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What is the function of thin layer in thin layer chromatography?
I just learned about the thin layer chromatography, which I think is basically the same as the paper chromatography.
I am wondering what is the function of the thin layer? I mean, what's special about thin layer, compared to normal chromatography paper? What mechanism/principle/property makes the thin layer special?

Thin layer chromatography uses a stationary phase which is alumina or silica. These phases are much more polar than paper. So they are more efficient than paper for detecting impurities in a given mixture.

In addition, to the above, TLC compares well to column chromatography as for the stationary phase is the same or at least similar. Column chromatography is a purification method of outmost importance in organic chemistry.

If you recall Tswett's original discovery of chromatography, he used a calcium carbonate column to separate the chlorophyll pigments (or different types of chlorophylls). Now calcium carbonate is a very fine powder and the permeation of liquid through a powder is very slow, especially if the column is long. Pressure drop across a column scales as 1/($d_p^2$), where $d_p^2$ is the particle diameter.
Paper chromatography was well known in the 1930s, it was relatively faster but you are limited by choice of stationary phases, just cellulose. In order to have "paper chromatography" with other stationary phases such as silica, alumina etc., a thin layer chromatography was invented. The main purpose was speed of separation for quality control of pharmaceutical products, as it is today!
So this "thin" in TLC is akin to a thin paper sheet, but made of different stationary phases. Note that if you use a thick layer, the capillary action will not be uniform, and the bands of analytes will be distorted badly.

I think you should add something on the line of my comment under the question. In fact, by TLC one can check the completion of a reaction as well as the purity of the crude, but very often also deduce what would be the best column for preparative purification purpose.


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Does smoking drugs degrade them?
Why doesn't smoking a drug degrade or destroy it?
A wide variety of psychoactive compounds ("drugs") are commonly consumed via smoking – e.g., nicotine, THC, amphetamines, cocaine, and DMT.  But in every case, the desired drug is present in the uncombusted source, and AFAIK the potency of any of these organic compounds is not enhanced by either heat or oxidation.  (Indeed, when possible users seem to prefer inhaling vaporized or powdered distillates to inhaling smoldering compounds.)
So how do these drugs survive smoking?  I'm guessing that either (a) the drugs resist oxidation at the combustion temperatures of their carrier compounds, or (b) some fraction does oxidize.  I'm also suspecting that the chemistry of "smoking" might be quite different from the chemistry of efficient combustion.

I'm pretty sure smoking does degrade and destroy the active components at least to some degree. I guess enough of it is vaporised to achieve some physiological effect.

Likely many of the psychoactive compounds are volatilized by the warm combustion air flowing over the raw material. So, from the raw tobacco in the cigarette that you are inhaling through, not from the actual burning part.

Why the downvotes though? It's a perfectly curiosity-based question.

@KanghunKim I agree. It is a perfectly good question and there is no reason to downvote.

Why should drugs behave differently, compared to non-drugs? Fire does not distinguish, if organic matter is drug or not. Depending on substance nature, some is oxidized, some is thermally converted, some evaporates or sublimes.

@Poutnik fire doesn’t distinguish what is burned. But controlled fire creates heat that does volatilise active compounds. The burning in a controlled way is merely a means to achieve that goal.

@matt_black You say what I say, just by different words, except I should use rather burning instead of fire.

Smoking volatilises active compounds but may also degrade them
Many active ingredients in drugs are volatile alkaloids readily absorbed through the mucus membranes in the mouth, nose, and lungs. Nicotine, THC, and heroin are good examples of varying legality.
The reason why smoking is often used to deliver them is that smoking is a convenient way to deliver the volatile active ingredients into the body. Sure, burning will degrade some of the active ingredients in a cigarette, but burning the tip of the cigarette causes hot air to be drawn through the unburnt portion of the device which releases the volatile active ingredients in the unburnt part. This is convenient, though dangerous as many of the less desirable contaminants are also created as a side effect of burning. This is why cigarettes kill: it isn't the desired nicotine but the side effects of burning and the other, undesirable, ingredients that are also released in the volatilisation process. And, burning may destroy some of the active ingredients (though most users will get enough for this not to matter much).
This is also why electronic cigarettes can deliver nicotine relatively uncontaminated by those other products and are, often, effective substitutes for getting nicotine into addicts far more safely than via burning sticks of tobacco (the process of controlled heating can volatilise relatively pure nicotine without the side effects of less controlled burning and the source can be far purer than tobacco leaf).
Similar considerations apply to other active ingredients.
So, the bottom line is that, yes, burning is usually bad but mostly because it is not a very controlled way to deliver active ingredients. We use it because it is convenient and easy despite its drawbacks.

Fire or burning does not distinguish if organic matter is a drug or not.
Depending on organic substance nature

some of it evaporates or sublimes,
the rest is thermally converted and finally eventually burnt.

Note that most things in this world do not burn instantly, and those that do are usually useless as drugs(imagine smoking aluminium powder). You need some time.
Therefore, e.g. if you smoke a cigarette for 3 minutes, the 3 minutes are just too short for 'all' the nicotine to burn away(= 'some' of the nicotine just gets vaporised and ends up in the smoker's body), and hence cigarettes are not available to those under 19 because that 'some' small amount of nicotine is still harmful to minors.

N.B. The age of majority where I live is 19(18 de facto partially; e.g. when it comes to duty, voting, driving and mature movies/games) and so is the minimum age for alcohol/cigarettes.

I know that the former is 18/the latter is 21 in the USA

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What is the "frequency of occurrence" referring to in this paper?
In Zhou et al.(2018) 22. 10.1016/j.geoderma.2018.02.015, table 1 of the different microplastics refer to a "frequency of occurence".

Why does the % frequency of occurrence add up to >100% ? What is the difference between frequency of occurrence and proportion ?
https://www.researchgate.net/publication/323593126_The_distribution_and_morphology_of_microplastics_in_coastal_soils_adjacent_to_the_Bohai_Sea_and_the_Yellow_Sea

Presumably it’s the percentage of sampling sites that contained any amount of the given size

You ask:

Why does the % frequency of occurrence add up to >100% ?

It's an aggregate binary (yes/no) statistic - "at the sampling sites [plural]" - which tells you how many times ("frequency") items in the 7 classes show up ("occur") in the total sample population (here, 120 samples from 53 sites, 6360 samples).
So, at 32% of the sites, we find pellets; and at 43.4% of sites, we find sponges. It doesn't tell us which of the sites in particular but it does yield information about overall prevalence of certain type of microplastics at the sites.
The reason the sum is not 100% is because the numbers are not normalized. All they tell us is that at 32% of the 53 sites, we found pellets. At 43.4% of the 53 sites, we found sponges. And so on for the remaining 5 classes.

What is the difference between frequency of occurrence and proportion ?

The former is "does something in this class appear here (yes or no, not how many);" the latter is "of the total number of things collected here, take the number of things in each class and divide that number by the total and multiply the result by 100%."
So we see that pellets appear at about a third of the sites (32%). However, they account for only a very small percentage of the overall amount of microplastics at only 0.1%. Foams and flakes show up at a lot of sites and comprise large amounts of the types of detritus at those sites.

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Why isn't there hydrofluoric acid in fluoridated water?
From what I've read, when water is fluoridated it's done so either by $\ce{NaF}$ artificially or $\ce{CaF2}$ naturally. If this is correct, shouldn't the reaction $\ce{CaF2 + H2O -> CaO + 2HF}$ occur? Does something else in the water just react more readily?

No, it can never occur. Tap water cannot or will not contain free hydrofluoric acid.

Fluoride ion is only present at about 1 part per million. Not much higher can lead to mottling of teeth. And CaO cannot be in water: it reacts vigorously with water, producing calcium hydroxide. So the solution is very dilute and hydrofluoric acid would be almost entirely dissociated, meaning it would be present in negligible concentrations.

@EdV I see, so because CaO would immediately react with water, form Ca(OH)2 which would in turn react with HF to form CaF2 again and overall this just amounts to basically nothing happening? Or does it not even manage to form the CaO in the first place?

Calcium fluoride is an ionic compound that has low solubility in water, so what little dissolves just gives calcium and fluoride ions.

Generally speaking, if you have a chain of reactions that amounts to nothing happening, it just doesn't happen in the first place. It is as simple as that.

Your proposed reaction is unlikely
That is because the combination $\ce{CaF2 + H2O}$ (under tap water conditions) is much more stable than $\ce{CaO + 2 HF}$. In other words, the electronic energy of the former is lower and the system preferably stays in this state. That does not mean, however, that the reaction does not occur! It just means that the concentration of the products is incredibly small and probably you would not be able to detect them at all.
Because of this, chemists tend to say that it does not take place, although that is technically incorrect.
HF does occur, but in tiny concentrations.
I think there is a more likely way how HF would form.
Let us look at the reactions that are necessary to have $\ce{HF}$ in the fluoridated water.
The first step is the solvation of the solid fluoride ($\ce{NaF}$ or $\ce{CaF2}$) in water: $\ce{NaF2(s) <=> Na+(aq) + F-(aq)}$ or $\ce{CaF2(s) <=> Ca+(aq) + 2 F-(aq)}$, respectively. Sodium and calcium ions are stable in aqueous solution, so we can neglect them in the next steps. Just imagine that they are still in the water, surrounded by water molecules.
The second and last step is the fluoride ion reaction with water. According to Wikipedia, the reaction $\ce{F- + H2O <=> HF + OH-}$ has a $pK_b$ value of 10.8. This means:
$$ pK_b = -lg(K_b) \Rightarrow K_b = 10^{-pK_b} $$
$$ K_b = \frac{[\ce{HF}] \cdot [\ce{OH-}]}{[\ce{F-}]} = 10^{-10.8}$$
So here we can see that in a solution containing only fluoride ions and inert counterions, the portion of hydrofluoric acid is incredibly low! As mentioned above, you would probably not be able to detect it.
Cunningham
Cunningham's Law worked here for me in the comments.

Generally speaking, if you have a chain of reactions that amounts to nothing happening, it just doesn't happen in the first place. It is as simple as that.

That is a severe misunderstanding of chemical reaction dynamics. In fact, the exact opposite is true: If you have a chain of reactions, all of them happen! The question is only to which extent, which I tried to answer above.
See here and here for more details.

Are you sure $\ce{[H2O]}$ enters the formula for the $K_b$?

Your doubt was justified, I looked it up and it doesn't. Thanks, I edit the answer accordingly.

The point is that is not occurrence at a microscopic level that we are looking for, but outcome. I think that Neretin comment is appropriate, in this thread.


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Selective reduction of aromatic nitro group
In the following compound, is it possible to selectively reduce the aromatic nitro group to an amine, as shown in the reaction?

I fear that the methods usually used to reduce aromatic nitro compounds, such as Fe/HCl or catalytic hydrogenation, will also reduce the aliphatic NO2. Is there any reaction that could achieve reduction of only the aromatic one? Or, is it possible to somehow protect the nitro group from reduction?

What does Skyrim do to user's nicks...

While I cannot find you a reference, I think SnCl.2H2O in refluxing EtOAc will do this selectively as described here  https://www.sciencedirect.com/science/article/abs/pii/S0040403901800411. Sodium hydrosulfite in EtOH is also a possibility


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Is a diluted aqueous solution of sodium hypochlorite more stable than a concentrated solution, all else being equal?
On the Clorox website, they recommend that you dilute bleach (NaOCl) in water prior to use for disinfection. They say that you should make a fresh solution each time you need to use it, adding: “Don’t save a diluted bleach solution as it will degrade over time into salt and water.”

I agree that the sodium hypochlorite in the diluted solution will eventually degrade, but so will the sodium hypochlorite in the stock solution in the bottle.
Per LeChatelier’s principle, doesn’t the rate of decomposition of a compound increase with increased concentration?  When I looked up some chemical manufacturers' literature on sodium hypochlorite, I found information that confirmed this:

And from another manufacturer:

If we compare two aqueous solutions of NaOCl, one that’s 5% and one that’s 0.1%, then under typical storage conditions, and all else being equal, wouldn’t the 0.1% solution be MORE stable (i.e., have a smaller proportion degrade over time) than the 5% solution? If so, that would mean that you should keep your diluted bleach solution for later use, as it will be stronger than if you make a new dilution with the concentrated stock solution (from the bottle). Yet that's the opposite of what Clorox recommends.
The only issues I can think of here are the following:
(1) If they assume that the diluted solution will be stored in a transparent spray bottle, then it won’t be protected from the light, and thus will degrade faster, when compared to the same dilution stored in an opaque container.
(2) If they assume that the dilution will be done with tap water that has minerals dissolved in it, then these metal ions will accelerate the degradation of the NaOCl, when compared to the same dilution done using distilled water.
But, ALL ELSE BEING EQUAL, using opaque containers, distilled water, and recommended storage conditions, would a 0.1% solution not be more stable than a 5% solution?

I do not see considering pH change.

pH is certainly different, bleach is manufactured with excess lye to stabilize it in storage. Also the manufacturer cannot know the conditions of storing the diluted solution, could be an open bucket in the sun, won't last a day like that. And... they want to sell more bleach. I keep a dilute bucket in a generally dark bathroom and it stays effective for weeks, I try not to contaminate it too much though, only clean wash rags or removing tannin stain from washed coffee/tea cups.

Everything else equal, dilute solutions of hypochlorite are more stable than concentrated. The reason is that a main reaction for the degradation is:
3 NaClO -> NaClO3 + 2 NaCl
Which is strongly dependent of the hypochlorite concentration.
But when you make a use-dilution everything else is not equal. As Poutnik points out, pH is of great importance. For maximum stability it should be at least above 11 and preferably around 13.
And as you mention yourself: minerals and trace metals will catalyze the degradation.
Many manufacturers of hypochlorite issues detailed information about stability, handling and safety of hypochlorite solutions, e.g.: OXY


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Stereochemical outcomes for hydrohalogenation when there is a carbocation rearrangement
The following reaction is given:

I want to be able to tell which of the following stereochemical outcomes are possible.

I am aware that there are other possible products than the ones shown here such as if the methyl shift did not occur or if the carbocation was placed on the lower carbon on the C=C. But for when the carbocation is on the top carbon in the C=C and then rearranged, I was thinking all 4 products given here are possible.
Here is my reason:
After the proton transfer, there can be a methyl shift. When the methyl shift occurs, either methyl group (wedged or dashed) on that top-most carbon could shift. Thus, if the wedged methyl shifted, the products A and B represent that process. If the dashed methyl shifted,  and C and D represent that process. This is what I mean:

Next, after the methyl shift, the carbocation is formed. Cl can attack from either side. This explains the difference between 1 and 2 (where the wedged methyl shifted) in terms of the position of Cl, as well as the difference between 3 an 4 (where the dashed methyl shifted) in terms of the position of Cl.

Thus, all four products A, B, C, and D are formed. Is this correct or not? Thanks.

Yes to your question. By "dilute HCl" I hope you don't mean hydroxyl solvent (H2O, MeOH, etc.) because tertiary chlorides will undergo solvolysis.

This is correct. Since the methyl rearrangement can occur at either side of the planar carbocation, and Cl- can attack either side of the resulting planar tertiary carbocation, all four possible diastereomers are formed with equal preference. Here, you don't need to consider 1,3-diaxial interactions of cyclohexane either since the reaction occurs at carbons 1 and 2.


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Understanding electrolysis w/ oxalic acid electrolyte, iron electrodes
I've been experimenting with electrolysis for rust removal, electroplating, de-plating, etc. Right now I have a rust removal setup with carbon steel electrodes and a solution of oxalic acid in distilled water as the electrolyte. I've never used oxalic acid with electrolysis before (I've never used any acidic electrolysis process for de-rusting before). It seems to be working well, and quickly.
The setup is:

Half a teaspoon (4-5g if I did the math right) of oxalic acid crystals dissolved in about 350 mL of distilled water. That's ~10% of the concentration recommended on the box. It seems pretty dilute compared to what I'm used to (for bleaching wood).
Carbon steel anode and cathode. Cathode initially had some rust (brown) on it.
Running at 6V (which yields ~300mA) to keep heat output low.
About 2cm distance between electrodes.

When I turned it on, a sufficient amount of current began to flow, which I was hoping for but wasn't sure (see below). After a while, the solution turned a sort of orange-brown color which I'm assuming is iron oxalate, or at least it looks like a similar color. When the process is turned off, it settles very quickly. The cathode is relatively clean compared to a sodium bicarbonate process.
But, I'm not really sure if I'm accomplishing anything beyond splitting up water and whatever happens when I soak rusty metal in oxalic acid without electrolysis.
So my question is: What is happening in the electrolytic process compared to soaking rusty iron in oxalic acid without electricity, and what's the mechanism behind it? Or am I just wasting electricity?

From here down isn't really important.
The path I took to get to this process (for context) was:

I've only been using alkaline processes for rust removal, usually with sodium bicarbonate + water as the electrolyte.
I came across a video of somebody instead using acetic acid + sodium chloride + water as the electrolyte. The main points in the video:

Described as using electrolysis to "boost the performance" of vinegar (which seems to be the case).
Table salt as an ion source (from my own research I gathered this is because acetic acid is a weak electrolyte).
Stated that advantage of acidic process is stuff tends to stay in solution or at least not stick to the electrodes as much (no idea why, but I'm assuming it's for the same reason that e.g. copper plating fails when the pH is too low).

I decided to try it, but I was out of vinegar. I decided to use oxalic acid, based on:

I had a box of crystals on my shelf.
It didn't really introduce any new elements to complicate things further.
I looked it up and found it dissolves into oxalate ions, then I just sort of guessed  that at some point "iron oxalate" would exist.  I looked that up and patted myself on the back because it actually turned out to be a real chemical (lol); but more importantly, it wasn't going to instantly kill me.
The internet said it was a strong electrolyte so I thought maybe I wouldn't need the sodium chloride (ended up being correct).
If I did need to add salt, I didn't know what the reactions would be but I figured (uneducated guessing) it'd probably involve some combination of sodium oxalate, HCl, maybe ferric chloride (?), maybe some chlorine gas, and I have the appropriate setup to deal with (and dispose of) all of those safely.
If any zinc got in the mix (it happens) I figured it'd end up as zinc oxalate, zinc chloride (?), neither being a cause for concern. This was also an uneducated guess, though.

I set everything up, turned it on, and to my pleasant surprise, there was indeed good current flow without having to add any sodium chloride (a relief since I didn't have to test my guesses). But I'm not really sure if I'm accomplishing anything, so I headed here.

Comments are not for extended discussion; this conversation has been moved to chat. (Feel free to carry on there.)


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What exactly is oxidation number with regards to [Fe(CN)6]-4
I read that $\ce{Fe}$ has an oxidation state of $+2$ in $\ce{[Fe(CN)6]^{4-}}$. Does that mean $\ce{Fe}$ gives away $2$ electrons? Doesn't $\ce{(CN)6}$ form coordination bonds with $\ce{Fe}$ meaning $\ce{Fe}$ should have a negative oxidation number?

Why should coordination bonds affect the oxidation state of iron?  E.g. the oxidation state of Cu^2+ and CuCl4^2- is the same.

Please use LaTeX to edit the chemical formula of compounds . See https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange/86#4726 especially section 3.1

TL, DR: The problem is with formal charge, not oxidation state; but there is no paradox with either when chemists consider a more complete picture of the bonding in transition-metal complexes.
Contrary to the premise in the question, coordinate-covalent bonding need not affect oxidation state. Consider the ammonia molecule, $\ce{NH3}$, accepting a proton to form $\ce{NH4^+}$. The nitrogen atom provides both electrons to bond with the additional hydrogen, but the electrons remain polarized towards the nitrogen like the original $\ce{N - H}$ bonds. As a result the nitrogen retains the same oxidation state, -3, in both $\ce{NH3}$ and $\ce{NH4^+}$. Similarly, iron in $\ce{Fe(CN)6^{4-}}$ has the same +2 oxidation state as in "bare" $\ce{Fe^{2+}}$.
What does change with coordinate-covalent bonding is formal charge. Opposite to the convention used for oxidation states, formal charging regards covalent bonds as nonpolar even where that is far from true. So when the nitrogen in ammonia uses its nonbonding pair to accept a proton, it picks up a positive formal charge. If we regard the incoming proton as bare, then that hydrogen atom takes the  opposing drop in formal charge from +1 to zero. If the proton is regarded as coming from a more usual acid, let us say $\ce{H3O^+}$, then a bond between oxygen and hydrogen is also broken and thus the oxygen in the original hydronium ion takes the formal drop from +1 to zero.
This is all well and good for the nonmetallic species given above, where only a single coordinate-covalent bond is formed and chemiats are comfortable with most nometal atoms having either one positive or one negative formal charge. With transition metal complexes, however, multiple coordinate-covalent bonds appear and the metal may take a multiple negative formal charge even though it's oxidation state is positive. Iron in $\ce{Fe(CN)6^{4-}}$ has an oxidation state of +2 but apparently a formal charge of -4. Negative formal charges on metal atoms, which hold their valence electrons only loosely even when neutral, is an uncomfortable situation which most chemist circumvent in one of two ways.
You're not going anywhere!
One possibility with relatively electronegative atoms bonded to the metal, like chloride ligands, the bonding might be considered essentially ionic. The relatively normal oxidation state is thus more relevant to the molecular structure than the problematic formal charge.
A taste of your own medicine
The second possibility is seen with many carbon-based ligands that have pi bonding, like cyanide ion and the most common organic ligands. In such cases the pi antibonding orbitals in the ligand can overlap the $d$ electrons on the metal and pi-electton transfer can occur. This weakens the pi bonding in the ligand but in exchange, creates metal-ligand pi bonding. Ligands with this property are called pi acceptors. If this is rendered as contributing Lewis structures, it is seen as a delocalization of the pi bonding:
$\ce{\overset{-}{Fe} - C#N <-> Fe=C=\overset{-}{N}}$
As the name "pi acceptor" implies, metal electrons in this interaction are regarded as coming from the metal, canceling out the sigma electrons coming from the ligand. As the overset charges above indicate, this removes the paradoxical negative formal charges from the metal and sends them to the more natural home of more electronegative ligand atoms.

Thank you for your time. Reading this cleared this doubt and other queries I had.

You may want to consider accepting this answer if you think it helps you. Doing so is proper etiquette and with the q closed you may not get anything much better. Stay safe!


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Pressure increase in a chamber with 1 ml of water heated to 200 °C
For safety reasons I want to know how much pressure will be generated in a sealed chamber with $1~\mathrm{ml}$ of water, pure nitrogen at $1.3~\mathrm{bar}$ at $22~\mathrm{^\circ C}$. This chamber will be heated to $200~\mathrm{^\circ C}$. What is the new pressure in the chamber?
Chamber volume is $582~\mathrm{cm^3}$.

With acceptable accuracy consider water vapour as ideal gas. It would provide the upper constraint. The real pressure will be somewhat lower..

See https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange/86#4730   especially section 3.3

For that temperature, initial volume of liquid water, and chamber volume, all the water will evaporate.

Related: Minimum amount of water in pressure cooker

$1\ \mathrm{ml}$ of water at $p_0=1.3\ \mathrm{bar}$ and $T_0=22\ \mathrm{^\circ C}$ corresponds to a mass of $m_{\ce{H2O}}=0.99779\ \mathrm g$ and an amount of water of $n_{\ce{H2O}}=0.055386\ \mathrm{mol}$ (obviously not rounded to significant digits).
$582\ \mathrm{ml}-1\ \mathrm{ml}=581\ \mathrm{ml}$ of nitrogen at $p_0=1.3\ \mathrm{bar}$ and $T_0=22\ \mathrm{^\circ C}$ corresponds to a mass of $m_{\ce{N2}}=0.8624364\ \mathrm g$ and an amount of nitrogen of $n_{\ce{N2}}=0.030786609\ \mathrm{mol}$.
Now you know $T=200\ \mathrm{^\circ C}=473.15\ \mathrm K$, $V=582\ \mathrm{ml}$, and the amount of substance $n$ for water and nitrogen. Using the ideal gas law
$pV=nRT$, you should get the partial pressures $p_{\ce{H2O}}=3.74\ \mathrm{bar}$ and $p_{\ce{N2}}=2.08\ \mathrm{bar}$. Thus, the total pressure is $p=5.82\ \mathrm{bar}$.
Better values can be obtained using so-called steam tables, which should give something like $p_{\ce{H2O}}=3.6711\ \mathrm{bar}$ and $p_{\ce{N2}}=2.0827\ \mathrm{bar}$. Therefore, the ideal gas law is a sufficient approximation for your purpose, especially considering the precision of your input data.


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Selective secondary amide/lactam hydrolysis
I am working on a synthesis for a compound (4-aminophenol with alkyl chain and lactam at carbon 2). The final step of the synthesis involves hydroylsis of the secondary amide at carbon 4. My problem is I can not find a way to hydrolyse the secondary amide at carbon 4 without hydrolysing the lactam (which I don't want). Does anyone have any suggestions for how I could selectively hydrolyse the secondary amide? Thank you.

Is it really a problem if both the amide and lactam are hydrolyzed? Look at what you have left.

I cannot think of a way of doing this selectively. Either follow @user55119's advice or adjust the route at an earlier stage.

How do you think that lactams are formed?@Waylander

@user55119 treatment of the fully hydrolysed material with a carbodiimide should give the lactam. I don't think either the phenol or aniline will interfere.

@Waylander: I was hoping to get a response from the OP but alas. Or esterify the hydrolysis product and heat the bis-aminoester.

Drawing on the comments of @user55119 above, I am going to to summarise the comments as an answer.
There is no way of selectively hydrolysing the acetanilide in the presence of the lactam.
It doesn't matter because the lactam can be readily reformed from the completely hydrolysed material by treatment with a carbodiimide such as 1-(3-Dimethylaminopropyl)-3-ethylcarbodiimide  hydrochloride. The free phenol and aniline are unlikely to interfere as they are less nucleophilic than the secondary amine and you are forming a [5] ring with the lactam.


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Why are holes appearing in this aluminum kitchen foil on my radiator?
I thought I would try to deal with dehumidified air from radiators by filling simple troughs of water placed on the top, which in theory would evaporate water as the radiator heated the air. I crafted three simple troughs from folded standard kitchen aluminium foil and sat them inside the two sides of the white-painted radiator, atop of the corrugated metal fins. After a few days where the evaporation rate seemed too low to be meaningful for my air-humidification aspirations, there were black splodges on the inside of the foil that I took to be mould. The water had by this time evaporated so I gave up on the idea and removed the troughs, only to notice numerous and at times significant holes where the aluminium has apparently corroded away. One was a channel of maybe 2 mm wide by 20 mm long.
In a previous revision of this question, there wasn't enough info to go on, so I've repeated the experiment. We have water from the gas boiler leading through copper pipes to a steel radiator painted white, heated to something like 50 °C. The aluminium tray is inserted in the top, and cold tap water from the hot tap in the bathroom is added to about halfway. The radiator cycled hot/cold two times a day. I'm in a heavy water area in the UK (Surrey) with significant calcium content in the water.
Below shows the experiment progression in pictures. Images 1 and 2 show the set up. Image three shows the tray at about 6 days. The black marks turned out to be the holes, and there are white deposits underneath the holes. The last two images show Day 7, the tray removed for observation looking at the back edge from the other side. There were no deposits on the outside of the foil. There were powdery, crusty white deposits on the radiator where the holes were.

I'm guessing the white powder is aluminium oxide, but I don't understand why the corrosion would be so pronounced simply because of 50 °C water temperatures.

Regular duty aluminum foil (0.001 inch thick) supposedly has about 1 pinhole per square foot; heavy duty Al foil (0.oo15 inches thick) has many fewer. Recycled aluminum is used in cheaper foils and might contain a few atoms of mercury or copper. Chloride ions in tap water are no friends of aluminum, nor is a low pH. Can we get some more information?

Chloride will corrode through Al foil overnight.  You could try distilled water ;  However, Al baking pans have a fairly long life .

Do the holes correspond to the level of the water?

I don't understand why this is closed, at least the focus is clear. What can corrode an aluminum foil containing tap water atop a radiator? Of course some kind of corrosion that I have observed as well. It might be even simply oxide, presumably passivation requires a continuous and solid surface that a thin cheap foil might not provide. Or the chlorine / chloride mentioned above. It is a sensible question with a possible answer.

@matt_black : Yes, the holes seem to be at the water line, created as the waterline receded with the evaporation.

@DavidCoombes Do you know what the main water pipes in the house are made of (copper, lead, plastic)?

@matt_black : All pipes are copper. Water is hard water, chlorinated to whatever UK standard is I guess. Certainly can't taste any chlorine. Radiator is old but in good condition, no signs of rust other than at bleed valve. Paint seems flawless inside the groove of the radiator. Is the proximity of the iron a factor, or is it just a matter of water + heat + trace chemicals inside the bath?

My speculation as the radiators are located by the window, some light induced photolysis may be occurring with metal rich dust particles settling on the surface of the water.
With any formed electron hole ($\ce{ h+)}$ acting on some $\ce{OH-}$ in the water may be transformed into the associated hydroxyl radical, $\ce{.OH}$. In the presence of say a phosphate ion, we could then have the phosphate radical which is acidic.
In this scenario, the attacked Al at the water line may be an aluminum phosphate salt. Other possible ions may similarly be radicalized.

Thanks for the theory. Radiator is on the side wall away from the window (front wall). It gets little direct sunlight in the winter as the sun is already setting by the time it's that side of the window. It's also in a household environment so would there be much to contribute to 'metal rich dust particles'?


text:
Representing units for concentration of a liquid (v/v)%
I'm doing an experiment which involves testing a particular solution in different concentrations. Since this is a liquid, the concentration should be represented in v/v % rather than w/v %. However, I'm not sure how to represent this in my research question. Suppose, these are the concentrations I'm testing: 20, 40, 60, 80.
Should I express it like (20%, 40%, 60%, 80% (v/v))
(or)
(20,40,60,80 (v/v)%)
(or)
20%(v/v), 40%(v/v), 60%(v/v), 80%(v/v)

Anything is fine as long you explicitly state what it is. You can even write one line in the experimental stating that all percentages are reported as volume/volume. Repeating v/v looks odd in the last choice.

The point is that v/v is one of the less clear way to express concentration as for everything is fine as far as you prepare the solutions but suddenly becomes a mess if one is seeking relation to other properties. Volumes aren't additive. I have no idea if parts per volume is a recommended way, but] at least it has a preparative connotation. Note that one can still use w/v. That would give unequivocally the amount of one component in a volume of the final solutions. Con is that one might need density of them, afterwards. It is all matter of protocols, glassware available, and purposes.

Nonconforming units such as "v/v %" and "w/v %" should not be used at all.

The use of v/v % depends on the field. For HPLC, all instruments are designed to use v/v % and this is a billion dollar industry. From a physical chemist's point of view, this is a bad practice.


text:
Redox reaction of cycloheptatriene ligand

The reaction I really don't understand is the second one which produces the cycloheptatriene ring with a positive charge. The only thing I can think of is that the hydrogen is being ejected and behaving as a nucleophile but that's atypical of hydrogen. How does it happen?

See tropilium ion.

Link to tropylium ion.

Nucleophilic or hydridic hydrogen is actually commonplace in organic chemistry. Examples include:

The Cannizzaro reaction whereby a single aldehyde is reacted with base to form an alcohol and a carboxylate salt, and the variant (cross-Cannizaro reaction) whereby formaldehyde is added and it is converted to formate, the other aldehyde being reduced to the alcohol.

Various reduction reactions involving sodium borohydride and lithium aluminum hydride, most commonly to reduce various carbonyl compounds to alcohols.

In both cases the hydrogen is in an environment that enriches it electronically, in the first case by the action of the base attacking the carbonyl group and in the second case by attachment to an electropositive atom. With cycloheptatriene the carbon ring is in effect also electropositive, because the cycloheptatrienyl ring becomes aromatic with a positive charge. Compare the resulting reactivity with 3-chlorocyclopropene as a source of chloride ion or, in the opposite sense of forming an aromatic anion, the acidity of cyclopentadiene.


text:
Does this coupled reaction actually happen (decomposition of calcium carbonate)
I found the following example of a coupled reaction to drive the decomposition of calcium carbonate. I get the calculation part of it, that the changes in Gibbs energy sum to a negative amount. But does this reaction actually happen when you place the reactants together?  Would it happen with nothing more than, say, a brief flame to ignite the coal?  Or does it require more complicated steps to make it happen?
From Chemistry LibreTexts 19.8 Coupled Reactions

Many chemicals' reactions are endergonic (i.e., not spontaneous ($\Delta{}G > 0$)) and require energy to be externally applied to occur. However, these reaction can be coupled to a separate, exergonic (thermodynamically favorable  $\Delta{}G < 0$) reactions that 'drive' the thermodynamically unfavorable one by coupling or 'mechanistically joining' the two reactions often via a share intermediate. Since Gibbs Energy is a state function, the  $\Delta{}G$  values for each half-reaction may be summed, to yield the combined  $\Delta{}G$  of the coupled reaction.
One simple example of the coupling of reaction is the decomposition of calcium carbonate:
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)} \;\;\;\;\;\;\; \Delta G^o = \pu{130.40 kJ/mol}$$
The strongly positive $\Delta{}G$ for this reaction is reactant-favored. If the temperature is raised above $\pu{837 ºC}$, this reaction becomes spontaneous and favors the products. Now, let's consider a second and completely different reaction that can be coupled ot this reaction. The combustion of coal released by burning the coal $\Delta{}G^\circ = \pu{−394.36kJ/mol}$ is greater than the energy required to decompose calcium carbonate ($\Delta{}G^\circ{} = \pu{130.40kJ/mol}$).
$$ \ce{C(s) + O2 <=> CO2(g)} \;\;\;\;\;\;\; \Delta{}G^\circ = \pu{-394.36 kJ/mol}$$
If reactions  19.8.1  and  19.8.2  were added
$$\ce{CaCO3(s) + C(s) + O2 <=> CaO(s) + 2CO2 (g)} \;\;\;\; \Delta{}G^\circ{} = \pu{-263.96 kJ/mol}$$
and then Hess's Law were applied, the combined reaction (Equation  19.8.3) is product-favored with $\Delta{}G^\circ = \pu{−263.96 kJ/mol}$. This is because the reactant-favored reaction (Equation  19.8.2) is linked to a strong spontaneous reaction so that both reactions yield products. Notice that the $\Delta{}G$ for the coupled reaction is the sum of the constituent reactions; this is a consequence of Gibbs energy being a state function:
$$ \Delta{}G^\circ = (\pu{130.40 kJ/mol}) + (\pu{-394.36 kJ/mol}) = \pu{-263.96 kJ/mol}$$

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

An other name for $\ce{CaO}$ than calcium oxide or quick lime is burnt lime; the process is known as calcination.

The question here is not entirely clear. Do you understand the concept of free energy change during a reaction? When it is negative, the reaction proceeds spontaneously to products. The details of the reaction are somewhat secondary, but you might speed it up (if just initially) by igniting the coal. The assumption is constant T and p, and no exchange of matter with the surroundings.

@BuckThorn  Yes that's exactly the point of the question--can you assume that a negative change in G means the reaction will actually take place.  And if not, then what makes coupled reactions coupled?

This might be helpful: https://iubmb.onlinelibrary.wiley.com/doi/full/10.1002/bmb.5 I think the answer is no, they are not coupled unless you have a way of channeling the free energy from one reaction to drive the second.

Those are not Coupled reactions; the burning of carbon simply supplies the energy to decompose the carbonate. Another way of looking at it would be if you were the fireman on an old steam locomotive would you prefer to shovel pure coal or a mix of coal and limestone into the firebox for several hours trying to keep the train on time. The reactions are independent except for the increased CO2 inhibiting carbonate decomposition.

Okay, thanks.  So what makes reactions coupled?  Also, I assume I would prefer pure coal, but I don't really see what you are getting at with that example.  Can you spell it out a little?

My idea of coupled reactions is when the overall reaction is broken into lower energy steps and the products of one step are the reactants of a subsequent step.  The best examples are cell respiration [Krebs cycle] and photosynthesis.  Many catalytic processes fit this description but are hard to analyse because of increased reaction rates and that the consecutive reactions are happening more or less together. One reaction that has always bothered me is the corrosion of Fe to rust, Fe2O3. This is usually written as a composite equation combining water and O2.  It probably isn't. [out of space]


text:
Do benzene and hexane form an ideal solution?
My teacher said that benzene and hexane do not form an ideal solution (with reference to Raoult's law) but did not explain any reason for it.
My approach:
Cyclic hydrocarbons (benzene here) have higher van der waal's forces of attraction than straight chain hydrocarbons (hexane here) because of shorter intermolecular distances and greater surface area. So when hexane goes in between benzene, there will be weaker intermolecular bonds, causing positive deviation from the Raoult's law.
My doubts:

Are the benzene-hexane interactions stronger than hexane-hexane interactions even though they are weaker than the benzene-benzene interactions? If so, is there still a net deviation from the Raoult's law?

Do the same results apply to other solutions of cyclic and straight chain hydrocarbons?

Better question could be, if deviation from ideal solution is negligible.  Strength of vdW bonds is hexane-hexane < hexane-benzene < benzene-benzene, unless there is strong deviation from the Raoult law.

Also, under My Approach, better jumping directly to Aromatics instead of Cyclics, as it's aromaticity here the more important feature of one component. Aromatics in solid states have quite a peculiar way of stacking (not as a "pile" BTW) and I guess some kind of residual interaction of the same nature can be present liquids, too. Also look at https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7099588/ about benzene interactions.

@Alchimista The source you mentioned says benzene-cyclohexane bonds are stronger. So that looks like a case of negative derivation. Am I right? But the question in my exam just said "hexane", so I think they were talking about the straight chain hexane only. I'm not sure how I could correlate the case of benzene-cyclohexane to this (I'm a class 12 student)

@Poutnik Even I thought that way at first, but that points to saying that the solution is an ideal solution. But my teacher said it isn't an ideal solution

No, it does not point to saying that. Saying C>A and C<B does not mean C=AVG(A,B)

@Balu yes hexane is n-hexane. My point was to consider benzene for what it is, not a cyclic hydrocarbon only as it is in your question. Its aromaticity also dictates its intermolecular interactions, ergo the link. For the cyclohexane mentioned therein, it has a BP almost coincident to that of benzene but it does not mean that not effecti takes place by mixing with benzene. You can look for the data to get a clear picture.

A picture is worth a kiloword (1000 words). Then you can write 2 kilowords explaining the picture. But data are paramount, so here is a picture (Ref 1) :

According to the picture, which has three complete lines and one partial line, the straight line going from 0,0 to 1,1 designates the molar composition of the liquid (the independent variable). Then the two curved lines above it are the connected dots of measurements of the vapor-phase composition at approximately 1 atmosphere pressure and the boiling point of the mixture (which is varying, but called constant because it is uniformly the boiling point). The data were obtained by boiling the mixtures and taking vapor samples in an equilibrium still (Ref 2).
All the colored dots lie above the straight line, indicating that the vapor is enriched in hexane, or the liquid is impoverished in hexane, so hexane-benzene attraction not only does not keep the hexane in the liquid, but forcing hexane into benzene seems to be somewhat unattractive.
The partial line (a few red dots between 0.9, 0.9 and 1,1, at ~5 atmospheres pressure is about 60ºC hotter than at 1 atm, and does not add anything more to the conclusion. It was shown that enrichment of the vapor (in hexane) ceased at about 3.5 mole % benzene (Ref 2). (Note than on the graph, 0,0 corresponds to all benzene, which has the higher boiling point (80.1º vs 69º for n-hexane)). It had been thought that hexane and benzene formed a constant-boiling mixture at ~2.5%  benzene in hexane, but this was disproven. It's just very difficult to get all the benzene out of n-hexane.
In a way, this suggests that there is an interaction between benzene and n-hexane, but only at very low concentrations of benzene, perhaps not well described by Van der Waals interactions. Maybe a little bit of benzene causes the n-hexane molecules to rearrange and use up their packing space more effectively.
Ref 1: http://www.ddbst.com/en/EED/VLE/Images/VLE%20Benzene;Hexane_002.png
Ref 2. https://pubs.acs.org/doi/abs/10.1021/ie50283a006


text:
The Mechanism of Termolecular Reactions
How does a termolecular elementary reaction happen under the law of mass action? My physical chemistry textbook only says details about bimolecular and unimolecular reactions with collision theory. I have searched some documents, but almost all of them didn't elaborate anything crucial, I mean, anything quantitative about its rate constant.
There is only one mentioning a result done by R.C.Tolman. It says we need to introduce a variable $\delta$ to measure the distance between the molecules (To be honest, I'm not sure what he really means). Anyway, the formula mentioned in the paper is:
$$k=4N^2_A\delta\sigma_{AB}\sigma_{BC}\sqrt{\frac{8RT}{\pi}}(\frac{1}{\sqrt{\mu_{AB}}}+\frac{1}{\sqrt{\mu_{BC}}})Pe^{-\frac{E_a}{RT}}$$
in which
$N_A$ is Avogadro number;
$\sigma_{AB}=\pi(r_A+r_B)^2$ and $\sigma_{BC}=\pi(r_B+r_C)^2$ are collision cross sections;
$R$ is universial gas constant;
$T$ is the temperature;
$\mu_{AB}=\frac{1}{\frac{1}{M(A)}\ +\frac{1}{M(B)}}$ and $\mu_{BC}=\frac{1}{\frac{1}{M(B)}\ +\frac{1}{M(C)}}$ are reduced molar masses;
$P$ is a probability factor;
$E_a$ is the activiation energy.
So $v=kc(A)c(B)c(C)$, according to the law of mass action.
But why? I don't have any idea to demonstrate it, and I cannot find the original papers of Tolman. I'd appreciate it if anyone professional at theory could lend me a hand. Any helpful or valuable thought is welcome.


text:
Unique notation to search for compounds in PubChem and in Chembl
I am not a chemist, I come from a computer science background. I am working on a project related to computer-aided drug design.
I have a list of chemical compounds in their SMILES notation, to which I am trying to retrieve some chemical properties. Some of these chemical properties exist in the Pubchem database, others exist in the Chembl database.
As an example:

Isoniazid (INH)'s SMILES in PubChem is : C1=CN=CC=C1C(=O)NN
https://pubchem.ncbi.nlm.nih.gov/compound/3767#section=Canonical-SMILES

Isoniazid (INH)'s SMILES in Chembl is : NNC(=O)c1ccncc1
https://www.ebi.ac.uk/chembl/compound_report_card/CHEMBL64/

Which means that if I were to use these SMILES as a key for my (automated) search, I will get chemical properties for different compounds.
Now, my question is: Is there some unique key, that exists in both databases (PubChem and Chembl) that I can use for my search?
Also, if possible, can anyone explain why these SMILES are different even for the same compound. As I was reading in a website, that these are canonical SMILES which I beleive are specific to each database.

InChi keys are unique (with very limited exceptions).
You could also pre-process your SMILES through some kind of standardisation code to get them all into the same format

@NotEvans. Depending on what you aim for, standard InChI/InChIKey (version 1.06, December 2020) may be good and efficient.  They may lead to confusion, too.  Because Amine Chadi stated a background different to the one of a «chemist by training», I sense potential (though not insurmountable) pitfalls using them and appended my answer.  Version 1.06+ possibly growths and improves by feedback.

There is not one, but there are multiple approaches to describe a molecule by a SMILES string. It is up to the developers of the programs (and databases) to define which one is used and to declare this in the documentation.  Thus, read the manual; if they do not state this explicitly, ask them.
One of the better approaches taken is, for example the one by OpenBabel; it offers you to write e.g., either a canonical, universal, or inchified SMILES string.
If SMILES you work with have origin in multiple sources adhering to more than one dialect, you may request e.g., OpenBabel to consolidate the representation you subsequently use.  Thus
obabel -:"C1=CN=CC=C1C(=O)NN" -ocan

yields to the canonical SMILES string NNC(=O)c1ccncc1.  In case you have a file listing many SMILES, consider a conversion like
obabel -ismi input.txt -ocan -O output_canonical.smi

Noel M O'Boyle wrote a paper about the methods to identify a canonical SMILES representation of (organic) structures which provides insight here, too.
Towards a Universal SMILES representation - A standard method to generate canonical SMILES based on the InChI in J. Cheminf. 2012; doi 10.1186/1758-2946-4-22.

A comment suggested to use InChI and their hashed form, InChI keys as a replacement of SMILES.  Their spread in databases (example NIST, or examples compiled by Warr) may be tempting and checking identity of entries based on InChI key appealing in terms of performance and collision resistance once they are in hand (e.g., Goodman et al.).
However, the discern of different compounds which are tautomers of each other currently is not possible using standard InChI/InChI key.  Take, for example, 2-hydroxypyridine and its tautomer, 2-pyridone.  Depending on the polarity of the solvent, one form may be favoured over the other; and while both structures differ, and are described by two different canonical SMILES, the standard InChI and InChIKey are the same for both:

because

«different tautomers have the same connectivity/hydrogen layer»
see 4.7. Standard InChI specifics, InChI trust's Technical FAQ by 2012-05-12

What can be done, e.g. using InChI trust's reference binaries (version 1.06/December 2020, repository), is to invoke the inclusion of a fixed H layer then leading to InChI
InChI=1/C5H5NO/c7-5-3-1-2-4-6-5/h1-4H,(H,6,7)/f/h7H
InChI=1/C5H5NO/c7-5-3-1-2-4-6-5/h1-4H,(H,6,7)/f/h6H

and
InChIKey=UBQKCCHYAOITMY-QDQILVOLNA-N
InChIKey=UBQKCCHYAOITMY-BRMMOCHJNA-N

As of now, the inclusion of a fixed H layer in InChI leads to a non-standard description (thus, there is no leading 1S).  Equally note that only the second hash block is affected to yield a different InChIKey.
Thus, in short depending on what you aim for, use InChI/InChIKey with caution.
References:
Goodman, J. M.; Platnev, I.; Thiessen, P.; Bolton, E.; Heller, S. R. InChI version 1.06: now more than 99.99% reliable in J. Cheminform. 13, 2021, 40; doi 10.1186/s13321-021-00517-z, open access.
Warr, W. A. Many InChIs and quite some feat in J. Comput. Aided Mol. Des. 29, 2015, 681–694; doi 10.1007/s10822-015-9854-3.

Thank you for the great informative answer.


text:
Dissociation of phosphorus pentachloride in gas syringe

Phosphorus pentachloride is introduced into an empty gas syringe which has a movable, tightly-fitting plunger. The gas is allowed to expand until equilibrium is reached at a temperature at which the phosphorus pentachloride partially dissociates.
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$

Which statements are correct?

The equilibrium pressure inside the syringe will be greater than atmospheric pressure.
When the plunger is pushed in the equilibrium adjusts to produce more $\ce{PCl5(g)}.$
The volume of gas in the syringe at equilibrium will be greater than if no dissociation had occurred.

The correct answer is 2 and 3. Option 2 can be explained using Le Chatelier's principle and I think option 1 is wrong because the pressure inside should be equal to the pressure outside when the reaction reaches equilibrium.
But why is option 3 correct? Assuming the plunger is entirely extended and held stationary until the reaction reaches equilibrium, I think the volume of gas will be smaller if no dissociation had occurred, because there will be less molecules, hence lower pressure to push against the plunger as soon as it is released. My explanation might be entirely wrong since I don't really get how the experiment is being conducted here.

The equation $3$ is correct because in the reaction, one gaseous molecule gets dissociated and produces a total of $2$ gaseous molecules. The number of gaseous molecules increases if the dissociation reaction proceeds. At a given pressure, the volume occupied by $2$ gaseous molecules is twice the volume occupied by $1$ gaseous molecule


text:
What factors have the greatest effects on the yield from steam distillation of ginger?
I have a modest home lab setup that I use primarily for steam distillation of plant matter to extract essential oils. My yields are typically at or below the low end of estimates given by different sources, but understandably so since I'm working with very small amounts (e.g. I recently got a 0.55% yield from 217g of thyme; cf. Holmes 2016, 2019 which both list typical yields between 0.67-1.0%).
However I recently tried distilling ginger which according to multiple sources typically has a fairly high yield (cf. Holmes 2016: 3.3-5.0%; Malle & Schmickl 2015: 1.9-2.6%; Möller 2018: ~4%). I used 378g of fresh ginger from a local grocery store, I chopped it coarsely, and filled the boiling flask with about 500mL distilled water. But after 4.5 hours, I ended up with <0.5mL of oil—a yield of ~0.1%.
After the first run, I even tried putting the ginger in a food processor to make it a bit finer and ran it again with another ~350mL water for 1.5 hours but didn't get enough oil for it even to separate.
In all I ended up with about 550mL of hyrdosol and ~0.25mL of essential oil.
This was the setup I used:

TL;DR: So my question is, what are the most likely factors that could have reduced the yield to such an extent?
I know fresh ginger from the grocery store may not have been the best choice for the purposes of extracting essential oils, so next time I'll try getting dried ginger intended for therapeutic uses, but other than that, I'm not sure what else to do differently next time.
Works cited

Holmes, Peter. 2016. Aromatica: a clinical guide to essential oil therapeutics, vol. 1: applications and profiles.
Holmes, Peter. 2019. Aromatica: a clinical guide to essential oil therapeutics, vol. 2: principles and profiles.
Malle, Bettina & Helge Schmickl. 2015. The essential oil maker's handbook. (Trans. by Paul Lehmann)
Möller, Kai. 2018. Tabletop distilling: how to make spirits, essences, and essential oils with small stills.

Nice setup. I would recommend a heating mantle. A question, did you ever tried simply putting the ginger on the boiler? This at least will shorten the path. But I have no experience on steam distillation except for a couple of time as student and long ago. The latter is a question rather than a suggestion.

@Alchimista I haven't tried that yet but am planning to next weekend. If it makes a huge difference I'll post an update.

@Martin-マーチン The sources don't go into a great amount of detail about the processes involved other than to say whether they used fresh vs. dried or simple vs. steam. Malle & Schmickl sometimes also provide information about when to collect the plant matter, but as I'm mainly using things you can find on Amazon or a grocery store, that's a bit out of my control.

You have not described the oil separation. Do you apply extraction/evaporation for oil/water separation? Or centrifugation of emulsion? Part of the essential oil can be dissolved or can form stable emulsion.

@Poutnik For separation I just let the oil/hydrosol settle in a separatory funnel and empty them into separate containers.

I remember my lab practice in 1985 of isolation of essential oil from cloves. After distillation, there was very fine and like stable oil emulsion. After extraction to CHCl3 and solvent evaporation I got about 1-2 mL of oil. Without extraction or centrifugation, it could help to dissolve enough NaCl, what may help in emulsion separation.


text:
Liquids with high specific heats?
I am wondering if there are any liquids other than water that have relatively high specific heat capacities. The tables of specific heats of common substances I have found do list ethanol (2.376 J/(g °C)), but I have not seen much else.

Liquid ammonia, as per https://www.engineeringtoolbox.com/specific-heat-capacity-d_391.html.

Liquid ammonia has the highest specific heat capacity for any liquid or solid. At room temperature, water is 2nd highest liquid. Edit: my reference is my environmental chemistry textbook by Stanley E. Manahan.

Which is why NH3 is used in regulating temperature aboard ISS (not to mention it makes a good propellant, should they run out of hydrazine ;-)


text:
Fugacity vs compressibility
What's the criterion for saying that the forces of attraction are dominant?
f < p or Z < 1
I ask this because there is a situation where
f < p and yet Z > 1 (below Boyle temperature)
if the criterion is fugacity, it is not so intuitive to say that the forces of attraction are dominant and the real gas has a molar volume greater than that of an ideal gas.
f: fugacity;
Z: compressibility factor;
p: pressure

Let $\varphi=\frac{f}{p}$, where f is the fugacity, and p is the pressure.
Then:
$$\ln \varphi = \int_0^p \big( \frac{Z-1}{p} \big)dp $$
[Source for above expression: Wikipedia article on fugacity]
Hence the fugacity coefficient, $\varphi$, at a certain $T$ and $p$, isn't determined by the compressiblity factor, $Z$, at that $T$ and $p$.  Instead, it's determined by integrating $Z$ with respect to $p$, at the given $T$, according to the above expression.
I.e., $\varphi$ is determined by all the values that $Z$ took on in bringing the gas up to the current pressure.  This makes sense, because it's directly related to the free energy of the gas.
Consider these three limiting cases, in which $Z$ has the specified value (or range of values) for the entire process (you should confirm these yourself, for your own understanding, by checking the above integral):
I. $Z=1 \implies \varphi=1$ ($f=p$)
II. $Z>1 \implies \varphi>1$  ($f>p$)
III. $Z<1 \implies \varphi<1$ ($f<p$)
One can explain the result you are seeing as a case in which $Z<1$ for a sufficient part of the integration such that $ \varphi<1$ ($f<p$) but, at the specific pressure you are considering, $Z>1$.
Physically, this means the following:

$Z>1$: Excluded volume effects dominate at the current pressure.

$\varphi<1$ ($f<p$): Attractive forces dominate over most of the range between $p=0$ and $p=\text{current pressure}$.

@ÉdersonD'MartinCosta  I added, at the end, a physical conclusion that corresponds to the mathematical conclusion.


text:
If bond types are in reality intermixed, how come different bonds form completely different structures?
According to the bond triangle, compounds don't exist as solely ionic or solely covalent, but rather have ionic, covalent, and metallic character to them. So each bond type is connected and similar in a way. However, when learning about the different structures each bond type forms as a solid, they seem to form vastly different structures. Ionic compounds form lattice structures, covalents have molecular and network structures, and metallics have metallic lattices. Why is this so? For example, why don't ionic compounds form molecules if they contain covalent character? Is it because there are sort of transition structures that are mixes of the different structures mentioned above?

The way you posed the question is surprisingly challenging. I know the answer but I am thinking how to tackle various point that you touched (inadvertently?) without falling into logical loops. One point is that you treat solid on the same basis, while a covalent solid can be regarded as a giant single molecule, removing one case. It is also possible that you wrote covalent solid erroneously meaning solid formed by an arrangement of covalent molecules. Still, even if this is the case, what I wrote above is valid. - I'm thinking about an answer that satisfies me first of all.

All humans are related; how come they don't form similar societies?

I'd say that your premise isn't all that useful to begin with. Not all ionics form solids, see ionic liquids; not all covalent bonds are restricted to discrete molecular structures, see diamond; not all metals are solids; etc. In addition, the term "solid" immediately hinges on the question, "At what conditions?"

Premise: in spite of my comment, which I leave posted for reference, I've decided that I was overthinking about the reasons that lead OP pose the question and that is better to answer to what is certainly asked.
However, after having tentatively edited the title to reflect what the text deals with, I must amend the question at two points:
1 - "compounds don't exist as solely ionic or solely covalent..." the subject should rather be bonds instead of compounds. Also the useful mantra "there are not fully ionic or fully covalent bonds" shall be used with more parsimony or at least more carefully, as it is true context given (phenomenologically it does not holds in biatomic hydrogen or for the C-C bond in ethane, etc...)***;
2 - about the various structures and  hierarchies listed by OP, what is relevant is that, when solid bulk materials are considered, ionic bonds leads to ionic lattices while covalent bonds can lead to both covalent lattices or to discrete molecules afterwards kept togheter by various forces which aren't strong enough nor have the nature of the aforementioned bonds.
Answer
We shall take as starting points that
1 - an element faced to other atoms might react to achieve an overall lower energy.
2 - Generally enough to be used here, atoms tend to fulfil the octet rule.
3 - The way they do so depends on their absolute electronegativity as well as on the electronegativity difference between them.
(Of course what determines electronegativity and why atoms with a filled octet are particularly stable can be the subject for two or more questions).
When the difference in electronegativity is high, typically between elements at the extremes of the periodic table periods, the octet is filled by a true electron transfer. In the simplest case, this leads to a couple of opposite charged ions (real and relatively stable particles, as far electric neutrality is satisfied they can individually exist in the proper conditions, for instance as solvated species in water) and the overall energy is further lowered by their electrostatic pairing.
Basically, an ion can be treated as a point charge with a spherical electrical field. There is no need to locate the extra electron of an elemental anion, or to know where the electron leaved a "hole" in the cation counterpart.
The two are attracted by the coulombic force and as such direction trivially matters only in the minimal entity they can form, kind of a molecule corresponding to the minimal formula.
An extented bulk made of ions just obey to electric neutrality and geometrical constraints that are somewhat "external", kinds of possible space-filling arrangements similar to those of a number of balls.
Indeed, in common conditions, ionic bonds hold togheter ionic crystals rather than molecules.
Conversely, when the difference in electronegativity isn't that high and when fulfilling the octet would require a transfer of too much electrons, the latter are instead shared in between the nuclei in a region of space that is indeed the bonding orbital. Although bonds are ultimately of electrical nature, the situation now does not involve electrostatic charges with their spherical fields, but defined directions as soon the involved element combines with multiple atoms.
Most molecules satisfy the octet rule within themselves, and the geometry constraints (VSEPR suffices, for this discussion) is what gives their shape.
The molecules, if eventually lead to a solid, arrange in crystals or amorphous bulks hold by various forces such as H-bond, permanent dipole interactions, or what often goes under the generic umbrella terms dispersion forces or Van der Walls forces. No need to go into detail, here. Those aren't chemical bond on the base of their strength and/or different nature. It is also quite matter of nomenclature and is usually easier to see why they are treated apart rather than writing down. The bulk so formed is called a molecular solid, and its arrangement its dictated by the shape and packing of the molecules and the predominant forces between them.
It can even be that molecules form dimers, eg by means of H-bonds, and these dimers then pack via dispersive forces; etc.
When covalent bonds are involved there is another possible outcome, of which C diamond is the typical example.
In that the geometry of the atoms allows for the building of a truly 3-D lattice. Note that - in the context of this thread - even this case is basically as the above one, just the bulk isn't a molecular solid but can be seen as a very very big 3-D macromolecule. It is called a covalent solid.
Finally, when we say that
bonds aren't solely ionic or solely covalent
we mean that, unless the molecule is highly symmetric about the bond under discussion, electron shared in a covalent bond aren't equally distributed between the bound atoms. In other words, the electron density is higher towards the element - or  the molecular moiety - having the higher electronegativity.
As such, most of the time "there are not solely covalent bonds" means that in reality the vast majority of covalent bonds are polar, ie there is a net electrical moment along the bond direction, the intensity of which depends on the electronegativity difference between the bound atoms/moieties.
For metals, I would say that the pictorial description of cations immersed in a sea of electrons can be usefully recalled. It suffices to explain while metal bonds are not directional and again atoms in a solid metal are arranged to fill the space as spheres do - with the addition that being those spheres positively charges, the interstitial volume is filled by a kind of "continuous" negative fluid, ie the electrons spherically distribute all around each atom and as such ultimately mobile through the bulk.
Leaving the alkali metals, as elements valence electrons are on less symmetrical orbital, increases the covalent bond character leading to less simple structure and specific direction between planes. Still, the bond is responsible for holding the whole bulk.

***at another level, which however has no phenomenological relevance, even a completely apolar covalent bond, ie one with no electrical dipole moment, can be thought as a hybrid of resonance of which some limiting forms have ionic character. In such a sense it might be regarded as supporting the idea that all covalent bonds have some ionic character. However, the true meaning is that we can regard A2 as a hybrid of the limiting forms A-A, A+ A-, and A- A+. The weight of the latter is certainly small if not negligible and especially has not direct consequences on the way A2   molecules build up their solid phase.
In a less symmetric case, AB (A-B; A+ B-), an electrical dipole moment would emerge. Still the whole entity would be neutral seen from a sufficient distance (that of the adiacent molecules) and as such there is no reason for an ionic lattice to form. At most the permanent dipole would dictate or influence how the molecules arrange in a molecular solid, when they do.

Thank you for your response. When you say there is a true electron transfer in an ionic bond, doesn't that mean that bond is 100% ionic and 0% covalent? I thought there is no such thing as a 100% ionic bond and that there will always be some sharing of electrons between the two particles. From that reasoning, there seems to be a strict cutoff for what counts as ionic and covalent. If, as the difference in EN increases, the ionic character increases, does the transition from directional bonding to nondirectional bonding happen at a specific point or is it gradual?

Somehow It is not that you need a transition from ionic to covalent. Either you have an ionic bond or covalent bonds, the latter more or less polar.. The words "more ionic", it is slang. That is the source of your doubt. Also don't mix a possible description of, say, an entity as NaCl in vacuum (might be that describing it as a hybrid of Na+Cl- and Na-Cl is good for something, i don't know) to what lead to a NaCl crystal. That's ionic for sure. Note also that you can't really have mixed directionality. Either something is directional, or it is not.

Otherwise put, there are no ions even in the most polar covalent bond.


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What dodecahedral molecule is Linus Pauling likely holding in this photograph? Does it have 40 carbon atoms?
The video Quasicrystals ; Prof. Daniel Shechtman ; Nobel Prize in Chemistry focuses on Professor Dan Shechtman but happens to include some photos of Linus Pauling who never believed in quasicrystals. From the video:

The leader of that group was Professor Linus Pauling, a two-time Nobel laureate. He was a very important figure and an idol of the American Chemical Society, and to his last day, he was standing on stages and published papers saying that "Daniel Shechtman is talking nonsense."

During this part of the video there is shown a photo of Linus Pauling holding some ball and stick molecular models. These appear to be dodecahedral arrangements (12 pentagons, 20 vertices) of possibly carbon atoms.
This photo is apropos because early quasicrystal indeed had dodecahedral structure.
But my question is about the molecule for which this is likely the model. I think I see a dodecahedron with one more ball attached to each of the 20 corners. All 40 balls are black, and I'm guessing that if the 20 atoms projected outwards were hydrogen and thus $\ce{C20H20}$ or dodecahedrane) then they would be white in this color photograph. Since all 40 atoms look the same and it's common to leave the hydrogens off when they are understood, I'm wondering if this was supposed to be $\ce{C40H60}$.
I have not found a name for such a molecule with a 20 atom dodecahedral core and 20 more attached to the vertices, the closest I've found here is Tetracontahydrobenzo[5,10]anthra[9,1,2-cde]dibenzo[kl,rst]pentaphene but of course it's flat and made of hexagons rather than pentagons.

Question: What dodecahedral molecule is Linus Pauling (likely to be) holding in this photograph? Does it have 40 carbon atoms?

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Using the mole ratio to figure out the molecular formula

It takes $\pu{8.0 cm^3}$ of aqueous sodium phosphate to fully react with $\pu{12.0 cm^3}$ of an aqueous solution of the nitrate of metal $\ce{T}$. If both solutions have a concentration of $\pu{1 mol dm^-3},$ what's the formula of the phosphate of metal $\ce{T}?$

The answer: As both solutions have the same concentration, the mole ratio is the same as the volume ratio which is $12:8$ or $3:2.$ Thus, the formula is $\ce{T3(PO4)2}.$

I get how the mole ratio is $3:2,$ but how do they deduce the formula as being $\ce{T3(PO4)2}$ from it? Isn't the mole ratio just a way of quantifying the coefficients of the chemical equation? How do they, from the mole ratio, figure out the molecular formula of the phosphate of the mystery metal $\ce{T}?$

Write reaction equations for various valences of M.

Metal T or metal M. Tomayto tomahto.

@KarstenTheis Metal T obviously, I have forgotten the unusual designation of a metal.

@Poutnik Speaking of unusual notations, keep in mind there are emojis used to teach math: YouTube — Emoji Maths Puzzle. I actually find it pretty effective for teaching (and horrible for IRL or field applications).

@andselisk Such puzzles are often very tricky  :-)

Very poor usage of notations. Usually, metals are denoted by the letter M.

@NilayGhosh Not necessarily, transition metals are often denoted with the letter T.

There is $0.008$ moles of $\ce{PO4^{3-}}$ ions. So the total number of negative charges is $3·0.008 = 0.024$ moles. This number must be equal to the number of positive charges. Let's check it.
As there is also $0.012 $ mol $\ce{T^{x+}}$, the total number of positive charges is : $\pu{x · 0.012 = 0.024}$ moles. As a consequence x = $2$, and the unknown ion $\ce{T^{x+}}$ is doubly charged, like $\ce{Ca^{2+}}$ or $\ce{Zn^{2+}}$ . And now we know the formula of the ions forming the precipitate. They are : $\ce{T^{2+}}$ and $\ce{PO4^{3-}}$.
In order to obtain a neutral precipitate (total charges = zero), it is necessary to combine at least $\ce{2 PO4^{3-}}$ and $\ce{3 T^{2+}}$. This would make $2·3 = 6$ positive charges neutralized by $3·2 = 6$ negative charges. The precicitate must have the formula $\ce{T3(PO4)2}$


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Determining which of these has the maximum S-O Bond Length?

S-O Bond length is maximum in?
(a)$\ce{SOBr_2}$
(b) $\ce{SOCl_2}$
(c) $\ce{SOF_2}$
(d)$\ce{SO(CH_3)_2}$

Initially I thought it would be (d) since if a Halogen was attached, the S-O bond would become more polar hence shorter. But on more pondering I realised that resonance might come into play as well and there would be no resonance in (d) hence that would make the S=O bond purely double bond so (d) gets ruled out.
I'm having trouble in determining between the other 3. More is the resonance effective between S and X (where X is the Halogen atom) lesser will be the double bond character of the S-O bond.
How do we determine which resonance is more effective? Is it just related to the sizes of the respective atoms?
Also, will the S-O bond also be purely double bond in $\ce{SOF_2}$ and it would have no resonance? A friend of mine told me about this and I wasn't able to get why.

I would appreciate if someone could help me with approaching such problems


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Where does Aluminum Hydroxide come from in a reduction of a ketone by LAH?
The experiment was done in a Laboratory. A ketone was reduced with LAH in THF under reflux.
At the end of the reaction, the protocol stated to add water until there is no more H2 released. The H2 might come from the reaction of water with unreacted LiAlH4 molecules.
What I don't seem to understand is that the protocol suggests adding acid to "dissolve aluminum hydroxide". I do not understand how lithium hydroxide can be formed in the first place and by which mechanism.

Ask yourself what is the species is formed after addition of hydride to the ketone. Is this species a base? What happens when water is added? Is the reaction mixture still basic?

$\ce{LiAlH4}$ is not formed by $6$ independent ions. But, for the present purpose, it reacts as if it was formed of $\ce{Li^+ + Al^{3+} + 4 H^-}$. The ions $\ce{Li+}$ and $\ce{Al^{3+}}$ do  not react with water. The only ion reacting with water is the hydride ion $\ce{H^-}$, and the equation is $$\ce{H^- + H2O -> OH- + 1/2 H2}$$ As each $\ce{LiAlH4}$ "contains" $4$ hydride ions, these $4$ $\ce{H^-}$  produce also $4$ $\ce{OH-}$ ions with water, according to : $$\ce{4 H^- + 4 H2O -> 4 OH- + 2H2}$$ and these $4$ $\ce{OH-}$ may combine with the metallic ions $\ce{Li+}$ and $\ce{Al^{3+}}$ so as to give the following final equation  : $$\ce{LiAlH4 + 4 H2O -> LiOH + Al(OH)3 + 2 H2}$$ The aluminum hydroxide $\ce{Al(OH)3}$ is not soluble in water, but it passes into solution by adding some acid, like $\ce{HCl}$,  according to $$\ce{Al(OH)3 + 3 HCl -> AlCl3 + 3 H2O}$$


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Is ozone paramagnetic?
This answer shows why ozone should be diamagnetic. Still, Wikipedia article says it is weakly paramagnetic, having positive magnetic susceptibility $χ = \pu{+6.7E−6 cm^3 mol^-1}.$ There's a paper discussing it [1], saying it's only a value reported for liquid $\ce{O3},$ but it's paywalled. What is the reason for such discrepancy?
Reference

Anantakrishnan, S. V. The Structure of Ozone—An Interpretation of Its Magnetic Susceptibility. Proc. Indian Acad. Sci. (Math. Sci.) 1947, 25 (6), 520. DOI: 10.1007/BF03172553.

The 1947 paper you cite concludes that "Calculation of magnetic susceptibility for the different valence bond
structures indicate that ozone should be diamagnetic, the susceptibility
being about -18 x 10^6 while reported experimental values show that it is
feebly paramagnetic in the liquid state, diamagnetic values being reported
for the gaseous state."

Basically when you think about such questions, you have to also think about the state of that compound/element and also the temperature. Ozone can be a gas, liquid, or a solid. This is how you can critically analyze the thoughts.

OP can read another open access article on ozone's magnetic behavior. It is dated but you can enjoy the original experiments. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1085222/

https://doi.org/10.1080/00268978300100581 might be of interest: "Ozone is a quite exceptional molecule in several respects; it is van Vleck paramagnetic, there is extreme antishielding of the nuclei and the dependence of both χ and σ on small changes of the geometry is fantastic."

It is also difficult to determine experimentally, since ozone is non-cooperative. See Derek Lowe on working with ozone and its derivatives. https://www.science.org/content/blog-post/things-i-won-t-work-peroxide-peroxides .

The wikipedia article about ozone says that it is diamagnetic, not paramagnetic, but I'm not sure if that's a recent edit or not.

@Bli231957 It's magnetic susceptibility confirms that it is a weekly paramagnetic substance.


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How would a low concentrated acetic acid react with highly concentrated KOH solution of equal volume?
If there are two solutions of 25mL volume and one contains 0.5M acetic acid and the other 1M KOH, then will every molecule of acetic acid convert into potassium acetate?
I know acetic acid is a weak acid, so it will not dissociate completely at a time. But when concentration of hydroxide ion increases due to the base it will undergo backward reaction with hydronium ion to form water. So amount of hydroxide ion will be more than hydronium ions.
But I am confused about whether the non-dissociated acetic acid molecules dissociate after addition of the base? If they do,why would they dissociate? If the existing hydroxide and hydronium ions can form water molecules to maintain the consistency of auto ionization constant, will there be any necessity for acetic acids to dissociate to supply more hydronium ions?
Another thing which is confusing me is that I used to know that acetic acid does not dissociate completely even if infinite amount of base is added to the solution. But as I have mentioned a stem at the beginning, based on that question I have seen many solution to such mathematical problems where they considered that all the acetic acid molecules will dissociate to form potassium acetate molecules. So will all the acetic acid molecules dissociate and form salts? Again salts remain ionized on aqueous solution so there remains a probability to undergo backward reaction to form acetic acid molecules.
I am really confised what happens in the reaction medium in such conditions.
Thank you.

Dissociated and non dissociated acetic acid react with $\ce{KOH}$ in the same way. This result does not depend on the concentrations of the initial acid and of the initial hydroxide. It will produce potassium acetate.
Now it is also right that the reaction is not total, as potassium acetate is slightly hydrolized. But this effect is rather weak. A solution $1$ molar in potassium acetate has a pH $9.37$. This corresponds to a residual $\ce{OH-}$ concentration equal to  $\ce{2.35·10^{-5}}$ mole par liter.
So out of a solution containing $1$ mole potassium acetate per liter, only $23.5$ micromole are decomposed into acid and hydroxide. This ins not zero of course, but it is rather small. $23.5$ parts per million are not transformed into potassium acetate. The purity of the obtained potassium acetate is $100$% - $0.00235$% = $99.99765$ %

As many comments have been written about my very first sentence, I think advisable to develop it in some detail. An acetic acid aqueous solution is made of a vast majority of unchanged $\ce{CH3COOH}$ molecules, plus a tiny percentage of ions due to the reaction
$$\ce{CH3COOH  <=> CH3COO- + H+ \tag{1}}$$ which could also be written  $$\ce{CH3COOH + H2O <=> CH3COO- + H3O+ \tag{2} }$$
This reaction is an equilibrium, with an equilibrium constant which is usually derived from equation $(1)$ : $K\pu{_a = \frac{[H^+][CH_3COO-]}{[CH_3COOH]} = 1.8 10^{-5} M}$. In a $1$ molar $\ce{CH3COOH}$ solution, the concentration $\ce{[H^+]}$ is $\sqrt{K\pu{_a}} = 4.24· 10^{-3}$ M. Whatever its numerical value, it means that this concentration $\ce{[H+]}$ is rather small, with respect to $1$ M.
However, if these rare $\ce{H+}$ ions are suddenly destroyed by adding some drops of a $\ce{KOH}$ solution, the concentration $\ce{[H+]}$ should fall to zero. It does not, as immediately Nature reacts, and enough $\ce{CH3COOH}$ molecules get dissociated according to ($1$) or ($2$)  to compensate for the sudden lost of $\ce{H+}$. As a consequence, the concentration of acetic acid decreases and the concentration of acetate ions increases. Apparently this reaction ($1$) or ($2$) is extremely fast, as it is impossible to follow its kinetics by spectrometric methods. Why is this reaction so fast ? It  is hard to explain, because it is a reaction from or between neutral substances. And usually reactions from or between neutral substances are not immediate, and their kinetics can be followed by spectrometric methods.
On the contrary, every chemist knows that reactions due to ions (like neutralizations) are so fast that their kinetics cannot be determined. This is why some chemists have emitted the idea that the added ions $\ce{OH-}$ should react directly with the most abundant substance in the solution, namely the molecule $\ce{CH3COOH}$ in a possible reaction $\ce{OH- + CH3COOH -> CH3COO- + H2O}$. The $\ce{H+}$ ions would not wait for the equilibrium to be re-adjusted.
But I would not fight for this idea, that I have read once in a manual. It is not the point of the present discussion. I would not abject if the reader prefers thinking that the added $\ce{OH-}$ react with the ions $\ce{H+}$ (or $\ce{H3O+}$) produced by the quick adjustment of reactions ($1$) or ($2$).

Can you please explain how undissociated acetic acid molecules react with potassium hydroxide? Thank you.

I object. The non-dissociated acetic acid dissociates, as soon as the dissociated part is eaten up by the alkali. Then it is eaten up as well. This happens about as fast as you can imagine, only faster.

Why would non-dissociated molecules dissociate as there is no scarcity of hydronium ion and kw is maintained? @Ivan Neretin

But there is scarcity of hydronium ion as it gets consumed by the alkali.

Isn't the existing amount of hydronium ions(auto ionization of water+initial ionization of acetic acid) enough to be consumed by alkali and maintain kw? Then why do acetic acid dissociate further?

There are two equivalent ways of explaining how acetic acid solutions react with $\ce{KOH}$. First way : the following equation occurs with undissociated acetic acid molecule : $\ce{CH3COOH + OH- -> CH3COO- + H2O}$. Second way, in three steps.

The existing amount of hydronium is exactly enough (read: not a little more than enough) before any of it is consumed. After some of it is consumed, it becomes less than enough.

Ohh sorry I forgot the amount of hydronium ions in acetic acid solution is extremely low

Low or not low (it is low indeed), that's not important. What I said is important.


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Fluorine detection in organic compounds
Many of you will know why I am asking this question. There is a lot of attention recently to the dangers of the so called Perfluoroalkyl and Polyfluoroalkyl Substances (PFAS). In essence organic fluoropolymers with very attractive commercial properties like repelling grease, water and dust particles. Making them very wide spread in consumer products and the industry as a whole.
Unfortunately they tend to mimic normal organic compounds in the body and interfere with enzyme chains and other metabolic processes in multiple ways. Their extreme stability and slow expulsion rate cause them to bioaccumulate and easily reach dangerous levels in humans and in the environment. Looking at the health effects (discovered so far) they might be on par, if not more dangerous than ingesting unstable radioisotopes or heavy metals.
Since those compounds and their products are too varied for standard lab tests to be practical. My questions is:
Is there a way to chemically detect fluorine presence in solids and liquids? I understand that is likely very difficult at room temperature, but maybe enough heat will break the fluorine-carbon bonds and could expose F to specific reactions that could be easily observed?
To clarify I am looking for a practical inexpensive method I can perform safely at home. I would like to test multiple things like food containers, pizza boxes, shoes, etc. Precision and scientific level of certainty is not an issue, just an indication is what I need. I might be able to perform the heating in some kind of securely sealed heat and pressure resistant contained that could be put into a microwave or other heat source with all the reagents inside.
I just want an indication of Florine present at more than trace quantity, not search for PFAS specifically (which is way more complex)
I have very basic chemistry knowledge, so please excuse any mistakes I could have made in my question.

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Unfortunately, your analytical question is not an easy one, and no test can be done safely at home or by using a handheld gadget. The problem is fluorine, and tiny amounts of it in the sample. Fluorine in organic compounds does readily not participate in analytical detection tests.
A straightforward method for the qualitative detection of fluorine in organic compounds requires fusion with an alkali metal (Na, K), followed by fluoride ion detection via zirconium alizarin reaction. A color change from violet to yellow indicates the presence of fluorine in an organic compound. Microgram quantities can be analyzed, but this reaction is not at all friendly and safe!
You wish to analyze cardboard, food containers, pizza boxes, shoes…there is no universal solution test. The analytical methods will be based on the substrate you wish to analyze. For example, shoe analysis will be very different from cardboard analysis in terms of sample preparation. Someone will need to develop an extraction method for each sample followed by gas chromatography-mass spectrometry.
Alternatively, you will need an elemental analysis technique. Ordinary inductively coupled plasma mass spectrometry is not great for fluorine (small mass, buried in noise in the mass spectrum, and fluorine does not want to lose an electron and go into a mass spectrometer). So fancy approaches are needed, like the one reported here High-Sensitivity Elemental Mass Spectrometry of Fluorine by Ionization in Plasma Afterglow
Sadly, nuclear magnetic resonance is fundamentally insensitive, and so is any infrared-based method. So both techniques are useless for parts per million until and unless someone is willing to let run one sample for 24 hours to collect the signals. So if you still wish to check the linings of food containers, a lot of sample preparation is needed for infrared measurement and an expensive spectrum library for matching spectra.
In short consult the manufacturer of food items and/or consult a professional analytical service.

Thanks a lot for the exhaustive reply @M. Farooq! I would like to clarify what do you mean by "Microgram quantities can be analyzed, but this reaction is not at all friendly and safe!"

Fusion with sodium or potassium results in converting the fluorinated organics to inorganic fluorides, which can then be detected via the wet chemical test described in the answer. But this is not something done outside a well equipped lab, with good working fume hood, PPE, etc.

Thinking about this, going forward with a fusion means dealing with sample preparation, careful method development and validation of results and methodology, in order to be assured of making valid measurements on purpose. This is what analytical chemists and chemical analysts deal with all the time and it is absolutely essential, though decidedly non-glamorous.

@BorisHamanov, Prof Ed already explained the fact that you cannot melt sodium with you sample outside a lab. I remember a very bad accident with my teacher in a university. She was demonstrating this fusion process and the hot burning piece of sodium flew from the small test tube and landed on her head. This is what I mean by not a safe test.

I am going to disagree with your NMR assessment here as $\ce{^19F}$ is a very strongly NMR-active nucleus on par with $\ce{^1H}$ and as fluorine is a single-isotope element. Furthermore, perfluorinated organic compounds have lots of fluorine atoms (as normal organic compounds have hydrogens) so they should be easily detected. However, NMR is not readily performed at home, I'll give you that.

@Jan Not only that bench top spectrometers «catch up» to equally record $\ce{^{19}F}$ (ex1, ex2), one may speculate the (per)fluoroorganic samples lesser soluble in «typical» solvents may be characterized/identified by ssNMR with larger spectrometers, too.

Long time ago Scientific American used to publish experiments for amatuers. They had an experiment to make an NMR using Earth's field. Brilliant people of 70s. My perspective is from an analytical chemist's point of view. Our interest is often trace analysis. NMR is inherently insensitive due to Boltzmann's distribution as compared to other spectrometric techniques like mass spectrometers where parts per billion detection is not a problem. Basically I meant NMR is nowhere near any chromatographic detector in terms of sensitivity and speed.


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Why do bubbles form in the reaction Zn(NO3)2 + Mg→ Mg(NO3)2 + Zn
What are these bubbles? What reaction is taking place?

Magnesium often forms bubbles in metal displacement reactions such as this one. See, for instance, this answer where a similar phenomenon is seen with copper as the displaced metal.
Magnesium reacts with liquid water to generate hydrogen, but only slowly when the water is pure due to the low solubility of the hydroxide that is also formed in this reaction ($\ce{Mg + 2H2O -> Mg(OH)2 + H2}$). But even a slightly acidic solution, such as that usually formed when a transition or post-transition metal salt is dissolved in water, dissolves the hydroxide and the hydrogen displacement then becomes faster and more noticeable.
For an introductory experiment, zinc would make a better reducing metal, displacing copper, or we can use the classic iron nail in a copper salt solution. These choices avoid the hydrogen displacement side reaction, and we see the metal displacement reaction more cleanly.


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Alkaline hydrolysis with KOH instead of NaOH. Why?
I really loved the series breaking bad. So I started reading about the science behind. I found this paper:
Harnisch, F., & Salthammer, T. (2013). Die Chemie bei Breaking Bad. In Chemie in unserer Zeit (Vol. 47, Issue 4, pp. 214–221). Wiley. https://doi.org/10.1002/ciuz.201300612
In Breaking Bad they used HF to dissolve a dead corpse. But in the scientific paper they say that it is not a good way to do it. They talk about some other chemicals, I can't remember them. But they also talked about NaOH.
In Dr. Stone they claim that sodium hydroxide is used to melt corpses by the Yakuza.
Reading a little on this topic I found a Wikipedia article. It is called Alkaline hydrolysis (body disposal). The article is about an alterntative method to burial or cremation. But there they use KOH.
I was just wondering: is there a scientific reason why they use KOH instead of NaOH?

KOH and NaOH are pretty much interchangeable.

Have a look at these articles: https://noahchemicals.com/blog/sicomparison-potassium-hydroxide-and-sodium-hydroxide .. http://info.soapwarehouse.biz/blog-0/bid/83765/Sodium-Hydroxide-vs-Potassium-Hydroxide-what-s-the-difference


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How does the electronic structure Hamiltonian scales with basis sets and the number electrons?
Given a chemical system with $N$ electrons and a basis set $S  = \{\phi_i\}$ with $K$ basis functions. If one to discretize the Hamiltonian with this basis sets, what is the dimension?
The electronic structure problem $H|\psi\rangle = E|\psi\rangle$ is usually said to be hard because it scales badly. I want to understand explicitly how does the dimension of $H$ scale in term of the problem size (the electrons and basis set in consideration)?
If someone can give me the detail to this exponential scaling (?) and maybe an explicit example then it would be greatly appreciated, but I would be happy to get any helpful information/pointer on this.

Well, to start with the obvious, with $K$ basis functions your wavefunction is going to be a $K \times 1$ column vector, and the Hamiltonian is going to be a $K \times K$ matrix. What scaling are you asking about, though? The word scaling on its own seems a bit vague to me. There is not only the size of the problem, but also the question of how long it takes to solve a problem of a given size (i.e. diagonalise $H$). Are you specifically asking about the relationship between $K$ and the number of atoms / number of electrons?

@orthocresol Sorry for not being detail in the question. Diagonalizing H can be done in $O(N^3)$ and potentially down to around $O(N^{2.37})$, so what make this problem hard must be the fact that $H$ scales exponentially with respect to the number of electrons, which in turn $K$ must be scales exponentially as well. Since as you mentioned, with $K$ basis function, the dim of $H$ is $K \times K$. So yes, what is the relationship between $K$ and the number of atoms or electrons?

I recommend to [edit] the question to include the details from your comment.