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(Beginner Question) Reaction Word Problems
I am having trouble understanding how to create equations which can then be balanced. For example, if I am told that a freshly scraped piece of aluminum reacts with oxygen from the air to form a protective coating - where does the unbalanced equation $\ce{Al_{(s)} + O_{2(g)}-> Al_2O_{3(s)}}$ come from? In that, where does the $\ce{O_3}$ come from in the product? Following a synthesis $\ce{A + B -> AB}$, how am I to know I get 3 oxygens within the product of aluminum oxide, and why can't I say $\ce{Al + O_2 -> AlO_2}$?
Then take the reaction of fresh scraped aluminum with hot water, where the products of reaction are aluminum hydroxide and hydrogen gas: $\ce{Al + H2O -> Al(OH)_3 + H_2}$. How do I actually determine that these are the products in the first place, and how do we get the $\ce{(OH)_3}$ in the reaction equation? Following a single replacement $\ce{A + BC -> AC + B}$, what is wrong with saying $\ce{Al + H2O -> AlO + H_2}$, and then balancing?
Sorry if this is preschooler stuff around here, but will appreciate any guidance.

It is like asking how to learn chemistry from reading few paragraphs. There is no such a shortcut to learn it. // See also Chemistry SE: resources-for-learning-chemistry

"Why Al2O3" — first, you need to know that aluminium usually comes with a 3+ charge, and oxygen usually with a 2− charge. Al2O3 is just the result of combining the two in an electronically neutral fashion. Why those charges, though? It's a result of their location in the Periodic Table, and their electronic configurations. This is very basic chemistry, if you haven't learnt it yet, I'm sure you will very soon, so just keep at it. And if you aren't taught this in whatever course you're doing, get yourself a textbook (it doesn't have to be a big fancy one to explain electronic configurations).

Thanks for the reply. So would the expectation be if only given a word problem "aluminum reacts with oxygen from the air" that I can read the periodic table, and look at aluminum and oxygen to discern their ion charges and determine that $Al_2O_3$ is the only logical configuration?

In the simple cases, yes, that's pretty much it. Sometimes it's more complicated. Explaining the more complicated stuff, though, is the purpose of a full chemistry course, as Poutnik's comment suggested.

Also could you recommend a good (free) beginner textbook to get these concepts down properly? I have never taken a chemistry course, and am trying to learn on my own from high school resources I am finding online. Edit: just saw Poutnik's link, going to check out that list of books. Thank you again.

Posting my findings for any other noobs that might be lurking and just starting out.
Special thanks to user orthocresol for guiding me where to start looking.
For aluminum reacting with oxygen from the air, the wording "forms a protective coating" suggests a synthesis reaction where aluminum and oxygen combine. The charge of the compound must be neutral, so this is how we determine the number of each atom in the product. Aluminum carries the +3, and oxygen the -2, so we need 2 aluminum atoms and 3 oxygen atoms to satisfy this neutrality. This leads to $\ce{Al + O_2 \rightarrow Al_2O_3}$, which balances to $\ce{4Al +3O_2 \rightarrow 2Al_2O_3}$
Likewise with the reaction with water, we can look at the form of the single replacement $\ce{A +BC \rightarrow AC +B}$ and look at $\ce{H_2O}$ as $\ce{H(OH)}$, we can find that $\ce{Al +H_2O\rightarrow Al(OH)_3 + H_2}$, when we consider the aluminum +3 charge, and hydroxide -1, so we need 3 hydroxide to 1 aluminum. And $\ce{H_2}$ as hydrogen gas is a diatomic molecule in nature so will normally always carry this form. This leads to the balanced equation $\ce{2Al +6H_2O\rightarrow 2Al(OH)_3 +3H_2}$

Yep, sounds like you got the basics down!


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Sodium Hydroxide additives?
We use a 10-15% site made sodium hydroxide solution to remove extruded aluminum residue in a boil out tank from steel tooling in an extrusion press. Our old nitrate/nitrite additive is no longer available. Does anyone know of a chemical additive that would increase the efficacy and usable life of our caustic solution? We use the caustic solution at an elevated temp in a tooling bath approximately 100 gallon batches. Without the additive, the life of the batch is around 3 weeks. With our old additive, the life was almost 2 months.

Why add some additive to a $10$% $\ce{NaOH}$ solution ? Such a solution reacts quickly and strongly with aluminum residues from steel tooling. And it will not react with the steel itself. So why ?

@Maurice Well, apparently it does "increase the efficacy and usable life of our caustic solution"... Anything can be made better.

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Why does aromatic sulfonation of 3-aminobiphenyl occur at the para position of the adjacent ring?
I know that the amino group $\ce{-NH2}$ is a strong activator, so the sulfonate group $\ce{-SO3H}$ should be directed at the ortho or para position. However, the sulfonation occurs at para position on the adjacent ring:

Why is that and what effect does the amino group have on the reaction?

See: https://chemistry.stackexchange.com/a/152874/17368

short answer - the amino group protonates making it electron withdrawing and deactivating that ring. For the longer answer see the link provided by @NilayGhosh

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Why is carbon monoxide non-Brönsted-basic in aqueous solution when it has a lone pair on carbon?
Carbon monoxide is not protonatable in aqueous solution; a quick Google search even leads to a paper that talks about using carborane acids, i.e. extreme superacids, to protonate the molecule.
However, there do not seem to be any reason for carbon monoxide to be non-protonatable in aqueous solution- for one thing, its HOMO is localised on the carbon, like the so-called "persistent" carbenes, and the latter are usually just too strong to exist unprotonated even in DMSO, let alone in water. Another problem is that other carbon-centered lone pairs are extremely strong bases as well(e.g. the methyl anion), unless it participates in resonance involving strongly electron-withdrawing groups(such as nitrile or triflyl- even fluorine fails), and I'm not really sure that the small oxygen-centered lobe of the HOMO of carbon monoxide (which is of charge-shift character) counts as "strongly electron-withdrawing".
My main question now follows- what is the reason for the extreme stability against protonation in aqueous solution of the carbon monoxide molecule? Is it simply because of its HOMO's low energy of -14eV (quoting NIST) or is it something else?
P.S. I'm NOT talking about the Lewis basicity of carbon monoxide here- the very reason that it easily complexes with transition metals is that transition metals have their own valence lone pairs to back-donate to the LUMO of carbon monoxide; this cannot happen with the proton, since the proton doesn't even have any electrons to start with, let alone valence lone pairs.

does HOMO mean highest occupied molecular orbital?

@DialFrost Yes, it is.

I think you're somewhat overestimating its low basicity. IIRC one can formylate aromatic compounds using CO and strong acid. If you wanted complete protonation, indeed something like carborane may be needed, though.

Not relevant to aqueous solution but maybe of tangential interest- the HCO+ ion is “often the most abundant cation in molecular clouds” https://www.frontiersin.org/articles/10.3389/fspas.2021.655405/full


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What is the chemical equation of citric acid titrated with sodium hydroxide?
What is the chemical equation of citric acid titrated with sodium hydroxide (titrant), given that the molecular formula of citric acid is
$$ \ce{C3H6O(COOH)3} $$
and the molecular formula of sodium hydroxide is
$$ NaOH $$
Citric acid being the acid and sodium hydroxide being the base
So, $$\ce{C3H6O(COOH)3 + NaOH -> ?}$$

Out of curiosity, does your source claiming "the molecular formula of lemon is $\ce{C36H32Cl4N6O4}$" also provides a molecular formula for, say, a cucumber, cat or a human?

i found the molecular here i am not exactly sure if its correct

That's a pigment, not a fruit. Also, $\ce{NaOH}$ is not a molecular formula, it's a formula unit. Now, what exactly you are asking about?

oh im very sorry about that, Im changing the question to ask about citric acid (as it is the main component of lemon)

The reaction of citric acid (Wikipedia) with sodium hydroxide could be written as
$$\ce{C3H6O(COOH)3 + 3 NaOH -> 3 H2O + C3H6O(COONa)3}$$
or in the ionic form
$$\ce{C3H6O(COOH)3 + 3 OH- -> 3 H2O + C3H6O(COO)3^{3-}}$$

Lemon has no formula. Lemon is a mixture of a lot of substances. Chemically speaking, lemon is a not a pure substance. Only pure substances have formula. Lemon is like all living beings. Living beings are not pure substances. They don't have formula. They are made of cells containing a membrane, a cytoplasm, a nucleus, and many other things which are all made of a lot of different substances.


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Using 3 layer ONIOM(M06-2X:PM7:UFF) over 2 layer ONIOM(M06-2X:AMBER) or 2 layer ONIOM(M06-2X:OPLS) to study enzymatic aldehyde reduction
I'm trying to model the reaction mechanism of an enzymatic steroid reduction. Experimental research proposes a large variety of intermediates, which I believe would each require separate parametrization through AMBER.
Using a PM7 medium-layer, I should be able to achieve a near-linear cost modelling of non-covalent interactions between the SUB/COF.
The reason for using AMBER over UFF is mostly those non-covalent interactions, from my understanding.
PM7, while not the best of levels of theory, should still be better than MM for getting accurate results from what papers I found.
The M06-2X functional is picked due to its performance in both non-covalent interactions and transition state energy calculations.
To note, the only paper that did a comparable study to mine was published 2021 May. Before that, people used (B3LYP/6-31G*:OPLS) for fairly low-accuracy results.
Would potential reviewers in the field take issue to using a deviation from existing methods if my results align with experimental data?

Is the meat of your question the final sentence? If it is, the answer is that you won't know until you submit the publication. The referees will/should be knowledgeable in the field and will offer their opinions.

It is the meat of it, yes. I'm trying to determine whether to invest in doing all the AMBER parametization, or focus on the already involved computation.

At the risk of being biased (I am an AMBER disciple), in general if there is an accepted standard you think the referees will lean towards (and your new method yields superior results), it can't hurt to front-run that and do the comparison now. It will probably come up at some point anyway and best to get it out of the way and establish your new benchmark.

I was working in that field and I am still not accustomed to all of the used abbreviations here. Better explain or introduce them. I'm, however, going with my gut here: Anything is better than B3LYP. Quite importantly will be justification and calibration; if you want to trailblaze, then start doing it right. (And also have a look at xtb, please.) || Anyway, this might have been a better fit at [mattermodeling.se] and future questions of this kind might get better attention/answers there. It's is a good resource in any case.


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State-of-the-art results on molecular vibronic energy calculation
Could someone kindly point me to the latest records in scale, accuracy, or timescale for molecular vibronic energy calculation?
I tried to start with Google Scholar: https://scholar.google.com/scholar?as_ylo=2022&q=%22vibronic+energy%22&hl=en&as_sdt=0,21
From, that link I have found a few papers or work in progress, e.g.:

Mátyus, Edit, and Dávid Ferenc. "Vibronic mass computation for the $EF$-$GK$-$H\bar{H}$$^1\Sigma_\mathrm{g}^+$ manifold of molecular hydrogen." arXiv preprint arXiv:2201.13205 (2022).
Talbot, Justin J., et al. "Dynamic signatures of electronically nonadiabatic coupling in sodium hydride: a rigorous test for the symmetric quasi-classical model applied to realistic, ab initio electronic states in the adiabatic representation." Physical Chemistry Chemical Physics 24.8 (2022): 4820-4831.

I am wondering if there is a faster way to know the latest records.

Comments are not for extended discussion; this conversation has been moved to chat.


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polymerization of 1,3,5-trichlorobenzene
I am interested in carrying out the polymerization of 1,3,5-trichlorobenzene, wondering if anyone had ever successfully performed this polymerization? I would think it might be possible to carry out this polymerization by applying the method of "nickel-catalyzed coupling of aromatic dihalides", which was used in the literature:
Wang, Y.; Quirk, R. P. Synthesis and Characterization of Poly(benzoyl-1,4-phenylene)s. 2. Catalyst Coligand Effects on Polymer Properties. Macromolecules 1995, 28 (10), 3495–3501. DOI: 10.1021/ma00114a001.

Here is a good example to begin with. You may need iodide and/or bromide instead of chlorides.

Also, would you put the correct reference of  "nickel-catalyzed coupling of aromatic dihalides"? I cant find it in 1995 issue 10 of Macromolecules as you mentioned it.

Now what is your question? idgi

@Karl I guess he wants to make this graphene-like structure in the picture. I sure won't work though.

Won't work because reasons I already told here: https://chemistry.stackexchange.com/questions/75948/what-would-you-call-the-polymerised-sheet-like-polymer-arising-from-fittig-react

@Mithoron  is it possible to use monomers such as 1,3,5-tris(chloromethyl)benzene,  1,3,5-Tribromobenzene, 1,3,5-Tris(bromomethyl)benzene, 1,3,5-Triiodobenzene, etc., to carry out the polymerization to make graphene-like sheet structures?

My point is there's no way to make such spontaneous organisation of matter that is not thermodynamically preferred. In your mind you imagine that molecules will bind as you like - in this case they won't.

So, unless you got some really weird enzyme made, you'll just get a mixture of every possible oligomer, i.e. some gunk.

Precisely what will happen: "gunk". ;) At low turnover, oligomers, then a rubbery substance, which finally cures into a solid, unstructured brick.

I had deja vu… so I am obliged to repeat myself: The first coupling will already be so far out of planarity, it is a biphenyl after all, that there would never be a 2D-structure. I believe with the third molecule the whole thing will start to coil. My best guess would be that there might be string-like oligomers.

On the technicality though. I don't think that a question that results in the answer (or along those lines) "this is impossible", warrants closure. Sure it is an unsatisfactory outcome, but I don't see the aspect of being actually off-topic.


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Explanation of the missing 1-s orbital electrons of carbon in the molecular orbital diagram of methane
Consider the molecular orbital diagram of methane, for example found here:
I would like to know what happens with the two 1s orbital electrons of carbon in the molecular orbitals it is forming. Is it that the sigma 1s* anti molecular orbital is having very high energy so it's not filled and hence the full figure actually would be like I have depicted below?

This (my figure) also tells intuitively why in the above shown figure on  starting electrons from the valence shells are getting filled in the sigma 2s bonding orbital first?
(Note : I don't want a deep explanation related to formation of possible combinations of bonding, antibonding orbitals on this case as I just know the knowledge sufficient for molecules of the form like $\ce{N2}$, $\ce{C2}$, $\ce{O2}$, etc.. I do not know about this polyatomic type, I was just intuitively and a bit logical trying to explanain it with respect to the MO diagrams of molecules which I do know. From that I am considering to explain this filling order.)

I have removed the embedding of an external image and tried to fix as much of the language issues I could comprehend. I could not find any copyright information of the image, so I did not include it here. If you want to include it, you should convert it to a format that is supported by the Stack Exchange Image hoster, upload it and cite it accordingly. I have left the image for those who are willing to click links.

I have also down-voted this question. Either you are interested in what is a complete and possible explanation, or you want to reaffirm your own believes. You have explicitly excluded the first, which makes this question utterly useless for most of our audience. Unfortunately what you think is incorrect on all levels of theory, which makes it also useless for most of our audience.

I understood i will fix the image and @Martin-マーチン what was incorrect on all levels of theory ? May you please tell ?

Please cite the original source and make sure you're company with their copyright. With "all levels of theory" I mean from all angles. Your theory is unfortunately incorrect and you need a deep explanation to understand why.

In order for atomic orbitals coming from two different atoms to produce a reasonable molecular orbital, they should have about the same energy. The energy of the 1s electrons of the carbon atom are rather low ($\pu{−37.8 MJ/mol}$), and much lower than the 1s electrons of the $\ce{H}$ atoms ($\pu{-1.31 MJ/mol}$). So their contributions to any molecular orbital are negligible. It looks as if they do not make any molecular orbital. More accurately, the C atoms's 1s orbital would become a nonbonding molecular orbital with nearly the same shape and energy as the C atom's 1s orbital.

May you answer this too regarding this "one thing i was thinking about was even though its very negligible but with which orbitals of hydrogen will the 1s of carbon combine to product molecular orbitals as we can observe that by itself all the four h 1s are foming bonding/anti with the upper orbitals (2s...) of carbons in the MOT diagram"?

$99.9999...99$ % of the four atomic orbitals $1s$ of $\ce{H}$ are forming molecular orbitals with the $2s$ and $2p$ of the carbon atom. The rest, $0.000..01$ % is forming a dim molecular orbital with the $1s$ of the carbon. As a result, this molecular orbital is not different from the $1s$ of the carbon atom.

You might not like the way the browser is presenting the post, but at the very least it is technically correct markup. That is in the sense that it will not break down with one of the next updates if ever they come. I know that this is unsatisfying, but that's the best on offer right now. In any case, MathJax would not have been necessary at all for this post.


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Centre of Invertion/Centre of Symmetry
I would need a detailed explanation of centre of symmetry in organic chemistry stereochemistry. I understand it is a point that conects 2 groups or atomos at same distance of a point. However, what gets me is about the achiral classification of molecule when one finds this center. Every time you find this centre in a molecule you can certainly say that the molecule is achiral? In any molecule and geometry? Please, explain if possible with pictures. Thanks if someone can help.

Welcome to ChemSE! You might want to start here. https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415

A reasonable question to start with is what it means to map a specific structure onto its mirror image. Before you read on, take some time to consider how to express this mathematically.
What is a mirror operation through the $xy$ plane? This takes a point $(x,y,z)$ to a point $(x,y,-z)$.
Also useful to consider what a rotation looks like. A simple 180-degree rotation about the $z$-axis takes a point $(x,y,z)$ to a point $(-x,-y,z)$.
So what is a center of inversion? A center of inversion means that you can take point to the point diametrically opposite through the center, and it should be the same. In other words $(x,y,z)\rightarrow(-x,-y,-z)$ is a symmetry operation.
But if you look closely, this is equivalent to a rotation about the $z$-axis followed by a mirroring over the $xy$ plane.
So what you have just said is that having a center of inversion means you can rotate and reflect the object and make it look like itself. Certainly, then, the mirror image is superimposable.
To be more specific, an object is achiral if it possesses an improper axis of rotation. The improper axis of rotation $S_{n}$ is a symmetry element of rotation about an axis equal to an amount $\frac{1}{n}$ of a full rotation followed by reflection over a plane perpendicular to that axis. As described above, inversion is an $S_{2}$ operation. Since the molecule possesses such an improper axis of symmetry, it is achiral.


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How does the size of a molecule subject to photo-isomerization affect it's isometric configuration?
When azobenzene is photo-excited to a higher energy configuration, the electron orbital take new energy levels and position's, changing the over all configuration of the molecule.
Now, if various phenyl rings were connected together by an azo bond (R-N=N-R’), what would be the resultant configuration after being photo-excited? In other words, how would the geometry of the new molecule change?
There will be several rotational and twisting energy barrier's imposed unto the molecule, resulting in I imagine really high levels of molecular strain. Would this molecule decompose explosively or would the reaction never happen at all?
I am using azobenzene as an example in this case, but any bond that is subject to this type of transformation would fall in the question

Do you know that diimide is R-N=C=N-R, quite different then azo? Do you mean to ask what happens if there are various substituents on the rings? Another issue is that names of compounds aren't capitalised.

Diimide bond's are classifed as R-N=N-R; not sure were you got that middle C in. In any case, the point of the question is very clear and grounded. No need for closing it, as I see two upvotes.

You see two closevotes and R-N=N-R is azo functional group. Diimide group is just what I wrote. Also this is pretty much a dupe of your earlier question https://chemistry.stackexchange.com/questions/153803/how-do-molecules-with-opposite-rotational-photoisomerization-bonds-behave

What exactly is "very clear and grounded"? For ex. "various benzene molecules" shouldn't perhaps be derivatives there? There's just one benzene. " how would the configuration result" what configuration?

I apologize If I came forward too hard. Haven’t had the best day today. I’ll edit the question now to see if these adjustments help. The other question you pointed is similar to this one but as you could see it was not answered nor commented on.

You would only have enough energy to isomerise one bond, but in general if molecule is long it may have each bond just a few degrees from the next so could be arched some what. Have a look at carotenes and related molecules in bacteriorhodopsin or photoactive yellow proteins (pyp) to see what the geometry look like before and after isomerisation.


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A problem with Miller indices
I know how to calculate Miller indices (hkl), but I have found a weird one I can not work out could you please help, I know your you have to expand the plane but after that I’m lost.

See also: https://chemistry.stackexchange.com/q/82937

Because of translational symmetry, every plane has a set of symmetry-equivalent parallel planes (infinitely many). To get the Miller indexes, you have to find the plane closest to the origin that does not pass through the origin. The example shows a plane that goes through the origin. If you translate it by -b, it will cross a at 1/2, b at -1 and c at -1. So it is the 2, -1, -1 plane, also known as -2, 1, 1.


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Why is aminolysis of esters never drawn with an initial protonation step?
I had noticed that the aminolysis of esters is always shown without an initial protonation step:
Mechanism for formation of amide by reaction of amine with ester
https://www.chemistrysteps.com/esters-reaction-with-amines-the-aminolysis-mechanism/
https://www.youtube.com/watch?v=1saLns0_h8M
It is also drawn the same way in Clayden's 2nd edition page 204.
I could be reading too much into this but I found it strange that this was the case because the reaction between ammonia and aldehydes/ketones to form imines can be acid catalysed, and in that instance the reaction mechanism is always drawn starting with protonation of the carbonyl oxygen, rendering the carbonyl carbon more electrophilic.
How is this reaction carried out in the lab? Is there something preventing acid catalysis? One may be inclined to say acidic conditions may cause the amine to be protonated, making it significantly less nucleophilic. But this isn't a problem for the formation of imines, as long as pH does not dip too low.
Of course, this could be overthinking. But I was just curious if there was something deeper.

Adding acid renders the amine much less nucleophilic

@Andrew thanks. But I thought that since formation of imines using ammonia and aldehyde/ketone can be acid catalysed, why not the case here?

You cannot know any mechanism for certain, but in any such question, the question is how would run an experiment to support or refute a hypothesis.
If all of the textbooks say it and assuming it's correct, the acid isn't involved in the rate determining step. I suspect that the amine is more likely to deprotonate your protonated ester on average than do an addition. The acid catalysis may be helpful in getting the tetrahedral intermediate to collapse.


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Clean a filler machine used to dispense epoxy resin
I am dispensing epoxy resin and at times epoxy hardener through a filler machine.  It needs to be cleaned after each run, but rather than do a complete tear down, I'd rather be able to complete a "soft clean" between like materials.  For example, doing a soft clean between two runs of different colors.  By "soft clean", I mean running a cleaner through the machine on a cycle mode where the cleaner is run through a 100+ cycles to essentially remove any standing material within any lines and pipe work on the machine.
I'd also prefer to use something to clean with that is not prone to fires or explosion.  I use things like acetone on tools and such but that can be pretty volatile.  I have soaked things in benzyl alcohol before, but that type of alcohol tends to soften/swell things like rubber o rings and such and I need to avoid that.
Does anyone have a cleaning fluid in mind that may work for me?
Thanks.

You might try a paint stripper, such as Klean-Strip. However, you'd need to determin a few things in your particular appliction.

Is it effective in removing thin layers of adhered epoxy (likely, yes)?
Is it effective in removing blobs of cured epoxy (likely, quite slowly).?
Does it damage seals and gaskets, or even metal parts?

As for flammability, less-flammable alternatives, such as chlorinated or otherwise substituted solvents, tend to be much more toxic and create difficulty when disposing of wastes. Personally, I can deal with care to avoid ignition more easily than working with toxic chemicals.


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What would be the major product of isobutyl alcohol + BF3 with benzene?
I'm having trouble understanding the mechanism for isobutyl alcohol +  BF3 with benzene. I cant figure out the major product

Two things to keep in mind:

$\ce{BF3}$ is a Lewis acid, and we would expect it to form a carbocation with an alcohol. The carbocation would then attack the benzene ring as an electrophile.

That carbocation might not be isobutyl cation. You should be able to see that such a cation would rearrange to the more stable, tertiary, tert-butyl cation with a facile hydride shift. You should identify the major product accordingly.

Carbocation rearrangement is a significant issue with Friedel-Crafts alkylation. F-C acylation tends to give a more controllable product.


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Can ammonium salts react with insoluble bases, such as copper(II) oxide?
Can ammonium salts react with insoluble bases, such as copper(II) oxide? Textbooks and preliminary searches on Google all yield answers that give reaction of ammonium salts with strong alkalis instead.

It can and it cannot, depending on what you mean by "react", the level of insolubility and basicity.

Ammonium salts act due hydrolysis as very weak acids:
$$\ce{NH4+(aq) <<=> NH3(aq) + H+(aq)}\tag{R1}$$
and as such can dissolve such bases:
$$\ce{2 H+(aq) + CuO(s) -> H2O(l) + Cu^2+(aq)}\tag{R2}$$
The above process can be somewhat supported by copper ions affinity to ammonia (for $n = 0\mathrel{..}3$):
$$\ce{[Cu(NH3)_n]^2+(aq) + NH3(aq) <=> [Cu(NH3)_{n+1}]^2+(aq)}\tag{R3}$$
In contrary to my original thought, it happens in considerably extent, as Wikipedia explicitly states (I have overlooked it before) it is soluble in ammonium chloride (credit to Oscar Lanzi to point it out).

Thank you! So will it be safe to say that this type of reaction is not feasible under standard lab conditions? To give some context, I'm actually asking in terms of Cambridge O Levels' Syllabus which loosely defines standard lab conditions as r.t.p. and about 1-1.5 hr.

Wikipedia shows that copper(II) oxide is soluble in ammonium chloride solution. In this respect it resembles metal(II) oxides of magnesium, manganese and zinc.
Dissolution of such oxides can occur by either of two mechanisms. With magnesium oxide and, to a lesser extent, manganese oxide, the oxide is a strong enough base to react with ammonium ions to bring the metal ions into solution and leave a mixture of $\ce{NH3}$ and $\ce{NH4^+}$ in the solution at equilibrium. The solution with this mixture of ammonia and ammonium ions is then slightly alkaline with the $\mathrm{pH}$ in the ammonium/ammonia buffer range. Copper and zinc oxides, instead, rely on forming a complex with the ammonia, leaving essentially this complex plus $\ce{NH4^+}$ ions in a solution that remains acidic.

Hmm, you are right about ammonium chloride.

Hi! Sorry, maybe I should generalize the reaction to just ammonium salts with insoluble bases. To give some context, in the Cambridge O level syllabus, ammonium salts can react with alkalis to form salt, water and ammonia gas. However, I am unsure if I can generalize this reaction to  include insoluble bases as well.


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Adhesion energy vs free energy of solvation
I am looking to compute the solvation free energy of Pt(111). Experimental data are still rare because the entropic term is hard to measure.
I have found a paper where they use a QM/MM approach to compute the free energy of solvation of Pt(111), they compare their result to the value of 0.32 J/m2 given in Singh, N., Campbell, C.T., 2019. ACS Catal. 9, 8116–8127.
However in this last paper they describe the value as "Water−Pt(111) Bond Energy (i.e., Adhesion Energy)." which in my understanding does not contains any entropic information.
I am confused that the first paper use the term of solvation free energy and compares it to a paper which only computes bond energy, am I missing something?


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Separation of very nonpolar mixtures using column chromatography
I am conducting a synthesis using Buchwald-Hartwig coupling. Below is the TLC result after nearly all consumption of limiting starting material. Th is 2,3,5-tribromothiophene, which is the limiting starting material in my reaction and reac. is the reaction mixture. The left is 254 nm UV radiation, and the right is stain using sulfuric acid/formaldehyde. The eluent is hexane only.

I don't have confidence about what materials does the spots correspond to and even whether the desired product is successfully synthesized or not. The chances are that some side reactions would occur. Therefore first I am considering separating two spots below the starting material and identify them with MS, IR or something. But Is this possible? Currently the eluent is only hexane but two spots are near the solvent front, and they are too close. Can I separate two spots in column chromatography? Even the third spot is also quite close to two spots that I want to separate.

In this situation, I would have attempted prep TLC to try to isolate some of the potential product...

I agree, prep TLC is the way to go

Given the separation, be careful not to overload the prep TLC, or you're going to need to do it again...

@Zhe I am a beginner in prep TLC, and some text says that use polar solvent. But in this case is it okay to use hexane?

Hexane would be my first try


text:
How to make a thin layer of crystalline barium titanate on copper substrate?
Is it possible to create a thin $\ce{BaTiO3}$ crystal layer on a copper plate by putting the plate in a beaker, pouring in heated hydrofluoric acid that has been saturated with $\ce{BaTiO3}$ powder and letting it cool?
An alternative method could be CVD, though it's more expensive.

There are hydrothermal methods, e.g., https://ceramics.onlinelibrary.wiley.com/doi/abs/10.1111/j.1151-2916.1991.tb06889.x and https://www.researchgate.net/publication/273241146_Preparation_of_Barium_Titanate_Thin_Film_on_Titanium-Deposited_Glass_Substrate_by_Hydrothermal-Electrochemical_Method, and alternate methods for nanocrystalline films: https://ceramics.onlinelibrary.wiley.com/doi/abs/10.1111/j.1151-2916.1991.tb06889.x . These seem a bit safer than working with heated HF.

I don't think your first mentioned method is possible due to a few reasons :

The hydrofluoric acid we're talking about is highly reactive and so, corrosive. It can etch and in fact, dissolve several metals including Cu and the fact that it is heated aggravates the situation leading to degradation of surface of copper rather than deposition.
As you've mentioned it as saturated, there may be some part undissolved suspended in the solution or settled below. By simply just immersing in the solution, there is a possibility that the particles settle out or be just suspended rather than adhering to the surface of the plate.
Even if the particles are deposited on the surface somehow, chances for achieving precise uniform crystalline film is very slim.

On the other hand, the second method though not pocket friendly it gets the job done. It even offers greater control over the film thickness,
composition and crystallinity.


text:
Why is this diagram depicting the molecular orbital (MO) basis for a back-side attack the way it is?
Consider:

The description of this image in my textbook is as follows:

In order to form a bond, the HOMO (the highest occupied molecular orbital) of one species must interact with the LUMO (the lowest unoccupied molecular orbital) of the other. Therefore, when the nucleophile (Nuc) approaches the alkyl halide to form a new bond, the nonbonding molecular orbital of the nucleophile (its HOMO) must interact with the empty s* antibonding molecular orbital associated with the C¬Br bond (its LUMO). Figure 9.1a shows that in a back-side attack, a bonding interaction (the interacting lobes are both green) occurs between the nucleophile and the larger lobe of the s* antibonding MO. But when the nucleophile approaches the front side of the carbon (Figure 9.1b), both a bonding and an antibonding interaction occur, so the two cancel each other and no bond forms. Therefore, an SN2 reaction is successful only if the nucleophile approaches the sp3 carbon from its back side.

For reference, the nucleophile in the example before this is a hydroxide ion, and the electrophile-leaving group compound
is $\ce{CH3-Br}$.
I suppose where I begin to be confused is why C and Br have an antibonding interaction in the first place. At least, that is what I get from the diagram. Am I interpreting it incorrectly? If my interpretation is correct, why do they have an antibonding interaction?
I was under the impression that C and Br had a normal sigma bond, but maybe it changes in the reaction mechanism? My textbook also says that Hughes and Ingold labeled this a concerted reaction from their experimental evidence, but the only way it makes sense to me at the moment is if a carbocation intermediate is formed.
I think the non-bonding and antibonding interaction is throwing me off. How does the interaction work in more detail? I feel like there is something obvious that I'm missing and it could be fairly simple, or that I am just overcomplicating things.

I suppose where I begin to be confused is why C and Br have an antibonding interaction in the first place. At least, that is what I get from the diagram. Am I interpreting it incorrectly? If my interpretation is correct, why do they have an antibonding interaction? I was under the impression that C and Br had a normal sigma bond

(Very loosely speaking) every bonding orbital is accompanied by an antibonding one. In this case, only the bonding orbital is filled; the antibonding orbital is initially empty, and that's why there is a covalent bond which can be talked about. (The C–Br bond order is given by (number of bonding electrons − number of antibonding electrons)/2 = (2 − 0)/2 = 1, right? — if you're not familiar with this, I suggest learning a bit about MO theory before coming back to this.)
The diagrams in the book are not intended to state anything about whether the orbitals are filled or not.

but maybe it changes in the reaction mechanism?

Well, yes, it does. If electrons from elsewhere (in this case, the nucleophile, or hydroxide ion) flow into the antibonding orbital, then the maths above changes: essentially, you start to break the C–Br bond because (again, loosely speaking) the number of antibonding electrons increases. And that's the whole point of the reaction, after all — the C–Br bond is being replaced with a C–OH bond.
(It's a bit of a simplification to talk about integer numbers of electrons, but it'll do for now.)

My textbook also says that Hughes and Ingold labeled this a concerted reaction from their experimental evidence, but the only way it makes sense to me at the moment is if a carbocation intermediate is formed.

The MO description doesn't tell you whether it's concerted (i.e. SN2) or stepwise (i.e. SN1). You need physical / experimental evidence to prove whether it's concerted or stepwise, and then the MO analysis follows on from that; not the other way around. The experimental evidence isn't provided here, so there's no way to evaluate this claim.

So I did learn about molecular orbital theory in my general chemistry courses, but it was rather rudimentary, and primarily covered MO diagrams for heteronuclear diatomic molecules, and even that was brief. I just had not seen an actual model depicting interactions involving antibonding except for a short introduction. But what you said about filling the antibonding orbital makes sense; as the antibonding orbital gets filled, the bond becomes more unstable, which helps.


text:
A question regarding excitation of electrons in atomic orbital
In Bohr's model of an atom, the formula used to find the energy between the 2 orbits and wavelength of emitted photon was valid only for single electron species like hydrogen.In the case of a multi-electron system like in the picture given above will the electron absorb a photon to go from 2s to 2p and also remitt a photon while dexciting from 2p to 2s.There are also elements like sulphur with two excited states thus showing variable covalency but how do the electrons not dexcite from higher energy orbital in a short time but give enough time gap to show two excitation states?Is the dexcitation and remission of photon a phenomenon which can only be seen when an electron goes from one shell to another like from n=1 to n=2 or can it also be seen when electron goes from orbitals and sub shells like 2s to 2p?Since there is an energy difference between the 2s and 2p sub shells there must be remission of photon on excitation but I did not find any online sources to verify this, so I need help.

You have merged all types of atomic theories. What is meant by two excited states and variable covalency.

Like how in sulphur an electron is excited from 3p to 3d and 3s to 3p thus showing two excited states

Imagine an isolated sulfur atom, it will be one excitation/emission at a time. One cannot have multiple excited states and multiple photon emission at the same time. Search Grotrian diagram of simple atoms first.

Is there something you are looking for aside from what I answered? Except for the long dead and totally obsolete and incorrect Bohr theory for the hydrogen atom (and atomic ions having only 1 electron), the 2p level is not the same energy as the 2s level, so a photon can be absorbed, if an electron is promoted from the 2s level up to the 2p level. And vice versa: a photon can be emitted if an electron drops down to the 2s level from the 2p level. In my answer, I show that yellow light from excited sodium atoms arises when electrons drop from the two 3p levels to the 3s level.

The OP asked

Is the dexcitation and remission of photon a phenomenon which can only be seen when an electron goes from one shell to another like from n=1 to n=2 or can it also be seen when electron goes from orbitals and sub shells like 2s to 2p?

A simple answer can be provided by examining the 3s and 3p levels of sodium atoms. As is well known, adding a little table salt (sodium chloride) to a flame causes the flame to emit fairly intense yellow light that is characteristic and diagnostic of sodium. This photograph shows the sodium emission from my homemade alcohol burner, with the fuel (70% isopropanol/30% water) ‘salted’ with two pinches of salt.

In the flame, sodium ions get their electrons back and some of the resulting sodium atoms get excited by energetic collisions. This results in some sodium atoms having electrons in their 3p levels, as per the following simplified Grotrian diagram:

Figure source: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html
An excited sodium atom cannot stay excited indefinitely: it must de-excite and return to the ground state. The next figure shows that sodium de-excites from the 3p levels to the 3s level by emitting the famous yellow sodium D lines:

Figure source: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html
How do we know that the yellow light is actually two slightly different colors (wavelengths) of light, as in the above figure? The answer is that we perform spectroscopy on the light: we send the light to an instrument that disperses light, similar to how a simple prism or grating disperses sunlight into a rainbow of constituent colors.
To do the necessary spectroscopy, we first need a source of excited sodium atoms. There are numerous ways to get excited sodium atoms and a particularly useful and convenient method uses a sodium hollow cathode lamp (HCL). My photograph below shows a sodium and potassium HCL with neon fill gas. (The red color is intense and is due to the neon, similar to a neon light.) The photograph shows the light being collected by a lens, to focus it on the end of an optical fiber at the left. Note in the first figure that the optical fiber also appears just to the left of the flame. So the optical fiber can be used to collect light from the flame, as in the first figure, or from the HCL, as in the figure below.

The other end of the optical fiber serves to provide light as the input to a spectrometer or spectrograph. The next figure shows my homemade echelle spectrograph, with the optical fiber at the upper left.

Input light from the optical fiber is collimated and sent to the echelle grating. Diffracted light from the echelle grating is then cross-dispersed by the prism and the camera records the resulting spectrum, called an echellogram. The next photograph is a composite of two echellograms I obtained with the HCL setup shown in the previous figure.

The pair of sodium D lines are indicated in the echellogram. All the rest is mostly due to neon, with potassium also contributing some emission lines.
Finally, the next composite photograph compares the sodium D lines obtained from echellograms with HCL and flame emission sources. The D lines are narrowest in the HCL and are somewhat broadened in the flame.

Added echellograms, to illustrate how they show spectral content.
White LED echellogram:

Shorter wavelengths are at left and, for a given grating order (“arc”) at bottom. This is a continuum spectrum: there are no atomic lines in an LED.
Echellogram of the sun:

showing Fraunhofer lines in absorption, e.g., H alpha (the red Balmer line), sodium D lines, the magnesium triplet, etc. This is a continuum spectrum with missing dark spots due to absorption of specific wavelengths: the Fraunhofer lines.
White compact fluorescent lamp:

showing mercury atomic emission lines and the red and green broad bands due to the rare earth-doped phosphors on the lamp’s glass. This is a continuum spectrum with some  mercury atom spectral lines.
A longer exposure echellogram of the sodium and potassium HCL emission, showing the very bright (very over-exposed) red neon spectral lines. I suppressed them in the previous composite photo by using a blue filter. This is a discrete line spectrum due to sodium, potassium and neon. Maybe also impurities: the HCL is old.

Sorry. I have not understood what are the eleven curved red lines, and the same number of double yellow lines. Why eleven, why curved, and why are they separated by the same distance ? I had always believed that in a spectroscope, the light ray is decomposed into one optical spectrum, and only one. It is even worse for the last picture, where the colored curves contains now bright spots. Please explain ! I am lost !

@Maurice In an ordinary spectroscope, things are exactly as you say: a long spectrum of intensity versus wavelength (or wavenumber). But the echelle spectrograph achieves high resolution by using an echelle grating in high orders and then cross-dispersing the otherwise overlapped orders. So it is kind of like having sentences on a page rather than one very long sentence. It would take many comments to explain this in detail, so I will add some echellograms of other light sources, e.g., LEDs, a white compact fluorescent lamp, etc. I will link to a physics SE answer as well.

@Maurice Here is a link to the answer by someone who knows what they are talking about: https://physics.stackexchange.com/a/515156/313612. The “curves” are grating orders. The “spots” are atomic emission lines, though maybe a couple are ion emission lines: I have not done line assignments yet. The non-spots are broad emissions of some sort. The curving is due to the diffracted rays traveling through different thicknesses of the prism (it is 50 mm on edge). One neat thing about this is I can remove the camera, put my right eye as close to the prism as possible and see the D lines! Exquisite!

@Maurice Maybe my answer at https://astronomy.stackexchange.com/a/49116/45954 will help clarify how echelle spectrographs are used for high resolution spectroscopic work.

I am sorry ! I do not understand the beginning of the beginning of all you are talking about. I don't know what is an echelle, a broad emission, a curving, a dekker, etc. I guess it has not too much to do with chemistry. And here we are supposed to discuss about chemistry

@Maurice Well, as a lifelong chemist, who did analytical laser-based spectroscopy and is now an emeritus chemistry professor, I think chemistry is a pretty broad field. Chemists use whatever tools they think most effective, regardless of whomever invented them, to address chemistry problems and solve them. People have been doing spectroscopic analysis since before Bunsen and Kirchhoff and they started off with simple prisms. In the case of Fraunhofer, he used a modified theodolite and a prism. I illustrated this answer by showing that the sodium D lines can be observed spectroscopically.

I perfectly green about all you are saying. And I perfectly know spectroscopic analysis before and after Bunsen and Kirchhoff, with prisms or diffraction gratings. I am even teaching it at the high school level. But I have never seen these "echelles", "dekker", high order diffraction, not to speak of cross dispersing the otherwise overlapped orders. Sorry !

@Maurice    Fair! So echelle gratings are simply gratings designed to work in high diffraction orders, to achieve high spectral resolution. But the high orders severely overlap, so cross-dispersion saves the day. No need to worry about that. A dekker is just a thing that allows a user to alter the length of a slit in a spectrometer or spectrograph. Please look at this link showing a little of what French expert Christian Buil does with his echelle spectrograph. No need to deal with details: it is simply beautiful science!


text:
Reaction between NaCl and NaHCO3
Does NaHCO3 (aq) react with NaCl(aq)? I don't think they should undergo any reaction but in the online database it is showing that they react to form Na2CO3 and HCl. Is it correct?

Can you provide a link to the database search result?

Use other online databases, unless you have misinterpreted the database info. Quoting the info would help.
Mixing $\ce{Na2CO3}$ and $\ce{HCl}$ leads to the possibly violent reaction releasing big volume of CO2.
$$\ce{Na2CO3(aq) + HCl(aq) -> H2O(l) + CO2(g) + 2 NaCl(aq)}$$
Be aware that $\ce{NaHCO3(aq)}$ and $\ce{NaCl(aq)}$ mean $\ce{Na+(aq)}$, $\ce{HCO3−(aq)}$ and $\ce{Cl−(aq)}$, all being present side by side in sea water and human blood.
$\ce{HCO3−(aq)}$ does not react with $\ce{Na+(aq)}$ nor $\ce{Cl−(aq)}$, but it does react with water via several acido-basic reactions:
\begin{align}
\ce{HCO3-(aq) + H2O(l) &<=> H2CO3(aq) + OH-(aq)}\\
\ce{HCO3-(aq) &<=> CO3^2-(aq) + H+(aq)}\\
\ce{H+(aq) + OH-(aq) &<=> H2O(l)}
\end{align}
The fact that there are present ions $\ce{Na+(aq)}$, $\ce{CO3^2−(aq)}$, $\ce{H+(aq)}$ and $\ce{Cl−(aq)}$ does not mean there is formed $\ce{Na2CO3}$ and $\ce{HCl}$. Significant presence of $\ce{Na2CO3}$ would mean the solution is strongly alkalic while the same for $\ce{HCl}$ would mean the solution is strongly  acidic.
It is neither of that, it is just mildly alkalic because of $\ce{HCO3-(aq)}$. $\ce{CO3^2−(aq)}$ and $\ce{H+(aq)}$ are there just in traces and are independent on $\ce{Na+(aq)}$ and $\ce{Cl−(aq)}$.

No reaction will occur for those two compounds.
They are both salts that will dissociate in water. The $\ce{HCO_3^-}$ ion will form a bland ph buffer, but nothing so extreme.
If any molecules of $\ce{HCl}$ will form, they will be instantly dissociated since hydrochloric acid is a strong acid and very soluble in water


text:
Concentration of copper ions in refining of copper metal during electrolysis
When refining copper, would there be a circumstance when the concentration of copper ions in the electrolyte drop? I only know that when refining copper the color of the solution should remain the same blue color.

If the copper to be refined is an alloy of copper plus some less electronegative metals, like zinc, both metals are oxidized and dissolved at the anode. But only copper is reduced at the cathode. So the solution contains more and more zinc. As a consequence, it contains less and less copper. And the solution becomes less and less blue.


text:
Why am I getting a triplet instead of a singlet on NMR for aryl HH2-group?
I am trying to interpret an $\ce{^1H}$-$\mathrm{NMR}$ spectrum on the molecule 4-amino-N-(2-chlorophenyl)benzenesulfonamide that I produced in a lab session. I have managed to assign all the hydrogen atoms to their peaks with reasoning but one, the 4-amino-group. The only peak left is a triplet, which has to be signal due to the 4-amino but this should be a singlet as there are no neighbouring hydrogen atoms. I have used software to generate a $\ce{^1H}$-$\mathrm{NMR}$ spectrum and it is coming up with the peak at the same position as my lab generated $\ce{^1H}$-$\mathrm{NMR}$, but as a singlet. Why is this triplet present in my lab $\mathrm{NMR}$ spectrum?
Is this to do with possible hydrogen bonding between the $\ce{NH2}$ and water? I do have a slight broad peak at $\pu{4.1 ppm}$.

Edit (02/04/22)
ChemDraw predictive NMR:

Lab generated NMR:
[

Er, I'm not sure if there's enough info in your question to properly answer it. It would be really, really helpful if you could attach the spectrum (an image of it) and your assignments, or if you can't, at least provide the shifts and couplings you extracted. Please specify the solvent as well. I will assure you, though, that you are not observing coupling between NH2 and water (observing intermolecular J-coupling is, for all intents and purposes, not possible, and I expect it's extra impossible if you have something which is exchanging rapidly, as is the case for NH and OH groups).

[...shifts, couplings, and integrals too, please. Also, how did you assign this? Did you use 2D NMR or are you just looking at the 1D multiplet structure? Lastly, how certain are you of your compound's purity? Sorry there are so many questions - but this is what I mean - I don't think it's possible to give a definitive answer without more details.]

I added some NMR spectra: 1) NMR generated from ChemDraw which shows a singlet at 6.06 ppm for the NH2 2) NMR generated from the lab shows a triplet at 6.6 ppm for the NH2.
The solvent used was d6-DMSO.
The shifts, couplings and integrals can be found on the second spectrum.
I did not use 2D NMR.
Purity could be a major factor, this analysis was part of a two-step synthesis and the intermediate is also shown in the second spectrum. My initial thoughts was that this triplet came from the intermediate CH3, but I'm assuming this would produce a peak much more upfield, not at 6.6?

Looks more like a minor product... how did you integrate the spectrum?

Also, ChemDraw NMR prediction is not as trustworthy as you may think it is -- especially for exchangeable protons.

Using a ruler is not only primitive, but also terribly inaccurate. In the old days, people used to print the spectrum on paper, cut out the peaks and then weigh them. That's primitive, but probably more accurate than a ruler. The more modern way would be to process the spectrum using software on a computer. Anyway, just eyeballing your spectrum, I'm pretty sure that the peak under question is not 2 protons, considering you have other peaks in the aromatic region which are larger and you've assigned as 1 proton.

Sorry, just updated the second spectrum and I've added the integration trace on.

Aha, ok; if you have those traces then a ruler is just fine. But my point stands - how can that be 2 protons?

Yeah, when I looked at the integration line, the peak at 6.6 ppm is definitely not 2 protons. But does that mean the peak at 9.8 ppm the NH2 then?

I think it's rather more likely that the amine NH2 is the very broad peak at 4 ppm. Typically, free amines/alcohols give broad peaks, sometimes to the point where you can't even see them in the spectrum. The one at 9.8 ppm is more likely to be the sulfonamide NH. Please do check your integrations again -- you can't have a peak that's 1 proton and corresponds to both H6 and H6', for example. That should integrate to 2 protons (i.e., everything should be multiplied by 2).

Yes, I just realised! Is this supposed to be 2H, d, H6 and H6' then 2H also for H7 and H7'?

Yeah that would make more sense to me. The peak at 9.8 would then look more like 1H (as you'd expect from the sulfonamide NH).

To understanding the OP's problem of corresponding $\ce{^1H}$-$\mathrm{NMR}$, it was nice exchange of comments between OP and orthocresol. I agreed of all advices  given by orthocresol. Yet I'm bothered by the large and significant signal at about $\pu{2.0 ppm}$, which was assigned as $\mathrm{DMSO}$-$d_5$ signal by OP. However, it is known fact that $\mathrm{DMSO}$-$d_5$ signal appeares around $\pu{2.5 ppm}$, which is exactly in the spectrum but errouniously assigned as signal of $\ce{CH3}$-group of intermediate with a question mark by OP. I donn't agree with these assignment of OP and my thoughts are given as follows:
I think OP's deacetylation experiment didn't work at all (only a trace of product appears). My reasoning is due to the appearence of two-proton doublet at around $\pu{7.7 ppm}$ due to the $\ce{H}$-7 and $\ce{H}$-7'. If acetamide group is removed, the lone pair of free $\ce{NH2}$-group is available and mesomeric effect would sheilded $\ce{H}$-7 and $\ce{H}$-7' positions (increasing electron density) more than that in phenyl nucleus. I'd expect that signal to be more at around $6.5$-$\pu{7.0 ppm}$. Therefore, I think the $\ce{^1H}$-$\mathrm{NMR}$ spectrum provided by OP is simply the unreacted starting material (acetamide intermediate) as I assigned the peak as follows:

I think the $\mathrm{NMR}$ signal due to acetamide proton $(\ce{CH3CONH}-)$ should be burried in $7.0$-$\pu{7.5 ppm}$ reagion. It is probably the sharp singlet around $\pu{7.25 ppm}$ to my best guess (comparision of integrations at $7.0$-$\pu{7.5 ppm}$ reagion and $7.5$-$\pu{8.0 ppm}$ reagion would give the confirmation).

Yup seems very sensible! I didn’t look too closely at the aromatic region, bit of a mess, but the DMSO peak is definitely off & I didn’t spot that.


text:
Separation methods
I put some iron nails in water, and obviously, those started oxidizing, making those hydrogen bubbles. Some days after, the water was orange, indicated that the FeO has just dissolved in it. I found some days after that the solution was oversaturated, and the insoluble FeO was at the bottom of the container. I need all the FeO, even the one solved. If I heat and evaporate the water, will I lose the FeO? Or will the water leave it alone to use? I think I would get steel 0.2, but I am not sure.

Corrosion works completely different then you think...

I would be very thankful if you can give me some investigation to learn about the topic

Well, there's plenty of resources on the internet. My concern is that I don't get what are you really asking about. Whatever "would get steel 0.2" is supposed to mean in particular?

I am asking if I can get FeO or Fe2O3 evaporating the water, because I have seen in some information sources that it will get a steel-looking compund, so I don't want to get some other thing, I want the ferrous or ferric oxide.

You just get rust, and should use better sources.

Like which one? The most important thing is to get some of thoose compounds. I have seen electrolysis, but I am not sure. Also heard about putting the iron in HCl. Can you tell me how can I get it fast and pure? (the most effective method)

The title is too vaque and not related to the post.

It is not FeO ( ferrous +2  iron), ferrous is unstable in the presence of air. And FeO is black as high temperature mill scale. You have a ferric ( +3) compound , likely some kind of hydroxide. The lab I worked in analyzed ( XRD) at least one "rust" sample a day for corrosion evaluation. The iron compounds were generally not easy to pin down to a specific compound using the Fink XRD Index.

So, that dust in the bottom of the recipient is ferric hydroxyde? (Fe(OH)3)?

If it is dark brown, the stuff is $\ce{Fe(OH)3}$. If it is black, it is $\ce{Fe3O4}$. If it is pale brown, it is $\ce{FeO(OH)}$.


text:
Why can't all chemist/IUPAC agree to make equilibrium constant as dimensionless?
I mean the gold book defines two terms as the equilibrium constant - the standard/thermodynamic equilibrium constant and generic equilibrium constant.
To me, only the thermodynamic one seems correct cause it is derived from thermodynamic factors or activities and Gibbs free energy. The other one just seems like a senseless ratio/pedagogical tool to teach this concept of "Law of Mass Action" which itself doesn't have any thermodynamic basis but only a kinetic one.
So, is the second one useful into some other parts of chemistry, if not why can't we do away with it.
Or is it because kinetic equilibrium without thermodynamic equilibrium is a serious thing.

Why do you think the law of mass action cannot be derived from thermodynamics factors ?

There are cases where the TD K looks like a senseless pedagogical tool looking good in textbooks, while not always applicable in real life.

It is both important when teaching chemistry and also in "real life" to understand that from kinetics we can get equilibrium constants. I have seen how using kinetics data for gold chloride complexes interconverting that we can calculate the stability constants for complexes such as AuCl4-.

if you ever use $\Delta G=-RT\ln(K_{eq})$ then $K_{eq}$ has to be dimensionless, which means that all equilibrium constants have to be dimensionless. It is a historical/convenient  thing that units are still given, and we implicitly assume that the $K$ are divided by 1 unit to make them dimensionless. The Gold book  definition gives them as dimensionless.

@porphyrin what you are saying is just the end results where we have already defined relation between chemical potential and activities (which are dimensionless), this just strengthen the point that we should do away generic equilibrium constant. I am looking for some strong counter arguments. One good one is by Karsten Theis below.

@Maurice if you know a thermodynamic derivation of law of mass action without defining chemical potential and activities. Please provide it. Afaik law of mass action is part of kinetic domain.

The three $K$'s of biochemistry
It is common to use $K$ with a unit in biochemistry. There are three main uses, and the unit customarily has the dimension of a concentration (e.g. $\pu{mol L-1}$).
$K_\mathrm{d}$: Dissociation constant, typically of a ligand bound to a biomolecule. You would say "the ligand has nanomolar affinity" to express that the dissociation constant is about $\pu{1e-9}$. This means that at nanomolar concentration, there would be a significant fraction of ligand-bound biomolecule (with the assumption that the ligand is always at higher concentration than the biomolecule).
$K_\mathrm{i}$: Inhibition constant, essentially the dissociation constant of the inhibitor binding to an enzyme. When the inhibitor concentration is equal to the inhibition constant, more than 50% is bound.
$K_\mathrm{m}$: Michaelis–Menten constant, a combination of three rate constants. This describes a steady state of an enzymatic reaction. When the substrate concentration is equal to $K_\mathrm{m}$, the enzymatic rate is half-maximal (compared to saturating concentrations of substrate).
In all three cases, you assume a bimolecular association event with no cooperativity.
Here is a snapshot from the Table 1. Recommended Symbols and their Units [1, p. 290]:

\begin{array}{lll}
\hline
\text{Symbol} & \text{Meaning} & \text{Customary unit} \\
\hline
K_\mathrm{i} & \text{inhibition constant (inhibition type unspecified)} & \pu{mol dm^-3} \\
K_\mathrm{iA} & \text{inhibition constant for A} & \pu{mol dm^-3} \\
K_\mathrm{ic} & \text{competitive inhibition constant} & \pu{mol dm^-3} \\
k_{ij} & \text{rate constant for step from}~E_i~\text{to}~E_j & \text{as}~k \\
K_\mathrm{iu} & \text{uncompetitive inhibition constant} & \pu{mol dm^-3} \\
K_\mathrm{m} & \text{Michaelis constant (or Michaelis concentration)} & \pu{mol dm^-3} \\
K_\mathrm{mA} & \text{Michaelis constant for A} & \pu{mol dm^-3} \\
K_\mathrm{s} & \text{substrate dissociation constant} & \pu{mol dm^-3} \\
K_\mathrm{sA} & \text{value of}~K_\mathrm{s}~\text{for substrate A} & \pu{mol dm^-3} \\
\hline
\end{array}

Relation to thermodynamic equilibrium constant
The relation is loose. For example, even if these are pH-dependent, you would not include the hydronium concentration in the scheme but instead quote the values at a specific pH. You would also combine distinct protonation states in a single constant (for examples for ATP).
Time to wash your hands
As a physical chemist, you might feel sullied by the muddy definitions above. However, they are part of the daily language in biochemical labs and in the pharmaceutical industry. This would be the "A" of IUPAC.
Logarithms and units
The argument and the result of a logarithm do not have units, but measurements do. Somewhere before taking a logarithm, you have to have a step that gets rid of the units. Here are some examples:
$$ \mathrm{pH} = -\log [\ce{H+}]\tag{1}$$
$$ \Delta G^\circ = − R T \ln(K)\tag{2}$$
$$ \ln \frac{k_2}{k_1} = \frac{E_A}{R} (\frac{1}{T_1} - \frac{1}{T_2})\tag{3}$$
Textbooks often are not explicit about what happens to the units. In (1), you have to divide the concentration by the standard state (unless you use activities instead of concentration). In (2), you "dropped" the units when calculating the equilibrium constant. In (3), you get away with rate constants that have units because they cancel out. It would be a fine convention to have equilibrium constants with units, and then divide them by $Q^\circ$, the reaction quotient for the standard state. So whether $K$ is defined as dimensionless or with units is not directly linked to thermodynamic or kinetic definition, it is a convention.
Kinetics vs thermodynamics
$K_\mathrm{m}$ by its nature is defined through kinetics, as is $K_\mathrm{i}$. For determining $K_\mathrm{d}$, however, you typically would do an equilibrium measurement. Combined with the (first-order) rate constant of dissociation, $k_\mathrm{off}$, you can calculate the (second-order) rate constant of association. For non-covalent complexes, this will be fast, near diffusion controlled, in many instances. So there is a connection between kinetics and thermodynamics, but it is not directly related to the choice whether $K$ has units or is dimensionless.
Reference

Nomenclature Committee of the International Union of Biochemistry (NC-IUB). Symbolism and Terminology in Enzyme Kinetics (Recommendations 1981). Eur J Biochem 1982, 128 (2–3), 281–291. DOI: 10.1111/j.1432-1033.1982.tb06963.x.

Other fields have chemistry have similar things, for example Henry's Law constants ($K_\mathrm{H}$) are generally reported with units because there is no single standard definition.

This answer seems to me like just a series of facts without an explantion of the chemistry. One should aim to understand chemistry rather than merely memorise it. I would be more happy if the answer showed the relationship between kinetics and the thermodynamic idea of an equilibrium.

@NuclearChemist I gave it a shot. However, I can not be responsible for your happiness.

@KarstenTheis what i understood from your answer was there are useful kinetic equilibrium without thermodynamic equilibrium where generic equilibrium constant will be useful. Am i correct ? I think it will be better if we distinguished purely kinetic equilibrium from thermodynamic ones (which by definition will be in kinetic equilibrium too).

@KarstenTheis to me the logarithm argument doesn't make sense. In fact it will point to flaws in our thermodynamic theory if some variable came under logarithm and all we have to say is look logarithm are unitless , so let's divide by something to make our variable unitless. That's just a bad theory/argument. It is unitless cause activities are unitless and activities are unitless cause they are defined w.r.t standard states.

@DeepakArya Here is a textbook appendix to ponder: https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470697733.app1

This letter in JChemEd and the reply also discusses the relationship between logarithms and units: https://pubs.acs.org/doi/pdf/10.1021/ed072p954

It is sloppy use of math. DeltaG = -RTlnKeq + RTlnQ  =  -RT [lnKeq -lnQ]  =  -RTln[Keq/Q].  The units of Keq and Q are the same so the expression is unitless.  At equilibrium Q=Keq so deltaG = 0. Under standard conditions All activities are defined as 1. so deltaG = deltaG[0] = _RTln[Keq/1]  still unitless.


text:
Why doesn’t ferrous sulfate appear orange in vinegar?
When I mix ferrous sulfate crystals with water, the solution almost immediately turns orange. However, when I mix ferrous sulfate with vinegar, it doesn’t change colors.
Why is this?

Welcome to the chemistry site of the StackExchange (SE) network.  Be aware that laconic, not elaborated questions are usually closed on the SE.
The site expects that you include the compact summary of your related current knowledge, involving your conclusion of searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself. Effort not shown may be considered as effort not done, possibly leading to the question closure. How do I ask a good question?.

What have you searched and found about salt hydrolysis, ferrous ion oxidation and how hydrolysis is affected by acids?

As Poutnik suggests, water contains a small proportion of dissolved oxygen, about $\pu{10 mg/L}$. These $\ce{O2}$ molecules oxidize quickly some $\ce{Fe^{2+}}$ ions, and produce some $\ce{Fe^{3+}}$ ions which have a brownish color when included in aqueous complexes like $\ce{[Fe(OH)]^{2+}}$ according to :
$$\ce{4 Fe^{2+} + O2 + 2 H2O ->  4 Fe^{3+} + 4 OH^- -> 4 [Fe(OH)]^{2+}}$$In vineggar, or in any acidic solution, the brown complex $\ce{[Fe(OH)]^{2+}}$ is not formed, because the oxidation equation is $$\ce{4 Fe^{2+} + O2 + 4 H^+ -> 4 Fe^{3+} + 2 H2O}$$ The ferric ion $\ce{Fe^{3+}}$ may be included in a water complex like $\ce{[Fe(H2O)6]^{3+}}$, which is not brownish.

Additionally, many impure or old lots of iron(II) sulphate already contain some iron(III) too, which happily hydrolyzes unless water is acidified.


text:
Products formed on the reduction of KMnO4 in aqueous solution
The Latimer diagram for $\ce{MnO^-_4}$ is given below:
$$ \ce{MnO^{-}_4 ->[0.56V] MnO^{2-}_4 ->[2.26V] MnO_2 ->[0.95V] Mn^{3+} ->[1.51V] Mn^{2+} ->[-1.18V] Mn}$$
Then I have calculated the standard reduction potential values for the convertion of $\ce{MnO^-_4}$ to $\ce{MnO^{2-}_4,MnO_2,Mn^{3+},Mn^{2+}and Mn}$.
The values are found to be :
$$ \ce{MnO^{-}_4 ->[0.56V] MnO^{2-}_4}$$
$$ \ce{MnO^{-}_4 ->[1.693V] MnO_2}$$
$$ \ce{MnO^{-}_4 ->[1.5075V] Mn^{3+}}$$
$$ \ce{MnO^{-}_4 ->[1.508V] Mn^{2+}}$$
$$ \ce{MnO^{-}_4 ->[0.74V] Mn}$$
So according to the $\mathrm{SRP}$ values the feasibility for the formation of respective products follows the order:
$$ \ce{ MnO_2 > Mn^{2+} > Mn^{3+} > Mn > MnO^{2-}_4 } $$
So my question is which is the favourable product formed on the reduction of $\ce{KMnO4}$ in acidic solutions. The values suggests that the formation of $\ce{MnO_2}$ is preferred over others. But it is not observed in experimental reactions. Most of the reaction suggests that $\ce{Mn^{2+}}$ is the favorable product.
For example:

In oxidation of toulene to benzoic acid, the total redox reaction is:
$$\ce{5 C6H5CH3 + 6 KMnO4 + 9 H2SO4 -> \\5 C6H5COOH + 14 H2O + 3 K2SO4 + 6 MnSO4}$$
In oxidation of ferrous ions to ferric ions, that is:
$$\ce{10FeSO4 + 2KMnO4 + 8H2SO4 -> \\5Fe2(SO4)3 + 2MnSO4 + K2SO4 + 8H2O}$$

And there are many more reactions that are known to us in which formation of $\ce{Mn^{2+}}$ takes place. So why there is such disprency between experimental and observed facts. Which other factors influences here?  Note: There are some questions related to this topic but none of them has an appropriate answer to this question.

Mn^2+ is preferred in strongly acidic solutions, MnO2 in neutral or mildly acidic/basic solutions and MnO4^2-  in strongly basic solutions. You may want to calculate also a chart of E=f(pH).

And finally a doubt is it not possible to reduce KMnO4 to Mn(III) salts or to Mn ?

What happened between getting the doubt and asking about it?

My first doubt was mentioned in question that MnO2 should form instead of Mn(+2) . Then I got a clarity on it that it also depends on the pH of the solution . Then my second doubt (which wasn't mentioned in the question ) " is it not possible to get Mn(III) salts or Mn by the reduction of KMnO4 in aq soln under any condition like strongly acidic or basic etc ? "

I have meant the second one.  What have you found before asking?

Let us continue this discussion in chat.

I think explanation in here may clear the air.

And here


text:
Importance of equilibrium constant
Here i am going to let out a doubt of mine which has been bothering me since high school about equilibrium constant. Let's say we have a reaction.
$A+B\rightarrow C+D$. When we say reversible reactions, we say that $A$ and $B$ react to form $C$ and $D$ and after that $C$ and $D$ again react back to form $A$ and $B$. When does this reacting back to form $A$ and $B$ occur?
Now,the equilibrium stage of a reaction. What's the importance of the equilibrium constant and why was the constant defined to be $\frac{[C][D]}{[A][B]}$. I have always struggled understanding why the equilibrium constant is defined in that way.

The reverse reaction occurs simultaneously to the forward reaction, if all products are present.
We say the reaction is at chemical equilibrium when rates of both reactions are equal and the net change is zero. The fact this equilibrium is dynamic and not static can be observed via isotopic exchange.
The equilibrium constant is defined that way (variants use partial pressures, molar fractions or activities), because, well, that expression is at the given temperature constant, assuming the equilibrium state.


text:
Explaining the process of boiling a liquid
I am confused by the thermodynamic definition of boiling. It is stated that boiling occurs when the vapor pressure of a liquid is the same as the ambient atmosphere. Now suppose we deal with an open reaction tube filled with water. It is an open system, connected to the outside air pressure of 1 atm (101325 Pa).
Now we heat the water to 373.15 K. The water will begin to boil at this temperature. There isn't any water vapor in the air yet though before boiling (at least not the equilibrium vapor pressure).
So basically, does that mean that a liquid boils when it's value for the vapor pressure function at the specific temperature reaches ambient pressure, and doesn't use equilibrium vapor pressure for its definition?
To clarify, let's look at another example. Water in an open system at 90 °C is exposed to 1 atm of dry air and, let's say, 0.5 atm of water vapor.
To boil, does the water need to overcome the 1 atm of air and the 0.5 atm of water vapor together, or only the 1 atm of air, because the 0.5 atm water vapor is the same species as the water?

B.P. means liquid  is able to push away surrounding gas to use the extra volume  for saturated vapor bubbles.// Is gas pressure caused by permanent gases only or also by present vapours?

Ask yourself this: If you placed into the liquid a narrow capillare as the output of gas source, what pressure must the gas have to form bubbles? Does play role the composition of surrounding gas or just it's pressure? // Now replace the external gas source device by the liquid itself as the vapor generator.

Hi Poutnik, thanks for the advice. If I understood correctly, the gas needs to push away all the gas, so 1.5 atm of pressure, regardless of the composition of the surrounding gas.

So basically, the water needs to overcome the ambient pressure, which consists of the air pressure and the water vapor pressure together.

Sure. Pressure caused by vapour is not different to pressure caused by other gases or by mechanical device.

On another note; is the equilibrium vapor pressure of a liquid at a certain temperature the same, regardless if there’s also air in the space over the liquid or just the vapor? For example at 90C equilibrium pressure is 70 kPa for water, is it any different if there was also air over the water or the same? I would guess it’s the same because air is an ideal gas?

Let us continue this discussion in chat.

At $\pu{90 °C}$ water vapor pression is $\pu{70100 Pa}$. Water will boil if the atmospheric pressure is $\pu{70100 Pa}$, which occurs at an altitude of about $\pu{3000 m}$. If this hot water is on a level with the sea and in an open system, it would not boil. It would be in equilibrium with a partial water pressure of $\pu{70100 Pa}$, and a partial pressure of air equal to $\pu{101325 Pa} - \pu{70 100 Pa} = \pu{31 225 Pa}.$

This has nothing to do with equilibration at the interface with the air above.  When water boils, essentially pure bubbles of water vapor are formed in the liquid below.  In order for these pure bubbles of water vapor to form, their pressure must be high enough to push back the surrounding water (which is essentially at the pressure of the gas above, aside from the small hydrostatic contribution at depth).  So the water vapor pressure must be at the pressure of the air above the liquid for bubbles to form at depth.


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Diffusion of particles across a plane surface

A relatively viscous material is coated with a thin film consisting of gold nanoparticles that can diffuse into the material. We can assume that all particles have a diameter of $\pu{25 nm}.$ Coating can be seen as a completely flat surface.
In an experiment performed at $\pu{30 °C},$ the nanoparticle concentration $c$ was determined at different distances $d$ from the coating after 60 days. The following data were obtained:
$$
\begin{array}{rr}
\hline
d/\pu{mm} & c/\pu{mol dm^-3} \\
\hline
1 & \pu{8.7E-12} \\
2 & \pu{5.8E-12} \\
4 & \pu{1.1E-12} \\
6 & \pu{7.5E-14} \\
8 & \pu{1.7E-15} \\
\hline
\end{array}
$$
In the same viscous material, equal number of latex nanoparticles with radii of $5$ and $\pu{10 nm}$ are added. How far from the point source do we have to go for the ratio (in concentration) between the small and the large latex particles to be $2$ or higher after $1,$ $10$ and $100$ days, respectively? The temperature is the same as in the first attempt.

I began by plotting $\ln c$ against $x^2$ to determine the diffusivity constant $D$ using the equation
$$c = \frac{n_0}{A\sqrt{\pi Dt}}\exp\left(\frac{-x^2}{4Dt}\right).$$
From this I got that the slope $-0.135$ and $D = \pu{3.57E-7}.$
However, I am unsure how to continue. At first I thought that I could use the same equation by finding the ratio $c(\pu{5 nm})/c(\pu{10 nm})$ and then solving for $x$ since both $A$ and $n_0$ would cancel out as they are equal.
But I think that there should be some other way in which I use both the radii and the temperature given as I don't see any other reason as to why they are provided.

@Maurice Please try to accomplish complete edits. Adding a single backslash and changing two braces to curly brackets when there are numerous other issues should not be the point of an edit. Keep in mind the post is bumped each time it gets edited, and such minuscular edits become annoying quickly.

You could try using the Stokes_Einstein law $D=k_BT/(6\pi\eta r)$ where $\eta$ is viscosity, $r$ radius. You have $D$ and $r$ from your data so you know how these are related, i.e. you know $k_BT/(6\pi\eta)$.

The average distance diffused is $\sqrt{4Dt}$ in two dimensions and this could also be useful.

@porphyrin I don't quite understand how using Stokes-Einstein's law would benefit me here, what exactly do I want from that equation? I would think that I somehow want to include the concentration ratio in order to solve the problem


text:
Observation of deflection in the Cathode ray experiment
In the Cathode Ray Experiment used to discover the electron,
(This is what my textbook says (Grade 11) so if there's any additional parameters that I do not know of I apologize)
The cathode ray tube has two electrodes and a evacuation. Under low pressures when high voltage is passed through the Electrodes, a stream of particles can be seen.
Its presence was further confirmed by perforating the anode and by adding a phosphorescent coating of Zinc Sulphide behind it.
My question is since the anode is perforated will we see deflection of electrons towards the edges of the perforations because the eletrons tend to flow towards the anode so i see no reason for it to pass through the holes instead of deflecting towards the perforation boundaries.
Diagram to visualise what i mean :

This is because electrons possess very high speeds in the cathode ray tube and cannot be stopped suddenly.

Yes, the field lines bend some near the holes. That is a small perturbation on the electron trajectory.

I gather you are talking about the canal ray experiment rather as shown in the picture. In the canal experiments, positive charges pass through the hole when there is a reasonable number of gas particles.
The electron was not discovered with the perforated cathode experiments. These tubes were highly evacuated. The "electrons" (unknown at that time), were studied with tubes like these (https://virtuelle-experimente.de/en/e-feld/geschwindigkeit/geschwindigkeit.php):

This virtual experiment shows that the single hole in this diagram is indeed positively charged and one may ask why electron just get absorbed in it? The key point that you remember is that the voltage difference, although it looks like a small battery, is actually several thousand volts.
If the acceleration voltage different between the anode and the cathode is $V$, you can determine the velocity of electrons $v$ as
$v = \sqrt{ \dfrac{2 V e}{m} }$
Now $e/m$ is a huge number for electrons, and $V$ is large, so you can imagine that the electrons are directionally moving with a very very high velocity. Note a negatively charged cylinder over the coil. That coil also keeps the electron in the centered path.

Your explanation is correct. But it is not an answer to the question. As far as I understand, the question was : Why do the electrons cross the holes of the anode and are not deflected so as to touch the edge of the hole ? In my opinion, this phenomena is due to another anode, situated further away.


text:
Why does the ν1+ν4 combination band for methane occur?
I am looking at the normal modes for methane and have attached a figure of them below:

The only IR active modes are $\nu_3$ and $\nu_4$ which I understand because the other two modes do not change the dipole moment of the molecule and hence do not interact with IR.
Now I have also found in my research that there exists a combination band $\nu_1 + \nu_4$.
I understand that combination bands are created when two different normal modes are excited by a given photon.
However, the $\nu_1$ mode of methane is IR inactive so how can a photon excite it to give the aforementioned combination band?

In a combination band you need to make the direct product and reduce it to its irreducible representation (irreps) and see if any symmetry species are ir active by looking for x, y, z in the point group table.  If you make the direct product in $T_d$ of $A_1 $ (totally sym. so changes nothing) and $T_3$ you get $T_3$ which is ir active.


text:
Why does my sea salt have a best by date?
Seems like it should be good for thousands of years, if not millions.

What kind of container is it in?

The best before date often also serves as a lot number. So if there was e.g. chemical contamination at the factory, they could withdraw all packages with best before date between X and Y.

You would not want to buy salt package 1000 years old, would you?

This table (no pun intended) from Morton Salt explains why table salt has a shelf life:
MORTON SALT EXPIRATION GUIDE
Essentially, salt is hygroscopic, which means it absorbs water from the air. Thus, unless it is stored under very dry conditions, the salt granules will clump together over time.
This clumping is inhibited by anticaking agents. According to Morton, in the presence of an anticaking agent, table salt has a shelf life of ~5 years. I.e., under typical storage conditions, it should last that long before it begins to clump. Without an anticaking agent, it's ~3 years.
Interestingly, even if caking weren't an issue, if the salt is iodized, and you want it to remain iodized, it still has a shelf life of ~5 years.  That's because, as the table explains, "iodine can dissipate over time."
Finally, as suggested by Karsten's comment, the cardboard container will also break down over time.

Culinary sea salt does not usually contain iodine or anti-caking agents. In the absence of iodization and preservation with antioxidants, the expiration date is merely a marketing gimmick.

@Farooq: As the link in this answer shows, culinary “sea salt” certainly can contain anti-caking agents.  And as the answer explains, the (potential very slight) deterioration that justifies the “best before” date is not caused by the anti-caking agent, but inhibitied by it.  Best before dates are never meant to be interpreted as “product expires now — discard it”, but just as “past this date, the manufacturer doesn’t promise that the quality will still be up to expectations”

slightly related: What component(s) of instant coffee make it deliquescent?

The container is glass and plastic. If there was an anti caking chemical, shouldn’t it have been listed in the ingredients? Best by date was 2010. My personal experience is that, like wine, it gets better with age.

@Mom'sBasement  (1) If an anti-caking agent isn't listed in the ingredients, and the manufacturer was accurate in their labeling, then there is none there.  (2) Some plastics can release odors into what they contain over time. (3) Storage conditions can vary widely, but it's possible their estimate was too conservative—why don't you contact the manufacturer and ask them their rationale for their best-buy date?

Salt does not expire. It does not decompose under normal conditions, and as you correctly assume, it could sit around for millions of years without undergoing any change. Sitting around for millions of years is actually exactly what it does as rock salt. However, while salt does not expire, other constituents of your salt package probably do; the expiration date is actually of the container itself.
Sea salt is different than table salt. It is less processed and does not have anticaking agents nor iodine compounds added to it. Also, while sodium chloride is hygroscopic, it is not that hygroscopic. It is rarely, if ever, used as a desiccating agent. Sea salt crystals tend to be larger and rougher than table salt particles and do not clump up as easily due to moisture. Therefore, iodine dissipation and bulking of salt particles are not as much of a concern as the other answer claims while ignoring that you explicitly specified "sea salt" instead of "table salt".

Try living on a boat with salt and ocean air and you will find that salt is VERY hygroscopic.  Doesn't care if it has anti-caking agents, is granules or sea salt, the moisture in the air on the ocean gets into almost all containers.  I use a snaplock glass container to get 6 months without clumping.  But to use the salt you have to open the container and that is enough time for the salt to grab moisture from the air.

"It is rarely, if ever, used as a desiccating agent." See Dehumidify Your Home ¶ Also: "I placed salt in a glass, covered it with enough water and left it overnight. To my surprise, the volume of water increased by around 20-25%, the glass is significantly more full." — Scientific Reason for Salt Solution Gaining Volume.

@boatcoder Being hygroscopic is usually assumed to be the property of material at usual room conditions, otherwise all soluble substances would be hygroscopic at 100% humidity. Table salt if being pure sodium chloride does not attract water until relative humidity  > 75% ( T dependent )


text:
Does electrophilic substitution of hydrogen by electrophiles occur on oxygen or on ring of phenol?
In some reactions such as C6H5ONa with alkyl halide, alkyl phenyl ether is formed, i.e, C6H5O-R in which hydrogen(here Na) is substitued by R. In other cases such as reaction of phenol with bromine water, 2,4,6-tribromophenol is formed (just like in most electrophilic substitution reactions, like kolbe schmitt reaction etc.) where hydrogn at ortho/para position is substituted.
So my question is why in first case R (alkyl group for alkyl halide) is not added to ring ortho/para as in case of friedel craft alkylation? Or in second case Br is not added to oxygen atom?
Is there any specific reason as to why in some cases the hydrogen of oxygen atom in phenol is substitued over the hydrogen atom at the ortho/para position of the ring?
Same is true in case of aniline, eg, in acetylation of aniline, CH3CO is added to nitrogen atom and not to the ring...
It would be really helpful if someone could clarify where my understanding is wrong. Thanks!
Edit:
My question arose out of this problem in which the reagent on the left is attacking the one on the right through its ortho position (Carbon labeled as 1 in picture) as per answer which is (A). I understand why it attacks the vinyl carbon (in second molecule) but not why it attacks through the ortho carbon and not through the oxygen atom.

It is ionic compound, so it is Ar-O-  + Cl-R -> Ar-O-R + Cl-.// It is rather nucleophilic substitution Cl by Ar-O

@Poutnik But why does this nucleophilic substitution not occur through the ortho/para position since the negative charge on oxygen is delocalised over the whole ring right? Why only through the oxygen atom? And moreover carbon with a partial negative charge should be a better nucleophile as per me...

In some sense, this  substitution is not about phenyl at all. It would happen with CH3ONa as well. It is not about what RX does to ArONa, but vice versa.

Benzene ring is a very poor nucleophile, similarly as the flat side of a knife is a very poor screwdriver.

@Poutnik I agree, but then benzene ring attack on CO2 molecule in kolbe schmitt reaction should not occur and instead oxygen should attack the CO2 but that doesn't happen right? My question is NOT why oxygen attacks the electrophile in some cases, my question is why oxygen does not ALWAYS attack the electrophile. In other words, why electrophilic addition occurs to the benzene ring...

That is different, the ring and CO2 are planar, CH arrangement in alkyl is tetragonal. Being good electro/nucleophil assumes being able to point focus, without sterical hindrance. That is not easily imaginable for SN2 and the ring acting as nucleophil.// Benzene ring is rather a substrate for electrophilic attack than being nucleofil.

@Poutnik But then why in other cases does the ring atack or in your words act as the substrate? Why does the addition of the electrophile NOT occur on the OXYGEN?

Let us continue this discussion in chat.

As to the product A in your Edit, it is what it is isolated after protonation of the anion formed by the Michael addition. For all we know, the oxygen of the enolate may add in a Michael fashion, but if so, it is likely formed reversibly.

A few rules of thumb:

The electrophile (an electron-poor alkene) is a "soft" electrophile, so we may expect it to be attacked by a soft nucleophile. The electron-rich C-C double bond on your left-side reactant is a soft nucleophile, the oxygen anion is not.

Carbonyls are thermodynamically favorable (relatively, of course). Nature has an opportunity here to form a carbonyl by attacking through the carbon, which would not be possible if the oxygen attacked.

This situation doesn't seem to have anything to do with phenol. The left-side reactant could be considered a diene, but not aromatic. I think in this scenario even the conjugation it has is irrelevant, and it is best viewed as an enolate. Phenol would probably behave differently, because maintaining aromaticity by attacking through the O would be thermodynamically preferred to forming a carbonyl and attacking through the C.

How do you differentiate between a hard and soft nucleophile/electrophile?

@AnkushNaskar That should probably be its own question, but you can familiarize yourself with [Hard-Soft Acid-Base Theory](https://www.practicallyscience.com/understanding-reactivity-with-hard-soft-acid-base-theory/#:~:text=Hard%2DSoft%20Acid%2DBase(,be%20large/polarizable%20and%20uncharged.)


text:
Why is the boiling point of fluorine lower than that of oxygen?
Fluorine boils at -188.1 °C and oxygen boils at -183 °C, but shouldn't $\ce{F2}$ boil after $\ce{O2}$?
Despite being electronegative elements, both are nonpolar molecules and posses dispersion forces as the only mean of intermolecular interactions. Yet, $\ce{F2}$ has more electrons than $\ce{O2}$, and should have stronger dispersion forces. Hence, $\ce{F2}$ should boil after $\ce{O2}$, but it doesn't. Why?

Obviously van der Waals forces are stronger for O2, perhaps because it is more polarizable than F2.

@Poutnik I didn't know that. Can you explain why?

It is quite well known polarizability of fluorine and fluoride is very low, due its high electronegativity and small size.

although there are no stable O4 molecules in liquid oxygen, O2 molecules do tend to associate in pairs with antiparallel spins, forming transient O4 units. from tetraoxygen

Bonding in there is completely different - the two molecules are hardly trivial. There's no point in comparing them like they were two elements of homologous series.

It would be tempting to argue that fluorine is so electronegative and holds its electrons so tightly that their polarizability is reduced, thus so are the dispersion forces in $\ce{F2}$. But upon further review, this does not stand up. We would expect nitrogen, being less electronegative than oxygen, to offer more polarizability still, yet molecular nitrogen boils at a temperature lower than both oxygen and fluorine (-196 °C). The real question is why $\ce{O2}$ boils higher than both neighboring diatomic molecules, $\ce{N2}$ and $\ce{F2}$.
Putting two and two together
What really distinguishes oxygen from its neighbors is its existence as a diradical, which arises from the degeneracy of its partially filled π molecular orbitals. This creates the possibility of interaction between unpaired electrons from different molecules.
Such an interaction is described, in terms of the magnetic properties of liquid oxygen, in this answer. Pairs of oxygen molecules tend to have "sticky collisions" in which they are indeed temporarily dimerized through bonding between their unpaired electrons. This counts as an attractive interaction that selectively occurs upon condensation and it is just enough to favor condensation at a slightly higher temperature than both neighboring, non-radical diatomics.
When one (and one) is enough
Another comparison is possible with the monoxides of carbon and nitrogen, versus $\ce{O2}$ and each other. Both $\ce{CO}$ and $\ce{NO}$ are weakly dipolar as well as having dispersion forces, but $\ce{NO}$ has the potential to dimerize like $\ce{O2}$ whereas $\ce{CO}$ does not. Here is how boiling points compare in this series:

Oxide
b.p./°C
Wikipedia page

$\ce{CO}$
−192
en.wikipedia.org/wiki/Carbon_monoxide

$\ce{NO}$
−152
en.wikipedia.org/wiki/Nitric_oxide

$\ce{O2}$
−183
en.wikipedia.org/wiki/Oxygen

Nitric oxide combines a weak dipole with the dimerization capability between molecules with an unpaired electron, forming a dimer. The tendency of nitric oxide to dimerize, though weak, is stronger than that in $\ce{O2}$, so this interaction produces a larger effect on boiling point with the nitrogen oxide.


text:
What solvent dissolves calcium nitrate but not ammonium nitrate?
I am experimenting with a precipitation reaction between ammonium sulfate solution and calcium nitrate. Calcium sulfate precipitates and can be filtered out easily.
I am wondering if there's a way I can separate other compounds from the solution. If I simply boil away water, I will be left with the mixture of all three solids: $\ce{NH4NO3},$ $\ce{(NH4)2SO4},$ and $\ce{Ca(NO3)2}.$
I suppose the easiest way to extract ammonium nitrate from the mixture would be to find a solvent that can dissolve every other compound except the ammonium nitrate. Is there a way to find solvents that won't dissolve ammonium nitrate, but will dissolve calcium nitrate, for example?
I found comprehensive tables with the data for aqueous solubility, but only a few references about specific substance's solubility/insolubility in non-aqueous media.
I do not have a background in university level chemistry, nor is this question for any publication purposes. I am purely asking from my curiosity.

Simultaneous obtaining calcium nitrate and ammonium sulphate is not possible.

Among other solvents that might prove possible, methanol is stated to have some solubility for NH4NO3 and Ca(NO3)2. (Ref: CRC Handbook) However, there is a very good reason to NOT use methanol with soluble metallic nitrates, especially when the nitrates are acidic (like NH4NO3). The mixture would be highly flammable and very likely susceptible to self-ignition if you stirred it too hard or heated it near any source of sparks or flame. The safety hazards are overwhelming. Any other replacement solvent that is flammable or contains carbon (even if not very flammable) could be a serious hazard.
However, there is a way to use water and get a fairly pure solution of ammonium nitrate. You can calculate the exact amount of NH4NO3 needed to react with a given amount of  Ca(NO3)2. You will, of course, need to know the exact concentration of your ammonium sulfate solution and the exact amount of water in the Ca(NO3)2 (because it is generally supplied as a tetra hydrate, but that varies, depending on whether you use a research grade or a technical (fertilizer) grade).
Now the problem is that poorly soluble calcium salts are generally more soluble in solutions of ammonium salts, and here we have one of the reactants and one of the products being an ammonium salt - you can't get away from it! Or can you? The reason Ca(SO4)2 is (somewhat) soluble in ammonium salts is that the acidity of the ammonium salt converts some of the SO4$^{2-}$ to bisulfate and some more of the calcium sulfate can dissolve. If you add NH4OH to reduce the acidity, you can eliminate the solvent power of the ammonium salt. In this way, you can complete the precipitation of the calcium sulfate.
There will be a price to pay: now the solution is going to smell like ammonia. However, if you do the addition of ammonia carefully, to about pH 7-8, the odor should be manageable. Use good ventilation. Filter off your calcium sulfate precipitate and evaporate your ammonium nitrate solution with minimal heat but plenty of ventilation.
Further purification of the calcium sulfate could be done by simple washing; the ammonium nitrate could be redissolved, filtered and evaporated to produce crystals as opposed to evaporation to dryness.

Both $\ce{Ca(NO3)2}$ and $\ce{NH4NO3}$ are highly soluble in water, but at 80°C or higher, $\ce{NH4NO3}$ is more soluble. (not too much higher, or the ammonium salt would decompose.)
What might be the easiest way to separate them is to evaporate water from the mixture at ~80°C until the $\ce{CaSO4}$ and most of the $\ce{Ca(NO3)2}$ precipitate, decant the $\ce{NH4NO3}$ solution, and evaporate water from it. This would still be impure if done rapidly, but performed slowly enough, the ammonium salt could be further purified by crystallization.