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Electrochemical double layer formation
I was revisiting some of my notes about the electrochemical double layer but I have some doubts, and even the textbook of Bard and Faulkner could not really help.
In electrochemical systems, at the potential of zero charge (PZC) there is no charge on the electrode surface and therefore no electrochemical double layer as I understand it (it is the maximum of the following graph, potential vs surface tension):

Using the PZC now as our reference point where there is no double layer, if we shift the applied potential on the electrode to more positive values than the PZC (assume there is no faradaic reaction, only capacitive effects) of course, a double layer forms. Now my question is:
If the applied potential is more positive than the PZC, does that mean, that the electrode is positively polarized due to the applied potential and anions are moving closer to the electrode surface? Or if the applied potential is more negative than the PZC the electrode is negatively polarized and more cations are near the electrode, or is a metal electrode generally positively polarized?
So does the “sign” of the electrode’s charge have something to do with the potential that we apply, or is the EDL independent of the applied potential (should be Helmholz assumption)?

Yes, if you apply a negative potential compared to the PZC the electrode will carry excess negative charge, vice versa for a positive potential compared to PZC. More details can be found in Bard Chapter 13 ( this part on excess charge and PZC is in section 2.2). The excess charge on your electrode is definitely dependent on the applied potential, a basic theory hinting at this is already Gouy Chapman, now of course with specific adsorption and different metal facets this becomes a very complicated issue.

Okay I will take a look at chapter 13. I was thinking it should be at earlier chapters in the book because it is a very important topic!!

I can also give you the points for a bounty if you answer the question because it helped me a lot

I feel like a complete answer that's nuanced would take me a real while to write and the comment is not answer worthy, so I'll just be happy that you have the resources to answer your question now.

Ok thank you for the effort tho.


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What is meant by 'as Cl2'
I feel like I'm over thinking this one, but think I need a confirmation (and quantified example) to put my confusion to bed. I'm more a mechanical than chemical background but trying my best.
I really get stuck on papers measuring concentrations 'as XXX' when the 'XXX' is only a component of the compound.
Example 1: Hypochlorous Acid (HOCl) has a weight of 52.457 g/mol, while chlorine weighs 35.45 g/mol. Thus, the chlorine weight fraction would be ~67.5%. As this would be measured as Cl2 would we then halve the above weight fraction as we only have a single molecule of chlorine.
Example 2: If we add 1mol (143g) of Calcium Hypochlorite to make 1-litre of solution.

We would create 2mol (105g) of Hypochlorous Acid, which includes 2mol of Cl (aka 1mol of Cl2) weighing 70.9g. As we are interested in chlorine, would we express this as a measurement of HOCl 70.9g/l "as Cl2"? Or do we consider the entire mass of hypochlorous acid?

In general, sentences like "The concentration of X as Y is z mg/L" should be interpreted to mean "If you extract all the X from 1 L of solution, then convert the X completely to Y (throwing away any unused atoms), you will have z mg."
In example 2, you have a 2 M solution of hypochlorous acid. We want to know the concentration of hypochlorous acid as chlorine. This means we take all the hypochlorous acid (2 mol), then convert it to chlorine, throwing away the unneeded atoms. We end up with 70.9 g of chlorine so the concentration of hypochlorous acid as chlorine is 70.9 g/L. The fact that chlorine is $\ce{Cl2}$ is not relevant here; there's no need to divide by an additional 2.
It's probably worth mentioning why people use these units. There are several basically equivalent bleaches: $\ce{NaOCl}, \ce{HOCl}, \ce{Ca(OCl)2}$, etc., which can sometimes interconvert based on pH and other factors. A solution with 74.5 g of $\ce{NaOCl}$ has the same bleaching power as one with 52.5 g of $\ce{HOCl}$, but the distinct numbers hide the equivalence. Indeed, the chemical tests used to measure the bleaching power probably cannot even tell what form the chlorine is in. So people invented a measuring system where both systems have the same numeric value: 35.5 g of bleach "as chlorine." This emphasizes that the two systems behave similarly and lets a test report one unambiguous result.
They could also have said "0.5 molar equivalents of $\ce{Cl2}$", and I'm sure most chemists would find that much more intuitive these days. But the chemists don't always get their way.
EDIT: I hate to edit an accepted answer, but after consulting some references on chlorine specifically, I'm pretty sure that this answer, while correct for the vast majority of "X as Y," might actually be wrong for chlorine. The problem is that when you dissolve chlorine in water, only half the chlorine in $\ce{Cl2}$ is active as bleach, with the other half getting converted to inactive chloride. It appears that the convention does not "count" the inactive half of the chlorine and therefore 74.5 g of $\ce{NaOCl}$ is considered equivalent to 71 g of $\ce{Cl2}$, not 35.5 g. (Here's an example from the Ohio EPA.)
So yeah, you probably shouldn't have listened to me. Sorry.

In clarification to your edit, would this depend on the form of chlorine being dissolved? The Ca(OCl)2 reaction posted in OP seems to retain all chlorine molecules as active HOCl.

Link from Ohio EPA gives me 404 error.

Let imagine there are 3 bank accounts, one with 10000 Euros, one with 10000 US dollars, one with 10000 UK pounds. Let assume - just illustratively - there is exchange ratio 5 dollars per 4 Euros  and 2 dollars per 1 UK pound. If there is requirement to express the money level "as dollars",
the first one is worthy "as 12500 dollars",
the third one "as 20000 dollars".
Back to chemistry, "as Cl2" is not limited to chlorine being a part of the considered compound. there is just expected some chemical equivalence, defined by chemical reactions.
$\pu{1 mol}$ $\ce{Cl2}$ is equivalent to $\pu{1 mol}$ of $\ce{HClO}$ or $\ce{ClO-}$ due the reactions
$$\ce{Cl2(aq) + H2O <=> H+(aq) +  Cl-(aq) + HClO(aq)}$$
$$\ce{Cl2(aq) + 2 OH-(aq) <=>> ClO-(aq) +  Cl-(aq) + H2O}$$
So if
$M_\ce{Cl2} \approx \pu{71 g mol-1}$
$M_\ce{HClO} \approx \pu{52.5 g mol-1}$
then $\pu{52.5 g}$ $\ce{HClO}$ means "as $\ce{Cl2}$"  $\pu{71 g} \ce{Cl2}$.
Mass-wise, it does not matter if we consider $\ce{Cl}$ or $\ce{Cl2}$). Amount-wise, $\ce{HClO}$ is equivalent to twice as much amount of $\ce{Cl}$.
If we consider reactions
$$\ce{2 Fe^2+(aq) + H2O2(aq) + 2 H+(aq) -> 2 Fe^3+(aq) + 2 H2O}$$
$$\ce{2 Fe^2+(aq) + Cl2(aq) -> 2 Fe^3+(aq) + 2 Cl-(aq)}$$
then $\pu{1 mol}$ of $\ce{Cl2}$ is equivalent $\ce{1 mol}$ of $\ce{H2O2}$, so $\pu{34 g}$ of $\ce{H2O2}$, if expressed "as $\ce{Cl2}$", means (approximately) $\pu{71 g}$ $\ce{Cl2}$.

The notation mentioned in the other thread ($\ce{NH3-N}$) means something little different: "nitrogen in ammoniacal form, expressed as elemental nitrogen.". The direct equivalent, if we e.g. analyzed solution containing chloride $\ce{Cl-}$, hypochlorite $\ce{ClO-}$ and chlorate $\ce{ClO3-}$, would be the analogical syntax $\ce{Cl-Cl}$, $\ce{ClO-Cl}$, $\ce{ClO3-Cl}$, or verbose equivalents "chlorine in chloride/hypochlorite/chlorate form.".
It is not used to express equivalence of different chemicals, but rather to express content of chemical element (or compound), present in various forms, in a unified way. Typically to make their sum easily.
Like in elemental analysis, the mass of total nitrogen in a waste water sample is the sum of ammoniacal+nitrite+nitrate+organic nitrogen.
$$m_\text{Total-N}=m_\ce{NH3-N}+m_\ce{NO2-N}+m_\ce{NO3-N}+m_\ce{Org-N}$$

If we consider various chloramins:
\begin{align}
\ce{1 NH2Cl} \equiv \ce{1 HOCl} \equiv \ce{1 Cl2}\\
\ce{1 NHCl2} \equiv \ce{2 HOCl} \equiv \ce{2 Cl2}\\
\ce{1 NCl3} \equiv \ce{3 HOCl} \equiv \ce{3 Cl2}
\end{align}
As
\begin{align}
\ce{NH_{3-n}Cl_n + n H2O &<=> NH3 + n HOCl}\\
\ce{HOCl + Cl- + H+ &<=>  H2O + Cl2}
\end{align}
If we consider  $\ce{NHCl2-Cl}$ as "chlorine in form of dichloramin,
$$\pu{1 mg L-1} \ce{NHCl2} \equiv \frac{2 \cdot M_\ce{Cl}}{M_\ce{NHCl2}} \pu{mg L-1} \ce{NHCl2-Cl}$$
But, if we consider  $\ce{NHCl2}$ "as $\ce{Cl2}$, or as active chlorine",
$$\pu{1 mg L-1} \ce{NHCl2} \equiv \frac{4 \cdot M_\ce{Cl}}{M_\ce{NHCl2}} \pu{mg L-1} \ce{Cl}$$

The former counts the equivalent mass concentration for the chlorine fraction contained in the substance.
The latter counts the mass concentration for the equivalent free/active chlorine.

Ok, that stirs the pot again. If it is possible to measure 'as X' without any content of 'X' makes it interesting indeed. I presume we need to know the original reaction we are equating to. I have only dealt with reactions from hypochlorous acid to different chloramines, how should I determine these (Monochloramine, Dichloramine and Nitrogen Trichloride) as Cl2 equivalents?

If you were expressing the amount of money in your wallet "as US dollars", would it come weird if you had no US dollars there?

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Ammonia (NH3) vs Ammonia Nitrogen (NH3-N)
I've been looking into chloramination (chlorine & nitrogen reactions) and noticed many papers either reference Ammonia (NH3) or Ammonia-Nitrogen (NH3-N), seemingly interchangeably.
Is the hyphenation simply nomenclature for "we are only interested in measuring the nitrogen component of ammonia" or does NH3-N present as a distinct compound in water treatment?
For example, if a paper provides a Cl2:N Ratio of NH3, is this equivalent to a Cl2:N ratio of NH3-N?
Note: Asking in parallel to my 'as xxx' question. Posting separately to avoid convolution.

$\ce{-}$ in $\ce{NH3-N}$ is just a hyphen, not a chemical bond, with meaning "nitrogen present in form of $\ce{NH3}$ or $\ce{NH4+}$". It can have both qualitative and quantitative meaning.
With rounded molar masses, $\pu{17(18) mg L-1}\text{ of }\ce{NH3(NH4+)} \equiv \pu{14 mg L-1}\text{ of }\ce{NH3-N}$.
Or $\ce{N-NH3}$, AFAIK the convention is not settled.
There can be literally any compound or material instead of $\ce{NH3}$ in the $\ce{X-N}$ syntax, containing $\ce{N}$. $\ce{NO2-N}$, $\ce{NO3-N}$, $\ce{Org-N}$ are very common too, e.g. in context of waste water analysis.
That syntax convention  is often used in elemental analysis, where are determined contents of particular forms of nitrogen, recalculated to element content.
The respective molar ratios are the same for ammonia and ammonia nitrogen, as there is 1 nitrogen atom in ammonia molecule. For mass ratios, they have to be multiplied by molar mass ratios.
The topic is related to the more general principle not limited to chemistry: quantities being expressed by  an equivalent quantity of something else.

E.g. spring scales, commonly used on farmer markets, measure an object weight = force. But, they are calibrated in the equivalent mass. So an object with weight $\pu{9.806 N}$ has this weight reported "as mass" $\pu{1 kg}$.
The content of phosphorus or potassium in solid fertilizers is for historical reasons often reported as $\ce{P2O5}$ and $\ce{K2O}$ content, even if there are no such oxides present. ($\ce{N}$ context is addressed by Nilay Ghosh).

+1. You can also represent phosphorus content in a fertilizer as P-content. Phosphorus fertilization rates often are given in terms of P, not P2O5. The proportion of P in P2O5 is used to calculate the equivalent quantity of P in a fertilizer: 2 P/P2O5 = (2 × 31 g/mol)/142 g/mol = 0.437. Thus, the percentage of P in a 12-24-12 fertilizer is: 24% P2O5 × 0.437 = 10.5% P. Although, most of the places still use P2O5 scales.

@NilayGhosh  Sure, that is obvious. The example was focusing on the mentioned general equivalence principle, not on particular chemistry application.

So interesting how equivalency terminology is almost colloquial depending on what is being measured. Like we seem to have "X-N" for nitrogen dominant compounds, but "as Cl2" for chlorine measurement. Thankyou though, answers are very helpful.

There are no formal requirement for terminology. It could be as well Cl-Cl, ClO-Cl, ClO3-Cl, or "as N2". It depends on circumstances and historical patterns.

I have one final query regarding the mass ratios of NH3 vs NH3-N. To my mechanical mind the proportionate mass of nitrogen in ammonia (14.007/17.031 = 0.8224). So, if I have a solution with NH3 of 10mg/l this would be (10*0.8224 =) 8.224mg/l NH3-N?

It is analogical to the question "If 100 g of the pie contains 30 g of sugar, how much sugar is in 500 g of the pie?"

Based on the context of question, the correct term for it is ammoniacal nitrogen. Most of the things are explained in @Poutnik's answer, so I am going to expand the last point. This term is also used in fertilizer domain. It used to refer a nitrogen source where most of the nitrogen will come from ammonia. The other form of nitrogen are nitrate-nitrogen and urea-nitrogen. It is slower acting than the nitrate form. It must be converted by soil acting bacteria to nitrate nitrogen before it can be absorbed by plant roots. It is not as easily lost from the soil by leaching.
Please refer to this article:

https://www.teagasc.ie/crops/soil--soil-fertility/fertiliser-types/nitrogen-n/
https://www.canr.msu.edu/news/understanding-the-forms-of-nitrogen-in-water-soluble-fertilizers-for-greenhouse-growers

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What are these plates in LixC6?
I'm completely new to chemistry. I can’t personally figure out the basic issues myself, so I’m asking for your help. I became interested in how lithium-ion batteries generally work. As a result, I came to what is the most used material for the anode. This is LixC6. But I don't understand what its structure means, i.e. How do we even understand this? Why are there 3 dies here? What is 'x' in LixC6? Can you explain please. If the question is not clear enough, then let me know, because I really don’t know much about chemistry and I can ask quite stupid questions.

References

Image Source - Electronic structure ‘engineering’ in the development of materials for Li-ion and Na-ion batteries. Link

The three "dies" (if you refer to the graphite planes on the left side of the picture that you provided) are just a convenient way to represent the graphite crystallographic lattice, see the picture below, from Wikipedia.

As for the x, my best guest would be that it represents a statistical fraction of Lithium ion presence. Id est in practice for a given carbon cycle ($\ce{C6}$) it is either 0 or 1, that is we either have $\ce{LiC6}$ or $\ce{C6}$. However, you can average this on your whole lattice. Hence if you have, say $\ce{Li_{0.5}C6}$, it would mean that half the vacancies in your graphite lattice are occupied with a lithium atom. Would you have $\ce{Li_{0.25}C6}$, it would be a quarter of the vacancies, etc.

EDIT: I found confirmation of this in [1], which also provided a figure similar to yours.

References
[1] DM1 - Le lithium dans les accumulateurs Li-ion (Centrale PSI 2015), Archive

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.

@Buttonwood, I must admit that I do not see the interest of your 'edit' here... I mean: is this not completely equivalent to write $\ce{C6}$ or C$_6$ (first one written with ce, second one with basic LaTeX command)? I get the point that in case of long equations, using mhchem would proove meaningful, but in more basic cases I simply do not see the point... Am I missing something here? (I may be completely misled)

It was for consistency in the representation and to ease of reading without JS enabled.


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When you do dehydrohalogenation of 1° alkyl halides, do you use a strong base or sterically hindered base?
I know when do you dehydrohalogenation of 2° alkyl halides, you can do either a strong base like KOCH2CH3 to get the most substituted product or a sterically hindered base like t-butoxide to get the less substitued product. However, I do not understand whether you would use the strong base or sterically hindered base for 1° alkyl halide because there are not two options here.

Simply put, by Saytzeff / Zaitsev rule, the favoured product in dehydrohalogenation reactions is the alkene with the greatest amount of alkyl groups connected to the doubly bound carbon atoms.
However in this case the base does not matter actually as both Saytzeff and Hoffmann alkenes will look same (except in the case carbocation's hydride shift / methyl shift takes place).
So you can use whatever base fits your budget!
Update: It seems that using a bulkier base will reduce the probability of getting the product of a nucleophilic displacement. It will then want to follow substitution and act as a nucleophile rather than as a base. Thus if you want ONLY elimination to take place the bulkier base is better as it offers higher steric hindrance. [inputs from comments]
Also you will have to note that even conditions make a difference in the products.
SOURCE: Haloalkanes, NCERT

Use of a sterically hindered base reduces the possibility of getting the product of nucleophilic displacement

It's misleading to state your choice as between a strong base and a sterically hindered base, as t-butoxide is in fact even stronger than ethoxide. Rather, you choose between two strong bases, one sterically hindered and one not (or less so).
To answer your question: yes, there is only a single possible product for the elimination reaction (hydride/alkyl shift isn't generally an issue as primary carbocations aren't formed in the first place). However, all bases are potentially also nucleophiles, and can also engage in substitution, in this case giving an ether instead of the desired alkene. If you want to minimize this reaction it would pay to use the more hindered base, as substitution is more sterically demanding than elimination and would be slowed down more by bulky substituents on the nucleophile/base.


text:
What is the ionic conductivity of the hydroperoxide ion?
Hydrogen peroxide is supposed to dissociate as
$$\ce{H2O2 <=> H+ + HO2-}$$
But I can't find a source for the conductivity, $\lambda$, of the  hydroperoxide anion $(\ce{OOH-}$ or $\ce{HO2-})$ on Google.

@HarikrishnanM If I'm reading that table right, 31% H2O2 solution has roughly $\Lambda_0 = 250$. But since the value of $\lambda_{H+}$ changes at 31% acc. to the paper, how do I get $\lambda_{HO2-}$ independently?

On reading closer, I think that table is only showing the acid's conductance after correcting for the H2O2-H2O, so I can't get my $\lambda$ anyway since they didn't record the total conductance including the solvent.

Well, I can't find $\lambda^0_{HO2^-}$. That's actually what I wanted.

Ohh I get it... will try and get something more as soon as possible.

By the way did you try the crc handbook of chemistry? It has all data there... the paper for your reference: https://cdnsciencepub.com/doi/pdf/10.1139/v58-108


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Why potassium carbonate breaks glass when heated?
I have been studying about desiccants for a while and testing some inorganic salts, including calcium chloride, sodium sulfate and potassium carbonate. When they get enough moisture from air, I dehydrate them by heating up in a borosilicate glass, and during the process everything looks fine for all, but with potassium carbonate the glass always breaks after 1 or 2 cycles, and I don't know why. It should be clear that this doesn't happen with the other salts even after more than 15 cycles so far, and it takes several minutes or even hours after the glass cools down (I didn't even watch the exact moment when this happens). I'm using an electric heater in a low setting temperature, taking about 15 minutes to reach the final temperature about 250 to 280 °C. I will be grateful for your answer and any suggestions of what material could be more appropriate for the task

How fast, in what way and to what T do you heat them? The best if you describe the whole procedure in details, otherwise the question itself may become yet another layer of the puzzle. // If hot enough, the carbonate may chemically damage the glass structure, potassium oxide getting incorporated in the borosilicate. Remember alkaline carbonates were used for ages to make glass. You may try rather iron based metallic vessel for it.

I am using an electric heater, it takes about 15 minutes to reach the set temperature of approximately 250 °C, and it doesn't break while heating up, but after cooling at least 20 minutes later

If the beakers are not made of borosilicate glass, it will react with the metal carbonate and break the glassware:

Fusion with an alkali metal carbonate, usually anhydrous sodium or potassium carbonate or their mixture, decomposes glass samples very effectively. This occurs above the melting point of the carbonate(s): $\ce{Na2CO3}$, 853°C; $\ce{K2CO3}$, 903°C; $\ce{KNaCO3}$, 712°C. If it is known that the glass does not contain boron, the fusion sample is fused with a mixture of sodium carbonate and borax to decrease the fusion temperature. The complex silicate network is broken down, and monosilicates are formed. Other anions present are released, and network modifiers form salts corresponding to the acid used finally to dissolve the cold fusion mixture. This reaction is accompanied by evolution of carbon dioxide.

Source: M.P. Syomov, O.V. Rakhimova, in Encyclopedia of Analytical Science (Second Edition), 2005

And even if it is borosilicate glass, it contains silica and will still react as you describe, though possibly more slowly.

But I am not reaching that high temperatures, the maximum I am using is 280 °C in the base of the hotplate of the electric heater


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How dangerous is Naphtha?
According to a safety datasheet of Naphtha, it's fairly dangerous stuff (including genetic defects). But, as they say, the dose makes the poison, just how much dose should I be worried about?
Currently, we use a silicone adhesion primer (SP-270), which is dissolved in Naphtha. We use it in a fume cupboard, brushing it onto small metal parts with a tiny brush. The Naphtha evaporates pretty quickly, before we remove the metal part from the fume cupboard. We don't wear gloves.
Questions:

Does this method of handling Naphtha sound unsafe?
How much exposure to Naphtha is necessary before I'm likely to experience any of the nasty effects? If I accidentally touch the brush, should I be worried? Or do I need to swim in it to get an effect?

Added clarification:
The datasheet does not discuss the level of exposure required to cause the listed effects. For example, the safety datasheet for Paracetamol says "Harmful if swallowed", but doesn't say how much you need to swallow before it's harmful. Adults can actually swallow up to 4g of it per day, but 24g is fatal. Perhaps there's something similar that can be said about Naphtha?

For the person downvoting: please can you explain how I can improve this question? Why is it worse than some of the other questions asking about the dangers of certain substances?

To put the risks of naphtha in perspective, the ingredients are fairly similar to the composition of the petrol (gasoline) you use in large quantities to power cars. The short term risks are its flammability in large quantities; the long term risks arise from repeated exposure to large amounts. The long terms risks involve regular repeated exposure which you can minimise with gloves and ventilation.

@DrMoishePippik - Read the question. I literally cite and link to the datasheet in the question. My question involves what's not mentioned in the datasheet. Perhaps I can explain it another way: The SDS for Paracetamol says "Harmful if swallowed". And yet adults can actually swallow up to 4g of it per day, but 24g is fatal. Perhaps there's something similar that can be said about Naphtha?

Naphtha and gasoline are synonymous enough that a report on the carcinogenicity of gasoline should be a good surrogate for one on naphtha. Et voici: https://cfpub.epa.gov/ncea/risk/recordisplay.cfm?deid=36176 [U.S. EPA. Evaluation of the Carcinogenicity of Unleaded Gasoline. U.S. Environmental Protection Agency, Washington, D.C., EPA/600/6-87/001 (NTIS PB87186151).]

This is a basic chemical safety and practice question not a personal medical question. it should be reopened.

On the practical side, if you are handling the substances in a fume hood (cupboard), wearing proper gloves, and out of precaution protective eyeware and a labcoat, you should be ok. More generally, somebody might respond by posting the dose response curve.

eg https://pubmed.ncbi.nlm.nih.gov/24179027/


text:
Iodine solution concentration
I am planning to make 1L of 0.005M iodine solution from KI and I2. I have found instructions that say to use 2 g of KI and 1.3 g of I2 but no explanation why exactly this particular amount. I also found that iodine solution concentration will be the same as the concentration of the added I2, again no explanation.
Could someone please explaine the calculations behind this? What and how affects the final concentration? Since there is a reaction happening (KI + I2 = KI3), I am a little confused.

Iodine has relatively poor solubility in water, but dissolves much more readily in an aqueous solution containing iodide ions because tri-iodide ions are formed. So the KI just facilitates dissolution of the iodine in the solution.

$\ce{I2}$ has a molar mass of $254$ g/mol. $\ce{KI}$ has a molar mass of $39 + 127 = 166$ g/mol.
Then $1.3$ g $\ce{I2}$ is $\frac{1.3}{254} = 5.12$ millimole $\ce{I2}$.
Also $2$ g $\ce{KI}$ is $\frac{2.0}{166} = 12.05$ millimole $\ce{KI}$
Now $\ce{I2}$ is nearly insoluble in water. But it is soluble in $\ce{KI}$ solutions, thanks to the following equation. If the previous amounts of $\ce{I2}$ and $\ce{KI}$ are dissolved in $1$ liter water, the $5.12$ millimole $\ce{I2}$ are totally transformed into $\ce{KI3}$. The obtained solution contains $\ce{5.12}$ millimole $\ce{KI3}$, $12.05 - 5.12 = 6.97$ millimole $\ce{KI}$, and $0.0$ millimole $\ce{I2}$, because of the reaction : $$\ce{KI + I2 <=> KI3}$$ This reaction has been considered as complete in the previous calculations for reason of simplicity. But in reality this reaction is an equilibrium, whose equilibrium constant is unknown. So the concentration of $\ce{I2}$ in the $\ce{KI3}$ solution is not zero. It has a small and unknown value. But this value has no importance and nobody is interested in knowing it. The only important parameter in such a solution is the total amount of $\ce{I2 + KI3}$. Why is it so ?
The reason is that such a solution of $\ce{I2}$ in $\ce{KI}$ is used for titrating reducing solutions like thiosulfate ion $\ce{S2O3^{2-}}$ (or the reverse rection). If an unknown solution of iodine in $\ce{KI}$ is titrated with a standardized solution of thiosulfate, the solution of iodine is slowly discolored by the reaction : $$\ce{S2O3^{2-} + I2 -> S4O6^{2-} + 2 I-}$$ In the beginning the solution of $\ce{I2 + KI}$ is dark violet and contains very few free iodine $\ce{I2}$. This iodine is quickly destroyed by the first drops of thiosulfate. But as soon as it is destroyed by thiosulfate, $\ce{I2}$ is regenerated by $\ce{KI3}$ thanks to the equilibrium $\ce{KI + I2<=>KI3}$. So progressively the added ions thiosulfate destroy all the iodine contained in $\ce{I2 + KI3}$. If we know that the unknown solution contains $\ce{I2 + KI3}$ in unknown amount, this amount is equal to the number of mole of added  thiosulfate, up to the endpoint of the titration. And this endpoint is defined by the entire discoloration of the solution. This phenomena can be defined and observed with a high precision : ± $1$ drop of thiosulfate. It is one of the more precise titrations of the whole domain of analytical chemistry.
So when $\ce{I2}$ in a $\ce{KI}$ solution, we can ignore that $\ce{I2}$ is partly in the form of $\ce{KI3}$. The titration with $\ce{S2O3^{2-}}$ proceeds as if iodine is simply dissolved in the solution.


text:
How do I explain that voltage decreases when increasing the concentration of the cathode in a voltaic cell?
I just finished a lab experiment in which I built a voltaic cell with copper (II) sulphate and zinc sulphate. My experiment was to identify the correlation between the concentration of copper (II) sulphate and the voltage generated.

I started with a 200 ml solution of 0.5 molar copper (II) sulphate and a 200 ml solution of 0.5 molar zinc sulphate solution (which is constant).
I measured the voltage of the cell at 0.5 M for both and I got about 1.088 V.
I added 1g of copper (II) sulphate anhydrous (crystal) to the copper (II) sulphate solution after every trial (which effectively increases the concentration of copper (II) sulphate (the cathode)) but I realized that each time I did that the voltage decreased. For context, I added 1g 18 times and I ended up with a voltage of 1.029 V.

So TLDR, increasing the concentration of copper (II) sulphate (the cathode) decreased the voltage of my voltaic cell instead of increasing it.
A few other students in my class did the same experiment (slight differences in methodology) and their voltage increased, which supports my original hypothesis of voltage increasing with concentration and also supports the Nernst equation (pretty sure).
Can anyone PLEASE help me understand why the voltage decreased when I increased the concentration of copper (II) sulphate?

There is possibility the activity coefficient of copper ions decreases faster than concentration increases, leading to decreasing of ion activity. Using concentrations instead of activities is just approximation, that is simple  enough for school, but diverges from reality. // Did the students use the same/similar concentration range? Did the voltage  with the given solution have any time trend?

There are basically 2 options, implying you have all  used the same or overlapping concentration range. 1/ it happened to all, so the activity thing is the reason. 2/ It happened just to you, so there is some error on your part we cannot remotely identify. // If there is possibility to repeat the experiment, use 10-50 times more diluted solution. 0.5 M is a lot.

What sort of salt bridge have you used ?

Anhydrous copper sulfate is a white powder, not a crystal (as you described)

How do you explain that your students obtain a result which is contrary to yours ?


text:
Why is sulfur more nucleophilic than nitrogen in thiourea?
In $\ce{S_N2}$ reaction using thiourea and alkyl halide, it is the sulfur atom rather than nitrogen atom that attacks alkyl halide. But nitrogen also has lone pair, so there is a chance that it acts as nucleophile. I know that nitrogen can push its lone pair to sulfur through resonance structure, but it is same in the case of urea. What makes the sulfur atom more nucleophilic than nitrogen atom in thiourea?

In a simplified analysis, I'm guessing the reduced pi orbital overlap of C=S versus C=O means resonance structures where the sulfur has a negative charge (and nitrogen has a positive charge) become more important for thiourea compared to urea.


text:
Does concept of degree of unsaturation not work for fullerene molecule?
Commonly known buckminsterfullerene has molecular formula C60 if we calculate it's degree of unsaturation by formulae, it comes out to be 61 and it is also known that this fullerene has 12 pentagon rings and 20 hexagon rings.
So, overall it has 32 rings. Also, degree of unsaturation can be thought of as number of pi bonds + number of rings for hydrocarbons, so the number of pi bonds calculated are 29 according to this concept
However, the number of pi bonds in fullerene have been found out to be 30. Why does degree of unsaturation not work for fullerene, what is it's actual degree of unsaturation?

Could you provide links as to where you are getting these "rules" from? My guess is that with fused rings you can forget them, but without seeing exactly how the terms you are using are defined its impossible to say.

It's not that your useless concept doesn't work, it's just that counting of rings is slightly more complicated.

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.03%3A_Calculating_Degree_of_Unsaturation#:~:text=Remember%2C%20the%20degrees%20of%20unsaturation,bond%20(2%20%CF%80%20bonds). degree of unsaturation is found using similar method for bicyclo compounds and the formula is given too

An alkane with 60 carbon atoms has the formula $\ce{C60H122}$. To get the degree of unsaturation of fullerene, you have to successively add dihydrogen to it until you have an alkane. For each double bond turned into a single bond, you need one dihydrogen. For each single bond you break (to remove rings), you need one dihydrogen. Looking at the formula of fullerene, you will need 61 dihydrogen molecules to turn it into an alkane, so the degree of unsaturation is 61.
If a ring system has N rings, it does not mean you need to break N bonds to open them all up. Here is an example with 4 rings:

I only need to break 3 bonds to get a non-cyclic alkane:

This is because the first cut breaks two rings at the same time.
EDIT:
This insightful comment made me think more about counting rings.

[Nicolau Saker Neto in the comments] Because C60 fullerene is so close to a sphere, I can't help but bring up this curiosity - the reason the formula is slightly off comes down to the fact that a 3D sphere missing even just a single infinitesimal point is topologically homeomorphic to a plane, but a complete 3D sphere is not, so there really is something special about that first cut.

If you contemplate the structure of naphthalene and norbornane, you will see that they are topologically equivalent. They both have two bridging carbons connected by three links. Yet most of us would probably count two rings for naphthalene and three rings for norbornane.

In the "flattened" view, it becomes clear why: In the top structure, we probably will not count the outer 10-membered loop, but in the bottom structure, we will typically count the outer 6-membered loop (and the two 5-membered rings). For the degree of unsaturation, you should only count "inner" rings in the flattened view of the molecule.
For fullerenes, this means counting one ring less, just as the analysis of breaking bonds to remove all the rings shows.

Because $\ce{C60}$ fullerene is so close to a sphere, I can't help but bring up this curiosity - the reason the formula is slightly off comes down to the fact that a 3D sphere missing even just a single infinitesimal point is topologically homeomorphic to a plane, but a complete 3D sphere is not, so there really is something special about that first cut.

@NicolauSakerNeto - I was trying to figure out how to express that in words. Thank you for putting it so clearly

@NicolauSakerNeto You can take a fullerene, grab one of the rings and expand it until the fullerene is flattened. When you do that, the ring you just expanded looks like the outer boundary of the ring system, which you wouldn't count usually. Here is an rough image of this, using a dodecahedron (C20): https://i.stack.imgur.com/zIA9j.png

@Andrew The same for adamantane: https://i.stack.imgur.com/kdDA9.png (3 rings or 4?)

Yes I understand that kind of drawing is possible. "Not counting the outer ring" is ultimately equivalent to taking an inflatable fullerene (or adamantane or dodecahedrane, etc.) wrapped in an elastic skin, squishing it against a plane to deform it into that shape, poking a hole in the front, then stretching the skin open and gluing the outer rim onto the plane. So you could say the act of planarizing it (which renders the unsaturation formula valid) "broke the outer ring" and you then decide not to count it. I can't seem to put in it better words.

@NicolauSakerNeto Good enough for me.


text:
Determining the concentration of ammonium nitrate by chemical means
I have several litres of ammonium nitrate solution that I do not know the concentration of, but is approximately 0.6M. I don't have precise enough instruments to determine it via its density. I titrated it with NaOH, since the ammonium ion is a weak acid, but it is too weak and the endpoint is impossible to discern clearly. I then realised of course I could just evaporate the water and weigh the solute, but the question still stuck with me: how could I determine its concentration by chemical means? The only other way I can think of is to measure its conductance and compare this value to existing data.

Try adding a precise amount to excess standard NaOH, carefully distilling off the ammonia and titrating the remaining NaOH with HCl.

If you have a good balance and a 100 ml volumetric flask you can determine the density of the solution directly.

@MaxW, I tried that but the values I got are way off -- less dense than pure water. The balance is only to 2 d.p. so I suspect it is not good enough.

As long as the solution is pure, your idea of making standardized $\ce{NH4NO3}$ solutions and comparing conductance should work.
Caveat: This works as long as conductivity vs. concentration is a monotonic function. For some other substances is not, e.g., for $\ce{H2SO4}$, which, at STP, has a distinct peak in conductance at ~30%, and then drops off, so that there are two points on the curve at which conductivity is the same, but concentration quite different.
This is also the case for $\ce{NH4NO3}$ above 8.0 m/mol kg-1, but, in the region in which you are interested, it should not be an issue.
However, there is another, simpler, rather old alternative: use a hydrometer. The density of $\ce{NH4NO3}$ is a monotonic function. And others have already done the work for you, charting density against concentration.
Eureka!

Thank you @DrMoishe. I have learned something new about monotonic functions!


text:
Why does the reaction "pause" when the battery is not connected to a circuit? (or does it?)
The general set up, In an alkaline battery, the negative electrode is zinc (Zn) and the positive electrode is manganese dioxide ($\ce{MnO2}$). The alkaline electrolyte of potassium hydroxide (KOH)
At the anode:
$$\ce{Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e−}$$
At the Cathode:
$$\ce{2MnO2(s) + 2H2O(l) + 2e− → 2 MnO(OH) (s) + 2OH−(aq)}$$
My question is that why does the reaction "pause" when the battery is not connected to a circuit?
For an example, If I were to have a container holding the electrolyte and I were to place the 2 metals in to the solution, I would expect sometime later that the reaction to be complete, meaning that all zinc and $\ce{MnO2}$ used up. I believe that the electrons would beable to find themselves in the position to be involved in the reaction. But for an unconnected battery, the reaction seems to stop? Why is this?

How could the electrons "find themselves in the position to be involved in the reaction" ? The electrons cannot go through the KOH solution. They must go to the cathode through a metallic wire.

@Maurice, Thanks for your reply. So I am not really understanding what happens. Suppose no wire again: I guess the reaction takes place at the anode as OH- ions float towards it and react with Zn. Then does the electrons cling to Zn and not float into the solution?

If no wire is joining the zinc plate and the $\ce{MnO2}$ plate, the whole setup would not be a cell or a battery. It wouldn't produce any anode or any cathode. The whole setup would simply be a chemical solution of $\ce{KOH}$ in contact with a zinc plate somewhere, and with a block of $\ce{MnO2}$ somewhere else. The only chemical effect that could happen is the possible reaction (not the half-reaction) between Zn and $\ce{KOH}$ somewhere, and independently also the possible reaction between $\ce{KOH}$ and $\ce{MnO2}$ elsewhere. Fortunately no such reaction happens between $\ce{KOH}$ and either zinc or $\ce{MnO2}$. If such a reaction would exist between zinc and $\ce{KOH}$, this reaction will soon consume the zinc in a couple of minutes before the battery would work. And the battery could not be sold: It would be used in a couple of minutes after construction. The zinc metal reacts (and produce electrons) only if there is an electric contact between the zinc and $\ce{MnO2}$. Without any contact, the zinc cannot emit electrons, because these electrons have nowhere to go.

Thanks again for the reply. (Sorry for my questions as I am not to good with chemistry) As metals don't know know the situation they are in so I guess the following happens: There are some reactions that take place at each metal but  it stops(maybe from the presence of electrons). But when the wire is connected the electrons can flow to the other metal solid, relieving the presence of electrons(or something) and restarting the reaction?

@Reuben At the atomic level, the reactions take place at each electrode, but the reverse reactions take place at the same rate. Once there is a wire connection, the anode has a way to get rid of electrons and the cathode has a source of electrons, so oxidation at the anode and reduction at the cathode are the net reactions. So the electrode half-reactions don't have a master switch, they keep on doing what they always do, but the availability of species drives the direction of reaction.

Also, if you swap out a half-reaction for another one (on the other side of the electrochemical series), sometimes the direction of electron flow will switch, and suddenly the old anode is the new cathode. Again, nobody has to tell the electrode to switch, it is simply driven by the availability of the species of the half-reaction.

In a battery, the zinc and manganese reagents are kept separate, and so the electrons can only flow when both parts of the battery are connected via an electric circuit.
If you take a look at the Construction section of the Wikipedia Alkaline battery article, you’ll see that a thin membrane divides each half-cell of the reaction, but that the electrons will be able to flow so long as a circuit connects both ends of the battery.


text:
Ether has +R effect nd CH3 has +I effect, if R effect is more effective than I effect then why is CH3COOH less acidic than CH3OCH2COOH
The More is the electron donating effect, the more will be the charge on carboxylate ion, the more will it be unstable and the less acidic it will be.

You aren't comparing $\ce{–CH3}$ with $\ce{–OCH3}$. You're comparing $\ce{–CH3}$ with $\ce{–CH2OCH3}$, as these are the groups actually attached to the carboxyl groups in your two acids. In the latter group the saturated  $\ce{CH2}$ function prevents delocalization of any electron pairs from the oxygen and bl9cks the proposed +R effect.

So what type of group is CH2OCH3 then?

Only the inductive effect can pass through a saturated atim, and the oxygen is electronegative. What does that suggest to you?

Actually I knew ethers have -I effect but, as u said that oxygen is not directly attached to the carboxyl group, and CH2 is also attached to it, so I was saying won't it donate the electrons to the carboxyl group, since O is farther away to -COOH it's inductive effect will decrease and CH2 is nearer and it's inductive effect will dominate!??

CH2 is not that good a donor, the more electronegative oxygen still dominates even one position removed.


text:
Reaction kinetics exercise for hydrogen iodide synthesis
The rate constant for the reaction of hydrogen with iodine is $\pu{2.45E-4 M-1 s-1}$ at 302 °C and $\pu{0.905 M-1 s-1}$ at 508 °C.
a. calculate the activation energy and Arrhenius preexponential factor for this reaction.
b. What is the value of the rate constant at 400 °C ?

I'm a bit confused because I get a negative value for the activation energy $E_a$
We have at 302 °C and 508 °C respectively
$$k_1 = \pu{2.45E-4 M-1 s-1}$$
$$k_2 = \pu{0.905 M-1 s-1}$$
We convert in Kelvin:
$$T_1 = 302 °C = 575.15 K $$
$$T_2 = 508 °C = 781.15 $$
The Arrhenius Equation is
$$\ln{k} = \ln{A} - \frac{E_a}{RT}$$
So we have
$$\ln{k_1} = \ln{A} - \frac{E_a}{RT_1}$$
$$\ln{k_2} = \ln{A} - \frac{E_a}{RT_2}$$
We want to solve for $E_a$. We subtract the second equation from the first:
$$\ln{k_1} - \ln{k_2} = - \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$
Solving for $E_a$, we get:
$$E_a = - R \times \frac{\ln{k_1} - \ln{k_2}}{\frac{1}{T_1} - \frac{1}{T_2}}$$
With $R = 8.314 J/mol K$, $T_1 = 575.15 K$, $T_2 = 781.15$, $k_1 = \pu{2.45E-4 M-1 s-1}$, $k_2 = \pu{0.905 M-1 s-1}$
(Link to WolframAlpha calculation)
I get $E_a = 1.48948 \times 10^5 J/mol$
Then using the Arrhenius equation with $k_1$ or $k_2$:
$$A = k_1 \times \exp{\frac{E_a}{RT_1}}$$
(Link to WolframAlpha calculation)
I get $A = \pu{8.26E9 M-1 s-1}$
But I'm confused, why the Activation Energy is negative ? Or did I miss something ? I checked all my calculations
EDIT: the error was due to a missing negative sign for $E_a = ...$. Now it makes more sense.

@Poutnik So how should I formulate the title ? "Why do I get a negative activation energy ?"

Instead of  $2.45 \times 10^{-4} M^{-1} s^{-1}$  ( $2.45 \times 10^{-4} M^{-1} s^{-1}$ ), try $\pu{2.45E-4 M-1 s-1}$  ($\pu{2.45E-4 M-1 s-1}$). // As a simplified rule, only symbols for variables, physical(chemical) quantities or physical constants are in italic, all the rest is upright.

@Poutnik Thank you very much for your help. The error was in fact due to the missing negative sign. Now it makes more sense. I get $E_a = 148957$ J/mol. This gives an Arrhenius Pre-exponential factor of $\pu{8.26E9 M-1 s-1}$. For b), can we simply use the same formula with $T = 400$ ° C $= 673.15$ K as temperature ? This would give $\pu{8.26E9 \exp{(−148957/(8.314×673.15))}}$ $= \pu{0.0228 M-1 s-1}$

@Poutnik Feel free to post an answer if you want (even just saying that it was due to the missing negative sign). I will accept it. Otherwise I will answer my own question later

It was nothing. But consider if "Always doublecheck signs in your calculations if you get weird results" is the good answer material.  :-)

A mistake must have happened in your calculations, because, when I do them, I obtain :
ln$k_1 = - 8.314$;
ln$k_2 = - 0.100$;
ln$k_1$ - ln$k_2$ = $-8.314 + 0.100 = -8.214$
$\frac{1}{T_1} - \frac{1}{T_2}$ = $ \frac{1}{575} - \frac{1}{781} = 1.76·10^{-3} - 1.97·10^{-3} = - 0.21 ·10^{-3} $ K$^{-1}$
$E/R = \frac{-8.214}{-0.21·10^{-3} \mathrm{K}^{-1}} = + 39 100$ K.
$E = 39100$K$ · 8,314$ $\frac{J}{mol·K} = + 325.2$ kJ/mol
It is a positive result.


text:
What is laboratory air?
Oftentimes in the literature, particularly for oxidation studies, the authors mention that experiments were performed in "laboratory air." Is there any distinction from laboratory air from regular, ambient air, or is this terminology only used to distinguish it from synthetic air (Only nitrogen and oxygen)?

"Laboratory air" probably means literally "laboratory air". Syntetic air would be explicitly mentioned, probably with explicit purity requirement.

Laboratory air is not a designated or special term. All it conveys to the experimentalist is that literally "laboratory" air was used for oxidation. This is perhaps a handwaving way of saying that we do not have the air analysis certificate in hand. We have no idea of water content, we have no idea of any trace impurity gases such as hydrogen chloride or ammonia or carbon dioxide. Normally when air is purchased, you know the purity of the gases. Laboratory compressed air is also contaminated with pump oils and hydrocarbons.

Just to clarify - it means "the air in the room at the time of the work"  and is the same air being breathed by the experimenter, right ?

@Criggie, We cannot say that absolutely. To split hairs further, many universities supply compressed air to research laboratories (stored air). So this may not be air that the experimenter is also breathing at the same time.


text:
Products formed on reaction of aqueous solution of iron(III) nitrate nonahydrate
Crystalline iron(III) nitrate nonahydrate, Fe(NO3)3.9H2O, has a very pale violet color. When added to water, the crystals dissolve to form a brown solution. Treatment of this brown solution with concentrated nitric acid yields a very pale violet solution while treatment with HCl yields a yellow solution.
I could identify that the brown solution is due to [Fe(OH)(H2O)5]2+, [Fe(OH)2(H2O)4]+, and on reaction with HNO3, we obtain [Fe(H2O)6]3+, but am unable to identify the yellow colored solution which is obtained on reaction with concentrated HCl.

It is probably [FeCl4]- .

See the answers to https://chemistry.stackexchange.com/questions/173730/why-do-we-have-to-prevent-the-hydrolysis-of-ironiii-nitrate


text:
Comparing reactivity of benzaldehyde and 2-butanone for nucleophilic addition
I was taught that, in general, aldehydes are more reactive than ketones because of two reasons: 1) less steric hindrance, and 2) alkyl groups on both sides of carbonyl group in ketones have a stronger +I effect and reduce the partial positive charge on the carbonyl carbon.
Here, even though the second compound is a ketone, it has small alkyl groups which would provide less hindrance than a bulky phenyl group in benzaldehyde. Also, there will be a resonance effect in benzaldehyde reducing the carbonyl carbon's positive charge more(?) than it would be reduced in butanone.
But my teacher said benzaldehyde would be more reactive, because "aldehydes > ketones in reactivity", and he also said that in butanone, electron donating and hindering groups are present on both sides of the carbonyl.
I would like a stronger argument (if there is one) for why the reactivity of benzaldehyde is more; comparing hindrance and electron donating behaviour like this doesn't seem concrete to me.
Any help is appreciated!

"there will be a resonance effect in benzaldehyde reducing the carbonyl carbon's positive charge more(?) than it would be reduced in butanone." That's what you got wrong.

@Mithoron so 2 alkyl groups have stronger electron donating effect (+I) than one phenyl ring's +R?

Phenyl has withdrawing effect. If you drew relevant structure, you'd see it only donates mesomerically to oxygen, not carbon.


text:
Using potassium thiosulfate instead of sodium thiosulfate in iodine clock reaction
I am familiar with the iodine clock reaction that uses sodium thiosulphate. Still, I'm wondering if I could use potassium thiosulphate instead, as it's the thiosulfate that reduces the Iodine.

I would not expect it to have a significant effect in your results. Check availabiltiy and price.

If you make corrections for different molar mass and ev. difference in crystal water, it should not matter.


text:
Why can't picric acid be made more acidic by adding nitro groups at the meta positions?
I understand that picric acid is acidic because of the $\ce{-NO2}$ electron withdrawing group, but why can't we add more of those on the meta positions to make it more acidic?
Is it because that $\ce{-NO2}$ shows less -R effect when it is on that meta positions?

We surely can add more nitro groups, albeit with great difficulty.

Would it be more acidic if more NO2 groups are added.

That is not immediately clear, nor are the experimental data readily available.

I think if more NO2 groups were added the repulsions between them would cause steric hindrance, because of which it would not longer be stabilized by resonance. Can this be a correct explanation?

Indeed, all NO2 groups will be turned somewhat sideways and the resonance partially broken. Whether this will overcome the increased electron withdrawing effect of five nitro groups, I don't know.

You can certainly make pentanitrophenol but it is way more harder to get the -NO2 group to occupy the last remaining m-places. Historically, it has been achieved previously in the early 1900s but the way to synthesis was not straightforward. They had to make many observations, understand the chemical behavior, analyze, assume things and then come to a conclusion. They have to start with halogen substituted aniline and phenol. They didn't know how the nitric acid will behave towards nitrated m-derivates of aniline and phenol and they understood that there was no way the extra -NO2 can be wedged in between o- and p- position. The only exception noticed was tetranitrophenylmethylnitramine discovered in 1889 which was obtained by direct action of fuming nitric acid on two different trinitrophenyldimethylanilines. They observed that the -NO2 group placed at the m-position is very mobile and very liable to substitutition.
Later they achieved tetranitrophenol by oxidizing oximes. In the same way, they subjected m-nitro and m,m-dinitromethylaniline with conc. nitric acid and proved that it is possible that the pentanitro- derivative is possible along with the tetranitro- derivative. Finally, came to m-nitro and m,m-dinitrophenol where they subjected them with conc. nitric acid and it formed the tetranitro and pentanitrophenol as they predicted (m.p 140 °C and 190 °C resp.).

Once they achieved this, they went ahead and achieved nitrating m-substituted phenol where the substituted group is something beside -NO2 group.
All of their findings can be found in the below journals:

On pentanitrophenylmethylnitramine and tetranitro and pentanitrophenol by Dr. J.J. Blanksma, 1901
Nitration of meta substituted phenols by Dr. J.J. Blanksma, 1906


text:
I need a good 3d drawing of ferropericlase
I'm not a scientist, I'm an artist and I need a good 3d drawing of ferropericlase, where can I find that?
Thank you!
Niki

try https://www.mindat.org/min-3161.html#autoanchor11

It's not clear what you mean by "3D drawing". Your tags refer to molecular structure and molecular orbital theory. Is it a drawing showing the schematic arrangement of atoms in a crystal of the substances what you are looking for?

The question is a bit fuzzy, however as you are in NYC, to get in touch with the group around Bart Kahr both interested in crystal growth/structure/properties as well as the visual arts might be fruitful, https://www.kahrlab.com/

Ferropericlase is a solid solution of $\ce{FeO}$ and $\ce{MgO}$ having a rocksalt structure. Your best bet is a rocksalt schematic with the cations randomly labeled $\ce{Fe^{2+}}$ and $\ce{Mg^{2+}}$.


text:
I get dizzy & light headed when go into store which sells helium ballons
I get dizzy when I go into a party store which sells helium ballons and need to get to fresh air outside ASAP. Am I sensitive to helium or why does this happen. I feel like I am going to fall down and very shaky on my feet while trying to get to door to leave the store.  Have had this happen at two party stores. If it is not helium sensitivity have not idea what causes it.

It's very hard to imagine a biochemical mechanism for helium sensitivity. Even if the correlation is real, it's vastly more likely you're making a wrong inference, and something other than helium is causing the effects - an unblinded qualitative statistical study with n = 2 samples isn't good enough evidence for that rather unusual hypothesis.

Shops that sell helium balloons also sell many other unusual items. How do you know the cause of your dizziness is helium?

Helium itself is inert, with no physiological clue to its presence. That has caused death of workers inside dirigibles and of these students. There is no smell, dizziness, or other sensation caused by helium, and that can be dangerous.
Here is another possibility: some businesses employ ultrasonic door openers and intrusion detection. Ultrasonic vibration can bother some people's middle ears, throwing off balance -- this happened to an acquaintance as we entered a department store. Walking back and forth, we quickly found the ultrasonic source. Is it possible that while filling balloons, an ultrasonic squeal is emitted? You could get or make a simple ultrasound converter to check... and have fun listening to bats, mice, and dolphins, too.


text:
Does hydrogen chloride gas react with metals?
I searched and couldn't find any reactions with $\ce{HCl (g)}$ as a reactant, only with the acid in the aqueous phase ($\ce{HCl (aq)}$). Do metals react with $\ce{HCl}$ gas to form hydrogen?

Umm... which metals -- Au? Cs? At what temperature -- 4.3 K? 1,273 K?

Any. My question is does it react or not in gaseous state.

Gaseous $\ce{HCl}$ do not react as an acid. For example, it does not react with calcium carbonate.

@Maurice, the OP was asking about reaction with metal, not acid/base reaction. However, it's too broadly worded to answer. Likely, it would react with Cs at 2,000 K, though I have no wish to test that. AS to whether H2 would be released, or form CsH, that's another question.

HCl gas would certainly react with any adsorbed water on the surface, and I'd guess that no surface exposed to the atmosphere is absolutely dry.

Worse yet, gaseous HCl will condense water from the atmosphere, and then...

I have found no definitive evidence for a reaction of hydrogen chloride gas with metals, but it can react with bases that do not require a redox reaction. This answer, citing References [1-3], discusses the anhydrous reaction between hydrogen chloride and ammonia. Combination of single molecules of $\ce{HCl}$ and $\ce{NH3}$ gives a hydrogen-bonded $\ce{NH4Cl}$ monomer rather than an ionic one; the proton is not fully transferred. The fully proton-transferred, ionic product is formed when this monomer is polymerized by further reaction.
Hartman et al. [4] describe the reaction of hydrogen chloride with solid sodium carbonate at 500°C, forming the familiar products seen in the aqueous reaction (sodium chloride, carbon dioxide, water). In this case the formation of water implies that the hydrogen chloride does not remain dry. Presumably it forms hydrogen-bonded monomers or oligomers with the water, similar to the ammonium chloride monomer described above. See also Ref. [3].
References

Robert A. Cazar, Alan J. Jamka, and Fu-Ming Tao (1998). "Ab Initio Investigation of Proton Transfer in Ammonia−Hydrogen Chloride and the Effect of Water Molecules in the Gas Phase", J. Phys. Chem. A., 102, 26, 5117–5123.

Richard J. Countess and Julian Heicklen (1973). "Kinetics of particle growth. II. Kinetics of the reaction of ammonia with hydrogen chloride and the growth of pariculate ammonium chloride", J. Phys. Chem. ,77, 4, 444–447.

Galina M. Chaban, R. Benny Gerber, and Kenneth C. Janda (2001). "Transition from Hydrogen Bonding to Ionization in (HCl)n(NH3)n and (HCl)n(H2O)n Clusters:  Consequences for Anharmonic Vibrational Spectroscopy", J. Phys. Chem. A, 105, 36, 8323–8332.

Miloslav Hartman, Karel Svoboda, Michael Pohořelý, Michal Šyc, Siarhei Skoblia, and Po-Chuang Chen (2014).
"Reaction of Hydrogen Chloride Gas with Sodium Carbonate and Its Deep Removal in a Fixed-Bed Reactor".
Ind. Eng. Chem. Res. , 53, 49, 19145–19158.
https://doi.org/10.1021/ie503480k


text:
Calculation of formation heat given molar heat capacities and reaction enthalpy

Consider the following reaction for the formation of liquid water at $\pu{298 K}:$
$$\ce{H2(g) + 1/2 O2(g) -> H2O (l)} \quad \Delta_\mathrm rH = \pu{-286 kJ/mol}$$
If molar heat capacities (at constant pressure) of $\ce{H2(g)},$ $\ce{O2(g)},$ $\ce{H2O(g)}$ are given as $39,$ $29$ and $\pu{75 J K^-1 mol^-1},$ calculate the heat of formation of $\ce{H2O(l)}$ at $\pu{348 K}.$ (Assume that molar heat capacities are independent of temperature in given range)

I tried solving this problem by taking $\Delta H = nC_p\Delta T$ and I added but I am getting a wrong answer. The data given is bit unusual (in thermodynamics we frequently work with enthalpy more). Any hints to the answer?
My problem is sourced from my prep material for JEE(Main).

Have you tried using the heat capacity of liquid water over the 50 C temperature range.

I am sorry for the late reply everyone, I was stuck in Cyclone Michaung for the past few days and could not check on anything. I had no internet for the past few days.

Regarding @ChetMiller's comment: I have not done that. Maybe I'll try and get back to you

@ChetMiller, nope I could not get the answer.

It seems there is an error in the task description. $\ce{75 J K-1 mol-1}$ is the approximate molar heat capacity of $\ce{H2O(l)}$  ( $\pu{4.18 J K-1 g-1 \cdot 18 g mol-1 \approx 75.2 J K-1 mol-1}$  ). The molar or specific heat capacity of $\ce{H2O(g)}$ is about a half of that. Considering the reaction related to the considered formation enthalpy involved $\ce{H2O(l)}$, it all points out some error.

I agree with @Poutnik's assessment.

You can apply the Hess law (Wikipedia, Libretexts), mere saying the change of enthalpy (or more generally any state variable) depends only on the starting and ending state, not on the path how the end state has been reached.
Applying it on the water formation enthalpies at different temperatures:
The enthalpy of liquid water formation at $\pu{298 K}$ is equal to

the enthalpy change to warm up respective elements from $\pu{298 K}$ to $\pu{348 K}$
plus the enthalpy of water formation at $\ce{348 K}$
plus the enthalpy change to cool down water back from $\pu{348 K}$ to $\pu{298 K}$.

Calculation:
\begin{align}
\Delta_\text{f} H_\ce{product}^{T_1} &= \Delta H_{\text{elements},T_1 \to T_2} + \Delta_\text{f} H_\ce{product}^{T_2} +  \Delta H_{\text{product},T_2 \to T_1}\\
\Delta_\text{f} H_\ce{product}^{T_1} &=  C_\text{elements} \cdot (T_2 - T_1) + \Delta_\text{f} H_\ce{product}^{T_2} +  C_\text{product} (T_1 - T_2)\\
\Delta_\text{f} H_\ce{H2O(l)}^{298} &=  (C_\ce{m,H2} + \frac 12 C_\ce{m,O2})(\pu{348 K} - \pu{298 K}) \\
&+ \Delta_\text{f} H_\ce{H2O(l)}^{348} +  C_\ce{m,H2O(l)} (\pu{298 K} - \pu{348 K})\\
- \pu{286 kJ mol-1} &=  (\pu{39 J K-1 mol-1} + \frac 12 \pu{29 J K-1 mol-1})(\pu{348 K} - \pu{298 K}) + \Delta_\text{f} H_\ce{H2O(l)}^{348} \\
&+  (\pu{75 J K-1 mol-1}) (\pu{298 K} - \pu{348 K})\\
- \pu{286 kJ mol-1} &=  (\pu{53.5 J K-1 mol-1} - \pu{75 J K-1 mol-1})(\pu{50 K} ) +
\Delta_\text{f} H_\ce{H2O(l)}^{348}\\
\Delta_\text{f} H_\ce{H2O(l)}^{348} &= -\pu{286 kJ mol-1} - \pu{1.075 kJ K-1 mol-1} \approx \pu{-287.1 kJ mol-1}
\end{align}


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15N NMR simulation with orca software
I would like to simulate J(15N-1H) couplings with orca. By default, the software uses the 14N isotope for NMR nitrogen simulation. In the program, in the xyz input, I added M 15 on the line of the N to declare the 15N isotope for the simulation. In the report, I can see it has been taken into account (mass =15) but in the coupling constant part of the report, nitrogen is refered to as 14N. Is there another way to set the 15N information onto the program?

Please include more information for your question, e.g. the input file. Your question won't be well received here as it is heavy on technical doing and very light on the actual chemistry. Once you've improved your question, I suggest migrating it to [mattermodeling.se]. Please do not cross post it there.

@Martin-マーチン This might fit (better) at Material Modeling but I think questions about simulation of chemical spectra are well suited for our site.

@Buck agreed, in general.


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Theoretic (QED) Atomic Ionization Energies Database?
The 2019 paper "QED tests with highly-charged ions" contains a survey of QED-calculated atomic ionization energies.  I can manually copy the values of interest therein, but it seem that theoreticians attempting to improve on algorithms used to calculate such things would benefit from a benchmark or gold standard database.  Does such a database exist?

Might the NIST database be of use to you ? https://physics.nist.gov/PhysRefData/ASD/ionEnergy.html

By "gold standard" I meant as a calculation gold standard -- in other words, QED-calculated figures, while they may not be the "gold standard" in the absolute (empirical) sense, they are in the sense that other algorithms should approach the QED-calculated figures and attempt to do so with fewer compute resources.


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Can SMILES name any possible chemical?
Similar: Does IUPAC nomenclature have the ability to name all organic compounds? - The answer to this question was no.
A relevant statement from the linked, though it is not explained:

The closest thing to an absolute method of describing a compound's structure is to have a table of positional data (XYZ coordinates) giving the relative positions between the atoms determined from X-ray/neutron diffraction. Any attempt at simplifying this data will be lossy, whether the structures are drawn (not too lossy) or named (very lossy).

Do SMILES, InChl, or any other chemical naming system have the capacity to name all possible chemical compounds? What would be required to demonstrate that a certain naming system has this property?

Not if the compound is unknown, e.g. a peak in a GC-MS analysis that does not match any known compound.

What is exactly the point of this question? Why not ask straight out what are the limitations of the systems. A few days ago someone asked whether they could be used to transmit information about orbital occupancy (answer is no).

These naming systems (similarly to any other but uniquely in the way they are encoded) are designed for exchange of particular information and are entirely utilitarian. Presumably they are updated if novel unforeseen structures should present themselves.

On the philosophical (scientific methodology) question: https://en.wikipedia.org/wiki/Inductionism


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Reagent for conversion of 1-phenylpent-1-en-3-one to 1-phenylpent-1-en-3-ol

This conversion was recently asked in the NSEC, an Indian (preliminary) chemistry olympiad for high schoolers and the official answer key stated LiAlH4 is the correct reagent to be used. However I have come across LAH reducing double bonds conjugated with benzene rings so I thought the double bond would be reduced if it was used. Therefore I assumed the birch reduction would be the appropriate method, but would that reduce the ring itself into some sort of diene? Or is that just for unsubstituted benzene?
The other options were the clemmenson (which would kill the carbonyl)
and H2/Pd-C (which should kill the double bond)

We had this come up recently and maybe someone smarter than me can find the discussion. The real world answer is that none of the four reagents offered will do this reliably. Modified borohydride is the way to go, either Luche reduction https://en.wikipedia.org/wiki/Luche_reduction   or sodium monoacetoxyborohydride https://pubs.acs.org/doi/10.1021/jo00284a046

Found it   https://chemistry.stackexchange.com/questions/177916/does-h2-pd-c-reduce-a-conjugated-ketone

Lanthanide modified LiAlH4 will also give the allylic alcohol  https://pubs.rsc.org/en/content/articlelanding/1986/p1/p19860001929

Thank you so much! A lot of times in these types of questions its more of which of the following is least wrong, more than which does the job effectively haha. It is difficult to teach us so much material in such a short time, so yeah nuance is lost sometimes.


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Spontaneity of irreversible reactions
I saw this reaction in my textbook:
$\ce{NH4Cl(s) -> NH3(g) + HCl(g)}$ given $\Delta{H}_{r} = \pu{176 kJ/mol}$ and $\Delta{S} = \pu{0.285 kJ/(mol \cdot K)}$
On a worksheet my teacher gave me, I saw the reverse reaction:
$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$ given $\Delta{H}_{r} = \pu{-176 kJ/mol}$ and $\Delta{S} = \pu{-0.285 kJ/(mol \cdot K)}$
Both reactions are said to occur at $\pu{298.15 K}$ and $\pu{1 atm}$ of pressure.
I know that when reversing the reaction, the sign of $\Delta{H}_{r}$ must be reversed as well. Doing the same for $\Delta{S}$ makes sense as well due to the state change.
The change in free energy in the first reaction is $\pu{+91 kJ/mol}$, whereas in the second it is $\pu{-91 kJ/mol}$.
Would it be correct to infer from this, that when an irreversible reaction is spontaneous, its reverse is nonspontaneous (under the same conditions) and vice versa?

Direction of spontaneity and the value of reaction Gibbs energy depends on partial pressures of gases.

Yes, it is correct to infer this, but only if the substances are in their standard state for which the values of $\Delta H$ and $\Delta S$ are given. For other temperatures and partial pressures, as mentioned by @Poutnik, you have to use $$\Delta G = \Delta G^\circ + RT \ln Q$$ where $Q$ is the reaction quotient.


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Regioselectivity for double elimination of dihalides
I came across a question shown below,

When predicting the product, I thought the internal alkyne would be preferred over the terminal alkyne as the former would be more substituted. However, the book states that the terminal alkyne should be the answer.
I was thinking that this could be due to the formation of the acetylide ion, which favors the terminal alkyne formation. Would this assumption be correct?
If that is the case, what is a way to form an internal alkyne using elimination?
Also, if the reagents were not excess $NaNH_2$ and just like 2 equivalence, would the internal alkyne be the product as the acetylide ion cannot form without three equivalence of sodium amide?
It would be great if I could receive some clarification over this!
Thank you :)

2-Butyne will be isomerized to 1-butyne with $\ce{NaNH2}$ and then deprotonated. The allene, 1,2-propadiene, is the intermediate. Water work up gives 1-butyne itself. It is called a "zipper reaction".


text:
Does Sugru (Formerol) decay underwater?
Sugru is described as waterproof and dishwasher-safe, but I still wonder if it will decay quickly (within months) if submerged in room-temperature tap water.
Aren't silicone rubbers susceptible to hydrolysis?

Mold can grow on silicone sealants (which is why they are sometimes mixed with mold suppressor) but they are otherwise a standard material for use in humid environments like kitchen sinks and bathrooms.

First, the silicone monomer does hydrolyze in water, causing crosslinking. However, the resultant crosslinked polymer is stable.
Second, silicone is, one might say, rabidly hydrophobic. For water to cause degradation, it would have to penetrate, but is confined to the surface. However, silicones are highly permeable to water vapor. so degradation might occur, depending on the formulation and additives.
As for the specifics for the proprietary formulation for Sugru, you'd need to test if it meets a specific need. It's ~50-75% silanes, the remainder talc and other substances, so that would would change its water resistance.

"the resultant crosslinked polymer is stable." -- so would it just become harder (less elastic)?

It is the process by which Sugru changes from a plastic form to an elastic solid.


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How does the Eschweiler Clarke reaction occur with sodium cyanoborohydride as a reducing agent?
I am specifically confused with 2 concepts.

How sodium cyanoborohydride acts as a reducing agent. I read that it acts as a source of hydride but I am not sure how this works. If anything, I assumed that the hydrogens bound to the boron atom would have a delta positive charge.

In the first step, the primary amine performs a nucleophilic attack on the carbonyl carbon of formaldehyde, generating an intermediate with a positive nitrogen and a negative charge on the formaldehyde oxygen. How does the reduction to the iminium intermediate occur? I read that the hydride attacks the formaldehyde carbon but why is that the case? Should the positive nitrogen not be a more attractive target? Or the negative oxygen attack first?

Essentially, I don't understand the logic behind it.

It probably boils down to proximity and five/six-membered rings in transition states and reactive intermediates.

What conditions do you consider the Eschweiler–Clarke? Typically sodium cyanoborohydride is not present. Instead, formic acid acts as the reducing agent.


text:
Is it possible to filter CO2?
I'm diving into the world of chemistry here, pondering a concept that might sound a bit out there. Can we actually filter out CO2 from a car's exhaust pipe and store it until it's safe to handle?
This idea is way beyond my beginner chemistry knowledge, but it's interesting, I'm curious: Are there methods in chemistry that could potentially filter and capture CO2 emissions from car exhaust? And if so, could we find a way to store this CO2 until it can be safely dealt with?
I'm also wondering about the challenges this kind of filtration might face. Are there chemical principles or reactions that could help trap CO2 while letting other gases through? And could this ever become a widespread solution for reducing vehicle emissions?
Can't wait for the thoughts, insights, or any existing research on CO2 filtration in car exhausts. Can we turn this idea into a mainstream reality, or is it still it still science fiction?
Thanks!

if you want to sequester CO2 from combustion, it is much more practical to do so at a stationary source such as a coal or gas-burning power plant than it is to do so in a vehicle.

Will always stay science fiction because of thermodynamics. For power stations where the heat is used for district heating, it might make sense.

It is possible as many things are. All is about scalability and money.
From $\pu{20 L}$ of gasoline/petrol is formed roughly $\pu{44 kg \ce{CO2}}$ and $\pu{18 kg}$ of $\ce{H2O}$. If captured by $\pu{80 kg}$ of $\ce{NaOH}$, it would form $\pu{106 kg \ce{Na2CO3}}$ and in total with the input $\pu{36 kg}$ of $\ce{H2O}$. Accordingly heavy supporting material and construction.
Cars would need quite a huge, heavy and expensive to maintain absorption unit.  It would also lead to big challenge for engine power. And I do not see anybody would buy such a car nor handle so big amount of material.

Trade off Pressure needed to push gases through  versus efficiency of absorption and related mass and volume.  // As an extreme example - imagine somebody would put a bunch of old cotton/linen clothes to the tubing of the engine exhaust, so the engine would need significantly higher pressure to blow exhaust gases out. Or that there is inserted some element with many narrow gas passes, again with the pressure drop on it.

My bad my keyboard is a bit funky, I was think something like this just to clarify!

It is a fake to pull money from pockets for false good feeling you did something good for environment. It may work, but for very small amount and flow rate of CO2. Not for more than 100 mL of gasoline, probably less. And with its small size, it cannot help itself but to allow to bypass of most of the exhaust gases.

Oh I see, It is possible, although it would be very exspensive and large is what im getting. I planning on entering this idea into a Young Entrepreneurs Challenge thing, I agree it needs refining, Is there anything possible to make it more efficient/affordable for the average consumer? Its been a decade since the idea was proposed hence I think there could be a space to make it better. Thanks alot for the help!

Rather, new ideas are seldom new ideas and if they were abandoned, there is good reason for it.

You need to compress the exhaust gas, remove all the water it contains, and then compress it further until the CO2 condenses, which you can fill into standard gas bottles. Every litre of fuel turns into roughly three times the amount of liquid CO2.
The fuel consumption of your car additionally increases drastically to run the machinery for separating and liquifying the CO2, even if you could do so with prefect effiency. Spoiler: you can't, meaning the fuel consumption increases even more.
And then you still have those gas bottles which you'll have to pay somebody to empty for you and press the CO2 into an empty oil field or something.
I recommend you just buy yourself a smaller car next time. Or an electric one.


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Assignment of NMR peaks of polystyrene
I’m having trouble identifying the peak around 3,75 ppm in the 1H-NMR spectrum of polystyrene measured at 400 Hz in CDCl3. Could someone please provide their perspective on it? I see it appear in a lot of polystyrene spectra but never know what it could be caused by. Thanks in advance!

Maybe it's from terminal groups.

I thought so too, just a bit confused which ones, and why would those peaks be so shifted.

There can be multiple causes of this kind of trace impurity in a 1H-NMR spectrum. For example trace solvents.
What I think is the most likely cause (it depends on the synthesis used), is that this peak is caused by a trace initiator or trace terminating agent. For example a brominated alkene chain.
I wouldn't worry about it too much as the integral is very small.

Got it! Thank you so much for your answer.


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Reading and adjusting pH of reverse osmosis water
I grow things for a living and regularly use pH meters + strips to read and adjust the pH of various nutrient and pesticide solutions. Oftentimes, I have a need for nearly naked water with the pH adjusted down to 5.5-5.75. I'm aware that taking the pH of naked water isn't possible with the meter or strips that I use so normally I add about 5% tap water to reverse osmosis water so I can get an accurate pH reading, then adjust the pH down with phosphoric acid to my desired level. Recently I've been trying to track down the source of a minor fungal contaminant and I have reason to suspect the tap water so I'm trying to eliminate it entirely. My question is this:
If I fill a bucket with pure reverse osmosis water, then add an amount of phosphoric acid to the RO that I think will get me close to my desired pH, will the added phosphoric acid put enough ions in the RO to get an accurate pH reading with a standard glass ball pH meter and/or pH strips?
I've already tried and it seems to work but I really have no way of double checking. Generally if I have reason to believe that my meter needs calibration (I try to stay on top of it but it's a shared tool and not everybody treats it right) then I just pull out my trusty pH strips and check. But pH strips also don't work in naked water so I'm not really able to blindly trust those either.
Thank you very much for your time.

What is "nearly naked water" ? Is it water obtained by reverse osmosis ? What is the pH of this original water? Is it not $5.6$ ? Usually demineralized water has a pH not far from $5.6$ when in contact with air.

RO water is usually passed thru a charcoal filter to remove the chlorine from city water. If the filter is not maintained I have found that the effluent water becomes acidic from HCl formed by reaction of the chlorine with carbon. To check the acidity of the water add a small amount of neutral KCl to a sample solution that is protected from the air [parafilm works] Then add Phosphoric acid to give the desired pH and adjust your water appropriately. If the water is originally acidic check the RO system. Also Phosphoric acid is a better buffer than any HCl from a faulty carbon filter. It is possible that addition of the phosphoric acid will be accurate enough. this can be determined by testing the finished water without and with KCl.
Calibration of the pH meter is important. For low acidity use a flowing reference electrode not a gel filled reference electrode.

Umm,.. "from HCl formed by reaction of the chlorine with carbon?" The activated charcoal might act as a catalyst, but direct reaction with dilute HCl seems improbable.

Reaction With CHLORINE Carbon is oxidized various chlorocarbons. Cl2 is reduced to chloride. I am not sure how the solution becomes acidic possibly a concurrent disproportional of Cl2. I know this from experience with RO systems. Systems on chlorinated water became acidic and bitter tasteing. Problem solved by an in line carbon filter before the RO to remove chlorine. [never checked for redox potential, should have]. RO systems need monitoring and maintenance usually neglected in home systems.


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How did scientists prove that electron configuration was true if we do not know exactly where electrons are?
In Wikipedia, there is an entry for electron configurations
which explains all about them, but I did not find from this article how they were discovered. If we take into account Schrodinger's quantum mechanical model and Heisenberg's Uncertainty Principle, then we can only know where electrons are most likely to be, not their location. Can someone clarify if that means that we know they are in a specific sublevel but not its exact location (i.e., it's in 3p somewhere...??). Or are the configurations just a model, to help us understand that electrons have different energies, as mentioned in a closed question (Determining electronic configuration)?

You can know the electron is in an orbital without knowing its spatial position. Kind of the point of quantum mechanics, actually.

What's this supposed to be, "what is this quantum mechanics all about?" question? You should be well aware what electron configuration (and orbital!) is before asking how it can be determined.

You may consider protecting yourself and vaccinate against microbes, even if you don't know  the microbe structure.

There's an either-or statement in your question that you can remove. An electron configuration is a simplified representation of an approximate model of the actual electronic structure. It is an abstraction that attempts to represent some relevant properties of the electrons in the atom. All electrons have the same rest mass, intrinsic angular momentum and charge, but their energy, the magnitude and orientation of angular momentum, and the orientation of intrinsic angular momentum (spin) can differ.
There is one wavefunction that describes the entire atom. Single electron atoms can be described approximately using an extension of the hydrogenic formula (obtained by applying the Born-Oppenheimer approximation to the Schrodinger equation for the system of electrons and nucleus and solving the resulting differential equation) by simply altering the nuclear mass. For multi-electron atoms, further approximations are necessary. Perhaps the simplest that can be made is to express the wavefunction as a product of single-electron wavefunctions. This is where electronic configurations come in. They can be interpreted as shorthand notation for a product of single-electron wavefunctions.
However this (the information in the electron configuration) only gets you so far.
Electrons are also indistinguishable fermions and the wavefunction of a multi-electron system is a fermionic wavefunction, which has particular anti-symmetric properties under electron exchange (if you put labels on the electrons in the wavefunction, swapping a pair changes the sign of the wavefunction). This requirement can be accommodated by writing the electronic wavefunction of a multielectronic atom as a sum of permutations of the product of single-electron spin-orbitals (see for instance Hartree-Fock theory and Slater determinants).
In addition, the electronic wavefunction is known to be complicated by electron-electron interactions. As soon as you add more electrons that hydrogenic formula loses accuracy, even if Slater showed that it can be used semi-empirically to make some very reasonable predictions. It's good enough for some of us, as an old teacher of mine used to say.


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Calculating partial pressure: Difficult cubic equation in bulky Chemical Equilibrium numerical
The problem, which I found in a popular book for Physical Chemistry, written by Neeraj Kumar, is fairly simple, being simply 2 equations and 2 variables, yet its simplification is a nightmare.

$\pu{5 atm}$ $\ce{AB(g)}$ and $\pu{0.5 atm}$ $\ce{BC(g)}$ are introduced in a reaction vessel and the following equilibriums are established:

$$\ce{AB(g) <=> A(g) + B(g)} \qquad K = \pu{10^{-7} atm}$$
$$\ce{BC(g) <=> B(g) + C(g)} \qquad  K = \pu{10^{-6} atm}$$

If the equilibrium partial pressure exerted by $\ce{B(g)}$ is around $10^{–n}$, determine the value of $n$.

We start of by writing the simultaneous equilibriums

$$10^{-7} = \dfrac{x(x+y)}{5-x}\tag{...1}$$
$$10^{-6} = \dfrac{y(x+y)}{0.5-y}\tag{...2}$$
Dividing the 2 gives us:
$$10^{-1} = \dfrac{x(0.5-y)}{y(5-x)}$$
This further simplifies to:
$$y=\dfrac{5x}{5+9x}\tag{...3}$$
We now have $y$ in terms of $x$, so we can substitute $\mathrm{(3)}$ in either $\mathrm{(1)}$ or $\mathrm{(2)}$, but substituting in $\mathrm{(1)}$ gives an easier simplification.
Substituting $\mathrm{(3)}$ in $\mathrm{(1)}$:
$$10^{-7} = \dfrac{x(x+\cfrac{5x}{5+9x})}{5-x}$$
This simplifies to give us:
$$10^{-7} = \dfrac{x(9x^{2}+10x)}{(5+9x)(5-x)}$$
It was the simplification of this cubic equation which gave me a problem.
Generally, approximations can be taken in such equations (by simply looking at the value of $K$, we know that $x$ will be a very small number) but here I don't see any. We can neglect $x$ in front of 5 in the $(5-x)$ term, but I doubt that makes the simplification of the cubic equation any easier.
On plugging the equation into Wolfram Alpha, it gives the following results which can be viewed here.
These are the results:

$x \approx 5 \times 10^{-4}$ is the only valid result in this case.
On plugging this value of $x$ into (3), we get:
y = $\dfrac{25 \times 10^{-4}}{5+9 \times 5 \times 10^{-4}}$ (we neglect $45 \times 10^{-4}$ in front of 5)
$$\implies y = 5 \times 10^{-4}$$
Therefore, the partial pressure of $\ce{B(g)}$ at equilibrium is $= x+y = 2 \times 5 \times 10^{-4} = 10^{-3}$
My only qualms are in solving the cubic we had initially:
$$10^{-7} = \dfrac{x(9x^{2}+10x)}{(5+9x)(5-x)}$$
Without Wolfram Alpha, how would we solve this cubic equation? Approximating $(5-x) \approx 5$ doesn't seem to help our case much. Is there any way to solve this without using a computer, simply using pen and paper?
P.S.: This problem was actually asked in 2013 to high school students in a written competitive exam called JEE, and in that exam, the students only have 5 to 6 minutes for each problem, so imagine the sheer skill required to solve such a difficult cubic equation and get the answer in that time-frame.

https://www.math.utah.edu/~wortman/1060text-tcf.pdf

Usually when the calculations are too tough JEE awards it as a bonus. One more thing, you have given only JEE and not specified whether it is Main or Advanced.

@HarikrishnanM I believe it was the the first year JEE Advanced was conducted.

Now coming to your question: I tried searching the archives on the JEE Adv 2024 website but could find none. https://jeeadv.ac.in/archive.html . It would be great if you could post the problem image.

A relatively straightforward solution is to start with your two equations:
$$10^{−7}=\frac{x(x+y)}{5−x}$$
$$10^{−6}=\frac{y(x+y)}{0.5−y}$$
In both cases, the variable in the denominator is negligible relative to the constant, so we simplify both:
$$10^{−7}=\frac{x(x+y)}{5}$$
$$10^{−6}=\frac{y(x+y)}{0.5}$$
If we divide both sides of the second equation by 10 to make the denominator an integer, we see that both equations are identical except for replacement of $x$ with $y$, which tells us that $x=y$. That allows us to substitute $x$ for $y$ to yield
$$10^{−7}=\frac{x(x+x)}{5}$$
$$\implies 5\times 10^{-7} = 2x^2$$
$$\implies x^2=25\times 10^{-8}$$
$$\implies x = 5\times 10^{-4}$$
$$\implies x+y=10^{-3}$$
$$\implies n=3$$

Nice! I've never known a book problem to require solving a cubic equation. There always seems to be some simplification that reduces the problem to a quadratic at most.


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Acidity of H4P2O7 and H3PO4
As given in pKa of H4P2O7 and H3PO4. The acidity of pyrophosphoric acid is clearly stronger than phosphoric acid. The two acid structures is given below.
Thought: Consider the pyrophosphoric acid dissociates first proton on the left, the stronger acidity might be attributed to the other phosphate group on the right. In other words, I think that the induction effect of the other P=O group results to O-H bond more polarized compared to H3PO4
Is it true in this case?

Have you checked whether the effect is not simply statistical? One molecule has 4 x OH, the other 3 x OH (correction!) that can dissociate. That alone can explain why one has a higher acidity. To figure out if that is the case, assume you were determining the pKa of a substance but were mistakenly using 4/3 x  the concentration that you thought you were. How would this affect the pKa determination?

Yes, your thought process is on the right track.
Reason: The enhanced acidity of pyrophosphoric acid (H₄P₂O7) compared to phosphoric acid (H₃PO₄) can be attributed to the presence of the additional phosphate group. The induction effect of the adjacent P=O group in pyrophosphoric acid makes the O-H bond more polarized, making it easier for the molecule to donate a proton, thereby increasing its acidity. This effect is a result of the electron-withdrawing nature of the phosphate group, which strengthens the acidity of the compound.


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Acidity of formic acid and 3-fluoropropanoic acid
I'm getting conflicting answers when I want to find out the acidity of 3-fluoropropanoic acid and formic acid.
I argue that 3-fluoropropanoic acid is stronger acid than formic acid as it has a -I group (although at a distance) whereas formic acid doesn't. However looking at the pKa table suggests 3-fluoropropanoic acid is weaker that formic acid!
Can anyone tell me what other factors have I missed?

Factors influencing the acid strength of carboxylic acids
For starters, it should have been considered that the acid strength of carboxylic acids depends on

electron releasing groups (which weaken the acid) / electron withdrawing groups (which strengthen the acid).

Number of such groups (higher number of ERGs will lead to weaker acids, while higher number of EWGs will lead to stronger acids)

The distance of the group from the carboxyl group (an EWG kept closer will lead to a stronger acid; an ERG kept further away will lead to a stronger acid.)

Inductive effect, due to the $\ce{-\color{red}{F}}$ group, is distance dependent, which means that the effect will become weaker as the group is placed further away. Check out the below example:

Apply a similar logic to the situation.
The reduction in $\mathrm{\color{red}{-I}}$ effect of fluorine is much more and because of that the formic acid turns out to be stronger.
Also, 3-fluoropropanoic acid has one EWG, but some ERGs before it [$\ce{\color{navy}{F}-\color{blue}{C2H4}-\color{magenta}{COOH}}$], and the alkyl group in between has some electron releasing tendency. There is competition between those two effects and $\ce{-\color{red}{F}}$ being further away from the COOH group leaving a $\mathrm{\color{green}{+I}}$ group in between acts as the differentiating factor.
Thus the order turns out in this fashion.

I would suggest you to read these paragraphs from our Chemistry reader:

Image sources:

https://shorturl.at/qzFSY (a Google image)

https://ncert.nic.in/textbook.php?lech2=3-5

This answer is derived from my learnings from NCERT, Class XII, Chemistry Volume II.

But in general, there is some EWG in 3-fluoropropanoic acid, however there is no EWG in formic acid. So shouldn't fluoropropanoic acid be slightly more acidic than formic acid? And how do we account for 'wee bit stronger' formally?

Question is edited to clarify. Also the quantifier 'wee bit' has been removed for prevention of confusion.

But we don't tell the same argument when we want to find the acidity of Chloracetic acid or fluoroacetic acid right? We never say the -CH2- group's inductive effect alters the acidity of the acid.

It depends on the context what factor we consider. In this case the halogen is attached to the compound as $\ce{X-CH2-COOH}$. Only the -I order matters in this case, as both are the same distance away.
I will upload a snapshot from my chemistry textbook that gives the complete order of all carboxylic acids. Read whatever factors I have said, understand those factors completely and then have a look at the order.

The edited answer should be able to satisfy all concerns.


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Steglich Esterification in Alkyl Gallates
Based on my research (reading some articles), some alkyl gallate syntheses used Steglich esterification instead of the usual esterification (Fischer esterification). The reason why Steglich esterification is used is because it allows the conversion of sterically demanding and acid-labile substrates.
However, I didn't understand which substrates are sterically demanding or acid-labile in alkyl gallate syntheses.
My question:
If I want to synthesize ethyl gallate, isobutyl gallate, and hexyl gallate, do I need to do Steglich esterification? or I can just do the usual esterification (Fischer esterification)?
Note:
Alkyl gallate synthesis uses alcohol and gallic acid to produce alkyl gallate

Who still uses Fischer? Well, there's bound to be some, but I doubt it was still the usual method for many decades.

As far as gallates (lowercase!) are concerned, gallic acid itself is too sensitive for heating it with H2SO4 - it can be oxidised even by air!

This paper appears from the abstract to be relevant  https://onlinelibrary.wiley.com/doi/abs/10.1002/recl.19510700310

Thanks for answering, I think it already answered my problem


text:
Why does a small decrease in [ATP] cause a large increase in [AMP]
I was wondering how the reaction catalysed by adenylate kinase and its equilibrium of approximately 1 leads to [AMP] changing greatly with a small change in [ATP]. The below information was provided but I fail to see how the percentage changes were achieved. If anyone could point me in the right direction for a mathematical/equilibrium calculation as to how these percentages might come to be I would be incredibly grateful.

$\ce{2ADP <=> ATP + AMP}$
The equilibrium constant for this reaction is given by $K_\mathrm{eq} = \dfrac{[\ce{AMP}] [\ce{ATP}]}{[\ce{ADP}]^2}$ and has a value of near 1.
Therefore $[\ce{AMP}]$ ~ $\dfrac{[\ce{ADP}]^2}{[\ce{ATP}]}$
As [ADP] rises during muscle contraction, it is rapidly converted into AMP and ATP, and because [AMP] concentrations are only ~2% of [ATP] a 10% decrease in [ATP] will result in a ~400% increase in [AMP].

[AMP] concentrations are only ~2% of [ATP]

If the product of AMP and ATP concentration is equal to the square of the ADP concentration, the ADP concentration is $\sqrt{0.02}$ of the ATP concentration, so 0.14 of the ATP concentration. In a well-energized cell, therefor, you could have concentrations of 5 mM, 0.7 mM, and 0.1 mM for ATP, ADP and AMP, respectively, for example.
If you use up (that is hydrolyze to ADP) 10 % of the ATP, the new ATP concentration would be 4.5 mM and the new ADP concentration would be 1.2 mM. With the concentration of AMP still at 0.1 mM, the system would be out of equilibrium:
$Q = \frac{\pu{4.5 mM} \cdot \pu{0.1 mM}}{\pu{1.2 mM}^2} = 0.28$
So, some of the ADP will react to form AMP and ATP. The approximate concentrations at equilibrium will be 0.21 mM, 0.97 mM and 4.4 mM for ATP, ADP and AMP, respectively. So the AMP concentration more than doubles when 10% of the ATP is hydrolyzed to ADP.
I'm not sure why the claim is an increase by 400%. In this example, I set the ATP concentration to 5 mM. This was arbitrary but does not affect the result because the problem scales linearly (e.g. if you have the system at equilibrium and dilute everything by a factor 10, it is still at equilibrium).

I believe what's being said here is that if you have a cell and it contains 1000 molecules of ATP, then there will be ~20 molecules of AMP (2% of 1000). If 100 molecules of ATP (10% of 1000) are used up they form 100 ADP. Using the equation you provided, you then get 50 AMP and 50 ATP from those 100 ADP. If there were only ~20 AMP to begin with, and you now have 20+50 = 70 AMP, then you have increased the amount of AMP by 70/20*100 = 350%. So not quite 400% (hence the ~400%) but close enough.
As a reference, you can see this paper where, they state discuss the ratio of ATP to AMP and ADP. I find it easier to think in terms of ratios instead of percentages for this case. From the paper:

Although produced in a few other reactions, the major cellular source of AMP appears to be the reaction catalysed by adenylate kinases, enzymes that catalyse the interconversion of the three adenine nucleotides (figure 1b). There is little change in free energy in this reaction, so it is readily reversible with an equilibrium constant close to 1 such that, at equilibrium, [ATP].[AMP]/[ADP]2 will be ≈1. If this reaction is indeed at or near equilibrium (which appears to be the case in most eukaryotic cells), and cells maintain an ATP : ADP ratio of 10 : 1 as discussed above, it follows that the ratio of ATP : AMP will be 100 : 1. In a cell with a fully charged ‘battery', the concentration of AMP will, therefore, be 100-fold lower than that of ATP and 10-fold lower than that of ADP.


text:
Basicity of pyridine and 2,6-di-tert-butylpyridine
Among pyridine and 2,6-di-tert-butylpyridine, which one would be more basic?
My guess was that since two bulky groups are substituted at ortho position in the latter, it will experience steric repulsions and hence it's lone pairs will easily be donated.
However, that doesn't seem to be the case.
Can someone please explain why steric factor does not lead to easy donation of lone pairs in this case?

https://en.wikipedia.org/wiki/Non-nucleophilic_base

Among pyridine and 2,6-di-tert-butylpyridine, which one would be more basic?

As Mithoron pointed out in the comment section, it's more fundamental than pretty obvious thing that large steric groups simply block access to atoms (and lone pairs). Thus it is fact that 2,6-di-tert-butylpyridine is a weaker base than pyridine because access to its lone pair is blocked by steric hindrance (Ref.1). According to Ref.1, the $\mathrm{p}K_\mathrm{a}$ of conjugate acid of 2,6-di-tert-butylpyridine in 50% aqueous ethanol at $\pu{25 ^\circ C}$ is 3.58 while that of conjugate acid of pyridine is 4.38. Brown and Kanner attributed this unusually weaker basicity  of 2,6-di-tert-butylpyridine to the steric strain involving the bound proton (Ref.1).
Keep in mind that the alkyl (electron donating) groups at appropriate positions would increase the basicity of pyridine by inductive effect. For example, the $\mathrm{p}K_\mathrm{a}$ of conjugate acid of 2,6-di-metylpyridine is 5.77 while that of conjugate acid of 2,6-di-isopropylpyridine is 5.34. However, the severe steric hindrance coursed by replacing the hydrogen atom in each of isopropyl group by a methyl group reversed this positive effect to other direction (Ref.1).
It is recommended that OP should read Ref.2 for additional information.
References:

Herbert C. Brown and Bernard Kanner, "2,6-Di-butylpyridine-An unusual pyridine base," J. Am. Chem. Soc. 1953, 75(15), 3865-3865 (DOI: https://doi.org/10.1021/ja01111a530).
Robert L. Benoit, Monique Fréchette, and Diane Lefebvre, "2,6-Di-tert-butylpyridine: an unusually weak base in dimethylsulfoxide," Canadian Journal of Chemistry 1988, 66(5), 1159-1162 (DOI: https://doi.org/10.1139/v88-190).