text:
Possible Products in the Addition To a Double Bond in a Ring

I don't understand why both these products should be different. I know hydrogenation is syn addition and both the hydrogens add from the same side, but why should the plane in which both the hydrogen atoms are present even matter?
Also, bond rotation is possible around sigma bond which makes both these products identical.
Please explain why the two products are different and where am I wrong with this logic?

You are right, these two are identical.

Those two are identical but the rotation of that sigma bond is not possible.

@orthocresol what is your logic of calling them identical?

Just flip the one on the left upside-down, it is the same

@orthocresol is it okay to flip molecules like that?

Of course, if you really don't like flipping the molecule you can always flip yourself and look at it from the bottom up instead of looking at it top down. Why don't you try using a molecular modelling set and make the two molecules. That will also show you why that bond can't be rotated.

That pretty much answers my question. Thank you.

The two molecules are not different, they are identical. You can prove this in a number of ways:

Build both with a molecular model kit and see if you can orient them in the same way (spoiler: you can).
Use your imagination and 3D-vision to rotate one of the two structures until it looks like the other one.
Look for decisive symmetry elements. In this case, there is a plane of symmetry traversing the molecule through the $\ce{C-Br}$ bond. Therefore, the molecule cannot be chiral. Therefore, two versions of it that differ only in stereochemical orientation must be identical.

I can only speculate why your source drew it that way. One possibility would be to stress that the hydrogenation is not only syn-selective but can also happen from both sides (even though it gives the same product here). That doesn’t change that they’re identical, though.

why should the plane in which both the hydrogen atoms are present even
  matter?

In this case it doesn't.  The author of the example is laying the groundwork for the future.
The area above and below the double bond is often referred to as a "face". In your molecule, the two faces are not equivalent. If you build two models of your compound and in both color the top of the double bond red and the bottom blue, you'll find there is no way to superimpose one model onto the other where the red face is superimposed on the blue face.  These two faces are therefore not equivalent, they are said to be enantiotopic (mirror images) in this molecule.
In this case, addition of (an achiral reagent such as) hydrogen to the two faces produces the same molecule.  In more complex molecules, or when using chiral reagents this will not be the case.
Consider the case where there is a methyl group attached to the methylene group adjacent to the $\ce{C-Br}$ carbon in your molecule.  Now the faces are different (diastereotopic) and addition of hydrogen to the top and bottom faces will produce two different products (methyl and bromine on the same side or opposite sides of the molecule).
The concept of faces is important in chemistry.  Always consider if the faces are equivalent, enantiotopic or diastereotopic. Depending on what you find, the products from attack at the two faces may be the same, enantiomers or diastereomers.  The concept of faces is introduced in your problem.


text:
Mechanism of a dephenylation reaction
I recently came across this reaction in my undergraduate labs. The experiment is taken from this article: J. Chem. Educ., 2008, 85, 413.
The condensation of 2-methoxynaphthalene with cinnamoyl chloride leads to the formation of 9-hydroxyphenalenone:

I think it's fairly clear that the first step is a Friedel-Crafts acylation. Subsequently to form the new C-C bond, there must be a Michael addition of some sort; however this would imply that the phenyl group leaves as "$\ce{Ph-}$". There are two things I'm not very sure about:

How does the phenyl group even leave?
How does the oxygen get demethylated? I am guessing that the mechanism should be similar to that with $\ce{BBr3}$, i.e. coordination of oxygen to the Lewis acid followed by nucleophilic attack of $\ce{Cl-}$ on the methyl carbon. Would that be right?

I looked up the original synthesis, which was reported quite a long time ago, in J. Org. Chem., 1941, 6, 558. The authors write that the reaction proceeds via the intermediacy of this compound:

which suggests that the initial nucleophile in the Michael addition is not the aromatic ring but rather the -OMe (or -OH) group. Unfortunately, that just makes things more confusing.

First, the methoxy- group can be demethylated by $\ce{AlCl3}$, beginning with $\ce{AlCl3}$ coordination to the ether oxygen, followed by an $\mathrm{S_N2}$ reaction:
$\hspace{3.6cm}$
$\hspace{0cm}$
The oxygen activates naphthalene's position 1, and the acid chloride will undergo a Friedel-Crafts acylation (as you noted):
$\hspace{.5cm}$
Being a strong Lewis acid, $\ce{AlCl3}$ can coordinate with the carbonyl oxygen. This activates position 4 of the $\alpha$-$\beta$ unsaturated carbonyl towards electrophilic addition:
$\hspace{.75cm}$
Electrophilic addition occurs at naphthalene's spatially adjacent position 8, the carbocation stabilized through the extensive conjugation of the molecule (as well as by the oxygen, as you noted):
$\hspace{1.75cm}$
The spatially adjacent phenyl group accepts the proton and is ejected as benzene:
$\hspace{1cm}$
Addition of water evolves $\ce{HCl}$ and precipitates $\ce{Al(OH)3}$, yielding the final product:
$\hspace{1.8cm}$

I came across ringo's detailed mechanism of orthocresol's query about dephenylation during a Friedel-Crafts (FC) acylation. The penultimate step in ringo's mechanism involves the dephenylation. My thought is why a sigma bond of a phenyl ring is abstracting a proton that should be lost readily on its own to effect aromatization? The following modification to this step of an otherwise cogent mechanism is offered. The FC reaction liberates the strong acid H+ AlCl4-.
First, cation 1 loses a proton to generate species 2. Ipso protonation of the phenyl ring via the π-framework provides resonance stabilized cation 3 that collapses to afford aluminum salt 4 and benzene. Aqueous acid work-up gives 9-hydroxy-1H-phenalen-1-one 5.

In step 3, shouldn't the carbocation formed after protonation be present at the ipso position as at ipso position, it would become a three degree carbocation while at the current position(depicted in the mechanism) it is a two degree carbocation, noting that the extent of conjugation in both cases is similar?

I thought that the position shown for the carbocation might be preferred so that the negative inductive effect of oxygen is minimised, but then again, I am not sure how effective is that factor so far away in the molecule

@YUSUF HASAN: Protonation of structure 2 at the ortho-position does lead to a stable cation many times over BUT the cation does NOT lead to dephenylation. When ipso-protonation occurs as in 2 with the positive charge at the ortho-position and delocalized, dephenylation occurs. Notice in structure 2 the tail of the red arrow is at the ipso end of the double bond meaning that the double bond should be considered polarized ipso-negative, ortho-positive.

I agree more with your proposed modification as it seems to provide stronger "motivation" for the dephenylation to occur. The concerted arrow-pushing starting from the oxygen anion coordinated to the $\ce {AlCl3}$ catalyst does seem to provide that "motivation". I suppose dephenylation may not actually be that absurd in this reaction due to the restoration of aromaticity in that particular ring of the tricyclic system and the ejection of the stable benzene ring. Furthermore, there is also intramolecular hydrogen bonding of the "1,3-diketone"-type in the final product.

The critical step is the formation of species 3. The rest is downhill.

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text:
Problems with creating sodium hydroxide from sodium (hydrogen) carbonate
If sodium hydrogen carbonate (baking soda, $\ce{NaHCO3}$) is heated to greater than $50~\mathrm{^\circ C}$, it will release carbon dioxide and water to form sodium carbonate:
$$\ce{2NaHCO3 -> Na2CO3 + H2O + CO2}$$
And when the sodium carbonate is heated further to about $850\ \mathrm{^\circ C}$, it will release more carbon dioxide and form sodium oxide:
$$\ce{Na2CO3 -> Na2O + CO2}$$
And according to Wikipedia:

Sodium oxide (SOX) is a chemical compound with the formula $\ce{Na2O}$. It is used in ceramics and glasses, though not in a raw form. It is the base anhydride of sodium hydroxide, so when water is added to sodium oxide $\ce{NaOH}$ is produced.

So it should be possible to produce sodium hydroxide from baking soda and water, with some heat. I heated some baking soda in a gas flame. I am not sure exactly what the temperature of the flame was, but it was enough for the stainless steel spoon that I used to glow red. (If it is useful, the gas where I live is approximately $50~\%$ hydrogen, $25~\%$ methane and the rest $\ce{CO2}$ and $\ce{CO}$, and it burns completely with a blue flame. I suspect this flame to be at least $1000\ \mathrm{^\circ C}$, based on available information about flame temperatures of different fuels.)
However, $\ce{NaOH}$ is produced industrially via the chloralkali process, so there must be problems with using this method. What are they, and are they significant if trying to produce a small amount of sodium hydroxide for a minor experiment?

Twoobservations.
First, the chlor-alkali process is surely cheaper, all the more so by producing other marketable products (chlorine, hydrogen).
Second, in the lab I would consider making the sodium hydroxide in solution by reacting sodium bicarbonate or sodium carbonate with calcium hydroxide.

For your first point, what do you mean? I used dry baking soda, heated it directly, and then poured it into a beaker. Then, I added water quickly, since I thought it might absorb carbon dioxide from the air.

Edited that out, misunderstood the first time.

You are correct: Taking baking soda and heating it to $850~\mathrm{^\circ C}$ will generate sodium oxide and liberate carbon dioxide and water. To generate sodium hydroxide, all you would need to do is wait for the heated soda to cool down and then add water. Beware! The reaction of sodium oxide and water is strongly exothermic. React the two carefully!
A red-glowing spoon is a certain indicator that the temperature of the flame was larger than $500~\mathrm{^\circ C}$. A blue flame is usually an indicator for complete combustion. In the context of bunsen burners, that is typically larger than $1000~\mathrm{^\circ C}$. I cannot say whether your gas flame reached that temperature or not.
Note that once you have created sodium hydroxide you should handle it carefully. Especially do not attempt drinking it and keep it off your hands. Proper safety equipment (goggles, gloves) would be preferable.

Concerning the industrial process, the question is usually not so much what the problems with a certain method are, but rather what the benefit of a certain method is — especially if that benefit relates to cash. Electrolysis of a sodium chloride solution is a very cheap process with sodium chloride and water both available at almost no cost and electricity also typically cheap. Acquiring sodium carbonate is probably already more expensive than the required sodium chloride and heating to $850~\mathrm{^\circ C}$ is certainly much more expensive than letting a current flow through the solution. Hence why the chloralkali process is preferred in industry.

That 850°C figure may be just some arbitrarily chosen temperature for "decomposition begins". Or perhaps what you saw was that Na2CO3 melts at 854°C but is listed as "decomposing" (i.e. there's more than just a simple phase change to contend with). But remember that thermal decomposition is an equilibrium reaction, not a phase change that occurs at a sharply defined temperature.
What I've found while similarly chasing down reaction data is https://www.solvay.us/en/binaries/HeatEffects_of_the_TronaSystem-237230.pdf that shows the equilibrium pCO2 at 850°C at something between 10 and 12mmHg. That's going to be a slow process, but at least it will occur. Now remember that the local pCO2 in your gas flame is going to be much higher than 400ppm-of-100kPa, and according to that table on page 2 you're going to need well in excess of 1200°C just for the reaction to progress forwards. Red heat isn't going to cut it here.

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text:
How to determine the equation for the reaction of europium with sulfuric acid based on experimental measurements?
I was given a set of results by a fellow researcher where he reacted $0.608~\mathrm{g}$ of europium ($M = 152$) with an excess of $\ce{H2SO4}$ and collected $144~\mathrm{cm^3}$ of $\ce{H2}$ gas at room temperature and pressure.
I was then asked to derive the equation for the reaction using these figures.
This is what I did.
I know that $1~\mathrm{mol}$ of gas is $24000~\mathrm{cm3}$.
I found out that $144~\mathrm{cm^3}$ is equivalent to $0.006~\mathrm{mol}$ of $\ce{H2}$.
Then, I did $0.608 / 152$ to get $0.004~\mathrm{mol}$.
This meant that europium and hydrogen gas were in the ratio $2:3$.
The equation I wrote was $$\ce{2Eu + H2SO4 -> Eu2SO4 + 3H2}$$
But I am wrong and I don't know where.

Your equation isn't balanced; the oxidation state for europium is +3 and not +1; and europium sulfate does not exist in the solution you describe.
From this link:

Reaction of europium with acids
Europium metal dissolves readily in dilute sulphuric acid to form solutions containing the very pale pink aquated Eu(III) ion together with hydrogen gas, $\ce{H2}$. It is quite likely that $\ce{Eu^3+}$(aq) exists as largely the complex ion $\ce{[Eu(OH2)9]^3+}$

$$\ce{2Eu(s) + 3H2SO4(aq) -> 2Eu^{3+}(aq) + 3SO4^{2-}(aq) + 3H2(g)}$$

With "excess sulfuric acid", sulfate ion does not exist either.  Free acid will protonate sulfate ions to give bisulfate, which is close to but not exactly a strong acid.  I would make bisulfate my anion and adjust coefficients accordingly.

@Oscar This is only true for highly concentrated solutions of sulfuric acid. It does not say which concentration was used, only that it was more than necessary to react europium fully.

Sulfate/bisulfate buffers around pH 2, thus about 0.01 M solvated ptotons.  At acid concentrations one would typically use bisulfate will predominate over free solvated protons.


text:
Is this gas in thermodynamic equilibrium?
Please consider the following graph of molar volume as a function of pressure for a fixed volume.

Can we say whether this system under thermodynamic equilibrium or not?
Update Thinking about this some more, the molar density vs pressure graph may be useful too.

Is this gas in thermodynamic equilibrium?

This is not a homework question. I got my BSc in Chemistry 20 years ago. I have not been working in a Chemistry related field since 2004 and somebody showed this to me to ask what I thought, being somewhat rusty I want to cross check with others

Did you really mean "for a fixed volume," or did you mean "for a fixed temperature?"  For a fixed volume, molar volume does not change if the number of moles does not change.

The volume within which the measurement is taken is of fixed size. The temperature and quantity/pressure vary. Think of this as taking a fixed volume sample and measuring the sample's temperature and pressure. The gas is a mixture and well mixed.

If that's the case then the molar density must be constant.  It is just the mass of gas (which doesn't change) divided by the volume of the container (which also doesn't change).

the mass of gas is changing as there is less gas at the lower temp & pressure

Where is the gas (i.e., matter)  going to?

Let us continue this discussion in chat.

I don't think "equilibrium" is a good word to use in your question.  We have no way of knowing if the gas is at equilibrium from the data you present.  A highly non-ideal gas that was exploding might give the same relationship.  A gas undergoing a rapid unimolecular reaction wouldn't really be at a true chemical equilibrium but could behave as an ideal gas from a pressure / temperature standpoint.
Did you really mean to ask if the gas was ideal?  An ideal gas follows $PV = nRT$, which can be re-written in several ways:

molar density $\frac{n}{V}$ as a function of pressure $P$: $$\frac{n}{V} = \frac{1}{RT}\;P$$, which is the equation of a straight line through the origin (bottom graph).
molar volume $\frac{V}{n}$ as a function of pressure $P$:  $$\frac{V}{n} = RT \frac{1}{P}$$, which is the equation of a hyperbola (top graph).

Your data is consistent with the gas being ideal (and isothermal).

So each sample represents a distinct fixed volume sampling from the test environment. The sample is returned after measurement. Some persons have hypothesised that the system is not equilibrated during the experiment (and argue that in other cases there is external energy being absorbed by the system at certain pressures). I am seeing - instead - what could be a phase change for a portion of the gas from monomolecular to dimolecular, but this theory can only be valid if the system itself is under Thermodynamic Equilibrium

We expect pV = constant for equilibrium (ideal gas law aka virial expansion to 1st order). Given this, doubling the pressure should halve the volume. Eyeballing the data points at 0.25 and 0.5 atm suggests such a relationship.

Yes, that is my rusty opinion too. And in fact if we look at the Molar Density (mol/L) instead of the Molar Volume (L/mol) we get - to my rusty eyes - something that looks suspiciously like a straight line. That kind of says TE to me. In other words this is necessary but is this sufficient to declare thermodynamic equilibrium?


text:
Combining acid dissociation constants to determine pH of diprotic acid
If I have a diprotic acid with $K_{\mathrm{a1}}$ and $K_{\mathrm{a2}}$ as the acid dissociation constants, why can't I calculate the final $\mathrm{pH}$ using $K = K_{\mathrm{a1}}\cdot K_{\mathrm{a2}}$ and the following net reaction?
$$
\ce{H2A + 2H2O ->  A^2- + 2\,H3O+} \\
K = \frac{4x^3}{I-x} \\
\mathrm{pH} = -\log_{10}[2x]$$
where $I$ is the initial concentration of $\ce{H2A}$. For instance $I$ could be $1\ \mathrm{M}$, and we could use $K_{\mathrm{a1}} = 5.9 \times 10^{-2}$ and $K_{\mathrm{a2}} = 6.4 \times 10^{-5}$ (constants for oxalic acid).
Solving it the standard way, one acid dissociation at a time, gives $\mathrm{pH} = 0.67$ whereas the combined way (above) gives $\mathrm{pH} = 3.9$. In what way is using this combined equilibrium constant neglecting the total hydronium concentration?

This should be broken into:
\begin{align}
\ce{H2A + H2O &-> HA- + H3O+}\\
\ce{HA- + H2O &-> A^{2-} + H3O+}
\end{align}
With equilibrium equations:
\begin{align}
K_\mathrm{a1}&=\frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}\tag1\label{Ka1}\\
K_\mathrm{a2}&=\frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]}{[\ce{HA-}]}\tag2\label{Ka2}\\
\end{align}
Let us solve for a general case. From logic and general knowledge we know the water dissociation constant, that the masses must balance, and that the charges must balance. These are shown here:
\begin{align}
K_\mathrm{w}&=[\ce{H3O+}]\cdot[\ce{OH-}]\tag3\label{Kw}\\
I&=[\ce{H2A}]+[\ce{HA-}]+[\ce{A^{2-}}]\tag4\label{Mass Balance}\\
[\ce{H3O+}]&=[\ce{OH-}]+[\ce{HA-}]+2[\ce{A^{2-}}]\\
&=\frac{K_\mathrm{w}}{[\ce{H3O+}]}+[\ce{HA-}]+2[\ce{A^{2-}}]\tag5\label{Charge Balance}\\
\end{align}
Using equations \eqref{Ka1} and \eqref{Ka2}, we can solve for $[\ce{HA-}]$ and $[\ce{A^{2-}}]$:
\begin{align}
K_\mathrm{a1} &= \frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}
&\Rightarrow &&
[\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}\tag6\label{HA-}\\
K_\mathrm{a1} \cdot K_\mathrm{a2} &= \frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]^2}{[\ce{H2A}]}
&\Rightarrow &&
[\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\tag7\label{A2-}\\
\end{align}
Substituting into \eqref{Mass Balance}:
\begin{align}
I &= [\ce{H2A}]+\frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}
     +\frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\\
  &= [\ce{H2A}] \cdot \frac{[\ce{H3O+}]^2 + [\ce{H3O+}]
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H3O+}]^2}\tag9\label{Sub Mass Balance}
\end{align}
Solving \eqref{Sub Mass Balance} for $[\ce{H2A}]$:
$$[\ce{H2A}]=\frac{I \cdot [\ce{H3O+}]^2}{[\ce{H3O+}]^2 + [\ce{H3O+}]
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}} \tag{10}\label{H2A}$$
Substituting \eqref{H2A} into \eqref{HA-} and \eqref{A2-}:
\begin{align}
[\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
\tag{11}\label{Sub HA-}\\
[\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{12}\label{Sub A2-}
\end{align}
Substituting equations \eqref{Sub HA-} and \eqref{Sub A2-} into equation \eqref{Charge Balance}:
\begin{align}
&&
[\ce{H3O+}] &= \frac{K_\mathrm{w}}{[\ce{H3O+}]}+\frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]+2K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\\
\Rightarrow &&
[\ce{H3O+}]^2 &= K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot ([\ce{H3O+}]^2+2K_\mathrm{a2}\cdot[\ce{H3O+}])}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{13}\label{Final Eq}
\end{align}
The above equation exactly calculates the $[\ce{H3O+}]$ for a solution of any diprotic acid. The same method used to solve for this equation can be applied to polyprotic acids as well. To make this into a more manageable equation, however, you may choose to assume $K_\mathrm{a1} \cdot K_\mathrm{a2}=0$ (but only in the case of weak acids).

As for the case you proposed, using $K_\mathrm{w}=1 \times 10^{-14}$, $I=1\ \mathrm{M}$, $K_\mathrm{a1}=5.9 \times 10^{−2}$ and $K_\mathrm{a2}=6.4 \times 10^{−5}$ in the equation $y=K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot (x^2+2K_\mathrm{a2})}{x^2 + x \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}-x^2$, and using a graphic calculator to find where the equation is equal to 0, we find $[\ce{H3O+}]=0.215508\ \mathrm{M}$ and $\mathrm{pH}=0.666537$.

I fail to see where the second equation of equation 4 comes from. Can anyone enlighten me?

Comes from charge balance.

I also have trouble understanding how you derived the second equation of equation 4. I couldn't get how it comes from charge balance. Can you please explain in more details. Thanks.

It is a way of counting how $\ce{H3O+}$ is made from the acids present in solution. 1 is created when $\ce{H2O}$ donates a proton, 1 is created when $\ce{H2A}$ donates a proton, and 2 are made when $\ce{H2A}$ donates two protons.

While your method is correct for determining the number of protons produced by double dissociation, it overlooks all the protons released by single deprotonation, which should contribute significantly to your overall hydronium concentration. To actually find the pH of a diprotic acid, you either assume that the first dissociation goes to completion or that the second dissociation is too weak to occur. In general, a large Ka1 value means that you can assume that the first disassociation goes to completion. From there, you can use the concentrations of [HA]- and [H]+ and create a new equilibrium equation to solve for the final concentration of H+. Here is an article on how to do that. On the other hand, if Ka1 is small, then you know that Ka2 will be even smaller, and thus will not have a significant effect on the overall pH. In this case, you solve the equilibrium equation assuming that the acid was monoprotic.

Well yes thank you I know how to do it the conventional way. In the example I gave, I did that.

no problem, just wanted to provide a resource for future readers.

Actually reading more closely, I think your statement "first dissociation goes to completion" isn't accurate. Many polyprotic acids have a small $K_{a1}$. In many cases, calculation based on the $K_{a1}$ is sufficient.

You are correct. When the first dissociation has a small Ka1, you should approximate the acid as monoprotic, and when Ka1 is large, you should treat it as a strong acid.


text:
Why isn't CH3OH basic if it contains OH?
I was asked to find the most basic solution from the following:

$\ce{K2SO4}$
  $\ce{CH3OH}$
  $\ce{Na2CO3}$
  $\ce{(NH4)2CrO4}$
  $\ce{BaCl2}$

I know that I have to identify the solution that contains the conjugate base of a weak acid.
Knowing this, I immediately eliminated $\ce{BaCl2}$ ($\ce{HCl}$), $\ce{K2SO4}$ ($\ce{H2SO4}$), and $\ce{(NH4)2CrO4}$ ($\ce{H2CrO4}$ — which I've heard is a strong acid).
This left $\ce{CH3OH}$ and $\ce{Na2CO3}$.
The correct answer is $\ce{Na2CO3}$ since $\ce{CO3^{2-}}$ is the conjugate base of $\ce{H2CO3}$.
However, why can the final answer not be $\ce{CH3OH}$, especially considering that $\ce{OH}$ is basic? Also, could $\ce{OH}$ be considered the conjugate base to a weak acid?

The corresponding acid would be $\ce{H3C-OH2+}$ which is about as strong as $\ce{HCl}$ and $\ce{H2SO4}$.

A pretty common mistake to make when you're first learning to identify acids and bases: you can't just assume that everything with an OH will be basic. I can rewrite water as $\ce{HOH}$ but that doesn't mean that water is strongly basic.

Come to think of it, $\ce{H2SO4}$ contains $\ce{OH}$ as well.

Unfortunately, chemistry is not as simple as going

Oh, I spotted $\ce{OH}$, it must be a basic compound!

I would argue that the vast majority of all compounds one could write with $\ce{OH}$ inside are either neutral or acidic and only a minority is basic.
Take, for example, sulphuric acid. As Ivan correctly mentioned in a comment, rather than writing the traditional $\ce{H2SO4}$, you could also write it as $\ce{O2S(OH)2}$ which lays more emphasis on the molecular structure (to hydroxy groups and two oxy groups all bound to sulphur). We all know that $\ce{H2SO4}$ is acidic, not basic, even though it has $\ce{OH}$ groups.
Or take alcohols such as methanol ($\ce{H3C-OH}$) or ethanol ($\ce{H3C-CH2-OH}$). These also feature $\ce{OH}$ groups, they are even commonly written that way, but are about as acidic or basic as water ($\ce{HOH}$) is.
So the question may be why did you get taught that $\ce{OH}$ is basic? Well strictly speaking, it is not the $\ce{OH}$ hydroxy group that is basic but the $\ce{OH-}$ hydroxide ion. Only those compounds that dissociate into some cation and a hydroxide ion can be considered basic. For that to happen, ‘some cation’ typically needs to be metallic, at least at your level. So recognise a metal bonded to $\ce{OH-}$ and you should most likely be right calling the compound basic.

Some final notes:

You should always check the directly corresponding conjugate acid. So when analysing $\ce{K2SO4, Na2CO3}$ and $\ce{(NH4)2CrO4}$, trace them back to $\ce{HSO4- , HCO3-}$ and $\ce{HCrO4-}$, respectively. That still leaves us with two rather strong acids but the third ($\ce{HCO3-}$) is even weaker.
The conjugate acid to methanol $\ce{H3C-OH}$ would be the methyloxonium ion $\ce{H3C-OH2+}$, which is a strong acid. This goes for pretty much all organic hydroxy groups.
The $\mathrm{p}K_\mathrm{a}$ of water — hydroxide’s conjugate acid — is $15.7$, so yes, that is a very weak acid.

$\ce{CH3OH}$ is a compound known as methanol. Unlike the sodium found in $\ce{Na2CO3}$, the $\ce{CH3}$ group found in methanol will not separate from the $\ce{OH-}$. This means that the only way that methanol can act as a base is for the oxygen to pick up another hydrogen from solution, making the compound $\ce{CH3OH2+}$.
On the other hand, sodium will readily dissociate from $\ce{CO3^{2-}}$. This is because sodium is a metal, and it will complete its octet when it loses an electron. Once dissociated, $\ce{CO3^{2-}}$ will act as a base and pick up protons, becoming either $\ce{HCO3-}$ or $\ce{H2CO3}$. Since when $\ce{CO3^{2-}}$ gets protonated its net charge decreases, this will act as the stronger base.


text:
Is this a phase change in a gas under Thermodynamic Equilibrium?

The above graph shows the molar density (i.e. n/V from PV=nRT) of a gas plotted against pressure. There are two regions with good fit to a linear relationship which I have coloured green and blue (though these are all from a single set of measurements. The visual distinction is just to highlight where the linear fit applies)
My rusty physical chemistry knowledge suggests that the regions where there is a linear fit are in Thermodynamic Equilibrium as the Ideal Gas law would appear to be holding. Given that we have two linear regions I would also conclude that there is a phase change and that the whole is actually in Thermodynamic Equilibrium when we account for the phase change (i.e. n is different between the two regions)
Is this naïve analysis based on a foggy recollection of Phys Chem lectures from 20 years ago correct or is there another explanation for this data (and if so what is the explanation)
Update
To clarify on the hypothesized phase change. The suggestion is that this phase change is a X + X <=> X2 phase change where X and X2 are both gaseous but the X2 form is only thermodynamically stable at lower temperatures (which correspond to the lower pressures). i.e. not the more common gas-liquid phase change

I don't think it necessarily suggests a phase change.  It just seems to me that, at the higher pressures, we are seeing non-ideal gas behavior setting in.

If it is a phase change as somebody else has hypothesized, it would be a phase change in a ~15-25% component of the gas mixture.

If there were a phase change (condensation), the molar density would be increasing faster, not slower, because liquid would be forming.

If it is a phase change (dimerization) then 1 mole will be turning into 0.5 moles so the molar density would be decreasing faster because the dimers are still a gas and assuming they are an ideal gas they will still (mostly) behave as points with mass and elastic collisions

You didn't say anything about a chemical reaction occurring.  A chemical reaction is not the same thing as a phase change.  A phase change involves a change in the state of aggregation of the matter.

It's not a chemical reaction - if the hypothesis I am trying to stress is correct it is rather a dipole association which can be viewed as a different gas phase, but still a gas phase


text:
What are the implications of having different conformations?
I have seen videos explaining what conformations are. I am curious about whether different conformations can give rise to slightly different properties. Particularly of interest is whether you can modify the smell of something by just rotating some bonds.

They are rotating all the time so you could say that compounds are described by their conformational ensembles. You could bind a molecule to manipulate conformations, but then you have complex of molecules not a single one.

That's an interesting question. To start off, what Mith said in the comments is absolutely right. Let's say that two conformations of a molecule have different smells (which is in fact not an absurd proposition by any means). The problem is that, because there is a very low barrier to the interconversion of the conformations, the individual conformations do not really hang around long enough to be individually detected; you can only really detect a weighted average of the conformations.
To give you an example, you can devise di-substituted cyclohexanes that have two predominant chair conformations. These can be investigated by proton NMR spectroscopy, since axial-axial coupling constants are larger than equatorial-axial or equatorial-equatorial couplings. From the integrals of the relevant peaks you can determine the relative proportions of both conformations.

However, if you somehow managed to synthesise a sample of the pure diequatorial conformer (the one in blue) and put it into a NMR tube, some of it would have converted into the diaxial conformer (the one in red) before you could even run your NMR. In fact, there's probably no way you can synthesise a pure conformer to begin with.
Likewise, if you replace the NMR machine with your nose, there's no way you can open a bottle of a perfume X, with conformation X1, and smell something, then open a bottle of the same perfume X with conformation X2 and smell another thing. They would both smell the same because conformers aren't isolable. (Well, not unless you are working in an argon matrix at temperatures near absolute zero or something like that!)

Mith then goes on to say that you can bind to a molecule to manipulate its conformation. That's the more interesting part, in my opinion at least! If you think about it, it's going to be very hard for us to devise some kind of surface that particularly favours one conformation over another. You'd need a very complex binding partner, with a very specific 3D shape, that "fits" with one conformation but not another conformation.
Does that ring a bell? It might, because nature has made those binding partners for us already. That previous paragraph is a perfect description of an enzyme: a protein that catalyses a biological reaction. Lysozyme is one of the most famous enzymes; it was the first enzyme whose structure was successfully solved by X-ray diffraction. It is also a great example of an enzyme that works by manipulating conformations.
Lysozyme binds to a series of carbohydrates in the peptidoglycan cell walls of some bacteria. Together with water, it hydrolyses the glycosidic bonds (a fancy name for an acetal) in these carbohydrates:

(image from Wikipedia)
I'm not going to go into the full mechanism of this, but the striking thing is that upon binding to lysozyme, the "D ring" (which is the NAM residue on the left-hand side of the diagram above) is distorted from its usual chair conformation into a half-chair conformation. This is because it "fits" better into the exact 3D conformation of the enzyme (not shown in the diagrams):

(image modified from Wikipedia)
This makes the ring more unstable and prone to hydrolysis (strictly speaking, more prone to nucleophilic attack by the aspartate-52 residue). Without this manipulation of the conformation, it's likely that the hydrolysis cannot proceed.
So, there you have it! One way in which the conformation of a molecule, upon binding to another molecule, affects its properties.

I really like when my ideas are so nicely used :)


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Temperature and reaction quotient
Consider an exothermic reaction. If we double the temperature at which the reaction is run, will the reaction quotient increase?
I know that an increase in Q relative to K causes the reaction to favor the reactants as the system tries to return to equilibrium.
If a reaction is exothermic, and we take heat to be a product, wouldn't an excess of heat cause the reaction to favor reactants?
The question I'm having trouble with is this one; apparently the answer is choice 2. I wonder if this is a typo or am I fundamentally misunderstanding something.

Just to clear things up first (I might be saying something you already know, but I just want to be sure): the intent of the question is not to ask you what happens to the reaction quotient after equilibrium is re-established.
If a system is originally in equilibrium, with $Q = K_1$, and then you disturb it such that the equilibrium constant changes to $K_2$, then after the system re-reaches equilibrium you will have $Q = K_2$. If you interpret the question as asking whether $Q$ has changed after equilibrium is re-established, then the question basically amounts to asking whether the equilibrium constants $K_1$ and $K_2$ are different.
What the question is asking is whether $Q$ is affected instantaneously by the disturbance. Re-reading your question I think you got it, so...

Anyway, heat is not a product.
The reaction quotient for that reaction is defined as
$$Q = \frac{a_{\ce{CO2}}}{(a_{\ce{CO}})^2} \approx \frac{p_{\ce{CO2}}}{(p_{\ce{CO}})^2}$$
and $K$ is defined as the value of $Q$ at equilibrium.
Nowhere does heat appear in the equation. The idea of using heat as a "product" is nothing more than a mnemonic to help you remember the direction in which the equilibrium shifts. As long as the pressures are fixed, the temperature does not affect the instantaneous value of the reaction quotient.
When you increase the temperature, what happens is that the equilibrium constant $K$ drops. Let's say that before changing the temperature $Q = K = 0.01$ (just an example). If you increase the temperature, $K$ might fall to $0.001$ but $Q$ at that instant will still be $0.01$. As the system re-equilibrates, $Q$ will drop to $0.001$.
So, it is not that $K$ remains constant and $Q$ increases. It is more like: $K$ drops and $Q$ remains constant (at least instantaneously).


text:
Microstates, molar mass, and degrees of freedom.
What is the relationship between the number of microstates and molar mass. The key says choices II and III are correct. If III is true, why isn't I also true? Also, if II is true, why isn't IV true as well?
I think that the greater the number of microstates, the greater the number of DoF. I figured that much out.
Why would NH3 have more degrees of freedom than NO2? Is it because NH3 has more attached atoms than NO2?
Is molar mass just a red herring in the first statement? NO2 has fewer bound atoms than NH3; it seems that the number of bound atoms affects the number of microstates more so than molar mass, because in III, where the two species are structurally identical, molar mass comes into play.
So when determining the number of microstates, is the number of attached atoms more important than molar mass?
I can't seem to find any good online resources relating molar mass with the number of microstates and degrees of freedom.


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Removing aluminum "corrosion?"
I have some aluminum alloy, possibly 6063 or 7050 (no clue really) which has corroded. The finish might be silver-anodized but I doubt it due to the high polish, could be urethane. The item is a fishing rod holder, used in a freshwater marine environment. Hardware is stainless and steel. Several screws and bolts need to be removed, and I was wondering if there was an acceptable reducing agent for the corrosion rather than busting driver bits and/or mangling threads.

Aluminum corrosion in most environment produces $\ce{Al2O3}$, corundum, which is extremely hard (9 on the Mohs scale, where 10 is diamond). Corundum is also quite resistant to most chemicals.
There are various products called "Aluminum Jelly", such as that from Loctite and from Henkel, which can remove some of the surface corrosion. They probably attack the aluminum at its point of contact with the oxide, rather than reacting with $\ce{Al2O3}$ itself. Regrettably, this probably means that though it might brighten the surface, the jelly will not substantially free the bolts.
The drill bits break when they encounter a mass of corrosion and suddenly seize, and this breakage can be minimized by running at slow speed with lots of lubricant. Applying gentle heat and cooling to the parts immersed in lubricant (e.g. Alumaslick or Vibra-TITE) allows it to penetrate better.
In any event, any threads in the aluminum have likely been ruined, so consider drilling oversize holes and re-tapping them (aluminum is easily tapped, but shavings tend to stick in the tool, so use lubricant and clear the tap frequently).


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Are the lone pairs in water equivalent?
I've read that the oxygen atom in water is $\mathrm{sp^2}$ hybridized, such that one of the oxygen lone pairs should be in an $\mathrm{sp^2}$ orbital and the other should be in a pure p atomic orbital.
First, am I correct about the lone pairs being non-equivalent?
Second, if so, does this have any significance in actual physical systems (i.e. is it a measurable phenomenon), and what is the approximate energy difference between the pairs of electrons?
Lastly, if it turns out the lone pairs are actually inequivalent, can this be reconciled with the traditional explanation (due to VSEPR theory) that oxygen is $\mathrm{sp^3}$ and the lone pairs are equivalent?

There are some recent discussions on this topic http://pubs.rsc.org/is/content/articlelanding/2015/rp/c4rp00245h/unauth#!divAbstract

You can observe the difference experimentally with photoelectron spectroscopy.

Water, as simple as it might appear, has quite a few extraordinary things to offer. Most does not seem to be as it appears.
Before diving deeper, a few cautionary words about hybridisation. Hybridisation is an often misconceived concept. It only is a mathematical interpretation, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will result from various factors, such as steric and electronic interactions, and furthermore interactions with the surroundings like a solvent or external field. The geometric arrangement will not be formed because a molecule is hybridised in a certain way, it is the other way around, i.e. a result of the geometry or more precise and interpretation of the wave function for the given molecular arrangement.
In molecular orbital theory linear combinations of all available (atomic) orbitals will form molecular orbitals (MO). These are spread over the whole molecule, or delocalised, and in a quantum chemical interpretation they are called canonical orbitals. Such a solution (approximation) of the wave function can be unitary transformed form localised molecular orbitals (LMO). The solution (the energy) does not change due to this transformation. These can then be used to interpret a bonding situation in a simpler theory.
Each LMO can be expressed as a linear combination of the atomic orbitals, hence it is possible to determine the coefficients of the atomic orbitals and describe these also as hybrid orbitals. It is absolutely wrong to assume that there are only three types of spx hybrid orbitals.
Therefore it is very well possible, that there are multiple different types of orbitals involved in bonding for a certain atom. For more on this, read about Bent's rule on the network.[1]
Let's look at water, Wikipedia is so kind to provide us with a schematic drawing:

The bonding angle is quite close to the ideal tetrahedral angle, so one would assume, that the involved orbitals are sp3 hybridised. There is also a connection between bond angle and hybridisation, called Coulson's theorem, which lets you approximate hybridisation.[2] In this case the orbitals involved in the bonds would be sp4 hybridised. (Close enough.)
Let us also consider the symmetry of the molecule. The point group of water is C2v. Because there are mirror planes, in the canonical bonding picture π-type orbitals[3] are necessary. We have an orbital with appropriate symmetry, which is the p-orbital sticking out of the bonding plane. This interpretation is not only valid it is one that comes as the solution of the Schrödinger equation.[4] That leaves for the other orbital a hybridisation of sp(2/3).
If we make the reasonable assumption, that the oxygen hydrogen bonds are sp3 hybridised, and the out-of-plane lone pair is a p orbital, then the maths is a bit easier and the in-plane lone pair is sp hybridised.[5]
A calculation on the MO6/def2-QZVPP level of theory gives us the following canonical molecular orbitals:

(Orbital symmetries: $2\mathrm{A}_1$, $1\mathrm{B}_2$, $3\mathrm{A}_1$, $1\mathrm{B}_1$)[6,7]
Since the interpretation with hybrid orbitals is equivalent, I used the natural bond orbital theory to interpret the results. This method transforms the canonical orbitals into localised orbitals for easier interpretation.

Here is an excerpt of the output (core orbital and polarisation functions omitted) giving us the calculated hybridisations:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   2. (1.99797) LP ( 1) O  1            s( 53.05%)p 0.88( 46.76%)d 0.00(  0.19%)
   3. (1.99770) LP ( 2) O  1            s(  0.00%)p 1.00( 99.69%)d 0.00(  0.28%)
   4. (1.99953) BD ( 1) O  1- H  2
               ( 73.49%)   0.8573* O  1 s( 23.41%)p 3.26( 76.25%)d 0.01(  0.31%)
               ( 26.51%)   0.5149* H  2 s( 99.65%)p 0.00(  0.32%)d 0.00(  0.02%)
   5. (1.99955) BD ( 1) O  1- H  3
               ( 73.48%)   0.8572* O  1 s( 23.41%)p 3.26( 76.27%)d 0.01(  0.30%)
               ( 26.52%)   0.5150* H  3 s( 99.65%)p 0.00(  0.32%)d 0.00(  0.02%)
 -------------------------------------------------------------------------------

As we can see, that pretty much matches the assumption of sp3 oxygen hydrogen bonds, a p lone pair, and a sp lone pair.
Does that mean that the lone pairs are non-equivalent?
Well, that is at least one interpretation. And we only deduced all that from a gas phase point of view. When we go towards condensed phase, things will certainly change. Hydrogen bonds will break the symmetry, dynamics will play an important role and in the end, both will probably behave quite similarly or even identical.
Now let's get to the juicy part:

Second, if so, does this have any significance in actual physical systems (i.e. is it a measurable phenomenon), and what is the approximate energy difference between the pairs of electrons?

Well the first part is a bit tricky to answer, because that is dependent on a lot more conditions. But the part in parentheses is easy. It is measurable with photoelectron spectroscopy. There is a nice orbital scheme correlated to the orbital ionisation potential on the homepage of Michael K. Denk for water.[8] Unfortunately I cannot find license information, or a reference to reproduce, hence I am hesitant to post it here.
However, I found a nice little publication on the photoelectron spectroscopy of water in the bonding region.[9] I'll quote some relevant data from the article.

$\ce{H2O}$ is a non-linear, triatomic molecule consisting of an oxygen atom  covalently bonded to two hydrogen atoms. The ground state of the $\ce{H2O}$ molecule is classified as belonging to the $C_\mathrm{2v}$ point group and so the electronic states of water are described using the irreducible representations $\mathrm{A}_1$, $\mathrm{A}_2$, $\mathrm{B}_1$, $\mathrm{B}_2$. The electronic configuration of the ground state of the $\ce{H2O}$ molecule is described by five doubly occupied molecular orbitals:
  $$\begin{align}
\underbrace{(1\mathrm{a}_1)^2}_{\text{core}}&&
\underbrace{(2\mathrm{a}_1)^2}_{\text{inner-valence orbital}}&&
\underbrace{
  (1\mathrm{b}_2)^2 (3\mathrm{a}_1)^2 (1\mathrm{b}_1)^2
       }_{\text{outer-valence orbital}}&&
\mathrm{X~^1A_1}
\end{align}$$
[..]
In addition to the three band systems observed in HeI PES of $\ce{H2O}$, a fourth band system in the TPE spectrum close to 32 eV is also observed. As indicated in Fig. 1, these band systems correspond to the removal of a valence electron from each of the molecular orbitals $(1\mathrm{b}_1)^{-1}$, $(3\mathrm{a}_1)^{-1}$, $(1\mathrm{b}_2)^{-1}$ and $(2\mathrm{a}_1)^{-1}$ of $\ce{H2O}$.

As you can see, it fits quite nicely with the calculated data. From the image I would say that the difference between $(1\mathrm{b}_1)^{-1}$ and $(3\mathrm{a}_1)^{-1}$ is about 1-2 eV.
TL;DR
As you see your hunch paid off quite well. Photoelectron spectroscopy of water in the gas phase confirms that the lone pairs are non-equivalent. Conclusions for condensed phases might be different, but that is a story for another day.

Notes and References

What is Bent's rule?
Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

Formal theory of Bent's rule, derivation of Coulson's theorem (Wikipedia).
Worked example for cyclo propane by ron.

A π orbital has one nodal plane collinear with the bonding axis, it is asymmetric with respect to this plane. A bit more explanation in my question What would follow in the series sigma, pi and delta bonds?
With in the approximation that molecular orbitals are a linear combination of atomic orbitals (MO = LCAO).
The terminology we use for hybridisation actually is just an abbreviation:
$$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$
In theory $x$ can have any value; since it is just a unitary transformation the representation does not change, hence
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 4\times\mathrm{sp}^3 \\
   &\leadsto 3\times\mathrm{sp}^2, 1\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}, 2\times\mathrm{p} \\
   &\leadsto 2\times\mathrm{sp}^3, 1\times\mathrm{sp}, 1\times\mathrm{p} \\
   &\leadsto \text{etc. pp.}\\
   &\leadsto 2\times\mathrm{sp}^4, 1\times\mathrm{p}, 1\times\mathrm{sp}^{(2/3)}
\end{align}
There are virtually infinite possibilities of combination.
This and the next footnote address a couple of points that were raised in a comment by DavePhD. While I already extensively answered that there, I want to include a few more clarifying points here. (If I do it right, the comments become obsolete.)

What is the reason for concluding 2 lone pairs versus 1 or 3? For example Mulliken has in table V the b1 orbital being a definite lone pair (no H population) but the two a1 orbitals both have about 0.3e population on H. Would it be wrong to say only one of the PES energy levels corresponds to a lone pair, and the other 3 has some significant population on hydrogen? Are Mulliken's calculations still valid? – DavePhD

The article Dave refers to is R. S. Mulliken, J. Chem. Phys. 1955, 23, 1833., which introduces Mulliken population analysis. In this paper Mulliken analyses wave functions on the SCF-LCAO-MO level of theory. This is essentially Hartree Fock with a minimal basis set. (I will address this in the next footnote.) We have to understand that this was state-of-the-art computational chemistry back then. What we take for granted nowadays, calculating the same thing in a few seconds, was revolutionary back then. Today we have a lot fancier methods. I used density functional theory with a very large basis set. The main difference between these approaches is that the level I use recovers a lot more of electron correlation than the method of Mulliken. However, if you look closely at the results it is quite impressive how well these early approximations perform.
On the M06/def2-QZVPP level of theory the geometry of the molecule is optimised to have an oxygen hydrogen distance of 95.61 pm and a bond angle of 105.003°. This is quite close to the experimental results.
The contribution to the orbitals are given as follows. I include the orbital energies (OE), too. The contributions of the atomic orbitals are given to 1.00 being the total for each molecular orbital. Because the basis set has polarisation functions the missing parts are attributed to this. The threshold for printing is 3%. (I also rearranged the Gaussian Output for better readability.)

Atomic contributions to molecular orbitals:
2:  2A1  OE=-1.039   is   O1-s=0.81 O1-p=0.03   H2-s=0.07   H3-s=0.07
3:  1B2  OE=-0.547   is   O1-p=0.63             H2-s=0.18   H3-s=0.18
4:  3A1  OE=-0.406   is   O1-s=0.12 O1-p=0.74   H2-s=0.06   H3-s=0.06
5:  1B1  OE=-0.332   is   O1-p=0.95

We can see that there is indeed some contribution by the hydrogens to the in-plane lone pair of oxygen. On the other hand we see that there is only one orbital where there is a large contribution by hydrogen. One could here easily come up with the theory of one or three lone pairs of oxygen, depending on your own point of view. Mulliken's analysis is based on the canonical orbitals, which are delocalised, so we will never have a pure lone pair orbital. When we refer to orbitals as being of a certain type, then we imply that this is the largest contribution. Often we also use visual aides like pictures of these orbitals to decide if they are of bonding or anti-bonding nature, or if their contribution is on the bonding axis.
All these analyses are highly biased by your point of view. There is no right or wrong when it comes to separation schemes. There is no hard evidence for any of these obtainable. These are mathematical interpretations that do in the best case help us understand bonding better. Thus deciding whether water has one, two or three (or even four) lone pairs is somewhat playing with numbers until something seems to fit. Bonding is too difficult to transform it in easy pictures. (That's why I am not an advocate for cautiously using Lewis structures.)
The NBO analysis is another separation scheme. One that aims to transform the obtained canonical orbitals into a Lewis like picture for a better understanding. This transformation does not change the wave function and in this way is as equally a representation as other approaches. What you loose by this approach are the orbital energies, since you break the symmetry of the wave function, but this is going much too far to explain. In a nutshell, the localisation scheme aims to transform the delocalised orbitals into orbitals that correspond to bonds.
From a quite general point of view, Mulliken's calculations (he actually only interpreted the results of others) and conclusion hold up to a certain point. Nowadays we know that his population analysis has severe problems, but within the minimal basis they still produce justifiable results. The popularity of this method comes mainly because it is very easy to perform. See also: Which one, Mulliken charge distribution and NBO, is more reliable?
Mulliken used a SCF-LCAO-MO calculation by Ellison and Shull and was so kind to include the main results into his paper. The oxygen hydrogen bond distance is 95.8 pm and the bond angle is 105°. I performed a calculation on the same geometry on the HF/STO-3G level of theory for comparison. It obviously does not match perfectly, but well enough for a little bit of further discussion.

NO  SYM    HF/STO-3G : N(O)     N(H2)      |      Mulliken : N(O)     N(H2)
1   1A1    -550.79     2.0014   -0.0014    |      -557.3     2.0007   -0.0005
2   2A1     -34.49     1.6113    0.3887    |       -36.2     1.688     0.309
3   1B2     -16.82     1.0700    0.9300    |       -18.6     0.918     1.080
4   3A1     -12.29     1.6837    0.3163    |       -13.2     1.743     0.257
5   1B1     -10.63     2.0000    0.0000    |       -11.8     2.000

As an off-side note: I completely was unable to read the Mulliken analysis by Gaussian. I used MultiWFN instead. It is also not an equivalent approach because they expressed the hydrogen atoms with group orbitals.
The results don't differ by much. The basic approach of Mulliken is to split the overlap population to the orbitals symmetric between the elements. That is a principal problem of the method as the contributions to that MO can be quite different. Resulting problematic points are occupation values larger than two or smaller than zero, which have clearly no physical meaning. The analysis is especially ruined for diffuse functions.
At the time Mulliken certainly did not know about anything we are able to do today, and under which conditions his approach will break down, it still is funny to read such sentences today.

Actually, very small negative values occasionally occur [...]. [...] ideally to the population of the AO [...] should never exceed the number 2.00 of electrons in a closed atomic sub-shell. Actually, [the orbital population] in some instances does very slightly exceed 2.00 [...]. The reason why these slight but only slight imperfections exist is obscure. But since they are only slight, it appears that the gross atomic populations calculated using Eq. (6') may be taken as representing rather accurately the "true" populations in various AOs for an atom in a molecule. It should be realized, of course, that fundamentally there is no such thing as an atom in a molecule except in an approximate sense.

For much more on this I found an explanation of the Gaussian output along with the reference to F. Martin, H. Zipse, J. Comp. Chem. 2005, 26, 97 - 105, available as a copy. I have not read it though.
Scroll down until the bottom of the page for the image, read for more information: CHEM 2070, Michael K. Denk: UV-Vis & PES. (University of Guelph) If dead: Wayback Machine
S.Y. Truong, A.J. Yencha, A.M. Juarez, S.J. Cavanagh, P. Bolognesi, G.C. King, Chemical Physics 2009, 355 (2–3), 183-193. Or try this mirror.

What is the reason for concluding 2 lone pairs versus 1 or 3?  For example Mulliken has in table V the b1 orbital being a definite lone pair (no H population) but the two a1 orbitals both have about 0.3e population on H.  http://scitation.aip.org/content/aip/journal/jcp/23/10/10.1063/1.1740588 Would it be wrong to say only one of the PES energy levels corresponds to a lone pair, and the other 3 has some significant population on hydrogen? Are Mulliken's calculations still valid?

@DavePhD Mulliken's analysis and calculations are still valid, one has to consider a few things though. For 1955(!) this was state of the art and as good as it gets. Nowadays we have more elaborate methods. (I did not have time to read the whole paper - but I assume it is one of the famous Mulliken analysis population papers.) 1. He uses SCF-LCAO-MO calculations, essentially HF. There is nothing wrong with that, but the orbitals will be slightly different from ones with correlated methods. 2. From what I can tell he uses a minimal basis, i.e. no polarisation functions. (cont.)

(cont.) 3. The analysis he presents is based on the canonical orbitals. Again, there is absolutely nothing wrong with doing that, we regularly do it with post-HF/DFT methods. We still have to understand, that we expect every atom to contribute to the molecular orbitals. So the assignment of "this is a lone pair" has always implied, that this is the largest contribution. The only reason that 1b1 can only be pure oxygen is, because hydrogen does not have any suitable functions to mix in. Today we include polarisation functions and we will see some H contribution for that orbital as well. (cont.)

(cont.) The population analysis of wave functions will always have some bias. There is neither a true nor the most reasonable separation scheme. We know today, that Mulliken's analysis is highly basis set dependent (it get's worse with larger basis) and often unreliable to produce consistent results with increasing accuracy of methods. A HF/STO-3G gives surprisingly good results though. If you look at the canonical orbitals I posted (I might amend that with actual numbers) you will see that there is H contribution in the in-plane lone pair (as we would expect). Mulliken still holds. (cont.)

(cont.) If we transform the canonical orbitals, we will arrive at a scheme that represents Lewis structures well. But this is also only one interpretation. Even in the localisation scheme I used we see that this lone pair has an occupation lower than two, hence there is still contribution from H. || Would it be wrong to say ... ? No, it would not. That does not necessary say it is correct. We know too little and we are stuck with interpreting. And as long as we do not have solid evidence against any of it, all representations need to be considered valid. (cont.)

(cont.) So in conclusion: There is still a lot to learn from Mulliken, parts of his analysis still hold as they did back then, while others are known to be too approximate to match other evidence. All in all we are just guessing elaborately and until we can actually measure the orbitals (lol) we will never know. Thank you for your comment and sorry for the long reply. (fin.)

@DavePhD I have extended the wall of text a little bit further, if you are still interested in that: it is notes 6 and 7.

Thanks for all the detail.  It seems that there is a long tradition going back to Pauling of considering one of the lone pairs to be the -32eV orbital (as an essentially pure s orbital of oxygen) and the other lone pair to be the highest energy orbital (as pure p).  http://pubs.acs.org/doi/abs/10.1021/ed064p124

I just wanted to add another publication without bumping the post again. (Thanks to @MAFIA36790 for finding it.) No rabbit ears on water. The structure of the water molecule: What should we tell the students? in J. Chem. Educ. 1987, 64 (2), 124. Or find it on a mirror.

You can check out the calculated molecular orbitals of water on Professor Zipse’s (LMU Munich) page. The surrounding ‘text’ in German need not interest you, just click on mo-number to access an image of the corresponding MO.
The lowest orbital with the lowest energy is, of course, oxygen’s core $\mathrm{1s}$-orbital. Next up we have an all-bonding orbital deriving from oxygen’s $\mathrm{2s}$, an all-bonding deriving from oxygen’s $\mathrm{2p}_x$ and a predominantly bonding one deriving from oxygen’s $\mathrm{2p}_z$. You could combine these three to give an almost $\mathrm{sp^2}$ type environment around oxygen with two $\mathrm{sp^2}$ orbitals involved in bonding to hydrogens and the third being an oxygen lone pair. (However note that $\mathrm{sp^2}$ is typically associated with $120^\circ$ while water has a bond angle of $105^\circ$.)
The final orbital is $\mathrm{2p}_y$ according to the way I defined the coordinates (diagram below). This is essentially π type with respect to the $\ce{O-H}$ bonds, so it is nonbonding (hydrogen cannot bond in a π manner). This represents the second lone pair of oxygen.
The two ‘lone pairs’ (remember that by an MO description all orbitals are delocalised across the entire molecule) are notably different from each other, that includes their energies. So calling them equivalent in an $\mathrm{sp^3}$ manner is wrong. $\mathrm{sp^2}$ is a description that would fit the calculated energies, but not so well with the geometry.

The bonding angle just doesn't fit sp², because that must be 120 degree.


text:
Most efficient way to purify household vinegar: distillation or freezing?
I am aware that there are better ways to purify acetic acid out of a water-acid mixture than these, but which one out of distillation (using heat) and fractional freezing (using cold) would gain a higher concentration of acetic acid?

With the correct equipment and technique distillation of the vinegar will produce glacial acetic acid however things can seem easier in theory than in practice. Freeze precipitation and centrifuge may also produce glacial acetic acid from vinegar but again this might not be so easy without access to the right equipment.

Related: http://chemistry.stackexchange.com/questions/2506/acetic-acid-freezing-distilation

@Ivan Neretin , the link suggest the acid be approximately 72% for fractional crystallisation . The household vinegar wouldnt be anywhere near this concentration?

That's just an approximation intended to explain how eutectics work. But anyway, typical household vinegar is of course less concentrated than needed for fractional freezing.

So distillation would be the most practical method? I've thought of converting the acetic acid to an insoluble acetate salt , filter off the water then react the dry acetate salt with an acid to form acetic acid at a high concentration which could then be purified easier.

@Ivan Neretin , what do think of the suggestion I posted above if a silver or basic beryllium acetate were formed by adding there carbonates to the household vinegar at the correct temp then cooled to precipitate?

I think that would be an awfully overcomplicated and expensive method (and dangerous too, in case of beryllium). Also, it probably wouldn't work anyway.

@ Ivan Neretin I agree it was just a theory. (:

In another Q&A we see " If you want to isolate acetic acid from a solution that is mostly water, I would distil the solution until the composition is mostly acetic acid, and then fractionally freeze it."  -  https://chemistry.stackexchange.com/questions/2506/acetic-acid-freezing-distilation

The internet sells 1 L of pure acetic acid in food quality for less than ten euros. ;)

There is no acetic acid/water azeotrope, so distillation can produce glacial (pure) acetic acid.

Freezing can do the same, once you've got a mixture more enriched with acid than eutectics.

Some experimental data after a 4 liter jug of 5% distilled white vinegar accidentally froze at -25°C:

The first 0.5 liter to melt out had an acetic acid concentration of about 13%, substantially above the starting concentration, and a pungent vinegar smell.
The next 1 liter to melt out had an acetic acid concentration of about 7%. The remaining ice was very porous and slushy, slow to drain and probably trapping lots of liquid, and helping explain the non-sharp separation between acetic acid and water.
The next 1 liter to melt was about 3% concentration, below the original.
The last-to-melt liquid measured at approximately 1% concentration, pH still measures about 3, but has almost no smell.

These concentrations, reported as mass percent, were estimated by measuring the solution density, which for acetic acid changes only slightly (e.g., 20% acetic acid is only 2.5% denser than water). Assuming a 0.1% absolute density measurement error (e.g., thermal compensation difference since the liquids were measured at near 0°C), there aren't a lot of significant figures in the percentages above.
The data above was batch gravity melting, followed by pouring off the liquid fraction. Particularly for the first pours, the melting is mostly on the exterior of the jug. The exact yield curves likely depend on the thermal profiles during both freezing and melting, the vessel shapes, and details like how the liquid is removed from the ice.
Unlike when freezing pure water, the bottom of the frozen vinegar jug wasn't visibly distended or ruptured, and actually showed no signs of significant volume change. At 20C, a 4 liter jug takes over 1 day to fully melt.
Overall, freeze distillation does appear straightforward to concentrate vinegar somewhat over normal household levels. Repeated freezing may improve the concentration further, although there's a freeze eutectic at 30% acetic acid / water (mol/mol), so reaching 100% glacial acetic acid would likely require a different approach.


text:
Reaction of alkenes (localised π bond) with nitrating mixture
The nitrating mixture for the nitration of benzene contains many strong electrophiles (e.g $\ce{H2SO4}$, $\ce{NO2+}$). All of these could, presumably, add across the π bond in a molecule containing a localised π bond. Do these additions occur and if so what is the major product?

http://pubs.acs.org/doi/abs/10.1021/ja01499a029 not sure about H2SO4/HNO3


text:
Why does sodium react with water to produce a hydroxide, while zinc produces an oxide?
Sodium belongs to the so-called active metals and produces a hydroxide on contact with water. Zinc produces an oxide. Why is it so?
I was given a hint: to read up about electrochemical series. I've read about it, and naturally zinc is way below sodium in the series. Sodium is more enthusiastic about giving away its electrons and forming positive ions.
Zinc is slightly below this bolded line in the series, but I'm afraid I do not fully understand what that means. "Hydrogen gas bubbled through an alkaline solution, shedding two electrons to produce water and neutralize the solution"?

Here's a page that might help me, I guess. I'm reading it now

The overall potential of an electrochemical cell is equal to the reduction potential for the substance being reduced plus the oxidation potential for the substance being oxidized. In other words, $ΔE_{rxn} = E°_{red} + E°_{ox}$. Now, all that we need to do is find a list of reduction potentials. From here, I found the following reduction potentials:
$$\ce{Na+(aq) + e- -> Na(s)}: E°_{red}=-2.71 V $$
$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH^{-}(aq)}: E°_{red}=-0.83 V $$
$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}: E°_{red}=-0.76 V $$
A redox reaction occurs between one species being oxidized (losing electrons) and another being reduced (gaining electrons). Since all of these values are reduction potentials, we can flip reactants and products to get oxidation potentials. The oxidation potential for a reaction is negative one times the reduction potential (due to conservation of energy) and since we possess sodium and zinc metal rather than ions, we want for these reagents to be oxidized. This gives us the following equations with oxidation potentials:
$$\ce{Na(s) -> Na+(aq) + e-}: E°_{ox}=2.71 V $$
$$\ce{Zn(s)->Zn^2+(aq) + 2e-}: E°_{ox}=0.76 V $$
Now, you might be wondering what any of this has to do with your original question. It turns out that a reaction is spontaneous (will occur) when the $ΔE_{rxn}>0$. Calculating $ΔE_{rxn}$ for the oxidation of sodium gives us $ΔE_{rxn} = E°_{red} + E°_{ox}=-0.83V + 2.71V=1.88V$, which is greater than zero, so this reaction will be spontaneous. On the other hand, if we calculate the cell potential for zinc being oxidized and water being reduced, we get a negative number, so this reaction is not spontaneous. In other words, sodium will readily reduce water, but zinc will not. Since zinc cannot directly reduce water, it will react with water to produce zinc hydroxide and hydrogen gas. The zinc will then decompose into zinc oxide and water. The reaction mechanism for this is given here:
$$\ce{Zn + 2 H2O → Zn(OH)2 + H2}$$
$$\ce{Zn(OH)2 → ZnO + H2O }$$
You might be wondering why this reaction occurs. It turns out that the standard electrode is:
$$\ce{2H+(aq) + 2e- -> H2(g) + e-}: E°_{red}=0 V $$
which means that while zinc cannot reduce $\ce{H2O}$, it can reduce $\ce{H+}$, reacting with the protons present due to the natural decomposition of water. After losing two electrons, the positively charged zinc will react with the hydroxide found in solution. This creates zinc hydroxide, which is an insoluble compound. Sodium will remain dissociated from the hydroxide, as sodium hydroxide is water soluble. From here, the decomposition of zinc hydroxide is spontaneous, producing water and zinc oxide.


text:
Percent composition by volume of the reaction H2 + Br2 <=> 2HBr

For the reaction $\ce{H2(g) + Br2(g) <=> 2HBr(g)}$ at a certain temperature, $K=1$. Determine the composition (in per cent by volume) of the equilibrium reaction mixture if the initial mixture consisted of $3$ moles of $\ce{H2}$ and $2$ moles of $\ce{Br2}$.

The answer given is $49.6\%$ of $\ce{H2}$, $29.6\%$ $\ce{Br2}$, $20.8\%$ $\ce{HBr}$.
I made an ICE table with following data:
$$
\begin{array}{c|ccc}
   *        & \mathbf I   & \mathbf C   & \mathbf E \\
\hline
 \ce{H2}    & 3   & -x  & 3-x \\
 \ce{Br2}   & 2   & -x  & 2-x \\
 \ce{HBr}   & 0   & 2x  & 2x \\
\hline
\end{array}
$$
Using this data I made the equation:
$$
1=\frac{4x^2}{(3-x)(2-x)}
$$
My work in solving it was:
$$
(3-x)(2-x)=4x^2 \\
3x^2+5x-6=0 \\
x=\frac{-5+9.85}{6}\text{ or } x=\frac{-5-9.85}{6}
$$
I used the answer $0.808$ for $x$. I calculated the volume percents as:
$$
\%~\ce{H2} = {3-x\over 5} = 43.84\% \\
\%~\ce{Br2} = {2-x\over 5} = 23.84\% \\
\%~\ce{HBr} = {2x\over 5} = 32.33\%
$$
But, these don't match the answer given. Where have I gone wrong?

There's a few spelling mistakes

What equation did you use for the equilibrium expression? You should have gotten a quadratic equation, not a linear one.

I got  quadratic equation but squared x are canceled each other

Please see my edits (click "edited just now") for examples of how to mark up your posts.

The $x^2$ terms should not have cancelled. Aside from the typo in the initial $\ce{HBr}$ column, your ICE table looks correct to me. Please add to your post the equilibrium expression you used to make the equation $6-3x-2x=0$.  See here for details of how to format equations, if needed.

Hmm. I get a different answer than the one given. Can you add into the question the volume percents that you calculated in your work?

Br:23.84%    H: 43.84%  Hbr:32.32%

This is problem from chemistry exercises book, N.L. Glinka, the problem #353

Your equation and your math appear correct to me: I also get $x=0.808$ from the quadratic equation, and calculate the same percentages.
As long as you are supposed to assume ideal gas behavior in the question, it appears to me that the answer key is incorrect.

authors: Музаффар Шакаров


text:
Why does two phenylhydrazine molecule react with glucose in osazone formation?
From Wikipedia:

It involves two reactions. Firstly glucose with phenylhydrazine gives glucosephenylhydrazone by elimination of a water molecule from the functional group. The next step involves reaction of one mole of glucosephenylhydrazone with two moles of phenylhydrazine (excess). First phenylhydrazine is involved in oxidizing the alpha carbon to a carbonyl group, and the second phenylhydrazine involves in removal of one water molecule with the formyl group of that oxidized carbon and forming the similar carbon nitrogen bond.

Why doesn't it first react with the formyl group and stop with just one phenylhydrazine? That is, why doesn't the nitrogen of phenylhydrazine attack the $\mathrm{sp^2}$ carbonyl carbon (a nucleophilic addition reaction)?
Why does it oxidize glucose first and then proceed?

This may help:https://www.youtube.com/watch?v=_-Ff4dsGkwI

As according to modern mechanism it has first reacted with formyl then goes for tautomers and to form an intermediate..which further form iminoketone ,as phenylhydrazine is in excess then it again react with the c=o formed on C2 and then with carbon first and give osazone .  And as further reaction stops by H- bonding between H and Nitrogen of each phenyl hydrazine which forms chelation and makes it stable.

This is very difficult to understand at the moment. I can't work out whether it is actually answering the question or not. Please consider rewriting it.

Three equivalents of phenylhydrazine are required to form an osazone. One acts as an oxidant. Look here: https://chemistry.stackexchange.com/questions/42702/if-glucose-can-react-with-phenyl-hydrazine-to-form-osazone-why-cant-it-react-wi


text:
What would make tracing paper clear?
This isn't much of a "chemistry" question, but this is the only category where I can get answer based on science. I already tried soaking it in cooking oil, which did not work (it only made the tracing paper even more slippery.)

I can tell you how to find something that will make tracing paper transparent, but I don't have an example to hand.
Tracing paper is partially opaque because it is "rough" and contains many small fibres of cellulose that scatter light. These scatter light effectively because of their size and because their refractive index is very different from air. If you can find a liquid which has a similar refractive index, that will make the paper much more transparent than in its natural state.
I'm sure their are some, but I don't know any right now.


text:
Why in thermodynamics, some reactions are taken at constant temperature and pressure? What is its significance?
To be more specific while deriving relation between cell emf and free energy, constant temperature and pressure  assumptions were taken. Why?

If you read my answer to this question: Why does the Gibbs free energy only correspond to non-expansion work?, you will see that
$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + đW_\text{add}$$
Of these three terms, only $đW_\text{add}$ is related to the electrochemical work. Therefore, in order to equate $\mathrm{d}G$ to $đW_\text{add}$, you have to assume constant $p$ and $T$ such that both $\mathrm{d}p$ and $\mathrm{d}T$ are equal to 0.
In general, this applies to chemical equilibria as well. Because of the above expression, we can say that $G$ is the "natural quantity" to use in systems with fixed pressure and temperature. If instead you had a system with fixed volume and temperature, the Helmholtz free energy would become much more relevant.

That being said, I'd add that fixed volume requires some sophisticated experimental setup, while fixed pressure comes for free.


text:
Glycogen Hydrolase
Can glycogen hydrolase cleave both glycogen and starch bonds?
My intuition says yes because they are both a 1-4(6) glycosidic linkages but I bet I am missing something.


text:
ZnO spontaneous combustion during transport
I found your contact information researching $\ce{ZnO}$ reactions. I was hoping you could please help me or orient me in the right direction to find an answer to my problem.
The company I work for produces brass alloys in a furnace and one of our byproducts is zinc oxide dust. We sell this zinc oxide to other manufacturers. The last shipment we sent on a truck combusted, it burned out of the plastic bag, burned the wooden pallet it was on and it burned a hole in the wooden bed on the truck.
This has only happened twice in the 30 year span of the company. We keep our zinc oxide for two weeks before we sell it, to make sure it has cooled down, we also use two plastic bags, closed off with tape to avoid contact with moisture.
Based on a past profile done on our material, some of the other elements found in our zinc oxide dust are: $\ce{H2O}$ 1.92%, $\ce{Cd}$ 0.13%, $\ce{Cu}$ 0.66%, $\ce{Fe}$ 0.17%, $\ce{Pb}$ 7.43%, $\ce{Mn}$ 0.01%, $\ce{K}$ 0.36%, $\ce{Na}$ 1.18%, $\ce{Ca}$ 0.06%, $\ce{Mg}$ 0.01%, $\ce{Al}$ 0.02%
Do you have any idea what could have caused this reaction? Can you please suggest a source I can look into for more information and answers? We don’t want to risk shipping this substance if there is a chance it will combust and put people in danger.

That is very strange... D0 you have any additional info? Also there was that much Pb?

For a problem of this magnitude might I suggest that you also hire somebody qualified to look into it - we are all more than willing to help but ultimately you shouldn't trust random people on the Internet who aren't privy to the actual situation etc. All we can offer are suggestions. (And personally I find it very strange too.)

Zinc hat a relatively low boiling point of 904°C. So it might be that the zinc oxide contained significant amounts of finely dispersed zinc, which can be pyrophoric in the presence of air (or probably even moisture). However, this is only a guess. If you want to find out the true reason for the accident, you should follow @orthocresol 's suggestion and hire a professional.


text:
Control the pressure of Hydrogen in a small container without the use of a pressure regulator
I had recently asked this question, and due to the comments and general consensus of safety concerns, I decided to re-ask the question, attacking the problem from another angle.
The Problem
I need to store a relatively large amount of Hydrogen in a small (12 cc) Aluminum container at about 1000 psi. The problem is not the generation of the hydrogen, nor the transfer of the hydrogen into the container at that high pressure, but how to extract the hydrogen out of the container without using a pressure regulator.
I am afraid that the container is too small for a pressure regulator to fit in it, and the whole device has to be small on the outside, so one cannot fit it on externally...
The output pressure I desire would be about 200 psi
An Idea: (I'm an Electrical Engineer, don't laugh!)
Emulate Vapor Pressure
Gasses like CO2, Butane, and Propane have relatively low vapor pressures, allowing them to form a liquid at reasonable pressures. At about 800 psi, CO2 will form a liquid, making it so that any container holding liquid CO2 will let out CO2 at that somewhat constant 800 psi of pressure until there isn't enough CO2 to keep itself in liquid form. Hydrogen, however, will not do this because it's vapor pressure curve is far too sensitive. It would have to be cooled down to some ungodly temperature, which is not an option:

I was wondering if I could somehow emulate this behavior by possible adding something to mix with the hydrogen and change it's vapor pressure. Is this even possible?
Edit:
Flow rate doesn't directly matter, but the more flow the better.

Do you want semi-constant output pressure, or semi-constant output flow? What kind of flow rate are you looking for? How constant does the flow rate need to be? Also, you really need to consider hydrogen embrittlement of aluminum and pick the best alloy.

Have you explored the range of inline micro pressure relief valves available, or even micro controllers/relief valves for pneumatic applications? Very small footprints.

@JonCuster Constant output pressure.

@long It seems to be that the tiny ones only handle tiny pressures. If you have an eye on one please send a link. But that was the first solution I was considering. I have looked far and wide.

Would something like H-Cube be interesting? They use water electrolysis to produce hydrogen and can go up to 100 bar.

@K_P Definitely interesting, but the hydrogen generation isn't the problem.

More like a suggestion than a solution.
First, understand what you are trying to achieve.
You are generating hydrogen at some rate and storing it at high pressure. From your previous question I understand that the pressure of your stored $H_{2}$ is only a very general approximation.
Your objective is now to extract $H_{2}$ in a controlled way without using regulator. This sounds absurd.
My complaints:

I am very sure that you do not have a pressure gauge to measure the pressure of your $H_{2}$ container. This is a serious design flaw.
When you will try to extract $H_{2}$ the flow rate will be really high.

Now to the possible solution:

Incorporate atleast one dual valve regulator. This will take care of both issues mentioned above. You will be able to measure container pressure, and  regulate flow-rate.

(But you do not want a such a regulator ! okay. What you can do is to get the smallest one possible and connect the regulator using 1/8 inch or 1/4 inch PTFE tubes and keep it separate. The regulator will then need to have custom input connector. It is not difficult to design one. Regulator to tube can be connected using all metal nut-ferrule connection which are safe upto 2000 psi or more. )
Well, your safety should be prime concern than the size of device I guess. In case oxygen leaks in and temperature at some local point is enough there is a chance of SERIOUS explosion amplified by the high pressure.
Some things about which I am concerned:

Do you evacuate the container before filling $H_{2}$ in ? If not, you are making a mixture of $H_{2}$ and $O_{2}$. Be careful !
The container of $H_{2}$ can handle high pressure or not. Have you tested it ? You can test it with $N_{2}$ or just air before using $H_{2}$.

If you need more information about custom connections then send a comment.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Competitive protonation between alkene and carbonyl

Given this compound in an acidic medium, what would get protonated first?
Would the double bond not get protonated preferentially, since oxygen is more electronegative than carbon and less willing to donate an electron pair for protonation?
I checked the $\mathrm{p}K_\mathrm{a}$ value for the conjugate acid of an  aldehyde, and it was around -10, which is pretty low; this suggests that the aldehyde would have a very low basicity.
Also carbocation had a $\mathrm{p}K_\mathrm{a}$ value of around -3.
Going by this information does it not make sense for a double bond to get protonated first?
On the other hand, the oxonium ion is resonance stabilised, which would indicate that it is protonated in preference to the double bonds.
Which reasoning is correct? And more importantly, in these situations how are we supposed to predict which factor dominates?

You'd need pKa of carbocation not alkene.

please don't spam bold+italics, it doesn't help the readability

@Mithoron  sorry about that,corrected.

@orthocresol oh,I thought they did, but sure.

Draw the mesomerics forms and select the part of the molecule in which this effect is the lowest then you'll have more electrons  there to attack the proton. If you have nothing else it is a good way to have a good idea.

@Shadock  okay, so  you mean that because of the conjugation of the double bond with the carbonyl oxygen,there would be more electrons on the oxygen atom to attack the proton?

No, the mesomeric effect if more important in this part so the electrons from the gamma carbon (from the carbonyl) to the oxygen are all more delocalized than the electrons on the bouble bond alone at the left of your molecule which also have a CH3 next to it which is donor of electrons. So the density of electron is higher in the double bond in the left than in the right of the molecule so I would prefer the left one. But I am not a specialist and well an other part of me would like to prefer a simple electrostatics interaction with the oxygen.

In the scheme below I have shown the three most probable protonations of the molecule in question. I have considered that a tertiary carbocation is typically more stable than a secondary one and that a resonance-stabilised carbocation is typically more stable than a non-stabilised one.

From left to right, let’s discuss the structures.

The one on the left is not good. The carbocation is only stabilised by being tertiary. It would take a very strong acid to sufficiently protonate in this way, or a good nucleophile that can capture the positive charge on the tertiary carbocation.
The one in the middle doesn’t look as bad initially, because it is resonance-stabilised. However, if you draw the other principal mesomeric structure, you arrive at a positively charged oxygen with a sextet — a very unstable thing to have. It is very improbable that protonation would occur like this.
There is another (undrawn) structure that would protonate the same double bond but on the other side. However, that carbocation would destroy the unsaturated ketone’s resonance and is therefore unfavored.
Finally, the protonated carbonyl. This is clearly the best cation, since it is an all-octet structure and the positive charge can be stabilised by resonance. Futhermore, if you draw the resonance structure, you realise that it localises the cation on a tertiary carbon — the best possible case.

Thus there should be no discussion; protonation of the carbonyl oxygen leads to the most stable cation and is therefore expected to prevail.
(Your lower $\mathrm{p}K_\mathrm{a}$ of carbocations is likely due to the fact that they can often deprotonate to form a double bond giving additional stabilisation.)

@ron I was sure that I had re-removed them, but ChemDraw again was more ‘intelligent’ than I was. Bugger. (Hopefully fixed.)

I am not saying it is a valuable answer because organic chemistry is not the field of chemistry in which I am the best. But doing what I did can help a bit as I said.
If you try a mesomeric form with the left bouble bond, which is possible in theory (everything is possible on a sheet of paper) you see that you need to make a carbon with a negative charge.
If you do the same in the second part of your molecule, you see that the gamma carbon take a positive charge and the oxygen the negative charge, which is correct about their electronegativity.
So the mesomeric forms I drew in the left of my sheet are more probable than those on the right.
So on your molecule you will have more a bouble bond between the two carbons on the left and more a "1.5" bond in the right and in the right the electrons are travelling between atoms and then they look less reactive (obviously none of them is static just these ones will move more than others). To get your proton you need electrons, and there are more of them in the left side then (which are not doing anything else).
So I would answer it is the bouble bond with react first.
I hope it can help you to think about that. And as I said I am not a specialist, I just propose a way to think.
And sorry in advance for my possible English mistakes ! :)


text:
Increasing C-alkoxycarbonylation of enolates
I am working with N-tosyl-2-piperidone and attempts to use the lithium enolate of this compound to perform an nucleophilic acyl substitution reaction on an several chloroformate esters in THF has lead to multiple products. The major side product of this appears to be the vinyl carbonate, formed by O-attack of the enolate on the chloroformate ester.

I have used the cyanoformate analogues with success, but benzyl cyanoformate is impressively expensive and if I can avoid synthesizing it (with sodium or potassium cyanide), I would prefer to do so.
Does anyone know of a procedure for increasing the C-addition product of a reaction like this?


text:
Does isopropyl alcohol have a similar affect as dipropylene glycol regarding absorption?
I am an aromatherapist and am making essential oil reed diffusers.  Do you know if isopropyl alcohol has a similar affect in absorption as dipropylene glycol?


text:
Half cell reactions for oxidation of water by acidified solution of potassium dichromate
From Chemguide:

I don't understand the first equation. The chromium atoms in the dichromate ion receive 6 electrons, and their oxidation state declines from 6+ to 3+. But where do the 14 hydrogen cations get their 14 electrons in order to form water with the seven oxygens?

Look closely at the oxidation state of all atoms in the equation.
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) + 6e- <=> 2Cr^3+ (aq) + 7 H2O (l)}$$
The oxidation state in chromium changes from +6 to +3. This is what the electrons in the equation are necessary for. Oxygen retains its oxidation state of -2 in both sides of the equation, as well as hydrogen which remains its +1 oxidation state.
In the other equation
$$\ce{O2 (g) + 4H+ (aq) +4e- <=> 2H2O (l)}$$
the oxygen gets reduced by changing its oxidation state form 0 to -2. Again for hydrogen, the oxidation state of +1 is retained.
The total equation for this process would therefore be
$$\ce{Cr2O7^2- (aq) + 14 H+ (aq) <=> 4Cr^3+ (aq) + 8H2O (l) + 3O2}.$$
From the standard potentials you can see, that this reaction may happen, it does not tell you anything about the conditions under which this reaction will happen. There are a lot more factors to consider.


text:
How can I recover PbNO3 from PbI2?
I'm a chemistry school teacher.
I want to know how to recover lead Nitrate from the ppt of PbI2.
I know I can do it using conc HNO3 but I do not know if NO or HI will form. Also I would need a guide on the conditions of the reaction.
I don't want to dispose the lead if I can avoid that.
Thank you

The reaction is given here.
$$\ce{3PbI2 + 8HNO3 ->[\Delta] 3Pb(NO3)2 + 3I2 + 2NO + 4H2O}$$

Lead(II) iodide react with nitric acid to produce  lead(II) nitrate,
  iodine, nitrogen monoxide and water. Nitric acid - 30% solution. The
  reaction takes place in a boiling solution.


text:
What is generally meant by distinct line spectrum?
In a single hydrogen atom the electron is excited to 6th orbit. The book says maximum 5 distinct spectral lines are possible when the electron comes to the ground state.
Looks like they have only considered those lines when the electron jumps from shell 6 to ground, shell 5 to ground, shell 4 to ground etc. But in my opinion we should also consider the lines produced while transfer of electrons from 6 shell to 5 shell etc. As the elctron can jump from shell 5 to shell 1 directly only when it has jumped from shell 6 to 5 earlier. That would lead to more number of spectral lines than just 5.
So, am i thinking something wrong here (the word "distinct"  seems to be the reason)?

Your intuition is right: all the transitions are possible.
But the language is a little vague and might just be a simplification.
The emission spectrum of hydrogen atoms contains a lot of lines which are often grouped into series (which are named Lyman, Ballmer and Paschen, for example). What groups the series is the lower energy level in the transition responsible for the line: the Lyman series are transitions to the lowest energy level; Ballmer is transitions to the second lowest level...). A good summary of all this is given here.
Your book may have been assuming you only counted the lines corresponding to the final drop to the ground state (the Lyman series) ignoring the intermediate drops to other energy levels. If not, it is clearly wrong as all the transitions are possible.


text:
Plastics and Cotton
Plastics melt on heating and Cotton burns.
When do substances burn with flame? Should they always contain volatile compounds for them to create the Flame?

Your generalisation about plastics is too simple. And volatile components are not required for flames to exist.
Many plastics don't burn at all easily. Bakelite, many epoxy resins or kevlar, for example, neither burn easily or melt. PVC melts but doesn't easily burn. Polyethylene and polystyrene both melt and burn.
Flames are created by gas-phase oxidation reactions (at least in air, even more things will burn in fluorine but you don't see that happening very often as not many labs or kitchens have access to fluorine). But plenty of things containing no volatile components can be made to burn with flames.
Coke or charcoal, for example, are mostly carbon but can be burned and, when burning, will have flames. But those flames are not made by volatilising components of the substance but consist of the burning of partial oxidation products (often carbon monoxide) created by surface reactions on the solid. The purer the carbon the fewer the number of volatile components but once the reaction has started, there is enough energy to keep the solid reacting directly with oxygen to generate gaseous volatiles like CO.
Candles, unlike coal, are made from mixtures of long chain hydrocarbons. These can be volatilised by moderate heat and will then burn in air as gases. This is why candles are easy to light compared to coal and also why they contain a wick which both starts the reaction and acts to wick the liquified hydrocarbons into the hotter zones of the flame where they can be vaporised and burned.
So while flames require volatile components, they can come from partial reactions in the substance and don't have to be present before the fire starts.

Burning is the result of a redox reaction between some sort of fuel source and oxygen in the air. Since cotton is a loosely packed substance, there is a lot of surface area for oxygen to react with it. In addition, cotton is mostly made up of cellulose, a compound that readily burns at low temperatures. On the other hand, plastics are made up of large hydrocarbons. Hydrocarbons are fuel sources; however, larger hydrocarbons require more energy to start combustion. Plastic should burn exothermically, but the energy required to start the process is too large for it to be ignited by a flame.


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Does glass adsorb methanol to an exceptional extent?
Does (borosilicate) glass adsorb methanol to an exceptional extent? We are doing vapor density measurements of organic solvents using a microbalance and a sealed glass sphere. Other polar molecules such as acetone present no problems, but we are getting anomalous results for methanol for some reason.
[Alternatively, does it absorb less than other solvents?]

What does 'exceptional' mean in this case?

@Lighthart Compared to other alcohols and ketones

Again, this cannot really be defined until you dictate what you mean by 'exceptional'.  I understand you are comparing it to other molecules, but how?

@Lighthart I want to know whether more molecules of methanol will stick to a bit of glass at a given temperature than (say) ethanol or acetone. Not sure how much simpler I can describe it.

Exceptional is subjective term, and can't be responded to sensibly.  But more/less can be discussed.  This is somewhat complex.  The isotherm is an equilibrium between adsorption and desorption.
Methanol is small, and when adsorbing to glass, presumably through hydrogen bonding, a single molecule will do very little to block other adsorption sites.  Other, larger, molecules might 'lay down' on the surface or have branched structures that may inhibit their peers from adsorbing, giving smaller surface coverage.  However, long linear molecules will tend to orient, and there is an energetic favorability to forming a monolayer for long linear alcohols (and thiols, carboxylates, amines, silanes, etc).
In these highly ordered linear systems there is an attractive force of roughly 0.2 kcal/mol per methylene unit.  Since methanol has the fewest number possible, if a methanol molecule desorbs it can (almost) freely leave the molecular layer, as opposed to longer chain molecules.  When you begin to discuss molecules that are more topologically diverse than simple linear chains, this attraction is significantly weakened, and as such those molecules may also (almost) freely desorb.
In essence the methanol isotherm has maximum kinetic lability, and as such the coverage will happen reasonably quickly, and be fairly complete.  At equilibrium I would expect methanol to have apparently higher coverage than any branched molecule, and before equilibrium, I would expect methanol to have more coverage than long-chain analogs.
Please note this is on a per molecular basis, and the mass equivalents of methanol will be smaller than any other molecule to be compared with (save water).

Thanks - sounds plausible. I will pass it on to the PhD student who discovered the anomalous results we got and see whether it might be testable.


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Burning Of Coal and Black Body Radiation
Coal burns with a red glow.
What is the ignition temperature of coal?
If it is T, and I take another substance which has melting point quite above T,
and heat it to T, Will it also emit red light?

Yes. Blackbody radiation is blackbody radiation regardless of what is emitting it (with some caveats that are material specific thrown in). After enough time around furnaces, you have a pretty good feel for what temperature emits what colors from just below 500C to 600C and slightly above.

Don't confuse the ignition temperature with the temperature of a burning coal. One is the temperature required to start the reaction, the other is the equilibrium temperature achieved when the substance is burning.
As for the colour of things at a given temperature, the colour will be broadly the same at the same temperature: this is black body radiation. For an interesting take on this in the context of movies, see this amusing video about why movies get it wrong on casting iron swords. A key point is that the glowing colour shown in most movies is far too cool (too orange red) for molten iron (which should be much whiter). Movies cheat and use aluminium which melts at less than 700 °C and glows orange. Iron needs ~1,600 °C and glows far whiter.