text:
How do you know if a bond according to VB theory consists of 2 "3 electron bonds" or one "fully paired" bond?
First of all, I want to discuss the whole thing strictly according to the VB - theory and not use any concepts of the MO - theory.
According to Linus Pauling, the "3 electron Bond" is the resonance of one electron between 2 half filled orbitals:
A| •B <-> A• |B
I use his representation: "A•••B".
I applied that Idea to Triplett Oxygen, which has one sigma bond and 2 x ½ π Bonds or "2 x 3 electron bonds": "O÷÷÷O" so it forms one sigma Bond between the two half filled 2py orbitals and two ½ π Bonds between

one fully occupied 2px orbital and one half filled 2pz orbital.
one half filled 2pz orbital and one fully occupied 2px orbital

Then I moved on to Peroxid with O2--.
I now have two options:

Starting with the following electron configuration: O: 2s² 2px² 2py¹ 2pz¹, O--: 2s² 2px² 2py² 2pz²

That must lead to two "3 electron bonds": "O:::O".

Starting with the following electron configuration: O-: 2s² 2px² 2py² 2pz¹, O-: 2s² 2px² 2py² 2pz¹

That must lead to one sigma bond: "O-O".
I know that it can't be option one because Peroxide is diamagnetic and probably more energetically favourable to form a single sigma bond than two ½ π Bonds, but is there another explanation for that?
I also tried to apply that on Dioxygenyl, and endet up with one sigma bond, one π bond and one ½ π Bond: "O=•••O":

But then I didn't have to face this question because I could only start with O and O+ or O+ and O.

I don't know why somebody downvoted without giving a reason, so I can't improve my question...

You can't, really.
Trihalide ions, $\ce{X3^-}$, are known for all halogens with stable isotopes ($\ce{F,Cl,Br,I}$). Conventionally they are regarded as having a two-elecron bond delocalized between two linkages, to wit:
$\ce{X-X\space\space :X^-\space\space <->\space\space X:^-\space\space X-X}$
But this apparently applies only for $\ce{Cl, Br, I}$. Braida and Hubertus[1] report that trifluoride ion, with the same formal valence structure as the others, has a stronger contribution from a delocalized three-electron bonding:
$\ce{F\therefore F^-\space\space \cdot F\space\space <->\space\space F\cdot\space\space F\therefore F^-}$
with the negative charge more delocalized here than in the "standard" structure that seems to apply to the heavier trihalide ions.
Both structures correspond to the same valence-bond arrangementcand thus simple valence bond theory cannot distinguish them. It's one area where molecular orbital theory is superior.
Reference

Benoit Braida and Phillipe C. Hubertus(2004). "What makes the F3- ion so special?" A breathing-orbital valence-bond ab initio study. _Journal of the American Chemical Society, 126(45), 14890-14898. https://doi.org/10.1021/ja046443a.

Oh ok thank you! I have to give a poster presentation on the binding in O2+, O2, O2-, O2-- in the 1st semester and thought it would be best to illuminate all of them once purely from the VB theory and then from the MO theory, at least then I have discussion material when I try to present it this way ;)

But can't VB theory predict bond strength or bond lenght somehow? I've read that a π bond is slightly weaker than a sigma bond, so shouldn't it be possible to calculate if it's rather a single sigma bond or two ½ π bonds somehow?

(I'm far from any calculations in quantum mechanics, that's more a superficial question)


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What is the minimum energy required to dessicate an ethanol+water azeotropic solution
I am wondering, how can we compute the minimum theoretical energy required to separate ethanol from a cold azeotropic water-ethanol mixture?
Are there industrial processes that are close to the theoretical performance? how much energy would be required in practice to dessicate azeotropic ethanol? Would this be just work or heat?

No energy is required to do this task. You just have to add enough desiccant agent like calcium chloride to eliminate the water.

@Maurice CaCl2 forms CaCl2 . 3 EtOH, so not really a good choice.

Industrial process uses water elimination by destillation away the ternary azeotrop with benzene.

After an initial distillation for purification and drying up to approx 95% of ethanol, drying with molecular sieves becomes more and more attractive (example).  You pass the solvent, remove the dry solvent from the granules by filtration, dry the molar sieves by heat and vacuum, and recycle the (re)activated cold molar sieves in the next batch.  So you may skip use of benzene and an additional distillation setup.

@Buttonwood Yes, that is true, I have forgotten to mention it.

Energy is required, but it is the very least of our concerns.

@Buttonwood thanks for your answer! This process of molecular sieves filtration, how much energy does it require when taking into account the recycling of the sieves? Is there some kind of figure/table of how many MJ/kg of ethanol are required?

@ChristianClaudel I don't know where to get the data. At lab scale, one evacuates a round bottom flask with an oil pump ($< \pu{0.1 mbar}$) /Schlenk line anyway in use and a heat gun at  $\approx \pu{200 ^\circ{}C}$; once done & cooled to ambient temperature, the content is poured into the liquid (static method). The dynamic method in (alternating) tube reactors with scales better (génie chimique/process chemistry), here regeneration purges the reactor bed with hot dry nitrogen gas over hours (sometimes heating jackets) typically without vacuum (for mechanical reasons of the reactors used).

The amount of heat needed to regenerate the sieves depends on the capacity of the molar sieves (one can adjust by porosity), and their loading (depends how much they adsorbed/how long they were used). And in a tube reactor, you need pumps to move the liquid across a packed bed (mechanics of fluids), but with flux not too high as you need some residential time to get the adsorption to advance well enough.

Just as an appetizer in addition to Kirk-Othmer (https://onlinelibrary.wiley.com/doi/10.1002/0471238961.1315120511210812.a01.pub2, https://onlinelibrary.wiley.com/doi/10.1002/0471238961.0418250913030315.a01.pub3) there is an open access publication in Industrial & Engineering Chemistry Research by Jović et al. which, though about dichloromethane instead of ethanol, gently taps into the topic «Adsorptive Water Removal from Dichloromethane and Vapor-Phase Regeneration of a Molecular Sieve 3A Packed Bed», Ind. Eng. Chem. Res. 2017, 56, 5042–5054 (https://doi.org/10.1021/acs.iecr.7b00433).

As the enthalpy change of mixing is negative and the entropy change of mixing is positive, I would expect the negatively taken change of Gibbs energy of mixing would be the minimal theoretical energy needed to separate them.

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How does the branched structure of glycogen, increase its ability to be hydrolysed?
"Glycogen branching is essential because it allows for increased water solubility and several sites to break it down; this allows for easy and quick glycogen utilization when it is broken down" (Daghlas and Mohiuddin 2022).
According, to my textbook "the extensively branched structure of glycogen fits its function: more free ends are available for hydrolysis".
Firstly, does hydrolysis not occur between covalently bonded monomers and not require 'free ends'?
Secondly, both quotes imply that a unbranched polymer consisting of a specific number of glucose monomers, would be harder to hydrolyse compared to a branched polymer consisting of the exact same number of glucose monomers. I do not understand why.
I do believe I understand that a single glucose molecule/monomer (shown below) has more polar groups (5 C—OH groups and 1 C—O bond for a single glucose molecule, versus 4 C—OH groups and 1.5 C—O bonds in a monomer of glucose in a maltose molecule, ultimately, there are 6 versus 5.5 oxygen atoms per glucose monomer), compared to a glucose monomer in a polysaccharide (such as maltose shown below), linked to another glucose monomer via a dehydration reaction (polar H2O molecule is lost, contributed by two glucose monomers).
Glucose:

Maltose:

However, the first quote stated branching increases water solubility. Would a branched glucose polymer not be as polar (soluble) as an unbranched glucose polymer of the same size? If we had two polymers, both made of three glucose molecules and one polymer had two 1-4 glyosidic linkages, like this:

Would it not have the same solubility as a polymer consisting of three glucose monomers, with one 1-4 and 1-6 glyosidic linkage? Similar to this:

Since both molecules would have the same number of oxygen atoms (3 glucose molecules - 2 H2O molecules) and hydroxyl groups and thus, both molecules would have the same polarity. This would apply to larger molecules of the same size, both with more branches and longer branches. I would expect unbranched and branched glucose polymers of the same size, to have the same number of polar groups.
I assume a more soluble molecule would attract water molecules 'better' thus, making its own hydrolysis 'easier'.
Finally, keep in mind I have only an introductory understanding of chemistry, as I am a 2nd year Biology student at Univeristy.

To your first question regarding the branching: visualize (or draw) a linear polymer of glucose, and compare it with a highly branched polymer. The purpose of glycogen hydrolysis is to get glucose molecules. Imagine you have a lot of enzyme molecules. How many glucose molecules can be produced "at once" from a linear polymer, and how many from a branched one? In a linear polymer, you have only two ends, and only two molecules can be produced at once. However, in a branched polymer, you have a lot of "free ends" so many glucose molecules can be produced in the same time.

I guess this question comes from not learning that enzymes need 'free ends' to speed up glucose hydrolysis.

Well, not all of them. Glycoside hydrolases can be classified as either endohydrolases or exohydrolases.

Ok, then I would assume an endohydrolase would be able to hydrolyse branched and unbranched sugar polymers of the same size, at the same rate.

Well, probably not. Imagine a large, very branched polymer. It is hard for an enzyme to access monomers that are located near the center of branching, because they are buried deeply in the branched structure.

So an endohydrolase would have more trouble hydrolysing glycogen for example, compared to an unbranched sugar polymer, because of the 1,6 glyosidic bonds, I would assume?

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How does burning natural gas produce nitrogen oxides?
I understand that the byproducts of burning methane to be carbon dioxide and water, and in low oxygen situations, carbon monoxide.
$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$
$\ce{2 CH4 + O2 -> 2 CO + 4 H2}$
Where does nitrogen come into this?  How does burning methane produce nitrogen oxides?

$\ce{N2 + O2 <<=>[high T] 2 NO}$

You are assuming the composition of natural gas only includes methane, although in reality it includes a variety of other gases like nitrogen in smaller quantities.

@Sam202 .... and in air used for burning in big quantities.

@Poutnik so just High temps?  Does an electric stove also produce NO?  Just trying to understand the politics going on in NY right now.

High as very high. Like in combustion engines or gas burners used for melting glass, to get the picture like  t > 1500 deg C. At low T like in stoves is the reaction unfavoured and mainly kinetically frozen.

@rtaft, An electric stove would be inefficient, both because temp is too low and surface area of burners is high, decomposing any NOx. An electric arc, though, is effective in producing NOx's, ozone and other (likely undesirable) gases. Only air is needed as input. See https://pubmed.ncbi.nlm.nih.gov/32119426/, for example.

Incomplete combustion of methane does not form H2. N2 and O2 react endothermically the reaction is favored at High temperature[combustion] and energy such as lightning in the atmosphere.  The original questioner seems reluctant to recognize that the burning of natural gas is happening at a high temperature in air that consists of 80 percent nitrogen gas.  Gas stoves and ovens in a non-ventilated room add small amounts of NOx and SO2 [from the odorant], some CO and a tremendous amount of CO2 [almost 4000 ppm]. It is a fossil fuel tho burning for energy is better than flaring or releasing it.

@Sam202 While natural gas direct from wells often contains small quantities of nitrogen and many other things, the natural gas that gets delivered rarely does as the cleaning process strips most of the minor components out leaving fairly pure methane (and sometimes ethane depending on the economics of separation).

Because air contains nitrogen
One of the problems with looking at a real world chemical reaction by looking at the reaction formula for the major reaction going on is that you miss the possibility of side reactions.
In this case it looks like you are reacting methane with pure oxygen. But very few things that burn methane use pure oxygen. Most devices that burn methane  (bunsen burners, gas cookers, boilers, some vehicles and plenty of other devices) use air not pure oxygen and air is ~80% nitrogen.
Nitrogen isn't that reactive under most normal circumstances. But it is a little reactive and can form nitrogen oxides if it gets hot enough. In low temperature flames the level is usually fairly minimal (like the flames in a gas cooker or a bunsen burner). In hotter flames the yield of nitrogen oxides (NOx is an often used all-encompassing term) can be significant. For example internal combustion engines have much hotter combustion than simple flames and tend to produce worrying amounts (though there are not many vehicles that use methane, but typical petrol (gasoline) engines produce a lot of NOx in the same way as a methane powered engine would). This is a big reason why most vehicles have to have catalytic converters.
Basic point: flames are hot and can encourage side reactions. If you are using air, those reactions will often involve nitrogen.

Here is a starting point for further research. If your fuel is coal or oil, nitrogen-containing compounds in the fuel contribute substantially to the NOx load. Natural gas also contains dinitrogen, so even burning it in pure oxygen would produce NOx. The amount of NOx depends on various parameters of combustion, not just the temperature.

@Karsten Raw natural gas sometimes contains small amounts of nitrogen but CO2 is usually higher. But the gas we use does not, especially if it has been liquefied fro transport. The production process usually leaves fairly pure methane. So the nitrogen content is really not relevant not relevant to this question.


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Why is the first electron affinity exothermic and successive ones endothermic?
I am having trouble wrapping my head around electron affinities. And the textbook explanations aren't very helpful.
So, the textbook says that the 1st electron affinity is generally exothermic. The reason given is that the electron added is strongly attracted by the effective nuclear charge. In contrast, the successive electron affinities are always endothermic. This is due to the additional energy needed to overcome electrostatic repulsion.
This leaves me wondering:

Why is so much energy released when an electron is added to a neutral atom? I understand that the more energy released, the more stable the product is, but why does it become more stable? And why is this seemingly independent of electronic configuration?

Why isn't the attraction between the electron and the nucleus a factor in the subsequent electron affinities? In the 1st electron affinity, the atom is neutral, and yet there is still attraction between the electron and the nucleus. In successive electron affinities, the ion is negatively charged, but won't there still be attraction?

Isn't there also repulsion to overcome in the 1st electron affinity? Sure, the atom is neutral, but won't the added electron be repelled by the valence shell electrons or something?

Thanks.

Forget atoms, they are complicated. Any neutral thing is attracted to any charge.

Why is so much energy released? It is not so much. It is less than atom ionisation energies.

@Ivan Neretin Really? What about a beam of neutrons in an electric field? It doesn't deflect because its charge is neutral.

I said "attracted to charge", not "affected by field". The latter is not true; if the field is uniform, then of course a neutral particle would not be affected by it. On the other hand, the former is true, because the field of a point charge is non-uniform, and the polarizability creates some net force. Sure enough, it is much smaller than the attraction between two opposite charges.

@Ivan Neretin. I see your point. But still, I don't see how this answers the question. Why is the first electron affinity exothermic? Why are -1 ions more stable than neutral atoms (since that is what's implied)?

It makes no sense to say that ions are more stable (or less stable) than neutral atoms. See, atoms can't turn into ions anyway. It is true, though, that an anion is more stable than a system of a neutral atom and a free electron far away.

Electron affinities are generally smaller than ionization energies.
$\pu{IE_{\ce{Na}} = 5.14 eV}$ but $\pu{EA_{\ce{Cl}} = 3.61 eV}$
If the electron clouds had been fully spherical symmetric and fully between the external electron and the nucleus, the effective nuclear charge would be for that electron zero.
But as individual orbitals, except $\mathrm{s}$ ones, are not spherically symmetric and as part of atomic electron probability density is behind the captured electron, the effective atom charge is slightly positive, so the electron can be bound. That is true for most of elements, except noble gases and some elements with full or half full orbital groups.

When an atom becomes an anion, it becomes a non-point-like negative charge. The same sign charges mutually repulse each other sooner than partial attraction due asymmetry or penetration of electron density decreases little the repulsion.

Yes, there is repulsion, but weaker than not fully screened off nucleus attraction, so there is still net attraction. Additionally, electrons partially repulse each other toward the nucleus.

Interesting, from the table, that the inert gases and Be, N, Mg and Mn, Zn, Cd, Yb, Hg, Pu and perhaps a few more, have negative heats of electron affinity. Those atoms won't hold onto a free electron! They would rather be neutral than negative.

@JamesGaidis If your review their positions in the periodic table, their electronic configuration involve full or half occupation of orbital groups, making from them kind of semi-noble gases in the electron affinity aspect.


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How can fragmentation result in a higher mass than parent ion?
I've collected MS/MS data on a (highly modified) peptide. While most of the data is fine, I've noticed that I have some daughter ions which are higher in mass than the parent ion (in this case, an increase from 879 m/z to 895 m/z).
My initial guess was that perhaps the parent ion was actually doubly charged, but the isotope increment by 1 m/z - so this means it's definitely singly charged, right? So how could CID fragmentation result in a higher mass?
For context, this is a Bruker "HCT" instrument, which uses an ion trap, and the data was collected from a direct flow injection in negative-mode ESI using methanol with no additives.
I was under the impression that concentrations in gas-phase are far too low for inter-molecular chemistry; in this case, however, the 895 peak is taller than all the rest!

Than the compound in there has higher molecular mass.

@Mithoron, this is not the typical case. Adducts are more common in ESI-MS. These adduct peaks appear at a higher mass than the original compound. Mohger, is your compound highly purified or not? m/z of 895 in positive mode is a contaminant from surfactants but you mention negative ion mode.

A very good reference for contaminants in MS is Interferences and contaminants encountered in modern mass spectrometry, Bernd O. Keller et al. in Analytica Chimica Acta (2008).

@AChem my molecule is fairly purified, but it's a shared instrument and there could be someone else's contamination. The surfactant idea is interesting: maybe there's a multiply-charged contaminant ion that's isobaric with the parent ion? Thanks for the reference, their spreadsheet looks incredibly useful :)

@mohger, this is why blanks in analytical chemistry are extremely critical. You can easily eliminate cross contamination "theory" if you run a solvent blank.


text:
What's the pH of a solution knowing only pKa, mass and volume? (no molecular mass known)
Is it possible to calculate the pH of a given solution knowing only the pKa (but not the structure of the acid, i. e. its molecular mass), its mass and the volume of the solution? E. g. What is the pH of a 400 ml solution with 9 g of mandelic acid (pKa = 3.39)? Suppose you don't know its structure or its molecular mass and you can't look it up online or someplace else. There's an online pH calculator that can do it (https://www.omnicalculator.com/chemistry/ph) where I only specified the pKa of the acid, the grams of acid in the solution and its volume but I don't know how the algorithm behind it works and I couldn't find anything else online.

Once the name is available, calculating molecular weight is trivial. Look it up. There is no mysterious calculation.

As I stated in the question body you cannot look it up online or someplace else. I didn't write the name of the acid in the website I used, only the pKa of the acid. I was wondering if there is a way of calculating the pH of the solution in the example only with the given data (pKa = 3.39, 400 ml, 9 g) and no extra knowledge.

Can you post a screenshot, where you are entering this information? Just putting in pKa will not lead to the correct pH. You do need molar concentration.

It seems the "custom" acid option in the calculator you linked has a molar mass default set to match $\ce{HCl's}$, since reverse-working the problem (using pH, Ka, V, m) leads to a molar mass of approximately $\pu{36.45g/mol}$. I don't believe it is possible to calculate pH with m, V, Ka alone.

It cannot be done. With the same pKa, acid A with the molar mass 1/2 of molar mass of acid B would have pH at the same mass concentration lower by 0.15.

Thank you very much. I thought it was impossible to solve this problem without adding extra data (you end up with a single equation with two variables), but the link convinced me otherwise. Being relatively new to chemistry I wasn't so sure and I didn't think of "reverse-engineering" the algorithm to see if it was assuming something wrong.

Unfortunately knowing only the pKa, the mass of the acid in the solution and the volume of the entire aqueous mixture it's not enough, one will always end up with an equation with two variables, which is impossible to solve. E. g. Assuming acid A and B have the same pKa, acid A with the molar mass 1/2 of molar mass of acid B would have pH at the same mass concentration lower by 0.15.
Using the "custom" option the online pH calculator assumes a molar mass of 36.45 g/mol (identical to HCl's).
I would like to thank everybody who took their time to read and comment this question.


text:
NH protons not appearing in H NMR spectrum of 3,4-dihydropyrimidone derivative (solvent: CDCl3)
My synthesized compound is 3,4-dihydropyrimidone derivative, which has 2 nitrogen in the ring. The $\ce{NH}$ protons are not appearing in the H NMR spectrum and the solvent used was deuterated chloroform $(\ce{CDCl3})$. I understand that the peak for $\ce{NH}$ proton tends to be broad, is there a possibility that it merges/overlaps with the neighboring peaks?

Likely very broad with the tautomrism. Have you done a D2O shake and rerecorded the spectrum?

Hi Waylander. Nope, I have not done a D2O shake as D2O is unfortunately not available in my undergrad lab. Thanks for the suggestion anyway!

ask your lab instructors for some.

Can you put a snap shot of the questioned NMR? So, we can see if water effecting the spectrum as Buttonwood suggested. My initial feeling was the same but I thought your sample is wet after workup.

Hi @MathewMahindaratne I have put up the NMR spectrum in my post

As indicated by @Waylander, your compound may undergo keto-enol tautomerism.  Beside a sample which may contain varying concentrations of three compounds (top row), the products of an intentional proton-deuterium exchange equally may be recognizable (bottom row), too:

Especially in humid climate/summer season, bottles of $\ce{CDCl3}$ catch water with each opening of the vial, the $\ce{H2O}$ peak in the NMR spectrum becomes more prominent.  Note, Wikipedia's property box reports for $\pu{20 ^\circ{}C}$ a solubility of $\pu{8.7 g}$ of chloroform per litre of water.  An additional potential cause is storage of NMR solvents in a fridge; if you lift the lid of the still cold bottles too early and draw the liquid with a Pasteur pipette (before the bottle warmed to ambient temperature), water merely condensates on them.
Possible solutions:

with a dry sample of your compound, seek a bottle of $\ce{CDCl3}$ which wasn't used yet
with a dry sample of your compound, seek an ampoule of $\ce{CDCl3}$ with enough solvent for one sample only
with a dry sample of your compound, seek access to a solvent which does not promote the proton-deuterium exchange.  Perdeuterated dimethylsulfoxide, DMSO-$d_6$ is an example.

The sequence represents an increase in financial investment per sample for the deuterated solvent.  On the other hand, the substitution of $\ce{CDCl3}$ by DMSO-$d_6$ may save time to record and subsequently interpret the NMR spectra.

Also worth adding activated molecular sieves to your bottles of CDCl3

@Waylander Molar sieves are an option if you can trust their origin & purity. If they are the ones shuttled around the lab and back to the bin for activation, not only to dry the solvents but got in contact with Bronsted acids (e.g., heterogeneous etherification) this may fall flat. Filtration over active basic Al2O3 for chromatography in a Pasteur pipette is an other option some senior chemists acquired the dexterity to set up (though to check the 25mg reaction, the plug below this column better is made of clean glass wool, not the cotton from chemists/drugstores like CVS & Walgreens).

Hi @Buttonwood , however the spectrum does not show any D2O peaks which can cause the proton deuterium exchange as you suggested. Is it possible that D in CDCl3 exchanges with the H of NH?

Hi @Waylander , however the spectrum does not show any D2O peaks which can cause the proton deuterium exchange as Buttonwood suggested. Is it possible that D in CDCl3 exchanges with the H of NH?

The D in CDCl3 is not exchangeable under normal conditions of 1H NMR

My synthesized compound is 3,4-dihydropyrimidone, which has 2 nitrogen in the ring. The $\ce{NH}$ protons are not appearing in the H NMR spectrum and the solvent used was deuterated chloroform $(\ce{CDCl3})$. I understand that the peak for $\ce{NH}$ proton tends to be broad, is there a possibility that it merges/overlaps with the neighboring peaks?

I know Buttonwood's answer is good enough to explain why you didn't see your $\ce{NH}$ peaks. I like first two of his solutions. However, as suggested in third solution, I wasn't a fan of using $\ce{DMSO}$-$d_6$ if I have a choice between $\ce{CDCl3}$ and $\ce{DMSO}$-$d_6$ because $\ce{DMSO}$-$d_6$ is highly hydroscopic than that of $\ce{CDCl3}$ (in addition to the financial setback).
When I analyzed your $\mathrm{^1H}$-NMR you have just posted, I have seen massive water peak around $\pu{1.56 ppm}$. If you have integrated that peak, you'd see the integration value is more than 3-protons of the real compound. That means your compound is really wet when you take the NMR. The following diagram shows ratios of water peak to TMS and residual $\ce{CHCl3}$ peaks in a typical $\ce{CDCl3}$ (with 99.8% $\ce{D}$ and 0.05% TMS) solvent bottle (Ref.1):

Assuming you have used similar $\ce{CDCl3}$ solvent bottle, your water peak is way bigger than expected. That means, you have either didn't dry your product (assuming this is crude product) or your $\ce{CDCl3}$ solvent bottle is really old absorbing way over water by time (as suggested by Buttonwood). Either way, this water is capable of exchanging acidic protons (similar to $\ce{D2O}$ shake) in the product so that equilibrium is moving water direction since presence of water is too much (Le Chatelier principle). Thus, you see only water molecules instead of two $\ce{N-H}$ peaks.
Other than that, your compound is pretty pure. If it is a solid, one more recrystallization would do the trick.
Note: To find out your sample is wet or your $\ce{CDCl3}$ solvent bottle is wet, just run a NMR of the solvent. If water peak is still more than TMS peak, it is really wet.
Late Edit: For my surprise, most of NMR of the derivatives of sought compound  have been done in $\ce{DMSO}$-$d_6$ solvent. Following is $\mathrm{^1H}$-NMR of your exact compound in $\ce{DMSO}$-$d_6$ (Ref.2). You may see both $\ce{N-H}$ peaks clearly. So, I think you should run another spectrum in $\ce{DMSO}$-$d_6$ to confirm it:

References:

Alexander L. Guzman and Thomas R. Hoye, "TMS is Superior to Residual $\ce{C}H\ce{Cl3}$ for Use as the Internal Reference for Routine $\mathrm{^1H}$-NMR Spectra Recorded in $\ce{CDCl3}$," J. Org. Chem. 2022, 87(2), 905–909 (DOI: https://doi.org/10.1021/acs.joc.1c02590).
Mohamed Ould M’hamed, Abdulrahman G. Alshammari, and O. M. Lemine, "Green High-Yielding One-Pot Approach to Biginelli Reaction under Catalyst-Free and Solvent-Free Ball Milling Conditions," Appl. Sci. 2016, 6(12), 431 (6 pages) (DOI:  https://doi.org/10.3390/app6120431).


text:
Passing protons through cation-permeable membranes
Protons are attached to a water molecule (making hydronium) in acidic solutions. If a container was split through the middle with a cation-permeable but anion impermeable membrane and, say, HCl was added to one side, what would happen?
I'd assume because of the concentration gradient, osmosis will cause hydronium ions to move to the less concentrated side until equilibrium is reached.
But that would also make charge very uneven on both sides, where one side is highly negatively charged and the other side is very positively charged. So, I am not sure osmosis alone could even lead to this state.
Considering the ions on either side are either electron deficient or have full shells, and have the same charge, I would not expect any gas to come out of solution.
If a negative electric current was applied to one side of the container, and a positive on the other, would that produce gas more efficiently than plain electrolysis? If it depends, then under what conditions?
Or am I wrong, and what will happen to the ions from the acid?

Non-zero electrostatic gradient would very quickly compensate the cation gradient.

Side note: There such membranes proton exchange membranes used for proton exchange fuel cells.

Without current you would establish an equilibrium situation in which a charge imbalance is created across the membrane interface that counteracts (via migration) diffusion of your protons through the membrane. This is what is described by the Donnan equilibrium.
You can pass current in such a system as well, you will find however that there's a potential drop (a resistance) across this membrane. This resistance is exactly (or more) what you would thermodynamically gain by having a higher proton concentration at one side. You sadly cannot cheat thermodynamics.
But will it work better? Depends.
The rate of the different electrochemical reactions is not only dependent on the thermodynamic driving force but also on other factors such as the kinetics and mass transport. If the rate is highly dependent on the proton concentration you might benefit from this.
People are working on such systems for water electrolysis, in which at the hydrogen side they want an acidic environment and at the anode an alkaline environment. These membranes are called bipolar membranes.


text:
Why are carbocations electrophiles
My lecturer taught that carbocations are electron-short, and are therefore positively charged. He then went on to say they are electrophiles, that is they accept electrons.
However, he taught last semester that electrons are donated for a substance to become a cation.
And this is very confusing.

It seems that you misunderstood your lecturer. If a compounds looses electrons it becomes a cation.

If you get conflicting information, it is mostly easy to verify it on independent offline on online sources, what should be preferred to asking. The latter leads to flooding internet with redundant questions and answers.

Your professor is right in his explanations. Consider the following example, you start with (CH${_3}$)$_3$CCl (tert-butylchloride) and then try doing a transformation where you want the Cl$^-$ to be replaced by some other group. It is known that if we try to remove Cl from a molecule, it would leave as Cl$^-$, meaning that the tert-butyl group that is now left behind is short of two electrons (one from the tert-butyl part, and one from chloride), in comparison to tert-butylchloride, which means that it will now have a positive charge, becoming tert-butyl cation.

Any cation is missing electrons.  Think of regular metal ions.  A sodium atom has 11 protons and 11 electrons.  The sodium ion ($\ce{Na+}$) has 11 protons and 10 electrons.  The ion has fewer electrons than protons, resulting in a positive charge.
The same is true for carbon.  Carbocations are electron-deficient, which is why they have a positive charge.
Some cations are stable (like $\ce{Na+}) and are not electrophilic.  However, carbocations are generally not stable, and are going to react with electron-rich species.  They are electrophiles.
Your confusion might be in the wording.  In order for a carbon atom to become a carbocation, it must donate electrons.  Once it is electron-deficient, it is now an electrophile.


text:
What do consecutive letters A and B in the 4th column of PDB file signify?
What do the consecutive A and B in the 4th column identify in a PDB file downloaded from rcsb.org?
The molecule name is 3CZ3 (consider the RNA molecule only). On visualizing it, A and B denote the two strands of a dsRNA molecule.  I always thought the 6th column (the one having E) is where the chain ID is defined.
ATOM      3  P  A  C E   1       4.436  40.379 -15.386  0.50129.95           P
ATOM      4  P  B  C E   1      15.290  -0.357  16.460  0.50123.26           P
ATOM      5  O1PA  C E   1       2.978  40.404 -15.652  0.50130.05           O
ATOM      6  O1PB  C E   1      14.336  -0.918  15.481  0.50123.33           O
ATOM      7  O2PA  C E   1       4.862  41.771 -15.118  0.50129.91           O
ATOM      8  O2PB  C E   1      16.239  -1.432  16.825  0.50123.24           O
ATOM      9  O5'A  C E   1       5.166  39.855 -16.710  0.50129.89           O
ATOM     10  O5'B  C E   1      14.451   0.058  17.755  0.50123.38           O
ATOM     11  C5'A  C E   1       5.056  40.545 -17.954  0.50129.97           C
ATOM     12  C5'B  C E   1      14.669  -0.584  19.010  0.50123.36           C
ATOM     13  C4'A  C E   1       6.424  40.853 -18.544  0.50130.09           C
ATOM     14  C4'B  C E   1      15.719   0.121  19.855  0.50123.10           C
ATOM     15  O4'A  C E   1       6.820  42.211 -18.203  0.50130.44           O
ATOM     16  O4'B  C E   1      17.036  -0.442  19.636  0.50123.30           O
ATOM     17  C3'A  C E   1       7.593  40.026 -18.018  0.50129.97           C
ATOM     18  C3'B  C E   1      15.991   1.586  19.561  0.50122.75           C
ATOM     19  O3'A  C E   1       7.657  38.724 -18.580  0.50129.64           O
ATOM     20  O3'B  C E   1      14.921   2.417  19.964  0.50121.76           O
ATOM     21  C2'A  C E   1       8.783  40.893 -18.422  0.50130.06           C
ATOM     22  C2'B  C E   1      17.243   1.792  20.405  0.50123.19           C
ATOM     23  O2'A  C E   1       9.124  40.767 -19.792  0.50129.87           O
ATOM     24  O2'B  C E   1      16.985   1.905  21.792  0.50123.34           O
ATOM     25  C1'A  C E   1       8.240  42.290 -18.111  0.50130.23           C
ATOM     26  C1'B  C E   1      17.990   0.488  20.131  0.50123.67           C
ATOM     27  N1 A  C E   1       8.694  42.824 -16.759  0.50130.14           N
ATOM     28  N1 B  C E   1      19.160   0.657  19.188  0.50124.11           N
ATOM     29  C2 A  C E   1       9.911  43.528 -16.654  0.50130.12           C
ATOM     30  C2 B  C E   1      20.401   1.113  19.686  0.50124.21           C
ATOM     31  O2 A  C E   1      10.611  43.722 -17.658  0.50130.10           O
ATOM     32  O2 B  C E   1      20.537   1.369  20.893  0.50124.22           O
ATOM     33  N3 A  C E   1      10.300  43.991 -15.435  0.50130.18           N
ATOM     34  N3 B  C E   1      21.445   1.262  18.825  0.50124.14           N
ATOM     35  C4 A  C E   1       9.545  43.780 -14.352  0.50130.19           C
ATOM     36  C4 B  C E   1      21.293   0.984  17.526  0.50124.09           C
ATOM     37  N4 A  C E   1       9.982  44.261 -13.181  0.50130.17           N
ATOM     38  N4 B  C E   1      22.357   1.153  16.733  0.50124.08           N
ATOM     39  C5 A  C E   1       8.309  43.068 -14.428  0.50130.11           C
ATOM     40  C5 B  C E   1      20.047   0.524  16.993  0.50124.23           C
ATOM     41  C6 A  C E   1       7.934  42.617 -15.634  0.50130.15           C
ATOM     42  C6 B  C E   1      19.022   0.378  17.850  0.50124.31           C
ATOM     43  P  A  G E   2       8.321  37.573 -17.683  0.50129.52           P
ATOM     44  P  B  G E   2      14.673   3.755  19.127  0.50120.94           P
ATOM     45  O1PA  G E   2       8.217  36.290 -18.413  0.50129.58           O

Your counting of columns possibly is influenced by awk* counting whitespace.  But the .pdb file format's count of columns takes columns as in characters on a punch chard.  In this perspective, the atom block is organized with 80 characters per line as fixed format (similar to the time when Fortran still was FORTRAN.)  Similar to typewriters, with an added enumeration on top, this looks as the following
         1         2         3         4         5         6         7         8
12345678901234567890123456789012345678901234567890123456789012345678901234567890
ATOM     32  N  AARG A  -3      11.281  86.699  94.383  0.50 35.88           N
ATOM     33  N  BARG A  -3      11.296  86.721  94.521  0.50 35.60           N
ATOM     34  CA AARG A  -3      12.353  85.696  94.456  0.50 36.67           C

To quote wwPDB details further:
COLUMNS        DATA  TYPE    FIELD        DEFINITION
-------------------------------------------------------------------------------------
 1 -  6        Record name   "ATOM  "
 7 - 11        Integer       serial       Atom  serial number.
13 - 16        Atom          name         Atom name.
17             Character     altLoc       Alternate location indicator.
18 - 20        Residue name  resName      Residue name.
22             Character     chainID      Chain identifier.
23 - 26        Integer       resSeq       Residue sequence number.
27             AChar         iCode        Code for insertion of residues.
31 - 38        Real(8.3)     x            Orthogonal coordinates for X in Angstroms.
39 - 46        Real(8.3)     y            Orthogonal coordinates for Y in Angstroms.
47 - 54        Real(8.3)     z            Orthogonal coordinates for Z in Angstroms.
55 - 60        Real(6.2)     occupancy    Occupancy.
61 - 66        Real(6.2)     tempFactor   Temperature  factor.
77 - 78        LString(2)    element      Element symbol, right-justified.
79 - 80        LString(2)    charge       Charge  on the atom.

So, let's overlay your file with the definition:
         1         2         3         4         5         6         7         8
12345678901234567890123456789012345678901234567890123456789012345678901234567890

ATOM      3  P  A  C E   1       4.436  40.379 -15.386  0.50129.95           P

This brings A you describe to position 17, altLoc.  And E (on position 22) to chainID, in line with your current use.
The two snippets copy the documentation Coordinate Section of wwPDB's documentation Atomic Coordinate Entry Format Version 3.3.  The root of this document is available open access as Protein Data Bank Contents Guide: Atomic Coordinate Entry Format Description, as version 3.3 (last revision by 2022-08-29).
* One noteworthy extension to awk to work with .pdb files is bioawk.  There might be some functional overlap with bioperl, and  biopython.

On visualizing it, A and B denote the two strands of a dsRNA molecule.

No, A and B denote two alternate conformations. For example, the dsRNA made of chain G and H almost has two-fold symmetry, and the protein binding to it has two-fold symmetry, so the RNA binds in either orientation. The occupancy for all the RNA atoms is 0.50 in each orientation (partially obscured in the pdb text file by running into the high B-factors in the next column).
The 3D viewers I tried ignore the second orientation when using a cartoon representation. I had to switch to ball-and-stick, and then I was able to visualize the second orientation in Jmol separately (with the command "display %B"). Alternatively, you can use the cartoon representation and use the command "configuration 2" to switch to the B-conformation.

I always thought the 6th column (the one having E) is where the chain ID is defined.

Yes, that's correct. The alternate conformation A of chain G makes a duplex with the alternate conformation A of chain H, and the alternate conformation B of chain G makes a duplex with the alternate conformation B of chain H.
Non-unit occupancies are a murky area in macromolecular crystallography, and software is often not fully capable of dealing with it.

You aren't kidding about "software is often not fully capable of dealing with it." I only heard about this in Spring 2022, so for ~20 years, Open Babel and Avogadro didn't know it. Avogadro2 should now add alternate conformations correctly in the latest releases.


text:
How heating of copper sulphate pentahydrate (blue vitriol) can be explained?

The chemical formula of copper sulphate pentahydrate, $\ce{CuSO4\cdot 5H2O}$ does not give information about its true structure because it's only an empirical formula.
It forms a network solid. Two units of sulphate ions are bonded with central metal ion $\ce{Cu(II)}$. But each $\ce{SO4^{2-}}$ ions are bonded to two $\ce{Cu^2+}$. Hence, the average number of sulphate ion per cupric ion is one. Similarly, the average number of water molecules per $\ce{CuSO4}$ unit is five. Out of which, one molecule is H-bonded.
Upon heating, first it loses two molecules of water then two more molecules of water and finally the remaining one in following steps:$^1$
$\underset{\underset{\text{(Crystal)}}{\text{Blue}}}{\ce{CuSO4\cdot 5H2O (s)}}\ce{\color{red}{->[63 ^\circ C][-2H2O]}}\underset{\text{Light blue}}{\ce{CuSO4\cdot 3H2O (s)}}\ce{\color{red}{->[109 ^\circ C][-2H2O]}}\underset{\text{Very light blue}}{\ce{CuSO4\cdot H2O (s)}}\ce{\color{red}{->[200 ^\circ C][-H2O]}}\underset{\underset{\text{(amorphous)}}{\text{Grey-white}}}{\ce{CuSO4 (s)}}$
1- Why removal of water molecules following this pattern: 2, 2, 1? Every hydrated salt follows a pattern, e.g., sodium carbonate decahydrate loses 9 molecules of water at once to form monohydrate.
2- In $\ce{CuSO4\cdot H2O}$, the water molecule must be hydrogen bonded but why the coordinated bonded water molecules are breaking bonds prior to H-bonded water? Coordinate bonds are much stronger than H-bonds even if we consider that one water molecule is making multiple H-bonds.
3- What can we say about the structure of $\ce{CuSO4\cdot 3H2O}$?
Reference

Page 766, Inorganic Chemistry (4E) By Catherine Housecroft, Pearson- 2012, ISBN: 9780273742784

https://chemistry.stackexchange.com/questions/71980/why-is-anhydrous-copperii-sulfate-white-while-the-pentahydrate-is-blue-even-t

https://chemistry.stackexchange.com/questions/110728/hydrate-decomposition-equation-steps

That's pretty much a duplicate of the second one.

Does this answer your question? Hydrate decomposition equation steps

No, as there are few more queries mentioned in this question and the answer ended with "In reality many assumptions suitable for estimation of melting points of molecular crystals, e.g. regarding the number of bonds and their types don't work well for ionic crystals ..."

Note that IPUAC's preferred spelling is "sulfate". IUPAC prohibits use of dative bonds (arrows) to represent coordination. This is not "polymeric structure", rather a network solid. The crystal structure of $\ce{CuSO4 · 3H2O}$ (bonattite) is freely available at multiple sources, e.g. see https://www.mindat.org/min-718.html.

@andselisk So in bonattite, the coordination no. of Cu(II) is 4.


text:
Density of dissolved and non-dissolved salts
I am doing a calculation, which needs to be 100% correct. My question is as follows; Does the density of dissolved salt differentiate from the density of non-dissolved salts?
My specific problem is a solution of 75% potassium formate in water. When the salt is dissolved, the salt molecule is broken down to K+ and COOH-. This, at least in my mind have to affect the density of the dissolved ions from the density of crystalline salt (non-dissolved). This in turn, would mean that a calculation of the exact density can not be calculated from the density of pure non-dissolved potassium formate (@ 1.91kg/l).
Is there any table I can look up the density of the ions directly, or does anyone have any tips to point me in the right direction? Are this difference negligible? Or are my hypothesis just plain wrong? All answers are appreciated.

Would [chemistry.se] be a better home for this question?

https://www.sciencedirect.com/science/article/pii/S0017931018353067 has the data and fits you seem to want.

Thanks a lot for the answers and the reference. Yeah, I wrote a little quick there, we are using formic acid in the production for the formate :) The 20-degree reference is perfect, specifically, I'm looking for the values at 22C. My wording while looking for an answer may be the reason I was struggling, it's not easy having just a bachelor. Anyways, thanks a lot for the quick reply. Now I know where to ask when I'm stuck

This is not quite an answer but an edit to the question. I can see you haven't registered to this site. Please register and edit your question. This will merge your account.

Poutnik has nicely explained in his answer what you have to do to get reliable data. Unfortunately, tabulated density data of potassium formate is hard to found. However, I have found two reliable datasets (Ref. 1 & 2) with very agreeable data sets at two extremes (densities of 5-25% solutions in Ref.1 and that of 50-78% solutions in Ref.2, which was mentioned by Jon Custer in his comment):

Now, since the graph has good agreement of data $(R^2 = 0.9971)$, and using the straightline equation, you can find any density of given solution. Note that the data is for solutions at $\pu{20 ^\circ C}$.
References:

Åke Melinder, Thermophysical Properties of Aqueous Solutions Used as Secondary Working Fluids," Ph.D. Thesis 2007, Royal Institute of Technology, Stockholm, Sweden (ISBN 978-91-7178-707-1).
Tao Wen, Meng Wang, Yi Chen, Weifeng He, and Yimo Luo, "Thermal properties study and performance investigation of potassium formate solution in a falling film dehumidifier/regenerator," International Journal of Heat and Mass Transfer 2019, 134, 131-142 (DOI: https://doi.org/10.1016/j.ijheatmasstransfer.2019.01.031).

Better than "density of dissolved salt" is using the term "density of the salt solution", which if a function of the particular solution concentration.
As there is no law of total volume conservation, the volume of final solution may differ (is usually little smaller for salts) from the sum of volumes of solvent and (bulk) salt to be dissolved.
The density of a particular salt solution cannot be calculated from basic principles. It has to be determined experimentally, or interpolated from already known tabulated solution density data.
For common salts, there are often published tabulated data of solution densities for given mass concentrations or mass percentages.
If no solution density data are available, you can use as a very rough first approximation the calculated density from the total masses and total volumes of the solvent and the (bulk) salt. BUT, it is highly recommended to measure density (or calculate it from the mass and the volume) at least for one or two solutions close to solution concentrations of your interest.
Then, you can use an interpolation function, available in numerical mathematical methods, to interpolate density data.
Look also on the related question  Why water volume went up by almost the same amount after adding salt?

There is also no salt molecule $\ce{HCOOK}$. It is just a summary formula, expressing stoichiometric ratio of ions  $\ce{K+}$ and $\ce{HCOO-}$  1:1 in a salt crystal or its solution.


text:
Identify two hydrocarbons using volume and density
Question:

A gaseous mixture with volume 16.8 dm3 contains two unsaturated neighboring hydrocarbons belonging to the same group. Density of the mixture was 14.4 times as hydrogen gas.
What could be the two hydrocarbons?
(a) $\ce{CH4}$, $\ce{C2H6}$
(b) $\ce{C2H2}$, $\ce{C3H4}$
(c) $\ce{C3H4}$, $\ce{C4H6}$
(d) $\ce{C3H6}$, $\ce{C4H9}$
(e) none of the above

I'm not really sure how to approach this question. The answer according to the mark scheme is b). What is the method to get this answer? Also, how can $\ce{C2H2}$ and $\ce{C3H4}$ be in the same group? One is an alkyne and the other a cycloalkene.

Read what you have written, the quoted task does not match your comment. $\ce{C2H2 means CH#CH}$, $\ce{C3H4 means CH3-C#CH}$, $\ce{C4H6 means CH3-CH2-C#CH}$

Some mixtures are too dense, and some aren't unsaturated.

@IvanNeretin All is allowed in incorrect answers. :-)

@Poutnik Fixed it. Oh right... I was thinking about cyclopropene for C3H4

The density of $\ce{H2}$ is such that $1$ mole, or $2$ g $\ce{H2}$ occupies a volume $22.4$ L at $0$°C at $1$ atm. Its density is $\pu{\frac{2 g}{22.4 L} = 0.0892 g/L}$.
Doing the same calculation with $\ce{C2H2}$ and $\ce{C3H4}$ gives densities equal to $\ce{1.1607}$ g/L for $\ce{C2H2}$ and $\pu{1.7857}$ g/L for $\ce{C3H4}$
The density of the gas mixture is $14.4 · 0.0892 = 1.2857$ g/L. This value is between the densities of $\ce{C2H2}$ and $\ce{C3H4}$. It means that these two gases can be mixed to give the observed density. So the choice b) can be an acceptable answer.
And indeed a mixture $80$% $\ce{C2H2}$ + $20$% $\ce{C3H4}$ has the observed density.
The solution c) contains too heavy hydrocarbons. The solution d) contains a non-existant compound ($\ce{C4H9}$)

But the question states the compounds have to be unsaturated so that answer can't be valid.

@matt_black. You are right. I edit my answer to take this remark into account.


text:
Relationship between grams of root and content in a tea
When we make a tea from a some herb, what exactly is the relation between the weight of the herb/root stepped in hot water and how many grams/milligrams are in the tea.
For context and giving a specific example:
After reading about side effects of using nutmeg, I am always checking for any spice or herb or root I use.
For a concrete example ginger I am reading that there is an upper gram limit that above which side effects appear.
I am trying to understand if this would also pertain to tea and if for example we take 20 grams of ginger root grated and put it in hot water would that mean that there are 20 grams of ginger (somehow) in the tea?
What is the relationship here?

From first principles, there is impossible to determine for specific cases relation between the dry "tee" mass,  the mass of dissolved components in infusion and the mass of active components. It requires extensive experimental work in the particular scenarios. // There is also unclear relation between amount of absorbed active components from a given mass of herb if used directly(e.g. added to food) or if used for infusion. There are too many variables there.

Like the previous answer, there isn't a way to know without performing some form of analysis. If someone else has done this sort of analysis, you could replicate the way they made their tea to have an idea of what yours might contain.
I think that there might be a deeper misunderstanding on your part.  Take ginger for example.  The ginger root is made up of hundreds (or probably thousands) of different chemical compounds. Some of these are the molecules that make up the cell walls (cellulose) and other fibres in root.  Some of these are the DNA and proteins that are part of any living thing. If the ginger is fresh, there will also be a lot of water.  So, 20 grams of ginger is not referring to 20 grams of one thing.
When you make a tea, you put the plant in hot water, and only some of the chemicals in the plant dissolve in water.  Which ones, and how much, depends on the plant, the temperature of the water and how much water vs plant material is used.
If you are concerned about potential health risks or side effects, you should look up reference material specific to teas made from those plants, not just the plants themselves.
I hope that helps.

It does help. When looking if there are any reports already about tea I found articles like this: https://senchateabar.com/blogs/blog/ginger-tea-benefits#:~:text=One%20cup%20of%20ginger%20tea,caused%20by%20bloating%20and%20cramps. here it says One cup of ginger tea contains an average of 250 milligrams of ginger but there is no reference of any lab analysis to back it up. This one https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4818021/ mentions that 1000 mg is equivalent to one teaspoon (5 g) of freshly grated ginger extract, has a reference but I can't access the full article

Do you think the second article is reliable i.e. they have followed the lab analysis you mention?

As far as I can see, the second reference is a reliable one, in that it is published in a peer-reviewed scientific journal. I don't have access to the article that they cite for the various equivalent doses of ginger, so I can't evaluate those.  The authors in the paper you refer to do point out that there are no real standards in regards to the active constituents of ginger.

Long story short, the answer is probably too complicated to answer without analysis of the specific tea.
The ingredients within the herb or root all have their own saturation levels. For example, the water will likely not be able to hold the same amount of sodium chloride when compared to sugar.
Different molecules also dissolve at different rates. Table salt is soluble in water, and so is limestone, but table salt dissolves much faster - partly due to its much higher saturation level.
One issue that makes it particularly complicated, however, is that these solutes affect each other. The chloride ions from the sodium chloride could, potentially, impact the solvation of potassium by limiting the rate at which it can be freed from its chloride ions.
And that's an oversimplification. There is actually a lot more to it than I just explained, but I hope my oversimplification can provide a little more clarity without requiring me to go too in depth.
TLDR; The most practical way to know would be to perform analysis of the tea you are curious about.

Does that mean that there is theoretically the possibility that if we have X grams of something stepped in hot water, we drink X grams of said thing when consuming the tea?

Only if X g of something is soluble in given volume of boiling water. "Soluble tea" is.


text:
What happens if an element has two different oxidation numbers on the same side of reaction?
I am taking General Chemistry 1 and I am studying redox reactions. I was looking at the following combustion reaction:
$$\ce{4CH3 + \underset{O: 0}{7 O2} \to \underset{O: -2}{4CO2} + \underset{O: -2}{6H2O}}$$
Under the reaction, I am labeling how the oxidation numbers for each oxygen atom. I noticed how the oxidation state on the left hand side of the reaction for O changes from 0 to -2 on the right hand side, even though there are now two sets of oxygen atoms in two different molecules.
What would happen if—disregarding the fact that O always has an oxidation of -2—the first molecule on the right hand side happened to have an O atom with an oxidation number of -2, but the second molecule happened to have an O with an oxidation number of -1? Is a situation like this possible in any reaction, with an element different than O? If so, how could the general change in oxidation number be described from the left hand side to the right hand side of the reaction?
EDIT:
I found a reaction that details what I'm confused about:
$$\ce{2 \underset{N: +5;O: -2}{N2O5}(g) \to 4 \underset{N: +4; O: -2}{NO2} (g) + \underset{O: 0}{O2} (g)}$$
Here, it is obvious that nitrogen is reduced from an oxidation number of $+5$ to an oxidation number of $+4$, but how would you describe how the oxidation number of oxygen changes? On one molecule, its oxidation number stays the same, but on another, the oxidation number has gone up—has O been oxidized, or has it not changed?
(I have now written this edit as a separate question).

First things first: your equation presumably is looking at the combustion of methane, and that should be $\ce{CH4}$, not $\ce{CH3}$. With that in mind, you might want to re-balance your equation.
Next, O doesn't always have an oxidation state of −2: peroxides, for example, exist, where oxygen is in an oxidation state of −1. That's actually useful to know, because I'll use it as an example later.
Lastly, to answer the actual question: sure, there are plenty of cases where you can't just describe the changes as one oxidation state to another oxidation state. I guess, you're asking specifically about something like:
$$\ce{A}^{(x)} + \ce{A}^{(y)} + \ce{O} \longrightarrow \ce{A}^{(z)}$$
where $\ce{A}$ is some element, $\ce{O}$ is some oxidising agent, parenthesised superscripts denote oxidation state, and $x < y < z$. (This isn't standard notation, so if someone wants to fix it please be my guest.) Off the top of my head, I can't think of anything that has this pattern;* but it's fairly common to have a reaction where $x < z < y$, i.e. the oxidation states get 'averaged' out. In this case you don't need another oxidising agent, and this has a special name: a comproportionation reaction, and essentially features the element $\ce{A}$ being both the oxidising and reducing agent at the same time. The Wikipedia page provides several examples.
The reverse reaction,
$$\ce{A}^{(z)} \longrightarrow \ce{A}^{(x)} + \ce{A}^{(y)}$$
with $x < z < y$ is a disproportionation reaction, the simplest being the decomposition of hydrogen peroxide:
$$\ce{2H2O2 -> 2H2O + O2}.$$

* The reason is probably because if you had a reaction like that, you could just split it into the sum of two redox reactions, one where you have the oxidation state change $x \to z$, and one with $y \to z$. In both cases this is a simple oxidation of the element $\ce{A}$.

I copied this equation from one of my chemistry lectures on redox reactions, which was covering combustion. Maybe I copied the molecule wrong, but why is it "wrong?" if I have a $\ce{CH3}$ there?

$\ce{CH3}$ does not exist as a usable substance. It is an intermediate entity that does not exist longer than $1$ microsecond. You cannot fill a bottle with $\ce{CH3}$. It will immediately dimerize and become ethane $\ce{C2H6}$.

How does it become ethane?

First it is a moot point if CH3 is a molecule or not. Think of it as the empirical formula of ethane C2H6. In the compounds CO2 and H2O oxygen has the oxidation number of -2; you seem to have understood that.
The decomposition of N2O5 requires that nitrogen gain an electron; this electron comes from oxygen the oxygen eventually is oxidized to O2 molecules, oxidation number zero [it is elemental oxygen]. The combined oxygen in NO2 is still minus 2 oxidation number, it has yet to react. NO2 can be decomposed further to N2 and O2 via a series of reactions. Write out the final equation.
The O2 molecule can react in steps to give intermediate negative oxidation numbers such as the superoxide ion, O2-, oxidation number -1/2 and the peroxide ion O2[minus 2], ox. no. minus one, but these are special compounds that are reactive.

The answer is part of the oxygen has been oxidized, as you can see from your oxidation state labels. We often see part of an element change oxidation state and the rest nit change or even change differently.
Such situations are handled by identifying whole molecules or ions as oxidizing or reducing agents. Here we may regard all of $\ce{N2O5}$ as an oxidizing agent, which when it takes electrons loses a virtual oxide ion:
$\ce{N2O5 +2e^- -> 2NO2 + O^{2-}}$
This ion is then the effective reducing agent, giving off the electrons that the nitrogen atoms in the $\ce{N2O5}$ capture.
$\ce{2O^{2-} -> O2 +4e^-}$
Thus the oxygen that is displaced from bonding with nitrogen is what gets oxidized; the oxygen that remains bound to nitrogen is redox-inert.

Sometimes acids can exist in dehydrated form.
SO3 is dehydrated H2SO4.  N2O5 is dehydrated 2 HNO3.
$\ce{N2O5 + H2O -> 2 HNO3}$
Now we see 3 Oxygens at -2, Nitrogen at +5, and Hydrogen at +1.
What this is saying is that Oxygen has a greater affinity for electrons than Nitrogen or Hydrogen.
HNO3 takes 1 electron away from Silver:
$\ce{2 (H +1 N +5 O3 -6) + Ag 0 -> Ag +1(N +5 O3 -6) + N +4 O2 -4 + H2 +2 O-2}$
and gives it to one of the NO3- moieties, because NO3- has a higher affinity for electrons than Silver.
We can write this reaction "short hand" as:
$\ce{2 HNO3 + Ag -> AgNO3 + NO2 + H2O}$.

How would you describe the way the oxidation number of Oxygen changes?

$\ce{2 N2O5 <-> 4 NO2 + O2}$
It makes perfect sense.  2 Oxygens went from -2 to 0.  4 Nitrogens went from +5 to +4.  That can only mean that, in spite of convention, N+5 (as NO3-) has a greater affinity for electrons than Oxygen.
It then becomes important to see where the equilibrium of the stated reaction lies.  Studying redox potential charts can be helpful in further understanding electron affinity beyond individual atoms.


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Phrasing for processes that "create" Entropy?
In many circumstances, in the real world, systems aren't "closed" in any meaningful sense.  So, when I burn a log in a fireplace, I can use words to artificially set up a kind of "log universe" that initially consists of nothing but log and some oxygen, and later consists of ashes and smoke and the air that that smoke is mixed with (which probably we should add back into our initial system) etc etc.  However, it seems like a lot of wordplay when I just want to indicate that the burning log is somehow causing entropy to increase in its immediate vicinity more than the un-burning log would have.  So, succinctly: "Burning a log creates entropy." or "Burning a log increases entropy." Is this the appropriate phrasing?  How do I succinctly but still correctly and understandably talk about processes that "create" entropy?

Increase is a better word. Entropy is a quantity. You cannot create a quantity out of nowhere. Think of creating mass (which implies from nothing) vs. increasing the mass of a liquid by adding more water in a pan. Increasing the mass still sounds better. I will not create meters if we stretch an elastic band and measure its new length.

@AChem so "Burning a log increases entropy." is comprehensible, and would generally elicit a "more-or-less, sure" from scientists?

I am not a nativer English speaker but "creating entropy" sounds awkward to me, although people might understand the meaning. Serious readers may not like it. Note the non-sense of order-disorder version of entropy in general chemistry and plenty of webpages, yes, there we created order and disorder, if entropy were disorder. This confusion arose because of mixing the concept of (information) entropy with Clausius' entropy. Many textbooks have corrected this notion due to one man's efforts, Dr. Frank L. Lambert. See his articles in the Journal of Chemical Education.

Entropy like temperature cannot be created but can be increased.

"Generate" or "create" are commonly encountered verbs in popular usage and in a technical context. What makes their use acceptable is that the entropy of system plus surroundings remains constant in reversible processes but increases in irreversible ones, whereas energy is always conserved and therefore constant (see for instance this post in Physics SE about the non-conservative nature of entropy). In a reversible process the entropy change of the system equals the opposite of the change in entropy of the surroundings. This is often described as entropy being "exchanged". When the entropy does not remain constant it can be said that it is "generated". Those terms should be interpreted with caution since they don't have an identical meaning to their common use. There is for instance no material substance that is being exchanged or created.
Exergonic is an adjective that describes a process carried out at constant T that results in a reduction in free energy and therefore an increase in total entropy for system plus surroundings. The term is used mainly in engineering circles and among thermodynamicists. The combustion of a log in a fireplace is an exergonic process.
Another related and  relatively new concept is "entransy", see for instance Wu and Guo (2014). As explained by  Kostic (2017):

The concept of new physical quantity, entransy G, represents functional-product of transferred or stored heat (Q or Q vh) as heat or ‘thermal charge’ and absolute temperature as intensive thermal potential (T), the way electrical energy represents functional product of transferred or stored ‘electrical charge’ and voltage as intensive electrical potential. The analogy is complete for ‘pure heat transfer’ without any work conversion or dissipation, referred here as ‘caloric processes’, where thermal heat is conserved, as is electrical charge, while both are degraded to lower respective potentials

However this is a new and potentially controversial concept used by some specialists, and I can't comment on how accepted it is.
References
(1) Wu, J.; Guo, Z. Entropy and Its Correlations with Other Related Quantities. Entropy 2014, 16 (2), 1089–1100. https://doi.org/10.3390/e16021089.
(2) Kostic, M. M. Entransy Concept and Controversies: A Critical Perspective within Elusive Thermal Landscape. International Journal of Heat and Mass Transfer 2017, 115, 340–346. https://doi.org/10.1016/j.ijheatmasstransfer.2017.07.059.


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Outer shell electrons
I'm trying to find the answers for how many free outer shell electrons that are present in Nickel chloride, I know that nickel has 2 outer shell electrons and chlorine 7.
But nickel chloride is made from one unit nickel and 2 units of chlorine, I am trying to work out how much of an increase is likely to occur in what is called Compton scatter, as this is dependant on the amount of free outer shell electrons (valence)
Quote "In Compton scattering, a photon interacts with a weakly bound outer shell electron. Instead of being completely absorbed as in the photoelectric interaction, in the Compton process the photon is deflected from its original direction and continues to exist but at lower energy"
So needing total unattached/weakly bound electrons in NiCl² please

You have mixed up all possible concepts, I am not sure what are you trying to ask.

Edited question, hopefully clearer?

It would probably be helpful if you will not as vague as to what possible concepts were all mixed up and as such a bit more informative

Nickel has indeed 2 electrons in the outer shell, but 10 valence electrons, as it is in the 10th Group of the Periodic Table. Its next lower shell has a similar energy level as its outer shell, so that there is a chemical interaction possible with all 8 electrons of that next shell.
Nickel(II)-chloride or NiCl2 is an ionic compound in which the 2 outermost electrons of the Nickel got transferred to the Chlorine atoms, leaving Nickel as well as Chlorine with a complete shell, the cheapest energetic state that can be reached in regular interaction those molecules. As the compound consists of ions which repel and attract each other, a crystal lattice is formed. In this lattice, the net force acting on each individual ion is the lowest possible.
As far as I am familiar with the Compton effect, I would say that it is not important in what kind of bond the electrons participate in, but how high the electron density in the matter is.
Unfortunately, I have no idea what you mean with increase. What part is increasing?
So in short, nickel(II) ions do have 8 outer shell electrons and chloride ions do, too.

I know studies have shown nickel chloride to have greater refraction of X-rays, than copper and I was wondering whether that the production of nickel chloride from cupronickel would actually increase the likelihood of Compton scatter, as it is my understanding this is dependant on the outer shell electrons that are unattached?


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Estimate scaled up rate of gas phase reaction
I have measured the conversion rate of a particulate-catalyzed gas phase reaction in a reactor. That is, I have a graph of how fast $\pu{1 mol}$ of catalyst converts $\pu{1 mol}$ of reactant into $\pu{1 mol}$ of product at STP. (It’s obviously not linear, assume that I have measured it over time enough that I have a very good time-product concentration graph.) I want to estimate just scaling this to a much larger space.
Say my reactor is $\pu{10 cm^3}$ and I want to scale to, say, $\pu{10 m^3}$ assuming that same concentrations of reactants and catalyst and STP).
Is it as simple as multiplying by the volume ratio?
My probably wrong intuition is that in scaling the average travel distance between reactions and (relatedly) the rate of rebound from the walls, which gives you the pressure, scales differently, akin to the way that weight does not scale with surface area.
I’m also slightly worried that a catalyzed reaction has a different non-obvious answer. I’m sure that this has a standard answer (and it’s probably just that it scales by volume ratio), but I’m checking my intuition. Thanks in advance.

I general, you can't just expect that process like that would be scaleable in any simple way, afaik.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

1 mole is like 1 kilogram (unit names) // 1 mol is like 1 kg (unit symbols). //  For options of formatting of the posted plain text, see as inspiration SE - help - formatting.

MAny things in chemical reactions do not scale with larger volumes or mass, so this isn't easy. For example, heat transfer. volume of stuff in a reactor scales with the cube of the size, but the surface area only scales with the square of the size (so heat will be far harder to manage heat transfer depends on the surface not the volume).


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Hydro distillation set up
Im trying to find out what the apparatus of this essential oil distillation set up is called (https://www.youtube.com/watch?v=yBOaFV7BSfw&t=76s), it looks like some sort of backwards clevenger apparatus with a conical drip catchment connected to condensers. Could Anyone tell me what this is called? Ive searched far and wide but cannot find anything like it, is it custom glass?
The video is different from this picture but they are both similar.
Any info would be fantastic, I have the copper pot and condensers, just need this piece.

The shape might be a bit particular, but by function, it is distillation separator, or specifically, an oil separator to perform a steam distillation.

Hot vapor (red arrow) travels up across the inner tube.  Condensed, it drops as liquid (blue arrows), some aside to the inner tube where the vapors rise, travels to the arm on the left into the cylinder.  Upon further cooling to room temperature, the components separate into inmiscible  liquids e.g., water (colorless, higher density, lower layer) and oil/non-water soluble liquid (here darker, lower density, top layer).  The more oil is captured by this receiver, the more condensed water enters back to the distillation (the tube from left bottom to centre/right, diagonally upward siphon).
There are distillation separators specifically designed to retain liquids less dense than water (e.g., the one your screen photo shows), or ones to collect liquids heavier/of higher density than water (Dean Stark receiver, same color scheme):

(a composite, reuses catalogue image by chemglass)
The distillation appears to be tilted, though this is neither necessary (the inner tube's wall for the uprising steam is higher than the tube to the left to divert the liquid to the left), nor helpful.  The vapors should rise up freely, vertically.  And instead of an Allihn condenser, using a Dimroth condenser were more efficient to retain the essential oil extracted.

When I watched your YouTube Clip, I realized the it is essentially a steam distillation setup. Again, Buttonwood has given good description for how it works. My intention here to make few suggestions:

OP's main concern was:

it looks like some sort of backwards Clevenger apparatus with a conical drip catchment connected to condensers.

That conical drip catchment is called Heckman head in distillation in micro-scale. I have attached a micro-scale distillation setup with Heckman head in right side to your distillation setup for demonstration. In your apparatus, the Heckman head was made in a larger scale (macro-scale) with horizontal modification (a tube) to move the distillated collection in the "well" (see the dip in the conical part of the Heckman Head in micro-scale setup) to a Dean-Stark type separator, which separate two immiscible liquids such as essential oil (lower density; in top layer) and water (higher density; in bottom layer) by gravity. According to the steam distillation principle, steam (high volume) takes some immiscible essential oil (only a little amount) as a solution below its boiling point. When condensed at the condenser, these two layers separate and can be collected separately (essentially by using simple distillation setup). Here, you are using large scale separation so that may need the a little complicated set up like one in display. According to the sought YouTube clip, the angle tube take excess steam down to the bottom of the Dean-Stark type separator, pushing essential oil layer upward (I didn't see why it needed, but good reason would be to reduce excess pressure).
Credit for image on the right hand side: McMaster University

Is the line art on the right hand side an illustration of the Survival Manual by Zubrick? Just because of the style of the drawing and because the more recent the edition of the helpful primer, the more consideration of small / microscale equipment by the author. And for the tube diagonal currently labeled "Excess water vapors condensed by air", I still think it is a siphon for condensed, then separated liquid water returning back to the still's round bottom flask.

@Buttonwood: I found the art online when checked for Heckman head. So, I'm sorry I didn't pay attention to where it came from. But I can try again and let you know. About  diagonal air condenser, I don't understand why it is necessary to be there. The way it has designed, it didn't save extra extract.

@Buttonwood: It was from here. But didn't have relevant lab manual available.

With the distillation up and running, condensed liquid is going to enter the wider cylinder to the left where the inmiscible phases eventually split.  The larger fraction will be water (which has to be drained automatically, as long as the distillation is running) carrying the smaller fraction of essential oil (here, the darker phase on top). This is why the diagonal siphon is needed, and why the separator (regardless if designed to retain liquids of higher, or lower density then the one of water/the carrier) have to be initially filled once with water/the carrier.


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Why does my sand dissolve in citric acid?
I soaked and washed white Darice decorative sand in distilled water.  I added food-grade citric acid to different distilled water, then I sprinkled in a few grains of the sand.   The sand bubbled and dissolved.
AFAIK, citric acid should not dissolve quartz sand.   Either

the sand is not quartz, or
the internet is wrong that citric acid cannot dissolve quartz sand.

If 2, then an equation would be appreciated.

It is (1).$\mathstrut$

It would be great if you listed the composition of this decorative sand. This will help in writing of a possible reaction equation between the "sand" and citric acid.

Bright white decorative sand is usually molten limestone or marble.

Did I say molten? It's crushed and ground, of course.

Your sand dissolves in citric acid because it is carbonate sand.
Not all sand is formed by quartz, even if it is usually implied in context of masonry or fish-keeping. There are sometimes used explicit respective expressions for "quartz sand" in some languages.
The term sand does not describe what the material is, but how the material looks like. It means solid, water insoluble "stoney" material, consisting of loose particles with the particular range of granularity $\pu{0.05 - 2 mm}$ (credit to David S).
Quoting the Wikipedia:

Sand is a granular material composed of finely divided mineral particles. Sand has various compositions but is defined by its grain size. Sand grains are smaller than gravel and coarser than silt.
Sand can also refer to a textural class of soil or soil type; i.e., a soil containing more than 85 percent sand-sized particles by mass.

There are beaches with bright, near white sand formed from shells. It is based on calcite/aragonite $\ce{CaCO3}$, soluble in acids while releasing gaseous carbon dioxide.
$$\ce{CaCO3(s) + 2 H+(aq) -> Ca^2+(aq) + H2O(l) + CO2(g)}$$
There are beaches and deserts with dark, near black sand formed from various volcanic rocks.

Sand is also used to define particle size. Gravel or Stones is above 2mm in size, Sand is 2mm - 0.05mm, Silt is 0.05mm - 0.002mm, and Clay is below 0.002mm.

Well, that is what is said in the quote and respective links. Thanks for the explicit values.


text:
Why is work done on the system in irreversible adiabatic compression greater than work done on the system in reversible adiabatic compression?
So, while reading about Quasi- Static process, I came across this question here:
Question about Quasi- static process
From the answers given there, what I understood is this: After irreversible adiabatic compression and expansion, even though the final pressure is same as initial pressure, final temperature of the gas is not the same as initial temperature. The final temperature will be more because more work has to be done by surroundings in compression than done by the gas in expansion.
My thoughts are: a) More work is done during compression because more weight is placed on the piston (two weights) during compression than expansion( only one weight is removed). So more work is transferred to the system during compression and then this work adds to system's internal energy, hence the temperature increase. Is this correct?
b) When one of the weights is reduced( for expansion), the pressure on the piston drops suddenly and the piston shoots up, so there will be temperature gradients inside the gas, which will give rise to entropy, hence making the process irreversible.
Is this correct? Thanks in advance!

Thanks for replying!But why? Why is the compression work done in irreversible adiabatic more than expansion work done in irreversible adiabatic compression? Is my reasoning in point a) correct or not?

The proper question is Why is work done on the system in irreversible adiabatic compression  > work done on the system in reversible adiabatic compression?  (and vice versa for expansion)

I think because in irreversible compression more weights have to be placed on the piston and in reversible process the pressure difference is infinitesimal.


text:
Why does cyclo heptyl carbocation is unstable?
When cyclo propane methyl carbocation under goes ring expansion it gives cyclo butyl carbocation, similarly when cyclo butyl methyl undergoes ring expansion it gives cyclo pentyl carbocation.  My question is why cyclo hexyl methyl do note under goes ring expansion?

Why would it?$,$

Let's put it this way: it doesn't expand, because it doesn't have a reason to. Smaller cycles, on the other hand, do have such a reason.

My question is why cyclohexylmethyl (carbocation?) do not undergoes ring expansion (cycloheptyl carbocation)?

Is above statement that you mean in your question? If so, it is easy to answer. If you look at ring stability, the lowest ring strain is given by cyclohexyl ring system by adopting to a chair conformation. Therefore, there is no such thing that thermodynamically more stable form will spontaneously transform to thermodynamically more unstable form (relatively speaking). According to Wikipedia, strain energy of cyclohexane is about $\pu{0.1 kcal mol-1}$ (not much different from methylcyclohexane), thus thermodynamically more stable than cycloheptane, the strain energy of which is listed as $\pu{6.2 kcal mol-1}$.
To support this phenomenon, I have solid experimental evidence: During my Ph.D. studies, I have observed Friedel-Crafts alkylation of toluene in details using tosic acid as a catalyst and alkenes including allyl and cinnamyl bromides, cycloalkenes, tosylates, etc. as the alkyl substrate (Ref.1). The frequent reactions using cycloheptene gives 100% 4-(1-methylcyclohenyl)-1-methylbenzene exclusively. No traces of 4-cycloheptyltoluene at all!
References:

Mathew P. D. Mahindaratne and Kandatege Wimalasena, "Detailed Characterization of p-Toluenesulfonic Acid Monohydrate as a Convenient, Recoverable, Safe, and Selective Catalyst for Alkylation of the Aromatic Nucleus," J. Org. Chem. 1998, 63(9), 2858–2866 (DOI: https://doi.org/10.1021/jo971832r).


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What is the IUPAC name of an ether if two alkyl groups attached to oxygen atom have functional groups attached to them?

Here, both alkyl groups attatched to oxygen atom have ketone functional group attatched to oxygen. Then, how will I write the IUPAC name?

Note, the structure is not well described as an ether. It is the (mixed) anhydride of acetic acid (left hand side of the formula), and propionic acid (right hand side).  The resources page includes a section about chemical nomenclature.

See: https://chemistry.stackexchange.com/q/59536/17368

Compounds consisting of two acyl groups bonded to the same oxygen atom are anhydrides. Symmetric anhydrides have identical acyl groups; mixed anhydrides have different acyl groups.
The corresponding subsections in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows:

P-65.7.1 Symmetric anhydrides
Symmetric anhydrides of monobasic acids, substituted or unsubstituted, are named by replacing the term ‘acid’ of an acid name by the class name ‘anhydride’.

P-65.7.2 Mixed anhydrides
Anhydrides derived from different monobasic acids are named by citing in alphabetical order the names of the two acids, substituted or unsubstituted, without the class name ‘acid’ followed by the class name ‘anhydride’ as a separate word.

Therefore, the name for the compound that is given in the question is acetic propanoic anhydride.

Nevertheless, your idea of “ketone groups attached to an ether oxygen” is not completely wrong. However, it is only used for cyclic anhydrides formed from two acid groups attached to the same parent hydride structure. Such names are called heterocyclic pseudoketones, for example

oxolane-2,5-dione (succinic anhydride)

furan-2,5-dione (maleic anhydride)

2-benzofuran-1,3-dione (phthalic anhydride)


text:
Application of Beer-Lambert Law to Absorptivity
The Beer-Lambert law gives a linear relationship between the concentration of a solute $c$ and the absorbance $A$, with absorbance defined as the logarithm of the ratio between the transmitted radiant power to the incident one:
$$A = \epsilon lc$$
$$A = -\log\left(\frac{I_t}{I_i}\right) = -\log\left(\frac{I_i - I_a}{I_i}\right) = -\log{\left(1 -\frac{I_a}{I_i}\right)} $$
In a simplified model of the effect of CO2 on the atmosphere I am playing around with, the absorptivity $a$ in the Stefan-Boltzmann law of radiation is used to quantify the aomunt of CO2 present, and I found that $a$ is defined as the ratio between the absorbed radiant power to the incident one.
$$j = a\sigma T^4$$
$$ a = \frac{I_a}{I_i} $$
I am wondering if it is plausible to say, from the Beer-Lambert law, that in this case the absorptivity is related to the concentration of CO2 in the atmosphere by the following, considering an atmosphere of uniform properties and thickness (which is certainly an overly ideal model). That is:
$$ -\log(1-a) = A \propto \ce{ [CO_2] }$$
P.S. I am not sure if this question is more suited for Physics Stackexchange, but since it's about the Beer-Lambert law, I'll post it here first.
Edit: as the commenters helpfully pointed out, my original definitions for the terms were awfully incorrect.

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

Absorbance $A$ of atmosphere, caused by $\ce{CO2}$, can be for the given relevant IR absorption wavelength $\lambda$ formulated as:
$$A(\lambda) = \int_{h_1}^{h_2}{(\epsilon(\lambda)c(h)\mathrm{d}h}$$
where

$A(\lambda)$ is absorbance at the given wavelength
$h$ is altitude.
$\epsilon(\lambda)$ is the $\ce{CO2}$ (mass/molar) extinction coefficient for the given wavelength
$c(h)$ is (mass/molar) $\ce{CO2}$ concentration at the given altitude.

Transmittance $T$ for the given wavelength is then $$T(\lambda) = \frac{I(\lambda)}{I_0(\lambda)}= 10^{-A(\lambda)}$$
The relative emissivity, referring to the emissivity of a black body at the given wavelength, is $\frac{I(\lambda,T)}{I_\mathrm{BB}(\lambda,T)}=1-T(\lambda) $
If we switch from the wav length to frequency by $\nu = \frac{c}{\lambda}$, then from Planck's law page we can use the relation of the Plancks' law and Stefan-Boltzmann law:

$$P = \int_0^\infty d\nu \int_0^{\tfrac{\pi}{2}} d\theta \int_0^{2\pi}d\phi \, B_\nu(T) \cos(\theta)\sin(\theta) = \sigma T^4$$
where $$\sigma = \frac{2k_\mathrm{B}^4\pi^5}{15c^2h^3}\approx 5.670 400 \times 10^{-8}\, \mathrm{J\, s^{-1} m^{-2} K^{-4}}$$ is known as the Stefan–Boltzmann constant.

and further

$$B_\nu(\nu, T) =\frac{ 2 h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 }$$

For the thermally emitted power at given frequency $\nu$, temperature $T$ and transmittance $T(\lambda=\frac{c}{\nu})$(Be aware of clash of quantity symbols):
$$dP(\nu,T,T(\lambda) =   d\nu (1-T(\lambda))\int_0^{\tfrac{\pi}{2}} d\theta \int_0^{2\pi}d\phi \, B_\nu(T) \cos(\theta)\sin(\theta) \\
=  d\nu (1-T(\lambda)) \int_0^{\tfrac{\pi}{2}} d\theta \int_0^{2\pi}d\phi \, \frac{ 2 h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } \cos(\theta)\sin(\theta) \\
=  d\nu (1-T(\lambda)) \int_0^{\tfrac{\pi}{2}} d\theta  \frac{ 4 \pi  h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } \cos(\theta)\sin(\theta)  \\
=  d\nu (1-T(\lambda)) \frac{ 4 \pi  h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 }  \int_0^{\tfrac{\pi}{2}} d\theta   \cos(\theta)\sin(\theta) \\
=  d\nu (1-T(\lambda)) \frac{ 2 \pi  h\nu^{3}}{c^2} \frac{1}{ e^{h\nu/(k_\mathrm{B}T)} - 1 } $$

I think your final equation is incorrect. The integral should be over the transmitted energy, not the absorbed energy, since this gives the total power emitted by the body beyond the absorbing layer.

@BuckThorn But it is over the transmitted energy, using the corrective factor (1-T(lambda)) for radiation of grey body at given wavelength/frequency). Unless I have missed something. // By other words, if a given air layer absorbs at given lambda 50% of IR, its emissivity at given lambda is 50% of emissivity of a black body. (referring to Kirchhoff's law of thermal radiation).


text:
How to find a method of determining whether an acid is strong or weak given only a formula?
I know that strong acids dissociate completely in water, and weak acids dissociate only partially. I searched for a method for finding if an acid is strong or weak given its formula, but the answer was a definitive, "You can't". My question is, what would be the best way of going about finding a pattern between strong and weak acids based on their formula?

Chemistry is semi-empirical science. There are few laws, many semi-empirical and empirical rules and many exceptions. There is no shortcut to learn it. First you have to memorize or search for the facts to know them. Later, when knowledge and experience comes, you will learn to see patterns in the facts, related to underlying known or unknown principles. It will allow you to reduce the data you need to remember or search for. But there will be always a lot of behavior that cannot be predicted from a simple rule set. Many facts can be predicted, but it may be inaccurate and very complicated.

https://en.wikipedia.org/wiki/Acid_dissociation_constant#Factors_that_affect_pKa_values

https://chemistry.stackexchange.com/questions/57574/is-there-a-formula-or-way-to-determine-whether-an-acid-is-a-strong-one-or-a-weak

https://chemistry.stackexchange.com/questions/105242/does-concentration-or-pka-define-acid-strength

The short answer is:  You can't.
The longer is answer is that there are some guidelines.
In most high-school and intro-level college general chemistry classes, students are taught that there are 6 common strong acids:  $\ce{HCl, HBr, HI, HNO3, H2SO4 and HClO4}$.  There are others, but these are the most common.  If the acid you are interested in is not one of these, it is likely to be a weak acid.
The only sure way to know is to look up the $\ce{Ka}$ or $\ce{pKa}$ of the acid.  Strong acids have $\ce{pKa's}$ below 0.  For example, hydrochloric acid has a $\ce{pKa}$ around -6, and nitric acid's is around -1.  Weak acids have $\ce{pKa's}$ greater than 0.
There might be some debate around where the cut-off line is between weak and strong acids; I've used zero here, but I've seen 1 used elsewhere.  That part isn't really important.


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Why sometimes normality is preferred over molarity?
I will participate in a practical course titled "Synthesis of Coordination Compounds." One of the materials needed to synthesize sodium trioxalatoferrate (III) is sulphuric acid with the concentration of 6 N (normality). Can someone explain the possible reasons the producers of H2SO4 with said concentration use normality instead of molarity for the unit of concentration? Can someone enlighten me on why equivalent concentration is sometimes used over molarity in lab products? Thanks.

Because some chemists are still mentally in 20th century?

For similar reasons why some people in scientific community still use imperial/"US" units.

Normality and equivalents are obsolete concepts today. Mole and molar units are the only units used today.
But these old units are still used when titrating a solution containing one (or several) unknown acids, like in waste waters or acidic rain for example, if nobody is able to do more measurements.
Let's consider an example. Suppose $250$ mL acidic waste water has required $36.2$ mL $0.1$ M (or $0.1$ N) $\ce{NaOH}$ to be neutralized. You can state that the sample of waste water contains $0.00362$ mol monoacid $\ce{HX}$ like $\ce{HCl}$, or $0.00181$ mol diacid $\ce{H2X}$ like $\ce{H2SO4}$, or some other value in case of a mixture. You cannot be more precise. Of course this "double" result cannot be printed in a report, and may be criticized by politicians, not aware of the problem.
But if you don't know the composition of acids in solution, you may state that the sample contains $0.00362$ equivalent of acid, once and for all. This a precise result.
And the acidic concentration (or normality) of your waste water can be calculated, even if you don't know which acid is in solution, and even if your solution is a mixture of different acids : $\ce{c = 0.00362 eq/0.25 L = 0.01448 eq/L = 0.01448 N}$. And this is an acceptable result to be published in a report or in front of non-chemists.

Normality is strongly deprecated, but not yet obsolete. Obsolete means not used anymore.

Normality is not obsolete at least in practice, it is used in various parts of the world including the USA/India, which means several billion people are familiar with this unit. Unfortunately, it has become a trend to bash historical ideas/notions and theories. Old does not mean that it is useless. Indeed, normality does cause a lot of conceptual difficults to beginning students but so does molarity, but once normality is understood along with its limitations, normality is/was not a bad concept after all. This unit is still used in industrial settings, pharma in some applications. All in all it is good to be familiar with it. Very soon we will start hearing that molarity is an old unit because we read that on the web and IUPAC website. Molarity is not going anywhere anytime soon!
The sole reason normality is useful is that all volumetric titration calculations are reduced to a very simple formula (CbVb=CtVt). Here b indicates the titrant in the burette and t indicates the substace being titrated... no thinking is required on the operator's part once the Standard Operating Procedure has been set for an analysis. Just plug and chug. This reduces the chances of mistakes or even forgetting the use of stoichiometric ratios. Other uses of normality are indicated by Maurice. You can titrate fruit juices without even knowing what acids are being titrated.
In your particular case, there is no specific reason to use normality but most likely the procedure was adapted from an older reference and they followed it verbatim. Normality use is being discouraged as concentration unit but it has not been wiped out from the face of the Earth.


text:
What does indicator "colour change" mean at equivalence and end point during titration
I've been trying to learn about indicators and titrations, but I keep on getting stuck on what "colour change" means in relation to the equivalence and endpoint, and I feel like I'm reading definitions which don't really support each other.
From libretext

The equivalence point is when the color changes most rapidly, not when the solution has changed color

And later

Indicators change color gradually at various pH...The long stretched color in the middle of the last line has equal intensity of BLUE and RED. If a solution has a color matching this, the pH would be the same as the pKai of the indicator

Which I understand as the equivalence point not so much referring to when the indicator changes colour itself, but when the ongoing colour change is most rapid. But because there seems to be a gradient of colour change within the range of the indicator, I was wondering what the "colour change" present at the equivalence point and endpoint refer to from some other sources on google. For example, from Khan Academy

an indicator is a weak acid or base that is added to the analyte solution, and it changes color when the equivalence point is reached

But one sentence later:

Endpoint: refers to the point at which the indicator changes color in an acid-base titration.

Given that both of equivalence and end point are said to have colour changes, I was wondering if someone would be able to help point out to me exactly what this "colour change" is referring to as I find myself really confused as to what the indicator is meant to measure. To my knowledge it's meant to measure equivalence point(?) as that is when pKa=pH, but would that be the initial colour change, rate of fastest colour change, or when you have a 50-50 blend of the colours of the acid and base form?
Further, in terms of Endpoint, would that be when there is an overwhelming concentration of one form of the indicator over the other such that only one colour is present, or would it be the initial change in colour?
Thanks in advance

The titration equivalence point is the point at the titration curve when there has been provided the equivalent amount of reactant.
The titration inflection point is the point at the titration curve where the curve is the most steep. It is very close to the equivalence point but generally not identical.
The titration end point is the point at the titration curve where the titration is stopped and the spent volume is taken to calculation.

It is usually related to the conventionally chosen color change or color appearance of the used titration indicator. E.g. at titration of strong acids by $\ce{NaOH}$ solution with methyl orange indicator, it is when red color turned to yellow.
For instrumental titration, the second derivative of being recorded titration curve may be used, detecting the inflection point.
Depending on the method and the used indicator, there may be smaller or bigger bias, caused by the difference between the end point and equivalence point.

The general approach is to choose such titration indicators, where the equivalence point falls with the indicator color change range.

Thanks so much for your explanation. Would there be the case where the methyl orange indicator is instead orange instead of red or yellow? So if the PkA of methyl orange is around 3.5, if equivalence point is around there would the solution be orange? If it is somewhere in between red and yellow, would one still say that the colour has "changed", or is that terminology more for complete colour changes?

Since the answer upvoting and accepting has been invented, there is no need for explicit thanks. But, they are, of course, appreciated.  :-)

MO range is usually listed as 3.1(red) - 4.4(orange). Note that due different eye sensitivity for different colors, the subjective middle color does not match the indicator pKa. Mixing red and yellow does produce orange. It is not like switching, it is transition.

End points are chosen in such a way that there is sharp visual edge between "not yet there" (MO not yet yellow, PhPh still clear) and "already there" (MO yellow, PhPh violet).