text:
Anhydrous calcium sulfate as a drying agent
I know that anhydrous $\ce{CaSO4}$ is used as drying agent.
I also know that anhydrous $\ce{CaSO4}$ (dead burnt plaster) is prepared by heating gypsum $\ce{CaSO4.2H2O}$ at a temperature above $\pu{393K}$, which is an irreversible process.
How does $\ce{CaSO4}$ perform the drying if it can't accept any $\ce{H2O}$ molecules? Is there some other mechanism here? Or do I have some wrong information fed in my brain?

How are you not answering your own question? Don't you think you just described the two opposite processes (water absorption and volatilization)?

I think your problem is with the idea of "irreversible" process.
The term irreversible here means that, at $\pu{393 K}$ and in the conditions where this is performed, the reaction
$$\ce{CaSO4.2H2O <=>  CaSO4 + 2 H2O}$$
is pushed towards products so much, that it can be considered "one-way" only:
$$\ce{CaSO4.2H2O -> CaSO4 + 2 H2O}$$
This is dependent on the temperature, and likely other conditions: for example, water exists as a gas at this temperature at a pressure of $\pu{1 atm}$ or lower. If you remove some of the water while heating (such as vacuum, or by circulating dry air around to get the water out), then you further push the equilibrium to the right (and make the backwards reaction impossible since water is no longer present, thus making the process irreversible).
Once you have your dry $\ce{CaSO4}$, it can be cooled down to room temperature (in a dessicator, to avoid it absorbing water). At temperatures below $\pu{393 K}$, the reaction will work backwards:
$$\ce{CaSO4 + 2 H2O -> CaSO4.2H2O}$$
Thus, your dry $\ce{CaSO4}$ will absorb water from the environment, becoming a drying agent. You may regenerate it by reheating it to $\pu{393 K}$, and so on and so forth.


text:
How can bromine water be used to distinguish between benzoic acid and salicylic acid?
I have come to learn that bromine water can distinguish benzoic acid from salicylic acid. I know that salicylic acid on reaction with bromine water forms 2,4,6-tribromophenol. I think that benzoic acid will form m-bromobenzoic acid. So with this respect how can bromine water distinguish the two compounds?

Salicylic acid is partially activated towards electrophilic aromatic substitution by the hydroxyl group, and slightly deactivated by the carboxylic acid. Overall, it is activated towards the substitution. Benzoic acid is deactivated by the carboxylic acid, so overall, it is deactivated towards substitution. Hence, salicylic acid will react without the need of a catalyst (and upon decarboxylation, the reaction will actually become faster), whereas benzoic acid needs a catalyst, such as $\ce{FeBr3}$, to react.
Hence, salicylic acid will react with bromine water, but benzoic acid won't. Physically, bromine water is orange due to $\ce{Br2}$, but if the bromine reacts, then the orange color will no longer be present. Thus, the solution will turn colorless upon contact with salicylic acid, and will stay orange in contact with benzoic acid. Furthermore, depending on the scale and concentrations, you may also observe the evolution of $\ce{CO2}$.

You can clearly distinguish Salicylic acid from Benzoic acid after treatment with $\ce{Br_2/H_2O}$. In Salicylic acid, when bromine water is added due to polarity of the modium, Oxygen atom of the $\ce{OH}$ group attached to the benzene ring becomes highly activated in Electrophilic aromatic substitution. It forms ortho-para substittued product and also ipso -substitution occurs where the $\ce{-COOH}$  group leaves the ring as $\ce{CO_2}$ and also $\ce{Br }$ group gets attached to the ortho position.So, in the polar medium, Salicylic acid reacts with bromine water to give 2,4,6-tribromophenol, and in you will get a white precipitate of the product formed in case of Salicylic acid.But in case of Benzoic acid, there can be a substitution at the meta position, or may not be  depending upon the condition of the reaction, but a white precipitate will not be formed in any way. Thus, bromine water can distinguish Salicylic acid from Benzoic acid by forming a white precipitate in the former case but not in the latter one.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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text:
For hydrogen like atom, find velocity from potential energy

PE of system is given by $$\mathrm{U(r) = \frac{-Ke²}{r^3}}$$ assume bohr model to be valid Find velocity

For this I used $\mathrm{\frac{-PE}{2} =KE}$ and then used $$\frac{mv^2}{2} =\frac{Ke^2}{2r^3}$$ and $$mvr=\frac{nh}{2\pi}$$
However, I am getting the answer as $$v=\frac{n^3h^3}{16Ke^2\pi^3m^2}$$  but the answer given has a factor of 24 rather than 16. I don't know where I'm going wrong.

Your question could really benefit from some formatting. As for the -PE/2 =KE part: where did you take that from? It looks suspiciously like the special case of virial theorem for Coulomb-like forces, which your force is not.

my concern is -PE/2 = KE i think it applies to usual atoms as there PE is proportional to 1/r. here it might change

The fault in your solution is the part where you assume  $\mathrm{\frac{-PE}{2} =KE}$ this is true only if the force follows inverse square law.
In this particular case we need to find the force first by using
$$F =\frac{-dU}{dr}$$
Which gives us
$$F =\frac{-3Ke^2}{r^4}$$
Now this force is central and provides centripetal force
$$\frac{mv^2}{r} =\frac{3Ke^2}{r^4}$$
This along with
$$mvr=\frac{nh}{2\pi}$$
gives us
$$v=\frac{n^3h^3}{24Ke^2\pi^3m^2}$$

The solution of the questions is given in steps in the figure.

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all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is there a terminology contradiction about whether the conjugate of a strong acid is a "weak base"?
(Please note, I am not simply asking, "Is the conjugate of a strong acid a weak base?"  I'm asking about the contradictory ways those terms seem to be used.)
I was revisiting strong/weak acids/bases to help some high school chemistry students and found a lot of apparent contradictions in the way that sources use the terms.  One is whether the conjugate acid of a strong base is considered a "weak acid".
I understand that the strongER the base, the weakER the conjugate acid, and all sources use the relative terms consistently.  The contradiction is in the use of the absolute terms "strong" and "weak".
There seem to be two conflicting conventions in use.  In Convention I, the conjugate of a strong base is indeed called a weak acid (whereas weak bases and weak acids can also be conjugates), so you have these conjugate pairings:
\begin{array}{cc}
\hline
\text{acid} & \text{base} \\
\hline
\text{strong} & \text{weak} \\
\text{weak} & \text{weak} \\
\text{weak} & \text{strong} \\
\hline
\end{array}
In Convention II, the conjugate of a weak acid is a weak base, and vice versa; the conjugate of a strong base is too weak to be considered an acid, and the conjugate of a strong acid is too weak to be considered a base:
\begin{array}{cc}
\hline
\text{acid} & \text{base} \\
\hline
\text{strong} & \text{not a base} \\
\text{weak} & \text{weak} \\
\text{not an acid} & \text{strong} \\
\hline
\end{array}
You can find examples of both just by googling "conjugate of a strong acid", "conjugate of a weak base", etc.  Many sources complete the sentence "conjugate of a strong acid [or base]" with "is a weak base [or acid]" (Convention I).  On the other hand many sources complete "conjugate of a weak acid [or base]" with "is a weak base [or acid]" (Convention II).
I just want confirmation different sources really are using these terms in ways that contradict each other, so I can pass that on to my students and move on.
So, which of the following is true?

There is no contradiction; I'm reading it wrong.  (This seems unlikely since you can literally google "conjugate of a weak acid" and get what seem to be multiple contradictory answers.)
One usage of these terms is correct, and the sources that say otherwise are using the terms incorrectly.  (In particular, the more reliable sources seem to skew towards saying that the conjugate of a weak acid is a weak base, i.e. Convention II.  Maybe the sources that say "the conjugate of a strong base is a weak acid" are wrong, and they're just garbling the (correct) statement that the strongER the base, the weakER the conjugate acid.)
There is no agreed-upon definition of "weak acid" and different sources use it in contradictory ways, but real chemists don't care because they just use Ka values.

To avoid horrible confusion, please be clear in your answer if you're voting for #1, #2, or #3, or something else not listed here.

Not really an answer but more of a comment from my side. As you already mentioned things like conjugate base stability or PKa are more of interest to most chemists. The question is, do you really need the classification for strong and weak acids? And why would you require that. Saying it fully or partially dissociates is much more useful or giving a pKa value perhaps. If this becomes problematic it's probably only because of exams or tests and then something you'd have to discuss with every teacher how he would define it as answer for an exam.

In real life you would probably use that also for strong acids like HCl if it's only 3% and sold somewhere, you'd say it's a relatively weak acid to calm down the people. If you write an article on how dangerous chemistry is you would call acetic acid a strong acid, I guess it depends but in any scientific manner I would not start arguing with stong and weak.

"I can't post the links to the sources because my reputation score is too low" Can you comment the links here so any other user can add them to your post?

You have so many problems with your reasoning that this question is too broad. You don't even really know the definition of strong acid, which has nothing to do with degree of dissociation.

@Mithoron You might want to take that up with Wikipedia, then, as (currently) their Acid strength page states: "A strong acid is one that completely ionizes (dissociates) in a solution".

@Mithoron and in fact almost every source defines a strong acid as one that dissociates completely in solution.  However, I'm adding this comment to clarify for anyone reading these comments, that the question was edited since it was originally posted, so the comments may not make much sense as it applies to the question above.

(Would rather have made this a comment than an answer but still new. So please don't hang me.)
Through high-school, I was also bugged by this. No, you are not going mad, there are many different views on this matter when you google it. The problem is that like many other scales humans create, the way we define 'strong' or 'weak' is a rather relative concept, especially when it gets to the mid-ranges. It is only a way for us to explain what we observe and get a handle on things.
One thing all chemists would agree on is that you should rather look at how strong an acid or base is relative to another (based on Ka or Kb values). In the end, these are the things you would actually use. However, as you need an answer in a school context I would suggest the following:
At school (at least where I come from) we mainly dealt with Bronsted-Lowry acids and bases. Here the strength of an acid depends on how easily it donates ("accepts" for bases) protons. If your students already know a bit about Ka values then anything with a Ka>1 can be considered (reasonably) strong: on a school level, this is usually limited to HI, HBr, HCl, HNO3, H2SO4 and sometimes HClO3 and HClO4. You might want to look at the relative strengths of these 'strong acids' (I seem to recall an exam question about why HI is a stronger acid than HCl (how stable are the ions) and why HF isn't considered a strong acid - the answer is H-bonding). Anything with Ka<1 can be seen as weak.
As to conjugates. Your statement in 2 about 'The stronger the Acid the weaker the conjugate base' and its inverse applies. Here you also work comparatively rather than absolute. You could try explaining it by saying if something gives away a proton easily (acid strength) why would it want to take that proton back later on? Mathematically this can be expressed by relating the Ka value to the Kb value of its conjugate Ka = Kw/Kb (Kw = 1e-14). You would notice that very few 'weak bases' (Kb<1) will give you a Ka>1, from there the confusion about "the conjugate of a weak base is a weak acid". eg. H2PO4^- (Ka = 6.23e-8) is neither a particularly strong base nor acid, but if it reacts as a base and forms H3PO4 this conjugate still only has a Ka = 7.52e-3 - which is still very weak compared to HI.

This comment is just to notify you that the question has been substantially revised/clarified since you first posted your answer. You may want to revise/expand it in light of the edits.

@ Bennett, I feel your pain. The first table you show is most unhelpful for students as they are trying to understand conjugate acid-base pairs. For what it's worth, I like to use a version of table 2 where I substitute, "Negligible acid/base" as the conjugate of a "Strong" (rather than "Not an acid/base"). I find this helpful as students understand the difference between an acid such as ammonium (weak acid, pKa = 9.25) and Na+ (negligible acid pKa = 13.8). Dissolving a salt containing the ammonium ion (at a reasonable concentration) will give a measurable effect on pH but the sodium ion will not. However, a hydrated sodium ion can, in extremis, act as an acid.
Although our definitions of "Strong" (and, therefore "Negligible") are somewhat arbitrary, students in most exams will need to understand and apply them, so they aren't going away.
Of your options, I believe that #2 is the closest (I feel that convention 1 is inherently misleading, at least to students with no schema). Having worked in research, I find that chemists still do use "Strong" acids and bases in the same way that we were taught in school, but with a bit more understanding and the ability to accept the grey areas.

That makes sense.  My concern is that there don't even seem to be any sources acknowledging that there are two conventions in widespread use.  Even the sources which use convention 2, don't caution you that you will run into convention 1 and that you should avoid being misled by it.
More generally, across a broad range of topics (not just chemistry), I think people don't realize how confusing it can be when the tutorial resources are misleading or wrong, because when experts read those resources, their brain automatically corrects it to what they know the author is trying to say.

Fair point. It amazes me that these concepts have been around for a while, yet many resources continue to explain them in confusing ways. Hopefully, by enough discussion, we can improve on the teaching methodology.


text:
Can a droplet of formalin evaporate in midair while falling?
I am designing an evaporator for use in an enclosed environment. Due to the hazardous nature of formaldehyde, I want my evaporator to be wick-free.
The environment in which the evaporator will be used is at a pressure of -10Pa to the outside atmosphere, has a humidity of 32%RH and a temperature that sits at around $36.5~ ^\circ \rm C$.
The basic design for my evaporator is a long tube with an induction coil wrapped around its length and an opening at the bottom for the vapour to escape. As the droplet makes it way down through the heated tube I am hoping that with enough heat it will vaporise and be mixed into the enclosed atmosphere.
The solution I am using is formalin 40.
http://www.strathclydenutrition.com/index.php/Page/product_page/formalin-401
The room in which the vapour is required has a large fan that circulates the air, though I don’t know what speed this runs at or if this will effect the air in my evaporator tube.
I have yet to develop a prototype of my design and short of gaining a thermodynamics or chemical engineering degree, I thought I’d ask the friendly folks at the Physics Stack Exchange if you could help me first.
Any links to external tutorials or articles would be greatly appreciated.


text:
Siloxane condensation mechanism
I want to formulate the condensation reaction mechanism for the formation of siloxane links.

I don't see how to express it as an addition-elimination reaction, because even though the silicone atom is positively polarized, it cannot form a fifth bond that would result from a nucleophilic attack by the other silanol. Is it a concerted reaction?
Any input is helpful, so no limit on conditions.
Any key publications?

Under what conditions? Acid or basic?

Note - silanols SiHnOH4-n such as drawn are highly unstable

@Waylander any conditions that you know the mechanism for :) Also, thanks for the note on stability; I understand that SiHnO4-n forms in situ from silyl chlorides.

If you google "Silanol condensation mechanism" one of the first hits is The Journal of Non-Crystalline Solids vol 100 (1998) 31-50 by CJ Brinker. The PDF is freely available and discusses acid and base-catalysed condensations at length


text:
Polyhedron with four capping ions
Let's say I have a trigonal prism in my crystal structure and this is further capped by two ions it's called a bicapped trigonal prism. If I have three ions it's called a tricapped trigonal prism. But what about four? I cannot find anything like quadcapped, fourcapped, fourfold capped trigonal prism. How would you name that?

By then, it would be too ugly to deserve a human-readable name. Nearly all huge polyhedra are like that.

After asking more people who already wrote reports with similar structures and a paper I found on the net the most convincing answer I could find was the term 'tetracapped'.
Source: 10.1002/zaac.201400143


text:
Why does pOH increase when pH decreases?
Let's say you add $\ce{HCl}$ to water. The $\ce{H+}$ ion concentration increases and that causes a decrease in $\mathrm{pH}$. But why would the $\mathrm{pOH}$ increase?
I can't see why added $\ce{H+}$ ions will decrease the $\ce{OH-}$ concentration? Is this because of the water auto-ionization? Is it because, according to Le Chatelier principle, when you add an amount of a component, the reaction will go in the other direction to counteract the disturbance, thus decreasing the concentration of $\ce{OH-}$ to form water?

Well, yeah, it may be thought of like that.

Increasing the concentration of $\ce{H+}$ increases the collisions of $\ce{H+}$ and $\ce{OH-}$ which can result in the formation of water.

At a particular temperature, the $K_{\text{eq}}$ for the following reaction (yes, it's the auto-protolysis of water):
$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$
will be constant. Note that $K_\mathrm{w}=K_{\text{eq}}\times[\ce{H2O}]=\ce{[H+(aq)][OH-(aq)]}$ is called the auto-protolysis constant of water. It is considered a constant because concentration of water $([\ce{H2O}])$ is assumed to be constant.
So, you can directly observe from here, that since $\ce{[H+(aq)][OH-(aq)]}=K_\mathrm{w}=\text{constant}$ at a particular temperature, if $\ce{[H+(aq)]}$ increases (i.e. $\mathrm{pH}$ decreases), $[\ce{OH-(aq)}]$ must decrease (i.e. $\mathrm{pOH}$ must increase).
Your interpretation of the Le Chatelier's principle is also correct. In fact, it is actually what is happening behind the scenes. When you add more acid (assuming a 100% ionized acid), the $Q$ value of the auto-protolysis of water increases, causing the reaction to shift backward. However, the backward shift is unable to decrease the concentration of $\ce{H+}$ ions as much as it decreases the concentration of $\ce{OH-}$ ions. See explanation below:
Consider adding $\pu{0.01M}$ $\ce{HCl}$ to pure water (at $\pu{25^\circ C}$) at $t=0$. This is what happens till achieving equilibrium at $t=t_0$:
$$
\begin{array}{ccc}
&\ce{H2O(l) &<=>& H+(aq)&+&OH-(aq)}\\\hline
-&-&&\pu{10^-7M}&&\pu{10^-7M}\\
t=0&-&&\pu{(10^-7+10^-2)M}&&\pu{10^-7M}\\
t=t_0&-&&(10^{-7}+10^{-2}-x)~\pu{M}&&(10^{-7}-x)~\pu{M}\\
\end{array}
$$
Now, if you actually solve for $x$ at $t=t_0$, you'll find $x$ to be of the order $10^{-8}$ or below. Notice that, for the $\ce{H+}$ ions, this means a decreases in concentration from $\approx\pu{0.01M}$ at $t=0$ to $\pu{0.0099999M}$ at $t=t_0$. This a hardly noticeable change. However, it is a catastrophe for the $\ce{OH-}$ ions, whose concentration would probably be reduced, by a factor of more than ten thousand, from the initial $\pu{10^-7 M}$! The change in $\mathrm{pOH}$ is, hence, much more pronounced.


text:
Is identifying a chemical equivalent to be able to produce it?
This question is a follow up of another thread, where we are discussing Theresa May's accusation of the Russian government which is supposedly behind the recent chemical attack in the UK.
Soon after the attack the UK implemented a chemical analysis which again supposedly shows that this compound can be produced by only one government in the world: the Russian government.
So, my question is, if the UK Chemical Weapons Institute can carry out such a chemical analysis and figure out what chemical weapon was used, does it mean that they themselves possess this technology?

On 14 March 2018, at UN Security Council Russia's representative Vasily Nebenzya claimed:

For the British specialists to be perfectly confident that this was a Novichok agent and not any other kind, they would need a control standard for proof. It [the substance used in the attack] must be compared to a control substance... They have a collection and they have the formula. In other words, if the UK is so firmly convinced this is Novichok, they have samples and formula and are capable of formulating it themselves.

Nebenzya in his speech refers to some professional chemist and I wonder what other professional chemists think of this.

Further reading:

A BBC article explaining the Novichok agent; the Wikipedia page about the same as well
Another BBC article detailing the timeline of political events.

The final proof of any structural determination is to make the compound and see if the properties are identical to the original.  You can make a structral assignment and publish it without this final step. It happens all the time. I would expect that with the array of modern analytics tools avaible that they are pretty sure what the agent is, but this does not mean they have previously prepared it.

It may just mean that they had a reference sample, or at least the results of analysis of one, to compare to. They need not have made the reference sample, nor be able (or willing) to do so.

@AndyW. So, for such a body as the UK Chemical Weapons Institute, the ownership of such a reference sample or results of analysis of one, is not enough to be able to produce it? Just in terms of chemical science.

Possibly. There are probably multiple labs that could work out and follow a synthesis pathway given a target structure. (Doing it safely is another matter!) But it's not clear to me if a specific compound has been identified, or just a member of a class of compounds. Also, looking at the wikipedia article on these agents, they are typically generated at the point of use by mixing two (safer) components. That's probably a substantially harder challenge than just making the compound itself.

Treaties don't prohibit the synthesis of chemical weapons, they prohibit their manufacture in quantity. And nerve agents are not particularly hard to make, just hard to handle. The word "produced" in the question creates ambiguity as what it really means is "produce on a large scale". Synthesising a small amount is something any chemist in a good lab could do, but it isn't the same as "producing" in the sense of the question.

@matt_black. Thank you, sir! In my question I rather mean the ability to synthesize this nerve agent.

@matt_black More specifically, the Chemical Weapons convention (of which both Russia and the UK are signatories) requires any site making more than 100g of a Schedule 1 material (roughly, chemical weapons or chemicals whose only known use is making chemical weapons) to be reported to the OPCW. This does allow them to make small amounts for research into antidotes, detection, etc.

In general it is much easier to identify a structure than it is to make it. Complex natural products are often extracted from biological sources and structurally characterised long before synthetic chemists find a way to synthesise them in a lab.

It is also worth noting that most nerve agents are not that hard to make: but they are hard to make without killing yourself and everyone else in your lab. Most competent military biological and chemical weapon labs will have that capability and will have used it to explore weapons their enemies might use so they can protect their armies from attack. This isn't prohibited by weapons conventions as they only mandate you don't produce the compounds in large quantities.

That would be a good and interesting question without the political fluff around it. Sure, it is there to give context, but it is absolutely unnecessary to talk about the chemistry issues involved. It doesn't even have anything to do with the nature of the compound. Have my down-vote.

I’ve just noticed that we have a question about nerve agents tagged with “everyday-chemistry”. I’m sure we can all wish that this never becomes the right tag.

We've gotten a lot of flags on this and some of the other answers.  I think that the political issues are strongly coupled to the question, and as such are difficult to factor out of the equation.  That being said, please keep opinions to a minimum.  Evaluating the science is okay, but evaluating the political motives is probably a slippery slope, and should be avoided if possible.

RE: ...that they themselves own this technology? / Own seems like  the wrong verb. Possess would be a better choice. So England and Russia could both possess the technology, but neither could "own" it exclusively.

To me the gist is that nerve agents are not particularly hard chemicals to make. What is hard is making them without killing yourself in the process. Then after you make it you still need to weaponize it. In other words you want to be able to kill someone else with it without killing yourself. Two overcome these two problems requires very significant technological expertise.

So, my question is, if the UK Chemical Weapons Institute can carry out such a chemical analysis and figure out what chemical weapon was used, does it mean that they themselves own this technology?

Ok, first.
Performing elemental chemical analysis does not give you a single idea if you can produce compound you analyzed. Knowing it structure does not guarantee this either. So, theoretically speaking the answer is no.
However, there are complications.
Novichok agent apparently belongs to the family of nerve agents that are toxic in extremely small concentration. Furthermore, they metabolize in the body, irreversibly binding to target proteins. Consequently, it is not likely to be possible to extract the toxic substance from the body, and to identify it one needs empirical data on the ways the compound is metabolized in the body (and even this does not give guarantees). This requires  access to measurable quantities of the agent in question and a dedicated effort to study its effect.
Structures claimed to belong to Novichok agent family belong to a known family of fluorophosphate nerve agents. Methods for synthesis of this class of compounds are known, so producing some of the compound should be possible having only the structures themselves, even though the first approach might be needlessly costly.
That said, nerve agents have structures similar to some insecticides, that are produced in quantities. It should be assumed that any developed country with moderately advanced agrochemical industry can produce chemical warfare agents in military relevant quantities if needed. In fact, it is a common opinion that former USSR chemical industry was designed with this idea in mind.
THAT said, use of such exotic compound is idiotic on its own and thus unlikely. There are countless possibilities to produce strong toxins using freely available consumer chemicals only.

-1 for the detour into politics at the end.

I really wonder, why people down vote a detailed and informative answer, made by a professional.

@Jacobian I really wonder why people wonder why people downvoted, when they explicitly say why they did it. More constructively, without the last paragraph, I'd probably reverse my vote.

@David Richerby. permeakra explicitely said why he did it, though deleted his comment. Besides, the main value of the answer obviously is not in the last two paragraphs.

Please see my comment under the question.  Thanks!

Something has just occurred to me that's completely independent of my earlier comments. I believe the theory is that the nerve agent was administered through the ventilation system of Skripal's car. Presumably, then, there was still some amount of the pure substance inside the car that investigators are analyzing directly, so they aren't only working with the products of reactions between the substance and the victims' bodies.

@DavidRicherby To my knowledge, nerve agents hydrolyze rather fast even in air and are fairly volatile in general. So I highly doubt they could.

No, the capability to identify a compound does not imply capability to synthesize that compound. Indeed, "analysis" and "synthesis" represent two branches of chemistry, and many chemists spend years trying to synthesize compounds of known structure.
Modern structure determination relies on roughly two steps. First is separation of sample mixtures into constituent parts, using various forms of chromatography. Next is analysis of chemical composition and bonding patterns using techniques such as IR spectroscopy and nuclear magnetic resonance spectroscopy, as well as mass spectrometry (MS), which weighs molecules (and their destructively ionized fragments) with extraordinary precision and dynamic range.
Advanced MS instruments can analyze very small amounts of collected samples, without special preparation, and at atmospheric pressure. Pure samples are not required for the instruments typically used in forensics and product testing (DART). When a pure sample is desired, one or more separation devices can feed the MS instrument in-line (LC-MS, GC-MS, MS-MS).
MS can measure molecular weight so precisely that the chemical formula can be deduced directly based on the fractional weights of the elements due to varying isotopic abundance. Often, a chemical formula is almost sufficient for unique identification. For example, the NIST Chemical WebBook contains only 4 entries with the same formula as the VX agent, all for VX or nearly identical compounds. The PubChem database contains only 1 (VX itself).
It is also possible to identify compounds using known chemical or physical interactions linked to a readout such as a color change. This principle is used in devices such as home pregnancy tests and lead test kits. The US military even has custom field test kits for some chemical weapons.
It should be noted that compounds introduced into an organism may be identified indirectly via metabolites found in blood or urine, even if the original compound is long gone. Again, VX agent is a suitable example.

Technically one could argue that a home pregnancy test does not involve a chemical reaction, or at least not the making or breaking of chemical bonds; instead I believe it's usually a form of immunoassay where antibodies labelled with coloured particles are transported along a test strip by capillary action and captured by other antibodies immobilised on the strip (https://en.wikipedia.org/wiki/Lateral_flow_test). Otherwise, great answer.

Advanced MS instruments can analyze very small amounts of collected samples, without special preparation, and at atmospheric pressure. || they do require a pure compound, though.

It should be noted that compounds introduced into an organism may be identified indirectly via metabolites found in blood or urine, even if the original compound is long gone | which, in turn, requires a prior empirical knowledge, i.e. a dedicated effort wih a compound obtained beforehand. Good luck with it.

@nekomatic Thanks! I changed the wording a bit so as not to inadvertently exclude ELISA.

@permeakra I think the theory is that the substance was administered through the ventilation system of Skripal's car. They could have obtained the pure compound from that.

With any material (or device) there are three distinct steps necessary to make a copy :

Be able to identify it by its behavior and interactions.
Be able to determine its structure.  Identification does not tell you something's structure, just its external behavior.
Develop a process to construct the material or device as a prototype at least.

All that has been done in this case is step (1) and arguably (2).  Step (3) is the biggie.  It could take years or decades to develop the manufacturing process.
In the case of extremely hazardous materials step (3) is even more complex than usual, as even more care than usual has to be exercised to avoid potential accidents.
In some cases you may be able to skip step (2) and get step (3) completed first, but that doesn't mean step (3) gets any easier.

So, my question is, if the UK Chemical Weapons Institute can carry out such a chemical analysis and figure out what chemical weapon was used, does it mean that they themselves own this technology?

This is a weapon and as such they may not say that they can prepare it, but may be able to.  They might even deny they can make it, but can.  I've no particular knowledge about this specific substance in this regard.

Novichok agent apparently belongs to the family of fluorophosphates, just like VX and other nerve agents. There are reasonably well established synthetic methods for this kind of work, so 3) is fairly simple.  Personally I have hard time with idea they were able to determine the agent. The extremely small amount of the agent means that there is no easy way to guess the exact structure of the agent. Guessing that it was a nerve agent is apparently simple, but finding exact structure is not.

@permeakra. Thanks for this clarification! It would be great, if you make a full answer.

@permeakra  There are a number of identifying techniques that require very little compound.  You can do proton and carbon NMR at 600 MHz with under one mg.  Less than 1 mg/ml concentration for mass spec as well, of course this is with some of the better instruments, but the UK government certainly has access to these.  Even I had access to them back in university as a freshman even.  Getting exact structure is less easy... , but you don't really need the exact structure to identify the agent, other similar structures may not be stable, etc....  so you can get a high confidence match.

@ttbek That works only if you do elemental analysis or have a reasonably pure compound. Or, in case of chromatomass at least have the original compound present as free molecules in the solution. Identifying compounds that irreversibly bound to proteins is an entirely different thing, and if they are dispersed in the body fluids... no.

@Chromato-mass in particular requires access to the source of the analyzed compound to obtain its mass-spectra beforehand.

I don't know what forms they were able to collect, I haven't followed the news, etc.... .  You don't need to have access to the source of the analyzed compound if it is already on file (beforehand can be far beforehand), and it most likely is.  It also isn't strictly true that you need it, it's just much more difficult without.  What do you mean by "elemental analysis," the term is fairly broad.  For instance, atomic mass spec would fall under it.  Some of the other techniques under that umbrella require much larger samples.  They also wouldn't return as clear cut results.

@ttbek >beforehand can be far beforehand | sure. But we are talking about what is a secret chemical warfare agent, i.e. the  data from the authors were never released to general public. Someone had to produce a bit and put it into mass-spectra (and ideally calibrate it with a given column)

@permeakra Hmm, I wasn't aware that it was all that secret... I thoughtit was in the sense that it is supposed to be, but is more like a "secret" in the way that a lot of now very well known encryption methods are "munitions."  I admittedly haven't followed the topic though.

Just chemistry here.

Chemists do not need to be already able to synthesize a molecule to identify it.
All they need is a reference, either as physical sample or as collection of chemical and physical data. In both cases is irrelevant how these were obtained.
There is no way to identify the producer of an isolated molecule. Multidimensional analysis (normally it works for complex samples with an inherent geographical signature such as oils and wines, to give examples) OR considering "markers" such as impurities and byproducts can possibly point to a specific way of production/producer.

If one is able to identify a molecule or has enough references to it,  then by replacing "one" with "a state, a government" it is surely possible to synthesis that idenfied molecule. It is just matter of effort, which can be big or relatively small.
There is not a straightforward "chemical" reason to trust one of the parts involved in the story.
I would also note that producing molecule X as chemical weapon (big scale, making enough heads,  etc...) and synthesising X at common lab scale are two different things.
A state can be unable to produce a warfare agent but will be unfair to the chemists of that nationality to deduce that they cannot synthesise that specific molecule.
Edit: of course the total synthesis of certain complicated molecules, think of some natural products, can be a tremendous challenge. A nerve agent developed in the seventies or so does not certainly fall in this category.
See also a comment by Matt Black on the same line of the end of my A.

You comment about multidimensional analysis  is spot on, but you need to explain that a bit more. It only works if you can obtain a "reasonably pure"  samples of the stuff being analyzed. A chemical weapon would leave on traces of itself and you'd never be able to obtain a "pure sample." So you'd never be able to use multidimensional analysis to prove who made a particular batch of the chemical.

@MaxW the entire paragraph is meant to say that you cannot identify the producer of an isolated (isolated in synthetic chemistry sense) molecule  The mentioning of multidimensional analysis and wines was to prevent debates. Obviously a chemical warfare agent extracted or recovered from a victim will be rather than impure too simple for multidimensional analysis. Thus I mentioned impurities  They can be easily added as well. But my answer is general. Even in possession of a sample made by state X, you cannot claim X is the producer solely based on chemistry alone.

There is really more than one question here: one is whether identifying a chemical requires the ability to make it; the other is whether the UK can reliably identify Russia as the manufacturer of the nerve agent.
The first question has a simple answer: there are plenty of known chemical structures that chemists don't know how to make, yet. This is common in natural products where chemicals are extracted from animals or plants and fully structurally characterised long before good mechanisms are found to synthesise them in a laboratory. So you don't need to be able to make something to know what it is.
But this is mostly irrelevant to the question of whether only the Russians can make Novichock agents. Not least because we know they have been made by others. But also because nerve agents are relatively simple chemicals and are easy to make (though not easy to make without killing everyone in the lab as a byproduct of the synthesis).
Chemical identification of a Novichock agent cannot possibly tell us where it was made: the chemistry can't tell us where the ingredients came from. The UK's portion down scientists have stuck to this line in statements.
This has led some to conclude that the evidence doesn't point to Russia. As does the fact that the alleged Russian work on the agents was never confirmed independently (other than the fact that others have confirmed that the recipes leaked by defectors actually work). But the allegation that this is some sort of false-flag conspiracy to damage Russia's image makes no sense. Firstly the development of nerve agents would be top secret and no government would blab about it in public. So we wouldn't expect the Russians to admit they had ever made or stockpiled the agents. Moreover, we would expect the western agencies to follow up the intelligence they had by both synthesising, testing and developing better ways to combat them. They wouldn't be doing their job if they hadn't. And we wouldn't be safe from hostile foreign powers.
So we should discount media simplifications that claim only the Russians could make them. But that isn't the point. We have good intelligence that they did research them and probably made stockpiles of them during the Cold War. So they have the expertise to handle them and use them. Some of the conspiracy theories allege that any competent chemist could have made them so why blame the Russians? But this ignores the issue of skill. Making a nerve agent isn't hard: not killing yourself in the process is very hard, deploying the agent is probably even harder. And it was the Russians who developed the agents and worked out how to deploy them. This matters, but conspiracy theorists have focussed on the ease of synthesis and ignored the extra expertise required to avoid death for all involved. They don't seem to understand chemistry: It takes large amounts of government investment to get that expertise.
So chemistry doesn't provide proof. But the chemical expertise to make and use the agents is held only by a very small number of players and the only one with a motive is Russia.
In short the key issue isn't chemistry but the expertise with particular chemicals. Plus the motivation to commit the act. The media should not claim that chemistry alone is definitive in fingering Russia, but the known-expertise plus history plus motivation makes a strong case.
Political appendix
Simple answers dealing with the chemistry out of context have already been widely appropriated to promulgate conspiracy theories about the Salisbury attack. Here is some of the context. Please ignore if you just want to stick with chemistry.
The real evidence that the Russians are the most likely suspect is a combination of the fact that they developed the agents (and therefore have the expertise required to work with them) plus the clear motivation they have to use them: it sends a strong signal that discourages defectors and opponents of the Russian government. They don't want to admit they did it but they want other defectors to know they did it. And they have form. A decade ago they used polonium 210 to kill another defector in London (unlike simple chemicals, polonium leaves a trace of radioactive debris that pointed very clearly to who used it as did the fact that only a very small number of countries can make it in the first place).

I think, Chemistry SE is not a good place for this kind of answer. 99% of what you are saying is about politics. But the question was asked with a clear idea to leave politics aside (just to avoid unnecessary debates).

@Jacobian I tried to put the key issues about chemical knowledge in context. Half the content is about what we can and cannot know based on chemistry and chemical expertise. The debate on this is full of selective quotation and misleading partial arguments so I put everything in context to avoid that possibility.

Please see my comment under the question.  Thanks!


text:
How to simulate phase separation in a discrete system using LAMMPS?
I would like to simulate phase separation in LAMMPS and need advice what kind of two particle systems and interactions will yield phase separation.
For example, a meta stable phase separation is achieved with a continuous phase flow model governed by Cahn–Hilliard equation, where $c$ is the concentration, $\mu$ is the chemical potential, and $\gamma$ is a constant:
$$\begin{align}
\frac{\partial c}{\partial t} &= D\nabla^2\mu&
\mu &= \left(c^3-c-\gamma\nabla^2 c\right)\\
\end{align}
$$

I tried to use a discrete two atom system in LAMMPS. The first type of the particles has an inter-LJ interaction and the second type has an inter-soft-potential interaction. Between the distinct two types there is also a soft potential:

Initial configuration (interleaved lattice of the two particles):

Steady state configuration at $T=0.1$ (reduced LJ units):

Configuration file:
units lj

timestep 0.001

dimension 2
boundary p p p
atom_style atomic
neighbor 0.3 bin
neigh_modify every 20 delay 0 check no

variable temp equal 0.1

lattice hex 0.5
region simbox block 0 50 0 25 -0.1 0.1
create_box 2 simbox

lattice hex 0.5
create_atoms 1 region simbox

lattice hex 0.5 origin 0.5 0.5 0
create_atoms 2 region simbox

mass 1 1
mass 2 1.0

velocity all create ${temp} 1234567 dist gaussian
pair_style hybrid lj/cut 2.5 soft 0.5
pair_coeff 1 1 lj/cut 2.0 0.8 4.0

pair_coeff 1 2 soft 0.5
pair_coeff 2 2 soft 0.5

fix 1 all nvt temp ${temp} ${temp} 0.1

dump 1 all atom 500 out/dump_temp_${temp}.lammpstrj
write_data data.all
thermo 500
run 500000

How do I make my system to look more similar to the continuous model suggested by Cahn and Hiliard? I want to get a more prominent phase separation like in the continuous model. I am not sure how to calibrate my system/interactions.
Is there a different simple system I could use to simulate phase separation?
If you think Matlab could be more suitable to simulate phase separation, please elaborate.

According to IUPAC (which I consider to be the authority on terminological issues) 'phase separation' is defined as: "The process by which a single solid or liquid phase separates into two or more new phases." Again, based on looking at your system, you have hard spheres sticking to each other in a simulation space larger than the system, comprising a single phase in vacuum.

@Bdrs I ended up using 2 particle system. But still looking for a more natural system that will experience phase separation.

Great! Try water/1-octanol. That's a popular one, and you should be able to find plenty of empirical data about it.

What is it with that first picture... it seriously messes with my head, no clue why though. Does anyone else have this feeling?

@Martin-マーチン what do you mean?

Somehow that colour combination gives me a headache. Absolutely not sure why...

The LAMMPS simulation here will rather produce phase glass phase transition and not a phase separation. see binary LJ from more details: http://www.sklogwiki.org/SklogWiki/index.php/Binary_Lennard-Jones_mixtures

Firstly, a couple of comments.
The Cahn-Hilliard model, for which you gave a snapshot, represents the two fluid phases on a much larger length scale than the atomic scale. It's a continuum model. So, whatever particle-based model you use, you would need an enormous system to get close to similar-looking pictures. The interfaces will always look more irregular, in a practical particle-based simulation. Don't expect to see nice circular droplets.
Secondly, even that Cahn-Hilliard simulation is not yet at equilibrium. The droplets are in the process of coalescing, and the final state will be a single large droplet of one phase in the other (or possibly the two phases will form slabs, with planar interfaces between them, it depends somewhat on the boundary conditions and dimensions of the surrounding box). The driving force is the reduction of the total interfacial area. This whole process takes time. Something similar should happen in the particle-based simulations, but because the length scales are so different, the time scales will also be different.
Given all that, your first choice of models may be suitable. However, you need to choose a temperature which is high enough for both the phases to be liquid. You haven't done this! The triple point temperature of 2D Lennard-Jones is $k_BT/\epsilon\approx 0.4$. At your temperature of $0.1$, the Lennard-Jones atoms will be trying to crystallize; in practice they will get "stuck" in irregular semi-crystalline semi-amorphous clumps. Nonetheless, if you raise the temperature above $0.4$, you might see liquid-liquid coexistence for this system. I can't be sure, and you may need to experiment.
I have some suggestions for a simpler approach, if all you want to do is see liquid-liquid phase separation using LAMMPS. Firstly, it is OK for both the fluids to be identical. For example, calling the two species A and B, the A-A and B-B interactions can both be Lennard-Jones. If you choose the A-B interaction to be different, and unfavourable compared to Lennard-Jones (for example, the completely repulsive Weeks-Chandler-Andersen version of Lennard-Jones), you should see phase separation.
The advantage of this approach is that the Lennard-Jones phase diagram is well known, making it fairly easy to choose a density and temperature at which both the phases will be liquid.
Even simpler, you can choose A and B to interact through a coarse-grained potential such as the dissipative particle dynamics DPD potential
$$
v(r) = \begin{cases} \frac{1}{2}a(1-r)^2
& r<1 \\ 0 & r>1
\end{cases}
$$
where the cutoff distance is taken as the unit of length.
LAMMPS has a DPD option, and you can find a perspective article on DPD
by Español and Warren J Chem Phys, 146, 150901 (2017)
also available as a preprint. There are also tutorials on DPD readily available on the internet.
DPD has the advantage of being very fast;
also it is sort-of intermediate between an atomic model and a continuum one,
since the particles in DPD really represent regions of fluid,
not individual molecules.
Once again, the A-A and B-B interactions can be made identical,
for example $a=25$, with a typical temperature $k_BT=1$
and a density $\rho=3$ in 3D; you would need to experiment for suitable parameters in 2D, but the precise values may not be crucial. The A-B interaction could be the same, but with a larger value of $a$, for instance $a=50$.
This approach has already been used to model fluid-fluid separation,
for instance in papers by Novik and Coveney
Phys Rev E, 54, 5134 (1996)
also available as a preprint,
and Phys Rev E, 61, 435 (2000)
also available as preprint.
Those papers are a bit involved,
and there are many complicated technical issues with
the DPD method.
It is not essential to go into those in depth.
If you are not essentially interested in the hydrodynamics,
it is quite satisfactory to use the DPD potential
in a standard molecular dynamics simulation with a thermostat of your choice.

I simulated the data from $T=0.01$ to $T=0.1$ only bellow $T=0.5$ a structure that resembles phase separation emerges.  For temperature higher than $T=0.4$ both particles have a continuous phase.

I don't quite understand your temperatures. You say that you saw phase separation below 0.5 but you did not see it above 0.4? Can you please clarify? More generally, it is hard to predict what you will see for a model which has not been studied before. I have suggested, in my answer, using the DPD potential, which has been studied before. Suitable parameters are suggested in the papers I cited, and you could also try the values I suggested. I suspect that there is a wide range of DPD parameters that will work. This may give you a better chance of seeing what you want to see.

it's a typo. There is a $T_c$. It's around 0.5

OK. The only other suggestion I have for that system is to make the 1-2 interaction more repulsive, instead of being just the same as the soft potential. For example WCA Lennard-Jones, which is fairly easy to do in LAMMPS: https://lammps.sandia.gov/threads/msg02715.html This would discourage the mixing more strongly. Maybe push $T_c$ up a bit? Perhaps worth trying at least.

if I have the 1-1,1-2, and 2-2 force fields couldn't I predict the parameters of the force fields at which I will see phase separation? Without simulating only using MATHEMATICA and intersection between the hyperplanes of the potentials?

Short answer is "no". You would need a way of predicting the thermodynamic free energy of each of the phases, as a function of temperature, pressure, composition, knowing just the interaction potentials. This is not a trivial task to do, and is a whole field of science in itself. Doing particle-based simulations (as you are doing) is actually one of the standard ways of finding out which phase is stable.

Could you suggest how to setup the LAMMPS input file with a DPD pair_style dpd T cutoff seed interaction? Namely, how to control the concentrations to observe phase separation.

I would expect to have phase separation bellow a critical temperature, is it true?

Usually it is expected at some temperature (unless crystallization happens first). Technically, the critical point is called a consolute temperature and is just at the top of the coexistence curve. As with all two-component systems, the phase separation temperature depends on the composition. See here for a schematic phase diagram (the UCST one). An advantage of the symmetrical models which I mentioned (where A-A is the same as B-B) is that the upper consolute point is bound to be at 50:50 composition.

In answer to your request about the LAMMPS input file, I'm not too good with detailed input information like that, you would do better to ask the LAMMPS community directly (I believe they have tutorials and a user guide too). Looking online, I see a tutorial here which seems to cover setting up a DPD binary mixture.

https://math.nist.gov/mcsd/savg/vis/concrete/index.html

I couldn't find parameter set for DPD to produce phase separation. BTW does DPD particles are considered active matter?

Well, I'm surprised you couldn't produce phase separation using appropriate DPD parameters for a two-component fluid, since the Novik/Coveney references I gave in my answer, as well as the online tutorial linked in my comment above, definitely study phase separation using DPD. As far as active matter is concerned, no, the DPD model itself is not active: you would need to add some extra driving forces.

I was able to run DPD and have a Cahn-Hillard like pattern. Español et al. in their perspective suggest one should consider to adapt the notion of DPD-like instead of continuum models. I am not sure how one can take a free energy and develop interactions that will resembles the continuum case.  For example, if we take the Cahn-Hillard model's free energy $F[c]=\int d^n x \left[\frac{1}{4}\left(c^2-1\right)^2+\frac{\gamma}{2}\left|\nabla c\right|^2\right]$ how one can develop a model like the DPD?

I'm glad that you managed to get a satisfactory pattern with DPD. As for your last question, that looks like a new question to me! Also, just a suggestion, it might be a better fit on https://physics.stackexchange.com/

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Does double replacement reaction only happen to aqueous reactants?
I was taught that double displacement reactions occur when reactants are aqueous. However, I came across this reaction
$$\ce{ZnO(s) + H2S (g) -> ZnS(s) + H2O (g)}$$
This is the reaction that is used to remove hydrogen sulfide that is naturally found in methane gas. This reaction indeed looks like a double replacement reaction. However, the reactants are not aqueous.
Additionally, this reaction happens at the temperature of $\pu{700 K}$. Does that mean if enough energy is given, double replacement reactions can occur without an aqueous medium? How so?

Source 1, Source 2

At high temperatures, it is likely that the outer layer of the solid reactants melts and the mobile ions in these molten outer layers react with one another. And if the reaction is exothermic, more of the solid reactants would melt and react as the outer layer reacts. The rate of reaction would increase quickly and soon, the double displacement would be complete before you know it.

Generally, reactions occur with liquids and gases (which encompasses solutions). This is because there is the fundamental problem of mass transport. When reactants react, they leave product in the wake of the reaction, which isolates the reactants from each other. For fully solid reactants, it means that more reactants have trouble finding each other. When reactions take place in liquid or a gas, there is the prospect of diffusion, which allows more reactant to find each other.
In the reaction you point out, hydrogen sulfide is a gas so it can easily diffuse to find more zinc oxide. The product zinc oxide might not form a perfect coating (i.e., it has holes in it) that allow the hydrogen sulfide to diffuse through this product to find more zinc oxide.
Other examples: when you burn a piece of wood, this is a solid reacting with a gas. But the combustion reaction is actually between vaporized wood and oxygen. The presence of a gaseous fuel is what allows good mixing between it and oxygen to produce a flame.
Finally, an example from a chemistry demonstration: solid barium hydroxide and solid ammonium chloride are mixed together and shaken. They reaction is quite endothermic and can freeze water on the outside of the reaction vessel. The trick here is that the barium hydroxide is usually the octahydrate. The hydrate provides enough water to create small amounts of solution for the reaction.

The barium Hydroxide/ammonium chloride system is a eutectic, not a double displacement.

I am referring to reactions in general, not just double displacements. @A.K.

First of all most of the literature will refer to double displacement as metathesis instead so that should help you search. But yes your reactant can be in any phase and perform a double displacement. However, mobility is a factor that is not considered in liquid or gas phases, but does impact solid phases. To make solid phases react at an observable rate, heat is typically required to increase mobility and thus effect a reaction.  One example is the synthesis of gallium nitride from gallium iodide and lithium nitride: $$\ce{GaI3_{(s)} + Li3N_{(s)} \rightarrow GaN_{(s)} + 3 LiI_{( s)}}$$

This is a very interesting reaction. If you're able to, please provide a procedure so that we can see we how the problem of mass transport is overcome.

https://pubs.acs.org/doi/pdfplus/10.1021/cm9901764


text:
Volume of orbitals
As I was learning about atomic structure, the lecturer made a seemingly dubious claim:

The volume of a p orbital is one-third that of the s orbital. Thus, inter-electronic repulsions are particularly significant in p orbitals.

She was explaining the lower first ionisation energy of oxygen compared to that of nitrogen and of course, she invoked the idea of inter-electronic, intra-orbital repulsion in her explanation. However, she then went on to ask why is it that we do not observe this trend when looking at $\ce {ns^{1}}$ and $\ce {ns^{2}}$ configurations and answered with the above claim.
I was rather intrigued by her claim as I've never heard it ever before and intuitively, I feel that it is wrong. Thus, I would like to clarify if this claim is indeed wrong?
I would also like to ask how can the volume of orbitals be determined?

It would most likely be impossible to come up with a reliable measure of the volume of an orbital.
Orbitals are defined over all space, so an integral like this
$$
\iiint\phi(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz
$$
is not guaranteed to be finite, and also doesn't really mean anything. Clearly, the orbitals are decaying, but are they decaying quickly enough to have this integral be finite? I would suspect they probably are because the hydrogen atom orbitals decay like $\exp(-r)$, and this integral from $0\rightarrow\infty$ is finite. Regardless, $\phi$, is an atomic orbital here, but the integral over the space of an atomic orbital, molecular orbital, or wavefunction doesn't mean anything physically that I am aware of. Only the integral over the magnitude squared.
One could think about picking a fixed distance from the nucleus, and then integrating over all space inside this distance from the nucleus, but this would always be arbitrary. I doubt that a $p$ orbital would ever be one third the volume of an $s$ orbital for a reasonable choice of distance.
Third, any measure of the volume of an orbital which gives a finite answer is going to be spoiled when you consider that the shape or orbitals is not unique in the sense that a unitary transformation of the molecular orbitals gives another set of valid molecular orbitals. These transformations will almost certainly not maintain the same volume within some distance of the nucleus or anything like this.
It is perhaps possible to quantify what this lecturer is getting at in a more concrete way, however. For instance, there are different schemes of creating localized molecular orbitals in the context of Hartree-Fock. The Edmiston-Reudenberg method of orbital localization maximizes the electron-electron repulsion. So, one could carry out this procedure to generate orbitals which look like the typical $s$ and $p$ orbitals we think about, then look at the result of this maximization procedure to verify that the electron-electron repulsion is larger in the $p$-like orbitals than the $s$-like orbitals. This might work, but because the shape of the orbitals can easily be changed, I doubt this measure has any chemical significance.
So to summarize, the lecturer's reasoning may not be strictly wrong, but she is being very loose with objects that are poorly defined, such as the volume of an orbital. I also explained why the volume of an orbital is poorly defined, and there is really no way to fix this completely.

As to the question she was actually trying to answer, inter-electronic repulsion in $p$ orbitals versus $s$ orbitals is completely unnecessary. Here is a graph of ionization energy with atomic number
It is quite clear that the dips in the graph correspond to time when we go from a half-filled shell to one extra electron (nitrogen to oxygen) or from a full shell to the first electron in an empty shell (beryllium to boron). The beryllium to boron case says this cannot possibly be due to inter-electronic repulsion being stronger in $p$ orbitals because there is only one electron in the $p$ orbital of boron. I would say most of the explanation of this phenomenon is answered in this question. Basically, it boils down to the fact that these electrons which are added to a new shell or half-full shell must be the opposite spin of the other electrons, so they are not stabilized by exchange, and are higher in energy despite the greater attraction to a more positively charged nucleus. Notice this effect decreases as you go down the periodic table. This is understood by noting that exchange is a somewhat distance-dependent phenomenon, so as the orbitals become larger, this loss of stabilization from losing exchange is not as competitive with the increased nuclear charge.

My take is, you could say that the average distance $r = |\vec{r}_1 - \vec{r}_2|$ between two electrons in the same p orbital is smaller than that between two electrons in the same s orbital. Hence the repulsion - which scales as $1/r$ - is greater for p electrons. Volume is indeed a poor choice of word.

@orthocresol I agree, but even this remains to be shown quantitatively. It may very well change as you go down the periodic table and so on. Also, the measure you suggest $|\vec{r}_1-\vec{r}_2|$ is the very thing which is minimized in the Edmiston-Reudenberg localization scheme as this does indeed maximize electron-electron repulsion. I guess the bigger point is that these orbitals aren't unique, so reasoning based off of any particular set of orbitals is always somewhat questionable.

Why do you claim that the integral above is not guaranteed to be finite? I think it is

@user1420303 Ya, it probably is, but the only thing that really must be finite is the integral over the square of that orbital, and there are certainly functions whose squares are convergent when integrated over all space, even though that is not true of the function. The most obvious example is $1/x$, which famously diverges and $1/x^2$ which converges. There is probably a physical argument that all atomic orbitals decay faster than $1/x$, but I don't know this is actually true since atomic orbitals are not strictly real things (in the physics sense).

Thanks for the insightful response. Rationalising the observations using the concept of exchange energy is much more convincing. Based on your explanatiok, why do we not see a dip going from ns1 to ns2 since the extra electron added is also not stabilised by exchange?

If they are solution to t.i.s.e. they should be. I would say that Fock operator is bounded, but as you said, maybe in a even broad sense they are not

@Tan Yong Boon I would say that is because the ns1 electron does not benefit from exchange, so there is no stabilization lost when that next electron is added, so the ionization energy increases because nuclear attraction just be larger than the electronic repulsion or else the electron wouldn't bind at all.

Another question I have is: Can the exhange energy and the inter-electron repulsion be quantitatively determined to show that it is indeed the inter-electron repulsion that is responsible for this phenomenon?


text:
Bond angles in NH3 and NCl3
The bond angle in a molecule is inversely proportional to the electronegativity of the surrounding atom if the central atom is same. This also happens with $\ce{NH3}$ and $\ce{NF3}$, as bond angle in $\ce{NH3 > NF3}$.
Why then is the bond angle in $\ce{NCl3 > NH3}$?

We will need to look at Bent's rule as well as steric effects. In the case of NH3 and NCl3, it is clear that steric effects play the dominant role. This is due to chlorine atoms being significantly larger than hydrogen atoms  and nitrogen being a small atom.

Possible duplicate of Why does NF3 have a smaller bond angle than NH3?

The bond angle of a molecule depends on several factors. We have to look at all the factors and then decide the result according to them.
The difference in bond angles in $\ce{NF3}$ and $\ce{NH3}$ is only determined by the electronegativity difference between the central atom and the bonded atom. As $\ce{F}$ is more electronegative than $\ce{H}$, in $\ce{NH3}$, bonding electron pair shifts more towards $\ce{N}$ than they shift in $\ce{NF3}$. So, the bond angle increases. There is no other effect at all in this case.
But in the case of $\ce{NCl3}$, you can notice there is a lone pair on the central $\ce{N}$ atom. Most importantly, there is an introduction of vacant low lying $\ce{3d}$ orbital of $\ce{Cl}$ where the lone pair of $\ce{N}$ can delocalise. This effect is known as $\ce{p\pi-d\pi}$ backbonding. So, due to this backbonding, which is quite strong, the bond between $\ce{N}$ and $\ce{Cl}$ adopts a significant double bond character.
Although the electronegativity difference is decreased in the case of $\ce{NCl3}$ and bond pair is shifted away from central $\ce{N}$, this introduction of double-bond character increase the repulsion between the bond pairs even more, and that's why the bond angle increases.
So, in case of $\ce{NF3}$ and $\ce{NH3}$ there was no vacant and valence d orbitals either in $\ce{F}$ or in $\ce{H}$. So, this backbonding doesn't occur there. But in $\ce{NCl3}$, the case is different and more factors determine the bond angle. As the backbonding is much stronger, this effect overall dominates over the electronegativity factor and make the bond angle increase.
Experimental evidence of this double bond character: In $\ce{NCl3}$, $\ce{N-Cl}$ bond length is 1.759 angstroms but $\ce{N-Cl}$ single bond length is 1.91 angstroms. This indicates the presence of some partial double bond character which shrinks the bond length.

Why 3d orbitals?  There could be pi back-donation with the orbitals combinations from the usual valence sibshells.

@Oscar Lanzi, I didn't get your point. Can you please specify which orbitals you are talking about??

Overlap between "nonbonding" electrons on chlorine and antibonding sigma orbitals, which weakens the sigma bonding but gives more bond energy with pi bonding.  It's a typical situation with molecular orbitals.

As Tan Yong Boon stated, it is Bent’s Rule with which we can explain the lower bond angle of $\ce{NF3}$ when compared to $\ce{NH3}$. The rule as stated by Henry Bent:

Atomic s character concentrates in orbitals directed towards electropositive substituents.

Fluorine is more electronegative that hydrogen, and the $\ce{N−F}$ bond would have greater p character than the $\ce{N−H}$ bond. And more s character leads to large bond angles. Thus, the bond angle is greater in $\ce{NH3}$ than in $\ce{NF3}$.
Now, consider $\ce{NCl3}$. Clearly, $\ce{Cl}$ atom islarger in size than the central atom, nitrogen. Hence the higher bond angle here is due to the steric crowding caused by $\ce{Cl}$ atoms.(More pronounced than the electrongativity of $\ce{Cl}$ atom). They repel eachother and hence bond angle increases. This is not the case with small $\ce{H}$ atoms around the nitrogen in $\ce{NH3}$. Therefore, the $\ce{Cl-N-Cl}$ bond angle of $\ce{NCl3}$ is greater than $\ce{H-N-H}$ bond angle of $\ce{NH3}$.

Some search on bond angle data for NCl3 and NH3.
Many places, it is mentioned that  bond angle of NCl3 > NH3 but no references are given. I looked around to find some experimental data on this.
To my search bond angle in NCl3 (107) < NH3 (107.8).
Here are the references for my analysis.
In case of NCl3:
(1) http://www.chm.bris.ac.uk/motm/ncl3/ncl3js.htm
(2) https://en.wikipedia.org/wiki/Nitrogen_trichloride#cite_note-2
In case of NH3 See book: Chemistry of Elements by N N Greenwood and Earnshaw (second edition)

Could you please type your answer in instead of posting a screenshot? Text on the image lacks readability, cannot be edited or parsed by the search engines and by the software used by visually impaired members.


text:
Is there a characterisation for conjugated dienes similar to cis-trans?

As shown above,we know there are two configurations possible for conjugated dienes.
Are they given any special names like we have cis and trans?
This came up because while studying the Diels Alder reaction, it occured to me that maybe only the 'cis' like dienes will go through the reaction.

Go through $E -Z $ nomenclature. Here it is not possible ,but in any double bonds where the attached groups on one side is different, that is very useful.

Yes, the terms "s-cis" (single bond cis) and "s-trans" (single bond trans) are often used to describe the geometrical relationship of the double bonds attached to the central single bond.  See this earlier answer  for more information as well as how those conformations affect the Diels-Alder reaction.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) supports various stereodescriptors that are used to denote specific conformers, in particular concerning conjugated dienes:

P-94.3.1.3 Stereodescriptors s-cis and s-trans
The spatial arrangement of two conjugated double bonds about the intervening single bond is described as ‘s-cis’ if ‘synperiplanar’ and ‘s-trans’ if ‘antiperiplanar’. This descriptor should not be applied to other systems such as N-alkyl amides (use ‘E/Z’ or ‘sp/ap’).


text:
Relative fraction of population of different enzyme conformations, knowing the rate constants
Say we have an enzyme that has multiple structural conformations, say A, B, C and D. Additionally, we say that the enzyme must go through all conformations, to get to one end to the other, i.e.
$$\large \ce{A <=>[k_{A-B}][k_{B-A}] B <=>[k_{B-C}][k_{C-B}] C <=>[k_{C-D}][k_{D-C}] D}$$
So there's no skipping any of these steps to go from A to D.
Given all the rate constants derived experimentally, and thus knowing the equilibrium constants for every step, is it possible to derive an overall equilibrium constant that describes which conformation the enzyme is most likely to be found, and perhaps more importantly, the relative fractions for every conformation?

Of course it's possible, are you asking how to do it in general or specific cases?

If we use the equilibrium constants:
$$\ce{A <=>[$K_1$] B <=>[$K_2$] C <=>[$K_3$] D}$$
And we call $[\text{E}]$ the total enzyme concentration, we have:
$$ [\text{A}] + [\text{B}] + [\text{C}] + [\text{D}] = [\text{E}]\label{a}\tag{1}$$
It's not hard to see that $\displaystyle K_1 K_2 K_3 = \frac{[\text{D}]}{[\text{A}]}$. From here, we have that $\displaystyle [\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3}$. We also get:
$$[\text{B}] = K_1[\text{A}] = \frac{[\text{D}]}{K_2 K_3}$$
$$[\text{C}] = K_2[\text{B}] = \frac{[\text{D}]}{K_3}$$
Hence, substituting these last three equalities in $\ref{a}$, we get:
$$ \frac{[\text{D}]}{K_1 K_2 K_3} + \frac{[\text{D}]}{K_2 K_3} + \frac{[\text{D}]}{K_3} + [\text{D}] = [\text{E}]$$
So:
$$[\text{D}] = \frac{K_1 K_2 K_3 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
From here we can also express the concentrations of the other species in terms of $[\text{E}]$:
$$[\text{A}] = \frac{[\text{D}]}{K_1 K_2 K_3} = \frac{[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{B}] = K_1 [\text{A}] = \frac{K_1 [\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
$$[\text{C}] = K_2 [\text{B}] = \frac{K_1 K_2[\text{E}]}{1 + K_1 + K_1 K_2 + K_1 K_2 K_3}$$
With these equations we may compare the concentrations of each species at equilibrium, thus we are able to determine which one is the highest based on the values for the equilibrium constants; and we can also obtain their mole fractions (divide each by $[\text{E}]$, which will eliminate this term from all the numerators).

Just to be clear, $[B] = K_1 [A]$ is just an algebraic construct, right? Because physically it wouldn't make sense to claim that $[B]$ is only dependent on this small part of the reaction.

Yes and no. In principle, at equilibrium $[A]$ can only convert to $[B]$, so its concentration at equilibrium is only dependent on $[B]$ at equilibrium. They are all related, you may also rewrite all the equations in term of any component, meaning that they all depend on each other.


text:
Van der Waals interaction and magnetic dipole dipole interaction
I have a conceptual question that bothers me.
From what I understand, van der Waals forces are the primary source of inter-molecular interaction. There are three different possible origins for van der Waals forces: permanent dipole-permanent dipole (Keesom/short range) forces, the permanent dipole-induced dipole (Debije/short range) interactions and induced dipole-induced dipole (London/long range) forces. (Here we talk about electric dipole.)
We also have magnetic dipole dipole interaction which is $\propto \frac{m_1m_2}{r^3}(3\cos^3\theta-1)$ and has quantum mechanical origins such as Pauli's exclusion principle and exchange interaction and etc.
My question is: since both magnetic and electric dipole dipole interactions are interpreted based on vector model mathematical formulation and these interactions happen at atomic level and they are of a similar range, in what cases we consider interaction between electric charge or a dipole, and magnetic dipole?
Or in general, can magnetic dipole-dipole interaction happen along with electric dipole-dipole interaction in a systems?
When I read text books on inter-molecular forces (Israelaschvili, Cosgrov, etc) non of them talk about magnetic interactions. I wonder why.

As far as I can understand, your question is: "Why aren't magnetic dipole-dipole forces also included alongside electric dipole-dipole forces in calculation of intermolecular forces?" is that right?

Yes,,,,,,,,,,,,

Magnetostatic interactions are much smaller than electrostatic interactions. For atomic and molecular systems, the magnitude of the magnetic potential energy is roughly $\alpha^2 \approx 5\times 10^{-5}$ times the magnitude of the electrostatic potential energy. Therefore magnetic dipole (and higher orders) are usually neglected.
The multipole expansions for the magnetostatic long-range potentials are formally equivalent to the usual electrostatic long-range, with $\mu_0/4\pi$ replacing the $1/4\pi\epsilon_0$ prefactor, where $\mu_0$ is the vacuum permeability.
References:
M. Lepers and O. Dulieu,  arXiv:1703.02833v2 [quant-ph] https://arxiv.org/abs/1703.02833v2
1 Z.C. Yan, J.F. Babb, A. Dalgarno, and G.W.F. Drake, Phys. Rev. A - At. Mol. Opt. Phys. 54, 2824 (1996).


text:
Long-term durability of linen vs polyamide vs polyester?
What is the comparative long-term durability of linen vs polyamide vs polyester when exposed to oxygen, but not sunlight?
I assume that linen, polyamides (Nylon) and polyesters will all oxidize and become brittle over long periods of time (decades). What is the comparative longevity of the three materials in this regard?


text:
How does a mostly non-polar molecule prevent a polar head from being pulled off by other polar molecules?
Linear-shaped molecules of a solute with a large non-polar section and a small polar head are not soluble in a polar solvent. Why is this? Why don't the attractive forces between a molecule of polar solvent and the small polar head of the solute molecule cause the polar head of the solute molecule to dissociate?
The typical explanation I find on google is that the non-polar section "outweighs" or "is greater than" the polar head, but I'm a little confused as to what that means.

The same way the Sun is powerless to rip our Moon away.  The Moon stays close by because the strength of Earth's gravity is great enough at this close distance to ward off the Sun.  Similarly, covalent bonds at close range are typically strong enough to prevent other molecules ripping them apart unless these also come into close contact.

During the disolution of a compound C  in a solvent S, intermolecular forces between molecules of C and between molecules of S are broken (which requires energy), and intermolecular forces between each molecule of C and several molecules of S around it are formed (which releases energy). As such, dissolution of C into S is only favorable if the released energy is greater than the required energy.
Let's consider octanol $\ce{CH3(CH2)7OH}$ in water. We need to break the hydrogen bonds between octanol molecules and the London dispersion forces in them to separate the molecules, and we need to break the hydrogen bond network in water to accommodate each octanol molecule. Then, we surround each octanol molecule by water molecules. Only the polar head ($\ce{OH}$) will be able to form new hydrogen bonds (and account for the broken ones), but the interaction between water and the hydrophobic tail is merely permanent dipole-induced dipole, which are weak compared to all the hydrogen bonds we had to break between the water molecules to be able to accomodate the large hydrophobic portion, so we're losing energy rather than gaining it. Thus, this will be poorly soluble in water.


text:
Why does increasing temperature for an endothermic reaction always increase equilibrium constant?
According to Le Chatelier's Principle, increasing temperature for an endothermic reaction shifts the reaction towards the products.
However, unless there are things that I am missing, I believe there are three ways to imagine this phenomenon, albeit they all arise from the same equation/concept.

Classic $\Delta G = \Delta H - T \Delta S$
$\Delta S_{\text{sys}} + \Delta S_{\text{sur}} = \Delta S_{tot\text{}}$
Van't Hoff Isochore: $\ln\left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

In the second case, it makes sense that an endothermic reaction would favor an increase in temperature because $\Delta S_{\text{sur}} = \frac{-\Delta H_{\text{sys}}}{T}$ assuming constant pressure. Thus an increase in temperature would mean an increase in total entropy change, indicating a more favorable change.
Van't Hoff's isochore also predicts that an increase in temperature favors the forward reaction of an endothermic reaction as well if you do out the math.
However, in the first case of the classic equation, the equation suggests that there is actually an entropy dependence rather than an enthalpy dependence of the gibbs free energy, which is related to the equilibrium constant by $-RT \ln(K) = \Delta G$.
It further confused me when considering that case 1 derives from case 2, and case 3 derives from case 1. I need help identifying my mistake or rectifying the confusion on why equilibrium is not entropy dependent.

The first equation, the corresponding change in G,  and the equilibrium constant is supposed to be related to the following two initial and final states:
State 1:  Pure reactants in separate containers at temperature T and pressure 1 bar
State 2:  Pure products in separate containers at temperature T and pressure 1 bar
The $\Delta G$ you are talking about refers to the change between these two thermodynamic equilibrium states.  But please note that these  thermodynamic equilibrium states that are not in chemical reaction equilibrium with each other.  So, to get the equilibrium constant for the reaction, you are using the $\Delta G$ for these two states which are not in chemical equilibrium.  To understand how this plays out, you need to review the derivation of the relationship between this $\Delta G$ (actually denoted $\Delta G^0$ to distinguish the specific states 1 and 2) and the equilibrium constant K.
Now, when you change the temperature, all three quantities in the equation for $\Delta G^0$ change:  T, $\Delta H^0$, and $\Delta S^0$, not just T.  This is why the Van't Hopf equation gives the correct result for the effect of temperature on the equilibrium constant, and the other approach you described does not.

Saying that $- RT \ln K = \Delta H - T\Delta S$ shows that the dependence of $\ln K$ to $T$ is related to $\Delta S$ is flawed since both terms are multiplied times $-T$. Hence:
$$ \ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$$
(which is part of the derivation of van't Hoff's equation, which results after derivation against $T$).
From here, you can see that the dependence of $\ln K$ with $T$ is associated with $\Delta H$, not $\Delta S$.


text:
Vibration motion at the imaginary frequency for TS of the Claisen rearrangment
I understand that the vibration along the reaction coordinate is different for the transition state , as the motion of the atoms takes TS towards products/reactants.
However, what would be the vibrational motion corresponding to the imaginary frequency of the Claisen rearrangement?

I started at PM7 niveau, where it looks like this:

and at the DFT niveau (DF-B97-D3/def2-SV(P)) it is not that different:

To summarize: Id say, it looks "as expected".
The transition state needs to connect both local minima. When you follow the animation of the normal modes above, you can see bond forming and bond breaking between either the two carbons on the left top or the carbon and the oxygen on the bottom right.

#p pm7 opt=(ts,noeigen,calcall)
8        0.416763000     -1.336451000     -0.208978000
6        1.497193000     -0.113742000      0.149606000
6       -0.772891000     -1.071748000      0.210613000
6       -1.509741000     -0.011936000     -0.327122000
6        0.829456000      1.060608000     -0.253868000
1        2.339023000     -0.475266000     -0.444938000
1        1.609073000     -0.346478000      1.210997000
6       -0.368126000      1.414823000      0.371361000
1       -0.923592000      2.282901000      0.026388000
1       -0.520920000      1.227340000      1.430860000
1        1.106186000      1.529032000     -1.191942000
1       -2.506791000      0.192973000      0.036038000
1       -1.388722000      0.264355000     -1.370246000
1       -1.103706000     -1.651275000      1.071122000

#p b97d3/def2SV/W06 opt=(ts,noeigen,calcall)
8       -0.556373000     -1.341873000     -0.246607000
6        1.267703000     -0.946693000      0.176479000
6       -1.313091000     -0.435583000      0.272443000
6       -1.504992000      0.814802000     -0.295637000
6        1.395538000      0.372483000     -0.314454000
1        1.717678000     -1.782296000     -0.375359000
1        1.166045000     -1.105900000      1.259763000
6        0.768640000      1.425447000      0.328791000
1        0.744341000      2.432184000     -0.112679000
1        0.518449000      1.360808000      1.397182000
1        1.705654000      0.503456000     -1.362984000
1       -2.163740000      1.551184000      0.190272000
1       -1.227963000      0.987609000     -1.343598000
1       -1.692271000     -0.594810000      1.314532000

That's true, it's exactly what you'd expect.
However, because of the bonds breaking and forming at the TS, would you account for the BSSE when computing the activation barrier for the reaction?

I never really saw someone using BSSE correction, if it is not directly about it. From what I could see, the common way is to use bigger basis sets.


text:
How does the redox of a battery happen?
I tried to understand how does a battery work. After doing some research, I knew chemical reactions inside are key. But when I came across the reactions inside a zinc carbon battery, almost all websites say "Zinc is oxidized and gives out two electrons".
Why does this reaction happen in the first place? Is it the result of the reaction between the electrolytes and the zinc? And I'm confused what exactly do electrolytes do because I saw they are described as "catalyst". I only understand they allow the flow of ions but prevents the flow of electrons.
Thank you and if what I say above is wrong, please correct me.

Hi! I'm afraid this would be too broad (i.e. too long) to explain in an answer. You should read up on online tutorials about the basics of electrochemistry. Redox are a series of chemical reactions that just tend to happen.

The zinc-carbon cell uses a zinc anode and a manganese dioxide cathode. It is manganese dioxide that takes part in the reaction, not carbon. Carbon is added to the cathode to increase conductivity and retain moisture(catalyst). It also contains a layer of ammonium chloride. Following is the overall reaction.
$$ \ce{Zn(s) + 2 MnO2(s) + 2 NH4Cl(aq) → Mn2O3(s) + ZnCl2 (aq) + 2NH3(aq)} $$
This is why the half reactions are as follows,
Anode: $\ce{2Zn -> Zn^{2+} + 2e–}$
Cathode: $\ce{2NH4+ + 2MnO2 + 2e– -> Mn2O3 + H2O + 2NH3}$
An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. Here, ammonium chloride solution acts as the electrolyte immobilized in the form of a paste. Basically its a chemical medium for flow of charge.

I was taught the second reaction instead as $\ce{NH4+ + MnO2 + e- -> NH3 + MnO(OH)}$ Where did you see the one you've written?

Wikepedia says the reaction which you told is when zinc chloride is used . > “If zinc chloride is substituted for ammonium chloride (i.e. replaces ammonium chloride) as the primary electrolyte, the anode reaction remains the same but the cathode reaction is:
$$\ce{MnO2(s) + H2O(l) + e− → MnO(OH)(s) + OH−(aq)}$$


text:
What is the difference between enthalpy of reaction and standard enthalpy?
What is the difference between enthalpy of reaction ($\Delta H$) and standard enthalpy ($\Delta H^\circ$)? I was told that the standard enthalpy of reaction is the change in heat when one mole of matter is transformed. I don't quite get what this one mole refers to in a chemical equation.
For example, I came across this equation in my textbook: $$\ce{N2 + O2 -> 2 NO \tag{ΔH = +181kJ/mol}}$$
Does the $\Delta H$ here mean the amount of energy absorbed during the formation of 1 mole of $\ce{NO}$? But since 2 moles of $\ce{NO}$ are produced, shouldn't it be $181\times 2 = 362$?

Be sure to take this [tour] to know about the community. Also use MathJax to format questions and answers.

Yes, the standard enthalpy of reaction ($\Delta_\mathrm{r}H^\circ$) is the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states. The only condition is that the participants have to be in their standard states, ie. usually at 1 bar pressure.
On the other hand, enthalpy of reaction ($\Delta_\mathrm{r}H$) is amount of energy absorbed or released at any condition. Considering your example,
$$ \ce{N2 + O2 -> 2NO} \qquad \tag{$H = \pu{+181kJ}$}$$
Here, $\Delta$H represents the amount of heat absorbed in the formation of 2 moles of $\ce{NO}$. For the formation of one mole of a substance from constituent elements, it is called enthalpy of formation($\Delta_\mathrm{f}H$)and if in standard states, standard enthalpy of formation.
$$ \ce{ 1/2 N2 + 1/2 O2 -> NO} \tag{$\Delta_\mathrm{f}H$}$$
As we can see, $\Delta_\mathrm{f}H$ = $\frac{1}{2}\Delta H$. Hence it should be $\Delta_\mathrm{f}H$= +90.5 kJ/mol.
Note: Unit of enthalpy is that of energy but unit of enthalpy for a reaction is of energy per unit mass or amount.

It would be worth saying that the units for entalpy are just $\text{kJ}$, whereas those of enthalpy of formation are $\pu{kJ mol^-1}$

Also, standard conditions also include $\pu{298 K}$.

No there is no condition on temperature as per IUPAC but commonly known is 298K. @ralk912 Refer standard states

Do you have a link for that?

I gave one in my answer as well as comment. Will update regarding units @ralk912

Interesting. I wonder why the IUPAC would decide to do that instead of just fixing a value. In any case, then the temperature should be given when speaking of standard states I think.

Yes, it is not very intuitive. But I was taught like that.

The IUPAC has proven themselves they're not usually intuitive. You were taught right.

Let us continue this discussion in chat.

Thanks for the help! I'm also wondering what the units are for standard enthalpy of reaction. So far I've seen kJ and kJ/mol. Can they be used interchangeably? Does the /mol part add anything?

Mol part indicates the enthalpy change per mol of the reaction. (If that makes sense) @chemquestion Unit for standard enthalpy change of a reaction is kJ/mol.(usual) Although we can use any combination of energy/mass or amount like kcal/mol


text:
How does water flows through pipes in cold countries?
This question occurred to me when I looked into the fridge, where the water had frozen due to very low temperatures. If the water freezes due to low temperatures (0 degrees Celsius), then how can it flow through pipes in the cold countries (especially near the north pole) where the temperature is almost always below 0 degrees Celsius, especially in the cold months?

Sure cold countries find ways to prevent this because they expect these cold weathers but it happened more than once to me that I came home from vacation and the water line was completely frozen because it wasn't used.

From the Wikipedia page on Frost Line

The frost line—also known as frost depth or freezing depth—is most commonly the depth to which the groundwater in soil is expected to freeze. The frost depth depends on the climatic conditions of an area, the heat transfer properties of the soil and adjacent materials, and on nearby heat sources. For example, snow cover and asphalt insulate the ground and homes can heat the ground (see also heat island). The line varies by latitude, it is deeper closer to the poles.

So they just dig the pipes in deep.

Canada: Can use table of estimates based on freezing index degree days.
  Ontario: Map of frost penetration depths for Southern

Ottawa: 1.8 metres (5.9 ft) [7][8]
Windsor: 1.0 metre (3.3 ft) [7][8]

Here's a  soil temperature vs depth graph.

Source

Source of picture? Also flow of water (hindering freezing) may also influence this.

In addition to @AvnishKabaj's answer, I'd like to add that in some places, people use thick pipes, and surround water pipes with more water to stop the water from freezing and then making pipes burst. If water freezes, it not only won't flow, it'll often cause pipes to burst due to the expansion of ice (between $4^o$ to $0^o$C due to air gaps in ice. See Water Pipes Burst)
Surrounding with water helps as water has a really high specific heat ($1$ $cal $ $kg^{-1}$ $K^{-1}$). As a result, a lot of heat needs to be lost before the water actually freezes. (Source: Concise Physics for Class X, Selina Publishers).
From the Wikipedia Article on Pipe Insulation,

Since some water pipes are located either outside or in unheated areas where the ambient temperature may occasionally drop below the freezing point of water, any water in the pipework may potentially freeze. When water freezes it expands and this expansion can cause failure of a pipe system in any one of a number of ways.
Pipe insulation cannot prevent the freezing of standing water in pipework, but it can increase the time required for freezing to occur—thereby reducing the risk of the water in the pipes freezing. For this reason, it is recommended to insulate pipework at risk of freezing, and local water-supply regulations may require pipe insulation be applied to pipework to reduce the risk of pipe freezing.

This is just another method by which water in pipes is prevented from freezing, and how it's flow is maintained in cold countries.


text:
Are allylic tertiary alcohols oxidized by the Jones' reagent via a classical carbocation intermediate?
I know the mechanism for Jones' oxidation is (picture from Organic Chemistry by Clayden):

and it proceeds through a chromate ester. And in general we understand that the Jones' reagent $\ce{H2CrO4}$ cannot oxidize tertiary alcohols.
However, in my book, a question asked to oxidize the following tertiary allylic alcohol:

The solution proceeded via a carbocation intermediate to eventually form 3-methylcyclohex-2-en-1-one.
I wish to ask if this a correct solution. Does this reaction really occur? That too in the way it is described? If so, why isn't it described in any other book, even as an exceptional case? (Most other books simply accept that tertiary alcohols cannot be oxidised) Or is this a by-product or a low-yield reaction?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q19; Alcohols, Ethers, Epoxides

See this post https://chemistry.stackexchange.com/a/134868/113297 . Tertiary allylic alcohols indeed undergo oxidation.

It makes sense. Jones's oxidation occurs in presence of a strong acid. Hence, the oxygen can get protonated and leave as a water molecule, giving rise to a tertiary carbocation. The charge can then delocalize, the secondary alcohol get formed, and this alcohol then undergo oxidation. The oxidation will probably drive the equilibrium towards the product.

It does make sense. I was kind of surprised that this hasn't been mentioned before in any other book. Like, I had certainly rote learned the fact that "three degree alcohols cannot be oxidised by Jones the boy" like a song in my brain :P

Well, in principle they can't, this is a secondary alcohol getting oxidized. Another thing that sometimes surprises people is that "ketones can actually be oxidized", via their enol tautomers.

Oh! That's interesting. Do you have an example link? (i think I might be knowing it already but am not sure)

Here is a study in the kinetics with $\ce{KMnO4}$. I can't think of a practical application off top of my head, I'd expect the reaction not to be very clean (the carboxylic acids/ketones formed can form enols themselves!). Also, the Baeyer-Villiger reaction is also an oxidation, but of course that's a different mechanism and off-topic (and well combustion is too, but even more off-topic).

Oh yes, the reaction is more common with enol ethers, of course, since milder conditions would be needed due to the higher concentration of the enol.

The oxidation may not go through a cation. The tertiary chromate ester may undergo a [3,3]-sigmatropic rearrangement to the secondary chromate ester and thence to the enone.

I actually like that one better

@ralk912: You should! Read this, Dauben and Michno, J. Org. Chem., 42, 682 (1977).