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[ [ "\nForty electrons is *tiny*. Even if we limit ourselves to just the valence electons, cyclohexane already has 36 electrons. Anything drug-like has *way* more electrons that 40. For example, [viagra](https://en.wikipedia.org/wiki/Sildenafil) has 178 valence electrons, and that's not necessarily a \"large\" drug. (Compare with [vancomycin](https://en.wikipedia.org/wiki/Vancomycin), for example.)\n\n\nEven if you're dealing with things inorganic compounds, where the total number of atoms in the formula unit is small, the properties of the material don't come from a single formula unit, but come from the interaction of a large number of atoms. -- That's an example of a more general principle. The important properties for most materials you use (including drugs) don't come the molecule in isolation, but come from the interactions of the molecule with other molecules, either of the same chemical or of different chemicals. To be accurate, all of those interactions need a system with much more than 40 electrons.\n\n\nThe 40 electron limit comes from the implict assumption here that you're talking about quantum mechanical calculations. QM calculations are rather computationally expensive, as you have to account for all the interactions of all the electrons with each other at all positions in their delocalized superposition. There's various tricks (like [DFT](https://en.wikipedia.org/wiki/Density_functional_theory)) which make the calculations for large numbers of electrons easier, but note that \"easier\" doesn't mean \"easy\". Even with DFT and other approaches, large systems take a lot of computer time to calculate accurately.\n\n\nThere are other approaches which don't suffer from the same limit as QM does, but they are able to make their gains in efficiency because they make approximations. For example, molecular mechanics approaches are able to simulate systems in the hundreds of thousands of atoms region. But they're able to do so because they don't actually calculate the position of electrons. Instead they treat the system \"classically\", and experimentally fit interaction potentials which approximate the underlying quantum effects. (For example, they don't exactly calculate the bond stretching potential, but instead approximate it as a harmonic one. That's \"close enough\" to the true bond stretching potential for the range of bond lengths typically seen in such simulations, but not 100% quantum mechanically accurate.) \n\n\nThere's many groups and companies which do use molecular mechanics and other similar approaches to inform their drug and material development process. The issue is that because the energetic potentials being used are only approximate the results from the simulation are also only approximate. Depending on what you're trying to simulate, the results of the simulation may or may not be accurate. As such, these simulations are treated mostly as a first step, to find potential leads/hypotheses, and then the scientists actually have to go into the lab and test the results to confirm.\n\n\n", "35" ], [ "\nAside from the computational power needed to simulate larger molecules, there is also a lack of knowledge about the exact mechanisms that some drugs could potentially use. Think for example of experiments in yeast or *Escherichia coli* cells, which are used to find new biochemical mechanisms that could be exploited for new drugs. Even though we already know a lot about those cells, it would be computationally very demanding to include all known proteins and mechanisms into any kind of simulation. Furthermore, even if we could do such a simulation, there would still be a whole lot of other proteins, genes and mechanisms which we don't really understand yet but which could very well provide new mechanisms that could be used for new drugs. For this reason we would still need (biological) experiments even if we had much greater computational power than we have today.\n\n\n", "8" ], [ "\nAttempt exists, lots of drugs are first generated randomly, those most likely to have useful effects given structural category and simulation (simulation of molecular interactions and of known cell molecular pathways and physiological interactions) are tested, in cell lines, animals, then human.\nAll that has costs and risks.\n\n\nReferences about cell simulation:\n\n\n* [Wikipedia: Metabolic network modelling](https://en.wikipedia.org/wiki/Metabolic_network_modelling)\n* [*Skip the Humans: Drug Discovery by Simulating Cells* by Adrienne Lafrance, the Atlantic, May 30, 2014](https://www.theatlantic.com/technology/archive/2014/05/virtual-clinical-trials-doctors-could-use-algorithms-instead-of-people-to-test-new-drugs/371902/)\n* R. Weinberg, M. Berkus, *Int. J. Biomed. Comput.* **1971,** *2*(2), 95-120. DOI: [10.1016/0020-7101(71)90025-0](https://doi.org/10.1016/0020-7101(71)90025-0); Mirrored at [deepblue.lib.umich.edu](https://deepblue.lib.umich.edu/handle/2027.42/33674) \n([pdf](https://deepblue.lib.umich.edu/bitstream/handle/2027.42/33674/0000186.pdf)).\n\n\n", "4" ] ]

[ [ "\nYou are most of the way to the reason yourself:\n\n\n$$\\begin{array}{|c|c|c|}\\hline\n\\text{name} & \\text{formula} & \\text{oxidation number of P}\\\\ \\hline\n\\text{phosphoric} & \\ce{H3PO4} & +5\\\\\n\\text{phosphorous} & \\ce{H3PO3} & +3\\\\\n\\text{hypophosphorous} & \\ce{H3PO2} & +1\\\\ \\hline\n\\end{array}$$\n\n\nThe \"hypo-\" prefix means \"beneath\" or \"less than\". Hypophosphoric acid, $\\ce{H4P2O6}$, has oxidation number of phosphorous as +4, which is in between phosphoric acid and phosphorous acid. Thus, \"hypophosphoric\" refers to \"beneath\" phosphoric (but above phosphorous) acid.\n\n\n", "9" ] ]

[ [ "\nIn order not to waste your resources I recommend that you specify the quantity that you need for the upcoming reaction.\n\n\nTo make a diluted quantity of $x$ liters you may use the equation $$c\\_1\\*v\\_1 = c\\_2\\*v\\_2 \\\\ v\\_2 = \\frac{c\\_1\\*v\\_1}{c\\_2} $$\n\n\nWhere $v\\_2$ is the volume that you want to use for the reaction, $v\\_1$ is the volume from the concentrated solution that you need to take, $c\\_1$ is the concentration of the concentrated solution, and $c\\_2$ is the concentration of the solution that you want to use.\n\n\nFor your solution you may use the weight percentages or convert them to molarities if you want to use volumetric lab glassware.\n\n\nTo convert the weight percentages to molarities, use the equation $$c = \\frac{x\\*\\rho}{m\\*(100-x)}$$\n\n\nWhere x is the weight percentage of the solution so for 5% you use the number 5 in the equation, m is the molar mass of the hydrochloric acid. To facilitate calculating the molar masses of different compounds I use the on-line tool [Molar Mass Calculator](http://www.webqc.org/mmcalc.php).\n\n\nTo get the best results I recommend using a volumetric flask and substituting the volume for the volumetric flask instead of the volume of the acid that you want to use. So the equation that want to use is $$v\\_{needed} = \\frac{c\\_{5\\%}\\*v\\_{volumetric}}{c\\_{13\\%}} $$\nYou then put the volume that you obtained in the volumetric flask and fill the rest with distilled water till the mark and you have thus obtained your solution.\n\n\n", "3" ] ]

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[ [ "\nThere are two really important properties of N2O that make it such an important greenhouse gas, both of which you have raised in your question but need some clarification. \n\n\n1. Infrared absorption - it's good to think in terms of gaps here. There's a lot of overlap between CO2, CH4 and (a frequently ignored GHG, but very important) water vapour in the wavelengths which they absorb. At the 4-5 micron and 7-8 micron range, N2O is extremely efficient at absorbing infrared radiation. These are important regions, precisely because there is no overlap with other GHGs in these regions. N2O is absorbing infrared very efficiently, and is doing so without \"competition\".\n2. Lifetime - you noted that the lifetime of N2O and CO2 are comparable, which is sort of true. N2O, as a stable, inert molecule that is well mixed in the atmosphere, has a very long lifetime of 120 years compared to some other GHGs (CH4, for example is 8 years, based on the fact that its removed by the hydroxyl OH radical). It's possible to determine this fairly accurately as the only sinks for N2O are its stratospheric photo-dissociation and reaction with O(1D) radical. CO2 by comparison has a lifetime that varies anywhere between 5 and 200 years. The comparable part comes from the fact that for calculations for global warming potential, CO2 is given an \"averaged\" lifetime.\n\n\nWith N2O, therefore, we have a very efficient infrared absorber at two \"windows\" at which there is no overlap with other GHGs, and a long lifetime. The Global Warming Potential is the time integrated radiative forcing for a 1 kg pulse emission of the compound, where the upper limit of the integration is time horizon (usually set at 100 yr), relative to the same quantity of the reference compound CO2. It becomes clear from this definition why a stable, inert molecule with a long lifetime and which is very efficient at absorbing at particular infrared wavelengths would have a greater global warming potential than CO2 (shorter lifetime, absorbing in \"windows\" shared by other compounds) on *a kg to kg* comparison. \n\n\n", "4" ], [ "\nAnother facet to the window factor mentioned in many other answers is the fact that the absorption of any species is blocked by itself - ie, the concentration of any gas already in the atmosphere is already absorbing its wavelengths, so the marginal addition of 1 more kg of that gas will have less IR to absorb. Because the concentration of CO2 is over 400ppm while the concentration of N2O is more than 1000x less at around 329ppb, a new kg of N2O has a lot less \"competition\" to absorb its IR than a new kg of CO2. For the same reason, the GWP of CO2 is decreasing over time as more CO2 builds up in the atmosphere.\n\n\n", "2" ] ]

[ [ "\nYes, there are many examples. For instance\n\n\n* The common reducing agents LithAl, $\\ce{LiAlH4}$, and sodium borohydride, $\\ce{NaBH4}$ (last one you might argue about)\n* $\\ce{[ReH9]^{2-}}$, see <https://en.wikipedia.org/wiki/Potassium_nonahydridorhenate>\n* Metal carbonyl hydrides, see <https://en.wikipedia.org/wiki/Metal_carbonyl_hydride>\n* More generally many organometallic compunds, e.g. $\\ce{(C5H5)\\_2ReH}$\n\n\nA quick google also shows up <http://www.ilpi.com/organomet/hydride.html>\n\n\n", "8" ] ]

[ [ "\nThis is but an urban legend. Glass does not get thicker at the bottom. (Sure, glass **is** fluid to some tiny extent, and it can be demonstrated, but not in this way.) It is just that the glass panes were initially manufactured with uneven thickness. Then the people who stuck the glass into the frames would orient the thicker side down, maybe because it is more stable this way, or because it feels right, or just to mess with us. Sometimes they didn't pay attention and turned it randomly.\n\n\nSupporting links: [1](https://physics.stackexchange.com/questions/55829/do-glass-panes-become-thicker-at-the-bottom-over-time), [2](http://io9.gizmodo.com/the-glass-is-a-liquid-myth-has-finally-been-destroyed-496190894).\n\n\n", "4" ] ]

[ [ "\n[Propyl](https://en.wikipedia.org/wiki/Propyl_group) is a simple hydrocarbon with three carbons in it. [Propane](https://en.wikipedia.org/wiki/Propane) is $\\ce{CH3CH2CH3}$.\n\n\nPropyl is the unit formed by attaching that chain to something else (which implies replacing one of the carbons with another bond). There are two ways to do this: one attaches the propyl unit using the end carbon on the chain (giving $\\ce{X-CH2CH2CH3}$); the other involves attaching the middle carbon to something else (giving $\\ce{CH3CHXCH3}$). The first is called (at least in older terminology) *normal*-propyl, *n*-propyl or $\\ce{n-Pr}$; the second is called *iso*-propyl, *i*-propyl or $\\ce{i-Pr}$. These reduce the amount of space required to write the structure.\n\n\nIn your compound helicene I'd guess that the propyls hanging off the rings are there to make the compounds far more soluble in organic solvents as flat aromatic structures tend to be fairly insoluble due to the way the pack in crystals.\n\n\n", "7" ] ]

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[ [ "\nExplanation of notation:\n\n\n$H$ is the enthalpy of the system.\n\n\n$\\Delta$ means change of, so $\\Delta H$ means change of the enthalpy\n\n\n[](https://i.stack.imgur.com/1U2uns.jpg)\n\n\nThis symbol means standard condition, standard condition is defined as a pressure of $100\\text{kPa}$ and reactants and products are in their standard state, or concentration of solutions are $1\\text{M}$. However, since LaTeX doesn't have this symbol, I'll substitute it for $^\\ominus$.\n\n\nEnthalpy changes:\n\n\n**Enthalpy of dilution** – The enthalpy change when a solution containing one mole of a solute is diluted from one concentration to another.\n\n\n**Enthalpy of** ($n\\text{th}$) **electron affinity** – The enthalpy change when $n$ electrons are added to one mole of gaseous atoms.\n$$\\ce{Li(g) + e-(g) -> Li-(g) +60\\ \\text{kJ}}$$\n$$\\ce{F(g) + e-(g) -> F-(g) +328\\ \\text{kJ}}$$\n\n\n**Enthalpy of** ($n\\text{th}$) **ionization** – The enthalpy change when $n$ electrons are removed from one mole of gaseous atoms. It is always positive.\n\n\n$$\\ce{Li(g) +520\\ \\text{kJ} -> Li+(g) +e-(g)}$$\n$$\\ce{He(g) +2372\\ \\text{kJ} -> He+(g) +e-(g)}$$\n\n\n**Enthalpy of lattice dissociation** – The enthalpy change when one mole of an ionic lattice dissociates into isolated gaseous ions.\nAn example for sodium chloride which have an enthalpy of lattice dissociation, also known as lattice energy, of $787\\ \\mathrm{kJ/mol}$\n$$\\ce{NaCl(s) +787\\ \\text{kJ} -> Na+(g) + Cl-(g)}\\qquad\\Delta H=787\\ \\text{kJ/mol}$$\n\n\n**Enthalpy of lattice formation** – The enthalpy change when one mole of solid crystal is formed from its scattered gaseous ions.\n\n\n$$\\ce{Na+(g) + Cl-(g) -> NaCl(s)} +787\\ \\text{kJ}\\qquad\\Delta H=-787\\ \\text{kJ/mol}$$\n\n\nWhich means Enthalpy of lattice dissociation$=-$Enthalpy of lattice formation\n**Enthalpy of hydration**($\\Delta\\_\\text{hyd}H^\\ominus$) – Enthalpy change when when one mole of ions undergo hydration.\n\n\n**Enthalpy of mixing** – The enthalpy change from a substance when mixed.\n\n\n**Enthalpy of neutralisation** – The enthalpy change when an acid is completely neutralised by a base.\n\n\nFor a strong acid, like $\\ce{HCl}$ and strong base, like $\\ce{NaOH}$, they disassociate almost completely $\\ce{Cl-}$ and $\\ce{Na+}$ are spectator ions so what is actually happening is $$\\ce{H+(aq) + OH-(aq) -> H2O(l) + 58\\ \\text{kJ/mol}}$$\n\n\nHowever, using a weak acid/base will have a lower enthalpy of neutralisation as normally most of the acid/base does not disassociate.\n\n\nFor example, mixing ethanoic acid and potassium hydroxide only has a enthalpy of neutralisation of $-11.7\\ \\text{kJ/mol}$\n\n\n**Enthalpy of precipitation** – The enthalpy change when one mole of a sparingly soluble substance precipitates by mixing dilute solutions of suitable electrolytes.\n\n\n**Enthalpy of solution** – Enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.\n\n\nEnthalpy of solution can be positive or negative as when a ionic substance dissolves, the dissolution can be broken into three steps\n\n\n1. Breaking of solute-solute attraction (endothermic)\n2. Breaking solvent-solvent attraction (endothermic), eg. hydrogen bonds, LDF\n3. Forming solvent-solute attraction (exothermic)\n\n\nAn example of a positive enthalpy of solution is potassium chlorate which has an enthalpy of solution of $41.38\\ \\text{kJ/mol}$\n\n\n**Enthalpy of** (Solid$\\rightarrow$Liquid: $\\Delta\\_\\text{fus}H^\\ominus$, Liquid$\\rightarrow$Solid:$\\Delta\\_\\text{freezing}H^\\ominus$, Liquid$\\rightarrow$Gas: $\\Delta\\_\\text{vap}H^\\ominus$, Gas$\\rightarrow$Liquid: $\\Delta\\_\\text{cond}H^\\ominus$, Solid$\\rightarrow$Gas: $\\Delta\\_\\text{sub}H^\\ominus$, Gas$\\rightarrow$Solid: $\\Delta\\_\\text{deposition}H^\\ominus$) – The enthalpy change from providing energy, to a specific quantity of the substance to change its state.\n$$\\Delta\\_\\text{fus}H^\\ominus=-\\Delta\\_\\text{freezing}H^\\ominus,\\Delta\\_\\text{vap}H^\\ominus=-\\Delta\\_\\text{cond}H^\\ominus,\\Delta\\_\\text{sub}H^\\ominus=-\\Delta\\_\\text{deposition}H^\\ominus$$\n**Standard enthalpy of atomization**($\\Delta\\_\\text{at}H\\_T^\\ominus$) – Change when a compound's bonds are broken and the component atoms are reduced to individual atoms at at $T^\\circ K$.\n$$\\ce{S\\_8 -> 8S}\\qquad\\Delta\\_{at}H^\\ominus=278.7\\ \\text{kJ/mol}$$\n\n\n**Standard enthalpy of combustion**($\\Delta\\_\\text{c}H\\_T^\\ominus$) – The enthalpy change which occurs when one mole of the compound is burned completely in oxygen at $T^\\circ K$ and $10\\ \\mathrm{kPa}$.\n$$\\ce{H2(g) +\\frac{1}{2}O2(g) -> H2O(g)}+572\\ \\text{kJ}\\qquad\\Delta\\_cH^\\ominus = -286\\ \\text{kJ/mol}$$\n\n\n**Standard enthalpy of formation**($\\Delta\\_\\text{f}H\\_T^\\ominus$) – Change in enthalpy during the formation of one mole of the compound from its constituent elements, with all substances in their standard states, and at a pressure of $100\\ \\mathrm{kPa}$ at $T^\\circ K$.\n\n\nIt can be calculated using Hess's law if the reaction is hypothetical. An example is methane, $\\ce{C}$ and $\\ce{H2}$ will not normally react but the standard enthalpy of formation of methane is determined by Hess's law to be $-74.8\\ \\text{kJ/mol}$\n$$\\ce{\\frac{1}{2}N2(g) +\\frac{1}{2}O2(g) -> NO(g)}\\qquad\\Delta\\_\\text{f}H^\\ominus=90.25\\ \\text{kJ/mol}$$\n\n\n**Standard enthalpy of reaction**($\\Delta\\_\\text{r}H^\\ominus\\_T$) – Enthalpy change that when matter is transformed by a chemical reaction at $T^\\circ K$ and $10\\ \\mathrm{kPa}$.\n$$\\ce{H2(g) +\\frac{1}{2}O2(g) -> H2O(g)} +572\\ \\text{kJ}\\qquad\\Delta\\_\\text{r}H^\\ominus = −572\\ \\text{kJ/mol}$$\n$$\\Delta\\_\\text{r}H^\\ominus=\\sum H^\\ominus\\_\\text{products}-\\sum H^\\ominus\\_\\text{reactants}$$\n$$\\Delta\\_\\text{r}H\\_\\text{forward}=-\\Delta\\_\\text{r}H\\_\\text{backwards}$$\nLaws:\n\n\nLe Chatelier's Principle – When an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change.\n\n\nYellow $\\ce{Fe^3+}$ reacting with colorless thiocyanate ions $\\ce{SCN-}$ to form deep red $\\ce{[Fe(SCN)]^2+}$ ions:\n$$\\ce{Fe^3+(aq) + SCN-(aq) -> [Fe(SCN)]^2+(aq)}$$\nWhen $\\ce{NH4Cl}$ is added, $\\ce{Cl-}$ reacts with $\\ce{Fe^3+}$ ions to form $\\ce{[FeCl4]-}$ ions. By the Le Chatelier's Principle, when $\\ce{Cl-}$ is added into a solution of deep red $\\ce{[Fe(SCN)]^2+(aq)}$ ions, the equilibrium will shift to $\\ce{Fe^3+(aq) + SCN-(aq)}$, turning the solution pale red.\n\n\nKirchhoff's Law – Enthalpy of any substance increases with temperature.\n\n\nHess's Law – Total enthalpy change of a chemical reaction is independent of the number of steps the reaction takes.\n\n\nHenry's Law – Amount of dissolved gas is proportional to its partial pressure in the gas phase.\n\n\nProcesses:\n\n\nConstant Entropy – Isoentropic\nConstant Pressure – Isobaric\n\n\nConstant Volume – Isovolumetric\n\n\nSide note: This is probably not a complete list as I may have missed some\n\n\n", "15" ], [ "\nMost of the chemistry happens in open containers. Open containers have constant pressure. At constant pressure work is done to carry out the reaction and also to \"push\" the air up (called as expansion work). This expansion work is $${\\Delta PV}$$. Thus some part of the heat leaks as expansion work without being converted into Internal Energy.\n\n\nSo we define a new term $${\\Delta H ~(Enthalpy) = \\Delta U + \\Delta PV}$$. It accounts for expansion work and internal energy. As the heat is used only for internal energy and expansion work, Enthalpy is equal to heat supplied.\nThere are various types of enthalpies depending upon Internal Energy is used to do what kind of work.\n\n\nPhase is something that is uniform in composition and physical state. It is similar to state of matter (solid, liquid and gas), but these is only one state of matter (solid carbon) but there can be many phases (graphite, diamond)\n\n\nEnthalpy of phase transition\n============================\n\n\nIt is the heat supplied at constant 1 atm pressure to bring about a change in phase. It can be $$\\ce{H2O (liquid) <=> H2O (gas) -> +44KJ}$$ or $$\\ce{C (graphite) <=> C (diamond) -> +4KJ}$$ Conversion of graphite into diamond is thermodynamically possible but kinetics does not allow it, as there are very strong bond to break and rearrange. This is true for every thermodynamical reaction.\n\n\n**Enthalpy of vapourisation** and **Enthalpy of Condensation**\n\n\n*Enthalpy of vapourisation* is the heat required to vaporise 1 mole of liquid into gas. (a phase change) Energy needs to be supplied and hence enthalpy is positive.\n\n\nAs Enthalpy is a state function, the reaction can proceed from the final state (gas) to initial state (liquid) by absorbing the same amount of energy. This heat absorbed at constant pressure is called as *enthalpy of condensation* and is numerically equal to negative of enthalpy of vapourisation.\n\n\n**Enthalpy of Fusion** and **Enthalpy of freezing**\n\n\nThe heat required to melt 1 gm of solid into liquid at a constant pressure of 1 atm is *enthalpy of fusion*. The enthalpy required for the reverse reaction is (liquid to solid) is called *enthalpy of freezing*. It follows the same lines as of above two enthalpies.\n\n\n**Enthalpy of sublimation** and **Enthalpy of vapour deposition**\n\n\nSome substances directly vaporise into gas, without forming liquid and liquid is not stable at any temperature and pressure. The enthalpy required to do so is *enthalpy of sublimation*. As enthalpy is a state function, the enthalpy required to deposit the gas into a solid is called *enthalpy of vapour deposition*.\n\n\nEnthalpy of atomic and molecular changes\n========================================\n\n\n**Enthalpy of ionisation** and *Enthalpy of electron capture*\n\n\nThe energy required to remove the first electron of one mole of gas at constant pressure is called *enthalpy of ionisation*. The energy required to remove the second electron is called *Second enthalpy of ionisation*.\nThe enthalpy of ionisation of 1 mole of gas at 273.15 Kelvin and 1 atm pressure is called \"Ionisation Energy\".\n\n\nThe enthalpy of capturing an electron by a neutral atom (neutral, not positive ion) is called *enthalpy of electron capture*. At 273.15 K and 1 atm, for 1 mole this energy is also called \"Standard electron gain Enthalpy\"\n\n\n**Enthalpy of disassociation** and **Enthalpy of bond formation**\n\n\nThe heat required to disassociate a molecule into 2 entities is called enthalpy of *bond dissociation* Eg $$\\ce{H2O -> H2 + O ~~~~ \\Delta H = +927 KJ}$$\n\n\nthe opposite of it is *enthalpy of bond formation* Eg. $$\\ce{H2 + O2 -> H2O}$$ Here the enthalpy of (H-O-H) is not equal to 2 times enthalpy of (O-H)\n\n\nfor this particular reaction, $$\\ce{H2O(gas) -> H + HO^. ~~~~\\Delta H = +428KJ }$$ and $$\\ce{OH^. -> H + O^. ~~~~ \\Delta H = +428 KJ}$$. Therefore total enthalpy is +927KJ. As Enthalpy of disassociation is dependent upon reaction itself $$\\ce{\\Delta H(O-H)~is~the~average~value~of~\\Delta H(O-H)~of~H-OH~,~\\Delta H(O-H)~of~CH3OH~etc.}$$\n\n\n", "6" ] ]

[ [ "\nThis isn't the most well-thought-out scenario, because if you dissolve $\\ce{KOD}$ in $\\ce{H2O}$ all the deuteriums will be lost immediately anyway.\n\n\n$$\\ce{KOD + H2O <=>> KOH + HOD}$$\n\n\nAssuming that the amount of $\\ce{KOD}$ is much less than the amount of $\\ce{H2O}$ (after all $\\ce{H2O}$ is the solvent) then the position of this equilibrium will lie very, very far to the right. So, at the end of the day, if you are using $\\ce{H2O}$ as the solvent, the use of $\\ce{KOD}$ won't have any effect. If you want to see any difference in deuteration, you have to at least use $\\ce{KOD/D2O}$. However, there is a second problem with this:\n\n\nIf you dissolve an amine in $\\ce{D2O}$, then all the amine hydrogens will be very rapidly replaced with deuteriums. It does not even matter mechanistically what the pathway is. Regardless of whether the initial product is $\\ce{RNH2}$, $\\ce{RNHD}$, or $\\ce{RND2}$, as long as there is $\\ce{D2O}$ in the mixture, all three will become $\\ce{RND2}$. The same can be said of dissolving them in $\\ce{H2O}$.\n\n\n\n\n---\n\n\nTo answer your questions:\n\n\n(1) No. In general, protonation can and will occur by any solvent molecule that happens to be close by. If you are running the reaction in $\\ce{D2O}$ solvent, then there is much, much more $\\ce{D2O}$ than there will ever be $\\ce{H2O}$ produced. So, protonation by $\\ce{H2O}$ is statistically close to impossible.\n\n\n(2) None of us can answer that, as we cannot presume what your teacher was thinking of when they set the question.\n\n\n(3) Depends on what your solvent is, see comments above on dissolving amines in $\\ce{D2O}$ or $\\ce{H2O}$.\n\n\n", "3" ] ]

[ [ "\nThose steps you mentioned are not merely critical steps, in glycolysis but are however the control points of the process and I would prefer to call them rate controlling steps. There is nevertheless one critical step which is believed to be the first committed step in glycolysis (not really sure if there are more).\n\n\n\n> \n> There are three irreversible steps in glycolysis, and the differences\n> between glycolysis and gluconeogenesis are found in these three\n> reactions. The first of the glycolytic reactions is the production of\n> pyruvate (and ATP) from phosphoenolpyruvate. The second is the\n> production of fructose-1,6-bisphosphate from fructose-6-phosphate, and\n> the third is the production of glucose-6-phosphate from glucose.\n> Because the first of these reactions is exergonic, the reverse\n> reaction is endergonic. Reversing the second and third reactions would\n> require the production of ATP from ADP, which is also an endergonic\n> reaction.\n> \n> \n> \n\n\nHaving said that, there are factors whih affect the process of glycolysis:\n\n\nEnzymes\n-------\n\n\n**Control of hexokinase**\n\n\nThe enzyme that catalyses the first reaction is hexokinase. The substrate of hexokinase is not necessarily glucose; rather, it can be any one of a number of hexoses, such as glucose, fructose, and mannose. Glucose-6-phosphate inhibits the activity of hexokinase; this is a control point in the pathway. Hexokinase is inhibited by high levels of its product, glucose-6-phosphate. When glycolysis is inhibited through phosphofructokinase, glucose-6-phosphate builds up, shutting down hexokinase. (**However, the liver contains a second enzyme that phosphorylates glucose-glucokinase**).\n\n\n**Control of phosphofructokinase**\n\n\nThe phosphorylation of fructose-6-phosphate is highly exergonic and irreversible, and phosphofructokinase, the enzyme that catalyses it, is the key regulatory enzyme in glycolysis. Phosphofructokinase is a tetramer that is subject to allosteric feedback regulation. fructose-2,6-bisphosphate an important allosteric activator of phosphofructokinase (PFK), the key enzyme of glycolysis; (it is also an inhibitor of fructose bisphosphate phosphatase (FBPase), which plays a role in gluconeogenesis.)\n\n\n**Control of Pyruvate Kinase**\n\n\nThe final step of glycolysis is also a major control point in glucose metabolism. Pyruvate kinase (PK) is allosterically affected by several compounds. ATP and alanine both inhibit it. (It would be wasteful to breakdown more glucose is there is already vast amounts of ATP). Alanine is essentially pyruvate with an amino group. (converted to pyruvate via a transaminase enzyme). Therefore, a high level of alanine indicates that a high level of pyruvate is already present, so the enzyme that would make more pyruvate can be shut down. Fructose-1,6-bisphosphate allosterically activates pyruvate kinase so that the incoming products of the first reactions of glycolysis can be processed.\n\n\nSubstrate\n---------\n\n\nThe reaction in which fructose-6-phosphate is phosphorylated to give fructose- 1,6-bisphosphate is the one in which the sugar is **committed to glycolysis**. Glucose-6-phosphate and fructose-6-phosphate can play roles in other pathways, but fructose-1,6-bisphosphate does not. **After fructose-1,6-bisphosphate is formed from the original sugar, no other pathways are available, and the molecule must undergo the rest of the reactions of glycolysis.**\n\n\nIt is frequently observed that control is exercised near the start and end of a pathway, as well as at points involving key intermediates such as fructose-1,6-bisphosphate. This is the summarised version of control factors of glycolysis:\n\n\n[](https://i.stack.imgur.com/D1QgS.png)\n\n\nRemarks\n-------\n\n\nTo sum up, the reason I explained your mentioned reactions not to be critical is because (with reference to reaction 1 and 3):\n\n\n* Even if hexokinase is inhibited, glycolysis can still take place via catalysis by glucokinase in liver.\n* There are other feeder pathways for glycolysis even if the glucose sugar is not present, the saccharides (maltose, lactose, trehalose, and sucrose, mannose and galactose) can still be used to produce fructose 6-phosphate.\n\n\ne.g $$\\ce{Fructose + ATP (Mg2+) → fructose 6-phosphate + ADP}$$\n\n\n* The last reaction would be bypassed by transaminase (is not important since the reaction would have been completed):\n\n\nSo these are major controlling steps, meaning glycolysis can still take place (if committed step is reached) though at reduced rates until optimum conditions are restored. The second part of your question can’t not answer it because it is a different concept. Hope this helps\n\n\nReferences\n\n\n1. Harpers illustrated Biochemistry\n2. Lehninger Principles of Biochemistry\n3. Biochemistry (Campbell and Farrell)\n\n\n", "6" ], [ "\nThe committed enzyme in glycolysis is considered to be PFK-1, and is under complex control. While the hexokinase/glucokinase steps are of course critical for getting glucose into the cell, the product, glucose 6-phosphate has multiple fates (on to glycogen or the pentose pathway), so is not unique to the glycolytic pathway. \n\n\nIn muscle, the key activator of PFK-1 is AMP, signalling a drop in the energy charge of the cell and therefore a need to switch on ATP generating pathways.\nIn the liver, the key activator is Fructose 2,6 bisphosphate, which rises in response to insulin.\n\n\n", "1" ] ]

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[ [ "\nThe reaction of $\\ce{NaSi}$ with water is both highly enthalpically and entropically favored—the reaction produces 5 molecules of $\\ce{H2}$ gas for every 2 $\\ce{NaSi}$ molecules that react, and the enthalpy of reaction is –175 $\\mathrm{kJ\\ mol^{-1}}$. The reverse reaction would require ludicrous amounts of energy and is likely not even kinetically feasible. \n\n\nIt would be especially pointless to attempt the reverse reaction when $\\ce{NaSi}$ is generated easily enough from $\\ce{Na}$ and silica gel. Liquid sodium-potassium alloy absorbed onto silica gel results in a fine black powder that spontaneously ignites in humid air. Heating $\\ce{Na}$ and silica gel to 165ºC with continual agitation produces a shelf stable, yet still highly reactive $\\ce{NaSi}$. When slowly heated to 400ºC, air stable $\\ce{NaSi}$ may be produced, though with less reducing power than those produced via the other two methods. It will, however, still react with water to produce $\\ce{H2}$ and $\\ce{ Na2Si2O5}$.$^{[1]}$\n\n\n\n\n---\n\n\n$^{[1]}$ [Alkali Metals Plus Silica Gel:  Powerful Reducing Agents and Convenient Hydrogen Sources](http://pubs.acs.org/doi/abs/10.1021/ja051786+)\n\n\n", "2" ] ]

[ [ "\nAn unit cell is a periodic structure. There are two types of unit cells: conventional unit cells and primitive unit cells. The conventional unit cell is the one that is used most often. Nevertheless, I suggest that you take a look at what is the difference between the two. There are plenty of resources that explain the difference. \n\n\nNow with that out of the way. Let's look at the unit cell of NaCl. According to [Chem Libre](https://chem.libretexts.org/LibreTexts/Howard_University/General_Chemistry%3A_An_Atoms_First_Approach/Unit_2%3A__Molecular_Structure/Chapter_4%3A_Ionic_Bonding/Chapter_4.1%3A_Ionic_Bonding), the unit cell is:\n\n\n[](https://i.stack.imgur.com/hJ7oc.png)\n\n\nRecall that ionic solids are held together by electrostatic forces. So what matters is that the charges in a unit cell cancel. In other words, we want the net charge to be 0. Let's count the number of sodium atoms and chloride atoms to see whether the charges cancel in this unit cell.\n\n\nAt the top plane, there are 4 chloride atoms that contribute $\\frac{1}{8}$ of an atom. There is also 1 chloride atom that contributes $\\frac{1}{2}$. On the other hand, there are 4 Na atoms that contribute $\\frac{1}{4}$ of an Na atom. Now, each Na atom has a 1+ charge whereas each Cl atom has a 1- charge. Therefore, at the top plane we have that:\n\n\n$$(+1)(4\\,\\cdot\\,\\frac{1\\,\\,\\text{Na atoms}}{4})+(-1)(4\\,\\cdot\\,\\frac{1\\,\\,\\text{Cl atoms}}{8})+(-1)(1\\,\\cdot\\,\\frac{1\\,\\,\\text{Cl atoms}}{2})=0$$\n\n\nSo we can see that the overall charge for the top layer is 0.\nThe same is true for the plane in the middle. Work through it. It is a good exercise.\n\n\nNow, lets look at the coordination number. What do we mean by coordination number? The coordination number is simply the number of nearest neighbors. That is all that there is to it. It doesn't matter whether a single sodium ion is in contact with 6 chloride ions. This doesn't imply that the sodium ion will end up with -5 negative charge. This is because even though the sodium ion is in contact with 6 chloride ions, each of these chloride ions is in contact with other sodium ions. As a result, the sodium ion does not really accumulate negative charge. \n\n\nWhat **determines** the coordination number is a bit more complicated. The coordination number is dependent on the size of each atom, the charges, and even the spin. One simple rule that helps you determine the coordination number of binary solids is the radius ration rule. I suggest that you take a look at it: <http://minerva.mlib.cnr.it/mod/book/view.php?id=269&chapterid=111>\n\n\n", "2" ] ]

[ [ "\nIf we talk about the reaction of *only* sulfur and potassium nitrate, then the reaction yields potassium oxide, sulfur dioxide and nitrogen.\n\n\n$$\\ce{4KNO3 + 5S -> 2K2O + 5SO2 + 2N2}$$\n\n\nHowever, when carbon is added in the form of charcoal, then the resulting mixture is called [gunpowder](https://en.wikipedia.org/wiki/Gunpowder). The overall chemical reaction of gunpowder is as follows:\n\n\n$$\\ce{2 KNO3 + S + 3 C -> K2S + N2 + 3 CO2}$$\n\n\n$$\\ce{10 KNO3 + 3 S + 8 C -> 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2}$$\n\n\nThere are plenty of questions in chem.SE related to gunpowder:\n\n\n1. [What is the oxidation mechanism of gunpowder?](https://chemistry.stackexchange.com/questions/35680/what-is-the-oxidation-mechanism-of-gunpowder)\n2. [What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder?](https://chemistry.stackexchange.com/questions/70734/what-causes-kno3-to-decompose-into-kno2-and-oxygen-in-gunpowder)\n3. [Why can sulfur act like a catalyst?](https://chemistry.stackexchange.com/questions/4173/why-can-sulfur-act-like-a-catalyst)\n4. [How to write the balanced equation of the reaction of potassium nitrate, carbon, and sulfur?](https://chemistry.stackexchange.com/questions/37684/write-the-chemical-equation-of-this-reaction/37686#37686)\n5. [What are the reaction products of sulfur and potassium nitrate?](https://chemistry.stackexchange.com/questions/42800/what-are-the-reaction-products-of-sulfur-and-potassium-nitrate)\n\n\n", "2" ] ]

[ [ "\nFirst things first: Don't mess with higher concentrations of hydrogen peroxide unless you are a trained and well-equipped chemist. We're talking \"steel-reinforced gauntlets\" here. This chemical is as volatile as nitroglycerine.\n\n\nAt 3% you have mostly de-ionized water and for good reason. Hydrogen peroxide is extremely, violently unstable, and is just as likely to oxidize anything else than what you actually intend to react it with.\n\n\nAnything above a concentration of 30% is very dangerous. Above 60% is suicidal in ill-equipped scenarios. 100% is a hypothetical, and is in essence a potential component of rocket fuel.\n\n\nSo, yes it will work. You may lose a hand, but it will work.\n\n\n", "3" ], [ "\nListerine Total Care Stain Remover AntiCavity Mouthwash contains 21% ethanol (from the label) and hydrogen peroxide (lower down on the ingredients list; I would hazard a guess of about 0.75 - 1.0% from the amount of foaming in use).\n\n\nThe product is stable for months - perhaps years. My bottle is way past its expiration, yet still foamable. Although the concentrations are not extreme, the long term stability suggests that H2O2 doesn't react significantly with ethanol, at least not at these concentrations.\n\n\nHigher concentrations are more reactive: On 16 July 1934, in Kummersdorf, Germany, a tank containing a mixture of hydrogen peroxide and ethanol exploded during a test, killing three people.\n\n\nFenton's Reagent, which is hydrogen peroxide with a catalytic amount of ferrous sulfate, is used to oxidize waste water containing organic compounds, but as indicated in a comment above, getting it to stop at acetic acid may be tricky.\n\n\n", "2" ] ]

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[ [ "\nI do not believe so.\n\n\nThe mechanism of decomposition of $\\ce{H2O2}$ is:\n\n\n$\\ce{H2O2(g) -> 2HO.(g)}$\n\n\n$\\ce{H2O2(g) + .OH(g) -> H2O(g) + HOO.(g)}$\n\n\n$\\ce{HOO.(g) + .OH(g) -> H2O(g) + O2(g)}$\n\n\n$\\ce{.OH(g), HOO.(g)}$ are both radical compounds. \n\n\nIf you take a look at the first step in the mechanism, it is the rate limiting step and it is the same as your first step in the mechanism, the photolysis step. Therefore if $\\ce{H2O2}$ undergoes partial photolysis all that will happen is that the sample will undergo decomposition quicker. \n\n\nAdditionally, your mechanism of oxidation has been proposed before. You can take a look at this [article](http://www.mdpi.com/2073-4344/6/4/50/htm), in the section Mechanism of Oxidation by Peroxides: Radical or not Radical, That is the Question, in the end of subsection 4.1 and 4.2 they refer to the mechanism you proposed. They end up proving that it proceeds through a concerted mechanism that requires the presence of a suitable metal catalyst. It involves oxidative insertion or some type of $\\ce{C-H}$ insertion by the metal followed by oxidation of the product of that reaction. \n\n\nBut say the generation of hydroxyl radicals does work out, the corresponding alcohol may not be the only product that forms. Hydroperoxides have a tendency of forming as well. This is because hydrogen peroxide tends to decompose into oxygen and water, the oxygen can interact with the alkyl radical to form hydroperoxides. Along with the fact that the alcohol formed will tend to undergo oxidation in the presence of hydrogen peroxide and hydroxyl radicals.\n\n\nAlso if you are still going to try the reaction, the bond dissociation energy of the $\\ce{O-O}$ single bond is around 146 kJ per mole. So the energy per molecule is $\\ce{2.433\\times10^{-19} J}$. So the maximum wavelength of light to ionize this is $\\ce{8.21\\times10^{-7} m}$. Your calculation is a little off, it is in the near infrared range (you may have forgot to convert from kilojoules to joules or something like that). This is why hydrogen peroxide is stored in dark bottles and told to not be put in the presence of light. \n\n\n", "7" ] ]

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[ [ "\nMonomers must contain a functional group at each side (at least bifunctional) and these functions have to react with each other (like amine and carboxylic acid to yield amide group or alcohol and carboxylic acid to yield ester group,...) .\n\n\n\n> \n> 1. We can start with two types of monomers that hold identical functional groups at each side.\n> \n> \n> \n\n\n[](https://i.stack.imgur.com/gXLLh.gif)\n\n\nOne example of that is the reaction of 1,6-diaminohexane with hexandioic acid to yield Nylon-6,6.\n\n\n\n> \n> 2. We can also start from one type of monomers with two different functional groups at each side.\n> \n> \n> \n\n\n[](https://i.stack.imgur.com/tcAb8.gif)\n\n\nOne example of that is the reaction of 11-aminoundecanoic acid to yield Nylon-11.\n\n\nThe condition that the reaction should give off a small molecule is no longer valid. The counter example is the formation of polyurethane by reaction of diisocyanates and diols without giving off small molecules. We prefer name this type of polymerization \"a step growth polymerization\" in opposite to \"chain growth polymerization\" (known before as polyaddition). \n\n\n", "2" ], [ "\nCondensation polymerization can be shortened by having two monomers which are bifunctional that can react giving off small molecules like $\\ce{H2O}$ or methanol.\n\n\nThis can be exemplified by amino acids which have a carboxyl group ($\\ce{-COOH}$) and an amine group ($\\ce{-NH2}$). When making the polymer the carboxylic group reacts with the amine group of the neighboring moleucle giving a dimer which is bonded via an amide bond ($\\ce{-CONH-}$). This dimer has an amine group on one side and a carboxylic group on the other making it susceptible for further reaction thus making a trimer and a tetramer... and so on.\n\n\n", "1" ], [ "\nCarbon-carbon double bonds are not necessary for condensation polymerisation. The reaction usually involves the elimination of water. When two different functional groups are present on a molecule a polymer results, for example with $\\omega$-hydroxyhexanoic acid repeated condensation yields a polyester,\n$$\\ce{HO(CH2)COOH \\rightarrow HO-(CH2COO)\\_n-H}$$ \n\n\nWith two different bifunctional molecules are used such as an amine and an acid, such as adipic acid and hexamethylenediamine, nylon can be made;\n\n\n$$\\ce{ HOOC(CH2)\\_4COOH + H2N(CH2)\\_6NH2 \\rightarrow HOOC(CH2)\\_4CONH(CH2)\\_6NH2 + H2O}$$\nand this is followed by further steps where the initial (acid -amine) product reacts with more adipic acid. \n\n\nPolyesters, polyamides, polyurethanes and silicones are made this way. If more that two functional groups are present insoluble, infusible polymers such as Bakelite are formed.\n\n\nThe kinetics do not follow that of chain reactions, there is no initiation or termination and the general mechanism is $\\ce{M\\_j + M\\_k \\rightarrow M\\_{j+k}}$. The kinetics can be analysed by assuming that only the end groups are important, thus the rate constant is assumed to be independent of molecular size then the rate is found to be bimolecular, (at least for acid catalysed reactions) i.e. $\\ce{d[A]/dt=-$k$[A][B]}$ for reacting species A and B.\n\n\n", "1" ] ]

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[ [ "\nActually, the initial theories before Lewis suggested that $\\ce{H+}$ is the cause of acidity. However, it soon turned up that an ion as small as the nucleus of hydrogen (you may simply call it a proton) can't be created in low energy reactions due to its high polarising power. So, $\\ce{H+}$ is though the cause of acidic nature in aqueous solutions, $\\ce{H+}$ never exists as $\\ce{H+}$ but as $\\ce{H3O+}$, $\\ce{H9O4+}$, etc. It is because the proton is heavily hydrated. Though we are not very sure of what is the actual hydrated form, we usually refer it to as $\\ce{H3O+}$ or $\\ce{H+ (aq)}$.\n\n\n", "22" ], [ "\nAccording to [Wikipedia](https://en.wikipedia.org/wiki/Hydronium#Structure), the actual structure of hydronium was discovered using infrared spectroscopy.\n\n\nIt should be noted that $\\ce{H+}$ notation is still useful for reaction balancing, since it's much easier to count atoms in $\\ce{H+}$ instead of $\\ce{H3O+}$, even if the former ion may not really exist.\n\n\n", "1" ], [ "\nAs stated [here](https://books.google.co.uk/books?id=Mtth5g59dEIC&pg=PA349), the notion that (pure) water disassociates into some kind of ions is actually very misleading. Instead of thinking of liquid water as existing in separate molecules, it is far better to consider it as a vast network of hydrogen and oxygen atoms, where bonds are not in a binary state of 'formed' or 'broken' only, but have continuously varying strength depending more or less on the distance. It is then easy to understand how a charge imbalance can be very rapidly transferred to another location by just slight shifts in the positions of the atoms in the network. Unfortunately, this process is still called \"proton transfer\" despite the fact that there is no actual movement of a single proton across a long distance.\n\n\n", "0" ] ]

[ [ "\nReact the enolate of pinacolone (t-Bu Methyl ketone) with methyl vinyl ketone. It will first add 1,4 then the resulting enolate will do an aldol on the pinacolone ketone to give 3-tBu-cyclohexenone. React this with Phenyl cuprate and, despite the steric hindrance, it will add Michael-wise to give you the desired product. \n\n\n", "1" ] ]

[ [ "\nYou can think of $\\ce{NH4OH}$ as a hydrate: $\\ce{NH3\\_{(aq)}}$; no need for complicated reactions interconverting the two.\n\n\nConcerning the problem you described, at these concentrations all subtleties are negligible, and you basically only need to calculate the *nominal* concentrations of $\\ce{NH4+}$ and $\\ce{NH3}$, and plug them into the Henderson-Hasselbach equation.\n\n\nYou start with $\\ce{0.5 M NH3}$ and $\\ce{0.5 M NH4Cl}$, and add $\\ce{HCl}$ to a total nominal concentration of $\\ce{0.15 M}$, without altering the total volume (the question is written ambiguously, but that's what it means, most likely). As $\\ce{HCl}$ is a very strong acid, it will convert all $\\ce{NH3}$ to $\\ce{NH4Cl}$ (actually $\\ce{NH4+}$, because we are in solution). Therefore your final situation is: $\\ce{(0.5-0.15)=0.35 M NH3}$ and $\\ce{(0.5+0.15)=0.65 M NH4+}$. This should give you a pH of about 8.98 (lower than the one you would have without the HCl, consistently with the addition of an acid).\n\n\nSo your main error above was that you dissociated $\\ce{HCl}$ (correctly), but then just added the resulting $\\ce{H+}$ to the total, forgetting that there was a ton of $\\ce{NH3}$ around that would react with it.\n\n\nAnother example is when you add a solution of some particular transition metal salt to a solution of a weak acid (could be iron sulfate + hydrogen sulfide for instance). Despite the fact that the salt solution isn't acidic at all, the pH drops considerably compared to the solution of the weak acid alone, simply because the metal and the anion form a very insoluble salt that precipitates, subtracting weak acid anions from the solution and forcing the previously non-dissociated acid to dissociate.\n\n\nIn general, my advice is stick to what you know, don't go looking for complications: if they did not teach you to do very exact/sophisticated pH calculations, they probably don't expect you to pull them out of a magic hat at the exam. And as my stoichiometry teacher used to say, always write down all species you have in solution.\n\n\n", "2" ], [ "\nWell, first of all, as you have written before, the main reaction is the following one,\n\n\n$$\\ce{NH3 + H2O <=> NH4+ + OH-}$$ \n\n\nIn addition, you know the ammonia constant: $K\\_\\mathrm{b}=10^{-4.75}$\n\n\n(Note that if the question does not have a mention about equilibrium, it is because you don't need use the equilibrium in order to make the calculations)\n\n\nSo, let's start.\n\n\n1. Initially you have: \n\n\n$$[\\ce{NH3}] = [\\ce{NH4+}] = \\pu{0.5 M}$$\n\n\nThen, just taking into account the previous reaction\n\n\n$$\\ce{NH3 + H2O <=> NH4+ + OH-}$$ \n\n\nyou can get the result easily:\n\n\n\\begin{align}\nK\\_\\mathrm{b} &= \\frac{[\\ce{NH4+}][\\ce{OH-}]}{[\\ce{NH3}]} \\\\ \n[\\ce{OH-}] &= K\\_\\mathrm{b} \\frac{[\\ce{NH3}]}{[\\ce{NH4+}]} \\\\ \n[\\ce{OH-}] &= K\\_\\mathrm{b} = \\pu{10^{-4.75} M}\n\\end{align}\n\n\nand finally the pH would results as\n\n\n$$\\mathrm{pH} = 14 - \\mathrm{pOH} \\overset{\\mathrm{pOH} = 4.75}{\\to} \\mathrm{pH} = 9.25$$\n2. You add $\\pu{0.15 M}$ $\\ce{HCl}$ (acid), where it reacts with, obviously, the ammonia (base) which is found at the equilibrium, so\n\n\n\\begin{array}{ccccc}\n& \\ce{&NH3 &+ &H2O &<=> &NH4+ &+ &OH-}\\\\\n&i) &0.5 & &0.15 & &0.5 & &/ \\\\\n&f) &0.35 & &/ & &0.65 & &/\n\\end{array}\n\n\nNow, you may observe that both concentrations have changed, so we could obtain, using the same strategy as before we did: \n\n\n$$K\\_\\mathrm{b} = \\frac{[\\ce{NH4+}][\\ce{OH-}]}{[\\ce{NH3}]} \\to [\\ce{OH-}] = K\\_\\mathrm{b} \\frac{[\\ce{NH3}]}{[\\ce{NH4+}]} = 10^{-4.75}\\frac{0.35}{0.65} \\to [\\ce{OH-}] = \\pu{10^{-5.01} M}$$\n\n\nTo conclude, the $\\mathrm{pH}$ is given by\n\n\n$$\\mathrm{pH} = 14 - \\mathrm{pOH} \\overset{\\mathrm{pOH}=5.01}{\\to} \\mathrm{pH} = 8.98$$\n\n\nObserve that the buffer turns into more acid as much more acid is introduced in it.\n\n\n", "1" ], [ "\nI mainly want to rectify one point that you are confused with and that all the other answers fail to point out explicitly. You say in the beginning that ammonia should react with water according to $(1)$.\n\n\n$$\\ce{NH3 + H2O <=> NH4+ + OH-}\\tag{1}$$\n\n\nObviously, this is an acid-base reaction. We can deduce where the equilibrium will be just by checking out the $\\mathrm pK\\_\\mathrm a$ values of the two Brønsted acidic species, water and ammonium:\n\n\n\\begin{array}{lc}\\hline\n\\text{compound} & \\mathrm pK\\_\\mathrm a\\\\ \\hline\n\\ce{H2O} & 14\\\\\n\\ce{NH4+} & 9.25\\\\ \\hline\\end{array}\n\n\nAs we can see, ammonium is more acidic than water by almost five logarithmic units. Therefore, it is safe to assume that in a solution of ammonia in water only a very neglegible amount of ammonia is protonated. Thus, when mixing a $\\pu{0.5M}$ nominal solution of ammonia (which is very close to an actual $\\pu{0.5M}$ solution) to a $\\pu{0.5M}$ solution of ammonium chloride, we can indeed just use the Henderson-Hasselbalch equation to shortcut the way to the $\\mathrm{pH}$ value:\n\n\n\\begin{align}\\mathrm{pH} &= \\mathrm pK\\_\\mathrm a + \\lg\\frac{[\\ce{NH3}]}{[\\ce{NH4+}]}\\\\\n&= 9.25 + \\lg 1\\\\\n&= 9.25\\end{align}\n\n\nWhen we then in the second step add $\\pu{0.15M}$ of a strong acid, we can simply identify the strongest base in solution and consider $\\pu{0.15M}$ of that base additionally protonated. The strongest base around to any appreciable extent is ammonia, so we now have a new concentration of $\\pu{0.65M}$ of ammonium and $\\pu{0.35M}$ of ammonia. Thus, our new $\\mathrm{pH}$ value becomes:\n\n\n\\begin{align}\\mathrm{pH} &= \\mathrm pK\\_\\mathrm a + \\lg\\frac{[\\ce{NH3}]}{[\\ce{NH4+}]}\\\\[0.5em]\n&= 9.25 + \\lg \\frac{\\pu{0.35M}}{\\pu{0.65M}}\\\\[0.5em]\n&= 9.25 + \\lg 0.54\\\\\n&= 9.25 - 0.269\\\\\n&= 8.98\\end{align}\n\n\n", "1" ] ]

[ [ "\n[SMARTS](http://www.daylight.com/dayhtml/doc/theory/theory.smarts.html) is deliberately designed to be a superset of SMILES. That is, any valid SMILES depiction should also be a valid SMARTS query, one that will retrieve the very structure that the SMILES string depicts.\n\n\nHowever, as a query language, SMARTS can be more general than SMILES is. For example, `CC` as a SMILES string depicts a single compound: ethane. As a SMARTS query, though, `CC` will match ethane, but will also match propane, acetic acid, cyclohexane, vancomycin, etc.\n\n\nThere's also SMARTS strings which are not valid SMILES strings. You list several in your question: `[#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8]` is not a valid SMILES representation, unless your SMILES parser is being particularly generous. Even if it is, `[#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8,#7]` is also a valid SMARTS query which should match your molecule, but not even a generous SMILES parser is likely to accept it.\n\n\nDespite being similar, SMARTS and SMILES are intended for fundamentally different things - the purpose of SMILES is to represent particular compounds, whereas SMARTS represents a query against a range of possible molecules, or an abstract description of a set of possible molecules. As such, they're not inter-convertible, even if the strings are literally identical.\n\n\nFor your *particular* case, yes, `CC1CC=CC(=O)C1` is both a valid SMILES and a valid SMARTS, but as a SMARTS query, it represents not just 5-methyl-2-cyclohexen-1-one, but also 5-propyl-2-cyclohexen-1-one and 3-hydroxy-5-butyl-6-amino-2-cyclohexen-1-one, as well as many others, all of which contain that substructure. The SMARTS viewer you link doesn't depict this explicitly, because it's implicit in the use of SMARTS that it's a substructure pattern for a broader class of compounds.\n\n\n", "9" ], [ "\nThilo asked a similar question on the `rdkit-discuss` mailing list, where Andrew Dalke chimed in with this response, which he gave me permission to post here. The answer uses the python-based rdkit library to give examples of converting between SMILES and SMARTS and other tasks.\n\n\nFrom Andrew Dalke:\n\n\nOn Apr 19, 2017, at 12:03, Thilo Bauer wrote:\n\n\n\n> \n> Is converting SMARTS to SMILES a \"lossless\" operation, or does one loose\n> information on doing so?\n> \n> \n> \n\n\nIt is obviously not lossless if you include terms that cannot be represented in SMILES.\n\n\n\n```\n>>> from rdkit import Chem\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"[C,N]\"))\n'C'\n\n```\n\nor which don't make sense as a molecule:\n\n\n\n```\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"c\"))\n'c'\n>>> Chem.MolFromSmiles(\"c\")\n[23:02:24] non-ring atom 0 marked aromatic\n\n```\n\nIt also loses some information which could be represented in SMILES:\n\n\n\n```\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"[NH4+]\"))\n'N'\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"C[N+]1(C)CCCCC1\"))\n'CN1(C)CCCCC1'\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"[12C]\"), isomericSmiles=True)\n'C'\n\n```\n\nDo be careful if you want to handle aromatic atoms and bonds:\n\n\n\n```\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"[#6]:1:[#6]:[#6]:[#6]:[#6]:[#6]:1\"))\n'C1:C:C:C:C:C:1'\n>>> Chem.MolToSmiles(Chem.MolFromSmarts(\"c=1-c=c-c=c-c=1\"))\n'c1=c-c=c-c=c-1'\n\n```\n\n\n> \n> Background:\n> I've got three different SMARTS strings representing the same structure\n> - at least when depicting it. Also all three strings result in the exact\n> same SMILES (see code and output below).\n> \n> \n> \n\n\nIt looks like you want SMARTS canonicalization.\n\n\nIn general this is hard, because SMARTS can include boolean expressions and recursive SMARTS.\n\n\nIf you limit yourself to patterns like `'[#6]-1=[#6]-[#6]...'`, with only atomic numbers and single/double/triple bonds, then I think RDKit will do what you want.\n\n\n[[CF: Andrew Dalke also had important answer-level commentary on R.M.'s answer, which I copy below.]]\n\n\n\n> \n> From chemistry stack exchange, an answer contributed by user R.M.:\n> \n> \n> SMARTS is deliberately designed to be a superset of SMILES. That is, any valid SMILES depiction should also be a valid SMARTS query, one that will retrieve the very structure that the SMILES string depicts.\n> \n> \n> \n\n\nExcept, that last clause isn't true. Try matching tritium against itself.\n\n\n\n```\n>>> from rdkit import Chem\n>>> mol = Chem.MolFromSmiles(\"[3H]\")\n>>> pat = Chem.MolFromSmarts(\"[3H]\")\n>>> mol.HasSubstructMatch(pat)\nFalse\n\n```\n\nFor hydrogens you must use '#1', because H in SMARTS means something different.\n\n\n\n```\n>>> pat2 = Chem.MolFromSmarts(\"[3#1]\")\n>>> mol.HasSubstructMatch(pat2)\nTrue\n\n```\n\nSMILES input under Daylight and most other toolkits gets normalized to the chemistry model, including aromaticity perception:\n\n\n\n```\n>>> mol = Chem.MolFromSmiles(\"C1=CC=CC=C1\")\n>>> pat = Chem.MolFromSmarts(\"C1=CC=CC=C1\")\n>>> mol.HasSubstructMatch(pat)\nFalse\n>>> pat2 = Chem.MolFromSmarts(\"c1ccccc1\")\n>>> mol.HasSubstructMatch(pat2)\nTrue\n\n```\n\nRDKit also does a small amount of additional normalization, or 'sanitization' to use the RDKit term. For example, it will convert \"neutral 5 coordinate Ns with double bonds to Os to the zwitterionic form\" (see GraphMol/MolOps.cpp):\n\n\n\n```\n>>> s = \"CN(=O)=O\"\n>>> mol = Chem.MolFromSmiles(s)\n>>> pat = Chem.MolFromSmarts(s)\n>>> mol.HasSubstructMatch(pat)\nFalse\n>>> Chem.MolToSmiles(mol)\n'C[N+](=O)[O-]'\n\n```\n\nI believe that the output SMILES from a toolkit, assuming that the SMILES doesn't have an explicit hydrogen, can be used a SMARTS which will match the molecule made from that same SMILES, by that same toolkit.\n\n\n", "4" ], [ "\nIn this case, the SMARTS and SMILES represent the same (sub)-structure, in particular because the SMARTS pattern explicitly defines the single atom type and connectivity for all atoms. So to answer your question, yes, this case is a lossless conversion. \n\n\nIn contrast, if the `O` could be an `O` or an `N`, the SMARTS would not catch that. And this both your SMILES and SMARTS would miss `N=C1C=CCC(C)C1`. However, you can write a SMARTS to match both `N=C1C=CCC(C)C1` and `O=C1C=CCC(C)C1`, but a SMILES will not.\n\n\nAlso, note that the SMILES could be written in many (equivalent) ways, e.g., `O=C1C=CCC(C)C1`\n\n\n", "1" ] ]

[ [ "\nFrom a mole balance, we find $C = n\\_x(t) + n\\_y(t) + n\\_z(t) = n\\_x(t) + 2n\\_y(t)$, where the second equality follows from the stoichiometry of the reaction. In particular, our boundary conditions are $C = n\\_x(0) = 2n\\_y(\\infty).$ \n\n\nWe assume that the reaction is first-order, so $n\\_x(t) = n\\_x(0)\\exp(-kt) = C\\exp(-kt).$ The two measurements give us the set of equations \n\\begin{align\\*} \n80D &= n\\_x(5) + 5n\\_y(5) &&= C\\exp(-5k) + \\frac{5}{2}C(1-\\exp(-5k)) \\\\\n100D &= 5n\\_y(\\infty) &&= \\frac{5}{2}C,\n\\end{align\\*}\nwhere $D$ is an undetermined proportionality constant for the titration, and the coefficient of 5 in $n\\_y$ accounts for the fact that 2 equivalents of $A$ react with $n\\_y$ and 3 equivalents with $n\\_z$. Dividing one equation by the other produces the final result $$\\frac{4}{5} = \\frac{2}{5}\\exp(-5k)+1-\\exp(-5k),$$ which can be solved for $k$ and hence $t\\_{1/2}$.\n\n\n", "1" ], [ "\nThis is quite a tricky problem, and I'm not sure I have it all. The amounts of $X,Y, Z$ at time $t$ are $X=X\\_0\\exp(-kt), ~Y=X\\_0(1-\\exp(-kt)), ~ Z=X\\_0(1-\\exp(-kt))$ and the measurement is volume A. At infinite time $X\\_\\infty=0$ and so using the ratios given $2Y\\_\\infty+3Z\\_\\infty=5\\cdot 100$ is equivalent to $2X\\_0$ (by adding up $X,Y,Z$ at $t=\\infty$) and at $t=5, $ $X\\_5+2Y\\_5+3Z\\_5=6\\cdot80$ is equivalent to $X\\_0(2-\\exp(-5k))$ so you should have all that you need if you now take ratios as $(500/480)=2X\\_0/(X\\_0(2-\\exp(-5k)))$\n\n\n", "0" ] ]

[ [ "\nThe international standard ISO 105-E01:2013 *Textiles* – *Tests for colour fastness* – Part E01: *Colour fastness to water* includes specifications for two different solutions:\n\n\n**Alkaline solution,** freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre: \n\n\n* 0.5 g of ʟ-histidine monohydrochloride monohydrate ($\\ce{C6H9O2N3.HCl.H2O}$);\n* 5 g of sodium chloride ($\\ce{NaCl}$);\nand either\n* 5 g of disodium hydrogen orthophosphate dodecahydrate ($\\ce{Na2HPO4.12H2O}$)\nor\n* 2.5 g of disodium hydrogen orthophosphate dihydrate ($\\ce{Na2HPO4.2H2O}$).\n\n\nThe solution is brought to pH 8 (± 0.2) with 0.1 mol/l sodium hydroxide solution.\n\n\n**Acid solution,** freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:\n\n\n* 0.5 g of ʟ-histidine monohydrochloride monohydrate ($\\ce{C6H9O2N3.HCl.H2O}$);\n* 5 g of sodium chloride ($\\ce{NaCl}$);\n* 2.2 g of sodium dihydrogen orthophosphate dihydrate ($\\ce{NaH2PO4.2H2O}$).\n\n\nThe solution is brought to pH 5.5 (± 0.2) with 0.1 mol/l sodium hydroxide solution.\n\n\n", "4" ] ]

[ [ "\nWater in the food your trying to cook has a certain boiling point. This boiling point is elevated with increase in pressure (since under high pressure, vapors tend to condense back). To cook the food properly, you'd want the water to stay as liquid as much as possible, even under high temperatures, due to the fact that liquid water has a high specific heat capacity, storing lots of heat energy for a given temperature and can transmit heat to food better. Under pressure, boiling point of water is increased, so that it still remains liquid. This helps cook the food faster than without pressure.\n\n\n", "2" ], [ "\nI think that you are talking about pressurized pots.\n\n\nIf this is the case I think looking at the ideal gas equation will give us a hint on what happens. $$pV = nRT$$ Where $p$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the number of moles of the gas, and [$R$](https://en.wikipedia.org/wiki/Gas_constant) is the gas constant.\n\n\nIn the pressurized pot the volume is assumed to be constant, and this is a valid assumption as the amount of water that is vaporized is small and could be neglected in order to get a more qualitative result, if however we want a more detailed result, then we wouldn't neglect any thing and we would use a real gas equation.\n\n\nAs we increase the temperature, the number of moles of the vapor increases and they both increase the pressure of the water vapor. The increased pressure is the key for the result that we will get shortly.\n\n\nIf we looked at the phase diagram (specifically at the liquid phase) we see that as we see that as we increase the pressure the boiling point of water (or liquid) increases and this allows to get to higher temperatures that is not achievable within normal pressure.\n\n\n\n\n\nIn short, heating pressure pot increases the pressure above the water which increases the boiling point of water and allows reaching of higher temperatures which speeds up cooking.\n\n\n", "1" ] ]

[ [ "\nWe speak in this case of multiple equivalence points. \n\n\nAs a concrete example, if we are titrating $\\ce{H2SO4}$ with $\\ce{NaOH}$, then the first equivalence point occurs when all the $\\ce{H2SO4}$ has reacted to form $\\ce{HSO4-}$, and the second equivalence point when all the $\\ce{HSO4-}$ has further reacted to form $\\ce{SO4^2-}$. If you started with 1 mole of acid in solution, then adding 1 mole of base gets you to the first equivalence point, and another mole (2 moles total) gets you to the second equivalence point. In the same vein, one might consider triprotic acids, which have three equivalence points, and so on. \n\n\nIncidentally, an interesting case to think about might be that of a titration between a diprotic acid and a dibasic base.\n\n\n", "3" ] ]

[ [ "\n**Definition of internal energy:** U is a function of state, representing the total kinetic and potential energy of the molecules. U = U(T,V)\n\n\n**First Law of Thermodynamics:**$$\\Delta U=\\delta Q+\\delta W$$where the symbol $\\Delta$ is used to represent the change in a (path-independent) function of state (like U) between an initial and final thermodynamic equilibrium state of a closed system and the symbol $\\delta$ is used to represent the change in a parameter that depends on the process path between and initial and final thermodynamic equilibrium state of a closed system. $\\delta Q$ is the heat added to the system over the path, and $\\delta W$ is the work done on the system over the path.\n\n\n**Relationship for the Work:**$$\\delta W=-\\int{P\\_{ext}dV}$$where, for both for reversible and irreversible process paths, $P\\_{ext}$ is the force per unit area exerted by the gas on the piston face, and, by Newton's 3rd law, the force exerted by the piston face on the gas. For an irreversible process path, the pressure typically varies with spatial location within the cylinder, so that the average gas pressure does not match $P\\_{ext}$ at the piston face. For a reversible process, the gas pressure is uniform within the cylinder, so $P\\_{ext}=P$ where P is the gas pressure calculated from the equation of state of the gas (such as the ideal gas law), based on the number of moles in the cylinder, the gas pressure in the cylinder, and the gas temperature in the cylinder.\n\n\n**Combining the First Law with the Relationship for Work:**$$\\Delta U=\\delta Q-\\int{P\\_{ext}dV}\\tag{rev and irrev processes}$$\n$$\\Delta U=\\delta Q-\\int{PdV}\\tag{rev processes}$$\n\n\n**Constant Volume ($V\\_i=V\\_f)$:**$$\\delta W=0$$\n$$\\Delta U=\\delta Q$$\n\n\n**Constant Pressure ($P\\_{ext}=P\\_i=P\\_f=P$):**$$\\delta W=-P\\_{ext}\\Delta V=-P\\Delta V$$\n$$\\Delta U=Q-P\\Delta V$$\nSo,$$\\Delta U+P\\Delta V=\\delta Q$$But, since P is constant,\n$$\\Delta U+\\Delta (PV)=\\delta Q$$\nThe definition of enthalpy is $$H\\equiv U+PV$$\nTherefore, for a constant pressure process (a process in which $P\\_{ext}$ over the entire process path is equal to the equilibrium pressures in both the initial and final equilibrium states of the system) $$\\Delta H=\\delta Q$$\n\n\n", "4" ], [ "\nI believe that motivating the definition of enthalpy via the Legendre transform *is* indeed useful, and would like to attempt to do that here.\n\n\nWe start with the differential form of the first law, $\\mathrm{d}U = T\\,\\mathrm{d}S - p\\,\\mathrm{d}V$. We see that $S$ and $V$ are the natural variables for $U$, in that the differential $\\mathrm{d}U$ takes a simple \"natural\" form. In addition, if we're considering constant-volume processes, then $\\mathrm{d}V = 0$, and $\\mathrm{d}U = T\\,\\mathrm{d}S = \\delta q\\_\\text{rev}$, which is even simpler to work with.\n\n\nMany processes, however, occur at constant pressure rather than at constant volume, and we'd like to work with a thermodynamic variable that undergoes the same simplification as does $U$ at constant $V$. This can be done by the Legendre transform\\*: $$\\mathrm{d}H \\equiv \\mathrm{d}(U+PV) = \\mathrm{d}U + \\mathrm{d}(PV) = T\\,\\mathrm{d}S - p\\,\\mathrm{d}V + \\mathrm{d}(PV) = T\\,\\mathrm{d}S + V\\,\\mathrm{d}P.$$ Adding the differential $\\mathrm{d}(PV)$ on either side of the first law leads to a new thermodynamic variable $H$ that is a natural function of $S$ and $P$, instead of $S$ and $V$. We call $H$ the enthalpy. It easily follows that, for a constant-pressure process, $\\mathrm{d}H = T\\,\\mathrm{d}S = \\delta q\\_\\text{rev}$.\n\n\n\n\n---\n\n\n\\*One must be careful about adding arbitrary differential elements and calling it a Legendre transform. To be precise, the reason that adding the differential $\\mathrm{d}(PV)$ works is because $P(V)$ is monotonic; equivalently, the inverse function $V(P)$ exists; equivalently, a value of $P$ defines a single value of $V$ and vice versa. The Legendre transform works only between variables satisfying such a criteria, for otherwise information is lost during the transformation. \n\n\nIncidentally, that $P(V)$ is monotonic comes from a variational treatment of the second law. The definition of enthalpy thus relies on the second law, even though it is commonly introduced after solely the first law.\n\n\n", "3" ] ]

[ [ "\nThe reaction over here is a $\\text{S}\\_{\\text{N}}(\\text{Ar})$ reaction. Normally, nucleophilic substitutions do not occur, but in presence of strong nucleophiles and electron withdrawing groups, it is possible.\n\n\nNow, coming to your reaction, let's see what really happens:\n\n\nConsider the resonation structures of 4-chloro-3-bromonitrobenzene:\n[](https://i.stack.imgur.com/3Fcf4.png)\nYou can see that the resonance structures produce a carbocation at the chlorine attached carbon, but not at the bromine attached one. Nucleophiles always tend to attack electrophilic centers, so it is the chlorine that will get replaced.\n\n\nContinuing with the reaction, we will get:\n[](https://i.stack.imgur.com/Clfrh.png)\n\n\nSo, the product would be **3-bromo-4-methoxynitrobenzene**.\n\n\n", "4" ], [ "\nThis reaction is difficult to occur because it is a nucleophilic substitution to a benzene ring but in the presence of a strong electron withdrawing group this reaction moves ahead.\nThe point here to be noted is that -OEt is a strong nucleophile and attacks that carbon position where there is higher positive charge and thereby replacing the halogen out there.[](https://i.stack.imgur.com/bduhE.jpg)\n\n\nAnd thereby we obtain the product....\nHope this was helpful\n\n\n", "2" ] ]

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[ [ "\nNo. The reason for this is not to be found in the excitations, but in the evaluation of the method, i.e. the working equations.\n$$%Introducing some shortcuts\n\\require{cancel}\n\\newcommand{\\op}[1]{\\mathrm{#1}} %\\op{H}\n\\newcommand{\\bracket}[2]{\\left\\langle#1\\middle|#2\\right\\rangle}\n\\newcommand{\\bra}[1]{\\left\\langle#1\\right|}\n\\newcommand{\\ket}[1]{\\left|#1\\right\\rangle}\n\\newcommand{\\order}[1]{^{(#1)}} %E\\_n\\order{1}\n\\newcommand{\\overlap}[3]{\\mathcal{#1}\\_{#2}\\order{#3}} %\\overlap{S}{m}{1}\n\\newcommand{\\integral}[3]{\\mathcal{#1}\\_{#2,#3}} %\\integral{V}{i}{j}\n$$\n\n\n### The variational principle\n\n\nIn short: a trial wave function always has a higher energy than the exact wave function.\n\n\nAssume we know the exact solutions to the Schrödinger equation, which form a complete basis and there is one wave function $\\Psi\\_0$ which leads to the lowest energy $E\\_0$.\n\\begin{align}\n\\op{H}\\Psi\\_i &= E\\_i\\Psi\\_i & \n\\{i \\in \\mathbb{N}, i\\geq 0\\}\\tag{1}\\label{schrodinger}\n\\end{align}\n\n\nWe choose the solutions to be orthonormal (boundary conditions).\n$$\\bracket{\\Psi\\_i}{\\Psi\\_j} = \\delta\\_{ij} = \n\\begin{cases}1, & i=j \\\\ 0, & i\\neq j \\end{cases}\n\\tag2\\label{orthonorm}$$\n\n\nThe trial wave function $\\Phi$ can be expressed as a linear combination of the complete set of wave functions.\n$$\\Phi = \\sum\\_i^\\infty a\\_i\\Psi\\_i\\tag3\\label{expansion}$$\n\n\nThe expectation energy of the trial wave function is given through\n$$W = \\frac{\\bra{\\Phi}\\op{H}\\ket{\\Phi}}{\\bracket{\\Phi}{\\Phi}}.\n\\tag4\\label{expen}$$\n\n\nWith $\\eqref{expansion}$, $\\eqref{orthonorm}$, and $\\eqref{schrodinger}$ we can expand and simplify.\n\\begin{align}\nW &=\n \\frac{\n \\sum\\_i^\\infty\\sum\\_j^\\infty a\\_i a\\_j \\bra{\\Psi\\_i}\\op{H}\\ket{\\Psi\\_j}\n }{\n \\sum\\_i^\\infty\\sum\\_j^\\infty a\\_i a\\_j \\bracket{\\Psi\\_i}{\\Psi\\_j}\n }\\\\\n &=\n \\frac{\n \\sum\\_i^\\infty\\sum\\_j^\\infty a\\_i a\\_j E\\_j\\bracket{\\Psi\\_i}{\\Psi\\_j}\n }{\n \\sum\\_i^\\infty\\sum\\_j^\\infty a\\_i a\\_j \\bracket{\\Psi\\_i}{\\Psi\\_j}\n }\\\\\n &=\n \\frac{\n \\sum\\_i^\\infty a\\_i^2 E\\_i\\bracket{\\Psi\\_i}{\\Psi\\_j} +\n \\sum\\_{j\\neq i}^\\infty a\\_i a\\_j E\\_j\\bracket{\\Psi\\_i}{\\Psi\\_j}\n }{\n \\sum\\_i^\\infty a\\_i^2 \\bracket{\\Psi\\_i}{\\Psi\\_j} +\n \\sum\\_{j\\neq i}^\\infty a\\_i a\\_j \\bracket{\\Psi\\_i}{\\Psi\\_j}\n }\\\\\n &=\n \\frac{\n \\sum\\_i^\\infty a\\_i^2 E\\_i\n }{ \n \\sum\\_i^\\infty a\\_i^2\n } & | -E\\_0\\\\\nW - E\\_0 &=\n \\frac{\n \\sum\\_i^\\infty a\\_i^2 (E\\_i - E\\_0)\n }{ \n \\sum\\_i^\\infty a\\_i^2\n } \\\\\nW - E\\_0 &=\n \\frac{\n \\sum\\_{i>0}^\\infty a\\_i^2 (E\\_i - E\\_0)\n }{ \n \\sum\\_{i>0}^\\infty a\\_i^2\n } + \\frac{a\\_0^2 (E\\_0 - E\\_0)}{a\\_0^2}\\tag{4'}\\label{notnegative}\\\\\nW - E\\_0 &\\geq 0\\tag5\\label{varprin}\n\\end{align}\n\n\nIn $\\eqref{notnegative}$ the last term is zero. Since $E\\_i > E\\_0$ (by definition) and $a\\_i^2$ is always greater or equal zero, the expectation value of the trial wave function must always have a higher energy than the exact solution of the Schrödinger equation, i.e. $\\eqref{varprin}$.\n\n\n### Configuration Interaction\n\n\nThe CI wave function is set up as a linear combination of determinants from a reference calculation. This reference calculation can in principle be any other method the only requirement is that it also forms a complete (orthonormal) basis. \n\nWhen Hartree-Fock is chosen, we know that we obtain a complete set and that this method itself is variational. The CI trial wave function can then be expressed as a linear combination of solutions of the HF wave function.\n$$\\Psi^\\mathrm{CI} = \\sum\\_i^\\infty a\\_i\\Phi\\_i = \na\\_0\\Phi^\\mathrm{HF} + \\sum\\_{i>0}^\\infty a\\_i\\Phi\\_i$$\nAn alternative expression of the above is using an \"excitation operator\" $\\op{T}$ to generate the higher energy determinants from the reference.\n$$\\Psi^\\mathrm{CI} = (1 + \\op{T})\\Phi^\\mathrm{HF}$$\n\n\nThe energy of the CI trial wave function can be evaluated with the Lagrange multiplier method, which requires, that the wave function is normalised (adding zero).\n$$L = \\bra{\\Psi^\\mathrm{CI}}\\op{H}\\ket{\\Psi^\\mathrm{CI}} \n - \\lambda\\left[\\bracket{\\Psi^\\mathrm{CI}}{\\Psi^\\mathrm{CI}}-1\\right]$$\n\n\nSince the boundary conditions did not change, the variational principle still holds.\n\n\n### Coupled Cluster\n\n\nBecause of the inherent complexity of the CC ansatz, one key element to evaluate to CC energy is missing: the requirement of the normalised CC trial wave function. \n\nWe generate the trial wave function with an exponential approach (rather than a linear) expanded into a Taylor series.\n\\begin{align}\n\\Psi^\\mathrm{CC} \n&= \\mathrm{e}^{\\op{T}}\\Phi^\\mathrm{HF}\\\\\n&= \\left[\\op{1} + \\op{T}\\_1 + \\left(\\op{T}\\_2 + \\frac12\\op{T}\\_1^2\\right)\n + \\left(\\op{T}\\_3 + \\op{T}\\_2\\op{T}\\_1 + \\frac16\\op{T}\\_1^3\\right)\n + \\cdots \\right] \\Phi^\\mathrm{HF}\\\\\n&= \\Phi^\\mathrm{HF} + \\left[\\op{T}\\_1 + \\left(\\op{T}\\_2 + \\frac12\\op{T}\\_1^2\\right)\n + \\left(\\op{T}\\_3 + \\op{T}\\_2\\op{T}\\_1 + \\frac16\\op{T}\\_1^3\\right)\n + \\cdots \\right] \\Phi^\\mathrm{HF}\\\\\n\\end{align}\n\n\nOr to shorten the journey\n$$\\Psi^\\mathrm{CC} = \\Phi^\\mathrm{HF} + \\sum\\_i^\\infty t\\_i \\Phi\\_i.$$\n\n\nOne can easily see what kind of mess it would become requiring the wave function to be normalised $\\bracket{\\Psi^\\mathrm{CC}}{\\Psi^\\mathrm{CC}}=1$, so the Lagrange ansatz won't work. Instead we choose the wave function to be orthonormal to the reference, which in the Hartree-Fock (complete orthonormal basis) takes away a lot of work.\n\\begin{align}\n\\bracket{\\Phi^\\mathrm{HF}}{\\Psi^\\mathrm{CC}}&= 1&\n\\implies\\quad \\bracket{\\Phi^\\mathrm{HF}}{\\Phi\\_i} &= 0\n\\end{align}\n\n\nWe can write the CC-Schrödinger equation as\n$$\\op{H}\\mathrm{e}^{\\op{T}}\\Phi^\\mathrm{HF} = \n E^\\mathrm{CC}\\mathrm{e}^{\\op{T}}\\Phi^\\mathrm{HF}$$\nand evaluate the energy as \n\\begin{align}\n\\bra{\\Phi^\\mathrm{HF}}\\op{H}\\mathrm{e}^{\\op{T}}\\ket{\\Phi^\\mathrm{HF}}\n &= E^\\mathrm{CC}\n \\bracket{\\Phi^\\mathrm{HF}}{\\mathrm{e}^{\\op{T}}\\Phi^\\mathrm{HF}}\\\\\nE^\\mathrm{CC} \n &= \\bra{\\Phi^\\mathrm{HF}}\\op{H}\\mathrm{e}^{\\op{T}}\\ket{\\Phi^\\mathrm{HF}}\\\\\nE^\\mathrm{CC} \n &= E^\\mathrm{HF} + \n \\bra{\\Phi^\\mathrm{HF}}\\op{H}\n \\ket{\\left[\\op{T}\\_1 + \\left(\\op{T}\\_2 + \n \\frac12\\op{T}\\_1^2\\right)\n + \\left(\\op{T}\\_3 + \\op{T}\\_2\\op{T}\\_1 + \\frac16\\op{T}\\_1^3\\right)\n + \\cdots \\right]\\Phi^\\mathrm{HF}}.\\\\\n\\end{align}\n\n\nFrom this we can see that the evaluation of the expectation value of the energy is significantly different from what is required for the variational principle $\\eqref{expen}$.\n\n\nWithout going further into details, the dropping of the requirement of a normalised trial wave function leads to not obeying the variational principle.\n\n\n### Additional considerations\n\n\nIt is often said that Full-CC is equivalent to Full-CI, which is a convenient half truth. It is extremely important that this gets more true with a better description of the reference wave function, i.e. with larger basis sets. Full-CI is exact at the complete basis set limit (for the electronic Schrödinger equation). This is not necessarily true for Full-CC. \n\n\nSince we are in computational chemistry hardly ever concerned about absolute energies, the non-variational nature of the method is by far outperformed by the \"infinite\" approach generating the \"excited\" determinants. Relative energies don't suffer significantly (compared with the size consistency and extensivity) under the non-variational approach. \n\n\n", "13" ], [ "\nMy previous answer was downvoted (-3!), but I maintain that in the particular case where there are only two electrons, CCSD is equivalent to CISD and Full-CI.\nIf the discussion was about asking if for a one-electron system Coupled Cluster gives an upper bound to the energy, everyody would agree that it is true. I agree with all that was said in the top answers for the general case, but with two electrons things are a bit special since there are no triples and quadruples. \n\n\n1) The CCSD equations can be rewritten in such a way to be solved with an intermediate Hamiltonian in the space of singly and doubly excited determinants.[1] In the special case where there are 2 electrons the space is the FCI space, and the equations of FCI, CISD and CCSD are strictly equivalent : the dressing matrix $\\Delta$ is zero so the effective Hamiltonian is equal to the Hamiltonian.\n\n\n2) I don't agree with the argument that the fact that the energy is computed by projection implies that the energy is non-variational. What makes the result usually approximate is the fact that the wave function is not an eigenfunction of the Hamiltonian. But when it is, computing the energy by projection is exact. In the special case where we have 2 electrons, the CISD is the Full-CI. So if one projects the equations on the singles and doubles, the Full space is considered and the FCI wave function is obtained (up to a normalization factor). In the FCI, the wave function *is* an eigenfunction of the Hamiltonian, so computing the energy by projection gives exactly the FCI energy (in a less numerically stable manner in practice).\n\n\n3) $\\Psi\\_{\\rm CCSD} = \\exp^{T\\_1+T\\_2} \\Phi\\_{\\rm HF} = (1 + (T\\_1 + T\\_2) + \\frac{1}{2}(T\\_1+T\\_2)^2 + \\dots) \\Phi\\_{\\rm HF}$\n\n\nWith only 2 electrons,\n\n\n$\\Psi\\_{\\rm CCSD} = (1 + T\\_1 + T\\_2 + \\frac{1}{2}T\\_1^2) \\Phi\\_{\\rm HF}$\n\n\nand $\\Psi\\_{\\rm FCI} = (1+T\\_1+T\\_2)$\n\n\nBoth $T\\_1^2\\Phi\\_{\\rm HF}$ and $T\\_2 \\Phi\\_{\\rm HF}$ span the same determinant space, so finding the $\\Psi\\_{\\rm CCSD}$ and the $\\Psi\\_{\\rm FCI}$ which minimize the energy will give the same solution in the basis of determinants.\nAll the additional constraints of Coupled Cluster are on the amplitudes of the Triples and Quadruples, which are absent here.\n\n\n\n\n---\n\n\n[1] Eigenvalue problem formulation of coupled-cluster expansions through intermediate Hamiltonians, I Nebot-Gil, J Sanchez-Marin, J.L Heully, J.P Malrieu, D Maynau, Chemical Physics Letters, Volume 234, Issues 1–3, 3 March 1995, Pages 45-49, <https://doi.org/10.1016/0009-2614(95)00026-Z>\n\n\nNote: In the paper the energies they obtain are slightly different from standard CCSD, but this is due to an approximation on the $T\\_1\\times T\\_2$ terms, which don't exist with only two electrons.\n\n\n", "6" ], [ "\nIn the most general case, CC can be understood simply as a prescription for a trial wave function (ansatz) that uses the exponential excitation operators. This ansatz can be then optimized variationally, and this is variational CC.[1] It is much more expensive than the common CC that uses another approximation (besides the ansatz), which linearizes the CC amplitude equations in a certain way. In most cases, this is a good approximation.\n\n\n\"Variational\" means that a method has some ansatz for the wave function which is optimized (varied) to get the minimal energy. This is then guaranteed to be greater or equal to the true energy by the variational principle. In this sense, the common CCSD is not variational even for two-electron systems: there is no energy that would be optimized in the procedure.\n\n\nPerhaps you could understand \"variational\" as a method that is in some way guaranteed to always give an energy greater or equal to the true energy. In general, the non-variational CC is not such a method. But for two-electron systems, the CCSD energy is exact, so technically, it is variational in this other derived sense. But as it is exact, which is a stronger statement, I see little sense in stressing the \"variational\" aspect. Also, I know of no method that would be variational in this second sense, but not in the first sense.\n\n\nReference \n\n\n1. [Troy Van Voorhis and Martin Head-Gordon. *J. Chem. Phys.* **2000,** *113*, 8873.](http://dx.doi.org/10.1063/1.1319643)\n\n\n", "5" ], [ "\nYes,\nin the particular case of 2 electrons, CCSD is strictly equivalent to CISD since there can't be any excitation higher that singles and doubles. CISD is the Full-CI of a 2-electron system, so CCSD is variational for 2 electrons.\n\n\n", "0" ] ]

[ [ "\nSo the example you present is of a poor question. It's a classic case of a \"best answer\" MCQ that requires you pick the answer that is most accurate, rather than an answer that is unambiguously right. (I *hate* these.) Alkanes are listed in most textbooks and guides to organic chemistry as a \"functional group.\" The reality is that it's hard to do chemistry on an alkane in most circumstances, so it's probably the least \"functional\" of the functional groups.\n\n\nThat said a side chain is an imprecise term. It's most consistently used in amino acid chemistry to refer to the differentiating moiety. But it's really used to indicate any moiety that hangs off of a common structure. What constitutes the side chain versus the primary structure is a function of context. A large polymeric structure can have side chains that consist of aromatic rings, while with a smaller structure consisting of a single benzene ring with some alkanes hanging off of it, the alkanes will be considered the \"side chain.\" There's no substantive difference between a side chain and a functional group except for emphasis in a discussion. \n\n\nSo in brief: Functional groups are the things you learned about like amines, amides, carboxylic acids, etc. Side chains are parts of a molecule that are considered to be \"hanging off\" the \"main\" part of the molecule, and are highly dependent on the context of the discussion. Sometimes these side chains are single functional groups, and sometimes they're much bigger structures. It really depends. But functional groups are always functional groups. Though in general, chemists tend to \"forget\" alkanes as functional. But as I indicated earlier, the question isn't all that great, so don't feel too bad if it confused you.\n\n\n", "5" ] ]

[ [ "\nI finally found the answer. \n\n\nIn temporary hardness, we have to remove $\\ce{Mg^2+}$ ions by precipitating it. In $\\ce{Mg(OH)2}$, $\\ce{Mg^2+}$ ions in water are present in the concentration that is cube root of the solubility product, but in $\\ce{Mg(CO3)}$ it is the square root of the solubility product. So, despite $\\mathrm{K\\_{sp}}$ of $\\ce{Mg(OH)2}$ being higher than $\\ce{Mg(CO3)}$, $\\ce{Mg^2+}$ ions in $\\ce{Mg(OH)2}$ are comparatively less than $\\ce{Mg^2+}$ ions in $\\ce{Mg(CO3)}$.\n\n\n", "14" ], [ "\nThe solubility data for $MgCO\\_3$ and $Mg(OH)\\_2$ are these: for $MgCO\\_3$ 0.01 g/100 mL cold H2O; for $Mg(OH)\\_2$ 0.0009 g/100 mL H2O @ 18C, 0.004 g/100 mL H2O @ 100C (CRC Handbook).\n\n\nThe published Ksp data are all over the place: for MgCO3, I have found $10^{-5}$, $6.8× 10^{-6}$, and $ 3.5 × 10^{-8} $(twice). For Mg(OH)2, in the same set of 4 documents, I have $ 1.8 ×10{-11}$ (twice), $1.6 × 10^{-12} $ and $5.6 × 10^{-12}$\n\n\nCalculating Ksp from the solubility data gives: for Mg(OH)2 @ 18C, $1.5 × 10^{-11}$ and @ 100C, $1.31 × 10^{-9}$. For MgCO3, Ksp calculates to $1.4 × 10^{-6}$.\n\n\nFrom the solubility data, Mg(OH)2 is by far less soluble than MgCO3. From the Ksp data, any way you look at them, the solubility product of Mg(OH)2 is smaller than for MgCO3, indicating a lower solubility. So, no conflict.\n\n\nThere is a complexity in comparing a solubility product of MgCO3 (which multiplies 2 concentrations) vs Mg(OH)2, which multiplies 3 concentrations. Not to mention the fact that solubilities and Ksp are given for hydrates of MgCO3, which I did not deal with. And you might consider that it is not we who are removing $Mg^{+2}$ ions from the temporary hard water, but that heating removes CO2 from the water, leading to production of Mg(OH)2 in solution and then precipitation. The loss of CO2 minimizes the possible precipitation of MgCO3, especially since it is more soluble at high temperatures, where the transformation is occurring.\n\n\n", "7" ] ]

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[ [ "\n\"The problem I face is I have been taught that the pure liquids aren't included in the expression of $\\mathrm{K}\\_{c}$ since they have a concentration of 1.\"\n\n\nLiquids have a constant concentration (really activity), so in general, it is not helpful to include them in the equilibrium expression. However, that does not mean that the value is 1. Pure water has a concentration of roughly $55.5\\ \\mathrm{M}$. You can absorb this concentration into the equilibrium constant, since it is a constant. Therefore,\n\n\n$$\\mathrm{K}\\_{w} = \\ce{[H+][OH-]} = K\\_{\\mathrm{c}}\\ce{[H2O]}$$\n\n\nEDIT:\n\n\nJust for the record, note that the autoionization of water is generally written:\n\n\n$$\\ce{H2O + H2O <=> H3O+ + OH-}$$\n\n\nsince we don't have naked protons in water.\n\n\nThis changes the equilibrium constant slightly:\n\n\n$$\\mathrm{K} = \\frac{\\ce{[H3O+][OH-]}}{\\ce{[H2O]}^{2}}$$\n\n\nbut $\\mathrm{K}\\_{w}$ is still the same.\n\n\n", "1" ] ]

[ [ "\nI believe this to be a simple typo in the book. In the [original image](https://i.stack.imgur.com/suJ5P.jpg) you'll see that in the surrounding schemes $<$ and $\\to\\text{Stability}$ are given. Only in this line the relations are reversed, while the stability indicator remains the same.\n\n\nHowever, since we're at it, we can at least try to put some values to the task. I'll adopt the same scheme used in the comparison of the [*t*-butyl cation and the benzyl cation](https://chemistry.stackexchange.com/q/74943/4945), or the [radicals](https://chemistry.stackexchange.com/q/26570/4945). I calculated the isodesmic reactions of the form in $\\eqref{isodesmic}$ at the DF-B97D3/def2-TZVPP level of theory.\n$$\\ce{ R+ + CH4 -> RH + H3C+ }\\tag{1}\\label{isodesmic}$$\n\n\nAccording to this we'll find the following order in decreasing stability: *t*-butyl, benzyl, allyl, (ethyl). Note that the ethyl cation is a non-classical cation, basically a proton coordinating to the π-bond of ethene; see also: [Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?](https://chemistry.stackexchange.com/q/29000/16683).\n\n\n\\begin{array}{llr}\n \\ce{R+} & \\ce{RH} & \\Delta G / \\pu{kJ mol-1}\\\\\\hline\n \\ce{H3C+} & \\ce{CH4} & 0.0 \\\\\n \\ce{[H2C=CH2]H+} & \\ce{H3C-CH3} & -197.9 \\\\\n \\ce{H2C=CH-CH2+} & \\ce{H3C-CH=CH2} & -259.3 \\\\\n \\ce{H5C6-CH2+} & \\ce{H2C-C6H5} & -343.0 \\\\\n \\ce{(H3C)3C+} & \\ce{HC(CH3)3} & -370.9 \\\\\\hline\n\\end{array}\n\n\n*(Side note: I was unable to remove a small imaginary mode from the isopropyl carbocation. That has almost no influence though.)*\n\n\n[](https://i.stack.imgur.com/U0bnn.jpg) [](https://i.stack.imgur.com/OmFJJ.jpg) \n\n[](https://i.stack.imgur.com/mBrw4.jpg) [](https://i.stack.imgur.com/embkP.jpg) \n\n[](https://i.stack.imgur.com/znnmH.jpg) [](https://i.stack.imgur.com/za1p0.jpg) \n\n[](https://i.stack.imgur.com/cBwKN.jpg) [](https://i.stack.imgur.com/dKkdJ.jpg) \n\n[](https://i.stack.imgur.com/r8Pcu.jpg) [](https://i.stack.imgur.com/xqOJU.jpg) \n\n\n(I won't attach geometries or absolute energies this time, because that would exceed the character limit.)\n\n\n", "11" ], [ "\nAs per Jerry March in *Advanced Organic Chemistry*, the stability order of the cations is benzyl > allyl > t-butyl as expected due to resonance. The ethyl cation does not even exist because it is so unstable that when it is tried to be made, the t-butyl cation is obtained instead. Same with methane and propane. \n\n\nA special case is the cation with cyclopropyl substituents wich is still more stable than the benzylic cation because the bended *p* orbitals stabilize the cation. \n\n\n", "3" ] ]

[ [ "\nYou likely refer to remission, especially if you would collect spectroscopic data with an [integration sphere](https://en.wikipedia.org/wiki/Integrating_sphere) at hand. More generally, you refer to [reflectance](https://en.wikipedia.org/wiki/Reflectance), which depends on the material, the selected wavelength, and the roughness of the top surface -- assuming the selected wavelength is not able to penetrate all across the reflective layer.\n\n\n[](https://i.stack.imgur.com/aVkxe.png)\n\n\n([source](https://en.wikipedia.org/wiki/Reflectance))\n\n\nMetals and non-metals (e.g. polymer lamp housings) may be covered by a layer of chromium, and then may be polished to obtain a very smooth surface; in contrast to other metals such a surface remains \"bright\", even when exposed to air.\n\n\nBeside the silver mirror already known from sugar chemistry, [mercury](http://www.marcmaison.com/architectural-antiques-resources/mercury-mirror) / almagam based mirrors were quite common till about 1800. The liquid nature of mercury renders this material still useful in the niche of [liquid mirror telescopes](https://en.wikipedia.org/wiki/Liquid_mirror_telescope). \n\n\nGlancing *briefly* over the *(n,k)* values of metals provided [here](https://www.photonics.com/EDU/Handbook.aspx?AID=25501) provides access to calculate the reflectance $R$ as the materials in question \n\n\n$$ R = \\frac{(n-1)^2 + k^2}{(n+1)^2 + k^2}$$\n\n\nMatching the experience to see much more often silver top-coated mirrors in the optics lab, than of other materials, silver is of higher reflectance than chromium:\n\n\n[](https://i.stack.imgur.com/Crm15.png)\n\n\n(The value at 10 µm equally provided by the referenc intentionally was omitted by mine.) \n\n\nSo why spokes are chromium plated, and not with silver? I speculate the much higher hardness of chromium (Mohs 8.5) than the one of silver (Mohs 2.5) contributes to the reasons why the former is used, and not the later. It remains a smooth surface. Maybe electrochemical reasons (chromium likely already is part of the steel / alloy the rims are made of; rather than silver) and the ease Ag would be tarnished by the environment (thinking about $\\ce{Ag2S}$) are other reasons, beside obvious cost in material and processing.\n\n\n", "13" ] ]

[ [ "\n[Sign conventions](https://en.wikipedia.org/wiki/Work_(thermodynamics)#Sign_convention) can make 'work' confusing, but the question here asks for the work done *by* the gas. When a gas expands, it's *doing* work (it's *losing* energy), so it's actually $W = + P \\, \\mathrm dV$, inverting all the inequality signs. So option B) is correct.\n\n\n", "3" ], [ "\nWork done *by* the gas and *on* the gas are two different things.\n\n\nIn your question, the volume is increasing and the pressure is decreasing.\n\n\nLet's consider some things before making an attempt to answer your question,\n\n\nFor simplification, consider the adiabatic process where $Q=0$.\n\n\nIf the external agent does positive work, the volume always decreases.\n\n\nBut in your question, as the volume is increasing, the external agent isn't the one doing the positive work, its actually the gas because only the gas can expand the system and still do positive work.\n\n\nIn the first law of thermodynamics,\n\n\n$\\mathrm dU=Q+W$\n\n\n$\\mathrm dU=Q-p\\,\\mathrm dV$, \n\n\nThe term $W$ represents the work done by the external agent.\n\n\nIf the external agent does positive work, volume decreases (i.e, $\\mathrm dV$ is negative => $W$ is positive) and the internal energy increases and vice versa.\n\n\nSo, the work done by the gas becomes $-(-p\\,\\mathrm dV)$, because for a reversible process the work done by external agent and the gas are equal in magnitude and opposite in sign. \n\n\nNow, as $\\mathrm dV$ is positive, work done *by* the gas in also positive.\n\n\nAs you can see, magnitude wise $W\\_3$ is the greatest and now we also know that the net positive work is done *by* the gas because the final volume is greater.\n\n\nThis should answer your question.\n\n\n", "1" ] ]

[ [ "\nMany commercial detergents are composed of pH-neutral surfactants like alkyl polyethylene oxides plus alkaline additives like sodium silicate or carbonate or borax, which help dissolve greasy materials better than acids. \n\n\nThere are some acidic detergents, like citric acid cleaners for stainless steel and other metals and toilet bowl cleaners. The acidic function is paramount, and the reduced surface tension is secondary, useful in providing better wetting.\n\n\nIt is instructive to consider what cleaning really means, and the individual steps that go on. First, water is the primary cleaning agent. It is the universal solvent, but not quite good enough or fast enough. So we add surface tension lowering agents - big improvement: oily materials are wetted and can be suspended better as the first step in removal. Alkaline additives then combine with polar surfaces to further displace oily materials, improving cleaning action. Acids don't help with oily materials much but can remove some contaminants better.\n\n\nDispersants for solids, like mud or clay, are typically anionic polymers, usually alkaline because the anion contains groups that will adsorb on polar surfaces and disperse the particles by electrostatic repulsion, assisting their removal.\n\n\nThere are classes of surfactants where the oleophilic group is anionic, like sulfonates or phosphonates, and these can be acidic if the cation is hydrogen or alkaline if an alkali metal, or nearly neutral if the cations are mixed. Another class of surfactants, called cationic, has the oleophilic group attached to nitrogen, and is typically somewhat acidic because the anion is the anion of a strong acid like HCl. \n\n\nSo you could optimize your cleaner by selecting a pH of water that cleans best; then add a specific surfactant to improve removal of your specific dirt - or the reverse. And consider interactions and whatever other additives might help.\n\n\n", "2" ] ]

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[ [ "\nIf we take small particles of a sparingly soluble salt, say AgCl, then one or two particles may dissolve in water and a saturated solution is obtained pretty soon. Upon adding extra AgCl to the solution precipitation of AgCl starts. At this stage, you'll find an equilibrium is established between the dissolved and the undissolved AgCl.\n\n\n$$\\ce{AgCl(s) <=> Ag+ (aq.) + Cl- (aq.)}$$\n\n\nAnd for this equilibrium if you're to find the equilibrium constant, let's call it $K\\_{sp}$, you get:\n\n\n$$ \\ce{K\\_{sp} = [Ag+] \\cdot [Cl-]} $$\n\n\n(Both the concentrations are the ones taken *at* equilibrium.)\n\n\nThis $K\\_{sp}$ is called solubility product. It's useful to predict whether or not more solute can be dissolved in your solution or not. We do this by comparing it to something which is called the ionic product.\n\n\n", "4" ], [ "\nThe solubility product is an equilibrium constant for the dissolution of a sparingly soluble but highly ionised electrolyte. It does not apply in principle to strong soluble electrolytes or to compounds that exist in an unionised state in solution, such as $\\ce{HgCl2}$. \n\n\nIn a reaction such as \n$$\\ce{AgCl <=>Ag^+ + Cl^-}$$ \nwhich must have reached equilibrium, which means that some solid is present together with the solution (and so the solution is saturated), the solubility product is \n$$\\ce{[Ag^+][Cl^-]=}K\\_s $$ \n\n\nIn the general case $\\ce{A\\_nB\\_M <=> nA + mB }$ then $\\ce{[A]^n[B]^m=}K\\_s$\n\n\n(The thermodynamic derivation of $K\\_s$ depends on using activities which may be replaced by concentration when the concentration is low as it is in the case of sparingly soluble / insoluble salts)\n\n\nThe solubility product illustrates a quantitative application of Le Chatelier’s principle. If, for example, the concentration of $\\ce{Cl^-}$ is increased by adding sodium chloride some AgCl is precipitated until equilibrium is re-established, i.e. $\\ce{[Ag^+]}$ is reduced until $\\ce{[Ag^+][Cl^-]=$K\\_s$} $. Similarly if $\\ce{[Ag^+]}$ in solution is reduced by forming a soluble complex, such as by adding ammonia, then more $\\ce{Cl-}$ will be produced as more solid AgCl dissolves until equilibrium is established. \n\n\nThe solubility product $K\\_s$ can be used to calculate the concentration of an ion in solution and hence its total amount in a solution. In the case where there is a common ion it can be used to determine how solubility varies with common ion concentration. Thus it can be used to answer questions such as 'How much $\\ce{Ag^+}$ is there in 100 ml of a solution of 0.1 M potassium chromate saturated with silver chromate'.\n\n\n", "1" ] ]

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[ [ "\nNo designations such as alpha-oryzanol, beta-oryzanol, etc. appear to be in use, and according to [this WebMD entry](http://www.webmd.com/vitamins-supplements/ingredientmono-770-gamma%20oryzanol.aspx?activeingredientid=770), as well as [this PubChem document](https://pubchem.ncbi.nlm.nih.gov/compound/6450219#section=Top), another name for gamma-oryzanol is simply \"oryzanol\". \n\n\nThere seems to be some ambiguity as to just what oryzanol, or gamma-oryzanol refers to. According to [Chemical Book](http://www.chemicalbook.com/ProductChemicalPropertiesCB5758140_EN.htm): \n\n\n\n> \n> Gamma Oryzanol is a mixture of 6 substances derived from rice bran oil, including sterols and ferulic acid, called Cycloartanyl Ferulate or Triterpene alcohol ferulate. \n> \n> \n> \n\n\nBut then other sources, including [this US National Institutes of Health](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4377434/) publication, consider gamma-oryzanol composed of four components: \n\n\n\n> \n> There are four main components of γ-oryzanols, namely, cycloartenyl ferulate, 24-methylene cycloartenyl ferulate, campesteryl ferulate, and sitosteryl ferulate [original publication: *Lerma-García M. J., Herrero-Martínez J. M., Simó-Alfonso E. F., Mendonça C. R. B., Ramis-Ramos G. Composition, industrial processing and applications of rice bran γ-oryzanol. Food Chemistry. 2009;115(2):389–404. doi: 10.1016/j.foodchem.2009.01.063* \n> \n> \n> \n\n\nFurther confusing the issue is that the PubChem article referenced above gives a single structure for gamma oryzanol (the same as that given in the OP), but does not include any physical properties that one would expect for a single pure compound (i.e. no melting point, solubility, vapor pressure data) although it does give the molecular weight. Again, PubChem lists \"oryzanol\" as a synonym and does not have any other oryzanol-named compound in it's database.\n\n\n**Summary TL/DR:** \n\nThe gamma in \"gamma-oryzanol\" doesn't refer to a specific structure or isomer and there is no corresponding compounds named alpha-oryzanol, beta-oryzanol, etc. Rather, the term gamma-oryzanol is also simply called \"oryzanol\" and it refers to a combination of several different moieties as described above.\n\n\n", "8" ], [ "\nSome works directly mention the alpha and beta designated substances – that were (incl. gamma) rather fractions obtained during first historic isolations:[1]\n\n\n\n> \n> (…) Oryzanol occurs in the unsaponifiable fraction of rice bran oil and is so named because it was first discovered in rice bran oil (*Orysae Sativa L.*) (Kaneko and Tsuchiya, 1954)[2] and contained a hydroxyl group. Oryzanol was originally considered a single compound but later was determined to be a mixture of ferulic acids esterified with normal sterols or triterpene alcohols, called **α-**, **β-** and **γ-oryzanol**, of which γ-oryzanol has been the most commonly mentioned. The sterol components of γ-oryzanol are primarily campesterol and sitosterol, and the triterpene alcohol components are cycloartenol and 24-methylene cycloartanol (…)\n> \n> \n> \n\n\nOther text sound like alpha a beta fractions were not oryzanol at all:[3]\n\n\n\n> \n> (…) During their studies, a number of lipid fractions were isolated from rice bran oil; the third (**gamma**) fraction isolated was **oryzanol**[2]. Its chemical composition was later identified as a mixture of \n> ferulic acid esters of various plant sterols (…)\n> \n> \n> \n\n\n(Hard to tell, the article[2] is in Japanese.)\n\n\nAs airhuff already said in his answer, **γ-oryzanol** (or just **oryzanol**) is a mixture of various sterol [ferulates](https://en.wikipedia.org/wiki/Ferulic_acid), which now usually are listed by their chemical names, however a couple of notable compendia[4,5] list the two major[6] formerly isolated constituents under names **oryzanol A** and **C** (no B, since it turned out to be a mixture of A and C[7]).\n\n\n\n\n\n* (I) **oryzanol A**, CAS 21238-33-5, cycloartenyl ferulate, \n\n9,19-cyclo-5α-lanost-24-en-3β-yl (2*E*)-3-(4-hydroxy-3-methoxyphenyl)prop-2-enoate\n* (II) **oryzanol C**, CAS 469-36-3, 24-methylenecycloartanyl ferulate, \n\n24-methylidene-9,19-cyclo-5α-lanostan-3β-yl (2*E*)-3-(4-hydroxy-3-methoxyphenyl)prop-2-enoate\n\n\nYour structure happens to be **oryzanol A**.\n\n\n\n\n---\n\n\nReferences:\n\n\n1. Huang, C. J. [Potential Functionality and Digestibility of Oryzanol as Determined Using in Vitro Cell Culture Models](https://digitalcommons.lsu.edu/gradschool_dissertations/261), Dissertation thesis, Louisiana State University and Agricultural and Mechanical College, **2003**.\n2. Kaneko, R.; Tsuchiya, T. [New Compound in Rice Bran and Germ Oils](https://doi.org/10.1246/nikkashi1898.57.526). *J. Chem. Soc. Jpn.* **1954**, *57*, 526–529.\n3. Wheeler, K. B.; Garleb, K. A. [Gamma Oryzanol—Plant Sterol Supplementation: Metabolic, Endocrine, and Physiologic Effects](https://doi.org/10.1123/ijsn.1.2.170). *International Journal of Sport Nutrition* **1991**, *1* (2), 170–177.\n4. O’Neil, M. J.; Heckelman, P.; Koch, C.; Roman, K. *The Merck Index*, 14th ed.; **2006**.\n5. Hill, R. A.; Makin, H.; Kirk, D.; Murphy, G. Dictionary of Steroids; CRC Press, **1991**.\n6. Xu, Z.; Godber, J. S. [Purification and Identification of Components of γ-Oryzanol in Rice Bran Oil](https://doi.org/10.1021/jf981175j). *J. Agric. Food Chem*. **1999**, *47* (7), 2724–2728.\n7. Shimizu, M.; Ohta, G. [Studies on the Constituents of Rice Bran Oil. V. Reexamination of Oryzanol-B](https://doi.org/10.1248/cpb.8.108). *Chem. Pharm. Bull.* **1960**, *8* (2), 108–111.\n\n\n", "3" ] ]

[ [ "\nSolid allotropes of pretty much *anything* are different phases. They have different crystal structures, and you can't have a single phase with two different crystal structures inside it.\n\n\nAs for the other criteria, some of them are sufficient but not necessary, and some are difficult to check. The distinct boundary must be there, but that's a knowledge that comes from them being two phases and not vice versa. Physical properties are not all that different, though you can tell one allotrope from another by density. (We used a solution of $\\ce{BaI2}$ to this end.)\n\n\nChemical properties are even more similar since the molecules are the same. I doubt that a purely chemical test can be devised. But then again, do we really need it?\n\n\n", "2" ] ]

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[ [ "\nElectricity in a wire submerged in water effectively turns it into an [electrochemical cell](https://en.wikipedia.org/wiki/Electrochemical_cell), and then it is no surprise that the anode gets oxidized pretty quickly. A modest potential of a few volts would suffice to oxidize any metal, even gold. This works for AC as well (a wire oxidizes during the positive half-wave and then does nothing during the other half-wave).\n\n\n", "4" ] ]

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[ [ "\nI actually take issue with the question (which was asked of you not by you), as I think this oversimplifies a molecule, that still isn't well understood in the first place.\n\n\nThat is why I find this question incredibly difficult to answer and why I find that [Nuclear Chemist's answer](https://chemistry.stackexchange.com/a/98947/4945) dangerous, because it propagates myths about bonding that have been disproved already.\n\n\nBut enough of the commentary, let's clear up some issues.\n\n\n\n> \n> \n> > \n> > Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\\ce{ClO2}$.\n> > \n> > \n> > \n> \n> \n> $\\ce{ClO2}$ has two $\\sigma$ bonds, one lone pair, two π bonds and one odd electron.\n> \n> \n> \n\n\nThere comes a time, where you should realise that hybridisation is not always a useful concept to understand bonding. This time is now.\n\n\nYour analysis of the bonding is not wrong, but maybe a bit too simple.\n\n\nThe reason for this is that the molecule has a very high symmetry, but it also has an unpaired electron. The problem here is that orbital symmetry cannot match with spacial symmetry, which means that hybridisation doesn't actually work here. \n\nA bit more precise: The electron density of α-spin electrons is different from that of β-spin electrons. The hybridisation of the orbitals to produce either electron density need to also be different. In the very least you would have to consider two different hybridisation schemes at once. \n\nAll possible states (that is independent of any hybridisation model) then will be in superposition to form the actual ground state. This ground state will have a complicated bonding situation.\n\n\nOf course you can use VSEPR theory to predict the shape of the molecule, that is a very good starting point. You will find out that the molecule is bent, and that matches the crystal structure. A bent molecule of the form $\\ce{B-A-B}$ always has $C\\_\\mathrm{2v}$ symmetry, it therefore always has π-orbitals. Whether those orbitals are essentially bonding or not, is an entirely different question. \n\nIt also does not tell you which atomic orbitals will be involved.\n\n\nOne obvious fault in the question is that there is no involvement of d-orbitals whatsoever in the molecules. The d-orbitals of chlorine are simply not accessible energetically; at least not in the way that it would form stable bonds. Anyone who is jibber-jabbering about *'octet-expansion'* has stopped learning about chemistry about 20 years ago. (There are plentiful discussions about this on chemistry.se alone.) \n\nAs long as that last part of the question is not meant as a trap (which would be even more awful), it alone shows that the exercise is not well-thought about.\n\n\nThe π-orbitals will therefore be entirely composed out of p-orbitals of the involved elements.\n\n\nIf you look close enough you will realise, that writing a valid Lewis structure of the molecule is not that hard, and you don't even need π-bonds in the first place. \n\nThese are solely dictated by symmetry.\n\n\nIn principle that molecule is of a similar system like allyl-systems (with two extra electrons), or ozone (with one more electron). You will need a better than rudimentary grasp at molecular orbital theory to understand it. \n\nLet's *assume* that things would be easier, and try tor roll with that for as long as possible. I have written [a couple of times](https://chemistry.stackexchange.com/search?q=user%3A4945+terminal+atoms+hybrid*) that terminal atoms can be (as a rule of thumb) regarded as having sp-orbitals.\nTherefore the two oxygen will use one sp-orbital to form a σ-bond towards chlorine, the other one is an in-plane-σ-lone-pair. That leaves for every oxygen one in-plane-p-lone-pair, which we will (in the spirit of the approximation) disregard as having any effect on bonding. There is one remaining out-of-plane-p-orbital per oxygen. These will, due to symmetry, participate in π-bonding. \n\nFrom VSEPR-theory we know the molecule is bent. The approximate hybridisation of the orbitals of chlorine that match that would be sp². Let's just assume that this is the case. Two of these will be the counterpart for the σ-bonds. The remaining will be either a lone-pair, or singly occupied. Already in this very simplified view, the problems start here. Let's further assume, that because it has higher s-character, the orbital will be lower in energy and therefore be doubly occupied. That leaves the odd electron in an out-of-plane-p-orbital, which will participate in π-bonding. \n\nIn this simplified and assumption riddled thought-experiment, the π-bonding will form three molecular orbitals pictured below.\n\n\n[](https://i.stack.imgur.com/fIC6A.png)\n\n\nEssentially the stabilisation from the π-bonds itself should be small, and could be described as [π-backbonding](https://en.wikipedia.org/wiki/Pi_backbonding), or is similar to [negative hyperconjugation](https://en.wikipedia.org/wiki/Negative_hyperconjugation); *maybe negative π-backbonding (!?)*.\n\n\nThis all was deduced from a very reduced and simplified set of theory; the reality is probably by far more complicated. **Some of the greatest minds of (theoretical) chemistry have been puzzled about $\\bf\\ce{ClO2}$**, and we have not even started to discuss [the three-electron bond they assumed in between](https://en.wikipedia.org/wiki/Chlorine_dioxide#Structure_and_bonding). Let's just say that this is not an undergrad study problem.\n\n\nI know this doesn't really answer your question, and this is not very satisfactory either, but this is science: It often gives you more questions than answers.\n\n\n", "20" ], [ "\nWhat I would suggest to you is to start with working out how many electrons (valence electrons) are present in the molecule.\n\n\nEach oxygen atom ($\\ce{O}$) brings six electrons and the chlorine ($\\ce{Cl}$) has seven electrons. This gives us a total of 19 electrons.\n\n\nWe should aim for having 8 electrons around each atom, keep in mind that the more heavy p block elements can expand above an octet.\n\n\nIf we look at the crystallography of chlorine dioxide ($\\ce{ClO2}$) (A.Rehr, M.Jansen, Investigations on Solid Chlorine Dioxide ($\\ce{ClO2}$)): Temperature-Dependent Crystal Structure, IR Spectrum and Magnetic Susceptibility, Inorganic Chemistry (1992) 31, (\\*) p4740-p4742) we find that the chlorine dioxide ($\\ce{ClO2}$) molecule is clearly bent. The O-Cl-O angle is 116.2 degrees. Now this is an interesting finding.\n\n\nNormally if we have a perfect $sp^3$ molecule such as methane ($\\ce{CH4}$) we get an angle of 109.5 degrees between the bonds. But in the case of ammonia ($\\ce{NH3}$) or water ($\\ce{H2O}$) the increased ability of the lone pairs to repel electron pairs decreases the angle.\n\n\nThus it looks like a distorted $sp^2$ to me, I think that the bonding is likely to be two lone pairs on each $\\ce{O}$, one lone pair on the $\\ce{Cl}$.\n\n\nTwo $\\sigma$ bonds each holding two electrons connecting oxygen ($\\ce{O}$) atoms to the chlorine ($\\ce{Cl}$).\n\n\nTwo $\\pi$ bonds each holding two electrons connecting oxygen ($\\ce{O}$) atoms to the chlorine ($\\ce{Cl}$).\n\n\nA $\\pi^{\\*}$ (antibonding) orbital will have one electron in it, think it would be interesting to do ESR on $\\ce{^{17}O}$ enriched chlorine dioxide ($\\ce{ClO2}$) and compare the results with ESR on normal chlorine dioxide ($\\ce{ClO2}$). This could allow us to get a better idea of where the unpaired electron is.\n\n\n", "2" ] ]

[ [ "\nRecently a research paper[[1]](https://doi.org/10.1126/sciadv.abc6688) has been published which claims to effectively convert *n*-alkanes into *n*-alcohols.\n\n\nIt proposes a quadruple relay catalysis for selective synthesis of *n*-alcohols from *n*-alkanes via the following reaction pathway:\n\n\n[](https://i.stack.imgur.com/giViG.jpg)\n\n\nThe reaction seems to get yields higher than 75%. The paper itself goes into much detail about each catalyst and its role.\n\n\nThe only catch here is that as this is a hydromethylation, you need to start with an alkane which has one carbon less in its chain than the required alcohol.\n\n\n\n\n---\n\n\n**Reference:**\n\n\n[[1]](https://doi.org/10.1126/sciadv.abc6688): Tang, X.; Gan, L.; Zhang, X.; Huang, Z. N-Alkanes to n-Alcohols: Formal Primary C─H Bond Hydroxymethylation via Quadruple Relay Catalysis. *Sci. Adv*. **2020**, 6 (47), eabc6688.\n\n\n", "2" ] ]

[ [ "\nThe issue is that reactivity is not an inherent property of a compound. It can be reactive towards specific types of reactions and completely inert to others. Acyl chloride reacting with phenol doesn't rely on phenol behaving as an acid, the type of reaction for which you would describe it as more reactive.\n\n\n", "3" ] ]

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[ [ "\nWrite down the actual (equilibrium) reaction that develops between carbonate ion and water when one proton is exchanged (only full-fledged strong acids or bases might exchange a second proton to any significant extent). Which conjugate acid/base appears? $\\ce {H2CO3/HCO3-}$ or $\\ce {HCO3-/CO3^{2-}}$? When you see which one it is, use that dissociation constant.\n\n\n", "1" ], [ "\nThe carbonate ion is the Conjugate base of the weak acid \n$\\ce{HCO\\_3^-}\\ (K\\_{a(HCO\\_3^-)}={5\\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\\ce{H\\_2CO\\_3}$ , so the solution of this problem as a solution of a monoprotic weak base:\n$\\ce{CO\\_3^{-2} + H\\_2O <=> HCO\\_3^- + OH^-}$\n$$\\ce{K\\_{b(CO\\_3^{-2})}}=\\frac{[OH^-][HCO\\_3^-]}{[CO\\_3^{-2}]} =\\frac{K\\_w}{K\\_{a(HCO\\_3^-)}}=\\frac{10^{-14}}{5\\times 10^{-11}}=\\ce{2\\times10^{-4}}$$\nNeglecting the $\\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\\ce{[OH^-]}={[HCO\\_3^-]}$,and little of the carbonate goes to bicarbonate,so$\\ce{[CO\\_3^{-2}]}={[CO\\_3^{-2}]\\_0}=0.5$ ,so thus\n $$\\dfrac{[OH^-]^2}{0.5}=K\\_{b(CO\\_3^{-2})}= \\ce{2\\times10^{-4}}$$\n$\\ce{[OH^-]}={10^{-2}}$ Which corresponds to $\\ce{pOH = 2}$ or $\\ce{pH = 12}$\n\n\n", "0" ] ]

[ [ "\nConsider what happens to your argument if I rewrite the equation in the following equivalent form:\n$$\\ce{A -> \\frac{3}{2}B +\\frac{1}{2}C +D}$$\nThat is, you write\n\n\n\n> \n> ... since rate of consumption of A is double the rate of reaction as can be seen from the stoichiometric coefficients\n> \n> \n> \n\n\nIn contrast, the rate of the reaction, in this case, is defined to be the rate of decay of $\\ce{A}$,\n$$-\\frac{\\mathrm{d}[\\ce{A}]}{\\mathrm{d}t} = k[\\ce{A}]$$\nWhen you look at it like this, it is more clear that the stoichiometry should not have an effect on the decay process as you claimed.\n\n\nFurthermore, if we integrate the equation above after separating by variables,\n$$\\int\\_{[\\ce{A}\\_0]}^{[\\ce{A\\_t}]}\\frac{1}{[\\ce{A}]}\\mathrm{d}[\\ce{A}] = \\int\\_0^t-k\\mathrm{d}t$$\nwe get,\n$$\\ln\\left(\\frac{[\\ce{A\\_t}]}{[\\ce{A}\\_0]}\\right)=-kt.$$\n\n\nIf we try to find the time at which the original concentration has halved, we get the formula you have written above which is,\n$$t\\_{\\frac12}=\\frac{\\ln(2)}{k}.$$\n\n\nThe important point here, and this isn't true for other reaction-orders, is that the half-life is independent of initial concentration. This is not what one would intuitively expect, and indeed it's not true in general, but it is characteristic of all exponential-decay processes.\n\n\nIndeed, this is a very deep result which doesn't tend to get emphasized enough in my opinion.\n\n\nIt is true that sometimes you can say that a reaction will be second order based on stoichiometric coefficients, but you are told this is first order, so that is probably the source of your confusion.\n\n\nSo, your textbook is right.\n\n\n", "3" ] ]

[ [ "\nThe statement you read that \"baking soda is a base\" comes from the fact that a solution of sodium bicarbonate (baking soda) and water has a pH of around $8.3$. \n\n\nHowever, sodium bicarbonate is *amphoteric* with respect to Brønsted–Lowry acid/base theory, which means that it can act as either an acid or a base. More specifically, the bicarbonate ion is *amphiprotic*, meaning that it can either donate a proton, acting as an acid, or accept a proton, making it act as a base. The reason that a sodium bicarbonate solution is slightly basic is that it's ability to take a proton from a water molecule is greater than its ability to donate a proton to a water molecule. In other words, the chemical reaction causing the solution to be basic is: \n\n\n$$\\ce{HCO3- + H2O <--> H2CO3 + OH-}$$ \n\n\nThe other equilibrium reaction at play, the reaction \"trying\" to make the system more acidic, is: \n\n\n$$\\ce{HCO3- + H2O <--> CO3^2- + H3O+}$$ \n\n\nAgain, in an otherwise pure aqueous system, it is the first reaction that dominates, and thus giving the observed slightly basic solution. \n\n\nNote that the $\\ce{H2CO3}$ produced in the first reaction is also part of yet another equilibrium process: \n\n\n$$\\ce{H2CO3 <--> CO2 ^ + H2O}$$ \n\n\nWhere the $\\ce{CO2}$ is liberated as a gas. This is the cause of the well-known \"fizzing\" when baking soda is added to vinegar (acetic acid). \n\n\n", "9" ], [ "\nSodium bicarbonate can indeed act as an acid. You just need a strong enough base with which it can react.\n\n\nTake 0.1 mol sodium hydroxide and add it to a 100 mL distilled water sample, then add 0.1 mol sodium bicarbonate. The pH goes down in the second step as the solutes are combined to make the almost-but-not-quite strong base sodium carbonate. Compare this reaction:\n\n\n$\\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$\n\n\nwith this one where we see the same proton transfer process:\n\n\n$\\ce{HCl + NaOH -> NaCl + H2O}$\n\n\n", "5" ] ]

[ [ "\nIf it were up to me, I would:\n\n\n1. Draw the nitrogen atom as if it were carbon but label it as \"N\".\n2. Assuming normal valence (e.g. an amine but not a nitro conpound), include the formally nonbonded electron pair as one of the substituents on nitrogen.\n\n\n", "3" ], [ "\nYou can just draw the nitrogen from the centre of the circle using a lesser \"wedge\" angle on the circle to indicate an eclipsed bond. \n\n\n", "0" ] ]

[ [ "\nBonds actually have negative potential energy. Think about it this way: in order to break a bond you have to expend energy so when a bond is created, energy must therefore naturally be released. Hence internal energy is lower.\n\n\n", "3" ], [ "\n\n```\nWhen stronger bonds are formed, *Ep* is bigger.\n\n```\n\nI take issue with this statement. When stronger bonds are formed, the internal energy becomes **more negative**, not just \"bigger\". It is this **decrease** in internal energy that makes the products' bonds so strong, since now the activation energy will be very large for a reaction to occur. \n\n\nThis link might help to explain: <http://www.rsc.org/Education/Teachers/Resources/cfb/energy.htm>\n\n\nMy favorite passage from this link is, looking at a one-step exergonic reaction: \"The height of the peak from the reactants side is a measure of the strength of the bonds in the reactants and the height of the peak from the products is a measure of the strength of the bonds in the product. Thus we can consider the bonds in the product to be stronger than the bonds in the reactants.\"\n\n\nIf PV constant, this means Ek is the same. Ep gets more negative. So naturally enthalpy decreases (negative sign).\n\n\n", "2" ], [ "\nWhen energy goes off from a molecule, we take the sign as negative - this implies that the sign of PE which we are considering in formation of stronger bonds is **MORE**, albeit **WITH NEGATIVE SIGN**, hence, it is going down in energy coordinate.\n\n\nThe magnitude of change in PE, with formation of stronger bonds is more, this will lower the Internal Energy.\n\n\n", "1" ], [ "\nThe change in enthalpy is negative in an exothermic reaction because energy is \"lost\" through the reaction (because there is more energy on the products side than on the reactants side).\n\n\nAnother way to think about this is by calculating the enthalpy before and after a reaction, for example - and this is a synthesis [and exothermic] reaction:\n\n\n$$\\ce{2H2 + O2 <=> 2H2O}$$\nThe ΔH [change in enthalpy] is -484 kJ (you can calculate this using standard enthalpy of formation numbers).\n\n\nYou can also think back to the definition of enthalpy - disorder: When two molecules that were 'freely' moving around, come together, there is less disorder [so the disorder in the system has decreased, and consequently is negative].\n\n\n", "-1" ] ]

[ [ "\nYup, you're forming sulfide.\n\n\nRender the thiosulfate ion as $(\\ce {S-}-\\ce {SO3-})$. Then at the cathode:\n\n\n$(\\ce {S-}-\\ce {SO3-})+2e^- \\rightarrow \\ce {S^{2-}}+\\ce {SO3^{2-}}$\n\n\nBoth products are basic, with sulfide ion being strongly so, and they will react further with protons; so the sulfide ion gives some hydrogen sulfide. Sulfide ions can also combine with silver ions, from the anode, giving a black precipitate. The other product, sulfite ion, can be reduced further, possibly forming still more sulfide ion/hydrogen sulfide.\n\n\nSo the reaction is not safe. But kudos for using a fume hood.\n\n\n", "3" ] ]

[ [ "\nHCl reacts with ammonia and removes it from the equilibrium above:\n\n\n$$\\ce{HCl + NH3 <=> NH4+ + Cl-}$$\n\n\ntherefore the equilibrium will shift to the left.\n\n\n", "4" ] ]

[ [ "\np-Anisaldehyde (PAA) stain\n==========================\n\n\n[This posting](http://www.ochemonline.com/TLC_stains) from www.ochemonline.com (which is available under a CC-By-SA-3.0 licence) provides the following information about the stain and its behavior. The information below is consistent with my personal experience using PAA stains.\n\n\n* Color: Range of colors on a light pink background (my experience was mostly blue spots)\n* Medium shelf life\n\n\nThe post on stains goes on to say:\n\n\n\n> \n> This stain is light and oxidation sensitive and will gradually turn pink/orange. The stain should be kept in an aluminum-covered jar while in use, and the excess should be kept cold and in the dark. Once the stain turns dark red, it should be discarded and made fresh again.\n> \n> \n> \n\n\nNow you have your answer - the stain should turn pink over time, it should produce (mainly) blue spots on a pink background, and you have a visual indicator of its shelf-life. Once it is dark red, it is done. To keep the stain around longer, wrap the jar in foil.\n\n\nBrown spots on white background\n===============================\n\n\nThe iodine stain is the only stain I know that does this for most compounds. \n\n\nNot all the spots are visible\n=============================\n\n\nPAA is principally an electrophile and stains nucleophilic compounds more strongly than other types of compounds.\n\n\nPersonal recommendation\n=======================\n\n\nI grew frustrated with the PAA stain because it degraded quickly and switched to the vanillin stain. The recipe is in the link posted, and reproduced below scaled down. Make small batches to avoid waste. You want to use the stain faster than it decomposes.\n\n\n\n> \n> Dissolve 3 g of vanillin in 50 mL of ethanol. Slowly add 0.5 mL of concentrated sulfuric acid.\n> \n> \n> \n\n\nThis stain is very sensitive to functional groups and produces a wider range of functional group specific colors than PAA. Vanillin (the structure of vanillin is simiular to PAA, but with an additional OH on the ring) is both an electrophile and a nucleophile and so reacts with a wider range of compounds.\n\n\nOchemonline indicates that it has the same \"medium shelf-life\" as PAA, but in my experience, PAA decomposed more quickly. I left my vanillin stain in an amber glass jar without foil wrapping in my hood at room temperature, and since I was doing lots of TLC every day, I used it faster than it degraded. PAA stain degraded much more quickly for me under these conditions, and I was throwing half of it out despite the number of TLC plates I was doing. If you are not doing TLC every day, then wrap in foil and refrigerate. \n\n\n\n> \n> Best reason to use vanillin: The awesome smell!\n> \n> \n> \n\n\nPoor spot quality\n=================\n\n\nThis main not be a problem with your stain.\n\n\n* Big streaky spots are indicative of too high concentration in your solutions.\n* Alternatively, you may need a more optimal solvent. For example, when doing amines on silica, you may want to add 1% triethylamine to your TLC solvent since silica is acidic. Amines and other basic compounds will streak less in these conditions.\n* Big streaky spots may also mean poorly resolved ocmpounds\n* Faint spots may mean poor compatibility with stain or that the concentration was too low.\n* If spots run together, perhaps your lanes are too close, the bottom of the TLC plate is not level, or your plate touched the side of the developing chamber.\n\n\n", "1" ] ]

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[ [ "\nIt depends on what the solution is saturated with, the degree of supersaturation, and what substance you are adding. \n\n\nIf you are adding more of the substance with which the solution is supersaturated, the addition will likely provide a \"seed\" for nucleation (precipitation of the solid). The same will happen for the addition of many substances, particularly those that have similar bonding structure in the solid phase as the substance that is supersaturated. \n\n\nAs you suggest, adjusting the temperature will probably be necessary in order to further supersaturate the solution. If you just want to achieve a particular degree of supersaturation, you could just cool the solution by the necessary amount. This is assuming that the compound that is supersaturated has a solubility that decreases with decreasing temperature (most, but not all, compounds behave this way). \n\n\nIf you want to increase the solution's supersaturation at a given temperature, say room temperature, then you should heat the solution so that it is no longer supersaturated, then dissolve more of the compound, then let it cool back to room temperature. Of course it may precipitate during cooling depending on the degree of supersaturation and the conditions described above.\n\n\n", "2" ], [ "\nYes, adding additional solute to a super saturated solution can be done without crystallization. \n\n\nAdding additional solute to a super saturated solution and preventing crystallization is dependent on temperature, degree of agitation, and rate of solute addition. Changing the temperature is usually the easiest way to control the degree of super saturation, but you can experiment with the other variables as well. The degree to which these three variables affect any given system is dependent on the identity of the system and the amount of impurities (generally you do a lot of work to minimize impurities before you create a supersaturated solution).\n\n\n", "2" ] ]

[ [ "\nLike most concepts that are not captured by a single number, nucleophilicity is not a simple (or at least, relatively simple) concept like acidity.\n\n\nNucleophilicity depends a lot on orbital alignment and interaction, in addition to thinking about charges. It's also confusing because nucleophility is not always married to the stability of the adduct. For example, iodide is a great nucleophile, but alkyl iodides are also highly reactive.\n\n\nIn your specific case, since the two nucleophiles have the same charge, you should look at other factors. Bromide is significantly larger than hydroxide, and it is a much better nucleophile, at least when we're considering carbon electrophiles. Indeed, we see bromide used in nucleophilic catalysis but not hydroxide.\n\n\n", "2" ], [ "\nBetter nucleophile, as per definition, are those which can donate their electron pair to (and form a bond with) a CARBON.\nAlthough, we could see usually, nucleophilicity and basicity (ability to donate LP to a proton) go side by side, there are cases where it is not true.\n\n\nSo, for OH-, it is better nucleophile than Br- in an aprotic medium, due to higher charge density on O atom. While when it come to basicity in aqueous medium, OH- ions are heavily hydrated due to extensive H-bonding, making it less likely to donate its LP to a Carbon, hence less basic than Br-.\n\n\n", "-2" ] ]

[ [ "\nFirstly, I will put across main points so that it will be easy to understand this complex control mechanisms of the enzymes involved in both processes:\n\n\n**Glycogen synthase control**\n\n\nControl of glycogen metabolism is effected via reciprocal **regulation of glycogen phosphorylase and glycogen synthase**. Thus, activation of glycogen phosphorylase is tightly linked to inhibition of glycogen synthase, and vice versa.\n\n\nBoth glycogen synthesis and breakdown are **exergonic** under the same physiological conditions. If both pathways operate simultaneously, however this is deemed to be **wasteful hydrolysis of UTP**. \n\n\nGlycogen phosphorylase and glycogen synthase therefore must be under stringent control such that glycogen is either synthesized or utilized according to cellular needs.\nRegulation involves both **allosteric control** and **covalent modification**, with the latter being under hormonal control.\n\n\n\n> \n> Glycogen synthase also exists in two distinct forms that can be\n> interconverted by the action of specific enzymes: active,\n> dephosphorylated glycogen synthase I (glucose- 6-P–independent) and\n> less active, phosphorylated glycogen synthase D (glucose-6-\n> P–dependent). Glycogen synthase, is activated by G6P.When there is\n> high demand for ATP (low [ATP], low [G6P], and high [AMP]), glycogen\n> phosphorylase is stimulated and glycogen synthase is inhibited, so\n> flux through this pathway favours glycogen breakdown.\n> \n> \n> \n\n\nWhy is this so?\n\n\n**ATP** also binds to the allosteric effector site, but in the T state, so that it **inhibits** rather than promotes the T-R conformational shift.\nHaving said that here is an illustration of control of glycogen phosphorylase activity;\n\n\n[](https://i.stack.imgur.com/HZtjE.png)\n\n\n\n> \n> The enzyme may assume the enzymatically inactive T conformation or the\n> catalytically active R form. The conformation of phosphorylase b is\n> allosterically controlled by effectors such as AMP, ATP, and G6P and\n> is mostly in the T state under physiological conditions. In contrast,\n> the modified form of the enzyme, phosphorylase a, is largely\n> unresponsive to these effectors and is mostly in the R state unless\n> there is a high level of glucose.\n> \n> \n> \n\n\n**Exactly this situation occurs in glycogen metabolism through the opposition of the glycogen phosphorylase and glycogen synthase reactions.**\n\n\nAnother point to note is that glycogen synthase is regulated by covalent modification through complex cyclic cascade. For more information, please see here: [Glycogen Metabolism](https://www.rpi.edu/dept/bcbp/molbiochem/MBWeb/mb1/part2/glycogen.htm)\n\n\nIf you need more clarification don't hesitate to put in comments. Hope it helps\n\n\nReferences\n\n\n1. Biochemistry (Grisham)\n2. Biochemistry (Voet and Voet)\n3. Harper’s Illustrated Biochemistry\n\n\n", "2" ] ]