text:
Intuitive explanation for depression of freezing point
Why does depression of freezing point takes place if a non-volatile solute is added to the solution?
For more details about the question checkout this link below
Depression of freezing point

Better not to mention the link as for it is totally unclear and closed because that.

Why is the entropy of mixing not a suitable enough explanation?

@JonCuster I think it is, as I just argued in my answer, but entropy is not a very intuitive concept for beginners.

@Zhe - or, better put perhaps, intuition is often not very useful and highly variable from person to person. I just hate questions asking for 'intuition' since they are really unanswerable.

Yes, for some folks saying that the Gibbs energy of reaction is concentration dependent and that the concentration of liquid water changes is quite intuitive. Or you could say K does not change, but the concentration of water does, so it is no longer at equilibrium. Short and to the point if you are familiar with the concepts.

Imagine a pure liquid. If part of it freezes, what happens to the rest of the solution? Well, it looks exactly the same since it's also pure, so nothing.
However, this is not true for the system you've described.
What happens if there's a solute mixed in liquid? When, now you're concentrating the solute in the remainder of the liquid, so you're creating an entropic barrier to freezing the liquid.
Freezing is a process with $\Delta H < 0$ and $\Delta S < 0$. This means that with $\Delta G = \Delta H - T\Delta S$, the process is spontaneous when $T$ balances the favorable enthalpy change against an unfavorable entropy change. This $T$ is the freezing point. However, notice that due to the above argument, $\Delta S$ is now more negative. This means that the same balance will be achieved with a smaller value of $T$, hence, a freezing point depression.

Here is a - hopefully intuitive  - explanation based on kinetics:
I start with a glass of water with ice cubes in it, sitting outside in winter on a day where the temperature happens to be equal to the freezing point of pure water. The system is at equilibrium, meaning that the rate of (solid) water molecules melting is equal to the rate of (liquid) water molecules freezing. Basically, some water molecules wiggle free from the surface of the ice, while some water molecules from the liquid gently bump into the surface of the ice and attach.
Now, I add a solute to the liquid water. The melting of the ice will not be affected because the ice is still made of pure water. However, some of the molecules of the liquid bumping into the surface of the ice can no longer attach because they are solute molecules and they don't fit. So the melting rate stays the same, while the freezing rate decreases (you could say because the concentration of liquid water got smaller by adding the solute).
As a net effect, the ice cubes melt, getting smaller. This means the freezing point of the solution is lower than that of pure water.

So your telling due to interaction between solute and solvent molecule depression for ex. (https://masterconceptsinchemistry.com/wp-content/uploads/2017/10/dissolivingsodim.jpg) but depression in freezing point is a colligative property it does not depend on interaction between solute and solvent molecule could you clarify


text:
Carboxylic Acid Nomenclature
According to my chemistry textbook, 2-hydroxy-1,2,3-propanetricarboxylic acid is represented as:

But 5-bromoheptanoic acid is given as:

In the former naming, propane is considered in the base chain while the COOH carbons are not. In the latter one,  the COOH carbons are considered in the base carbon chain.
On checking NCBI Pubchem, I found the following:
https://pubchem.ncbi.nlm.nih.gov/compound/16217547#section=Top
https://pubchem.ncbi.nlm.nih.gov/compound/5-Bromovaleric_acid#section=Top
Why are the carbons of carboxylic acids excluded in the former name but included in the latter?

Related: IUPAC name for citric acid

I don't remember the specific recommendations but taking pentane or pentanoic ac. as parent compound would be rather messy. Keep in mind that except for reporting a new isolated molecule following recommendations is very important but less crucial than what it might seem to beginners (assuming you are). Perhaps both examples adhere to recommendations and/or others will answer. I just want to point out that you should not over think if you can univocally name a structure (or draw from name).

@Alchimista Yeah, thanks. Found that dicarboxylic acids (such as glutaric acid; structure: https://en.m.wikipedia.org/wiki/Glutaric_acid) have the COOH carbon considered in the base name in most cases but in tri/more carboxylic acids, this rule doesn't follow.

According to IUPAC Nomenclature rules, the carboxylic acid functional group carbon is included in the base aliphatic chain if one or two carboxylic acid functional groups are present and the standard suffix '-oic acid' is used. For three or more carboxylic acid groups in an aliphatic chain, the carboxylic acid carbons are excluded and the suffix 'carboxylic acid' is used instead.
Suffixes such as carbonitrile and carbaldehyde (When is the carbaldehyde suffix correct?, an example for cyclic compounds) are used similarly to denote the exact positions of the functional groups to avoid ambiguity.

This response is beautifully specific to the question, but its apparent credibility suffers by not citing the IUPAC rule which forms the basis of the answer.

When identical groups are present such that neither can be preferred over the other then, such nomenclature is preferred to emphasise on the equivalence of the groups such as in the one we have here. We cannot name in such a way that includes the three groups together. So, we name it that way instead.
As such, the rules of IUPAC Nomenclature are themselves arbitrary therefore, it is just something that is the way it is. For more information you might like to see https://old.iupac.org/publications/books/principles/principles_of_nomenclature.pdf, Page 34 onwards.
Edit: I almost forgot that a compound can have more than one name, even by the IUPAC nomenclature rules. As such, the names used commonly as IUPAC names are the preferred ones.

source: Stack Exchange

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sort: votes


text:
How to calculate the pH of a solution given the concentration and pKa's of solutes?
For example, if I have a solution of 50mM acetlycholine (pKa = 4.5), how would I go about calculating the pH of the solution?
Thanks

You wouldn't. In situations like this, you would typically have some other solutes at much greater concentrations, and it is them that will ultimately define the resulting pH. Otherwise, go for the Ostwald’s dilution law.

A pKa of 4.5 would be typical for a carboxylic acid group, but I don't see any functional groups that look like it (there is a quaternary amine and an ester group, neither one have acid/base chemistry in water). What else is in the solution, and what is the solvent?

I am assuming in your example that there is only one solute. If you have more than one then ionisation is mutually suppressed and things are a bit more complicated.
A useful expression that you can derive from an ICE table for a weak acid is:
$\qquad \qquad\qquad\qquad\qquad\rm pH=\dfrac{1}{2}[pK_a-\log{c}]$
Where c is the concentration of the acid which we assume to be just about equal to the equilibrium concentration.
This is reasonable for $\rm pK_a$ values between 4 and 10.
Putting in the numbers:
$\rm pH=\dfrac{1}{2}[4.5-(-1.3)]$
$\rm \underline{pH=2.9}$
Acetylcholine has a quaternary nitrogen so I can only assume that, given that the $\rm pK_a$ value is correct, the deprotonisation occurs at a different site on the molecule.

How did you get log(0.05) = -0.7?

Oops ! Sorry about that. Should be -1.3. Have corrected answer.

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text:
Relation between freezing point depression and vapor pressure
My lab manual defines the freezing point as "The temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase." It goes on to explain why adding a non-volatile solute to a solvent, which lowers the vapor pressure compared to that of the pure liquid, lowers its freezing point.
What does the freezing point, which concerns liquid and solid phase, have to do with vapor pressure? If I do an experiment in a gas-tight syringe containing liquid and solid only, there will be no equilibrium with the vapor phase, but the freezing point will still change.

I beg to differ - there is always an equilibrium vapor pressure over a solid or liquid. Using the vapor pressure provides a way to compare the solid to the liquid (using the vapor as an intermediary). The solid and liquid will only be in equilibrium when the vapor is in equilibrium with both of them.

Note that @JonCuster's excellent comment here is an application of the zeroth law of thermodynamics.

A freezing point is a temperature at which solid and liquid phases are in equilibrium with each other. In equilibrium the chemical potentials of each component in all phases present are equal. It means that if there are three phases in the system, solid, liquid and gas, then the chemical potential should be equal in all the three. As a consequence, the vapor pressure of a given component over the liquid phase should equal that over the solid phase. But if there are only solid and liquid phases, I see no need to involve gas. The basic derivation, which I was taught, of the freezing point depression of liquid solvent A (when the solute is insoluble in solid A) starts with $$\mu_\mathrm{A}(\mathrm{solid})=\mu_\mathrm{A}(\mathrm{solution})=\mu_\mathrm{A}(\mathrm{pure\,liquid})+RT\ln{x_\mathrm{A}}$$ - no gases here.
I might miss something here, but my guess is that your lab manual is just trying to be more obscure than it ought to be (no offence to its authors intended). While the expression for $\mu_\mathrm{A}(\mathrm{solution})$ itself can be derived from the Raoult's law, which requires gases, that's slightly different, although related, story. Provided that the expression for $\mu_\mathrm{A}(\mathrm{solution})$ is already known, I can see no sound reason for inviting the vapor pressures to explain cryoscopy.


text:
Do substances dissolved in a solvent oxidise?
If a substance is dissolved in a solvent does it still oxidise? Does it depend on weather the solvent can dissolve oxygen?

There are very many reagents that will oxidise substances in solution

The rate would be slower if oxygen does not dissolve in the solvent, but at the interface the solute would still be in contact with the atmosphere (assuming that's the source of the oxygen).

Not only can things be oxidized in solution, but in some cases (such as ferrous chloride) exposure to air can contaminate your solution with oxidation products.  Such solutions should be stored in a way that prevents or at least retards air from getting in (e.g., taping around the lid).


text:
Pyrimidine chemistry
Why C4 position in pyrimidine ring is more reactive then C2 position in nucleophilic aromatic substitution and Pd chemistry? Should not be the C2 position the most reactive because of the effect of the 2 N atoms?

The SNAr bit is already asked here: https://chemistry.stackexchange.com/q/80995/16683 I think the Pd coupling bit would be good as a stand-alone question.

That q does not have an answer with a positive score.  So it is not "really" answered.

@OscarLanzi I never said it was "answered", merely "asked" ;) and arguably, asking both questions together might be too broad.

It is true, it might be a lot to ask in one go but, I asked them toghether because on every books or paper they explicitly say that since in the SNAr reaction, C4 is the preferred electrophilic site of reaction, it is the same for Pd chemistry. But I am very confused.

I thought that the presence of the 2 Nitrogens would have made the C2 position more electrophilic and not the other way around. Again, if you use a negatively charged nucleophile it goes on C4, if you use a neutral one, it goes on the C4 first. So, it does not really depend on the nucleophile. Might it depend on the presence of the orbital on the N preventing some how the reaction t C4? and it is the same for Pd chemistry: if you run a Pd coupling on the C4 it goes in 1 h, on the C2 lack of reactivity or anyway harsh conditions.


text:
Comparing the number of oxidation states of Vanadium and Arsenic
USNCO 2004 Question 45 states:

Which element exhibits the greatest number of oxidation
  states in its compounds?
(A) Ca (B) V (C) Cu (D) As

I ruled out Ca, as I know it only exists as +2. I then drew electron configurations for V, Cu and As. I ruled out Cu as it had half filled 4s and full 3d orbitals, also because I've only ever seen copper in a +2 oxidation state.
However, I wasn't able to differentiate between V and As. They both have 3 unpaired electrons, albeit As has an extra full 3d orbital.
I was tempted to go with As, as being less metallic than V, it is more likely to have negative oxidation states, but the answer states Vanadium has the most oxidation states.
Is there any way I can work this question out by looking at electron configurations, or will I have to memorize the oxidation states?

V and As are evenly matched - both on all states between -3 and 5. BTW I'd expect you heard about Cu +1 compounds, it goes up to +4, actually and there's also exotic -2.

@Mithoron I saw on a Wiki page that V and As do have the same number of oxidation states, but is there any way I can work out which have more 'common' states which I think USNCO is asking for?

Well, I wager V has more "common" ones, just like tr. met. do - unpaired electrons aren't a problem for them.


text:
How do three-electron bonds work in MO theory?
When we're taught MO theory in basic chemistry, a rule-of-thumb hammered into our heads is that an orbital takes up to 2 electrons with opposite spins. Sigma or pi bonds are represented by stable bonding orbitals. Orbitals sometimes have just the one electron (radicals) but are much more stable with the pair.
While this justifies the occurrence of one-electron bonds, like that of the dihydrogen cation, it does not smoothly explain the occurrence of three-electron bonds like in nitric oxide and dioxygen. If we draw a straightforward MO diagram for nitric oxide with the "two electrons per orbital" rule, we end up with a lone electron in a antibonding pi orbital. "Antibonding" seems to imply the lone electron is sitting at its host atom, but that's not the case. Where does this lone electron go? Does it "merge" with one of the bonding sigma or pi orbitals and if so how do we know which one? Is this just a shortcoming of MO theory?

There's nothing inherently wrong with an electron sitting in an antibonding orbital. All "antibonding" means is that when you put an electron in it, the bond order decreases. It doesn't necessarily say anything about where the electron is - it is still delocalised over both atoms. And no, you can't merge two orbitals.

Yes, I'm aware that electrons can sit in antibonding orbitals. We often see them "cancelling" out bonding orbitals (e.g. why dihelium doesn't form). But here's a case of an unpaired electron sitting in an antibonding orbital that is known to participate in bonding.

Mmhmm. All that means is that the "antibonding-ness", if you will, is less than the "bonding-ness". And so you have a weak(er) bond.

Well, we can use NO as an example. https://mgh-images.s3.amazonaws.com/9781111779740/508967-10-99EEI1.png We can ignore the bottom bit, because it's not relevant. There's a σ orbital with 2 electrons, but the corresponding σ* orbital at the very top is unfilled, so that makes one σ bond between N and O. For the π orbitals, there are 4 bonding electrons and 1 antibonding electron, so we have a (4-1)/2 = 1.5 π bond. Altogether we have a 2.5-bond between N and O.

So how can we know where the electron goes in the molecule? What makes NO have a three-electron bond instead of just normal two-electron bonds plus a lone electron in an antibonding orbital pulling the nuclei apart? Or does MO theory simply not account for such detail?

"What makes NO have a three-electron bond instead of just normal two-electron bonds plus a lone electron in an antibonding orbital pulling the nuclei apart?" There is no difference. "how can we know where the electron goes in the molecule?" you need to draw out the MO. It looks like this: https://chemistry.stackexchange.com/a/66654/16683 the middle one in the row labelled 8.


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Photochromism time of colour change
I know there are chemical pigments/dyes which produce an immediate reversible colour change in UV or direct sunlight. Is there a way to postpone the colour change so that it does not happen immediately but rather over a period of 30 minutes for example? Would it have to do with the concentration of dye or is it the chemicals themselves which have different colour change times.

Older photochromic glasses (the ones that went grey in sunlight) were pretty slow in changing, particularly when cold. I have no idea what changed over the years to speed them up.

@jonCuster they probably found better chemicals to use.


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number of optical isomers of a cyclic molecule

What is the number of optical isomers for this compound?
Also, there seems to be no clear-cut boundary between geometrical isomers and optical isomers. Referring to a post in StackExchange,
is (RS)- and (RR)-form of 1,2-dichlorocyclohexane considered optical isomers or geometrical isomers? The isomers exist due to restriction of rotation caused by the ring, you can also called the two geometrical isomers, can't you?

Please specify how you intend to isomerize the drawn molecule: If you exchange only the halogens amongst themselves, you have the same situation as in the methane derivative CFClBrI. If you allow exchange with the hydrogens, it gets complicated.

@TAR86, he's talking about stereoisomerism, not structural/constitutional isomerism. One has to predict number of optical isomers of the given compound, without playing around with the basic skeleton.

I think that you are over thinking. As enantiomers do not superimpose each other obviously they haven't the same geometry. When discussing enantiomerism you are already further. It is just that at a first glance they might look the same, but enantiomers certainly have different geometry. You can call them "geometrical isomers with optical activity" but "enantiomers" do it better.

The question for this molecule is heavy constrained because the molecule is essentially rigid and there is no meaningful way to "swap" the stereochemistry at any individual centre. The carbon core is the same as adamantine and it isn't called that for nothing. However the overall molecule has the same stereochemistry as the (equally rigid and tetrahedral) methane so generates just two enantiomers with 4 different halogens.

Dammit editor. Adamantane not -ine.

1-Bromo-3-chloro-5-fluoro-7-iodoadamantane would be an interesting example of central tetrahedral chirality without the actual (physical) center. Exchanging halogen atoms will result in only two stereoisomers.
Due to its rigidity, its stereochemistry is related to tetrasubstituted tetrahedrane (tricyclo[1.1.0.02,4]butane; or rather its imaginary geometrical model, because it would be very strained and unstable), whose stereochemistry is related to bromo(chloro)fluoro(iodo)methane.

To your additional question: It depends how (far) you'd like to go with removal of the rotation restriction, anyway the resulting compounds will already not be original 1,2-dichlorocyclohexane.

I mean yes, (1R,2R)-1,2-dichlorocyclohexane and (1R,2S)-1,2-dichlorocyclohexane can probably be considered geometrical isomers (in addition to optical), they can even be named trans-1,2-dichlorocyclohexane (which has two enantiomers, the other one being (1S,2S)) and cis-1,2-dichlorocyclohexane; but this isomerism is not caused by rotation restriction, but by different arrangements on chiral tetrahedral carbons.
But note that geometrical isomerism is usually used in relation with double bond cis/trans isomerism; and together with optical isomerism are both deprecated terms.


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What is the maximum concentration of HNO3 (nitric acid) which may be handled outside of a hood?
Commercial stock solutions of HNO3 (nitric acid) are usually 70% w/w, and according to several MSDSs (e.g. here), it is dangerous to inhale it:

Symptoms/effects after inhalation : Irritation of the respiratory tract. Dry/sore throat. Corrosion of the upper respiratory tract.
    Coughing. FOLLOWING SYMPTOMS MAY APPEAR LATER: Respiratory difficulties. Possible
    inflammation of the respiratory tract. Risk of lung edema. Blue/grey discolouration of the skin

What I want to know is what is the maximum concentration that 70% HNO3 can be diluted to, which may be safe to work outside of a hood? Are there guidelines/credible sources for such information?
Please note, this is not a medical question, it is a question of safety practice in a chemistry lab.

Why would you need to work outside the hood?  If you can, work with chemicals in the hood.

What he said! If possible keep it in a hood at all times.

And if it isn't possible to keep it in a hood you almost certainly need to rethink what you are trying to do. (Yes, my lab regularly uses small amounts of IPA outside of a hood to clean optics and vacuum equipment, but we sure would not use similarly small quantities of nitric acid.)

I can see why it may be really dangerous for your throat to use fuming HNO3, but normal conc. nitric acid shouldn't be, unless you sniff it on purpose.

This is a good question!  Probably everyone agrees that 0.0000001 mol / L of nitric acid in water is not particularly hazardous to handle outside the fume hood, and that concentrated nitric acid is very dangerous and must be handled in a fume hood.  Where is the in-between point?

I feel like the commenters are missing that the OP is not asking about the the amount of concentrated acid that can be handled outside the hood, he is asking about the concentration of acid that can be handled outside the hood.

That's right, @Curt F., and I'll try to elaborate: my work is usually with low concentrations of 1% nitric acid, and the experienced students say it's not a problem to work with it outside of the hood. We prepare the 1% from 70% and this IS done in the hood, always. This is the background, and as Curt F. Put it nicely: what would be the cut-off point within this range?

Sounds like you are referring to inhalation risk and damage. Safety may depend on ventilation in the room, how close you approach the solution, and both the volume (or surface area), temperature and concentration, how you handle the solution, to what purpose it is put. If you are working in a lab ask someone associated with the lab in charge of safety.

I want to guess that this is mainly determined by the relevant legislation of whichever country you’re in, particularly your country’s equivalent of Germany’s MAK (maximal working-space concentration). Because these are set to be well below any truly hazardous limit.


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Calculating dissociation degree from pH and Ka without given initial acid concentration
Is it possible to calculate the dissociation degree (DD) from pH and Ka without directly been given the initial acid concentration?
I found the following formula:
$$\text{DD} = \frac{K_\mathrm{a}}{10^\mathrm{-pH}}$$
Is this true?

Briefly: yes, it's possible for any polyprotic acid, using pH and Ka only; no, the formula for DD seems to be incorrect. If you read German, the Wikipedia article covers it all: https://de.wikipedia.org/wiki/Dissoziationsgrad. If not, I can put some sort of an answer based on that in a couple of hours.

I can't read german, but thanks. I got this equation from ostwalds law and pH = 1.2pka - 1/2log ca. However, this is probably only valid when the acid is considered very weak?

No, it's only valid for any monoprotic acid and weak polyprotic acids.

Even though the degree of dissociation $α$ depends both on the nature of the dissolved electrolyte (e.g. acid) and the concentration, for the homogeneous medium it is possible to determine $α$ by $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ only, without any auxiliary information such as initial concentration since $\mathrm{pH}$ is a function of concentration.
I would suggest you not to remember certain formulas, since they have been derived only for particular systems and cannot be universally applied. Instead, use the law of mass action, especially when it comes to multi-component systems and multi-step dissociating electrolytes.
The following is heavily inspired by the German Wikipedia article for the degree of dissociation. Consider all $n$ steps of the dissociation of the $n$-protic acid $\ce{H_nA}$:
$$
\begin{align}
\ce{H_nA + H2O &<=> H_{n-1}A- + H3O+}&\quad &K_{\mathrm{a}1} \tag{1}\\
\ce{H_{n-1}A- + H2O &<=> H_{n-2}A^2- + H3O+}&\quad &K_{\mathrm{a}2} \tag{2}\\
&\ldots &&\ldots\\
\ce{H_{n-(m-1)}A + H2O &<=> H_{n-m}A^m- + H3O+}&\quad &K_{\mathrm{a}m} \tag{$m$}\\
&\ldots &&\ldots\\
\ce{H2A^{(n-2)-} + H2O &<=> HA^{(n-1)-} + H3O+}&\quad &K_{\mathrm{a}(n-1)} \tag{$n-1$}\\
\ce{HA^{(n-1)-} + H2O &<=> A^n- + H3O+}&\quad &K_{\mathrm{a}n} \tag{$n$}
\end{align}
$$
where $n\in\mathbb{N}$, $m\in\mathbb{N}$, $m < n$; $K_{\mathrm{a}m}$ – $m$-th dissociation constant. Dissociation degree $α$ for any component of the composition $\ce{H_{n-m}A^m-}$ formed at the $m$-th step is by definition
$$α_m = \frac{[\ce{H_{n-m}A^m-}]}{c_0}$$
where $c_0$ – initial concentration:
$$
\begin{align}
c_0 &= [\ce{H_nA}] + [\ce{H_{n-1}A-}] + \ldots + [\ce{H_{n-m}A^m-}] + \ldots + [\ce{HA^{(n-1)-}}] + [\ce{A^n-}]\\
  &= \sum_{m=0}^{n}{[\ce{H_{n-m}A^m-}]}
\end{align}
$$
Each concentration in this set can be found recursively from the dissociation constant:
$$K_{\mathrm{a}m} = \frac{[\ce{H_{n-m}A^m-}][\ce{H3O+}]}{[\ce{H_{n-(m-1)}A^{(m-1)-}}]}$$
$$\implies$$
$$
\begin{align}
[\ce{H_{n-m}A^m-}] &= \frac{K_{\mathrm{a}m}[\ce{H_{n-(m-1)}A^{(m-1)-}}]}{[\ce{H3O+}]}\\
  &= \frac{K_{\mathrm{a}m}K_{\mathrm{a}(m-1)}[\ce{H_{n-(m-2)}A^{(m-2)-}}]}{[\ce{H3O+}]^2}\\
  &\ldots\\
  &= \frac{\prod_{i=0}^m{K_{\mathrm{a}i}}[\ce{H_{n-m}A^{m-}}]}{[\ce{H3O+}]^{n-m}}
\end{align}
$$
so that for $α_m$ after rearranging sums of the products the following universal equation  of two arguments ($K_\mathrm{a}$ and $\mathrm{pH}$) emerges:
$$α_m = \frac{[\ce{H3O+}]^{n-m}}{[\ce{H3O+}]^n + K_{\mathrm{a}1}[\ce{H3O+}]^{n-1} + K_{\mathrm{a}1}K_{\mathrm{a}2}[\ce{H3O+}]^{n-2} + K_{\mathrm{a}1}K_{\mathrm{a}2}\cdot\ldots\cdot K_{\mathrm{a}n}}\cdot\prod_{i=0}^{m}{K_{\mathrm{a}i}}$$
For the monoprotic acid $\ce{HA}$ ($n = 1$, $m=0$) the formula becomes much simpler:
$$α_0 = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + [\ce{H3O+}]} = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + 10^{-\mathrm{pH}}},$$
and if the acid is very weak ($K_{\mathrm{a}} \ll 10^{-\mathrm{pH}}$, e.g. $K_{\mathrm{a}} \ll \text{ca.}~10^{-7}$), only then you can use your formula:
$$α_0 \approx \frac{K_{\mathrm{a}}}{10^{-\mathrm{pH}}}$$

Pragmatically I think that this also means that pH < 5 or pH > 9 to avoid complications due to autodissociation of water (assuming that you want 1% accuracy...)

The formula seems fine, depending on how DD is defined.
The calculation is based on the Henderson-Hasselbalch equation (also known as the Buffer eq.):
pH = pKa + log ([Base]/[Acid]) => [B]/[A] =10^(pH-pKa)
The ratio [B]/[A] is independent of the total concentration.
The ratio B/A can also be expressed as the fraction of the total concentration present as Protonized acid (Ya):
pH = pK + log ((1-Ya)/Ya), where Y goes from 1 (all acid protonized) to 0 (all acid deprotonized)
A Bjerrum plot is often used to illustrate Ya as a function of pH

And is this only valid if you say that the acid is very weak? So doesn't dissociate very much?

It is valid for all acids, but (very) strong acids will for all practical purposes be completely dissociated (Ya=1). But it is only valid for diluted solutions. When the concentration is too high the calculated result will deviate.

The equation isn't valid for dilute solutions of a weak acid where the autoionization of water would need to be considered.


text:
Consequence of interactions between filled orbitals
It is known that the interaction between filled orbitals from the perspective of molecular orbital theory is overall destabilising because antibonding MOs are more antibonding than bonding MOs are bonding. Consider a large molecule (e.g. $\ce {I2}$). Due to the presence of multiple net destabilising interactions between the filled inner shell atomic orbitals (e.g. $\ce {1s, 2s, 2p}$ ...), wouldn't its formation be almost impossible since the single stabilising interaction due to the valence atomic orbitals would be small, compared with the total destabilisation from all the interactions between the inner shell orbitals, which are in fact, very numerous?

Indeed, the single stabilising interaction of the valence orbitals is small, but the total destabilization is smaller yet. See, all those 1s, 2s... are awfully small, so from their own POW they are awfully far from each other.

When overlap is zero, the MOs don't have bonding or antibonding character. The overlap is not zero, but is tiny enough that it might as well be zero.

@orthocresol Are you trying to say that for the iodine molecule, the overlap integral S for 1s-1s interaction is almost negligible, hence even though there is an interaction, the bonding and antibonding interactions are equally bonding and antibonding respectively?

Yes, and that's what Ivan was trying to say too, I believe.

@orthocresol Actually, I just read Molecular QM by Atkins & Friedman (4th ed.) concerning this topic and they mentioned in the text that interactions between compact AOs and diffuse AOs can be neglected. So I suppose that the case of compact AOs would be 1s orbitals interacting in iodine molecule. And this can be explained by the overlap integral S being small. What about "diffuse AOs", is the idea also S being small?

I don't know. I think I'd need to see the context, and I'm not sure whether it's referring to the overlap between two compact AOs, or whether it's referring to the overlap between one compact AO and one diffuse AO. Right now, I'm too lazy to look in my 5th ed MQM though. Sorry. But in general, yes, if S is small then the "interaction" will be small.

On p. 261, Atkins & Friedman (2005) mention two criteria when considering the significance of orbital interactions. On the first criterion, they write:

First, to participate significantly in bond formation, atomic orbitals must
  be neither too diffuse nor too compact. In either case, there would be only
  weak constructive or destructive overlap between neighbouring atoms, and
  only feeble bonds would result. It follows that in Period 2, ($\ce {1s,1s}$)-overlap can be largely neglected in comparison with ($\ce {2s,2s}$)-overlap, for $\ce {1s}$ orbitals are too compact to have significant overlap with each other. Indeed, it is generally safe, for qualitative discussions at least, to consider only overlap between orbitals of the valence shell, for only these orbitals are neither too compact nor too diffuse to have significant overlap.

This means that it is likely that for the $\ce {I2}$ molecule I mentioned in the question, the destabilisation from the interaction of filled inner shell orbitals (e.g. $\ce {1s, 2s, 2p}$ etc.) is probably very insignificant. As orthocresol writes in the comments, this can also be related to the overlap integral $\ce {S}$.
In orthocresol's answer here, we note that the reason behind antibonding MOs being more antibonding than bonding MOs are bonding is that the value of the overlap integral is not negligible (i.e. $\ce {S}$ is not much less than $\ce {1}$ such that $\ce {S}$ $\approx$ $\ce {0}$). However, in the case of the $\ce {1s-1s}$ overlap in the $\ce {I2}$ molecule, the orbitals may be too compact to interact such that $\ce {S}$ may actually become negligible. Thus, the antibonding and bonding MOS are destabilised and stabilised to the same extent respectively. Thus, these filled orbital interactions of the inner shell orbitals do not constitute much destabilisation.
Reference
Atkins, P.; Friedman, R. Molecular Quantum Mechanics (4th ed.). Oxford University Press, 2005.


text:
Guidelines on qualitatively drawing MOs
I would just like to clarify the guidelines of drawing MOs formed from interactions between AOs. Strangely, this topic is rarely touched on in most books on molecular orbital theory. The following are some examples of what I consider to be guidelines:

When there is an in phase combination of atomic orbitals, i.e. the lobes in contact are of the same phase, the two lobes would merge into one bigger lobe.
When there is an anti phase  combination of atomic orbitals, i.e. lobes of different phase, the two lobes would seem to be "pushed away" from each other. The electron density seems to shift away from the internuclear region. For example, in the case of a p orbital, the lobe involved in the anti phase combination would shrink in size while the other lobe would increase in size.
For bonding MOs, the lobes closer to the more electronegative atom must be necessarily larger. For antibonding MOs, the lobes closer to the less electronegative atom must necessarily be larger.

Is there a well-established set of such guidelines which beginners in MO theory can follow when drawing resultant MOs?

I use JD Lee's Concise Inorganic Chemistry from page 93 onwards with special emphasis on page 97 and 98

The problem is an extension of drawing a function of one spatial variable, say x. Consider an in phase orbital to be positive, an anti phase orbital to be negative. Assume the orbitals have same occupancy (integrated density), then at some point there has to be a node in the summed orbitals, a point with zero density. Cancellations will occur elsewhere to different extents depending on orbital shapes. Extend this line of thought to 3d.


text:
Drawing bonds and delocalised electrons in condensed formulae
The $\LaTeX$ mhchem package provides commands for drawing bonds within condensed chemical formulae. Some of them have a clear allocation to covalent bonds while others have not. Single, double, and triple bonds can be drawn in a very easy and convenient way and, of course, delocalised $\pi$-electrons in condensed formulae like $\text{H}_2\text{C}=\text{CH}-\text{CH}=\text{CH}_2\;\;$of 1,3-butadiene can easily be drawn by dashed lines, too (see image below). However, the following four general formulae remain incomprehensible with respect to their application.
Does anyone know semi-structural (condensed) formulae of chemical compounds in which such chemical bonds (dashed lines/arrows) are drawn? Or are they just a "nice to have feature" provided by the package?


text:
Does gas chromatography follow the "like interacts with like" rule of thumb?
The "like interacts with like" rule of thumb works well when there are polar and non-polar molecules that are mobile (liquid-liquid extraction, membrane vs cytosol, questions of solubility).

Polar molecules interact with polar molecules rather than with non-polar molecules because the interactions are stronger.
Non-polar molecules "hang out" with non-polar molecules because that
way, they don't disrupt the stronger interactions between polar
molecules.

The latter is sometimes called the hydrophobic effect when
   water is present as a solvent. Self-assembly of detergents into
   micelles, folding of proteins, formation of lipid bilayers are all
   rationalized by "like interacts with like".
In a previous post, the question was asked why polar molecules have lower retention times than non-polar molecules when running gas chromatography with a non-polar column. One of the answers invoked "like interacts with like" as rationalization.
I have two questions:

Do polar molecules have lower retention times than non-polar molecules on a non-polar GC column (to make it comparable, the molecules should have similar boiling points)?
Does the "like interacts with like" rule of thumb apply to gas chromatography, given that there are no interactions in the gas phase and there is no hydrophobic effect in the stationary phase (i.e. the non-polar stationary phase interacting with the polar analyte would not disrupt any polar-polar interactions)?

In the post I referred to, the question started with a convincing statement:

I remembering reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.

This would suggest non-polar molecules have the weaker interaction and elute first, even on a non-polar column.

I think it's helpful here to consider what the stationary phase looks like for GC.

For open tubular columns, I found wall-coated (WCOT), support-coated (SCOT) and porous layer (PLOT). For WCOT and SCOT, the stationary phase is a liquid (at least at operating temperature), which I did not know.

Also important is the polarity of the stationary phase, right?

Yes, the question is about a non-polar stationary phase. Like with like would suggest that non-polar analytes stick more, while priniciples of IMF suggest that polar analytes stick more (induced dipole: dipole forces vs. dispersion forces).

"I remembering reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules." You seem to be basing your post on this statement, which looks incorrect to me. By definition, polar molecules will interact with each other via their dipoles. This means that for any non-polar molecule to interact with a polar molecule, it must be strong enough to break the polar-polar interaction. It's not, because dipole-dipole are stronger IMF. That's why non-polar will interact with each other and leave the polars to themselves.

In many scenarios, you are correct. Here, the polar molecules would be in the gas phase (by themselves) and in diluted solution (in the stationary phase). I'm not sure they get a chance to interact with each other, unless they form immiscible droplets in the stationary phase. If they don't, there would be no occasion to break dipole-dipole interactions.

@SteffX Maybe asking for similar boiling points was not the best idea, maybe I should have asked for same molar mass. This way, all the analytes would have similar Van der Waals interactions. For example, take nonane, dibutyl ether and 1-octanol (Mr = 128, 130, 130). If like interacts with like, nonane should have the highest retention time on a non-polar column, but the retention indices (900, 873, 1071) show that 1-octanol interacts stronger. On a polar column (indices 900, 972, 1558) this is even more pronounced, which is expected from "like with like".

The following is based on comparing retention indices, using two main sources:

Practical retention index models of OV-101, DB-1, DB-5, and DB-Wax
for flavor and fragrance compounds, DOI: 10.1016/j.lwt.2007.07.007
Linear retention indices in gas chromatographic
analysis: a review; DOI: 10.1002/ffj.1887

Difference in retention are due to differences in interaction with the stationary phase:

When
  considering the average gas linear velocity as a constant
  throughout an analysis, the solutes should spend an
  identical period of time in the mobile phase to elute from
  the column. To this extent, only the differences in the
  time spent in the stationary phase are responsible for the
  solutes distinct retention time values.[2]

The two properties of the solute that determine time spent in the stationary phase are vapor pressure (of the solute in the stationary phase) and activity:

In general, the affinity of a solute for the stationary phase
  depends on its vapour pressure and the activity coefficient
  of the solute in that phase.[2]

To summarize, the relevant interactions are between the analyte and the stationary phase. There are no analyte:analyte interactions (not in the vapor phase and not while in solution with the stationary phase). If the stationary phase is non-polar, the retention time will be determined by non-polar:non-polar interactions in the case of a non-polar analyte, and by polar:non-polar interactions in the case of a polar analyte. The question is which of these interactions is stronger.
It is difficult to find two analytes, one polar and the other non-polar, with all other "things" being equal, to address the first part of question. Looking at retention index models based on boiling point and octanol:water partition alone1, you can compare analytes with the same boiling point and with different polarities (based on the partition coefficient).
For polar columns like DB-WAX, it is very clear that polar analytes have longer retention times, i.e. bind more tightly compared to non-polar analytes with the same boiling point.
For non-polar columns like DB-5, there is very little dependence of retention times on polarity, and the model is non-linear (highest retention time for medium polarity, lower retention times for high or low polarity).
One example of actual data is a comparison between eugenol (polar) and tetradecane (non-polar). Eugenol has a boiling point of 254 deg Celsius and a pKow of 2.73, while tetradecane has a similar boiling point of 253 deg Celsius and a pKow of 7.22. The retention index for eugenol is 1360, while that for tetradecane by definition is 1400 (14 carbon atoms). So the more polar eugenol has a slightly lower retention index compare to the non-polar tetradecane on the non-polar column. (Data from 1).

For GC, the most significant factor affecting separation resolution is differences in vapor pressures, boiling points if you will, between analytes.  While column phase selection, oven temperature programming, detector type and temperature, and other factors will have an effect on the efficiency of separation, IMFs are not generally important for the majority of separations.  The "like interacts with like" principle does apply to GC such that in the case you describe, nonpolar analytes will spend more time interacting with the stationary phase while polar will have little to nearly zero interaction and elute first.  The separation of the various nonpolar compounds will, holding all the other column conditions constant, "boil down" to varying vapor pressures.
Edited for clarification:
To begin, I think it beneficial to keep in mind that the question you pose is sufficiently clear we are dealing with polar analytes using a non-polar GC column.  Regarding proprietary column selection, temperature programming, and the choice of various other method parameters, these things are necessarily important in an efficient and successful separation (generally called method development) but they do not do much to answer the principle questions asked about whether 1) polar compounds are less retained on a non-polar column, and 2) does "like with like" apply to GC.  To that end, I will again address these two primary questions and then try to clarify another issue brought into the discussion that does not bear directly on answering the questions as asked.
The most direct answer to your first question is simply, yes, polar analytes do have shorter retention times than non-polar compounds when using a non-polar column because polar compounds have little interaction (partitioning) with the column's stationary phase.  The polar molecules of interest have little affinity for the non-polar stationary phase, i.e. hydrophobicity.  Had it been a polar stationary phase column then IMFs would be significant.  You mention making things comparable by considering compounds of similar boiling points which is precisely why a ramping temperature program is most beneficial for efficiently resolving such a separation.  This too, however, is a method development consideration and thus an example of the applied science of gas chromatography taking advantage of pure science principles.
Considering your second question, this, too, is most directly answered with another yes - the "like with like" concept does apply here, but as you point out, this is a rule of thumb, and as such, a conceptual aid not strict principle.  The qualification about hydrophobic effects in the stationary phase you make in your wording of the question actually explains, as well as I could, what is happening.  Also, as @Blaise points out, the statement with which you end your original question referencing a previous user (I [remember] reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.), is in error: the interaction between polar and non-polar molecules is not stronger than between two non-polar molecules.  Perhaps what this person meant is the interaction between two polar molecules is stronger, and relatively much stronger, than interactions between two non-polars.
With regard to your request I clarify why IMFs are not generally important (in this case), it must be remembered that the analytes in GC are not in "pure liquid form" but, rather, in the gas phase and they are completely surrounded by inert carrier gas molecules - little chance for IMFs to be of significance.  The polar molecules to be resolved do not spend a significant time partitioning into the stationary phase so they cannot form immiscible droplets as you brought up.
With respect to @SteffX, your question is about GC specifically and is not a question of what is the best separation method to use.  I take your question to be about using GC (on small sample sizes, right or wrong on my part) to effect a separation for analytical purposes, not as a preparatory separation like distillation.  Also, your original question(s) are with regard to gas chromatography in general and make no reference to column packing, bore, or length (other than being of a non-polar stationary phase) and I tried to answer in general terms.  However, since capillary GC columns were brought into the discussion, I think it deserves clarifying that, in fact, most capillary column separations are not isothermal.  Without going too deeply into something you were not asking about, it should be noted that isothermal temperature separations are a simpler and therefore preferred method development as long as you have good analyte resolution; the more common practice is to program a temperature gradient to resolve similar boiling point compounds - part of your original Question 1.  Lastly, Van der Waals force(s), a subject for much definitional discussion in its own right, have little importance to your polar analytes at GC conditions, and are moot almost before the sample hits the injector.
I hope this has been helpful and, in my opinion, I think you already have a good understanding of this.

Could you clarify why you say IMFs are not generally important? I'm confused because you also say the differences in vapor pressure are important, but these depend on the strength of IMFs in the pure liquid form of the analytes. Also, it seems to me that column phase selection, which you say is important, would affect IMFs between the stationary phase and the analyte.

I tried to post a reply here but I was too wordy for this section.  I am editing my original answer to reflect your request.  Still new here - Thanks.

I respectfully disagree with @Quip_Boy. If you want to separate compounds based on their vapor pressures, you can just carry out a distillation, which is also based on a gradient in temperatures along the column, whereas a capillary column used in GC is kept at the same temperature all the way in an oven. So there must be something else in the process.
You are quite right citing "like interacts with like", but interaction between 2 non-polar molecules is more than escaping polar molecules (hydrophobic effect). Non-polar molecules can form an attraction because of the Van der Waals force. The Van der Waals force is weak but it can add up a carbon-carbon chain (with the necessary hydrogen atoms, of course) just like with a zipper(1).
(1) i.e. each "hook" on one side of the zipper forms a small attraction to a "hook" on the other side, but altogether, the whole zipper is very strong.

I agree that non-polar molecules interact via Van der Waals force (temporary dipole). However, a polar molecule would interact with a non-polar molecule via dipole:induced dipole interactions, which are supposedly stronger.


text:
Roothaan matrix equation equation orbital coefficients
To go from the Hartree_Fock
$f\psi{_i}=\epsilon\psi{_i}$
to the  Roothaan equation
$FC=SC\epsilon$.
equation we expand the orbitals as
$\psi{_i}=\sum C_{\mu i}\phi_i$
But for Helium atom we have just one orbital function so the coefficient $C_{\mu i}$ should be a vector not a matrix.
My question is  what is the interpretation of the matrix coefficient $C$ in the Roothaan equation  $FC=SC\epsilon$ for the helium atom since we've got just one spatial orbital function?

The $\mathbf{C}$ matrix will always be $K\times K$, where $K$ is the size of the finite basis set you are using. That is to say, with a basis set of size $K$, you will produce exactly $K$ MOs. This means you will never wind up with $\mathbf{C}$ being a vector. In the example you linked, they use a basis set of size 2, so they get 2 MOs for helium and thus the $\mathbf{C}$ matrix is $2\times 2$. While helium has only one occupied spatial orbital, it will have $K-1$ virtual orbitals depending the size of the basis set you use to solve the restricted Roothaan-Hall equations.

So  what you mean is that if $\mathbf{C}$ is  $2\times 2$ matrix, one column of the matrix will be the coefficients of the occupied spatial orbital(the one with smaller $\epsilon$ ) and the other coefficients of a virtual excited state ?

@amiltonmoreira exactly. The only way you wouldn't wind up with a $\mathbf{C}$ matrix per se is if you only used a single basis function, though you could obviously still treat it as a $1\times 1$ matrix.


text:
vapor pressure of one part of binary compound
Let's say we have a binary compound AB in an enclosed container, and let's say there's a tendency to form B-vacancies so that B atoms/molecules escape and form a vapor. Let's assume there's enough of AB that the pressure saturates before AB$_{1-x}$ decomposes too much. Would the vapor pressure of B be the same as for the pure element B?

Usually the formation of vacancies is not accompanied by vapor formation. Now, unless the AB compound is in equilibrium with solid B, then, no, the vapor pressure of B over AB will not be the same as B over B.

@JonCuster - for quite a few oxides, for example, gaseous oxygen evolves when oxygen vacancies form, so I wouldn't say that simultaneous vapor and vacancy formation is a rare case

@voffch - indeed, for oxides that is the case (and is exploited by various oxide sensors since the vacancy concentration affects the conductivity). I usually deal more with metal alloys, where losing constituents to the gas phase is not an issue. Your answer below can be applied to the more general case.

No. The equilibrium vapor (gas) pressure is defined by the corresponding equilibrium. When it's just the element $\ce{B}$ that evaporates, the equilibrium is $$\ce{B(liquid)<=>B(gas),}$$ and the thermodynamics of evaporation (or sublimation if $\ce{B}$ is solid), such as the Clausius–Clapeyron equation, gives you the equilibrium $p-T$ relationship.
When you have nonstoichiometric compound $\ce{AB_{1-x}}$ which loses $\ce{B}$, the process is quite different. Let's imagine purely hypothetical $\ce{AB}$, in the crystal structure of which neutrally charged $\ce{B}$ vacancies, $\ce{V^x_B}$, can be formed. Then, our hypothetical equilibrium between the gas phase and defects in solid can be written in Kröger–Vink notation as $$\ce{B^x_{B}<=>V^x_{B} + B_{gas}}.$$ In this case, until $\ce{AB_{1-x}}$ decomposes too much, the equilibrium constant $K=\frac{p_B\cdot[\ce{V^x_{B}}]}{[\ce{B^x_{B}}]}$ and its temperature dependence define the "partial pressure of $\ce{B}$ - temperature - composition" relationship for $\ce{AB_{1-x}}$.
Please note that the example above is oversimplified for clarity. In the real-world applications of defect chemistry, point defects in nonstoichiometric solids are often charged, and they tend to interact with each other, complicating the equilibria.


text:
What are good examples of questions that make college students think about chemistry concepts?
I am teaching the second semester of general chemistry at the college level. The course tends to get bogged down in calculations, and sometimes we forget about the big picture and what is going on at the atomic level. What are some good questions that would make students think deeply about the following topics:

Intermolecular forces
Kinetics of chemical reactions
Equilibrium and direction of reactions
Acid/base reactions
Redox reactions (electrochemistry)

Perhaps this would be better on meta.

How do I move it to meta.

If it gets closed mod can do it if he agrees.

@Mithoron Meta is for discussion, support and feature requests for the site. Won't this post be better as a community wiki such as Resources for learning Chemistry.

@AvnishKabaj Thing is that's a list of links to other questions. Not unlike https://chemistry.meta.stackexchange.com/questions/3472/the-giant-list-of-duplicates. Well, there were similar posts on main so I guess it may be better here, but I can't say it's particularly good fit for the site.

Well, it's something like https://chemistry.stackexchange.com/questions/37425/what-are-the-most-extreme-chemicals

Intermolecular forces
Are there contradictions about intermolecular forces in the postulates of kinetic theory?
Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?
Kinetics of chemical reactions
Does a first order reaction really get completed only at infinity?
Equilibrium and direction of reactions
What does it mean to shift equilibrium?
At equilibrium are the rates of the forward and backward reactions equal to zero?
Acid/base reactions
Is the behavior of a dilute strong acid the same as the behavior of a weak acid?
Redox reactions (electrochemistry)
Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?


text:
How do I know crystal structures from formula?

Give an explanation why $\ce{MgF2}$ and $\ce{CaF2}$ adopt different structure types.

For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae,  on how I determine this?

You don't. That being said, a post factum explanation is somewhat possible.

Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.

I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.

Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.

Suggestion:  The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. https://www.ccdc.cam.ac.uk/Community/initiatives/cspblindtests/).  Which is why I'm going to replace the link set by you by the one pointing to the sixth one  -- equally published as open access article on Acta Cryst B.

@Buttonwood Good catch, thank you for the edit!

You knew better about USPEX, and me a bit about the contests.  It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.

You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:

Mg(II) and Ca(II) are different cations
Their fluorides form different structures (e.g. the anion is the same)
(you should know) Mg and Ca pretty much always are in +2

So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:

The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
The course covered simple ionic structures (sphere packing)
Sphere packing is only affected by radius
Mg, Ca have different radii. Both are "big"
F is "small"
The reason the structures are different is because Mg, Ca have different radii
(Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]

By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.


text:
Is it possible to convert a combustion reaction directly into an electrochemical one?
I'm fairly inexperienced in chemistry so forgive me if this question is silly but, why is it not possible to convert a combustion reaction to an electrochemical one directly and harvest the electrical energy instead of the thermal energy? For example in my chemistry textbook it says:
$$\ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g) + energy}$$
The book then says: "Even though none of the reactants or products in this reaction is ionic, the reaction is still assumed to involve a transfer of electrons from carbon to oxygen"
my question is, why is it necessary for us to burn fuel in say combustion engines if technically electrons are being transferred, is it possible to create a new engine that will take say one of the hydrocarbons found in gasoline plus oxygen and instead somehow turn the generated energy to the electrical form directly without burning it? Perhaps some complicated engine that uses a cycle similar to the ATP cycle

It's not that easy to do that, but it is in principle how a fuel cell works.

I'll say that any reaction which is spontaneous and violent can be transformed into one that is controlled and electrochemical. Only in some cases you have to take more detours than in others, and depending on the specifics, you still lose some energy into heat.

As @Zhe states, there are methane fuel cells: https://www.futurity.org/fuel-cell-methane-1910432/ . At room temperature, though, methanol is easier to use.

Expanding the previous comments: yes, you need a fuel cell to convert the "chemical" energy to electricity. The basic concept is so old (since 19th century) that even Wikipedia has more or less decent articles on that. Theoretical efficiency of fuel cells surpasses that of combustion engines. Why are we still not surrounded by fuel cells? Let's take the SOFC prototype from @DrMoishePippik 's comment as an example. The materials are expensive (cobalt, rare-earths), working temperatures are still high, long-term stability not well-understood, etc. Meanwhile, it is cheaper just to burn fuel.

Thanks everyone, I didn't know anything about fuel cells before seeing the comments.


text:
What is the radius of the atoms?

In a recipient there is an unknown monoatomic gas that occupies a volume of $\pu{230 cm3}$, at $\pu{300 K}$ and $\pu{1 atm}$. It is known that the atoms occupy a volume of $2\times10^{-4}$. Find the radius of the atoms.

I used the ideal gas law to calculate the number of moles, then I calculated the number of atoms, then divided the volume occupied by the atoms by the number of atoms to find the individual volume of each atom, then applied the formula for the volume of a sphere to find the radius, the problem is that I found 92x10^-10 and the correct answer is 1,25x10^-10, can anybody spot the mistake? Thank you.
Here are the maths I did:

Your first error is for $n$. The value you have for R is wrong. $R= 82.057338(47)\ \mathrm{cm}^3 \mathrm{atm}\  \mathrm{K}^{−1} \mathrm{mol}^{−1}$.

Your right, but its still not enough to get to the right answer.

Leo - Well fix that part of the math and work the correction through the problem...

… and always write the values with the correct units and carry the units through the calculations.

I haven't dived into your math, but I think there are 3 things that are confusing:

intermediate calculations, even though you weren't asked to do any;
missing units and inconsistent notations (e.g. base 10 is not always shown);
ambiguous phrase "atoms occupy a volume of $2\times 10^{-4}$."

I always find it cleaner and more productive to derive the formula first, and only then plug in the numbers minding the proper units. As for the volume issue, I tend to believe that $2\times 10^{-4}$ means the fraction of the volume occupied by gas atoms and it is not the absolute value expressed in cubic centimeters. In fact, in this case you get the correct answer and you don't need to use gas volume ($\pu{230 cm3}$) at all.
Check this out. Radius of an atom $r_\mathrm{a}$ is indeed can be found from the volume of an atom $V_\mathrm{a}$
$$r_\mathrm{a} = \sqrt[3]{\frac{3V_\mathrm{a}}{4π}}.$$
At the same time
$$V_\mathrm{a} = \frac{V_\mathrm{atot}}{N}$$
where $V_\mathrm{atot}$ – volume of all gas atoms depicted with spheres; $N$ – number of the atoms that can be found via Avogadro's number $N_\mathrm{A}$ and amount $n$:
$$N = nN_\mathrm{A}.$$
Finally, you are allowed to use ideal gas law to determine the amount $n$:
$$n=\frac{pV}{RT}$$
where $p$ – pressure; $V$ – volume of the gas (macro scale); $R$ – the molar gas constant; $T$ – temperature. Now, lets's put all the things together assuming that  $V_\mathrm{atot}/V = 2\cdot 10^{-4}$:
$$
\begin{align}
r_\mathrm{a} &= \sqrt[3]{\frac{3V_\mathrm{atot}RT}{4πpVN_\mathrm{A}}}\\
  &= \sqrt[3]{\frac{3\cdot2\cdot10^{-4}\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{300 K}}{4\cdot 3.14\cdot\pu{101325 Pa}\cdot\pu{6.022e23 mol-1}}}\\
  &= \pu{1.25e-10 m}
\end{align}
$$
Now that's the value that actually makes sense:)

Thank you, my units were really mesy and I got confused if the volume occupied by the atoms was a pure volume or a fraction of the volume, thank you again.

No prob, keep it up:) By the way, as you are probably using $\mathrm{\LaTeX}$ judging from the image you posted, I'd suggest to use siunitx package. It helps to format units and numbers according to SI and ISO standards and is very versatile.


text:
Must a sample of a deuterated compound be isotopically pure for kinetic isotope experiments?
I'm trying to do a project on the primary kinetic isotope effect on the dehydroxylation of $\alpha-FeOOH$, goethite. The goethite consists of edge-sharing double octahedra of 3 oxyanions and 3 hydroxyanide. The double octahedra are joined at corners by the oxyanion or hydroxide, leaving empty channels across which hyroxides and oxyanions form hydrogen bonds.

The dehydroxylation mechanism is believed to consist of proton diffusion from the bulk of the material to the surface (likely facilitated by a series of protonations down the hydrogen-bonding channels), then binding of hydrogens to surface hydroxyl group, forming water desorbing. The proton diffusion is believed to be the rate-determining step.
I thought I could lend more evidence to whether this is the case by performing the dehydroxylation on a deuterated compound and a compound of natural isotopic abundance. However, I cannot find the deuterated compound for sale and will have to synthesize it. The most facile synthetic method given time constraints involves potassium hydroxide. We could use deuterated hydroxide, but that would be more expensive. Grey et. al.** were able to do this synthesis with the undeuterated hydroxide, but they didn't report an H/D ratio.
All the papers I've come across on kinetic isotope effect experiments work under the assumption that their deuterated compounds are isotopically pure. It may just be a matter of convention, but would I actually need to do this experiment on an isotopically pure sample? Could I use a sample with, say, an isotope ratio halfway between pure deuterated and natural abundance? Would the calculated kinetic isotope effect (defined as hydrogen rate constant/deuterium rate constant) then be twice what I measure with this partially-enriched sample?
*Surface and Bulk Reactivity of Iron Oxyhydroxides - A Molecular Perspective
Xiaowie Song
Umea University
**2H MAS NMR Studies of Deuterated Goethite (α-FeOOD)
Kathryn E. Cole,†,§, Younkee Paik,†,§, Richard J. Reeder,‡,§, Martin Schoonen,‡,§ and, and Clare P. Grey*,†,§
The Journal of Physical Chemistry B 2004 108 (22), 6938-6940
DOI: 10.1021/jp0486090

There are two approaches to measuring kinetic isotope effects.
One is to have two completely separate, isotopically pure systems, one heavy, and one light. In each system you can measure the rate of the reaction however you want.  You don't have to be able to measure the rate in an "isotopically resolved" way, just measure the bulk rate.  Each of your systems is isotopically pure, so the rate of each system is 100% reflective of the isotope composition of that system.
The other way is to have a single system with variable/unknown isotopic composition.  You measure the rate of your reaction, but you have to be able to measure (e.g.) the rate of depletion of each substrate isotopomer separately.  (Or the rate of formation of each product isotopomer, but that sounds very difficult in your system.)  For example, if you had a mixture of $\ce{CH4}$ and $\ce{CD4}$ and you were interested in the kinetic isotope effect of some kind of combustion reaction of those gases with oxygen, you could measure the ratio of $\ce{CD4}$ to $\ce{CH4}$ over time, e.g. using mass spectrometry.  It would increase as the $\ce{CH4}$ is preferentially depleted by the combustion reaction.  You could also measure the rates of formation of $\ce{H2O}$ and $\ce{D2O}$.  The rate of formation of $\ce{H2O}$ would be much higher than that of $\ce{D2O}$.  (The exact numerical values here would depend strongly on the nature of your combustion system, e.g. temperature, catalysts used, etc.)
So iff you can measure $\ce{FeOOH}$ and $\ce{FeOOD}$ separately in the same system, then you don't need an isotopically pure material.  I'm not an expert in these types of measurements, but it sounds very hard to me.  Probably if you don't already know how to do it, it will be cheaper just to buy $\ce{KOD}$ and get on with the synthesis of an isotopically pure material so you can use the first approach.


text:
Why do you use acetic acid instead of water to create your standard riboflavin solutions?
I must create my own lab and I was going to use water, however, a similar experiment online says that acetic acid should be used to create my standard riboflavin (vitamin B2) solutions.
Why is that?

Please provide more information, for instance links to relevant protocols that might clarify the origin of the differences. Of the cuff, I'd say solubility though.

https://www.stellarnet.us/wp-content/uploads/StellarNet-Exp3_Fluor_Riboflavin.pdf

From Sigma product information:

The flavins are light sensitive and unstable in alkaline solutions.  The dry solid is stable to diffuse light, but is highly photolabile in solution, especially in alkaline solutions. Neutral and acidic solutions are stable in the dark (3% decomposition per month at 27 °C at  pH 6.0).  Acidic solutions may be sterilized by autoclaving.

So making a slightly acidic solution probably helps to prevent decomposition. If you dissolve it in pure water without a buffer, the pH is not well defined.


text:
Ionic compound having highest solubility in water
Which ionic compound has highest solubility in water? I can find CsBr having highest solubility with 1230 g/L at 25 °C.
Note: compounds like ethanol are soluble to any extent in water, but they are covalent, not ionic.

It would be interesting to split in the two cases of molar solubility and mass solubility, though the latter is easier to find data on directly.

I could be cheeky and add Ionic Liquids to the list, which often mix with water at any ratio. Maybe clarify that you are talking about solids at standard conditions.

@TAR86 Darn, I was just a bit late with my edit! I'll leave that sidenote in my answer anyway.

Caesium salts are unapologetically ionic, and they typically have quite high mass solubilities in many solvents, including water.
Assuming organic ions are allowed, caesium acetate ($\ce{H3CCO2^-Cs+}$) in particular has a remarkably high solubility of 9451 g/kg water at −2.5 °C, increasing to 13 455 g/L water at 88.5 °C.
Caesium formate ($\ce{HCO2^-Cs+}$) is also quite soluble, with a solubility of 4880 g/kg water at 20 °C, resulting in 2.56 L of solution with a density of 2.297 g/mL (reference, .docx file). However, its solubility increases much faster with increasing temperature (J. Chem. Soc., Trans., 1922, 121, 1837-1843). At 100 °C, it reaches an outstanding value of 20 071 g/kg water! That's 11.6 molal, or roughly 20-25 mol/L assuming the density doesn't change too much. This saturated solution is 67.7% cesium formate by number of moles, which means more than two caesium ions and two formate ions per molecule of water.
I believe I have read somewhere that caesium formate is the record holder for highest mass solubility in water (evidently only at high temperatures). If this is not true, then I can scarcely believe it will be topped by much.
Tangentially, Ivan mentions Clerici's solution, which is actually a mixture of thallium(I) formate and thallium(I) malonate. The mixture doesn't count (though the individual components are quite soluble themselves), but it interesting to analyse. Apparently 300 g of each compound will dissolve in 40 g of water without saturating it room temperature (ref), giving a lower bound to their combined solubility of 15 000 g/kg water. This value rises with heating, and is the only way I can see to beat the mass solubility of caesium formate.
For further entertainment, I recommend these two solubility tables with a large number of entries (1, 2). The second one can be ordered by solubility at different temperatures. It's interesting to see the variety of cations and anions which can be combined to display extreme mass solubility.
Edit: If room-temperature ionic liquids are allowed, then it is quite likely some of them are miscible with water in any proportion, which is effectively "infinite solubility in water". Something as simple as ethylammonium nitrate ($\ce{C2H5NH3+NO3-}$) may suffice.

If they are ionic, the state of matter does not matter.  +1.

The following data is compiled from [1, pp. 4-44, 5-167]:
Table 1. Selected solubility values of the inorganic compounds with significant ionic character at $25~\mathrm{^\circ C}$.
$$
\begin{array}{lc}
\hline
\text{Formula} & \text{Solubility in water}/\pu{g L-1}\\
\hline
\ce{CsF} & 5730\\
\ce{SbF3} & 4920\\
\ce{LiClO3} & 4587\\
\ce{Pb(ClO4)2} & 4405\\
\ce{ZnCl2} & 4080\\
\hline
\end{array}
$$
Solubility of antimony(III) trichloride $\ce{SbCl3}$ is $9870~\mathrm{g~L^{-1}}$ at $25~\mathrm{^\circ C}$, but technically it's not an ionic compound.
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97.

I've seen antimony trichloride before in tables, but it is apparently very easily hydrolysed, so perhaps it shouldn't be counted either way.

@NicolauSakerNeto Yep, you are right, and the same probably goes for $\ce{ZnCl2}$. I also omitted $\ce{ZnBr2}$ for similar reason (and it's covalency, too).

There is not going to be a single definitive answer, primarily because of a wide gray zone surrounding the domain of ionic compounds. Besides, as Nikolau noted, the question is ambiguous.
If you want mass concentration, then look at $\ce{InI3}$ which claims a whopping $13100~\mathrm{g/L}$. Pity that it is probably ionic in name only, judging by the solubility in non-polar solvents. Well, then look at those mentioned by andselisk, though the ionic nature of some of them is also debatable, and then at the thallium formate (a component of Clerici solution) with $\sim5000~\mathrm{g/L}$.
If you want molar concentration, then the question is still ambiguous (are we looking at molarity or molality?), and the pretty strong contenders are $\ce{NaOH}$, $\ce{BeF2}$, $\ce{LiClO3}$.
So it goes.

True, $\ce{InI3}$ is weird, but it's definitely not ionic and the reference for the solubility value dates back to 1940s or something:)

We can do better than that.  Ammonium nitrate = 1500 g/L at 20°C.


text:
Major product for aldol condensation between acetophenone and acetaldehyde
Crossed aldol condensation is done between acetaldehyde and acetophenone by using NaOH as the base. So, here we can get 4 possible products. But which one would be the major?
I know that acetaldehyde will act as electrophile to which addition will occur. But what will act as nucleophile to give major product: acetaldehyde or acetophenone?
I don't have any solid reasoning whether acetaldehyde enolate or acetophenone enolate will give major product with acetaldehyde.

The so called 'cross aldol condensation' between a ketone and an aldehyde is actually called 'claisen-schmidt condensation'

And for the answer... it's generally easier for a nucleophile to attack on an aldehyde as compared to a ketone (steric and polar factors both can be used to explain that)

All the reactions leading to the 4 possible products are reversible so you will get the thermodynamic product.

@Carrick i know that neucleophile attack on aldehyde is better. But my question is which is better nucleophilic aldehyde or acetophenone enolate.

@Waylander But which out of the 4 is thermodynamic that is more stable product.

The product derived from acetophenone enolate

@Waylander can you explain why is that major? The reason for that enolate being more stable?

That product is the major because the extended conjugation is energetically more favourable.

Ha, when you think the Bordwell or Evans $\mathrm pK_\mathrm a$ tables have the answer but they don’t: no value for acetaldehyde. (:

I would expect the aldehyde to be the major electrophile in this reaction. The reason is that the steric demands for a nuelophile coming close to the carbonyl carbon are lower than for the same nucelophile coming close to a ketone.
I would expect that the enolate formed from the reaction of the methyl phenyl ketone (acetopheone) to be slightly more stable than the one formed from ethanal. This is because the enolate can be delocalised into the benzene ring.
If we were to form the enolate from ethanal then if this was to attack the carbonyl carbon of acetophenone then we would get a tertary alcohol as the product. The major aldo product in this reaction will be a secondary alcohol formed by the action of the enolate from acetopheone on the ethanal.


text:
A wrong trivial name?
Isopropyl chloride should have been called secondary propyl chloride, right? The first iso must be isobutyl chloride, methinks.

Trivial names can't be wrong by definition.

Oh, it must be a matter of common acceptance, right?

Yeah, that's right.

In this case sec-propylchloride and isopropylchloride coincide.  The prefix “iso” has several interpretations, but generally means that a terminal carbon (methyl group) is moved down the chain by one, which for propylchloride means that a methyl group is attached to what was carbon 1 (the one with the chloride).
So you could call it either sec-propylchloride or isopropyl chloride, neither would be wrong, but it is desired to standardise names when possible, and isopropyl was chosen.  AFAIK the choice of iso- over sec- was arbitrary, but as Neretin says, trivial names can’t be wrong by definition.  I suspect, although I don’t know, that iso was chosen because it was known that the molecule was similar to propylchloride or had the same formula (thus an isomer) before the actual structure was known.
One should be aware that “iso” is sometimes also used in a looser sense to refer to any member of the class of isomers of a compound.

I follow your point; yet i feel if we standardise the prefixes 'iso' and 'neo',  "isoalkane" can be  generalized as the isomer having  one additional methyl group on the second carbon and "neoalkane" having two methyl groups on the second carbon; also  substituted alkanes like  "isoalkyl halide" & neoalkyl halide"have halogen at the distant end of the parent chain. To make this point more convincing, first 'neo' alkane is neopentane and first neoalkyl chloride is neopentyl chloride;  first iso alkane is isobutane and so first isoalkylchloride must be isobutylchloride.

As a teacher, i used to explain students that way and ask them to remember isopropyl to be an exception; then they are quite convinced.

Isopropanol and isopropyl chloride are not exceptions if you make the slight adjustment that “iso” compounds have the methyl at the far end moved down by one to the second-to-last carbon, rather than the second carbon.

You can think of “sec-“ and “tert-“ as adjusting the functionalised (business end) of the molecule, and “iso-“, “neo-“ as adjusting the other end.  A good example is sec-isoamyl borane (siamyl)₂BH, the bulky reducing agent where “amyl” means 5-carbons. “Sec” moves the boron up from C1 to C2, and “iso” moves the last carbon C5 from being attached to C4 down to C3.


text:
Acidic, Basic, and Neutral Medium
So I'm new to this concept and I can't find any proper explanation to this. What does it mean for a reaction to occur in an acidic, basic, or neutral medium?

It's just a matter of pH of reaction mixture. As far as redox is concerned in high pH is "milder".

So does that mean that the redox reaction in lower ph will be more vigorous?

Not necessarily. Often there's different reaction pathway if difference in pH is big.

Description of reactions being conducted in "acidic, basic, or neutral medium" are rather qualitative terms which don't add anything until and unless more details are provided. The first clue is that the concept of pH is dedicated to purely aqueous solutions. All it indicates that an acid or a base was present during the course of a reaction. Neutral usually implies that either the pH was brought close to 7 or no acid or base was intentionally added. pH can be defined for organic solutions but it is more interesting to physical chemists rather than synthetic chemists and it is very hard to quote a pH value when organic solvents are present.
For example titration of iron (II) and potassium permanganate is conducted in a highly acidic medium. This is accomplished by adding a small amount of ~ 2 M sulfuric acid. It does not proceed quantitatively in a neutral medium.

Why do we call it an acidic medium instead of acidic solution?
This is just a guess, but redox reactions can take place in solution, on surfaces, in the vapor phase, underground, in pastes found in batteries etc. So to have a single word instead of listing all the different environments, we just say the redox reaction takes place in a "medium" that is acidic, neutral or basic.
There are similar collective terms for other situations in chemistry. If you want to talk about reactants, product, catalysts and spectators at the same time, you can refer to them as "species". If you want to talk about atoms, molecules and ions at the same time, you can refer to them as "particles".


text:
What exactly is a reducing equivalent and what does it do?
I've encountered this term in the context of cellular metabolism, but I can't seem to find an explanation of what a reducing equivalent is or why it is named this way. Wikipedia was not really helpful, and most other online sources seem to use the same phrasing (which I am struggling to parse).
Is a reducing equivalent something that is itself reduced, or is it something that causes reduction (thereby being itself oxidized in the process)? The Wikipedia article uses the expression "which transfer the equivalent of one electron" which has left me all the more confused.
I would be most grateful for an explanation or links to material I could read to clarify this for myself.

The number of "equivalents" of a reagent refers to how much of the reagent there is relative to another reactant of interest. To use a concrete example, if we are interested in the $\mathrm{S_N2}$ transformation $$\ce{CH3Cl -> CH3Br},$$ a possible way of bringing this about would be to treat the starting material (methyl chloride) with a certain amount of bromide ion. If we carry out the reaction on 0.5 mol of methyl chloride, in theory we would only need 0.5 mol of bromide ion, because of the stoichiometry of the reaction:
$$\ce{CH3Cl + Br- -> CH3Br + Cl-}$$
In this case, 0.5 mol of bromide ion would be referred to as being "1 equivalent" of bromide (relative to methyl chloride, which is the reactant of interest).
We could also say that for the complete combustion of hexane, we would (theoretically) need "9.5 equivalents" of oxygen gas:
$$\ce{2C6H14 + 19O2 ->12CO2 + 14H2O}$$
In the case of a redox half-reaction, you usually have electrons in either the reactants or the products:
$$\ce{Fe^3+ + e- -> Fe^2+}$$
(this highly simplified reaction is relevant in, e.g., cytochrome c's role as an electron carrier in oxidative phosphorylation). Under biological conditions, this is rather unrealistic, as electrons don't actually exist in aqueous solution, nor is there any electrolysis going on. So, you need something else to act as a supplier of electrons, i.e. a reducing agent, such as NADH.
$$\ce{NADH + H+ -> NAD+ + 2H+ + 2e-}$$
In this case, because 1 mol of NADH provides 2 mol of "electrons", you could say that NADH acts as two reducing equivalents. Here, NADH is acting as a reducing agent, and is itself oxidised to NAD. [I don't agree with the Wikipedia page about hydride ion being one reducing equivalent; it should be two.]


text:
Structure of (CH3COO)3Fe
I'm a grade 12 student trying to comprehend MOT & LFT in Co-ordination compounds.
For one of our salt analysis(practicals), the test for acetate ion includes the addition of Neutral Ferric Chloride to salt solution.
This results in the formation of the above Co-ordination compound.
I've been obsessed, and I'm probably at the brink of insanity trying to comprehend it's structure via MOT/LFT. I have seen many Lewis structures but I'd really know to know it in terms of MOT. I assume it's octahedral in shape, with the 2 oxygen atoms of the carbonyl group acting as ligands; But I'm not sure, hence the assumption.
Thanks!

Have you seen this ?

https://link.springer.com/article/10.1007%2FBF01393501

@YUSUF HASAN yes I have, but it seems to be of no avail.

Related: https://chemistry.stackexchange.com/questions/43187/what-is-the-product-of-the-chemical-reaction-between-phenol-and-ferric-chloride/43216?r=SearchResults#43216


text:
Is Sulfuric hydrofluoric acid a real thing?
I saw a video on social media of a person disappearing into air when exposed to Sulfuric hydrofluoric acid vapours. I don't have much knowledge in chemistry. Is this possible combination.

I saw a video of a dude made of liquid metal, he used to stab people with swords made of his own hands.

Yes, it's a real compound commonly known as Fluorosulfuric acid. Not the strongest among superacids, but is definitely corrosive. However, that thing you saw would be called a witchery 500 years ago; these days it's called BS of fake news (as most of the social media is:) ).

There is an actual compound called fluorosulfuric acid, $\ce{HSO3F}$ as mentioned by @andselisk. It is one of the strongest acid known. It hydrolyzes slowly to sulfuric acid and hydrofluoric acid. It is highly toxic and corrosive in nature. If it comes in contact with human skin, it will eat your skin right away leaving no trace and so it will help you to disappear happily. Anyway, jokes aside, fluorosulfuric acid disappearing people is absolute nonsense. Don't try any experiment with fluorosulfuric acid. It is highly toxic and I highly recommend to stay away from it.


text:
Flow rate of the fuel consisting of ethanol and water burning in air of a given flow rate
I hope that someone could give me some hints or tips on how to approach this question, even though these types of questions are usually closed off as homework.

A burner is run by a fuel mixture of $\pu{90 mol.\%}$ ethanol and $\pu{10 mol.\%}$ water. Assume a complete combustion with air ($\pu{79 mol.\%}$ $\ce{N2}$, $\pu{21 mol.\%}$ $\ce{O2}$). The whole process takes place at $\pu{100 kPa}$.
Calculate the inlet flow rate of the fuel given that the flow rate of air is $\pu{100 mol s-1}$ and that the excess of air is $20\%$.

The answer key only gives the answer $\pu{6.48 mol s-1}$ of the fuel without explaining the calculations.
My attempt at the question was:
$$\ce{C2H5OH + H2O + 3O2 -> 2CO2 + 4H2O}$$
Knowing that we have $20\%$ excess in air, we could calculate the theoretical moles needed.
$${\text{fractional excess} = (n_\mathrm{feed} - n_\mathrm{theo})/n_\mathrm{theo}}$$
$$\to n_\mathrm{theo} = \pu{83.3 mol}$$
I then used the molar ratio from the reaction to calculate how much ethanol and water was needed. I just don't see how the answer becomes $\pu{6.48 mol s-1}$. Any help is greatly appreciated!

A burner combusts fuel. What takes part in such reactions?

Also, the question looks like just stoichiometry. Why did you include a heat exchanger etc?

@EashaanGodbole I added the water in the reaction since they stated in the problem that the fuel consisted of both ethanol and water, although I know that a combustion reaction does not include water in the reactants. The diagram of the system was given, there are other questions which involve the heat exchanger specifically.

The inlet water should not be included in the balanced chemical reaction formula, and you balanced the formula incorrectly.  Your first step should be to get that right.

I think you overcomplicate the problem. There are many variables you don't need and many components that are duds (nitrogen, water, pressure). What important is that we burn ethanol:
$$\ce{C2H5OH + 3 O2 -> 2 CO2 + 3 H2O}$$
so that the ratio between ethanol and oxygen is $1:3$. I have taken chemistry engineering classes long time ago and I don't remember exact notations, so I'd rather denote molar flow rate with an n-dot notation:
$$\dot{n} = \frac{\mathrm{d}n}{\mathrm{d}t}$$
where $n$ – amount; $t$ – time. Since fuel is only 90% alcohol, then
$$\dot{n}(\text{fuel}) = \frac{\dot{n}(\ce{C2H5OH})}{x(\ce{C2H5OH})}$$
where $x$ denotes molar fraction. From the combustion reaction
$$\dot{n}(\ce{C2H5OH}) = \frac{\dot{n}(\ce{O2})}{3}$$
and for the air the following is fulfilled:
$$\dot{n}(\ce{O2}) = x(\ce{O2})\frac{\dot{n}(\text{air})}{1 + α}$$
where $α$ is the partial excess of air. Now we can put everything together and plug in the numbers:
$$
\begin{align}
\dot{n}(\text{fuel}) &= \frac{x(\ce{O2})\dot{n}(\text{air})}{3x(\ce{C2H5OH})(1 + α)}\\
  &= \frac{0.21\cdot\pu{100 mol s-1}}{3\cdot 0.9\cdot (1 + 0.2)}\\
  &= \pu{6.48 mol s-1}
\end{align}
$$
One hell of a burner.

You calculated 83.3 moles of air reacting with the ethanol.  Since air is 21% oxygen, this means that 83.3 x 0.21 = 17.5 moles of oxygen react with the ethanol.  Since each mole of ethanol reacts with 3 moles of oxygen, the number of moles of ethanol that react is 17.5/3=5.833 moles ethanol.  Since the fuel mixture is 90% ethanol, there are 10 moles of fuel mixture for every 9 moles of ethanol.  So the number of moles of fuel mixture are 5.833 x 10/9=6.48 moles of fuel mixture.


text:
Difference between adiabatic and nonadiabatic tunneling for molecule
What is the difference between adiabatic and nonadiabatic tunneling for a molecule in a laser-field?
Does nonadiabatic tunneling necessarily relate to the Conical intersection or curve-crossing?

there is a short answer here https://chemistry.stackexchange.com/questions/100458/what-is-the-difference-between-non-adiabatic-coupling-and-diabatic-coupling

Can you provide a reference for these terminologies? I have never heard of non-adiabatic or adiabatic tunneling. I have seen things similar to what I would guess this is using both diabatic and adiabatic representations. Generally though, nonadiabatic almost always has to do with conical interesections.

Hi here is one of the references..https://aip.scitation.org/doi/10.1063/1.463981

Look at D. Yarkony, Chem. Rev., 2012, 112 (1), pp 481–498 for a thorough review of the subject.

This is very far from a complete answer, especially w.r.t the part of your question regarding "laser fields". After doing some reading, nonadiabatic tunneling is conceptually the same as adiabatic tunneling except that one allows for coupling to another adiabatic state.
Adiabatic tunneling is the phenomenon where a particle described by a wavefunction is propagating on a potential and begins with less energy than the height of the barrier it approaches. One finds, however, that there is amplitude of the wavefunction on the other side of the barrier despite the particle having insufficient energy to cross the barrier.
Now, in physics and chemistry, nonadiabatic processes typically describe those processes where it is a bad approximation to consider only a single potential energy surface. That is, both surfaces may be adiabatic states, but these states are coupled in some manner. In general, there will always be some small coupling between states that are not too far apart in energy.
Nonadiabatic tunneling just describes the situation that one does not neglect this coupling between states. To a first approximation, nothing will change and one will find the same amplitudes as a function of position and time. If you consider the presence of the higher-energy adiabatic state as a perturbation of the Hamiltonian of the original particle, then the zeroth order picture will just be adiabatic tunneling. The next term will be something like the case where the particle is either reflected or transmitted over the barrier, but it does so by first "visiting" the higher-energy adiabatic state.
Now, with that description in mind, I think it is reasonable to say that nonadiabatic tunneling does not necessarily have to be involved with with conical intersections or curve crossing. Nonadiabatic tunneling is just regular tunneling which allows other states to couple with the original state. Now, of course, conical intersections and curve crossing could very well come into play in nonadiabatic tunneling because this would just be another form of coupling one could introduce.

As an aside, the description I give where the higher-energy state is a perturbation on the original system is probably not a good one in general as one could imagine very strong coupling between the potentials (as would be the case with a conical intersection).


text:
Favorskii rearrangment
Why does the alkoxide anion dont act as a nucleophile in the fevorskii rearrangement?
It can displace halide as halide is alpha to carbonyl and carbonyl decreses the energy of c-halide lumo

It can but because intremolecule reactions are much faster then intermolecular reactions you get mostly the rearrangement product.

SN2 reactions on secondary alkyl halides is not that fast.


text:
How is a NMR spectrum obtained?
I am reading about NMR, and from what I'm understanding it should give information on the transition energies in the spectrum of the nuclear spin in a magnetic field.
What I don't understand is how this information is accessed during the experiment.
The NMR measurement is usually described as: a magnetic field $H_0$ polarizes the sample along a certain direction; then a short pulse of an auxiliary magnetic field alters the direction of the magnetization, which then starts to precess around $H_0$. The changing magnetization induces a current the coils of the NMR machine which is measured and gives the precession frequency. The magnetization gradually aligns with $H_0$, leading to a diminishing current intensity. The decrease in intensity gives the nuclear relaxation rate.
A first question is: is the above description of the measurement correct?
A second question is: How do we extract the information on the energy levels $E_m$, or their separation, from the precession frequency? If not from the precession, how is the energy spectrum obtained?

The description is simplified, but actually fairly accurate. You can find all the details in a specialised NMR book (other general texts usually do not go into much detail).
Quantum mechanically, you can show that the precession frequency (Larmor frequency), denoted $\omega_0$, is related to the energy difference between the up and down spin states of a spin-1/2 nucleus. The energy levels are
$$\begin{align}
E_\alpha &= +\frac{1}{2}\hbar\omega_0 \\
E_\beta &= -\frac{1}{2}\hbar\omega_0 \\
\end{align}$$
and so the transition occurs at the frequency $|E_\beta - E_\alpha|/\hbar = \omega_0$, so if we can identify how much precession is happening at the frequency $\omega_0$, that tells us how strong the corresponding signal should be in the spectrum.
The initial data that you get is in the form of a free induction decay (FID), which plots the amount of detected magnetisation against time. The process of extracting the frequencies from the raw time-domain data is accomplished by a Fourier transform.
Feel free to ask if you would like any part of this to be elaborated. I intentionally did not go into great detail, partly because that is the role of a textbook, but also partly because it can get pretty complicated very quickly.
(There is also a complicated sign convention to do with the definition of $\omega_0$, and different books define it differently. Occasionally the definition even depends on the sign of the gyromagnetic ratio $\gamma$. I would advise to not worry too much about plus/minus signs.)

Many thanks for your answer! Before accepting I would like your confirmation on the following. A problem I had earlier was how to extract the frequencies from the macroscopic magnetization, $M=\sum_i Tr{\left[ I_i\hat\rho \right]}$, but your mention of the FID and its fourier transform made me realize that, in the time evolution of that average, one should get factors of $exp(-\frac{it E_m}{\hbar})$ coming from the energy eigenstates, and it is from those factors we should get the oscillations in the FID (and by Fourier tr. , the spectrum). Is this how it works?

@tbt, the spectrometer measures the $x$- and $y$-magnetisations (like in your equations, $M_x = \operatorname{Tr}(I_x\rho)$ and $M_y = \operatorname{Tr}(I_y\rho)$). We then fuse these two quantities together into a complex magnetisation $M = M_x + \mathrm i M_y$, which evolves in time $\sim \exp(i\omega t)$, or $\exp(iEt/\hbar)$ (in NMR usually everything is expressed in frequencies, so we prefer the former ;) ). So you are absolutely right!

Thank you again, orthocresol. The $i$ in my formula was meant to indicate the sum over the $N$ atoms in the sample, not the spatial direction (which i left implicit) but thanks for the clarification anyways. I have accepted your answer:)