text:
pH responsive polymer which is immiscible in water?
Can anyone suggest a pH-responsive polymer, which is immiscible in water (polar solvent) but miscible in oil (e.g vegetable oil)?
I want to study the change in the density of such polymers when they are subjected to changing pH.
Edit:
I'm having a biphasic system of oil & water, and want to study the viscosity changes of oil with time!
For this, I was earlier thinking of having a polymer in the oil phase itself and somehow triggering (changing) its state of polymerization (density) by changing the pH of the underlying aqueous solution! (remember it's a biphasic system of oil & water).
But after reading some more papers and getting help from the underlying comments, I think it's not possible to have a polymer in an aprotic phase and then be able to study its response towards changing pH!
So now I have thought to change the experiment to synthesis a polymer at the biphasic boundary (surface) of the water and oil phase. This polymer must be soluble in the oil phase! Hence the product of interfacial polymerization must diffuse into the oil phase.

pH is a quantity that makes no sense in an aprotic solvent.

And what is that polymer supposed to do? "responsive" could mean it turns green. ?!?

You could've read the very next line, where I talk about density variation with pH.

pH change doesn't mean anything in an aprotic solvent, but I am only talking about pH with respect to the polymer. I understand that reading and understanding is not a part of Chemistry 101, but you might've learned that in grade 1!

There's no such thing as "pH with respect to the polymer". If you're referring to the protonation state of the polymer, then you may [edit] that in. By the way, it's not immediately obvious that your two sentences are linked; my suggestion would be to make that clearer. It could very well be that you are interested in studying the density of polymers that respond to "pH" (or protonation state) in other ways. More clarity usually means you get more chance of a helpful answer; I'm trying to help you here.

Actually I'm not a student of polymer science, but I feel it's a very nice area to explore since it has a huge lot of customization capabilities! I refer to "pH with respect to polymers" because I was recently reading a research paper titles "pH-responsive polymers" DOI: 10.1039/c6py01872f

Please have a look at the edits! Thanks


text:
Unusual Result in Kinetics Lab
The question: In the lab described below, why did the decolorization reaction described, which should be first order in each of the two reactants, go most slowly when the reactants were most concentrated?
As part of the introduction to our kinetics unit, I asked my students for factors that they thought would influence the amount of time it takes a reaction to occur.  Not surprisingly, they came up with a variety of things, including temperature and the concentration of reactants.  I set up a simple lab to compare temperature and concentration conditions based on the reaction of crystal violet with sodium hydroxide.  Later in the unit we will come back to the reaction and analyze its kinetics more carefully, but for the moment, I just wanted my students to do a more basic analysis.
The students did trials for 4 conditions, and each involved the same quantity of reactants: 25 ml of 0.10 mM crystal violet and 25 ml of 1.0M sodium hydroxide.  The reaction was performed by adding the crystal violet to a flask, adding water to dilute, adding the sodium hydroxide to the flask, stirring, and timing the reaction until it no longer showed visible purple.  (Yes, that's a sloppy endpoint, but it was sufficient for this activity's purposes.)
Trial 1: 25 ml crystal violet, 150 ml room temp distilled water, 25 ml sodium hydroxide (a 1:8 dilution of stock reactants)
Trial 2: 25 ml crystal violet, 150 ml hot distilled water, 25 ml sodium hydroxide
Trial 3: 25 ml crystal violet, 50 ml room temp distilled water, 25 ml sodium hydroxide (a 1:4 dilution of stock reactants)
Trial 4: 25 ml crystal violet, 0 ml distilled water, 25 ml sodium hydroxide (a 1:2 dilution of stock reactants)
The results:

The hot reaction went much faster than the cold.
The 1:4 dilution went faster than the 1:8 dilution.  (~6 minutes vs 9 minutes, respectively)
The 1:2 dilution went much, much more slowly than any other. (~16 minutes)

Result #3 is the anomaly.  It involved the most concentrated solutions and so should have been the fastest of the 3 room temperature reactions, but it was far and away the slowest.
Things I've considered:

The reaction isn't significantly endo- or exothermic, so the solution
didn't have a different temperature due to its minimal mass.  (All of
the reactions except for the one with hot water added were within a
couple of degrees of each other.)
The students didn't make a mistake in execution - this was observed by multiple groups over multiple trials.
The difference was not due to the sloppy endpoint of the reaction.  Though the reduced volume meant that a given amount of remaining crystal violet would be most purple in this trial, the difference in color was enormous and obvious.
The reverse reaction never becomes significant with such a massive excess of sodium hydroxide.
New idea: The unusual result is the only trial without distilled water - maybe the water is contaminated or something like that, but in a way that is promoting the reaction?  I was excited when I thought of this (I actually popped up out of a dead sleep with it in my head,) but it seems unlikely, I used the exact same container of water an hour earlier to make each of the stock solutions...

I've wracked my brain, but I can't figure out a reasonable explanation for what we observed.  Any suggestions would be welcome.

Have you tried with still more concentrated solution ? For example with no dilution at all ? Or with 2.0 M or 4.0 M solution ?

@Maurice The unusual result is essentially no dilution - the only dilution that occurs is the mixing of the two stock solutions at in equal volume, so that the reaction mixture is half of the stock's concentration for each.  I didn't run the trial with higher concentrations.

Don't you think there is a different reaction or a new equilibrium at high concentrations of NaOH ?

@Maurice It's entirely possible.  I will test on Monday.

I hadn't realized that I left this hanging.  I did run a series of tests the next week and could not replicate the anomalous result.  I honestly have no idea what happened - it must have been a contaminant of some sort.  I selected @Andrew 's answer for its good explanation of the general experiment and analysis of the situation.

Your analysis of the expectation is correct, and we can look at it more quantitatively in this way:
Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of:
$$-\frac{d[\ce{CV}]}{dt}=k[\ce{HO-}][\ce{CV}]\approx k[\ce{HO-}]_{initial}[\ce{CV}]$$
For simplicity, we'll define a $k'$ as $k'=k[\ce{HO-}]_{initial}$ and treat the reaction as pseudo-first-order.
The half-life of a first-order reaction is given by $t_{1/2}\approx \frac{0.7}{k}$, so for the concentrated reaction we have $t_{1/2}=\frac{0.7}{k'}$.
For the 1:4 reaction, the initial hydroxide concentration is halved, so we have a pseudo-first-order rate constant of $0.5k'$ and $t_{1/2}=\frac{0.7}{0.5k'}$ or 2 times the half-life of the concentrated reaction.
Likewise, the half-life of the 1:8 reaction is twice that of the 1:4 reaction and four times that of the 1:2 reaction, so it's $t_{1/2}=\frac{0.7}{0.25k'}$.
Let's say that the concentrated reaction needs 6 half-lives for the concentration to get to the point where the color disappears. Since the 1:4 reaction starts at half the initial CV concentration, it only needs 5 half-lives. But the half-lives are each twice as long, so it should take a time equivalent to 10 half-lives of the concentrated reaction, so 10/6=1.7 times as long.
For the 1:8 reaction, the same analysis says that it will take 4 half-lives, but each is 4 times as long as that of the 1:2 reaction, so it takes 16 equivalent half lives, meaning 16/6=2.7 times as long.
The only instance in which the more dilute reaction will be faster is if the concentrated reaction requires less than two half-lives to be observed to be complete. That can't be the case here or there would be no color at the start of the 1:8 reaction.
So we must look for an alternative explanation for your observations. 0.5 M NaOH is a very concentrated solution, and it is possible that other reactions are occurring besides the intended one. To test that, I would follow Maurice's suggestion and run the reaction at a wider range of base concentrations. It would also be preferable to have the same concentration of CV in every reaction, so only one variable is involved. To do this, used a fixed volume of (NaOH + water) in each reaction.
UPDATE: This paper1 reports doing this reaction at much lower hydroxide ion concentrations and also notes that the reaction rate decreases with increasing ionic strength, which could be at least a partial explanation for why your observed rate at high concentrations is lower than expected.
1 Felix, L.D. Kinetic Study of the Discoloration of Crystal Violet Dye in Sodium Hydroxide Medium. J Chem Appl Chem Eng Vol: 2 Issue: 1

Yes, typically the reaction is done in a spectrometer cell over the course of an hour or so.  At the end the solution is still purple.  In order to get decolorization in a reasonable time I raised the [NaOH] substantially.  It is usually treated as pseudo-first order in order to find the order of CV and the concentration of NaOH is varied to find its order.  We'll be doing that later in the unit. (This is the most common kinetics lab that high schoolers do these days - I was trying to expand its use as an intro to the unit.)  The ionic strength idea sounds promising.  I'll test on Monday.

All the reaction have different concerntration at start . As you have taken so much excess NaOH it would remain constant. We could look it as you have a solution of 50 ml with 0.05 mM violet red and 0.5M NaOH . You dilute the N times and then to have to compare rates. Threotically we could find that but when we say discolouration in the solution does it mean the reaction has been 100% complete? Actually there is a particular concerntration at which we will see discolouration . Lets say that concerntration is 0.001mM of violet red.
Time taken in completion of reaction (assuming the reaction is 1st order with respect to violet red and also note that pseudo rate constant depends on intial concertration of NaOH)= $\frac{ln(\frac{intial\,contretaion}{final\,concentration})}{k(concentration\,of\,NaOH)}$ = $\frac{ln(\frac{0.05*1000}{n})}{(k*0.5/n)}$=$\frac{2n*ln(\frac{50}{n})}{k}$=F(n)
Here k is the kinetic rate of the 2nd order reaction.
F'(n) =  2ln($\frac{50}{n}$) -2
As you can the derivative of F is greater than zero till 50/e and after that less than zero
(You could assume the complete reaction concerntration be anything but assume it be constant)
So whatever concentration you choose the function comes out that time of completion of reaction increases(F'(n) may come out be negative at the very start though) till a particular value of n and after that decrease as n increase(n is directly proportional to dilution)
If we assume the kinetic 2nd order law to be completely valid we can find the concerntration of violet red at which our eyes think the reaction has been completed.Yeah for that you will need to conduct many experiments.
The result is not unusual and its perfectly fine. If you find the rate of the reaction at intial points I m sure you will find it to be greatest in most concerntrated solution.

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Is it ok or needs more modifications?

I'm struggling to see how this explains the event.  In each case we used the same number of moles of reactant.  The order of the reaction is first in NaOH and first in CV, but even if I'm wrong about that and it's one or the other, the most concentrated solution should react fastest at any particular number of remaining CV moles.  If it's always fastest and the reactions all consume equal quantities of CV, how could it do anything other than finish first?

Exactly what i m saying is that rate depends on concerntration , and from a concerntration(which depends on how much we diluted the intial mixture) we have to reach a particular concerntration. If u add less water more moles need to react because u have to reach one particular concerntration which means if less water implies less moles left at the end(intial moles were the same but final moles are not same so we cant say anything about time).

Discolouration does not mean the reaction is 100 percent complete but it means it has reached a stage where your eyes cannot detect the colour which will happen at a fixed concerntration.U r assuming 100 percent reaction.If the reaction was 100 percent complete it would infinite time not mere 10 or 15 minites

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text:
Why is there a big jump in first ionisation energy between sodium and potassium?
I know the trend in group 1 is that ionisation energy decreases down the group due to an increase in atomic radius and more energy levels are added so more shielding, but I'm not sure why there's such a big jump between sodium and potassium.
My textbook doesn't explain it very clearly for me but says the d orbital doesn't shield the outer electrons very well explaining why K to Cs has similar values, but potassium's 3d orbital is not yet filled, so I'm not sure why it's value is so much lower.
I'll put the values from my textbook in to help explain what I mean.
Li=513, Na=496, K=419, Rb=403, Cs=376. (in kJ/mol)
The gap is much more obvious when you look at a graph.
Sorry if I haven't made it very clear but any help would be appreciated. Thanks

Wikipedia has plotted the graph you mentioned above (in terms of eV) and it seems normal. Also check out This graph

We also have Rb above the average of K and Cs.  The stepwise nature of the graph seems to correlate with the subshells filled by the preceding period: Li -> s, Na and K -> s and p, Rb and Cs -> s, p and d.

@aniruddhadeb the Wikipedia graph does not include lithium.  If that element were included the drop from Na to K would look like a very sharp step because Li is only a little above Na.  This is what the OP is asking, and there is a similar though less pronounced effect centered around Rb.

For Alkali metals, the First Ionisation Energy is equal to the work function $ (\phi) $ of the metal. As metallic character increases down the group, thus $\displaystyle  \phi_{Li} > \phi_{Na} > \phi_{K} > \phi_{Rb} > \phi_{Cs} $. Hence Ionisation Energies are different.

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text:
Oxide caused by running a saucepan with aluminium handle through the dishwasher
Having run a saucepan with an aluminium handle through the dishwasher, the handle is producing a dark grey dusty residue, which I assume is an oxide of aluminium. The dishwasher was run with a typical dishwashing detergent as well as some drying agent (potentially).
The residue can be scoured off, leaving the shiny aluminium below. Having been scoured, the oxide doesn't seem to return.
What reaction is responsible for this? Obviously the high temperatures of the dishwasher is one factor, but what are the others?

Interesting question, but I don't think its oxide, as aluminium has a white oxide

@HahaHahaha pure aluminium oxide is white in colour - Most probably the compound is aluminium oxide with some other impurities, giving it a grey colour. It could also be sodium aluminate, as most dish washing liquids contain sodium Lauryl sulphate or some sodium based foaming agent

Aluminium used for cooking plate is not pure aluminium. It is an alloy, which contains Manganese, Silicium or other metals. If aluminium is oxidized, the impurities will not be oxidized and remain in the aluminium oxide layer. This makes a grey color.

@Andrew, aluminum (well, the Al2O3 coat) is attacked by alkali much more quickly than by most acids (except HF). Machine dish-washing detergents are highly alkaline, containing alkali metal carbonates and hypochlorites to saponify fats.

Hypochlorites are added to remove food stains/sterilize rather than to saponify fats. Hypochlorite is such a bad choice anyway due to its corrosive nature on heating elements and metals. Hope they are completely replaced by oxygen bleaches.

@DrMoishePippik - HF does not attack Al2O3 particularly well, requiring a boiling HF solution to make much of a dent. I can leave a piece of sapphire in room-temperature HF all day long without noticing anything happening.

@JonCuster, the thin, somewhat porous layer of Al2O3 on aluminum exposed to air is attacked much more rapidly... though I admittedly don't work much with nasty HF. https://blogs.sciencemag.org/pipeline/archives/2004/03/03/things_i_wont_touch_1

Does this answer your question? Why dishwashers are corrosive to aluminum?


text:
In reversible processes in dynamic equilibrium why do we only see the products and not the reactants?
If both forward and backward reaction rates are equal why is it that only the products are visible to a naked eye?

Not really. It just depends on the reaction conditions. Consider the following reaction:
$$\ce{N2O4 <=> 2NO2}$$
Here’s a picture showing this equilibrium in action:

One of the reasons you can’t see the reactant here is because it is colourless, but this gives a pretty good example regarding visual equilibrium changes.
this is a good answer giving another example of a equilibrium that can be seen visually.

Why do you think so ?
As I have missed the other answer, this is just a complement to it.
In the equilibrium:
$$\ce{ 2 NO2 <=> N2O4}$$
both are reactants and products at the same time.
If you formally consider products on the right side, then you see by the naked eyes the red-brown reactant.
What is or is not seen depends solely on the properties and amount/concentration of the substance.


text:
Lattice enthalpy and ionic character
As we move down in sulphates of group 2 their solubility decreases because there is a drastic decrease in hydration enthalpy as compared to decrease in lattice enthalpy.
Does the decrease in lattice enthalpy mean decrease in ionic character?
But according to Fajans' rules the ionic character should increase or covalent character must decrease  down the group.
Kindly explain this.

AFAIK there is no such thing as "fazan rule", so I corrected it to "Fajans' rules".

Lattice enthalpy is given by the Born-Lande equation:
$${\displaystyle E=-{\frac {N_{A}Mz^{+}z^{-}e^{2}}{4\pi \epsilon _{0}r_{0}}}\left(1-{\frac {1}{n}}\right)}$$
This seems very complicated, but removing some constants, we can reduce it to say:
$$E \propto \frac {z^+z^-}{r_0}$$
here, $z^+$ and $z^-$ are charge on cation and anion and $r_0$ is distance to closest ion.
Coming to your question, In the sulphate series of Alkaline earth metals, the charge remains the same for all members: what changes is the distance to closest ion, which is smaller for small ions such as $\ce{Be^2+}$, thereby causing their lattice energies to be high.
The decrease in lattice enthalpy may not neccessarily mean a decrease in ionic character. If two lattices have similar values of $r_0$, then a higher lattice enthalpy for one of them may imply larger ionic character due to large charges on either of the atoms.

Lattice energy is not at all related to ionic character.
Ionic character is given by Fajans’ rule, whereas lattice enthalpy depends upon a whole set of reactions involved in the Born–Haber cycle. For instance:

Compound
Lattice energy (kJ/mol) [Ref 1]
Ionic character (%) [Ref 2]

KF
801
82.5

LiF
1009
86.7

KI
632
73.7

LiI
734
65.0


text:
Counting the number of conjugated pi bonds in a complex aromatic molecule
So for multiple bonds to be conjugated, they must be in an alternating
double bond – single bond – double bond, etc arrangement. However, do they need to be continuous? In my textbook, they count the number of conjugated multiple bonds like this:

However, for other molecules such as 9,10-Bis(phenylethynyl)anthracene :

Would I count the number of conjugated multiple bonds as a total of 13 or as 7? In other words, do the bonds need to be continuous to be counted as conjugated pi bonds?

The total number of conjugated double bonds is 13: these bonds form groups of 7,3,3 conjugated bonds each in the anthracene and benzene rings respectively.

The problem is in the geometry of the molecule. Are the two external phenyl groups coplanar with the anthracen central group ?

There's 15 conjugated pi bonds there and two perpendicular in triple bonds. What is the point of this question though? Illustration from book is for specific context - dependence of color of compounds with linear group of bonds. Other compounds behave quite differently.


text:
Is the solution isotonic, hypertonic, or hypotonic?
The problem is shown below. I don't understand how can the tonicity of a solution be determined if the solution has multiple solutes dissolved in it.

Do I take the total sum of the mass of solutes and divide it by the volume of the solution in both outside and inside of the cell (total concentration) and determine the osmosis direction based on whichever is greater? Or should it be done in another way?
If it could be of use, here are the questions that were asked below the picture and their answers. This question was taken out of a test-bank for the 8th edition of the Campbell biology textbook.

Osmotic pressure for non-electrolytic solutes is given by
$$\pi = CRT$$
where $C$ is the effective concentration of all the solutes. In our case, with multiple solutes, we simply add all their concentrations to obtain the effective concentration. This gives us
$$
\begin{align}
\pi_\mathrm{cell} &= 0.05RT\\
\pi_\mathrm{environment} &= 0.03RT
\end{align}
$$
Since $\pi_\mathrm{environment} < \pi_\mathrm{cell},$ The environment solution is hypotonic.

Thanks for the answer, but the problem states that the membrane is permeable to fructose and glucose but not to sucrose; The 2 simple sugars should balance their concentration by diffusion only, and sucrose's is balanced by osmosis. Therefore, I speculate that the tonicity of the solution should be determined based on only the solutes which cannot pass through the membrane; Needless to say, osmosis wouldn't occur if the membrane was permeable to all solutes. Could this be the case?


text:
How to manage complex equilibria when studying a mecanism?
Consider a complex multistep reaction mechanism with several intermediates. There are multiple possible pathways as well as equilibria in solution. One wants to study the reaction mechanism using computational chemistry.
It is easy to calculate the $\Delta G$ and $\Delta G^\ddagger$ of individual reactions, but how do you take in consideration all the possible equilibria?
For example, consider an aldol reaction in presence of water (taken from Wikipedia):

Acetone can be either as a ketone, a protonated ketone, an enol or an enolate. This means that there are 16 possible pairs (although only 4 which make sense). These will yield different intermediates, which can dehydrate in different ways (ex. by protonation of the hydroxyde first or not, through a ketone or an enol or an enolate...)
In this case, many possibilities could be eliminated as too high in energy just by general knowledge of chemistry, but if you don't know enough about the system, how do you tackle this problem? How do you handle all the equilibria possible?

Unfortunately, you'll have to determine whether or not some combinations can be eliminated. Equilibria are on a potential energy surface also only points with a low barrier connecting them. If the paths following out of these yield different products, which are similarly stable, you'll have to consider both. I suggest you look for computed reaction mechanisms and try following their argumentation there. The unsatisfactory answer is: You have to calculate all paths, before you can determine the most likely one.

If you know what the final equilibrium products are or may be, and you know how much mass you start with, either as elements or molecules, you can calculate the Gibbs energy of the products as a function of (unknown) concentrations (assuming constant temperature and pressure) and then minimize the Gibbs energy with respect to those concentrations. You have to constrain the solution to conserve all of the elements that you began with, and constrain it to have all concentrations be non-negative.


text:
Chemical balance problem
The ammonium chloride dissociates at 317 ° C according to the reaction:
$$\ce{NH4Cl (s) -> HCl (g) + NH3 (g)}$$
reaching a total pressure of 500 mm Hg in equilibrium in a closed container. What will be the total pressure in the equilibrium when a gaseous hydrogen chloride at 500 mm Hg of pressure is introduced into a closed vessel together with the ammonium chloride?
What I have calculated has been:
In the first balance.
$$\ce{Kp=pHCl*pNH3}$$
$$\ce{pHCl=pNH3=P/2}$$
$$\ce{Kp=(p^2/4)=(500/760)^2/4=0,108}$$
And in the second balance, when is added 500 mm Hg of HCl
$$\ce{Kp=p(2p)^2=4p^3}$$
$$\ce{pT=p + 2p=3p}$$
And now I don't know what else to do.
Sorry if the formulas are not well placed

In the second balance, where did the $\rm2p$ come from?

In the beginning, both pressures of $\ce{NH_3}$ and $\ce{HCl}$ are $250$ Torr = $0.329$ atm. So that the equilibrium constant is $K_p$ = ($0.329$ atm$)^2$ = $0.108$ atm$^2$
After addition of $500$ Torr $\ce{HCl}$, the total pressure of $\ce{HCl}$ should be $750$ Torr = $0.986$ atm. But it is not so high, as a small amount $x$ of $\ce{HCl}$ and $\ce{NH_3}$ must react to maintain the equilibrium constant. One may write that :
$$K_p = (0.329 - x)(0.986 - x)= 0.108 \space \text{atm}^2$$
The solution of this equation is :  $x = 0.18$. So the final pressures of $\ce{HCl}$ and $\ce{NH_3}$ are :
$$p(\ce{NH_3}) = 0.329 - 0.108 = 0.211 \space \text{atm}$$
$$p(\ce{HCl}) = 0.986 - 0.108 = 0.878 \space \text{atm}$$
The total pressure is $0.211 + 0.878 = 1.089 \space \text{atm}$

Thank you very much, your response has been very helpful, the only thing that the value of x gives 0.193.


text:
Substituted Diene Preparation from Aniline / Novel Cyclic Imine Preparation
I have recently been studying the following mechanism recommended by another user from the following DOI: https://doi.org/10.1016/j.tet.2004.06.026. In this paper, the following mechanism is proposed (I am only focusing on  the proposed mechanism in EtOH)

Please note that the author of the paper did not show the intermediate prior to keto-enol tautomerization. This provides a valuable insight into plausible reactions that can be derived from this mechanism. Below is the intermediate prior to keto-enol tautomerization (to restore aromaticity).

It is clear that the α-hydrogens next to the carbonyl are crucial in restoring aromaticity. If, however, these hydrogens were replaced with alkyl groups, the final product would be quite different, as outlined in the following reaction (DOI: https://doi.org/10.1002/ejoc.200701009):

While the author of the paper did not include all of the steps of the reaction, it is clear, that the alkyl groups are preventing the restoration of aromiticity via tautomerization. I drew out a new mechanism that could be applied to substances other than phloroglucinol in one of my previous questions. I have since confirmed this using SciFinder. Here is the general mechanism:

It is clear that the driving force behind this type of reaction is the lone pair on the oxygen atom which can allow not very powerful electrophiles to attack the aromatic ring in a concerted fashion in which all charged intermediates are prevented by the formation of the carbonyl. Now to my question. Does this mechanism seem applicable to deprotonated aniline? This would theoretically result in the formation of a primary imine in accordance to the following theoretical mechanism I have proposed:

I would have expected to find a reaction very similar to this on SciFinder; however, I did not come across anything involving substituted anilines and alkyl halides directly. I did, however, find a very similar reaction that involved substituted nitrosobenzene instead of aniline and Grignard reagents. A very similar product is also formed with the only difference being an oxime is formed instead of an imine. Here is the DOI: https://doi.org/10.1246/bcsj.48.3727. The reaction is shown below.

The paper does not go into much detail about the specific mechanism about this step in the reaction (mostly steps afterward). It does not seem to me, however, that this reaction is following the same mechanism as my proposed reaction. The Grignard reagent may help deprotonate the nitrosobenzene; however, in this case, there is no longer an electrophile, so so the mechanism is probably not the same. Is there any obvious reason why this newly proposed mechanism would not work (I do not think polymerization of the product will be an issue), as I would have expected this to already have been attempted, however, it has not.

Your DOI link didn't work for me. Here is what I sent you: doi.org/10.1016/j.tet.2004.06.026 Structure 1c above has the double bond in the wrong position. It is in conjugation with the aromatic ring. (See the earlier 1c in the paper). 1d may arise by direct SN2' alkylation. In the Experimental, they evaporate solvent in vacuo. If they heated the flask, 1d may arise by Claisen rearrangement of the SN2 O-alkylation product.

Oh, I did not see that Claisen rearrangement possibility in the last question... sorry.

@user55119: The reference is: Tetrahedron 2004, 60(32), 6807-6812 (https://doi.org/10.1002/ejoc.200701009). Only abstract is available through internet, if you are not in campus computer.

@Mathew: I  was referring to the link in the second line of question that appears to have been updated. I have full access.

I now see the Claisen rearrangement! I am curious, however, because it is the second elementary step of the reaction wheater or not the reaction would proceed forwards without the Claisen rearrangement. Maybe it would be too thermodynamically unfavorable. I included a better example of the mechanism from the third picture, which does not follow a Claisen rearrangement but seems to have a similar mechanism. The only issue is that a stabilizing factor could be the keto-enol tautomerization, so I'm still not sure wheater or not this could be applied to substituted phenol and aniline as I proposed


text:
Orbital overlap in E2

In order to overlap, orbital lobes must have the same sign of the wavefunction i.e. they must have the right symmetry. In case 2 in the image above, can therefore E2 elimination not occur?

What is the difference between your case 1 and case 2? I mean, it's clear that you've changed your colouring scheme; but the molecule can't see and doesn't know about your colouring scheme. What about the molecule is different such that you coloured it differently?

I thought that the molecule knows about the sign of the wavefunction at each orbital lobe and only shaded lobes can overlap with shaded lobes and similarly for unshaded lobes. Are you saying that in general we can flip the sign of the wavefunction of each lobe in any given orbital independently of all other orbitals in the molecule?

Ok, I think I might have used a bad phrasing. A better analogy is probably the MOs of H2. If you "shade in" both 1s orbitals with the same colour (sign), then the bonding MO is just $\mathrm{1s_A + 1s_B}$, and the antibonding is $\mathrm{1s_a - 1s_B}$. But if you choose to shade them differently, it doesn't change the fact that you have a bonding MO and an antibonding MO. It doesn't even change the final form of the MOs. It only changes the sign that you write: bonding = $\mathrm{1s_a - 1s_B}$, antibonding = $\mathrm{1s_a + 1s_B}$.

So, are you saying that in general we can flip the sign of the wavefunction of each lobe in any given orbital independently of all other orbitals in the molecule?

Yes, pretty much, although I'm not beating around the bush just to be annoying. I care more that you understand why this is the case, rather than just knowing that it is the case.

Sorry, I can't speculate at all why this is allowed. Could you please explain why?

That goes back to my second comment: you are mixing orbitals to form other orbitals (in H2 you have 1s & 1s -> σ & σ*, here we have σ & σ* -> π & other bonds). This mixing takes place regardless of the sign of the orbital you ascribe: it's just that if you flip your relative orbital signs, then your mathematical description of the mixing needs a minus sign instead of a plus (or the other way round). Nothing changes from the molecule's point of view.


text:
Polymers which are miscible in oil
Can someone suggest me a list of polymers or some paper where I may find polymers that are soluble in oils like silicon oil, mustard oil, coconut oil, bean oil or petroleum ether, etc.?

There are many types of polymers, from elemental sulfur to  polyborazylenes. Be more specific.

Can you suggest to me how to look for these types? I don't want to single out some properties, since my research demands a holistic picture of things, rather than just simply looking at a certain set of polymers!
I would be very grateful if you would share some of your knowledge with me! Thank you

Plastics Design Library series include tabulated parameters for the variety of polymers.
One of such parameters is a PDL number rating from 0 to 9:
0: solvent dissolved disintegrated
1: decomposition
2: severe distortion; oxidizer and plasticizer deteriorated
…
9: highest resistance, no change
In the table below I assembled data on thermoplastics chemically unstable towards various oils with PDL numbers below 3:
$$
\begin{array}{lrcll}
\hline
\textbf{Polymer} & \boldsymbol{T/\pu{°C}} & \textbf{PDL} & \textbf{Oil} & \textbf{Resistance Note} \\
\hline
\text{CAB} & 20 & 0 & \text{spearmint oil} & \text{Limited Resistance} \\
\hline
\text{LDPE} & 20 & 1 & \text{aniseed oil} & \text{Not resistant; tensile strength at yield and elongation at break greatly reduced} \\
\hline
\text{HDPE} & 20 & 2 & \text{camphor oil} & \text{Not resistant} \\
\hline
\text{PP} & 100 & 2 & \text{two-stroke engine oils} & \text{Unsatisfactory/severe effect} \\
  & 20 & 2 & \text{turpentine oil} & \text{Not resistant} \\
  & 100 & 2 & \text{vaseline oil} & \text{Unsatisfactory/severe effect} \\
\hline
\text{LLDPE} & 20 & 1 & \text{aniseed oil} & \text{Not resistant; tensile strength at yield and elongation at break greatly reduced} \\
\hline
\text{PS} & 22 & 2 & \text{clove oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{lemon peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{orange peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{pine needle oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{spearmint oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
\hline
\text{SAN} & 23 & 2 & \text{citronella oil} & \text{Severe attack; softened in few hrs} \\
  & 23 & 2 & \text{clove oil} & \text{Severe attack; softened in few hrs} \\
  & 22 & 2 & \text{lemon peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 52 & 2 & \text{pine needle oil} & \text{Severe attack; softened in few hrs} \\
  & 23 & 2 & \text{spearmint oil} & \text{Severe attack; softened in few hrs} \\
\hline
\text{ABS} & 52 & 2 & \text{pine needle oil} & \text{Severe attack; softened in few hrs}  \\
  & 23 & 2 & \text{spearmint oil} & \text{Severe attack; softened in few hrs} \\
\hline
\end{array}
$$
Acronyms
$$
\begin{array}{ll}
\text{CAB} & \text{Cellulose Acetate Butyrate} \\
\text{LDPE} & \text{Low Density Polyethylene} \\
\text{HDPE} & \text{Polyethylene} \\
\text{PP} & \text{Polypropylene} \\
\text{LLDPE} & \text{Linear Low Density Polyethylene} \\
\text{PS} & \text{Polystyrene} \\
\text{SAN} & \text{Styrene Acrylonitrile Copolymer} \\
\text{ABS} & \text{Acrylonitrile Butadiene Styrene}
\end{array}
$$
References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; Plastics Design Library; William Andrew: Norwich, N.Y, 2012.

I'm very thankful for your guidance! I will surely be referring to this book from now onwards! You may want to look at this question, which I did post before, in case you need some more insights towards the research I'm working on!
https://chemistry.stackexchange.com/questions/126078/ph-responsive-polymer-which-is-immiscible-in-water

Hey, I find an Alphabetical List of Exposure Media at the end of the book. What should be my strategy to detect which of these media are immiscible in water and will tend to float on water (i.e lower density)? Does this follow any trend, or are they only a few which would follow any such rule, or else I have to look at all of them individually? Thank you for all the help!

@AnshumanSinha I suspect you have to look it up (miscibility) or even determine the density yourself experimentally for your samples as it might slightly deviate from one manufacturer to another.


text:
Does the prefix Ortho have some significance?
Are orthoboric acid and boric acid one and the same thing or we can not write H3BO3 as B(OH)3?

The title and the body are poorly related. The question in the body has already been answered: Why is boric acid written as B(OH)3 and not H3BO3?; Why is arsenous acid denoted H3AsO3?; Basicity of boric acid …

…and the question from the title is addressed here: Origin of “ortho”, “meta”, “para”; What do the prefixes meta, ortho, pyro mean in inorganic chemistry?.

Well what I mean is that my professor said this specifically that orthoboric acid is H3BO3 and boric acid is B(OH)3...so is that right?

The reference given by Andselisk does not bring an answer to the question about the origin of the prefix "ortho" in inorganic nomenclature.
The prefix "ortho" is added before the name of an acid $H_aX_bO_c$ if first $b = 1$ and if it corresponds to the highest possible number of $OH$ groups attached to the central X atom. For example $H_4SiO_4$ is the orthosilicic acid, $H_3BO_3$ is the orthoboric acid. $H_4CO_4$ is the orthocarbonic acid (which only exists in esters). $H_3PO_4$ is the orthophosphoric acid (where the prefix is often skipped). $H_3AsO_4$ is the orthoarsenic acid. The use of the prefix does not depend on the oxidation number of $X$. For example, if $P$ and $As$ are taken with a lower oxidation number, one obtains $H_3PO_3$ as the orthophosphorous acid, and $H_3AsO_3$ as the orthoarsenous acid.
The prefix "meta" is added to the name of an acid, if its formula is formed by removing one molecule $H_2O$ from the formula of the ortho-acid. $H_3BO_3$ - $H_2O$ = $HBO_2$ and this is the metaboric acid. $H_3PO_4$ - $H_2O$ = $HPO_3$ which  is the metaphosphoric acid. Repeating this process, one gets :$H_2CO_3$ is the metacarbonic acid (often written without the prefix meta). $H_2SiO_3$ is the metasilicic acid. As with ortho-acids, the use of "meta-"  does not depend of the oxidation number. $HPO_2$ is the metaphosphorous acid. $HAsO_2$ is the metaarsenous acid.
It should also be mentioned that removing one $H_2O$ out of two molecules of an ortho- acid gives a pyro- acid. Example :
$2 H_3PO_4 - H_2O = H_4P_2O_7$ and this is the pyrophosphoric acid. $H_4As_2O_7$ is the pyroarsenic acid. This prefix pyro- is even used for acids existing only under the ortho- definition. For example, a pyro derivate from the sulfuric and sulfurous acids $H_2SO_4$  and $H_2SO_3$ exist. They are the pyrosulfuric acid $H_2S_2O_7$. and the pyrosulfurous acid $H_2S_2O_5$
It should also be mentioned that the prefixes ortho-, meta-, and pyro- are never used with acids having already a prefix like hypo- and per-.
All these rules have been edited in the middle of the 19th century.

I admit the link I provided wasn't the most relevant one, thank you for pointing this out. I corrected my comment and included the link that IMO fits better. Either way, the question appears to a be a duplicate.


text:
Why is chromate coloured but tungstate colourless?
I have to try and work this out using the tetrahedral molecular orbital diagram for $\ce{[WO4]^2-}$ but I am confused as both are $\mathrm d^0$ right?

I think this is a great question about CT complexes $\ce{[M^{VI}O4]^2-}$ and can be further extended with examples for sulfate $\ce{SO4^2-}$ and molybdate $\ce{MoO4^2-}$ (both colorless) and uranate $\ce{UO4^2-}$ (yellow).

@andselisk why is this the case? Is it to do with LMCT?

What doesn't? The majority of metal complexes are colored precisely because of LMCT.

@andselisk does that mean [WO4]2- does not have LMCT bands then because it appears colourless? Sorry I'm confused.

LMCT bands may lie not only in visible, but also UV or NIR range.

@andselisk - This is getting off topic of the question, but I don't think it's accurate to say that the "majority" of metal complexes are colored because of LMCT. Intensely colored ones like dichromate and permanganate are, but many transition metal complexes are weakly colored because of $d \rightarrow d$ transitions, not LMCT. I think the majority of common complexes that students encounter are likely $d \rightarrow d$.

@Andrew Yeah, I gave it a second thought and you are probably correct, it's just that I primarily encountered LMCT-type complexes in my research. Thanks for tuning in and correcting me.

https://chemistry.stackexchange.com/questions/39829/how-can-the-intense-color-of-potassium-permanganate-be-explained-with-molecular?noredirect=1&lq=1 https://chemistry.stackexchange.com/questions/27/color-of-chromate-and-permanganate

Both complexes (and many more such as permanganate, sulphate, …) share the same general MO scheme which I am going to shamelessly copy from my older answer:

Figure 1: Qualitative MO scheme of a tetrahedric complex with σ and π bonding between metal and ligands. Double vertical lines represent electron pairs.
The intense colour of permanganate and chromate is due to LMCT transitions: ligand-to-metal charge transfer. Essentially, an electron from the HOMO (a $\mathrm{t_1}$ orbital that can be thought of a ligand-centred p-type orbital) is excited to the LUMO (an $\mathrm{e}$-type orbital that can be thought of one of the metal’s empty d orbitals) by absorbing a photon from the visible range; we then see the complementary colour, meaning that chromate absorbs blue light and looks yellow to our eyes.
The energies of LUMO and HOMO – or rather, the energy difference between the two – determines the wavelength of the absorbed photon and thus the colour that we see. For both permanganate and chromate, the wavelength is well within the visible spectrum resulting in the strongly coloured complexes. In cases such as sulphate or tungstate, the absorption wavelength is in the ultraviolet range which cannot be perceived by human eyes; therefore, the solution appears uncoloured.
Without computer calculations, it is probably impossible to accurately predict the ultimate wavelength but note that such an absorption is practically always present if the corresponding orbitals are available although usually in the ultraviolet range. An obvious counterexample would be any zinc complex in which the metal has a $\mathrm d^{10}$ configuration and the LUMO is thus the much further removed $\mathrm a_1$ orbital; therefore, zinc complexes are typically colourless and do not display such an LMCT band.


text:
Carbonyl chloride from stabilized chloroform treated with hydrogen peroxide
In our lab, chloroform contains 0.5–1% ethanol to prevent it from producing carbonyl chloride when exposed to UV light.
Will the following reaction occur using such chloroform?
$$\ce{CHCl3 + H2O2 -> HCl + H2O + COCl2 ^}$$

I can't find any evidence that carbonyldichloride (phosgene) is formed this way. However, this paper discusses the reaction of chloroform with hydrogen peroxide. It is behind a paywall so I can only see the abstract  https://www.tandfonline.com/doi/abs/10.1080/09593339009384930?journalCode=tent20

The source is from here: https://zh.wikipedia.org/wiki/%E5%85%89%E6%B0%94    It's unavailable in English.


text:
Can Methylene Blue Detect Spoilage in Food?
Initially, there is already a product which is a sensor based on Methylene Blue, which can detect changes in oxygen levels in a packaging. However, as our thesis project, we decided to take on this and make it more simple and replicable.
So this is our concern. We decided to try and replicate the "blue bottle experiment", which is Glucose and Sodium hydroxide mixed to form an alkaline solution and then Methylene Blue will be added. At first it will be blue, but once it settles, it will turn clear. When the solution is shaken, it will turn blue again, and then turn clear once again when settled.
In theory, if a strip of filter paper was dipped into this "blue-bottle" solution, will it turn blue once it detects high levels of oxygen? (This is provided that the filter paper will remain moist)
ps. the reason why i'm asking for professional opinions and answers is that we cannot afford to buy expensive materials if we don't know if it will work in the first place.

This is an excellent idea. And you could keep these paper strips in small closed plastic bags. What should be known is the sensitivity of the test, which I don't know.

The test is incredibly sensitive to oxygen as normally performed. It seems you have two tasks: reducing sensitivity, and keeping the test paper moist without touching the food. Good luck with your experiments, and please report back here with your own answer.


text:
Why does SiH4 have unusually large bond angles?
The bond angles of $n > 2$ hydrides are usually around $90^\circ$, such as $\ce{PH3, H2S},$ $\ce{H2Se},$ $\ce{H2Te}.$ I understand that this is from increased $\mathrm{p}$ character in the bonding with these compounds.
However, for $\ce{SiH4}$ and $\ce{GeH4}$ the bond angles are actually quite close to $109.5^\circ$. Why do $\ce{Si}$ and $\ce{Ge}$ not typically use unhybridized $\mathrm{p}$ orbitals, and instead use closer to $\mathrm{sp^3}$ hybrid orbitals for bonding?

Tech note: please don't use ^\text{o} in place of a degree sign ^\circ, don't enclose text in \ce{…}, and keep in mind that labels such as atomic orbitals are recommended to be upright (to avoid confusion with other variables). If you haven't seen yet, or need to refresh your memory, visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Because $\ce{SiH4}$ is forced to use its $s$ orbital for four bonds and the other elements can keep that orbital relatively inert?

Related: https://chemistry.stackexchange.com/questions/64513/why-is-the-h%e2%88%92c%e2%88%92h-bond-angle-in-%d1%81hloromethane-larger-than-109-5/121709?r=SearchResults#121709

And https://chemistry.stackexchange.com/questions/70981/why-is-sulfur-unhybridized-in-dimethyl-sulfide/70993?r=SearchResults#70993

Basic symmetry suggests the bond in trivalent hydrides won't be the same as a perfect tetrahedron: there are three bonds and a lone pair around the central atom. But there are 4 equivalent bonds and nothing else to interfere with symmetry in silicon tetrahydride. Why would you expect them not to be the same?

Imagine how to arrange things on a sphere. For 4 particles surrounding a single particles, the best arrangement is a tetrahedron - which has a 109.4712 tetrahedral angle. You are letting the hybrid model cloud your geometric intuitive understanding.

In hydrides that contain two or three hydrogens connected to a central atom such as $\ce{PH3}$ or $\ce{H2Se}$, the number of bonding partners does not exceed the number of available p orbitals on the central atom. In an extreme application of Bent’s rule, the most ideal bonding situation would be to have pure p-type orbitals form the required bonds and at least one lone pair reside entirely in an s-type orbital. This allows stabilisation of the s lone pair (as an s orbital’s energy is lower than a p orbital’s one) and better bonding (as p orbitals are directional and thus offer greater overlap with hydrogen; furthermore, they are higher in energy which better matches hydrogen’s orbital energy and again allows for greater bonding energy).
This ideal situation already falls apart in the second period where the bond angles of ammonia and water are unusually large; this is because the central atoms are too small to accomodate two or three bonding partners at $90^\circ$ angles so the bond angles are increased in a delicate balance until the lowest overall energy is obtained. The results are hybrid orbitals with a nonzero s component.
In $\ce{AH4}$ molecules, the situation is different and worse yet again. There are not enough p orbitals to accomodate four bonds so the s orbital has to help. Symmetrically, the optimal arrangement is a tetrahedron and this just happens to almost perfectly match an $\mathrm{sp^3}$ arrangement. As silane (and methane, germane, etc.) has four identical substituents, there is no energy to be gained by reducing symmetry so the ultimate geometry is that of an ideal tetrahedron whose bond angles are $109.5^\circ$.

The geometry depends on the nature of the doublets around the central atom. If the doublets are all bonding, like in $\ce{CH4},$ $\ce{SiH4}$ and $\ce{GeH4},$ the geometry of the molecule is tetrahedral, and the angles between the bonds are 109.5°. If one or two of these doublets are non-bonding, they are more cumbersome than covalent bonds. They repel the bonding electrons, and the angles between these bonding electrons are smaller. That is the case for $\ce{NH3}$ and $\ce{PH3}$, which have one pair of non-bonding electrons. It is even worse for $\ce{H2O},$ $\ce{H2S},$ $\ce{H2Se}$ and $\ce{H2Te},$ which have two pairs of non-bonding electrons.

This is an especially outdated view on these types of molecules; the old myth about lone pairs needing more space.


text:
Are there reactions operated at high temperature at industrial level?
Temperature is an important parameter when you deal with a chemical reaction, first of all $\Delta G$ and $\Delta G^\circ$ depend on temperature, and as a consequence the spontaneity of a reaction depends on temperature. Second, also the equilibrium constant $K$ depends on temperature so, it is possible to modify the reaction yield by varying it.
Finally, the kinetic constant $K_{V}$, and so the reaction speed, are related to temperature. Said that, I was wondering if at industrial level, there are some important reactions which are operated at high temperature (I mean at least more than $\pu{40 °C})$ in order to obtain at least one, or more of the effects I've written (or some other that I don't know). Of course, I also include reactions such as electrolysis and electrochemistry reactions.

The temperature of all production reactions in industry is chosen by economic consideration of the combination of kinetics and equilibrium.  Other examples are ammonia synthesis and oxidation of cyclohexane to produce adipic acid.

The Haber-Bosch process for producing ammonia works at both high temperature and pressure for both kinetic and thermodynamic reasons.

If you wanna talk about electrochemistry, write it in title, not last sentence! BTW what industry thinks as high temps is at least order of magnitude higher.

The cracking reactions transforming heavy fuel, tar, domestic or heavy oils into lighter molecules like gasoline, kerosene or white spirits, they are all carried out at temperatures higher than 200 °C. The electrolysis of alumina to produce metallic aluminium is done at about 960 °C. The production of iron from ore and charcoal is done in a blast furnace at 1400 °C


text:
How to check whether intramolecular hydrogen bonding takes place in cyclic compounds?
How can we confirm or have a clear idea whether hydrogen bonding is taking place in substituents of cyclic compounds? For example, it takes place in ortho-nitrophenol, but it doesn't take place in ortho-flurophenol despite the fact that the nitro group and the fluorine are present at the same position and both are highly electronegative.

Nitro and fluorine are not really similarly placed despite being at the same position.  The former has a six-member ring when you make the hydrogen bond, the latter only a five-member ring.  Given the constraints involved with placement on an aromatic ring, even in the ortho case, the six-membered ring is easier to form.

Computational chemistry is a great way to study whether hydrogen bonding is significant or not

From a structure diagram alone it is hard to determine these interactions. The reason for this is that important structural parameters are stretched or shortened in order to flatten the molecule for a 2D drawing.
In a first order approximation a molecular modelling kit based on balls and sticks could already be very helpful. It'll let you approximate the 3D structure and get a 'feeling' (also known as chemical intuition) about where to expect close contacts, or what bonds are strained, etc.; just to play around with it will give you a better visualisation of the shapes of molecules in general.
More advanced is using computers. While this might seem intimidating at first, it is actually something you can do at home. I've previously written extensively about how to do these identifications in this Q&A: How to identify hydrogen bonds and other non-covalent interactions from structure considerations? There are quite a few programs out there that let you build ball and stick models of compounds using tabulated bond lengths. That is probably already enough to identify likely positions for these interactions. Many of these molecular editor programs can run optimisations on a quick but low level, too. My personal favourite is Molden (very lightweight), but Avogadro does the job equally well, and probably has a nicer user interface. (Geoff, if you read this, we are all still hoping for a major update.)
Obviously you can increase the level at which you are running the calculations; for more complex structures you may need to. A lot of this can still be done with a home computer. You can start out with semi-empirical models, of which I recommend the fantastic open-source xtb, go on to ab-initio wave function methods (esp. HF), climb Jacob's ladder (AIP Conference Proceedings 2001, or see these slides by Perdew via helper.ipam.ucla.edu), and finally make your way to the various post-HF methods. (the last one you probably don't want to do on a laptop computer.) In this context I would like to point you towards Orca.
For the two molecules you mentioned, I've done the analysis with xtb and Multiwfn.

With these two molecules it is still quite simple, because they are flat (mostly at least; they are of course not static). Here is a model of the 3D structures with the hydrogen bond measured. For ortho-fluorophenol it is 2.25 Å and for ortho-nitrophenol it is 1.73 Å. You can see a clear difference already.

And here are the non-covalent interactions from Multiwfn analysis (Option 20, 2, visualised with ChemCraft at contour value 0.2 and in range ±0.05):

The red coloured blobs correspond to repulsive interactions, while the green and blue coloured values correspond to attractive interactions. As you can see, there is no hydrogen bonding in ortho-fluorophenol, but there is some in o-nitrophenol.


text:
Deriving rate law from a reaction mechanism for a chain reaction using steady-state approximation
I am trying to derive the overall rate equation for the photocatalysed halogenation (chlorination) of an alkane that proceed via this 4 step chain reaction:

$\ce{Cl2->2Cl^. \quad(1) initiation}$
$\ce{Cl^. +RH->R^. + HCl \quad(2) propagation}$
$\ce{R^. + Cl2-> RCl + Cl^. \quad(3) propagation}$
$\ce{R^. + Cl^. ->RCl \quad(4) termination}$

I have calculated the differential rate equations for $\ce{R^.}$ and $\ce{Cl^.}$ and set them to zero at steady state.
[The free radical dot symbols are omitted from this point onward for brevity]
This gives me $$\ce{For Cl ,2k1[Cl2]-k2[Cl][RH] +k3[R][Cl2] -k4[R][Cl]=0 \quad(5)}$$
$$\ce{For R, k2[Cl][RH] -k3[R][Cl2] -k4[R][Cl]=0 \quad(6)}$$
Adding these two equations,
$$\ce{[Cl]=\frac{k1}{k4}\frac{[Cl2]}{[R]}}$$
Then I tried substituting this into equation (6), which gives me a quadratic equation. The solution of the quadratic equation is of the form $$\ce{[R]=constant +\sqrt{\mathrm{constant [RH]}}}$$
The overall rate law from this expression is a very complicated equation.
This looks wrong because the rate law for radical halogenation is usually of the form $\ce{k[RH]\sqrt{\mathrm{[Cl_2]}}}$ (Wikipedia)
Where am I going wrong?

The initiation rate may not depend on [Cl2]( for high enough [Cl2]), but on irradiation level only. i.e being a reaction of zero order .

In the book that Wikipedia cites (Advanced Organic Chemistry: Reaction Mechanisms
By Reinhard Bruckner, ISBN 9780080498805), they have a different set of assumptions. They are saying the formation of chlorine radicals proceeds by a fast equilibrium. Then, they say that the organic radical is at steady state, ignoring reaction (4) given by the OP (the termination reaction).

Where am I going wrong?

I'm not sure why your steady state approximation is less appropriate than that of the textbook. I'm sure that there is some experimental data (and some fitted rate constants) that would speak to that. Also, the appropriate approximation depends on the initial conditions.

[Comment from Poutnik] The initiation rate may not depend on [Cl2]( for high enough [Cl2]), but on irradiation level only. i.e being a reaction of zero order.

The derivation in the book is for the thermal reaction, but the OP asked about the photocatalyzed reaction, so there is a discrepancy there.

But I don't understand why using one termination reaction (R and Cl) instead of another termination reaction (Cl and Cl, which would give the equilibrium) would change the rate law. The text book's rate law is $\ce{\sqrt{[Cl2]}[RH]}$ whereas the rate law that I am getting is $\ce{\sqrt{const.+[RH]}[Cl2]}$. (Also this is a question from a book, so it must be a well known problem)

@ShoubhikRMaiti That is not the only difference. For your set of equations, [Cl] depends on [R]. For the assumptions in the textbook, it doesn't, so it becomes possible to get a simple expression for the steady state of [R], and for the rate of product formation.

The mechanism of the gas phase halogenation reactions is discussed in some detail in a paper by Benson and Bus[1]. In the following, I submarised section II ( mechanism and rate laws).
The mechanism of the reaction thermal reaction is

$\ce{X2 + M -> 2X^.  + M  \;\; k_1 \quad(1) }$
$\ce{2X^.  + M  -> X2 + M  \;\; k_{-1} \quad(-1) }$
$\ce{X^. + RH -> R^. + HX \;\; k_2 \quad(2) }$
$\ce{R^. + HX  -> X^. + RH \;\; k_{-2} \quad(-2)} $
$\ce{R^. + X2-> RX + X^. \;\; k_3  \quad(3) }$
$\ce{X^. + RX -> R^. + X2 \;\; k_{-3}   \quad(-3) }$
$\ce{R^. + X^. (+ M) ->RCl (+ M) \;\; k_{4}       \quad(4) }$
$\ce{R^. + R^. (+ M) -> R2 (+ M)   \;\; k_{5} \quad(5)}$

Using the stationary state approximation for the intermediates $R^.$ and $X^.$,
and neglecting the back reaction 4 which is generally unimportant[[1]] you find
$ \ce{\frac{|R|}{|X|} = \, \theta \; = \frac{k_2 |RH|}{\,k_3 |X_2|\,} \left(1 + \frac{k_2 |HX|}{k_3 |X_2|}\right)^{-1}}$
$ \ce{|X|_{ss} = \left(\frac{k_1 |M||X_2|}{k_{-2}|M| + k_4 \theta + k_5 \theta^2 }  \right)^{1/2}
}$
This equation results in
$ \ce{-\frac{d|X_2|}{dt} = k_1|RX||X|_{ss} \left( 1 + \frac{k_{-2}|HX|}{k_{3}|X_2|}\right)^{-1}} $
Depending on which of the postulated termination reactions predominates, the authors obtain three expressions for the velocity equation. Thus in case I,
$ \ce{-\frac{d|X_2|}{dt} = k'|RX||X_2|^{1/2} \left( 1 + \frac{k_{-2}|HX|}{k_{3}|X_2|}\right)^{-1}} $
that for low concentrations of HX reduces to
$ \ce{-\frac{d|X_2|}{dt} = k'|RX||X_2|^{1/2} } $
The equations for a photocatalized reaction are similar to the ones above but you must substitute $k_1$ by $I_a/|M||X_2|$ where $I_a$ is the specific rate of absorption of light.
[1]: S. W. Benson, and J. H. Buss, Kinetics of Gas Phase Halogenation
Reactions, J. Chem. Phys. 28, 301 (1958); http://doi.org/10.1063/1.1744111


text:
Acid-Temperature coagulation of milk
I was making coffee with a cup of boiling milk which was 2 days old. I used instant coffee powder and some sugar for the making of it. Tried to make a very hot coffee but usually prefer medium hotness. When I mixed the hot milk with coffee powder, I immediately observed coagulation.
Then, I decided to use an unopened milk packet and boiled it the same way. But this time I used a relatively more roasted coffee powder with same amount of common sugar. Again I observed coagulation but the particles were smaller.
I know that milk is a colloid and can be coagulated by using an acidic compound.
The casein starts coagulating at pH 4.6,
But I don't think that coffee is acidic enough to lower the pH. As in my second experiment, I've used a lesser acidic coffee powder (more roasted) but still observe coagulation.

In fresh dairy milk, is there any relation between titratable acidity and heat coagulation?
Does hydrogen ion concentration determine the coagulation?
Is it the coffee power or the milk or the temperature?
How to avoid this effect?

Coagulation of milk is a result of casein micelles clumping together. This is favoured by higher temperatures and also by favour a pH of around 4.6 as this is the isoelectric point. At this isoelectric point, the amino acids within the casein micelle are electrically neutral. This means there is less electrostatic repulsion between the casein micelles, facilitating coagulation Furthermore, the addition of acid means that the kappa casein, i.e. hairs on the outside of the casein molecule which stabilise it, become protonated. This protonisation causes them to destabilise, making it easier for the micelle to fall apart and clump together with other micelles that have fallen apart. Relatively High temperatures also increase reaction rates, which overall speeds up the aforementioned processes as well.
Even if the coffee does not reduce the pH by much, any change in pH which leads to the pH being closer to the isoelectric point (4.6) of the casein micelle will increase the rate of the coagulation process.
Using old milk also increases coagulation, as milk naturally coagulates due to rennet, an enzyme, cleaving the kappa-casein, causing the casein molecules to destabilise.
To avoid coagulation during heating, perhaps try heating the milk gently. The ideal temperature at which milk coagulates is about 43ºC, so perhaps staying below that could work. This source also provides tips on how to avoid coagulation while heating milkAlternatively, you have yourself some cheese to enjoy :)


text:
Ozonolysis: mechanism for reductive workup of ozonide using Zn/Acetic Acid/Water
What is the mechanism for the reductive workup of an ozonide using zinc/acetic acid/water? Many sources online show the mechanism using DMS instead:
https://www.khanacademy.org/science/organic-chemistry/alkenes-alkynes/alkene-reactions-tutorial/v/ozonolysis-1
https://www.youtube.com/watch?v=PcO49xCb4Gs

Use a pair of electrons on Zn metal the same way that one uses an unshared pair on dimethyl sulfide.

@user55119 can metals act as nucleophiles?

@user55119 so I tried the mechanism using zinc as the nucleophile. Would you get ZnO as the product in addition to the carbonyl compounds?

@user55119 In acid? It looks like typical reduction with hydrogen in statu nascendi.

ZnO or likely zinc acetate ultimately.


text:
Acetone-resistant clear plastic
Is there a plastic that's clear, acetone resistant, and ideally sold in sheet stock that's ~1/8" or 1/4" thick?  Both acrylic and polycarbonate are out, they turn cloudy on contact with acetone.  I tried a piece of PETG and it was better, but still turned fairly cloudy after a few seconds.  I don't need continuous service, but I need it to still look decent if it gets wiped down with acetone.

Would glass work? If you want the more durable plastic properties, maybe laminate glass onto acrylic/polycarb? If your piece only needs to be tablet-sized, buying a glass tablet screen protector is super cheap. For something bigger, could epoxy

I think the only real option for this is a clear plastic sheet that's had a chemical resistant thin film applied to the faces.  One commercially available example is Makrolon AR2, it's polycarbonate with some proprietary coating that's listed as having >24hrs resistance to acetone and a few other chemicals.

According to Palstics International's Chemical Resistance Chart, the A-rated plastics (no solvent attack) towards acetone are:

ECTFE (Halar®): transparent films available
Fluorosint® PTFE: white
HDPE: transparent films available
Nylon®, Type 6/6: white
PP: clear sheets available
PPS: opaque white
PTFE: white

Among these, polypropylene appears to be the only plastic that satisfies all criteria.

+1 iPP can be made to crystallise in structures smaller than optical wavelenght. Its done with comonomer contents and/or tacticity defects in a small percent range and/or a lot of nucleating agents.

From what I can tell, clear polypropylene is biaxially oriented polypropylene, it looks like it gets stretched in 2 directions while being formed.  I think this process is mostly used for thinner packaging material, something a few mil thick.  I wasn't able to find anything in sheet form that's an 1/8" or 1/4" thick.

@Shawn You are correct, getting optically clear or even just transparent PP 1/8" sheet isn't trivial. If semi-transparent white PP sheet isn't an option, then one should probably look further for less common proprietary plastics.

Alternatively, cast your own sheet (or straight your close-to-final piece) using a mold from a chemical resistant 2 part epoxy. Usually, difunctional or polyfunctional amine hardeners confer chemical/temperature resistance, causing a polymerization in a 3D cross-linked network (it also means saying goodbye to thermoplasticity).
Google search - customize it for the TLD/geo area you live in.
Example of a result that works for me (haven't use it yet, but I had a chat with a sale representative and confirmed that it maintains its clear/mechanical resistance when using with acetone after the cure).

It sounds like you're looking for a plastic material that is clear, acetone-resistant, and available in sheet stock around 1/8" or 1/4" thickness, especially in the context of epoxy hardeners. In this case, you might consider using PTFE (polytetrafluoroethylene), commonly known as Teflon. PTFE is known for its chemical resistance, including resistance to acetone. It's transparent, maintains clarity, and is available in various thicknesses, including sheets.
When working with epoxy hardeners like used in epoxy flooring, having a clear, acetone-resistant barrier like PTFE can be essential for maintaining a clean and visible workspace. Ensure to check specific product specifications and compatibility before use.
Best of luck with your project!


text:
exchange integral at large distances
I want to ask a question about the exchange integral $\ce{H_{AB}}$ -
If I consider an exchange integral $\ce{H_{AB}}$ the following exchange integral can be evaluated:

so $\ce{S_{AB}}$ is $0$ at $\ce{R = \infty}$
but in this presentation I was shown today, at large values of R, $\ce{H_{AB}}$ is preportional to the overap integral $\ce{S_{AB}}$.
Surely at large distances and towards infinity, as $\ce{S_{AB}}$ goes to zero, $\ce{H_{AB}}$ also goes to zero too.
Where does the preportionality
$$H_{AB} \propto S_{AB}$$
come from then using these approimations when R is large?

Great question @vik1245. Would you be interested in committing to a brand new stack exchange dedicated towards computational modeling? https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=NzE1YTZmNTZhOGE1N2EwYWYzZGM2NWQyYTFiNzQ5ZDI5MWQyM2EwNzA5OWY5OWY2MDIzYzY2ZTMxNjJlYzIwMSLiKdOhPeuMkh2ODtejwcgH38HFpQq64gQljgKUrZrZ0. I wonder if you might be interested in committing to it?

The author of the presentation is talking about large, but not yet very large/infinite distance here. Note the annotations:
$$
\frac{e^2}{4 \pi \epsilon R} = k/R
$$
when $R \rightarrow \infty$: slowly to zero

$$
S_{AB}
$$
when $R \rightarrow \infty$: to zero

$$
\left< 1s_A\left| k/r_A \right| 1s_B\right>
$$
when $R \rightarrow \infty$: rapidly to zero

So there is some regime somewhere, in which the change of the entire expression is dominated by the change in $S_{AB}$. The verification of the correctness of the different notations is apparently left to the reader ...


text:
What is the difference between cyanato, cyanido and cyano? When should i use which?
I understand that cyanido is the newly adopted naming in place of cyano but what is cyanato? Can anyone please give examples to illustrate the difference?

Cyano- is related to the neutral CN group. Cyanido- is related to the CN- anion. Cyanato- is related to the CNO- anion.


text:
How do we weigh one C atom?
As far as I get it, we did weigh one carbon (12,6-isotope) atom. Then we said that electrons have nearly zero weight and protons weight nearly the same as neutrons. Hence we took the weight of that one atom and divided it by twelve and this way we got the mass of a proton/neutron:
$w_\mathrm {n,p} = 1.6726 \cdot 10^{-24}\ \mathrm g$
Because that isn't easy to calculate with, we introduce a number which multiplied with the weight of a proton/neutron gives us one gram:
$N_\mathrm A = w_\mathrm {n,p}^{-1} = 1\ \mathrm{mol}$
So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that:

How did we actually measure the weight of one carbon atom in the first place?
Why is the molar mass not precisely 1 and 12?

1.) did you check wikipedia? Its explained thoroughly there. 2.) nuclear binding energy, $E=mc^2$

Your equation for the Avogadro constant is not correct.

This is a question worth debating and discussion. First of let us realize the fact that we are used to looking at relative atomic weights or masses (if someone wants to be too accurate). No Table lists the absolute atomic masses of atoms in the periodic table, i.e., through out the history we have to assumed either the weight of hydrogen atom is exactly 1.0000, or oxygen-16 had a weight of 16.0000 or lately the weight of carbon-12 atom is exactly 12.0000 units.  So let us forget the picture that someone actually made a measurement, took a single atom of hydrogen or carbon and weighed by some fancy means. These are all man made references and numbers. We could have taken the weight of C-12 as 100, and all the atomic masses would change proportionally.

So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that: How did we actually measure the weight of one carbon atom in the first place?
  Why is the molar mass not precisely 1 and 12?

See above. You have to choose a reference. Fix hydrogen to 1 or carbon as 12 exactly but not both. Now you may ask how these numbers came up? You have to look at the question historically, when Avogadro's number was not used. If you understand German, have a look at Handbuch der Anorganischen Chemie by Abegg. There is long history how the number 12  got associated with carbon.
https://archive.org/details/a2handbuchderano03abeguoft/page/12;

You must realize that one proton free, lonely, has not the same mass as when included in a nucleus. A proton looses a small part of its mass when entering a nucleus. This mass loss is transformed into energy through the Einstein law : E = mc2.


text:
Do amphoteric cations exist?
I know that anions such as $\ce{CH3COO-}, \ce{OH-}, \ce{CN-}$ are basic in nature, since they are conjugate bases of weak acids. Similarly, cations such as $\ce{NH4+}, \ce{H3O+}$ are acidic in nature , since they conjugate acids of weak bases.
However, we also have another case whereby, anions such as $\ce{HSO3-}, \ce{HS-}$ are amphoteric in nature. This is due to the fact that their conjugate acids are polyprotic acids. Thus, one can have the two reactions:
$\ce{HS- + H2O <=> S^2- + H3O+}$ and $\ce{HS- + H2O <=> H2S + OH-}$
Since, for this case, we got amphoteric anions from polyprotic acids, is there a possibility that we can do the same for polyprotic bases so as to get amphoteric cations? i.e. Can we have a conjugate acid of a polyprotic base, which can be regarded as an amphoteric cation? I haven't seen any such examples of amphoteric cations yet, hence I am asking this question.

Polyamines. Or basic aminoacids.

$\ce{H3O+}$ and $\ce{H3S+}$ are technically amphoteric (as well as likely many others), though they're (always?) extremely weak bases. I wonder if there are more natural examples which are limited to few atoms.

Or CaOH+,  as Ca(OH)2 has 2 dissociation constants.

What about singly protonated DABCO? (ie 1,4-diazabicyclo[2.2.2]octane). It is cationic, and the neutral conjugate base has no additional acidic protons. With one proton, it is weakly basic (pKa of conjugate acid is ~3) and weakly acidic (pKa ~ 9).

Thing is, almost any cation can be protonated. Only the smallest ones may be to unstable to have minimum on PES. Another thing is that "amphoteric" tends to be used for stuff easily / commonly acting as acid and base, so what exactly you'd like to get here?

Your bolded text isn't entirely clear: the examples you cited, like $\ce{HS-}$, already satisfy this requirement. For example, $\ce{HS-}$ is the conjugate acid of $\ce{S^2-}$, which is a polyprotic base. I assume you want this polyprotic base to be a neutral species, and not a negatively charged species. But the other examples given in the comments already suit that perfectly well. A molecule with multiple amine groups is the simplest and most common example.

@Poutnik: Isn't Ca(OH)2 a strong base? No dissociation constant?

Diamines, triamines, etc. There are commercial polyamines that are used as thickeners.

@James Gaidis Pkb1 = 1.37, pKb2 = 2.43
https://en.wikipedia.org/wiki/Calcium_hydroxide

Yes, for example lysine.
Lysine has two amino groups and one carboxyl group. When dissolved in aprotic solvents, all of these will be uncharged so the carboxyl group will be $\ce{-COOH}$ and the two amino groups $\ce{-NH2}$. When dissolved in water, these three groups have distinct $\mathrm pK_\mathrm a$ values of $2.15, 9.16$ and $10.67$. Depending on the acidity of the solution, the following (acid-sided) equilibriums occur:

When dissolved in water, the carboxyl group’s proton will immediately be removed and used to immediately protonate the ε-amino group (the far end). However, the α-amino group is still more basic than water and thus will be partially protonated; meaning that the right-hand equilibrium will be happening predominantly if lysine is dissolved in water. Dissolved lysine will be a mixture of uncharged lysine and lysine cation.
When an acid such as hydrochloric acid is added, at first the α-amino group is completely protonated and then the left-hand equilibrium occurs predominantly as more and more lysine cation is converted to lysine dication.
If a base is added (such as sodium hydroxide), the ε-amino group’s proton can also be removed yielding lysine anion; I have not shown that equilibrium in the scheme above.
Lysine cation thus fits your question as it is an amphoteric cation, the conjugate acid of a dibasic compound.


text:
Optimization completed on the basis of negligible forces
I am performing relaxed scan in Gaussian (B3LYP/6-31G(d,p)). There are two small organic molecules, one of which is a radical, the other has a benzene ring.
I'm trying to understand the optimization output. Specifically, the line Optimization completed on the basis of negligible forces. I understand that these forces are the maximum and RMS forces above.
Some examples of "negligible forces" from my output file:
        Item               Value
Maximum Force            0.000004
RMS     Force            0.000001
Maximum Displacement     0.019917
RMS     Displacement     0.004747

        Item               Value
Maximum Force            0.000002
RMS     Force            0.000000
Maximum Displacement     0.002231
RMS     Displacement     0.000490

        Item               Value
Maximum Force            0.000002
RMS     Force            0.000001
Maximum Displacement     0.003132
RMS     Displacement     0.000805

However, some small values like
        Item               Value
Maximum Force            0.000002
RMS     Force            0.000001
Maximum Displacement     0.000666
RMS     Displacement     0.000149

are followed only by Optimization completed. So what does "negligible" mean? How small do they need to be to say they are negligible? Or is this connected to at least one NO in convergence criterion section?
Next question, how reliable is optimization when such message pops up? If it's unreliable, what do I do about this? Would altering the initial geometry fix it when in only happens in a couple of scan steps?

Please list the displacements (Max and RMS) for each set of forces you've reported. Also, the forces you've reported are very small.  You should look at the convergence criteria for Gaussian (under opt in the manual).

I'd love to bounty this question.  OP, please update the question and include the displacements for the corresponding forces you report. Also, it would help to know what level of theory you're using as well as what type of system is leading to this behavior.

@Anna Your energies should be fine but your frequencies could be affected. I've seen responses to specific cases and situations but never a comprehensive answer.  Time to bounty up.

Related and at least explains the basic distinction between "optimization completed" and "Optimization completed on the basis of negligible forces": https://gaussian.com/faq3/  Essentially, the negligible forces occurs when the displacement criteria aren't below Gaussians thresholds, but the force criteria are both two orders of magnitude smaller than the threshold.

@Tyberius The FAQ addresses converged forces and displacements in the optimization but unconverged displacements in the frequency.  How about when the optimization has forces converged but not displacements?

@LordStryker it's kind of tucked in there, but in the convergence disagreements section, the two bullet points state how define convergence in Gaussian

@Tyberius Right, but how does this affect the frequency computation? I've heard cautionary tales from others in the past about this but never had to deal with it myself until recently.

@LordStryker I see, I misinterpreted your last comment. I'm not certain, I'd be interested to see if the there are known cases where the approximate Hessian used in a typical geom opt is bad and how the quality of the stationary point is related to convergence criteria used.

@LordStryker, your profile says you like taking things to the CBS limit. Have you considered supporting the Materials Modeling Stack Exchange? In the "example questions", many of them are basis set related, and Kirk Peterson is there. Kirk Peterson is a professor that did his PhD under Dunning, and optimized most of the Dunning basis sets on Basis Set Exchange.

According to the Gaussian Frequently Asked Questions:

In general, a stationary point is found when one of two criteria is met for a point in an optimization calculation:

All four values listed in the output are smaller than the indicated thresholds.
The Maximum Force and RMS Force are two orders of magnitude smaller than the thresholds shown, regardless of the values of the displacements.

You indicate three molecules where the second criteria activate - the forces are really, really small. So the optimization gives up.
In your final case, you can see that the displacements are also small, so the first criteria activate.
You then want to know how "reliable" the geometries are when the forces are small, but the displacements are not. That depends a lot on your needs. The optimization quits because such small forces imply a very flat potential energy surface and it may take a long time before "complete" convergence. My guess is that you still have some imaginary frequencies, but that might not matter depending on your needs.
It's worth pointing out that many quantum chemical programs have different criteria for declaring an optimization "finished." (In other words, your calculation in a different program might be declared done because of negligible energy and force/gradient change.)
If you want to push for a true minima, you should take that optimized geometry, run a force / frequency calculation to get an accurate Hessian, and save the result to a checkpoint, then re-start the optimization using the saved Hessian:
%Chk=myfile
# B3LYP/6-31G(d,p) OPT=ReadFC Freq Geom=AllCheck Guess=Read

Of course, that means probably breaking your relaxed scan calculation into a set of separate steps. (Honestly, this is how I usually do relaxed scans anyway - a set of separate jobs with each stepped geometry.)
One other very important point. If you're doing a scan of intermolecular interactions with B3LYP, you really should be using a dispersion correction such as EmpiricalDispersion=GD3BJ. I often see incorrect repulsive potential energy curves from B3LYP without dispersion corrections.


text:
Using the electrical potential created by the electron transport chain to power electrical systems
My friend and I were discussing the extremely low energy density of modern batteries and read that food has a much higher energy density. Thus, we were considering the possibility of using biological processes to extract energy from food and use that to create an electric potential or do work in some other way.
The electron transport chain creates an electrochemical gradient and thus and electric potential using protons. If some sort of system could be devised to mimic/perform glycolysis and the citric acid cycle and also mimic the ETC membrane and necessary proteins, is it feasible/is there already research on using glucose and cellular respiration (or a similar biological process) to power electrical systems? How would this relate to fossils fuels and what scale could this be useful at?

There are fuel cells that run on alcohols such as ethanol or methanol and on glucose.
That said, the energy density in current Li-ion cells is quite high - a 18650 cell holds ~30,000 joules, about 9 kCal. By comparison, a  "biochemically" powered shotgun shell (well, the nitrate and carbon of gunpowder can come from bio sources) packs ~2,000 J.


text:
Comparing different stereochemistry terms
What is the difference between:

Meso compound and an optically inactive compound
Chiral compound and an optically active compound

For all examples i have seen, chiral compounds were optically active and meso were optically inactive so is there any difference between them?

Meso is only a subtype of optically inactive compounds and chirality doesn't guarantee noticeable optical activity.

"A meso compound is an optically inactive member of a set of stereoisomers, at least two of which are optically active."  See: The Meaning of Meso", J. Chem. Educ., 2008, 85, 441.

What was the reason behind all these downvotes??

Meso compounds have a chiral centre , but they also have a plane of symmetry making them optically inactive....or u can say meso compounds are optically inactive due to internal compensation.
Whereas enantiomers taken  together ie. formation of racemic mixture (50:50) is optically inactive as a whole due to external compensation.(They can be optically active individually)
Therefore chirality isn't the only defination for the optical activity of a compound, there are reasons such as symmetry to account for as well.


text:
What is the equivalence of 40 ppt to pg/mL?
Please help me converting 40 ppt to pg/mL. I have tried an online calculator and done it manually and there are differences. I`m aware that ppt can be used as parts per trillion or per thousand but I don´t understand where the difference is coming from. The calculator is  http://www.endmemo.com/sconvert/pg_mlppt.php
Thanks,

You wrote that you did it manually. Could you please edit your question and show what you've done so far? That wound help to get an idea where the problem arises...

Check out http://picotrillion.com/

By manually I mean thinking automatically 40 pg/mL is equal to 40 ppt . Thanks someone just confirmed that the online converter is broken so my calculation is ok.

You are in luck. The converter is broken, so it gave you the wrong answer:

It does fine with ppm and ppb, but conversions to and from ppt seem to be broken. A trillion can mean ten to the twelve (conventional), a lot (informal), or ten to the eighteen (dated, British) according to lexico.com.


text:
How can AuCl3 exist?
The reduction potential for $\ce{Au^3+ + 3e- -> Au}$ is $1.503 \text{ V}$. The reduction potential of $\ce{Cl2 + 2e- -> 2Cl-}$ is $1.36 \text{ V}$. So, the $E^o$ for $\ce{2Au^3+ + 6Cl- -> 2Au + 3Cl2}$ is $0.143 \text{ V}$.
If the reaction $\ce{2Au^3+ + 6Cl- -> 2Au + 3Cl2}$ has a positive E standard ($E^o$ = $0.143 \text{ V}$), then how is it possible for $\ce{AuCl3}$ to exist in aqueous solution without $\ce{Au^3+}$ oxidizing $\ce{Cl-}$ to $\ce{Cl2}$?

There is not considered $\ce{Au^3+ 4 Cl- <-> [AuCl4]-}$.

$\ce{AuCl3}$ does not exists. It is an empirical formula for $\ce{Au2Cl6}$ (https://doi.org/10.1107/S0365110X58000694).

Gold(III) chloride does not exist as $\ce{AuCl3}$. According to Wikipedia, the name gold trichloride is a simplification of the name, which is referring to the empirical formula, $\ce{AuCl3}$. The X-ray defraction studies revealed that the chemical exists as a chloride-bridged dimer, $\ce{Au2Cl6}$ (Ref.1):

In $\ce{Au2Cl6}$, each gold center is square planar, and the bonding in $\ce{Au2Cl6}$ is considered somewhat covalent (Wikipedia). Yet, your suggestion is somewhat true, since a recent research has shown that $\ce{Au2Cl6}$ is susceptible to decompose to metalic $\ce{Au}$ and $\ce{Cl2}$ gas at high temperatures or expose to $\ce{e-}$-beam irradiation (Ref.2).
Also, another research group (Ref.3) has shown that:

Gold chloride, supplied as $\ce{HAuCl4}$ solution, is rapidly adsorbed on the silicate surfaces, the Au(III) is reduced to metallic gold, and gold particles grow on the surface. SEM images show agglomerates of gold unevenly distributed on the surface of the silicates, including in some areas forming agglomerates, especially on quartz and feldspar.

References:

E. S. Clark, D. H. Templeton, C. H. MacGillavry, "The crystal structure of gold(III) chloride," Acta Cryst. 1958, 11, 284-288 (https://doi.org/10.1107/S0365110X58000694).
Keita Kobayashi, Ryo Kitaura, Katsuhiro Sasaki, Kotaro Kuroda, Takeshi Saito, Hisanori Shinohara, "In Situ Observation of Gold Chloride Decomposition in a Confined Nanospace by Transmission Electron Microscopy," Materials Transactions 2014, 55(3), 461-465 (https://doi.org/10.2320/matertrans.MD201308).
Sima Mohammadnejad, John L. Provis, Jannie S. J. van Deventer, "Reduction of gold(III) chloride to gold(0) on silicate surfaces," Journal of Colloid and Interface Science 2012, 389(1), 252-259 (DOI: 10.1016/j.jcis.2012.08.053).

There are other III chlorides that form the same dimeric structure, such as molten aluminium (III) chloride.

Not really the same dimer.  Aluminum (III) chloride has tetrahedral coordination not square planar.  Blame it on a different electronic structure.


text:
Strength of Hydrogen Bond and Angle
This link has a short paragraph on the very first page, which says: "There are two useful criteria to evaluate the presence and the strength of a H-bond. The first is to look at the distance between the H and the A atom, which should be shorter than the sum of their Van der Waals radii. A second method suggests to evaluate the angle formed by X-H∙∙∙A, centered on the H atom: the closer this angle is to 180°, the stronger the H-bond is"
I wish to understand why the part in bold is true, that is, what is the angle dependence (qualitative or quantitative) of the strength of hydrogen bonds.
To add to the above discussion, in an introductory biology course at university, we were also given the following example -

Well, once again, why is it so?
I did some research, didn't really find much, but came across an interesting paper that may be worth a read if anyone's really interested. Attaching it here!
Thanks in advance!

Empirical data
If you analyze crystal structure data for short intermolecular distances between hydrogen atoms and potential acceptor atoms, you will find a distribution of these angles. The simplest explanation for this distribution is that there is a favorable interaction that is strongest for the most commonly observed angles. The fit between empirical data and computational results is fairly good, e.g. https://www.researchgate.net/publication/242862021_Distribution_of_hydrogen_bond_angles_in_molecular_crystals
Dipole-dipole interactions
If you treat hydrogen bonds as pure dipole-dipole interactions (Q-Y ... H-X), you expect the bond to be directional. After all, you maximize the interaction by close approach of the negative partial charge on the acceptor and the positive partial charge on the donor (i.e. the hydrogen atom in a polar bond). However, for this model the angular dependence should be symmetrical, i.e. there should be a similar decrease in bonds strength when you rotate the donor or the acceptor away from a colinear arrangement. The "rules" for hydrogen bonds say otherwise. It is more important that the angle at the hydrogen atom is maximal (Y ... H-X) than that the angle at the acceptor atom is maximal (Q-Y ... H).
Covalent contribution
There is a covalent contribution to the hydrogen bond that explains why a linear arrangement around the hydrogen atoms is favored. The strongest hydrogen bonds are those where hydrogen bond and covalent bond with hydrogen become interchangable ("symmetric hydrogen bonds"). In his 2002 review, Steiner wrote:

All hydrogen bonds can be considered as incipient proton transfer reactions, and for strong hydrogen bonds, this reaction can be in a very advanced state.

There is some controversy whether the hydrogen bond has covalent character, and some question how to describe it theoretically (as a 3 center 4 electron bond?). There is, however, strong evidence from NMR that there is indeed a covalent contribution because nuclei couple in a through-bond fashion across hydrogen bonds (e.g. this review).

It has direct mechanical analogy.
If you pull a tree branch weakly, it remains more or less in its original direction. If you pull it strongly, it points towards you.
By other words, if an object is pulled by 2 non colinear forces, it has tendency to move itself to make them colinear, as the net force makes it to do so.
If one of forces is much weaker, the effect is weak.

I can't give you a definitive answer, since I've not studied this myself, but here is what my physical intiution says:
Let's use the following labels:
Oxygen atom in the same molecule as the hydrogen: O'
Oxygen atom with which the H is hydrogen bonding, i.e., the oxygen on the other molecule:  O
As you know, a hydrogen bond forms because of the attraction between the partial positive charge on the H and partial negative charge on the O (it can also be an S or an N, but in your illustration it's an O). The greater the partial positive charge the O sees when interacting with the H, the stronger the bond.  And the region around the H in the direction opposite O' (i.e., that furthest from O') is the most positive.
For an illustration of this, take a look at the diagram labeled "Water charge distribution" here: "http://www1.lsbu.ac.uk/water/water_molecule.html
On the LHS of your diagram, the O is interacting with the side of the H furthest from O', i.e., the most positive side of the H. Hence the hydrogen bond is relatively strong.  On the RHS of your  diagram, the O is interacting with a side of the H that is closer to O', which thus has a slightly higher electron density, making the H less positive when seen from that direction, which in turn yields a somewhat a weaker hydrogen bond.

A shorter way of saying the same is that in the bent structure the oxygen atoms are repelling one another more strongly than in the linear structure.

note that the heteroatom donor and the H-bond acceptor atom both contain a negative dipole. They repel each other.
When you draw out a diagram of the electrostatic interactions in question, you will notice that the magnitudes of this (-)(-) repulsion are minimized when the negative dipoles are at a maximum possible distance from one another—or in other words, if you were to imagine a circle with the H-atom at the center, the (-)(-) repulsion is minimized when the two negative dipoles are on the exact opposite points of the circle.
With this in mind, you can imagine that when the angle is not 180 degrees, the H-bond is weaker, as the magnitude of (-)(-) repulsion is stronger than it might otherwise be, resulting in the H-bond being  "pulled apart" more.


text:
GTOs of higher quantum numbers?
I recently realised that the STO-nG basis functions were fit to 1s/2s/3d STOs of various exponents, instead of being fit to STOs of higher principal quantum numbers (such as 2s/3p orbitals etc.).
Has a Gaussian expansion of STOs of higher quantum numbers (for example, 2s or 3p STOs) been performed, and are there any GTO integration schemes for Cartesian GTOs of higher principal quantum numbers?

Good question. I gave you a +1. If you have interest in basis sets, you might be interested in committing to the Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=NzE1YTZmNTZhOGE1N2EwYWYzZGM2NWQyYTFiNzQ5ZDI5MWQyM2EwNzA5OWY5OWY2MDIzYzY2ZTMxNjJlYzIwMSLiKdOhPeuMkh2ODtejwcgH38HFpQq64gQljgKUrZrZ0. Some questions so far: "Is there any hybrid of Gaussians and plane waves that is accepted as mainstream for materials modeling?" "How does a complete basis set remove Pulay forces?" "Is it practical to model [next comment]

"Is it practical to model materials using Gaussians?", "Are plane wave basis sets re-optimized for each material?", "Why are there no correlation consistent basis sets for potassium that are generally contracted?"

I recently realised that the STO-nG basis functions were fit to 1s/20/3d STOs of various exponents, instead of being fit to STOs of higher principal quantum numbers (such as 2s/3p orbitals etc.).

If I understand what you are saying, this is incorrect (although feel free to clarify if I am not understanding the question).
Here is the paper documenting STO-nG for Na-Ar: https://doi.org/10.1063/1.1673374. The first sentence of the abstract:

"Least‐squares representations of the 3s and 3p Slater‐type atomic orbitals by a small number of Gaussian functions are presented."

However, it is true that they are expanded in terms of 1s and 2p Gaussians, which leads to your next question

Are there any GTO integration schemes for Cartesian GTOs of higher principal quantum numbers?

I am not aware of any codes that take advantage of higher principal quantum numbers. There likely isn't any need to.
The Molecular Electronic Structure Theory book (Helgaker et al), has this to say about STO-3G (page 290-291):

"Note the ... absence of a radial node in the STO-3G $2s$ orbital. In calculations using this basis, the $2s$ radial node will arise from a linear combination of the STO-3G 1s and 2s orbitals. The absence of a node in the STO-3G orbital is therefore not a problem."

These nodes arise from higher principal quantum numbers. Note that (almost) all basis set formats do not include principal quantum numbers. (A few do, but they are materials-oriented).

Thanks for your answer! I was simply wondering if any basis set out there fit 2s basis functions to the 2s STO with radial component r e^-ζr, or the 3p basis function to the 3p STO with radial component r^2 e^-ζr.

@Benjamin Pritchard: I gave this answer a +1.  If you have interest in basis sets, you might be interested in committing to the Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=NzE1YTZmNTZhOGE1N2EwYWYzZGM2NWQyYTFiNzQ5ZDI5MWQyM2EwNzA5OWY5OWY2MDIzYzY2ZTMxNjJlYzIwMSLiKdOhPeuMkh2ODtejwcgH38HFpQq64gQljgKUrZrZ0 Some questions so far: "Is there any hybrid of Gaussians and plane waves that is accepted as mainstream for materials modeling?" "How does a complete basis set remove Pulay forces?" "Is it practical to model [next comment]

"Is it practical to model materials using Gaussians?", "Are plane wave basis sets re-optimized for each material?", "Why are there no correlation consistent basis sets for potassium that are generally contracted?"

Has a Gaussian expansion of STOs of higher quantum numbers (for example, 2s or 3p STOs) been performed

I am unaware of such fitting. Typically, basis sets using higher angular momenta (cc-pV$n$Z, def2-$n$ZVP, pc-$n$) are fitted on total energies or correlation energies. See https://www.basissetexchange.org/ for examples and references.

Are there any GTO integration schemes for Cartesian GTOs of higher principal quantum numbers?

Yes, there are. I have seen the following approach in action before:

hand-crafted algorithms are used for common, low-angular-momentum cases
more general schemes, but still optimized schemes for higher cases
a catch-all, not very efficient, unspecific algorithm above a certain threshold

You have quite good knowledge of basis sets. If you have interest in basis sets, you might be interested in committing to the Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=NzE1YTZmNTZhOGE1N2EwYWYzZGM2NWQyYTFiNzQ5ZDI5MWQyM2EwNzA5OWY5OWY2MDIzYzY2ZTMxNjJlYzIwMSLiKdOhPeuMkh2ODtejwcgH38HFpQq64gQljgKUrZrZ0 Some questions so far: "Is there any hybrid of Gaussians and plane waves that is accepted as mainstream for materials modeling?" "How does a complete basis set remove Pulay forces?" "Is it practical to model [next comment]

"Is it practical to model materials using Gaussians?", "Are plane wave basis sets re-optimized for each material?", "Why are there no correlation consistent basis sets for potassium that are generally contracted?"