text:
How to calculate how much weak acid would dissociate in water?
So I know water has a pH of 7. If we add 0.1 mol of a strong acid inside 1 liter of water, it WILL fully dissociate into 0.1 mol of A- and 0.1 mol of H+ thus making the pH of the solution 1.
But what if we add 0.1 mol of a weak acid HA, let's say acetic acid, in 1 liter of water? How would I know how much of this acid would dissociate in water and what would be the pH of the solution?
Using the found information that only 1% of the acid would dissociate, the pH of solution would be 2. Knowing that the pKa of acetic acid is 4.75, I calculated using Henderson-Hasselback that only 1% of acetic acid would be deprotonated.
But, to calculate the pH, firstly I needed to know how much of it dissociated. So how can I know in first place how much of a weak acid would dissociate in water?

The amount dissociated and the pH are one and the same quantity. In other words the $\ce{H+}$ cations come from the dissociation.

First, you should be given $K_\mathrm{a}$ of the weak acid. If you know it, you can set up the expression for $K_\mathrm{a}$:
$$\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}= K_\mathrm{a}$$
Let $x$ be the amount of the weak acid dissociate at equilibrium. Thus, at equilibrium, $[\ce{H+}] = [\ce{A-}] = x$
and $[\ce{HA}]$ will be its initial concentration minus $x$. This can be derived from an ICE table as well. The equilibrium expression then becomes:
$$\frac{x^2}{[\ce{HA}]_i-x}= K_\mathrm{a}$$
We can assume $[\ce{HA}]_i-x = [\ce{HA}]_i$ because $x$ would be quite small compared to $[\ce{HA}]_i$ since it is a weak acid. Then you get the following expression:
$$\frac{x^2}{[\ce{HA}]_i} \approx K_\mathrm{a}$$
You can then solve for $x$, which would be how much of the weak acid dissociates given the $K_{a}$
One important thing to note is that this approximation will not work well with acids that have a relatively high Ka or there is a low initial concentration. So you should check that $$[HA]_i >> [H^+] >> [OH^-]$$ If not, you cannot simplify the expression and would instead have to solve it as a quadratic.

I would add in a compact way there is assumption of this simplification that $c_0 \gg [\ce{H+}] \gg [\ce{OH-}]$. So generally, after getting pH, it is good to check if these conditions are met.

@M.L: Your $x$ can't be an integer. So, I change it accordingly. I'd like you to check it that you agree or not. I upvoted regardless.

Thanks. x is generally not supposed to be an integer. It is just how much the concentration of the weak acid changes at equilibrium based on the Ka given (If this is what you meant). As for the first comment, I agree that it is important to check that because sometimes Ka might be large enough for the approximation to be inaccurate. I will make this correction.


text:
"Scrubber" for storing HCl
I'm in need of hydrochloric acid for cleaning parts before electroplating, but I'd rather not deal with the HCl fumes given off by it, as I'd rather keep my workshop contents from rusting.
Given that HCl basically from what I've understood escapes any container it's put in, I had the idea of putting the container inside a bigger bucket and putting something in the bottom of the bucket to react with the fumes give off by the hydrochloric acid.
My first idea was either sodium bicarbonate or lye in solid form, would these react with it or do I need to have the base in liquid form? I'd rather not have my hydrochloric acid container covered in a lye solution.

Marble or limestone scraps likely would serve as well, and you could even make a retaining wall with polyethylene netting, so that the cleaning bucket would be well below the level of surrounding limestone chips. Also, keep the opening just large enough tpo admit the parts. Definitely don't use lye, which is perhaps more dangerous than HCL, as well as hygroscopic!

Inexpensive [about US\$0.30 / liter] marble or limestone scraps, $\ce{CaCO3}$, likely would serve well. You could make an outer bucket with polyethylene netting, so that the pickling bucket would be well below the level of surrounding limestone chips. Coarse netting would not work with powders. Make the neck just large enough to admit the parts.
Better would would be to draw air continuously from above the container through a scrubber stack filled with marble or limestone chips, and possibly vent outside.
Definitely don't use lye, $\ce{NaOH}$, which is perhaps more dangerous than $\ce{HCL}$, as well as hygroscopic! You'd wind up with a puddle of caustic lye solution at the bottom of the outer bucket. And $\ce{NaHCO3}$ would also make a mess, though non-toxic.
BTW, if using large amounts, be careful that a spill into the outer container doesn't release so much $\ce{CO2}$ as to displace air and asphyxiate someone.

We're talking maybe 5 liters here of HCl at 30ish % at the most. My idea would be to seal both containers, and just make the outer bucket say 20 liters or so. I'll have to think about a scrubber stack instead, that way I could theoretically at least make a water trap with some suitable solution in it.

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text:
Why doesn't aromatic substitution add at multiple positions in one reaction?
When you do aromatic substitutions, you usually see the major products with one substituent added. My question is why can't you add multiple substituents in one reaction? For example, the image below shows nitrobenzene reacting with $\ce{Br2}$ and $\ce{FeBr3}$ to add $\ce{Br}$ to the meta position. Why can't the reaction continue and add to the other meta position and thus make 1,3-Dibromo-5-nitrobenzene?

Do we get even this?  I have read elsewhere that F-C requires a fairly active ring, and a nitro group could be too deactivating for even one substitution.

@Oscar Lanzi: Yes it is possible with $\ce{Fe}$ powder.

The rate of Aromatic substitution depends upon the activity of the aromatic system, because when the collision happens the aromatic system has to donate electrons to an electrophile.
In the example above you have used nitro-benzene, which is very strongly deactivated due to the $\ce{-NO2}$ group and thus the ortho and para positions are completely blocked. The meta position of nitro-benzene is not good but also isn't as bad as the para and ortho positions, in terms of electron density. So, we can say that the meta position of nitro-benzene is just a smidgen less electron rich than normal benzene, and thus first bromination takes place. But, after the addition of one bromine the ring becomes even more electron deficient which just makes further bromination even harder and slower. The reaction slows down to such an extent that it practically ceases to proceed.
On the contrary, if you would use a strongly activated aromatic system like phenol or aniline. It would provide such electron density to the para and ortho positions that they would undergo complete bromination, forming a 2,4,6-tribromo product. The electron density provided by $\ce{-OH}$ and $\ce{-NH2}$ is simply so high that it overcomes the electron deficiency created by the added bromines.

During a electrophilic aromatic substitution, it is always possible to have multiple substitutions during one reaction. However, your example is not the ideal one for a discussion. As noted, Oscar Lanzi has questioned even aromatic ring of nitrobenzene is active enough to give even one $\ce{Br}$ substitution. To clear that, I have found a reliable reference: According to Ref.1, it is possible to brominate nitrobenzene if you have used $\ce{Fe}$ powder in the place of $\ce{FeBr3}$ and at high temperature:

Once brominated, the benzene nucleus is even more deactivated by the newly added $\ce{Br}$ atom than it was with nitro group, which is a strong deactivating group (c.f., the relative rate of reactivity of benzene and chlorobenzene is $1.0$ and $0.033$). This phenomena prevented the additional bromination of the ring even with a stronger catalyst.
One of the best example for the electrophilic aromatic multi-substitution is bromination of phenols and anilines because of the presence of strongly activating groups on the ring such as $\ce{OH}$ in the phenols and $\ce{NH2}$ in the anilines.
Many monohydric phenols react rapidly with bromine in polar solvents such as water to yield solid bromination products, all most all of them are multi-substituted (Ref.2). All the vacant ortho- and para-positions are usually substituted. For example, phenol itself yields an immediate quantitative precipitate of 2,4,6-tribromophenol (Wikipedia):

It is said that controlled partial halogenation of phenols can be effected by using the theoretical amount of halogen dissolved in a less polar solvent such as carbon disulphide and chloroform (Ref.2). However, according to my experience in organic chemistry laboratory, even this reaction gives a significant amounts of minor products with more than one $\ce{Br}$ substitution. The best method to achieve this task is blocking the activating group, in this case, $\ce{OH}$ group. In that way, the blocking group not only significantly deactivate the ring, but also provide enough steric hindrance, which prevent substitution at both ortho-positions.
For instance, the bromination of phenyl acetate would gives you exclusively para-substitution (4-bromophenyl acetate). However, you may need to have a prior arrangement to remove $\ce{HBr}$ formed during the bromination in this case. That's because acetyl protecting group is acid sensitive. For example, the bromination of phenyl acetate have given almost exclusively the para-isomer in quantitative yield, when the reaction is performed in the presence of sodium forms of zeolites X and Y, which have removed $\ce{HBr}$ formed during the reaction by a process of cation exchange (Ref.3).
All of all apply to aniline as well. For example, 2,4,6-tribromoaniline can be prepared by treating bromine water with aniline in a solution of acetic acid or dilute hydrochloric acid (Wikipedia).

Other than halogenation reaction of above types, other example for electrophilic aromatic multi-substitution is Friedel-Crafts alkylation reaction. Once alkylated, the ring would get activated more than it was before the first alkylation (alkyl is activating group). Thus, it would further undergoes second (and/or third) alkylation. Foe comparison, the relative rate of benzene and toluene are $1.0$ and $24.5$. That's the reason, it take long way to mono-alkylate benzene ring by first doing Friedel-Crafts acylation reaction and then reducing carbonyl group to methylene group by an appropriate method.
References:

John R. Johnson, C. G. Gauerke (Checked by Frank C. Whitmore and T. Otterbacher), "m-Bromonitrobenzene (benzene, 1-bromo-3-nitro-)," Organic Syntheses, Coll. Vol. 1, p.123 (1941); Vol. 8, p.46 (1928)(DOI: 10.15227/orgsyn.008.0046)(PDF).
P. W. G. Smith, A. R. Tatchell, In Organic Chemistry for General Degree Students, Volume 2: Aromatic Chemistry; Pergamon Press, Ltd: Oxford, United Kingdom, 1969, "Chapter V: Phenols," pp. 79-104 (ISBN-13: 978-0080129471).
Keith Smith, Ping He, Ashley Taylor, "Selective para-bromination of phenyl acetate. Under the control of zeolites, bases, acetic anhydride or metal acetates in the liquid phase," Green Chem. 1999, 1, 35-38 (DOI: https://doi.org/10.1039/A808242A).

Source for relative rates

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Why is tetrachlorocuprate (II) yellow if it is high-spin?
From my knowledge of transition metal complexes, tetrahedral complexes generally have less splitting (compared to octahedral complexes) because they only involve four ligands are they aren't placed directly near any d-orbitals (unlike octahedral complexes). Furthermore, chloride is a weak-field ligand and therefore forms high-spin complexes. Both of these facts point to tetrachlorocuprate absorbing light of low energy (red to orange) and therefore being blue in colour (as blue is complementary to red in the colour wheel).
However, it is a yellow complex, suggesting that it absorbs high energy purple or violet light. I can't make any sense of this. Even if I factor in Jahn-Teller distortion, it shouldn't surpass that of hexaaquacuprate (II) which is blue in colour. Does this have to do with de-excitation? If so, what makes tetrachlorocuprate different from all other complexes?

I think it is a very complicated story. I think there is no single species but a mixture. See Stability and Electronic Spectra of the Copper(II) Chloro Complexes in
Aqueous Solutions, M. A. KHAN and M. J. SCHWING-WEILL in Google Scholar.


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Stability of Carbanions
For comparing stability of carbanion, we use the fact that "negative charge is stable on more electronegative atom." So, The stability order is, {electronegative order: $\mathrm{sp \gt sp^2 \gt  sp^3}$} and therefore, $\ce{HC#C^- \gt H2C=C^-H \gt CH3-C^-H2}$.
My doubt is that, we generally say, carbanion is stable when there is less electron density on $\ce{C^-}$.
In $\ce{HC#C^-}$, $\ce{C}$ is $\mathrm{sp}$ hybridized that means it is more electronegative, so it will pull electrons more as compared to other two. So, electron density order will be reverse of above order. And, $\ce{HC#C^-}$ will be least stable.
By, my thought, answer is completely opposite.
Please help me to get out of this doubt and please correct me if I said something wrong.

Read this.

I don't understand your doubt. In term of Electronegativity you have to interpret it as the CH is pulling from the C(-).

s-Orbitals are lower in energy than p-orbitals. The same reasoning applies to hybrid orbitals. This may help. https://chemistry.stackexchange.com/questions/124896/c%e2%88%92h-bond-length-in-organic-compounds/124909#124909

$\mathrm{sp}$ hybridized carbanions are more stable as the lone pair of electrons is placed in an orbital consisting of higher $\mathrm s$-character. In general, electrons are more stable in $\mathrm s$ orbitals rather than in the $\mathrm p$ orbitals of the same shell. This can be verified by the Aufbau principle.
Now, when we say $\mathrm {sp}$ hybridized species are more electronegative than $\mathrm{sp^2}$ hybridized species, we mean that the $\mathrm{sp}$ orbital are more accomplished in holding an electron within them when compared to their $\mathrm{sp^2}$ counterparts.
As electrons are more stable in an $\mathrm{sp}$ orbital, electrons are pulled more towards them rather than the $\mathrm{sp^2}$ orbitals, which makes us say that $\mathrm{sp}$ orbitals are more electronegative than $\mathrm{sp^2}$ orbitals.
The question you are asking is similar to asking if a negative charge is more stabilized on an oxygen atom or on a nitrogen atom.

Can the downvoter take a moment to describe what is wrong with this answer? I am always open for feedback. :)

@Nisaragbhavsar do you want to say that in case of first carbocation, the lone pair remains in an sp orbital and hence close to nucleus and least instability ,not that what OP thinks that sp carbon is more electronegative so it pulls electrons of it's bond with other carbon more.  but Both sp and sp(2) carbons have same tendency to attract electrons of carbon -carbon bond

The stability works on the following principle:

More spread out the electron density, more stable is the compound.

As you can see, sp hybridized carbon atom pulls the electron density to a larger extent, thereby, effectively spreading out the charge.
Can you also think of why this is the reverse for cations?
In the case of carbocations, the sp hybridized carbon atom pulls electron density causing even more positive density on the carbocation.

Would the downvoter care to explain the reason?

Yeah, it seems like someone has mass downvoted all answers to this question. My answer has also been downvoted without stating a reason.

@DatBoi and @ Nisarg Bhavsar. Thanks for your reply. But ☹ I think you didn't understand my doubt. My doubt is that, In HC---C- , C- is sp hybridized so it is more electronegative. So due to high electron pulling capacity C - will have more electron density . Since , more electron density on Carbanion decreases it's stability . So, HC ---C- will be least stable. This is my thought. please help. I am getting totally opposite stability order

@The Explorer You are thinking about it in the opposite sense. Assume if we had acetylene and ethene. Which of them would be more reactive towards a base?? Obviously acetylene because $\ce{#}$ is more electronegative than a $\ce{=}$. What you need to understand is that the $\ce{#}$ will pull the electrons from the carbon and not the other way around.

@TheExplorer Stop! I suggest you rather ask yourself, what does it even mean for a hybridised orbital to pull away electron density.? The answer you will get will be that in a sp orbital, electron density is closer to the nucleus and thus the least instability.


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Hard water filtration
Why does hard water have to pass through the cation exchange resin first followed by the anion exchange resin? Why can't it be the other way around?

Is better to release bicarbonates as CO2 in catex (  cation exchange resin )  in H cycle or to replace them by anex ?

To answer your question, what if we did utilize the method you propose?
Before going into the details, it should be noted that the Ion-Exchange method is used to remove hardness of water at an industrial scale. The principal reason that this method is preferred over any other hardness removal process is that it has a unique advantage:
All exchanges take place in an aqueous medium in the form of ions alone.
Keeping that in mind, let's come back to the adjustment you suggested. In the Anion Exchanger, the usual reactions are
$$\ce{R'OH- +Cl^- -> R'Cl- + OH-}$$
$$\ce{2R'OH- + SO4^2- -> R'2 SO4^2- + 2OH-} $$
$$\ce{2R'OH- + CO3^2- -> R'2 CO3^2- + 2OH- }$$
Given that you still haven't removed the cations, the ensuing process would result as follows:
$$\ce{Ca^2+ + OH- <=>> Ca(OH)2} \   \\K_{-1}=\frac{1}{K_{spCa(OH)_2}}$$ and $$\ce{Mg^2+ + OH- <=>>Mg(OH)2}\\K_{-1}=\frac{1}{K_{spMg(OH)_2}}$$
Note that $K_{spMg(OH)_2}=1.8\cdot10^{-11}$ and  ${K_{spCa(OH)_2}}=6\cdot 10^{-6} $
. Further, due to the high concentration of $\ce{OH-}$ ions in the medium, the equilibrium shifts forward, meaning that a high amount of salt is precipitated and deposited within the exchanger, also called scaling. This is detrimental to the industry from any perspective.
Now, let's go back to the part where we discussed advantage of the catex. Do you get it now?
In simpler terms, the fundamental phenomenon behind this process (ION EXCHANGE) is challenged and its efficacy is vastly downgraded.
For future references:

Detailed exchange process
$K_{sp}$ values


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How exactly does copper react with hydrogen peroxide, salt and citric acid?
Context
I have no knowledge in chemistry. I am looking for information about etching circuit boards with citric acid instead of hydrochloric acid because it seems it could be a safer, easier way as citric acid is way safer to handle and not that toxic and harmful.
I am struggling to find much information online about the process since it is not as popular as other etchants.
The process
It seems to work really well if I mix 100ml of 3% hydrogen peroxide solution in water with quite a lot of citric acid powder(4-6 teaspoons) and a bit of salt(around 1 teaspoon), heat everything in a microwave oven to a warmer temperature(not boiling, around 70-80° C) so everything dissolves and then while it's still warm I put the copper clad in it and this seems to work just as good as hydrochloric acid and hydrogen peroxide in room temperature - etches 10x15cm PCB in 15-20 minutes.
Concerns/questions
I have done it a few times in well ventilated room, but I have no knowledge about what exactly happens in the reaction and if it is dangerous in any way, so I would like some insight into what exactly might be happening, what materials are produced and how safe/toxic they are.

See also: What results from the neutralization of copper citrate with Na2CO3

See also: Copper Citrate handling and disposal.  Also, this page has a list of compounds and their disposal requirements (suggestions).  Interestingly, copper citrate is the only copper compound indicating that small quantities can go down the drain flushed with lots of water.  Not sure what "small" means, however.

"Small" amounts means much less than 1 mole. Maybe amounts like a couple of millimoles.

The acids, hydrochloric or citric, have a common property : they produce $\ce{H+}$ in water. These ions are necessary to react with copper and hydrogen peroxide $\ce{H2O2}$. Indeed metallic copper reacts with hydrogen peroxide $\ce{H2O2}$ and an acid in the following way : $$\ce{Cu + H2O2 + 2 H+ -> Cu^{2+} + 2 H2O}$$ So whatever the acid chosen, hydrochloric or citric, the copper will be dissolved the same way. With citric acid, the dissolution should be slower. But it leads to the same final result : metallic copper will be destroyed. The only difference is the final composition of the liquid. With HCl, the final liquid is made of $\ce{Cu^{2+}}$ and $\ce{Cl-}$ ions, which may be partially recombined. With citric acid, the final liquid is made of citrate ions, which will make a complex with $\ce{Cu^{2+}}$. Citral ions from citric acid will make a small change of color, but has no other effect. Both solutions are equally dangerous, or equally non-dangerous if you prefer, because copper solutions are always toxic. You should take care of the obtained copper solutions in the same way, whatever their compositions.
Personally I would recommend using citric acid instead of hydrochloric acid. It is less corrosive, and has nearly the same effect on copper + $\ce{H2O2}$.

Thank you very much for the answer. This is indeed what I have observed. Hydrochloric acid is a very nasty thing to have around in comparison with citric acid and I would love to not have to deal with it. The time issue seems to be very easily solved with increased temperature. I have read/heard about adding salt(NaCl) to speed things up in this process and did so as well. Is there any logic in this? Could you comment on the role of salt here? Also any comments on the amounts/proportions of these ingredients?

Regarding "copper solutions are always toxic", what about Copper Citrate?  It is sold as a dietary supplement, just google for "copper citrate supplement".  Won't the OP's process produce a copper citrate solution?  Does that make it safe(er)?

@KJ7LNW "The dose makes the poison".


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Are high spin complexes and spin free complexes the same thing?
While solving some Coordination Compounds problems, I came across a problem that asked

Select which complex is high spin or spin free octahedral complex.

The term 'spin free' sounds like a complex with no unpaired electrons and I know high spin complexes are those which have unpaired electrons. But the question was was single correct and this got me confused.
Do the terms high spin and spin free bear the same meaning?

The answers seem to show that, at least in the context of inorganic coordination chemistry, these terms are used synonymously. However, as you noted, in most other contexts, "spin-free" is generally taken to mean "free of spin" or "no spin dependence": see https://pubs.acs.org/doi/10.1021/j100842a004 and https://aip.scitation.org/doi/abs/10.1063/1.466508

Spin Free Complex means that the electrons are free to spin!
High spin complexes are the ones with Weak Field Ligands which aren't able to pair up electrons of the central ion.
Hence the 2 terms are usually used synonymously.
On the same lines Low Spin Complexes and Spin Paired Complexes are used synonymously!

Is this a regional dialect? I do find sources where "spin-free" is used in this way, but at least in US-English, "X-free" would mean something that lacks, or is free of, "X", see guilt-free, fat-free, worry-free.

I am not really sure if it's taken from a dialect but that's what it's used for in chemistry. I 100% agree that is confusing, high spin complex is a much better word to use.

@Tyberius Asbestos free cereals

i dont think Spin Free Complex means that the electrons are free to spin, i think it means there are orbitals with spin momentum (since electrons cant technically spin)

In a paper by the renowned chemist, F. Albert Cotton,
"Magnetic Investigations of Spin-free Cobaltous Complexes. III. On the Existence
of Planar Complexes"
they investigate "spin-free" complexes which the authors defines as complexes with three unpaired electrons. Following this definition of spin-free complexes does not necessary imply high-spin, as in the case of some octahedral Cr(III) complexes, which have three unpaired electrons that can occupy the $ t_{2g}$ energy level, making it low spin.
Side Note:
This definition of spin-free may be how Al. Cotton defined it for his paper; this is not a claim that it is representative of the entire group. Additionally, this paper is from the 1960s so contemporary definitions may differ.


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TD-DFT transition dipole moment vector does not align with experimental data
I have been modelling excited states of some molecules with similar stucture using TD-DFT theory, however, I have encountered an issue that sometimes transition dipole moment does not align with the direction of experimental. For example, for molecule 1 the right direction of a transition dipole moment is with B3LYP functional and CAM-B3LYP functional results in a dipole moment in an opposite direction (the results were independent of the basis set). For molecule 2 the B3LYP functional yielded worse results than CAM-B3LYP. What could be the issue? I have done explicit geometry optimizations and frequency calculations (no imaginary frequencies were found) for each functional and basis set.

A possible error could be the identification of the excited state involved in the transition.  It can happen that excited states switch places if there are other excited states energetically close by. Another case are systems with charge-transfer states, which typically have very different energies in Cam-B3LYP and B3LYP.

Thank you very much for the comment, @HansWurst. However, it is known from the experiments for sure that it is really the S1 state that we are searching for. $S_0->S_1$  and $S_0->S_2$ transitions differ in about 100 nm, so, I believe, the switch of states is not the case here (as the involvement of the molecular orbitals also states that the states should be correct), however, I am not sure about the charge transfer involvement in the system. Any other types of the issues that could be the case?

I am not familiar with measurements of the transition dipole moments but there shoudn't be an "absolute" direction since it is an off diagonal element of an hermitian operator. The negative and positive direction should be equivalent since we can only determine the magnitude square of $|\vec \mu_{if} \cdot \vec E| = |-\vec \mu_{if} \cdot \vec E|$ but I am not an expert when it comes to this so i could be wrong. Perhaps someone else who is more familiar with this topic can provide more help.

@HansWurst, thank you very much for the insights!


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Differing number average molecular weight results for 1HNMR, MALDI and GPC of a polymer
I have been given 1HNMR, MALDI-ToF-MS and GPC spectra for a laboratory synthesized sample of polycaprolactone.
The number average molecular weight (Mn) obtained from their analysis differed significantly though:
Mn(1HNMR) = 3186,41 g/mol
Mn(MALDI) = 4751,64 g/mol
Mn(GPC) = 6750,81 g/mol
I tried to find in the literature an explanation for that pattern but found only that 1HNMR is the more reliable one for being quantitative source, while MALDI can present inconsistencies due to chain breakage during ionization and that GPC might present wrong molecular masses if the calibration is not done correctly (in this case, calibration was done with PS). I am also considering that the analysts made no mistakes in the generation of these spectra, as it was performed by a professor in laboratory course on the subject.
So, I wanted to ask, does anyone have an insight as to why those three techniques are presenting such inconsistent results for this sample?
(Also, this is my first question here, so any remarks on question posing are also welcome)
Thanks for the help!

Do not assume that a calibration exists. This said, if you go technique by technique you will find pros and cons, as you succinctly listed abive5. For details let's wait for a true polynerist.

Above read "perfect calibration".


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retrosynthesis gamma-hydroxy ketone
One of the sections about retrosynthesis in Warren's book addresses how to access $\gamma$-hydroxy carbonyl compounds like ZM181.  This is again one of the examples to advice the reader to check the results for plausibility: while epoxides often may be a suitable building block, here, the intermediate would react further with the ester present to yield a lactone (second line) instead:

In the edition accessible to mine, Warren apparently does not detail out if an alternative approach with chloro- or bromoethanol (third line) to yield ZM181 were viable.
Question: Would be the approach of the third line, chloroethanol and ethyl acetoacetate to yield ZM181, still viable outside academia, too?
I'm aware the intramolecular transesterification to yield a five-membered cycle needs to be considered as a possible product for both approaches.  On the other hand, I think either route could be favoured by choice of the reaction conditions, i.e.

complete single deprotonation of ethyl acetylacetonate in dry THF by one equivalent $\ce{NaH}$ (not two), then followed by the addition of the expoxide may ease the formation of the lactone. The reasoning of $\ce{NaH}$ is to ensure all of the starting material is deprotonated once with the irreversible escape of $\ce{H2}$. Or,

the partial, equilibrium based deprotonation of the $\beta$-dicarbonyl compound by $\ce{K2CO3}$ in a mixture of THF/water, in presence of the bromo ethanol already present, might reduce the concentration of alcoholate and thus lower chances of the formation of the lactone.  This line of thought accounts for the different $pK_a$ of the reagents present, e.g., water could protonate free alcoholate, too.

Reference: Stuart Warren Organische Retrosynthese, Teubner, Stuttgart 1997.

I don’t think the second and third lines are realistically different - you’d need a base present in both so if the lactone is favourable , it would form in both cases.
Practically, depending on what the next step in the synthesis is, the lactone might be fine (eg can still add a nucleophile in, or reduce it down)

Also the 1,3-dicarbonyl likely exists (to a large extent) as the enolised form, which may protect the lactone from forming immediately after the addition.

The only advantage I see of the bromoethanol route is that you could put a protecting group on the -OH and prevent lactone formation that way

For a moment, I thought running the reaction of the second line in dry THF would indeed yield the lactone. Contrasting to this -- given the different pKa of the beta-dicarbonyl compound and of an alcohol -- I think now, running the reaction of the third line with $\ce{K2CO3}$ in a mixture of water and THF could reduce a the concentration of free alcoholate to trigger the intramolecular transesterification.  But still it could be an uphill battle (already because intramolecular reactions are favoured over intermolecular ones, and the formation of a cycle of five atoms, etc.).

What is the [tag:green-chemistry] aspect of this? The bromide looks hazardous enough to me to kick this out of the category.

@Martin-マーチン Maybe today there is an other nucleofuge / other C2 precursor (halogen free) which gained much popularity.

Don't get me wrong, I think it's a viable question well asked. I just don't see how the tags fit. The question is essentially about a synthesis; not retrosynthesis, because that part was already done. And forgive me, but I don't see any aspect of green chemistry mentioned here. I'd use completely different tags, but I wanted to ask, maybe you just would like to clarify this.

@Martin-マーチン My interpretation of your comment is that you suggest to retract the retrocynthesis tag (regardless if this inspired the question) an a reformulation.  Perhaps in lines of this: «Aiming for ZM181, oxirane is not suitable because the intermediate forms a lactone (second line) if the reaction takes place in dry THF.  Possibly chloro ethanol / bromo ethanol could yield the desired product under equilibrium conditions of $\ce{K2CO3}$ and a mixture of THF/water (third line). Question: is there a halogen-free C2 precursor leading to the target?»


text:
How can I convert cathodic energy efficiency to overall energy efficiency?
Im working on calculation which involve the energy efficiency of producing formate by CO2 electroreduction.
I wonder how to convert the reported cathodic energy efficiency (71%) of papers e.g (1) this paper or (2) this paper into an overall energy efficiency.
I want to assume that water is oxidized at an anode. If needed we can plug in an assumed efficiency for the anode reaction.
I am relatively new to electrochemistry, and trying to learn so forgive me if this is trivial, but I can barely find the term "cathodic energy efficiency" online for me to learn this.

If the efficiency is not $100$%, it means that the rest of the electric energy is lost in heat by the Joule effect.

Thank you. Is there a way to estimate what is the loss of energy on the anode side for these kind of papers?

@Maurice I think the <100% cathode energy efficiency they report of producing the desired product (formate)  is also result of the competing HER reaction, and not just heat.

You may be right, after all. Why not ?


text:
Neutralisation of weak acid by base
I'm still working on my concepts related to chemical equilibrium. I came across this question:

10mL of a weak acid HA is 20%
dissociated in water. This solution is completely neutralised by 10 mL
of $10^{-3}$ M of NaOH.

Is [$\ce{HA}$] = $10^{-3}$ M or is [$\ce{H^+}$] =$10^{-3}$?
Why is it wrong to say that since NaOH completely dissociates, and solution is neutral, [$\ce{OH^-}$] =$10^{-3}$ M= [$\ce{H^+}$]?
Here, [.] denotes concentration.

If your weak acid is neutralized by $\ce{NaOH}$, the result does not depend on the percentage of dissociation before neutralization. Here your initial acid is $\pu{10^{-3} M}$, as you stated. But the initial concentration of $\ce{H+}$ is $20$% of this value, so : $\ce{[H+] = 2·10^{-4}} M$

Neutralised is ambiguous here. In aqueous acid/base solutions neutralized usually means that the pH was adjusted to 7.0. It could also mean that as many moles of base were added as there were moles of acid. In a titration this is referred to as the equivalence point. The final pH will be greater than 7 due to the reaction $$\ce{A- + H2O <=> HA + OH-}$$

@MaxW you're correct; here the equivalence point is where the equivalents of the acid and base will be equal. The hydrolysis will yield a solution whose ph is $$ph =7+\frac{1}{2}\biggr[pka + log[HA]\biggr]$$

@MaxW  This is why I got confused. Now I understood it. Thanks a lot!

As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage.

Is $\pu{[}\ce{HA}{] = 10^{−3} M}$ or is $\pu{[}\ce{H+}{] = 10^{−3}}$?

$\pu{[}\ce{HA}{] = 10^{-3}}$.
Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base:
$$N_1V_1=N_2V_2 $$  Since $V_1=V_2$, therefore, $\pu{[}\ce{HA}{]=[}\ce{NaOH}{]}$.

Why is it wrong to say that since $\ce{NaOH}$ completely dissociates, and solution is neutral, $[\ce{OH−}]$ $\pu{= 10^−3 M = [}\ce{H+}]$?

The secret lies in the name of the concept itself. Chemical equilibrium. To attain an equilibrium state, all favorable chemical reactions are first carried out. The remaining products/ions constitute the equilibrium state. Coming back to the question, the neutralisation reaction between $\ce{HA}$ and $\ce{NaOH}$ is carried out first:
$$\ce{HA + NaOH ->NaA +H2O}$$
The resultant ions constitute the equilibrium, which, are none unless you consider the hydrolysis of the resultant salt of a weak acid/strong base.

Since your weak acid is neutralized to the end point by a strong base, $\ce{NaOH}$, the final reding does not depend on the percentage of dissociation of the weak acid before neutralization. Suppose your ionization of the weak acid is as follows:
$$\ce{HA + H2O <=> H3O+ + A-} \tag1$$
Thus at the end of neutralization:
$$\text{amount of $\ce{NaOH}$ needed} = \text{amount of $\ce{HA}$  at equilibrium} + \text{amount of $\ce{H3O+}$ at equilibrium}$$
Since volumes of both solutions are $\pu{10 mL}$ each:
$$[\ce{HA}] + [\ce{H3O+}] = [\ce{NaOH}] = \pu{10^{-3} M}$$
Here, $[\ce{HA}]$ and $[\ce{H3O+}]$ concentration of each at the equilibrium. My argument is when you get $\pu{10 mL}$ of aqueous solution of a weak acid $\ce{HA}$, all three species of equation $(1)$ is in equilibrium, regardless of its ionization percentage. Therefore, sum of $[\ce{HA}]$ and $[\ce{H3O+}]$ is equal to $\pu{10^{-3} M}$, which is $[\ce{OH-}]$ since $\ce{NaOH}$ has completely dissociated in an aqueous solution.
Note that, the percentage is given to calculate the $K_\mathrm{a}$ of the acid. In this case (your highlighted question is not completed),
$$K_\mathrm{a} = \frac{0.2 \times 0.2}{1-0.2} = 0.05$$


text:
Is this analogy right?
Is my logic correct in this case ?
Since the formal charge represents the charge the element possess in a covalent compounds, is it right to say that formal charge is the equivalent term for covalent bonds as is oxidation number is to ionic bonds ?

Oxidation number applies to covalent compounds just as well.

But that is an assumption, right ? Covalent compounds don't generally carry charges to possess oxidation number.

No, that's a definition. We define oxidation numbers so as to be meaningful for covalent compounds too. They do not necessarily have any physical meaning, though. In particular, they are not real charges.

The oxidation number is the charge that one atom would have if all the covalent bonds of the molecule (or of the ion) were transformed into ionic bonds, with a negative charge on the most electronegative atom of the bond, and a positive charge on the least electronegative atom.

So if I'm correct formal charge is the actual charge possesed by the atom in a covalent compound ?

@Parvathy neither this is correct

@Parvathy. Not really. For example, in $\ce{CH4}$, no atoms are charged. The bonds are all covalent. But the electronegativities are $2.1$ for $\ce{H}$, and $2.5$ for $\ce{C}$. So there is a partial (or formal)  positive charge $\delta +$ on the $\ce{H}$ atom, and a negative $\delta -$ on the Carbon atom. When calculating the oxidation number of C you add all the partial (or formal) charge, as if they were ionic charges. And you find $-4$ for the Carbon atom, and $+1$ for each hydrogen atom.

Your analogy isn't correct. As @Ivan pointed out in the comments, oxidation numbers apply to covalent compounds as well as they do to ionic ones.
You calculate oxidation numbers assuming that the more electronegative atom takes up both the electrons in a bond. Formal charge, on the other hand, is determined assuming no difference in electronegativity.
Take the carbon atom in methane as an example. It forms four $\ce{C-H}$ bonds; none of them are polar enough to be called ionic. Yet we consider this bond to be perfectly ionic and assign a "charge" of $-4$ to carbon. This is its oxidation number.
When you calculate its formal charge, though, you ignore the (slight) difference in electronegativity between carbon and hydrogen. Carbon, then, happily ends up with the four electrons it started with. Its formal charge, therefore, is zero.
Note that neither of these numbers has any physical meaning.
Can you figure out the formal charge of each atom in the nitrate ion and ozone?

Thank you for clearing my doubt ! It's 1 for O₃ and 0 for the oxygen of NO₃⁻ and -1/3 for the Nitrogen,  if I'm not wrong.

@Parvathy https://en.m.wikipedia.org/wiki/Formal_charge


text:
How many CO2 is emitted (in kg) by using 1 m3 of LNG?
How much CO2 in kg is emitted by using 1 cubic meter of LNG? It is used for a dry booth after painting products.
Unfortunately, this is all information I have. Please note the assumptions for answering this question.

are you talking about burning LNG? And are you measuring a cubic metre of the gas or the liquid (as the product is often measured by the volume of the gas at STP not by the volume of the very cold liquid)?

Using LNG ( liquid natural gas) in small scale is quite unusual, as it demands cryogenic devices. I would expect rather LPG ( liquid propan/butan gas ) or PNG ( pressurized natural gas ).

STP (to simplify a little) means room temperature and pressure. The point being that LNG is normally distributed as a gas (it is basically the same as the natural gas distributed in domestic and industrial piped gas and consists mostly of methane). LNG is never delivered as a liquid to customers: it is transported as a liquid in very large refrigerated tankers but is regasified and injected into the normal gas pipelines used to distribute natural gas.

Also worth noting the possible confusion with LPG which consists mainly of propane or butane both of which can be liquefied by pressure alone and don't require cooling. This makes them far easier to transport efficiently in cylinders for small scale use.

Thank you for your help. I will ask some additional questions to the user of this LNG.

@matt_black Thanks Matt. Using LNG for restored NG is quite confusing. As liquid nitrogen is frequently used e.g. in cryocenters, I have thought perhaps LNG can be delivered as well. But I would agree it would be a big security risk becoming a NG bomb.

Density of gas, assuming ideal behaviour, is $\rho=\frac{pM}{RT}$ .
With assumption the natural gas is pure methane, burning follows the reaction:
$$\ce{CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)}$$
Therefore the mass of $\ce{CO2}$ created by burning $\pu{1 m3}$ of natural gas at $\pu{1 atm}$ and $\pu{25 ^{\circ}C}$ is
$$m_\ce{CO2} = \frac{pM_\ce{CH4}}{RT} \cdot V \cdot \frac{M_{\ce{CO2}}}{M_{\ce{CH4}}} =  \frac{p}{RT} \cdot V \cdot M_{\ce{CO2}}\approx \frac{\pu{e5 Pa} \cdot \pu{1 m3} \cdot \pu{0.044 kg/mol}}{\pu{8.314 JK-1mol-1} \cdot \pu{298 K}} \approx \pu{1.78 kg}\ \ce{CO2}$$


text:
How does addition of AgCl affect concentration of iodide in KI solution?
In a mixture of $\ce{AgCl(s)}$ and initial $\pu{0.10 M}$ $\ce{KI}$, what is the $[\ce{I-}]$ after the reaction proceeds?
Disclaimer: This is not a graded assignment, but rather a problem I'm trying to figure out for practice purposes from my textbook.

Assume the reaction goes to completion. AgI is highly insoluble in water so it is completely removed - there is no equilibrium.

The reaction goes to completion quantitatively, that is to say to all the significant figures that the KI concentration is known. There will be a equilibrium and a minuscule amount of  $\ce{I-}$ left in solution. $\ce{[I-]{\mathrm{inital}} \gg [I-]{\mathrm{final}}}$ // NOTE: The problem ignores the possibility that the AgI coats the AgCl particles thus cutting off the reaction.

I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it.
Given the reaction:
$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}$$
there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is to say that if the initial concentration of $\ce{I-}$ is $\pu{0.10 M}$, then the final concentration of $\ce{Cl-}$ will be  $\pu{0.10 M}$. (This is a significant figure simplification. The $\ce{[I-]}$ isn't given to 10 decimal places.)
Now Wikipedia gives the $K_\mathrm{sp}$ of silver chloride as $1.77 \times 10^{-10}$ and for silver iodide as $8.52 \times 10^{−17}$.
Since the final concentration of $\ce{Cl-} = \pu{0.10 M}$, the final concentration of silver is given by:
$$\ce{[Ag+]} = \dfrac{K_\mathrm{sp, AgCl}}{\ce{[Cl-]}} = \dfrac{1.77 \times 10^{-10}}{0.10} = \pu{1.77\times 10^{-9} M}$$
So the final concentration of $\ce{I-}$ is given by:
$$\ce{[I-]} = \dfrac{K_\mathrm{sp, AgI}}{\ce{[Ag+]}} = \dfrac{8.52 \times 10^{−17}}{1.77 \times 10^{-9}} = 4.8\times10^{-8}$$
Since the $\ce{[I-]}$ starts out as $\pu{0.10 M}$ and ends up as $\pu{4.8 \times 10^{-8} M}$, it is better to say that the reaction proceeds quantitatively rather than completely.

The OP asked the question again, and proposed the following solution from his book.

My textbook says to solve it in the following way:
1) Ignore the 0.10 M KI since it doesn't matter.
2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its sp.
sp=1.77×10−10=^2
=1.3×10^−5
This is the Ag+ concentration
3) Substitute calculated Ag+ concentration into equilibrium for sp of AgI.
sp=8.51×10^−17=[Ag+][I−]
8.51×10^−17=[1.3×10^−5][I−]
[I−]=6.4×10^−12

The book answer is wrong!!
The crux of the matter is what limits the final concentration of silver. Point (2) in the book answer is wrong, which makes point (1) wrong also.
The equilibrium for the $\ce{AgCl}$ precipitate depends on both the concentrations of both $\ce{Ag+}$ and $\ce{Cl-}$. (We're ignoring the activity/concentration factor for the ions.)
Given the reaction:
$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}\tag{1}$$
The reaction is an exchange reaction. The AgCl(s) dissolves via:
$$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}\tag{2}$$
and then KI precipitates via:
$$\ce{Ag+(aq) + I-(aq) <=> KI(s)}\tag{3}$$
When the reaction ends, then the $\ce{K+}$ cations left in solution must be balanced by the same concentration of anions in the solution. The anions in this case could be either $\ce{Cl-}$ or $\ce{I-}$. However due to the much greater insolubility of $\ce{AgI}$ we know that $\ce{[Cl-] \gg [I-]}$. So the solution must end up (to two significant figures), $\ce{\pu{0.10 M} K+}$ and $\ce{\pu{0.10 M} Cl-}$ with traces of $\ce{Ag+}$ and $\ce{I-}$.
For the book answer the final concentration of $\ce{Ag+}$ is $\ce{1.3\times 10^{−5}}$. Since the final $\ce{Cl-}$ is $\pu{0.10 M}$ that means that
$$\mathrm{K'_{sp,AgCl}} =\ce{[Ag+][Cl-]}= (1.3\times 10^{−5})(0.1) = 1.3\times 10^{−6}$$
The acknowledged value for $\mathrm{K_{sp,AgCl}} = \pu{1.77\times 10^{-10}}$. Thus Point (2) in the book answer must be wrong.


text:
variable oxidation states
which of the statements is true:

looking at the electronic configuration of an element, can the possible oxidation states/oxidation number of an elemnet be predicted.
for example, the valence electronic configuration of nitrogen is 2s^2 2p^3. From this can we know all the possible oxidation states exhibited by the element nitrogen.
                            (OR)

2.We should first know all the compounds nitrogen can form, and then find the oxidation number of nitrogen in each compound, and then give all the possible oxidation states nitrogen can exhibit.

The oxidation numbers cannot be predicted from the structure of the atom. They are experimentally determined. Not much chance you predict that while nitrogen can get all integer oxidation numbers from $−3$ to $+5$, its neighbor oxygen can only have $−2$, $−1$, $0$, $+1$, $+2$ in known compounds. And then come fractional oxidation states...

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


text:
Is any element base or acid?
I have seen classification of Mg+2 or any other cation or anion for that matter as an acid or a base, but I have never seen an element labelled as an acid or a base. Isn't there a possibility of Mg+2 being the conjugate acid of Mg a base? If not,why so?

Actually, most elements are neither.

An acid is a donor of proton. So no element or no ion of any element can donate a proton. They cannot give what they don't have. A base in an acceptor of protons. Here it is more feasible, Any atom or atomic ion can in principle accept a proton, provided there is a strong enough donor to give it a proton.

@Maurice That's only Bronsted definition and even here there is one - hydrogen.

@Mithoron Can you guide better,please?

Almost everything can act as acid or base depending on reaction and definition of acid and base. You'd be looking for Usanovich's definition, I guess.

Well any element cannot be called an acid but we surely call Na a base in organic reactions.

Short answer is: Yes and no. It depends on what acid-base theory is applied.

Redox reactions like $\ce{Mg -> Mg^2+ + 2 e-}$  are not usually considered acid-base reactions, but  over-generalized acid-base theories like Usanovich acid-base theory treat them as a special case of acid-base reactions.
Within this theory, $\ce{Mg}$ is relatively strong Usanovich base, $\ce{Mg^2+}$ is very weak Usanovich acid.

If we consider typical reactions by classical Broensted-Lawry theory:
$\ce{H2SO4 + H2O -> HSO4- + H3O+},$
or $\ce{NH3 + H3O+ -> NH4+ +H2O}$,
or by another classic - Lewis theory:
$\ce{NH3 + BF3 -> H3N-BF3}$,
then atoms involved in acid-base interactions have the same oxidation number before and after the reaction.
$\ce{Mg^2+(aq)}$ is the conjugate acid to the conjugate base $\ce{MgOH+ (aq)}$.
$$\ce{[Mg(H2O)6]^2+ + H2O <=> [Mg(H2O)5(OH)]+ + H3O+}$$
$\ce{MgOH+ (aq)}$ is is then the conjugate acid to $\ce{Mg(OH)2 (aq)}$.
$$\ce{[Mg(H2O)5(OH)]+ + H2O <=> [Mg(H2O)4(OH)2] + H3O+}$$
$\ce{\pu{Mg}}$ has in all 3 species the oxidation number +2.

What acid-base reaction you consider redox reaction ?

In Usanovich's theory, all of them, see comment under q.^

Usanovich theory is rather a rare animal and is not widely used or even known.

Well, yeah, but it's the answer to OP's last question.

@Mithoron But it should be the other way . Redox reactions are not usually considered acid-base reactions. As in the Usanovich theory Redox reactions are special cases of acid-base reactions, not the opposite.

Let us continue this discussion in chat.


text:
Am I visualizing/understanding atomic form factor correctly?
I'm having difficulty visualizing the concept of atomic form factor or atomic scattering factor as it has been presented in a chemistry course on X-ray crystallography that I'm taking. At the moment, I visualize it as such:

(Please note that the question marks in the diagram above are included to communicate my confusion.)
Am I visualizing this correctly? Is there any other way for me to interpret these diagrams? The course doesn't dive into the mathematics of atomic form factor (and neither does our accompanying text), so I don't have much to go on in developing a mental image of the phenomenon at play here--and I strongly suspect that the mental image that I do have is wrong since I can't find any evidence to support it.

I don't think we need a visualization here in the first place. That being said, I'd picture the right image as big as the left one, only much more faint.

In your illustration, there seems to be a little «almost step».  But more importantly, if displaying a heavy scatterer (iron, Fe) next to a weaker one (C), you should put a bit more emphasis on the descent for iron.  This is one of the reasons why the atomic position of the heavier atoms are easier to attribute in the Fourier map, than the lighter ones.
Take a look at IUCr's illustration from their freely accessible on-line dictionary:

(credit)
Maybe your school has the printed or / and electronic version of the IT (the International Tables of Crystallography) which goes into further detail (e.g., listing $f$ and $f'$ for the elements, obviously dependent on the wavelength of radiation used during the diffraction experiment).
Because currently many schools' libraries may restrict access and borrow the books in person, note that IUCr moderates a dedicated section of teaching pamphlets, too.  Topics include e.g., crystal packing and symmetry.

I agree with Ivan, your coloured circles are misleading. The factor is a ratio, a number, and is scattering relative to a single electron's scattering and as such is independent of any radial distance $r$ in the (spherical) atom. It depends on the angle that scattering is observed at and also on the total number of electrons.  $f=\int_o^\infty U(r)\sin(\mu r)/(\mu r) dr$ where $U(r)=4\pi r^2\rho(r)$ and $\mu=4\pi\sin(\theta)/\lambda$ making $f$ after integration only a function of $\mu$ which as you see depends on the scattering angle. The integration removes any dependence on radial distance; $\rho(r)$ is the electron density at $r$.  As $f$ depends on the number of electrons in an atom, it is larger for heavier atoms, i.e. more electrons more scattering.

There are two major differences between carbon and iron in a diffraction experiment. First, iron has more electrons. Second, most electrons in iron (18 of 26) are tightly bound inner electrons, so they are close to the nucleus. On the other hand, most of the electrons in carbon are valence electrons, so its electron density is more diffuse.
If you need an analogy or a mental image, iron is like a peach and carbon is like a strawberry. You should not have different mental images of the objects depending on wavelength - it is the same object. If you need some hard data, you can Fourier-transform the form factor to get the radial electron density of the atoms of the two elements.


text:
Make pyridine from furfural
Furfural $(\ce{C5H4O2})$ has been used as a feedstock for synthesis. Furfurylamine $(\ce{C5H7NO})$ is made from furfural.
Given that industrial manufacturing ammonia is matured, how about combining furfural, ammonia and hydrogen to directly synthesize pyridine? Is the following reaction realizable?
$$\ce{C5H4O2 + NH3 + H2 -> C5H5N + 2 H2O}$$
Both furfural and pyridine are significant organic compounds. If one compound can be made from the other, the cost of the other can drop. So far is how pyridine itself is being manufactured remained an industrial secret? Substituted pyridines are covered in many articles and online information.
Wikipedia pyridines production indicates that

Chichibabin synthesis makes dihydropyridine, a pyridine derivative from acrolein, acetaldehyde and ammonia and byproducts are observed.
Bönnemann cyclization of acetonitrile and two molecules of acetylene gives 2-methylpyridine, a substituted pyridine. Following dealkylation gives pyridine.

In essence, substituted or derived pyridines were first made in multisteps and they are further processed to the parent pyridine.
Another possible route would be using the aza variety of the Achmatowicz reaction on furan.

Hydrogen replaces Ts() and removing of hydroxyl and alkyl make the parent pyridine.
( removing the hydroxyl makes the parent pyridine.

Still, substituted furfurylamines $(\ce{C5H7NO})$ is needed at the beginning. In other words, my question was about if as few steps as possible making the parent pyridine from a prevalent starting material such as furfural is realizable.

This approach tends to give 3-hydroxypyridines

$$\ce{C5H4O2 + NH3 -> C5H5NO + H2O}$$ 2 or 3 or 4-hydroxypyridines do not matter if the hydroxypyridine can be hydrogenated to pyridine and water. $$\ce{C5H5NO + H2 -> C5H5N + H2O}$$

if your first step is to make furfurylamine then treat with hydrogen peroxide and HCl  https://www.researchgate.net/publication/293244971_Preparation_of_3-hydroxypyridine_from_2-aminomethylfuran_with_hydrogen_dioxide_as_oxidant

3-hydroxypyridine behaves much like phenol so good luck with trying to reduce it by hydrogenation.

That is one reason I want to ask... I can be flawed and wrong in lines of thought.

Wikipedia has a good summary of known industrial processes for producing pyridine. It is not an industrial secret. And there is probably little scope for a new process.

I do not know whether the following is a real answer or not but Wikipedia 3-methylpyridine article shows

3-Methylpyridine is produced industrially by the reaction of acrolein with ammonia:
:$$\ce{2 C3H3O + NH3 -> 3-CH3C5H4N + 2 H2O}$$ This reaction also affords substantial amounts of pyridine.

Wikipedia Nicotinonitrile article

A colorless solid, it is produced by ammoxidation of 3-methylpyridine:[2]. Nicotinonitrile is a precursor to the vitamin niacin.

In other words, the question could become one of separating pyridine from 3-Methylpyridine.


text:
How ph affects aesculune fluorescence?
I am facing a ween problem. I am trying to find out, how would look like the graph for fluroscence of aesculine (intensity of fluoroscence on Y axis and pH on X axis).
I know, that increasing the pH causes a corresponding increase in the maximum fluorescence intensity from 77.74 (units) in the acidic sample to 146.13 at neutral pH and 232.69 at alkaline pH.
But, can I apply this on aesculine?

You should measure the fluorescence vs that from a a standard compound excited at the same wavelength (with same absorbance) and observed at the same wavelength. If you look in the literature there are many standards and also compounds whose fluorescence yields are known. Look also for measuring relative fluorescence yields rather than absolute yields as these are v tricky. In any experiment you must ensure that you correct for any differences in the amount of light absorbed as this can be a source of large errors.

A pH dependence is an indication that a pH buffer should be used in the sample matrix before the instrumental analysis.

I have no idea, you are talking about...

Keep also in mind that coumarins are labile at $\mathrm{pH} \ge 12$.

You should investigate several pH and study the effect of time on the fluorescence spectra. One cannot quote single values without mentioning the wavelengths of excitation and emission.
pH can alter the state of  ionization and it can also cause hydrolysis of the sugar unit.
You need to collect three different types of spectra at each pH, say from pH 2 to 12.
At each pH, first measure the absorption spectrum at those pH values. This is the easiest experiment. Do you see a difference in absorption spectra at different pH values? Note the trends.  If yes, pH is affecting either the structure (decomposing) or the state of ionization.
Next, collect the excitation spectrum at a fixed emission wavelength (you have to estimate that). The excitation spectra should look similar to the absorption spectrum in shape.
Once you know the correct excitation wavelength at each pH, collect proper emission spectrum by using the lamba max of excitation.
So the answer is that be ready to see the changes in fluorescence spectra since there is no "the" fluorescence spectrum of a given molecule in a solution. You have to specify the solvent, pH, and excitation wavelength for experimental measurements.


text:
Why is hydrated beryllium ion acidic?
The beryllium ion, $\ce{Be^{2+}}$, forms the aquo complex $\ce{[Be(H2O)4]^{2+}}$. According to LibreTexts, this complex is acidic in solution: $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$
It is not clear to me why it should be acidic. I read here that "the hydrolysis happens because the $\ce{Be−O}$ bond is very strong and so in the hydrated ion this weakens the $\ce{O−H}$ bonds hence there is a tendency to lose protons."
I have two questions:

What is meant by a "very strong" $\ce{Be-O}$ bond? Is it stronger than those in other metal aquo complexes? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?
Why does a strong $\ce{Be-O}$ make the $\ce{O-H}$ bond weaker? I know that oxygen forms a coordinate bond with metal cation so the electrons in the $\ce{O-H}$ bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O-H}$ bonds more polar. Why does a polar bond have to be a weaker one?

All aquo cations are more acidic then water. How could they be not, when H2O is bound to Lewis acid?

What is meant by a "very strong" Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?

You got that absolutely right.
$\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length and hence increasing the bond strength.

[...]the electrons in the O−H bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O\bond{−}H}$ bonds more polar.

You struck at the right point. Let's look at the original question.
I'm sure you must know that a stronger acid has a higher tendency to release $\ce{H+}$ ions. This means that the electron pair should be drawn away from $\ce{H}$ atom in the $\ce{O\bond{−}H}$ bond. Thus, a polarized bond makes it easier for the H+ ion to 'escape' from the molecule.

Why does a polar bond have to be a weaker one?

It is not appropriate to answer this question in definitive terms. The strength of a bond is defined by its bond length, not its polarity. To address the confusion you're having, look back at the context of the question. $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$
We're talking about a compound with polar O-H bonds dissolved in water. Water has a high dielectric constant, and can easily break ionic/polar bonds since it liberates a high amount of solvation energy. This is because of its high dielectric constant.
So, are polar bonds weaker than non-polar bonds in water? Sure, I can see your line of thinking. But, always remember that ionic bonds can easily be broken by water, but that doesn't mean that ionic bonds are weaker than non-polar bonds.


text:
n-Chlorobutane addition to benzene by Friedel-Crafts mechanism
I recently had an exam question, pictured below, regarding a Friedel-Crafts alkylation of benzene. The question wants to know what would form as the major product. I responded that the second option, sec-butylbenzene, would form as the major product. This response was marked incorrect. I discussed the item with the professor, who told me that the correct answer (last option) was that "more than one of these products" would form.
I located a textbook example, also pictured below, which lists the sec-butyl product as the major (49% yield), and the n-butylbenzene as the minor (29% yield). This sums to 78% of the yield. I wholly agree that more products will form, but I see no evidence to suggest that they could be considered a majority product.
I have searched Reaxys and ACS for clarifying literature, but have not found anything helpful. Is anyone aware of literature which would offer experimental data to support either position, or perhaps another way that I should be thinking about this reaction?

It’s not clear to me what your selected answer was and what answer the instructor considered correct.

My selected answer was the sec-butyl option, the one with the filled in dot. The professor said that the last option, "More than one of these products" was correct.

It says major product in the question, so I’d agree with the rearranged sec-butyl option.
(I would say that this seems less about the chemistry, which you seem to understand just fine, and more about the semantics of how the question is worded)

It says "major product (s)", which kinda hints at a possibility that multiple products can all be considered major at once. If not for that detail, I'd say that sec-butyl is the only correct answer.

Even though butylbenzene is considered minor in this case, would it be fair to say that it is a secondary major product when taking the other miscellaneous products into account? That is the only other way I can rationalize the answer

If you deleted out the "major product(s)" part of the question, what would people answer?


text:
Separating plant extract from ethanol solution
I have to do an experiment which involves getting the extract of certain leaves. I had decided to macerate it in ethanol (70%) solution and then to evaporate the solvent using water bath/rotary evaporator.
Due to the pandemic, I couldn't get access to labs to conduct this experiment. Is there any other way to evaporate the solvent at home without the need for water bath/rotary evaporator?
This extract will be used for testing antibacterial effects. So, I'm kinda concerned about simply heating up the solution because it may destroy the necessary compounds. Would it work if I try to heat the solution in lower temperatures?
Thank you,

I would use a combination of vacuum and water bath.

How much liquid are we talking about? Ethanol will just evaporate when leaving it open, so you could just leave it open outside somewhere so your flat is not covered in ethanol fumes.

Free evaporation or using a water bath ?
Both ethanol and water evaporate at any temperature, just with the exponential-like rate. ( Washed clothes will dry up even at freezing temperatures ).
If living in area with currently cold weather, consider using evaporation on the central heating radiator, parhaps partially isolated.
Free evaporation could be a safe bet, but slow evaporation can support oxidation of dissolved substances, so I would try both free and heated ways.
I remember that in my young-chemist days, I had bought a water stream vacuum pump, based on the Bernoulli equation.

Setting up a fan blowing over the top of whatever you are storing your extract in would be more than enough to evaporate any sort of alcohol/water mixture. It might take a while, but unless you have another option then it could be worth trying.


text:
What pressure does liquid nitrous oxide need at −45 °C?
What pressure does liquid nitrous oxide need at −45 °C to remain liquid? How can I calculate its pressure required for other temperatures?

Rather read from phase diagram then calculate, unless you find curve fitting equation.

@Mithoron Would you point me to a phase diagram to read this from? I could really not find a decent one.

I also found a bad one...

«Nitrous oxide: Saturation properties and the phase diagram», Ferreira, Lobo, The Journal of Chemical Thermodynamics, 41, 2009, 1394-1399, doi 10.1016/j.jct.2009.06.017, seems to cover the range you need here.

If such a diagram was not available, or for other compound, one could use  the knowledge of the boiling point, the evaporation molar enthalpy (or its approximation by the empirical rule ( forgot the rule name ) saying evaporation molar entropy change is 90 J/K/mol and the Clausius-Clapeyron equation.) Then it is possible to compute the approximation of the curve p_vap  = f(T)

Using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I found the saturation point of nitrous oxide at equilibrium for a temperature of $T=-45.000\ \mathrm{^\circ C}$ at a pressure of $p=783110\ \mathrm{Pa}=7.8311\ \mathrm{bar}$;
for $T=-78.000\ \mathrm{^\circ C}$, it is $p=182290\ \mathrm{Pa}=1.8229\ \mathrm{bar}$.

Thank you very much! I was about to ask if there may be a (freely accessible) online database/calculator where one could get this info and I just found this REFPROP version 10 is selling for $325 here: https://www.nist.gov/srd/refprop I guess you have purchased that, right? Would you mind looking up one more value - pressure for -78°C? Thanks!

Looks like there is an alternative to REFPROP - it is called CoolProp. Not very user friendly though. Website: http://www.coolprop.org/


text:
Finding reaction rate constant for second order equation in Wolfram Mathematica
I’m working in Mathematica for my master’s project and I can’t work out how to find $k$ in the second order rate equations where I have a concentration vs time graph for $\ce{AHA}:$
$$\mathrm{rate} = k[\ce{AHA}][\ce{H}].$$
I have the starting concentrations for $\ce{AHA}$ and $\ce{H}.$

What is the integrated form of the second order equation?

Are the starting concentrations [AHA] and [H] different or same? This information is important to determine which integrated rate equation to use.

@Ele If you are using Mathematica a lot, you may find Mathematica.SE useful.

If your question is about Mathmatica, you should  show us the math you want to work out in Mathematica. If you don't know what the mathematical description of your problem is, you should show us the data and tell us where you got stuck.

The general second order equation is
$$\frac{\mathrm d[A]}{\mathrm dt}=-k[\ce{A}][\ce{B}]\tag{1}$$
with $[\ce{A}]_0,$ $[\ce{B}]_0$ the initial amounts. At a time $t$ there are $[\ce{A}]_0-x$, and $[\ce{B}]_0-x$ remaining if amount $x$ reacts. Thus
$$\frac{\mathrm dx}{\mathrm dt}=k([\ce{A}]_0-x)([\ce{B}]_0-x).\tag{2}$$
You should be able to integrate this from $0\to x$ by using partial fractions or directly using computer algebra. The result is
$$\ln\left(\frac{[\ce{A}]_0-x}{[\ce{B}]_0-x}\right)=k([\ce{A}]_0-[\ce{B}]_0)t+\ln\left(\frac{[\ce{A}]_0}{[\ce{B}]_0}\right),\tag{3}$$
so that you can plot (as Maurice has pointed out) the left hand side vs $t$ and obtain the rate constant $k$ from the slope $k([\ce{A}]_0-[\ce{B}]_0)$ in units $\pu{dm^3 mol^-1 s^-1}.$

Here's one way you could get from eqn. (2) to eqn. (3) in porphyrin's answer using Mathematica.  I made extensive use of the Part function, whose shorthand is "[[ ]]", to pull out the desired subexpression from each answer.
E.g., Part[x+y, 1], which means "take the first part of the expression (x+y)", can be expressed as (x+y)[[1]], and returns "x"). But you could, alternately, just do copy and paste.
[I used "y" in place of porphyrin's "x" just because I'm used to x being an independent, rather than dependent, variable.]
ans1 = DSolve[{y'[t] == k (Ao - y[t])*(Bo - y[t]), y[0] == 0}, y[t],
  t]
ans2 = ans1[[1, 1, 2]]
ans3 = Reduce[
  y == ans2 && Ao > 0 && Bo > 0 && t >= 0 && y >= 0 && k > 0, k]
ans4 = ans3[[2, 2, 2, 4]]
ans5 = Reverse@Simplify[MultiplySides[ans4, t]][[1, 1, 1]]
ans6 = ans5 /. {(Ao - y) -> A, (Bo - y) -> B}
ans7 = ans6 /. {A -> "[A(t)]", B -> "[B(t)]"}

And here is what it would look like.  You would just need to plot the LHS of ans6 vs. t; a linear regression would give you the slope, which is the rate constant.  Note that "Log" in Mathematica is the natural log (ln), not the base-10 log:

If the intent is to fit actual experimental data, a direct nonlinear fit is generally going to be much more accurate than linear regression to a plot of the log unless proper weighting is performed. Generally speaking, linear regression for this application is considered very outdated

@Andrew  On the contrary, the opposite is true, for both theoretical and practical reasons. THEORETICAL: It's always better to linearize the data if you can because there exist, for linearized data, algorithms that can can always find the global minimum (the true best fit). With non-linear data, you are only certain to find a local minimum; whether you've found the global minimum depends on how clever/lucky you've been in providing starting values for the fitting parameters. PRACTICAL: Experimental data is typically measured data, and is thus typically heteroscedatic rather than homoscedatic.

...Specifically, it typically has variability that is proprotional to its magnitude.  To properly build models for such data, it is necessary to minimize the relative, rather than absolute, size of the residuals. Yet none of Mathematica's fitting functions (Fit, LinearModelFit, FindFit NonLinearModelFit) offer such minimizations.  [And I suspect this is true of most other symbolic mathematica systems as well.] Instead, they only offer minimzations based on the absolute size of the residuals (such as least squares), which give poor fits for data in the former category.

...And the nice thing is that, when you linearize by taking the logarith of the data and apply least squares, you are minimizing the absolute size of the logs of the residuals, which is equivalent to minimizing the relative size of the residuals themselves, thus achieving an appropriate minimization for heteroscedatic data whose variability is proportional to its magnitude.  [Also: While I put the need, when doing non-linear fits, to specify good starting values under "THEORETICAL", it could go here as well, since it's a pain to find those values; doing a linear fit makes that unnecessary.]

As your comments demonstrate, data fitting is a large topic, but here are some brief replies: 1) if you can't be sure you have the global min of a nonlinear fit, you need better data 2) linear least squares suffers from disproportional weighting of points farther from the center point 3) without knowledge of the specific measuring system, there's no a priori reason to believe that absolute residuals of logs are any more accurate. The better way to deal with variable error is to include multiple y points for each x and do the nonlinear fit to all of those values simultaneously.

@Andrew  My replies to your replies: #1 is overly glib; data can be difficult or expensive to obtain, and often we have to work with what we have.  #2:  That disproportional weighting is trivial compared to the disproportionality that typically occurs when you have a large range of measured values (say, when measuring concentrations over a few orders of magnitude). #3: Sure, one can't say anything absolutely a priori.  But when your measured values vary over a significant range (say, a few orders of magnitude or more), it would be an unusual measurement device whose measurement error...

....doesn't increase with the magnitude of the value being measured.  You may find these two papers informative:  https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2017SW001669?scrollTo=references
https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2635088
Finally, it won't,by itself, address heteroscedacity to  "include multiple y points for each x and do the nonlinear fit to all of those values simultaneously". Rather, after measuring multiple y values, you'd need to determine the variance for each x, and weight the fit according to that variance (more variance -> less weight).

In addition, there's the issue of model utility.  Typically, if your ind. var. ranges over a few orders of mag., it's far more useful to minimize relative errors rather than absolute errors.  E.g., suppose you want to predict the safe use limit for cables with cross-sectional areas ranging from 1 mm^2  to 1,000 mm^2 (~1 mm dia. to ~ 35 mm dia.).  It would be far more useful to have a model that was optimized to give predictions with min. rel. errors rather than min. absolute errors; in the latter case the predicted errors for the smallest cables could be multiple times the predicted values.

Try to report first the logarithm of the concentration of the reagent versus the time $t$. If it gives perfectly aligned points, going downwards, the kinetics is first order, and the slope of the line is the rate constant $k_1$ in $\pu{s^-1}.$
If the points are not aligned on the logarithmic curve, try to report the inverse $1/c$ of the concentration of the reagent versus time. If the points are aligned, and going upwards, it means that the kinetics is second order. The slope of the line is the rate constant $k_2$ in $\pu{L mol^-1 s^-1}.$
If none of these attempts are satisfied, the kinetics is probably described by a hypothesis of an intermediate state.


text:
Why does beryllium have an exceptionally high hydration energy?
You would expect solubility of  Group $2$ fluorides to increase down the group, as lattice energy plummets much more sharply than hydration energy does. For the most part, this is true: $\ce{BaF2}$ is more soluble than $\ce{SrF2}$, which is more soluble than $\ce{CaF2}$, and so on. However, $\ce{BeF2}$ is several times more soluble than even $\ce{BaF2}$. Why is that?
Beryllium forms the complex $\ce{[Be(H2O)4]^{2+}}$ in water. According to Wikipedia, $\ce{M-O}$ distance expectedly increases down the group and enthalpy of solvation becomes less and less negative, but beryllium aquo complex is the black sheep:
\begin{array}{c|c|c}
 & \ce{[Be(H2O)4]^{2+}} & \ce{[Mg(H2O)6]^{2+}} & \ce{Ca^{2+}(aq)} & \ce{Sr^{2+} (aq)} & \ce{Ba^{2+} (aq)}\\
\hline
\ce{M-O distance (pm)} & 167 & 209 & 242 & 263 & 281\\
\ce{\Delta H- solvation (kJ mol^{-1})} & 2494 & 1921 & 1577 & 1443 & 1305
\end{array}
I found these solubility figures from JD Lee's Concise Inorganic Chemistry, Fifth Edition:
\begin{array}{c|c|c}
 & \ce{BeF2} & \ce{MgF2} & \ce{CaF2} & \ce{SrF2} & \ce{BaF2}\\
\hline
\ce{solubility (grams solute per 100 g water} & \ce{very soluble} & 0.008 & 0.0016 & 0.012 & 0.12
\end{array}

So are you asking why Be has a high hydration energy or are you asking why BeF2 is more soluble than BaF2? I did a bit of researching through a few books and, it appears that there are conflicting beliefs over the relative solubility. Can you confirm the  solubility data?

Huh. Although my teacher said that the solubility follows a clear trend: $\ce{BeF2 > BaF2 > SrF2 > CaF2 > MgF2}$, the trend of solubility according to JD Lee is more here-and-there

@C_Lycoris I am asking why BeF2 is more soluble than BaF2. I surmised it was because of beryllium's high hydration enthalpy.

I believe that the high hydration enthalpy is due to the points correctly stated by you. Furthermore, I find that the general trend is for the solubility to increase down the group when the size of the cation is near about that of the anion. [The lattice enthalpy usually dictates the solubility in this case] for eg: $$\ce{NaF<KF<CsF}$$. While it is true that the Hydration energy dominates the solubility in this case, I have yet to find a satisfactory reason for this.

The problem with solubility of ionic compounds is that it hinges on two different factors [Hydration enthalpy and lattice enthalpy], which further hinge on 2 other factors:[size of cation and anion]. Another factor that comes into play is Fajan's polarization rule. All this culminates into varying, often confusing trends.
Ultimately, what decides solubility is ΔG= ΔH-TΔS. That's the surefire way of comparing solubility, but given that this data is not always available during exams, it isn't very useful.


text:
Viscosity at CMC
What happens to viscosity at critical micelle concentration? Does it increase or decrease abruptly, and if so, what's the reason behind it?
I'm aware of changes in other properties such as conductivity, concentration etc, but not viscosity.

Do you mean as the concentration of the monomer increases (at constant T)?

No, I mean at the critical micelle concentration, when the individual ions of the soap molecules cluster together

Yes, sorry, never mind my comment. It turns out the viscosity often increases.

whats the reason behind it? i tried making a comparison with gases, where the viscosity increases with the radius of the gas, and also with density... here, the "radius" is increasing, due to micelle formation and the "density" (which i think is analogous here to concentration) is decreasing... how do we decide which factor dominates?

One of the more fundamental theories of solution viscosity (by Einstein) predicts the following for dilute solutions of perfect spheres in a Newtonian fluid:
$$\frac{\eta}{\eta_0} = 1 +2.5\phi \tag{1}$$
where $\eta_0$ is the viscosity of the pure solvent,  $\phi$ is the volume fraction of solute (here: aggregating detergent) Note we are ignoring higher order terms that describe the dependence on micelle-micelle interactions (e.g. on the second virial coefficient).
The important point of Eq. 1 is that the viscosity is independent of the size of the spheres, or the composition (provided they are rigid, not flexible). So according to this theory, if the volume occupied by the micelles is not different from that occupied by the monomers, you should not see a change in viscosity. But typically you do, and that means the volume fraction is changing, at least according to this theory.
How? Maybe through hydration. The theory does not say whether hydration plays a role, but hydration can alter the effective volume occupied by the spheres, if you include the hydration layer as part of the micelle volume.
Most likely the theory is not really accurate and molecular shape also plays a role. However that would require a more complex theory to address.
It turns out that viscosity often goes up at the CMC (with formation of micelles).


text:
Flame Test of Alkali Metals and Alkaline Earth Metals
The flame test colors for alkali metal and alkaline earth metal salts are well known, but what about the metals themselves? Do they also give rise to the same colors (eg, brick red for Ca, pale green for Ba etc), or does it differ? And if so, why?

Yes, the metals, when sent in a flame produce the same colors as their salts. In fact, the metallic ions do not produce any color when sent in a flame. It is only the neutral atom that produce the famous yellow double line of the sodium. This means that a tiny amount of the $\ce{Na+}$ and $\ce{Cl-}$ ions are first transformed into neutral $\ce{Na}$ and $\ce{Cl}$ atoms. And the color of the flame is due to the deexcitation of these rare atoms being first electronically excited in the flame. So it means that the following processes are occurring successively in the flame:$$\ce{NaCl(s) -> Na(g) + Cl(g)}$$ $$\ce{Na + heat -> Na^*}$$ $$\ce{Na^* -> Na + h\nu }$$ The famous yellow light of the $\ce{Na}$ atom is due to the return of the outer electron from the level $4s$ to $3p$ of the sodium atom, if my memory is good. Nothing to do with the $\ce{Na+}$ ion.

Why does caesium, a pretty large atom, need high-energy blue light to get excited but lithium, a small atom whose nucleus tugs strongly at the outermost electrons, need low-energy red light?

The visible light emitted by these atoms is not always the return of the electron to the lowest level. It is even rarely the case. It may be the passage from a very high excited level to a not so high excited level.

@RayBradbury and what we see is the energy of the jump, not the energy of the levels themselves.

I suggest changing the dot to an asterisk.

@RayBradbury   In addition to the comment by Maurice, there is another reason the colour you see has little to do with the energy of different orbitals: you only see lines in the visible but many emission lines are UV or IR so you usually see only a small selection of the actual range of emission lines.


text:
IUPAC name of H[AuCl4]
Is $\ce{H[AuCl4]}$ tetrachloroauric [III] acid or hydrogen tetrachloroaurate[III]? As per the rules I have studied the latter name is correct. Then why am I getting the former name mentioned everywhere on the internet as the IUPAC name, although Wikipedia does mention the other name (hydrogen tetrachloroaurate[III]) though not as the IUPAC name?

Well, HCl can be considered both hydrogen chloride (in gaseous form) and hydrochloric acid (especially when it is in the aqueous form). However, this acid is a solid. I would go with chloroauric acid. That is the widely used name.

The authoritative source Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (Red Book) lists $\ce{H[AuCl4]}$ as an example of a salt in the subsection IR-4.4.3.4 Generalized salt formulae [1, pp. 61–62].
Further, introduction to the section IR-8 Inorganic Acids and Derivatives underlines that IUPAC nomenclature is established from composition and structure, not properties, hence the word "acid" should not be used in systematic names [1, p. 124]:

Certain inorganic and simple carbon-containing compounds are commonly given non-
systematic or semi-systematic names containing the word ‘acid’. Examples are boric acid or
orthoboric acid, metaboric acid, phosphoric acid, diphosphoric acid, cyclo-triphosphoric
acid, catena-triphosphoric acid, dithionous acid, peroxodisulfuric acid or peroxydisulfuric acid, etc. These names are unique in modern nomenclature in that, interpreted literally, they describe a particular chemical property of the compounds in question. Systematic names are otherwise based solely on composition and structure.
All such acids may also be given structure-based systematic names using principles
already described in preceding chapters on substitutive and additive nomenclature, so in that respect the ‘acid’-containing names are superfluous. Furthermore, many species which
would be classified as acids based on their chemical properties are never named as such,
e.g. aqua ions such as hexaaquaaluminium(3+), and hydrides and derivatives such as
ammonium, hydrogen sulfide (sulfane), etc. The term ‘acid’ is thus not used consistently.
Based on these considerations, the use of the word ‘acid’ in any new name in inorganic
nomenclature is discouraged. However, a number of the existing ‘acid’ names are so
commonly used (sulfuric acid, perchloric acid, etc.) that it would be unrealistic to suggest replacing them altogether by systematic alternatives.

According to Wikipedia, $\ce{H[AuCl4]}$ exists in aqueous solutions as square planar $\ce{AuCl4-}$ anion, which is also retained upon crystallization as tetrahydrate, but there is a rapid
exchange of all protons in the unit cell [2]. For compounds of this class where the structural aspects of protons are either not essential or cannot be precisely established, a hydrogen name can be used. From the section IR-8.4 HYDROGEN NAMES [1, pp. 134–136]:

An alternative nomenclature for hydrogen-containing compounds and ions is described here. The word ‘hydrogen’, with a multiplicative preﬁx if relevant, is joined (with no space) to an anion name formed by additive nomenclature and placed within appropriate enclosing marks (see Section IR-2.2). This construction is followed (again with no space) by a charge number indicating the total charge of the species or structural unit being named (except for neutral species/units).
Hydrogen names are useful when the connectivity (the positions of attachment of the
hydrons) in a hydron-containing compound or ion is unknown or not speciﬁed (i.e. when
which of two or more tautomers is not specified, or when one does not wish to specify
a complex connectivity, such as in network compounds).
[…]
Examples:
[…]

$\ce{H4[Fe(CN)6]}$
tetrahydrogen(hexacyanidoferrate)
$\ce{H2[PtCl6]·2H2O}$
dihydrogen(hexachloridoplatinate)—water (1/2)

[…]
The strict deﬁnition of hydrogen names proposed here is meant to eliminate such
confusion by imposing the requirements:
(i) that ‘hydrogen’ be attached to the rest of the name,
(ii) that the number of hydrogens must be speciﬁed by a multiplicative preﬁx,
(iii) that the anionic part be placed in enclosing marks, and
(iv) that the charge of the total structure being named is speciﬁed.
Hydrogen names constructed in this way cannot be mistaken for other types of name.

To sum it all up, I think the IUPAC name for $\ce{H[AuCl4]}$ is hydrogen(tetrachloridoaurate).
Note that new recommendations also suggest using chlorido instead of chloro for anionic ligands. Also, in names formal oxidation states are enclosed in parentheses, not square brackets.
Wikipedia might not always mention IUPAC names, and even those which are listed as such should not be automatically trusted. Also, acceptance of systematic names in chemical community is inertial; not every systematic name is immediately replacing the traditional one which everyone is familiar with. I suspect this is the case as to why chloroauric acid still gets a pass.
References

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.
O’Reilly, D. E.; Peterson, E. M.; Scheie, C. E.; Williams, J. M. Nuclear Magnetic Resonance of the Aquated Proton. II. Chloroauric Acid Tetrahydrate. Phase Transitions and Molecular Motion. J. Chem. Phys. 1971, 55 (12), 5629–5635. DOI: 10.1063/1.1675731.


text:
Why transition metal chloride complex have lower reduction potential than metal aqua ions?
Why transition metal chloride complex have lower reduction potential than metal aqua ions? For example:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &-> Au(s)} &\quad E^\circ &= \pu{+1.52 V}\tag{1}\\
\ce{[AuCl4]-(aq) + 3 e- &-> Au(s) + 4 Cl-} &\quad E^\circ &= \pu{+0.93 V}\tag{2}
\end{align}
$$
Is it because the chloride complex formation constant is very high (higher than aqua ions) so the reduction of $\ce{Au^3+}$ is less favored?

Welcome to Chemistry.SE! Please use MathJax for writing chemical equations: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

The  anion  $\ce{AuCl4−}$ is  the  most  known  complex  of  gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we are dealing with:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &<=> Au(s)} &\quad E^\circ_\mathrm{cathode} \tag1\\
\ce{[AuCl4]-(aq) + 3 e- &<=> Au(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} \tag2
\end{align}
$$
If we substract the equation $(2)$  from the equation $(1)$, we get:
$$\ce{Au^3+(aq) +  4 Cl- (aq) <=> [AuCl4]-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag3$$
The equation $(3)$ is the reaction for the formation of $\ce{AuCl4−}$ with formation constant $K_f$. If you applied the Nernst equation for the equation $(3)$, then at $\pu{25 ^\circ C}$:
$$E^\circ_\mathrm{cell} = \frac{RT}{nF} \ln Q = \frac{RT}{nF} \ln K_f = \frac{0.0592}{n} \log K_f \tag4$$
This last solution of the equation $(4)$ (where $n = 3$) is because, when at equilibrium $Q = K$. If we know the $K_f$, we could have find the $E^\circ_\mathrm{cell}$, and hence $E^\circ_\mathrm{anode}$ (which is for $\ce{AuCl4−}$). Unfortunately, I couldn't find $K_f$ for $\ce{AuCl4−}$ ion formation. Thus, I'd show how big $K_f$ for $\ce{AuCl4−}$ by applying the given values of $E^\circ_\mathrm{cathode} = \pu{+1.52 V}$ and $E^\circ_\mathrm{anode} = \pu{+0.93 V}$ on the equation $(4)$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f \ \Rightarrow \log K_f = \frac{n}{0.0592} E^\circ_\mathrm{cell} = \frac{3}{0.0592} (1.52 - 0.93) = 29.9$$
$$\therefore \ K_f = 10^{29.9} = 7.91 \times 10^{29}$$
It is also in similar way you can find the $E^\circ$ for redox half-reaction of $\ce{PtCl4^2−}$ since we know its formation constant ($K_f = 1.0 \times 10^{16}$):
$$
\begin{align}
\ce{Pt^2+(aq) + 2 e- &<=> Pt(s)} &\quad E^\circ_\mathrm{cathode} = \pu{1.18 V} \tag5\\
\ce{[PtCl4]^2-(aq) + 2 e- &<=> Pt(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} = ? \tag6
\end{align}
$$
If we substract the equation $(6)$  from the equation $(5)$, we get:
$$\ce{Pt^2+(aq) +  4 Cl- (aq) <=> [PtCl4]^2-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag7$$
From equation $(4)$ where $n = 2$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f = \frac{0.0592}{2} \log( 1.0 \times 10^{16})  = 0.474$$
Since, $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.474 V}$, and $E^\circ_\mathrm{cathode} = \pu{1.18 V}$,
$$E^\circ_\mathrm{anode} = 1\pu{1.18 V} - \pu{0.474 V} = \pu{0.706 V}$$
$$\therefore \ \ce{[PtCl4]^2-(aq) + 2 e- <=> Pt(s) + 4 Cl- (aq)} \quad E^\circ = \pu{0.706 V}\text{ (calculated)} $$
This value is in good agreement with the experimental value, which is given as $\pu{0.755 V}$ in electrpchemical series of CRC Handbook of Chemistry and Physics.
Source of formation constant of platinum complex ion

This difference of redox potentials is a consequence of Nernst's law. The second equation is nothing else as the first one, where the concentration of the $\ce{Au^{3+}}$ in a solution of $\ce{AuCl4^{-}}$ is extremely low. This sort of comparison could be used to calculate the equilibrium constant of the equilibrium $$\ce{Au^{3+} + 4 Cl- <-> AuCl4^{-}}$$
Here, Nernst's law can be written :  $$\ce{Au -> Au^{3+} + 3 e-}$$
$$\ce{E = 0.93 V = 1.52 V + \frac{0.059 V}{3} . log [Au^3+]}$$ so that the concentration $\ce{[Au^{3+}]}$ in $\ce{AuCl4^-}$ is given by  $$\mathrm{log}\ce{[Au^{3+}] = \frac{3}{0.059 V}(0.93 V - 1.52 V) =  - 30}$$ This means that in a $1$ molar solution of $\ce{AuCl4^-}$, the concentration of the ion $\ce{Au^{3+}}$ is only $\pu{1E-30M}$. But the redox reaction happening in $\ce{AuCl4^{-}}$ is the same redox reaction that occurs between $\ce{Au}$ and $\ce{Au^{3+}}$. Just the concentration of $\ce{Au^{3+}}$ is extremely low in $\ce{AuCl4^-}$ solutions.

Thank you. With this part of the last period "the reduction of Au is less favourited" I meant exactly what you have written (my fault, I was not very clear). If chloride complex's Kf is higher, than the concentration of free Au3+ is very low. Thank you again for answering.

$\pu{1E-30M}$ is effectively nothing. Reduction of  aquo and chlorido complex have only the product in common.


text:
How to use anodic and cathodic overpotentials to calculate electrolytic energy efficiency?
I am trying to calculate the energy efficiency for $\ce{CO2}$ reduction to formate, coupled to water oxidation based on literature data.
$$\ce{CO2 + 2e- + 2H+ -> HCOOH}$$
$$\ce{H2O -> 2H+ + 1/2O2 + 2e-}$$
I hope someone can help me to understand how to combine the information that is presented in the papers so I can calculate an overall energy efficiency of the reaction.
E.g. things that are given:

cathodic overpotential
anodic overpotential

The energy efficiency should eventually be equivalent to
$\mathrm{EE = \frac{Combustion\;energy\;of\;Formate}{Invested\;electrical\;energy}}$

More on the site's notation: https://chemistry.stackexchange.com/help/notation and https://chemistry.meta.stackexchange.com/q/86/73265

Note that there are additional potential differences, affecting the efficiency, that are related to non equlibrium system state due nonzero current. Like electrolyte potential gradient and electrode double layer potential shift due electrode reaction kinetics.


text:
How to calculate molar concentration from equilibrium equations with no initial concentration given
Consider the following equations:
\begin{align}
\tag{1}
  \ce{CO2 (g) + H2O (l) &<=> H2CO3 (aq)} &
  K_\ce{CO2} &= 10^{-1.47}\\
\tag{2}
  \ce{H2CO3 (aq) &<=> H+ (aq) + HCO^-_3 (aq)} &
  K_1 &= 10^{-6.35}\\
\tag{3}
  \ce{HCO^-_3 (aq) &<=> H+ (aq) + CO^-_3 (aq)} &
  K_2 &= 10^{-10.33}
\end{align}
I am also told atmospheric $\ce{CO2}$ is at $\pu{10^{-3.5} atm}$ and the solution is pure water.
So far I have been able to find the equilibrium equations and have got the concentration of $\ce{H2CO3}$ to be $\pu{10^{-4.97}}$. However, from this I do not know where to go in order to find the concentration of bicarbonate.


text:
Maximum number of Spectral Lines "A better quantum model shows that there will be n^2 transitions"?
I was parsing the following post What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state? and came across with the reply from @porphyrin.
cite : "lease note that these answers are only true in the Bohr model, a better quantum model shows that there will be $^2$ transitions. The orbitals are now labelled with increasing orbital angular momentum, ℓ=0,1,2,3..."
Please, does someone can point me out to references or more extended explanations, from where does the $^2$ come from?
Update
The general formula for calculating number of spectral lines is (n2−n1)(n2 −n1+1)/2; which "merely" relate to triangular numbers n(n+1)/2. I wonder about $n^2$ since the sum of two consecutive triangular numbers give a square number; what does the (n-1)n/2 constituent part describe? Is this due to Pauli exclusion Principle; Bose/Einstein Vs Fermi/Dirac distribution?
Really appreciate.
PS: I don't have 50 reputation to comment directly

Phys. chem. textbooks cover this as well as specialist books on spectroscopy and, of course, books on quantum mechanics.

Dear @porphyrin thanks for the reply, I have updated my question. Please could you provide me with references or articles explaining the idea; I'm lost with precise keywords..

@Xavier These type of questions can easily become "rabbit holes" unless you are comfortable with the basic model systems of quantum mechanics. Do you know the quantum mechanical treatment of angular momentum and the treatment of the hydrogen atom and selection rules ?

@HansWurst, thanks :) - I'm used to live in rabbit hole questions. I feel confortable. Yesterday night I finally end up in the Zeeman Effect and start to read about selection rules.. does it have something to do with $n^2$