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What you understand by "Order of reaction"?
When studying Kinetic chemistry (and most of the Physical chemistry topics), we often find definitions based upon the mathematical formalism. For example, Order of reaction is often defined as the sum of the partial order of a Rate law.
But is there a way to define it without using the mathematical formalism? I mean, if we think microscopically, how would we define Order of reaction?

See Wikipedia article Rate Equation -- Gist the order of the reaction is the sum of the product's coefficients is the reaction is an elementary reaction.  If not an elementary reaction, then the sum of the exponents from the rate equation.

Out of curiosity, what is "kinetic chemistry"? I know there is physical chemistry, and thermodynamics and kinetics are the branches within; "kinetic chemistry" is something new to me. And if you mean kinetics, then the above-mentioned Wikipedia articles pretty much answers your question.

I'm voting to close this question as off-topic because sometimes, you just gotta invoke math in order to understand things in the mathematical sense.

Historically, the concept of reaction order came from Ostwald's famous Lehrbuch of 1890s. An abridged translation is available in Internet Archive as "Outlines of General Chemistry". In this book, surprisingly, when a single reactant undergoes a transformation he does not call it a first order reaction. He designated rate of reactions controlled by two species as reaction of the second order.

Reaction of the Second Order. We come to a second kind of chemical process, regulated by a new law, when in its course two substances
  undergo change of quantity. We must again assume that the action is
  proportional to the concentration of each present.
Reactions of Higher Order. We must now consider reactions in which three or more substances take part.

So roughly, you can think of reaction order as the number of species in a chemical reaction which control the rate of reaction.
Coming to modern times, you have fractional reaction orders, and zero order and so on. For understanding you will need to see the concept of elementary and overall reaction orders.
A reaction may be occurring in several steps. In such cases, you might end up in fractional exponents. However, once you are able to know the steps of that reaction, you will see that those elementary reactions don't have fractional exponents. You can then interpret, each elementary reaction in the original Ostwald's way.


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Chirality and number of sterioisomers
I have recently read that the number of sterioisomers in a compound having n chiral centres is $2^n$. (That is, if no meso form is present)
The below question asks for the number of sterioisomers in the product obtained and the product obtained has 3 chiral centres, so it should be $2^3$ = 8, but the answer given is two.
There is no plane of symmetry.
Do I not consider the carbon attached to the benzene ring when looking for chirality?
Original question:

Solution:

The formula is of no use here. Simply think about where the configuration of which carbon(s) in the final product can change in the course of the reaction(by looking at the mechanism), which carbons will maintain a constant configuration, and then count manually the number of ways the configuration of the variable carbon(s) can vary

Is the Grignard reagent truly optically active?  I doubt it.

How can you calim that a Grignard reagent can be optically pure?

I believe you are misreading the question. Rather than asking for the number of stereoisomers the product can have (this is where 2^n comes in), it is asking for the number of stereoisomeric products formed. Note that that the two products are diastereomers, which is a type of stereoisomeric relationship. It's not looking for any fancy calculation; just for the related products.
In a Gringard mechanims, the Gringard reagent can attack the carbonyl on either the re or si face, which in turn, can yield an R chiral center or an S center at the location of attack. This lends to the diastereotpic products.

The inversion takes place at a single carbon inbetween, so 2^1 = 2. Knowing that the mechanism takes place at the carbonyl carbon, the inversion takes place at one point only. It will not form several stereoisomers, only two at the reaction centre.

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Help with molecular dynamics and computational modelling to check validity of hypothesis regarding SN2 reaction mechanisms
Polar aprotic solvents have a less hindered electronegative atom pointing out, namely oxygen, fluorine, and nitrogen, all of which have lone pairs besides having a partial negative charge. Now, all the molecules inside a reaction mixture are moving at great speeds. Not all nucleophiles will attack the carbon at 180° to the leaving group since this provides the best overlap between the nucleophile's lone pair and the C–X σ* antibonding orbital. Other nucleophiles running in random directions can maybe bounce off the solvent molecules’ oxygens’ exposed lone pair right into the antibonding orbital, thereby increasing the number of molecules that successfully overlap in some unit time. The nucleophiles are also negatively charged which increases the repulsion while 'bouncing off'. This is illustrated in the picture attached. I am aware of the present explanation which suggests that oxygen or fluorine can solvate the corresponding cations of the nucleophile and it makes sense. This is just another reason that I had thought of(there can be more than one favorable reasons for phenomena?). The number of solvent molecules will obviously outnumber the substrate and nucleophiles, the ones which are roaming around idly can maybe aid this acceleration ( vis-a-vis the repulsion hypothesis) and increase the rate doubly (not doubly but slightly more than the previously anticipated increased rate).

There is a paper that I am currently working on regarding this. I have been made to realize that I'm going to need a molecular dynamics software to check if this phenomenon is statistically more or less likely. However, I'm currently in high school, hence I do not know anything about the same. Can someone please recommend a software and guide me through the basics? Is this something that I can independently work out, or will I need some assistance from a university( or professors)? How much time and work will it take to simulate?

I'm voting to close this question as off-topic because you're asking us to write at least a Master's thesis.

@ToddMinehardt what no? I'm just asking you to recommend me a software and if this is something I'll be able to do myself. this can be as short as blah blah ( software name) and yes you can or no you cannot

@ToddMinehardt what does closing do? I think my answer is still helpful although I agree the question is way to difficlt

@CodyAldaz closing means that in its current form, the question is not a suitable one to be answered on SE. For example, if a question is too broad i.e. asks too much at once, then this is simply not the best place to do it (better to sit down and talk to somebody about it). This is interesting, but almost certainly too broad. It is basically planning a research project, plus asking for a guide on software, which takes tens or hundreds of pages depending on its complexity. I think you too would know that your answer, while definitely helpful, only skims the surface of what could be done. :)

@ToddMinehardt and orthocresol, the user says "I'm currently in high school, hence I do not know anything about the same."  Therefore, do we have to close the question, or is there something we can do to save its life?

Your hypothesis concerning a "bouncing" contribution to the reaction kinetics would be incredibly difficult to observe or quantify even with the best supercomputers.
The likelihood of a molecule to react in a manner like the SN2 reaction is actually an incredibly rare event at the molecular time scale (femtoseconds 10^-15 s). This may sound contradictory since these reactions are commonplace and occur readily under appropriate conditions (e.g. a few hours heating). However, the two timescales are separated by over 15 orders of magnitude. Many reactions which do approach the transition state can also revert back to the reactant, or in other words abort. So you'd have to run the incredibly long simulations thousands of times in order to see only a small number of reactions. This is the meaning of the pre-factor in the Arrhenius rate  equation
$k = Aexp(\frac{-\Delta G^\ddagger}{k_B T})$
Nevertheless, some of your intuition is correct. A molecule would indeed require momentum along the transition vector in order to overcome the highest energy point (the transition state) and pass over to the product. Otherwise it would return to reactant. Some of this velocity may naturally come from bouncing off a nearby solvent molecule.

Extra Information:
The topic concerning the relationship of transition states and molecular dynamics is actually very complex and a major challenge in the field of chemistry.  The challenge essentially boils down 1) the size of the system and 2) the time length of the simulation.
Fortunately for most practical cases we can simply ignore dynamical effects and study isolated because it is known that solvent interactions are generally weak and reactions closely follow the minimum energy paths.Therefore, the free energy at the transition state ($G^{\ddagger}$) can be approximated in solution using the Harmonic approximation and implicit solvent models like PCM. If you'd like to see how some of these calculations are done, please check out my GitHub on the growing string method (ref 1). These are much simpler to model and actually give a lot of information. Usually the transmission coefficient is assumed to be 1. or 0.5.
If you're interested in some more advanced considerations of the relationship between dynamics and reaction barriers I'd recommend the following references: ref 2,3
References

"Single-ended transition state finding with the growing string method" Journal of Computational Chemistry 2015, 36, 601–611
"QM/MM Protocol for Direct Molecular Dynamics of Chemical Reactions in Solution: The Water-Accelerated Diels–Alder Reaction" J. Chem. Theory Comput. 2015, 11, 12, 5606-5612
https://cims.nyu.edu/~eve2/string.htm

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Mechanism of Possible Acid Catalysed Ring Opening
Here's a problem I'm stuck with -

The first step looks like the usual addition of hydroxylamine to the ketone, resulting in two diastereomeric oximes (this is probably what (A) is). Further, in the second step, the acid can protonate the hydroxyl group of the oxime, or catalyse ring opening.  I don't see how to decide which one occurs.
Moreover, the final product doesn't contain three and four membered rings at all - and has a cyanide group! I wonder how that happened. Please help me figure out the mechanism for the conversion, thank you!

This looks like a variatiant of the Beckmann fragmentation

The first problem is that the reactant has eleven carbons and the product ten. A search of Chem. Absts. does not locate the reactant. The product is formed by acid-catalyzed fragmentation of the oxime of camphor via a second order Beckmann reaction. https://pubs.acs.org/doi/pdf/10.1021/ja01320a046 @Waylander


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Is there any effect on bonds formation and reactants due to a reaction in adiabatic calorimeter
Is there any effect on reactants and bond formation due to a reaction going on a adiabatic calorimeter.We know enthalpy of a reaction should be measured at constant temperature but in adiabatic calorimeter every second the temperature of the chemical solution keeps on changing.

How did you plan to measure reaction enthalpy, if temperature in  calorimeter was constant ? Ideally the constant temperature is meant for the environment, so temperature of calorimeter does not have externally caused drift.

@Poutnik You can measure reaction enthalpy at constant T using isothermal heat flow calorimetry.

Actually, that is an industry-standard type of calorimetry.

@theorist Well, it could be, but it would have indeed an interesting setting and would not be adiabatic. Also I am not sure anyone using such an equipment would ask the above question.


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Connection between irreducible representations and electronic states in diatomic molecules
I am trying to understand the connection between irreducible representations and molecular electronic states.
To explain the problem: I found some potential energy curves (PECs) for $\ce{N2+}$ in the form of irreducible representations such as $\mathrm{A_g}$, $\mathrm{A_u}$, $\mathrm{B_{2g}}$, ...
However, in the literature, I normally find the PECs in the form of $\Sigma$, $\Pi$, $\Delta$... states. How do we connect these two pictures? I am not an expert in quantum chemistry calculations. A simple explanation will be highly helpful.

In the linear $C_{\infty \mathrm{v}}$ and $D_{\infty \mathrm{h}}$ point groups there are two notations for the irreducible representations which are equivalent, in that
$$\begin{align}
\mathrm{A_1} &\equiv \Sigma^+ \\
\mathrm{A_2} &\equiv \Sigma^- \\
\mathrm{E_1} &\equiv \Pi \\
\mathrm{E_2} &\equiv \Delta \\
\mathrm{E_3} &\equiv \Phi \\
&\,\,\vdots
\end{align}$$
For $C_{\infty \mathrm{v}}$, these are the actual irreps; for $D_{\infty \mathrm{h}}$ you simply need to tack on a g or u label to denote the symmetry with respect to inversion. See e.g. the $D_{\infty \mathrm{h}}$ character table found at http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=1001&option=4 which contains both notations.


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What would be the aldol condensation products in B
In this question (13) I figured out A,C were wrong and D would be correct choice but I am stuck at finding out products formed in  B and also why it would be wrong?
Please help.

When you have a mixture of two aldehydes under aldol conditions it is always possible to get self-condensation of one of the aldehydes.

But is is given equimolar mixture? I think may be in in B the para CH3 can lose H and become carbanion and attack acetophenone giving aldol product. Is the logic right?

B is the correct answer, but not for the reasons you have given.
4-Me-benzaldehyde cannot self-condense, and is a far more reactive electrophile than acetophenone.  Acetophenone has a pKa of around 16 so will readily undergo base-catalysed aldol reaction with the aldehyde. This is the only reaction that will occur. All the other sets of reactants shown are likely to undergo self-condensation as well as cross-condensation giving a mixture of products.


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The amount of heat energy required to raise the temperature of 1g of helium at NTP from T1 K to T2 K is?

The amount of heat energy required to raise the temperature of 1g of helium at NTP from $T_1 \pu{K}$ to $T_2 \pu{K}$ is?

My confusion is, that as the gas is at NTP, therefore, it is an isochoric process as well as isometric process.

There are at least two possible answers to this question. It depends on the conditions of the heating process. The gas can be heated at constant volume or at constant pressure. In the first case, Cv = 3R/2 must be used. In the second case, Cp = 5R/2  must be used.

Just tell me if this is right

A gas is at constant pressure if at NTP as NTP means 1 atm

Also at constant volume because NTP means 22.4 l for 1 mole

Also pls look into this working             if we consider both isobaric and isochoric   then delta U=delta Q -delta W so change in work is zero so this should be U=Q which means

n Cv delta T= ncp delta T

And this is wrong

Pls tell where I am wròng

Well. A gas is at constant pressure if the pressure does not change in the container with the time. This pressure may be 1 atm, but it may have any other value, provided this value does not change during your experiment or your measurement.
A gas at constant volume is a gas inside a cloud vessel, where the temperature or itse composition may change. This property has nothing to do with the volume 22,4 L/mol at STP.
Now, if the volume of a gas is constant, there is no work done on or by this gas, and Delta U = Delta Q

What is cloud vessel?

The pressure is 1 atm because it is at NTP       if this isn't the idea then why NTP is used in the question and what does it mean ?

From thermodynamics,
$\Delta U = Q - W$
$\Delta U = Q - \Delta(PV)$
$\Delta(PV) + \Delta U = Q - \Delta(PV) + \Delta(PV)$
We know that, $\Delta(PV) + \Delta U = \Delta H$
$\therefore$ assuming constant pressure, $\Delta P =  0$,
$\Delta H = Q$
$mC_p\Delta T = Q$
From a differential perspective, for the case of a gas at NTP (1 atm), we can use the definition of $Q=m_{g}\int^{T_2}_{T_1}C_p(T)dT$. Where $m_g$ is the mass of gas, $C_p(T)$ is the heat capacity (at constant pressure) of the gas at a given temperature  (heat capacities for gases are widely reported in by institutions like NIST); generally in a form like $C_p(T) = a + bT + cT^2 + dT^3 ...$


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How to create black precipitate in clear liquid
I am not a chemistry person at all, so my apologies if my question seems dumb.
I am working on the set for a short film and in one of the scenes we would like the main actor to look at a sample of water, drop something into it and watch a black precipitate form. Clouding the water.
I have looked online to see what reaction would yield the desired result, however, my chemistry knowledge is very basic.
Could somebody here help me out and let me know what I could use? Safely.
I understand that we need to handle all chemicals with care.

Could you just have the actor drop in some dark food coloring? Do you actually need solids to be formed that can separated out?

Too bad you can’t have a green precipitate and green screen in the black in post-processing. ;-)

@EdV Recalling the old, tube-based TV and their color knobs, I speculate one may record with a non-green precipiate in front of a green screen, and then «simply» alter the color channels in the PAL / SECAM color space.  Because changing colors selectively via huemap on .png with ImageMagick works so fine (e.g. https://www.imagemagick.org/discourse-server/viewtopic.php?t=20280).

@Buttonwood Thanks for that link and information! I was just joking about the green screening, but barium sulfate gives an excellent white precipitate, starting from colorless solutions, and it could be flipped to black fairly easily with software post-processing.

@EdV Indeed, I understood the comment as friendly.  Re colorless / white BaSO4, I recall a lecture by Hervé This (a French chemist active to popularize science) stating «...and as soon you use yellow light instead of day light, you see the albumen («egg white», in French) turning into egg yellow, without [further] cooking, or chemicals» (because egg yolk literally translates into «egg yellow» in French).

I can think of iron(II) sulfide($\ce{FeS}$). There are many ways to make iron(II) sulfide but one reaction procedure will suit your purpose. It uses iron(II) sulfate and lime sulfur(calcium polysulfide) to form a clear black precipitate of iron(II) sulfide with a clear supernatent. See this e-book excerpt.
You can use other sulfides as well. See this picture (full picture here):

From left to right: silver sulfide ($\ce{Ag2S}$, black), lead sulfide ($\ce{PbS}$, black), nickel sulfide ($\ce{NiS}$, black-brown), copper sulfide ($\ce{CuS}$, dark indigo). These are formed by reacting $\ce{Na2S}$ with corresponding metal ions. If you want to fully make your solution black, then your options would be to form silver sulfide and nickel sulfide. You could also make your solution black by forming elemental mercury solution but that would be too toxic for your purpose.
Other ways include adding iodide to excess amount of bismuth(III) nitrate  in dilute nitric acid but be careful not to overdo as excess concentration of iodide will make solution orange-red(See here).
If you do not want any reaction to happen then just dump manganese(IV) dioxide in water. They are black and insoluble.

Many transition metal sulfides are brown/black so they're enticing, but most of their water-soluble salts start out already coloured, sometimes quite intensely, and they could impart that colour momentarily to the aqueous sulfide solution. Iron(II) sulfate has a relatively light colour, though.

The reaction with Iron sulfate could be a good choice but the solution will be greenish so I don't know if the user is okay that the water will have this tint because he said a clear liquid. I would not use mercury for these reactions, they should then disposed of it in special containers.

@GM see updated answer.

+1 that's better, the only cons is that by experience iron sulphate will give some greenish nuance to the solution

What about managanese? I think the sulfate is almost colorless and the sulfide is fairly dark.

@barbecue you can use whatever you want as long as they impart black color.

I would suggest an iodine starch reaction as shown in this video. This reaction is safer compared to the use of other heavy metals that must always be disposed of properly.
In this case, you have to dissolve first starch in water (for instant soaking bread in water and then filtering) or buying potato starch powder. The water will be still transparent but some starch will be dissolved in the solution.
You could try to add some ethanol to the solution this should improve the chance that the starch will precipitate.
The solution to be poured in should be made of iodine (I2) and an iodide for example potassium iodide (KI) (this solution can be found with the commercial name Lugol's solution). When you add this solution you will see appearing some dark spot the precipitate will not so evident if the solution is not saturated so you might want to try to maximize the amount of starch dissolved.

Furthermore is not actually black but a dark blue, however, I don't think you will notice the difference. Once finished, it can be disposed of in diluted water in the sink.

Chemically related (because the same compound is formed) is the iodine clock (e.g., https://www.youtube.com/watch?v=xLCDJ0m_qrk) with the added delay (movie speak: suspense) between mixing almost colorless liquids and the formation of the dark matter.

Would this be a precipitate though?

@JMac No, both forming the iodine-starch complex equally will only cloud the colorless aqueous solution, and not form a solid precipitate.

@Buttonwood the precipitate will not be very evident if the solution is not saturated, but you can try to add ethanol to facilitate the precipitation (I wrote it now in the answer)

+1 for ingredients being (relatively) safe and widely available

A classic lab experiment would be the formation of black lead(II) sulfide by mixing potassium sulfide with lead(II) acetate in an aqueous solution:
$$\ce{Pb(CH3COO)2(aq) + K2S(aq)  -> PbS(s) + 2 CH3COOK(aq)}$$
Safety precautions must be taken due to neurotoxicity of soluble lead(II) salts.
In place of $\ce{K2S}$ one can also use aqueous solution of ammonium sulfide or hydrogen sulfide.

Because of the considerable hazard of lead solutions, it may be prudent to avoid it altogether. A silver nitrate solution would give a similar effect, though ammonium sulfide may not work for this specific metal. I suspect there are better options still, but nothing comes to mind right now.

@NicolauSakerNeto Correct, silver would do too. Bismuth as well. I picked lead simply as the cheapest option assuming one would need a substantial amount of the precipitate to occur for the good footage.

Can someone tolerate the smell of hydrogen sulfide? TBH I can't ;)

@NilayGhosh Actors are paid to smell things that are even worse than this. Besides, I added $\ce{H2S}$ solution as the last option;)

I was planning to put condition infront of OP, "will tell only if you project me as the actor", but you ruined my plan ;)

Folks might not like their skin turning black if the silver solution gets on their skin.

I know fountain pen's are less often used nowadays, but you could drop some droplets of their ink into water and than take a macro photo / video:

(screenphoto from here)
There are plenty of companies in the field (e.g., Waterman, Caran d'Ache, etc.) offering ink.  You might be surprised how little ink is necessary for this, e.g., just while rinsing the nib and the inner tank of your fountain pen.  And you are not limited to black, or even a single color.

Yes, it would work, but mixing a clear or lightly colored liquid would be more  surprising....that is the magic of chemistry,

@AJKOER Like pushing a white shirt into the yellowish aqueous solution of leuco-indigo which, just brought back into the air slowly becomes jeans-blue. Or the dark red by mixing the colorless solutions of $\ce{FeCl3}$ and $\ce{NH4SCN}$, «suddenly» disappearing by addition of $\ce{NaF}$ as in https://www.youtube.com/watch?v=Q4OfkdyMLRQ or with a nail: https://www.youtube.com/watch?v=2cknYNl5v6Y.  Who knows what they have at their disposition.

The question does request a precipitate though.  I don't think ink droplets water are forming a precipitate.

@JMac True, this does not yield a solid, crystalline precipiate one could separate e.g., by gravity filtration.  It is clouding the water, though.

This is not a chemistry answer, but it is not a comment either?  Please remove if it doesn't fit here.
Since you are not familiar with chemistry, why risk it?
I would recommend just dripping dark food coloring or ink into the water, and using post-processing to either replace the drop with another shot of plain water dripping into the beaker, or modifying the color.  There is likely to be a very clear cutoff (especially if filmed parallel to the surface of the water) and a limited time range.  This could be done without expensive software or special skills.

@Ciero The thread becomes fuzzy.  But indeed, CGI is advanced enough to simulate the ink droplets in water; a tutorial post by 2017 (https://www.blendernation.com/2017/04/20/tutorial-ink-dropping-water-simulation/) for the freely available blender.  There even is a relevant entry on blenderSE, https://blender.stackexchange.com/questions/135244/black-lines-in-smoke-render/135253#135253 for a «chemistry free» version.

I wasn't even suggesting simulation, merely composing a water drop falling into a glass with an ink drop falling into a glass.  Worst case this can be no special effects whatsoever, just a cut between the two shots at the moment of impact


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In free radical substitution of propane with halogens, how do you know which carbon ends up with an unpaired electron after fission of C-H bond?
In free radical substitution of propane with a halogen (eg- chlorine), how do you know which carbon out of the three ends up with an unpaired electron after fission of C-H bond?

This image shows methane, so there is only one possibility since there is only one carbon, but what about alkanes like propane with multiple carbons? How do we know which carbon atom ends up with the unpaired electron and why?

On which one is the radical most stable?

@Waylander The middle carbon. Because it would be surrounded by alkyl groups?

You are correct

This is to do with the relative stabilities of the free radicals produced. Here is a schematic diagram showing the two possibilities for propene (similar mechanism, also with two free radical possibilities like propane):

As you can see, a more stable and less stable free radical is produced. The stability depends on how many carbons are attached to the carbon with the unpaired electron. In the diagram, the free radical where two carbons are attached is more stable than the free radical with one carbon attached.
As a general rule, the stability will be greatest in tertiary free radicals, then in secondary, and least stability in primary.
This is due to the inductive effect of carbons bonded to the carbon with the unpaired electron:

Carbons bonded to the free radical carbon are able to 'move' some of their electron density over to the free radical carbon, stabilising it. This is due to something called sigma conjugation, which we needn't go into.
But in summary, both free radicals are formed, as seen in the first diagram. Both free radicals go onto generate products, but much more come from the more stable species. Therefore, in fact both carbons possible in propane end up with the unpaired electron, but the route involving the secondary free radical species is more successful and gives the major product. The route involving the primary intermediate gives a minor product.

In free radical substitution, a free radical is formed that is an electron deficient species and is stabilised by electron donating species bonded to atom on whose orbital it is present. So, electron donating groups that is +I ,+M increase chances of formation of free radical on it and presence of electron withdrawing species (-I,-M, Backbonding,by Cl) reduce chances. Hyperconjugation is also applicable here as this reaction satisfies its conditions, Carbon atom (sp3 hybridised)having atleast 1 Hydrogen atom directly bonded with carbon having a vacant p orbital or bonded with Carbon that is bonded with sp2 hybridised atom. Concept is that: Free radical or any chemical species forms  where is most stable or follows the most stable path.


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What does the value of the pH meter mean when there aren't any hydronium ions in the substance being tested?
What does the value of the pH meter mean when there aren't any hydronium ions in the substance being tested?
How does the pH meter show a value which fluctuates as non-aqueous solvents are being tested? I thought that the pH meter only measured hydronium ions. Is there any way that this value actually means something in terms of acidity and basicity?

Please be more specific, include the substance and solvent in question and also tell us what pH readings you are getting. If a purely hypothetical question, please edit it as such.

If you put a glass electrode into a non polar solvant, it will not give any stable values. The measured pH changes erratically form to +1.68 to -4.05, then to +2.34, then to  -0.65 or to +3.19, then to - 6.95, then to any values you may think of. It will not stay stable. Try ! You will be surprised !

It may very depend on solvent. For nonpolar solvents like n-hexane, it may wildly fluctuate, driven by random electrostatics ( I remember an old vacuum tube based pH meter in a high school we had to be 1.5-2 m away and not moving to have a steady value. I guess PVC floor did not help. ) For polar ionizable solvents like alkohols, it may get more or less stable, even if strange value without particular interpretation.

Nowadays you have to differentiate between pH measurements with non-glass sensors and with glass sensor. In the typical pH meter with a glass sensor, there could be many things that could go wrong resulting in biased results. In general, some electrodes may be sensitive to interfering ions such as $\ce{Ag+}$. So some potential could be generated also by other ions not only hydronium ions. However, the most common thing is that you are actually reading biased measurements because you are perturbing the status of your instrument.
Interaction of the solvent with the materials of your pHmeter
First thing you have to consider is the interaction of the solvent with the materials of your pH meter. You can easily damage some part of it or dehydrate the glass membrane of your electrode.
Solubility of the electrolyte in the junction
Then you have to bear in mind that in your glass pHmeter there is a junction that leaks an electrolyte for forming the saline bridge if this electrolyte is not soluble in your solvent the "circuit" will not close and you will have a wrong measurement.


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Why was Avogadro's number chosen to be the value that it is?
I understand that the mole is an extremely useful way of measuring and understanding atomic quantities. I did find some history of the number in its wikipedia article, but I did not find an answer to the question "Why is the number chosen to be what it is?".
I understand that it is not necessary to know the history of the number in order to apply it, but it would seem to me that the number wasn't chosen arbitrarily. Why is it, then, that Avogadro's number was chosen as $6.022\times10^{23}$?

Have you tried wikipedia ? BTW, current definition is exactly $\pu{6.02214076e23 mol^-1}$ ( A. constant = A, number with the unit )

I have, although there is some history of the number, I didn't find anything that really satisfies my curiosity. My question still being "Why is the number chosen to be what it is?", and what I think is more important is the second question "How do chemists approximate atomic mass?".

It is supposed on chemistry SE site you explicilty mention what you have tried and failed, so responders do not try to advice you what you have already done.

See also Dalton unit All what is needed is patience, thinking about terms to be searched and following links about the topic.

If you know why the mole is important, then that's all there is to it, really. The mole is chosen to have a certain significance in that it relates atomic masses to macroscopic masses (if you don't get that, I suggest re-reading its definition). Ignoring any quibbles over the recent redefinition of the mole, the actual number 6.022e23 just so happens to be the number of particles that make up a mole, and that number doesn't have any real significance that wasn't already captured in the mole definition.

The number was arbitrarily chosen back in the day so that 12g of carbon-12 are exactly one mole.

https://chemistry.stackexchange.com/questions/32443/what-is-the-correct-value-of-the-avogadro-constant-and-how-was-it-derived

https://chemistry.stackexchange.com/questions/60767/effects-of-changing-avogadros-constant

Does this answer your question? Why is the definition of the mole as it is?

Why was Avogadro's number chosen to be the value that it is?

Your question implies that you already know that it was a choice rather than something derived from first principle. There are some numbers that are derived from first principles. In math, $\pi$ and $e$ are not a choice, but can be derived from their properties. In physics, the fine structure constant is not a choice, but would be measured to get the same value by aliens on a different planet (at least as far as we can tell).

[...] Avogadro's number [...]

To be correct, it is actually Avogadro's constant $N_A$ that we use more often in chemistry (for identities such as $k_B = R / N_A$, where $k_B$ is the Boltzmann constant and R the universal gas constant). Avogadro's constant has the dimension (in the SI unit system) of one divided by amount of substance. So the number in front of the unit will change as you change the unit, just like 30 minutes is the same as 0.5 hours even though 30 is not the same as 0.5.

I also understand that it is not necessary to know the history of the number in order to apply it. But it would seem to me that the number wasn't chosen arbitrarily.

The value of Avogadro's constant is linked to the choices that were made when setting up the metric system. There is one definition, that of the meter, that is linked to our planet (so aliens on another planet would have no reason to choose it). The meter was chosen to be approximately 1/10,000,000th of the quadrant of the Paris meridian (i.e distance from Paris to South Pole to North Pole to Paris divided by 40,000,000). Then, they made an artifact to define the meter exactly. Next, the kilogram was chosen to be approximately the mass of 1/1000th of a cubic meter of water (at some temperature). It is no coincidence that the density of water is very close to 1 kg/L. Again, they made an artifact to define the kilogram exactly.
Once the kilogram was defined, the mole was chosen to contain the same number of particles as (pick your favorite, they are all approximately correct at this point) one gram of protons or neutrons, 16 g of oxygen atoms or 12 g of pure carbon-12 atoms.
How are the definition of kilogram and mole related to Avogadro's constant?
Avogadro's constant tells us how many particles there are per mole of particles. The mole was defined with reference to the gram and fundamental constants such as the mass of a proton or of a single carbon-12 atom, so once these choices are in place, Avogadro's constant is no longer a choice.
Relationship to current definitions of units
The SI units (including meter, kilogram and mole) are defined in a different way today, but their values are still the same up to many significant digits. So you can still trace back the choice of the Avogadro constant to choices made when defining the meter, the kilogram and the mole.

Nitpick: a mole of protons weighs a quite measurable amount more than 1 gram. That approximation is off by close to 1%. The difference between using C-12 vs O-16 would be quite a bit less.

From the practical historical perspective chemists and physicists needed a number to use as a conversion factor from Daltons to grams to perform stoichiometric reactions.
In the beginning, Dalton proposed to use the mass of the Hydrogen and assign it to a value of 1 Dalton without knowing how many grams correspond to 1 Dalton. Scales use grams, not daltons so scientist needs a conversion factor. To do that we could take a gram of hydrogen and count how many atoms are there this would return what is called Avogadro constant. In "theory" this will give the same result also if we take 16 grams of Oxygen (that has 8 protons and 8 neutrons so roughly 16 Daltons) and the results should be the same (of course in practice we would have to take 32 grams of molecular oxygen $\ce{O2}$ because atomic oxygen is not stable. In practice is not like this because there are other factors related to the fact that as Einstein noted mass is also related to energy. Physicists however used Oxygen-16 for a while, then they team up with IUPAC and decided to use 12 grams of Carbon-12. The values of Avogadro costant changed a bit but they finally now after the 2019 redefinition of the SI base units it has been fixed to a value.
This is a table form this article (Jensen, William B. "Why Has the Value of Avogadro’s Constant Changed over Time?." Journal of chemical education 87.12 (2010): 1302-1302).
Table 1. Example Experimental Values for Avogadro's Constant over Time*

|------+-----------+-----------|
| Year | Author    |  Nᴬ/10^23 |
|------+-----------+-----------|
| 1908 | Perrin    |       6.7 |
| 1917 | Mullikan  |     6.064 |
| 1929 | Birge     |    6.0644 |
| 1931 | Bearden   |     6.019 |
| 1945 | Birge     |   6.02338 |
| 1951 | DuMond    |   6.02544 |
| 1965 | Bearden   |  6.022088 |
| 1973 | Cohen     |  6.022045 |
| 1987 | Deslattes |  6.022134 |
| 1994 | Basile    | 6.0221379 |
| 2001 | De Bièvre |  6.021339 |

* A more complete list can be found in ref (3).

Was the choice of the element arbitrary?
No, in fact, the choice of the element was based on some practical considerations initially was preferred oxygen because the way they measured the weight was using ratios between the compounds citing IUPAC publication:

Berzelius published a series of articles on atomic weight measurements
  in which he reported actual weighings along with results from other
  chemists on an oxygen scale i.e., value of O = 100. Berzelius felt
  that oxygen was a good reference material because it combined with
  most elements (in inorganic chemistry) and many molecular and atomic
  weights could be measured directly. The drawback to hydrogen as a
  reference was that it combined directly with a limited number of
  elements only and the atomic weight must be evaluated in comparison
  with oxygen. As the oxygen to hydrogen ratio could never be determined
  with absolute accuracy, any errors in this ratio would be carried over
  into all other values of atomic weights.

When finally they start using mass spectrometry for determining weights they switched to carbon:

In April 1957 at the bar in the Hotel Krasnapolski in Amsterdam, Nier
  suggested to Mattauch that the 12C = 12 mass scale be adopted because
  of carbon's use as a secondary standard in mass spectrometry63. Also,
  12C = 12 implied acceptable relative changes in the atomic weight
  scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for the
  16O = 16 scale (which would not acceptable to chemists).
  Enthusiastically, Mattauch made a worldwide effort in the late 1950s
  to publicize the 12C = 12 scale and obtain the physicist's approval,
  while Wichers obtained the chemist's approval.

That is a good explanation. How is it that one "counts" how many atoms are in the original one gram of Hydrogen though?

@LukePaluso Actually most of times is using indirect measurement (e.g. current) but this is another question...

You gave a glimpse into the history, but it is really more convoluted than how you describe. Dalton's Law of Partial Pressures dates to 1803, and his first publication on the atomic masses was in A New System of Chemical Philosophy, published 1808, and a supplement published in 1810. Mendeleev's periodic table was published in 1869. The existence of isotopes was first suggested in 1913 by the radiochemist Frederick Soddy. So there is a tremendous about of scientific advancement wrapped up in all of this.

Also to be excessively fussy it is the Avogadro constant, not the Avogadro number.

@MaxW thanks for your feedback yes indeed I tried to summarize, I think it's important to understand that is indeed a convention but it linked to historical and technological advances and some practical aspects, the history I quite complex but in the IUPAC publication the user can have a good summary.

@MaxW thanks for pointing it out I'll correct the answer

Please understand that I'm trying to point out that science hasn't moved in a straight line from the Greek notion of the four elements being earth, air, fire and water to today's perspective. Rather it has been a meandering cow path between the two.

Let us continue this discussion in chat.

@GM The last line of the table from J. Chem Ed, "2001 | De Bièvre |  6.021339", has a typo in the value.  "2" should be inserted as the thousandths digit.  Their mistake not yours.  See https://ieeexplore.ieee.org/document/918199

Initially the Avogadro was defined as the number of atoms contained in 1 g Hydrogen. Later on, it was understood that Hydrogen can contain various amounts of Deuterium. So that, instead of referring to the fluctuating H atom, it was decided that the Avogadro would be related to the weight of another atom. There had been long discussions about the choice of this atom. Oxygen-16 and Carbon-12 were proposed. After discussion, the atom that is 12 times heavier than the H atom was chosen, namely the Carbon-12 atom. So the Avogadro number was up to recently,  the number of Carbon-12 which is contained in $12.000$ g $\ce{C-12}$. But one or two years ago, it was decided that the Avogadro will not be related to any atom, and that it is a given number, which cannot be changed later on if some not yet known technical difficulty would happen in the future.

Is it the case that all atom's atomic mass is measured relatively to a gram of Hydrogen using experimentation?

The atomic masses are always determined by experimentation. They are usually measured by mass spectrometry, with respect to Carbon-12.

Other answers have covered it adequately. But there's one other way in which Avogadro's constant is not completely arbitrary. It's an order-of-magnitude measure of the quantity of particles that can sustain sentience on an Earth-type planet. 12 grams of carbon-12 is a sample of matter that's roughly on the same scale as us humans.
I'm serious here. How many atoms does it take to build a self replicating molecule? How many molecules per a cell that's capable of forming multicellular organisms? How many cells per a sentient lifeform? The answers to those questions clearly have a lower bound. Avogadro's constant is the measure of minimal required complexity of sentience. We don't have a lot of reference points on all three - DNA/RNA (citation needed), eucaryotic cell, human. That said, if there was a simpler way to build a cell, or a cheaper way to sustain a central nervous system of 15 billion neurons, evolution would've probably already stumbled upon it, what with the humongous amount of thermal noise it had to work with.
In that way, those aliens on a different planet from Karsten Theis' answer, were they to come up with a concept similar to mole, might have the their coefficient of proportionality roughly in the same ballpark.
See also: Kolmogorov complexity.

Thanks for this out-of-left-field analysis. It is also possible that aliens would have decided not to have the concept of amount of substance at all; in fact, there are some humans who think it is a superfluous part of the SI system, see e.g. https://physics.stackexchange.com/questions/174541/why-is-the-mole-amount-of-substance-a-dimensional-quantity


text:
What is the materials science behind thermal diffusion in multispectral camouflage?
I've been interested in infrared and near-infrared camouflage (called multispectral camouflage) for a couple of days now, and I've found a video or two of certain companies that have created fabrics that both reject outside thermal and near-infrared radiation, and allow for scattering of infrared and near-infrared radiation originating from the body as body heat.
It's been three days and I've been searching for the materials science behind the camouflages, but I can't find anything relevant.
So, what materials/substances do you think they are using for this type of camouflage and why?

What do you mean by "reject" and "scattering"? Is the aim for a body to appear black when illuminated  by an IR lamp? And you can only scatter outside light. Thermal emission always occurs in all directions anyway.

The aim is both to take incoming thermal (IR) radiation, and thermal energy from body heat, and when the material radiates the IR energy it sends it off into arbitrary directions, not just toward the observer, lowering absorbed heat by the observer. In this sense the external heat (for instance from an IR lamp) is not very reflected by the material; it should appear like normal background when illuminated, but also take internal body heat and "diffuse" it in all different directions, again, giving out  a heat signature that is similar or equal to the background.

You seem to be missing the point. Except polished metal, hardly any material reflects (like a mirror) IR. And any warm body, including a piece of camouflage, radiates thermal IR into all directions. You can add an insulation layer on the inside to keep your body heat, but I think that´s not the idea here.

The wiki article suggests that Peltier devices are used to spatially modulate the thermal output across an object to break up an infrared image. Is that kind of what you mean?

What would make sense is a material that absorbs NIR (making it look black in an IR searchlight), and does not absorb FIR, because a low absorbance generally means a low emittance at the same wavelenght. Could that be it?

No in the video they have a graphic showing incoming IR radiation being deflected and emitted (body heat) IR radiation being sent in arbitrary directions. The video also clearly shows the fabric is thin and flexible, so I doubt physical peltier plates would be involved. What I'm thinking is one of two things; one, that they made a fiber or coating that scatters IR energy, or two, a fiber or coating exhibits the Peltier effect; using up the body heat to produce electric current to drive a peltier cooler. Any thoughts?

New development; Rayleigh scattering. Do you think they found a material that scatters infrared radiation?


text:
What is nowadays called silazane?

Example from the Recommendations 2013. Left: Alleged disilazane ; Right: My version

Clearer example without obvious flaws in alternative structure. Left: Alleged disilazane ; Right: My version
I am confused with the contemporary IUPAC nomenclature of organic compounds. According to Recommendations 93 (available at acdlabs.com), silazanes are heterogeneous hydrides where both terminal atoms are silicon atoms (R-5.1.4.2). The same can be concluded from the general rule on heterogeneous hydrides (R-2.2.3.2). However, in the introductory chapter of Recommendations 2013 (which is freely available at RSC Publishing site) on page 82 the compound given in the image is given two names, with the least priority (non-PIN) variant being 1,1′-(2-ethenyl-1,2,3-trimethyldisilazane-1,3-diyl)bis(ethan-1-one). I am confused by this name, as, by rule R-2.2.3.2, terminal atoms should be lowest in the Hantzsch–Widman order, which preserves unambiguity of the structure. From this rule, I expect disilazane to be something like the structure on the second image (although I acknowledge that the tetravalent nitrogen must be positively charged, which changes the name automatically).
TL;DR So the question is:

Do Recommendations 2013 change something about naming heterogeneous hydrides? And if so, what is now called silazanes?

With view on both the wikipedia entry (https://en.wikipedia.org/wiki/Silazane) and by a commercial supplier (https://www.sigmaaldrich.com/chemistry/chemistry-products.html?TablePage=16245452) I would like to hint that the second formula features a tetravalent N the aforementioned examples (and the first example by you) do not.  I thought «disila» then indicates the presence of «two atoms of silicon» as in the right hand example and HMDS (https://en.wikipedia.org/wiki/Bis(trimethylsilyl)amine) though.

I know about the tetravalent N, I addressed it in the question. I will add another pair of structures for my point to be clearer; it is just that this is example from the Nomenclature book.

The Recommendations 2013 are not bug-free, in fact: see, for example, the 2nd structure on page 75.

It seems this name should not be in the blue book. It is an error.
The following instruction is what Corrections to Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013 has given for your problem:

Page 82, P-15.3.2.4.1, example 4 on this page.
  Delete 1,1′-(2-ethenyl-1,2,3-trimethyldisilazane-1,3-diyl)bis(ethan-1-one).

As Igors Mihailovs (OP) pointed out in comment section to the question, the Recommendations 2013 are not error-free. You may see how nany errors have been corrected in this web page. For example, page 75 has 4 errors in that page alone!

That error list on the web page struck me like a lightning bolt :-0

I actually had another subquestion about that structur, as to why there is locant 1 cited in the "ethane-1-one", but multiple corrections on the page You cited confirmed this is ANOTHER bug in this name.


text:
Why is it that when I add yolk to vinegar and olive oil, it doesn't create a stratification?
Why is it that when I add yolk (which mostly consists of minerals, vitamins, lipids and proteins) to vinegar and olive oil, it doesn't create a stratification?
I'm really confused about this because if I only add vinegar to olive oil it actually does create a stratification. Could it be because the proteins somehow react with the fat in the olive oil or what's going on?

Please see “Mayonnaise” in wiki.

With the egg yolk and mechanical whipping, you're making a more stable emulsion than without the egg yolk.

The egg yolk contains a lot of lecithin, which is a phospholipid with 1 fatty acyl replaced with phosphatidylcholine.
It acts like emulsifier, what is used in industrial food production.


text:
Can ferrocene type sandwich compounds be made with fluorene?
Ferrocene will be well known to made members of this site. It consists of two cyclopentadienyl anions sandwiching an iron-II ion.
During my first year organic chemistry course (about 40 years ago) I remember a friend suggesting that the fluorene anion had a similar stability and it might be possible to make ferrocene type compounds with an iron-II ion sandwiched between two fluorene anions.
A group of us persuaded a sympathetic professor to let us have a go, and as I recall we prepared the anion by adding lithium aluminium hydride to solution of fluorene in ether cooled with dry ice then adding iron-II chloride. The iron chloride immediately went into solution, so some form of complex was formed, but we couldn't get much further as on warming to room temperature we got a hideous mess of many different products and we didn't have time to analyse it thoroughly.
So my question is whether these types of fluorene compound are known, or alternatively whether it is known that they cannot be prepared? I have attempted to Google this, but there is such a vast literature on ferrocene type compounds that I have not been able to find anything relevant.

You mean sandwich compounds in general? They sure are.

I rather doubt you added LiAlH4 to a solution of flourene in acetone. The reaction between LiAlH4 and acetone is immediate and violent. Have you misremembered?

@Waylander It was 40 years ago so I have quite possibly misremembered. Maybe the solvent was ether?

Ether is much more likely, or THF

@Waylander the guy who set it up for us kept an eye on us to make sure we didn't harm ourselves or anyone else :-)

I doubt the reagent was LiAlH4 as that is a powerful reducing agent and would not give the anion of fluorene, possibly it was lithium hydride. When were you at Cambridge?

Yes there are, many in fact.  One of the examples is having a complex consisting of a fluorene molecule and of cp altogether with iron, described by Decken in Acta Cryst. E (doi: 10.1107/S1600536804028077):

(source)
You are not limited to complexation with iron, though.  Already by the title «Synthesis and characterisation of Ti, Cr, Mo and W bis(fluorene) complexes» by Cloke et al. in J. Organomet. Chem. (source, paywall) you may see that there is a lot of variation possible by either varying the metal, or the second molecule to build the complex.

Thanks :-) Although I'm a bit surprised that is a neutral fluorene molecule. I assumed any compound would be with the fluorenyl anion. Do you know if that type of compound exists?

@JohnRennie There's lots of kinds of sandwich compounds, this one's still rather "classical" ;) Still, Wikipedia art. shows one of an anion derivative https://en.wikipedia.org/wiki/Fluorene

@JohnRennie What the image depicts is only a selection of the structure model, the ($\eta^5$-Cyclopentadienyl)($\eta^6$-9H-fluorene)iron(II) cation.  There equally is a hexaphosphate anion, $\ce{PF6^-}$ in the model, too.  If you access the publication as .html (https://journals.iucr.org/e/issues/2004/12/00/hg6105/index.html), the fourth button in the left hand column, just below «cif» opens an interactive 3D view (direct: http://publcif.iucr.org/cifmoldb/gui/cifjmol.php?cifid=hg6105) -- and this tetragonal bipyramide easily is seen.

@JohnRennie Regarding the fluorenyl anion alone, formally obtained by giving away one of the two H's at C-9, there equally are complexes with metal cations much less nicely «piled» as ferrocene and the example shown in the answer, too.  Ad hoc an example as Mg salt (https://doi.org/10.1524/ncrs.2012.0243) which may be seen interactively with the remaining hydrogen atoms in the CCDC data base here: https://www.ccdc.cam.ac.uk/structures/Search?Doi=10.1524%2Fncrs.2012.0243&DatabaseToSearch=Published

Thanks again. It looks as if the negative charge is not delocalised as it is in the cyclopentadienyl anion i.e. the metal is coordinated to the C9 carbon not to the centre of the C5 ring. We had anticipated the charge would be delocalised so you'd get a direct analogue of ferrocene.


text:
C60 Fullerene Polymers?
Upon scouring Wikipedia for info about fullerenes to research, I came across this paragraph in the page on Buckminsterfullerene (C60):

Under high pressure and temperature, repeated [2+2] cycloaddition between
    C60 results in a polymerized fullerene chains and networks. These polymers
    remain stable at ambient pressure and temperature once formed, and have
    remarkably interesting electronic and magnetic properties, such as being
    ferromagnetic above room
    temperature.[43]

This sounded like a very intriguing topic, and I instantly got hooked. However, the source that it cited was "Chapter 14: The group 14 elements" from some inorganic chemistry textbook, and contained little more information than the quote above. Upon performing an internet search, I found little related research, and no online journal that both published the research and was one that I had access to. If anyone has more information about this mysterious polymer or it's variations, properties, or related research, I would be grateful.
Note that I'm not referring to the dumbbell-shaped dimer of C60, which I found plenty of research on. However, all answers and contributions about the polymeric form are greatly appreciated.

Interesting reaction! Here is what I found: Dimerization and Fusion of Two C60 Molecules by Kaur et al.


text:
How do we explain mutations, in the deterministic world of chemical reactions?
I'm not a chemist, neither a molecular biologist. But I'm fascinated with science.
The way I understood, DNA is a macro-molecule which is made of atoms. And in reproduction, either mitosis or meiosis then mating, or in RNA creation mutation can happen. And mutation means a letter molecule in the chain of DNA can become another letter.
From the other side I know that by combining a known set of reactants, we get the same products, ALL the times. In other words, chemical reactions are deterministic and no matter how much we try, H2 + O results in water, and not in solphuric acid for example.
At this point I'm stuck. And I can't find an answer. How mutation happens in DNA while it's the outcome of chemical reactions and those reactions are deterministic?

You know that short wavelength UV, i.e., UV-C, can damage DNA and kill viruses, bacteria and cause cancer, right? And radiation and cosmic rays can cause damage to biological systems. So it is not perfect clockwork, even ignoring competitive reactions, etc.

Chemical reactions aren't deterministic. Side reactions happen all the time.

The topic is much larger than a scope for a single answer. But staying at general, low level chemical principles, it can be pinned down to several mechanisms leading to possible mutations:

Chemical reactions are not deterministic. Generally, on molecular level, there is less than 100% probability the desired and expected reaction happens, and more than 0%  probability it does not happen, or other reaction happens.
A molecule may react with other molecule in more than one way, with alternative reactions having their probability to form  "by-products".
A molecule may react with similar, but still native molecule, that it is not supposed to react with to form desired product.
A molecule may react with a molecule that should not be present, coming e.g. from pollution, medical treatment or poisoning.
Some foreign molecules may not react directly, but can affect DNA replication or synthesis of RNA or proteins.
A molecule may be broken by ionizating irradiation like UV, X-ray or  $\gamma$ photons, high energy subatomic particles or ions from the natural body radioactivity ($\ce{^{14}C}$, $\ce{^{40}K}$), from natural radioactive background ($\ce{U}$, $\ce{Ra}$, $\ce{Rn}$), from pollution (radioctive elements from coal burning ), medical treatment, radiation incidents or even poisoning.

$TAR86 Aromatics from coal are not radioactive, but heavy elements in coal ash are. Radioactive pollution for coal burning is much higher than pollution from nuclear plants.

Mutations are changes in the genomic sequence of an organism and more broadly, per the Wikipedia, viruses and extrachromosomal DNA. Mutations can occur at the source (original genome) or during replication (generation of product genome). DNA may be damaged in many ways. In addition to spontaneous reactions there can be all sorts of reactive chemicals (such as reactive oxgyen species or ROS) in a cell and external radiation sources that can cause unwanted changes to the chemical structure of the original DNA. Changes to the DNA can cause difficulties replicating it with high fidelity. In addition, the replication system is not foolproof, despite error-checking mechanisms.
Many reactions that occur in the cell are predictable, in that they have been studied and are often well understood, and in that the probability that they should occur under specific conditions (for instance in vitro - "in a test tube") have been well characterized. A cell is a very complex soup of different proteins, carbohydrates, lipids, ions, complexed metals, metabolites, and water. All these molecules can impact the chemical stability of DNA and the replication system. This opens up the door for a lot of potential wrenches in the mechanism, but only if the molecules gain physical access to the DNA or the replication system. In addition, cells have built-in repair mechanisms, but, like humans, they are not foolproof. However, estimates are available of the frequency with which they fail.


text:
Why does a bonding molecular orbtial increase stability?
When two atomic orbitals come together, they fuse to form a molecular orbital. There is a large probability that the electrons will spend time between the nuclei. This will reduce the nuclei - nuclei repulsion and hence increase stability.
The part that I do not get is how do the electrons reduce nuclei-nuclei repulsion? Is it because the electrons in the middle attract the positive nuclei. This creates an opposing force to the electrostatic repulsion? Is this the case and if not what is actually causing the stability?

The forces involved are electrostatic, nuclei repelling one another and electrons similarly, but also being attracted to the opposite nuclear charges. Also there is kinetic energy of the electrons (for simplicity we consider the nuclei to be stationary) which form the total energy of the molecule.
The simplest molecule to consider is $H_2^+$. The interactions have potential energy arising from Coulomb interactions $$\displaystyle \displaystyle V(r) = -\frac{e^2}{r_1}-\frac{e^2}{r_2}+\frac{e^2}{R}$$
where $r_1,r_2$ are the separation of the electron from the nuclei and $R$ the separation of the nuclei themselves. In addition there is kinetic energy to consider and this has the form $p^2/2m$ where $p$ is the momentum (operator). Combining these terms the Schroedinger equation is formed and in doing so some integrals have to be solved.
Some of these have classical counterparts and are Coulomb type integrals, but one type, the 'resonance' or 'exchange' integral, is purely quantum mechanical. It is difficult to describe in words but it expresses the possibility of the electron moving from the vicinity of one proton to that of the other. All these integrals are positive but add together according to the signs in the potential.  The upshot of this is that the total energy reaches a (negative) minimum as the separation between nuclei is increased from a very small value, and becomes zero when the nuclei are well separated.
The electron has kinetic energy as well as potential energy so besides the potential energy effect of reducing the repulsive interaction between nuclei, by virtue of having increased density between them, the positive kinetic energy can counteract this effect.
To find out which is most important the Virial Theorem (as result from classical mechanics) can be used. In the special case of a diatomic molecule the average kinetic energy is given by $\langle T\rangle=-E-R(dE/dR)$ and the potential energy $\langle V\rangle=2E+R(dE/dR)$ where $E$ is the total energy at $R$ the internuclear separation.
The plot below shows that it is the minimum in the potential energy that caused the chemical bond to be formed not the smaller minimum in the kinetic energy which is positive anyway. Thus in forming a bond the kinetic energy increases (destabilising) but the potential energy decreases by more and this stabilises the bond.  This is shown below where $\rho$ is bond length in reduced units and $E_I$ is the ionisation energy. $E_0$ is the minimum total energy.


text:
Writing a organic reaction in different form
Can I write this reaction
$$\ce{A + B ->[\ce{X}] C + D ->[\ce{Y}] E + F}$$
as
$$\ce{A + B ->[\ce{(1) X}][\ce{(2) Y}] E + F}$$
Can I write any organic reaction in this way or are there any conditions/rules?
Textbooks sometimes mention this form and sometimes they do not. What I have seen is that they mention mostly oxidation or reduction reactions this way but other reactions like addition are always presented in long form only.
This answer describes a little bit but my question is a little different.

It is depend on the reaction. For example, the given two reactions you have sighted are related to each other. It is a example of alkene to alcohol conversion using oxymercuration followed by demercuration with $\ce{NaBH4}$. The overall reaction is electrophilic addition of water to the alkene following Markovnikov's rule (when the nucleophile is water). The first reaction gives an organomercury intermediate, which is not important here for authors to mention. Yet, it is necessary to have a second reaction to convert carbon-mercury bond of the intermediate to carbon-hydrogen bond to get the product of interest. However, it is important to know (and is a common knowledge) that this intermediate should be isolated before treating it with $\ce{NaBH4}$ for following reasons:

One of the byproduct of reaction (1) is acetic acid, which violently react with $\ce{NaBH4}$.
One of the reactant of reaction (1) is water, which also react with $\ce{NaBH4}$.
The other reactant of reaction (1) is $\ce{Hg(OAc)2}$, which also react with $\ce{NaBH4}$.

Nonetheless, any chemist following this reaction sequence knows what hidden behind the sequence. It is also important to know that most organic reactions are written without byproducts, which are not important to the next step. To my understanding, it is okay to write two reaction sequence (even three or more) in a single reaction using steps (1) and (2) and more, as long as consequent intermediates are not important. For example, one of your intermediates $\ce{C}$ and $\ce{D}$ has to be a byproduct. But, if $\ce{C}$ and $\ce{D}$ give $\ce{E}$ and $\ce{F}$, respectively, reacting with $\ce{Y}$, you should show that in your sequence for readers benefits. I you are interesting only on $\ce{E}$ and $\ce{F}$ (together or you have a better separating method for the mixture), you can write it however you want it. The ball is always at your corner, but more explanatory way is always better.
Notes:
Your first reaction doesnt make sense unless $\ce{Y}$ is a catalyst. If not, you should draw it as tri-component reaction:
$$\ce{A + B + X -> C + D}$$
If it is a catalyst, then you can write it as:
$$\ce{A + B ->[\text{cat. } \ce{X}] C + D}$$
An common example for this is:
$$\ce{CH3CO2H + CH3CH2OH ->[\text{cat. } \ce{H2SO4}] CH3COOCH2CH3 + H2O}$$


text:
Breaking bonds with magnetic fields
I was wondering if we could break a bond by applying a big magnetic field on a molecule. The magnetic would just cause the spin of cores of the atoms making the molecule to be aligned , raising the energy of molecular bonding orbital in the same energy of the antibonding molecular orbital allowing an electron to hop from the molecular bonding orbital to the molecular antibonding orbital destroying the molecule.

You'll need a field bigger than just big.

Well, aromatic systems tend to align in a strong magnetic field. LC systems. You can make a fragile molecule where this alignment ruptures a bond. Why anybody would want to do that is beyond my imagination. ;)

Extreme magnetic fields can do interesting things (though not turn to steel while attempting time travel, if you're catching my drift) but "cause the spin of cores of the atoms" is, at very least, not a proper wording.

cause the spin of the cores of atoms aligned.


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How to define axes if all rotation axes pass through the same number of atoms?
I was wondering how one can define the axes for a molecule with several equal axes of rotation, all of which pass through the same atom. I know that the $z$-axis is usually defined as the principal axis, or the axis passing through the most atoms if none are equal. But in the case of, say, the $D_{2h}$ molecule below (assuming all R are the same), how do you define the axes? All 3 axes of rotation pass through only one atom, the central Cu.

What axes do you mean in "define the axes"? The principal one? If so, then you should already know the answer as you mentioned the point group.

What's to define? The rotation axes? We don't define them; we just find them. Or do you want to know where to put the $z$ axis? It doesn't matter.

In this molecule you can put the Z axis in any of three orthogonal directions  for the $C_2$ axes  and then choose one to make this the principal axis. When you look at the point group this will be based on just one of these axes ( which can vary between different textbooks) and so you need to check the notation say $\sigma(z, y) $ is the same as you get. If not move the principal axis and try again. Frequently, Z is chosen to point out of the plane of a molecule if it has one.

In response to andselisk, how do I know the principal one in this case since they're all c2? Finding the rotation axes is clearly straightforward, but my question was how I assign which of them is the z axis and hence the principal in this particular case?

And while I understand Ivan Neretin saying it doesn't matter, this is in the context of revision for an exam where almost all the past paper questions on this require you to show the axes

@atbm While often the assignment of the point group may be formalized in a flow diagram like the one on wikipedia (https://en.wikipedia.org/wiki/Molecular_symmetry#Point_groups_and_permutation-inversion_groups) maybe there are previous exams / is a reasoning in your lecture notes why one of the three $C_2$ axes conventionally gains seniority. You may find a kind of counting the $C_2$ axes by primes, though: $C'_2, C''_2$, etc. (e.g., https://symotter.org/tutorial/benzene)


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How to find relative lowering of vapour pressure from phase diagrams in terms of temperature?
Problem

Answer

 C

My question
Does a tie rule exist for solution solvent phase diagram to express relative lowering of vapour pressure in terms of temperature?
The phase diagram is the one provided as an explanation to colligative properties of binary solutions with one component non volatile.
Edit: Is there any relation between relative lowering of vapour pressure; relative elevation of boiling point and relative depression of freezing point ??
I think I found (  ∆p/p  = 1.23 ∆Tb/Tb) by differentiation of Clausis- Clayperon eqn and Troutons law; but am unsure.
If this is true? Can we approximate 1.23 as 1 ??( 1.23 was ∆Svap/R and ∆Svap  was put as 10.3)
Also does there exist a like rel for freezing point?

Please don't use Dropbox or any third-party cloud storage for uploading essential parts of the question. Instead, input the information as text or an image directly. Check your spelling and grammar. Also, please add your thoughts/argumentation.

There is some problem with my internet. So Dropbox is the only option. I tried to add directly five times but failed. I am new to stack exchange and therefore I am unaware of its many features.

I corrected the formatting and uploaded the illustration for you. I suspect you are using a mobile StackExchange app; if so, please switch to the web browser — the app is known for uploading issues, problems with rendering math and hasn't been updated in years.


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If D₂O (Deuterium Oxide) in large quantities, is exposed to explosive amounts of Neutrons or Protons, can it possibly go Nuclear?
If you exposed large amounts of Protons or Neutrons to Deuterium Oxide, could it possibly explode or implode.
Im basing this off a Chemical Simulation, in which simulated Protons are exposed to simulated D₂O, and a large energetic explosion occurs.
In the other scenario, simulated Neutrons are exposed to simulated D₂O, and an extremely energetic (it can overcome proton energy explosions, meaning it will consume the deuterium faster than the protons could) implosion and causes a gravitational wave that sucks all simulated material around it into the core, where it is vaporized and destroyed versus a proton bomb causing matter to explode outwards.
Is any of this even realistic? In one scenario, simulated Deuterium acts like a liquid nuclear fission material, but in real life... its heavy water...
This is all from a computer simulation, I doubt any of it is realistic, but I've never heard of a Deuterium nuclear reaction, Im just curious

What kind of "computer simulation" is this supposed to be? What is "go nuclear"? Temperature, pressure? And that part about a gravitational wave is just complete nonsense, sorry. Are you writing a novel? I´m sure it could be funny, but sounds to me like there is no science in your fiction.

Out of curiosity, what program and input parameters were used for such "simulation"?

I think, the current question offers (too) little detail for a reasonable answer on ChemSE.  Speaking of the later, possibly PhysSE is more suitable for this type of question. ChemSE here typically deals with reactions with atom's electrons, rather than their nuclei.  Thirdly, «nuclear» ... of course you may start with heavy water $\ce{D2O}$ (there equally is $\ce{HDO}$ ...) to convert some of the nuclei into (than radioactive) tritium.  But it is not spectactular efficient (https://en.wikipedia.org/wiki/Tritium#Deuterium).

Deuterium plus a proton makes 3He, a stable isotope. Deuterium plus a neutron makes tritium which decays by beta emission. Neither will 'go nuclear' whatever that means.

This question is a typical non-sense, based on nothing else than imagination, and fear of scientific discoveries. But well ! Maybe it is related to the process of fusion of deuterium nuclei in a plasma to produce Helium nucleus at a temperature about one million degrees in the so-called H bomb. Maybe the applicant thinks a deuterium nucleus can be obtained from deuterium oxide. It cannot, of course. And protons or neutrons are no use here. it is a typical delirium.

@Maurice - sustaining D-D fusion (in, e.g., a Tokamak) requires more like 400-500 million K.

I'm voting to close this question as off-topic because it is so poor in terms of science, it is also a question which makes very little sense in my mind as a nuclear worker (nuclear chemist)

@NuclearChemist Makes sense, I was just curious because it was stated as if it was real.

In a sense, atoms are "going nuclear" all the time, forming radioactive isotopes under the influence of high-energy particles in the upper atmosphere.  In perhaps the best-known example, the bombardment of nitrogen with high-energy neutrons may generate carbon-14, which subsequently enters organic matter before it slowly decays (half-life = 5700 years), providing a basis for dating said organic matter.
Carbon-14 is only one of many similarly generated radioisotopes.  A list of such cosmogenic nuclides is found here.

Deuterium is a stable combination of a proton and a neutron, but it could accept and bind a proton, forming helium-3 or a neutron, forming tritium. Neither reaction would be explosive, in the sense of liberating enough energy to cause some other nuclear reaction. You might be able to make a small explosion as the sudden heat boils something very fast.
Now, if you could mash two deuterium atoms together, you would get a very hot helium-4, which 50% of the time ejects a proton to give tritium, and 50% of the time ejects a neutron to give helium-3. The heat liberated is significant (2-3 MeV), but not enough to cause another nuclear reaction. Once in a million times, the product is helium-4 and a very energetic gamma ray (about 24 MeV). The pressure required is enormous.
Some fission reactors use heavy water as a moderator because deuterium slows down neutrons without absorbing them much, so if you think about it, D2O would tend to extinguish a nuclear reaction rather than adding to it.


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Is the total ionic-dipole bond (hydrogen bonds) acting on an ion stronger than the lattice energy holding the ions together?
When magnesium chloride dissolves in water, its $\Delta H^⦵_\mathrm{sol}$ is negative.
Since $\Delta H^⦵_\mathrm{sol} = \Delta H^⦵_\mathrm{hyd} - \Delta H^⦵_\mathrm{latt},$ this means $-\Delta H^⦵_\mathrm{hyd} > - \Delta H^⦵_\mathrm{latt}.$
Also, the value of $\Delta H^⦵_\mathrm{hyd}$ depends on strength of ionic-dipole bond, the stronger the bond, the more negative the value is.
Therefore, is the total ionic-dipole bond (hydrogen bonds) acting on an ion stronger than the lattice energy holding the ions together, in the case of magnesium chloride (I am aware for some salts, it is the reverse)?

It means that the energy of the interactions gained (ion with some number of solvent molecules) is larger than the energy of interactions lost (solvent-solvent in solution and ion-ion in the lattice). It does not mean that a single ion-ion pair interaction in the solid is weaker than a single solute-solvent pair interaction.


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what is the physical significance of ∆H=∆U for a chemical reaction?
In case of few chemical reactions $\Delta n=0$,so according to the equation $\Delta H= \Delta U+\Delta nRT$ change in enthalpy equals change in internal energy. But what does this in essence actually mean? Like physically what does this indicate and do we get any additional information from that an equation? I got this query seeing the following equation of formation of hydrogen chloride.
$\ce{H2 +Cl2 =2HCl}$ with $\Delta n=0$.
Like can it be associated to why the reaction is zero order?

$\Delta n = 0$, not $\Delta T =0$. I guess you are taking isothermal condition

This equation holds for a reaction involving ideal gases at constant T. Under these conditions, $\Delta (pV) = \Delta n RT$. Assume now that a gas phase reaction occurs at constant pressure and temperature. Then  $\Delta (pV) = p\Delta V$. This is equal to the negative of the pressure-volume work done by the system during the reaction since $w = -p\Delta V$. Therefore, at constant p and T, $\Delta H$ is equal to the change in energy of the system minus the pV work it did. This is consistent with the fact that we equate the enthalpy change for such a process at constant p and T with the heat exchanged, that is, $\Delta H = \Delta U -w = q_p$. Now if $\Delta n =0$ no expansion occurs at constant p and T, and so $w=0$ and $\Delta H = \Delta U = q$.

So if ∆n≠0 and so w≠0, in case of a chemical reaction like Haber's process of ammonia formation, on what is the work being done? Like is  it used for breaking bonds? I just want to understand for what are we using the extra energy of ∆(PV)

Here the work is pV work (gas expanding or contracting against the surroundings). The making/breaking of bonds results in a change in thermal energy which either increases the T of the system or dissipates into the surroundings. If a reaction proceeds at constant T and p, and the reaction changes n, then the volume also has to change. That means work was done.

Note that work doesn't have to be done. You might carry the reaction out in an isochoric bomb. In that case the likely outcome is a change in temperature and or pressure, in addition to the change in n. If the reaction is isothermal and isochoric, then if n changes you should see a change in p.

Thank you very much bro. I understood it but there are still few doubts of mine regarding constant T,P and how can number of moles change with volume in reality ( not the equation). Could you please let me know where I can study this as I couldn't find anything when I looked up for this.

How do you think a internal combustion engine in a car works? It has a piston into which fuel is discharged and then ignited, expanding and thereby doing work which is used to propel the vehicle forward.

Wikipedia is not necessarily the best place to look if you want a clear explanation, but it usually provides good summaries/introductions, so try this: https://en.wikipedia.org/wiki/Internal_combustion_engine

In addition to standard introductory physical chemistry textbooks, which usually provide a solid description of chemical thermodynamics, I can recommend that you look at other posts on this site, for instance this recent one: https://chemistry.stackexchange.com/questions/131470/t%e2%88%86s-implies-that-heat-flows-which-way


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Reaction of 1,2-diketones with periodic acid
$\ce{HIO4}$ can oxidise alcohols. But does $\ce{HIO4}$ undergo any reaction/act as oxidant with 1,2-diketones?

Hint: https://en.wikipedia.org/wiki/Malaprade_reaction

I have just noticed that HIO4 / H5IO6 , even if they are periodic acids, are not periodic acids. :-)

@Poutnik Be aware that $\ce{NaIO4}$ equally is capable to shrink your cellulose and cotton based bills (well, except in Australia with their polymer AU$).  Strong et al., «How To Shrink Paper Money: A Macroscopic Demonstration of the Malaprade Reaction», J. Chem. Educ. 2019, 96, 6, 1199-1204, doi 10.1021/acs.jchemed.8b00951

@Poutnik I didn't got that joke. Can you explain? :-/

@Nilay Ghosh periodic(1) = per-iodic versus periodic(2) = regularly repeating :-). E.g. PVC has periodic(2) occurance of Cl atom at every other C atom, even if there is no bound periodic(1) acid. Funny thing was I had originally read the title as periodic(2), partly because in my native language iodine/iodide are jod/jodid.

Yes it does, under basic conditions to give two carboxylic acids, according to this paper by Bunton & Shiner J. Chem. Soc., 1960, 1593-1598  here
Abstract of the paper:

The oxygen atom of the acetone from oxidation of pinacol and 2-methyl- propane-1,Z-diol by periodic acid comes from the l,Z-diol, in agreement with the suggestion that an intermediate is formed by electrophilic attack of the periodate. In contrast to this, the additional oxygen atom of acetic  acid, formed by oxidation of biacetyl by periodate in basic solution, comes from the periodate. The mechanism is a nucleophilic attack by periodate upon the carbonyl-carbon atom,  and does not seem to involve the hydrates of the ketone as intermediates. In the oxidation of methylacetoin by periodic acid the oxygen of the acetone comes from the hydroxyl group of the hydroxy- ketone, and the additional  oxygen atom of the acetic  acid comes from the periodate. The mechanism  probably  involves  nucleophilic attack by the periodate  ion  upon the carbonyl  group, and co-ordination of the hydroxyl group on to the iodine atom of the periodate.


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How many non-cyclic isomers exist for C4H7Br?

The compound Q has the formula $\ce{C4H7Br}$ and contains a double bond. How many non-cyclic isomers exist for Q? Should include cis-trans and any other form of isomers as well.

Homework. Add a list with all those you have found to your question, then we give hints as to what is missing.

I can count at least 12...

Also duplicate of this


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Ammonia decomposition on platinum surface
I was given a problem I have to solve, unfortunately I cant wrap my head about it. I've searched the page for answers but they doesnt seem to fulfill my needs, so i'm asking for little guidance.
The problem:

When the decomposition of ammonia is carried out on platinum surface at 1000°C the
  hydrogen binds strongly on the surface. Prove the that the rate is
  equal to $\frac{dp(NH_3)}{dt}=-k\frac{p(NH_3)}{p(H_2)}$.

I dont understand why we have to divide with the partial pressure of hydrogen. In my understanding the rate is $\frac{dp(NH_3)}{dt}=-kp(NH_3)$. I know I have to use the fact that hydrogen binds strongly at the surface, but I'm lost. Can you please explain to me how should I derive this, or just hint me to the right direction.
I lack of good understanding on this topic, so sorry for this novice question.
What I've searched:
Why is decomposition of ammonia gas on quartz a 1st order reaction?
What is the molecularity of the RDS of a zero order complex reaction?
Thank you for your time.

Search for reaction kinetics on adsorptive surfaces, like Wikipedia - Reactions_on_surfaces. There are competitive adsorption processes, affecting the reaction kinetics.

Thanks to Poutnik's comment, I arrived to the answer and I would like to share it.
$$\ce{2NH_3 ->N_2 + 3H_2}$$
$$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{N_2}=\frac{K_{N_2}p(N_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{H_2}=\frac{K_{H_2}p(H_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
Since the hydrogen binds strongly on the surface, we can safely determine that $K_{H_2}>>1$, wich means the denominator $1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)\approx K_{H_2}p(H_2)$. So the surface coverage of ammonia can be written as $\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$. By definition $v=-\frac{dp(NH_3)}{dt}=k_0\theta_{NH_3}=k_0\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$, defining $k=k_0\frac{K_{NH_3}}{K_{H_2}}$ we arrive at the desired equation.
Feel free to edit.


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Derivation of the isothermal Joule-Thomson coefficient
The definition of the Joule-Thomson effect is:
$$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$
And this is defined in an isenthalpic process, i.e. $dH=0$.
My book shows the derivation of the isothermal Joule-Thomson coefficient ($\varphi$) using the cyclic rule:
$$\left(\frac{\partial T}{\partial P}\right)_H\times\left(\frac{\partial P}{\partial H}\right)_T\times\left(\frac{\partial H}{\partial T}\right)_P=-1$$
Using this cyclic rule,
$$\mu=\left(\frac{\partial T}{\partial P}\right)_H = -\frac{\left(\frac{\partial H}{\partial P}\right)_T}{\left(\frac{\partial H}{\partial T}\right)_P} = -\frac{\left(\frac{\partial H}{\partial P}\right)_T}{C_P}$$
And therefore
$$\left(\frac{\partial H}{\partial P}\right)_T = -\mu C_P = \varphi$$
I understand the mathematics of this derivation, but I don't understand it on a conceptual level.
$\mu$ was derived in an isenthalpic process, with $dH=0$. It owes it existence to the fact that there is no enthalpy change. So how can we now use this variable when $dH\neq 0$?

This question may be a more fundamental misunderstanding on my behalf of how thermodynamics equations are derived. Often I see an equation derived under the assumption that some variable is held constant, but then the equation applied when that variable is not constant any more. The above example is just one where I think this discrepancy is obvious.

"Often I see an equation derived under the assumption that some variable is held constant, but then the equation applied when that variable is not constant any more." If you give some examples.....

@Zenix let's focus on the current question, and then maybe more examples will not be necessary.

Your last equation is correct but for a different experiment and that is one carried out at constant temperature in a calorimeter. The derivative is the heat absorbed per unit difference in pressure  at constant T. This  change in conditions is generally the case when moving between different forms of these equations.

Are you familiar with the equation $$dH=C_pdT+\left[V+T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

When you write a total differential such as
$$dH = \left(\frac{\partial H}{\partial T}\right)_p dT + \left(\frac{\partial H}{\partial p}\right)_T dp \tag{1}$$
you are applying the methods of differential geometry, so in reality the answer to your question lies in the applicability of these methods in thermodynamics (the rest being "math"), something which statements such as "so-and-so is a state function" implicitly justify. One can write a total or exact differential of a state function, as for the enthalpy in the equation above. This equation can be interpreted as follows: small (differential) changes in p and T, which are orthogonal dimensions (in the sense that they can be varied independently), additively cause a linearly proportional differential change in the function H. In the differential limit, the surface of H looks like a plane. The partial derivatives describe the slope of the plane in the orthogonal dimensions.
The cyclic rule can be derived from the above equation by taking the partial derivative wrt one of the independent variables while holding H constant.
$$ 0 = \left(\frac{\partial H}{\partial T}\right)_p \left(\frac{\partial T}{\partial p}\right)_H + \left(\frac{\partial H}{\partial p}\right)_T \tag{2}$$
What does it mean here to hold H constant? It means we are looking for an isenthalpic path on the enthalpy surface, from the initial point at which we computed the partial differentials of the surface wrt T and p, in direction $\left(\left(\frac{\partial T}{\partial p}\right)_H dp, dp\right)$, where the partial differential $\left(\frac{\partial T}{\partial p}\right)_H$ also happens to be given (thanks to the geometry of the problem) by
$$ \left(\frac{\partial T}{\partial p}\right)_H = -\frac{\left(\frac{\partial H}{\partial p}\right)_T }{\left(\frac{\partial H}{\partial T}\right)_p} \tag{3}$$
Alternately, consider the line resulting from intersection of a horizontal isenthalpic plane $c(T,p)=H_0=H(T_0,p_0)$ and the plane $s(T,p)$ tangential to the surface $H$ at the point $(T_0,p_0,H_0)$, the tangential plane given by $$s(T,p) = H_0 + C_p \Delta T + \varphi \Delta p$$
where
$$C_p=\left[\left(\frac{\partial H}{\partial T}\right)_p\right]_{(T_0,p_0)}$$
$$\varphi=\left[\left(\frac{\partial H}{\partial p}\right)_T\right]_{(T_0,p_0)}$$
are the partial derivatives of $H$ evaluated at $(T_0,p_0)$, and $\Delta T = T-T_0,~ \Delta p = p-p_0$. Solving for the intersection line by setting $s(T,p)=c(T,p)$ gives
$$T = -\frac{\varphi}{C_p} p + T_0 + \frac{\varphi}{C_p} p_0 +\frac{c-H_0}{C_p}$$
The slope of the intersection line can be recognized to be the same as $\left(\frac{\partial T}{\partial p}\right)_H$ given by Eq. (3). The geometric nature of the problem and relation between the different derivatives should then be clear.

So how can we now use this variable when dH≠0?

A Joule-Thompson coefficient $\mu$, like any other thermodynamic (or state) properties, is strictly valid at the conditions under which it is determined (it may have a broader useful range depending on how much it varies with T and p and the tolerated error). This is different from the question of the mathematical accuracy of the relationships used to derive the properties.    $\mu$ is derived at a specific state defined for a pure substance by a specific point (T,p) and as such is a fixed property of the substance at that point.

Ok, so the fact that $\mu=\left(\frac{\partial T}{\partial P}\right)_H$ was found at constant enthalpy does not require for it to be valid only at constant enthalpy. We used only constant enthalpy to simply the equation so we can practically find what $\left(\frac{\partial T}{\partial P}\right)_H$ is. A simpler analogy would be finding the intercept in something like $y=2x+c$. You put $x=0$ and $y=3$ (e.g., some experimentally measured value, just like $\mu$ is) to find $c=3$. But then this is correct for all $x$, not just $x=3$. Is this the general idea?

Yes, a geometric interpretation in terms of intersecting planes is possible. I added this to the answer.

Thanks, I now understand. Regarding your last paragraph on the validity, so it is correct to say that it's strictly valid at the conditions for which it was derived (eg $dH=0$), and an approximation for all other conditions, yet often the approximation is good enough? For example, in the derivation of the $PVT$ relations in adiabatic expansion, we often use $C_p$ and $C_v$. Yet, $dP \neq 0$ and $dV \neq 0$. So how does this work out? Or should this be a new question?

The thing is that $dH$ is not a property at a particular point, H is (or rather H relative to its value at some reference state). dH is just a small change in H at that point, and is for all purposes so small as to not affect the properties (state) of the system.

Ok, so even if $dH \neq 0$, because it is a differential, $H$ itself is infinitesimally close to whatever new $H$ we have due to $dH \neq 0$, so for all practical purposes it doesn't matter. This can be generalised to my other question, for example with adiabatic expansion. Even though $C_p$ was derived with $dP = 0$, the "new" $P$ when $dP \neq 0$ is close enough to the "old" $P$, therefore it works.

Which then leads me to think - is there a case when the differential is sufficiently large to change the system? Can this be somehow measured or quantified?

No, it should never fail because in general one deals with "state functions" and their derivatives evaluated at fixed points, and state functions are well behaved. It is possible for those derivatives (not the differentials) not to behave smoothly however, for instance at phase transitions some properties such as heat capacities can blow up (see eg https://en.wikipedia.org/wiki/Lambda_point) and crash.

A differential such as dH is by definition an infinitely small change in H - it is actually a mathematical abstraction.

Let us continue this discussion in chat.


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Protons and osmotic potential
In a hypothetical enclosed membrane only permeable to water (including hydronium), where inside the pH is lower than the outside, would water move inside?

At the end, both solutions will be equally concentrated in H+, provided the two solutions contains only water and an acidic aqueous solution

If the water is pure on both sides, it will have the same pH (assuming the same temperature and pressure).  Thus the only way to create a difference in pH would be through the presence of conjugate acids and/or bases other than $\ce{H_2O}$, $\ce{H^+}$ and $\ce{OH^-}$.
And since you've specified that your membrane is permeable only to water, these conjugate acids and bases would necessarily act as osmolytes.  I.e., their presence would lower the chemical potential of the water, causing the water to move towards the direction with the higher osmolyte activity.  So whether the water moves one direction or the other depends on the relative activities of the conjugate acids and/or bases you have used to create the pH difference.

No, because osmosis is not an electrical phenomenon, but rather a chemical one. The concentration of protons would diffuse across the membrane if the fluid, I'm assuming which is water, will be allowed to interact with the other side of the membrane. However, water will not seek to increase the pH of an acidic solution within a membrane the same way it will to dilute concentrations of ions. The resulting balance of pH across the membrane will be as a result of the water to water interaction across the membrane and not because of the pH being imbalanced.

Osmosis is not a chemical reaction.

@AndrewKovács It is not, but not everything in Chemistry is about reactions.


text:
What is the solubility of d-Glucose in water as a function of temperature?
I can't find solubility of d-Glucose in water as a function of temperature.
I.e. a function that can tell me that at 35 Celsius the maximum amount of d-Glucose that can be dissolved in 100g water is X

I presume that this publication has the answer, but I don't have access to it.

Alves et al (2007) Solubility of d-Glucose in Water and Ethanol/Water Mixtures
https://pubs.acs.org/doi/abs/10.1021/je700177n

Have you used Google Scholar (https://scholar.google.com) ? The very first result is your desired answer. https://pubs.acs.org/doi/abs/10.1021/je700177n.

I don't have access to that publication

It is a linear function of temperature (upto 60 C) as per this paper:
Solubility of d-Glucose in Water and Ethanol/Water Mixtures, . Chem. Eng. Data 2007, 52, 6, 2166-2170


text:
Unit cell in a hexagonal closed pack structure
My textbook says that there are only 14 possible 3D lattices(Bravais Lattices).
One of these 14 lattices is a hexagonal lattice in which only a primitive variation is given to be possible.
But then I look at hexagonal closed packed structured lattice and I can't figure out how I could possibly make that lattice using primitive hexagonal unit cell. Am I missing something here?
Source- NCERT class 12th part 1

Imagine two identical sheets of paper (say, A4). They are of the same shape, but the symbols you write on them can be very different. Same thing here. Atoms are your symbols. You may put them in the unit cell in a myriad of ways.

Indeed, hcp is not a primitive Bravais lattice. It has a 2-atom basis.