text:
Acidic strength order among hydroxytetralones
I can understand in 1 the conjugate base has more stability for better resonance but in 2 and 3 the effect of resonance cannot be considered. How can we compare the inductive effect in such case?

In the future, please use a competent title specific to your problem, apply proper punctuation and capitalization, name the compounds and make sure the numbering and wording you are using are traceable.

Let's look at some experimental $\mathrm{p}K_\mathrm{a}$ values of known hydroxytetralones and related compounds (Ref.1):
$$
\begin{array}{l|c|c|r}
\text{Entry} & \text{Compound} & \text{Ionic strength} & \mathrm{p}K_\mathrm{a} \\
\hline
1& \text{phenol} & 0 & 9.98 \\
2& \text{phenol} & 0.100 & 9.78 \\
3& \text{8-Hydroxy-1-tetralone} & 0 & 11.31 \\
4& \text{8-Hydroxy-1-tetralone} & 0.100 & 11.14 \\
5& \text{6-Hydroxy-1-tetralone} & 0 & - \\
6& \text{6-Hydroxy-1-tetralone} & 0.100 & 7.74 \\
7& \text{2-Hydroxyacetophenone} & 0 & 10.26 \\
8& \text{2-Hydroxyacetophenone} & 0.100 & 10.07 \\
9& \text{3-Hydroxyacetophenone} & 0 & 9.19\text{ (Ref.2)} \\
10& \text{4-Hydroxyacetophenone} & 0 & 8.05\text{ (Ref.2)}  \\
11& \text{4-Hydroxyacetophenone} & 0.100 & 7.87 \\
\hline
\end{array}
$$
It should be noted that the properties of 8-Hydroxy-1-tetralone, which is similar to that of 2-Hydroxyacetophenone except that the carbonyl oxygens is restricted to the free rotate. In contrast, after the proton transfer, the carbonyl oxygen of the anion of 2-Hydroxyacetophenone is free to rotate away from the anionic oxygen (conjugate base of the acid). Nonetheless, in these compounds, the hydrogen of hydroxyl group may be intramolecularly hydrogen bonded to the carbonyl oxygen.
A comparison of p- (Entries 10 & 11; mesomeric) and m-substitutions (Entry 9; pure inductive) in phenol, which are included for completeness, shows the well known enhancement from mesomeric interaction of p-groups. Such effect is clearly evident in the entry 6. However, the effects important to this question is the interactions of groups ortho to each other such as those in entries 3, 4, 7, and 8. The authors of Ref.1 adequately put is as:

Organic theory is notably inadequate in the quantitative sense for correlating substituent effects and particularly so for o-substitution. which in the present study involves the formation of intramolecular hydrogen bonds.

The complexity of the problem is apparent from a consideration of $\mathrm{p}K_\mathrm{a}$ values in the above Table. Whereas p-substitutions of relavent carbonyl groups in phenol (or in the tetralone structure) increases the acidity by about $2$ $\mathrm{p}K$ units (due to mesomeric interactions), the corresponding substituénts in the o-position produce again two molecules of similar structure (entries 3, 4, 7, and 8) but with acid strengths less than that of phenol. Keep in mind that the complete theoretical treatment of the effects of substituents on the acid dissociation equilibrium will require consideration of both the acid and anion (after proton transfer). In the acid, variation in the strength of intramolecular hydrogen bonding must be a major factor in the o-carbonyl substitution.

References:

Lawrence B. Magnusson, Clarence Postmus, Jr., and Carolyn A. Craig, "Coördination in Solutions. I. Acid Strengths of Phenol Derivatives in Water," J. Am. Chem. Soc. 1963, 85(12), 1711-1715 (ODI: https://doi.org/10.1021/ja00895a001).
The $\mathrm{p}K_\mathrm{a}$ for 3-hydroxyacetophenone is from: F. G. Bordwell and Glenn D. Cooper, "Conjugative Effects of Methylsulfonyl and Methylthio Groupings," J. Am. Chem. Soc. 1952, 74(4), 1058–1060 (ODI: https://doi.org/10.1021/ja01124a057).


text:
Can I convert copper(II) acetate or carbonate to copper(II) sulfate?
I am a high school student who does chemistry as a hobby, and as such I really don't know very much. I managed to make some copper(II) acetate and copper(II) carbonate, and now I'd like to convert it into copper(II) sulfate (mostly for the purpose of electroplating).
I do have sulfuric acid and I understand that I could react the copper compounds with it to make copper sulfate. However, the sulfuric acid I have claims to have metal inhibitors, and I don't know if that will impede the reaction.
It would be really interesting to me if I could take elemental sulfur and use it in the conversion process, because I do have a good amount of that.

I doubt the metal inhibitors will stop this reaction

Or you could test plating directly with copper acetate... it might work as well or better than the sulfate for shiny plating.

https://icandochemistry.com/2021/05/22/reactions-to-make-copper-sulfate/

$1$) How have you prepared copper acetate, and copper carbonate ? $2$) Why don't you use copper acetate for plating copper as suggested by Moishe Pippik ? $3$) Forget about using elementary sulfur. It is difficult to transform it into sulfate in a non-industrial way.

Practical approach
Arguably, the easiest way is to trade/exchange your chemicals — copper(II) salts and elemental sulfur — for copper(II) sulfate, which should be readily available for purchase online or from a gardening store as it's considered both fungicide and fertilizer. Since you are going to use the sulfate for electroplating, it doesn't matter much whether you get an anhydrous salt or a hydrate. This way you will save energy, time, resources and will be aware of the purity.
Although it's perfectly doable in a chemical lab, I would advise against conversion of elemental sulfur to sulfate as it requires a harsh conditions and a strong oxidant (hydrogen peroxide, chlorine/bromine water) and is not only time-consuming and arduous, but is also a bit dangerous, and given the availability of vitriol, the risk is just not worth it. At all.
For educational purposes you can use sulfuric acid, the chemistry is pretty straightforward. If you need to isolate a relatively pure product, you will need to crystallize the product $(\ce{CuSO4.5H2O})$ from the saturated aqueous solution, which implies losses, so make sure you have enough starting material (copper(II) salts) as well as a heating plate, preferably with a stirrer.
Synthesis from copper(II) carbonate
I doubt you have pure $\ce{CuCO3}$ because it cannot be obtained directly by exchange reactions in aqueous solutions between $\ce{Cu(II)}$ salts and soluble carbonates: due to hydrolysis you always get hydroxocarbonates $(\ce{Cu2(OH)2CO3}$ or $\ce{Cu3(OH)2(CO3)2}).$
Anyway, careful addition of sulfuric acid to either carbonate or hydroxocarbonate should yield in solution of copper(II) sulfate:
$$\ce{CuCO3(s) + H2SO4(aq) -> CuSO4(aq) + CO2(g) + H2O(l)}\tag{R1}$$
Synthesis from copper(II) acetate
Since both acetate and sulfate are readily soluble in water, you might need to remove acetate to facilitate crystallization of sulfate.
This can be relatively laborious depending on how you can do this.

Convert acetate to insoluble copper(II) hydroxide by adding any water-soluble inorganic base (except, of course, ammonia):
$$\ce{Cu(OAc)2(aq) + 2 NaOH(aq) -> Cu(OH)2(s) + 2 NaOAc(aq)}\tag{R2}$$
The precipitate is usually quite amorphous and will be hard to filter.
To make it more crystalline, try to warm up the solution a bit (do not exceed 80 °C) and then cool it down, say, in a fridge, to 4 °C.
After decantation, washing the hydroxide from remaining acetate with deionized water on the filter and transferring it to a glass vessel, you can add sulfuric acid:
$$\ce{Cu(OH)2(s) + H2SO4(aq) -> CuSO4(aq) + 2 H2O(l)} \tag{R3}$$

Alternatively, you may try to thermally decompose acetate to copper(II) oxide by heating it in the air above 300 °C:
$$\ce{Cu(OAc)2(s) + 3 O2(g) ->[\Delta] CuO(s) + 3 H2O(g) + 2 CO2(g)} \tag{R4}$$
Then, treat the cooled down oxide powder with the acid:
$$\ce{CuO(s) + H2SO4(aq) -> CuSO4(aq) + H2O(l)} \tag{R5}$$

Note on inhibitors
Acid inhibitors are designed to prevent corrosion of construction materials such as carbon/stainless steel, aluminum and copper metals and alloys and should not affect the aforementioned reactions.

source: Stack Exchange

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text:
Why does hydrogen bonding cause a broad stretch in IR spectroscopy?
So hydrogen bonding is an intermolecular forces. I'm assuming that the hydrogen bonding strength varies between molecules and some molecules won't undergo hydrogen bonding causing different bond lengths and stretch frequencies.

Or that the intermolecular distances vary, unlike in a molecular bond where they're relatively stable (although, consider molecules like ammonia, which flips back and forth, as in a MASER).

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text:
Does a free radical localized at one atom in a compound make sense or could it be?
In most of our cases, a dot that represents a radical is always considered affiliated to an atom in a molecule, especially when we talk about a reaction mechanism.
However, Levine's Quantum Chemistry [Chap 15.8 - Localized MOs] tells me that

the Hartree–Fock wave functions of nonclosed-shell electronic states
are, in most cases, linear combinations of a few Slater determinants
[for example, see (10.44) and (10.45)], and the above localization
procedure does not apply to the open-shell orbitals in these wave
functions. Thus, in a molecule in an excited electronic state with an
open-shell configuration, the electrons in the incompletely filled MOs
are delocalized over much of the molecule.

Does that mean a radical is actually delocalized in the whole molecule and that the localized image of a radical is inappropriate? If so, why do we still use it to explain some chemical mechanisms?

Well, for starters, radicals are rarely excited.

Is that section of Levine's book specifically about restricted Hartree-Fock? In general for open shell systems the RHF solution is not a single determinant, and this might be what is being discussed. But we need a little more context

Is that to say radicals are actually in their ground states but with a high energy? In spite of this, I think the key point is not excited but open-shell. Radicals do have open-shell electronic configurations. @Mithoron

I understand what you said. But it's pointed out that the localized MOs, which the section talks about, do not apply to open-shell systems, which means electrons in these molecules are delocalized. So the question at the top was asked. Do you have any idea? @IanBush


text:
How to name ether with ketone and carboxyl groups?
Carboxylic acid is the main functional group due to higher priority, but since it's an ether, what's the proper way to name the following compound?

My attempt: 2-(1-methyl oxo propyl ether) ethanoic acid.

You have already found out that the carboxylic acid group is the principal characteristic group. Therefore, the ether group has to be expressed as a prefix. The corresponding rule in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows.

P-63.2.2.1.1 Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. These compound prefixes require the numerical multiplying prefixes ‘bis’, ‘tris’, etc.

In this case, the substituent prefix name for the $\ce{R'-{}}$ group is 3-oxobutan-2-yl and thus the substituent prefix name for the $\ce{R'-O-{}}$ group is (3-oxobutan-2-yl)oxy. Therefore, the complete name for the compound that is given in the question is [(3-oxobutan-2-yl)oxy]acetic acid.

Note that locants are omitted for parent compounds when all substitutable hydrogen atoms have the same locant. Therefore, the preferred IUPAC name is [(3-oxobutan-2-yl)oxy]acetic acid and not 2-[(3-oxobutan-2-yl)oxy]acetic acid.

Thanks but can you just explain 1) why did you add "2-yl" and 2) is it 3 oxo instead of 2 oxo because you went lower numbering to the location of point where the ether connects

@Chuck butan-2-yl is an alkanyl-type substituent group with the free valence at position 2, not to be confused with the alkyl-type substituent group butyl with the free valence at position 1. It's 3-oxobutan-2-yl and not "2-oxobutan-3-yl" because the free valence gets a low locant first.


text:
Polycyclic or heterocyclic compound naming
What is the order and name of the following polycyclic compound?
I am studying heterocyclic chemistry to teach some notions to my high school students even if it is outside curriculum (to give them advantages over those who don't know them at their age) and I am planning to make my own notes about the subject. However, I am not sure of the method of naming these classes of compounds, and after consulting the internet, I saw there are SEVERAL ways to name them (some of them confusing for me, specially the notation with [x.y.z.t...] notation. I can not understand where the numbers come from and the preference order between rings...
Remark: What are the rules to name polycyclic compounds?
References to name polycyclic and 3d organic compounds would be useful too.

What do you mean by "order"?

There should be some stereochemistry, too.

@Loong The numbering sequence ordering and the notation. I tried to explain it with my new formatted question...Let me know if you find it useful now. I do want to explain it to my students...

@riemannium You mention "heterocyclic" several times (text, title, tag, and alt text of the image); your example, however, is not heterocyclic.

Maybe start here: https://iupac.qmul.ac.uk/BlueBook/P2.html#25 I'm not sure it's possible to summarise several sections of the Blue Book in an answer, though.

related https://chemistry.stackexchange.com/questions/169432/iupac-naming-of-lysergic-acid-diethylamide-lsd-heterocyclic-core-and-its-locan

@Loong, heterocyclic does not mean that there are several rings? I thought so...

@riemannium heterocyclic compounds

@Loong Then, how is named the category of an organic compound with several merged rings like the one of mine?

@riemannium According to the IUPAC Blue Book, they are called "fused ring systems".

First of all, it sometimes helps if you draw the structure correctly according to the IUPAC recommendations for orientation of fused ring systems (see Subsection P-25.3.2.3 for details).

Then you should start with the corresponding hydrocarbon parent component that has the maximum number of noncumulative double bonds (called mancude ring system). We will handle the saturation later.

Check if the mancude ring system is on the list of retained traditional names (like naphthalene) in Subsection P-25.1.1. If the whole structure is not on the list, check if there is at least a part of it. In this case, you should find indene.

Next you can do the fusion with a cyclobutane ring.
Numbers and letters [1,2-f] are used to indicate the side where the fusion occurs and the relative orientation of the fused ring system. The direction of the numbers of the attached component (e.g. ‘1,2’ or ‘2,1’) corresponds to the direction of the lettering of the peripheral sides of the parent component (i.e. a, b, c, etc.).
The corresponding Subsection P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), reads as follows:

(…)
Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (including sides whose locants are distinguished by letters, for
example, 2a,3a) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)

In this case the first two positions of cyclobutane (‘1,2’) are fused with the sixth side of indene (‘f’). That gives us cyclobuta[1,2-f]indene. However, all sides of cyclobutane look the same, so no indication of the position and orientation is actually necessary. Therefore, we can ignore this part and just keep cyclobuta[f]indene. Nevertheless, we will need these positions later.
We are still working with a mancude ring system, so we now have 1H-cyclobuta[f]indene.

Next you can do the fusion with a cyclopentane ring.
Remember that numerical locants (‘1,2’) were used for the first-order attached component cyclobutane. For the next fusion, we need to indicate the opposite side (‘3,4’).
The locants of the second-order attached components cyclopentane (‘1',2'’) are primed to contrast with those of the first-order attached components cyclobutane (‘3,4’). The two sets of locants are separated by a colon. This gives us cyclopenta[1',2':3,4]cyclobuta[1,2-f]indene. Again, all sides of cyclopentane look the same, so no indication of a position and orientation is actually necessary. Therefore, we can ignore this part and just keep cyclopenta[3,4]cyclobuta[1,2-f]indene.

Finally, you have to restore the saturation. Since the compound is completely saturated, no locants are needed. This gives us the complete name tetradecahydrocyclopenta[3,4]cyclobuta[1,2-f]indene.


text:
When does Van't Hoff's reaction rate rule not work?
I have a problem answering a question in my lab report.
The question is about the empirical van't Hoff's rule - what does it say and what are its limitations?
I managed to find online that the reaction speed doubles when you increase the temperature by 10°C, and that somewhat matches our experiment.
Now, the question is when does it not work. I found only a few answers but they seemed complex (they were either written in a confusing way or it's because I'm not a native English speaker).
I found this: "The rule is an approximation that works best when temperatures approximate those under which the reaction normally occurs." #
But what does it mean, "temperature under which the reaction normally occurs"?
Also, related to the last question, should van't Hoff's rule work for a reaction between sodium thiosulphate and sulphuric acid?

https://chemistry.stackexchange.com/questions/22153; https://chemistry.stackexchange.com/questions/66233

The van't Hoff's empirical rule says:

The reaction rate typically raises 2-4 times with the temperature increase by $\pu{10 ^\circ C}$.

The range of the rule corresponds to typical range of reaction activation energies. The higher the reaction activation energy is, the higher is the coefficient of the rule.
The rule you have quoted is from a medical dictionary. Biochemical reactions have in average rather lower activation energies and for this specific domain the value $2$ may generally about fit. But note that the value for a particular reaction is a real number.
It is closely related to the Arrhenius equation (the simplest equation for the dependence of a reaction kinetic constant on temperature):
$$k = A \exp{( - \frac{E_\mathrm{a}}{RT})}.$$

We can see in the below equation the relation of the van't Hoff rule coefficient and the reaction activation energy:
$$C = \frac{ \exp{(- \frac{{E_\mathrm{a}}}{{R(T+\Delta T)}}})}{\exp{(- \frac{{E_\mathrm{a}}}{{RT}}} )}=\exp{\left(\frac{E_\mathrm{a}}{RT} - \frac{E_\mathrm{a}}{R(T+\Delta T)}\right)}=\exp{\left(\frac{E_\mathrm{a}\Delta T}{RT(T+\Delta T)}\right)}$$

$C$ is the ratio of reaction kinetic constants.
$E_\mathrm{a}$ is the reaction activation energy.
$T$ is the lower absolute temperature of measurement.
$\Delta T$ is the measurement temperature difference.

If $\Delta T = \pu{10 K}$,  $C$ is equal to the van't Hoff rule empirical coefficient.
$E_\mathrm{a}$ can be computed from $C$, $T$ and $\Delta T$ this way:
$$\ln{C} = \frac{E_\mathrm{a}\Delta T}{RT(T+\Delta T)} $$
$$E_\mathrm{a} = RT\ln{(C)} \left(\frac {T+\Delta T}{\Delta T}\right)$$
For $T= \pu{298.15 K}$ and $\Delta T = \pu{10 K}$: $E_\mathrm{a} \approx  \pu{76.4 \ln{(C)} kJ mol-1}$
For $C = 2$: $E_\mathrm{a} \approx  \pu{76.4 \ln{(2)} kJ mol-1} \approx \pu{52.9 kJ mol-1}$
For $C = 4$: $E_\mathrm{a} \approx  \pu{76.4 \ln{(4)} kJ mol-1} \approx \pu{106 kJ mol-1}$
Therefore, the coefficient of the van't Hoff rule is about 2 for reaction activation energies around $\pu{50 kJ mol-1}$.

So the rule is that the reaction time raises around 2 times when the temperature increases by 10°C, but, the higher the activation energy of the reaction, the larger the increase? That actually makes sense.
How would one calculate the theoretical activation energy of a reaction between sodium thiosulphate and sulphuric acid?

See the answer update.

Ok, but if I know the reactants and the products, how would I determine the activation energy of the reaction?
The question I need to answer is if van't Hoff's rule works for a reaction between $\ce{Na2S2O3}$ and $\ce{H2SO4}$. I guess I need to see if that reaction's activation energy is around 50 kJ/mol.
Based on the experiment I would guess it's lower than 50 kJ/mol.

The activation energy is an experimental value. I doubt you find it written, so you have to determine it experimentally, obtaining the reaction speed ratio at 2 different temperatures. But you need not to calculate the activation energy at all. Enough is to compare reaction speed at temperatures different by 10 K.

Oh ok, that's what I did. I wasn't sure if that was enough. I will mark your answer as correct. Thank you for taking your time to help.


text:
Flash point of naptha and propanol
I filled a Zippo lighter with 2-propanol and it lit up easily when the ambient temperature was around 6-7 degrees Celsius.
It was impossible to light it up when the temperature was around 3-4 degrees Celsius. Then I put a drop of lighter fluid (naphtha) on the wick and it lit up first try. After the drop of naphtha was burned the lighter was heated up a bit and then it also sparked easily with the propanol.
The reported flash point of propanol is 12 degrees Celsius while the flash point of naphtha is 30-38 degree Celsius.
How can this be? Shouldn't be the flash point of naphtha lower than propanol recording to my "experiment" ?

Flash point relates to a open pan or pool of liquid with no wick.
The definition of flash point is the lowest temperture of a combustable liquid at which it will catch fire when a flame is presented above the liquid.
When you have a wick present then it is much more easy to burn a fuel, even if the bulk of the fuel is below the flash point it is possible to burn it at the wick. For example a parafin lamp normally has a wick, you light the wick with a match and then the parafin burns.
So I think that the flash point issue is not so important, I am a bit surprised. Normally the fuel for a zippo is a version of gasoline (US) or petrol (UK) which does not have to have the fuel additives or a given octane rating. Also when you give the flash point of naphtha keep in mind.
"All naphtha is equal but some naphthas are more equal than others".
There are lots of different grades of naphtha, they will vary a lot in their properties (including their flash point). For example if we take the kerosene fraction then the different kerosenes have a wide range of different flash points.

Thanks for your answer. It makes sense to me that using a wick can reduce the actual flash point but shouldn't this be true for both substances to the same degree? Or is is possible that the wick supports the flash point of naphtha more than propanol?

The lighter fluid is not from Zippo just a cheap version from a supermarket. Is it possible that it is made of compounds which have a so much lower flash point than the reported 30 degrees?

Interesting because I had a similar issue trying to run a Coleman catalytic heater with kerosene.  This lead to some checking of flash points and petroleum distillate definitions.
The distillate "cuts" run from gasoline (flash point -23 C) to Coleman fuel, "white gas", or light naphtha (flash point - 4C) to kerosene (flash point + 38C).
2 propanol flashes at + 12 C. Zippo lighter fluid flashes at "< 6.5 C"$^1$. Hexane flashes at - 22 C. Butane flashes at - 60 C.
So, just as kerosene worked for a warm catalytic heater but not for a cold one, 2 propanol only works when your lighter is "warmed up".
This is because, below the flash temperature, the fluid will not produce enough vapor pressure (as a percentage of air) for the spark to ignite it.  This is called the lower flammability (or explosion)  limit.
$^1$ Zippo lighter fluid MSDS sheet

Do you have a source for the flash point of -4°C for light naphtha? I can't find this anywhere but this would indeed explain it.

@flappix naphtha is an archaic term meaning any distillate fraction above gasoline up to kerosene and diesel.  "Light" naphtha is the lower boiling cut, heavy naphtha is kerosene.  Coleman fuel is light naphtha.  MSDS sheets on anything you buy shoud have flashpoint information.

@flappix more reading here: petroleum ether, Standards subsection, and here: naphtha


text:
Acidic strength of nitrophenols
I've seen the pKa values of Nitrophenol as follows:-

pKa of o-nitrophenol= 7.23
pKa of p-nitrohenol= 7.14

so p-nitrophenol is more acidic than p-nitrophenol

But both phenoxide ions are stabilized by -I effect and resonance.
the only difference is accumulation of negative charge on o-nitrophenolate ion, which would make it more stable anion, which means o-nitrophenol is more acidic
But the data says that p-nitrophenol is more acidic than o-nitrophenol.
What gives p-nitrophenol extra stability for it to be more acidic than o-nitrophenol?

There is a typo somewhere here: so p-nitrophenol is more acidic than p-nitrophenol.

I have recently answer a question about some $\mathrm{p}K_\mathrm{a}$ value discrepancies on different hydroxytetralone derivatives. As Oscar Lanzi pointed out in his answer that the major reason why o-isomer is less acidic is acid groups' capability to have (strong) hydrogen-bond with available electron donor. For example, for tetralone, it is the carbonyl oxygen of o-ketone group. Thus, the stability of the conjugate base does not solely determine the acidity of the parent compound.
Let's look at some experimental $\mathrm{p}K_\mathrm{a}$ values of known nitrophenols and related compounds (Ref.1):
$$
\begin{array}{l|c|c|r}
\text{Entry} & \text{Compound} & \text{Ionic strength} & \mathrm{p}K_\mathrm{a} \\
\hline
1& \text{phenol} & 0 & 9.98 \\
2& \text{phenol} & 0.100 & 9.78 \\
3& \text{2-nitrophenol} & 0.100 & 7.05 \\
4& \text{3-nitrophenol} & 0.100 & 8.18 \\
5& \text{4-nitrophenol} & 0.100 & 6.96 \\
6& \text{2-Hydroxybenzaldehyde} & 0.100 & 8.14 \\
7& \text{3-Hydroxybenzaldehyde} & 0 & 9.02\text{ (Ref.2)} \\
8& \text{4-Hydroxybenzaldehyde} & 0.100 & 7.45 \\
9& \text{6-Hydroxy-1-tetralone} & 0.100 & 7.74 \\
10& \text{2-Hydroxyacetophenone} & 0 & 10.26 \\
11& \text{2-Hydroxyacetophenone} & 0.100 & 10.07 \\
12& \text{3-Hydroxyacetophenone} & 0 & 9.19\text{ (Ref.2)} \\
13& \text{4-Hydroxyacetophenone} & 0 & 8.05\text{ (Ref.2)}  \\
14& \text{4-Hydroxyacetophenone} & 0.100 & 7.87 \\
\hline
\end{array}
$$
The compounds were chosen as the hydrogen of o-hydroxyl group may be intramolecularly hydrogen bonded to the electro-donating substitution group. In nitro compounds, it is $\ce{N-O}$ group, In benzaldehyde compounds, it is oxygen of $\ce{CHO}$ group, and In acetophenone compounds, it is oxygen of $\ce{CO-CH3}$ group.
A comparison of p- (Entries 5, 8 & 13; mesomeric) and m-substitutions (Entries 4, 7, and 12; pure inductive) in all phenol derivatives, have shown the well known enhancement from mesomeric interaction of p-groups and inductive interaction of m-group effects, increasing the acidity in both cases (increasing the stability of the conjugate base). However, as in the previous question, the effects important to current question is the interactions of groups ortho to each other such as those in entries 3, 6, and 11. As the authors of Ref.1 correctly put it, the unusual correlating substituent effects particularly by the o-substitution should be due to the formation of intramolecular hydrogen bonding. Thus, it is safe to say that the strength of hydrogen-bonding in parent molecule play a significant role in acid strength compared to the stability of conjugate base achieved by the same substitution.

References:

Lawrence B. Magnusson, Clarence Postmus, Jr., and Carolyn A. Craig, "Coördination in Solutions. I. Acid Strengths of Phenol Derivatives in Water," J. Am. Chem. Soc. 1963, 85(12), 1711-1715 (ODI: https://doi.org/10.1021/ja00895a001).
The $\mathrm{p}K_\mathrm{a}$ for 3-hydroxyacetophenone is from: F. G. Bordwell and Glenn D. Cooper, "Conjugative Effects of Methylsulfonyl and Methylthio Groupings," J. Am. Chem. Soc. 1952, 74(4), 1058–1060 (ODI: https://doi.org/10.1021/ja01124a057).

There may be hydrogen bonding between the hydroxyl hydrogen and a nitro group oxygen in the ortho isomer, which would stabilize the proton. But 0.09 $\pu{pK_a}$ unit could be within experimental error.

The nitro-group in the one case will be distorted out of plane, where in the other it isn't. Any reasoning based on the given two values is a mere guess in my opinion though.

Agreed, especially when you have elephants on the weighing scale and they are so nearly cancelling out that a mouse tips the balance.


text:
How does tetraethyl lead prevent knocking?
My understanding of knocking is that in a spark ignition engine as the fuel is being compressed it heats up.  The heat causes some of the oxygen to break up into oxygen radicals the $\ce{O^.}$ radicals react with the fuel grabbing parts of the fuel and creating radicals. These radicals react with molecular Oxygen thus releasing oxygen radicals to go on and continue the chain reaction.
I can understand how adding ethanol to fuel would fix this as $\ce{O^. + H3CCH2OH → H2O + H3CCHO}$ would mop up the oxygen radicals. But how does tetraethyl lead fix this?
Or is my understanding of knocking and cold flame just wrong?

https://www.britannica.com/science/tetraethyl-lead

Related: Why would tellurium + sodium hydroxide have worked as a good anti-knock gasoline additive (if it wasn't so smelly)?; see also the comment by @ToddMinehardt for the mechanism

About tellurium - related dramatised reading or listening: https://episodictable.com/tellurium/

Engine combustion reactions are complex but Tetraethyl Lead (TEL) slows thermal combustion
It is important to understand what causes knocking in internal combustion engines. In diesel engines the heat cause by compression is what ignites the fuel. They typically use linear hydrocarbon fuels as they are more likely to auto-combust under compression. Petrol (gasoline) engines are designed differently and the combustion is initiated by a spark. Combustion initiated by compression or heat is undesirable as the engine no longer controls when the ignition happens. Typical fuels contain more branched chain, aromatics and unsaturated hydrocarbons.   If early combustion does happen it can damage the engine and is called "knocking".
The important issue with all antiknock agents is to inhibit early combustion caused by compression and heat. And this is exactly what TEL has been observed to do.
And it wasn't discovered from theory. It was found to work by experimenting with a large variety of alternatives (and, unfortunately, economics, not safety, dominated the choice over other possibilities).
But what is the mechanism? This is harder to determine as there are a large variety of radical reactions possible in air/fuel mixtures. It isn't likely that the key step is the production of oxygen radicals (dioxygen is already a diradical under normal atmospheric conditions). But it is reasonable to assume that highly reactive hydrocarbon radicals play a part (the radicals produced from the linear hydrocarbons in diesel fuel would be more reactive than those from petrol/gasoline fuel). And it is also reasonable that TEL would "mop up" some of the potential radicals produced from heat and compression in a standard spark engine. Inhibiting these, even to a small extent, seems to be enough to reduce the propagation speed of those unwanted early side reactions so that knocking does not initiate early combustion.
BTW, though ethanol does work, it takes about 10% of it in the fuel. Better keep that for drinking and develop other, better and non-toxic, agents.


text:
Why is the boiling point of a iodoalkane greater than that of the corresponding fluoroalkane?
Why is the boiling point of ethyl iodide (b.p. 72 °C) higher than the boiling point of ethyl fluoride (b.p. −37 °C) even though the dipole–dipole interaction should be better in $\ce{C2H5F}?$ Is it because of higher molecular mass of iodine?

Related - Comparison of boiling points of n-propyl chlorine and ethyl iodide

The molecular weight difference is significant.

Because given enough polariz-able electrons dispersion forces beat polarity. And most things have enough polariz-able electrons.

The way in which intermolecular forces (IMF) are taught can sometimes be confusing.  Students are generally taught that dispersion forces (London dispersion, van der Waals) are the weakest IMF, with by dipole-dipole interactions being stronger and then hydrogen bonding being the strongest.
This is true for individual interactions, but since properties like melting point and boiling point are affected by the sum of all the interactions between molecules, the situation is not as straightforward as that.
As a rule of thumb, when two molecules have similar molar masses (about ${\pm}$10 g/mol) to each other, the more polar molecule will have the higher boiling point.
For example, consider ethane, fluoromethane and iodomethane.

Ethane
Fluoromethane
Iodomethane

Formula
$\ce{C2H6}$
$\ce{CH3F}$
$\ce{CH3I}$

Molar mass (g/mol)
30.07
34.03
141.94

Dipole moment (Debye)
0
1.847
1.620

Boiling Point (${^o}$C)
-89
-78.4
42

If we compare ethane and fluoromethane, they have similar molar masses, so the dispersion interactions for each molecule can considered similar.  That being the case, the difference in boiling point results from the presence of dipole-dipole interactions.
However, if we compare fluoromethane to iodomethane, the difference in the molar masses (really the difference in size between F and I) is the determining factor.  Even though $\ce{CH3F}$ is more polar than $\ce{CH3I}$, it has a lower boiling point.
It's important to realize as well that the difference in polarity between similar haloalkanes is not as dramatic as students are led to think.

$\ce{CH3F}$
$\ce{CH3Cl}$
$\ce{CH3Br}$
$\ce{CH3I}$

Molar mass (g/mol)
34.03
50.49
94.94
141.94

Dipole moment (Debye)
1.847
1.870
1.810
1.620

Boiling point (${^oC}$)
-78.4
-24
3.56
42

References
Boiling points: Wikipedia
Dipole moments: https://cccbdb.nist.gov/diplistx.asp


text:
The edge case of an edge case: Vapor emissions of the descaling agent Sulfamic acid
I am wondering about the following scenario:
What happens if ready bought sulfamic acid descaling (usually 15% cont. + $\ce{H2O}$) is vaporized or at least steam develops while very briefly descaling the steam wand of an espresso machine? Is there any real danger to the respiratory system with this acid in a liquid solution, even if it is already diluted? When would $\ce{SO2}$ or sulfuric acid be produced?
The material safety sheet does not warn of any real dangers if the solution becomes airborne. Pure sulfamic acid can produced $\ce{SO2}$ around 300 °C but while descaling its mixed with water.
A possible product through hydrolysis seems to be ammonium bisulfate, which in turn can thermally decompose above 120 °C to sulfur dioxides.
How much harmful $\ce{SO2}$ can there even be produced if we are only starting out with an already diluted solution and we are looking at two different reactions with additional byproducts. $\ce{SO2}$ furthermore reacts in water(vapor) to sulfuric acid which has a much higher density than air and also has a much higher boiling point than water. Is there even enough "left" for a meaningful amount of anything dangerous to be created?
I admit my problem is sadly further amplified by the fact, that the concerning espresso machine has a steam generator(circulation heater) which probably runs hotter than the regular ~140°C of an espresso machine boiler.

According to the Handbook of Chemistry and Physics, sulfamic acid $\ce{NH2SO3H}$ is decomposed at about $200°$C in $\ce{NH3}$ and $\ce{SO3}$, which produces immediately ammonium bisulfate $\ce{NH4HSO4}$ with atmospheric water. And sulfamic acid is very soluble in water : one part acid in $2$ parts water at $100°$C.
As a consequence, you do not risk any absorption when heating solutions before $200°$C. And at higher temperatures, you will probably never get to the dangerous limit of $4$ g/m3, specially with espresso machines. All you risk is being touched by micro bubbles of ammonium bisulfate solutions


text:
Why does the chlorine radical attack the carbon-hydrogen bond in methane as apposed to the chlorine - carbon bond in a molecule of chloromethane?
During the propagation step in a free radical substitution for the production of haloalkanes, a chlorine radical attacks the carbon-hydrogen bond in chloromethane. Even though this feels right, and I understand why, I couldn't find much information on the interaction between the Chlorine-carbon bond and the chlorine radical. Why doesn't it break that bond and make $\ce{Cl2}$ and a methyl radical? Is it because the chlorine carbon bond is much stronger than the carbon-hydrogen bond?

The reaction is exothermic. The C-H bond of methyl chloride is attacked by  chlorine atoms. Higher chlorination products are formed. Take a look at either of the first two items here.


text:
How to make a small zinc-air battery?
Background:
I want to make a very small (20-30mg, 3 x 3 x 1 mm) primary battery cell, for powering a device that will go on a bumblebee. It will only need a capacity of about 300μWh (~1J). I'll need a peak current of about 2.5mA (although this is 'negotiable'). Ideally it needs to provide 1.8V - 3.3V (but obviously can place in series, to get voltage up).
Plan:
I thought a zinc-air battery would be ideal, as it only needs to be a primary cell; high energy density; simple (so easy to make and miniaturise); doesn't require complicated equipment.
Following a fairly standard approach: the initial version will just be sandwich of zinc foil (anode, ~20μm thick), copper foil (catalyst & negative terminal, ~10um thick), and sodium hydroxide as the electrolyte.
I need a separator and a semipermeable enclosure...
Question 1: For the separator, would a standard PE Battery Separator work? (I don't want something that will dissolve or react with the hydroxides). I think this is what is used in li-ion batteries (needs to be something I can buy).
Question 2: At least some of the enclosure needs to be oxygen permeable - what material should I consider (to keep hydroxides in, but allow ingress of oxygen)? (ideally not too exotic - need something I can actually buy!).
Question 3: What modifications would you suggest to make this more capacious or powerful? (e.g. I wonder about replacing the zinc foil with zinc powder suspended in an electrolyte gel, to increase surface area, for example). I guess I could switch to a dual-electrolyte battery - but I'm trying to keep it simple (especially given how small it is). I also wonder about whether increasing the surface area of the 'cathode' catalyst?
Question 4: Feel free to answer with wider suggestions of other chemistries that I might consider that are similarly simple, and easy to build/use.

On a bumblebee ?

A SR4165w / 337 cell weighs ~100 mg (https://www.mouser.com/ProductDetail/Seiko-Instruments/SR416SW?qs=wUXugUrL1qxKS1AC4Wuieg%3D%3D) and is ~5 mm dia x 1.65 mm. Perhaps removing most of the metal shell could bring it down to fly-weight?

@DrMoishePippik: I'm currently about to use this 70mg battery (https://www.sii.co.jp/en/me/datasheets/ms-rechargeable/ms412fe-5/). As you've spotted, the metal case is a large part of the weight (also the discharge rate means I'll need to combine it with a 100uF cap probably). Yours is a good suggestion: I've considered modifying these batteries before, but not tried it. Maybe lithium batteries aren't a good start for that (they won't like oxygen getting in).

@Maurice: I'm putting a bluetooth (BLE) SoC on the bee: It receives and stores packets [and received signal strength] sent by high-gain, rotating, bluetooth transmitters. Reading off this data later, we can triangulate the path the bee took.

BTW, would a passive corner reflector work, triangulated by multiple light-sources and cameras? It would need to be NIR or visible light, though, due to size limit, limiting tracking to line-of-sight, and blocked by foliage. sigh

@DrMoishePippik - yes! Here's our paper about it. We've also done as you described, with 6 camera systems arranged around the nest & reconstructed the 3d path of the bee during learning flights. Unfortunately the signal strength suffers from the radar equation's 1/d^4 which means that beyond about 30-40m the signal is very hard to detect over false positives. I'm finding low energy bluetooth (esp with the high-gain antenna on the transmitter) has a range of 100s of metres (if there's not too much in the way).

@lionfish, congrats on that fine paper, and thank you for forwarding the link! Best wishes on the "RFID tags". Hmmmm... charge from a piezo transducer when they flap wings? ;-) Just kidding.

@DrMoishePippik That's been done: piezo tranducer to get power from bees paper - a few uW though - so I think a 1mm^2 solar cell would probably be more use!
On topic: I've ordered a Celgard single layer PP diaphragm (2500) as a separator, some zinc foil, copper foil and stainless steel foil [I couldn't easily find pure iron foil] and some NaOH. Will report back what works. I've not solved what semi-permeable membrane to use - so any advice on that would be really appreciated.


text:
Can stainless steel "soap" bars actually work or is it just deceptive marketing?
The Wikipedia — Stainless steel soap article sounds highly skeptical, although it does include a possible mechanism involving the chromium in the stainless steel binding with the sulfur compounds and then being rinsed away. It seems to me that if that's what happens, you would quickly run out of chromium in the outer surface of the stainless steel, and it would no longer be effective. Seems like snake oil to me.  Has the effectiveness of the SS soap been proved from a perspective of chemistry?

Well it works ok for garlic. Have you tried it?

Did you compare it to just rinsing (without the stainless steel)?

But the real question is whether one would expect it to work based on the chemistry.

It "works ok"? What it would even supposed to do, adsorb stuff? Any real smell absorber needs to have large surface area, for starters.


text:
What is the best way to deprotonate a methyl group?
I am attempting to deprotonate the methyl group on 2-methylbenzo[d]thiazol-6-ol (hydroxyl group is protected with a silyl group) in order to attach a phosphate group in the form of diethyl chlorophosphate. I have tried n-BuLi, LDA, NaH in DMF and THF at -78, 0, and 25 degrees Celsius and at reflux but none of these trials has resulted in any reaction apart from NaH in DMF at reflux, which resulted in the silyl protecting group falling off and unconvincing NMR evidence of the phosphate attaching. I need to attach the phosphate to allow a Horner-Wadsworth-Emmons reaction in the next step. Are there any other options I can explore?

Have you tried running the deprotonation and quenching with D2O? Your problem maybe with the chlorophosphate electrophile rather than with the deprotonation conditions.

It has been done with the parent substrate: https://pubs.acs.org/doi/abs/10.1021/ja00069a016

Alternatively, you may consider an Arbuzov approach. Brominate (NBS) the methyl group and react the bromide with $\ce{(EtO)3P}$ as done here. It has all been done on the parent substrate.

BTW: Welcome to ChemSE!

The deprotonation of 1a has been conducted with LDA/THF and subsequently converted to the phosphonate.[1] Alternatively, bromide 3a has been prepared[2] and converted[3] to phosphonate 2a via an Arbuzov reaction. It also appears that your desired compound 2b has been prepared by the Arbuzov route from 3b. [4]

G. Cai, N. Bozhkova, J. Odingo, N. Berova, K. Nakanishi, J. Am. Chem. Soc., 1993, 115, 7192. (The preparation of compound 11 (2a here) in the Experimental Section of this paper omits the addition of $\ce{(EtO)2POCl}$ (?)).

T. Kotani, et al., https://www.jstage.jst.go.jp/article/cpb1958/45/2/45_2_297/_article

L. Jiang, et al., RSC Adv., 2016, 6, 17318.

Higuchi, Makoto, et al., Chem. Absts. Link not available.


text:
unexpected ozonolysis products: short chain alkanes (methan, ethane, propane) as reaction products when ozone reacts with organic matter
My question is which molecule or functional group reacts with ozone to yield short-chain alkanes?
Some background: I use ozone produced by an ozone generator with the aim to clean my vacuum instrument to reduce the blank for trace gas analysis (methane, ethane,...). The idea was that ozone would react with any labile molecules that are attached to the surfaces of my instrument. The hope was that the reaction products of these background organics with ozone can be pumped away with the vacuum and afterwards the surface would be clean and my blank is lower. Unfortunately, I observed an unexpected effect and found a large suite of short-chain alkanes (C1-C3) and also ethene and propene that desorb for hours from my surfaces (especially when liquid water is present). I was very surprised as reactions with ozone usually lead to O-containing functional groups while alkanes are not typical reaction products of "ozonolysis". Would it be possible that methyl- and ethyl-radicals are produced in an intermediate step which then abstracts an H atom from other molecules to finally yield alkanes?
Any help is highly appreciated!

Why do you think it worked at all?

What is "my vacuum instrument" and how did you analyze and quantify gases?

@JochenSchmitt I would definitely [edit] in this information to the question.

@andselisk I tried to keep my question short so more details of "my vacuum instrument" can be found in this publication: http://www.atmos-meas-tech.net/7/2645/2014/. In short, it is a GC-IRMS method where we measure gases extracted from ice core samples, thus air from past times. The sample size is small with around 150 g of ice which has ca. 15 mL STP air and the concentration of the trace gases is very low in the range of 0.5 ppb for ethane. So I am dealing with tiny amounts of substance and thus contamination is a big issue.


text:
What has caused a plateau in the lead up to the equivalence point of a conductometric titration?
In a titration of HCl against NH3OH, conductivity had increased in a linear fashion (think y=x), and before it approached the equivalence point, there was approximately 0.8mL of 'plateau'. Before logging (based on 'event' - mL of HCl titrated), I had ensured that the values were not fluctuating significantly for at least 10 seconds; the plateau appeared across 7 titrations. I can't find any literature which provides reason for this. TIA.

Are you adding NH4OH to an HCl solution, or HCl to NH4OH?  There is some information on NH4OH being not being nearly as conductive as HCl solution, because it exists in equilibrium with NH3 + H2O.

HCL was added to NH4OH

Add the graph please.

In an aqueous ammonia solution there is an equilibrium of
$\ce{NH3 + H2O  << NH4+  + OH-}$
This solution has much lower conductivity than HCl titrant being added.
HCl reacts with free OH-
$\ce{NH3 <-> NH4+ + OH-  ->}$
 $\ce{OH- + H+ Cl- ->  H2O + Cl-}$
Net reaction:
$\ce{NH3 + H+ Cl- -> NH4+ + Cl-}$
Conductivity increases as HCl is added.
The "plateau" may be occurring in much the same manner as a weak acid titration because, essentially, ammonia buffers the reaction of HCl and OH -.  The final neutralization of remaining ammonium hydroxide creates less of a net charge change
$\ce{NH4+ OH- + H+ Cl- -> NH4+ + Cl- + H2O}$
Notice there is no net charge change here.
One might conclude the "endpoint" is only there when conductivity resumes its rise as still more HCl is added.

Thankyou for the response! I did find some literature which subtly suggested that would be the case, however, there were some inconsistencies between sources so I figure I would ask here. Have a great day!


text:
Is change in enthalpy heat supplied at constant internal or external pressure?
From what I understood, we introduced enthalpy to work with reactions that were carried out at constant external pressure. But then I came across this formula (for ideal gas):
∆H=∆U + p∆V
where  p∆V=∆n(RT).
But doesn't the p in the second formula represent the external pressure? So how are we using pv=nRT which is only valid if the pressure we are dealing with is the internal pressure of the gas?

By Newton's 3rd law, the force per unit area exerted by a gas on its surroundings is equal in magnitude to the pressure the surroundings exerts on the gas. In this sense, the "internal pressure" is always equal to the external pressure.  In an irreversible process, however, the" internal pressure" is not determined by the ideal gas law (or other real gas equation of state).  Did you think it was?

Yes, I did..so how is it determined?

The ideal gas law is valid only for thermodynamic equilibrium states.  In rapid irreversible deformations, viscous stresses also contribute to the compressive stresses in a gas.  So, unless the stress on the gas is specified based on external considerations, say specified external pressure, the compressive stress in the gas at the piston face must be solved for using the complicated partial differential equations of fluid dynamics combined with the partial differential energy equation within the gas.

Even if the change occurs irreversibly for the ideal gas mixture, the work is still $\int{P_{ext}dV}$.  The equation you wrote assumes that the initial and final states of the gas are thermodynamic equilibrium states at pressure p, and that the external pressure is also $P_{ext}=p$ over the entire change.  In that case, the work done by the gas on its surroundings, even if the change occurs irreversibly, is $W=P_{ext}\Delta V=p\Delta V$.

When a work $p$∆$V$ occurs in thermodynamic calculations, whatever the ± sign of the work, the pressure $p$ is the pressure of the force doing the work. When a work $w$ is done on a sample of gas whose volume decreases, the pressure $p$ to introduce in the work formula is the outer pressure $p_{ext}$ . On the contrary, when the work $w$ is done for example by the heated inner gas so that its volume increases by ∆$V$, in this case the force doing the work comes from the inner gas, and the working pressure is the inner pressure $p_{int}$.
Summary : Whatever the ± sign of the work,

when the work $w$ is positive (when the volume is decreasing), the pressure $p$ included in $p$∆$V$ is the inner pressure  $p_{int}$
when the work $w$ is negative (when the volume is increasing), the pressure $p$ included in $p$∆$V$ is the outer pressure $p_{ext}$

But in all the books that I have come  across, it was always specifically mentioned that in ∆u=q+w, work done is always take as -∫(p external )dv

Yes. Work done on the system is always obtained using the external pressure


text:
Can acyl chloride give Wurtz reaction?
Since the mechanism involves formation of free radical so is it possible to yield a ketone from wurtz reaction of acyl chloride
Such as CH3COCl reacts with Na in dry ether so can it yield CH3COCH3

No, and if someone tried it could end up quite badly. And ketones themselves react with sodium... Also don't write posts like were scribbling.

@Mithoron if we speak ok theoritical grounds only , and take 1 mol of each quantity then ?

The answer is: Boom! Or theoretically ist could be: Bang!

Acid chlorides are completely different entity compared to alkyl chlorides. You cannot expect both give similar (or closely related) products when reacted with metallic sodium. But, interestingly there are a reaction between acid chloride and sodium or potassium metal, yet it is not a boom or bang (sorry Martin!) according to reference 1. Ref. 1 states that:

Sodium reacts with benzoyl chloride in anhydrous ether solution to give ethyl benzoate as the main product. This reaction, with other metals,
was reported by Varvoglis (Ref.2). However, sodium or potassium reacts with benzoyl chloride in dry xylene to give benzoic anhydride and sodium chloride.

Under the same conditions in dry xylene, the reaction of p-toluenesulfonyl chloride with potassium or sodium behave differently to give disulfone instead of anhydride (Ref.1).
Lucky for Martin (:-)), these mixtures are not safe enough if you don't know what you are doing. For instance, if you mix propionyl chloride and diisopropyl ether, there would be a bang even without sodium or potassium metal (Ref.3). Just for your information:

Mixtures of propionyl chloride with diisopropyl ether are very unstable. The exothermic reaction giving isopropyl propionate and isopropyl chloride
has a $\Delta H_\pu{25 ^\circ C}$ of $\pu{-10 kcal mol-1}$ and a $\Delta G_\pu{25 ^\circ C}$ of $\pu{-11 kcal mol-1}$ , and can proceed spontaneously.

The reaction also has catalyzed by some metal salts such as $\ce{ZnCl2}$ in very dilute conditions (even about $\pu{4 ppm}$ levels). Thus, as you have seen, you may not allowed to react any acid chlorides in any ether of choice in the presence of alkali metals.
Edit:
In a recent publication suggested that the reaction of benzoyl chloride in THF/benzene in the presence of magnesium metal has undergone to form the corresponding diketone, $\ce{PhC(O)C(O)Ph}$ (Ref.4).

References:

Irwin A. Pearl, Theodore W. Evans, and William M. Dehn, "Reactions of Sodium and Potassium on Acid Chlorides," J. Am. Chem. Soc. 1938, 60(10), 2478–2480 (ODI: https://doi.org/10.1021/ja01277a054).
According to the authors of Ref.1, Varvogli has treated acid chlorides in absolute ether with a large number of metals. Of all the metals tried, only iron and zinc were effective, and these yielded ethyl esters of the respective adds. On the basis of this work, potassium and sodium probably would have been predicted to be without action: G. A. Varvoglis, "Uber die Einwirkung von Metallen auf Samechlorid," Ber. 1937, 70, 2391-2396.
W. M. B. Könst, "The hazard of a propionyl chloride-diisopropyl ether mixture," Journal of Hazardous Materials 1981, 4(3), 291-298 (ODI: https://doi.org/10.1016/0304-3894(81)80007-6).
S. V. Maslennikov, V. P. Sergeeva, I. V. Spirina, and S. V. Kazakina, "Oxidation of Magnesium with Benzoyl Chloride in a Mixture of Tetrahydrofuran and Benzene," Russian Journal of General Chemistry 2002, 72(5),  678-680 (ODI: https://doi.org/10.1023/A:1019583513232).

It's mostly the difference in reactivity between aromatic and aliphatic acyl chlorides. Still, I think reaction with pure chloride could be overly even for aromatic one and you overestimate this reaction with ether. Pretty interesting nonetheless.


text:
Calculating error in acid-base titration
You perform a titration of an acid (let's say $\pu{20 ml}$ $\ce{CH_3COOH}$) with a base (let's say $\pu{0.5 M}$ $\ce{NaOH}$). The burette is marked at $\pu{0.10 ml}$ intervals. From what I read in a previous thread, the largest source of error in acid-base titrations comes from endpoint determination (i.e. adding one drop too much or little). However, I want to investigate if this is the case and compare this error to the indicator error. Below are my reasoning that I need help to correct and fill in.
One drop has approximately a volume of $\pu{0.05 ml}$. In the example titration given above, adding one extra drop would lead to overestimating the amount of substance of $\ce{CH_3COOH}$ with $\pu{0.05ml} \cdot \pu{0.5 M} = \pu{0.05E-3} \cdot \pu{0.5 mol} = \pu{2.5E-5 mol}$. This would increase the concentration with $\frac{\pu{2.5E-5 mol}}{\pu{0.020 l}} = \pu{0.00125 M}$. This error appears very small.
For the phenolphthalein indicator, we only know the effective pH range ($8.3$ - $10.0$ according to this source) with two significant figures. Furthermore, the pH value of the solution of $\ce{CH_3COOH}$ and $\ce{NaOH}$ (which is obtained after the titration) depends on the concentration of $\ce{CH_3COOH}$. Let's say the $c_{\ce{CH_3COOH}} = \pu{0.1 M}$. When this reaction
$$\ce{CH3COOH + OH^- ⇌ CH3COO^- + H2O}\tag1\label{eq:one}$$
is in equilibrium, the concentration of $\ce{CH_3COOH}$ and $\ce{OH^-}$ will have decreased with $x$:
$$x/(0.1-x)^2 = 1.7\times10^9$$
The pH of this solution will thus be $8.9$ (two significant figures). If instead $c_{\ce{CH_3COOH}} = \pu{0.2 M}$ then the pH of the solution at the equivalence point would have been $9.0$ (two significant figures). This already presents a difficulty since we don't know $c_{\ce{CH_3COOH}}$ before the titration. Combined with the broad effective range of the phenolphthalein indicator (which starts to turn pink already at pH $8.3$), the indicator error appears larger to me.
Let's investigate this claim. Assume that $c_{\ce{CH_3COOH}} = \pu{0.1 M}$. At the equivalence point, the pH will be $8.9$. Assume that you saw the faintest colour of pink when the pH was $8.3$ and you stopped titrating. In this solution, we have $c_{\ce{OH^-}} = 10^{-(14-8.3)}$. The equilibrium expression for reaction \eqref{eq:one} is:
$$\frac{x}{((0.1-x) \cdot c_{\ce{OH^-}})} = 1.7 \times 10^9,$$
where $x$ is the decrease in the concentration of $\ce{CH_3COOH}$ and $\ce{OH^-}$ compared to the initial concentrations. The initial concentration of $\ce{OH^-}$ is $10^{-(14-8.3)} + 0.09997\ldots = \pu{0.09997\dots M}$. Thus one would determine $c_{\ce{CH_3COOH}} = \pu{0.09997\dots M}$ instead of $c_{\ce{CH_3COOH}} = \pu{0.1 M}$, which is an error of $\pu{0.000027 M}$. Apparently, based on this example, the indicator error is less than the error of adding one drop too much. If this conclusion generalizable?
So if the endpoint determination error is the biggest source of error, the error in the above titration should be $\pu{0.00125 M}$. It seems like the error in titration should be larger.

Well, assume you are the error. Are you sure you can read the burette the same way every time? How accurate is the burette anyway (without reading errors)? What about the volume of the solution you are going to determine?

A "standard drop" is usually considered as 0.03 mL, with a pointy plastic(=hydrophobic) burette ending even 0.02 mL. // It seems you have not considered the 2nd indicator error, i.e. titration of the indicator itself ( which is assumed very small too, but worthy to estimate it.).

@Martin yes, I was also thinking that after these calculations. So then errors due to endpoint determination and indicator error are smaller compared to the error sources you listed?

@Poutnik ok that will then lower the error of adding one drop too much if the volume is lower than 0.05 ml. What is the 2nd indicator error? I tried googling it but found nothing useful. Could you show me how to calculate that?

The indicator is an additional acid or base, added to the titrated system, that changes its color by being neutralized to the respective conjugate base or acid. Typical amount is 1-2 drops of 0.1 % indicator solution(water, ethanol). The exact degree of neutralization is difficult to quantify due subjective nature of color shift detection. But the rough estimate is possible. // Quick calculation for 1 drop of 0.1 % Phenolphthalein gives it is equivalent to roughly 0.001 mL of 0.1 M NaOH, i.e. 1/30 of the standard 0.03 mL drop. for the first color appearance, perhaps 1/10 of that ).

Thanks @Poutnik. So are the indicator error, 2nd indicator error and endpoint determination error smaller than those Martin listed?

See also http://www.titrations.info/pipette-burette about accuracy of calibrated glassware.

Your analysis is overestimated, I think. I think the human error is the biggest. My suggestion is, if you are serious about errors, do the titration 10+ times and do the statistical analysis.


text:
What is the goopy substance that forms before CaCO3 precipitates?
I do a lab where we mix a solution of sodium carbonate with a solution of calcium nitrate. As expected, calcium carbonate (white solid) precipitates out. However, if the concentration is high enough, when the solutions are first mixed a translucent, goopy substance forms. Continuing to stir that substance for another minute or two always produces the white solid.
The goopy substance does not form if the solutions are dilute enough (~2 g or less of CaCO3 forming in ~100 mL of water) but always forms if the solutions are more concentrated (~4 g or more of CaCO3 forming in ~100 mL of water).
I'm trying to figure out what that goopy substance is.
Also, it only forms once the two solutions are mixed. (So it's not as if it's just the calcium nitrate being hygroscopic). My best guess is it's some form of hydrated calcium carbonate, where the ions haven't settled into their most stable crystal lattice set and still have water molecules trapped between them, but I'm not sure.

Have a look at an open access article, "Pre-nucleation clusters as solute precursors
in crystallisation" it discusses calcium carbonate system in detail. See the link: https://pubs.rsc.org/en/content/articlehtml/2014/cs/c3cs60451a. Your hypothesis seems plausible but proving it can be a nightmare because it will require too many advanced techniques.

A better scientific replacement for the word "goopy" is gelatinous precipitate.


text:
The Shielding of valence p Electrons of Bromine Atoms
So I was studying Slater's Rule and this question came up on libretexts
The electronic configuration of Br would be $1s^22s^22p^63s^23p^64s^23d^{10}4p^5$
So according to my calculations, the Screening constant would be $2+8+(8 \times 0.85)+10+(\underline6 \times 0.35) = 28.9$ but the answer given is $2+8+(8 \times 0.85)+10+( \underline4 \times 0.35) = 28.2$
Is the answer given in libretexts wrong?

I dont think my answer is wrong
An electron in the valence p subshell has 4 p-neighbours and 2 s-neighbours so it should be (4+2)(0.35) = 60.35 as I understand the slater's rules

You seem right. // alternative Libretexts and Wikipedia

Thanks for the confirmation :)

In my understanding, they confused the groups of orbitals of the same kind like np versus the Slater rule specific groups, merging ns and np orbitals.


text:
Open MSDS (GHS/CLP) database API by CAS#
I am looking for an open API to a database that provides GHS/CLP classifications.
I am aware this is a duplicate, however the post in question never provided an answer since, as far as I can tell IUCLID 6 does not have a publically accessible API or I never found it (despite it being called a public API ? Im a bit confused on this but chances are I simply missed the API)

https://github.com/TomLent1/GHScrunch Could be helpful

Khoi Van wrote find_sds, a Python-based implementation to collect SDS based on the CAS registry number submitted to the command line.  The lookup then considers ChemBlink, VWR, Fisher Scientific, TCI Chemicals, ChemicalSafety, and Fluorochem as sources for the pdf.  It is a project with a permissive GPL3 license.

Unfortunately this seems to download PDFs directly which is not what im looking for right now since I would love to avoid parsing PDFs but if all fails this might be a nice alternative, thank you!

@Julien "as sources for the pdf", yeah.  I see the author offers an management/inventory program; following the documentation, there is a cross-link for the demo of the inventory. Search (left magnifying glass), then addition of a new sample (+ next to erlenmeyer) opens a pan, entering 109-99-9 and read from suppliers fetches

the strings of GHS codes and hazard statements and symbols.  With code analysis / contact with the authors, one possibly could extract the algorithm used to provide a "standalone" version (enter the CAS, report the codes back to the CLI). The example with CAS 109-99-9 is about tetrahydrofurane.


text:
in a mass spectrometer why do the ions inside the acceleration chamber not interact with each other and disturb the reading
This would be in both time of flight and deflection mass spectrometers, and this would be when you are trying to find the relative abundance of isotopes of an element.

There is high enough vacuum for ions to be far enough apart for mutual force negligible to external EM force.

There is a force from the other ions $\sum_{i \ne j} e_ie_j/r_{i,j}^2 $ and one from the accelerating electric field also $e_jE_0$. Combined these limit the resolution. The force between ions can be limited by the ionising conditions used.

Space charge effects are well known, as are conditions and techniques to minimize the effect on the data. Lots of papers out their to dig into details.


text:
How to determine the unitary transformation from atomic orbitals to symmetry-adapted atomic orbitals of Ethylene
I am trying to find the unitary transformation matrix U from the atomic orbitals $|\mathbf{\chi}\rangle$ of ethylene to the symmetry-adapted atomic orbitals $|\mathbf{\chi}'\rangle$ of Ethylene, i.e $|\mathbf{\chi}'\rangle = |\mathbf{\chi}\rangle \mathbf{U}$.
Now, I've concluded that $\Gamma = A_g + B_{1g} + B_{2u} + B_{3u}$ for the hydrogen atomic orbitals $\{\chi^H_i\}_{i=1}^4$ and my line of thinking is now that given the ordering of the AOs :$|\mathbf{\chi}\rangle = [\chi_{1s}^{H1}$, $\chi_{1s}^{H2}$, $\chi_{1s}^{H3}$, $\chi_{1s}^{H4}$
$\chi_{1s}^{C1}$, $\chi_{2s}^{C1}$, $\chi_{1s}^{C2}$, $\chi_{2s}^{C2}$
$\chi_{2py}^{C1}$, $\chi_{2py}^{C2}$, $\chi_{2pz}^{C1}$, $\chi_{2pz}^{C2}$, $\chi_{2px}^{C1}$, $\chi_{2px}^{C2}$] the 14x14 unitary matrix $\mathbf{U}$ should be of the form:
\begin{equation}
\mathbf{U} = \begin{bmatrix}
A & 0\\
0 & I
\end{bmatrix}, \text{ } A = \frac{1}{2}\begin{bmatrix}
\vdots & \vdots & \vdots & \vdots\\
A_g & B_{1g} & B_{2u} & B_{3u}\\
\vdots & \vdots & \vdots & \vdots\\
\vdots & \vdots & \vdots & \vdots
\end{bmatrix} = \frac{1}{2}\begin{bmatrix}
1 & 1 & 1 & 1\\
1 & -1 & 1 & -1\\
1 & 1 & -1 &-1\\
1 & -1 & -1 & 1
\end{bmatrix} ,
\end{equation}
and the rest of  $\mathbf{U} \text{ is diagonal of 1.}$
In this way, to my understanding, the hydrogen atomic orbitals will be symmetry-adapted after transformation by $\mathbf{U}$ but do I also need to consider the carbon orbitals? I.e is the rest of $\mathbf{U}$ not as simple as I believe or is $\mathbf{U}$ found by some other process?


text:
How to use Graham's Law to solve for time
A question from one of my GenChem practice exams:

"A 3.0 L sample of helium gas was placed in a container filled with a porous membrane.  Half of the helium gas effused through the membrane in 24 hours.  A 3.0 L sample of oxygen gas was placed in an identical container.  How many hours will it take for half of the oxygen gas to effuse through the membrane?"

(correct answer: 68 hours)
Seeing effusion here, one of the first things that came to mind was the formula from Graham's Law:
$$\ce{\frac{rate of effusion of A}{rate of effusion of B} = \sqrt{\frac{m_\text{A}}{m_\text{B}}}}$$
But that, nor the original formula involving proportionality, has anything to do with time. I thought that maybe velocity could be manipulated to be related to time, but we don't know anything about the distance that has been covered by each molecule. How would you approach a problem like this?

Any rate implicitly involves time. In your case, volumetric rates of effusion $(L/h)$ are relevant.  Distance over time is not relevant here, volume over time is.

Let:
A represent He
B represent $\ce{O2}$
$$\frac{\dot{V}_A}{\dot{V}_B}=\frac{\frac{V_A}{t_A}}{\frac{V_B}{t_B}}=\sqrt{\frac{M_B}{M_A}}$$
Solving for $t_B$:
$$t_B=t_A\left(\frac{V_B}{V_A}\right)\sqrt{\frac{M_B}{M_A}}$$
Substituting values:
$$t_B=24\left(\frac{1.5}{1.5}\right)\sqrt{\frac{32}{4}}$$
$$t_B=\pu{68h}$$

Why would you use molar masses in this case?

@Mailbox Graham's law says that the rate of effusion of a gas is inversely proportional to the square root of its density.  Equal volumes of gas at the same T and P contain equal numbers of gas molecules, so the rate of effusion is also inversely proportional to the square root of its molecular weight under said conditions.


text:
Solving for Partial Pressure given first-order decomposition
From my GenChem practice exam:

Consider the following reaction:
$\ce{C4H8 (g) → 2C2H4 (g)}$
The first-order decomposition of cyclobutane to two molecules of
ethene has a rate constant of $\ce{9.20\times10^{-3} s^{-1}}$.  When an initial $\ce{2.26}$ moles of cyclobutane were allowed to decompose in a $\ce{3.00 L}$ container
for $2$ mins, the total pressure in the container reached $\ce{6.50 atm}$. What
is the partial pressure of the cyclobutane at this time?

(correct answer: 1.29 atm)
The first thing I noticed was that when the reaction is complete, there will be two moles of products for every one mole of products that existed prior to the reaction.
The first thing I tried doing was finding the concentration of cyclobutane at $t = 120\ce{\,s (2 min)}$. This was easy enough if we use the first-order integrated rate law:
$$
\ce{[C4H8]_t = [C4H8]_0}\cdot e^{-kt} \\
\implies \ce{[C4H8]}_{t=120\text{ s}} = \left( \ce{\frac{2.26 mol}{3.00 L}} \right) \cdot e^{-(\ce{9.20\times 10^{-3} s^{-1}})(120\text{ s})} \\
= \ce{0.250 M}
$$
But where do I go from here?

The first thing OP has tried was the correct step, by finding the concentration of cyclobutane using the first-order rate law:
$$
\mathrm [A]_t = [A]_0 e^{-kt}$$
where $\mathrm [A]_t = [\ce{C4H8}]_t$ when $t = \pu{120 s}$, and $\mathrm [A]_0 = [\ce{C4H8}]_t = \pu{\frac{2.26}{3} mol L-1} = \pu{0.753 mol L-1}$ when $t = 0$.
Thus, at $t = \pu{120 s}$:
$$
\ce{[C4H8]_t = [C4H8]_0}\cdot e^{-kt} \ \implies \ \ce{[C4H8]}_{t = \pu{120 s}} = \pu{0.753 mol L-1} \cdot e^{-(\pu{9.20 \times 10^{-3} s-1})(\pu {120 s})} \\
= \pu{0.250 mol L-1}
$$
Therefore, the amount of $\ce{C4H8}$ remaining in the container is $\pu{0.250 mol L-1} \times \pu{3 L} = \pu{0.750 mol}$.
Meantime, above finding reveals that (according to the decomposition equation) $\pu{(2.26-0.750) mol}$ of $\ce{C4H8}$ has been decomposed to give $2 \times \pu{(2.26-0.750) mol = \pu{3.02 mol}}$ of $\ce{C2H4}$. Thus, the mole fraction of $\ce{C4H8}$ $(\chi_\ce{C4H8})$ in the container is:
$$\chi_\ce{C4H8} = \frac{\pu{0.750 mol}}{\pu{(3.02 + 0.750) mol}} = 0.199$$
Thus, partial pressure of $\ce{C4H8}$ in the container $(p_\ce{C4H8})$ is:
$$p_\ce{C4H8} = \chi_\ce{C4H8} \cdot P_\text{Tot} = 0.199 \cdot \pu{6.50 atm} = \pu{1.29 atm}$$
Note: I like the Mailbox's answer using the ratios. The reason to provide this answer is to show how to use the theory to solve the problem.

Thanks for posting this! I tried using mole fractions, but I ended up getting it wrong and I didn't know why. Seeing your answer helped me see why!

Found it.
So, after 2 minutes, the concentration of cyclobutane is around $0.250$ M, which means that around $0.750$ moles of cyclobutane is left in the reaction. If $0.750$ moles of cyclobutane are left, this means that $1.50$ moles of cyclobutane have reacted, so $2\times1.50 = 3$ moles of $\ce{C2H4}$ are left. So, at this time, a total of about $3.75$ moles of molecules, including both products and reactants, exist in the container.
Given this information, we can solve for the initial pressure of the gaseous mixture:
$$\frac{P_{t = 0}}{2.26\text{ mol}} = \frac{6.50\text{ atm}}{3.75 \text{ mol}} \\
\implies P_{t = 0} = 3.92\text{ atm}
$$
Finally, using the initial pressure, we can solve for cyclobutane's partial pressure at $t = 120$ seconds:
$$\frac{3.92\text{ atm}}{2.26\text{ mol}} = \frac{P_{t = 120 \text{ s}}}{0.75 \text{ mol}} \\
\implies P_{t = 120\text{ s}} = 1.30\text{ atm}
$$


text:
If everything tends to entropy, why does water spontaneously evaporate?
Knowing that the second law of thermodynamics states that everything tends to entropy, why does water and other volatile substances spontaneously evaporate, getting energy to do so from the surface they were touching? Wouldn't that unobey said law, as the substance would get more energetic and the surface it was previously touching less energetic, decreasing the entropy?
P.S.: Im sorry if this question is too basic or of I dont understand something someone says, as Im still in grade ten and not so knowledgeable about chemistry and physics.

"Everyting tends to entropy" has no scientific  meaning. The sentence "tend to entropy" has no meaning. Entropy is a state function. Nothing can "tend to a state function".

Evaporation or water (or anything else, for that matter) certainly increases entropy.

There's actually two factors at play here, which are being confused. The relevant formula is,
$$ \Delta G = \Delta H - T\Delta S$$
The total energy of a system is a combination of its entropy (S) and its enthalpy (H).  Entropy can be seen as the net disorder, the total amount of possibilities that every component in the system has (position, velocity, etc.).  Enthalpy is effectively the energy inherent in all of the chemical bonds that have formed.
As a liquid H2O has a high degree of enthalpy, from all of hydrogen bonds between molecules.  This is why evaporation makes the "surface it was previously touching less energetic".  Breaking these bonds took energy, cooling the system.
As a gas there are fewer bonds between molecules, so the enthalpy is worse.  However, the gas molecules can move in all directions. The more kinetic energy they have the faster they can move and the more possibilities they have, favouring the entropy (the $T\Delta S$ in the above equation). Whether H2O is a liquid or gas therefore depends on the temperature.  The boiling point (100 Celsius) is the point where H2O as a gas is overall lower in free energy/$\Delta G$ than H2O as a liquid.
As an aside, a system in general will always seek to maximize entropy, because that's the statistically most likely outcome. That's what's meant by the second law of thermodynamics.

The second law of thermodynamics actually talks about the total (net) entropy of a system. The water body and the surface that it is in contact with constitutes a system.
When water evaporates taking the energy from the contacted body, it's entropy surely increase with a accompanying decrease in the entropy of that surface. But the entropy increase in water is significantly greater than the decrease in the entropy of the surface. This is because water is a liquid and the entropy change it can have is greater due to the characteristic properties of a typical liquid as compared to the change in entropy of the solid surface (solid being already low in entropy).
The result is that the total entropy change of the system of both is positive that is it's entropy has increased.


text:
Do leaked potassium hydroxide from AA alkaline batteries, eventually become potassium bicarbonate?
I understand that liquid leaked from AA alkaline batteries is an aqueous solution of potassium hydroxide (which is highly alkaline). It becomes potassium carbonate (which is less alkaline than potassium hydroxide but still relatively alkaline) once exposed to air (or, more precisely, to carbon dioxide).
I also understand that potassium carbonate becomes potassium bicarbonate (which is mildly alkaline) once exposed to air (or, more precisely, to carbon dioxide).
I have two questions:

Then, does it mean that an aqueous solution of potassium hydroxide from AA alkaline batteries will first become potassium carbonate and then potassium bicarbonate, if they are left untouched and exposed to air (as in normal life)?

If potassium carbonate, which was created by exposing potassium hydroxide to carbon dioxide, were exposed to water, would it become an aqueous solution of potassium hydroxide (which would perhaps be more dilute than the original)? Similarly, if potassium bicarbonate were exposed to water, would it become an aqueous solution of potassium hydroxide (which would perhaps be even more dilute than the original)?

I am asking this question, basically to know if it is safer to wait longer (i.e., for potassium hydroxide to become potassium bicarbonate) before I clean leaked liquid from AA alkaline batteries.
I would appreciate your inputs. Apologies if the questions sound nonsensical; chemistry is not my strong area.

The scenario has 2 paths:
If the diffusion driven reaction with aerial carbon dioxide is faster than solution drying up, it follows the full reaction scheme:
$$\ce{KOH(aq) ->[+ CO2] K2CO3(aq) ->[+ CO2] KHCO3(aq)}$$
But the above path is not very probable for AA cell leakage cases, as the solution usually dries up before the formation of bicarbonate is completed. It then stops at achieved progress of formation of potassium carbonate/bicarbonate:
$$\ce{KOH(aq,s) ->[+ CO2][- H2O] K2CO3(s) (+ KHCO3(s))},$$
as dry potassium carbonate does not react with carbon dioxide any more.

In solution, $\ce{K2CO3(aq)}$ partially hydrolyzes, forming minor concentration of $\ce{KOH}$.
$$\ce{CO3^2-(aq) + H2O(l) <=> HCO3- + OH-(aq)}$$
$\ce{KHCO3(aq)}$ hydrolyzes too, but negligibly, forming minor concentration of $\ce{KHCO3}$ and negligible traces of $\ce{KOH(aq)}$.
$$\ce{HCO3-(aq) + H2O(l) <<=> H2O(l) + CO2(aq) + OH-(aq)}$$

It is always good to neutralize strong acids/bases by weak bases/acids, not strong ones. Otherwise we figuratively "release a wedge by another wedge". It is good practice to keep handy in a lab neutralization solutions of 2 % acetic acid (at home use diluted vinegar) and 2 % sodium bicarbonate (baking soda).

Thank you very much for your reply. Re: your first answer, which is generally more common: potassium hydroxide becoming potassium bicarbonate before the solution drying up, or stopping at potassium carbonate? It might depend on the amount of liquid and the external environment, but is it possible to find a general theory that explains the relative speed of drying up and the process to potassium bicarbonate? (In my layperson view, an environment that fastens drying up seems also to fasten the exposure of potassium hydroxide solution to the air, so the relationship seems not inverse)

I guess you may overthink it. Just wipe the leakage or dried up residue by clothes or paper napkin/tissue, soaked by diluted vinegar and then water. // General theory says it depends on particular scenario, geometry and conditions. E.g. higher temperature means faster evaporation but slower CO2 dissolution, supporting the "dry path".

Thank you for the explanation!

Your question $1$ is correctly treated. Exposed to air, potassium hydroxide solutions become slowly potassium carbonate solutions, then potassium bicarbonate solutions. But this operation is not reversible. In question $2$, potassium carbonate solutions or potassium bicarbonate solutions will never spontaneously be transformed into potassium hydroxide solutions, diluted or not. And yes, in next paragraph, it is better to wait a long time before throwing away the liquid coming out of an AAA alkaline battery. Of course, it is much quicker if this alkaline solution is neutralized by any acidic solution (like hydrochloric acid) before getting rid of it.

Sorry for the late reply, but thank you for the answer!


text:
3-Bromo-2-propylbutanal vs 2-(1-Bromoethyl)pentanal, which one is correct name according to IUPAC nomenclature standards and why?

In above example, What should should be the main chain. According to the IUPAC nomenclature standards whether it's name should be - (a) 3-Bromo-2-propylbutanal (b) 2-(1-Bromoethyl)pentanal

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

You have already found the principal characteristic group, i.e. the aldehyde group that is expressed as suffix ‘al’. And you have already included the principal characteristic group in the parent structure according to Rule P-44.1.1.
Next, the principal chain has the greater number of skeletal atoms according to Rule P-44.3.2; i.e. the principal chain is the longest chain. Therefore, the principal chain is pentanal and thus the complete name of the compound is 2-(1-bromoethyl)pentanal.


text:
What's the point of hydrogen ions in a PEM fuel cell?
I understand in a fuel cell, that oxidation occurs at the anode where hydrogen gas is split into hydrogen ions and electrons, the former of which travel through the electrolyte. Then, at the cathode, oxygen combines with hydrogen ions and electrons and is reduced. What I don't understand is the point of the hydrogen ions. Some answers I've gotten from Google:

To complete the circuit (how? I'm not entirely sure how circuits work as I haven't studied physics, but I don't get how they complete the circuit, especially considering that the electrons and hydrogen ions are travelling in the same direction while having opposite charges?)
To combine with oxygen for reduction (this kinda makes sense; however, this might be a dumb question, but why can't the oxygen just be reduced into something else without using H+ ions?)

I guess the question I'm asking in a more broad sense is: why are ions important in fuel cells (and more broadly, galvanic cells).
I'm only a high school student, so although I've searched quite a bit to try to find answers, I haven't been able to and hence why I'm asking here.
Thanks in advance everyone!

Keep in mind that every local element has to be in about balance of overall charge. Otherwise it would extremely quickly reach such a potential that would stop any current.  Or it would cause the opposite current, electric sparcles, lightning or coulombic explosion. All depending on how much you would be determined to ignore such balance requirement.

None of your questions are dumb.
The electrons travel from the anode to the cathode when looking at the outside system (the wire who is connecting the two). The hydrogen ions move from the anode to the cathode. So you see that the negatively charged electrons move in the outside circuit to meet again with oxygen and protons at the cathode. By saying "completing the circuit" what is important to take away is that there is no charge accumulation, otherwise said the charge shouldn't pile up somewhere. This wouldn't work as charges repel each other and it would take more and more energy to keep piling them up. In a classical electric circuit electrons go around in loops, never piling up somewhere, as such "completing the circuit". (It's more nuanced than this, but not important for your question).
Oxygen can be reduced to something else without using H+, it can be reduced to the superoxide anion. It is a possibility, also reducing to hydrogen peroxide (with some hydrogen ions) can happen. But when reducing together with H+ it forms a very stable H2O and as such it is energetically better (get more 'electricity' for your buck).
Lastly, to finish the general remark at the end, ions are important in electrochemical reactions because we don't just need to make the reaction work. We also need to combat this charge buildup. We often add a "supporting electrolyte" to improve the conduction of the solution. The ions will then behave similar to electrons in a wire and will push each other slightly to reduce the charge build-up. If you did not have a lot of ions, then you will quickly get a lot of charge in one place and it will take a while for it to get from one electrode to the other.
Is this somewhat clear?

Perfect, I think I understand; your remark about how completing the circuit is essentially just preventing charge accumulation makes a lot of sense

TL;DR
It is both of what you found. To create water, you need hydrogen and oxygen in some form. The design of a fuel cell requires electrons to be moved and in exchange, some other charge has to move, here it is provided by the protons.
The lengthy explanation:
If we just combine oxygen and hydrogen and ignite it, it would (violently) react to water. It does this because it is an exothermic reaction. The interesting thing about a fuel cell is that we can control the flow of the energy that is set free during this reaction. We do not want to have just sudden heat. So at this point, a fuel cell is really just burning hydrogen with extra steps. But no matter how these steps look like, in the end you need both reaction partners. So yes, it is about "combine with oxygen for reduction".
Now to the "complete the circuit" part, and here it is important how the extra steps of burning hydrogen look like. In order to control the reaction, we now separate the gases with a membrane that can only be passed by hydrogen cations (protons). But molecular oxygen does not form a stable product with just protons, so we give the oxygen a few more electrons such that we have $O^{2-}$ and voilá, this is a nice reaction that can take place: $2 H^+ + O^{2-} \rightarrow H_2O$. Now we have to get the extra electrons for the oxygen from somewhere, and we have to put the spare electrons from the hydrogen somewhere, and again this is a perfect match. Since we blocked the direct way with the membrane, we use a wire to transfer the electrons from one place to another. We can plug in electrical load here which is driven by the energy that would normally be released by burning hydrogen (minus some losses). So you see, you need the proton to complete the circuit. No ions, no electric current. No electric current, no reaction.
About your other questions:
Yes you can use many different ions. This is what happens in batteries. A fuel cell is simply speaking a battery with certain specifications. Here is a list with many actually used combinations: Electric battery - Wikipedia
In any galvanic cell, you are creating an imbalance of charge due to the electrons pushed from one to another cell. This imbalance causes the reaction to stop at a certain point (very fast). The charge difference is then so large that it outweighs the reaction enthalpy and nothing happens anymore. To compensate, charges have to be moved from one part of the cell to another. A classic is to move sulfate ions via a salt bridge, because they don't participate in the reaction but can compensate the charge.


text:
What two processes occur that dictate the microstructure of a polymer?
Firstly, I'm not 100% sure on what a microstructure is as our lecturer didn't explain it all too well. From what I've gathered, its the backbone structure of the polymer chain - but I may be wrong.
In relation to the question, I have a thought that chain-walking is a process that dictates the microstructure but am unsure as to what else could be.
The example I am using is {OSSO}-ZrBn2 to polymerize propylene with Zr and Hf being the metal centre options as shown below.
Based on the catalyst, what processes would dictate the microstructure of the polymer?

You are wrong. https://en.wikipedia.org/wiki/Microstructure https://en.wikipedia.org/wiki/Crystallization_of_polymers

What is Bn?    ...

@Poutnik the image of the catalyst used has been added to the question

@Mithoron Can you provide any more clarification instead of just posting links? I'm a bit lost on what you want me to look at here.

@Karl I've completed the lecture course and there was a total of one slide on microstructures where they didn't give an explanation for it :/

What kind of lecture was that exactly?

The catalyst and reaction conditions dictate the backbone structure. Tacticity, number and size of side chains, Mw, Mw distribution, ... The backbone structure, together with "processing parameters" (temperature profile, flow, ...) influences what crystalline microstructure you get. Now what exactly is your question?

Let us continue this discussion in chat.

Nothing to chat about. You, and very probably your teacher, are using the term "microstructure" for something that is not usually called microstructure. Judging from that you're supposed to interpret the results of CP-MAS experiments, this is some quite advanced lecture. So please get clear with the nomenclature, and then update your questions so it makes sense. And also tell us what info about the polymerisation of propylene with these catalysts you get from the data presented. Otherwise I'm inclined to vote "close" because this is a homework question without any visible effort by the OP.


text:
Experiment to crush soda can by air pressure
You put a can with water at its base on a hot plate and allow the water to heat. Once steam is visible from the opening on top of the can, you quickly flip the can upside down into a bowl of water with gloves or tongs. The can will be crushed by the atmospheric pressure.
I know by heating the can, we boiled the water inside it. The process of boiling turned the water into vapor. And since the water vapor molecules are much more spread out than the water molecules, they take more space and are forcing the molecules of air out from the can.
And when we put the can in the cold water, we suddenly cooled it. That cooling caused the water vapor in the can to condense, creating a partial vacuum. Because of that, the pressure outside of the can became much greater than the pressure inside, and that pressure difference crushed the can.
However, my point of confusion is...
How may someone calculate the air pressure of which the can is crushed before or after impact?

Here is a demo without boiling water (at least they are not showing or discussing it): https://youtu.be/lT3PJ0RY4oY And here is a demo with boiling water: https://youtu.be/Wz5_Cq3aoec?t=8

Anywhere between 0 and 1

How may someone calculate the air pressure of which the can is crushed before or after impact?

The outside air pressure is one atmosphere. The pressure in the can is one atmosphere as long as there is a way for air to come in and out, and as long as you heat the can gradually.
After the can is crushed, the gas pressure in the can is one atmosphere again (once water has had time to flow in and out).
When you submerge the can in the ice bath, the gas pressure inside the can depends on the remaining volume and the remaining amount of water in the gas phase.
If we assume that condensation of the water is fast and the liquid water is slow to enter the can and the can has not been crushed yet, you would be able to estimate the pressure inside as the vapor pressure of water at the current temperature of the gas. The temperature is difficult to estimate (it is $\pu{100 ^\circ C}$ when the water is boiling, and $\pu{0 ^\circ C}$ on the surface of the ice water). If we just assume room temperature, the vapor pressure is 0.0231 atm, i.e. 1/50th of the outside ambient pressure.
You get a sense of the pressure difference by looking at the volume of the can before and after crushing. Ideally, you could also measure how much water made it into the can, testing the hypothesis that entry of water is slow compared to the crushing of the can (either process will decrease the vapor volume).