text:
Partial pressure of gases in a container with a piston
I've been having problems with the following problem.

Figure 1.

A container with a piston like that shown in Figure 1 is filled with $0.10$ $mol$ of Ar and $1.00$ $mol$ of water (liquid and vapor). The temperature of the entire container is $87 °C$ and the total pressure is $1$ $bar$. It is assumed that the volume of the liquid or cup is negligible with respect to that of the gas, and that the temperature of water and Ar is always maintained at $87 °C$. In this initial situation, the partial pressure of Ar is $A$ $bar$, the partial pressure of water vapor is $B$ $bar$, the volume of the gas is $C$, and the number of moles of liquid water is $D$ $mol$.

The piston is then pulled back quickly to fix the gas volume to $15.8$ $L$. At the moment when the piston
is pulled, the partial pressure of Ar decreases to $E$ $bar$ and the partial pressure of water vapor
decreases to $F$ $bar$. In this situation, the water will boil, because the total pressure of the gas is
lower than the saturated vapor pressure of water at $87 °C$, which is $0.6226$ $bar$. The partial pressure of water vapor increases
due to the boiling until the boiling eventually stops. When the boiling stops, the number of moles of
liquid water is $G$ $mol$. After that, evaporation proceeds until equilibrium is reached. At
equilibrium, the number of moles of liquid water is $H$ $mol$.

Calculate the appropriate numerical values for blanks $A$ - $H$.

The reason I am having trouble solving the problem is the fact that there seems to be too many unknown variables: the initial volume of the container $(V_0)$, the partial pressures of Ar $(p_{Ar,1})$ and water $(p_w)$ and the number of moles of liquid water $(n_{w,l})$. I've come up with the following formulas but I am unable to get any further.
$\ce{p_{total}=p_{Ar,1} + p_{water}=1 bar}$
$\ce{n_{total}=n_{Ar,1} + n_{water,g}=0.10 mol + (1 mol-n_{water,l})}$
$\ce{p_{Ar,1}=\frac{n_{Ar}*R*T}{V_0}}$
$\ce{p_{w,1}=\frac{n_{water}*R*T}{V_0}}$
$\ce{V_0=\frac{n_{total}*R*T}{p_{total}}}$

$B=\pu{0.6226 bar}$ as the water-water vapour system would be in equilibrium.
Now,
$$
\begin{align}
A+B&= \pu{1 bar}\\
A&= \pu{0.3774 bar}
\end{align}
$$
For initial volume of the container,
$$
\begin{align}
P_{\ce{Ar}_i}V_i&=n_{\ce{Ar}_i}RT\\
V_i&=\frac{n_{\ce{Ar}_i}RT}{P_{\ce{Ar}_i}}\\
C&=\pu{\frac{0.1 \times 0.08314 \times 360}{0.3774} L}\\
C&=\pu{7.93 L}\\
\end{align}
$$
For initial moles of water vapour,
$$
\begin{align}
P_{\ce{H2O{(v)}}_i}V_i&=n_{\ce{H2O{(v)}}_i}RT\\
n_{\ce{H2O{(v)}}_i}&=\frac{P_{\ce{H2O{(v)}}_i}V_i}{RT}\\
n_{\ce{H2O{(v)}}_i}&=\pu{\frac{0.6226 \times 7.93}{360 \times 0.08314} L}\\
n_{\ce{H2O{(v)}}_i}&=\pu{0.165 mol}\\
\end{align}
$$
Now,
$$
\begin{align}
n_{\ce{H2O{(v)}}_i} + n_{\ce{H2O{(l)}}_i} &=\pu{1 mol}\\
n_{\ce{H2O{(l)}}_i} &= (1 - n_{\ce{H2O{(v)}}_i}) \pu{mol}\\
D &= \pu{ (1 - 0.165) mol}\\
D &= \pu{ 9.835 mol}\\
\end{align}
$$
Now as the volume is doubled by Boyle's law we can say that the partial pressure of argon and water vapour will decrease to half of original.
Thus,
$$
E=\frac{A}{2}=\pu{0.1887 bar}\\
F=\frac{B}{2}=\pu{0.3113 bar}\\
$$
Now as the partial pressure of water vapour is less than the vapour pressure of water at that temperature, water will evaporate to make it equal.
Now as the temperature is same thus the vapour pressure of water is also same as initial. Thus only change in system is it's volume which was doubled. So we can conclude that twice amount of water vapour as compared to initial conditions will be required to attain this pressure.
Thus,
$$
\begin{align}
n_{\ce{H2O{(v)}}_f}&=2 \times n_{\ce{H2O{(v)}}_i}\\
n_{\ce{H2O{(v)}}_f}&=\pu{0.33 mol}\\
\end{align}
$$
But as the total moles of water is conserved,
$$
\begin{align}
G=H=&(1-n_{\ce{H2O{(v)}}_f}) \pu{mol}\\
G=H=&\pu{0.67 mol}\\
\end{align}
$$

Thank you! It seems like I didn't realize that I could calculate the volume of the gases from the number of moles of Ar alone.


text:
Why would electrode be positively charged in electrolysis?
From my understanding, a battery provides a driving force to pull electrons from the anode, thus the anode becomes positively charged.
Suppose the anode is copper, the process should be as follows:
$\ce{Cu->Cu^{2+} +2e-}$
However, the copper ions should be dissolved into the electrolyte after this process, and only neutral copper atoms are present in the electrode, so why would it be positively charged and attract anions?

Copper may work as an anode if and only if the cathode is made of a metal whose redox potential is higher than $0.34 V$. like metallic silver $\ce{Ag}$ in a solution of $\ce{AgNO3)$

Neutral atoms turn into cations according to the half-reaction you wrote. What is confusing about it? Anions are not a part of the process at all.

I frequently get confused by the terms cathode and anode when they are used without specifying where they are being used. Electrochemists have to juggle words that are very similar.
In an active cell, the electrode dissolves and positive CAT-ions leave the AN-ode and leave it negative so it can push an electronic current (electrons go from - to +) thru an external system while the electric current flows the other way (from + to -). But in a passive cell (one that is operated upon by an external electromotive force, it's the reverse: the electrode made negative, called the CAT-hode, attracts the CAT-ions, while the electrode made positive, called the AN-ode, attracts AN-ions.
The solution is to visualize the process pictorially, without words, then apply the words carefully, like labels on a jar of chemicals. I'm going on at length to demonstrate as many of the confusing terms as I can remember.
The question to ask is "What is the first process - what is the initiating agent?" Is it something spontaneous that you are trying to explain, or is it something you are causing and have to describe it?
So your question is: why would it (the copper anode) be positively charged and attract anions?  You say "battery provides a driving force to pull electrons from anode". The implication is that the cell is passive; it is operated on by an external voltage. So far, everything makes sense.
In the very next sentence, "Suppose the anode is copper...", you show copper actively (arrow -->) dissolving, which makes it negatively charged. Now copper is low in the electrochemical series, but an aggressive medium can corrode it - let's not get too specific. This makes sense too.
The "paradox" is that the (active) copper anode is now negative while in the preceding situation, the (passive) copper anode is positive. But there is no paradox - you have changed the system from passive to active. The term anode is the same, copper is the same, so your terminology is correct - in each case, separately. But the last paragraph asks 1. why positive, while implying 2. active copper dissolution. There's the problem.
I would say, blame it on the terminology; it's complex, and you have to visualize first and name it afterward, rather than name it and then call up a picture of what's going on.

Try to remember that, in the $\ce{Zn/Cu}$ cell, like in all other electrochemical devices, whatever the working type (cell or electrolysis), the following sentences may be memorized :

the anode is always where oxidation happens (transforming $\ce{M}$ into $\ce{M^{z+}}$ ion), so that the cathode is always where reduction happens.

the zinc electrode is always the negative pole, and the copper plate is always the positive pole. The + sign can be printed on the copper plate, and "minus" on the zinc plate

Depending on the outer connections, each plate can be anode or cathode, and oxidation and reduction can occur on either plates. But remember : Oxidation = Anode.
Some more explanations. When working in electrolysis, $\ce{Zn}$ is still the negative pole, but Zn is reduced from $\ce{Zn^{2+}}$ ion into metallic $\ce{Zn}$. So Zn is working as a cathode. Simultaneously, the copper plate is still the positive pole. But $\ce{Cu}$ is oxidized into metallic copper and copper is the anode.

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text:
Understanding the pH scale for edible food!
I'm working on a website where I want to provide the pH level of plants! Although i don't even know what this is (I have zero knowledge in chemistry), I have to provide it! What I want to do is provide the pH spectrum on the website so that it would point to the correct pH level of a specific plant.
Let's use the following hypothetical examples from the database I was given:

Peas:  preferred pH : 6.0 - 7.5
Beans  preferred pH : 6.0 - 6.8

At this stage I want to point out that I'm only working with edible plants, vegetables, fruits etc.
Also here is a reference image that would be ideally used on the website:

From the image that was given above, I want to provide:
1(One pH value) : An arrow to point out the preferred pH of a plant.
2 (Range of pH values): an arc highlighting the range of the preferred pH of a plant.

Questions:

Why is there a range in the preferred pH of a plant?
Is it possible for an edible food to have a negative pH value? Answer: NOT POSSIBLE
Is a pH level let's say 20 dangerous to consume (what is the minimum and maximum values that can be safely consumed)?

Negative pH is hardly ever encountered outside of a chemical lab, and pH 20 is physically impossible. Now to the point. You might want to know that pH is not just some arcane scientific quantity; no, it has a very common-sense, everyday meaning.

....which is????

The measure of acidity, of course. Can you survive on a diet of lemons alone? Not for long, I guess. Well, that's the lower bound.

thanks @IvanNeretin ! It makes a lot of sense now! Now I have an answer to one of my questions

Beside that it is hard to find a sense on what you are asking and what you want to do, it is even unclear what is the pH "preferred by edible plants". Are you referring to the pH of the soil where they grow best, to the pH of the raw lymph, to the pH of the descending lymph, to the pH of the juice by pressing? All can be different. It smells about strange pH scales of food, lemon bicarbonate therapy and nonsense like that.

@Alchimista thanks for your reply, but I'm afraid I cannot answer that! I literally have no idea! I will have to contact the person that created the dataset!

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text:
Is the cannizaro reaction possible for a molecule having hydrogens attached to sp2 hybridised carbons?
As known benzaldehyde can undergo a cannizaro reaction in which a phenyl group is an electron donating group, so the reaction similar to below  (of prop-2-enal) (cannizaro) should occur for similar molecule 'either having all hydrogen connected to SP2 hybridised carbon (eg: pent2,4dienal) or the carbon which is SP3 hybridized connected to double bond should not have any hydrogen connected to it (eg: 4,4,4triimethylbutanaldehyde)' should show cannizaro.
Major product in basic medium should be decided by the below reaction (Cannizaro Reaction) and not aldol condensation. Am I correct?

I don't think that acid base reaction can occur with concentrated base in this case.
This may be helpful
Bond energy of $\ce{C-H}$ bond in methane $\pu{415 kJ/mole}$.
Bond energy of $\ce{C-H}$ bond in ethene $\pu{444 kJ/mole}$.

Your question isn't about "hydrogens next to sp2 carbons": benzaldehyde has $\ce{C_\mathrm{sp^2}-H}$ bonds as well, but you already know what's going on with benzaldehydes. Your question is really about α,β-unsaturated aldehydes, isn't it?

@orthocresol Yes. I have tried to express my thoughts in question as much as I can and I am sorry because I didn't know acrolien will undergo hazardous reaction (and wasted much time of community) and will not represent my doubt clearly.

According to The Hazardous Substances Databank entry for acrolein  here

An extremely violent polymerization reaction of acrolein results from contact with alkaline materials such as sodium hydroxide, potassium hydroxide, ammonia and amines. Mixing sodium hydroxide and acrolein in a closed container caused the temperature and pressure to increase.

The next compound in the series, crotonaldehyde, certainly also polymerises in alkaline conditions. I am inclined to think this is general. There are just too many other pathways available for these reactive molecules under the strongly basic conditions of Cannizaro, not least retro-aldol.

I think the retro aldol is the more likely outcome of creating the enolate. You then have pivalaldehyde plus acetaldehyde. Pivaldehyde gives Cannizzaro but acetaldehyde will react all over the place.


text:
Mass percentage of limonene in a sample
I realize Chemistry Stack Exchange is not for homework and this will seem incredibly simple to many of you, but I have received no help elsewhere. I would really appreciate any support:

Limonene is obtained in the oil of lemon feel and it is the only alkene present.

$\pu{0.500 g}$ of the oil reacted with exactly $\pu{30.6 cm3}$ of a solution of bromine dissolved in cyclohexane with a concentration of $\pu{0.200 mol dm-1}$.
Calculate the percentage by mass of limonene in the oil.
Give your answer to an appropriate number of significant figures.
Assume that there is nothing else in the oil that react with bromine.

What I have done:
I tried working out the moles of $\ce{Br2}$ from these data (using the concentration and volume) and got $\pu{0.00612 mol}$. Assuming $\ce{Br2}$ is in the same ratio as limonene the moles will also be $\pu{0.00612 mol}$. However, this gives yield of $\pu{0.832 g}$ for limonene, which surely can’t be possible given the starting mass. Have I miscalculated the moles of bromine because it is dissolved in cyclohexane? Is there anything I am missing?

How many moles of bromine can be additioned to one mole of limonene?

You are almost there: you have a factor of two problem, i.e., there are two double bonds in limonene.

Thank you for the help, seems so simple now!

If you calculated correctly, there should be 83.4% limonene in your sample.

@Eliza: Please do not attach word problem as an image. Please type it in.


text:
How to maximize shelf-life of freeze-dried food beyond 25 years? Can this exceed 100 years?
Many manufacturers will sell and advertise emergency preparedness food packs to still taste "great" 25-30 years into the future (or at least maintain a consistent taste over this period). For instance, they will cook eggs, lasagna or vegetables (anything really), then place it into a freeze-drier unit, which eliminates the water via sublimation. They then place the food product into a mylar or aluminium pouch (sometimes polyurethane), insert an oxygen dissector pouch in it, then instruct the customer to store it in a cool (less than 16 degrees Celsius or 60 Fahrenheit) and dry environment for long term storage.
What process happens after 25-30 years which limits its shelf life? It is the accumulation of oxygen diffusing within the bag over time? It is the oxygen pouch that reaches it absorption limit? Is moisture sneaking in? Is the organic structure of the food break down due to normal thermodynamic forces over time (increase in entropy)?
What if one wanted to increase the shelf-life to say 50 years, maybe even 100 years. What food break down processes will be encountered as the limiting factor, that would need to be countered?
What other tricks (other than mylar pouch, oxygen desiccator pouch, freeze-drier) can I use to maximize the shelf-life in excess of 100 years?

In 50 years, 3D printed foods will be common, so all that is needed is feedstock and the printer recipe. Yum! ;-)

I'm not sure that the issue is technical rather than marketing/demand.

@JonCuster Are you suggesting that no changes are required to the formula employed by the manufacturers to achieve 100 years shelf-life. Thus, is it purely a marketing decision that they write 25 years instead of 100 years on the packaging?

I think oxygen will diffuse throught the foil into the bag and the food will oxidize.
In addition, as long as we are near room temperature, there can be thermal degredation.
What could you do to improve the stability? Thicker foil to decrease oxygen diffusion.  Create a Vacuum around the package which prevents oxygen from reaching the foil.
Store it at low temperatures to prevent thermal degradation.
The global seed vault stores seed at -18°C for 50 years at least.
https://en.wikipedia.org/wiki/Svalbard_Global_Seed_Vault

@Anon21 - no, I think there has been no need to push the technology (or prove an actual lifetime in excess of 25 years) since there is not sufficient demand. Canned food, stored under reasonable temperatures, is still good well past the date printed on the can.

@JonCuster I don't suppose, by any chance, that you would have any suggestion on how to "prove" (with hopefully a high degree of confidence) that some food, under some specified storage condition, could survive 100 years, without waiting 100 years?

@Anon21 - like other similar questions (semiconductor reliability, spacecraft, other high reliability applications), one has to do accelerated aging studies as well as proving that you are accelerating the right processes. That can take a lot of work.

@JonCuster So if I were to assume that 'everything has been taken care of' using a "new process" (vacuum, triple-thick aluminium foil, etc), then the only process remaining is likely to be thermal degradation of the food sample... then all one has to do is raise the temperature and measure which specific foods "self-detoriate" at which speeds and temperatures, then commit to only using those foods which are identified, by those tests, to last more than 100 years to prepare the meals? I cannot think of anything else than thermal degradation at this point - am I missing anything?


text:
Discrepancy when calulating mol weights with ChemSketch and Python RDKit
I am an IT guy with little knowledge of Chemistry, so please bear with me...
For context: I am working with a system that has a lot of structures, their .mol files and some metadata (like mol weights) in a database. Due to an error, some of the metadata was lost and needs to be recalculated. Since there are a lot of structures, I need to automate this.
Originally, the mol weights were calculated using the Freeware version of ACD ChemSketch. Since I did not find a way to call ChemSketch and have it return the mol weight or otherwise automate ChemSketch itself, I looked at RDKit for Python as an alternative.
In my tests, I noticed that ChemSketch and RDKit produced slightly different results when calculating the mol weights. I used m = Chem.MolFromMolFile('amisalin.mol') to load the file and then MolWt(m) respectively ExactMolWt(m).
Example mol file amisalin.mol:

  -ACD/LAB-

 18 17  0  0  0  0  0  0  0  0  0 V2000
   18.3905   -6.6892    0.0000 C   0  0  0  0  0
   19.4429   -6.0748    0.0000 C   0  0  0  0  0
   18.3905   -8.0661    0.0000 C   0  0  0  0  0
   17.2059   -5.9960    0.0000 C   0  0  0  0  0
   20.5930   -6.7427    0.0000 N   0  0  0  0  0
   19.4429   -4.2757    0.0000 O   0  0  0  0  0
   17.2059   -8.7560    0.0000 C   0  0  0  0  0
   16.0211   -6.6892    0.0000 C   0  0  0  0  0
   21.7398   -6.0748    0.0000 C   0  0  0  0  0
   16.0211   -8.0661    0.0000 C   0  0  0  0  0
   22.5401   -6.5380    0.0000 C   0  0  0  0  0
   14.8742   -8.7309    0.0000 N   0  0  0  0  0
   23.6965   -5.8762    0.0000 N   0  0  0  0  0
   24.8434   -6.5380    0.0000 C   0  0  0  0  0
   23.6965   -4.0709    0.0000 C   0  0  0  0  0
   25.9934   -5.8762    0.0000 C   0  0  0  0  0
   24.8434   -3.4124    0.0000 C   0  0  0  0  0
   21.0845  -12.6663    0.0000 Cl  0  0  0  0  0
  1  2  1  0  2  0  0
  1  3  2  0  1  0  0
  1  4  1  0  1  0  0
  2  5  1  0  2  0  0
  2  6  2  0  2  0  0
  3  7  1  0  1  0  0
  4  8  2  0  1  0  0
  5  9  1  0  2  0  0
  7 10  2  0  1  0  0
  9 11  1  0  2  0  0
 10 12  1  0  2  0  0
 11 13  1  0  2  0  0
 13 14  1  0  2  0  0
 13 15  1  0  2  0  0
 14 16  1  0  2  0  0
 15 17  1  0  2  0  0
  8 10  1  0  1  0  0
M  END

Results:

Average molar weight
Exact molar weight

Chemsketch
271.78628
271.14514

RDKit
271.7919999999999
271.145140004

As you can see, the results are slightly different. Since the input data is identical, I assume this is due to some internal differences when calculating the weights, possibly due to rounding (?).
My question now is: Is the difference negligible? Does anyone know where it comes from? Is one of the results "correct" / more accurate than the other?

What is the intended use of the .mol files?  Often, it they are used as a descriptor of the structure, how the atoms are bound together and having an idea about the molecular mass is a nicety, but not a must have.  For example, if the .mol files «anyway» are rewritten later into an other format to perform subsequent (e.g., quantum chemical) computations.

And re automation of GUI programs with no programmatic API; I occasionally use PyAutoGui in Python. It works better if the program already uses short-cuts (e.g., Ctrl + C) than if you have to define the clicks of the «simulated mouse» by the pixel coordinates of your screen (this is then is less portable ...).  A demo (starts around 11:00 mn:ss) by the developer is here; basic-enough to be part of his book «Automate the boring stuff» in chapter 20.

Thank you for your very helpful comments and answer. I had also considered using pyautogui but went another way since I'd ideally like my solution to run "headless". However, thanks to your help, I'm now equipped to discuss this with the users. If the discrepancy shouldn't be acceptable to them, as an alternative, I also found that the most recent version of openbabel yields almost exactly the same results for average mol weight as ChemSketch.

For the use, the installation of OpenBabel should follow reading the current documentation (readthedocs.io with its many examples; openbabel.org still often points to the elder version 2.3. Overall, the installation of OpenBabel requires much less disk space than the sum of (miniconda Python + RDKit) and already may suffice here. About its interfaces for being called from a program, see e.g., (here).

Yes, I noticed and was surprised by that too. I built the most recent OpenBabel version from source and successfully integrated it with Python using pip/conda.

In first place, you must clarify with the users of the database the purpose of the entries about molecular weight because average molecular weight and exact molecular weight differ in their meaning and subsequent use.  It may be that the documentation of the database provides this information.  Later, once this target is set, use either one of the program to compute the entries for all compounds of this database to ensure consistency in the data.  RDKit's current release 2021_03_1 is just some weeks old and probably the easier route to cover eventually all entries in your database.

Molecules consist of atoms (like oxygen, nitrogen, etc).  For some of them, say nitrogen, despite they are all nitrogen, there might be more than one form and these differ slightly in their mass per individual atom (isotopes).

The mass of these isotopes is published.  Equally, biannually, there are updated public records about the probability that an atom of e.g., nitrogen either is $\ce{^{14}N}$ (about 99.6%) or $\ce{^{15}N}$ (about 0.4%).

These probabilities are equally valid for molecules, if you have 1000 molecules each containing one atom of nitrogen, than 4 of them should contain the heavier isotope $\ce{^{15}N}$.  Similar there are such probabilities about other atoms, like carbon, oxygen, etc., which are all independent of each other.  Among carbon atoms, for example, 98.9% are $\ce{^{12}C}$, and 1.1% are $\ce{^{13}C}$, respectively.  (Except in radiochemistry / nuclear chemistry.)

Exact molecular weight is used, e.g. in databases eventually used for e.g., mass spectroscopy.  An analysis of your sample will recognize these different probabilities and may yield different, separate signals about the molecules with $\ce{^{14}N}$ from the ones about $\ce{^{15}N}$.  Thus exact molecular weight is about molecules of one isotopic composition.
Say our sample were $\ce{(CN)2}$ of two carbon atoms and two nitrogen atoms per molecule, than there would be four mass signals for all combinations of the two isotopes of carbon and nitrogen possible.  Their intensity  correlate with these probabilities.  Exact molecular weight is a monoisotopic mass, typically to consider only the mass of the molecules consisting with the most frequently observed isotopes per element, i.e., only about $\ce{^{12}C}$ and $\ce{^{14}N}$.  Depending on the intended analysis, it may be worth to retain six (or even more) decimals.

Average molecular weight now accounts not only for the molecules consisting of the most frequently seen isotopes per element, but for any combination of the elements with each other.  These weights are are averaged by the probability that these combinations occur, e.g that there is a molecule $\ce{(^{13}C^{14}N)2}$ consisting only of the heavier carbon isotope and the lighter nitrogen isotope among all $\ce{(CN)2}$ molecules of your sample.  Average molecular weights are typically used in to set up a reaction in the lab, to know how many grams of compound A shall react with how many grams of compound B.  Depending on the scale of the reaction and the needs of your clients, to retain two decimals of this value may suffice.

The pace these IUPAC publishes the data about these isotopes is independent from the one the programs are updated by the developers, and later by the users. This contributes why an old program still may include outdated values.

The sort of answer that you wish you could upvote twice. Solid facts, clearly explained, appropriate level of detail!


text:
chemistry 2 final about organic chemistry asap pls

I need this as soon as possible please. Any help would be appreciated

We are not here to solve your homework without you starting at least some effort. We help those who try to solve their problem, do some attempt, and fail half way. We will help them go ahead. But we will not do all the job and you nothing !


text:
What is the difference between an Aryl Halide and a Vinyl Halide?
Are aryl halides a type of vinyl halide? From my knowledge, when a Halogen atom is directly bonded to a doubly bonded carbon atom it is a Vinyl halide. When we look at halogen atom directly bonded to a benzene ring we see a similar thing. So are these terms never correlated? Also is Phenylic Halide a term which exists?

You are pretty much asking whether a benzene ring is a double bond. Well, in some ways they are similar, but in many ways they aren't; I'm sure you can come up with examples. The same applies to aryl / vinyl halides.

Ooh that is a good way of putting it and I do realise the stupidity in my question now, thanks.

No, aryl halides are not a type of vinyl halides.
A compound is called an aryl halide when it contains a halogen group directly attached to an aromatic system. For example,

Whereas a compound is called a vinyl halide when it contains a halogen group directly attached to a doubly bonded carbon atom. For example,

The double bonds in an aromatic system are not the same as the aliphatic double bonds. They differ quite a bit in terms of electron density and thus there derivatives behave very differently is many circumstances.
The term phenylic halide doesn't exist but a very similar term phenyl halide exists which is a subset of aryl halides family.
An aryl halide can be called a phenyl halide when the aromatic system of the compound is specifically the phenyl system($\ce{-C6H5}$). For example,


text:
Dissolution of CO2 in water
The following question has troubled me for a while.

For gases that are slightly soluble in water, there is a proportional relationship between the partial
pressure, $P$, and the mole fraction, $x$, of the gas molecules dissolved in water (Henry's law):
$\ce{P = k_H*x}$
A container ($V$ = $570$ $mL$) is filled with water ($V_l$ = $500$ $mL$) and pressurized with $CO_2$ gas ($P_0$ = $50$ $atm$), before it is allowed to stand at $10^\circ$C until the vapor–liquid equilibrium is established.
Calculate the pressure of $CO_2$ $[atm]$ in the container and the amount of $CO_2$ $[mol]$ dissolved in the water. The Henry coefficient of $CO_2$ for water at $10^\circ$C is $k_H$ = $0.104 × 10^4$ $atm$, and we will consider that the reaction of $CO_2$ in water can be ignored.

I did the following calculations but I am unsure with the outcome.
$\ce{n_{total}\approx n(H_2O) = \frac{\rho V}{M(H_2O)} = 27.753 mol}$
$\ce{x = \frac{P_0}{k_H} = 0.048077}$
$\ce{n(CO_2)_{dissolved} = x*n_{total} \approx 1.3343 mol}$
$\ce{P_1 = \frac{nRT}{V} = \frac {1.3343*0.0831451*1.01325*283}{0.570} = 55.8 atm}$
Shouldn't the final pressure be smaller than the initial pressure ($50$ $atm$)?

yes, the problem is wanting you to split the $\ce{CO2}$ initially introduced into the container between the gas and liquid phase.

Assumptions:

Reaction of carbon dioxide with water is neglected.
Vapour pressure of water is negligible.
Volume of solution does not change on dissolution of carbon dioxide.
Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2{(aq)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$

Initially no $\ce{CO2}$ is dissolved in water and so initial moles of $\ce{H2O{(l)}}$ and $\ce{CO2{(g)}}$ can be calculated as follows:
$$
n_{{\ce{H2O{(l)}}}_i} = \frac{\rho V}{M} =\pu{\frac{500 \times 0.9975}{18} mol}= \pu{27.71 mol}\\
n_{{\ce{CO2{(g)}}}_i} = \frac{P_iV}{RT} = \pu{ \frac{50 \times 0.07}{0.0821 \times 283} mol}= \pu{0.15 mol}\\
$$
Now after the equilibrium is settled some amount of carbon dioxide will dissolve in water according to Henry's law.
Final pressure in the container can be written as:
$$
P_f= \frac{n_{{\ce{CO2{(g)}}}_f}RT}{V}\\
$$
And by Henry's law we can say,
$$
P_f= K_H \times X_{\ce{CO2{(aq)}}}
$$
Thus from both these equations we can conclude:
$$
K_H \times  \frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}} = \frac{n_{{\ce{CO2{(g)}}}_f}RT}{V}\\
$$
We also known that,
$$
n_{{\ce{CO2{(g)}}}_f} = n_{{\ce{CO2{(g)}}}_i} - n_{{\ce{CO2{(aq)}}}_f} \\
$$
So,
$$
K_H \times  \frac{n_{\ce{CO2(aq)}_f}}{n_{\ce{H2O(l)}}} = \frac{(n_{{\ce{CO2{(g)}}}_i} - n_{{\ce{CO2{(aq)}}}_f})RT}{V}\\
$$
Solving this we get,
$$
n_{\ce{CO2(aq)}_f} = \pu{0.135 mol}\\
$$
This correspondingly means:
$$
n_{{\ce{CO2{(g)}}}_f}= \pu{0.015 mol}\\
P_f= \pu{5 atm}\\
$$
The final pressure is less than initial pressure as expected. The mistake you were making was implicitly considering $P_f=P_i$ when finding the mole fraction of carbon dioxide in water, which clearly isn't the case.


text:
Can fluorine ever have a positive oxidation state?
I know fluorine is the most electronegative element but can we humans ever synthesize fluorine in a positive oxidation state like +1?

What element would have a negative oxidation state to balance that +1?

There has been a previous discussion: https://chemistry.stackexchange.com/questions/62397/does-fluorine-in-fno3-have-1-oxidation-number

There is possible to have a ion $\ce{F+}$ in gaseous or plasmatic state, so yes, it is possible to have fluorine in oxidation state +I.

As noted in the referenced question, fluorine is not in the +1 oxidation state in hypofluorous acid (which, incidentally, is the only hypohalous acid that has been isolated) nor in any other compound where its bonds give it a complete octet.  Even where the fluorine has a positive formal charge, attaching it to less electronegative atoms will lead to an oxidation state of -1.
However, it is possible for fluorine to show a zero oxidation state and for a fluorine atom to apparently show a positive charge.  As one might expect, this occurs in one of the most unusual compounds in all of chemistry.
Striking gold -- and a lucky 7
That compound is "gold heptafluoride", $\ce{AuF7}$, whose unusual structure is reflected in the IUPAC name: instead of gold(VII) fluoride, it is officially called difluorinegold(V) fluoride. It is more accurately formulated $\ce{AuF5\cdot F2}$. It has been isolated at liquid nitrogen temperatures [1].
Himmel and Riedel [2] give calculations showing that unlike other heptafluorides, not all seven fluorine atoms are bonded to the central (gold) atom.  Rather, the gold is bonded octahedrally to five individual fluorine atoms and a difluorine ($\ce{F2}$) ligand, the latter attached end on and rotated so that in the most stable conformation (according to most of the several optimization schemes used), the remote fluorine atom is eclipsed with one of the other ligands as if attracted to it (a, below, picture from [2]).  The staggered arrangement one might expect (b) is instead a transition state in this fluxional molecule.  If we assume that the individual fluorine ligands are negatively charged as usual, this interaction indicates that at least the remote fluorine atom, labeled $\text{F}_b$, is positively charged.

And indeed, Table 2 of the paper shows the following charge distribution:

Thus the relatively electronegative gold atom, its effective electronegativity further enhanced by the other fluorine ligands, actually imparts positive charge to both fluorine atoms in the molecular fluorine ligand, as if there were a more electronegative moiety even than atomic fluorine.  Note the significantly greater portion of positive charge on the remote fluorine, the one which wheels around to attract one of the fluoride ligands in the eclipsed conformation.
To sum up, difluorine gold (V) fluoride has:

A difluorine ligand, apparently the first known case of thus

A gold-fluorine moiety, either $\ce{AuF5}$ or $\ce{AuF6}$ depending on how one views the structure, that is effectively more electronegative than fluorine itself

A resulting positive charge on some fluorine atoms.

References
1.
Timakov, A. A.; Prusakov, V. N.; Drobyshevskii, Y. V. (1986). "Gold heptafluoride". Dokl. Akad. Nauk SSSR (in Russian). 291: 125–128.
2.
Daniel Himmel and Sebastian Riedel, "After 20 Years, Theoretical Evidence That 'AuF7' Is Actually AuF5‚F2", Inorganic Chemistry 46 (13), 5338-5342 (2007).

Cool, but 0 isn't positive, and I'm pretty sure positive values are possible for fluorine.

Right.  I mention (partial) positive charge as distinct from oxidation state.  If we are looking at species where the fluorine has its full octet, we can get +1 oxidation state (with the usual convention) only by having it bound by pairwise bonds to two fluorine atoms.  Gold heptafluoride has fluorine in one such bond, not two.  Ergo oxidation state 0 even though there is a positive partial charge.

But after a bit of searching I found only theoretical study of $\ce{F4^+}$ https://www.researchgate.net/publication/255755579_Investigation_of_heterodimeric_and_homodimeric_radical_cations_of_the_series_F2O2_F2Cl2_Cl2O2_F4_and_Cl4

I think it's just if isolated cation can hold together, though. Not all that different from F+ in plasma - there's probably no counterion for such cation.

The reference does mention the possibility of matrix isolation of its identified thermochemically stable species including $\ce{F4^+}$.  Such a matrix would presumably have to include some counterion.

Well, matrix makes counterions largely irrelevant, I guess. I imagine there could even be no specific counterion in some cases, but I dunno much about it.


text:
Explain on the basis of valence bond theory that tetracyanonickelate(II) ion with square planar structure is diamagnetic
Answers on the internet suggest that tetracyanonickelate ($\ce{([Ni(CN)4]^{2-}}$) with square planar structure is diamagnetic because $\ce{CN-}$ is a strong field ligand and causes pairing, but my book says that VBT does not distinguish between strong and weak field ligands. So can anyone explain the logic using VBT?

According to VBT, for $\ce{[Ni(CN)_4]^{2-}}$, because it is a square planar geometry, it will experience $\ce{dsp^2}$ hybridization. The following process and result will be the following:

The diagram clearly shows that using VBT, we get that $\ce{[Ni(CN)_4]^{2-}}$ is diamagnetic. However, in general, VBT is not used often to describe these bonds, and more and more complex theories of bonding have been developed, although all involve approximations/assumptions. One of them is crystal field theory which is what the internet answers were suggesting.
In terms of crystal field theory. $\ce{Ni^{2+}}$ has a $\ce{d^8}$ configuration, and the crystal field splitting diagram is square planar. Since $\ce{CN-}$ creates large splitting, only four orbitals will be filled in the following diagram:

We will find that only $\ce{d_{yz}}$, $\ce{d_{xz}}$, $\ce{d_{z^2}}$, and $\ce{d_{xy}}$ will be filled as $\ce{d_{x^2-y^2}}$ is too high in energy. Since there are 8 electrons, all of the 4 orbitals will be fully filled and this makes it diagmagnetic according to crystal field theory.

please see https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-for-transition-metal-complex

It is dsp2 because that is the hybridization of square planar geometries according to VBT. If it is tetrahedral in the case of $\ce{[Nicl_4]^{2-}}$, it will undergo sp3 hybridization.

@Mithoron I am not sure if you saw my comment but see this https://chemistry.stackexchange.com/questions/58012/are-there-square-planar-complexes-with-sp2d-hybridization

And you saw Martin's comment there? VBT isn't what you think - it's part of class of computational methods - about as good approach as MOT and the calculations say there's almost always only negligible mixing of p and d orbitals - no actual hybridisation of d and p orbital together.

@Mithoron I know, that is what my second paragraph explains that VBT is not generally used to describe these bonds and so on. I know there are other theories like MOT, CFT, and Ligand Bond Theory (Which I posed a question about that hasn't been answered). It's just the question asked about how would VBT be used to model the bonding in $\ce{Ni(CN)_4}^{2-}$

"VBT is not used often to describe these bonds, and more and more complex theories of bonding have been developed, although all involve approximations/assumptions. One of them is crystal field theory" - that's just not true and you still didn't get my point. CFT is about as outdated as what you think is VBT, but is rather so called VSEPR - bunch of ancient ideas, already outdated in last century.

Sure you can say that but what does my understanding of the outdatedness of theory has to do with my answer and the question? Anyways, more importantly about the correctness of my statements: I watched a few lectures from UC Irvine to make that statement you said is not completely true. But if you can point out what is wrong there, please let me know. I just want to be correct in what I say.

@M.L has already provided the solution as per VBT, but that answer seems to have some flaws when it comes to CFT (looks like he has corrected the mistakes now). So, I would like to take some time to answer this question using CFT.

According to crystal field theory, the degeneracy of the orbitals of a same orbital is lost in presence of a ligand and the pattern in which they are distorted depends upon the nature of the ligand. The spectrochemical series helps us understand the nature of a ligand.
In general there are two types of ligands, strong field and weak field. Strong field ligands have a greater power to distort the degeneracy of the sub-orbitals whereas weak field have a weaker power of distortion.
As per the spectrochemical series, $\ce{CN-}$ is a strong field ligand and thus creates a greater distortion in the degeneracy of $\mathrm d$ orbitals.
The central metal ion here is the $\ce{Ni^2+}$ with its $\mathrm d^8$ electronic configuration.
A $\mathrm d^8$ system prefers a square planar approach of ligands if the ligands are strong field. In such a condition the the degeneracy is disrupted and the energy diagram of the $\mathrm d$ orbitals is as below:

Now when trying to fill this diagram with electrons we have to take into consideration that an electron in the $\mathrm d_{x^2-y^2}$ orbital can pair up with the electron in $\mathrm d_{xy}$ in case of a strong field ligand. This is because the extent of splitting done by a strong field ligand is so high that it can overcome the energy gained due to pairing of electrons.
Thus the energy diagram filled with electrons looks as following:

Now the ligands donate their lone pairs in the $3\mathrm d_{x^2-y^2}$, $4\mathrm s$, $4\mathrm p_x$ and $4\mathrm p_y$ sub-orbitals.
In reality, no hybridization takes place but if forced to say the hybridization of this complex, we can say $\mathrm{dsp}^2$ because of the orbitals involved in the $\sigma$ bonds between the metal and ligands.
As you can see the energy diagram does not contain any unpaired electron and thus the complex is diamagnetic.


text:
Does the appearance of turbidity upon addition of water to an organic solvent mixture imply the presence of nonvolatiles?
I was attempting to ascertain the suitability of GCMS to analyze a solvent mixture recovered from a solgel process. The filtered mixture contains heptane and denatured ethanol as major components and small amounts of HMDS and other impurities (various hydrocarbons in the heptane plus methanol in the ethanol). I spiked the mixture with deionized water and biphasicity ensued. The upper organic layer was clear but the lower aqueous layer was very white and turbid. Does this imply the presence of nonvolatile organics that precipitated, making the mixture unsuitable for GCMS analysis?


text:
What is calcined lawsonite?
For context, I am planning to try and use this sample to make a ceramic glaze.
I recently collected a sample of what I think is lawsonite, which has the chemical formula $\ce{CaAl_2Si_2O7(OH)_2*H_2O}$. After collecting it I calcined it in an electric (oxidation) kiln to around 2016 °F. Assuming that what I collected is indeed lawsonite, what happened to it? I'm assuming that all the water "evaporated", leaving some combination of $\ce{CaO},\,\ce{Al_2O_3},$ and $\ce{SiO_2}$.

It probably decomposed.

@NilayGhosh what does that mean?

It means lawsonite decomposed at high temp, leaving behind compounds you mentioned:  CaO, Al2O3 and SiO2. The water present as water of crystallization  evaporated.

@NilayGhosh Okay so we have $$\ce{CaAl2Si2O7(OH)2H2O}=\ce{CaO}+\ce{Al2O3}+2\ce{SiO2}+2\ce{H2O}$$ so I guess $$\ce{CaAl2Si2O7(OH)2H2O}+\text{heat(2016 °F)}\to \ce{CaO}+\ce{Al2O3}+2\ce{SiO2}.$$
Does that sound accurate?

Yes. $\ce{~~}$.


text:
What functional group is this in this specific molecule?
I know without the nitrogen this molecule would probably be considered a crown ether. What exactly would the nitrogen/oxygen functional groups be named as in this molecule shown below? Any help would be greatly appreciated.

replacement nomenclature will always work.

Well, compound is substituted hydroxylamine, not ether.

Thanks, just wanted to get an idea of what it would be called to look up relevant research on functional groups like this.


text:
Why 9BBN's rings don't migrate during hydroboration-oxidation?
My organic chemistry textbook (Clayden, Warren, Greeves) says this:

If we use the $\ce{HBR2}$ reagent $\ce{9-BBN}$, then only the
non-cyclic substituent formed in the hydroboration reaction will
migrate, selectively giving us the product we want.

Why can't the original substituents of $\ce{9-BBN}$ migrate? It seems to me that there is nothing stopping the rings of $\ce{9-BBN}$ from migrating to the oxygen in the $\ce{-OOH}$ attached to the Boron.

I suspect the ring strain raises the energy of the transition state enough that those migrations are disfavored.

I agree. 9-BBN is a pretty rigid molecule, which is exactly why is it used


text:
Methanol-water mix freezing points - clarify contradictory online data > 60% w/w?
Extending an exploration of windshield washer-fluid choices to freezing points of methanol-water mixes, all web-sources located agreed to within a couple of degrees from 0-60% w/w, yet above this proportion (and below 100%), repeatedly came across conflicting data (which fell into three groups). (Yes, the disagreements are all outside the range of winter driving on this planet - except perhaps for around Vostok Station...)

w/w
various graphs by eye
various tables
engineering toolbox table
all˚C

60%
- 71
- 71
- 71
˚C

70%
-102
-137
- 82
˚C

80%
-116
-143
- 87
˚C

90%
-108
-146
- 90
˚C

100%
- 96
- 98
- 98
˚C

examples of sources for above data:
http://www.methanol.org/wp-content/uploads/2016/06/FreezingPointsMethanol-WaterSolutions.pdf    graph
https://patents.google.com/patent/WO2011153147A1/en table (~ 1/3 way into the document)
https://www.engineeringtoolbox.com/methanol-water-d_987.html        table
Thanks for any comment.

Different source data, different data processing, different errors made.

Thanks. Your emphasis on "different" inspired a new search strategy which led to a satisfactory answer to the question.

From Solid-Liquid Phase Diagram of the System Methanol-Water by G. A. Miller and D. K. Carpenter in J. Chem. Eng. Data 1964, 9, 3, 371–373 (July 1, 1964) [https://doi.org/10.1021/je60022a017]

"A solution of methanol and water that is methanol-rich tends to form
a glass when cooled below the melting point. The liquid is very
viscous at such temperatures (below -100° C.) because of the presence
of hydrogen-bonded polymers. Because it is difficult to induce
crystallization, the melting point is determined from a warming curve.
Much of the wide divergence of melting point data for the composition
range of 55 to 95 mole per cent* methanol results from the use of
cooling curves."

* by a back-of-a-napkin calculation (assuming 32.042 g/mol MeOH and 18.015 g/mol H2O) the mole percentage mentioned is approximately 68% to 97% methanol (w/w).
Further - successful - online research was inspired by a comment; it appears the failure to find information with common search terms was due to a language difference. It seems specialists (presumably physical chemists) use "solid-liquid equilibrium" or "solid-liquid phase diagram" in documents which may completely exclude the term "freezing-point" - such as the publication quoted above which answers the original question.


text:
Relation between the number of hybridised orbitals and the number of sigma bonds
March's Advanced Organic Chemistry (8th Ed.), has in its first chapter several tantalizing statements of the following variety;

Boron has only three valence electrons available to form bonds, hence the valence of 3. Any hybridization model must take this into account.

If ethene is examined... each carbon has three σ bonds, one to
each of the three atoms. Therefore, sp2 orbitals are used to form those three bonds.

When carbon is connected to another carbon atom by a triple bond, as in ethyne... each carbon is connected to only two other
atoms by a σ bond and hence uses sp hybridization.

The key is to understand that an atom forms two σ bonds for sp hybridization and three σ bonds for sp2 hybridization.

From reading that section, I  understood the author's technique of determining hybridisation as;

You call those bonds that form when there's maximum overlap between the bonding orbitals "sigma bonds" , and such initial bonds that form while using the author's method, are arranged so that they do so at max overlap(the atoms are arranged in space so as to have such a configuration).
You make a bond each between the central atom and the others this way, and the remaining orbitals, if any contain electrons left over, interact to form pi bonds(and they're not sigma bonds because in the simple molecules examined, these leftover electrons aren't in orbitals that overlap terribly well. So you make a name for them-pi bonds- and say they interact in the way they do).

Now, all these statements seem to border on stating that the number of sigma bonds formed and the number of hybridised orbitals are the same, and that kind of makes sense to me. By the method of determining hybridisation (detailed above as I understood it), $\pi$-bonds seem to be defined to be those that're formed from unhybridised orbitals.
But the author never makes such a statement. And it's a chemistry textbook. Cue the alarm bells in my head.
On looking it up, there isn't any statement on the internet of that sort, either, except for this Quora thread where the general consensus seems to be that yeah, pi bonds can be formed from hybridised orbitals. But I don't get how.
Is there any such rule? If not, why?

Yes, in general pi-bonds are formed from unhybridised orbitals, because they are formed by sideways overlap of p-orbitals, at least in the usual organic structures we see. There are problems with this model if you look at unusual compounds like benzyne, where the third bond is formed by sp2-sp2 overlap. Whether we can call that a pi bond is debatable.

That answers it, thanks. I've understood the hybridisation thing correctly, right?

Yes, that understanding of hybridisation mostly works for first row of main group (B,C,N,O). When you move to other parts of the periodic table (d-block, or 2nd row main group like P,S) it becomes very complicated, and it's not always possible to use the hybridisation model.

Hybridisation is just a mathematical description. The statement about benzyne by @ShoubhikRMaiti is not correct, at least very incomplete and from an experimental organic chemist point of view. Bonds do not form by orbital overlap. Bonds are what we observe; orbitals is what we use to describe these, and overlap is our interpretation. This unfortunately carries multiple ways of looking at things.

@Martin: hmm. I know these models are just constructs. That said, I just needed clarification on how the theory itself worked, which I trust is how he verified it as being?

@Martin-マーチン Why is it not correct? In LCAO model bonding orbitals do form by overlap (or constructive interference) of atomic orbitals, so I don't see the issue with using orbital overlap to describe bonds.


text:
How do lyophilic colloids reduce surface tension?
It's given in my textbook that the surface tension of lyophilic/hydrophilic solutions is lower than water.
But my intuition says, since in lyophilic solutions the interaction between a lyophilic colloid and water molecules would be more than it would be in between two water molecules, then the surface tension would increase. But my textbook says otherwise. What am I missing?


text:
Comparing acidic strengths between benzylammonium ion and phenol
Question

Compare the acidic stengths between benzylammonium ion and phenol.

I first tried to remove the H+ ion from both of them and tried to compare the relative stabilities of resulting conjugate bases, but I am not sure, which one would be more stable.
The first one would be a neutral compound and the second one would have resonating structures on which the negative charge gets mostly delocalised to carbon.
I have seen somewhere on the internet that benzylammonium ion has a $\mathrm{pK_a}$ value of 9.33 while phenol has a $\mathrm{pK_a}$ value of 10. If this data is correct, then how do we explain the acidity order?

Nitrogen being electronegative pulls the electrons which destabilize the ring Whereas in o- acts as an electron donating stablize the ring so it is more stable

You could as well take an orange and a banana and ask why one winds up heavier then another.

@Mithoron I don't understand your point. Doesn't this question make sense? To the downvoter: And I would also like to know the reason for downvote.

You can't compare two "different" compounds when there are many factors and some of them contradict each other. In such cases we only have the data to look upon. It's true that many hypothesis can be made from the data, but those are nothing but humans being humans and acting like they know what is going on. In short: Data drives theory(many times wrong) and not the other way around.

@NisargBhavsar But we can come up with an explaination for the observation. So, that's what I am looking for. What's wrong with that?

The problem is that, the explanation isn't going to serve anyone. Not even you.

And even if someone gave an explanation it will boil down to the fact that the the ammonium ion's inductive effect is dominant over the the resonance in the phenoxide ion "as per the data".  Sitting on a chair with armrest we can't predict which is more favourable a carbanion with -I of ammonium ion and -M of benzene or an oxygen anion with -M of benzene.

Have you considered first comparing say the pKa of benzylamine and benzyl alcohol?

@BuckThorn I just did. So benzylamine has a pKa of 8.82 and benzyl alcohol has a pKa of 15.4 . Even in this case, I don't know the reason behind this. I would appreciate if answerer explains this case too. But why should this be related to the original question?

When you pick benzylamine and benzyl alcohol you narrow the source of the difference to electronegativity mostly (removing much of the influence of the aromatic ring), and yet retain the relative ordering of the pKas.

What is the source of your pKa, particularly benzylammonium. It is surprisingly similar (no coincidence I think) to that of benzylamine in the Wikipedia: https://en.wikipedia.org/wiki/Benzylamine

@BuckThorn I just googled it and this is the first result I got. Of course I knew that it was not reliable. I couldn't find any good source. Also this link shows an experimental value of 8.82 for benzylamine.

The link to 8.82 is for benzyl amine in 60% aqueous dioxane, not water, so it does not directly compare to the pKa for phenol, or the 9.33/9.34 for benzyl amine. But it does suggest that solubilizing the amine with a water-compatible organic solvent increases the acidity of the ion by a factor of 3.3, which is quite large. Thus solvent effects are as important as some inductive effects!

The acid strength of each compound can be explained, but the acidity order is much more difficult to compare, because the two compounds are only remotely connected. It is misleading to conclude that the mere presence of a phenyl group somehow connects these molecules. The similarity of the pK$_a$s is likely a coincidence.
The question needs a clear definition of pK$_a$. We can ask what the pK$_a$ of phenol is, or the pK$_b$ of phenoxide. When we ask for the pK$_a$ of benzyl amine, we want the pK$_a$ of benzylammonium ion, not the pK$_a$ of benzyl amine going to C$_6$H$_5$CH$_2$NH$^-$. There is a bit of flexibility (or sloppiness) in the terminology here, and also a conflict in the values presented: the question quotes 9.33 for benzylammonium ion (I found 9.34 - no big deal), but the comment by the OP gives 8.82 for benzyl amine. The values should be identical because we know what the OP means. It appears that 8.82 is incorrect.
The difference between these molecules is far more important than any similarities. For example, one is a neutral molecule that you can put into a bottle with 6 x 10$^{23}$ more; the other is a charged piece of a molecule that you can hardly bring next to another one.
The phenyl ring has an enormous effect on the acidity of phenol, due to the resonance effect. It stabilizes the oxygen anion by more than 5 orders of magnitude, compared to, e.g., methanol (pK$_a$ = 15.5) or benzyl alcohol (15.4) or ethanol (15.9).
(Interesting that phenylethanol (pK$_a$ = 14.81) is more acidic than benzyl alcohol. This suggests that the methyl group in ethanol is more electron-donating than the benzyl group in benzyl alcohol; this agrees with the discussion in #3 below, for phenethylammonium ion.)
The phenyl ring has a much smaller effect on the amine molecule-ion; no resonance, just inductive. The pK$_a$ of benzylammonium ion (9.34) can be compared with, e.g.:

methylammonium ion (pK$_a$ = 10.66) Replacing the phenyl of benzylammonium ion with H reduces the K$_a$ by a factor of 21.

This is equivalent to saying that benzyl amine is less basic (pK$_b$ = 4.66) than methylamine (pK$_b$ = 3.34); therefore the protonated benzyl amine will give up an added proton more readily than methylamine. We can conclude that methyl is more electron-donating than benzyl.

ethylammonium ion (pK$_a$ = 10.71) Replacing the phenyl of benzylammonium ion with a methyl group reduces the K$_a$ by a factor of 23. Thus, ethylamine, like methylamine, is more basic than benzyl amine because methyl is more electron-donating than phenyl. (But we knew that.)

Phenethylammonium ion has a pK$_a$ = 9.73. Putting another CH$_2$ group between phenyl and nitrogen reduces K$_a$ by a factor of 2.5. This indicates that a benzyl group is less electron-donating than a methyl or ethyl group, in agreement with the comparison of the inductive effect between benzylammonium vs methyl- or ethylammonium ions and benzyl alcohol vs phenylethanol.

Benzyl amine and phenol are the neutral actors in this drama, and might react nicely in a one-to-one ratio. What would be the pH in solution? The graph shows the ratio of ion concentration to neutral molecule concentration for phenol and benzyl amine, separately, over the pH range of 5 to 9. They show contrasting slopes which cross at pH 7.33. At this pH, the ionic fraction of each compound is 0.21% of the whole - it doesn’t matter what concentration you started with if everything is soluble (benzyl amine is very soluble, but phenol, mp 43 C, is soluble in water slightly less than ~0.1 M in water at room temperature). Just add the same number of moles of each neutral compound to get 0.21% reaction and pH 7.33. What a coincidence! Practically the same pH as pure water.

I sincerely thank you for your efforts into this answer. The values that I gave were just based upon a google search and so I knew that these are not reliable values :).

@Light Yagami: You're welcome; my pleasure. It was a good question - it stretched the imagination.


text:
Why is the spectroscopic entropy of carbon monooxide higher than that of nitrogen?
While solving a question set, I found a question with this data table (at $\pu{298K}$):
$$
\begin{array}{lrrr}
\hline
\text{Substance} & T_\mathrm b/\pu{K} & S^⦵_\mathrm{cal}/R & S^⦵_\mathrm{spec}/R \\
\hline
\text{Nitrogen} & 77.35 & 18.3 & 18.30 \\
\text{Carbon monoxide} & 81.61 & 18.6 & 19.22 \\
\hline
\end{array}
$$
The first part of the question asked why $\ce{CO}$ has higher $S_\mathrm{spec}$ (spectroscopic entropy) than $S_\mathrm{cal}$ (calorimetric entropy). I know that this is due to the residual entropy of $\ce{CO}$.$^*$
However, the next part of the question asked why the value of spectroscopic entropy of $\ce{CO}$ is larger than that of $\ce{N2}$. The question supplied the information that the rotational reduced temperature ($\theta_T$) was same for both molecules (I think that means the rotational partition function $q_\mathrm{rot}$ will be same for both); and that the contribution to entropy from the vibrational modes to entropy is negligible in both cases.
Now the spectroscopic entropy can be written in terms of the contribution of each degree of freedom:
$$S_\mathrm{spec}=S_\mathrm{trs}+S_\mathrm{rot}+S_\mathrm{vib}+S_\mathrm{elec}$$
The translational entropy contribution depends on the translational partition function $q_\mathrm{trs}$ , which depends on the mass of the molecule. $\ce{CO}$ and $\ce{N2}$ has almost the same mass (28.01 u), so I don't think there would be any difference in $S_\mathrm{trs}$.  The rotational and vibrational terms are assumed to be same in the question. The electronic term should also be zero in both cases, as the electronic partition function would be 1 for both. (I think the excited states are all inaccessible at room temperature).
So where does the difference in the calculated entropy come from?
I feel like the answer is something simple that I have missed.
$^*$ For the concept to discern calorimetric entropy from spectroscopic entropy, see for example this Chemistry LibreText; an literature example e.g., in  1958JAmChemSoc1054.

One contribution is the number of identical conformations of a molecule that can be generated by symmetry operations, which is described by a symmetry number $\sigma$. In the high temperature limit
$$S_\mathrm{rot} /R = 1 + \ln\left( \frac{8\pi^2IkT}{h^2\sigma} \right) $$
A larger symmetry number decreases the entropy. This number is 2 for nitrogen and 1 for carbon monoxide. That alone can account for $\log(2)=  0.69 \times R$. The larger symmetry number for nitrogen decreases the entropy by this amount.
In addition the moments of inertia (inversely proportional to the rotational temperature) of the two molecules are not really identical (it is ~5% larger for $\ce{CO}$), but the difference accounts for only ~$0.05 \times R$.


text:
Absorption Curves Normalization
I have absorption curves for three species that are nominally the same concentration that I need in order to calculate spectral overlap. Part of that process is dividing the curve through by the concentration as a normalization process. The curves of two species are a factor of 10 lower than the other (not necessarily a problem), but that is true even out away from where the peaks from these species are. All curves were taken on the same spectrometer. It seems to me like maybe someone changed out the blank solvent in-between data taking (a possibility in a shared lab). Is there a way to normalize this prior to calculating spectral overlap?

Are the data recorded on a two-beam instrument, where sample and blank are simultaneously present?  If so, the raw spectral data accessible to you (prior baseline correction, peak pick, etc.) may already be the result of subtracting the signal of the blank from the signal of your sample (= no recovery).  Note, some spectrometer's software allow the user to define (i.e., to change) which of the two paths is about the sample, or the blank; an additional potential pitfall in a shared lab.  Suggest: Add an illustration of the curves here, clarify the aim to "calculate the spectral overlap".

Otherwise, if you may normalize by maximum intensity (most intense peak in each spectrum = 1.00 a.u.) and then set the three spectra in one diagram in common.  Then, it it might have been a good thing that each sample's concentration was similar to each other.

I am calculating spectral overlap for resonant energy transfer reasons which unfortunately means I can't divide through by the maximum value. It is a two-beam instrument, and I've just realized that I saved the original data file type in addition to the csv. I'll check if that has the needed information to get the raw curve back out. Thank you

Ok, it is not for "looking nice" (to advertise the location of the bands) in first place, but for a FRET. A look at this overview regarding normalizing, and FRET em particular may be helpful.  Sadly, though, it is not a walk-through including the .cvs data.

@MJAFort do you really need the normalisation if the species had the same concentration? Because if the answer is yes, I can not see at this moment how one would proceed, - at all - with different concentrations.

@Alchimista I am suspicious that there is something gone wrong due to the fact that the absorption far away from the peaks from these species is a factor of 10 higher as well as the peaks due to the species. I guess I was trying to figure out if there were normalization processes that are commonplace but I was unaware of, especially given all the rules surrounding the calculation of the spectral overlap integral. Nominally all solutions in this case were 1mM

Isn't it possible to check for the absorption of that one compound? To see if it should have somewhere a zero Abs between peaks? It might be genuine? Molecules might have very different absorption coefficients...


text:
Why is a sodium chloride molecule stable?
The ionization energy of sodium is $\pu{5.139eV}$. This is the energy absorbed when a neutral sodium atom is stripped of its outermost electron. The electron affinity of chlorine is $\pu{3.62eV}$. This is the energy released when an electron is added to a neutral chlorine atom to form the chlorine anion.
Now suppose we have a system containing exactly one neutral sodium atom and exactly one neutral chlorine atom infinitely far apart so that the systems potential energy is zero. If we wish to make a $\ce{NaCl}$ molecule from our system, we need to put $\pu{5.139eV}$ of energy into the system to ionize the sodium atom. Afterwards, we place the electron we have just pulled off the sodium atom onto the chlorine atom upon which we get back only $\pu{3.62eV}$ of energy. So the total energy of the system has increased by $\Delta E =\pu{5.139eV -3.62eV= 1.519eV}$.
If the total energy of the system has increased after performing the previous supposition, then why does the $\ce{NaCl}$ ionic bond occur? Surely it is less stable than having the two neutral atoms separate from each other? My understanding up till now has been that molecules form when the total energy of the system is lower than the total energy of the sum of its constituents?
Is this all because an individual $\ce{NaCl}$ bond is not stable on its own but it is stable when its within a crystal lattice?
Any help on this issue would be most appreciated!

Sodium chloride molecule is not much of a thing at all. Then again, you might want to consider that the ions are not infinitely far apart.

A similar thing with potassium and fluorine: https://chemistry.stackexchange.com/a/128771/79678. If you had two well separated atoms of K and F, an energy input would be required to get potassium ion and fluoride ion. But that is just one piece of the whole energetics. Same for sodium and chlorine. If you boil NaCl, there are some actual NaCl molecules in the vapor, for what it is worth.

As you can see from Maurice's answer, you've underestimated the amount of energy required to produce NaCl molecules from pure Na and Cl2.  But essentially, you're on the right track: What you've shown is that molecular NaCl is unstable...which is why NaCl's equilibrium state is not the molecular form, but the ionic form.  Once you account for the energy you get from forming the ionic crystal, the net energy change is highly negative.

In short you are right just did not considered that when the ions get togheter the potential energy is lowered. Every X+Y- is more stable than the two taken with separation. Though it can be crucial the energy required for the two be formed,as per the answer of Maurice. But hold on the idea that molecules form when the total energy of the system is lower than the sum...... It is correct. So normally its is the lattice formation, though NaCl is certainly a possible molecular thing.

The lattice energy numbers for NaCl  (a whole mole) were addressed in a recent answer to another question: https://chemistry.stackexchange.com/questions/138455/what-exactly-is-lattice-energy/151973#151973  The "lattice energy" for a single pair of ions would be significantly less.

What you have calculated is the energy required to make two separated ions out of two separated $\ce{Na}$ ad $\ce{Cl}$ atoms. When they are separated, these ions do not make a molecule or a crystal of salt. Separated ions attract one another and are able to release a huge amount of Coulomb energy when approaching each other.
Anyhow your calculation should still take into account the fact that chlorine atoms do not actually exist. They must be produced by breaking a $\ce{Cl2}$ molecule. And this requires some energy. You may also introduce the sublimation energy of sodium atoms


text:
small negative frequencies in DF T optimizations with organic molecules containing CF3 groups
When optimizing by DFT medium-sized organic molecules containing aromatic CF3 groups (for example, adducts involving takemoto's catalyst), the frequency calculation on optimized structures have always small negative frequencies involving CF3 groups (small rotations), no matter which program (gaussian or orca), which functional (b3lyp, m06-2x, pbeh-3c or b97-3c), which grid (ultrafine, grid6), which base (double or triple-zeta, with or without diffuse or polarization terms) or convergence criteria.
The only solution is to manually rotate the groups a bit and launch another optimization (with smaller step) and another frequency calculation, over and over (sometimes I finally find a true minimum after 3 or 4 optimization/frequency cycles, sometimes after 7 or 8).
This procedure is quite time-consuming, considering the number of frequency calculations involved.
Anyone has ever faced a similar problem? Are there any tips or tricks able to solve (or mitigate) this problem?
thanks

The Matter Modeling site may be a good place to ask.


text:
SrSO4 Solubility in Acid
$\ce{SrSO4}$ is insoluble in water. Will it be more soluble in an acid solution? I know that if the anion is the conjugate base of a weak acid, it'll be more soluble. Does $\ce{SO4^2-}$ count as the conjugate base to $\ce{H2SO4}$ (strong acid) or $\ce{HSO4-}$?

Even HSO4- is not really a weak acid.

Wait so would it be more or less soluble in acid

Wikipedia says "slightly soluble in acids" and yes, sulfate ion is conjugate base of HSO4-

BaSO4, that is even less soluble than SrSO4, is known to be somewhat soluble in concentrated H2SO4(particular data not handy).

In analytical chemistry books, $\ce{SrSO4}$ is said to be soluble in boiling concentrated $\ce{HCl}$ and $\ce{HNO3}$ solutions. On the other hand, $\ce{BaSO4}$ is said to be soluble in concentrated sulfuric acid, because of the formation of both $\ce{HSO4^-}$ ion and of the complex  $\ce{H2[Ba(SO4)_2]}$, which is decomposed by dilution. Ref.; R. Delaby, Analyse qualitative minérale, Masson, Paris, 1950 (in French).


text:
Does the amplitude of the RF pulse affect the NMR signal?
I know that the NMR signal depends on the static magnetic field strength $ B_0 $ (by Boltzmann equation) and its gyromagnetic factor. While NMR occurs when the RF pulse frequency matches the equation:
$$ \nu = \frac{g}{h}\mu_{n}B_{0} $$
what about its amplitude? Does the amplitude of the RF pulse affect the NMR signal? For example, does it affect the signal strength?

Please be a little more clear. You write about "signal amplitude" and "pulse amplitude" in different sentences, but sometimes its not 100% clear which one you mean (such as immediately after the equation).

V. approx explanation. The rf pulse is at 90$^\text{o}$ to the static field $B_0$ and has the effect of moving the magnetisation (sum of spin vectors) away from alignment along $B_0$ (z-axis normally) into the xy plane where it rotates and cuts through detector coils and induces FID signal. The FID can only be detected with a magnetisation component in the xy plane. The size of the rf pulse's magnetic field $B_I$ is proprotional to angle $\theta$ moved away from z-axis as $B_I\sim \tan(\theta) $ and so via magnetisation's component in xy plane (arrow on axis in sketch) to the size of the FID.


text:
Competition between aromatic sulphonation and nitration
Say you take a substituted benzene* and react it with oleum and nitric acid. Would it result in a nitrated ring, a sulphonated ring, or some sort of a combination?
Thinking about it in terms of the first step, the stronger base of $\ce{SO3}$ and $\ce{HNO3}$ seems likelier to get protonated first, making the preferred protonated electrophile (say it's A) less reactive than the other possible intermediate (say that one's B). But then the next step, the r.d.s, requires a pretty strong, reactive electrophile to attack the ring, which would make B the probable winner.
How do I decide between these factors, if such a general assumption can be made?

*I assumed a substituent wouldn't make a difference in this comparison. The answer, if so, would be more widely applicable than if the question were just about benzene. If there are differences, a note on them would be appreciated.

100% dry nitric acid is not possible and thus the oleum would dissolve in that water to form sulphuric acid. And it is widely know that a mixture of sulphuric acid and nitric acid causes nitration.

Wikipedia says: "Oleum is a harsh reagent, and is highly corrosive. One important use of oleum as a reagent is the secondary nitration of nitrobenzene. The first nitration can occur with nitric acid in sulfuric acid, but this deactivates the ring towards further electrophilic substitution. A stronger reagent, oleum, is needed to introduce the second nitro group onto the aromatic ring."

@NisargBhavsar: great reasoning there. But does the added water always suffice to turn all the SO3 into sulfuric acid? If not, there's a chance for sulfonation to occur.

@NilayGhosh: thanks, that explains how nitration works. Isn't sulfonation possible, though?


text:
Bond lengths in Fullerenes
I came across this about fullerenes-

All carbon atoms are equal and the undergo sp2 hybridisation. Each carbon atom forms three sigma bonds with three other carbon atoms. The remaining electron at each carbon is delocalized in molecular orbitals which in turn give aromatic character to the molecule. This ball-shaped molecule has 60 vertices and each one occupied by one carbon atom and it also contains both single and double bonds with C-C distances of 143.5 pm and 138.5 pm respectively.

It speaks of fullerenes being aromatic, which would imply that the bond lengths in fullerene must be equal. I am aware of the fact that resonance won't mean that bond lengths be equal in all cases, for example, I found this about naphthalene-

The molecule is planar, like benzene. Unlike benzene, the carbon–carbon bonds in naphthalene are not of the same length. The bonds C1−C2, C3−C4, C5−C6, and C7−C8 are about 1.37 Å (137 pm) in length, whereas the other carbon–carbon bonds are about 1.42 Å (142 pm) long.

I know that such a case in naphthalene is due to dissimilar conjugation as it behaves as a benzene ring joined to a diene. Speaking of graphite, as far as I know, all bond lengths are equal in a layer due to conjugation that is quite comparable to benzene and even better than naphthalene (resonance energy per atom). But I would like to know exactly what happens in a fullerene molecule.

What you see is what you have: all bonds are conjugated to a certain extent, but not as perfectly as to be called aromatic.

For this question you can easily find much material by simply searching for "fullerenes bond length"., "fullerenes and aromaticity", "fullerenes vs graphite" and so on. Or simply "fullerenes". The main difference should be already in pure geometry, as there are pentagon rings and, but I can be wrong, segments leading to a regular hexagon cannot lead to a regular pentagon.

You'll find a ton of more or less confident claims that fullerenes are aromatic. Just dismiss them as nonsense.

Well yes, I guess I got it, obviously hexagonal and pentagonal rings would mean dissimilarity and there will have to be different bond lengths.


text:
higher viscosity base oil instead of better lubrication results in worse
UPDATE:
The problematic "symptom" that I try to fix, is the static friction of two sliding plastic parts.
This is about a keyboard key switch that was made during the 80s. The problem/issue is that after I applied Grease #1, then when the slider is pressed under a small angle, it gets strong binding issues (i.e. it is stuck and needs a lot more force to move it) and it also exhibits a substantial "stick-slip" after it starts moving, i.e. irregular dynamic friction, which instead of a steady velocity and a smooth sliding it slides in a starting-stopping way.
When I applied Grease #2, the aforementioned issues got greatly reduced (not completely eliminated the static friction though), resulting in a much better and smoother sliding.
However, this is contrary to what I have expected. On paper, Grease #1 is much more viscous than Grease #2. And according to the consensus saying how much viscosity matters, it should be the #1 that should make it smoother and not the other way around.
These 2 different greases have a big difference in their base oil viscosity at 40C. Their Techincal Data Sheets claim that grease #1 is 9500 mm2/s viscous and grease #2 is 1500 mm2/s viscous at 40C.
Another important difference is this:

Grease #1 is a grease with a Polydimethylsiloxane (PDMS) "silicon oil" as Base Oil, and an "inorganic" Thickener (probably silica)
Grease #2 is a grease with a Polyalphaolefin (PAO) "synthetic hydrocarbon" as Base Oil, and Silica as a Thickener.

After a simple study of the basics of Tribology and by reading various articles on the subject, I have came to know that viscosity is regarded as the number one factor for lubricating two sliding surfaces and that higher viscosity results in lower static and dynamic friction and more importantly eliminates the unwanted stick-slip phenomenon*. The higher the viscosity the better.
According to the above, for my use case, grease #1 should be much better at both reducing friction and eliminating stick-slip sliding. However, when applied and tried empirically the exact opposite is true. #2 is actually lubricating where #1 is probably making the friction and the stick-slip even worse than without applying any grease at all.

Why this happens instead? Is the viscosity rated in the TDS wrong? Is there another explanation?
What property exactly of #2 is responsible that provides the desired lubricating  properties, which #1 lacks?
Is there another property of a lubricant (grease) that would be better for this use case?

Use case: Two plastic parts sliding against each other. They are parts of a keyboard switch. The plastics are Nylon (the Slider) and the second (Housing) is probably ABS. More information: https://deskthority.net/wiki/Alps_SKCL/SKCM_series (the parts are the Slider and the Shell)
UPDATE:
Some relevant and helpful information:
The switch was made during the 1980s, therefore this means that they must have used oils/lubricants/greases that used the technology that was available at that time. The switch's Slider came pre-applied with a grease/spray/solid/whatever from the factory, which strongly suggests that the manufacturer knew that without it, the slider would not slide smoothly. And this is a fact that have been verified, if this is cleaned off, the switch gets a high static friction.
One more important thing: Another grease, Grease #3**, actually worked  very well with this switch and completely eliminated any static friction, even at obtuse angles. It has Polyalphaolefin (PAO) as Base Oil and Silica as thickener, however it is extremely viscous. It is considered a heavy "dampening" grease instead of a "lubricating" one, which makes the sliding slower, but also makes it very precise and smooth. These kind of greases are used with binoculars, microscopes, rotary dials and controls that need a very precise and smooth movement, and they are regarded as highly desirable qualities for these use cases.
My guess is that the factory applied grease/solid/whatever, is probably a "dampening" one too. Switches in a good condition from that series, have that precise and smooth slide, which is a characteristic behavior that differs significantly from the behavior of other switches applied with usual lubricants.
Technical details of the greases:
Grease #1 Technical Data Sheet: https://www.oks-germany.com/product_downloads/pdf/PI/en/PI_OKS_1110_110645_EN.pdf
Grease #2 Technical Data Sheet: https://www.oks-germany.com/product_downloads/pdf/PI/en/PI_OKS_477_113696_EN.pdf

Footnotes:
*. "The test results show that, by increasing the viscosity of the lubricant, the potential for stick slip is greatly reduced." - How to Measure, Prevent, and Eliminate Stick-Slip and Noise Generation with Lubricants
**. Grease #3 Technical Data Sheet: NyoGel 767A "Dampening" grease

Are you sure slider is nylon? Acetal (aka POM, Dupont trade name is Delrin) is generally preferred for that type of application and is typically not lubricated as the plastic itself has a low coefficient of friction and self-lubricates.But if it's definitely nylon, then that's irrelevant.

No, I'm not sure if the slider is nylon, it is what "the consensus" of the vintage keyboard community says. However, I am certain that it is not a self-lubricating one, precisely because this particular switch is a version that was made during the 80s, which were typically factory pre-lubricated and precisely by the fact that the manufacturer actually made a newer version, in beginning of the 90s, that is different material which is self-lubricating and thus needs no lubrication. I don't know how to tell what exact kind of material it is, are there any ways that I can use for that?

The effectiveness of lubricants depends on a number of factors. If the objects are in motion, hydrodynamic lubrication is important.
For static friction, adhesion of the lubricant to the surface is more important. For that reason, motor oil has antiwear additives to prevent metal-to-metal contact. Since the plastic parts in a keyboard are initially at rest before sliding, static friction predominates.

I think your answer is making a lot of sense. The greases in fact have completely different Base Oils. Grease #1 has Polydimethylsiloxane (PDMS), which AFAIK is a "silicon oil" and grease #2 has Polyalphaolefin (PAO) i.e. synethetic hydrocarbons. Do the hydrocarbons adhere that much better in the surface? Is that what is causing the  different results?
You are spot on about the static friction, as this is exactly what I am trying to "fix" with the lubrication. I will update my original question to be more specific. Please check it out and give your opinion.

The time scale of viscosity measurement is so much faster than the time scale of your switch operation that there may be only a slight connection. The temperature difference is another big effect. Although your switch may move rapidly when it does move, there will be long periods of no movement, when the grease can reorganize to bad effect.
Grease #1 has an inorganic thickener (could be silica); Grease #2 has a silicate thickener (could be some clay). Grease #1 had an entry that identified a very small amount of oil separation after some period. My thought is that a viscosity measurement is run on a homogeneous mix, and oil separation implies a non-homogenization after a period of standing, generally in the direction that the more fluid ingredient separates out while the thicker part gets thicker - to the point where it could develop a yield stress, i.e., a strength resembling a solid structure, hence the stick part of the behavior. Once the structure is broken, the more fluid part comes into play. This effect might take place only on a very small thickness level, e.g., when two parts are pressed together, squeezing the lubricant to very thin dimension. This would not be a part of technical specifications, but could well be a performance issue.
Grease #2 did not have an oil separation entry, and did have a very low pour point (-20$^o$C). Grease #1 had no drop point, indicating that it was so viscous as to have a structure, not a flow.
Very high viscosity oils by themselves could be good lubricants for your switch, but a three-dimensional thickener like silica, by not preventing oil separation, might contribute to excess thickening, a yield stress that locks up to give stick. A silicate (clay is layer-like, not three-dimensional) thickener that does not exude oil could remain in place homogeneously indefinitely and be slippery.
Since no long-term effects have been noted for silicone or petroleum oils on the switch materials, I feel there is no softening or material degradation to consider. But the performance test you require is specific to your need, and you probably need to devise your own test regime that measures slipperiness after a prolonged rest time with a pressure between two objects made of the very materials of the switch - although, if my hypothesis about the oil separation and yield stress formation is correct, it should probably happen between any two smooth objects.
Technical terms like rheopexy or thixotropy are not applicable, in my mind, because I think the problem is with time-dependent non-homogeneity. But non-Newtonian would be OK - maybe most fluids are non-Newtonian at some level.
Some petroleum greases are thickened with calcium or lithium soaps. A big difference is the temperature of the dropping point. This is probably not your concern. The lithium soaps, even at low concentration, exert powerful thickening and a slipperiness without gelling. Some of these (e.g., white lithium grease) are higher cost and come in little tubes for electrical use; others come in 55 gallon drums, with varying amounts of added lithium soap. One of these might provide the adhesion and slip without the stick.


text:
silyl ketene acetal vs Benzene, which compound has a higher nucleophilicity?
With the use of Lewis acid Aluminium Trichloride, which one would have a stronger nucleophilicity? Silyl Ketene Acetal (CAS:  31469-15-5) or a phenyl group?

Reacting with what electrophile?
In general the silyl ketene acetal will be more reactive, after all the Mukaiyama reaction can be run in Toluene  Mukaiyama Reaction wikipedia page

α,β-unsaturated carbonyl.
yeah I think SKA will be more reactive. Thank you for sharing.

If you are happy with the answer please accept it


text:
Can cis-trans isomers have same InChI?
From all we know only tautomers should get the same InChI key, or, in case of the second block, if there are lots of stereo centers. However these cases are just collisions, they still have different InChI. I have found a class of molecule pairs, with four real-world examples in ChEBI and PubChem, where both web input pages of ChEBI and PubChem generate identical InChI and keys, despite there being a different E-Z configuration on a double bond. And these are also entries in the said databases. My question:
Is this normal/expected? Is it a bug in both software, which part of it?
The pairs:

5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)-1-benzopyran-1-ium-3-yl 6-O-[(2Z)-3-(3,4-dihydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside and 5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)-1-benzopyran-1-ium-3-yl 6-O-[(2E)-3-(3,4-dihydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside

2-(3,4-dihydroxy-5-methoxyphenyl)-5,7-dihydroxychromenium-3-yl 6-O-[(2Z)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside and 2-(3,4-dihydroxy-5-methoxyphenyl)-5,7-dihydroxychromenium-3-yl 6-O-[(2E)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside

5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)chromenium-3-yl 6-O-[(2Z)-3-(4-hydroxyphenyl)prop-2-enoyl]--β-D-glucopyranoside and 5,7-dihydroxy-2-(4-hydroxy-3,5-dimethoxyphenyl)chromenium-3-yl 6-O-[(2E)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside

5,7-dihydroxy-2-(4-hydroxy-3-methoxyphenyl)chromenium-3-yl 6-O-[(2Z)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucopyranoside and 5,7-dihydroxy-2-(4-hydroxy-3-methoxyphenyl)chromenium-3-yl 6-O-[(2E)-3-(4-hydroxyphenyl)prop-2-enoyl]-β-D-glucoside-glucopyranoside

Addendum: I also confirmed with two different SDF files that the current inchi-1 Linux binary generates the same InChi.
P.S. This is a pair of minimal molecules:

Both have the InChi InChI=1S/C11H10O3/c1-14-8-11(13)7-4-9-2-5-10(12)6-3-9/h2-8H,1H3/p+1. If the methyl is removed, leaving a =OH+ group at the end, both isomers get different InChi.

Observation: I started with the .sdf of all four pairs and converted them into .smi (OpenBabel 3.1.1, as repackaged for Debian 11/bullseye, branch testing).  Pairwise, the SMILES are different.  Note, converting the .sdf  into .inchi (again same OpenBabel version),  e.g. by obabel -isdf 2A_ChEBI_75708.sdf -oinchi -O 2A.inchi yielded this warning information: *** Open Babel Warning  in InChI code   petunidin 3-O-(6-O-(Z)-4-coumaroyl-beta-D-glucoside) :Proton(s) added/removed 1 molecule converted.  (Then, pair wise, inchis equate each other.).  Maybe there is some sanitizing issue.

I also get Warning (Proton(s) added/removed) structure #1 from INCHI-1. Do all these software (web input at ChEBI, PubChem, OpenBabel) use the same InChi library?

Indeed, it could be that there is one reference implementation with a license permissive enough to be copy-pasted into OpenBabel (perhaps a question for Geoff Hutchison?).  Could be this strives RDKit, too.

I vaguely recall a presentation (either OpenBabel, or RDKit UGM) about the difficulties to represent and later sanitize the structure representation as strings.  But I think, this was more around metal organic compounds (their sometimes less classical bonding), and  tautormerism (e.g., 2-hydroxy pyridine vs. 2-pyrididone).  From there I got the term of «sanitizing strings».

I think there is the possibility that this effect is caused by a bug in the tautomer code, in that the double bond is replaced by a single bond, because the tautomer code thinks this is an enol. One hint is that the effect goes away if you remove the OH-groups on the dihydroxyphenyl moiety in the 4 pairs. But the oxonium also plays a role, as it is necessary for this.

@RalfStephan : I think you're right: the issue is very likely to be the OH in para to the vinyl substituent. That is common to all your examples. I don't think the oxonium matters. If the software generates a quinone-like tautomer, it will indeed make an enol on the original enone side, probably losing the stereochemistry. I had a similar issue at work, and I had to change the settings of the Inchikey calculation, to avoid having unwanted loss of stereochemistry. So in essence my guess is: long-range tautomerism forming a quinone. Easy to test: draw the quinone and calculate its Inchi.

Here's the SMILES I mean: C1=CC(=O)C=C/C/1=C\C=C(/O)\C=[OH0+]C

Professor Jonathan M Goodman gave an explanation on the InChi mailing list:
https://sourceforge.net/p/inchi/mailman/inchi-discuss/thread/B7D99356-8B55-43BE-8C2A-B8CC3EB37799%40cam.ac.uk/#msg37289766
In summary, the given class of molecules represents a special case where the current InChi algorithm (1.06) produces less than useful results in order to being able to handle all other cases satifyingly. Prof Goodman hints at work being done in organometallics which may affect this behaviour. So, it's not a bug in the implementation of the algorithm, it is the algorithm itself that needs improvement.
The linked message also contains a smaller minimal case, confirming that it is the hetero-ion together with the keto-enol configuration causing the phenomenon.

Another aspect of the side effects of the tautomerism code is that structures are changed when converting. I wanted to convert the europinidine structure COc1c(O)c(O)[c]c(-c2[o+]c3c(c(OC)[c]c(O)[c]3)[c]c2O)[c]1 to InChI, this gives InChI=1S/C17H14O7/c1-22-13-5-9(18)6-14-10(13)7-12(20)17(24-14)8-3-11(19)16(21)15(4-8)23-2/h3-7H,1-2H3,(H3-,18,19,20,21)/p+1 which looks like a pseudo-tautomer but I don't believe that change is possible, at all. Of course it breaks matching a naive SMARTS pattern aimed at anthocyanidins...


text:
Where does the residue accumulate in the boiling point composition curve?
My teacher was teaching the boiling point composition curve for fractional distillation and provided the diagram shown below. It shows that the residue starts accumulating at the point on the vaprous line just above the liquidus line (above $χ_{A},χ_{B}$). Note that $χ_{A},χ_{B}$ don't represent the coordinates, they represent the liquid composition of the two components $A$ and $B$. Similarly, $y_{A}, y_{B}$ represent the vapour composition of the vapour in equilibrium with the liquid with composition $χ_{A},χ_{B}$.
But why does residue accumulate at that point? My teacher gave no explanation why.

Let's start with the mixture $x_A$, which is made of 85% A + 15 % B. When heating this liquid, the first bubble vaporized appeared at a temperature just a little above $T'_A$, as given by the horizontal line. And the composition of the vapor is nearly pure $A$, with $1$% $B$. As a consequence, the liquid is loosing some $A$. its volume decreases, but as it does not loose $B$, its composition changes from $15$% $B$ to maybe $20$% $B$. As it has more B, it is necessary to heat the liquid to a higher temperature. The composition of the liquid is a point following the lower curve upwards to the right hand side of the graph. Meanwhile the composition of the vapor slightly changes and contains a little bit of $B$.
When the volume of the liquid has decreased by more than one half, the remaining liquid contains more $B$ than $A$. Its composition is $20$% $A$ and $80$% $B$. The boiling point of this mixture has increased : it is now about half way between the boiling temperatures of the pure $A$ and $B$ stuff. And the vapor gettin out of this liquid can be obtained in drawing a horizontal line from the point $0.8$ on the lower curve. This line cuts the upper line $y$ at a point $y_A$ = $85$%, and $y_B$ = $15$%. It is possible to continue heating the rest of the liquid. The amount of liquid is smaller and smaller, and the vapor has a composition following the upper $y$ curve.
Now, if you have a hot vapor containing the same proportion ($15$% B , $85$% A), and if you coli it down, the first drop of liquid happening at the beginning of the condensation has a composition corresponding the th horizontal line passing at $y_A, y_B$. So this liquid droplet contains $80$% $B$ and $20$% $A$.


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Properties of stilbene isomers
(E)-stilbene may be separated by crystallization from (Z)-stilbene and the side product of the Wittig reaction, triphenylphospine oxide (TPPO), by using absolute ethanol. My question is why? (E)-stilbene and (Z)-stilbene are isomers of each other, why (and how) do they differ in polarity? If both are of low polarity, why is (Z)-stilbene soluble in ethanol, but not (E)-stilbene? Also why is (Z)-stilbene the more polar isomer of the two and has a lower $R_f$ value in TLC?
An initial reasoning of mine was that (Z)-stilbene were more hindered than the (E)-isomer, and thus less stable. Yet, this doesn't make sense to me.

Please pay attention to your spelling: it’s stilbene, not stillbene. You get it correct half the time. Stereochemical descriptors like E and Z should be italicised, placed in parentheses, and connected to the main name by a hyphen: so the correct names are (E)-stilbene and (Z)-stilbene.

Neither E nor Z stilbene are very polar. But their polarities must differ if their geometries are different and they are very different.

With the $\ce{C=C}$ double bond of the stilbenes as geometric reference, you may think of the phenyl rings as substitutents increasing the electron density of the former.  This creates a small, locally constraint, dipole moment.  Dipoles like dipoles differ from a scalar (like e.g., temperature) that they have a direction, which is why I drew little arrows in the simplified representation of either isomer in the illustration below:

In the case of (E)-stilbene, these vectors oppose each other; for symmetry reasons, they may cancel out each other as if there were no vector present.  This contrasts to the case of (Z)-stilbene where the sum of the two vectors does not vanish; symbolized (the doodle is a concept, not at exact scale) by the dashed arrow: there is a non-zero molecular dipole moment.
The drawing of (Z)-stilbene's configuration however does not state (much) about the molecule's energetically preferred conformation(s).  To lessen intramolecular steric hindrance of the hydrogens of the two phenyl rings, the following spatial arrangement is more likely to happen:

Note, regardless how the phenyl rings rotate around the bond between $\ce{C_{ar}-C_{sp^2}}$, the overall (molecular) dipole moment will not be affected from these local variations.
Equally note that to achieve an efficient recrystallization of the three constituents of your mixture, the isomers of stilbene and TPPO, you use a mixture of ethanol/water heated to reflux then allowed to cool slowly to room temperature.  The addition of water is to attenuate the solution capacity of ethanol for each of the three compounds so that one preferentially crystallizes out.  This is why this purification method is called fractional crystallization.

Also worth noting that the factors influencing propensity to crystallise easily are far more complex that just the dipole moment if the isomer. In this case one isomer is much flatter than the other and likely to have far larger pi-pi interactions between neighboring molecules' phenyl rings. Explanations are not going to be simple!

@matt_black True, crystallization of (organic) matter depends on many factors in addition to «just» dipole moments. Rigidity and shape of the molecules equally contribute to this, too. One of the well-known in the field is Kitaigorodski (e.g.,«Molecular Crystals and Molecules», one of the references mentioned here). One may argue about the wording to describe the complexity known and unknown parameters bring to this, yet there may be many ...

Thank you very much @Buttonwood, if you allow me though, in a specific step of this lab, we get trans/cis isomers and the side product TPPO, so basically two solids and a liquid, to get the cis isomer, can we just used filtration to get the liquid which is the cis isomer, and then we are left with two solids, one is polar and the second is not which is easy to seperate because of the difference in polarity?

I assume the specific step you describe is after performing the reaction, extraction against dichloro methane and water, drying the organic phase, and gentle evaporation of dichloro methane.  If so, I would not simply filter off the liquid, even if pure (Z)-stilbene is liquid (at room temperature) while the other two chemicals are solids, because (Z)-stilbene may equally act as a solvent for them.  Recrystallization of this mixture from a bit of ethanol, from boiling down to an ice bath (slow cooling, not crushing precipitation) helps you to concentrate (E)-stilbene.

By means of atom economy, assuming your Wittig is like benzaldehyde + benzyltriphenylphosphonium chloride + a base, it may be more efficient to create a stilbene, e.g., by reaction of benzaldehyde + the Grignard of benzylbromide, and a subsequent dehydration.  (Perhaps refluxing the raw product in toluene with a grain of iodine increases the percentage of (E)-stilbene (if necessary).) Depends, of course, of what is at your disposition, but basically, in [PPh3-CH2-Ph+][Cl-], there is only one Ph in your wanted, but three Ph in the by-product.

Alternatively, you might consider disolving the raw material and «filter» it across a plug of silica. TPPO will not travel well, and thus may be removed by this (relatively) easily. (A prior check with TLC and authentic samples guides you through the selection of the organic solvent.) Then concentrate the filtrate (contains basically both stilbenes only) to dryness, and recrystallize from, say boiling ethanol slowly cooled to r.t.. Keep the flask in the ice bath for completion of crystallization; then wash the crystals of (E)-stilbene filtered off (Buchner) with little of ice-cold ethanol.