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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.03%3A_Acid-base_reactions_a_la_Brnsted

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. The older Arrhenius theory of acids and bases viewed them as substances which produce hydrogen ions or hydroxide ions on dissociation. As useful a concept as this has been, it was unable to explain why NH , which contains no OH ions, is a base and not an acid, why a solution of FeCl is acidic, or why a solution of Na S is alkaline. A more general theory of acids and bases was developed by Franklin in 1905, who suggested that the plays a central role. According to this view, an acid is a solute that gives rise to a (positive ion) characteristic of the solvent, and a base is a solute that yields a (negative ion) which is also characteristic of the solvent. The most important of these solvents is of course H O, but Franklin's insight extended the realm of acid-base chemistry into non-aqueous systems as we shall see in a later lesson. In 1923, the Danish chemist J.N. Brønsted, building on Franklin's theory, proposed that In the same year the English chemist T.M. Lowry published a paper setting forth some similar ideas without producing a definition; in a later paper Lowry himself points out that Brønsted deserves the major credit, but the concept is still widely known as the Brønsted-Lowry theory. These definitions carry a very important implication: , and . As a very simple example, consider the equation that Arrhenius wrote to describe the behavior of hydrochloric acid: \[HCl \rightarrow H^+ + A^–\] This is fine as far as it goes, and chemists still write such an equation as a shortcut. But in order to represent this more realistically as a proton donor-acceptor reaction, we now depict the behavior of HCl in water by in which the acid HCl donates its proton to the acceptor (base) H O. "Nothing new here", you might say, noting that we are simply replacing a shorter equation by a longer one. But consider how we might explain the alkaline solution that is created when ammonia gas NH dissolves in water. An alkaline solution contains an excess of hydroxide ions, so ammonia is clearly a base, but because there are no OH ions in NH , it is clearly not an Arrhenius base. It is, however, a Brønsted base: In this case, the water molecule acts as the acid, donating a proton to the base NH to create the NH . The foregoing examples illustrate several important aspects of the Brønsted-Lowry concept of acids and bases: Brønsted (1879-1947) was a Danish physical chemist. Although he is now known mainly for his proton donor-acceptor theory of acids and bases (see his ), he published numerous earlier papers on chemical affinity, and later on the catalytic effects of acids and bases on chemical reactions. In World War II he opposed the Nazis, and this led to his election to the Danish parliament in 1947, but he was unable to take his seat because of illness and he died later in that year. (1874-1936) was the first holder of the chair in physical chemistry at Cambridge University. His extensive studies of the effects of acids and bases on the optical behavior of camphor derivatives (specifically, how they rotate the plane of polarized light) led him to formulate a theory of acids and bases similar to and simultaneously with that of Brønsted. There is another serious problem with the Arrhenius view of an acid as a substance that dissociates in water to produce a hydrogen ion. The hydrogen ion is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an excess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a , H O . In a sense, H O is acting as a base here, and the product H O is the conjugate acid of water: Owing to the overwhelming excess of \(H_2O\) molecules in aqueous solutions, a bare hydrogen ion has no chance of surviving in water. Although other kinds of dissolved ions have water molecules bound to them more or less tightly, the interaction between H and H O is so strong that writing “H ” hardly does it justice, although it is formally correct. The formula H O more adequately conveys the sense that it is both a molecule in its own right, and is also the conjugate acid of water. The equation "HA → H + A " is so much easier to write that chemists still use it to represent acid-base reactions in contexts in which the proton donor-acceptor mechanism does not need to be emphasized. Thus it is permissible to talk about “hydrogen ions” and use the formula H in writing chemical equations as long as you remember that they are not to be taken literally in the context of aqueous solutions. Interestingly, experiments indicate that the proton does not stick to a single H O molecule, but changes partners many times per second. This molecular promiscuity, a consequence of the uniquely small size and mass the proton, allows it to move through the solution by rapidly hopping from one H O molecule to the next, creating a new H O ion as it goes. The overall effect is the same as if the H O ion itself were moving. Similarly, a hydroxide ion, which can be considered to be a “proton hole” in the water, serves as a landing point for a proton from another H O molecule, so that the OH ion hops about in the same way. According to the Brønsted concept, the process that was previously written as a simple of a generic acid HA ("HA → H + A )" is now an in its own right: \[HA + H_2O \rightarrow A^- + H_3O^+\] The idea, again, is that the proton, once it leaves the acid, must end up somewhere; it cannot simply float around as a free hydrogen ion. A reaction of an acid with a base is thus a ; if the acid is denoted by AH and the base by B, then we can write a generalized acid-base reaction as \[AH + B \rightarrow A^- + BH^+\] Notice that the of this reaction, \[BH^+ + A^- \rightarrow B + AH^+\] is also an acid-base reaction. Because simple reactions can take place in both directions to some extent, it follows that transfer of a proton from an acid to a base must necessarily create a new pair of species that can, at least in principle, constitute an acid-base pair of their own. In this schematic reaction, base is said to be to acid , and acid is conjugate to base . The term means “connected with”, the implication being that The table below shows the conjugate pairs of a number of typical acid-base systems. We can look upon the generalized acid-base reaction as a competition of two bases for a proton: If the base H O overwhelmingly wins this tug-of-war, then the acid HA is said to be a . This is what happens with hydrochloric acid and the other common strong "mineral acids" H SO , HNO , and HClO : Solutions of these acids in water are really solutions of the ionic species shown in heavy type on the right. This being the case, it follows that what we call a 1 M solution of "hydrochloric acid" in water, for example, does not really contain a significant concentration of HCl at all; the only real a acid present in such a solution is H O ! The second of these statements is called the . It means that although the inherent proton-donor strengths of the strong acids differ, they are all completely dissociated in water. Chemists say that their strengths are "leveled" by the solvent water. A comparable effect would be seen if one attempted to judge the strengths of several adults by conducting a series of tug-of-war contests with a young child. One would expect the adults to win overwhelmingly on each trial; their strengths would have been "leveled" by that of the child. Most acids, however, are able to hold on to their protons more tightly, so only a small fraction of the acid is dissociated. Thus hydrocyanic acid, HCN, is a weak acid in water because the proton is able to share the lone pair electrons of the cyanide ion CN more effectively than it can with those of H O, so the reaction \[HCN + H_2O \rightarrow H_3O^+ + CN^–\] proceeds to only a very small extent. Since a strong acid binds its proton only weakly, while a weak acid binds it tightly, we can say that Strong acids are "weak" and weak acids are "strong." If you are able to explain this apparent paradox, you understand one of the most important ideas in acid-base chemistry! This is just a re-statement of what is implicit in what has been said above about the distinction between strong acids and weak acids. The fact that HCl is a strong acid implies that its conjugate base Cl is too weak a base to hold onto the proton in competition with either H O or H O . Similarly, the CN ion binds strongly to a proton, making HCN a weak acid. The fact that HCN is a weak acid implies that the cyanide ion CN reacts readily with protons, and is thus is a relatively good base. As evidence of this, a salt such as KCN, when dissolved in water, yields a slightly alkaline solution: CN + H O → HCN + OH This reaction is still sometimes referred to by its old name ("water splitting"), which is literally correct but tends to obscure its identity as just another acid-base reaction. Reactions of this type take place only to a small extent; a 0.1M solution of KCN is still, for all practical purposes, 0.1M in cyanide ion. In general, the weaker the acid, the more alkaline will be a solution of its salt. However, it would be going to far to say that "ordinary weak acids have strong conjugate bases." The only really strong base is hydroxide ion, OH , so the above statement would be true only for the weak acid H O. The only really strong bases you are likely to encounter in day-to-day chemistry are alkali-metal hydroxides such as NaOH and KOH, which are essentially solutions of the hydroxide ion. Most other compounds containing hydroxide ions such as Fe(OH) and Ca(OH) are not sufficiently soluble in water to give highly alkaline solutions, so they are not usually thought of as strong bases. There are actually a number of bases that are stronger than the hydroxide ion — best known are the O and the NH , but these are so strong that they can rob water of a proton: O + H O → 2 OH NH + H O → NH + OH This gives rise to the same kind of leveling effect we described for acids, with hydroxide ion as the strongest base in water. Hydroxide ion is the strongest base that can exist in aqueous solution. The most common example of this is ammonium chloride, NH Cl, whose aqueous solutions are distinctly acidic: NH + H O → NH + H O Because this (and similar) reactions take place only to a small extent, a solution of ammonium chloride will only be slightly acidic. From some of the examples given above, we see that water can act as an acid CN + H O → HCN + OH and as a base NH + H O → NH + H O If this is so, then there is no reason why "water-the-acid" cannot donate a proton to "water-the-base": This reaction is known as the . Chemists still often refer to this reaction as the "dissociation" of water and use the Arrhenius-style equation H O → H + OH as a kind of shorthand. As discussed in the previous lesson, this process occurs to only a tiny extent. It does mean, however, that hydronium and hydroxide ions are present in any aqueous solution. Other liquids also exhibit autoprotolysis with the most well-known example is liquid ammonia: 2 NH → NH + NH Even pure liquid sulfuric acid can play the game: 2 H SO → H SO + HSO Each of these solvents can be the basis of its own acid-base "system", parallel to the familiar "water system". Water, which can act as either an acid or a base, is said to be : it can "swing both ways". A substance such as water that is amphiprotic is called an . As indicated here, the hydroxide ion can also be an ampholyte, but not in aqueous solution in which the oxide ion cannot exist. It is of course the amphiprotic nature of water that allows it to play its special role in ordinary aquatic acid-base chemistry. But many other amphiprotic substances can also exist in aqueous solutions. Any such substance will always have a conjugate acid and a conjugate base, so if you can recognize these two conjugates of a substance, you will know it is amphiprotic. For example, the triplet set {carbonic acid, bicarbonate ion, carbonate ion} constitutes an amphiprotric series in which the bicarbonate ion is the ampholyte, differing from either of its neighbors by the addition or removal of one proton: If the bicarbonate ion is both an acid and a base, it should be able to exchange a proton with itself in an autoprotolysis reaction: \[HCO_3^– + HCO_3^– \rightarrow H_2CO_3 + CO_3^{2-}\] Your very life depends on the above reaction! CO , a metabolic by-product of every cell in your body, reacts with water to form carbonic acid H CO which, if it were allowed to accumulate, would make your blood fatally acidic. However, the blood also contains carbonate ion, which reacts according to the reverse of the above equation to produce bicarbonate which can be safely carried by the blood to the lungs. At this location the autoprotolysis reaction runs in the forward direction, producing H CO which loses water to form CO which gets expelled in the breath. The carbonate ion is recycled back into the blood to eventually pick up another CO molecule. If you can write an autoprotolysis reaction for a substance, then that substance is amphiprotic.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.05%3A_Specific_Heat_Calculations

Water has a high capacity for absorbing heat. In a car radiator, it serves to keep the engine cooler than it would otherwise run. As the water circulates through the engine, it absorbs heat from the engine block. When it passes through the radiator, the cooling fan and the exposure to the outside environment allow the water to cool somewhat before it makes another passage through the engine. The specific heat of a substance can be used to calculate the temperature change that a given substance will undergo when it is either heated or cooled. The equation that relates heat \(\left( q \right)\) to specific heat \(\left( c_p \right)\), mass \(\left( m \right)\), and temperature change \(\left( \Delta T \right)\) is shown below. \[q = c_p \times m \times \Delta T\nonumber \] The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by \(\Delta T = T_f - T_i\), where \(T_f\) is the final temperature and \(T_i\) is the initial temperature. A \(15.0 \: \text{g}\) piece of cadmium metal absorbs \(134 \: \text{J}\) of heat while rising from \(24.0^\text{o} \text{C}\) to \(62.7^\text{o} \text{C}\). Calculate the specific heat of cadmium. The specific heat equation can be rearranged to solve for the specific heat. \[c_p = \dfrac{q}{m \times \Delta T} = \dfrac{134 \: \text{J}}{15.0 \: \text{g} \times 38.7^\text{o} \text{C}} = 0.231 \: \text{J/g}^\text{o} \text{C}\nonumber \] The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures. Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that \(60.0 \: \text{g}\) of water at \(23.52^\text{o} \text{C}\) was cooled by the removal of \(813 \: \text{J}\) of heat. The change in temperature can be calculated using the specific heat equation: \[\Delta T = \dfrac{q}{c_p \times m} = \dfrac{813 \: \text{J}}{4.18 \: \text{J/g}^\text{o} \text{C} \times 60.0 \: \text{g}} = 3.24^\text{o} \text{C}\nonumber \] Since the water was being cooled, the temperature decreases. The final temperature is: \[T_f = 23.52^\text{o} \text{C} - 3.24^\text{o} \text{C} = 20.28^\text{o} \text{C}\nonumber \]

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/02%3A_The_Hydrosphere/2.04%3A_Chemical_budgets_of_oceanic_elements

To the extent that the composition of the ocean remains constant, the rate at which any one element is introduced into seawater must equal the rate of its removal. A listing of the various routes of addition and removal, together with the estimated rate of each process, constitutes the budget for a given element. If that budget is greatly out of balance and no other transport routes are apparent, then it is likely that the ocean is not in a steady state with respect to that element, at least on a short time scale. It is important to understand, however, that short-term deviations from constant composition are not necessarily inconsistent with a long-term steady state. Deviations from the latter condition are most commonly inferred from geological evidence. The major input of elements to the oceans is river water. Groundwater seepage constitutes a very small secondary source. These were considered the only sources until the 1970’s, when the existence of hydrothermal springs at sites of seafloor spreading became known. There are presently no reliable estimates of the magnitude of this source. Pollution represents an additional input, mainly dissolved in river water, but also sometimes in rain and by dry deposition. Routes of removal are formation and burial of sediments, formation of evaporite deposits, direct input to the atmosphere by sea-salt aerosol transfer associated with bubble-breaking, and burial with sediments, either in interstitial water or adorbed onto active surfaces. Reaction with newly-formed basalt associated with undersea volcanic activity appears to be an important removal mechanism for some elements. The major elements undergoing steady-state dynamic change in the ocean are connected with biological processes. The key limiting element in the development of oceanic biomass is phosphorus, in the form of the phosphate ion. For terrestrial plant life, nitrogen is more commonly the limiting element, where it is taken up in the form of the nitrate ion. In the ocean, however, the ratio of the nitrate ion concentration to that of phosphorus has been found to be everywhere the same; this implies that the concentration of one controls that of the other. The source of nitrate ion is atmospheric N , which is freely soluble in water and is thus always present in abundance. The conversion of N to NO is presumed to be biologically mediated, probably by bacteria. The constancy of the NO /P concentration ratio implies that the phosphorus concentration controls the activity of the nitrogen-fixing organisms, and thus the availability of nitrogen to oceanic life. Photosynthetic activity in the upper part of the ocean causes inorganic phosphate to be incorporated into biomass, reducing the concentration of phosphorus; in warm surface waters, phosphate may become totally depleted. A given phosphorus atom may be traded several times among the plant, animal and bacterial populations before it eventually finds itself in biodebris (a dead organism or a fecal pellet) that falls into the deep part of the sea. Only about 1% of the phosphorus atoms that descend into deeper waters actually reach the bottom, where they are incorporated into sediments and permanently removed from circulation. The other 99% are released in the form of soluble phosphate, which is eventually brought to the surface in regions of upwelling. An average phosphorus atom will undergo one cycle of this circulation in about 1000 years; only a few months of this cycle will be spent in biomass. After an average of 100 such cycles, the atom will be removed from circulation and locked into the bottom sediment, and a new one will have entered the sea with river or juvenile water. Phosphorus is unique in that its major source of input to the oceans derives ultimately from pollution; in the long term, this represents a transfer of land-based phosphate deposits to the oceans. About half of the phosphorus input is in the form of suspended material, both organic and inorganic, the latter being in a variety of forms including phosphates adsorbed onto clays and iron oxide paticles, and calcium phosphate (apatatite) eroded from rocks. These various particulates are known to dissolve to some extent once they reach the ocean, but there is considerable uncertainty about the rates of these processes under various conditions. The major sink for oceanic phosphorus is burial with organic matter; this accounts for about two-thirds of the phosphorus removed. Most of the remainder is due to deposition with CaCO . A minor removal route is through reaction with Fe(II) formed when seawater attacks hot basalt, and in the formation of evaporite deposits. However, there is more phosphorus in evaporite deposits in the Western U.S. than in all of the ocean, so it is apparent that the long-term phosphorus budget is still not clearly understood. Carbon enters the ocean from both the atmosphere (as CO ) and river water, in which the principal species is HCO . Once in solution, the carbonate species are in equilibrium with each other and with H O , and the concentrations of all of these are influenced by the partial pressure of atmospheric CO . The mass budget for calcium is linked to that of carbon through solubility equilibria with the various solid forms of CaCO (mainly calcite). During photosynthesis, C is taken up slightly more readily than the rare isotope C . Since the rate of photosynthesis is controlled by the phosphate concentration, the C /C ratio in the dissolved carbon dioxide of surface waters is slightly higher than in the ocean as a whole. Observations of carbon isotope ratios in buried sediments have been useful in tracking historical changes in phosphate concentrations. The ratio of carbon to phosphorus in sea salt is about eight times greater than the same ratio measured in organic debris. This implies that in exhausting the available phosphate, the living organisms in the upper ocean consume only 12.5% of the dissolved carbon. Even this relatively small withdrawal of carbon from the carbonate system is sufficient to noticeably reduce the partial pressure of gaseous CO in equilibrium with the ocean; it has been estimated that if all life in the ocean should suddenly cease, the atmospheric CO content would rise to about three times its present level. The regulation of atmospheric CO pressure by the oceans also works the other way: since the amount of dissolved carbonate in the oceans is so much greater than the amount of CO in the atmosphere, the oceans act to buffer the effects of additions of CO to the atmosphere. Calculations indicate that about half of the CO that has been produced by burning fossil fuel since the Industrial Revolution has ended up in the oceans. This is of course the major carbonate species in the ocean. Although it is interconvertible with CO and is thus coupled to the carbon and photosynthetic cycles, itself can neither be taken up nor produced by organisms, and so it can be treated somewhat independently of biological activity. In this sense the only major input of into the oceans is river water. The two removal mechanisms are formation of CO \[H^+ + HCO_3^– \rightarrow H_2O + CO^2\] and the (biologically mediated) formation of \(CaCO_3^:\) \[Ca^{2+} + HCO_3^– \rightarrow CaCO_3 + H^+\] Since the pH of the oceans does not change, \(H^+\) is conserved and the removal of \(HCO_3^–\) by biogenic secretion of \(CaCO_3\) can be expressed by the sum of these reactions: \[Ca^{2+} + 2 HCO_3^– \rightarrow CaCO_3 + CO_2 + H_2O\] whose reverse direction represents the dissolution of the skeletal remains of dead organisms as they fall to lower depths. The upper parts of the ocean tend to be supersaturated in CaCO . Solid CaCO , in the form of calcite, is manufactured by a large variety of organisms such as foraminifera. A constant rain of calcite falls through the ocean as these organisms die. The solubility of increases with pressure, so only that portion of the calcite that falls to shallow regions of the ocean floor is incorporated into sediments and removed from circulation; the remainder dissolves after reaching a depth known as the lysolcline. At the present time, the amounts of carbonate and Ca supplied by erosion and volcanism appear to be only about one-third as great as the amount of calcite produced by organisms. As the carbonate concentration in a given region of the ocean becomes depleted due to higher calcite production, the lysocline moves up, tending to replenish the carbonate, and reducing the amount that is withdrawn by burial in sediments. Organic residues that fall into the deep sea are mostly oxidized to CO , presumably by bacterial activity. Calcium is removed from seawater solely by biodeposition as CaCO , a process whose rate can be determined quite accurately both at the present and in the past. As is explained above, the upper part of the ocean is supersaturated in calcite but the lower ocean is not. For this reason, less than 20% of the CaCO that is formed ends up as sediment and is eventually buried. The main questions about the calcium budget tend to focus on the rates and locales at which dissolution of skeletal carbonates occurs, and on how to interpret the various kinds of existing carbonate sediments. For example, the crystalline form aragonite is less stable than calcite, and will presumably dissolve at a higher elevation. The absence of aragonite-containing pteropod shells in deeper deposits seems to confirm this, but in the absence of rate data is it difficult to know at what elevations these particular organisms originated. The data in the above Table indicate that at the present time there is a net removal of calcium from the oceans. This is due to the rise in sea level since the decline of the more recent glacial epoch during the past 11,000 years. The additional water has covered the continental shelves, increasing the amount of shallow ocean where the growth of organisms is most intense. Over the more distant past (25 million years) the calcium budget appears to be well balanced. Evidence from geology and peleontology indicates that the salinity, and hence the chloride concentration of sewater has been quite constant for about 600 million years. There have been periods when climatic conditions and coastal topography have led to episodes of evaporite formation, but these have evidently been largely compensated by the eventual return of the evaporite deposits to the sea. The natural input of chloride from rivers is about 215 Tg/yr, but the present input is about half again as great (Table 7 on page 33), due to pollution. Also, there are presently no significant areas where seawater is evaporating to dryness. Thus the oceanic chloride budget is considerably out of balance. However, the replacement time of Cl in the oceans is so long (87 million years) that this will probably have no long-term effect. Although sodium is tied to chloride, it is also involved in the formation of silicate minerals, the weathering of rocks, and in cation exchange with clay sediments. Its short-term budget is quite out of balance for the same reasons as is that of chloride. On a longer time scale, removal of sodium by reaction with hot basalt associated with undersea volcanic activity may be of importance. Considerably more sulfate is being added to seawater than is being removed by the major mechanisms of sediment formation (mainly CaSO and pyrites). The natural river input is 82 Tg of S per year, while that due to pollution is 61 Tg/yr from rivers and 17 Tg/yr from rain and dry deposition. This element is unusual in that its river-water input is balanced mostly by reaction with volcanic basalt; removal through biogenic formation of magnesian calcite (dolomite) accounts for only 11% of its total removal from the ocean. The present-day magnesium budget seems to have been balanced for the past 100 million years. However, most of the extensive dolomite deposits were formed prior to this time, so the longer-term magnesium budget is poorly understood. The potassium budget of the ocean is not well understood. The element is unusual in that only about 60% of its input is by rivers; the remainder is believed to come from newly formed undersea basalt. The big question about potassium is how it is removed; fixation by ion-exchange with illite clays seems to be a major mechanism, and its uptake by basalt (at lower temperatures than are required for its release) is also believed to occur. About 85% of the silicon input to the oceans comes from river water in the form of silicic acid, H SiO . The remainder probably comes from basalt. It is removed by biogenic deposition as opaline silica, SiO · H O produced mainly by planktonic organisms (radiolaria and diatoms). Unlike the case for CaCO , the ocean is everywhere undersaturated in silica, especially near the surface where these organisms deplete it with greater efficiency than any other element. Because opaline silica dissolves so rapidly, only a small fraction makes it to the bottom. The major deposits occur in shallower waters where coastal upwelling provides a good supply of N and P nutrients for siliceous organisms. Thus over half of the biogenic silica deposits are found in the Antarctic ocean. In spite of the fact that dissolved silica has the shortest replacement time (21,000 years) of any major element in the ocean, its concentration appears to have been remarkably constant during geological time. This is taken as an indication of the ability of siliceous organisms to respond quickly to changes in local concentrations of dissolved silica. This element is complicated by its biologically-mediated exchange with atmospheric nitrogen, and by its existence in several oxidation states, all of which are interconvertible. Unlike the other major elements, nitrogen does not form extensive sedimentary deposits; most of the nitrogen present in dead organic material seems to be removed before it can be buried. Through this mechanism there is extensive cycling of nitrogen between the shallow and deep parts of the oceans. The real difficulty in constructing a budget for oceanic nitrogen is the very large uncertainty in the rates of the major input (fixation) and output. Both of these processes are biologically mediated, but little is known about what organisms are responsible, where they thrive and how they are affected by local nutrient supply and other conditions. )

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Membrane_Potentials

Membrane potential is what we use to describe the difference in voltage (or electrical potential) between the inside and outside of a cell. Without membrane potentials human life would not be possible. All living cells maintain a potential difference across their membrane. Simply stated, membrane potential is due to disparities in concentration and permeability of important ions across a membrane. Because of the unequal concentrations of ions across a membrane, the membrane has an electrical charge. Changes in membrane potential elicit action potentials and give cells the ability to send messages around the body. More specifically, the action potentials are electrical signals; these signals carry efferent messages to the central nervous system for processing and afferent messages away from the brain to elicit a specific reaction or movement. Numerous active transports embedded within the cellular membrane contribute to the creation of membrane potentials, as well as the universal cellular structure of the lipid bilayer. The chemistry involved in membrane potentials reaches to many scientific disciplines. Chemically it involves molarity, concentration, electrochemistry and the Nernst equation. From a physiological standpoint, membrane potential is responsible for sending messages to and from the central nervous system. It is also very important in cellular biology and shows how cell biology is fundamentally connected with electrochemistry and physiology. The bottom line is that membrane potentials are at work in your body right now and always will be as long as you live. The subject of membrane potential stretches across multiple scientific disciplines; Membrane Potential plays a role in the studies of Chemistry, Physiology and Biology. The culmination of the study of membrane potential came in the 19th and early 20th centuries. Early in the 20th century, a man named professor Bernstein hypothesized that there were three contributing factors to membrane potential; the permeability of the membrane and the fact that [K+] was higher inside and lower on the outside of the cell. He was very close to being correct, but his proposal had some flaws. Walther H. Nernst, notable for the development of the Nernst equation and winner of 1920 Nobel Prize in chemistry, was a major contributor to the study of membrane potential. He developed the Nernst equation to solve for the equilibrium potential for a specific ion. Goldman, Hodgkin and Katz furthered the study of membrane potential by developing the Goldman-Hodgkin-Katz equation to account for any ion that might permeate the membrane and affect its potential. The study of membrane potential utilizes electrochemistry and physiology to formulate a conclusive idea of how charges are separated across a membrane. Differences in concentration of ions on opposite sides of a cellular membrane produce a voltage difference called the membrane potential. The largest contributions usually come from sodium (Na ) and chloride (Cl ) ions which have high concentrations in the extracellular region, and potassium (K ) ions, which along with large protein anions have high concentrations in the intracellular region. Calcium ions, which sometimes play an important role, are not shown. In discussing the concept of membrane potentials and how they function, the creation of a membrane potential is essential. The lipid bilayer structure of the cellular membrane, with its lipid-phosphorous head and fatty acid tail, provides a perfect building material that creates both a hydrophobic and hydrophilic side to the cellular membrane. The membrane is often referred to as a mosaic model because of its semi-permeability and its ability to keep certain substances from entering the cell. Molecules such as water can diffuse through the cell based on concentration gradients; however, larger molecules such as glucose or nucleotides require channels. The lipid bilayer also houses the Na /K pump, ATPase pump, ion transporters, and voltage gated channels, and it is the site of vesicular transport. The structure regulates which ions enter and exit to determine the concentration of specific ions inside of the cell. Animals and plants require the breakdown of organic substances through cellular respiration to generate energy. This process, which produces ATP, is dependent on the electron transport chain. Electrons travel down this path to be accepted by oxygen or other electron acceptors. The initial electrons are obtained from the breakdown of water molecules. The hydrogen build up in the extracellular fluid leaving a gradient. As per membrane potentials, when there a gradient, the molecules flow in the opposite direction. In this case, hydrogen flows back into the cell through a protein known as ATP synthase which creates ATP in the process. This action is essential to life because the number of ATP created from each glucose increases drastically. Chemical disequilibrium and membrane potentials allow bodily functions to take place. Transport proteins, more specifically the 'active' transport proteins, can pump ions and molecules against their concentration gradient. This is the main source of charge difference across the cellular membrane. The following points should help you to understand how membrane potential works Human nerve cells work mainly on the concept of membrane potentials. They transmit chemicals known as serotonin or dopamine through gradients. The brain receives these neurotransmitters and uses it to perform functions. Check out this YouTube video if you want to know more about how the Na /K pump and how the membrane potential works. www.youtube.com/watch?v=iA-Gdkje6pg The calculation for the charge of an ion across a membrane, The Nernst Potential, is relatively easy to calculate. The equation is as follows: (RT/zF) log([X] /[X] ). RT/F is approximately 61, therefore the equation can be written as (61/z) ln([X] /[X] ) The only difference in the Goldman-Hodgkin-Katz equation is that is adds together the concentrations of all permeable ions as follows (RT/zF) log([K ] +[Na ] +[Cl ] /[K ] +[Na ] +[Cl ] ) The charges are equal on both sides; therefore the membrane has no potential. 2)There is an unbalance of charges, giving the membrane a potential. 3) The charges line up on opposite sides of the membrane to give the membrane its potential. 4) A hypothetical neuron in the human body; a large concentration of potassium on the inside and sodium on the outside. 1. List the following in order from highest to lowest permeability. A , K , Na 2. Which of the following statement is NOT true? 3. What would be the equilibrium potential for the ion K be if [K ] = 5mM and [K ] =150mM? 4. True or false: At resting membrane potential, the inside of the membrane is slightly negatively charged while the outside is slightly positively charged. 1. 2. Answer is true; membrane potential exists in neurons and is responsible for action potential propagation in neurons. 3. E = (61/z) log([K ] /[K ] ) = (61/1) log([5mM]/[150mM]) = z=1 4. The resting membrane potential is negative as a result of this disparity in concentration of charges.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Degree_of_Unsaturation

There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of \(\pi\) bonds and/or cyclic rings. In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated. Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s). Degree of Unsaturation (DoU) is also known as . If the molecular formula is given, plug in the numbers into this formula: \[ DoU= \dfrac{2C+2+N-X-H}{2} \] As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C H the number of actual hydrogens needed for the compound to be saturated is 8 . The compound needs 4 more hydrogens in order to be fully saturated . Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C H Cl, there is one less hydrogen compared to ethane, C H . For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C H N compared to C H . Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C H OH, matches the number of hydrogens in ethane, C H . The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination. Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, 3 double bonds. What is the Degree of Unsaturation for Benzene? The molecular formula for benzene is C H Thus, DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds. However, when given the molecular formula C H , benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene. 1. (a.) (E (b.) (c.) (d.) 2. (a.) (b.) (c.) (d.) 3. (a.) (b.) (c.) (d.) 4. (a.) (b.) (c.) (d.) 5. (a.) (Remember-a saturated molecule only contains single bonds) (b.) (c.) ( (d.)

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Lasers/Gas_Lasers

In these lasers the lasing medium is made-up of one or a mixture of gases or vapors. Gas lasers can be classified in terms of the type of transitions that lead to their operation: atomic or molecular. The most common of all gas lasers is the helium-neon (He-Ne) laser. The presence of two atomic species (helium and neon) in this gas laser might suggest that the medium is made of molecules, but these two species of atoms do not form a stable molecule. In fact, all inert atoms like helium, argon, krypton, etc. (those in the last column of the Periodic Table) hold tightly to their own electron clouds and seldom form a molecule or react with other atoms (hence the name: inert). In the He-Ne laser the transition that produces the output light is an atomic transition. Gas lasers that employ molecular gas or vapor for their lasing medium use molecular transitions for their lasing operation. Molecular transitions tend to be more complex than atomic ones. As a consequence the laser light produced by molecular lasers tends to have a wider and more varied collection of properties. Examples of some common molecular gas lasers are carbon monoxide (CO), carbon dioxide (CO2), excimer, and nitrogen (N2) lasers. The He-Ne laser was the first continuous wave (cw) laser invented. A few months after Maiman announced his invention of the pulsed ruby laser, Ali Javan and his associates W. R. Bennet and D. R. Herriott announced their creation of a cw He-Ne laser. This gas laser is a four-level laser that use helium atoms to excite neon atoms. It is the atomic transitions in the neon that produces the laser light. The most commonly used neon transition in these lasers produces red light at 632.8 nm. But these lasers can also produce green and yellow light in the visible as well as UV and IR (Javan's first He-Ne operated in the IR at 1152.3 nm). By using highly reflective mirrors designed for one of these many possible lasing transitions, a given He-Ne's output is made to operate at a single wavelength. He-Ne lasers typically produce a few to tens of mW (milli-Watt, or \(10^{-3}\) W) of power. They are not sources of high power laser light. Probably one of the most important features of these lasers is that they are highly stable, both in terms of their wavelength (mode stability) and intensity of their output light (low jitter in power level). For these reasons, He-Ne lasers are often used to stabilize other lasers. They are also used in applications, such as holography, where mode stability is important. Until the mid 1990's, He-Ne lasers were the dominant type of lasers produced for low power applications - from range finding to scanning to optical transmission, to laser pointers, etc. Recently, however, other types of lasers, most notably the semiconductor lasers, seem to have won the competition because of reduced costs. The above energy level diagram shows the two excited states of helium atom, the 2 3S and 2 1S, that get populated as a result of the electromagnetic pumping in the discharge. Both of these states are metastable and do not allow de-excitations via radiative transitions. Instead, the helium atoms give off their energy to neon atoms through collisional excitation. In this way the 4s and 5s levels in neon get populated. These are the two upper lasing levels, each for a separate set of lasing transitions. Radiative decay from the 5s to the 4s levels are forbidden. So, the 4p and 3p levels serve as the lower lasing levels and rapidly decay into the metastable 3s level. In this way population inversion is easily achieved in the He-Ne. The 632.8 nm laser transition, for example, involves the 5s and 3p levels, as shown above. In most He-Ne lasers the gas, a mixture of 5 parts helium to 1 part neon, is contained in a sealed glass tube with a narrow (2 to 3 mm diameter) bore that is connected to a larger size tube called a ballast, as shown above. Typically the laser's optical cavity mirrors, the high reflector and the output coupler, form the two sealing caps for the narrow bore tube. High voltage electrodes create a narrow electric discharge along the length of this tube, which then leads to the narrow beam of laser light. The function of the ballast is to maintain the desired gas mixture. Since some of the atoms may get imbedded in the glass and/or the electrodes as they accelerate within the discharge, in the absence of a ballast the tube would not last very long. To further prolong tube lifetime some of these lasers also use "getters", often metals such as titanium, that absorb impurities in the gas. A typical commercially available He-Ne produces about a few mW of 632.8 nm light with a beam width of a few millimeters at an overall efficiency of near 0.1%. This means that for every 1 Watt of input power from the power supply, 1 mW of laser light is produced. Still, because of their long operating lifetime of 20,000 hours or more and their relatively low manufacturing cost, He-Ne lasers are among the most popular gas lasers. Another commonly used gas laser is the argon-ion laser. In these lasers, as in the He-Ne the lasing transition type is atomic. But instead of a neutral atom, here the lasing is the result of the de-excitations of the ion. It takes more energy to ionize an atom than to excite it. By the same token, more energy can be obtained from the de-excitation of the ion. So, doubly (Ar++) and singly ionized (Ar+) argon atoms can radiate shorter wavelength light than could the neutral argon atom, Ar. Because of this, argon-ion lasers can produce uv light with a wavelength as short as 334 nm. In addition, these lasers can produce much more power than He-Ne lasers. Argon-ion lasers typically range in output power from one to as much as 20 W. At the higher power levels their output is multi-mode, i.e. contains several distinct wavelengths. Some of these wavelengths are: Because of these two reasons, high power and multicolor output, argon-ion is one of the most commonly used lasers in laser light shows, as well as in a variety of applications. The make-up of a typical argon-ion laser is very similar to a He-Ne's, but with a few slight differences. First, these lasers are much larger in size. A typical Ar++ laser tube is about one meter long, as compared to just 20 cm for that of a He-Ne. Second, the optical cavity of these lasers is built external to the tube. This is partly because of the high power operation of the laser and partly because such external arrangement allows for the use of optional wavelength selection optics within the optical cavity. A prism or a diffraction grating located just before the high reflecting mirror selects only one of the lasing transitions for amplification within the cavity; other wavelengths are deflected out of the resonant cavity. In this way these ion lasers can operate in a so called single mode. With this arrangement the two mirrors holders on opposite sides of the laser tube are typically attached together with an invar rod for thermal stabilization (Invar is a steel alloy that contains nickel). Its most valued property is that it expands and contracts very little when its temperature changes. As a result, when the laser's temperature changes as it heats up due to the large electric current within the electromagnetic pump discharge, the optical path length, and therefore the modal character of the laser output, remains relatively unchanged. Finally, because of their high power argon-ion lasers require active cooling. This is most commonly accomplished by circulating water, either directly from tap or from commercially produced chillers, in closed coils that surround the plasma (a gas of charged ions) tube and parts of the electric power supply. Some of the lower powered argon-ion lasers are just air cooled using a fan, which makes them less cumbersome to use. Another type of ion laser, the krypton laser, operates very much the same as the argon-ion laser. To take advantage of all the colors available in both argon and krypton lasers, manufacturers make argon-krypton ion lasers by using a suitable mixture of these two gases. The mixed gas lasers are very useful for entertainment applications because, in addition to many colors, they can also produce a "white" beam. (Why is the word "white" in quotations?) In both of these lasers the gaseous medium is made-up of molecules, which in addition to electronic energy levels of atoms also have both molecular vibrational and rotational energy levels. The vibrational energy levels are similar to finer spaced ladder rungs that span two rungs of the electronic energy levels. The rotational levels are still more finely spaced rungs that span the vibrational rungs! In these gas lasers the lasing transitions occur among the vibrational levels, typically belonging to different electronic levels. Specifically, \(CO_2\) lasers can generate an output wavelength from about 9 micro-meters (mm, or microns) to about 11 microns (1 micron is one millionth of a meter, or 1000 nm.) These outputs generally contain many closely spaced wavelengths, if the laser is used for high power output. But for more wavelength specific applications the optical cavity of the laser is designed to amplify just one or a few of the vibrational radiative decay lines. The wavelength range for the CO laser is lower, from about 5 to 6 mm. Another feature of these gas lasers that make them one of the most versatile of all gas lasers is that they can be made to operate over a large range of power outputs, either in a pulsed or cw mode. The \(CO_2\) laser, in particular, ranges in cw power from few Watts to kWs, making these lasers ideal for many industrial applications including welding and drilling. Molecular vibrations for \(CO_2\), a linear molecule, are shown in the figure below. Other combinations of these are possible but these three are fundamental. There are different varieties of \(CO_2\) lasers that flow fresh gases through the resonant cavity area in order both to remove heat and to provide lots of gas to achieve high laser powers. For these lasers in the cw mode powers can reach as high as 100 kW. These intense laser beams are essentially tremendous invisible "heat" beams that can cut through thick pieces of metal and are used extensively in industrial applications. We mention two interesting tidbits about these lasers. First, since glass in not very transparent to IR light, the mirrors are actually made of special crystalline materials that are transparent to the IR. Second, recall that IR light is invisible to our eyes and so special precautions are needed to protect people working around these lasers. It turns out that although these lasers can easily cut through metal, they cannot pass through a thin sheet of clear plexiglass, and so often these systems are housed in a plexiglass shell to block any stray reflected IR light. Other types of gas lasers include the nitrogen laser (N2), excimers, copper-vapor, gold-vapor, and chemical lasers. Of these the excimer lasers and the chemical lasers are the most different from the ones we have already discussed above. Similar to \(CO_2\), \(CO\), and \(N_2\) lasers, these gas lasers also use molecular transitions for their lasing operation. What makes them especially different is that the molecular gas used for these lasers has no ground state! Typically these molecules include an atom belonging to the inert gas family (argon, xenon, krypton) and one from the halide group (chlorine, fluorine, and bromine). The inert gas atoms (also known as the rare gases) do not want to interact with any other atoms. On the other hand, the halide gases are highly reactive. Still, they cannot bond with the inert gases to form a molecule. But when sufficient energy is provided to these atoms they bind together in a short-lived excited state that soon (few nanoseconds) decays back into the original two separate atoms (i.e. the molecule dissociates). Because of this rapid molecular dissociation, these lasers obtain population inversion just by excitation alone! In fact, the word excimer is short for "excited dimer," although most excimer lasers do not use two identical atoms as a strict dimer would. The excimer molecules are created from a mixture of inert gases along with one of the halides. Typically a few percent of Ar, Kr and Xe are mixed with a few percent of a halide to form excimer molecules: ArF, KrF , and XeF. The other 90% of the gas mix consists of other inert gases such as He and Ne which act simply as a buffer and do not take part in the reaction. A large electric pulse is often used for the excitation and formation of the excimer molecule. The rapid decay of the short-lived molecule then leads to a very short laser pulse lasting 10 - 100 ns (10-8 - 10-7 s). So, another unusual feature of the excimers is that they do not require an optical amplifier. They are very efficiently formed in the reaction with an efficiency of around 30% so that the gain is extremely high. A high reflector and a glass (really quartz - why?) window is sufficient for laser light production. This means that only about 4% of the light is reflected back into the cavity at the front window, but the gain is so high that only a single pass through the cavity is needed to produce lots of uv light. Typically about 1 Joule of energy is in a 10 ns pulse, so that the pulse power is 1J/10ns = 100 MW. If this power were steadily produced it would be equivalent to powers generated by large power plants. However, only about 1 - 100 pulses are produced per second, so that the average power produced is about 1 - 100 W. Because of the highly reactive nature of the halide gas used in these lasers, excimers are not very easy to operate. The halides tend to be very corrosive and therefore add a great deal to the operational cost (as well as the danger!) of these lasers. But still these lasers are very much in use because their output wavelength is in the UV, from 350 nm down to as low as 193 nm; all with a good deal of power. In these lasers the energy of excitation comes from a chemical reaction that takes place in the medium itself. In this sense, then, chemical lasers are self-pumped. Even more interesting is the operation of one of the most common of these lasers, the hydrogen-fluoride laser that operates in the IR at 4.6 microns. The added intrigue is due to the chain reaction that takes place to excite the laser molecule, \(\ce{HF}\). A mixture of hydrogen (molecular) gas, \(\ce{H2}\), and fluorine gas, \(\ce{F_2}\), is subjected to an electric discharge to start the chain reaction, which results in the production of a hydrogen-fluoride molecule in an excited vibrational level (excited state is denoted by a starred superscript: \(HF^*\)) and the dissociative production of H or F for the next reaction: \[\ce{F + H_2 \rightarrow HF^{*} + H}\] \[\ce{F2 + H \rightarrow HF^{*} + F}\] \[\ce{F + H2 \rightarrow HF^{*} + H}\] \[\ce{F_2 + H \rightarrow HF^{*} + F}\] etc. So, the medium in these lasers is used up as a fuel to generate their laser light. Therefore, practical limitations aside, the power of these lasers depends on the amount of the chemical (gas volume) that is used in the laser. Because of their conceivably limitless power output, these lasers have been studied mostly for their military applications. The most famous of these is the US Army's Mid-Infrared Advanced Chemical Laser, MIRACL, located at White Sands Missile Range, New Mexico. This is a formidable chemical laser that has been developed as part of the Department of Defense's efforts for its Strategic Defense Initiative (SDI). MIRACL is the US's most powerful laser. It operates in a band from 3.6 to 4.2 microns producing megawatts of CW output for as long as 70 seconds. For its fuel it burns ethylene, \(C_2H_4\), with nitrogen trifluoride, NF3. The resulting free fluorine atoms combine with deuterium gas that is injected to this burning fuel to form deuterium fluoride molecules (DF), which ultimately provide the laser light. In this laser the beam within the optical cavity is about 21 cm high and 3 cm wide, resulting in a 14 cm2 output beam. For testing purposes this laser is used with an aiming and focusing telescope system called the SEALITE Beam Director (SLBD) which was first developed by the Navy. The moving part of SLBD that is capable of fast rotations and high accelerations weighs about 18,000 pounds. Its telescope can focus the beam to any target that is located within a range of a minimum of 400 meters up to infinity. Jay Newman and Seyffie Maleki ( )

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Introduction_to_Catalysis

Catalysis is the ability of some species to rapidly speed up the rate at which a chemical reaction proceeds. For historical reasons, the discipline is normally split into two sub-catagories; homogeneous (homo = same, geneous = phase) and heterogeneous (hetero = different). Homogeneous catalysis is concerned with catalysts that are in the same phase as the chemical reactions they are speeding up. These reactions are normally in the liquid phase and include all of biology's enzymes. While the majority of homogeneous catalysis is in the liquid phase there are gas phase and solid phase homogeneous catalytic reactions. Heterogeneous catalysts have a catalyst that is in a different phase. This type of catalysis is responsible for the vast majority of 'bulk' chemicals that are produced each year that go into making all the things we take for granted around us such as plastics, and are also extensively used for refining oil in gasoline. This chapter focuses on heterogeneous catalysis and specifically how and why they work at all. The key concept in catalysis, and indeed all of chemistry, is the tension between the thermodynamics of a particular reaction which tells you whether something happen, and the kinetics which tells you something will happen. This tension in the world around you is everywhere i.e. thermodynamics says the diamond on an engagement ring should turn into graphite, a more stable allotrope of carbon but the kinetics is such that you would be waiting a very, very long time before it happened! Catalysis works by giving chemical reactions that happen a way to happen. The standard way to think of a chemical reaction happening is to imagine the energy of a molecule as being represented by a z co-ordinate while the reaction 'proceeds' in the x-y plane. If this sounds confusing, just think of the energy as being like the height of land features like mountains relative to their position. A molecule in this landscape would be a place where you can't go downhill without having to go uphill first, like a crater lake. Even though on a global scale, a crater lake should flow down to the sea, the local conditions around it mean that it would have to first flow uphill, which we could postulate doesn't happen on a regular basis. This analogy helps to visualise what is going on in a chemical reaction: the atoms involved when they're joined in one fashion are in an energy 'lake' and aren't able to change, but given the right conditions, such as temperature or photons, they are able to get over the energy barrier and flow down to another energy 'lake' which has the atoms joined in another fashion. This energy barrier is called the activation energy. In our analogy, imagine drilling a hole from one lake to another lake, since the water doesn't have to go 'up and over', it can flow easily to another position. This is exactly what a catalyst does, it doesn't change the total energy of the system or where the system will eventually end up (the thermodynamics dictate this), it just provides an easier way for the system to get there and hence speeds it up (it increases the rate). This analogy paints a picture that helps you remember the text-book definition of a catalyst: a substance which increases the rate at which a thermodynamic equilibrium is obtained by lowering the activation energy of the reaction pathway. That is what a catalyst does but it doesn't give us any idea of how. It is only useful in deciding whether you are observing catalysis in action, the mechanism of how heterogeneous catalysts work is far more subtle.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Anabolism/Pentose_Phosphate_Pathway

The pentose phosphate pathway is the major source for the NADPH required for anabolic processes. There are three distinct phases each of which has a distinct outcome. Depending on the needs of the organism the metabolites of that outcome can be fed into many other pathways. is directly connected to the pentose phosphate pathway. As the need for glucose-6-phosphate (the beginning metabolite in the pentose phosphate pathway) increases so does the activity of . The main molecule in the body that makes anabolic processes possible is NADPH. Because of the structure of this molecule it readily donates hydrogen ions to metabolites thus reducing them and making them available for energy harvest at a later time. The PPP is the main source of synthesis for NADPH. The pentose phosphate pathway (PPP) is also responsible for the production of Ribose-5-phosphate which is an important part of nucleic acids. Finally the PPP can also be used to produce glyceraldehyde-3-phosphate which can then be fed into the TCA and ETC cycles allowing for the harvest of energy. Depending on the needs of the cell certain enzymes can be regulated and thus increasing or decreasing the production of desired metabolites. The enzymes reasonable for catalyzing the steps of the PPP are found most abundantly in the liver (the major site of gluconeogenesis) more specifically in the cytosol. The cytosol is where fatty acid synthesis takes place which is a NADPH dependent process.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.08%3A_Molecular_Orbital_Theory

Make sure you thoroughly understand the following essential ideas The molecular orbital model is by far the most productive of the various models of chemical bonding, and serves as the basis for most quantiative calculations, including those that lead to many of the computer-generated images that you have seen elsewhere in these units. In its full development, molecular orbital theory involves a lot of complicated mathematics, but the fundamental ideas behind it are quite easily understood, and this is all we will try to accomplish in this lesson. This is a big departure from the simple and models that were based on the one-center orbitals of individual atoms. The more sophisticated model recognized that these orbitals will be modified by their interaction with other atoms. But all of these models, as they are generally called, are very limited in their applicability and predictive power, because they fail to recognize that distribution of the pooled valence electrons is governed by the totality of positive centers. Chemical bonding occurs when the net attractive forces between an electron and two nuclei exceeds the electrostatic repulsion between the two nuclei. For this to happen, the electron must be in a region of space which we call the . Conversely, if the electron is off to one side, in an , it actually adds to the repulsion between the two nuclei and helps push them away. The easiest way of visualizing a molecular orbital is to start by picturing two isolated atoms and the electron orbitals that each would have separately. These are just the orbitals of the separate atoms, by themselves, which we already understand. We will then try to predict the manner in which these atomic orbitals interact as we gradually move the two atoms closer together. Finally, we will reach some point where the internuclear distance corresponds to that of the molecule we are studying. The corresponding orbitals will then be the of our new molecule. To see how this works, we will consider the simplest possible molecule, \(\ce{H2^{+}}\). This is the hydrogen molecule ion, which consists of two nuclei of charge +1, and a single electron shared between them. As two H nuclei move toward each other, the 1 atomic orbitals of the isolated atoms gradually merge into a new molecular orbital in which the greatest electron density falls between the two nuclei. Since this is just the location in which electrons can exert the most attractive force on the two nuclei simultaneously, this arrangement constitutes a . Regarding it as a three- dimensional region of space, we see that it is symmetrical about the line of centers between the nuclei; in accord with our usual nomenclature, we refer to this as a . There is one minor difficulty: we started with two orbitals (the 1 atomic orbitals), and ended up with only one orbital. Now according to the rules of quantum mechanics, orbitals cannot simply appear and disappear at our convenience. For one thing, this would raise the question of at just what internuclear distance do we suddenly change from having two orbitals, to having only one? It turns out that when orbitals interact, they are free to change their forms, but there must always be the same number. This is just another way of saying that there must always be the same number of possible allowed sets of electron quantum numbers. How can we find the missing orbital? To answer this question, we must go back to the wave-like character of orbitals that we developed in our earlier treatment of the hydrogen atom. You are probably aware that wave phenomena such as sound waves, light waves, or even ocean waves can combine or interact with one another in two ways: they can either reinforce each other, resulting in a stronger wave, or they can interfere with and partially destroy each other. A roughly similar thing occurs when the “matter waves” corresponding to the two separate hydrogen 1 orbitals interact; both in-phase and out-of-phase combinations are possible, and both occur. The in-phase, reinforcing interaction yields the that we just considered. The other, corresponding to out-of-phase combination of the two orbitals, gives rise to a molecular orbital that has its greatest electron probability in what is clearly the antibonding region of space. This second orbital is therefore called an orbital. When the two 1 wave functions combine out-of-phase, the regions of high electron probability do not merge. In fact, the orbitals act as if they actually repel each other. Notice particularly that there is a region of space exactly equidistant between the nuclei at which the probability of finding the electron is zero. This region is called a , and is characteristic of antibonding orbitals. It should be clear that any electrons that find themselves in an antibonding orbital cannot possibly contribute to bond formation; in fact, they will actively oppose it. We see, then, that whenever two orbitals, originally on separate atoms, begin to interact as we push the two nuclei toward each other, these two atomic orbitals will gradually merge into a pair of molecular orbitals, one of which will have bonding character, while the other will be antibonding. In a more advanced treatment, it would be fairly easy to show that this result follows quite naturally from the wave-like nature of the combining orbitals. What is the difference between these two kinds of orbitals, as far as their potential energies are concerned? More precisely, which kind of orbital would enable an electron to be at a lower potential energy? Clearly, the potential energy decreases as the electron moves into a region that enables it to “see” the maximum amount of positive charge. In a simple diatomic molecule, this will be in the internuclear region— where the electron can be simultaneously close to two nuclei. The bonding orbital will therefore have the lower potential energy. This scheme of bonding and antibonding orbitals is usually depicted by a molecular orbital diagram such as the one shown here for the . Atomic valence electrons (shown in boxes on the left and right) fill the lower-energy molecular orbitals before the higher ones, just as is the case for atomic orbitals. Thus, the single electron in this simplest of all molecules goes into the bonding orbital, leaving the antibonding orbital empty. Since any orbital can hold a maximum of two electrons, the bonding orbital in H is only half-full. This single electron is nevertheless enough to lower the potential energy of one mole of hydrogen nuclei pairs by 270 kJ— quite enough to make them stick together and behave like a distinct molecular species. Although H is stable in this energetic sense, it happens to be an extremely reactive molecule— so much so that it even reacts with itself, so these ions are not commonly encountered in everyday chemistry. If one electron in the bonding orbital is conducive to bond formation, might two electrons be even better? We can arrange this by combining two hydrogen atoms-- two nuclei, and two electrons. Both electrons will enter the bonding orbital, as depicted in the Figure. We recall that one electron lowered the potential energy of the two nuclei by 270 kJ/mole, so we might expect two electrons to produce twice this much stabilization, or 540 kJ/mole. Experimentally, one finds that it takes only 452 kJ to break apart a mole of hydrogen molecules. The reason the potential energy was not lowered by the full amount is that the presence of two electrons in the same orbital gives rise to a repulsion that acts against the stabilization. This is exactly the same effect we saw in comparing the ionization energies of the hydrogen and helium atoms. With two electrons we are still ahead, so let’s try for three. The dihelium positive ion is a three-electron molecule. We can think of it as containing two helium nuclei and three electrons. This molecule is stable, but not as stable as dihydrogen; the energy required to break He + is 301 kJ/mole. The reason for this should be obvious; two electrons were accommodated in the bonding orbital, but the third electron must go into the next higher slot— which turns out to be the sigma antibonding orbital. The presence of an electron in this orbital, as we have seen, gives rise to a repulsive component which acts against, and partially cancels out, the attractive effect of the filled bonding orbital. Taking our building-up process one step further, we can look at the possibilities of combining to helium atoms to form dihelium. You should now be able to predict that He cannot be a stable molecule; the reason, of course, is that we now have four electrons— two in the bonding orbital, and two in the antibonding orbital. The one orbital almost exactly cancels out the effect of the other. Experimentally, the bond energy of dihelium is only .084 kJ/mol; this is not enough to hold the two atoms together in the presence of random thermal motion at ordinary temperatures, so dihelium dissociates as quickly as it is formed, and is therefore not a distinct chemical species. The four simplest molecules we have examined so far involve molecular orbitals that derived from two 1 atomic orbitals. If we wish to extend our model to larger atoms, we will have to contend with higher atomic orbitals as well. One greatly simplifying principle here is that only the orbitals need to be considered. Inner atomic orbitals such as 1 are deep within the atom and well-shielded from the electric field of a neighboring nucleus, so that these orbitals largely retain their atomic character when bonds are formed. For example, when lithium, whose configuration is 1 2 , bonds with itself to form Li , we can forget about the 1 atomic orbitals and consider only the σ bonding and antibonding orbitals. Since there are not enough electrons to populate the antibonding orbital, the attractive forces win out and we have a stable molecule. The bond energy of dilithium is 110 kJ/mole; notice that this value is less than half of the 270 kJ bond energy in dihydrogen, which also has two electrons in a bonding orbital. The reason, of course, is that the 2 orbital of Li is much farther from its nucleus than is the 1 orbital of H, and this is equally true for the corresponding molecular orbitals. It is a general rule, then, that the larger the parent atom, the less stable will be the corresponding diatomic molecule. All the molecules we have considered thus far are ; they are made up of one kind of atom. As an example of a molecule, let’s take a look at a very simple example— lithium hydride. Lithium hydride is a stable, though highly reactive molecule. The diagram shows how the molecular orbitals in lithium hydride can be related to the atomic orbitals of the parent atoms. One thing that makes this diagram look different from the ones we have seen previously is that the parent atomic orbitals have widely differing energies; the greater nuclear charge of lithium reduces the energy of its 1 orbital to a value well below that of the 1 hydrogen orbital. There are two occupied atomic orbitals on the lithium atom, and only one on the hydrogen. With which of the lithium orbitals does the hydrogen 1 orbital interact? The lithium 1 orbital is the lowest-energy orbital on the diagram. Because this orbital is so small and retains its electrons so tightly, it does not contribute to bonding; we need consider only the 2 orbital of lithium which combines with the 1 orbital of hydrogen to form the usual pair of sigma bonding and antibonding orbitals. Of the four electrons in lithium and hydrogen, two are retained in the lithium 1 orbital, and the two remaining ones reside in the σ orbital that constitutes the Li–H covalent bond. The resulting molecule is 243 kJ/mole more stable than the parent atoms. As we might expect, the bond energy of the heteronuclear molecule is very close to the average of the energies of the corresponding homonuclear molecules. Actually, it turns out that the correct way to make this comparison is to take the geometric mean, rather than the arithmetic mean, of the two bond energies. The geometric mean is simply the square root of the product of the two energies. The geometric mean of the H and Li bond energies is 213 kJ/mole, so it appears that the lithium hydride molecule is 30 kJ/mole more stable than it “is supposed” to be. This is attributed to the fact that the electrons in the 2σ bonding orbital are not equally shared between the two nuclei; the orbital is skewed slightly so that the electrons are attracted somewhat more to the hydrogen atom. This , which we considered in some detail near the beginning of our study of covalent bonding, arises from the greater electron-attracting power of hydrogen— a consequence of the very small size of this atom. The electrons can be at a lower potential energy if they are slightly closer to the hydrogen end of the lithium hydride molecule. It is worth pointing out, however, that the electrons are, on the average, also closer to the lithium nucleus, compared to where they would be in the 2 orbital of the isolated lithium atom. So it appears that everyone gains and no one loses here! The molecules we have considered thus far are composed of atoms that have no more than four electrons each; our molecular orbitals have therefore been derived from -type atomic orbitals only. If we wish to apply our model to molecules involving larger atoms, we must take a close look at the way in which -type orbitals interact as well. Although two atomic orbitals will be expected to split into bonding and antibonding orbitals just as before, it turns out that the extent of this splitting, and thus the relative energies of the resulting molecular orbitals, depend very much on the nature of the particular orbital that is involved. You will recall that there are three possible orbitals for any value of the principal quantum number. You should also recall that orbitals are not spherical like orbitals, but are elongated, and thus possess definite directional properties. The three orbitals correspond to the three directions of Cartesian space, and are frequently designated , and , to indicate the axis along which the orbital is aligned. Of course, in the free atom, where no coordinate system is defined, all directions are equivalent, and so are the orbitals. But when the atom is near another atom, the electric field due to that other atom acts as a point of reference that defines a set of directions. The line of centers between the two nuclei is conventionally taken as the axis. If this direction is represented horizontally on a sheet of paper, then the axis is in the vertical direction and the axis would be normal to the page. These directional differences lead to the formation of two different classes of molecular orbitals. The above figure shows how two atomic orbitals interact. In many ways the resulting molecular orbitals are similar to what we got when atomic orbitals combined; the bonding orbital has a large electron density in the region between the two nuclei, and thus corresponds to the lower potential energy. In the out-of-phase combination, most of the electron density is away from the internuclear region, and as before, there is a surface exactly halfway between the nuclei that corresponds to zero electron density. This is clearly an again, in general shape, very much like the kind we saw in hydrogen and similar molecules. Like the ones derived from -atomic orbitals, these molecular orbitals are σ ( ) orbitals. orbitals are cylindrically symmetric with respect to the line of centers of the nuclei; this means that if you could look down this line of centers, the electron density would be the same in all directions. orbitals, we get the bonding and antibonding pairs that we would expect, but the resulting molecular orbitals have a different symmetry: rather than being rotationally symmetric about the line of centers, these orbitals extend in both perpendicular directions from this line of centers. Orbitals having this more complicated symmetry are called π ( ) orbitals. There are two of them, π and π differing only in orientation, but otherwise completely equivalent. The different geometric properties of the π and σ orbitals causes the latter orbitals to split more than the π orbitals, so that the σ* antibonding orbital always has the highest energy. The σ bonding orbital can be either higher or lower than the π bonding orbitals, depending on the particular atom. If we combine the splitting schemes for the 2 and 2 orbitals, we can predict bond order in all of the diatomic molecules and ions composed of elements in the first complete row of the periodic table. Remember that only the of the atoms need be considered; as we saw in the cases of lithium hydride and dilithium, the inner orbitals remain tightly bound and retain their localized atomic character. Carbon has four outer-shell electrons, two 2 and two 2 . For two carbon atoms, we therefore have a total of eight electrons, which can be accommodated in the first four molecular orbitals. The lowest two are the 2 -derived bonding and antibonding pair, so the “first” four electrons make no net contribution to bonding. The other four electrons go into the pair of bonding orbitals, and there are no more electrons for the antibonding orbitals— so we would expect the dicarbon molecule to be stable, and it is. (But being extremely reactive, it is known only in the gas phase.) You will recall that one pair of electrons shared between two atoms constitutes a “single” chemical bond; this is Lewis’ original definition of the covalent bond. In C there are two paris of electrons in the π bonding orbitals, so we have what amounts to a double bond here; in other words, the bond order in dicarbon is two. The electron configuration of oxygen is 1 2 2 . In O , therefore, we need to accommodate twelve valence electrons (six from each oxygen atom) in molecular orbitals. As you can see from the diagram, this places two electrons in antibonding orbitals. Each of these electrons occupies a separate π* orbital because this leads to less electron-electron repulsion (Hund's Rule). The bond energy of molecular oxygen is 498 kJ/mole. This is smaller than the 945 kJ bond energy of N — not surprising, considering that oxygen has two electrons in an antibonding orbital, compared to nitrogen’s one. The two unpaired electrons of the dioxygen molecule give this substance an unusual and distinctive property: O is c. The paramagnetism of oxygen can readily be demonstrated by pouring liquid O between the poles of a strong permanent magnet; the liquid stream is trapped by the field and fills up the space between the poles. Since molecular oxygen contains two electrons in an antibonding orbital, it might be possible to make the molecule more stable by removing one of these electrons, thus increasing the ratio of bonding to antibonding electrons in the molecule. Just as we would expect, and in accord with our model, O has a bond energy higher than that of neutral dioxygen; removing the one electron actually gives us a more stable molecule. This constitutes a very good test of our model of bonding and antibonding orbitals. In the same way, adding an electron to O results in a weakening of the bond, as evidenced by the lower bond energy of O . The bond energy in this ion is not known, but the length of the bond is greater, and this is indicative of a lower bond energy. These two dioxygen ions, by the way, are highly reactive and can be observed only in the gas phase.

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Enkephalins (penta-peptides) have recently been discovered as regulators of nerve impulses involving pain in the brain. Apparently these peptides act as natural analgesics (pain-killers) and their action mimics that of morphine and other opiates. At present it is thought that the morphine-like effects are due to aromatic side chains on phenylanine and tyrosine which mimic a similar sturcture on morphine. Apparently, it does not make much difference whether the enkephalin contains methionine or leucine at the acid end of the peptide. The primary structures are: methionine-enkephaline: tyr - gly - gly - phe - met leucine-enkephalin: tyr - gly - gly - phe - leu QUES. Write the peptide structure for methionine-enkephalin. Enkephalin - Met -

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The Hope diamond, a form of carbon, is a very expensive piece of jewelry—currently worth about $350,000,000. A graphite pencil, comprised partly of carbon, can be purchased for less than a dollar. Both items contain carbon, but there is a big difference in how that carbon is organized. The diamond was formed under very different reaction conditions than the graphite, so it has a different heat of formation. A relatively straightforward chemical reaction is one in which elements are combined to form a compound. Sodium and chlorine react to form sodium chloride. Hydrogen and oxygen combine to form water. Like other reactions, these are accompanied by either the absorption or release of heat. The is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states. The standard conditions for thermochemistry are \(25^\text{o} \text{C}\) and \(101.3 \: \text{kPa}\). Therefore, the standard state of an element is its state at \(25^\text{o} \text{C}\) and \(101.3 \: \text{kPa}\). For example, iron is a solid, bromine is a liquid, and oxygen is a gas under those conditions. The standard heat of formation of an element in its standard state by definition is equal to zero. The \(\Delta H^\text{o}_\text{f} = 0\) for the diatomic elements, \(\ce{H_2} \left( g \right)\), \(\ce{N_2} \left( g \right)\), \(\ce{O_2} \left( g \right)\), \(\ce{F_2} \left( g \right)\), \(\ce{Cl_2} \left( g \right)\), \(\ce{Br_2} \left( g \right)\), and \(\ce{I_2} \left( g \right)\). The graphite form of solid carbon is in its standard state with \(\Delta H^\text{o}_\text{f} = 0\), while diamond is not its standard state. Some standard heats of formation are listed in the table below.

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Fatty acids are merely carboxylic acids with long hydrocarbon chains. The hydrocarbon chain length may vary from 10-30 carbons (most usual is 12-18). The non-polar hydrocarbon alkane chain is an important counter balance to the polar acid functional group. In acids with only a few carbons, the acid functional group dominates and gives the whole molecule a polar character. However, in fatty acids, the non-polar hydrocarbon chain gives the molecule a non-polar character. The most common fatty acids are listed. Note that there are two groups of fatty acids--saturated and unsaturated. Recall that the term unsaturated refers to the presence of one or more double bonds between carbons as in alkenes. A saturated fatty acid has all bonding positions between carbons occupied by hydrogens. The melting points for the saturated fatty acids follow the boiling point principle observed previously. Melting point principle: as the molecular weight increases, the melting point increases. This observed in the series lauric (C12), palmitic (C16), stearic (C18). Room temperature is 25 C, Lauric acid which melts at 44 is still a solid, while arachidonic acid has long since melted at -50 , so it is a liquid at room temperature. Note that as a group, the . The reason for this phenomenon can be found by a careful consideration of molecular geometries. The tetrahedral bond angles on carbon results in a molecular geometry for saturated fatty acids that is relatively linear although with zigzags. See graphic on the left. This molecular structure allows many fatty acid molecules to be rather closely "stacked" together. As a result, close intermolecular interactions result in relatively high melting points. On the other hand, the introduction of one or more double bonds in the hydrocarbon chain in unsaturated fatty acids results in one or more "bends" in the molecule. The geometry of the double bond is almost always a cis configuration in natural fatty acids. and these molecules do not "stack" very well. The intermolecular interactions are much weaker than saturated molecules. As a result, the melting points are much lower for unsaturated fatty acids.

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As useful and appealing as the concept of the is, it raises a somewhat troubling question that we must sooner or later face: what is the nature of the in which the shared electrons are contained? Up until now, we have been tacitly assuming that each valence electron occupies the same kind of atomic orbital as it did in the isolated atom. As we shall see below, his assumption very quickly leads us into difficulties. Consider how we might explain the bonding in a compound of divalent beryllium, such as beryllium hydride, BeH . The beryllium atom, with only four electrons, has a configuration of 1 2 . Note that the two electrons in the 2 orbital have opposite spins and constitute a stable pair that has no tendency to interact with unpaired electrons on other atoms. The only way that we can obtain two unpaired electrons for bonding in beryllium is to promote one of the 2 electrons to the 2 level. However, the energy required to produce this excited-state atom would be sufficiently great to discourage bond formation. It is observed that Be does form reasonably stable bonds with other atoms. Moreover, the two bonds in BeH and similar molecules are completely equivalent; this would not be the case if the electrons in the two bonds shared Be orbitals of different types, as in the "excited state" diagram above. These facts suggest that it is incorrect to assume that the distribution of valence electrons that are shared with other atoms can be described by atomic-type , and orbitals at all. Remember that these different orbitals arise in the first place from the interaction of the electron with the electrostatic force field associated with the positive nucleus. An outer-shell electron in a bonded atom will be under the influence of a force field emanating from positive nuclei, so we would expect the orbitals in the bonded atoms to have a somewhat different character from those in free atoms. In fact, as far as valence electrons are concerned, we can throw out the concept of atomic orbital altogether and reassign the electrons to a new set of that are characteristic of each molecular configuration. This approach is indeed valid, but we will defer a discussion of it until a later unit. For now, we will look at a less-radical model that starts out with the familiar valence-shell atomic orbitals, and allows them to combine to form whose shapes conform quite well to the bonding geometry that we observe in a wide variety of molecules. First, recall that the electron, being a quantum particle, cannot have a distinct location; the most we can do is define the region of space around the nucleus in which the probability of finding the electron exceeds some arbitrary value, such as 90% or 99%. This region of space is the . Because of the wavelike character of matter, the orbital corresponds to a standing wave pattern in 3-dimensional space which we can often represent more clearly in 2-dimensional cross section. The quantity that is varying (“waving”) is a denoted by ψ ( ) whose value varies from point to point according to the wave function for that particular orbital. Orbitals of all types are simply mathematical functions that describe particular standing-wave patterns that can be plotted on a graph but have no physical reality of their own. Because of their wavelike nature, two or more orbitals (i.e., two or more functions ψ) can be combined both in-phase and out-of-phase to yield a pair of resultant orbitals which, to be useful, must have squares that describe actual electron distributions in the atom or molecule. The and orbitals that you are familiar with are the most convenient ones for describing the electron distribution in isolated atoms because assignment of electrons to them according to the usual rules always yields an overall function that predicts a spherically symmetric electron distribution, consistent with all physical evidence that atoms are in fact spherical. For atoms having more than one electron, however, the , basis set is only one of many possible ways of arriving at the same observed electron distribution. We use it not because it is unique, but because it is the simplest. In the case of a molecule such as BeH , we know from experimental evidence that the molecule is linear and therefore the electron density surrounding the central atom is no longer spherical, but must be concentrated along two directions 180° apart, and we need to construct a function having these geometrical properties. There are any number of ways of doing this, but it is convenient is to use a particular set of functions (which we call ) that are constructed by combining the atomic and functions that are already familiar to us. You should understand that ; it is merely a that combines the atomic orbitals we are familiar with in such a way that the new (hybrid) orbitals possess the geometric and other properties that are reasonably consistent with what we observe in a wide range (but certainly not in all) molecules. In other words, hybrid orbitals are abstractions that describe reality fairly well in certain classes of molecules (and fortunately, in much of the very large class of organic substances) and are therefore a useful means of organizing a large body of chemical knowledge... but they are far from infallible. ; it is merely a that combines the atomic orbitals we are familiar with in such a way that the new (hybrid) orbitals possess the geometric and other properties that are reasonably consistent with what we observe in a wide range (but certainly not in all) molecules. This approach, which assumes that the orbitals remain more or less localized on one central atom, is the basis of the theory which was developed in the early 1930s, mainly by . "In December 1930 Pauling had his famous 'breakthrough' where, in a rush of inspiration, he ' '. This flurry of calculations would eventually become the first of Pauling's germinal series of papers on the nature of the chemical bond. ' ', Pauling would recall. " Although the hybrid orbital approach has proven very powerful (especially in organic chemistry), it does have its limitations. For example, it predicts that both H O and H S will be tetrahedrally coordinated bent molecules with bond angles slightly smaller than the tetrahedral angle of 109.5° owing to greater repulsion by the nonbonding pair. This description fits water (104.5°) quite well, but the bond angle in hydrogen sulfide is only 92°, suggesting that atomic orbitals (which are 90° apart) provide a better description of the electron distribution about the sulfur atom than do hybrid orbitals. The hybrid orbital model is fairly simple to apply and understand, but it is best regarded as one special way of looking at a molecule that can often be misleading. Another viewpoint, called the , offers us a complementary perspective that it is important to have if we wish to develop a really thorough understanding of chemical bonding in a wider range of molecules. Hybrid orbitals are constructed by combining the functions for atomic orbitals. Because wave patterns can combine both constructively and destructively, a pair of atomic wave functions such as the - and - orbitals shown at the left can combine in two ways, yielding the shown. From an energy standpoint, we can represent the transition from atomic - and -orbitals to an hybrid orbital in this way: Notice here that 1) the total number of occupied orbitals is conserved, and 2) the two hybrid orbitals are intermediate in energy between their parent atomic orbitals. In terms of plots of the actual orbital functions we can represent the process as follows: The probability of finding the electron at any location is given not by , but by , whose form is roughly conveyed by the solid figures in this illustration. Returning to the example of BeH , we can compare the valence orbitals in the free atoms with those in the beryllium hydride molecule as shown here. It is, of course, the between the hydrogen-1 orbitals and the two lobes of the beryllium -hybrid orbitals that constitutes the two Be—H "bonds" in this molecule. Notice that whereas a single -orbital has lobes on both sides of the atom, a single -hybrid has most of its electron density on one side, with a minor and more spherical lobe on the other side. This minor lobe is centered on the central atom (some textbook illustrations don't get this right.) As far as the of the molecule is concerned, the result is exactly the same as predicted by the VSEPR model (although hybrid orbital theory predicts the same result in a more fundamental way.) We can expect any central atom that uses -hybridization in bonding to exhibit linear geometry when incorporated into a molecule. We can now go on to apply the same ideas to some other simple molecules. In , for example, we start with the boron atom, which has three outer-shell electrons in its normal or ground state, and three fluorine atoms, each with seven outer electrons. As is shown in this configuration diagram, one of the three boron electrons is unpaired in the ground state. In order to explain the trivalent bonding of boron, we postulate that the atomic - and - orbitals in the outer shell of boron mix to form three equivalent hybrid orbitals. These particular orbitals are called , meaning that this set of orbitals is derived from one s- orbital and two p-orbitals of the free atom. This illustration shows how an -orbital mixes with two orbitals to form a set of three hybrid orbitals. Notice again how the three atomic orbitals yield the same number of hybrid orbitals. Boron Trifluoride BF is a common example of hybridization. The molecule has plane trigonal geometry. Let us now look at several tetravalent molecules, and see what kind of hybridization might be involved when four outer atoms are bonded to a central atom. Perhaps the commonest and most important example of this bond type is methane, CH . orbitals of carbon mix into four hybrid orbitals which are chemically and geometrically identical; the latter condition implies that the four hybrid orbitals extend toward the corners of a tetrahedron centered on the carbon atom. Methane is the simplest hydrocarbon; the molecule is approximately spherical, as is shown in the space-filling model: This shows how an orbital on each of two two carbon atoms join (overlap) to form a carbon-carbon bond, and then the remaining carbon orbital overlaps with six hydrogen 1s orbitals to form the ethane molecule. If lone pair electrons are present on the central atom, these can occupy one or more of the orbitals. This causes the molecular geometry to be different from the coordination geometry, which remains tetrahedral. In the ammonia molecule, for example, the nitrogen atom normally has three unpaired p electrons, but by mixing the 2 and 2 orbitals, we can create four -hybrid orbitals just as in carbon. Three of these can form shared-electron bonds with hydrogen, resulting in ammonia, NH . The fourth of the hybrid orbitals contains the two remaining outer-shell electrons of nitrogen which form a non-bonding lone pair. In acidic solutions these can coordinate with a hydrogen ion, forming the ammonium ion NH . Although no bonds are formed by the lone pair in NH , these electrons do give rise to a charge cloud that takes up space just like any other orbital. In the molecule, the oxygen atom can form four orbitals. Two of these are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding. The observed H-O-H bond angle in water (104.5°) is less than the tetrahedral angle (109.5°); one explanation for this is that the non-bonding electrons tend to remain closer to the central atom and thus exert greater repulsion on the other orbitals, thus pushing the two bonding orbitals closer together. Hybridization can also help explain the existence and structure of many . Consider, for example, electron configurations of in the compounds in the illustrations below. The (top row) is another structure derived from zinc and chlorine. As we might expect, this ion is tetrahedral; there are four chloride ions surrounding the central zinc ion. The zinc ion has a charge of +2, and each chloride ion is –1, so the net charge of the complex ion is –2. At the bottom is shown the electron configuration of atomic zinc, and just above it, of the divalent zinc ion. Notice that this ion has no electrons at all in its 4-shell. In zinc chloride, shown in the next row up, there are two equivalent chlorine atoms bonded to the zinc. The bonding orbitals are of sp character; that is, they are hybrids of the 4s and one 4p orbital of the zinc atom. Since these orbitals are empty in the isolated zinc ion, the bonding electrons themselves are all contributed by the chlorine atoms, or rather, the chloride ions, for it is these that are the bonded species here. Each chloride ion possesses a complete octet of electrons, and two of these electrons occupy each sp bond orbital in the zinc chloride complex ion. This is an example of a , in which the bonded atom contributes both of the electrons that make up the shared pair.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photosynthesis/Photosystem_II/Photosystem_II_2

Photosystem is the form of pigments on the thylakoid membrane1. It collects energy over the wavelengths and concentrates it to one molecule which uses the energy to pass one of its electrons on to a series of enzymes1. Photosystem II occurs with two series of enzymes followed by Photosystem I in order to create energy for a plant1. In Photosystem II which also called water- plastoquinone oxidoreductase, the generated hydrogen ions help to create a proton gradient that is used by ATP synthase to generate ATP, and the transferred energized electrons are used to reduce 2NADP+ to 2NADPH. Photosystem II is the first membrane protein complex in oxygenic photosynthetic organisms in nature. It produces atmospheric oxygen to catalyze the photo-oxidation of water by using light energy. It oxidizes two molecules of water into one molecule of molecular oxygen. The four electrons removed from the water molecules are transferred by an electron transport chain which is formed hydrogen ions and molecular oxygen to plastoquinone2. By obtaining these electrons from water, photosystem II provides the electrons for all of photosynthesis to occur4. Photosystem II is composed of 20 subunits such as D1, D2, CP43, CP47, and PsbO3. Subunit D1 (beta-carotene, quinine and manganese center) reacts in the center of protein and binds Chlorophyll P680 and pheophytin, and Subunit D2 reacts in the center Protein. D1 and D2 form the core of this membrane protein3. D1 (colored in red) is homologous to the L subunit of the bacterial photosystem where as D2 (colored in blue) is homologous to the M subunit of the bacterial photosystem3. Chlorophylls is bounded by D1 and D2 and colored in green in the Figure A shown below3. CP43 binds with manganese center and CP47 appears in Photosystem I3. Last, PsbO (colored in purple) occurs in Manganese center to stabilize Protein. These subunits contains 99 cofactors and coenzymes; “35 chlorophyll a, 12 beta – carotene, two pheophytin, three plastoquinone, two heme, bicarbonate, 25 lipid and seven n-dodecyl – beta – D – maltoside detergent molecules, the six components of the Mn4Ca cluster, and one Fe2+ and two putative Ca2+ ion per monomer”1. Chlorophyll absorbs light4, Beta – carotene absorbs photoexcitation energy4, and heme contains iron4. Pheophytin is transferred an electron from P680 which is formed of 2 chlorophylls that absorb light at the wavelength of 680nm4. It is a primary electron acceptor and contains chlorophyll with the Magnesium replaced by two protons5. Then the electron is transferred to Plastoquinone (PQ) at QA site then QB site4. Plastoquinone can be one or two electron acceptor or donor from Photosystem II to the cytochrome bf complex in mobile intra-thylakoid membrane5. The arrival of a second electron at QB site with the uptake of two protons produces PQH24. When Plastoquinone is fully reduced to PQH2, it is called Plastoquinol. Therefore, the overall reaction for Photosystem II is shown below; 2PQ + 2H2O -> O2 + 2PQH2 (3) When the electron is transferred from P680 to Phephytin, a positive charge is formed on P680+ which is a strong oxidant that extracts electrons from water at manganese center5. Manganese center is the oxygen evolving center (OEC) and the site of water oxidation. It includes 4 manganese ions, a calcium ion, a chloride ion, and a tyrosine radical5. It is the core of this redox center because it has four stable oxidation states such as Mn2+, Mn3+, Mn4+, and Mn5+.5 Each time the P680 is excited and an electron is kicked out, the positively charged pair extracts an electron from the manganese center5. The manganese center is oxidized one electron at a time so it requires four steps to complete the oxidation. A tyrosine residue is not shown participates in the proton electron transfers, therefore; the structures are designated S0 through S4 to indicate the number of electrons removed6. We know there are five different oxidation states because of S0 through S4. When S4 is attained, an oxyzen molecule is released and two new molecules of water bind. The site of plastoquinone reduction is on the stroma side of the membrane6. The manganese complex is on the thylakoid lumen side of the membrane6. For every four electrons harvested from water, two molecules of PQH2 are formed extracting four protons from the stroma6. The four protons formed during the oxidation of water are released into the thylakoid lumen6. This distribution of protons across the thylakoid membrane generates a pH gradient with a low pH in the lumen and a high pH in the stroma6. The oxygen evolving complex of photosystem II contains Mn4, a redox-active tyrosine, and Ca2+/Cl- ions, but its molecular structure has not yet determined8. However, by looking at Figure B above, the point group for Photosystem II can be determined as C2 with a metal, Mn7. The Figure B describes an oblique surface-rendered view of the 3D structure of the C. reinhardtii supercompex6. The supercomplex is dimeric, therefore; it is found to be C2 point group symmetric containing two sets of subunits6. The primary emphasis of the Raman study in Photosystem II is on the low frequency range from 220 to 620 (cm-1)8. The low frequency region is examined for both S1 and S2. The Raman spectra of Photosystem II in the S1 state represents a few unique low-frequency bands that do not represent in S2 state8. This indicates that it is coordinated by two H2O or OH-. The Raman Mn-depleted Photosystem II and Photosystem II in the S2 are almost the same8. This indicates that the S1 state of the Manganese has a near infrared electronic transition from the resonance enhanced Raman scattering can be induced8. Photosystem II which is a part of Photosynthesis is one of the protein complexes. It has been the focus on many studies as a major biological energy source for life on the earth. This process requires water to obtain the electrons in order to provide the electrons for all of photosynthesis.

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Write equations that show NH as both a conjugate acid and a conjugate base. One example for NH as a conjugate acid: \(\ce{NH2- + H+ ⟶ NH3}\); as a conjugate base: \(\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l)\) Write equations that show \(\ce{H2PO4-}\) acting both as an acid and as a base. Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid: Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry acid: Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base: Show by suitable net ionic equations that each of the following species can act as a Brønsted-Lowry base: What is the conjugate acid of each of the following? What is the conjugate base of each? H O, O ; H O , \(\ce{OH^-}\) ; H CO , \(\ce{CO3^2-}\); \(\ce{NH4+}\), \(\ce{NH2-}\); H SO , \(\ce{SO4^2-}\); \(\ce{H3O2+}\), \(\ce{HO2-}\); H S; S ; \(\ce{H6N2^2+}\), H N What is the conjugate acid of each of the following? What is the conjugate base of each? Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations: The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. HNO (BA), H O(BB), H O (CA), \(\ce{NO3- (CB)}\); CN (BB), H O(BA), HCN(CA), \(\ce{OH^-}\) (CB); H SO (BA), Cl (BB), HCl(CA), \(\ce{HSO4- (CB)}\); \(\ce{HSO4- (BA)}\), OH (BB), \(\ce{SO4^2- (CB)}\), H O(CA); O (BB), H2O(BA) \(\ce{OH^-}\) (CB and CA); [Cu(H O) (OH)] (BB), [Al(H O) ] (BA), [Cu(H O) ] (CA), [Al(H O) (OH)] (CB); H S(BA), \(\ce{NH2- (BB)}\), HS (CB), NH (CA) Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations: What are amphiprotic species? Illustrate with suitable equations. Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H O. State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species: State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species. amphiprotic: \(\ce{NH3 + H3O+ ⟶ NH4OH + H2O}\), \(\ce{NH3 + OCH3- ⟶ NH2- + CH3OH}\); \(\ce{HPO4^2- + OH- ⟶ PO4^3- + H2O}\), \(\ce{HPO4^2- + HClO4 ⟶ H2PO4- + ClO4-}\); not amphiprotic: Br ; \(\ce{NH4+}\); \(\ce{AsO4^3-}\) Is the self ionization of water endothermic or exothermic? The ionization constant for water ( ) is \(2.9 \times 10^{-14}\) at 40 °C and \(9.6 \times 10^{-14}\) at 60 °C. Explain why a sample of pure water at 40 °C is neutral even though [H O ] = 1.7 × 10 . is 2.9 × 10 at 40 °C. In a neutral solution [H O ] = [OH ]. At 40 °C, [H O ] = [OH ] = (2.910 ) = 1.7 × 10 . The ionization constant for water ( ) is 2.9 × 10 at 40 °C. Calculate [H O ], [OH ], pH, and pOH for pure water at 40 °C. The ionization constant for water ( ) is 9.614 × 10 at 60 °C. Calculate [H O ], [OH ], pH, and pOH for pure water at 60 °C. x = 3.101 × 10 M = [H O ] = [OH ] pH = -log 3.101 × 10 = −(−6.5085) = 6.5085 pOH = pH = 6.5085 Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: pH = 3.587; pOH = 10.413; pH = 0.68; pOH = 13.32; pOH = 3.85; pH = 10.15; pH = −0.40; pOH = 14.4 What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely? What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52? [H O ] = 3.0 × 10 ; [OH ] = 3.3 × 10 Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See below Figure for useful information. Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See for useful information. [H O ] = 1 × 10 ; [OH ] = 1 × 10 The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10 at 25 °C. What is the concentration of hydroxide ions in the rainwater? The hydroxide ion concentration in household ammonia is 3.2 × 10 at 25 °C. What is the concentration of hydronium ions in the solution? [OH ] = 3.1 × 10 Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidic solution. Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basic solution. The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH , which causes the solution to be basic. Use this list of important industrial compounds (and ) to answer the following questions regarding: CaO, Ca(OH) , CH CO H, CO HCl, H CO , HF, HNO , HNO , H PO , H SO , NH , NaOH, Na CO . The odor of vinegar is due to the presence of acetic acid, CH CO H, a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in a 1- aqueous solution of this acid. [H O] > [CH CO H] > \(\ce{[H3O+]}\) ≈ \(\ce{[CH3CO2- ]}\) > [OH ] Household ammonia is a solution of the weak base NH in water. List, in order of descending concentration, all of the ionic and molecular species present in a 1- aqueous solution of this base. Explain why the ionization constant, , for H SO is larger than the ionization constant for H SO . The oxidation state of the sulfur in H SO is greater than the oxidation state of the sulfur in H SO . Explain why the ionization constant, , for HI is larger than the ionization constant for HF. Gastric juice, the digestive fluid produced in the stomach, contains hydrochloric acid, HCl. Milk of Magnesia, a suspension of solid Mg(OH) in an aqueous medium, is sometimes used to neutralize excess stomach acid. Write a complete balanced equation for the neutralization reaction, and identify the conjugate acid-base pairs. \(\underset{\large\ce{BB}}{\ce{Mg(OH)2}(s)}+\underset{\large\ce{BA}}{\ce{HCl}(aq)}⟶\underset{\large\ce{CB}}{\ce{Mg^2+}(aq)}+\underset{\large\ce{CA}}{\ce{2Cl-}(aq)}+\underset{\:}{\ce{2H2O}(l)}\) Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO ) , a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO with CuO. What is the ionization constant at 25 °C for the weak acid \(\ce{CH3NH3+}\), the conjugate acid of the weak base CH NH , = 4.4 × 10 . \(K_\ce{a}=2.3×10^{−11}\) What is the ionization constant at 25 °C for the weak acid \(\ce{(CH3)2NH2+}\), the conjugate acid of the weak base (CH ) NH, = 7.4 × 10 ? Which base, CH NH or (CH ) NH, is the strongest base? Which conjugate acid, \(\ce{(CH3)2NH2+}\) or (CH ) NH, is the strongest acid? The strongest base or strongest acid is the one with the larger or , respectively. In these two examples, they are (CH ) NH and \(\ce{CH3NH3+}\). Which is the stronger acid, \(\ce{NH4+}\) or HBrO? Which is the stronger base, (CH ) N or \(\ce{H2BO3-}\)? triethylamine. Predict which acid in each of the following pairs is the stronger and explain your reasoning for each. Predict which compound in each of the following pairs of compounds is more acidic and explain your reasoning for each. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign. Rank the compounds in each of the following groups in order of increasing acidity or basicity, as indicated, and explain the order you assign. Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F or CN , is the stronger base? See . The active ingredient formed by aspirin in the body is salicylic acid, C H OH(CO H). The carboxyl group (−CO H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a 0.001- aqueous solution of C H OH(CO H). \(\ce{[H2O] > [C6H4OH(CO2H)] > [H+]0 > [C6H4OH(CO2)- ] ≫ [C6H4O(CO2H)- ] > [OH- ]}\) What do we represent when we write: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)?\] Explain why equilibrium calculations are not necessary to determine ionic concentrations in solutions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibrium calculations necessary as part of the determination of the concentrations of all ions of some other strong electrolytes in solution? Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base. Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base in water directly proportional or inversely proportional? Explain your answer. What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak acid? What two common assumptions can simplify calculation of equilibrium concentrations in a solution of a weak base? Which of the following will increase the percent of NH that is converted to the ammonium ion in water (Hint: Use LeChâtelier’s principle.)? The addition of HCl Which of the following will increase the percent of HF that is converted to the fluoride ion in water? What is the effect on the concentrations of \(\ce{NO2-}\), HNO , and OH when the following are added to a solution of KNO in water: The equation for the equilibrium is: What is the effect on the concentration of hydrofluoric acid, hydronium ion, and fluoride ion when the following are added to separate solutions of hydrofluoric acid? The equation for the equilibrium is: Why is the hydronium ion concentration in a solution that is 0.10 in HCl and 0.10 in HCOOH determined by the concentration of HCl? This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO H exists primarily as HCO H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H O ] produced by the stronger acid. From the equilibrium concentrations given, calculate for each of the weak acids and for each of the weak bases. CH CO H: \(\ce{[H3O+]}\) = 1.34 × 10 ; [CH CO H] = 9.866 × 10 ; ClO : [OH ] = 4.0 × 10 ; [HClO] = 2.38 × 10 ; [ClO ] = 0.273 ; HCO H: [HCO H] = 0.524 ; \(\ce{C6H5NH3+ : [C6H5NH3+]}\) = 0.233 ; [C H NH ] = 2.3 × 10 ; From the equilibrium concentrations given, calculate for each of the weak acids and for each of the weak bases. NH : [OH ] = 3.1 × 10 ; [NH ] = 0.533 ; HNO : \(\ce{[H3O+]}\) = 0.011 ; [HNO ] = 1.07 ; (CH ) N: [(CH ) N] = 0.25 ; [OH ] = 4.3 × 10 ; \(\ce{NH4+ : [NH4+]}\) = 0.100 ; [NH ] = 7.5 × 10 ; Determine for the nitrite ion, \(\ce{NO2-}\). In a 0.10- solution this base is 0.0015% ionized. Determine for hydrogen sulfate ion, \(\ce{HSO4-}\). In a 0.10- solution the acid is 29% ionized. \(K_\ce{a}=1.2×10^{−2}\) Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid: Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid: For which of the following solutions must we consider the ionization of water when calculating the pH or pOH? Even though both NH and C H NH are weak bases, NH is a much stronger acid than C H NH . Which of the following is correct at equilibrium for a solution that is initially 0.10 in NH and 0.10 in C H NH ? is the correct statement. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 in HCO H and 0.10 in HClO. Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 in HNO and 0.120 in HBrO. [H O ] = 7.5 × 10 Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 in CH NH and 0.10 in C H N ( = 1.7 × 10 ). Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 in NH and 0.100 in C H NH . [OH ] = \(\ce{[NO4+]}\) = 0.0014 Using the values in , place \(\ce{Al(H2O)6^3+}\) in the correct location in . Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in and . \(\ce{\dfrac{[H3O+,ClO- ]}{[HClO]}}=\dfrac{(x)(x)}{(0.0092−x)}≈\dfrac{(x)(x)}{0.0092}=3.5×10^{−8}\) Solving for gives 1.79 × 10 . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H O ] = [ClO] = 1.8 × 10 [HClO] = 0.00092 [OH ] = 5.6 × 10 ; \(\ce{\dfrac{[C6H5NH3+,OH- ]}{[C6H5NH2]}}=\dfrac{(x)(x)}{(0.0784−x)}≈\dfrac{(x)(x)}{0.0784}=4.6×10^{−10}\) Solving for gives 6.01 × 10 . This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: \(\ce{[CH3CO2- ]}\) = [OH ] = 6.0 × 10 [C H NH ] = 0.00784 [H O ] = 1.7× 10 ; \(\ce{\dfrac{[H3O+,CN- ]}{[HCN]}}=\dfrac{(x)(x)}{(0.0810−x)}≈\dfrac{(x)(x)}{0.0810}=4×10^{−10}\) Solving for gives 5.69 × 10 . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [H O ] = [CN ] = 5.7 × 10 [HCN] = 0.0810 [OH ] = 1.8 × 10 ; \(\ce{\dfrac{[(CH3)3NH+,OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{(0.11−x)}≈\dfrac{(x)(x)}{0.11}=7.4×10^{−5}\) Solving for gives 2.85 × 10 . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [(CH ) NH ] = [OH ] = 2.9 × 10 [(CH ) N] = 0.11 [H O ] = 3.5 × 10 ; \(\ce{\dfrac{[Fe(H2O)5(OH)+,H3O+]}{[Fe(H2O)6^2+]}}=\dfrac{(x)(x)}{(0.120−x)}≈\dfrac{(x)(x)}{0.120}=1.6×10^{−7}\) Solving for gives 1.39 × 10 . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are: [Fe(H O) (OH) ] = [H O ] = 1.4 × 10 \(\ce{[Fe(H2O)6^2+]}\) = 0.120 [OH ] = 7.2 × 10 Propionic acid, C H CO H ( = 1.34 × 10 ), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698- solution of C H CO H? White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm , what is the pH? pH = 2.41 The ionization constant of lactic acid, CH CH(OH)CO H, an acid found in the blood after strenuous exercise, is 1.36 × 10 . If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution? Nicotine, C H N , is a base that will accept two protons ( = 7 × 10 , = 1.4 × 10 ). What is the concentration of each species present in a 0.050- solution of nicotine? [C H N ] = 0.049 The pH of a 0.20- solution of HF is 1.92. Determine for HF from these data. The pH of a 0.15- solution of \(\ce{HSO4-}\) is 1.43. Determine for \(\ce{HSO4-}\) from these data. \(K_\ce{a}=1.2×10^{−2}\) The pH of a 0.10- solution of caffeine is 11.16. Determine for caffeine from these data: The pH of a solution of household ammonia, a 0.950 M solution of NH is 11.612. Determine for NH from these data. \(K_\ce{b}=1.77×10^{−5}\) Determine whether aqueous solutions of the following salts are acidic, basic, or neutral: Determine whether aqueous solutions of the following salts are acidic, basic, or neutral: acidic; basic; acidic; neutral Novocaine, C H O N Cl, is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is 7 × 10 . Is a solution of novocaine acidic or basic? What are [H O ], [OH ], and pH of a 2.0% solution by mass of novocaine, assuming that the density of the solution is 1.0 g/mL. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134- solution of H CO , a diprotic acid: No calculations are needed to answer this question. [H O ] and \(\ce{[HCO3- ]}\) are equal, H O and \(\ce{HCO3-}\) are practically equal Calculate the concentration of each species present in a 0.050- solution of H S. Calculate the concentration of each species present in a 0.010- solution of phthalic acid, C H (CO H) . \(\ce{C6H4(CO2H)2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H4(CO2H)(CO2)-}(aq) \hspace{20px} K_\ce{a}=1.1×10^{−3}\) \(\ce{C6H4(CO2H)(CO2)}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{C6H4(CO2)2^2-}(aq) \hspace{20px} K_\ce{a}=3.9×10^{−6}\) [C H (CO H) ] 7.2 × 10 , [C H (CO H)(CO ) ] = [H O ] 2.8 × 10 , \(\ce{[C6H4(CO2)2^2- ]}\)3.9 × 10 , [OH ] 3.6 × 10 Salicylic acid, HOC H CO H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838. \[\ce{CH3CO2C6H4CO2H}(aq)+\ce{H2O}(l)⟶\ce{HOC6H4CO2H}(aq)+\ce{CH3CO2H}(aq)\] The ion HTe is an amphiprotic species; it can act as either an acid or a base. Explain why a buffer can be prepared from a mixture of NH Cl and NaOH but not from NH and NaOH. Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H PO and a salt of its conjugate base NaH PO . Excess H O is removed primarily by the reaction: \(\ce{H3O+}(aq)+\ce{H2PO4-}(aq)⟶\ce{H3PO4}(aq)+\ce{H2O}(l)\) Excess base is removed by the reaction: \(\ce{OH-}(aq)+\ce{H3PO4}(aq)⟶\ce{H2PO4-}(aq)+\ce{H2O}(l)\) Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH and a salt of its conjugate acid NH Cl. What is [H O ] in a solution of 0.25 CH CO H and 0.030 NaCH CO ? \(\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5}\) [H O ] = 1.5 × 10 What is [H O ] in a solution of 0.075 HNO and 0.030 NaNO ? What is [OH ] in a solution of 0.125 CH NH and 0.130 CH NH Cl? [OH ] = 4.2 × 10 What is [OH ] in a solution of 1.25 NH and 0.78 NH NO ? What concentration of NH NO is required to make [OH ] = 1.0 × 10 in a 0.200- solution of NH ? [NH NO ] = 0.36 What concentration of NaF is required to make [H O ] = 2.3 × 10 in a 0.300- solution of HF? What is the effect on the concentration of acetic acid, hydronium ion, and acetate ion when the following are added to an acidic buffer solution of equal concentrations of acetic acid and sodium acetate: What is the effect on the concentration of ammonia, hydroxide ion, and ammonium ion when the following are added to a basic buffer solution of equal concentrations of ammonia and ammonium nitrate: What will be the pH of a buffer solution prepared from 0.20 mol NH , 0.40 mol NH NO , and just enough water to give 1.00 L of solution? pH = 8.95 Calculate the pH of a buffer solution prepared from 0.155 mol of phosphoric acid, 0.250 mole of KH PO , and enough water to make 0.500 L of solution. How much solid NaCH CO •3H O must be added to 0.300 L of a 0.50- acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.) 37 g (0.27 mol) What mass of NH Cl must be added to 0.750 L of a 0.100- solution of NH to give a buffer solution with a pH of 9.26? (Hint: Assume a negligible change in volume as the solid is added.) A buffer solution is prepared from equal volumes of 0.200 acetic acid and 0.600 sodium acetate. Use 1.80 × 10 as for acetic acid. What is the pH of a solution that results when 3.00 mL of 0.034 HCl is added to 0.200 L of the original buffer? A 5.36–g sample of NH Cl was added to 25.0 mL of 1.00 NaOH and the resulting solution diluted to 0.100 L. Which acid in is most appropriate for preparation of a buffer solution with a pH of 3.1? Explain your choice. To prepare the best buffer for a weak acid HA and its salt, the ratio \(\dfrac{\ce{[H3O+]}}{K_\ce{a}}\) should be as close to 1 as possible for effective buffer action. The [H O ] concentration in a buffer of pH 3.1 is [H O ] = 10 = 7.94 × 10 \(\dfrac{\ce{[H3O+]}}{K_\ce{a}}=1\) \(K_\ce{a}=\dfrac{\ce{[H3O+]}}{1}=7.94×10^{−4}\) Which acid in is most appropriate for preparation of a buffer solution with a pH of 3.7? Explain your choice. Which base in is most appropriate for preparation of a buffer solution with a pH of 10.65? Explain your choice. For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio \(\dfrac{\ce{[OH- ]}}{K_\ce{b}}\) that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH ] is [OH ] = 10 = 10 = 4.467 × 10 . Which base in is most appropriate for preparation of a buffer solution with a pH of 9.20? Explain your choice. Saccharin, C H NSO H, is a weak acid ( = 2.1 × 10 ). If 0.250 L of diet cola with a buffered pH of 5.48 was prepared from 2.00 × 10 g of sodium saccharide, Na(C H NSO ), what are the final concentrations of saccharine and sodium saccharide in the solution? The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is: What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C H NO , a diprotic acid; = 8.5 × 10 , = 3.39 × 10 ) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added? Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid. At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example. Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH. Why can we ignore the contribution of water to the concentrations of H O in the solutions of following acids: but not the contribution of water to the concentration of OH ? In an acid solution, the only source of OH ions is water. We use K to calculate the concentration. If the contribution from water was neglected, the concentration of OH would be zero. We can ignore the contribution of water to the concentration of OH in a solution of the following bases: 0.0784 C H NH , a weak base 0.11 (CH ) N, a weak base but not the contribution of water to the concentration of H O ? Draw a curve for a series of solutions of HF. Plot [H O ] on the vertical axis and the total concentration of HF (the sum of the concentrations of both the ionized and nonionized HF molecules) on the horizontal axis. Let the total concentration of HF vary from 1 × 10 to 1 × 10 . Draw a curve similar to that shown in for a series of solutions of NH . Plot [OH ] on the vertical axis and the total concentration of NH (both ionized and nonionized NH molecules) on the horizontal axis. Let the total concentration of NH vary from 1 × 10 to 1 × 10 . Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 barbituric acid ( = 9.8 × 10 ) with 0.100 KOH. The indicator dinitrophenol is an acid with a of 1.1 × 10 . In a 1.0 × 10 - solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).

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This page describes ways of hydrolyzing esters - splitting them into carboxylic acids (or their salts) and alcohols by the action of water, dilute acid or dilute alkali. It starts by looking at the hydrolysis of simple esters like ethyl ethanoate, and goes on to look at hydrolyzing bigger, more complicated ones to make soap. Technically, hydrolysis is a reaction with water. That is exactly what happens when esters are hydrolyzed by water or by dilute acids such as dilute hydrochloric acid. The alkaline hydrolysis of esters actually involves reaction with hydroxide ions, but the overall result is so similar that it is lumped together with the other two. The reaction with pure water is so slow that it is never used. The reaction is catalyzed by dilute acid, and so the ester is heated under reflux with a dilute acid like dilute hydrochloric acid or dilute sulfuric acid. Here are two simple examples of hydrolysis using an acid catalyst. First, hydrolyzing ethyl ethanoate: . . . and then hydrolyzing methyl propanoate: Notice that the reactions are reversible. To make the hydrolysis as complete as possible, you would have to use an excess of water. The water comes from the dilute acid, and so you would mix the ester with an excess of dilute acid. This is the usual way of hydrolyzing esters. The ester is heated under reflux with a dilute alkali like sodium hydroxide solution. There are two advantages of doing this rather than using a dilute acid. The reactions are one-way rather than reversible, and the products are easier to separate. Taking the same esters as above, but using sodium hydroxide solution rather than a dilute acid: First, hydrolyzing ethyl ethanoate using sodium hydroxide solution: and then hydrolyzing methyl propanoate in the same way: Notice that you get the sodium salt formed rather than the carboxylic acid itself. This mixture is relatively easy to separate. Provided you use an excess of sodium hydroxide solution, there will not be any ester left - so you don't have to worry about that. The alcohol formed can be off. That's easy! If you want the acid rather than its salt, all you have to do is to add an excess of a strong acid like dilute hydrochloric acid or dilute sulfuric acid to the solution left after the first distillation. If you do this, the mixture is flooded with hydrogen ions. These are picked up by the ethanoate ions (or propanoate ions or whatever) present in the salts to make ethanoic acid (or propanoic acid, etc). Because these are weak acids, once they combine with the hydrogen ions, they tend to stay combined. The carboxylic acid can now be distilled off. This next bit deals with the alkaline hydrolysis (using sodium hydroxide solution) of the big esters found in animal and vegetable fats and oils. If the large esters present in animal or vegetable fats and oils are heated with concentrated sodium hydroxide solution exactly the same reaction happens as with the simple esters. A salt of a carboxylic acid is formed - in this case, the sodium salt of a big acid such as octadecanoic acid (stearic acid). These salts are the important ingredients of soap - the ones that do the cleaning. An alcohol is also produced - in this case, the more complicated alcohol, propane-1,2,3-triol (glycerol). Because of its relationship with soap making, the alkaline hydrolysis of esters is sometimes known as . Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photosynthesis/Photosystem_II/Photosystem_II_3

Photosystem II is crucial to life as we know it. This process is the only natural process capable of forming O2 from water and sunlight (Siegbahn, 2009).This capability is used to convert light energy to chemical energy in plants. This process also provides the energy to drive the conversion of carbon dioxide into the oxygen critical to animal life. Animals also derive benefit from chemical energy stored in the form of sugars within the cells of the plants through their consumption. This may also occur indirectly via the consumption of other animals who have previously consumed these plants. The overall process is divided into light reactions which need light to occur, and dark reactions, also known as the Calvin Cycle, which do not directly need light. (Campbell, 2005) Electrons and protons are passed from the light reactions to NADPH which allows for their transport to the Calvin Cycle. (Campbell, 2005). At the heart of this indispensable system is the protein p680 which drives the light reaction, and thereby the entirety of the process (Campbell, 2005). At the heart of p680 is the Oxygen Evolving Complex (OEC), shown below, which is a cluster of inorganic atoms. The precise form of the OEC continues to elude researchers, and many models have been proposed. The Dangler Model (Siegbahn, 2009) has been selected for this assignment because it gives the best opportunity to delve into the symmetrical aspects of the OEC. This model of the OEC consists of four manganese atoms, four oxygen atoms, and one calcium atom. A chlorine atom becomes involved in the reactions but is not directly a part of this cluster. (Vrettos, 2002) Figure 1: Oxygen Evolving Complex. (simplified† from distorted cubic layout) Seeing the oxygen evolving complex arranged in the manner of the Dangler Model provides an excellent opportunity to assess the symmetry of this structure, at least locally. Knowledge of the symmetry is of importance when predicting or interpreting the results of IR and Raman Spectroscopy. Using the simplified depiction of the OEC in figure 1 it can be seen that the cluster is of a cubic shape with one magnesium stretching away from a corner (Siegbahn, 2009) the symmetry is as follows: The complex without the blue Mn would be C3v With the blue Mn the complex is C1 The entire protein also has C1 symmetry Through knowing the structure and employing the character tables, the IR and Raman spectroscopic results can be predicted for this complex. In the event that the simplified version of the distorted cubic structure were are to be the precise layout of the OEC, we could can expect to see the following. Through an understanding of the basics of inorganic chemistry and looking at ongoing research we can see the pivotal role inorganic chemistry plays in the chemistry of life. Looking at a periodic table with each atom marked as to whether it is essential to all investigated species yields 5 elements in the which reside in the d-block, metals. The same table shows an additional 5 metals being essential to at least one biological species (Schore, 2007) Through this example, and likely many, many more, it becomes obvious that there is no distinct chasm that inorganic and organic chemistry can not bridge. Life is a powerful and diverse process rooted in chemistry and as such takes advantage of opportunities not provided by the typical carbon, hydrogen, oxygen, nitrogen arsenal of organic chemistry. †: The Dangler Model does not predict all atoms for the OEC to be equidistant, not for all the angles to perfectly fit the 90º required of a perfect cube. It can be described as distorted cubic. By representing it as a simple cube, there is access to a greater variety of symmetries to evaluate. The image is a simplified depiction of that found in the Siegbahn paper in the references section.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.11%3A_Real_and_Ideal_Gases

The behavior of a molecule depends a lot on its structure. Two compounds with the same number of atoms can act very differently. Ethanol \(\left( \ce{C_2H_5OH} \right)\) is a clear liquid that has a boiling point of about \(79^\text{o} \text{C}\). Dimethylether \(\left( \ce{CH_3OCH_3} \right)\) has the same number of carbons, hydrogens, and oxygens, but boils at a much lower temperature \(\left( -25^\text{o} \text{C} \right)\). The difference lies in the amount of intermolecular interaction (strong \(\ce{H}\)-bonds for ethanol, weak van der Waals force for the ether). An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas needs to completely abide by the kinetic-molecular theory. The gas particles need to occupy zero volume and they need to exhibit no attractive forces whatsoever toward each other. Since neither of those conditions can be true, there is no such thing as an ideal gas. A is a gas that does not behave according to the assumptions of the kinetic-molecular theory. Fortunately, at the conditions of temperature and pressure that are normally encountered in a laboratory, real gases tend to behave very much like ideal gases. Under what conditions then, do gases behave least ideally? When a gas is put under high pressure, its molecules are forced closer together as the empty space between the particles is diminished. A decrease in the empty space means that the assumption that the volume of the particles themselves is negligible is less valid. When a gas is cooled, the decrease in kinetic energy of the particles causes them to slow down. If the particles are moving at slower speeds, the attractive forces between them are more prominent. Another way to view it is that continued cooling of the gas will eventually turn it into a liquid and a liquid is certainly not an ideal gas anymore (see liquid nitrogen in the figure below). In summary, a real gas deviates most from an ideal gas at low temperatures and high pressures. Gases are most ideal at high temperature and low pressure. The figure below shows a graph of \(\frac{PV}{RT}\) plotted against pressure for \(1 \: \text{mol}\) of a gas at three different temperatures—\(200 \: \text{K}\), \(500 \: \text{K}\), and 1000 \: \text{K}\). An ideal gas would have a value of 1 for that ratio at all temperatures and pressures, and the graph would simply be a horizontal line. As can be seen, deviations from an ideal gas occur. As the pressure begins to rise, the attractive forces cause the volume of the gas to be less than expected and the value of \(\frac{PV}{RT}\) drops under 1. Continued pressure increase results in the volume of the particles to become significant and the value of \(\frac{PV}{RT}\) rises to greater than 1. Notice that the magnitude of the deviations from ideality is greatest for the gas at \(200 \: \text{K}\) and least for the gas at \(1000 \: \text{K}\). The ideality of a gas also depends on the strength and type of intermolecular attractive forces that exist between the particles. Gases whose attractive forces are weak are more ideal than those with strong attractive forces. At the same temperature and pressure, neon is more ideal than water vapor because neon's atoms are only attracted by weak dispersion forces, while water vapor's molecules are attracted by relatively strong hydrogen bonds. Helium is a more ideal gas than neon because its smaller number of electrons means that helium's dispersion forces are even weaker than those of neon.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.05%3A_Kinetics_of_Reactions_in_Solution

Make sure you thoroughly understand the following essential ideas: The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions. Most of the added complications of kinetics and rate processes in liquid solutions arise from the of the liquid phase. In a typical gas at atmospheric pressure, the molecules occupy only about 0.2 per cent of the volume; the other 99.8 percent is empty space. In a liquid, molecules may take up more than half the volume, and the "empty" spaces are irregular and ever-changing as the solvent molecules undergo thermal motions of their own. In a typical liquid solution, the solvent molecules massively outnumber the reactant solute molecules, which tend to find themselves momentarily (~10 sec) confined to a "hole" within the liquid. This trapping becomes especially important when the solvent is strongly hydrogen-bonded as is the case with water or alcohol. When thermal motions occasionally release a solute molecule from this trap, it will jump to a new location. The jumps are very fast (10 - 10 sec) and short (usually a few solvent-molecule diameters), and follow an entirely random pattern, very much as in . Consider a simple bimolecular process A + B → products. The reactant molecules will generally be jumping from hole to hole in the solvent matrix, only occasionally finding themselves in the same where thermal motions are likely to bring them into contact. The process can be represented as \[A + B \rightarrow \{AB\} → \text{products}\] in which the \(\{AB\}\) term represents the caged reactants including the and the activated complex. Contrast this scenario with a similar reaction taking place in the gas phase; the molecules involved in the reaction will often be the only ones present, so a significant proportion of the collisions will be \(A\)-\(B\) encounters. However, if the collision should fail to be energetically or geometrically viable, the reactant molecules fly apart and are unlikely to meet again anytime soon. In a liquid, however, the solute molecules are effectively in a constant state of collision — if not with other reactants, then with solvent molecules which can exchange kinetic energy with the reactants. So once an A-B encounter pair forms, the two reactants get multiple whacks at each other, greatly increasing the probability that they will obtain the kinetic energy needed to kick them over the activation hump before the encounter pair disintegrates. The encounter pair model introduces some new rate parameters: \[\ce{A + B <=>[k1,k_{-1}] {AB} -> products}\] The first step is an equilibrium between the reactants outside and inside the solvent cage. The rate constants \(k_1\) and \(k_2\) reflect those relating to diffusion of molecules through the solvent; their values are strongly dependent on the viscosity (and thus the temperature) of the solvent. (Note that \(k_1\) is a second-order rate constant, while \(k_2\) is first-order.) is the transport of a substance through a concentration gradient; that is, from a region of higher concentration to one of lower concentration. Think of the way the color of tea spreads out when a tea bag is immersed in hot water. Diffusion occurs because random thermal motions are statistically more likely to move molecules out of a region of higher concentration than in the reverse direction, simply because in the latter case fewer molecules are available to make the reverse trip. Eventually the concentrations become uniform and equilibrium is attained. As molecules diffuse through a liquid, they must nudge neighboring molecules out of the way. The work required to do this amounts to an activation energy, so diffusion can be thought of as a kinetic process with its own rate constant and activation energy. These parameters depend on the sizes of the solute and solvent molecules and on how strongly the latter interact with each other. This suggests two important for reactions in solution. For water at room temperature, is typically 10 -10 dm mol s and is around 10 -10 dm mol s . Given these values, > 10 s implies diffusion control, while values < 10 s are indicative of activation control. Several general kinds of reactions are consistently very "fast" and thus are commonly found to be diffusion-controlled in most solvents: Gas-phase rate constants are normally expressed in units of mol s , but rate constants of reactions in solution are conventionally given in mol/L units, or dm mol s . Conversion between them depends on a number of assumptions and is non-trivial. For example the formation of I from I atoms in hexane at 298 K has = 1.3×10 dm mol s . The most famous of these is one of the fastest reactions known:\[H^+ + OH^– → H_2O \nonumber \] for which = 1.4×10 dm mol s at 298 K. Polar solvents such as water and alcohols interact with ions and polar molecules through attractive dipole-dipole and ion-dipole interactions, leading to lower-energy solvated forms which stabilize these species. In this way, a polar solvent can alter both the thermodynamics and kinetics (rate) of a reaction. If the products of the reaction are markedly more or less polar than the reactants, solvent polarity can change the overal thermodynamics (equilibrium constant) of the reaction. Nowhere is this more apparent than when an ionic solid such as salt dissolves in water. The Na and Cl ions are bound together in the solid through strong coulombic forces; pulling the solid apart in a vacuum or in a nonpolar solvent is a highly endothermic process. In contrast, dissolution of NaCl in water is slightly exothermic and proceeds spontaneously. The water facilitates this process in two important ways. First, its high dielectric constant of 80 reduces the force between the separated ions to 1/80 of its normal value. Secondly, the water molecules form a around the ions (lower left), rendering them energetically (thermodynamically) more stable than they were in the NaCl solid. In the same way, a reaction whose mechanism involves the formation of an intermediate or activated complex having a polar or ionic character will have its activation energy, and thus its rate, subject to change as the solvent polarity is altered. As an example we will consider an important class of reactions that you will hear much about if you take a course in organic chemistry. When an aqueous solution of a strong base such as KOH is added to a solution of -butyl chloride in ethanol, the chlorine is replaced by a hydroxyl group, leaving -butyl alcohol as a product: This reaction is one of a large and important class known as that are discussed in most organic chemistry courses. In these reactions, a species that possesses a pair of non-bonding electrons (also called a or ) uses them to form a new bond with an — a compound in which a carbon atom has a partial positive charge owing to its bonds to electron-withdrawing groups. In the example here, other nucleophiles such as NH or even H O would serve as well. In order to reflect the generality of this process and to focus on the major changes that take place, we will represent this reaction as Extensive studies of this class of reactions in the 1930's revealed that it proceeds in two activation energy-controlled steps, followed by a simple dissociation into the products: In step , which is rate-determining, the chlorine leaves the alkyl chloride which becomes an intermediate known as a ("cat-ion"). These ions, in which the central carbon atom lacks a complete octet, are highly reactive, and in step the carbocation is attacked by the hydroxide ion which supplies the missing electron. The immediate product is another cation in which the positive charge is on the oxygen atom. This is unstable and rapidly dissociates ( )into the alcohol and a hydrogen ion. The reaction coordinate diagram helps us understand the effect of solvent polarity on this reaction. Polar solvent molecules interact most strongly with species in which the electric charge is concentrated in one spot. Thus the carbocation is stabilized to a greater degree than are the activated complexes in which the charge is spread out between the positive and negative ends. As the heavy green arrows indicate, a more polar solvent will stabilize the carbocation more than it will either of the activated complexes; the effect is to materially reduce the activation energy of the rate-determining step, and thus speed up the reaction. Because neither the alkyl chloride nor the alcohol is charged, the change in solvent polarity has no effect on the equilibrium constant of the reaction. This is dramatically illustrated by observing the rate of the reaction in solvents composed of ethanol and water in varying amounts:

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.07%3A_Successive_Approximation

An approximation is often useful even when it is not a very good one, because we can use the initial inaccurate approximation to calculate a better one. A good example of this occurs in where it was necessary to solve the equation \[\text{0}\text{.0200}=\frac{\text{(1 + }x\text{)}x}{\text{(1 }-\text{ 2}x\text{)}^{\text{2}}} \nonumber \] The conditions of the problem suggest that is so small that it can be ignored. 1. Accordingly we can approximate \[ 1 + x \approx 1 \approx 1 – 2x \nonumber \] from which we obtain the approximate result (called the first approximation), : \[\text{0}\text{.0200 }\approx \text{ }\frac{\text{1}x_{\text{1}}}{\text{1}^{\text{2}}} ~~~ \text{or} ~~~ x_{1} \approx 0.0200 \nonumber \] Although for some purposes this is a sufficiently accurate result, a much better approximation can be obtained by feeding this one back into the formula. If we write the formula as \[x=\frac{\text{0}\text{.0200 (1 - 2}x\text{)}^{\text{2}}}{\text{1 + }x} \nonumber \] we can now substitute = 0.0200 on the right-hand side, giving the second approximation: \[x_{\text{2}}=\frac{\text{0}\text{.0200 (1 - 2 }\times \text{ 0}\text{.0200)}^{\text{2}}}{\text{1 + 0}\text{.0200}}=\frac{\text{0}\text{.0200 }\times \text{ 0}\text{.96}^{\text{2}}}{\text{1}\text{.0200}}=\text{0}\text{.0181} \nonumber \] If we repeat this process, a third approximation is obtained: \[ x_{3} \approx 0.0182 \nonumber \] in exact agreement with the accurate result obtained from the quadratic formula in the example. With practice, using this method of successive approximations is much faster than using the quadratic formula. It also has the advantage of being self-checking. A mistake in any of the calculations almost always leads to an obviously worse approximation. In general, if the last approximation for differs from the next to last by less than 5 percent, it can be assumed to be accurate, and the successive-approximation procedure can be stopped. In the example just given, \[\frac{x_{\text{2}}\text{ }-\text{ }x_{\text{1}}}{x_{\text{1}}}=\frac{\text{0}\text{.0181 - 0}\text{.0200}}{\text{0}\text{.0200}}=\frac{-\text{0}\text{.0019}}{\text{0}\text{.0200}}\approx -\text{0}\text{.10}=-\text{10 percent} \nonumber \] so a third approximation was calculated. This third approximation was almost identical to the second and so was taken as the final result.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity/Aromaticity

The adjective "aromatic" is used by organic chemists in a rather different way than it is normally applied. It has its origin in the observation that certain natural substances, such as cinnamon bark, wintergreen leaves, vanilla beans and anise seeds, contained fragrant compounds having common but unexpected properties. Cinnamon bark, for example, yielded a pleasant smelling compound, formula C H O, named cinnamaldehyde. Because of the low hydrogen to carbon ratio in this and other aromatic compounds (note that the H:C ratio in an alkane is >2), chemists expected their structural formulas would contain a large number of double or triple bonds. Since double bonds are easily cleaved by oxidative reagents such as potassium permanganate or ozone, and rapidly add bromine and chlorine, these reactions were applied to these aromatic compounds. Surprisingly, products that appeared to retain many of the double bonds were obtained, and these compounds exhibited a high degree of chemical stability compared with known alkenes and cycloalkenes ( ). On treatment with hot permanganate solution, cinnamaldehyde gave a stable, crystalline C H O compound, now called benzoic acid. The H:C ratio in benzoic acid is <1, again suggesting the presence of several double bonds. Benzoic acid was eventually converted to the stable hydrocarbon benzene, C H , which also proved unreactive to common double bond transformations, as shown below. For comparison, reactions of cyclohexene, a typical alkene, with these reagents are also shown (green box). As experimental evidence for a wide assortment of compounds was acquired, those incorporating this exceptionally stable six-carbon core came to be called "aromatic".

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.03%3A_The_Equilibrium_Constant

The constancy of the ratio of of one isomer to the of the other at a given temperature is characteristic of all gaseous equilibria between isomers, i.e.,of all reactions of the general type \[\text{A}\text{ }({g})\rightleftharpoons \text{B}\text{ }({g})\label{1} \] The constant ratio of concentrations is called the and is given the symbol \(K_c\). For reactions of the type given by Equation \(\ref{1}\) the equilibrium constant is thus described by the equation \[K_{c}=\frac{[\text{ B }]}{[\text{ A }]}\label{2} \] where, by convention, the concentration of the product B appears in the numerator of the ratio. If, for some reason, we wish to look at this reaction in reverse, \[\text{B}\text{ }({g})\rightleftharpoons \text{A}\text{ }({g}) \nonumber \] then the equilibrium constant is denoted as the reciprocal of the constant given in Equation \(\ref{2}\). \[K_{c}=\frac{[\text{ A }]}{[\text{ B }]} \nonumber \] In general the equilibrium constant \(K_c\) and also differs based on the substances involved. Examples illustrating this behavior are given in Table where the experimentally determined equilibrium constants for various cis-trans isomerization equilibria are recorded at various temperatures. Note that the equilibrium changes as the temperature and the composition of the molecule changes. When we turn our attention to more complex equilibrium reactions, we find that the relationship between the concentrations of the various species is no longer a simple ratio. A good demonstration of this fact is provided by the dissociation of dinitrogen tetroxide, N O . This compound is a colorless gas, but even at room temperature it dissociates partly into a vivid red-brown gas, NO , according to the equation \[\text{N}_{2}\text{O}_{4}\text{ }({g})\rightleftharpoons \text{2NO}_{2} \text{ }({g})\label{5} \] If 1 mol N O contained in a flask of volume 1 L is heated to 407.2 K, exactly one-half of it dissociates into NO . If the volume is now increased, the ratio of [NO ] to [N O ] does not remain constant but increases as more dissociates. As shown in Table , if we increase the volume still further, even more dissociation occurs. By the time we have increased the volume to 10 L, the fraction of N O molecules dissociated has increased to 0.854 (i.e., to 85.4 percent). Obviously the situation is now not quite so straightforward as in the previous example. Nevertheless there is a simple relationship between the equilibrium concentrations of the reactant and product in this case too. We find that it is the quantity \[\frac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]} \nonumber \] rather than the simple ratio of concentrations, which is now constant. Accordingly we also call this quantity an equilibrium constant and give it the symbol . Thus for Equation \(\ref{5}\) is given by the relationship \[K_{c}=\frac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]} \nonumber \] where again by convention the product appears in the numerator. It is easy to check that actually is a constant quantity with the value 2.00 mol/L from the data given in Table . Thus if we take the result from line , we find that when 1 mol N O is placed in a 10 L flask at 407 K, 0.854 mol dissociate. Since from Equation \(\ref{5}\) each mole which dissociates yields 2 mol NO ,there will be \[\text{0.854 mol N}_{2}\text{O}_{4}\times \frac{\text{2 mol NO}_{2}}{\text{1 mol N}_{2}\text{O}_{4}}=\text{1.708 mol NO}_{2} \nonumber \] present in the reaction vessel. There will also be (1 – 0.854) mol = 0.146 mol N O left undissociated in the flask. Since the total volume is 10 L, the equilibrium concentrations are \[[\text{ NO}_{2}]=\frac{\text{1.708 mol}}{\text{10 L}}=\text{0.1708 mol/L} \nonumber \] and \[[\text{ N}_{2}\text{O}_{4}]=\frac{\text{0.146 mol}}{\text{10 L}}=\text{0.0146 mol/L} \nonumber \] Accordingly \[K_{c}=\frac{\text{1.708 mol/L}\times \text{ 0.1708 mol/L}}{\text{0.0146 mol/L}}=\text{2.00 mol/L} \nonumber \] In exactly the same way, if we use the data from line a in Table , we find \[K_{c}=\frac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]}=\frac{\text{(1.00 mol/L}\text{)}^{2}}{\text{0.5 mol/L}}=\text{2.00 mol/L} \nonumber \] You can check for yourself that lines and also yield the same value for . When 2 mol N O gas is heated to 407 K in a vessel of volume 5 dm , it is found that 0.656 of the molecules dissociate into NO . Show that these data are in agreement with the value for of 2.00 mol dm given in the text. Many equilibrium problems can be solved in a fairly standardized fashion in three stages. Calculate the amount of each substance transformed by the reaction as it comes to equilibrium, i.e., the amount of each reactant by the reaction and the amount of each product by the reaction. Stoichiometric ratios derived from the equation must always be used in these calculations. In this particular example we note that 0.656 of the original N O dissociates. Since 2 mol was used, a total of 0.656 × 2 = 1.312 mol N O is . The amount of NO is accordingly \(n_{\text{NO}_{2}}=\text{1.312 mol N}_{2}\text{O}_{4}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol N}_{2}\text{O}_{4}}=\text{2.624 mol NO}_{2}\) Use the amounts calculated in the first stage to calculate the amount of each substance present at equilibrium. Dividing by the volume, we can obtain the equilibrium concentrations. Since 1.312 mol N O dissociated out of an original 2 mol, we have (2 – 1.312) mol = 0.688 mol N O left. The equilibrium concentration of N O is thus \([\text{ N}_{2}\text{O}_{4}] = \dfrac{0.688\text{ mol N}_{2}\text{O}_{4}}{\text{ 5.00 dm}^{3}} = \text{0.1376 mol dm}^{-3}\) Since no NO was originally present, the amount of NO present at equilibrium is the amount produced by the dissociation, namely, 2.624 mol NO . Thus \([\text{ NO}_{2}]=\dfrac{\text{2.624 mol NO}_{2}}{\text{5.00 dm}^{3}}=\text{0.525 mol dm}^{-3}\) It is usually worthwhile tabulating these calculations, particularly in more complex examples. Note that a quantity in the column headed Amount Produced indicates that a given substance (such as N O in this example) has been . There is less of that substance when equilibrium is reached than was present initially. In the third stage we insert the equilibrium concentrations in an expression for the equilibrium constant: \(K_{c}=\dfrac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]}=\dfrac{\text{0.525 mol dm}^{-3}\times \text{ 0.525 mol dm}^{-3}}{\text{0.1376 mol dm}^{-3}}=\text{2.00 mol dm}^{-3}\)

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/05%3A_Gases/5.06_The_Kinetic_Molecular_Theory_of_Gases

The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles. The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates: Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8, we explain how this theory must be modified to account for the behavior of real gases. Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a (impulse) from molecules to the walls (Figure \(\Page {2}\)). The to the wall perpendicular to \(x\) axis as a molecule with an initial velocity \(u_x\) in \(x\) direction hits is expressed as: \[\Delta p_x=2mu_x \label{6.7.1}\] The , a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume. \[f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{6.7.2}\] The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency. \[P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{6.7.3}\] At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace \(u_x^2\) in the expression above with the average value of \(u_x^2\), which is denoted by \(\overline{u_x^2}\). The overbar designates the average value over all molecules. The exact expression for pressure is given as : \[P=\dfrac{N}{V}m\overline{u_x^2} \label{6.7.4}\] Finally, we must consider that there is nothing special about \(x\) direction. We should expect that \(\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}\). Here the quantity \(\overline{u^2}\) is called the defined as the average value of square-speed (\(u^2\)) over all molecules. Since \(u^2=u_x^2+u_y^2+u_z^2\) for each molecule, \(\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}\). By substituting \(\dfrac{1}{3}\overline{u^2}\) for \(\overline{u_x^2}\) in the expression above, we can get the final expression for the pressure: \[P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{6.7.5}\] Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules. Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the kinetic energy of the molecules of a gas \((\overline{e_K})\), which can be represented as \((T)\) \[\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{6.7.6}\] where \(N_A\) is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by \(N_A\): \[N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{6.7.7}\] where \(M\) is the molar mass of the gas molecules and is related to the molecular mass by \(M=N_Am\). By rearranging the equation, we can get the relationship between the root-mean square speed (\(u_{\rm rms}\)) and the temperature. The rms speed (\(u_{\rm rms}\)) is the square root of the sum of the squared speeds divided by the number of particles: \[u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{6.7.8}\] where \(N\) is the number of particles and \(u_i\) is the speed of particle \(i\). The relationship between \(u_{\rm rms}\) and the temperature is given by: \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{6.7.9}\] In this equation, \(u_{\rm rms}\) has units of meters per second; consequently, the units of molar mass \(M\) are kilograms per mole, temperature \(T\) is expressed in kelvins, and the ideal gas constant \(R\) has the value 8.3145 J/(K•mol). The equation shows that \(u_{\rm rms}\) of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones. At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed. The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed (\(v_{\rm av}\)) root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)). particle speeds average speed (\(v_{\rm av}\)), root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)) Use Equation 6.7.6 to calculate the average speed and Equation 6.7.8 to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving. The average speed is the sum of the speeds divided by the number of particles: \[v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s\] The rms speed is the square root of the sum of the squared speeds divided by the number of particles: \[v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s\] The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence \(v_{\rm m}=6.0\) m/s. The \(v_{\rm rms}\) of the particles, which is related to the average kinetic energy, is greater than their average speed. At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction, as you will learn in Chapter 14 "Chemical Kinetics". This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure \(\Page {1}\). Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure \(\Page {1}\) were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases. We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. The temperature of a 4.75 L container of N gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the temperatures and volume effect of increase in temperature Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas. A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the a. no change; b. no change; c. no change; d. no change; e. decreases; f. decreases; g. decreases Kinetic-Molecular Theory of Gases: The behavior of ideal gases is explained by the . Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same . The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a , in which some molecules have higher or lower speeds (and kinetic energies) than average.

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Precipitation and volatilization gravimetric methods require that the analyte, or some other species in the sample, participates in a chemical reaction. In a direct precipitation gravimetric analysis, for example, we convert a soluble analyte into an insoluble form that precipitates from solution. In some situations, however, the analyte already is present in a particulate form that is easy to separate from its liquid, gas, or solid matrix. When such a separation is possible, we can determine the analyte’s mass without relying on a chemical reaction. A particulate is any tiny portion of matter, whether it is a speck of dust, a globule of fat, or a molecule of ammonia. For particulate gravimetry we simply need a method to collect the particles and a balance to measure their mass. There are two methods for separating a particulate analyte from its matrix. The most common method is filtration, in which we separate solid particulates from their gas, liquid, or solid matrix. A second method, which is useful for gas particles, solutes, and solids, is an extraction. To separate solid particulates from their matrix we use gravity or apply suction from a vacuum pump or an aspirator to pull the sample through a filter. The type of filter we use depends upon the size of the solid particles and the sample’s matrix. Filters for liquid samples are constructed from a variety of materials, including cellulose fibers, glass fibers, cellulose nitrate, and polytetrafluoroethylene (PTFE). Particle retention depends on the size of the filter’s pores. Cellulose fiber filter papers range in pore size from 30 μm to 2–3 μm. Glass fiber filters, manufactured using chemically inert borosilicate glass, are available with pore sizes between 2.5 μm and 0.3 μm. Membrane filters, which are made from a variety of materials, including cellulose nitrate and PTFE, are available with pore sizes from 5.0 μm to 0.1 μm. For additional information, see our earlier discussion in this chapter on filtering precipitates, and the discussion in of separations based on size. Solid aerosol particulates are collected using either a single-stage or a multiple-stage filter. In a single-stage system, we pull the gas through a single filter, which retains particles larger than the filter’s pore size. To collect samples from a gas line, we place the filter directly in the line. Atmospheric gases are sampled with a high volume sampler that uses a vacuum pump to pull air through the filter at a rate of approximately 75 m /h. In either case, we can use the same filtering media for liquid samples to collect aerosol particulates. In a multiple-stage system, a series of filtering units separates the particles into two or more size ranges. The particulates in a solid matrix are separated by size using one or more sieves (Figure 8.4.1 ). Sieves are available in a variety of mesh sizes, ranging from approximately 25 mm to 40 μm. By stacking together sieves of different mesh size, we can isolate particulates into several narrow size ranges. Using the sieves in Figure 8.4.1 , for example, we can separate a solid into particles with diameters >1700 μm, with diameters between 1700 μm and 500 μm, with diameters between 500 μm and 250 μm, and those with a diameter <250 μm. Filtering limits particulate gravimetry to solid analytes that are easy to separate from their matrix. We can extend particulate gravimetry to the analysis of gas phase analytes, solutes, and solids that are difficult to filter if we extract them with a suitable solvent. After the extraction, we evaporate the solvent and determine the analyte’s mass. Alternatively, we can determine the analyte indirectly by measuring the change in the sample’s mass after we extract the analyte. For a more detailed review of extractions, particularly solid-phase extractions, see . Another method for extracting an analyte from its matrix is by adsorption onto a solid substrate, by absorption into a thin polymer film or chemical film coated on a solid substrate, or by chemically binding to a suitable receptor that is covalently bound to a solid substrate (Figure 8.4.2 ). Adsorption, absorption, and binding occur at the interface between the solution that contains the analyte and the substrate’s surface, the thin film, or the receptor. Although the amount of extracted analyte is too small to measure using a conventional balance, it can be measured using a quartz crystal microbalance. The measurement of mass using a takes advantage of the piezoelectric effect [(a) Ward, M. D.; Buttry, D. A. , , 1000–1007; (b) Grate, J. W.; Martin, S. J. ; White, R. M. , , 940A–948A; (c) Grate, J. W.; Martin, S. J. ; White, R. M. , , 987A–996A.]. The application of an alternating electrical field across a quartz crystal induces an oscillatory vibrational motion in the crystal. Every quartz crystal vibrates at a characteristic resonant frequency that depends on the crystal’s properties, including the mass per unit area of any material coated on the crystal’s surface. The change in mass following adsorption, absorption, or binding of the analyte is determined by monitoring the change in the quartz crystal’s characteristic resonant frequency. The exact relationship between the change in frequency and mass is determined by a calibration curve. If you own a wristwatch, there is a good chance that its operation relies on a quartz crystal. The piezoelectric properties of quartz were discovered in 1880 by Paul-Jacques Currie and Pierre Currie. Because the oscillation frequency of a quartz crystal is so precise, it quickly found use in the keeping of time. The first quartz clock was built in 1927 at the Bell Telephone labs, and Seiko introduced the first quartz wristwatches in 1969. Particulate gravimetry is important in the environmental analysis of water, air, and soil samples. The analysis for suspended solids in water samples, for example, is accomplished by filtering an appropriate volume of a well-mixed sample through a glass fiber filter and drying the filter to constant weight at 103–105 C. The microbiological testing of water also uses particulate gravimetry. One example is the analysis for coliform bacteria in which an appropriate volume of sample is passed through a sterilized 0.45-μm membrane filter. The filter is placed on a sterilized absorbent pad that is saturated with a culturing medium and incubated for 22–24 hours at 35 ± 0.5 C. Coliform bacteria are identified by the presence of individual bacterial colonies that form during the incubation period (Figure 8.4.3 ). As with qualitative applications of precipitation gravimetry, the signal in this case is a visual observation of the number of colonies rather than a measurement of mass. Total airborne particulates are determined using a high-volume air sampler equipped with either a cellulose fiber or a glass fiber filter. Samples from urban environments require approximately 1 h of sampling time, but samples from rural environments require substantially longer times. Grain size distributions for sediments and soils are used to determine the amount of sand, silt, and clay in a sample. For example, a grain size of 2 mm serves as the boundary between gravel and sand. The grain size for the sand–silt and the silt–clay boundaries are 1/16 mm and 1/256 mm, respectively. Several standard quantitative analytical methods for agricultural products are based on measuring the sample’s mass following a selective solvent extraction. For example, the crude fat content in chocolate is determined by extracting with ether for 16 hours in a Soxhlet extractor. After the extraction is complete, the ether is allowed to evaporate and the residue is weighed after drying at 100 C. This analysis also can be accomplished indirectly by weighing a sample before and after extracting with supercritical CO . Quartz crystal microbalances equipped with thin film polymer films or chemical coatings have found numerous quantitative applications in environmental analysis. Methods are reported for the analysis of a variety of gaseous pollutants, including ammonia, hydrogen sulfide, ozone, sulfur dioxide, and mercury. Biochemical particulate gravimetric sensors also have been developed. For example, a piezoelectric immunosensor has been developed that shows a high selectivity for human serum albumin, and is capable of detecting microgram quantities [Muratsugu, M.; Ohta, F.; Miya, Y.; Hosokawa, T.; Kurosawa, S.; Kamo, N.; Ikeda, H. , , 2933–2937]. The result of a quantitative analysis by particulate gravimetry is just the ratio, using appropriate units, of the amount of analyte relative to the amount of sample. A 200.0-mL sample of water is filtered through a pre-weighed glass fiber filter. After drying to constant weight at 105 C, the filter is found to have increased in mass by 48.2 mg. Determine the sample’s total suspended solids. One ppm is equivalent to one mg of analyte per liter of solution; thus, the total suspended solids for the sample is \[\frac{48.2 \ \mathrm{mg} \text { solids }}{0.2000 \ \mathrm{L} \text { sample }}=241 \ \mathrm{ppm} \text { solids } \nonumber\] The scale of operation and the detection limit for particulate gravimetry can be extended beyond that of other gravimetric methods by increasing the size of the sample taken for analysis. This usually is impracticable for other gravimetric methods because it is difficult to manipulate a larger sample through the individual steps of the analysis. With particulate gravimetry, however, the part of the sample that is not analyte is removed when filtering or extracting. Consequently, particulate gravimetry easily is extended to the analysis of trace-level analytes. Except for methods that rely on a quartz crystal microbalance, particulate gravimetry uses the same balances as other gravimetric methods, and is capable of achieving similar levels of accuracy and precision. Because particulate gravimetry is defined in terms of the mass of the particle themselves, the sensitivity of the analysis is given by the balance’s sensitivity. Selectivity, on the other hand, is determined either by the filter’s pore size or by the properties of the extracting phase. Because it requires a single step, particulate gravimetric methods based on filtration generally require less time, labor and capital than other gravimetric methods.

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The following table provides critical values for \(G(\alpha, n)\), where \(\alpha\) is the probability of incorrectly rejecting the suspected outlier and is the number of samples in the data set. There are several versions of Grubb’s Test, each of which calculates a value for where is the number of suspected outliers on one end of the data set and is the number of suspected outliers on the opposite end of the data set. The critical values for given here are for a single outlier, , where \[G_\text{exp} = G_{10} = \frac {|X_{out} - \overline{X}|} {s} \nonumber\] The suspected outlier is rejected if is greater than \(G(\alpha, n)\).

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.26%3A_Osmotic_Pressure

Suppose we have a solution of sugar in water separated by a membrane from a sample of pure water. The membrane is porous, but the holes are not large enough to allow sucrose molecules to pass through from one side to the other while still being large enough to allow water molecules to pass freely through them. In such a situation, water molecules will be hitting one side of the membrane more often than the other. As a result, water molecules will move more often from right to left through the membrane in the animation than they will in the reverse direction. There will thus be a net flow of water from the compartment containing pure solvent, through the membrane, into the compartment containing the sucrose. This is another example of the tendency for moving molecules to become more thoroughly mixed together. The following movie shows this process, called . Osmosis occurs when two solutions of different concentrations are separated by a membrane which will selectively allow some species through it but not others. Then, material flows from the less concentrated to the more concentrated side of the membrane. A membrane which is selective in the way just described is said to be . Osmosis is of particular importance in living organisms, since most living tissue is semipermeable in one way or another. In the movie, we have depicted a membrane which is selective purely because of the pore size. In biological systems, the semipermeability relies on a set of solute transporters and channels. The cell membrane is formed of a with polar head groups facing out, and nonpolar hydrocarbon tails in the middle of the membrane. The consequence is that charged and polar substances cannot cross the membrane. Aquaporins are membrane proteins which allow water, but no other molecule, not even H O to pass through. For other solutes and ions, there exist specific transporters, some which allow a solute to diffuse down a natural gradient, and others which actively pump ions or other solutes in or out of the cell. These transporters, pumps and channels can be gated and regulated as well, allowing a cell to respond to varying osmotic conditions. A simple demonstration of osmosis is provided by the behavior of red blood cells. If these are immersed in water and observed under the microscope, they will be seen to gradually swell and to finally burst, as seen in the video below. Osmotic flow occurs from the surrounding water into the more concentrated solutions inside the cell. If the blood cells are now immersed in a saturated solution of NaC1, osmosis occurs in the opposite direction, since the solution inside the cell is not as concentrated as that outside. Under the microscope, the blood cells can be seen to shrink and shrivel. In medical practice, any solution which is to be introduced into the blood must take the possibility of osmosis into account. Normal saline, a solution of 0.16 NaCl (0.16 mol NaCl per dm of solution) is always employed for intravenous feeding or injection, because it has the same concentration of salts as blood serum. The video below shows what happens when a red blood cell is immersed in water and osmosis occurs. The tendency for osmotic flow to occur from a solvent to a solution is usually measured in terms of what is called the osmotic pressure of the solution, symbol Π. This osmotic pressure is not a pressure which the solution itself exerts but is rather the pressure which must be applied to the solution (but not the solvent) from outside in order to just prevent osmosis from occurring. A simple method for measuring the osmotic pressure is shown in Figure 2. The wider end of a funnel-shaped tube is covered with a membrane. The tube is filled with solution and placed in a container of the solvent. The height of the solution above the solvent increases until a maximum value is reached. The osmotic pressure is then the pressure exerted by the column of a solution of height where ρ is the density of the solution and is the gravitational acceleration. \[\Pi = \rho gh \nonumber \] Experimentally the osmotic pressure is found to obey a law similar in form to the ideal gas law and hence easy to remember: \[\Pi V = nRT \label{1} \] where is the amount of solute in volume of solution. In practice it is more useful to have Eq. (1) in terms of the concentration of solute / . Accordingly we rearrange it to read \[\Pi = n * V^{-1}RT \nonumber \] or \[\Pi = cRT \label{2} \] Equation \(\ref{2}\) is a very useful relationship since it means that we can find the concentration of any solution merely by measuring its osmotic pressure. This, in turn, allows us to find the molar mass of the solute. Suppose we have a solution in which a known mass of solute is dissolved in a known volume of solution. By measuring the osmotic pressure of this solution, we can find the concentration of solute and hence the amount of solute in the total volume. Since we already know the mass of solute, the molar mass follows immediately. A solution of 20.0 g of polyisobutylene in 1.00 dm of benzene was placed in an osmometer, similar to the one shown in Figure 2, at 25°C. After equilibrium had been obtained, the height was found to be 24.45 mm of benzene. Find the average molar mass of the polymer. The density of the solution is 0.879 g cm . We must first find the osmotic pressure from the height with the formula \[\Pi = \rho gh \nonumber \]. In doing this, it is most convenient to convert everything to SI base units. \[\begin{align} & \Pi =\rho gh=\text{0}\text{.879}\frac{\text{g}}{\text{cm}^{\text{3}}}\text{ }\times \text{ 9}\text{.807}\frac{\text{m}}{\text{s}^{\text{2}}}\text{ }\times \text{ 24}\text{.45 mm} \\ & \text{ }=210.8\frac{\text{g}}{\text{cm}^{\text{3}}}\frac{\text{m}}{\text{s}^{\text{2}}}\text{mm }\times \text{ }\frac{\text{1 kg}}{\text{1000 g}}\text{ }\times \text{ }\left( \frac{\text{100 cm}}{\text{1 m}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{1 m}}{\text{1000 mm}} \\ & \text{ }=\text{210}\text{.8}\frac{\text{kg m}^{\text{2}}}{\text{m}^{\text{3}}\text{ s}^{\text{2}}}=\text{210}\text{.8}\frac{\text{kg m}}{\text{s}^{\text{2}}}\text{ }\times \text{ }\frac{\text{1}}{\text{m}^{\text{2}}}=\text{210}\text{.8 N m}^{-\text{2}}=\text{210}\text{.8 Pa} \\ \end{align} \nonumber \] Knowing Π, we can now calculate the concentration by rewriting \[\Pi = cRT \nonumber \] as \[\text{c}=\frac{\Pi }{RT}=\frac{\text{210}\text{.8 }\times \text{ 10}^{-\text{3}}\text{ kPa}}{\text{8}\text{.3143 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ }\times \text{ 298}\text{.15 K}}=\text{8}\text{.50 }\times \text{ 10}^{-5}\text{ mol dm}^{-\text{3}} \nonumber \] For 1 dm of solution therefore, \[n_{solute} = 8.50 × 10^{–5} \text{mol} \nonumber \] while \[m_{solute} = 20 \text{g} \nonumber \] Thus \[M_{\text{solute}}=\frac{\text{20 g}}{\text{8}\text{.50 }\times \text{ 10}^{-\text{5}}\text{ mol}}\text{2}\text{.35 }\times \text{ 10}^{\text{5}}\text{ g mol}^{-\text{1}} \nonumber \] Note: An average polyisobutylene molecule thus has a molecular weight of close to a quarter of a million! Such a molecule is made from units, each with a molar mass of 56.10 g mol . An average polyisobutylene chain is thus \[\frac{\text{2}\text{.35 }\times \text{ 10}^{\text{5}}\text{ g mol}^{-\text{1}}}{\text{56}\text{.10 g mol}^{-\text{1}}}=\text{4189} \nonumber \] units long.In other words the chain length is over 8000 carbon atoms in this sample of polymer.

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In both developed and developing nations, indoor air pollution poses a greater health risk than outdoor air pollution. According to the World Health Organization (WHO) and other agencies such as the Environmental Protection Agency (EPA), indoor air generally contains higher concentrations of toxic pollutants than outdoor air. Additionally, people generally spend more time indoors than outdoors, hence, the health effects from indoor air pollution in workplaces, schools, and homes are far greater than outdoor. Indoor pollution sources that release gases or particles into the air are the primary cause of indoor air quality problems in homes. Inadequate ventilation can increase indoor pollutant levels by not bringing in enough outdoor air to dilute emissions from indoor sources and by not carrying indoor air pollutants out of the home. Outdoor air enters and leaves a building by infiltration, natural ventilation, and mechanical ventilation. In infiltration, outdoor air flows into the house through openings, joints, and cracks in walls, floors, and ceilings, and around windows and doors. In natural ventilation, air moves through opened windows and doors. Air movement associated with infiltration and natural ventilation is caused by air temperature differences between indoors and outdoors and by wind. Finally, there are a number of mechanical ventilation devices, from outdoor-vented fans that intermittently remove air from a single room, such as bathrooms and kitchen, to air handling systems that use fans and duct work to continuously remove indoor air and distribute filtered and conditioned outdoor air to strategic points throughout the house. The rate at which outdoor air replaces indoor air is described as the air exchange rate. When there is little infiltration, natural ventilation, or mechanical ventilation, the air exchange rate is low and pollutant levels can increase. High temperature and humidity levels can also increase concentrations of some pollutants. There are many sources of indoor air pollution in any home (Figure \(\Page {1}\)). These include combustion sources such as oil, gas, kerosene, coal, wood, and tobacco products; building materials and furnishings as diverse as deteriorated, asbestos-containing insulation, wet or damp carpet, and cabinetry or furniture made of certain pressed wood products; products for household cleaning and maintenance, personal care, or hobbies; central heating and cooling systems and humidification devices. Pollutants causing indoor air pollution can also originate from outside sources such as radon, pesticides, and outdoor air pollution. Radon is a naturally occurring radioactive gas produced from the decay of uranium in rock. If a building/home is constructed in an area with uranium containing rock, the gas can seep through the foundations and accumulate in basements. Exposure to radon can cause lung cancer. The relative importance of any single source depends on how much of a given pollutant it emits and how hazardous those emissions are. In some cases, factors such as how old the source is and whether it is properly maintained are significant. For example, an improperly adjusted gas stove can emit significantly more carbon monoxide than one that is properly adjusted. Some sources, such as building materials, furnishings, and household products like air fresheners, release pollutants more or less continuously. Other sources, related to activities carried out in the home, release pollutants intermittently. These include smoking, the use of unvented or malfunctioning stoves, furnaces, or space heaters, the use of solvents in cleaning and hobby activities, the use of paint strippers in redecorating activities, and the use of cleaning products and pesticides in house-keeping. High pollutant concentrations can remain in the air for long periods after some of these activities. Risks from indoor air pollution differ between less industrialized and industrialized nations. Indoor pollution has a greater impact in less industrialized nations where many people use cheaper sources of fuel such as wood, charcoal, and crop waste among others for cooking and heating, often with little or no ventilation. The most significant indoor pollutant, therefore, is soot and carbon monoxide. In industrialized nations, the primary indoor air health risks are cigarette smoke and radon. Combustion pollutants are gases or particles that come from burning materials. In homes, the major source of combustion pollutants are improperly vented or unvented fuel-burning appliances such as space heaters, wood stoves, gas stoves, water heaters, dryers, and fireplaces. The types and amounts of pollutants produced depends on the type of appliance, how well the appliance is installed, maintained and vented and the kind of fuel it uses. Common combustion pollutants include: which is a colorless, odorless gas that interferes with the delivery of oxygen throughout the body. Carbon monoxide causes headaches, dizziness, weakness, nausea and even death. Average levels in homes without gas stoves vary from 0.5 to 5 parts per million (ppm). Levels near properly adjusted gas stoves are often 5 to 15 ppm and those near poorly adjusted stoves may be 30 ppm or higher. which is a colorless, odorless gas that causes eye, nose and throat irritation, shortness of breath, and an increased risk of respiratory infection. Average level in homes without combustion appliances is about half that of outdoors. In homes with gas stoves, kerosene heaters or un-vented gas space heaters, indoor levels often exceed outdoor levels. which refers to fine particles that forms in smoke when wood or other organic matter burns. from wood smoke In addition to particle pollution, wood smoke contains several toxic harmful air pollutants including benzene, formaldehyde, acrolein, and polycyclic aromatic hydrocarbons (PAHs). Secondhand smoke is a mixture of the smoke given off by the burning of tobacco products, such as cigarettes, cigars or pipes and the smoke exhaled by smokers. Secondhand smoke is also called environmental tobacco smoke (ETS). Exposure to secondhand smoke is sometimes called involuntary or passive smoking. Secondhand smoke, classified by EPA as a Group A carcinogen, contains more than 7,000 substances. Secondhand smoke exposure commonly occurs indoors, particularly in homes and cars. Secondhand smoke can move between rooms of a home and between apartment units. Opening a window or increasing ventilation in a home or car is not protective from secondhand smoke. Radon is a radioactive gas that results from the natural decay of and found in nearly all rocks and soils. Elevated radon levels have been found in every state. Radon is in the atmosphere and can also be found in ground water. The national average for radon in outdoor air is 0.4 picocuries per liter (pCi/L), while the average for indoor air is 1.3 pCi/L. Any building can have high levels of radon, including new and old homes, well-sealed and drafty homes, office buildings and schools, and homes with or without basements. Radon gas can get into buildings through cracks in solid floors and walls, construction joints, gaps in suspended floors and around service pipes, cavities inside walls, the water supply and building materials. Testing is the only way to know if your home has elevated radon levels.EPA recommends fixing your home when the radon level is at or above 4 pCi/L. EPA estimates that about 21,000 lung cancer deaths each year in the U.S. are radon-related. Exposure to radon is the second leading cause of lung cancer after smoking. For most people, radon is the single greatest environmental source of radiation exposure. EPA recommends that all homes and schools be tested for radon. For smokers, the risk of lung cancer is heightened due to the combined effects of radon and smoking. are usually not a problem indoors, unless mold spores land on a wet or damp spot and begin growing. Molds have the potential to cause health problems. Molds produce allergens (substances that can cause allergic reactions) and irritants. Inhaling or touching mold or mold spores may cause allergic reactions in sensitive individuals. Allergic responses include hay fever-type symptoms, such as sneezing, runny nose, red eyes, and skin rash. Allergic reactions to mold are common. They can be immediate or delayed. Molds can also cause asthma attacks in people with asthma who are allergic to mold. In addition, mold exposure can irritate the eyes, skin, nose, throat, and lungs of both mold-allergic and non-allergic people. Symptoms other than the allergic and irritant types are not commonly reported as a result of inhaling mold. are emitted as gases from certain solids or liquids. VOCs include a variety of chemicals, some of which may have short- and long-term adverse health effects. Concentrations of many VOCs are consistently higher indoors (up to ten times higher) than outdoors. VOCs are emitted by a wide array of products numbering in the thousands. Organic chemicals are widely used as ingredients in household products. Paints, varnishes and wax all contain organic solvents, as do many cleaning, disinfecting, cosmetic, degreasing and hobby products. Fuels are made up of organic chemicals. All of these products can release organic compounds while you are using them, and, to some degree, when they are stored. EPA's Office of Research and Development's "Total Exposure Assessment Methodology (TEAM) Study" (Volumes I through IV, completed in 1985) found levels of about a dozen common organic pollutants to be 2 to 5 times higher inside homes than outside, regardless of whether the homes were located in rural or highly industrial areas. TEAM studies indicated that while people are using products containing organic chemicals, they can expose themselves and others to very high pollutant levels, and elevated concentrations can persist in the air long after the activity is completed. VOCs are emitted by a wide array of products used in homes including paints and lacquers, paint strippers, cleaning supplies, varnishes and waxes, pesticides, building materials and furnishings, office equipment, moth repellents, air fresheners, and dry-cleaned clothing. VOCs evaporate into the air when these products are used or sometimes even when they are stored. Volatile organic compounds irritate the eyes, nose and throat, and cause headaches, nausea, and damage to the liver, kidneys and central nervous system. Some of them can cause cancer. is a mineral fiber that occurs in rock and soil. Because of its fiber strength and heat resistance asbestos has been used in a variety of building construction materials for insulation and as a fire retardant. Asbestos has also been used in a wide range of manufactured goods, mostly in building materials (roofing shingles, ceiling and floor tiles, paper products, and asbestos cement products), friction products (automobile clutch, brake, and transmission parts), heat-resistant fabrics, packaging, gaskets, and coatings. Asbestos fibers may be released into the air by the disturbance of asbestos-containing material during product use, demolition work, building or home maintenance, repair, and remodeling. In general, exposure may occur only when the asbestos-containing material is disturbed or damaged in some way to release particles and fibers into the air. Three of the major health effects associated with asbestos exposure are: · lung cancer · mesothelioma, a rare form of cancer that is found in the thin lining of the lung, chest and the abdomen and heart · asbestosis, a serious progressive, long-term, non-cancer disease of the lungs Take steps to help improve your air quality and reduce your IAQ-related health risks at little or no cost by: Usually the most effective way to improve indoor air is to eliminate individual sources or reduce their emissions. Ventilating: Increasing the amount of fresh air brought indoors helps reduce pollutants inside. When weather permits, open windows and doors, or run an air conditioner with the vent control open. Bathroom and kitchen fans that exhaust to the outdoors also increase ventilation and help remove pollutants. Always ventilate and follow manufacturers’ instructions when you use products or appliances that may release pollutants into the indoor air. Central heaters and air conditioners have filters to trap dust and other pollutants in the air. Make sure to change or clean the filters regularly, following the instructions on the package. The humidity inside can affect the concentrations of some indoor air pollutants. For example, high humidity keeps the air moist and increases the likelihood of mold. Keep indoor humidity between 30 and 50 percent. Use a moisture or humidity gauge, available at most hardware stores, to see if the humidity in your home is at a good level. To increase humidity, use a vaporizer or humidifier. To decrease humidity, open the windows if it is not humid outdoors. If it is warm, turn on the air conditioner or adjust the humidity setting on the humid Get a quick glimpse of some of the most important ways to protect the air in your home by touring the Indoor Air Quality (IAQ) House. Room-by-room, you'll learn about the key pollutants and how to address them. Interactive Version Text Version  ( ) US Environmental Protection Agency

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There are many other types of solids, or other important qualities that distinguish solids, like catalysis, porosity, etc. We haven't described minerals or ceramics very well, or superconductors, magnetic materials, etc. For now, we'll just describe two more types: polymers and nanomaterials. Polymers are familiar and important from life, because we use them for almost everything. Nanomaterials are less familiar, but are starting to be used in commercial products although we don't always understand them too well. are also called , which means big molecules. Polymers are usually long chains made of small molecules covalently bonded together. The important properties of the polymer come from the "intermolecular" or non-covalent forces that determine how the polymer chains fold up or tangle together. Most important biomolecules (like DNA, proteins and starch) are polymers. We also make a wide variety of polymers, usually starting with hydrocarbon molecules from oil, that are used in clothing, plastics, construction materials, etc. Many different molecules and reactions can be used to make polymers. The properties of the polymer will often depend on the structure, which is usually amorphous or disorganized. If the macromolecules are more orderly and rigid, the structure may be more crystalline and less flexible, with a higher melting point (because the molecules can pack together well with relatively strong intermolecular forces). If the macromolecules are very different sizes and have irregular shapes, then the polymer might be more flexible. Sometimes covalent bonds are added between the molecules, which can also make the material tougher. (For example, rubber is made by stronger by heating with sulfur, which can connect the chains together.) are very small solids, like extremely fine powders only smaller. They have dimensions measured in nanometers, or 10 meters. At this scale, they are too small to have all the normal properties of solids, because they have so much edge. For example, nanomaterials have bigger band gaps than the same material in bigger chunks. The smaller the particles, the bigger the band gap. (Molecules have bigger HOMO/LUMO gaps than solid materials; nanoparticles are in between molecules and normal solids.) People are pretty excited about nanoparticles (they're popular!) but it's good to remember that we don't really know much about their health effects yet, except to say that they are complicated and need more study. I don't recommend running out and buying the new nano-shampoo or nano-silver handwipes or whatever you might encounter unless it's noticeably superior to the non-nano alternative. Nanostructures and nanoparticles can be natural or manufactured. For example, most birds that look blue don't have blue pigment, they have nanostructures on their feathers that scatter light. These days, most electronic devices are based on nanostructures also: each individual transistor in a computer chip is nanosized, and it may not be possible to make them much smaller than they are now without too many size-based complications.

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Table 16.3.1 , at the bottom of this appendix, gives the proportion, , of the area under a normal distribution curve that lies to the right of a deviation, \[z = \frac {X -\mu} {\sigma} \nonumber\] where is the value for which the deviation is defined, \(\mu\) is the distribution’s mean value and \(\sigma\) is the distribution’s standard deviation. For example, the proportion of the area under a normal distribution to the right of a deviation of 0.04 is 0.4840 (see entry in in the table), or 48.40% of the total area (see the area shaded in Figure 16.3.1 ). The proportion of the area to the left of the deviation is 1 – . For a deviation of 0.04, this is 1 – 0.4840, or 51.60%. Normal distribution curve showing the area under a curve greater than a deviation of +0.04 ( ) and with a deviation less than –0.04 ( ). When the deviation is negative—that is, when is smaller than \(\mu\)—the value of is negative. In this case, the values in the table give the area to the left of . For example, if is –0.04, then 48.40% of the area lies to the left of the deviation (see area shaded in Figure 16.3.1 . To use the single-sided normal distribution table, sketch the normal distribution curve for your problem and shade the area that corresponds to your answer (for example, see Figure 16.3.2 , which is for ). This divides the normal distribution curve into three regions: the area that corresponds to our answer (shown in ), the area to the right of this, and the area to the left of this. Calculate the values of for the limits of the area that corresponds to your answer. Use the table to find the areas to the right and to the left of these deviations. Subtract these values from 100% and, voilà, you have your answer.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/06%3A_Equilibrium_Chemistry/6.13%3A_Additional_Resources

The following experiments involve the experimental determination of equilibrium constants, the characterization of buffers, and, in some cases, demonstrations of the importance of activity effects. A nice discussion of Berthollet’s discovery of the reversibility of reactions is found in The following texts provide additional coverage of equilibrium chemistry. The following papers discuss a variety of general aspects of equilibrium chemistry. Collected here are a papers that discuss a variety of approaches to solving equilibrium problems. Additional historical background on the development of the Henderson-Hasselbalch equation is provided by the following papers. A simulation is a useful tool for helping students gain an intuitive understanding of a topic. Gathered here are some simulations for teaching equilibrium chemistry. The following papers provide additional resources on ionic strength, activity, and the effect of ionic strength and activity on equilibrium reactions and pH. For a contrarian’s view of equilibrium chemistry, please see the following papers.

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Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new   access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ).   and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. .     .

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There are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction). The ocean is a unique example of a system with , or multiple states of solubility equilibria working simultaneously. Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H CO ). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions \(\ce{(HCO3- )}\), which can further ionize into more hydrogen ions and carbonate ions \(\ce{(CO3^2- )}\): \[\ce{CO2}(g)⇌\ce{CO2}(aq) \nonumber \] \[\ce{CO2}(aq)+\ce{H2O}⇌\ce{H2CO3}(aq) \nonumber \] \[\ce{H2CO3}(aq)⇌\ce{H+}(aq)+\ce{HCO3-}(aq) \nonumber \] \[\ce{HCO3-}(aq)⇌\ce{H+}(aq)+\ce{CO3^2-}(aq) \nonumber \] The excess H ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure \(\Page {1}\)). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world’s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years. Learn more about ocean and how it affects other marine creatures. Slightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO , FeS, and Ca (PO ) in HCl because their basic anions react to form weak acids (H CO , H S, and \(\ce{H2PO4-}\)). The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Chatelier’s principle. Of particular relevance to us is the dissolution of hydroxylapatite, Ca (PO ) OH, in acid. Apatites are a class of calcium phosphate minerals (Figure \(\Page {2}\)); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca (PO ) OH and dissolved Ca , \(\ce{PO4^3-}\), and OH ions: \[\ce{Ca5(PO4)3OH}(s)⟶\ce{5Ca^2+}(aq)+\ce{3PO4^3-}(aq)+\ce{OH-}(aq) \nonumber \] When exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid: \[\ce{PO4^3-}(aq)+\ce{H3O+}⇌\ce{H2PO4^2-}+\ce{H2O} \nonumber \] \[\ce{PO4^2-}(aq)+\ce{H3O+}⇌\ce{H2PO4-}+\ce{H2O} \nonumber \] \[\ce{H2PO4- + H3O+ ⇌ H3PO4 + H2O} \nonumber \] Hydroxide ion reacts to form water: \[\ce{OH-}(aq)+\ce{H3O+}⇌\ce{2H2O} \nonumber \] These reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Chatelier’s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF . They function by replacing the OH ion in hydroxylapatite with F ion, producing fluorapatite, Ca (PO ) F: \[\ce{NaF + Ca5(PO4)3OH ⇌ Ca5(PO4)3F + Na+ + OH-} \nonumber \] The resulting Ca (PO ) F is slightly less soluble than Ca (PO ) OH, and F is a weaker base than OH . Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information. As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca (PO ) F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure \(\Page {3}\)). Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater. When acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid: \[\ce{H3O+}(aq)+\ce{CO3^2-}(aq)⟶\ce{HCO3-}(aq)+\ce{H2O}(l) \nonumber \] Calcium hydrogen carbonate, Ca(HCO ) , is soluble, so limestone and marble objects slowly dissolve in acid rain. If calcium carbonate is added to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation: \[\ce{2H3O+}(aq)+\ce{CO3^2-}(aq)⟶\ce{H2CO3}(aq)+\ce{2H2O}(l) \nonumber \] (Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved: \[\ce{H2CO3}(aq)⟶\ce{CO2}(g)+\ce{H2O}(l) \nonumber \] These reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved. Calculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH) in a solution with [Mg ] = 0.10 and [NH ] = 0.10 . Two equilibria are involved in this system: To prevent the formation of solid Mg(OH) , we must adjust the concentration of OH so that the reaction quotient for Equation (1), = [Mg ,OH ] , is less than for Mg(OH) . (To simplify the calculation, we determine the concentration of OH when = .) [OH ] can be reduced by the addition of \(\ce{NH4+}\), which shifts Reaction (2) to the left and reduces [OH ]. Solid Mg(OH) will not form in this solution when [OH ] is less than 9.4 × 10 . When \(\ce{[NH4+]}\) equals 0.19 , [OH ] will be 9.4 × 10 . Any \(\ce{[NH4+]}\) greater than 0.19 will reduce [OH ] below 9.4 × 10 and prevent the formation of Mg(OH) . Consider the two equilibria: \[\ce{ZnS}(s)⇌\ce{Zn^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=1×10^{−27} \nonumber \] \[\ce{2H2O}(l)+\ce{H2S}(aq)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26} \nonumber \] and calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 in Zn and saturated with H S (0.10 H S). \[\ce{[H3O+]}>0.2\:M \nonumber \] ([S ] is less than 2 × 10 and precipitation of ZnS does not occur.) Therefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions. Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na S O , called hypo) to form the complex ion \(\ce{Ag(S2O3)2^3-}\) ( = 4.7 × 10 ). The reaction with silver bromide is: What mass of Na S O is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of \(\ce{Ag(S2O3)2^3-}\)? Two equilibria are involved when AgBr dissolves in a solution containing the \(\ce{S2O3^2-}\) ion: In order for 1.00 g of AgBr to dissolve, the [Ag ] in the solution that results must be low enough for for Reaction (1) to be smaller than for this reaction. We reduce [Ag ] by adding \(\ce{S2O3^2-}\) and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na S O is needed to provide the necessary \(\ce{S2O3^2-}\). Because 5.33 × 10 mol of AgBr dissolves: Thus, at equilibrium: \(\ce{[Ag(S2O3)2^3- ]}\) = 5.33 × 10 , [Ag ] = 9.4× 10 , and = = 4.7 × 10 : When \(\ce{[S2O3^2- ]}\) is 1.1 × 10 , [Ag ] is 9.4 × 10 and all AgBr remains dissolved. Thus, 1.00 L of a solution prepared from 1.9 g Na S O dissolves 1.0 g of AgBr. AgCl( ), silver chloride, is well known to have a very low solubility: \(\ce{Ag}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)\), = 1.6 × 10 . Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed: \(\ce{Ag+}(aq)+\ce{2NH3}(aq)⇌\ce{Ag(NH3)2+}(aq)\), = 1.7 × 10 . What mass of NH is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of \(\ce{Ag(NH3)2+}\)? 1.00 L of a solution prepared with 4.81 g NH dissolves 2.0 g of AgCl. We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Chatelier’s principle. For example, one way to control the concentration of manganese(II) ion, Mn , in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion: \[\ce{Mn(OH)2}(s) ⇌ \ce{Mn^2+}(aq)+\ce{2OH-}(aq) \hspace{20px} K_\ce{sp}=\ce{[Mn^2+,OH- ]^2} \nonumber \] This could be important to a laundry because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn ion while increasing the amount of solid Mn(OH) in the equilibrium mixture, as predicted by Le Chatelier’s principle. What is the effect on the amount of solid Mg(OH) that dissolves and the concentrations of Mg and OH when each of the following are added to a mixture of solid Mg(OH) in water at equilibrium? The equilibrium among solid Mg(OH) and a solution of Mg and OH is: \[\ce{Mg(OH)2}(s) ⇌ \ce{Mg^2+}(aq)+\ce{2OH-}(aq) \nonumber \] (a) The reaction shifts to the left to relieve the stress produced by the additional Mg ion, in accordance with Le Chatelier’s principle. In quantitative terms, the added Mg causes the reaction quotient to be larger than the solubility product ( > ), and Mg(OH) forms until the reaction quotient again equals . At the new equilibrium, [OH ] is less and [Mg ] is greater than in the solution of Mg(OH) in pure water. More solid Mg(OH) is present. (b) The reaction shifts to the left to relieve the stress of the additional OH ion. Mg(OH) forms until the reaction quotient again equals . At the new equilibrium, [OH ] is greater and [Mg ] is less than in the solution of Mg(OH) in pure water. More solid Mg(OH) is present. (c) The concentration of OH is reduced as the OH reacts with the acid. The reaction shifts to the right to relieve the stress of less OH ion. In quantitative terms, the decrease in the OH concentration causes the reaction quotient to be smaller than the solubility product ( < ), and additional Mg(OH) dissolves until the reaction quotient again equals . At the new equilibrium, [OH ] is less and [Mg ] is greater than in the solution of Mg(OH) in pure water. More Mg(OH) is dissolved. (d) NaNO contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg and OH . (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.) (e) The addition of solid Mg(OH) has no effect on the solubility of Mg(OH) or on the concentration of Mg and OH . The concentration of Mg(OH) does not appear in the equation for the reaction quotient: \[Q=\ce{[Mg^2+,OH- ]^2} \nonumber \] Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of , and no shift is required to restore to the value of the equilibrium constant. What is the effect on the amount of solid NiCO that dissolves and the concentrations of Ni and \(\ce{CO3^2-}\) when each of the following are added to a mixture of the slightly soluble solid NiCO and water at equilibrium? (a) mass of NiCO ( ) increases, [Ni ] increases, \(\ce{[CO3^2- ]}\) decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO ; (d) mass of NiCO ( ) increases, [Ni ] decreases, \(\ce{[CO3^2- ]}\) increases; (e) mass of NiCO ( ) decreases, [Ni ] increases, \(\ce{[CO3^2- ]}\) decreases Several systems we encounter consist of multiple equilibria, systems where two or more equilibria processes are occurring simultaneously. Some common examples include acid rain, fluoridation, and dissolution of carbon dioxide in sea water. When looking at these systems, we need to consider each equilibrium separately and then combine the individual equilibrium constants into one solubility product or reaction quotient expression using the tools from the first equilibrium chapter. Le Chatelier’s principle also must be considered, as each reaction in a multiple equilibria system will shift toward reactants or products based on what is added to the initial reaction and how it affects each subsequent equilibrium reaction.

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Time_Dependent_Quantum_Mechanics_and_Spectroscopy_(Tokmakoff)

Note: These pages were automatically transcribed and have not been proofread. Please visit https:\\tdqms.uchicago.edu for the most recent versions of this material. These notes are meant as a resource for chemists that study the time-dependent quantum mechanics, dynamics, and spectroscopy of molecular systems. They are derived from my lectures in graduate quantum mechanics that focus on condensed phase spectroscopy, dynamics, and relaxation. The content originated from many sources. In large part this includes the references for the course, as noted in the notes and readings. I would like to thank a number of colleagues and prior instructors who were the source of content that guided my preparation of several of these lectures, including Bob Silbey, Keith Nelson, Troy Van Voorhis, Bob Field, John Ross and Graham Fleming. I also want to thank Anne Hudson, Peter Giunta, and Tanya Shpigel for their assistance preparing the notes over the years, and the Department of Energy and National Science Foundation for their ongoing support of my research in this area.

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are familiar to you from everyday life, because acids are sour things, like vinegar and lemons. might be less familiar to you. Common bases in regular life are baking soda and antacids that people take if they are having stomach trouble. Also, many soaps are basic, although this is just because of how they are made, not necessary to make them soapy. Bases make things feel slippery, and they taste bitter. They can be used to clean greasy things. Common acids and bases will react with each other and when they react completely, the products are usually a salt solution that isn't sour or bitter. This is called . Acid-base neutralization reactions are what make most cakes fluffy, because sometimes these reactions generate a gas that makes holes in the cake. It is important to know that there are many different definitions of acid and base. This page describes the simplest and most specific definition. Practicing chemists use broader definitions that stretch the general concept to a variety of situations that are important in more advanced chemistry. This description describes acid-base reactions in water. (It works a little bit differently in other solvents, but you don't need to think about that too much until you study more advanced chemistry). An acid is an electrolyte (strong or weak) that produces H ions when it dissolves in water. Hydrogen ions are also called protons, because a hydrogen nucleus is just a proton (unless it is a heavier isotope, but this is rare). Acids are sometimes called "proton donors" meaning they give away protons, but this is not a very good word, because the protons are pulled away by the solvent, not dropped by the solute. Examples of strong acids include HCl and H SO , which are called hydrochloric acid and sulfuric acid. But sulfuric acid has 2 protons bound to sulfate ion, and only one comes off completely (just like the weak electrolytes discussed ), so a solution of sulfuric acid will have hydrogen ions, bisulfate ions, and sulfate ions in solution. Just like in the case of precipitation reactions, if a base is added, both protons might come off completely and react with the base. Acids are called , , etc. depending on how many acidic protons they have. HCl, acetic acid (vinegar, CH COOH) and nitric acid (HNO ) are monoprotic acids. (Acetic acid has other protons, but only the last one is acidic.) Sulfuric acid and many others are diprotic acids. A base is an electrolyte (strong or weak) that produces hydroxide ions when dissolved in water solution. This could be because it is a hydroxide salt, like NaOH, or because it takes hydrogen ions from water, leaving hydroxide behind. A good example of this is ammonia, NH , which is sometimes used in house cleaning products. Ammonia reacts with water to make ammonium hydroxide (but only a little bit, ~1% of the ammonia reacts): \[NH_{3}(aq) + H_{2}O(l) \rightarrow NH_{4}^{+}(aq) + OH^{–}(aq)\] In general, bases react with hydrogen ions. This is how neutralization happens. The acid produces hydrogen ions, and the base produces hydroxide ions. These react together to make water. The anion that came from the acid and the cation are left, so if you evaporate the water, you would get a salt. The general reaction looks like this: \[X^{–}(aq) + H^{+}(aq) + M^{+}(aq) + OH^{–}(aq) \rightarrow H_{2}O(l) + X^{–}(aq) + M^{+}(aq)\] Overall, the reaction is: \[H^{+}(aq) + OH^{–}(aq) \rightarrow H_{2}O(l)\] Thus, the hydrogen ions, which makes acids acidic, are consumed, and the hydroxide which makes bases basic is also consumed, and if the moles of acid and base are equal, only neutral water and a salt is left. (Actually, it is a little bit more complicated than this if the acid or base is weak. The solution will only really become neutral when the moles are equal if both are strong.) Q: How do acid-base reactions make cake fluffy? A: Usually cakes include an acidic ingredient (this varies) and sodium bicarbonate, a base. When they react, the proton from the acid is transferred to the bicarbonate, making the weak acid carbonic acid. Carbonic acid is the product of an reaction between carbon dioxide and water. This reaction can be reversed, or carbonic acid can decompose into water and carbon dioxide. Especially at the high temperatures inside a baking cake, this decomposition will happen, and produce carbon dioxide gas. The pressure of the hot gas will form bubbles inside the cake, making it fluffy. In the previous section ( ), instead of having hydroxide react with hydrogen ions to form water, the acid base reaction made carbonic acid from protons and bicarbonate. In general, a base is something that will bind tightly to a proton. Bicarbonate and carbonate ions are bases, and so are sulfide ions. Both of these reactions can produce a gas, either carbon dioxide or hydrogen sulfide. In the lab, sodium bicarbonate is usually used to neutralize acid spills. When it reacts with acid it produces bubbles, so it's easy to see when the reaction finishes. Most chemists agree that acid-base reactions are combination reactions without redox. This is a much more general definition than described here, but after you read about , go over these examples and convince yourself that they all fit that definition. What other examples can you think of that also fit this definition? are strong/weak electrolytes that produce H ions when dissolved in water. A notable property of acids is that they have a sour taste. are strong/weak electrolytes that produce hydroxide ions (OH ) when dissolved in water. They are also known to have a bitter taste as well as a slippery or soapy texture. is a type of reaction in which acids react with bases to form a salt solution (water and a salt), usually. (18 min, skip the last 3 min, optional: skip the first 4 min)

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. In addition to these individual basis; please contact What is the relationship between mechanical work and energy? Does a person with a mass of 50 kg climbing a height of 15 m do work? Explain your answer. Does that same person do work while descending a mountain? If a person exerts a force on an immovable object, does that person do work? Explain your answer. Explain the differences between electrical energy, nuclear energy, and chemical energy. The chapter describes thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy. Which form(s) of energy are represented by each of the following? Describe the various forms of energy that are interconverted when a flashlight is switched on. Describe the forms of energy that are interconverted when the space shuttle lifts off. Categorize each of the following as representing kinetic energy or potential energy. Are the units for potential energy the same as the units for kinetic energy? Can an absolute value for potential energy be obtained? Explain your answer. Categorize each of the following as representing kinetic energy or potential energy. Why does hammering a piece of sheet metal cause the metal to heat up? 3. Technically, the person is not doing any work, since the object does not move. 11. The kinetic energy of the hammer is transferred to the metal. How much energy (in kilojoules) is released or stored when each of the following occurs? Calculate how much energy (in kilojoules) is released or stored when each of the following occurs: A car weighing 1438 kg falls off a bridge that is 211 ft high. Ignoring air resistance, how much energy is released when the car hits the water? A 1 tn roller coaster filled with passengers reaches a height of 28 m before accelerating downhill. How much energy is released when the roller coaster reaches the bottom of the hill? Assume no energy is lost due to friction. 1. 5. 250 kJ released In some simple case, when a reaction only involves solids, liquids or liquid solution, we can say that \[Zn_{(s)} + 2HCl_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Cl^−_{(aq)} + H_2(g)\] When 3.00 g of zinc metal is added to a dilute HCl solution at 1.00 atm and 25°C, and this reaction is allowed to go to completion at constant pressure, 6.99 kJ of heat must be removed to return the final solution to its original temperature. What are the values of q and w, and what is the change in internal energy? Calculate the amount of work done against a pressure of 1.0 atm when 4.0 mol of acetylene are allowed to react with 10 mol of O at 1.0 atm at 20°C. What is the change in internal energy for the reaction? Heat implies the flow of energy from one object to another. Describe the energy flow in an a. exothermic reaction. b. endothermic reaction. When a thermometer is suspended in an insulated thermos that contains a block of ice, the temperature recorded on the thermometer drops. Describe the direction of heat flow. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. State whether each process is endothermic or exothermic. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. Determine whether each process is endothermic or exothermic. Is Earth’s environment an isolated system, an open system, or a closed system? Explain your answer. Why is it impossible to measure the absolute magnitude of the enthalpy of an object or a compound? Determine whether energy is consumed or released in each scenario. Explain your reasoning. The chapter states that enthalpy is an extensive property. Why? Describe a situation that illustrates this fact. The enthalpy of a system is affected by the physical states of the reactants and the products. Explain why. Is the distance a person travels on a trip a state function? Why or why not? Using erive a mathematical relationship between and . Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C. Using Table 9.6.1, how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal. Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K). In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy. How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change? A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in Table 9.5.1 and solubility rules to identify the salt. Using the data in Table 9.8.2, calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)? A 3.00 g sample of TNT (trinitrotoluene, C H N O ) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the Δ of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the Δ of TNT? = × (molar mass) For Cu: = 58 J; For Al: = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: = 0.385 and 0.897 (g•K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al. 4.7 kJ Δ = −0.56 kJ/g; based on reaction with AgNO , salt contains halide; dividing Δ values mass of salts gives lithium bromide as best match, with −0.56 kJ/g. = 43.1°C; the combustion reaction is \[4C_7H_5N_3O_{6(s)} + 21O_{2(g)} \rightarrow 28CO_{2(g)} + 10H_2O_{(g)} + 6N_{2(g)}\] with \[Δ_f^οH (TNT) = −65.5\; kJ/mol\] ​ Based on the following energy diagram, a. write an equation showing how the value of Δ could be determined if the values of Δ and Δ are known. b. identify each step as being exothermic or endothermic. Based on the following energy diagram, a. write an equation showing how the value of Δ could be determined if the values of Δ and Δ are known. b. identify each step as being exothermic or endothermic. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state? In their elemental form, A and B exist as diatomic molecules. Given the following reactions, each with an associated Δ °, describe how you would calculate ΔH for the compound AB . \( \begin{matrix} 2AB & \rightarrow & A_{2} + B _{2} & \Delta H_{1}^{o}\\ 3AB & \rightarrow & AB_{2} + A _{2}B & \Delta H_{2}^{o} \\ 2A_{2}B &\rightarrow & 2A_{2} + B _{2} & \Delta H_{3}^{o} \end{matrix} \) ​ Methanol is used as a fuel in Indianapolis 500 race cars. Use the following table to determine whether methanol or 2,2,4-trimethylpentane (isooctane) releases more energy per liter during combustion.   a. Use the enthalpies of combustion given in the following table to determine which organic compound releases the greatest amount of energy per gram during combustion. b. Calculate the standard enthalpy of formation of 1-ethyl-2-methylbenzene. Given the enthalpies of combustion, which organic compound is the best fuel per gram? 1. 2. a. To one decimal place Octane provides the largest amount of heat per gram upon combustion. b, ​​ Using , calculate for each chemical reaction. a. 2Mg(s) + O (g) → 2MgO(s) b. CaCO (s, calcite) → CaO(s) + CO (g) c. AgNO (s) + NaCl(s) → AgCl(s) + NaNO (s) Using , determine for each chemical reaction. a. 2Na(s) + Pb(NO ) (s) → 2NaNO (s) + Pb(s) b. Na CO (s) + H SO (l) → Na SO (s) + CO (g) + H O(l) c. 2KClO (s) → 2KCl(s) + 3O (g) Calculate for each chemical equation. If necessary, balance the chemical equations. a. Fe(s) + CuCl (s) → FeCl (s) + Cu(s) b. (NH ) SO (s) + Ca(OH) (s) → CaSO (s) + NH (g) + H O(l) c. Pb(s) + PbO (s) + H SO (l) → PbSO (s) + H O(l) Calculate for each reaction. If necessary, balance the chemical equations. a. 4HBr(g) + O (g) → 2H O(l) + 2Br (l) b. 2KBr(s) + H SO (l) → K SO (s) + 2HBr(g) c. 4Zn(s) + 9HNO (l) → 4Zn(NO ) (s) + NH (g) + 3H O(l) Use the data in to calculate for the reaction Sn(s, white) + 4HNO (l) → SnO (s) + 4NO (g) + 2H O(l). Use the data in to calculate for the reaction P O (s) + 6H O(l) → 4H PO (l). How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas? How much energy is released or consumed if 10.0 g of N O is completely decomposed to produce gaseous nitrogen dioxide and oxygen? In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction: Calculate ΔH for this reaction. Is the reaction exothermic or endothermic? Would you expect heat to be evolved during each reaction? How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate.   a. −1203 kJ/mol O b. 179.2 kJ c. −59.3 kJ −174.1 kJ/mol −20.3 kJ −34.3 kJ/mol Cl ; exothermic Δ = −2.86 kJ CaCl : −4.6 kJ; K CO , −0.65 kJ; mixing, −0.28 kJ Determine the amount of energy available from the biological oxidation of 1.50 g of leucine (an amino acid, Δ = −3581.7 kJ/mol). Calculate the energy released (in kilojoules) from the metabolism of 1.5 oz of vodka that is 62% water and 38% ethanol by volume, assuming that the total volume is equal to the sum of the volume of the two components. The density of ethanol is 0.824 g/mL. What is this enthalpy change in nutritional Calories? While exercising, a person lifts an 80 lb barbell 7 ft off the ground. Assuming that the transformation of chemical energy to mechanical energy is only 35% efficient, how many Calories would the person use to accomplish this task? From how many grams of glucose would be needed to provide the energy to accomplish this task? A 30 g sample of potato chips is placed in a bomb calorimeter with a heat capacity of 1.80 kJ/°C, and the bomb calorimeter is immersed in 1.5 L of water. Calculate the energy contained in the food per gram if, after combustion of the chips, the temperature of the calorimeter increases to 58.6°C from an initial temperature of 22.1°C.

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The introductory part of the organic chemistry course has three major modules: (structure), (conformational analysis), and (chemical reactions). An understanding of the first two is crucial to an understanding of the third one. The rest of the organic chemistry course will be largely spent on studying different kinds of reactions and mechanisms. A summary of key concepts follows. 1. MOLECULAR ARCHITECTURE - Basic principles of molecular structure. a) Atomic structure b) Orbitals and hybridization c) Covalent bonding d) Lewis structures and resonance forms e) Isomerism, structural and geometric isomers f) Polarity, functional groups, nomenclature systems g) Three-dimensional structures, stereochemistry, stereoisomers. 2. MOLECULAR DYNAMICS - Basic principles of molecular motion The focus is on conformations and their energy relationships, especially in reference to alkanes and cycloalkanes (conformational analysis). a) Steric interactions b) Torsional strain and Newman projections c) Angle strain in cycloalkanes d) Conformations of cyclohexane and terminology associated with it 3. MOLECULAR TRANSFORMATIONS - This is the part that comprises the bulk of organic chemistry courses. It is the study of chemical reactions and the principles that rule transformations. There are three major aspects of this module. In organic chemistry I we will focus largely on the first two, and leave the study of synthetic strategy for later. a) Reaction mechanisms - Step by step accounts of how electron movement takes place when bonds are broken and formed, and the conditions that favor these processes (driving forces). An understanding of the basic concepts of thermodynamics and kinetics is important here. b) Energetics - An account of energy changes and/or requirements that take place during the course of a transformation. c) Synthetic strategy - An understanding of the specific chemistry of functional groups and how it can be used to design molecules with specific structural features.

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A biomolecular nucleophilic substitution (S 2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group. It is possible for the nucleophile to attack the electrophilic center in two ways. The following diagram illustrates these two types of nucleophilic attacks, where the frontside attack results in retention of configuration; that is, the product has the same configuration as the substrate. The backside attack results in inversion of configuration, where the product's configuration is opposite that of the substrate. S 2 Experimental observation shows that all S 2 reactions proceed with inversion of configuration; that is, the nucleophile will always attack from the backside in all S 2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic center, and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism. The S 2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an R enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the S enantiomer. Conversely, if the substrate is an S enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the R enantiomer. In conclusion, S 2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are and substrates that are . If the configuration is the substrate, the resulting product will be . Conversely, if the configuration is the substrate, the resulting product will be .

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Make sure you thoroughly understand the following essential ideas which have been presented above. The term was first used in the seventeenth century; it comes from the Latin root -, meaning “sharp”, as in , vinegar. Some early writers suggested that acidic molecules might have sharp corners or spine-like projections that irritate the tongue or skin. Acids have long been recognized as a distinctive class of compounds whose aqueous solutions exhibit the following properties: is a natural dye found in certain lichens. The name is of Scandinavian origin, e.g. (color) + (moss) in Icelandic. "Litmus test" has acquired a meaning that transcends both Chemistry and science to denote any kind of test giving a yes/no answer. The first chemistry-based definition of an acid turned out to be wrong: in 1787, Antoine Lavoisier, as part of his masterful classification of substances, identified the known acids as a separate group of the “complex substances” (compounds). Their special nature, he postulated, derived from the presence of some common element that embodies the “acidity” principle, which he named , derived from the Greek for “acid former”. Lavoisier had recently assigned this name to the new gaseous element that Joseph Priestly had discovered a few years earlier as the essential substance that supports combustion. Many combustion products (oxides) do give acidic solutions, and oxygen is in fact present in most acids, so Lavoisier’s mistake is understandable. In 1811 Humphrey Davy showed that muriatic (hydrochloric) acid (which Lavoisier had regarded as an element) does not contain oxygen, but this merely convinced some that chlorine was not an element but an oxygen-containing compound. Although a dozen oxygen-free acids had been discovered by 1830, it was not until about 1840 that the hydrogen theory of acids became generally accepted. By this time, the misnomer was too well established a name to be changed. The root comes from the Greek word οξνς, which means "sour". The key to understanding acids (as well as bases and salts) had to await Michael Faraday's mid-nineteenth century discovery that solutions of salts (known as electrolytes) conduct electricity. This implies the existence of charged particles that can migrate under the influence of an electric field. Faraday named these particles (“wanderers”). Later studies on electrolytic solutions suggested that the properties we associate with acids are due to the presence of an excess of in the solution. By 1890 the Swedish chemist Svante Arrhenius (1859-1927) was able to formulate the first useful theory of acids: "an acidic substance is one whose molecular unit contains at least one hydrogen atom that can dissociate, or , when dissolved in water, producing a hydrated hydrogen ion and an anion." Strictly speaking, an “Arrhenius acid” must contain hydrogen. However, there are substances that do not themselves contain hydrogen, but still yield hydrogen ions when dissolved in water; the hydrogen ions come from the water itself, by reaction with the substance. There are three important points to understand about hydrogen in acids: S sulfide ion You will find out in a later section of this lesson that hydrogen ions cannot exist as such in water, but don't panic! It turns out that chemists still find it convenient to pretend as if they are present, and to write reactions that include them. The name has long been associated with a class of compounds whose aqueous solutions are characterized by: Just as an acid is a substance that liberates hydrogen ions into solution, a yields when dissolved in water: NaOH → Na + OH Sodium hydroxide is an because it contains hydroxide ions. However, other substances which do not contain hydroxide ions can nevertheless produce them by reaction with water, and are therefore also classified as bases. Two classes of such substances are the metal oxides and the hydrogen compounds of certain nonmetals: Na O + H O → [2 NaOH] → 2 Na + 2 OH NH + H O → NH + OH Acids and bases react with one another to yield two products: water, and an ionic compound known as a . This kind of reaction is called a . This "molecular" equation is convenient to write, but we need to re-cast it as a net ionic equation to reveal what is really going on here when the reaction takes place in water, as is almost always the case. H + + + OH + + H O If we cancel out the ions that appear on both sides (and therefore don't really participate in the reaction), we are left with the net equation H + OH → H O ( ) which is the fundamental process that occurs in all neutralization reactions. Confirmation that this equation describes all neutralization reactions that take place in water is provided by experiments indicating that no matter what acid and base are combined, all liberate the same amount of heat (57.7 kJ) per mole of H neutralized. In the case of a weak acid, or a base that is not very soluble in water, more than one step might be required. For example, a similar reaction can occur between acetic acid and calcium hydroxide to produce : 2 CH COOH + Ca(OH) → CH COOCa + 2 H O If this takes place in aqueous solution, the reaction is really between the very small quantities of H and OH resulting from the dissociation of the acid and the dissolution of the base, so the reaction is identical with Equation 1: H + OH → H O If, on the other hand, we add calcium hydroxide to pure liquid acetic acid, the net reaction would include both reactants in their "molecular" forms: 2 CH COOH + Ca(OH) → 2 CH COO + Ca + 2 H O The “salt” that is produced in a neutralization reaction consists simply of the anion and cation that were already present. The salt can be recovered as a solid by evaporating the water.

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. In addition to these individual basis; please contact Plot molar volume versus molar mass for these substances. According to Meyer, which would be considered metals and which would be considered nonmetals? Meyer found that the alkali metals had the highest molar volumes, and that molar volumes decreased steadily with increasing atomic mass, then leveled off, and finally rose again. The elements located on the rising portion of a plot of molar volume versus molar mass were typically nonmetals. If we look at the plot of the data in the table, we can immediately identify those elements with the largest molar volumes (A, B, F) as metals located on the left side of the periodic table. The element with the smallest molar volume (E) is aluminum. The plot shows that the subsequent elements (C, D) have molar volumes that are larger than that of E, but smaller than those of A and B. Thus, C and D are most likely to be nonmetals (which is the case: C = sulfur, D = phosphorus). What happens to the energy of a given orbital as the nuclear charge of a species increases? In a multielectron atom and for a given nuclear charge, the experienced by an electron depends on its value of . Why? The electron density of a particular atom is divided into two general regions. Name these two regions and describe what each represents. As the principal quantum number increases, the energy difference between successive energy levels decreases. Why? What would happen to the electron configurations of the transition metals if this decrease did not occur? Describe the relationship between electron shielding and on the outermost electrons of an atom. Predict how chemical reactivity is affected by a decreased effective nuclear charge. If a given atom or ion has a single electron in each of the following subshells, which electron is easier to remove? The electrons of the 1 shell have a stronger electrostatic attraction to the nucleus than electrons in the 2 shell. Give two reasons for this. Predict whether Na or Cl has the more stable 1 shell and explain your rationale. Arrange K, F, Ba, Pb, B, and I in order of decreasing atomic radius. Arrange Ag, Pt, Mg, C, Cu, and Si in order of increasing atomic radius. Using the periodic table, arrange Li, Ga, Ba, Cl, and Ni in order of increasing atomic radius. Element M is a metal that forms compounds of the type MX , MX , and MX , where X is a halogen. What is the expected trend in the ionic radius of M in these compounds? Arrange these compounds in order of decreasing ionic radius of M. The atomic radii of Na and Cl are 190 and 79 pm, respectively, but the distance between sodium and chlorine in NaCl is 282 pm. Explain this discrepancy. Are shielding effects on the atomic radius more pronounced across a row or down a group? Why? What two factors influence the size of an ion relative to the size of its parent atom? Would you expect the ionic radius of S to be the same in both MgS and Na S? Why or why not? Arrange Br , Al , Sr , F , O , and I in order of increasing ionic radius. Arrange P , N , Cl , In , and S in order of decreasing ionic radius. How is an isoelectronic series different from a series of ions with the same charge? Do the cations in magnesium, strontium, and potassium sulfate form an isoelectronic series? Why or why not? What isoelectronic series arises from fluorine, nitrogen, magnesium, and carbon? Arrange the ions in this series by What would be the charge and electron configuration of an ion formed from calcium that is isoelectronic with The 1 shell is closer to the nucleus and therefore experiences a greater electrostatic attraction. In addition, the electrons in the 2 subshell are shielded by the filled 1 shell, which further decreases the electrostatic attraction to the nucleus. Ba > K > Pb > I > B > F The sum of the calculated atomic radii of sodium and chlorine is 253 pm. The sodium cation is significantly smaller than a neutral sodium atom (102 versus 154 pm), due to the loss of the single electron in the 3 orbital. Conversely, the chloride ion is much larger than a neutral chlorine atom (181 versus 99 pm), because the added electron results in greatly increased electron–electron repulsions within the filled = 3 principal shell. Thus, transferring an electron from sodium to chlorine decreases the radius of sodium by about 50%, but causes the radius of chlorine to almost double. The net effect is that the distance between a sodium ion and a chloride ion in NaCl is than the sum of the atomic radii of the neutral atoms. Plot the ionic charge versus ionic radius using the following data for Mo: Mo , 69 pm; Mo , 65 pm; and Mo , 61 pm. Then use this plot to predict the ionic radius of Mo . Is the observed trend consistent with the general trends discussed in the chapter? Why or why not? Internuclear distances for selected ionic compounds are given in the following table. If the ionic radius of Li is 76 pm, what is the ionic radius of each of the anions? What is the ionic radius of Na ? Arrange the gaseous species Mg , P , Br , S , F , and N in order of increasing radius and justify your decisions. Identify each statement as either true or false and explain your reasoning. Based on electronic configurations, explain why the first ionization energies of the group 16 elements are lower than those of the group 15 elements, which is contrary to the general trend. The first through third ionization energies do not vary greatly across the lanthanides. Why? How does the effective nuclear charge experienced by the n electron change when going from left to right (with increasing atomic number) in this series? Most of the first row transition metals can form at least two stable cations, for example iron(II) and iron(III). In contrast, scandium and zinc each form only a single cation, the Sc and Zn ions, respectively. Use the electron configuration of these elements to provide an explanation. Of the elements Nd, Al, and Ar, which will readily form(s) +3 ions? Why? Orbital energies can reverse when an element is ionized. Of the ions B , Ga , Pr , Cr , and As , in which would you expect this reversal to occur? Explain your reasoning. The periodic trends in electron affinities are not as regular as periodic trends in ionization energies, even though the processes are essentially the converse of one another. Why are there so many more exceptions to the trends in electron affinities compared to ionization energies? Elements lying on a lower right to upper left diagonal line cannot be arranged in order of increasing electronegativity according to where they occur in the periodic table. Why? Why do ionic compounds form, if energy is required to form gaseous cations? Why is Pauling’s definition of electronegativity considered to be somewhat limited? Based on their positions in the periodic table, arrange Sb, O, P, Mo, K, and H in order of increasing electronegativity. Based on their positions in the periodic table, arrange V, F, B, In, Na, and S in order of decreasing electronegativity. 5. Both Al and Nd will form a cation with a +3 charge. Aluminum is in Group 13, and loss of all three valence electrons will produce the Al ion with a noble gas configuration. Neodymium is a lanthanide, and all of the lanthanides tend to form +3 ions because the ionization potentials do not vary greatly across the row, and a +3 charge can be achieved with many oxidants. 11. K < Mo ≈ Sb < P ≈ H < O The following table gives values of the first and third ionization energies for selected elements: Plot the ionization energies versus the number of electrons. Explain why the slopes of the and plots are different, even though the species in each row of the table have the same electron configurations. Would you expect the third ionization energy of iron, corresponding to the removal of an electron from a gaseous Fe ion, to be larger or smaller than the fourth ionization energy, corresponding to the removal of an electron from a gaseous Fe ion? Why? How would these ionization energies compare to the first ionization energy of Ca? Which would you expect to have the highest first ionization energy: Mg, Al, or Si? Which would you expect to have the highest third ionization energy. Why? Use the values of the first ionization energies given in Figure 7.11 to construct plots of first ionization energy versus atomic number for (a) boron through oxygen in the second period; and (b) oxygen through tellurium in group 16. Which plot shows more variation? Explain the reason for the variation in first ionization energies for this group of elements. Arrange Ga, In, and Zn in order of increasing first ionization energies. Would the order be the same for second and third ionization energies? Explain your reasoning. Arrange each set of elements in order of increasing magnitude of electron affinity. Arrange each set of elements in order of decreasing magnitude of electron affinity. Of the species F, O , Al , and Li , which has the highest electron affinity? Explain your reasoning. Of the species O , N , Hg , and H , which has the highest electron affinity? Which has the lowest electron affinity? Justify your answers. The Mulliken electronegativity of element A is 542 kJ/mol. If the electron affinity of A is −72 kJ/mol, what is the first ionization energy of element A? Use the data in the following table as a guideline to decide if A is a metal, a nonmetal, or a semimetal. If 1 g of A contains 4.85 × 10 molecules, what is the identity of element A? Based on their valence electron configurations, classify the following elements as either electrical insulators, electrical conductors, or substances with intermediate conductivity: S, Ba, Fe, Al, Te, Be, O, C, P, Sc, W, Na, B, and Rb. Using the data in Problem 10, what conclusions can you draw with regard to the relationship between electronegativity and electrical properties? Estimate the approximate electronegativity of a pure element that is very dense, lustrous, and malleable. Of the elements Al, Mg, O , Ti, I , and H , which, if any, would you expect to be a good reductant? Explain your reasoning. Of the elements Zn, B, Li, Se, Co, and Br , which if any, would you expect to be a good oxidant? Explain your reasoning. Determine whether each species is a good oxidant, a good reductant, or neither. Determine whether each species is a good oxidant, a good reductant, or neither. Of the species I , O , Zn, Sn , and K , choose which you would expect to be a good oxidant. Then justify your answer. The general features of both plots are roughly the same, with a small peak at 12 electrons and an essentially level region from 15–16 electrons. The slope of the plot is about twice as large as the slope of the plot, however, because the values correspond to removing an electron from an ion with a +2 charge rather than a neutral atom. The greater charge increases the effect of the steady rise in effective nuclear charge across the row. Electron configurations: Mg, 1 2 2 3 ; Al, 1 2 2 3 3 ; Si, 1 2 2 3 3 ; First ionization energies increase across the row due to a steady increase in effective nuclear charge; thus, Si has the highest first ionization energy. The third ionization energy corresponds to removal of a 3 electron for Al and Si, but for Mg it involves removing a 2 electron from a filled inner shell; consequently, the third ionization energy of Mg is the highest. insulators: S, O, C (diamond), P; conductors: Ba, Fe, Al, C (graphite), Be, Sc, W, Na, Rb; Te and B are semimetals and semiconductors. Mg, Al, Ti, and H I is the best oxidant, with a moderately strong tendency to accept an electron to form the I ion, with a closed shell electron configuration. O would probably also be an oxidant, with a tendency to add an electron to form salts containing the oxide ion, O . Zn and Sn are all reductants, while K has no tendency to act as an oxidant or a reductant. see above question to tease out

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Case_Studies/Blood_Glucose_Test

With a couple of million people in the U. S. with diabetes, it is necessary to have a simple, specific test for the concentration of glucose in the blood. A rapid test is needed to manage the levels of insulin in diabetes mellitus. If not enough insulin is present, the blood glucose may be very elevated. On the other hand if too much insulin is present, the glucose levels are too low. Blood glucose levels are now measured by a procedure based upon the enzyme . Since an enzyme is used, it is very specific for only D-glucose, and will not be subject to interferences from other molecules in the blood. Glucose is a , which means that it can be oxidized. The enzyme glucose oxidase catalyzes the oxidation of beta-D-glucose to D-gluconic acid. The alpha-D-glucose is rapidly converted to the beta form so that all of the glucose is measured at one time. Diatomic oxygen from the air is the oxidizing agent acting upon the glucose reducing agent. During the reaction the ring opens and the aldehyde on carbon # 1 is converted to the acid, D-gluconic acid. At the same time the oxygen in the presence of water is converted to hydrogen peroxide. So far all of the chemicals are colorless, so you would not be able to see the reaction taking place. Therefore another step is needed to produce a color. Several methods of detection are possible, but in most cases, the hydrogen peroxide reacts with a second color producing chemical. An example is something called o-Toluidine or 2-methylaniline which reacts with the hydrogen peroxide using an enzyme called peroxidase to produce a color forming chemical. The concentration of the glucose can be related to the intensity of color produced. The more the intensity, the higher the concentration of glucose. A simple color chart can be used to "read" the concentration of the glucose.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.01%3A_Chemical_Potential_Energy

Gunpowder was originally developed by the Chinese in the ninth century , primarily for rockets. This material is composed of charcoal, sulfur, and saltpeter (potassium nitrate). The explosive reaction that occurs involves the conversion of the charcoal to carbon dioxide, with the potassium nitrate providing the extra oxygen needed for a rapid reaction. Although sulfur was included to stabilize the product, gunpowder is still highly explosive. Two basic types of energy exist: potential energy and kinetic energy. is stored energy. It has not yet been released, but is ready to go. is the energy of motion. It causes work to be done through movement. Energy is the capacity for doing work or supplying heat. When you fill your car with gasoline, you are providing it with potential energy. is the energy stored in the chemical bonds of a substance. The various chemicals that make up gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car. The release of that energy does two things: some of the potential energy is transformed into work, which is used to move the car; at the same time, some of the potential energy is converted to heat and makes the car's engine very hot. The energy changes of a system occur as either heat or work, or some combination of both. Dynamite is another example of chemical potential energy. The major component of dynamite is nitroglycerin, a very unstable material. By mixing it with diatomaceous earth, the stability is increased and it is less likely to explode if it receives a physical shock. When ignited, the nitroglycerin explodes rapidly, releasing large amounts of nitrogen and other gases along with a massive amount of heat.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Absolute_Configuration_R-S_Sequence_Rules

To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature. The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and, as such, is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer: However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of , although different for the two enantiomers of a chiral molecule,at the same temperature, be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes. The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S. Consider the first picture: a curved arrow is drawn from the highest priority ( ) substituent to the lowest priority ) substituent. If the arrow points in a counterclockwise direction ( when leaving the 12 o' clock position), the configuration at stereocenter is considered ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,( when leaving the 12 o' clock position) then the stereocenter is labeled ("Rectus" → Latin= "right"). The is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. ( )-2-Bromobutane ( )-2,3- Dihydroxypropanal Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules: First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number. When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated. Remember that If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority. If the chains are similar, proceed down the chain, until a point of difference. : an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl. If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to. A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below: Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above. Keep in mind that priority is determined by the point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant. After all your substituents have been prioritized in the correct manner, you can now name/label the molecule or i) If it is it is ii) if it is it is . Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S). : remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models. Are the following R or S?

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Solubility_-_What_dissolves_in_What%3F

A solution is a homogeneous mixture in which the solute particles are so small that they cannot be filtered out and they do not scatter light. For a solution to form, there needs to be the correct combination of enthalpic (intermolecular forces) and entropic effects to enable the solvent and solute particles to interact strongly with each other without decreasing the entropy of the system. In this section we will closely examine the effects of enthalpy, but we will not spend too much time discussing the effects of entropy. To understand the effects of intermolecular forces, we need to go all the way back to bonds. A bond is defined as the sharing of electrons by two atoms. Bond polarity occurs when two atoms share electrons unequally because of a difference in the strength of attraction for the shared electrons by the nuclei of the two different atoms. The strength of attraction scale that is most commonly used is electronegativity (EN). There are three common categories of bonds, based on the difference in electronegativity (∆EN) values of the two bonding atoms: There are approaches to the task of determining into which category a given bond should be placed. Some methods are more rigorous than others, but often the rigorous methods are too complicated to be used quickly and simply. You should find a method that you like and that serves your needs sufficiently and stick with it until you are asked to use a more rigorous method. The simplest method is that a bond between a metal and a nonmetal is ionic, whereas a bond between two nonmetals is covalent. There are two common conventions used to show the polarity of a bond: δ The concept of molecular polarity is most easily applied to small molecules, but it can be applied to larger molecules with the understanding that larger molecules may have multiple “identities.” The first task in determining molecular polarity is to identify whether a substance is comprised of ions or of molecules. Individual small ions are not polar, because they do not have a charge separation that leads to the existence of a positive pole and a negative pole. Thus, if you identify a compound as consisting of ions (either monatomic or small polyatomic), you are no longer concerned about molecular polarity of that substance. To determine the molecular polarity of a compound that contains only covalent bonds, you need to consider three aspects of the molecule: Taking these three aspects into account, you should be able to determine if a small molecule is polar or nonpolar. Some helpful that are true: a) If the central atom(s) in a molecule has at least one lone electron pair, the molecule will be polar. NH is polar, and so is O b) If the molecule contains only nonpolar bonds and there are no lone pairs on the central atom(s), the molecule will be nonpolar. All C H molecules are nonpolar c) If all of the atoms attached to the central atom are of the same element and there are no lone pairs on the central atom, the molecule is nonpolar. CCl and CO are nonpolar d) If the atoms attached to the central atom are of different elements, the molecule is polar. CH Cl and OCS are polar e) If a molecule of a compound contains at least one N, O, F, or Cl atom, the molecule is usually polar, unless c) is true. CH CH OH, HF, CH NH , and CH Cl are all polar f) When molecules get large enough, they might have polar sections and nonpolar sections. A general rule of thumb for carbon-containing compounds is that molecules with 4/5 or fewer C atoms per polar bond can be considered to act as if they are polar. Thus, molecules with 4/5 or more C atoms per polar bond can be considered to act as if they were nonpolar. CH CH OH is considered polar; CH CH CH CH CH CH CH OH is considered nonpolar; CH CH CH CH OH is right on the polar/nonpolar divide g) All single atoms are nonpolar. Most, but not all, molecules of elements are nonpolar. If you have a collection of independent atoms, molecules, or ions, these particles will always exert forces of attraction and repulsion among themselves as they more closely approach each other. The repulsive forces tend to be ignored, but they do exist and come about mainly because of the repulsion of the negatively-charged electron clouds surrounding each particle (even for cations, although the positive charges do lead to repulsion with other cations and the nucleus of other atoms.) A great deal of time is spent studying the attractive forces among independent particles. These forces are what enable molecules and atoms to gather together and remain in a condensed phase as liquids and solids. Collectively, these attractive forces are commonly known as intermolecular forces (IMFs). They are the forces of electrostatic attraction that exist between separate particles. Along with the common definition that bonds occur molecules and IMFs occur molecules, there are two closely related factors that distinguish bonds and IMFs: These statements are generalizations, and there are certainly exceptions, but they hold true for most common substances. The simplest way to categorize IMFs is to classify the particles that are interacting as one of the following: From these three particle types, you can describe five different types of interparticle interactions: (All particles exert and experience London forces, but the forces among nonpolar particles are London forces) The only remaining interaction would be ions interacting with ions, but that is an ionic bond. There is, however, a sixth IMF. It is a case of extreme dipole/dipole interactions: The only bonds that are polar enough to result in a large enough partial positive charge on a small atom are: δ δ δ δ δ δ F—H, O—H, and N—H bonds. The only atoms that have a large enough EN value to result in a large enough partial negative charge on a small atom are: δ δ F, O, and N This discussion ignores entropy and focuses on enthalpy, but entropy definitely plays a role. That discussion is for later. A solution is a homogeneous mixture in which the solute particles are so small that they cannot be filtered out and so dispersed throughout the mixture that they do not scatter a beam of light that travels through the mixture. The probability of one substance dissolving/mixing in/with another substance to form a solution depends on whether or not the newly formed IMFs are as energetically favorable as the currently existing IMFs. Thus, the mutual solubility of two substances depends mainly on the substance with the IMFs that involve the largest charge separation. In other words, there is a hierarchy of what will mix: Ionically-bonded solids will form a solution (dissolve) in a solvent only if the solvent can provide very strong ion/dipole interactions. Usually water is the only material that can dissolve inorganic salts to any great extent, although some inorganic salts dissolve slightly in methanol and DMSO. You can find tables of solubility rules for common inorganic salts in water. One of these rules is that almost all compounds containing Group IA cations are soluble. Organic salts have a reasonable solubility in many polar organic solvents, but are generally much more soluble in water than in these organic solvents. The non-organic counter cations are usually Na , K , or NH , all of which generally form soluble salts. The non-organic counter anions are usually Cl , NO , and SO , all of which generally form soluble salts. Water will form a solution (mix or dissolve) with another covalently-bonded substance only if: the molecules of that substance can form hydrogen bonds with the water molecules. -The molecules of the other substance do not need to be able to form hydrogen bonds among themselves when in the pure form, but they must be polar molecules and contain an N, O, or F atom that has a partial negative charge. Acetone is an example of such a substance. The ability of a molecule to form hydrogen bonds with water molecules does not guarantee that the molecule will dissolve in water. If there is a portion of the molecule that is nonpolar, this nonpolar section interferes with the water molecules’ ability to form hydrogen bonds among themselves. Thus, the larger the nonpolar section, the less soluble the substance is in water. A general rule of thumb is the : Molecules with fewer than 4-5 C atoms per hydrogen bonding group will likely be soluble in water. Molecules with more than 4-5 C atoms per hydrogen bonding group will likely be insoluble in water. The water solubility of molecules with 4-5 C atoms per hydrogen bonding group is best determined by experiment or by a search of the literature. For instance, the following compounds, all of which are polar and can form hydrogen bonds with water, contain four C atoms per molecule: butanoic acid is soluble with water in all ratios; 1-butanol is reasonably soluble in water; and diethyl ether is so slightly soluble in water that it is considered insoluble. the molecules of that substance react with water molecules to form ions, which then dissolve in the water because of ion/dipole interactions. - The six strong acids (HCl, HBr, HI, HNO , HClO , and H SO ) are the most common examples. All of these acids have K values greater than 10. (Remember, the K value tells you how likely it is for the acid to react with water, and a K greater than 1 means that the reaction is product-favored at equilibrium) All molecular weak acids and molecular weak bases react with water to some extent to create ions, but these weak acids all have K values that are less than 1 and these weak bases all have K values that are less than 1. Thus, all these reactions are considered reactant-favored, and for purposes of predicting water solubility, we ignore the fact that some ions are formed. Remember that the intact weak acid or weak base molecules may themselves dissolve in water through hydrogen bonding. (See section b) (i) above. The common adage “Like dissolves like” works well for most compounds. Polar compounds tend to mix well with other polar compounds. The IMFs involved can be hydrogen boding, dipole/dipole, and London forces. Nonpolar compounds tend to mix well with other nonpolar compounds, with London forces being the only IMF involved. With a few notable exceptions (CCl , for instance), organic compounds containing chlorine and bromine and iodine do have a dipole moment, and are therefore polar molecules. These molecules cannot, however, form hydrogen bonds with water, nor are the halogen atoms present as ions. Thus, halogenated organic compounds do not dissolve well in water. Because most organic compounds have at least one small nonpolar section on their molecules, some polar organic compounds mix well with nonpolar organic compounds. Solubility in aqueous acid solutions and/or aqueous base solutions is a confusing concept, mainly because of the way that chemists talk about the process. The statement, “Check the solubility of benzoic acid in 3 M NaOH” may be clear to a practicing chemist, but it is misleading to most beginning students. The statement would be much clearer if it were written, “Check the water solubility of the organic anion formed when benzoic acid reacts with the OH ion in an aqueous 3M NaOH solution.” The same clarity, or lack thereof, is true for statements about solubility in aqueous acid solutions. The order of events when an aqueous solution of a base is added to an organic compound that can act as an acid is: i) The organic compound reacts with the OH ion, donating an H ion to the OH ion. ii) The products of this reaction are water and the conjugate base of the organic compound. If the original organic compound was a neutral molecule, the conjugate base will be a water-soluble anion. The order of events when an aqueous solution of an acid is added to an organic compound that can act as a base is: i) The organic compound reacts with the acid, accepting an H ion. ii) The products of this reaction are the conjugate base of the acid, and the conjugate acid of the organic compound. If the original organic compound was a neutral molecule, the conjugate acid will be a water-soluble cation. If no acid/base reaction occurs between the aqueous acid/base and the organic compound, then the organic compound remains as a neutral substance, and maintains its original solubility (or lack thereof ) in water.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.26%3A_Cross-Linking

The formation of covalent bonds which hold portions of several polymer chains together is called . Extensive cross-linking results in a random three-dimensional network of interconnected chains, as shown in the figure. As one might expect, extensive cross-linking produces a substance which has more rigidity, hardness, and a higher melting point than the equivalent polymer without cross-linking. Almost all the hard and rigid plastics we use are cross-linked. These include , which is used in many electric plugs and sockets, , which is used in plastic crockery, and epoxy resin glues. Below is a video of the formation of Polyurethane Foam. Polyether polyol, a blowing agent, which adds a gas to the mixture to produce a foam, silicone surfactant, and a catalyst is mixed with a second liquid contains a polyfunctional isocyanate. The polyol and the polyfunctional isocyanate react to form polyurethane - a very hard substance when dried. The general reaction is shown below: In the reaction in the video, each R group has multiple isocyanate groups; the reactants are polyfunctional. Thus there is a high degree of cross-linking in the polyurethane. This causes the foam to become rigid after cooling.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Background_of_Alkenes

This is an introductory page about alkenes such as ethene, propene and the rest. It deals with their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity. Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are: You can work out the formula of any of them using: C H The table is limited to the first two, because after that there are isomers which affect the names. All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula. For example, with C H , it isn't too difficult to come up with these three structural isomers: There is, however, another isomer. But-2-ene also exhibits geometric isomerism. The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH groups on either end of the molecule locked either on one side of the molecule or opposite each other. These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides). Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below. The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons. Alkenes are virtually insoluble in water, but dissolve in organic solvents. We just need to look at ethene, because what is true of C=C in ethene will be equally true of C=C in more complicated alkenes. Ethene is often modeled like this: The double bond between the carbon atoms is, of course, two pairs of shared electrons. What the diagram doesn't show is that the two pairs aren't the same as each other. One of the pairs of electrons is held on the line between the two carbon nuclei as you would expect, but the other is held in a molecular orbital above and below the plane of the molecule. A molecular orbital is a region of space within the molecule where there is a high probability of finding a particular pair of electrons. In this diagram, the line between the two carbon atoms represents a normal bond - the pair of shared electrons lies in a molecular orbital on the line between the two nuclei where you would expect them to be. This sort of bond is called a sigma bond. The other pair of electrons is found somewhere in the shaded part above and below the plane of the molecule. This bond is called a pi bond. The electrons in the pi bond are free to move around anywhere in this shaded region and can move freely from one half to the other. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.03%3A_Enthalpy

Both heat and work represent energy transfer mechanisms. As discussed previously, the term "heat" refers to a in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a than its surroundings. Similarly, work refers to the transfer of energy that does not involve temperature differences. Hence, work, like energy, can take various forms, the most familiar being mechanical and electrical. The unit of work is Joules. Work, like heat, exists only when energy is being transferred. When two bodies are placed in thermal contact and energy flows from the warmer body to the cooler one, this process is called “heat”. A transfer of energy to or from a system by any means other than heat is called “work”. To further understand the relationship between heat flow (\(q\)) and the resulting change in internal energy (\(ΔU\)), we can look at two sets of limiting conditions: reactions that occur at constant volume and reactions that occur at constant pressure. We will assume that work is the only kind of work possible for the system, so we can substitute its definition \(w=-P\Delta V\) into the first Law of Thermodynamics (\(\Delta U = q + w\)) to obtain the following: \[ΔU = q − PΔV \label{5.2.5} \] where the subscripts have been deleted. If the reaction occurs in a closed vessel, the volume of the system is fixed, and ΔV is zero. Under these conditions, the heat flow (often given the symbol q to indicate constant volume) must equal ΔU: \[\underset{\textrm{constant volume}}{q_{\textrm v}=\Delta U} \label{5.3.3} \] At constant volume, no \(PV\) work can be done, and the change in the internal energy of the system is equal to the amount of heat transferred from the system to the surroundings or vice versa. Many chemical reactions are not, however, carried out in sealed containers at constant volume but in open containers at a more or less constant pressure of about 1 atm. The heat flow under these conditions is given the symbol \(q_p\) to indicate constant pressure. Replacing q in Equation \(\ref{5.3.3}\) by \(q_p\) and rearranging to solve for q , \[\underset{\textrm{constant pressure}}{q_{\textrm p}=\Delta U+P\Delta V} \label{5.3.4} \] Thus, at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Because conditions of constant pressure are so important in chemistry, a new state function called enthalpy (H) is defined as \[H =U + PV \nonumber \] At constant pressure, the change in the enthalpy of a system is as follows: \[ΔH = ΔU + Δ(PV) = ΔU + PΔV \label{5.3.5} \] Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: \(ΔH = q_p\). This expression is consistent with our definition of enthalpy, where we stated that enthalpy is the heat absorbed or produced during any process that occurs at constant pressure. At constant pressure, the change in the enthalpy of a system is equal to the heat flow: \(ΔH = q_p\). The molar enthalpy of fusion for ice at 0.0°C and a pressure of 1.00 atm is 6.01 kJ, and the molar volumes of ice and water at 0°C are 0.0197 L and 0.0180 L, respectively. Calculate ΔH and ΔU for the melting of ice at 0.0°C. : enthalpy of fusion for ice, pressure, and molar volumes of ice and water : ΔH and ΔU for ice melting at 0.0°C : A Because 6.01 kJ of heat is absorbed from the surroundings when 1 mol of ice melts, q = +6.01 kJ. When the process is carried out at constant pressure, q = q = ΔH = 6.01 kJ. B To find ΔU using Equation \(\ref{18.11}\), we need to calculate Δ(PV). The process is carried out at a constant pressure of 1.00 atm, so C Substituting the calculated values of ΔH and PΔV into Equation 18.11, ΔU = ΔH − PΔV = 6010 J − (−0.0017 J) = 6010 J = 6.01 kJ At 298 K and 1 atm, the conversion of graphite to diamond requires the input of 1.850 kJ of heat per mole of carbon. The molar volumes of graphite and diamond are 0.00534 L and 0.00342 L, respectively. Calculate ΔH and ΔU for the conversion of C (graphite) to C (diamond) under these conditions. ΔH = 1.85 kJ/mol ΔU = 1.85 kJ/mol If ΔH for a reaction is known, we can use the change in the enthalpy of the system (Equation \(\ref{5.3.5}\)) to calculate its change in internal energy. When a reaction involves only solids, liquids, liquid solutions, or any combination of these, the volume does not change appreciably (ΔV = 0). Under these conditions, we can simplify Equation \(\ref{5.3.5}\) to ΔH = ΔU. If gases are involved, however, ΔH and ΔU can differ significantly. We can calculate ΔU from the measured value of ΔH by using the right side of Equation \(\ref{5.3.5}\) together with the ideal gas law, PV = nRT. Recognizing that Δ(PV) = Δ(nRT), we can rewrite Equation \(\ref{5.3.5}\) as follows: \[ΔH = ΔU + Δ(PV) = ΔU + Δ(nRT) \label{5.3.6} \] At constant temperature, Δ(nRT) = RTΔn, where Δn is the difference between the final and initial numbers of moles of gas. Thus \[ΔU = ΔH − RTΔn \label{5.3.7} \] For reactions that result in a net production of gas, Δn > 0, so ΔU < ΔH. Conversely, endothermic reactions (ΔH > 0) that result in a net consumption of gas have Δn < 0 and ΔU > ΔH. The relationship between ΔH and ΔU for systems involving gases is illustrated in Example \(\Page {2}\). For reactions that result in a net production of gas, ΔU < ΔH. For endothermic reactions that result in a net consumption of gas, ΔU > ΔH. The combustion of graphite to produce carbon dioxide is described by the equation \[C_{(graphite, s)} + O_{2(g)} → CO_{2(g)} \nonumber \] At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔU for the reaction? : balanced chemical equation, temperature, pressure, ΔH, and molar volume of reactant : ΔU : A In this reaction, 1 mol of gas (CO ) is produced, and 1 mol of gas (O ) is consumed. Thus Δn = 1 − 1 = 0. B Substituting this calculated value and the given values into Equation \(\ref{5.3.7}\), To understand why only the change in the volume of the gases needs to be considered, notice that the molar volume of graphite is only 0.0053 L. A change in the number of moles of gas corresponds to a volume change of 22.4 L/mol of gas at standard temperature and pressure (STP), so the volume of gas consumed or produced in this case is (1)(22.4 L) = 22.4 L, which is much, much greater than the volume of 1 mol of a solid such as graphite. Calculate \(ΔU\) for the conversion of oxygen gas to ozone at 298 K: \[3O_2(g) \rightarrow 2O_3(g). \nonumber \] The value of \(ΔH\) for the reaction is 285.4 kJ. 288 kJ As Example \(\Page {2}\) illustrates, the magnitudes of ΔH and ΔU for reactions that involve gases are generally rather similar, even when there is a net production or consumption of gases. Enthalpy is a state function, and the change in enthalpy of a system is equal to the sum of the change in the internal energy of the system and the PV work done. Enthalpy is a state function whose change indicates the amount of heat transferred from a system to its surroundings or vice versa, at constant pressure. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU.

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You may be wondering, why should neutral molecules attract each other at all? If the molecules are polar, the explanation is fairly obvious. When two polar molecules approach each other, they can arrange themselves in such a way that the negative side of one molecule is close to the positive end of the other:   The molecules will then attract each other because the charges which are closest together are opposite in sign. (This behavior is very similar to the attraction between two bar magnets placed end to end or side by side with the north poles opposite the south poles.) Forces between polar molecules which arise in this way are called . The existence of dipole forces explains why polar molecules have higher boiling points and melting points than do nonpolar molecules. In the following table, we compare the boiling points of several pairs of molecules. In each pair, one molecule is polar and the other is nonpolar, but otherwise they are as similar as possible. The polar substance always has the higher boiling point, indicating greater attractive forces between separate molecules, that is, larger intermolecular forces.

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Make sure you thoroughly understand the following essential ideas: The periodic table in the form originally published by Dmitri Mendeleev in 1869 was an attempt to list the chemical elements in order of their atomic weights, while breaking the list into rows in such a way that elements having similar physical and chemical properies would be placed in each column. At that time, nothing was known about atoms; the development of the table was entirely empirical. Our goal in this lesson is to help you understand how the shape and organization of the modern periodic table are direct consequences of the atomic electronic structure of the elements. To understand how the periodic table is organized, imagine that we write down a long horizontal list of the elements in order of their increasing atomic number. It would begin this way: H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca... Now if we look at the various physical and chemical properties of these elements, we would find that their values tend to increase or decrease with in a manner that reveals a repeating pattern— that is, a . For the elements listed above, these breaks can be indicated by the vertical bars shown here in color: ... To construct the table, we place each sequence in a separate row, which we call a . The rows are aligned in such a way that the elements in each vertical column possess certain similarities. Thus the first short-period elements H and He are chemically similar to the elements Li and Ne at the beginning and end of the second period. The first period is split in order to place H above Li and He above Ne. The "block" nomenclature shown above refers to the sub-orbital type (quantum number l, or s-p-d-f classification) of the highest-energy orbitals that are occupied in a given element. For =1 there is no block, and the block is split so that helium is placed in the same group as the other inert gases, which it resembles chemically. For the second period ( =2) there is a block but no block; in the usual "long form" of the periodic table it is customary to leave a gap between these two blocks in order to accommodate the blocks that occur at =3 and above. At =6 we introduce an block, but in order to hold the table to reasonable dimensions the blocks are placed below the main body of the table. : Each column of the periodic table is known as a . The elements belonging to a given group bear a strong similarity in their chemical behaviors. In the past, two different systems of Roman numerals and letters were used to denote the various groups. North Americans added the letter to denote the -block groups and for the others; this is the system shown in the table above. The the rest of the world used for the -block elements and for the others. In 1985, a new international system was adopted in which the columns were simply labeled 1-18. Although this system has met sufficient resistance in North America to slow its incorporation into textbooks, it seems likely that the "one to eighteen" system will gradually take over. Chemists have long found it convenient to refer to the elements of different groups, and in some cases of spans of groups by the names indicated in the table shown below. The two of these that are most important for you to know are the and the . The properties of an atom depend ultimately on the number of electrons in the various orbitals, and on the nuclear charge which determines the compactness of the orbitals. In order to relate the properties of the elements to their locations in the periodic table, it is often convenient to make use of a simplified view of the atom in which the nucleus is surrounded by one or more concentric spherical "shells", each of which consists of the highest-principal quantum number orbitals (always - and -orbitals) that contain at least one electron. The shell model (as with any scientific model) is less a description of the world than a simplified way of looking at it that helps us to understand and correlate diverse phenomena. The principal simplification here is that it deals only with the of the - and -blocks, omitting the - and -block elements whose properties tend to be less closely tied to their group numbers. In particular, the number of outer-shell electrons (which is given by the rightmost digit in the ) is a major determinant of an element's "combining power", or . The general trend is for an atom to gain or lose electrons, either directly (leading to formation of ) or by sharing electrons with other atoms so as to achieve an outer-shell configuration of s p . This configuration, known as an , corresponds to that of one of the noble-gas elements of Group 18. The above diagram shows the first three rows of what are known as the — that is, the - and -block elements only. As we move farther down (into the fourth row and below), the presence of -electrons exerts a complicating influence which allows elements to exhibit multiple valances. This effect is especially noticeable in the transition-metal elements, and is the reason for not including the d-block with the representative elements at all. Those electrons in the outmost or valence shell are especially important because they are the ones that can engage in the sharing and exchange that is responsible for chemical reactions; how tightly they are bound to the atom determines much of the chemistry of the element. The degree of binding is the result of two opposing forces: the attraction between the electron and the nucleus, and the repulsions between the electron in question and all the other electrons in the atom. All that matters is the net force, the difference between the nuclear attraction and the totality of the electron-electron repulsions. We can simplify the shell model even further by imagining that the valence shell electrons are the electrons in the atom, and that the nuclear charge has whatever value would be required to bind these electrons as tightly as is observed experimentally. Because the number of electrons in this model is less than the atomic number , the required nuclear charge will also be smaller. and is known as the . Effective nuclear charge is essentially the positive charge that a valence electron "sees". Part of the difference between and is due to other electrons in the valence shell, but this is usually only a minor contributor because these electrons tend to act as if they are spread out in a diffuse spherical shell of larger radius. The main actors here are the electrons in the much more compact inner shells which surround the nucleus and exert what is often called a shielding or " " effect on the valence electrons. The formula for calculating effective nuclear charge is not very complicated, but we will skip a discussion of it here. An even simpler although rather crude procedure is to just subtract the number of inner-shell electrons from the nuclear charge; the result is a form of effective nuclear charge which is called the of the atom. The concept of "size" is somewhat ambiguous when applied to the scale of atoms and molecules. The reason for this is apparent when you recall that an atom has no definite boundary; there is a finite (but very small) probability of finding the electron of a hydrogen atom, for example, 1 cm, or even 1 km from the nucleus. It is not possible to specify a definite value for the radius of an isolated atom; the best we can do is to define a spherical shell within whose radius some arbitrary percentage of the electron density can be found . When an atom is combined with other atoms in a solid element or compound, an effective radius can be determined by observing the distances between adjacent rows of atoms in these solids. This is most commonly carried out by X-ray scattering experiments. Because of the different ways in which atoms can aggregate together, several different kinds of atomic radii can be defined. Distances on the atomic scale have traditionally been expressed in Ångstrom units (1Å = 10 cm = 10 m), but nowadays the picometer is preferred: 1 pm = 10 m = 10 cm = 10 Å, or 1Å = 100 pm. The radii of atoms and ions are typically in the range 70-400 pm. A rough idea of the size of a metallic atom can be obtained simply by measuring the density of a sample of the metal. This tells us the number of atoms per unit volume of the solid. The atoms are assumed to be spheres of radius in contact with each other, each of which sits in a cubic box of edge length 2 . The volume of each box is just the total volume of the solid divided by the number of atoms in that mass of the solid; the atomic radius is the cube root of . Although the radius of an atom or ion cannot be measured directly, in most cases it can be inferred from measurements of the distance between adjacent nuclei in a crystalline solid. This is most commonly carried out by X-ray scattering experiments. Because such solids fall into several different classes, several kinds of atomic radius are defined. Many atoms have several different radii; for example, sodium forms a metallic solid and thus has a metallic radius, it forms a gaseous molecule Na in the vapor phase (covalent radius), and of course it forms ionic solids such as NaCl. is the effective radius of ions in solids such as NaCl. It is easy enough to measure the distance between adjacent rows of Na and Cl ions in such a crystal, but there is no unambiguous way to decide what portions of this distance are attributable to each ion. The best one can do is make estimates based on studies of several different ionic solids (LiI, KI, NaI, for example) that contain one ion in common. Many such estimates have been made, and they turn out to be remarkably consistent. Many atoms have several different radii; for example, sodium forms a metallic solid and thus has a metallic radius, it forms a gaseous molecule Na in the vapor phase (covalent radius), and of course it forms ionic solids as mentioned above. We would expect the size of an atom to depend mainly on the principal quantum number of the highest occupied orbital; in other words, on the "number of occupied electron shells". Since each row in the periodic table corresponds to an increment in n, atomic radius increases as we move down a column. The other important factor is the nuclear charge; the higher the atomic number, the more strongly will the electrons be drawn toward the nucleus, and the smaller the atom. This effect is responsible for the contraction we observe as we move across the periodic table from left to right. The figure shows a periodic table in which the sizes of the atoms are represented graphically. The apparent discontinuities in this diagram reflect the difficulty of comparing the radii of atoms of metallic and nonmetallic bonding types. Radii of the noble gas elements are estimates from those of nearby elements. is always smaller than the neutral atom, owing to the diminished electron-electron repulsion. If a second electron is lost, the ion gets even smaller; for example, the ionic radius of Fe is 76 pm, while that of Fe is 65 pm. If formation of the ion involves complete emptying of the outer shell, then the decrease in radius is especially great. is in a class by itself; having no electron cloud at all, its radius is that of the bare proton, or about 0.1 pm— a contraction of 99.999%! Because the unit positive charge is concentrated into such a small volume of space, the of the hydrogen ion is extremely high; it interacts very strongly with other matter, including water molecules, and in aqueous solution it exists only as the H O . are always larger than the parent ion; the addition of one or more electrons to an existing shell increases which results in a general expansion of the atom. An is a sequence of species all having the same number of electrons (and thus the same amount of electron-electron repulsion) but differing in nuclear charge. Of course, only one member of such a sequence can be a neutral atom (neon in the series shown below.) The effect of increasing nuclear charge on the radius is clearly seen. Chemical reactions are based largely on the interactions between the most loosely bound electrons in atoms, so it is not surprising that the tendency of an atom to gain, lose or share electrons is one of its fundamental chemical properties. This term always refers to the formation of ions. In order to remove an electron from an atom, work must be done to overcome the electrostatic attraction between the electron and the nucleus; this work is called the of the atom and corresponds to the exothermic process \[M_{(g)} → M^+_{(g)} + e^–\] where \(M_{(g)}\) stands for any isolated (gaseous) atom. An atom has as many ionization energies as it has electrons. Electrons are always removed from the highest-energy occupied orbital. An examination of the successive ionization energies of the first ten elements (below) provides experimental confirmation that the binding of the two innermost electrons (1 orbital) is significantly different from that of the =2 electrons.Successive ionization energies of an atom increase rapidly as reduced electron-electron repulsion causes the electron shells to contract, thus binding the electrons even more tightly to the nucleus. Ionization energies increase with the nuclear charge Z as we move across the periodic table. They decrease as we move down the table because in each period the electron is being removed from a shell one step farther from the nucleus than in the atom immediately above it. This results in the familiar zig-zag lines when the first ionization energies are plotted as a function of Z. This more detailed plot of the ionization energies of the atoms of the first ten elements reveals some interesting irregularities that can be related to the slightly lower energies (greater stabilities) of electrons in half-filled (spin-unpaired) relative to completely-filled subshells. Finally, a more comprehensive survey of the ionization energies of the main group elements is shown below. Some points to note: Formation of a negative ion occurs when an electron from some external source enters the atom and become incorporated into the lowest energy orbital that possesses a vacancy. Because the entering electron is attracted to the positive nucleus, the formation of negative ions is usually exothermic. The energy given off is the electron affinity of the atom. For some atoms, the electron affinity appears to be slightly negative, suggesting that electron-electron repulsion is the dominant factor in these instances. In general, electron affinities tend to be much smaller than ionization energies, suggesting that they are controlled by opposing factors having similar magnitudes. These two factors are, as before, the nuclear charge and electron-electron repulsion. But the latter, only a minor actor in positive ion formation, is now much more significant. One reason for this is that the electrons contained in the inner shells of the atom exert a collective negative charge that partially cancels the charge of the nucleus, thus exerting a so-called shielding effect which diminishes the tendency for negative ions to form. Because of these opposing effects, the periodic trends in electron affinities are not as clear as are those of ionization energies. This is particularly evident in the first few rows of the periodic table, in which small effects tend to be magnified anyway because an added electron produces a large percentage increase in the number of electrons in the atom. In general, we can say that electron affinities become more exothermic as we move from left to right across a period (owing to increased nuclear charge and smaller atom size). There are some interesting irregularities, however: When two elements are joined in a chemical bond, the element that attracts the shared electrons more strongly is more . Elements with low electronegativities (the metallic elements) are said to be . Moreover, the same atom can exhibit different electronegativities in different chemical environments, so the "electronegativity of an element" is only a general guide to its chemical behavior rather than an exact specification of its behavior in a particular compound. Nevertheless, electronegativity is eminently useful in summarizing the chemical behavior of an element. You will make considerable use of electronegativity when you study chemical bonding and the chemistry of the individual elements. Because there is no single definition of electronegativity, any numerical scale for measuring it must of necessity be somewhat arbitrary. Most such scales are themselves based on atomic properties that are directly measurable and which relate in one way or the other to electron-attracting propensity. The most widely used of these scales was devised by Linus Pauling and is related to ionization energy and electron affinity. The Pauling scale runs from 0 to 4; the highest electron affinity, 4.0, is assigned to fluorine, while cesium has the lowest value of 0.7. Values less than about 2.2 are usually associated with electropositive, or metallic character. In the representation of the scale shown in figure, the elements are arranged in rows corresponding to their locations in the periodic table. The correlation is obvious; electronegativity is associated with the higher rows and the rightmost columns. The location of on this scale reflects some of the significant chemical properties of this element. Although it acts like a metallic element in many respects (forming a positive ion, for example), it can also form hydride-ion (H ) solids with the more electropositive elements, and of course its ability to share electrons with carbon and other -block elements gives rise to a very rich chemistry, including of course the millions of organic compounds.

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When a donates one or more electrons, its goes up, and the resulting species is capable of reaccepting the electrons. That is, the oxidized species is an . For example, when copper metal dissolves, the copper(II) ion formed can serve as an oxidizing agent: Similarly, when an oxidizing agent such as silver ion is reduced, the silver metal can donate an electron, serving as a reducing agent: This is analogous to what we observe in the case of . For every oxidizing agent, there corresponds some reducing agent, and for every reducing agent, there corresponds an oxidizing agent. \(\Page {1}\): Selected Redox Couples Arranged in Order of Decreasing Strength of Oxidizing Agent. An oxidizing and reducing agent which appear on opposite sides of a half-equation constitute a . Redox couples are analogous to conjugate acid-base pairs and behave in much the same way. . Thus the strong oxidizing agent F produces the weak reducing agent F . Conversely, the strong reducing agent Li corresponds with the weak oxidizing agent Li . This type of relationship is demonstrated in Table \(\Page {1}\), where selected redox couples have been arranged in order of increasing strength of the reducing agent. As in the case of acids and bases, this table of half-equations can be used to predict which way a will proceed. The reactions with the greatest tendency to occur are between strong oxidizing agents from the upper left and strong reducing agents from the lower right. If a line is drawn from oxidizing agent to reducing agent for such a reaction, it will have a slope. Reactions with little tendency to occur involve the weak oxidizing agents at the lower left and weak reducing agents at the upper right. In such cases a line from oxidizing to reducing agent has an slope. When the slope of the line is not far from horizontal, the oxidizing and reducing agents are similar in strength and their reaction will go only part way to completion. Predict whether iron(III) ion, Fe , will oxidize copper metal. If so, write a balanced equation for the reaction. A line from the oxidizing agent Fe to the reducing agent Cu is downhill, and so the reaction will occur. The balanced equation is the sum of the two half-equations adjusted to equalize the number of electrons transferred. \(\ce{2Fe^{3+} + 2e^{–} -> 2Fe^{2+}}\) \(\underline{\text{Cu} \rightarrow \text{Cu}^{2+} + \cancel{2e^–}}\) \(\ce{2Fe^{3+} + Cu -> 2Fe^{2+} + Cu^{2+}}\) Note that the half-equation for Cu is reversed from that in Table 1 because the reactant Cu is a reducing agent. All half-equations in Table \(\Page {1}\) have the oxidizing agent on the left. Use Table 11.15 to find a reagent that will oxidize H O (other than F or Cl , which were mentioned earlier). We must find an oxidizing agent from which a line to the reducing agent H O has a downhill slope. The only possibilities are MnO or CrO . The latter reacts extremely slowly, but aqueous permanganate solutions decompose over a period of weeks. The concentration of MnO decreases because of the reaction

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Unit cells need not be cubes, but they be parallel-sided, three-dimensional figures. A general example is shown in Figure \(\Page {1}\). Such a cell can be described in terms of the lengths of three adjacent edges, , and , and the angles between them, α, β, and γ. Crystals are usually classified as belonging to one of seven crystal systems, depending on the shape of the unit cell. These seven systems are shown in the image below. The simplest is the cubic system, in which all edges of the unit cell are equal and all angles are 90°. The tetragonal and orthorhombic classes also feature rectangular cells, but the edges are not all equal. In the remaining classes, some or all of the angles are not 90°. The least symmetrical is the triclinic, in which no edges are equal and no angles are equal to each other or to 90°. Special note should be made of the hexagonal system whose unit cell is shown in Figure \(\Page {2}\). It is related to the two-dimensional cell encountered as the second example of a 2D crystal lattice structure, in that two edges of the cell equal and subtend an angle of 120°. Hexagonal crystals are quite common among simple compounds, like quartz, seen here below.

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There is more to the structure of DNA than just the primary sequence of nitrogenous bases. Secondary structure also plays a crucial biochemical role. Each DNA molecule consists of two nucleotide chains wrapped around each other in a double helix and held together by hydrogen bonds. This hydrogen bonding involves only the nitrogenous bases. Each of the purine bases can hydrogen bond with one and only one of the pyrimidine bases. Thus adenine can hydrogen bond with thymine and guanine with cytosine, as shown in Figure \(\Page {1}\). Note that in both cases there is an exact match of hydrogen atoms on the one base with nitrogen or oxygen atoms on the other. Note also that the distance from sugar linkage to sugar linkage across each of the base pairs in Figure \(\Page {1}\) is almost exactly the same. This explains why only these two combinations occur in DNA. Other combinations (i.e., adenine-cytosine) are not nearly so favorable energetically. The spacing between base pairs is 340 pm, and there are 10.5 base pairs in one full turn of the helix. The two nucleotide chains in the double helix are said to be to each other. Because of the exact pairing of the bases we can always tell the sequences of bases in the one chain from that in the other. Thus if the first six bases in one chain are AGATCC, we know that the first six bases in the other will be TCTAGG. Both chains are therefore alternative representations of the same information. If one or two bases become misplaced in either strand, this can be recognized because of mismatching with the complementary strand. Repair enzymes can then correct the sequence of bases along the incorrect strand. A final point to make is that the two strands are . This means that one strand, from bottom to top is going from the 5' carbon to the 3' carbon, while the complimentary strand is going 3' to 5' from bottom to top. This double-helix model for DNA was first suggested in 1953 by James D. Watson (born 1928) and Francis Crick (1916 to 2004). It was an important milestone in the history of science, since it marked the birth of a new field, molecular biology, in which the characteristics of living organisms could at last begin to be explained in terms of the structure of their molecules. In 1962 Crick and Watson shared the Nobel Prize with M. F. H. Wilkins, whose x-ray crystallographic data had helped them to formulate their model. Rosalind Franklin(1920-1958) who performed the x-ray crystallography experiments did not win the Nobel Prize, as they are not awarded posthumously, but should be included in any discussion on the discovery of the double helix. A fascinating account of this discovery, which does not always put the author in a favorable light, can be found in Watson’s book “The Double Helix."

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In addition to , there is another class of molecules which does not fit easily into of . For these molecules it is possible to draw which obeys but which is unsatisfactory in other ways. A simple example of such a molecule is ozone, an unusual form of oxygen, whose molecular formula is O . Like the oxides of nitrogen, ozone is important in a discussion of atmospheric-pollution problems, but for the moment we will confine ourselves to its structure. We can draw two Lewis diagrams for O , both of which obey the octet rule: Structure 1 suggests that there is an O―O single bond on the left and an double bond on the right side of the molecule. Structure 2 suggests the opposite placement of the double bond. However, it seems unlikely that electrons should be able to distinguish left from right in this way, and experimental evidence confirms this suspicion. Both bonds are found to have the same length, namely, 128 pm. This is intermediate between the double-bond length of 121 pm in O and the O―O single-bond length of 147 pm in H O . In other words the structure of O is somehow intermediate in character between the two structures shown. On a mathematical level, we can satisfactorily account for the properties of ozone by regarding its structure as a of the two structures shown above, the term hybrid having exactly the same sense as for hybrids. We then obtain an electron probability distribution in which both bonds receive equal treatment and are intermediate in character between double and single bonds. Such a structure is called a and is indicated in one of the following ways:   The term resonance and the use of a double-headed arrow, , are both unfortunate since they suggest that the structure is continually oscillating between the two alternatives, so that if only you were fast enough, you could “catch” the double bond on one side or the other. One can no more do this than “catch” an hybrid orbital instantaneously in the form of an or a orbital. The most important example of resonance is undoubtedly the compound benzene, C H , which has the structure   The circle within the second formula indicates that all C—C bonds in the hexagon are equivalent. This hexagonal ring of six carbon atoms is called a . Each carbon-carbon bond is 139 pm long, intermediate between the length of a C―C single bond (154 pm) and a double bond (135 pm). Whereas a double bond between two carbon atoms is normally quite reactive, the bonding between the carbons in the benzene ring is difficult to alter. In virtually all its chemical reactions, the ring structure of benzene remains intact. Even when a molecule exhibits resonance, it is still possible to . Any bonds which are intermediate in character can be treated as though they were single bonds, though perhaps a bit fatter. On this basis one would predict that the ozone molecule is angular rather than linear because of the lone pair on the central oxygen atom, with an angle slightly less than 120°. Experimentally the angle is 117°. In the same way each carbon atom in benzene can be expected to be surrounded by three atoms in a plane around it, separated by angles of approximately 120°. Again this agrees with experiment. All the atoms in C H lie in the same plane, and all bond angles are 120°. Write resonance structures to indicate the bonding in the carbonate ion, CO . Predict the O—C—O angle and the carbon-oxygen bond length. We must first write a plausible Lewis structure for the ion. Counting valence electrons, we have a total of 4(from C) + 3 × 6(from O) + 2(from charge) = 24 electrons There are 4 octets to be filled, making a total of 4 × 8 = 32 electrons. We must thus count 32 – 24 = 8 electrons twice, and so there are 4 shared pairs. Since there are only three oxygen atoms, one must be double bonded to the carbon atom:   Two other equivalent structures can also be drawn, and so the carbonate ion corresponds to the following resonance hybrid:   Since the carbon has no lone pairs in its valence shell, the three oxygens should be arranged trigonally around the carbon and all four atoms should lie in a plane. As we saw in the previous chapter, the C—O single-bond length is 143 pm, while the double-bond length is 122 pm. We can expect the carbon-oxygen distance in the carbonate ion to lie somewhere in between these values. Experimentally it is found to be 129 pm.

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1. Ionic and covalent compounds are held together by electrostatic attractions between oppositely charged particles. Describe the differences in the nature of the attractions in ionic and covalent compounds. Which class of compounds contains pairs of electrons shared between bonded atoms? 2. Which contains fewer electrons than the neutral atom—the corresponding cation or the anion? 3. What is the difference between an organic compound and an inorganic compound? 4. What is the advantage of writing a structural formula as a condensed formula? 5. The majority of elements that exist as diatomic molecules are found in one group of the periodic table. Identify the group. 6. Discuss the differences between covalent and ionic compounds with regard to a. the forces that hold the atoms together. b. melting points. c. physical states at room temperature and pressure. 7. Why do covalent compounds generally tend to have lower melting points than ionic compounds? Answer 7. Covalent compounds generally melt at lower temperatures than ionic compounds because the intermolecular interactions that hold the molecules together in a molecular solid are weaker than the electrostatic attractions that hold oppositely charged ions together in an ionic solid. 1. The structural formula for chloroform (CHCl ) was shown in Example 2. Based on this information, draw the structural formula of dichloromethane (CH Cl ). 2. What is the total number of electrons present in each ion? a. F b. Rb c. Ce d. Zr e. Zn f. Kr g. B 3. What is the total number of electrons present in each ion? a. Ca b. Se c. In d. Sr e. As f. N g. Tl 4. Predict how many electrons are in each ion. a. an oxygen ion with a −2 charge b. a beryllium ion with a +2 charge c. a silver ion with a +1 charge d. a selenium ion with a +4 charge e. an iron ion with a +2 charge f. a chlorine ion with a −1 charge 5. Predict how many electrons are in each ion. a. a copper ion with a +2 charge b. a molybdenum ion with a +4 charge c. an iodine ion with a −1 charge d. a gallium ion with a +3 charge e. an ytterbium ion with a +3 charge f. a scandium ion with a +3 charge 6. Predict the charge on the most common monatomic ion formed by each element. a. chlorine b. phosphorus c. scandium d. magnesium e. arsenic f. oxygen 7. Predict the charge on the most common monatomic ion formed by each element. a. sodium b. selenium c. barium d. rubidium e. nitrogen f. aluminum 8. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. a. \(_4^9X^{2+} \) b. \(_1^1X^-\) c. \(_8^{16}X^{2-} \) 9. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. a. \(_3^7X^+ \) b. \(_9^{19}X^-\) c. \(_{13}^{27}X^{3+}\) 5. a. 27 b. 38 c. 54 d. 28 e. 67 f. 18 9. a. Li, Li , 2nd period, group 1 b. F, F , 2nd period, group 17 c. Al, Al , 3nd period, group 13 1. What are the differences and similarities between a polyatomic ion and a molecule? 2. Classify each compound as ionic or covalent. 3. Classify each compound as ionic or covalent. Which are organic compounds and which are inorganic compounds? 4. Generally, one cannot determine the molecular formula directly from an empirical formula. What other information is needed? 5. Give two pieces of information that we obtain from a structural formula that we cannot obtain from an empirical formula. 6. The formulas of alcohols are often written as ROH rather than as empirical formulas. For example, methanol is generally written as CH OH rather than CH O. Explain why the ROH notation is preferred. 7. The compound dimethyl sulfide has the empirical formula C H S and the structural formula CH SCH . What information do we obtain from the structural formula that we do not get from the empirical formula? Write the condensed structural formula for the compound. 8. What is the correct formula for magnesium hydroxide—MgOH or Mg(OH) ? Why? 9. Magnesium cyanide is written as Mg(CN) , not MgCN . Why? 10. Does a given hydrate always contain the same number of waters of hydration? 7. The structural formula gives us the connectivity of the atoms in the molecule or ion, as well as a schematic representation of their arrangement in space. Empirical formulas tell us only the ratios of the atoms present. The condensed structural formula of dimethylsulfide is (CH ) S. 1. Write the formula for each compound. a. magnesium sulfate, which has 1 magnesium atom, 4 oxygen atoms, and 1 sulfur atom b. ethylene glycol (antifreeze), which has 6 hydrogen atoms, 2 carbon atoms, and 2 oxygen atoms c. acetic acid, which has 2 oxygen atoms, 2 carbon atoms, and 4 hydrogen atoms d. potassium chlorate, which has 1 chlorine atom, 1 potassium atom, and 3 oxygen atoms e. sodium hypochlorite pentahydrate, which has 1 chlorine atom, 1 sodium atom, 6 oxygen atoms, and 10 hydrogen atoms 2. Write the formula for each compound. a. cadmium acetate, which has 1 cadmium atom, 4 oxygen atoms, 4 carbon atoms, and 6 hydrogen atoms b. barium cyanide, which has 1 barium atom, 2 carbon atoms, and 2 nitrogen atoms c. iron(III) phosphate dihydrate, which has 1 iron atom, 1 phosphorus atom, 6 oxygen atoms, and 4 hydrogen atoms d. manganese(II) nitrate hexahydrate, which has 1 manganese atom, 12 hydrogen atoms, 12 oxygen atoms, and 2 nitrogen atoms e. silver phosphate, which has 1 phosphorus atom, 3 silver atoms, and 4 oxygen atoms 3. Complete the following table by filling in the formula for the ionic compound formed by each cation-anion pair. 4. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. a. zinc and sulfur b. barium and iodine c. magnesium and chlorine d. silicon and oxygen e. sodium and sulfur 5. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. a. lithium and nitrogen b. cesium and chlorine c. germanium and oxygen d. rubidium and sulfur e. arsenic and sodium 6. Write the empirical formula for each compound. a. Na S O b. B H c. C H O d. P O e. KMnO 7. Write the empirical formula for each compound. a. Al Cl b. K Cr O c. C H d. (NH ) CNH e. CH COOH Answers 1. a. MgSO b. C H O c. C H O d. KClO e. NaOCl·5H O 3. 5. a. Li N b. CsCl c. GeO d. Rb S e. Na As 7. a. AlCl b. K Cr O c. CH d. CH N e. CH O 1. Name each cation. 2. Name each anion. 3. Name each anion. 4. Name each anion. 5. Name each compound. 6. Name each compound. 7. Name each compound. 8. Name each compound. 9. Name each compound. 7. a. rubidium bromide b. manganese(III) sulfate c. sodium hypochlorite d. ammonium sulfate e. sodium bromide f. potassium iodate g. sodium chromate 1. For each ionic compound, name the cation and the anion and give the charge on each ion. 2. For each ionic compound, name the cation and the anion and give the charge on each ion. 3. Write the formula for each compound. 4. Write the formula for each compound. 5. Write the formula for each compound. 6. Write the formula for each compound. 7. Write the formula for each compound. 8. Write the formula for each compound. 9. Write the formula for each compound. Conceptual Problems 1. Name each acid. 2. Name each acid. 3. Name the aqueous acid that corresponds to each gaseous species. 4. For each structural formula, write the condensed formula and the name of the compound. a. b. 5. For each structural formula, write the condensed formula and the name of the compound. a. b. 6. When each compound is added to water, is the resulting solution acidic, neutral, or basic? 7. Draw the structure of the simplest example of each type of compound. 8. Identify the class of organic compound represented by each compound. a. b. CH CH OH c. HC≡CH d. e. C H NH f. CH CH=CHCH CH g. h. 9. Identify the class of organic compound represented by each compound. a. b. c. d. e. f. CH C≡CH g. h. 1. Write the formula for each compound. 2. Write the formula for each compound. 3. Name each compound. 4. Name each compound.

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Make sure you thoroughly understand the following essential ideas: Chemical change is guided and driven by energetics, but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products. The energetic aspects of change are governed by the laws of thermodynamics (the "dynamics" part of this word is related to the historical origins of the field and is not a part of dynamics in the sense of these lessons.) Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more and the system is said to be in . The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need. Thermodynamics points the way and makes it possible...but it says nothing about how long it will take to get there! The stoichiometric equation for the reaction says nothing about its By mechanism, we mean, basically, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products: The inner workings of the black box are ordinarily hidden from us, are highly unpredictable and can only be inferred by indirect means. Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements. The thermodynamics of these reactions are all similar (they are all highly exothermic), but their (their kinetics and mechanisms) could not be more different. Careful experiments, carried out over many years, are consistent with the simplest mechanism: a single collision between the two reactant molecules results in a rearrangement of the bonds: One might be tempted to suppose that this would proceed in a similar way, but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br. The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively. What is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation. Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed: \[\color{red} \text{rate} = \dfrac{\Delta \text{concentration}}{\Delta \text{time}} \label{2-1}\] For a reaction of the form A + B → C, the rate can be expressed in terms of the change in concentration of any of its components: \[\text{rate} = \dfrac{-\Delta[A]}{\Delta t} \label{Eq1a}\] \[\text{rate} = \dfrac{-\Delta[B]}{\Delta t} \label{Eq1b}\] \[\text{rate} = \dfrac{\Delta[C]}{\Delta t} \label{Eq1c}\] al \(\Delta t = t_2 – t_1\): \[Δ[A] = [A]_2 – [A]_1 \label{2-2}\] Notice the minus signs in Equations \(\ref{Eq1a}\) and \(\ref{Eq1a}\); the concentration of a always with time, so Δ[A] and Δ[B] are both negative. Since negative rates does not make much sense, in order to make the rate come out positive. Consider now a reaction in which the coefficients are different: \[A + 3B → 2D \] It is clear that [B] decreases three times as rapidly as [A], so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient: \[\text{rate} = \dfrac{-\Delta[A]}{\Delta t}= \dfrac{-\Delta[B]}{3 \Delta t} = \dfrac{+\Delta[D]}{2\Delta t}\] Each of the above quotients is a legitimate expression of the rate of this particular reaction; they all yield the same number. Which one you employ when doing a calculation is largely a matter of convenience. For the oxidation of ammonia \[4 NH_3 + 3O_2 \rightarrow 2 N_2 + 6 H_2O \nonumber\] it was found that the rate of formation of N was 0.27 mol L s . : Because of the way this question is formulated, it would be acceptable to express this last value as a negative number. Most reactions slow down as the reactants are consumed. Consequently, the rates given by the expressions shown above tend to lose their meaning when measured over longer time intervals Δ . Thus for the reaction whose progress is plotted here, the actual rate (as measured by the increasing concentration of product) varies continuously, being greatest at time zero. The of a reaction is given by the slope of a tangent to the concentration-vs.-time curve. Three such rates have been identified in this plot. An instantaneous rate taken near the beginning of the reaction (t = 0) is known as an (label here). As we shall soon see, initial rates play an important role in the study of reaction kinetics. If you have studied differential calculus, you will know that these tangent slopes are whose values can very at each point on the curve, so that these instantaneous rates are really defined as \[rate = \lim_{\Delta t \rightarrow 0} \dfrac{-\Delta[A]}{\Delta t}\] However, if you does not know calculus, just bear in mind that the larger the time interval Δ , the smaller will be the precision of the instantaneous rate. Instantaneous rates are also known as differential rates. The relation between the rate of a reaction and the concentrations of reactants is expressed by its . For example, the rate of the gas-phase decomposition of dinitrogen pentoxide \[2N_2O_5 → 4NO_2 + O_2\] has been found to be directly proportional to the concentration of \(N_2O_5\): \[\text{rate} = k [N_2O_5]\] Be very careful about confusing equilibrium constant expressions with those for rate laws. The expression for \(K_{eq}\) can always be written by inspecting the reaction equation, and it contains a term for each component (raised to the appropriate power) whose concentration changes during the reaction. For this reaction it is given by \[ K_{eq} = \dfrac{[NO_2]^2 [O_2]}{[N_2O_5]^2}\] In contrast, the expression for the rate law generally bears no necessary relation to the reaction equation, and must be determined . More generally, for a reaction of the form \[n_A A + n_B B + ... → products\] the rate law will be \[rate = [A]^a[B]^b ... \] Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant will depend on the exponents of the concentration terms in the rate law. To make this work out properly, if we let \(p\) be the sum of the exponents of the concentration terms in the rate law \[p = a + b + ...\] then will have the dimensions (concentration /time). The of a rate law is the sum of the exponents in its concentration terms. For the decomposition with the rate law k[ ], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. We can also say that the reaction is "first order in ". For more complicated rate laws, we can speak of the overall reaction order and also the orders with respect to each component. As an example, consider a reaction \[A + 3B + 2C → \text{products}\] whose experimental rate law is \[rate = k[A] [B]^2\] We would describe this reaction as means that the rate is independent of the concentration of a particular reactant. However, enough C must be present to allow the equilibrium mixture to form. The rate of oxidation of bromide ions by bromate in an acidic aqueous solution \[6H^+ + BrO_3^– + 5Br^– → 3 Br_2 + 3 H_2O \nonumber\] is found to follow the rate law \[rate = k[Br^–,BrO_3^–,H^+]^2\] What happens to the rate if, in separate experiments, In order to determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form A → products in which the rate is – [A]/ , we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A]. Use the tabulated experimental data to determine the order of the reaction \[2 N_2O_5 → 4 NO_2 + O_2 \nonumber\] The ideal gas law can be used to convert the partial pressures of \(N_2O_5\) to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of \([N_2O_5]\) and \([N)2O)5]^2\). It is apparent that the rates are directly proportional to \([N)2O)5]^1\), indicating that this is a first-order reaction. When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found. We then plot the five initial rates of consumption of \(N_2O_5\) as a function of its molar concentration. As before, we see that these rates are directly proportional to \([N_2O_5]\). The slope of this plot gives the value of the rate constant. = (5.2 × 10 ) [N2O5] mol L s A study of the gas-phase reduction of nitric oxide by hydrogen \[2 NO + 2 H_2 → N_2 + 2 H_2O \nonumber\] yielded the following initial-rate data (all pressures in torr): initial rate (torr s–1) Find the order of the reaction with respect to each component. In looking over this data, take note of the following: : Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide. : Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen. The rate law is thus \[rate = k[NO]^2[H_2]\] It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is \[A + B + C → \text{products}\] and we need to find the order with respect to [B] in the rate law. If we set [B] to 0.020 M and let [A] = [C] = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively the one in which we are interested.

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The Lanthanides consist of the elements in the f-block of period six in the . While these metals can be considered transition metals, they have properties that set them apart from the rest of the elements. Lanthanides (elements 57–71) are fairly abundant in the earth’s crust, despite their historic characterization as . Thulium, the rarest naturally occurring lanthanoid, is more common in the earth’s crust than silver (4.5 × 10 % versus 0.79 × 10 % by mass). There are 17 rare earth elements, consisting of the 15 lanthanoids plus scandium and yttrium. They are called rare because they were once difficult to extract economically, so it was rare to have a pure sample; due to similar chemical properties, it is difficult to separate any one lanthanide from the others. However, newer separation methods, such as ion exchange resins similar to those found in home water softeners, make the separation of these elements easier and more economical. Most ores that contain these elements have low concentrations of all the rare earth elements mixed together. Like any other series in the periodic table, such as the or the , the Lanthanides share many similar characteristics. These characteristics include the following: The commercial applications of lanthanides are growing rapidly. For example, europium is important in flat screen displays found in computer monitors, cell phones, and televisions. Neodymium is useful in laptop hard drives and in the processes that convert crude oil into gasoline (Figure \(\Page {1}\)). Holmium is found in dental and medical equipment. In addition, many alternative energy technologies rely heavily on lanthanoids. Neodymium and dysprosium are key components of hybrid vehicle engines and the magnets used in wind turbines. As the demand for lanthanide materials has increased faster than supply, prices have also increased. In 2008, dysprosium cost $110/kg; by 2014, the price had increased to $470/kg. Increasing the supply of lanthanoid elements is one of the most significant challenges facing the industries that rely on the optical and magnetic properties of these materials. The transition elements have many properties in common with other metals. They are almost all hard, high-melting solids that conduct heat and electricity well. They readily form alloys and lose electrons to form stable cations. In addition, transition metals form a wide variety of stable , in which the central metal atom or ion acts as a Lewis acid and accepts one or more pairs of electrons. Many different molecules and ions can donate lone pairs to the metal center, serving as Lewis bases. In this chapter, we shall focus primarily on the chemical behavior of the elements of the first transition series. Similarly, the Lanthanides have similarities in their electron configuration, which explains most of the physical similarities. These elements are different from the main group elements in the fact that they have electrons in the f orbital. After Lanthanum, the energy of the 4f sub-shell falls below that of the 5d sub-shell. This means that the electron start to fill the 4f sub-shell before the 5d sub-shell. The of these elements were primarily established through experiments (Table \(\Page {1}\)). The technique used is based on the fact that each line in an emission spectrum reveals the energy change involved in the transition of an electron from one energy level to another. However, the problem with this technique with respect to the Lanthanide elements is the fact that the 4f and 5d sub-shells have very similar energy levels, which can make it hard to tell the difference between the two. Another important feature of the Lanthanides is the , in which the 5s and 5p orbitals penetrate the 4f sub-shell. This means that the 4f orbital is not shielded from the increasing nuclear change, which causes the atomic radius of the atom to decrease that continues throughout the series. One property of the Lanthanides that affect how they will react with other elements is called the basicity. Basicity is a measure of the ease at which an atom will lose electrons. In another words, it would be the lack of attraction that a cation has for electrons or anions. For the Lanthanides, the basicity series is the following: La > Ce > Pr > Nd > Pm > Sm > Eu > Gd > Tb > Dy > Ho > Er > Tm > Yb > Lu In other words, the basicity decreases as the atomic number increases. Basicity differences are shown in the solubility of the salts and the formation of the complex species. Another property of the Lanthanides is their characteristics. The major magnetic properties of any chemical species are a result of the fact that each moving electron is a micromagnet. The species are either diamagnetic, meaning they have no unpaired electrons, or paramagnetic, meaning that they do have some unpaired electrons. The diamagnetic ions are: La , Lu , Yb and Ce . The rest of the elements are paramagnetic. The metals have a silvery shine when freshly cut. However, they can tarnish quickly in air, especially Ce, La and Eu. These elements react with water slowly in cold, though that reaction can happen quickly when heated. This is due to their electropositive nature. The Lanthanides have the following reactions: The size of the atomic and ionic radii is determined by both the nuclear charge and by the number of electrons that are in the electronic shells. Within those shells, the degree of occupancy will also affect the size. In the Lanthanides, there is a decrease in atomic size from La to Lu. This decrease is known as the . The trend for the entire periodic table states that the atomic radius decreases as you travel from left to right. Therefore, the Lanthanides share this trend with the rest of the elements. The color that a substance appears is the color that is reflected by the substance. This means that if a substance appears green, the green light is being reflected. The wavelength of the light determines if the light with be reflected or absorbed. Similarly, the splitting of the orbitals can affect the wavelength that can be absorbed (Table \(\Page {4}\)). This is turn would be affected by the amount of unpaired electrons. Each known Lanthanide mineral contains all the members of the series. However, each mineral contains different concentrations of the individual Lanthanides. The three main mineral sources are the following: In all the ores, the atoms with a even atomic number are more abundant. This allows for more nuclear stability, as explained in the Oddo-Harkins rule. The Oddo-Harkins rule simply states that the abundance of elements with an even atomic number is greater than the abundance of elements with an odd atomic number. In order to obtain these elements, the minerals must go through a separating process, known as separation chemistry. This can be done with selective reduction or oxidation. Another possibility is an ion-exchange method. The abundance of elements with an even atomic number is than the abundance of elements with an odd atomic number. The pure metals of the Lanthanides have little use. However, the alloys of the metals can be very useful. For example, the alloys of Cerium have been used for metallurgical applications due to their strong reducing abilities. The Lanthanides can also be used for ceramic purposes. The almost glass-like covering of a ceramic dish can be created with the lanthanides. They are also used to improve the intensity and color balance of arc lights. Like the Actinides, the Lanthanides can be used for nuclear purposes. The hydrides can be used as hydrogen-moderator carriers. The oxides can be used as diluents in nuclear fields. The metals are good for being used as structural components. The can also be used for structural-alloy-modifying components of reactors. It is also possible for some elements, such as Tm, to be used as portable x-ray sources. Other elements, such as Eu, can be used as radiation sources.

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Plants store glucose as the polysaccharide starch; the cereal grains (wheat, rice, corn, oats, barley) as well as tubers such as potatoes are also rich in starch. Starch can be separated into two fractions-- . Natural starches are mixtures of amylose (10-20%) and amylopectin (80-90%). forms a colloidal dispersion in hot water whereas amylopectin is completely insoluble. The structure of amylose consists of long polymer chains of glucose units connected by an linkage. Starch - Amylose shows a very small portion of an amylose chain. All of the monomer units are alpha -D-glucose, and all the alpha acetal links connect C #1 of one glucose and to C #4 of the next glucose. As a result of the bond angles in the acetal linkage, amylose actually forms a spiral much like a coiled spring. See the graphic below, which show four views in turning from a the side to an end view. Amylose in starch is responsible for the formation of a deep blue color in the presence of iodine. The iodine molecule slips inside of the amylose coil. Iodine - KI Reagent: Iodine is not very soluble in water, therefore the iodine reagent is made by dissolving iodine in water in the presence of potassium iodide. This makes a linear triiodide ion complex with is soluble that slips into the coil of the starch causing an intense blue-black color.

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In the first half of the nineteenth century it was generally believed that reactions of organic compounds proceeded with minimal structural change. This tenet simplified the elucidation of the numerous substitution, addition and elimination reactions that characterized the behavior of common functional groups. However, subsequent discoveries showed that nature was not always so obliging, leaving chemists and chemistry students to grapple with the possibility of deep seated structural change occurring during certain reactions. A large number of these structural rearrangements are triggered by intermediates incorporating positively charged or electron deficient atoms, which in the case of carbon are carbocations. Two such examples, already noted, are the and forced . This chapter will describe and discuss other cases of this intriguing group of transformations.

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In mass spectrometry (MS), we are interested in the mass - and therefore the molecular weight - of our compound of interest, and often the mass of fragments that are produced when the molecule is caused to break apart. There are many different types of MS instruments, but they all have the same three essential components. First, there is an ionization source, where the molecule is given a positive electrical charge, either by removing an electron or by adding a proton. Depending on the ionization method used, the ionized molecule may or may not break apart into a population of smaller fragments. In the figure below, some of the sample molecules remain whole, while others fragment into smaller pieces. Next in line there is a mass analyzer, where the cationic fragments are separated according to their mass. Finally, there is a detector, which detects and quantifies the separated ions. One of the more common types of MS techniques used in the organic laboratory is . In the ionization source, the sample molecule is bombarded by a high-energy electron beam, which has the effect of knocking a valence electron off of the molecule to form a . Because a great deal of energy is transferred by this bombardment process, the radical cation quickly begins to break up into smaller fragments, some of which are positively charged and some of which are neutral. The neutral fragments are either adsorbed onto the walls of the chamber or are removed by a vacuum source. In the mass analyzer component, the positively charged fragments and any remaining unfragmented are accelerated down a tube by an electric field. This tube is curved, and the ions are deflected by a strong magnetic field. Ions of different mass to charge (m/z) ratios are deflected to a different extent, resulting in a ‘sorting’ of ions by mass (virtually all ions have charges of z = +1, so sorting by the mass to charge ratio is the same thing as sorting by mass). A detector at the end of the curved flight tube records and quantifies the sorted ions. Below is typical output for an electron-ionization MS experiment (MS data below is derived from the Spectral Database for Organic Compounds, a free, web-based service provided by AIST in Japan. The sample is acetone. On the horizontal axis is the value for m/z (as we stated above, the charge z is almost always +1, so in practice this is the same as mass). On the vertical axis is the relative abundance of each ion detected. On this scale, the most abundant ion, called the , is set to 100%, and all other peaks are recorded relative to this value. For acetone, the base peak corresponds to a fragment with m/z = 43 - . The molecular weight of acetone is 58, so we can identify the peak at m/z = 58 as that corresponding to the , or . Notice that there is a small peak at m/z = 59: this is referred to as the . How can there be an ion that has a greater mass than the molecular ion? Simple: a small fraction - about 1.1% - of all carbon atoms in nature are actually the C rather than the C isotope. The C isotope is, of course, heavier than C by 1 mass unit. In addition, about 0.015% of all hydrogen atoms are actually deuterium, the H isotope. So the M+1 peak represents those few acetone molecules in the sample which contained either a C or H. The fragmentation of molecular ions into an assortment of fragment ions is a mixed blessing. The nature of the fragments often provides a clue to the molecular structure, but if the molecular ion has a lifetime of less than a few microseconds it will not survive long enough to be observed. Without a molecular ion peak as a reference, the difficulty of interpreting a mass spectrum increases markedly. Fortunately, most organic compounds give mass spectra that include a molecular ion, and those that do not often respond successfully to the use of milder ionization conditions. Among simple organic compounds, the most stable molecular ions are those from aromatic rings, other conjugated pi-electron systems and . Alcohols, ethers and highly branched generally show the greatest tendency toward fragmentation. The mass spectrum of dodecane illustrates the behavior of an unbranched alkane. Since there are no heteroatoms in this molecule, there are no non-bonding valence shell electrons. Consequently, the radical cation character of the molecular ion (m/z = 170) is delocalized over all the covalent bonds. Fragmentation of C-C bonds occurs because they are usually weaker than C-H bonds, and this produces a mixture of alkyl radicals and alkyl carbocations. The positive charge commonly resides on the smaller fragment, so we see a homologous series of hexyl (m/z = 85), pentyl (m/z = 71), butyl (m/z = 57), propyl (m/z = 43), ethyl (m/z = 29) and methyl (m/z = 15) cations. These are accompanied by a set of corresponding alkenyl carbocations (e.g. m/z = 55, 41 &27) formed by loss of 2 H. All of the significant fragment ions in this spectrum are even-electron ions. In most alkane spectra the propyl and butyl ions are the most abundant. The presence of a functional group, particularly one having a heteroatom Y with non-bonding valence electrons (Y = N, O, S, X etc.), can dramatically alter the fragmentation pattern of a compound. This influence is thought to occur because of a "localization" of the radical cation component of the molecular ion on the heteroatom. After all, it is easier to remove (ionize) a non-bonding electron than one that is part of a covalent bond. By localizing the reactive moiety, certain fragmentation processes will be favored. These are summarized in the following diagram, where the green shaded box at the top displays examples of such "localized" molecular ions. The first two fragmentation paths lead to even-electron ions, and the elimination (path #3) gives an odd-electron ion. Note the use of different curved arrows to show single electron shifts compared with electron pair shifts. The charge distributions shown above are common, but for each cleavage process the charge may sometimes be carried by the other (neutral) species, and both fragment ions are observed. Of the three cleavage reactions described here, the alpha-cleavage is generally favored for nitrogen, oxygen and sulfur compounds. Indeed, in the spectra of 4-methyl-3-pentene-2-one and N,N-diethylmethylamine the major fragment ions come from alpha-cleavage. The complexity of fragmentation patterns has led to mass spectra being used as "fingerprints" for identifying compounds. Environmental pollutants, pesticide residues on food, and controlled substance identification are but a few examples of this application. Extremely small samples of an unknown substance (a microgram or less) are sufficient for such analysis. The following mass spectrum of cocaine demonstrates how a forensic laboratory might determine the nature of an unknown street drug. Even though extensive fragmentation has occurred, many of the more abundant ions (identified by magenta numbers) can be rationalized by the three mechanisms shown above. Plausible assignments may be seen by clicking on the spectrum, and it should be noted that all are even-electron ions. The m/z = 42 ion might be any or all of the following: C H , C H O or C H N. A precise assignment could be made from a high-resolution m/z value (next section). Molecules with lots of oxygen atoms sometimes show a small (2 m/z units greater than the parent peak) in their mass spectra, due to the presence of a small amount of O (the most abundant isotope of oxygen is O). Because there are two abundant isotopes of both chlorine (about 75% Cl and 25% Cl) and bromine (about 50% Br and 50% Br), chlorinated and brominated compounds have very large and recognizable M+2 peaks. Fragments containing both isotopes of Br can be seen in the mass spectrum of ethyl bromide: A common fragmentation pattern for larger carbonyl compounds is called the : The mass spectrum of 2-hexanone shows a 'McLafferty fragment' at m/z = 58, while the propene fragment is not observed because it is a neutral species (remember, only cationic fragments are observed in MS). The base peak in this spectrum is again an acylium ion. When alcohols are subjected to electron ionization MS, the molecular ion is highly unstable and thus a parent peak is often not detected. Often the base peak is from an ‘oxonium’ ion. Other functional groups have predictable fragmentation patterns as well. By carefully analyzing the fragmentation information that a mass spectrum provides, a knowledgeable spectroscopist can often ‘put the puzzle together’ and make some very confident predictions about the structure of the starting sample. The mass spectrum of an aldehyde gives prominent peaks at = 59 (12%, highest value of in the spectrum), 58 (85%), and 29 (100%), as well as others. Propose a structure, and identify the three species whose values were listed Quite often, mass spectrometry is used in conjunction with a separation technique called gas chromatography (GC). The combined GC-MS procedure is very useful when dealing with a sample that is a mixture of two or more different compounds, because the various compounds are separated from one another before being subjected individually to MS analysis. We will not go into the details of gas chromatography here, although if you are taking an organic laboratory course you might well get a chance to try your hand at GC, and you will almost certainly be exposed to the conceptually analogous techniques of thin layer and column chromatography. Suffice it to say that in GC, a very small amount of a liquid sample is vaporized, injected into a long, coiled metal column, and pushed though the column by helium gas. Along the way, different compounds in the sample stick to the walls of the column to different extents, and thus travel at different speeds and emerge separately from the end of the column. In GC-MS, each purified compound is sent directly from the end of GC column into the MS instrument, so in the end we get a separate mass spectrum for each of the compounds in the original mixed sample. Because a compound's MS spectrum is a very reliable and reproducible 'fingerprint', we can instruct the instrument to search an MS database and identify each compound in the sample. The extremely high sensitivity of modern GC-MS instrumentation makes it possible to detect and identify very small trace amounts of organic compounds. GC-MS is being used increasingly by environmental chemists to detect the presence of harmful organic contaminants in food and water samples. Airport security screeners also use high-speed GC-MS instruments to look for residue from bomb-making chemicals on checked luggage.

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Prefixes are often used for decimal multiples and submultiples of units. Often, the symbols are used together with units. For example, MeV means million electron volts, units of energy. To memorize these prefixes is not something you will enjoy, but if you do know them by heart, you will appreciate the quantity when you encounter them in your reading. You can come back here to check them in case you forgot. The table is arranged in a symmetric fashion for convenience of comparison. Note the increments or decrements by thousand 10 or 10 are used aside from hecto (centi) and deca (deci). When you want to express some large or small quantities, you may also find these prefixes useful. Greek prefixes are often used for naming compounds. You will need the prefixes in order to give a proper name of many compounds. You also need to know them to figure out the formula from their names. The common prefixes are given in this Table. Note that some of the prefixes may change slightly when they are applied to the names. Some of the examples show the variations. Note also that some names are given using other conventions. For example, \(\ce{P4O6}\) and \(\ce{P4O10}\) are called phosphorus trioxide and phosphorus pentoxide respectively. These are given according to their empirical formulas.

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By the end of gen chemistry, calculating of different metals should be pretty familiar. Here’s what you do. Take a typical compound – \(FeCl_3\), for instance. Treat every bond between the metal and a different atom as if it were an ionic bond. That means the more electronegative elements (like chlorine, say, or oxygen) bear negative charges, and the less electronegative element (such as the metal) bears the positive charge. If the compound is neutral, the sum of the oxidation states also has to be neutral. (If the compound has a charge, you adjust the oxidation states accordingly so that their sum equals the charge). Now here’s a fun exercise. Try applying the same rules to carbon. It’s going to feel a little bit weird. Why? Because there are two key differences: Two answers. So unlike metals, which are almost always in a positive oxidation state, the oxidation state of carbon can vary widely, from -4 (in CH4) to +4 (such as in CO2). Here are some examples. (Don’t forget that this is called a “formalism” for a reason. The charge on the carbon is not +4 or –4. But the oxidation state formalism helps us keep track of where the electrons are going, which will come in handy very soon). With an understanding of how to calculate oxidation states on carbon, we’re ready for the next step: understanding in the oxidation state at carbon, through reactions known as (where the oxidation state is increased), and (where the oxidation state is reduced). More on that next time.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/Cell_Diagrams

Cell notation is shorthand that expresses a certain reaction in an electrochemical cell. Recall that standard cell potentials can be calculated from potentials E for both oxidation and reduction reactions. A positive cell potential indicates that the reaction proceeds spontaneously in the direction in which the reaction is written. Conversely, a reaction with a negative cell potential proceeds spontaneously in the reverse direction. Cell notations are a shorthand description of voltaic or galvanic (spontaneous) cells. The reaction conditions (pressure, temperature, concentration, etc.), the anode, the cathode, and the electrode components are all described in this unique shorthand. Recall that oxidation takes place at the anode and reduction takes place at the cathode. When the anode and cathode are connected by a wire, electrons flow from anode to cathode. A typical arrangement of half-cells linked to form a galvanic cell. Using the arrangement of components, let's put a cell together. One beaker contains 0.15 M Cd(NO ) and a Cd metal electrode. The other beaker contains 0.20 M AgNO and a Ag metal electrode. The net ionic equation for the reaction is written: In the reaction, the silver ion is reduced by gaining an electron, and solid Ag is the cathode. The cadmium is oxidized by losing electrons, and solid Cd is the anode. The anode reaction is: The cathode reaction is: Using these rules, the notation for the cell we put together is: Cd (s) | Cd (aq, 0.15 M) || Ag (aq, 0.20 M) | Ag (s) Give us feedback on this content: Assign Concept Reading Assign just this concept or entire chapters to your class for free. Edit this content PREV CONCEPT Electrolytic Cells Standard Reduction Potentials NEXT CONCEPT Referenced in Given the following information, provide the appropriate electrochemical cell notation for the following reaction:ZnSO4(aq) + Mn(s) Zn(s) + MnSO4(aq)assuming all solutions are at 1.0M, 1.0 atm and 298 K Zn (s) | Zn || Mn | Mn (s), Mn (s) | Mn || Zn | Zn (s), Zn | Zn (s) || Mn (s) | Mn , or Mn | Mn (s) || Zn (s) | Zn Which of the following conditions are considered standard when writing electrochemical cell notation?a. 1 liter of solution volumeb. 1 atmosphere of pressurec. 1.00 molar solution concentrationd. 298 kelvin temperature a, b and d, a, c and d, a. b and c, or b, c and d What is the cell notation for a voltaic cell with the following equation? Pb2+(aq) + Cd(s) → Pb(s) + Cd2+(aq) Pb | Pb2⁺|| Cd2⁺| Cd, Pb2⁺ | Pb || Cd | Cd2⁺, Cd | Cd2⁺|| Pb2⁺ | Pb, or Cd | Pb2⁺|| Pb | Cd2 Boundless. “Electrochemical Cell Notation.” Boundless Chemistry. Boundless, 21 Jul. 2015. Retrieved 11 Apr. 2016 from www.boundless.com/chemistry/...tion-513-3688/

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Regio_and_Stereoisomerization_in_Polymers

Symmetrical monomers such as ethylene and tetrafluoroethylene can join together in only one way. Monosubstituted monomers, on the other hand, may join together in two organized ways, described in the following diagram, or in a third random manner. Most monomers of this kind, including propylene, vinyl chloride, styrene, acrylonitrile and acrylic esters, prefer to join in a head-to-tail fashion, with some randomness occurring from time to time. The reasons for this regioselectivity will be discussed in the synthetic methods section. If the polymer chain is drawn in a zig-zag fashion, as shown above, each of the substituent groups (Z) will necessarily be located above or below the plane defined by the carbon chain. Consequently we can identify three configurational isomers of such polymers. If all the substituents lie on one side of the chain the configuration is called isotactic. If the substituents alternate from one side to another in a regular manner the configuration is termed syndiotactic. Finally, a random arrangement of substituent groups is referred to as atactic. Examples of these configurations are shown here. Many common and useful polymers, such as polystyrene, polyacrylonitrile and poly(vinyl chloride) are atactic as normally prepared. Customized catalysts that effect stereoregular polymerization of polypropylene and some other monomers have been developed, and the improved properties associated with the increased crystallinity of these products has made this an important field of investigation. The following values of Tg have been reported. The properties of a given polymer will vary considerably with its tacticity. Thus, atactic polypropylene is useless as a solid construction material, and is employed mainly as a component of adhesives or as a soft matrix for composite materials. In contrast, isotactic polypropylene is a high-melting solid (ca. 170 ºC) which can be molded or machined into structural components. ),

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Heterogeneous_catalysis

According to Surface adsoprtion theory heterogeneous catalysis has five stages: The rate at which reactants will diffuse to the surface will be influenced by their bulk concentration and by the thickness of the boundary layer. Bonds are formed as the reactant(s) are adsorbed onto the surface of the catalyst. The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface. Bonds form between the atoms and molecules on the surface Bonds are broken as the product(s) desorb from the surface. The products are then desorbed from the surface of the catalyst. The contact process is the current method of producing sulfuric acid in the high concentrations needed for industrial processes. Platinum used to be the catalyst for this reaction; however, as it is susceptible to reacting with arsenic impurities in the sulfur feedstock, vanadium(V) oxide (\(V_2O_5\)) is now preferred In the contact process vanadium [V] oxide (V O ) is solid whereas the reactants SO and O are gaseous. \[2SO_2 (g) + O_2 (g)\overset{V_2O_5(s)}\rightleftharpoons 2SO_3 (g)\] More detailed method: \[2V_2O_5(s) + 2SO_2(g)\rightleftharpoons 2SO_3(g) + 2V_2O_4(s)\] \[2V_2O_4(s) + O_2 (g) \rightleftharpoons 2V_2O_5 (s)\] Therefore: \[2SO_2(g) + O_2(g)\rightleftharpoons 2SO_3(g)\] The original Haber–Bosch reaction chambers used as the catalyst, less expensive iron-based catalyst, which is still used today \[N_2 (g) + 3H_2 (g) \overset{Fe(s)}\rightleftharpoons 2NH_3(g)\] The reaction mechanism, involving the heterogeneous catalyst, is believed to involve the following steps:

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Organic_Acids_and_Bases/Organic_Bases

This page explains why simple organic bases are basic and looks at the factors which affect their relative strengths. For A'level purposes, all the bases we are concerned with are primary amines - compounds in which one of the hydrogens in an ammonia molecule, NH , is replaced either by an alkyl group or a benzene ring. All of the compounds we are concerned with are derived from ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water. Ammonia in solution sets up this equilibrium: An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left. The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water. The strengths of weak bases are measured on the pK scale. The smaller the number on this scale, the stronger the base is. Three of the compounds we shall be looking at, together with their pK values are: Remember - the smaller the number the stronger the base. Comparing the other two to ammonia, you will see that methylamine is a stronger base, whereas phenylamine is very much weaker. Two of the factors which influence the strength of a base are: Methylamine has the structure: The only difference between this and ammonia is the presence of the CH group in the methylamine. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions. Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion: In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there isn't any way of spreading the charge. To summarize: Taken together, these mean that methylamine is a stronger base than ammonia. The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of the other aliphatic primary amines are very similar to methylamine. For example: An aromatic primary amine is one in which the -NH group is attached directly to a benzene ring. The only one you are likely to come across is phenylamine. Phenylamine has the structure: The lone pair on the nitrogen touches the delocalized ring electrons . . . . . . and becomes delocalized with them: That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalized electrons will be attracted towards it, but the intensity of charge around the nitrogen is nothing like what it is in, say, an ammonia molecule. The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalisation. That means that the delocalization would have to be disrupted if the phenylamine acts as a base. Delocalization makes molecules more stable, and so disrupting the delocalization costs energy and will not happen easily. Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalization - this means that phenylamine is a very weak base indeed.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Prepartion_of_Alkenes

This page looks at ways of preparing alkenes in the lab by the dehydration of alcohols. This is a simple way of making gaseous alkenes like ethene. If ethanol vapor is passed over heated aluminum oxide powder, the ethanol is essentially cracked to give ethene and water vapor. \[ CH_3CH_2OH \overset{Al_2O_3}{\longrightarrow} CH_2=CH_2 + H2O\] To make a few test tubes of ethene, you can use this apparatus: It wouldn't be too difficult to imagine scaling this up by boiling some ethanol in a flask and passing the vapor over aluminum oxide heated in a long tube. The acid catalysts normally used are either concentrated sulfuric acid or concentrated phosphoric(V) acid, H PO . Concentrated sulphuric acid produces messy results. Not only is it an acid, but it is also a strong oxidizing agent. It oxidizes some of the alcohol to carbon dioxide and at the same time is reduced itself to sulfur dioxide. Both of these gases have to be removed from the alkene. It also reacts with the alcohol to produce a mass of carbon. There are other side reactions as well Ethanol is heated with an excess of concentrated sulfuric acid at a temperature of 170°C. The gases produced are passed through sodium hydroxide solution to remove the carbon dioxide and sulfur dioxide produced from side reactions. The ethene is collected over water. \[ CH_3CH_2OH \overset{conc. H_2SO_4}{\longrightarrow} CH_2=CH_2 + H2O\] The concentrated sulfuric acid is a catalyst. Write it over the arrow rather than in the equation. This is a preparation commonly used at this level to illustrate the formation and purification of a liquid product. The fact that the carbon atoms happen to be joined in a ring makes no difference whatever to the chemistry of the reaction. Cyclohexanol is heated with concentrated phosphoric(V) acid and the liquid cyclohexene distils off and can be collected and purified. Phosphoric(V) acid tends to be used in place of sulphuric acid because it is safer and produces a less messy reaction. You have to be wary with more complicated alcohols in case there is the possibility of more than one alkene being formed. Butan-2-ol is a good example of this, with no less than three different alkenes being formed when it is dehydrated. Butan-2-ol is an example to illustrate the problems. It is important that you understand it so that you can work out what will happen in similar cases. When you dehydrate an alcohol, you remove the -OH group, and a hydrogen atom from the next carbon atom in the chain. With molecules like butan-2-ol, there are two possibilities when that happens. That leads to these products: The products are but-1-ene, CH =CHCH CH , and but-2-ene, CH CH=CHCH . In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene. Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds, follow the link in the box below. Which isomer gets formed is just a matter of chance. Dehydration of butan-2-ol leads to a mixture containing: Jim Clark ( )

https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/10.E%3A_Liquids_and_Solids_(Exercises)

In terms of their bulk properties, how do liquids and solids differ? How are they similar? Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid. In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids? In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases? They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast, a gas will expand without limit to fill the space into which it is placed. Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape. What is the evidence that all neutral atoms and molecules exert attractive forces on each other? All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature. Open the to answer the following questions: Define the following and give an example of each: The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid? Why do the boiling points of the noble gases increase in the order He < Ne < Ar < Kr < Xe? The London forces typically increase as the number of electrons increase. Neon and HF have approximately the same molecular masses. Arrange each of the following sets of compounds in order of increasing boiling point temperature: (a) SiH < HCl < H O; (b) F < Cl < Br ; (c) CH < C H < C H ; (d) N < O < NO The molecular mass of butanol, C H OH, is 74.14; that of ethylene glycol, CH (OH)CH OH, is 62.08, yet their boiling points are 117.2 °C and 174 °C, respectively. Explain the reason for the difference. On the basis of intermolecular attractions, explain the differences in the boiling points of –butane (−1 °C) and chloroethane (12 °C), which have similar molar masses. Only rather small dipole-dipole interactions from C-H bonds are available to hold -butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the Cl-C bond; the interaction is therefore stronger, leading to a higher boiling point. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses. The melting point of H O( ) is 0 °C. Would you expect the melting point of H S( ) to be −85 °C, 0 °C, or 185 °C? Explain your answer. −85 °C. Water has stronger hydrogen bonds so it melts at a higher temperature. Silane (SiH ), phosphine (PH ), and hydrogen sulfide (H S) melt at −185 °C, −133 °C, and −85 °C, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds? Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules. The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of F is greater than that of O. Consequently, the partial negative charge on F is greater than that on O. The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O. Under certain conditions, molecules of acetic acid, CH COOH, form “dimers,” pairs of acetic acid molecules held together by strong intermolecular attractions: Draw a dimer of acetic acid, showing how two CH COOH molecules are held together, and stating the type of IMF that is responsible. Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix. What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together: H-bonding is the principle IMF holding the DNA strands together. The H-bonding is between the \(\mathrm{N−H}\) and \(\mathrm{C=O}\). The density of liquid NH is 0.64 g/mL; the density of gaseous NH at STP is 0.0007 g/mL. Explain the difference between the densities of these two phases. Identify the intermolecular forces present in the following solids: (a) hydrogen bonding and dispersion forces; (b) dispersion forces; (c) dipole-dipole attraction and dispersion forces The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration. Rank the motor oils in order of increasing viscosity, and explain your reasoning: Although steel is denser than water, a steel needle or paper clip placed carefully lengthwise on the surface of still water can be made to float. Explain at a molecular level how this is possible: (credit: Cory Zanker) The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of “skin” at its surface. This skin can support a bug or paper clip if gently placed on the water. The surface tension and viscosity values for diethyl ether, acetone, ethanol, and ethylene glycol are shown here. You may have heard someone use the figure of speech “slower than molasses in winter” to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature. Temperature has an effect on intermolecular forces: the higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid; the lower the temperature, the lesser the intermolecular forces are overcome, and so the less viscous the liquid. It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this. The surface tension and viscosity of water at several different temperatures are given in this table. (a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason. At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm? Refer to for the required information. Water rises in a glass capillary tube to a height of 17 cm. What is the diameter of the capillary tube? 9.5 × 10 m Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change? Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change? The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise. What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container? Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate? We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids. Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime? What is the relationship between the intermolecular forces in a liquid and its vapor pressure? The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases. What is the relationship between the intermolecular forces in a solid and its melting temperature? Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day? As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures. Carbon tetrachloride, CCl , was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl . When is the boiling point of a liquid equal to its normal boiling point? When the pressure of gas above the liquid is exactly 1 atm How does the boiling of a liquid differ from its evaporation? Use the information in to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa. approximately 95 °C A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced? Explain the following observations: (a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water. The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why. Explain why the molar enthalpies of vaporization of the following substances increase in the order CH < C H < C H , even though all three substances experience the same dispersion forces when in the liquid state. Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids. Explain why the enthalpies of vaporization of the following substances increase in the order CH < NH < H O, even though all three substances have approximately the same molar mass. The enthalpy of vaporization of CO ( ) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS ( ) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers. The boiling point of CS is higher than that of CO partially because of the higher molecular weight of CS ; consequently, the attractive forces are stronger in CS . It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO . A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible. The hydrogen fluoride molecule, HF, is more polar than a water molecule, H O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain. Ethyl chloride (boiling point, 13 °C) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride. The thermal energy (heat) needed to evaporate the liquid is removed from the skin. Which contains the compounds listed correctly in order of increasing boiling points? How much heat is required to convert 422 g of liquid H O at 23.5 °C into steam at 150 °C? 1130 kJ Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.) Titanium tetrachloride, TiCl , has a melting point of −23.2 °C and has a Δ = 9.37 kJ/mol. (a) 13.0 kJ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them. From the phase diagram for water, determine the state of water at: What phase changes will take place when water is subjected to varying pressure at a constant temperature of 0.005 °C? At 40 °C? At −40 °C? At low pressures and 0.005 °C, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At 40 °C, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At −40 °C, water goes from a gas to a solid as the pressure increases above very low values. Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm. What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning. From the phase diagram for carbon dioxide, determine the state of CO at: (a) liquid; (b) solid; (c) gas; (d) gas; (e) gas; (f) gas Determine the phase changes that carbon dioxide undergoes as the pressure changes if the temperature is held at −50 °C? If the temperature is held at −40 °C? At 20 °C? Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of 65 atm and a temperature of 20 °C. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature. Dry ice, CO ( ), does not melt at atmospheric pressure. It sublimes at a temperature of −78 °C. What is the lowest pressure at which CO ( ) will melt to give CO ( )? At approximately what temperature will this occur? (See for the phase diagram.) If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer. Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry. Is it possible to liquefy nitrogen at room temperature (about 25 °C)? Is it possible to liquefy sulfur dioxide at room temperature? Explain your answers. Elemental carbon has one gas phase, one liquid phase, and three different solid phases, as shown in the phase diagram: What types of liquids typically form amorphous solids? Amorphous solids lack an ordered internal structure. Liquid materials that contain large, cumbersome molecules that cannot move readily into ordered positions generally form such solids. At very low temperatures oxygen, O , freezes and forms a crystalline solid. Which best describes these crystals? (e) molecular crystals As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid? (d) amorphous Explain why ice, which is a crystalline solid, has a melting temperature of 0 °C, whereas butter, which is an amorphous solid, softens over a range of temperatures. Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range. Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances: (a) SiO , covalent network; (b) KCl, ionic; (c) Cu, metallic; (d) CO, molecular; (e) C (diamond), covalent network; (f) BaSO , ionic; (g) NH , molecular; (h) NH F, ionic; (i) C H OH, molecular Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances: (a) CaCl , ionic; (b) SiC, covalent network; (c) N , molecular; (d) Fe, metallic; (e) C (graphite), covalent network; (f) CH CH CH CH , molecular; (g) HCl, molecular; (h) NH NO , ionic; (i) K PO , ionic Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid: X = metallic; Y = covalent network; Z = ionic Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid: X = ionic; Y = metallic; Z = covalent network Identify the following substances as ionic, metallic, covalent network, or molecular solids: Substance A is malleable, ductile, conducts electricity well, and has a melting point of 1135 °C. Substance B is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072 °C. Substance C is very hard, does not conduct electricity, and has a melting point of 3440 °C. Substance D is soft, does not conduct electricity, and has a melting point of 185 °C. A = metallic; B = ionic; C = covalent network; D = molecular Substance A is shiny, conducts electricity well, and melts at 975 °C. Substance A is likely a(n): (b) metallic solid Substance B is hard, does not conduct electricity, and melts at 1200 °C. Substance B is likely a(n): (d) covalent network solid Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell. The structure of this low-temperature form of iron (below 910 °C) is body-centered cubic. There is one-eighth atom at each of the eight corners of the cube and one atom in the center of the cube. Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell. What is the coordination number of a chromium atom in the body-centered cubic structure of chromium? eight What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum? Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom? 12 Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom? Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 Å. (a) 1.370 Å; (b) 19.26 g/cm Platinum (atomic radius = 1.38 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum. Barium crystallizes in a body-centered cubic unit cell with an edge length of 5.025 Å (a) 2.176 Å; (b) 3.595 g/cm Aluminum (atomic radius = 1.43 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum. The density of aluminum is 2.7 g/cm ; that of silicon is 2.3 g/cm . Explain why Si has the lower density even though it has heavier atoms. The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12). The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space? Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer. In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS. A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer. What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions? Co O A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound? A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? Explain your answer. In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is TlI. Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closest-packed array with all of the octahedral holes occupied: Li, Na, Be, Ca, or Al? What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium? 59.95%; The oxidation number of titanium is +4. Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure. As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation. Both ions are close in size: Mg, 0.65; Li, 0.60. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of Si for Al . Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. What is the formula of the compound? One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound? Mn O NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is 4.880 Å. Thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å. Calculate the ionic radius of TI . (The ionic radius of I is 2.16 Å.) 1.48 Å A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge. What is the spacing between crystal planes that diffract X-rays with a wavelength of 1.541 nm at an angle of 15.55° (first order reflection)? 2.874 Å A diffractometer using X-rays with a wavelength of 0.2287 nm produced first-order diffraction peak for a crystal angle = 16.21°. Determine the spacing between the diffracting planes in this crystal. A metal with spacing between planes equal to 0.4164 nm diffracts X-rays with a wavelength of 0.2879 nm. What is the diffraction angle for the first order diffraction peak? 20.2° Gold crystallizes in a face-centered cubic unit cell. The second-order reflection (n = 2) of X-rays for the planes that make up the tops and bottoms of the unit cells is at = 22.20°. The wavelength of the X-rays is 1.54 Å. What is the density of metallic gold? When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted. These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64 Å. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first-order diffraction? 1.74 × 10 eV

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Halogenation_Alkanes

This page describes the reactions between alkanes and cycloalkanes with the halogens fluorine, chlorine, bromine and iodine - mainly concentrating on chlorine and bromine. \[CH_4 + 2F_2 \rightarrow C + 4HF \] In the presence of a flame, the reactions are rather like the fluorine one - producing a mixture of carbon and the hydrogen halide. The violence of the reaction drops considerably as you go from fluorine to chlorine to bromine. The interesting reactions happen in the presence of ultra-violet light (sunlight will do). These are photochemical reactions that happen at room temperature. We'll look at the reactions with chlorine, although the reactions with bromine are similar, but evolve more slowly. Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane. The original mixture of a colorless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. All of the organic products are liquid at room temperature with the exception of the chloromethane which is a gas. If you were using bromine, you could either mix methane with bromine vapor , or bubble the methane through liquid bromine - in either case, exposed to UV light. The original mixture of gases would, of course, be red-brown rather than green. One would not choose to use these reactions as a means of preparing these organic compounds in the lab because the mixture of products would be too tedious to separate. The mechanisms for the reactions are explained on separate pages. You would again get a mixture of substitution products, but it is worth just looking briefly at what happens if only one of the hydrogen atoms gets substituted (monosubstitution) - just to show that things aren't always as straightforward as they seem! For example, with propane, you could get one of two isomers: If chance was the only factor, you would expect to get three times as much of the isomer with the chlorine on the end. There are 6 hydrogens that could get replaced on the end carbon atoms compared with only 2 in the middle. In fact, you get about the same amount of each of the two isomers. If you use bromine instead of chlorine, the great majority of the product is where the bromine is attached to the center carbon atom. The reactions of the cycloalkanes are generally just the same as the alkanes, with the exception of the very small ones - particularly cyclopropane. In the presence of UV light, cyclopropane will undergo substitution reactions with chlorine or bromine just like a non-cyclic alkane. However, it also has the ability to react in the dark. In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane. This can still happen in the presence of light - but you will get substitution reactions as well. The ring is broken because cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal value of about 109.5° when the carbon makes four single bonds. The overlap between the atomic orbitals in forming the carbon-carbon bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken. Jim Clark ( )

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Organic_Acids_and_Bases/Organic_Acids

This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths. For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water. An acid in solution sets up this equilibrium: A hydroxonium ion is formed together with the anion (negative ion) from the acid. This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid. If you write it like this, you must include the state symbols - "(aq)". Writing H implies that the hydrogen ion is attached to a water molecule as H O . Hydrogen ions are always attached to something during chemical reactions. The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left. The strengths of weak acids are measured on the pK scale. The smaller the number on this scale, the stronger the acid is. Three of the compounds we shall be looking at, together with their pK values are: Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pK of 10.00, and ethanol is so weak with a pK of about 16 that it hardly counts as acidic at all! In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X": So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths? Two of the factors which influence the ionization of an acid are: In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case. Ethanoic acid has the structure: The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH COO . You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond. To understand why this is, you have to look in some detail at the bonding in the ethanoate ion. Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen. In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them This leads to a delocalised pi system over the whole of the -COO group, rather like that in benzene. All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms. So where is the negative charge in all this? It has been spread around over the whole of the -COO group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as: The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid. Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group. When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C H O . Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring. This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion. Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this: But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram). If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge. That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic. In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge. What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent. However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic. Ethanol, CH CH OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed: This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form. Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all. You might think that all carboxylic acids would have the same strength because each depends on the delocalization of the negative charge around the -COO group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion. In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids: Remember that the higher the value for pK , the weaker the acid is. Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid. The methanoate ion (from methanoic acid) is: The only difference between this and the ethanoate ion is the presence of the CH group in the ethanoate. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions. Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions. The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid. The acids can be strengthened by pulling charge away from the -COO end. You can do this by attaching electronegative atoms like chlorine to the chain. As the next table shows, the more chlorines you can attach the better: Trichloroethanoic acid is quite a strong acid. Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO end and so strengthening the acid. The effect is there, but isn't as great as you might expect. Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO end. Here is what happens if you move a chlorine atom along the chain in butanoic acid. The chlorine is effective at withdrawing charge when it is next-door to the -COO group, and much less so as it gets even one carbon further away.

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Heating a home is becoming more and more expensive. The decision to use gas, oil, electricity, or wood can be multi-faceted. Part of the decision is based on which fuel will provide the highest amount of energy release when burned. Studies of thermochemistry can be very useful in getting reliable information for making these important choices. When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of \(1 \: \text{mol}\) of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of a balanced equation: \[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) + 890.4 \: \text{kJ}\nonumber \nonumber \] The equation tells us that \(1 \: \text{mol}\) of methane combines with \(2 \: \text{mol}\) of oxygen to produce \(1 \: \text{mol}\) of carbon dioxide and \(2 \: \text{mol}\) of water. In the process, \(890.4 \: \text{kJ}\) is released and so it is written as a product of the reaction. A is a chemical equation that includes the enthalpy change of the reaction. The process in the above thermochemical equation can be shown visually in the figure below. In the combustion of methane example, the enthalpy change is negative because heat is being released by the system. Therefore, the overall enthalpy of the system decreases. The is the enthalpy change for a chemical reaction. In the case above, the heat of reaction is \(-890.4 \: \text{kJ}\). The thermochemical reaction can also be written in this way: \[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ}\nonumber \nonumber \] Heats of reaction are typically measured in kilojoules. It is important to include the physical states of the reactants and products in a thermochemical equation, as the value of the \(\Delta H\) depends on those states. Endothermic reactions absorb energy from the surroundings as the reaction occurs. When \(1 \: \text{mol}\) of calcium carbonate decomposes into \(1 \: \text{mol}\) of calcium oxide and \(1 \: \text{mol}\) of carbon dioxide, \(177.8 \: \text{kJ}\) of heat is absorbed. The process is shown visually in the figure above (B). The thermochemical reaction is shown below. \[\ce{CaCO_3} \left( s \right) + 177.8 \: \text{kJ} \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber \nonumber \] Because the heat is absorbed by the system, the \(177.8 \: \text{kJ}\) is written as a reactant. The heat of reaction is positive for an endothermic reaction. \[\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = 177.8 \: \text{kJ}\nonumber \nonumber \] The way in which a reaction is written influences the value of the enthalpy change for the reaction. Many reactions are reversible, meaning that the product(s) of the reaction are capable of combining and reforming the reactant(s). If a reaction is written in the reverse direction, the sign of the \(\Delta H\) changes. For example, we can write an equation for the reaction of calcium oxide with carbon dioxide to form calcium carbonate: \[\ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \rightarrow \ce{CaCO_3} \left( s \right) + 177.8 \: \text{kJ}\nonumber \nonumber \] The reaction is exothermic and thus the sign of the enthalpy change is negative. \[\ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \rightarrow \ce{CaCO_3} \left( s \right) \: \: \: \: \: \Delta H = -177.8 \: \text{kJ}\nonumber \nonumber \]

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Solids as well as liquids exhibit vapor pressure. If you hang out a wet cloth in winter, the cloth first freezes as hard as a board, but after sufficient time all the ice evaporates and the cloth becomes soft and dry. Another solid which exhibits evidence of a vapor pressure is para-dichlorobenzene, C H Cl , which is used for mothballs. The fact that you can smell this solid from across the room means that some of its molecules must have evaporated into the air and entered your nose. Indeed, given a year or so, moth crystals will evaporate completely. Still another example is dry ice, CO ( ), whose vapor pressure reaches atmospheric pressure at –78.5°C. Consequently CO ( ) sublimes, forming vapor without passing through the liquid state. Unlike ordinary ice, it remains dry, making it very convenient as a refrigerant. \(\Page {1}\) Like the vapor pressure of a liquid, the vapor pressure of a solid increases with temperature. This variation is usually presented in combination with other curves in a , such as that for water in accompanying figure. The term in this context is used to distinguish solid, liquid, and gas, each of which has its own distinctive properties, such as density, viscosity, etc. Each phase is separated from the others by a boundary. In the phase diagram for water, the variation of the vapor pressure of ice with temperature is shown by the line . As might be expected, the vapor pressure of ice is quite small, never rising above 0.006 atm (0.61 kPa). The vapor pressure of liquid water is usually much higher, as is shown by the curve . The point on this curve corresponds to the critical point, and the vapor pressure is the critical pressure of water, 218.3 atm (22 MPa). The curve stops at this point since any distinction between liquid and gas ceases to exist above the critical temperature (this state is called a ). The other end of this vapor-pressure curve, point , is of particular interest. At the temperature of point (273.16 K or 0.01°C) both ice and water have the same vapor pressure, 0.006 atm. Since the same vapor is in equilibrium with both liquid and solid, it follows that all three phases, ice, water, and vapor, are in equilibrium at point . This point where all 3 phases exist at the same time is therefore referred to as the . Line remains to be discussed. This line describes how the melting point of ice varies with pressure. In other words, it includes temperatures and pressures at which solid and liquid phases are in equilibrium. Note that the line is not exactly vertical. Point corresponds to a slightly lower temperature than point . This means that, as we increase the pressure on ice, its freezing point decreases. Again we can draw on everyday experience for an example of this behavior. A person on a pair of skates standing on ice exerts his or her whole weight on the ice through the thin skate blade. The very high pressure immediately under the blade causes some ice to melt, affording lubrication which enables the skater to glide smoothly over the ice. It should be emphasized that point is not the normal freezing point of water. When we measure the normal freezing point of water (or any other liquid), we do so at standard atmospheric pressure in a container open to the air (part of the next figure). In this container we have not only ice and water in equilibrium with each other but also The total pressure on the contents of this container is 1 atm (101.3 kPa) and its temperature is exactly 273.15 K (0.00°C). As far as liquid and solid are concerned, this corresponds to point in the phase diagram. If we could now pump all the air out of the container so that only pure water in its three phases was left, we would have the situation shown in part . Ice, water, and pure water vapor, but no air, now occupy the container. The pressure has dropped from the 1.00 atm of the atmosphere to the vapor pressure (0.006 atm). Because of this decrease in pressure, the melting point of ice increases slightly and the new equilibrium temperature is 0.01°C. This corresponds to the triple point of water, point in the phase diagram. Because the triple point is more readily attained (one need not be at sea level) and also more reproducible, it is now used as a primary reference temperature for the international thermodynamic scale of temperature. In SI units the temperature of the triple point of water is defined as 273.16 K.

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Though petroleum consists mainly of normal and branched alkanes, other classes of compounds also occur. The next best represented are the , which are characterized by at least one of carbon atoms. If a hydrocarbon chain is to be made into a ring, a new C—C bond must be formed between carbon atoms at the end of the chain. This requires that two hydrogen atoms be removed to make room for the new bond. Consequently such a ring involves one more C—C bond and two less C—H bonds than the corresponding normal alkane, and the general formula for a cycloalkane is C H . The most common cycloalkane in petroleum is methylcyclohexane, which has the projection formula   A ball-and-stick model of one of the conformations of methylcyclohexane is shown in the figure below. Note that the ring of six carbon atoms is puckered to permit tetrahedral arrangement of bonds around the carbon atoms. Also shown in the figure is cyclopentane, C H , which contains a five-membered ring. Virtually all the cycloalkanes in petroleum contain five- or six-membered rings similar to those illustrated.

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A good example of a nonpolar lipid is the neutral fat glycerol tristearate. This most-common form of animal fat serves as a storehouse for energy and as insulation against heat loss. On a molecular level it is constructed from three molecules of stearic acid and one of glycerol: (1) Notice that for each stearic or other fatty acid molecule that combines with one of the —OH groups of glycerol, a molecule of water is given off, and so the reaction is a condensation. It turns out that a great many important biological molecules are put together by condensation reactions during which water is given off. The reverse of Eq. (1), in which water reacts with a large molecule and splits it into smaller pieces, is called . By carrying out hydrolysis living organisms can break down molecules manufactured by other species. The simple building blocks obtained this way can then be recombined by condensation reactions to form structures appropriate to their new host. By contrast with the glycerol tristearate found in animals, vegetable fats contain numerous double bonds in their long hydrocarbon chains. This polyunsaturation introduces “kinks” in the hydrocarbon chains because of the barrier to rotation and the 120° angles associated with the double bonds. Consequently it is more difficult to align the chains side by side (see Figure \(\Page {1}\) ), and the unsaturated fats do not pack together as easily in a crystal lattice. As was true with alkanes, chain length also determines whether a fat is liquid or solid, and where the melting point occurs. Most unsaturated fats (like corn oil) are liquids at ordinary temperatures, while saturated fats (like butter) are solids. Vegetable oils can be converted by hydrogenation to compounds that are solids. This process involves adding H catalytically to the double bonds: Hydrolysis of fats [the reverse of Eq. (1)] is important in the manufacture of soaps. It can be speeded up by the addition of a strong base like NaOH or KOH, in which case the reaction is called . Since saponification requires that the pH of the reaction mixture be high, the fatty acid that is produced will dissociate to its anion. When glycerol tristearate is saponified with NaOH for example, sodium stearate, a relatively water-soluble substance and a common soap, is formed. The ability of soaps to clean grease and oil from soiled surfaces is a result of the dual hydrophobic-hydrophilic structures of their molecules. The stearate ion, for example, consists of a long nonpolar hydrocarbon chain with a highly polar —COO group at one end. The hydrophobic hydrocarbon chain tries to avoid contact with aqueous media, while the anionic group readily accommodates the dipole attractions and hydrogen bonds of water molecules. The two main ways that the hydrophobic portions of stearate ions can avoid water are to cluster together on the surface or to dissolve in a small quantity of oil or grease (see Figure \(\Page {2}\) ). In the latter case the hydrophilic heads of the soap molecules contact the water outside the grease, forming a structure known as a . Since the outsides of the micelles are negatively charged, they repel one another and prevent the grease droplets from recombining. The grease is therefore suspended (emulsified) in the water and can be washed away easily. Natural soaps, such as sodium stearate, were originally made in the home by heating animal fat with wood ashes, which contained potash, K CO . Large quantities are still produced industrially, but to a considerable extent soaps have been replaced by detergents. This is a consequence of the undesirable behavior of soaps in hard water. Calcium, magnesium, and other hard-water cations form insoluble compounds when combined with the anions of fatty acids. This produces scummy precipitates and prevents the soap molecules from emulsifying grease unless a large excess is used. Detergents such as alkylbenzenesulfonates (ABS) and linear alkylbenzenesulfonates (LAS) have structures very similar to sodium stearate except that the charged group in their hydrophilic heads is —SO attached to a benzene ring. The ABS detergents also have methyl (CH ) groups branching off their hydrocarbon chains. Such molecules do not precipitate with hard-water cations and therefore are more suitable for machine washing of clothes. The LAS detergents replaced ABS during the mid-nineteen-sixties when it was discovered that the latter were not biodegradable. They were causing rivers and even tap water to become covered with detergent suds and foam. Apparently the enzymes in microorganisms that had evolved to break down the unbranched hydrocarbon chains in natural fats and fatty acids were incapable of digesting the branched chains of ABS molecules. LAS detergents, though manufactured by humans, mimic the structures of naturally occurring molecules and are biodegradable.

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Another useful set of techniques for separating mixtures is called . Perhaps the simplest of these techniques to describe is paper chromatography, an example of which is shown in the video below. Three substances are applied to a strip of chromatography-grade paper (the of this experiment). As the liquid level rises and meets the spots, the sample partially dissolves in the liquid (the because it is moving) and travels up the plate within the solution. Different substances will travel different distances along the plate. The distance that a substance will travel depends on how strongly it adheres to the stationary phase (a process called ) versus how much time it spends dissolved in the mobile phase. The more a substance adsorbs, the less it dissolves and the less it moves along the plate. The pink and blue spots at the end of the video are examples of substances highly adsorbed to the stationary phase. The less a substance adsorbs, the more it dissolves and the farther it travels, such as the yellow on the far left. The process is continued until a good separation is created. In this manner, a mixture of substances may be separated such as the middle sample, which was originally green but separated into blue and yellow. Notice that while it was not initially obvious that the middle spot contained both substances, this fact is clear after performing paper chromatography. All forms of chromatography work on the same general principle as paper chromatography. There is always a stationary phase which does not move and a mobile phase which does. The various components in the mixture being chromatographed separate from each other because they are more strongly held by one phase or the other. Those which have the greatest affinity for the mobile phase move along the fastest. The most important form of chromatography is or vapor-phase chromatography. A long column is packed with a finely divided solid whose surface has been coated with an inert liquid. This liquid forms the stationary phase. The mobile phase is provided by an inert , such as He or N , which passes continuously through the column (seen below), among the solid particles. A liquid sample can be injected into the gas stream at the and vaporized just before it enters the tube. As this sample is carried through the by the slow stream of gas, those components which are most soluble in the inert liquid are held up, while the less-soluble components move on more rapidly. The components thus emerge one by one from the end of the tube, into the detector. In this way it is possible to separate and analyze mixtures of liquids which it would be impossible to deal with by distillation or any other technique. The development of chromatography is one of the major revolutions in technique in the history of chemistry, comparable to that which followed the development of an accurate balance. Separations which were previously considered impossible are now easily achieved, sometimes with quite simple apparatus. This technique is particular essential to the science of biochemistry, in which complex mixtures are almost always encountered. In the field of environmental chemistry, chromatography has helped us separate and detect very low concentrations of contaminants like DDT or PCB (polychlorinated biphenyls). The major drawback to chromatography is that it does not lend itself to large-scale operation. As a result it remains largely a laboratory, rather than an industrial, technique for separating mixtures.

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Altogether there are 20 amino acids which commonly occur in all organisms. Under most circumstances amino acids exist as zwitterions and have the general formula R represents a group called a which varies from one amino acid to another. One aspect of amino acid structure not obvious in the above representation is the stereochemistry. Notice in the above image the carbon marked with an α, has 4 different groups attached to it, and thus amino acids are One of the most familiar chiral objects are hands, and therefore chirality is often thought of in terms of handedness. Below are the two enantiomers of the amino acid alanine, L-alanine and R-alanine, so you can see the difference. How do you tell the difference between the two enantiomers? While most chemists use the R and S priority system to distinguish between enantiomers, many biochemical compounds, including amino acids and sugars use the D and L system. This is based upon a historical method for determining enatiomers using glyceraldehyde. Glyceraldehyde is a chiral three carbon sugar. with one form classified as D and the other as L. Amino acids and sugars can be synthesized from glyceraldehyde. The enantiomer syntheiszed from the D-form of glyceraldehyde are also labeled D, while those from the L-form, L. To determine if an amino acid is L or D, look at the α carbon, so that the hydrogen atom is directly behind it. This should place the three other functional groups in a circle. Follow from OH to to H , or . If this is in a counterclockwise direction, the the amino acid is in the L-isomer. If this order is in the clockwise direction, the amino acid is a D-isomer. Try this trick with the two models of alanine. If you assigned priority and used the R,S system, you will find that most amino acids are S-isomers. There is one exception however, which is cysteine. the sulfur in the R group gives it priority over the carboxylic acid group. This it is an R-isomer in the R,S system, but an L-isomer in the D,L system. Structures of the 20 different R groups are given in Figure \(\Page {1}\), where they are shown as part of a polypeptide chain. The figure also shows the three-letter code and one-letter code often used to identify amino acids in a polypeptide chain. Although each amino acid side chain has its own individual properties, it is useful to divide them into several categories. In Figure \(\Page {1}\) this has been done on the basis of how strongly hydrophilic or hydrophobic they are. On the extreme left of the figure are the six most hydrophilic side chains. All are polar, and four are actually ionic at a pH of 7. Next in the figure are the six most hydrophobic side chains. These are all large and nonpolar and contain no highly electronegative atoms like nitrogen and oxygen. The remaining eight side chains either contain small nonpolar groups or groups of fairly limited polarity, and are therefore not very strongly hydrophilic or very strongly hydrophobic. Whether a side chain is hydrophilic or hydrophobic has considerable influence on the conformation of the polypeptide chain. Like the hydrocarbon tails of fatty acid molecules, hydrophobic amino acid side chains tend to avoid water and cluster together with other nonpolar groups. In a globular protein like trypsin for example, most hydrophobic residues are found among twisted chains deep within the molecule. Hydrophilic side chains, by contrast, tend to occupy positions on the outside surface, where they contact the surrounding water molecules. Hydrogen bonds and dipole forces attract these water molecules and help solubilize the globular protein. Were all the nonpolar R groups exposed to the aqueous medium, the protein would be much less soluble.

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The term lipid applies to any water-insoluble substance which can be extracted from cells by organic solvents such as chloroform, ether, or benzene. Two major categories may be identified. have molecular structures which contain no electrically charged sites, few polar groups, and large amounts of carbon and hydrogen. They are similar to hydrocarbons in being almost completely insoluble in water, and so they are said to be (from the Greek, meaning water-hater). On the other hand, consist of molecules which have polar groups (such as —OH) or electrically charged sites at one end, and hydrocarbon chains at the other. Since polar or charged groups can hydrogen bond to or electrostatically attract water molecules, one end of a polar lipid molecule is said to be (water-loving). Lipids with a polar and nonpolar end are sometimes called amphipathic lipids, because one end is hydrophilic, while the other is hydrophobic. Such substances often form structures which bury hydrophobic surface, while exposing hydrophilic surface to water. Some typical structures of both types of lipids are shown in Figure \(\Page {1}\).

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is the amount of 3D space a substance or object occupies. In the photo above, the same volume of water (50 mL) is shown in each of the beakers. As you noticed, the 50 mL looks radically different from beaker to beaker. What causes this variation? Volume is a derived unit, depending on 3 quantities: length, width, and height. Therefore, though the beakers each contain the same volume, their differing lengths, widths, and heights makes that volume look deceptively different.   The most commonly used derived units are those of volume. As we have already seen, calculation of the volume of an object requires that all 3 dimensions are multiplied together (length, width, and height). Thus the SI unit of volume is the cubic meter (m ). This is rather large for use in the chemical laboratory, and so the cubic decimeter (dm ) or cubic centimeter (cm , formerly cc) are more commonly used. The relationship between these units and the cubic meter is easily shown: \[\text{1 dm} = \text{0.1 m} \nonumber \] \[\text{1 cm} = \text{0.01 m} \nonumber \] Cubing both sides of each equation, we have \[\text{1 dm}^\text{3} = \text{0.1}^\text{3} \text{m}^\text{3} = \text{0.001 m}^\text{3} = \text{10}^{-\text{3}} \text{m}^\text{3} \nonumber \] \[\text{1 cm}^\text{3} = \text{0.01}^\text{3} \text{m}^\text{3} = \text{0.000 001 m}^\text{3} = \text{10}^{-\text{6}} \text{m}^\text{3} \nonumber \] Note that in the expression dm the as well as the base unit. A cubic decimeter is one-thousandth of a cubic meter, not one-tenth of a cubic meter. Two other units of volume are commonly encountered in the chemical laboratory — the liter (l) and the milliliter (ml — one-thousandth of a liter). The liter was originally defined as the volume of one kilogram of pure water at the temperature of its maximum density (3.98°C) but in 1964 the definition was changed. The liter is now exactly one-thousandth of a cubic meter, that is, 1 dm . A milliliter is therefore exactly 1 cm . Because the new definition of liter altered its volume slightly, it is recommended that the results of highly accurate measurements be reported in the SI units cubic decimeters or cubic centimeters, rather than in liters or milliliters. For most situations discussed in this online textbook, however, the units cubic decimeter and liter, and cubic centimeter and milliliter may be used interchangeably. Thus when recording a volume obtained from laboratory glassware calibrated in milliliters, you can just as well write 24.7 cm as 24.7 ml.

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According to the central dogma of molecular genetics, DNA is the genetically active component of the chromosomes of a cell. That is, DNA in the cell nucleus contains all the information necessary to control synthesis of the proteins, enzymes, and other molecules which are needed as that cell grows, carries on metabolism, and eventually reproduces. Thus when a cell divides, its DNA must pass on genetic information to both daughter cells. It must somehow be able to divide into duplicate copies. This process is called . Given the complementary double strands of DNA, it is relatively easy to see how DNA as a molecule is well structured for replication, as is show in Figure \(\Page {1}\). Each strand serves as a template for a new strand. Thus, after DNA is replicated, each new DNA double helix will have one strand from the original DNA molecule, and one newly synthesized molecule. This is referred to as semiconservative replication. A rather complex mechanism exists for DNA replication, involving many different enzymes and protein factors. Let us consider some of the more important aspects of DNA replication. First, the double strand needs to be opened up to replicate each template strand. To do this, a set of proteins and enzymes bind to and open up the double helix at an origin point in the molecule. This forms replication forks, points where double stranded DNA opens up, allowing replication to occur. A helicase enzyme binds at the replication forks, with the function of further unwinding the DNA and allowing the replication fork to move along the double strand as DNA is replicated. Another enzyme, DNA gyrase, is also required to relieve stress on the duplex caused by unwinding the double strand. Further, single strand binding proteins are needed to prevent the single strands from reforming a double strand. Another essential enzyme in this initiation phase is primase, which creates an RNA primer on each single strand of DNA to begin replication from. All of these initial functions are necessary to prepare the DNA for the main enzyme which builds then new strands, DNA polymerase. Multiple polymerase enzymes exist, but for the moment we will DNA polymerase III, the main DNA polymerase in . DNA polymerase III catalyzes the reaction by which a new nucleotide is added to a growing DNA strand. That reaction is seen in Figure \(\Page {2}\). The DNA polymerase enzymes need a free 3' OH group in order to begin synthesizing a new strand, which explains the necesity of the RNA primer, which gives a 3'OH group for DNA polymerase III to start from. This leads to another constraint on DNA polymerase III. One strand, the can be polymerized continuously since the new strand being created goes 5' to 3' from the replication fork, but since the original strands are anti-parallel, the other strand, the is going in the wrong direction for polymerization. In this case, the polymerzation reaction starts away from the replication fork and works back toward it. This means that the lagging strand is synthesized in disconnected segments, known as Okazaki fragments, instead of continuously. Later, another DNA polymerase, in the case of , DNA polymerase I, removes RNA primers and fills in the missing discontinuities. Then, enzyme, DNA ligase, connects breaks between 3'OH groups and 5' phosphate groups in the newly synthesized strands that exist due to these discontinuities. While the enzymes of this process differ in eukaryotes, they fulfill similar mechanisms. Even with this complexity of this process, DNA polymerase III is able to add new nucleotides at a rate of 250-1,000 nucleotides per second. A number of advantages of the double-stranded structure held together by hydrogen bonds is evident in the process of replication. Complementary base pairing insures that the two new DNA molecules will be the same as the original. The large number of hydrogen bonds, each of which is relatively weak, makes complete separation of the two strands unlikely, but one hydrogen bond, or even a few, can be broken rather easily. The helicase portion of the replication complex can therefore separate the two strands in much the same way that a zipper operates. Like the teeth of a zipper, hydrogen bonds provide great strength when all work together, but the proper tool can separate them one at a time.

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As the name suggests, polysaccharides are substances built up by the condensation of a very large number of monosaccharide units. Cellulose, for example, is a polymer of β-glucose, containing upwards of 3000 glucose units in a chain. Starch is largely a polymer of α-glucose. These two substances are a classic example of how a minor difference in the monomer can lead to major differences in the macroscopic properties of the polymer. Good-quality cotton and paper are almost pure cellulose, and they give us a good idea of its properties. Cellulose forms strong but flexible fibers and does not dissolve in water. By contrast, starch has no mechanical strength at all, and some forms are water soluble. Part of the molecular structure of cellulose and starch are shown in Figure \(\Page {1}\). Cellulose and starch are different not only in overall structure and macroscopic properties. From a biochemical point of view they behave so differently that it is difficult to believe that they are both polymers of the same monosaccharide. Enzymes which are capable of hydrolyzing starch will not touch cellulose, and vice versa. From a plant’s point of view this is just as well since cellulose makes up structural material while starch serves as a storehouse for energy. If there were not a sharp biochemical distinction between the two, the need for a bit more energy by the plant might result in destruction of cell walls or other necessary structural components. Bacteria, protozoa, termites, some cockroaches, and ruminant mammals (cattle, sheep, etc.) are capable of digesting cellulose. Ruminant mammals, termites and cockroaches themselves do not produce cellulase, the enzyme which breaks down cellulose, but rather, maintain a symbiotic relationship with bacteria in their guts which do breakdown the cellulose for digestion. Most organisms, including humans, are not capable of digesting cellulose, either through their own enzymes or through a symbiotic relationship with an organism which can. If our digestive enzymes could hydrolyze cellulose, humans would have available a much larger food supply. Quite literally we would be able to eat sawdust! It is possible to hydrolyze cellulose in the laboratory either with strong acid or with cellulase, the enzyme used by bacteria. So far, however, such processes produce more expensive (and less tasty!) food than we already have available.

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Until comparatively recently, it was believed that the noble gases "had no chemistry". The observation that dioxygen reacts with platinum hexafluoride lead N. Bartlett to attempt a similar reaction with Xe in the early '60's. His attempt was prompted by his recognition that the first ionization enthalpy of Xe was almost identical to that of O (going to O ). The compound actually formed with xenon was subsequently found to have a more complicated structure, but nevertheless, the way was paved for an exhaustive investigation of the chemistry of the noble gases. The known compounds with fluorine and oxygen are listed in Table 21-2: A detailed study of this topic is not necessary for chem 242. There are also a few compounds containing a Xe-C bond, for example: Krypton forms only the unstable KrF . Radon probably has a chemistry more extensive than xenon, but because of its radioactivity, has not been very much studied.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al)/05%3A_The_Electronic_Structure_of_Atoms/510%3A_Coulombs_Law

As seen in the section on , the energy of an electron in an isolated hydrogen atom is determined by the principal quantum number . The reason for this is not exactly the same as the reason that the energy of the particle in a box depended on a whole number , however. In an atom an electron has both (like the particle in a box) and potential energy. The electron’s potential energy is a result of the attractive force between and . When unlike charges (one negative and the other positive) attract each other, or like charges (both positive or both negative) repel each other, governs the force between them. According to this law the force of attraction or repulsion varies between the charges. Suppose two particles, one with a charge of +1 μC (microcoulomb) and the other with a charge of –1 μC are placed 1cm apart. The force of attraction between these two charges is found to be 90 N (newton), about the same force as gravity exerts on a 20-lb weight. If the distance between the charges is now multiplied by a factor of 100 (increased to 1 m), then the force of attraction between the two charges is found to be by a factor of 100 , i.e., by a factor of 10 . Consequently the force decreases to about the weight of a grain of sand. Another hundredfold increase in the distance (to 100 m or about 100 yd) reduces the force of attraction by a further factor of 10 , making the force virtually undetectable. As the above example shows, electrostatic forces of attraction and repulsion are very strong when the charges are close, but drop off fairly quickly as the charges are separated. If the separation is very large, electrostatic forces can often be neglected. This behavior resembles the more familiar example of attraction and repulsion between the poles of magnets and also is akin to the force of gravity in the solar system. Magnetic forces and gravitational forces follow an inverse square relationship as well. A second part of Coulomb’s law states that the force is proportional to the of each charge. In the above example if one charge is doubled (to ±2 μC), the force is likewise doubled, while if charges are doubled, the force is multiplied by . Coulomb’s law is summarized by the equation \[F=k\frac{Q_{1}Q_{2}}{r^{2}}\] where is the force, and are the charges, and is the distance between the charges. The constant has the value 8.988 × 10 N C . To see this equation in action, check out the following : Ed Vitz (Kutztown University), (University of

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/09%3A_Covalent_Bonding/9.08%3A_Coordinate_Covalent_Bond

Remember when you were younger, and were told to share your favorite toy with your brother, sister, or friend? You probably didn't want to share, but did anyway. It likely turned out that you had more fun playing with the toy together than if you had kept it to yourself. Atoms often share electrons with other atoms that have nothing to contribute to the situation; the end result is a new structure. Each of the covalent bonds that we have looked at so far has involved each of the atoms that are bonding contributing one of the electrons to the shared pair. There is an alternate type of covalent bond in which one of the atoms provides both of the electrons in a shared pair. Carbon monoxide, \(\ce{CO}\), is a toxic gas that is released as a byproduct during the burning of fossil fuels. The bonding between the \(\ce{C}\) atom and the \(\ce{O}\) atom can be thought of in the following procession: At this point, a double bond has formed between the two atoms, with each atom providing one of the electrons to each bond. The oxygen atom now has a stable octet of electrons, but the carbon atom only has six electrons and is unstable. This situation is resolved if the oxygen atom contributes one of its lone pairs in order to make a third bond with the carbon atom. The carbon monoxide molecule is correctly represented by a triple covalent bond between the carbon and oxygen atoms. One of the bonds is a , a covalent bond in which one of the atoms contributes both of the electrons in the shared pair. Once formed, a coordinate covalent bond is the same as any other covalent bond. It is not as if the two conventional bonds in the \(\ce{CO}\) molecule are stronger or different in any other way than the coordinate covalent bond.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.03%3A_Electrolysis_of_Brine

Three important chemicals, NaOH, Cl , H , can be obtained by an aqueous NaCl solution (brine). This forms the basis of the . The diaphragm cell (also called a Hooker cell) in which the electrolysis is carried out is shown schematically in Figure \(\Page {1}\). At the cathode, water is : \[\ce{2H2O + 2} e^{-} \ce{ -> H2 + 2OH^-} \nonumber \] Chlorine is produced at the anode: \[\text{2Cl}^{-} \rightarrow \text{Cl}_2 + 2e^{-} \nonumber \] Thus the overall reaction is \[\text{2H}_2\text{O}(l) + \text{2Cl}^{-}(aq) \rightarrow \text{H}_2(g) + \text{Cl}_2(g) + \text{2OH}^{-}(aq)\label{3} \] Since the H ( ) and Cl ( ) might recombine explosively should they come in contact, the cathode must be entirely surrounded by a porous diaphragm of asbestos. Hence the name of this type of cell. Both the H ( ) and Cl ( ) produced in Eq. \(\ref{3}\) are dried, purified, and compressed into cylinders. Fresh brine is continually pumped into the cell, and the solution which is forced out contains about 10% NaOH together with a good deal of NaCl. [Remember that the spectator ions, Na ( ), are not included in a net ionic equation such as Eq. \(\ref{3}\). H O is allowed to evaporate from this solution until the concentration of the solution reaches 50% NaOH, by which time most of the NaCl has crystallized out and can be recycled to the electrolysis. The NaOH is sold as a 50% solution or further dried to give crystals whose approximate formula is NaOH•H O. The considerable effort required to concentrate the NaOH solution obtained from diaphragm cells can be avoided by using mercury cells. The cathode in such a cell is mercury, and the cathode reaction is \[\text{Na}^{+}(aq) + e^{-} + \text{xHg}(l) \rightarrow \text{NaHg}_x(l) \nonumber \] The sodium metal produced in this reaction dissolves in the liquid mercury, producing an . The liquid amalgam is then transferred to an- other part of the cell and reacted with water: \[\text{NaHg}_x(l) + \text{2H}_2\text{O}(l) \rightarrow \text{2Na}^{+}(aq) + \text{2OH}^{-}(aq) + \text{H}_2(g) + \text{xHg}(l) \nonumber \] The 50% sodium hydroxide solution produced by this reaction contains no sodium chloride and can be sold directly, without being concentrated further. Up until 1970, however, chlor-alkali plants using mercury cells did not have adequate controls to prevent losses of mercury to the environment. About 100 to 200 g mercury was lost for each 1000 kg chlorine produced-apparently a small quantity until one realizes that 2 500 000 kg chlorine was produced by mercury cells day during 1960 in the United States. Thus every 2 to 4 days 1000 kg mercury entered the environment, and by 1970 sizable quantities were being found in fish. Since 1970 adequate controls have been installed on mercury cells and most new alkali plants use diaphragm cells, but the very large quantities of mercury introduced into rivers and lakes prior to 1970 are expected to remain for a century or more.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/07%3A_Technique_Summaries/7.26%3A_Melting_Points

Load the sample by jabbing the open end of a capillary tube into a pile of the sample. With closed end down, drop the tube down a long hollow tube so that it hits the benchtop and packs the sample into the closed end of the tube. Load the sample to a height of \(2\)-\(3 \: \text{mm}\). Place a sample into a slot in the MelTemp. Turn the dial to begin heating. Heat at a medium rate to \(20^\text{o} \text{C}\) below the expected melting point. Then heat . ( Record the temperature where the first droplet of liquid is seen (there is movement in the tube). Record the second temperature when the entire sample liquefies (the entire sample changes from opaque to transparent). Record a melting range, e.g. \(120\)-\(122^\text{o} \text{C}\). Attach the sample to a thermometer with a tiny rubber band, positioning the sample flush with the bottom of the thermometer. Insert the sample into a Thiele tube, so that the sample is near the middle of the tube.

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/13%3A_States_of_Matter/13.20%3A_Phase_Diagram_for_Water

A specific consistency of snow is required to make the best snowballs. Dry snow can be tightly pressed, and will form snowballs because the higher pressure causes the snowflakes to melt somewhat. However, when you release the pressure, the snow goes back to a more solid form and the flakes no longer stick together. Ideally, instead, the snow needs to be a little bit wet so that the particles will stick together. Water is a unique substance in many ways. One of these special properties is the fact that solid water (ice) is less dense than liquid water just above the freezing point. The phase diagram for water is shown in the figure below. Notice one key difference between last section's general phase diagram, and the above phase diagram for water: in water's diagram, the slope of the line between the solid and liquid states is negative rather than positive. The reason is that water is an unusual substance, in that its solid state is less dense than the liquid state. Ice floats in liquid water. Therefore, a pressure change has the opposite effect on those two phases. If ice is relatively near its melting point, it can be changed into liquid water by the application of pressure. The water molecules are actually closer together in the liquid phase than they are in the solid phase. Refer again to water's phase diagram (figure above). Notice point \(E\), labeled the . What does that mean? At \(373.99^\text{o} \text{C}\), particles of water in the gas phase are moving very, very rapidly. At any temperature higher than that, the gas phase cannot be made to liquefy, no matter how much pressure is applied to the gas. The \(\left( P_\text{C} \right)\) is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid. For water, the critical pressure is very high, \(217.75 \: \text{atm}\). The critical point is the intersection point of the and the critical pressure.

https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/The_Octet_and_Other_Stable_Groups

Lewis considered the "group of two" to be of greatest importance for understanding molecules and chemistry. However, he had also described the "group of eight", which just said that many atoms gain, lose or share electrons until they have 8 valence electrons. But he later said that the "group of eight" was less fundamental than the "group of two". This may be in part because Langmuir invented the term "octet" to replace "group of eight" and tried to force many compounds to fit it, even if they didn't seem to. Lewis recognized that not all stable compounds follow the . Although it usually works pretty well for elements in the p-block, the transition metals usually follow the "18-electron rule". You can see that the rule matches the periodic table: hydrogen and helium want 2 electrons, p-block elements want 8, d-block elements 18, etc. The essence of Lewis' theory is that stable compounds can be predicted and understood using what are now called . These show the arrangement of valence electrons in a molecule. For stable molecules, it is usually possible to draw a structure in which electrons are shared so that every atom has its octet (8 electrons), either by adding, losing, or sharing electrons.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.18%3A_Atomic_Weights_of_the_Elements

The atomic weight of any isotope of an element is referenced to C, which is assigned an exact atomic weight of 12. The atomic weight of an element, therefore, is calculated using the atomic weights of its isotopes and the known abundance of those isotopes. For some elements the isotopic abundance varies slightly from material- to-material such that the element’s atomic weight in any specific material falls within a range of possible value; this is the case for carbon, for which the range of atomic masses is reported as [12.0096, 12.0116]. For such elements, a conventional, or representative atomic weight often is reported, chosen such that it falls within the range with an uncertainty of \(\pm 1\) in the last reported digit; in the case of carbon, for example, the representative atomic weight is 12.011. The atomic weights reported here—most to five significant figures, but a few to just three or four significant figures—are taken from the IUPAC technical report (“Atomic Weights of the Elements 2011,” , , 1047–1078). Values in ( ) are uncertainties in the last significant figure quoted and values in [ ] are the mass number for the longest lived isotope for elements that have no stable isotopes. The atomic weights for the elements B, Br, C, Cl, H, Li, Mg, N, O, Si, S, Tl are representative values.

https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/09%3A_Titrimetric_Methods/9.07%3A_Additional_Resources

The following set of experiments introduce students to the applications of titrimetry. Experiments are grouped into four categories based on the type of reaction (acid–base, complexation, redox, and precipitation). Additional experiments emphasizing potentiometric electrodes are found in . For a general history of titrimetry, see the following sources. The use of weight instead of volume as a signal for titrimetry is reviewed in the following paper. A more thorough discussion of non-aqueous titrations, with numerous practical examples, is provided in the following text. The sources listed below provides more details on the how potentiometric titration data may be used to calculate equilibrium constants. The following provides additional information about Gran plots. The following provide additional information about calculating or sketching titration curves. For a complete discussion of the application of complexation titrimetry see the texts and articles listed below. A good source for additional examples of the application of all forms of titrimetry is

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The focus of this chapter is on methods in which we measure a time-dependent signal. Chemical kinetic methods and radiochemical methods are two examples. In this section we consider the technique of flow injection analysis in which we inject the sample into a flowing carrier stream that gives rise to a transient signal at the detector. Because the shape of this transient signal depends on the physical and chemical kinetic processes that take place in the carrier stream during the time between injection and detection, we include flow injection analysis in this chapter. (FIA) was developed in the mid-1970s as a highly efficient technique for the automated analyses of samples [see, for example, (a) Ruzicka, J.; Hansen, E. H. , , 145–157; (b) Stewart, K. K.; Beecher, G. R.; Hare, P. E. , , 167–173; (c) Valcárcel, M.; Luque de Castro, M. D. , Ellis Horwood: Chichester, England, 1987]. Unlike the centrifugal analyzer described earlier in this chapter (see ), in which the number of samples is limited by the transfer disk’s size, FIA allows for the rapid, sequential analysis of an unlimited number of samples. FIA is one example of a continuous-flow analyzer, in which we sequentially introduce samples at regular intervals into a liquid carrier stream that transports them to the detector. A schematic diagram detailing the basic components of a flow injection analyzer is shown in Figure 13.4.1 . The reagent that serves as the carrier is stored in a reservoir, and a propelling unit maintains a constant flow of the carrier through a system of tubing that comprises the transport system. We inject the sample directly into the flowing carrier stream, where it travels through one or more mixing and reaction zones before it reaches the detector’s flow-cell. Figure 13.4.1 is the simplest design for a flow injection analyzer, which consists of a single channel and a single reagent reservoir. Multiple channel instruments that merge together separate channels, each of which introduces a new reagent into the carrier stream, also are possible. A more detailed discussion of FIA instrumentation is found in the next section. When we first inject a sample into the carrier stream it has the rectangular flow profile of width shown in Figure 13.4.2 a. As the sample moves through the mixing zone and the reaction zone, the width of its flow profile increases as the sample disperses into the carrier stream. Dispersion results from two processes: convection due to the flow of the carrier stream and diffusion due to the concentration gradient between the sample and the carrier stream. Convection occurs by laminar flow. The linear velocity of the sample at the tube’s walls is zero, but the sample at the center of the tube moves with a linear velocity twice that of the carrier stream. The result is the parabolic flow profile shown in Figure 13.4.2 b. Convection is the primary means of dispersion in the first 100 ms following the sample’s injection. The second contribution to the sample’s dispersion is diffusion due to the concentration gradient that exists between the sample and the carrier stream. As shown in Figure 13.20, diffusion occurs parallel (axially) and perpendicular (radially) to the direction in which the carrier stream is moving. Only radial diffusion is important in a flow injection analysis. Radial diffusion decreases the sample’s linear velocity at the center of the tubing, while the sample at the edge of the tubing experiences an increase in its linear velocity. Diffusion helps to maintain the integrity of the sample’s flow profile (Figure 13.4.2 c) and prevents adjacent samples in the carrier stream from dispersing into one another. Both convection and diffusion make significant contributions to dispersion from approximately 3–20 s after the sample’s injection. This is the normal time scale for a flow injection analysis. After approximately 25 s, diffusion is the only significant contributor to dispersion, resulting in a flow profile similar to that shown in Figure 13.4.2 d. An FIA curve, or , is a plot of the detector’s signal as a function of time. Figure 13.4.4 shows a typical fiagram for conditions in which both convection and diffusion contribute to the sample’s dispersion. Also shown on the figure are several parameters that characterize a sample’s fiagram. Two parameters define the time for a sample to move from the injector to the detector. Travel time, , is the time between the sample’s injection and the arrival of its leading edge at the detector. Residence time, , on the other hand, is the time required to obtain the maximum signal. The difference between the residence time and the travel time is \(t^{\prime}\), which approaches zero when convection is the primary means of dispersion, and increases in value as the contribution from diffusion becomes more important. The time required for the sample to pass through the detector’s flow cell—and for the signal to return to the baseline—is also described by two parameters. The baseline-to-baseline time, \(\Delta t\), is the time between the arrival of the sample’s leading edge to the departure of its trailing edge. The elapsed time between the maximum signal and its return to the baseline is the return time, \(T^{\prime}\). The final characteristic parameter of a fiagram is the sample’s peak height, . Of the six parameters shown in Figure 13.4.4 , the most important are peak height and the return time. Peak height is important because it is directly or indirectly related to the analyte’s concentration. The sensitivity of an FIA method, therefore, is determined by the peak height. The return time is important because it determines the frequency with which we may inject samples. Figure 13.4.5 shows that if we inject a second sample at a time \(T^{\prime}\) after we inject the first sample, there is little overlap of the two FIA curves. By injecting samples at intervals of \(T^{\prime}\), we obtain the maximum possible sampling rate. Peak heights and return times are influenced by the dispersion of the sample’s flow profile and by the physical and chemical properties of the flow injection system. Physical parameters that affect and \(T^{\prime}\) include the volume of sample we inject, the flow rate, the length, diameter and geometry of the mixing zone and the reaction zone, and the presence of junctions where separate channels merge together. The kinetics of any chemical reactions between the sample and the reagents in the carrier stream also influ-ence the peak height and return time. Unfortunately, there is no good theory that we can use to consistently predict the peak height and the return time for a given set of physical and chemical parameters. The design of a flow injection analyzer for a particular analytical problem still occurs largely by a process of experimentation. Nevertheless, we can make some general observations about the effects of physical and chemical parameters. In the absence of chemical effects, we can improve sensitivity—that is, obtain larger peak heights—by injecting larger samples, by increasing the flow rate, by decreasing the length and diameter of the tubing in the mixing zone and the reaction zone, and by merging separate channels before the point where the sample is injected. With the exception of sample volume, we can increase the sampling rate—that is, decrease the return time—by using the same combination of physical parameters. Larger sample volumes, however, lead to longer return times and a decrease in sample throughput. The effect of chemical reactivity depends on whether the species we are monitoring is a reactant or a product. For example, if we are monitoring a reactant, we can improve sensitivity by choosing conditions that decrease the residence time, , or by adjusting the carrier stream’s composition so that the reaction occurs more slowly. The basic components of a flow injection analyzer are shown in Figure 13.4.6 and include a pump to propel the carrier stream and the reagent streams, a means to inject the sample into the carrier stream, and a detector to monitor the composition of the carrier stream. Connecting these units is a transport system that brings together separate channels and provides time for the sample to mix with the carrier stream and to react with the reagent streams. We also can incorporate separation modules into the transport system. Each of these components is considered in greater detail in this section. The propelling unit moves the carrier stream through the flow injection analyzer. Although several different propelling units have been used, the most common is a , which, as shown in Figure 13.4.7 , consists of a set of rollers attached to the outside of a rotating drum. Tubing from the reagent reservoirs fits between the rollers and a fixed plate. As the drum rotates the rollers squeeze the tubing, forcing the contents of the tubing to move in the direction of the rotation. Peristaltic pumps provide a constant flow rate, which is controlled by the drum’s speed of rotation and the inner diameter of the tubing. Flow rates from 0.0005–40 mL/min are possible, which is more than adequate to meet the needs of FIA where flow rates of 0.5–2.5 mL/min are common. One limitation to a peristaltic pump is that it produces a pulsed flow—particularly at higher flow rates—that may lead to oscillations in the signal. The sample, typically 5–200 μL, is injected into the carrier stream. Although syringe injections through a rubber septum are possible, the more common method—as seen in —is to use a rotary, or loop injector similar to that used in an HPLC. This type of injector provides for a reproducible sample volume and is easily adaptable to automation, an important feature when high sampling rates are needed. The most common detectors for flow injection analysis are the electrochemical and optical detectors used in HPLC. These detectors are discussed in and are not considered further in this section. FIA detectors also have been designed around the use of ion selective electrodes and atomic absorption spectroscopy. The heart of a flow injection analyzer is the transport system that brings together the carrier stream, the sample, and any reagents that react with the sample. Each reagent stream is considered a separate channel, and all channels must merge before the carrier stream reaches the detector. The complete transport system is called a . The simplest manifold has a single channel, the basic outline of which is shown in Figure 13.4.8 . This type of manifold is used for direct analysis of analyte that does not require a chemical reaction. In this case the carrier stream serves only as a means for rapidly and reproducibly transporting the sample to the detector. For example, this manifold design has been used for sample introduction in atomic absorption spectroscopy, achieving sampling rates as high as 700 samples/h. A single-channel manifold also is used for determining a sample’s pH or determining the concentration of metal ions using an ion selective electrode. We can also use the single-channel manifold in Figure 13.4.8 for an analysis in which we monitor the product of a chemical reaction between the sample and a reactant. In this case the carrier stream both transports the sample to the detector and reacts with the sample. Because the sample must mix with the carrier stream, a lower flow rate is used. One example is the determination of chloride in water, which is based on the following sequence of reactions. \[\mathrm{Hg}(\mathrm{SCN})_{2}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \: \mathrm{HgCl}_{2}(a q)+2 \mathrm{SCN}^{-}(a q) \nonumber\] \[\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{SCN})^{2+}(a q) \nonumber\] The carrier stream consists of an acidic solution of Hg(SCN) and Fe . Injecting a sample that contains chloride into the carrier stream displaces thiocyanate from Hg(SCN) . The displaced thiocyanate then reacts with Fe to form the red-colored Fe(SCN) complex, the absorbance of which is monitored at a wavelength of 480 nm. Sampling rates of approximately 120 samples per hour have been achieved with this system [Hansen, E. H.; Ruzicka, J. , , 677–680]. Most flow injection analyses that include a chemical reaction use a manifold with two or more channels. Including additional channels provides more control over the mixing of reagents and the interaction between the reagents and the sample. Two configurations are possible for a dual-channel system. A dual-channel manifold, such as the one shown in Figure 13.4.9 a, is used when the reagents cannot be premixed because of their reactivity. For example, in acidic solutions phosphate reacts with molybdate to form the heteropoly acid H P(Mo O ). In the presence of ascorbic acid the molybdenum in the heteropoly acid is reduced from Mo(VI) to Mo(V), forming a blue-colored complex that is monitored spectrophotometrically at 660 nm [Hansen, E. H.; Ruzicka, J. , , 677–680]. Because ascorbic acid reduces molybdate, the two reagents are placed in separate channels that merge just before the loop injector. A dual-channel manifold also is used to add a second reagent after injecting the sample into a carrier stream, as shown in Figure 13.4.9 b. This style of manifold is used for the quantitative analysis of many analytes, including the determination of a wastewater’s chemical oxygen demand (COD) [Korenaga, T.; Ikatsu, H. , , 301–309]. Chemical oxygen demand is a measure of the amount organic matter in the wastewater sample. In the conventional method of analysis, COD is determined by refluxing the sample for 2 h in the presence of acid and a strong oxidizing agent, such as K Cr O or KMnO . When refluxing is complete, the amount of oxidant consumed in the reaction is determined by a redox titration. In the flow injection version of this analysis, the sample is injected into a carrier stream of aqueous H SO , which merges with a solution of the oxidant from a secondary channel. The oxidation reaction is kinetically slow and, as a result, the mixing coil and the reaction coil are very long—typically 40 m—and submerged in a thermostated bath. The sampling rate is lower than that for most flow injection analyses, but at 10–30 samples/h it is substantially greater than the redox titrimetric method. More complex manifolds involving three or more channels are common, but the possible combination of designs is too numerous to discuss. One example of a four-channel manifold is shown in Figure 13.4.10 . By incorporating a separation module into the flow injection manifold we can include a separation—dialysis, gaseous diffusion and liquid-liquid extractions are examples—in a flow injection analysis. Although these separations are never complete, they are reproducible if we carefully control the experimental conditions. Dialysis and gaseous diffusion are accomplished by placing a semipermeable membrane between the carrier stream containing the sample and an acceptor stream, as shown in Figure 13.4.11 . As the sample stream passes through the separation module, a portion of those species that can cross the semipermeable membrane do so, entering the acceptor stream. This type of separation module is common for the analysis of clinical samples, such as serum and urine, where a dialysis membrane separates the analyte from its complex matrix. Semipermeable gaseous diffusion membranes are used for the determination of ammonia and carbon dioxide in blood. For example, ammonia is determined by injecting the sample into a carrier stream of aqueous NaOH. Ammonia diffuses across the semipermeable membrane into an acceptor stream that contains an acid–base indicator. The resulting acid–base reaction between ammonia and the indicator is monitored spectrophotometrically. Liquid–liquid extractions are accomplished by merging together two immiscible fluids, each carried in a separate channel. The result is a segmented flow through the separation module, consisting of alternating portions of the two phases. At the outlet of the separation module the two fluids are separated by taking advantage of the difference in their densities. Figure 13.4.12 shows a typical configuration for a separation module in which the sample is injected into an aqueous phase and extracted into a less dense organic phase that passes through the detector. In a quantitative flow injection method a calibration curve is determined by injecting a series of external standards that contain known concentrations of analyte. The calibration curve’s format—examples include plots of absorbance versus concentration and of potential versus concentration—depends on the method of detection. Calibration curves for standard spectroscopic and electrochemical methods are discussed in and in , respectively and are not considered further in this chapter. Flow injection analysis has been used to analyze a wide variety of samples, including environmental, clinical, agricultural, industrial, and pharmaceutical samples. The majority of analyses involve environmental and clinical samples, which is the focus of this section. Quantitative flow injection methods have been developed for cationic, anionic, and molecular pollutants in wastewater, freshwaters, groundwaters, and marine waters, three examples of which were described in the previous section. Table 13.4.1 provides a partial listing of other analytes that have been determined using FIA, many of which are modifications of standard spectrophotometric and potentiometric methods. An additional advantage of FIA for environmental analysis is the ability to provide for the continuous, in situ monitoring of pollutants in the field [Andrew, K. N.; Blundell, N. J.; Price, D.; Worsfold, P. J. , , 916A–922A]. : Adapted from Valcárcel, M.; Luque de Castro, M. D. , Ellis Horwood: Chichester, England, 1987. As noted in Chapter 9, several standard methods for the analysis of water involve an acid–base, complexation, or redox titration. It is easy to adapt these titrations to FIA using a single-channel manifold similar to that shown in [Ramsing, A. U.; Ruzicka, J.; Hansen, E. H. , , 1–17]. The titrant—whose concentration must be stoichiometrically less than that of the analyte—and a visual indicator are placed in the reagent reservoir and pumped continuously through the manifold. When we inject the sample it mixes thoroughly with the titrant in the carrier stream. The reaction between the analyte, which is in excess, and the titrant produces a relatively broad rectangular flow profile for the sample. As the sample moves toward the detector, additional mixing oc- curs and the width of the sample’s flow profile decreases. When the sample passes through the detector, we determine the width of its flow profile, \(\Delta T\), by monitoring the indicator’s absorbance. A calibration curve of \(\Delta T\) versus log[analyte] is prepared using standard solutions of analyte. Flow injection analysis has also found numerous applications in the analysis of clinical samples, using both enzymatic and nonenzymatic methods. Table 13.4.2 summarizes several examples. : Adapted from Valcárcel, M.; Luque de Castro, M. D. , Ellis Horwood: Chichester, England, 1987. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of phosphate provides an instructive example of a typical procedure. The description here is based on Guy, R. D.; Ramaley, L.; Wentzell, P. D. “An Experiment in the Sampling of Solids for Chemical Analysis,” , , 1028–1033. As the title suggests, the primary focus of this chapter is on sampling. A flow injection analysis, however, is used to analyze samples. The FIA determination of phosphate is an adaptation of a standard spectrophotometric analysis for phosphate. In the presence of acid, phosphate reacts with ammonium molybdate to form a yellow-colored complex in which molybdenum is present as Mo(VI). \[\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+12 \mathrm{H}_{2} \mathrm{MoO}_{4}(a q) \leftrightharpoons \: \mathrm{H}_{3} \mathrm{P}\left(\mathrm{Mo}_{12} \mathrm{O}_{40}\right)(a q)+12 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \nonumber\] In the presence of a reducing agent, such as ascorbic acid, the yellow-colored complex is reduced to a blue-colored complex of Mo(V). Prepare the following three solutions: (a) 5.0 mM ammonium molybdate in 0.40 M HNO ; (b) 0.7% w/v ascorbic acid in 1% v/v glycerin; and a (c) 100.0 ppm phosphate standard using KH PO . Using the phosphate standard, prepare a set of external standards with phosphate concentrations of 10, 20, 30, 40, 50 and 60 ppm. Use a manifold similar to that shown in , placing a 50-cm mixing coil between the pump and the loop injector and a 50-cm reaction coil between the loop injector and the detector. For both coils, use PTFE tubing with an internal diameter of 0.8 mm. Set the flow rate to 0.5 mL/min. Prepare a calibration curve by injecting 50 μL of each standard, measuring the absorbance at 650 nm. Samples are analyzed in the same manner. 1. How long does it take a sample to move from the loop injector to the detector? The reaction coil is 50-cm long with an internal diameter of 0.8 mm. The volume of this tubing is \[V=l \pi r^{2}=50 \mathrm{cm} \times 3.14 \times\left(\frac{0.08 \mathrm{cm}}{2}\right)^{2}=0.25 \mathrm{cm}^{3}=0.25 \mathrm{mL} \nonumber\] With a flow rate of 0.5 mL/min, it takes about 30 s for a sample to pass through the system. 2. The instructions for the standard spectrophotometric method indicate that the absorbance is measured 5–10 min after adding the ascorbic acid. Why is this waiting period necessary in the spectrophotometric method, but not necessary in the FIA method? The reduction of the yellow-colored Mo(VI) complex to the blue-colored Mo(V) complex is a slow reaction. In the standard spectro-photometric method it is difficult to control reproducibly the time between adding the reagents to the sample and measuring the sample’s absorbance. To achieve good precision we allow the reaction to proceed to completion before we measure the absorbance. As seen by the answer to the previous question, in the FIA method the flow rate and the dimensions of the reaction coil determine the reaction time. Because this time is controlled precisely, the reaction occurs to the same extent for all standards and samples. A shorter reaction time has the advantage of allowing for a higher throughput of samples. 3. The spectrophotometric method recommends using phosphate standards of 2–10 ppm. Explain why the FIA method uses a different range of standards. In the FIA method we measure the absorbance before the formation of the blue-colored Mo(V) complex is complete. Because the absorbance for any standard solution of phosphate is always smaller when using the FIA method, the FIA method is less sensitive and higher concentrations of phosphate are necessary. 4. How would you incorporate a reagent blank into the FIA analysis? A reagent blank is obtained by injecting a sample of distilled water in place of the external standard or the sample. The reagent blank’s absorbance is subtracted from the absorbances obtained for the standards and samples. The following data were obtained for a set of external standards when using to analyze phosphate in a wastewater sample. What is the concentration of phosphate in a sample if it gives an absorbance of 0.287? Figure 13.4.13 shows the external standards calibration curve and the calibration equation. Substituting in the sample’s absorbance gives the con- centration of phosphate in the sample as 36.1 ppm. The majority of flow injection analysis applications are modifications of conventional titrimetric, spectrophotometric, and electrochemical methods of analysis; thus, it is appropriate to compare FIA methods to these conventional methods. The scale of operations for FIA allows for the routine analysis of minor and trace analytes, and for macro, meso, and micro samples. The ability to work with microliter injection volumes is useful when the sample is scarce. Conventional methods of analysis usually have smaller detection limits. The accuracy and precision of FIA methods are comparable to conven- tional methods of analysis; however, the precision of FIA is influenced by several variables that do not affect conventional methods, including the stability of the flow rate and the reproducibility of the sample’s injection. In addition, results from FIA are more susceptible to temperature variations. In general, the sensitivity of FIA is less than that for conventional methods of analysis for at least two reasons. First, as with chemical kinetic methods, measurements in FIA are made under nonequilibrium conditions when the signal has yet to reach its maximum value. Second, dispersion dilutes the sample as it moves through the manifold. Because the variables that affect sensitivity are known, we can design the FIA manifold to optimize the method’s sensitivity. Selectivity for an FIA method often is better than that for the corresponding conventional method of analysis. In many cases this is due to the kinetic nature of the measurement process, in which potential interferents may react more slowly than the analyte. Contamination from external sources also is less of a problem because reagents are stored in closed reservoirs and are pumped through a system of transport tubing that is closed to the environment. Finally, FIA is an attractive technique when considering time, cost, and equipment. When using an autosampler, a flow injection method can achieve very high sampling rates. A sampling rate of 20–120 samples/h is not unusual and sampling rates as high as 1700 samples/h are possible. Because the volume of the flow injection manifold is small, typically less than 2 mL, the consumption of reagents is substantially smaller than that for a conventional method. This can lead to a significant decrease in the cost per analysis. Flow injection analysis does require the need for additional equipment—a pump, a loop injector, and a manifold—which adds to the cost of an analysis. For a review of the importance of flow injection analysis, see Hansen, E. H.; Miró, M. “How Flow-Injection Analysis (FIA) Over the Past 25 Years has Changed Our Way of Performing Chemical Analyses,” , , 18–26.

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. In addition to these individual basis; please contact 1. Balance each chemical equation. a. KI(aq) + Br (l) → KBr(aq) + I (s) b. MnO (s) + HCl(aq) → MnCl (aq) + Cl (g) + H O(l) c. Na O(s) + H O(l) → NaOH(aq) d. Cu(s) + AgNO (aq) → Cu(NO ) (aq) + Ag(s) e. SO (g) + H O(l) → H SO (aq) f. S Cl (l) + NH (l) → S N (s) + S (s) + NH Cl(s) 2. Balance each chemical equation. a. Be(s) + O (g) → BeO(s) b. N O (g) + H O(l) → HNO (aq) c. Na(s) + H O(l) → NaOH(aq) + H (g) d. CaO(s) + HCl(aq) → CaCl (aq) + H O(l) e. CH NH (g) + O (g) → H O(g) + CO (g) + N (g) f. Fe(s) + H SO (aq) → FeSO (aq) + H (g) 3. Balance each chemical equation. a. N O (g) → NO (g) + O (g) b. NaNO (s) → NaNO (s) + O (g) c. Al(s) + NH NO (s) → N (g) + H O(l) + Al O (s) d. C H N O (l) → CO (g) + N (g) + H O(g) + O (g) e. reaction of butane with excess oxygen f. IO F(s) + BrF (l) → IF (l) + Br (l) + O (g) 4. Balance each chemical equation. a. H S(g) + O (g) → H O(l) + S (s) b. KCl(aq) + HNO (aq) + O (g) → KNO (aq) + Cl (g) + H O(l) c. NH (g) + O (g) → NO(g) + H O(g) d. CH (g) + O (g) → CO(g) + H (g) e. NaF(aq) + Th(NO ) (aq) → NaNO (aq) + ThF (s) f. Ca (PO ) F(s) + H SO (aq) + H O(l) → H PO (aq) + CaSO •2H O(s) + HF(aq) 5. Balance each chemical equation. a. NaCl(aq) + H SO (aq) → Na SO (aq) + HCl(g) b. K(s) + H O(l) → KOH(aq) + H (g) c. reaction of octane with excess oxygen d. S (s) + Cl (g) → S Cl (l) e. CH OH(l) + I (s) + P (s) → CH I(l) + H PO (aq) + H O(l) f. (CH ) Al(s) + H O(l) → CH (g) + Al(OH) (s) 6. Write a balanced chemical equation for each reaction. a. Aluminum reacts with bromine. b. Sodium reacts with chlorine. c. Aluminum hydroxide and acetic acid react to produce aluminum acetate and water. d. Ammonia and oxygen react to produce nitrogen monoxide and water. e. Nitrogen and hydrogen react at elevated temperature and pressure to produce ammonia. f. An aqueous solution of barium chloride reacts with a solution of sodium sulfate. 7. Write a balanced chemical equation for each reaction. a. Magnesium burns in oxygen. b. Carbon dioxide and sodium oxide react to produce sodium carbonate. c. Aluminum reacts with hydrochloric acid. d. An aqueous solution of silver nitrate reacts with a solution of potassium chloride. e. Methane burns in oxygen. f. Sodium nitrate and sulfuric acid react to produce sodium sulfate and nitric acid. Please be sure you are familiar with the topics discussed in Essential Skills 2 ( ) before proceeding to the Numerical Problems. a. What is the formula mass of each species? a. ammonium chloride b. sodium cyanide c. magnesium hydroxide d. calcium phosphate e. lithium carbonate f. hydrogen sulfite ion b. What is the molecular or formula mass of each compound? a. potassium permanganate b. sodium sulfate c. hydrogen cyanide d. potassium thiocyanate e. ammonium oxalate f. lithium acetate 1. What is the mass percentage of water in each hydrate? a. H AsO ·5H O b. NH NiCl ·6H O c. Al(NO ) ·9H O 2. What is the mass percentage of water in each hydrate? a. CaSO ·2H O b. Fe(NO ) ·9H O c. (NH ) ZrOH(CO ) ·2H O 3. Which of the following has the greatest mass percentage of oxygen—KMnO , K Cr O , or Fe O ? 4. Which of the following has the greatest mass percentage of oxygen—ThOCl , MgCO , or NO Cl? 5. Calculate the percent composition of the element shown in bold in each compound. a. Sb b. I c. Al O d. H O 6. Calculate the percent composition of the element shown in bold in each compound. a. H O b ReO c. H O d. Fe O 7. A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound? 8. The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds. 9. What is the mass percentage of water in each hydrate? a. LiCl·H O b. MgSO ·7H O c. Sr(NO ) ·4H O 10. What is the mass percentage of water in each hydrate? a. CaHPO ·2H O b. FeCl ·4H O c. Mg(NO ) ·4H O 11. Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates? 12. Which contains the greatest mass percentage of sulfur—FeS , Na S O , or Na S? 13. Given equal masses of each, which contains the greatest mass percentage of sulfur—NaHSO or K SO ? 14. Calculate the mass percentage of oxygen in each polyatomic ion. a. bicarbonate b. chromate c. acetate d. sulfite 15. Calculate the mass percentage of oxygen in each polyatomic ion. a. oxalate b. nitrite c. dihydrogen phosphate d. thiocyanate 16. The empirical formula of garnet, a gemstone, is Fe Al Si O . An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula? 17. A compound has the empirical formula C H O, and its formula mass is 88 g. What is its molecular formula? 18. Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula? 19. How many moles of CO and H O will be produced by combustion analysis of 0.010 mol of styrene?   20. How many moles of CO , H O, and N will be produced by combustion analysis of 0.0080 mol of aniline?   21. How many moles of CO , H O, and N will be produced by combustion analysis of 0.0074 mol of aspartame?   22. How many moles of CO , H O, N , and SO will be produced by combustion analysis of 0.0060 mol of penicillin G?   23. Combustion of a 34.8 mg sample of benzaldehyde, which contains only carbon, hydrogen, and oxygen, produced 101 mg of CO and 17.7 mg of H O. a. What was the mass of carbon and hydrogen in the sample? b. Assuming that the original sample contained only carbon, hydrogen, and oxygen, what was the mass of oxygen in the sample? c. What was the mass percentage of oxygen in the sample? d. What is the empirical formula of benzaldehyde? e. The molar mass of benzaldehyde is 106.12 g/mol. What is its molecular formula? 24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO and 17.0 mg of H O. a. What is the mass of oxygen in the sample? b. What is the mass percentage of oxygen in the sample? c. What is the empirical formula of salicylic acid? d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula? 25. Given equal masses of the following acids, which contains the greatest amount of hydrogen that can dissociate to form H —nitric acid, hydroiodic acid, hydrocyanic acid, or chloric acid? 26. Calculate the formula mass or the molecular mass of each compound. a. heptanoic acid (a seven-carbon carboxylic acid) b. 2-propanol (a three-carbon alcohol) c. KMnO d. tetraethyllead e. sulfurous acid f. ethylbenzene (an eight-carbon aromatic hydrocarbon) 27. Calculate the formula mass or the molecular mass of each compound. a. MoCl b. B O c. bromobenzene d. cyclohexene e. phosphoric acid f. ethylamine 28. Given equal masses of butane, cyclobutane, and propene, which contains the greatest mass of carbon? 29. Given equal masses of urea [(NH ) CO] and ammonium sulfate, which contains the most nitrogen for use as a fertilizer? 1) What is the relationship between an empirical formula and a molecular formula 2) Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition. a. What is the formula mass of each species? a. 53.49146 amu b. 49.0072 amu c. 58.3197 amu d. 310.177 amu e. 73.891 amu f. 81.07 amu b. What is the molecular or formula mass of each compound? a.158.034 amu b. 142.04 amu c. 27.0253 amu d. 97.181 amu e. 124.1 amu f. 65.99 amu 1. To two decimal places, the percentages are: a. 5.97% b. 37.12% c. 43.22% 2. To two decimal places, the percentages of water are: a. 20.93% b. 40.13% c. 9.52% 3. % oxygen: KMnO , 40.50%; K Cr O , 38.07%; Fe O , 30.06% 4. % oxygen: ThOCl2, 5.02%; MgCO3, 56.93%; NO2Cl, 39.28% 5. To two decimal places, the percentages are: a. 66.32% Br b. 22.79% As c. 25.40% P d. 73.43% C 6. a. 61.98% Br b. 34.69% Cs c. 59.96% C d. 21.11% S 7. Cr O . 8. Empirical Formulas 9. To two decimal places, the percentages are: a. 29.82% b. 51.16% c. 25.40% 10. What is the mass percentage of water in each hydrate? a. 20.94% b. 36.25% c. 32.70% 11. NiSO · 6H O and CoCl · 6H O 12. FeS 13. NaHSO 14. Calculate the mass percentage of oxygen in each polyatomic ion. a. 78.66% b. 55.17% c. 54.19% d. 59.95% 15. a. 72.71% b. 69.55% c. 65.99% d. 0% 16. The empirical formula of garnet, a gemstone, is Fe Al Si O . An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula? No, the calculated mass percentage of silicon in garnet is 16.93% 17. C H O 18. Empirical Formula: C Cl Molecular Formula: C Cl 19. How many moles of CO and H O will be produced by combustion analysis of 0.010 mol of styrene? Moles of CO2: 0.08 mol CO Moles of H2O: 0.04 mol H O 20. How many moles of CO , H O, and N will be produced by combustion analysis of 0.0080 mol of aniline? Mole of CO2: 0.048 mol CO Mole of H2O: 0.028 mol H O Mole of N2: 0.004 mol N 21. How many moles of CO , H O, and N will be produced by combustion analysis of 0.0074 mol of aspartame? Mole of CO2: 0.104 mol CO Mole of H2O: 0.666 mol H O Mole of N2: 0.0074 mol N 22. How many moles of CO , H O, N , and SO will be produced by combustion analysis of 0.0060 mol of penicillin G? Mole of CO2: 0.096 mol CO Mole of H2O: 0.054 mol H O Mole of N2: 0.060 mol N Mole of SO2: 0.060 mol SO 23. a. 27.6 mg C and 1.98 mg H b. 5.2 mg O c. 15% d. C H O e. C H O 24. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO and 17.0 mg of H O. a. What is the mass of oxygen in the sample? 70.4mg b. What is the mass percentage of oxygen in the sample? 61.70% c. What is the empirical formula of salicylic acid? C H O d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula? C H O 25. hydrocyanic acid, HCN 26. Calculate the formula mass or the molecular mass of each compound. a. 130.1849 amu b. 60.1 amu c. 158.034 amu d. 323.4 amu e. 82.07 amu f. 106.17 amu 27. To two decimal places, the values are: a. 273.23 amu b. 69.62 amu c. 157.01 amu d. 82.14 amu e. 98.00 amu f. 45.08 amu 28. Cyclobutene 29. Urea Please be sure you are familiar with the topics discussed in Essential Skills 2 before proceeding to the Conceptual Problems. Please be sure you are familiar with the topics discussed in Essential Skills 2 before proceeding to the Numerical Problems. 1. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass. 2. Calculate the molecular mass or formula mass of each compound. 3. Calculate the molecular mass or formula mass of each compound. 4. Calculate the molar mass of each compound. a.     c. d. e. 5. Calculate the molar mass of each compound. a. b. c. d. 6. For each compound, write the condensed formula, name the compound, and give its molar mass. a. b. 7. For each compound, write the condensed formula, name the compound, and give its molar mass. a. b. 8. Calculate the number of moles in 5.00 × 10 g of each substance. How many molecules or formula units are present in each sample? a. CaO (lime) b. CaCO (chalk) c. C H O [sucrose (cane sugar)] d. NaOCl (bleach) e. CO (dry ice) 9. Calculate the mass in grams of each sample. a. 0.520 mol of N O b. 1.63 mol of C H Br c. 4.62 mol of (NH ) SO 10. Give the number of molecules or formula units in each sample. a. 1.30 × 10 mol of SCl b. 1.03 mol of N O c. 0.265 mol of Ag Cr O 11. Give the number of moles in each sample. a. 9.58 × 10 molecules of Cl b. 3.62 × 10 formula units of KCl c. 6.94 × 10 formula units of Fe(OH) 12. Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I )? 13. What is the total number of atoms in each sample? a. 0.431 mol of Li b. 2.783 mol of methanol (CH OH) c. 0.0361 mol of CoCO d. 1.002 mol of SeBr O 14. What is the total number of atoms in each sample? a. 0.980 mol of Na b. 2.35 mol of O c. 1.83 mol of Ag S d. 1.23 mol of propane (C H ) 15. What is the total number of atoms in each sample? a. 2.48 g of HBr b. 4.77 g of CS c. 1.89 g of NaOH d. 1.46 g of SrC O 16. Decide whether each statement is true or false and explain your reasoning. 17. Complete the following table. 18. Give the formula mass or the molecular mass of each substance. 19. Give the formula mass or the molecular mass of each substance. 1. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass. 2. Calculate the molecular mass or formula mass of each compound. 3. Calculate the molecular mass or formula mass of each compound. 4. Calculate the molar mass of each compound. a. 153.82 g/mol b. 80.06 g/mol c. 92.01 g/mol d. 70.13 g/mol e. 74.12 g/mol 5. Calculate the molar mass of each compound. a. 92.45 g/mol b. 135.04 g/mol c. 44.01 g/mol d. 40.06 g/mol 6. For each compound, write the condensed formula, name the compound, and give its molar mass. a. C H O , Valeric Acid, 102.13 g/mol b. H PO Phosphorous acid, 82 g/mol 7. For each compound, write the condensed formula, name the compound, and give its molar mass. a. C H NH Ethylamine, 45.08 g/mol b. HIO Iodic acid, 175.91 g/mol 8. Calculate the number of moles in 5.00 × 10 g of each substance. How many molecules or formula units are present in each sample? a. 5.37 × 10 mol b. 3.01 × 10 mol c. 8.80 × 10 mol d. 4.04 × 10 mol e. 6.84 × 10 mol 9. Calculate the mass in grams of each sample. a. 47.85 grams b. 384.52 grams c. 536.57 grams 10. Give the number of molecules or formula units in each sample. a. 7.83x10 molecules b. 6.20x10 molecules c. 1.60x10 molecules 11. Give the number of moles in each sample. a. 1590.8 moles b. 6011.3 moles c. 115244.1 moles 12. Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I )? 2.61 x10 molecules 13. What is the total number of atoms in each sample? a. 2.60x10 atoms b. 1.01x10 atoms c. 1.09x10 atoms d. 2.41x10 atoms 14. What is the total number of atoms in each sample? a. 5.9x10 atoms b. 2.8x10 atoms c. 3.31x10 atoms d. 8.15x10 atoms 15. What is the total number of atoms in each sample? a. 3.69x10 atoms b. 1.13x10 atoms c. 8.54x10 atoms d. 3.50x10 atoms 16. a. False, the number of molecules in 0.5 mol Cl2 are the same amount of molecules in H2 b. False, the number of molecules in H2 is 2 x (6.022 x10^23) H atoms c. True, 2 H (1.01 amu) + 1 O (16.01) = 18.0 amu d. True, C6H6 -> 12(6) + 1(6) = 78 amu 17. Complete the following table a. 0.39 b. 2.36x10^23 c. 7.08x10^23 d. 482.8 e. 1.71x10^24 f. 8.55x10^24 g. 7932.7 h. 148.3 i. 5.36x10^26 j. 46938.5 k. 425.7 l. 1276.98 m. 126.5 n. 7.77x10^23 o. 5.44x10^24 p. 0.14 q. 8.27x10^22 r. 1.65x10^24 s. 34358 t. 406.8 u. 1.23x10^2 18. Give the formula mass or the molecular mass of each substance. 19. Give the formula mass or the molecular mass of each substance. Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems. 12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess. a. 2.46 g barium(s) plus 3.89 g bromine(l) in water to give barium bromide b. 1.44 g bromine(l) plus 2.42 g potassium iodide(s) in water to give potassium bromide and iodine c. 1.852 g of Zn metal plus 3.62 g of sulfuric acid in water to give zinc sulfate and hydrogen gas d. 0.147 g of iron metal reacts with 0.924 g of silver acetate in water to give iron(II) acetate and silver metal e. 3.142 g of ammonium phosphate reacts with 1.648 g of barium hydroxide in water to give ammonium hydroxide and barium phosphate 13. Under the proper conditions, ammonia and oxygen will react to form dinitrogen monoxide (nitrous oxide, also called laughing gas) and water. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants. a. 24.6 g of ammonia and 21.4 g of oxygen b. 3.8 mol of ammonia and 84.2 g of oxygen c. 3.6 × 10 molecules of ammonia and 318 g of oxygen d. 2.1 mol of ammonia and 36.4 g of oxygen 14. When a piece of zinc metal is placed in aqueous hydrochloric acid, zinc chloride is produced, and hydrogen gas is evolved. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants. a. 12.5 g of HCl and 7.3 g of Zn b. 6.2 mol of HCl and 100 g of Zn c. 2.1 × 10 molecules of Zn and 26.0 g of HCl d. 3.1 mol of Zn and 97.4 g of HCl 15. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced. a. NaI(aq) + Cl (g) → NaCl(aq) + I (s); 1.0 mol of NaCl b. NaCl(aq) + H SO (aq) → HCl(g) + Na SO (aq); 0.50 mol of HCl c. NO (g) + H O(l) → HNO (aq) + HNO (aq); 1.5 mol of HNO 16. Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced. a. AgNO (aq) + CaCl (s) → AgCl(s) + Ca(NO ) (aq); 1.25 mol of AgCl b. Pb(s) + PbO (s) + H SO (aq) → PbSO (s) + H O(l); 3.8 g of PbSO c. H PO (aq) + MgCO (s) → Mg (PO ) (s) + CO (g) + H O(l); 6.41 g of Mg (PO ) 17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.) a. KClO (s) \(\underrightarrow {\Delta} \) KCl(s)+O (g);2.14 g of KClO produces 0.67 g of O b. Cu(s) + H SO (aq) → CuSO (aq) + SO (g) + H O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide c. AgC H O (aq) + Na PO (aq) → Ag PO (s) + NaC H O (aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate 18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction? 19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction? 20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting. A + 3B → 2C 21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol. a. Write a balanced chemical equation for this reaction. b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide. 22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction: NaClO (s) + Cl (g) → ClO (aq) + NaCl(aq) a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO ? b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide. 23. The reaction of propane gas (CH CH CH ) with chlorine gas (Cl ) produces two monochloride products: CH CH CH Cl and CH CHClCH . The first is obtained in a 43% yield and the second in a 57% yield. a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion? b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane? c. Use the actual percent yield to calculate how many grams of each product would actually be obtained. 24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions: \[ \ce{2PaI_5 (s) ->[\Delta] 2Pa (s) + 5I_2 (s)} \nonumber \] a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized? b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction? c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation? 25. Aniline (C H NH ) can be produced from chlorobenzene (C H Cl) via the following reaction: C H Cl(l) + 2NH (g) → C H NH (l) + NH Cl(s) Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia. a. Which is the limiting reactant? b. Which reactant is present in excess? c. What is the theoretical yield of ammonium chloride in grams? d. If 4.78 g of NH Cl was recovered, what was the percent yield? e. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess. 26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following: a. 9.36 × 10 formula units of NaCl b. 8.5 × 10 mol of Br c. 3.7 × 10 g of NaCl 12. Write a balanced chemical equation for each reaction and then determine which reactant is in excess. a. Ba(s) + Br (l) = BaBr (aq) Reactant in excess: Br b. Br (l) + 2KI(s) = 2KBr + I Reactant in Excess: Br c. Zn + H SO = ZnSO + H Reactant in excess: H SO d. Fe + 2 AgC H O = Fe(C H O ) + 2 Ag Reactant in excess: AgC H O e. 2 (NH ) PO + 3 Ba(OH) = 6 NH OH + Ba (PO ) Reactant in excess: Ba(OH) 13. The balanced chemical equation for this reaction is 2NH + 2O → N O + 3H O a. NH b. NH c. O d. NH 14. a. Excess: HCl b. Excess: HCl c. Excess: HCl d. Excess: Zn 15. a. 150 g NaI and 35 g Cl b. 29 g NaCl and 25 g H SO c. 140 g NO and 27 g H O 16. a. 2 AgNO (aq) + CaCl (s) = 2 AgCl(s) + Ca(NO ) (aq); Mass of Reactant: 212.34g AgNO3 and 69.4 g CaCl2 b. Pb(s) + PbO (s) + 2 H SO (aq) = 2 PbSO (s) + 2 H O(l); Mass of Reactant: 1.3g Pb, 1.5g PbO , 1.2g H SO c. 2 H PO (aq) + 3 MgCO (s) = Mg (PO ) (s) + 3 CO (g) + 3 H O(l); Mass of Reactant: 6.17 g of MgCO , 4.78g of H PO 17. a. 80% b. 30% c. 35.7% 18. 81% 19. 45%. 20. (x/molecular mass of B) x (2/3) x (molecular mass of C) 21. a. CO + 2H → CH OH b. 58.28% 22. 2 NaClO (s) + Cl (g) → ClO (aq) + NaCl(aq) a. 11.9 g Cl b. (0.7458g= Grams of ClO2 = (x/(90.44 g/mol)) x (2/2) x (67.45 g ClO2/ 1 mol ClO2 23. a. 2.24 g Cl b. 4.95 g c. 2.13 g CH CH CH Cl plus 2.82 g CH CHClCH 24. a. 4.14mg Pa b. 82.1% c. 0.0162 g PaI5 25. a. chlorobenzene b. ammonia c. 8.74 g ammonium chloride. d. 55% e. \(Theoretical \, yield \, (NH_4Cl) = {mass \, of \, chlorobenzene \, (g) \times 0.92 \times \times 53.49 \, g/mol \over 112.55 \, g/mol} \) 26. a. 6619.19g Cl b. 6.77 x 10 g Cl c. 2.52 x 10 g Cl Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems. 17. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.) a. KClO (s) \(\underrightarrow {\Delta} \) KCl(s)+O (g);2.14 g of KClO produces 0.67 g of O b. Cu(s) + H SO (aq) → CuSO (aq) + SO (g) + H O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide c. AgC H O (aq) + Na PO (aq) → Ag PO (s) + NaC H O (aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate 18. Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction? 19. A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction? 20. Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting. A + 3B → 2C 21. Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol. a. Write a balanced chemical equation for this reaction. b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide. 22. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction: NaClO (s) + Cl (g) → ClO (aq) + NaCl(aq) a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO ? b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide. 23. The reaction of propane gas (CH CH CH ) with chlorine gas (Cl ) produces two monochloride products: CH CH CH Cl and CH CHClCH . The first is obtained in a 43% yield and the second in a 57% yield. a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion? b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane? c. Use the actual percent yield to calculate how many grams of each product would actually be obtained. 24. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions: \[ 2PaI_5 (s) \underrightarrow {\Delta} 2Pa (s) + 5I_2 (s) \] a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized? b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction? c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation? 25. Aniline (C H NH ) can be produced from chlorobenzene (C H Cl) via the following reaction: C H Cl(l) + 2NH (g) → C H NH (l) + NH Cl(s) Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia. a. Which is the limiting reactant? b. Which reactant is present in excess? c. What is the theoretical yield of ammonium chloride in grams? d. If 4.78 g of NH Cl was recovered, what was the percent yield? e. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess. 26. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following: a. 9.36 × 10 formula units of NaCl b. 8.5 × 10 mol of Br c. 3.7 × 10 g of NaCl 17. a. 80% b. 30% c. 35.7% 19. 45%. 21. a. CO + 2H → CH OH b. 58.28% 23. a. 2.24 g Cl b. 4.95 g c. 2.13 g CH CH CH Cl plus 2.82 g CH CHClCH 25. a. chlorobenzene b. ammonia c. 8.74 g ammonium chloride. d. 55% e. \(Theoretical \, yield \, (NH_4Cl) = {mass \, of \, chlorobenzene \, (g) \times 0.92 \times \times 53.49 \, g/mol \over 112.55 \, g/mol} \)

https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Biochemical_Cycles/Biochemical_Cycles

Plants such as trees and algae undergo the photosynthesis reaction where carbon dioxide and water in the presence of sunlight are converted to organic materials and oxygen. An important reverse reaction occurs in the water: Fish use metabolism where oxygen and organic materials - other small fish or algae - as food is converted to carbon dioxide, water, and energy. Bacteria in water, as well as land, also undergo metabolism and use oxygen and decompose organic wastes as food to convert to carbon dioxide, water, and energy. By products in the decomposition of organic waste are nitrates and phosphates. The major natural biochemical cycles include the , , and cycles. They are presented in brief in this graphic.   The overall health of a body of water depends upon whether these factors are in balance. Municipal sewage systems are now doing a better job of removing most of the organic waste products in the discharge water, but some organic waste still enters the streams and lakes. If an excess amount of organic waste is present in the water, the bacteria use all of the available oxygen in the water in an attempt to decompose the organic waste. The amount of organic waste in water is represent by a chemical test called BOD - Biological Oxygen Demand. The concentration of oxygen is measured in a water sample at the beginning of the test and again after five days. The difference between the oxygen concentrations represents the amount of oxygen consumed by the bacteria in the metabolism of the waste organics present. In situations where eutrophication occurs, the natural cycles are overwhelmed by an excess of one or more of the following: nutrients such as nitrate or phosphate, or organic waste. In the first case under aerobic conditions (presence of oxygen), the natural cycles may be more or less in balance until an excess of nitrate and/or phosphate enters the water. At this time the water plants and algae begin to grow more rapidly than normal. As this happens there is also an excess die off of the plants and algae as sunlight is blocked at lower levels. Bacteria try to decompose the organic waste, consuming the oxygen, and releasing more phosphate and nitrate to begin the cycle anew. Some of the phosphate may be precipitated as iron phosphate to remove the soluble form from the water solution. In the second case under anaerobic conditions (absence of oxygen), as conditions worsen as more phosphates and nitrates may be added to the water, all of the oxygen may be used up by bacteria in trying to decompose all of the waste. Different bacteria continue to carry on decomposition reactions, however the products are drastically different. The carbon is converted to methane gas instead of carbon dioxide, sulfur is converted to hydrogen sulfide gas. Some of the sulfide may be precipitated as iron sulfide. Under anaerobic conditions the iron phosphate in the sediments may be solubilized into solution to make it available as a nutrient for the algae which would start the growth and decay cycle over again. The pond may gradually fill with undecayed plant materials to make a swamp.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Book3A_Medicines_by_Design/03%3A_Drugs_from_Nature_Then_and_Now/3.05%3A_TestingI_II_III

To translate pharmacology research into patient care, potential drugs ultimately have to be tested in people. This multistage process is known as clinical trials, and it has led researchers to validate life-saving treatments for many diseases, such as childhood leukemia and Hodgkin's disease. Clinical trials, though costly and very time-consuming, are the only way researchers can know for sure whether experimental treatments work in humans. Scientists conduct clinical trials in three phases (I, II, and III), each providing the answer to a different fundamental question about a potential new drug: Is it safe? Does it work? Is it better than the standard treatment? Typically, researchers do years of basic work in the lab and in animal models before they can even consider testing an experimental treatment in people. Importantly, scientists who wish to test drugs in people must follow strict rules that are designed to protect those who volunteer to participate in clinical trials. Special groups called Institutional Review Boards, or IRBs, evaluate all proposed research involving humans to determine the potential risks and anticipated benefits. The goal of an IRB is to make sure that the risks are minimized and that they are reasonable compared to the knowledge expected to be gained by performing the study. Clinical studies cannot go forward without IRB approval. In addition, people in clinical studies must agree to the terms of a trial by participating in a process called informed consent and signing a form, required by law, that says they understand the risks and benefits involved in the study. Phase I studies test a drug's safety in a few dozen to a hundred people and are designed to figure out what happens to a drug in the body—how it is absorbed, metabolized, and excreted. Phase I studies usually take several months. Phase II trials test whether or not a drug produces a desired effect. These studies take longer—from several months to a few years—and can involve up to several hundred patients. A phase III study further examines the effectiveness of a drug as well as whether the drug is better than current treatments. Phase III studies involve hundreds to thousands of patients, and these advanced trials typically last several years. Many phase II and phase III studies are randomized, meaning that one group of patients gets the experimental drug being tested while a second, control group gets either a standard treatment or placebo (that is, no treatment, often masked as a "dummy" pill or injection). Also, usually phase II and phase III studies are "blinded"—the patients and the researchers do not know who is getting the experimental drug. Finally, once a new drug has completed phase III testing, a pharmaceutical company can request approval from the Food and Drug Administration to market the drug. Scientists are currently testing cone snail toxins for the treatment of which health problem? How are people protected when they volunteer to participate in a clinical trial? Why do plants and marine organisms have chemicals that could be used as medicines? What is a drug "lead?" Name the first marine-derived cancer medicine.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Book3A_Medicines_by_Design/04%3A_Molecules_to_Medicines/4.04%3A_Transportation_Dilemmas

Scientists are solving the dilemma of drug delivery with a variety of other clever techniques. Many of the techniques are geared toward sneaking through the cellular gate-keeping systems' membranes. The challenge is a chemistry problem—most drugs are water-soluble, but membranes are oily. Water and oil don't mix, and thus many drugs can't enter the cell. To make matters worse, size matters too. Membranes are usually constructed to permit the entry of only small nutrients and hormones, often through private cellular alleyways called transporters. Many pharmacologists are working hard to devise ways to work not against, but nature, by learning how to hijack molecular transporters to shuttle drugs into cells. Gordon Amidon, a pharmaceutical chemist at the University of Michigan-Ann Arbor, has been studying one particular transporter in mucosal membranes lining the digestive tract. The transporter, called hPEPT1, normally serves the body by ferrying small, electrically charged particles and small protein pieces called peptides into and out of the intestines. Amidon and other researchers discovered that certain medicines, such as the antibiotic penicillin and certain types of drugs used to treat high blood pressure and heart failure, also travel into the intestines via hPEPT1. Recent experiments revealed that the herpes drug Valtrex and the AIDS drug Retrovir also hitch a ride into intestinal cells using the hPEPT1 transporter. Amidon wants to extend this list by synthesizing hundreds of different molecules and testing them for their ability to use hPEPT1 and other similar transporters. Recent advances in molecular biology, genomics, and bioinformatics have sped the search for molecules that Amidon and other researchers can test. Scientists are also trying to slip molecules through membranes by cloaking them in disguise. Steven Regen of Lehigh University in Bethlehem, Pennsylvania, has manufactured miniature chemical umbrellas that close around and shield a molecule when it encounters a fatty membrane and then spread open in the watery environment inside a cell. So far, Regen has only used test molecules, not actual drugs, but he has succeeded in getting molecules that resemble small segments of DNA across membranes. The ability to do this in humans could be a crucial step in successfully delivering therapeutic molecules to cells via gene therapy.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Chemistry_of_Cooking_(Rodriguez-Velazquez)/09%3A_Spices/9.08%3A_Seasoning_and_Flavoring

Many ingredients are used to enhance the taste of foods. These ingredients can be used to provide both seasoning and flavoring. Knowing how to use seasonings and flavorings skillfully provides cooks and bakers with an arsenal with which they can create limitless flavor combinations. Flavoring and seasoning ingredients include wines, spirits, fruit zests, extracts, essences, and oils. However, the main seasoning and flavoring ingredients are classified as herbs and spices. Knowing the difference between herbs and spices is not as important as knowing how to use seasonings and flavorings skillfully. In general, fresh seasonings are added late in the cooking process while dry ones tend to be added earlier. It is good practice to under-season during the cooking process and then add more seasonings (particularly if you are using fresh ones) just before presentation. This is sometimes referred to as “layering.” When baking, it is difficult to add more seasoning at the end, so testing recipes to ensure the proper amount of spice is included is a critical process.

https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/DNA/DNA_Structure/DNA_History

DNA is the primary agent for all genetic material. This is common sense but it was not always that way. In the early 1900's many people thought that must be the genetic material responsible for inherited characteristics. One of the reasons behind this belief was the knowledge that proteins were quite complex molecules and therefore, they must be specified by molecules of equal or greater complexity (i.e. other proteins). DNA was known to be a relatively simple molecule, in comparison to proteins, and therefore it was hard to understand how a complex molecule (a protein) could be determined by a simpler molecule (DNA). What were the key experiments which identified DNA as the primary genetic material? The overall conclusions from these experiments was that there was a "transforming agent" in the the heat treated type I bacteria which transfomed the live mutant (rough) type II bacteria to be able to produce type I capsule polysaccharide. This "transforming agent" could have been the DNA or the protein or something completely different. T2 is a virus which attacks the bacteria . The virus, or , looks like a tiny lunar landing module. The viral particles adsorb to the surface of the E. coli cells. It was known that some material then leaves the phage and enters the cell. The "empty" phage particles on the surface cells can be physically removed by putting the cells into a blender and whipping them up. In any case, some 20 minutes after the phage adsorb to the surface of the bacteria the bacteria bursts open (lysis) and releases a multitude of progeny virus. If the media in which the bacteria grew (and were infected) included P labeled ATP, progeny phage could be recovered with this isotope incorporated into its DNA (normal proteins contain only hydrogen, nitrogen, carbon, oxygen, and sulfur atoms). Likewise if the media contained S labeled methionine the resulting progeny phage could be recovered with this isotope present only in its protein components (normal DNA contains only hydrogen, nitrogen, carbon, oxygen and phosphorous atoms). Phage were grown in the presence of either P or S isotopic labels. 1) were infected with S labeled phage. After infection, but prior to cell lysis, the bacteria were whipped up in a blender and the phage particles were separated from the bacterial cells. The isolated bacterial cells were cultured further until lysis occurred. The released progeny phage were isolated. Where the S label went: 2) were infected with P labeled phage. The same steps as in 1) above were performed. Where the P label went: The material which was being transfered from the phage to the bacteria during infection appeared to be mainly DNA. Although the results were not entirely unambiguous they provided additional support for the view that DNA was the "stuff" of genetic inheritance.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Thermosetting_vs._Thermoplastic_Polymers

Most of the polymers described above are classified as thermoplastic. This reflects the fact that above Tg they may be shaped or pressed into molds, spun or cast from melts or dissolved in suitable solvents for later fashioning. Because of their high melting point and poor solubility in most solvents, Kevlar and Nomex proved to be a challenge, but this was eventually solved. Another group of polymers, characterized by a high degree of cross-linking, resist deformation and solution once their final morphology is achieved. Such polymers are usually prepared in molds that yield the desired object. Because these polymers, once formed, cannot be reshaped by heating, they are called thermosets .Partial formulas for four of these will be shown below by clicking the appropriate button. The initial display is of Bakelite, one of the first completely synthetic plastics to see commercial use (circa 1910). A natural resinous polymer called lignin has a cross-linked structure similar to bakelite. Lignin is the amorphous matrix in which the cellulose fibers of wood are oriented. Wood is a natural composite material, nature's equivalent of fiberglass and carbon fiber composites. A partial structure for lignin is shown here: ),

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Free_Radicals

In , a (more precisely, a ) is an , , or that has or an , and therefore may be seen as having one or more "dangling" . With some exceptions, these "dangling" bonds make free radicals highly towards other substances, or even towards themselves: their molecules will often spontaneously or if they come in contact with each other. Most radicals are reasonably stable only at very low concentrations in inert media or in a vacuum. A notable example of a free radical is the hydroxyl radical (HO•), a molecule that is one atom short of a water molecule and thus has one bond "dangling" from the oxygen. Two other examples are the molecule ( ), which has two dangling bonds; and the superoxide (• ), the oxygen molecule with one extra electron, which has one dangling bond. In contrast, the ( ), the ( ) and the ( ) are not radicals, since the bonds that may appear to be dangling are in fact resolved by the addition or removal of electrons. Free radicals may be created in a number of ways, including synthesis with very dilute or rarefied reagents, reactions at very low temperatures, or breakup of larger molecules. The latter can be affected by any process that puts enough energy into the parent molecule, such as , heat, electrical discharges, , and chemical reactions. Indeed, radicals are intermediate stages in many chemical reactions. Free radicals play an important role in combustion, , polymerization, chemistry, , and many other chemical processes. In living organisms, the free radicals and and their reaction products regulate many processes, such as control of vascular tone and thus blood pressure. They also play a key role in the intermediary metabolism of various biological compounds. Such radicals can even be messengers in a process dubbed redox signaling. A radical may be trapped within a or be otherwise bound. Until late in the 20th century the word "radical" was used in chemistry to indicate any connected group of atoms, such as a methyl group or a carboxyl, whether it was part of a larger molecule or a molecule on its own. The qualifier "free" was then needed to specify the unbound case. Following recent nomenclature revisions, a part of a larger molecule is now called a functional group or substituent, and "radical" now implies "free". However, the old nomenclature may still occur in the literature. The first organic free radical identified was triphenylmethyl radical. This species was discovered by Moses Gomberg in 1900 at the University of Michigan USA. Historically, the term in radical theory was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons; however, they can be observed as radicals in when broken apart by irradiation with energetic electrons. In chemical equations, free radicals are frequently denoted by a dot placed immediately to the right of the atomic symbol or molecular formula as follows: Radical use single-headed arrows to depict the movement of single electrons: The cleavage of the breaking bond is drawn with a 'fish-hook' arrow to distinguish from the usual movement of two electrons depicted by a standard curly arrow. It should be noted that the second electron of the breaking bond also moves to pair up with the attacking radical electron; this is not explicitly indicated in this case. Free radicals also take part in and as . Chain reactions involving free radicals can usually be divided into three distinct processes. These are , , and . The formation of radicals may involve breaking of covalent bonds , a process that requires significant amounts of energy. For example, splitting H into 2H has a Δ ° of +435 kJ/mol, and Cl into 2Cl has a Δ ° of +243 kJ/mol. This is known as the homolytic , and is usually abbreviated as the symbol Δ °. The bond energy between two covalently bonded atoms is affected by the structure of the molecule as a whole, not just the identity of the two atoms. Likewise, radicals requiring more energy to form are less stable than those requiring less energy. Homolytic most often happens between two atoms of similar electronegativity. In organic chemistry this is often the O-O bond in species or O-N bonds. Sometimes radical formation is , presenting an additional barrier. However, propagation is a very . Likewise, although do exist, most species are electrically neutral. Radicals may also be formed by single electron of an atom or molecule. An example is the production of by the . Early studies of organometallic chemistry, especially tetra-alkyl lead species by F.A. Paneth and K. Hahnfeld in the 1930s supported heterolytic fission of bonds and a radical based mechanism. Although radicals are generally short-lived due to their reactivity, there are long-lived radicals. These are categorized as follows: The prime example of a stable radical is molecular (O ). Another common example is nitric oxide (NO). Organic radicals can be long lived if they occur in a conjugated π system, such as the radical derived from ( ). There are also hundreds of examples of radicals, which show low reactivity and remarkable thermodynamic stability with only a very limited extent of π . Persistent radical compounds are those whose longevity is due to around the radical center, which makes it physically difficult for the radical to react with another molecule. Examples of these include Gomberg's triphenylmethyl radical, Fremy's salt(Potassium nitrosodisulfonate, (KSO ) NO·), nitroxides, (general formula R NO·) such as TEMPO, TEMPOL, nitronyl nitroxides, and azephenylenyls and radicals derived from PTM (perchlorophenylmethyl radical) and TTM (tris(2,4,6-trichlorophenyl)methyl radical). Persistent radicals are generated in great quantity during combustion, and "may be responsible for the oxidative stress resulting in cardiopulmonary disease and probably cancer that has been attributed to exposure to airborne fine particles." Diradicals are molecules containing two radical centers. Multiple radical centers can exist in a molecule. Atmospheric oxygen naturally exists as a diradical in its ground state as triplet oxygen. The low reactivity of atmospheric oxygen is due to its diradical state. Non-radical states of dioxygen are actually less stable than the diradical. The relative stability of the oxygen diradical is primarily due to the spin-forbidden nature of the triplet-singlet transition required for it to grab electrons, i.e., "oxidize". The diradical state of oxygen also results in its paramagnetic character, which is demonstrated by its attraction to an external magnet. Radical alkyl intermediates are stabilized by similar physical processes to carbocations: as a general rule, the more substituted the radical center is, the more stable it is. This directs their reactions. Thus, formation of a tertiary radical (R C·) is favored over secondary (R HC·), which is favored over primary (RH C·). Likewise, radicals next to functional groups such as carbonyl, nitrile, and ether are more stable than tertiary alkyl radicals. Radicals attack double bonds. However, unlike similar ions, such radical reactions are not as much directed by electrostatic interactions. For example, the reactivity of nucleophilic ions with α,β-unsaturated compounds (C=C–C=O) is directed by the electron-withdrawing effect of the oxygen, resulting in a partial positive charge on the carbonyl carbon. There are two reactions that are observed in the ionic case: the carbonyl is attacked in a direct addition to carbonyl, or the vinyl is attacked in conjugate addition, and in either case, the charge on the nucleophile is taken by the oxygen. Radicals add rapidly to the double bond, and the resulting α-radical carbonyl is relatively stable; it can couple with another molecule or be oxidized. Nonetheless, the electrophilic/neutrophilic character of radicals has been shown in a variety of instances. One example is the alternating tendency of the copolymerization of maleic anhydride (electrophilic) and styrene (slightly nucleophilic). In intramolecular reactions, precise control can be achieved despite the extreme reactivity of radicals. In general, radicals attack the closest reactive site the most readily. Therefore, when there is a choice, a preference for five-membered rings is observed: four-membered rings are too strained, and collisions with carbons six or more atoms away in the chain are infrequent. Carbenes and nitrenes, which are diradicals, have distinctive chemistry. Spectrum of the blue flame from a butane torch showing excited molecular radical band emission and Swan bands A familiar free-radical reaction is combustion. The oxygen molecule is a stable diradical, best represented by ·O-O·. Because spins of the electrons are parallel, this molecule is stable. While the ground stateof oxygen is this unreactive spin-unpaired (triplet) diradical, an extremely reactive spin-paired (singlet) state is available. For combustion to occur, the energy barrier between these must be overcome. This barrier can be overcome by heat, requiring high temperatures. The triplet-singlet transition is also "forbidden". This presents an additional barrier to the reaction. It also means molecular oxygen is relatively unreactive at room temperature except in the presence of a catalytic heavy atom such as iron or copper. Combustion consists of various radical chain reactions that the singlet radical can initiate. The flammability of a given material strongly depends on the concentration of free radicals that must be obtained before initiation and propagation reactions dominate leading to of the material. Once the combustible material has been consumed, termination reactions again dominate and the flame dies out. As indicated, promotion of propagation or termination reactions alters flammability. For example, because lead itself deactivates free radicals in the gasoline-air mixture, tetraethyl lead was once commonly added to gasoline. This prevents the combustion from initiating in an uncontrolled manner or in unburnt residues (engine knocking) or premature ignition (preignition). When a hydrocarbon is burned, a large number of different oxygen radicals are involved. Initially, hydroperoxyl radical (HOO·) are formed. These then react further to give organic hydroperoxides that break up into hydroxyl radicals (HO·). In addition to combustion, many reactions involve free radicals. As a result many plastics, enamels, and other polymers are formed through radical polymerization. For instance, drying oils and alkyd paints harden due to radical crosslinking by oxygen from the atmosphere. Recent advances in radical polymerization methods, known as living radical polymerization, include: These methods produce polymers with a much narrower distribution of molecular weights. The most common radical in the lower atmosphere is molecular dioxygen. Photodissociation of source molecules produces other free radicals. In the lower atmosphere, the most important examples of free radical production are the photodissociation of nitrogen dioxide to give an oxygen atom and nitric oxide (see eq. 1 below), which plays a key role in smog formation—and the photodissociation of ozone to give the excited oxygen atom O(1D) (see eq. 2 below). The net and return reactions are also shown (eq. 3 and 4, respectively). In the upper atmosphere, a particularly important source of radicals is the photodissociation of normally unreactive chlorofluorocarbons (CFCs) by solar ultraviolet radiation, or by reactions with other stratospheric constituents (see eq. 1 below). These reactions give off the chlorine radical, Cl•, which reacts with ozone in a catalytic chain reaction ending in Ozone depletion and regeneration of the chlorine radical, allowing it to reparticipate in the reaction (see eq. 2–4 below). Such reactions are believed to be the primary cause of depletion of the ozone layer (the net result is shown in eq. 5 below), and this is why the use of chlorofluorocarbons as refrigerants has been restricted. Free radicals play an important role in a number of biological processes. Many of these are necessary for life, such as the intracellular killing of bacteria by phagocytic cells such as granulocytes and macrophages. Researchers have also implicated free radicals in certain cell signalling processes, known as . The two most important oxygen-centered free radicals are and . They derive from molecular oxygen under reducing conditions. However, because of their reactivity, these same free radicals can participate in unwanted side reactions resulting in cell damage. Excessive amounts of these free radicals can lead to cell injury and death, which may contribute to many diseases such as cancer, stroke, myocardial infarction, diabetes and major disorders. Many forms of cancer are thought to be the result of reactions between free radicals and DNA, potentially resulting in mutations that can adversely affect the cell cycle and potentially lead to malignancy. Some of the symptoms of aging such as atherosclerosis are also attributed to free-radical induced oxidation of cholesterol to 7-ketocholesterol. In addition free radicals contribute to -induced damage, perhaps more than alcohol itself. Free radicals produced by are implicated in inactivation of in the . This process promotes the development of . Free radicals may also be involved in , senile and drug-induced , , and . The classic free-radical syndrome, the iron-storage disease , is typically associated with a constellation of free-radical-related symptoms including movement disorder, psychosis, skin pigmentary abnormalities, deafness, arthritis, and diabetes mellitus. The of aging proposes that free radicals underlie the itself. Similarly, the process of mito suggests that repeated exposure to free radicals may extend life span. Because free radicals are necessary for life, the body has a number of mechanisms to minimize free-radical-induced damage and to repair damage that occurs, such as the , , and . In addition, play a key role in these defense mechanisms. These are often the three vitamins, , and and . Furthermore, there is good evidence indicating that and can act as antioxidants to help neutralize certain free radicals. Bilirubin comes from the breakdown of ' contents, while uric acid is a breakdown product of . Too much bilirubin, though, can lead to , which could eventually damage the central nervous system, while too much uric acid causes . or ROS are species such as , , and and are associated with cell damage. ROS form as a natural by-product of the normal metabolism of and have important roles in cell signaling. has been found to form in sunlight, and therefore may be associated with cell damage as well. This only occurred when it was combined with other ingredients commonly found in sunscreens, like titanium oxide and octyl methoxycinnamate. In most fields of chemistry, the historical definition of radicals contends that the molecules have nonzero spin. However in fields including spectroscopy, chemical reaction, and astrochemistry, the definition is slightly different. Gerhard Herzberg, who won the Nobel prize for his research into the electron structure and geometry of radicals, suggested a looser definition of free radicals: "any transient (chemically unstable) species (atom, molecule, or ion)". The main point of his suggestion is that there are many chemically unstable molecules that have zero spin, such as C , C , CH and so on. This definition is more convenient for discussions of transient chemical processes and astrochemistry; therefore researchers in these fields prefer to use this loose definition. Free radical diagnostic techniques include:

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Sigmoidal kinetic profiles are the result of enzymes that demonstrate g. cooperativity refers to the observation that binding of the substrate or ligand at one binding site affects the affinity of other sites for their substrates. For enzymatic reactions with multiple substrate binding sites, this increased affinity for the substrate causes a rapid and coordinated increase in the velocity of the reaction at higher \([S]\) until \(V_{max}\) is achieved. Plotting the \(V_0\) vs. \([S]\) for a cooperative enzyme, we observe the characteristic sigmoidal shape with low enzyme activity at low substrate concentration and a rapid and immediate increase in enzyme activity to \(V_{max}\) as \([S]\) increases. The phenomenon of cooperativity was initially observed in the oxygen-hemoglobin interaction that functions in carrying oxygen in blood. Positive cooperativity implies allosteric binding – binding of the ligand at one site increases the enzyme’s affinity for another ligand at a site different from the other site. Enzymes that demonstrate cooperativity are defined as allosteric. There are several types of allosteric interactions: (positive & negative) homotropic and heterotropic Figure 1: Rate of Reaction (velocity) vs. Substrate Concentration. Positive and negative allosteric interactions (as illustrated through the phenomenon of cooperativity) refer to the enzyme's binding affinity for other ligands at other sites, as a result of ligand binding at the initial binding site. When the ligands interacting are all the same compounds, the effect of the allosteric interaction is considered homotropic. When the ligands interacting are different, the effect of the allosteric interaction is considered heterotropic. It is also very important to remember that allosteric interactions tend to be driven by ATP hydrolysis. The degree of cooperativity is determined by Hill equation (Equation 1) for non-Michaelis-Menten kinetics. The Hill equation accounts for allosteric binding at sites other than the active site. \(n\) is the "Hill coefficient." When n < 1, there is negative cooperativity; When n = 1, there is no cooperativity; When \(n > 1\), there is positive cooperativity \[ \theta = \dfrac{[L]^n}{K_d+[L]^n} = \dfrac{[L]^n}{K_a^n+[L]^n} \label{1}\] where Taking the logarithm of both sides of the equation leads to an alternative formulation of the HIll Equation. \[ \log \left( \dfrac{\theta}{1-\theta} \right) = n\log [L] - \log K_d \label{2}\] Currently, there are 2 models for illustrating cooperativity: the concerted model and the sequential model

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Write the chemical formula and Lewis structure of the following, each of which contains five carbon atoms: There are several sets of answers; one is: (a) C H (b) C H (c) C H What is the difference between the hybridization of carbon atoms’ valence orbitals in saturated and unsaturated hydrocarbons? On a microscopic level, how does the reaction of bromine with a saturated hydrocarbon differ from its reaction with an unsaturated hydrocarbon? How are they similar? Both reactions result in bromine being incorporated into the structure of the product. The difference is the way in which that incorporation takes place. In the saturated hydrocarbon, an existing C–H bond is broken, and a bond between the C and the Br can then be formed. In the unsaturated hydrocarbon, the only bond broken in the hydrocarbon is the π bond whose electrons can be used to form a bond to one of the bromine atoms in Br (the electrons from the Br–Br bond form the other C–Br bond on the other carbon that was part of the π bond in the starting unsaturated hydrocarbon). On a microscopic level, how does the reaction of bromine with an alkene differ from its reaction with an alkyne? How are they similar? Explain why unbranched alkenes can form geometric isomers while unbranched alkanes cannot. Does this explanation involve the macroscopic domain or the microscopic domain? Unbranched alkanes have free rotation about the C–C bonds, yielding all orientations of the substituents about these bonds equivalent, interchangeable by rotation. In the unbranched alkenes, the inability to rotate about the \(\mathrm{C=C}\) bond results in fixed (unchanging) substituent orientations, thus permitting different isomers. Since these concepts pertain to phenomena at the molecular level, this explanation involves the microscopic domain. Explain why these two molecules are not isomers: Explain why these two molecules are not isomers: They are the same compound because each is a saturated hydrocarbon containing an unbranched chain of six carbon atoms. How does the carbon-atom hybridization change when polyethylene is prepared from ethylene? Write the Lewis structure and molecular formula for each of the following hydrocarbons: (a) C H (b) C H (c) C H (d) C H (e) C H (f) C H Write the chemical formula, condensed formula, and Lewis structure for each of the following hydrocarbons: Give the complete IUPAC name for each of the following compounds: (a) 2,2-dibromobutane; (b) 2-chloro-2-methylpropane; (c) 2-methylbutane; (d) 1-butyne; (e) 4-fluoro-4-methyl-1-octyne; (f) -1-chloropropene; (g) 5-methyl-1-pentene Give the complete IUPAC name for each of the following compounds: Butane is used as a fuel in disposable lighters. Write the Lewis structure for each isomer of butane. Write Lewis structures and name the five structural isomers of hexane. Write Lewis structures for the isomers of \(\mathrm{CH_3CH=CHCl}\). Write structures for the three isomers of the aromatic hydrocarbon xylene, C H (CH ) . Isooctane is the common name of the isomer of C H used as the standard of 100 for the gasoline octane rating: (a) 2,2,4-trimethylpentane; (b) 2,2,3-trimethylpentane, 2,3,4-trimethylpentane, and 2,3,3-trimethylpentane: Write Lewis structures and IUPAC names for the alkyne isomers of C H . Write Lewis structures and IUPAC names for all isomers of C H Cl. Name and write the structures of all isomers of the propyl and butyl alkyl groups. Write the structures for all the isomers of the –C H alkyl group. In the following, the carbon backbone and the appropriate number of hydrogen atoms are shown in condensed form: Write Lewis structures and describe the molecular geometry at each carbon atom in the following compounds: Benzene is one of the compounds used as an octane enhancer in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: \[\ce{3C2H2 ⟶ C6H6}\] Draw Lewis structures for these compounds, with resonance structures as appropriate, and determine the hybridization of the carbon atoms in each. In acetylene, the bonding uses hybrids on carbon atoms and orbitals on hydrogen atoms. In benzene, the carbon atoms are hybridized. Teflon is prepared by the polymerization of tetrafluoroethylene. Write the equation that describes the polymerization using Lewis symbols. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) \(\mathrm{CH=CCH_2CH_3 + 2I_2⟶ CHI_2CI_2CH_2CH_3}\) (b) \(\ce{CH3CH2CH2CH2CH3 + 8O2 ⟶ 5CO2 + 6H2O}\) Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product. 65.2 g Acetylene is a very weak acid; however, it will react with moist silver(I) oxide and form water and a compound composed of silver and carbon. Addition of a solution of HCl to a 0.2352-g sample of the compound of silver and carbon produced acetylene and 0.2822 g of AgCl. Ethylene can be produced by the pyrolysis of ethane: \(\ce{C2H6⟶C2H4 + H2}\) How many kilograms of ethylene is produced by the pyrolysis of 1.000 × 10 kg of ethane, assuming a 100.0% yield? 9.328 × 10 kg Why do the compounds hexane, hexanol, and hexene have such similar names? Write condensed formulas and provide IUPAC names for the following compounds: (a) ethyl alcohol, ethanol: CH CH OH; (b) methyl alcohol, methanol: CH OH; (c) ethylene glycol, ethanediol: HOCH CH OH; (d) isopropyl alcohol, 2-propanol: CH CH(OH)CH ; (e) glycerine, l,2,3-trihydroxypropane: HOCH CH(OH)CH OH Give the complete IUPAC name for each of the following compounds: (a) (b) (c) Give the complete IUPAC name and the common name for each of the following compounds: (a) (b) (c) (a) 1-ethoxybutane, butyl ethyl ether; (b) 1-ethoxypropane, ethyl propyl ether; (c) 1-methoxypropane, methyl propyl ether Write the condensed structures of both isomers with the formula C H O. Label the functional group of each isomer. Write the condensed structures of all isomers with the formula C H O . Label the functional group (or groups) of each isomer. HOCH CH OH, two alcohol groups; CH OCH OH, ether and alcohol groups Draw the condensed formulas for each of the following compounds: MTBE, Methyl -butyl ether, CH OC(CH ) , is used as an oxygen source in oxygenated gasolines. MTBE is manufactured by reacting 2-methylpropene with methanol. (a) (b) 4.593 × 10 L Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) propanol is converted to dipropyl ether (b) propene is treated with water in dilute acid. Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) \(\mathrm{CH_3CH=CHCH_3+H_2O⟶CH_3CH_2CH(OH)CH_3}\) ; (b) \(\mathrm{CH_3CH_2OH⟶CH_2=CH_2+H_2O}\) Order the following molecules from least to most oxidized, based on the marked carbon atom: Predict the products of oxidizing the molecules shown in this problem. In each case, identify the product that will result from the minimal increase in oxidation state for the highlighted carbon atom: (a) (b) (c) (a) (b) (c) Predict the products of reducing the following molecules. In each case, identify the product that will result from the minimal decrease in oxidation state for the highlighted carbon atom: (a) (b) (c) Explain why it is not possible to prepare a ketone that contains only two carbon atoms. A ketone contains a group bonded to two additional carbon atoms; thus, a minimum of three carbon atoms are needed. How does hybridization of the substituted carbon atom change when an alcohol is converted into an aldehyde? An aldehyde to a carboxylic acid? Fatty acids are carboxylic acids that have long hydrocarbon chains attached to a carboxylate group. How does a saturated fatty acid differ from an unsaturated fatty acid? How are they similar? Since they are both carboxylic acids, they each contain the –COOH functional group and its characteristics. The difference is the hydrocarbon chain in a saturated fatty acid contains no double or triple bonds, whereas the hydrocarbon chain in an unsaturated fatty acid contains one or more multiple bonds. Write a condensed structural formula, such as CH CH , and describe the molecular geometry at each carbon atom. Write a condensed structural formula, such as CH CH , and describe the molecular geometry at each carbon atom. (a) CH CH(OH)CH : all carbons are tetrahedral; (b) \(\ce{CH3C(==O)CH3}\): the end carbons are tetrahedral and the central carbon is trigonal planar; (c) CH OCH : all are tetrahedral; (d) CH COOH: the methyl carbon is tetrahedral and the acid carbon is trigonal planar; (e) CH CH CH CH(CH )CHCH : all are tetrahedral except the right-most two carbons, which are trigonal planar The foul odor of rancid butter is caused by butyric acid, CH CH CH CO H. Write the two-resonance structures for the acetate ion. Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures: Write two complete balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. (a) \(\ce{CH3CH2CH2CH2OH + CH3C(O)OH⟶CH3C(O)OCH2CH2CH2CH3 + H2O}\): (b) \(\ce{2CH3CH2COOH + CaCO3⟶(CH3CH2COO)2Ca + CO2 + H2O}\): Yields in organic reactions are sometimes low. What is the percent yield of a process that produces 13.0 g of ethyl acetate from 10.0 g of CH CO H? Alcohols A, B, and C all have the composition C H O. Molecules of alcohol A contain a branched carbon chain and can be oxidized to an aldehyde; molecules of alcohol B contain a linear carbon chain and can be oxidized to a ketone; and molecules of alcohol C can be oxidized to neither an aldehyde nor a ketone. Write the Lewis structures of these molecules. Write the Lewis structures of both isomers with the formula C H N. What is the molecular structure about the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion, (CH ) NH ? What is the hybridization of the nitrogen atom in trimethyl amine and in the trimethyl ammonium ion? Trimethyl amine: trigonal pyramidal, ; trimethyl ammonium ion: tetrahedral, Write the two resonance structures for the pyridinium ion, C H NH . Draw Lewis structures for pyridine and its conjugate acid, the pyridinium ion, C H NH . What are the geometries and hybridizations about the nitrogen atoms in pyridine and in the pyridinium ion? Write the Lewis structures of all isomers with the formula C H ON that contain an amide linkage. Write two complete balanced equations for the following reaction, one using condensed formulas and one using Lewis structures. \[\ce{CH3NH2 + H3O+ ⟶CH3NH3+ + H2O}\] Write two complete, balanced equations for each of the following reactions, one using condensed formulas and one using Lewis structures. Ethylammonium chloride is added to a solution of sodium hydroxide. Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in . CH CH = CHCH ( ) + Cl \(⟶\) CH CH(Cl)H(Cl)CH ( ); 2C H ( ) + 15O \(⟶\) 12CO ( ) + 6H O Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in . Identify any carbon atoms that change hybridization and the change in hybridization during the reactions in . the carbon in CO , initially at , changes hybridization to in CO

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For general purposes it is useful to consider temperature to be a measure of the kinetic energy of all the atoms and molecules in a given system. As temperature is increased, there is a corresponding increase in the vigor of translational and rotation motions of all molecules, as well as the vibrations of atoms and groups of atoms within molecules. Experience shows that many compounds exist normally as liquids and solids; and that even low-density gases, such as hydrogen and helium, can be liquified at sufficiently low temperature and high pressure. A clear conclusion to be drawn from this fact is that intermolecular attractive forces vary considerably, and that the boiling point of a compound is a measure of the strength of these forces. Thus, in order to break the intermolecular attractions that hold the molecules of a compound in the condensed liquid state, it is necessary to increase their kinetic energy by raising the sample temperature to the characteristic boiling point of the compound. The following table illustrates some of the factors that influence the strength of intermolecular attractions. The formula of each entry is followed by its formula weight in parentheses and the boiling point in degrees Celsius. First there is molecular size. Large molecules have more electrons and nuclei that create van der Waals attractive forces, so their compounds usually have higher boiling points than compounds made up of smaller molecules. It is very important to apply this rule only to like compounds. The examples given in the first two rows are similar in that the molecules or atoms are spherical in shape and do not have permanent dipoles. Molecular shape is also important, as the second group of compounds illustrate. The upper row consists of roughly spherical molecules, whereas the isomers in the lower row have cylindrical or linear shaped molecules. The attractive forces between the latter group are generally greater. Finally, permanent molecular dipoles generated by polar covalent bonds result in even greater attractive forces between molecules, provided they have the mobility to line up in appropriate orientations. The last entries in the table compare non-polar hydrocarbons with equal-sized compounds having polar bonds to oxygen and nitrogen. Halogens also form polar bonds to carbon, but they also increase the molecular mass, making it difficult to distinguish among these factors. The melting points of crystalline solids cannot be categorized in as simple a fashion as boiling points. The distance between molecules in a crystal lattice is small and regular, with intermolecular forces serving to constrain the motion of the molecules more severely than in the liquid state. Molecular size is important, but shape is also critical, since individual molecules need to fit together cooperatively for the attractive lattice forces to be large. Spherically shaped molecules generally have relatively high melting points, which in some cases approach the boiling point. This reflects the fact that spheres can pack together more closely than other shapes. This structure or shape sensitivity is one of the reasons that melting points are widely used to identify specific compounds. Notice that the boiling points of the unbranched alkanes (pentane through decane) increase rather smoothly with molecular weight, but the melting points of the even-carbon chains increase more than those of the odd-carbon chains. Even-membered chains pack together in a uniform fashion more compactly than do odd-membered chains. The last compound, an isomer of octane, is nearly spherical and has an exceptionally high melting point (only 6º below the boiling point).

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Write the following isotopes in hyphenated form (e.g., “carbon-14”) Write the following isotopes in nuclide notation (e.g., " \(\ce{^{14}_6C}\) ") For the following isotopes that have missing information, fill in the missing information to complete the notation For each of the isotopes in Question 21.2.3, determine the numbers of protons, neutrons, and electrons in a neutral atom of the isotope. Write the nuclide notation, including charge if applicable, for atoms with the following characteristics: Calculate the density of the \(\ce{^{24}_{12}Mg}\) nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 × 10 cm and is spherical in shape. What are the two principal differences between nuclear reactions and ordinary chemical changes? The mass of the atom \(\ce{^{23}_{11}Na}\) is 22.9898 amu. Which of the following nuclei lie within the band of stability? Which of the following nuclei lie within the band of stability? Write a brief description or definition of each of the following: Which of the various particles (α particles, β particles, and so on) that may be produced in a nuclear reaction are actually nuclei? Complete each of the following equations by adding the missing species: Complete each of the following equations: Write a balanced equation for each of the following nuclear reactions: Technetium-99 is prepared from Mo. Molybdenum-98 combines with a neutron to give molybdenum-99, an unstable isotope that emits a β particle to yield an excited form of technetium-99, represented as Tc . This excited nucleus relaxes to the ground state, represented as Tc, by emitting a γ ray. The ground state of Tc then emits a β particle. Write the equations for each of these nuclear reactions. The mass of the atom \(\ce{^{19}_9F}\) is 18.99840 amu. For the reaction \(\ce{^{14}_6C ⟶ ^{14}_7N +\, ?}\), if 100.0 g of carbon reacts, what volume of nitrogen gas (N ) is produced at 273 K and 1 atm? What are the types of radiation emitted by the nuclei of radioactive elements? What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios? What is the change in the nucleus that results from the following decay scenarios? Many nuclides with atomic numbers greater than 83 decay by processes such as electron emission. Explain the observation that the emissions from these unstable nuclides also normally include α particles. Why is electron capture accompanied by the emission of an X-ray? Explain how unstable heavy nuclides (atomic number > 83) may decompose to form nuclides of greater stability (a) if they are below the band of stability and (b) if they are above the band of stability. Which of the following nuclei is most likely to decay by positron emission? Explain your choice. The following nuclei do not lie in the band of stability. How would they be expected to decay? Explain your answer. The following nuclei do not lie in the band of stability. How would they be expected to decay? Predict by what mode(s) of spontaneous radioactive decay each of the following unstable isotopes might proceed: Write a nuclear reaction for each step in the formation of \(\ce{^{218}_{84}Po}\) from \(\ce{^{238}_{92}U}\), which proceeds by a series of decay reactions involving the step-wise emission of α, β, β, α, α, α, α particles, in that order. Write a nuclear reaction for each step in the formation of \(\ce{^{208}_{82}Pb}\) from \(\ce{^{228}_{90}Th}\), which proceeds by a series of decay reactions involving the step-wise emission of α, α, α, α, β, β, α particles, in that order. Define the term half-life and illustrate it with an example. A 1.00 × 10 -g sample of nobelium, \(\ce{^{254}_{102}No}\), has a half-life of 55 seconds after it is formed. What is the percentage of \(\ce{^{254}_{102}No}\) remaining at the following times? Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the Pu present today will be present in 1000 y? The isotope Tl undergoes β decay with a half-life of 3.1 min. If 1.000 g of \(\ce{^{226}_{88}Ra}\) produces 0.0001 mL of the gas \(\ce{^{222}_{86}Rn}\) at STP (standard temperature and pressure) in 24 h, what is the half-life of Ra in years? The isotope \(\ce{^{90}_{38}Sr}\) is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life. Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of \(\ce{^{99}_{43}Tc}\). What is the age of mummified primate skin that contains 8.25% of the original quantity of C? A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87. A laboratory investigation shows that a sample of uranium ore contains 5.37 mg of \(\ce{^{238}_{92}U}\) and 2.52 mg of \(\ce{^{206}_{82}Pb}\). Calculate the age of the ore. The half-life of \(\ce{^{238}_{92}U}\) is 4.5 × 10 yr. Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this Pu was the capture of neutrons by U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7 × 10 years ago? A \(\ce{^7_4Be}\) atom (mass = 7.0169 amu) decays into a \(\ce{^7_3Li}\) atom (mass = 7.0160 amu) by electron capture. How much energy (in millions of electron volts, MeV) is produced by this reaction? A \(\ce{^8_5B}\) atom (mass = 8.0246 amu) decays into a \(\ce{^8_4Be}\) atom (mass = 8.0053 amu) by loss of a β particle (mass = 0.00055 amu) or by electron capture. How much energy (in millions of electron volts) is produced by this reaction? Isotopes such as Al (half-life: 7.2 × 10 years) are believed to have been present in our solar system as it formed, but have since decayed and are now called extinct nuclides. Write a balanced equation for each of the following nuclear reactions: Write a balanced equation for each of the following nuclear reactions: transuranium How does nuclear fission differ from nuclear fusion? Why are both of these processes exothermic? Both fusion and fission are nuclear reactions. Why is a very high temperature required for fusion, but not for fission? Cite the conditions necessary for a nuclear chain reaction to take place. Explain how it can be controlled to produce energy, but not produce an explosion. Describe the components of a nuclear reactor. In usual practice, both a moderator and control rods are necessary to operate a nuclear chain reaction safely for the purpose of energy production. Cite the function of each and explain why both are necessary. Describe how the potential energy of uranium is converted into electrical energy in a nuclear power plant. The mass of a hydrogen atom \(\ce{(^1_1H)}\) is 1.007825 amu; that of a tritium atom \(\ce{(^3_1H)}\) is 3.01605 amu; and that of an α particle is 4.00150 amu. How much energy in kilojoules per mole of \(\ce{^4_2He}\) produced is released by the following fusion reaction: \(\ce{^1_1H + ^3_1H ⟶ ^4_2He}\). How can a radioactive nuclide be used to show that the equilibrium: \[\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)\] is a dynamic equilibrium? Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave? Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-131 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β emission. If a hospital were storing radioisotopes, what is the minimum containment needed to protect against: Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous? Given specimens uranium-232 (\(t_{1/2} = \mathrm{68.9 \;y}\)) and uranium-233 (\(t_{1/2} = \mathrm{159,200\; y}\)) of equal mass, which one would have greater activity and why? A scientist is studying a 2.234 g sample of thorium-229 ( = 7340 y) in a laboratory. Given specimens neon-24 (\(t_{1/2} = \mathrm{3.38\; min}\)) and bismuth-211 (\(t_{1/2} = \mathrm{2.14\; min}\)) of equal mass, which one would have greater activity and why?