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Acidity of alcohols and basicity of amines
I've been studying acids and bases in organic chemistry. My textbook says that alcohols are very weak acids, therefore they cannot react with strong bases. the conclusion i formed from this was: very low acid / base strength leads to not being able to react with stronger bases / acids. But then, while reading the chapter about Amines, i read that aromatic amines (such as aniline) ONLY react with strong acids. Therefore my question is: why is it that very weak acids cannot interact with strong bases, however very weak bases can ONLY interact with strong acids?

Some very is more very than some other very.

Does ethanol react with RMgX?

There exist several sorts of weak acids. They can be weak, very weak and very very weak acids.
Weak acids react with weak and strong bases.
Very weak acids do not react with weak bases. They do react with strong bases.
Very very weak acids do not with weak and with ordinary weak bases.
Ethanol is such a very very weak acid. It does not react significantly with strong bases. I only reacts with sodium metal in a redox reaction.
Aniline is a weak base, but not a very very weak base. It reacts with strong acids.

I think I understand now. Alcohols are very very weak acids so they do not react with weak bases at all, and their reactivity to strong bases is negligible. Since my textbook said that alcohols do not react with strong bases, I thought that they could interact with weaker ones (if A is not B then it is C). Is it fair to conclude that if something does not have significant reactivity to strong bases, then as a consequence it will NOT neutralize a weak base?
(sorry if the questions seem trivial, I am a high school student barely scratching the surface of chemistry)

@mral The essential condition is what "does not react" mean quantitatively. The threshold is a subjective, deliberate parameter. If fraction $\pu{e-3}$ reacted, it reacts. What about $\pu{e-4}$?  $\pu{e-6}$? $\pu{e-10}$? $\pu{e-15}$?  Any molecule?

The weak base $\ce{NH3}$ with amount concentration $\pu{1 M}$ has $\mathrm{pH} \approx 14 -  4.75/2 \approx 11.63$. If I added a small amount (wrt $\ce{NH3}$) of a  very weak acid $\ce{HA}$ with $\mathrm{p}K_\mathrm{a} = 12.63$, it would have $\pu{\approx 9 %}$ in dissociated form $\ce{A-(aq)}$, complemented by $\ce{NH4+(aq)}$. So very weak acids do react with weak bases. // Similarly, Aniline would react with acetic acid too. $\ce{Ph-NH3+}$ has pKa 4.63, acetic acid 4.75.  In solution with equimolar concentration, both pairs would have roughly equal concentration of respective acid/base.

Qualitative attributes like strong, weak, very weak, very very weak (acid/base) have no widely accepted agreement/definition what it exactly means quantitatively. As consequence, different sources may give different boundaries between them.
That is the reason why you have ended with different evaluation for acids and bases.
In context of water solutions, acids have the characteristic acidity parameter $\mathrm{p}K_\mathrm{a}$. It is equal to $\mathrm{pH}$ when $\pu{50 \%}$ of the acid is dissociated.
\begin{array} {|l|c|}
\hline \mathrm{pH} & \text{% of acid dissociation} \\
\hline \mathrm{p}K_\mathrm{a} - 3 & \approx \pu{0.1 \%} (\frac{1}{1001}) \\
\hline \mathrm{p}K_\mathrm{a} - 2 & \approx \pu{1 \%} (\frac{1}{101}) \\
\hline \mathrm{p}K_\mathrm{a} - 1 & \approx \pu{9 \%} (\frac{1}{11}) \\
\hline \mathrm{p}K_\mathrm{a} & \pu{50 \%} (\frac{1}{2}) \\
\hline \mathrm{p}K_\mathrm{a} + 1 & \approx \pu{91 \%} (\frac{10}{11}) \\
\hline \mathrm{p}K_\mathrm{a} + 2 & \approx \pu{99 \%} (\frac{100}{101}) \\
\hline \mathrm{p}K_\mathrm{a} + 3 & \approx \pu{99.1 \%} (\frac{1000}{1001}) \\
\hline
\end{array}
Mutual reactions of acids with bases have to be assessed in sense what $\mathrm{pH}$ wrt the acid $\mathrm{p}K_\mathrm{a}$ is set by the base excess.
In context of Broensted-Lawry acid-base theory, bases accept H+ ion:
$$\ce{B(aq) + H+(aq) <=> BH+(aq)\text{(conjugate acid)}}$$
where $\ce{B}$ may be a molecule or a ion.
If the base has the basicity constant $\mathrm{p}K_\mathrm{b}$ then its conjugate acid has the acidity constant (at $\pu{25 ^\circ C}$)
$\mathrm{p}K_\mathrm{a, conj}$ = 14 - $\mathrm{p}K_\mathrm{b}$
We can then compare the relative strength of the given acid and the conjugate acid of the given base. (That applies on a solvent too, if it can act as a base like water)
For the stoichiometric ratio of the given acid and base, we can estimate the degree of neutralization.
\begin{array} {|c|r|}
\hline \mathrm{p}K_\mathrm{a,conj} -  \mathrm{p}K_\mathrm{a} & \text{Neutralization degree}   \\
\hline 8 & \pu{0.01 \%}  \\
\hline 6 & \pu{0.1 \%}  \\
\hline 4 & \pu{1 \%}  \\
\hline 2 & \pu{9 \%}  \\
\hline 0 & \pu{50 \%}  \\
\hline -2 & \pu{91 \%}  \\
\hline -4 & \pu{99 \%}  \\
\hline -6 & \pu{99.9 \%}  \\
\hline -8 & \pu{99.99 \%}  \\
\hline
\end{array}

Well, weak acids are even worse with weak bases.  So try logic from 2 or 3 points rather than one.
The alcohol proton is very weakly acidic, pKa 17, but much more so than hydrogen covalently bonded to alkane, around pKa 50.
A weaker base, such as ammonia, pKa 9.3, will not touch it, but a very strong base, such as lithium diisopropyl amide, pKa 35, will remove it.

pKa 9.3 for ammonia $\ce{NH3}$, without mentioning it is the acidity constant of ammonium $\ce{NH4+}$ , may be confusing for the OP.

To add to the existing answers, you can make the statement in your textbook more correct by adding "even":

Alcohols are very weak acids, therefore they cannot react with even strong bases.

Without that word, it was somewhat misleading, naturally leading to your question.

Your question employs the Bronsted theory, where acids are proton donors, and bases are proton acceptors.  Similar to e.g., a farmers' market with sellers and buyers, acid and base can action only in presence of the other.  pKa values are a convened scale to sort them according to the  likelihood they exchange goods (here, the theory deals with protons) in a set of parameters (temperature, solvent, ...).
Mayr's scales of nucleophilicity and electrophilicty numbers of reagents (link to the project / project's database) illustrate «trading criteria» can equally be applied elsewhere.


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What happens to the air pressure above water as its heated to and past boiling?
If I put a pressure sensor inside an airtight container half-filled with water, and heat the water, what will the pressure sensor read as the temperature increases? Would it linearly increase (black line in graph) or sharp incline only after 100C (orange line) or something else? Could one tell the temperature of the water from this measurement?

The curve looks like the yellow line.
This experience is rather famous. It shows the critical point of water. Let's see how it goes.
If the temperature increases, the inner pressure increases a lot inside. At $110$°C, the pressure is about $2$ bars. The mass  of the water decreases. But its volume does not change much in the liquid phase, because of the liquid dilatation.
At $180$°C, the pressure is about $10$ bar. The density of the gas increases ($0.0079$ g/mL) and the density of the liquid decreases to $0.87$ g/mL
At $312$°C, the pressure becomes $100$ bars. The liquid density is as low as $0.71$ g/mL. The gas density has increased to $0.046$ g/mL.
The difference density decreases when increasing the temperature.
When the temperature is at $374.8$°C, the pressure is $217$ bars. And both liquid and vapor have the same density $0.31$ g/mL. At this point, suddenly, the limit between the two phases disappears!! There is just one phase inside the container, which can be described as neither a liquid nor a gas, or both simultaneously. This phase is known as the critical phase, and it is both a liquid and a gas. The point $212$ bars, $374.8$°C is called the critical point of water.
At higher temperatures, the water is supercritical, with just one phase.

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Ni etching using nitric acid
I etched Ni film (200nm) coated on diamond substrate using 30% nitric acid. It worked well and I was able to remove the Ni in 1 minute. A few things I noted that I'm hoping to understand why they happened and if I could do something about it.

flakes of Ni were left of the diamond. The current experiment was an etch test and part of the Ni was coated by photo resist. I don't want to etch any longer because I already have significant under etching in the patterned area. Would changing the concentration of HNO3 help? My primary Etchant is 70% HNO3 which I diluted to 30% because from what I've read 30% HNO3 etches Ni better than 70% (that seems counterintuitive to me so maybe it's wrong).

The area with the Ni flakes seem to have etched the diamond. The areas are all 360nm deep holes instead of being 200nm islands which surprised me. I know diamond has solid solubility in Ni but that only happens at temperatures above 800C. The etching was done at room temperature. Is it possible to get such high temperature on the sample surface during etching? Is there any way to avoid it?

Thank you for your help.

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Planarity of rings
Can a 7 or less than 7 membered homocyclic ring with unsaturated C atoms and fully conjugated system, lose planarity? The smallest homocyclic non-planar ring that I know is cyclooctatetraene which is a 8C compound.

Planarity is not a boolean value.

Only cyclopentane is planar, all other saturated rings are at least puckered.

@Martin-マーチン You probably meant cyclopropane. And I think OP is asking about fully conjugated ring systems...

@orthocresol yes, yes I did. And I guess you are right about OP's intentions.

@IvanNeretin The mathematician in me says planarity is very much a boolean value' for 3 vectors a, b and c in 3D they are coplanar iff a. b x c is zero. You can talk about small deviations from planarity yes, especially if you quantify them in some way, but the property of planarity itself is a yes/no question.

@IanBush Good point. Well, since the real-world numbers are known with some uncertainty, they never exactly equal anything, in particular zero. Hence the strict mathematical definition of planarity is useless to us, and we resort to using the approximate version, which is where my first comment applies.

@IvanNeretin Hmmm, different worlds. I'm a computationalist. I can almost always tell you exactly when my atoms all lie in a plane. Possibly the reason behind me wanting to keep the concept to planarity "pure". I would still rather you say something like "with deviations from planarity measured as ".

Setting aside rings with saturated carbon atoms, it appears that the cyclobutadiene-dication ring can be puckered despite being formally aromatic according to the $4n+2$ rule. Puckering may be favored by dispersion of the charge which, in the planar geometry, is highly concentrated in the four-carbon ring. From the abstract of Firme et al.[1]:

The cyclobutadiene dication was not experimentally characterized to the date. However, some of its derivatives were. Most of them have planar geometry, but tetramethylcyclobutadiene dication has a nonplanar geometry according to ab initio calculations. ... The planar cyclobutenyl dication derivatives have bond order of chemical bonds in the ring close to unity and relatively small electronic density in the ring. The puckered cyclobutadiene dication and its puckered derivatives have relatively high electronic density in the ring.

Reference

Firme CL, Antunes OA, Esteves PM. Electronic nature of planar cyclobutenyl dication derivatives. J Phys Chem A. 2007 Nov 22;111(46):11904-7. doi: 10.1021/jp075869j. Epub 2007 Oct 19. PMID: 17948974.

I read somewhere that similar thing can happen with analogous dianion.


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Question about ozonolysis of alkene/alkyne

I'm having trouble with this problem.  I do not see how any of the options could produce the given products on ozonolysis.

We're performing reductive ozonolysis, so products should be either aldehydes or ketones, but there are carboxylic acids in the products too.
To form the first given product, there needs to be a double bond 3 carbons away from the triple bond, only option a) satisfies this but it has a double bond in between which will also ozonolyse.
Can anyone tell me what I'm doing wrong?
The answer given in the key is a)

@Grimm I have added an answer.  If it answers your question please click the green tick next to the answer. It takes effort to write an answer. So please acknowledge that. Also you get +2 :)

We can use reverse engineering and process of elimination to arrive at the right answer.
All the options have a triple bond. So let us name the terminal alkyne carbons with number 1 and 2. With 2 being the Carbon on the rightmost end.
Let us represent the second and third products as follows : R1-COOH and R2-COOH where R1 = CHO and R2= H.
So we get R2-COOH from carbon number 2 of the terminal alkyne carbons. To get R1-COOH there needs to be a double bond at the carbon that is attached to carbon number 1.
Therefore we can rule out options (B) and (C).
Now to further reduce our options, the option D would produce a CO2 since there is a carbon which has two double bonds. Since there is no CO2 in the products we can conclude that the correct option is (A)


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How is voltage generated and maintained by a battery?
I was sent here by the folks on the physics stack exchange as they thought my question was more chemistry-related, but I am only a 10th grader and don't have a very strong chemistry background, so please be patient if I am not very proficient in expressing my question.
I was trying to figure out why parallel connected batteries don't add voltages, and the exceedingly common answer was that batteries act as sources of fixed potentials, not fixed fields. So my next step was to figure how they maintained a fixed potential - I discovered this answer which explains how batteries maintain a constant potential difference.
So what I understood was the field created by the batteries grows in strength as more charges are spewed out of the cathode. When this field becomes strong enough, it stops the redox reactions occurring in the battery. This causes the potential difference to become constant as there is no more charge accumulation.
I have my doubts about this explanation - I thought potential difference had nothing to do with how many charges there were present, but rather that the battery added some amount of energy into electrons coming out of the cathode which created a potential difference that was used to do work in a circuit.
At this point I am confused about how a potential difference is created by a battery - is my past understanding correct or does charge density truly create voltage? If charge density does create voltage, what is the point of adding energy to electrons inside a battery (if the chemical reactions in a battery even "add" energy to the electrons)?
If charge density does indeed create voltage, how would that explain stuff like voltage drops accross resistance?
Note: I want to know what happens at the atomic level, not analogies.

Delete whatever you have been reading on the internet and random websites about electrochemistry, including Q&A in physics from your mind. They can easily drive any sane mind crazy from rambling "answers". I understand that you are in grade 10 and your curiousity is appreciated. Only read stuff which has been written by reliable authors, i.e., published books or research papers for concepts. Can you access journals? There is a beautiful article, "Potentially Confusing: Potentials in Electrochemistry" DOI: 10.1021/acsenergylett.0c02443. There are so many potentials.

@AChem Perhaps later you might add a short answer that tells how Volta, long before the discoveries by Dalton, Davy, Faraday, etc., devised his condenser electroscope and used it to find the polarity of charges on the poles of his voltaic piles. ;-)

@EdV, So far no student or teacher has ever questioned the obvious.

The battery (for simple example think of galvanic cell) starts operating through spontaneous chemical reactions happening at electrodes. The electrons move between the electrodes depending upon the feasibility of reaction. The driving force here is provided from chemical work which is apparently non-conservative.This process continues until an equilibrium is reached , irrespective of the connected external circuit.

@An_Elephant, If I remove the word "apparently" in your text, that is, if I write "from chemical work that is not conservative" instead of "from chemical work that is apparently not conservative", is there anything wrong?

@HEMMI No , it won't be wrong.

The electric scalar potential $\phi$ is conservative and is a single-valued quantity except for the overall constant value.
There are some places of high potential values and some places of low potential values.  A positively charged particle can gain some kinetic energy if the particle starts from a high potential place, but continuous gain of kinetic energy is impossible only from electric scalar potentials because it is conservative. So I am curious about the origin of the non-conservative force of the battery.

The reason of voltage not being additive for batteries of parallel cells has nothing to do specifically with electrochemistry. It is shared behaviour among all parallel voltage sources, regardless of the nature of the source. This part is physics.

@HEMMI Electrostatic forces are conservative and electrostatic forces are not always equivalent to electric force.

For example , in a galvanic cell, the thermal motion and electric field comes to an equilibrium. The former is the one of which I am talking about. Although the origin is ultimately EM force, but on a large scale (and hence the emphasis of apparently) it appears to be non-conservative and a bit messy. I'm sorry for previous mistakes.

Voltage is kind of function of length of the path covered along the wire so multiple batteries in parallel must not add to more voltage.

I think it's important that you understand what a battery is and how it works from your question.
A battery is a portable galvanic cell.
It depends on the type of galvanic cell, but I'll give you an example of a Zinc-Copper battery. I don't understand what you mean by the addition of energy to electrons.
On one half cell is the anode. This is the site of oxidation. In this one, you have zinc ions in solution and the anode is made of zinc metal.
On the other half cell you have the cathode. This is the site of reduction. You have copper ions in solution and a copper cathode.
Linking these two half cells, you have two circuits: the internal circuit and the external circuit.
Internal circuit contains a salt in solution. Let's say KNO3. External circuit is just wires. Zinc is of course the stronger reductant, so it is oxidised like so:
Zn(s) --> Zn2+(aq) + 2e-
That makes a charge difference, so the internal circuit spews respective ions (NO3- in this case) to balance the charge in solution.
The electrons flow through the wire  to the cathode. They come into contact with Cu2+ ions and make solid copper metal. Also, potassium ions are shot into solution to balance charge.
Also, in regards to potential difference and charge density. Voltage is related to how many electrons pass through the wire in a span of time. Voltage is also the potential difference.
Stronger oxidants tend to have a larger E degree value than stronger reductants. Below hydrogen, they have negative values. So, the larger the distance between reductant and oxidant, the greater the potential difference.
I don't get what you mean by charge density. The charge on either side of the cell remains the same. In the wire, it makes more sense to talk about rate of oxidation and electron flow than charge density.
At least, that's what my textbook says.
Hopefully that answers your question.
You should do more research and find more reliable sources.

Rather, a battery(1) can be, among other meanings, a set of connected batteries(2), where a battery(2) is a portable galvanic cell. For the case of  car batteries of 6 or 12 cells, here the battery is not a  portable cell and their cells are not really portable. These are fruits of not keeping meaning of words.

Intrinsic Voltage is not determined by current flow altho it can affect operating voltage. An electrolyte carries current it is not spewing ions, it may or not be nonreactive. In you example it is imperative to prevent copper ions from contacting the Zn metal.  To maintain a constant voltage the cell must have a low internal resistance and the change in activities of the anode and cathode reaction chemicals must be kept as small as possible over the useful life. Parallel connection simply increases the amount of chemicals in the cell with no effect on the voltage.


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Mixing Boric Acid, Sodium Borate and alcohol
This is for a liquid detergent that I am developing - when I dissolve boric acid (doesn't fully dissolve though) and then mix it with dissolved sodium borate( also doesn't fully dissolve in water), settling occurs after sometime.
How do firstly dissolve these chemicals and how do I prevent the settling of these after sometime at room temperature?

Do you respect solubility of substances?

Please explain...I don't understand

Explain what you do not understand. Every substance has its solubility limit. If you provide enough water not to reach the limit, the substance can be fully dissolved. Otherwise, it cannot be. // It could be temporarily dissolved at higher temperature, but the raise of solubility with temperature very much depends on the substance. E.g. solubility  of NaCl is almost T independent, while for KNO3 it raises progressively steeply. And it would precipitate back when cooled down to original temperature.

So I must increase the amount of water when dissolving?

You must know the thing from everyday life. Take table salt. AFAIK there can be dissolved about 33 g in 100 mL of water. If you provided less than 100 mL for 33 g of salt, it will not fully dissolve, no matter how much you stir. // At 20 deg C, there can be dissolved 4.72 g boric acid/100 mL, or 2.5 g sodium tetraborate decahydrate(borax)/100 mL // See  https://en.wikipedia.org/wiki/Solubility_table

You have not said what mass of solids and what volume of water you have used.  If you do not say exactly  what you did, we cannot say what you did wrong or could do better.

Thank you for the response. I recently dissolved 150grams in 750ml of warm water. In another instance, I dissolved 75grams of Boric acid in 500ml of water, and added 15grams of Sodium Hydroxide. I am begining to see the light - thank you!

Before you go further with boric acid and sodium borate, you should know the characteristic properties of them first. Boric acid has the molecular formula $\ce{H3BO3}$ and has a triangular planner structure: $\ce{B(OH)3}$ with the central boron having only six electrons (not fulfilling Octet rule and hence a Lewis acid). According to this article:

Its physical appearance is white or almost white crystalline powder or can call it colorless. It is soluble in water and in ethanol (96 per cent), and freely soluble in boiling water to give acidic solution. For example, 3.3% aqueous solution of boric acid shows $\mathrm{pH}$ of $3.8$–$4.8$. Boric acid is also soluble in glycerol (85 per cent).

Thus, for instance, you may use heat to dissolve boric acid well. Yet, you may need to know this: $\mathrm{p}K_\mathrm{a}$ of boric acid is $8.92$–$9.24$ that is sensitive to temperature, ionic strength, and concentration (Ref.1). Accordingly,

The $\mathrm{p}K_\mathrm{a}$ varies with concentration because of polymerization above $\pu{0.02 M}$. Boric acid reacts reversibly with alcohols, especially 1,2-diols including carbohydrates, with carboxylic acids, thiols, and amines. These esters/adducts, are also Lewis acids with lower $\mathrm{p}K_\mathrm{a}$ values. Boric acid can stabilize some materials while catalyzing the degradation of others.

Thus, the settling you experience must be due to the polymerization of boric acid solution:

Finally, boric acid and/or borate can easily undergoes esterification with alcohol functions (Ref.2). For example,

References:

Antonio Lopalco, Angela A. Lopedota, Valentino Laquintana, Nunzio Denora, Valentino J. Stella, "Boric Acid, a Lewis Acid With Unique and Unusual Properties: Formulation Implications," Journal of Pharmaceutical Sciences 2020, 109(8), 2375-2386 (ODI: https://doi.org/10.1016/j.xphs.2020.04.015).
Supachok Tanpichai, Farin Phoothong, and Anyaporn Boonmahitthisud, "Superabsorbent cellulose-based hydrogels cross-liked with borax," Scientific Reports 2022, 12, 8920 (12 pages) (ODI: https://doi.org/10.1038/s41598-022-12688-2).

Thank you for enlightening me.


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What does the constant mean in this equation for the energy levels of an atom?
My textbook, Silberberg & Amateis, Chemistry: The Molecular Nature of Matter and Change (9th ed.), gives the following equation for calculating the energy level of an atom:
$$E = \pu{-2.18E-18 J}\left(\frac{Z^2}{n^2}\right)$$
Where $Z$ is the charge of the nucleus.
They never explained what the constant term that is multiplying the (Z/n)^2 term means, or where it comes from. Where does this come from? It looks similar to the Rydberg equation, but I still can't see its relation to it exactly.

Well, if you figure out how to compute the energy associated with a particular wavelength of light, you can check for yourself if it is Rydbergs equation.

Rydberg formula is but an empirical observation, it doesn't explain anything either. The real explanation comes from quantum mechanics.

Essentially, the equation conveys how the energy of a one-electron atom varies with factors such as the nuclear charge and electronic orbit number.
$-2.18\times10^{-18}\ \mathrm J$ is an empirically obtained factor, but you can also derive it from first principles using Bohr's atomic model.
$$2.18\times10^{-18}\ \mathrm J=\frac{(k_\mathrm ee^2)^2m_\mathrm e}{2\hbar^2}$$
Where $k_\mathrm e$ is Coulomb's constant, $e$ is the electronic charge, $m_\mathrm e$ is the mass of an electron, and $\hbar$ is the reduced Planck constant.
You can find the derivation on Wikipedia: https://en.wikipedia.org/wiki/Bohr_model
Intuitively, you can think of $-2.18\times10^{-18}\ \mathrm J$ as the energy of a hydrogen atom system with an electron in its first orbit. Here $Z=1$ and $n=1$.


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Does high pressure reverse reaction between zinc and sulfuric acid?
When zinc is added to sulfuric acid, it undergoes a chemical reaction that generates hydrogen gas and zinc sulfate. Can this reaction be reversed by applying pressure to the products converting them back into zinc and sulfuric acid?
$$\ce{Zn(s) + H2SO4(aq) <=> ZnSO4(aq) + H2(g)}$$
Are there any sources claiming this reaction can be controlled by pressure at all?

It should even be possible to get a very crude estimate for what kind of pressure would be needed by assuming that the pressure dependence of the Gibbs free energy of reaction is solely due to the hydrogen gas (while also assuming it's an ideal gas), and then figuring out at what pressure the reaction free energy reaches zero. As Poutnik mentions, it'll be very high, probably tens of thousands of atmospheres at least.

A classical example for this is Fe. For zinc the enthalpy is too big.

And there's another, more immediate, issue: why would it reduce Zn and not sulfur?

I am new to chemistry and found out this suggestion using chatgpt. We could say it was his idea.

It's a chatbot so it has no ideas. Just rips off of stuff it found and pretends to be someone.

As a very rough estimation of pressure when the hydrogen redox potential equals zinc standard redox potential, we can use the extrapolation of the Nernst equation:
\begin{align}
E^\circ_\ce{Zn/Zn^2+} &= E^\circ_\ce{H2/H+} + \frac{\pu{0.059 V}}{2}
\log\left(\frac{[\ce{H+}]^2}{p_{\ce{H2}}}\right) \tag{1}\\
\log(p_\ce{H2}) &= \left(E^\circ_\ce{H2/H+} -  E^\circ_\ce{Zn/Zn^2+}\right) \frac{2}{\pu{0.059 V}} \tag{2}\\
p_\ce{H2} &= 10^{\left(E^\circ_\ce{H2/H+} -  E^\circ_\ce{Zn/Zn^2+}\right) \frac{2}{\pu{0.059 V}}} \tag{3}\\
  &= \pu{10^{\frac{2\times\pu{0.76 V}}{\pu{0.059 V}}} atm} \\
  &\approx \pu{5.8E25 atm}
\end{align}
This extrapolating estimation is not realistic, being far out of validity scope of the Nernst equation. Even pressure many orders lower would make the system very different, everything solid, including hydrogen. $\pu{E25 bar}$ would cause an universal nuclear fusion.
As Loong has noted, the Solar core pressure is $\pu{2.5E11 bar}$. If protons had been fusable directly like deuterium is…
But it gives the clear message the pressure cannot help in reaction reversal.

Any equilibrium calculation involving ordinary compounds that gives a pressure of $10^5$ bar or more is probably not realistic.
Chemical bonds can deform and rearrange under such pressures and thus create different compounds we would consider exotic. One example we have seen on this site involves "sodium chloride" in which the bonding rearrangement leads to stoichiometries other than the normally "obvious" $\ce{NaCl}$ [1].
Reference

Zhang, W.; Oganov, A. R.; Goncharov, A. F.; Zhu, Q.; Boulfelfel, S. E.; Lyakhov, A. O.; Stavrou, E.; Somayazulu, M.; Prakapenka, V. B.; Konôpková, Z. (2013). "Unexpected Stable Stoichiometries of Sodium Chlorides". Science. 342 (6165): 1502–1505. arXiv:1310.7674. Bibcode:2013Sci...342.1502Z. doi:10.1126/science.1244989. PMID 24357316. S2CID 15298372.


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Stability of hexazine
I read about a hypothetical compound called hexazine on Wikenigma. It's a 6-membered ring all of whose atoms are nitrogen and they form alternate single and double bonds just similar to benzene. But it also says that this would be highly unstable. How can it be so much unstable when it's structure resembles so much to that of benzene which is exceptionally stable. Has it something to do with nitrogen's compact size?

Related: https://chemistry.stackexchange.com/questions/53864/ring-compounds-of-non-carbon-atoms

The decomposition to 3 molecules of nitrogen gas would be highly exothermic

Well, chemistry is like that: all elements are different, even if they are neighbors. Size is certainly not an explanation: one N to another N is the same as one C to another C.

In addition you might ask what happens when you place 6 N atoms in bonding proximity in a vaccuum, that is, what is (or are) the most stable structures of N molecules containing 6 atoms.

It's hard to beat that NN triple bond. You can make N clusters but they are not rings: https://arxiv.org/ftp/arxiv/papers/1805/1805.00835.pdf

But why does benzene not decompose spontaneously into three acetylene molecules ? The reverse equation is possible for acetylene being transformed into benzene. Why is $\ce{N_2}$ never transformed into hexazine ?

Related: "Azidoazide azide" at  acs.org or Wikipedia

Should't hexazine be aromatic? May be that could provide some stability.

Notice relative bonding energies in simple, double and triple bonds of carbon and nitrogen. Get the benzene stabilization energy due aromaticity. Evaluate and conclude.

How can it be so much unstable when it's structure resembles so much to that of benzene which is exceptionally stable.

Benzene is not exceptionally stable to start with. As you substitute more and more CH groups with N in benzene, the enthalpy of formation (already positive for benzene) becomes larger and larger (e.g. 1,3,5 triazene). A positive enthalpy of formation tells you how much energy is released when the molecule turns back into elements. (It does not tell you about the activation energy needed for the process, so a molecule can exist even if a positive enthalpy of formation).
Playing around with the different unsubstituted species on MolCalc, it seems that the predicted enthapies of formation are higher when two nitrogen atoms are directly next to each other. In hexazine, there would be 6 of those pairs.
There are no examples of pentazine or hexazine. There are different tetrazines, but 1,2,3,4 tetrazine only exists fused to benzene (in the picture below, compounds (1), (4), and (5) have not been observed yet).

Source: https://www.sciencedirect.com/science/article/pii/B9780080965185001386
Like the comments of the question already say, bonding in elemental carbon is very different from that in elemental nitrogen. Similarly, substituting N for CH in ethane yields hydrazine, a rocket fuel. So while the Lewis structure look the same, many properties change dramatically when you substitute N for CH.

Hydrazine isn't explosive... unless you mix it with oxidant, but so is ethane. There are plenty of nitrogen rich explosives, though, like tetrazene explosive

@Mithoron Point well taken, I removed that bit. Ethane has a negative enthalpy of formation, while hydrazine has a positive enthalpy of formation. Thus, the latter can be used as a monopropellant in thrusters operating in outer space, e.g. here.

An aromatic five-nitrogen ring is known as the pentazolate ion, $\ce{N_5^-}$. Among several documentations of this anion, which is isoelectronic with the cyclopentadienyl anion, the Wikipedia article identifies an isolated salt with moderately good thermal stability:

In 2017, white cubic crystals of the pentazolate salt, $\ce{(N5)6(H3O)3(NH4)4Cl}$ [the article and cited reference render the anion first in the formula] were announced. In this salt, the $\ce{N_5^-}$ rings are planar. The bond lengths in the ring are 1.309 Å, 1.310 Å, 1.310 Å, 1.324 Å, and 1.324 Å.[1] When heated, the salt is stable up to 117 °C, and over this temperature it decomposes to ammonium azide.[1]

Crystal packing presumably stabilizes the anion in the presence of protic species that would surely react with it in solution.
Cited Reference

Zhang, Chong; Sun, Chengguo; Hu, Bingcheng; Yu, Chuanming; Lu, Ming (26 January 2017). "Synthesis and characterization of the pentazolate anion cyclo-N5ˉ in (N5)6(H3O)3(NH4)4Cl". Science. 355 (6323): 374–376. Bibcode:2017Sci...355..374Z. doi:10.1126/science.aah3840. PMID 28126812. S2CID 206651670.

Cool reference! From the editorial introduction: "The flip side of the robust stability of N2 is the instability of any larger molecules composed exclusively of nitrogen. These molecules nonetheless remain enticing targets for explosive and propellant applications."

$\ce{N3- and N5+}$ also exist, see here. They aren't aromatic, though (not even a ring).

@karsten the WP article says chemists trued to synthesize $\ce{N5^+N5^-}$ for use as rocket fuel. That apparently failed, but we do have $\ce{LiN5}$ synthesized wt high pressure. Might that work with an appropriate pentazenium compound?

I can't tell, I'm not a rocket scientist. ;-)


text:
Derivation of Flory's Model for Linear Condensation of Polymers
I'm having some trouble understanding a part of Flory's model for linear polycondensation reactions. In his 1936 paper, he derives an expression for the probability of the existence of a particular $x$-mer as

$$\Pi_x = xp^{x-1}(1 - p)^2\tag{2}$$

Later, he states the following:

In order to locate the $x$ value which gives a maximum in $\Pi_x,$ let
$$\partial\Pi_x/\partial x = (1 - p)^2(p^{x - 1} + xp^{x - 1}\ln p) = 0$$

Not being the most mathematically savvy person, can someone explain to me how he got to this expression?
Reference

Flory, P. J. Molecular Size Distribution in Linear Condensation Polymers. J. Am. Chem. Soc. 1936, 58 (10), 1877–1885. DOI: 10.1021/ja01301a016.

"Not being the most mathematically savvy person" - How much do you know about differentiation? Does https://www.mytutor.co.uk/answers/5998/A-Level/Maths/How-do-you-differentiate-a-x/  Make any sense? Do you know the Product Rule for differentiation?

@Justus Have a look at my answer and if it answers your question, please click on the tick mark next to the answer. You also get +2 points on doing this :)

Given $\Pi_{x}$= $*^{−1}$ *$(1−)^2$
$\partial$ refers to partial differentiation.
$\frac{\partial \Pi_{x}}{\partial x}$ means differentiating $\Pi_{x}$ with respect to x.
Therefore $(1−)^2$ is a constant when differentiating with respect to x.
So during differentiation the constant stays as a constant multiplier.
Now see that there are two terms involving x, therefore we need to use the product rule of differentiation.
$\left(\frac{\partial \Pi_{x}}{\partial x}\right)$ =  $(1−)^2$ * ($^{−1}* \frac{\partial x}{\partial x} + x*\frac{\partial p^{x-1}}{\partial x}$)
$\frac{\partial x}{\partial x}$ = 1 and for calculating $\frac{\partial p^{x-1}}{\partial x}$ look at exponentiation rule of differentiation as suggested by @Ian Bush in the comments.
Therefore we finally have :
$\left(\frac{\partial \Pi_{x}}{\partial x}\right)$ =  $(1−)^2$ * ($^{−1}* 1 + x*ln(p)*p^{x-1}$)
And since he is trying to locate the  value which gives a maximum in $\Pi_{x}$, he equates the equation to 0.
$\left(\frac{\partial \Pi_{x}}{\partial x}\right)$ =  $(1−)^2$ * ($^{−1}* 1 + x*ln(p)*p^{x-1}$) = 0


text:
Can you write oxidation states with negative Roman numerals?
As I understand, Roman numerals are used for oxidation states, whereas Arabic numerals are used for the charge of full ions, with some ambiguity allowed when a metal centre is all but ionised, such as in hexaaqua compounds.
But I have never seen a Roman numeral with a negative sign in front of it.  Would I write the oxidation state of sulphur in hydrogen sulphide (H$_2$S) as (-II)?

The Romans would surely look at it with raised eyebrows.

@FurrierTransform Can you clarify if you intend to use the Roman numerals in chemical names, such as iron(II) sulfate, iron(III) sulfate?  So far, I interpreted your question could be about reaction equations - maybe departing from an erroneous assumption

@IvanNeretin TIL supercilium for eyebrow.  Too sad TLL did not yet advance to letter S, but it lists cilium (lemma 3:1057) mesosupercilium (lemma 8:854).

According to IUPAC guidelines, roman numerals are used to denote oxidation states when used in the name of a compound. This is used when the cationic group can have varying oxidation states. For example, $\ce{FeSO4}$ is written as iron (II) sulphate, and $\ce{Fe2(SO4)3}$ is written as iron (III) sulphate. Although anionic groups can also have variable oxidation states, we omit mentioning the oxidation state in the name of the compound. Instead, we use standard prefixes to identify the oxidation states. For example, potassium oxide ($\ce{K2O}$), potassium peroxide ($\ce{K2O2}$), and potassium superoxide ($\ce{KO2}$) have different oxidation states for oxygen: -2, -1, and -1/2 respectively. But instead of mentioning this explicitly, we denote these with different standard names.

Have a look on mhchem syntax which one is encouraged to deploy in the body of questions, answers, and comments for chemical formulae and equations.  If used well, the results have share the professional appearance (e.g., italic fonts are used, but not on symbols of chemical elements), and a syntax easier to read and maintain.

It would be good to cite the specific IUPAC guideline to which you refer

Wikipedia's list of known (whole-number) oxidation states uses Roman numerals only for positive ones. Zero and negative numbers are represented with Arabic numerals and the negative ones with of course a minus sign.

I certainly only tend to use them for the cation as well.  However, I am comparing the various oxidation states of sulphur, with the upper bound, sulphur(IV), as in sulphate, being non-controversial.  But I'm not sure how to write the lower bound.  It would seem weird to mix notation. But also weird to incorrectly label the sulphur with a +6 charge.

I think that @NilayGhosh has answered my question:  You do use negative Roman numerals, it's just less commonly needed and therefore less commonly seen, but would still be correct in situations where Roman numerals are the correct notation. If you post this, I will accept. Pending any further input.

@FurrierTransform I have converted my comments into answers.

Oxidation state of an anion gets fixed automatically once we have specified the oxidation state of the associated cation.

Converting my comment into answer upon OP's request.

Basically the other answer is correct. Metals tends to hold positive charge and they achieve multiple oxidation state. So, in order to differentiate them, we use roman numerical in their names to avoid confusion and simultaneously providing the O.S. of metal which is always a useful information. For e.g iron(II) oxide and iron(III) oxide In few occasion, metal can achieve negative when the entity holds a negative charge, thus becoming an anion. For anions, we use specific suffixes for different  oxidation state. So, in this case we don't need roman numbers. For e.g. Disodium tetracarbonylferrate. We don't need to mention (-II) since it is redundant and in theory, roman numerals doesn't exist for 0 and negative numbers. But some publication(example) and textbooks still adopts the writing style for convenience sake(at least in formulae).
Here is a good article on this topic:https://pubs.acs.org/doi/pdf/10.1021/ed2001347

Just to clarify, we usually denote negative anion oxidation states with different suffixes, which allows the oxidation state of the relevant atoms, like sulphur in sulphate, to be figured out (VI here).  However, if one wished to refer to the oxidation state of a specific atom, where that atom is not just an ion with a charge $\pm n$, like S in H$_2$S, or F in F$_2$O, than a negative Roman numeral would be appropriate, just weird and not commonly seen.


text:
Can anyone help me with the derivation for the formula of entropy?
I have this formula in the chapter of thermodynamics:
$$\Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) + n  R \ln\left(\frac{V_2}{V_1}\right),$$
where $\Delta S$ is entropy, $n$ moles, $C_v$ molar specific heat capacity at constant volume, and $R$ universal gas constant.
I'm curious how it got derived. Can anybody please derive this formula for me?

What have you tried ? Have you learn carnot cycle ?

Review the guide How to ask. Not following the guidance may lead to objections, question down-voting or even question closure.

@An_Elephant  I would like to give some context here. We have thermodynamics in both physics and chemistry. I was studying entropy topic in chemical thermodynamics where i encountered this formula which was given in my textbook without derivation. And yes i have learned the carnot cycle in thermodynamics which is given in my physics textbook.

For an ideal gas, you can derive it this way:
\begin{align}
\Delta S &= \int_{T_1,V_1}^{T_2,V_2}{\frac{\delta Q_\mathrm{rev}}{T}} \\
&= \int_{T_1}^{T_2}{\left(\frac{\delta Q_\mathrm{rev}}{T}\right)_V} + \int_{V_1}^{V_2}{\left( \frac{\delta Q_\mathrm{rev}}{T} \right)_T }\\
&= \int_{T_1}^{T_2}{\left(nC_V \frac{\mathrm{d}T}{T}\right)_V} +
\int_{V_1}^{V_2}{\left(\frac{p \cdot \mathrm{d}V}{T}\right)_T} &( \delta Q_\mathrm{rev})_T = (p.\mathrm{d}V)_T\\
&= nC_V \int_{T_1}^{T_2}{\left( \frac{\mathrm{d}T}{T}\right)_V} +
nR \cdot \int_{V_1}^{V_2}{\left(\frac{ \mathrm{d}V}{V}\right)_T} &(p=\frac{nRT}{V})\\
&= nC_V \ln {\left(\frac{T_2}{T_1}\right)} +
nR \ln {\left(\frac{V_2}{V_1}\right)} \\
\end{align}
If we consider an isobaric reversible thermal expansion instead, it would be:
\begin{align}
\Delta S &= nC_V \ln {\left(\frac{T_2}{T_1}\right)} +
nR \ln {\left(\frac{V_2}{V_1}\right)} \\
&= nC_V \ln {\left(\frac{T_2}{T_1}\right)} +
nR \ln {\left(\frac{T_2}{T_1}\right)}\\
&= n(C_V + R)  \ln {\left(\frac{T_2}{T_1}\right)} \\
&= nC_p \ln {\left(\frac{T_2}{T_1}\right)}
\end{align}

For an ideal gas undergoing a reversible process, the 1st law of thermodynamics tells us that $$cU=nC_vdT=dq-PdV=dq-\frac{nRT}{V}dV$$Dividng both sides of this equation by T then gives:  $$\frac{dq}{T}=nC_v\frac{dT}{T}+nR\frac{dV}{V}$$Each term in this equation is an exact differential, with the left side representing dS.  Therefore integrating, we obtain:  $$\Delta S=nC_v\ln{(T_2/T_1)}+nR\ln{(V_2/V_1)}$$


text:
How to make bonds thicker in ChemDraw?
I often see molecules depicted using ChemDraw with all or selected bonds visibly thicker than those used with the style defined with ACS-1996 template.
For example, see the PDF handouts for Baran's laboratory group seminars created with ChemDraw, for instance, Biotech: A High Stakes Table for Science (Garrido-Castro, 2022) (PDF):

Another example would be an illustration produced in Meng et al. [1]:

Where can I find the relevant parameters for the bond thickness control in ChemDraw?
Reference

Meng, G.; Lam, N. Y. S.; Lucas, E. L.; Saint-Denis, T. G.; Verma, P.; Chekshin, N.; Yu, J.-Q. Achieving Site-Selectivity for C–H Activation Processes Based on Distance and Geometry: A Carpenter’s Approach. J. Am. Chem. Soc. 2020, 142 (24), 10571–10591. DOI: 10.1021/jacs.0c04074.

Did you try "File -> Document Settings … -> Drawing" to set your own value, or "File -> Apply  Document Settings from" to use a predefined one?

ChemDraw stores styles as *.cds files (while templates, about structures, glass ware, etc as *.ctp).  They are in a binary format Windows stores in a folder of .../ChemOffice20XX/ChemDraw/ChemDrawItems with a varying string of ChemOffice20XX - depending on the version of ChemDraw installed.
Anyone can create a style file from scratch.  Or can adjusting one already provided (ACS, RCS, ...) by ChemDraw and save it as a new style.  Phil Baran's choice to use a non-serif font Arial (?) but bold arguably is not that unique.  It equally is seen for example in the retrosynthesis books by Warren and Wyatt, or in Classics in Total Synthesis by Nicolaou and Snyder.  Hence, if you want a formula similar like

and don't get in touch with any of the authors / the author's groups mentioned, you can play with the parameters similar to the ones the .gif outlines below.

I think there is a recommendation to cite references on chemistry.se in the pattern of ACS journals, but do not recall if there equally is dedicated "use this (or any other specific) style to format structure formulae" rule here.  Hence it is not surprising, authors on chemistry.se use a style a sketcher offers by default, which is not too personalized.
Book references:
Warren, S.; Wyatt, P. Organic Synthesis: The Disconnection Approach. Wiley, 2008, ISBN 978-0-470-71236-8 (link to the publisher's landing page).
Nicolaou, K. C.; Snyder S. A. Classics in Total Synthesis II: More Targets, Strategies, Methods. Wiley, 2003, ISBN 978-3-527-30684-8 (link to the publisher's landing page).


text:
Can carbocations exist in a nonpolar solvent?
I have seen in organic chemistry books that the solvents they used in reactions are either protic or aprotic, both polar, but can carbocation exist in a non polar solvent? I believe it cannot exist.

You're overestimating solvent effect here, but it's there. Related thing is solvatochromism.

The one here: https://chemistry.stackexchange.com/questions/9212/can-a-organic-compounds-such-as-hydrocarbons-contain-an-ionic-bond should be just fine.

As described by Amogh, we would not expect carbocations, or ions in general, to be dissolved in nonpolar solvents except in very special circumstances such as cryptand chelation of some cations. Stable carbocations would likely be too bulky and irregularly shaped for such a situation to work for them.
What we can do, however, is form some carbocations in nonpolar solvents and recover them as precipitated salts. Breslow et al. synthesized several ions based on the cyclopropenyl cation ring in this manner. The synthesis of these ions under non-superacidic conditions attests the aromatic stabilization of this otherwise relatively strained and high charge-density ring system.
The example given here (often given as a textbook problem) involves forming the unsubstituted $\ce{C3H3^+}$ cation by reaction of 3-chlorocyclopropene ($\ce{C3H3Cl}$) with antimony pentachloride ($\ce{SbCl5}$) [1]. The precursors are each separately soluble in cold carbon tetrachloride; when combined these solutes form the aromatic cation through the Lewis acid $\ce{SbCl5}$ extracting the chloride from $\ce{C3H3Cl}$. The resulting salt, $\ce{[C3H3^+][SbCl6^-]}$, is of course insoluble in the carbon tetrachloride and is thereby quantitatively recovered.
Reference

Breslow, R.; Groves, J. T. (1970). "Cyclopropenyl Cation. Synthesis and Characterization". J. Am. Chem. Soc. 92 (4): 984–987. doi:10.1021/ja00707a040.

Meh, as I mentioned in earlier comment, you could make a soluble salt.

I'm aware of the hydrocarbon ions, but the answer does not specify that they could be dissolved in a nonpolar solvent.

While you can always assume there to be some reversible reaction between the neutral molecule and its carbocation, in the case of non-polar solvents the equilibrium is so far towards the neutral molecule that the concentration of the carbocation can be assumed to be non-existent. This is because carbocation formation is quite endergonic (non-spontaneous), which can only be compensated by the energy released during solvation by a polar solvent.


text:
Why does Methanol have a higher surface tension than Ethanol?
I am researching the surface tension of different alcohols and found something odd: the surface tension of Methanol was higher than that of Ethanol, which doesn't make sense to me. My understanding is that surface tension depends on the strength of intermolecular forces--Both alcohols have a hydrogen bond, but Ethanol has a stronger london-dispersion force(this is evidenced by the boiling/melting point), so shouldn't Ethanol have a higher surface tension? Well, I know that this is wrong, but I can't find a good explanation to why Methanol has a higher surface tension.
The only theory I currently have in mind is that the LDF of Ethanol somehow cancels out the strength of its hydrogen bond, but I am not sure if this is correct.

Why just methanol  over ethanol? Why not benzene over toluene? Did you consider the strengths of H-Bonding? That must be overwriting your london-dispersion force argument.

As I pointed out in my comment, the strengths of hydrogen-bonding may play a role in difference in the surface tensions between ethanol and methanol. Methanol is a smaller molecule compared to ethanol and certainly has stronger H-bonding comparatively.
This is because methanol has only one carbon atom in its main carbon chain but ethanol has two. This extra methylene group in ethanol in addition to methyl group of methanol $(\ce{CH3CH2})$ would be able to form London Forces with methyl groups in ethanol molecules, making the formation of hydrogen bonding within the molecules comparatively less prominent, thus can be considered less strong than that in methanol. In methanol, the strongest inter-molecular force between molecules are only hydrogen bonding.
Aside hydrogen bonding, the difference between activity coefficients of methanol and ethanol has playing bigger role according to published data (Ref.1). According to Ref.1:

The study of the influence of the activity coefficient on the surface tension behavior is of particular importance. [...] Taking the activity coefficient into account leads to a substantial decrease in the surface tension (except for the $\ce{KCl}$ electrolyte and the case of the $\ce{NaCl}$ electrolyte where the activity coefficient significantly increases the surface tension at high salinity).

The activity coefficient of ethanol is $\gt 1$ while that of methanol is $\lt 1$ in ethanol-methanol binary system, according to Ref.2:

Perhaps the most interesting of these three binaries is the methanol-ethanol system, which is generally assumed to be ideal. Figure 4 shows that the components when present in low concentration do not behave ideally. The activity coefficients of methanol at low methanol concentrations are less than 1.00, and the coefficients for ethanol at low ethanol concentrations are greater than 1. To support all of above arguements, the ref.3 states clearly that:
The behavior of ethanol-methanol binary system is far from ideal because both the hydrogen bond donating and accepting strengths are seem not to be identical.

References:

Philippe Leroy, Arnault Lassin, Mohamed Azaroual, and Laurent André, "Predicting the surface tension of aqueous 1:1 electrolyte solutions at high salinity," Geochimica et Cosmochimica Acta 2010, 74(19), 5427-5442 (DOI: https://doi.org/10.1016/j.gca.2010.06.012).
H. H. Amer, R. R. Paxton, and Matthew Van Winkle, "Methanol-Ethanol-Acetone: Vapor-Liquid Equilibria," Ind. Eng. Chem. 1956, 48(1), 142–146 (DOI: https://doi.org/10.1021/ie50553a041).
K. Nilavarasi, Thejus R. Kartha, and V. Madhurima, "Evidence of anomalous behavior of intermolecular interactions at low concentration of methanol in ethanol-methanol binary system," Spectrochimica Acta Part A: Molecular and Biomolecular Spectroscopy 2018, 188(1), 301-310 (DOI: https://doi.org/10.1016/j.saa.2017.07.015).


text:
Will charcoal (carbon) oxidize in the air?
Wikipedia — Carbon sequestration claims that burying charcoal into the soil offsets $\ce{CO2}$ and thus reduces the greenhouse effect:

In the soil, the biochar carbon is unavailable for oxidation to $\ce{CO2}$ and consequential atmospheric release.

Is it a must to cover charcoal/biochar with the soil to prevent the air and rain water exposure to avoid oxidation, i.e. offset carbon dioxide? Would it otherwise oxidize to $\ce{CO2}$ under normal conditions (no fire) if left above the ground?

The spontaneous oxidation of charcoal by air at room temperature happens spontaneously. But it is so slow that it can be forgotten. The half-life of the reaction iis much much longer than the duration of the universe.

slightly related (especially Sir Poliakoff's video "Charcoal in Liquid Oxygen (slow motion)") Why doesn't carbon fiber overwrapping in LOX catch fire? (watch this video first)

So to capture carbon this way you don't necessarily need to bury biochar, you just need to find a meaningful usage for it, correct? And what about water? Can charcoal oxidation be accelerated by regular water contact (like rain/river water)?

Related: https://chemistry.stackexchange.com/questions/59884/at-what-temperature-does-charcoal-burn


text:
Influence of (activated) charcoal on water's pH
Does immersing (a) charcoal / (b) activated charcoal (both clean, no ash) into distillate water influences water's pH? If yes - in which direction - acidic or alkaline?
If there is an influence - is this influence reduced with each new immersion into new water (i.e. charcoal gets rinsed)?

Yes, the pH can change depending on source of carbons.

What are you actually trying to do or achieve in # 3?

Create activated charcoal at home easily.

The vinegar story is not right. Activated charcoal is difficult to make as home. You need very high temperatures (like 900 C)  and an inert atmosphere.

@AChem - I've read that there are 2 ways to activate carbon - a physical one (which indeed involves very high temperatures) and a chemical one. Here is an article that describes "How To Make Activated Charcoal With Vinegar" (search for that phrase there)... Is it disinformation?

Yes complete disinformation.

Pure water exposed to air is never neutral with pH 7, but has pH about 5.6 due traces of dissolved carbon dioxide, which is weakly acidic. Tap water is pH neutral just accidentally.

Active charcoal/carbon alone is indifferent with respect to water $\mathrm{pH}$.
But what may matter is what has been adsorbed on the active surface and can be released to water, or adsorbed from water (ammonia, acetic acid, carbon dioxide etc).
$\mathrm{pH}$ of pure destilled water is very unstable, affected by dissolved impurities from solid surfaces or air. The major factor is aerial $\ce{CO2}$ forming equilibrium $\mathrm{pH} \approx 5.6$, making water slightly acidic. These effects may overrule the effect the active charcoal.

As far as I know, pH will not be affected.
Carbon is not soluble in water and does not react with it under normal conditions. Hence why carbon can be used for filtration systems. Activated carbon is chemically the same.
And activated carbon that you buy is very unlikely to be contaminated with acids.
Method of production does not matter. Though I'm not sure that vinegar would make activated carbon. If the "activated" carbon is still wet with vinegar, then obviously pH will decrease.
Hopefully that answers your question.

Carbons (not necessarily activated carbon) can change water pH. They can have acidic functional groups bonded to their surface.

Define carbons. Are you talking about organic acids? In that case, yes. But they aren't C(s). Are you talking about acids adsorbed onto impure carbon? Yeah, I talked about that.

Carbons are carbons (solid carbonized stuff). They are not organic acids.

@AChem - I'm interested in charcoal/biochar (completely washed from ashes) - will it influence pH?

@AChem - and if carbons/charcoal/biochar do change water pH - in what way? Will it become more acidic or more alkaline?

It will depend on the treatments. A properly made activated carbon should not change water pH, but all other types (home-made) will not be neutral. These carbons have "surface" groups, which can change water pH.

@AChem - you write: "these carbons have "surface" groups, which can change water pH" - but how can they change the pH value? Will it become more acidic or more alkaline?


text:
Choice of Basis for solution space of a Hamiltonian
I'm quite new to the field of Quantum chemistry, but I can't wrap my head around the following postulate of Quantum Chemistry, stated as Postulate 3 in [1]:
Postulate 3: In any measurement of the observable associated with the operator A, the only values that will ever be observed are the eigenvalues $a_n$ which satisfy the eigenvalue equation $$A\psi_n = a_n \psi_n.$$
My first thought was, that this was just a little bit fuzzy, because one result in QC is, that energies, as eigenvalues of the Hamiltonian, don't exists continiuosly but only discrete, so we would need restrict the set of possible $\psi_n$ to a set of basis vectors for the space of eigenfunctions of A. Otherwise we could use any linear combination of eigenvectors and thus realise any (energy) eigenvalue.
This is for example done, if we take a look at the 1 dimensional particle in the box, which is usually provided with the basis set $B_1 = \{\psi_n\}_{n \in \mathbb{N}}$ with
\begin{equation}
    \psi_n(x) = \begin{cases}\left(\frac{2}{L}\right)^{1/2} \sin\left(\frac{n\pi x}{L}\right) & 0\leq x \leq L \\ 0 & else\end{cases}.
\end{equation}
According to the postulate above, the choice of the basis is quite important as the corresponding energy values to the wave functions $\psi_n$ are the only ones that could be observed. However, and here is my problem, the choice of this basis set for the 1D particle in the box seems arbitrary to me, as we could also set
\begin{equation}
    \psi'_1 = \frac{1}{\sqrt{2}}(\psi_1 + \psi_2) \text{ and } \psi'_2 = \frac{1}{\sqrt{2}}(\psi_1 - \psi_2),
\end{equation}
so that $B_2 = \{\psi'_1, \psi'_2, \psi_3, \psi_4,\dots\}$ would also be an (orthonormal) basis set for the same space as $B_1$.
The important implication would now be, that the corresponding energy values for $B_1$ and $B_2$ are different and hence, depending on the choice of such a basis set, the system could be in different (discrete) states.
To me this seems as a contradiction, as I would expect that the energy states of the system should be independent of that mathematical framework, i.e. every mathematical formulation should yield the same physical properties. This leads me to the conclusion, that I made a mistake somewhere or I did understand something wrong. Any help would be much appreciated.
[1]: D. A. McQuarrie, “Quantum Chemistry,” Oxford University Press, Oxford, 1983.

The expectation value of the energy, defined by $\langle E \rangle = \langle \psi | \hat{H} | \psi\rangle$, is not the same as the value that would be obtained if the energy were to be measured. Expectation values are continuous, and are linear combinations of eigenvalues. The eigenvalues, which are the values you would measure, are discrete.

@orthocresol What I meant was that, if $\psi_1$ and $\psi_2$ are eigenvalues of $H$ then so is any linear combination of them both. Hence the choice of the basis set of these $\psi_n$ is crucial, if that makes sense.

@student7481 Your above comment is garbled. If ψ1 and ψ2 are eigenvectors of a given matrix then a linear combination of those eigenvectors is also an eigenvector of the original matrix if and only if the corresponding eigenvalue is the same for both cases, i.e. they correspond to degenerate states. The linear combination of two non-degenerate evecs is in general not an eigenvector of the original matrix.

You are right, that was also the error.

If you can algebraically solve the Schroedinger differential equation then generally $n$ quantised solutions are  obtained and are determined by the potential energy function you use and the boundary conditions. If you cannot solve the equation algebraically then using a basis set of other functions, such as sin/cos, gaussians etc, then the solution can be found by adding these functions to form a basis set of $m$ functions and $m$ can increase until sufficient accuracy is obtained for the $n^{th}$ solution to the Schroedinger equation. Generally $m \gg n$ because the basis set functions are not solutions to your problem but will approximate to it if sufficient number of terms are used. See  'Solution of the Schrödinger Equation for One-Dimensional Anharmonic Potentials' J. Chem. Educ. 2011, 88, 7, 929–931 for examples.

This is actually something that had me stuck for a long while as an undergraduate, so let me see if my resolution helps you. I don’t want to get into any philosophically sticky propositions about wavefunction collapse or the interpretation of quantum measurements, but suffice to say that this postulate can be presented in a rather pragmatic form using spectroscopy. In fact, McQuarrie shows that spectroscopic transitions arising from the coupling of the electric dipole of a molecule with the electromagnetic field of light cause the coefficients of a wavefunction to change. If you begin in some initial state, then the probability to transition to a final state is proportional to the transition dipole moment integral between the states. This result is called Fermi’s Golden Rule of time dependent perturbation theory. But more importantly, in the process of deriving this result you find that only when the frequency of the electromagnetic radiation is on resonance (matches energy) with the gap between the energy eigenvalues of the states will you cause any transitions to occur. This is a very practical example of why the eigenvalues of the Hamiltonian have such a prominent role in quantum measurement. For less friendly cases like the double slit experiment, we somewhat have to throw our hands up because we cannot write down an explicit Hamiltonian for the problem. In other words, we really have no mechanism to describe why we observe the results that we do. But whenever we can write out a Hamiltonian explicitly we get results like we see with Fermi’s Golden Rule. I hope this was helpful to you.

The postulate could have been stated more clearly as

Postulate 3: In any measurement of the observable associated with the operator A, the only values that will ever be discretely observed are the individual eigenvalues $a_n$, each corresponding to an eigenstate described by wavefunction $\psi_n$, which satisfy the eigenvalue equation $$A\psi_n = a_n \psi_n.$$

One consequence being that degenerate eigenstates (those sharing the same eigenvalue) can be arbitrarily combined to generate an observable state with the same observable eigenvalue, but orthogonal states (differing in eigenvalue) cannot be combined to generate pure observable states.

Hat tip to @IanBush for the enlightening comment to the OP.

Another consequence is that when a system exists as a superposition the act of observation perturbs the system when not in an eigenstate and results in collapse of the wavefunction (selection of eigenstates).


text:
Why this preparation of Alkyne by dehydrohalogenation is done in two steps?
My teacher was explaining preparation of alkyne by dehydrohalogenation of Alkane dihalide. Actually the preparation is carried out in two steps. In the first step an alkane dihalide is dehydrohalogenated with alcoholic KOH and then the formed alkene halide is again dehydrohalogenated with Sodamide (NaNH2). Given that sodamide is a more strong dehydrohalogenating agent than alcoholic potassium hydroxide, why can't we carry out it in a single step with sodamide itself ? Why is preparation of Alkynes by dehydrohalogenation is carried out in two steps ?

It may be a question of price. Sodamide is more expensive than $\ce{KOH}$

Consider the other pathways open to the dihalide starting material with sodamide and ammonia.

What Waylander said - this wouldn't work, because sodium amide is nucleophilic.

@Mithoron So The starting material cannot be Sodamide ?

It is not just that sodamide is nucleophilic. Any deprotonation that occurs generates ammonia which is more nucleophilic so you have two pathways that will give amines.

@Waylander I can't understand what you are saying

Consider the possible reactions of sodamide with the dihalide:
1 - Sodamide acts as a base deprotonating the dihalide forming the double bond by elimination of halide anion giving the intermediate alkenyl halide that you want. This is stable to the reaction conditions, but the other product of this is a molecule of ammonia - we'll come back to this.
2 - Sodamide acts as a nucleophile displacing one of the halides (and then possibly going on to displace the other halide giving the aziridine). This pathway is irreversible, costing you yield and complicating your purification of the product. Remember that molecule of ammonia? It is a better nucleophile than sodamide and can do this pathway too, giving rise to multiple amine derivatives of the original dihalide.

Not sure about aziridine ring formation being "irreversible", especially in the presence of nucleophiles.

Why is a ammonia a better nucleophile than NaNH2 ?

The mechanism of the first dehalogenation is by KOH extracting a proton, a rearrangement to a vinyl chloride, and one of the chlorides as a leaving group (as a concerted E2 elimination).
Since the first proton extraction is stabilized by 2 electron withdrawing Cl, KOH is adequate for this step, less expensive, and more easily controlled$^1$.
The second proton extraction to form the alkyne requires a much stronger base, NaNH2, which is far more reactive.

One thing you don't want is something out of control, like a run-away polymerization or overheating of a large batch.

This is why sodamide could be added cautiously, in a second step.
There may also be a higher chemical cost to using 2 NaNH2 instead of KOH/NaNH2, but this would be off-set by saving a synthetic step and worth investigating if safe.
Alkyne is known to be synthesized in one step with 2 NaNH2 in anhydrous ammonia.
$^1$ which is helpful when they tell you, OK, go out into the plant and make 5000 gallons of it.

Do you have an example of the prep of an alkyne from a vicinal dihalide with only sodamide, rather than teaching materials claiming it works?

Not to mention, since when E2 has carbanion?

@Waylander Here is another reference for vicinal dihalides.

@Mithoron you are right.  The carbanion would not form as a separate intermediate.  E2.

@RobertDiGiovanni   Does the Chem. Rev. referenced discuss using NaNH2? I can only see the abstract

Vicinal dihalide reference in comments, please see subsection 3 in table of contents.

@RobertDiGiovanni  The dehydrobromination in ref. subsection 3 is run with KOtBu or DBN/KOtBu, not sodamide


text:
"Best" way of referencing chemicals in databases
What is considered the "best" way of referencing chemicals in a database ? As far as I understand CAS nr. is a specific type of hash whereas something like smiles stores structural information too. Is there any possible overlap between identifiers (I know there is for SMILES) ? If so which one has the least overlap ? Im not quite sure I understand exactly the purpose of each of these

TL,DR: If your database is about organic molecules with less than 1024 atoms per entry, each described in the .sdf file format, and your structure-related identifier is not used to recover the underlying structure, I recommend non-standard InChI keys assigned with the optional fixed H layer.  This is a 27 character long hash  with high collision resistance.
The CAS number is a book keeping register number of a chemical compound; neither the first, nor the only attempt to index chemicals in a database - in the particular case, to offer an eventually machine readable tag for Chemical Abstracts and subsequent products.  Outside the ecosystem of the American Chemical Society, it equally is used by other databases and catalogues of chemical suppliers, in many of the about 20k chemicals indexed in Wikipedia' property boxes.  Its use is advantageous e.g. because different to the systematic name, or generic names, CAS number(s) assigned to a chemical do not change in natural languages and scripts (English, Japanese, Arabic, etc.).  Though the formatting is known (see e.g., here), and a small public sub set of about 500k entries (see here), eventually, the assignment of CAS numbers is up to the ACS.
There is no way to predict the  CAS number for a new structure, nor to reconstruct a chemical structure with the CAS registry number in hand.  Because the present form of your question does not specify further what chemicals your database covers: different crystal structures of the same compound get different CAS RN, e.g., 1317-80-2 about rutile, and 1317-70-0 about anatase - yet both about titanium dioxide, $\ce{TiO2}$.  This variation, described as polymorphism, equally is observed for organic materials.

SMILES is a notation to retain some chemical information (see for example Wikipedia's article, and Daylight's tutorial). Accepted for file input and output e.g., by openbabel, or RDKit it is possible to simplify a 3D structure of a molecule into such a string, and recover a reasonable 3D structure from this string. (The input 3D structure need not to match the reconstructed 3D structure in the absolute atomic positions, but typically does for the relative spatial arrangement of the atoms.)
While you have the advantage to assign a SMILES string to a structure on your computer, independent from a curator of a database, the SMILES string's complexity increases with the complexity of the structure they describe. There is no fixed string length of SMILES strings.  If the structures are "complex enough", different implementations may assign one structure different SMILES strings, too.  An example for this is pyridine; it can be described in a kekulized form with alternating single and double bonds C1=CC=NC=C1 (e.g., Marvin) and C1=CN=CC=C1 (e.g., ChemDraw), or in a form where all atoms concerned are described as equals in an aromatic system, hence as c1ccncc1 (e.g., OpenBabel). Sometimes, the kekulized form has an advantage, in other case it does not.  Further, openbabel alone allows the export of canonical, universal, and inchified SMILES - equally see Noel O'Boyle's paper about this.  Hence if you elect SMILES as a chemical reference in a database, test your tools to assign SMILES (e.g., with cipvalidationsuite) before deployment and document how it fits into your intended workflow.

A younger approach is the assignment of International Chemical Identifiers (InChI) by an algorithm curated by the InChI trust.  Similar like a SMILES string, the InChI string is a simplified  representation of the chemical structure submitted in layers (Hill formula, connectivity of skeleton atoms, charges, stereochemistry, ...) and it is possible to recover one plausible 3D structure from this string - for examples with their executables.  Further on, one can assign the InChI key a string of fixed length (27 ASCII characters), though (checksum like practically) you are no longer can retrieve the original structure submitted to the algorithm.  Hence, InChI keys can represent an computational affordable approach to reference chemicals in databases.  There are suppliers and databases which describe their chemicals both by SMILES, as well as InChI / InChI key.
Note this approach equally has some caveats:

For now, the algorithm to assign standard InChI (and InChI key)
focuses on small molecules with less than 1024 atoms (Heller et al.).
For larger molecules (current limit 32766 atoms, e.g. polymers) you have
to assign InChI not yet considered as standard (as by current release 1.06,
distributed since December 2021).

The present default algorithm to yield standard InChI (and standard
InChI key) does not discern tautomers.
A consequence is e.g., 2-hydroxypyridine (Oc1ccccn1) and its tautomeric
pyridone (O=c1cccc[nH]1) both yield standard InChI string
InChI=1S/C5H5NO/c7-5-3-1-2-4-6-5/h1-4H,(H,6,7) (note 1S!) and
standard InChI key UBQKCCHYAOITMY-UHFFFAOYSA-N.
The assignment with the optional (additional) fixed H layer however
yields for 2-hydroxypyridine
InChI=1/C5H5NO/c7-5-3-1-2-4-6-5/h1-4H,(H,6,7)/f/h7H (note the now
absent S, and added sub string following f) and InChI key
UBQKCCHYAOITMY-QDQILVOLNA-N vs.
1/C5H5NO/c7-5-3-1-2-4-6-5/h1-4H,(H,6,7)/f/h6H and
UBQKCCHYAOITMY-BRMMOCHJNA-N for the isomeric pyridone.  Consider the
second approach if your database might or indeed does contain such
compounds and document your choice for future users of the database.

Beside InChI trust's executables for Windows, or Linux, your operation system / tool of preference
might/might not reflect the current version 1.06.  The CLI executable
inchi_main provided by DebiChem
for example is stuck in version 1.03 (represents the state of the art
by 2010) because InChI trust later adopted a bespoken license,
underwent/undergoes reorganization after the death of the lead developer.
Hence, depending on the route you take, some of the more recent corrections
and improvements might not be available to you, or you compile the
executable with the source code
shared by the InChI trust.
In case of DebiChem, inchi_main already includes the optional assignment
of the fixed H layer, hence can discern tautomers. However, it processes
only .sdf of the elder V2000 format, and not .sdf V3000 (See e.g.,
Biovia's definition
of this file format.  To some degree a conversion between the two dialects
is possible, e.g. thanks to openbabel).

Literature references:
O'Boyle, N. Towards a Universal SMILES representation - A standard method to generate canonical SMILES based on the InChI. J. Cheminform. 2012, 4, 22. doi 10.1186/1758-2946-4-22, open access
Hanson, R. M; Musacchio, S.; Mayfield, J. W.; Vainio, M. J.; Yerin, A.;  Redkin, D. Algorithmic Analysis of Cahn–Ingold–Prelog Rules of Stereochemistry: Proposals for Revised Rules and a Guide for Machine Implementation. J. Chem. Inf. Model. 2018, 58, 1755-1765. doi 10.1021/acs.jcim.8b00324 and https://github.com/CIPValidationSuite/ValidationSuite
Heller, S.; McNaught, A.; Stein, S.; Tchekhovskoi, D.; Pletnev I. InChI - the worldwide chemical structure identifier standard. J. Cheminform. 2013, 5, 7. doi 10.1186/1758-2946-5-7, open access.
Pletnev, I.; Erin, A.; McNaught, A.; Blinov, K.; Tchekhovskoi, D.; Heller, S. InChIKey collision resistance: an experimental testing. J. Cheminform. 2012, 4, 39. doi 10.1186/1758-2946-4-39, open access.


text:
Determine if particles are aluminium oxide
I am getting a white dust in my car, coming from the air vents and landing all over, mostly on the dashboard. After searching about this, it seems that it could be caused by the fins of the air conditioner evaporator oxidizing (for whatever reason).
I recently bought the car (used), and I before I claim fixing/compensation from the dealer I want to have proof that it is in fact aluminium oxide, or else I expect that they will try every other alternative (=cheaper) explanation possible without getting to the root cause (seen this sooo many times).
So now I'm wondering if there is any reasonably cheap way to test and verify that the dust particles are aluminium oxide?

Check and see if your car has an interior cabin air filter (not the one in the engine).  Sometimes they are in places like the glove box and are difficult to change.  The previous owner may have neglected to do this.

XY problem: What difference would the material make to getting the problem fixed? Even if it were aluminum oxide, not likely, either the dealer resolves it -- or not. Could be they used a power-washer on an aluminum part, e.g., to remove smoke smell from previous owner, and it corroded a bit. Whether the surface releases dust from Al2O3 or remains of detergent, the dealer would still have to fix it the same way.

@DrMoishePippik In theory, yes. In the real world, things get more complicated. I've had numerous situations where I have a strong suspicion of the cause of a problem, but of course the responsible party will try various cheaper methods of "fixing" it and claim "Now it's fixed!" only to have the problem still remain. This creates unnecessary time, work, expense and irritation. I find it's often better to find proof of the root cause of a problem beforehand to avoid ending up with various "fixes" that might even cause new problems or make the original one worse...

Aluminium forms a protective oxide layer that is strongly associated with the metal. Therefore, I doubt what you are seeing is actually aluminium oxide. Air probably wouldn't be able to disperse if.
In the case that I am wrong, though, there are several ways you can go about identifying whether or not the substance is aluminium oxide.
The first thing I would try is to drop in some water. Does it dissolve? If it does, definitely not aluminium oxide.
Firstly, since you wanna use this for insurance you wanna have it absolutely foolproof. You can subject a sample to mass spectrometry analysis. The downside is that it is expensive and you need to find a lab willing to analyse your sample.
Secondly, which is much more crude, you can subject it to metals and measure whether a reaction will occur or not. This can pinpoint where it is on the reactivity series, thus identifying the compound. For example, it may react with magnesium but not manganese. And this doesn't prove there is an oxide. It just proves it is an aluminium compound.
Research the reactivity series and redox reactions.
Thirdly, and this is also expensive, you can measure the melting point.
There are definitely other methods to doing this.
But I'm not sure this is a good idea in the first place. I don't think running a chemical analysis is a good way to make an insurance claim, but there is your answer.


text:
Which pKa should I use in this Henderson/Hasselbalch rearrangement?
I have a question about the following formula:
$$Mu/Pu = \frac{1+10^{pHm - pKa}}{1+10^{pHp- pKa}}$$
This formula can be found in the following paper: Begg, E. J., & Atkinson, H. C. (1993). Modelling of the passage of drugs into milk. Pharmacology & Therapeutics, 59(3), 301–310. https://doi.org/10.1016/0163-7258(93)90072-l.
$Mu/Pu$ defines the ratio of an unbound drug between mother's milk $Mu$ and plasma $Pu$. $pHm$ and $pHp$ are the pH values for mother milk and plasma, and are $7.2$ and $7.4$ respectively.
In the project I'm using this formula for, I want to calculate the $Mu/Pu$ for various complex drugs. An example of this is the drug eplerenone, with the chemical formula $\ce{C24H30O6}$.
This drug has various pKa's, ranging from -4.2 to 16.44 (Source: DrugBank | Powering health insights with structured drug data. (March 6, 2023). DrugBank. https://drugbank.com/). I'm not very well versed (anymore) in this subject, and don't know which pKa to use.
Most likely (to me) seems the pKa which is the closest to the pH of milk/plasma, but it could also be the pKa at which the relevant drug will be neutral.
Experimental data of emtricitabine, another drug with the formula $\ce{C8H10FN3O3S}$, shows for example an pKa of 2,65, according to the same source.
Can somebody help me with this question?

Does this drug migrate between blood plasma and milk  in neutral or ionic form ? There would be some distribution constant for this migration-active form and then 2 equlibriums between active and non-active form in each environment. Depending on that, higher pH in one environment can both favour or disfavour the drug presence there. // Perhaps take 2 pKa for anion-neutral and neutral-cation, and then take which is closer to biological pH.


text:
Reversible path for Entropy change on crystallisation
I am trying to re-learn thermodynamics but have come stuck with this question so am seeking help.
Two moles of supercooled water at $265$ K freeze to give ice at the same temperature. Using $\Delta H_f = 6.008 $ kJ/mol and $C_p(ice) =36.8$ J/mol/K at $273$ K, calculate $\Delta H$ and find a reversible route to calculate $\Delta S$ for this process.
(a) I tried to calculate the enthalpy change assuming that $C_p$ is constant over the small temperature range; $\Delta H_{265}=-(\Delta H_f+\Delta T C_p)$, where $\Delta T =-8$ which for two moles gives $-11.43$ kJ/mol and is the answer given.
(b) I'm not sure how to calculate the entropy change. $\Delta S=2\Delta H_{265}/265 = -43.12$ J/K would seem to be the entropy change due to the phase change in crystallizing from the supercooled water but as this is a natural process it is presumably under irreversible conditions. The same point could be reached by calculating the entropy change in crystallizing at $273$ K and then the change from $273 \to 265$ as $\Delta S = -\Delta H_f/273 + C_P\ln(265/273)$ which for two moles gives $-41.8$ J/K which is close to the answer given ($-41.76$ J/K) but how could this be a reversible pathway? Additionally does this not  ignore the heat used in initially taking the water from $265$ to $273$, so I think that my reasoning and answer is wrong.

Heating the water to 273 K so it is no longer supercooled, crystallizing it there and cooling the resulting ice back to 265 K can surely be done reversibly. Now to the point: which substance your $C_p$ refers to?

Apologies, I missed it in writing, $C_p$ refers to ice, that of water is much larger.

Good. Then my point about the reversible path is still valid.

@Ivan Neretin, Thanks, but If the reversible path is from heating water at 265 to water at 273, then crystallisation at 273 then cooling the crystal back to 265, how do I calculate the entropy for the first part, i.e. warming water from 265 to 273. (the $C_p$ of water is not given) Why is this a reversible path as opposed to fixing a heat bath to the system at 265 and proceeding reversibly to crystallise at the same temperature?

Crystallizing at 265 K would be a jump from some elevation, and we need a smooth walk. As for Cp of water, just take the standard value.

@opa About Cp of water not given, you do know the energy ratio  1 cal/1 J.

@Poutnik well, yes,  it is 75.6 J/mol/K as we are using SI units, but Its not given in the question so I suppose that the answer must not need it, which takes me back to my initial questions.

A task description need not contain all info (not limited to numerical data) that is needed. It must contain all info that must not be missing to be able to resolve the task. OTOH, some tasks contain info that is not needed to confuse resolvers.

The path that you have carried out is perfectly fine. If we add that both processes take place at constant pressure named $P_1$, then we can also calculate the heat exchanged by $Q =n \Delta h$
\begin{align}
  \ce{H2O}(P_1, T_1, \pu{liquid}) &\rightarrow \ce{H2O}(P_1, T_2, \pu{liquid})
  \hspace{1 cm} (\pu{Step 1}) \\
  Q_1 = nc_\pu{p} (T_2 - T_1) = 588.8 \; \pu{J} &\hspace{0.5 cm}
  \Delta S_1 = n c_\pu{p} \ln\bigg(\frac{T_2}{T_1}\bigg) = 2.1890 \; \pu{J/K} \\
  \ce{H2O}(P_1, T_2, \pu{liquid}) &\rightarrow \ce{H2O}(P_1, T_2, \pu{solid})
  \hspace{1 cm} \pu{(Step 2)} \\
  Q_2 = n(-\Delta_\pu{fus} H) = -12016 \; \pu{J} &\hspace{0.5 cm}
  \Delta S_1 = n\bigg(\dfrac{-\Delta_\pu{fus} H}{T_2}\bigg) = -44.0147 \; \pu{J/K} \\
\end{align}
and indeed you get $\Delta S_\pu{sys} = \Delta S_1 + \Delta S_2 = -41.2681 \; \pu{J/K}$.
Since we are treating with changes in state functions, $\Delta H$ and $\Delta S$ will be the same for the process carried in a reversible manner, as well as irreversible. Now, one way of stating a reversible path, is for example:

Mechanically reversible process of heating of water from $T_1$ to $T_2$ at constant pressure.
Thermally reversible process of crystallization of water from the liquid state to the solid state.

To be honest, this doesn't say much, so we are going to do some additional calculations to understand what is happening. Indeed, the process is irreversible and it has to do with how the heat is exchanged between the surroundings
Step 2 The second step is fine since the heat given to the surroundings is $Q_\pu{2,surr} = -Q_2 = 12016 \; \pu{J} $, and if we imagine it as a heat reservoir at $T_2$
\begin{equation}
  \Delta S_\pu{2,sys} + \Delta S_\pu{2,surr} = -44.0147 \; \pu{\dfrac{J}{K}} +
  \pu{\dfrac{12016 \; J}{273 \; K}} = (-44.0147 + 44.0147)
  \; \pu{\dfrac{J}{K}} \rightarrow
  \boxed{\Delta S_\pu{2,universe} = 0 \; \pu{\dfrac{J}{K}}}
\end{equation}
Step 1 Here lies the problem. The heat given by the surroundings is $Q_{2,surr} = -Q_2 = -588.8 \; \pu{J}$. For example, if this one is a heat reservoir at $T_2$, the 2nd law states
\begin{equation}
  \Delta S_\pu{1,sys} + \Delta S_\pu{1,surr} = 2.1890 \; \pu{\dfrac{J}{K}} -
  \pu{\dfrac{588.8 \; J}{273 K}} \rightarrow
  \boxed{\Delta S_\pu{1,universe} \; 0.032222 \; \pu{\dfrac{J}{K}} >0}
\end{equation}
We can do better if we split the heating in multiple steps, so that the heat reservoir can gradually accompany the system. In this case
\begin{equation}
  \Delta S_\pu{1,universe} = \Delta S_\pu{1,sys} + \Delta S_\pu{1,surr} =
   2.1890 \; \pu{\dfrac{J}{K}} - \sum_{k=1}^N \dfrac{Q_\pu{k,surr}}{T_\pu{k,surr}}
\end{equation}
E.g., two heat reservoirs that give $(588.8/2) \; \pu{J} = \pu{294.4 J}$ to the system, the first one at $\pu{265 K}$ and the second one at $\pu{273 K}$. The results are the following:

So now after all of this we can say something better: a reversible path for this process, is that of a heat exchange with the surroundings in the heating step, so that the surroundings is only infinitesimally at a higher temperature than the system. When $N\rightarrow \infty$ a reversible process may be approached.

Part (b) The supercooled liquid is not in equilibrium with the solid and therefore $\Delta G$ is not zero, so you cannot calculate the entropy for the phase change from $\Delta H(265)/265$. You can't make the supercooled liquid from the solid phase without melting it and allowing it to supercool again. (You lose the heat released on crystallisation to the lattice vibrations of the solid and then to the surroundings in the case of constant T.)
The only path remaining is then to start with the heat of fusion and correct for the change in temperature using the heat capacity of ice, as you show in your question. ( I presume the very slight difference in values is due to a typo some where in the calculation)


text:
How to write a rate expression when one reactant is not soluble?
For a chemistry experiment, I would like to determine the rate expression for the reaction between zinc and hydrochloric acid. Given that zinc is not soluble in the water, would I use a "false" concentration of sorts and divide the moles of zinc added by the volume of the hydrochloric acid, or is one simply not able to write a rate expression for this reaction?

Consider if the rate is meaningful: you can determine the degree of the equation, but the actual rate would be dependent on external factors, e.g., Is the zinc a single sphere, a nanometer thick sheet, etc.? What removes hydrogen bubbles that quickly remove Zn from contact with HCl? Gravity? Stirring?

Trying to evaluate reaction rates where three different phases are involved (solid, liquid, gas) is like opening the Pandora's box.

(1) About the "false" concentration Hello. In general, a rate law may (among other magnitudes) include some measure of the concentration of all the participants, in this case
$$ \ce{Zn(s)} + \ce{2 HCl(aq)} \rightarrow \ce{ZnCl_2 (aq)} + \ce{H_2 (g)} \tag{1} $$
As you said correctly, zinc is not soluble on this medium. However, there is no such thing of "concentration of Zinc" or "concentration of a solid", because this is a pure solid, and there is no sense of talking about a measure of concentration of a pure solid phase. Mathematically this is an advantage for you, because the activity of a pure solid phase, is by definition 1, i.e. $ a_{\ce{Zn}} = 1 $ always.
(2) About the possibility to write down a rate law You can, sometimes easily sometimes not, track the concentration of a certain species in this solution. Here may come your knowledge in analytic chemistry for ions in solution for example.
If you can guarantee that only Eq. (1) takes place, this will be enough, and by stoichiometry you can find the moles of all the other species consumed/generated within time. After that, you can perform a mole balance and try to relate the rate of change of any concentration to the rate of reaction. However, finding a general equation that relates the set of experimental values to a rate law $ r $ that depends on to the concentration of $ \ce{H^+} $, of $ \ce{Cl^-}$, the partial pressure of $ \ce{H_2} $, the temperature $ T $, etc., is a matter of art. Good luck in your experiment!

In principle, the rate law should include anything that changes the rate, or specify that parameters that were kept constant (temperature, for example). So you could say you investigated the rate law a room temperature and ambient pressure.

[OP] Given that zinc is not soluble in the water, would I use a "false" concentration of sorts and divide the moles of zinc added by the volume of the hydrochloric acid, or is one simply not able to write a rate expression for this reaction?

You would have to do some experiments to see what about the zinc affects the rate. Surface area would be a first guess. It is unlikely that the quantity you suggest would be directly linked to the speed of reaction. I don't think the reaction slows down when you add more hydrochloric acid to a block of zinc (but your "false" concentration would be lower).

[from the comments] What removes hydrogen bubbles that quickly remove Zn from contact with HCl? Gravity? Stirring?

If you stir, you would have to do an experiment whether the type of stirring and the speed of stirring affects the rate (or keep it constant and specify it when you state the rate law and constants, see above).

For a chemistry experiment, I would like to determine the rate expression for the reaction between zinc and hydrochloric acid.

The first rate laws you learn about are often zero, first or second order in a reactant. When it comes to heterogeneous reactions, you can expect more complicated behavior (which might look like zero or first order in same part of the multi-dimensional parameter space). A classic example is enzyme kinetics, where the reaction looks first order in substrate at lower concentration and zero order at higher concentration, with a rate law that is not a simple power law.


text:
If the 2 electrons of a hydrogen molecule are more likely to be found between the 2 hydrogen atoms, why does this diagram says otherwise?

So I am reading about bonding and  antibonding molecular orbitals and i can't understand how does the probability diagram show that the electron density in the sigma 1s molecular orbital is greatest between the two positively charged nuclei.
As i read this diagram [(b) in the picture],the probability of finding the 2 electrons in a hydrogen molecule is higher around each hydrogen atom nucleus. But molecular theory says that the electron density is the highest between the 2 nuclei.
The diagram shows that the highest ψ^2 values are on the hydrogens nuclei and not between them.
So what am i getting wrong?

This is a consequence of plotting in 2D and focusing on the wave function to emphasize the additive nature at the expense of clarity. It might help to compare the atomic 1s wave function to its probability as a function of radius, which is the wavefunction multiplied by a spherical volume element, so that even though $\psi^2$ is maximum at $r=0$, the maximum probability is actually at a distance r>0 as shown here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html.

Perhaps what is meant is not that the maximum of the probability lies there but that the integrated probability over that region is a maximum.

What should be plotted, perhaps, is the difference between the probability density of two isolated hydrogen atoms and two bonded hydrogen atoms. That would make the situation clearer.

The shape of the orange blob is highly affected by the chosen contour level. In the image below, if you choose a contour level of 0.04, you get a oval much like the one shown. If you choose a contour level of 0.2, you get a dumbbell shape. If you choose a contour level of 0.4, you get two spheres around the atoms, which does not look connected at all.

It turns out that the electron density of two bonded atoms is very similar to the electron density of the two individual atoms, especially for atoms with inner electrons. The difference, called the deformation density, gives you an idea how the electron density "shifted" with respect to that of the unbound atoms. Unfortunately, I did not find a deformation density for dihydrogen in a quick internet search. Maybe someone can calculate it, though.


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Does Propylene Glycol cause motor oil to coagulate?
It is well known that ethylene glycol when mixed with motor oil will cause catastrophic engine failure.  The mix coagulates to a brown-mud that fails to lubricate.  But, will a Propylene Glycol / motor oil mix also cause a failure?
The few articles I have been able to find refer to glycol/oil mix.  The ethylene seems to be implied. It appears the glycol reacts with the additives in the motor oil.  At the micron level, this mixture is very abrasive.  I am not finding any info on how Propylene Glycol reacts with engine oil.
I heated a Propylene Glycol/motor oil mixture to boiling.  It never coagulated.  When it cooled, the mixture separated.  I repeated the same with ethylene glycol/oil mix.  That mixture failed to coagulate as well.  Again, when it cooled, the mixture separated.
Oil pumps generally run in the 20-30psi range.  I doubt pressure is needed to cause the coagulation as the oil pan is typically full of mud when an engine fails.
I'm wondering why I can't get the mixtures to coagulate.  Does anyone know if Propylene Glycol/motor oil coagulate?

The oil/coolant mixtures sat for a week.  Both versions had separated.  When agitated, the mixtures would quickly separate.  The ethylene version showed signs of mud at the interface layer.  The propylene version was very dark in color.  Very close the same shade as the oil, but with no sign of mud.
When the ethylene version was heated (not to boiling) and agitated, it quickly emulsified into thick mud.  When it cooled the emulsion held.
When the propylene version was heated and agitated it showed signs of emulsification but when the agitation was removed it would quickly separate.  When cooled the separation was more pronounced.  It did not coagulate the oil.

Both the EG and PG are a 50/50 mix with water.
Update: A week later:
The ethylene glycol (EG) is still fully emulsified.  It is showing signs of separation but is still fully incorporated.
The propylene glycol mixture has coagulated to mud.  There is a layer of oil sitting on thick layer of mud with no sign of any unincorporated PG.
Both ethylene glycol and propylene glycol do coagulate when mixed with motor oil.

Motor oils are carefully designed mixtures that lubricate engines under operating conditions. Adding other ingredients is unlikely to improve their performance under those hot conditions in an operating engine. But the problem is unlikely to be illuminated by simply mixing the extra ingredients: you need to see how the new mix would function in an operating engine with all the friction and heat. Your experiment does not provide much insight into what happens in a real engine.

Have you tried intense mixing? Some non miscible liquids form quite stable  gel-like mixtures, especially with air, if mixed really well and impurities are present.

@Poutnik I had thoughts the activity in the oil pump may be causing an  emulsification. As far as I can tell, once in the mud form, it no longer seperates.  In the non-mud form, coolant and oil will seperate.  I did vigorously agitate the boiling mixture, and got no change.

M experience is the violent agitation of a running engine causes the thickening. At that time the engine is also pretty much ruined.


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What does it mean when a molecule is described as "optically stable"?
In this question about "Why don't trigonal S and P compounds undergo inversion at room temperature?", phosphines, sulfoniums and sulfoxides are described as "optically stable", while ammonia is not. From this question and elsewhere I infer that "optical stability" in this context refers to the racemization rate of the molecules being very slow.  Is there a precise definition of "optical stability"?

related https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/ejoc.202100816

Optical stability is a qualitative property rather than a quantitative one. If you want to describe the rate of racemization quantitatively, you can measure the Gibbs free energy of activation for the transition from one enantiomer to the other enantiomer. The higher the activation energy, the more optically stable the compound is.