text:
Is gas solubility independent of pressure?
My physical chemistry textbook poses this question:

Prove the statement that an alternative way to express Henry’s law of gas solubility is to say that the volume of gas that dissolves in a fixed volume of solution is independent of pressure at a given temperature.

My proof is thus:
$P_2 = Kx_2$, where $P_2$ is the partial pressure of a solute, $K$ is the Henry's law constant, and $x_2$ is the mole fraction of solute dissolved in solution. By the ideal gas law,
$\frac{n_2RT}{V} = Kx_2 \rightarrow V = \frac{n_2RT}{Kx_2}$.
Therefore, V depends on temperature, not  pressure. Regardless of if my proof's correct, I'm missing part of the picture here. Intuitively, I want believe increasing pressure leads to an increase in dissolved gas. If you have a beaker of water and some gas in a metal box, and then you decrease the volume of that container, like a piston, at constant temperature, wouldn't more gas dissolve?

read Dr Pippiks answer

You state, it seems that, logically, "increasing pressure leads to an increase in dissolved gas." Your logic is not wrong... an increased mass of gas dissolves.
The confusion is because here, Henry's law is stated in terms of the volume of gas. Since increasing pressure makes the gas occupy less volume, it has higher density. As a hypothetical example:

At STP, i.e., 1 atm, Gas X has a density  ${1 g/l}$. It dissolves in water at the ratio ${1 l_{Gas X}/l_{H2O}}$, so 1 gram of Gas X is absorbed in a liter of water.
At 10 atm, Gas X has a density  ${10 g/l}$. Still, just one one liter dissolves in a liter of water. But that one liter is 10 grams of Gas X at that higher pressure, so ten times the mass of Gas X dissolves in a liter of water.

Of course, this assumes a perfect gas, and that there is no chemical reaction with the liquid. For example, $\ce{CO2}$ in water may form clathrates, throwing off a linear relation. The reference cited provides corrections to Henry's law to account for empirical results.

Let n be the number of moles of solute dissolved in $n_{solvent}$ moles of solvent (in dilute solution).  Then, $$x=\frac{n}{n_{solvent}}$$So, $$P=\frac{n}{n_{solvent}}K$$or$$K=n_{solvent}\frac{P}{n}=n_{solvent}\frac{V}{RT}$$So, the volume of ideal gas at temperature T that dissolves is $$V=\frac{KRT}{n_{solvent}}$$

Yes, from the OP, I believe that is understood. However, the question from that intuitive approach implies concern about the increased mass, rather than constant volume.


text:
Speed of unmixing between polar and nonpolar solvent
I have put water and silicone oil (viscosity 100 cSt) together into a jar. Then I colored the water blue to make identifying the interfaces easier. When I shake it heavily, a kind of silicone+water "foam" builds up, which only unmixes very slowly. Even when I only shake it slightly (as for the image below), so that a few major bubbles of water enter the silicone, the bubbles won't pop back into the bulk of water for ages.

I think I remember, from doing chemistry in the past (I am not a chemist), a mixture of chloroform and water unmixes very fast, while a mixture of water and kerosene did not (i.e. performed like my recent silicone oil experiment). I first thought a lack of difference in density (because the silicone seems pretty close to 1 g/cm^3) hinders unmixing, but while chloroform has indeed much higher density than water, kerosene also has significantly lower density than water. So I think it must be something else than density. But it can't be surface tension either, can it. Because nonpolar solvents always have lower surface tension than water (lack of hydrogen bonding), don't they.
So what makes one pair of polar/nonpolar solvent unmix fast while the other builds up foam? I think that must be very important for a chemist during washing after synthesis, but I can't seem to find a suitable search term.

https://www.sciencedirect.com/science/article/abs/pii/S0920410521016521

https://en.wikipedia.org/wiki/Emulsion

@Mithoron: unfortunately I can't read the science direct article due to the paywall. But the contained keyword "demulsification" led me to this article which I found helpful: https://petrowiki.spe.org/Oil_demulsification

So my preliminary explanation for my problem child is threefold: 1) small density difference between the two solvents keeps the droplets from coming close to each other due to gravity, 2) once the (water) droplets have come close enough, further approaching is inhibited by still comparatively high viscosity (of the silicone oil) 3) internal surfactant effects of the silicone keeps the droplets from coalescing.

In an inspiration that the surfactant effects of silicone oil might come from the "etherlike" siloxane groups which pair with the hydrogen atoms of water to form hydrogen bonds, I put acetone into the mixture, in the hopes that its molecules would straddle inbetween water and siloxane groups and so weaken the surfactant action. And indeed, the interliquid "foam" resolved much faster then (estimated factor 100). Only disadvantage is, the water fraction has become a little blurry now (where it was rather clear before), possibly because some of the silicone got emulsified into it now.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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text:
Why wont Al(OH)3 form in this reaction?
I am in grade 11. We were learning qualitative salt analysis and my teacher wrote a reaction on board - $$\ce{Al^3+ +H2O\rightarrow[Al(H2O)6]^3+}.$$
Now as far as structure of the complex is concerned i get it that $\ce{H2O}$ is a ligand and $\ce{H2O}$ donates lone pair to form that complex and $\ce{H2O}$ acts as Lewis base. My question is why wont the following reaction be possible -
$$\ce{Al^3+ + H^+ OH^- \rightarrow Al(OH)3 + H+_{(aq)}}$$
I have a limited knowledge of coordination chemistry. So am I missing something?

The equations written by Aleph are all wrong, mainly because they are not stoichiometrically correct. They should have been written  $$\ce{Al^{3+} + 6 H2O -> [Al(H2O)6]^{3+}}$$ $$\ce{Al^{3+} + 3 H2O -> Al(OH)3 + 3 H^+}$$ Anyway, the first reaction produces a complex hydrated cation. And the second equation ignores the complexation of the aluminium ion. In reality, as Poutnik mentioned,  the ion $\ce{Al(H2O)6^{3+}}$ produced in the first equation is successively deprotonated according to the series of three equations $$\ce{[Al(H2O)6]^{3+} + H2O -> [Al(H2O)5(OH)|^{2+} + H3O+}$$ $$\ce{[Al(H2O)5(OH)|^{2+} + H2O -> [Al(H2O)4(OH)2|^{+} + H3O+}$$ $$\ce{[Al(H2O)4(OH)2|^{2+} + H2O -> [Al(H2O)3(OH)3| + H3O+}$$ The last product is sometimes simplified as $\ce{Al(OH)3}$

It is rather
$$\ce{[Al(H2O)_n(OH)_{6-n}]^{(n-3)+}(aq) + H2O(l) <=> \\
<=> [Al(H2O)_{n-1}(OH)_{7-n}]^{(n-4)+}(aq) + H3O+(aq)}.$$ Coordinated water molecules become more acidic than free ones.

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Why aren't lanthanum, praseodymium, osmium, and mercury not on the same energy level?
The following question is from "The Official Guide to the HiSET Exam, Second Edition 2nd Edition", published by McGraw Hill, 2nd edition (March 25, 2016).2016 text, p. 75 (ISBN10: 1259640795).
Along with the question, students are given a periodic table (going up to element 109, Mt / meitnerium) that only contains element symbols (without atomic numbers or any other numbers).

Which group of elements has valence electrons in the same energy
level?
A.  B, Si, As, Te
B. Ga, As, Se, Kr
C. H, Li, Na, K
D. La, Pr, Os, Hg

The correct, given answer is B. I get this as all elements are on energy level 4.
But why is choice D not correct?  To my understanding, all of these elements are on energy level 6 (and all seem to have valence electrons on 6th level).
Is the book incorrect or am I missing something?
I'm good with all the other chemistry questions and can explain them to the students.  I'm stuck on this one and text fails to explain the answer.  Again I think I understand why B works, but don't understand fully why D can't.  Both answers feature elements each a part of the same respective horizontal row or energy level / period.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Lanthanides are "f-block" elements, filling the 4f shell in series.
Mercury and Osmium have filled their f shells with 14 electrons.
Valence electrons are not always the outermost.

I had edited the answer to include the electronic configurations of the elements and explain what was the valence shell actually by Hund's rule. It was rejected though and I have submitted yet again. For those in the privileged team, please do know that this is quite relevant only and not deviating from the original interest of DrMoishe Pippik. I don't seem to find a reason why my edit should be irrelevant.

@HarikrishnanM, sorry, I'd neither approved nor rejected it. If you'd like, make it an additional answer, shedding more light on this!

@HarikrishnanM please write a new answer. Your edit guess further than the original.

@Martin-マーチン even if I write my new answer most of the points will match DrMoishe Pippik's answer except for the electronic configs added in. I don't seem to have a reason for my answer to be unique and it's better off I edited Dr Moishe Pippik's answer as 2 answers with only EC added is gonna be redundant. And given ChemSE's strict intolerance to that, I don't wish to risk getting bad remarks from everyone.

@HarikrishnanM, No risk - if your answer is unappreciated, delete it. But get credit yourself for making this more clear!

$\ce{Hg}$ : $\ce{[Xe]}$ $\ce{(4f)^14 (5d)^10 (6s)^2}$
$\ce{Os}$ : $\ce{[Xe]}$ $\ce{(4f)^14 (5d)^6 (6s)^2}$
$\ce{Pm}$ : $\ce{[Xe]}$ $\ce{(4f)^3 (6s)^2}$
$\ce{La}$ : $\ce{[Xe]}$ $\ce{(5d)^1 (6s)^2}$

Over here, we see that in Hg, Os and La 5d shell is the last filled shell and in Pm 4f is filled last by Hund's rule.
Thus Pm is in a lower energy level than the others.
Moral of the story is that valence electrons are not always the outermost.
Sources:

Clark's Tables: Science Data Book. India: Pearson Education, 2006.

This is very helpful.  When looking at a more detailed periodic table that includes electron configurations I noticed that the final shell for the 4 elements never changed (always 6s^2) compared to a contrasting row like row 4 where we have Ga, As, Se, Kr  (Gallium, Arsenic, Selenium, Krypton) where the final shell does change in number (4p^1, 4p^3, 4p^4, 4p^5).  Intuitively I had wondered if that might be a clue--but couldn't conceptualize it--as I was assuming valence electrons were always of the outmost shell.  Now I know otherwise.  MUCH appreciated.

@Sail2DeepBlue - do remember, always follow Hund's rule. Whatever comes last in Hund's rule is the valence orbital.

For the person who downvoted - answer follows IUPAC conventions now.


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Does acetone react with acidic precipitate product in this reaction?
In this reaction procedure, It seems like there is a mistake in step 3 which would result in a 0% yield. From step 3, it looks like everything is fine, its performing the acid catalysed hydrolysis of Nitriles to form a carboxylic acid, and the reaction is maintained in its acidic conditions meaning the carboxylic group would be neutral and will precipitate in the aqueous condition. Could this mistake be when we wash the acid precipitate with acetone?
Does acetone react with mesitylacetic acid?
Step 1: Chloromethylmesitylene 3. Mesitylene 2 (200 g, 1.66 moles), 1 L of
concentrated hydrochloric acid (1 L), and formaldehyde solution (63 mL, 0.84
mole) were placed in a round bottomed flask. Hydrogen chloride was introduced
below the surface of the mixture through a gas inlet tube. The mixture was stirred
vigorously and heated to keep the temperature of the mixture at 55-60 °C for 5.5
hours. At the halfway point, additional formaldehyde solution (63 mL, 0.84 mole)
was added. After the mixture had been cooled to room temperature, it was
extracted with ether (3 x 300 mL). The combined ether extracts were washed
successively with water, dilute sodium hydroxide, and brine, dried over sodium
sulfate, and filtered. The ether was removed and the residue was distilled under
reduced pressure (130–131°/22 mm) to give chloromethylmesitylene 3 (155–170
g); the distillate solidified to a crystalline mass (mp 37-38 °C).
Step 2: Mesitylacetonitrile 4. A solution of sodium cyanide, (66.2 g) in water (110
mL) and ethanol (160 mL) was heated at reflux until all the sodium cyanide had
dissolved. Chloromethylmesitylene 3 (152.5 g, 0.90 mole) was then added
slowly, and the heating was continued for 3 hours. The mixture was allowed to
cool to room temperature and extracted with ether (3 x 300 mL). The organic
extracts were washed with water, dried over sodium sulfate and filtered. After
removal of the ether, the residue was distilled under reduced pressure (160–
165°/22 mmHg) to give mesitylacetonitrile 4 (128–133 g) which solidified upon
standing.
Step 3: Mesitylacetic acid 1.
Concentrated sulfuric acid (750 mL) was cautiously added to water (900 mL).
When the mixture had cooled to about 50 °C, mesitylacetonitrile 4 (127 g, 0.80
mole) was added. The mixture was heated at reflux. A precipitate formed during
heating. The contents of the flask were cooled and poured into ice water (3 L).
The precipitate was collected on a Büchner funnel, washed well with acetone and dried in a vacuum desiccator to give mesitylacetic acid 1 (123 g), mp 163–166
°C.

I think it unlikely there will be a reaction with the acetone if the product 1 is quenched into ice water first. If you are concerned about the possibility chill your acetone wash solution before use.


text:
Electrical properties of glycol / water mixtures
I'm looking for a reference on electrical properties (specifically, conductivity) of mixtures of water / ethylene glycol. Specifically, I'm interested in

How does the electrical conductivity of the mixture depend on the volume fraction of the glycol? Can I use standard formulas (which I also struggle to find) for liquid mixtures? Are there such formulas?
If there are such formulas, is there a standard reference for the electrical properties of ethylene glycol?
How does the conductivity of the mixture depend on temperature? Does the rule of thumb 2%/°C that is used for water stand also for water/glycol mixtures?

Try the CRC handbook of physics and chemistry.

https://iopscience.iop.org/article/10.1088/1742-6596/142/1/012059/pdf
Would highly suggest to check this paper.

Disclaimer: I am no conductivity modelling expert so take what follows with a grain of salt. Also do not hesitate to comment this answer, and I will try to improve it accordingly.

Your question consists in six(!) questions which I will try to answer below.
1. How does the electrical conductivity of the mixture depend on the volume fraction of the glycol?
You may have a look at the figure below, from Loskutov et al.[1], for water / ethylene mixtures at 25°C.

2. & 3. Can I use standard formulas (which I also struggle to find) for liquid mixtures? Are there such formulas?
I must admit my lack of knowledge on this very topic, but the authors in [1] propose a formula for conductivity introducing a physical quantity they called excess electrical parameter, however noting that "excess electrical conductivity is not a universal parameter, so it is introduced for general consideration of different characteristics in a
single scale".
You may also want to have a look at references [2], into which the author propose two different curve fittings (exponential and quadratic models). However, I fear that these equations are purely experimental. Furthermore, in your case (binary mixture without additional salt), this would lead to R=0 in their formula, for which their fit is not that great (probably due to the actual shape of the conductance shown in the figure above).
Additionally, although I did not find many paper on water / glycol binary mixtures, I did find quite a lot of publications on ternary mixtures, adding NaCl, CaCl$_2$, carbon nanotubes, and whatnot (see [4] for instance). In the latter cases, determinations of conductivities are also very much empiric.
4. If there are such formulas, is there a standard reference for the electrical properties of ethylene glycol?
I honestly do not know what you mean by standard reference. Do you mean a standard setup and / or procedure for measuring it? Like a set of temperature / pressure / purity / cell dimension / apparatus / frequency / voltage? Do you rather refer to the measured conductivity under such conditions / with such a procedure? Both?
5. & 6. How does the conductivity of the mixture depend on temperature? Does the rule of thumb 2%/°C that is used for water stand also for water/glycol mixtures?
Regarding the influence of temperature on said conductivity, I found nought but [3], which studies the influence of temperature on the RF conductivity (1 MHz) and dielectric constant of glycol / water mixtures.
Do you have any reference for this "rule of thumb" that you mention?

References
[1] Molecular Structure of an Ethylene Glycol–Water Solution at 298 K, Loskutov et al. (2018), DOI: 10.1134/S003602441902016X
[2] Electrical Conductivity Measurements for the Ternary Systems of Glycerol/Sodium Chloride/Water and Ethylene Glycol/Sodium Chloride/Water and Their Applications in Cryopreservation, Chen et al. (2009), DOI: 10.1089/bio.2009.0001
[3] RF Conductivity and Dielectric Relaxation Studies on an Ethylene Glycol-Water Mixture, Saha et al. (1999), DOI: 10.1088/0022-3727/32/7/011
[4] Electrical Conductivity Measurements for the Ternary Systems of Glycerol/Sodium Chloride/Water and Ethylene Glycol/Sodium Chloride/Water and Their Applications in Cryopreservation, Chen et al. (2009), DOI: 10.1089/bio.2009.0001

Thanks a lot for your answer.
Across sources for e.g. monoethlyene glycol I found values in disagreement with each other. Typically, they would rate the conductivity to approx $1 \mu S / cm$ but that's not always true.

For the "rule of thumb": https://en.wikipedia.org/wiki/Conductivity_(electrolytic)#Conductivity_versus_temperature

In the reference [1], the conductivity of water is quite high (1.5 $\mu S/cm$). Could I expect a lower curve in case the conductivity of water is lower?

@G.Gare, the wikipedia page that you mentioned also states that "This value is broadly applicable for most salts at room temperature.", so I guess it may be quite bold of an asumption to believe that it holds for non-salt substances  such as glycol. For your remark on water conductivity: the latter depends on water purity (purer water will have a lower conductivity), so I would guess your expectation is right.

thanks for your comments. What I mean with "lower curve" is: shall I expect that the "bell effect" or the excess conductivity is also, proportionally, loweR?

@G.Gare, I am sorry, but I have absolutely no idea... Intuitively, I would tend to say that both phenomena are uncorrelated: the bell effect is likely caused by water / glycol interactions, while the "baseline" conductivity of not-so-pure water is ionic in essence. So I would think that using lower conductivity water should only translate the curve down but not scale it down (i.e. flatten it). However this is pure speculation on my side, and I would advise checking this experimentally if possible.


text:
Fischer Esterification and Dean Stark
Why is it Impossible do use a Dean Stark Montage if our reactives are more volatile than the mixture (H20 + Toluene for example)

Because the reactants distill over into the toluene phase of the collection arm, where they cool down and there is no catalytic acid so the reactions stops. The result is a concentrated toluene solution of the reactants in the collection arm.


text:
HOMO and LUMO energy levels diagrams
Good time of day. I want to draw the same beautiful pictures in my work as I will attach below taking from some articles. Are there any programms or scripts which can do this? I would be glad if someone shared their scripts.

Thank you for your time!


text:
Synthesizing "Methyl 2-(2-oxopyrrolidin-1-yl)-2-phenylacetate"
I am a chemistry student, here I mapped out the steps one might take to synthesize Methyl2-(2-oxopyrrolidin-1-yl)-2-phenylacetate, a piracetam dirivitive. Please add any steps I might be missing and correct any mistakes. Thank you so much To synthesize Methyl 2-(2-oxopyrrolidin-1-yl)-2-phenylacetate (C16H15NO3) from Pyrrolidine (C4H9N) and phenylacetic acid (C8H8O2), here is the detailed step-by-step synthesis process:

Protection of Pyrrolidine:
    a. Dissolve Pyrrolidine (C4H9N) in an appropriate solvent, such as dry ether or dichloromethane (DCM).
    b. Add a protecting group to the Pyrrolidine's nitrogen atom, like the carbobenzyloxy (Cbz) group. Typically, benzyl chloroformate is used as the Cbz protecting agent.
c. Cool the mixture to 0°C and add triethylamine (TEA) dropwise to react with the benzyl chloroformate.
    d. Allow the reaction to stir at room temperature for a few hours.
    e. Monitor the reaction using thin-layer chromatography (TLC) until completion.
    f. Work up the reaction by washing it with an aqueous solution, such as saturated sodium bicarbonate, followed by water, and then dry it over a drying agent like MgSO4.
    g. Filter and concentrate the mixture under reduced pressure to obtain Cbz-protected Pyrrolidine.
Activation of Phenylacetic acid:
    a. Dissolve phenylacetic acid (C8H8O2) in an appropriate solvent, like DCM.
    b. Add an activating agent such as oxalyl chloride or thionyl chloride along with a base such as TEA or pyridine in order to form Phenylacetyl chloride.
    c. Stir the mixture at room temperature until completion, which can be monitored by TLC.
Coupling Reaction:
    a. To the solution of Cbz-protected Pyrrolidine from the first step, slowly add the Phenylacetyl chloride solution from the second step.
    b. Add an appropriate base, like TEA or DIPEA (N,N-diisopropylethylamine), to the mixture.
    c. Stir the reaction mixture at room temperature for a few hours and monitor completion using TLC.
Deprotection:
    a. Add a deprotecting agent such as Palladium on carbon (Pd/C) to remove the Cbz group.
    b. Hydrogenate the mixture under a hydrogen atmosphere while stirring for a few hours.
Oxidation:
    a. Filter and concentrate the crude product obtained in Step 4.
    b. Dissolve the crude product in an appropriate solvent, like DCM or ethyl acetate.
    c. To this solution, add an appropriate oxidizing agent like Dess-Martin periodinane (DMP) or Jones reagent (Chromium trioxide in acetic acid).
    d. Stir this reaction at room temperature for a few hours until completion (as monitored by TLC).
Purification:
    a. Quench the oxidation reaction with an aqueous solution such as saturated sodium bicarbonate or diluted brine solution.
    b. Separate layers and dry the organic layer with MgSO4.
    c. Filter and evaporate under reduced pressure to obtain the crude product.
    d. Purify the crude product using either column chromatography or crystallization.

I am confused by the reaction you are doing with the phenylacetyl chloride. You add CBZ- protected pyrrolidine and then hydrogenate. 1- why bother with protecting then deprotecting the pyrrolidine?  2 - this will give the pyrrolidinyl amide not substitution at the 2 position of the phenylacetate

You copy-pasted a recipe (source is an AI?) and are asking us to double check it. Not really how the site works.

No it's not AI. I read a couple different studies on the synthesis of similar molecules including Piracetam,  Phenylpiracetam,  and Methylphenidate and kind of mixed them up in a way that seems logical.  If you are referring to the photograph, that is from an app called King Draw Chemical structure. I have no idea if that's considered AI or not.

I would start from the commercially available Methyl alpha-chlorophenylacetate (CAS 7476-66-6), react this with pyrrolidine and then oxidise the product with RuCl3/NaIO4 as reported Sharma & Ganesh  here
Alternatively you could react Methyl alpha-chlorophenylacetate with the Methyl 4-aminobutyrate hydrochloride (also commercially available) in the presence of tertiary base to displace chloride then heat to cyclise to the pyrrolidone.


text:
IUPAC name of the following bicyclic / spiro compound
We have the given compound

It's IUPAC naming is more sinister than things seem.
It's not a proper bicyclic compound, or a proper spiro cpmpound either.

I could interpret the spiro compound in red, though I doubt it does much, as well as the $-CHO$ functional group to be the special suffix, carbaldehyde, but not much else. It seems very odd to have a spiro compound, inside a bicyclic compound, making this compound feel very awkward.
Is this a tricyclic compound? If so, what is it's IUPAC name?

The IUPAC name is tricyclo[6.2.0.01,4]decane-5-carbaldehyde. In general, for complex structures like this the IUPAC name is automatically generated, usually using software like ChemDraw or ChemAxon.


text:
Does Adsorption Depend On Shape Of The Adsorbent?
I was reading about the Freundlich adsorption isotherm and when I read the equation I was quite confused. It states that
$$x/m=k(p)^{1/n}$$
So basically x is mass of the adsorbate adsorbed, m is mass of the adsorbate at pressure p, and k and n are constants, so basically the shape or surface area of the adsorbent do not come into play.
But would it happen like that? I mean, if we take a solid sphere and a cube of same mass and material but different surface area would both have same amount of gas adsorbed?
I have just completed jee adv chem syllabus so if possible please try to keep the answer simple without involving heavy terminologies.

k and n might depend on the species involved (they are constants for those particular species).

Please use better formatting if you post again. See the help pages


text:
"Fireblob" in KO₂ and PCl₅ reaction
The reaction between $\ce{KO2(s)}$ and $\ce{PCl5(s)}$ leads to the strange formation of a black "fireblob" which eventually turns orange.
This is demonstrated in this YouTube video (screenshots below) but no complete explanation is given.
How is the fireblob formed?

Here's my best guess at the play-by-play.
$\ce{PCl5}$ is a solid with a relatively low melting point (160 °C) and high volatility (sublimes at 167 °C). When the solid $\ce{KO2}$ gets dropped on the $\ce{PCl5}$, the solids don't really interact much (two solids in contact only need to touch on as little as three points, after all). However, there is some amount of $\ce{PCl5}$ vapor surrounding the solid, and some of that accidentally bumps onto the surface of the $\ce{KO2}$ above it. Reaction of the vapor and the surface is almost immediate, ultimately forming $\ce{KCl}$ and $\ce{P4O10}$, though the detailed process involves a number of intermediates such as oxychlorides and is likely fairly complicated. It also evolves a prodigious amount of heat.
Well, that heat quickly gets absorbed by the $\ce{PCl5}$, which evaporates more, feeding the reaction further. The temperature rises enough that the $\ce{PCl5}$ begins to melt locally and can spread over the surface of the $\ce{KO2}$. There's no going back now. The temperature rises very quickly, likely to ~600-800 °C, above the melting point of $\ce{KO2}$ at 560 °C. The $\ce{KO2}$ also melts, and any reaction products of the surface of the crystals which could be a physical barrier towards reaction are now mixed into the liquid, exposing fresh $\ce{KO2}$ for reaction. The heat and mixing eventually spread over the entire sample of $\ce{KO2}$, forming a large molten blob.
Let's pause here for a bit. $\ce{KO2}$ is bright yellow, and $\ce{PCl5}$ is an even lighter colour, so why is this blob black? It's difficult to say exactly what chemistry is going on such a short timespan and high temperature because these are rather out-of-equilibrium conditions, but it's quite likely that one way or another (either as a reaction intermediate, or from the thermal decomposition of $\ce{KO2}$, or straight up thermal excitation), some electrons are being knocked loose from the superoxide anions ($\ce{O2^-}$) and becoming free in the bulk of the molten droplet. Free electrons are quite short-lived in most chemical systems, but in certain situations they can be stabilized briefly. The classic example is solutions of free electrons in liquid ammonia, whereby adding a tiny speck of a silvery-white metal into a clear colourless liquid immediately generates a deep black-blue solution.
But the reaction isn't done, there's still $\ce{KO2}$ inside that molten blob, and it continues to react faster and faster. The temperature rises to as much as around 1000 °C, at which point black-body radiation starts to pump out visible photons, mostly in the red-orange part of the spectrum. The glow is due to this physical process, rather than directly due to a chemical change. As it heats further the reactions accelerate, but we avoid a finite-time singularity because the rate at which photons carry away the reaction energy increases as $T^4$. This creates a very strongly limiting factor for further temperature increases, so from the colour of the blob, it likely doesn't go further than roughly 2000 °C.
By this stage, the reaction is starting to peter out - almost all the $\ce{KO2}$ (the limiting reagent) has been consumed. As the last bits react, the temperature of the surface lowers to the point that the black-body photons dip back into the infrared, and the red-orange glow becomes weaker and weaker. The drop becomes black again, showing the last few free electrons. These eventually react with additional $\ce{PCl5}$, and the black colour disappears. Interestingly, the droplets generally begin glowing again, but this is not because they are significantly reheating - the interior of the droplet was still at 1000+ C, but the visible photons being radiated from the inside were being absorbed by the free electrons on the surface (it's black because it absorbs light, after all). Now that the free electrons are completely gone, the opaque medium becomes transparent, and the photons generated inside shine through. This quickly cools off the interior, and now finally, with no additional energy to be generated, everything cools down. No more reactions, just the almost-colourless remnants, still hot enough to be molten, forming transparent droplets of mostly $\ce{KCl}$ which will eventually solidify.


text:
Electronic configuration and symmetry labels of 3d$^7$ ions
In 3d$^7$ ions, e.g. Co$^{2+}$, the ground state is $^4$F, and it contains different levels due to the crystal field, such as $^4T_{1g}$, $^4T_{2g}$ and $^4A_{2g}$. Can anyone tell me what exactly these symbols mean? I need to know the exact configuration, not just "the even/odd symmetry of wavefunction". Thanks in advance!

Chris, you need to check some material on the applications of molecular symmetry in Chemistry and crystal field theory. You can find an introduction in many Physical or Inorganic Chemistry textbooks. The notation relates to the symmetry of the central ion wavefunction  and the electronic states of the same multiplicity you could build considering the valence electrons and the electronic configurations. Maybe someone can give you further directions.

These appear to be a mixture of Term Symbols and symmetry species in Oh point group. The superscript 4 could be spin multiplicity  $2S+1=4$ with 3 electrons where $S=1/2$.  The orbitals $T_{1g}$ etc have the symmetry properties given by their row in the point group.

These are labels for irreducible representation in the Oh point group:
http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=904&option=4
T and A are standard labels for triply and singly degenerate representations, the g denotes gerade (=even under inversion) and 1 and 2 are just labels. Functions in a specific irreducible representation transform in specific ways under the 48 symmetry operations in the Oh point group.

Thanks but as I said in the question, I need the exact configuration. For example, $^4$A${2g}$ for 3d7 means 4 electrons on t${2g}$ orbitals and 3 on e$g$ orbital. Same for the $^4$T${2g}$, but what exactly is the difference?


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If the substituents are distinguishable, is it possible to prioritize them using the CIP rules?
In the R/S stereochemical nomenclature,
when we try to  determine the priority of substituents around a chiral carbon, are there specific examples where the ranking cannot be definie under the CIP priority rules?
(for instance, where the preference between the 2nd and 3rd positions cannot be determined)?"

My questions:

If the substituents are distinguishable, is it possible to prioritize them using the CIP rules?；
Even if the chemical formulas are identical substituents, can they be properly ranked by the CIP rule if they are distinguishable?
If so, why.

At least, as an example of "isomeric" relationship between substituents, it seems that we can rank by RS ranking in the following figure.

Note: In the above figure, the upper left is a 2-propyl group and the upper right is a 1-propyl group.
In the above figure, the 2-propyl group on the upper left and the 1-propyl group on the upper right are both represented only as C3H7- in the composition formula, but if we arrange the atoms following the carbon (deficit) directly connected to the chiral carbon atom in order of increasing atomic number;
-Three atoms, (C,C,H) in the 2-propyl group on the left side,
In the 1-propyl group on the right, three atoms, (C,H,H).
The CIP rule seems to determine the higher rank of the 2-propyl group on the left.
In this sense, in the above example, the branching structure is different, so we can use the CIP rule.
However, for example what if the two of the substituents is just trans and one is cis relation?
Anyway, are there any examples that,  under the existing rules known as CIP, despite various efforts, we cannot rank?
References:(Written in Japanease)
https://detail.chiebukuro.yahoo.co.jp/qa/question_detail/q12289227327

Blue Various: Per your request, I offer the following examples. Consider the D-pentoses ribose 1, xylose 4 and arabinose 5. Each one has stereodescriptors in uppercase letters (R/S) at $\ce{C2, C3 and C4}$. These carbon atoms also find themselves in chiral environments. They are both stereogenic and chirotopic after Mislow and Siegel. Now imagine the reduction of these pentoses to their respective pentitols, ribitol 2, xylitol 3 and arabitol 6. In the case of ribitol, $\ce{C2 and C4}$ retain uppercase descriptors while $\ce{C3}$ has a lowercase descriptor, "s". $\ce{C3}$ remains stereogenic but it is no longer chirotopic as $\ce{H-C3-OH}$ lies in a mirror plane. Because
CIP rule 1a precedes rule 5, the priorities at $\ce{C3}$ are O>R>S>H and are, necessarily, in lower case. If the stereochemistry at $\ce{C3}$ is inverted, a new stereoisomer is formed--namely, xylitol 3--that still still has $\ce{HO-C3-H}$ lying in a mirror plane. $\ce{C3}$ now has the descriptor "r" and is also stereogenic and achirotopic.
Inversion of stereochemistry at $\ce{C3}$ in arabitol 6a forms arabitol 6b. They are identical by 180o rotation about an axis perpendicular to the page. $\ce{C3}$ in arabitol 6 is non-stereogenic and chirotopic, which leads to the absence of a stereodescriptor.

If you use "Search on Chemistry" at the top of the page and enter user:55119 CIP, you will find related topics that I have previously addressed on ChemSE.


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Systematic method to determine D/L notation of substances other than amino acids and glyceraldehyde
The fundamental rules of D/L notation, as I understand it, are as follows;

In D/L notation, a molecule is named by mapping it to glyceraldehyde.
The D/L notation is based on the stereo configuration of d-glyceraldehyde, and compounds that can be made without disrupting this stereo configuration are designated as D-forms, and their mirror-isomers are designated as L-forms.
Here, the distinction between whether the reference substance, d-glyceraldehyde, is a d-body (d is lowercase-letter) or an l-body can be determined experimentally from polarization experiments.

My questions:

But can "without disrupting this stereo configuration" in the abovementioned  be done under a uniform rule?
If there is such a uniform rule, what exactly is it?
Why is S-3 hydroxyacyl-CoA, for example, L-body?
As another example, for what reason was the L-body of thalidomide treated as an L-body?

As a "local rules" applicable only to amino acids, there seems to be a known method of distinguishing between D and L bodies in the following manners.
(In practice, perhaps only alpha amino acids can be distinguished?)
'Local rules' for amino acids;

(1)H for H
(2)OH for NH2
(3)COOH for CHO
(4)R for CH2OH

But then, why is S-3 hydroxyacyl-CoA an L-body?
Although a simple mapping of β-hydroxyacyl-CoA to glyceraldehyde would be as follows;

(1)Corresponding H to H
(2)OH to OH
(3)CH2CO-S-CoA to CHO
(4)R for CH2OH

But, the above method of mapping different atomic groups seems quite arbitrary.
Why do you map them in this way? Is there any logic that justifies such a correspondence?


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What is the **minimum** externally applied voltage for an electroplating setup?
I am trying to plate a piece of iron jewelry by using a silver anode and a silver nitrate solution. I am confused on how to calculate the voltage needed. Since silver is oxidized at the anode and then reduced at the cathode, is the cell potential zero?
Or am I supposed to calculate cell potential as (potential of silver oxidation) + (potential of iron reduction)? This doesn't make sense to me, because as far as I understand it, iron isn't reduced at the cathode. Please help me understand.

@Poutnik thank you for the reply, but I am trying to figure out the minimum cell potential needed, in terms of electrode potentials. I have revised my question for clarity.

Better to say the minimal, externally applied voltage.
There no minimum voltage unless you mean the trivial zero voltage. In initial stage, the iron spontaneously replaces silver ions, so there is no potential barrier there. In later stages, there are two silver electrodes and there is no potential barrier either.
The minimum practical voltage is any nonzero voltage when you are able to reach the desired result during the given time interval.
Primary is controlling current density, that should be not more than (ballpark value) $\pu{10 mA/cm2}$. Too high values could lead to rough surface. The voltage for it is the best to be determined experimentally.

Look also for recipes for silver electroplating solutions, as it may be the key to surface quality.
A patent for cyanide-free electroplating solution recipe is:

The formula of non-cyanide silver electroplating solution consists of: Silver Nitrate 40 g/L, Sulfothiorine 230 g/L, sodium sulphite anhydrous 99.3 80 g/L, ammonium acetate 30 g/L, Potassium ethanoate 30 g/L, boric acid 20 g/L, nicotinic acid 1 g/L. The pH value of plating solution is 5-6.

Note that Sulfothiorine is the old good sodium thiosulfate pentahydrate $\ce{Na2S2O3 . 5 H2O}$.
There is many more recipes.

Another option is electrolysis-less plating by chemical silver deposition, many recipes available too.

If you want to deposit a silver layer that is bright and adherent, you should not electrolyze a solution containing simple silver ions. Because the electrolysis of silver ions produce a rough deposit of silver that does not adhere perfectly on the cathode. To get a bright shiny silver layer,  a complexing agent should be added in the solution, like cyanides or EDTA solutions.

"In initial stage, the iron spontaneously replaces silver ions, so there is no potential barrier there." Can you explain this please?

@Maurice I am well aware of it, the optimal composition of solutions for electroplating is 1/4 of chemistry, 1/4 of alchemy, 1/4 of art and 1/4 of secrets. That is why I have mentioned the electroplating solutions, perhaps not strongly enough.

Iron immersed in a silver ion solution will spontaneously immersion plate silver. There is a potential but the cells are shortcircuited.  The purpose of the chelating or complexing agents is to lower the activity of silver ions, ie. change the voltage, to prevent immersion plating. If done properly the electroplating gives more even plating and possibly a better shine. There is a lot more chemistry involved when one thinks about it.


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Conjugate addition of alkyl groups to α,β-unsaturated Aldehyde
why the conjugate addition of an alkyl or other organic group to an α,β-unsaturated aldehyde is not available while ketones can? (for example of addition of lithium diorganocopper reagent, R2CuLi)

Quoting M. Mauduit et al in Comprehensive Organic Synthesis (2nd Ed, 2014) from  here

α,β-Unsaturated aldehydes represent certainly the most challenging Michael acceptor concerning the asymmetric copper-catalyzed conjugate addition of organometallic nucleophiles. This is mainly because of their higher reactivity that allows the undesired formation of 1,2-adducts as well as the aldol by-product.

Further reading of this article reports strategies to allow 1,4 addition.


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Can geometry of a molecule change when the spin-orbit coupling is taken into account?
Cross-posted on MMSE.
Is there any evidence that spin-orbit coupling (SOC) leads to significant changes in atomic structure? Using DFT, I am getting different relaxed lattice constants when I take SOC into account compared to when I neglect it. Is this a bug or a feature? Are there any known materials where SOC leads to structural distortion?

Have a look at Pekka Pyykkö's work. I recommend starting with his great 2012 review, Relativistic Effects in Chemistry: More Common Than You Thought - check out section 4.6.

If there are two  states with different symmetry so they do not interact, then you introduce an operator that mixes them the total energy should remain the same with one state  increasing the other lowering in energy and locally form an avoided crossing which should, I suppose,  change the geometry at the crossing point.

Sure, eg IrO2 is doing it, but it is doing the other way:SOC is needed for higher symmetry models, and the lack of SOC leads to distortion

@NicolauSakerNeto I remember seeing it and it is great, but now paywalled. I assume we (now) have to pony up?

@OscarLanzi You can find it on ResearchGate here

I'm closing this question as it was cross-posted to [mattermodeling.se]: https://mattermodeling.stackexchange.com/q/11949/

"I am getting different relaxed lattice constants when I take SOC into account and when I neglect it."

That sounds correct. I would be surprised if the lattice constants remained precisely the same after taking into account spin-orbit coupling.

"Are there any known materials where SOC leads to structural distortion?"

All of them. Unless you're able to show an example in which the lattice constants obtained from theory remain precisely the same after introducing spin-orbit coupling into the model.


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Deviation from ideal gas behavior
I know that gases deviate from ideal behavior at low temperature and high pressure
But what if we increase the pressure of a gas by increasing the temperature keeping the volume constant. Would the gases still show deviation?
I think they shouldn't because when we increase the temperature, we increase their kinetic energy and make attractive forces between the molecules more negligible, at the same time, we don't change the volume so the volume of the gas molecules should remain negligible as compared to the volume of the container.

More and more molecules will spend time colliding, forming for fraction of time an effective dimer or adduct, before them spring away. At very high temperatures, molecules start to break or, together with atoms, start self-ionization.

It is true, though, that gases at higher T tend to behave "more ideally".


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Acid dissociation constant from conductometric data

$\ce{0.007 M}$ aq. solution of anilinium hydrochloride had molar conductivity equal to $\pu{119.4 S cm2 mol-1}$, which became $\pu{103 S cm2 mol-1}$ when a few drops of aniline were added to the solution. Molar conductivity of $\ce{HCl}$ in the same ionic concentration was found to be $\pu{413 S cm2 mol-1}$. Find out the hydrolysis constant of anilinium hydrochloride.

I knew that $K_\text{h}=\dfrac{[\ce{PhNH2}][[\ce{H3O+}]}{[\ce{PhNH3+}]}$ and I tried to find out degree of hydrolysis $h=\dfrac{Λ_m^c}{Λ_m^∞}$ and hence $K_\text{h}$ but that attempt miserably failed. Can anyone please explain and give the correct method to go about this?

More details, from the comments:

I took the equilibrium,
$\ce{PhNH2 + HCl <=> PhNH3+ + Cl-}$
So I guessed that x=0.007 and found concentration, $K_h=0.007$. This was blatantly incorrect.
But I am not able to proceed after writing down the reaction.

Further comment:

$h=\frac{Λ^c_m}{Λ^∞_m} = \frac {103}{119.4} = 0.86$ isn't it?
Thus $K_h = \frac {c \cdot h^2}{1-c} = \pu{3.7E-2}$
But this is incorrect...
Answer is given as $\pu{2E-5}$

Deleting obsolete comments. // Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// No MathJax in titles.

You can write  $\pu{2E-5}$ as $\pu{2E-5}$  and $\ce{PhNH3+}$ as $\ce{PhNH3+}$

It’s either aniline hydrochloride or anilinium chloride. Otherwise, you’d have too many hydrogen atoms.

@Karsten the question said it and thus the word 'anilinium hydrochloride'. Ig its either $\ce{PhNH2+ HCl-}$ or $\ce{PhNH3+ Cl-}$

PhNH2.HCl=aniline hydrochloride, PhNH3+ Cl- = anilinium chloride, PhNH3+ . HCl = anilinium hydrochloride . The last one is wrong.

The molar conductivity $\Lambda_\ce{AHCl hydr}=\pu{119.4 S cm2 mol−1}$ is equivalent to the partially hydrolyzed hydrolyzed anilinium chloride, according to the hydrolysis constant.
$$\ce{PhNH3+(aq) + Cl-(aq) + H2O(l) <=>[K_\text{h}] PhNH32(aq) + Cl-(aq) + H3O(aq)}$$
$$K_\text{h} = \frac{[\ce{PhNH2}][\ce{H3O+}]}{[\ce{PhNH3+}]}$$
The molar conductivity $\Lambda_\ce{HCl}=\pu{413 S cm2 mol−1}$ is equivalent to the fully hydrolyzed anilinium chloride.
$$\ce{PhNH3+(aq) + Cl-(aq) + H2O(l) -> PhNH32(aq) + Cl-(aq) + H3O(aq)}$$
The molar conductivity $\Lambda_\ce{AHCl}=\pu{103 S cm2 mol−1}$ is equivalent to the not hydrolyzed hydrolyzed anilinium chloride, as free aniline suppresses hydrolysis by increasing pH and shifting the equilibrium.
$$\ce{PhNH3+(aq) + Cl-(aq) + H2O(l) <-[PhNH2] PhNH32(aq) + Cl-(aq) + H3O(aq)}$$
Therefore, the degree of hydrolysis
$$h= \frac{\Lambda_\ce{AHCl hydr} - \Lambda_\ce{AHCl} }{ \Lambda_\ce{HCl} - \Lambda_\ce{AHCl}} =    \frac{\pu{119.4 S cm2 mol−1}  - \pu{103 S cm2 mol−1} }{  \pu{413 S cm2 mol−1}   - \pu{103 S cm2 mol−1}}\approx 0.0529$$
$$K_\text{h} = \frac{[\ce{A}][\ce{H3O+}]}{[\ce{AH+}]}\approx \frac{[\ce{A}]^2}{[\ce{AH+}]}=\frac{c_{\ce{AHCl}}^2\cdot h^2}{c_{\ce{AHCl}}(1-h)}=\frac{c_{\ce{AHCl}}\cdot h^2}{1-h}=\\
=\frac{\pu{0.007 mol L-1}\cdot 0.0529^2 }{ 1 - 0.0529}\approx\pu{2.07E-5 mol L-1}$$

In summation, molar conductivity of hydrolysed - molar conductivity of non hydrolysed = how much of the anilinium hydrochloride is actually being hydrolysed, and the molar conductivity of HCl - molar conductivity of non-hydrolysed anilinium hydrochloride = concentration of the conjugate acid of aniline. Thus, you got $h$ from there and then went on to find $K_h$. Thank you so much @Poutnik!

This answer is more about the chemistry and how the conductivity relates to the ions present, and does not show a numerical solution. For that, check the answer by Poutin.
Here is the acid/base reaction without the chloride counter ion:
$\ce{PhNH3+ <=> PhNH2 + H3O+}$
The number of ions on the left and on the right is the same. However, the hydronium ion is a much better conductor, so the dissociation reaction changes the conductivity.
Chloride, on the other hand, is present at the same concentration in all three solutions, so it is not responsible for the changes in conductivity.
The part of the question I did not get at first is the addition of aniline. Aniline is neutral, so its concentration does not change the conductivity directly. However, as you can see in the equation above, it can react with hydronium ions to form anilinium ion. This will lower the conductivity of the mixture (ignoring dilution).
If you assume that the addition of the aniline drops the hydronium concentration sufficiently that the conductivity reflects 100% anilinium and chloride, then the way Poutnik calculated the equilibrium constant is sound. You can check after the fact that the approximation is valid by calculating the pH and the ratio of anilinium to hydronium before and after the addition of aniline.

aniline is slightly basic in nature having $pK_b=9.38$ and thus on addition of acid it will accept the proton donated by HCl thus forming anilinium hydrochloride (that's what the question says). So by common ion effect the hydrolysis is suppressed. At first when Poutnik said it I realised this fact and this answer gives me more clarity into this.


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Multidimensional Chemical Space - nD coordinate system
I have a dataset of many organic compounds.
I would like to describe them using a set of coordinates.
A 2D example of what I want to do is chemplot, this package lets you create a 2D coordinate system to distinguish the chemicals and thus it creates a 2D chemical space.
It does that using some kind of decomposition technique (PCA, t-SNE, Umap) applied on the molecular representation that is said to be n-Dimensional.
Now what I want is actually the n-Dimensional representation of the molecules and not the simplified mapping on 2D space.
I read on the article that the procedure they use is the following

ChemPlot uses the RDKit library to convert SMILES and InChI notations into ECFPs with a bit-vector length of 2,048 bits and radius of 2 adjacent atoms. After each molecule is converted, for all molecules, the bits that contain only 0 s or only 1 s are removed from the bit-vectors. The remaining number of bits determines the total number of dimensions and they are used as the input for the dimensionality reduction phase.

I would like some help on how to reproduce this process in python.
The 2 steps I would like help on are:

How to get the 2048 bits vector (with radius of 2 atoms)
How to clean the dimensions that only have 0s or 1s

This is an open source package, well documented, using other packages (for classification and parameter reduction, e.g. mordred) that are also open source.
This means you can download their python code and modify it, instead of reproducing it. All you have to do is to output the 2048 bits vector, and the shortened vectors. You would have to find the function where these are generated (or the function calls), and add a line to output the intermediate data you are interested in.
The details on how to do that are not chemistry, but computer programming, so I'll stop here.


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Enzymatic mechanism of Lipase
When lipase catalyses the hydrolysis of a triglyceride (oil/fat) which bond does it break and why?
What about in the case of saponification (triglyceride + NaOH) in the presence of lipase?
Is the same bond broken in both cases?
If during esterification, alcohol gives H and acid gives OH, then during hydrolysis/saponification is the
O—COR the bond broken by lipase?

As shown here https://proteopedia.org/wiki/index.php/Lipase  the O-COR bond is broken

As an Organic chemistry/Biochemistry student, you must understand when bond making and bond breaking reaction occur in vitro or in vivo conditions, it's always involved the weakest bond. For example, weakest bond in a ester hydrolysis is bond between carbonyl carbon and oxygen attached to corresponding alkyl group $(\ce{R-C(=O)-OR'})$, which would break during the hydrolysis regardless of the conditions. Difference between ester hydrolysis by $\ce{NaOH}$  and lipases is no specificity when used $\ce{NaOH}$ (e.g., both 1,3- or 2-lincages in glyceride will be break to give glycerol and fatty acids as products), but could have high specificity when used lipases.
The ref.1 gives a general mechanism of hydrolysis by a lipase:

Figure 1. Mechanism of lipase catalysis (From Ref.1)

Their (lipases') catalytic triad is composed of serine, histidine and aspartate or glutamate, which is also found in serine proteases. The mechanisms involved during lipase catalysis are illustrated in Fig. 1. In the first step, serine is activated by deprotonation, for which histidine and aspartate are required (Fig. 1a). Consequently, the nucleophilicity of the hydroxyl residue of serine is enhanced and attacks the carbonyl group of the substrate forming an acyl-enzyme intermediate (Fig. 1b). The presence of an oxianion hole contributes to the stabilization of charge distribution and reduction of the ground state energy of the tetrahedral intermediate. The deacylation step in Fig. 1c is shown which is controlled by the electronegativity of the molecules populating the interface. In this process a nucleophile (e.g. $\ce{H2O}$ or monoglyceride) attacks the acylated enzyme leading to product release and regeneration of the catalytic site.

References:

P. Reis, K. Holmberg, H. Watzke, M.E. Leser, and R. Miller, "Lipases at interfaces: A review," Advances in Colloid and Interface Science 2009, 147–148, 237-250 (DOI: https://doi.org/10.1016/j.cis.2008.06.001).


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How to convert molar volume to volume?
This seems like a simple question that people must have asked already, but I can't find an answer online.
I'm not a chemist, but need to use a code that deals with thermodynamics.  I am trying to use an input data file from an older code but have to convert it to the form the current code uses.  I'm having problems understanding how to deal with the volume term.
In the previous code, there's molar volume in cm3 mol-1.  In the current code, I need to input volume in J bar-1.  I know I can do a simple units conversion by multiplying the molar volume value by 0.1.  However, I have an uncomfortable feeling that there's something I'm not understanding.
Do I need to worry about an equation of state to do things properly?  I'll be dealing with solids (not liquid or gas) at a room pressure and temperature but don't want to make any simplifying assumptions.  Thank you for any tips.

If you have given

the molar volume $V_\text{m}$ in $\pu{cm3 mol-1}$,
the molar amount in $\pu{mol}$,

you get the volume as $V = n \cdot V_\text{m}\ [\pu{cm3}] = 0.1 \cdot  n \cdot V_\text{m}\ [\pu{J bar-1}]$.
As $$\pu{J bar-1} = \dfrac{\pu{N m} }{ \pu{e5 Pa}} = \dfrac{\pu{N m3}}{ \pu{e5 N}} = \pu{e-5 m3} = \pu{10 cm3} = \pu{10 mL}$$
Note that this is the first time I have even seen to express volume in such a unit.

I had to do a similar thing while deriving the formula for molar conductivity from conductivity. My rep was too low to comment and thus I could not answer before @Poutnik


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is alkyl halide turning into alkene reduction or oxidation?
Is the reaction of 2-iodobutane turning into but-2-ene reduction or oxidation reaction?

HI is being removed in the reaction. It is $\beta$-elimination. I really am not able to see any changes in the oxidation numbers also.
$\ce{CH3-CH(I)-CH2-CH3}$ will undergo dehydrohalogenation.
Only rearrangement in molecular structure takes place.

Why was my answer unaccepted? Like it was accepted long ago and now all of a sudden its unaccepted

I guess only OP knows why. Well, a reason could be that technically ox states do change, even if sum stays the same. BTW did you check out the site chat? For example this room or the main chatroom


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Determination of ground state energy that has the same L and S values
Determination of the energy of a multielectron atom in the ground state can be estimated using the atomic term symbol which involves the values L, S and J. If an atom has the same L and S values then the determination of the ground state is.......
Select one:
a. According to Hund's rule, for orbitals less than half full, the ground state is for the highest
b. According to Hund's rule, for orbitals filled more than half full, the ground state is for the lowest J
c. According to Pauli's rule, for orbitals filled less than half full, the ground state is for the lowest J
d. According to Pauli's rule, for orbitals filled more than half full, the ground state is for the highest J
e. According to Hund's rule, for orbitals filled less than half full, the ground state is for the lowest J
The Hund's rule is used to determine the ground state of an atom. I knew for the maximum multiplicity of S lies lowest in energy. And the term with the largest L value is lowest in energy, but the thing is, when L and S were in the same value, the one that determine this is the J value which is the highest J value for half filled, and lowest J value for >half filled
Considering the question, i have no idea which were the answer, the a or b? or am i wrong throughougth? can you give me some resources to help me understand this more?

I'm not sure you stated Hund's rules properly. Stated briefly they are (a) max number of unpaired electrons lowest energy (b) for same spin multiplicity, largest orbital ang mom usually lowest in energy (c) less that half filled level, lowest J usually lowest in energy.


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Thermal conductivity and diffusivity of diatomite (also known as diatomaceous earth or kieselguhr)
Does anyone know where can I find measured thermal conductivity and thermal diffusivity of diatomite for different temperatures? I'm looking for the temperature dependency for temperatures higher than 300K. The upper temperature limit shall be at least 1000K, but more is better. All I have found is thermal conductivity in "Touloukian, Y. S., et al. "Thermophysical properties of matter - The TPRC data series. Volume 2. Thermal conductivity - nonmetallic solids. (1971)". I'm still missing data on thermal diffusivity though, and I'd like to see more data on thermal conductivity, if available.
Any help will be appreciated.

As a granular material the thermal diffusivity will depend greatly on things like grain size, shape, and packing density (how much you compress it). For a specific application you likely need to determine it experimentally to get a good answer.

@JonCuster, Your statement is true -- but since similar diatoms have similar shells (tests), http://www.microscopy-uk.org.uk/mag/indexmag.html?http://www.microscopy-uk.org.uk/mag/artfeb05/cbdiatoms.html, there likely would be much less variation between samples with the same origin, as opposed to ground powders, such as MgO, that have not been sieved. Results of a few tests (no pun intended) might apply to an entire source of diatomaceous earth.

Cross-posting is discouraged on the Stack Exchange network. As there are duplicates on Earth Science (https://earthscience.stackexchange.com/q/25763) and Matter Modeling (https://mattermodeling.stackexchange.com/q/11965), I am closing this question.


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Quantitative Method Comparison & Experimental Error
I'm a Y1 Chem Undergrad and I standardised hydrogen peroxide using two different methods (both redox titrations) in the lab. This was done to compare the 2 different methods and their reliability. I am unsure how to quantitatively evaluate/compare them as there is no 'theoretical' H2O2 conc., just the experimental values you get.
The methods give [H2O2] conc. = 12x10^-3, 9x10^-3 (gL^-1). I thought I might be able to calculate the percentage error if I assumed the mean was the 'theoretical' value (gives ~22%) but this seems like a bit of a cop-out to me. Random errors for the titrations are 3.2% and 0% respectively.
Is there a statistical technique you are aware of that might allow me to quantitatively compare my results?
Many thanks,
J

There is something called as the paired t-test that compares the two means and tells us if they belong to the same population or not.

Assuming you’ve done multiple runs of each titration, you can work out the 95% confidence interval and the standard deviation. This can be done in excel. It would also be worth quantifying the errors from each instrument (the burette, pipette, etc.). That being said, if you already know that the random errors are different, doesn’t that already tell you one result is more reliable than another?


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Radioactivity of Promethium
In school I was taught while looking at the difference between lanthanides and actinides that all of the actinides were radioactive, while only promethium among the lanthanides was radioactive. The question that struck my mind was 'why only promethium'?
On further research I also found out that technetium has a similar behaviour, being located in the midst of stable elements. Why does this happen?


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Why is expected bond energy the *geometric* mean of two pure bonds?
Suppose we are calculating the electronegativity of a bond A—B. In my textbook (Chemical Principles) calculation of electronegativity, they calculate electronegativity in terms of a variable difference in bond energies $\Delta$, which they define as $$\ce{\Delta = (A-B)_{actual} - (A-B)_{expected} }$$
and that discrepancy is used to calculate electronegativity. Fair. But I'm confused by their formula for the expected energy of $\ce{A-B}$: $$\ce{Expected A-B bond energy = \sqrt{\ce{(A-A bond energy) \times (B-B bond energy)}} }$$
which is the geometric mean of the two. But why the geometric mean, and not, say, an arithmetic or harmonic mean? Why even use a mean at all? What's the intuitive rationale for this?

Relevant but not really an answer: https://en.wikipedia.org/wiki/Combining_rules

See also en.wikipedia.org  Electronegativity#Methods_of_calculation

Yea, like these articles are relevant because of the similar kind of formulae used? As for the answer, I think it is because bond energy is dependent on the orbital overlap. As for the GM, it's used to get a balanced contribution from each atom as the bond strength is proportional to the product of the combining factors. GM is taken when things are multiplied and so yea, we get it

Am I right everyone?

If pushed and threatened with violence I would probably wave my hands very vigorously about second order perturbation theory grossly approximated, possibly via the closure approximation. But really I don't think there is a right answer in the sense that the underlying exact theory behind chemistry is very, very difficult and often insoluble, so as such quite often we end up using things which just empirically work - and this is one of them.

@IanBush no answers here?

@Poutnik......?

You can actually used arithmetic mean instead of geometric mean. Mean (or average) of 2 different value are designed to be somewhere in between the value.
For example, let $X=3$ and $Y=7$, the means are
$$HM = \frac{1}{2}(\frac{1}{3} + \frac{1}{7}) = 4.2$$
$$GM = (3\times 7)^{\frac{1}{2}} \approx 4.58$$
$$AM = \frac{3+7}{2} =5$$
There are other types of mean that can also be use (i.e., root mean square).
Now, I believe that the electronegativity that you are referring to are Pauling's electronegativity. I think the main reason that Pauling pick $GM$ over the other is due to the fact that energy may come in many order of magnitude. Let's say $X=1$ and $Y=10000$
$$HM = \frac{1}{2}(\frac{1}{1} + \frac{1}{10000}) = 20000/10001 \approx 1.9998$$
$$GM = (1\times 10000)^{\frac{1}{2}} =100$$
$$AM = \frac{1+10000}{2} =5000.5$$
If you see this, which of the 3 means ($HM$, $GM$, $AM$) is more  in the middle. It would be more intuitive to consider $GM$ to be more in the middle in this case.


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Alcohol preparation process
Currently Studying in high school, I encountered a reaction in organic chemistry. I got a reaction in which alcohol(ethanol) was dehydrated to form unsaturated hydrocarbon(ethene) and water. I wanted to know that can the opposite of this also occur? That is can we form alcohol from unsaturated hydrocarbon? If yes, what is the process to do so?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

See also chem.libretexts.org - Case_Study - The_Manufacture_of_Ethanol_from_Ethene

There is a reaction.
The reaction which is listed by you is called the dehydration reaction. It is used to prepare alkenes.
$\ce{CH3-CH2-OH ->[conc. H2SO4]CH2=CH2}$
If you are from India you will study this in Hydrocarbons in class XI and Alcohols, Phenols and Ethers from class XII (The reverse reaction is based on this chapter).
It is the same catalyst, only the reaction is reversed.
By Le Chatelier's principle, the byproduct $\ce{H2O}$ is removed by the concentrated acid used, thus increasing yield of alkene or alcohol (depending on what you want to prepare).
SOURCES: NCERT - Chemistry, Classes XI and XII


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Constituents of LaI2
An ionic solid $\ce{LaI2}$ shows electrical conduction due to presence of which anions and cations?

Note that questions looking like homework/test/exam/lab tasks, posted without significant attempts to understand/resolve them, are explicitly off-topic on this site and will be closed.
// Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  downvoting or closure.

You will never find out, short of knowing its formula.

It is an electride, actually having a chemical formula of $\ce{La^{3+}[(I^−)_{2}e^{−}]}$

So it is neither due to cations nor anions, after all.

As indicated by Nilay, the electrical conductivity does not come from atomic ions but from electrons that are transfered from lanthnum into a delocalized conduction band rather than to the iodine. According to Wikipedia, lanthanum also forms a monoiodide ($\ce{LaI}$) and a dihydride ($\ce{LaH2}$) with similar ionic/metallic mixed bonding. All these compounds have lanthanum(III), with the balancing negative charge from the nonmetal combined with conduction electrons (e.g. $\ce{LaI2}$ has two iodide ions plus one conduction electron per lanthanum(III) ion). Both the iodides and the hydride can be combined with more of the nonmetal to form $\ce{LaI3}$ or $\ce{LaH3}$, with loss of electrical conductivity as the additional nonmetal atoms capture the otherwise-conduction electrons.


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Universal indicator solution in electrolysis
Sometimes universal indicator solution, or whatever acid-base indicators, is added into an electrolytic cell to indicate the change in pH at electrodes over time.
I wonder, if universal indicator solution will get electrolyzed in the set-up, leading to errors in the experiment.
My lecturer commented that universal indicator solutions are generally stable and do not involve in reactions. Moreover, we can observe that the products we get match what we have expected, so universal indicator solution has not been involved in the electrolysis.
But then, I am still confused. For example, we don't know if the substances in the indicator are electrolyzed to produce the same products expected.
Can somebody provide me with some ideas?

I guess the relative acidity of the solution is the same if you add the indicator before or after electrolysis. It means that the indicator is not electrolyzed. Have you tried these two approaches ? Is there any significative difference ?

It really depends on the stability of your indicator under the electrochemical conditions you apply in your cell, like Poutnik has mentioned. And actually it is a good remark that you mention the formation of the to be expected products from electrolyzing your indicator. If you do the charge balance at a half-reaction, you will see you need to consume or produce some ions. In watery solutions this is typically hydroxide or protons, so a pH indicator is typically not really telling you what you're forming.
Besides the stability of the indicator, the main issue I would have throwing an indicator in my solution is the possibility of it to adsorb on the electrodes. So even if the indicator is completely stable, you will get deviation in your current-potential behavior.

Products being obtained and indicators being changed are two independent processes. There is no universal answer if the indicator can be reduced or oxidized, as it depends on the indicator and the potentials on either of electrodes.
Some indicators may be vulnerable to Kolbe electrolysis
$\ce{2 R-COO- ->[- 2 CO2][- 2e-]-> R-R}$
and generally to oxidation at anode, as there may be present reactive oxoradicals or atomic chlorine.


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Effect of Steric Hindrance in Pinacol-Pinacolone Rearrangement
In the acid catalysed rearrangement of vicinal diols to carbonyl compounds which is pinacol-pinacole rearrangement,
The migratory order of the groups is H>Aryl>Alkyl
My question is that how aryl groups take precedence over alkyl groups when the aryl groups cause steric hindrance.

For an aryl group, what if the ipso-carbon is bound to both carbons in question at same time in the transition state for rearrangement? What becomes of the positive charge?


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Breakdown of addition polymers
We know that condensation polymers can be broken down by hydrolysis with the elimination of small water molecules.
Then what about addition polymers? How can they be broken down into their monomers?
I think everything has their own ways to get built up, and a way to be broken down.

Hydrolysis with addition of water molecules. With addition polymers situation is analogous - reaction is the same as polymerisation, but backwards.

If addition polymer is polythene $\ce{(CH2CH2)_n}$, the back reaction is not exactly the contrary of the polymerization, because the polymer is a long chain like in an alcane. Water is no use and the pyrolysis produces a lot of smaller alkanes and alkenes. Of course ethene $\ce{C2H4}$ is produced, but several other simple hydrocarbons are also produced, with a number of $\ce{C}$ atoms different from $2$.


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Stereochemical relationship of molecules including double bond configuration and absolute configuration
Are (S,E)-4-isopropoxypent-3-en-2-ol and (R,Z)-4-isopropoxypent-3-en-2-ol both enantiomers and diastereomers? I am a bit confused because they would be diastereomers when only looking at the configuration of the double bond or enantiomers when just looking at the chiral center. However, it doesn't make sense that molecules can be diastereomers and enantiomers at the same time. Could someone explain? Thanks.

You don't separate the stereochemical entities. Do they have the same atom connectivity? YES! Are they enantiomers? NO! Then they are diastereomers.

Diastereomers, by definition, are stereoisomers that are not enantiomers, so you're right they can't be both.
I think you might be confused because you are trying to apply a simplified heuristic (opposite configuration of double bond = diastereomer, opposite configuration at chiral center = enantiomer) outside of its proper scope. This works well for molecules with a single stereocenter, but for slightly more complex cases you have to step back and think about the meaning of the terms for a moment.
Two molecules are enantiomeric when they are mirror images of one another, that is, if they are identical in every respect, except the configuration around their chiral center(s). (R)-4-isopropoxypentan-2-ol and (S)-4-isopropoxypentan-2-ol are enantiomers, because their single chiral center contains the exact same set of four substituents (H, OH, methyl and 2-isopropoxypropyl), just in reverse orders. However, your pair of molecules have only three substituents (H, OH and methyl) in common; the fourth substituent at C2 is (E)-2-isopropoxypropen-1-yl in the first compound and (Z)-2-isopropoxypropen-1-yl in the second, which are isomeric but clearly distinguishable groups. (R)-4-isopropoxypentan-2-ol and (S)-4-ethoxypentan-2-ol aren't enantiomers, and neither are these molecules.
They are, however, still stereoisomers of one another, so they must be diastereomers.


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How to display atomic charges from MOPAC calculation?
MOPAC (http://openmopac.net/) is nice semiempirical code. It also calculates charges on atoms.
Is there a free GUI to display them please ? What I know, Avogadro diplays vibrations, JMol orbitals...but which GUI display charges on atoms ?

How do you want to display these? As numbers overlayed on the atoms, or as atom colors, shades, etc? You can color surfaces in Jmol: https://jmol.sourceforge.net/docs/surface/#Color

Maybe better to link to the source: https://github.com/openmopac/mopac

Both Jmol and Avogadro can color atoms by their partial charges provided that you have a way of passing that information to the GUI. MOPAC has a special output format to pass orbital and partial charge information to Jmol by using the GRAPHF keyword [http://openmopac.net/manual/graph.html], which produces a *.mfg output file. If you open up the *.mfg file in Jmol, you can visualize the MOPAC partial charges by right-clicking on the molecular display and going through the menus: "Color"->"By Scheme"->"Partial Charge".
To display the partial charges as numbers, go to the Jmol console (also in the right-click menu or File->Console drop-down menu) and type the commands "select all" to select all of the atoms and then "label %P" to display partial charges as the labels of the selected atoms.

@MiroIliaš If this answered your question please check the checkmark to accept it.


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Does CsBr have BCC lattice or cubic?
Everywhere I see on the internet, it says BCC, but my professor still says "it is cubic lattice, no matter which book/website might tell you otherwise".
The argument was that for deciding which lattice it is, we only see the atoms/ions which make up the lattice, in this case Br, and they're only present at the corners of the cube, which seems correct.
I haven't seen this being said anywhere else, and every site says CsBr is BCC, considering the Cs at body center while deciding the structure.
I do understand its just a cube with Br at corners and Cs in the middle, but I want to know what the correct terminology here.

It is cubic lattice, and calling it BCC is an abuse of language. In the real BCC lattice the cube's center is equivalent to its corner.

https://en.wikipedia.org/wiki/Cubic_crystal_system#Caesium_chloride_structure

Lattice and crystal structure are two different objects. A lattice exists in the geometrical space and is formed by lattice points characterized by having the same geometrical properties; on these bases are defined the Bravais lattices. Crystal structures are obtained by the convolution of a lattice with a motif (one or more atoms with a mutual geometrical relationship); a structure exists in real space and is formed by atoms. Don't confuse atoms with lattice points.
CsBr crystallizes in the Pm-3m space group; "P" means that the underlying lattice is primitive, not body-centered. The motif in this case is given by two atoms, Cs and Br. Cs atoms are located at the vertexes of the cube, the same positions as the P lattice, whereas Br is at the center (but you can also describe the structure in the opposite way, with Cs at the center and Br at the vertexes). The important thing is that atoms at the center and those at vertexes are chemically different; for this reason, the lattice is P and not I (=bcc). If all atoms were Cs, you would have a bcc structure.
Cubic P and cubic I (=bcc) are both cubic lattices, but with different symmetry operations.


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Why did this synthesis of Aspirin fail?
I tried to synthesize acetylsalicylic acid(aspirin) by the method below. However, I definitely failed and got something unknown.
The method used was:

Add 2.5 g of salicylic acid(about 2494 mg), 3.0 ml of acetic anhydride, and 5 drops of concentrated(85%) phosphoric acid to a 50-mL Erlenmeyer flask.
Heat the flask with a hot water bath at about 70°C for 5 minutes. The solution should be transparent after the reaction.
Add 2 mL of distilled water to the flask. After a minute, take off the flask from the water bath and add 20 ml of distilled water to the flask. Then put the flask in an ice bath and get crystals.
Collect the aspirin crystals by vacuum filtration. Use a Buchner funnel and dry filter papers. Scrape out the solid in the flask with a glass rod and dump on the filter paper. Rinse it with 5 mL of distilled water.
Dry the aspirin in 110°C oven for 5 minutes. I get about 2556 mg of aspirin.
Take 1 g of aspirin and follow a process similar to 1-5. Add 15 mL of diethyl ether and 1 g of aspirin to a 50-mL Erlenmeyer flask. Then heat the reaction with a hot water bath at about 50°C. Take off the flask and add 15 mL of petroleum ether to the flask. Put the flask in an ice bath and get white needle shaped crystals. Collect the crystals by vacuum filtration. Rinse it with petroleum ether. Finally dry the crystals in 110°C oven. (Recrystallization)

When I took them out after 5 minutes, they was slightly less dry. (now I get about 311 mg) I wanted to use a melting point apparatus, so I continued heating a little more. However, it melted...

It seems that new substances have been formed instead of aspirin. Especially, at the point where there were filter holes, black powder like something created. I wonder what these could be.
When I was synthesizing aspirin (at method 3), crystals formed unexpectedly because the temperature of the hot water bath slowly decreased. However, since the solution was already transparent enough, I believed it would not have a significant impact. Additionally, the aspirin was not finely powered but rather in chunk form. But I also believed it would not be a major issue.

Some SWAG: 1. There was dirt in the apparatus. 2. Chemicals were impure (phosphoric acid, for example, may attack a paper lid-seal, carbonizing it). 3. At some point the reactants overheated. For example, in step 1 of the exercise, heat may be evolved. It might be necessary to add them slowly, with cooling, until mixed.

What is the meaning of the first sentence in § 6 ? I copy it here in italics : Take 1 g of aspirin and follow a process similar to 1 - 5 ? Do we have to add acetic anhydride and phosphoric acid to this aspirin like in paragraph 1?


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Which of the following two IUPAC names would be accurate according to 1993 IUPAC rules?
I'm not looking for answers that are based on 2013 IUPAC recomendation based rules. But rather the old 1993 rules of IUPAC.
The two names are :
1.) Penta-1,2-dien-5-ol
2.) Penta-3,4-dien-1-ol
would we start numbering from the carbon attached to the main functional group [i.e.,secondary suffix] which would give us locants at 1,3 and 4 or just simply follow the lowest locant rule giving us locants 1,2 and 5 ?
Again, I'm specifically looking for name that be accurate on basis of 1993 IUPAC rules and not the PIN.
What rules to keep in mind while naming this ?

Concerning numbering of locants according to the recommendations of 1979, 1993, as well as 2013, the lowest locant is given first to the principal group that is named as suffix (here: -ol) and then to unsaturation (ene/yne).
Thus, the correct name according to the 1993 and 2013 recommendations is penta-3,4-dien-1-ol. (The 1979 recommendations had a different placement of locants in the name.)
See also this example given in the 1993 recommendations: pent-4-en-2-ol


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using either 2-propanol or isopropyl alcohol for writing a report
I'm writing a report based on testing I've done on a material. I'm wondering if there is any convention on using different nomenclature for isopropanol. Which way of writing this molecule is consider the most accepted or the most "correct?"

The answer depends on the audience. For theoretical, more nomenclature inclined chemists, I would use the preferred IUPAC name propan-2-ol,  but technicians would probably use isopropanol or isopropyl alcohol.

Many compounds can have more than one systematic IUPAC name. A compound may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC.
However, “isopropanol” is one of the rare cases of a widespread name that is actually considered wrong and not just not preferred.

Designations such as isopropanol, sec-butanol, and tert-butanol are incorrect because there are no hydrocarbons isopropane, sec-butane, and tert-butane to which the suffix "-ol" can be added; such names should be abandoned.

– Nomenclature of Organic Chemistry (The IUPAC 'Blue Book'), Sections A, B, C, D, E, F, and H. Oxford: Pergamon Press. 1979.
Isopropyl alcohol, however, is correct but not preferred.

Isopropyl alcohol, sec-butyl alcohol, and tert-butyl alcohol are, however, permissible (see Rule C-201.3) because the radicals isopropyl, sec-butyl, and tert-butyl do exist.

– ibid.
The preferred name in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is propan-2-ol.


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How can standard enthalpies of fusion and vaporization add up to that of sublimation when these occur at different temperatures?
In Atkins' Physical Chemistry, the following diagram appears in the chapter on thermochemistry:

The definition of standard enthalpy change is also given:

The standard state of a substance at a specified temperature is its pure form at 1 bar.

My doubt is regarding the temperature at which these changes occur. Say the fusion occurs at the melting point, vaporization at the boiling point. According to the definition of standard enthalpy change, these need to occur at a fixed temperature. How can we add the standard enthalpies associated with the two to obtain the standard enthalpy of sublimation, which occurs at a different temperature and may not be possible at the standard $1\pu{bar}$ pressure? Shouldn't the heats associated with changing the temperature of the system be included in the equation?
The phase diagram of a general solid-liquid-gas may be represented by the following:

According to the diagram, solid, liquid, and gas phases can only exist together at the triple point, leading me to the conclusion that the equation $\Delta_\text{fus}H^\ominus+\Delta_\text{vap}H^\ominus=\Delta_\text{sub}H^\ominus$ is only valid at the triple point.

(...)''can only exist together'' in equilibrium ''at the triple point'', but who says there's got to be equilibrium?

@Mithoron thank you for the comment. Say the fusion occurs at the melting point, vaporization at the boiling point, and sublimation, which may not be possible at standard pressure, occurs at a different temperature. According to the definition of standard enthalpy change, these need to occur at a fixed temperature. Shouldn't the heats associated with changing the temperature of the system be included in the equation?

Say the standard temperature were closen as the triple point.  Would you be comfortable with the explanation then?

@ChetMiller I think changing the definitions set by IUPAC is just bypassing the problem and not really solving it.

That is not what I asked.

@ChetMiller, to answer your question, no. I wouldn't be comfortable with that definition because triple point of different substances is different, and setting the standard as the triple point of water wouldn't work everywhere.

I'm not talking about that as the standard temperature.  I'm asking whether you think that that, at the triple point, the heat of fusion plus the heat of vaporization is equal to the heat of sublimation.

@ChetMiller that was answered in the question itself. Yes, I do think that the equation works at the triple point.

You can certainly have liquid and vapor at equilibrium at temperatures other than the atmospheric boiling point.  And you can certainly have solid and vapor at equilibrium at temperatures less than the triple point.  In establishing the standard enthalpy change for vaporization at a temperature less than the atmospheric boiling point (say 25 C), the vapor state is hypothetical, and an extrapolation of ideal gas behavior from the equilibrium vapor pressure to 1 bar.  Something similar (extrapolation) is done with the heat of fusion to estimate the heat of fusion at 1 bar.

According to your answer, the formula adding standard enthalpies of fusion and vaporization to that of sublimation is an approximation derived from extrapolation. However, as far as I know, this formula is exact. I think this requires further explanation. Do you know where I can read about all the extrapolation?

Do you think that water vapor at 25 C and 1 atm. is a thermodynamic equilibrium state of water?  See Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics.

Yes, it is definitely in equilibrium with liquid water, and thanks for the reference.


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Which of the following might turn NaNO2 into N2?

Which of the following might turn $\ce{NaNO2}$ into $\ce{N2}$?
A. $\ce{NaCl}$ B. $\ce{NH4Cl}$ C. $\ce{HNO3}$ D. $\ce{H2SO4}$

The answer is B and we have
$$\ce{NH4Cl + NaNO2 = NaCl + 2H2O + N2↑}.\tag1$$
How can we solve this multiple choice question without knowing $(1)$?

To get elemental nitrogen for Sodium Nitrite you need to reduce the nitrite. Only B contains a reasonable reducing agent, the ammonium ion. So it must be B.

The ability to answer questions requires knowledge of facts or principles leading to them.  That is valid in chemistry, in science and all human activities.

You can't. Or rather, you can conclude that (B) is the only compound here containing N in negative oxidation state, which may lead to some beneficial consequences... but that very way of thinking only develops when you know a good deal of chemistry, and in particular a great many reactions, most likely including (1).

To get N2 from NaNO2, NaNO2 must be reduced right, So we need a reducing agent which oxidizes itself and would reduce NaNO2
Now look at the options
Let's see HNO3, the oxidation state of Nitrogen is already $+5$ and since $+5$ is the highest oxidation state, there is no chance of HNO3 to get oxidized further. So it cannot reduce NaNO2.
The same analogy goes with H2SO4 ( S is already in its highest oxidation state $+6$ ), Also Na+ cannot get oxidised further also.
NH4Cl can get oxidised as N's oxidation state is in $-3$, This could be a good way to guess the answer for a multiple choice question, You will also not able to guess between two reducing agents, you should know the reaction anyways.
But as Poutnik did state in comments - "The ability to answer questions requires knowledge of facts or principles leading to them. That is valid in chemistry, in science and all human activities", You would need the knowledge of the reaction to answer such questions.

Wonderful idea! The problem was probably intended to examine knowledge about redox, instead of any particular reaction.


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Justifying that Standard Entropy of a Reaction is a Constant of Temperature
One of the parts in a question I'm trying to solve for an assignment goes like this:

A key issue in the Hydrogen Economy is Hydrogen Storage. This can be restored if a substance $X$ can bind to $H_2$. The binding should be strong enough so that $XH_2$ is a liquid or solid. Necessarily, $X$ must also be a liquid or solid.
Consider the chemical reaction at constant pressure:
$ X(l) + H_2(g) \rightarrow XH_2 (l) $
Write a relation between $\Delta G^{\circ}(T)$ and $\Delta S^{\circ}$ for this reaction. Take $\Delta S^{\circ}(T)$ to be a constant of temperature, equal to $ -85 kJ/mol $. Justify this approximation.

(and then there are more parts)
I know how to write the relation ($\Delta G^{\circ}(T)$ =  $\Delta H^{\circ}(T)$ - T$\Delta S^{\circ}$ since $S$ is constant), but how do I justify the approximation?
I'm confused, because the value given seems to be the one given by Trouton's rule - but if you use Trouton's rule, then entropy of $X$ and $XH_2$ are both approximately equal, so $\Delta S^{\circ}(T) \approx -S_{H_2}$. But entropy of $H_2$, being a gas, would massively change with temperature, and would definitely not be $-85 kJ/mol.$
Does that mean the approximation is wrong, or am I thinking in the wrong direction?

Entropy must have unit J/K, respectively molar entropy J/K/mol.  Additionally, respective Delta S and delta H are either both functions of T, either considered both constant, as they are linked by the difference of heat capacity of reactants and products.

Further parts of the question require calculation of delta H at different temperatures, so they don't expect us to take that as a constant. How, then, do I justify the constant delta S?


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Hydrolyse Phosphonate diester to free phosphonic acid under relatively mild conditions?
I want to hydrolyse a phosphonic acid diethylester to the corresponding free phosphonic acid. In the literature, it is described, that usually strongly basic or acidic conditions are utilized to achieve this.
The Hydrolysis of Phosphinates and Phosphonates: A Review.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC8150351/

Alternatively one can utilize ester cleavage by TMSBr and water workup.
In my case, however, multiple complications occur. R is a complex substituent, which will not tolerate too acidic conditions. In addition, when performing a reaction under basic conditions, the problem is, that base like NaOH or KOH would introduce a very polar species to the mixture in which the product can not separated off easyly from the free phosphonic acid that would be for sure also very polar. (I know, that one possible way get rid of the excess would be the utilization of ion exchange, but from my experience  this is an extremely time consuming process when your product as well your impurities are UV inactive.) In addition to that, reaction control would be problematic due to the fact, that the product is likely to co-elute with the base, therefore the reliable reaction control by LCMS, would be problematic, also because the product is not carrying UV active groups. I am therefore wondering if there are possible other strategies to transform the diester into the acid preferably in a fashion, that does not create a problematic purification situation and work in comparison to TMSBr with milder conditions that are not too acidic.

Do you have any control over what the ester group is? Can you make it something other than Et/Me?

The Et group is a consequence of the precursor used in the previous step, because it contains it. I thing however, it would be possible to exchange Et to a different group before hydrolysis. I did however not try this yet and Methyl is however the only one that should hydrolyse more readily.

I am wondering, if it would not be an idea worth testing to just try a strong organic base like DBU, DBN or a similar one in water, water/methanol mixture. And to see if the product will appear. I would also think to apply microwave conditions could be an good idea to counteract the weaker base's reaction speed in comparison to alkaline bases. In addition, if lucky, an easy DCM water extraction would be enough, to remove the excess base.

Is it possible to swap the phosphonate alkyl group for something that could be removed under entirely different conditions? Dibenzyl phosphonate that could be removed by hydrogenation or diallyl with Pd+Nucleophile?

How about a resin-bound strong organic base?

Resin bound strong organic bases will likely permanently bind to the final product. Observed this on strong anionic exchangers in comparison to weak anion exchangers.

would your substrate stand hydrogenation?

I did not try hydrogenation out yet. Coming back to the benzyl alcohol protection: Introducing this to the precursor Is not so easy however. Transesterrication of ethyl to benzyl alcohol is problematic because it can not be evaporated and high excess is usually used for transesterrication. Maybe a workaround would be to do first all the reactions that form R and than try to replace ethyl by benzyl alcohol, because due to a charge of the product, it could be seperated off from  by anion exchange at this point. But I am not sure if transesterrication of ethyl to benzyl would be easy.

Maybe cyanoethyl protection may also be a way to achieve the free phosphonic acid by soft basic conditions. It is usually cleaved by ammonia or triethylamine. I would need to test, if this group would survive the conditions that deliver the R group.

How many steps back do you have to go to change the starting phosphonate?

The phosphonate utilized 2 steps back determines the structure I want to hydrolyse to the phosphonic acid. The first reaction undergoes basic conditions, the second one water free acidic.

Can you run that step with dibenzyl phosphonate instead of diethyl? You probably have to make it from methyl phosphonic dichloride, as I don't think it is commecially available

I think this would definitely work. The benzyl alcohol group should stay intact during the acidic key reaction step that would form the protected ester and even be better than ethyl in terms of stability. However the trouble would be an access route to the correct phosphonate that contains the benzyl alcohol residues and a third needed functionality. I imagine a route starting from dibenzyl phosphite which is affordable to form the necessary building block in 2 subsequent reactions. This would be manageable as well because these 2 reactions would have fairly good yield.

...I would however need to think and validate, that this additional effort is really unavoidable, if hydrolysis would be achievable by fairly soft conditions I would not be pushed to invest the extra time for these steps. Also, the product must be stable against hydrogenation which I don't know yet, so trying this would be a little bit of a gamble, but it's definitely an interesting thought if the other strategies will not lead to success, especially because Pd/Charcoal should be easy to be seperated off.

@Waylander I checked the dibenzyl idea as a precursor. It seems that I purified the correct structure, but until know it did not react in the predicted way. I will try different conditions to check again on this, but the deprotection step indeed works fast and completely The tlc spot at RF- 0.5 dissapears after Pd/C treatment and a new baseline spot appears.


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How ring expansion is possible here?
Recently I have came across this question.

I went with the 3rd option as ring expansion cannot occur due to the formed stable 3 degree carbocation. But the answer key claims 4th option. How can product through the ring expansion be the major product  here ? As the 3 degree carbocation is stable, and if ring expansion happens, a 3 degree carbocation to 2 degree carbocation transition should occur (although in the end 3 degree carbocation would form) as I have written in the mechanism for this pathway here:

But this pathway has to undergo activation barrier for the 3 degree to 2 degree carbocation transition. So how would this product be major product ? If the starting compound is a 4 membered or 3 membered ring, then the ring stability may be more dominating factor than the stability of the carbocation, but in this case it is a quite stable 5 membered ring.
Links:
Ring expansion in cyclo compounds
Carbocation rearrangement with ring expansion

All steps are reversible until you get the thermodynamically most stable product.

@MathewMahindaratne Yes. But isn't the product through ring expansion a minor product ?

If it's minor would depend on conditions of reaction.


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High-temperature interactions with salts on nichrome wire
I'm working on creating aluminium-doped zinc oxide (AZO) thin films, but instead of the usual glass substrate, I want to try this on a fine nichrome wire.
I'm using a mix of zinc acetate $\ce{Zn(CH3COO)2·xH2O}$ and aluminium nitrate $\ce{Al(NO3)3·xH2O}$ in water. The zinc acetate concentration is $\pu{0.25 M},$ and I'm tweaking the $\ce{Al}/\ce{Zn}$ atomic ratio from $0$ to $10\,\%$. This mix gets nebulized and sprayed onto the nichrome wire, heated up to $\pu{450 °C},$ and then I anneal it at $\pu{600 °C}$ in an $\ce{N2}/\ce{H2}$ atmosphere.
Here's where I need your insight: I'm a bit concerned about how the nichrome wire will hold up under this process. We're talking about spraying salts onto a heated wire. Do you think this could be too harsh and cause corrosion? And how might this affect the AZO coating itself?

Many questions. $1$) Why are you working with such a diluted solution ($0.25$ M in zinc acetate) ? Zinc acetate saturated solution is about $1.5$ M.  $2$) Why spraying your solution on a wire ? Why not on a plate ? $3$) Why not dipping the wire or the plate into a solution before heating it up ? $4$) Heated in vacuum, zinc acetate is supposed to sublimate. Don't you think this sublimation may destroy your coating? $5$) What will you do or measure with your coating after it has been done?

Of more concern than corrosion damage might be inclusion of some of the Nichrome metals in the ZnO, an additional dopant. Nichrome is good to ~1000°C or higher, https://asrichards.com/heater-wire/. A Nichrome heater under borosilicate glass or quartz would be my choice.

@Maurice Good critical questions!
The choice of the process is mainly about minimizing reagents and maintaining control over the film. The classic approach would be to create sol-gel AZO and dip-coat, but this adds reagents. Just mixing the salts into a bath to cover NiChrome results in a uneven and incomplete coat.
Choice of a wire is downstream engineering, we're most interested in getting AZO electrical properties, low ion bleed and high-ish conductivity.
The true test of this is if we're able to get a solid AZO/salty insulator/AZO interface that behaves like known memristors.

@DrMoishePippik Thanks for this, looking into the effect of co-dopants would be a good idea. And honestly, if I could find a reliable source of fine borosilicate (coated) wire - I'd do it in a heartbeat - Inconel/Nichrome wire is a compromise!

Oh, isn't that what a fiber optic cable is?

@ThomasPluck, fiber optic cable is made from very pure glass, not necessarily high temperature softening point, and the glass may be solid or hollow core, but most definitely does not surround a wire. There may be a parallel cable for support (and powering repeaters).

@DrMoishePippik Yeah, I was more thinking fused-silica with a simple buffer like this: https://www.edmundoptics.com/f/buffered-fiber-optics/12870/