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[ [ "\nIt is very easy to purify the amino compounds from the reaction mixture of reduction of nitro- to amine- group. Firstly, we will do workup after completion of the reaction by extraction method with organic solvents either with DCM or with ehtylacetate, then organic layer wlii be separated, and then dried with anhydrous sodium sulphate and the solvent will be evaporated under reduced pressure by applying rotary evaporator. Then after making slurry by adding little amount of silica gel we will pack slurry in Column-chromatography silica gel (100-200 mesh) by using eluent as Ethylacetate:Hexane (20% to 60%), the amino- group compond will be purify. This is so because nitro- group is less polar compared to anino- group. So, nitro- compound will come first from column chromatograhy and amino- compound will come letter in pure form.\n\n\n", "1" ] ]

[ [ "\nI don't have a fireplace or wood stove, so I don't have a practical technique that I know works to prevent soot buildup on glass, but I can answer some of your questions about glass and soot. To start, I should add a caveat here: while many fireplaces use tempered glass windows, wood stoves generally have glass-ceramic windows to withstand the increased heat. As far as surface properties go for this, glass-ceramic isn't dramatically different from normal glass.\n\n\n\n> \n> does each glass (especially my one) have pores with a size that starch\n> particles can fit in? I read somewhere that starch has a diameter\n> between 2-170 µm depending on the type.\n> \n> \n> \n\n\nWhile special techniques exist for making porous glasses, normal glass is not porous and float glass (the standard technique for fabricating flat glass panes) produces quite smooth pieces even without polishing ([hundreds of nm peak-to-peak](http://www.delta-technologies.com/Products.asp?C=13)), so probably whole starch particles are not getting stuck in the surface of the glass and the surface roughness probably has little impact on soot adhesion in this case.\n\n\n\n> \n> I'm making a fire where I burn wood! Why wouldn't the starch simply\n> burn and create itself the soot?\n> \n> \n> \n\n\nUnless the flame is right up to the glass, it probably doesn't get hot enough to really burn starch (if any is even there). That site didn't compare it directly with not applying starch so it's not clear that it makes any difference.\n\n\n\n> \n> How does soot building work? I mean, why does the soot stick so hard\n> at the window? After all it's just little particles flowing around and\n> landing on the window. (I know it's called burning in, but how does\n> the heat work on a already burned particle?)\n> \n> \n> \n\n\nSoot is basically the deposition of incomplete combustion products from a flame. A wood flame appears yellow or orange because tiny particles of mostly carbon are incandescing. There is insufficient oxygen to fully combust the carbon in the fuel to $\\ce{CO2}$, so the combustion results in carbon molecules in various states of oxidation from small polymers to elemental carbon. These species are really only stable as gases within the heat of the flame itself and will thus readily deposit onto cold surfaces, like the window glass. Many modern wood stoves seem to have \"[air-wash](http://www.woodheat.org/maintenance.html)\" systems that flow air past the inside of the window to reduce how much soot can actually reach the surface.\n\n\nWhat is actually deposited as soot is a [complex mixture](http://pubs.rsc.org/en/Content/ArticleLanding/2010/CP/b924344e#!divAbstract), but what is emitted from the flame is related to how much air is available and the temperature of the flame (the two are also connected with each other). If the oxygen supply is increased, more complete combustion is favoured so less carbon species that can deposit as soot are formed. The temperature that is attained influences the composition of soot particles: higher temperatures produce more aliphatic and less aromatic compounds. The practical significance is that reducing the flame intensity by limiting oxygen can greatly increase the production of soot. (and may have some effect on how well it adheres to the glass, though I wasn't able to find anything concrete to that effect)\n\n\nFrom the non-scientific sources I glanced through, it seems to be well known that excessive soot buildup can be due to wet fuel or low airflow and that burning a very hot fire can help make soot easier to remove. As for the corn starch, it may indeed be leaving some deposit that hinders soot adhesion (there are commercial cleaners that claim to leave a film behind that makes soot easier to remove), the act of applying it may be removing existing soot that's too small to see but new soot adheres readily too, or it may do nothing. Perhaps you might want to try applying corn starch to half of a freshly-cleaned window to see if there is a noticeable difference.\n\n\n", "9" ], [ "\nExcellent question. One I've been pondering myself. I moved to Norway 3 years ago. Before then, I'd never had a wood burning stove in any of the places I had lived in. Now, our main sources of heat are wood burning stoves with electrical unit for the coldest times. So, it has been a bit of a learning experience for me and one that I've found fascinating, having had a Biology & Chemistry tertiary education (as well as having slight pyromaniac tendencies!).\n\n\nTo answer your question...\nI've found that soot deposition on the glass doors is dependant on wood composition. More resin means more soot on the door. It also makes the soot difficult to remove. Pine is the worst, closely followed by fir. Hard wood is the best. My personal favourites are beech and elm but both can be difficult to come by. Birch and willow are what I use mostly. Although I've had a bit of a windfall - one of our old elms was blown down in this year's autumnal storms. It is a bit of a shame as I really like the tree but I'll be glad to make use of it. I tend to use a mixture of coniferous and hard woods. We have an excess of the former. They burn very easily and quickly so are good at the start. Hard wood comes later once I've got a good collection of embers and the oven has heated up properly. but I digress... \nStarch is combustible so, after the coating applied and fires have been burnt the chemical structure will be have changed. The process is called The Maillard Reaction. You'll find plenty of articles about that online. The products of the Maillard reaction tend to be brown in colour so the starch film would need to be very thin or you would see it. They also tend to have a hydrophylic nature. If there are indeed pores then these will end up as pockets of hydrophylic Maillard reaction products. Coniferous resin is mainly composed of hydrophobic compounds. (I found this fascinating paper that gives details of them and their structure. Also discusses high temperature treatment products. See text around Fig. 2.1 in \"Chemical Investigations of the Organic Embalming Agents\nEmployed in Ancient Egyptian Mummification\" <https://research-information.bristol.ac.uk/files/34498395/396674.pdf>) Hydrophobic and hydrophylic repel each other, think oil in water. So, if this theory of pores and starch holds true then it will the mutual repulsion that is most likely to be the factor that influences the soot deposition rate. I have to say though - I do doubt it because starch is combustible. There will be plenty of oxygen and heat to facilitate that.\n\n\nAs stated elsewhere, high airflow / oxygen availability prevents soot build up. That is because the resin combustion reaction is allowed to go to full completion. That is why, as you've noticed, a roaring fire at a later time clears the glass. It basically finishes the job.\n\n\nDelivering the appropriate airflow is a fine art, one I've yet to master. It dependant on air pressure and wind as well as how wide the stove's vents are. My step father just burns with his vents full open. He has done this all his life so perhaps he is right. However, imho his method wastes a lot fuel and heat as the wood burns very quickly and draught sucks a lot of the air out of the room which is replaced by cold air from outside. So, I try to find the balance.\n\n\nFor cleaning, I use a hoover with plastic scraper to start (reduces mess), then a steam lance (gets most of it off) and finally oven cleaner (removes the most hard stuck deposits). That works a treat.\nI like the idea of using the ash. Thank you!\n\n\nBy the way - I got to this forum post via google. It was highly ranked. So thanks for asking and thanks for your update.\n\n\n", "3" ] ]

[ [ "\nThe light and heat are created by the exothermic nature of the reaction. If you look up \"Born Haber cycle sodium chloride\" you will see information on why the reaction is exothermic.\n\n\nThe reaction \"needs help\" because of the fact that the sodium will likely be coated with a layer of sodium oxide or hydroxide that will not allow chlorine to collide with the sodium metal. Even then, with pure sodium metal the reaction will be faster if heated as any reaction needs activation energy to proceed. Adding water causes an exothermic reaction that can provide activation energy for the reaction with chlorine.\n\n\n", "3" ] ]

[ [ "\nThe reason behind the bitterness of most fruits is due to presence of Tannin in them.\n\n\nTannin is a naturally occurring polyphenol found in plants, seeds, bark, wood, leaves and fruit skins. \n\n\nTannin tastes bitter,dry and astringent and you can feel it specifically on the middle of your tongue and the front part of your mouth. Besides the peels of fruit, we can feel bitterness in other food products as well like,\n\n\nHigh-Tannin Foods =\n Tea Leaves\nWalnuts, Almonds and Nuts with Skins\nDark Chocolate\nCinnamon, Clove and other spices\nPomegranates, Grapes \nQuince\nRed Beans\n\n\nThe bitterness and astringency are caused by tannins binding with proteins in our saliva. The chemical constituent of tannins are: gallic acid , ellagic acid and their derivatives(present in hydrolyzable tannins) , catechin and its derivative (present in condensed tannins) \n\n\n\n\n\n\nAs for your specific question about the the pomegranate fruit pith, it contains ellagitannins, pelargonidin, punicalin, punicalagin, anthocyanins, cyanidin, ellagic acid which makes it bitter.\n\n\n", "9" ] ]

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[ [ "\nHere is the structure of cyanogen\n\n\n\n\n\nEach carbon has 4 valence electrons and each nitrogen has 5, for a total of ((4x2)+(5x2)) 18 electrons as you said. Again, as you noted, 6 electrons are involved in sigma bonds, the 2 C-N sigma bonds and the one C-C sigma bond. Each nitrogen also has a lone pair of electrons and that accounts for (2x2) 4 more electrons. That leaves 8 electrons to fill the 4 pi bonds. Just like acetylene with 2 pi bonds, the two carbon-nitrogen triple bonds in cyanogen also have two pi bonds, with each pi bond holding 2 electrons.\n\n\n", "5" ], [ "\nTo visualize ron's answer, here is the valence orbital scheme of cyanogen. Note, that this is the canonical molecular orbital picture, which does not use hybridisation. From this depiction, the stated fact from Geoff is also clearly visible. Due to delocalisation the carbon-carbon bond is significantly shorter than a normal single bond. The right hand side orbitals are taken from a DF-BP85/def2-SVP calculation. (Since I forgot to add that to the picture itself, the horizontal dotted line separates unoccupied from occupied orbitals.) \n\n\n\n\n", "3" ] ]

[ [ "\n\n> \n> Why and how do enthalpy and $E\\_\\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?\n> \n> \n> \n\n\nOlefins are less stable than their corresponding alkane analogues. For example, when ethene is hydrogenated to ethane roughly 33 kcal/mol of heat is given off because ethene has a higher energy content than ethane.\n\n\nWhy is this, why is a molecule with a double bond higher in energy than a molecule without a double bond?\n\n\nThere are two equivalent ways to describe the orbitals in ethene. One is the \"double bond\" description where a pi and sigma bond exist together side by side. We know that a pi bond is not as strong as a sigma bond due to poorer overlap with the pi bond compared to the the sigma bond. It only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about the double bond, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane. Alternatively one can view ethene as a two-membered ring, no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring. It is easy to see that such a ring system would contain a significant amount of strain.\n\n\nWhether you use the pi bond or two-membered ring approach to describe olefinic double bonds, both lead to the conclusion that olefins are destabilized (higher in energy content) due to poor overlap or ring strain (which in itself is really a reflection of poor overlap). Because of this destabilization, alkenes will generally have lower activation energies and more heat will be given off when they react compared to alkanes.\n\n\n**Example:**\nLet's consider the bromination of ethene and ethane\n$$\\begin{align}\n\\ce{C2H4 + Br2 &-> BrH2C-CH2Br}\\\\\n\\ce{C2H6 + Br2 &-> 2 CH3Br}\n\\end{align}$$\n\n\nThe [heats of formation](http://www2.ucdsb.on.ca/tiss/stretton/database/organic_thermo.htm) of ethene, bromine and 1,2-dibromoethane are 52.5, 0 and -37.8 kJ/mol respectively; the reaction is **exothermic** by 90.3 kJ/mol. On the other hand, in the case of ethane, the heats of formation are -84.7, 0 and (2x) -37.4 kJ/mol; this reaction is **endothermic** by 10.3 kJ/mol. The olefinic reaction is significantly favored for the reasons discussed above.\n\n\n", "3" ], [ "\nEthane will release more heat per mole but ethene should produce a higher temperature. The extra heat released by ethane goes into heating up more water which has a high heat capacity.\n\n\nAs for which has lower activation energy, ethyne is dangerous to store because it can explode so I am assuming ethene is faster to react because of some trend. I also worked with someone who was researching possible fuels for a ramjet and his group chose ethene because it was the best compromise between safety and speed of combustion. I do not know why the trend ethane>ethene>ethyne for reaction rates is as it is though.\n\n\n", "1" ], [ "\nEthene produce a high temperature. Ethane release more heat per mol. The extra heat released by ethane goes into heating up more water which has a high heat temperature.\n\n\n", "-1" ] ]

[ [ "\nThe problem is that neither of these nucleophile are hard. Vanilla hydrides such as potassium hydride or sodium hydride release very soft hydride ions. Hydrogen only has one proton. Giving it two electrons to stabilize as in the case of the hydride anion is a tall order. Note the proton to electron ratio: 1 to 2. Even carbanions which are unstable species don't have such a ratio. \n\n\nTherefore, the hydride ion has a massive ionic radius and is easily polarized, making it an extremely soft base. Electrophilic carbon atoms in carbonyl groups are usually hard(er). \n\n\nPlus the 1s orbital of hydrogen is too small to effectively overlap with anything other than another hydrogen. Is your textbook written by Clayden? I distinctly remember Clayden's remark about hydrogen's 1s orbital with regard to its anion's nucleophilicity. \n\n\n\n\n---\n\n\nUsual reagents used for hydride delivery/reduction of carbonyl compounds (and related) include LAH, NaBH4, and NaBH3CN, in order of decreasing strength. As far as I know, the lithium ion in LAH plays an integral role in its mechanism of reduction by bonding with the carbonyl oxygen of carboxylic acids. \n\n\nAlso, these aforementioned reducers are in order of decreasing strength because they are listed in order of decreasing partial negative charge on their hydride ions. For example, the nitrile group in sodium cyanoborohydride withdraws electron density from the hydrogens. Hence the reason it is the weakest of the three aforementioned reducers. NaBH3CN will only act on strongly polarized iminium ions rather than plain imines or carbonyl compounds. \n\n\n", "2" ] ]

[ [ "\nIf you want sodium hydroxide for soap making you should just buy some: it is a cheap and common chemical. This is because there are effective industrial processes to make it in vast quantities. \n\n\nTrying to make it yourself is not easy and will cost you far more effort than just buying it. The least of your problems will be what happens to contaminants like the hexacyanoferrate (which is present in such low concentrations it isn't really significant). Have a look at the way industry makes bulk hydroxide and chlorine (e.g. [here](http://www.essentialchemicalindustry.org/chemicals/chlorine.html#membrane_cell)); it isn't as simple as just electrolysing brine. All the processes involve sophisticated ways to separate the products produced at the different electrodes. You don't get decent quantities of either product if you don't maintain this separation (one method uses flowing mercury; another expensive ion-exchange membranes). In addition to this, uncontrolled electrolysis is dangerous as it produces toxic and explosive gases. You really don't want to try it at home.\n\n\nSince the most basic components for an effective reaction will cost you far more than the couple of dollars a large bottle of pure sodium hydroxide will cost, buy the chemical and ignore the electrolysis as a route to what you want. \n\n\n", "3" ], [ "\nElectrolize salt (NaCl) water without a proton exchange membrane is more likely to produce NaOCl. You need the membrane to keep chlorine and sodium hydroxide from combining to form bleach. The membrane may be made from a polymer which only allows positive sodium ions to pass through it. \n\n\nSaftey note. The whole process also creates dangerous potentially deadly chlorine gas if not vented properly. \n\n\nIts probably safer and cheaper to buy your sodium hydroxide from a commercial source as the producer is likely able to capture revenue from selling the sodium hydroxide as well as chlorine, and hydrogen that are made during the process, while buying their electricity at lower than residential rates. \n\n\nEdit, I am not sure what will happens to the sodium ferrocyanide. The worse case is it is also broken down, and that at above room tempurature, cyanide gas will evolve at the same electrode as the chlorine gas, underscoring the need for proper ventilation.\n\n\n", "1" ] ]

[ [ "\nYes, we can definitely see atoms in modern microscopes. I'll try not to get deep into the subject of \"seeing\", but I will note that most people say they are seeing when they use their eyes to observe objects. That is, they are using an optical system (eye) consisting of a lens and a detector (retina) and some image processing carried out by a computer (brain). Thus, any instrument which uses some sort of detector and a computer to turn the signals of that detector into an image we feel comfortable looking at qualifies to me as \"seeing in a microscope.\" If you're comfortable with that then read on.\n\n\nMy favorite microscope which is capable of seeing atoms is the transmission electron microscope (TEM). Using a high quality TEM, we can easily resolve objects around an angstrom in size, which is the distance between atoms. To image a single atom, you also need to have a sample which is only a single atom thick at some point. This is tough to accomplish but can be done. Today, images of graphene provide arguably the best \"single atom\" TEM images. A seminal paper is [here](http://www.nature.com/nmat/journal/v10/n3/fig_tab/nmat2964_F1.html) (paywall). You can also search the internet for graphene images such as [this](http://www.er.esi.nagoya-u.ac.jp/rescwe/kaiseki/photos/GrapheneHRTEM.jpg).\n\n\nIn the links above, the images show the shapes of atoms as little blobs, and to a large extent the blobs are more a description of the microscope's ability to focus rather than the shape of the atom itself. However, the atomic force microscope (AFM) excels at seeing the electrons at the surface of an atom. With AFM, researchers are even able to [image the chemical bonds of molecules sitting on a substrate](http://www.sciencemag.org/content/342/6158/611). In this case, the image shows the bonds as rods and the atom centers are not really visible. Chemistry textbooks often show the \"shapes\" of atoms by drawing the electron orbitals, so in a sense these rods are the shapes of these atoms too.\n\n\nThis just points up that the \"shape\" of atoms is a non-trivial subject, which is very dependent on how you look at them.\n\n\n", "10" ], [ "\nWe can't \"see\" individual atoms with light. The wavelength of visible light is too large to resolve individual atoms, but that doesn't mean we can't *detect* individual atoms using various methods. \n\n\nX-rays and beams of electrons both have about the right wavelengths to detect atoms and some microscopes based on either can produce images that resolve to show atomic positions. Transmission electron microscopes can \"see\" individual atoms in some circumstance. More importantly, the structure of crystals can be resolved by interpreting the diffraction of x-rays from the crystal lattice, providing detailed analysis of the 3-dimensional structure of crystals. This is almost a routine way of determining the structures of molecules as complex as proteins. In a sense this is \"seeing\" atoms (more strictly it is detecting the electron density). But it is a very indirect way of \"seeing\" and, strictly, you are detecting using a large assemblage of atoms not just one at a time.\n\n\nThere are modern techniques that allow *individual* atoms to be detected and even manipulated. These are based on the use of very small needles with atom-sized tips. Atomic force microscopy or scanning tunnelling microscopy and other variants can use the controlled movement of these fine needles to measure the force of interaction between the atoms on the tip of the needle and the molecules or atoms on a surface. So what you are \"seeing\" is the atomic forces generated by the electron clouds in individual atoms the they interact with something else that is also atom-sized.\n\n\nFor example this image shows a sheet of graphene:\n[](https://i.stack.imgur.com/lt5AI.jpg)\n\n\nAnd this image shows a single molecule of pentacene (with a model of the structure shown below the actual image):\n\n\n[](https://i.stack.imgur.com/c7ZrA.jpg)\n\n\nThere is even an amusing [animated video](https://www.youtube.com/watch?v=oSCX78-8-q0) by IBM (who invented the atomic force microscopy technique) showing a story told by manipulating the positions of carbon monoxide molecules on a surface.\n\n\nThese techniques have even been used to understand reaction mechanisms by observing what happens to individual molecules.\n\n\nSo we can \"see\" molecules at atoms by using devices that can measure atomic forces very precisely. In that sense we can see what they \"look\" like at least from the point of view of other atoms interacting with them.\n\n\n", "5" ], [ "\nSince the nucleus is concealed within the much larger electron cloud, the shape of an atom is the shape of its orbitals; the 3D pattern of bonds in crystals, etc. So, if we could somehow see one without using light, an atom would \"look like\" its electron wavefunction, its surrounding standing-wave of probability.\n\n\n* [H orbitals wave function](https://upload.wikimedia.org/wikipedia/commons/e/e7/Hydrogen_Density_Plots.png): a 2D probability plot\n* [Gallery of orbitals](https://i.stack.imgur.com/8aDOv.jpg \"orbitals galleryquot;\"): three-D, the hard and shiny version!\n\n\nIf I understand it right, the wavefunction axes of an individual atom in space have uncertain direction. Therefore unless we somehow probe an atom and \"collapse\" its orientation, the above wave functions should be seen as smeared-out spherically.\n\n\nOn the other hand, everyone in the 1950s knew that [atoms look like this](https://www.google.com/search?q=atoms%20atomic%201950s&num=100&source=lnms&tbm=isch&sa=X). \n\n\nThe common \"three-planet solar system\" image has ended up deeply embedded in American pop-culture. It's wrong, and should be replaced by various plots of 2D wavefunctions, but just try to convince the public! Heh, also, radio waves don't come out of the tips of broadcasting towers. And, falling raindrops don't have sharp points. Instead they're spheres!\n\n\n:)\n\n\n", "2" ] ]

[ [ "\nThere are different strategies for synthesizing superheavy elements. \n\n\nTwo of the main strategies are referred to as \"hot fusion\" and \"cold fusion\". \n\n\nThe first two reactions are good examples of the \"hot fusion\" strategy. I will focus on these two reactions, as the other two are much older and I don't think they are good examples of current thinking. \n\n\nExperimentally, a very heavy element target (the Es or Bk in the first two reactions) is bombarded by an ion beam of the second element. The target element is choosen on the basis of being very heavy, availible and not rapidly decaying. \n\n\n$^{48}\\_{20}\\ce{Ca}$ is choosen for a very special reason. This is not ordinary calcium. The most common isotope of calcium has only 20 neutrons, but this isotope has 28!. It is extremely neutron rich, espeically compare to other relatively stable isotopes in this mass-range of the periodic table. $^{50}\\_{22}\\ce{Ti}$ is also very neutron rich. \n\n\nWhy are neutron rich nuclei chosen? There are two reasons for this:\n\n\nFirst, confining many protons in a small space causes electrostatic repulsion. As atomic number increases, the stable nuclei deviate further and further from a 1:1 neutron:proton ratio, because having more neutrons keeps the protons further apart and decreases electrostatic repulsion energy. \n\n\nSecondly, you will notice the asterisk (\\*) in the 1st, 3rd and 4th equations. This denotes the [compound nucleus](http://encyclopedia2.thefreedictionary.com/Compound+Nucleus) which initially forms in the fusion reaction. It is an extremely ($<10^{-21}$s) short lived intermediate state, not the ground state of a nucleus, but more like a highly excited state. It's formation is not considered creating a new element. It is critical to the strategy of forming an element that the intermediate state loss energy through neutron \"evaporation\". Notice that in the second equation, the step of the neutrons evaporating is shown, while the intermediate state is not shown. Generally speaking, in the hot fusion strategy, 3-5 neutrons will evaporate to cool the intermediate state. \n\n\nFor further reading see:\n\n\n[Future of superheavy element research: Which nuclei could be synthesized within the next few years?](http://arxiv.org/abs/1207.5700) and\n\n\n[The discovery of the heaviest elements](http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.72.733) Rev. Mod. Phys. 72, 733\n\n\n", "8" ], [ "\nThe stability of heavy nuclei is a function of both atomic number and atomic mass. Therefore, an appropriate number both of protons and of neutrons in the target nucleus must be attained, or else vanishingly short lifetimes can be expected. (See the Wikipedia article on, e.g., the [island of stability](https://en.wikipedia.org/wiki/Island_of_stability \"island of stability\"), and links therein.) The 'source' nuclei are chosen to provide proton and neutron counts that are anticipated to yield heavy nuclei with sufficiently long lifetime so as to be measurable.\n\n\nOf course, it's not as simple as just choosing the right total number of protons and neutrons in your starting nuclei, as you're not guaranteed that all of the neutrons and protons will stay put in the synthesized nucleus, nor that you won't have rapid radioactive emissions that knock you away from your desired composition. For example, in the various [syntheses of dubnium reported at Wikipedia](https://en.wikipedia.org/wiki/Dubnium \"syntheses of dubnium reported at Wikipedia\"), various numbers of neutrons were ejected as part of the nuclear synthesis 'reaction'.\n\n\n", "4" ] ]

[ [ "\n\n> \n> Surprisingly, I learned that there are also usages for orbitals g,h,i\n> and even j.\n> \n> \n> \n\n\nActually, the letter \"j\" is not used, so it is s, p, d, f, g, h, i, k, l, etc.\n\n\nThe higher angular momentum orbitals do enter the domain of science, due to excited states of atoms. Transitions to and from excited states are observable through atomic spectroscopy.\n\n\nFor example there is the article [Microwave spectroscopy of Al I atoms in Rydberg states: D and G terms](http://iopscience.iop.org/0953-4075/36/24/006) \n\n\n", "17" ], [ "\nOne use of orbitals beyond f is in computational chemistry to construct basis sets. It's important to remember that orbitals are entirely a mathmatical construct that chemists and physicists have found useful in conceptualizing chemical properties and so we can use the idea of orbitals outside of their normal context of framing the periodic table. To develop a nearly complete basis set for certain elements, one can use a large collection of orbitals of different angular momentum. As an older example, [Nesbet, Barr, and Davidson](http://www.sciencedirect.com/science/article/pii/0009261469801016) attempted to compute the energy of a neon atom using a basis set that included up to i orbitals. \n\n\nAs an interesting side note, I found a nice visualization of the orbitals (really the spherical harmonics, but its essentially synonymous in this case) up to n on [Sylvain Paris's](https://people.csail.mit.edu/sparis/sh/index.php?img=64) MIT page.\n\n\n", "8" ], [ "\nThe letters beyond $f$ can also appear in higher orbital angular momentum states, as, $^1G$ in a low spin $d^2$ complex.\n\n\n", "2" ] ]

[ [ "\nLet's do it step by step.\n\n\n1. $C = 2$. The reason is that you have two components: $\\ce{HBr}$ and $\\ce{C4H6}$. The third compound, $\\ce{BrC4H7}$, is simply a combination of both of them. Remember what is $C$: It's the minimal number of components needed to describe all phases in the system. Since $\\ce{BrC4H7}$ can de described as $1\\times\\ce{HBr} + 1\\times\\ce{C4H6}$, it is not an independent component.\n2. $P=1$ or $P=2$. It depends on how much the reactions has progressed. Basically, you have one phase on each side of the reaction. On the left, you have one gas phase that has a mixture of both $\\ce{HBr}$ and $\\ce{C4H6}$. On the right, you one one liquid ([?](http://www.chemspider.com/Chemical-Structure.4516556.html)) phase: $\\ce{BrC4H7}$. So here are the possibilities:\n\t1. The reaction has not started yet. All you have in the one gas phase. In that case, $P=1$ and $F=2-1+2=3$.\n\t2. The reaction is going as written, but not completed yet. In this case not all gas is consumed yet, but liquid exists. Then, $P=2$ and $F=2-2+2=2$.\n\t3. The reaction is finished. All gas is now liquid. Then we're back to $P=1$ and $F=2-1+2=3$.\n\n\nSo what happened in stage 2? Why did you lose a degree of freedom? Well, it depends on what what the driving force for the reaction. For example: Why did gaseous $\\ce{HBr}$ and $\\ce{C4H6}$ combine to form liquid $\\ce{BrC4H7}$? Was it because of cooling or heating? In this case, the lost degree of freedom is $t$. Was it pressure? Then the lost degree of freedom is $P$.\n\n\n$X$ is a bit problematic. There is not enough information given in the question. In my opinion, $X$ was fixed to begin with: I assume that once the reaction is complete, none of the reactants remain. Thus their ratio is fixed to 1:1 and $X$ is constrained. Had it not been the case, you would also get differing $P$ in stage three because you would have two phases, not one. \n\n\n", "3" ] ]

[ [ "\nAll of the paths given by the OP can lead to discovery of catalysts. Usually, it is a combination of strategies and an accumulation of knowledge from different sources. I will give two examples, ammonia synthesis and enzymatic reactions.\n\n\n**Ammonia synthesis**\n\n\n$$\\ce{N2(g) + 3H2(g) <=> 3NH3(g)}$$\nThe synthesis of ammonia from the elements is one of the largest scale reactions nowadays, and it provides most of the fixed nitrogen (i.e. other than elemental) for human food globally (some nitrogen is fixed by bacteria, especially in roots of legumes). In this sketch, I am using \"Ammonia Synthesis Catalysts: Innovation And Practice\" by Liu Huazhang as one of the sources. There is also a wikipedia article on one of the players, Alwin Mittasch, that I used.\n\n\nFritz Haber figured out that high pressure and low temperature would give the best yield for the reaction (as it is exothermic and the number of gas molecules decreases). This provided the \"negative criteria\", point 4 made by the OP: the catalyst has to be stable at high pressure. He developed a circulating process to enrich the product and received a patent for it in 1908. Haber and Bosch, working for the BASF company, brought this to industrial scale in the early 1900s, resulting in the Haber-Bosch process. To find a catalyst, they screened thousands of potential substances (strategy 3 given by the OP). They found an efficient osmium catalyist, but osmium was to expensive for commercial use.\n\n\nThe next improvement was a combination of serendipity and rational followup. Mittasch and Wolf, both working for Bosch, were key players. Wolf tried a magnetite from Gallivare, Sweden, that had been \"sitting on a shelf\" as a catalyst, and it showed promise. There were earlier observations that iron is a good catalyst, but pure iron did not work. Mittasch systematically followed up, figuring out how to reproduce the result by making a catalyst from $\\ce{Fe3O4, K2O, CaO, Al2O3 and SiO2}$. \n\n\nThis catalyst, with minor modifications, is still used today. Haber and Bosch won the Nobel prize in 1918, and Ertl won the Nobel prize in 2007 for elucidating the molecular mechanism of the catalysis. Research is ongoing to find other paths to nitrogen fixation (see e.g. [here](https://en.wikipedia.org/wiki/Nitrogen_fixation#Ambient_nitrogen_reduction)), using other catalysts inspired by biological nitrogen fixation or different precursors (e.g. nitrogen and water for photocatalytical synthesis of ammonia).\n\n\n**Enzymatic catalysis**\n\n\nEnzymes have been developed by trial and error in the course of evolution. It is thought that often enzymes catalyzing on particular reaction led to related enzymes that catalyze related reactions.\n\n\nA rational way to find a protein that catalyzes a given reaction is to make a transition-state analog. Then, you raise antibodies against that analog. In theory, the antibodies should act as catalysts for the desired reaction, and often they do (catalytic antibodies). This might be the best example of strategy 1 given by the OP.\n\n\nIn 2018, Frances Arnold was awarded a Nobel prize for directed evolution. She was able to mutate known enzymes (or more generally proteins) to catalyze reactions that have no known counterpart in biological systems (for example making carbon-silicon bonds). The process is rational (as in having a plan) but involves randomization and screening steps.\n\n\n\n> \n> How are catalysts for chemical reactions found?\n> \n> \n> \n\n\nThe answer is all of the above. \n\n\n", "5" ] ]

[ [ "\nThree pieces of information from the comments:\n\n\n1. It fizzed and disappeared in vinegar.\n2. It had a yellow-orange color in blue flame.\n3. It's salty (note that it's usually a bad idea to taste random stuff you find on your floor).\n\n\nThe only white crystalline material that is a plausible candidate for being there and fizzes in acetic acid (ie vinegar) is calcite - $\\ce{CaCO3}$. It's also a mineral that has 'retrograde solubility' meaning it dissolves more readily in cold water. This is probably how it got to the water in the first place. This is also why it precipitates on your electric kettle. It is the same stuff.\n\n\nThe salty stuff that burns yellow-orange is obviously $\\ce{NaCl}$, aka table salt. It also probably dissolved with the vinegar as well. Why is the salty taste not quite like table salt? Well, first of all it is mixed with calcite. But, calcite has no taste. It's possible that there is some $\\ce{KCl}$ or $\\ce{MgCl2}$ in there as well. They are slightly more bitter than $\\ce{NaCl}$. This is the stuff they use in low sodium salt.\n\n\nThere's another possibility - it could be any of the epsom salt family, $\\ce{MgSO4.$n$\\,H2O}$.\n\n\nAs to would this be dissolved in the first place, you can only speculate. You say that it came from a table that was outside. Did people eat there? Could people had spilled salt on it? Is the airborne dust in your area contain a lot of calcite? What about your basement? Could someone have spilled some salts on the floor, maybe even years ago?\n\n\n", "5" ] ]

[ [ "\n\n> \n> Is it crystallization?\n> \n> \n> \n\n\nYou are correct. The main difference is that sand is crystalline and glass is not—it is amorphous.\n\n\nThe main component (> 95%) of common yellow sand is quartz (the mineral whose composition is SiO2). Note that not all sand is quartz. There are white sands containing calcite (CaCO3) and black sand (containing various heavy minerals). But the most common sand is indeed quartz sand: SiO2.\n\n\nGlass, the type you see in your everyday life, on the other hand, is not composed of pure SiO2. It has a bunch of other additives such as Na, K, B, and others. This is done to modify the properties of the glass and make it more suitable for human use. It doesn't matter much though for our discussion.\n\n\nSo if they are made of the same thing, why the difference? The answer is cooling rate. If you cool molten SiO2 slow enough, the atoms have enough time to organize themselves into a crystalline structure. In the case of pure SiO2, this is a network of SiO4 tetrahedra: One silicon atom surrounded by four oxygens. If it cools too fast, then the crystalline structure does not form. It may be completely amorphous, or form into a sub-microscopic array of SiO2 crystals in various structures (CT-opal for example).\n\n\nWhat determines the cooling rate? Well, in the case of glass it is a matter of minutes. You've seen glass making: The glass is molten and very quickly it solidifies to a solid. In contrast, most of the quartz sand you're seeing is actually broken fragments of rocks called granite. This type of rock has abundant quartz in it, and it forms deep underground (as in 10s of kilometers) at very slow cooling rates. While a glass maker can take his glass and let it cool in the atmosphere or in water, molten silicate magma (\"glass\") deep in the Earth is surrounded by rocks that are in the hundreds of degrees. This slow cooling facilitates crystallization of the SiO2 into quartz rather than glass. How slow is this? At least tens of years, more commonly hundreds or even thousands of years. This is much slower than the seconds and minutes in glass making.\n\n\nSlow crystallization of SiO2 in the Earth is not the only process. There are actually processes that can take molten silicate magma and cool it very rapidly. This commonly happens in volcanic settings: A volcano throws molten magma (again, made mostly out of SiO2) and throws it into the atmosphere where it can cool very rapidly. It's even faster if the magma erupts into water. The result is volcanic glass: sometimes called obsidian.\n\n\nThe difference in the visual appearance is actually not due to the presence or absence of a crystalline structure. It's mostly related to grain size. If you take window glass and shatter it so it's all small grains, it will look just like sand. On the other hand, if you take quartz, and make it big and smooth (just like window glass), it will look like this:\n\n\n \n\n(By Parent Géry (Own work) [Public domain], via Wikimedia Commons)\n\n\n", "27" ], [ "\nTake sand, melt it, let it cool and you have (sort of) glass.\n\n\nGlass is an amorphous solid. It appears to be solid, but is, in reality, a viscous liquid. \n\n\nSand is essentially the same, although broken up into very small pieces. Although sand could contain quartz (crystallized) fragments, it doesn't change what we observe about the sand.\n\n\n\n> \n> How could it result in so much visual differences?\n> \n> \n> \n\n\nGlass is clear since the sand it is made from has chemicals added to lower the melting point and to make it transparent (by removing some impurities). Sand is composed of everything in the earth, so it is not as transparent as glass. Also the small particles deflect light so it is difficult to see a pattern through sand.\n\n\n", "0" ] ]

[ [ "\nI guess that it actually was a silver cleaner, not one for gold or precious stones.\n\n\nThe latter mostly contain nonionic surfactants, which are made by the reaction of water-insoluble long chain alkanols (C12 - C15, from natural fatty acids) with oxirane. These washing agents dissolve well in water and help to remove grease and fingerprints from the jewelery. Their action isn't much different from that of a dishwasher for manual cleaning.\n\n\nThese cleaners don't smell the way you described and the car retailer probably used similar cleaners for the car.\n\n\nSilver cleaners are a different league. The do contain the same surfactants (washing agents), but in order to actually remove the tarnish (silver sulfide) from the silverware, they do also contain a bit of acid (typically sulfuric acid) and **thiourea**, which probably is responsible for the smell.\n\n\nRepeated shampoeing probably is the best method to get rid of the thiourea without ruining the carpet.\n\n\n", "2" ], [ "\nHomemade Carpet Fresh\nI would recommend a home made dry cleaning using diatomaceous soil compound. This will eliminate the problem of toxins and erosion of the metal interior under the carpet. Use some DAP mixed with your favorite dry detergent and vacuum it up. If plaster of Paris is not available basically any spackle will work. Mix 1/1\n\n\n", "-1" ] ]

[ [ "\nThere are a few things that influence the properties of polymers:\n\n\n* Molecular weight of the chains, i.e. how many monomers are joined in each chain, can drastically alter the polymer's properties. For example, for polyethylene, UHMWPE has extremely long chains and produces a very strong material suitable for bullet-resistant vests, whereas LLDPE has much shorter chains that produce a weaker material that is much more ductile (think about how easy it is to stretch a plastic shopping bag or the like).\n* Another factor that affects a polymer's properties is branching. Branching is when a polymer is not made up of nice straight chains, but chains that occasionally fork off into several different chains. The difference between LLDPE and LDPE is that LDPE has a lot more branching. This keeps the chains from packing as nicely so LDPE isn't as strong.\n* The other big factor (other than adding another monomer to make a copolymer) is the introduction of crosslinking. Crosslinking is where different chains are connected to one another, often by addition of a molecule that can bond to a chain at both ends, e.g. rubber vulcanization crosslinks the different chains with linear sulfur chains. Natural rubber is very different than what most people think; it's a soft sticky material that doesn't hold its shape. The polyisoprene chains it's composed of aren't crosslinked and can easily slide past one another, but once crosslinked, the chains are connected and aren't free to move as much. Depending on the degree of crosslinking, you can go from something somewhat soft, like a bouncy ball, to something very rigid, like a car tire.\n\n\nHydrogels are not simple linear polymers. While they're mostly made from a single monomer, small additions of related molecules are used to control how they polymerize. N.B. These apply generally to silicones, other polymers may have other branching/termination processes. If the monomer only has two sites at which it can connect to other monomers, naturally the only structure that can be formed is a linear chain. One can control the length of the chains by adding a small amount of a terminator—a monomer that can only make one connection. If a terminator reacts with a growing chain, it stops the growth as it can't connect to anything else. The more terminator added, the shorter chains are, on average. Similarly, if you add a monomer that has more than two reactive groups, branching and crosslinking will occur. Often, a non-crosslinked polymer is supplied as a liquid, then a crosslinking agent is added to harden the polymer. [Dow Corning](http://www.dowcorning.com/content/discover/discoverchem/cure-systems.aspx) has a great primer on the actual chemistry involved.\n\n\nWhat makes a hydrogel special is that it's a polymer *network* that encapsulates water, i.e. it's an extensively crosslinked polymer that is hydrophilic. If the polymer was just a series of individual chains, it would just dissolve in water since nothing is holding the different chains together. When crosslinked, water can get in, but the mesh of polymer is linked together enough to hold its shape, though it may swell to accommodate the water. Depending on the degree of crosslinking, more or less water can fit within the structure. A less linked polymer can fit more water since there are fewer crosslinks resisting, but is less rigid when hydrated for the same reason (think Jell-O). A more crosslinked polymer won't absorb as much water, but will be much more rigid (like a contact lens). Contact lenses have to balance holding their shape, oxygen permeability, surface hydrophilicity, etc. so there are many reasons why, despite being based on the same primary monomer, two different brands may have a different blend of the other stuff that lets one control the hydrogel's properties.\n\n\n", "3" ], [ "\nWhen a polymer is produced, the final result does **not** only rely on what the monomers were. For example, there is \"extended\" polystyrene and \"extruded\" polystyrene. Or there are polystyrene sulfonate chemical polymers and etc.\n\n\nSince nowadays, we need polymers to do many things for us (sheathing, sheltering, and sometimes even heat conductivity), we usually tend to modify their structure to make them more temperature-resistant, harder, stronger and ...; and that is because researches on new polymers are not always cheap or successful.\n\n\nLet's explain with an example. Imagine you have $10000$ ethylene monomers that you will need to make a polymer from. Each of the PE polymers require $100$ ethylene molecules to form. Since you need this thing to be a little bit stronger, you would pressurize it, heat it, liquify it etc. So the result is a set of well-arranged polymer molecules [You can imagine lines of pupils standing in a school's yard, waiting to go to their classes] They are resistant to minor temperature rises and can be used in places where hard polymers are needed. (actually, not yet, but you are currently imagining :))\n\n\nOn the other hand, Imagine the raw \"plastic\" that is industrially produced with $10000$ monomers and the polymers are made of 100 ethylene molecules. This time it seems as if the pupils are dispersed through the school yard, though in tact with each other. You can easily conclude that this polymer is *potentially* weaker and is more vulnerable to temperature differences.\n\n\nThe first example is called \"extruded polyethylene\" and the 2nd one is called \"expanded polyethylene\". Notice that the monomers are the same, but the result is different in as big of a range that includes usage to specific heat capacity.\n\n\nThis is not the only case. You remember that I mentioned PS sulfonate. Sulfur, in here, is added to polystyrene to make chain-like \"links\" to polymers. For example, imagine the lines of pupils again. These sulfonates act as some older students that hold the pupils still and do not let them escape. These chains make polystyrene sufonate ideal for being an agent for dying clothes (since it becomes resistant to dissolution).\n\n\nIn your case, Hydrogel monomers are the same, but as I described above, either **the how of their production** or **additional materials that modify the behavior of the Hydrogel** or both are different.\n\n\nSorry, I didn't find any appropriate applicable images without license and I hope the description is not that vague. Hope I've helped.\n\n\n", "2" ], [ "\nI don't konw much about contact lenses. But for hydrogel, the reason may be as followed:\n\n\nConventional synthesized hydrogel is chemical crosslinked polymer network. Except the monomer, another reagent named crosslinker with multi-functionality such as the MBAA(N,N'-Methylenebisacrylamide)is needed. Then the swelling ratio or the water content of a hydrogel is strongly affected by the relative amount of crosslinker. Higher the crosslinking density, lower the water content. \n\n\n", "0" ] ]

[ [ "\nYour synthesis is unlikely to become a huge commercial success ;)\n\n\nMethylchloride is a technical product, but usually made by reaction of methanol with hydrogen chloride:\n\n\n$$\\ce{CH3OH + HCl -> CH3Cl + H2O}$$\n\n\nThis works nicely without side products, because\n\n\n* $\\ce{CH3Cl}$ has a low solubility in water (around $\\mathrm{5\\,g\\cdot L^{-1}}$)\n* under standard conditions, $\\ce{CH3Cl}$ is a gas (bp around $\\mathrm{-28\\,°C}$)\n\n\nThe radical halogenation of methane will usually lead to various halogenation products and thus is not a good idea. The same is true for the similar reaction of ethane.\n\n\nIn addition, I fail to see how the reaction of ethanol and carbon dioxide will lead to glycerol.\n\n\nYou should further take into account that there's plenty of glycerol available as a side product in the production of *biodiesel*.\n\n\nFrom a commercial and ecological point of view, methane therefore is better converted to cabon monoxide and hydrogen by steam reformation. Further hydrogen may be obtained from a subsequent shift reaction and the resulting carbon dioxide may be reacted with gycerol to yield 1,2-glycerol carbonate.\n\n\n", "2" ] ]

[ [ "\nThere isn't a simple redefinition that would work in all cases, and this is probably why nobody does this. Of course, you COULD define such a unit for a given system by defining it as a function of temperature, but it would be applicable only to the specific system.\n\n\nAs you noted, the physics behind the temperature dependence of a material's heat capacity lies in the quantum states which become available for excitation at different energies. Consider an insulator which stores heat primarily in the lattice vibrations. As you increase the temperature, more vibrational modes are accessible. This is often described using the Debye model which gives the heat capacity as such:\n\n\n$C\\_V = k(\\frac{T}{\\theta})^3$\n\n\nwhere $C\\_V$ is the heat capacity, $k$ is a constant, $T$ is the temperature in Kelvin, and $\\theta$ is a constant pertinent to a specific solid.[1] \n\n\nHowever, a metal can also store a significant amount of heat in the motions of electrons since conduction electrons are not bound to specific sites. Heat capacity for the electron 'gas' is often described like this:\n\n\n$C\\_V' = k' \\frac{T}{\\theta'}$\n\n\nwhere $C\\_V' \\neq C\\_V$, $k' \\neq k$ and $\\theta' \\neq \\theta$.[2] Depending on $\\theta$ and $\\theta'$ for a given material, at low temperatures (often cryogenic) the electron contribution can be the dominant mode of storing heat in the solid. Thus, if you chose a unit which was sensible for an insulator, it wouldn't apply to a conductor (and vice versa). \n\n\nEven within a specific material, the heat capacity on either side of a phase transition is often different since the phase transition has changed the quantum states available in the material. Therefore, your new unit would become discontinuous at phase transitions and would need a different mathematical expression for each segment of the heat capacity function, not to mention a different parameterization for each material. Hence, no real gain.\n\n\n[1] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 106, eq. 47b\n\n\n[2] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 193, eq. 37. Note that folks don't actually use the symbol $\\theta'$ in this case, but choosing so makes the explanation above clearer.\n\n\n", "5" ] ]

[ [ "\nAspartame is one of the most well studied food additives there is. It seems to be controversial in the same way that evolution is—scientific research is overwhelmingly on the side of the substance being safe to consume, even at very high levels, but a Google search on it brings up things like \"[Aspartame is, by Far, the Most Dangerous Substance on the Market that is Added To Foods](http://aspartame.mercola.com)\" (from a site trying to sell natural health products). A lot of this comes perhaps from the naturalistic fallacy: aspartame is artificial and thus unsafe. A huge number of regulatory authorities have deemed it safe to consume: FDA, Health Canada, EFSA, FSA, etc.\n\n\nThere is also a good deal of bad biochemistry thrown about: aspartame is metabolized into the two amino acids aspartic acid and phenylalanine, as well as methanol. There have been some hypotheses that aspartic acid intake could induce some amount of toxicity, but experimental evidence is lacking. Methanol, of course, is well known to be toxic. The key is, that aspartame is very sweet and only tiny amounts are added to food. So even drinking enough diet cola to hit the FDA's acceptable daily intake of aspartame (which works out to ~7.5 L), the quantity of methanol consumed is only around 300 mg, and things like fruits are larger sources of methanol in people who drink sane amounts of diet soft drinks.\n\n\nThe one condition in which aspartame is known to do harm at reasonable intake is in people who have phenylketonuria (PKU). This might be what you're thinking of by \"faults in the vital proteins of their body\". PKU is a genetic disease caused by a defective gene for the enzyme phenylalanine hydroxylase which results in an inability to metabolize phenylalanine correctly; thus, consuming phenylalanine from any source is hazardous to sufferers of the disease.\n\n\n", "8" ] ]

[ [ "\nPure water can form two hydrogen bonds per molecule. The enthalpy of these hydrogen bonds is about 23 kJ/mol of water. You have to give water a very good reason to rip apart its hydrogen bond network, and the interactions water experiences with the organic compounds often aren't enough, even considering increased entropy, to make this favorable. \n\n\n", "9" ], [ "\nWater is not a bad solvent for all organic molecules. In fact many organic molecules *are* soluble in water. In general the solubility of a substance in a given solvent is determined by the strength of the intermolecular forces between the solvent and solute. \n\n\nThe predominant intermolecular force in water is hydrogen bonding and so molecules with groups which can hydrogen bond generally dissolve well in water. These groups can have acidic hydrogens of their own such as alcohol or amine groups which can h-bond to water both through their hydrogens and through the lone pairs on their oxygen and nitrogen atoms respectively. Also other groups which have lone pairs on electronegative atoms such as carbonyl groups can h-bond with water and so will generally dissolve, although not as well as alcohols or amines. All of these groups are generally termed to be hydrophilic.\n\n\nThere are also hydrophobic groups which tend not to bond well with water and so do not tend to dissolve. These are generally groups which are non-polar and interact primarily through Van der Waals forces. Therefore they do not form strong interactions with water (because water primarily interacts through h-bonding) and so do not dissolve. These groups tend to be apolar, the most common example being alkyl groups.\n\n\nSo in answer to your question, many large organic molecules with predominant hydrophobic groups do not dissolve in water whereas smaller molecules with hydrophilic groups do tend to dissolve.\n\n\nThis questions has a more detailed discussion of the properties of water as a solvent: [Why is water \"the universal\" solvent?](https://chemistry.stackexchange.com/questions/21898/why-is-water-the-universal-solvent)\n\n\n", "8" ] ]

[ [ "\nAmides will react with three equivalents n-butyllithium to give nitriles via geminal lithium oxyimide intermediates ([*J. Org. Chem.*, **1967**, *32* (11)](http://pubs.acs.org/doi/abs/10.1021/jo01286a075)). n-butyllithium is a strong enough base to deprotonate the molecule twice at nitrogen and once alpha to the carbonyl group. Grignard reagents may well behave similarly.\n\n\nI do not think this is a particularly useful transformation though because n-butyllithium is a rather indiscriminate reagent. There are a wide range of other methods available - see [this paper](http://www.researchgate.net/profile/Sangeeta_Jha/publication/260273495_Dehydration_of_Amides_to_Nitriles_A_Review/links/0c960530724add6933000000) if you're interested.\n\n\nIn the special case of a N,O-dimethylhydroxyamide, you can [make ketones using either.](http://en.wikipedia.org/wiki/Weinreb_ketone_synthesis)\n\n\n", "3" ], [ "\nFor a N,N-alkyl, a N,N-diaryl or a N-alkyl-N-aryl amide that what will happen is that the grignard will react attacking the carbonyl carbon. This will form R''C(NR2)(R')-OMgX from RMgX and R''-CONR2.\n\n\nWhat we normally do is to treat the product with dilute acid, this will then convert the initial additon product into either a ketone or an aldehyde.\n\n\nOne common method of making aldehydes is to react Grignards and other similar reagents with DMF, then treating the intial product with aqueous acid.\n\n\n", "1" ] ]

[ [ "\nThere is a detailed consideration of this issue in [Hydrogen Bonding of the Ammonium Ion](http://www.sciencedirect.com/science/article/pii/002228609408381Q#) Journal of Molecular Structure 345 (1995) 77-81.\n\n\nFirst, the try to be clear about what they mean by \"hydrogen bond\":\n\n\n\n> \n> For the purposes of the discussion, a hydrogen\n> bond is taken to be a specific short-range interaction\n> in which the N-H bonds within the\n> ammonium ion are significantly affected by the\n> surrounding ions, and not simply electrostatic\n> attraction.\n> \n> \n> \n\n\nThey conclude that ammonium ions rotate too freely in water for significant hydrogen bonding to exist. \n\n\nIn gas phase calculations, they find that one water molecule hydrogen bonds to one ammonium ion (in other words, the dimer is hydrogen bonded), based upon the lengthening of the N-H covalent bond, but as you add 2, 3, and 4 water molecules, the N-H bond length returns to the free ammonium N-H bond length and the O --- H length increases.\n\n\n", "6" ] ]

[ [ "\nYou should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent. \n\n\nAlso let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\\ce{HI}$) lack of stability; it's relatively easy to ionize its proton due to the weak hydrogen-iodide bond. \n\n\nOn the other hand, the high nucleophilicity of iodide ion *in water* reflects the high rate of reaction between iodide ion and some electrophilic atom (other than hydrogen). \n\n\nThe relative lack of basicity of the iodide anion in water is precisely what makes it a good nucleophile. Its lack of affinity toward protons - i.e. the protons of water - physically frees $\\ce{I^-}$ to attack electrophilic atoms (not limited to carbon). There is little hydrogen bonding between iodide ion and water. \n\n\nOn the other hand, the relative basicity and thus strong hydrogen-bond forming capabilities of $\\ce{F^-}$ makes it a poorer nucleophile in water. Rates of reaction between fluoride anion and some electrophilic atom are slowed because the fluoride anion is trapped (strongly solvated) by water molecules. And there will also be a sizable number of fluoride anions that have completely abstracted a proton from water to make $\\ce{HF}$, which isn't nucleophilic at all. \n\n\nIn a polar aprotic solvent, however, where there are no positively polarized hydrogens for a nucleophile to hook onto; $\\ce{F^-}$ becomes a much stronger nucleophile. Actually, $\\ce{F^-}$ becomes a better nucleophile than $\\ce{I^-}$, likely due to $\\ce{F^-}$'s greater charge density. \n\n\n", "16" ], [ "\nThe bond between I and H is weak so I- is a weak base. I- is a strong nucleophile because it is polarizable, making it faster for its orbitals to overlap with the electrophile. Remember that basicity is a thermodynamic concept and nucleophilicity is a kinetic concept. \n\n\nI have been told that I- can be added to reactions as a catalyst because it is a good nuclephile and a good leaving group. Leaving group capability seems to be inversely related to basicity. \n\n\n", "1" ] ]

[ [ "\nThe term peptide bond refers to amide bonds. The 21 amino acids in insulin's A-chain are covalently linked by 20 amide bonds. The 30 amino acids in insulin's B-chain are linked by 29 amide bonds. The two polypeptides are covalently linked by disulfide bonds, which are not amide bonds.\n\n\nThe 3D structure of proteins is a complex topic and a field of intense research. It's worth pointing out though that even polypeptides (with no disulfides or other covalent linkages) have a more complex 3D structure than you might expect. I suggest you look into [alpha-helices](http://en.wikipedia.org/wiki/Alpha_helix) and [beta-sheets](http://en.wikipedia.org/wiki/Beta_sheet) as an entry into the topic.\n\n\nTook some info from the [Beta Cell Consortium](http://www.betacell.org/content/articleview/article_id/8/).\n\n\n", "4" ], [ "\nAside from covalent bonds (amide and disulfide), the sturcture of a protein is determined by hydrogen bonds, [salt bridges](http://en.wikipedia.org/wiki/Salt_bridge_(protein_and_supramolecular)), and less specific interactions such as hydrophobic and hydrophilic effects. Hydrophobic portions of the protein chain tend toward the interior of the folded protein and hydrophilic regions to the exterior, in aqueous solution.\n\n\n", "2" ] ]

[ [ "\n**Most definitely yes.**\n\n\nI took multiple statistics classes as an undergraduate and they have helped me considerably. (I can argue fairly that it helped me get tenure.)\n\n\nA few suggestions from my own work:\n\n\n* **Statistical design of multivariate experiments**: Many scientists think you should test one variable at a time. No. I taught a grad student (in another group) about optimization theory and proper design and three months later, a great paper came out.\n* **Data mining and exploration**: Many chemistry data sets are extremely complicated, with cross-correlated factors. Understanding how to draw empirical inferences, find the most important factors, etc. is *enormously* helpful across a wide range of chemistry.\n* **Principal component analysis**: An emerging area in spectroscopy and imaging is taking complicated spectra and performing PCA to help with quantifying chemical components in mixtures and following everything from cancer growth to materials design.\n* **Multivariate regression and ANOVA**: My group has created numerous quick statistical screens using multivariate surrogates for complex simulations or experiments.\n* **Cross-validated analysis**: Again, the understanding of training and test sets, bootstrapping, etc. is very useful.\n\n\nNow, you also mention SDEs and Markov processes. These are crucial in chemical kinetics, including areas like reaction-diffusion systems. Monte Carlo methods are also highly used in studying dynamical processes.\n\n\nThe list would be enormous. Basically, I think it's a great skill to have and you'll use it throughout your career.\n\n\n", "7" ] ]

[ [ "\nIn a 1979 paper entitled [Copper-catalysed reactions of Grignard reagents with epoxides and oxetane](http://dx.doi.org/10.1016/S0040-4039(01)86190-6), Huynh, Derguini-Boumechal, and Linstrumelle report that the copper-catalysed ring opening of oxetane by various Grignard reagents gives the alcohols \"in acceptable yields\":\n\n\n[](https://i.stack.imgur.com/jmSaJ.png)\n\n\n$$ \\begin{array}{cccc}\n\\ce{R} & \\ce{X} & \\text{Solvent} & \\text{Yield (%)} \\\\\n\\hline\nn\\text{-}\\ce{C4H9} & \\ce{Cl} & \\text{Ether} & 75 \\\\\n\\ce{C6H5} & \\ce{Br} & \\text{THF} & 52 \\\\\n\\ce{CH2=CH-CH2} & \\ce{Br} & \\text{Ether} & 50 \\\\\n\\ce{(CH3)2C=CH-CH2} & \\ce{Cl} & \\text{THF} & 50\n\\end{array} $$\n\n\n", "4" ] ]

[ [ "\nMethanol isn't particularly toxic in and of itself, although it's no walk in the park.\n\n\nIf methanol flowed through the body without being broken down, it would cause roughly the same kind of harm as ethanol, i.e. intoxication.\n\n\nThe real culprit is one of its metabolic products, methanoic acid, also known as formic acid.\n\n\nTo understand how formic acid, present as the formate ion, is toxic, we look to [Wikipedia](http://en.wikipedia.org/wiki/Methanol):\n\n\n\n> \n> Formate is toxic because it inhibits mitochondrial cytochrome c oxidase, causing the symptoms of hypoxia at the cellular level, and also causing metabolic acidosis, among a variety of other metabolic disturbances.\n> \n> \n> \n\n\n**Edit:** As DavePhD points out, an intermediate product in this process is formaldehyde, or methanal. While formaldehyde is also toxic, it is rapidly metabolized to methanoic acid.\n\n\n**Reedit:** The deeper, more historical reason that this happens is that methanol isn't readily available in nature, meaning that few species have developed biochemical tools to deal with it. There simply hasn't existed an evolutionary pressure to deal with methanol.\n\n\n", "32" ], [ "\nThe enzyme [alcohol dehydroganase](http://en.wikipedia.org/wiki/Alcohol_dehydrogenase) converts the methanol to formaldehyde in the body. Formaldehyde is then converted to formic acid.\n\n\nFormaldehyde can cause blindness before being converted to formic acid, while formic acid causes acidosis as Williham Totland points out.\n\n\nSee [Biochemical Aspects of Methanol Poisoning](http://thetruthaboutstuff.com/pdf/(116)%20Cooper%201962%20Biochemical%20Aspects%20of%20Methano%20Poisoning.pdf) for more information.\n\n\n", "19" ], [ "\nI just wanted to point out that everything is toxic in the right proportion. THe problem is with the dose. I think you could drink a bit of methanol without getting particulary harmed. Do not do it, obviously.\n\n\nThe LD50 is very low at 5g/kg.\n\n\nEdit: **Nothing to do with methanol**, but it is a good infographic \n\n\n\nSource: <http://www.compoundchem.com/2014/07/27/lethaldoses/>\n\n\nAs a comparison with ethanol, **I believe** (not sure) that the oxidation product of these two compounds, ie, acetic acid and formic acid are not equally reactive. The former (metabolite of ethanol) is less reactive than the latter (metabolite of methanol). At least from an organic point of view this makes sense. But maybe there is another more biological reason.\n\n\n", "11" ], [ "\nOld thread but...\n\n\nMethanol toxicity is attributed to formic acid.\n\n\nBut formic acid itself is a \"low-toxicity\" food additive E 236 (authorized in animal feed only though, at ~1%, afaik), which is a bit confusing.\n\n\nI assume formic acid derived from methanol/formaline metabolism has different fate-distribution-bioavailability than ingested formic acid. Or it's simply a matter of dosage.\n\n\n", "1" ] ]

[ [ "\n**TL;DR:** Don't taste pure elements. They either taste of nothing, taste foul, or kill you. Or all of the above.\n\n\n**Edit:** Also, for clarification, elements don't have a taste. Taste is a biological reaction to chemical interactions happening with an element, not a property of the element itself. So, like everything else in chemistry, the taste is decided by the electron configuration of the element.\n\n\n\n\n---\n\n\nTaste as we know it happens in chemical reactions between our taste buds and the chemical species in question. Most of the taste is actually governed by the nose, where very specific receptors interact with very specific chemicals to send very specific signals to the brain, the chemistry of which is, while not trivial, also not interesting. It's your basic, run of the mill presumptive test.\n\n\nIn the taste buds, however, the receptors look for certain specific *characteristics* of the species being tasted. As an example, acids are sour, indicating that our sour taste buds are actually pH meters (although there might be more to it than this).\n\n\nSimilarly, most sugars have similar characteristics, that can be detected by a relatively simple chemical detector.\n\n\nIf, however, you go around tasting elements, you'll find two very interesting things.\n\n\nFirst, you'll find that you're dying from all kinds of poisonings. At a rough count, *all* of the periodic table is poisonous in its pure form, with only a few exceptions (carbon, oxygen, hydrogen, nitrogen, silicon, titanium, noble gases, gold and platinum, possibly a few others).\n\n\nSecond, you'll find that the pure elements don't taste much at all, if anything. Taste is a completely biological concept, and these elements simply aren't found in nature, meaning that the tongue and nose have no frame of reference.\n\n\nMost metals will taste metallic, due to tarnishes forming on their surface, mostly oxides, which all have the familiar, metallic taste. The non-metals are mostly fairly reactive, and taste absolutely foul as they react to the air in your moist mouth.\n\n\nGases like chlorine and fluorine may react and form chloride and fluoride, which taste sour, and then dissolve your mouth and nose, and then kill you.\n\n\nThe more radioactive metals will seem to taste of blood, but that's actually just acute radiation poisoning setting in, actually filling your mouth with blood. If they didn't kill you, they'd probably taste metallic like the rest of the metals, but they do, so they don't.\n\n\nFinally, the alkali metals like lithium and its friends will taste bitter, also kill you.\n\n\n**\"But wait!\"**, I hear you scream, \"What of, say, lithium in pharmaceuticals?\"\n\n\nWhen elements like lithium are ingested, they are done so in a rather more controlled fashion. You don't get a brick of pure, elemental lithium and told to go crazy when you go crazy, but rather, delivered in pills, in conjunction with a delivery agent, in this case, lithium citrate.\n\n\nMost elements have some biological role or other, but if they are in the kind of quantity you could taste, they are most likely poisonous, and in many cases, fatally so.\n\n\n", "7" ], [ "\nI don't know how many variables determine the taste but we recently found out that materials can have completely different properties just by the amount. We know that with different structures of atoms materials get completely different like diamond and graphit but, we know at least since the discovery of graphene that the periodic table might be much more complicated than we had imagined. \n\n\nEvery element gets completely different properties with just the quantity of atoms. If you take for example a really small amount of an element with a spoon it glows yellow, then you take with an other spoon a bigger amount and it glow red, then you take a bigger amount it glows blue. Currently it doesn't make any sense why this happens since you don't change the atomic structure of the material.\n\n\nIn the future the periodic table will get a 3rd dimension with every element becoming different with just the amount. This is not about the structure of the atoms or something, just the amount. \n\n\nFor the colors of matrials we have the wave lenght of the light as a reference but for the taste there are to much variables since we taste with our nose and our tongue.\n\n\n<http://periodictable.com/Posters/PlaceMatBackSide3000.JPG>\n\n\n<http://periodictable.com/Posters/LetterSizeBackSide1500.JPG>\n\n\n", "-1" ] ]

[ [ "\nIndeed, [Prof. Alán Aspuru-Guzik](http://aspuru.chem.harvard.edu) has placed his [\"clean energy project\"](https://cepdb.molecularspace.org) database on the web in a searchable form.\n\n\nAt the moment, it covers ~2.3 million compounds.\n\n\n\n> \n> Use of the data requires citation of the appropriate literature sources. The data in this database is released under the [Creative Commons Attribution ShareAlike](http://creativecommons.org/licenses/by-sa/3.0/) license and therefore any academic publications associated with the CEPDB should be appropriately [cited](http://www.molecularspace.org/about-us/how-to-cite).\n> \n> \n> \n\n\nAs you point out, the database only covers HOMO and LUMO orbital eigenvalues. While this may not be the most useful for experimental applications (i.e., you really want ionization potentials and first excited state energies or electron affinities) it should be a helpful \"first screen\" to find interesting lead compounds.\n\n\n", "3" ] ]

[ [ "\nLet's reword it:\n\n\n1. Traces of oil and grease are distilled off at temperatures below their boiling points.\n2. The temperature in the frying pan is above 100 °C.\n3. The food does contain significant amounts of water.\n\n\nThis sounds like the conditions for steam distillation, except that you're collecting the mixture of water and water-insoluble material everywhere in the kitchen and not in a flask.\n\n\n", "5" ] ]

[ [ "\nThe lowest energy MO is : $\\psi\\_1 = (1/2)(\\phi\\_A+ \\sqrt2\\phi\\_B+\\phi\\_C)$.\n\n\nThen comes the MO: $\\psi\\_2 = (1/\\sqrt2)(\\phi\\_A- \\phi\\_C)$.\n\n\nThe highest energy MO is $\\psi\\_3 = (1/2)(-\\phi\\_A+ \\sqrt2\\phi\\_B-\\phi\\_C)$.\n\n\nWhere $\\phi$ denotes the atomic orbital $1s$ on each hydrogen atom. $A$, $B$ and $C$ denote hydrogen atoms, where $B$ is the central one.\n\n\n", "5" ] ]

[ [ "\nAs you have already figured out that we have to use the ideal gas law $PV= nRT$, which is you have modified as $$\\frac{n}{V}= \\frac{P}{RT}$$\n\n\nTo find the density, we can further modify the equation as,\n\\begin{align}\n\\frac{m}{MV} &= \\frac{P}{RT}\\\\\n\\frac{\\rho}{M}&= \\frac{P}{RT}\\\\\n\\rho &= \\frac{PM}{RT}\\\\\n\\end{align}\n\n\nNow we just have to replace the parameters with their respective values.\n\n\nThe value of temperature should be converted to Kelvin as $(0 + 273)~\\mathrm{K} = 273~\\mathrm{K}$.\n\n\nLikewise, the value of gas constant $R$ should be chosen as $0.8206~\\mathrm{L~atm~K^{-1}{mol}^{-1}}$.\n\\begin{align}\n\\rho &= \\frac{1~\\mathrm{atm} \\cdot 44.09562~\\mathrm{g~mol^{-1}}}{{0.8206~\\mathrm{L~atm~K^{-1}{mol}^{-1}} \\cdot 273~\\mathrm{K}}}\\\\\n&=1.9683~\\mathrm{g/L}\\\\\n&=0.0019683~\\mathrm{g/cm^3}\\\\\n\\end{align}\n\n\n", "4" ] ]

[ [ "\n**TL;DR:** Water is incredibly easy to get and work with.\n\n\n\n\n---\n\n\nIn full\n-------\n\n\nWater is a good solvent for polar compounds[citation needed], and the reasons for this are laid out pretty well by John Snow, but that's not really what makes it the universal solvent.\n\n\nInstead, a series of other, incidental properties makes it a popular choice:\n\n\nAvailability\n------------\n\n\nThere's a *lot* of water[citation needed]. It *literally* falls from the sky. This means that for most applications, it's the absolutely cheapest solvent available.\n\n\nReactivity\n----------\n\n\nWater is a reasonably stable molecule. Few things you can dissolve in water are expected to react significantly with it. It won't readily burn or combust, even in vapour form, and its high heat capacity and prevalence (see above) means that it can effectively douse most runaway reactions that might occur, such as fires (there are times when adding water to a fire is a bad idea, especially perhaps in a lab, but even those times the problem can be solved by adding *a lot* of water).\n\n\nToxicity\n--------\n\n\nUnlike some other common solvents, water is not toxic[citation needed], and is not a bio- or eco-hazard. This means that unused solvent can be disposed of without special precautions, and if the solute can be disposed of trivially, the solvent can as well.\n\n\nPurity and stability\n--------------------\n\n\nWater is everywhere[citation needed], giving it another unique property: Unlike many other solvents, water won't absorb atmospheric water vapour, decreasing its purity as a matter of course. (It'll still exchange water with the atmospheric vapour, of course, but it doesn't significantly change its make-up.)\n\n\nDistilled 100 % water will remain more or less pure unless something is actively added to it, which, while not unique, is useful and somewhat uncommon. Water molecules will also not spontaneously decompose under most sane conditions.\n\n\nOperating temperatures\n----------------------\n\n\nWater is liquid at a wide range of temperatures, stretching from 0°C to 100°C. Under laboratory conditions of ~1 atm and 20-25°C, water is a well-behaved and practical liquid.\n\n\nFurthermore, while the liquid phase is available at a wide range of temperatures, the solid and gaseous states are not out of reach when using even extremely basic equipment. Cooling water to well below freezing, say -18°C, or to boiling, are both trivial activities, using equipment that is not only available in every lab, but even most homes.\n\n\nIn short...\n-----------\n\n\nWater isn't always the best tool for the job, but it's almost always a good or great one, at least when working with polar compounds. A common alternative to water is ethyl alcohol, which has similar polar properties and operating temperatures, but ethanol does have weaknesses in the other categories mentioned here, i.e. its reactivity, its absorption of water vapour, and its price.\n\n\n", "22" ], [ "\nTo directly address where the phrase comes from:\n\n\n\n> \n> Water is called the \"universal solvent\" because it dissolves more substances than any other liquid. -[USGS](http://water.usgs.gov/edu/solvent.html)\n> \n> \n> \n\n\nWhat are ideal qualities of solvent?\n------------------------------------\n\n\nThe strength of a solvent can be attributed to the strength of its intermolecular forces like london forces, dipole-dipole forces, ion-induced dipole, and hydrogen bonding. These are forces of attraction and repulsion. Solvation occurs when a molecule is surrounded by the solvent, so when there are strong intermolecular forces, stronger solvation occurs. \n\n\n\n\n\nThe hydrogen bond\n-----------------\n\n\nWater makes use of the hydrogen bond, a type of intermolecular force experienced when hydrogen is attracted to the electronegative atoms nitrogen, oxygen, or fluorine. Hydrogen bonding is the strongest intermolecular force.\n\n\n\n\n\nHydrophobia\n-----------\n\n\nWater is a great solvent for hydrophilic molecules but hydrophobic molecules are, by definition, not easily disturbed by water. This makes a clear exception to the \"Universal Solvent\" idea. I haven't read any literature claiming it to actually be the *best* solvent, but it does work well in many chemical situations.\n\n\n", "21" ], [ "\nJust to provide an alternative answer:\n\n\nConsider a [solvent miscibility table](https://www.erowid.org/archive/rhodium/pdf/solvent.miscibility.pdf) like the one linked.\n\n\nWhat is the least miscibile solvent? Water! Water is the worst solvent. Water is immiscible with 17 out of 30 of the other listed solvents. \n\n\nThere are 6 solvents in the table which are miscible with all the other solvents: ethanol, acetone, tetrahydrofuran, n- & iso-propanol and dioxane. \n\n\nWhy is water such a bad solvent? Hydrogen bonds. Water molecules are strongly hydrogen bonded to other water molucules. To dissolve something else, these hydrogen bonds need to be broken. \n\n\nFor another prespective, look across the periodic table. What does water stably dissolve? Does water dissolve gold, silver, copper, tin, lead, aluminum, and many other metals and metaloids like antimony, silicon, and germanium, [as mercury does](http://www.tandfonline.com/doi/abs/10.1080/10408347408542677?journalCode=batc19#preview)? \n\n\nThere is nothing universal about water, except that there is a lot of it on Earth and we need it to live. \n\n\n", "17" ], [ "\nSince water is polar as you note, and \"like dissolves like\" remains a good principle, I think the answer is that lots and lots of substances are polar. Most (ionic) salts are, for example. Even substances that are polar covalent are at least slightly polar and so somewhat soluble (and with agitation, more soluble). So you have to be really non-polar (e.g. hydrocarbons, \"oils\") to not be soluble in water.\n\n\n", "12" ], [ "\nIt's worth noting some of the history behind the term \"universal solvent,\" and why it is used even though water isn't truly *universal* or even necessarily the most versatile solvent available.\n\n\nBefore chemistry existed, and for that matter before science existed as the practice we would recognize today, alchemy was an important protoscience. A number of important breakthroughs and refinements came about due to the work of alchemists, and for all that it often got buried in mysticism (and, at times, fraud), alchemy began the research into the chemical properties of matter. Chemistry developed largely in reaction to alchemy, rejecting the mystical underpinnings, and applying the scientific method.\n\n\nWithin alchemy, there was a belief in the existence of a \"universal solvent,\" which would be part of the process for producing the philosopher's stone and/or transmuting lead into gold. Numerous alchemists claimed to know how to make it, while many others considered producing it to be one of their major, overarching goals, the most difficult or secret step in obtaining whatever powers they believed they could achieve through alchemy.\n\n\nObviously, no chemical with the properties described for this universal solvent has ever been found. However, because of the critical use of water as the default solvent in chemical work, it has often been referred to with the title. Other answers describe better than I can why water is a particularly useful solvent.\n\n\n", "11" ] ]

[ [ "\nNaming complex chemical compounds in a straightforward systematic fashion is usually not very easy. And often there are trivial names, that existed way before this systematic naming scheme was introduced. Especially when it comes to naturally occurring compounds like steroids. The IUPAC defines [steroids](https://goldbook.iupac.org/S06005.html) as the following:\n\n\n\n> \n> Naturally occurring compounds and synthetic analogues, based on the cyclopenta[a]phenanthrene carbon skeleton, partially or completely hydrogenated; there are usually methyl groups at C-10 and C-13, and often an alkyl group at C-17. By extension, one or more bond scissions, ring expansions and/or ring contractions of the skeleton may have occurred. Natural steroids are derived biogenetically from triterpenoids. \n> \n> [](https://i.stack.imgur.com/q1V7T.png) \n> \n> (source: [iupac.org](https://goldbook.iupac.org/SVG/S06005.png)) \n> \n> \n> \n\n\nThere is even a quite lengthy article about the nomenclature of steroids: [1] [G. P. Moss, *Pure & App. Chem.* **1989**, *61* (10), 1783-1822.](http://dx.doi.org/10.1351/pac198961101783) \n\n\nThe name Deoxycorticosterone acetate is itself derived from a trivial name. The underlying principle compound is [corticosterone](http://en.wikipedia.org/wiki/Corticosterone) and its systematic steroid name is 11β,21-dihydroxypregn-4-ene-3,20-dione, see [1] Table 2. It can be derived from [cortisol](http://en.wikipedia.org/wiki/Cortisol), which has an additional hydroxy group in position 17. The systematic name therefore is (11β),17,21-trihydroxypregn-4-ene-3,20-dione.\n\n\nYou assigned correctly, that the prefix de(s)oxy- refers to losing an oxygen containing functional group. In this case a hydroxy group at position 11. The systematic name for 11-deoxycorticosterone would therefore be 21-hydroxypregn-4-ene-3,20-dione.\n\n\nFinally the acetate refers to an acetyl ester group at position 21 (in the sidechain). Interestingly the systematic name for cortisol acetate is 21-*O*-acetylcortisol. Since it has already a trivial name, see [1], 3S-4.3:\n\n\n\n> \n> When the steroid acid has a trivial name that already specifies the hydroxyl group that is esterified, the prefix must be *O*-acyl-.\n> \n> \n> \n\n\nApplying the same logic, the systematic name for Deoxycorticosterone acetate should be 21-*O*-acetyldeoxycorticosterone. Which makes little to no sense. A more systematic approach is described again in [1], 3S-4.3 \n\n\n\n> \n> *Esters of steroid alcohols* \n> \n> Esters of steroid alcohols are named by the appropriate steroid substituent group followed by that of the acyloxy group in its anionic form. The steroid name is formed by replacing the terminal -e of the hydrocarbon name by -yl, -diyl, etc. and inserting this before the locant with a Greek letter to designate position and configuration. When necessary locants should be used with the anionic part of the name. Examples: [...] \n> \n> The prefix form acyloxy-, with the appropriate locants, is used if there is a preferred functional group as the suffix (e.g. an acid or lactone). Examples: [...]\n> \n> \n> \n\n\nSo in this case since you have a ketone, which is probably more important, I would choose the second part of this rule. Hence the systematic name becomes 21-acetoxypregn-4-ene-3,20-dione.\n\n\nAs you can see, naming is quite a big, often confusing subject and sometimes the origins are not clear at all. And your analysis was completely correct in every way. Trivial names are often like the vocabulary of a specific language - if you don't know it, you have to look it up.\n\n\n\n\n---\n\n\n*Additional fun fact*: \n\nThe name acetate is a [retained name](http://en.wikipedia.org/wiki/Retained_name) which can be used in a [semisystematic](https://goldbook.iupac.org/S05601.html) way.\n\n\n", "7" ], [ "\n\n> \n> What is the law behind naming \"corticosterone\"?\n> \n> \n> \n\n\nThere is no law, since the name indicate where the compounds are produced. Corticosterone and corticosteroid are synonyms for a groups of compounds (steroids) that are [produced in the cortex of the adrenal glands](http://en.wikipedia.org/wiki/Corticosterone). [Cortisone](http://en.wikipedia.org/wiki/Cortisone) and [cortisol](http://en.wikipedia.org/wiki/Cortisol) refer to 2 of the corticosteroids. \n\n\n", "4" ] ]

[ [ "\nThe Friedel-Crafts alkylation is another example of an electrophilic aromatic substitution. I've drawn the mechanism below showing the formation of cyclohexylbenzene. The reaction is typically run in the presence of an acid and the aluminum chloride serves as a catalyst to enhance the electrophilic properties of the proton, at the end of the reaction the aluminum chloride has been regenerated.\n\n\n\n\n\nIn the insert box I've shown that when the reaction is carried out with n-propylchloride isopropyl benzene is produced, not n-propylbenzene. A carbocation intermediate is generated in thr Friedel-Crafts alkylation and will rearrange to a more stable carbocation if possible. In the cyclohexene case, rearrangement of the intermediate carbocation just regenerates the same cyclohexyl carbocation, so the rearrangement is \"invisible\" in this case (unless we label the starting cyclohexene).\n\n\n", "6" ], [ "\n**MECHANISM OF ALKYLATION**: At first, electrophile methyl carbonium ion ($\\ce{CH3}$) is formed from $\\ce{CH3Cl3}$ and $\\ce{AlCl3}$, which, at the second step, attracted by $\\pi$-electron of the benzene ring to form sigma complex. At last step, the approach of nucleophile like, $\\ce{AlCl4-}$, one proton is given out from sigma complex to form stable product toluene\n\n\n", "0" ] ]

[ [ "\nTo solve this problem, you need to know the pKas of carbonic acid.\n\n\nThe pKas are 6.35 and 10.33.\n\n\nIf you add 1 equivalent of acid to $\\ce{Na\\_2CO\\_{3}}$, the product will be $\\ce{NaHCO\\_{3}}$ (sodium bicarbonate).\n\n\nIn this case, you are going 0.4-.377 = 0.023 moles past the point of 1 equivalent. \n\n\nI would model the final solution as 0.354 moles of $\\ce{HCO\\_{3}^-}$ and 0.023 moles $\\ce{H\\_2CO\\_{3}}$ , and use the pKa of 6.35 and the Henderson-Hasselbalch equation to calculate pH.\n\n\nThe proposed answer 10.77 is incorrect. \n\n\n", "1" ] ]

[ [ "\nThe boiling point elevation fomula assumes that all particles are equal, only the number of particles in the solution matters. \n\n\nSo I would use:\n\n\nm = (0.2 + (3)(0.8))(molality of $\\ce{MgCl\\_2}$) + (0.2 + (2)(0.8))(molality of $\\ce{NaCl}$)\n\n\nTo find molality of $\\ce{MgCl\\_2}$ and molality of $\\ce{NaCl}$, consider a 1kg sample of the solution. There are 95g $\\ce{MgCl\\_2}$, 117g $\\ce{NaCl}$ and 788g water. \n\n\nCalculate the number of moles $\\ce{MgCl\\_2}$ in 95g $\\ce{MgCl\\_2}$, and divide this number by 0.788kg. This is the molality of $\\ce{MgCl\\_2}$ in the solution. \n\n\nRepeat for $\\ce{NaCl}$\n\n\n", "1" ] ]

[ [ "\nYou have multiplied the mass of the sample, 1.50g, by temperature change and heat capacity. \n\n\nHowever, the water provides most of the heat for the reaction. \n\n\nThe total mass of the solution is 1.50g + 35.0g = 36.5g. \n\n\nYou should be multiplying 36.5g by the temperature change and heat capacity.\n\n\nThen, you need to consider how many moles 1.50g KCl is. \nDivide the change in enthalpy of the solution by the number of moles of KCl to determine the molar heat of solution of KCl. \n\n\n", "3" ], [ "\nHere's the calculation, step by step:\n\n\n$$q\\_\\mathrm{cal} = 36.5\\cdot 4.18\\cdot (-2.4) = \\pu{-366 J}$$\n$$q\\_\\mathrm{rxn} = -q\\_\\mathrm{cal} = \\pu{366 J}$$\n$$n(\\ce{KCl}) = \\frac{\\pu{1.50 g}}{\\pu{74.55 g mol-1}} = \\pu{0.0201 mol}$$ \n\n$$\\frac{\\pu{366 J}}{\\pu{0.0201 mol}} = \\pu{18.209 J mol-1} = \\pu{18.2 kJ mol-1}$$\n\n\n", "0" ] ]

[ [ "\nThe answer above only shows one of the two possible isomers that could be formed by elimination from the tertiary carbocation intermediate. \n\n\n\n\n\nI suspect that both isomeric cyclohexenes with tri-substituted double bonds would probably be produced in roughly equal amounts, along with a smaller amount of the exo-methylene compound.\n\n\n**Edit:** Thanks to Bon and Yolo123 for their comments\n\n\nElimination towards the isopropyl group is also a possibility, but two factors argue against this being the major elimination pathway. First there is a statistical factor; there is only proton on the central isopropyl carbon available for elimination, whereas the are 4 (2+2) protons available alpha to the carbocation center in the 6-membered ring. Second, if you look at a model and consider that the proton to be eliminated must be lined up with the carbocation p-orbital you see that each methylene group adjacent to the carbocation has one hydrogen well lined up for elimination, whereas to properly line up the isopropyl proton results in some adverse steric interactions between the isopropyl methyl hydrogens and the ring methylene hydrogens. \n\n\nMy guess is that the gem-dimethyl group in the cyclohexane ring will not favor one in-ring elimination pathway over the other - my model doesn't show them exerting any adverse steric effect one way or the other.\n\n\n", "2" ] ]

[ [ "\nI think the most widely-used approach at the moment comes from Roi Baer and Leeor Kronik and others, e.g.\n\n\n* [*Phys. Rev. Lett.* (2010) 105, art. 266802](http://dx.doi.org/10.1103/PhysRevLett.105.266802)\n* [*Phys. Rev. B* (2011) 84, art. 075144](http://dx.doi.org/10.1103/PhysRevB.84.075144)\n\n\nBasically the idea is to tune the range-separation parameter $\\gamma$ between short-range and long-range electrostatic effects in a range-separated hybrid functional. You find a match such that Koopmans' theorem holds.\n\n\nThe result is an \"optimally-tuned\" range-separated hybrid DFT functional (OT-RSH).\n\n\nIt's not perfect, for example, it violates size consistency: [*J. Chem. Phys.* (2013) 138, 204115](http://dx.doi.org/10.1063/1.4807325)\n\n\n", "5" ] ]

[ [ "\nThe difference in electronegativity between copper (1.9) and magnesium (1.3) is the key difference. \n\n\nSince copper is more electronegative than magnesium its electronegativity is much closer to that of carbon (2.55). This results in carbon-copper bonds being less polarized and more covalent than carbon-magnesium bonds. The electrons in a carbon-copper bond are better shared between the two atoms, whereas the electron distribution in a carbon-magnesium bond places more electron density on carbon.\n\n\nAs a result of this difference in electron density on the carbon atoms the carbon portion of a cuprate is a softer nucleophile, while the carbon portion of a Grignard is a harder nucleophile. These difference in nucleophilicity produce different reactivity patterns for the two reagents. For example, the softer cuprate will undergo 1,4 addition to an $\\alpha,\\beta$-unsaturated carbonyl compound (a softer electrophile), while the Grignard adds directly to the carbonyl (which is a harder electrophile).\n\n\n", "11" ] ]

[ [ "\nThe difference between THF and epoxides is perhapse bigger than you might think. Due to the three-membered-ring of an epoxide, a lot of energy is 'stored' in ring strain. This destabilizes the stucture and is a driving force for a Grignard reaction (or any substitution on an epoxide). \n\n\nThe THF is a five-membered-ring, which is far more stable. This bond is simply an ether, that happens to be in a cyclic structure. Therefore it also reacts like an ether: not easily attacked by nucleophiles.\n\n\n", "7" ] ]

[ [ "\nThe upper graph is just the most simple way to visualize the effect of a catalyst on a reaction $\\ce{S -> P}$:\n\n\n**The activation energy is lowered**.\n\n\nThe activation energy for the reaction in that direction is the difference of the energies of the starting material $S$ and a transition state $TS^\\#$. \n\n\nSince it is the same starting marterial in the presence or absence of the catalyst, the **energy of the transition state** is different.\n\n\nCan the same transition state have two different energies - just through the remote distance magical action of a catalyst located somewhere? Probably not!\n\n\nIt is much more plausible that - in the absence and presence of a catalyst - two completely different (different in structure and different in energy) transition states are involved.\n\n\nExactly this situation is described in the second graph!\n\n\nThe catalyst \"reacts\" with the starting material, either by forming a covalent bond, by hydrogen bonding, etc. and thus opens a different reaction channel with **different intermediates and a transition state** not found in in the non-catalyzed reaction.\n\n\nIn the end, the same product $P$ is formed and the catalyst is regenerated, but this doesn't mean that the catalyst wasn't heavily involved in the reaction going on.\n\n\n", "7" ], [ "\n1# Curve 1 is just for a simple case where it's showing the comparison between a theoretical pathway for a catalyzed and non-catalyzed reaction but it has an another significant which I will discuss in part two. \n\n2# How many of these curves can occur: Infinite (Theoretically) but in practical case you have to consider all types of possible elementary reaction in your model. An elementary reaction is a reaction which can occur in one step or the reactant, product and TS has well defined position on potential energy surface. For example if you have a simple ketone hydrogenation reaction, you can get an overall energy profile for gas phase reaction but on a catalytic surface there are lots of possibilities. At first the ketone and hydrogen will adsorb on the surface and then 1st hydrogen addition can occur either on C of the ketone group or on the hydrogen of the ketone group. They are intuitively main reaction pathway but there can be much more. \n\n[](https://i.stack.imgur.com/O1WW4.jpg) \n\nIn the above image I was working on LA hydrogenation and you can see there are two different pathway for the hydrogenation and there might be some other more pathway. And if you want to capture the real chemistry you have to make sure you are including all important intermediates. So your energy profile may look like this\n[](https://i.stack.imgur.com/R6pTI.jpg)\nHere the picture becomes quite complicated with lots of elementary reaction. To determine the overall activation energy of a catalyzed reaction scientists follow two methods \n\nmethod-1: Energetic span theory: Given by Shaik et al. I am adding the link to that paper here,\n<http://pubs.acs.org/doi/ipdf/10.1021/ar1000956>\n\n\n2) Apparent activation barrier: This is an age old method. Here the sensitivity of turn over frequency (TOF) with the change of temperature is measured and defined as apparent activation barrier.\n$$E\\_a=-R\\frac{\\delta~ln(TOF)}{\\delta(\\frac{1}{T})}$$ \n\n\nNow you can plot the first E profile with your gas phase activation energy and apparent activation energy of the catalyzed reaction.\n\n\n", "3" ], [ "\n\n> \n> A reaction involving more than one elementary step has one or more intermediates being formed which, in turn, means there is more than one energy barrier to overcome. In other words, there is more than one transition state lying on the reaction pathway. As it is intuitive that pushing over an energy barrier or passing through a transition state peak would entail the highest energy, it becomes clear that it would be the slowest step in a reaction pathway. However, when more than one such barrier is to be crossed, it becomes important to recognize the highest barrier which will determine the rate of the reaction. This step of the reaction whose rate determines the overall rate of reaction is known as rate determining step or rate limiting step. The height of energy barrier is always measured relative to the energy of the reactant or starting material. Different possibilities have been shown in [figure 6](https://upload.wikimedia.org/wikipedia/commons/thumb/6/6d/Reaction_Co-ordinate_Diagrams_for_reactions_with_0%2C_1%2C_2_intermediates.png/500px-Reaction_Co-ordinate_Diagrams_for_reactions_with_0%2C_1%2C_2_intermediates.png)\n> \n> \n> \n\n\nFrom the link you gave on energy profile i found this. Basically there is no **\"net change\"** in $E\\_{a}$ . In the first graph in question the curve which is at the top is the representing how the reaction proceeds without a catalyst and the one below is of the reaction with catalyst. In the second graph there are 4 intermediate reactions/transition states before the product is formed. Hence you have 4 peaks of different activation energy as shown. \n\n\nA catalyst basically provides alternate routes to a reaction by forming the **\"transition/metastable/intermediate state** at a lower energy than the original reaction. You will study later that this has to do with basic Collision theory and molecular dynamics. \n\n\nIn the given graphs a catalyst can make the **R->P** through just one route where only one transition happens (as in the graph 1) or for 4(or even more) transition states (as in the graph 2).\n For one set of conditions(P,T,V etc) I am guessing the number of these peaks will be unique. It is also possible that for a certain catalyst and reactant the number of peaks / intermediate states is constant.\n\n\n", "2" ] ]

[ [ "\nUnfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding.\n\n\nYour explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp2 hybrid orbitals, giving rise to two σ bonds to the neighbouring carbons and one σ bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. *D*6h, three of them are bonding and three of them are antibonding. In the π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.)\n\n\n[](https://i.stack.imgur.com/30E9Z.jpg)\n\n\nThe nitrate ion $\\ce{NO3-}$ is what I like to call a [Y-aromat](https://chemistry.stackexchange.com/q/32354/4945) since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene. \n\nThe resonance structures, however, look much more complicated.\n\n\n[](https://i.stack.imgur.com/Th3Jh.jpg)\n\n\nIn quite good approximation, you can regard the nitrogen as having three sp2 hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen σ bonds. Each oxygen has one lone pair in the in-plane p orbital and one lone pair in the sp hybrid. The remaining out-of-plane p orbitals of nitrogen and oxygen build up the delocalised π system. This time there are only four orbitals. Because of symmetry constraints, i.e. *D*3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.\\* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene. \n\n\n[](https://i.stack.imgur.com/Vqc2x.jpg)\n\n\nSince the carbonate anion $\\ce{CO3^{2-}}$ is isoelectronic to nitrate, approximately the same orbital picture will result.\n\n\n\n\n---\n\n\n\\* Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another. \n\nFor example, the E\" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $\\ce{N-O}$ bonds and antibonding towards the other. (It is also bonding with respect to a $\\ce{O\\bond{~}O}$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that. \n\n\n[](https://i.stack.imgur.com/ovpOp.jpg)\n\n\n", "18" ] ]

[ [ "\nAs with every process, dissolving can be described by *the* thermodynamical equation:\n\n\n$$\\Delta G = \\Delta H - T \\Delta S$$\n\n\nWhich the process being spontaneous if $\\Delta G$ is negative.\n\n\nWe have two components here, enthalpic and entropic. The enthalpic component $\\Delta H$ is a constant of the salt in question. The entropic component $\\Delta S$ can usually be considered positive because a highly ordered crystal lattice is broken to give dissolved, much less ordered fragments — hence why for most salts they dissolve easier in hot water.\n\n\nIf the solubility *decreases* with higher temperatures, then $\\Delta S$ must be negative. The salt will then only dissolve if $\\Delta H$ is negative enough to give an overall negative result. This seems to be the case for $\\ce{Na2SO4}$\n\n\nIt is impossible to predict $\\Delta S$ and $\\Delta H$ *a priori* without any computer quantum calculations. One has to resort to experiments to determine whether a salt will dissolve better or worse in the heat.\n\n\n\n\n---\n\n\nWhile I have used the term ‘salt’ most of the time, the same is true for molecules. Take sucrose: I was asked in a physical chemistry lab to guess with my colleague whether sugar would dissolve endothermicly or exothermicly. We decided, that the degree of order be a lot higher in dissolved sugar due to a high amount of water surrounding it rather tightly, that the entropy must decrease, so that the enthalpy has to be negative and sugar has to dissolve exothermically. Hence sugar should dissolve more easily in cold water, too.\n\n\nAccording to a comment by Nicolau Saker Neto, our reasoning was false and sucrose displays [increasing solubility at higher temperatures](http://chestofbooks.com/food/science/Experimental-Cookery/The-Solubility-Of-The-Sugars.html). Apparantly the testing faculty member approved of it, though.\n\n\n", "6" ] ]

[ [ "\nMany different carbon ions exist in the gas phase. \n\n\nIn [Electronic spectra of linear carbon anions](http://www.sciencedirect.com/science/article/pii/S0301010497003583), spectra of $\\ce{C\\_2^-}$, $\\ce{C\\_3^-}$... $\\ce{C\\_10^-}$ are discussed. \n\n\nDianions such as $\\ce{C\\_7^{2-}}$ have also been observed.\n\n\nIn the solid phase, aluminum carbide and berylium carbide have more than 50% ionic character according to the Handbook of Refractory carbides and nitrides, and correspond to $\\ce{C^{4-}}$. Calcium carbide, which could be considered a salt of acetylene, corresponds to $\\ce{C2^{2-}}$ rather than $\\ce{C^{2-}}$ \n\n\n", "5" ] ]

[]

[ [ "\nThe principle that \"heavier nuclei are more stable\" is a very crude approximation. Theoretical explanations of nuclear stability are complex. While the transuranium elements become progressively more unstable with increasing atomic mass, this trend is not predicted to continue indefinitely; there is an island of stability around copernicium and flerovium, and the as-of-yet unsynthesized copernicium-291 has been predicted to have a half-life of 1200 years or so. You may be interested [this paper](http://iopscience.iop.org/1742-6596/420/1/012001/pdf/1742-6596_420_1_012001.pdf) for details; figure one is a nice illustration of how stability is affected by neutron/proton number towards the end of the periodic table.\n\n\n", "5" ], [ "\nIf the ratio of neutrons to protons in the nucleus is too high, the nucleus is unstable to beta decay. A neutron emits an electon and becomes a proton. \n\n\nIf the ratio of neutrons to proton is too low, the nucleus is unstable to positron decay and/or electron capture. \n\n\nBut how high/low is too high/low, and why does this occur?\n\n\nWe need to look at the [Semi-empirical mass formula](http://en.wikipedia.org/wiki/Semi-empirical_mass_formula).\n\n\n$$E = vA -sA^{2/3} - c\\frac{Z^2}{A^{1/3}} - a\\frac{N-Z}{A} - \\delta (A,Z) $$\n\n\nWhere $E$ is the binding energy of the nucleus, $N$ is number of neutrons, $Z$ is number of protons, $A$ is $N + Z$, and $v$, $c$, $s$ and $a$ are empirical constants. \n\n\nOnly the $c$ (coulomb) and $a$ (asymmetry) terms involve the relative number of protons and neutrons. \n\n\nThe asymmetry term is zero when the number of protons and neutrons are equal, and relates to the Pauli exclusion princple (neutrons and protons being fermions).\n\n\nExcept for the coulomb term (that protons electrostatically repel each other), the ideal ratio of protons to neutrons would be 1 to 1. Electrostatic repulsion of protons explains why somewhat more neutrons than protons is favorable. \n\n\n", "4" ], [ "\n\"The heavier the atom, the more unstable it gets, right?\" Wrong. Nickel-62 is the \"most stable\" isotope, measured by binding energy per nucleon. Since in the core of massive stars, it is subject to photodissociation, Fe-56 is often considered to be the \"most stable\" isotope. This discrepancy shows that context matters. Take a look at the graph of binding energy/nucleon vs nucleon number <https://en.wikipedia.org/wiki/Nuclear_binding_energy#/media/File:Binding_energy_curve_-_common_isotopes.svg> You can see that the graph is nearly flat around ~±10 nucleons of Fe-56, so there's no value in making broad statements, imho. Even if unlike the OP, the statement was accurate, the devil is in the details (context, environment, what is meant by \"stability\" , etc.). Why did anyone bother to resurrect this old question, I wonder? Heavier nuclei are neither or both more and less stable than lighter nuclei. It isn't (imho) a \"very crude approximation\", it is simply wrong. A more accurate \"crude\" approximation is to say that low mass (not weight, for God's Sake!) elements tend to release less and less fusion energy up to approximately iron, and also past iron, the tendency is a slow increase in the fission energy released. (but details matter, especially the exact balance between the various forces). You simply can not predict the stability of an isotope by using only its mass (at least, not below ~ 250 or 300 amu).\n\n\n", "1" ] ]

[ [ "\n\n> \n> [For silver], why don't the electrons in the 4d orbital move to the 5p orbital?\n> \n> \n> \n\n\nIn the case of silver, this is likely because the energy separation between the 4d and 5s orbitals are quite large ([absorbing in ultraviolet region, which is why silver appears white/silver](https://physics.stackexchange.com/questions/72368/why-are-most-metals-gray-silver)), so the energy separation between 4d and 5p would be even larger.\n\n\nIn general, when you write down the electron configuration you are writing down the **ground state**, i.e. the most stable configuration. While it is possible in principle to move electrons from 4d to 5p, it will not give you the most stable configuration as moving that electron requires a large amount of energy. \n\n\nFor inferring the likely valence states, you can only use orbitals with unpaired electrons, and/or move the electrons that are not separated by large energy differences between different orbitals to make hybrid orbitals with unpaired electrons. In this case, as you cannot move 4d electrons to 5p (which would make some kind of spd hybrid orbital), you can only have a valence state of 1 from the unpaired electron in 5s.\n\n\n\n> \n> Is it possible for any elements [to move their electrons between $n$d and $(n+1)$p orbitals]?\n> \n> \n> \n\n\nYes, but only for elements nearer the top of the periodic table (and not at the very top, as you will then not have any accessible d orbitals). Period 3 and period 4 (I think) would be most likely to do this. For example, phosphorus can have a valence state of 5, and sulfur can have a valence state of 6, both by involving their accessible 3d orbital, to make hybrid $\\ce{sp^3d}$ and $\\ce{sp^3 d^2}$ orbitals respectively. \n\n\n\n> \n> When the f orbital is [completely] filled with 14 electrons, should we take it in consideration to find the valence states?\n> \n> \n> \n\n\nIn general, no. When the f block is completely filled, the electrons in the f orbital are generally assumed to be core-like, i.e. stable, or low in energy, therefore will have a large energy gap to the other electrons in 6s, 5d and 6p (or 7s, 6d and 7p), and do not tend to participate in bonding/valency. There are possibly one or two exceptions, but I cannot recall them right now. \n\n\n\n> \n> When Z=58, can you show me the electronic [configuration] and how do we form the valence state(s)?\n> \n> \n> \n\n\nZ = 58 is Cerium, which has the following electronic configuration: $\\ce{[Xe] 4f^1 5d^1 6s^2}$. \n\n\nOn paper, it will appear that due to the diffuseness of the orbitals 4f, 5d and 6s, it is probably unusual to form covalent compounds to cerium, and you would expect a valence state of 3 (if you only use the 5d and 6s electrons), or a valence state of 4 (if you also use in addition the 4f electron).\n\n\n[It was found the 4f is actually near the energy of 5d and 6s, giving rise to a variable electron structure and a valence state change of 3 to 4 when cerium is cooled or compressed](http://en.wikipedia.org/wiki/Cerium#Physical_properties). Moreover, it is possible to form [organocerium compounds with a valency of 3](http://en.wikipedia.org/wiki/Organocerium_chemistry). By the way, this is not something that you can deduce simply by looking at the electronic configuration. You have to find this out empirically, using experiments!\n\n\n", "1" ] ]

[ [ "\nApart from titanium dioxide, the active species in sunscreens are organic compounds from different substance classes, such as cinnamates, benzophenones, benzimidazoles, and triazines.\n\n\nThe do absorb in the 280-315 (UVB) and/or 315-380 nm (UVA) range. This \"uptake\" of energy results in an electronic excitation, the compounds undergo a transition from the $S\\_0$ (ground) state to an excited $S\\_1$ state.\n\n\nThe excited singlet state ($S\\_1)$ may \n\n\n* deactivate radiationless via *internal conversion* (**IC**), the energy of the photon is released as heat or\n* undergo *intersystem crossing* (**ISC**) to a triplet state (usually $T\\_1)$ with different spin multiplicity\n\n\nThe latter state then might again loose its energy via **IC** or undergo a chemical transformation, through which the suncreen might be turned to a species that not longer is able to absorb UV radiation.\n\n\n", "4" ], [ "\nJust to add, Zinc (like Titanium) is another UV filter that reflects UV light rather than absorbing it. I believe ever other UV filter, at least those approved in the U.S., absorb the UV rays.\n\n\n", "0" ] ]

[ [ "\nIf I understood what strong means, then graphene has high strength because of the 2-D layer structure of carbon atoms. The strength can also be attributed to its **$sp^{2}$** hybridisation.\n\n\nAny crystalline material will have a lattice structure. When we mean strength of a material its usually the ability to break apart the bond between two lattice atoms. But the observed strength is usually 2 or 3 orders lower than theoretically predicted strength due to **flaws (voids, dislocations etc)**. But when graphene has a near flawless 2-D structure we can reach its theoretical strength. Though its is a very expensive process to actually realize **\"flawless\"**. Even with flaws its strength is really high compared to steel.\n\n\nSo its essentially the **2-D one atom layer structure** that can be responsible for its high strength.\n\n\nOne important thing is though graphene has high strength and elasticity (ability to get stretched) it is relatively brittle (fracture toughness ~4 MPa$\\sqrt{m}$).\n\n\n", "4" ], [ "\nImagine carbon as some sort of wool.\nGraphite would be just some wool without any pressure on it.\nDiamond would be a wool ball with a structure and a shape.\nGraphene would be a textile knitted with the strings of wool.\n\n\nWool is soft, a wool ball is more durable to pressure and a textile could carry some weight.\n\n\nNow imagine graphene as a canvas fabric knitted with the finest strings the wool has and all strings are equally long. The pressure would be splitted equally on each of this strings and you wouldn't have a weak point. You would carry an elephant with it.\n\n\n", "1" ] ]

[ [ "\nI would have to say \"f\" overwhelmingly stands for \"fundamental\" (not \"fine\"). \n\n\nAn early (1915) reference is \"Some Recent Discoveries in Spectrum Series\" Astrophysical Journal, vol. 41, p.323 \n\n\n\n> \n> Within each of these three systems there are, in general, four different *types* of series, known as \"principle\", \"sharp\" (second subordinate), \"diffuse\" (first subordinate), and \"fundamental\" (Bergmann). These names are not always very appropriate; for instance the \"sharp\" series is often diffuse, the \"fundamental\" may be composed of the faintest lines in the spectrum, but this is not vital.\n> \n> \n> \n\n\nThe fundamental series was discovered by Arno Bergmann in 1907, and is sometimes called the Bergmann series. It was named \"fundamental\" by William M. Hicks, I beleive in [A Critical Study of Spectral Series. Part I: The Alkalies H and He](http://www.jstor.org/stable/90988) (1911): \"F is the symbol for the new series (fundamental) as P.S.D. stand for those already known\" (page 57) and \"It was my first impression that this would be the same for all the alkalies and the .987 of the earlier discussion. It would be the basis of a fundamental series for each\nelement, and the letter F which has been attached to it had its origin in this idea.\" (page 94).\n\n\nThe names existed before orbital theory. There is no need to investigate f-block elements to observe excitations to f-orbitals. Electrons of any elment can be excited to an f-orbital. In fact, Bergmann discovered the series in the sodium spectrum. What was orginially known as the [sharp series](http://en.wikipedia.org/wiki/Sharp_series), were transition to s-orbitals, the primary series transitions to p-orbitals, etc. \n\n\nSee [The Origin of the s, p, d, f Orbital Labels](http://www.che.uc.edu/jensen/W.%20B.%20Jensen/Reprints/137.%20s,%20p,%20d,%20f.pdf) for additional information. \n\n\n", "6" ] ]

[ [ "\nInterestingly, the first paper from Pauling on electronegativity [\"The Nature of the Chemical Bond. IV. The Energy of Single Bonds and The Relative Electronegativity Of Atoms\"](http://dx.doi.org/10.1021/ja01348a011) published in 1932 gives a different scale with H = 0.0 and F = 2.0!\n\n\nLater, he changed to 2.1 for H and 4.0 for F, e.g. <http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-01.html>\n\n\nIf we look through his [book](http://books.google.com/books?id=L-1K9HmKmUUC&q=electronegativity+scale#v=snippet&q=electronegativity%20scale&f=false), on p. 90 we find the following text:\n\n\n\n> \n> Following these there are given the values obtained by adding 2.05 to them...\n> \n> \n> \n\n\nThe resulting numbers are then rounded to one decimal, from H = 2.05 to H = 2.1 and F = 4.0.\n\n\nHis notes are online (e.g., [here](http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-29.html)) but there's no obvious explanation I can find yet.\n\n\nMy ***hunch*** is that he started with H = 0.0, and realized that the alkali metals would be negative numbers. The [shifted scale](http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-29.html) gave Li ~ 1.0 and F ~ 4.0, which seem like \"nice\" numbers.\n\n\nI tried to hunt down the reason for the move to 2.2 as the standard for hydrogen. The first reference seems to be by Huggins ([*J. Am. Chem. Soc.*, **1953**, 75 (17), pp 4123–4126](http://dx.doi.org/10.1021/ja01113a001)) but the paper itself doesn't really explain the reasoning.\n\n\n[Wikipedia](http://en.wikipedia.org/wiki/Electronegativity#cite_note-Allred-5) suggests a paper by Allred, but that cites the Huggins paper for the 2.2 number.\n\n\nPresumably more could be found with a very careful search through the papers of Pauling and Huggins, but I've never heard any clear explanations and the documents online don't give them either.\n\n\n", "5" ] ]

[ [ "\n**Short summary - HSAB theory has it's limits of applicability, and this is a good example.**\n\n\nMore recent discussion of acid-base theory center on the frontier orbital concept, e.g. W. B. Jensen, [*The Lewis Acid-Base Concepts* (1980) pp. 112-155.](https://books.google.com/books?id=RrREAAAAIAAJ&q=The+Lewis+Acid-Base+Concepts&dq=The+Lewis+Acid-Base+Concepts&hl=en&sa=X&ei=l2mTVO23GqHbsAT12ICgBw&ved=0CB8Q6AEwAA)\n\n\nIn this concept, an \"acid-base\" reaction creates new HOMO-LUMO orbitals of the product. So it's about energy matching between the acid (empty LUMO) and base (filled HOMO) (and also symmetry and orbital overlap). It's still very much a Lewis picture, since you have the electron pair on the base and an electron acceptor on the acid.\n\n\nFor example, if we react $\\ce{NH3 + H+ -> NH4+}$ we combine the HOMO of $\\ce{NH3}$ (i.e., a lone pair) with an empty LUMO on $\\ce{H+}$ to create the adduct.\n\n\nIMHO, this is a nice concept, since the orbital energy match between Al and O is good, you get covalent character to the resulting product. (Now you *also* get ionic character, but that's a different question.)\n\n\n", "6" ], [ "\nThe Orbital Molecular Theory explains it very well; the symmetry and orbital overlap defined by the wave functions gives a complete approach. I add some thoughts to the question qualitatively:\n\n\nThe concept of covalent and ionic bonds must be taken carefully when talking of acidity, a property normally refereed to aqueous solution where other interactions (hydration enthalpy) take place.\n\n\nThe $\\ce{Al^3+}$ ion is very polarizing, and has a high charge and low ionic radius leading to high effective nuclear charge. In solution the cation must have many shells of hydration, coordinating or polarizing water molecules and attaching $\\ce{OH-}$ groups in order to compensate the positive charge, thus increasing the net amount of available $\\ce{H+}$.\n\n\nSimilarly, carbon dioxide is a gas and it can be seen as a separated phase on a soda drink (bubbles); so first it must be dissolved in water. The dissolution of a $\\ce{CO2 (g)}$ molecule in water increases the acidity, $\\ce{[H+]}$, as can be tasted when leaving a glass of soda open in equilibrium with air for a while. I copy here the equations found at [this link](http://ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Carbonate/CO2.html) which explains why carbon dioxide is acidic. If you follow the link, it has a brief explanation:\n\n\n$\\ce{CO2 (g) <=> CO2 (aq)}$\n\n\n$\\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)}$\n\n\n$\\ce{H2CO3 (aq) <=> HCO3- (aq) + H+ (aq)}$ \n\n\n$\\ce{HCO3- (aq) <=> CO3^2- (aq) + H+ (aq)}$\n\n\n", "0" ] ]

[ [ "\nI would say no, it is only a hydrogen bond, not a halogen bond. \n\n\nTo be a halogen bond, the halogen atom must accept electron density from the other member of the bond. If the other member of the bond is a hydrogen atom bonded to a more electronegative element, I don't see have the halogen atom could be an acceptor of electron density. \n\n\nIf you can get access see \"[Halogen Versus Hydrogen](http://www.sciencemag.org/content/321/5891/918.short)\" Science Vol. 321 no. 5891 pp. 918-919. Footnote 7 there explains: \"A halogen bonding donor is a species that contains an electrophilic halogen that can become a member of a halogen bond. In the literature on halogen bonding, the reader should be alert to how the words 'donor' and 'acceptor' are used. In a complex RX---B, RX is the halogen bond donor but the electron acceptor (Lewis acid); B is the electron donor and halogen bond acceptor (Lewis base).\"\n\n\n", "3" ] ]

[ [ "\nAn electron being at/in the nucleus results in the [Fermi Contact Interaction](http://en.wikipedia.org/wiki/Fermi_contact_interaction).\n\n\nThis interaction can be observed experimentally through NMR, EPR, and electron capture. \n\n\nIn [NMR of paramagnetic compounds](http://en.wikipedia.org/wiki/Paramagnetic_NMR_spectroscopy), chemical shifts can be far outside the usual diamagentic range. The contribution to chemical shift of the electron density at the nucleus is referred to as the \"contact shift\".\n\n\nFor more information on Fermi contact, see \"Origin and meaning of the Fermi contact interaction\" Theor Chim Acta (1988) 73:173-200.\n\n\n", "7" ], [ "\nIf you do a (simple) calculation of a quantummechanical rotator in a spherical potential — i.e. calculate where an electron would be in the vicinity of an nucleus — you will find that the eigenfunction $\\left | \\Psi\\_1 \\right \\rangle$ corresponding to the lowest energy $E\\_1$ is dependent only on the distance of the electron from the nucleus ($\\left | \\Psi\\_1 \\right \\rangle = f(r)$ — use spherical coordinates because it makes your job a million times easier!) — which gives us a sphere like the s-orbitals are often drawn.\n\n\nIf you carry on, you realise that $\\left | \\Psi\\_1 \\right \\rangle$ is more or less a decaying exponential function, so its value is highest for $r = 0$. The probability of an electron being at a specific point is typically given as $\\left \\langle \\Psi\\_1 \\middle | \\Psi\\_1 \\right \\rangle$ and evaluates as $\\Psi^2\\_1$ in this case. Since the function has only one phase, the highest $\\Psi\\_1^2$ value is — *in the nucleus*.\n\n\nSo the nucleus already carries the highest single probability of having the electron right inside it. It’s not even ‘special’ for the electron to be in the nucleus.\n\n\nHowever, check Dave’s answer for the physical implications, which I didn’t even know about!\n\n\n", "3" ] ]

[ [ "\nNoble gases usually do not form strong bonds between their atoms - it takes a fair amount of energy to dimerise them into [excimers](https://en.wikipedia.org/wiki/Excimer), but those are short-lived excited molecules. Thanks to excitation, shells of the atoms aren't closed and they react, but very quickly they lose energy and become separate atoms.\n\n\nWith exception of weakly bound $\\ce{Xe2^+}$ cation, [stable compounds](https://en.wikipedia.org/wiki/Noble_gas_compound) created from noble gases (mainly xenon) only feature bonds with other elements, typically fluorine or oxygen.\n\n\nAs mentioned in the comments, you may have heard about detection of so called [van der Waals molecules](https://en.wikipedia.org/wiki/Van_der_Waals_molecule) of helium, which aren't \"true\" molecules, but very weakly bound pairs of atoms. In fact helium vdW dimer may have weakest bond even among them, and it was an achievement to observe it.\n\n\n", "32" ], [ "\nThe reason why noble gases form molecules is because of London forces, which are very weak. The electrons in noble gas atoms are on the move all the time, so there will be a moment when they will spend more time in an area, which generates a temporary dipole and allows for interaction with another atom.\n\n\nAn addendum regarding noble gas *compounds* in general: the first noble gas compound synthesised was $\\ce{XePtF6}$. When $\\ce{[O2+][PtF6-]}$ was discovered by the oxidation of dioxygen with platinum hexafluoride, it was hypothesised that $\\ce{XePtF6}$ could be made via the oxidation of xenon with platinum hexafluoride, because of the similar ionisation energies of dioxygen and xenon.\n\n\n", "7" ] ]

[ [ "\nPer Wikipedia, we have this laboratory method:\n\n\n$$\\ce{NH4Cl(aq) + NaNO2(aq) → N2(g) + NaCl(aq) + 2 H2O (l)}$$\n\n\nAlso per Wikipedia, this method generates small quantities of impurities of $\\ce{NO}$ and $\\ce{HNO3}$, which can be eliminated by passing the gas through potassium dichromate $\\ce{(K2Cr2O7)}$ dissolved in sulfuric acid $\\ce{(H2SO4)}$ \n\n\nBoth ammonium chloride and sodium nitrite are reasonably available chemicals, and the other products (table salt and water) are undeniably safe. Sulfuric acid might a bit tougher to get depending on where you live, and potassium dichromate will be the most difficult given that the use of hexavalent chromium is discouraged (and you should research procedures to reduce the $\\ce{Cr2O7^{-2}}$ to $\\ce{Cr2O3}$). But otherwise, this looks like the best approach.\n\n\n", "13" ], [ "\nYou could try a simple two step process. Take some air, burn some carbon-rich material in it until no further burning occurs. Then pass the remaining gas through an alkaline solution to remove the carbon dioxide. The remaining gas will be more than 95% nitrogen.\n\n\nAnd you get this with no expensive chemicals. \n\n\n", "10" ], [ "\nOne more classic method for $\\ce{N2}$ preparation in the lab that hasn't been mentioned yet is oxidative dehydrogenation of ammonia to nitrogen, typically with the help of transition metal oxides, e.g.:\n\n\n$$\\ce{2 NH3 + 3 CuO ->[\\pu{500 .. 550^\\circ C}] 3 Cu + N2 ^ + 3H2O ^} \\tag{1}$$\n\n\nReaction with copper(II) oxide is considered canonical and is often listed in textbooks and lab manuals, for example *A Text-book of Practical Chemistry* [1, p. 6] from 1921 describes this process as follows:\n\n\n\n> \n> Nitrogen is obtained by passing air first through ·880 ammonia and then over red-hot copper turnings. The oxygen of the air is absorbed by the copper and the resulting copper oxide is at once reduced back to copper by the ammonia. The frequent renewals of the contents of the copper tube are thus entirely avoided.\n> \n> \n> \\begin{align}\n> \\ce{2 Cu + O2 & = 2 CuO},\\\\\n> \\ce{2 NH3 + 3 CuO & = N2 + 3 H2O}\n> \\end{align}\n> \n> \n> The gas should be washed with sulphuric acid (1 : 4) to free from excess of ammonia, and dried by strong sulphuric acid. The use of dilute acid to free it from ammonia is to avoid the danger of a great rise in temperature should a large excess of ammonia come over.\n> \n> \n> \n\n\nMethod of obtaining nitrogen *via* thermal decomposition of $\\ce{NH4NO2}$ is criticized as a rather inconvenient one that may potentially result in an explosion.\n\n\n\n> \n> Small quantities of nitrogen may be prepared by carefully heating a solution of ammonium nitrite (sodium nitrite and ammonium chloride in molecular proportions). [...] The method is not recommended as it is difficult to control, and the nitrogen is generally contaminated with oxides of nitrogen, especially when the action becomes violent.\n> \n> \n> \n\n\nInstead, one can utilize \"volcano demo\" ([thermal decomposition of ammonium dichromate](https://en.wikipedia.org/wiki/Ammonium_dichromate#Reactions)) as somewhat safer alternative:\n\n\n$$\\ce{(NH4)2Cr2O7 ->[\\pu{170 .. 190^\\circ C}] Cr2O3 + N2 ^ + 4 H2O ^} \\tag{2}$$\n\n\nThese days various sources provide slightly deviating techniques for reaction type (1). Other metal oxides such as $\\ce{TiO2}$, $\\ce{V2O5}$, $\\ce{MnO2}$, their binary and ternary mixtures on silica- and alumina-supported substrates doped with rare-earths (Y, mostly) are capable of selective catalytic oxidation at lower temperatures [2, 3].\n\n\n### References\n\n\n1. Hood, G. F.; Carpenter, J. A. A Text-Book of Practical Chemistry; J. & A. Churchill, 1921. ([Google Books](https://books.google.com/books?id=uQ4KAAAAIAAJ&printsec=frontcover&dq=A%20Text-book%20of%20Practical%20Chemistry%20By%20George%20Frederic%20Hood,%20John%20Arthur%20Carpenter&hl=en&sa=X&ved=0ahUKEwijkf_9nubWAhUU24MKHe8dBfAQ6AEIKDAA#v=onepage&q&f=false)).\n2. Li, Y.; Armor, J. N. *Applied Catalysis B: Environmental* **1997**, 13 (2), 131–139 DOI: [10.1016/S0926-3373(96)00098-7](http://www.sciencedirect.com/science/article/pii/S0926337396000987).\n3. Gang, L.; van Grondelle, J.; Anderson, B. G.; van Santen, R. A. *Journal of Catalysis* **1999**, 186 (1), 100–109 DOI: [10.1006/jcat.1999.2524](http://www.sciencedirect.com/science/article/pii/S0021951799925241).\n\n\n", "5" ], [ "\nThere is one incredible simple, cheap and easy way to make nitrogen gas with near 99 percent purity if you don't need it compressed. I've used this method in my chemistry experiments when I needed just modest amounts of nitrogen or when I need to do reactions in the absence of reactive oxygen.\n\n\nSimply buy handwarmers from Walmart or many other sources. Use a heavy duty, large resealable bag and fill it as best you can with regular, uncontaminated air. Drop the hand warmer in there and it will absorb almost all oxygen, leaving you with about 1 percent \"all other gases\" that are naturally in the atmosphere. Personally, I glue an airtight small rubber tube (a hose from aquarium bubblers, also found at Walmart or any pet store) then clamp it shut with a small but strong spring clamp (although other clamps will due). I've found the hand warmer can dry out before it loses its capacity to absorb O2, so I cut one end open, seal it with laundry clamps (just to keep the contents from spilling out, making a mess) and spray it LIGHTLY now and then with mist from a stray bottle. However, if your not worried about the carbon, iron powder, and table salt contained in these packets (all very harmless) you can cut the packet open and just dump it in the bottom of the bag, and spray as described above once in a while. You would be surprise how long they last and how many bags full of nitrogen you can get from just one hand warmer packet. If you're not worried about adding excess heat to your O2 purged experiment, just put the handwarmer in the bag containing the enclosed beaker. Otherwise, you can purge air out of a different bag, vessel by inserting the hose, and squeezing the nitrogen filled bag.\n\n\n", "5" ], [ "\nAccording to [Wikipedia](http://en.wikipedia.org/wiki/Nitrogen#Production) you can generate pure nitrogen by heating metal azides, which gives out \nnitrogen and the metal. Sodium azide is [not really expensive](http://www.sigmaaldrich.com/catalog/product/sial/s2002).\n\n\n$$\\ce{2NaN\\_3 ->2Na + 3N\\_2}$$\n\n\nHowever, as you may expect, this may not be a very safe reaction as the first reaction that came to my mind when you asked for a reaction with large yield of nitrogen gas is that of TNT, though that would not give a pure yield as you have required. Azides are also explosive. Sodium azide is [used in car airbag and airplane escape chutes](http://en.wikipedia.org/wiki/Sodium_azide#Automobile_airbags_and_airplane_escape_chutes) where this exact reaction is used to quickly inflate these safety features. The metallic sodium formed is then further reacted to less dangerous sodium compounds. \n\n\n", "4" ], [ "\nTo generate nitrogen you could mix potassium nitrate, hydrochloric acid, and zinc.\nIt would generate nitrogen and the byproducts will be zinc chloride, ammonium chloride, and potassium chloride.\nThe nitrogen would be mixed with some HCl vapors, which you could remove through bubbling it in water.\n\n\nThe combustion of dinitrogen monoxide can also produce nitrogen.\n\n\n", "1" ], [ "\n\n\n\n\n\n\n\n\n\n\n**Want to improve this post?** Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.\r\n \r\n \n\n\n\n\n\n\nThermal decomposition of ammonium nitrite:\n$$\\ce{NH4NO2 -> N2 + H2O}$$\n\n\n", "-1" ], [ "\nAmmonium Dichromate reacted with heat produces only one gaseous product, Nitrogen. FIlter through a bubbler to get particles out of it. See Wikipedia under Ammonium Dichromate for the reaction equation and stoichiometry.\n\n\n", "-1" ] ]

[ [ "\nThe difference between a weak base and a strong base is that a strong base is completely converted to ions in a given solvent, whereas a weak base is not. Although this depends on the actual basicity of the solute, it is also very dependent on the solvent involved.\n\n\nIn water, $\\ce{NH3}$ can react to form $\\ce{NH4+}$ and $\\ce{OH-}$. However, most of the ammonia remains as $\\ce{NH3}$. This is due to the difference in p*K*as:1\n\n\n$\\ce{NH4+}$ = 9.2\n\n\n$\\ce{H2O}$ = 15.7\n\n\nBecause the ammonium ion is more acidic than neutral water, very little $\\ce{NH4+}$ will be present at equilibrium.\n\n\nIn contrast, the relevant p*K*as for the $\\ce{HCl}$ solution are:\n\n\n$\\ce{NH4+}$ = 9.2\n\n\n$\\ce{HCl}$ = -8\n\n\nThis tells us that $\\ce{HCl}$ is much more acidic than the ammonium ion, and therefore essentially all of the ammonia will be protonated when dissolved in $\\ce{HCl}$. As a result, it is considered a strong base in this case.\n\n\n1[Evans' p*K*a tables](http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf)\n\n\n", "1" ], [ "\nI'm guessing because ammonium chloride $\\ce{(NH4Cl)}$ is ionic and ammonium hydroxide $\\ce{(NH4OH)}$ is not. You simply get more dissolution with the former.\n\n\n", "0" ] ]

[ [ "\nFirst off, gold *does* react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the [other answer](https://chemistry.stackexchange.com/a/21813/5017)\n\n\nThe reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).\n\n\nA nice summary of some work by Jens K. Norskov can be found here:\n<http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490>\n\n\n\n> \n> In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms. \n> \n> \n> \n\n\nI think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions.\n\n\nYou can, of course, react gold with ***aggressive*** reagents like [aqua regia](http://en.wikipedia.org/wiki/Aqua_regia), a 3:1 mix of $\\ce{HCl}$ and $\\ce{HNO3}$.\n\n\nIf properly done, the product is $\\ce{HAuCl4}$ or [chloroauric acid](http://en.wikipedia.org/wiki/Chloroauric_acid).\n\n\n", "25" ], [ "\nRelativistic effects account for gold's lack of reactivity. Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand. \n\n\nSince the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.\n\n\nIf you'd like to see the formulas and math behind this (it's not all that complicated) [see here](https://chemistry.stackexchange.com/questions/16633/why-is-gold-golden/16637#16637). Also note that similar arguments explain [mercury's anomalous properties](https://chemistry.stackexchange.com/questions/13667/why-does-mercury-have-low-melting-and-boiling-points/13671#13671).\n\n\n", "17" ] ]

[ [ "\nA chemist would say that all 3 methyl groups are equivalent (identical by symmetry). Therefore they all produce the same chain, and that chain is the longest chain in the molecule. Consequently, since the methyl groups are all the same, it doesn't matter which one you pick.\n\n\n", "7" ] ]

[ [ "\n$$\\ce{A~ + :\\!\\!B -> \\ \\ A\\!\\leftarrow:\\!\\!B}$$\n\n\n$$K\\_f = \\dfrac{\\ce{[A\\!\\leftarrow:\\!\\!B]}}{\\ce{[A][B]}}$$\n\n\nSee sections 4.1 and 4.2 in the following reference for more information:\n\n\n<http://goicoechea.chem.ox.ac.uk/teaching/abse/abse_lecture2.pdf>\n\n\nBut $K\\_f$ is specific to an acid-base pair. \n\n\nA more universal scale is formed by the [Gutmann–Beckett method](http://en.wikipedia.org/wiki/Gutmann%E2%80%93Beckett_method) by observing the phosphorous NMR spectrum of triethylphosphine oxide as it interacts with a sample lewis acid. \n\n\nAn Acceptor Number (AN) scale is formed with AN=0 corresponding to triethylphosphine oxide in hexane (P chemical shift of 41), and AN = 2.21 (d-41), where d is the chemical shift of triethylphosphine oxide interacting with the sample. The greater AN is, the more acidic the compound. \n\n\n", "9" ] ]

[ [ "\nB has more branching and this means that molecules of B pack less efficiently and so there are fewer Van der Waals forces between them and so B has the lowest boiling point. On the other hand A is a linear molecule and so packs efficiently, has stronger VdWs forces and so has the highest boiling point. As @DavePhD's data shows $\\ce{B < C < A}$\n\n\n", "4" ], [ "\n$A$ n-hexane, boiling point 69 degrees C. \n\n\n$B$ 2,3 dimethyl butane, boiling point 58 degrees C. \n\n\n$C$ 3-methyl pentane, boiling point 63 degrees C. \n\n\n$B<C<A$\n\n\n", "4" ] ]

[ [ "\nHydrogen bonds often have a significant covalent component. This can be seen in many aspects of the hydrogen bond: it is directional, unlike Coulombic interactions, and the bond length is generally significantly shorter than predicted by Coulombic models. There is great variation in the covalent character of hydrogen bonds depending on their environment, with e.g. bond lengths between 2.2 and 4.0 Angstroms known.\n\n\nOther intermolecular interactions have significant covalent components; there is nothing special about hydrogen per se but it is the most important example.\n\n\n", "3" ] ]

[ [ "\nThis is really a cheminformatics problem. It's a reflection of the chirality or atomic stereochemistry about the atom you mention.\n\n\nThere are multiple ways of specifying chiral information in 2D and 3D.\n\n\nAs mentioned by [Dave](https://chemistry.stackexchange.com/a/21742/5017) SMILES uses a clockwise and counter-clockwise from left to right in the string.\n\n\nIn 2D, we usually use wedge and hash notation. The problem is that some users use wedges and hashes for graphical appearance and not for stereochemical information (e.g., pseudo-3D look). So file formats also add atom parity to indicate the *actual* stereochemistry of atoms.\n\n\nThe \"parity\" idea has to do with the relationship between the atoms as they are ordered in the file, vs. the stereochemistry around the central atom. As indicated in the other answer, these are enantiomers, even though they use the same \"@@\" SMILES stereochemical indicator:\n\n\n* $\\ce{O[C$@@$](F)(Cl)Br}$\n* $\\ce{O[C$@@$](F)(Br)Cl}$\n\n\nNote that the atom ordering is different, so the \"parity\" will be different. For your example, $\\ce{O[C$@@$](F)(Cl)Br}$ I get \"1\" for the MDL parity.\n\n\nFrom the MDL specification (Appendix A: Stereo Notes):\n\n\n\n> \n> Number the atoms surrounding the stereo center with 1, 2, 3, and 4 in order of increasing atom number (position in the atom block) (a hydrogen atom should be considered the highest numbered atom, in this case atom 4). View the center from a position such that the bond connecting the highest-numbered atom (4) projects behind the plane formed by atoms 1, 2, and 3.\n> \n> \n> Sighting towards atom number 4 through the plane (123), you see that the three remaining atoms can be arranged in either a clockwise or counterclockwise direction in ascending numerical order.\n> \n> \n> \n\n\n\n\n\nFor more details, you might want to look at the [Open Babel documentation](http://openbabel.org/dev-api/group__stereo.shtml), [OpenSMILES](http://opensmiles.org/) or the [MDL specification](http://accelrys.com/products/informatics/cheminformatics/ctfile-formats/no-fee.php).\n\n\n", "4" ], [ "\nIn SMILES notation, the order of the remaining three atoms corresponds to clockwise order when viewed in the direction looking from the first listed atom toward the carbon center. \n\n\nSo you have written $\\ce{O[C@@](F)(Cl)Br}$. Looking in the direction from the O atom toward the C atom, the clockwise order of the remaining atoms is F, Cl, Br. \n\n\nYou could write the other enantiomer as $\\ce{O[C@@](F)(Br)Cl}$.\n\n\n", "3" ], [ "\nThe molecule is not 2 dimensional, it is 3 dimensional, in this case tetrahedral. Since the vertexes have different CI priorities, the molecule will spin one way in space, this is visualized by placing the lowest priority atom in the back and assigning priorities to the remaining. This spin causes rotation of polarized light and other effects. \nIf you are familiar with chirality, which is an extension of Group Theory, it would seem to answer the questions you are having, unless I totally misunderstood what it is you are asking. \n\n\n", "1" ] ]

[ [ "\nIt is important to make clear any requirement that a variable be constant for each equation you write. \n\n\nTo go from\n\n\n$$\\mathrm{d}G = V\\,\\mathrm{d}p - S\\,\\mathrm{d}T$$\n\n\nto\n\n\n$$\\mathrm{d}G = V\\,\\mathrm{d}p,$$ \n\n\nyou would need to assume temperature is constant. \n\n\nAnd to go from\n\n\n$$\\mathrm{d}G = V\\,\\mathrm{d}p$$\n\n\nto\n\n\n$$\\Delta G = V(p\\_2 - p\\_1)$$\n\n\nyou would need to assume volume is constant. \n\n\nIf temperature and volume are both constant, the pressure is constant also. \n\n\nSo your final equation is only saying: when temperature, pressure and volume are constant, there is no change in Gibbs free energy. \n\n\n", "7" ], [ "\nWhen you write $\\mathrm{d}G=V\\,\\mathrm{d}P-S\\,\\mathrm{d}T,$ you are talking about the differential change in $G$ between two ***closely neighboring thermodynamic equilibrium states***, one at $(T,P)$ and the other at $(T+\\mathrm{d}T,P+\\mathrm{d}P).$\n\n\nSo, when you integrate this equation, you must integrate over a path comprised to a continuous sequence of thermodynamic equilibrium states. Over such a path, at constant temperature, the molar volume $V$ changes with pressure $P,$ and this needs to be taken into account in the integration.\n\n\n", "5" ] ]

[ [ "\nOnly 1 in 10,000,000 water molecules dissociate. \n\n\nHI, HBr and HCl fully dissociate, and I-, Br-, and Cl- and are much better leaving groups. \n\n\n", "7" ], [ "\nWater dissociating into H+ and OH- isn't the same concept as free OH- leaving. A more accurate comparison would be forming NaOH by dropping Na metal in water, because in that case, OH- leaves water to become part of a new molecule as opposed to just ionizing in solution for a second before it becomes full water again.\n\n\nAlcohols are very capable of SN1. Water is the leaving group as it is obviously protonated first. Benzyl alcohol to benzyl chloride, t-butanol to t-butyl bromide, etc.\n\n\n", "1" ] ]

[ [ "\nSugar contains a lot of hydroxyl groups. So due to hydrogen bonds, sugar can readily dissolves in water.\n\n\n", "1" ] ]

[ [ "\nIf you don'y want to use maleated polyolefin (the best coupling agent for this composite), you can use oxidized polyethylene, ethylene propylene rubber, ethylene-co-glycidyl methacrylate or polyolefin elastomer. \n\n\n", "1" ] ]

[ [ "\n\n> \n> We can share the women, we can share the wine\n> \n> \n> We can share what we got of yours, cause we done shared all of mine\n> \n> \n> \n\n\n\"Jack Straw\" Grateful Dead\n\n\nBut seriously now, the most important thing in the [Princeton.edu article](http://www.princeton.edu/%7Eachaney/tmve/wiki100k/docs/Coordinate_covalent_bond.html) (which explains at the bottom that its content comes from Wikipedia) is\n\n\n\"**The distinction from ordinary covalent bonding is artificial**\"\n\n\nDoes $\\ce{H2}$ come from a hydride and a proton or from two hydrogen atoms? The reason a molecule is stable is independent of a particular path by which it is made.\n\n\nFor $\\ce{CO}$, it is only in your mind that oxygen brought more wine (electrons) to the party, but what really matters is that there is enough wine (electrons) at the party for everyone to be satisfied.\n\n\n$\\ce{CO}$ is isoelectronic with $\\ce{N2}$ and $\\ce{CN-}$; the reason for bonding is the same for each.\n\n\n", "17" ], [ "\nElectronegativity affects [coordinate covalent bonds](https://en.wikipedia.org/wiki/Dipolar_bond); indeed atoms which accept electrons should be \"electron hungry\" enough to take them. However it stems mainly from that the acceptor doesn't have complete electron shell, thus acts as [Lewis acid](https://en.wikipedia.org/wiki/Lewis_acids_and_bases) and donor acts as base.\n\n\n", "6" ] ]

[ [ "\nSome of your electron configurations are wrong.\n\n\nXenon has 18 $N$ electrons ($s+p+d$ sublevels full) and eight $O$ electrons ($s+p$ sublevels full). Radon has 18 $O$ and 8 $P$. Erbium does not have all the $f$ subleblvels full and ends up with $N^{30}O^8P^2$.\n\n\nYou may want to look up the [Aufbau Principle](https://en.wikipedia.org/wiki/Aufbau_principle). It's imperfect but more accurate than the shell-by-shell ordering you assume. Basically, when you have more than one electron the repulsions between electrons raises the energy of sublevels that have higher angular momentum and don't get close to the nucleus, and then they overlap with lower angular momentum orbitals of what you thought were higher shells.\n\n\n", "2" ] ]

[ [ "\nA simple example can be the disproportionation of chlorine when bubbling a current of chlorine gas $\\ce{Cl2}$ into a solution of $\\ce{NaOH}$ (containing $\\ce{OH^-}$ ions):\n$$\\ce{Cl2 + 2 OH^- -> ClO^- + Cl^- + H2O}$$\nHere the chlorine atoms have the oxidation number equal to zero in the reactants, and get the final oxidation number $-1$ in $\\ce{Cl^-}$ and $+1$ in $\\ce{ClO^-}$. Here one of the $2$ original $\\ce{Cl}$ atoms oxidises the other one from $0$ to $+1$. In this operation it is reduced fro $0$ to $-1$.\n\n\nThe final solution is the famous bleach, used for cleaning dirty surfaces.\n\n\n", "-1" ] ]

[ [ "\nThe addition of a Grignard to a carbonyl doesn't form an alcohol—not directly at least. It forms an alkoxide ion (i.e. the conjugate base of an alcohol).\n\n\nOnly when that reaction is complete, is water (or some other acid, ammonium chloride is used quite regularly) added to the solution. The water protonates the alkoxide ion to form the alcohol, and also destroys any remaining Grignard. This step is usually referred to as 'protic workup' or just 'workup'.\n\n\nSo the Grignard never sees any alcohol.\n\n\nThe same is true of the carboxylic acid formation (assume you're referring to Grignard addition to carbon dioxide). It doesn't form a carboxylic acid straight away, it forms a carboxylate ion.\n\n\n", "11" ] ]

[ [ "\nYours thoughts seem to be correct. [Imidazole](https://en.m.wikipedia.org/wiki/Imidazole) pKa is 14.5 and 6.8. [Pyrrole](https://en.m.wikipedia.org/wiki/Pyrrole) pKa is 16.5. You do have a resonance form for the imidazole ring. Additionally, nitrogen is electron withdrawing. The localized negative charge on deprotonated pyrrole will be more basic.\n\n\nBoth nitrogens in imidazole can be protonated, hence 2 pKa values.\n\n\n", "1" ] ]

[ [ "\nYes they can, but they have higher specificity for esters. Here is part of the abstract of a [paper](https://academic.oup.com/peds/article/18/2/93/1545211) using directed evolution to increase the activity:\n\n\n\n> \n> A double mutant F207S/A213D gave the highest molecular activity of 1.1 min−1 for the amide, corresponding to a 2-fold increase compared with that of the wild-type lipase. A structural model of the lipase indicated that the mutations occurred at the sites near the surface and remote from the catalytic triad, but close to the calcium binding site. This study is a first step towards understanding why lipases do not hydrolyze amides despite the similarities to serine proteases in the active site structure and the reaction mechanism and towards the preparation of a general acyl transfer catalyst for the biotransformation of amides.\n> \n> \n> \n\n\nNote that the abstract is a bit self-contradictory, first saying that they increased the activity two-fold, then saying lipases have no amide hydrolysis activity. It is possible that a turnover of $\\pu{1.1 min-1}$ is considered not active, considering the much faster turnover for the lipase reaction. The introduction, which has lots of references, clarifies this.\n\n\nI found this in a search for \"lipase specificity amide\" on google scholar. I did not look for the specific substrate you had in mind because that feels like giving out fish instead of hints on fishing.\n\n\n", "6" ] ]

[ [ "\nThe absolute values of internal energy $U$ or enthalpy $H$ (which could be more useful for water vapour) at a single state point are meaningless. It is only the difference between two different state points that matter. Thus, the value for a single state point can be set to any arbitrary value. **Therefore, this arbitrary meaningless value cannot be directly measured.** Many handbooks set the arbitrary state point so that the values of these properties are positive for most liquid or gas states. For water, the triple point is selected as the reference state, where the internal energy $U$ of saturated liquid is assigned a zero value.\n\n\nThat's why *all the laws you know depend on knowing the changes that happened that led to forming this vapour at that temperature.* Heating a mass of $m=1\\ \\mathrm{kg}$ of water from the triple point with $U=0$ to a new state with $T=423\\ \\mathrm K$ and $p=10^5\\ \\mathrm{Pa}$ would give you $U=2582.7\\ \\mathrm{kJ}$ and $H=2776.3\\ \\mathrm{kJ}$.\n\n\nFurthermore, you want to directly measure the energy that is *contained as heat.* You should remember that internal energy $U$ is a state quantity or state function because it describes an equilibrium state of a thermodynamic system, regardless of how the system has arrived in that state. For a closed thermodynamic system,\n$$\\Delta U=Q+W$$ where is $Q$ amount of heat transferred to the system and $W$ is work done on the system (according to the sign convention used by ISO and IUPAC) provided that no chemical reactions occur. Thus, simply by looking at the value of $U$, you cannot know how much was supplied by heat. You also cannot know how much heat can be extracted from it. Again *all the laws depend on knowing the changes.*\n\n\nHowever, since the internal energy $U$ describes an equilibrium state of the thermodynamic system, regardless of how the system has arrived in that state, you do not need to know the exact path taken from the reference state to the observed state; you just need to assume a possible path. So instead of measuring $U$ directly, you can measure the values for a set of thermodynamic parameters (like mass $m$, temperature $T$, and pressure $p$ in the example above) that describe the state of the thermodynamic system. Then you can calculate the change in internal energy $\\Delta U$ for a system that changes from the reference state to the new state with the observed parameter values assuming any possible path.\n\n\nThere are precalculated values (so-called steam tables) that already did this job for you. So you can simply measure the values for a set of thermodynamic parameters and then look up or interpolate the corresponding value for the internal energy $U$.\n\n\n", "2" ] ]

[ [ "\n\n> \n> I'd like to learn how to approach more complex problems. Is there a\n> standard set of steps to apply, perhaps iteratively?\n> \n> \n> \n\n\nEvery electrolyte has its own decomposition potential and a stability window. One can carefully design mixtures and study their electrolytic behaviour, for example, what happens if we electrolyze an aqueos solution of transition metal salts. There are no standard set of steps or iterations, although for simple ionic cases, people look up the electrode potential table to predict to products of electrolysis. For example, in higher classes one might study the electrolysis of KCl and KI in water. What products do we expect under standard conditions? These can be answered systematically. Electrolysis products of random mixtures like soda are difficult to predict, because once you have organic compounds, they start reacting at the electrodes via radical reactions.\n\n\nFor the conceptualization of electrolysis, take binary mixtures which represent soda such as a very dilute phosphoric acid and citric acid solution. If you electrolyze this mixture at the decomposition potential of water, you will the see expected products, hydrogen and oxygen.\n\n\nLet us say you have water and tons of fructose in it (this is why sodas are bad!), it will almost be a non-conductor of electricity. Simple underivatized sugars are not good electrolytes. On the other hand, carefully controlled (pH adjusted) solutions of sugars can be oxidized. Search electrolysis of carbohydrates.\n\n\nAgain think about the electrolysis carbon dioxide saturated water; this solution will also be a poor conductor of electricity. Still since we will have plenty of water molecules, and using the right decomposition potential, hydrogen and oxygen will be the main electrolysis products.\n\n\nIf you are still curious, search voltammetry of mixtures.\n\n\n", "0" ], [ "\nFructose and $\\ce{CO2}$ are covalent solute molecules. They are not modified by electrolysis when dissolved in a solution of phosphoric acid. Fructose, for example, does not produce ions in solution : it will never be decomposed (by electrolysis) partially into fragments, or totally into $\\ce{C, H2}$ and $\\ce{O2}$. Water is a solvent, made of covalent molecules. It can be electrolyzed.\n\n\n", "0" ] ]

[ [ "\nThe local coupling constant is typically given in units of energy, such as electronvolts (eV), because it represents the strength of the coupling between the electronic and vibrational degrees of freedom in a system (which is thus an 'energy-related' quantity). The Holstein Hamiltonian, which is used to describe the interaction between electrons and phonons in a system, includes the atom displacement term, which represents the coupling between the electronic state $P$ and the vibrational state $\\lambda$, with the coupling constant $g^\\lambda$ being the coefficient in front of this term for unit equality in the equation.\n\n\n", "2" ] ]

[]

[ [ "\n**Because reality, not theory, dominates chemistry**\n\n\nOzone, by the way, has three oxygen *atoms* not molecules.\n\n\nBut while many theoretical structures are possible and look OK in different bonding theories, that isn't how chemists work out the real structure. Most bonding theories are too weak to make accurate structural predictions for many \"difficult\" structures.\n\n\nWhen it was realised that benzene, for example, was a ring some proposed it was essentially the same as cyclohexatriene, with alternating double and single bonds. But observations of the crystal structure showed that, despite simple bonding theories ideas, the bond lengths were all equal. Better theories were developed to explain the actual structure.\n\n\nOzone, in some theories, could be a 3 membered ring or, indeed, a variety of other structures. But it isn't. Microwave spectroscopy show it to be a bent molecule as does the fact is has a dipole moment. That's what is is, theory be damned. The central angle is about 116° and the central oxygen has a net positive charge withe the terminal oxygens having net negative charges but bond lengths between those of a single and double bond.\n\n\nThere are theories that can account for this, but the predictions of simpler theories are useless. What we *observe* the structure to be is far more important than theory.\n\n\n", "2" ] ]

[ [ "\nFor our organic chemistry lab program we have developed GC program for one of the separation labs and we use alkane homologues as the analyte mixture to be separated. The chromatogram give a baseline separation on shorter (3-m column) Carbowax-20M packed column. Thus, for curiosity, I checked for separation of alkene isomers on similar column. Sure enough, I found one separating $\\ce{n-C16H32}$ alkene isomers:\n\n\n[](https://i.stack.imgur.com/R4Ox5.jpg)\n\n\nThe column is $\\pu{300 m}$ Garbowax-2OM capillary column operating at $\\pu{96 ^\\circ C}$ with $\\pu{0.15 MPa}$ $\\ce{N2}$ mobile phase (Ref.1). As indicated in the diagram, the method has even separated most of $cis/trans$ isomers at same position.\n\n\n\n\n---\n\n\n**References:**\n\n\n1. Ladislav Soják, Ján Krupčík, and Jaroslav Janák, \"Gas chromatography of all $\\ce{C15}$-$\\ce{C18}$ linear alkenes on capillary columns with very high resolution power,\" *Journal of Chromatography A* **1980**, *195(1)*, 43-64 (ODI: <https://doi.org/10.1016/S0021-9673(00)81542-3>).\n\n\n", "2" ] ]

[ [ "\nIf a reaction is second order in $\\ce{A}$ and $\\ce{B}$, with $\\ce{[A]\\_0 ≠ [B]\\_0}$, and with a rate law given by $\\pu{d[A]/dt = d[B]/dt = k[A][B]}$, the integrated law is : $$\\ce{\\frac{1}{[B]\\_0 - [A]\\_0}ln{ {\\frac{[A]\\_0 [B]}{[A][B]\\_0}}}} = kt$$\nAs it is pretty long to establish this law, I recommend that you open a book on chemical kinetics like A. A. Frost and R. G. Pearson, Kinetics and Mechanism, John Wiley & Sons, New York, 4th ed. 1965, p. 16.\n\n\n", "2" ] ]

[ [ "\nYes, a milkshake is heterogeneous.\n\n\n* milk is heterogeneous all on its own; [fat globules](https://www.compoundchem.com/2018/06/02/milk/) in the milk are a different phase than water solutions in the milk. (Curds vs. whey).\n* ice cream adds two additional heterogeneities: it contains microscopic ice crystals, and ice is a different phase than both water and fat globules. It also contains tiny air bubbles, and this is a fourth phase.\n\n\n", "6" ], [ "\nHomogenity/heterogenity is dependent on evaluation conditions.\n\n\nA mixtures is heterogenous if both these conditions are true:\n\n\n* Local variability of mixture properties are above the threshold of\n\t+ property value resolution AND\n\t+ mixture application requirement.\n* Spacial granularity of the above variability is above the threshold of\n\t+ spacial resolution AND\n\t+ mixture application requirement.\n\n\nOtherwise, a mixture is homogenous.\n\n\nIf values vary less than we notice or care \n\nOR if values vary within smaller distance than we notice or care \n\nTHEN the mixture is homogenous.\n\n\nEvery mixture can be considered homogenous or heterogenous at certain ( sometime extreme) conditions.\n\n\n* If there are floating evenly distributed potatoes in water, observed by a telescope with resolution $\\pu{1 m}$, it is homogenous mixture.\n* If there is an old favourite red T-shirt, a \"16-Windows(3.x)-colors man\" may say it is still homogenously colored, even if his wife disagrees.\n\n\n", "2" ], [ "\nMilk is a [colloid dispersion](https://en.wikipedia.org/wiki/Colloid), i.e. a\n\n\n\n> \n> \"substance consisting of microscopically dispersed insoluble particles is suspended throughout another substance.\"\n> \n> \n> \n\n\n(source: English [Wikipedia](https://en.wikipedia.org/wiki/Colloid))\n\n\nAs such, visible light is scattered when it passes this medium; equally known as [*Tyndall effect*](https://en.wikipedia.org/wiki/Tyndall_effect). In case you get your hands on a bright light source, you then can monitor the optical path across this medium. It needn't be a laser as in this video [recording](https://www.youtube.com/watch?v=t1d_B7_zWiQ) about a *very dilute* milk:\n\n\n[](https://i.stack.imgur.com/pGzQI.png)\n\n\n(screen photo of the mentioned video, 1:28 min into the recording)\n\n\n*Addition after a comment:*\n\n\nReplication with «true milk» requires a brilliant light source passing an aperture *narrow enough*. On the other hand, because of the high concentration of particles with a dimension about light's wavelength, scattering occurs so frequently that the beam fuzzes out rapidly: (For these two *transmission mode* observations, equally note color of the milk and glare (and apparent color) of the glass in the second run with milk which is less prominent for the experiment with tap water in the experiment.)\n\n\n[](https://i.stack.imgur.com/tS6oX.png)\n\n\nSetup: daylight-white light of a LED torch (focus to infinity) passes a flat bottom glass, perpendicular to the surface (aperture: $\\pu{3 mm}$ diameter, thickness glass wall: $\\approx \\pu{2 mm}$). Path length $\\approx \\pu{10 mm}$ for either water (first experiment), or fresh milk (second experiment; 1.5% fat, purchased less than $\\pu{12 h}$ earlier, which passed both [Pasteurization](https://en.wikipedia.org/wiki/Pasteurization) and [homogenization](https://en.wikipedia.org/wiki/Homogenization_(chemistry)). *Homogeneous* hereby is not about a discern between solution (e.g., $\\ce{NaCl}$ dissolved in water), or collide (e.g., fat dispersed in milk), but refers to a narrowed (i.e., adjusted) [particle size distribution](https://en.wikipedia.org/wiki/Particle-size_distribution) of the *fat globules suspended* in milk after passing homogenization for which [Wikipedia](https://en.wikipedia.org/wiki/Milk) reports\n\n\n\n> \n> an average diameter of two to four micrometers [of unhomogenized cow's milk] and with homogenization, [an] average [of] around 0.4 micrometers.\n> \n> \n> \n\n\n($\\pu{1 µm} = 1 \\times \\pu{10^-6 m}$, and $\\pu{1 nm} = 1 \\times \\pu{10^-9 m}$; so $\\pu{0.4 µm} \\approx \\pu{400 nm}$. For comparison: average diameter of a human hair $\\approx \\pu{100 µm}$ ([ref](https://en.wikipedia.org/wiki/Orders_of_magnitude_(length)#100_micrometres)), wavelength of visible light $\\approx 250\\ldots\\pu{700 nm}$.)\n\n\n", "1" ] ]

[ [ "\nThe important thing is to remember that for a formal redox half-reaction, oxidizing H2 to H+,\nthere is relation of the standard reaction Gibbs energy and the standard redox potential\n\n\n$$\\Delta\\_r G^\\circ = -nFE^\\circ .$$\n\n\nThe lower is the potential, the stronger(thermodynamically, not necessarily kinetically) is the respective reductant and weaker the oxidant. And vice versa.\n\n\nSo $\\ce{Mn(s)}$ is the strongest reductant, $\\ce{Pb(s)}$ the weakest one.\n\n\nSo $\\ce{Mn^2+(aq)}$ is the weakest oxidant, $\\ce{Pb^2+(aq)}$ the strongest one.\n\n\nFrom above must be clear that the only spontaneous reaction is :\n\n\n$$\\ce{Mn(s) + Fe^2+(aq) -> Mn^2+(aq) + Fe(s)}$$\n\n\n", "1" ], [ "\nYou can also state that, when comparing two redox couples, the metal belonging to the redox couple with the most negative values will react. Here, comparing the three metals $\\ce{Mn, Fe, Pb}$, the metallic $\\ce{Mn}$ will always react and produce $\\ce{Mn^{2+}}$, because of its highest negative value ($E° = - 1.18$ V) and $\\ce{Pb}$ will never react: its redox potential is not negative enough.\n\n\n", "0" ], [ "\nOne of the easiest ways is to draw the reactions in the direction they occur. For example, considering:\n\n\n\n> \n> $$\\ce{Mn + Fe++ -> Mn++ + Fe}$$\n> \n> \n> \n\n\n$$\\ce{Mn++ + 2e- -> Mn - 1.18 V Does not want 2 electrons}$$\n\n\nCan be viewed as:\n\n\n$$\\ce{Mn -> Mn++ + 2e- + 1.18 V glad to give them away}$$\n\n\n \n$$\\ce{Fe++ + 2e- -> Fe. - 0.45 V happy without electrons, but not as much as Mn}$$\n\n\nThen just add: 1.18 + (-0.45) = + 0.73 V positive, it will occur spontaneously. Electrons flow toward Fe++.\n\n\nDo this with all the others, only A is positive. The sign (positive or negative) of the *sum* of the 2 half reactions shows whether or not it is spontaneous. This is how batteries work.\n\n\nMost importantly, direction of electron flow is *relative* to the to metals involved, with most positive to most negative (as written!) producing the highest voltage.\n\n\n", "-1" ] ]

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[ [ "\nRecall the definitions of molality and molarity, it is mol solute / kg of solvent and mol solute/ Vol of solution in (L). It is not kg of solution in molality. Instead of memorizing a plug and chug formula, would not it be far better to understand the conversion and do it yourself?\n\n\nI quote an example from General Chemistry by Ebbing and Gammon (2016). Read the strategy:\n\n\n\n> \n> Molality to Molarity: An aqueous solution is 0.273 m\n> $\\ce{KCl}$. What is the molar concentration of potassium chloride,\n> $\\ce{KCl}$ ? The density of the solution is 1.011 x 10$^3$ g/ L.\n> \n> \n> Problem Strategy You are asked to determine the molarity of the\n> solution, which is the moles of solute $(\\ce{KCl})$ per liter of\n> solution (volume of $\\ce{H2O}$ and $\\ce{KCl}$. The molality tells you that the solution\n> contains 0.273 mol $\\ce{KCl}$ in $1 \\mathrm{~kg}$ of\n> water. If we determine the volume of solution that contains $0.273\n> \\mathrm{~mol} \\ce{KCl}$, we should be able to calculate the\n> molarity. Because the solution is made up of only $\\ce{KCl}$ and\n> water, the solution mass is the mass of 0.273 mol\n> $\\ce {KCl}$ plus the mass of 1.000 kg of water. Knowing\n> the total mass of solution, we then can use the solution density to\n> convert from the mass of solution to the volume of solution. The\n> molarity is obtained by dividing the moles of solute by the volume of\n> solution in liters.\n> \n> \n> \n\n\n**EDIT:** Now that the OP posted the text, we can discuss the hints.\n\n\nThe equation $$V=\\frac{1000+mM\\_2}{d}$$\n\n\nis derived from mass balance principle. Keeping the above discussion in mind,\n\n\n$$Total\\hspace{0.15cm}mass\\hspace{0.15cm}of\\hspace{0.15cm}solution, M=mass \\_{solvent} + mass\\_{solute}$$\n\n\nThis is in grams. Now recall the mass volume relation from density. Here $V$ is the volume of solution *not* the solvent.\n\n\n$$d=\\frac{M}{V}$$ or $$M=dV$$\n\n\nThese hints should be enough to derive the desired equation.\n\n\n", "3" ], [ "\nThe equation given is\n$$V = \\frac{1000 + m \\cdot M}{d} \\pu{cm3}$$\n\n\nIt only works if you know to enter the quantities' numerical values after expressing them in the conventional units (g for mass, mol/kg for molality, g/mol for molar mass and e.g. g/mL for density). With these specifications, the expression gives the volume of a solution containing 1000 g of solvent.\n\n\nAnother way of writing this, using the modern abbreviation for molality, $b$, and mass, $m$:\n\n\n$$V\\_\\mathrm{solution} = \\frac{m\\_\\mathrm{solvent} + m\\_\\mathrm{solvent} \\cdot b\\_\\mathrm{solute} \\cdot M\\_\\mathrm{solute}}{d\\_\\mathrm{solution}}$$\n\n\nThe numerator is the mass of the solution as sum of the mass of solvent and solute. In the cited equation, the 1000 reflects that molarity has units of $\\pu{mol kg-1}$ but the molar mass conventionally has units of $\\pu{g mol-1}$.\n\n\n", "1" ] ]

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[ [ "\nThis arises from the quantum mechanical properties of the angular momentum operator (*vide infra*). Since the operators for angular momentum and translational energy are different, there is no reason to expect a similarity between the form of the eigenvalues.\n\n\n\n\n---\n\n\n### Spectra of the angular momentum operators\n\n\nFirst we note the classical relationship between the angular momentum $L$ and the rotational kinetic energy\n\n\n$$E\\_\\mathrm{rot} = \\frac{1}{2}I\\omega^2 = \\frac{L^2}{2I} \\tag{1}$$\n\n\nsince $L = I\\omega$, with $I$ being the moment of inertia and $\\omega$ being the angular velocity. By analogy, the quantum mechanical Hamiltonian for rotational motion is\n\n\n$$\\hat{H} = \\frac{\\hat{L}^2}{2I} \\tag{2}$$\n\n\nwhere $\\hat{L}^2$ is the operator for (the square modulus of) orbital angular momentum. Because $L$ is a vector, we have\n\n\n$$\\hat{L}^2 = \\hat{L}\\_x^2 + \\hat{L}\\_y^2 + \\hat{L}\\_z^2 \\tag{3}$$\n\n\nwhere $L\\_x$ is the operator for the projection of orbital angular momentum onto the $x$-axis, and so on. The classical angular momentum is given by $\\vec{L} = \\vec{r} \\times \\vec{p}$; by expanding the cross product we can obtain the equations $L\\_x = yp\\_z - zp\\_y$, $L\\_y = zp\\_x - xp\\_z$, and $L\\_z = xp\\_y - yp\\_x$. Using these relations as well as the classical commutation relationship $[x,p\\_x] = [y,p\\_y] = [z,p\\_z] = \\mathrm{i}\\hbar$, we can derive the commutators of all four operators in equation $(3)$:\n\n\n$$\\begin{align}\n[L\\_x,L\\_y] &= \\mathrm{i}\\hbar L\\_z; & [L\\_y,L\\_z] &= \\mathrm{i}\\hbar L\\_x; & [L\\_z,L\\_x] &= \\mathrm{i}\\hbar L\\_y; \\tag{4} \\\\\n[L^2,L\\_x] &= 0; & [L^2,L\\_y] &= 0; & [L^2,L\\_z] &= 0 \\tag{5}\n\\end{align}$$\n\n\nConventionally, angular momentum eigenstates are chosen to be simultaneous eigenstates of $L^2$ and $L\\_z$: since $[L^2,L\\_z] = 0$, it is guaranteed that there is a complete set of states which are eigenstates of both operators. We denote these states by $|\\lambda,\\mu\\rangle$ where $\\lambda$ and $\\mu$ are related to the eigenvalues of these states:\n\n\n$$\\begin{align}\nL^2|\\lambda,\\mu\\rangle &= \\lambda\\hbar^2 \\tag{6} \\\\\nL\\_z|\\lambda,\\mu\\rangle &= \\mu\\hbar \\tag{7}\n\\end{align}$$\n\n\nIn this paradigm, we then define the raising and lowering operators $L\\_+$ and $L\\_-$ (the meaning will become clear in due time)\n\n\n$$L\\_\\pm \\equiv L\\_x \\pm \\mathrm{i}L\\_y \\tag{8} $$\n\n\nNow consider the effect of $L\\_\\pm$ on any arbitrary state $|l,m\\rangle$ which is simultaneously an eigenstate of $L^2$ and $L\\_z$. Since $L^2$ commutes with both $L\\_x$ and $L\\_y$, we have\n\n\n$$\\begin{align}\nL^2L\\_\\pm|\\lambda,\\mu\\rangle &= L\\_\\pm L^2|\\lambda,\\mu\\rangle \\\\\n&= L\\_\\pm \\lambda\\hbar^2|\\lambda,\\mu\\rangle \\\\\n&= \\lambda\\hbar^2 L\\_\\pm|\\lambda,\\mu\\rangle \\tag{9}\n\\end{align}$$\n\n\nor in other words, $L\\_\\pm|\\lambda,\\mu\\rangle$ is also an eigenstate of $L^2$ with the same eigenvalue $\\lambda\\hbar^2$ as $|\\lambda,\\mu\\rangle$ itself. Also, using the commutators in equation $(4)$,\n\n\n$$\\begin{align}\nL\\_z L\\_\\pm |\\lambda,\\mu\\rangle &= (L\\_zL\\_x \\pm \\mathrm{i}L\\_zL\\_y)|\\lambda,\\mu\\rangle \\\\\n&= [(\\mathrm{i}\\hbar L\\_y + L\\_xL\\_z) \\pm \\mathrm{i}(L\\_yL\\_z - \\mathrm{i}\\hbar L\\_x)]|\\lambda,\\mu\\rangle \\\\\n&= [\\hbar(\\mathrm{i}L\\_y \\pm L\\_x) + (L\\_x \\pm \\mathrm{i}L\\_y)L\\_z]|\\lambda,\\mu\\rangle \\\\\n&= [\\pm\\hbar(L\\_x \\pm \\mathrm{i}L\\_y) + (L\\_x \\pm \\mathrm{i}L\\_y)L\\_z]|\\lambda,\\mu\\rangle \\\\\n&= L\\_\\pm(L\\_z \\pm \\hbar)|\\lambda,\\mu\\rangle \\\\\n&= L\\_\\pm(\\mu \\pm 1)\\hbar|\\lambda,\\mu\\rangle \\\\\n&= (\\mu \\pm 1)\\hbar L\\_\\pm |\\lambda,\\mu\\rangle \\tag{11}\n\\end{align}$$\n\n\nwhich tells us that $L\\_\\pm|\\lambda,\\mu\\rangle$ is also an eigenstate of $L\\_z$, but with a *different* eigenvalue: the new eigenvalue is $(\\mu \\pm 1)\\hbar$. The raising operator (for example) therefore has the effect of conserving the total angular momentum $L^2$, but increasing the $z$-projection of this angular momentum $L\\_z$.\n\n\nHowever, for this to make physical sense, there has to be a limit to how far this can go: or else we will have no maximum value of $L\\_z$, even for a given value of $L^2$. Note that since $L\\_x^2$ and $L\\_y^2$ are both real and non-negative (which follows from the Hermiticity of $L\\_x$ and $L\\_y$), equation $(3)$ sets a maximum limit on $L\\_z$:\n\n\n$$L\\_z^2 \\leq L^2 \\tag{12}$$\n\n\nMathematically, the only way this can be fulfilled is if there is a \"highest\" state $|\\lambda,\\mu\\_\\mathrm{max}\\rangle$ which the raising operator turns into *zero*:\n\n\n$$L\\_+|\\lambda,\\mu\\_\\mathrm{max}\\rangle = 0 \\tag{13}$$\n\n\nNow consider\n\n\n$$\\begin{align}\nL\\_-L\\_+|\\lambda,\\mu\\_\\mathrm{max}\\rangle &= (L\\_x - \\mathrm{i}L\\_y)(L\\_x + \\mathrm{i}L\\_y)|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= [L\\_x^2 + L\\_y^2 + \\mathrm{i}(L\\_xL\\_y - L\\_yL\\_x)]|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= [L^2 - L\\_z^2 + \\mathrm{i}(\\mathrm{i}\\hbar L\\_z)]|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= (L^2 - L\\_z^2 - \\hbar L\\_z)|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= (\\lambda\\hbar^2 - \\mu\\_\\mathrm{max}^2\\hbar^2 - \\mu\\_\\mathrm{max}\\hbar^2)|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\tag{14}\n\\end{align}$$\n\n\nHowever, we already know that $L\\_+|\\lambda,\\mu\\_\\mathrm{max}\\rangle = 0$. When $L\\_-$ acts on zero, it simply returns zero, and so the \"eigenvalue\" in equation $(14)$ must be equal to zero:\n\n\n$$\\begin{align}\n\\lambda\\hbar^2 - \\mu\\_\\mathrm{max}^2\\hbar^2 - \\mu\\_\\mathrm{max}\\hbar^2 &= 0 \\\\\n\\lambda &= \\mu\\_\\mathrm{max}^2 + \\mu\\_\\mathrm{max} \\\\\n&= \\mu\\_\\mathrm{max}(\\mu\\_\\mathrm{max}+1) \\tag{15}\n\\end{align}$$\n\n\nFundamentally, that is where the term comes from. There are several steps now to close some loopholes. I will not go through them in detail because it is far too long, but I will outline what is needed.\n\n\n1. You need to show that for orbital angular momentum the value of $\\mu\\_\\mathrm{max}$ is restricted to non-negative integer values; we define this quantity to be $l$ and we can replace the quantum number $\\lambda$ with $l$ because $\\lambda$ is related to $\\mu\\_\\mathrm{max}$, which is in turn equivalent to $l$. The eigenvalues of the total angular momentum operator $L^2$ are $\\lambda \\hbar^2 = \\mu\\_\\mathrm{max}(\\mu\\_\\mathrm{max}+1)\\hbar^2 \\equiv l(l+1)\\hbar^2$.\n\n\nNote that for *orbital* angular momentum only integer values of $l$ are allowed; half-integer values are not allowed. This condition can only be derived by solving the differential equations corresponding to the eigenvalue equations of the angular momentum operators (see further reading). For example, you need to solve this:\n\n\n$$\\begin{align}\nL\\_z f(x,y,z) &= (xp\\_y + yp\\_x)f(x,y,z) \\\\\n&= \\left(-x\\mathrm{i}\\hbar\\frac{\\partial}{\\partial y} - y\\mathrm{i}\\hbar\\frac{\\partial}{\\partial x}\\right)f(x,y,z) = \\mu\\hbar f(x,y,z) \\tag{16}\n\\end{align}$$\n\n\nHalf-integer values for spin are allowed because the exact form of the spin angular momentum operators are not known. However, they obey the same commutation relations as the orbital angular momentum operators, so the rules governing them are the same except for the constraint on integer values.\n2. You need to show that the allowed values of $\\mu$ are integers between $-l$ and $l$ respectively. We have already shown that $\\mu$ increases in integer steps, and that $\\mu \\leq l$, so what remains is to show that $\\mu \\geq -l$: the proof is exactly analogous to that in equations $(13)$ through $(15)$, except that you need to consider $L\\_+L\\_-|\\lambda, \\mu\\_\\mathrm{min}\\rangle$.\n\n\nConventionally the quantum number $\\mu$ is renamed to $m\\_l$ and the eigenstates are labelled with $|l,m\\_l\\rangle$. So now we have shown that $l$ is the maximal value of $m\\_l$, and that $m\\_l$ ranges from $-l$ to $l$ in integer steps, as we expected\n3. The total angular momentum $J$ is the sum of the orbital and spin angular momenta, i.e. $J = L + S$. For the rigid rotor system where you ignore spin, $J = L$ and hence we can simply replace $L$, $l$, and $m\\_l$ with $J$, $j$, and $m\\_j$: we therefore have $J^2|j,m\\_j\\rangle = j(j+1)\\hbar^2|j,m\\_j\\rangle$.\n4. Finally, the *energies* are then given by\n\n\n$$\\begin{align}\nE &= \\langle j, m\\_j|H|j,m\\_j \\rangle \\\\\n&= \\left< j,m\\_j \\middle| \\frac{J^2}{2I} \\middle| j,m\\_j \\right> \\\\\n&= j(j+1)\\frac{\\hbar^2}{2I} \\tag{17}\n\\end{align}$$\n\n\nwhere $j$ is a non-negative integer (i.e. $j = 0,1,2, \\ldots$).\n\n\n\n\n---\n\n\n### Further reading\n\n\nChapter 4 of [*Molecular Quantum Mechanics*, 5th ed., Atkins](https://global.oup.com/academic/product/molecular-quantum-mechanics-9780199541423?cc=gb&lang=en&).\n\n\nChapter 7 of [*The Physics of Quantum Mechanics*, Binney & Skinner](https://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf) for an alternative derivation.\n\n\n", "13" ] ]

[ [ "\nI guess (from having heard a few lectures about solid state inorganics, no real experience on the subject) that complex crystalline structures that form a polycrystalline surface just don't fit together sufficiently, crystallographically, with a second surface. Even if both are polished flat down to the atomic scale.\n\n\nThat is especially true if they are not just metallic, but have some of covalent character bonds in them too. The chance that evem a tiny fraction of domains on both sides would by chance fit together is paractically zero.\n\n\nFor the two pieces to form a continuous phase, a lot of surface atoms have to move around, and they're not going to do that if the jump to the next energetically favourable position is more than one or two atomic diameters.\n\n\n", "1" ] ]

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[ [ "\nThe prostaglandin molecule you have given has four stereogenic centers: Three on ring substitutions and one in a one of long side chains $(\\ce{-CH(OH)(CH2)4CH3})$. Since none have other symmetric factors such as plain of symmetry, usual numbers of stereo isomers of the molecule are $2^4 = 16$. However, both long chain substitutions contain disubstituted double bonds, each of which can exist as $E$- and $Z$-isomers ($cis/trans$). Thus, each of the original 16 isomers can exist as four other srereoisomers: $EE$-, $EZ$-, $ZE$-, and $ZZ$-isomers. For example: $(2S,3S,4S,3''R)$- isomer can exist as $(2S,3S,4S,3''R)$-$(2'E,1''E)$-, $(2S,3S,4S,3''R)$-$(2'E,1''Z)$-, $(2S,3S,4S,3''R)$-$(2'Z,1''E)$- and $(2S,3S,4S,3''R)$-$(2'Z,1''Z)$-isomers, all four of which are stereoisomers with optical activity. Thus, you may find total of 64 optical active isomers in given prostaglandins.\n\n\n", "1" ], [ "\n$1)$ Cis-trans isomers are also stereoisomers. They are called as diastereomers.\n\n\n$2)$ Consider, 1-bromo-3-methylcyclohexane. It shows cis-trans isomerism and each of the isomers(that is cis and trans) have enantiomers.\n\n\n$3)$ Again, consider the above example. 1-bromo-3-methylcyclohexane has $4$ stereoisomers, each of them is chiral.\n\n\nThe compound you have given have 16 stereoisomers *excluding* cis-trans isomers.\n\n\n", "1" ] ]

[ [ "\nSteam (actually, water vapor) above a rapidly boiling can be *hotter* than the *average* temperature of the liquid because the situation is *not* in equilibrium.\n\n\nMacroscopically, consider that in [nucleate boiling](https://en.wikipedia.org/wiki/Nucleate_boiling) the greater part of water in the container is *not* boiling; only in certain parts of the vessel has the water been superheated, and then flashes into vapor. \"When the temperature difference is [sufficient], isolated bubbles form at nucleation sites and separate from the surface... [at a greater temperature difference] the vapor escapes as jets or columns which subsequently merge into slugs of vapor.\"\n\n\nThose regions that formed bubbles were *above* the boiling point of water. Though the temperature of the bubble drops due to rapid expansion, it must still be **above 100 °C** or it would condense before reaching the surface (That oscillation of bubbles forming and collapsing can be observed at an incipient boil , i.e., [liquid entrainment](https://www.youtube.com/watch?v=N1yZwRcQSZw) stage of boiling.). Therefore, the vapor immediately released from the bubble can be hotter than the body of the liquid, which can be verified by putting a thermometer in the spout of a steaming kettle and in the liquid.\n\n\nOn a microscopic level, consider the fastest moving particles, i.e, the hottest, are the ones that escape. Of course, that is on *average*, since there can be areas in the liquid that are [superheated](https://en.wikipedia.org/wiki/Superheating) and have not yet boiled.\n\n\nNote, though, that in film boiling, the [Leidenfrost effect](https://en.wikipedia.org/wiki/Film_boiling) maintains a layer of exceedingly hot **water vapor** against the surface of the container, *below* the level of the water! Of course, that is *not* liquid water, but it is below the surface of the water... so the answer to your question depends on the rate of boiling and where in the container you measure the temperature.\n\n\n", "2" ], [ "\nIn addition to the nonequilibrium factors described well by DeMi8she Pippik, we may also consider the behavior when the water is not pure. Assuming we have set up nucle-ation sites and other factors so that the temperature no longer differs measurably from equilibrium, that equilibrium temperature will, in most practical cases, tend to rise as boiling proceeds and the composition of the liquid inevitably changes.\n\n\n* With nonvolatile solutes, the frugality and therefore vapor pressure of the water will drop with the composition shifting to more solute. In the case of sugar solutions in water, this effect should be familiar to anyone who cooks candies or jams.\n* With volatile solutes such as ethanol, the boiling starts off with predominantly the volatile solute at lower than the boiling point os the water alone, and the temperature rises as the liquid composition shifts towards more of the less volatile water component. Here the manufacturing of distilled liquor comes to mind.\n\n\n", "0" ] ]

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[ [ "\nThe black color is due black [copper(II) oxide](https://en.wikipedia.org/wiki/Copper(II)_oxide):\n\n\n\n> \n> It can be formed by heating copper in air at around 300–800°C: $\\ce{2 Cu + O2 -> 2 CuO}$\n> \n> \n> \n\n\nPossible range of temperature seems quite wide.\n\n\nAppearing and intensity of brass blackening is combination of temperature and duration. It can be formed faster with growing temperature due faster kinetics of formation, but getting unstable at very high temperature due thermodynamic decomposition.\n\n\nAs organic substances are not reportedly present (above natural trace background), carbon from pyrolyzed organic matter is not considered.\n\n\n", "7" ], [ "\nA metallurgist would guess at the temperature with a metallographic examination (grain size, twins, etc.) and hardness tests. (It would anneal to some degree depending on temperature.)\n\n\nThe first step would be to determine the brass/bronze alloy. Common wrought brass products are 70:30 or 60:40, and interestingly the 70:30 is most yellow. The 60:40 picks up a pinkish cast.\n\n\n", "4" ], [ "\nThe blackening of copper alloys (e.g., basic brass is 33% zinc with 67% copper) is due to the formation of black oxide of copper (the process is known by the trade name [*Ebonol C*](https://www.jacksonplating.co.uk/ebonol-c.php): Accordingly, if the copper alloys is containing 65% or more of copper, a black oxide treatment can be applied to alloy surface to blacken by converting the copper to cupric oxide $(\\ce{CuO})$. According to [Wikipedia](https://en.wikipedia.org/wiki/Black_oxide), the temperature required to blacken brass alloy is about $\\pu{400 ^\\circ F} \\ (\\pu{204 ^\\circ C})$.\n\n\nIt is also noteworthy to mentioned here this somewhat cool educational demonstration. According to this [Illini website](http://matse1.matse.illinois.edu/metals/g.html):\n\n\n\n> \n> [...] penny coins, starting in 1983, were made of zinc with a thin layer of copper plated on the surface. If these coins are heated, the zinc will diffuse into the copper layer, producing a surface alloy of zinc and copper. These alloys are brasses. Not only does the zinc change the properties of copper, but also the color of the brasses changes with zinc content - reaching a golden yellow color at around 20% zinc and golden at 35-40% zinc. Copper also oxidizes when heated in air, producing a black layer of copper oxide $(\\ce{CuO})$. Thus when heated, there is a competition between the rate of oxidation (making the surface black) and the rate of diffusion (making the surface a golden-yellow color).\n> \n> \n> \n\n\nYou may use this demonstration to find out the exact temperature when the coin starts blackening (hopefully)!\n\n\n", "3" ] ]

[ [ "\nTry 3D pin and ball visual model from home available items, creating respective 3D models according to the Fisher projections of your image. Then try to superimpose them, rotating the whole models or the bonds. If you succeed, you will know why. If not, you will know they are not identical.\n\n\nYou can formally rotate the bottom part of molecules around the central $\\ce{C-C}$ bond to have the side methyl on the bottom. Then it is clear it is the case of the same symmetry as for optically inactive [meso tartaric acid](https://en.wikipedia.org/wiki/Tartaric_acid). Both stereogenic centers mutually cancel each other. The molecule has a plane of symmetry and therefore is not optically active, as it's mirror image can be superimposed with the original.\n\n\n[](https://i.stack.imgur.com/GqVJMm.png)\n\n\nThe suggested turning does not change configuration. Mirror images of your case and of meso tartaric acid can be turned around the bonds to become original image, what disqualifies them from being enantiomers.\n\n\n*With bonds properly rotated, the upper molecule half is the mirror image of the lower half. If you mirror the whole molecule, the upper half becomes the lower half and vice versa, leading to the identical molecule. That is revealed if you turn the mirror image by half turn clockwise.*\n\n\n\n\n---\n\n\nWith methyls on top and the bottom, the enantiomers would be those isomers with H and Br atoms alternating sides. They form mirror twins that cannot be superimposed. Again, they have the same symmetry as optically active isomers of [tartaric acid](https://en.wikipedia.org/wiki/Tartaric_acid).\n\n\n\n\n---\n\n\n\n> \n> If I consider flipping the molecule that would result in every pair of enantiomers being identical.\n> \n> \n> \n\n\nI have suspicion you take the charts as 2D flat structures, what is wrong. [Fischer projection](https://en.wikipedia.org/wiki/Fischer_projection) on your image is convention to display 3D bond structure, with the vertical bonds arranged in a way of bending top and bottom away from you, while horizontal bonds are stretching toward you. Like if there was a horizontal wooden log behind and you tried to wrap the molecule around it. Remember four C bonds are not rectangular in a single plane, but are pointing to the corners of a tetraedr.\n\n\n", "1" ] ]

[ [ "\nWater triple point is $\\pu{+0.01 ^\\circ C}$ at about $\\pu{610 Pa} \\ll \\pu{100 kPa}$ and water does sublime at freezing temperature. But slowly. I assume $\\pu{100 mmHg}$ is a practical threshold for sublimation to have a possibly sufficient rate. There is no need to fit atmospheric pressure. Liquid evaporation occurs with vapor pressure below external pressure too.\n\n\nPhase diagrams are about equilibrium.\n\n\nAt temperatures below their melting point, solids sublimate until the vapor pressure reaches the solid-gas line. Higher temperature means faster sublimation with higher vapor pressure. If the triple point has too low a pressure, we can use only a low maximum vapor pressure which may not be practical for larger quantities.\n\n\n", "4" ] ]

[ [ "\nI assume OP is wondering why always showing carbonyl carbon protonation instead hydroxyl carbon protonation in the mechanism of Fischer esterification. As I have mentioned this in another answer, those reactions are reversible and protonation happens in both oxygens. yet reaction would go through to give the thermodynamically most stable product, which in this case the final ester.\n\n\nThen, why protonation go through carbonyl carbon? Carbonyl carbon of acid group is $\\mathrm{sp^2}$ hybridized and hence, when oxygen is protonated, carbon is capable of stabilizing the positive charge on oxygen by resonance, although this act may migrate the positive charge to unfavorable carbon atom. Yet, since that same carbon (previously carbonyl carbon) already has another oxygen attached to it, which can stabilize the unfavorable positive charge on carbon by mesomeric effect (giving one of the lone pairs to make the third resonance). The higher the participation of the third resonance form (the higher the ability of $\\ce{OH}$ group to release electrons), the more stable the protonated form of a carbonyl group (the larger the basicity of carbonyl group).\n\n\nIf the initial protonation has gone to the hydroxyl oxygen (instead of carbonyl oxygen), there is no way to stabilize the positive charge by any mesomeric effects (neither by first and second resonance). Thus, the protonated form is very unstable and reversed reaction (deprotonation) occurs very quickly. On the other hand, in a protonated carbonyl oxygen of a carboxylic acid, the first and third resonance forms are identical (both having $\\ce{HO^+=C}$), which gives an extra stabilization ($\\mathrm{p}K\\_\\mathrm{a}$ of protonated acid is about $-6$).\n\n\n", "2" ], [ "\nYour mentioned points are somewhat correct. Look in acidic medium electrophilicity of carbonyl increase, secondly nucleophile(like alcohol) get consume due to protonation. For the reaction to happen we maintain a particular particular pH(4.5-5) at which both factors are maintained so that not much consumption of electrophile takes place along with increase in electrophilicity of carbonyl.\n\n\n", "0" ] ]

[ [ "\nThankfully, 2 liters is not alot. Muriatic acid is used to clean bricks and concrete. A local mason or contractor might happily take it for free.\n\n\nBut, if you love a good science experiment, start by being safe with googles and have a baking soda solution ready in case there is a spill. Use a large plastic tub for containment and you are ready to work. Make it outdoors or in a well ventilated area. Outdoors is far better.\n\n\nStart by diluting a portion of the 20% muriatic 1 to 10 with water. \n\nAdd acid to water. Concerns about heat evolution are well founded. You won't get Cl2 gas, but you will get HCl gas during neutralization, which is only slightly less worse if inhaled. The chemistry is fairly simple:\n\n\n$$\\ce{NaHCO3 84g + HCl 36 g -> NaCl + H2O + CO2 fizz}$$\n\n\nYou can make a dilute [baking soda](https://en.m.wikipedia.org/wiki/Sodium_bicarbonate) solution as well and add slowly with stirring. Just keep adding until it no longer fizzles, then test with pH paper. Repeat.\n\n\n", "11" ], [ "\n$1$ liter $\\ce{HCl}$ $20$% weighs $1.101$ kg, and contains $0.20 · 1101$ g/($36.45$ g/mol) = $6.041$ mol $\\ce{HCl}$. $2$ liters $\\ce{HCl}$ $20$%, contain $12.082$ mol $\\ce{HCl}$. To neutralize this acid with baking soda ($\\ce{NaHCO3}$), you will need $12.082$ mol baking soda, which weighs $12.082 · 84.02 g = 1015$ g $\\ce{NaHCO3}$. To neutralise the same amount of acid with lime (impure $\\ce{CaCO3}$), you will need $6.041$ mol $\\ce{CaCO3}$, which weighs $604.1$ g lime.\nThe reaction will be quickly finished. The obtained final solution may be thrown away with the waste waters. And I repeat, no toxic chlorine gas will be produced.\n\n\nYou may also replace baking soda or lime by a metal like iron or aluminium. But the reaction will be slow, especially at the end, when $90$% of the acid will have reacted. It may last more than one hour. Just to get an idea, you will need $6.041$ mol of metal, which weighs $338.3$ g with iron, and $163.1$ g with aluminium.\n\n\n", "7" ], [ "\nYou are so lucky as to have found 2 liters of toilet bowl cleaner (concentrated). There used to be a variety in my grocery store that was 20% HCl, but now it is only 10%, so now I have to use twice as much. The commercial materials contain a little surfactant to make some foam, and usually some blue color.\n\n\nHowever, you can make do with the clear stuff, and don't really need the surfactant; if you want to, you can add that to the toilet water separately.\n\n\nAs a first try, pour about 10-20 mL of the 20% acid into your toilet bowl and scrub around with the usual toilet brush. Then flush. Depending on how often your toilet needs cleaning, your hydrochloric acid will last about 2-3 years. Consider it a blessing. When it's gone, you may consider buying a gallon from the hardware store.\n\n\nNow, if it really gets on your nerves to have a generic toilet bowl cleaner in your house, you could use twice as much, twice as often, and make it go away four times faster.\n\n\nBTW, I have a gallon of 31% HCl that I use - but I dilute it to 15% for toilet bowl cleaning. I keep the jug (tightly stoppered) in a plastic bucket, in a place where there is plenty of ventilation (just in case of leakage). So, you do need to be careful, but aggressive chemicals can be useful if well contained.\n\n\n", "4" ], [ "\nSome of the answers above go into a lot of good detail as to the chemistry behind the answer, but I want to get to the point and avoid extraneous detail.\n\n\n\n\n---\n\n\nGradually pour chalk or baking soda into it, stir and repeat until it stops reacting.\n\n\nCongratulations, you've neutralised your acid and now have a salty solution of water and sodium or calcium chloride (depending on whether you went with chalk or baking soda respectively). This can be poured down the drain no bother.\n\n\nJust make sure you do it outside so the hydrogen doesn't build up and you do it slowly enough that the heat doesn't cause a problem.\n\n\n", "3" ] ]