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[ [ "\nTollen's reagent (ammoniacal silver nitrate) reacts with terminal acetylenes to form the silver acetylide which precipitates out of solution\n\n\n$$\\ce{RC#CH + AgNO3 + NH4OH -> RC#CAg v + NH4NO3 + H2O}$$\n\n\nIn the case of acetylene itself, since both ends of the molecule are (if you will) terminal acetylenes, a bis-silver acetylide salt is formed and precipitates out of solution (see page 3, item 6b in [a PDF from www.sakshieducation.com/](http://www.sakshieducation.com/EAMCET/QR/Chemistry/Jr%20Chem/13.OC%20Alkynes.pdf) or [via the Internet Archive](https://web.archive.org/web/20200204140114/http://www.sakshieducation.com/EAMCET/QR/Chemistry/Jr%20Chem/13.OC%20Alkynes.pdf))\n\n\n$$\\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg v + 2NH4NO3 + 2H2O}$$\n\n\nTollen's reagent is not very stable, therefore it needs to be freshly prepared before use. The reagent has a number of other uses in chemistry (test for aldehydes and α-hydroxy ketones), anatomical staining, and silver mirroring. [This Wikipedia article](http://en.wikipedia.org/wiki/Tollens'_reagent) provides a nice, concise overview.\n\n\n", "21" ], [ "\nTerminal hydrogens in acetylene are acidic and are easily removed to form acetylide ions, and in this case, elementary silver will be displaced from the ammoniacal silver nitrate and will form silver mirror on the surface of the reacting vessel.\n\n\n", "-3" ] ]

[ [ "\nYou could call ethyl alcohol \"ethyl hydroxide\" analogously to \"sodium hydroxide\". But it is a poor analogy, since (unlike NaOH) ethyl alcohol is not an ionic compound, does not dissociate in water to form ethyl cation and hydroxide anion, and does not raise the pH when dissolved in water.\n\n\n", "10" ], [ "\nMany people confuse between the three forms of the OH group.\n\n\nWhen the OH has a negative charge, it is called hydroxide, and it forms ionic compounds with cations. An example is sodium hydroxide, which contains discrete Na+ ions and OH- ions. When dissolved in water, it separates into Na+ ions and OH- ions (solvation not mentioned for simplicity)\n\n\nWhen the OH forms a covalent bond (different from the ionic bond above) with any other group, it is called a hydroxy group. An example is ethanol, H3C-CH2-OH, the oxygen in the OH forming a covalent bond with the carbon in the ethyl group. When you dissolve ethanol in water, it does not separate into C2H5+ and OH- ions like the NaOH would.\n\n\nThe final form of the OH is called a hydroxyl radical. Imagine removing an electron from a negatively charged hydroxide ion to make it charge-neutral. In this form it is very, very reactive, as it lacks an electron to achieve the stable form of hydroxide (like how a halogen behaves).\n\n\n", "4" ], [ "\nWell, I don’t understand what you are trying to ask. \n\n\nBut the $\\ce{OH}$ you are bothered about can be thought exactly as that of $\\ce{OH}$ of water ($\\ce{H2O}$) \n\n\n$\\ce{C2H5-OH}$ in the same way as $\\ce{H-OH}$\n\n\n", "2" ], [ "\nAs other answers point out, the hydroxyl group in an alcohol is covalently bonded to carbon and thus is not the same as a hydroxide ion.\n\n\nBut don't assume that the hydroxyl group lacks any basic character. It can still act as a Lewis base towards strong acids like $\\ce{HCl}$, facilitating nucleophilic substitution reactions between alcohols and these strong acids. The hydroxyl group in an alcohol can also act as a Lewis base towards the $\\ce{-COOH}$ function in a carboxylic acid, enabling the formation of esters from such acids. You should encounter these properties when you study nucleophilic reactions, including ester formation, in organic chemistry.\n\n\n", "2" ], [ "\nAlcohols contain a hydroxyl functional group. This is different from the hydroxide ion, $\\ce{OH-}$.\n\n\nBy way of example, the formula for the alcohol present in alcoholic drinks, ethanol or ethyl alcohol, is $\\ce{C2H5OH}$. If you want to emphasize the hydroxyl functional group, you could also write $\\ce{C2H5-OH}$. To symbolize any alcohol, you could use $\\ce{R-OH}$. As you can see, this is different from $\\ce{OH-}$ because there is no formal charge on the hydroxyl group and it is connected to the remainder of the molecule with a covalent bond.\n\n\nIn terms of chemical reactions, you can find similarities in the reactivity of water and alcohols. Many reactions of water and of alcohols involve proton (hydrogen ion, $\\ce{H+}$) transfer. Water can lose a proton (to make hydroxide, $\\ce{OH-}$), and alcohols can as well (to make an alkoxide, $\\ce{R-O-}$). Both can also gain a proton (to make hydronium $\\ce{H3O+}$ or protonated alcohol, $\\ce{R-OH2+}$, respectively). So the alcohol is analogous to water, and the alkoxide is analogous to hydroxides in some ways.\n\n\nThis is illustrated below (top, alcohol; bottom, water; left, deprotonated; center, as is; right, protonated):\n\n\n[](https://i.stack.imgur.com/xo17Z.png)\n\n\n", "1" ] ]

[ [ "\nThe Fresnel equations, <http://en.wikipedia.org/wiki/Fresnel_equations>, are used to calculate reflectivity of transparent substances based on the **index of refraction** of the liquid vs. that of air or other medium. There are complications introduced by conductivity, and *solids* may display anisotropic reflectivity, see <http://www.hindawi.com/journals/isrn/2012/732973/>.\n\n\n*Dispersion staining* is a technique for staining microscopic specimens depending on the change in the index of refraction with the color of light (dispersion), see <http://en.wikipedia.org/wiki/Dispersion_staining>. This uses the shininess of substances in a liquid, often an organic immersion oil, to help distinguish or identify them.\n\n\nNote that **metallic reflection** is based on a \"sea\" of electrons bouncing back an inverted wave, and differs from refraction. Liquid metals such as mercury are shiny for that reason, but I do not know if there are sufficiently conductive organic liquids that would have metallic reflection.\n\n\n", "2" ] ]

[ [ "\nThe **electron affinity** of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.\n\n\nIt is the tendency of an atom to gain electron. It is numerically equal to the negative of electron gain enthalpy.\n\n\nAlthough fluorine has the highest electronegativity, chlorine has the highest electron affinity and this is because the considerable repulsion in the tightly packed $2p$ subshell of fluorine.\n\n\nWhile **electron gain enthalpy** is enthalpy change when an isolated gaseous atom attain an electron to form a monovalent gaseous anion.\n\n\nChlorine ($\\ce{Cl}$) has more negative electron gain enthalpy than fluorine the reason is same as above.\n\n\n\n\n---\n\n\nSource:\n\n\nWikipedia and chemistry reader\n\n\n", "1" ] ]

[ [ "\nAtoms can be created and destroyed. For example, in nuclear fusion two or more atoms make one atom. In nuclear fission, one atom makes two or more atoms.\n\n\n\n> \n> how did the atoms form after the Big Bang? \n> \n> \n> \n\n\nThe theory of this process in known as [Big Bang Nucleosynthesis](http://en.wikipedia.org/wiki/Big_Bang_nucleosynthesis)\n\n\n\n\n\nStarting from protons and neutrons, light elements formed. Later, in stars, nuclei up to iron formed through nuclear fusion. Nuclei heavier than iron formed from supernovae by the [r-process](http://en.wikipedia.org/wiki/R-process) and asymptotic branch giant stars by the [s-process](http://en.wikipedia.org/wiki/S-process). \n\n\n\n> \n> Also, does this mean that the number of atoms in our universe has remained the same since then? \n> \n> \n> \n\n\nNo. The number constantly changes due to fusion, fission, supernovae, creation of neutron stars, black holes and possibily [quark stars](http://en.wikipedia.org/wiki/Quark_star). \n\n\n\n> \n> Also, what happens as the universe expands? Does the atoms just scatter or are new atoms being created?\n> \n> \n> \n\n\nCurrently, this process does not create or destroy atoms. However, according to the theory of [accelerating expansion](http://en.wikipedia.org/wiki/Accelerating_universe), there is a senario referred to as the \"[big rip](http://io9.com/5919193/the-big-rip-theory-says-the-universe-could-end-in-tears)\" in which all atoms would be destroyed. \n\n\n", "8" ] ]

[ [ "\nDepending on how long the battery has been sitting empty even adding electrolyte might not help. \n\n\nYour best bet is to fill the battery with distilled water and put it on a charge, charge it for a couple of days if possible.\n\n\nGet yourself a hydrometer (used for checking the specific gravity of your electrolyte) and check the water's gravities. After you charge the battery for awhile check them again. If the gravities start to rise, and the liquid is reasonable clear stop charging the battery and let it sit 12-24 hours. (If however the water turns cloudy/grey the battery is scrap get a new one). \n\n\nAfter the battery has sat and cooled down, empty the water into a container, and neutralize with soda ash, or baking soda. BE CAREFUL even though what you put it was water, what will come out might/will be corrosive. Dispose of properly. \n\n\nNow go to your local auto parts store and buy a box of battery electrolyte its usually sold as 1.28 spg, fill the battery with electrolyte re-charge until your gravities come up, and voila, use just re-conditioned a battery.\n\n\nBe sure to wear rubber gloves / goggles or face shield and any other appropriate safety gear. Remember that when a battery charges it releases hydrogen sulfide gas which is explosive so no open flames sparks etc.... \n\n\nOf course the battery may only last your a couple of months or another year or so, and it may not even come back at all. But I have used this procedure with great success to revive old batteries....and squeeze a bit more life out of them. \n\n\nJeff\nIndustrial battery specialist for 30 years.\n\n\n", "2" ], [ "\nObviously handling sulfuric acid is hazardous. If all the acid leaked out, I would get a new battery, otherwise the acid will leak out again and possibly hurt someone or damage the vehicle. \n\n\nIf there is no acid, certainly adding water will not help. If you do add acid, the concentration of acid needs to be correct. \n\n\nLead-acid batteries do not contain pure sulphuric acid, but acid dilute with water. The concentration of acid can increase over time due to electrolysis of the water to hydrogen and oxygen gases. If the concentration of acid is too high (solution density above 1.19 g/ml), adding water to dilute the acid is beneficial. \n\n\n", "1" ] ]

[ [ "\n1) Because $-\\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\\Delta G \\le w\\_{non PV}\\quad (\\text{constant T and P, closed system})$$\nwhere the equality sign holds for reversible changes.\n\n\nHere's proof:\n$$\\begin{align\\*}\nH &= U + P V&\\text{(definition of enthalpy)}\\\\\ndH &= dq + dw + d(PV) &\\text{(because dU = dq + dw)}\\\\\ndG &= dH - T dS - S dT &\\text{(because G = H - TS)}\\\\\n&=dq + dw + d(PV) - T dS -S dT &\\\\\n&= dq + dw + d(PV) - T dS & \\text{(when the change is isothermal)}\\\\\n&= dw\\_{rev} + d(PV) &\\text{(if reversible,}\\ dq\\_{rev} = T dS\\text{)}\\\\\n&= dw\\_{electrical,rev} + dw\\_{expansion,rev} + P dV + V dP & \\text{(partitioning the work)}\\\\\n&= dw\\_{electrical,rev} + V dP & (dw\\_{expansion,rev} = - P dV)\\\\\n&= dw\\_{electrical,rev} & \\text{(if pressure is constant)}\\\\\n\\end{align\\*}$$\n\n\n2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.\n\n\n", "5" ] ]

[ [ "\nSo you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is. \n\n\nFirst, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and **heat absorbed by the calorimeter**. If energy is conserved, $$q\\_{hot} + q\\_{cold} + q\\_{calorimeter} = 0$$\n\n\nOne thing that's confusing you is that last term, $q\\_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!\n\n\nNow, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. **It also gives the approximate temperature change for the calorimeter** if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.\n\n\nThe heat capacity of the calorimeter is $$C = \\frac{q\\_{calorimeter}}{\\Delta T\\_{calorimeter}}$$\nYou know $\\Delta T\\_{calorimeter}$ from your time vs temperature graph. You know $q\\_{calorimeter}$ from the energy conservation equation above; $q\\_{calorimeter} = -q\\_{hot}-q\\_{cold}$. Therefore you know $C$. You should know how to get $q\\_{hot}$ and $q\\_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?\n\n\n", "3" ], [ "\nThe point is that heat is not just stored in the water, but also in the calorimeter sturcture. Heat that is unaccounted for just considering the temperature and amount of water must have been taken up by the calorimeter itself.\n\n\n", "1" ] ]

[ [ "\nFrom what I gathered in the comments you confused a few terms. Both compounds are monodentate, i.e. they can only coordinate one metal at a time. However, pyridine usually donates one electron pair, while pyrrole can donate three (to one metal) — I haven’t seen a structure where pyrrole does that, though.\n\n\nThe reason is pretty simple: Pyrrole is much like the cyclopentadienyl anion except that one carbon atom is replaced by nitrogen. If it wants to coordinate in any way, it will have to do that like a slice of bread from its side, forming a $\\eta ^5$ complex. Much like ferrocene in the picture below.\n\n\n\n\n\nPyridine, on the other hand, has no hydrogen on the nitrogen atom; instead it has a lone pair pointing outwards, but still in the plane of the ring. This allows pyridine to donate two electrons in a $\\sigma$-like fashion to a metal centre, $\\eta ^1$-style. Of course, pyridine could also perform a sandwich-like coordination much like benzene sometimes does. However, the direct two-electron $\\sigma$ donation will be much stronger and therefore largely outweigh a sandwich-type donation in most compounds.\n\n\n", "11" ] ]

[ [ "\n$\\Delta U = q + w$ and $dU = dq + dw$\n\n\nThis is the [first law of thermodynamics](http://en.wikipedia.org/wiki/First_law_of_thermodynamics). It is not an inequality.\n\n\n$dU\\leq TdS-PdV$ is correct when there is only PV work. It is incorrect when there is other work such as electrical or gravitational.\n\n\n\n> \n> the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.\n> \n> \n> \n\n\nThe PdV term alone represents the reversible work, the TdS term represents reversible heat\n\n\n\n> \n> Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \\leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?\n> \n> \n> \n\n\nthe system must lose internal energy in a spontaneous change\n\n\n", "1" ] ]

[ [ "\nAs long as there is less than 100% relative humidity, the ice will sublime to some extent.\n\n\nJust in the past week it has been in the -12 °C to -5 °C range in my area, and the snow and ice on the ground did sublime, on the order of 1 cm per day, more in the sun less in the shade. \n\n\nFor quantitative information see \"[Sublimation from a seasonal snowpack at a continental,\nmid-latitude alpine site](http://snobear.colorado.edu/Markw/Research/hood99.pdf)\" Hydrol. Process. 13, 1781-1797 (1999).\n\n\n", "10" ], [ "\nOn cold days, the wet clothes dry due to the water in the clothes that freezes, then sublimates, to create water vapour that escapes the clothes, therefore drying it.\n\n\n", "-1" ] ]

[ [ "\nWe've got:$$G\\_{(p,T)}= U+pV-TS$$\nWhere G is Gibbs free energy, U is the thermodynamic system's internal energy, p is pressure, V is volume, T is temperature and finally, S in entropy. In a chemical reaction, we can change the equation above to have:$$\\Delta{G}\\_{(p,T)}= \\Delta{U} +p\\Delta{V}- T\\Delta{S}$$\nWhich is then \"interpreted\" to:$$\\Delta{G}\\_{(p,T)}= \\Delta{H} - T\\Delta{S}$$\nThis is what learners like me learn during the high school. Note that Gibbs free energy (measured in here) works well when deciding about the spontaneity of processes **that occur in constant temperatures and pressures**.\n\n\nSo if you're to use the equation, know that you oblige the poor poor system to be in a constant temperature! Think about it. If you're being asked a question from things you've studied, you *will* definitely run into phrases like `STP` or `RTP` in the question content. There is, however, a \"formula expansion\" to where we have variable temperatures; but of course, that wouldn't use the equations above.\n\n\nIf it's still vague for you, you can comment me.\n\n\n", "1" ], [ "\nWe can not calculate the exact amount of G (Gibbs free energy), only we can calculate change of that,so we must put temperature change on this equation:\n$$\\Delta G = \\Delta H - \\Delta TS $$\n\n\n", "1" ], [ "\nThe temperature must remain constant so it is the value both at the start and the end. I will try to explain this simply as, from your question, I suspect that you have just been introduced to $G$ with the equation $G=H-TS$.\n\n\nThe Gibbs free energy helps to indicate whether a process/reaction will be spontaneous at conditions where the pressure and temperature remain constant. This is useful because these are the conditions that occur in lab when we do an experiment. As I am sure you know the Gibbs free energy must decrease for a spontaneous process; in other words $\\Delta G<0$.\n\n\nNotice that the equation $G=H-TS$ does not involve $\\Delta G$ is only gives a relation between Gibbs free energy and enthalpy and entropy. It is clearly useful to consider changes in $G$ so lets say $G$ changes by an infinitesimally small amount: $\\mathrm{d}G$ (this basically means the same as $\\Delta G$ but the change is much smaller) we can change the expression so that $\\mathrm{d}G$ is written in terms of infinitesimal changes in entropy temperature and enthalpy:\n\n\n$$(G+\\mathrm{d}G)=(H+\\mathrm{d}H)-(T+\\mathrm{d}T)(S+\\mathrm{d}S)$$\nExpanding the brackets in the last expression:\n$$G+\\mathrm{d}G=H+\\mathrm{d}H-TS+S\\mathrm{d}T+T\\mathrm{d}S+\\mathrm{d}S\\mathrm{d}T$$\nI will now rearrange to make the following step clear:\n$$\\mathrm{d}G+[G=H-TS]+\\mathrm{d}H+S\\mathrm{d}T+T\\mathrm{d}S+\\mathrm{d}S\\mathrm{d}T$$\nNotice the $G=H-TS$ in the equation (highlighted in square brackets) – this can now cancel because it’s the same as having $G$ on each side. This leaves:\n$$\\mathrm{d}G=\\mathrm{d}H+S\\mathrm{d}T+T\\mathrm{d}S+\\mathrm{d}S\\mathrm{d}T$$\nThe last term is the product of two infinitesimally small changes (very very small numbers) so is basically zero and can be ignored. This leaves:\n$$\\mathrm{d}G=\\mathrm{d}H+S\\mathrm{d}T+T\\mathrm{d}S$$\nNow, remember towards the start of my answer I said that this is useful at a constant temperature and pressure. Well, if temperature is constant $\\mathrm{d}T=0$ so the $S\\mathrm{d}T$ term vanishes leaving:\n\n\n$$\\mathrm{d}G=\\mathrm{d}H-T\\mathrm{d}S$$ and for larger changes: $$\\Delta G =\\Delta H -T\\Delta S$$\n\n\nIt is crucial that you know, and hopefully now understand after my explanation, that the above equation is **only** valid at a constant pressure and temperature.\n\n\nSo in short $T$ is **the constant** value for temperature.\n\n\n", "1" ] ]

[ [ "\nThe molecule length value you are using is the problem. It is more like 300 pm, which is the van der Vaals diameter and includes regions where electron density is significant. You have calculated an internuclear distance, but the nuclei of one molecule won't come close to those of another molecule because of electron-electron repulsion.\n\n\nThis will change your answer by a factor of 8.\n\n\nSee Finney, J. L. The water molecule and its interactions: the interaction between theory, modelling, and experiment. *J. Mol. Liq.* **2001,** *90* (1-3), 303–312. [DOI: 10.1016/S0167-7322(01)00134-9](https://doi.org/10.1016/S0167-7322(01)00134-9).\n\n\n", "8" ] ]

[ [ "\nIf a reaction has just one step, it's named an *elementary reaction*. In this reaction all of reactant directly participate on reaction and effectively collide with each other. So, the order of reaction corresponds to its coefficient, but if a reaction is not a elementary reaction, there isn't any reason for colliding all reactants with each other. For example, consider the following overall reaction:\n\n\n$$\\ce{A + B + D -> F}$$\n\n\nAssume the corresponding mechanism is constituted of two elementary reactions:\n\n\n$$\\ce{A + B -> C}$$\n$$\\ce{C + D -> F}$$\n\n\nAs shown above, there is no need to collide A with D or A with C ... Then, in these type of reactions coefficients are not corresponded to order of reaction . \n\n\n", "9" ] ]

[ [ "\nThere are two other widely used acid/base theories that apply here: [Brønsted–Lowry theory](http://en.wikipedia.org/wiki/Br%C3%B8nsted%E2%80%93Lowry_acid%E2%80%93base_theory) and [Lewis theory](http://en.wikipedia.org/wiki/Lewis_acids_and_bases).\n\n\nA Brønsted base is defined as a proton acceptor. Technically, $\\ce{CuO}$ is capable of reacting with $\\ce{H+}$ to form hydroxide ions and eventually $\\ce{H2O}$. Although $\\ce{CuO}$ is insoluble in water (as you said), this reaction can occur in other solvents or in a biphasic system (where the proton is in solution and the $\\ce{CuO}$ is present as an undissolved solid).\n\n\nA Lewis base is defined as an electron donor. Generally, this means it has a lone pair of electrons that is available for reactions. The oxygen atom in $\\ce{CuO}$ has lone pairs of electrons available.\n\n\n$\\ce{CuO}$ is considered a base because it fits these two descriptions.\n\n\n", "8" ], [ "\nAn additional way to consider it: $\\ce{CuO}$ is defined as a base in the Lux-Flood theory because it is an oxide ion donor; it donates oxide to oxide acceptors ([Lux-Flood acids](http://en.wikipedia.org/wiki/Lux-Flood_theory))\n\n\nAn example of a Lux-Flood acid-base reaction involving $\\ce{CuO}$ is the high pressure reaction with carbon dioxide to give $\\ce{CuCO\\_3}$ ([source](http://onlinelibrary.wiley.com/doi/10.1002/zaac.19744100207/abstract)).\n\n\n", "5" ], [ "\n$\\ce{CuO}$ is a basic oxide and when it reacts with water it gives $\\ce{Cu(OH)2}$ which is a base as it gives $\\ce{Cu^2+}$ and $\\ce{OH-}$ ions when dissolved in water.\nSo technically $\\ce{CuO}$ is a base rather a basic oxide.\nGenerally all metallic oxides and hydroxides are bases.\n\n\n", "2" ], [ "\nI checked with an expert and they said Copper Oxide is considered to be \"sparingly soluble\". And is considered to be an Arrhenius base.\n\n\nSo I think that might solve the conundrum. It's not considered insoluble.\n\n\nFrom what I can tell, Al2O3 (Aluminium Oxide), is considered to be insoluble, and it's said that it thus wouldn't form Hydroxide ions in water. So Al2O3 is not an Arrhenius base.\n\n\nAnother technicality is\n\n\n<https://www.khanacademy.org/science/chemistry/acids-and-bases-topic/acids-and-bases/a/arrhenius-acids-and-bases> \"Note that depending on your class—or textbook or teacher—non-hydroxide-containing bases may or may not be classified as Arrhenius bases.\"\n\n\nHence <https://freechemistryonline.com/arrhenius-concept-of-bases.html> \"a base was defined as a compound containing OH groups which can become hydroxide ions when the base is dissolved in water.\" and \"The Arrhenius definitions do not include such compounds as ammonia (NH3) and calcium oxide (CaO)\" and \"CaO is termed a basic oxide but not a base\"\n\n\nSo on that definition of Arrhenius base, where it has to contain an OH group, Copper Oxide wouldn't be considered an Arrhenius base.\n\n\n", "0" ] ]

[ [ "\nIt might take some time, but I would use vinegar.\n\n\nThe acid in vinegar is strong enough to dissolve the calcium carbonate, but it is not strong enough to dissolve metals (unless the metal reacts with water).\n\n\n", "2" ], [ "\nI recommend [sulfamic acid](https://en.wikipedia.org/wiki/Sulfamic_acid) with formula $\\ce{H3NSO3}$. The reaction is irreversible, no dangerous compound:\n\n\n$$\\ce{2 H3NSO3 + CaCO3 -> Ca(H2NSO3)2 + CO2 + H2O}$$\n\n\n", "1" ] ]

[ [ "\nIf you want to track where the energy is going, do a first-law analysis (compute heat and work for a given change in internal energy or enthalpy). \n\n\nIf you want to predict whether a process occurs spontaneously or not, or you want to study an equilibrium, the second law is the tool you need. It's convenient to apply the second law by looking at free energy changes, since they let you focus on the system rather than the entire universe. \n\n\nWhen we're interested in experimental internal energies or enthalpies we're tracking energy transfers. When we're interested in the Gibbs free energy we're usually answering questions about process spontaneity, or equilibrium conditions, or maybe we want to compute non-PV work. Different tools for different jobs.\n\n\n\n\n---\n\n\nLet me edit some of my comments below into my answer, because the comments are scrolling off the page and they make it clearer why you can't really track energy transfers just by looking at the Gibbs free energy:\n\n\nThe Gibbs free energy *is* an energy in the sense that it is the part of the energy that is \"free\" to do useful work. It does indeed have units of energy, and $\\Delta G$ can be used to compute the maximum possible nonexpansion work.\n\n\nIt is *not* an energy in the sense that it is not conserved in a process, the way internal energy and enthalpy are. $G$ is derived partly from entropy. It's quite possible that $G\\_\\text{system} + G\\_\\text{surroundings}$ isn't constant in a process.\n\n\nIf you have an isothermal process, $\\Delta G= \\Delta H−T\\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD's answer gives a concrete example of that.\n\n\n", "6" ], [ "\nImagine you have a divided insulated container. On one side is helium gas and on the other side is neon gas. Both sides are uniformly at standard temperature and pressure. \n\n\nYou remove the divider.\n\n\nWhat happens?\n\n\nThe gases mix, because there are greatly more microstates we would consider \"mixed\" compare to not mixed, statisitically. \n\n\nEntropy (S) increased, and in the ideal gas approximation there is no change in temperature or pressure (but there is a significant change in entropy of the system). \n\n\nGibbs so-called \"free energy\" decreased solely because entropy increased.\n\n\nNo energy went anywhere. \n\n\n", "4" ], [ "\nI think the other answers here are pretty good, but it seems to me that the main point of confusion is over what it means for energy to be conserved.\n\n\nYou know that total energy is conserved, but you also know that it can take different forms. Gibbs Free energy is just one form of energy. When Gibbs free energy decreases, it goes somewhere - either as heat released by the system, or into increasing the entropy of the system. Those are really the only two options, because this is how Gibbs energy is defined.\n\n\nFor your other questions:\n\n\n\n> \n> why is this even useful?\n> \n> \n> \n\n\nYour thoughts at the end are correct - it is useful because it lets us determine spontaneity, and also the amount of energy available to do \"useful\" work - useful in the sense that we can control its transformation to other forms of energy.\n\n\n\n> \n> If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) \n> \n> \n> \n\n\nWe track enthalpy even though it is the thermodynamic potential for constant P & S systems because at constant T & P, in the absence of non-PV work, it is also equal to the heat produced or absorbed by the system.\n\n\nIn the lab that might not seem like a big deal, but if you are designing a steam engine, air conditioning system, or a chemical reactor, being able to predict the amount of heat that is absorbed or lost during a transition is very useful!\n\n\n", "2" ] ]

[ [ "\nWhile enthalpy, internal energy, and Gibbs's energy are all types of energy, they are not all measuring the same quantities. Internal energy is the energy of the molecules in your system. Enthalpy takes into account the internal energy plus whatever work is being done by or on your system. And Gibbs's energy is the maximum amount of non-*PV* work that you can get from your system. While these three quantities are all in units of energy, they are not just different terms for the total energy. Since enthalpy and Gibbs's energy involve work, and hence exchange of energy with the surroundings, these quantities will be different than just the internal energy of the system alone.\n\n\n", "2" ] ]

[ [ "\nThe $K\\_{sp}$ values are the way to go .. the other information you mentioned are just some convenient rules of thumb for predicting relative solubilities of compounds when you don't know the $K\\_{sp}$ values. Also, the 4th choice in your list .. radium .. probably also belongs on that list, since it's the next alkali earth element after barium, but it's included because it's a radioactive element, and people don't commonly use or encounter it.\n\n\n", "2" ] ]

[ [ "\nIn the first example, the 4(2+2), if you were ABSOLUTELY going to adhere to significant figure rules, then 20 would be the correct answer, though I don't believe many people would fault you for saying 16. In your second case, 2.0(0.5+0.5), the appropriate answer here would indeed be 2.0. \n\n\nIn any case, significant figures are not really valuable when referring to purely mathematical equations like you did. We use them in order to avoid making errors when it comes to the accuracy of a result, in other words, significant figures are there so that your answer is not more specific than your measurement was. In mathematics, this is not an issue, as you are not making any measurements. So, while you COULD use significant figures in a situation like this, it really is not quite the appropriate situation in which to do so.\n\n\n", "1" ] ]

[ [ "\nNot sure what kind of explanation you are looking for, but \n\n\n[Electrode Dynamics, A. C. Fisher, Oxford Chemistry Press, 1996, ISBN: 019855690X](https://global.oup.com/academic/product/electrode-dynamics-9780198556909?cc=us&lang=en&)\n\n\nis a great concise explanation of overpotentials among other things.\n\n\n", "2" ], [ "\nI have always thought that corrosion textbooks give some of the most concise and enlightening overviews of electrochemistry. One of my favorites is:\n\n\n\n> \n> *Corrosion and Surface Chemistry of Metals*, by D. Landolt, EPFL Press, 2007, ISBN: 0849382335\n> \n> \n> \n\n\nThis book will explain to you how to interpret polarization diagrams. The emphasis, of course, is on corrosion phenomena, but the same type of reasoning applies to any electrochemical system. The author does a great job of breaking down all the sources of overpotential at an electrode and shows many polarization diagrams along with the discussion. It really is a great book and I highly recommend it!\n\n\n", "1" ] ]

[ [ "\nI found the following p*K*a values for glycine and alanine from the CRC Handbook of Chemistry and Physics via [Sigma-Aldrich](http://www.sigmaaldrich.com/life-science/metabolomics/learning-center/amino-acid-reference-chart.html):\n\n\nGlycine $\\ce{COOH}$: 2.34 \n\nAlanine $\\ce{COOH}$: 2.34\n\n\nGlycine $\\ce{NH3+}$: 9.60 \n\nAlanine $\\ce{NH3+}$: 9.69\n\n\nFirst, although the values for the carboxylic acid are identical as reported, you would see a difference if you found a value reported with enough precision. The carboxylic acid is not immune to changes in the structure of the rest of the molecule.\n\n\nSecond, the difference between 9.60 and 9.69 is pretty small. In most practical situations, the two amino acids would behave identically with regard to the $\\ce{NH3+}$ p*K*a.\n\n\nThe fact that the magnitude of the difference is larger for the $\\ce{NH3+}$ p*K*a values can likely be attributed to proximity. In this case, the relevant proton is 2 bonds away from the $\\alpha$ carbon, whereas the proton in the $\\ce{COOH}$ group is 3 bonds away.\n\n\n", "7" ] ]

[ [ "\nYour teacher is right. For example, at room temperature, the pressure of water vapor over a pond or lake might be around 25 torr; that's less than atmospheric pressure, but the water vapor is undeniably a gas.\n\n\nMaybe your confusion stems from a belief that gases must be produced by boiling liquids (since a liquid must have a vapor pressure equal to the external pressure before it begins to boil). But a liquid can also become a gas through evaporation, no matter what the external pressure is.\n\n\nIn **evaporation**, molecules escape from the liquid's surface and become a gas. \n\nMolecules in the liquid are in constant motion, and they frequently collide with each other. Every now and then one will get knocked off the surface of the liquid by the others. This can happen at any temperature or pressure the liquid exists at.\n\n\nIn **boiling**, bubbles of gas form throughout the liquid. This can only happen when the temperature is high enough to make the vapor pressure equal to the external pressure.\n\n\n", "2" ], [ "\nNo. Suppose we had a container containing nothing but water. Suppose it is only half full, so there is empty space. **There is no pressure from other gases.** If we heat this container enough, the water will boil and eventually turn completely into a gas. \nIn short, clearly other things influence the state of matter in everyday life besides pressure from atmospheric gases. In this example, I illustrate how water evaporating in a vacuum is determined by temperature. I think this is a good example because it involves just one substance and doesn't bring into question how to think about mixed gases and partial pressures. \n\n\nThere is an extremely useful equation that connects these relationships beautifully: \n\n\n$PV = nRT$\n\n\nP stands for pressure, V for volume, T for temperature, n for the number of molecules, and R is a constant. \n\n\nThis equation basically says that\n*any change in pressure or volume to a closed system of gas will result in a corresponding change to the temperature of the system.* \n\n\nThus, as long as $P\\_1V\\_1\\neq P\\_2V\\_2$, you can always think about a change in volume or pressure as a change in temperature for a closed system. \n\n\n", "1" ], [ "\nIf you know Dalton's law of partial pressure that \"in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases\"\nAnd since pressure is non-negative quantity then:\n\n\nP(total)=P1 + P2 + ... + Pn\n\n\nSO the pressure of gas in atmosphere in fact, cannot be greater than atmospheric pressures . \n\n\n<http://en.wikipedia.org/wiki/File:Dalton%27s_law_of_partial_pressures.png>\n\n\nAnd also I am not saying that gas stored in container cannot have pressure greater than atmosphere . The best example is your LPG cylinder.\n\n\n", "0" ] ]

[ [ "\n\n\n\nAll credit to Zhang et al. \"[Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy](http://www.sciencemag.org/content/342/6158/611)\" Science Vol. 342 no. 6158 pp. 611-614.\n\n\nYes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded.\n\n\nIt is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms.\n\n\nScanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds.\n\n\n", "34" ], [ "\nNo I don't think that is possible.\n\n\nIt's all about the **attractions** *(the wanting of a full octet)* between the atoms, to be honest we don't even have the technology to view the actual atoms itself (if we do then we are at the initial stages). So these \"bonds\" that you are talking about are just imaginary connections between atoms so people don't get confused or overwhelmed. \n\n\nTake an ionic bond, one of the strongest bonds out there, we know it is caused by strong electrostatic attractions between the atoms in which one completely takes an electron or gives it up. If there was like a connection between them it would look almost surreal. It would make more sense if they are beside each other.\n\n\nNow for the covalent bond, in a covalent bond atoms are sharing electrons, in this case the electron clouds merge, this maybe hard to grasp but that is just how this works. That is why for most covalent bonds we look at the Lewis Structures.\n\n\nNow take a look at an **ionic** \"bond\".\n\nTake a look at the **covalent** \"bond\".\n\n\n\nCheck out the **double** and **triple** \"bond\". - Bad Image\n\n\n\nBy looking at the above picture you can see that a **triple** \"bond\" is just two more electrons. \n\n\nSo in reality to make these drawings of atoms easy, we have came up with a **bond**. There might be \"bonds\" but one has to note that they are **attractions** instead. Here I gave you a couple examples, mostly 2D but if you want to really get the idea you should take a look at the 3D structures that include everything. But you have to also note that we don't know how far our current depictions are true, I just gave you the most plausible ones, so don't believe everything as a fact.\n\n\nTo sum it up: You Will Not See The Bonds (Not the sticks and ball kind). Though you may see the attractions. \n\n\n", "-2" ] ]

[ [ "\nBriefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways.\n\n\n**Entropy is a measure of the number of different ways a state with a particular energy can be realized**. Specifically, $$S=k\\ln W$$ where $k$ is Boltzmann's constant and $W$ is the number of equivalent ways to distribute energy in the system. If there are many ways to realize a state with a given energy, we say it has high entropy. Often the many ways to realize a high entropy state might be described as \"disorder\", but the lack of order is beside the point; the state has high entropy because it can be realized in many different ways, not because it's \"messy\". \n\n\nHere's an analogy: if energy were money, entropy would be related to the number of different ways of counting it out. For example, there, there are only two ways of counting out two dollars with American paper money (2 1-dollar bills, or 1 two-dollar bill). But there are five ways of counting out two dollars using 50-cent or 25-cent coins (4 50-cent pieces, 3 50-cent pieces and 2 quarters, and so on). You could say that the \"entropy\" of a system that dealt in coins was higher than that of a system that dealt only in paper money. \n\n\nLet's look the change in entropy for a reaction $\\rm A\\rightarrow B$, where A molecules can take on energies that are multiples of 10 energy units, and B molecules can take on energies that are multiples of 5 units. \n\n\nSuppose that the total energy of the reacting mixture is 20 units. If we have 3 molecules of A, there are 2 ways to distribute our 20 units among energy levels with 0, 10, and 20 units:\n\n\n\n\n\nIf we have 3 molecules of B, there are 4 ways to distribute 20 units among energy levels with 0, 5, 10, 15, and 20 units:\n\n\n\n\n\nThe entropy of B is higher than the entropy of A because there are more ways to distribute the same amount of energy in B than in A. Therefore, $\\Delta S$ for the reaction $\\rm A\\rightarrow B$ will be positive.\n\n\n", "11" ], [ "\n\n> \n> what would better describe entropy verbally?\n> \n> \n> \n\n\nWe can also use \"measurement of randomness\" or \"amount of chaos\" or \" energy dispersion\"\n\n\n\n\n---\n\n\nInitially in 1862, Rudolf Clausius asserted that thermodynamic process always \"admits to being reduced to the **alteration** in some way or another of the arrangement of the constituent parts of the working body\" and that internal work associated with these alterations is quantified energetically by a measure of **entropy** change.\n\n\nBut later after few years Ludwig Boltzmann translated word **alteration** from Rudolf Clausius' assertion to **order and disorder** in gas phase molecular systems.\n\n\nBut if you see latest books they use concept of **energy dispersion** instead of order or disorder to explain entropy. \n\n\n\n\n---\n\n\nSource: [Wikipedia](http://en.wikipedia.org/wiki/Entropy_(order_and_disorder))\n\n\nAlso have look at [physics S.E.](https://physics.stackexchange.com/a/33385/46239)\n\n\n", "4" ], [ "\nI will take a crack at this although I admit that this topic sometimes confuses me as well. Here is how I like to think about entropy. Consider a box containing equal amounts of gases A and B. So, we have a fixed volume and fixed number of molecules. Let us also isolate the box from its surroundings so that it has a fixed energy as well. We have just created a microcanonical ensemble (constant NVE). The ensemble consists of every possible configuration of molecules having the same NVE. Now, if we were able to step back and take a broad view of the ensemble, we would observe that the vast majority of boxes contain a rather bland homogeneous mixture of gases A and B. In fact, they would be indistinguishable for all practical purposes. Let us count the number of boxes in this state and call the number $W\\_1$. Continuing with the box counting, we find that there are only a handful of distinguishable boxes left, but they are quite interesting! One box, for instance, might have all of the A molecules crowded together in one corner and all of the B molecules in another corner! This one box can be labelled $W\\_2$. We start to understand that some configurations of molecules will be extremely improbable because they represent such a small fraction of the total number of possible configurations. Boltzmann quantified this relationship as $S=k \\log W$. If we plug $W\\_1$ into this equation we get a high value for the entropy, $S$, because $W\\_1$ is so large. On the other hand, if we plug in $W\\_2$ the entropy we get is very low.\n\n\nSo, to sum up, entropy is really the measure of how likely a given system configuration is when compared to all of the possible configurations. In this sense, I would argue that you could say a messy room has a higher entropy than a clean room because there are so many more ways a room could be considered \"messy\" than \"clean\".\n\n\n", "2" ] ]

[ [ "\nCarbon nanotube black is highly absorbent in and outside the visible spectrum, see <http://www.extremetech.com/extreme/186229-its-like-staring-into-a-black-hole-worlds-darkest-material-will-be-used-to-make-very-stealthy-aircraft-better-telescopes>.\n\n\nOf course, that does not make it ideal for energy conversion, as it would also need at least the following qualities to be practical.\n\n\n1. A good conductor of heat to the heat-exchanger surface below\n2. An enduring finish, resistant to abrasion, heat and environmental degradation\n3. Economical, both in material cost and application\n4. Safety in use and in manufacture\n\n\n", "4" ] ]

[ [ "\nThe key to the answer is what kind of oil do you have, there are differences. In general, oils with big amount of double bonds MAY oxidize by oxygen from air. The process is radical, UV-induced, and very slow. Some other oils, with no or too little double bonds do not oxidize. and remain mostly same.\n\n\nThe process of oxidation (drying) initially produces hydroperoxides, later able to serve as a source of more free radicals, leading eventually to inter-molecular linkages. I found contradicting notions for what these linkages consist of (it is either oxygen bridges or direct links similar to polymers), but the end process if well known and used in varnishes. However, varnishes has extra components: they are usually diluted by some solvent to reduce viscosity and always contain a metal catalyst to accelerate drying of the oil, that also is usually heated for prolonged period of time, again, to increase the speed of drying.\n\n\nThe considerations above ignore obvious process of evaporation of light impurities and possibility of bio-degradation (I'm not aware of bio-degradation processes for pure fatty oils, but there are bacteria that live in almost undiluted toluene and there are bacteria that thrive on petrol, so I'm open to this idea).\n\n\n", "3" ], [ "\nSome of the lighter constituents could evaporate. There are a lot of minor components in olive oil, and they are largely responsible for the fragrance and flavor. As these evaporate, the fragrance and flavor would change. It is unlikely that they would all evaporate though, as many of these components likely have high boiling points/low vapor pressures.\n\n\nThe oil itself would not evaporate, but it would not solidify either. At most you might notice a thickening of the oil.\n\n\nOver long periods of time the oil may react with oxygen to form various oxidation products. This process is so slow, however, that it is definitely not a concern for olive oil you might leave out in your kitchen. The oil will start tasting and smelling stale long before significant oxidation occurs.\n\n\n", "3" ], [ "\nVegetable oils do lose some lighter constituents through evaporation, particularly substances responsible for aroma, but the primary degradation pathways, making it rancid, are through oxidation, hydrolization and microorganism metabolism; see <http://www.ivannikolov.com/how-cooking-oil-goes-rancid/> and <http://en.wikipedia.org/wiki/Rancidification>.\n\n\nSome tree-nut oil, in particular, quickly develops a noticeable off-taste. BHT, BHA and other additives (often applied to packaging rather than to the food itself) help decrease oxidative rancidity.\n\n\nThat said, vegetable oil used for lubrication in old machine parts certainly does thicken to the consistency of wax, though I do not know how much of that is due to evaporation vs. degradation.\n\n\n", "2" ] ]

[ [ "\nThis is really a physics question but I will answer it anyway.\nI'm not sure how much subatomic physics you know so I will give two different versions.\nSimply put, a neutron decays to form a proton and electron.\n$$\\ce{n -> p+ + e-}$$\nThis explains why the proton number increases by one to form Xenon.\n\n\nMore properly a down quark inside a neutron decays to form an up quark, an electron and an electron antineutrino as a result of weak nuclear force interactions.\n$$\\ce{d -> u + e- + \\overline{\\nu}\\_{e}}$$\n\n\nThis causes the neutron ($\\ce{udd}$) to turn into a proton ($\\ce{uud}$) leading to the increase in atomic number.\n\n\nThis type of beta decay is know as $\\beta^{-}$ decay. There is also $\\beta^{+}$ decay which is a similar process where a proton becomes a neutron, positron and an electron neutrino:\n$$\\ce{^23\\_12Mg -> ^23\\_11Na + e+ + \\nu\\_{e}}$$\n$$\\ce{u -> d + e+ + \\nu\\_{e}}$$\n\n\n<http://en.wikipedia.org/wiki/Beta_decay> $%edit$\n\n\n", "5" ], [ "\nWhen we talk about an unstable atom that will decay by emitting $\\alpha$, $\\beta$ or $\\gamma$ radiation, the nucleus is the thing that's unstable, right?\n\n\nAs [Wikipedia](http://en.wikipedia.org/wiki/Beta_particle) puts it,\n\n\n\n> \n> An unstable atomic nucleus with an excess of neutrons may undergo β− decay, where a neutron is converted into a proton, an electron and an electron antineutrino (the antiparticle of the neutrino)\n> \n> \n> \n\n\nYou can forget the anti-matter part. Just remember than in $\\beta$ decay there is a neutron from the nucleus that is being \"devolved\" into eminently a proton and an electron. The electron leaves the atom with high velocity and energy, but the proton remains. So, for each atom that a $\\beta$ particle is released from, we have:\n\n\n$$\\ce{^A\\_zM -> ^A\\_{z+1}M+}$$\n\n\n", "3" ], [ "\nEssentially, a neutron is converted to a proton, releasing an electron in the form of beta radiation. That is why the mass stays constant but the number of protons increases by 1.\n\n\n", "0" ] ]

[ [ "\nI actually think this is an unfair question, unless you were given some additional information, or you were specifically instructed to memorize specific ionization energies, rather than learning the trends. The trend for ionization energies is that they increase from left to right across rows, and from bottom to top in columns. Based on only that information, I can't see how one should distinguish between $O$ and $Br$ ... yes, it happens to be true that $O$ has a higher IP than $Br$, but how are you supposed to tell? Suppose the example had used $S$ instead of $O$, and $I$ instead of $Br$? Those elements have the same relative positions in the table, but in that case, it would be $I$ that has the larger ionization energy.\n\n\nSo, in this case I think your confusion was quite understandable .. I would never construct a test question like this unless the student had some additional information. It's not even true that you can always use the rule about them increasing from left to right along rows .. in the first row, $N$ actually has a slightly higher IP than $O$, because there is a penalty for disrupting the half-filled p-shell in $N$, which is anomalously stable. You can see the periodic trends [here](http://www.iun.edu/~cpanhd/C101webnotes/modern-atomic-theory/ionization-energy.html).\n\n\n", "4" ], [ "\nThe larger that atoms get, the outer electrons experience whats called a [shielding effect](https://en.wikipedia.org/wiki/Shielding_effect).\n\n\nElectrons that are closer to the nucleus have a stronger attraction than the electrons further away. Because the electrons in oxygen experience less shielding, they have a stronger affinity.\n\n\nOxygen's stronger electron affinity contributes to the observed [electronegativity](https://en.wikipedia.org/wiki/Electronegativity) of the atom.\n\n\n\n> \n> On the most basic level, electronegativity is determined by factors like the nuclear charge (**the more protons an atom has, the more \"pull\" it will have** on negative electrons) and the number/location of other electrons present in the atomic shells (the more electrons an atom has, **the farther from the nucleus the valence electrons will be**, and as a result **the less positive charge they will experience**—both because of their **increased distance from the nucleus**, and because the other **electrons in the lower energy core orbitals will act to shield the valence electrons** from the positively charged nucleus).\n> \n> \n> \n\n\n", "0" ], [ "\nOk so I thought about the question to and I thought that Bromine should have a higher first ionization energy but then if you really think about it, the answer would be oxygen. Look at it this way. Oxygen desperately needs two electrons. And Bromine needs only one more. So if you look at it, it would be harder to lose another electron when you already lost two and it would be easier to lose one more if you are Bromine. And also if you look at a basic ionization energy trend, it goes to the top right corner.\n\n\nImagine you were broke, would you be willing to pay to get something even though you know that you need money? No like Oxygen.\n\n\nThink of Bromine as an investor who needs a little bit more $ for an achievement, Bromine would be more likely to sacrifice its electron or to invest in something like an investor. \n\n\nReally bad analogies but hopefully you get the idea.\n\n\n", "-1" ] ]

[ [ "\n\n> \n> Did any chemical reaction occur between the coconut oil and H2O2 to make a new compound?\n> \n> \n> \n\n\nMost likely, **no chemical reaction occurred**.\n\n\nWhether any reaction occurred depends on the conditions you used when mixing the oil and hydrogen peroxide. How long did they mix for? What temperature was the mixture at? Did you add anything else to the mixture?\n\n\nHydrogen peroxide is used in the production of epoxidized vegetable oils. However, the reaction is performed for prolonged times at elevated temperatures with concentrated hydrogen peroxide in the presence of a catalyst.\n\n\nI am guessing that you just poured the two together at room temp, then soon after used the mixture to wash your hands. Compared to the commercial reaction, you are missing time, heat, and a catalyst, so it is very unlikely that any reaction occurred.\n\n\n\n> \n> My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?\n> \n> \n> \n\n\nOils are composed of triglyceride esters of fatty acids, not fatty acids themselves. So you would not see any peroxylauric acid.\n\n\n\n> \n> It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)\n> \n> \n> \n\n\nIt felt less greasy because it was less greasy: you added water to it (3% hydrogen peroxide is $\\ce{H2O2}$ in water). You probably made an emulsion of oil and water and used that to wash your hands (kind of like using soap).\n\n\n", "5" ] ]

[ [ "\nFirstly, in water Co(II) is not exclusively $\\ce{[Co(H2O)6]^2+}$, there is also a small amount of tetrahedral $\\ce{[Co(H2O)4]^2+}$, 0.08% at room temperature increasing substantially at high temperature. [Canadian Journal of Chemistry, 1980, 58(14): 1418-1426](http://www.nrcresearchpress.com/doi/abs/10.1139/v80-223).\n\n\nAccording to Greenwood's \"Chemistry of the elements\" 2nd edition page 1131, the more polarizable the ligand, the more the tetrahedral form of Cobalt (II) is preferred since fewer ligands are required to neutralize the metal ion's positive charge. \n\n\n", "3" ], [ "\nI don't know whether it's correct or not but what I think is due to steric factors .\n\n\nGiven the SCN- is pretty large ligand as compare to H2O .\n\n\nAlso there might be some additional stabilisation in water as ligand due to intra atomic hydrogen bonding. \n\n\n", "1" ] ]

[ [ "\n1) Yes and no ... the d-d orbital overlap will almost certainly have pi-orbital character (angular node along bond axis), and it could reasonably be considered \"backbonding\" because the electron density is going in the canonically \"unexpected\" direction (i.e. away from positively charged metal center). On the other hand, chemists (particularly inorganic chemists) usually restrict the phrase \"pi-backbonding\" to refer to back donation from the d-orbitals of a neutral or ionic metal center into anti-bonding orbitals of a ligand, typically an olefin or a carbonyl ligand.\n\n\n2) Yes .. that's essentially correct ... d-orbitals tend to be rather diffuse, and so can overlap with lobes of diffuse orbitals on ligands (or counter-ions, as with the metal sulfides). In traditional metal-ligand pi-backbonding, lobes the d-orbitals overlap with lobes of an unoccupied anti-bonding orbital of the ligand. In the example given above, they overlap with lobes of unoccupied d-orbitals on the sulfur. For a given system, unoccupied, or virtual, orbitals tend to be much more diffuse than occupied orbitals.\n\n\n3) Those diagrams are just intended to capture the qualitative aspects of the bonding interaction, such as the symmetries of the orbitals involved. Remember, the orbitals are 3-dimensional shapes that describe the probability of finding an electron with a particular set of quantum numbers in a given region of space. Depictions like the one in your post are helping to visualize the orbital by assuming a fairly tight \"tolerance\" for the probability. For example, those images might depict the surfaces that enclose 75% of the total probability density. If you expand that to 90%, or even 99%, the lobes maintain the same basic shapes but get much larger, which might make the overlap seem more significant, but could also be confusing for other reasons.\n\n\n4) I think it is quite well-written, but I suppose it would depend on the broader context in which it was being presented. I guess it is presented at an inappropriately high level for an introductory chemistry textbook, for example.\n\n\n", "5" ] ]

[ [ "\nThe problem rises when we are mistaken about what a property is. \n\n\n\n> \n> The different types of matter can be distinguished through two components: composition and properties. [UCDavis](http://chemwiki.ucdavis.edu/Analytical_Chemistry/Chemical_Reactions/Properties_of_Matter)\n> \n> \n> \n\n\nDo a search about chemical property, and you get:\n\n\n\n> \n> a property or characteristic of a substance that is observed during a reaction in which the chemical composition or identity of the substance is changed. [Dictionary.com](http://dictionary.reference.com/browse/chemical+property)\n> \n> \n> A chemical property is any of a materials properties that becomes evident during a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity. Simply speaking, chemical properties cannot be determined just by viewing or touching the substance; the substance's internal structure must be affected for its chemical properties to be investigated. However a catalytic property would also be a chemical property. [Wikipedia](http://en.wikipedia.org/wiki/Chemical_property)\n> \n> \n> Any **characteristic** that can be determined only by changing a substance's molecular structure.\n> [Boundless](https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/introduction-to-chemistry-1/physical-and-chemical-properties-of-matter-28/physical-and-chemical-properties-of-matter-181-1817/)\n> \n> \n> \n\n\nWhat's a physical property?\n\n\n\n> \n> A physical property is any property that is measurable whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe its transformations or evolutions between its momentary states. Physical properties are often referred to as observables. **They are not modal properties.** - [Physical property in wikipedia](http://en.wikipedia.org/wiki/Physical_property).\n> \n> \n> \n\n\nBasically, a physical property is what that doesn't deal with the composition of matter; but other factors, such as it's physical state. \"Physical\" isn't at its literal meaning.\n\n\nTable salt is never pure $\\ce{NaCl}$; because simply there's nothing pure in the nature. It's *mainly* composed of Sodium Chloride crystals. But if, and only if, we consider it to be only made from $\\ce{NaCl}$, we are talking about its chemical composition and therefore, it must be a chemical property.\n\n\nI understand that sometimes there are wrong questions for us to answer, so if composition is to be a property, it **must** be a chemical one.\n\n\n", "6" ], [ "\nFirst, table salt is composed of $\\ce {NaCl}$ wich is different from $\\ce {Na}$ & $\\ce{Cl}$. \n\n\nSecond, actually table salt isn't exclusively composed of $\\ce {NaCl}$ depending of the soure but contains also a fair amount of $\\ce {KCl}$\n\n\nThird, composition isn't what you can call a \"property\". I mean, it's a definition, compound containing only $\\ce {NaCl}$ or other $\\ce {KCl}$ is table salt that's all. \n\n\n", "0" ] ]

[ [ "\nWell, one important thing to keep in mind is that atomic and molecular orbitals are not \"real\" things, they are mathematical constructions that are useful in helping understand and predict how electrons behave in atoms and molecules. They are like a set of rules that tell us all of the ways in which it is *possible* for the electrons with a particular set of quantum numbers to behave. For example, a H-atom in the ground state has only one single electron in a $1s$ orbital (quantum numbers: $n=1$, $l=0$, $m\\_l=0$), but we also know that if the H-atom absorbs some energy, the electron can end up in a higher electronic state, for example the $3\\mathrm p\\_z$ orbital (quantum numbers: $n=3$, $l=1$, $m\\_l=0$). \n\n\nNow, when you are dealing with molecular orbital theory, you are talking about the *possible* ways in which electrons in atomic orbitals can combine in a covalent interaction. So it's not that the two atomic orbitals combine \"constructively and destructively at the same time\", it's just that those are the two ways in which its *possible* for the two orbitals to combine: in the region where they overlap, they can either add their amplitudes, or subtract them. The \"constructive\" case where they add produces an molecular orbital that is lower in energy (that is to say, an MO that describes a state where the electrons will have lower energies) than the \"destructive\" case, so by the aufbau principle, electrons will occupy the lower energy \"bonding\" orbital first. In a case like $\\ce{H2}$, where there are only two electrons, only the bonding orbital is occupied, but because the Pauli exclusion principle tells us we can only have two electrons in a given orbital, if you add another electron (as in $\\ce{He2+}$), it will occupy the higher-lying \"anti bonding\" orbital.\n\n\n", "6" ] ]

[ [ "\nAromaticity is a electronic property of a molecule, not a description of its shape. An aromatic molecule contains a circular system of conjugated π bonds and p orbitals that contains 4n+2 electrons, where n is an integer.\n\n\nBenzene is aromatic: it has 6 electrons in a circular conjugated system (4n+2=6, n=1).\n\n\nCyclooctatetraene is not aromatic: it appears to have a circular conjugated system, but it contains 8 electrons (4n+2=8 would require n=3/2). This molecule actually adopts a non-planar geometry to avoid [*antiaromaticity*](http://en.wikipedia.org/wiki/Antiaromaticity). \n\n\nCyclohexane is not aromatic, even though it is circular: it does not have a conjugated electron system.\n\n\n", "6" ], [ "\nAs @Kyle said aromaticity is an electronic property, not a description of shape and so circular would not be an appropriate description, particularly seeing as these molecules are not at all circular (benzene is hexagonal, for example).\n\n\nThe name 'aromatic' is a historical quirk. Supposedly it was used because many of the chemically aromatic substances know at the time happened to have notable aromas although wikipedia seems to [dispute this](http://en.wikipedia.org/wiki/Aromaticity#History). Several other reputable sites like [this](http://hyperphysics.phy-astr.gsu.edu/hbase/organic/aromatic.html) and [this](http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/react3.htm#rx8) do support this view though so I'm not sure what the right answer is (or even if anybody knows).\n\n\n", "2" ] ]

[ [ "\nIf you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed. \n\n\nThe reaction that you propose - \n\n\n$\\ce{Cl- + H2O -> HCl + HO-}$ \n\n\nis highly thermodynamically unfavorable. \n\n\nWe can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a **much** (as in almost a trillion trillion times) stronger acid than water. \n\n\nGiven that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water. \n\n\nSo *even* if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH. \n\n\n\n\n---\n\n\nAs a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a **lot** bigger). \n\n\n$\\ce{Cl- + H2O <<=> HCl + HO-}$ \n\n\n", "22" ], [ "\nIt does form HCl, the only question is how much? Dissociation constant of hydrochloric acid is ([reference](http://depts.washington.edu/eooptic/links/acidstrength.html)):\n\n\n$$K\\_\\mathrm{a} = \\frac{[\\ce{H+}][\\ce{Cl-}]}{[\\ce{HCl}]} = \\pu{1.3e6 M} \\tag{1}$$ \n\n\n$$K\\_\\mathrm{w} = [\\ce{H+}][\\ce{OH-}] = \\pu{1e-14 M^2} \\tag{2}$$\n\n\nIn neutral water, $[\\ce{H+}] = \\pu{1e-7 M}$.\n\n\nSo in $\\pu{1 M}$ $\\ce{NaCl}$, assuming neutral pH and substituting $[\\ce{H+}] = \\pu{1e-7 M}$ in equation (1), we find $\\pu{1e-13 M}$ $\\ce{HCl}$ is present.\n\n\nNot very much, but still billions of molecules in a 1 liter sample.\n\n\nIf you want to calculate rigorously, instead of assuming neutral pH solution, to the above two equations you would add two more:\n\n\n* Electrical neutrality requirement:\n$$\\ce{[Na+] + [H+] = [Cl-] + [OH-]}$$\n* Number of chlorine atoms is conserved:\n$$\\ce{[HCl] + [Cl-] = [Na+]}$$\n\n\nAnd you would have 4 equations and 4 unknowns ($\\ce{[HCl]}$, $\\ce{[Cl-]}$, $\\ce{[H+]}$, and $\\ce{[OH-]}$), with $\\ce{[Na+]}$ being known from the amount of $\\ce{NaCl}$ added to the water.\n\n\n\n\n---\n\n\nThe above is a simplification of the real situation. \n\n\nAs explained in *Essentials of Chemistry* by Hessler and Smith at page 327, which even though it is over 100 years old I still think is quite accurate:\n\n\n\n> \n> in an aqueous solution of sodium chloride [are] eight distinct things: the ions hydrogen, hydroxyl, sodium, and chlorine, and the undissociated substances: water, sodium chloride, sodium hydroxide and hydrochloric acid.\n> \n> \n> \n\n\n", "13" ] ]

[ [ "\nThe first thing is that it has to be going from liquid to solid because the substance is being cooled. Like tap water put in the freezer: liquid to solid.\n\n\nOn that note, the average kinetic energy goes down (directly assumed because that's what temperature is, and it's going down.)\n\n\nWhich leaves us with two answer choices: latent heat absorbed or latent heat released, where latent heat is the heat of a (thermodynamic system) at a constant temperature (like a phase change!)\n\n\nIf you're gaining energy, particles are going to have more velocity, like those virtual diagrams of gases with fast moving balls. Since we're going to a solid, losing energy makes sense because particles can't move as fast and are trapped in a fixed location (like an ice cube, which has a definite cuboid shape).\n\n\nSo liquid to solid, latent heat because the temperature is constant, and losing energy to become a solid - B is the answer.\n\n\n", "5" ] ]

[ [ "\n[Here's a cool example](http://en.wikipedia.org/wiki/Hydrothermal_vent) that should help illustrate the correct answer to your question. As for *why* it works that way, you might think about it in terms of bubble formation .. in order for a liquid to boil, it needs to form bubbles. Is it harder or easier to do that when the ambient pressure is high?\n\n\nAs for the calculation, you can use the Clausius-Claperyon equation to calculate the correction as described [here](http://en.wikipedia.org/wiki/Boiling_point). You will need some values for $P\\_0$, $T\\_0$ and $\\Delta H\\_{vap}$; you can use 1 atm, 373 K and 40.68 kJ/mol.\n\n\n", "2" ] ]

[ [ "\nActually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)\n\n\nThe correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):\n\n\n$$\n\\mu\\_i=(\\frac {\\partial H}{\\partial n\\_i})\\_{p,S,n\\_{j\\neq i}}\n$$\n\n\nSo the total exact differential is expressible as you had originally expected:\n\n\n$$\ndH=TdS+VdP+\\sum\\_i\\mu\\_idn\\_i=(\\frac {\\partial H}{\\partial S})\\_{p,n\\_i}dS+(\\frac {\\partial H}{\\partial p})\\_{S,n\\_i}dS+\\sum\\_i(\\frac {\\partial H}{\\partial n\\_i})\\_{p,S,n\\_{j\\neq i}}dn\\_i\n$$\n The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.\n\n\nSorry for any confusion my initial errors may have caused you. Hope this clears it up.\n\n\n", "1" ] ]

[ [ "\nFor starters, it might be fun to grow different kinds of crystals in shallow dishes and watch them grow in real time under a microscope. Either a low melting solid, melted and resolidified, or growing from a saturated solution. Also, looking at how a metal surface is etched by acid using a microscope would be fascinating.\n\n\n", "2" ] ]

[ [ "\nThis is a good question, but you may have gotten off on the wrong track by thinking about the electronic shell structure (although what you have written seems correct, if a bit old-fashioned). The relevant phenomenon here is the relative abilities of the metal ions to oxidize (or remove electrons from) each other. For your specific example, the copper ion is a better oxidizing agent than the iron atom. Therefore it is able to take two electrons away from the iron atom, oxidizing it and leaving the iron 2+ ion behind. \n\n\nAs for *why* this happens, at your current stage of education the best answer is probably \"because iron is easier to oxidize than copper\". You can look up the electrochemical activity series to find out how various metals relate to each other in terms of how easy they are to oxidize. Copper is one of the harder ones to oxidize, but if you were to mix copper metal with silver ions, you would end up with silver metal and copper ions, because silver is even harder to oxidize than copper.\n\n\nDoes that help?\n\n\n", "2" ] ]

[ [ "\nThere are two types of materials I'm aware of that produce light in the way I think you want: two-photon fluorescence and photon upconversion. Two-photon fluorescence is basically the same as normal fluorescence but two photons are absorbed simultaneously, emitting a single photon when returning to the ground state. [Photon upconversion](http://en.wikipedia.org/wiki/Photon_upconversion) similarly involves the absorption of multiple photons, but sequentially (by various mechanisms), followed by emission of a single photon.\n\n\nTwo-photon fluorescence is mostly seen used for [two-photon excitation microscopy](http://en.wikipedia.org/wiki/Two-photon_excitation_microscopy) and useful dyes tend to be big conjugated organic systems. Materials used for photon upconversion are generally lanthanide-doped nanoparticles and they've found many uses in applications traditionally served by quantum dots and organic fluorophores. They're both particularly useful in medical imaging as IR can penetrate tissue deeply, but [upconverting nanoparticles](http://onlinelibrary.wiley.com/doi/10.1002/anie.201005159/full) have a big advantage in that two-photon fluorescence requires very high excitation intensity (pretty much just lasers) as both photons must be absorbed simultaneously, but upconversion processes are sequential and can thus occur at lower intensity. Though, for some applications, the intensity requirements for two-photon absorption are useful as one can play tricks with focusing such that only a small region receives enough excitation light to emit, making imaging small areas possible.\n\n\nAs for a database of compounds, Thermo-Fisher has a [catalogue of two-photon dyes](https://www.thermofisher.com/ca/en/home/references/molecular-probes-the-handbook/technical-notes-and-product-highlights/fluorescent-probes-for-two-photon-microscopy.html), but I don't know that there are a lot of commercial manufacturers of upconverting nanoparticles, though Sigma-Aldrich formerly carried up converting nanocrystal under the Sunstone brand.\n\n\n", "4" ], [ "\nI am not aware of any materials that have this property. There are certainly molecules that emit fluorescence in the IR, when irradiated with shorter wavelength light, but I can't think of any that would emit in the IR after being irradiated in the IR in a manner that can be called \"luminescence\". Luminescence describes the specific phenomenon where a photon of a particular wavelength had excited a molecule to a higher-energy electronic state, the molecule then relaxes to a lower energy state, with emission of a second photon, which is always of longer wavelength than the original photon. Because molecular electronic transitions are typically in the UV-visible region of the spectrum, the phenomenon of luminescence is generally restricted to those spectral regions.\n\n\nOne particular issue with IR luminescence is that it would be pretty difficult to tell apart from normal thermal IR emission ... luminescence refers specifically to light that is emitted outside of the thermal spectrum of an object. That means that an object would need to be pretty cold initially, in order to distinguish luminescence in the IR region. And again, since the emitted wavelength is necessarily longer than that of the absorbed wavelength, a molecule that was luminescent under IR irradiation would emit deeper in the IR than where it absorbed, making all of this harder.\n\n\nHaving said that, there are certainly molecules with electronic transitions that absorb in the IR, and if you got them cold enough and used an appropriate spectrometer, you might be able to observe IR luminescence from them. It's something that would be restricted to pretty highly specified research labs though.\n\n\n", "1" ], [ "\n*In addition to Michael D. M. Dryden's excellent answer:*\n\n\nFreshly prepared **graphene oxide** seemingly shows two-photon absorption in the IR and deactivation by fluorescence, as described in an 2014 article in \n*Nature Science Reports*. ([DOI](http://dx.doi.org/10.1038/srep06090))\n\n\nFig. 2D in that article shows a linear correlation of the emission intensity at $\\lambda\\_{\\textrm{em}} = \\textrm{680 nm}$ with the square of the intensity of the excitation at $\\lambda\\_{\\textrm{exc}} = \\textrm{1140 nm}$.\n\n\nFig. 2C shows that the emission wavelength can be tuned by variation of the excitation wavelength!\n\n\n", "1" ] ]

[ [ "\nOkay, I think I found an answer. It's a set of four papers which were published consecutively in *Physical Review* 36 (1930). Interestingly, each was submitted on June 23rd, 1930, which makes me wonder about the back story.\n\n\nThe first comes from General Motors Corporation Research Laboratories in Detroit ([Kettering](http://en.wikipedia.org/wiki/Charles_F._Kettering), Shutts, and Andrews). The other three papers are out of Johns Hopkins (Donald Andrews, from Chemistry, and Yates, from Mathematics). I suppose Andrews could have graduated from Hopkins and found that GMCRL was working on the same thing; or perhaps he had a dual appointment.\n\n\nAs a side note, there appears to be [a patent suggesting that L. W. Shutts invented shock absorbers](http://www.google.com/patents/US1797799), so it makes immediate sense that he would propose the spring law as an explanation.\n\n\nSo Donald Andrews (or at least Johns Hopkins) is the obvious link between GMCRL and Robert C. Yates. Did Andrews know Yates? Was someone about to be scooped? The answer to *that* is probably beyond the scope of this question.\n\n\n", "3" ] ]

[ [ "\nYour definition of enantiomer is correct. \n\n\nThis simple further classification might be helpful. Let's take 2 molecules with the same molecular formula. These molecules are either constitutional (or structural) isomers or stereoisomers. If they have different structures (or connectivity patterns) and are separated by an energy barrier, then they are constitutional isomers. If they have the same structure or connectivity pattern, then they are stereoisomers. \n\n\nAmong stereoisomers there are 3 classes that we can place a pair of molecules into. Take any two stereoisomers and compare them, they are either\n\n\n1. the same (2 identical molecules)\n2. mirror images of one another (enantiomers)\n\n\nIf they are not the same or enantiomers, then they are\n\n\n3. different, and we call them diastereomers.\n\n\n", "3" ], [ "\nOne additional thing to add to @ron's answer is that *diastereomers* have different chemical properties .. they can be distinguished using conventional spectroscopic techniques, or even by measuring thermodynamic parameters (at least in principle .. sometimes the differences are very small and hard to resolve). They can also have different chemical reactivities. Apart from their different optical activities, *enantiomers* have identical chemical properties ... that is to say, you have to do an experiment that exploits their chiral natures to tell them apart.\n\n\nAnother point is that diasteromers always have at least two chiral centers ... a molecule with a single chiral center will have two enantiomers. A molecule having two chiral centers will have two enantiomers and two disasteromers. There is a nomenclature convention that can be used to index the various choices: each stereocenter is labeled \"R\" or \"S\" depending on whether the components are arranged clockwise (R) or counter-clockwise (S) around it, according to a conventional order of preference. In the case of a molecule with two stereo centers, the (R,R) and (S,S) stereoisomers are mirror images, and therefore are enantiomers, and the (R,S) and (S,R) stereoisomers are not mirror images, and therefore are *usually* diastereomers. As pointed out by @ron in the comments, if the molecule is highly symmetrical, the (S,R) and (R,S) forms may actually be indistinguishable, which would be a \"meso\" compound.\n\n\n", "1" ] ]

[ [ "\nOne thing to try would be just to build the two isomers and compare their total energies .. that would at least show which position was preferred thermodynamically (i.e. more stable). It sounds like you are more interested in the kinetic consideration though (i.e. which reacts faster) ... in that case, looking at the relative energies of the transition states will give you some basis for rationalizing the preference. I am not sure how good MM is at dealing with transition states, however ... you would need to rely rather heavily on your chemical intuition to find the correct structures, and even then, the bond angles and bond lengths may be distorted to such an extent that your forcefield is not well parameterized to deal with them.\n\n\n", "2" ] ]

[ [ "\nIf a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral (achiral).\n\n\nThe all hydrocarbon [2]catenane will be achiral. The molecule contains a plane of symmetry; picture a plane that contains one of the rings and bisects the other ring. In the following image, you can see that the plane of the screen contains the ring on the right and bisects the ring on the left.\n\n\n![A [2]catenane](https://i.stack.imgur.com/Gt8Vj.gif)\n\n\nYou might say that if we replaced the circles with cyclohexane rings, because of their chair-like shape, the plane of symmetry is lost. That is true, but you can flatten one of the cyclohexane rings and the plane of symmetry is restored. Remember, if any thermally accessible conformation of a molecule is achiral, then the molecule is achiral.\n\n\nSometimes it helps to build a model or use a Newman projection to help with the stereochemical analysis. Here is such a projection for your chloro-[2]catenane. Note that it still possesses a plane of symmetry (a plane perpendicular to the screen). \n\n\n\n\n\nIf we go further and add another chlorine atom, but now to the back ring, then the molecule becomes chiral — just like 1,3-dichloroallene. \n\n\n\n\n---\n\n\nRegarding:\n\n\n\n> \n> The compounds attached seem to be enantiomers. I don't see how by rotation they could be super-imposable.\n> \n> \n> \n\n\nLook at the bottom figure in my second drawing. I've taken your interlocking 5- and 6-membered catenane and rotated the 6-membered ring a bit further along (a conformational isomer; remember, as I mentioned above, if a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral [achiral]). The plane of the screen now contains the 5-membered ring and it bisects the cyclohexane ring. The plane of the screen also contains the methyl group. The plane of the screen is a symmetry element of this molecule; therefore this molecule is achiral.\n\n\n", "13" ] ]

[ [ "\nTwo of the methods are:\n\n\nAtomic Absorption Spectroscopy - similar to the flame test, but is usually quantitative. The sample is heated (in a flame or furnace) and the absorbance at a defined wavelength is measured against standardized solutions. It will work for any atom if you use a tunable laser.\n\n\nICP/MS (Inductively Coupled Plasma/Mass Spectrometry) - a solution of the compound is ionized and passed into a mass spectrometer. Some atoms (especially halogens) cannot be properly ionized and passed into the mass spectrometer. Oxygen and nitrogen (and a few others) would be difficult to quantify since it may come from impurities in the sample.\n\n\n", "3" ] ]

[ [ "\nWhat you are forgetting is that the reaction will also involve hydrogen.\n\n\n$\\ce{C6H6 + 3 H2 -> C6H12}$\n\n\nIn principle, the $\\Delta H$ for $\\ce{H2}$ should be 0, but in practice, that may not be true with a force field. The energies from molecular mechanics are seldom accurate enough to compute reaction energies. They are parameterized on a set of molecules, but the errors are usually large, both systematic and random errors.\n\n\n", "6" ] ]

[ [ "\nYes, you can define chemical potentials as partial molar enthalpies, or partial molar internal energies, or partial molar Helmholz free energies. (Sorry for my initial confusion on this point ... see also my answer on your other question on this topic, [here](https://chemistry.stackexchange.com/questions/23565/how-does-the-chemical-potential-allow-for-open-systems-to-be-considered/23567#23567)).\n\n\nThe definitions as partial molar enthalpies and partial molar internal energies are not particularly useful, since they involve working at constant entropy, which is practically impossible, but they are still mathematically valid. \n\n\nThe definition of chemical potential as partial molar Helmholz free energy at constant temperature and volume may actually have some practical applications, (e,g. perhaps for reactions that evolve gases in closed containers? although I have never come across any. \n\n\n", "2" ] ]

[]

[ [ "\n*Your question is almost impossible to answer - it simply is too broad.*\n\n\n1. Excitation spectra of fluorescent organic compounds typically aren't sharp lines but consist of one (or more) bands. Even if the maximum of the excitation spectrum isn't located at $\\lambda = \\textrm{365 nm}$, **monochromatic** irradiation at that wavelength will excite **a lot** of different compounds (and even compound classes) which then deactivate by fluorescence.\n2. Your light source does not emit monochromatic radiation! Hand lamps used to inspect TLCs typically have coated low-pressure mercury lamps. While the maximum of the radiation density might be centred around 360 nm, the full with at half maximum is - roughly estimated - about 30 nm or more. This means that there's even more compounds that might get excited by the emission of the hand lamp.\n\n\nIn summary, there's no exhaustive answer to your question, but just to name some compounds that are potential candidates:\n\n\n\n\n\n* Diphenylacetylene (**1**)\n* 1,6-Diphenylhexatriene (DPH, **2**)\n* 4',6-diamidino-2-phenylindole (DAPI, **3**)\n* 6,8-difluoro-7-hydroxy-4-methylcoumarin (DIFMU, **4**)\n\n\n", "1" ] ]

[ [ "\nYes this happens fairly regularly. It is a deviation from ideal behavior associated with the property of partial molar volume. For an ideal solution, the total volume always equals the summed volumes of the pure components, but ideal behavior neglects several important facts about real molecules, to wit, that they may have different shapes, and may also experience stabilizing or destabilizing interactions with each other.\n\n\nIt's fairly easy to see how molecular shapes and sizes could affect this property. Macroscopic objects experience similar effects. Suppose I have a bucket \"full\" of golf balls and a bucket full of sand, if I work carefully, I can pour quite a lot of sand into the bucket of golf balls because the smaller size of the sand grains allows them to fit into the gaps between the golf balls. Another example: Suppose you have a 10 liter bucket of golf balls and a 10 liter bucket of tiny cubes packed tightly together. If I then try to pour those two buckets into a 20 liter container simultaneously, it almost certainly will not be able to fit all of the balls and cubes, because they will not pack as efficiently in the 20 L container, costing some space. These are extreme examples, but illustrate one way that different shapes can play a role.\n\n\nDifferences in intermolecular interactions can play an even bigger role. If you are dissolving salt in water, the individual ions making up the salt crystal can be solvated very efficiently by the polar water molecules, which experience a strong attractive interaction with the ions. (Recall that the O and H-atoms in water have partial negative and positive charges, respectively.) This results in the solution having a significantly smaller volume than expected based on the summed volumes of the pure salt and water used to make the solution. \n\n\nPartial molar volume is not too hard to measure experimentally. For something like aqueous sodium chloride it can be determined in a fairly straightforward way from precise measurements of the densities of a series of solutions with increasing concentrations. However it is worth noting that the volume and mass measurements used for the calculation must be very precise, yielding measured densities with 5 or more significant digits. This requires careful use of volumetric glassware and precision balances, and the temperature of the solutions must be kept constant throughout.\n\n\nFor purposes of demonstrating the qualitative effect, you could just use water and ethanol (or isopropyl alcohol), which also have a significant negative excess volume. (see the first graph on the wikipedia page for partial molar volume for details). \n\n\nFor comparison you could use mixtures of benzene and toluene, which behave almost ideally with respect to volume effects. \n\n\n", "13" ], [ "\n1.Volumes do not add when you mix things together because the intermolecular forces of a mixture are different than the ones in the pure substances. The way this is handled is by using a property of mixtures called the \"partial molar volume.\" \n\n\n2.There are a couple of reasons why volumes of solution are not additive. One important one is that some chemical reaction occurs. The second is especially important if one component is water. Liquid water has an open structure, that is, the water molecules even in the liquid form molecule sized structures of lower density than the mean. The addition of ionic and/or polar solutes can \"collapse\" this structure resulting in an increased density. To complicate matters, the solute may also have structure that may be smaller than or greater than the average density of the solid. This will result in an expansion (or contraction) when the solute is dissolved.\n\n\nThere are no good theoretical models, of a general nature, to predict or estimate whether a solvent/solute combination will have a positive or negative \"excess\" volume of solution. \n\n\n3.When $50\\mathrm{mL}$ of water and $50\\mathrm{mL}$ of ethanol are combined in a $100\\mathrm{mL}$ graduated cylinder, the resulting mixture has a volume of $\\approx 97\\mathrm{mL}$.\nIntermolecular interactions between water molecules in a pure sample involve hydrogen bonding, though the details of these complex interactions are not completely understood. Some models predict transient ice-like structures within liquid water, where the local density of the sample is temporarily diminished. Ethanol and water molecules are also attracted to each other through hydrogen bonding. The ethanol likely interrupts the transient ice-like structures present in liquid water, resulting in a slight contraction of the sample.\n\n\n", "9" ] ]

[ [ "\nCheck out the van der Waals equation wikipedia page for a start ... it gives an explanation of one of the most common modifications of the ideal gas law that allows for modeling of deviations from ideality in the way you describe. This approach is quite nice in my opinion, because it includes two coefficients (a and b) that account for specific deviations for a given gas ... a scales with the strength of intermolecular interactions, and b accounts for the actual volume of the gas molecules.\n\n\nYou should be able to use that information to do what you want to do regarding Charles Law. The graph on the page shows isotherms, but you should be able to generate the isobar curves that you are after with a little bit of work.\n\n\n", "1" ] ]

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[ [ "\nThe textbook statement you gave (\"solvent is always the one present in larger quantity\") is simply a convention that makes it easier for us to talk to each other about mixtures of substances. We have other words for other types of mixtures. For example, in your ethanol-water example, you are correct in stating that it doesn't make much sense to talk about solute and solvent. So, we call these two liquids \"miscible\". They mix together in all proportions. A good demonstration that captures the essence of different types of mixtures is the phase diagram of phenol-water:\n\n\n \n\n\nHere we have a feature that is called a \"miscibility gap\". There is a region, defined by the rainbow shape in the diagram above in which phenol and water separate into two distinct phases. If we had pure water and added just a little bit of phenol to it at room temperature, we would find that it would dissolve into the water up to about 7 or 8 wt%. So, phenol would be the solute and water the solvent. Adding more than this amount pushes the mixture into the miscibility gap so that now the phenol and water would mostly separate into two separate phases. Let's say we add enough phenol to get us to 40 wt%. If we heat the two phase mixture up to 67 C then it will suddenly become miscible. Now only one phase exists and we can add as much phenol as we want to it without having to worry about two phases forming again. There is no solute or solvent here. Just a mixture of two liquids that exist in a single phase. \n\n\n", "11" ], [ "\nI think you may be emphasizing the term \"universal solvent\" a bit too hard. Water is \"universal\" as a solvent simply because lots of molecules are at least somewhat polar and so dissolve in polar water. But otherwise I'd say your textbook is right: the solvent is generally in greater quantity than the solute by convention.\n\n\n", "2" ], [ "\nI think you may be putting too much significance on the semantics of some mnemonics that are perhaps phrased more absolutely than they are intended to be used. That is particularly true with the phrase that \"water is the universal solvent\", which is not really a scientifically significant statement at all. Certainly gasoline, or mineral oil, will not dissolve in water, even in very small quantities, so why would it be considered \"universal\" in the sense you seem to be taking it?\n\n\nTo answer your question, the better rule of thumb is to use the statement that the solvent is always the component present in larger quantity .. but even that is not correct in all cases. For example in a multiphase extraction, there can be multiple solvents ... components that are polar will dissolve in the aqueous phase, so for those components, water is the solvent .. while components that are non-polar will dissolve in the non-aqueous phase, which will be the solvent for those components.\n\n\nA final point is that for something like a 49-51% mixture of ethanol and water, the concept of a solvent becomes largely academic ... is the ethanol dissolved in the water, or is the water dissolved in the ethanol? The answer is: both, and neither .. they are miscible liquids, so it's a two component mixture, rather than a solution. The physical properties of a mixture depend rather strongly on the properties of both components, while the physical properties of a solution are more likely to be dominated by the properties of the solvent.\n\n\n", "2" ] ]

[ [ "\nI figured out the answer. If there is a lone pair of electrons belonging to nitrogen, oxygen, or fluorine, then it is possible to bond to water. Therefore, it is can accept a hydrogen. If there is a hydrogen bonded to one of the three elements listed above, then it able to donate. \n\n\n", "6" ] ]

[ [ "\nA better way of thinking about it, without getting too bogged down in details, is that the plant absorbs the energy of photons and uses it to drive chemical processes - ultimately the synthesis of sugar out of carbon dioxide and water - that are energetically \"uphill\", and would not happen spontaneously without the energy from the photons. The energy from the photons is used by the plant to move electrons from one place to another, but no electrons are actually generated or consumed in the process ... photosynthesis is a cycle, so you always end up back where you started, with some molecules of $\\ce{CO2}$ and $\\ce{H2O}$ having been consumed (along with a lot of light energy), and some extra molecules of sugar and $\\ce{O2}$ having been produced.\n\n\nThe places where the photonic energy is used to mobilize the electrons are protein complexes called photosystem I and photosystem II, which are concentrated in the chloroplasts of green plants. The details of how this happen are still not completely understood, but a LOT of progress has been made fairly recently using fancy equipment like laser spectrometers that can measure physicals processes that occur on timescales shorter than a nanosecond. This has allowed researchers to map the progress of the electron as it moves from one part of the protein complex to another, following absorption of the photonic energy. \n\n\nDoes that help at all?\n\n\n", "3" ], [ "\nNo, photons of visible light don't ionize molecules during photosynthesis - electrons are transferred via [redox](https://en.wikipedia.org/wiki/Redox) reactions in [electron transport chain](https://en.wikipedia.org/w/index.php?title=Electron_transport_chain) - it seems to me that fragment of the article was maybe a little misleading.\n\n\n", "2" ], [ "\nIn some steps, a red light photon has enough energy (~0.5 e.v.) to raise an electron to a higher, metastable orbit (activating it), but might not actually remove the electron from the molecule (see <http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120072/bio13.swf::Photosynthetic+Electron+Transport+and+ATP+Synthesis> (very clear), <http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html> and <http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html>).\n\n\n", "1" ] ]

[ [ "\nThe same mechanism discussed in [my answer to your earlier question](https://chemistry.stackexchange.com/questions/23500/regioselectivity-of-alpha-halogenation-of-ketones/23504#23504) on acid catalyzed halogenation alpha to a carbonyl still applies here. For reference, here is the same [link](http://chemwiki.ucdavis.edu/Organic_Chemistry/Reactivity_of_Alpha_Hydrogens/Alpha_Halogenation) supplied earlier that shows the mechanism for this acid catalyzed halogenation.\n\n\nIn that earlier answer I said that the enol that \"produces the more stable double bond (the one that is more highly substituted)\" is the enol that will form and react nucleophilically with the halogen to produce the alpha-halo carbonyl product. In light of your new question, the statement that more highly substituted double bonds are more stable needs some qualification. \n\n\nThe closer in electronegativity a substituent is to carbon, the more it will stabilize a double bond. Consequently, the more alkyl groups attached to a double bond, the more stable the double bond (barring large steric effects). On the other hand, fluorine destabilizes a double bond. It is so electronegative that it prefers to bond with orbitals high in p character. When forced to bond to a double bond $\\ce{sp^2}$ orbital (that is enriched in s character) destabilization results (see this earlier [fluorobullvalene](https://chemistry.stackexchange.com/questions/14057/what-is-the-conformer-distribution-in-monosubstituted-fluoro-bullvalene) example for more on why fluorine \"hates\" to be on a double bond).\n\n\nBased on its electronegativity, chlorine should be somewhere between fluorine and an alkyl group in terms of its ability to stabilize a double bond. If we compare the heats of formation of 1-butene (-0.1 kJ/mol) and trans-2-butene (-11.2 kJ/mol) we see that placing the methyl group on the double bond stabilizes the molecule by 11.1 kJ/mol. If we compare the heats of formation of allyl chloride (-5.6 kJ/mol) and trans-1-chloropropene (-12 kJ/mol) we see that the placing the chlorine group on the double bond stabilizes the molecule less, only 6.4 kJ/mol.\n\n\nThese differences mean that we will have much more of the enol **without** the chlorine on the enol double bond, then we would have of the enol without the methyl group on the enol double bond. So we have more of the unsubstituted enol in the chlorine case **and** the enol with the chlorine on the double bond is much less nucleophilic (less reactive) due to the electron withdrawing inductive effect of the chlorine substituent. These two factors combine to make the second alpha-chlorination take place on the unsubstituted side of the molecule.\n\n\nAs to steric effects, they probably don't play a significant role. One metric commonly used to compare \"steric size\" is the cyclohexane [A value](http://en.wikipedia.org/wiki/A_value). The A value for chlorine is 0.43, much smaller than the A value for methyl (1.7), suggesting that a chlorine substituent is smaller than methyl substituent.\n\n\n", "1" ], [ "\nMost probably due to steric reasons and the electron withdrawing effect that the halogen has on the substituted methylene group.\n\n\n", "0" ] ]

[ [ "\nThe chemistry you have posted is for non-hydraulic cement, which requires air drying for precisely the reasons you enumerated. The kind of cement that sets underwater uses silicates and various oxides (primarily calcium, aluminum and iron oxides). The basic chemical process that is occurring is hydration, whereby the mixture of oxides and silicates incorporates several water molecules directly into the crystal lattice. The fact that the setting process both uses water, and also produces a structure that is insoluble and robust to attack by water, is what makes underwater setting of hydraulic cement possible.\n\n\nWith regard to teaching your high school class, it's probably better to stick to the lime cycle for non-hydraulic cement as far as detailed chemical reactions are concerned. One interesting point that you can tell them about hydraulic cement is that its chemistry is still not completely understood, and even the molecular crystal structure was only very recently sorted out by [MIT scientists in 2009](http://cee.mit.edu/cee-in-focus/2011/spring/cement-structure).\n\n\n", "9" ] ]

[ [ "\nThe acidic and basic catalyzed alpha-halogenation of a carbonyl proceed through an enol and an enolate intermediate respectively. The difference in regioselectivity arises because of the different process used to arrive at the enol/enolate in the acidic and basic cases.\n\n\nIn the acidic case, 1) the carbonyl oxygen is protonated and then 2) a proton is removed from the alpha carbon to form the enol. [See here](http://chemwiki.ucdavis.edu/Organic_Chemistry/Reactivity_of_Alpha_Hydrogens/Alpha_Halogenation) for a good picture of this mechanism. In this process, the reaction is an equilibrium so thermodynamic control will lead to formation of the most stable enol. Therefore, a proton will be removed from the more highly substituted alpha-carbon to produce the more stable double bond (the one that is more highly substituted). \n\n\nWhen the reaction is run under basic conditions, the process is no longer a thermodynamically controlled equilibrium, but rather proceeds under kinetic control. The same link used above also shows the mechanism of the base-catalyzed alpha-halogenation. The rate determining step is removal of an alpha-proton by the base. The lowest energy proton to remove is the one on the least substituted carbon. Remember that a primary carbanion is more stable than a secondary carbanion which is more stable than a tertiary carbanion, so removing a primary proton will produce the most stable carbanion-enolate. Hence, in the base catalyzed case the proton will be removed from the least susbstituted alpha carbon so to produce the most stable carbanion-enolate intermediate.\n\n\nTo summarize the acid-catalyzed halogenation proceeds under thermodynamic control to produce the most highly substituted (most stable double bond) enol. Whereas the base-catalyzed halogenation is kinetically controlled and produces the least substituted (most stable carbanion) enolate anion.\n\n\n", "6" ], [ "\nRon's answer is not completely true. \n\nYes, the least substituted α-hydrogen is the more acidic one and the least hindered one (kinetic control) and the more substituted enolate is the most stable (thermodynamic control). What is not completely true is that acidic conditions favour thermodynamic control because it is reversible but that basic conditions are limited to kinetic control. \n\nWhether or not basic conditions afford kinetic or thermodynamic control depends on the deprotonation conditions. Hence, the use of a reversible base (with associated *p*Ka less than that of the carbonyl compound, *i.e.* 18-25) such as NaOH will promote the thermodynamic enolate, as will a protic solvent such as H2O. Even a very strong base such as LDA does not guarantee kinetic control, because any trace of proton or unreacted ketone will cause equilibration to the thermodynamic enolate, even at –78°C over time. \n\nThe real reason for the obtention of the least substituted halide from an unsymmetrical ketone under basic conditions is that the molecular halogen (*e.g.* Cl2) reacts instantly with the kinetic enolate as it forms, trapping it in a sense, and that reaction between a powerful nucleophile and a powerful electrophile is apparently faster than the enolate equilibration. Some of the reason for this is that Cl2 in OH– is actually OCl– which acts as both base and electrophile, so that a second collision is not required. \n\nThis property is the reason for the success of the haloform reaction, which occurs at the CH3 end of methyl ketones. The first halogenation facilitates the second deprotonation (by virtue of lowering the *p*Ka) and the second halogenation facilitates the third. \n\n\n", "4" ] ]

[ [ "\nThe standard reaction enthalpy you have describe is only for a pure substance if you are talking about a standard enthalpy of formation. Otherwise the equation wouldn't be valid for reactions that have more than one product, right? In the specific case of a standard enthalpy of formation, you are talking about the enthalpy change associated with the creation of one mole of a pure substance *from its component elements in their standard states.* This gets at your second question ... the zero on the enthalpy scale is established by convention (and this is the reason why the definition of standard enthalpy of formation works) ... the standard enthalpy of formation of an element in its standard state is DEFINED to be zero. \n\n\nNow, you can estimate the standard reaction enthalpy for any chemical reaction by subtracting the summed standard enthalpies of formation of the reactants from the summed standard enthalpies of formation of the products, taking into account the appropriate stoichiometric coefficients from the balanced chemical equation. That process is similar to what you have written, but not precisely the same, since you omitted the stoichiometry information (and the chemical equation, for that matter). The standard enthalpy estimate obtained will pertain to reactions carried out at the given conditions .. standard enthalpies are defined for 1 atm pressure but variable temperature, so you may need to correct tabulated standard enthalpies of formation to match the given reaction conditions.\n\n\nOne last point is that enthalpy is absolutely defined as a thermodynamic quantity .. it is the internal energy plus the product of pressure and volume, $H=U + pV$. The issue is that the internal energy of a substance is hard to define absolutely .. are you going to include the binding energies of the core electrons to their atoms? How about the strong forces holding the nuclei together? Thus for chemical processes it almost always makes sense to deal with just changes in enthalpy, since most of the really low level stuff I mentioned doesn't change between reactants and products. I think that may have been part of what you were getting at with your second question.\n\n\n", "1" ] ]

[ [ "\nWhether or not oxygen can reach oxidation states higher than $\\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\\mathrm{-III}$ and $\\mathrm{+V}$ but would also be restricted to <insert arbitrary number here>.\n\n\nThe actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\\ce{XeF6, NF3}$ or $\\ce{SF6}$ to $\\ce{XeO4, NO3-}$ or $\\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\\mathrm{+II}$ on oxygen given our current knowledge.\n\n\nFurthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. [$\\ce{Na2S}$](https://en.wikipedia.org/wiki/Sodium_sulfide), [$\\ce{Na2Se}$](https://en.wikipedia.org/wiki/Sodium_selenide), [$\\ce{Na2Te}$](https://en.wikipedia.org/wiki/Sodium_telluride) and [$\\ce{Na2Po}$](https://en.wikipedia.org/wiki/Sodium_polonide) are all known and in all of these compounds the chalcogen has an oxidation state of $\\mathrm{-II}$.\n\n\n", "2" ], [ "\nYou need to review the concepts surrounding atomic shell structure and quantum numbers, and how they relate to the organization of the periodic table, I think. Briefly, since oxygen is in the second row of the periodic table, we know that its valence electrons occupy the second orbital shell, corresponding to orbitals with principal quantum number $n=2$. Recall also that for a given principal quantum number $n$, only sub-shells with orbital angular quantum number $l$ up to $n-1$ are allowed. Therefore, the valence shell of osygen only includes s ($l=0$) and p ($l=1$) orbitals. On the other hand, the other members of group 16 will by definition have higher principal quantum numbers, which means that their valence shells include d ($l=2$) orbitals. It is these orbitals that can be used to rationalize the existence of higher oxidation states for those elements, for example by the $sp^3d$ and $sp^3d^2$ hybridization schemes in VSEPR (valence shell electron pair repulsion) theory. \n\n\nNow, that explains why heavier elements in group 16 can have higher (i.e. more positive) oxidation numbers than +2. Your question as stated appears to have an error however, since it says that those elements \"exhibit only +2,+4,+6\" ... that's not technically correct .. for example there are compounds where sulfur has an oxidation state of -2 (e.g. $H\\_2S$). However I don't think it matters for the answer to your question here.\n\n\n", "-1" ] ]

[ [ "\nNear room temperature, considering solutions of water and sulphuric acid, the maximum conductivity is at about 30% sulfuric acid. For pure water or pure sulfuric acid the conductivity is orders of magnitude lower. \n\n\nIn pure water, there are only $10^{-7}$M of $\\ce{H+}$ and $\\ce{OH-}$ to carry current. \n\n\nWhen a relatively small amount of sulfuric acid is added, it dissociates fully to $\\ce{H+}$ and $\\ce{HSO4-}$. For example if you increase the concentration of sulfuric acid from 0M to 0.01 M, the number of ions in the solution increase by a factor of $10^5$. \n\n\n", "1" ] ]

[ [ "\nThe crystal structure shows a strangely distorted molecule with only $C\\_\\mathrm{i}$ symmetry. Values have quite a wide range, $\\angle(\\ce{CCH})\\approx 100-120^\\circ$. See below for all values and xyz coordinates to read in with a molecular viewer. Source:\n [R. Kahn, R. Fourme, D. André and M. Renaud, *Acta Cryst.* **1973**, *B29*, 131-138](http://dx.doi.org/10.1107/S0567740873002074).\n\n\nA DF-BP86/def2-TZVPP calculation (gas phase, 0 K) gives a highl $D\\_\\mathrm{3d}$ symmetric structure with \n\\begin{array}{ll}\n\\angle(\\ce{CCH\\_{ax}}) & =109.1^\\circ,\\\\\n\\angle(\\ce{CCH\\_{eq}}) & =110.3^\\circ,\\\\\n\\angle(\\ce{CCC}) &=111.5^\\circ,\\\\\n\\angle(\\ce{H\\_{eq}CH\\_{ax}}) &=106.4^\\circ.\\\\\n\\end{array}\n Also Bond length differ very slightly (and impossibly to determine at room temperature and pressure), with \n\\begin{array}{ll}\n\\mathbf{d}(\\ce{CC}) &=1.537~\\mathrm{\\mathring{A}},\\\\\n\\mathbf{d}(\\ce{CH\\_{ax}}) &=1.105~\\mathrm{\\mathring{A}},\\\\\n\\mathbf{d}(\\ce{CH\\_{eq}}) &=1.103~\\mathrm{\\mathring{A}}.\\\\\n\\end{array} \nThe tiny differences can be explained with [Bent's rule](https://chemistry.stackexchange.com/q/15671/4945). \n\nIn principle the assignment Goeff already gave you is absolutely valid, especially if you take the dynamic nature of the molecules into account that they have at standard temperature and pressure. \n\nAlso see xyz coordinates for visualisation below.\n\n\n\n\n---\n\n\nAppendix\n========\n\n\nCrystal structure \n\n\n\\begin{array}{lr}\nR(1-2) &1.528 \\\\ \nR(1-3) &1.519 \\\\ \nR(1-4) &0.884 \\\\ \nR(1-5) &1.146 \\\\ \nR(2-6) &1.055 \\\\ \nR(2-7) &0.929 \\\\ \nR(2-10) &1.521 \\\\ \nR(3-8) &1.100 \\\\ \nR(3-9) &0.944 \\\\\nR(3-11) &1.521 \\\\\nR(10-12) &1.519 \\\\\nR(10-13) &1.100 \\\\\nR(10-14) &0.944 \\\\\nR(11-12) &1.528 \\\\\nR(11-15) &1.055 \\\\\nR(11-16) &0.929 \\\\\nR(12-17) &0.884 \\\\\nR(12-18) &1.146 \\\\\nA(2-1-3) &110.4 \\\\\nA(2-1-4) &106.7 \\\\\nA(2-5) &105.8 \\\\\nA(1-2-6) &106.6 \\\\\nA(1-2-7) &118.2 \\\\\nA(1-2-10) &111.3 \\\\\nA(3-1-4) &103.0 \\\\\nA(3-1-5) &114.0 \\\\\nA(1-3-8) &110.6 \\\\\nA(1-3-9) &116.9 \\\\\nA(1-3-11) &112.4 \\\\\nA(4-1-5) &116.7 \\\\\nA(6-2-7) &109.5 \\\\\nA(6-2-10) &110.6 \\\\\nA(7-2-10) &100.5 \\\\\nA(2-10-12) &112.4 \\\\\nA(2-10-13) &109.1 \\\\\nA(2-10-14) &100.4 \\\\\nA(8-3-9) &106.9 \\\\\nA(8-3-11) &109.1 \\\\\nA(9-3-11) &100.4 \\\\\nA(3-11-12) &111.3 \\\\\nA(3-11-15) &110.6 \\\\\nA(3-11-16) &100.5 \\\\\nA(12-10-13) &110.6 \\\\\nA(12-10-14) &116.9 \\\\\nA(10-12-11) &110.4 \\\\\nA(10-12-17) &103.0 \\\\\nA(10-12-18) &114.0 \\\\\nA(13-10-14) &106.9 \\\\\nA(12-11-15) &106.6 \\\\\nA(12-11-16) &118.2 \\\\\nA(11-12-17) &106.7 \\\\\nA(11-12-18) &105.8 \\\\\nA(15-11-16) &109.5 \\\\\nA(17-12-18) &116.7 \\\\\n\\end{array}\n\n\n\n```\nC 2.329589000 3.011988000 -0.044238000\nC 3.630888000 2.632672000 0.661249000\nC 1.367908000 1.836044000 -0.049671000\nH 2.524936000 3.113740000 -0.900292000\nH 1.933922000 3.936128000 0.506802000\nH 3.388552000 2.469096000 1.674854000\nH 4.348450000 3.220000000 0.605369000\nH 1.030575000 1.607424000 0.972471000\nH 0.592407000 1.921052000 -0.581309000\nC 4.247092000 1.383956000 0.049671000\nC 1.984112000 0.587328000 -0.661249000\nC 3.285411000 0.208012000 0.044238000\nH 4.584425000 1.612576000 -0.972471000\nH 5.022593000 1.298948000 0.581309000\nH 2.226448000 0.750904000 -1.674854000\nH 1.266550000 -0.000000000 -0.605369000\nH 3.090064000 0.106260000 0.900292000\nH 3.681078000 -0.716128000 -0.506802000\n\n```\n\nComputed (DF-BP86/def2-TZVPP) \n\n \n\n\n\n```\nC -1.270112000 -0.733300000 0.229506000\nC 0.000000000 -1.466599000 -0.229506000\nC -1.270112000 0.733300000 -0.229506000\nH -1.328386000 -0.766944000 1.332131000\nH -2.167276000 -1.251277000 -0.145050000\nH 0.000000000 -1.533888000 -1.332131000\nH 0.000000000 -2.502555000 0.145050000\nH -1.328386000 0.766944000 -1.332131000\nH -2.167276000 1.251277000 0.145050000\nC 1.270112000 -0.733300000 0.229506000\nC 0.000000000 1.466599000 0.229506000\nC 1.270112000 0.733300000 -0.229506000\nH 1.328386000 -0.766944000 1.332131000\nH 2.167276000 -1.251277000 -0.145050000\nH 0.000000000 1.533888000 1.332131000\nH 0.000000000 2.502555000 -0.145050000\nH 1.328386000 0.766944000 -1.332131000\nH 2.167276000 1.251277000 0.145050000\n\n```\n\n", "3" ], [ "\nIn the chair form of cyclohexane, the carbon atoms and the bonds around them are almost perfectly tetrahedral.\n\n\nSo the C-C-H angles will be almost exactly 109.5 degrees. (Note that while you defined the bond midpoint, the angle will be the same regardless of whether it's the midpoint of the bond or the neighboring carbon atom itself.)\n\n\n", "2" ] ]

[ [ "\n*You have already mostly answered the question yourself ;)*\n\n\nA heating block, made from a material with a high thermal conductivity, helps to keep (**a lot of**) samples distributed over a larger area (several vials, Eppendorf tubes, etc.) at exactly the same temperature even if the primary heat source is punctual or just a layer of heating wires at the bottom of the instrument.\n\n\n", "4" ] ]

[ [ "\nYes, a common synthesis of benzotriazole starts with o-phenylenediamine as shown in the following figure.\n\n\n\n\n\n[Image Source](http://en.wikipedia.org/wiki/Benzotriazole#Synthesis)\n\n\nThe first step in the reaction involves straightforward diazotization of one of the amino groups (exactly like what's done in the first part of the [Sandmeyer reaction](http://en.wikipedia.org/wiki/Sandmeyer_reaction)) to produce the *ortho*-amino diazonium ion. Instead of eliminating nitrogen as in the Sandmeyer case, the molecule loses a proton and then the diazomium ion internally captures the *ortho*-imino group to yield the triazole product.\n\n\nThe reaction is not reversible, the benzotriazole is very stable and the intermediate diazonium ion is relatively high energy making the back reaction unlikely under normal conditions. The reaction will also work with substituted *o*-phenylenediamines as long as the substituent doesn't interfere with the diazotization step. So for example, alkyl- or halo-substituted *o*-phenylenediamines will produce the corresponding alkyl- or halo-benzotriazoles.\n\n\nIntramolecular cyclization to produce a triazole will only occur if the amino groups in the starting diamine are ortho to one another. If *m*- or *p*-phenylenediamines or non-*ortho* naphthyl diamines are used, then products resulting from intermolecular azo coupling will occur. See an article by Freeman [[1](https://doi.org/10.2741/4093)] for some examples.\n\n\n### Reference\n\n\n1. Freeman, H. Aromatic Amines: Use in Azo Dye Chemistry. *Front Biosci* **2013**, 18 (1), 145. DOI: [10.2741/4093](https://doi.org/10.2741/4093). (Open Access)\n\n\n", "6" ] ]

[ [ "\nI suspect that the reason you see a decay in the cell potential of $\\ce{Zn/1M~ NaCl/Cu}$ with increasing temperature is due to increased corrosion of zinc. In $\\ce{NaCl}$ solutions the anodic corrosion reaction is dissolution of $\\ce{Zn}$ as $\\ce{Zn^{2+}}$ and the cathodic reaction is reduction of dissolved oxygen to form an insoluble zinc hydroxide ([Ichiro Suzuki, *Corrosion Science* **1985**,*25* (11), 1029-1034](http://dx.doi.org/10.1016/0010-938X(85)90070-8)). The insoluble hydroxide passivates the metal surface. So, you have two effects which would contribute to EMF decay: 1) increased $\\ce{Zn^{2+}}$ concentration and 2) passivation of the electrode. \n\n\nOne last thing which may or may not be helpful to you - if you are planning on utilizing the EMF from this cell to drive any kind of current you must account for the hydrogen overpotential on the copper electrode. It takes a lot of extra voltage to make hydrogen on pretty much any metal except platinum. You may find that you have a cell that looks good at open circuit (i.e. high potential with no current flow) only to see it crumble to nothing when you try connecting it to a load.\n\n\n", "3" ] ]

[ [ "\nNot really sure what you are after here, but maybe this will help ... artificial flavors and colors are generally present at very low concentrations, dissolved in the water. \"Dissolved\" means that the individual additive molecules are completely surrounded by water molecules, which are interacting with it to stabilize it in its dissolved form. Since their concentration is low, each such additive-water aggregate is pretty far away from any others, so what you have is not very different from water, in a chemical sense at least. Note also that the water molecules around a given additive molecule are not static .. they are constantly exchanging with the \"bulk\" pure water solvent around them.\n\n\nIn some cases, an additive may not be very water-soluble ... in such a case a surfactant solubilizer molecule may be added to help the additives dissolve in the water. The surfactant molecules have one end that likes to interact with the insoluble additives (the hydrophobic end), and one end that likes to interact with water (the hydrophilic end). They aggregate themselves into larger spherical structures called micelles, where the hydrophobic parts of the molecules are protected on the inside (along with the additive molecules), and the hydrophilic parts are on the outside, exposed to the water. These micellar aggregates can be large enough to scatter light, and so drinks with these additive-solubilizer combinations will often appear cloudy. An example of this is the citrus flavoring in sodas like Mountain Dew and Squirt ... in those cases, brominated vegetable oil (BVO) is added as a surfactant to dissolve the citrus flavor molecules. \n\n\n(Aside: BVO is not exactly good for you, and several soft drink manufacturers have said that they will eliminate it from their formulations, which means maybe I can start drinking Fresca again)\n\n\n", "2" ] ]

[ [ "\nCarboxyl groups do share some properties with hydroxy and carbonyl groups such as the ability to engage in hydrogen bonding but in terms of chemical reactivity they are very different.\n\n\nThe reason for this lies in the structure of the carboxyl group. Instead of the carbonyl and the hydroxyl groups being completely independent of each other, there is an orbital interaction between the three $p$ orbitals on the carbon and the two oxygens which creates a $\\pi$ system across the three atoms that tends to stabilise the group, particularly when it is deprotonated. (I couldn't find any decent pictures of these $\\pi$ orbitals but if anyone can please edit them in)\n\n\nThe most obvious property that carboxyl groups have which hydroxyl and carbonyl groups do not display (hydroxyls do slightly but nowehere near as much as carboxyls) is acidity. This high acidity is caused by the fact that the carboxylate ion which forms when carboxyl groups donate a proton is stabilised by the electron delocalisation $\\pi$ system. This can also be expressed using resonance structures as shown below.\n\n\n\n\n\nThe strength of the acidity can be seen in the $pK\\_a$ values of carboxylic acids compared to alcohols.$$Ethanoic~acid~pK\\_a = 4.74$$$$Ethanol~pK\\_a = 16$$\n\n\nRemebering that $pK\\_a$ has a logarithmic scale we can see that carboxylic acids are over $10^{10}$ times more acidic than their respective alcohols.\n\n\nAn excellent introduction to many other properties and reactions of carboxylic acids can be found here: <https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/crbacid1.htm>\n\n\n", "3" ] ]

[ [ "\nThe only volume that is important in the denominator of your equation is the total volume of solution; neither the volume of solute or solvent are explicitly relevant.\n\n\nPragmatically, there is not likely to be much of a volume change when taking a concentrated solution and diluting it.\n\n\nSo with reference to your question, you should add enough solvent to get the proper final volume. Explicitly, add the 25 mL of stock solution and 'dilute to mark'-- that is add enough solvent so the total solution is 50 mL. The added solvent might not be exactly 25 mL to make this final volume.\n\n\n", "4" ], [ "\nYou only add the $\\ce{25 mL}$ water. After all you want to dilute to a certain molarity. You start from a 10 M solution and want to go to a 5M solution. Neglecting the volume of the dissolved compound you simply add another equivalent of solvent to dilute by a factor 2, i.e. $\\ce{25 mL}$ water in this case.\n\n\n", "0" ] ]

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[ [ "\nDi-tert-butyl dicarbonate melts around room temperature (m.p.=23°C). It does not decompose at this or even slightly higher temperatures. For example, it is typically purified by distillation under reduced pressure at temperatures up to around 65°C. At higher temperatures it will decompose to isobutene, t-butyl alcohol and carbon dioxide.\n\n\n\n\n\nIn any case it is OK to melt di-tert-butyl dicarbonate and use it between room temperature and up to around 40-60°C. Just make sure that when you open the bottle and use it that you do so in a dry environment as water will decompose di-tert-butyl dicarbonate.\n\n\n", "4" ] ]

[ [ "\n\n> \n> Why is dG=dH-TdS?\n> \n> \n> \n\n\nIt isn't. That would only be true at constant temperature. In general:\n\n\n$$\\mathrm{d}G = \\mathrm{d}H - T\\,\\mathrm{d}S - S\\,\\mathrm{d}T$$\n\n\n", "6" ], [ "\nIf you write $\\mathrm{d}G = \\mathrm{d}H - T\\,\\mathrm{d}S$, you drop off one term of the differentiation of $G$ which is $S\\mathrm{d}T$.\nBy definition $G = H - TS$. So, $$\\mathrm{d}G = \\mathrm{d}H - T\\,\\mathrm{d}S-S\\,\\mathrm{d}T$$\nThen, you should write $\\mathrm{d}H = T\\,\\mathrm{d}S + V\\,\\mathrm{d}p$ (not as you have written $\\mathrm{d}H = S\\,\\mathrm{d}T + V\\,\\mathrm{d}p$ ).\n\n\nFinally, $$\\mathrm{d}G = T\\,\\mathrm{d}S + V\\,\\mathrm{d}p - T\\,\\mathrm{d}S - S\\,\\mathrm{d}T$$\n $$\\mathrm{d}G = V\\,\\mathrm{d}p - S\\,\\mathrm{d}T$$\n\n\n", "6" ] ]

[ [ "\nI don't understand what you mean by saying \"if you don't make the substitution you do not get zero\". The relevant point is that, for ideal gases, $$p = nRT/V,$$ so $p/T$ is constant with respect to $T$. Therefore $$\\frac \\partial{\\partial T}\\left(\\frac pT\\right)=0$$ for an ideal gas, whether or not you use the substitution to simplify the process.\n\n\nIf you don't use the substitution, you need to use the quotient rule to do the partial derivative properly, and you should still end up getting zero if you do it correctly. Could that be where your confusion is coming from?\n\n\n", "2" ] ]

[ [ "\nIn simple molecules the intermolecular forces are, in order of increasing strength, dispersion forces, permanent dipole interactions and hydrogen bonding (which has significant covalent character but is generally considered to be an intermolecular force).\n\n\nBy contrast giant covalent repeating structures such as diamond and $\\ce{SiO2}$ are not molecular in the same sense as they can theoretically be infinitely large. Therefore they do not really have intermolecular forces but they are simply held together by covalent bonds between the atoms in the structure.\n\n\n", "1" ], [ "\nMolecules are made of fixed numbers of atoms joined together by covalent bonds, and can range from the very small (even down to single atoms, as in the noble gases) to the very large (as in polymers, proteins or even DNA).\n\n\nThe covalent bonds holding the molecules together are very strong, but these are largely irrelevant to the physical properties of the substance. Physical properties (melting points, boiling points, solubility in water,...) are governed by the **intermolecular forces** - forces attracting one molecule to its neighbors - van der Waals attractions or hydrogen bonds.\n\n\nMolecular substances tend to be gases, liquids or low melting point solids, because the intermolecular forces of attraction are comparatively weak. You don't have to break any covalent bonds in order to melt or boil a molecular substance.\n\n\nThe value of the melting or boiling point will depend on the strength of the intermolecular forces. The presence of hydrogen bonding will lift the melting and boiling points. **The larger the molecule the more van der Waals attractions are possible** - and those will also need more energy to break.\n\n\n", "1" ], [ "\n**Two factors matter: the nature of the bonds and the scale of the interactions**\n\n\nThe simple case is that, for many giant covalent substances, all the bonds are strong *covalent* bonds. Diamond, where all the carbons in the substance are connected to each other by strong covalent bonds, is a good example.\n\n\nThe more complex case relates to how strong the weak interactions between individual molecules (van der Waals forces) are in molecules of different size. Iodine, a substance containing small molecules containing two iodine atoms, is volatile because it has weak forces between the molecules. But polyethylene, a polymer is a fairly solid and non volatile substance, contains long strands of alkane-like chains held together by the same forces.\n\n\nBut the strength of those weak forces between the components making up the substance, depend on the surface area of the components. Iodine molecules have a small surface area, so the forces are weak; polyethylene chains have a large surface area, so the attractive forces are strong. This is also why geckos can walk up walls: same forces at play but the structure of their fingers and toes creates huge surface area maximising the overall strength of the weak interactions.\n\n\nThe surface area also explains why the shape of the molecule matters. Compact molecules have smaller forces than their isomers of the same size that are less compact. So the compact 2,2 dimethyl butane (sometimes called neopentane, C(CH3)4) is far more volatile than its isomer pentane, a 5-carbon long chain.\n\n\n", "0" ], [ "\nIn covalent molecules, there’s a theory that the bigger the molecule, the higher the melting point and boiling point. \n\nHowever, the main reason why giant covalent structures have high melting and boiling point is that it is the strong covalent bond that must be overcome in order to melt or boil the giant covalent structure. \n\nLet me use graphite as an example, although there is weak intermolecular bonds between the layers of carbon atom (which is actually graphene). However, breaking the weak intermolecular bonds between layers of carbon atom will only allow layers of carbon atom to slide over each other, which make graphite slippery and soft. In order to melt or boil graphite, you have to break the strong covalent bonds. \n\nIn contrast, for simple molecular structures, the reason why it have low melting and boiling point is because it is the weak intermolecular forces that must be overcome in order to melt or boil it.\n\n\nI understand it’s so complicated, but as long as you review it for a few times, you will remember it.\n\n\n", "-1" ] ]

[ [ "\nThe mass scale has changed over time, largely due to different isotopes of the \"baseline.\" Not surprisingly, there's a good [Wikipedia](http://en.wikipedia.org/wiki/Atomic_mass_unit) article on the matter.\n\n\n\n> \n> In the 20th century, until the 1960s chemists and physicists used two different atomic-mass scales. The chemists used a \"atomic mass unit\" (amu) scale such that the natural mixture of oxygen isotopes had an atomic mass 16, while the physicists assigned the same number 16 to only the atomic mass of the most common oxygen isotope (O-16, containing eight protons and eight neutrons). However, because oxygen-17 and oxygen-18 are also present in natural oxygen this led to two different tables of atomic mass. The unified scale based on carbon-12, $\\ce{^12C}$, met the physicists' need to base the scale on a pure isotope, while being numerically close to the chemists' scale.\n> \n> \n> \n\n\nIn short, [Dalton](http://en.wikipedia.org/wiki/John_Dalton) suggested $\\ce{^1H}$ as the basis of the mass scale, but [Ostwald](http://en.wikipedia.org/wiki/Wilhelm_Ostwald) pushed later for $\\ce{^16O}$. **Unfortunately, no one knew about isotopes yet.**\n\n\nThe problem was that physics stuck to isotopically pure O-16 for their mass scale, and chemistry used \"oxygen\" (i.e., with some O-17 and O-18 present). There's 0.028% difference between those scales. That may not sound like much, but with molecules of ~200-500 amu, it starts to add up.\n\n\nUsing C-12, it's isotopically pure (making physics happy) very close to the previous chemistry \"O-16 + O-17 + O-18\" scale.\n\n\nFrom [IUPAC](http://www.iupac.org/publications/ci/2004/2601/1_holden.html):\n\n\n\n> \n> In April 1957 at the bar in the Hotel Krasnapolski in Amsterdam, Nier suggested to Mattauch that the $\\ce{^12C}$ = 12 mass scale be adopted because of carbon's use as a secondary standard in mass spectrometry. Also, $\\ce{^12C}$ = 12 implied acceptable relative changes in the atomic weight scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for the $\\ce{^16O}$ = 16 scale (which would not acceptable to chemists). Enthusiastically, Mattauch made a worldwide effort in the late 1950s to publicize the $\\ce{^12C}$ = 12 scale and obtain the physicist's approval, while Wichers obtained the chemist's approval.\n> \n> \n> \n\n\n", "15" ] ]

[ [ "\nThe glass transition temperature depends on the cooling rate and molecular weight distribution and **could be influenced by additives**:\n\n\nMost liquid plasticizers are low molecular weight organic materials with a glass transition temperature in the range of $125-225^0\\text{C}$. When this kind of additives is added to an organic polymer, the result is a weight averaging of the glass transition temperature between that of the polymer and that of the plasticiser if they are miscible. The reduction of the the glass transition temperature of a polymer by the plasticiser is usually about $5^0\\text{C}$ per weight percent of plasticiser.\n\n\nImpact modifiers ( a kind of additives) can affect the impact resistance of a polymer. Impact modifiers are key additives for\nincreasing flexibility and impact strength\nto meet physical property requirements of\nrigid polymers. All impact modifiers are elastomeric or\nrubbery in nature, with a lower modulus\nthan the host polymer. The dispersed\nrubber phase acts to absorb or dissipate the\nenergy of impact in order to stop craze or\ncrack propagation. \n\n\n", "1" ] ]

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[ [ "\nThere is no reaction since the electrons will only flow one way in the wire. When you connect the zinc metal to the copper metal with a wire, there is a voltage potential generated between them, just like in a thermocouple.\n\n\nFrom [Wikipedia](http://en.wikipedia.org/wiki/Thermocouple):\n\n\n\n> \n> Any junction of dissimilar metals will produce an electric potential related to temperature. \n> \n> \n> \n\n\nIn this case, we don't care about the temperature effects. \n\n\nWhen a salt bridge is placed between the solution, then the electrons are carried by the anions in the solution from one electrode to the other electrode. \n\n\nSo the zinc metal releases electrons when it dissolves, which travel through the wire easily, to the copper. The $\\ce {Cu^{2+}}$ in solution grabs the electrons from the copper electrode, adding copper metal to the electrode. Since there is now more negative ions in this solution, the anions move across the bridge to the zinc electrode. The anions attack the zinc metal, causing the zinc to loose electrons when it dissolves. \n\n\nI'm not sure what starts the reaction, nor why the zinc dissolves easily.\n\n\n", "4" ], [ "\nElectron move from zinc to copper as zinc is more reactive metal than copper and thus more readily loses the electron to turn into zn2+ ion . Thus electrons move from zinc to copper\n\n\n", "-3" ] ]

[ [ "\nI have never worked with potassium personally, but have experience with other alkali metals. You are spot on about potassium superoxide. I have direct knowledge of an explosion that occurred when potassium metal was stored for an extended period of time in mineral oil. I would therefore be extremely cautious if you intend to store it in this manner. Factors such as length of storage time, permeability of your container seals and storage temperature can all affect the amount of superoxide that forms. Based on the evidence I have seen, it is not safe to assume that there will be no transport of oxygen through a mineral oil barrier. The only way I would work with this metal is in an inert glovebox, period. That is my own preference, of course, and I always tend to err on the side of caution. \n\n\n", "6" ], [ "\nThe following excerpt regarding the management of alkali metals is from: \n\n\nManagement of time-sensitive chemicals (I): Misconceptions leading to incidents by Jim Bailey, David Blair, Lydia Boada-Clista, Dan Marsick, David Quigley, Fred Simmons and Helena Whyte\n\n\n\"Slow reactions of alkali metals and their alloys with oxygen to form oxides and superoxides have been well documented\\*. Even when these metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react. Potassium, under these circumstances, forms yellow or orange coating that can explode or catch fire upon cutting. NaK, a eutectic mixture of potassium and sodium, can undergo a similar reaction. Lithium stored under dry nitrogen can react with the nitrogen to form a nitride. Formation of the nitride is autocatalytic and can eventually lead to autoignition of the lithium. Fires of alkali metals are extremely dangerous and difficult to extinguish.\"\n\n\n* Management of time-sensitive chemicals (II): Their identification, chemistry and management,Jim Bailey,David Blair,Lydia Boada-Clista,Dan Marsick,David Quigley,Fred Simmons,Helena Whyte: **Chemical Health and Safety**, Volume 11, Issue 6, November–December 2004, Pages 17–24\n\n\nI've come to the realization that if you plan to keep chemicals around you need to know all the risks and provide effective management including a plan for periodic inspection.\n\n\n", "3" ], [ "\nI have used potassium metal in glass containers stored in mineral oil for 30 years. You are correct that there is residual oxygen in mineral oil that will \"rust (oxidize)\" the metal, but I have never seen the yellow/orange superoxide layer that some have described. If you need it unreacted, then it needs to be stored so as to not come in contact with oxygen. If you are using it to remove residual oxygen from a glove box, then you probably don't want to even open it until you get ready to use it and then open it in the glove box. If you are using it to demonstrate periodic reactivity, then storing it in mineral oil in a sealed glass container has worked well for me over hundreds of demos for three decades with no issues, even with potassium that has been stored this way for three or more years.\n\n\n", "2" ] ]

[ [ "\nThanks for the edit - I see now what you meant.\n\n\nThe short answer for why we \"need\" Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does **not** necessarily decrease. Internal energy is the [thermodynamic potential](http://en.wikipedia.org/wiki/Thermodynamic_potential) (wiki link) for constant $S$ and $V$ - this makes sense if you think about it:\n\n\nIf the internal energy decreases, then heat must be released by the system, and/or work must be being done on the surroundings. However, real spontaneous processes happen where heat is absorbed and work is not done on the surroundings - the reason is that if the net entropy increase is greater than the enthalpy decrease, the process will still be spontaneous.\n\n\nThe equation for $\\mathrm{d}U$ includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does.\n\n\nThe other reason we \"need\" Gibbs energy is that it also tells us how much \"thermodynamically useful\" work we can get out of a process.\n\n\nMathematical Explanation:\n\n\nThe two inequalities that you started with only represent spontaneous process under very specific conditions:\n\n\n$$\\mathrm{d}U < 0$$\n\n\nis true for spontaneous conditions only when $S$ and $V$ are constant. The reason for this is that we define spontaneity as the direction in which total entropy of the universe increases - it is at the root of our concept of \"spontaneous\" vs. \"nonspontaneous.\"\n\n\nFor example, to arrive at this expression for $\\mathrm{d}U$, we start with the [Clausius inequality](http://en.wikipedia.org/wiki/Clausius_theorem):\n\n\n$$\\mathrm{d}S - \\frac{\\mathrm{d}q}{T} \\geq 0$$\n\n\nThis is just a way of saying that the net change of the entropy of the universe must be greater than or equal to zero for any process, and that the net entropy change of the universe is given by the change for the system ($\\mathrm{d}S$) plus the change in surroundings, $-\\frac{\\mathrm{d}q}{T}$.\n\n\nTo turn this into something useful, we make a restriction: constant volume. This lets us assume no $pV$ work, and we can therefore write:\n\n\n$$\\mathrm{d}S - \\frac{\\mathrm{d}U}{T} \\geq 0$$\n\n\nand\n\n\n$$T\\,\\mathrm{d}S \\geq \\mathrm{d}U$$\n\n\nsince $\\mathrm{d}U = \\mathrm{d}q$ at constant $V$.\n\n\nIf we then assume no entropy change for the *system*, we can also write:\n\n\n$$\\mathrm{d}U \\leq 0$$\n\n\nIn other words, the restriction of constant $S$ and $V$ are *built-in* into the inequality. You can substitute in other thermodynamic variables to get $\\mathrm{d}U$ in terms of $T$ and $p$, but it doesn't tell you anything about spontaneity under conditions other than constant $S$ and $V$.\n\n\nGibbs energy is useful because the inequality $\\mathrm{d}G < 0$ is true for spontaneous processes under constant $T$ and $p$ conditions, which as you mentioned, are easy to achieve in a lab.\n\n\nThe other thermodynamic potentials are derived in similar ways - by holding different variables constant and seeing what happens when change in entropy is greater than zero.\n\n\n", "10" ] ]

[ [ "\nThe article by Martin Chaplin 'Water's Hydrogen Bond Strength' (arXiv:[0706.1355](https://arxiv.org/abs/0706.1355) **[cond-mat.soft]**, no evidence of peer review) is discussing how changes to the properties of water would alter life. \n\n\nIf the hydrogen bonding in water was stronger, there would be less ionization (less $\\ce{H+}$ and less $\\ce{OH-}$).\n\n\nIf the hydrogen bonding in water was weaker, there would be more ionization (more $\\ce{H+}$ and more $\\ce{OH-}$).\n\n\n", "4" ] ]

[ [ "\nThe statement in the title question \"internal energy [can] always be expressed as the product of the constant volume heat capacity and the change in temperature\" would only apply to a [calorically perfect gas](http://en.wikipedia.org/wiki/Perfect_gas#Thermally_and_calorically_perfect_gas).\n\n\nIf you instead said \"an infinitesimal change in internal energy can be expressed as the product of the constant volume heat capacity and an infinitesimal change in temperature\", that is only true for a thermally perfect gas. \n\n\nFor real gas, the heat capacities depend upon both temperature and pressure.\n\n\n", "4" ], [ "\nFor a real gas (and for single phase liquids and solids of constant composition), the effect of temperature and molar volume on molar internal energy is given by:\n$$dU=C\\_vdT-\\left[P-T\\left(\\frac{\\partial P}{\\partial T}\\right)\\_T\\right]dV$$where, in general, Cv is a function of both temperature and pressure. Note that, for an ideal gas, the term in brackets is zero and Cv becomes dependent only on temperature.\n\n\n", "2" ] ]

[ [ "\nHot oxidizing acids, such as nitric acid od sulfuric acid, are the more canonical methods to separate gold from less rare metals.\n\n\n\n> \n> \n> > \n> > My solution I assume contains a large amount of silver.\n> > \n> > \n> > \n> \n> \n> \n\n\nIf so, shouldn't you have noticed a precipitate of silver chloride ($\\ce{AgCl}$) under these conditions?\n\n\n", "2" ], [ "\nI would have used aqua regia but was trying to leave the foils solid. I should have used just nitric acid.I would hate to waste the dissolved metals in the HCL. Wanted to recover the silver and copper.Am I correct to assume if I added the nitric acid the silver chloride would participate.\n\n\n", "0" ] ]

[ [ "\nI'd doubt that exclusive nitration of toluene in para position is possible in homogenous solution.\n\n\nBut imagine to stuff a toluene molecule into a very narrow \"tube\". This tube will block any attack at C-2 (and C-3), while a nitrating agent may approach C-4.\n\n\nExactly this has been done in the vapour phase nitration of toluene with nitric acid over a zeolite ([ZSM-5](http://en.wikipedia.org/wiki/ZSM-5)). See: Kalbasi, R.; Ghiaci, M.; Massah, A. Highly selective vapor phase nitration of toluene to 4-nitro toluene using modified and unmodified H3PO4/ZSM-5. *Applied Catalysis A: General* **2009,** *353* (1), 1–8. [DOI: 10.1016/j.apcata.2008.10.013](https://doi.org/10.1016/j.apcata.2008.10.013).\n\n\n", "7" ], [ "\nI think It can be done. But it's a rather long process.First prepare nitrobenzene from benzene by nitration.Then, convert it to aniline by adding a reducing agent(preferably Sn/Hcl).Then, Convert the NH2 group to NHCOCH3 by adding ethanoic anhydride in acetic acid(Which is a weaker activating group than NH2 but a stronger one as compared to CH3).Then, Add the methyl group by friedel-crafts alkylation. Then, convert NHCOCH3 back to NH2 by hydrolysis.Then, convert aniline to diazonium chloride by adding NaNo2 in Hcl. Finally convert diazonium chloride to Nitro group by Adding NaNo2/Cu. U get p-Nitrotoluene. But, I am not too sure this is acceptable.\n\n\n", "-1" ] ]

[ [ "\nLet's do a simple generic example. We have a reaction:\n\n\n$$\\ce{2A <=>[k\\_1][k\\_{-1}] B}$$\n\n\nExperimentally we determine that the rates of the forward and reverse reactions are:\n\n\n$$\\text{Rate}\\_1=k\\_1[\\ce{A}]^2$$\n$$\\text{Rate}\\_{-1}=k\\_{-1}[\\ce{B}]$$\n\n\nAt equilibrium:\n\n\n$$\\text{Rate}\\_1=\\text{Rate}\\_{-1}$$\n$$k\\_1[\\ce{A}]^2=k\\_{-1}[\\ce{B}]$$\n\n\nSince the rate constants are **constants** we can combine them into the equilibrium constant to generate the law of mass action:\n\n\n$$K\\_{eq}=\\frac{k\\_1}{k\\_{-1}}=\\frac{[\\ce{B}]\\_{eq}}{[\\ce{A}]\\_{eq}^2}$$\n\n\nNow, if $k\\_1=2k\\_{-1}$, we have a value of the equilibrium constant, which gives us the relationship between $[\\ce{A}]\\_{eq}$ and $[\\ce{B}]\\_{eq}$ that defines equilibrium:\n\n\n$$K\\_{eq}=\\frac{k\\_1}{k\\_{-1}}=\\frac{2k\\_{-1}}{k\\_{-1}}=2$$\n$$K\\_{eq}=\\frac{[\\ce{B}]\\_{eq}}{[\\ce{A}]\\_{eq}^2}=2$$\n$$2{[\\ce{A}]\\_{eq}^2}=[\\ce{B}]$$\n\n\nIn this case, only concentrations of $\\ce{A}$ and $\\ce{B}$ that satisfy $2{[\\ce{A}]\\_{eq}^2}=[\\ce{B}]$ constitute equilibrium. Other combinations of concentrations are not at equilibrium and therefore the forward and reverse rates are not equal. \n\n\nLet's see how this works. An equilibrium situation is $[\\ce{A}]=1\\text{ M}$ and $[\\ce{B}]=2\\text{ M}$ because:\n\n\n$$\\frac{[\\ce{B}]\\_{eq}}{[\\ce{A}]\\_{eq}^2}=\\frac{2}{1^2}=2$$\n\n\nThe rates are:\n\n\n$$\\text{Rate}\\_1=k\\_1[\\ce{A}]^2=2k\\_{-1}[\\ce{A}]^2=2k\\_{-1}(1^2)=2k\\_{-1}$$\n$$\\text{Rate}\\_{-1}=k\\_{-1}[\\ce{B}]=k\\_{-1}(2)=2k\\_{-1}$$\n\n\nHowever, what if $[\\ce{A}]=[\\ce{B}]=2\\text{ M}$?\n\n\n$$\\frac{[\\ce{B}]\\_{eq}}{[\\ce{A}]\\_{eq}^2}=\\frac{2}{2^2}=\\frac{1}{2}\\ne K\\_{eq}$$\n$$\\text{Rate}\\_1=k\\_1[\\ce{A}]^2=2k\\_{-1}[\\ce{A}]^2=2k\\_{-1}(2^2)=8k\\_{-1}$$\n$$\\text{Rate}\\_{-1}=k\\_{-1}[\\ce{B}]=k\\_{-1}(2)=2k\\_{-1}$$\n\n\nThe concentrations, when plugged into the mass action expression, do not match the relationship defined by the equilibrium constant. More importantly, **the rates are no longer equal**. In this example, the rate of the forward reaction is now 4 times the rate of the reverse reaction. This is how equilibrium is reestablished. The forward reaction is going faster, but as it does so, $\\ce{A}$ is consumed (slowing down the forward reaction) and $\\ce{B}$ is produced (speeding up the reverse reaction) until the rates are equal again. \n\n\nWe can even do a bit of algebra to figure out what the concentrations of $\\ce{A}$ and $\\ce{B}$ will be once equilibrium is reestablished. Let's have $x$ represent the change in concentration of $\\ce{B}$, so the change in the concentration of $\\ce{A}$ is $-2x$.\n\n\n$$K\\_{eq}=2=\\frac{[\\ce{B}]}{[\\ce{A}]^2}=\\frac{2+x}{(2-2x)^2}=\\frac{2+x}{4-8x+4x^2}$$\n$$2(4-8x+4x^2)=2+x$$\n$$8-16x+8x^2=2+x$$\n$$6-17x+8x^2=0$$\n$$x=0.44695..., 1.6781...$$\n\n\nIn our case, we must use $x=0.44695$ since the other value would lead to a negative concentration for $\\ce{A}$. Thus:\n\n\n$$[\\ce{A}]\\_{eq}=2.00-2(0.44695)=1.11$$\n$$[\\ce{B}]\\_{eq}=2.00+0.44695=2.45$$\n\n\n", "4" ] ]

[ [ "\n* Let’s start with the last question: Calculate $\\Delta\\_\\text{mix}G$ when the pressures are equal in both compartments before mixing. Most physical chemistry textbooks treat this case (including Atkins), and the expression of $\\Delta\\_\\text{mix}G$ is:\n$$\\Delta\\_\\text{mix}G= nRT(x\\_1\\ln x\\_1 +x\\_2\\ln x\\_2)$$ In this exercise, $1$ denotes hydrogen, $2$ denotes nitrogen $n=6\\ \\text{mol}$, $x\\_1=\\frac{2}{6}$ and $x\\_2=\\frac{4}{6}$. So, by substituting these values in the above expression, we find: $$\\Delta\\_\\text{mix}G= 6\\ \\text{mol}\\times 8.314\\ \\frac{\\text{J}}{\\text{mol}\\cdot\\text{K}}\\times 298\\ \\text{K}\\times(\\frac{2}{6}\\ln\\frac{2}{6} +\\frac{4}{6}\\ln\\frac{4}{6})= -9462\\ \\text{J}$$ As $\\Delta\\_\\text{mix}G<0$ at constant temperature and pressure, this means that the mixing of two ideal gases is a spontaneous process .\n\n\n\t+ Let’s now treat the first question with two different pressure in the two compartments. The initial Gibbs energy is $$G\\_\\text{i}=n\\_1(\\mu\\_1^0 +RT\\ln p\\_1)+n\\_2(\\mu\\_2^0 +RT\\ln p\\_2)$$ Where $n\\_1=2\\ \\text{mol}$, $n\\_2=4\\ \\text{mol}$, $p\\_1=2\\ \\text{atm}$, $p\\_2=3\\ \\text{atm}$. $$G\\_\\text{i}=2\\ \\text{mol}\\times(\\mu\\_1^0 +RT\\ln2)+4\\ \\text{mol}\\times(\\mu\\_2^0 +RT\\ln3)$$\n\tWhen the partition is removed, and each gas occupies the volume $V\\_1+V\\_2$.\n\tThe partial pressure of hydrogen is $$p^\\prime\\_1=\\frac{2\\ \\text{mol}\\times RT}{V\\_1+V\\_2}$$\n\tThe partial pressure of nitrogen is $$p^\\prime\\_2= \\frac{4\\ \\text{mol}\\times RT}{V\\_1+V\\_2} $$ On the other hand: $$V\\_1=\\frac{2\\ \\text{mol}\\times RT}{2\\ \\text{atm}}=1\\frac{\\text{mol}}{\\text{atm}}\\times RT$$ $$V\\_2=\\frac{4\\ \\text{mol}\\times RT}{3\\ \\text{atm}}=\\frac{4}{3}\\frac{\\text{mol}}{\\text{atm}}\\times RT$$ i.e. $$V\\_1+V\\_2=\\frac{7}{3}\\frac{\\text{mol}}{\\text{atm}}\\times RT$$ \n\tSo $$p^\\prime\\_1=\\frac{2\\ \\text{mol}\\times RT}{(7/3\\ \\text{mol/atm})RT}=\\frac{6}{7}\\ \\text{atm}$$ $$p^\\prime\\_2= \\frac{4\\ \\text{mol}\\times RT}{(7/3\\ \\text{mol/atm})RT}=\\frac{12}{7}\\ \\text{atm} $$ The final Gibbs energy is $$G\\_\\text{f}=n\\_1(\\mu\\_1^0 +RT\\ln p^\\prime\\_1)+n\\_2(\\mu\\_2^0 +RT\\ln p^\\prime\\_2)$$ $$G\\_\\text{f}=2\\ \\text{mol}\\times(\\mu\\_1^0 +RT\\ln\\frac{6}{7})+4\\ \\text{mol}\\times(\\mu\\_2^0 +RT\\ln\\frac{12}{7})$$ The Gibbs energy of mixing is the difference between $G\\_\\text{f}$ and $G\\_\\text{i}$ \n\t$$\\Delta\\_\\text{mix}G= 2\\ \\text{mol}\\times RT\\ln\\frac{6/7}{2} +4\\ \\text{mol}\\times RT\\ln\\frac{12/7}{3}$$ $$\\Delta\\_\\text{mix}G= 2\\ \\text{mol}\\times RT\\ln\\frac{6}{14} +4\\ \\text{mol}\\times RT\\ln\\frac{12}{21}$$ $$\\Delta\\_\\text{mix}G= 2\\ \\text{mol}\\times 8.314\\frac{\\text{J}}{\\text{mol}\\cdot\\text{K}}\\times 298\\ \\text{K}\\times(\\ln\\frac{3}{7} +2\\times\\ln\\frac{4}{7})= -9744\\ \\text{J}\\approx{-9.7\\ \\text{kJ}}$$ \n\tIn this case, the value of $\\Delta\\_\\text{mix}G$ is the sum of two contributions: the mixing itself and the changes of pressure of the two gases to their final total pressure.\n\n\nI hope it’s clear now.\n\n\n", "4" ] ]

[ [ "\nUsually with the *reaction mechanism* the sequence of events (electron transfers) is indicated, while the term *reaction pathway* usually refers to the reaction coordinate diagram associated with the reaction, i.e. the change in energy with an extra notion on the transition states.\n\n\n", "6" ], [ "\nAs far as my knowledge is concerned when we talk about Reaction Mechanism then we mean the Steps through which the reactants are passed and reach to the Product. Say for example the making of water from Hydrogen and Oxygen which involves 40 steps and at the end we get what we know H2 +1/2 O2 giving H2O\nAnother example can be the chlorination of Methane to form CCl4.\nWhile in Reaction pathway we talk about the inside of the reactions involved like talking about electrons movement .\n\n\n", "0" ], [ "\nNo, Khair Muhammabd. Go through the reaction mechanism. Actually, the reaction mechanism is about sequence and detailed understanding of elementary reactions(that are here called elementary steps in the term of Composite Reaction) for conversion of reactants to the desired products.\nIt is accompanied with the formation of Reaction Intermediate at each elementary step(except last one). Where the overall reaction can be obtained by adding all elementary reactions together. So as the overall rate of reaction can be obtained by adding the rates of all elementary steps.\nBut what happens, every composite reaction has one elementary step (Mostly 1st step) that is carried out very fast as compared to other steps. Say, T1=1sec for 1st step. and T2=0.0000000001 sec reaction.\nSo the total time of the composite reaction will almost be equal to first step (T(total)=T1) while T2 is neglected. Basically the first step is called Rate dertermining step(RDS).\nSo overall reaction rate will be taken equal to the rate of RDS.\nR(r\\*n)=R(RDS).\nWhile I know nothing about Reaction Pathways, still going through it.\n\n\n", "-1" ] ]

[ [ "\nTo complement entropid's answer, let's look at the formulation of $U$ that is slightly more useful:\n\n\n$$\\begin{align}\nU &= q + w \\\\\n\\mathrm{d}U &= \\mathrm{d}q + \\mathrm{d}w \\\\\n&= C\\,\\mathrm{d}T - p\\,\\mathrm{d}V\n\\end{align}$$\n\n\nAt constant volume, $\\mathrm{d}V = 0$, so $p\\,\\mathrm{d}V = 0$\n\n\n$$\\begin{align}\n\\mathrm{d}U &= C\\_V\\,\\mathrm{d}T - 0 \\\\\nC\\_V &= \\frac{\\mathrm{d}U}{\\mathrm{d}T}\n\\end{align}$$\n\n\nHowever, this notation assumes that $U$ is only a function of $T$, as in $U(T)$. It is not. In this version, $U$ is a function of $V$ and $T$: $U(V, T)$. Note that we do not need $p$ as a variable, since it is dependent only on $V$ and $T$ as well - for an ideal/perfect gas $p(V,T)=\\frac{nRT}{V}$. Thus, we need partial derivative notation so that we know we are not trying to take the total derivative of $U$:\n\n\n$$C\\_V = \\left(\\frac{\\partial U}{\\partial T}\\right)\\_{\\!V}$$\n\n\n", "3" ], [ "\nThe difference in **internal energy** ($\\mathrm{d}U$) is a function of many different variables:\n\n\n$$\\mathrm{d}U = T\\,\\mathrm{d}S - p\\,\\mathrm{d}V \\!$$\n\n\nTherefore you must specify – via a partial derivative – with respect to which of the variables you are doing the derivative, otherwise you would need a *total derivative*, which is not what the equation requires.\n\n\n", "3" ] ]

[ [ "\nThe turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)\n\n\nIt might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? Most people would express how fast the factory works in terms of \"widgets per hour\". It's the same with enzymes - it's most natural to describe how fast the enzyme works in terms of \"turnovers per second\". It's just that the \"turnover\" is implied and is viewed as \"dimensionless\", so it's left off, leaving you simply with \"s-1\" as a unit. \n\n\n(Note that even though the \"turnover\" is left off, it's still *very* important to know it's there, and to know what, exactly, one \"turnover\" is. In complicated reactions there is sometimes different reasonable ways to count what \"one turnover\" is, and different definitions could result in a 2, 3, 4, 5, etc. fold difference in value. Those vexing numeric differences become easy to explain once you realize *this* researcher is using the formation of one molecule of this product as \"one turnover\", but *that* researcher is using the disappearance of one molecule of that substrate as \"one turnover\".)\n\n\nNaively, one could invert the ratio to come up with a value in time, rather than in 1/time. For example, if a factory makes 5 widgets per hour, it takes them 0.2 hours to make a widget ... sort of. One issue you run into is that that's the time *on average*. In a stochastic process like enzyme turnover, it's highly unlikely that any single turnover will take exactly the time specified. Some will take longer, some will be quicker. Only for a large number of turnovers in aggregate does that time value have meaning. Saying that \"a turnover takes 5 s\" gives a false impression about individual turnovers in a way that saying \"this enzyme works at 0.2 turnovers per second\" doesn't.\n\n\nAdditionally, in the case of multi-step reactions, saying that \"a turnover takes 5 s\" may be incorrect. The entire pathway may take much more that that. It's just that once the pathway has reached steady state it's only the rate limiting step which takes 5 s. For example, take a pathway of 20 steps, 19 of which take 4 s each, but one which takes 5 s. It's going to take (19\\*4 + 5) = 81 seconds for any given substrate molecule to start at the beginning of the pathway and make it to the end. However, once the pathway is full and each step is handing off products to the next step, the pathway will be producing product at a rate of 1 every 5s. (This, of course, doesn't apply to enzymes where all reactions take place at the same active site, but it's a consideration in the general case, which explains why 1/s is the preferred unit for reaction rates.)\n\n\n", "9" ], [ "\nYou were actually describing the turnover frequency of a catalyst, and frequency has the unit hertz which is 1/s. Turnover number is a dimensionless value: number of molecules of reagent that a catalytic site can convert before becoming inactivated.\n\n\n", "3" ], [ "\nAs you say this represents \"conversions per unit of time\". \n\n\nThe mathematical expression for \"per time unit\" is 1/(time unit), and seconds is a particular time unit. \n\n\n", "2" ], [ "\nVmax=kcat \\* ([E total])\nIf the units for Vmax are M/s i.e. s-1 and you divide Vmax by the concentration of total enzyme which has units in M, the Ms will cancel out, leaving you with 1/s or s-1\n\n\n", "0" ] ]

[ [ "\nThe shielding model works as follows:\n\n\n ([source](http://en.wikipedia.org/wiki/Effective_nuclear_charge))\n\n\nSo it is a mixture of attraction and pushing from protons and non-valence electrons, in accord with the [Coulomb's law](http://en.wikipedia.org/wiki/Coulomb%27s_law#The_law). The prediction of the effective core charge is not an easy task - you will have to use quantum chemistry methods (e.g. [Hartree-Fock](http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method)) to predict those, like its done in [this reference](http://scitation.aip.org/content/aip/journal/jcp/38/11/10.1063/1.1733573).\n\n\nI think its best viewed as a pragmatic model to boil down the higher, not very graphic, quantum theories, to one number, the effective nuclear charge:\n$$ Z\\_\\text{eff} = Z - S$$\n\n\n", "3" ], [ "\nIts actually Coulombic repulsion between electrons and attraction between electron and protons.\n\n\nIt can be said shelding if you are considering the last electron. But if u consider any electron in the lower shells the electron in the upper shells will push the electron towards the nucleus and it won't be called shilding.\n\n\n", "0" ] ]

[ [ "\n[Potassium chlorate](http://en.wikipedia.org/wiki/Potassium_chlorate) is a source of oxygen. After heating, it decomposes to $\\ce{O2}$ and $\\ce{KCl}$:\n$$\\ce{4 KClO3 → KCl + 3 KClO4}$$\n$$\\ce{KClO4 → KCl + 2O2}$$\n\n\nThe gummy bear is mainly composed of sugar and other carbohydrates. Those carbohydrates will react with oxygen, [combustion](http://en.wikipedia.org/wiki/Combustion#Chemical_equations) occurs. For example, glucose will react in this manner:\n$$\\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O}$$\n\n\nIf there is any material present which does not burn, such as $\\ce{H2O}$, the temperature will not rise as high. For gummy bears the reaction works spectacularly because they are mainly carbohydrates (>70%).\n\n\nAn apple, for example, has only ~13% carbohydrates – unless you dry it, of course. On the other hand, [this video](https://www.youtube.com/watch?v=xXts1AlpR_g#t=36) on YouTube is an example of how sugar itself reacts violently with potassium chlorate.\n\n\n", "11" ] ]

[ [ "\n$\\ce{N2H2}$ is a very unstable molecule, prone to disproportionation into hydrazine and nitrogen. $\\ce{N-N}$ bonds except one in $\\ce{N2}$ molecule are unstable and generally unfavorable. So, no there are no polymers with nitrogen as a sole component of the main chain. \n\n\nSeveral classes of compounds of 3-row elements demonstrate tendency for oligomerization, namely $\\ce{(HP)\\_{n}}$ derivatives usually exist as a ring. It is possible to obtain metastable at room temperature plastic sulfur, containing linear sulfur chains terminated by free radicals. \n\n\nHowever, generally when in search of easily polymerizing molecule that produce main chain containing elements other than carbon, one usually have to look into compounds producing chains with two elements. The most common and known one is formaldehyde $\\ce{CH2=O}$, producing $\\ce{HO-(CH2-O)\\_{n}-H}$ chains. A less known, but interesting example is [hexachlorotriphosphazene](http://en.wikipedia.org/wiki/Hexachlorophosphazene), the source material for production of [Polyphosphazenes](http://en.wikipedia.org/wiki/Polyphosphazene).\n\n\nIn general, metastable double bonds are rare for elements other than carbon (sometimes with nitrogen or oxygen partners), so usually reactions intended to produce such bonds, say, $\\ce{Si=Si}$ usually end with olygomer instead, typically of ring structure. Moreover, single homoelemental bonds are usually much weaker for elements other than CHNOB, so typically when a chains containing other elements is occurred, it is usually hetero-atomic chain, produced by means other than olygomerization of double-bond containing monomer. For example, an extremely wide-used family of polymers (polysiloxanes, or [silicones](http://en.wikipedia.org/wiki/Silicone)) is produce by hydrolisys of dialkyldicholorsilanes $\\ce{R2SiCl2}$.\n\n\n", "4" ], [ "\nTo my knowledge, diazene (diimine, $\\ce{N2H2}$) does not polymerize but simply decomposes (to hydrazine and nitrogen). Even in the polymerization azobenzene-substituted vinylethers (consider azobenzene as diazene in with both hydrogen atoms are replaced by phenyl rings), the $\\ce{N=N}$ double bond does not participate in the polymerization.\n\n\nBut you are right, a lot of **copolymers** consists of monomers in which the polymerization does not take place at $\\ce{C=C}$ bonds.\n\n\nThink in the reaction of $\\ce{\\alpha,\\omega}$-diols\n\n\n* with phosgene ($\\ce{COCl2}$) to polycarbonates\n* with diisocyanates to polyurethanes\n\n\nor the reaction of dianhydrides with diamines to polyimides.\n\n\n", "3" ], [ "\nYou're right, the monomer needn't be an alkene. Non-alkenes with double bonds can also form addition polymers. In fact, the monomer needn't always have a double bond either! Surprised? Take a look at this:\n\n\n\n> \n> [](https://i.stack.imgur.com/ytVhg.png)\n> \n> \n> \n\n\n(Image source: [chemwiki.ucdavis.edu](http://chemwiki.ucdavis.edu/@api/deki/files/7295/Nylon_6_formation.png))\n\n\n", "2" ] ]

[ [ "\nFirstly, $\\ce{NaBH4}$ is not a Lewis acid because it cannot accept any more electrons. Aqueous $\\ce{NaBH4}$ is a commonly used reducing agent often for reduction of carbonyls to their respective alcohols.\n\n\nThe mechanism of Lewis acid catalysis depends on the reaction but often Lewis acids are used to generate cationic electrophiles such as in the Friedel-Crafts reaction. \n\n\n", "1" ] ]

[ [ "\nI think that the second explanation is the correct one. A simple experiment can prove it: if you place a porous cover over the pure solvent, its vapour pressure will not change. In alternative, you can place a few corks and observe the same result (no variation in the vapour pressure). In both cases surface sites are blocked, but the vapor pressure will obviously remain unaltered. Therefore the reason cannot be that the solute molecules are hindering the solvent molecule escape by occupying the solution surface. \n\n\nThe force that is driving the solvent molecules to the vapour phase is the difference between the entropy of the liquid and the vapour phases: when the molecule escape the liquid there is an entropy gain. Adding a non-volatile solute will increase the entropy of the liquid phase without having an effect on the entropy of the vapour phrase (given that the solute is non-volatile). This will lower the difference in entropy between the two phases, which is the driving force that make vaporisation happen, therefore there will be a lower number of solvent particles in the vapour phase and a consequent lower vapour pressure.\n\n\nTextbooks like *Chemistry: The Molecular Nature of Matter and Change* (Silberberg, Amateis) or *Physical Chemistry for the Biosciences* (Chang) provides the same explanation. In addition, there is a short paper/article on this topic[[1](http://dx.doi.org/10.1021/ed075p787)] where the author highlights how only very few freshman chemistry texts give an adequate explanation of this phenomenon by making use of a entropy-based argument.\n\n\n     1 [Vapor Pressure Lowering by Nonvolatile Solutes, G. D. Peckham, J. Chem. Educ., 1998, 75 (6), 787](http://dx.doi.org/10.1021/ed075p787) \n\n\n", "11" ], [ "\nThis is similar to adsorption on a surface: van der Waals and other forces (e.g. hydrogen bond) bind the solute to the solvent, just as a gas may be bound to a solid. The lowest energy state depends on the strength of the solute-solvent bonds (adhesion) and the solute-solute bonds (cohesion) as opposed to random molecular motion causing evaporation. See <http://en.wikipedia.org/wiki/Van_der_Waals_force> and <http://www.chem.purdue.edu/gchelp/solutions/eboil.html>\n\n\n", "0" ] ]

[ [ "\nFirstly, copper would have to be in the 4+ oxidation state for there to be $\\ce{Cu(SO4)2.5H2O}$, so this isn't a realistic question.\n\n\nBut, no it won't be 5M, because you need to know the final volume of the solution to calculate molarity. If you add 345.55 grams of just about anything I can think of to 200 mL of water, the volume will increase substantially. So not only will the 90 grams of hydrate water add to the volume of the final solution, generally you need to consider how all added substances contribute the final volume. \n\n\nIn some cases, such as magnisium sulfate, a small amount added to water can actually decrease the volume.\n\n\n", "2" ], [ "\nJust to be clear, I don't think you can actually make a 5M solution of copper sulfate, but I'll ignore solubility for this answer.\n\n\nWhen you make a solution of a given concentration from a solid you generally dissolve the solute (your copper sulfate pentahydrate) in a relatively small amount of water and then dilute it to the final volume. Here, you would dissolve your 249g (1 mole) of copper sulfate pentahydrate in sufficient water to dissolve it but less than 200mL total, then carefully add enough water to reach a solution volume of 200mL, yielding a 5M solution. \n\n\nThe water of hydration would indeed become part of the solution. If instead you began this process with anhydrous copper sulfate and used 1 mole of that (159.61g) then you would dissolve it in enough water to make the solution but less than 200mL total then carefully add enough to reach your final solution volume and concentration, 200mL at 5M.\n\n\nThe difference here would be that you would have to add 90g of additional water to prepare a solution from the anhydrous compound as compared to the pentahydrate.\n\n\nIf you prepare a solution by adding a fixed mass of solute to a fixed volume of water, you are never going to make the desired concentration and volume.\n\n\n", "2" ], [ "\nI realize this may not be what you are asking, but this is worth pointing out. The most common source for error in solution concentration comes from neglecting to account for the mass of the hydrate. If you weight out an amount of a hydrate, and use the molecular mass of the anhydrous, your final concentration will be erroneous. For example, 5g of $\\ce{CuSO4 \\dot{} 5 H2O}$ will not yield the same as 5g $\\ce{CuSO4}$ (anhydrous). When diluted to a volume of 1 liter:\n\n\n$$\\frac{\\ce{5g\\,CuSO4.5H2O}}{1} \\frac{\\ce{1mol\\,CuSO4.5H2O}}{249.685\\ce{g}}\\frac{1}{\\ce{1L}}=0.02\\ce{M\\,CuSO4}$$\n\n\n$${\\frac{\\ce{5g\\,CuSO4}}{1} \\frac{\\ce{1mol\\,CuSO4}}{159.6086\\ce{g}}\\frac{1}{\\ce{1L}}=0.03\\ce{M\\,CuSO4}}$$\n\n\n", "1" ], [ "\nOf course, the water in the hydrate affects the final concentration.\n\n\n* Molar mass of $\\ce{CuSO4.5H2O}$ is $\\pu{249.5 g/mol}.$\n* Amount of substance is\n$$n = \\frac{345.55}{249.5} = \\pu{1.385 mol}$$\n* Concentration:\n$$C = \\frac{1.385}{0.2} = \\pu{6.925 mol/L}$$\n\n\nBut, this is only an illustrative example. **It's not real.** The solubility of $\\ce{CuSO4.5H2O}$ is $\\pu{320 g/L}$ at $\\pu{20 °C}.$\n\n\n", "0" ], [ "\nThe molar mass of copper (ii) sulfate penta-hydrate is $\\pu{249.69 g/mol}$ and that for anhydrous copper(ii) sulfate is $\\pu{159.61 g/mol}$. Thus, in $1$ mole of the hydrated substance there are $\\pu{159.61 g}$ of copper (ii) sulfate and $5 \\times \\pu{18.00 g}$ (i.e. $\\pu{90.00 g}$) of molecules of water of crystallization. \n\n$\\pu{5M}$ copper (ii) sulfate solution means $5$ moles of the substance is contained in $1$ liter of solution. The mass of copper (ii) sulfate in $\\pu{5M}$ of the substance is $5 \\times \\pu{159.61 g/L} = \\pu{798.05 g/L}$. \n\nThis means that to prepare a $\\pu{5 M}$ copper sulfate solution, weigh out $\\pu{798.05 g}$ of the anhydrous substance and dissolve in $\\pu{1 L}$ volumetric flask with distilled or de-ionized water, making up to volume by filling up to the thin ring mark at neck of flask.This can be scaled down by a factor of $5$ by dissolving $\\pu{159.61 g}$ of the substance in $\\pu{0.2 L}$ ($\\pu{200 mL}$) solution, using a $\\pu{200 mL}$ volumetric flask.\n\n\nNow if the penta-hydrate is to be used to achieve a concentration of $\\pu{5 M}$ copper sulfate solution, from the previous we see that $\\pu{159.61 g}$ copper sulfate is contained in $\\pu{249.69 g}$ of the hydrated copper sulfate. Therefore,\n$\\pu{798.05 g}$ of copper sulfate is contained in $(249.69/159.61) \\times \\pu{798.05 g} = \\pu{1248.45 g}$ of the pent-hydrate, i.e. dissolve $\\pu{1248.45 g}$ of the hydrated salt in $\\pu{1L}$.\n\n\nOr if a smaller quantity is desired, say $\\pu{200 mL}$ solution: weigh $(249.69/159.61) \\times \\pu{159.61 g} = \\pu{249.69 g}$ of the penta-hydrate and dissolve in $\\pu{200 mL}$ solution. Care must be taken to add water in small quantities so as not to exceed the ring graduation at neck of flask and the lower meniscus of solution should align with the mark.\n\n\n", "0" ] ]

[ [ "\nWell, *very* long hydrocarbons *can* become twisted around each other. This is definitely known in the polymer literature. Consider some branched polyethylene with ~1,000 repeat units. It's not too hard to imagine that multiple chains can become entangled in much the same way as shoelaces or strings when you mix a lot of them in a box.\n\n\nYou are also right that with large molecules, there are more sites for intermolecular interactions. Consider that there is a \"combination\" effect with lots of sites. That is, to pull two very long linear chains apart, you need to break lots of interactions.\n\n\n\n\n\nNotice that pulling one atom on the bottom away from its neighbor on the top chain will require the movement of *all* atoms on the bottom chain. So it's slightly harder to separate two long chains, not only because there are so many interactions, but also because of the combination. \n\n\nPut simply, the difficulty in separating the two chains is slightly higher than simply the sum of the interactions.\n\n\n", "1" ] ]

[ [ "\nYou have to look which forces act where. You have a force that tries pull the liquid up the capillary, which stems from the gain in energy due to adhesion. At the same time, however, gravity acts against this force. In equilibrium, both forces have the same value:\n\n\n$$F\\_\\text{g} = F\\_{\\text{surf}}$$\n$$ mg=\\rho Vg= \\rho R^2\\pi h g=2\\pi R\\sigma\\_s \\cos(\\varphi)$$\n\n\nWhich leads to the formula presented by Babounet in another answer:\n\n\n$$ h = \\frac{2\\sigma\\_s\\cos(\\varphi)}{R\\rho g}$$\n\n\nSo, to boil it down, if you lower the radius $R$, you lower the force due to gravity $F\\_g$, but the force due to surface energy $F\\_\\text{surf}$ does not fall as fast, which leads to a higher height $h$.\n\n\n\n([source](http://www.engineeringarchives.com/les_fm_capillaryeffect.html), $W=F\\_g$ here, naming isn't perfect, $F\\_{\\text{surf}}$ is not drawn into) \n\n\n", "4" ], [ "\nA meniscus is formed by the opposing forces of adhesion between the fluid and the walls of the container drawing the fluid up and gravity pulling the fluid down. The adhesive forces are proportional to the diameter of the tube while the gravitational effect due to the liquid's weight is proportional to the square of the diameter. This results in the greater height of the meniscus for narrower tubes. \n\n\n", "3" ], [ "\nAccording to the Jurin's Law \n\n\n$ h = \\frac{2\\gamma\\cos(\\theta)}{r\\rho g} $\n\n\nWith : \n\n\n$h$ the height of the fluid \n\n$\\gamma$ the surface tension \n\n$\\theta$ the contact angle with the tube \n\n$r$ the tube radius \n\n$\\rho$ the liquid density \n\n$g$ the gravitational constant\n\n\nSo, the $h$ is inversly proportional to $r$. In other words, the larger the tube, the lower the liquid climbs.\n\n\nNote that this law is only valid if $r < \\lambda\\_\\text{c}$ with $\\lambda\\_\\text{c} = \\sqrt{\\frac{\\gamma}{\\rho g}} $\n\n\n", "1" ] ]

[ [ "\nThough the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly. The [Wikipedia article for cyanoacrylates](http://en.wikipedia.org/wiki/Cyanoacrylate#Properties) shows the polymerization more clearly than I can easily explain in words. The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool provides a large number of sites for the glue to cure.\n\n\nWhen I read your question I was curious; I had never heard of this before. I searched the internet and was only able to find one or two examples of people actually trying this successfully, and both of them looked dubious. I've spilled superglue on cotton shirts and pants any number of times and while it cures almost instantly, it has never caught my shirt on fire. \n\n\nI decided to try it myself, and I soaked a cotton ball in superglue to see what would happen. The glue cured extremely rapidly, producing very irritating (colorless) fumes, and the mass of glue and cotton got warm, but it was nowhere near hot enough to ignite - my tap runs hotter than the glue got. There may be a particular type of glue that cures significantly more exothermically, but I'm inclined to think that this is an internet hoax, and [Popular Science agrees](http://www.popsci.com/diy/article/2012-09/gray-matter-playing-fire).\n\n\n", "17" ], [ "\nCyanoacrylates include methyl 2-cyanoacrylatecommonly sold under the trade names \"Super Glue\".In general, cyanoacrylate (consists of monomers of cyanoacrylate molecules) is an acrylic resin that rapidly polymerises in the presence of water (specifically hydroxide ions), forming long, strong chains, joining the bonded surfaces together. Because the presence of moisture causes the glue to set, exposure to normal levels of humidity in the air causes a thin skin to start to form within seconds, which very greatly slows the reaction. Because of this cyanoacrylate is applied thinly, to ensure that the reaction proceeds rapidly and a strong bond is formed within a reasonable time.\n\n\n\n\n\nSo to sum up, in order to start the reaction off some water is normally needed.. damp things stick better/quicker than dry ones and the glue goes hard faster on a humid day.\n\n\nIn cotton wool, which is made of cellulose, a polymer of sugar molecules, there are lots and lots of hydroxy (-OH or alcohol groups), which can start the reaction in the same way as the water does, only because there are lots of them they can start many more reactions at once.\n\n\nSince the reaction gives out heat, the cotton bud therefore gets hot (and as it becomes hotter so the reaction goes faster etc), and it may get hot enough to catch fire.\n\n\n", "11" ] ]

[ [ "\nYou are certainly thinking in the right direction. Fluorine strongly prefers the axial position (for reasons described [here](https://chemistry.stackexchange.com/questions/18427/why-f-replaces-axial-bond-in-pcl%E2%82%85/18544#18544)), so at low temperature, the isomer that you've drawn on the right (2 axial and 2 equatorial fluorines) is **the only isomer present**. As a result we have an $\\ce{A2B2X}$ spectrum. The 2 equivalent equatorial fluorines (spin = 1/2) split the phosphorous signal into a triplet. Each line of this triplet is further split into a triplet by the 2 equivalent axial fluorine's. BTW, the same reasoning that explains fluorines axial preference, also explains why generally $\\ce{J(P-F)\\_{equatorial}}$ > $\\ce{J(P-F)\\_{axial}}$\n\n\nAt room temperature, [pseudorotation](http://en.wikipedia.org/wiki/Pseudorotation) is rapid (more thermal energy is available and the barrier to pseudorotation can be surmounted) and all fluorines become equivalent on the nmr timescale. Therefore, at room temperature and above a quintet is observed in the $\\ce{^{31}P}$ nmr spectrum.\n\n\n", "2" ] ]