text:
How can I make AgNO3 more basic?
Is there anything that I can add to the AgNO3 solution that will make it more basic without forming a precipitate such as reacting with OH or HCl?
Please be more specific and give an example.

You are requesting to be more specific but could you explain more of this context, why do you want to increase the pH of the silver salt? What is the experimental requirement?

Because I am trying to then react to the solution of AgNO3 with NaCl to check if a more basic solution could affect the amount of precipitate formed because there will be a change in activation energy. I also used HNO3 solution to make the solution of AgNO3 more acidic and it worked so I wanted to see if there will be a difference with a basic solution. Do you also think if I used complexation of ammonia, would that give us a strong base and would that work for my experiment?

The reaction of AgNO3 with NaCl is $quantitative$ as long as the silver ions are in solution (as long as the pH is acidic ,say < 6). By definition, a basic solution implies the presence of -OH ions in water, so silver ions will react with base and NaCl (both). There is no way to circumvent this problem. There is no point in spending time on this problem.

Just one last question for your valuable comments. I want to do a catalyst in my experiment, so then I decided on doing acid-base catalysis and the problem is that I won't be able to do a base as a catalyst (would [Cu (NH3)4(H2O)2] 2+ work or does it also has OH ions?) so in conclusion, do you have any idea for any other way to add a catalyst or if not even a variable apart from temperature and concentration that would probably result in a higher amount of precipitate. I know these are a lot of questions to ask but I have an essay due very soon and you're comments are very useful. thanks!!!

I still do not understand your research topic. What is the topic? Every alkaline solution in water has -OH ions. That is it. You cannot avoid them by any means.

If you dissolve silver nitrate $\ce{AgNO_3}$in water, and add some NaOH solution to make it basic, it will destroy your substance, and produce a brown precipitate of silver oxide $\ce{Ag_2O}$ : $$\ce{2 AgNO_3 + 2 NaOH -> Ag_2O + 2 NaNO_3 + H_2O}$$
But instead of using a hydroxide like $\ce{NaOH}$, you may add some ammonia solution $\ce{NH_3}$. In this case you will produce some $\ce{Ag_2O}$ in the very beginning, but this dark precipitate will soon be dissolved by more ammonia, and a nearly colorless solution of diamminosilver(I) is obtained : $$\ce{2 Ag^+ + 2 NH_3 + H_2O -> Ag_2O + 2 NH_4^+}$$$$\ce{Ag_2O + 4 NH_3 + H_2O -> 2 [Ag(NH_3)_2]^+  +  2 OH^-}$$

That is a bad idea to mix Ag+ with ammonia.

Continuing Poutnik's comment, there is always a threat of forming silver nitride. Also for context, see: https://chemistry.stackexchange.com/questions/43441/is-tollens-test-safe.

@Poutnik and Nilay Ghosh. If you try to evaporate this solution containing diamminesilver ions, silver nitride may be formed, which is explosive. But if you don't evaporate, you can use this solution without danger to produce nice silver mirrors on glass, I have done it with all my students in the lab, without any difficulty. You just have to add some glucose and a drop of NaOH, then to heat at 40°C (not more). You obtain a nice silver mirror. That is the way all mirrors were prepared in the19th and in the beginning of the 20th century. Now they are made in condensing aluminium vapors.

I look at the formation of silver nitride closely.  My take on a photoactivation path, change in NH3 concentration, fostering reverse reaction to Ag2O, activation by light to Ag/Ag2O, a powerful visible light photocatalyst, introduction of the .NH2 radical forming AgNH2, 2 AgNH2 --> Ag2NH + NH3 followed by AgNH2 + Ag2NH --> Ag3N + NH3.

@Maurice I am well aware of the purpose of that solution. But OP did not express the purpose and these solutions are not safe to store. They are recommended to be sanitised by HCl. Ag3N is formed if ammonia concentration is about 1.5 M or above, At cca 0.75 M it is not reportedly formed.


text:
Why are Buckminsterfullerene (C60) solutions pink/purple?
A solution of Buckminsterfullerene $\ce{(C60)}$ is shown below:

According to the April 8th Wikipedia page on Buckminsterfullerene $\ce{(C60)}$:

Solutions of pure C60 have a deep purple color which leaves a brown residue upon evaporation. The
    reason for this color change is the relatively narrow energy width of the band of molecular levels
    responsible for green light absorption by individual C60 molecules. Thus individual molecules
    transmit some blue and red light resulting in a purple color. Upon drying, intermolecular
    interaction results in the overlap and broadening of the energy bands, thereby eliminating the blue
    light transmittance and causing the purple to brown color change.[36]

Now, I'm not too bright in the field of quantum chemistry/electrochemistry, so this doesn't click with me very well. Does anyone have a good way to explain in layman's terms why the $\ce{C60}$ solutions appear pink/purple? I'm very curious why they experience such a definite color absorption when they are made solely of carbon atoms, while most organic dyes have a host of other heteroatoms and functional groups. Also, why do they exhibit a very specific absorption only when in solution? Any and all contributions are greatly appreciated.

https://en.wikipedia.org/wiki/Stacking_(chemistry)

For starters, that wp section is terribly written. "band of molecular levels" is balderdash as much as the term "transmittance".

@Karl Yeah, I knew it couldn't have just been my less-than-perfect electrochem skills that were confusing me.

All pi systems have an absorbance due to the possible electronic excitation from the HOMO to the LUMO. The larger the pi system, the lower the energy, particle in a box and so on. Rings a bell? Benzene in the UV, azulene is blue, beta carotene yellow, etc.

related https://chemistry.stackexchange.com/questions/27609/iodine-color-change-in-different-complexes https://chemistry.stackexchange.com/questions/60055/pi-electron-stacking-how-does-it-work

@Karl Ah, Thanks Karl! That makes more sense. So the particular pi orbital arrangement for C60 allows for a specific green absorption when in solution, and then I guess pi stacking in the crystalline form blurs the line?

In the solid, the molecules are close enough to each other that energy emitted by one can be absorbed by another instead of leaving the solution. Vibrations also have an effect and are different in the solid. Regarding heteroatoms and UV/vis absorption: Pentacene and tetracene are just multiple fused benzene rings (no heteroatoms) in a flat molecule and strongly colored.

@TAR86 I guess there are a few other dye molecules with only carbon. Thanks for the tip!

The solid form of many coloured species appears different to that in dilute solution. Some of this may be due to $\pi-\pi$ interaction in the solid or dielectric effect causing solvent shifts when dissolved, but more likely is due to the huge absorption that occurs in a solid compared to solution. This changes how we perceive the colour as many more wavelengths are completely absorbed than in dilute solution and only relatively few wavelengths reflected back to us. Solid C60 appears as a fine black powder.

@porphyrin Oh, duh! I guess I just was overthinking it. Thanks!
BTW, you might want to consider posting that comment as an answer. ;)

source: Stack Exchange

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

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text:
Converting Chlorine Dioxide to Free Residual Chlorine
Our water treatment facility used Chlorine Dioxide to disinfect the produced drinking water. The analysis results indicate concentrations of 0.2 mg/l of chlorine dioxide at the source. Can we analys residual chlorine in the distribution network? We dont have water quality standards for chlorine dioxide, only residual chlorine. Can I calculate the concentration of chlorine in the water from the concentration of chlorine dioxide?

Some chemistry, first add H2O2 and NaOH:

$\ce{2 ClO2 + H2O2 + 2 NaOH -> 2 NaClO2 + O2 (g) + 2 H2O }$

Source: https://chemiday.com/en/reaction/3-1-0-1114
Continue adding H2O2, as:

$\ce{2 NaClO2 + 2 H2O2 ->  2 NaClO + 2 O2 (g) + 2 H2O }$
$\ce{2 NaClO + 2 H2O2 -> 2 NaCl + 2 O2 (g) + 2 H2O }$

which implies a net reaction of:

$\ce{2 ClO2 + 5 H2O2 + 2 NaOH -> 2 NaCl + 5 O2 (g) + 6 H2O }$

So, slowly add H2O2 until the bubbling ceases. Collect the volume of oxygen and convert to moles. Divide by 2.5 to determine moles of ClO2 and your chlorine therefrom.
Here is a related reference, Titrimetric and photometric methods for determination of hypochlorite in commercial bleaches, which cites the use of hydrogen peroxide and acidic permanganate.

Please have a look at Is ChemiDay a reliable enough source (for inorganic reactions) to be cited on our site?

The action of H2O2 on NaOCl has to be one of the better-known reactions.  The reaction mechanics is actually more complex than many are aware, but the final products are not in dispute.

Also, do not employ concentrated H2O2, dilute solutions are safer especially in forming normal dioxygen (not singlet oxygen).

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what catalyst to use for the reaction between AgNO3 and NaCl?
What is the best catalyst to use for the reaction between AgNO3 and NaCl that would result in a change in activation of energy so therefore more precipitate formed. Please give me the ways I can catalyze this reaction with the methodology on how can I carry it out. please be more specific.

This is not a reaction that needs a catalyst.

No catalyst is needed for this reaction.

Probably not but I want to try a catalyst and see if there will be a difference.

Even if there was a catalyst, it would not affect the amount of precipitate, as catalysts do not affect equilibrium.

The reaction of chloride ion with silver ion is 1:1 and highly quantitative. It is one of the best studied reactions inorganic chemistry. You cannot carry this reaction in basic/alkaline medium by any means. There is no need of any catalyst. This was worked out at least 120 years ago by people who were interested in atomic weight determinations. Theodore Richard got Nobel prize on this subject.

I upvoted, as this is a great answer! I still suspect this is, however, just a camouflage question to extract a 'solution' to a prior inappropriate question.

No need to be paranoid. There are plenty of useless and potentially dangerous questions here posted from all over the world. We don't need to play Sherlock Holmes.

I apologize for the paranoia, as I thought I am seeing either a curious innocent non-chemist wanting to learn chemistry by starting with the synthesis to a high explosive, or someone not at all innocent persistently trying to do the same thing, either scenario likely does not end well, in my assessment.

so by what you mean by the answer is that no other variable could increase the amount of precipitate between Ag and Cl except for concentration?

No variable can increase the amount of precipitate. It is 1:1. For each mol Cl,  1 mol AgCl is precipitated.

I appreciate your valuable comment and everyone else comment that tried to help.


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Indicator error in acid-base titration
If we value a 20 mL solution of 0.25 M acetic acid with a 0.1M solution of NaOH
The end point of the titration is reached when 50 mL of NaOH is added from the burette. At this point, the pH of the solution is 8.81
Since the pH at the end point is 8.81, phenolphthalein, whose pKa is 9.7, has been used as an indicator.
I need to calculate the error associated with using this indicator, because as the turn does not exactly coincide with the pH at the end point, there must be some error.

Phenolphthalein is a fine indicator for this titration, but the answer you seek depends on just how well the titration was done. At the titration end point, the phenolphthalein should be the faintest pink you can see, as distinct from the colorless solutions, and it fades quickly because of carbon dioxide dissolving in the swirled solution. So, if the end point was really pink, the end point error would be larger than it would otherwise have been. Practice helps a great deal.

Also, have a look at this nice answer by @MaxW : https://chemistry.stackexchange.com/a/88650/79678.

The error is not related to the choice of the indicator. It is related to the number of drops, needed to be sure that you have obtained the equivalent point.

@Maurice - RE: "The error is not related to the choice of the indicator." -- That is not true. There is a definite bias in the analysis due to the difference between the pH at the equivalence point and the pH at which the indicator is noticed to change. It would be very unusual for the two pH values to be equal.

@MaxW. You are right. I have not been  precise enough. I should have said that when titrating acetic acid with NaOH using phenolphtalein as an indicator, the main error is not related to the choice of the indicator. It is in the doubt about plus or minus one drop of NaOH added

Or half a drop, as @MaxW  said somewhere else here (and I can’t find it.) Anyway, that is what we used to teach.

For future titration practices, you should never use the buret until the end. Most high quality burets (Class A) are limited to 50 mL (max) or even 25 mL. If it is a made up exercise by a teacher, they should not promote this approach either. You will have far more error from the incorrect volume measurements rather than the choice of your indicator. Also be careful with reporting concentrations: 0.1 M NaOH is not the same as 0.10 M NaOH.
Anyway, one cannot determine the indicator error from the given data. Practically, you would plot a pH titration curve on a graph paper. Determine the equivalence point (= the true point at which the titration is stoichiometrically complete) from the graph. Lets say, you get 40.56 mL (last digit estimated). With an indicator you get an end-point at 40.60 mL. The end-point is the point where the indicator "hints" you the end of the titration. Now the excess concentration of acetic acid calculated via the end-point is your indicator error.


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What are the cheapest basis sets able to model intermolecular dipole interactions inside a metal complex and give good relative energies, with DFT?
As a example, If I have a square planar complex of a cation with both a weakly polar ligand (X) and strongly polar ones, like ammonia and water I expect a bit of stabilization due to formation of a hydrogen bond between ammonia and water, if they are in adjacent positions:
   NH3
   │
X─ M² ─OH2
   │
   X

What would not happen if they were in opposite positions:
     X
     │
NH3─ M² ─OH2
     │
     X

I think it is a bit tricky to model that, as intermolecular interactions are long-range, and most small basis sets approximations start to break down in such ranges. I imagine it's necessary at least some diffuse and polarization functions to start seeing such interactions. So my guess is that one needs to start at least with 6-31+G(d) level or higher. But I'm not sure. If you need do do a large number of such calculations, would it be possible to go lower than that, and still get reliable results, at least in a relative way, say, finding the first structure has lower energy than the second, even if neither value matches experimental data very well?

The proper nomenclature is called cis vs trans. See for example cis-platin. Although not clear, it seems your question is asking if the trans conformer will have any hydrogen bonding in the trans conformation. The answer would then be clearly no.

Therefore, you should be able to easily use all types of DFT and basis set combinations that you wish. If you're worried about long-range interactions consider long-range corrected functionals, and dispersion. Dispersion is practically free so it's usually a good idea to include it.

Thank you, Cody. Trying to make clear my question, I'm not interested just in this particular compound. The square planar complex example has only two possible configurations (cis/trans), what makes it trivial. But in complexes with higher number and  a more diverse set of ligands, it's not trivial. I'm interested in a fast method to ordering a large number of such combinations by relative strength of the interligand interactions.

You don’t tell anything about your available ressources, but if you have the capacities, go for something like Tight-DLPNO-CCSD(T)/def2-QZVPPD//TPSS-D3(BJ)/def2-TZVP.

The resources are small-scale. Desktop machine, 8 cores/16GB RAM.

I don’t know your X but if it is not too large give it a try or try to find reliable downscaling hints from that point on. Maybe some Grimme and Neese papers might give you a good start into reliable energies at acceptable costs.


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What is the relative rate of diffusion of ammonia to hydrogen chloride, both in gaseous states?
I encountered this problem during an assignment:

I answered it as C, but the assignment grader marked it as wrong and indicated that A is correct.
I asked another instructor about it and gave me a different answer: D.
I am confused on how to approach this problem given that I have received different answers from what I consider reputable sources. How do I proceed to solve this?

Hint: Lighter gases diffuse faster than heavier gases!

I suggest you to better change the institution, if possible ;)

@Zenix Having administered multiple choice exams to many general chemistry students over the years, the occasional bad question happens. Example of an inadvertantly bad question: Which compound (of the listed ones on the exam) has the most polar bonds? Some students will look for the listed compound that had bonds that are most polar. That was the intent of the question. But others will count the number of polar bonds per compound and rank the listed compounds that way. Not unreasonable,  actually.

No answer is correct, but "A" is  closest to the correct answer.  Heavier molecules diffuse slower than lighter ones because the same amount of thermal energy has to move more mass.  However, this relation (diffusion rate is inversely proportional to the square root of molecular weight) actually leads to 44 seconds not 64.  The professor who told you "D" probably misunderstood that in the second chamber there is HCl (acidic) instead of ammonia (basic).

I am getting 44s, still not 60s..

It should definitely be slower.  Actualy I got just 60 s at first ... because I mistook the molecular weight of HCl for that of Cl2.  Maybe they also did so.  Good catch, will edit.

@Zenix Yeah, they forgot the square root in Graham’s law. But it was really a logic problem!


text:
Effect of phenyl and vinyl substituents on Acidity of carboxylic
Recently I came across this statement in my textbook:

Direct attachment of groups such as phenyl or vinyl to the carboxylic
  acid, increases the acidity of corresponding carboxylic acid contrary
  to the decrease expected due to resonance effect.

On researching online, I found this statement as an answer:

Although we expect a decrease in acidity due to the resonance effect
  of phenyl and vinyl groups, but because of the greater
  electronegativity of the the $\ce{sp^2}$ hybridized carbon which is
  directly attached to the carboxylic carbon.

My Thoughts
Please keep in mind that the carbonyl carbon(here carboxylic carbon) is also $\ce{sp^2}$ hybridized. So $\ce{sp^2}$ hybridized carbon is directly attached to the carboxylic carbon. Therefore there is no electronegativity difference between the two carbons and as a result of this only +R effect should work and direct attachment of groups such as phenyl or vinyl to the carboxylic acid should not increase the acidity of corresponding carboxylic acid rather they should decrease its acidity due destabilization (due to electron donating nature vinyl and phenyl group by means of resonance) of conjugate base so formed.
The question
Kindly explain it theoretically by giving plausible reasons as to why the statement in my textbook is true?
Note: avoid quoting pKa values

There is sp3-sp2 versus sp2-sp2 difference of electronegativity  differences.

I didn't get you?

sp2  carbon of e.g. acrylic acid  has bigger electronegativity than sp3 carbon of e.g. acetic acid, what leads to the former causing lower carboxyl electron density, compared to the latter, and bigger carboxyl acidity.

Look at the carboxylate ($\ce{O-C-O}$) part of the molecule.  It is an analogue of the allyl system where we have replaced the two terminal carbons with oxygens.  There are 4 pi electrons in the carboxylate part of the ionized acid, so the HOMO is $\ce{\psi_{2}}$ of the allyl system.  Note that in this orbital there is a node at the carboxylate carbon.  This node effectively disconnects substituents attached to this carbon in a pi or resonance sense. Hence any resonance interaction between a substituent attached to the carboxylate carbon and the carboxylate fragment is minimized.

The title of this question is incomplete, and too general. Please add citations for the statements you are referring to.

Could you suggest one please , I'm unable to come up with good one

When the cases of carboxilic acids are considered, we must note that the conjugation of $\ce{COO-}$ group is so strong in itself, that it's conjugation with the rest of the compound is rendered relatively weak. In such cases, the inductive effect is dominant.
Let us take the example of benzoic acid. The carbon to which $\ce{COOH}$ is attached shall be named as the key-carbon $\ce{C}$1. The conjugate base would be $\ce{PhCOO-}$, and the +I(Inductive effect) of $\ce{COO-}$ group will be dominant on $\ce{C}$1. Any group attached to this compound will further increase or decrease the acidic strength of the compound as a whole, depending upon the attached group's contribution towards electron density at $\ce{C}$1.
For example, in case of para-nitrobenzoic acid, electron density at $\ce{C}$1 decreases due to $\ce{NO2-}$'s strong -R (Mesomeric) Effect(partial positive charge is developed on $\ce{C}$1 which is stabilised by $\ce{COO-}$'s +I). Hence it turns out to be a strong acid.
The opposite case is that of para-methoxybenzoic acid. The electron density at $\ce{C}$1 increases due to +R of $\ce{OCH3}$ group and makes the conjugate base unstable, hence acid weaker. This can be cross checked by the $\mathrm{p}K_\mathrm{a}$ values of above mentioned two compounds as well.
Now coming to the compounds mentioned in the question, which are phenyl and vinyl, one must note that both of these would reduce the charge density on $\ce{C}$1 because of the $\mathrm{sp^2}$ hybridised Carbon atom which contributes to the overall electro-negativity of these two groups.
Long story short, the resonance effect between $\ce{COO-}$ and the rest of the compound is suppressed by a relatively stronger Inductive effect which works mutually between $\ce{COO-}$ and the rest of the compound.


text:
Trying to synthesize and crystallize iron (iii) ammonium sulfate
I've been following instructions (from here, the first method) to synthesize iron (iii) ammonium sulfate and crystallize it. The first attempt was a fail, I spilled most of my solution and what I managed to save crystallized green from a dark yellow-brown solution. I tried again, first reacting the iron (in the form of filings) with the sulfuric acid, then adding the ammonium nitrate dissolved in water and allowed to warm to room temp. There was no immediate reaction when I added the amm. nitrate but the solution changed color slowly over the next few days to a dark green-yellow. When crystals finally started forming, they were green. NH4Fe(SO4)2•12H2O should crystallize out as a pink/purple color, whereas the iron (ii) compounds crystallize green, with the regular sulfate being blue-green and Mohr's salt being a darker green. I believe I got the latter.
Basically I'm confused as to why I seem to have gotten iron (ii) ammonium sulfate rather than the iron (iii) variant. I'm going to try again for iron (iii) but with sulfates or iron (ii) and ammonium to start, I just had ammonium nitrate on hand. I can put up pictures of the crystals that I got from any of these solutions, if it might help identify. My plan is to just keep an eye on them. Mohr's salt is stable on air, so I'll watch to see if it degrades over time.
Edit: I noticed the crystals turning brown so clearly I only got iron (ii) sulfate and it's oxidizing to iron (iii) sulfate. I redissolved them in about 25% sulfuric acid and I'm going to recrystallize. To be fair the procedure for the method I used wasn't super clear.

Did all the iron get dissolved ? As there can be ongoing parallel dissolving of iron by Fe(III). ( FeCl3 is used for Cu dissolution in electronic. )

I used a slight excess of iron and after filtering there was only an expected small amount left in the filter. Maybe what went wrong was the amm. nitrate I used wasn't pure so although my weight was correct I was also weighing the impurities.


text:
Why does orbital optimization remove spin contamination?
I've come across numerous articles asserting that orbital optimized methods, where minimization of the energy is achieved through a series of unitary rotations (achieved by exponentiated single-electron excitations from the ground state reference) of the reference determinant, are exceptionally good at removing spin contamination. I've seen the numbers they report, and I believe them! I get the gist that orbital optimization allows us to overcome the error introduced by the UHF orbitals, but nowhere can I find the actual explanation of why these rotations effectively cure spin contamination.
Both references and explanations warmly welcomed.

Could you please list some papers where you saw the numbers?

For example, Neese, Schwabe, Kossmann, et al. J. Chem. Theory Comput. 2009, 5, 3060–3073.

Or Bozkaya, J Chem Phys. 135, 224103 (2011)

I agree with the statement that FCI has no spin contamination, however, it's not necessarily true that if one reference is closer to FCI than other, it has less spin contamination (which is why UHF can produce better energies but lose spin symmetry). One way to think about it is by using the Thouless Theorem to relate any two non-orthogonal references
$$| \tilde{\Phi} \rangle = e^{T_1} |\Phi\rangle$$
where $\langle \tilde{\Phi} | \Phi \rangle \neq 0$ and the single-excitation operator is $T_1 = \sum_{ai} t_{a}^{i} a_a^+ a_i$ where $a$ denotes unoccupied MOs and $i$ denotes occupied MOs. As long as the references are not of different symmetries or spin multiplicities, they are equivalent up to an orbital rotation. This is how orbital optimization works - by changing $T_1$, I can move over the entire manifold of different references. For instance, an optimization condition of $T_1 = 0$ produces something called Brueckner orbitals. This also means that I can rotate from RHF to UHF or to something in between like ROHF. If you read some of the ROHF papers by Scuseria et al. for instance, you can use this framework to define references with spin constraints, such as preserving $\langle S^2 \rangle$ by applying a penalty for deviations from $S = 0$ or $S = 1$. That will easily produce orbitals that are both a good description and free (for the most part) of spin contaminatination.
However, that's all to do with the reference. The paper you've cited is about applying orbital-optimized MP2 which essentially rotates an existing reference and then applies MP2. The reason why is probably works so well is that the convergence of MP2 is very sensitive to the choice of reference. So, by rotating first, you obtain a more suitable reference and then perturbation theory gets you much closer to FCI than usual, hence, decreasing spin contamination for the reason you noted. It could be cool to compare orbital-optimized MP2 to CCSD. The term $e^{T_1}$ is built into the cluster operator $T = T_1 + T_2$ which is why CCSD (and all CC methods) do not exhibit this sensitivity to the choice of reference, thus you would expect that they would have the same or lower spin contamination right off the bat.

Upon further thought, I've summarized it as follows. If anyone comes along with a better explanation, they'll get the cake!
Spin contamination is a fictitious effect arising from not achieving the full CI solution to the Schrödinger equation. For, had we found the exact solution, then the solution we found would be an exact eigenfunction of the ${\hat{S}}{}^2$ operator, as $[\hat H, {\hat{S}}{}^2]=0$.
In a general sense, orbital optimization can be viewed as simply allowing for more correlation to be recovered on top of whatever other improvements you've applied to your initial Hartree-Fock reference. So, by recovering more correlation, you've brought your solution closer to FCI. In doing so, you've brought your solution closer to a solution that contains no spin contamination.
However, there seems to be a stronger statement that could be made. This type of correlation seems especially potent in removing spin contamination.


text:
Why is the experimental pH of vinegar (acetic acid) so much higher than expected?
I bought some sodium hydroxide on Amazon and went about titrating solutions I make with it.
As I'm an amateur chemist at home, I have limited tools. A pad of paper, some 5% acetic acid vinegar, and some pH strips.
I assumed the pH of vinegar was 2.5 ± 0.5.
Making a solution of 4g NaOH powder in 100g of (tap) water, I expected a solution of 1 mol/L. (NaOH 40g/mol)
I went about finding the ratio of concentrations, working by factors of 10 and then refining, that pH strips would indicate around 7.
At my great surprise, this turned out to be around 0.9:1 NaOH 1M to vinegar (about 20mL:22mL – the margin of error being around a factor of 1.1)
A vinegar pH of 2.5 suggests an H+ concentration of 3.2 mM; and so that that ratio would be, $1\mathrm M/3.2 \mathrm{mM} = 312:1$.
So experimentally, $\mathrm{[OH^-]}$ is about 350 times lower than expected. I'm trying to figure out why that would be.
Am I missing something obvious here? I'm looking for a mistake but can't find one.
Here are my thoughts:
Bought NaOH might not be pure
It would take a "purity" of 0.25% to explain such a difference. Unlikely.
NaOH might not dissociate completely experimentally
NaOH is a strong base. pKb would need to be extra high to explain a factor of 350.
Calculation error
I've checked and rechecked my math, everything seems to hold up. 22mL of vinegar brings 70µmol [H+], 20 mL NaOH solution brings 50µmol [OH-].
Measurement error
I've done the experiment twice. Same kind of values. Adding 80% of vinegar volume leaves the paper unmistakably dark. Adding 120% of equilibrium volume leaves the paper yellow-ish. 90% to 110%: green-ish. To measure quantity added, I use a scale with a resolution of ±0.01g. (Measurements were taken ±0.5g)
Bad pH paper
It would take really bad paper to make a pH 9 liquid appear to be pH 7, wouldn't it? A factor of 350 is about $log_{10}(350) \approx 2.5$ off on the log scale. Unlikely. Besides, aren't the indicator substances pretty standard?
Acidic tap water
I live in a big city, I've read the water analyses. Looks like the water is around pH 6.5. I don't think at this scale it would change much.
Wrong chemistry/theoretical background
It's possible a 4% w/v solution of NaOH doesn't have a pH near 14. Or that mixing it with 1.1 parts of 5% vinegar doesn't make a pH of 7. I have not yet found the basis for that. At equal quantities [H+] and [OH-] in room temperature water, I don't see a reason for the weak acid to be a factor.
So I'm pretty much stumped. Why does my solution have a pH 11.5 instead of 14? Is there an obvious chemical concept I've missed?

pH of vinegar is irrelevant to your calculations. It is not the amount of H+, it is the amount of CH3COOH that matters. Related: https://chemistry.stackexchange.com/questions/60407/why-will-a-strong-acid-neutralize-as-much-base-as-a-weak-acid/60409#60409

I see. So the pH of 2.5 is not equal to all the potential H+ it can ever release because that's what a weak acid is, right?

Okay yes, looks like that's pretty much it. So 5% acetic acid (0.83M) and 0.95 NaOH pretty much neutralize each other at a 0.9:1 ratio. - I'd welcome suggestions or edits to make this question more useful for others

The ration  0.9:1 has no meaning  ! When you neutralize acetic acid with NaOH, you not only destroy the H+ ions, as these ions are pretty rare is the solution. Above all NaOH destroys the non dissociated acidic molecules, which are much more abondant in the solution. As consequence, if you want to make calculations about the neutralization of CH3COOH you should forget about the pH. You should simply consider that the number of moles of NaOH is equal to the total number of moles of CH3COOH (dissociated or not)


text:
Transformation of virial coefficients from molar volume to pressure
I have the following virial equation for a real gas:
$Z = \frac{PV_m}{RT} = 1 + \frac{B}{V_m} + \frac{C}{V_m^2}$
I would like to transform the coefficients $B$ and $C$ so the equation is expressed in terms of pressure:
$Z = \frac{PV_m}{RT} = 1 + B'P+ C'P^2$
My book says that the transformations are:
$B' = \frac{B}{RT}$ and $C' = \frac{(C-B^2)}{(RT)^2}$
How are these derived? I can get the first one by:
$Z(\frac{1}{V_m}) = Z(P)$ and only considering $B$:
$1 + \frac{B}{V_m} = 1 + B'P$,
$\frac{BP}{RT} = B'P$,
$\frac{B}{RT} = B'$.
For the second one I also consider C:
$1 + \frac{B}{V_m} + \frac{C}{V_m^2}= 1 + B'P + C'P^2$,
$\frac{BP}{RT} + \frac{CP^2}{(RT)^2}= \frac{B}{RT}P + C'P^2$,
$\frac{CP^2}{(RT)^2}= C'P^2$,
$\frac{C}{(RT)^2}= C'$.
This result (mine) differs to the one given by the book. They have an additional $-B^2$ term in the numerator. Where is that term coming from?

Should not it be rather $\frac{pV_\mathrm{m}}{RT}$ ,or   $\frac{pV}{nRT}$  ?
As either $pV = nRT$ either $pV_\mathrm{m} = RT$
Perhaps you follow a notation from a book, as I have seen this once, but it is wrong, IMHO.

@Poutnik yes, corrected. Doesn't matter - the problem still stands.

In the 3rd equation from a bottom you make substitution V_m=RT/P, but it is  true only if both B=0 and C=0
Try equation indexing by usage  \tag{N} at the end of equation Matjax expression. Like
$$X=Y+Z \tag{1}$$ -->
$$X=Y+Z \tag{1}$$

We want to take this:
$$Z = \frac{PV_m}{RT} = 1 + \frac{B}{V_m} + \frac{C}{V_m^2}+…\tag{1}$$
And transform it into this:
$$Z = \frac{PV_m}{RT} = 1 + B'P+ C'P^2+…\tag{2}$$

We begin by expressing (1) in terms of $\rho_m = 1/V_m$:
$$Z=1+B\rho_m+C\rho_m^2+…\tag{3}$$
We then multiply (2) and (3) by $RT\rho_m$ to get (4) and (5), respectively. Remember that $ZRT\rho=ZRT/V_m =P$, because $Z=PV_m/RT$.
$$RT\rho_m Z = P = RT\rho_m + RT\rho_mB'P+ RT\rho_mC'P^2+…\tag{4}$$
$$RT\rho_m Z= P = RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…\tag{5}$$
Next, we use (5) to substitute for $P$ in (4):
$$P = RT\rho_m + RT\rho_mB'(RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…) +$$
$$ RT\rho_mC'(RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…)^2+…\tag{6}$$
Collecting all the $\rho_m^2$ terms from (5) and (6):
$$RTB=(RTB')(RT)$$
Rearranging leads to the desired $$B'=\frac{B}{RT} \tag{7}$$
Next, collecting all the $\rho_m^3$ terms from (5) and (6):
$$RTC=(RTB')(RTB)+(RTC')(RT)^2$$
Simplifying, rearranging, and making use of (7) leads to the desired
$$C' = \frac{C-B^2}{(RT)^2}\tag{8}$$

There is a derivation here too: https://physics.stackexchange.com/questions/414386/virial-expansion-in-terms-of-pressure


text:
Mechanism of electron delocalisation from the other end of alicyclic chain
When one adds $\ce{Br2}$ and $\ce{CH2Cl2}$ to 5-Methylhex-5-enol, this mechanism follows;

Also,in adding $\ce{Br2}$ to 2,2-Dimethylhexa-3,4-dienol, this follows;

What is the mechanism here, in which the electrons from a hydroxyl on the other side of the C-chain is delocalised to the other side?

I guess you could call it an intramolecular cyclisation

Is there anything particular about the OH group?

I am not aware of some significant naming regarding that


text:
Nitrocellulose (flash paper) synthesis without Sulfuric Acid?
According to the Wikipedia page on Nitrocellulose:

The process uses nitric acid to convert cellulose into cellulose nitrate and water:
$$\ce{3HNO3 + C6H10O5 ->[H2SO4] C6H7(NO2)3O5 + 3H2O}$$
  The sulfuric acid is present as a catalyst to produce the nitronium ion, $\ce{NO2+}$. The reaction is first order and proceeds by electrophilic substitution at the C−OH centers of the cellulose.[8]

Since I have been on quarantine, I have lost access to the lab and therefore must carry out reactions from the comforts of my home. I have some sulfuric acid drain cleaner that is sufficient for many at-home applications. However, it consists of only about 93% $\ce{H2SO4}$, and contains a lot of dark dyes. I am trying to synthesize nitrocellulose paper from letter printer paper and envelopes, for the purpose of developing a self-destructing letter that burns away when the letter is returned to the envelope after being removed. However, the dyes in the sulfuric acid severely discolor the paper, and make it stand out considerably in a pile of mail, defeating the purpose of secrecy. So, at least until the quarantine is lifted, I may need an alternative catalyst to Sulfuric acid to synthesize the nitronium ions in vitro. Does anyone have an alternative to $\ce{H2SO4}$, or a way to eliminate the dyes without distillation? (because I broke my condenser column, and can't use the lab's)
Update: I'm still curious about my question, but it might not even be legal to send nitrocellulose envelopes through the mail. One spark and you could end up with flaming mail processing machines. Yikes!

I think it is quite a dangerous hobby for quarantine, I would avoid going to a hospital at this moment.

The KNO3 process looks interesting. Just remember when nitrating to keep the temperature under control or your experiment will likely flash as it cooks… In the USA, making energetic materials for your own use is permitted—nitrocellulose is commonly shipped by mail. So long as it’s wet it’s inert.

An interesting photolysis experiment, to be performed in small amounts only, as you may actually produce a higher yield of an energetic than you expect with available reagents.
Dissolve KNO3 in dilute H2O2 (or, just water as any released electrons via photolysis can also scavenge the nitrate radical) with a bit of B-12 Vitamin (or, try substituting a B-complex).
Soak cotton in the mix. Flatten out the treated cotton and place it at the center of a plastic bowl lined with aluminum foil.
Place in sunlight until the color vanishes (less than a couple of hours).
Let the treated cotton dry in the dark (no Al foil contact).
Test lite the product.
You may wish to compare the results to only KNO3 treated cotton.
Chemistry:
The B-12 photocatalyst produces electron holes (h+), which react with water creating hydroxyl radicals.
$\ce{H2O2 + h+ -> H+ + .OH }$
The latter reacts with NO3- to form the nitrate radical which may attack the cotton.
$\ce{NO3- + .OH -> .NO3 + OH- }$
Similarly, the H2O2 may be attacked by electrons also released from the photocatalyst. The one-electron reduction of the hydrogen peroxide can also produce the hydroxyl radical as follows:
$\ce{H2O2 + e- -> .OH + OH- }$
A not favored reaction resulting in associated loss of the nitrate radical:
$\ce{ .NO3 + e- -> NO3- }$
You may wish to test the results from several photolysis samples, some of which were repeatedly subject to the B-12/H2O2/KNO3 sunlight treatment.
Does anyone wish to know which is a more powerful nitrating agent in practice, strong nitric acid or the transient nitrate radical? It may be time to discover!
To assist, here is an interesting source, 'Nitrate radicals and biogenic volatile oxidation, mechanisms, and organic aerosol', to quote:

Abstract. Oxidation of biogenic volatile organic compounds (BVOC) by the nitrate radical (NO3) represents one of the important interactions between anthropogenic emissions related to combustion and natural emissions from the biosphere. This interaction has been recognized for more than 3 decades, during which time a large body of research has emerged from laboratory, field, and modeling studies.

Here is a mention of a non-photolysis creation path (but, my preference would be nitrate in a reaction mix of copper with HOCl in the presence of graphite):

The nitrate radical (NO3) arises from the oxidation of nitrogen dioxide (NO2) by ozone (O3) and occurs principally in the nighttime atmosphere due to its rapid photolysis in sunlight and its reaction with NO (Wayne et al., 1991; Brown and Stutz, 2012). The nitrate radical is a strong oxidant, reacting with a wide variety of volatile organic compounds, including alkenes, aromatics, and oxygenates as well as with reduced sulfur compounds. Reactions of NO3 are particularly rapid with unsaturated compounds (alkenes) (Atkinson and Arey, 2003).

A possible important implication of the above, in the current proposed experiment, is not to overdue the photolysis (assuming that it is, at all, successful, else investigate non-photolysis paths).

I do not think it will be possible to form nitrate esters of cellulose without sulfuric acid present, I am sure that you need the strong bronsted acid to make the reaction work. But I think that there is another issue here, your safety / health / the law on energetic materials.
I hate to have to say this, but I do not think it is safe to make energetic materials using nitric acid at home. You should also consider for a moment the question of "is it legal" to make the energetic material at home.
Keep in mind years ago some "home chemists" would make explosives / fireworks and it was regarded as "boys will be boys". But nowadays the "boys in blue" (Police) are very keen to chase after people who make explosives / things which go bang etc.
I think that the distillation of sulfuric acid at home does not sound like a good idea, I think that you would need to work in a good fume hood. Most school labs would be unlikely to have the equipment required to do it.
With all due respect I think you should choose another chemical project to pass the time while you are on "lockdown".

Oh, yeah... It might not even be legal to send nitrocellulose through the mail. you could end up with flaming mail processing machines. yikes.

In many cases, sodium hydrogensulfat can be used instead of sulfuric acid. For example, a mixture of sodium chloride and sodium hydrogensulfat produces gaseous HCl at high temperature as if it was sulfuric acid plus sodium chloride.


text:
How to calculate the volume of HNO3 needed to produce a specific concentration?

How much $\pu{cm^3}$ of $\ce{HNO3}$, $\omega_{\ce{(HNO3)}} = \pu{63}\%$ with density $\ce{HNO3}$ of $\pu{1.4 g cm-3}$ do we need to create $\pu{500 cm^3}$ of solution whose $c = \pu{0.2 mol dm-3}$?

I tried to solve it and got following answer:
\begin{align}
V_\ce{(HNO3)} &= ? \\
ω_\ce{(HNO3)} &= 63\% \\
\rho_\ce{(HNO3)} &= \pu{1.4 g cm-3} \\
V_\mathrm{(solution)} &= \pu{500 cm3} \\
c &= \pu{0.2 mol dm-3} \\
c &= \frac{n_\ce{(HNO3)}}{V_\mathrm{(solution)}}  \\
n_\ce{(HNO3)} &= c \times V_\mathrm{(solution)} \\
n_\ce{(HNO3)} &= \pu{0.2 mol dm-3} \times \pu{0.5dm^3}\\
n_\ce{(HNO3)} &= \pu{0.1 mol}\\
m_\ce{(HNO3)} &= \pu{6.3 g}\\
m_\mathrm{(solution)} &= \frac{m_\ce{(HNO3)} \times 100\%}{\omega_\ce{(HNO3)}}\\
m_\mathrm{(solution)} &= \frac{\pu{6.3 g} \times 100\%}{63\%}\\
m_\mathrm{(solution)} &= \pu{10 g}\\
\rho_\ce{(HNO3)} &= \frac{m_\ce{(HNO3)}}{V_\ce{(HNO3)}}\\
V_\ce{(HNO3)} &= \frac{m_\ce{(HNO3)}}{\rho_\ce{(HNO3)}}\\
V_\ce{(HNO3)} &= \frac{\pu{6.3 g}}{\pu{1.4 g cm-3}}\\
V_\ce{(HNO3)} &= \pu{4.5 cm^3}
\end{align}
Calculating mass of solution here wasn't needed. Also, my friends did the same exercise and they got answer: $\pu{7.14 cm^3}$
This is new lesson that we are taking and I quite don't understand it. Our teacher wrote these as formulas:
\begin{align}
\omega_\mathrm{solute} &= \frac{m_\mathrm{solute}}{m_\mathrm{solution}}\\
m_\mathrm{solution} &= m_\mathrm{solute} + m_\mathrm{solvent}\\
\rho &= \frac{m_\mathrm{solution}}{V}\\
c_\mathrm{supstance} &= \frac{n_\mathrm{supstance}}{V_\mathrm{solution}}
\end{align}
Are these formulas correct? If I made mistake, please point it out. I don't understand this part at all.

since ρ = Msolution / v ,
it should be 10g and not 6.3 divided by 1.4 in your last step
Hope this helps!

That's the thing. Teacher did another exercise as I did it up there (kinda similar, not same) and that's why I can't understand this whole thing. Once, she used formula that I wrote down in formulas section, second time she used formula from up there...

What was the other formula taught?

Formula written in first lesson was p = m(solution) / v (V of what????)
The other formula is same as mine
p = m(HNO3) / V (according to thins V should be used of solution, but in other case she used V of solute)
That's what makes me not understand this. I'll translate other exercises and edit my post in 10 minutes

Here is a picture of notepad in which I wrote it. I copy pasted what she wrote and translated question: https://imgur.com/a/fLy0lU4
W = concentration in percentage
R = density

What's $\omega$?

What's supstance?

Supstance in this situation would be HNO3. Omega is percentage of solute in solution (if in 100g of solution there is 50g of HNO3 that means that HNO3 is 50% of that solution)

@Zenix When you edit questions like these, please make sure to also remove the deprecated homework tag and give them a proper title.

@Martin - マーチン ♦Actually am newbie. Will remember in future...

Frankly I think that you did so many individual calculations that you lost track of what you were doing. Obviously the amount (moles or grams) of concentrated acid used equals the amount of acid in the dilute solution. Let's use the subscript $\mathrm{i}$ for the initial concentrated solution and the subscript $\mathrm{f}$ for the subscript for the final solution.
$$c_\mathrm{i}\times V_\mathrm{i} = c_\mathrm{f}\times  V_\mathrm{f}\tag{1}$$
Rearranging
$$V_\mathrm{i} = \frac{c_\mathrm{f}\times  V_\mathrm{f}}{c_\mathrm{i}}\tag{2}$$
Now $V_\mathrm{f}$ has units of $\pu{cm^3}$, which is fine for a volume measurement. But $c_\mathrm{f}$ has units of $\pu{mol \cdot dm^3}$ but the concentrated acid is given as $63\%$ by weight. So we need to convert the concentration acid to $\pu{mol \cdot dm^3}$ so that the units cancel.
$$c_\mathrm{i} = \frac{0.63 \times \pu{1.4 g/cm^3} \times \pu{1000 cm^3/dm^3}}{\pu{63 g/mol}} = \pu{14.0 mol/dm^3}\tag{3}$$
Now going back to equation 2
$$V_\mathrm{i}
= \frac{c_\mathrm{f}\times V_\mathrm{f}}{c_\mathrm{i}}
= \frac{\pu{0.2 mol/dm^3}\times \pu{500 cm^3}}{\pu{14.0 mol/dm^3}}
= \pu{7.1 cm^3}\tag{4}$$

Thank you, I mostly understand what are you saying. I know I did a lot of calculations but that's how we are encouraged to work. We actually never learned this approach (using initial and final variables). Thank you again for introducing me to this new approach and I'll definitely research it and try to understand it a bit more.


text:
Are clathrates and interstitial compounds the same thing?
In the book$\ce{^{[1]}}$,  it's given that clathrates are compounds in which gases are entrapped within cavities of crystal lattices of certain organic or inorganic compounds. Interstitial compounds are formed when small atoms occupy positions in the interstices of the metal lattice. So I'm not sure what is the difference between the two? Also it's given that only Helium forms interstitial compounds and Argon, Krypton and Xenon forms clathrates. Could anyone help me understand the difference between the two?
Reference

Pradeep, New Course Chemistry, Class XII, Vol. 1

Clathrate compound :

A clathrate is a chemical substance consisting of a lattice that traps or contains molecules. The word clathrate is derived from the Latin clathratus (clatratus), meaning ‘with bars, latticed’. Traditionally, clathrate compounds are polymeric and completely envelop the guest molecule, but in modern usage clathrates also include host–guest complexes and inclusion compounds. According to IUPAC, clathrates are inclusion compounds "in which the guest molecule is in a cage formed by the host molecule or by a lattice of host molecules."

The famous is the methane in ice clathrate $\ce{4 CH4 \cdot 23 H2O}$
Another case are clathrates of noble gases, found in hunting for noble gas compounds.
Many clathrates are stable just at low temperatures and/or high pressures.
Interstitial compound :

An interstitial compound, or interstitial alloy, is a compound that is formed when an atom with a small enough radius sits in an interstitial “hole” in a metal lattice. Examples of small atoms are hydrogen, boron, carbon and nitrogen. The compounds are industrially important.

So in some loose sense, the former has more or less stable non metallic lattice, the latter has stable, metallic lattice.


text:
Does the reaction of 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one with NaOH involve ring contraction?

What is the product and mechanism when 2‐hydroxy‐6‐methyl‐5‐nitrocyclohepta‐2,4,6‐trien‐1‐one is reacted with
  $\ce{NaOH}$?

The only reagent used is $\ce{NaOH}$ so I suppose it abstracts the acidic hydrogen from the $\ce{-OH}$ group. A nucleophilic attack at the carbonyl carbon is also possible, as it could form a tropylium like cation. But I am not sure as to how a ring contraction occurs. The final answer given is 3-methyl-4-nitrobenzoic acid. How does the formation of an acid group and the ring contraction take place?

It may be worthwhile drawing the dicarbonyl tautomer and thinking of a possible C-H deprotonation, and also considering the mechanism of the Favorskii rearrangement.

Electrocyclization would get you to the Favorskii intermediate. Alternatively,  the @Waylander diketone may undergo a benzylic acid rearrangement.

Drawing on the suggestion of @user55119 I propose the mechanism shown below.
Whether the first intermediate bicyclic [4.1.0] heptanone arises from an intramolecular electrocyclic process or an intramolecular aldol is unclear but either seems possible. Once the highly strained bicyclic [4.1.0] heptanone intermediate is formed, the reaction proceeds with OH- attack on the carbonyl in a process analogous to a Favorskii mechanism here, with the second intermediate anion stabilised by the nitro group which also directs the direction of ring opening. Finally loss of OH- in a manner similar to base-catalysed aldol and driven by re-aromatisation gives the observed product.

I no longer have Scifinder access to see if this is a literature example. If anyone who does would care to post the reference, please do.


text:
Conception and manufacture of the Avogadro Project's (nearly) perfect silicon sphere (and, what is “atomic feeling”?)
The Avogadro Project, an effort prior to the recent redefinition of Avogadro's constant to design a nearly perfect sphere of silicon-28, apparently relied on the fine fingers of optician Achim Leistner - at least according to the Wikipedia:

Achim Leistner is an Australian optician of German origin.3 During
  his retirement, he was asked to join the Avogadro project to craft a
  silicon sphere with high smoothness, as automated machining does not
  match his precision.3
In addition to precision instruments, Leistner uses his hands to feel
  for irregularities in the roundness of the sphere.3 The research
  team has called his extraordinary sense of touch "atomic feeling".3

I suppose this is largely a set of questions about machining and metrology, but materials science is important too. Also, because of the relevance to the redefinition of $N_\text{A}$ I thought it appropriate for this site. With all due respect for Achim Leistner's amazing fingers:
How was this sphere conceived and crafted? For instance: how was the size of the sphere decided? How was the material selected? What is the concentration of defects in the silicon that allows assumption that the lattice is perfect? How is this degree of quality guaranteed? Why can't "automated machining" match the precision of (assuming the alternative) "machining by hand"? What is this purported "atomic feeling"? Is this PR talk, to make the manufacture of the sphere seem more mysterious?

If this were a PR move, I would say it was a terrible one. Words like "feel", "believe", "hope" and so on are usually poorly received in STEM community. A better word would be "sense" or "detect", but then it raises the question of the mechanism of achieving such sensitivity or detection limit.

The new definition of Avogadro number has nothing to do with Achim Leistner's performance. His merit is to be able to get as near as possible to the Avogadro number with his sphere. But even if he had failed in his attempt, the Avogadro would not have been modified.

@andselisk Struck me as very odd, too, especially in our day and age. There is such a thing as craftmanship, but I don't see how a sense of touch would be useful to determine sphericity.

@Maurice Evidently I have the story somewhat backwards, I'll make an edit.

First thing I think we have to differentiate between checking the roundness  and the roughness (that in the article seems it is called smoothness) these two properties of the surface necessitate different tools. Regarding the roundness I doubt that the atomic feeling of Leistner can do much in fact if you look at this video it is clear that the roundness is measured with a CMM machine (even though in the publications is reported a special interferometer).
Regarding the roughness, indeed the human tactile sense has a great sensitivity to roughness in an article appeared in Nature the authors found that humans can distinguish patterns with a wavelength of 760 nm and amplitude of only 13 nm. But from the new scientist article it seems that “rough spots” in the sphere stick out no more than 0.3 nanometres so basically I don't think that his atomic touch, in the sense of measuring roughness, was enough for this task.  Working in surface metrology I would guess that is not so easy to measure the roughness at a nanometric level in the whole sphere in a curved surface! So if we want to take "the atomic feeling" literally what we could suppose is that he used his tactile sense to guide the roughness sampling with more accurate instruments, for instance, some interferometric technique.

Why can't "automated machining" match the precision of (assuming the
  alternative) "machining by hand"?

In the polishing phase is not only a matter of precision but rather on adapting to the changes of surface. Expert polishers are incredibly well paid, they know how to modulate the polishing pattern and force to solve the specific problem of the surface (e.g. scratches valleys etc. etc.) nowadays automated polishing still can not adapt quickly to the chaotic behaviour of the surface.
I quote the article  Towards an automated polishing system: capturing manual polishing operations:

Some automated solutions have already been proposed to assist or
  replace human operator. However, these solutions typically lack
  flexibility and dexterity that are provided by human operators. For
  example, some of the polishing skills that are particularly
  challenging to automate include rapid reasoning and decision making
  based on visual inspections, and fast adjustment of the polishing
  patterns, e.g. when a surface defect is identified

So I would rather think that with "atomic feeling" they are just referring to the impressive, not well-defined skills set that an expert polisher has matured in his career to reach "atomic" polishing results!
Or maybe not?
Looking at the answer of @Buttonwood it seems that the roundness error was +- 28 nm and the diameter around 10 cm that would be a signal with an amplitude of 56 nm and very long wavelength of around let's say 2 cm, that would fit into the range in the Nature article that I linked (even if so long wavelength would indeed necessitate another study...)so maybe somehow he was able to detect this imperfection on the roundness as well for guiding the polishing? In the meanwhile, at PTB it seems they found a way to replace the atomic feeling of Achim, as shown in this video.

Short answer:  The success of the project benefitted from many contributors.
According to BIPM and its Avogadro project, there are two spheres (reference).  The page equally refers to a publiction about the 2011 $\ce{^{28}Si}$ atom count (paywall) and the report in 2017 (here, open access).  From the later paper, you may identify the origin of the two spheres in the Czochralski crystal obtained initially:

as well as where samples were taken to determine the amount of carbon, oxygen, nitrogen, and boron (loc. cit., table 1), the molar mass of silicon (table 2), concentrations of point defects in the lattice (table 3).  Equally, the analysis took into consideration the $\ce{SiO2}$ layer naturally formed if the sample is exposed to air, as well as adsorbed water; for both, the upper limit of layer width as well as the mass are provided.  As an example, the deposit of oxygen on the spheres was mapped by X-ray photon spectroscopy (XPS):

The article describes that the diameter of the spheres was assessed by a sphere interferometer

in vacuum at $20\,^\circ{}\mathrm{C}$ within a tolerance of $\pm 3\,\pu{mK}$ to allow the statement

The measured diameter was the apparent diameter, which is not corrected for the phase shift due to the surface layers. The mean apparent diameter was 93 723.723 61(61) μm.

As for imperfections, this property equally was mapped:

With papers titled like The self-weight deformation of an x-ray interferometer the dedicated issue of Metrologia 48(2), April 2011 shed some light on monitoring the quality of the samples and the used tools for the analyses.
Additions:
A video of channel veritassium is dedicated to the reference spheres, too.  The section starting at 7:00 min focusses on carving and polishing them (8:11 min, at 8:19 min likely showing Leistner himself).
A now open-access publication by Fujii et al. equally briefly describes the stages to carve-out the spheres:

Starting from the ingot, a section is cut.  A lathe is used to obtain a sphere already with an shape accuracy of $1\,\pu{mm}$.  To smoothen the surface, these spheres are lapped with alumina particles in water with finer and finer grains; eventually (fine lapping) within a margin of $100\,\pu{nm}$.  To polish the surface, collodial alumina and eventually titanium dioxide yielded a shape deviation of less than $50\,\pu{nm}$ and a surface roughness below $0.2\,\pu{nm}$.  This publication explicitly refers to Leistner's work including publications authored by him (example, example) and then established protocols easing to manufacture such samples reliably.
PTB continues both further development of the interferometers (here) and the deployment of these spheres as substitute of the old standards (example).  (As of writing, the mass standard's group seeks collaborators for Si spheres (reference, perma).)  Comments by Maurice and GM suggest the advancement of mechanical engineering and this new definition of the kilogram might offer smaller mass references which do not need to be as large and heavy as $1\,\pu{kg}$.

+1 Thanks for the link, I could not find any publication related to the making of the sphere! Quite disappointing!

@GM Reading both the English and the French wikipedia entry about Leistner, even if the publication by Bartl et al. is more of technical nature, I notice that there is a section of acknowledgments; e.g. for the Russian lab to purify the Si, the German IKZ for growing the crystal. But later on, there only is «We wish to thank [....] R
Meeß and his colleagues for the manufacturing of the spheres, our directors for their advice and generous financial support, and all our colleagues for their daily work.»  Quite contrary to my expectation, Leistner's work is not mentioned at all.

Rudolf Meeß works at PTB here is the link to the manufacturing division that he's guiding https://www.ptb.de/cms/en/ptb/fachabteilungen/abt5/fb-55/ag-556/grinding.html

@GM Interestingly, sub-group 5.41 works on / with sphere interferometers (https://www.ptb.de/cms/en/ptb/fachabteilungen/abt5/fb-54/ag-541/kugelinterferometer541.html) and department mass 1.1.1 still seeks further collaborators for additional isotopically enriched Si spheres (https://www.ptb.de/cms/en/about-us-careers/careers/bms-stellen/bms-stelle.html?tx_jobmodul_pi1%5Bjob%5D=3223&tx_jobmodul_pi1%5BlistBackPid%5D=11489&cHash=d5af8c68fa533e496032881c998a6720) | In a youtube video in a channel «veritassium», I meanwhile found some hints about carving the spheres, too.

https://www.youtube.com/watch?v=pouZ3oWIQCk

Indeed Ächim Leistner is not an academic so maybe it does not appear in the publication but It's strange that in the issue of Metrology 48 there is no contribution on the making of the sphere. But from the video PTB seems be able to produce them for Avogadro project without the "atomic feeling" of Ächim. Maybe the skilled artisan was replaced by a CNC at the end?

@GM Note #1: I did not find a reference which uses a diaresis / trema / umlaut in M Leistner's name. | Note #2: Similar to a publication, if M Leistner contributed in a significant way to the success of the project, than the natural way is to include him; he did - without him was no access to the spheres.  You equally thank the gents solving the crystal structures even if your paper is about Maoecrystal (10.1021/jacs.6b06623), don't you? | Note #3: Maybe now, helping to replicate (reproducible research -> production) the specimen. Still Leistner was in the first team, and deserves credit.

Hi, I am not saying Achim Leistner does not deserve credit I am just trying to justify the fact that is not cited. I there is not a motivation for the fact that is not cited it means that did not make the work, is that what you think?

@GM  One thing that still puzzles me is the insistence on attempting to create a 1 kg prototype, rather than choosing whatever mass might be most convenient or result in the most accurate estimate. From this New Scientist article: "The mass of each sphere matches that of the Australian copy of the kilogram." Why, say, didn't they aim to make a 50 g sphere, or perhaps 10 such spheres?

@BuckThorn I speculate practical reasons contributed to stay with 1 kg.  What would be the diameter of a 50 g sphere of Si (the 1 kg one of about handy 9.4 cm)?  Given the grinding and polishing is depicted in the video as manual work (at the time), given the incertainties of all measurements eventually leading to the point: this is 1 kg.  | On the other hand, I now start to read Maurice' earlier comment as this: this here is both conceptual (a device-independent) advancement, and the mechanical / practical one to realize the concept.

I suspect that the budget was generous, although the milling/machining equipment might not have been tailor built for the Avogadro project. I agree that handling a ball of 1 kg mass might be easier than a smaller one (because of the size, not the mass). But they could have aimed for just roughly 1 kg - bringing it to an approximately exact 1 kg mass must have been a significant additional and perhaps unnecessary expense, but perhaps not?

It doesn't mean your other points are not correct: it is a mind-blowing technological feat.

@BuckThorn I'm just reading an other publication no-longer behind a paywall about this project and some of the details like one of the X-ray interferometers used to eliminate all kind of interferences indeed is mind-blowing.

@BuckThorn  my understanding they choose one Kg because the scope was to redefine the kilogram, they wanted to provide a method to create your own reference species not copying the reference K but making it from Avogadro constant and Silicon.

@GM It is human to want a round number, but the important parts appear to be (1) to make a (nearly) perfect sphere and accurately measure its diameter (2) to measure the mass as accurately as possible and (3) to measure the atom packing from crystallographic information. The actual mass or diameter seems unimportant, provided you can determine them accurately. Naturally, if it's no additional effort, then it's a bonus to get to exactly 1 kg


text:
What actually happens to the volume of a gas in an isolated system?
In an isolated system with adiabatic walls in which gas is filled what exactly happens? How do we define the volume of a gas here? Is volume of 1 mole of gas  present in that inelastic container the same as the volume of say 5 moles of the same gas in that container?

Sure.  What did you think?

By definition, gasses fill the volume of their container.

Do you think, if there were originally 5 moles and 4 moles were spent, there would be 1 mol in the 1/5 of the container volume, and rest would be vacuum ?

@Poutnik, that would break the second law of thermodynamics !

@Edison Yau Thinking is not bound by laws of thermodynamic.  More precisily, thinking is, but its conclusions are not :-)

What I actually thought was since the gas is confined to the system(container) shouldn't the volume occupied by 5 moles and 1 mole of the same gas be the same?

It is. Why it should not be ? It does not matter, if the system is isolated or not. Gases always occupy all volume they can, as fast as they can.

@Poutnik is correct if you want the simple reason to this, it's actually related to the dispersion of the energy, specifically the potential energy stored in the intermolecular forces. It's related to entropy.

The volume of a gas is based on the pressure, temperature and number of moles.
According to the ideal gas law :
$PV = nRT$
Where P is the pressure, V is the volume, n is the number of moles, R is a gas constant and T is the temperature.
In your case, the volume of both containers is the same, thus the volume of both gases are equal. However, assuming the temperature of the system is constant, the pressure that the gases exist must be different :
V, T, R = Constants -> $ n \propto P$
Therefore, the pressure of the container with 5 moles is five times of that of the one with 1 mole.
To actually make a fair comparison of the volumes of both moles of gases, the pressure that they are in need to be the same. So, if 5 moles of the gas is placed in a container that has 5 times the volume of the one with one mole, the pressure would be the same but the volume would be different (5:1).
Here is a helpful video for you to understand the concept: https://www.youtube.com/watch?v=BxUS1K7xu30


text:
Adiabatic flame temperature of hydrogen peroxide calculation
I will first demonstrate my cursory method for evaluating the adiabatic flame temperature of H2O2 at atmospheric pressure.
$$\ce{H2O2 -> H2O + 0.5O2}$$
At $\pu{1200 K}$,
\begin{align}
h_\ce{H2O}^\pu{1200 K} - h_\ce{H2O}^\pu{298 K} &= \pu{34.506 kJ/mol}\\
h_\ce{O2}^\pu{1200 K} - h_\ce{O2}^\pu{298 K} &= \pu{29.761 kJ/mol}\\
\Delta_f H^\pu{298 K} &= \pu{-241.826 kJ/mol}
\end{align}
Sources: $\ce{H2O}$: https://janaf.nist.gov/tables/H-064.html;  $\ce{O2}$: https://janaf.nist.gov/tables/O-029.html
At this temperature,
$$\text{Energy released} = 241.826 - 34.506 - 0.5 \times 29.761 = \pu{192.44 kJ/mol}$$
For Adiabatic flame temperature, the energy released should be equal to zero which is not so in this case (I am aware that the process is iterative). However, multiple sources indicate the flame temperature to be around $\pu{1000 K}$. A screenshot from RPA (a rocket propulsion software) is one such source.

The software obviously takes into account other products but, even as a first estimate my calculation shouldn't have been so off from the software calculation. I would appreciate any help in pointing out the mistake in my calculation.

I have tried to fix your markup, I hope I didn't screw it up; please check. If you want to know more, please have a look here and here. I personally don't really understand your question, maybe it is an option for you to explain why you did these calculations and how you arrived at these formulae.


text:
Reactivity of Ammonium
I was going through some displacement reactions when this struck me:

What is the position of the ammonium cation in the metal reactivity series?

I can find other positive ions like carbon and hydrogen in my textbook but not ammonium. I tried to look it up in google but the search results were not up to the mark.
I could get some comparison at this site. From there, it seems that ammonium will be above zinc, lead and iron.
This post also claims its position but does not refer it Do copper and ammonium hydroxide react?.
I know 'ammonium in the metal reactivity series' sounds a bit strange but what I basically want is its position among the metals and the positive ions. Can anyone here help me with this?

Metal reactivity series is for metals. I don't really think you can find carbon there. (Hydrogen is a special case.) Some metals reduce others and thus displace them from their salts. But you won't get any metal, if you reduce ammonium.

It is possible, with the right conditions, to electrolytically produce ammonium amalgam, analogous to sodium amalgam. But analogies only go so far.

@ivan you can sort of find carbon in a reactivity series -- for high temperature oxidation.  Note how the entropy effect of forming gaseous oxides from a solid element causes the carbon to become a stronger reducer at higher temperatures, cutting through the other elements instead of occupying a fixed rank.  Many metallurgical processes use this property.

@Oscar Yeah, sure. However, when people mention "the metal reactivity series", they typically don't mean that one.

The reference given about Ammonium Hydroxyde is not valid, because Ammonium hydroxyde does not exist. It is a solution of ammonia in water. All reactions attributed to ammonium hydroxide are reactions of ammonia $\ce{NH3}$. Ammonium hydroxyde is a molecule that has invented in the 19th century to explain why NH3 reacts so violently with water. But if both ions $\ce{NH4^+}$ and $\ce{OH-}$ are present in solution, they must react by protolysis producing ammonia $\ce{NH3}$ and water, leaving only one or two percent of the initial ions in solutions. See later on.

But the labels $\ce{NH4OH}$ have been maintained because caretakers and other buyers of this solution preferred this name, and said that the ammonia solution does not clean blocked pipes in the beginning of the 20th century. See : "why we all are using a non-existant substance, NH4OH, M. Cosandey, CHIMIA 2021, 75, 6/7, 679.

Let's compare ammonium cation NH4+ and the neutral molecule, NH4 with other metals.
As Ed V pointed out, with a mercury cathode and low concentration of H+, NH4+ can be reduced to NH4 electrolytically (also with sodium amalgam), which forms an amalgam that is stable for a short period of time: the mercury bubbles up, then collapses, evolving NH3 and H2. Ammonium amalgam was discovered in 1808 by Humphry Davy and Jöns Jakob Berzelius.
https://en.wikipedia.org/wiki/Amalgam_(chemistry)
A recent (2016) paper http://scripts.iucr.org/cgi-bin/paper?xk5030 gives the crystal ionic radius of ammonium as falling between 140 and 154 pm, depending on the anion; this compares nicely with the atomic radius of mercury which is 150 - 155 pm, so an alloy is not inconceivable.
Metallic ammonium (NH4) has been considered for some off-world chemistry in a 1954 paper. The phase transition of mixed crystals of hydrogen and ammonia to metallic ammonium was found (by calculation) to take place at pressures almost certainly less than 250 kilobars, suggesting that Neptune and Uranus contain large deposits of metallic ammonium.
http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&bibcode=1954MNRAS.114..172B&letter=0&classic=YES&defaultprint=YES&whole_paper=YES&page=172&epage=172&send=Send+PDF&filetype=.pdf
But on our world, the reaction of a metal like iron with a high concentration of NH4+ cation would produce H2, not NH4. The reaction could be examined on the basis of the pH, or concentration of H+ ions. As it turns out, the potentials are simply related to the concentration of H+ ion, which depends on the anion.
A 0.1M ammonia solution (anion = OH-) has a pH of 11.122 (http://www.chem.science.unideb.hu/Solutionsmonobasic.pdf). A solution of NH4Cl (anion = Cl-) gives a pH of 5.13 (https://socratic.org/questions/calculate-the-ph-of-the-following-aqueous-solutions).
When we displace from a solution of ammonium salts, we wind up displacing hydrogen. You could imagine that adding iron or zinc to a solution of NH4+Cl- could evolve neutral NH4 which immediately decomposes to NH3 plus a hydrogen radical H., which then dimerizes to H2, and the NH3 immediately redissolves in the water. Far better to think of the NH4+ dissociating in the aqueous solution before reacting with the iron or zinc, in other words, the reaction with the metal is a reaction of H+ (or, actually, hydrated H+, or hydronium ion) after dissociation of NH4+ in the presence of H2O.

Thanks a lot! Can you give me an example (in form of a reaction) for the last paragraph?

Not this: Fe + 2 NH4Cl + nH2O --> FenH20++  + 2 NH4  +  2Cl-  +nH2O, but rather:

Not this: Fe + 2 NH4Cl + n H2O --> Fe++  + 2 NH4 + 2 Cl-  + n H2O, but rather:     Fe + 2 NH4Cl + n H2O  -->  Fe + 2 NH4+ + 2 Cl- + n H2O  --> Fe + 2 NH3 + 2 H+ +2 Cl- + n H2O (which reflects the acidity of ammonium salts) --> Fe++ + 2 NH3 + H2 + n H2O --> Fe++ + 2NH4+ + H2 + 2 OH- + (n-2) H2O (reflecting the basicity of NH3).    Or, remember that NH4+ dissociates to NH3 + H+, and simplify to Fe + 2H+ --> Fe++ + H2. It's simpler because we have no evidence of indication of NH4 ever appearing on this scene, but we can measure the presence of H+ (or H3O+) - you know what I mean.


text:
Is the first step of most the accepted mechanism of Perkin's reaction correct?
If I am not wrong this is the most accepted mechanism of perkin's reaction:

In the first step, is the acetate ion basic enough to abstract proton from acetic anhydride?
If yes, I have a counter arguement. It is well known that acetic anhydride doesn't answer iodoform test despite having an acetyl group. It is mainly because, abstraction of proton is not possible as it would produce a unstable carbanion (because of cross conjugation).
If OH- cannot abstract the proton iodoform rxn. , how could acetate ion which is a weaker base abstract the same proton in Perkin's reaction?

AcO- in AcOH is a different base to AcO- in aqueous.

anyway it is weaker than OH- ,Isn't it? If it can abstract the proton then OH- should also be able to.

It is weaker than OH-  IN WATER

Exactly ,@waylander ,  and @ anonymous ever heard about alcoholic KOH reagent, and it's use in elimination?

@Waylander Are you telling AcO- in AcOH is stronger than OH- in AcOH? If yes, could you provide an explanation? And  yes have heard about alcoholic KOH.

@Waylander: if I understand it correctly, in the aqueous medium, ethanoic acid is readier to donate its proton to a hydroxyl, which makes sense. Did you mean to say this is not the case in pure ethanoic acid? What exactly makes it different?


text:
Are these two reactions the same?
I read the book an it said when $\ce{NH4Cl}$ dissolves in the water, the reaction  of $\ce{NH4+}$ will be
$$\ce{NH4+ <=> NH3 + H+},$$
but instead, the first reaction came to my mind was
$$\ce{NH4+ + H2O <=> NH4OH + H+}$$
Are these two reaction the same? And if they are the same, does that means that $\ce{NH3} = \ce{NH4OH}$ in the water?

I'd say NH4OH is not much of a thing. You may assume it is always fully dissociated. But some (maybe a major part) of NH3 exists as just NH3.

Thanks guys. Forgive my foolishness.


text:
Nernst equation and equilibrium
I am currently stuyding electrochemistry, and recently I've stumbled upon a problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate, but I obtained an unrealistic result and I became confused.
What I first did was write the half-reactions involved within the voltaic cell:
Anode (-): 5Fe2+(aq) ⇌ 5Fe3+(aq) + 5e- ε01= -0.77 V
Cathode (+): MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) ε02= 1.5 V
Notice the fact that I obtained the first reaction, the oxidation, by reversing the equation for the reduction of Fe3+.
By combining both half-cell reactions, one can obtain the equation for the whole process:
5Fe2+(aq) + MnO4-(aq) +  8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) E0= 0.73 V
A positive standard potential will yield a negative value for the Gibbs free energy, so the reaction is product-favoured. The problem is concerned with calculating the concentration at equillibrium, corresponding to the equivalence point of our titration, so the overall Gibbs free energy will be 0, just as the non-standard potential, which we can calculate using the Nernst equation (considering 298.15 K of temperature and 1 atm of pressure). I'm also assuming that the reaction quotient is equivalent to the equilibrium constant, since we are at the equivalence point.
\begin{align*}
    E=E^{0}-\frac{0.0592}{5}\log{K}
\end{align*}
\begin{align*}
    E=0, E^{0}=\frac{0.0592}{5}\log{K}
\end{align*}
I then obtained the mathematical expression for the equilibrium constant:
\begin{align*}
    K=10^{\frac{5E^{0}}{0.0592}}
\end{align*}
I then entered the values, and calculated K (substantially larger than 1):
\begin{align*}
    K=4.523*10^{61}
\end{align*}
However, this is the hurdle I have yet to overcome. I didn't quite understand what activities are in the Nernst equation (afaik they are not equilibrium concentrations), so I assigned a formula for K based on the equilibrium reaction:
\begin{align*}
    K=\frac{[Fe^{3+}]^{5}[Mn^{2+}]}{[Fe^{2+}]^{5}[MnO4^{-}][H^{+}]^{8}}
\end{align*}
The problem mentions the following concentrations for the titration:
Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
H2SO4: 1N (probably 0.1M for H+)
KMnO4: 1N (0.2M MnO4- following the same steps)
In my conception, the activities of the reagents in the equilibrium constant were their initial concentrations, and the activities of the products were their equillibrium concentrations. I then made an ice table, considering a concentration of x. By replacing those values into the equilibrium constant, I obtained:
\begin{align*}
    K=\frac{5x^{5}*x}{0.02^{5}*0.2*0.1^{8}}
\end{align*}
X was supposed to be the concentration of Fe^{2+}, however the answer I got is definitely not the right one.
\begin{align*}
    x=6726501.235
\end{align*}
Where did I go wrong?

When you computed $\frac{RT}{nF}$, did you copy over the correct number of zeros?

What do you mean? I always used 0.0592. As for the other concentrations, I copied those correctly as well.

$\frac{(8.3141\ \mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}) 298\ \mathrm{K}}{96485\ \mathrm{C}\ \mathrm{mol}^{-1}} \neq 0.0592\ \mathrm{J}\ \mathrm{mol}\ \mathrm{C}^{-1}$

We actually are supposed to do more than that. The original Nernst equation uses the natural logarithm of the constant, as opposed to the generalized formula which uses the log function. It's just for the purpose of simplifying things. What you wrote would be roughly equal to 0.0257, which, when divided by loge, yields 0.05919 which can be safely approximated to 0.0592. More precise calculations would lead to about 0.059156973, but that is really "simplifiable" to 0.0592. You can also use 0.059 when doing rough calculations.

problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate,

There are several things which may be problematic in your approach.

For the stated problem you do not need to invoke Nernst equation. Simply use the balanced equation.
When potentiometric titration curve are to be estimated, it is far more convenient would use the half-cell instead of a full reaction. I am assuming the redox electrode is dipping in iron solution.

\begin{align*}
    E=E^{0} -\frac{0.0592}{5}\log{[Fe^{2+}]/[Fe^{3+}]}
\end{align*}
where $E^{0}$ is the iron half cell value. You should not calculate equilibrium constants from potentiometric titrations.
See page 17 for a full example Potentometric Titration clculation

I do not know when people will stop teaching normality. It is obsolete (18th century concept). Your normality conversions are incorrect.

The problem mentions the following concentrations for the titration:
  Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
  H2SO4: 1N (probably 0.1M for H+)
  KMnO4: 1N (0.2M MnO4- following the same steps)

1 N sulfuric acid will be half the molarity =0.5 M
1 N permanganate will be 1/5th the molarity = 1/5 M
Check the concept of normality in your textbook.

Apologies for misusing normal concentrations. However, how is applying the equation you've mentioned in your second point different from basically applying the Nernst equation for the anodic half cell? And why is it [Fe2+]/[Fe3+] and not the opposite, [Fe3+]/[Fe2+], since Fe3+ is the product obtained via the oxidation of Fe2+. Also, should i replace the values in the square brackets with the activities, or with the equilibrium concentrations?

No problem, normality is a part of Indian syllabus. When you use negative sign in Nernst equation before the log term, the reduced form comes in the numerator. I remember it by a word NERO = Negative sign reduced over oxidized. If you use a positive sign, PORE = Positive is oxidized over reduced.

What I didn't understand, though, and the textbook failed to explain that clearly, is the difference between activities and molar concentrations. I never quite actually understood what are activities and how do they differ from molar concentrations, nor how are they used. When working with the Nernst equation, why are we allowed to compute the activities of the reactants and the equilibrium concentrations of the products? This is the stuff that really baffled me and I'd be highly appreciative if you could clarify it for me.

@user89030, please post this as a separate question so that it gets more attention. Also search the Chemistry stack exchange (previous question), I am sure it must have been addressed before.


text:
Vapour Pressure of A Solution
I study in class 12, and have a doubt in vapour pressure that my school teacher  and my friends are not able to clarify.
My doubt is the reason why the partial vapour pressure of each individual component of a solution is not equal to their respective standard vapour pressures($P°$)?
My reasoning is that consider you have a pure liquid A in a closed container, the vapour pressure of it shall be equal to $P°_A$. Now when you add another volatile liquid B, a solution should be formed; however, the equilibrium of the reaction
$$\ce{A_{liq} <=>  A_{vap}}$$ should not be affected since the concentration of reactants and products remains the same since B is different from A. Eventually, the reaction of B must also reach equilibrium, independently, and thus the partial pressure of B  must be equal to $P°_B$ .
However, this does not seem to be the case. Could someone please help in clarifying my doubt through the viewpoint of equilibrium(Mathematically if possible) and not the standard explanation that the surface of the solvent get's blocked by the particles of the solute, hence the vapour pressure reduces.
Regards,
Arun

I assume you are trying to apply Dalton's law of partial pressures on liquids? Also have you seen Raoult's law? In ideal gases, the molecules are independent of each other. They do not "know" the presence of  each other. The situation for liquid is very different, things are so close to each other that you cannot neglect intermolecular interaction.

@M.Farooq Thanks for your response sir...however, I am afraid that I still have not entirely understood your response. Are you trying to say that the adhesive forces between molecules of A and B in the solution will be different compared to the cohesive forces between pure A molecules and pure B molecules, which would have existed if they had been present separately, which is in turn affecting the individual partial pressures of each of the components of the solution? I would greatly appreciate a response to this...

They keyword you should be searching is the molecular origin of the Raoult's law. See, by definition of the Raoult's law (from Wiki) "It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture." What you suggest is exactly $against$ this law. In your proposition, you are neglecting the mole fraction of component A and B in the liquid state. Even in the gases, the total pressure is their partial pressure times their mole fraction, so why change for liquids.

@M.Farooq Yes sir, I know the definition of Raoult's law, however, I am unable to understand the reasoning behind it for we have not been taught the detailed derivation, just the fact that the partial pressure is proportional to the mole fraction . I would appreciate if you could read through my reasoning in my original question posted(not the comments) and highlight where I am right and where I am wrong(The main part being whether the equilibrium of reaction involving ' A' would be affected). Thank You

@M.Farooq Also to add on, I do agree that the partial pressure of a gas depends on it's mole fraction in vapour phase. However, I am unable to wrap my head around how it depends on the mole fraction of the component in the solution.

You can try to convince yourself by an approach called $reductio$ $ad$ $absurdum$. In this way, you assume whatever you are saying is true. Once you assume it is true, you deduce results out of it. If your proposition were false, the results are $absurd$. The biggest incorrect result is that you propose that mole fractions are $not$ needed. Would this match the experimental result?

Another way be rigorous is to derive Raoult's law from fundamental principles, without knowing that it exists. Search Derivation of Raoult's law from statistical thermodynamics. See how the mol fractions enter into this equation. This will be the most rigorous way.

Let us continue this discussion in chat.

@M.farooq rate of volatility can't be same for all liquid when we use the term partial fraction we generally amount of x in big y, this pressure can't be equal to pure vapor pressure of A.

Rauolts law and dalton law are equal just a term difference there they use total pressure instead of vapour pressure

This is where your logic fails:

since the concentration of reactants and products remains the same since B is different from A

As soon as you mix two components you change the concentration of both, you effectively dilute all components.
Otherwise your intuition is correct. The gases can often be assumed to act independently. This is true if the solution is ideal (that is, assuming interactions between all molecules in solution are roughly the same). Then each component equilibrates independently with its vapor. However, the vapor pressure of each component is in general not the same as it would be if it were pure. Instead, for an ideal solution the partial pressure of each component follows Raoult's law $p_i=p^*_i\chi_i$, where $p^*_i$ is the partial pressure of the pure component and $\chi_i$ is the mole fraction in solution.
The total pressure above the solution is then the sum of the partial pressures of the components (following Dalton's law, where $\chi_i^g$ is the mole fraction in the gas phase):
$$ \begin{align} p &= \sum_i p_i = \sum_i \chi_i^g p \\ &= \sum_i \chi_i p^*_i \end{align} $$


text:
What will happen if I trap an ozone producing device inside a small airtight area?
I'm currently building a UVC light, and right now I'm working on figuring out how to place it in an aluminum reflector housing to cast the beam of light effectively. It's creating a lot of ozone, and I'm worried about the longevity of the aluminum reflector it's going to be placed inside, as I know ozone oxidizes and corrodes metals. If I make this airtight with a lens on the front, will the ozone production cease after the oxygen inside converts? Will any dangerous conditions occur inside if it's not ventilated? Will ventilation create more ozone to the structure, or less?

I don't think you will have any problems with your aluminium because most of the aluminium usually is already passivated by a layer of aluminium oxide ($\ce{Al2O3}$), this depends actually on the alloy that you are using but if you buy normal aluminium most of the time has this behaviour. If the enclosure it's watertight all the oxygen will be converted in ozone, and when the oxygen will run out the reaction will stop, if the enclosure has some leakages you will continue to create ozone so you will have to operate the lamp in a ventilated environment.
If the environment is watertight and the UV lamp is generating a lot of heat you could have a substantial increase in pressure. This could cause your window to break or explode.

One problem I see (other than creating a very toxic high concentration of ozone) is the ozone itself quickly reacts with organics, including lubricants, tubing,.. used in the machine producing it. Per a source, to quote:

Ozone is an effective sterilant due to its ability to oxidise most organic matter...
During ozone sterilisation, ozone breaks down into reactive species, including hydroxyl radicals and atomic oxygen. Because of the strong oxidising nature of ozone, polymers must be resistant to oxidation. Polymers and medical devices should also be resistant to high relative humidity levels (> 80%), which are required for ozone to be effective. Consequently, materials should be resistant to oxidation and moisture.

As such, in the presence of significant water vapor, a possible corrosion problem with ozone attacking the OH- ion (formed in the dissociation of water), to quote a reference:

The first reaction that takes place is accelerated ozone decomposition by a type of initiator. This can be an OH-molecule, see reaction 1:
1: $\ce{O3 + OH- -> •O2- + HO2• }$
This radical has an acid/ base equilibrium of pKa = 4,8. Above this value, this radical no longer splits, because it forms a superoxide radical, see reaction 2:
2: $\ce{HO2• -> •O2- + H+ (pKa = 4,8) }$
Radical chain-reaction
3: $\ce{O3 + •O2- -> •O3- + O2 }$
4: $\ce{•O3- + H+ -> HO3• (PH < ≈ 8) }$
The OH-radicals that have formed react with ozone according to the following reaction mechanism:
5: $\ce{OH• + O3 -> HO4• }$
6: $\ce{HO4• -> O2 + HO2• }$
During the last reaction, HO2• radicals are formed, which can start the reaction all over again (see reaction 2). As a result, a chain-reaction develops, which is maintained by so-called promoters.

Source: https://www.lenntech.com/library/ozone/reaction/ozone-reaction-mechanisms.htm#ixzz6J8fJIVgr .
The latter HO2• is highly acidic (see ‘Radical-Enhanced Acidity: Why Bicarbonate, Carboxyl, Hydroperoxyl, and Related Radicals Are So Acidic’) and could attack metal, like Aluminum forming Al(OH)3 or hydrated Al2O3.
Bottom line, I suspect, in time this will prove to be a bad idea, and could result in significant harm to the machine, and any leakage of ozone is problematic, given its potential health issues, to quote a government source:

When inhaled, ozone can damage the lungs. Relatively low amounts can cause chest pain, coughing, shortness of breath and throat irritation. Ozone may also worsen chronic respiratory diseases such as asthma and compromise the ability of the body to fight respiratory infections.

And a UV (ozone) cleaner is a nice (and pretty standard) lab device to get residual organic crud off of surfaces you want to be clean...


text:
What chemical reactions could ancient people use to propel projectiles?
Are there reactions other than burning of gunpowder that ancient people could perform that would produce gas from non gaseous substances, so that it could have been used for propulsion of projectiles had they thought of it? An example would be CaO + H2O reaction, as upon mixing in a closed container this mixture would pressurize the container, and at some pressure the cord tying the projectile to the chamber would rip and projectile would be propelled

How would those folks even know about pressure in the first place? It was only a few hundred years ago that people realized that the atmosphere had weight.

@EdV you are right, but you don't need to know what is pressure to use it!

@GM But why even have the idea of pressurizing some container and using it to propel something? There was Greek fire, too, and the  Greeks were very intelligent, but they apparently did not invent rockets. Also, Hero’s engine demonstrated the use of steam, though they probably did not realize what steam was.

My point is that modern chemists, with 20-20 hindsight, could MacGyver a pre-historic cannon or rocket, but it would have taken too many big conceptual and practical innovations for ancient peoples to have had. Especially if they had no written language, etc.

@ Ed V: Most of people out side USA don't know who's MacGyver ! :-)

@MathewMahindaratne Yes, especially now, we have to be thankful for small blessings! ;-)

Guess I missed the part where the OP’s question started with “Are there reactions other than burning of gunpowder”. Of course, there are actual baking soda mines, burnt wine and the white cliffs of Dover, but not much to go on. Maybe we do not hear of ancient rockets and cannons because their genius inventors were early winners of what we now call the Darwin Awards! ;-)

@MathewMahindaratne  you'd be surprised. I've see MacGyver DVD sets sold in European countries. Hollywood's reach is far and wide.

The answer depends on what do you mean for ancient people. There are a lot of reactions that could produce gas for ejecting projectiles. The most common was the mixture of sulphur, potassium nitrate $\ce{KNO3}$ and carbon that is known as gun powder and has been reported to be used since the IX century in China.
If you want some less efficient and not used reaction,  you could think of using methane explosion extracted from some swamp. Or some reaction between calcium carbonate and acetic acid. Like this one between sodium hydrogen carbonate:
$$\ce{NaHCO3 + CH3COOH -> NaCH3COO  + CO2 + H2O}$$

One of the earliest use of Black Powder was crude large-bore handheld guns (early China). And, actually much more, to quote a source:

Other Song military applications of gunpowder included primitive hand grenades, poisonous gas shells, flamethrowers and landmines. The first artillery pieces were rocket tubes made from hollow bamboo shoots, but these were soon upgraded to cast metal

Cannon tube + water in a sealed container + cannonball.
Add heat.
How far the cannonball goes depends in part on the strength of the container.

I suppose you are referring to the steam cannon. Could you please elaborate?


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T∆S implies that heat flows which way?

From Schroeder's Introduction to Thermal Physics.
Parts $(a)$ and $(b)$ are quite simple, and we find that $ \Delta H =-2808\ \mathrm{kJ}$ and $\Delta G =-2879\ \mathrm{kJ}$, so by $\Delta G = \Delta H -T\,\Delta S $, we find that $T\,\Delta S = 71\ \mathrm{kJ}$ at room temperature.
What I absolutely cannot understand is how $(c)$ and $(d)$ work. It intuitively makes sense that entropy is increasing. Assuming $1$ mole of glucose, we start the reaction with $7$ moles of reactants and end with $12$ moles of products. Entropy obviously increases, as we can arrange these molecules in many more ways. However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment. But this isn't true in real life – metabolism reactions obviously release heat. What's going on here?

The fact that the book is using $\Delta G$ as the relevant state variable for free energy implies this is being carried out at constant T and p.  So, proceeding:

"However, this implies that we are getting usable work from increases in entropy...."

Yes, that's correct.  The positive $\Delta S$ lowers $\Delta G$

"However, this implies that we are getting usable work from increases in entropy – meaning that the muscles are somehow absorbing heat energy from the environment."

No, that's incorrect.  The heat flow here is purely due to $\Delta H$, and it is flowing out of the system.
What you're probably thinking is that, at constant T (which we have here), $$\Delta S = \frac{q}{T},$$
and thus, since $\Delta S >0 $,  q must be $> 0$ as well, i.e., that the  positive $\Delta S$ means heat is flowing into the system.
This is wrong.  It is not the case that $\Delta S = \frac{q}{T}$.  Rather:
$$\Delta S = \frac{q_{rev}}{T},$$
and this reaction is not happening reversibly.  Instead,  $\Delta S$  is determined from the difference between $S_{products}$ and  $S_{reactants}$, and each of these individual values are calculated from:
$$S(T') =\int_{0}^{T'} dS= \int_{0}^{T'}\frac{\text{đ}q_{rev}}{T}$$
[For more on this, see my answer here:  Infinite Increase in Entropy when Energy added to Absolute Zero ]
In summary, in a chemical reaction at constant pressure (and with no non-pV work),
$$\text{đ}q_{sys}=dH_{sys}$$
Thus it is the sign of $\Delta H$, not $\Delta S$, that determines the direction of heat flow.
Finally, the answer to (d) is that, as thermal energy is generated by the reaction, it heats up the system, and the 2nd law of thermodynamics says that thermal energy naturally flows from hotter to cooler regions.  So there is heat flow into the surroundings until the reaction mixture returns to the temperature of the surroundings.

Let's start from the famous expression $\ce{\Delta G° = - zEF}$. In the case of glucose, it means that the combustion of glucose in a galvanic cell produces 2879 kJ/mol. It is more than the heat that could be released if the reaction does not produce electricity (2808 kJ/mol). In other words, of all the energy available from this reaction (2879 kJ), a part of it (71 kJ) is used to increase the entropy of the system due to the chemical reaction, and this part will not be converted into heat. The heat released (2808 kJ/mol) will be used for heating the system, then for heating the environment and increasing its entropy, if the heat can get out of the system. This amount of heat is slightly smaller than the total available energy $\ce{\Delta G°}$.