text:
Finding pH after dilution?

Assume that $\ce{pKa}$ is 4.76
Mix $\pu{50 mL}$ of $\pu{0.1M}$ $\ce{HA}$ and $\pu{50 mL}$ $\pu{0.1M}$ $\ce{NaA}$. This is solution (i). Calculate concentrations of $\ce{HA}$ and $\ce{A-}$, measure the $\ce{pH}$.
Then, mix $\ce{10 mL}$ of solution (i) and $\ce{90 mL}$ of water. Calculate $\ce{[HA]}$ and $\ce{[A-]}$, and measure the $\ce{pH}$.

How do I approach these calculations? I have 13 of these to do, these are the first two.
Equation is:
$$\ce{pH} = \ce{pKa}+\log{\left(\frac{\ce{[A-]}}
 {\ce{[HA]}}\right)}$$
For the first one, I got,
$$4.76 + \log{\left(\frac{0.1}{0.1}\right) = 4.76}$$
But I don't know if that's correct. How would I approach the subsequent problem when water is added?

I have edited your post with MathJax. You can learn how to use MathJax to write chemical equations here: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

Moreover I changed A to (i), so that we don't confuse with A anion.

The solution (i) is a weak acid and it's conjugate base type of buffer, the $pK_a$ value matches that of acetic acid. Although the $pH$ of a buffer remains unchanged upon reasonable dilution, $10 \pu{mL}$ of solution (i) and $90 \pu{mL}$ of water will have quite an impact. Have you tried writing the ionic equilibrium equation of the dissociation of weak acid $\ce{HA}$? You can probably apply common-ion effect for the concentration of $\ce{H+}$ (due to water).

@Tyberius: I think you're neglecting Berry Holmes' comment above. Dilution is not always inconsequential to the pH, especially when concentrations get quite low. As my analytical chemistry teacher once put it, if you squeeze a lemon into a lake...
@notiarogers: Tyberius already showed you that you miscalculated the formal concentrations of $\ce{HA}$ and $\ce{A^-}$. Don't skip any steps: $50 \space mL$ of $0.1 \space M$ solution contain $50 \cdot 0.1 = 5 \space mmol$ of substance; the final volume is $100 \space mL$, therefore the final concentration is $5 \space mmol / 100 \space mL = 0.05 M$.
The formula used above (for the calculation of the pH of an aqueous buffer solution made from a weak acid + its salt with a strong base) is only an approximation. Its main assumption is that the concentrations of acid $\ce{HA}$ and anion $\ce{A^-}$ are equal to the nominal ones (0.05 M, as explained above). In reality, when you dissolve $\ce{HA}$ in water, some of it already dissociates (forming some $\ce{A^-}$), and when you dissolve $\ce{NaA}$ in water, it dissociates to $\ce{Na^+}$ and $\ce{A^-}$, and some $\ce{A^-}$ already reverts to $\ce{HA}$ (which, BTW, is why the former solution is slightly acidic and the latter is slightly basic). So things are quite a bit more complicated than it seems: even your initial solutions don't contain 'only' $\ce{HA}$ or 'only' $\ce{A^-}$.
Now, it may be that whoever is asking you to do this calculation wants you to stick to such approximation, and in fairness the final concentration after dilution is probably still large enough, unless you want a very precise result.
In case they want you to do a more complicated calculation, you can write an equation for each equilibrium, one for each mass balance and one for the charge balance. Your equation is the dissociation equilibrium of $\ce{HA}$: $$\ce{K_A=\frac {[H^+] \cdot [A^-]}{[HA]}}$$
You can add the dissociation equilibrium of water: $$\ce{K_w=[H^+] \cdot [OH^-]}$$The mass balance of all acid-related species is:$$\ce{[HA] + [A^-] = C_{HA} + C_{NaA}}$$ where $\ce{C_HA}$ is the nominal concentration of $\ce{HA}$, i.e. 0.05 M as calculated above, and $\ce{C_{NaA}}$ is the nominal concentration of $\ce{NaA}$, i.e. also 0.05 M in this case.
The charge balance is: $$\ce{[H^+] + [Na^+] = [OH^-] + [A^-]}$$
And as $\ce{[NaA]}$ is a strong electrolyte, we know that: $$\ce{[Na^+] = C_{NaA}}$$
If you put all these equations together and eliminate all unknowns except $\ce{[H^+]}$, you get a 3rd degree equation in $\ce{[H^+]}$ and known constants alone, which you can solve (probably best numerically) to find the pH.
The exact solution would look very different from the approximate one you were using, which was essentially:$$\ce{[H^+] = K_A \cdot \frac{C_{HA}}{C_{NaA}}}$$
I can tell you that the pH after your 10-fold dilution is only slightly higher than 4.76 (it's 4.763). You would have to dilute 100-fold to start seeing some difference (4.79); and frankly that looks to me like a minor difference for most practical applications.
Note: we are neglecting the difference between activity and concentration; but I guess we have seen enough maths for one day...

Equal amounts of the acid and its corresponding base are mixed. The concentration of the acid will be equal to its corresponding base, i.e. $0.05 \text{ M}$. When the concentration of the acid is equal to the concentration of its corresponding base, $pH = pk_a$. Thus the $\text{pH} = 4.76$.
Dilution of the solution will decrease the concentration of the acid, as well as, of its corresponding base. The concentrations will be $[\text{acid}] = [\text{base}] = 10*0.05/100 = 0.005 \text{ M}$.
The ratio between the acid and the base will not change. Thus, the pH will still be at $4.76$.


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Purify gas produced after reactions with sulfuric acid
Say, I preheat 50ml of 98% $\ce{H2SO4}$ and proceed with adding 25ml of 85% $\ce{HCOOH}$.
The result of which is:
$$\ce{H2SO4 + HCOOH -> CO + H2O}$$
Sulfuric acid will dehydrate formic acid producing $\ce{CO}$, however, gas could still be contaminated with vapor. What is the correct way of removing these impurities?

Pass the gas mixture through Anhydrous Calcium Chloride or Silica Gel. Calcium Chloride is quite hygroscopic and absorbs water vapors from the gas mixture. Alternatively you can use Silica gel, which adsorbs water due to its nature as a good adsorbent.
Edit: To get rid of the acids present in the vapor, bubble the mixture through a concentrated base solution. Acids would react with the base, while $\ce{CO}$ being a neutral oxide, simply doesn't react and can be collected.

It'll absorb water vapor, but there should be some vapor from the acids still present, I guess.

@Halo I have made the edit :)

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How does aqueous ammonia react with hypochlorous acid in water?
Here is what I have so far:
$$\ce{NH3 + HOCl -> NH2Cl + H2O}$$
Since ammonia reacts with hypochlorous acid to form the chloramine $\ce{NH2Cl}$.
But I don't know if this is correct.

This is correct according to the Wikipedia entry on Chloramine

Yes, aqueous ammonia will react with hypochlorous acid to form chloramine gas as suggested by @Waylander. You can also check this Reddit post.
I am assuming this is an exercise question. Do not try to attempt this reaction if you are not an experienced chemist as it can also produce deadly byproducts like chlorine gas, hydrazine, dichloroamine and probably nitrogen trichloride. Chloramine is itself poisonous.

+1 once when was in graduate school one of the campus custodians told me that he had tried to put ammonia and chlorine bleach together in his cleaning solution and it knocked him out. I promptly got religion about household chemicals.

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How to determine the mass of ammonium nitrate needed and enthalpy of solution?
I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.
I have the formula:
$$Q=ms\Delta T$$
I looked up the enthalpy of solution for ammonium nitrate and found it to be $\pu{25.41 kJ/mol}$.
I also have data that if I add $\pu{15 g}$ of $\ce{NH4NO3}$ the temperature will decrease by $\pu{25 ^\circ C}$, if I add $\pu{10 g}$ the temperature will decrease by $\pu{20 ^\circ C}$, and if I add $\pu{5 g}$ the temperature will decrease $\pu{13 ^\circ C}$.
Clearly then, to get my solution to be $\pu{3 ^\circ C}$, I would need to add a mass somewhere in between $10$ and $\pu{15 g}$.  In the lab I could just add random amounts between until the solution settled on $\pu{3 ^\circ C}$. Shouldn't there be a mathematical way to solve this rather than by random guessing?
I tried to use $Q=ms\Delta T$ by substituting the enthalpy of solution (in J) for $Q$, the specific heat of water for $s$, and $\pu{20 ^\circ C}$ for $\Delta T$ (since I am going from $23$ to $\pu{3 ^\circ C}$). However, solving $$\pu{25410 J} = m (\pu{4.186 J// ^\circ C})(\pu{20 ^\circ C})$$ gives me $\pu{303.66 g}$ which is clearly nowhere near the $\pu{10 - 1 g}$ that is sufficient to change the temperature.
How can I accurately determine the mass using a formula?

@Pritt There is always a space between the value and the unit. You can use \pu{5 g} for easier and correct typesetting.

@Martin-マーチンThanks for telling me this. I wasn't aware of that.

@Martin-マーチンAnd how did you write that syntax for \pu{5 g} with greying background? Moreover, how is that yellow highlight background made in the questions?

@Pritt you can include pieces of code in your comments by enclosing them within back-ticks. So the code gets shaded like this. See Jeff Atwood's answer.

The yellow highlighting is the quote option, you use that to add chunks of text copied from external references; it can be achieved by using > symbol before the text.

I am trying to figure out how to determine the mass of ammonium nitrate needed to reduce the temperature of solution from $\pu{23 ^\circ C}$ (measured room temperature at start) to $\pu{3 ^\circ C}$, once it dissolves in water.

The result will depend on the amount of water where you're dissolving the salt. For instance, the data you present refers to different solvent mass. And by the way, for your temperature difference ($\pu{20^\circ C}$) the problem is already solved: a $\pu{20^\circ C}$ temperature decrease, will require $\pu{10 g}$ of the salt.
It is important to distinguish the mass of the salt from the mass of the solvent.
\begin{align}
Q &= m_\mathrm{salt} \times \Delta_\mathrm{solution} H^\circ_\mathrm{salt}\\
Q &= m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T\\
\therefore m_\mathrm{salt} &= \frac{m_\mathrm{water} \times C_{p(\mathrm{water})} \times \Delta T}{\Delta_\mathrm{solution} H^\circ_\mathrm{salt}}
\end{align}
Hope this helps.


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What is the difference between an assay and an extemporaneous preparation?
What is the difference between an assay and an extemporaneous preparation?
Can an extemporaneous preparation be labelled as the BP mixture?

Usually most extemporaneous preparations are mixtures, solutions, creams, emulsions etc. and formulations are available in the British Pharmacopoeia and others

An assay is an investigative (analytic) procedure for qualitatively
  assessing or quantitatively measuring the presence, amount, or
  functional activity of a target entity (the analyte).

E.g percentage purity of a given substance, viscosity, bulk density, total organic content etc.
The analyte can be a drug, a biochemical substance, or a cell in an organism or organic sample.
Extemporaneous preparation refers to the preparation (mixing, assembling, packaging and labelling) of a medicinal product by a pharmacist (or licensed practitioner) according to prescription requirements for a specific patient . Extemporaneous preparations  are made in  relatively small quantities  for short-term supply and use.
For example: Magesium trisilicate mixture can be extemporaneously prepared from the following:

Magnesium trisilicate
Light magnesium carbonate
Sodium Carbonate
Concentrated Peppermint emulsion
Double Strength Chloroform water

(Magnesium trisilicate mixture is an oral suspension containing 5% w/v of these excepients in suitable vehicle.)
Guidelines of extemporaneous preparations can be found in the Pharmacopoeia.
Where you want to determine the content of say one of the excepients , is assaying. For example you may wish to experimentally find the actual content of magnesium in the mixture, you may perform titration with a suitable standard solution, then you can actually be able to calculate its percentage weight in volume


text:
Is rotation/flip of skeletal structural formula of Sildenafil/Claim 6 okay?
I want to show visually the similarities between Sildenafil (Viagra) and the molecule on the right:

Is it allowed to flip the skeletal structural formula and rotate bond strings of the molecule on the right to better show the similarities like I did it in the following figure?

The molecule on the right is flipped at the horizontal line and the bottom left C$_2$H$_5$ is rotated outwards. In both cases, the structure is the same, but I stumbled on things like enatiomeres and optical isomerism, and now I'm confused. I red "Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008)" partially, which seems to allow transformations like tried above, but as non-chemist, I'm still unsure.

Hi @Pi.Lilac, sorry for the downvotes earlier, your choice of the word Viagra made an anti-spam system think you were a spammer. That's been corrected now, and welcome to Chemistry.SE :)

yes, free rotation about single bonds not part of rings, but don't rotate groups about double or triple bonds or change relative positions in rings etc.


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Tert-butoxide in SN2 reactions
Can Tert-butoxide displace bromide via SN2 reaction? I read that Tert-butoxide is bulky and has steric hindrance, even with primary substrates. How is this reaction (picture below) possible?
Source: Levy, Arrow pushing in Organic Chemistry

Well, either substitution or elimination can occur. In case of bromomethane, theres no question of whether eliminations can occur or not. So tert-butoxide did the expected thing, it substituted the bromine. By the way, its not that hindered to not approach methyl carbons.

This seems interesting.
The $(CH_3)_3CO^-$ is a strong nucleophile and we observe here methyl bromide is being attacked and there is no point of elimination. There has to substitution in this case.
The bulkier the nucleophile is the more probable is the elimination product due to steric hindrance. In majority of the cases Tert-butoxide undergoes elimination reaction.
The further factor includes temperature at which the reaction is being carried out.
We can show this relation with $$\delta G  = \delta H - T \delta S$$
Hope it was helpful.


text:
Why does gaseous HCl not change dry blue litmus paper to red?
I am doing a practical in lab, in which HCl is to be evolved. I found that when blue litmus is near the evolved gas, it doesn't turn red. But, when litmus is dipped in water and then brought near the HCl gas it does change to red. Why is this happening?

Dry $\ce{HCl}$ gas doesnt furnish $\ce{H+}$ ions. Moist $\ce{HCl}$ however, can.

Also see: https://chemistry.stackexchange.com/questions/2838/how-does-the-litmus-ph-indicator-work

This is because gaseous HCl  does not dissociate to give up protons and chloride ions.
But, in aqueous solutions HCl does dissociate and exist in ionic form:
$$\ce{H2O(l) + HCl(aq) -> H3O+ (aq) + Cl−(aq)}$$
Thus it acts as a typical Brønsted acid (proton donor) in an aqueous environment. In this case, water is gaining a proton $\ce{(H+)}$, so it is a base, while HCl is giving one away, so it is an acid.
Therefore it is able to turn the blue litmus paper red (a behaviour typical of acids).
For related topic please see: How does the litmus pH indicator work? as suggested above in comments.


text:
What is the directive influence of the tert-butyl group in electrophilic aromatic substitution?
We know that hydrocarbon side-chains on a benzene ring influence electrophilic substitutions by directing the incoming electrophile to attack at ortho and para positions. The following hyperconjugations are shown to explain directive effect of toluene:

However, the situation was different when I tried to analyse for tert-butylbenzene. Tert-butylbenzene has no $\alpha$-hydrogens so the above mechanism cannot be extended to this case. Does the following mechanism work?

Does tert-butylbenzene have any directive influence at all? If we consider only inductive effects, then the incoming electrophile may attack at the meta or para positions (ortho being crowded).

And yes the later canonical structures do enjoy a lot a hyperconjugation. It might be possible to form a tertiary carbanion in that case.

But still forming 3° carbanion on its own is difficult isn't it ?

What you've drawn in your bottom figure is an example of carbon-carbon hyperconjugation. The t-butylbenzene and the charge separated structure drawn to its right are resonance structures (so you should use a double-headed resonance arrow to connect them).  Additional resonance structures with the negative charge delocalized into the ortho and para positions of the benzene ring can also be drawn, as well as additional structures using the other methyl groups attached to the central t-butyl carbon.  Based on the carbon-carbon hyperconjugation argument we would expect that

the t-butyl group would be an ortho-para director and
that the t-butyl group would accelerate the rate of electrophilic aromatic substitution compared to benzene itself.

Generally, carbon-carbon hyperconjugation is a small effect (smaller than carbon-hydrogen hyperconjugation) and not that commonly invoked.  The problem that I have with using carbon-carbon hyperconjugation arguments in this case is that our resonance structures involve a high energy methyl cation.
Backing up to toluene for a moment, there are two arguments why the methyl group is an activating ortho-para director. You’ve already outlined the hyperconjugative argument where electron density is shifted from a $\ce{C-H}$ bond to the aromatic ring using pi orbitals. However, in addition to this pi transmission argument there is also electron transfer from the methyl carbon to the aromatic ring inductively through the sigma system. The methyl group is roughly $\ce{sp^3}$ hybridized while the aromatic carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow through the sigma bonds from $\ce{sp^3}$ to $\ce{sp^2}$ carbons. So both inductive (sigma) and hyperconjugation (pi) arguments can be used to explain the electron donating ability of a methyl group.
Here is some data on the relative rates for the electrophilic nitration of several aromatic compounds:
benzene=1, toluene=24, t-butylbenzene=15.7 (reference, p. 1060)
We see that t-butylbenzene undergoes electrophilic aromatic nitration slightly faster than benzene and slightly slower than toluene - and the effect is very small.  Further, it is also reported (same reference) that nitration of t-butylbenzene produces ortho, meta and para products in a 12 : 8.5 : 79.5 ratio. Since the bulky t-butyl group will sterically block the ortho position to some extent, the fact that the para isomer predominates and there is more ortho than meta isomer is consistent with the t-butyl group acting as an activating ortho-para director.
Small effects are often difficult to explain, but I suspect that the carbon-carbon hyperconjugation effect (pi transmission) is a small contributor at best here.  Rather, the inductive sigma transmission effect (electron flow from the roughly $\ce{sp^3}$ central t-butyl carbon to the aromatic  $\ce{sp^2}$ carbon) is how the t-butyl group donates electron density and stabilizes the electrophilic aromatic substitution transition state.  In fact, we could even argue that since carbon is more electronegative than hydrogen, the 3 methyl groups surrounding the central t-butyl carbon pull some electron density away from it compared to the 3 hydrogens surrounding the methyl carbon.  This would make the t-butyl carbon slightly less electron releasing than a methyl group in accord with observed relative rates.

So, the mechanism I suggested is valid I guess. Although I think that this cannot be extended to triethyltoluene or tributyltoluene. Do you have any ideas on what happens there?

Yes, the resonance stabilization you suggested is valid.  Why wouldn't it extend to your other examples as well?

Does this work https://i.stack.imgur.com/5bwYl.png

Also forgive me for adding an extra carbon on that carbocation by mistake

Yes, except I'd use a double-headed resonance arrow rather than a reaction arrow.

Does the mechanism shown work? I'm not really sure about it though.

It is not a mechanism, it is resonance.

Thanks for pointing that out. Is it correct?

Yes it is correct, except I'd use a double-headed resonance arrow rather than a reaction arrow.

@ron In last, you said that inductive effect is at major work here instead of Carbon-carbon hyperconjugation. But inductive effect shall not support para product.


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Zinin Reduction of 1,3-dinitrobenzene
Remember asking my chemistry teacher this question in 1979/80: in the reduction of 1,3-dinitrobenzene with hydrogen sulphide/ aqueous ammonia, why is only one nitro group reduced?. He said that the reaction must be driven by the electon-depleted ring (two nitro groups); as one nitro-group becomes an amino, electron density in the ring increases, by definition; therefore, the reaction stops.
This is almost certainly true; but, is there a more fulsome explanation? Attempts to find the mechanism, on the net, have established that it has not yet been elucidated.
Any thoughts; theories; ideas; or, is this a worthy Ph.D. project for an ambitious young student?

Strongly related: https://chemistry.stackexchange.com/questions/34066/selective-nitro-reduction-of-poly-nitro-compounds/53698#53698

https://chemistry.stackexchange.com/questions/68817/selectivity-of-zinin-reduction

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text:
What happens with ammonia ions in tap water after they are separated from chlorine?
Tap water is increasingly treated with chloramine which is composition of chlorine and ammonia. Chlorine can be removed by catalytic activated carbon and ammonia ions are then left. What happens with them in the water? Do they bind with something or are they left in ionic form? How quickly they evaporate at room temperature?

By "chlorine", do you mean chloride ions or elemental chlorine? The former are not removed by activated carbon, and the latter is simply not there.

@IvanNeretin The former are indeed removed if you have a charcoal filter.

Chloramines (mono-, di-, and tri-) are used as a disinfectant for drinking water treatment (the source water, not the tap water) - this process is called combined chlorine. Upon contact with pathogens, organics, etc., chloramine is reduced to chloride (not chlorine), ammonia, and hydroxide. However, this half reaction is for alkaline conditions. The pKa of ammonia is around 9.4, so around circumneutral pH, some ammonia will exist as ammonium, the protonated state.
So, will activated carbon (AC) remove the products? First, we need to clarify something about AC - it is not catalytic, it is a physical adsorption removal process. AC will remove chloride ions, but it will not remove ammonium. Ammonia is quite volatile and will move from the water to the gas phase if e.g., a pitcher of water is left at room temperature. If you wanted to remove ammonium as well you could use something called biologically activated carbon (BAC). One last note, chloride will not "bind" will AC, it will be adsorbed through electrostatic forces.


text:
Why does silver oxide form a coordination complex when treated with ammonia?
Part of the preparation of Tollens' reagent occurs according to the formula:
$$\ce{Ag2O (s) + 4 NH3 + 2 NaNO3 + H2O → 2 [Ag(NH3)2]NO3 + 2 NaOH}$$
I wonder how exactly the oxygen gets "unstuck" from the silver under the influence of ammonia. I've read about coordination complexes, so I understand what $\ce{[Ag(NH3)2]NO3}$ is, but I'm puzzled at what exactly happens to pull that oxygen apart from $\ce{Ag2O}$
I've done very little exploration of the issue, so this may be too basic a question - in which case just please give a hint on what to read up.
I found this (now deleted) comment:

$\ce{NH3}$ dissolves silver oxide because the nitrogen has a pair of electrons that it can share with the silver forming a diammine complex. It has nothing to do with pH adjustment. – A.K. Jan 20 '16 at 17:17

Chemguide states this:

This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate.

I'm stashing this for future reading: Mechanism of Tollen's test and Purpose of ammine in Tollen's reagent

It's quite elementary (perhaps). Silver (and gold) are known to show two main coordination numbers in its compounds, namely, $2$ and $4$. To start, we can consider $\ce{[Ag(OH)2]-}$ to be our initial complex as in aqueous solutions it can be thought of to be existing in an equilibrium with $\ce{Ag2O}$. See footnote [1].
A quick glance at the order for the strength of ligands tells us that $\ce{NH3}$ is a stronger ligand than $\ce{OH-}$. What I now intend to do is to compare the thermodynamic stability of $\ce{[Ag(OH)2]-}$ with that of the diammine complex $\ce{[Ag(NH3)2]+}$.
The strength of ligands is directly proportional to the thermal stability of a complex so yes, the diammine complex is more stable.
A similar case is seen with copper(I). Cyanide complexes of copper(I) are known to be extremely stable and hence it prefers forming a complex with cyanide over other ligands.
[1] Nevertheless, saying this may be highly controversial because Ellingham diagrams tell us that $\ce{Ag(OH)}$ being unstable gets converted into $\ce{Ag2O}$ which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist.

How do you quantify the "strength" of a ligand? Different ligands bind to different metals with different affinities, and the usual way to measure this affinity is by the formation constant of the complex $$\ce{M + n L <=> ML_n}; \qquad K_\mathrm d = \frac{[\ce{ML_n}]}{[\ce{M}][\ce{L}]^n}.$$ Your argument then becomes tautological, because it is essentially now "it is more thermodynamically stable because it is more thermodynamically stable".

Some thoughts on the issue received via email from Jim Clark, the author of the Chemguide website. I hope these thoughts will prove helpful to readers, thus I post them here (with the author's permission).

I don’t really see the problem here.  The silver oxide is formed in the reaction by adding hydroxide ions to silver(I) ions.  Instead of precipitating silver(I) hydroxide, it dehydrates (either totally or partially - I don’t know) to silver(I) oxide.  But as with any other insoluble compound, there will be an equilibrium between the solid compound and its ions in solution - albeit strongly to the solid side of the equilibrium.
When you add ammonia solution, it complexes with silver(I) ions to give diamminesilver(I) ions which are soluble.  The equilibrium position of that reaction is well to the complex side.
So you can think of this in two ways.
The ammonia reacts with silver(I) ions in the solid silver(I) oxide and goes into solution.  The oxide ions react with water to produce hydroxide ions.  Or...
The ammonia reacts with the tiny amount of silver(I) ions already in solution, and the precipitation equilibrium shifts to replace them.  The solid therefore goes back into solution producing the hydroxide ions that were added originally, but with the silver ions now mainly complexed.
The first way is not only simpler, but also, of course, applies to any solid silver(I) oxide you might have - for example a silver(I) oxide tarnish on some item of silver which can be removed by wiping it with ammonia solution (a source of both ammonia to complex the silver ions and water to react with the oxide ions.

Actually both comments perhaps, to quote:
"because Ellingham Diagrams tell us that Ag(OH) being unstable gets converted into Ag2O which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist"
"NH3  is a stronger ligand than OH−"
Assuming a starting Ag2O presence from the reaction, some Ag+ ions from limited solubility, and noting, "a ppt of AgOH appears and dissolves in excess ammonia",  see https://books.google.com/books?id=uwJbSFRIqyAC&pg=PA37&lpg=PA37&dq=radical+attack+on+silver+Ag+%2B+%E2%80%A2OH++--%3E+AgOH&source=bl&ots=jG9-2NG9x3&sig=V9nVefopAXXKP3lFnxQYnJriG14&hl=en&sa=X&ved=2ahUKEwjasar_g7ffAhWlhOAKHQEDA_4Q6AEwGHoECAcQAQ#v=onepage&q&f=false :
2 AgOH = Ag2O + H2O
both comments may be right.
Further, in the presence of visible light leading to the photocatalyst Ag2O/Ag, which per research out of P.R. China (published 2011 in the ‘Chemistry, a European Journal’), which has been cited as a powerful stable visible light photocatalyst capable of generating electrons (e-) and holes (h+) (see https://onlinelibrary.wiley.com/doi/abs/10.1002/chem.201101032 ):
Ag2O/Ag + hv --> e− + h+
With electrons:
Ag+ + e- --> Ag
And, in the presence of active electron holes, any OH- created via:
XOH = X+ + OH-   (where X = H or NH3 or Ag)
could be converted into •OH radical:
OH- (aq) + h+ --> •OH (aq)
which may attack Silver metal directly (see https://pubs.acs.org/doi/abs/10.1021/acs.jpca.7b08081?src=recsys&journalCode=jpcafh and superacidity related redox couples at http://beta.chem.uw.edu.pl/people/WGrochala/Ag2+_mechanism.pdf, and more generally radical attacks on metals in their lower valence states at https://srd.nist.gov/NSRDS/NSRDS-NBS-65.pdf , where one could postulate the half cell reactions: Ag = Ag+ + e- and •OH + e- = OH- with the net reaction below):
Ag + •OH (aq) --> AgOH (aq)
2 AgOH = Ag2O + H2O
Or: AgOH (aq) = Ag+ (aq) + OH- (aq)  (based on some limited solubility of Ag2O)
producing the oxygen free Silver ion to complex with NH3 (or OH-, depending on the respect ion concentration) in a photo cyclic reaction. Note: the hydroxyl radical can be largely scavenged at high ammonia concentration (creating •NH2), so this path is pH sensitive.


text:
Ionic radius of Al[3+] compared to K[+], Ca[2+], S[2-], and Cl[-]
I came across this question:

Which of the following is bigger?
$$\ce{K+, Ca^2+, Al^3+, S^2-, Cl-}$$

Now, I know that the radius decreases as the period progresses to the right and increases as we go down on the P.T. So, I get to this:
$$\ce{S^2- > Cl- > K+ > Ca^2+}$$
Also, I know that $\ce{Al^3+ < Al^0}$ and that $\ce{Al^0 > S^0}$. But I can't figure out how to connect that to the inequality above.

$\ce{Al{3+}}$ is a 2nd group element, and should be even smaller than the rest.

Think of $\ce{Sc^3+}$; it should be easy to compare it to $\ce{Al^3+}$ and also to the rest.

Well $Sc^{3+}<Sc$ and $Al^{3+}<Al$. Of course $Sc>Al$. If $Sc^{3+}>Al^{3+}$, then yes, I can compare them. And then $Al^{3+}$ is the smallest. But why does $Sc>Al\rightarrow Sc^{3+}>Al^{3+}$ hold?


text:
What's considered a "higher" level of theory in quantum mechanics calculations?
Comparing wavefunction-based methods and density-based methods for quantum mechanical calculations, how is "rank" determined?
For example, in wavefunction methods, CCSD(T) is considered a higher level than MP2, which is a higher level than HF.

What about if I'm comparing MP2 to B3LYP? Or if I add dispersion
corrections to the DFT method to use something like B3LYP-D3BJ?
What if I just want to compare different DFT functionals? Is hybrid higher than pure? Is B3LYP higher than PBE0?

DFT doesn't fit well into the scheme, because it uses different approach comparing to post-HF methods. Personally I would consider it betwee  HF and MP2, for all common functionals. Functionals themselves are not 'better' or 'worse' because all of them are semiempirical.

A higher level of theory in QM calculations is always the one the anonymous reviewer of your publication claims is superior to the one you chose.

@ToddMinehardt, that's funny. By the way, I noticed that you're a data scientist who once did theory. Can I ask how the transition went?

@halcyon - It was a discontinuity, and there were/are pluses and minuses. I'd be happy to slate a chat with you in a chem.SE "room" as time permits.

@ToddMinehardt oh god I cried. Kinda true though :(

For single reference wavefunction based methods, the comparison is relatively straightforward within method "families". This is due to the nature of how these methods are formulated, usually involving a very long series that has been terminated at some point to make the calculation feasible.
You start with Hartree-Fock, which ignores electron correlation.
After that you can add perturbative corrections of increasing order (this will be the MP2, MP3, MP4, MPn family), do something called configuration interaction, truncated at increasing levels of excitation (CIS, CISD, CISDT,...,FCI), or something called the coupled cluster methods, again truncated at some level of excitation (CCSD, CCSDT, CCSDTQ,...,FCI).
In these cases the most straightforward interpretation of "higher level" would mean a less aggressive truncation of the series that includes more terms. In case of CI and CC, this almost always results in more accurate results, and both converge to the same limit, where all possible terms are included and correlation effects are exactly calculated. (of course there are errors from other approximations) The MPn perturbation series has no such strong guarantees, and this is one of the reasons why pretty much noone uses anything beyond MP4 these days. (even MP4 is seldom used)
For DFT you have something called Jacob's ladder, but things are generally more murky, as most DFT functionals have many empirical (fitted) parameters, and there is no straightforward way of comparing them.
In general, the only way to compare them, is to take a big database of either experimental or calculated (via CCSD(T) or similar high level method) data, and compare the average/maximum errors.

If you want to solely know about DFT functionals, read about Jacob's ladder of functionals. There's a good recent review of a slew of DFT functionals:  http://dx.doi.org/10.1063/1.4948636
I do not think hybrid functionals are higher in theory. All that is being done is mixing exact exchange with other functionals that have already been established. But I think some people may argue that is not the case.
Another thing that you may want to look at are the "plus methods": DFT + GW, DFT + U, DFT + DMFT. These methods add a term to the Hamiltonian to correct errors or capture behavior that cannot be observed with traditional DFT.
Most comparisons between DFT and Post-Hartree Fock Methods are done by comparing calculation results. And this comparison is not very straightforward. Usually physical values (e.g. binding energy, bond lengths) are calculated by different method and compared to experimental values. Usually, however, wavefunction methods are deemed to be more accurate or consistent. So they are often used as benchmarks for DFT calculations.

Jacobs ladder is highly subjective though, since DFT-Functionals mostly tend to do one thing pretty good and another thing horribly bad.

In general, GGA yields better results than LDA. The problem is that most "higher level" functionals such as meta-GGA are not very general. In some systems, they will be very good. But at some, they will underperform. The journal in my answer brings attention to this. It also brings attention to new meta-GGA functionals that are generally better than GGA. Hence the significance of the journal

Aside from the used method, you should also pay attention to the used basis set. Generally, the larger the basis set the better the results. (This is strictly speaking not true, as just having a "large basis set" does not guarantee the additional basis functions are helpful. But usually basis sets are optimized to achieve maximum quality with a given number of basis functions.)
When comparing methods of different levels of theory, you should use a basis set of same quality. This will ensure the differences are really due to the method. In extreme cases a cheap method with a large basis set may give better results than a higher level method with a small basis set.
However, one should be careful with the "basis set of same quality" formulation, especially for DFT where thing are generally more fuzzy.

I've seen plenty of examples where smaller basis sets with DFT actually can perform better than larger ones.  The reason for this is because DFT is not a convergent theory unlike ab initio approaches.

I work more with the ab initio approaches, so I don't know too much about DFT. But I can image this very well. However, I would also expect that in these examples not all molecular properties are better described at the same time?


text:
Spectroscopic transition frequency
I have difficulty understanding the first sentence of the Introduction in the paper linked below. Can someone explain to me what is spectroscopic transition frequency and thermal bath?

The question addressed here is how the modulation of the spectroscopic transition frequency Ω(t) by fluctuations of the thermal bath is related to the observed spectral line shape.

http://pubs.acs.org/doi/abs/10.1021/jp5081059

Hello and welcome to Chemistry.SE. It's a good idea to take the short [tour], and visit the [help] if you have any further questions about how this site works. Note that I added the text from the paper that you referred to in your question (feel free to change my [edit] if I got it wrong). It is much easier to read that way, and plain links are not searchable for people who may have the same question in the future. Best of luck with your question!

@user44545 - please don't link to illegal repositories of papers. If there is a particular passage/diagram that you wish to refer to, just take a screenshot/copy the text, with a link back to the literature on the publishers website.

In an isolated molecule /atom the intrinsic width of the transition is limited by the lifetime of the upper level and this can be huge ( seconds) and so the line-width can be minuscule by time-energy 'uncertainty'. If the decay is exponential the spectral line shape is Lorentzian by Fourier Transform theory.
In a medium, the surrounding solvent or gas molecules repeatedly interact with our victim molecule by collisions and by electric and magnetic dipoles etc, and do so  in a random fashion and thus cause fluctuation in its energy levels. These changes are then observed in the transition energy and so broaden the observed transition in the ensemble of molecules. So far this is quite general. This paper examines one model of this, linear + quadratic coupling and the quadratic part is presumably why the peak is split - but I only skimmed the paper so you will have to check this.  The detailed maths is just the technical part, ignore this and try to get to the basic physics first.
(The term 'thermal bath' is just jargon for all the random interactions from solvent/buffer gas surrounding the victim molecule, 'modulation' just means changes, so the first sentence could read 'We describe how random interactions with a solvent or buffer gas changes the observed spectroscopic line shape ....')


text:
What is Fjord Region?
I'm studying molecular motors for a project of mine and I'm trying to understand what a Fjord region is.
It's sort of labeled in the image below but I'm not entirely sure which part.
I would like a suggestion of a book or article that may help me understand what it is?

Here's another source with some descriptions, probably not as good as in the answer though.

Fjords and bays
Fjord is a term originally developed by chemists studying polycyclic aromatic hydrocarbons, along with the term bay.
The descriptions aren't rigidly defined in the same way as stereochemical descriptors, but were designed to be 'self-explanatory', following the real world uses of the words1

Bay: "a broad inlet of the sea where the land curves inwards."
Fjord: "a long, narrow, deep inlet of the sea between high cliffs, as in Norway, typically formed by submergence of a glaciated valley."

In an attempt unify the usage of these terms, IUPAC issued a set of guidelines for their usage in 2014.2

Some historical context
The earliest publication of these terms was given in the references above by Harry Heaney, Keith Bartle, Denny Jones, and Peter Lees who were working at the now non-existent Institute of Technology (Bradford, UK). They were studying triphenylene, and trying to find descriptions for the various environments present.

Circumnavigating triphenylene, the British crew sighted two essentially different kinds of H's; and Derry Jones spontaneously called one type "bay" and the other "peninsular." They published these descriptive "self-explanatory" names,40 but not without some apprehension. In fact, Professor Jones confessed to having dreamed that referees rejected their manuscript on account of its "bay" and "peninsular" nomencla ture

Reference:
1 Spectrochim. Acta 1966,22,pp941-951; Trans. Farad.  Soc. 1967,63,pp2868-287
2 J. Polycyclic Aromatic Hydrocarbons, 2015, 35, pp 161-176

The arrows make a big difference here in the understandability of the IUPAC drawings. The original drawing in the question lacks an arrow, and is therefore ambiguous on what exactly is the "fjord region". But honestly, these terms seem like an attempt to be far too clever, and I'm not a big fan.

@CodyGray Seems more like they were trying to give a nomenclature that was easily understood just by inspection.  Doesn't seem like it worked particularly well; but I don't think the point was to "be clever", someone was just clearly thinking about geographic features when they decided to name them.


text:
Do electric charges increase activation barriers?
The activation energy governs the rate of reaction, and is thought to be equal to the energy needed to break the bonds needed to initiate the reaction. But if we have a reaction like:
$$\ce{A +B -> C}$$
And another like:
$$\ce{A+ + B+ -> C^2+}$$
Would the former have a lower activation energy since there is no repulsion between the molecules? Assume the same bonds are being broken in each case.

The very example of two reactions related in such a way, especially with the requirement of same bonds being broken, looks highly improbable and artificial. That being said, if such an example can be found, your reasoning should apply.

What I meant in general is what is the definition of activation energy?
I've only been introduced to it in the form of the energy required to break bonds, but if we consider a reaction like the one with Iodide ions and S2O8 2- ions ( I think ) , then the normal explanation given is that the reaction does not proceed without a catalyst due to the high activation energy required to overcome the repulsion of like charges.

@Saad based on your comment, the example you chose seems to obscure your real question. Also, to get Ivan to see your comment, you have to include  "@Ivan Neretin" in your comment (you can only include one reference like this in a comment and it will only ping someone if they have made a comment on that particular post).

Some of the answer to your question is given here, https://chemistry.stackexchange.com/questions/34909/why-are-rates-of-reaction-faster-for-ionic-compounds-than-for-covalent-compounds/71587#71587. The rate constants are affected by the charges on the molecules and so we may assume that this is equivalent to affecting the activation energy.

Here is a specific example from biochemistry. In the hydrolysis of ATP, the first step is the attack of a hydroxide anion at one of the phosphates. The triphosphate portion of ATP has multiple negative charges at neutral pH, so this reaction is very slow.
When catalyzed by an enzyme, the enzyme often binds to a magnesium cation in such location that it shields some of the negative charge. This makes the reaction proceed faster.

Do electric charges increase activation barriers?

Yes, they do. And there are various strategies to lower the activation barrier in these cases. One is providing a counter ion as described above. Another is to change the charge via acid or base catalysis, i.e. protonating or deprotonating one or the other reactant to decrease electrostatic repulsion.

Would the former have a lower activation energy since there is no repulsion between the molecules?

Yes, and it would be sufficient to remove one of the charges in the latter to lower the activation barrier because electrostatic repulsion requires both particles to be charged.


text:
How do I lower the surface tension of water without using a detergent?
For many years I've used a Rainbow vacuum, which uses a reservoir of water as the filter.   While I'm quite pleased with it, I've found that when trying to vacuum cement dust or wood ash not all of the  particles are captured in the water.   I believe a remedy to this would be to lower the surface tension of the water.  Yes, I can always be sure to use hot water, but I'd like to go further.  Obviously I can't use a suds-producing detergent because the vacuum produces lots of turbulence in the reservoir.
What commonly available chemicals could I use that would give me the greatest reduction of surface tension? So far I've tried to research trisodium phosphate, sodium carbonate, and a product called Haemo-Sol which is a non-ionic surfactant... but I can't find surface tension numbers for any of these much less the concentration to use.
The manufacturer suggests adding a tablespoon of vinegar, which I think is
rather lame.
Can anyone make a good guess or point me in the right direction?

Reduction of surface tension is a most interesting science however currently friccohesity has superseded the surface tension experimental approach.The trimethylsulfoxonium iodide (TMSOI), methyltrioctylammonium chloride (MTOAC) have largely reduced the surface tension which was measured with survismeter. Pl consult the following paper-
Int. J. of Thermodynamics ISSN 1301-9724, Vol. 11 (No. 4), pp. 181-186, December 2008. 'Solvodynamics of Benzene and Water Phases by DTAB, MTOAC,TMSOI and Orcinol Studied with Interfacial Tension, Surface Tension and Viscosity Measured with Survismeter'

Your goal is to reduce the surface tension of the water so that it does not support the formation of large bubbles or inhibit the whetting of the dirt particles.
When the bubbles are small, the dust inside will be more likely to make contact with the liquid surface of the bubble within the lifetime of the bubble. Also, when it does make contact, it will be much more likely to irreversibly adhere to the water if you have reduced the surface tension.
The word "surfactant" is a blend of "surface active agent"$^1$, and this is what you need to interfere with the surface tension of your water.  I believe that you may have written off detergents too hastily. I think the best solution will be to add a small amount of laundry detergent, or possibly non-foaming carpet cleaner. Each of these products are designed to foam to a lesser degree than, say, dishwashing detergent or most hand soaps.
I suspect that you will have to experiment a bit, but that something on the order of a single cc per liter may be sufficient to increase the effectiveness of dust removal without causing excessive suds. Your goal should be to see how much detergent you can add without adversely affecting your vacuum, and decide whether this has resulted in a decreased throughput of dust particles.
As to your title question, "How do I lower the surface tension of water without using a detergent?", what I'm really saying is 1) that most anything that does what you want could probably be deemed a detergent and 2) that a detergent is likely the most practical solution to your problem.
1) Surfactants and Interfacial Phenomena, Fourth Edition, Rosen and Kunjappu, John Wiley and Sons, Inc., 2012

Vinegar does reduce the surface tension quite a bit. If you put a tablespoon or two of vinegar in a bowl and then run water into it, the bubbles that form on the surface persist much longer than they do if there's no vinegar present.
I suggest reconsidering your rejection of the vinegar recommendation.

Beyond that, airhuff's right -- you're searching for something that will preferentially accumulate at the air-water interface. Air is far less polar that water, which means that you need a chemical that has both polar and nonpolar characteristics. No small inorganic solutes possess appreciable nonpolar regions, and once you tack on a nonpolar tail of any substantial length, you're back under the umbrella of detergents/surfactants.
The foaming potential is absolutely a function of concentration, as airhuff says, so it's a matter of finding the right number of drops per gallon to achieve the balance of properties you're seeking.

UPDATE, 22 Oct 2023: As a practical example, Shin and Liu (2004) controlled the void size in bubble-microtemplated electrodeposited copper foams by tuning the acetic acid (vinegar) concentration.

Amway makes a product called L.O.C. It is somewhat of a detergent, However, it has been sold as a product that "makes water, wetter". It has been added (as I have been told) to chemicals in crop dusters. It allows the chemicals to spread over the leaves, giving greater surface area of coverage because of loss of surface tension.
Product: L.O.C.™ Multi-Purpose Cleaner (click "Ingredients") and MSDS

That's just an ordinary detergent. The main ingredient is sodium lauryl ether sulfate (a.k.a. SLES or laureth sulfate), one of the two most ubiquitous synthetic detergents. The other ingredients are also common in detergent formulations. Amway is just good at marketing this as something revolutionary.

Basically, the surface tension of water is originated when the water molecules which are dipolar, and their polar interactions exist due to the residual forces which exist at the surfaces. The dipoles in the bulk are counterbalanced by the dipoles of the water molecules placed nearby at a Lennard Jones distance and hence the global potential energy is higher as no residual negative or positive charges exist in the bulk. The dipoles are highly vibrant causing rotational and vibrational motions or oscillations leading to translations motions. Thereby at air and water interfaces, these dipoles become most active and the vibrant activities of the dipoles which remain at air phase are not counterbalanced contrary to the activities of the dipoles which are in contact of the other water molecules. Thus, the downward pull of the molecules is generated that causes a tension or surface tension. Thus, the dipolar forces of water molecules at the air-water interface differs from the dipolar forces which are with bulk water molecules. Stronger dipolar activities develop higher surface tension. Thus, the dipolar forces which face air having different gases and particulate matters are not strongly adhered to either whereas the dipolar forces which face similar water molecules become stabilized. The dipoles exposed towards air are active as these are not optimized and hence create tension at interface, but they could not enter the bulk due to stronger optimization which cause exceptionally stronger friction. So remained confined to the surface causing the surface tension. This difference of the dipolar forces causes friccohesity chemistry. The surface tension creation is example of friccohesity chemistry because the frictional forces and cohesive forces remain operative during the surface tension creation. Now any molecules which weaken the dipolar forces reduce the surface forces. For example, apart from surfactant, the proteins, alcohols, dendrimers, webdrimers, graphene, graphene oxide, adhesives, fluorescent dyes, gums, aloe vera, phytochemicals, decrease the surface tension of the water.

A really informative answer. Welcome to the site :) I have a suggestion - wouldn't it be better to split the answer into paragraphs for easy reading?

Please edit for clarity, grammar, and style: "The dipoles are highly vibrant causing rotational and vibrational motions or oscillations leading to translations motions" is not only incorrect in the physical sense, for example.

Surface tension and viscosity work hand in hand, there is hairline to distinguish cohesion and adhesion of solvent with self or with ingredient molecules. The hydrogen bonded water with dipolar interactions generate tendency to cohere the energy at surface as dipoles remain unengaged generating friccohesity resonating energy transfer (FRET). The air does not constitute any fixed surface unlike a molecules so these dipole cohere at water surface. However, these dipoles are balanced in the bulk with no tendency to cohere as the similar water molecules are present. But when different molecules like drug, detergent, antioxidant-curcumin, are dispersed in water, then the water dipoles cohere with these molecules noted as different molecules. The water dipoles before cohering to different molecules get  disrupted the hydrogen bonding which was attained due to coherence of the similar water molecules. Thus the coherence or cohesive forces with different molecules is noted as adherence or adhesion which resist rotational,vibrataional, and translational motions of the different molecules could be noted as ingredient molecules when the nanoformulations are made. Thus an adhesion generates a frictional force that increases shear or viscosity and decreases a cohesion or the surface tension. Therefore the surface tension and viscosity work hand in hand, increasing or decreasing either surface tension or viscosity, a new and novel interface noted as friccohesity is originated and measured with survismeter. Genesis of friccohesity is Fric is derived from frictional and the cohesity from cohesive forces. Truly the chemistry of friccohesity most fabulously generate the molecular economy which resolve the surface tension and viscosity related chemical processes like cosmetics, electroplating, coating, thin film. For these a continuity and monolayer adhesion is essential. The continuity is attained by moderating both the surface tension and viscosity. These all the issues are explained in the book.
https://www.cambridgescholars.com/product/978-1-5275-9363-3

Ethanol, ionic liquids, dyes, protein, reduce surface tension of water, however these chemicals increase viscosity so it is better to deal with friccohesity. The surface tension and viscosity determinations need density data measurements but friccohesity does not. The deepest mechanistic changes are accurately resolved by friccohesity. Acetic acid, sodium acetate and others reduce the surface tension of water.


text:
Nomenclature: equivalent atoms by single locant
Sometimes the IUPAC name is too verbose despite the structure's high symmetry. If we could use only the first locant of the ("topologically symmetrically") equivalent set of locants, and designate all of them by some symbol (e.g. xallsym, which is too long, and Greek letter sigma has some meanings already, so I will use Greek letter stigma), then we could have

1,3,5-trinitrohexahydro-s-triazine vs
1ς-trinitrohexahydro-s-triazine or
Nς-trinitrohexahydro-s-triazine

naphthalene-1,4,5,8-tetrol vs
naphthalene-1ς-tetrol

naphthalene-1,2,4,5,8-pentol vs
naphthalene-1ς,2-pentol

naphthalene-1,2,3,6,7-pentol vs
naphthalene-1,2ς-pentol

1,5(1,2,3),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane vs
1ς(1,2,3),2ς(1,4),3ς(1,2)-nonabenzenabicyclo[3.3.1]nonaphane

1,2,3,4,5,6-hexachlorobenzene or
hexachlorobenzene vs
1ς-hexachlorobenzene (the name is not convenient in this case)

1,2,3,4,5,6-hexachlorocyclohexane vs
1ς-hexachlorocyclohexane

1,1,2,2,3,3,4,4,5,5,6,6-dodecachlorocyclohexane or
perchlorocyclohexane vs
1ς,ς-dodecachlorocyclohexane
(not convenient in this case, but mentioned the special feature, listing two identical sets. Imagine 1ς means 1,2,3,4,5,6; 1ς,ς would mean 1,2,3,4,5,6,1,2,3,4,5,6)
Wrong usage can lead to contadictions/confusion, e.g. 1ς-trichlorobenzene would be illegal, because benzene carbon 1 is equivalent with all other carbons, total 6. But it is distantly similar problem like for with name like 1,2,3-hexachlorobenzene.
Is anyone aware of such feature in any chemical nomenclature?

"topologically" isn't appropriate here, as this is geometry; topology is a different thing altogether. Also your "pentol" is bad example.

Mithoron, could you explain what bad do you see on the pentol example? The "stigma" apples to the single locant immediately before the exponent, of course; and to the parent structure without any substituent that are defined by the prefixes.

gilleain, it means "all equivalent atoms", so 1* on cyclooctane means all eight carbons, so you cannot use this "stigma" for the mentioned set of positions.

Mithoron, geometry is much less appropriate term here, since you can draw cyclohexane as non-regular distorted hexagon, but the atom equivalence does not change.


text:
Is DNA a base inside the cell?
DNA is deoxyribonucleic acid because it has phosphate groups in every one of it's nucleotides.
However, in aqueous solutions protons will ionize, leaving phosphates alone as negative groups, i.e., lewis bases.
In this case, which is seen in all of our cells, is DNA structure actually the conjugate base of the theoretically described crystal form of DNA?

Hello and welcome to Chemistry.SE. If you have any questions about the site, feel free to visit the [help]. Best of luck with your interesting question!

In nuclei DNA as coniugated base is bound to histons which work as coniugated acid.

Related: Why are lactate and lactic acid used synonymously in biochemistry?

In a solution of Hydrochloric acid, almost all of  $\ce{HCl}$ exists as $\ce{Cl-}$ which is a base. Would you call it Hydrochloric base instead?

By the way it's very related to pKa concept as I now understood it. DNA should be reduced to a bulk of phosphate groups for simplicity, of which the pKa value is ~3. Since we know that physiologic pH is ~7, ionized form of phosphate groups must have ten thousand times more concentration than the non-ionized form. That's how we know DNA is actually what makes the environment acidic. Otherwise I was thinking we might have very well defined DNA in it's ionized form, as DNB. But understanding why this would be wrong depends on understanding the pKa concept. So thanks @orthocresol.

The pH is 7, it's not really an acidic environment. In terms of charge it's simply balanced by a whole host of positively charged stuff, like histones (which Mithoron mentioned), but also cations like Mg2+, etc. The name DNA is, to some extent, a misnomer; DNA as it exists in the body is not an acid, since it exists mostly in the form of its conjugate base.

(As for whether it's different from the crystal structure, I am not sure whether Franklin crystallised DNA as the acid or as the salt, would have to look it up later. However you are certainly quite correct in saying that DNA exists as the conjugate base)

related https://chemistry.stackexchange.com/questions/24791/why-dna-is-negatively-charged-and-what-makes-it-so


text:
Water vapor equilibrium calculations
The question had two parts.
The first part said:
Calculate the amount of water required to saturate a vessel of volume 82 L at 300 K given that the vapor pressure of water is 38 torr at 300 K.
The solution went like this:
Since we have to saturate the vessel, we have to find the minimum number of moles that will occupy the given volume (that is what I understood).
Now, applying the gas equation
$$PV=nRT$$
$$\frac{1}{20}*82 = n * 0.82 * 300$$
$$n = \frac{1}{6};  w=n * M = 3 gm$$
This part is fine.
Now the second part:
If 5 gm of water is introduced in the vessel (initially evacuated), then, calculate the amount of water remaining.
So, my teacher said that, 3 gm will evaporate and 2 gm will remain. But, 3 gm will evaporate for 82 L. So, when I pour 5 gm of water, shouldn't the volume decrease?

You are correct that there will initially be a volume of 82L - 5cc. Then 3cc evaporates, so the volume of the container is 82 L - 0.002 L. Will this make a difference in your calculation that will show up in a 2 significant figure answer?

@airhuff then, in the first question also, the volume should be 82-0.003. right?

But the point for the second question, is would it matter if you just ignored the effect of the volume of the liquid water? In the first question there is no liquid water left because it all evaporated and you have nothing but water vapor. Back to the second question: 82L - 0.002L = 81.998L, which rounds to 82L (note that you are only given 2 significant figures in the problem), and for that matter it rounds to 82.00L! So, is the volume of the small amount of water insignificant compared to the size of the container in this question (hint, hint), such that you can just ignore it?

You are not supposed to write "Thanks" or "Help is appreciated" or any statements of weaknesses in your post. And fix that Vap0r in the title. [I edited his post, apparently he rejected it]


text:
Can the enthalpy change of allotrope transformation can be done with Hess cycle?
My A- level textbook has some points about the uses of Hess cycle. For example: indirectly determining the lattice enthalpy, vaporisation enthalpy etc.
But, my book also says it is possible to determine the enthalpy change of allotrope transformation (eg. graphite to diamond); which I find hard to understand how. I have understood (and drawn the cycle) how to use Hess cycle for calculating lattice enthalpy and heat of vaporisation. But don't know how to do it for allotrope transformation. Is it really possible?

Explain you downvote.

Can I direct you to Jim Clark's excellent book on A level calculations where the example you mention is covered?
If I remember correctly using combustion data for both allotropes allows the Hess cycle to be triangulated and the enthalpy of transformation from one to the other form to be calculated.

It would help if you be more elaborate on how the calculations are done, so people won't have to go to see the book. External references should be the additive, not the gist of your answer.

It will be better if you just shoe all of us the method here and do mention the link.

When would you mention the link? @Jim_Pearce


text:
Finding the rate constant of a first order reaction

The reaction
$\ce{A + B -> AB}$
is 1st order with respect to A and zero order with respect to
  B. The reaction is begun with the initial concentration of both reactants at $0.100 \,\text{M}$. After $1.5$ hours the concentration of
  B has dropped to $0.060 \,\text{M}$. What is the approximate value of the specific rate (reaction rate) constant for this reaction?
(A) $0.15$ hr$^{-1}$
(B) $0.34$ hr$^{-1}$
(C) $0.61$ hr$^{-1}$
(D) the specific rate constant cannot be determined unless the
  mechanism of the reaction known.

My thoughts: As the reaction is first order with respect to A and 0th order with respect to B, we can write the rate law as $\frac{d[A]}{dt} = -k[A]\to [A] =[A_0]e^{-kt} = 0.1e^{-kt}.$ Now, because A and B are in a 1:1 mole ratio, the concentration of A after 1.5 hours is also $0.06 \,\text{M},$ thus, we solve for $k$ as follows:
$$0.06=0.1e^{-k\cdot 1.5}$$
$$\to \ln(0.6)=-k\cdot 1.5$$
$$\to k = -\ln(0.6)/1.5\approx 0.34$$
Thus, the answer I obtained is B. However, the correct answer given is C. Is there any error in my work, or is the answer key wrong?

You're correct. There's a key mistake. Sounds like you could teach the author of the book you're referring :)


text:
Oxidizing and Reducing agent in redox reaction

In the oxidation-reduction reaction
$\ce{Sn^{4+} + 2 Fe^{2+} -> 2 Fe^{3+} + Sn^{2+}}$
(A) $\ce{Sn^{4+}}$ is the oxidizing agent and $\ce{Fe^{2+}}$ is the
  reducing agent.
(B) $\ce{Sn^{4+}}$ is the reducing agent and $\ce{Fe^{2+}}$ is the
  oxidizing agent.
(C) $\ce{Sn^{4+}}$ is the reducing agent and $\ce{Fe^{3+}}$ is the
  oxidizing agent.
(D) $\ce{Fe^{3+}}$ is the oxidizing agent and $\ce{Sn^{2+}}$ is the
  reducing agent.

My thoughts: Because $\ce{Sn^{4+}}$ is getting reduced, and $\ce{Fe^{2+}}$ is getting oxidized, then $\ce{Sn^{4+}}$ is therefore an oxidizing agent, and $\ce{Fe^{2+}}$ is a reducing agent, which leads me to believe that A is the right answer. However, the answer key says D. I note that (D) is true for the reverse reaction, but why would A not be true then?

Yea, D lists the products not the reactants. Sometimes the answer key is wrong and A is right ;)

If I were you @DisplayName, I'd avoid using that book.


text:
Why does ethanol have a health rating of [2] in NFPA 704 whereas methanol, which is far more toxic, has a rating of only [1]?
Why Ethanol has the blue health hazard rating of 2 in NFPA 704 where as both Acetone and Methanol which are far more toxic than Ethanol have a rating of only 1?    (Click on each link to see the NFPA diamond)

Ethanol
Methanol
Acetone

it's the "fire diamond", F in NFPA states for fire hazard. Anyway, the health info (blue) difference is strange.

I believe that the rating is dependent on toxicity tests.
The acute oral toxicity LD50 value of acetone is 5800 mg/kg
The acute oral toxicity LD50 value of methanol is 5628 mg/kg
The acute oral toxicity LD50 value of ethanol is 3450 mg/kg
Low LD50 values means that only a small amount of chemical is needed to kill someone. According to these values, you would need to ingest a larger amount of acetone or methanol to die.I am not sure how valid these toxicology test are. Fun fact: fructose is deadlier than methanol according to LD50 values.
Furthermore, there appears to be some variation in the values as well. Wikipedia lists the LD50 value of ethanol as 7,060 mg/kg.
Maybe this question can be addressed at Biology SE.

LD 50 is a rotten metric for the relative toxicity of methanol and ethanol. This might be because methanol kills you slowly and its acute toxicity isn't the one that matters.


text:
Is acidified dichromate(VI) capable of oxidizing hydrogen sulfide to sulfur dioxide?
My textbook outlines reductive behaviour of $\ce{H2S}$ with the reaction
$$\ce{3H2S + K2Cr2O7 + 4H2SO4 -> 3S + Cr2(SO4)3 + K2SO4 + 7H2O}.\label{rxn:QR1}\tag{R1}$$
This reaction oxidizes $\ce{S^2-}$ to $\ce{S}$. But why sulfur is not oxidized further to $\ce{SO2}$ according to
$$\ce{H2S + K2Cr2O7 + 4H2SO4 -> SO2 + Cr2(SO4)3 + K2SO4 + 5H2O}?\label{rxn:QR2}\tag{R2}$$
The reaction seems stoichiometrically plausible, yet likely requires highly concentrated oxidants. I know \eqref{rxn:QR1} happens at low concentration of $\ce{H2SO4}$. Maybe using highly concentrated $\ce{H2SO4}$ would do the work for \eqref{rxn:QR2}. What are the prerequisites for \eqref{rxn:QR2} to occur?

All reactions are possible. H2S can be oxidized to sulfur, sulfite, or sulfate.

Well, permeakra's answer suggests me it's quite hard to produce SO2 in this case!

I never said it's easy. In fact, I think, permeakra is right. SO2 will be produced as a minor product, if at all. Still, it will be produced at least in some cases, hence the reaction is possible, like I said.

But I thought why not $\ce{SO2}$

While the reaction is stechiometrically correct, it is hard to perform.

In excess of $\ce{H2S}$ sulfur will be produced.
In excess of oxidant sulfate will be produced.

And exact balance is impossible to achieve.
That said, $\ce{SO2}$ reacts with  $\ce{H2S}$ finally producing sulfur. This reaction, just like many inorganic reactions, is an unholy mess of many different processes with many products obtained simulationaly and the only way to produce understandable and reproduceable results is to simplify the situation. One of the way to simplify the situation is to use a vast excess of one of the reagents.

Have you seen the second answer? It says H2S will also be produced in this reaction. will it?

Actually the first reaction is possible at normal conditions.
$$
\begin{align}
\ce{3H2S + K2Cr2O7 + 4H2SO4 → 3S + Cr2(SO4)3 + K2SO4 + 7H2O}   & \tag{R1}\\
\end{align}
$$
Now, if you observe the two reactions, you see that there is 7 moles of water molecule in 1st reaction and 5 moles of water molecule in 2nd reaction. So, the 2 moles of water must somehow reacts with sulfur to produce sulfur dioxide.
$$
\begin{align}
\ce{3S + 2H2O ->[400 °C] 2H2S + SO2}   & \tag{R2}\\
\end{align}
$$
You observe that 2 moles of hydrogen sulfide gas is produced which get cancelled from each side and give the 2nd reaction.
$$
\begin{align}
\ce{3H2S + K2Cr2O7 + 4H2SO4 &-> (3S + 2H2O) + Cr2(SO4)3 + K2SO4 + 5H2O}    \tag{R1}\\
\ce{3H2S + K2Cr2O7 + 4H2SO4 &-> (SO2 + 2H2S) + Cr2(SO4)3 + K2SO4 + 5H2O}  \tag{From R2}\\
\ce{H2S + K2Cr2O7 + 4H2SO4 &-> (SO2) + Cr2(SO4)3 + K2SO4 + 5H2O}   & \tag{R3}\\
\end{align}
$$

I changed the alignment of the bottom equations to make them more legible. Hope you agree with the change.

@BuckThorn I'd probably go even further and retag '(From R2)' to something shorter of a label, and also replace full reactions with the net ionic equations. That way it's arguably easier to track the difference and there will be no horizontal scroll bar. But this is probably for the author to decide.

Three critics : 1)The equations are not readable. 2) Potassium is missing. 3) H2S appears on both sides in equation (2).


text:
Terminal Alkyne in Suzuki Coupling
I've done a very similar reaction to the one below but by TLC I observe only starting materials. It's super clear that nothing else is going on. Conditions: K3PO4, Pd(dppf)Cl2, dioxane/water, heat to 100 deg.C.

What thoughts do you have on why this doesn't work?

I imagine the alkyne chelating to the Pd and potentially 'poisoning' it? I currently lack the words and technical lingo to put this into a formal explanation.
A competing Sonogashira perhaps? Unlikely given the aqueous conditions.
Any other issues?

I don't see the alkyne interacting with the Pd being an issue. Any competing Sonogashira will be dependent on your choice of catalyst, base and solvent.

@Waylander Assume fairly ordinary conditions: dioxane/water, K3PO4 as a base and a slight excess of the boronic acid. Heat to 100 deg. C. I've actually done a similar reaction and did not see any coupling product. By TLC it was literally just starting material. What would you propose as the issue?

@Waylander I've updated the reaction to more closely align with what I did. I can't show the exact structure for confidentiality reasons. NOTE: I have carried this out in the lab to no avail.

What catalyst did you use?

Try Fu's standard 'wet' Suzuki conditions of tricyclohexyl phosphine, Pd(dba)2Cl2, K3PO4, in dioxane/water as described in Tet. Letts (2001) 4295

@Waylander Pd(dppf)Cl2.DCM

Is transmetallation able to outcompete a $\beta$-migratory insertion?

If that's the problem, maybe just try a very large protecting group on the terminal alkyne...

If TLC is the only analysis we have, it may be difficult to determine exactly what is happening (or not happening). Other things that might be happening include homocoupling (biaryl formation), proteo-deborylation, and debromination. These may be hard to discern by TLC alone.
If there is no reaction of starting materials this might be due to:

Failure to reduce Pd(II) precatalyst to active Pd(0) species: your catalytic cycle never initiates. Consult chapter 1 of Negishi's Handbook of Organopalladium Chemistry for Organic Synthesis. Consider adding a catalytic amount of exogenous reducing agent, changing the solvent to EtOH/PhMe/H2O, increasing the temperature, switching to a Pd(0) catalyst system, or a system that is more easily reduced to Pd(0) like Pd(OAc)2 + xs phosphine ligand.
Stalling in the catalytic cycle: there is insufficient energy to get over the activation barrier in the oxidative addition, transmetallation, or reductive elimination step. Try increasing the temperature.
Catalyst destruction by oxidation: did you degas your system with N2 or Ar?

Alkyne poisoning of the homogeneous Pd catalyst seems unlikely since there are plenty of examples with Sonogashira coupling of terminal alkynes involving dppfPdCl2 that work just fine.


text:
Precision Balance (1mg) vs. Analytical Balance (0.1mg)
I'm in the lucky position to building my own home lab and I'm looking to get a nice balance, but I don't want to waste money on something I won't use though.
I'm currently about to start second year chemistry (Organic) and I plan to take at least a few courses beyond this as time allows (I already have my Comp Sci degree and a full time job and I've been in and out of grad school for Mathematics and Computer Science). So this is just a hobby.  As far as what I'll be doing I honestly don't know chemistry well enough to say exactly, but I think it'd be fun to synthesize various molecules you come across and see what kind of purity I could get after running the experiment a few times.
I probably won't ever get much past 4th year undergrad or 1st year graduate level knowledge since I can only take one night class at a time.
I understand that analysis and p chem have a decent amount of use for sub-mg level accuracy, but how often do these situations come up in practice?
Should I spend the extra on a 0.1mg balance or just go for a 1mg balance?  This decision won't affect future lab purchases much more likely just the timeline of the purchases. Eventually I'd like to get additional useful instruments as I start needing them like a centrifuge and/or autoclave. But I probably won't ever get a FT-IR for example.

What's the cost differential you're looking at?

Around $800. So 0.1 mg balance would be about 2x price.

Either of these balances will require some care to set up. It will need to be on a solid surface away from any drafts, etc. However, neither would have the more expensive setup and maintenance requirements of say a $\pu{0.001mg}$ balance. And in either case you can take care of the cleaning and calibration yourself, particularly as it's just a hobby with no certification requirements.
I would seriously consider looking at used equipment from a place like LabX (I have no affiliation whatsoever). I would hesitate to purchase a used $\pu{0.001mg}$ balance because that is a very sensitive piece of equipment, but a $\pu{0.1mg}$ analytical balance; my lab always bought that sort of thing used.
You can probably get a good used $\pu{0.1mg}$ Mettler for example for the same or less than a new $\pu{1mg}$ balance and there is really no downside. It's easy to imagine you wanting to weigh out a few mg of some reagent and having the ability to do that with the extra accuracy of at least a $\pu{0.1mg}$ balance would probably be desirable.

Thanks for the link.  LabX looks great.  I plan to get much of my equipment used, but I'm fine with buying a new balance.  I was looking at the Ohaus AX series.

I'm glad it was helpful. I really can't speak to Ohaus directly, but I knew other labs that used them and I never heard complaints (and lab people do like to complain about troublesome equipment ;) Best of luck.

Analytical balances require special care in setup (e.g. air current shielding) and some maintenance so if you think the environment is going to be a bit more "rough and tumble" it might not be worth the money. Given the price differential, I think the prudent move would be to purchase the precision balance.


text:
Which resonance structure to use for writing curved arrow mechanisms?
If a structure with delocalized electrons participates in a reaction, which resonating structure do you use to write down the mechanism? Is using either equivalent? Is it preferred to use the resonance hybrid? If so, what are the rules of using curved arrows for partial bonds?

Generally, you can use either resonance form, and no, don't use the hybrid, that makes it impossible to draw mechanisms

@orthocresol point made. Usually, try the reactions with both resonance forms and see which one actually works out.

I nominate this for reopening. This is a good question commonly asked by beginning students of organic chemistry and has an unequivocal answer.

Any resonance structure can be used. Why? Resonance structures represent multiple views of the same compound, so it must be possible to start with any valid resonance structure. Because resonance structures consider different positions of electrons, the sets of arrows needed to describe the mechanisms of different resonance structures will be different.
While it's possible to use any resonance structure in a mechanism, selecting one resonance structure can be used to improve communication or emphasize particular property/reactivity of a compound.
As an example, below is a figure describing the formation of an enolate and its attack on an aldehyde (aldol reaction) using different resonance structures. For clarity, all of the curved arrows are describing reactions, not the relationships between resonance structures.

We know that carbonyls such as 1 have an important resonance structure which places a negative charge on oxygen and leaves a positive charge on carbon (2). While 1 is the most important contributor, 2 emphasizes the carbocationic nature of the carbonyl carbon. I like this structure, because I know some reactions of carbocations: SN1 and E1. This helps explain to me why the alpha-position is acidic. Deprotonation of 2 next to the carbocation, just as in E1, gives enolate 3. Thinking of the carbon-oxygen pi-bond as a leaving group, deprotonating 1 at the alpha- position gives the same enolate 3 but by a different set of arrows compared to 2 to 3.
Note that there is a third way (not shown) to describe the same deprotonation where a base deprotonates the alpha-hydrogen and the electrons are simply localized on the alpha-carbon to give structure 4.
Structures 3 and 4 are resonance structures, and again these can both be used to describe further chemistry. The blue set of curved arrows from 3 to 5 contains one additional electron pair movement to the red set from 4 to 5. Both are valid descriptions of the attack on an aldehyde, but emphasize different aspects of enolate reactivity.
Keep in mind that bond-line drawings, hybrid drawings, and curved arrows are just simple representations of molecules. Thankfully they work very well and allow us to reduce the complexity of electronic structure to a manageable level.
Finally, as @orthocresol suggests in the comments, using hybrid drawings is difficult. This is because bond-line drawings represent electron pairs as localized and curved arrows work best when showing the action of electron pairs.


text:
Why do helium balloons expand in volume as they go higher?
I realize as balloons go higher, the atmospheric pressure decreases, doing less to counteract the force of the gas particles pushing against the inner walls of the balloon. But at the same time, doesn't the outside temperature decrease, causing the gas particles in the balloon to lose energy and have less impactful collisions and a lower velocity? Is the reason why the volume nonetheless increases because the pressure outside decreases faster than the temperature? Or am I overthinking this?

I didn't know that balloons expanded during the fly because of thermodynamics, and I didn't know how high they can fly, but a rapid search tells that a partially unfilled regular balloon can fly until an altitude of around $\pu{25 km}$.
Now, $\pu{25 km}$ means that it reaches the first part of the stratosphere, with temperatures of $\pu{-60 ^\circ C}$, that gradually increase until $\pu{0 ^\circ C}$ at $\pu{50 km}$.
As for the pressure, it goes from around $\pu{40 mmHg}$ to $<\pu{1 mmHg}$ in the range $25\text{–}\pu{50 km}$.
If you try a $pV=nRT$ calculation on these data, you see that the gas is already at around 10 times its initial volume with $\pu{40 mmHg}$ pressure and a $\pu{213 K}$ temperature, and that at the $\pu{50 km}$ point the volume is increased 700 times!
Also: while the trend of the pressure is quite logical, that of temperature is caused by complex interations (eg: sun rays that heat particles). You can find this image quite interesting:

You are exactly correct that it is a matter of atmospheric pressure decreasing at a rate great enough to overcome the contraction due to decrease in temperature. On a nice, clear, dry 25°C day at sea level, atmospheric pressure decreases by about 12% per km, where the air temperature decreases by about 3% per km.
This is very similar to the process that allows convectional clouds to expand with height and the resulting thunderstorms to form. If an air mass started to rise then contract due to cooling, that would be the end of that convection event.

Thanks to the ideal gas law, the volume of the balloon doesn't really depend on the fact that the gas inside is helium in particular. So we should get the same volume if we fill the balloon with ordinary air and hoist it into the atmosphere mechanically instead of letting it rise by its own buoyancy.
In this modified thought experiment, the density of the gas inside the balloon is exactly the same as the density of the surrounding air, assuming that both pressure and temperature are equalized.
If the volume of the balloon stayed the same, this would require that the atmosphere has the same density at all altitudes. But then there would need to be infinitely much of it, which is absurd in conflict with observations: It is well known that the density in space approaches zero, and since the atmosphere is not a liquid, it has no distinct upper edge, so its density must decrease gradually as we move up.
Also, if the density were the same everywhere, air at sea level would not be compressed by the weight of the atmosphere above us, which is also absurd.
(Consider what happens if we compress a portion of air with a piston -- say, a bicycle pump. Both pressure and temperature will increase -- but temperature does not rise so fast that the volume will stay constant!)


text:
Why are lithides not known?
In the last few decades, many alkalides - anions of alkali metals - have been synthesised. The most famous is undoubtedly that of sodium: $\ce{[Na(\text{2.2.2-cryptand})]+Na-}$, but the alkalides $\ce{K-}$, $\ce{Rb-}$, and $\ce{Cs-}$ are all known. However, $\ce{Li-}$ is not.
James Dye writes in a 1984 review1 that for the product of the reaction
$$\ce{M(s) + N(s) + L(s) -> [ML]+N-(s)}$$
to be thermodynamically stabilised ($\ce{M}$, $\ce{N}$ are metals, and $\ce{L}$ is the macrocyclic ligand), several criteria must be met:

Small lattice energies for $\ce{L(s)}$, $\ce{M(s)}$, and $\ce{N(s)}$ so that the enthalpies of sublimation will not be too large.
Low ionization energy of $\ce{M}$.
High electron affinity of $\ce{N}$.
Large complexation energy of $\ce{M+}$ by $\ce{L}$.
Large lattice energy of $\ce{[ML+]N-}$ (which depends mainly on the interionic separation).

From these given factors, I suppose the only possible explanation is the larger heat of sublimation of lithium. However, lithium is still known to form electrides in solution - compounds of the form $\ce{[ML+]e-}$. In order for these to be formed, the sublimation energy of $\ce{Li}$ still has to be overcome.
Is there a thermodynamic reason why lithides, $\ce{Li-}$, have not yet been made?
Reference

Dye, J. L. Electrides, negatively charged metal ions, and related phenomena. Prog. Inorg. Chem. 1984, 32, 327–441. DOI: 10.1002/9780470166338.ch4.

This page tells me that transition metal lithides have been observed, although my level of knowledge isnt sufficient to explain what they're saying: http://pubs.acs.org/doi/abs/10.1021/jp952638g?journalCode=jpchax&

The context of that article is vastly different (computational study on discrete MLi molecules, which aren't $\ce{M+Li-}$), so it's not relevant to this question.

Perhaps some of these will prove of limited interest: doi: 10.1021/ja056314+; doi: 10.1002/cphc.201200805; doi: 10.1021/jp8040023; doi: 10.1146/annurev.ms.23.080193.001255; doi: 10.1063/1.2393225.

The more interesting question is how sodides can form. Sodium electron affinity is ~52 kJ/mol, but heat of vaporisation of sodium metal is 72 kJ/mol.

One barrier is the greater tendency of lithium to form partially covalent bonds than heavier alkali metals.  This is a good thing for organic chemists, since this stabilizes organolithium compounds and enhances the solubility of lithium chloride as a chloride ion source (see this answer).  But it's not so hot for forming lithides.  Covalent bonding to more electronegative elements, meaning to practically all other elements in this case, oxidizes the intended lithide ion.  In effect the cryptand in an intended lithide is converted into one of those organolithium compounds.
Linear Christmas, in the comments to the question, hints at a way to beat this tendency.  Two of the references given there, https://doi.org/10.1021/ja056314+ and https://doi.org/10.1002/cphc.201200805, involve theoretical investigations of lithides built with pyrrole-based cryptands.  Such cryptands resist covalent bonding to lithide (or other nucleophilic) anions because this would destroy the aromatic of the ring and require electrons to overlap into a high-lying, strongly antibonding orbital.  These compounds, or counterparts with heavier alkalide ions, draw interest for their expected nonlinear optical (NLO) properties, making them synthetic targets.  So, stay tuned.
Polyatomic lithide species have been characterized in the gas phase, the simplest being the linear $\ce{Li3^-}$ ion and the rectangular $\ce{Li4^-}$ ion (both geometries given are for ground states). See Alexandrova et al [1].
Reference

Anastassia N. Alexandrova, Alexander I. Boldyrev, Xiang Li, Harry W. Sarkas, Jay H. Hendricks, Susan T. Arnold, Kit H. Bowen; "Lithium cluster anions: Photoelectron spectroscopy and ab initio calculations". J. Chem. Phys. 28 January 2011; 134 (4): 044322. https://doi.org/10.1063/1.3532832.

Not too complicated of an answer - don't make a mountain out of a mole-hill. This is simple nuclear stability. You can just google "Nuclear Band of Stability" and it explains the necessary balance between protons, neutrons and electrons. Li-1 would have 3 protons, 4 neutrons, and 4 electrons. Crowded around such a small nucleus, the electrons would experience too much repulsion which would cause at least one to be ejected from the cloud. Additionally, 3 measly protons does not suffice for another electron, at a greater distance, experiencing electron repulsion, to experience an even greater force by effective nuclear charge.

If you didn't notice, electron affinity of lithium is positive and $\ce{Li-}$ anion perfectly possible. https://en.wikipedia.org/wiki/Electron_affinity_(data_page)

Downvoting but not delete-voting because wrong answers should stay but clearly be labelled as wrong.


text:
Mixing water and oil in microgravity outside of a container
What would happen if I  where to mix water and oil in a vacuum in microgravity  where the temperatures  were low enough so that the oil and water remain liquids also, the mixture is not inside of a container.
 I would expect the water and oil two separate into two blobs, but wouldn't the London dispersion forces hold the two blobs together?

https://www.google.com/amp/lightyears.blogs.cnn.com/2011/10/18/oil-and-water-in-microgravity/amp/

An experiment similar to this was carried out using a free-fall drop shaft to provide the microgravity environment in Japan in 2000.
From the abstract of the publication$^1$:

An experimental study was performed to obtain the detailed information needed to provide a deep understanding of the combustion process and the secondary atomization of an oil-in-water emulsion droplet. The oil-in-water emulsion, which consisted of n-hexadecane as a base fuel, distilled water, and a trace of surfactant was tested. Photographic observation and temperature measurements were made of the suspended emulsion droplet during the heating-up and combustion processes under microgravity. The primary attention was toward the phase separation in the droplet, and the time histories of droplet temperature and the amount of water in the droplet, during the period of time prior to disruptive microexplosion. The results showed that the separation of the base fuel and water as well as their agglomeration and coalescence occurred with the lapse of time. The increase in the droplet temperature resulted in phase separation, and the formation of a single water droplet enveloped by a shell of the base fuel, prior to the microexplosion.

Interestingly, the phase separation was attributed to the increase in droplet temperature. It's my guess that the heating increased the rate of separation, but that given time the final result of "the formation of a single water droplet enveloped by a shell of the base fuel" would have taken place, and I think this would be the result of the situation described in the question; that van der Waals forces between the oil and water would prevent them from separating into two separate balls of liquid. The stronger hydrogen bonds of water were likely the reason for water forming on the inside of the droplet, and I would expect the same to happen in the case described in the question.
Summary, TL;DR:
Based on the results of the drop-shaft microgravity experiment described above, I would expect similar results for an experiment as described in the question. There should be an inner ball of water held together by hydrogen bonding, surrounded by an oil shell held together, and to the surface of the water, by the weaker van der Waals forces.
1) "Water-coalescence in an oil-in-water emulsion droplet burning under microgravity", Proceedings of the Combustion Institute, Volume 28, Issue 1, 2000, Pages 985-990.

The "trace of surfactant" gives a bias towards a large interface, favouring the described core-shell structure.

Absolutely, but I think this is just a matter of how quickly the preferred structure is achieved. Not knowing the concentration of "trace" makes this a bit speculative, but I think it's unlikely that something other than the core-shell structure would form in the absence of the tract of surfactant. I guess it's conceivable that it is holding the 2 phases together, but I suspect that they would be held together by van der Waals (London) forces as proposed in the OP and your answer. By the way, nice catch in your answer bringing up the pressure issue.

The core-shell structure will be stable if the interfacial tension is less than the surface tension of water minus 0.59 ($=1-2^{2/3}$) times the surface tension of the oily phase (assuming they have the same volume), if my reasoning is correct.

Your reasoning, that there are still London dispersion forces between the oil and the water, is correct. However the contact area will either be small, because you still have to deform the ideal spherical phase boundaries of each phase, or the phase with higher surface tension will surround itself with the other phase (core-shell droplet, see other answer). It depends on the relative values of the surface/interfacial tensions.
(It does not work in vaccuum, though. The water will boil until it freezes. Just use a big enough container filled with air instead.)

Wouldn't the air affect the forces holding the two together? That is to say, there would be forces between the liquids and the air that would make it more preferable for them to seperate.

@Te55eract Not measurably.


text:
Are there alternative ways to find n or number of e- in problems like the one pictured?
I need help understanding how we are to come up with 12 e- because according to the directions, we are supposed to focus on any element, but it doesn't tell us what to look for. I see how carbon has a -2 value in the reactants side and a +4 on the products side. the difference is 6 e- and we are suppose to multiply that by the number of carbons.
But is there another way to calcualte to find 12 e- ?
Couldn't we add the charges from the reactants and products after multiplying them respectively with the subscripts? For instance: -2 X 2 = -4 or 4 e- and +4 X 2 = 8 or 8 e- and so 4 e- + 8 e- = 12 e- ?
Isn't this a valid substitute than the explanation below? Because the directions below ask us to look for the difference in any of the elements. But this doesn't explain how to calulcate for hydrogen. But in the explanation I prepared above does, doesn't it?
But then again, this doesn't explain how we can calculate for oxygen, which has a 0 charge in it's own gas phase and a -2 in water and -2 in carbon dioxide.
So what would be another way to calculate for -12 e- ?

In this case, we've to calculate the oxidation states of carbon in the reactant and product side separately, the difference between those two and finally multiply it with the number of carbon atoms. While this is may be a seemlingly simple calculation, things can get uber-tricky because of the mammoth number of exceptional cases. Let's start with the basics; here's a gift for you:
$${\begin{array}{|l|c|} \hline
\pu{Element} & \pu{Oxidation Number} & \pu{Remarks} \\\hline
\pu{Alkali Metals} & \pu{+1} & \pu{Always} \\
\pu{Alkali Earth Metals} & \pu{+2} & \pu{Always} \\
\pu{Flourine} & \pu{-1} & \pu{Always} \\
\pu{Halogen Family} & \pu{-1} & \pu{Generally} \\
\pu{Oxygen Family} & \pu{-2} & \pu{Generally} \\
\pu{Hydrogen} & \pu{+1} & \pu{Generally} \\\hline
\end{array}}$$
Whoa! I see three exceptional cases already (remarked "generally" in the table) and things haven't started yet. What's up with them?
Oxygen: Oxygen doesn't always exist as an oxide ion $(\ce{O^2-})$ in its compounds, it can often-times exist as

Peroxide $(\ce{O_2^2-})$: Oxidation state in this case is $-1$. Popular examples include $\ce{H2O2}$, $\ce{Na2O2}$, $\ce{BaO2}$, etc.
Superoxide $(\ce{O_2^-})$: Oxidation state is $-0.5$. Common examples $\ce{KO2}$, $\ce{RbO2}$, $\ce{CrO2}$
Ozonide ($\ce{O3-}$): $\frac{-1}{3}$ oxidation state; for instance in $\ce{KO3}$.

Halogen atoms (except flourine): When $\ce{Cl, Br or I}$ combine with elements of higher electronegativity ($\ce{O and F}$), then these halogen atoms can show positive oxidation numbers; commonly $\pu{+1, +3, +5, +7}$. In $\ce{BrF5}$, that's an inter-halogen compound, $\ce{Br}$ is in $+5$ oxidation state.
Hydrogen: In case of metal hydrides,  it can show $-1$ oxidation state. As in $\ce{NaH}$.
Phew, er any more exceptions? You dare ask? Metals. Wait! Aren't metals electro-positive!? Yes, they are, but in some cases they can show zero or even negative oxidation numbers. For example in $\ce{CsAu}$, gold shows a $-1$ oxidation state. Let's bring in coordination-compounds: nickel in $\ce{[Ni(CO)4]}$ finds itself in zero oxidation number.
There are plenty of them, I'm saving them for another day.

Let's try calculating things you've asked.
For ethanol ($\ce{C2H6O}$) we don't know the oxidation state of the carbon atom. Why don't we assume it to be $x$. Now, the oxidation state of a compound is either zero (if it's a neutral species) or it's equal to the charge it bears. Is ethanol neutral? Yes it is. So the individual oxidation numbers of elements (times the number of elements bearing that oxidation state) should sum up to be zero.
$$2x+6(+1)+1(-2)=0$$
Solving, we get $x=-2$. Doing the same thing for carbon-dioxide, we can get $x=+4$. So a carbon atom changed its oxidation state from $-2$ to $+4$, that's a jump of $+6$. And how many carbon atoms were doing this? Two carbons. That gives us $n=12$.


text:
AP Chemistry Practice exam
It's starts off the question with "Propanoic Acid, $\ce{C2H5COOH}$, is an organic acid that is a liquid at room temperature. An incomplete Lewis diagram for the propanoic acid molecule is provided in the box below. Complete the diagram, showing how the remaining atoms in the molecule are arranged around the carbon atom marked with an asterisk (*). Your structure should minimalize formal charge and include any lone pairs of electrons."
I don't really know the best way to do A. For B I said $\text{sp}^2$ hybridization. For Ci, I said intermolecular forces and dipole dipole forces. For Cii, I said dipole dipole forces. Double check me please.

Oxygen forms two bonds. Where will it be placed? Think carefully.

Didn't it go with carbon? The asterisk one.

Would oxygen go into the right side of the carbon with the asterisk, then another oxygen right next to it?

For part a, I assume that the question is asking you to complete the structure of the propanoic acid. Carboxylic acids always have a $\ce{-COOH}$ functional group as follows:
https://upload.wikimedia.org/wikipedia/commons/thumb/b/b5/Carboxylic-acid.svg/1200px-Carboxylic-acid.svg.png
As you can see, the R in propanoic acid is the ethyl ($\ce{CH3CH2}$) group.
For part b, the carbon is indeed sp2 hybridised as the carbon in question has 3 bond pairs.
For part c(i), the types of intermolecular forces present will be weak London Dispersion Forces, Hydrogen Bonding (due to $\ce{-OH}$) and Dipole-Dipole Forces.
As for part c(ii), both butanoic acid and propanoic acid contain the same functional groups, except that butanoic acid contains an extra $\ce{CH3}$. Hence, the same types of intermolecular forces will be present. However, the strength of London Dispersion Forces in butanoic acid will be stronger than that in propanoic acid since butanoic acid has a larger and hence more polarizable electron cloud. As a result, butanoic acid has a higher boiling point than propanoic acid.