text:
When pH increases, does the effect of increased [OH-] or reduced [H+] control precipitation of CaC2O4
The image below shows how oxalic acid dissociates and forms calcium oxalate ($\ce{CaC2O4}$ or $\ce{Ca(OOC-COO) as written below}$).

When the pH increases, does this result in more or less $\ce{Ca(OOC-COO)}$ forming?
I can think of two factors:

As $\ce{pH}$ increases, $\ce{OH-}$ increases. Greater $\ce{OH-}$ means more of the $\ce{Ca^{2+}}$ will react with $\ce{OH-}$, forming $\ce{Ca(OH)2}$. So the reaction in the third line below will shift to the left, resuling in less $\ce{Ca(OOC-COO)}$.
As $\ce{pH}$ increases, $\ce{H+}$ decreases. Less $\ce{H+}$ results in the reactions (in line 1 and 2) shifting to the right. Hence, there will be more $\ce{(-OOC-COO-)}$ and so more $\ce{Ca(OOC-COO)}$.

These give opposing results. Which one outweighs the other, and why?

The original question from by textbook:
Drinking strong antacids to relieve indigestion produces kidney fluid that is more alkaline than usual.
Are kidney stones more likely to occur under these conditions?

Yes, as at a higher pH a higher concentration of oxalate ions will form.
Yes, as both calcium and oxalate ions are bivalent.
No, as the presence of hydrogen oxalate ion (HOOCCOO–) means that a buffer is present.
No, as the higher concentration of hydroxide ion (OH–) will remove calcium ion by forming calcium
hydroxide.

You seem to have the specifications for the problem and your work mixed together. Would you first state the problem as posed to you and then your assumptions/work?

Do you mean just add the textbook problem? If so, I have now edited it in. Thank you.

In the body we'll assume that:

There is some low level of $\ce{Ca^{2+}}$ present in the blood or urine.
The pH isn't nearly high enough to precipitate out $\ce{Ca(OH)2}$.
Under "normal" bodily conditions calcium oxlate doesn't precipitate out either so the overall oxalate concentration is low.

Let's let $x$ be the concentration of all the oxalate species. By knowing $x$, the pH and the two pka's we can calculate the exact concentrations of all three oxalate species.
$x = \ce{[HOOC-COOH] + [HOOC-COO^-] + [^-OOC-COO^-] }$
Now the pKa values for oxalic acid are far enough apart that at most two of the three oxlate species will have an appreciable concentration.
For normal bodily functions only $\ce{[HOOC-COO^-]}$ and $\ce{[^-OOC-COO^-]}$ will be the only significant species. Since only a couple of significant figures will be of interest, we can use the Henderson–Hasselbalch equation to calculate the relative concentration of the two species to reasonable accuracy. (The essence is that the third species has a negligibly small concentration.)
pH = $\text{pK}_{a2} + \text{log}(\dfrac{\ce{[^-OOC-COO^-]}}{\ce{[HOOC-COO^-]}})$
Thus as the pH increases relatively more $\ce{[^-OOC-COO^-]}$ is formed and less $\ce{Ca^{2+}}$ will be soluble.
But normal blood is alkaline already. Thus there would be a negligibly small amount of $\ce{[HOOC-COO^-]}$ in the blood to start with since the $\text{pK}_{a2} = 4.2$. So calcium oxalate will be unlikley to ppt in the blood.
However the kidneys make urine which has a pH from 4.5 to 8. So if the pH in the urine changes as much as from 4.5 to 8 then there could be a greater tendency to ppt calcium oxalate in the urine.
ANSWER
(4) is out since if body was that alkaline your tissue would dissolve.
(3) is out since you can see from the HH equation specified that the presence of another buffer would effect the absolute pH but not the relative oxalate species at that pH.
(2) is true, but the answer doesn't note any effect of pH. So the answer doesn't really address the problem.
Thus the answer must be (1).

Thank you for the answer. Could you please explain the significance of this line "x=[HOOC−COOH]+[HOOC−COO−]+[−OOC−COO−]" It looks to be the sum of each of the reactants except Ca2+. Is that right, and if so, why are they being summed?

Ok, tried to expand that point. The simplification doesn't make any real difference for a computer program which can solve for all three species concentrations iteratively in a fraction of a second, but using the HH equation makes hand calculations a lot simpler than solving a cubic equation. Note that pka's and Ksp only have two significant figures.

At a higher pH a higher concentration of oxalate ions will form (Le Chatelier's principle). Thus the formation of kidney stones is more likely.
If calcium hydroxide precipitates in your body you're most likely dead.
The solubility product of $\ce{Ca(OH)_2}$ is $5.5 \cdot 10^{-6}$. The calcium concentration in blood is about $3 \cdot 10^{-3}~\pu{mol\cdot l^{-1}}$. It needs a pH above 13 to precipitate calcium hydroxide in blood.

Thank you for the answer. Would you please be able to point out why "a higher pH a higher concentration of oxalate ions will form" (which does make sense) outweighs: a higher pH = more [OH-] = Ca2+ being used up to form Ca(OH)2 = reaction in 3rd line shifting to the left = less calcium oxalate or kidney stones.

Added the pH required to precipitate $\ce{Ca(OH)_2}$ from blood.


text:
When removing a reactant and equilibrium shifts left, doesn't this make an excess of the 2nd reactant, forcing equilibrium back to the right?
In the reaction:$$\ce{Ca^{2+}(aq) + ^-OOC-COO^- <=> Ca(OOC-COO)(s)}$$
If $\ce{Ca^{2+}(aq)}$ is removed, the reaction shifts to the left to produce more $\ce{Ca^{2+}(aq)}$ to compensate.
But when the reaction shifts to the left, isn't a side effect that more $\ce{^-OOC-COO^-}$ is also produced (as when $\ce{Ca(OOC-COO)(s)}$ dissociates it produces BOTH $\ce{Ca^{2+}(aq)}$ and $\ce{ ^-OOC-COO^- }$, and not $\ce{Ca^{2+}(aq)}$ by itself?
So if there is more $\ce{^-OOC-COO^-}$, won't the reaction want to shift to the right?
So wouldn't these two (i.e. wanting to shift left and shift right) counteract and the reaction not shift anywhere? I think there is something about Le Chateliers's that I have misunderstood, so please point it out if you see it.

The thing to be careful about with Le Chatelier's principle is that it relates to equilibria :

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium.

Le Châtelier's Principle
Now when you say in the above "But when the reaction shifts to the left, isn't a side effect  ..." note you can't use Le Chatelier's principle here again so fast because the system has not reached an equilibrium yet, it is just moving towards one.
It is likely that the next thing you will study is how to actually calculate the new equilibrium conditions. This gives usually a deeper understanding of how changes in various conditions make the reaction move forward/backward establishing new equilibrium conditions.

It's quite likely that it will overshoot as the OP suggests. However, it will gradually reach equilibrium.

According to Wikipedia, regarding the Law of Mass Action:

In chemistry, the law of mass action is the proposition that the rate
  of a chemical reaction is directly proportional to the product of the
  activities or concentrations of the reactants. It explains and predicts behaviors of solutions in dynamic equilibrium. Specifically, it implies that for a chemical reaction mixture that is in equilibrium, the ratio between the concentration of reactants and products is constant.

So the key with respect to your question is that the equilibrium will shift in a manner that is proportional to the product of the reactants.
However, I think your specific example needs a bit of a clarification.  As written, you are suggesting removing only positively charged species, plucking out just the $\ce{Ca^2+}$ ions without adding more cations or also removing anions.  If you did this through some sort of ion-exchange technique, you would have to consider whether the cation replacing $\ce{Ca^2+}$ would also react with your other reactant or if it would purely be a spectator ion.  The discussion of the Law of Mass Action above assumes the latter.

Calcium oxalate is relatively unsolvable. So there will be a $\text{K}_{\text{sp}}$ such that
$\text{K}_{\text{sp}}$ = $\ce{[Ca^{2+}] [^-OOC-COO^-]}$
So if the calcium oxalate is dropped into a sodium chloride solution, then
$\ce{[Ca^{2+}] \approx [^-OOC-COO^-]}$
It isn't quite true because a bit of $\ce{HOOC-COO^-}$ will form. Let's ignore this factor which will be small. Let's define $x$ as:
$x = \sqrt{\text{K}_{\text{sp}}} = \ce{[Ca^{2+}]_i = [^-OOC-COO^-]_i}$
where "i" stands for initial. Now we take out our magic tweezers and remove half the calcium ions and twice that number of chloride ions. After all the magic tweezers aren't magic enough to leave the solution charged.
But now more $\ce{Ca^{2+}}$ must dissolve as well as more $\ce{^-OOC-COO^-}$. But we know that
$0.5x = \ce{[^-OOC-COO^-]_2 - [Ca^{2+}]_2}$
and that
$\text{K}_{\text{sp}}$ = $\ce{[Ca^{2+}]_2 [^-OOC-COO^-]_2}$
So we have two equations with two unknowns and the new equilbrium conditions can be solved.

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text:
Simple experiment to do at home that involves acids and bases?
so long story short I am in the IB Diploma and I am required to come up with an experiment to investigate pretty much anything to do with chemistry. I'm thinking about doing mine on investigating how temperature affects the pH value of a solution, but my teacher said it was too simple.
Basically, is there any reaction I can make at home with simple chemicals(vinegar, baking soda, salt, juice, etc.) that would produce a meaningful experiment and involves me measuring the pH level? Thank you guys so much.

Maybe mixing boric acid (maybe made from borax) and glycerine or other polyol? https://en.wikipedia.org/wiki/Boric_acid Question looks too broad unfortunately.

There will be no measurable impact on pH at temperatures you can generate at home. How about something like measuring how accurate the percentage of store bought hydrogen peroxide is? Simple titration.

Measure the pH+ of various foods, and make a spreadsheet listing them.
You might also measure pH+ change with time in a food, e.g. apple juice or bread dough, with a few mg of yeast sprinkled in.

How about home made indicators? There are plenty of plant based colorants that can change color on pH change. You can make it easy way ( listing colors in acidic, neutral, basic  solutions) or make it more difficult, eg actually comparing in what pH the color changes (hint: it is generally not exactly pH 7)

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text:
Regioselectivity in the formation of an imidazo[1,2-a]pyrimidine
In the heterocycle synthesis below (taken from Org. Process Res. Dev. 2006, 10 (3), 398–402), a 5,6-fused heterocycle (an imidazo[1,2-a]pyrimidine) is formed by treating 2-amino-4-(trifluoromethyl)pyrimidine with the diethyl acetal of bromoacetaldehyde, with the cited paper claiming to get a 25:1 selectivity for 5 (minimal amounts of 5a also produced were able to be easily removed).

Ref: Org. Process Res. Dev. 2006, 10 (3), 398–402.

There are (as ever with heterocycle syntheses) a few different mechanisms one could draw, however, no matter which way round one draws the mechanism, the major product is always the one that arises from what appears to be the less favoured intermediates.
By less favoured intermediates, what I mean to say is that to arrive at the major product, necessarily a cation ends up closer to the electron withdrawing trifluoromethyl group than it would do if the minor product was formed.
Given that all the steps are essentially reversible, is the product distribution arising from a purely ground state thermodynamic preference for 5 over 5a, or is there a mechanism that I'm completely neglecting (sorry for the lack of pictures, I'm not able to ChemDraw right now).

I propose the route in the following scheme. I don’t see any unfavourable intermediates. In short, the first step would be formation of a Schiff base with the extracyclic amino group and the aldehyde. This — in one of many orientations as shown below — places the bromide nicely so that one of the pyrimidine nitrogens can attack it nucleophilicly. Upon displacement, an intermediate nitrogen-centred cation is generated; loss of a proton and switching of electrons regenerates aromaticity and gives the desired product.

Scheme 1: Proposed reaction scheme. The attacking nitrogen $\mathrm{sp^2}$ orbital in step 2 is drawn explicitly.
We can draw structures of the Schiff base intermediate that would allow either of the pyrimidine nitrogens to attack. However, the top one experiences a stronger $-I$ effect by the neighbouring $\ce{CF3}$ group hence its electron density and nucleophilicity should be reduced. The para-nitrogen does not experience this inductive effect since it greatly weakens with greater distance.
The ortho-nitrogen’s lone pair is indeed affected since inductive effects affect both σ and π bonds while mesomeric effects mainly affect π electrons only.

What about a concerted mechanism?

@Martin-マーチン Concerted mechanism for what? The entire addition? Don’t know, can’t think of one. The nucleophilic substitution/elimination? Why not, never sold on the details.

@Jan. Thanks for the answer. I had considered that mechanism, but I feel a little uncomfortable using the nitrogen LP like that without making use of the ring in some way (plus admittedly I had a feeling the DMAP pathway was more likely due to reactions of similar systems- this of course doesn't help get to the product).

@Martin-マーチン - I'd be interested to see the concerted (pericyclic?) mechanism. I'm struggling to see how

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text:
Difference between positional isomerism and metamerism (organic chemistry)
I don't want the definition of positional isomerism and metamerism but want what differentiates the two types of isomerism.
I know metamers have different alkyl groups on either side of it, but doesn't that just mean the position of the functional is changing? Hence shouldn't it be a positional isomer.

If functional group carbon is part of principal chain then metamers are also positional isomers. In case of polyvalent functional group like sulphoxide($\ce{>S=O}$), secondary amine ($\ce{-NH-}$) etc. structure formed by change in structure of alkyl groups will be metamers only.  For example, diethyl amine ($\ce{Et2NH}$) and methyl propyl amine ($\ce{MeNHPr}$) are metamers.

Actually metamerism comes under positional isomerism. It's like this "metamerism is a subset of positional isomerism". When a compound has a functional group as a part of its chain then we call  positional isomerism as metamerism to specify that the functional group is a part of the chain.


text:
pH of aqueous solution of HCl of low concentration

What is the pH of $10^{-8}~\mathrm{M}$ $\ce{HCl}$ solution in water?

My attempt:

pH = $-\log(10)^{-8}$ = 8

But this is wrong because it should be acidic. Where have I gone wrong?

It is known that the equilibrium constant for the following reaction is $10^{-14}$.
$$\ce{2H2O(l) <=> H3O+(aq) + OH-(aq)}$$
That means,
$$\frac {[\ce{H3O+(aq)}][\ce{OH-(aq)}]} {[\ce{H2O(l)}][\ce{H2O(l)}]} = 10^{-14}$$
where the concentration of water $[\ce{H2O(l)}]$ is assumed to be $1$.
Initially, we have $[\ce{H3O+(aq)}] = 10^{-8}$.
Then, assuming that $x$ moles of $\ce{H3O+(aq)}$ is generated. An equal amount of $\ce{OH-(aq)}$ must also be generated:
$$(10^{-8}+x)(x) = 10^{-14}$$
Solving for $x$ gives $x = 9.51 \times 10^{-8}$.
So, the total concentration of $\ce{H3O+(aq)}$ is $1.051\times10^{-7}$.
Therefore, the pH is $6.978$.

The problem with your approach
Let the initial molarity of the hydronium ion be $a$.
Solving $(a+x)(x) = 10^{-14}$, if $a$ is well above $10^{-5}$, gives $x \approx 0$.
The approximation that the pH is equal to $-\log{{\left([\ce{HCl(aq)}]\right)}}$ is only accurate for normal concentrations.

I'll quibble. ...where the concentration of water $[\ce{H2O(l)}]$ is assumed to be $1$. just isn't true. The assumption is that the concentration of water is a constant and is thus ignored in the autopyrolysis equation for water. Water is about 55.5 molar of course.

Here is another approach:
Suppose we have a 10^-8 M HCl solution.
We have:
Mass balance: [H3O+] = [Cl-] = 10^-8 (M).
Charge balance: [H3O+] = [Cl-] + [OH-].
Autoprotolysis of water: [H3O+] * [OH-] = 10^-14.
We multiply the charge balance with [H3O+] and get:
[H3O+]^2 = [H3O+] * [Cl-] * [H3O+] * [OH-] = [Cl-] * [H3O+]  + 10^-14
=>
We insert the value of [Cl-] and get:
[H3O+]^2 = 10^-8 * [H3O+]  + 10^-14
=>
[H3O+]^2 - 10^-8 * [H3O+]  - 10^-14 = 0
We solve the equation for [H3O+] and get:
[H3O+] = 1,0512*10^-7
=>
pH = 6.98 ≈ 7

Caution!
This answer uses a simplification! It assumes that the concentration of $\ce{H3O+}$ as contributed by $K_\mathrm{w}$ is constant at $10^{-7}~\mathrm{M}$. This is incorrect. In reality, an additional equation must be introduced and the resulting formula is cubic rather than quadratic. Unfortunately, solving a cubic equation is very non-trivial and I am not able to do it. One possible cubic equation with which the final value should be calculateable is the following:
$$x^3 + K_\mathrm{a}x^2 - (K_\mathrm{a}c_0)x - K_\mathrm{a}K_\mathrm{w} = 0\\
x = [\ce{H3O+}]$$
Due to the approximation, this answer is off the actual value. The other answer uses a different simplification which arrives at a result closer to the actual value. I’m leaving this post here for posteriority, because it still serves to exemplify how complicated the calculation must be at least and how easily a pitfall is encountered.
Original answer
You used the simplified equation $(1)$ wherein acid is a strong, monoprotic acid. It encompasses a few shortcuts that are only valid for reasonably concentrated strong acids. Using it for an acid this dilute will break a lot of the assumptions used. To clarify, I will use the long method.
$$\mathrm{pH} = -\lg c_0(\text{acid})\tag{1}$$
Upon dissolution of any acid (using hydrogen chloride as an example, but any will work) equilibrium $(2)$ will be found.
$$\ce{HCl + H2O <=> H3O+ + Cl-}\tag{2}$$
This equilibrium can be adequately described by the acidity constant $K_\mathrm{a}$ which is defined as shown in equation $(3)$ (note that where concentrations are implied, activities are actually used. The solvent does not show up because its activity is $1$).
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{Cl-}]}{[\ce{HCl}]}\tag{3}$$
The most important simplification present in $(1)$ is that it assumes $c_0(\ce{H3O+}) \ll c_0(\ce{HCl})$ or — equally valid — $c_\mathrm{initial}(\ce{H3O+}) \ll c_\mathrm{final}(\ce{H3O+})$. In that case, it can be equated that $[\ce{H3O+}] \approx [\ce{Cl-}]$ which is used to reduce the number of independent variables and solve the equation. This will not be the case in this calculation with $c_0(\ce{HCl}) = 10^{-8}~\mathrm{M}$ and therefore we need to explore other methods. What we can say is shown in $(4)$.
$$\begin{align}[\ce{Cl-}] + [\ce{HCl}] &= c_0(\ce{HCl})\\
c_0(\ce{H3O+}) + [\ce{Cl-}] &= [\ce{H3O+}]\end{align}\tag{4}$$
$c_0(\ce{H3O+})$ is known from the autodeprotonation of water $(5)$ from which follows that $c_0(\ce{H3O+}) = 10^{-7}~\mathrm{M}$.
$$\begin{align}\ce{2 H2O <=> H3O+ + OH-}&&K_\mathrm{w} = 10^{-14}\end{align}\tag{5}$$
Now, finally, we have enough information to solve the equation.
$$\begin{align}K_\mathrm{a} &= \frac{(10^{-7} + [\ce{Cl-}])[\ce{Cl-}]}{c_0(\ce{HCl}) - [\ce{Cl-}]}\tag{6.1}\\[0.8em]
K_\mathrm{a} &= \frac{10^{-7}\times [\ce{Cl-}] + [\ce{Cl-}]^2}{c_0(\ce{HCl}) - [\ce{Cl-}]}\tag{6.2}\\[0.8em]
K_\mathrm{a} c_0(\ce{HCl}) - K_\mathrm{a} [\ce{Cl-}] &= 10^{-7}[\ce{Cl-}] + [\ce{Cl-}]^2\tag{6.3}\\[0.6em]
0 &= [\ce{Cl-}]^2 + (10^{-7} + K_\mathrm{a})[\ce{Cl-}] - K_\mathrm{a} c_0(\ce{HCl})\tag{6.4}\\[0.6em]
[\ce{Cl-}]_{1/2} &= \frac{-K_\mathrm{a} - 10^{-7} \pm \sqrt{(10^{-7} + K_\mathrm{a})^2 + 4K_\mathrm{a} c_0(\ce{HCl})}}{2}\tag{6.5}\end{align}$$
Plugging in all the values (taking $K_\mathrm{a}$ from Wikipedia) and realising that only the positive square root makes sense, this is our result:
$$\begin{align}[\ce{Cl-}]_{1} &= \frac{-10^{5.9} - 10^{-7} + \sqrt{(10^{-7} + 10^{5.9})^2 + 4 \times 10^{5.9} \times 10^{-8}}}{2}\tag{6.6}\\[0.8em]
[\ce{Cl-}]_1 &= \frac{-10^{5.9} - 10^{-7} + \sqrt{(10^{-7} + 10^{5.9})^2 + 4 \times 10^{-2.1}}}{2}\tag{6.7}\\[0.6em]
[\ce{Cl-}]_1 &= 10^{-8}\tag{6.8}\end{align}$$
Thankfully, this equates to full deprotonation as the theory predicts. In the simplified context $(1)$, we just assume that complete deprotonation occurs. We can plug the value of $[\ce{Cl-}]$ into equation $(4)$ to get the final result:
$$\begin{align}c_0(\ce{H3O+}) + [\ce{Cl-}] &= [\ce{H3O+}]\tag{4}\\[0.4em]
10^{-7}~\mathrm{M} + 10^{-8}~\mathrm{M} &= 1.1 \times 10^{-7}~\mathrm{M}\tag{7}\\[0.4em]
\mathrm{pH} = -\lg [\ce{H3O+}] &= 7\lg 1.1 = 6.959\tag{8}\end{align}$$
You experience a very slight decrease in $\mathrm{pH}$ but the principal contributor of $\ce{H3O+}$ ions is still the autoprotonation of water.

For pH of an aqueous solution we should consider -log of total [H+] in the solution. In this case Jan's explanation is very good. For quick result find out the H+ concentration from solute and then from H+ ion concentration from solvent. H+ ion concentration from water is generally less than 10^-7 due to common ion effect.

@narendrakumar Yes, we can call that the condensed essence stemming from the reations between equations $(3)$ and $(4)$.

This is wrong!! If $$\ce{[H^+] =} 1.1\cdot10^{-7}$$ then $$\ce{[OH^-]} = 0.91\cdot10^{-7}$$ and $$\ce{[H^+] - [OH^-] =} 1.9\cdot10^{-8}$$.

@MaxW Here’s the promised second comment. My maths is not good enough to deal with the cubic equation that I end up with when doing things properly. Therefore, I have added a note to the beginning of the answer.

Jan - the solution is not a qubic equation but rather a quadratic equation.

@MaxW I urge you to share. I notice your answer to the other question (which is linked to this one as a dupe) does not go into detail much.

Just added solution to that one.

@MaxW Ah. You used the simplification that deprotonation is complete. Which is true for all practical intents and purposes, but I attempted to go ‘all the way’ and account for any residual $\ce{HCl}$. I believed it might not be fully irrelevant at these concentrations.

Yes, I see your point. If $\text{K}_a$ is not much greater than $\text{K}_w$ then you'll end up with a cubic equation which is a mess to solve by hand. I'd also worry about the other end. I'd doubt that 12 molar HCl is completely dissociated. $1.00\cdot10^{-8}$ HCl would be completely dissociated.

How is there a cubic eqn?  I get only a quadratic.

@OscarLanzi To be honest, it is too long ago, I cannot recall it fully. And I also certainly threw away the scrap paper I did the rough calculations on. But the gist is that there is not only the obvious equilibrium described by $K_\mathrm{a}$ that needs to be considered but also the one described by $K_\mathrm{w}$ and if I plug all those together in the hopes of reducing the number of independent variables I end up with a cubic equation.


text:
Dipole-quadrupole polarizability term for induced dipole
For describing an induced dipole, I have usually seen the following equation,
$$
P_{i} = \alpha_{ij}E_{j} + \frac{1}{2}\beta_{ijk}E_{j}E_{k}
$$
where $P_{i}$ is the $i^{\text{th}}$ component of the induced dipole moment, $\alpha$ is the (dipole) polarizability and $\beta$ is the first (dipole) hyperpolarizability.
However, I recently read a paper with quadrupole-coupled terms coming in, like the dipole-quadrupole polarizability as shown below:
$$
P_{i} = \alpha_{ij}E_{j} + \frac{1}{3}A_{ijk} \frac{dE_{j}}{dr_{k}}+ \frac{1}{2}\beta_{ijk}E_{j}E_{k}
$$
where in the second term, $A_{ijk}$ is the dipole-quadrupole polarizability term. This equation is introduced for the case of polar molecules in the condensed phase, and in the related paper it is for water.
It appears that the contribution of dipole-quadrupole polarizability is rather small, but maybe larger than the hyperpolarizability.
Could someone explain or refer some literature so as to describe

the importance of the cross-coupled terms like the dipole-quadrupole polarizability, and
if there are some cases where the cross-coupled terms (the second term in the second equation) become negligible?

Reference: Batista, E. R.; Xantheas, S. S.; Jónsson, H. Molecular multipole moments of water molecules in ice Ih. J. Chem. Phys. 1998, 109 (11), 4546-4551. DOI: 10.1063/1.477058.

This question has been around for more than a year without an answer, so probably you have figured it out by now! But anyway, here goes. The standard literature references in this field are Theory of Molecular Fluids vol 1 by CG Gray and KE Gubbins (Clarendon Press 1984) and The Theory of Intermolecular Forces by Anthony Stone (Oxford University Press 2013) and these books explain the various terms in your equation. The dipole-quadrupole polarizability $A_{i,jk}$ may vanish by symmetry (e.g. for centrosymmetric molecules and isolated atoms) but does not vanish for a molecule like water. The corresponding term in your equation is also proportional to the local electric field gradient, so the importance of this term (relative to the other terms) depends on some of the details of the surroundings. In Appendix C of Gray and Gubbins they state

For the cases we consider in this book the induction $A$ terms are negligible

but clearly there might be some cases in which they should not be neglected.

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text:
Term used to describe scalar coupling over a single bond?
I know that coupling on NMR over two bonds is referred to as geminal coupling and over three bonds is referred to as vicinal coupling. However is there a term for coupling over one bond (such as a carbon-13 bonded to a fluorine?

As far as I know it's usually called "one-bond coupling". Like DSVA I've never heard of any other special term. For example: http://u-of-o-nmr-facility.blogspot.co.uk/2009/01/weak-one-bond-or-multiple-bond.html

FWIW: geminal and vicinal are generic chemistry terms and not unique to NMR at all. The technically correct way to describe coupling is to use the J descriptor; but common expressions include directly bonded and especially one-bond coupling

I've never heard a special term but you can describe it as $^1 J$ or $^1 J_{CF}$ coupling using the systematic nomenclature.

Strictly speaking, the IUPAC convention for J specifies that the nucleus of higher mass should come first; hence this would be J(FC). Not sure that this convention is strictly adhered to in literature though


text:
Hydrogen atom transfer between enantiomers of rac-1-phenylethanol

(Taken from Dalton Trans., 2012, 41, 13423–13428. DOI:10.1039/c2dt31781h)
I have difficulties understanding the hydrogen transfer in the image above. I always thought that hydrogen transfers followed the following formula:
$$\ce{X^. + H-Y -> X-H + Y^.}$$
I can't really see the connection between the picture and the formula. Is it the whole ethanol group that is transferred?

The hydrogen transfer is between the enantiomers. The dashed hydrogen is transferred in the picture.

Is there any more context provided as to how the racemisation occurs? Hydrogen atom transfer via a radical process seems plausible to me since the resultant hydroxybenzyl radical is quite stabilised, but you need something to take the hydrogen atom, since C-H bonds don't just fall apart and re-form spontaneously.

I think, a deprotonation/reprotonation mechanism would be even more plausible than implying radical formation ($420~\mathrm{kJ/mol}$). (CC @orthocresol)

Your image appears to be from this source: http://pubs.rsc.org/en/Content/ArticleHtml/2012/DT/c2dt31781h  It's good practice to cite or link to such source material when you use it in a question (or answer).

@Jan Imo a radical process would be more selective for the C-H bond (benzylic C-H ~370 kJ/mol) over the O-H bond (~440 kJ/mol) whereas an ionic deprotonation/reprotonation would happen at the O-H preferably. I don't know, though, that's just what I remember from lectures :) Of course, I'm not suggesting homolytic cleavage, I was thinking more of a reversible hydrogen atom abstraction by something else.

@CurtF. Sorry don't know much about this site yet, but you are correct i should have cited it porpperly.

Did you continue reading the paper? :) The source is important and is what was needed to answer the question!
I'm just going to quote from it.

In DKR [dynamic kinetic resolution] a chemical catalyst is added in order to racemize the enantiomers, ensuring that all the substrate can be utilized. [...] Highly successful catalysts are often Ru(II) complexes [...] The metal complex catalyses racemization by a hydrogen transfer mechanism (Scheme 2).

So, the supposed hydrogen transfer occurs via coordination to a metal, followed by β-hydride elimination. Readdition of the metal hydride across the opposite face of the resultant ketone (acetophenone in this case) produces the other enantiomer.

I feel like a complete idiot... sorry i wasted your time. I'll double check any further source based questions in the future. Thank you

No problem, it was an interesting question.


text:
Do Bases and Acids actually react with each other?
Let's say we have $5$ $\ce{HCl}$ molecules and we dump them into a container full of water. It's a strong acid so all $5$ $\ce{H+}$ ions would react with water and create $\ce{H3O+}$. It reacts with the water, right?
Now, we add $5$ $\ce{NaOH}$ molecules down there. They all react with some water molecules of their own as well and create $5$ $\ce{OH-}$. So now we have $5$ $\ce{H3O+}$ and $5$ $\ce{OH-}$. Those two don't react with each other, correct? They just react with the water molecules and balance each other out, giving us a neutral solution.
I am unsure of this. Is this true? Thanks in advance.

They certainly react with each other: $$\ce{H3O+ + OH- -> 2H2O}$$

Hmm ... so they react separately with water molecules and then the products react with each other?

Yes. The steps would be: $$\ce{HCl + H2O -> H3O+ + Cl-}$$ $$\ce{NaOH -> Na+ + OH-}$$ $$\ce{H3O+ + OH- -> 2H2O}$$

Thank you. But is it then wrong to say that acids and bases balance each other out in a solution and thereby make it neutral? Provided we have an equilibrium

That question doesn't make sense.

Hahah, ok. I was always taught in school that a pH of 7 means that the concentration of $ OH^- $ and $ H_3O^+ $ is identical.

That is true.$ $

So is it wrong to say that during a titration, when an equilibirum has been reached, that the acid and base have 'balanced each other out'? Or is it more correct to say all available $ H^+ $ have reacted with all available $ OH^- $?

pH  7 does not mean that the concentration of $\ce{OH-}$ and $\ce{H3O+}$ is identical. pH is defined as $$\mathrm{pH = -\log{[H_3O^+]}}$$ The amount of autoionisation of water varies with temperature hence so does the pH. Neutrality is defined as having $\ce{[H3O+] = [OH-]}$, not as pH = 7.

Pretty sure that's what I said ... or meant

$$\ce{NaOH + HCl -> NaCl + H2O}$$
$$\ce{Na+ + Cl- + OH- + H+ -> Na+ + Cl- + H2O_{(l)}}$$
Both sides $\ce{Na+}$ ans $\ce{Cl-}$ cancel out remaining reaction
$$\ce{H+ + OH- <=> H2O_{(l)}}$$
Any strong acid and strong base reaction the remaining reaction is hydrolization of water.


text:
Understanding Chemical Potential (Li-ion)
Can anyone help me to understand what the "Chemical Potential" is? I can see that in batteries the electrical energy is stored as "chemical potential", according to this:

Li-ion batteries, Na-ion batteries, and Mg-ion batteries, reversibly
convert between electrical and chemical energy via redox reactions,
thus storing the energy as chemical potential in their electrodes.

So what is chemical potential? According to Wikipedia, it is defined as:

In thermodynamics, chemical potential, also known as partial molar
free energy, is a form of potential energy that can be absorbed or
released during a chemical reaction or phase transition

Which is quite straight forward, in a way. However, it doesn't really make me understanding it entirely, therefore I would like to ask if there is anyone that can help me understand this. I'm always trying to use analogies and therefore I would like to ask the following question:

Can the chemical potential be seen like the latent heat, where the
energy is stored as bonds and, in order to break the bonds, one needs
to add extra energy.

Or is it more complex than this? Please see below a sketch with what I imagine happens inside. Did I understand it right?

It's hard for me to understand what keeps the energy there. Any help counts!
Thank you!

You're talking about two very different things, thermodynamic potential and electrode potential.

Can you please explain which is which?

Well, you should be able to guess... or simply google it https://en.wikipedia.org/wiki/Thermodynamic_potential https://en.wikipedia.org/wiki/Electrode_potential


text:
Purifying/Recrystallising N-Iodosuccinimide
I'm planning on an Iodination reaction of a 2-aminothiazole derivative. Specifically, 2-aminothiazole, with two Boc groups on the amine.

What practical advice do you have for recrystallising NIS? Literature recommends dissolving in hot dioxane and then adding $\ce{CCl4}$ to precipitate out the NIS.
If I do not have access to $\ce{CCl4}$ would $\ce{CHCl3}$ suffice?
Do NIS and NBS (I've previously used NBS for this reaction in $\ce{CHCl3}$) have similar solubilities in $\ce{CHCl3}$?

Recrystallization is finicky. Maybe it'll work, maybe it won't. Try just lowering the temperature first. The reference I saw was -20 degrees C.

Purification of Laboratory Chemicals (a generally accepted reference book for methods of purification of common chemicals) appears to agree with your procedure:

N-Iodosuccinimide [516-12-1] M 225.0, m 200-201 °C. Crystallise it from dioxane/CCl4. It iodinates arenes in triflic acid. [Olah et. al. J. Org. Chem. 1993, 58, 3194, Beilstein 21/9 V 544.]

As you point out, carbon tetrachloride is difficult if not impossible to get hold of now, so the methodology will not work.
The most obvious thing to try would be the method for the purification of NBS, however NIS is not tolerant to recrystallisation from water and from previous attempts doesn't achieve anything.
Depending on what you're doing with the NIS, I've often had success in simply dissolving in solvent and washing with sodium thiosulfate to remove any liberated iodine. Drying and evaporation then gives cleaner, though probably not pure NIS.
Also note that NIS should be stored in the dark at reduced temperatures to prevent decomposition (which is noted by the appearance of an orange/brown decoloration).

I've seen this entry before in the book, and I've always wondered why use $\ce{CCl4}$. Surely its only purpose is as a poor solvent for NIS, but which is still good for the main impurity, molecular iodine. If so, then precipitation of pure NIS from a dioxane solution by adding hexane should work just as well.


text:
Starch gelatinization, retrogradation and disgestability
Over on the sub-reddit on prebiotics we're a bit stumped on figuring out the process for turning raw potato starch (mainly containing R2 resistant starch) into R3 via retrogradation/gelatinization. The best thing we have found so far on the topic is this article, which is quite a bit too technical for us (e.g. what's melting enthalpy?) and doesn't fully answer our questions.
I myself had a haphazard go at it, boiling water with potato starch for an extended period of time (sorry, don't have exact ratios or times) and then leaving it in the fridge for 3 days. The end result was large chunks of a milky-clear substance of soft-gummybear consistency (can be picked up and lightly squeezed without loosing form or oozing) that will not disolve in water. My personal question is: what is the substance described here and what's could be assumed its digestibility?
For the r/prebiotics reddit in general, the question is what process should be be used by a layman to turn raw potato starch powder into either gelatinized and or retrograded starch (assuming here they're different). I presume the variables we need are water/starch ratio, mixing technique, heating temperature, lenght of heating, cooling temperature and cooling time.
Thank you in advance and for your time.


text:
Reagents to synthesise butanedioic acid from 1,2-dibromoethane
Why can't we use $\ce{HCN (g)}$ and $\ce{HCl (aq)}$ to convert 1,2-dibromoethane into butanedioic acid?
I found out that the answer is $\ce{KCN (aq/alcoholic)}$ and $\ce{H2SO4 (aq)}$, I don't understand the differences between these reagents and their reactions.

Related: Viability of proposed carbon-carbon bond-forming reaction

In the first step — aside from the fact that nobody wants to work with gaseous $\ce{HCN}$ due to its toxicity — using $\ce{HCN}$ wouldn’t work. The reaction in question is a bimolecular nucleophilic substitution. For that, you need to have a nucleophilic lone pair that is capable of attacking and it has to be on carbon (otherwise we would arrive at a diisonitrile). However, in hydrogen cyanide there is no free lone pair on carbon; it would need to get deprotonated before it can attack. Rather than using the toxic, gaseous $\ce{HCN}$ and deprotonating it, it is much quicker to just immediately start from an ionic compound such as $\ce{KCN}$ wherein the cyanide is already present as $\ce{^-C#N}$ anions ready for nucleophilic attack.
$$\ce{Br-CH2-CH2-Br + 2 CN- -> NC-CH2-CH2-CN + 2 Br-}\tag{1}$$
In your second step, it really doesn’t matter which acid you use. In fact, many problem sets wouldn’t even specify the acid, they would just write what I did in equation $(2)$. Thus, you should get the same points for using $\ce{HCl}$ as you would for using $\ce{H2SO4}$. You should use a strong acid, though; acetic acid will not do the trick.
$$\ce{NC-CH2-CH2-CN ->[H+/H2O][-2 NH3] HOOC-CH2-CH2-COOH}\tag{2}$$

Would $\ce{NAOH(aq)}$ and $\ce{K2Cr2O7/H2SO4(aq)}$ work because of the $\ce{OH}$ having a lone pair?

@rudraksh Hydroxide does not work. Cyanide provides two extra carbons to give the 4-carbon chain.


text:
How/where can I find a reliable atomic radii table?
I am currently working on a periodic table to include it in a experimental chemistry book and I would like to include non-bonded atomic radii data so it would be easier for the students to learn about periodic trends and quickly determine if an atom is more likely to produce a hard or a soft Lewis acid-base.
The problem is that every reference I looked have different values for the same atom. For instance, see the results for Hydrogen and Gold non-bonded atomic radii below:

In this site we have: 0.53 (H) / 1.74 (Au) Å
In RSC site, we have: 0.32 (H) / 1.30 (Au) Å
This aricle (Slater, 1964 - see the DOI below): 0.25 (H) / 1.35 (Au) Å
Finally, in Wikipedia (which is the result Google will return if you type the search query "hydrogen/gold atomic radius") is: 0.53 (H) / 1.66 (Au) Å

These differences are observed in most elements. I do not know which one is more reliable. In my point of view, variations of 0.20 angstrons are too big in atomic scale, aren't they?
What do you think about what cause those variations? Are they significant? Could someone help me to find a reliable source for atomic radii data?
Maybe those variations occur due to statistical deviations from the method used to measure them, but could it be so big (about 20%)?
Thanks!
Slater, 1964. DOI: 10.1063/1.1725697

Forget it, there is no such thing. Atomic radii are inherently arbitrary and uncertain. Also, welcome to Chem.SE.

There have been several attempts for systematic van der Waals and covalent radii sets using quantum mechanics and crystallography recently. But I find your particular request for "atomic" radii a bit unclear. Do you want VdW or covalent radii?

@IvanNeretin I see... I think it is so because the electron cloud has undefined edges. So do you think I should stick to one source or don't use a atomic radii data at all?

@GeoffHutchison I want to use these values in a periodic table for a didactic book, as I mentioned before. I am looking for the best atomic radii data for studying periodic trends related to size and consequently to Zef, such as, lanthanidic compression and hard and soft Lewis acid-base. I know that VdW is the best for noble gases since they aren't bonded  to each other (except for Xe and Kr in some compounds). But I am not sure about what "radius" is best for the other elements: non-bonded, ionic or covalent?

Yes, exactly. Stick to some respectable source, remember that atomic radii are uncertain, maybe add a note to that effect.


text:
Unknown salt formed from electrolysis of urine
I am not sure what I was expecting to happen with this but I created a circuit to electrolyze various household liquids. My electrodes are copper — or at least claim to be, I think they might just be zinc wires with some copper-colored coating.
Now I'm trying it on urine, which seems to eat through the wire rather rapidly. It produces some sort of gas with tiny bubbles. When these bubbles come into contact with the other electrode, they turn gray and fall to the bottom of the container. This gray material is almost green and that makes me think it is a copper salt rather than zinc. What are the bubbles and what is this salt?
Also, as a side question, how can I isolate this salt without potentially exploding whatever is in the bubbles?
If it's important: I'm using ~6 V, but I don't know the current.

Just to be sure I understand: both electrodes are copper, and the bubbles that form at one electrode migrate to the other.  When they come in contact with the other electrode, a gray/green solid is formed.  On which electrode does the solid form?

The main components of urine are water and then urea.  Water makes up 90% to 95% of urine with urea making up about 50% of the remainder.  It's likely that your reaction is mainly the hydrolysis of water.  In that case, you are forming hydrogen gas at cathode and oxygen gas at the anode.  Also, some $\ce{NaCl}$ will be electrolyzed and form $\ce{Cl2}$.
As you stated, you are likely oxidizing copper, probably making $\ce{CuCl2}$.  Most zinc salts that you could be forming in your reaction would be white.  If urea is being electrolyzed, then you would also be forming nitrogen and carbon dioxide at the anode.  It's possible that at least some of the migrating bubbles contain $\ce{Cl2}$ gas which then could form solid $\ce{CuCl2}$.
You could do a quick test of some aspects of my answer by observing the hydrolysis of water using your system.  Just conduct your procedure exactly as you do for urine, but instead of urine use water.  The main salt present in urine is $\ce{NaCl}$, whose concentration varies quite a bit but is on the order of 3 g/L.  So just conduct your experiment on a 3 g/L $\ce{NaCl}$ in water solution and compare your results to the urine experiment.


text:
Is there a pH at which all proteins are negatively charged? How can it be determined?
I know that depending on the amino-acid-composition of the peptide the distance to the isoelectric point determines the charge. Is there a universal point at which all proteins are definitely negatively charged?

$14%Ha, not even misusing MathJax here!$

That depends highly on what you mean by "all." Since acid-base chemistry is equilibrium based, there is in principle always some of the less preferred species around even though it might be quite small. At the very least, there is a nonzero probability of finding the disfavored species, in your case the neutral species.

Proteins will consist primarily of the 20 proteinogenic amino acids. Among those, lysine and arginine are positively charged at neutral $\mathrm{pH}$ while aspartate and glutamate are negatively charged. All proteins also have an amino terminus which is positively charged at neutral $\mathrm{pH}$ and an acid terminus which is negatively charged.
We don’t need to care about the acidic groups; they are already negatively charged. But if we want a $\mathrm{pH}$ level where the protein is definitely negatively charged, we want to deprotonate all positively charged groups.
The amino groups of the N-terminus and lysine have a $\mathrm{p}K_\mathrm{a}\approx 10$. Thus, raising the $\mathrm{pH}$ value to $\mathrm{pH}\ 12$ should get them fully deprotonated. The guanidinium group of arginine is a little less acidic; $\mathrm{p}K_\mathrm{a} \approx 13.6$ (approximated from the value of unsubstituted guanidinium; value found on Wikipedia). We therefore need to raise the $\mathrm{pH}$ a little bit more for certain deprotonation. If we go to the absolute limits of aquaeous $\mathrm{pH}$ of larger than $14$ (i.e. more than $1~\mathrm{M}\ \ce{NaOH}$) we should get these protonated. However, the two logarithmic units one typically assumes would mean $100~\mathrm{M}\ \ce{NaOH}$ — not possible. We can get the $\mathrm{pH}$ to around $15$, and then that’s the end of it. The fully deprotonated form should be the principal species in solution there, though.
Bear in mind that these harsh conditions are also able to hydrolyse the peptide bonds. Thus you may just destroy your protein while getting there.


text:
Should chemical compounds and elements be capitalized?
In the case of a name of a person, the first letter is written in capital letters.

Should the first letter of name of a chemical compound or element be written in capital letters?

Related: Why do people often capitalize element names?

The names of chemical compounds and elements should be capitalized if they appear at the beginning of a sentence or in a title - that is, they are treated just like any other common noun.
For example, a title:

Why I Don't Like Zinc

or a sentence:

Boron is my favorite element.

Within a sentence:

We used boron and zinc in the experiment.
Vinegar contains acetic acid.

The symbols for chemical elements are always capitalized, no matter what:

We combined $\ce{As}$ and $\ce{W}$ to make a new alloy.

In the 1979 and 1993 IUPAC recommendations, example names were actually written with a capital initial letter. This was not necessarily wrong since most examples started on a new line. However, this practice has been abandoned in the 2013 recommendations in order to ensure that names of organic compounds are not considered as proper nouns.

Capitalising (some) words in a title is a matter of taste, not orthography.

@Karl That's like saying "Whether it's 'colour' or 'color' is a matter of taste, not spelling." Capitalization is literally orthography: the way things are written down.

@DavidRicherby Your point being?

@Karl My point being that your point is completely wrong.

@DavidRicherby I have no idea what your first post had to do with mine, but https://en.wikipedia.org/wiki/Capitalization#Titles

@Karl You said that which words get capitalized in titles is not orthography. I am pointing out that this is incorrect: orthography is exactly the set of "rules" for how to write things down, including things like which letters get capitalized.

I'm sorry, but the diverse, incoherent and arbitrary "rules" by which some English speaking writers, editors, publishers uppercase first letters in titles, and others don't, is not exactly orthography. As opposed to rather widely recognised rulesets like "British English" or "American English" on general spelling.

Thank Todd Minehardt for your good answer, I just have a last complement question, What about substitutions on the nitrogen of an amide and similar cases of other types of substitutions? Should that atom be capitalized?

@Another.Chemist - Can you give a particular example?

@ToddMinehardt i.e. glucosamine (N-acetyl)-6-sulfatase

"N" is an element symbol, always uppercase.

One addition to Todd's correct answer is that of trade names.  The canonical example of Teflon being capitalized,  the name being property of DuPont.  https://en.m.wikipedia.org/wiki/Polytetrafluoroethylene
Although Teflon is not the chemical name,  the distinction is added to avoid possible confusion.

And complicated cases, such as acetylsalicylic acid, which is "Aspirin" in countries such as Canada, where Beyer holds a trademark, and "aspirin" in countries such as the US and UK, where the trademark lapsed and it's now considered to be a nonsystematic name.


text:
Deriving Thermodynamics equations confusion
Please explain:
I have taken up to Cal 3 so I understand deriving but I guess because of how it is used in context of this class I am not understanding. In the thermodynamics equation: ( as an example) what is the difference in using dU = Cvdt VS. $\Delta$U=Cv$\Delta$T. This may come across as a stupid question but I just don't understand, maybe because I have not had much practice, but would I need to be deriving something? Another example, I was asked to derive the Maxwell-Boltzmann constant, but in the book it shows the derivation and its literally just plugging in different equations;however, when I say that I don't mean its like dx/dt = v and then I substitute in v for that part but rather substituting in full equations.
EDIT: I'm sorry my question seems unclear, but I am asking it to the best of my ability. I guess all I want to know is what deriving an equation in physical chemistry means. If I derive x^2 it is 2x. Well as I have seen, if I derive the RMS equation, instead of actually (dx/dt)'ing anything, I am just plugging in different parts of different equations. Pretty ridiculous I get down voted for asking a question, unclear or not. The whole purpose of this forum is to help communicate issues with one another, not be rude.

Well it seems I entirely misunderstood what you were asking and answered a different question. In light of this I think it's quite fair that others voted to close as unclear. I tried to guess what you meant but in truth, I shouldn't be forced to guess. Your new edit does help to clarify. And the problem with your thinking is: a derivation does not necessarily involve derivatives. It simply means manipulating the maths somehow to prove a theorem or result. In this respect it means exactly the same thing as a proof. See Wikipedia: Formal proof.

Furthermore the construction "If I derive $x^2$ it is $2x$" is quite uncommon and is, to the best of my knowledge, incorrect. Even though the first derivative of $x^2$ is $2x$, you don't derive $x^2$; you differentiate it to get $2x$. So, this is a terminology issue.

yes.. thank you! It is clear I just did not understand the difference between the two words. Thanks again. Also, so does the dU in the example I provided mean taking the derivative of U, or just what you said about infinitesimal change? @orthocresol

The derivative of U with respect to something else, say T, is (dU/dT). This may loosely be understood to be the same thing as a ratio of infinitesimal changes: the ratio of dU to dT when some process happens. It doesn't make sense to say "take the derivative of U": you need something to differentiate it with respect to, since a derivative has dU over d(something). When you "take the derivative of $x^2$" the implicit assumption is that you are taking it with respect to $x$ which is why dx is in the denominator: $$\frac{\mathrm{d}(x^2)}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}( x^2) =2x$$

$\mathrm{d}U = C_V\,\mathrm{d}T$ relates an infinitesimal change in temperature to an infinitesimal change in internal energy.
$\Delta U = C_V \Delta T$ relates a finite change in temperature to a finite change in internal energy.
To get from infinitesimal to finite you have to add up, or sum, a huge number of infinitesimal changes. This isn't done with a sum, but rather an integral (which acts like a sum for infinitesimal quantities) from an initial state to a final state. Since both $U$ and $T$ are state functions you can do this quite safely without worrying about intricacies of path functions.
$$\begin{align}
\mathrm{d}U &= C_V\,\mathrm{d}T \tag{1} \\
\int_\text{initial}^\text{final}\mathrm{d}U &= \int_\text{initial}^\text{final}C_V\,\mathrm{d}T \tag{2}
\end{align}$$
The limits of integration on the left are simply the initial internal energy and the final internal energy. You can call it $U_1$ and $U_2$ respectively. Likewise the limits of integration on the right are $T_1$ and $T_2$
$$\begin{align}
\int_{U_1}^{U_2}\mathrm{d}U &= \int_{T_1}^{T_2} C_V\,\mathrm{d}T \tag{3}
\end{align}$$
In general, $C_V$ is a function of $T$. However, to get to your desired equation, you have to assume that $C_V$ is independent of $T$ over the temperature range you are interested in, i.e $C_V$ is a constant. That allows you to take it out of the integral much like you would any other constant.
$$\begin{align}
\int_{U_1}^{U_2}\mathrm{d}U &= C_V\int_{T_1}^{T_2} \mathrm{d}T \tag{4} \\
U_2 - U_1 &= C_V(T_2 - T_1) \tag{5} \\
\Delta U &= C_V\Delta T \tag{6}
\end{align}$$
So, yes, there is a difference. For the finite change you made an extra assumption that $C_V$ doesn't vary over the temperature range. For the infinitesimal change the temperature range is infinitesimally small - hence $C_V$ cannot vary and the assumption is not necessary.


text:
What can I use as a fuel for an engine with carbon dioxide as the oxidizer?
On Mars carbon dioxide is abundant. It is plausible to utilize a fuel in an engine that can use carbon dioxide as an oxidizer? I know this is a strange inquiry, but could a engine be made to use $\ce{CO2}$ in the martian atmosphere the same way most engines consume $\ce{O2}$ on Earth?

Usually when we say something "burns", it's being oxidized.  In the case of carbon dioxide only the oxygen can be oxidized, by displacing it as the element; that requires fluorine or a sufficiently powerful fluorinating agent.
Carbon dioxide supports combustion, acting as the oxidizer instead of being oxidized, with some active metals such as magnesium.  Thus
$\ce{2 Mg + CO_2 -> 2 MgO + C}$
Such reactivity is why special reagents are needed to extinguish burning active metals.
Addendum:
The comments indicate that the OP is interested in burning fuel on Mars.  Absent any active metals like magnesium in the oxide-rich and still water-bearing environment on Mars, it is not practical to use carbon dioxide for combustion.  Better is to bring some water along or extract it from underground, electrolyze it with solar power, and re-burn the hydrogen and oxygen.

How come this is no longer accepted?  I assumed that there had been a better answer but I find that none exists.  Thank you.

I need to see if there are other chemical ways this can be done. Now that it has been reopened I will accept it if no one else answers. My question originally had grammars issues, but @A.K. thanks, fixed it. Maybe it will now start getting up votes.

I like the suggestion to think outside the box and use fluorine as an oxidizer.

Tempting, but how?  I assume you can electrolyze fluorides with solar power, but are these known on Mars?

You would probably have to brink tanks of fluorine to Mars with you.


text:
Is Zirconium 3+ ion more stable than Zr?
Is $\ce{Zr^{3+}}$ more stable than the atom, and why if yes?

And what is your question supposed to mean? Do we need to make a filter for "stable" in titles for automatic flagging as unclear?

Not sure how else I can ask: Is Zirconium 3+ ion more stable than Zr? Stable as in "electronic configuration". Is that really that unclear from the ion formula?

"Stable as in "electronic configuration"." is meaningless. You don't know the context in which you want to ask so there's no question.

You can't compare a neutral atom to an ion. That doesn't make sense between you're not comparing two systems that have the same components.

I cannot tell that to the teachers who designed the course. I am expected to explain which is more stable. This is the assignment. Sorry if this not satisfactory to your high standards, but I have no choice in formulating the question or answering it in any other terms than expected.

I think your main confusion lies in the statements "most common" and "most stable".  These statements are not equivalent.
The "most common" statement implies the conditions found in nature, meaning the different conditions where it is found or where it was formed.
The term "most stable" requires that you specify a set of conditions explicitly.  In the presence of one set of conditions one for will be the most stable, and under another set of conditions another form will be most stable.


text:
What happens to the molecules of a liquid when it evaporates?
Let's say I spill liquid on some fabric or on a impermeable material like a countertop. Eventually, the liquid stops being a liquid and we say it has "dried", but what happened to the molecules? Did they change phase? How were the conditions for phase change met? Assume we are at room temperature and the liquid is something that doesn't naturally boil or freeze at that temperature or pressure (like water).
EDIT: I'm aware this is called evaporation, but what happens to the molecular structure during this process? What causes molecules to leave the liquid?

Evaporation....

The molecules don't change their structure during evaporation. It is only that they spread further apart. This is associated with the 2nd law of thermodyamics: entropy (basically) always increases. Entropy is the measure of disorder or possible arrangements of a system of particles.
Molecular changes are reactions. Phase changes, e.g. evaporation etc. are merely changes in the "nearest-neighbor" distance between particles.
There may be a very slight change in average bond angles etc. (e.g. when water freezes into a meta-stable state) but all bonds remain intact for a phase change.
Coffee and wine are mixtures of tons of different chemicals but the primary substance is water, so they follow the phase-transition of water very closely, i.e. same $T$ and $p$.
To give you some perspective, gases (e.g. water vapor) are generally 1,000 times less dense than liquids/solids (e.g. water/ice).

There is no change in molecular structure. A phase change is a physical, not chemical process. To understand why evaporation occurs, you need to realize that every substance has a vapor pressure. This is the pressure exerted by that substance's vapors when it is in thermodynamic equilibrium with its condensed form in a closed system. Substances with high vapor pressures, also called volatile substances, evaporate more quickly than those with low vapor pressures. Consider the illustration below:

This illustration shows a liquid coming into thermodynamic equilibrium with its vapor, only made possible by the stopper in the flask. Remove it, and it is no longer a closed system. As vapor leaves the flask, more liquid will vaporize in order to maintain equilibrium, but as the process continues there is eventually no liquid left to replace the vapor as it leaves.
This is why it's important to put the caps back on your gasoline canisters after you're done using them, and how you're able to dry your clothes on a clothes line.

Suppose that there are some molecules dissolved in the solvent, say a dye from coffee dissolved in water. These dye molecules usually have a far higher boiling point than water does and so have a lower vapour pressure, often effectively zero, at room temperature. Thus water evaporates at a far greater rate than they do and so the solution 'dries' leaving the dye behind. The dye is chemically unchanged by not being dissolved in a solvent. If the solution was, say, alcohol & water then as both have a large vapour pressure at room temperature both will evaporate and nothing will be left.
The water (solvent) evaporates due to the difference in vapour pressure at the liquid surface and the average value in a room. The liquid will evaporate trying to establish equilibrium, at which point the rate of evaporation will equal the rate of condensation of water vapour in the air entering the liquid. Usually equilibrium is never met and so all the liquid evaporates. However, on hot very humid days you may have noticed that it often takes far longer for a puddle to evaporate than on less humid days.
Not all the molecules in a liquid have the same energy, they follow a (Boltzmann) distribution of energies, many have low energy and fewer higher energy than the average. (Only the average energy is constant at constant temperature). Of these higher energy molecules (achieved by random jostling of molecules in solution) some of them are close to the surface and randomly are on the right trajectory to escape into the gas phase. As some of the most energetic molecules leave the liquid this will cool. (It will therefore receive heat from the surface it is sitting on and from  molecules in the air (including water molecules), but if these processes are slow the liquid will be measurably cooled). You can feel the cooling effect of evaporation after stepping out of a hot shower.

Water (or any liquid in general) consists of particles. Electrostatic forces hold the particles together, but they can still move and slide over each other - which is what makes it a liquid.
Their energy depends on the temperature. The higher the temperature, the more vigorously they move and collide and slide over each other. Through collision, they can gain or lose energy, as well as change direction.
Eventually, by random chance, some particles on the surface has enough speed in the correct direction to leave the surface, overcoming the electrostatic forces that used to hold the particle to the water.
This particle has escaped, and we say that it has evaporated.


text:
What is the relationship between equilibrium concentrations in two similar experiments?

Two experiments were performed involving the following equilibrium. The temperature was the same in both experiments.
  $$\ce{H2 (g) + I2 (g) <--> 2HI (g)}$$
In experiment A, $\pu{1.0 M}~\ce{H2}$ and $\pu{1.0 M}~\ce{I2}$ were initially added to a flask and equilibrium was established. In experiment B, $\pu{2.0 M}~\ce{HI}$ was initially added to a second flask and equilibrium was established. Which of the following statements is always true about the equilibrium concentrations?

$\ce{[H2]}$ equals $\ce{[HI]}$ in experiment A.
$\ce{[HI]}$ equals $\ce{2[H2]}$ in experiment A.
$\ce{[HI]}$ in experiment A equals $\ce{[HI]}$ in experiment B.
$\ce{[HI]}$ in experiment A equals 2 $\ce{[I2]}$ in experiment B.

Can someone explain why the answer is 3. Doesn't that mean 2 and 4 are true as well?

@airhuff Your edit was not only completely superfluous, but also harmful. Please pay attention to [meta-tag:buzzwords], and don't introduce them after they have been already removed. Unfortunately I noticed after I spent too much time to polish that thing, that I had done that already. Also note, that the abbreviation for concentration should be all set in math mode, in some browsers it looks terrible otherwise.

@Martin-マーチン Yea, agreed. I'd been a member for about 2 weeks at the time I made the edit and have no idea why I thought that was an improvement. Rookie mistake ;)

Note that you are not given the equilibrium constant so you do not know the absolute equilibrium concentration of any species at equilibrium.  This eliminates 2 and 4.
Since all conditions and total concentration of reactants + products are the same between the two experiments, you do know that the concentration of any species in experiment A is equal to that species in experiment B.  Again, only at equilibrium, and you don't know to what degree that equilibrium lies in terms of absolute concentrations.

Thank you! I don't have that deep of an understanding so could you please expand? I don't get if the stoichiometric ratio has anything to do with the calculation of HI or any other species in this problem. Or do you need to use ICE charts?

I hope I can explain. Note that the reaction is written with a 2-way arrow, meaning the reactants and products are in a state of equilibrium.  What is the concentration of any species at equilibrium? You can't tell without more information so that wasn't the point of the problem. The point is that it doesn't matter if you start with only the compounds on the right side of the equation or the left.  At equilibrium they will come to the same concentration either way.  That's how you know C is right; [HI]=[HI], it doesn't matter how you got there.  I hope I understood you and I hope that helped!


text:
Calculating relative molecular mass given mass of solute and solvent, as well as freezing point depression, but no cryoscopic constant?
Find RMM of anthracene given that when 0.813g is dissolved in 45g of benzene. freezing point of depression is 0.51 degrees celsius.


text:
η1 or sigma complex involving cyclopentadienyl anion
When $\ce{\eta^1-C5H5-}$ acts as a ligand, does the donor carbon become $sp^3$ hybridized, and would this be a sigma donor through sigma orbitals? Or, is it $sp^2$ where the lone pair is donated through a pi orbital, making it a sigma donor through a pi orbital?
The second option seems much more likely as the ligand changes the $\ce{C}$ attached to metal randomly with time through rearrangements.
Also, does "sigma complex" and "pi complex" only differentiate the hapticity? as $\ce{\eta^1-C5H5-}$ is a sigma complex and $\eta^3$, $\eta^5$ are pi complex.

Even though depictions of $\eta^1$-coordination might suggest a $\mathrm{sp^3}$-carbon donating, there is little evidence, that it actually is. Taking the example you have presented in your answer from Casey et. al.[1] I have run a geometry optimisation to obtain molecular orbitals at the DF-BP86/def2-SVP level of theory. The following geometry parameters around the coordinating carbon already suggest only little pyramidalisation:
\begin{array}{lrr}\hline
\text{Parameter} & \text{Experimental [1]} & \text{DF-BP86/def2-SVP}\\\hline
\mathbf{d}(\ce{Re-C^1}) & 236.0 & 240.2\\
\mathbf{d}(\ce{Re-H})   & 288.7 & 282.6\\
\mathbf{d}(\ce{Re-C^2}) & 317.8 & 321.3\\
\mathbf{d}(\ce{Re-C^5}) & 315.6 & 315.6\\
\angle(\ce{Re-C^1-H})   & 114.3 & 100.7\\
\angle(\ce{Re-C^1-C^2}) & 110.8 & 109.5\\
\angle(\ce{Re-C^1-C^5}) & 108.5 & 106.4\\
\angle(\triangle[\ce{C^1C^2C^5}]\ce{-H}) &
                        122.8   & 139.1\\
\text{RMSD} & 0.0 & 0.1\\\hline
\end{array}
We know that crystal structures are not really that accurate with protons, hence I'd put a little more trust into the calculation at this point. The paper does not give a value for any protons.
Here is a picture of the calculated structure.

When we look at the canonical MO we will see that all (occupied) π-obitals of the Cp-ligand are still there. Obviously they are distorted, but clearly recognisable.

The second orbital is the HOMO-1, which is the main contribution for the coordination, and the last is the HOMO (click for large). The only s-contribution to the HOMO-1 is from the carbon of the $trans\text{-}\ce{CO}$-ligand (and some hydrogen).
In output numbers, missing from 1 are minor contributions (<0.05):

 Alpha occ 81 OE=-0.276 is C39-p=0.19 C42-p=0.12 C41-p=0.12 C43-p=0.11 C40-p=0.11
 Alpha occ 87 OE=-0.185 is C39-p=0.26 C42-p=0.23 C41-p=0.22 Re24-d=0.08 C27-s=0.05
 Alpha occ 88 OE=-0.153 is C43-p=0.32 C40-p=0.32 C41-p=0.13 C42-p=0.12

When we localise the orbitals with the Natural Bond Orbital (NBO) analysis, we find a lone-pair at $\ce{c^1}$ (blue/orange) which donates into the anti-bonding $\ce{Re-^{$trans$}CO}$ orbital (yellow/red):

The composition of the lone pair orbital is

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 36. (1.27157) LP ( 1) C 39            s(  8.77%)p10.40( 91.23%)d 0.00(  0.00%)

The low occupancy is normal for donor-acceptor-complexes with some covalent character.
TL;DR Most $\eta^1$ complexes are in first approximation π-coordination complexes like the definition suggests[2]. This may change significantly when moieties are altered.

Further thoughts:
Wikipedia makes one important statement (it rarely does, but here it is true)3:

Molecules with polyhapto ligands are often fluxional, also known as stereochemically non-rigid.

You can expect that in $\eta^1$-ligands the metal can easily migrate from one carbon to another, as the MO show. I would expect some kind of pseudo-rotation.
Ref.

Charles P. Casey, Joseph M. O'Connor, William D. Jones, Kenneth J. Haller. Organometallics 1983, 2 (4), 535–538. (link via doi, pdf via researchgate.net) Crystallographic data via CCDC number: 1118829.
η (eta or hapto) in inorganic nomenclature
Wikipedia: Hapticity and fluxionality

Casey and coworkers in Organometallics 1983, 2, 535:

http://www.ilpi.com/organomet/cp.html
It's $sp^3$

It is really not that easy. Just because you found a Lewis structure that displays it like an sp3 carbon, does not mean that this is actually true. There is a lot of flexibility in these complexes and I very much doubt that there is much s character in the co-ordination bond. Sorry, but I have to down-vote your answer.


text:
Do argon have a high energy?
Do argon have a large energy density? And what level is it on? Is it weak or have a density like gasoline. How much energy can argon store and what noble gas have the highest energy density?

Argon is hardly a palatable substitute for gasoline.

In terms of chemical reactions, Argon is not useful a fuel, as it doesn't react easily with any other elements - that's why it's called a noble gas.  Compounds with Argon are very rarely seen.
Gasoline is a mixture of compounds of carbon and hydrogen.  These react easily with other chemicals, particularly oxygen, and these reactions allow it to be used as a fuel.
So Argon is useless as a conventional chemical fuel because it hardly reacts with any elements.


text:
Why do post-Hartree-Fock methods fail to predict the direction of the dipole moment of carbon monoxide?
In carbon monoxide the dipole moment (negative to positive) points towards the oxygen, as I explained it in How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?
A calculation using density functional approximation, the level of theory is BP86/def2-QZVPP, yields the correct direction:
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.19~\mathrm{D} &&
\text{Direction:}~\longrightarrow
\end{align}

 ---------------------------------------------------------------------
 Center     Atomic      Atomic             Coordinates (Angstroms)
 Number     Number       Type             X           Y           Z
 ---------------------------------------------------------------------
      1          6           0        0.000000    0.000000   -0.648987
      2          8           0        0.000000    0.000000    0.486740
 ---------------------------------------------------------------------
[...]
 Dipole moment (field-independent basis, Debye):
    X=              0.0000    Y=              0.0000    Z=              0.1907  Tot=              0.1907

Using Møller-Plesset Perturbation Theory of second order, i.e. MP2/def2-QZVPP, we obtain
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.30~\mathrm{D} &&
\text{Direction:}~\longleftarrow
\end{align}

 ---------------------------------------------------------------------
 Center     Atomic      Atomic             Coordinates (Angstroms)
 Number     Number       Type             X           Y           Z
 ---------------------------------------------------------------------
      1          6           0        0.000000    0.000000   -0.648450
      2          8           0        0.000000    0.000000    0.486337
 ---------------------------------------------------------------------
[...]
 Dipole moment (field-independent basis, Debye):
    X=              0.0000    Y=              0.0000    Z=             -0.3002  Tot=              0.3002

There is no improvement going to other post-Hartree_Fock methods, as the following table indicates. A positive value means that the dipole is directed towards the carbon; the used basis set is in all cases def2-QZVPP.
\begin{array}{lr}
\text{Method} & \mathbf{q}(\overrightarrow{\ce{CO}})\\\hline
\text{Experimental}^1 & 0.11\\\hline
\text{HF} & -0.14\\\hline
\text{MP2}  &   -0.30\\
\text{MP3}  &   -0.22\\
\text{MP4(SDQ)} & -0.27\\
\text{MP4(SDTQ)//MP4(SDQ)} &  -0.27\\
\text{CCSD} &   -0.25\\
\text{CCSD(T)//CCSD} & -0.25\\
\text{CISD} & -0.22\\
\text{QCISD} & -0.26\\
\hline
\text{BP86} &    0.19\\
\text{PBE0} &    0.12\\
\text{M11}  &    0.06\\
\text{B3LYP} &   0.10\\
\hline
\text{B2PLYP} & -0.07\\
\hline
\end{array}
This failure probably already starts with Hartree-Fock. Where does the deficiency lie in the HF description in this particular case? And why is this deficiency not corrected by post-HF methods? Even the double hybrid functional B2PLYP cannot predict the direction of the dipole moment correctly.

According to Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. the value can be found in J. S. Muenter. J. Mol. Spectrosc. 1975, 55, 490. (I don't have access to that publication.)

Sample Input for Gaussian 09

#p MP2/def2QZVPP
scf(tight) opt(verytight,maxcycle=100)
symmetry(loose)
gfinput gfoldprint iop(6/7=3)

carbon monoxide

0 1
C   0.0 0.0 1.135
O   0.0 0.0 0.0

FWIW, QTAIM ELF analysis appears to refute the common writing of CO as having a conventional triple bond. See chat conversation and specifically this result. If it's a triple bond, it seems it's got a lot of charge-shift character to it.

Full CI results for CO. That's pretty much definitive.

The question itself is void, as all methods with the exception of Hartree-Fock predict the direction of the dipole moment correctly. Deathbreath found a Full CI calculation in Jeremy P. Coe, Daniel J. Taylor, and Martin J. Paterson. J. Comp. Chem. 2013, 34 (13), 1083-1093 of which luckily an open version exists (arXiv:1301.4904 [physics.chem-ph]). The following passage is taken from the latter:

The groundstate dipole moment of carbon monoxide, although fairly small, when calculated using HF strikingly has the incorrect sign compared with experimental results. Previous work has suggested that the accuracy of the dipole calculation
  depends on the amount of correlation accounted for.14  The bond length (2.1316 Bohr) and the experimental dipole value (0.122 Debye) are taken from Ref. 15. The positive value for the dipole here signifies a polarity of $\ce{C^−O^+}$.
  With a cc-pVDZ basis, two frozen core orbitals and a cut-off value of $c_\mathrm{min} = 5 \times 10^{−3}$, we see in Fig. 1 [omitted] that the MCCI method, starting from close to the incorrectly signed result of the HF single SD, quickly reaches a correctly signed value which converges at around half of the FCI value.

14 G. E. Scuseria, M. D. Miller, F. Jensen, and J. Geertsen. J.
  Chem. Phys. 1991, 94, 6660.
15 J. S. Muenter. J. Mol. Spectrosc. 1975, 55, 490.

The problem indeed is just a technical one, resulting from analysing the wrong density. Gaussian 09 by default only analyses the SCF density (see population, density). In order to obtain values for the post-HF methods, the density=current keyword needs to be specified. The updated table therefore becomes:
\begin{array}{lr}
\text{Method} &
\mathbf{q}(\overrightarrow{\ce{CO}})/\mathrm{D}&
\mathbf{d}(\ce{CO})/\mathrm{10^{-10}m}\\\hline
\text{Experimental}^{15} & 0.122 & 1.128\\\hline
\text{HF} & -0.14 & 1.102\\\hline
\text{MP2}  &   0.26 & 1.135\\
\text{MP3}  &   0.08 & 1.118\\
\text{MP4(SDQ)} & 0.11 & 1.118\\
\text{MP4(SDTQ)//MP4(SDQ)} & \text{n.a.} \\
\text{CCSD} &   0.08 & 1.124\\
\text{CCSD(T)//CCSD} & \text{n.a.}\\
\text{CISD} & 0.05 & 1.118\\
\text{QCISD} & 0.08 & 1.127\\
\hline
\text{BP86} &    0.19 & 1.135\\
\text{PBE0} &    0.12 & 1.122\\
\text{M11}  &    0.06 & 1.119\\
\text{B3LYP} &   0.10 & 1.124\\
\hline
\text{B2PLYP} & 0.15 & 1.129\\
\hline
\end{array}

Sample Input for Gaussian 09 for analysis
(requires the addition of %chk=mp2qzvpp.chk to the sample input of the question)

%oldchk=mp2qzvpp.chk
#p MP2/def2QZVPP
Geom=check
Guess=(Read,Only)
Density=(Check, Current)

dens-curr

0 1

I apologise for getting everyone worked up over a stupid beginner mistake.

I like the fact that your ‘stupid beginner mistake’ gained you seven question and four answer upvotes as of now ;D


text:
Predicting the product of bromination of acetanilide

I figured the answer would be option (B) as the group attached to the benzene ring is ortho- and  para- directing.
I thought $\ce{FeBr3}$ would be used in electrophilic aromatic substitution.
However, the model answer to the question is (D).
Is this correct? If so, what makes it possible?

Related: http://chemistry.stackexchange.com/questions/61123/bromination-of-acetanilide

This is a classic $\alpha$-bromination of a carbonyl. You can achieve this with an electrophilic source of bromine like in the reaction conditions provided or something like N-bromosuccinimide.
The reason you're not getting the EAS product is because that's a much higher energy pathway since you need to temporarily break aromaticity. Given more time, heat, and bromine, you can probably get the ortho- and para- products, but you will still $\alpha$-brominate on the amide first.

There are two competing mechanisms.

The first one is the electrophilic aromatic substitution which you expected:

The rate-determining step is presumably the first step, where a relatively unstable arenium ion is formed.

The second one is the carbonyl alpha-substitution reaction:

I failed to identify the rate determining step because of the coordination of the Lewis acid $\ce{FeBr3}$.

Therefore, the second mechanism occurs at a higher rate compared to the first mechanism, so more 2-bromo-N-phenylacetamide is produced per unit time, which makes it the major product.

You should have the Lewis acid $\ce{FeBr3}$ coordinate to the carbonyl oxygen before tautormerising. And the amide group is meta-directing, not ortho/para afaik. Apart from that, this is practically spot-on.

@Jan AFAIK, the amide is meta-directing when the carbon in the amide is attached to the benzene, and ortho/para-directing when the nitrogen is attached to the benzene instead...

@Jan is it alright now? The coordination is amazing.

Curly arrows in second step are wrong.

@orthocresol I have fixed it now.

Step 3 as drawn (i.e., the actual bromination) is most likely the r.d.s.

@Zhe I think the bromine, under the presence of the catalyst, has already been significantly polarized up to that step, lowering the r.d.s.

Yes, but that's still the intermolecular step, and you're competing with what? Enolization?

Also, I'm not criticizing your answer. I already up-voted it. :)

Not sure why you drew an ortho substitution for the EAS example, wouldn't para be far more abundant due to the steric blocking by the acetyl group?


text:
Differences between galvanic cells and Li-ion?
There are clearly some differences between the classical galvanic cell and the Li-ion batteries, but if one looks at the schematics, they looks a bit similar. Please see below:
a) The galvanic cell:

and
b) Li-ion

So I know that in theory there are two fundamental differences between the galvanic cells and Li-ion batteries:

In LIB, the Li ions intercalate in the structure of the anode and cathode
In LIB, there is solid-state mass diffusion of ions.

This is easy to understand, in theory, but what happens in fact in the galvanic cell? Why isn't there any diffusion? Is it because it is not concentration driven (there are two species instead of one)? But if there is no diffusion, how are they moving? In the galvanic cell, are the ions diffusing on the surface of the electrodes instead of intercalation in the structure?
Can anyone help me to understand better the differences between the two cells?
Thank you!


text:
Do hydrogen atoms not bound to carbon also appear in 1H NMR spectra?
I would like to know if hydrogens that are not bound to a carbon (such as in groups $\ce{-OH}$ and $\ce{-NH2}$) would lead to the presence of a singlet peaks in the NMR spectra or not, and if we have to count or omit them in the integration of the spectrum. I thought I was sure about it but lately I noticed that it depends on the spectra.

Yes, with a but.
First the yes: they do appear, albeit at slightly different chemical shifts as, of course, the environment around a proton in $\ce{-CH2\!\;\!-}$ is slightly different to the electronic environment around the proton in $\ce{-NH2}$. The $\ce{N-H}$, $\ce{O-H}$, and $\ce{C-H}$ bonds all have slightly different electronegativies.
Now for the but: the thing that more significantly distinguishes $\ce{N-H}$ and $\ce{O-H}$ from $\ce{C-H}$ is their lability in protic solvents. Exchange of labile protons in the sample molecule with deuterons (i.e. deuterium ions, $\ce{^2H+}$) in the solvent can can remove the proton peak from the spectrum (this exchange is an equilibrium process, but as the solvent is in large excess the equilibrium lies over towards full deuteration of the sample's labile groups. This can actually be useful in identifying peaks: first run a sample in, for example, $\ce{CDCl3}$, then add a couple of drops of $\ce{D2O}$ to your NMR tube, give it a shake, and re-run the spectrum -- your labile protons will have vanished.
Edit 1: Furthermore, as @user1573870 points out in the comments, exchange of protons (exhanging protons for protons, rather than exchanging protons for deuterons),  on "the NMR timescale" can lead to broadening, and the resulting peak can be, while technically present, very difficult to spot. I'd suggest anyone doing NMR do background reading on that, because it's both useful and interesting (and I need to do some, because I'm out of practice with NMR).
Edit 2:To directly answer @Chewie's question whether "we have to count or omit them in the integration of the spectrum", you only count each of the peaks that are present. Using the example of ethanol, as asked in the comments, $\ce{CH3CH2OH}$ should give three peaks:

The $\ce{CH3}-$ peak should integrate to show three protons,
The $\ce{-CH2}-$ peak should integrate to show two protons, and
The $\ce{-OH}$ peak should integrate to show one proton.

If due to exchange with the deuterated solvent the $\ce{-OH}$ peak is not present, or if the $\ce{-OH}$ peak is 'there' but so broad as to be invisible to you (or so broad that it is difficult to integrate accurately), then you would only give the integrals for the other two peaks, and those two peaks should give integrals with a ratio equivalent to their respective number of protons, i.e. 3:2. If the $\ce{-OH}$ peak is not there, its proton does not just get thrown in with one of the other peaks/integrals (unless, sneaky proton, it just happened to have the same integral as another peak, and was hidden behind it, but for something as simple as $\ce{CH3CH2OH}$ that is unlikely).
The punchiest way to say all of thise would be that one does not "integrate spectra"; one integrates peaks.

Also, such labile protons (e.g. acidic ones) do often give rise to a broadened peak due to having exchange rates on the NMR time scale. In extreme cases, this can broaden the peak to an extent that you might even overlook it.

Thanks! And so regarding the integration of the spectrum, if for example the molecule is ethanol C2H6O, I will count only 5 H on the integration, right?

@Chewie Sorry that I didn't address that explicitly. I'll add that in.

Even if exchange is slow on the nmr timescale, proton peaks may be broadened due to quadrupolar relaxation.  This is often the case with $\ce{N-H}$ protons. See here for some background.

Thank you very much @owjburnham! So if the -OH gets deuterated we won't be able to tell the ethanol from ethanal, that's a shame...

@Chewie Ethanal and ethanol would each have a $\ce{-CH3}$ peak, albeit in different places due to different environments; however, the other peak in each spectrum would be radically different: in the case of ethanol, an alkane $\ce{-CH2}$ around 4 ppm, shifted 'upfield' by proximity to the $\ce{OH}$, and in the case of ethanal (acetaldehyde) an aldehyde $\ce{C-H}$ at around 10ppm. Furthermore, the ethanol $\ce{-CH2}$ would integrate to 2, while the ethanal $\ce{C-H}$ would integrate to 1.

@Chewie These pictures may make things clearer. First, ethanol: http://www.chemguide.co.uk/analysis/nmr/nmrethanol.GIF .
And here, ethanal: https://chem242.wikispaces.com/file/view/nmracetaldehyde.gif/62855114/800x581/nmracetaldehyde.gif

This website is quite a good one to predict the spectrum of a given molecule: http://www.cheminfo.org/flavor/structuralAnalysis/NMR/Tools/1H_Prediction/index.html

Peaks from exchangeable protons do not just "vanish". They reappear as a shifted water (HDO) peak, at a position between the original and a native water peak in the same solvent. And it all depends on the relative molar concentrations of sample and water.

@Karl My impression / gut instinct is that it depends on miscibility. In e.g. d$_3$-acetonitrile, in which the $\ce{D2O}$ being shaken is totally miscible, then yes. Whereas in, say, d-chloroform, I feel like (the vast majority of) the $\ce{HDO}$ produced would not show up (assuming that you had enough solvent in your tube that the new water layer was above the area interrogated by the spectrometer).
Edit: The manager of U. Ottawa NMR seems to agree with me for $\ce{CDCl3}$ at least
http://u-of-o-nmr-facility.blogspot.co.uk/2007/10/proton-nmr-assignment-tools-d2o-shake.html

@owjburnham Obviously. The water solubility in chloroform is still 8g/l, though. That can easily translate into a molar concentration similar to your actual sample, if the latter has a somewhat higher molar mass.


text:
What is the number of hydrate isomers of chromium(III) chloride hexahydrate?
Assuming $\ce{CoCl3·6H2O}$ is an octahedral complex, the answer should be four as there can be $0, 1, 2$ or $3$ $\ce{Cl-}$ ligands inside the coordination sphere. However, numerous sources say the number of isomers is three discarding the case with three $\ce{Cl-}$ ligands inside the coordination sphere.
Only Loong's answer and another answer in a linked question say otherwise acknowledging the existence of the isomer with three $\ce{Cl-}$ ligands within the inner sphere. Every other place simply states that only three isomers have been experimentally discovered.
Also, in coordination isomerism too, in the case where both spheres become uncharged is ignored.
What is the reason?

Regarding the coordination isomerism question, the reason why it's not considered as an isomer in the case of two uncharged complexes, is that there is no ionic attraction. The two coordination spheres should act as individual ions and hence participate in the ionic sphere. But in this case, where the two spheres carry no charge, there isn't a force, namely the electrostatic force, which binds the two spheres.
Your question on CrCl3.6H2O is a very debatable topic and there hasnt been a unanimous decision that has been put forth by all the authors. Some say there are 3 or rather most of them say 3, only one author, JD Lee says that there are 4.
The reason given by the authors to support there claim of only 3 isomers is not very concrete, but it still majority of the authors believe that it is true. The H2O that exists outside the coordination sphere is the water of crystallization. The water of crystallization is believed to be present within the voids created by an anion and a cation. Now if the isomer [CrCl3(H2O)3] 3H2O should exist then it must have these voids. But here there isn't any anion in the ionic sphere, in fact it is believed that there isn't any ionic sphere. Therefore this isomer doesn't exist and hence cannot be the 4th isomer.
Now, it's also believed that the H2O is actually held to the central metal atom (Cr) with the help of dipole forces and that the voids are not needed. So that would mean that this exists and since it is theoretically an isomer, it should also be an isomer practically.
That brings us to to this very important question, what is the significance of the water in the ionic sphere? Now most books say that it's the water of crystallization of that compound. So if the water of crystallization must be present in the voids so that they can be removed by physical processes like heating, the 3H2O wont be an isomer anymore, but just another compound that exists with the same molecular formula.

I have a copy of 5th edition- J. D. Lee, which says that 3 isomers are known. These complexes are formed when $\ce{CrCl3}$ dissolves in water. Came across an article where it is written that: If diethyl ether is added to the solution then a fourth complex $\ce{[Cr(H2O)3Cl3]}$ can be formed (without the water of crystallization).


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Adding non stoichiometric amount of NH3 to copper solution

Consider the following reaction: $$\ce{Cu(H2O)_n^2+ + 4NH3 -> Cu(NH3)4^2+}$$
The amount of ammonium hydroxide added to your penny solution was not controlled (i.e. you did not add
“stoichiometric” amounts). Discuss what would happen to its absorbance value, and explain why, if the amount
added was too low and too high. (Assume the mark on the volumetric flask was not overshot in either scenario.)

I was hoping to get a second opinion on what happens to the absorption in the case where you add too much $\ce{NH3}$ and the case where you add too little.
I think that if I had added not enough $\ce{NH3}$ that some of the copper would not turn into $\ce{Cu(NH3)4^2+}$ and thus the absorption reading would be lower.
As for adding too much... I think it would be okay to add too much. I don't think it would affect the absorption.

Related: Why is ligand substitution only partial with copper(II) ions and ammonia?

Why does copper(II) sulfate react with ammonia?

Both your conclusions are absolutely correct.
I'll point out that you'll never get "all" the $\ce{Cu^{2+}}$ to the amine complex. You're trying to get greater than 99% of the $\ce{Cu^{2+}}$ to the amine complex. Because 99% and 99.9% and 99.99% would have virtually the same absorption reading.
To give even more detail, in mildly acid solution the copper will be in a octahedral complex $\ce{Cu(H2O)6^{2+}}$ which is a light blue. As you add ammonia at first $\ce{Cu(OH)2}$ forms, which is then dissolved as the various ammine complexes form.
$\ce{Cu(H2O)_6^2+ + NH3 <--> Cu(NH3)(H2O)5^2+}$
$\ce{Cu(NH3)(H2O)5^2+ + NH3 <--> Cu(NH3)2(H2O)4^2+}$
$\ce{Cu(NH3)2(H2O)4^2+ + NH3 <--> Cu(NH3)3(H2O)3^2+}$
$\ce{Cu(NH3)3(H2O)3^2+ + NH3 <--> Cu(NH3)4(H2O)2^2+}$
The tetraammine complex, $\ce{Cu(NH3)4(H2O)2^2+}$, is a much darker blue than the corresponding complex with water molecules alone. Thus colorimetry is much more sensitive measuring the tetraammine complex that the water complex.
The tetraammine complex has shorter Cu-N bonds than the C-O bonds. Thus the four ammine groups are one plane and the two water molecules at the apexes of the square bipyramidal complex.
Hopefully now the point about not converting all the copper to the tetraammine complex is clear. The various ammine complexes are in equilibrium with each other. Given that $\ce{[NH3] >> [Cu^{2+}]}$  then one can imagine the overall reaction as
$\ce{Cu(H2O)_6^2+ + 4NH3 <--> Cu(NH3)4(H2O)2^{2+}}$
In extremely concentrated solutions of ammonia the final two ammine groups will attach.
$\ce{Cu(NH3)4(H2O)2^{2+} + NH3 <--> Cu(NH3)5(H2O)^2+}$
$\ce{Cu(NH3)5(H2O)^{2+} + NH3 <--> Cu(NH3)6^2+}$


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Bombarding molecules with electrons
I have been reading a book titled "High Voltage Engineering". It seems to me that in this book it is considered that in a gas when they collide an electron with a molecule, the molecule is stationary and the electron is not. This is because the $v_\text{molecule} \ll v_\text{electron}$.
What do you think about this?

Welcome!  Is this in the context of particle accelerator?  If so they are likely accelerating the low-mass electron to a high velocity rather than accelerating the molecule.  From a standpoint of relativity (think Galileo, not Einstein) it would seem that it is ambiguous and irrelevant which is moving and which is stationary.

Hi, thanks for the welcome. The subject focuses on electric discharges in gases, they effectively plan to bombard a gas with electrons to generate ionization  by collision of their molecules.

So, I think all they are saying is that the molecular movement under the experimental conditions (i.e. temperature, pressure)  is negligible compared to the high-velocity electrons bombarding the "stationary" (in the bulk sense) gas.  In this case, the velocity in which the molecules collide can be approximated as the velocity of the electrons alone.

A molecule moves with thermal speed. An electron having enough energy to ionize a molecule (that is, at least few eV) moves much faster than that.

Exactly, but could you assume something like that or not?

As @IvanNeretin says, yes, it has to be so in order for ionization of the gas to happen.  In other words, if that weren't true, nothing would happen, electrons would just meander through the gas molecules and no reaction/ionization would take place.

Room temperature is about 23 meV (milli-electron Volts). One can easily make 100keV or higher energy electrons. Clearly the electrons will be moving much faster than the gas molecules. They are even moving much faster than a 100keV ion. (Also, remember that the rest energy of an electron is 511keV, so that 100keV electron should be treated relativistically).

@JonCuster Interesting point of view, recommend some bibliography for this topic?

Almost any intro plasma physics book will cover the relative mobilities of electrons vs ions. Look at the simplest system - a hydrogen plasma. The electron has roughly 1/2000th the mass of a proton. If electrons and ions are in thermal equilibrium, the electrons will have a velocity 40x that of the ions. Heavier ions will have even slower relative velocities. And, that assumes thermal equilibrium, which generally does not hold for plasmas.

@JonCuster ty :)


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experimentally determine oxidation state of Iron compound
I need to make several Iron compounds but I am having a bit of trouble finding a decent way to determine the oxidation state of Iron.
I'm starting with a Iron(II) compound that I try to oxidize using several methods. I can use raman to determine the compound but that requires me to create the compound I want to make and check that spectrum against my experimental sample. This is quite a bit of work and I was hoping someone could point me towards a better solution.
I can make an EDS but since my compounds usually contain carbon or oxygen as counter ion so EDS isn't very accurate. I don't (currently) have access to an XPS and I don't have unlimited access to a mass spectrometer (I have to travel quite a bit to get to one).
Can someone help me find a better way to determine the oxidation state??

I worked on a project years ago where we needed determine iron(II) and iron(III) in a lot of samples in a hurry.  It was a UV-VIS absorbance technique.  I think it was a form of phenanthroline (ortho?) that complexed with the iron(II) to give the colored species.  There was then a reagent added to reduce iron(III) to iron(II), and one measured that and calculated iron(III) by difference.  At least I've maybe given you some search terms. I'll try to find how we did that but no promises.  Again, it was fast, cheap and easy, so long as you had a UV-VIS spectrometer...do you have one?

I might be able to get one but phenanthroline is known to change color depending on the metal and its oxidation state. I'll try to use a solid state uv-vis and see if I can see any changes between an iron(II) and iron(III) complex but I am working with nanoparticles so I'm not very hopefull.

Yes, I believe what you said is part of the reason we were able to differentiate oxidation states;  the iron(II) absorption max was distinct from that of iron(III).  For one project in particular this distinction in the final analysis was absolutely critical.  Since you are interested I'll try to get the method reference.  I'm pretty sure it was in Standard Methods (for the Examination of Water and Wastewater).

I still haven't made it into the lab to get a copy of our method, but I thought these links might be interesting regarding the theory and technique behind the VIS absorption with phenanthroline methods for iron determination.  The biggest problems discussed are regarding total iron determination when some digestion is required to get the solid into solution, which doesn't seem to be applicable to your case. I don't have good (e.g. free) access to the full articles. 1) https://www.ncbi.nlm.nih.gov/pubmed/22349079   2) https://www.jstor.org/stable/23349093?seq=1#fndtn-page_scan_tab_contents

Here is a link to the full Standard Methods for the Examination of Water and Wastewater method 3500-Fe for determining ferrous, ferric and total iron in aqueous solutions.
Here is an excerpt describing the principle of the procedure:

"Principle:
  Iron is brought into solution, reduced to the ferrous
  state by boiling with acid and hydroxylamine, and treated with
  1,10-phenanthroline at pH 3.2 to 3.3. Three molecules of
  phenanthroline chelate each atom of ferrous iron to form an orange-red
  complex. The colored solution obeys Beer’s law; its intensity is
  independent of pH from 3 to 9. A pH between 2.9 and 3.5 insures rapid
  color development in the presence of an excess of phenanthroline.
  Color standards are stable for at least 6 months."

In this method, ferrous iron is measured directly and ferric is determined by difference between the total iron (measurement also described by method) and the ferrous iron.  Obviously, when the ferrous iron concentration approaches the total the uncertainty in the the ferric iron concentration becomes large.

In all the (physical) experimental literature I've read, at least for biomolecules (usually enzymes), the technique of choice is Mossbauer spectroscopy. Here is a good starting point.

One downside is the requirement that the sample be in the solid phase.