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[ [ "\nThe electronic energy is lowered when bonding occurs, therefore the energy levels become more stable. Stable just means they have a lower energy level.\nSo if we have two 1s orbitals they can combine to form a single sigma bond which has a lower energy than the components separately.\n\n\n\n", "-1" ] ]

[ [ "\nIf the reaction till \"then\" was a solution of, say 500 mL, than the addition of the next liquid – from your writing, it reads like it were methanol – should be equally just 500 mL. (And 500 divided by 500 were ... just 1).\n\n\nMaybe adding *more* than one part of methanol would no longer lead to (this much / this clean) precipitation intended, as methanol alone is quite a good solvent of your precipitate\\*), too.\n\n\n\\*) Intentionally leaving open, if the precipitate is an impurity to be removed, or really refers to the product *to be isolated*. You have the literature reference.\n\n\n", "1" ] ]

[ [ "\nYour argument is valid, though I believe the scenario you describe is not within the range of applicability of Ostwald's law. Allow me to recast your example in a more obvious form. \n\n\nYou have a solution of weak base. You add a stronger base. The increase in hydroxide concentration leads, by Le Chatelier's principle, to a lower extent of dissociation.\\*\n\n\nThis is beyond the scope of Ostwald's law, which tacitly assumes the lack of what amounts to a common-ion effect. Ostwald's law holds if the solvent which you're using to dilute your solution does not interfere (or interferes only weakly) with your electrolyte's solubility; the added solvent must serve to decrease the concentration of your electrolyte. Evidently, then, adding a stronger base will interfere with the solubility of your weak base, and is not governed by Ostwald's law.\n\n\n\n\n---\n\n\n\\*There is perhaps some subtlety here. As an example, consider the salt $\\ce{MX}$, which dissociates weakly into $\\ce{M+}$ and $\\ce{X-}$. The intuition behind the greater degree of dissociation is that (1) upon dilution, the equilibrium constant for dissociation doesn't change, whereas (2) the concentration of each species decreases. Because dissociation produces more ions than was originally present, this always leads to $Q < K$, and more dissociation will occur. \n\n\nIf we add a salt $\\ce{MY}$, then, we are introducing two effects: (1) decreased concentration of $\\ce{X-}$ and (2) increased concentration of $\\ce{M+}$. Because these effects pull the equilibrium in opposite directions, we can't conclusively determine what exactly will happen. Going back to your example, if we add a slightly stronger base, then we get exactly these two contrasting effects, and the change in the degree of dissociation is indeterminate. In contrast, if we add a much stronger base, then certainly (2) dominates and the degree of dissociation is reduced.\n\n\n", "4" ] ]

[ [ "\nJoh,\n\n\nPlease let me know if you referred to this compound:\n\n\nAlso, if this is your compound, please find the name.\n\n\n[](https://i.stack.imgur.com/jwEjS.jpg)\n\n\n", "4" ], [ "\n[ChemExper](http://www.chemexper.com) has a free-to-use structure search, which would usually give you a name and access to safety information from the supplier. Caveat: this will necessarily only work for commercially available and indexed chemicals.\n\n\nIf you have access to university resources or similar, you may be able to use SciFinder or Beilstein.\n\n\n", "3" ], [ "\nThere are two altogether different aspects to this question.\n\n\nFirst, you may want to look for your structure in the databases which contain known molecules. This aspect is covered (partially) by **TAR86**.\n\n\nSecond, you may want to know the name of a molecule of which you have a drawing, and which may or may not be known, or even exist in nature. In this case you may resort to some automatic naming engine, which are built into many popular molecular drawing programs, and also available [online](https://web.chemdoodle.com/demos/iupac-naming/). What's the point of knowing a name (but not the properties) of some strange molecule is another question. If you intend to use it to search for the molecule in [some database](http://www.chemspider.com/StructureSearch.aspx), then look again: maybe the said database comes with an interface for search by structure, and you don't need the name in the first place.\n\n\n", "3" ], [ "\nI would suggest visiting www.organic-chemistry.org/chemicals/structuresearch.htm\n\n\n[](https://i.stack.imgur.com/wAvAg.png)\n\n\nIt's a pretty good website, I must say. Simply stretch and draw carbon bonds, replace atoms of carbon with other heteroatoms, or even add double and triple bonds. It usually gives you the name of the compound as well as sources to buy it upon clicking the *Exact Search* button. \n\n\nHappy structure searching!\n\n\n", "3" ] ]

[ [ "\nYou need to narrow down your question. Currently, the potential solutes include NaOH (inorganic salt), dextrose (organic neutral compound), and choline chloride (organic quaternary ammonium salt) on one hand, and on the other hand a wide range of potential solvents. It is unlikely a single database will (ever) cover such a broad range of possibilities.\n\n\nNevertheless, there are some resources that may be useful to consult. For example solubility tables [like this](https://en.wikipedia.org/wiki/Solubility_table) or the properties tables for some of the compounds in wikipedia, like [here](https://en.wikipedia.org/wiki/Choline). If you know both solvent, and solute, NIST-IUPAC's solubility database [here](https://srdata.nist.gov/solubility/sol_detail.aspx?sysID=62_223) may be worth to visit.\n\n\nIf your institution / library provides access to Reaxys/Beilstein (a commercial database by Elsevier), you may find solubility data in the category of *physical properties*. This database covers both organic and inorganic (including metal organic) compounds and will provide you both the numerical value, as well the primary reference.\n\n\n", "2" ] ]

[ [ "\nYou could proceed from either end member of the olivine solid series and yield $\\ce{SiO2}$ as you suggest.\n\n\nHowever, I'd consider mechanisms that have been researched in the course of studying so-called mineral sequestration in addition to what you've written, especially considering the energy requirements you propose: the mineral sequestration reactions are (slightly) exothermic and would thus also yield heat as a by-product which I suspect would be desirable in a cold place like the Moon. That said, the reactions are slow at standard conditions and can be accelerated at the cost of supplying energy. How that balance works in your scenario is ultimately up to you, of course.\n\n\nAdditionally, carbon dioxide is a reactant (but needs to be in a supercritical state, which costs energy), which you'd get from human respiration and other organic sources \"for free,\" which relieves you of having to have a lot of hydrogen gas on-hand.\n\n\nThe idea with mineral sequestration is to allow [supercritical carbon dioxide](https://en.wikipedia.org/wiki/Supercritical_carbon_dioxide) to react with certain minerals to yield carbonates that are stable over some long time period:\n\n\n\n> \n> Mineral carbonation reactions are known to geologists and occur spontaneously on geological time scales. For example, the reaction of $\\ce{CO2}$ with common mineral silicates to form carbonates like magnesite or calcite is exothermic and thermodynamically favored.\n> \n> \n> \n\n\nAn example is:\n\n\n$$\\ce{Mg2SiO4 + 2CO2 -> 2MgCO3 + SiO2}$$\n\n\n\n> \n> (which) illustrates the transformation of forsterite, which is the end member of the common silicate mineral olivine. One ton of olivine can dispose of approximately two-thirds of a ton of $\\ce{CO2}$. Again, the reaction is\n> exothermic and releases 90 kJ/mole of $\\ce{CO2}$. \n> \n> \n> \n\n\nIn summary: There's nothing wrong with your chemistry or geology and there is more than one way to yield silica from mafic/ultramafic minerals.\n\n\nThe reference I have quoted above and cited below also details the process in general terms for minerals such as olivine and serpentine, and illustrates schematics for implementing such schemes on industrial scales, which might also be of interest to you.\n\n\n[Reference](http://www.netl.doe.gov/publications/proceedings/01/carbon_seq/6c1.pdf) from the National Energy Technology Laboratory within the DOE:\n\n\n\n> \n> Goldberg, P. ,Chen, Z. Y. ,O’Connor, W. ,Walters, R., and Ziock, H. $\\ce{CO2}$ Mineral Sequestration Studies in US. *Technology*. 1 (1): 1–10 (2000).\n> \n> \n> \n\n\n", "9" ], [ "\nThe previous answer is not really helpful because you would need huge amounts of CO2 which are not available on the moon. You need 2 moles of CO2 to generate 1 moles of SiO2. You simply do not have that amounts of CO2.\n\n\nFurthermore, assuming you do someone get that CO2, you need to physically separate the Mg and Fe carbonates from the silica. One way would be dissolving the carbonates in acid, which is also not available on the moon. Another way would be to melt the thing and separate the two immiscible carbonate and silicate liquids by centrifuging. This will require very high pressure to keep the carbonate in, otherwise, this thing will decarbonate and just make olivine again.\n\n\n\n> \n> If the forsterite is pure enough, could that be used to make decent\n> clear glass?\n> \n> \n> \n\n\nNo. Forsterite has a melting point above 1800 °C. It also has the problems I mentioned in my previous answer to you:\n\n\n[Can glass be made with anorthite?](https://chemistry.stackexchange.com/q/67403/8083)\n\n\nThe problem of quench crystals forming in your liquid once you cool the forsterite is much worse than what I described for anorthite in the answer given in the link. You will have to find a way to cool it down even faster, and I doubt you can do it. It's also probably going to crack and break when cooling down.\n\n\nThen there's the question of if you can even make it pure enough, but the answer for that will be in your other question here:\n\n\n<https://earthscience.stackexchange.com/q/10110/725>\n\n\n\n> \n> Could the fayalite be processed into iron, water, and silicon dioxide...\n> \n> \n> \n\n\nYes. It still doesn't solve the problem of forsterite being a terrible glass material.\n\n\n**EDIT**\n\n\nThere's a recently published paper:\n\n\nSchleppi, J., Gibbons, J., Groetsch, A. et al. J Mater Sci (2019) 54: 3726. <https://doi.org/10.1007/s10853-018-3101-y>\n**Manufacture of glass and mirrors from lunar regolith simulant**\n\n\nIt is open access. You might find it informative.\n\n\n", "5" ] ]

[ [ "\nIf I assume you refer to 1,2-cylohexanediols, the two isomers to consider were (*cis*), and (*trans*)\n\n\n[](https://i.stack.imgur.com/2czj4.png)\n\n\nAssuming the cyclohexane ring as a plane, an (axial, equatorial) orientation of the two O-substitutents is puts them both on the same side of this reference plane; the relative orientation of the two substituents is (*cis*):\n\n\n[](https://i.stack.imgur.com/tont9.png)\n\n\nThis contrasts to the case of the (*trans*)-configuration, where the two substituents may be either both in axial, or (likely preferred by thermodynamics) both in equatorial orientation:\n\n\n[](https://i.stack.imgur.com/uIBDo.png)\n\n\nFrom the later picture, taking into account *i*) the relative orientation of the two hydroxyl groups towards each other as well in respect to the cyclohexane moiety and *ii*) the (assumed) conformational preference for this form over the conformer with two axial oriented hydroxyl groups, I would assume the two individual vectorial contributions along $\\ce{C -> O}$ are better lined up than in the instance of the (*cis*)-configuration to yield a larger (global) dipolar moment.\n\n\n", "4" ] ]

[ [ "\n**The clue is the moles in the dilute solution is the same as the moles of the concentrated stock solution - it has to be constant.**\n\n\nThe method I have used is (therefore) correct. \n\n\n", "2" ] ]

[ [ "\nWith the same reasoning as provided [here](https://chemistry.stackexchange.com/questions/71553/geometrical-isomerism-cis-trans-in-trans-2-fluoro-3-methylpent-2-ene/71560#71560), I recommend that you name this compound (2 *E*)-2-bromobut-2-ene.\n\n\nUsing (*cis*/*trans*) may be useful around an isolated double bond with only two substitutents. While in the example presented by you the two methyl groups are on the same side of the double bond, they are not the only ones to be considered here. Triple and quadruple substituted dienes should be named in accordance to the CIP rules only.\n\n\n**Edit by 02-Jan-2019:**\nAgreeing with a comment by @Loong, the previously suggested name, \"(*E*)-bromo-2-butene\", was replaced by \"(2 *E*)-2-bromobut-2-ene\". This edit removes simultaneously the ambiguity about the position of the bromine, as well as the double bond. The interested is referred to the illustrated examples of rule R-7.1.2 of the corresponding Blue Book [here](https://www.acdlabs.com/iupac/nomenclature/93/r93_626.htm).\n\n\n", "6" ] ]

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[ [ "\nThis is a titration problem - not necessarily a buffer problem. While you have $0.0055$ moles of $\\ce{NH3}$, you have $0.01$ moles of $\\ce{H}$ from $\\ce{HCl}$ dissociating. $0.01 - 0.0055 = 0.0045$ moles of $\\ce{H}$ left. Take H/total L to find M. Take the $-\\log$ of $\\ce{[H]}$.\n\n\n", "3" ] ]

[ [ "\nI'm fairly sure the states $\\require{\\begingroup} \\begingroup \\newcommand{\\ket}[1]{|#1\\rangle} \\newcommand{\\braket}[1]{\\langle #1 \\rangle} \\ket{i}$ and $\\ket{f}$ are eigenstates of the unperturbed Hamiltonian $\\hat{H}\\_0$. Therefore, by virtue of the Hermiticity of $\\hat{H}\\_0$, they are necessarily orthonormal\n\n\n$$\\braket{i|f} = \\delta\\_{ij}$$\n\n\ntherefore if we define $\\hat{V'} = \\hat{V} + k$ where $\\hat{V}$ is the perturbation Hamiltonian and $k$ is some arbitrary constant\n\n\n$$\\begin{align}\n\\braket{i|\\hat{V'}|f} &= \\braket{i|\\hat{V}|f} + \\braket{i|k|f} \\\\\n&= \\braket{i|\\hat{V}|f} + k\\braket{i|f} \\\\\n&= \\braket{i|\\hat{V}|f}\n\\end{align}$$\n\n\nif $\\ket{i}$ and $\\ket{f} \\endgroup$ are different states. If they are the same state, then you are calculating the transition probability from one state to itself - not a very useful concept.\n\n\n", "6" ] ]

[ [ "\nUsing the Henderson-Hesselbach equation is a good idea. As you want the buffer's pH to be acidic, then using the $\\ce{NH3/NH4+}$ buffer is not a good idea, as that would be basic, so we should use the $\\ce{C5H5N/C5H5NHCl}$. Then, by Henderson-Hesselbach, we have that $\\ce{5.05 = pH = pKa + \\log \\left(\\frac{[C5H5N]}{[C5H5NH+]}\\right)}\\implies \\ce{5.05 - pKa = 5.05 - 5.23 = \\log \\left(\\frac{[C5H5N]}{[C5H5NH+]}\\right)} \\implies \\ce{\\frac{[C5H5N]}{[C5H5NH+]} = 10^{-0.18}\\implies \\frac{[C5H5N]}{[C5H5NH+]} \\approx 0.66\\approx \\frac 23}.$ Thus, we want a buffer solution where the ratio of the base to the acid is approximately 2/3, which answer choice C satisfies.\n\n\n", "1" ] ]

[ [ "\n$\\displaystyle \\ce{x\\_{solute}} = \\ce{\\frac{n\\_{solute}}{n\\_{solute} + n\\_{solvent}}}$\n\n\nThus, we have that $\\displaystyle\\ce{x\\_{solute}} + \\ce{x\\_{solvent}} = \\ce{\\frac{n\\_{solute}}{n\\_{solute} + n\\_{solvent}}} + \\ce{\\frac{n\\_{solvent}}{n\\_{solute} + n\\_{solvent}}} = \\ce{\\frac{n\\_{solute} + n\\_{solvent}}{n\\_{solute} + n\\_{solvent}}} = 1,$ so $\\ce{x\\_{solute} + 0.974 = 1\\implies x\\_{solute} = 0.026}$ \n\n\n", "1" ] ]

[ [ "\nSr(NO3)2 + Na2SO4 = SrSO4 + 2NaNO3\n\n\nSo, here we have, 1 mol Sr(NO3)2 reacts with 1 mol of Na2SO4 by unitary method, you can say, 0.003 mol Sr(NO3)2 reacts with 0.003 mol of Na2SO4.\nBut checking the reactants we have 0.003 moles Na2SO4 and 0.014 moles Sr(NO3)2\n\n\nSo only 0.003 moles of each reacts with the other to leave out any excess in either of the reactants, as in this case, Sr(NO3)2.\n\n\nWhen 0.003 mol reacts of Sr(NO3)2, we have $$0.014 (initial) - 0.003 (used) = 0.011mol$$ of Sr(NO3)2 left.\n\n\n", "0" ] ]

[ [ "\nLet us read Keeler and Wothers's *Chemical Structure and Reactivity* carefully. (It's really a great book. I wish I had used it in the past.) We excerpt the relevant paragraphs that detail the points you mention, and highlight particularly important statements regarding screening. All page numbers are taken from the second edition.\n\n\n\n\n---\n\n\n*Point 2, on Slater's rules.*\n\n\n\n> \n> [Energy ordering:] (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) ...\n> \n> \n> 3. If the electron being considered is in an ns or np orbital, then electrons in the next lowest shell (i.e. that with (n-1)) each contribute 0.85 to $S$. Those electrons in lower shells (i.e. (n-2) and lower) contribute 1.00 to $S$.\n> 4. If the electron being considered is in an nd or nf orbital, all electrons below it in energy contribute 1.00 to $S$ [the shielding constant]. (p. 262)\n> \n> \n> \n\n\n*Point 3, on Z$\\_\\text{eff}$ and d-block contraction.*\n\n\n\n> \n> ...there is a downwards kink in the [s-block] orbital energy on moving from the 3s to the 4s [orbitals]. [...] This difference is due to the filling of the 3d orbitals between calcium and gallium. The effects of all the extra protons in the nucleus is to cause an additional lowering in the energy of the valence orbitals for the p-block elements in Period 4. However, the effect is not really drastic: the 4s and 4p orbitals in gallium do not seem to have experienced the full effect of the ten extra protons. We interpret this by saying that **the effect of the extra protons on the orbital energies of the 4s and 4p has, to a large extent, been cancelled out by the electrons that have been added in the 3d orbitals**.\n> \n> \n> We could rephrase all of this by looking at the problem from the perspective of effective nuclear charges rather than orbital energies. The increase in Z$\\_\\text{eff}$ for gallium is slightly greater than might have been expected by comparison with boron and aluminium because of the filling of the d-block. (p. 266)\n> \n> \n> \n\n\n*Point 4, on orbital energies.*\n\n\n\n> \n> ...between scandium (Sc) and copper (Cu) electrons are being added to 3d orbitals, and **such electrons form an effective screen for the 4s electrons**. To put it another way, as far as the 4s electrons are concerned, on moving from one element to the next between scandium and copper, the effect of the extra proton is largely cancelled out by the addition of the electron to the lower 3d orbital. Consequently the energy of the 4s falls rather slowly. (p. 261)\n> \n> \n> The outer electron in potassium occupies the 4s AO, rather than the 3d. This is because the 4s orbital penetrates to the nucleus more effectively than does the 3d, resulting in the energy of the 4s being lower than that of the 3d. [...] ...both the 3s and the 3p have subsidiary maxima close in to the nucleus, but the first maximum for the 3s is very much closer in. The 3d has no such subsidiary maxima, and therefore is much less penetrating than 3s or 3p.\n> ...**the electrons in the 4s do not screen the 3d particularly well since much of the electron density from the 4s is further out from the nucleus than that from the 3d**.\n> \n> \n> From scandium onwards the energies of both the 4s and 3d AOs drop steadily: this is simply the result of the increase in nuclear charge not being quite offset by the increase in electron-electron repulsion. Put another way, **the electrons in the 3d sub-shell do not screen one another particularly well**. Although both AOs fall in energy, the 4s falls less steeply, which can be explained by noting that **this AO is quite well screened by the 3d electrons, not least as these are in a lower shell**. (p. 627)\n> \n> \n> \n\n\n\n\n---\n\n\nWe see that Keeler and Wothers are at least self-consistent regarding points 2, 3, and 4: 3d electrons are effective at shielding 4s and 4p electrons and not effective at shielding 3d electrons. I believe, therefore, that the contradiction lies in point 1, which should be corrected as follows:\n\n\n\n> \n> d-block electrons shield poorly **s or p electrons in the same shell**\n> \n> \n> \n\n\nI suspect the statement that d-block electrons shield poorly is taught in reference to the concept of penetration. If we look at plots of radial probability density for the 3s, 3p, and 3d orbitals, we might naively conclude, on the basis of mean distance from nucleus, that 3d electrons are closest to the nucleus, which means that they should shield 3s and 3p electrons well. In reality, because of penetration, this is not the case. Thus we conclude that 3d electrons shield (s or p electrons in the same shell) poorly, and the misconception arises if we try to extend this poor shielding to 4s electrons and the like.\n\n\n", "4" ], [ "\nIf u see the radial probability distribution function of nd orbital and that of ns orbital u will notice small humps are more in case of s orbital than that of d orbital. Because of this reason s orbital are more 'penenetrating' towards nucleus than that of a d orbital. So the screening constant value for a d orbital is much less compared to that of s orbital.\n Again if we consider the angular probability distribution function then we can see that d orbital are very much distributed in space compared to that of a s orbital which is spherically symmetric. Due to this reason d orbital registers a lower value of screening constant.\n\n\n", "0" ] ]

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[ [ "\nThe alkoxide and the hydroxides aren't good leaving groups. Consider an alcohol, the $\\ce{OH}$ group never leaves on its own. Oxygen donates a lone pair to the hydrogen of a hydronium ion (considering it to be in an aqueous solution). The water molecule now attached is a good leaving group (oxygen has a positive charge).\n\n\nNucleophilicity does not determine the suitability of a leaving group. A group is said to be a good leaving group when it can leave as a relatively stable ion. For example: iodine is a very good leaving group for an $\\mathrm{S\\_N2}$ reaction.\n\n\n", "5" ], [ "\nBoth alkoxides and hydroxides are not great leaving groups, but since we are just comparing their leaving group character...\n\n\nFirst of all, leaving group tendency does not depend on nucleophilicity. Rather, it depends on basicity. Both of these are often confused to be the same thing. Yes, it's true that coincidentally both of these, in most cases, suggest similar things, but they are not the same. Let us compare the two for OH and OR.\n\n\nNucleophilicity:\nOH is a better nucleophile than OR since it has more mobility(due to its small size)than OR. A good nucleophile should be small in order to attack sites easily.\n\n\nBasicity:\nThis does not depend on mobility. It only depends on electron donating tendency (inversely proportional). Here, OR is a weaker base, hence it is a better leaving group\n\n\n", "3" ], [ "\nYou are confusing hyperconjugation effect on a carbocation with basicity. Basicity mainly depends on the following factors in order of significance: Atom → Resonance → Inductance → Orbital. Atom is the same (oxygen), there is no resonance, so inductance is key. Carbon is more electronegative than H, so OR will better stabilize a negative charge, thus OR is a better leaving group stable weak base.\n\n\n[](https://i.stack.imgur.com/9L3Qr.png)\n\n\n", "3" ] ]

[ [ "\nIn order to understand an answer to this you need to be able to differentiate between system and surroundings \nThe system is the reactant mixture , now for an endothermic reaction the temperature of the reactant mixture will increase , however the temperature of the surroundings will decrease since heat is lost by the surroundings to the system , let's assume that an endothermic reaction took place in a test tube , if we put our hand around the test tube our palm will feel cooler since heat was transferred from our palm , however that heat was transferred to the reaction mixture hence the temperature of the mixture was raised . Hence this increase in temperature means we have altered the equilibrium conditions and thus excess heat needs to be lost therefore the eq will shift towards the exothermic reaction . However a decrease in temperature means the eq needs to gain heat which is in the form of an endothermic reaction \n\n\n", "2" ] ]

[ [ "\nConsidering you have chosen a $\\ce{Zn}$ concentration cell:\nThe cell representation is:\n$$\\ce{Zn(s)/ZnSO4(M1)||ZnSO4(M2)/Zn(s)}$$\n\n\nAnodic reaction puts more $\\ce{Zn^2+}$ ions in the solution:\n$$\\ce{Zn(s) -> Zn^2+(aq) + 2e-}$$\nTo balance these extra $\\ce{Zn^2+}$ ions salt bridge pulls some anions into anodic compartment.\n\n\nCathodic reaction removes $\\ce{Zn^2+}$ ions from cathodic compartment:\n$$\\ce{Zn^2+(aq) + 2e- -> Zn(s)}$$\nTo make up this equivalent amounts of cations are pulled off from the salt bridge.\n\n\nSo activity of $\\ce{SO4^2-}$ is not changed in either of the solutions. It is only those of $\\ce{Zn^2+}$ ions in both the compartments, which undergoes a change leading to redox reaction.\n\n\n", "1" ] ]

[ [ "\nYes, based on what we can see in the video, your guess appears to be correct: as the propane-filled bottle warmed up, just enough propane evaporated to keep the pressure inside the bottle equal to the [equilibrium vapor pressure](https://en.wikipedia.org/wiki/Vapor_pressure) of the liquid propane.\n\n\n[According to the video](https://youtu.be/mnZ_Rk6kD9M?t=270), the ambient temperature outside at the time it was recorded was \"about 45 °F\", or about 7 °C. Using the formula given [here](https://en.wikipedia.org/wiki/Propane_(data_page)#Vapor_pressure_of_liquid), I calculate the vapor pressure of propane at that temperature to be about 4400 mmHg, or about 590 kPa or about 5.9 bar.\n\n\nMeanwhile, according to [this page](http://hypertextbook.com/facts/2000/SeemaMeraj.shtml), the pressure inside a warm can or bottle of Coke can reach at least 380 kPa, or about two thirds of the vapor pressure of propane on a cold day. As the bottles are certainly designed with a considerable safety margin, to make sure that they won't burst even if handled carelessly or slightly damaged, it's not surprising that they can easily withstand the pressure of the propane in the video.\n\n\nBTW, this is exactly how [aerosol spray cans](https://en.wikipedia.org/wiki/Aerosol_spray) work: they contain a mixture of the liquid being sprayed and a propellant substance (quite often propane) that has a boiling point at 1 atm only slightly below room temperature (or, equivalently, that has an equilibrium vapor pressure only slightly above 1 atm at room temperature). Thus, as the can is drained, the partial boiling of the propellant maintains the pressure inside the can at the propellant's vapor pressure, which is high enough to propel the spray out of the nozzle, but not so high that it would require an excessively sturdy and expensive can to contain it.\n\n\n\n\n---\n\n\nAs for what would happen at higher temperatures, at room temperature (i.e. 25 °C), the vapor pressure of propane would be about 7100 mmHg or 950 kPa (according to the formula, or about 7600 mmHg or 1000 kPa according to the table, which seems to be taken from a different source). According to [this random forum post](http://chemistry.mdma.ch/hiveboard/acquisition/000426482.html), the small ½ liter Coke bottles used in the video can apparently withstand *at least* 180 psi, or 1250 kPa, so the propane-filled bottle *probably* wouldn't burst even at room temperature (unless it happened to be damaged or otherwise particularly weak). If the temperature was raised to, say, 45 °C (113 °F, a *very* hot day), the vapor pressure of the propane would rise further to about 1500 kPa, which just *might* be enough to make the bottle fail. Also, elevated temperatures will soften and weaken the plastic somewhat, making failure more likely.\n\n\nIn any case, in my personal experience, the weakest point of such bottles seems to be the relatively thin and weak cap, which would likely fail at some point before the bottle itself did. I wouldn't be surprised if that was by design, to make the typical failure mode relatively safe and predictable.\n\n\n", "22" ], [ "\nThe person in the video said it was about $\\pu{45^oF}$. At this temperature propane has a vapor pressure of about 8 atm, so that is the pressure that built up inside the Coke bottle. Once this pressure had built up, it quit boiling. \n\n\nThe fact that he could still fairly easily squeeze the bottle suggests that it was probably not very close to bursting from the pressure, and I very much doubt that even a hot day would have caused enough pressure for it to burst. I once saw the aftermath of a similar experiment with a bottle like that in which the lucky-to-be-alive idiot had used liquid nitrogen to blow it up. The bottle had badly deformed and stretched out before bursting. \n\n\n**Bonus Answer: Why it didn't work** \n\nSince propane is less dense than, and immiscible with, water (or Coke), it just floated on top and was the first thing to pour out when the bottle was inverted. If he had held his glove-covered hand over the opening and inverted it so that the propane was on top inside the bottle, rapidly expanding in a confined space, it would have at least had a chance of working.\n\n\n", "12" ] ]

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[ [ "\nBecause an oxygen ion is $\\ce{O^2-}$ and a chlorine ion $\\ce{Cl^-}$, you know that the salts have to be of the form $\\ce{M\\_xO\\_y}$ and $\\ce{M\\_xCl\\_{2y}}$.\n\n\nLet $o$ be the atomic mass of oxygen, $c$ of chlorine and $m$ of the unknown metal M. The problem states that\n\n\n$$0.6(xm+yo)=xm$$\n\n\nwhich implies\n\n\n$$0.6yo = 0.4xm \\implies m = \\frac{3y}{2x}o$$\n\n\nIf you recall some formulas for salts, you know that $x$ can only be 1 or 2 (e.g. $\\ce{M\\_5O\\_3}$ makes no sense). This means that $m$ is a multiple of $\\frac34o$ which is (approximately) 12. So we only need to check the elements with an atomic weight which is a multiple of 12:\n\n\n* 12: Carbon (C) isn't a metal\n* 24: Magnesium (Mg) is, and you can check the oxide/chloride satisfy the equations\n* 36: n/a\n* 48: Titanium (Ti) checks out\n* 60: n/a\n* 72: n/a\n* 84: Krypton (Kr) isn't a metal\n\n\nand so on. Even if the atomic weight corresponds to a metal, you can verify the value of $y$ doesn't match with the oxide that's formed.\n\n\n", "2" ] ]

[ [ "\nYour misunderstanding appears to be some confusion between stoichiometry and equilibrium concentrations. The chemical equation describes the **stoichiometry** or the ratio of the species in the reaction. In other words, each time the reaction happens, the equation describes how many of each reactant are consumed and how many are produced. In your example:\n\n\n$$\\ce{HCO3- + H2O <=> H2CO3 + OH-}$$\n\n\nEach time this reaction occurs in the forward direction, one bicarbonate anion and one water molecule react to form one carbonic acid molecule and one hydroxide anion. In the reverse direction, one carbonic acid molecule and one hydroxide anion react to form one bicarbonate anion and one water molecule. \n\n\nSince this reaction is reversible, the equilibrium constant tells you something about **the extent** of the reaction. In other words, what portion of the overall system is sitting on the reactant side and what portion sis sitting on the product side. Put another way, the system is at equilibrium when the rates of the forward and reverse reactions are equal. These rates are dependent on the concentrations of the species on the reactant and product sides of the equation and on rate constants.\n\n\n$$\\mathrm{rate\\_{forward}=rate\\_{reverse}}\\\\\nk\\_\\mathrm{f}[\\ce{HCO3-}][\\ce{H2O}]=k\\_\\mathrm{r}[\\ce{H2CO3}][\\ce{OH-}]$$\n\n\nThe equilibrium constant is the ratio of the rate constants:\n\n\n$$K=\\frac{k\\_\\mathrm{r}}{k\\_\\mathrm{f}}=\\frac{[\\ce{H2CO3}][\\ce{OH-}]}{[\\ce{HCO3-}][\\ce{H2O}]}$$\n\n\nThe equilibrium constant is thus a law of mass action expression for the situation where the forward and reverse rates are equal. At this state, both forward and reverse reactions are occurring following the stoichiometric ratios, but the concentrations of each species has reached a steady state.\n\n\n", "2" ] ]

[ [ "\nI believe that the word you are looking for is a pH-neutral compound, or simply a neutral compound.\n\n\n", "1" ], [ "\nI think your best bet would be neutral salts. This term describes compounds such as $\\ce{NaCl, KNO3, CaBr2, CsClO4}$. This [page](http://www.science.uwaterloo.ca/~cchieh/cact/c123/salts.html) from the University of Waterloo gives a brief table of ions that when paired produce a neutral salt.\n\n\n", "1" ] ]

[ [ "\nYour supposition that gases do not have equal access to teh surface does not seem to me to be correct. \n\n\nAt the surface physisorption and chemisorption can occur. The former is due to inter-molecular interaction with the gas molecule and the surface molecules so clearly will depend on the nature of both of them. In chemisorption a chemical bond is formed. The physisorption occurs at longer range and has a smaller potential well (as measured by potential energy vs distance from surface) than chemisorption. (See Atkins & DePaula 'Physical Chemistry' for a figure.)\n\n\nThe gases approach the surface at random and the number of collisions of each will be in proportion to their partial pressure. What happens on collision depends, as mentioned above, on the strength of the intermolecular interaction of gas and surface and so depends on what they are. At higher temperature the collision energy is greater than at lower ones and so one would expect less physisorption as the collision energy will be greater than that of most available physisorption sites.\n\n\nIt is also possible for multiple layers to be formed (as in BET isotherm) and for a physisorbed molecule to diffuse around on the surface until it leaves again (by random thermal activation) of finds a more stable site or becomes chemisorbed. \n\n\nAs to your particular case we can only surmise but $\\ce{CO2}$ is far more polarisable than methane so should have the larger dispersion forces so larger induced dipole - induced dipole intermolecular interaction and so preferentially occupy the surface. Both gases have transient dipoles due to molecular vibrations of which the $\\ce{CO2}$ should be the larger. This may also be important.\n\n\n", "1" ], [ "\nYou are talking about a porous solid. This is very different to a flat surface where adsorption of all molecules can occur without any barrier. To be adsorbed in the pores, the molecules need to be transported from the gas bulk to inside the pores, this process can be very slow depending on the diffusion rate.\n\n\nBy measuring an adsorption isotherm, you just get the amount adsorbed **at equilibrium**, that is, after all the molecules entered inside the pores.\n\n\nIf your solid has very narrow pores, say, [micropores](https://en.wikipedia.org/wiki/Microporous_material), then the size of the pore is very near the size of your molecules, and adsorption will be slow. In your case, [the size of the $\\ce{CO2}$ and $\\ce{CH4}$ is very different](https://en.wikipedia.org/wiki/Kinetic_diameter), $\\pu{0.33 nm}$ for $\\ce{CO2}$ and $\\pu{0.38 nm}$ for $\\ce{CH4}$. So the diffusion inside the pores is much more restricted for $\\ce{CO2}$, then it is expected to be slower. Probably your sample have many narrow micropores that result in this behavior.\n\n\n", "0" ] ]

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[ [ "\nThis is a case of salt hydrolysis of a salt of weak acid and strong base\n$$\\ce{HIO + NaOH -> NaOI + H2O}$$\n$$\\ce{NaOI -> IO^- + Na^+}$$\n$$\\ce{IO^- + H2O <=> HIO + OH^-}$$\n\n\nYou can calculate the final concentration of $NaOI$ to be $0.06M$ (denote this by $c$). \n\nLet $h$ be the hydrolysis constant. \n\nThus in the equilibrium state, $[IO^-] = (c-ch)M$, $[HIO] = (ch)M$ and $[OH^-] = (ch)M$. \n\nNow $$ k\\_h = \\frac{[OH^-][HIO]}{[IO^-]} = \\frac{(ch)(ch)}{(c-ch)} = \\frac{ch^2}{(1-h)}$$\nHere $k\\_h$ is the hydrolysis constant. As the hydrolysis constant is small hence we can approximate $(1-h)\\approx 1$. \n\nTherefore $$h = \\sqrt{\\frac{k\\_h}{c}}$$\nSo $$[OH^-] = ch = \\sqrt{ck\\_h}$$\nAs $k\\_h = \\frac{k\\_w}{k\\_a}$\nThus $$[OH^-] = \\sqrt{c\\frac{k\\_w}{k\\_a}}$$\nAs $[H^+] = \\frac{k\\_w}{[OH^-]}$\nSo finally we can write $$[H^+] = \\sqrt{\\frac{k\\_{w}k\\_{a}}{c}}$$\nThus $$pH = -\\frac{1}{2}log(\\frac{k\\_{w}k\\_a}{c}) = \\frac{1}{2}(-log(k\\_w)-log(k\\_a)+log(c)) = \\frac{1}{2}(pk\\_w + pk\\_a + log(c))$$\nPutting the values of $k\\_a$, $k\\_w$ and $c$, we have\n$$ pH = \\frac{1}{2}(14 + 10.64 - 1.22) = 11.71$$\n\n\n", "-1" ] ]

[ [ "\nAs Zhe explains in the comments, the answer is **D**. Even though small changes in pressure are not expected to significantly affect the chemical potential of a solid, it is a function of pressure:\n\n\n$$\\mu(\\mathrm{s}) = \\mu^\\circ(\\mathrm{s}) + \\int\\_{p^\\circ}^{p}V\\_\\mathrm{m}\\,\\mathrm dp \\qquad T = \\mathrm{const}$$\nThis means that a small change in the chemical potential of the solid also contributes to the total free energy change of the system when the pressure is altered. Usually this response is ignored as $V\\_\\mathrm{m}$ is orders of magnitude greater for gases than for condensed phases.\n\n\n", "3" ], [ "\n**Answer:**\n\n\nWell it's about the amount of moles of gases on left and the right sides of the equal/equilibrium sign\n\n\nfor A, it is 3 moles vs 2, so then if we apply pressure it shifts to the right, to make this more visual, imagine a tower with 3 blocks stacked and another tower stacked 2 blocks high, if we were place them side by side and push them down with a plate, obviously the tower with the 3 blocks is going to be compressed ***First***\n\n\nfor B, it is 4 vs 2 moles \n\n\nfor C, it is 2 vs 1 moles\n\n\nfor D, it is 2 vs 2 moles\n\n\nfor E, it is 3 vs 2=3 moles (with solid)\n\n\nFrom the list, we can see that in choice D, we have the same amount of moles on both sides. Adding pressure or decreasing them wouldn't affect in which ***direction*** it would go to. \n\n\nYou said that you wanted help between choice D and E, but again if we look at the problem, for choice E, we have a tower stacked 3 blocks high vs 3+solid blocks high, so that means that the one with the 3+solid blocks is higher and thus means unbalance\n\n\nto convert that into chemistry terms, because you have an imbalance of moles of gases, that means that if you apply pressure, you are going to be disrupting the side with 3+solid moles as compared to 3 moles. \n\n\nSo\n\n\nyou want to have choice D because there is no ***imbalance*** of moles of gases\n\n\nAlso I understand that you are wondering about the solid being there, I believe it is not choice D because we have the same amount of moles but the definition of an equilibrium is that it goes ***both*** forward and back so then that implicates that the Z turns into products as well. So by adding pressure you actually favor one side more because either adding or decreasing pressure means that it becomes harder for Z to become a gas or to turn into a solid depending on the imbalance created. Z will eventually become a gas at one point and the added pressure or decreased pressure might make that harder. Z turns into gas molecules at one point in time, so your going to have more than 3 moles gas on left side, it does that naturally because definition of equibrium, but if you apply pressure it becomes harder to do that because RXN shifts right\n\n\n", "0" ] ]

[ [ "\nIn general, the solubility of a sparingly soluble salt having a basic anion will increase with the addition of a strong acid. Given: \n\n\n$$\\ce{X+A- <--> X+ + A-}$$ \n\n\nwhere $K\\_{sp}$ is small and A- is a weak base, the addition of a strong acid will form HA, driving the reaction to the right and increasing $K\\_{sp}$. \n\n\nIn the specific case of \n\n\n$$\\ce{CaF2 <--> Ca^2+ + 2F−}$$ \n\n\nthe addition of $\\ce{HCl}$ gives \n\n\n$$\\ce{CaF2 + 2HCl <--> Ca^2+ + 2Cl- +2HF}$$ \n\n\ndriving the reaction to the right, thus increasing the solubility of $\\ce{CaF2}$. Note that $\\ce{HF}$ itself is a weak acid, having a pKa of $3.17$, and will be in equilibrium with $\\ce{H+}$ and $\\ce{F-}$.\n\n\n", "4" ] ]

[ [ "\nYou are right about the equilibrium existing because both freezing and melting occur at the same rate. \n\nImagine ice and water kept in a thermos flask under normal atmospheric pressure. You will observe that there is no change in the mass of both the phases. \n\nBut there is actually a lot of activity going on. Some molecules of ice enter the liquid phase and some molecules of water enter the ice phase. As both these processes occur at the same rate you don't see any change in the mass of both the phases. \n\nIn fact such equilibriums exist for all pairs of solids in their own liquids and gases above their own liquids at particular temperatures. \n\nIn your case I think there is a mechanism to extract heat from the system which shifts the equilibrium towards liquid changing to solid ie freezing.\n\n\n", "2" ], [ "\nThe flat regions are where the solid is changing into liquid and liquid into gas at the higher temperature. These regions persist whenever there are still two phases in equilibrium. The heat necessary to cause the melting used to be called the Latent Heat of Fusion but nowadays is called the enthalpy of fusion (or melting) and similarly for vaporisation.\n\n\nThe solid is held together by a cohesive or lattice energy, generally called intermolecular energy (potential energy with attractive and repulsive parts) which has to be overcome for the solid to melt, and the heat absorbed by the solid is used to do this which means that the temperature does not rise until all the solid has melted. A liquid also has intermolecular forces but weaker than in the solid and thus has increased entropy. This increase in entropy has the effect of lowering the Gibbs free energy.\n\n\n( The change of enthalpy on fusion can be calculated from the definition of enthalpy $H=U+PV$. Between the two phases, liquid to solid, the change is \n\n$\\Delta H = U\\_l-U\\_s +P(V\\_l-V\\_s)$\nbut as the change in volume between liquid and solid is small we can ignore the *PV* work term and so \n$\\Delta H = U\\_l-U\\_s $. \nUsing the heat capacity allow this to be evaluated since $U=U^{\\mathrm{o}} +\\int\\_0^T C\\_p(T)dT$ \nwith the result that\n$ \\Delta H = U\\_l^{\\mathrm{o}}-U\\_s^{\\mathrm{o}} +\\int\\_0^T C\\_p(T)\\_l-C\\_p(T)\\_s dT$.\n\n\nIn vaporisation the *PV* term ignored in the solid-liquid change has to be added and becomes $PV\\_g$ for the gas and which for an ideal gas is *RT* )\n\n\n", "2" ] ]

[ [ "\nThe amount of increase is dependent on the activation energy. For a very low activation energy, there will be almost no change (this is what you obtained by removing $E\\_a$ and $T$). For a high activation energy, the change will be quite large. For example, with an activation energy of $110\\mathrm{\\frac{kJ}{mol}}$, you the rate will increase by about $16$ times. $$\\frac{\\exp({\\frac{-E\\_a}{R\\*320}})}{\\exp({\\frac{-E\\_a}{R\\*300}})}=\\frac{\\exp({\\frac{-111000\\text{J}}{8.314\\*320}})}{\\exp({\\frac{-111000\\text{J}}{8.314\\*300}})}=\\frac{7.6\\times10^{-19}}{4.7\\times10^{-20}}=16.14$$\n\n\nSo there really isn't an \"in general\" answer. However, many text books give a crude estimate that the rate will double when the temperature is increased by $10$. This isn't in general the case, as it depends on the temperature range (a change from $300$ to $320$ is more significant that a change from $600$ to $620$) and the the value of the activation energy. \n\n\n", "5" ] ]

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[ [ "\nThe alkyl groups mentioned could be primary, secondary or tertiary\nThe last one's will not be going by Wurtz reaction (these will rather undergo elimination to give alkenes); only primary & secondary can.\nEven here, we can have a mixture of products:[](https://i.stack.imgur.com/LQQWH.jpg)\n\n\n", "1" ] ]

[ [ "\nI think choice D, *All of the above* may be the correct answer. All three pairs are structural isomers, with different functional groups, thus functional group isomers (FGIs). \n\n\n[This brief overview](http://www.chemguide.co.uk/basicorg/isomerism/structural.html) uses a carboxylic acid and ester pair as an example of FGI, which validates choice B. It also cites an aldehyde $\\ce{CH3CH2(C=O)H}$ and a ketone $\\ce{HCH2(C=O)CH3}$. If the methyl group is swapped out for an amino group (not called a nitrile), we have your compounds $\\ce{NH2CH2(C=O)H}$ and $\\ce{HCH2(C=O)NH2}$. Not only should this pair be FGIs by analogy with the aldehyde-ketone, but they satisfy the definition of functional group isomers: Structural isomers (same atoms, different connectivity) that have different functional groups. This validates choice A.\n\n\n[Wikipedia](https://en.m.wikipedia.org/wiki/Structural_isomer?wprov=sfla1) cites your exact ether-alcohol pair as example of FGIs, validating choice C.\n\n\nThus, *all of the above* must be the correct answer. If you are in a class, I would ask your instructor about this problem.\n\n\n", "3" ] ]

[ [ "\nBenzene diazonium chloride is stable upto 5 degree Celcius, while in Lassaigne's test we need to heat the organic compound with sodium till red hot, before breaking it by plunging into cold water. All the nitrogen will be liberated before it finds a chance to combine with Na and form any NaCN, which is responsible for the test ultimately.\n\n\n", "4" ] ]

[ [ "\nBoth your reactions are equivalent. Here’s why. We start from $(1)$.\n\n\n$$\\ce{4 OH- -> O2 + 2 H2O + 4 e-}\\tag{1}$$\n\n\n(I have omitted the phase descriptors for clarity.) In any chemical reaction, we can add or subtract spectators as we see fit; this is akin to mathematical equations where we can add or substract expressions as we see fit as long as we do the same to both sides. For example, I want to add four protons to $(1)$ to give $(2)$.\n\n\n$$\\ce{4 OH- + 4 H+ -> O2 + 2 H2O + 4 e- + 4 H+}\\tag{2}$$\n\n\nIf this were now a mathematical equation, I could combine (or divide) parts as I see fit. In a chemical reaction, we need to be a bit more careful. But as long as our focus is not on proton transfers, we can combine anything basic enough with acidic protons. Here, you can combine the four hydroxides and four protons on the reactant side of $(2)$ to give $(3)$.\n\n\n$$\\ce{4 H2O -> O2 + 2 H2O + 4 e- + 4 H+}\\tag{3}$$\n\n\nNow, we notice that we again have spectators. Two water molecules turn up both on the reactant and on the product side (an additional two are unique to the reactant side). We can remove them like we did the protons before, giving us $(4)$. I will also perform some rearranging.\n\n\n$$\\ce{2 H2O -> 4 H+ + O2 + 4 e-}\\tag{4}$$\n\n\nIt so turns out that $(4)$ is your second suggested reaction. It is the same equation as $(1)$ as I have just shown; the difference is merely in the type of water molecules present.\n\n\nNow which one should you use? Notice that one reaction consumes hydroxide while the other produces protons (formally! The consumption of hydroxide is equivalent to the production of protons). Thus, if your reaction medium is basic, you should use $(1)$ for your redox reaction. If, however, your reaction mixture is acidic, $(4)$ is more appropriate. In short, choose the one with the same charged particle as is used in the other half-reaction to make your life simpler.\n\n\n", "4" ], [ "\nYou are looking for a reaction where water is oxidized, i.e. loses electrons. In the chemical equation, the electrons must appear on the opposite side as $\\ce{H2O}$, since electrons are products of an oxidation half-reaction. \n\n\nOn the other hand, any time we write $\\ce{X + e-}$ we are depicting a reduction of $\\ce{X}$. Which of your proposed reactions shows water being oxidized (and not reduced)?\n\n\n", "0" ] ]

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[ [ "\nOne of the reasons to take Carbon-12 could be that it's hard to get your hands on one mole of pure hydrogen (and to weigh it). Carbon-12 is a (room temperature) solid which is relatively easy to produce, keep pure, store and weigh.\n\n\n", "2" ] ]

[ [ "\nSurely1. At the end of the day, it is just the matter of supplying enough energy: the homolysis requires approximately 4.5 eV per hydrogen molecule, while for heterolysis more than 17 eV per molecule must be provided.\n\n\n\n\n---\n\n\n1) G. Busca, A. Vaccari, Heterolytic dissociation of hydrogen on high-temperature methanol synthesis catalysts, Journal of Catalysis, 108(2), 1987, 491-494, <http://dx.doi.org/10.1016/0021-9517(87)90198-9>.\n\n\n", "11" ] ]

[ [ "\nBecause the client is ultimately interested in removing the barium by precipitation, gravimetric analysis is likely a suitable solution. This can be done in a few hours in a lab with basic equipment like a filter flask, a drying oven, an analytical balance and the appropriate reagents. \n\n\nYou can take advantage of the fact that barium sulfate is very insoluble in water by adding some saturated solution of sodium sulfate to a known volume of a sample of the water. Then just filter out the barium sulfate onto some pre-weighed filter paper, dry, weigh, and calculate the mass of barium (from the measured mass and the molecular weight of barium sulfate) per volume of water. You should repeat the analysis to be sure you added enough sodium sulfate solution to precipitate all of the barium. \n\n\nIf there is little or no precipitate found, you have a couple options. Firstly, you can optimize the gravimetric analysis by using large volumes of sample with a small filter and a good analytical balance. Also be sure to pre-filter the sample prior to adding the sodium sulfate in case there are suspended particulates, which would get measured just as if they were barium. \n\n\nIf a gravimetric analysis does not provide the detection limit required by the client, the next best options are atomic absorption spectroscopy (AAS) and inductively coupled plasma - atomic emission spectroscopy (ICP-AES). As these analyses can be done very cheaply (~\\$10-$30 per sample) you may actually want to do this to begin with. The down side of outsourcing is that the turnaround time could be weeks, or the price could quadruple if you need results fast.\n\n\n", "1" ] ]

[ [ "\nI think a good place to start is with the idea that that $K\\_{eq}$ will not change in value unless the temperature changes (see the second answer [here](https://chemistry.stackexchange.com/questions/49225/why-is-kc-not-affected-by-change-in-pressure) for a good explanation). With this in mind we can say that statement **III** is false because the temperature is not being changed. \n\n\nStatement **II** is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.\n\n\nStatement **I** can be shown to be true via Le Chatelier's principle by the same logic as Statement **II**: for the reaction to shift towards the product, the rate of formation of product must increase. \n\n\nTherefore, only Statement **I** is true. \n\n\nYou might wonder how $K\\_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K\\_{eq}$ is the ratio of the forward and reverse rate constants, *not* the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change. \n\n\n", "1" ], [ "\nFew links that are relevant : \n\n[chemguide](http://www.chemguide.co.uk/physical/equilibria/change.html) \n\n[chemistry.purdue.edu](https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Reaction_Quotient.htm) \n\n[wikipedia](https://en.m.wikipedia.org/wiki/Le_Chatelier%27s_principle) \n\n\nAs given in the first link **$K\\_{eq}$ will change only if you change temperature. What changes here is reaction quotient, $Q$.**\n\n\n$Q$ is calculated just like $K$ : $$\\frac{(concentration\\, of\\, products)^a}{(concentration\\, of\\, reactants)^b}$$ where a and b are the respective stoichiometric coefficients. \n\nIt is clear that when you will decrease the volume of the container you will increase the concentration of both the reactants and products. But $Q$ will decrease as $\\ce{O\\_2}$ in the reactants is cubed while both the products are squared. \n\n\nIf the system has to attain equilibrium $K$ has to be equal to $Q$ thus $Q$ has to increase which means that the concentration of the reactants has to decrease and that of the products has to increase. \n\nTherefore the reaction shifts in the forward direction and the rate of the reaction increases. \n\nThus only I is right.\n\n\n", "0" ], [ "\n**concept:**\n\n\nYou should base it on the amount of moles on both sides\n\n\n\n\n---\n\n\nso if you wanted to figure it out u know that the side with the more moles is going to be affected i.e. if you decrease the pressure the side with the more moles of *gas* is going to decrease and shift the other way\n\n\nto clarify:\n\n\n1) You said that I is true but it isn't because we have to understand that only temperature changes the rate of the reaction so for rate of reaction we don't care about the decrease in volume. \n\n\n2)For equilibrium concentration of reactants, we know that there is 5 moles of gases on the left side and there is four moles on the right side with heat, because you are not chaining the heat, when you decrease the volume that means that you are forcing the moles from the left to the moles on the right side. \n\n\n3)For $K\\_{eq}$ we will go with your explanation of why it is true. But then we can rule out number 1. But it seems to me that your textbook either has a typo that tried to say \"which of these are not true\" or that you have not read the problem correctly. \n\n\n", "-1" ], [ "\nMind well, equilibrium constant $K\\_c$ as well as $K\\_p$ depends only on the temperature, they remain unaffected even though pressure is changed, or concentration of reactant or product is changed. \n\n\n", "-1" ] ]

[ [ "\nOne possibility: Starch/iodine solution + sodium thiosulfate. The starch/iodine solution goes from dark blue/black to clear when sodium thiosulfate reduces the iodine.\n\n\n", "0" ] ]

[ [ "\nStarting with $$dH=TdS+VdP\\tag{1}$$H is a function of T and P, so$$dH=\\left(\\frac{\\partial H}{\\partial T}\\right)\\_PdT+\\left(\\frac{\\partial H}{\\partial P}\\right)\\_TdP\\tag{2}$$Substituting Eqn. 2 into Eqn. 1,\n$$\\left(\\frac{\\partial H}{\\partial T}\\right)\\_PdT+\\left(\\frac{\\partial H}{\\partial P}\\right)\\_TdP=TdS+VdP\\tag{3}$$But, from the **definition of $C\\_p$**, $$C\\_p\\equiv \\left(\\frac{\\partial H}{\\partial T}\\right)\\_P \\tag{4}$$Substituting Eqn. 4 into Eqn. 3 gives:$$C\\_pdT+\\left(\\frac{\\partial H}{\\partial P}\\right)\\_TdP=TdS+VdP\\tag{5}$$Solving for dS then gives the desired relationship.\n\n\n", "4" ], [ "\nFirstly note that the total differential of $S(p,T)$ is \n\n\n$$\\require{begingroup} \\begingroup \\newcommand{\\md}{\\mathrm{d}} \\newcommand{\\pdiff}[3]{\\left(\\frac{\\partial #1}{\\partial #2}\\right)\\_{\\!#3}} \\md S = \\pdiff{S}{p}{T}\\,\\md p + \\pdiff{S}{T}{p}\\,\\md T$$\n\n\nso your question amounts to evaluating the two partial derivatives.\n\n\nFor the first partial derivative, I think one simple possibility is to note that $G = H - TS$ and hence $S = (H - G)/T$. Therefore\n\n\n$$\\begin{align}\n\\pdiff{S}{p}{T} &= \\pdiff{}{p}{T}\\frac{H - G}{T} \\\\\n&= \\frac{1}{T}\\left[\\pdiff{H}{p}{T} - \\pdiff{G}{p}{T}\\right] \\\\\n&= \\frac{1}{T}\\left[\\pdiff{H}{p}{T} - V\\right]\n\\end{align}$$\n\n\nusing the equation $\\md G = V\\,\\md p - S\\,\\md T$ which implies $(\\partial G/\\partial p)\\_T = V$.\n\n\nIn case you're wondering where the First Law comes in: you need the First Law to derive that equation for $\\md G$, via $\\md U = \\md q + \\md w = T\\,\\md S - p\\,\\md V$.\n\n\nThe second partial derivative is straightforward: under constant pressure,\n\n\n$$\\md S = \\frac{\\md q\\_\\mathrm{rev}}{T} = \\frac{C\\_p\\,\\md T}{T} \\Longrightarrow \\pdiff{S}{T}{p} = \\frac{C\\_p}{T} \\endgroup$$\n\n\n", "2" ] ]

[ [ "\nThe stoichiometric coefficients $\\nu\\_i$ are arbitrary anyway; the only thing that matters is the ratio of them. $\\ce{3A -> 2B}$ and $\\ce{6A -> 4B}$ both describe the same physical process occurring (unless it is implying something about mechanisms, but in this context of thermodynamics, that's not the case). Clearly, even for a single unchanging system, its value of $\\xi$ will vary depending on how you define the stoichiometric coefficients. So, I wouldn't expect $\\xi$ to have any physical meaning. Personally (in my limited knowledge of thermodynamics) I just view it as a useful, but ultimately arbitrary, definition.\n\n\n[In the same vein, I would say that the magnitude of the equilibrium constant $K$ doesn't have any physical meaning either, since it also depends on how you define the stoichiometric coefficients. But I digress.]\n\n\n", "6" ], [ "\nSuppose that there is a reaction at constant temperature and pressure $\\ce{A <=> B }$ then the free energy change is given by considering the changes in the chemical potential $\\mu\\_A,~\\mu\\_B$ and number of moles $dn\\_A,~ dn\\_B$ as reaction proceeds, i.e. $dG=\\mu\\_Adn\\_A+\\mu\\_Bdn\\_B$. The quantity $dn\\_A$ is negative and $dn\\_B$ positive. \n\n\nThe extent of reaction $\\xi$ (unit mole) is defined so that it is zero at the start of a reaction (all reactants) and 1 when one mole of reactants has converted into products. When $\\ce{A <=> B }$ the change in extent of reaction is $d\\xi = dn\\_B=-dn\\_A$ and so $dG=(\\mu\\_B-\\mu\\_A)d\\xi$.\nThe reaction will proceed until the change in free energy is zero, $\\left (\\frac{\\partial G}{\\partial \\xi} \\right)\\_{T,p} = \\mu\\_A-\\mu\\_B=0$ or $\\mu\\_A=\\mu\\_B$.\n\n\nAs an example of using $\\xi$, the chemical potential for a perfect gas is, $\\mu =\\mu^{\\mathrm {o}}+RT\\ln(p)$ \nthen\n$$ \\left(\\frac{\\partial G}{\\partial \\xi} \\right)\\_{T,p} =\\mu\\_B^{\\mathrm {o}}-\\mu\\_A^{\\mathrm {o}}+RT\\ln \\left(\\frac{p\\_B}{p\\_A} \\right)$$\n\n\nOften the (confusing) notation $$\\Delta\\_rG =\\left(\\frac{\\partial G}{\\partial \\xi} \\right)\\_{T,p}$$ is used, and is confusing because in thermodynamics using $\\Delta$ usually means a simple difference not a derivative. At equilibrium \n$$\\Delta\\_rG = \\left(\\frac{\\partial G}{\\partial \\xi} \\right)\\_{T,p}=0 $$\nand so the familiar expression $\\Delta G= -RT\\ln(K\\_p) $ is obtained where the equilibrium constant is $K\\_p= (p\\_B/p\\_A)\\_{eq}$ \n\n\nNotes:\nThe extent of reaction does simplify some calculations but the same results are obtained if $dn\\_A, ~dn\\_B$ etc. are used.Then it is necessary to express all the changes in $dn\\_i$ terms of say $dn\\_A$ and then calculate $\\left(\\frac{\\partial G}{\\partial n\\_A} \\right)\\_{T,p}$ etc.\n\n\nIn the definition of chemical potential the pressure is understood to be divided by 1 bar so that $\\ln(p)$ is dimensionless.\n\n\nSome authors definitions of $\\xi$ refer to the change from the initial to equilibrium amount.\n\n\n", "4" ] ]

[ [ "\nMy own experience with Luche conditions is that they are fickle, and as likely to give you a mixture of 1,2 and 1,4 reduction products as not. For complete reduction of an enone I would go with Lithium triethyl borohydride to reduce the enone to ketone then sodium borohydride for ketone to alcohol. You can equally use catalytic hydrogenation for the enone reduction to ketone.\n\n\n", "1" ] ]

[ [ "\nMan! you must have got impure sample of Mg(OH)2\nMoreover, Mg(OH)2 being sparingly soluble may not be finding a way for direct titration, so the method suggested in the problem is a process we call as BACK-TITRATION. This you have to follow i all such situations, where direct titration might not, for any of the reasons, be possible.\n\n\nBecause all 1 g Mg(OH)2 is not pure, you can expect incorrect answer if you are directly comparing the moles of H+ & OH- ions.\n\n\nWould this be fine?\n\n\n", "4" ] ]

[ [ "\nAs pointed out in the comments, your question is logically problematic. The only way to practically demonstrate that something is impossible is to show that every conceivable attempt fails. This is one reason that theory is used, because otherwise you could spend an infinite amount of time trying different experiments and still fail to show it's impossible. To contradict a general statement, all that is required is one counter-example. However, there are plenty of real world examples where the efficiency (1 - work out/heat in) is demonstrated to be far less than 100%, and there are (of course) no known counter-examples. See <https://en.wikipedia.org/wiki/Heat_engine> for a lot of examples. Contrast the statement of the 2nd Law with the statement that \"Man cannot fly\". It took tens of thousands of years to build a machine which enabled man to fly (1961) see <https://en.wikipedia.org/wiki/Human-powered_aircraft#First_flights>, but once it had been documented the statement became false. \n\n\n", "0" ] ]

[ [ "\nFirstly I would like to expand a bit on what a colligative property is, in order to bring more understanding and also be able to apply this to a named physical property.\n\n\n**Colligative properties** depend mainly on the number of particles in a solution.\n\n\nThe values of the colligative properties are approximately the same for equal\nconcentrations of different constituents in solution regardless of the species or chemical nature of the constituents. Examples are osmotic pressure, vapor pressure lowering, freezing point depression, and boiling point elevation (as you have noted).\n\n\nFurthermore, physical properties of substances are not just limited to the colligative nature. In fact a number of properties have been defined:\n\n\n* **additive** (depend on the total contribution of the atoms in the molecule or on the sum of the properties of the constituents in a solution e.g molecular weight).\n* **constitutive** (depend on the arrangement and partly on the number and kind of atoms within a molecule e.g interfacial characteristics).\n* other thermodynamic physical properties are known (**extensive, intensive**)\n\n\n\n\n---\n\n\nHaving gained this background, it is now evident that **surface tension** (a force that pulls the molecules of the interface together) is dependant on the arrangement and also on the nature of atoms or molecules , therefore is more of a **constitutive** property rather than a **colligative** property.\n\n\nI have included a screenshot to better visualise the phenomenon:\n\n\n[](https://i.stack.imgur.com/0lI01m.jpg)\n\n\nHope this helps.\n\n\n", "7" ] ]

[ [ "\nDo not forget solubility effects involving the hydroxide product. Magnesium is plenty electropositive enough to decompose water, only to passivate with a hydroxide that has limited solubility (you can detect the presence of some reaction by decanting the liquid and then adding phenolphthalein, or by carefully looking for a few hydrogen bubbles during the reaction). Put magnesium in methanol, which is less protic than water; the methoxide is less passivating and you can actually bring the alcohol to a boil! See <https://www.google.com/url?sa=t&source=web&rct=j&url=https://m.youtube.com/watch%3Fv%3DNMfs3e9OdZQ&ved=0ahUKEwjlnLOxpPDWAhVDKiYKHVFwB5wQwqsBCCgwAA&usg=AOvVaw2TaAZjVq0m8EHL-JluhxGn>.\n\n\n", "1" ], [ "\nLikely, it's caused by ionization that act on water molecules through coulombic reaction. A sodium/potassium ,alloy which is liquid at room temperature, is good example of a alika reaction and shows the effects in question clearly and in useful for testing this idea. \nExperiments have been done to confirm this under argon atmosphere and when delivered by a syringe with a well-defined amount of the metal alloy(Na/K) it gives an consistent unoxidized surface,leading to controlled explosions and taking out possibility of side reactions being the initiator of such high rate of reaction. This was a observed with use of high speed cameras to follow the process with a 100 μs time resolution. High ionization then exceeds the stability of the molecules themselves which is the Rayleigh instability limit. So yes it seems ionization plays a part.\n\n\nThe atomic radius(alkaline ions') may take part through the ratio to waters molecules radius. The breakdown is know as Coulomb fission and is defined as the ratio between the droplet Coulomb self-energy and twice its surface energy. The surface tension of the Na/K drop can be estimated to be coulombically unstable for radius greater than 5 angstroms.\n Hence the largest molecules can not exceed that(slightly larger than the reacting water molecules one layer deep).The reaction been equated to a capacitor going off with electron coming from the surface alkaline ions to water.This can be seen here <http://marge.uochb.cas.cz/~jungwirt/paper263.pdf>\n\n\n", "0" ] ]

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[ [ "\nOperational electrochemical cells have $\\Delta G\\lt0$\n\n\nAnd,\n$$\\Delta G=-nFE$$\n$$\\Delta G=\\Delta H-T\\,\\Delta S$$\n\n\nKnowing the entropy change, enthalpy change could be calculated\n\n\nor\n\n\n$$\\Delta H=-nFE+nFT\\frac{\\mathrm d\\Delta G}{\\mathrm dT}\\quad\\text{at contant }p$$\n\n\n", "1" ] ]

[ [ "\nIn general it is very difficult to estimate the stability of a nucleus based on the number of protons and neutrons it contains, as it is an interplay of different forces (repulsion between protons and the Strong force). However, a crude estimate can be made based on these empirical rules:\n\n\n1. There are no stable nuclei with $Z>83$ (Also no stable nuclei for Technetium ($Z=43$) and Promethium ($Z=61$)).\n2. The stability of a nucleus depends on the ratio $$\\eta=\\frac{A-Z}{Z}$$\nthere are no stable isotopes with $\\eta<1$, except for $^1$H and $^3$He. Stable nuclei with relatively small atomic number have $\\eta \\gtrsim 1$, while nuclei with larger $Z$ need more neutrons to stabilize the nuclear charge, for instance $^{206}\\_{\\,\\,82}$Pb has $\\eta=\\tfrac{124}{82}\\approx1.51$.\n3. Most stable nuclei have an even number of protons and neutrons, there are only a few stable nuclei that have both an odd number of protons and neutrons.\n4. Nuclei that have a number of protons or neutrons that equals the so-called *magic numbers* **2, 8, 20, 58, 50, 82, 126** are especially stable. Nuclei that have both a magic number of neutrons and protons (such as $^4\\_2$He, $^{16}\\_{\\,\\,8}$O, $^{40}\\_{20}$Ca, $^{48}\\_{20}$Ca, and $^{208}\\_{\\,\\,82}$Pb) are exceptionally stable.\n\n\nIn your example, $^{40}\\_{20}$Ca belongs to this special category.\n\n\n", "1" ] ]

[ [ "\nFor most physical properties of substances, it's useful to know how those properties depend on other parameters. For example, the various [coefficients of thermal expansion](https://en.wikipedia.org/wiki/Thermal_expansion) (CTEs) indicate how rapidly the dimension(s) of a substance change when the temperature is changed by a given amount. A common CTE used when working with materials is the [*linear CTE*](https://en.wikipedia.org/wiki/Thermal_expansion#Linear_expansion):\n\n\n$$\n\\alpha\\_\\mathrm L = {1\\over L} {dL \\over dT}\n$$\n\n\nWhen thermal expansion occurs, if the temperature change is relatively small the amount of thermal expansion is roughly proportional to the original length $L$. So, the definition of $\\alpha\\_\\mathrm L$ divides the length dependence $dL/dT$ by $L$ to yield a parameter that is approximately constant within a particular temperature range.\n\n\nA similar rationale underlies the definition of the [molar conductance](https://en.wikipedia.org/wiki/Molar_conductivity) (also called the 'molar conductivity') as:\n\n\n$$\n\\Lambda\\_\\mathrm m = {\\kappa \\over c}\n$$\n\n\nThe conductivity $\\kappa$ of an electrolyte is a strong function of the concentration $c$ of the electrolyte of interest. However, for electrolytes containing only strong acids and/or bases (or salts of these), $\\kappa$ is nearly a linear function of $c$ over a *very* wide concentration range, and thus $\\Lambda\\_\\mathrm m$ for these electrolytes is nearly constant.\n\n\n**Thus, for species that show this strong linearity, defining $\\mathbf{\\Lambda\\_m}\\equiv\\kappa / c$ is useful because a single $\\mathbf{\\Lambda\\_m}$ value can be employed to calculate the absolute conductivity $\\mathbf\\kappa$ across a wide range of electrolyte concentrations $c$**, as:\n\n\n$$\\kappa = \\Lambda\\_\\mathrm m c$$\n\n\n", "2" ] ]

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[ [ "\nYou have certainly understood the main idea.\n\n\nIn general the boiling point of a substance is determined by the way the molecules come together. The closer they are, the bigger are the intermolecular forces making it harder for one molecule to leave the entire group.\n\n\nIn this particular case the molecules are similar, since both substances have a hydroxyl group, thus being to able to create H-bonds. But in the case of 1-butanol, the molecule is linear while 2-butanol is not. This stereochemical difference affects the way the molecules attract each other. The linear molecules are able to come closer, thus creating stronger bonds or in other words raising the boiling point, because a higher amount of energy has to be applied in order for one molecule to obtain enough, to leave the liquid phase.\n\n\n", "9" ], [ "\nAs it is known that more the packing efficiency more will the molecules interact with one another and hence we require a large amount of energy to overcome the intermolecular force and as a result the boiling point increases.\nIf we see the two molecules i.e. butan-2-ol and butan-1-ol, we see that butan-1-ol is linear and butan-2-ol is branched. As the branching increases there is limited interaction between the molecules, and thus the hydrogen bonding in butan-2-ol is not effective due to the branching thus its boiling point is lowe than butan-1-ol.\n\n\n", "0" ] ]

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[ [ "\n1. Standard electrode potential can only be measured experimentally by comparing the half cells, in your case $\\ce{Cr^{VI}/Cr^{III}}$ and $\\ce{I2/I-}$ with a standard hydrogen electrode or a calomel electrode. The net standard electrode potential will be obtained by subtracting those of cathode-anode. You can find the standard electrode potential of most half cells in your textbook or online.\n2. See in your reaction, two moles of $\\ce{Cr^{VI}}$ turns to $\\ce{Cr^{III}}$. So $n = 2 \\times (6 - 3) = 6$. You can try out the same thing with the other half cell. The answer will be the same.\n\n\n\n\n---\n\n\n", "1" ] ]

[ [ "\nFirstly carrots are sources of carotenoids (precursors to vitamin A or provitamins)\n\n\nIts not really a chemical reaction per se, Vitamin A (Vitamin A is a nutritional term for retinol and related compounds with its biologic activities) is a fat soluble vitamin that is mainly stored in adipose tissue when in excess due to the hydrophobic nature hydrocarbon carotenoids.\n\n\nHowever some is metabolised through various mechanisms e.g through cellular retinoid binding proteins and others.\n\n\na typical carotenoid:\n\n\n[](https://i.stack.imgur.com/r3ndd.png)\n\n\nThese compounds contain carotenoid pigments.\n\n\nMost carotenoids are stored in adipose tissue. Lutein and zeaxanthin are specifically accumulated in the pigment layer of the retina, and cause a yellowish appearance of eyes when over-consumed.\n\n\nSimilarly over acculumulation of pigments in layer of skin can produce a yellowish hue which may signify vitamin A toxicity (or hypervitaminosis A).\n\n\nHowever in my practice I have seen very few cases of such toxicity.\n\n\n", "1" ] ]

[ [ "\nThe energy required to break the first $\\ce{H-O}$ bond in water is $\\pu{493 kJ/mol}$. For comparison, the energy required to break the first $\\ce{H-C}$ bond in methane is only $\\pu{435 kJ/mol}$. As a point of reference for just what these numbers mean in terms of bond strength, we can think about how long it takes for methane to be destroyed in the atmosphere. The first step of the destruction of atmospheric methane is to break that first $\\ce{H-C}$ bond; because this is so difficult to do (again, requiring $\\pu{435 kJ/mol}$) it takes an average of about 9 years for a molecule of methane to be destroyed! And the energy required to break the $\\ce{H-O}$ bond in water to form hydroxide and hydronium ions is even greater. \n\n\nI realize that my analogies involve very different processes, but the underlying issue in both cases (atmospheric destruction of methane and self ionization of water) is bond strength. I hope that this gives some insight as to why the self-ionization of water is as low as it is. Don't hesitate to ask for any clarifications in the comments.\n\n\n", "1" ] ]

[ [ "\nNucleophilicity of an ion does not solely depend on the number of lone pairs it has. The solvent in which the ions are dissolved matters, too.\n\n\nIf you are using a polar protic solvent, then flouride ion would behave as a very bad nucleophile (it would be solvated in the case of water; water will form hydrogen bonds with it). \n\n\nAmong the nucleophiles you have stated, $\\ce{R3C}$ anion would be the best nucleophile in a polar protic solvent. However, if you use a polar aprotic solvent, flouride ion would behave as a very good nucleophile (because aprotic solvents do not solvate anions). \n\n\nRemember, nucleophilicity is a *kinetic concept* and basicity is a *thermodynamic* one. Your reasoning for comparing the nucleophilicities was based somewhat on basicity of the ions. (This would work well in an aprotic solvent.)\n\n\nIn very layman terms:\n\n\n* **Nucleophilicity**: How good is something at reacting (doesn't concern stability of the ion)\n* **Basicity**: How badly something would want to react (concerns stabilities)\n\n\n", "1" ] ]

[ [ "\nDefine basicity. For Brønsted-Lowry, it's all about $\\ce{H+}$ and the conjugate bases. In Brønsted theory, acceptance of a hydrogen ion is what a base does. If you didn't know before, you should know that the hydrogen halides acid strength is $\\ce{HF < HCl < HBr < HI}$. So for the conjugate bases, you just flip the sign $\\ce{F- > Cl- > Br- > I-}$. \n\n\nYou way overthought the question. You should also think about where you got this idea about \"electron donation\". (Lewis acid-base theory deals with electron *pair* donation/acceptance but there's a huge difference between electron donation and electron pair donation.) You claim for some species $\\ce{X-}$, that if $\\ce{X- -> X^0 + e-}$ easily, then $\\ce{X-}$ is a stronger base (than some $\\ce{Y- -> Y^0 + e-}$ which is less easy). (By 'easily', assume I mean in the thermodynamic (energetic) sense)\n\n\nWell, take a look at that and ask yourself whether it is relevant to Brønsted acid-base theory. I say no, it's not. Whether it is true or not, it is not relevant to the base strength as defined by Brønsted-Lowry (at least it has no clear, direct relevance).\n\n\n", "4" ] ]

[ [ "\n1. A methyl-group is not t-butyl, so it isn't all that crowded.\n2. There are two ortho-positions.\n3. There is also the possibility of an ipso-attack with the nitronium donor attacking the position of the methyl group, then shifting to the ortho-position.\n\n\nGiven all that without any orientation effect, the ratio between ortho and para should be around 3/1 - 2/1. It is commonly observed around 3/2, so the crowding does have an effect.\n\n\n", "15" ], [ "\nThere are three effects: sterics, electronics and statistics. \n\n\nYes, there is more steric crowding in the ortho position, but a methyl group isn't that big. In addition, electronic effects are stronger in ortho position than in para, since it is much closer to the +I substituent. And last but not least: there are two ortho positions which can react, but only one para-position, which can have a rather big influence if reactivity towards ortho and para position is similar.\n\n\n", "8" ], [ "\nOrtho nitrotoluene is a major product while para nitrotoluene is a minor product .\nReasons:\nIn case of reactivity:\nOrtho product is more reactive comparing with para product because electron density is more enriched at ortho position compared to the para position.\nIn case of stability:\nPara product is more stable because is away from substituent group so it does not affected by steric hinderence as the ortho position I'd more steric hindered.\n\n\n", "0" ] ]

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[ [ "\nTap water contains calcium and magnesium ions; more so if the water is hard. These ions can bind to soap over time and make an insoluble soap curd.\n\n\nIs the sponge clean? If not, another reason is that the surfactants in the soap that make the foam might be forming micelles around whatever grime is in your sponge, so that there aren't many left to make the foam.\n\n\n", "4" ] ]

[ [ "\nProtons and neutrons are both [baryons](http://en.wikipedia.org/wiki/Baryon), subatomic particles composed of three quarks. [Quarks](http://en.wikipedia.org/wiki/Quark) are smaller fermionic particles that make up protons, neutrons, and other [hadrons](http://en.wikipedia.org/wiki/Hadron).\n\n\nElectrons are [leptons](http://en.wikipedia.org/wiki/Lepton), fermions that are not composed of quarks.\n\n\nQuarks have mass, spin, charge, and (confusingly) color charge. Color charge is a quantum property that quarks have, and all hadrons need to have a net zero color charge. \n\n\nThe proton is composed of three quarks: two up quarks and one down quark. Each up quark has a spin of $+\\frac{1}{2}$, a mass of $2.3~\\text{MeV/c}^2$, and a charge of $+\\frac{2}{3}$. Each down quark has a spin of $-\\frac{1}{2}$, a mass of $4.8~\\text{MeV/c}^2$, and a charge of $-\\frac{1}{3}$. The total charge of the proton is $2(+\\frac{2}{3})+(-\\frac{1}{3})=+1$. The total spin of the proton is $2(+\\frac{1}{2})+(-\\frac{1}{2})=+\\frac{1}{2}$. The masses of the quarks do not add up because most of the proton's mass is in its confinement energy.\n\n\nWe can do the same analysis for a neutron, which is composed of one up quark and two down quarks. The charge of the neutron is $(+\\frac{2}{3})+2(-\\frac{1}{3})=0$. The total spin of a neutron is $(\\frac{1}{2})+2(-\\frac{1}{2})=-\\frac{1}{2}$. Again the masses do not add up. \n\n\nProtons and neutrons are different because the are composed of different quarks. However... enough physics. Now for the nuclear chemistry. \n\n\n\n> \n> Interconversion of protons, neutrons, and electrons\n> \n> \n> \n\n\nThe proton is a [stable particle](http://en.wikipedia.org/wiki/Proton#Stability) with a lifetime at least $2.1\\times 10^{29}$ years.\n\n\nHowever, protons can be converted into neutrons (and electron neutrinos) by electron capture:\n\n\n$$\\ce{p+ + e- -> n +\\nu}\\_e$$\n\n\nThis requires an investment of energy because the rest mass of the neutron is higher than the rest mass of a proton by $1.29~\\text{MeV/c}^2$.\n\n\nFree neutrons are [not stable](http://en.wikipedia.org/wiki/Neutron#Beta_decay_and_the_stability_of_the_nucleus), undergoing beta decay to a proton, electron, and electron antineutrino with a halflife of about 10 minutes.\n\n\n$$\\ce{n -> p+ + e- +\\bar{\\nu}}\\_e$$\n\n\nMost importantly, and fortunately, most neutrons bound in nuclei are stable. The protons and neutrons in a nucleus can be treated quantum mechanically analogously to electrons from several atoms bound in a molecule. There are not a collection of independent particles that do not interact. The nucleons occupy quantum mechanical energy levels. Neutrons can only decay if the proton it would become can occupy a lower energy level, which is usually not possible except for those radioisotopes which tend to undergo beta decay, thanks to the Pauli Exclusion Principle! Thus, the bound neutron is stabilized.\n\n\n", "9" ] ]

[ [ "\nAccording to the [IUPAC gold book](http://goldbook.iupac.org/V06597.html) a van der Waals force is:\n\n\n\n> \n> The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces.\n> \n> \n> \n\n\nHydrogen bonding is a type of dipole-dipole interaction, so it would fit the definition of a van der Waals force.\n\n\nThe way I think of it is: van der Waals forces are anything that make a gas non-ideal, since that's how they were originally discovered and defined.\n\n\n", "14" ], [ "\nThere are two definitions of a hydrogen bond in the Gold Book: [hydrogen bond (1994)](https://goldbook.iupac.org/terms/view/H02899) and [hydrogen bond (1999)](https://goldbook.iupac.org/terms/view/HT07050). The later stresses both the ~~covalent~~ *bond* nature ~~and the electrostatic *interaction* component~~, which ~~both~~ *can* make it fall outside of the [van der Waals forces](https://goldbook.iupac.org/terms/view/V06597) as of the 1994 definition.\n\n\nEDIT: Answer edited following the remarks made by @tomij below. See also the interesting discussion below [his answer](https://chemistry.stackexchange.com/a/24159/5135) as to the context (historical and experimental) leading to the introduction of the notion of van der Waals forces/interactions.\n\n\n", "1" ] ]

[ [ "\nUranium is found in low-levels in **all** rocks and soil. Radon is a gaseous radioactive decay product of uranium. As the uranium undergoes radioactive decay, radioactive radon is generated and **trapped** in the rocks that contain the uranium. \n\n\nThe earthquake theory involving radon suggests that prior to the actual quake, there is some subterranean movement where rocks are crushed, soil is uncompacted and the trapped radon is released producing a pre-quake spike in radon concentration. \n\n\nSo radon presents the following attributes:\n\n\n* it is present in all rocks and soil\n* if a rock is broken or if soil is disturbed, radon will be released\n* it is gaseous and air currents / thermal gradients will carry it up to the earth's surface producing a detectable plume\n* it is radioactive, this makes detecting small amounts or small changes in radon concentration relatively routine due to well-developed and very sensitive methods for detecting and accurately measuring radioactivity.\n* radon has a very short radioactive half-life, a bit under 4 days. Being a gas and having a short half-life is very useful in terms of measuring radon emissions. If a radon emission spike occurs, the gas will dissipate quickly and after about 10 half-lives (40 days) normal background levels of radioactivity will return.\n\n\n", "10" ] ]

[ [ "\nIf the enzyme is staturated with substrate, the reaction rate will be $k\\_2$.\n\n\n1450 moles per second, per mole of enzyme. \n\n\n87,000 moles per minute, per mole of enzyme.\n\n\n87,000,000,000 micromoles per minute, per mole of enzyme.\n\n\nBased upon \"50,000 kDa\" (which looks like a typo, 50 kDa sounds more reasonable), one mole of enzyme is 50,000,000 grams or 50,000,000,000 milligrams.\n\n\nSo 87/50 = 1.74 units per milligram.\n\n\nThen for when product is staturated, use $k\\_{-1}$ instead of $k\\_2$.\n\n\n", "2" ] ]

[ [ "\nThere is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another. For example, in some elimination reactions the E1 and E2 pathways can operate in competition with one another. An activation energy is associated with each of these 3 reaction pathways. Whichever pathway has the lowest activation energy will be the major pathway followed. By changing solvent, reaction temperature, relative strength of the nucleophile, relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other. \n\n\n\n> \n> Typically the leaving group for E1cb is poor (like -OH or -OR) but why\n> must this be the case?\n> \n> \n> \n\n\nThe E1 mechanism involves ejecting a leaving group in its first step, while the E1cb mechanism involves removing a proton in its first step. Let's consider how changing the leaving group can shift an elimination reaction towards one pathway or the other. To a first approximation, changing a leaving group will not affect how hard or easy it is to remove the proton. So it is reasonable to assume that the activation energy for the E1cb process doesn't change as we vary our leaving group. If we use a better leaving group (make our leaving group more stable) that means that we have made ejecting the leaving group a lower energy pathway and the E1 process will become more favorable relative to the other elimination mechanisms. If we change our leaving group to one that is an extremely **poor leaving group** (make our leaving group less stable), then ejecting it becomes a higher energy process and the E1 reaction becomes **less competitive** with the other reaction pathways. Saying this last sentence differently, when we use a poor leaving group we raise the activation energy for the E1 mechanism. Since the rate of the E1cb process is not affected by the leaving group, its rate remains unchanged. Consequently, a poor leaving group will disfavor the E1 process making the E1cb process more competitive. If the leaving group is bad enough, we can disfavor the E1 process so much that we wind up pushing our reaction all the way over to the E1cb side.\n\n\n", "12" ] ]

[ [ "\nPrecipitation is usually a single or double displacement reaction, and refers to the formation of an insoluble, solid salt from soluble ions. For instance, silver bromide, $\\ce{AgBr}$, is insoluble in water. However, it is made up of $\\ce{Ag+}$ and $\\ce{Br-}$ ions, which we can introduce in the form of soluble salts:\n\n\n$$\\ce{LiBr(aq) + AgNO3(aq) -> LiNO3(aq) + AgBr(s)}$$\n\n\nIn an aqueous solution, lithium bromide and silver nitrate underwent a reaction, forming lithium nitrate and silver bromide. In the products, silver bromide is denoted with the state symbol (s), which stands for \"solid\". This reflects its insolubility in water and hence its tendency to \"precipitate\".\n\n\nThe net ionic equation is when we only write the ions that participate in the reaction. The ions that do not participate are called **spectator ions**. For example, in the reaction above, only $\\ce{Ag+}$ and $\\ce{Br-}$ participated in precipitation. Thus, the net ionic equation for the reaction above is:\n\n\n$$\\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$\n\n\nOne way of deriving this net ionic equation is to realise that salts such as $\\ce{LiBr}$ and $\\ce{AgNO3}$ do not exist as molecules in water, but rather dissociate to give their constituent ions:\n\n\n$$\\begin{align}\n\\ce{LiBr(s) &->[dissolution] Li+(aq) + Br-(aq)} \\\\\n\\ce{AgNO3(s) &->[dissolution] Ag+(aq) + NO3-(aq)}\n\\end{align}$$\n\n\nHowever, $\\ce{AgBr}$ is insoluble and does not dissociate into ions. If we go back to the first equation and replace all the soluble salts with their constituent ions, we get\n\n\n$$\\ce{\\underbrace{Li+(aq) + Br-(aq)}\\_{from LiBr} + \\underbrace{Ag+(aq) + NO3-(aq)}\\_{from AgNO3} -> \\underbrace{Li+(aq) + NO3-(aq)}\\_{from LiNO3} + \\underbrace{AgBr(s)}\\_{does not dissociate}}$$\n\n\nThe spectator ions which do not take part in the reaction are $\\ce{Li+}$ and $\\ce{NO3-}$, as they are unchanged in this reaction. Subtracting them from both the left- and right-hand sides, we obtain the net ionic equation:\n\n\n$$\\ce{Ag+(aq) + Br-(aq) -> AgBr(s)}$$\n\n\nIf we take this idea and apply it back to $\\ce{CaCO3}$, we first need to come up with soluble sources of the ions $\\ce{Ca^2+}$ and $\\ce{CO3^2-}$. A possible choice is $\\ce{CaCl2}$ and $\\ce{Na2CO3}$, although that is hardly the only option.\n\n\nThe overall equation may be written as\n\n\n$$\\ce{CaCl2(aq) + Na2CO3(aq) -> 2NaCl(aq) + CaCO3(s)}$$\n\n\nand the net ionic equation as\n\n\n$$\\ce{Ca^2+(aq) + CO3^2-(aq) -> CaCO3 (s)}$$\n\n\nwhere $\\ce{Na+}$ and $\\ce{Cl-}$ are the spectator ions.\n\n\nFinally, here are some web pages you might find helpful:\n\n\n<http://www.occc.edu/KMBailey/Chem1115Tutorials/Net_Ionic_Eqns.htm>\n\n\n<http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Reactions_in_Aqueous_Solutions/Precipitation_Reactions>\n\n\n", "5" ] ]

[ [ "\nThe most important factor is the nucleophilicity versus basicity of the attacking species.\n\n\nTaking the trimethylcarbenium cation as a classic example:\n\n\n\nSterically hindered bases such as lithium diisopropyl amide (LDA) will favour elimination because it is physically much easier for them to remove a small proton from the outside of the cation than to reach the carbon atom at the centre surrounded by three methyl groups and form a new bond there. On the other hand, small species such as cyanide and water will favour substitution because because it is physically easier for them to reach the central carbon.\n\n\nAdditionally the strength of the base influences which reaction takes place. LDA is a strong base and the methyl hydrogens are quite acidic means that the elimination reaction is much more likely. However, weak bases such as iodide and water will favour substitution because they are less basic and so they are less likely to abstract an acidic proton from one of the methyl groups.\n\n\nIt should also be noted that the nature of the solvent has a strong effect both on the stability of the cation but also on the nucleophilicity of the attacking species. This page has some discussion on the effects of solvents on nucleophilicity and the website also has good discussions of elimination and substitution mechanisms.\n\n\n<http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx1.htm#hal1>\n\n\n", "3" ] ]

[ [ "\nEach hemoglobin molecule has 4 subunits (4 hemes each having one iron atom). Each time a molecule (such as oxygen or CO) binds to one subunit, it changes the binding properties of the other subunits. This is referred to as positive \"[cooperative binding](http://en.wikipedia.org/wiki/Cooperative_binding)\" in the case of hemoglobin, because binding one molecule makes the other sites bind more easily. \n\n\nInitially, considering hemoglobin with no CO or oxygen bound, CO competes with oxygen for the same binding sites (ligating one of the iron atoms). CO has a binding constant 300 times lower than oxygen, meaning a factor of 300 lower CO concentration will result in an equal percent of bound protein.\n\n\nConsidering CO has reached a level such that 2 out of the 4 sites can have CO bound, the remaining 2 sites would still provide enough oxygen carrying capacity for a person to live if the sites were function as normal; however, the 2 remaining sites then bind oxygen too tightly and don't release it.\n\n\nOther iron-containing proteins such as myoglobin can also bind CO and are also inhibited.\n\n\nSee [Pharmacokinetics and Mechanisms of Action of Carbon Monoxide](http://www.bvsde.paho.org/bvsacd/cd51/monoxide/cap05.pdf) for more infomation. \n\n\n", "12" ] ]

[ [ "\nYou are correct that the first step with both reagents involves the formation of the trichloromethyl anion ($\\ce{CCl3^{-}}$). This anion can eject $\\ce{Cl^{-}}$ and irreversibly form dichlorocarbene ($\\ce{:CCl2}$). [Carbenes](http://en.wikipedia.org/wiki/Carbene) are very reactive intermediates, reactive enough to add to a double bond and form a strained 3-membered ring (a cyclopropane).\n\n\nThe reaction you've shown is said to be stereospecific. That is, you started with cis-2-butene and wound up with a cis-dimethyl-dichlorocyclopropane; the cis olefin geometry is preserved in your cyclopropane product. This indicates that the reaction involved singlet dichlorocarbene. Carbenes can exist as singlets or triplets. As the following figure indicates, singlet carbenes add stereospecifically (your case), triplets do not. The \"slow spin flip\" step in the triplet carbene reaction is slow enough that rotation can occur about what was the olefinic bond and this rotation gives rise to cis and trans cyclopropane isomers.\n\n\n\n\n\n", "7" ], [ "\nI think that both reactions involve carbene as common intermediate which then is the species involved in the subsequent cyclopropanation reaction.\nThese are the two equations which lead to the formation of the carbene:\n$$\\ce{KOH + CHCl3 -> :CCl2 + KCl + H2O}$$\n$$\\ce{Cl3CCO2Na -> :CCl2 + NaCl + CO2}$$\nSubsequently, the alkene undergoes cyclopropanation by dichlorocarbene.\nIn the second case, heat is required in order to promote decarboxylation, which is entropically favored.\n\n\n", "7" ], [ "\n\n> \n> It will be the conjugate base of trichloromethane that is involved. \n> \n> \n> \n\n\nOnly as the initial intermediate! $¸\\ce{CHCl3 + OH- ->Cl3C^- + H2O}$\n\n\nThe carbanion will lose chloride and yield dichlorocarbene $\\ce{Cl3C- -> Cl- + Cl2C:}$\n\n\nThe latter then adds to the $\\ce{C=C}$ double bond.\n\n\n\n> \n> The same is true of the carboxylate.\n> \n> \n> \n\n\nNo! In this case, decarboxylation (loss of $\\ce{CO2}$) is the first step. The fate of the carbanion is however as described above.\n\n\n", "3" ] ]

[ [ "\nA redox reaction is any reaction in which oxidation and reduction are occurring. The word redox is a portmanteau of REDuction-OXidation. So, if we can correctly identify reduction and oxidation, we can identify redox reactants. In practice, this can be a little challenging, so I will focus first on the second part of your questions (definitions) and then go back to the reaction.\n\n\n\n> \n> Ionization is any process that produces ions. \n> \n> \n> \n\n\nThe conversion of a hydrogen atom to a hydrogen ion by loss of an electron is an ionization. \n$$\\ce{H -> H+ + e-}$$\nSo to is the reaction of hydrochloric acid and water, which produces hydronium ions and chloride ions.\n$$\\ce{HCl + H2O -> Cl- + H3O+}$$\n\n\n\n> \n> Oxidation is a loss of electrons.\n> \n> \n> \n\n\nThe ionization of hydrogen atoms to hydrogen ions is an oxidation.\n$$\\ce{H -> H+ + e-}$$\n\n\nSo to is the conversion of carbon monoxide to carbon dioxide:\n\n\n$$\\ce{2CO + O2 -> 2CO2}$$\n\n\nNote that this type of reaction is what gave us the word oxidation - compounds are reacting with oxygen and atoms end up with more oxygen atoms bound to them. \n\n\n\n> \n> Reduction is a gain of electrons.\n> \n> \n> \n\n\nThe ionization of chlorine atoms to chloride ions is a reduction.\n\n\n$$\\ce{Cl + e- -> Cl-}$$\n\n\nSo too is the conversions of acetylene to ethane:\n\n\n$$\\ce{C2H2 + 2H2 -> C2H6}$$\n\n\nNot all oxidations and reductions are ionizations. There is a simple mnemonic in English for remembering oxidation and reduction: OIL RIG (Oxidation Is Loss Reduction Is Gain).\n\n\n\n> \n> Oxidation and reduction happen together. \n> \n> \n> \n\n\nIn principle we can write an equation that shows only oxidation:\n\n\n$$\\ce{H -> H+ + e-}$$\n\n\nbut oxidation cannot happen on its own. That electron that was lost needs to be picked up by something else. This reaction that shows only oxidation is called a \"half-reaction\" and needs to be coupled with a reduction half-reaction to show a valid chemical reaction:\n\n\n$$\\ce{H -> H+ + e-}$$\n$$\\ce{Cl + e- -> Cl-}$$\n$$\\ce{H + Cl ->HCl}$$\n\n\nIn the other examples I gave, both processes are occurring. In the oxidation of carbon monoxide, the oxygen molecules are being reduced.\n\n\nThe surest way to identify oxidation and reduction (and thus redox) is to compare the oxidation states of the reactants and the products. Determining oxidation states/numbers is more complex that I want to put here, so start with the [Wikipedia article](http://en.wikipedia.org/wiki/Oxidation_state#Determining_the_oxidation_state_or_number).\n\n\nFor the carbon monoxide example:\n\n\n\n```\n reactants products\nC +2 +4\nO (bonded to C) -2 -2\nO2 0 -2\n\n```\n\nThe carbon atoms are being oxidized (oxidation number increased) and the oxygen atoms in $\\ce{O2}$ are being reduced (oxidation number decreases). The oxygen atom already attached to carbon did not change its oxidation number.\n\n\n", "7" ], [ "\nA redox reaction involve transfer of electrons between species and the term redox comes indeed from the combination of terms \"oxidation\" and \"reduction\". \n\n\nThe species which is oxidized loses electrons whereas the species which is reduced (with higher oxidation potential) gains electrons. An oxidant (oxidating agent) is a species that can oxidize other species, whereas reducing agents can be oxidized, thus transferring electrons to other species (subsequently reduced). While being reduced, the oxidation number of a species decreases and vice versa, oxidation implies an increase in the oxidation number. \n\n\nOne example of everyday life redox reaction is rusting, which consists in oxidation of iron by oxygen thus forming $\\ce{Fe(III)}$:\n\n\n$$\\ce{4Fe + 3O2 -> 2Fe2O3}$$\n\n\nAnother example is the cellular respiration, a complex pathway of oxidative degradation of sugars producing energy (stored as chemical energy in the acid anhydride bonds of ATP). Although the process is quite complex involving several enzymes and many steps, the overall reaction is quite simple: \n$$\\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}$$\n\n\n", "1" ] ]

[ [ "\nAs far as I know there isn't a specific term for that point. \n\n\nIf you have a one-step first order reaction\n\n\n$$\\rm A \\xrightarrow{k\\_1} B + C, \\quad \\text{rate} = k\\_1[A]$$\n\n\nthat can run backwards as a second order reaction\n\n\n$$\\rm B + C \\xrightarrow{k\\_{-1}} A, \\quad \\text{rate} = k\\_{-1}[B][C]$$\n\n\nand the rates of those reactions are equal, you can call that point \"equilibrium\". \n\n\nYou can't really plot the rates separately the way you have unless you have a common concentration, e. g. plot a change in concentration $x$ where $\\rm [A] = [A]\\_0 - x$, $\\rm [B] = [B]\\_0 + x$, and $\\rm [C] = [C]\\_0 + x$.\n\n\n", "2" ] ]

[ [ "\nIon selective electrodes are a group of electrodes that are very different from each other. \n\n\nThe common bits are, a selective membrane, a \"sensing\" electrode and a reference electrode to measure against. \n\n\nThe details of the membrane and sensing electrode depend highly on what selective electrode you are trying to build. \n\n\nHere is an old patent on a Cl selective electrode for reference:\n<http://www.google.com/patents/US3740326>\n\n\nRegarding getting accurate results with a DIY: this really depends on which electrode you are trying to build. In the Cl example, I don't see any reason why a DIY would perform worse.\n\n\n", "2" ] ]

[ [ "\n$\\ce{Co[(NH3)\\_4Cl2]}$ can exist as cis and trans isomers as shown in the following figure.\n\n\n\n\n\n[image source](http://chemistry.tutorvista.com/inorganic-chemistry/coordination-compound.html)\n\n\nIf a molecule contains a plane of symmetry it **cannot** be optically active or chiral.\n\n\nIn the cis isomer there is a plane of symmetry containing the $\\ce{Cl-Co-Cl}$ atoms, therefore this isomer cannot be optically active. With the trans isomer there are several planes of symmetry (one contains the $\\ce{Co}$ atom and the four $\\ce{NH3}$ groups, and there are four more planes of symmetry containing the $\\ce{Cl-Co-Cl}$ atoms and either containing or bisecting the pairs of amino groups), therefore the trans isomer is also optically inactive.\n\n\n", "8" ], [ "\nThe 6 ligands are at the 6 vertices of an octahedron.\n\n\nThere will be cis and trans isomers, but neither isomer is optically active.\n\n\nFor the trans isomer, there will be a central axis containing Cl-Co-Cl, and the 4 N atom lie in a plane of symmetry, bisecting the Cl-Co-Co line segment.\n\n\nFor the cis isomer, there is a 90 degree Cl-M-Cl angle, with the line segment Cl-Cl being an edge of the octahedron. The plane perpendicularly bisecting the Cl-Cl segment is a plane of symmetry. The plane passes through the Co and 2 of the N atoms as well.\n\n\n", "7" ] ]

[ [ "\nThe logic is wrong. Lower $\\mathrm{pH}$ (actually you mean $\\mathrm{p} K\\_\\mathrm{a}$ here) does not mean more polar.\n\n\nIt is more complicated to decide the polarity then you think. For reverse phase HPLC, the more polar and more soluble compound come out first. For reverse phase HPLC, it is common to have acid (TFA for example) in the eluent. It this case, caffeine can be protonated and become much more polar because it is ionic now. Solubility is another factor to consider. So if you want to predict the order, you have to provide more information.\n\n\n", "3" ] ]

[ [ "\nIodine is having 53 electrons and cesium is having 55 electrons in general. Iodine ion can be made by adding or by removing electrons from it. Here you added two electrons to iodine and made it an iodine ion in which number of electrons of iodine ion became equal to the number of electrons of cesium but the composition of nucleus(number of protons) is different in both iodine ion and cesium ion. Obviously cesium is having more number of protons than iodine, so cesium nucleus can hold it's electrons more closer in cesium than iodine nucleus in an iodine ion which results in larger radius of iodine ion than cesium.. \nSame number of electrons doesn't mean that their radii are same. Same logic is applicable for lithium ion and lithium.. Lithium ion is having less electrons and more protons compared to lithium.. So radius of lithium is larger than radius of lithium ion\n\n\n", "2" ] ]

[ [ "\nThe key issue here that we are comparing excited states. So let's begin by defining Singlet and Triplet Excited States.\nSinglet and Triplet Excited States: A singlet or a triplet can form when one electron is excited to a higher energy level. In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired). In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in the illustrative Figure below. Singlet, doublet and triplet is derived using the equation for multiplicity, $2S+1$, where $S$ is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up ($s = +1/2$) or spin down ($s = -1/2$). If we were to calculated the S for the excited singlet state, the equation would be $2[(+1/2) + (-1/2)]+1 = 2(0)+1 = 1$, therefore making the center orbital in the figure a singlet state. If the spin multiplicity for the excited triplet state was calculated, we obtain $2[(+1/2) + (+1/2)]+1 = 2(1)+1 =3$, which gives a triplet state as expected.\n\n\n\n\n\nIllustrative Figure: Spin in the ground and excited states The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. \\*\\*\\*For this reason, the lifetime of the triplet state is longer than the singlet state by approximately 104 seconds fold difference.\\*\\*\\*The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.\n\n\n", "10" ], [ "\nIf you consider stability simply in terms of energy and *not* lifetime, then a triplet state is definitely more stable than a singlet state, i.e., a triplet state is of lower energy than a singlet excited state. \n\n\nThis can be understood when we look at the energy contributions in a singlet and a triplet. Consider a simple case with two electrons labeled as $1$ and $2$ and two spin orbitals $a$ and $b$. In the singlet state electron $1$ occupies orbital $a$ with $1/2$ spin and electron $2$ occupies orbital $b$ with $-1/2$ spin. In the triplet state, $1$ occupies $a$ and $2$ occupies $b$, both with the same spin. The energy of the singlet state is given as:\n\n\n$E\\_S = h\\_{11}+h\\_{22}+J\\_{12}$\n\n\nAnd the energy of the triplet state is given as:\n\n\n$E\\_T = h\\_{11}+h\\_{22}+J\\_{12}-K\\_{12}$\n\n\nHere $h\\_{nn}$ is a contribution to energy irrespective of the spin of the electron. $J\\_{nm}$ is the coulomb term which exists between every unique pair of electrons. $K\\_{nm}$ is the exchange term, which only exists for an electron pair of the same spin. The exchange term has no easy classical interpretation. But importantly, it is a positive number, which implies that $-K\\_{nm}$ always decreases the energy.\n\n\nTherefore, $E\\_S$ is greater than $E\\_T$, meaning a triplet state is more stable compared to a singlet state.\n\n\n", "4" ], [ "\n\n> \n> Is triplet state stable or singlet state ??\n> \n> \n> \n\n\nIt depends upon the spin system (atom, molecule, etc.).\n\n\nDiamagnetic compounds have singlet ground states, so for many compounds the ground states are singlets.\n\n\nExamples of triplet ground states include dioxygen. [Singlet dioxygen](http://en.wikipedia.org/wiki/Singlet_oxygen) is a excited state.\n\n\nFor dihydrogen, the singlet state is the ground state, but at room temperature, there is an equilibrium concentration of 75% triplet state and 25% ground state. Near zero kelvins, the singlet state dominates, but the energy difference between the states is so small that, in accordance with Boltzmann statistics, at room temperature the states are populated in proportion to the number of microstates, the triplet having 3 and the singlet having 1. But in this spin system, the spins are proton spins not electron spins.\n\n\n[Methylene](http://en.wikipedia.org/wiki/Methylene_(compound)) also has a triplet ground electronic state and singlet excited state.\n\n\n\n> \n> How can an atom stay long time in a triplet state though the electrons have same spin?? \n> \n> \n> \n\n\nFirstly, electronic \"triplet state\" doesn't exactly mean the electrons have the same spin. The three possibilities referred to by the word \"triplet\" are 1. both spins are up 2. both spins are down and 3. $1/\\sqrt2$(up,down + down,up)\n\n\nIn any case, the lifetime of any excited state is not fixed even for a particular spin system. Mechanisms for the system to transition to the ground state need to be considered. In the gas phase the excited state may be long lived but it could be many orders of magnitude less in a liquid solvent. \n\n\nJust to give an example, for atomic helium in the triplet 1s2s state, the lifetime to decay by radiation to the 1s1s singlet state is 7859 seconds.\n\n\n", "1" ], [ "\nStability is about the energy level which is a thermodynamic factor while lifetime is a kinetic factor. That is why they are not necessarily agree with each other.\nTriplet live longer simply because it is kinetically harder for the excited electron to jump back to the original orbital and at the same time flip its spinning state. It is not really connected to the stability of it. Stability is more complicated depending many other aspects.\n\n\n", "0" ] ]

[ [ "\n\n> \n> However if the single or linear combinations of Slater determinants in\n> a same configuration are the zeroth order functions (eigenfunctions of\n> the unperturbed atomic Hamiltonian), how the fact that S2 and L2\n> commute with the atomic Hamiltonian makes the Slater determinants be\n> the eigenfunctions of the atomic Hamiltonian??\n> \n> \n> \n\n\nActually, you go it right, but let us spell this out step-by-step. First, forget about angular momentum operators, since you do not need them to understand that a single Slater determinant is indeed *not* an eigenfunction of the atomic Hamiltonian (as you called it). That is it, you are right. A Slater determinant is an eigenfunction of the unperturbed Hamiltonian which describes a system of independent electrons, but not of the exact one.\n\n\nNevertheless, a Slater determinant can be used as a trial wave function in a [variational procedure](http://en.wikipedia.org/wiki/Variational_method_(quantum_mechanics)). That is the whole point of the *[Hartree-Fock method](http://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method)*: a Slater determinant $\\Phi$ is not an eigenfunction of the atomic Hamiltonian, but we could evaluate it energy $\\left\\langle \\Phi \\mid H \\mid \\Phi \\right\\rangle$ using [Slater rules](http://en.wikipedia.org/wiki/Slater%E2%80%93Condon_rules), and consequnetly, we could minimize it to find an upper bound to the ground state energy. The resulting Slater determinant obtained by minimizing the energy is indeed only an *approximation* to the ground state wave function.\n\n\nNote also that the exact wave function (which is an eigenfunction of atomic Hamiltonian) can be expressed as a linear combination of Slater determinants for the various possible electronic configurations, not just a single Slater determinant. And this is the theoretical basis for [configuration interaction (CI)](http://en.wikipedia.org/wiki/Configuration_interaction) method.\n\n\n\n\n---\n\n\nNow back to all this business with angular momentum operators. A single Slater determinant is *not necessarily* an eigenfunction of neither $\\hat{L}^2$ nor $\\hat{S}^2$. However, as you mentioned, the Hamiltonian indeed commutes with all these operators (in non-relativistic approximation), and thus, the exact wave function is also an eigenfunction of $\\hat{L}^2$ and $\\hat{S}^2$ (as well as $\\hat{L}\\_z$ and $\\hat{S}\\_z$). To be more precise, the exact wave function can be chosen to be a simultaneous eigenfunction of all these commuting operators $\\hat{H}, \\hat{L}^2, \\hat{S}^2, \\hat{L}\\_z, \\hat{S}\\_z$. And since it is really desirable, we require the same from out trial wave function: we would like it to be an eigenfunction of these angular momentum operators.\n\n\nSo we want to construct our trial wave function being an eigenfunction of angular momentum operators $\\hat{L}^2, \\hat{S}^2, \\hat{L}\\_z, \\hat{S}\\_z$, but a single Slater determinant would not necessarily do it, so we form what is called a spin-adapted [configuration state function (CSF)](http://en.wikipedia.org/wiki/Configuration_state_function) - a linear combination of Slater determinants which is an eigenfunction of angular momentum operators. \n\n\nNote, however, that this is not always possible.\nFor instance, for the case of unrestricted determinants using a linear combination would not help in making the trial wave function an eigenfunction of $\\hat{S}^{2}$.\n\n\n", "6" ] ]

[ [ "\nOne way to understand this is if you think of air as bunch of atoms and molecules moving around in random fashion. The speed at which they move is measured by the temperature of air. (This is true even if we are not talking about air, but we don't need to bother about that here.) So when we say higher temperature we mean that the speed at which these particles move is higher. Furthermore the faster they move the less likely they are to stick together. \n\n\nNow water molecules like to stick to each other and when too many of them are together they \"condense\". But if you were to increase the temperature of the air (and hence the water molecules), they end up moving around faster and less likely to stick together. This way you can add more water molecules for the same amount of air.\n\n\nThe other thing to think about is pressure. In some sense it describes how crowded the place is. The higher the pressure, the more crowded it is. So when you increase the pressure, you increase the chances that the water molecules will stick together and hence \"condense\".\n\n\n\n\n---\n\n\nAnother way to think about this is through [Le Çhatelier's principle](http://en.wikipedia.org/wiki/Le_chatelier%27s_principle), which states that when a system in equilibrium (especially chemical equilibrium) is disturbed it tries to adjust in a way which minimizes the disturbance. \n\n\nSo now think of a box with some water and air. If you leave it long enough the system will reach an equilibrium, wherein no more water will evaporate or condense. But say now you increase the temperature of the air, the system will try to readjust so as to reduce the change in temperature. To do this it will cause the water to evaporate (as evaporation requires/consumes heat) and thereby reducing the change in temperature.\nSimilarly when you increase the pressure which is similar to decreasing the volume of the box, the air tries to account for this by condensing out some of the water (since liquid water occupies less volume than gaseous water), thereby reducing the effective decrease in volume seen by the air.\n\n\n\n\n---\n\n\nYou may wonder why other particles in air do not \"condense.\" They do, but only at much lower temperatures and/or higher pressures.\n\n\n", "6" ] ]

[ [ "\nIt \"transitions from one polymorph to another\"\n\n\nIt \"transitions to another polymorph\"\n\n\nIt \"transition to polymorph beta\"\n\n\nIt \"transforms to another polymorph\"\n\n\nIt \"transforms to polymorph beta\"\n\n\nI like \"transitions\" better than \"transforms\", but I've seen both\n\n\n", "4" ] ]

[ [ "\nIf you have a molecule or ion with multiple sites that can be protonated (as you do here), and the sites can interact with each other (as they can here), the affinity for protons drops rapidly as more sites are protonated. \n\n\nThe proton affinity of the ion in your second reaction is a couple of orders of magnitude less than in your first reaction because of this effect.\n\n\nPiling a third proton on there would be far less favorable than that---oxalic acid is a weak acid, and you're forcing it to act as a base by making it accept a proton1. You should stop with the two reactions you have.\n\n\nIt's really the same thing with hypochlorite ion; putting one proton on is fine (you get HClO, which is a weak acid). Putting a second proton on is tough, because you'd be making an acid act as a base.\n\n\n\n\n---\n\n\n1. This kind of thing *does* happen, though such protonated acids will quickly lose their extra proton to the surrounding water molecules. If you're using glacial acetic acid as a solvent, your acid ion is $\\rm CH\\_3COOH\\_2^+$ instead of $\\rm H\\_3O^+$, \nand $\\rm CH\\_3COO^-$ instead of $\\rm OH^-$.\n\n\n", "7" ] ]

[]

[ [ "\nYes there are. In Vogels famous textbook on practical organic chemistry there is a section about the cooling effect of certain salt (mixture) solutions in water.\n\n\n\n> \n> If ice is temporarily not available, advantage may be taken of the\n> cooling effect attending the solution of certain salts or salt\n> mixtures in water. Thus a mixture produced by dissolving 1 part of\n> NH4C1 and 1 part of NaNO3 in 1-2 parts of water causes a reduction in\n> temperature from 10 to — 15 °C to — 20 °C; 3 parts of NH4C1 in 10\n> parts of water from 13 to -15°C; 11 parts of Na2S2O3.5H2O in 10 parts\n> of water from 11 to - 8 °C; and 3 parts of NH4NO3 in 5 parts of water\n> from 13 to — 13°C\n> \n> \n> \n\n\n(Taken from Vogels 5th edition, p. 70)\n\n\n", "5" ] ]

[ [ "\n\n> \n> I know that electrons revolve around nucleus in their orbits.\n> \n> \n> \n\n\nYou should \"unknow\" that, because that is the failed [Bohr Model](http://en.wikipedia.org/wiki/Bohr_model) of the atom. \n\n\nThe are no fixed orbits for electrons in atoms. \n\n\nThe Bohr Model has been superceded by the [Schrodinger model](http://www.pha.jhu.edu/~rt19/hydro/node3.html).\n\n\nThe atom is described in terms of probabilty density: for each point in space the probability that an electron will be at that point (when the atom is in a given state such as 1s, 2s, 2p, etc). \n\n\n\n> \n> my question is \"how the orbitals are located in the orbits?\n> \n> \n> \n\n\nThere are no orbits\n\n\n\n> \n> Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits??\n> \n> \n> \n\n\nThe shapes you seen in drawings that are spherical, dumb bell, etc. are meant to show regions of higher electron probabilty density. For example, someone might draw a sphere on the basis that there is a 90% probablity of the electron in that orbital being within the sphere at a given moment. However, in reality, for example in the 1s state, the electron could be anywhere in space. Other states, such as 2s, have nodes, with the electrons possibly being anywhere except at the [nodes](http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Electronic_Orbitals#Radial_and_Angular_Nodes). \n\n\nSo think of the shapes as showing where the electrons in that orbital are often, but remember that they can really be anywhere excpet at the nodes. \n\n\n", "4" ] ]

[ [ "\n$\\mu\\_i=\\mu\\_i^o+RT ln(a\\_i)$ is the [definition of activity](http://goldbook.iupac.org/A00115.html) ($a\\_i$).\n\n\nTherefore, the [standard state](http://goldbook.iupac.org/S05925.html) for [standard chemical potential](http://goldbook.iupac.org/S05908.html) ($\\mu\\_i^o$) is necessarily part of the definition of activity.\n\n\nFor a pure liquid or solid, the standard state is the pure substance at 1 bar.\n\n\nFor a liquid solvent, the standard state is also the pure liquid at 1 bar.\n\n\nFor a gas, the standard state is the fictitious state of an ideal gas at 1 bar.\n\n\nFor solutes, there are multiple conventions, for example the fictitious state of the solute having a molality of 1 mole/kilogram and exhibiting the behavior it does at infinite dilution. \n\n\n$\\Delta G^o = \\sum\\_i \\nu\\_i \\mu\\_i^o$ by definition.\n\n\nSo there is no one standard state for $\\Delta G^o$, because the various substances may have different standard states.\n\n\n", "2" ], [ "\nSo is the point of contention here \"how does the standard state for the activities relate to the standard state for the Gibbs free energy change\"?\n\n\nTo see how choice of standard state for an activity relates to the standard state chosen for the chemical potential, write the equation as\n\n\n$$\\mu\\_i = \\mu\\_i^\\circ + R T \\ln a\\_i = \\mu\\_i^\\circ + R T \\ln \\frac{\\gamma\\_i c\\_i}{c\\_i^\\circ}$$\n\n\nwhere the standard states are consistent.The standard state is being defined so that\nwhen $\\gamma\\_i c\\_i = c\\_i^\\circ$, $\\mu\\_i = \\mu\\_i^\\circ$. If we're using molarities, we can choose $c\\_i^\\circ = 1\\ M$. \n\n\nNow suppose you want to define a standard state that's different from 1 in the units you're using for concentration; e. g, suppose we have a reaction that includes $\\rm H^+$ and we want to define the biological standard state ($^\\oplus$) as $c\\_{\\rm H^+}^\\oplus = 10^{-7}\\ M = 10^{-7} c\\_{\\rm H^+}^\\circ$, while using molarities as our concentration unit. \n\n\nWe can compute a new chemical potential:\n\n\n$$\\begin{array}{rcl}\\mu\\_{\\rm H^+} &=& \\mu\\_{\\rm H^+}^\\circ + R T \\ln \\dfrac{\\gamma\\_i c\\_{\\rm H^+}}{c\\_{\\rm H^+}^\\oplus}\\\\\n&=& \\mu\\_{\\rm H^+}^\\circ + R T \\ln \\dfrac{\\gamma\\_i c\\_{\\rm H^+}}{10^{-7} c\\_{\\rm H^+}^\\circ}\\\\\n&=& (\\mu\\_{\\rm H^+}^\\circ + 7 R T\\ln 10) + R T \\ln \\dfrac{\\gamma\\_i c\\_{\\rm H^+}}{c\\_{\\rm H^+}^\\circ}\\\\\n&=& \\mu\\_{\\rm H^+}^\\oplus + R T \\ln \\dfrac{\\gamma\\_i c\\_{\\rm H^+}}{c\\_{\\rm H^+}^\\circ}\\\\\n&=& \\mu\\_{\\rm H^+}^\\oplus + R T \\ln a\\_{\\rm H^+}\n\\end{array}$$\n\n\nNow at unit activity $\\gamma\\_i c\\_i = c\\_i^\\circ$ and $\\mu\\_i = \\mu\\_i^\\oplus$. Your chemical potential (and your Gibbs free energy changes) are now defined in terms of this new standard state. *And note that the chemical potential at this new standard state ($\\mu\\_{\\rm H^+}^\\oplus$) is indeed temperature dependent*. To convert between the two standard state Gibbs free energies in this particular case,\n\n\n$$\\Delta G^\\oplus = \\Delta G^\\circ + 7 \\nu R T \\ln 10$$\n\n\nwhere $\\nu$ is the stoichiometric coefficient of $\\rm H^+$ in the reaction.\n\n\n*TL;DR: The standard state you choose for the concentration at unit activity becomes the standard state for the standard state Gibbs free energy change.*\n\n\n", "2" ] ]

[ [ "\nIn the equation that you stated initially, water would be omitted from the equilibrium constant expression.\n\n\nYour second question (about esterification that you posted in a comment to one of the answers) is extremely interesting.\n\n\nI have given it some thought: in the standard case of esterification water should be included because it is responsible for causing the reverse process ( hydrolysis of ester formed).\n\n\nBut in the case of zeolites, which by virtue of their structure, probably act as shape selective catalysts..ester hydrolysis by the water would be affected. (in some cases, I have read that selectivity towards ester formation is nearly 100%) In a case such as this, one can safely remove water from the equilibrium constant expression.\n\n\n(I am not be a 100% correct, please excuse me, since I am just a student myself. Perhaps, someone else can provide a better/clearer explanation)\n\n\n", "2" ], [ "\nNo. If the water condenses it is effectively removed from the reaction equilibrium. Just as with solids that precipitate.\n\n\n", "1" ] ]

[ [ "\n$F^2/\\sigma(F^2)$ is basically the signal-to-noice ratio, a simple measure of how well your crystal diffracts. It is hardly related to $R\\_{int}$ at all. Indeed, to calculate $R\\_{int}$ you need to have the whole dataset, otherwise it just would not make sense. As for $F^2/\\sigma(F^2)$, you have it as soon as you've collected a little bit of data.\n\n\n", "3" ], [ "\n$R\\_{int}$ considers in this summation symmetry relevant reflections, *e.g.* in your equation the term $\\Sigma|F^2-F\\_{mean}^2|$ takes an intensity of reflection and subtracts from this number the mean intensity of all its symmetry equivalents. If redundancy of your data collection is high (the case for all area detectors), then with this number you can tell how close in intensity are the equivalents, and thus how good is your crystal. \n\n\nAnother measure of quality is $R\\_{\\sigma}$\n$$R\\_{\\sigma}=\\frac{\\Sigma[{\\sigma}(F^2)]}{\\Sigma|F^2|}$$\nIt tells you how close are the estimated standard deviation of equivalents (*e.g.*: were all equivalents collected with the same precision?)\n\n\n", "1" ] ]

[ [ "\n$\\ce{HF}$ reacts with glass ($\\ce{SiO2}$).[1]\n\n\n$$\\ce{SiO2 + 4HF -> SiF4 + 2H2O}$$\n\n\n$\\ce{SiF4}$ is not a solid that consists of vertex-connected tetrahedra like $\\ce{SiO2}$ but is a gas at room temperature. Technically, $\\ce{HF}$ is not a solvent since in this case it reacts with the glass vessel.\n\n\nAccording to Spierings:[2]\n\n\n\n> \n> $\\ce{HF2-}$ ions are adsorbed on surface silanol groups, the $\\ce{HF}$ molecules on vicinal silanol groups and $\\ce{H+}$ on surface bridging oxygens in siloxane units. [...] These are transformed into surface groups such as $\\ce{\\bond{#}Si-F}$ and $\\ce{\\bond{#}Si-O-SiF3}$. The adsorption of $\\ce{HF}$ and $\\ce{HF2-}$ increases the electronic density on the bridging oxygen in the siloxane unit. This in turn makes these oxygens more basic, so more $\\ce{H+}$ ions are adsorbed, which leads to more siloxane bonds being broken per time unit, i.e. a kind of catalytic effect. [...] The catalytic action of $\\ce{H+}$ ions on breaking siloxane bonds also occurs in the dissolution of glasses in acidic and weakly alkaline solutions. \n> \n> \n> \n\n\n1. Hollemann, A. F.; Wiberg, E. *Hollemann–Wiberg: Lehrbuch der Anorganischen\nChemie,* 34th ed.; de Gruyter: Berlin, 2007; Vol. 102.\n2. Spierings, G. A. C. M. Wet chemical etching of silicate glasses in hydrofluoric acid based solutions. *J. Mater. Sci.* **1993,** *28* (23), 6261–6273. [DOI: 10.1007/BF01352182](https://doi.org/10.1007/BF01352182).\n\n\n", "8" ] ]

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[ [ "\nWhen you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as \"spending more time\". Electrons are less described as a dot but as a smearing (electron density).\nChlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say \"the electron spends more time at F/Cl\".\n\n\nThis thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: <http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory>\nBasically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.\n\n\n", "4" ], [ "\nOn the question: 'What does \"spend more time\" mean here?', from a theoretical vantage, why explore approximations that are based on the time-independent Schrödinger equation?\n\n\nPerhaps instead, this could be viewed as an application of the nonrelativistic version of the time-dependent Schrödinger equation for the wave function in position space of a single particle (a shared electron) subject to a potential (from an electric field from the respective cases of the chlorine and fluorine atoms).\n\n\nConceivably (for those more of an expert in this field), the product could be a theoretically based graphic depiction of the expected position over time. Visually, does the shared pair of electrons spend more time nearer the chlorine atom with HCl? In the case of the HF molecule, does the shared electrons spend more time nearer the fluorine atom?\n\n\nAs an example of one of the more readable discussions on the time-dependent Schrödinger equation in a position basis, complete with graphs, please see, for example,[this source](https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation). Here is a good educational video describing the probabilistic nature of the quantum realm [at here](https://www.youtube.com/watch?v=O6g-7rUgrdg) and [also this](https://www.youtube.com/watch?v=7jY5Q6u65uo). A related video [is found here](https://www.youtube.com/watch?v=_DXHrp6-LZI).\n\n\n", "-3" ] ]

[ [ "\nThat is not a quartet, but probably two overlapping signals. You always have to pay attention to the signal ratios and not just the number of split signals. The ratio here is 2:2:2:1, that is not a quartet ratio. A quartet would be 1:3:3:1.\n\n\nThere is a certain distortion in strongly coupled systems, the \"roof effect\" you can also see in this spectrum. This distorts the ratios a bit from the ideal ones, but not so much as to explain the ratio in the pseudo-quartet here.\n\n\n", "5" ], [ "\n1. Does the IR have any stretches near ~1700? This would be a great way to rule in (or out) a carbonyl. It looks like there is one from the above graph, and you have a broad peak ~ 11 ppm, which should stand out as a very specific functional group.\n2. Mad Scientist is correct, that is not a quartet at 7 PPM, but is a doublet and a triplet partially overlapped.\n3. Here's your hint: you have TWO oxygen-containing functional groups and an aromatic system (It looks like you already knew that from your sketches in the photo). If the aromatic system contains 4H, and is there are two functional groups, it is likely a di-substituted benzene ring. Which of the three possible types (ortho, meta, para) could give rise to two doublets and two triplets in the splitting?\n\n\nP.S. I believe the spectrum is correctly referenced. The tiny peak near 7.26 ppm is likely chloroform from the solvent - it is much smaller than the other peaks and chloroform is the solvent of choice for most organic NMR.\n\n\n", "3" ], [ "\nLooking at your picture you also seem to have annotated an IR peak at around $\\mathrm{1700~cm^{-1}}$ which is indicative of the carbonyl group. Together with the IR peak at $\\mathrm{3238~cm^{-1}}$ and the very broad NMR peak far downfield this supports your supposition that there is a carboxylic acid present.\n\n\nLooking at the molecular formula you now only have one $\\ce{O}$ and one $\\ce{H}$ unaccounted for so it must be a phenol group. As mentioned by Mad Scientist the pseudo-quartet on the NMR spectrum is in fact an overlapping doublet and triplet. You have two doublets and two triplets in total so you should think about which of ortho, meta and para substitution would give you this splitting pattern.\n\n\n", "2" ], [ "\nI once had a similar example in which benzene was used as solvent but the spectrum was not referenced to its peak at 7.16 ppm. If the quartet is really a quartet you might think about the meaning of that singlet at 7.26 ppm.\n\n\n", "1" ] ]

[ [ "\nThere's something called the [\"solvent leveling effect\"](http://en.wikipedia.org/wiki/Leveling_effect). In short, there's a lowest pKa in a particular solvent, based on the basicity of the conjugate base.\n\n\nIn water, you're limited by $\\ce{OH-}$. So given the same concentration of $\\ce{HI}$ and $\\ce{HCl}$ in water, they will indeed have the same pH.\n\n\nIn other solvents, you may have a difference between acids, depending on the pKa.\n\n\n", "3" ], [ "\nNo, not all strong acids are the same strength\n\n\nThere are more than just 6 strong acid. \n\n\nThe acid dissociation constant ($K\\_a$) quantifies the [strength of an acid](http://depts.washington.edu/eooptic/links/acidstrength.html). pK is negative log base 10 of $K\\_a$. \n\n\nTrifluoromethanesulfonic acid pk <-13\n\n\nPerchloric acid pK = <-9 \n\n\nHI pK = -9.3\n\n\nHBr pK = -9\n\n\nHCl pK = -6\n\n\nsulfuric acid pK = -3\n\n\nchloric acid pK = -1\n\n\nnitric acid pK = -1\n\n\n", "2" ] ]

[ [ "\n[$\\ce{H2S}$](http://en.wikipedia.org/wiki/Hydrogen_sulfide) has a mostly covalent character. You can decide for it by the crude way textbooks give: Calculate the difference in electronegativity to decide whether it's a non-polar covalent compound, a polar one or has ionic characters.\n\n\nFor the dissociation part (which is usually endothermic overall) it'll need more \"ionic character\". And it *does* dissociate in water, resulting in a weak acid named sulfhydric acid. (The spelling varies with citable sources) Your textbook likes to think about it as not dissociable, only because it's a **weak acid**.\n\n\nAddendum:\n---------\n\n\nThe \"ionic character\" plays its role of a rule a good one as explaining acid 'strength'. However, it acts almost flawlessly in the same period, but has exceptions when it gets to groups. As Dave mentions, $\\ce{HF}$ is weaker than for instance, $\\ce{HCl}$ and that's because **the atomic radius changes more dramatically than electronegativity difference**. (See [here](http://drholly.typepad.com/ask_me_a_chemistry_questi/2005/12/why_isnt_hf_a_s.html) ) But, in the $\\ce{H2S}$ molecule, this exception hasn't occurred. Dave pointed out a good example of when the rule isn't working that well, but as I said, we can ignore the other factors in this very case.\n\n\n", "2" ] ]

[ [ "\nYou may be familiar with the equilibrium that exists between a gem-diol and the corresponding carbonyl compound, as shown in the figure below. The carbonyl double bond is very strong, so in most cases the equilibrium lies far to the carbonyl side.\n\n\n\n\n\nThe same type of equilibrium exists in the case of bis- and tris-amino compounds where all of the amino groups are attached to the same carbon atom. In the case of such a tris-amino compound (see figure) the imino analogue of a carbonyl is generated, but since a third amino group is present, another equilibrium between the imino compound and a nitrile also exists. Just like in the carbonyl case, the nitrile triple bond is very strong. So when a tris-amino compound is generated an equilibrium with a very stable nitrile will be set up and the equilibrium will (usually) lie far to the nitrile side. Most tris-amino compounds prefer to exist as nitriles. If the amino group is fully alkylated (all of the amino hydrogens are replaced by alkyl groups), then this equilibrium cannot occur and such alkylated tris-amino compounds should be isolable.\n\n\n", "13" ] ]

[ [ "\nDibromohydrocinnamic acid can eliminate in a number of different ways as shown in the following diagram. I'm not sure what conditions lead to cinnamic acid, but I know it can happen. You have recently been asking questions about E1 and E2 eliminations, so my guess is that you are running a well-known lab experiment that generates *cis*- and *trans*-bromostyrene and allows the student to study the effect of reaction conditions on the mechanism of an elimination reaction.\n\n\n\n\n\nUsually, the experiment starts with the bromination of *trans*-cinnamic acid. The bromination is run under conditions that lead to trans-addition of bromine so that only the *erythro* ([see here for diagrams and mechanism](http://users.ipfw.edu/tahmassd/chm265/fall%202006/bromination%20procedure.htm)) or (*S*,*R*) [and its mirror image (*R*,*S*) enantiomer] dibromocinnamic acid diastereomer is formed. The other dibromohydrocinnamic acid diastereomer, the *threo* isomer [(*R*,*R*) and its (*S*,*S*) enantiomer], which can only be formed by cis-addition of bromine is not produced.\n\n\nIf you build a model or draw a Newman projection it will make it easier to understand what follows. \n\n\nUsing the *erythro* [(*S*,*R*) and (*R*,*S*)] diastereomer obtained from the above bromination, running the elimination reaction in a solvent like acetone (a solvent that won't stabilize a carbonium ion as much as protic solvents like water or methanol) you drive the mechanism towards the E2 elimination side. With your model or drawing, put the carboxylate group (the base has removed the acidic carboxylic acid proton to produce the corresponding carboxylate anion) and the bromine on the next carbon in an anti-periplanar arrangement (you've discussed anti-periplanar in your earlier questions on the E2 mechanism) and see how in a one-step, concerted elimination of both carbon dioxide and bromide you wind up with the ***cis***-bromostyrene isomer. \n\n\nIf you ran the reaction in a protic solvent (like water or methanol) that stabilizes carbonium ions, you push the reaction towards the E1 mechanism involving loss of the bromine adjacent to the benzene ring to produce a stabilized benzylic carbonium ion. The stable carbocation ion has an appreciable lifetime and will rotate so 1) the soon-to-be-departing carboxylate group is lined up with the carbocation p-orbital on the adjacent carbon **and** 2) that the benzene ring and remaining (bulky) bromine are as far apart as possible (the lowest energy conformation). Elimination of carbon dioxide from this conformation produces ***trans***-bromostyrene.\n\n\nTypically, the student is given several test tubes and the reaction is run under a variety of conditions where solvent and reagents are changed and then the amounts of *cis*- and *trans*-bromostyrene are measured by nmr. From this information the student can see how solvent and reagent can shift the mechanism. See this [link](http://www.ebah.com.br/content/ABAAAA_A0AK/journal-of-chemical-education-decarboxylative-eliminationof-2-3-dibromo-3-phenylpropanoic-acid) for a bit more detail on the actual experiment along with some good Newman projections.\n\n\nBTW, the lab should smell nice since the *trans*-bromostyrene produces a hyacinth-like odor.\n\n\n", "7" ] ]

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