text:
Conversion of cyclohepta-1,3-diene to cyclohepta-1,3,5-triene
In the landmark synthesis of tropinone by Willstatter in 1901, an interesting step involves the use of $\ce {Br_2}$, quinoline to convert cyclohexa-1,3-diene to cyclohexa-1,3,5-triene. In the scheme below, 7 is first converted into cyclohepta-1,3-diene via a Hofmann elimination. Further reaction with $\ce {Br_2}$, quinoline yields 8. I am quite puzzled as to how this transformation occurs. Wouldn't addition of bromine to the double bonds occur if bromine is reacted with cyclohepta-1,3-diene?

It's probably 1,4-addition and later elimination creating two bonds.

@Mithoron 1,4 addition to the diene and then elimination to form the two bonds? Did I get your right?

Yep, pretty simple, but it may be a bit tricky to get it to add bromine selectively in 1,4 manner.

@Mithoron I could not find an example of the bromination of cyclohepta-1,3-diene but the related case of buta-1,3-diene gives a 1:1 mixture of 1,2 and 1,4-bromination see this SE answer https://chemistry.stackexchange.com/questions/118945/does-1-3-butadiene-in-bromine-water-do-addition-twice  and   https://chem.libretexts.org/Courses/Purdue/Purdue%3A_Chem_26605%3A_Organic_Chemistry_II_(Lipton)/Chapter_11.__Addition_to_pi_Systems/11.1%3A_Electrophilic_Addition/11.1.4_Addition_to_Conjugated_Dienes

I think the answer is here: https://synarchive.com/syn/303. The reagents are slightly different though and I am not sure how at -5°C, there is a 82% yield of the 1,4 adduct because the 1,2 adduct is usually favoured at lower temperatures I believe.

@M.L Interesting! That is some very old work, I wonder how they are sure they have one di-bromoproduct and not a mixture?

Thanks to @Mithoron, @Waylander and @M.L for your insightful comments! Willstatter's synthesis is described in greater detail on synarchive, as shared by M.L. Indeed, as Mithoron pointed out, 1,4-addition to the diene formed after Hofmann elimination occurs. Subsequent elimination with the base quinoline yields the triene, as depicted below.

I have grave doubts whether this actually gives an 82% yield of the 1,4 addition product. It is very old work and I doubt that the researchers could distinguish between the 1,4 and 1,2 addition products. It probably does not matter, kick the 1,2 dibromo product hard enough (and quinoline at 150C is hard enough) and it will give the di-elimination to the tri-ene.

@Waylander Hmmm... Could you elaborate on what you mean by "kick the 1,2 dibromo product hard enough"? I'm guessing your suggesting the triene is the thermodynamic product of this sequence of reactions and that sufficiently high temperature would favour it?

Exactly so. With the 1,2-addition product the pathway to the triene may well pass through a bromodiene intermediate (bromine on an alkenyl carbon) which is harder to eliminate to the triene. However, quinoline at 150C is sufficiently forcing that it will eventually eliminate.


text:
Stability of Microemulsions
Unlike emulsions, microemulsions are thermodynamically stable. Its stability is often explained by entropy changes brought about by dispersing liquid in another liquid, however this can't be the whole story behind its stability since dispersing liquids also happens in regular emulsions and they aren't stable thermodynamically.
I am not sure if I am correct, but I think that stability of microemulsions is due to the interactions between molecules on liquid interphase and emulsifier being better (stronger) than between molecules in the bulk of the liquid droplet in similar way when liquid spreads on the surface of the solid forming no contact angle with the surface. If interactions between emulsifier and surface molecules are more favourable than surface molecules have lower potential energy compared to the bulk and there is tendency to increase surface area between molecules and emulsifier or in other words there is tendency to create emulsion.
I wasn't able to find this explanation anywhere online though, but I think it has sense. What are your thoughts?

I think they are more like a true solution of a supramolecular complex, when normal emulsions are rather two phases only somewhat protected from separating.

source: Stack Exchange

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

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How do I crystallize Sodium Acetate w/o a crystal?
I made some sodium acetate. Definitely not pure or anything close. But I just did it as a quick experiment. Just some vinegar and baking powder.
When I search how to make $\ce{NaOAc}$ crystals, people say put crystal in it. I understand how that works but I don't have crystals in the first place.
As for the metal button, I don't have one either. I'm left with this kinda thick opaque liquid. I want to crystallize it. Can you help me with that?
Reiteration
I boiled the solution for so long, but I still didn't get any crystal formation. I left it in room temperature and tried to freeze it for more than a day with no luck. I've evaporated so much water that it is became very thick.
What I'm trying to achieve is, getting crystal to fully crystallize the salt to then dry it and make a powder.

@Mithoron The OP says he got some crystals with freezing it. Does that work? If so, why don't he use the crystals to further crystallize the rest? I'm not really sure how that works but I left it in the freezer for long time and it is still liquid.

@Mithoron can you open this question? It's not a duplicate. Sure, the other user asks the "same question" but his conditions are different. He is in a different situation. Thus, the answers he get is different than what I need. I hope you can re-open this

can you edit the question to differentiate it from the other question? Include anything that can make this question different from the other.

Evaporate a bit of your solution until crystals appear. Then you will have a crystal, or several crystals thereof. Some of those crystalline structures may be a mix of sodium acetate and sodium bicarbonate (this latter is what makes baking soda). You might as well suspend a thin thread of cotton into a glass filled with your solution, and let it evaporate. Crystals will form around the thread, but that shall take a long time, so be prepared to be patient. If you want to save time, depart from a concentrate solution of your salts (by boiling it until it becomes a bit milky).
https://esplora.org.mt/wp-content/uploads/2020/04/Salt-Crystals-Website.jpg

The thread method is very cool. I didn't know that was possible. I tried to reduce (evaporate) my solution a bit more. It got to a point where it's very very viscous. A very thin white layer were left along the walls of the pan where there were solution before. I doesn't look like a crystal to make because it's very smooth and thin powder like. I tried to freeze it but nothing happens

One thing we should have checked beforehand is whether or not sodium acetate is either  hygroscopic or, worse yet, deliquescent. Whichever the case, crystals may never form naturally (by evaporation at room temperature) because the salt keeps absorbing moisture from the air (1st case) or dissolving itself in the very moisture is absorbs.

@RooTenshi Check this out. It seems that sodium acetate is deliquescent.
https://en.wikipedia.org/wiki/Sodium_acetate

Hmmm. I will need to understand further before asking for more advice. Thank you for guiding me.

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


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Physical origin of induction and dispersion energies
I am reading this book on non-covalent interactions. Chapter 1 states that there are 3 types of non-covalent interactions:

Electrostatic interactions, which are just the coulomb interactions between two static charge distributions. (either repulsive or attractive)
Induction effects that arise from the distortion of a particular molecule in response to the electric field of its neighbors. (always attractive)
Dispersion interactions, which are due to the correlation of the motion of electrons in different molecules. (always attractive)

The chapter uses perturbation theory. I am not very familiar with it, but intuitively it does make sense. The perturbation of the Hamiltonian due to the second molecule is given by (15) (but I imagine that one could generalize the treatment to another part of the same molecule, if one was interested in intramolecular interactions for big molecules, rather than intermolecular interactions . . . right?):

So the electrostatic interaction comes out by evaluating the expectation value of $H'$ with respect to $\Psi_{00}^{AB}$, which is just the product of the ground state wave functions of the individual molecules. The second-order terms, instead, are obtained by "mixed" states, where one or both the molecules are not in their ground state:

which is split into:

My question has to do with induction and dispersion. From the subscripts of (1.7,1.8,1.9) it seems that these are the mathematical representations of the description I gave above using only words. That is, induction terms are the ones where I'm considering a molecule in its ground state, and the other in an excited state. Dispersion terms instead have to do with the "mixing" of states in which both molecules are excited. It's not entirely clear to me to what extent this is true, and to what extent we are just applying arbitrary categorization to quantum effects. In particular for the induction term: it seems that (1.7,1.8) is allowing each molecule to deviate from its ground state wave function/density (the focus is always DFT) only in the context of its interaction with the other molecule . . . shouldn't induction have to do with the effect that electrons on A feel due to the ground-state density of B?
The other thing that confuses me is the nomenclature usually adopted here. I have always heard using the word "dispersion", and never "induction", to describe the corrections one has to implement in DFT to account for long-range intermolecular interactions (such as the D3 correction by Grimme). Other chapters of the same book do the same, and always talk about "dispersion". But given what I wrote above it would seem to me that corrections should include both induction and dispersion . . . I can't imagine any way of differentiating the two, by the way. So when we are only talking about "dispersion", is it just for brevity, or am I missing something? (According to the equations above, both dispersion and induction are after all just the second-order corrections to the energy, so essentially they are the same thing . . . )

If you reduced this to just one question, and moved it to Matter Modeling Stack Exchange there might be more of an interested audience  there.


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At what pressure will ambient atmospheric air liquify?
At what pressure will ambient atmospheric air liquify? (Ambient Earth atmosphere at say five foot above sea level. From any specific point on Earth, since it would vary from point to point due to differing compositions at different locations of course.)
I realize that would never happen naturally on earth, anywhere above sea level, of course. Rather, I am asking about artificially pressurized ambient air that is stored under pressure.
(Sorry, not sure if this should be categorized as a chemistry or physics question since it's really both at once.)

Nitrogen , oxygen ,argon ?

Read up on phase diagrams and supercritical fluids to answer that question.

Air cannot be liquified at temperature naturally occuring on Earth.
More exactly, oxygen cannot be liquified above -118.6°C, nitrogen above -146.9°C.
Things are little more complicated as air is a mixture, but you can get the picture.
At extreme pressures like tens or hundreds gigapascals, air would solidify.


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When we are looking for volts vs SHE, is it that the potential of the element is compared to hydrogen potential?
I'm new to understanding this aspect of electrochemistry and only just now refreshing my memory of high school chemistry for a specific purpose. If I've worded the question or anything else incorrectly, don't hesitate to help me adjust.
This one link I'm reading on Corrosion has a standard EMF Series table with a column for reaction and a column for volts vs SHE. Under reaction we have for hydrogen $\ce{2H+ + 2e- -> H2}$ and then under volts vs SHE 0.0000.
I understand the reaction from high school chem 32 yrs ago (wow). Then, I look down to one of the two elements I'm interested in, Mg... $$\ce{Mg^2+ + 2e- -> Mg}$$ and its redox potential w.r.t SHE is -2.372 V.
The other element which I can't seem to find hard data on is carbon in the form of 4k carbon filament. I have a table that puts it in the range of +0 to +0.2. So, if Mg and carbon filament are used as two electrodes and connected into a circuit, theoretically with the right electrolyte and surface area I could get a potential of 2.5 volts?
The current (μAs) is what I'm trying to understand next and I'm sure I'll have more questions about that, but I just want to make sure I have this first part right first.
Thank you

See also policy-on-amirite-questions

Your title has got a mistake. You should remove the word "atom", because  there is no monoatomic hydrogen in a hydrogen half-cell. There are only hydrogen molecules $\ce{H2}$ (or ions). So your last words might be :  ...is compared to hydrogen potential !

I'm quite over my head in the electrochem section... quite obviously. But definitely needing help. Jump to the comments below and see [from @RobMcN] comment from @ToddMineHardt. I've got a chat room open and a little direction in conduction experiements and formulating.... formulas would be fantastic.

The answer is yes, you will get a potential difference of 0.2-(-2.37) = 2.57 V, if Mg and carbon filament are used as the electrode materials and connected into one circuit. The third digit after the decimal is removed, because different resources give slightly different values. From our own experiments using glassy carbon, its open circuit potential is -0.07 V w.r.t. Ag/AgCl (sat) reference, which in turn is 0.197 V w.r.t. SHE. Therefore, the potential of glassy carbon is 0.197-0.07=0.127 V w.r.t. SHE, close to your quoted value of 0 to 0.2 V for 4k carbon filament. Note that SHE implies pH=0. In practice, people often use RHE as the reference point, which has a pH-dependent potential. This is convenient because experiments done at any pH values can be converted to a difference w.r.t. RHE reference.
Noticed that your question title is also a question. "When we are looking for volts vs SHE, is it that the potential of the element is compared to hydrogen atom's potential?" Strictly speaking, the potential is the redox potential of a half electrochemical reaction. By comparing two potentials, you are comparing the degrees of ease or difficulty to lose or gain electrons between two half reactions. In a sense, this is related to work functions or Fermi levels of different species. The half reactions with negative potentials can lose electrons more easily than the ones with positive potentials. Or conversely, the half reactions with positive potentials can gain electrons more easily than the ones with negative potentials. Therefore, when you are looking at volts vs. SHE, you can comparing the redox of the element vs. the $\ce{H2 / H+}$ (or sometimes written as $\ce{H+ / H2}$) redox pair.

Please use MathJax and mhchem for the whole expression: \ce{...}. Other uses will break in different browsers and for different users and while it might look fine for you, it looks wrong for many others.

@RobMcN - Please leave comments in the comments section.

[From RobMcN] Thank you @Quantum That was truly helpful.
I am doing an experiment with a few battery cells made up of 99.8% Mg rods and 4k carbon filament under various loads and getting such different results that I am trying to understand. Results are being posted there. Help in understanding this would be greatly appreciated.


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Would wood turn into charcoal faster if it is burned in a pressurized environment?
It is very easy to make charcoal from firewood and there are many tutorials about this online. However they all state that the process of turning firewood into charcoal can take atleast 3 to 4 hours. That is quite a long time.
I was wondering whether this time span could be shortened by making the charcoal inside a pressurized container like a pressure cooker instead. A pressure cooker cooks food about 30 percent faster than conventional methods like steaming, boiling, and braising. So maybe a pressure cooker can also produce charcoal 30 percent faster. Pressure cookers also use 50 to 75 percent less energy due to shorter cooking times.
So would using something like a pressure cooker turn wood into charcoal faster?

If wood had higher temperature when pressurised at otherwise similar conditions, it would form charcoal faster. But it should not be pressurised air, it would burn brighter.

@Poutnik Do you think it would also produce better quality charcoal with a higher energy content? I'm assuming that would also be the case because the wood would be exposed to only a limited amount of air while inside the container. But correct me if i'm wrong.

In case of the cooker, pressure serves to increase boiling point. Effect of pressure in charcoal case is unclear. It needs just right amount of air being provided to support the process but not to burn wood.  Another option would be external heating.

Charcoal doesn't form when wood burns: it forms when wood is heated in a low oxygen environment (conventional methods pack wood so that the outer wood burns to provide heat while restricting the oxygen available so the inner wood creates charcoal). So, if you could recreate those conditions, it might work but a pressure cooker might not be the best way. Perhaps a vacuum chamber?

Increased pressure will increase the rate at which wood burns, probably due to the increased concentration of oxygen.  Li, He, Zhang and Wang compared the rate at which wood burns at high and low altitude [1], and found that "...for the same burning surface area, the burning rate, radiation heat flux and flame temperature at the high altitude was lower than those at the low altitude."
However, as pointed out by @matt_black, this won't work for charcoal production, since it requires a slow burn in a low-oxygen environment.  This is needed to remove the water and the volatiles, so that the wood undergoes pyrolysis to form charcoal, instead of combustion to form ash.
Additionally, note the reason a pressure cooker increases cooking rate is entirely different from the reason wood burns faster at increased pressure.  A pressure cooker is for food that is immersed in water.   The temperature of water is limited by its boiling point, which is higher within the higher pressure of a pressure cooker.
[1] Zhen-hua Li, Yaping He, Hui Zhang, Jian Wang.  Combustion characteristics of n-heptane and wood crib fires at different altitudes.  Proceedings of the Combustion Institute, Volume 32, Issue 2, 2009, Pages 2481-2488, ISSN 1540-7489, https://doi.org/10.1016/j.proci.2008.06.033.

Wood burning doesn't create charcoal. Heating wood in a low oxygen environment does. So I don't think this would work.

@matt_black  Thanks!  I'll update my answer based on this.


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Can Surface Energy Be Negative?
In some sources I read, it is written that surface energy can't be negative. However, this doesn't really make sense since surface energy should be negative if molecules on interphase have lower potential energy compared to the bulk.
If molecules on the surface experience interactions or bonds more favourable compared to bulk,surface energy should be negative. This doesn't happen often though, so surface energy is positive for most interphases.

See:  https://www.nature.com/articles/nmat1336a

Thanks. So, in multicomponent systems, surface energy can be negative.

@DarioMirić You can make it into a self-answer if you want.

Would be helpful to give a couple of examples with negative surface energy.

In general scientific literature it is said that surface energy of solids is always positive but there is an unsaid assumption being made while doing so. The assumption being that we are talking about a pure single component system.
The more accurate way to say this would be,
The surface energy of all pure single component solids is positive.
Now coming to the question, "Can surface energy be negative?". The answer is yes!
This can happen as you pointed out, if molecules on interphase have lower potential energy compared to the bulk. Now, this lowering of potential energy might happen due to some chemical interactions at the interface, with some other component which is present in the system.
This has also been experienced in practice like in the article, Interface Stresses and Their Effects on the Elastic Moduli of Metallic Multilayers(1).

The appearance of negative interface stresses and negative interface
energies in Table II seems surprising at first sight. However, the
negative interface energy $\gamma_{111}$ for the Pt/Ni system can be understood
if we introduce the idea that the interface energy of a semicoherent
heterophase boundary contains a chemical component, which arises from
the chemical inhomogeneity in the interface region, and a structural
component due to the distortions associated with misfit dislocations.
(See Ref. 19 for details. ) The chemical component should be negative
for all chemically abrupt interfaces in miscible systems with negative
mutual heats of solution. ' If it is also larger in absolute value
than the structural component, a negative semicoherent interface
energy is obtained. Negative interface stresses may also appear on
boundaries with positive interface energy, because the sign of the
second term in Eq. (3) can be either positive or negative. Following
these arguments we expect a loose correlation of the interface
stresses and energies with the mutual heats of solution, which is
present in our results on the (100) boundary plane as well as on the
(ill) boundary plane, as seen in Table II

Reference:
(1)   Gumbsch, P.; Daw, M. S. Interface Stresses and Their Effects on the Elastic Moduli of Metallic Multilayers. Phys. Rev. B Condens. Matter 1991, 44 (8), 3934–3938. https://doi.org/10.1103/physrevb.44.3934.

Yes, thank you. So, I got this one right.


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What exactly is a neutralisation reaction?
I personally understood a neutralisation reaction as a reaction for which the pH after the reaction gets closer (or becomes) 7 (neutral pH). But the definition in my textbook says that :

A neutralisation reaction is a reaction during which an acid reacts with a base to form a salt and water only.

I understand the part :

A neutralisation reaction is a reaction during which an acid reacts with a base

since, given that the acid is an H+ donor and the base and H+ acceptor, they will react to form water and bring the pH back to 7. But why should they react to form water and a salt only ? The reaction of metal carbonates with acids also produce carbon dioxide gas along with a salt and water and if metal carbonates are also bases, then when they react with acids, aren’t these also neutralisation reactions? Furthermore, even if metals are not bases, they will react with acids to form a salt and hydrogen gas, which will also cause H+ concentration to decrease and pH to become closer to 7. So why isn’t  this also a neutralisation reaction?
Can someone please give a better definition of a neutralisation reaction and clarify which of the above is or is not a neutralisation reaction, and a reason for why? Thanks

Neutralization is not limited to water solutions and pH changes. See definitions of acids and bases by Arrhenius, Broensted-Lawry, Lewis and Usanovich. The most used are the middle two.

There are multiple definitions of acid and base.  If one picks the one by Brønstedt and Lowry, than indeed the acid is the proton donor, the base the proton acceptor; and the transfer of the proton from the acid to the base is the neutralization reaction.
However, e.g. hydrochloric acid isn't just water and $\ce{H+}$, and lye isn't only water and $\ce{OH^-}$.  For charge compensation, the former solution equally contains $\ce{Cl^-}$, and the later $\ce{Na+}$ ions.  In this neutralization reaction yields water, $\ce{H2O}$.  Simultaneously, $\ce{Na+}$ and $\ce{Cl^-}$ however remain untouched and in the solution which (at equivalence point of acid and base) is like a neutral aqueous solution containing (dissociated) salt.
In the case of neutralizing e.g., an aqueous solution $\ce{HCl}$ with $\ce{Na2CO3}$, the product of neutralization is not water, but (formally) $\ce{H2CO3}$.  It is one of the properties of carbonic acid that this one is in a chemical equilibrium
$$\ce{H2CO3 <=>>[H2O] CO2 ^ + H2O}$$
where the subsequent decomposition releasing gaseous $\ce{CO2}$ is favoured.  Like the earlier case, $\ce{Na+}$ by the base and $\ce{Cl^-}$ of the acid are not affected by the neutralization reaction, nor alter the pH value of the solution.  They remain in aqueous solution.

Note, it is by convention that the equivalence point for the neutralization of $\ce{OH^-}$ by a strong base (e.g., $\ce{NaOH}$) with $\ce{H^+}$ by a strong acid (e.g., $\ce{HCl}$) in water at ambient conditions with $c(\ce{OH^-}) = c(\ce{H^+}) = \pu{10E-7 mol/L}$ (measured quantities) is set as $\text{pH} = 7$ (a derived scale).  It is special case though; for the neutralization of e.g., a weak acid (e.g., acetic acid), the equivalence point need not be at $\text{pH} = 7$.  In case of acids like $\ce{H2SO4}$ and $\ce{H3PO4}$, there are multiple equivalence points, too (link to a program to predict them).

But is the reaction of metals with acids also a neutralisation reaction?

The answer addresses the theory by Brønstedt/Lowry (like the one by Arrhenius with focus on the proton, regardless if $\ce{H^+}$, or $\ce{H3O^+}$ is a better description for the reaction in aqueous solutions). $\ce{2HCl + Zn ->[H2O] H2 + ZnCl2}$ however is redox reaction because electrons are exchanged (cf. change of the oxidation numbers). Among other theories, the one by Lewis tracks electrons. But since there aren't Zn molecules, redox reaction still seems to fit better.


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Q-TOF vs Q-orbitrap MS data
I recently had the opportunity to analyze the same (intact protein) sample on an Agilent QTOF and a Thermo Q-Exactive, and noticed that in addition to having a moderately higher resolution the orbitrap appeared to generate much "cleaner" deconvoluted data, as in many apparent adduct/variant/glycoform peaks that were present in the QTOF data were completely absent for the orbitrap data.
What is the source of this difference? Could the orbitrap be simply not detecting these other species or the QTOF generating these other species as artifacts?

I’m no expert in MS, but is this an unexpected result? They’re different kinds of instrumentation, and much of the product literature for Orbitrap systems focuses on cleaner baselines and better resolutions for intact proteins

(Only user-perspective about MS, rather than method development:) higher resolution reads like different a priori design features and parameter settings of the MS experiment yield results of better quality (like low resolution bench-top vs. sector field or tandem MS-MS high-res MS, or using MSes by different vendors).  Deconvolution, on the other hand, is about intelligent signal processing of already recorded raw data.  So, there may be a blend of reasons.

@NotEvans It's not without tradeoffs though. I know that higher resolution comes at the cost of slower scanning speed, but could cleaner baseline also be coming at the cost of non-detection of lower abundance species?

Deconvolution is very noise sensitive and we have no clue how the black-boxes software enchance resolution. They will never tell.

You might get  decent response if you add more details about deconvolution. What deconvolution method are you using? What is meant by cleaner? Does it mean higher signal-to-noise ratio?

Check https://chemistry.stackexchange.com/questions/161823/q-tof-vs-q-orbitrap-ms-data for more details

Lots of differences between TOF instruments and Orbis.  Orbitraps are Fourier Transform mass analyzers, so the longer you let the Orbi "scan" a packet of ions, the cleaner the signal becomes (signal averages towards the positive, noise averages towards zero).  That's where the FT deconvolution comes in.
The drawback to the Orbitrap is with duty cycle.  While the Orbi is "filled" and measuring those ions, there's still a beam of ions coming from the source.  They either get utilized by another mass analyzer/component (i.e. an HCD cell for MS/MS fragmentation in parallel to the Orbitrap acquiring the MS1), or get thrown out entirely.  If you're measuring a complex sample and using liquid chromatography before the MS, that means you're losing out on analyzing an ever-changing beam of ions.

Welcome to SE Chemistry, but I apologize because there are a couple of conceptual problems in the answer above. Noise does not average to zero in time average...this is a persistent myth. David Traficante has addressed this beautifully in his myth-buster article "Time averaging. Does the noise really average toward zero?" https://doi.org/10.1002/cmr.1820030203

Secondly, Fourier deconvolution is wildly noisy. It goes haywire with a small amount of noise even in a clean synthetic signal... thus they would not be using Fourier deconvolution methods and produce a better "deconvolved" spectrum even after applying window functions. Mostly likely they are using more sophisticated deconvolutions such as maximum entropy deconvolution, which is far better than Fourier deconvolution.

The OP never bothered to follow up and explain more. I wish he had added more information about his data.


text:
2-Pyridone tautomer ratio
The following excerpts from different textbooks about keto–enol tautomerism in pyridones seem to be contradictory. Could someone explain which one is right?
MARCH’S ADVANCED ORGANIC CHEMISTRY pg-103 states that:

In many heterocyclic compounds in the liquid phase or in solution, the keto form is more stable, although in the vapor phase the positions of many of these equilibria are reversed. For example, in the equilibrium between 4-pyridone (149) and 4-hydroxypyridine (150), 149 is the only form detectable in ethanolic solution, while 150 predominates in the vapor phase. In other heterocycles, the hydroxy-form predominates. 2-Hydroxypyridone (151) and pyridone-2-thiol
(153) are in equilibrium with their tautomers, 2-pyridone 152 and pyridine 2-thione 154, respectively. In both cases, the most stable form is the hydroxy tautomer, 151 and 153.

Now for the final few lines I think the result is same for both solvents that's why he didn't mention anything about it.
Organic chemistry 2e by Clayden, pg-729  states that:

In fact, 2-hydroxypyridine prefers to exist as the ‘amide’ because that has the advantage of a strong C=O bond and is still aromatic. There are two electrons in each of the C=C double bonds and two also in the lone pair of electrons on the trigonal nitrogen atom of the amide. Delocalization of the lone pair in typical amide style makes the point clearer

but doesn't mention any solvents.

Wikipedia states that :  X-ray crystallography shows that the hydrogen in solid state is closer to the nitrogen than to the oxygen (because of the low electron density at the hydrogen the exact positioning is difficult), and IR-spectroscopy, which shows that the C=O longitudinal frequency is present whilst the O-H frequencies are absent

That's an improvement, although you could have cleaned up the awkward new lines when copying from a PDF, as well as the reference numbers in March which are not really helpful when the associated references aren't provided. I've done that for now. Can you add in book titles, editions, and page numbers please. A proper citation should, at the very least, contain sufficient information to find the quoted text. Finally, on-topic, I suspect that the Clayden extract may be a bit simplified. These things can be quite solvent-dependent, AFAIK, and March says as much.

March didn't mention anything about which solvent is used in the tautomerism I asked, same with clayden. But clayden's reasoning seems correct to me though I don't know if March has errors

"2-Hydroxypyridone (151) and pyridone-2-thiol (153)" are wrong in the book. They should be 2-Hydroxypyridine (151) and pyridine-2-thiol (153).

Ig that is a small error of i and o and could be noticed from diagram, but yeah that doesn't answer my doubt

Why not start with the corresponding entry in Wikipedia as suggested in the comment by @Maurice?  There, it discerns the solid state

«The predominant solid state form is 2-pyridone. This has been confirmed by X-ray crystallography which shows that the hydrogen in solid state is closer to the nitrogen than to the oxygen (because of the low electron density at the hydrogen the exact positioning is difficult), and IR-spectroscopy, which shows that the C=O longitudinal frequency is present whilst the O-H frequencies are absent.»

from the one of the liquid/in solution:

«The tautomerization has been exhaustively studied. The energy difference appears to be very small. Non-polar solvents favour 2-hydroxypyridine whereas polar solvents such as alcohols and water favour the 2-pyridone.
The energy difference for the two tautomers in the gas phase was measured by IR-spectroscopy to be $2.43$ to $\pu{3.3 kJ/mol}$ for the solid state and $\pu{8.95 kJ/mol}$ and $\pu{8.83 kJ/mol}$ for the liquid state.»

Don't forget $\Delta{}G = -RT \ln{}K$ when accessing the primary references both sections point to, too.
There equally are databases about keto-enol equilibria.  The authors of Tautobase for example deposit the extracted data into a public repository on GitHub.  1680 pairs may comfortably read and searched e.g., by (sub)structure with DataWarrior.  In the present case, the equilibrium tabulated lists for 2-pyridone in water an estimated $\log{}K = 3.5$ (while the gaseous state is listed with $\log{}K = -0.4$):

Note related earlier posts about this topic here on chemistry.se, like this, or this, too.
References:
Wahl, O.; Sander, T. Tautobase: An Open Tautomer Database. J. Chem. Inf. Model. 2020, 60, 1085-1089; doi 10.1021/acs.jcim.0c00035.
Sander, T.;  Freyss, J.; von Korff, M.; Rufener, C. DataWarrior: An Open-Source Program For Chemistry Aware Data Visualization And Analysis. J. Chem. Inf. Model. 2015, 55, 460-473; doi 10.1021/ci500588j.

March states that "In other heterocycles, the hydroxy-form predominates. 2-Hydroxypyridone and pyridone-2-thiol are in equilibrium with their tautomers, 2-pyridone and pyridine 2-thione , respectively. In both cases, the most stable form is the hydroxy tautomer." By which I think he means that solvent has no effect on this as in the line before he stated how solvent affected the other tautomerism and in this he said that it dominates

The same database (again, estimated for an aqueous solution) lists for pyridine-2-thiol vs. the thione a $\log{}K = 4.6$ (i.e., thione is the favoured form, too), while pyridine-2-amine is favoured as amine (listed as $\log{}K = -6.5$). Maybe, on occasion, the database's listing of $\log{}K$ - while text books tend to use $pK_a = - \log{}K$ - might pass unnoticed. In line with the database and against «2-Hydroxypyridone and pyridone-2-thiol [...] the most stable form is the hydroxy tautomer» is e.g., Baran's yearly hetchem lecture (video, 25:47).

Doesn't that video say that keto form is more stable? why you're saying hydroxy or did I get you wrong

Likely you got me wrong when I used a quote of yours.

Didn't get you sir, the video you linked says keto form better due to c=o etc..as mentioned in clayden so which one is right, it didn't mention about solvents ig

One factor to consider is the role of hydrogen bonding.
In the gas phase at low pressure there is no hydrogen bonding because the molecules are too isolated; then the hydroxyl form with a pure cyclic conjugation = maximum aromaticity is more stable. In the neat liquid or in a polar solvent, the keto form will have stronger hydrogen bonding due to the carbonyl group being more electron-rich, thus a better proton-acceptor, and the N-H hydrogen with a positive charge on nitrogen acting as a better proton donor for the bond.

If its given in a question where solvent is not mentioned (I have never seen solvent being given when asked to compare keto-enol forms) then which one to mark, I mean which is the general state where we consider tautomerism?

«Neat liquid» is synonym to «the pure compound itself» (i.e., no solvent added); and then (for 2-hydroxypyridine vs 2-pyridone), the form of the pyridone is dominant over the other.


text:
What is the dimension of molar mass?
I was told that the number of moles is ‘dimensionless’. Does this mean that the dimension of molar mass is simply mass and hence the unit g/mol can be expressed as equal or proportional to the dalton?

The mole   (1 mol ) is 1 of 7 base units of SI system for many years  (m, kg, s, A, K, cd, mol).

The quantity “amount of substance” is not dimensionless and it shall not be called “number of moles” just as the quantity “mass” shall not be called “number of kilograms”.

I think we use 'number of moles' as an informal way of saying it, I've seen chemistry lecturers/students and books use the term any times.

The definition of 'amount of substance' acknowledges 'number of moles' as a deprecated term. It is no longer acceptable, even though it is still very popular in a few reunions of this world.

In some sense, any number alone is dimensionless. Even the Avogadro number is dimensionless, as the opposite to the Avogadro constant.

According to ISO ISO/IEC 80000 Quantities and units, BIPM The International System of Units (SI), IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book), NIST Guide for the Use of the International System of Units (SI) etc., the quantity amount of substance is not dimensionless. Amount of substance and mass are two of the seven ISQ base quantities; the corresponding SI base units are the mole and the kilogram, respectively.
The quantity molar mass (quantity symbol: $M$) for a pure sample is defined as
$$M = m/n$$
where $m$ is mass and $n$ is amount of substance.
The dimension of the molar mass is
$$\dim M = \mathsf{M}\;\mathsf{N}^{-1}$$
The coherent SI unit for molar mass is ‘kilogram per mole’ (unit symbol: $\mathrm{kg/mol}$).
(The usually used unit is gram per mole ($\mathrm{g/mol}$) rather than kilogram per mole ($\mathrm{kg/mol}$).)

From the IUPAC Gold Book:

The mole, symbol $\pu{mol},$ is the SI unit of amount of substance. One mole contains exactly $\pu{6.02214076E23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_\mathrm{A}$, when expressed in the unit $\pu{mol^-1}$ and is called the Avogadro number.

So a mole is a short notation for a certain amount of molecules/atoms etc. In this respect, it is similar to a e.g. a dozen which is equal to 12 elementary entities.
The unit of the molar mass is $\pu{g mol^-1}$. So, one could argue that indeed the unit Dalton, for which you also find a definition in the Gold Book, and $\pu{g mol^-1}$ are equal:
$$\frac{\pu{g mol^-1}}{\pu{Da}}=\frac{\pu{g}}{\pu{6.022E23}\times\pu{1.661E-24 g}}=1$$

would we consider the dimension of g/mol to have equal dimension to mass in that case? I understand the idea of using dozen but a mole is specifically a unit of molecules or atoms, where a dozen is similar to just saying a grouping of  12 of any objects.

@user1007028, you could use the unit $\ce{mol}$ for anything, such as cars, houses, planets... you just usually only have too small amounts to make this reasonable. You could drop the unit altogether also in chemistry, you then would have to work with very small numbers (masses) or large numbers (amounts). The unit is a very handy helper. Between $\ce{Da}$ and $\ce{g mol^{-1}}$, the latter spells out more explicitly how the mass is normalized.

Using the current official definition, you should not substitute a dimensionless number for the unit mole. A dozen is equal to 12, but a mole is not equal to the Avogadro number in the SI unit system. Thus, g/mol and Da are not equal in the SI system, and the given equation is incorrect. However, anyone is free to switch to another unit system; they just have to explicitly state it. (Quantum physicists for example like using the Hartree atomic units, which are all dimensionless.)

I was told that the number of moles is ‘dimensionless’.

According to the official definition, mole is a unit that has a dimension. This is counterintutitive perhaps given its definition ("the amount containing $\pu{6.02214076E23}$ particles"), but is has a dimension nonetheless.

Does this mean that the dimension of molar mass is simply mass and hence the unit g/mol can be expressed as equal or proportional to the dalton?

The dalton is set as 1/12 of the mass of the 12-carbon isotope, so it has dimensions of a mass. As measurements get more accurate, the value will change a bit over time. The mass of a particle and the molar mass of the pure substance made of those particles is proportional. The ratio of g/mol to Da, using current measurements, is 0.99999999965(30) times the Avogadro constant. So g/mol and dalton have different dimensions and can not be equal.
To say they are proportional does not make sense because both of them are units, not measurements. So while it makes sense to say that the mass and volume of samples of gold are proportional (at certain temperature and pressure), and call the proportionality factor the density of gold, it makes no sense to say that the units mL and g are proportional.

I should have said 'differ by a factor of a pure number' such as 1kg = 1000kg, they need to have the same dimension on each side for this to be true.

Yes, I acknowledge that your and @Loong's answer follow the current official interpretation of the definition of the mole, and mine does not. Anyway, the paper you linked by Klaus Schmidt-Rohr is a very good treatise on this issue (thank you for sharing) and I fully agree with his reasoning.


text:
H+ or H3O+ in mitocondrial electronic transport chain
Just learnt that $\ce{H+}$ (aq) is equal to $\ce{H3O+}$ and now I know why.
I'm thinking about cell processes that involve $\ce{H+}$ transport. I wonder if that $\ce{H+}$ is $\ce{H3O+}$ too, like mitochondrial electron transport chain.

$\ce{H3O+}$, written as $\ce{H3O+}$

Already at the level of composition, $\ce{H^+}$ and $\ce{H3O^+}$ are not equal to each other.  In addition, given the high charge density $\ce{H+}$ would carry, $\ce{H3O^+}$ and «higher forms» of hydronium ions are more likely in aqueous solutions.

This gif might help you understand how $\ce{H+}$ is transferred.

First of all, a proton is highly unstable, because it is an ionized H atom. Therefore, it always wants to find a potential well to be stablized (just like electrons want to find a potential well on neuclei; however, note that the two potential wells are inverted due to opposite signs). A proton is often found to bind to lone-pair electrons ":", e.g. the lone pair of a $\ce{H2O}$ molecule, to give $\ce{H3O+}$. Neutralization reaction between HCl and $\ce{NH3}$ in water can be thought of as HCl being dissociated, with the proton leaving the lone pair on $\ce{Cl-}$ and being transferred to the lone pair on water forming $\ce{H3O+}$, and then to $\ce{NH3}$ forming $\ce{NH4+}$. Using gas phase data as a tentative reference, $\ce{H2O}$ and $\ce{NH3}$ have proton affinities of 697 and 854 kJ/mol (see https://en.wikipedia.org/wiki/Proton_affinity_(data_page)). Therefore, eventually proton prefers to bind to $\ce{NH3}$ than $\ce{H2O}$.
However, in water the proton may be shared between two water molecules, which in turn are solvated by a shell of other water molecules. Therefore, it is not accurate to say "$\ce{H+}$(aq) is equal to $\ce{H3O+}$". If you read the Wiki page on Hydronium https://en.wikipedia.org/wiki/Hydronium, you will see the following statement relevant to your question: "Three main structures for the aqueous proton have garnered experimental support: The Eigen cation, which is a tetrahydrate, $\ce{H3O+(H2O)3}$; the Zundel cation, which is a symmetric dihydrate, $\ce{H+(H2O)2}$; and the Stoyanov cation, an expanded Zundel cation, which is a hexahydrate: $\ce{H+(H2O)2(H2O)4}$. Spectroscopic evidence from well-defined IR spectra overwhelmingly supports the Stoyanov cation as the predominant form. For this reason, it has been suggested that wherever possible, the symbol $\ce{H+}$(aq) should be used instead of the hydronium ion."
Next about $\ce{H+}$ transport mechanism, we can look at how protons are transported in proton-exchange membranes (PEMs) such as Nafion. There are essentially three types of mechanisms: Grotthus mechanism, vehicle mechanism, and surface mechanism. The following figure is taken from a book titled "Polymer Science: A Comprehensive Reference - Volume 10 Polymers for a sustainable environment and green energy." Grotthus mechanism involves consecutive hop of proton and turn of water molecules so that the proton being transported migrates between neighboring water molecules. In this case, imagine a team of soccer players standing in a line, which does not have to be straigt line. Each soccer player has two arms each holding one soccer standing for hydrogen atom. Each has two legs standanding for two lone-pair orbitals (two arms and two legs are equal to eight valence electrons and each limb equal to two valence electrons). Then a soccer is passed from one player to another, where each player uses only one of his/her leg to take the ball and then pass to the next one in the line. (Note this picture is simplified - the soccer passed to the next player could be the one on one of the two hands.) During this process, the hop and turn movements are involved. This is how you can understand Grotthus mechanism. In contrast, vehicle mechanism involves a proton-carrying molecule to do all the work instead of relying on a team work. These two mechanisms take place in the aqueous region between Nafion surfaces. A third mechanism called surface mechanism takes place on the surface of Nafion in the boundary region, which takes advantage of the lone pairs on the terminal sulfonic acid units.
With that background said, how protons are transported in cell processes depends on where it takes place, either through the cell membrane or within a cell. You have to be more specific about the question. It can be imagined that due to the presence of water, protons bind to the lone pairs of water molecules, and therefore both Grotthus mechanism and vehicle mechanism are possible. In addition, the surface of some cell structures may have lone pairs as well, which may faciliate the surface mechanism. In any case, protons do not exist by themselves but bind to lone pairs and undergo transport from lone pairs to lone pairs. Note that between the transfer from one lone pair to another lone pair, there should be an energy barrier separating the two potential wells. However, protons are very light and may display some degree of wave character, and therefore, they may undergo quantum tunneling to realize transfer without needing to fully climb up the energy barrier. Hope this helps.

First of all, a proton is highly unstable, because it is an ionized H atom.  Rather an ionized H atom is highly unstable, because it is a proton.

:-) Just one way to understand it :-)

$\ce{H+}$ (aq) is equal to $\ce{H3O+}$

It would be more accurate to say $\ce{H+ (aq)}$  is equivalent to $\ce{H3O+ (aq)}$. This acknowledges that the two species have different mass, and that it does not matter how to write them because they are surrounded by plenty of water anyway, and the water reacts with the species, leading to a highly dynamic situation.

I wonder if that $\ce{H+}$ is $\ce{H3O+}$ too, like mitochondrial electron transport chain.

There are several ways how protons are thought to be transported in the electron transport chain:

As part of redox half-reactions like $\ce{2H+ + 2e- + Q -> QH2}$. If this half reaction occurs on one side of the membrane and the reverse occurs on the other side, you get a net transport of 2 protons. The protons travel neither in the form of $\ce{H+}$ or $\ce{H3O+}$, but rather covalently attached to the Q electron carrrier. This is well-established for complex III.

As part of protonation and deprotonation reactions, either involving water or other acids and bases. Complex I has a "water wire" that enables multiple such steps. It also has several amino acid side chains that get protonated and deprotonated. For example, histidine can gain a proton on one side of its ring and donate a second proton on the other side of its ring. In complex IV (cytochrome oxidase), the acid/base reactions are thought to be driven by electron transfer leading to local electrical fields.

Protons can tunnel between proton carriers. This is invoked when the distances between proton carriers are too large to form hydrogen bonds.

You might talk of $\ce{H3O+}$ transfer for mechanism 2) involving water molecules. Otherwise, it makes more sense to use the term proton transfer.
Some sources:

Complex IV  https://www.pnas.org/content/95/22/12747
Complex I https://www.biorxiv.org/content/biorxiv/early/2021/04/16/2021.04.16.440187.full.pdf
Complex III https://en.wikipedia.org/wiki/Q_cycle


text:
Why does temperature increase rate of electrolysis even though resistivity of electrodes increase?
My experiment is just the electrolysis of iron rust at different temperatures. It consists of a rusted iron cathode and an iron anode in a salt water solution. From my data, there is an approximately 0.032A increase in current per degree Celsius.
I understand that by increasing temperature, you increase the conductance of the electrolyte and thus the current and rate, but I don't quite understand how this increase in conductance outweighs the increase in resistivity of electrodes.
It would be good if there are any equations that describe the relationship between conductivity/conductance/current in salt water and resistivity of iron/iron rust, and temperature.
Thanks in advance

Resistivity of electrodes  has negligible importance here.

The short answer as to why the rate of electrolysis increases with temperature likely relates to elevated efficiency. However depending on the nature of the cell, there can be an offset in the cell's corresponding longevity.
In case of the electrolysis of molten sodium chloride, for example, the ions are free to migrate to the electrodes of an electrolytic cell. At a high temperature the process is more efficient, in essence similar to the functioning of a good electrolyte.
Electrode surface area is also an important feature impacting the rate of reaction. To the extent that increased kinetic energy positively impact the electrodes performance, elevating temperature could be positive. However, in time this can be offset by an increase in electrode deterioration.
More technically per this source is a rise of efficiency from heating may be related to an electrode's particular overpotential, to quote:

Under conditions of electrolysis, the cell is operating away from its equilibrium (reversible) potentials determined from Thermodynamics. Certain electrode reactions are very fast and depart very little from the equilibrium potential. Such reactions are frequently referred to as reversible (see Figure 1). Other electrode reactions are inherently slow and require a potential, E, significantly greater in magnitude than the equilibrium potential to achieve a reasonable current density. This potential is called the overpotential, η (=E–Ee), and the electrode is the said to be polarized. Such reactions are referred to as irreversible (Figure 2). As overpotential is increased in magnitude (more negative for cathodic processes, more positive for anodic processes) current density increases, typically exponentially at high overpotentials. The relationship between current density and electrode potential is the subject of electrode kinetics.

And also, to answer the resistivity part of your question as altering the electrode coating likely elevates resistance but can address overpotential:

For example, in chlorine cells anode design has been revolutionized in the 1960s by the use of coated titanium electrodes (so-called dimensional stable anodes or DSAs) as replacements to carbon. The coating, based on ruthenium oxide, valve metals, precious metals and transition metals, gave significant reductions in overpotential (< 50 mV for chlorine generation). Such materials are widely used for many commercial electrolyses.

Lastly, in the particular case of a galvanic cell, the total chemical based source of electrons can be discharged sooner with a rise in temperature based on a corresponding increase in the reaction rate, a known function of temperature.
So some statistical curve fitting can be performed, but the time to failure of the cell for various reasons should also be noted at higher temperatures.


text:
How can we find the normalised wave function for this particle?

A particle moving on the x-axis has a probability of $1/5$ for being in the interval $(-d-a,-d+a)$ and $4/5$ for being in the interval $(d-a,d+a)$, where $d \gg a$.
The normalised wave function for the "left" interval is $\phi_-$ and for the "right" interval is $\phi_+$. What is the normalised wave function $\phi_x$ for the particle.

From Atkins' Physical Chemistry; Chapter 7 Quantum Mechanics, International Edition; Oxford University Press, Madison Avenue New York; ISBN 978-0-19-881474-0; p. 234:

It's always possible to find a normalisation constant N such that the probability density become equal to $|\phi|^2$

This means that:
$$\begin{align}
\int_{-d-a}^{-d+a}|\phi_-|^2 \,\mathrm{d}x &= \frac{1}{5} \tag{1} \\
\int_{d-a}^{d+a}|\phi_+|^2 \,\mathrm{d}x &= \frac{4}{5} \tag{2}
\end{align}$$ $$$$
Since $d \gg a$, $$|\phi_-|^2 = \frac{1}{5 \cdot 2a}$$ and  $$|\phi_+|^2 = \frac{4}{5 \cdot 2a}$$
Also we can say $\phi=c_1\phi_-+c_2\phi_+$, so $$\phi \cdot \phi^*=|\phi|^2$$
$$\implies|\phi|^2=|c_1\phi_-|^2+|c_2\phi_+|^2+2c_1c_2^*\phi_-\phi_+^*$$.
They have written the solution as $\phi = (1/\sqrt{5})\phi_-+ (2/\sqrt{5})\phi_+$. I'm not able to understand how they came to this result.

This problem can be thought of as a linear combination of atomic orbitals $\phi_-$ and $\phi_+$ to molecular orbital $\phi$ with broken symmetry (i.e. LCAO-MO and $c_1 \neq c_2$). The answer to it can be figured out as follows.
As stated in the conditions, the normalized atomic orbitals are $\phi_-$ and $\phi_+$ for the left and right intervals centered at $-d$ and $+d$, respectively. Since they are normalized, the integration of probability density of atomic orbitals in eqns. 1 and 2 should be equal to 1 for each.
When you integrate the probability density of the total wave function shown in the last equation, you don't need to consider the complex form. In addition, the first term can be integrated within $[-d-a,-d+a]$ to $c_1^2\int|\phi_-|^2 \,\mathrm{d}x = c_1^2 = 1/5$, the second term can be integrated within $[d-a,d+a]$ to $c_2^2\int|\phi_+|^2 \,\mathrm{d}x = c_2^2 = 4/5$, and the third term is integrated to zero due to the absence of overlap. Note that for simplicity, the open intervals $(-d-a,-d+a)$ and $(d-a,d+a)$ are changed to closed intervals $[-d-a,-d+a]$ and $[d-a,d+a]$, as the integration in open and closed intervals should lead to the same result (see Integrating on open vs. closed intervals on Mathematics.SE). This gives $c_1=1/\sqrt5$ and $c_2=2/\sqrt5$, which in turn means $\phi=(1/\sqrt5)\phi_- + (2/\sqrt5)\phi_+$.
The solution indicates that the total wave function has a constructive combination of the two $\phi_-$ and $\phi_+$ orbitals. This type of solution can be seen in the ground-state broken-symmetry solution of $\ce{H2}$ due to non-dynamic electron correlation, as the two H atoms are stretched to a bond length longer than the Coulson-Fischer point, where the two energy curves obtained from restricted and unrestricted (symmetric and broken-symmetry) wave functions start to bifurcate from each other. In addition, you can imagine that a node may appear in the lowest excited state, which means that the total wave function for that state has a destructive combination of the two atomic orbitals.
Of course, this problem is a simplified version of the practical problem because in reality there is an overlap between the two atomic orbitals unless the interatomic distance is stretched to very long where the overlap asymptotically approaches zero. However, I don't think the problem is aimed to teach about electron correlation or overlap but is used to familiarize students with LCAO-MO.


text:
Reaction between anilinium chloride with ferric chloride
I have a question about reaction between aniline and $\ce{FeCl3}$. I supposed that the following reaction proceeds with formation of brown precipitate:
$$\ce{3C6H5NH2 + FeCl3 + 3H2O -> \underset{brown ppt.}{Fe(OH)3} \downarrow + 3C6H5NH3^+Cl^-}$$
But if we firstly acidify the solution, it’s colorless:
$$\ce{C6H5NH2 + HCl -> C6H5NH3^+Cl^-}$$
If we then add $\ce{FeCl3}$, solution gives pale green color. What substance determines the color if anilinium chloride is colorless?
Here’s the video of experiment. After addition of aqueous solution of $\ce{FeCl3}$ to aniline, there’s no formation of precipitate ($\ce{Fe(OH)3}$ or polymer), only color changes.
Also in book Practical Organic Chemistry for biology students, there’s the following information: o-toludine in such reaction gives green color, p-toludine gives brown color, α-Naphthyl amine gives green ppt.

This looks more like an inorganic reaction. It would be an organic reaction, for instance, if it were a $\ce{FeCl3}$ catalyzed Friedel-Crafts reaction.

Note the other precipitate. Iron hydroxide or iron oxide hydrate is brown in color. On the other hand, in the second reaction, aniline is oxidised to polyaniline and iron(III) chloride is reduced to iron(II) chloride giving pale green solution. Here is the reaction performed electrolytically: https://www.researchgate.net/figure/Scheme-1-Schematical-presentation-of-the-initiation-of-ANI-polymerization-when_fig3_26446343

@NilayGhosh here’s the video of experiment in a test tube. I’m not sure, maybe precipitate will fall out in other conditions, different from this video?

The answer is FeCl2, which per Wikipedia:

FeCl2 crystallizes from water as the greenish tetrahydrate, which is the form that is most commonly encountered in commerce and the laboratory. There is also a dihydrate. The compound is highly soluble in water, giving pale green solutions.

You apparently have a reverse reaction observed in the normal REDOX reaction system (think Photo-Fenton with oxygen exposure acting as a stand-in for $\ce{H2O2}$, or more precisely, via the light assisted formation of $\ce{.HO2}$ or Hydroperoxyl radical):

$\ce{ Fe+++ + e- -> Fe++}$

In particular, I point to Eq (3.4) per this Science Direct source:

$\ce{Fe+++ + .HO2 -> Fe++ + O2 + H+}$

This is just part of classic Fenton chemistry, albeit, I suspect that the organic here also actually nicely promotes solvated electron activity as well.

What is oxidized, if Fe is reduced from 3+ to 2+? Thanks.

According to Wikipedia, aniline is oxidized by $\ce{FeCl3}$, producing  the polymer $\ce{-(C6H5-NH)_n  -}$

This the reverse reaction in a REDOX system, recycling ferric to ferrous. It occurs normally in a REDOX system in a so-called Photo-Fenton reaction. I agree with the comment from Quantum namely: "This looks more like an inorganic reaction".  Is it possible that light is a causative agent here? Note, for clarity, I have added my comments here into my answer.

https://youtu.be/xZZpGSpGHSg - here’s the video of experiment. After addition of aqueous solution of FeCl3 to acidified aniline, there’s no formation of precipitate (Fe(OH)3 or insoluble polymer —(C6H5–NH)n—), only color changes. Maybe it’s caused by acidified medium?

Ann Ivanov, thank you! You just supported my Fenton chemistry argument.

AJKOER, what is the role of aniline in this reaction?Why’s aniline required?

@AJKOER, Also in book Practical Organic Chemistry for biology students there’s the following information: o-toludine in such reaction gives green color, p-toludine - brown color, α-Naphthyl amine - green ppt, that’s results are different. I’m not sure, but I think, not only Photo-Fenton is observed. What’s your opinion?

Reference: by J Cheng · 2016 · Cited by 33 — Intermolecular C–H Quaternary Alkylation of Aniline Derivatives Induced by Visible-Light Photoredox Catalysis · Abstract · Figure 1 · Scheme 1. at https://pubs.acs.org/doi/10.1021/acs.orglett.6b02179 .

Reference: An enantioselective and branch-regioselective α-allylation of N-methyl anilines under dual palladium/photoredox catalysis is described. Readily available N-methyl anilines are used as formal “hard” alkyl nucleophiles without preactivation at https://pubs.acs.org/doi/10.1021/acscatal.0c00871 .


text:
IUPAC name of benzoyl chloride

I came up with two possible names: benzoyl chloride and benzene carbonyl chloride. I believe both names are correct.
Wikipedia says that the preferred IUPAC name is benzoyl chloride:

Can someone verify it?

Please note that chemical names should generally not be capitalised, unless occurring at the beginning of a sentence / in a title. There's no need to capitalise the 'b' in 'benzoyl chloride': in fact, it's wrong to capitalise it.

@orthocresol noted.

@orthocresol maybe capitalize them in a language where nouns generally are capitalized, like "Deutsch".

@OscarLanzi Justifying improper chemical names using fitting rules from an arbitrary language is a great idea. From now on, I'm going to name my benzoyl chloride half-Russian, half-Japanese with a colon in between and peppered with leetspeak: б3н30ил:塩化. Isn't it beautifully useful?

Just don't assume the rules of English are universal.

@OscarLanzi How so? Are there any non-universal rules for capitalization of common nouns, which the chemical names are?

The relevant rule for the numbering in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-65.5.1 Acyl halides from suffix acids
P-65.5.1.1 Acid halides in which hydroxy groups of all acid groups expressed as the suffix denoting the principal characteristic group (carboxylic, sulfonic, sulfinic, selenonic, etc. acids) have been replaced by halogen atoms ($\ce{F}$, $\ce{Cl}$, $\ce{Br}$, and $\ce{I}$) are named by citing the name of the acyl group (see P-65.1.7) followed by the name(s) of the specific class(es) as a separate word(s), in alphabetical order, each preceded by a multiplicative prefix, as needed.
The names formyl, acetyl, benzoyl, oxalyl, and oxamoyl are retained as preferred
prefixes.

So yes, the name benzoyl chloride is the preferred IUPAC name.
The suggested name "benzene carbonyl chloride" is not in accordance with IUPAC nomenclature. The similar name benzenecarbonyl chloride is a correct systematic name. However, it is not the preferred IUPAC name since "benzoyl" is the preferred prefix.
Also note that the retained name benzoic acid is the preferred IUPAC name, although the systematic name is benzenecarboxylic acid.


text:
Question on Graham's law of diffusion
Identical balloons are connected through identical valves with two identical cylinders. One cylinder contains gaseous helium and the other gaseous nitrogen. Both the gases may be assumed ideal, and both the cylinders weigh equally. Which balloon will be inflated faster when the valves are opened, and why?
(a) The nitrogen balloon will be inflated faster because nitrogen is a heavier element, and so the molecules of nitrogen have greater momentum and thus force the balloon to expand at a greater rate.
(b) The helium balloon will be inflated faster, because helium is a lighter element and so its atoms move faster and can get into the balloon at a greater rate.
(c) The helium balloon will be inflated faster because the helium must be at higher pressure and hence the gas will be forced into the balloon at a greater rate.
(d) It will depend on whether the gases have to flow up or down to enter the balloon. Helium being lighter than air as compared to nitrogen will rise faster than nitrogen. Therefore, if the balloons are at the top of the cylinders, the helium balloon will be inflated faster; and if they are at the bottoms, the nitrogen balloon will be inflated faster.
The above question is from a well known book, the answer given there is option C.
We know that diffusion rate is directly proportional to pressure and PV=nRT (n is number of moles), here in this case number of moles of 'He' will be greater than 'N', R is constant in both cases and volume is given constant but there is no information about temperature in the question, so how can we say that Pressure of 'He' would be greater than 'N'? Also, rate of diffusion is inversely proportional to root to mass, so how option B is wrong?

As for temperature, since nothing is said, it is appropriate to assume it is constant for both tanks. Valves on a tank of gas hold back pressure. Diffusion is a slow process that is generally controlled (if at all) by porosity. You fill up a balloon in 10 seconds with significant pressure; then the balloon deflates over a week by diffusion. The inflated balloon B will DE-flate faster than inflated balloon A because He diffuses faster than N2 thru latex (mylar would take many weeks).

The key words are IDENTICAL HARDWARE and IDENTICAL WEIGHT. Since He is a lighter molecule there must be more He molecules in the tank and a higher pressure. But that doesn't mean it will flow faster. The chemistry is solved now we need a chemical engineer to work out Reynolds numbers viscosities, velocities, J-T effects, etc. It's a stupid question! simply ask : ''Which tank has more molecules and why?"1

additional comment this has nothing to do with Graham's Law of diffusion. That is appropriate when there are no external forces and the motion is random not directed.

Be aware there are tricky tasks where not all info is there to help you. Some may be  provided with intention to confuse or mislead you.

It is overcomplicating the answer to assume it has anything to do with diffusion
Framing this as a "diffusion" problem is a distraction which is why it is hard to see the key features if you start with the assumption diffusion has anything to do with it.
The basic setup is simple. The starting position is two cylinders of gas with the same mass of gas. The key feature here is that, if the two sources have the same mass, there are about 7 times more moles of helium than of nitrogen (helium atoms being about 7 times less weighty than nitrogen molecules). This implies, given basic gas laws, that the pressure in the helium cylinder is far higher than the pressure in the nitrogen cylinder (same volume, but more moles of gas, pressure must be higher). So, if an identical valve is opened, the helium will flow much faster than the nitrogen. We don't even need to calculate how much faster as it is obvious that the flow will be much faster. Diffusion doesn't come into it at all as this is just basic gas law stuff.
The other provided answers are designed to distract from this basic idea with irrelevant complications. C is the correct answer.


text:
Number of chiral centers in Proline
I think it's two, but I am not able to access/find any information on the net that supports my claim or justifies why it's wrong.
One chiral center is the obvious carbon attached to $\text{-(COOH)}$ group. I think another one is the secondary Nitrogen because it has three different groups attached to it + a lone pair. Moreover, it's also a part of a cyclic ring structure so that should slow down Nitrogen Inversion.
What's the actual answer to the above mentioned question and why?

I think you would need stronger obstacle to affect nitrogen  inversion.

From the conceptual perspective applying (CIP)-rules, you are right; beside the carbon atom both $\alpha$ to nitrogen and the carboxylate, the nitrogen may be either up or down the plane of the cycle -- and given the pattern of substitution (including the free electron pair on $\ce{N}$), there could be diastereomers of proline.
However in praxi there are not.  Embedded in a conformational flexible  five membered cycle (with little cyclic strain, compared to cycles of four or three atoms) with single bonds only, the two forms of the free electron pair up or down interconvert into each other so rapidly, that this averages out like there were no stereogenic centre present.  (Build molecular models and try this manually.)  This is reason the Wikipedia article states as general rule

«For a compound that would otherwise be chiral due to a nitrogen stereocenter, nitrogen inversion provides a low energy pathway for racemization, usually making chiral resolution impossible.»
source

(For free ammonia at room temperature, the same page states 30 billion flips per second.)  Thus for proline, count only one stereogenic centre.

Note, rigid cycles may prevent this flips to occur (e.g., Tröger bases).


text:
Does the spin of electron in d half-filled d-orbital changes while forming low spin complex?
Recently I learned how VBT explains the formation of coordination complexes. But I am stuck at a point that seems hard to digest for me.
Consider a metal ion having an electronic configuration as Noble Gasd4 ns np nd.
Now if the ligand nearby is a strong field, it makes the (n-1)d4 configuration as follows:

Now I don't understand why one of the electron of d orbital paired with another in the presence of strong field ligand, and even the spin of electron got changed. How is that possible?

The short answer is yes, the spin state of the system must change, and you are right to question how that could happen, given that spin changes are generally considered "forbidden". I put forbidden in quotation marks, because these reactions in reality happen quite readily despite the prohibition.
As for "how does this happen?", the spin-state transition (termed an "inter-system crossing") is typically enabled by spin-orbit coupling, which is a complicated quantum phenomenon. To get acquainted with some of the basic principles first, I recommend reading up on the topics "spin-forbidden reactions" and "inter-system crossing."

Oh, thanks, sounds really interesting. Looking forward to study more.

Can u please provide me a source for the detailed information about Spin Forbidding Reactions and Inter System Crossing.

These topics should be covered an any advanced physical chemistry book that you have access to. If you don't have access to a textbook, you can get a lot of info free online using those search terms. however, since you've just started VBT, you may want to focus on increasing your general understanding of physical chemistry before diving into the advanced topics.

@SamyakMarathe https://chemistry.stackexchange.com/q/37303/4945 there is a list with many resources. There are free ones on the Internet available.

One such example is $\ce{[Mn(CN)$_6$]^3-}$ with $\mathrm{3d^4}$ electrons. The electrons of the six CN ligands populate the $\mathrm{d^2sp^3}$ hybridized orbitals giving octahedral geometry. That's how it can be understood with the valence bond theory (VBT). To answer your question, you can switch to a different theory. You can imagine the five d orbital split into two groups, i.e., $t_{2g}$ and $e_g$, according to crystal field theory (CFT). Then the $\mathrm{3d^4}$ electrons are filled into the $t_{2g}$ orbitals. Usually people write it as spin-up and spin-down in the leftmost instead of the rightmost box, and then two unpaired electrons in the next two boxes of the $t_{2g}$, where they can be either both spin-ups or both spin-downs. You can just treat this as a result of aufbau principle. I know this does not directly answer your question about spin flip. However, it provides an alternative way of understanding so that you would not have that question.
Please note that there are different types of theories being applied to understand the bonding in coordination compounds, including VBT, CFT, and molecular orbital (MO) theory, etc (https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/07%3A_Coordination_Chemistry_II_-_Bonding/7.01%3A_Theories_of_Electronic_Structure). Each theory looks at the problem in hands from different perspectives. In a way, it is like several blind men trying to tell what an elephant looks like by touching different parts of its body in the well-known Indian parable. Need to point out every theory has its own limitations. As scientists trained after modern MO theory has been established, we would prefer to rely on the more advanced MO theory but VBT and CFT are still simple, elegant, and useful. Note that even MO theory cannot explain it all and sometimes runs into problems, when we have to resort to even higher level of theories that include electron correlations and/or relativistic effects. At the moment, different theories can be used together to gain a unified understanding, as long as their interpretations do not contradict with each other.

Ligand Field Theory does not require the concept of hybridisation. The splitting you have mentioned also does not work with hybridisation as symmetry is not conserved für hybrid orbitals. The answer also does not actually answer the main question.

To stay within the parable: if the blind man touch all of the elephant they should come to the same conclusion. Well, these are humans so nothing is guaranteed. With valence bond (VB) theory and molecular orbital (MO) theory it is different: they are, given the same basis, congruent. The result is the same. So they do have also the same limitations, if they are complete. Also note that there is nothing simple about VB theory. Please also understand that Crystal Field theory is somewhat outdated and has been combined with MO theory to form ligand Field theory many years ago.

All theories mentioned above including VB, CF, and MO were discussed within their applications to coordination compounds. If gone out of that, VB can be very complicated and as powerful as MO theory (see e.g. https://pubs.acs.org/doi/pdf/10.1021/ar030162a), but its application to coordination compounds is simple and elegant. For that same reason, the MO theory applied to coordination compounds is actually Ligand Field theory. Please also understand that Crystal Field theory is still being taught at the college level in the US, see the latest edition of Chemistry the Central Science.

To see some supports for the discussions above, please see: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Inorganic_Coordination_Chemistry_(Landskron)/07%3A_Coordination_Chemistry_II_-_Bonding/7.01%3A_Theories_of_Electronic_Structure (three is no room for this link in the previous comment and therefore it is put here). Hope all these discussions are helpful for the one who asked the question.


text:
Why do silver halide crystals in suspension, develop a surface charge
This is a question related to analog photography. Photographic films have a suspension of silver halide crystals in them, and these crystals have adsorptive properties which are used in this application. I have read that when there are $\ce{Ag+}$ and $\ce{Br-}$ ions in solution along with a crystal of $\ce{AgBr}$, $\ce{Br-}$ ions preferentially bind to the surface of the crystal relative to $\ce{Ag+}$ ions. This creates a net negative charge on the silver halide crystal which plays a role in adsorptive properties.
I want to know why $\ce{Br-}$ ions preferentially bond to the silver halide crystals.
I thought it might have something to do with the Frenkel defect, because of which the $\ce{Ag+}$ ions move from the surface to inner interstitial sites, thus creating a negative charge on the surface. But this would mean that silver halide crystals attract $\ce{Br-}$ ions not $\ce{Br-}$ ions.


text:
lowest melting point of alkali salts
I want to make the alkali metals sodium and potassium through electrolysis of molten salts. However, I only have the ability to heat stuff up to about 260 degrees Celsius.
What would be the lowest melting point salts of sodium and potassium that could be used for this purpose?

Instead of halides, consider hydroxides A eutectic of sodium hand potassium hydroxides melts at 170°C. See:
Sergei Devyatkin. "Interaction of Oxides and Molten Alkalis, Products of Reaction and Application", Sustainable Industrial Processing Summit & Exhibition, Volume 7: Ionic Liquids & Energy Production, Edited by Florian Kongoli, Flogen, 2015, p. 237.

Is there a similar substance that exists as a pure substance? (not a mixture). I want pure sodium or pure potassium.

@user16791137 And you somehow didn't think about eutectic of two or more sodium salts?

@mithoron I am finding your suggestion hard to get by searching, so if you know something do tell. It does seem that NaCl+NaOH does not go below 260°C because the eutectic composition has too much NaOH, and I can't imagine any other binary more likely to succeed.

Maybe sodium tetrachloroaluminate would work.

It would. But I hesitate, because electrolyzing it could deposit aluminum instead of the alkali metal.

https://en.wikipedia.org/wiki/Molten-salt_battery#Na-NiCl2

So, put it in as your answer. I am done, I do not want to be dinged about moving this to chat.


text:
How is O3 GHG gas created by human activity causing climate change?
Please help me undestand these.
Greenhouse gases from human activity directly are: $\ce{CO2}$, $\ce{CH4}$, $\ce{NO2}$, $\ce{O3}$, and the $\ce{F}$-gases.
Wikipedia agrees: https://en.wikipedia.org/wiki/Greenhouse_gas

It is clear that $\ce{CO2}$ and $\ce{NO2}$ is mainly from burning fossil fuel. Solution to decrease climate change effects: burn less (preferable zero) fossil fuel

CH4 is mainly coming agriculture and landfills. Solution to decrease climate change effects: less meat, less landfill (recycle, waste incineration)

F-gases are used in refrigerator, spray, various cooling devices. Solution to decrease climate change: no sprays, using diffent gases in cooling devices.

But where is $\ce{O3}$ (ozone) coming from? I know it is varios layer of the stratosphere, and protects us from UV. But what kind of human activity makes the $\ce{O3}$ in such amount that $\ce{O3}$ is classified as GHG (Greenhouse gas?) How we can prevent generating $\ce{O3}$, in that way we can decrease $\ce{CO2}$, $\ce{CH3}$, $\ce{NO2}$ and $\ce{F}$-gases?
Also I don't understand if $\ce{O3}$ is a GHG, then why the percentage in GHG is missing from most statistics, including inourworldindata.org?
https://ourworldindata.org/greenhouse-gas-emissions
(See this chapter: By gas: how much does each contribute to total greenhouse gas emissions?)
While other reseaches show as a major GHG, but there is no percentage:
https://www.acs.org/content/acs/en/climatescience/greenhousegases/whichgases.html
(Seek for the graph at this text: Source: Figure 2, FAQ 2.1, IPCC Fourth Assessment Report (2007), Chapter 2)
Could you help me understand the $\ce{O3}$ role in this story?
Thank you!

Ozone doesn't last long enough in the atmosphere to be a serious contributor to warming. Our biggest environmental issues with it are not having enough of it in the upper atmosphere (the legacy of CFCs) and having too much in the lower atmosphere where it is an irritant contributing to smog (caused by eg traffic pollution).

Ozone is typical component of the smog of the Los Angeles type, where car emissions in pre-catalyst era were combined with UV light. Even with catalyst, it takes time to reach the catalyst operational temperature. I agree its GHG contribution is rather minimal.

The contribution of ozone to global warming is complex and there is no strong consensus on its impact
The problem with thinking about ozone as a greenhouse gas (alongside, for example, carbon dioxide and CFCs) is that it isn't something human activity created directly or something where we just need less of it.
First, we want more ozone in the upper atmosphere where it is a barrier to UV reaching the earth's surface (the ozone hole was caused by overuse of CFCs which were banned because they severely reduced the amount in the upper atmosphere). The ozone hole has not yet completely healed despite the ban in the 1980s.
Second, ozone is a short lived component of the atmosphere as it is very reactive, so it is hard to see that it can be a big contributor to warming (and would we choose to minimise it if we could given the problem of its protective value in the upper atmosphere?)
But some researchers have argued it does make a notable contribution in the lower atmosphere, but indirectly. There doesn't appear to be a consensus on how big this effect is (and the Wikipedia reference is old and obsolete). The argument for ozone being a notable contributor is based on the following. Hydrocarbon pollution in the lower atmosphere (often from vehicle emissions) leads to a variety of undesirable reactions some of which lead to the production of ozone (as well as many other irritating components of smog). We really don't want too much smog or ozone in the lower atmosphere because it is bad for health. Some have estimated that it also adds to the warming caused by hydrocarbon emissions (exacerbating the warming potential of methane, for example).
It is hard to judge the estimates of its contribution to warming not least because they rely on models of complex reactions caused indirectly by other pollutants. Also, the big issue with emissions leading to ozone are not its contribution to warming but its contribution to pollution which causes direct harm to people in the short term. In fact regulations around emissions has been striving to reduce those emissions since before we started worrying about global warming. And, many countries have sharply reduced them (this is a major reasons why most western countries insist on catalytic converters in their vehicles). We should reduce ozone pollution by reducing the other emissions that cause it and we have been doing that for decades.
I would argue that ozone is essentially irrelevant to global warming. We should strive to reduce it in the lower atmosphere even if we were not worried by global warming. So even if we can't agree on how big its contribution to warming is (which the literature isn't clear on) we should be reducing it as much as we can for more direct reasons.
And, even if we wanted to report its contribution to warming, the best place to account for it is to add it to the contribution of other emissions (eg methane) rather than to account for it separately as we don't directly emit it from anything.


text:
Will electrolysis of molten sodium hypochlorite work
I have an anhydrous mixture of sodium hypochlorite and sodium chlorate that was obtained by boiling bleach. If I were to melt this mixture and perform electrolysis on it, would sodium metal be produced, or would it immediately react with one of the chemicals in the mixture?

Wow, you'd sooner make it from salty soup. That idea is wrong on many levels.

"The anhydrous compound is unstable and may decompose explosively."

No. Sodium hypochlorite decomposes at 101°C.

Boiling down Chlorine Bleach composed of sodium hypochlorite (and sodium chloride owing to its industrial creation from the action of chlorine on NaOH) results in mostly just NaCl and a touch of NaClO3.
It is the first step in a generally inefficient path to chlorate (see, for example, discussion here).
Sodium chlorate liberates upon heating oxygen gas (see Wikipedia) leaving Sodium chloride.
So, expect the same products associated with the electrolysis of Sodium chloride. In particular: Sodium metal and chlorine gas can be obtained with the electrolysis of molten sodium chloride, while the electrolysis of aqueous sodium chloride yields hydrogen and chlorine, with aqueous sodium hydroxide remaining in solution. Source, see this educational reference.


text:
At physiological pH, why isn't the amide of glutamine protonated?
If the amide in the R group of glutamine has a pKa of 15, then pKa is certainly greater than physiological pH. We'd expect it to be pronated, then, but basically all diagrams of glutamine at physiological pH show it just as...an amine with NH2 and C=O.
Why wouldn't the amide be protonated? If so, where would the protonation site be — would it be an NH3⁺ as if it were an amine?

You think into the wrong direction. What is the $\mathrm{p}K_\mathrm{b}$ value of the amide? Or the $\mathrm{p}K_\mathrm{a}$ of the corresponding acid? Also, are you aware that the lone pair on the nitrogen is in conjugation with the carbonyl and what this means for its accessibility for protonation?

pKa 15 belongs to amide as an acid , you want to protonate it as a base. That is a different acid/base pair.

See also: https://chemistry.stackexchange.com/questions/45249/what-test-can-be-used-to-differ-amide-and-amine/46286?r=SearchResults&s=3|19.0845#46286

because it is completely non-basic

There are two equilibria, one between the deprotonated amide (one hydrogen, negative charge) and the amide (two hydrogens, neutral), and the other between the amide (two hydrogens, neutral) and the protonated amide (three hydrogens, positive charge). The major species is the one in the middle.
If you try hard to protonate an amide in aqueous solution, the proton might attach to the oxygen instead of the nitrogen, see e.g. https://pubs.acs.org/doi/10.1021/jo962063z.
A $\mathrm{p}K_\mathrm{a}$ of 15 means that the amide is a very weak acid and will not lose its proton even at very basic pH, i.e. the species on the left will not form in aqueous solution. To give an example using other acids and bases: acetic acid, with a $\mathrm{p}K_\mathrm{a}$ of 3.75, will be deprotonated at pH 7. One the other hand, the ammonium ion, with a $\mathrm{p}K_\mathrm{a}$ of about 9, will be protonated at pH 7. So if the pH is lower than the $\mathrm{p}K_\mathrm{a}$, the major species is the protonated one (of the equilibrium you are looking at). If the pH is higher than the $\mathrm{p}K_\mathrm{a}$, the major species is the deprotonated one.

If the amide in the R group of glutamine has a pKa of 15, then pKa is certainly greater than physiological pH. We'd expect it to be pronated, then, but basically all diagrams of glutamine at physiological pH show it just as...an amine with NH2 and C=O.

It is protonated. If it acted as an acid, it would lose one of its two protons, and aquire a negative charge.

Why wouldn't the amide be protonated? If so, where would the protonation site be — would it be an NH3⁺ as if it were an amine?

The $\mathrm{p}K_\mathrm{a}$ for $\ce{R-CO-NH3+}$ is on the order of -0.5, so it would not be the major species at neutral pH (or in aqueous solution of any pH, for that matter). Instead, it would act as an acid, yielding the amide. As mentioned above, if an amide does get protonated, it might either become O-protonated or N-protonated, depending on the specific amide and the conditions.


text:
Low temperature sodium salt eutectic
I have a need for a sodium salt that melts below 265 degrees celsius, which will be exposed to liquid sodium metal. The anion is not important, only that it has a sodium cation. I have done a lot of research and the nearest thing that I have found is sodium hydrogen sulfate. However, this specific salt will not work for what I need, because of a side reaction, namely that between Na metal and NaHSO4. The only option left that I can think of is a eutectic, consisting of only sodium salts (so no potassium hydroxide or the like). Are there any sodium salt eutectics that have a melting point of less than 265 degrees celsius?

As an additional keyword to consider, ionic liquids (or, IL) is a field with rising interest e.g., in catalysis and electrochemistry. NIST's database #147 compiles some information (including melting points) though it is up to the user to evaluate if these components are compatible with the chemistry intended (since Na may engage in redox reactions, e.g. by electrochemical window, etc.): https://ilthermo.boulder.nist.gov/

WP does not say sodium bisulfate melts anywhere near 265C. Rather it decomposes at 315C to sodium disulfate (Na2S2O7) which remains solid. In any event a eutectic based on NaOH seems better anyway.

Calculation from the FACT thermodynamics package suggests a mixture of about 80 mol% NaOH and 20 mol% NaI (sodium iodide) will melt below 250°C.

Source

Hm, NaOH plus sodium metal?

@Karl The halide component should read Na and then a capital eye, not Na and then a small ell. Unfortunately this confusion in nomenclature is with us unless we change the symbol for iodine from I to J. Sprechen sie Deutsch (iodine -> Jod)?

Ja, it's especially bad with lousy font sets where capital "i" and regular "L" are indistinguishable. But I meant this from the original question above ".. which will be exposed to liquid sodium metal."

As for keeping the metal out of the hydroxide, you do need to keep the temperature low. The old Castner process for getting Na(liq) from molten NaOH required a tight temperature range to keep the electrolyte liquid but limit sodium solubility. Adding the iodide salt to lower the melting point would have actually helped.

Why are you answering this OP all the time? He's asking about the same thing really and I think he's violating homework policy - has a task and wants us to do his research.


text:
Exploiting 8-fold symmetry of ERI tensor for building Coulomb and Exchange matrices
I'm trying to write a restricted Hartree Fock code in Fortran that reads in a file of zeroth-iteration 1 and 2 electron integrals (FCIDUMP format) and uses them to do the SCF procedure from an initial guess of a density matrix. The thing I'm struggling with is the 8-fold symmetry of the 2-electron integrals and how I can leverage it to construct my Coulomb and Exchange matrices efficiently.
FCIDUMP files use the symmetries of 1 and 2 electron integrals to save space, i.e. since $H_{core,p,q}$ = $H_{core,q,p}$, it will only print matrix elements of $H_{core}$ where $p \geq q$. Thus, reading in the matrix elements from FCIDUMP makes a lower triangular matrix which must be symmetrized along the diagonal to get the $H_{core}$ matrix. Easy enough:
h_core = e1ints
do q =1,norb
    do p =1,q
        h_core(p,q) = h_core(q,p) !lower triangular to symmetric
    end do
end do

Where norb is the number of spatial orbitals (ie the dimension of the Fock matrix is norb*norb) and e1ints is the lower triangular matrix I've constructed from FCIDUMP. Of course, it's more memory-efficient to keep the matrix lower-diagonal, but since $H_{core}$ must be added to the Fock matrix eventually, I don't think there's any way around this.
The same printing methodology holds true for the 2 electron integrals, where the symmetry is (pq|rs) = (pq|sr) = (qp|sr) = (qp|rs) = (rs|pq) = (rs|qp) = (sr|qp) = (sr|pq). Where I'm getting tripped up is a) how much more time and memory symmetrizing a "lower triangular" 4-dimensional electron repulsion integral (ERI) tensor will take and b) generally thinking about how to code the following across eight-fold symmetry instead of two-fold:
The general formula for the elements of the Coulomb and Exchange matrices, respectively, are:
$J_{pq} = \sum_{r,s=1}^{norb} (pr|qs)P_{rs}$
and $K_{pq} = \sum_{r,s=1}^{norb} (pr|sq)P_{rs}$
where $P$ is the density matrix. Since of the $norb^4$ elements of the full ERI tensor, I only have $\frac{\frac{norb*(norb+1)}{2} *(\frac{norb*(norb+1)}{2}+1)}{2}$ of them (for a nxn triangular matrix, there are $\frac{n(n+1)}{2}$ nonzero matrix elements). How can I efficiently use this symmetry to construct the $J$ and $K$ matrices without making the ERI tensor huge by symmetrizing it? Since the tensor is not used explicitly in the construction of the Fock matrix, I don't think it makes sense to symmetrize it - I'm just at a loss for how to get around this. There are codes on GitHub that deal with FCIDUMP files, but after many hours of searching, I unfortunately haven't been able to find one that uses these files in the way I need to.

Suggest: by content and purpose of your question, consider to move your question to the sibling site of https://mattermodeling.stackexchange.com/.

For a given two electron integral work out whih elements of the Hamiltonian it contributes to and then add in each of those contributions. You are then finished with that integral - no need to store the full rank 4 tensor, just (at most) the rank 2 Hamiltonian.

BTW do you mean (pq|rs) in your expression for your Coulomb integral?

There's an entire tag for FCIDUMP on Matter Modeling SE (see the link given by @Buttonwood) then click "tags" then "FCIDUMP".


text:
Why is my Povidone-Iodine 10% Solution losing it's color which disappeared, 2 tsp mixed with 6 oz distilled water, sitting out for 3 days?
I was directed to use this method of 2 tsp povidone mixed with 6 oz distilled water for mouth gargling, and also either using the mixed solution in a neti pot on the nose, or using the mixed solution in a spray bottle on the nose.
?1.  I do not know if I should have added salt to make simple saline out of the distilled water.
?2.  The bright brown color disappeared in an open measuring cup mixture of distilled water + povidone after sitting out uncovered on the counter for 2 or 3 days.  Do I have to have a lid on the mixture?  Or do I need to create smaller portions of the mixture so I only prepare what I need in each treatment for only one treatment at a time? And all the original povidone needs to stay in the original bottle unless needed for one specific treatment at a time?  Thank you.

Be aware  that personal medical advices are explicitly off-topic on chemistry SE site. Users may address chemical aspects, but will avoid any medical context. It is also possible the question will be closed due its medical relation. It is best to ask about in a pharmacy or at a pharmacologist.

According to this site

Povidone iodine is inactivated or becomes unstable in the presence of sodium thiosulphate, heat, light or alkaline pH. Once opened – use immediately. Discard immediately after first use.

Note the instability to light. Do not store this open-topped or in a clear glass bottle. You will need to make up your gargle solution fresh each time.

Protection of the product against day light is the reason why these bottles including iodine solutions are either stored in containers of dark brown glass, or of at least dark and opaque plastic (example).

An iodine solution gets discolored if left uncovered after more than one day. And there are at least two reasons : 1) Iodine is slightly volatile and slowly evaporates from any aqueous solution (but this is not the main reason). 2) The French chemist Louis Pasteur has shown in $1880$ that air contains plenty of organic spores, pollen, dust, bacteria and pathogenic agents that are continuously falling down on any aqueous surface, slowly producing polluted water. All this organic material has reducing properties : it will soon reduce brown $\ce{I2}$ to colorless iodide ions $\ce{I-}$. Two or three days is enough to discolor your solution.