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An Index Poisoning Case Involving Dichlor and Carbonic Acid
I'm currently working an unusual case of poisoning, with Sodium Dichloro-S-Triazinetrione.

In my case, I need to know the expected reaction with carbonic acid.
I'm also curious if anyone could give me an opinion as to the suspected, expected chemistry of an air dried sample, after a partial reaction of this material has occurred. In other words, a hard pellet of dichlor was almost completely dissolved in carbonic acid, and then fished out of the patients cup.
My C13 NMR analysis has yielded the following signal, using D2O as a solvent. Here's the NMR scan of the material after it reacted with carbonic acid and was air dried:

I'm no longer in a chemical field of work, but I do have a background in chemistry. However, I'm working on an unusual poisoning case. As far as I can tell, this is also an index case, with no other examples in the medical literature. The patient* essentially made a drink, by accident, with carbonic acid and dichlor, yielding active chlorine, but the air dried the remaining "pellet" of the poison material which is the mystery. That post reaction, partially reacted material is what was sent through the NMR.
*The patient in this case lost his voice for about 3 months, and then suffered a permanent change in his digestive system.
Reference

The Effect of Cyanuric Acid and Other Interferences on Carbonate Alkalinity Measurement by John A. Wojtowicz, JSPSI, 1, 1, 2001(link)

Carbonic acid (pKa 6.35) (https://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s31-appendix-c-dissociation-consta.html) is not strong enough to protonate sodium dichlorocyanuric acid pKa 3.75 (https://pubchem.ncbi.nlm.nih.gov/compound/Dichloroisocyanuric-acid#section=Decomposition) so you are dealing with the properties of the sodium salt.

I have made some edits, removed some extra, chatty words and made the question fit for this site. Hope you don't mind. If you want, you can revert back the edit. 2. It is very difficult to obtain pure carbonic acid as it quickly decompose to carbon dioxide and water. See: https://en.wikipedia.org/wiki/Carbonic_acid

No problem ... It's basically just carbonated mineral water with a little flavoring, a tiny amount of ethanol, no sugar or sweetener. Link to reagent solution - https://www.instacart.com/products/3308584-the-mountain-valley-sparkling-water-key-lime-twist-pure-fine-33-8-fl-oz

I don't know if such symptoms can be provoked by the ingestion of a cup of tea amount of unknown concentration, but why hypochlorite and/or Cl2 poisoning/burning of the tissues has been ruled out?

I would guess that the larger peak is unreacted dichlor and the smaller is cyanuric acid (reported at ~150 ppm), ie dichlor that has lost its Cl atoms, which is the expected (slow) reaction with water.

I agree with alchemista that the toxic agent is likely HOCl. One consequence of the carbonic acid would be maintain a high HOCl/ClO- ratio which means much more potent oxidizing activity

The most likely scenario here seems to be a minor variant of bleach poisoning. Dichlor is primarily used as a UV-stable slow-release source of HOCl for swimming pool disinfecting. When dissolved in water, it is slowly hydrolyzed to HOCl and cyanuric acid.
In this scenario, the dry pellet recovered from the drink would be a mix of unreacted or singly dechlorinated dichlor along with the product cyanuric acid. No information is provided about the time that the dichlor was in water, but given that cyanuric acid has a single 13C NMR peak at ~150 ppm (see https://spectrabase.com/spectrum/7YT6cO2Flau), it appears that the bulk of the material is still at least singly chlorinated. Nonetheless, it would only take a small amount of HOCl to cause significant damage. Household bleach is typically only about 5-10% sodium hypochlorite (NaOCl) and is toxic even after further dilution.
In the dichlor/carbonic acid mix, the carbonic acid would further act to keep the hypochlorous acid in the protonated form ($\ce{HOCl}$) rather than the much less reactive hypochlorite ($\ce{ClO-}$) that dominates in household bleach.


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Does geminal dibromide reacts with sodamide to form internal alkyne or terminal alkyne?
I am currently studying the topic of synthesizing alkenes from double dehydrohalogenation of vicinal and geminal dihalides.

This is a reaction problem I got from an organic tutor on YouTube and he was explaining how $\ce{NaNH2}$ will favor the production of terminal alkynes on the contrary to $\ce{KOH}$ that will favor internal alkynes. He didn't go into the mechanism  for this one but drew the product only which will be 1-butyne.
I wanted to understand the mechanism behind it but couldn't find an exact problem on internet. Could someone please explain to me what is the mechanism and why can it not favor the internal part of the chain as it produces a more stable alkyne ?

Welcome to chem.SE. I have edited the question. Make sure to give proper punctuation and capitalization.

oooh i see. so it depends if its occurring at high temperature or low temperature right?

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How can we say that one reaction is faster than another if two conflicting factors are at play in both of them
While learning about the preparation of alkyl halides from alcohols by reacting them with hydrohalic acids, it was mentioned that the order of reactivity of different alcohols is Tertiary > Secondary > Primary and also that since the reaction takes place via nucleophilic substitution, the reactivity order also varies as $HI > HBr > HCl$ , as $I^-$ is a stronger nucleophile than $Br^-$ and so on. My question is that, if we consider two reactions, one of a secondary alcohol with HI and another of a tertiary alcohol reacting with HCl, then which of the two will be faster and why?
I tried to come with an answer and I think it may be that the Rate Determining Step is the one where the carbocation is formed and thus the tertiary alcohol will undergo the reaction faster in all cases. Pls let me know if this is right or wrong.

General trends like the ones you mention are useful. But they should not be used to generally quantify an effect. Therefore, mixing two trends the way you suggest makes no sense.

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Would a PNG geyser work on biogas?
I am trying to understand if a water heater designed to run on piped natural gas (PNG) could be powered by biogas without any modifications. Typically, these geysers have a battery-powered ignition and appropriate nozzles to generate flames. From a chemistry viewpoint, my two questions are:

Would biogas ignite with the same ignition setup as PNG's? I am not sure about the right terms but thinking of the time required for a spark to start the flames on PNG versus biogas.

Would the same nozzles hold well or do the hole diameters need to be adjusted to achieve greater (or lower) gas flow to generate the same degree of heat?

Both PNG and biogas are mostly methane, but the latter contains significantamount of CO2 ( typically 25-50% ), so it would need less air for optimal combustion.

One problem would be the acid gases.  Allegedly, it is possible to construct a DIY device that removes acid gases from biogas using, say, kitchen-grade sodium bicarbonate.

You might find producer gas (https://www.vedantu.com/chemistry/producer-gas) is similar enough to biogas that a simple gas orifice interchange to producer gas would suffice -- or not. Perhaps a local HVAC company could help on specifics.


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what are the differences between reversible cells and rechargeable cells?
My chemistry textbook says daniel cells are reversible, but not rechargeable.
Why is it that all reversible cells are not rechargeable?
Are all rechargeable cells reversible? why/why not?

Good question. I think the reason is essentially degree rather than kind: rechargeable batteries are designed to be discharged and recharged many times, with minimal deterioration in performance. But non-rechargeable cells, such as the Daniell cell, can be reversed but not brought back to nearly original state.

when I researched a bit more on the internet, I found it is because when Daniel cell is allowed to recharge, reduction of zinc ions to zinc does not take place in the anodic compartment, but evolution of hydrogen takes place. But where did the hydrogen come from?

@AbenPhilip Water contains hydrogen.

Just as Poutnik said. Also, the cell is Daniell, not daniel.

Moreover reversible has subtle meaning. Often, although not in chemistry, reversible and rechargeable are used as synonyms.

Per Wikipedia on Electrolytic Cells, to quote:

Galvanic cells compared with electrolytic cells
In an electrolytic cell a current is passed through the cell by an external voltage, causing an otherwise nonspontaneous chemical reaction to proceed. In a galvanic cell the progress of a spontaneous chemical reaction causes an electric current to flow. An equilibrium electrochemical cell is at the state between an electrolytic cell and a galvanic cell. The tendency of a spontaneous reaction to push a current through the external circuit is exactly balanced by an external voltage that is called a counter electromotive force or counter e.m.f. so that no current flows. If this counter voltage is increased the cell becomes an electrolytic cell and if it is decreased the cell becomes a galvanic cell.[1]

I do also, operationally, agree with Aben Philip comments, where apparently, attempting to recharge (aka, chemically reverse) a galvanic cell, may undesirably further introduce hydrogen gas. The problematic $\ce{H2}$ gas evolution can result in bursting and the usual safety concerns working with hydrogen (as in an explosion hazard).
More technically from a electrochemistry perspective, per another source, rechargeable batteries are again distinctly claimed again to an "Electrolytic Cell", characterized as having non-spontaneous oxidation-reduction reaction versus spontaneous for a galvanic cell. But further, in addition, they may also have a semipermeable membrane governing the flow of ions, which are not found in a galvanic cell.
I hope this helps.


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What is the LUMO and HOMO in and O2 diatomic
In the molecular orbital (MO) diagram for the $O_2$ diatomic, the $2π_g^*$ contains two unpaired electrons. This is certainly the HOMO, as it is the highest molecular orbital with electrons. My question is, is the LUMO of this molecule the $6σ_u^*$ (which is definitely unoccupied)? or is the LUMO the $2π_g^*$?
On the one hand, it feels very weird to have both the LUMO and the HOMO be the same orbital, but it also feels like you should be removing electrons from the HOMO, and adding them to the LUMO, but you both add and remove electrons to the  $2π_g^*$.
In other words, what happens to the LUMO, when an orbital is partially filled.

In the typical way these terms are used, the LUMO can't be the same as the HOMO. Since each of the pi spatial orbitals already have one electron, they are not unoccupied, so they can't be the LUMO. In the sense of singly occupied spin orbitals, you could say that these pi spin orbitals are the LUMO, but its fairly nonstandard to refer to spin orbitals when discussing HOMO/LUMO.


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On what basis is the rate of Sn1 reaction dependent
Does the rate of reaction depend on the stability of the  1st carbocation formed just after the leaving group leaves or the rearranged carbocation's stability? for eg

Here if i compare the stability of initially formed carbocation then the order should be C>D>B>A here in B more hyperconjugation should bring about more stability hence the order which I wrote just now. But if we compare the overall stability after rearrangement of carbocation then order should be C>D>A>B please suggest what should I compare

What do you mean by "rearranged carbocation"? Also, why do you think there will be more hyperconjugation with B than with A?

In general, as a first order approximation, the rate of the reaction is dependent on the activation energy, which is often a higher lying transition state. Reactive intermediates, like carbocations, are stable structures, at least more than transition states.

@Snijderfrey Rearranged carbocation means that initially for the formation of carbocation in part (A) Cl will be removed giving a primary carbocation. This happens in both (A) and (B) so I thought more stable initial carbocation (rate determining step is formation of carbocation) will require less activation energy.

@Snijderfrey I think hyperconjugation depends on the number of "alpha" hydrogens present next to carbocation. Initially (before the carbocation (A) shifts to tertiary carbon by methyl shift) (b) has more "alpha" hydrogens than (A). And** hyperconjugation is a more  important step for determining stability than inductive effect**

@Martin-マーチン Any suggestions? for the order that I mentioned?

Hyperconjugation is not limited to C-H bonds.

yes i know but its not required here

There is reverse hyperconjugation also like in Ph-CCl3 etc

@Snijderfrey Please elaborate different types of hyperconjugation as I can recall only 2

The rate of a SN1 reaction depends solely on the stability of the rearranged carbocation and also on the leaving group tendency. The better the leaving group, the faster will be the reaction.
Here, the correct order would be D>C>A>B.
In C, $\ce{-OCH3}$ is present which is an EWG. So, it won't stabilize the carbocation. Hence, D>C.
And between A and B, A will be a secondary carbocation whereas B will be a primary carbocation after the leaving group leaves. So, A>B.

I think in methoxy the +M effect outweighs the -I effect, especially in regard towards a cation.


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Why are camphor vapours taken in nitrogen gas?
I have read some examples of "solid-in-gas" solutions in several books. These examples include iodine vapours, camphor vapours etc.
However, one fact that has stuck me is that nearly every book mentions "Camphor vapours in N2" instead of simply saying air. If it's simply "iodine vapours in air", why the distinction for camphor?

Vapour in gas is not solid in gas, but mixture of gases.
$\ce{N2}$ is not "simply air". Camphor in $\ce{N2}$ means they use nitrogen instead of air, probably to prevent camphor vapour oxidation  by aerial oxygen.


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Does protein phosphorylation add a phosphate group, or does it add a phosphoryl group? Wikipedia seems to have conflicting info on it
Does protein phosphorylation add a phosphate group, or does it add a phosphoryl group? Wikipedia seems to have conflicting info on it
Wikipedia on Phosphorylation, https://en.wikipedia.org/wiki/Phosphorylation  says  "In chemistry, phosphorylation of a molecule is the attachment of a phosphoryl group.....A phosphoryl group should not be confused with a phosphate group....The phosphoryl group plays the central role in phosphorylation...."  And it mentions Protein Phosphorylation.
Wikipedia on Protein Phosphorylation , https://en.wikipedia.org/wiki/Protein_phosphorylation   "Protein phosphorylation is a reversible post-translational modification of proteins in which an amino acid residue is phosphorylated by a protein kinase by the addition of a covalently bound phosphate group. "
Note- I am going to post an answer based on some really good info from Mithoron.

Note-Gotta credit Mithoron for this and if he posts an answer i'll accept his answer, as he deserves the rep for this answer.
There are two parts to this question

Are there phosphate groups in ATP

and

Is a phosphate group transferred.from the ATP , or is a phosphoryl group transferred from the ATP.

And also, you have to know the difference between a phosphate group and a phosphoryl group.
A phosphoryl group has three oxygens. A phosphate group has four oxygens.
https://en.wikipedia.org/wiki/Phosphoryl_group

Now you can look at ATP compared to ADP. There is no question that the difference is a phosphoryl group. Not a phosphate group.

I’ll use the general term phosphus compound to refer to something that may be a phosphate group and may be a phosphoryl group.
So when you look at ATP or ADP, do you draw lines to include four oxygens, or three oxygens. It’s arbitrary but that makes the difference to whether you would see phosphate groups, or phosphoryl groups.
But if you look at the first phosphor compound, and the O to the left of the P. That O is actually part of Adenosine.
Adenosine has that O, (though plus an H)

In answer to Are there phosphate groups in ATP.. One could maybe arguably(and not a good argument), consider there to be. if one draws the lines like that such that it considers the O from the adnosine to be part of the phosphate groups. Though(and to say phosphate groups would be a bad case, because), that O is from the adenosine. It’s Adenosine + three phosphoryl groups.
Phosphorylation involves the transfer of a phosphoryl group and not a phosphate group. And we know this because the thing that is transferred has three oxygens, not four oxygens.
Why then is it called triphosphate?
the naming "adenosine triphosphate" may have come from the fact that it can be formed by a condensation reaction between adenosine and a triphosphate group.

So ATP = adenosine + triphosphate - H - OH (i.e. minus water).
So, it might arguably be a bit of a misnomer, but well... it's stuck. (besides that perhaps apparently it should apparently be called adenosyl.... A follow up q I had was..Could you look at ATP as adenosine minus OH with triphosphate minus H. ? In which case, ATP could be seen legitimately as having the phosphate groups and that not being a misnomer. So the oxygen could legitimately be seen as part of the phosphor compounds.. Whether the oxygen came off the adenosine or off the triphosphate when they reacted, is perhaps arbitrary?
If you mean us counting the atoms on paper, then yeah, it is quite arbitrary whether we take the OH away from adenosine or the triphosphate. But in the biosynthesis of ATP, (without checking one could bet,), the oxygen comes from adenosine, i.e. it's (adenosine - H) + (triphosphate - OH).


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How to correctly use the avogadro number?
Avogadro constant, $N_A$ is equal to $\frac{6.02214076 \cdot 10^{23}}{mol}$. There is no unit of particle or anything in $N_A$ (nothing like $N_A = \frac{6.02214076 \cdot 10^{23} \ce {particles}}{mol}$). So like, can I Is it correct if I say
$$N_A=\frac{6.02214076 \cdot 10^{23}\ce{atom}}{mol}$$
if I'm doing stoichiometry in terms of atoms or
$$N_A=\frac{6.02214076 \cdot 10^{23}\ce{molecule}}{mol}$$
if I'm doing math in terms of molecules.
For example, $\ce{H2O}$ has the molecular mass of approximately $18.015u$ or $\frac{18.015u}{\ce{molecule}}$. I believe saying that the molecular mass of water is $\frac{18.015u}{molecule}$ is also a correct way of saying it. Though, I'd appreciate if someone can confirm this. If I want to convert from the molecular mass to molar mass, mathematically, it would be like such:
$$18.015u \cdot N_A = \frac{18.015u \cdot 6.022 \cdot 10^{23}}{\pu{mol}} = \frac{\pu{18.015g}}{mol}$$
or, I believe I can write like like such too:
$$\frac{18.015u}{\ce {molecule}} \cdot N_A = \frac{18.015u}{\ce{molecule}} \cdot \frac{6.022 \cdot 10^{23} \ce {molecule}}{mol}$$
The unit of molecule here will cancel out, leaving me with
$$\frac{18.015u \cdot 6.022 \cdot 10^{23}}{mol}$$
or since $6.022 \cdot 10^{23} \cdot 1u = 1$g,
$$\frac{18.015u \cdot 6.022 \cdot 10^{23}}{mol} \cdot \frac{1g}{6.022 \cdot 10^{23}u} = \frac{18.015g}{mol}$$
So my question is, can I add the "unit" particle in the $N_A$ formula like I showed before ($N_A = \pu{\frac{6.02214076 \cdot 10^{23}\ce{atom}}{mol}}$ or  $N_A = \pu{\frac{6.02214076 \cdot 10^{23} molecules}{mol}}$, etc)? If so, wouldn't I just changed the definition of $N_A$ or what $N_A$ equals to? Which I believe would be mathematically incorrect. Maybe I'm missing something. Also, can I say that the molecular mass of $\ce{H2O}$ is $\frac{\pu{18.015u}}{\pu{molecule}}$ instead of just $\pu{18.015u}$. I just hope to do the stoichiometry with the units right. I appreciate any help.

It goes as far back as to concept of abstract integer numbers, abandoning binding integers to real objects. No 3 rabbits, no 3 carrots, no 3 eggs, just 3. No Na atoms/mol, no Na ions/mol, no Na molecules / mol, just Na/ mol.

"There is no unit of particle or anything"... Although there is no a SI unit, the counting of particles is certainly within the definition of Amount of substance. Without particles, there is no Avogadro constant. The unit does not exist in my opinion just because it would be redundant (it does not require any definition or standard as counting objects seems a natural skill. Indeed positive integers were called natural number). What you did is what we have in mind.

@Alchimista so it's still correct then? I can add and treat atoms, molecules, ions, etc in the avogadro number and molecular mass?

@Mohammadmuazzamali I think that if it helps you why not? Remember that it is unit with no dimensions, it is just us counting. If I would be your teacher it will be fine for me. Otherwise just think of it, and don't write it down, if this thread is of practical relevance for your exercises.

Is it correct if I say ...

There are two questions you have to consider:

Do you obtain the correct answer using this notation?
Are you using formally correct technical language?

The answer to the first question depends on whether you understand what you are doing - no matter which notation you decide to use. The answer to the second question is no. "Molecules", "Particles", "Carbon", "Atom" are all not part of the official SI units, so you would not use this notation in a formal document. In fact, "u" is not even part of the SI unit system (It is one of the "Non-SI units whose values in SI units must be obtained experimentally", https://www.bipm.org/en/publications/si-brochure).
Adding annotations to units in general
Introductory textbooks sometimes add annotations to units as a way to explain stoichiometry and unit conversion. Here is one example:
$$\require{cancel}
0.75~\cancel{\mathrm{mol}~\ce{C3H8}} \times \frac{3~\cancel{\mathrm{mol}~\ce{CO2}}}{1~\cancel{\mathrm{mol}~\ce{C3H8}}} \times \frac{6.022 \times 10^{23}~\ce{CO2}~\text{molecules}}{\cancel{\mathrm{mol}~\ce{CO2}}} $$ $$= 1.4\times 10^{24}~\ce{CO2}~\text{molecules}$$
The formal way to write this would be:
$$\begin{align}
N_\mathrm{A}  &= \pu{6.02214076E-23 mol-1} \\
n_\ce{C3H8}   &= \pu{0.75 mol} \\
\nu_\ce{CO2}  &= 3 \\
\nu_\ce{C3H8} &= 1 \\[10pt]
N_\ce{CO2}    &= n_\ce{C3H8} \cdot \frac{\nu_\ce{CO2}}{\nu_\ce{C3H8}} \cdot N_\mathrm{A} \\
              &= \pu{0.75 mol} \cdot \frac{1}{3} \cdot \pu{6.02214076E-23 mol-1} \\
              &= \pu{1.35E24}
\end{align}$$
Here, the annotations are added to the quantity symbols (not to the units), and the units are the official SI units. Obviously, the formal way of writing it is more abstract, and in this case requires more space.
On the other hand, the formula used is quite expressive, and would apply no matter what the specific quantities are:
$$N_\ce{CO2} = n_\ce{C3H8} \cdot \frac{\nu_\ce{CO2}}{\nu_\ce{C3H8}} \cdot N_\mathrm{A}$$
It would roughly translate into: To get the number of carbon dioxide particles that are formed from a certain amount of propane, you take the amount of propane, multiply by the ratio of stoichiometric coefficients and by Avogadro's constant. Notice that the way the textbook shows the problem solution, it never mentions the names of the quantities that are used, but instead describes them using units and additional annotation only. Notice also that you would not "cancel out" the symbol $\ce{C3H8}$ because that would make the formula less meaningful.


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Dark smudge at the TLC solvent front
I have recently started using TLC in my graduate work to test for the cleavage of a cyclic oligo-adenylate in an enzymatic reaction. The sample I'm spotting on my plate is in an aqueous solution buffered with 25mM Tris. I am using a 20x20cm fluorescent silica plate as my stationary phase and my mobile phase is 20mM Ammonium Bicarbonate in 30% water and 70% ethanol. I am allowing the migration of the solvent to occur at room temperature in a chamber pre-equillibrated with the running buffer (The running buffer was added to the chamber and allowed to sit for > 1.0 h to maximize the vapor content of the chamber). These conditions are consistent with the literature that I have read dealing with cyclic oligo-adenylates.
I am visualizing my plate with a short-wavelength (254nm) UV lamp that causes the plate to fluoresce and analytes cast shadows. When I visualize my plate I see shadows for some of my analytes, but I also have a fat band at the solvent front (see attached photo). My questions are:

Is this dark band a problem? I don't think it is obscuring anything important, but I would feel more confident if it wasn't there.

Any ideas about what could be causing this band?

3.Any suggestions about getting rid of the band?
Like I said, I am pretty new to this whole TLC thing so I would be extremely grateful for any assistance.

Your mobile phase is too strong and the analytes (solutes) are traveling with the solvent. In other words, Rf is almost 1 and there is no retention on the plate for most of the solutes in your mixture.

The dark smudge are the major sample component spots, all joined together in a line. You want it to have RF somewhere in interval 0.3 - 0.7 for better spotting impurities/side products.  Your stationary phase may be too little active. Try temperate it in an oven at 105-120 Deg V to get lower Rf. Alternative is experimenting with increasing ethanol content in mobile phase.

Or you just have only a major component. But anyway seems that you have a too strong eluent / not enough active plate indeed. What I think you took wrong is the fact that you consider the most intense spots as an annoyance while (hopefully) they are the major component(s).


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Calculating the number of isoprenes in Lanosterol
Hey guys I'm having trouble understanding the structure of lanosterol regarding the isoprenes in the structure shown below:

The first isoprene I think is where the 3rd carbon would be at the OH. But I thought that carbon 3 of isoprene doesn't have any groups sticking out- only carbon 2 with the methyl group sticking out. Can isoprene units have hydroxyl groups too?
Also from looking at the image above, there also appears to be another isoprene unit right in the middle of lanosterol in blue- but it looks like a straight 5 carbon chain unlike your typical isoprene unit with a methyl group sticking out somewhere. Why is this? Thanks!

Squalene 1 is produced in vivo by reductive coupling of two farnesyl pyrophosphate (FPP)  units. FPP is ultimately formed from acetate through the agency of the mevalonic acid pathway. Squalene is enantioselectively epoxidized to form squalene epoxide 2. In vivo acid-promoted cyclization leads to carbocation 3a with stereocontrol at the newly formed centers. Note that the red mechanism arrow in structure 2 leads to an anti-Markovnikov addition creating vicinal quaternary carbons at $\ce{C8 and C14}$. At this juncture all of the isoprene units are intact.
The conformation of the chain at $\ce{C20}$ in 3a is critical because it determines the stereochemistry at this center in lanosterol 4. The hydride at $\ce{C17}$ in structure 3b migrates to $\ce{C20}$ while the $\ce{C13}$ hydride shifts to $\ce{C17}$. The continuity of the original isoprene units is now broken as the $\ce{C14}$ methyl (methide) group migrates to $\ce{C13}$ and the $\ce{C8}$ methyl group moves to $\ce{C14}$. Finally, a proton is lost from $\ce{C9}$ to form the tetrasubstituted double bond of lanosterol. The facial integrity of the four migrating groups is maintained throughout the rearrangements. The migration of the two methyl groups destroys the integrity of the isoprene rule within lanosterol itself.


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Using a pH meter to measure the presence of water in liquid ammonia?
I intend to do some experiments with building an ammonia-water gas adsorption refrigerator similar to a Crosley Icyball Wikipedia - Icyball. When recharging the icyball, it is important to not overheat the 'hot' side (containing water and ammonia) when driving off the ammonia. Too much heat will cause water to also evaporate from the hot side, and condense on the cold side along with the ammonia, compromising the efficiency of the device.
While experimenting with the device, I would like to be able to measure the concentration of water/ammonia within the two chambers of the device. The operating pressure would nominally be around 2,000 kPa, temperature range ~-10 to +50 degrees C. Other than water and ammonia, the only other gases present would be an inert/insoluble gas such as helium or hydrogen.
Would using a pH meter be a reasonable way of determining the concentration of water/ammonia in such a system? Or is there a more practical way of measuring water/ammonia concentration?

pH measurement will not be applicable for several reasons. I would suggest measuring physical parameters, like thermal conductivity,   heat capacity, relative permitivity, refraction index etc.

pH is usually measured with a glass electrode. This electrode gives a voltage depending on its constitution. If dipped in dilute aqueous solutions, this voltage is proportional to $\ce{[H^+]}$. It can be used to determine the concentration $\ce{[H^+]}$. But if this electrode is not dipped into a dilute aqueous solution, it gives a voltage which is related to the concentrations in an unknown way. When the aqueous solution is concentrated, it is possible to introduce activity coefficients, which are able to be transformed into sensible concentration values. But if the electrode is dipped into a non-aqueous solution, the measured tension will have no relation with any concentration. It may even be unstable, and change without any reason from positive to negative values.

Speculation: Perhaps the autoprotolysis of ammonia (formally $\ce{2 NH3 <=> NH^+_4 + NH^-_2}$) in addition to the one of water renders the situation even more complex.

@Buttonwood. You may be right. This autoprotolytic reaction should increase the conductivity of the ammonia solution. Is it the case ? I don't know.

The potential complication is inferred from the Birch reduction of arenes in ammonia (though e.g., Phil Baran's group worked on electrochemical alternatives easier to scale without this solvent) and speculation only. Possibly US American farmers know/know better how to check if liquid ammonia is dry, or/and those active in the business of industrial refrigeration with ammonia.


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Why are ligand orbitals always lower in energy than the metal orbitals in the ligand field theory?
The below image has been used to represent the ligand field theory for transition metals:

(Source: https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/ligand.php)
Also, other similar images are available on other sites/books which portray the LFT.
My doubt is that by looking at these images, it seems that the ligand orbitals are always lower in energy than the Metal orbitals, is there any specific reason for this?

Perhaps you could provide an example of some specific compounds in which this is not the case? I think you'll have a hard time finding very many. Common ligand atoms are generally pretty small with low energy orbitals, eg O, N, F, Cl, C, S, P, so in a generic scheme, assuming ligand orbitals are low seems appropriate. One compound that seems likely to go against the trend is $\ce{TiCl4}$, but I'm not sure the exact numbers.


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Is it appropriate to say "solid-in-gas solution" and "liquid-in-gas solution"?
A number of books mention "solid-in-gas" and "liquid-in-gas" types of gaseous solutions.
However, for something to dissolve in a gas, it must be in its vapour phase. So, for instance, why wouldn't iodine vapours in air be a gas-in-gas solution?
Or, for that matter, since water vapour is gaseous water, why wouldn't water vapour in air be a gas-in-gas solution?
One reason I can think of is that we probably mention the main state these substances are in at room temperature.  But then, water vapour also exists at room temperature (though water is obviously liquid).
How do I understand it?

Can you give the titles and authors of such books?

I actually found a paper using the term "gaseous solution", from 1928. https://pubs.acs.org/doi/pdf/10.1021/ja01391a006

@Karl a gas mixture is a solution. It is just trivial so the explicit term is unusual, and became desuete if ever was common. IUPAC reserves the term to solutions in a liquid or a solid,for the above reason. A mixture of gas is always one phase.

@Alchimista IUPAC reserves the term for liquid or solid cases, ergo a gas mixture is NOT a solution. qed ;)

Solid/liquid-in-gas solution is factical nonsense, being communication shortcut for a vapour of solid/liquid in a gas.
Solid sublimates/liquid evaporates into space not occupied by a condensed phase. Gas has nothing to do with that.
Mixture of gases is and is not solution, depending on the point of view.
It is a solution because it is a homogenous fluid mixture.
It is not a solution, because air/major gas does not participate in bringing the solid/liquid to gaseous phase at all. There is no action of dissolving, tearing a molecule from the condensed phase by molecules of gas. The vapour would be there even if there was no other gas at all. They just share common space, the gas being a bystander.
We do not say air is solution of oxygen and minor gases in nitrogen. There is no difference between "permanent gases" and vapours here.

You must look at the phases in your system. Whenever there is only one phase, but there are two or more chemical species, then you have a solution.
Beside this, all the rest is semantic. Some can be regulated by IUPAC, some not.
Refer to air as a solution would sound unusual. A gas mixture - unless reacting to give a liquid or solid product - is always one phase, so the term gaseous solution isn't even considered by the IUPAC's definition of solution: https://goldbook.iupac.org/terms/view/S05746.
About "solid in gas" you are correct, once in the gas phases the iodine or naphthalene vapours aren't solid anymore.
However, note that the same applies to sugar or salt in water. You can certainly speak of a solid (dissolved) in water although the actual solution obviously has no actual solid anymore, and even no salt but ions. In this case, the use is common because it is what operationally we do, we take a spoon of salt and dissolve it, even in the kitchen, and so we speak.
"Solid in gas (or liquid)" as it is and without context would mean smoke (or suspension), so we are not dealing with solutions.
I hope this is clear enough for you to understand that you were somehow over thinking the subject.

I like you answer, perhaps you could add that the term "gaseous solution" is pretty outdated. And I would like to point out all gases are 100% miscible (unless they react strongly with each other).

@Karl I will, as OP pose a side question about gas in gas.


text:
Why use diamond-like carbon instead of diamond?
Diamond-like carbon is an amorphous form of carbon, made of many small crystallites, mixing up the polytypes of diamond (most commonly the cubic lattice diamond, but also its lonsdaleite polytype with a hexagonal lattice and other forms). It is applied as a coating to lend some of the qualities of diamond to an object.
So if the goal is to give the object diamond-like qualities, why not just coat it with regular diamond instead of DLC?

The balance of cost and effect matters. If we had a good way of producing thin diamond films on surfaces (and it worked to give the expected properties) then it might be worth doing. But current technology (and economics) has techniques that give DLC which has most of the desired properties. We don't know how to cheaply create perfect films of pure diamond and they might not be better anyway.

How would you apply "real" diamond to a manufact? With a ring?

You balance cost of manufacture and return of investment for the anticipated use.
If you read about the methods of production of diamond-like carbon (DLC), e.g., on Wikipedia, e.g.

«DLC is typically produced by processes in which high energy precursive carbons (e.g. in plasmas, in filtered cathodic arc deposition, in sputter deposition and in ion beam deposition) are rapidly cooled or quenched on relatively cold surfaces.»
credit to Wikipedia's article

you may infer that DLC may offer a smooth continuous coating of a surface at a layer thickness you may tune by adjusting the process parameters.  Said Wikipedia article illustrates this with a cobblestone coverage:

On the other hand, how would you do with diamonds?  You would need to find them or rely on synthetic diamonds and then to break and crash these into fine grains.  Not all directions in a diamond crystal are of equal hardness (keyword cleavage planes), however diamonds are the most hard naturally occurring material.  Even if you arrive this, crushing mineral yields to some dispersion of the grain sizes, so by sieves or floatation, you have to narrow this size distribution.  But even then, the surface coating would be like icing a cake with shots (or sprinkles), potentially leaving some areas of the surface exposed without this coating.  Instead, the sharp edges of the isolated diamond shots likely would be abrasive like a fine sanding paper.

In some cases the continuous nature of the DLC film is so important that the far worse properties compared to diamond particles are unimportant. Being non-porous would be one;  (in-plane) thermal conductivity would be another.  Polycrystalline diamond CVD-grown in situ is a step closer to diamond but for growth the substrate has to be able to handle the high temperatures needed, and for good diamond growth these can get too high for many materials

Why do you think that would be preferable? A perfect crystal is rather fragile - diamonds are hard, but they also have perfect cleavage - one good hit and you have two diamonds (or a bunch of shards). We only use monocrystals when we have to - like for microchips. For material strength, they're usually really poor - you want many grain boundaries that halt the propagation of cracks, and you certainly don't want perfect cleavage.
It's also hard to imagine how you would coat anything with a single crystal of diamond in the first place. How would the atoms fit together? Using many grains allows you to nicely coat a surface, mostly without having to think too much about its shape - heck, if you look around yourself, we do that all the time with pretty much all the materials we use for strength - concrete and asphalt, stone walls and brick walls, metal/ceramic coatings...

Since DLC is pretty much a physical vapor deposition (PVD) thing, I will address this question from this perspective.
The simple answer to your question is: So far we cannot deposit diamonds in PVD processes (at least I have not heard about it). If you read into the literature of DLC, a measure of success for the coatings is the average hybridization, or in other words, how close the bonding structure gets to the sp3 bonds in proper diamonds. Reaching almost pure sp3 bonds in a DLC coating is a goal that many research groups try to reach. If we could deposit diamond properly, we would. Diamond would certainly be better than all alternatives for some problem, and would be used as such.
Chemical Vapor Deposited (CVD) diamonds are a thing, as a comment mentioned. I have seen a grown CVD diamond wth my own eyes for use as a window in a spectrometer. But CVD requires very high process temperatures that many substrates cannot endure. PVD on the other hand can be done at much lower temperatures.

Disregarding costs aspects, diamonds makes a good noble cathode, where I am assuming there is an intended electrochemical use for the coated object (else, this question is on aesthetics).
Generally, for the anode and cathode in the special case of an electrochemical cell, it is known that one can greatly accelerate galvanic based corrosion reaction, for example, other things held constant, by having a small area anode relative to a much larger area cathode.
Here is a supporting answer provide on Research Gate, to quote:

The influence of cathode/anode area ratio is a key factor to the rate of galvanic corrosion in seawater. It can be explained that the anode current is always equal to the cathode current when the galvanic corrosion occurs, and anode area is smaller, the anode current density is greater. That is, the corrosion rate of anode metal is greater. Increasing Sc/Sa accelerated galvanic corrosion rate of high potential difference coupling, the fitting curves of the galvanic current by experiments will show the parabola shape. When anode density is greater than corrosion rate will be higher. Galvanic corrosion rate linear growth with the Sc/Sa in seawater indicates, the galvanic potential shift and driving voltage decreases.

However, if we are talking about battery technology, we are in more complex arena of surface modifications. To quote a Science Direct source:

Electrode surface modifications generally prolong the lifetime and improve the safety of Li–ion batteries. Stabilization of electrode surfaces and the interphases reduces the extent of side reactions, thus suppressing impedance growth, capacity fade, and the release of significant heat and gases which cause safety risk of thermal runaway. Although significant progress has been achieved, particularly during the past decade, the field of electrochemical and chemical surface modification of battery electrodes is still wide open for further exploration. Novel synthetic approaches for electrode surface functionalization and preparation of thin-layer coatings are desired. However, the processing and manufacturing approach of electrode coatings must be facile and scalable at industrial level, as the main bottleneck for the large-scale implementation of Li–ion batteries today are the high cost of cathode materials and processing.4,6 Another practical issue is the stability of the surface modification, for example, the coating, which tends to degrade over time and cycling. Degradation is typically accelerated under harsh chemical and electrochemical operating conditions (e.g., high or low temperature, high voltage, and fast cycling rate). Further fundamental understandings of operational principles of the interphases, the synergetic interplay between their components, and how they may be modified for improved long-term cycling are necessary.

So, some knowledge of the intended purpose of the application for the diamond electrodes would appear to be important as to which is actually better as, for example, diamond-like carbon (an amorphous form of carbon) may not be suitable due to stability/degradation concerns for the targeted application.


text:
Racemization on nucleophilic substitution by OH- ion
Here's the question I'm stumped by -

According to the answer key, i) is the answer but why can't it be iv)? There is a chiral carbon in compound c) CH3(C2H5)CHCH2Br , so why won't the product (2-Methyl-1-butanol show enantiomers?

Points to consider:

In said substitution, $\ce{OH^-}$ is the nucleophile (the agent), and $\ce{Br^-}$ the nucleofuge leaving the molecule.

While both a) and c) contain a stereogenic centre, only in the case of substitution a) this happenz on this very stereogenic centre.

Thus the answer key points to option i).

As pointed out by @Snijderfrey's comment, the question equally aims to reflect on the mechanism of the reaction.  None of the formulae in the question explicitly states if the stereogenic centres in molecules a) or c) are in (R), or (S)-configuration one may convey with the wedge notation. Of course, an individual stereogenic centre is either one; describing the molecules of the sample (i.e., an ensemble) maybe this was left out because it is a mixture of (R) and (S) of these molecules (special case: racemic mixture), or omitted because the sample was not characterized accordingly.  In the later case, a wavy bond may highlight this:

Typically, the presence of a stereogenic centre eventually leads to optical activity of the material which then is either dextrorotatory or laevorotatory.  However for this question, your focus should be the correct assignment of (R), or (S) in the context of the underlying mechanism.
The effect eventually recorded in the lab depends on multiple parameters; the presence of an additional stereogenic centre does not need to be synergistic.  This then leads to circular dichroism spectroscopy used indeed for structure elucidation.  Among publications easier to understand about this topic in the Journal of Chemical Education are the ones by Urbach, Thomson, and Andrews.
References:
Andrews S. S.;  Tretton J. Physical Principles of Circular Dichroism. J. Chem. Educ. 2020, 97, 4370-4376, doi 10.1021/acs.jchemed.0c01061.
Urbach, A. R. Circular Dichroism Spectroscopy in the Undergraduate Curriculum. J. Chem. Educ. 2010, 87, 891–893, doi 10.1021/ed1005954.
Thomson P. I. T. Is That a Polarimeter in Your Pocket? A Zero-Cost, Technology-Enabled Demonstration of Optical Rotation. J. Chem. Educ. 2018, 95, 837–841, doi 10.1021/acs.jchemed.7b00767.

Since c) has a stereocenter, it would have optical activity right? If it doesn't show dextrorotatory and laevorotatory forms, why?

I agree with this answer. However, the question is a little hard to answer without knowing if the starting material is a racemic mixture. I suspect the question also aims that you think about the reaction mechanism, $S_N1$ vs. $S_N2$, and its implication for stereochemistry.


text:
Formula for collision frequency and mean free path
We were studying collision frequency and mean free path today, and in it, we got the equations of collision frequency as:
$$Z=\sqrt 2 \pi \sigma^2v_{rms}N$$
where $N$ is the number density, equal to $n/V$.
Then, I looked online for more information, but surprisingly, I got completely different formulae for $Z$ itself.
One formula was $Z=\frac1{\sqrt 2}\pi\sigma^2v_{avg}N^2$. Another was $Z=\sqrt 2\pi\sigma^2v_{avg}\frac{N} V$.
Similarly, the formula for mean free path given was $\lambda= \frac{V_{rms}} Z$. So, changing the value of $Z$ would obviously change the value of $\lambda$ as well.
Also, all these formulae in all the different websites appear to come with really complex mathematics and physics which I don't really understand, so if anyone could give a simpler explanation, that would be nice.

This has to do with collisions between like or unlike molecules as well as a factor of $\sqrt{2}$ from going from average velocity to RMS. See McQuarrie $\it{Statistical, Mechanics}$ (1976), p. 371-372.

If you don't mind, could you yourself not answer?

I am struggling with the same question and didn't find any reliable explanation.


text:
Why the electrode that is the electron source on a x-ray tube is called cathode?
In electrochemistry, the electrode that lose electrons and oxidizes is called anode. Why on an x-ray tube the electrode that lose electrons is called cathode?

In an electrolytic cell, the cathode is negative: it supplies the electrons to reduce electroactive species. In a voltaic cell, the anode is negative.

Because it loses electrons.

You're contradicting yourself...

@Mithoron Why am I contradicting myself?

@TitoMorennoPonteseSouza Can you re-phrase your query? The title says cathode, the text says anode! What is your exact question?

In solution or in vacuum (i.e. not in the wire), cations travel to the cathode and anions travel to the anode. Electrons in the vacuum are like anions (negatively charged), so they are emitted from the cathode and travel to the anode. However, in the wire, electrons travel from the anode back to the cathode.
The definitions are similar for a cathode ray tube (old style TV or oscilloscope). Here, the electrons creating the image are generated in the glowing cathode. They never make it to the anode but fly through it to reach the screen.

https://upload.wikimedia.org/wikipedia/commons/b/b0/Cathode_ray_Tube.PNG

In BOTH electrochemistry and X-ray tube context:

The electrode that "looses electrons" toward the external circuit is called the anode.
The electrode that "looses electrons" toward the system (an electrolytic cell or an X-ray tube) is called the cathode.

The point is that the direction of "electron loosing" is opposite in these two cases.

There is one exception in cathode-anode electrode marking, taken rather from the practical engineering point of view than theoretical electrochemical one.
For the case of rechargeable galvanic cells ( like Li-Ion, Ni-Cd/MH, lead acid etc.), the cathode/anode is permanently assigned to the electrode being the cathode/anode at the galvanic stage - discharging. This was probably chosen to avoid confusion, because from the electrochemical point of view, anode-cathode marking mutually switch the place and relative potentials when switching between discharging and charging.
By other words, cell researchers and developers approach secondary cells -- in the electrode naming context -- like if they were just primary cells and if recharge ability was just secondary detail. I.e. cathodes are for them the more positive electrodes.
Therefore, in the Li-Ion cells engineering context, the graphite-lithium intercalate electrode is always the anode and the mixed-oxide electrode is always the cathode, regardless of fact that it is the opposite during charging in the electrochemical context.

Poutnik, The marking on rechargeable batteries is just a commercial practice (for the everyday user) not an electrochemical convention. Imagine, it took 150 years for electrochemists to sort this out what will happen to the common user. This is why manufacturers will also color code the wires. Scientifically, the fixed sign is wrong. Electrochemically, the labels must change during use and recharge as you stated.

@M.Farooq  Not overlapping domains feels fine with not compatible own term definitions. I have explicitly commented it is engineering  point of view, so saying it is wrong is - wrong, as it is not using electrochemical definition. It is not commercial practice for everyday user either - they do not care. It is for cell developers and researchers. There regularly refer to electrodes in their papers as the anode/cathode without further explanation what they mean by it, because it is clear to them and to the targetted readers.


text:
Does the molecule LiF3 exist?
Can LiF3 exist? (1 lithium and 3 fluorines), and if so, what would the Lewis dot diagram look like?

Related: https://chemistry.stackexchange.com/questions/27920/why-is-nacl3-possible

https://chemistry.stackexchange.com/questions/7800/can-fluorine-act-as-the-central-atom-in-interhalogen-compounds

Lithium with an oxidation state of +3 would seem beyond the pale, but all the halogens are known to form trihalide ions. Thus in principle one could imagine a salt, $\ce{Li^+(F3^-)}$.
However, trihalide ions are generally not stable in salts except with a bulky, essentially nonpolarizing cation or at GPa-level pressures; even caesium does not impart a perfectly symmetrical triiodide ion structure. Therefore, the proposed lithium trifluoride, with a small and relatively polarizing cation, would be expected to decompose to $\ce{LiF + F2}$.

Perhaps at high pressures the trifluoride could be a stable compound, however there are certain technical difficulties involved in such an investigation.

Agreed. I chose not to open that can of worms, partly because at least with caesium other (more fascinating) factors enter the picture. See Zhu et al..

I wonder, if a Lewis adduct of LiF and F2 would be possible in gas phase. AuF5 is strong enough to make it, so maybe monomeric LiF also is...

@mithoron I would not bet on it, AuF5 is a uniquely strong Lewis acid. F2 coordination seems not to be reported with other Lewis acids such as SbF5 or BF3.


text:
The higher energy to transform Cu(s)/Cu^2+(aq) is not balanced by its hydration enthalpy
The book says that "the higher energy to transform Cu(s)/Cu^2+(aq) is not balanced by its hydration enthalpy". I also read that when we add energy of sublimation and ionization energy , their sum is not compensated by the hydration enthalpy.

are we adding both the ionization energies or just one?

2)and how is heat of sublimation + heat of ionization = heat of hydration

and if the (2) point is true then V,Cr,Ni they all should follow the statement that Cu is following about its energy being not compensated

If i am all wrong about the concepts, please make me understand the statement and why other elements that i have mentioned are not having the same uncompensated total energy by hydration energy.
Thanks.

Both ionization enthalpies are needed.

$\Delta H_\mathrm{atom} + \Delta H_\mathrm{ion} + \Delta H_\mathrm{hydr} \ne  0$.
The process is either exothermic either endothermic. It can be about energy neutral just incidentally.

As Q2 premise is false .....

The whole formal thought process is this:
$$\ce{Cu(s) ->[$\Delta H_\mathrm{atom}$] Cu(g) ->[$\Delta H_\mathrm{ion}$] Cu^2+(g) ->[$\Delta H_\mathrm{hydr}$] Cu^2+(aq)}$$


text:
Can addition of a "neutral" salt like NaCl assist pH test strip measurement accuracy in low ionic strength/poorly buffered solutions?
After a web-search, it seems ISAs (Ionic Strength Adjusters) are offered commercially to assist pH electrode/probe measurements. Presumably meant to increase conductivity (without affecting pH significantly) to make a reading more stable/accurate. Commonly KCl, presumably since it's present already for electrode fill/storage.
Would increasing the ionic strength, of say tap water, with KCl or NaCl also help the rather different mechanism of measurement by pH test strip color change be more accurate?
(Again from web-searching, see there are "low-ion" pH test strips offered for sale, but cannot tell from the description how they work - except possibly they may have to be read while submerged into a provided solution.)


text:
Spilled denat. alcohol (5% methanol)
I spilled a small amount of denatured alcohol (~3ml) on my skin and table while prepping microscope slides for springtails. The bottle says its 95% ethanol, with unspecified amounts of isopropanol and methanol. Wiped it up immediately with a paper towel, washed my hands with soap and water, scrubbed the table with water and wiped it down again. I'm a little bit worried, first about skin contact, and second about fumes. I had been using the denatured alcohol in small amounts for about an hour previously while preparing slides (small drops pipetted onto glass). I didn't realize it had methanol until after the fact, if I had known I would not have used it or at the very least been more careful (and perhaps not leaned over the microscope with my nose uncomfortably close to the slide...)
I called poison control and they said it should be fine but the MSDS indicates you should seek immediate medical attention so I'm not sure.

The dose makes the poison and your dose was very small.

I wonder what the poison control center is good for. Your behavior was unnecessary - or perhaps necessary due to lack of knowledge -  and they should know that.

Did you compiled the MSDS of the chemicals to be used before working with the chemicals in question?  Did you asked your colleague/supervisor ahead of time about the chemicals to learn about the potential hazard and to set this into context (quantity of the chemicals, possible pathways of intoxication [skin, inhalation, ...] and how to limit the exposure)? When in doubt, ask and get familiar ahead of time.

the antidote to methanol is ethanol. And the risk from methanol in this mixture would be very small anyway unless you drank it.

According to researchgate.net  Legal-limits-for-methanol-contents(fast search, not pretending to be an authoritative source), alcoholic drinks have the methanol content limit 7 g / L of 100% ethanol ( US/AU/NZ). For EU it is 2-13.5 g/L of 100% ethanol, depending on the drink class.
IF we take the US value 7 g/L, it is close to 1% of methanol in ethanol.
Fear of having skin contaminated by 3 mL of 5% methanol is not justified. It could have methanol content equivalent roughly to some 40 mL of low quality 40% spirit and you have not even drunk it.
Remember the quote "The dose makes the poison.". Your dose was very small.
Note that ethanol denaturation additives are unbearable or dangerous at intentional drinking of such ethanol, but safe at accidental skin contamination.

The table is form http://dx.doi.org/10.3390/pr8121609, which seems to be a peer-reviewed journal. So while researchgate in itself might not be an authoritative source, the journal is likely to be that. Also, honey spirits do not have a limit in the EU. Also: Why is Denaturated alcohol made poisonous?


text:
Toughening of PMMA
For an experimental design with UV-activation I have moulded some UV-transparant (from ~250 nm up) PMMA sheets that I then cut and drill to my final mould. As expected, the PMMA sheets fracture easily and can only used for several experiments. To overcome this I would like to toughen the PMMA while maintaining its UV-transparancy. What would be a good additive for this?

Are you absolutely restricted to PMMA?


text:
Is configuration interaction able to produce degenerate ground states?
I am new to multi-determinant theories. What motivates me in asking this question is the following observation: Consider the Dyson orbital of a $\ce{CO2}$ molecule whose axis is parallel to the $z$ axis defined as
$$
\textrm{Dy}(x)
  = \int \mathrm{d}x_1 \ldots \mathrm{d}x_{N-1} \,
    (\Psi^+(x_1,\ldots,x_{N-1}))^* \, \Psi^0(x_1,\ldots,x_{N-1}, x)
$$
where $x$ is space-spin coordinate, $\Psi^0$ and $\Psi^+$ are the total wavefunctions of the neutral and cation, respectively.
Now, if you calculate the neutral and cation wave functions using Hartree-Fock method and then use them in the equation above, you will get one Dyson orbital with mostly $\pi$-symmetry such that its lobes are oriented in a certain direction, e.g. in $x$ direction. But there is a clear imbalance, so to speak, here, isn't there. The formula of the Dyson orbital above does not have any spatial preference and the total densities of both $\ce{CO2}$ and $\ce{CO2^+}$ are axially symmetric around the molecules' axis, so why would their Dyson orbital necessarily have a spatial preference?
I am thinking it might be due to the fact that I am using just one determinant to represent the wave functions. If this were the true reason, then would using multi-determinant methods can give me $m$-fold degenerate GS (especially for $\ce{CO2^+}$) and hence there will also be $m$ Dyson orbitals?

I have no answer for the Dyson orbital part as I couldn't find an openly available, easy to understand definition of it. So it might very well be possible, that they are ill-defined for systems with degenerate ground states, they are approximations themselves, interpretation tools, etc. That would require study of the theory. What I can tell you is, that by its definition, full CI will produce the exact expectation energy in the given basis set. In a complete basis, the wave function will also be exact. That implies that theoretically CI can reproduce a degenerate ground state. [...]

So technically you can produce two wave functions and plug them into the formalism you have written. The orbital you obtain will be as meaningless as the orbitals you have plugged in though. In the CI formalism, orbitals are not optimised, they are simply the basis. There are better techniques to describe the multi-reference character of molecules than CI; have a look at CASSCF and RASSCF. However even with those orbitals remain a mathematical approach, and since $x$ is a space-spin coordinate, you'll always end up with one Dy(x). From that I'd expect it's more an illustrative approach.

What about truncated CI, e.g. CISD, is it known to be able to produce degenerate GS?

I am not a pure chemist people either (my background is physics), so I don't have any clue how the concept of Dyson orbital is perceived within chemical community. But in atomic and molecular physics, $Dy(x)$ usually appears in the consideration of ionization process. As you may see from the equation above, that the hand-waving meaning of $Dy(x)$ is that it represents the difference between neutral and cation GS's.

By the way, I don't agree that Dyson orbital is meaningless as it is unique, unlike ordinary orbitals. To see why this is so, imagine when one hypothetically has the exact wavefunctions for neutral and cation. The equation above clearly guarantees that Dyson orbital so calculated will be unique.

@Martin-マーチン It is better not to interpret Dyson orbitals in the same way like molecular  orbitals that form a basis set. Dyson orbitals normally occur when modeling photo-ionization processes. It can be interpreted as describing an orbital from which an electron is emitted upon ionization. The overlap of the Dyson orbital with scattering states contributes to the ionization probability of a photo-ionization process. I don't know if this is the background here but that is one application where Dyson orbitals are used. The calculation of the Dyson orbital happens afterwards, not during the CI.

@HansWurst, exactly. In my research, I will come to a point where I need to be able to compute the Dyson orbital, which is then to be plugged into a sort of overlap integral. The integral is roughly an overlap between Dyson orbital with a function that is somehow related to scattering states. The ionization probability is then proportional to this overlap integral.


text:
How to relate equilibrium constant and equilibrium conversion?
For the following gas phase reaction:
$$ A\leftrightarrow B $$
The concentration equilibrium constant ($K_c$) can be expressed as:
$$ K_c = \frac{c_B}{c_A} = \frac{\frac{F_B}{\vartheta}}{\frac{F_A}{\vartheta}} = \frac{F_{A0}X_{eq}}{F_{A0}(1-X_{eq})} = \frac{X_{eq}}{1-X_{eq}} $$
Where $c_i$ is the concentration and $F_i$ is the molar flow rate of component i ($F_{A0}$ is the initial molar flow rate of A, assuming no B is present initially), $\vartheta$ is the volumetric flow rate, and $X_{eq}$ is the equilibrium conversion of the reaction.
Following the same methodology for the following gas-phase reaction:
$$ A \leftrightarrow 2B $$
$$ K_c = \frac{c_B^2}{c_A} = \frac{(\frac{F_B}{\vartheta})^2}{\frac{F_A}{\vartheta}} = \frac{(2F_{A0}X_{eq})^2}{\vartheta F_{A0}(1-X_{eq})} = \frac{4F_{A0}X_{eq}^2}{\vartheta (1-X_{eq})} $$
Would this be correct?

Well, the second reaction is most likely not chemically possible, but if it was, you'd be correct.
One nice thing is you can phrase the second using the first.
$K_{c,2}=\frac{4F_{A0}X_{eq}^2}{\vartheta (1-X_{eq})}=\frac{4F_{A0}}{\vartheta}4K_{c,1}$

A and B aren't meant to be the same things in the two reactions.

If so, then the calculations seem correct.


text:
Inhibitors of hydration
Superglue (Cyanoacrylate) polymerizes via hydration, and using an activator such as a borax solution speeds up that process. I need to mix these 2 without them reacting. (Or at least not noticeably.)
So, I was wondering if there are any inhibitors that can stop this reaction at least temporarily. I'm not well versed in chemistry, so if I got a concept or anything wrong with my question, please tell me. Thanks.

Cyanacrylate polymerisation is initiated by single water molecules. No "hydration" takes place whatsoever.

Googling "borax superglue" does not lead to any sensible results. Borax certainly does not act as an activator there.

Inhibitors for cyanacrylate polymerisation are drying agents, or substances that catch the active ionic end group of the growing polymeric anion and transform it into something that doesn't (or only much slower) react with cyanacrylate.

There are inhibitors and "stabilizers" such as acids. You might experiment with acetic acid (of course, "glacial", not as a water solution), or $\ce{FeCl3}$, which would make the polymer dark or opaque. Another inhibitor is pyrogallol and boric acid, which likely would be compatible borates.
The reference above also cautions that an excess of inhibitor can prevent polymerization, so test various inhibitors and various concentrations to see what works for your intended use.
If you do find a good inhibitor, to help others, please add an answer here or at least a comment on what worked in that application.


text:
Why is 1,3-dimethyl-cyclohexane a meso compound?

So i was trying to find the plane of symmetry in 1,3-dimethyl-cyclohexane in chair conformation, but then i realised the plane is present only at a particular orientation [see figure]. So why do we consider this to be a meso compound?

Main point: symmetry of a molecule does not depend on the orientation. Practical: one orientation / sketch suffices. Draw the axis / plane and then let it move together with the molecule...

It does depend on the conformation, though.

You need to learn how to properly draw substituents on cyclohexane. See Drawing Cyclohexane.

1,3-dimethylcyclohexane is not exactly a compound, but three different compounds, depending on the configuration of the chiral centers. Two of them (RR and SS) are enantiomers of each other, and the third one (RS) is meso.
Why meso?
Well, because it has no enantiomeric forms which couldn't be transformed to each other by single-bond rotations. That's what is important, and the symmetry plane is just one possible evidence of this situation. It is sufficient, but not necessary.
We can put it differently: you seem to know full well that all conformations are in fact one compound, since you can't separate them. Also, at least one of the conformations is surely a meso compound, since it has a mirror plane. Then what about any other conformation, possibly having no mirror plane? Is it a different compound? No.

On a side note, mirror plane is not necessary at all, and Wikipedia gives us an example:

A meso isomer need not have a mirror plane. It may have an inversion or a rotoreflexion symmetry such as S4. For example, there are two meso isomers of 1,4-difluoro-2,5-dichlorocyclohexane but neither has a mirror plane, and there are two meso isomers of 1,2,3,4-tetrafluorospiropentane (see figure). In fact, a meso compound may have no symmetry in some conformations...


text:
What material properties are unpredictable (not reproducible) but can be measured consistently and at a low cost?
I'm looking for a process to create a material which has some easy to measure properties.
These properties should be consistent over a long period.
It should be very hard (expensive) to predict/reproduce a material which results in the same measurable properties. Every material product should generate an unique measurement result and impossible (or very expensive) to product a second product with the same measurement result.
What process, material and/or measurement could be used?
Some context:
The goal would be block-chain backed physical cash currency.
The process should result in a 'coin' (material product).
The 'coin' can be 'read' (measurement of some over time consistent material properties) resulting in a 'coin-id' (measurement).
The producer of the 'coin' reads the coin-id and adds a block to the chain which contains the 'coin-id' and his signature of the 'coin-id' and spends some (crypto) currency value. The physical coin represents the spend (crypto) currency value. For ease of use the signature could be attached to the coin (bar or qr-code).
The blockchain also contains the certificates of the coin producers.
The coins can be exchanged in the physical world without changes on the blockchain.
A receiver of a coin can read (measure) the unique coin-id and scan the signature.
A receiver knows the certificates of the coin producers. A receiver can check the veracity by checking if the coin-id is signed by a known coin producer certificate.
Addendum
Let the cost to produce a coin be c.
Let the probability of a duplicate measurement be p.
Let the value represented by the coin be v.
Because the represented value should be a lot bigger than the cost of a coin. The value should be a factor f bigger than the cost.
$v=cf$
The minimal value to make forgery too expensive:
$v=\frac{c}{p}$
The maximal p should be:
$p=\frac{c}{v}$
So given a $c=0.1$\$ and $v=100$\$
then the maximum $p=\frac{0.1}{100}=0.001$
Or put otherway around: given a more realistic forgery probability of $p=10^{-10}$
Maximum $f=\frac{1}{p}$, $f=10^{10}$
So a coin given $c=0.1$\$ could have a maximum value of $v=0.1*10^{10}=1.000.000.000$\$

Such material could be reverse engineered and reproduced.

@Poutnik Entropy can be stored in the form of information, although it has none.

@Karl It can be, of course. But it would be more about material arrangement than material itself. The "coin" could contain some hologram with blockchain code or some public encryption key. The coin itself can be a common coinage alloy.

@Poutnik Of course it could be reverse engineered. Likewise, md5 can be broken, given enough effort. The question (as always) is "Where is the money". If breaking a thing costs too much, it is effectively unbreakable.

@IvanNeretin I have meant it in context of a single, not structured, reproducibly managed material, as I have not at the time thought about heterogenous structured options.

A good red wine. Everyone knows that it is good, but it is impossible to make another bottle just like it. Measuring would be easy - just send me the bottle.

Process to create: sexual reproduction;  resulting material: fingerprints of the offspring (or retina patterns).

@KarstenTheis To get rich quick, I recommend the wine here. Cheers, and merry Christmas etc.!

Impossible for a homogeneous material, however ...
basically every nonhomogeneous material fits your description. Say the pattern of microphase separation in a copolymer. Or the arrangement of filler particles in a composite.
You just have to think of something that is not only easy to make, but also easy to measure, i.e. take a digital photo of (microscopes and MRT or µCT scanners are out, I guess):
Pour resin of say four colours into a round bin, so you have four differently-coloured pie pieces. Now take a fork and run it through the bin a few times, perhaps in a "random" fashion (rotation, speed, direction, duration).
Let resin set, and you get a disk ("coin") with a pattern that is impossible to regenerate, even if someone stole the random number, which you delete immediately after use, that initialised the fork movement. They'd get something that looks somewhat similar, but clearly distinguishable. If they tried ten thousand times, they might get one or two that look similar enough to fool your algorithm, but that forgery is uneconomic, so you're safe.
Of course someone could forge those coins by printing a photographed pattern onto a coin, just like you can photocopy a dollar note.
The bigger problem I see is to make sure the "measurement" never creates a false negative (negative==forged) outcome, and that it doesn't require you to save a multi-megabyte dataset for every single coin.

I thought about the pattern of crystals formed by cooling of a liquid metal (zinc?)

Good idea. They would keep the database of patterns of all released coins and any non-genuine coin would not match.

@IvanNeretin Yea, but that's not very persistant. Put ten coins in your wallet, and they will be non-recognisable after a week.

Not if you cover them with transparent plastic. That's better, but not very persistent either. But wait, I have an even better idea...

The resin coin meets the requirement for processing; cheap and impossible (or at least the expensive) to reproduce. The difficulty will indeed be the 'measurement' to read the unique coin value. Your points are valid, the more unique a description (measurement), the higher the cost of forgery, but also the larger the 'unique value'. But if a handheld device could reduce the information of the optical pattern using an cryptographic hash function, the storage would be less.

@Hubert A hash doesn't work,  where any tiny difference (a scratch or sth.) makes the comparison fail, and you don't know why. You want a compression algorithm for the actual pattern, so you can quantify the difference between the thing on the blockchain and the picture of the coin in your hand.

I believe the nuclear weapons monitor folk have used glitter in epoxy as parts of seals before (with photographs). You could combine that with illumina-style coloured-dot imaging.

Like your answer but concerned about robustness of the material. As such, I provided my alternate suggestion.

The OP asks: What process, material and/or measurement could be used?
Answer: Snowflake maker, dihydrogen monoxide, photograph (with metric ruler included). "Snowflake maker" is a vapor condenser from steam onto a cooler surface or simply from the vapor.

It is well-known that no two snowflakes are alike. The unscientific media (Ref 1) propose the opposite, but the scientific consensus is that the probability of finding two identical snowflakes is zero (Ref 2).
An advantage of Snowflake Bitcoin is that it requires a physical repository (a bank, but not a snowbank) with a temperature low enough to prohibit melting and sublimation (which would allow the snowflakes to grow or diminish). Antarctica might not be cold enough; the actual snowflakes might have to be stored in outer space, but that cost could be prohibitive. An alternate bank could be physical storage of the actual photographs, or digital images with a very high resolution.
Ref 1. https://www.nbcnews.com/id/wbna16759121
Ref 2. https://www.loc.gov/everyday-mysteries/meteorology-climatology/item/is-it-true-that-no-two-snow-crystals-are-alike/

Very seasonal, but the idea was that you cannot copy them, that excludes photographs, on paper or digital. And a currency that you cannot remove from the back does not need be represented as physical objects.

The photos are only the proof of ownership; the snowflakes are the product and cannot be duplicated, but are stored in the (very cold) bank. The snowflakes can be stored at great expense - for a time. The measurements are unique and irreproducible; the metric ruler can be augmented with another at 90 degrees, and  various angles can be specified, as well as opacities and reflectance. Trust in the bank grows over time; the numbered photos serve as the substitute coin and are traded. Like serially numbered dollar bills vs gold in Fort Knox. No counterfeiting possible!

Serially numbered bills have been counterfeited for centuries. If the copy is perfect, there is no way of knowing if a specific note is original. Even if you have a second one with the same number, you only know that you have a problem, but no solution.  A picture that has no representative in the bank would obviously be a fake, but for that you can also store a copy of the photo, or a hash of the digital picture. No need to keep the snowflake.

Gotta combine the snowflakes with NFTs!

The cold temperature would make handeling and exchanging the coin difficult. Is there an other crystallisation process that results in room temperature crystall-flakes?

@Hubert: Very good idea! I hadn't thought of materials other than ice/snow because there have been no claims of non-similarity of crystals. Many organic materials sublime, so if you catch them early, there will be many small crystallites forming. I have done this with dibenzofuran mp ~82C. Some materials sublime, but appear to wet the collector (or spread so thin, which is similar) and simply grow into a lump. But maybe there is more similarity between crystals because of higher temperatures - closer to some equilibrium.

As to what process, material and/or measurement could be used, the same that exists for minting metal alloy coins.
The material could be composed of said four (or more) alloys with the coin displaying an ID. The latter can be a mathematical function of associated chemical and electrical properties of said coin.
For example, an alloy of Ni, Sn, Fe and Zn may work based on the there actual (or predicted) respective variations in an anodic index (see this Table).
So, a simple chemical (like density) and electrochemical measurements (electrical resistance) may provide in conjunction with engraved ID an authenticity check.
This concept differs from government issued coins in the respect they are crafted for corrosion resistance (to increase circulation life span) while these coins are minted for identification purposes primarily,
Coins are known historically to be robust instrument for currency, and said coins can also be made completely virtual by mathematical fitting algorithms (applied to a provided data string) based on actual coins with specified varying compositions, from which the ID is generated.
Note: this answer complies with the need to be measured consistently over time at a low cost.

Very clever to ingrave the ID so it will be a part of the material properties.

What now stops me from reproducing a single coin with the same ID and alloy composition? Also highly accurate density measurements are expensive, and accurately measuring the electrical resistance of a coin is basically impossible.

Karl: Alas, experimenting with actual produced coins will likely suggests needed specifics. By the way, how does a vending machine (accepting coins for payment) actually determine a fake coin? Idea: Take advantage of existing technology!


text:
Why do CCl4 and CH4 have the same bond angle?
I was reading about the molecular shape of compounds. I learned that the electronegativity of the central atom and the terminal atom in a molecule both play a role in determining bond angle. In $\ce{NH3}$ and $\ce{NF3}$, $\ce{F}$ having higher electronegativity than $\ce{H}$, $\ce{NF3}$ has a smaller bond angle compared to $\ce{NH3}$.
Applying the same logic, it was expected that $\ce{CCl4}$ would have a smaller bond angle than that of $\ce{CH4}$. Surprisingly, I found that both of them have the same bond angle. Why is this the case?

A molecule is very much a 3-dimensional thing. Unless you have its model before your eyes (maybe only as a mental image, but it has to be a good one), you totally can't move further.

You have 4 atoms of one element, singly bonded to carbon, so they get, on average, as far apart as possible, which means tetrahedrally around the central carbon: https://chemistry.stackexchange.com/a/118202/79678.

A tetrahedron is a tetrahedron no matter how small - with apologies to Dr. Seuss…

@melanieshebel When applying MathJax to formulas, why not to use mhchem \ce{}? It makes generally cleaner MJ code and keeps symbols upright as expected.

The molecules $\ce{CCl4}$ or $\ce{CH4}$ contain six angles $\ce{Cl-C-Cl}$ or six $\ce{H-C-H}$. Why should one of these angles be different from any other one ?

The OP puts more emphasis about the similarity of an optimal arrangement either four chlorine, or four hydrogen atoms around a carbon atom rather than if these four (within a set) all arrange the same way.

In $\ce{NH3}$ and $\ce{NF3}$, $\ce{F}$ having higher electronegativity than $\ce{H}$, $\ce{NF3}$ has a smaller bond angle compared to $\ce{NH3}$.

Both of these compounds have a lone pair on the central atom. So the bound electrons and the lone pair (if you are using the simple "electrons pair up" model) compete for space.

Applying the same logic, it was expected that CCl4 would have a smaller bond angle than that of CH4.

All electrons around carbon are involved in bonding, so all four pairs are the same. To apply the electronegativity argument, you should compare the distinct bond angles in $\ce{CH2F2}$ or in $\ce{CH2Cl2}$.

Welcome to Stack exchange chemistry.
Consider for a moment what is known as the isolobal concept, there are a series of atoms and groups which all present the same types of orbitals (or at least close to identical orbitals) and the number of electrons.
Consider for a moment a methane molecule, if we were to break a C-H bond then the carbon atom would only have seven valance electrons. The carbon in an alkane such as methane has rehybridized its orbitals to give us four sp^3 orbitals. These are arranged in a tetrahedron around the carbon.
A covalent bond is formed by sharing the one electron in the sp3 orbital of the carbon with an atomic orbital from another atom that has the right geometry to overlap with the sp3 orbital. The sp3 orbital has two pear-shaped lobes, one is large and one is small. These have opposite signs of the wavefunction.
A hydrogen atom in the ground state has a single electron in an s orbital, this is a sphere-shaped orbital that can interact with the sp3 orbital to form both a bonding and an antibonding orbital. We will only concentrate in this answer on the bonding orbitals.
The sphere-shaped s orbital has the right geometry to interact with the sp3 orbital and it can result in the formation of an occupied (2 electrons in it) bonding orbital between the carbon and the hydrogen. This will be a sigma bond (single bond)
If we change to chlorine, then the outermost orbital (for the valence electrons) of the atom has also rehybridized to give us four sp3 orbitals. Three of these are occupied with two electrons while one in an isolated chlorine atom only has one. The orbital with only one electron can interact with the sp3 orbital on the carbon (bearing only one electron) to form two new molecular orbitals. One is antibonding and one is bonding.
If the bonding orbital between the carbon and the chlorine is occupied with two electrons then we have a bond. The C-Cl and C-H bonds will be different in length. But the angle between them will be dictated by the arrangement of the sp3 orbitals around the carbon atom.
If you still do not understand it then I would suggest that you fall back to VSEPR theory. As it is the festive season go and grab an orange and four cocktail sticks. Stab them into the orange in such a way that they are the greatest angle apart. You should find that the tips of them form a triangle-based pyramid (tetrahedron). It will not matter if you put grapes on the points of the cocktail sticks to represent hydrogen atoms in the methane. Or apples to represent the chlorine atoms in carbon tetrachloride. You will still have the same arrangement of the atoms in your model.
You can then hang it on the tree as a decoration or pull it apart and eat the fruits. When I can not lay my hands on my molecule modeling kit made of plastic balls and straws I tend to grab oranges and then draw atoms with a marker pen on the skin.

you have tried to explain it with M.O theory.But my ques. wasn't based on that. It was actually based on the e.n of the terminal atoms of the two molecules which play a significant role in bond angle.And you didn't even raise a word on that.

The electronegativity does not matter: see my first comment and the linked tetrahedral figure. Ammonia is polar because it has three hydrogens attached and a lone pair of electrons. Nitrogen trifluoride is also polar and for the same reason. The electronegativity differences determine the respective dipole moments. But with 4 H atoms around the central carbon, they are, on average, as far apart as possible, hence the tetrahedral geometry, and methane is non-polar. Same principle for carbon tetrachloride. Simple as that.

The electronegativity is not relevent to the shape of the molecule, I went through two methods of predicting the shape of the molecule.


text:
Oxidation of sulfuric acid to Marshall's acid and its half cell reaction

$\ce{H2O2}$ can be prepared by successive reactions:
$\ce{2NH4HSO4 -> H2 + (NH4)2S2O8}$
$\ce{(NH4)2S2O8 + 2H2O -> 2NH4HSO4 + H2O2}$
The first reaction is an electrolytic reaction and second is steam
distillation. What amount of current would have to be used in first
reaction to produce enough intermediate to yield $\pu{102g}$ pure
$\ce{H2O2}$ per hour. Assume efficiency 50%.

I was solving this question on Faraday's Laws Of Electrolysis when I stumbled across a conceptual flaw of mine. I realized that the sulfate anion in the first reaction is at a +6 Oxidation State and so is Marshall's acid, as far as what I know of Redox Reactions and their balancing we look at the number of electrons exchanged and thus formulate the half-cell reaction.
However, this logic of mine failed in the above question as the oxidation state of the central atom is unchanged, which makes me wonder how to calculate the valency-factor/N-factor and correspondingly the equivalent weight. I know this is a conceptual shortcoming of mine and that the aforementioned logic is very 'methodical' per se, which is why it fails.
If someone could please point out where I am going wrong it'd be highly appreciated.

It is oxidation state of oxygen, not sulphur, that is changing.

Using image of text is highly discouraged. Please, type the text instead of pasting a photo now onwards. Images cannot be properly indexed by the search engines.

Forget about the equivalent weights. Work with moles, and only with moles. Look how it goes.
$102$ g $\ce{H2O2}$ is $\ce{\frac{102 g}{34 g/mol} = 3.00 mol H2O2}$. The production of $1$ mol $\ce{H2O2}$ requires $1$ mol $\ce{(NH4)2S2O8}$. Now we will show that the production of $1$ mol $\ce{(NH4)2S2O8}$ requires $2$ moles electrons. Before doing this, we will first show that the ion $\ce{HSO4-}$ from $\ce{NH4HSO4}$ is at least partly decomposed into the following two ions according to the following equation :  $$\ce{HSO4^- <=> H+ + SO4^{2-}}$$ One of these ions ($\ce{H+}$) is reduced in $\ce{H2}$ at the cathode, and the other one ($\ce{SO4^{2-}}$) is oxidized at the anode according to : $$\ce{2H+ + 2 e- -> H2}$$ $$\ce{2 SO4^{2-} -> S2O8^{2-} + 2 e-}$$ This shows that $2$ electrons are needed to produce $1$ mole $\ce{(NH4)2S2O8}$, and later on $1$ mole $\ce{H2O2}$.
Then Faraday law gives you the intensity I needed to produce $3.00$ mol $\ce{(NH4)2S2O8}$ in $1$ hour = $3600$ s.
It is :
$$\pu{I = \frac{3.00 ~mol~·~2~·~96500~ As/mol}{3600~ s} = 160 A}$$
This value is obtained if the yield is $100$%. As the yield is $50$%, the intensity must be twice the previous value. This is $320$ A.

In the event that this question is actually more than just a theoretical exercise, the general electrolysis half-reactions, as cited by Maurice in my opinion, may not actually be in accord with this particular's complex reaction system per a review of the possible underlying mechanics.
More precisely, here is a suggested overview of reaction mechanics that support my comment:

$\ce{H2O = H+ + OH-}$

$\ce{Electrolysis of HSO4- => .H + .SO4-}$

$\ce{NH4+ = H+ + NH3}$

$\ce{.H + .H = H2}$

$\ce{.H + .SO4- = HSO4-}$

$\ce{.SO4- + .SO4- = S2O8^{2-} }$

$\ce{ 2 NH4+ + S2O8^{2-} = (NH4)2S2O8 }$

Also, a slow reaction, introducing a powerful radical:

$\ce{.SO4- + H2O = .OH + H+ + SO4^{2-} }$

$\ce{HSO4- +.OH = H2O + .SO4- }$

So, while one may claim seemingly that only that 2 electrons are required, my analysis as outlined above which is subject to kinetics, suggests some possible reversed reactions resulting in reduced efficiency.
As a result, not surprisingly, more than 2 electrons may actually be required to produce the single mole of $\ce{(NH4)2S2O8}$ in an experiment, and as such, I would recommend qualifying Maurice analysis with the words "at least", if one is writing up this experiment to account for observed results. More interesting is the provided statement "Assume efficiency 50%", which appears supportive of my take on the reaction system.

Hmm intensive anodic oxidation at high current density is not friendly for producing hydrogen radicals.

Albeit, the somehow formed sulfate radical anion, per this source Eq (8) at  https://acp.copernicus.org/articles/18/2809/2018/acp-18-2809-2018.pdf is the cited path to $\ce{S2O8^{2-}}$  via a self reaction with the sulfate radical anion. Interesting, the slow action of H2O on $\ce{.SO4^{-}}$ per Eq (7) does introduce the $\ce{.OH}$ radical. Also, I am unclear if any formed NH3 is converted into the $\ce{.NH2}$ radical.

OK, I have edited my thread to correct for the formation of the sulfate radical. Per the electrolysis of HSO4- to H+ + e- + .SO4- with the implication in concentrated solutions the formation of S2O8(2-) as confirmed in this reference https://www.toppr.com/ask/en-us/question/during-an-electrolysis-of-conc-h2so4-perdisulphuric-acid-h2s2-o8-and-o2-form-in-equimolar/ .


text:
Lennard-Jones Potential repulsion by nucleus nucleus or Pauli repulsion?
I read different things, some textbooks say that the repulsive part of the lennard jones potential is due to the repulsion of the nuclei when they are too close, others say it is due to the pauli repulsion between the electrons.
Or can I imagine that the pauli repulsion creates an antibonding sigma orbital, which means that the electron density between the nuclei is no longer so high and therefore the nuclei repel each other?

Some say that new kids are found in cabbage patches, and some say they are brought by a stork. Then you grow up and have your first course in Quantum Mechanics, and things come out in a very, very different light.

@IvanNeretin So both approaches are permissible? Electron repulsion by pauli repolsion or/(and?) repulsion between the nuclei?

Electrons repel each other by their charge, too .

If you learn nothing else from this post and ensuing discussion, learn this: it's one guy - John Lennard-Jones - hence the hyphen.

it isn't even clear clear that the best way to think about Pauli "repulsion" is as a force.

The short answer is: the Lennard-Jones potential is a completely heuristic model, and the short-range part of the potential is not physically meaningful except in the sense that it becomes very repulsive very quickly. Pauli repulsion is known, also heuristically, to decay as an exponential, so using a very steep polynomial works reasonably well. As to whether it is modelling the nuclear repulsion or the Pauli repulsion, the answer is both. Also, these two effects are closely related when discussing intermolecular interactions because exchange has the effect of de-shielding electrons and hence enhancing nuclear repulsion.

To be more explicit, the Lennard-Jones potential,
$$
V(r)=4\epsilon\left[\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^{6}\right]
$$
aimed to provide a simple functional form that reproduced the interactions between neutral gas molecules. In short, any potential that is repulsive at short-range, has a minimum at some intermediate distance, and goes to a constant as $r\rightarrow\infty$ is a heuristic model for the medium-range to long-range interactions between nonpolar molecules. What I mean by medium-range is something like twice the diameter of the molecule in question.
The special thing that the Lennard-Jones potential does is get the correct asymptotic form of the attractive interaction. Namely, it is known from the London Dispersion formula (and other ways) that  the dispersion interaction decays as $r^{-6}$. Not only that, but the induced dipole interaction between two nonpolar molecules also decays as $r^{-6}$.
So, this brings us to the short-range part of the potential. As you mention, there are two effects which result in short-range repulsion: nuclear-nuclear repulsion and so-called Pauli repulsion due to the electron exchange. Note that classically, the reason exchange results in repulsion is that it de-shields electrons in the internuclear region and hence results in increased nuclear repulsion.[1]
In fact, many theories predict that exchange repulsion decreases exponentially with distance, so using a polynomial distance dependence is somewhat arbitrary. Due to this, there are other potentials that choose to use an even steeper polynomial which goes as $r^{-14}$, which tends to give somewhat better results in simulations.
Now the question is why does the Lennard-Jones potential choose the specific polynomial of $r^{-12}$? The answer is rather funny. Back in the day when computers were very slow and floating point operations were very expensive, one needed to be very mindful of how expensive calculations were. Now, calculating exponents can be rather expensive. To calculate $(\sigma/r)^6$ requires a division and six multiplies. It would be a shame to have to do even more multiplication operations for the short-range part of the potential, which is the least physically meaningful part anyways. Hence, the exponent 12 was chosen because it is simply the product $(\sigma/r)^6\cdot (\sigma/r)^6$. Hence, one can save many floating point operations by doing this.
Note that this story about choosing the exponent 12 for computational speed is mentioned on the wikipedia page and I have been told this story by multiple scientists I trust, but I haven't looked closely enough at the original paper to know if this is really what Lennard-Jones had in mind.

References:
[1]: Rackers, J. A., & Ponder, J. W. (2019). Classical Pauli repulsion: An anisotropic, atomic multipole model. The Journal of chemical physics, 150(8), 084104.

"To calculate (σ/r)6 requires a division and six multiplies". No it doesn't: rsq = r * r; r4 = rsq * rsq; r6 = r4 * rsq; V = s/r6. In general any integer power takes approximately log to the base 2 of that power .multiplies. But your general point standsl


text:
How to prevent Sodium Hydroxide (Catalyst) from reacting with the Formic Acid (product) after electrolysis is done
So,
I am trying to create Formic Acid from PET and to break the ester bonds before electrolysis, I am planning on using water and sodium hydroxide. I was told that if I take this Plastic-Sodium Hydroxide-Water solution and electrolyze it with my gold diode, I should be able to get formic acid, as the Sodium Hydroxide is supposed to act as the catalyst. My only concern is that once the Formic Acid is formed, it would get neutralized by the Sodium Hydroxide catalyst. How do I prevent this?

You cannot prevent it in the described scenario.

It will also react with the terphthalic acid product

Yes, you may be able to reduce the lose due to a neutralization reaction with a carefully constructed electrolysis cell.
Try following, for example, the prototype provided in the electrolysis of aqueous $\ce{NaCl}$ as a source of $\ce{NaOH}$ along with $\ce{H2}$ and $\ce{Cl2}$ gas. In particular, per this educational reference, note the use of a diaphragm to limit product mixing.
To quote the source on actual products collected for commercial use:

The three products of the electrolysis of concentrated sodium chloride solution have important uses in the chemical industry:

hydrogen is used as a fuel and for making ammonia

chlorine is used to sterilise water supplies, and to make bleach and hydrochloric acid

sodium hydroxide is used to make soap, paper and bleach

Note, during and after the electrolysis, the diaphragm limits further unwanted reactions.
See if you can make a related design structure work in your case.


text:
Normalized absorbance/transmittance definition
In the absorption spectroscopy we can calculate transmittance $T$ of a given sample by comparing how the intensity of the incident beam $I_{0}$ is decreased with the distance (Lambert-Beer Law).
$A = -\ln{\frac{I}{I_{0}}}=\alpha L$
where $A$ is absorbence, $T = 1 -A$, $\alpha$ is an absorption coefficient and $L$ is the distance.
But I encountered something called Normalized transmittence I don't really understand how can I get it from measured data. Normalized how? It appears in scientific articles but I couldn't find any good source explaining it really.
I read that I should divide my measured $I$ by reference intensity (it can fluctuate so it is measured along the way too). So the normalized absorbance would be $A = -\ln{\frac{I}{I_{0}^2}}$ or just by definition $A = -\ln{\frac{I}{I_{0}}} $
Can anyone help me please?
The spectra then look like this:

[https://www.researchgate.net/figure/Hydrogen-cyanide-H-13-C-14-N-2-3-rotational-vibrational-band-spectrum-obtained-by_fig4_242279131]

This was posted by you and got an answer that you accepted: https://physics.stackexchange.com/q/685025/313612. Is there something that is specifically not being addressed?

I’m voting to close this question because it's a cross-posted question from https://physics.stackexchange.com/questions/685025

Is cross-posting a question on multiple Stack Exchange sites permitted if the question is on-topic for each site?

Recall the more familiar definition of light transmittance
$T = \frac{I}{I_o}$, many instruments quote percent transmittance as
$\%T = 100\frac{I}{I_o}$
Normalized transmittance is nothing but the maximum measured transmittance set to unity. This is easy to accomplish. Suppose you have 2000 data points (y-values) in your spectrum collected as a function of wavelength. Locate the maximum value of transmittance, say it was 95% or 0.95 in your data set (depending what was collected experimentally). Now divide the entire y-values with 95.65% or 0.9565. With that your maximum, transmittance will be exactly equal to 1  and the rest of the values will be less than 1. This process is called normalization.
It is just for convenience when you are comparing a lot of spectra.

Upvoted, but it is not clear why the OP first posted at physics SE, accepted an answer and then came here. Modern double beam-in-time instruments are rather good, so double beam-in-space spectrometers are not necessarily preferred for source noise performance.

@EdV, I think the OP was confused how to normalize the experimental data. I did not know it was posted twice.

Yes, that is what I figured. But the HCN IR spectrum was apparently used to calibrate a laser system, based on what I find when searching. In the old days, a pulsed laser might have been used to pump a tunable dye laser and then pulse-to-pulse fluctuations in the pump laser’s output, typically a few percent, had to be normalized out. Missing and runt pulses were especially annoying and experimental data had to be corrected for those. I used to use a window comparator to ensure collection of data from non-defective laser pulses.