text:
Complete combustion of mixtures
I am working on my scholarship exam practice (around high school and first year university level). I could actually solve the problem below but my solution is pretty time consuming. It would be appreciated if I could find other aspects in solving this.

Exactly $\pu{4.32 g}$ of oxygen gas was required to completely burn a $\pu{2.16 g}$ sample of a mixture of methanol and ethanol. How many moles of ethanol are contained in the sample? (Answer: $\pu{0.04 mol})$

What I did is setting $x$ and $y$ as the mass of methanol and ethanol, respectively. Then I formed 2 equations. First one was $(x + y = \pu{2.16 g})$ and another was the equation $(3x/64 + 3y/46 = \pu{0.135 mol})$ from the number of moles of oxygen gas, derived from two complete combustion equations. And then I solved simultaneous equations. I am wondering if this was the only way to solve this problem. If you think differently, please let me know.

Yes, for two variables you need two equations. You have some choices in how you define the variables and equations. Namely do you want to have the variables be masses or moles.

Good news is you are in right track and set up two equations correctly. However, I think the calculations would get easier if you have chosen your units of variables in  $\pu{mol}$. Also,  you may not needed to do more calculations since the answer is in $\pu{mol}$. Thus, your two equations should be:
$$32x + 46y = 2.16 \;\text{.... (1), and}$$
$$\frac{3}{2}x + 3y = 0.135 \;\text{.... (2)}$$
Simplify equation (2) as:
$$x + 2y = 0.090 \;\text{ and }\therefore x=0.090-2y $$
Substitute $x$ in equation (1):
$$32(0.090-2y) + 46y = 2.16 $$
$$(64-46)y = 32 \times 0.090 - 2.16 = 0.72$$
$$\therefore y = \frac{0.72}{18} = 0.04$$

I can think of another solution, which is somewhat more tedious, but as long as you asked to think differently, here it is: you can utilize the fact that both components are alcohols and share the same empirical formula $\ce{C_yH_{2y+2}O}$, $y\in (1,2)$.
This allows to find the unknown amount from the average molar mass $\bar{M}$ of the mix:
$$\ce{C_yH_{2y+2}O + 3/2$y$ O2 -> y CO2 + $(y + 1)$ H2O}$$
Let's denote methanol with 1, ethanol with 2.
Then amount of ethanol $n_2$ in the mix of amount $n$ can be found via its mole fraction $x_2$:
$$n_2 = x_2n \label{eqn:1}\tag{1}$$
Mole fraction of a component in the binary mix can be expressed in terms of molar masses:
$$x_2 = \frac{\bar{M} - M_1}{M_2 - M_1} \label{eqn:2}\tag{2}$$
Total amount of the mix can also be found from the total mass $m$ as well as reaction stoichiometry, expressing $n$ via the amount of oxygen:
$$n = \frac{m}{\bar{M}} = \frac{2}{3\cdot y}n(\ce{O2}) = \frac{2\cdot m(\ce{O2})}{3\cdot y\cdot M(\ce{O2})} \label{eqn:3}\tag{3}$$
Now, $y$ can be found by solving an equation with one unknown from the combination of average molar mass' definition and \eqref{eqn:3}:
$$
\begin{cases}
\bar{M} = y\cdot M(\ce{C}) + (2\cdot y + 2)M(\ce{H}) + M(\ce{O}) \\
\displaystyle\bar{M} = \frac{3\cdot y\cdot m\cdot M(\ce{O2})}{2\cdot m(\ce{O2})}
\end{cases}
$$
or
$$y\cdot M(C) + (2\cdot y + 2)M(H) + M(\ce{O}) = \frac{3\cdot y\cdot m\cdot M(\ce{O2})}{2\cdot m(\ce{O2})} \label{eqn:4}\tag{4}$$
Plugging in the numbers and solving \eqref{eqn:4}, one finds $y = 1.81$ and $\bar{M} = \pu{43.41 g mol-1}$.
From \eqref{eqn:2} $x_2 = 0.81$; from \eqref{eqn:3} $n = \pu{4.98e-2 mol}$.
Finally,
$$n_2 = 0.81\cdot \pu{4.98e-2 mol} = \pu{4.03e-2 mol}$$


text:
Reaction Mechanism with a Rate Determining Step
Just a quick question on how to deal with multiple intermediaries in a proposed reaction mechanism.
Here is the overall reaction:
$$\ce{I-(aq) + OCl-(aq) -> IO-(aq) + Cl-(aq)}$$
Here is the proposed mechanism:
$$
\begin{align}
\ce{OCl- + H2O &<=>[$k_1$][$k_{-1}$] HOCl + OH-} & &\text{(Fast equilibrium)} \\
\ce{I- + HOCl &->[$k_2$] HOI + Cl-} & &\text{(Slow/RDS, non-reversible)} \\
\ce{HOI + OH- &->[$k_1$] H2O + IO-} & &\text{(Fast)}
\end{align}
$$
A few things to note, the question does indeed label both the first reaction going forwards and the third reaction going forwards as $k_1$. It also mentions that we should treat the concentration of water as being constant.
Here is my question:
I know how to write the rate law for the Rate Determining Step as $k_2[\ce{I-}][\ce{HOCl}]$ and then I know I need to use the equilibrium expression in the first reaction to isolate concentration of $[\ce{HOCl}]$ as
$$[\ce{HOCl}] = \frac{k_1[\ce{OCl-}]}{k_{-1}[\ce{OH-}]}$$
but then I am confused about how to isolate the second intermediary $(\ce{OH-})$ so I can get the proper rate law. Should I use the third reaction?

You should use steady state on intermediate species HOI, i.e. $\ce{d[HOI]/dt=0=\cdots }$ and the rate of product formation is $\ce{d[IO^-]/dt=k_3[HOI][OH^-]}$. Use these two equations then the equilibrium to get $\ce{OCL^-}$  in the final answer.

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text:
What is the electronic structure of Mn in KMnO4 and K2MnO4?
I know that is is d3s hybrid. But in my book it is unclear which orbitals participate in pi bond with O. I think it should be both p and d orbitals of Mn that do.

Sigh... please see https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-for-transition-metal-complex . Can the hybridisation tag be retired?

OK, I'll bite.  What is a d3s hybrid?  C'mon man!

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text:
Given solubility of sodium carbonate, find mass of hydrate sodium carbonate dissolved?
I am working on a problem from my scholarship exam practice (MEXT). I am not quite sure if I got it right. The question was provided with multiple choices (a) to (f) below:

The solubility of sodium carbonate in $\pu{100 g}$ water is $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$. How many grams of the hydrate $\ce{Na2CO3∙10H2O}$ can be dissolved in $\pu{100 g}$ of water at $\pu{22 ^{\circ}C}$? (a) $\pu{0.556g}$; (b) $\pu{0.762g}$; (c) $\pu{9.27g}$; (d) $\pu{67.5g}$; (e) $\pu{81.7g}$; (f) $\pu{117g}$

I honestly do not know how to solve this problem. My best guess is to find the moles of sodium carbonate that can be dissolved in water first (from $\pu{25.0 g}$). And use that moles equate to the moles of hydrate form (1:1). Then find the mass of hydrate form by multiplying its molecular weight and my answer would be $\pu{67.5g}$. There is no answer key provided so I would like to hear other opinions and get some advice.

To answer this kind of problem, it'd be better starting with elimination approach. It is given the solubility of sodium carbonate in $\pu{100 g}$ of water as $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$.  By that, you know, solubility of its hydrate should be more than $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$ because the mass of hydrate is including additional mass of water as well (as your calculation shows). Therefore, you can eliminate answers (a), (b), and (c) directly, because they are less than 25 and are not correct. Your calculations shows it must be greater than $\pu{67.5 g}$ (therefore, answer (d) is incorrect as well). It should be greater than $\pu{67.5 g}$, because you forgot to include additional amount of salt, which would dissolve in 10 equivalents of water in that added salt: Molar mass of $\ce{Na2CO3}$ is $\pu{106 gmol^{-1}}$ while Molar mass of $\ce{Na2CO3.10H2O}$ is $\pu{286 gmol^{-1}}$. This means, each $\pu{mol}$ of hydrate has $\pu{186 g}$ or $\pu{186 mL}$ of water. That is a lot and we need to consider that as well in this question. Now we have only two answers to be considered: (e) and (f).
To do the calculations, suppose $x\; \pu{g}$ of $\ce{Na2CO3.10H2O}$ would dissolve in $\pu{100 g}$ of water at $\pu{22 ^{\circ}C}$. Thus,
$$\text{amount of }\ce{Na2CO3}\text{ in the solution} = \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
$$\text{and, amount of }\ce{H2O}\text{ in the solution} = \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
From given data for anhydrous salt, we can conclude that upon saturation:
$$\text{The amount of }\ce{H2O}\text{ in the solution} = 4\times \text{amount of }\ce{Na2CO3}\text{ in the solution}$$
$$\therefore \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g} = 4 \times \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
Solve for $x$:
$$\left(4 \times 106 - 180\right)x = 28600 \; \text{and therfore, } x=\frac{28600}{244}=117.2$$
Therefore your answer is (f) ($\pu{117 g}$).

Your calculation would be good, if you did not forget that hydrates contain water.
The solubility of the anhydrous carbonate is 25 g in 100 g of water.
Both carbonate forms, once diluted, are the same compound, forming the same ions plus eventually releasing water.
So in case of the hydrate, the ratio carbonate:water must be 1:4 as well.
$M_{\ce{Na2CO3}}=106\mathrm{~g/mol}$.
$M_{\ce{Na2CO3. 10 H2O}}=286\mathrm{~g/mol}$.
Using $x$ grams of the hydrate + $100$ grams of water
produces solution of $x \cdot 106/286 \mathrm{g}~\ce{Na2CO3} $
in $100 + x \cdot 180/286~ \mathrm{g}$ of water.
I will leave the rest on you to give you the honour of some effort, as this question is rather the homework class of questions.
See Homework


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Electron-donating trends for ligands bound to metal complexes
Imagine an inorganic complex with a bunch of terminal ligands. All else the same, would $\ce{F-}$ or $\ce{Br-}$ ligands be more electron-donating and why?
Of course, in terms of $\mathrm{p}K_{\mathrm{a}}$, $\ce{F-}$ is more basic than $\ce{Br-}$, predominantly due to the smaller size of the $\ce{F-}$ anion. I have also read in multiple places that the $\mathrm{p}K_{\mathrm{a}}$ trends can be used to describe Lewis acidity/basicity trends, although it's not immediately clear to me why this is true. If one takes the prior statement at face-value, then it appears that that $\ce{F-}$ is more Lewis basic and, by the very definition of a Lewis base, is more electron-donating than $\ce{Br-}$. However, why would $\ce{F-}$ be more electron-donating if its electronegativity is higher than $\ce{Br-}$? These two points seem paradoxical.
For full context, I am trying to understand why $\ce{F-}$ would be better at stabilizing higher oxidation states of metals in inorganic complexes than $\ce{Br-}$, which based on the literature appears to be related to the electron-donating strength of the ligands (with more electron-donating ligands increasing the stability of higher oxidation states).


text:
Relationship between equilibrium constants of a multi-step reaction
If a reaction proceeds in steps, is the equilibrium constant of every step taken into account to arrive at the final constant? Or can we just calculate equilibrium constant for the overall reaction with reactants on one side and the products on the other?

It might help if you provide a concrete example. In general, you can compute an equilibrium constant between any reagents or products, irrespective of the number of intervening steps in the mechanism, as long as they are in equilibrium.

To give a specific example, let's take this set of reactions:
$$\tag{1}\ce{A + B <=> I_1}$$
$$\tag{2}\ce{I_1 + C <=> I_2}$$
$$\tag{3}\ce{I_2 <=> I_3 + D}$$
$$\tag{4}\ce{I_3 <=> E + F}$$
To get the net reaction, we add up the equations and cancel the intermediates:
$$\ce{A + B + C <=> D + E + F}$$
When the system is a equilibrium, all the reactions are at equilibrium. You can write the equilibrium constant expression based on the net reaction:
$$K_\text{net} = \frac{[\ce{D}][\ce{E}][\ce{F}]}{[\ce{A}][\ce{B}][\ce{C}]}$$
You would arrive at the same expression if you take the product of the individual equilibrium constants (and cancel):
$$K_\text{net} = K_1 K_2 K_3 K_4 =\frac{[\ce{I_1}]}{[\ce{A}][\ce{B}]} \times \frac{[\ce{I_2}]}{[\ce{I_1}][\ce{C}]} \times \frac{[\ce{I_3}][\ce{D}]}{[\ce{I_2}]} \times \frac{[\ce{E}][\ce{F}]}{[\ce{I_3}]}$$
The concentrations of the intermediates will not be zero because they are in equilibrium with the other species. They are intermediates in the sense that we care a bit less about them since they don't appear in the net equation of interest. However, they are part of the mass balance and may well be among the major species.


text:
How is the aromaticity in graphene different from the aromaticity in benzene?
The web page Aromaticity in Graphene and other 2-D Systems begins:

I. Graphene
While the σ-bonding in graphene is assumed to be a rigid honeycomb framework built out of two-center two-electron (2c-2e) C-C σ-bonds, the π-bonding is supposed to be delocalized. However, according to the adaptive natural density partitioning (AdNDP) analysis and the electron sharing indices, graphene is aromatic, but its aromaticity is different from the aromaticity in benzene, coronene, or circumcoronene. Aromaticity in graphene is local with two π-electrons located over every hexagon ring. (emphasis added)
II. BC3 Honeycomb Epitaxial Sheet
[...]

Update: The wording is similar to what is written in Is Graphene Aromatic? Popov, BozhenkoEmail &  BoldyrevEmail (2012), Nano Research, 5, (2), pp 117–123 (open access here):

ABSTRACT
We analyze the chemical bonding in graphene using a fragmental approach, the adaptive natural density partitioning method, electron sharing indices, and nucleus-independent chemical shift indices. We prove that graphene is aromatic, but its aromaticity is different from the aromaticity in benzene, coronene, or circumcoronene. Aromaticity in graphene is local with two π-electrons delocalized over every hexagon ring. We believe that the chemical bonding picture developed for graphene will be helpful for understanding chemical bonding in defects such as point defects, single-, double-, and multiple vacancies, carbon adatoms, foreign adatoms, substitutional impurities, and new materials that are derivatives of graphene.

Question: How is the aromaticity in graphene different from the aromaticity in benzene? It is likely that those familliar with the topic will already understand the explanation in the block quote, but is it possible to find a simpler if less accurate way to explain the difference, in a similarly compassionate way to how this answer was extremely helpful to my question about double bonds and chromophores?

Might not be quite so simple - this "adaptive natural density partitioning (AdNDP) analysis and the electron sharing indices" stuff is very specialised. I've never even heard of them.

@orthocresol the word "simple" never crossed my mind ;-)

I'm skeptical about their conclusions for circumcoronene.

@orthocresol I think I've found the source for the wording, see update.

@A.K. (ditto)..

I might be too naive but at the end it means that while in benzene there is a sextet delocalised in a ring, in graphene the delocalisation over the neighbours results in just two pi electrons delocalised but pinned onto a single ring. The other are described by extended orbitals and are those responsible for the in plane electrical conductivity. I am aware that worded this way might even sound trivial.

You'll probably get some mileage from this wikipedia comment: " In fused aromatics, not all carbon–carbon bonds are necessarily equivalent, as the electrons are not delocalized over the entire molecule. The aromaticity of these molecules can be explained using their orbital picture. Like benzene and other monocyclic aromatic molecules, polycyclics have a cyclic conjugated pi system with p-orbital overlap above and below the plane of the ring." https://en.wikipedia.org/wiki/Aromaticity#Characteristics_of_aromatic_(aryl)_compounds

In their paper in Nature materials (2007), Fasolino et al. note that the bonds in free hanging graphene are not all the same lengths and the graphene is rippled. The paper, titled “Intrinsic ripples in graphene”, is freely available [here]. (https://arxiv.org/pdf/0704.1793.pdf). Don’t know if this helps any.

@EdV oh that's interesting I'll dig into that right away! The distinction between rippling and buckling is important as comments on and the answer to Are all stable Xenes (graphene-like 2D honeycomb sheets) buckled? mention.

Simpler explanations, can by their nature be less accurate than a detailed description, but are helpful in beginning to understand things.  Getting to the heart of your question:
“How is the aromaticity in graphene different from the aromaticity in benzene? . . .  but is it possible to find a simpler if less accurate way to explain the difference,”
Perhaps a productive way to approach your question [portion given above] is to consider the molecular structure of coronene and circumcoronene while contrasting them to the interpretation of aromaticity of graphene given in your cited article.
 circumcoronene
 CORONENE
Graphene from article cited in question.
According to the computational method employed “adaptive natural density partitioning method” [we can choose to agree or not agree with their interpretation as indicated by a comment to your post by @A.K.], the aromaticity of graphene is centralized within the set of hexagonal rings.  To quote the authors “Aromaticity in graphene is local with two π-electrons delocalized over every hexagon ring.”  The salient point here being the “delocalization” is within and isolated to or restricted to [my words] a six-membered ring.
As discrete molecules, the aromaticity in coronene and circumcoronene, as depicted the PubChem structures, is delocalized throughout the entire discrete molecule and not restricted to a discrete set of hexagonally bonded carbon atoms.
In other words, it comes down to electron density distribution.  In graphene, according to the authors of the article cited, the areas of highest electrons density are within the center of the six-membered ring while in coronene and circumcoronene the electron density is more uniformly distributed over the entire molecule.
Ref:
https://pubchem.ncbi.nlm.nih.gov/compound/coronene
https://pubchem.ncbi.nlm.nih.gov/compound/Circumcoronene

Wow, thank you very much for your insightful answer! It will take a few days for me to get back to the point where understood as much as I did, and then read through carefully, but I can get a general idea now.

Looking at the question only by comparing  Benzene versus Graphene is not as helpful as thinking about aromaticity in coronene vs graphene.

getting closer... :-)


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Why does lemon juice reduce the "fish" odor of sea food — specifically fish?
In studying amines, I read that lemon juice is also used to wash fish because it reacts with the amines on and in the fish to convert the amines to its salt, just reducing the "fishy smell".
I was thinking that it must be the citric acid in lemon juice that somehow reacts with the amines to convert them to their salts.
What is the general chemical reaction of citric acid with amines in fish, and why does this reaction actually reduce the "fishy" smell?

It's all about an acid-base reaction: amines are basic. And the amines found in fish are quite volatile.
Addition of acid forms non-volatile salts (eg: aminium citrates)

There are two cases, both related to acid-base reactions. Both are also partial reasons why so many fish recipes use lemon juice.
Fish, especially sea fish, naturally contain trimethylamine-N-oxide $\ce{(CH3)3N-O}$ that, after death, gets enzymatically reduced to trimethylamine $\ce{(CH3)3N}$, the source of the ammonia-like fish odour.

Trimethylamine N-oxide is an osmolyte found in saltwater fishes, sharks, rays, molluscs, and crustaceans. It is considered a protein stabilizer that may serve to counteract urea, the major osmolyte of sharks, skates and rays. It is also higher in deep-sea fishes and crustaceans, where it may counteract the protein-destabilizing effects of pressure.TMAO decomposes to trimethylamine (TMA), which is the main odorant that is characteristic of degrading seafood.
TMAO is biosynthesized from trimethylamine, which is derived from choline.

In acidic environments, TMA forms water-soluble, non-volatile, odourless trimethylammonium salts, which eliminates the volatile smelly trimethylamine.
$$\ce{(CH3)3N + H+ <=>> (CH3)3NH+ }$$
It is the direct analogy to the smelly volatile ammonia and odourless ammonium:
$$\ce{NH3 + H+ <=>> NH4+}$$

Another source of fishy smell/taste, especially of some sweet water species like carp or catfish, is geosmin, a polycyclic alcohol, being a derivate of decahydronaphthalene(decaline). It is released after the death of some Streptomyces or cyanobacteria cells which gives a muddy or earthy taste, and the human smell sensors are extremely sensitive to it (5 ppt).
Geosmin is also responsible for the smell of the soil after rain, of some beetroots and even of drinking water of surface origin. There is speculation the nose sensitivity for geosmin is an evolutionary response to the need for efficient searching for water.
Geosmin reportedly breaks down in acidic environments.


text:
Thermodynamics behind organic chemical reactions?
I took organic chemistry a few years ago at my university, and although I went through the motions to pass the exams, I didn't understand the thermodynamic reasoning behind why the reactions actually occurred. For instance, I memorized that Claisen Condensation reactions occur between two esters in the presence of a strong base, but I don't fully grasp WHY the reaction happens in the first place.  Why are the products more stable than the reactants in cases like these? Does this reaction always occur whenever these components are in a mixture, or is there a sort of catalyst to trigger it? I feel similarly confused for many other organic reactions (e.g., Aldol condensations, Fischer esterification, Sn2, etc). I suppose I'm wondering if there is a general thermodynamic reasoning I'm missing in organic chemistry.

Generally speaking, we would refer to the concept of the Gibbs free energy change (https://en.wikipedia.org/wiki/Gibbs_free_energy) of reaction when talking about the thermodynamic driving force of reactions. It essentially relates to the 2nd Law of Thermodynamics, which tells us that the total entropy of the universe always increases. This is the essential idea behind all reactions, not just those in organic chemistry.

Alkoxides form ester enolates that add to electrophilic carbonyls of neutral esters with loss of alkoxide to form beta-ketoesters. Because beta-ketoesters are much more acidic than the conjugate acid of the alkoxide, they are deprotonated by the alkoxide which drives the reaction to completion. Thus, the reaction is at least stoichiometric in base. In your example, 2 beta-keto esters will form. The other one is where  the red and blue esters switch places.

Whether or not the product is "more stable" would really depend on the sense in which you consider its reactivity. The $\mathrm{p}K_\mathrm{a}$ of a $\beta$-keto ester is about 11, so in that sense, it is much more reactive than either of the reactants. Once the $\beta$-keto ester is formed it is quickly deprotonated. It remains predominantly in this form because the anion is so stable, and the reaction is incapable of reversing. Thermodynamics does not always determine the result of an organic reaction. Many times, kinetics is the deciding factor.
The Claisen condensation is catalyzed by the base, as it is recovered in the end, and though I am sure it occurs spontaneously without it, I doubt it happens very often. I don't think ethyl acetate would be as popular a solvent as it is if it quickly and spontaneously converted to ethyl acetoacetate.

I wouldn't say the Claisen is base-catalysed, one equivalent of base is consumed in the reaction (to form the enolate of the product).

I suppose that’s fair, but another equivalent of alkoxide is generated ($\ce{OR'^-}$).

Yes but no; if you check the mechanism very carefully (took me a while to do), overall the transformation is $$\ce{2RCH2COOR' + OR'- -> RCH2C(O)C(R)^-CO2R' + 2ROH}$$ The first equiv of base is used to deprotonate the ester starting material. After addition of enolate into ester, the tetrahedral intermediate kicks out one equiv of base, so we are at net zero base consumption/production here. However, you then need one equiv of base to deprotonate the b-ketoester.


text:
Ordering the lines obtained in the H line spectrum based on energy they possess
I am trying to do the following question which asks me to calculate the wavelength of the green line in the visible spectrum of H. The authors claim that the green line is the second most energetic line but I believe that's a misprint because wavelength of green is second largest so energy should be second least. Am I right?

The Balmer series [https://en.wikipedia.org/wiki/Hydrogen_spectral_series#Balmer_series_(n%E2%80%B2_=_2)] is in the visible region of the spectrum, so this is presumably the group of transitions you should examine. However there is no line which corresponds to green colored light

@NightWriter I know from that we get that $n_final=2$ but to figure out $n_initial$ we have to use the relative energy of the green line in the spectrum. I'm pretty sure there is a green line in the spectrum :)

Ok but then I'm confused about the question because the hydrogen spectral series (emission spectrum) does not contain a green line (see the source I linked to).

@NightWriter Eh? Transition from $n = 4$ to $n = 2$ corresponds to $λ = \pu{486 nm}$, which is quite green to me (looks more like cyan on the image, but still).

@andselisk Oh, ok. I would have called that cyan.

@NightWriter The line having wavelength $486.1$ nm is the green line

@NightWriter Ha, I've just edited this into my comment:)

I would agree with that argument but it seems like potentially a problem with abuse of language. The "green" line is more energetic than the red line at 656 nm.


text:
CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions
I have been stuck on a chemistry problem for a long time now and if anyone here can help me I would be eternally grateful.

40. Du tillsätter $\pu{100 ml}$ $\pu{0,01 M}$ $\ce{Na2SO4}$-lösning i $\pu{100 ml}$ $\pu{0,02 M}$ $\ce{CaCl2}$-lösning. Det bildas en $\ce{CaSO4}$-fällning. Hur mycket av $\pu{0,01 M}$ $\ce{Na2SO4}$-lösning måste ännu tillsättas för att den bildade $\ce{CaSO4}$-fällningen just och just ska upplösas helt? $K_\mathrm{s}(\ce{CaSO4}) = \pu{2,5e-5 M}$. $(t = \pu{25 °C})$ (2p)
a. $\pu{0,43 l}$
b. $\pu{0,48 l}$
c. $\pu{0,53 l}$
d. $\pu{0,58 l}$

Translation: We add 100 ml of 0.01 M $\ce{Na2SO4}$-solution to 100 ml of 0.02 M $\ce{CaCl2}$-solution. A precipitate of $\ce{CaSO4}$ is formed. What is the (minimal) volume of 0.01 M $\ce{Na2SO4}$-solution that needs to be added to the mix for the $\ce{CaSO4}$ precipitate to be just dissolved completely?
The problem then lists the solubility constant, $K_\mathrm{s}$, for $\ce{CaSO4}$ as $2.5\cdot 10^{-5}$.
The only way I can think of solving this myself is by looking at the added $\ce{Na2SO4}$-solution as just water and using the $K_\mathrm{s}$ value for $\ce{CaSO4}$ to see what concentration of $\ce{CaSO4}$ is possible in just water. Then calculate the amount of moles of precipitate that was actually formed to see how much water I would need to add to the mixture:
$$K_\mathrm{s}(\ce{CaSO4}) = \pu{2.5e-5 M^2} = [\ce{Ca^2+}][\ce{SO4^2-}]$$
$$\therefore [\ce{Ca^2+}] = [\ce{SO4^2-}] = \sqrt{\pu{2.5e-5 M^2}} = \pu{5e-3 M}$$
$$C = \frac{n}{V} \to n = CV$$
Amount of precipitate formed:
$$n(\ce{CaSO4}) = \pu{0.02 M} \cdot \pu{0.1 dm^3} = \pu{0.002 mol}$$
Required total volume of water to dissolve $\pu{0.002 mol}$ of $\ce{CaSO4}$:
$$V = \frac{n}{C} = \frac{\pu{0.002 mol}}{\pu{5e-3 M}} = \pu{0.4 dm^3}$$
This is wrong, but I can't figure out the right way to go about this. The answer surely has to do with how well $\ce{CaSO4}$ is dissolved in a $\ce{Na2SO4}$-solution as opposed to just water.

When all of the $\ce{CaSO4}$ is dissolved completely, all the calcium ions originate from the $\pu{100 mL} \; \ce{CaCl2}$ solution, so it is easy to determine the amount:
$$n_\ce{Ca^2+} = \pu{100 mL \times 0.02 mol/L = 2 mmol}$$
To get the total sulfate amount, we have to consider what is present initially and what we add when we add a volume of $V_\text{add}$ of $\ce{Na2SO4}$.
$$n_\ce{SO4^2-} = \left(\pu{100 mL} + V_\text{add}\right) \times \pu{0.01 mol/L}$$
From this, we can figure out the concentration:
$$c_\ce{Ca^2+} = \frac{\pu{2 mmol}}{ V_\text{add} +\pu{200 mL}} $$
$$c_\ce{SO4^2-} = \frac{\pu{(100 mL} + V_\text{add}) \times \pu{0.01 mol/L}}{ V_\text{add} +\pu{200 mL}} $$
The product has to be equal to $K_s$:
$$K_s = \frac{\pu{2 mmol}}{ V_\text{add} +\pu{200 mL}}  \times  \frac{\pu{(100 mL} + V_\text{add}) \times \pu{0.01 mol/L}}{ V_\text{add} +\pu{200 mL}}$$
This gives a quadratic equation for $V_\text{add}$, giving you the numeric result upon solving it. The answer will be approximate because it assumes that volumes are additive and that activities are independent of spectator ion concentrations.

I like the fact that you added to two caveats that make the answer approximate. We don't explicitly state the assumptions frequently enough here.

This is an exercise where I see a mismatch between the level of description of the chemistry and the level of math required to solve it. You might learn more chemistry if the question had been "What volume 10 mM  $\ce{Na2SO4}$would it take to dissolve 2 mmol $\ce{CaSO4}$? Compare that to the volume of pure water it would take to dissolve the same amount of calcium sulfate."

Nice observation, I agree. The chemistry problem is buried under so much math that any appreciation of the chemistry is lost. Students really need the problems to reinforce the chemistry so they can  develop better chemical intuition.

Totally agree with @MaxW on the assumptions. Nice response. Agreed on the other comment as well. This would we a nicer problem as: (1) set up the equilibrium expression in terms of the added volume $V$ and (2) state any assumptions.

As you've already noted, there are 3 substances of interest here: $\ce{Ca^{2+}}, \ce{SO4^{2-}}, \ce{H2O}$.
Let $V$ be the additional volume of sodium sulfate solution. Then:
The total volume is $V + 0.2\ \mathrm{L}$.
The total amount of sulfate is $(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})$.
The total amount of calcium is $(0.1\ \mathrm{L}) (0.02\ \mathrm{M})$.
We have then the following relationship:
$$\ce{[Ca^{2+}][SO4^{2-}]}=\frac{(0.1\ \mathrm{L}) (0.02\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\frac{(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\leq K_{\mathrm{sp}} = 2.5\times 10^{-5}$$
Now, it's just algebra to solve for the threshold value of $V$, keeping in mind that there are two solutions for $V$ because this is a quadratic equation. In this case, the correct solution is the positive root:
$$V \approx 0.48\ \mathrm{L}$$
The mistake you made is that the calcium and sulfate ion concentrations are definitely not equal, since the additional volume of solution contains only sulfate and not calcium.

Goggle translate from Swedish yields a slightly different translation.

You add $\pu{100 ml}$ of $\pu{0.01 M} \; \ce{Na2SO4}$ solution in $\pu{100 ml}$ of $\pu{0.02 M} \; \ce{CaCl2}$ solution. A $\ce{CaSO4}$ precipitate forms. How much of $\pu{0.01 M}\; \ce{Na2SO4}$ solution has yet to be added to completely dissolve the $\ce{CaSO4}$ precipitate formed? $K_\mathrm{s}\ \ce{(CaSO4)} = \pu{2.5e−5 M}. (t =\pu{25 ^{\circ}C})$

The key point being that the problem is looking for the "extra" amount of the $\ce{Na2SO4}$ solution to be added, not including the initial $\pu{100 ml}$.
To start let's convert the volumes to liters. So $\pu{0.100 L}$ of $\pu{0.01 M} \; \ce{Na2SO4}$ solution is added to $\pu{0.100 L}$ of $\pu{0.02 M} \; \ce{CaCl2}$ solution.
Let's let $V_e$ be the extra volume of the $\pu{0.01 M} \; \ce{Na2SO4}$ solution, which needs to be added.
We know that for the product of the final concentrations, $\ce{[Ca^{2+}]}_f$ and $\ce{[SO4^{2-}]}_f$, is equal to the $K_\mathrm{sp}$.
$K_\mathrm{sp} = \ce{[Ca^{2+}]_f \cdot [SO4^{2-}]_f }$
Now the amount of  $\ce{Ca^{2+}}$ is $\pu{0.1 mol/L} \times \pu{0.02 L} = \pu{0.002 mol}$. There is $\pu{0.2 L}$ of solution initially to which  $V_e \; \pu{L}$ of additional $\pu{0.01 M} \; \ce{Na2SO4}$ solution is added. So the final volume is $\left(0.2 + V_e\right)$ liters. Thus,
$\ce{[Ca^{2+}]}_f = \dfrac{0.002}{0.2+V_e}$
The amount of  $\ce{SO4^{2-}}$ is $(0.1+V_e)\times0.01$ moles. There is $0.2$ liters of solution initially to which  $V_e$ liters of additional $\pu{0.01 M} \; \ce{Na2SO4}$ solution is added. So the final volume is again $\left(0.2 + V_e\right)$ liters. Thus:
$\ce{[SO4^{2-}]}_f = \dfrac{(0.1+V_e)\times0.01}{0.2+V_e}$
Now making the substitutions:
$K_\mathrm{sp} = \dfrac{0.002}{0.2+V_e} \times \dfrac{(0.1+V_e)\times0.01}{0.2+V_e}$
Solving the quadratic equation yields $V_e \approx \pu{0.483 L}$


text:
while using Henry's law does it matter if the solvent is volatile or not?
I am studying about binary solutions (with liquid solvents) and came across Henry's law. This is how it's stated in my textbook:

Henry's law states that at a constant temperature the solubility of a
  gas in a liquid is directly proportional to the partial pressure of
  the gas present above the surface of the solution.

The way it's been stated does not say if the liquid solution is volatile or not.
So, while using this law, does it matter if the liquid solvent is volatile or non volatile?

The partial pressure is irrespective of any other component of the system in the gaseous phase. All we're doing is establishing that the equilibrium $\ce{A(g) <=> A(aq)}$ is a function only of temperature and the identify of $\ce{A}$.

@Zhe so since the partial pressure is irrespective of any other component of the system thus it does not really matter if the solvent is volatile or not.?

That depends on what you mean by "does it matter?" It does not matter for the purposes of Henry's law, but that fixes the solvent. If the solvent is different, the constant of proportionality is different. So actually, my previous state is not as accurate as possible: given a fixed $\ce{A}$ and fixed solvent, the equilibrium is only a function of temperature.

@Zhe actually i meant to ask the question for a fixed solute and solvent. i should have mentioned it in the question.

In that case, I think you have your answer. :)

The main importance with Henry's law is that the solute be very dilute

@CharlieCrown yeah there's always a risk of it turning into raoult's law :D


text:
Basic strength of nitrogen
What is the order of basic strength of nitrogens in the compound given in the figure?
$$\ce{
H2\overset{.\!\!.}{\underset{(γ)}{N}}-\overset{.\!\!.}{\underset{(β)}{N}}H-\overset{\overset{\huge O}{||}}{C}-\overset{.\!\!.}{\underset{(α)}{N}}H2
}$$
The most basic is the γ-N since its lone pair can't go in conjugation. My doubt is between α and β nitrogens, whose lone pair can go in conjugation.

For comparison between the alpha and beta nitrogen atoms, you may want to refer to the alpha effect: https://en.m.wikipedia.org/wiki/Alpha_effect

Tang Yong Boon certainly is in the correct direction (please refer to his comment).
I'll just wrap things up:
$$\gamma > \beta > \alpha $$
$$\mathrm{(Basic \ strength)}$$
Reasoning:

The $\gamma$ nitrogen is unquestionably the strongest basic site.
$\beta$-nitrogen follows, as its tendancy of lone pair donation is greatly enhanced due to the presence of an adjacent (ie. alpha wrt. itself) electron donating group ($\ce{-NH2}$), otherwise known as the alpha effect. Some explainations include destabilization and elevated energy of the ground state, resulting in lower activation barrier.
The $\alpha$-nitrogen is best called an acidic site.


text:
Can anyone help me with this question on vibrational modes of free base porphyrin (C20H14N4)?

4 a. The structure of free-base porphyrin $(\ce{H2P}$, chemical formula: $\ce{C20H14N4})$ is shown below. Given that this molecule belongs to the $D_\mathrm{2h}$ point group (character table given above) identify the symmetries (irreps) for all the vibrational modes of this planar polyatomic molecule;

(i) Identify the number of vibrational modes in $\ce{H2P}$. [3 marks]
(ii) Work out the reducible representation for the vibrational modes. [6 marks]
(iii) Use the reduction formula to identify the (irreps) for all the vibrational modes. [6 marks]
(iv) Identify which of the vibrational modes are Raman-active and which are IR-active. [4 marks]

In particular I'm getting stuck with figuring out the unmoved atoms when applying symmetry operations. I know i need to calculate gamma vib in order to use the reduction formula but I cannot picture how to apply the symmetry operations to calculate unmoved atoms for each symmetry class (aside from E) because it is a more complicated structure than I'm used to dealing with (h20, benzene etc)

One thing to think about in the question: is the free base porphyrin really D2h symmetric?

If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.
Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.


text:
Making a soapy (bubble) solution
There are fun and practical applications of soapy solutions for bubble making.  An example of a practical one is detection of gas leaks from fittings and unions.
Given various household products, what would have the best combination of wetting agents or surfactants for bubble creation at temperatures above 0C and up to 40C?  For example, hand soaps, dish soaps, etc. realizing that many products vary in ingredients and design intent.  For example, most laundry detergents are designed to have reduced foaming.
What common household chemicals or products will make effective soap bubble solutions?
Addendum:
1. The following article describes creating soap bubble mixtures:  https://www.scientificamerican.com/article/bring-science-home-best-bubbles/
2. It mentions the use of glycerin and/or corn-syrup as a thickening agent.  "Glycerin and corn syrup form weak bonds with the water molecules and slow down the evaporation process, thus improving the life span and durability of the bubble. Glycerin makes stronger, longer-lasting bubbles, but corn syrup is often substituted in bubble solutions because it is cheaper."  (This is consistent with comments for the question.)
A significant remaining question is what detergent product might perform better over others, or what detergent product characteristics might perform better over others.  Obviously one without anti-foaming agents.

More commercial soap less water. For instance at least  2:1 vol using concentrated dish soap. Some add glycerol but I don't know the physical reason in details so just a comment.

The wiki on glycerol provides little explanation, "Glycerol is added to bubble mixture for blowing soap bubbles. It creates longer-lasting bubbles by thickening the mixture."

Glycerol simply lowers the vapor pressure of the water thus slowing evaporation.


text:
Stability of the alkenyl and phenyl cations
One reason often cited for why alkenyl and phenyl halides do not undergo nucleophilic substitution by the $\ce {S_N1}$ mechanism is that the formation of the alkenyl and phenyl cations involve a prohibitively large activation energy since they are rather unstable. In fact, this reason has been discussed in various posts on this site; see for example, here and here. Recently, I chanced upon some calculated energies for these cations in the gas phase and it seems to suggest that they are not as unstable as various textbooks and educators claim. The following table is taken from p. 300 of Carey and Sundberg (2007):

From the data above, we clearly see that the alkenyl carbocation is more stable than the primary carbocation although still significantly less stable than the secondary carbocation. Perhaps, we should be cautious when we discuss the stabilities of these carbocations because some sources seem to suggest that these alkenyl carbocations are so unstable that they are even less stable than primary carbocations. One thing that is consistent with most sources that the phenyl carbocation is the least stable of the lot.
There seems to be a cut-off point of stability, after which the $\ce {S_N1}$ reaction proceeds with much difficulty. This "cut-off point" seems to be between the stability of the alkenyl carbocation and the secondary carbocation. Is my interpretation correct?
Reference

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.


text:
Molecular orbitals of the enolate ion
The reactivity of the enolate ion is most frequent explained with reference to its system of $\pi$ molecular orbitals. The image below taken from Wikimedia Commons depicts this system of $\pi$ molecular orbitals. $\ce {C}$-alkylation is favoured when the orbital interaction with the $\ce {HOMO}$ of the $\pi$ system is dominant while $\ce {O}$-alkylation is favoured when the electrostatic interaction with the negative charge built up on the oxygen atom is dominant. This is the case because the coefficient of the $\ce {HOMO}$ is the largest on the terminal carbon atom while being more electronegative, oxygen naturally bears greater negative charge.

The explanation provided by Fleming (2010) on p. 160 is that in the lowest-lying $\pi$ molecular orbital, the oxygen atom exerts the strongest polarising effect on the electron distribution while for the second highest $\pi$ molecular orbital, the polarisation is in the opposite direction. I understand that these MO diagrams are grounded in quantum chemistry theory but is there any way that we could intuitively understand why the coefficient of the $\ce {HOMO}$ is indeed the largest on the terminal carbon?
Update
After discussion with some friends, I have more thoughts I would like to share. There is this connection between resonance structures and MO diagrams. For example, the MO diagrams of the allyl cation and allyl anion feature a node in the second MO, informing us that there is no sharing of the charge with the central carbon atom. This information is also reflected when we draw the two resonance structures. Can we perhaps, extend this connection a little further and say that the first MO for this enolate $\pi$ system reflects the charge distribution in the resonance structure with negative charge localised on the oxygen atom while the second MO reflects the charge distribution in the resonance structure with the negative charge localised on the carbon atom?

Image source
Reference
Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference ed.). United Kingdom : Wiley, 2010.

Not directly relevant to the question, but regarding the issue of C vs O alkylation, conventional HSAB theory is not correct,
https://doi.org/10.1002/anie.201007100

Not sure if I answer your question so I leave a comment: In $\pi_2$, there is a node between C and O due to the opposing signs of the wavefunctions. You won't find any electrons here and the oxygen can't pull electrons on the other side of the node due to the deconstructive interference of the waves, hence why the polarization is in the opposite direction towards the terminal carbon.

This is not directly related, but you might want to give my answer a read: How to rationalise with MO theory that CO is a two-electron donor through carbon? Similar reasoning can be applied explaining the coefficient of pi2 at carbon.

One possible way to intuit the result is to remember that all of the orbitals are normalized. Thus, the contribution of a single AO to the three MOs must add up to 1, and the contribution of the three AOs in each MO must also add up to 1. (This is a bit of an oversimplification, since the normalization is of the square of the wave function, but conceptually it's the same idea). So if a particular AO contributes a lot to one MO, it is proportionally less represented in the others. Also, the other AO's have to contribute less to the first MO in order for it to have the right magnitude.
In MOs composed of AOs from atoms of differing electronegativity, the lower energy MOs have a larger contribution from the more electronegative atom (and therefore smaller contribution from the less electronegative atoms). The higher energy MOs must therefore have a smaller contribution from the electronegative atom.
As a very simple first example, consider the $\sigma$ and $\sigma^*$ MOs of a polar diatomic molecule such as HF. The valence F AO is much lower in energy than the hydrogen orbital, so the bonding MO has a large contribution from F and very small contribution from H. In fact, the MO is nearly indistinguishable from the fluorine AO, as evidenced by the bond being nearly completely ionic. In order for the normalization to be correct, the antibonding orbital must be the reverse - a very small contribution from F and large contribution from H.
A slightly more complicated example is a triatomic like H2O. Here is a drawing of the lowest three valence MO's for XH2, when X is more electronegative than H:

(This is a portion of Figure 7.3 of Albright et al. Orbital Interactions in Chemistry).
You can see that again the electronegative atom (O) is overrepresented in the lowest energy MOs, at the expense of the H atomic orbitals. Therefore, in the high energy orbitals, we have to compensate with the reverse, an overrepresentation of the H orbitals (as seen in the 3a1).
Now getting to your question of C-C-O, again the O is most electronegative, so it contributes most to the lowest energy orbital. That means it has less to contribute to the higher energy orbitals, where the carbon orbitals are the bigger contributors.
This overly simplistic approach is harder to apply to the difference between the two carbons, for which you really have to incorporate the actual math of the atomic orbital contributions, but maybe it at least helps with the concepts.

Thanks for the answer. Your point on the normalisation of the orbital coefficients is helpful and leads us closer to the answer. However, I am interested as to why exactly does the largest amplitude of the 2nd MO lie on the terminal carbon. Maybe, this simply cannot be figured out based on intuition...

Is it intuitive to think about denisty decreasing near nodes? Consider that if it were three equivalent atoms, there would be a node through the central atom in $\pi_2$. Even though the node is shifted a bit here, the density still tapers off approaching the node.  I'd also add that the basic math of combining AOs to make MOs is not super complicated. If it's something you're really interested in, it's worth learning it right.

Minor point of correction: The depicted orbital scheme uses four orbitals, there would be one missing.

@Martin-マーチン - Are you referring to the three MOs from H2O that I've shown? You're correct that four AOs are involved, but since I've only used the  $\sigma_g$ combination of the H orbitals, there are really only three orbitals being combined - the oxygen $p_z$ and $s$ and the H2 $\sigma_g$. (To complete the MO diagram for H2O, there is one MO that is the oxygen $p_y$ and two that result from combination of oxygen $p_x$ with H2 $\sigma^*_u$)


text:
Unusual oxidation numbers for alkaline earth metals?
Are there compound of alkaline earth elements showing unusual oxidation numbers? For the sake of the question, every oxidation number different from +2 counts as unusual, most exciting would be oxidation numbers greater than +2.

This answer is an extension to @Ian Bush answer. Not only magnesium, but every group 2 element has a lower oxidation state of +0(beryllium, magnesium) and +1(for all AE metals lighter than radium).

Be(0)

A 2016 paper gives an insight to the existence of a zero valent beryllium complex compound. Previously the stability of Be(0) compounds had been computationally predicted; the referenced paper reports the first actual synthesis.

[...]The team added a single ligand i.e CAAC = cyclic
(alkyl)(amino)carbene to $\ce{BeCl2}$ in benzene solution to give
$\ce{(CAAC)BeCl2}$, and obtained the final product,
$\ce{[Be(CAAC)2]}$, by reducing the beryllium with KC8 in the presence
of a second equivalent of CAAC ligand.

Be(I)

High resolution infrared emission spectra of beryllium monohydride and
monodeuteride have been recorded. The molecules were generated in a
furnace-discharge source, at 1500 °C and 333 mA discharge current,
with beryllium metal and a mixture of helium and hydrogen or deuterium
gases.(source)

Mg(I)

It has been described in @Ian Bush's answer. For more information, refer to these sources(1 and 2). Some examples are $\ce{Mg2RuH4, Mg3RuH3, and Mg4IrH5}$ having the $\ce{Mg-Mg}$ bond and magnesium diboride containing the metastable $\ce{Mg2^{2+}}$ ion(Credit @Oscar Lanzi).

Mg(0)

This more recent (2021) addition has been obtained as an organic complex containing the $\ce{Na2Mg2^{2+}}$ cluster by Rösch et al. The authors also report a compound containing the mixed $\ce{Mg(0)}$-$\ce{Mg(I)}$ species $\ce{Mg3^{2+}}$.

Ca(I)

A Ca(I) Sandwich Complex $\ce{[(thf)3Ca(μ-C6H3-1,3,5-Ph3)Ca(thf)3]}$ is being described in this 2010 paper.

Sr(I)

The high-resolution infrared spectrum of gas-phase $\ce{SrF}$ was
obtained in emission with a Fourier transform spectrometer.[...]
(Source)

Ba(I)

3 isotopes of barium in $\ce{BaF}$ (Source).  In [2018] Wu and Lerner (source) reported a barium(I) in a complex graphite intercalation compound.

Ba(III, IV, V)?

Various studies have suggested that elements can assume oxidation states outside their usual range under GPa-level pressures. Luo et al. (2018) have performed such calculations with barium fluorides and predict the possibility of $\ce{BaF3,BaF4,BaF5}$. These proposed compounds await experimental verification.

I did not had to google each and every element to search for lower oxidation state. The wikipedia article of oxidation element gave a list of all posible O.S of element. Information regarding unusual O.S of any element can be found in the footnote and thus only clicked the relevant elements for the answer.

How does $\ce{MgB2}$ contain magnesium-magnesium bonds?  When I look at sketches of the structure it looks like there are boron-boron and boron-magnesium bonds (the latter possibly having ionic character to make up the electron deficiency in the boron) instead.

See this reference for the ionic bonding.

Yes, you are right @Oscar. I will edit my answer.

Tganks @nilay.  The quoted reference does not, however, mention any magnesium(I) species.  It renders magnesium fully ionized with two positive charges.  Most of the compensating negative charge in the normal state is in metallic bonding, but switches to one-half metallic, one-half ionic Mg-B bonding in the superconducting state.

In 2018 a graphite intercalation compound was reported showing evidence for barium(I).  Feel free to roll back if you find this inadequate.

@Oscar no worries. You always learn something new everyday : )

Thanks, and stay safe.  How to incorporate the reference in hidden form like you did for the others?

@Oscar I used the option hyperlink to embed links into the text.

Does not work for me possibly b/c using a mobile device cannot see the option.  Thanks.

@Oscar I always use my pc if I want to write an answer. It gives me the option to insert links, pictures and make my text bold, italics etc. If you want to write answer in mobile, then use the stack exchange app but I cannot guarantee as I heard(ortho said to someone) that the app is crappy and full of bugs.

FYI I add a new reference for Mg(0).

Magnesium(I) compounds are known, but they are of the form $\ce{[Mg-Mg]^{2+}}$ rather than a bare $\ce{Mg+}$ - so more like Mercury than Sodium. See  https://pubs.rsc.org/en/content/articlelanding/2011/dt/c0dt01831g#!divAbstract, and in a way this furthers the similarity between Mg and Zn. Similar compounds are also possible for the heavier metals.

"...the similarity between Mg and Zn." I think this is overplayed. In the +2 oxidation state you see some divergence, e.g. in the crystal structure of the oxides and basicity of the hydroxides. They are due to the zinc ion having lower-energy acceptor orbitals making it a stronger Lewis acid. In lower oxidation states where Lewis acidity is not so important, the Mg-Zn similarity is stronger.


text:
Will lead bind to dissolved polymer (Kapton) monomers?
We are attempting to determine the concentration of lead impurities in Kapton. Essentially we need to extract lead in the tape polymer Kapton (structure seen below);

The amount of lead in the tape is expected to be extremely low.  I plan on dissolving the Kapton into its monomers using a mixture of NaOH and hydrazine(N2H4). The process of then extracting lead ions from solution is figured out. However, I have to be sure that approximately 100% of the lead which is in the Kapton is dissolved in solution and not bonded to the Kapton monomers.
Will the oxygens on the monomers bond with any lead ions and therefore inhibiting any further reaction with that lead atom? If so is there a way to separate the lead atoms from the Kapton monomers?
Since we actually add lead separately as we are looking at the isotopic ratio, another solution to this issue would be if the lead binds to the Kapton only rapidly and is then replaced with another lead atom. This would mean that the ratio of added lead and lead impurity are equal in the solution as well and in the Kapton monomers. Is this a valid assumption over a certain time period?
Any comments would be helpful. Thank you.


text:
What acids will Sodium Lauroyl Methyl Isethionate yield after hydrolysis?
As far as I understand, after hydrolysis Sodium Lauroyl Methyl Isethionate will yield sodium, lauroyl and methyl alcohols, and isethionic acid. Meaning lauroyl is referring to the fatty acid alcohol rather than to the acid itsIf.
But I have read that this is compound is defined as "the sodium salt of the methyl lauric acid ester of isethionic acid", which sounds like it would actually yield lauric acid rather than the alcohol.
So which is it? Will it yield lauric acid after hydrolysis or lauroyl alcohol? And how can I tell based off the name?
I think I may be mixed up in my understanding of salts versus esters, since I definitely have more experience with the latter. Most of the information I have found on the nomenclature of salts is not comprehensive enough to discuss complex salts such as this one. If you are aware of such a reference, please share.

Isethionates are esters of long-chain aliphatic carboxylic acids ($\ce{C8-C18}$) with isethionic acid (2-hydroxyethanesulfonic acid) or salts thereof, such as ammonium isethionate or sodium isethionate (Wikipedia). Therefore, hydrolysis of any of them may yields more than one acid.

Is there any way to tell based on the name which acids might be yielded?

Since you said it is a Sodium Lauroyl derivative, it is already a salt of isethionaic acid (sulfonate). Upon hydrolysis, it release hydroxy terminal of isethionate. So, you got salts of both acids (assuming base hydrolysis).

Lauroyl not lauryl - it's acyl not alkyl.


text:
What will be the form of coulomb and exchange operators for electron m in an n-electron molecule?
I know that the Fock operator for electron $m$ in $n$-electron molecule would be:
$$\hat{f}\!(m)u_i(m) = \varepsilon_iu_i(m)$$
$$\hat{f}\!(m) \equiv -\frac 1 2 ∇_m^2 - \sum_α\frac{Z_α}{r_{ma}} + \sum_{j = 1}^n\left[\hat{j}_j(m) - \hat{k}_j(m)\right]$$
And the Coulomb operator according to Wikipedia is given by:
$$\hat{J}_{\!j}(1)f(1) = f(1)\int\left|\varphi_j(2)\right|^2\frac{1}{r_{12}}\mathrm dr_2$$
My understanding of this equation is that we are finding the interaction of $f(1)$ with the electrons of $j$-th orbital (we take interaction with electron 2 and twice that result, hence $2J$ is used in the expression for Fock operator), and $r_{12}$ is the separation between our electron in $f(1)$ and 2nd electron in $j$-th orbital.
This is the expression for exchange operator for electron $1$ in $j$-th orbital with an arbitrary function '$f$'. What would be the expression of Coulomb and exchange operators for electron $m$? Will we consider the $1$st electron of $(m/2)$-th orbital? Or will we simply put $m$ in place of $1$?
Or have I understood the whole thing incorrectly?

Remember that the electrons are indistinguishable. Also, you can find an expression for the exchange operator in terms of a permutation operator in Szabo's Modern Quantum Chemistry.

Electrons in the same orbital are indistinguishable, right?
So, that only is my doubt. For mth electron, will we consider the 1st electron of (m/2)th orbital, and use the same notation, or would the expression be different?

All electrons are indistinguishable. We write the expressions using a dummy index, usually 1, but we can't really label the electrons or say that 2nd electron is in some particular orbital.


text:
How to calculate the equivalent mass of Na2S2O3?
Let us consider the following reaction
$$\ce{I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6}$$
Now, in order to calculate the equivalent mass of $\ce{Na2S2O3}$, first I need to calculate it's $n$-factor which turns out to be $0.5$ because the oxidation state of $\ce{S}$ in $\ce{Na2S2O3}$ is $+2$ whereas in $\ce{Na2S4O6}$ it is $+2.5$.And the $n$-factor of $\ce{I2}$ is $1$. So, the equivalent mass of $\ce{Na2S2O3}$ is
$$\frac{\text{molecular mass}}{0.5}$$
And Iam confused with the following half-reactions:
$$\ce{I2 +2e->2I-}$$
$$\ce{2S2O3^{2-}->S4O6^{2-} +2e}$$
I appreciate any help in order to solve a problem using milliequivalents?

The first and third reactions are not balanced in atoms.

N-factor Is the change of atom oxidation state multiplied by number of atoms changing this state.
Therefore, n-factor of $\ce{I2}$ is $2$, n-factor of $\ce{Na2S2O3}$ is $1$.
So, equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass, for molecular iodine it is half of its molar mass.

excuse me: Is the equivalent mass of $\ce{KIO3}$ equal 5.$$\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$$

Well, this is somewhat tricky. For the mentioned reaction, releasing iodine, the equivalent mass is M/5. But if the iodine is just an intermediate product to be titrated by Na2S2O3, the equivalent mass is M/6. As in the former, the oxidation number change is by 5, in the latter by 6.


text:
Is the -OCH3 group bulky?
Our teacher told us that $\ce{OCH3}$ is a less bulky group when compared to other bulky groups like $\ce{CH3}$, $\ce{NO2}$ etc. Is this true? If yes, why?

In organic chemistry bulkiness isn't just a matter of geometry; one can find the volume of the group using vdW radii, but it's a crude approximation.
Quantitatively steric bulk relies on energy measurements and can be described with a cyclohexane A-value — energy required for a group to switch between axial and equatorial positions in cyclohexane.
From the table of A-values methoxy group is indeed less bulk in comparison to nitro and methyl groups:
$$
\begin{array}{lc}
\hline
\text{Substituent} & \text{A-Value}/\pu{kcal mol-1} \\
\hline
\ce{-OCH3} & 0.6 \\
\ce{-NO2} & 1.1  \\
\ce{-CH3} & 1.7 \\
\hline
\end{array}
$$


text:
Justify this statement whether correct or incorrect . He2^+ has the same energy as two isolated He atoms
My explanation : how can a stable molecule be at the same energy level as two isolated inert gas atoms?? Is this even possible?

I cannot confirm the energy is the same, but if it was, what would make it unstable ? Note also the thermodynamic stability as measure of Gibbs energy difference and kinetic stability as measure of reaction activation energy.

This statement is not well-formed and cannot be justified as correct or incorrect.

I'm trying to figure out if its supposed to be He$_{2}^{+}$, He$^{2+}$, 2He$^{+}$, or what in the title...

$\text{He}_2^+$ would not have the same energy as two isolated He atoms.


text:
Is temperature double-counted in the Gibbs free energy equation?
Complete amateur here. When looking at one form of the equation deriving the change in Gibbs free energy for a given chemical reaction, the terms change in enthalpy, and the negative product of temperature (in kelvin) and change in entropy:
$$\Delta G = \Delta H - T\,\Delta S$$
This seems to double count how temperature will influence the spontaneity of the reaction if temperature is already considered when deriving entropy, $S$.
Any guidance would be appreciated.

So what? Yes, $\Delta S$ is a function of temperature, and moreover, so is $\Delta H$, and then $\Delta G$ is exactly what you said it is. "Double counting" is not a thing of any slightest importance, much like double occurrence of the letter "S" in the text of your question.

Do you think that the temperature has to change in order to get a change in enthalpy or entropy of a reacting mixture?

One reason we write equations this way is because all of the parameters on the right-hand-side can be measured or computed independently.
Consider an analogous (but more intuitive) equation, the Stokes-Einstein relation for the diffusion coefficient $D$ of a particle in a liquid:
$$D = \frac{k_\mathrm{B} T}{6\pi\eta r}$$
where $\eta$ is the dynamic viscosity, $r$ is the radius of the particle, and $k_\mathrm{B}$ is Botzmann's constant. Here we see an explicit dependence of $D$ on $T$, but there is also an implicit dependence on $T$ through $\eta$ and possibly also $r$. Is this a problem? No, assuming the equation holds and we can determine the parameters accurately at each $T$ of interest. Whether that is a simple thing is another question.
Also, it is fair to wonder to what extent the parameters are fundamentally independent, or, whether by digging into the gory details of atomic structure of matter, we might not find $r$ and $\eta$ to be connected in some other way. But that is another question.

Even if the final temperature of the products is the same as the initial temperature of the reactants, you can still have a change in entropy, enthalpy, and Gibbs free energy of products relative to the reactants.  This is the result of making and breaking chemical bonds and not the result of a temperature change. Entropy and enthalpy are not just functions of temperature (and pressure), but also chemical makeup.  In such a well-defined case, there would certainly not be any double counting of the temperature.

Ah, thank you! It only seemed that since temperature was always considered (to my very limited knowledge) when measuring the entropy, micro-states, etc. that multiplying by such a factor would be almost redundant or have some sort or ^2 squared relationship.


text:
Can entropy in a closed system remain constant when you heat a liquid?
This is in regards to an exam question that I did not understand and got wrong.
The question was: When you heat acetone from 200 K to 400 K, what happens to the entropy?
Answers: Increases, decreases, more info needed
I put more info needed, but this was incorrect. My logic was that if you have a sealed container completely filled with acetone (there is no gas or empty space in the container) and you heat it, there are no new states that the liquid particles can occupy so there is no change in entropy. Is this reasoning incorrect?

Technically, that's right: there are no new states. But the particles distribute more uniformly among the existing states.

Exactly, as the higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly.

Poutnik's answer gives an explanation for the increase in Entropy from a classical thermodynamic point of view. I would like to add the statistical mechanical perspective, which is what you based you logic on.
So entropy in statistical mechanics is interpreted as the randomness in the distribution of particles in various microstates.
$S = k_B \ln \Omega$
where $\Omega$ is the number of microstates.
When the temperature of the system is raised, the energy of the system increases. This leads to more availability of higher energy microstates, and thus the particles are more randomly distributed (since there are more microstates available to them).
I would like to know where your assumption "there are no new states that the liquid particles can occupy" came from.
PS: I think the question in the exam was intended to make you think about entropy in thermodynamic terms, but there is also a way to look at it statistically.

The quantitative differential classical thermodynamic definition of entropy is $$\mathrm{d}S=\mathrm{d}Q/T$$
Therefore providing heat $Q$ to the acetone increases its entropy as
$$\Delta S=\int_{T_\mathrm{1}}^{T_\mathrm{2}} \frac{C(T)}{T}\cdot \mathrm{d}T$$
where $C$ is the heat capacity of the system.
The statistical thermodynamics definition has an alternative definition to $$S=k_\mathrm{b} \cdot \ln \Omega$$ as $$S=-k_\mathrm{b} \cdot \sum_\mathrm{i} p_\mathrm{i} \cdot \ln p_\mathrm{i}$$ as the sum over all states.
The higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly.
In such a case, even for the same number of states, more evenly distributed probabilities at higher temperatures lead to higher entropy.


text:
How can I check the sodium content of food?
What is an easy (i.e., home project) way to measure sodium in food?
For example, the nutritional label of a tahini brand that I often consume shows very little sodium (around 10mg/100g). I suspect the sodium content is much higher.
What is the chemical way to measure the salt content?

Sodium is a difficult ion to estimate at home.
Since most sodium salts are water-soluble, colorimetric or spectroscopic methods are often employed, both of which I assume are not within arms' reach.
Some alternatives:

Speaking of sodium estimation in foodstuff, it is usually determined by Mohr's
method which involves the precipitation titration of $\ce{Cl-}$ ions. This is done under the assumption that sodium is mainly present as sodium chloride. Additionally, it will also register $\ce{Cl-}$ from other chlorides, and hence may give an elevated reading. You might be able to do this at home. Here is a page
to help you with the process.
It is done in the presence of chromate ion ($\ce{CrO4^2-}$), and $\ce{Ag+}$ ions are added gradually. This precipitation happens initially:
$$\ce{Ag+ + Cl- <=> AgCl(s)}$$
When the solution runs out of $\ce{Cl-}$ ions, precipitation of the red-brown $\ce{Ag2CrO4}$ begins:
$$\ce{Ag+ + CrO4^2- <=> Ag2CrO4(s)}$$
This signals the end point.
There are sodium ion meters, you can get them for ~300$.
There is a volumetric estimation[1] for sodium directly, but it is not expedient to do it. It is based on the precipitation of sodium zinc uranyl acetate.

Conclusions
IMO it is difficult to do this as a home project, as it will require stuff that is not usually found at home to do it precisely.

References
[1] J. Am. Chem. Soc. 1931,  53, 9, 3288-3291 DOI: 10.1021/ja01360a011

Thank you! This is quite amazing. Do you think that the assumption that in food, most  $\ce{Cl-}$ originates from table salt can be justified?


text:
Will heating of KClO3 be considered disproportionation?
In all examples of disproportionation I have seen, there is one particular atom whose oxidation state is both increasing and decreasing to give rise to different products
For example, on heating phosphorous acid
$$\ce{4 H3PO3 -> 3 H3PO4 + PH3}$$
the oxidation state of phosphorous changes from +3 to +5 and -3.
However, in the case of heating $\ce{KClO3}$
$$\ce{2 KClO3 -> 2 KCl + 3 O2}$$
the oxidation state of oxygen changes from -2 to 0, while that of chlorine changes from +5 to -1.
So, will such a reaction be called a disproportionation since its the same molecule giving rise to two products, or will it be simply a redox reaction?

Thermal decomposition of potassium chlorate is not disproportionation, just a redox reaction.
Disproportionation refers to the same element acting both as oxidizing agent and a reducing agent, resulting in compounds which contain the same element in different oxidation states.
On the other hand, if you consider preliminary decomposition stage involving perchlorate formation, then it is a disproportionation reaction:
$$\ce{4 K\overset{+5}{Cl}O3 ->[\pu{400 °C}] 3 K\overset{+7}{Cl}O4 + K\overset{-1}{Cl}}$$
Also, synthesis of $\ce{KClO3}$ is a disproportionation reaction:
$$\ce{6 KOH + 3 \overset{0}{Cl}_2 -> 3 H2O + 5 K\overset{-1}{Cl} + K\overset{+5}{Cl}O3}$$


text:
Resonance forms of deprotonated hydroquinone
What resonance forms stabilize deprotonated hydroquinone?
Deprotonated hydroquinone looks like this:

Possible resonance forms include:

1 and 2 look reasonable: they don't carry multiple charges. My question is: how much do these two structures contribute to the stability of deprotonated hydroquinone? I am thinking that they should be much higher energy than the structure on top, because the negative charge is on C, which is not nearly as electronegative as O.
Can this be quantified (by and for amateur chemists)?

Per Michael Lautman's suggestion, here are partial charges as calculated in Avogadro:

The pattern expected from the resonance forms above, with more negative charge on carbons 2, 4, and 6, is not there.
Why is C4 -0.20?
For comparison, deprotonated cyclohexane-1,4-diol:

Resonance forms can be determined by following conjugation, which is the alternation of single bonds with p or pi electrons.  Follow the conjugation.

Yeah I know, thanks. I updated the question.

You can use modeling software to quantify the partial charges on each atom.   On windows, check out ArgusLab, or Avogodro for mac.

@michael partial charges will not reproduce anything usable about the resonance, except an overall superposition. Avogadro only produces charges based on force fields, not the most reliable for an anion anyway.

@Martin-マーチン, what do you suggest instead?

You can use electronic structure calculations to determine the approximate wave function, then superimpose wave functions of Lewis form to match this; basically a natural resonance theory from nbo. Or you can try ab initio Valence bond techniques (but I'm not familiar with those). Localising orbitals will probably at least give you the dominant form, which I would assume is the one you have drawn.


text:
Could sodium in ethanol reduce amides to amines?
Strong reducing agents like $\ce{LiAlH4}$ can reduce amides to amines. Sodium in ethanol (as a source of protons) is also a very strong reducing agent. Could it be therefore used to, for example, turn formamide into methylamine?

Yes, that is used as a pathway for synthesis of amines from amides, albeit rarely.
I quote from the abstract of a communication that reports the same: (emphasis mine)

The reduction of amides to their corresponding amines has been used extensively in the synthesis of many drugs and their intermediates. This reaction has usually been carried out by means of metal hydrides (especially lithium aluminium hydride) or by catalytic hydrogenation over various catalysts. Metal and alcohol has been used for the reduction of ketones to their corresponding hydroxy compounds and oximes to their corresponding amines, but the reduction of amides under these conditions, although previously reported, is not well documented in major works of organic chemistry. This procedure is specially useful for industry since it is safer, less expensive, requires less time and provides higher yields (65-80 per cent) when compared with lithium aluminium hydride or catalytic reduction methods.

The emphasized text is written in a slightly confusing language but nonetheless confirms that amides can be reduced to amines using $\ce{Na/ROH}$, otherwise known as Bouveault-Blanc reduction.

References:
Bhandari, K. , Sharma, V. L. and Chatterjee, S. K. (1991), ChemInform Abstract: A Convenient Method for the Reduction of Amides to Their Corresponding Amines.. ChemInform, 22: no-no. DOI: 10.1002/chin.199118077


text:
Are there any crystals that are theoretically possible, but haven't yet been made?
Are there any crystals (that exist at standard temperature and pressure) that are theoretically possible, but haven't yet been made?
For example, the predicted requirements for temperature, pressure, materials, etc. for crystallisation to occur are too difficult to achieve.

You might want to clarify what you mean by "crystals". Are you talking about specific crystal symmetries or space groups (e.g. body-centered-cubic or P422), or are you talking about any chemical substances in crystal form? -- If the latter, a large number of organic molecules (and even proteins!) will crystallize at standard temperature and pressure, and there's a practically unbounded number of them we've yet to produce (let alone crystallize).

This feels like asking if there are any integers we haven't written down yet.

Is the "perfect" crystal at absolute zero with zero absolute entropy theoretically possible?

When we are talking about the prediction of materials, it is generally about the structure and corresponding free energy, not the crystallization process. That being said there are a huge number of inorganic materials that predicted to be stable yet no one has synthesized them.
There are computational databases like Materials Project that contain automatically generated (theoretical) structures and their estimated free energy, solubility, magnetic properties, bandgap, stability compared to competing phases, etc.

As an addition, at much smaller scale than the Materials Project -- so far about 3k entries -- COD's sister, Theoretical Crystallography Open Database, http://www.crystallography.net/tcod/.

Arguably, there are more such crystals than those which have been made. It is just that nobody bothered to write them all down, because hey, what's the point.
A tiny minority of those theoretical crystals haven't been made for the reasons you envisioned, that is, because the conditions of crystallization are hard to achieve. Here by "conditions" we mean pressure because temperatures that are relevant to solid state chemistry can't be much of a problem. Slowly and painstakingly, one by one, these crystals are being dragged into existence. Oganov's discoveries are relatively recent examples. Metallic hydrogen is still waiting for its turn.
A vast majority, however, haven't been made for an entirely different reason: not because the conditions are hard to reach, but because there are no such conditions. If you want to make them, you have to invent some ingenious workaround ways.
You surely heard of graphite and diamond, and of the amount of effort it takes to turn the former into the latter, and that it's been successfully done. Now, what about this guy?

Diamond does have a place (albeit far away on the phase diagram) where it is stable. Fullerenes don't. I can claim with a good deal of confidence that you won't find the conditions where they are thermodynamically stable, not even in the areas where we haven't reached yet. However, they have been made. Others are still waiting:

(more can be found at https://en.wikipedia.org/wiki/Allotropes_of_carbon)
So it goes.

"Now what about this guy?" however should not lead to the impression that pure fullerenes -- like $\ce{C60}$ likely depicted would not crystallize "well enough" to perform a diffraction study.  See https://pubs.rsc.org/en/Content/ArticleLanding/RC/2004/RC040455#!divAbstract or http://www.chem.msu.su/rus/lab/phys/cryschem/liter/pap_5_eng.pdf, for example.


text:
What zeolite types are typically used to absorb CO and CO2 gases at maximum zeolite absorption temperature?
Me and my friends are doing a project involving carbon monoxide and carbon dioxide as products to be filtered, we are using zeolites and would like to know the zeolite type typically used to filter $\ce{CO}$ and $\ce{CO2}.$

Welcome to Stack Exchange! It's better to take a moment to describe what you know based on information you have read or searched already. Stack Exchange shouldn't be your "first stop". If you have identified at least some zeolite materials that can do this as proof of principle, you should mention it in your question. Thanks!

NASA uses lithium hydroxide and silver oxide to remove CO2, but I'm not sure about the effectiveness for CO. Why zeolites? https://space.stackexchange.com/a/32177/12102 and also this PDF https://ttu-ir.tdl.org/bitstream/handle/2346/74151/ICES_2018_180.pdf?sequence=1&isAllowed=y

My apologies, uhoh. I have researched the question however I was unable to find information on zeolites that would typically be used to filter CO and CO2. However thank you for the information on Lithium Hydroxide and Silver Oxides!

@uhoh The idea of using zeolithes is to have an easily reversible (by simply changing the pressure) adsorption. It's a well introduced method for gas separation. LiOH absorbs CO2, and you need an energy-intensive chemistry to get it back out again.

@Karl I see what you mean!


text:
Software or library for creating images of ball and stick models?
Is there any code library (in any programming language) or software capable of producing images of ball and stick diagrams?
Example of an image of a ball-and-stick model for Ammonia:

(Image source)
Preferences:

It should be capable of producing images programmatically through a code library. Any mainstream programming language (C, Java, Python, etc.) is acceptable.
It should be able to take chemical descriptions as input to produce images of ball and stick diagrams as output.
It should be free and open source.

I know that it is possible to create these models using Mathematica's ChemicalData built-in symbol, but I am looking for something that can integrate better into my existing code base (hence the preference for free and open source libraries in mainstream programming languages).

You should try  JSmol,  it is free and easy to use.  Have a look at one of the many Wikipedia pages that use this for drawing molecules.

For general open source solutions:

Avogadro - Molecular editor GUI. Has a Python API.
Jmol - Chemical structure viewer. Fully scriptable using a custom scripting language. It also has a JavaScript-only version named 'JSmol', allowing models to be displayed in web browsers.
ghemical - Has a GUI for editing molecules, but no API.

For a $\LaTeX$ solution, see this question on the $\TeX$ Stack Exchange site: Help creating solid cylinders along a path in Tikz for crystal lattice diagram
Other solutions:

If you have access to Mathematica (e.g. on a Raspberry Pi running Raspbian [which comes with a free copy of Mathematica]), you can use Mathematica's ChemicalData built-in symbol to create ball and stick models.

I would also mention QuteMole:

QuteMol is an open source (GPL), interactive, high quality molecular
  visualization system. QuteMol exploits the current GPU capabilites
  through OpenGL shaders to offers an array of innovative visual
  effects. QuteMol visualization techniques are aimed at improving
  clarity and an easier understanding of the 3D shape and structure of
  large molecules or complex proteins.


text:
Molar absorption coefficient for potassium hydrogen phthalate
I would like to draw a comparison of the sensitivity between several spectrometers. Therefor I measure the absorption of a serial dilution of potassium hydrogen phthalate (KHP). The idea is to compare the theoretical absorption with the measured absorption.
To calculate the theoretical absorption I need the molar absorption coefficient of KHP at 254 nm wavelength. Unfortunately I haven't found anything till now.
Does anyone know where I can find this coefficient?

Why use KDP? Benzene or naphthalene and many other stable compounds have well known extinction coefficients vs wavelength.

@porphyrin Maybe the manufacturer has some documented arcane calibration procedure involving KHP that has to be followed for accreditation, but this just my speculation.

I use KHP because it's used for calibration.

According to Burke et al. (R.W. Burke, E.R. Deardorff, O. Menis, Liquid Absorbance Standards, J. of Research of the National Bureau of Standards - A, Physics and Chemistry, 76A, Sept. - Oct. 1972, 469-482), solutions of KHP have "been used as a spectral standard in the comparative evaluation of spectrophotometers" (p. 478). But they say that absorbance is pH dependent and 262 nm is the shortest wavelength they apparently used. So I would do what @porphyrin suggested, assuming that degree of freedom.

I would imagine that you could just use one of the spectrophotometers you have available to determine the molar absorptivity constant experimentally. After all, the molar absortivity constant is simply the constant which equates the terms (those being Absorbance and Concentration) which are already proportional. So, using something like the beer-lambert law...
$$A = \epsilon b C$$
...make a solution of a known concentration, measure its absorbance at your wavelength of interest, and solve for $\epsilon$.
$$\epsilon = \frac{A}{b C}$$
You might consider doing this a number of times to get an average value and thus minimize the random error associated with your constant.

This is a very good answer, but it can be even better with a little tweaking! From the reference I gave in my comment above, $\epsilon$ is roughly 1400 L/mole cm at 254 nm. A 0.5 mM KHP solution in water gives an absorbance of about 0.7. So how about suggesting that the OP carefully make a KHP solution of about 0.5 mM, measure the pH and then do as you say in your answer? Averaging a few absorbances is right. And then OP can experimentally get the actual $\epsilon$ for use in calibrations. They also need to measure pH to see if anything unexpected is happening. Upvoting now!

I get the logic behind noting the pH of your solutions, as Absorbance can change as a function of pH, but I don't know how necessary that piece of data is if the system we're investigating is the actual spectrophotometer instead of the chemistry of the sample.
Actually, now that I'm thinking about it, I'm not even sure if you need the $\epsilon$ value if you're comparing machines; since $\epsilon$ is constant then just rearrange BL such that $\frac{A_1}{C_1} = \frac{A_2}{C_2}$ and use the same solution such that $C_2 = C_1$. That's all you need to compare results I would suspect.

Well, the OP says she wants to compare the theoretical and experimental absorbances, so the theoretical one depends on just what the molar absorptivity happens to be. If this was just a comparison of two spectrometers, then no issue.

Ya, I guess I'm not entirely tracking what she means when she says 'theoretical absorbance'. Like, if $\epsilon$ is determined experimentally, then 'theoretical absorbance' just means the correlation associated with the machine from which you determined $\epsilon$, unless there's a first-principles method for determining $\epsilon$ that I'm not familiar with. And at that point, you don't need $\epsilon$ since you have $A_i$ of the spectrophotometer and ideally $C_i$ as well.


text:
Proton NMR chemical shift of water peak in different solvents
Why does the water peak appear at different chemical shift values (ppm) in different solvents in proton NMR spectra? For example, the water peak in DMSO-d6 appears at nearly 3.33 ppm, but the same moisture peak in $\ce{CDCl3}$ appears at 1.56 ppm.
What is the reason behind it? Why should the same species ($\ce{H2O}$) give rise to two different chemical shift values?

Consider also water in addition to the solvents you mention:

water-d2 (or, $\ce{D2O}$) is an extensively H-bonding polar solvent (dielectric constant $\epsilon = 79$)
$\ce{DMSO-d6}$) is polar ($\epsilon= 47$) but aprotic (no-H-bond donation ability)
chloroform-d is an apolar aprotic solvent ($\epsilon=4.81$)

Now consider the state of a trace amount of $\ce{H2O}$ in each of these solvents, giving rise to an NMR signal:

in $\ce{D2O}$ the trace $\ce{H}$ can exchange and forms $\ce{HDO}$ which retains an extensive H-bond network. $\ce{H}$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $\ce{H}$ carrying a significant positive partial charge. The chemical shift is $\approx \pu{4.7 ppm}$
in DMSO-d6, $\ce{H}$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $\approx \pu{3.33 ppm}$
in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $\ce{H}$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $\approx \pu{1.56 ppm}$

The following table shows shifts for residual water $\ce{H}$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):
$\begin{array}{c}
\begin{array}{c|c|c}
\text{solvent} &\text{shift(ppm)}&\epsilon\\\hline
\ce{C6D6 }&0.40 & 2.28\\
\text{toluene-d8} &0.43 &2.38\\
\ce{C6D5Cl} &1.03 &5.69\\
\ce{CD2Cl2 }&1.52 &9.08\\
\ce{CDCl3} &1.56 &4.81\\
\ce{CD3CN}  &2.13 &36.64\\
\text{THF-d8} &2.46 &7.52\\
 \ce{(CD3)2CO} &2.84 &21.01\\
 \ce{(CD3)2SO}  &3.33 &47\\
\text{TFE-d3} &3.66&8.55 \\
\ce{CD3OD} &4.87 &32.6\\
\end{array}
\end{array}$
Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
Note also that the top three have an additional effect on water shifts due to the aromatic ring current.

Reference
[1] Fulmer et al. Organometallics 2010, 29, 2176–2179

Why do shifts change anyway? You might consider that different solvents may (de-)stabilize certain conformers, or result in different interactions? An example from a previous life (sadly a long time ago, and I can't recall the exact detail or even if NMR was reported in the paper I was following) is that in certain 3,4,5-trisubstituted pyridines (with non identical C3 and C5 substituents) H-2 and H-6 have similar shifts in d-CDCl3 (so it looks like an integral of 2) but they are clearly separated in d6-DMSO.


text:
How to determine polarity of components in TLC

Estimate how a TLC-analysis would look like for the following reaction when you take a TLC at: a) the start of the reaction, b) after 65% conversion and c) after full conversion of A. (You may assume that A-C separate on the TLC-plate)

I am having a very difficult time understand how you can determine which compound is more polar by just looking at the structure. I don't know what to look at or what to prioritize.
When attempting to answer this question I assumed that B would be more polar (why: because it contains a nitrogen which is more electronegative than oxygen). So B should be closest to the baseline and A should be further away. What I then thought was that C should be in between A and B because the reaction reacts something polar and something less polar. This was not correct. Our professor drew the TLC-analysis like this:

And I don't really understand how I can determine that the product is more polar than both of the reactants. Is it because it contains both an oxygen and a nitrogen?
I would truly appreciate it if someone could give me some general rules or steps to follow to easily determine the polarity of different compounds.

I'd just like to clarify "because it contains a nitrogen which is more electronegative than oxygen". No. Oxygen as an element is in fact more electronegative than nitrogen.  In this case, however, two O-C bonds are present, where the electronegativity difference is much lower as compared to N-H bond. This is why B is more polar than A.

An almost crazy exercise (without knowing details about to whom and when is given). Also don't think of polarity as a full explanation. See answer by @Waylander

I am assuming this is TLC run on silica. Silica is a mildly acidic surface so basic compounds such as amines do not come off the baseline unless you use a very polar eluent with added base (e.g. ammonia/triethylamine). Starting material B is a basic amine. Starting material A is an epoxide and fairly non-polar (kinda similar to an ether) and will run high in most systems, certainly any system where B comes off the baseline. Your product C is more basic (tertiary v secondary), this is the strongest effect. The second non-polar aromatic group raises the likelihood that it will run above B, but this is a weak effect. The alcohol is a polar group and will also pull it down thus the sum of these effects is that it runs lower
This missing information is the eluent. From what what is given in the question it is clear that the eluent is something like DCM/MeOH/NH3 (that would be my first attempt to separate such basic components). As B is in excess what you are really looking for is loss of A. When you can no longer see any A by TLC, the reaction can be worked up.
The groups that make for low running are those that interact strongly with the silica surface: basic amines, acids, alcohols. This is a general rule of thumb, but in my experience trying to guess where a multi-functioned molecule is going to run is quite a lottery. I do not believe that it is a safe assumption that a product similar to C would always run below a SM similar to B.


text:
Gibbs free energy of activation and activation energy
What the difference between Gibbs free energy of activation and activation energy

This question has the appropriate title for this related question, with the answers addressing both questions: https://chemistry.stackexchange.com/questions/15417/what-is-the-correct-definition-of-the-gibbs-free-energy-of-activation?rq=1
I don't know if that makes it a duplicate.

Activation energy is a broad term which may refer to the energy barrier of a reaction, whether calculated as enthalpy or Gibbs energy or whatever. The post linked by @KarstenTheis explains this pretty well.


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Calculating dew point and bubble point of gas mixtures
How do you calculate approximately the triple point and critical point of any given CH-based gas mixture?
I have developed a software which uses iterations to find bubble point and dew point of any CH-based gas mixture. It uses Peng-Robinsson EOS and Van Der Vaals mixing algorithm.
When calculating the dew point pressure for a certain temperature, it's difficult to know the starting range for the iteration. First I used False-Position method in finding the root, but now I have switched to Bisection. Although it's slower, it's more safe as it's not affected by function derivatives.
What would be needed is as approximation of the range where the liquid/vapour phases are found. Setting the iteration's range to start below triple point pressure, or above critical point pressure will cause the iteration to diverge and never find a root.

Would iterations work with Simplex?

What do you use for the liquid phase?

@MaxW I don't know much about Simplex. Just that it's an optimisation algorithm. I have to look it up.

@ChetMiller I have calculated the fugacity to a large degree according to this https://cheguide.com/bubble_p_flash.html

OK.  So you are assuming an ideal liquid solution.

In the picture above, the C is the critical point. There seems to be some easy methods to calculate the critical temperature, but not so with the critical pressure.
For bubble point it would give me a good indication of where the max range is.
But I realised that the critical temperature point won't help me finding the min and max range for dew point calculation. As you can see in the picture, the point M (max temperature point) would be a good indication of where the max range is for dew point.
Anyway I found a much simpler solution. I consider the bubble point and dew point calculation functions as a black box.
As I'm only seeking the bubble point / dew point for a specified temperature, I step the possible pressure range (1 kPa - 10 MPa), with steps of 1 kPa. Then I calculate the numerical derivate of using current step and previous step.
This function derivate is consistent and negative in the range of the function domain where a bubble or dew point value is valid.
Outside the function domain the derivate is close to zero (+-1E-15).
So this gives me a good starting point for the root finding algorithm.
I also changed from False Position method to Bisection method, as it's less sensitive to derivates.
I hope this will help someone else with similar problem.


text:
Does peroxide ion have any unpaired electrons?

According to the MOT diagram there are no unpaired electrons but in an exam I gave the answer key said it has 2 unpaired electrons. I can also find some images on google where it has 2 unpaired electrons.

Which one is correct and why?

@santimirandarp I believe, in the above MO, the BO is 2.

@Random, peroxide is diamagnetic. So the provided MO is correct. Oxygen ($\ce{O2}$) has 2 unpaired electrons, but $\ce{O2^2-}$ hasn't.

The MO diagram you show is correct. $O_2^{2-}$ has a bond order of 1 (equivalent to a single bond between the two O atoms) as the antibonding $\pi^*$ orbitals contain 4 electrons and this cancels the $\pi$ orbitals' 4 bonding electrons. See the Mo diagrams for the different $O_2$ species below:

https://www.chem.uci.edu/~lawm/Ch%205%20Solutions.pdf
The Lewis diagram is incorrect. It shows the O atoms with an expanded octet (not allowed for period 2 elements by the rules of Lewis diagrams) and it shows the bond between the two O atoms as a double, rather than the measured single O-O bond length:

http://www.wiredchemist.com/chemistry/data/oxygen-selenium-compounds
It also shows the peroxide ion as having unpaired electrons, when it is actually diamagnetic. It should look like this:

https://en-journal.org/ViewImage.php?Type=F&aid=489334&id=F1&afn=142_EN_24_4_325&fn=en-24-325-g001_0142EN


text:
Elimination of hydroxyl group while deprotecting acetal
Regarding this reaction, other than de-protecting the acetal, will we have an elimination reaction for the alcohol or no?

Base is of no relevance here. It depends entirely on how strong the acid conditions are. It is easy to select conditions that will deprotect without elimination though you could easily form the hemi-acetal of the ketone.

You can't eliminate hydroxyl groups with base

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As I said in the comments, it entirely depends on the conditions. It is easy to select mild conditions to cleave the dioxolane. Elimination would require very strong forcing conditions (concentrated acid and heating). The first site of protonation is the carbonyl, so you would form the hemi-acetal of the product long before elimination or substitution occured at the primary alcohol. If you were foolish enough to heat this with cH2SO4 then you could get elimination of the primary alcohol and generate quite a mess.


text:
Why is Barium Phosphate soluble in dil HCl?
Why is Barium Phosphate soluble in dil HCl
While Barium Sulphate is not?

What do you know about the acid-base equilibria? More to the point, what dissolved anion is in equilibrium with solid barium phosphate and what happens to it as you add an acid?

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Many phosphates are not soluble in water at standard temperature and pressure, except for the sodium, potassium, rubidium, cesium, and ammonium phosphates, which are all water-soluble. As a rule, the hydrogen and dihydrogen phosphates are slightly more soluble than the corresponding phosphates (Wikipedia). For example, based on Wikipedia Solubility Table, the solubility of $\ce{BaHPO4}$ is $\pu{0.013 g}$ in $\pu{100 mL}$ at $\pu{20 ^{\circ}C}$, and a CDC.gov Report states that:

The water solubility of barium compounds increases with decreasing $\mathrm{pH}$.

Nevertheless, the solubility studies of the phosphates of strontium, barium, and magnesium have been done previously (Ref.1). Alkaline earth phosphates as a whole have been reviewed as well (Ref.2).
That said, let's look at phosphates behaviour with $\mathrm{pH}$. The three $\mathrm{p}K_\mathrm{a}$ values for phosphoric acid are 2.16, 7.21, and 12.32. Thus, aqueous phosphate exists in four forms (Wikipedia):

In strongly basic conditions, the phosphate ion ($\ce{PO4^3-}$) predominates.
In weakly basic conditions, the hydrogen phosphate ion ($\ce{HPO4^2-}$) is prevalent.
In weakly acidic conditions, the dihydrogen phosphate ion ($\ce{H2PO4-}$) is most common.
In strongly acidic conditions, trihydrogen phosphate ($\ce{H3PO4}$) is the main form.

When dilute $\ce{HCl}$ (say $\pu{3 M}$ and hence $\mathrm{pH} \approx -0.48$) is added to solid $\ce{Ba3(PO4)2}$, following reactions will occur with aqueous hydronium ions:
$$\ce{Ba3(PO4)2(s) + nH2O(l) <=>  3Ba^2+(aq) + 2PO4^3-(aq)}$$
$$\ce{PO4^3-(aq) + H3O+(aq) ->  HPO4^2-(aq) + H2O(l)}$$
$$\ce{HPO4^2-(aq) + H3O+(aq) ->  H2PO4-(aq) + H2O(l)}$$
$$\ce{H2PO4-(aq) + H3O+(aq) ->  H3PO4(aq) + H2O(l)}$$
Because, $\mathrm{pH}$ of the acid solution  is lower than the $\mathrm{p}K_\mathrm{a1}$ of the $\ce{H3PO4}$ acid, $\ce{Ba3(PO4)2}$ solid would slowly dissolve to give $\ce{H3PO4(aq)}$ and $\ce{BaCl2(aq)}$ in the solution (Channel Le Chatelier's principle).
However, this phenomenon would not happen with $\ce{BaSO4}$ solid because $\mathrm{p}K_\mathrm{a1}$ of $\ce{H2SO4}$  is about $-3$, which is much lower than the operating $\mathrm{pH}$. Therefore, since $\mathrm{p}K_\mathrm{a2}=1.99$ for $\ce{H2SO4}$, it is possible to form  $\ce{Ba(HSO4)2}$, yet its solubility is said to be low as well ($\approx \pu{0.015 g}$ in $\pu{100 mL}$ of water at $\pu{25 ^{\circ}C}$).
References:

L. E. Holt, Jr., J. A Pierce, C. N Kajdi, “The solubility of the phosphates of strontium, barium, and magnesium and their relation to the problem of calcification,” Journal of Colloid Science 1954, 9(5), 409–426 (https://doi.org/10.1016/0095-8522(54)90029-X).
R. W. Mooney, M. A. Aia, “Alkaline Earth Phosphates,” Chem. Rev. 1961, 61(5), 433–462 (DOI: 10.1021/cr60213a001).

Minor point.  Is there actually $\ce{(NH4)3PO4}$, or must ammonium phosphate be an acid salt?  The normal phosphate ion "should be" strongly basic enough to displace $\ce{NH3}$.

@Oscar Lanzi: Honestly, I did not know what is the $\mathrm{p}K_\mathrm{a}$ of $\ce{(NH4)3PO4}$. But I found this infor for $\ce{(NH4)H2PO4}$: A solution of stoichometric monoammonium phosphate is acidic (pH 4.7 at 0.1% concentration, 4.2 at 5%).