Adding more solid to a solid/gas equilibrium $$\ce{CaCO3(s)<=> CO2(g) + CaO(s)}$$

In this equation, what happens if you increase the amount of one of the solids while there are existing quantities of everything else?

According to the calculation for the equilibrium constant, equilibrium is only dependent on the concentration of $\ce{CO2}$, so it technically should not change anything, right? However, if you decrease the pressure of the $\ce{CO2}$ enough, then shouldn't there be a point where the equilibrium constant can no longer be met because there aren't enough moles of $\ce{CO2}$?

Additionally, at this point, shouldn't adding more $\ce{CaCO3}$ change the point of equilibrium?

The concentrations of the solids change only negligibly with temperature or other reaction conditions and so are involved in the equilibrium only as constants. The amount of solids present does not change the concentration of each solid in its crystal structure. Therefore the equilibrium of the reaction is written as:

$\mathrm{K = constant}\times \ce{[CO2]}$

If one of the reactants ($\ce{CaCO3, CaO, CO2}$) is not present, there is no equilibrium.


Symmetry representations I'm Having a real struggle trying to understand symmetry representations matrices and character tables. Is there anyone who would be kind enough to help me out here? I understand it is probably a little broad but nevertheless, I will list some areas that if anyone could shed some light on I would be very appreciative.

1) What is an irreducible representation? My book just starts going into $2\times 2$ matrices like:
\begin{equation}
\left( \begin{array}{cc}
1 & 0\\
0& 1\end{array}
\right)
\end{equation}
Saying that we can represent the effect of symmetry operations as such! When the operation actually affects the molecule by changing an atom then the diagonal of the matrix swaps round seemingly !!

2) Leading on from the first question, what is a reducible representation? Why the distinction?

3) What is a basis? My book introduces them via vectors down each of the cartesian axes and then a "set" of them together (why) to give four basis's for a $C_2v$.

4) I can use the reduction formula to see how many irreducible representations are contained within a particular reducible representation, but what am I doing!

Let's start with the basics:

A representation is a set of matrices that fulfill the multiplication table for the point group.

So I might be able to use a set of 3x3 matrices, since of course I can define any of the symmetry operations over Cartesian (x,y,z) space. e.g.:

$$E = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array}
 \right) \;
i =
\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \end{array}
 \right)
$$

But usually, I can fulfill the multiplication table with much simpler (smaller) matrices.

An irreducible representation is a set of matrices that are as small as possible.
A reducible representation is a set of matrices that can be decomposed into a linear combination of irreducible representations. This can occur by multiple techniques, including block-diagonalization.

The distinction exists, because for chemical point groups, there are usually a finite number of irreducible representations (given in character tables), but of course there are an infinite number of reducible representations.

The power of symmetry and group theory is that any function of a molecule, like orbitals, vibrations, etc. can be described using a representation. Thus, we usually label orbitals and vibrations by their corresponding irreducible representation.

I'll have to hunt down my copy of Ogden, but usually people use the Cartesian axes or $p_z$, $p_x$, and $p_y$ to illustrate the different irreducible representations. That is, a basis is some function (like the Cartesian axes or an atomic orbital) used to generate a representation. We look at how the basis is transformed under the different symmetry operations and write down the matrix or characters from those transformations.

(I try not to use that term because students find it confusing and different guides and texts use subtly different definitions for it.)

The reduction formula (or algorithm) is a way of decomposing the reducible representation (which is really a big matrix) into a combination of the different irreducible representations, but only needing to act on the characters of the representations.

That is, I don't need to actually use the reducible matrices themselves. Using the reduction formula, I can get the linear combination of irreducible representations only using multiplication of characters of the representations. It's much easier to do the reduction formula than to figure out how to transform a large reducible matrix.


Validating a GC instrument / column Are there any generic rules / heuristics to identify a good GC vs a bad one?

The context is that during manufacturing-troubleshooting I often get shown GCs of various fine chemicals and I often wonder whether a problem is truly process-related or merely just an artifact of a bad GC operator / column / instrument etc.

Are there any benchmarks or signs to look for that will give a hint that a particular GC wasn't very reliable etc?

For example I've added one such GC graph below. Are there things too look for? e.g. I suppose non flat baselines or clamped off peaks are obviously bad? Merging peaks too? What else?

For any instrument that quantifies via integration of a peak (GC, NMR, etc.), sample concentration (peak height) is important especially with low signal/noise ratios. Obviously if the peak isn't in the sweet spot - so to speak - (where its area is significantly above the baseline, but not too big that it saturates the detector) the quantification may be unreliable. Also peak shape is a consideration in integration. If the peak is too broad, much of the area under it can be obscured by the baseline. On GC/MS or GC/FID in particular I used to have problems with peaks 'tailing', that is a broad lingering asymmetry after the peak max which would often overlap with species with greater retention times. No es bueno for quantification. Baking out the column, finding a better heating profile/method usually did the trick.

Optimal methods are going to vary widely based the instrument you use, the solvent, what components you have in your mixture, there relative amounts and concentration etc.. A lot of trial and error. As a general rule though typically when I had 2 components that were close in rt, but wouldn't resolve, I'd move to a method with a much slower ramp for the column temp, upping the carrier gas if possible on your equipment might help, too.


Why does the first lithium stationary state have n = 9/4? The ionization energy of $\ce{He+}$ is $19.6 \times 10^{-18}~\mathrm{J~atom^{-1}}$. What is the energy of the first stationary state ($n=1$) of $\ce{Li^{2+}}$?

Since the question specifically states that $n=1$, I used the formula

\begin{align}
E &= -2.178 \times 10^{-18} \cdot \frac{Z^2}{n^2}.\\
E &= -2.178 \times 10^{-18} \cdot \frac{3^2}{1}\\
E &= -1.96 \times 10^{-17}~\mathrm{J}
\end{align}

But the correct answer is supposed to be $-4.14 \times 10^{-17}~\mathrm{J}$ and many people are telling me to use $n=\frac{9}{4}$ which gives the right answer but I don't know why I am supposed to set $n$ at $\frac{9}{4}$.

I think your calculation is correct. There is an error in this textbook.  All you have to do is to switch  between $\ce{Li^{2+}}$ and  $\ce{He^{+}}$. And 'll prove my point:

$\ce{Li^{2+}}$ and  $\ce{He^{+}}$ are hydrogen-like atoms.

For these species, the ionization energy is given by the equation:
$$E_\mathrm{i}=E_\mathrm{i}(\ce{H})\times Z^2$$

For $\ce{Li^{2+}}$:
\begin{align}
E_\mathrm{i}(\ce{Li^{2+}})&=E_\mathrm{i}(\ce{H})\times Z^2\\
19.6 \times 10^{-18}&=E_\mathrm{i}(\ce{H})\times 3^2\\
E_\mathrm{i}(\ce{H})&=2.18 \times 10^{-18}  \,\mathrm {J/atom}\\
E_\mathrm{i}(\ce{H})&=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}}\\
  &= 13.6 \,\mathrm{eV}
\end{align}

$13.6 \,\mathrm{eV}$ is well-known to be the first ionization energy of hydrogen. So, my assumption is true.
To solve the problem after correction:
 The ionization energy of $\ce{He^{+}}$ is given by the equation:
\begin{align}
E_\mathrm{i}(\ce{He^{+}})&=\frac{E_\mathrm{i}(\ce{Li^{2+})}\times Z^2(\ce{He^{+})}}{Z^2(\ce{Li^{2+})}}\\
E_\mathrm{i}(\ce{He^{+}})&=8.72\times 10^{-18}\,\mathrm {J/atom}
\end{align}
The energy of the first stationary state:
$$E_\mathrm{i,1}=E_\infty-E_1$$
But, $E_\infty=0$, we find:
$$E_1=-E_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$


Is the half-full rule and full rule followed in the 6th and 7th periods? Is the half-full rule and full rule followed in the 6th and 7th periods? (Note: Half-full rules is Hund's rule)

Example: What is the correct electron configuration?
$$\ce{W = [Xe] 6s^2 4f^{14} 5d^4}$$ doesn't use half full rule, or
$$\ce{W = [Xe] 6s^1 4f^{14} 5d^5}$$ uses half full rule.

Short answer: no! The first known exception to occur in the periodic table is in period 5:  niobium’s ground state electronic configuration is

  $\ce{Nb: [Kr] 5s^1 4d^4}$

which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.

Let’s look at Wikipedia’s list of exceptions to Madelung’s rule. In period 4, the only exceptions are $\ce{Cr}$ and $\ce{Cu}$, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for Ni.)

In period 5, the exceptions to Madelung’s rule are classified in three groups:

$\ce{Mo}$ and $\ce{Ag}$: they are the analogues of Cu and Cr
$\ce{Pd}$ is $\ce{5s^0 4d^10}$: a different type of application of the explanation that “fully filled shells are particularly stable”, because in this case two electrons from the s shell were pulled into the d shell. Not what you usually learn.
$\ce{Nb}$ is $\ce{5s^1 4d^4}$,  and $\ce{Ru}$ is $\ce{5s^1 4d^7}$: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.

Period 6 contains further inconsistencies, because of the introduction of $\ce{f}$ orbitals into the mix.


How do technicians treat spilt mercury and bromine? As we know, mercury and bromine are in liquid state at room conditions and they are hazardous. The technicians should have ways to treat them if they are spilt.

  How can we safely treat spilt $\ce{Hg}$ and $\ce{Br}$?

I have thought of using dry ice to cool the mercury down into a solid, which can then be easily removed. But I do not think that this is really practical. Maybe some chemical reactions can help to somehow convert the them into a solid in order to remove them.

Any reputable chemical supplier will normally sell a mercury spill kit, and anyone working with mercury these days should have one. (Either a commercially prepared one, or an equivalent one that's assembled from individual supplies.) Generally speaking, they contain some sort of absorbent/adsorbent to pick up mercury, and a chemical which will react with the mercury and keep it from volatilizing. Sulfur is a common choice (recommended by the EPA), as mercury sulfide is pretty darn inert, as well as being very insoluble in water. (Also, "flowers of sulfur" does double duty as an absorbent.)

After the cleanup, the absorbent is collected (by sweeping - vacuuming can spread mercury vapors), packaged securely, and sent to a hazardous waste disposal company.

As elemental bromine is less widely used than elemental mercury, designated spill kits for bromine are less widely available, but the procedure is generally the same: spread a chemical to react with the bromine and render it more inert, then spread an absorbent to contain the liquid. Elemental bromine is rather reactive (an oxidizer), so a number of common reducing agents can be used. A 5% solution of sodium thiosulfate is a common choice (1) (2).

Again, if you're working with elemental bromine, your lab should have a spill response kit already assembled, with sufficient pre-prepared sodium thiosufate and absorbent to handle the amount of bromine you normally deal with.

The previous was for general information purposes only. If you are looking to work with elemental mercury or bromine, consult with your institution's chemical safety officer before purchasing for more information and restrictions which may apply in your area. Your institution likely also has chemical safety and spill response training, which will inform you of current best practices in handling mercury and bromine spills.

As always with accident handling, the most important thing is not to put yourself in greater danger by attempting to fix it. If the size or extent of the spill is such that you're unsure you can handle the spill safely, the best course of action is to evacuate the premises and let the hazardous materials team do the cleanup. (They'll likely use the same techniques of reacting and absorbing the material, but they'll be wearing more extensive personal protective equipment while doing it.)


Would I use PBr3 or HBr in this synthesis? In the above reaction, I know that I have to use either $\ce{HBr}$ or $\ce{PBr3}$ and then $\ce{NaCN}$, DMF, $\ce{H+}$ and $\ce{H2O}$. However, I'm just confused on when to use $\ce{HBr}$ or $\ce{PBr3}$. How is the stereochemistry controlled?

As @permeakra has said neither reagent will get you the stereochemistry you want in the product. The HBr route goes through a carbocation which scrambles the stereochemistry completely.

PBr3 will invert the stereochemistry, then SN2 -CN addition will invert it again, giving you the wrong product.

You need something that turns the -OH group into a leaving group while preserving the stereochemistry, so that the CN can displace it by a SN2 process. This will invert the stereochemistry giving you the product, after hydrolysis, you want.

I suggest forming a sulfonate, either tosylate, triflate or (my choice) mesylate. This should be displaced smoothly by cyanide under the DMF conditions you refer to.


How can you find the concentration of copper(II) ions in a solution of 0.5 mole copper(II) chloride and 0.5 mole of copper(II) sulfate? The question I'm trying to answer is below and the answer is C.

I tried finding the total moles of copper(II) ions first by using mole ratios for copper(II) sulfate and copper(II) chloride, obtaining an answer of 0.42 moles. Then, I used $n=cv$ and obtained the concentration, using the volume in the question in litres, of 0.84, which is incorrect. Now, I'm sure my method incorrect but I can't think of another way to this.

1 mole of copper(II) sulphate contains 1 mole of copper ($\ce{Cu^{2+}}$) and one mole of sulphate ($\ce{SO_4^{2-}}$). Or, if you want, 1 mole of copper, 1 mole sulphur, and 4 moles of oxygen.

This is by the definition of the mole as the unit of measurement for the amount of substance.

From this it follows that 0.5 moles of copper(II) sulphate contain 0.5 moles of copper. Similarly 0.5 moles of copper(II) chloride also contain 0.5 moles of copper. Together these are 1 mole of copper.

1 mole of copper ($\ce{Cu^{2+}}$) in $\mathrm{0.5\,l}$ water gives us a concentration of $\mathrm{2\, mol\,l^{-1}}$


CH2 group in alkyl glucoside gives two separate HNMR signals Zoom in:

It seems like the CH2 next to the sugar gives two different NMR signals which is very interesting. Could someone explain me why?

Without the sugar, those two hydrogen atoms are enantiotopic, since substitution of those protons would lead to different enantiomers. Achiral measurements still will appear identical though.

Once the sugar is attached, the two hydrogens are diastereotopic, and are chemically and magnetically different. Especially so close to the chiral environment of the sugar, these two hydrogen see enough of a difference in their local environments that their shifts are different.


Why are acids usually depicted in fiction as green? In cartoons, etc., acid is generally shown as vividly green, as in this scene from the Simpsons:

We even have the colour "acid green".

But when I used to use acids in chemistry class, they weren't bright poisonous green. They were mostly colourless, as, for example, this picture of sulfuric acid from Wikipedia:

So where does this idea of green acid come from? Are there acids which are actually green? Or perhaps acids are green when highly concentrated?

The green is the color of bile which usually accompanies gastric acid.  Acid  in its original context referred to a bitter taste.  Bitter tasting compounds, such as vinegar, is what led to the modern chemical definition.  The chemical definition is less likely to have influenced the populace than the experience associated the somewhat bitter, acidic, taste of gastric acid accompanied by bile.


Which temprature is taken in Gibbs free energy equation, of the system or the surrounding? Which temprature is taken, the temprature of the system or the temprature of the surrounding? I think it should be the temprature of the surrounding. This is because if we take the temprature of the system then change in Gibbs free energy would always be zero as then, change in enthalpy of the system would always be equal to change in entropy of the system multiplied by temprature of the system.

The Gibbs free energy is a property of a system, so you should use that system's temperature when calculating it.  The Gibbs free energy is the amount of work that could be extracted from the system in a isothermal, isobaric process: i.e. the temperature and pressure of the system won't change.  Thus Gibbs energy has to do only with the temperature of the system, not of the surroundings.

There is a related but distinct variable called exergy or sometimes "availability".  This variable is given the symbol $B$, and is also a thermodynamic state function.  It is defined as $\Delta B = \Delta H - T_0 \Delta S$, where $T_0$ is the temperature of the surroundings.

Exergy represents the maximum amount of work that could be extracted from a system when that system is brought into thermal equilibrium with the surroundings, for example using Carnot engines.  Thus exergy indicates work obtainable through a combined (a) isothermal change in Gibbs free energy at temperature $T \neq T_0$ and (b) a temperature change occurs when Carnot engines are used to transfer heat from the system at $T$ to the surroundings at $T_0$ until thermal equilibrium is obtained.


Why doesn't gold corrode? As the electronic configuration of gold is [Kr] $4d^{10} 4f^{14} 5s^2 5p^6 5d^{10} 6s^1$, the valence shell is not completely full. Why doesn't gold  corrode?

You can look it up on James Huheeys book, gold's few properties have been qualitatively explained by relativistic theory. Due to that and high nuclear charge acting on the electron, the single electron in $6s$ is actually attracted to the nucleus strongly, thus the electron is not available to react that easily and thus gold does not corrode.


Number of hydrate isomers of CrCl3.6H2O Assuming ocahedral complex, The answer should be 4 as there can be 0,1,2,3 Cl- inside the coordination sphere. But, in many places the case with 3 Cl inside the coordination sphere is not present, answer provided is 3.

Loong's answer says otherwise. Only this answer and another answer in a linked question acknowledge the existence of the 3 Cl form. Every other place simply states that only 3 forms have been experimentally discovered.

Also, in coordinatiom isomerism too, in the case where both spheres become uncharged is ignored.

What is the reason?

Regarding the coordination isomerism question, the reason why it's not considered as an isomer in the case of two uncharged complexes, is that there is no ionic attraction. The two coordination spheres should act as individual ions and hence participate in the ionic sphere. But in this case, where the two spheres carry no charge, there isn't a force, namely the electrostatic force, which binds the two spheres.

Your question on CrCl3.6H2O is a very debatable topic and there hasnt been a unanimous decision that has been put forth by all the authors. Some say there are 3 or rather most of them say 3, only one author, JD Lee says that there are 4.

The reason given by the authors to support there claim of only 3 isomers is not very concrete, but it still majority of the authors believe that it is true. The H2O that exists outside the coordination sphere is the water of crystallization. The water of crystallization is believed to be present within the voids created by an anion and a cation. Now if the isomer [CrCl3(H2O)3] 3H2O should exist then it must have these voids. But here there isn't any anion in the ionic sphere, in fact it is believed that there isn't any ionic sphere. Therefore this isomer doesn't exist and hence cannot be the 4th isomer.

Now, it's also believed that the H2O is actually held to the central metal atom (Cr) with the help of dipole forces and that the voids are not needed. So that would mean that this exists and since it is theoretically an isomer, it should also be an isomer practically.

That brings us to to this very important question, what is the significance of the water in the ionic sphere? Now most books say that it's the water of crystallization of that compound. So if the water of crystallization must be present in the voids so that they can be removed by physical processes like heating, the 3H2O wont be an isomer anymore, but just another compound that exists with the same molecular formula.


Regioselectivity of electrophilic addition of Halogen and Water (Halohydrin) on Alkenes? I am currently studying for an organic chemistry exam and I just couldn't come up with a solution to the following question:
Consider the reaction of 1-Propene + Br2 with water as solvent.
Why will the nucleophilic water molecule add on C-2 and rather than on C1 of the Bromonium-Ion?
The answer seems to be, because in the transition state the bonding between Br and C will dissociate quicker than the bonding between O and C will form, which leads to a positive charge on that C-Atom in the transition state and therefore water will add on the more stable "carbokation-like" carbon.
My question is: why does the Br-C bonding dissociate quicker than the O-C bonding forms? (I read that in the pearson organic chemistry textbook). Does it have anything to do with the Hammond-postulate? Or are there any other thermodynamic or kinetics reasons? So I am basically searching for an explanation why one bond would dissociate quicker than a different bond would associate, because I guess that's the reason why there is a partial positive charge that would then be more stable on the higher substituted carbon...
Unfortunately I haven't found an answer on the internet yet and also not in other textbooks.
Maybe you can help :) Thanks!

This is not a 100% answer, but it was too detailed to fit in the comment section.
Following up from the comment, here's a reference from Inorganic Chemistry by Morrison and Boyd:

There is evidence, of a kind we are not prepared to go into here, that these compounds are not formed by addition of preformed hypohalous acid, $\ce{HOX}$, but by reaction of the alkene with, successively, halogen and water.

Halogen adds (Step 1) to form the halonium ion; this then reacts, in part, not with the bromide ion, but with water (Step 2) to yield the protonated alcohol.
Propylene gives the chlorohydrin in which chlorine is attached to the terminal carbon. This is typical behaviour for an unsymmetrical alkene; orientation follows markovnikov's rule, with positive halogen going to the same carbon that the hydrogen of a protic reagent would. Now, this orientation would be perfectly understandable if the intermediate were an open carbocation: the initial addition of halogen yields the more stable carbocation - secondary, in the case of propylene. But the stereochemistry indicates that the intermediate is not an open carbocation, but a cyclic halonium ion.

And that's all. Quite anticlimatic, as I was expecting a reason for the regioselectivity. Most other references and papers I have seen online also do not have a concrete explanation for the regioselectivity, and they resort to hand-wavy explanations involving resonance for the same. This reaction mechanism is also quite similar to Oxymercuration-demercuration, whose regioselectivity is again explained by the "three resonating structures". I could not find diagrams of these resonating structures, and while this question asks the same, the resonance structures or the intermediate are not properly explained in the answer.
My best suggestion for you would be to accept the explanation in your textbook as that is adequate for now. Until someone more knowledgable posts a better answer with references, that should suffice.


Analysis of salts using color change when heated In order to detect basic radicals one of the preliminary test done is dry test tube heating. My book says that on doing so salts that would appear black when hot and cold are supposed to be Mn, Co, Ni or Cu salt. But does it refer to white salts turning black during heating or salts of any colour including black when it says that they appear black after heating?

No when cold they can be of different colours as can be seen from the following table:


Why can't the molecularity be determined from the rate law? Basically can someone explain why the molecularity cannot be determined from the rate law?

The coefficients only correspond to the rate law coefficients if the reaction is elementary. Elementary reactions are ones where the reagent(s) combine through a mechanism implied by the reaction. These are simple. Many reactions aren't as simple and have multiple steps.

Check out this link, example two. The reaction is actually comprised of two reactions where the slow step determines the rate coefficients.

As you can see, you cannot just read the equation and get the coefficients. One can only determine whether a reaction is elementary (i.e. find its rate law coefficients) by experimentation.


Why is the silver salt ammoniated to form the Tollen's reagent? When doing a distinguishing test between aldehydes and ketones using Tollen's reagent, is it necessary to convert $\ce{Ag+}$ ions into $\ce{Ag(NH3)2+}$ in order to oxidize the aldehyde?

Yes, it is important to convert your silver ion to diamminesilver(I) complex. If you do not add ammonia to silver nitrate, the silver ion is reduced so quickly that colloidal silver metal would appear and the solution would become a black, cloudy liquid and you won't get the silver mirror when it reacts with aldehyde.
It is also important that you maintain the pH of the reaction. During the preparation of Tollen reagent, you will see the use of sodium hydroxide. This is very important for maintaining the pH. During the whole reaction, the pH is between 8-10 and it goes even faster if the pH is beyond 10. The reaction also becomes more sensitive. This has been explained in details in this journal by going through the exact mechanism of the reaction.


Do colligative properties depend on the number of solute molecules? Colligative properties are properties that are affected by the number of solute molecules dissolved in the solution. True or false?

I think the statement is false, and the correct one should use "number of solute particles". I asked my TA and he said the molecule version is true, although I show him an example that $\ce{H2CO3}$ can dissolve water in many ways to create different sets of miles for $(\ce{CO3^2-},$ $\ce{HCO3-},$ $\ce{H+}).$ In this case, one mole of molecules $(\ce{H2CO3})$ can result in different total numbers of particles.

However, my TA said that $\ce{CO3^2-},$ $\ce{HCO3-},$ and $\ce{H+}$ can be considered as different molecules. I still don't satisfy with his answer though.

Could someone explain it for me, whether I am right or wrong.

It's not good notation on the TAs part. IUPAC defines a molecule as having to be electrically neutral. I see what they are intending, but solute particles would be more correct. This is why, as you noted, compounds that dissolve into ions produce a greater effect than the number of original molecules would suggest and molecules that aggregate in solution have a smaller effect. Equations for colligative properties depend on the van't Hoff factor, which essentially converts from the original number of molecules to the number of distinct particles.


Hydrogen storage using activated carbons On Wikipedia, a possibility of using activated carbons to store hydrogen is reported. According to the Wikipedia page, this solution can be used to store around 5 wt% by cooling down the hydrogen to -196$\;^\circ$C at 1 bar. I am wondering if the storage system has to be kept always at this temperature?

Short answer: yes, if you want to store the same amount of hydrogen and on condition that you do not mess with pressure.

Long answer: I'll skip some intermediate physicochemical derivations for brevity. The phenomenon we are dealing with is called adsorption. Perhaps, the simplest theoretical explanation of adsorption was given by Langmuir. Langmuir adsorption isotherm can be written as $$\theta = \frac{Kp}{1 + Kp},$$ where $\theta$ is a fraction of the surface of adsorbent (solid) covered by the adsorbate (gas) and $K$ is an equilibrium constant. It is seen that at a constant temperature the more pressure you apply to the system, the more surface is covered (i.e. the more gas is adsorbed).

But what about the temperature dependence? Since $K$ is an equilibrium constant, its temperature dependence is defined by the van't Hoff equation. Also, $K$ is directly proportional to the pressure required for a certain amount of gas to be adsorbed. In other words, when $\theta$ is constant, $\ln{K}+\ln{p}=\mathrm{const}$, transforming our van't Hoff into something like $$\left(\frac{\partial\ln{p}}{\partial(1/T)}\right)_\theta=\frac{\Delta_{ad}H^○}{R}.$$ Here, $R$ is the gas constant, and the sign of $\Delta_{ad}H^○$ - adsorption enthalpy - is negative, for adsorption is exothermic. Consequently, the slope of $\ln{p}$ vs $1/T$ line is negative. In plain English, the equation above reads like this: in order to obtain the same amount of adsorbate $\theta$ you should either decrease the temperature or increase the pressure.

If you're that interested in the topic, you may also want to check out how the sorption pumps work.


Why do electrodes need to share an electrolyte in a lemon battery? From what I understand of a lemon battery, there are two electrodes one zinc, and one copper. The electrodes are then both placed separated in the same electrolyte, in this case a lemon. The Zinc dissolves into the electrolyte oxidizes and loses 2 electrons. Those 2 electrons stay on the zinc electrode. On the copper electrode 2 h+ combine with 2 electrons from the copper and then form h2. the h2 bubbles up and leaves the electrolyte. The extra electrons from the zinc then move to the copper when there is a connection between the anode and cathode. This reaction continues until the zinc is too dissolved or if there is not enough h+ in the electrolyte to take electrons from the copper.

If my understanding is correct (please correct me if I'm wrong), then why can't each electrode have is own electrolyte, completely separated. For example the anode in half a lemon and the cathode in the other half. Both half's separated by air.

Like @Pritt said, the circle has to be closed, 2 electrons are traveling from one half-cell to the other, means 2 negative charges are missing, the half-cell needs to replace it by other 2 negative ions or send 2 positive charges, since charges cannot exist separately.


Why is it possible to separate NaOH from certain compounds using acetone? Is it due to the aldol reaction that happens? But the product of that reaction is still soluble in acetone.

Apparently if you have a solid mixture of 6-ethoxycarbonyl-3,5-diphenyl-2-cyclohexenone with NaOH and Na2CO3 impurites, and you dissolve them in acetone, after which you centrifuge, the solid impurites will settle out.

Why is this?

Ask yourself what kind of reaction happens?
What glassware have you used to remove impurities?

By controlling pH and polarities you are able to separate compounds because of different affinities between them and solvents. You change molecular solvatation of your analyte and impurities. Then by controlling boiling point, using a rotavap, you may be able to separate two phases that are miscible. Or by making impurities insoluble you can remove by mechanical operation such as centrifugation. Such post-treatments are very common in chemistry and they scale up very well.

So, there is no chemical reaction (except acid/base reaction if any) there are physical reactions (solvatation, phase extraction, phase change: evaporation, condensation, crystallisation and sedimentation).


compression set of tin catalyzed silicone better than platinum? Platinum silicone is more dimensionaly stable, it shrinks less when it cures.

However, i just need silicone which has good compression set (small change by compression), and perhaps, once they are cured, platinum and tin catalyzed silicone become the same material - silicone elastomer. Perhaps that tin silicone, having shrunk, is more resistant to compression set?

Does tin silicone have better compression set? Are the highest quality o rings made from tin silicone?

Tin catalyzed silicone lasts for a couple of years and platinum catalyzed lasts for 10 years, after which they lose their elasticity, becoming brittle and splitting. So while they may have equal properties for the first year, After a while the Platinum silicone will perform better. Using less catalyst for longer cure times also increases durability, and vacuum treating the silicone prior to casting and heat treatng it afterwards increases it's compression set performance.


What does the first $1$ mean in $\ce{Li}_2 1 ^1\Sigma_g^+: (1s\sigma_g)^2(1s \sigma_u)^2(2s \sigma_g)^2$? I am trying to learn the nomenclature behind molecular states and in (Demtröder, 2010) I came across this expression:
 $$\ce{Li}_2 \color{red}{1} ^1\Sigma_g^+: (1s\sigma_g)^2(1s \sigma_u)^2(2s \sigma_g)^2$$
now I can explain the origin of every term except the $1$ which I have put in red. What does this 1 mean?

The conventional labelling of molecular states starts at X for the lowest state and A, B, C for higher ones and a, b, c, for equivalent states of the same symmetry but different multiplicity; see Herzberg spectroscopy books for many examples. These labels, however, reflect the order in which the states were discovered and so do not reflect their relative energies. It is becoming more common to precede a molecular state's symmetry label by an integer e.g. $1~^1\Sigma _g^+,~ 2~^1\Sigma _g^+,~3~^1\Sigma _g^+$ to reflect the energy ordering of states with the same symmetry. This notation reflects the way atomic levels of the same symmetry are labelled as $1$s, $2$s, $3$s, etc.

As an example, the conventional notation for the first four electronic states of  $\ce{C2}$ in order of energy is

X $^3\Pi_u$ ;  a $^1\Sigma_g^+$ ; b $^1\Pi_u$ ; A $^3\Pi_g$

which becomes

$1~ ^3\Pi_u$ ;  $2 ~^1\Sigma_g^+$ ; $1 ^1\Pi_u$ ; $3~ ^3\Pi_g$

In atoms the notation for a set of transitions starting from the 3p level in sodium is  Na $3~ ^2P_{3/2}$ .


Aqueous Chemistry: Ranking Ion Concentration I am trying to rank the concentration of $\ce{PO4^3-}$, $\ce{NO3^-}$, and $\ce{Na+}$ that remain in a reaction. The reaction occurs when 100mL of 1.0M $\ce{Na3PO4}$ solution is mixed with 100mL of 1.0M $\ce{AgNO3}$ soln. I have written the equation as:

$\ce{Na3PO4(aq) + 3AgNO3(aq) -> 3NaNO3(aq) + Ag3PO4(s)}$

From this equation I find that there are 0.3mol of $\ce{Na+}$ by 0.1L * 1.0M * 3mol Na, and
0.1mol of $\ce{PO4^3-}$ by: 0.1L * 1.0M * 1mol $\ce{PO4^3-}$ ,
and finally 0.3mol of $\ce{NO3-}$ by 0.1L * 1.0M * 1mol $\ce{NO3-}$.

By these calculations, the concentration of the nitrate ion and sodium ion should by equal. However, the correct answer is that the most concentrated ion is sodium followed by nitrate followed by phosphate. Could someone please explain where I'm going wrong.

Thanks!

Let's look at the overall reaction:

$\ce{Na3PO4(aq) + 3AgNO3(aq) -> 3NaNO3(aq) + Ag3PO4(s)}$

As you can see $\ce{Na^+}$ and  $\ce{NO_3^-}$ are spectators ions, they don't participate in the main reaction. Their concentrations, taking into account the dilution, are:

$\ce{[Na^+]= \frac {3 \times 1.0 \times 100}{200}}= 1.5 \ce{M}$.

${[\ce{NO_3^-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

The net ionic equation is:

$\ce{PO4^{3-}(aq) + 3Ag^+(aq) ->  Ag3PO4(s)}$

The initial concentration of phosphate and silver ions taking into account the dilution are:

${[\ce{PO_4^3-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

${[\ce{Ag^+}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

According to the stoichiometry of the reaction: 3 moles of silver ions react with one mole of phosphate ion. This means that silver ion is a limiting reactant. It will completely react to form the precipitate of silver phosphate. On the other hand, the concentration of phosphate at the end of the reaction will be:
${[\ce{PO_4^3-}]= 0.5-\frac{0.5}{3}=\frac{2\times 0.5}{3}=0.33 \ce{M}}$.

In order to determine exactly the concentration of silver ion, we have to use the solubility product of silver phosphate:
${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$


Aryl halides by Hunsdiecker reaction I wanted to know wether we can prepare aryl halides by Hundiecker reaction or not. if so then what is its mechanism i know the mechanism for alkyl halides. please refer these links and please try to explain with reason.

http://www.organic-chemistry.org/abstracts/literature/918.shtm

http://www.organic-chemistry.org/abstracts/literature/449.shtm

There might be more recent publications, but an old review by Robert G. Johnson and Robert K. Ingham, Chem. Rev., 1956, 56, 219-269, gives some directions.

Based on early attempts with benzoic acid that date back to the time around 1840, is was long believed that the silver salts of aromatic carboxylic acids do not undergo a Hunsdiecker reaction at all. The authors write:

  In no case was bromobenzene obtained as a result of the
  action of bromine on silver benzoate. The product of Peligot and Bunge was a monobromobenzoic acid; apparently Kekule’s experiment was completely unsuccessful.

However, it later turned out that the inital verdict wasn't quite true.

Benzoic acids bearing electron-withdrawing substituents, such as $\ce{Cl}$ or $\ce{NO2}$ can be converted to their corresponding bromoarenes in a Hunsdiecker reaction. The authors of the review denote:

  Thus, the three isomeric nitrobenzoic acids (as silver salts) were
  converted to the corresponding bromides in excellent yields (ortho, 95 per cent; meta, 89 per cent; para, 79 per cent). The yields from the chlorobenzoic acids were less satisfactory, but sufficiently high to indicate the efficacy of the chloro group.


Why are allenes chiral? How is 1-bromo-3-chloropropa-1,2-diene ($\ce{BrHC=C=CHCl}$) chiral? It doesn't have any carbon that is connected to four different groups.

Asked and answered, but I think one thing that's missing is that allenes are not planar like alkenes or alkynes are. You can refer to this question for an explanation.

The dihedral angle between the two halogens is 90 degrees (ideally). Here's an animation hopefully providing a better view of the 3D structure:

This results in the two mirror images being non-superimposable upon each other, which means the molecule satisfies the criterion for chirality. If you rotate the right-hand molecule by $90^\circ$ such that the blue and white atoms at the top coincide, the green and white atoms will not match each other.

In general, any allene with the formula $\ce{R^1R^2C=C=CR^3R^4}$ will be chiral as long as $\ce{R^1}\neq\ce{R^2}$ and $\ce{R^3}\neq\ce{R^4}$. In this case we have $\ce{H} \neq \ce{Br}$ and $\ce{H} \neq \ce{Cl}$, so the compound in the question is chiral.

If $\ce{R^1} = \ce{R^2}$ or $\ce{R^3} = \ce{R^4}$, then the compound will possess a mirror plane and will therefore be achiral.


Can acid mine drainage form acid rain? Can acid mine drainage form acid rain?

Since normal acid rain is formed via combustion of fossil fuels, the sulphur dioxide and nitrogen dioxide are all in the gas phase, whereas in acid mines the sulphuric acid is in solution.

Here's the reaction for acid rock drainage if it helps:

$$\ce{FeS2 + 3.75 O2 + 0.5 H2O → Fe^3+ + 2 SO4^2- + H+}$$

I don't think acid rain formation from liquid acid mine drainage is very likely.  The reason is that in acid rock drainage (a.k.a. acid mine drainage or AMD), the acidity is caused by sulfuric acid.  Sulfuric acid, $\ce{H2SO4}$ is much less volatile than water.  It simply won't appreciably evaporate from bulk liquids at normal conditions on the Earth.

There is one condition where acid mine drainage could possibly contribute to the formation of acid rain: if there were lots of aerosols formed from acid mine drainage waters.  If there were an AMD-laden stream or creek that flowed over a waterfall or was otherwise agitated enough to form very small droplets and splashes, some sulfuric acid might travel to the atmosphere in the small aerosol droplets.  That could contribute to acid rain, although a more likely effect would be a much more local (tens to hundreds of feet) acidification of the environment.

However, even then, I suspect that sulfur from coal combustion would be the dominant source of airborne sulfuric acid in most regions.  In the United States, coal consumption is roughly one gigaton per year, leading to about 20 megatons of sulfur emissions to the atmosphere.


Which of these would most likely be a strong electrolyte? An ionic compound
A covalent compound
A polar compound
A nonpolar compound

Previously, I have been taught that polar compounds can be either strong or weak electrolytes. I also learned that ionic compounds are generally strong. Therefore, my guess would be a, an ionic compound. However, I am not entirely sure of my answers. Could you please explain to me how I could confirm and get the right answer? Thank you!

Yes. You are right. Ionic compound are strong electrolytes if they are soluble in water. $\ce{CaF2, BaSO4}$ are ionic compounds, but they are not soluble in water. I would not consider them as strong electrolytes. On the other hand, ionic compounds are a part of the bigger family of the polar compounds. But polar compounds are not always strong electrolytes. $\ce{CO, H2S, HF, H2O}$ are polar, but they are not strong electrolytes.


Why does my coffee creamer have titanium dioxide in it and is it safe? From the manufacturers website of a coconut french vanilla coffee creamer:

  Ingredients: ORGANIC COCONUT MILK (WATER, ORGANIC COCONUT CREAM), DRIED CANE SYRUP, NATURAL FLAVORS, COLORED WITH TITANIUM DIOXIDE, DIPOTASSIUM PHOSPHATE, CARRAGEENAN, GUAR GUM.

My guess, from looking at the abstract of the report (which is behind a paywall), "Titanium Dioxide Nanoparticles in Food and Personal Care Products" is that the the metal is simply used for the (white) coloring. Is this all it is used for?

From the paper abstract:

  ... human exposure analysis to $\ce{TiO2}$ through foods identified children as having the highest exposures because $\ce{TiO2}$ content of sweets is higher than other food products, and that a typical exposure for a US adult may be on the order of 1 mg Ti per kilogram body weight per day.

Is there any potential risk (toxicity) to consumption of $\ce{TiO2}$ at these levels?

$\ce{TiO2}$ is is the most widely used white pigment out there, and it is considered to be non toxic. You'll find it used as food coloring commonly in candies, but in this instance  it's in your coconut sweetener. Aside from food, it's used as a primary pigment in paint (the more $\ce{TiO2}$ generally the higher the quality of paint, because it results in higher opacity), solar cells, toothpaste, and as you've mentioned, sunscreen.

The only health concerns I found were from inhalation

  Titanium dioxide dust, when inhaled, has been classified by the International Agency for Research on Cancer (IARC) as an IARC Group 2B carcinogen, meaning it is possibly carcinogenic to humans. The findings of the IARC are based on the discovery that high concentrations of pigment-grade (powdered) and ultrafine titanium dioxide dust caused respiratory tract cancer in rats exposed by inhalation and intratracheal instillation. The series of biological events or steps that produce the rat lung cancers have also been seen in people working in dusty environments. Therefore, the observations of cancer in animals were considered, by IARC, as relevant to people doing jobs with exposures to titanium dioxide dust.


Equivalence ratio explained To let you know where I'm coming from: I haven't studied chemistry since high school. I am now studying something else, but I have a dataset coming from your field and I would like some help in understanding just the very very basics of what it is.

What I have is $\ce{NOx}$, nitric oxide and nitrogen dioxide in an engine exhaust and the equivalence ratio. It's from the dataset ethanol in the package SemiPar in R, and the original reference is Brinkman (1981). You'll find a figure at the bottom.

What I want: I want to be able to, extremely briefly, explain what this is describing.

My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.

Is this correct? Would you be able to provide a better explanation? Imagine you're trying to explain it to a five-year-old, that's about how much of this I know!

If you've read this far, please also help me with appropriate tags for the question. I'm not sure what to choose.

Reference:

Brinkman, N.D. (1981). Ethanol fuel – a single-cylinder engine study of efficiency and exhaust emissions. SAE transactions Vol. 90, No 810345, 1410–1424.

You are correct.  There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$.  This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that are in the fuel.  For fuel that are pure octane, the combustion reaction is:

$$\ce{\frac{25}{2}O2 + C8H18 -> 8CO2 + 9H2O}$$

Which means 12.5 moles of oxygen are required for every mole of octane.  For 12.5 moles of oxygen you need about 60 moles of air because air is only 21% oxygen.  That works out to 1740 grams of air (~29 grams per mole).  One mole of octane is 114 grams.  So the theoretical, or stoichiometric ratio of air to fuel is 1740/114 or about 15.

Most motor fuels are not pure octane.  The additives complicate the math a bit, but the stoichiometric ratio is around 14 for many fuels.

The equivalence ratio is just the actual air-to-fuel ratio divided by the theoretical.

  My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.

You said the second part backwards.  If the ratio is below one, there is too little air (i.e. too much fuel), and if it above one, there is too much air (not enough fuel).  So incomplete combustion should happen more at lower equivalence ratios.

Also, the exhaust temperature will be maximized for equivalence ratios close to 1.  Otherwise, there are extra reactants, either fuel or air, which remain after combustion and absorb some of the combustion energy, resulting in a lower temperature increase.  Since temperature is important for NOx formation, I figured it was worth mentioning.


How could mass increase when sulfuric acid is added to calcium carbonate? Context: I was doing an experiment in class that investigates the rate of mass loss over time when $\pu{20 mL}$ of sulfuric acid is added to $\pu{5g}$ of calcium carbonate. I did the experiment 3 times, each experiment involves mixing four, $\pu{20 mL}$ different concentration of sulfuric acid ($\pu{0.5 M}$, $\pu{1.0 M}$, $\pu{1.5 M}$, and $\pu{2.0 M}$) into $\pu{5g}$ of calcium carbonate.

Hypothesis: If concentration increase, the rate of mass loss should increase too

Result: Everything went out just fine, however, in each of the 3 experiments, when I add the $\pu{2.0 M}$ sulfuric acid to the calcium carbonate, the mass INCREASES (mass still loss overtime).

Could anybody please explain how could this be happening? As they react, carbon dioxide is produced and there should only be a mass loss. Also, this only occurs with the $\pu{2.0 M}$ sulfuric acid, the other concentrations work out just fine.

Could the electronic scale wasnt working properly? but this only happened with $\pu{2.0 M}$ acid.

There are several contributing factors to this outlier observation. First of all, when scientists did mass based experiments, even hundreds of years ago, they recorded mass up to four decimal places. The mass of CaCO3 should have been recorded to 0.1 mg for example 5.0612 g. When you say that mass increases, we do not know the details of mass change. In short, never record mass with one significant figure.

Secondly, how did you measure volumes of 20 mL acid? If you were using cylinders, they are never accurate. There is significant error in volume transfer. With a relatively viscous solution such as 2M H2SO4, if you transfer 20 mL, the transferred volume via cylinder is usually less than 20 mL.

This all about good laboratory practice. Coming to chemistry: when CaCO3 reacts with H2SO4, if forms an insoluble salt called calcium sulfate (CaSO4). With concentrated 2 M sulfuric acid, once it comes in contact with CaCO3, it may form a coating of CaSO4, which would slow down the rate of reaction. Did you see some unreacted solid mass?

Also conc. sulfuric acid is hygroscopic.


Difference between Epimers and Diastereomers My lecture notes describe epimers as compounds which differ by
configuration at only one carbon and gives the example of D-erythrose and D-threose as being epimers.

However isn't that the same thing as diastereomers? My understanding is that epimers are supposed to be some kind of subset of diastereomers, however I am not sure how or why. I tried searching the web but I got more confused with one site saying that carbohydrates in open form can have two different chirality centres and still be epimers.

This scheme I just drew up specifically for you should answer your question.

Diastereomers are stereoisomers that are not enantiomers of each other. That includes conformers (geometric isomers that derive from single bond rotation; usually interconverting rapidly) and atropisomers (under which I would subsume E/Z isomers; they derive from hindered rotation around a typically single bond and are separable) and anything with differeces in asymmetric carbons.

If you have exactly one asymmetric carbon with inverted stereochemistry, then the diastereomer is a special one: an epimer. Anomers (e.g. α- and β-D-glucose) should be considered special cases of epimers.

Let me use the menthol isomers (image taken from Wikipedia, where a full list of authors is available) to clarify the concept:

Two vertically aligned structures are always enantiomers. Two structures that are not vertically aligned are diastereomers.

(+)-Isomenthol is (+)-menthol’s C4-epimer, while ($-$)-neoisomenthol is (+)-menthol’s C1-epimer. And ($-$)-neomenthol is (+)-menthol’s C2-epimer. All of these are also diastereomers.

(+)-Neomenthol is not an epimer of (+)-menthol. It is still a diastereomer, however.


Why there were changes in molar mass of lithium? I am curious why there were changes in molar mass of lithium in the past.

The values were like 9 to 10 before 1830. But suddenly after 1830, the range of molar masses seems similar to the molar mass that is used now.
I thought there might be some technical issues in the past. But, from 1830, the values were consistent and stable. So, I cannot assume there were lack of development in equipments and technics.

So, my questions are:

1. why the values were around 9-10 before 1830, not around 6.941?

2. why the values suddenly became similar to the molar mass which is used now?

This doesn't necessarily completely answer the question, but it at least moves closer to the root of the problem. From the Wikipedia article that MaxW included in his comment:

  Brande also described some pure salts of lithium, such as the chloride, and, estimating that lithia (lithium oxide) contained about 55% metal, estimated the atomic weight of lithium to be around 9.8 g/mol (modern value ~6.94 g/mol)

So it seems to stem from an erroneous estimation of the percent composition of lithium oxide. If we go to the source (Brande's manual of chemistry:https://books.google.com/books/reader?id=zkIAAAAAYAAJ&printsec=frontcover&output=reader&pg=GBS.PA190), we find that this error likely stems from the fact that lithium oxide was difficult to obtain and so very limited experimentation had been done to characterize it.


t-butylbenzene in highly acidic medium Why would the following mechanism be valid for the conversion of t-butylbenzene to benzene in highly acidic medium followed by an aqueous workup?

The main problem I have with this mechanism is the formation of the secondary doubly allylic carbocation. Why not go for the tertiary, inductively and doubly allylic stabilized carbocation?

But I guess the problem with going down this path it becomes hard to see how the t-butyl group might leave.

As you can see, the book's mechanism is entirely dependent on having the secondary allylic carbocation.

All 6 positions on the benzene ring are being protonated (the frequency of protonation at the different sites is related to the stability of the carbocation generated) in acidic media, so all of the ions you suggested are being formed.  Protonation at 5 of these positions can only regenerate starting material when one of the two protons at the initial site of protonation is removed (an identity reaction, we could make this pathway visible by running the reaction in deuterio-acid + $\ce{D2O}$).  Protonation of the benzene ring at the carbon bearing the substituent provides a choice.  Either the proton can again be removed or the stable t-butyl cation can be ejected.  All of these things are happening in the reaction, but only removal of the t-butyl group is visible.


Synthesis of an epoxide from 1,2-diol Why is there no formation of epoxide when the two OH-groups are on different planes? I do understand the epoxidation mechanism. However, I would be grateful if someone cleared the specific requirements for epoxidation.

Thionyl chloride first reacts with the alcohol to form an alkyl chloro sulfite,which gives various stereochemical products as shown here.

Extending this to the current question , alkyl chloro sulfite 2 is formed.${Cl^-}$ attacks in a ${S_N^2}$ mode to give 3.

Base deprotonates 3 to give 4.

Alkoxide and chloride in 4 are in correct trans configuration for a ${S_N^2}$.This results in 5(B in your scheme).

If we take C in your question , and follow the scheme above, 9  is formed.In 9 alkoxide and chloride are not in correct configuration for ${S_N^2}$.Therefore
C in your question does not undergo epoxidation.


SN2 on compound containing halogen and carbonyl I was given the following question on a sample exam for my organic chemistry course:

I am quite certain that it would happen in a SN2-type reaction considering that

the nucleophile is $\ce{CN-};$
the solvent is DMF.

However, I'm not sure what electrophilic center would be attacked first. I'd appreciate any explanation about why the nucleophile would choose one carbon over the other (if that were the case). The question also states that there are 2 product formed, so could it be possible that an elimination reaction is happening as well?

If it attacks the carbonyl carbon, then you form an unstable tetrahedral intermediate with a negative formal charge on the oxygen. Unless you treat the compound with water so that the oxygen can remove it's formal charge, what will happen is that the reaction goes backwards where the CN group is kicked off.
However, if the CN attacks the Cl group, then you can have a substitution where you get a stable compound finally.


Finding the percent of calcium carbonate in commercial antacid by using back titration I'm an IB Chemistry student and I had to do a back titration lab where 1g of a powdered antacid tablet were mixed with 50mL 1M HCl, and the mixture was titrated using 0.2 M NaOH; the NaOH was to neutralize the leftover HCl that hadn't reacted with the antacid. The active ingredient in the antacid is CaCO3.

It took me about 40mL of NaOH to titrate the leftover HCl to endpoint; 40 mL at 0.2 mol/L translate to 0.008 moles of NaOH. In the equation for reaction between NaOH and HCl the ratio of the two reactants is 1:1; therefore there were 0.008 mol of unreacted HCl in the mixture.

There were 50mL of HCl (1 mol/L) in the mixture originally; 50mL * 1 mol/1000mL = 0.05 mol HCl total. Subtracting from this the molar quantity of unreacted (leftover) HCl yields: 0.05 - 0.008 = 0.042 mol HCl that did react with the CaCO3 in the antacid.

The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl --> CaCl2 + CO2 +H2O
Therefore, molar ratio between CaCO3 and HCl is 1:2. 0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.

That's a problem - i can't be having 210% composition by mass (the sample was 1g; I calculated 2.1 g CaCO3 present). So somewhere along the line i made a mistake, but I've checked my math so I don't know if it was a calculation error or if my method is dead wrong? Can anyone help me with this?

In chemical analysis usually do three times the same experimet, so you can minimize experimental errors.

After that, in order to find where is the error you can check:

Has everyone in you class use the same HCl and NaOH solutions (taken from the same bottle i.e.?

The antiacid tablets are all the same (given by your teacher)? If don't, could you post the composition?

Have you measured correctly the volumes used? To obtain that 210%, it seems tha you have used around 25 mL of HCl solution instead 50 mL.


Why is NH4OH used in Polarimetric analysis? Why is a sugar solution pre-treated with ammonium hydroxide before analyzing? And why is it specifically ammonium hydroxide instead of sodium hydroxide or calcium hydroxide?

Do you see a potential corresponding to the sugars, or to their Schiff bases, or to said imines' trimers?  Perhaps ammonia gives you a nice pH buffer, as opposed to alkali hydroxides that will wildly swing pH and allow Cannizzaro disproportionation of reducing sugars' aldehyde groups at high pH. Ammonia does not reduce at the cathode.


Index of Hydrogen Deficiency for para-Toluenesulfonylchloride In para-Toluenesulfonylchloride, there are 5 pi bonds and 1 ring. This suggests a IHD of 6. However, the actual IHD is 4. How come?

Is it because the formula IHD = rings + pi bonds does not apply to compounds containing sulfur? If not, what formula do I use? I have this formula:

$IHD = \frac{2C + N - X - H + 2}{2}$

Using this formula yields a IHD of 4.

You have to start from the general formula for the degree of unsaturation
$$DU = 1 + \frac{1}{2} \sum n_i(v_i-2),$$
where $n_i$ is the number of atoms with a valency of $v_i$.

For para-toluenesulfunylchloride ($\ce{C7H7SO2Cl}$) you will therefore arrive at
\begin{align}DU &= 1 + \frac{1}{2}\Big[n_\ce{C}(4-2) + n_\ce{H}(1-2) + n_\ce{S}(6-2) + n_\ce{O}(2-2) + n_\ce{Cl}(1-2)\Big]\\
&=1+\frac12\Big[7(4-2) + 7(1-2) + 1(6-2) + 2(2-2) +1(1-2)\\
&=1 + \frac12\Big[14 -7 +4 +0 -1\Big]\\
&=6,
\end{align}

because sulfur uses all six valence electrons in this case.


Introduction to Chemistry Textbook - Theory based TL:DR conditions below

Highschool to first year undergraduate level
Theory based(Experimentation not wanted)
General - Giving a good introduction to all or most areas of Chemistry
Rigorous - Professional formatting - Not a casual textbook

I am looking for a textbook for someone(me) who has done no chemistry, but isn't too basic as it progresses. What are some of the better textbooks for starting to learn chemistry? I don't know any specific fields, so I would like to get some exposure to a number of different topics. I am a third year Mathematics student, so I am not worried about it being too concise, I am willing to reread areas one hundred times if need be.

One note: I am not interested in doing experimentation at all, and would just like to learn theory, as I have (sometimes) embarrassingly bad knowledge of the field as I didn't take chemistry in highschool(which I greatly regret).

Highschool through to the end of first year undergraduate level preferably, or alternatively multiple textbooks by the same author of successive difficulty.

Chemistry by Zumdahl is very good. I teach General Chemistry I and this is the book we use. I still reference it frequently. You will want the instructor solutions to accompany it. I let students check this out for classroom use, but you should have it for yourself at least. (I saw for the first time this semester a version that is loose leaf; this is great since it is sort of big. We have a custom version that takes out parts not covered. There is also a nice ebook.) This book is lovely, except for electron affinity.
For introductory chemistry, Zumdahl and DeCoste wrote Introductory Chemistry: A Foundation. We use this for Chem 101. Also very good, but a little too elementary for my taste and there is not as many topics discussed. But this may be what you want. You are probably somewhere between these two.

The next two are probably more rigorous than you would like, but worth mentioning.

Principles of Instrumental Analysis by Douglas A. Skoog is outstanding, but probabbly more advanced than you would like.
Physical Chemistry - Atkins. This book is very dense and is definitly rigorous. This will take an undergrad 2-3 semesters to get through. This may be a little overkill. You will actually see many general chemistry textbooks take exercises from this book. Really need the solutions bad for this book, it is that hard. A great place to find exam questions.


Specific heat problems I am in need of help for my understanding in specific heat capacity. To be honest, I understand the lesson effectively, However, just a few days ago, I've encountered two similar yet different-in-approach problems. Here are the problems:
1.) A sample of gold was dropped to a glass of water. The water had a given initial temperature that was higher than the gold. Eventually, a final temperature for both elements were given. The problem was asking for the mass of the gold?

The solution showed to me that the heat of the gold (qgold) released is equal to the heat of water (qwater) absorbed.

qgold = -qwater

Now, I can see that the glass which served as the container of the water was not included in the equation. As far as I can get the logic of this, the second problem seemed to have a different approach but this time the container is involved in the heat transfer. Now this is the second problem:

2.) An unknown sample with a given mass and given initial temperature was dropped to a can with water. The can and the water had a similar initial temperature, but the two had a different mass. The final temperature of all elements present in this problem was given. The problem was asking for the specific heat of this unknown sample. The thing is, this is the part where I am a bit confused because the solution given to me was this:

q of unknown substance = q of the can + q of water

Now, what I really want to ask is this: When do I consider including the heat of the container of the system in this type of problem, as it is very situational in this case? My professor told me it's about material thing. However, I seem to not deem my professor's explanation to be sufficient, for I know a better explanation could be given.

I hope I would get a very comprehensive reply as soon as possible.

Thank you!

There are often some unstated simplifying assumptions in questions of this kind. For example, in both cases, no account is taken of heat transfer to the surroundings (the air?). Unless the system is isolated, one expects that eventually the system would reach the temperature of the surroundings. However, it is hardly worth stating this explicitly: our common experience is that we can neglect the transfer of heat to the surroundings, on the timescale of the experiment, because it happens comparatively slowly.

In the examples you give, some similar assumptions may apply. If you have a metal can, perhaps not very thick (low overall thermal capacity) and of high thermal conductivity, it's a reasonable assumption that it comes to thermal equilibrium with the contained liquid fairly quickly. On the other hand, for a glass, with lower thermal conductivity, it may be a reasonable approximation that no heat is transferred to it on the timescale of the experiment.

It is, as you say, situation-dependent: it depends on the material properties of the container. It would be most helpful if problems of this kind made it clear what you should assume. Alternatively you may be able to deduce what to assume, from the data provided (or not provided). Or indeed, perhaps it is expected that you can make the right assumptions, knowing something about the materials concerned.


Is heat an extensive or intensive property? Is heat extensive or intensive property?

I surfed the Internet for the same and found quite different answers. Like, heat is not a property, so it is neither. One said that since heat depends on the amount of substance, it is extensive. I thought about the latter one as well.

So, my question could be primarily divided into:

1. Is heat a property of matter?

2.1. If yes, is it extensive or intensive? Why?

2.2. If no, why?

In thermodynamics, heat is not a function of state but a process function. So for the question

  Is heat a property of matter?

the answer is no.

  ... is it extensive or intensive?

Since the concept of intensive and extensive properties only applies to physical properties that are a function of state, heat is neither intensive nor extensive.

As Chester said in the comments, it is worth noting that heat (as well as work) applied to a system can cause changes in the thermodynamic equilibrium state of the system.


What's the relation between molarity and molality? I am trying to derive the relation between the molarity and molality of a solution

Mole concept $\displaystyle n = \frac{M'}{M''}$

Here, ($n$) stands for number of moles
$\ce{(M^{'})}$ stands for mass of substance in gram
$\ce{(M^{''})}$ stands for molecular/atomic mass of substance

Let $V_{\text{sol}}= $ volume of solution
Molarity (M) $= n/V_{\text{sol}}$

The above equation of molarity ($M$) shows the concentration of solution of substance, i.e. ($M$) is the ratio of number of moles ($n$) with respect to it's volume ($V_{\text{sol}}$)

let $W$ = mass of solvent in kilogram
molality $(m) = n/W$

The above equation of molality ($m$) shows the concentration of solution of substance
i.e. ($m$) is the ratio of number of moles ($n$) with respect to mass of it's solvent ($W$)

What is the relation between molarity ($M$) and molality ($m$)?

Molarity ($M$) = $n$/$V_{\text{sol}}$
Molality ($m$) = $n$/$W$

I'm confused on how to relate the $W$ with $V_{\text{sol}}$.

$W$ and $V_{\text{sol}}$ can be related via the density of solution. We have, mass of solution = mass of solute + mass of solvent.
Hence,
$$ M_{\text{sol}} = W + M’ $$

Let us assume the density of the solution to be $\rho$. Since, density is equal to mass over volume, we have,
 $$ \rho=  \frac{M_{\text{sol}}}{V_{\text{sol}}} = \frac{W + M’}{V_{\text{sol}}} $$
Thus, you have relation between $W$ and $V_{\text{sol}}$.


Does an electron's spin change when being excited? My preconception is that when you fill an empty orbital by exciting an electron (with a photon), the empty orbital should be filled by an electron with the same spin. However, I've seen examples of that happening and not happening. For example, on this University of Calgary
web page there is an image of the electron spin being flipped:

The lowest energy transition is that between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) in the ground state.  The absorption of the EM radiation excites an electron to the LUMO and creates an excited state. The more highly conjugated the system, the smaller the HOMO-LUMO gap, DE, and therefore the lower the frequency and longer the wavelength.

Does it just not matter that much energetically?

The picture shows a triplet excited state returning to the ground state by emitting a photon, i.e. phosphorescence. This can only happen if there is also an interaction that couples angular momentum change with the transition, such as spin orbit coupling.
The electron has two quantum numbers; the spin $S=1/2$ but is not the spin of the electron that is changing but the other quantum number 'z' , 'azimuthal' or 'magnetic' spin quantum number $m_z=\pm 1/2$.


Electrophilic Addition Reaction When HBr reacts with ethene it proceeds by the attack of proton on Ethene to give Ethyl carbocation and then attack of nucleophilic bromide ion to give Bromoethane.

My question is why does the proton react with Ethene earlier than the bromide ion? Is a proton more reactive than bromide ion or is there some other reason?

The mechanism of the addition of hydrogen bromide to ethene is called electrophilic addition. The region of highest electron density in ethene is the double bond, where the π electrons are. Hydrogen bromide acts as an electrophile toward the π electrons of the double bond. The proton of hydrogen bromide is positively polarized and thus electrophilic. (Obviously, the negatively charged bromide ion is not electrophilic.) Hence, the first step is an acid-base reaction in which hydrogen bromide donates a proton to ethene, forming a carbocation.


Does isopropyl chloride undergo SN1 or SN2 with sodium acetate? What's the reaction of isopropyl chloride with sodium acetate? (The solvent is acetic acid)
Is it SN1 or SN2?

My thoughts:

It's not that difficult to approach isopropyl chloride's carbon, for there isn't much steric crowding. It's just a 2 degree alkyl halide with two methyl groups. This favours SN2.
The carbocation formed by the 2 degree alkyl halide, here in this case, is stabilised by the +I effect of two methyl groups and also six hyperconjugative structures. This should favour carbocation formation, hence SN1.
Acetic acid is a polar protic solvent, and can solvate sodium acetate.
Acetate ion isn't a great nucleophile (or is it?). I'm not sure, here. I know that acetic acid is a weak acid, so acetate ion is a strong base but I believe its nucleophilicity is suppressed by solvation.
Chlorine isn't an amazing leaving group.

So, does SN1 take place or SN2? Could someone also give some experimental data to support their reasoning?

Here's something I found, it says that both SN1 and SN2 can occur. Now which one occurs here, is probably decided by the nucleophile? I'm not sure. Anyway, thought I'd put this table up here for reference:

Source:
Organic Chemistry - Clayden - Oxford University Press - 1st edition - Page 424

Edit: Someone in the comments section said that you can't predict which mechanism is major, and have to look for experimental facts. So, for the sake of the completeness, where are the facts? Could someone please post the relevant details as an answer?

Actually in these type of reactions, where the reactant can both possibly undergo $\mathrm{S_N1} $ and $\mathrm{S_N2} $, the reactions actually proceeds through a mixture of both type of reaction mechanism. In these type of reactions, rate is related as $$\text{Rate} = k_1 [\ce{R}] + k_2 [\ce{R}][\ce{X-}]$$ So, there is also the stereochemically inverted product and also racemised product.
Fraction of inverted product is $k_2/(k_1+k_2)$ and the fraction of racemised product is $k_1/(k_1 +k_2)$. But the actual values of $k_1$ and $k_2$ is experimentally determined and probably very difficult to determine by logical approach only.


Understanding the CIP configuration for cyclic substituents with hetero-oxygen atoms As per CIP sequence rules, the topmost $\ce{C}$ on the left gets 1st priority. Moving on to the right side of the double bond, we have:

Top $\ce{C}$: ($\ce{C}$,$\ce{O}$,$\ce{O}$,$\ce{H}$)
Bottom $\ce{C}$: ($\ce{C}$,$\ce{O}$,$\ce{O}$,$\ce{H}$,$\ce{C}$...)

Thus bottom $\ce{C}$ gets priority one on the right side.

Now, since 1st priorities are opposite, it's an E isomer. But the solution manual states that it's a Z isomer. Why is my method incorrect?

CIP rules for cyclic substituents are different. See
this question for the rules. (The rules are given in the question itself).

This will be the hierarchical digraph for the right side of the molecule.

The atoms encircled are the ghost atoms. From the digraph one can see why the compound is Z.


Is an amorphous solid a single phase? Amorphous solid possess short-range order but no long-range order (e.g. amorphous silica, the short-range order is originated from the tetrahedral bonding between Si and O). Is it possible to have multiple domains with different short-range orders in one amorphous solid?

If so, should we treat this solid as a single phase or multiple phases? Can we detect the existence of multiple domains?

Relative concept: polyamorphism

(It seems that polyamorphism implies that amorphous phases can be distinct from each other.)

Amorphous silica, https://en.wikipedia.org/wiki/Structure_of_liquids_and_glasses

It is difficult to give a general answer. A glass is an amorphous solid that may change its composition in a continuous way from one point to another. When melting brown bottles and colorless bottles in the oven of a glass factory, you may obtain a non homogeneous mixture, with regions more or less colored. It is obviously one and the same phase all over the mixture. But as the mixture is not homogeneous, its composition is not the same everywhere.


Is there a better equation than the Nernst equation for low concentrations? The Nernst equation can be applied to the whole cell reaction or a half-cell reaction. Consider the half cell reaction:
$$\ce{Cu^{2+} + 2e^- -> Cu(s) \quad \quad E^0 = +0.337~\rm{V}}$$

The Nernst equation expressed in base 10 log at 25 °C is:

$$E = E^0 - \frac{0.1596}{2}\log_{10} \frac{a_{\ce{Cu^2+}}}{a_{\ce{Cu}}}$$

But $a_{\ce{Cu}} = 1$ by definition, and for low concentrations of $\ce{Cu^2+}$ where $a_{\ce{Cu^{2+}}} \ll 1$ then: $\ce{[Cu^2+]} \approx a_{\ce{Cu^2+}}$.

But when $\ce{[Cu^2+]} < 1$ then the log will be negative so:

$$\lim_{\ce{Cu^2+}\to 0} (0.337 - 0.0798 \times \log_{10} [\ce{Cu^2+}]) = \infty $$

This doesn't seem realistic. For instance $\ce{[Cu^2+] = 1.0\times 10^{-1,000,000}}$ is fine mathematically, but that sort of concentration doesn't make any sense chemically.

Thus what is the limit of the half cell potential as the $\ce{[Cu^2+] -> 0}$?

Edit-1 - My gut reaction to this is that you can't really measure the activity of a single ion. If the ion of interest is low in concentration compared to the ionic strength in general, then it would seem that the activity of that ion should go to zero as its effect on the total overall ionic strength goes to zero.

The Nernst equation does not need a replacement for low concentrations. Yes the limit is $\infty$, which essentially means you can't realistically have zero concentration of anything. We may think of absolutely pure water as an idealized situation. Well, as soon as it gets in touch with copper, some $\ce{Cu^2+}$ ions instantly jump into the solution, and from that point on, you have a certain non-zero concentration of them (though probably pretty low), and the equation works just fine.


What is the meaning of an oxide being 'Higher' or 'Lower'? To put it simply, many books use the term 'Higher Oxide' or 'Lower Oxide'.

What does this mean?

I think it generally means that the average oxidation state of the metallic element in the oxide is higher or lower. You can also say that the mass percentage of oxygen in the oxide is higher or lower.


Can you calculate the properties of a substance based solely on its atomic properties? I'm trying to write some software that I can use to determine, roughly, what the physical properties of a pure substance are. I know I could just use a database of the known properties of each element, but that doesn't do much for me when I go to create molecules and combine substances. A database that lists how each element will react to another is likely impossible, so I got to thinking... there has to be a mathematical way of determining these properties!

For example, I understand that there are trends on the periodic table that show that the stronger the force between the molecules is, the higher the melting point will be. I also understand that we can calculate how strong that force is based on what we know about the atom. So why can't I find any information relating the two in some mathematical way? The same goes for color, hardness, malleability, etc. Do we really only know these values through testing? We know the values but we don't know why they are what they are?

I know this is deeply complex topic and may not even be computational possible. In those cases, I'm trying to do as much as I can to approximate.

For atomic and molecular-level properties (bond, angle, torsion energies, etc), quantum mechanics provides the equations for solving for these properties. However, there are a number of approximations that are employed, especially for larger, more complex molecules.

For bulk properties of a substance, thermodynamics and statistical mechanics provide equations for solving for a variety of material properties (critical properties, phase equilibria, various types of enthalpy, and so on). Read up on equations of state.

There is also the option of molecular simulation. In this case you are solving relatively simple, well-established equations (eg, Newton's equations of motion in molecular dynamics simulations) repeatedly in order to compute the (usually) equilibrium properties of a substance of interest.

In all cases, the accuracy varies, and validating results by comparing various methods is considered best practice. While there are "purists" out there who believe that all parameters in a model should be derived from first principles alone, in practice models are often fit to experimental data (or even data from quantum mechanical calculations) to improve accuracy.


Fluorescence versus phosphorescence Not in an atomic aspect, I learned that fluorescence is the emission of light from substances until the source of light disappears (there is no emission in the dark), on the other hand, phosphorescence, due to a blocking of electrons on their way back to normal state, is the emission of light from substances even without any source (emission in the dark).

If it is so, then why do we call glowing objects in the dark as fluorescent?

If it is so, then why do we call glowing objects in the dark as fluoroscent?

I think this is just marketing: most people have heard the term fluorescence, while phosphorescence or luminescence are largely unknown terms.

While a delayed emission from a singlet state is possible, most glow-in-the-dark items do in fact show phosphorescence, i.e. emission from a tripet state.


How is 2D spectroscopy able to show quantum coherent transport through networks? For wave-like (quantum coherent) energy transfer in networks (eg. propagation of excitation in photosynthetic protein complexes of algae or FMO complex in plants) 2D electronic (photon echo?) spectroscopy is used (eg. in this article). I have read an article (unfortunately I can't cite it here) about how it works, but I didn't understand it. I know that the off-diagonal peaks in the spectrum mean some kind of coherence in the network but I don't know why. I also don't know what kind of coherence it means. What are meant by electronic eigenstates? What are the so called 'quantum beatings'? It would be very much appreciated if somebody would be so kind as to make these thing clear.

I'm going to focus on one question here. You really have several and it'd be better if you asked these as separate questions on the site.

First off, 2D spectroscopy involves using multiple pulses, just like multi-dimensional NMR experiments. There are a variety of techniques but in all cases, you're sending some sort of pump pulse which excites the system (e.g., electronic or vibrational levels) and then probing the system over time.

The resulting data is inherently two-dimensional. You know the absorption of the system along one axis, and the emission (or second absorption) along another axis. Often, you can watch this 2D data over time, giving you even more information.

Now how does this show coherence? Let's say you have some complex system. You take a UV/Vis electronic absorption spectra and you see two peaks.

Now normally, you'd think of this system as having two distinct absorptions. Maybe it's a metal complex and you see a ligand-metal or metal-ligand charge transfer band and a d-d transition at longer wavelength.

But perhaps instead, some of the intensity in a peak comes from energy transfer. This can particularly be the case in biology or materials systems, where there are multiple photosynthetic reaction centers with ever-so-slightly different energies. So you excite one center, but then energy dumps into another one (e.g., at lower energy).

By using two-dimensional spectroscopy, that will show up as an off-diagonal peak. You'll excite at one wavelength, but it fluoresces at another one!

That's an example of coupling between oscillators (either vibrational or electronic). Now there are multiple ways that energy transfer could occur. Usually, this happens essentially as a diffusive random walk - all phase information is lost.

In some cases, the wave packet transfers from one state to another, but retains its phase. That's coherent.

In 2D photon echo spectra, you'll see oscillations in the peak heights when coherent transfer occurs, thus "beats."


Nitration of Aromatic Compounds I am told by Dr. Klein that in the nitration of benzene, hydrogen sulfate anion acts as the base and deprotonates the sigma complex.

My question is why would the hydrogen sulfate anion be doing the deprotonation? We've just established that sulfuric acid is a stronger acid than nitric acid; hence sulfuric acid is able to protonate nitric acid to some extent. Now are we expected to believe that the conjugate base of a stronger acid is stronger than the conjugate base of a weaker acid? Why not have nitrate anion do the deprotonation? It is a stronger base than hydrogen sulfate anion.

It's not advisable to consider these types of reaction schemes in a slavishly literal fashion. In reality, there are multiple proton transfer equilibria occurring simultaneously, and the reactions will generally be quite rapid. While it's logical and technically correct to depict the most basic of the molecules (among those present in sufficiently large concentration) performing the proton abstraction, it's also of literal practical interest or importance in many cases. I suspect the reason that $\ce{HSO4-}$ was chosen is to emphasize that the sulfuric acid is actually catalytic, so it should be regenerated in the course of the depicted reaction mechanism. Of course, water is produced as a byproduct, so the sulfuric acid will not exist primarily in its neutral molecular form.


Does Methyl Iodide (CH3I) react with water (H2O) and if it does what would be its byproducts? Does Methyl Iodide ($\ce{CH3I}$) react with water ($\ce{H2O}$) and if it does what would be its byproducts?

A project I am designing requires a reaction with methyl iodide and a polar compound, so I was wondering if methyl iodide reacts with water.

Methyl iodide is hydrolyzed by water in an $\ce{S_{N}2}$ reaction ($\ce{S_{N}1}$ is disfavored by the instability of the primary methyl cation), which yields methanol and hydroiodic acid as products.

$$\ce{CH3I + H2O <=> CH3OH2^+ + I-}$$
$$\ce{CH3OH2^+ + H2O <=> H3O+ + CH3OH}$$

A possible side reaction is the formation of iodine and methane by the redox reaction

$$\ce{CH3I + H3O+ + I- <=> CH4 + I2 + H2O}$$

However, this reaction becomes only important when a large concentration of methanol pushes the hydrolysis equilibrium to the educt side. It is negligible in pure aqueous solution (reference).


Writing a system of differential equations with the given arbitrary reactions We have the following elementary reactions:
\begin{align}
&\overset{k_1}{\longrightarrow}A\overset{k_2}{\longrightarrow}{B+C}\\
&C \overset{k_4}{\longrightarrow} 2D \overset{k_5}{\longrightarrow}{C}\\
&B \overset{k_3}{\longrightarrow}\\
&D \overset{k_6}{\longrightarrow}
\end{align}
Let $\alpha(t), \beta(t), \gamma(t), \delta(t)$ denote the concentrations of $A, B, C, D$, respectively.
\begin{align}
\frac{\mathrm d}{\mathrm dt}\alpha(t) &= k_1 - k_2\alpha(t)\\[3pt]
\frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t)\\[3pt]
\frac{\mathrm d}{\mathrm dt}\bigl(2\delta(t)\bigr) = 2\frac{\mathrm d}{\mathrm dt}\delta(t) &= k_4\gamma(t)-k_6\delta(t)^2 \quad \text{or} \quad \frac{\mathrm d}{\mathrm dt}\delta(t) = \frac{k_4}{2}\gamma(t) - \frac{k_6}{2}\delta(t)^2\\[3pt]
\frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_5\delta(t)^2 - k_4\gamma(t) + k_2\alpha(t)\\
\end{align}
Is my systems of equations correct?

$\displaystyle\frac{\mathrm d}{\mathrm dt}\alpha(t) = k_1 - k_2\alpha(t)$ is correct.

Just to show you the way, here are the next equations:

$$
\begin{align}
\frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t) \\
\frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_2\alpha(t) - k_4\gamma(t) + k_5\delta(t)^2 \\
\frac{\mathrm d}{\mathrm dt}\delta(t) &= 2k_4\alpha(t) - k_6\delta(t) - k_5\delta(t)^2
\end{align}
$$


What is the molecular structure of xenon hexafluoride? Xenon hexafluoride, $\ce{XeF6}$, can be obtained by reaction of xenon with excess fluorine gas.

What is the gas-phase structure of the molecule?

What experimental evidence is there to support it, and is there any theory which allows for an prediction of the structure?

[Note that solid $\ce{XeF6}$ has multiple polymeric forms, and is hence excluded from the scope of this question.]

TL;DR Xenon hexafluoride has a fluxional structure in the gas phase, with multiple rapidly interconverting conformers. The three most important conformers have $C_\mathrm{3v}$, $O_\mathrm{h}$, and $C_\mathrm{2v}$ symmetries. The minimum energy conformer is probably $C_\mathrm{3v}$.

Experimental evidence
The structure of xenon hexafluoride ($\ce{XeF6}$) has always been controversial; it is a famously strong oxidising agent, so experimental studies of it have been hampered by difficulties in the isolation and storage of pure samples.
Early studies indicated that the structure of $\ce{XeF6}$ showed deviations from octahedral symmetry. Gas-phase electron diffraction (dating from 1968) showed that the structure was approximately, but not exactly, octahedral: the experimental diffractogram differed from the theoretical diffractogram calculated for a hypothetical octahedral molecule.1 The authors considered a wide range of possible geometrical distortions, and found that $C_\mathrm{2v}$ or $C_\mathrm{3v}$ structures were relatively consistent with the spectrum obtained, with the latter being the most probable configuration.2 These could be interpreted in terms of a stereochemically active lone pair:

If we start with an undeformed octahedron we can characterize the deformation to the equilibrium structure in terms of the site of the repulsive lone pair [...] Symmetry considerations would seem to dictate that the lone pair be centered on the corner, face, or edge in the equilibrium structure. [...] If the lone pair protrudes through the center of a face, a $C_\mathrm{3v}$ structure results [...]  If the lone pair points toward an edge a $C_\mathrm{2v}$ configuration is induced.

This approach has stood the test of time, and is reproduced in multiple recent inorganic chemistry textbooks.3–5
At roughly the same time, experiments with molecular beams of $\ce{XeF6}$ were carried out by Falconer et al. These beams were not deflected in an inhomogeneous electric field, and the authors derived an upper bound of $\pu{0.03 D}$ for the electric dipole moment of a rigid structure.6
This was occasionally presented as evidence for an undistorted structure, but the calculations did not take into account the possibility of a highly fluxional process, and later Bernstein and Pitzer showed that depending on the barrier height, dipole moments up to $\pu{0.6 D}$ could be plausible.7 On top of that, the distorted conformers were not expected to have large dipole moments, as the dipole moment of the lone pair would oppose the net dipole moments of the $\ce{Xe-F}$ bonds.
Vibrational spectroscopy offers a unique possibility to eliminate, or at least reduce, the influence of dynamic processes, since the timescales involved are extremely short. In 1972, Claassen et al. carried out infrared (IR) and Raman analysis of $\ce{XeF6}$ at various temperatures.8 A group theoretical analysis leads to the prediction that an octahedral molecule should possess two IR-active bands, one $\mathrm{t_{1u}}$ stretch and one $\mathrm{t_{1u}}$ bend. However, the IR spectra at both room temperature and $\pu{4 K}$ (in an argon matrix) show at least three stretches alone. Similar inconsistencies with an octahedral structure were observed in the Raman spectra. However, the extremely broad peaks precluded a conclusive assignment to a specific distortion.
Based on all the experimental data, Pitzer and Bernstein concluded that $\ce{XeF6}$ could be described by a pseudorotational model with the potential minimum most likely being a $C_\mathrm{3v}$ structure.9 Following in the footsteps of Bartell and Gavin,2 they attributed this deformation to a second-order Jahn–Teller effect, which we will now turn to.

Theories of bonding
Clearly, traditional VSEPR theory cannot satisfactorily explain the structure of $\ce{XeF6}$. Often, the structure is described as having a stereochemically active but fluxional lone pair, which is generally accurate; however, VSEPR cannot be used to predict the region of space which the lone pair occupies.
We could reject the 2c-2e bonds utilised in VSEPR theory, and move instead to 3c-4e bonds, which works for the lower fluorides $\ce{XeF2}$ and $\ce{XeF4}$, and is briefly explained here. Essentially, each $\ce{F–Xe–F}$ interaction is modeled as a linear 3c-4e interaction. However, this model predicts a regular octahedral structure, which is again incorrect.
As far as simple bonding theories go, the best explanation is given by qualitative molecular orbital theory. A MO scheme for octahedral $\ce{XeF6}$ can be constructed from Xe 5s, Xe 5p, and F 2p orbitals (neglecting π-type interactions). It is similar to the one for $\ce{SF6}$ (see here), except that there are two more electrons:

This opens up the possibility of a second-order Jahn–Teller distortion, and I have described the mathematical basis of this elsewhere. By distortion to a lower symmetry, the HOMO and LUMO are allowed to mix, which leads to a stabilising interaction. The distortion should be mediated by a vibrational mode with symmetry $\mathrm{A_{1g} \otimes T_{1u} = T_{1u}}$, which does indeed lead to a $C_\mathrm{3v}$ structure. The Jahn–Teller effect, as well as the application to the case of $\ce{XeF6}$, is described in the textbook by Albright et al.10 I have reproduced one diagram from the book (p 364):

Equation 14.4 shows the HOMO-LUMO mixing upon lowering of symmetry; equation 14.5 depicts the particular linear combination of the traditional $2\mathrm{t_{1u}}$ orbitals that transforms as $\mathrm{a_1}$ in the lowered symmetry; and equation 14.6 shows the interaction of the $2\mathrm{a_{1g}}$ orbital with this linear combination.

Computational studies
Computational studies of $\ce{XeF6}$ have generally been plagued by a strong dependency on the level of theory and on the basis sets used, due to the very small energy differences between the relevant conformers. There is still a lack of agreement on the exact details of the potential energy surface.
Kutzelnigg et al. provided the first thorough in silico investigation of xenon hexafluoride, along with a number of valence-isoelectronic species.11 They concluded that the balance between an undistorted $O_\mathrm{h}$ structure and a distorted $C_\mathrm{3v}$ structure is governed by competing effects. Distortions are found to be present when nonrelativistic Hartree–Fock theory is used, and the addition of polarisation functions (e.g., Xe 4f) to the basis set favours the $C_\mathrm{3v}$ structure. However, relativistic effects as well as electron correlation both favour the $O_\mathrm{h}$ structure. In the case of $\ce{XeF6}$, these lead to very small differences between the two structures.12
Very sophisticated coupled-cluster calculations (CCSD(T)-F12b) were carried out by Peterson et al., and led to the conclusion that the $C_\mathrm{3v}$ structure was the ground state.13 The eight equivalent $C_\mathrm{3v}$ structures (an octahedron has eight faces) underwent interconversion via $C_\mathrm{2v}$ transition states $\pu{1.08 kcal/mol}$ higher in energy. The perfect $O_\mathrm{h}$ structure was found to be a saddle point $\pu{1.80 kcal/mol}$ above the ground state, with a threefold degenerate imaginary vibrational frequency.
However, the accuracy of these calculations was called into question by Cheng et al., who found that the $O_\mathrm{h}$ structure was in fact a local minimum (i.e. no imaginary frequencies).14 According to the authors, the imaginary frequency found by Peterson et al. arises due to basis set effects in the CCSD(T) triples correction.
All in all, computations broadly support the fact that $\ce{XeF6}$ is dynamically distorted from $O_\mathrm{h}$ symmetry. However, the exact details have yet to be ironed out, and given that the energies involved are so close to each other, "an unbalanced calculation may fortuitously give the right answer" (in Kutzelnigg's words11).

Notes and references
1. Gavin, R. M., Jr.; Bartell, L. S. Molecular Structure of XeF6. I. Analysis of Electron‐Diffraction Intensities. J. Chem. Phys. 1968, 48 (6), 2460–2465. DOI: 10.1063/1.1669470.
2. Bartell, L. S.; Gavin, R. M., Jr. Molecular Structure of XeF6. II. Internal Motion and Mean Geometry Deduced by Electron Diffraction. J. Chem. Phys. 1968, 48 (6), 2466–2483. DOI: 10.1063/1.1669471.
3. Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 4th ed.;  Pearson Education, Inc.: Upper Saddle River, NJ, 2012.
4. Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford University Press: Oxford, U.K., 2014.
5. Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th ed.;  Pearson Education, Inc.: Upper Saddle River, NJ, 2014.
6. Falconer, W. E.; Büchler, A.; Stauffer, J. L.; Klemperer, W. Molecular Structure of XeF6 and IF7. J. Chem. Phys. 1968, 48 (1), 312–318. DOI: 10.1063/1.1667921.
7. Bernstein, L. S.; Pitzer, K. S. Electric field deflection of molecules with large amplitude motions. J. Chem. Phys. 1975, 62 (7), 2530–2534. DOI: 10.1063/1.430833.
8. Claassen, H. H.; Goodman, G. L.; Kim, H. Spectral Observations on Molecular XeF6: Raman Scattering and Infrared, Visible and Ultraviolet Absorption in the Vapor and in Matrix Isolation. J. Chem. Phys. 1972, 56 (10), 5042–5053. DOI: 10.1063/1.1676986.
9. Pitzer, K. S.; Bernstein, L. S. Molecular Structure of XeF6. J. Chem. Phys. 1975, 63 (9), 3849–3856. DOI: 10.1063/1.431880.
10. Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. Orbital Interactions in Chemistry, 2nd ed.; John Wiley & Sons, Inc.: Hoboken, NJ, 2013.
11. Kaupp, M.; van Wüllen, Ch.; Franke, R.; Schmitz, F.; Kutzelnigg, W. The Structure of XeF6 and of Compounds Isoelectronic with It. A Challenge to Computational Chemistry and to the Qualitative Theory of the Chemical Bond. J. Am. Chem. Soc. 1996, 118 (47), 11939–11950. DOI: 10.1021/ja9621556.
12. Interestingly, $\ce{RnF6}$ is predicted to be octahedral even at the SCF level, i.e. ignoring electron correlation. This is a result of the strong relativistic effects. On the other hand, $\ce{ClF6-}$ is octahedral simply because of electron correlation effects; the energy gained by distortion is insufficient. Together, these clearly show that even the second-order Jahn–Teller approach described earlier is incomplete, as it fails to explain other species valence isoelectronic with $\ce{XeF6}$.
13. Peterson, K. A.; Dixon, D. A.; Stoll, H. The Use of Explicitly Correlated Methods on XeF6 Predicts a C3v Minimum with a Sterically Active, Free Valence Electron Pair on Xe. J. Phys. Chem. A 2012, 116 (39), 9777–9782. DOI: 10.1021/jp3084259.
14. Cheng, L.; Gauss, J.; Stanton, J. F. Relativistic coupled-cluster calculations on XeF6: Delicate interplay between electron-correlation and basis-set effects. J. Chem. Phys. 2015, 142 (22), 224309. DOI: 10.1063/1.4922112.