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Alpha particles and nitrogen triiodide
I've been reading Wikipedia's article on $\ce{NI_3}$ (which doesn't contain an $\ce{I_3^-}$ ion, so "nitrogen III iodide" would probably be a better name for it) and getting very confused. I'll start by distinguishing the decompositions$$\begin{align}2\ce{NI_3}&\to\ce{N_2}+3\ce{I_2},\\8\ce{NI_3NH_3}&\to5\ce{N_2}+6\ce{NH_4I}+9\ce{I_2},\end{align}$$since $\ce{NI_3}$ is often isolated or stored in the form $\ce{NI_3NH_3}$. Here are the two things I'm confused about:

"Nitrogen triiodide was first characterized by Raman spectroscopy in 1990 when it was prepared by an ammonia-free route", despite at least one paper on experiments with nitrogen triiodide dating to 1958. I assume previous experiments were instead by the ammonia route.
Now for the real mystery. Apparently, alpha radiation can trigger such a decomposition (according to Bowden 1958, so presumably it's the second reaction). Someone has earned a "citation needed" for adding, "several attempts to replicate this action have failed".

So here are my questions:

How confident are we that alpha radiation has the claimed effect?
What mechanism, be it purely theoretical or empirically mandated, has been proposed for such a decomposition? (See also my discussion below with @IvanNeretin of its attribution to the high KE of alpha particles.)

You have a rather distorted perspective of the whole "chemical vs nuclear" picture, as if looking at it from the wrong end of a telescope. Electronegativity of He does not matter. Charge on He does not matter (well, almost). Helium chemistry is not a part of the picture. What matters is the energy of an alpha particle, and the energy is really, really big, and even bigger than that.

@IvanNeretin Are you saying the mechanism is that the alpha's multiple-MeV KE allows it to break covalent bonds? If so, does alpha radiation cause many materials to decompose?

Yes, very much so. It is just that your wording, while technically correct, is a massive understatement, as in "High energy of a nuclear explosion may result in property damage, if performed indoors". Indeed, radiation of any type would gradually destroy many materials, and naturally, those which tend to decompose by themselves are the most vulnerable.

@IvanNeretin OK, but how specifically do we expect energy to be redistributed among the nuclei and electrons? Will it primarily go to specific particles and thus trigger a particular reaction route, or are just about all options likely, dependent on just how the radiation strikes the target? In particular, do we expect the energy to be first delivered to an atom or a covalent bond?

A typical chemical bond is about 2-4eV. A Typical alpha particle has about 5,000,000eV KE (https://en.wikipedia.org/wiki/Alpha_decay). You are asking "If somebody drops a nuke on my house can you tell me where the bricks will land"

@IanBush We know where the bricks land; we know the products. It's more that I'm asking, "if non-identical bricks are linked by cement that obeys very complicated bond-specific quantum-mechanical rules, are there nontrivial constraints on how energy is dissipated?" A negative answer is far from obvious, if only because not only do alpha particles not routinely destroy everything, but doubt has been expressed this example even happens at all.

https://chemistry.stackexchange.com/questions/107151/what-chemical-explosives-detonate-when-exposed-to-alpha-particles-or-nuclear-fis

Under what conditions could NI3 be used as an alpha detector?

@JamesGaidis It lacks essential attributes like easy operations, stability and recovery.


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Why are colloidal sol particles of acidic dyes negatively charged and basic dyes positively charged, when it should be the other way round?
From NCERT Class 12 Chemistry; Sri Aurobindo Marg, New Delhi, India; January 2019; p. 136:

Colloidal solutions of basic dyes like methylene blue are positively charged whereas colloidal solutions of acid dye stuffs (eosin and Congo red) are negatively charged.
The charge on the sol particles is due to one or more reasons, due to preferential adsorption of common ions from solution and/or due to formulation of electrical double layer.

A base releases $\ce{OH-}$ ions, so it should preferably adsorb the same ions from water and become negatively charged. Why do basic dyes become positively charged then?
Similarly, acidic dyes release $\ce{H+}$ ions, so they should become positively charged, but that doesn't happen.

Bases rather capture H+(aq), as the Broensted-Lawry definition is wider than the Arrhenius one. E.g. NH3 does not release OH-. OH- is created indirectly by water autodissociation when H+(aq) is being depleted, forming NH4+(aq).

source: Stack Exchange

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Data on Vapour Pressures of Silicone Oils
I am trying to find specific information on vapour pressure of silicone oils with different viscosities which I have used in some experimental work. Is there a chemical databook where this information can be found, or is estimated?
The vapour pressures should differ for oils of different viscosities because their molecules will be of different molecular weights.
The data sheet I found online from the manufacturer for 2 cSt silicone oil merely states that the oil has vapour pressure which can be measured in hectopascals.  This is not specific enough as that could still be several hundreds of pascals and I need to quantify how low it is.

Since each manufacturer uses a proprietary mix of silicones, there is no general rule. You'll need to rely on manufacturer's data. Some companies do give actual vapor pressure at specific temperatures, e.g.,Silicone 704: "Vapour pressure @ 25°C  1x10-9 mbar".
Why not contact the manufacturer directly, instead of asking a third party?

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A dense solute for oils that's easy to acquire
I'm working on a crafts project that is basically putting oil, water and glue in a bottle together with glitter just so you look at it. I want to see how it looks if the density of the oil is very close to that of the water-glue mixture.
The oil I'm using is baby oil, which sits on top of the water, so I need to increase its density. If I had wanted to increase the density of the water, I would just have added salt. Now I was thinking if there might another dense solute that I could dissolve in the oil to make the density do up, but I found that search engines interpreted my search terms in the wrong way and I couldn't find any.
That's why I'm here: to find a suitable solute. Originally I was thinking of powders, like the example of salt in water, but a dense liquid would also work. It would need to be something that's readily available commercially, and preferably not too expensive or dangerous.
Are there any dense solutes for oil that match my criteria?

Why can't you buy a denser oil?

@J.Ari densities are rarely listed on household products. I also probably wouldn't be able to get it as close to the density of the water (based mixture) as I want (which I might not be able to with a solute either, but at least I can try)

You can try powdering any fat soluble vitamin from vitamin tablets (i.e. Vitamin E is a fat soluble vitamin) and see if that works. I think your safest approach is to do some research on oil products by finding the Safety Data Sheet (SDS) for the product online and noting the densities listed there.

And for even denser organic solvents, try bromoform. Iodine-substituted solvents tend to be somewhat unstable, though denser yet.

The @Maurice answer is OK-ish, but there is a better substance to mix with oil to get the desired density:
https://en.wikipedia.org/wiki/Tetrachloroethylene
(also called perchloroethylene and is also known by few other names)

It is denser (~ 1.6) so less of it will be needed
It is WAY less toxic than dichloromethane
It is WAY less volatile than dichloromethane
It is WAY less bad as an environment pollutant.
It is cheaper.
It is widely used as a dry cleaning agent, so it is commercially available and has more favorable regulatory status.

This substance is generally what you smell when you enter a dry cleaning shop. A cooperative dry cleaning worker may give or sell you the small amount you probably need (likely ~200g if I understand your idea).
p.s. be warned that organic substances expand/shrink more than water when their temperature changes. Your oily fraction will tend to float when hot and sink when cold. This is how lava lamps work.

I found a place that sold 250mL of this for a reasonable price. I placed my order and will be back after I have tried it!

Also, regarding the warning, that might actually be nice bonus. If for example the body heat from your hand could make it float, that would be absolutely awesome.

You can add drop by drop some heavy liquid like dichloromethane $\ce{CH2Cl2}$, which is heavier than water (density $1.336$). This substance will easily mix with any oil. Each drop added will increase a bit the density of the organic phase. When enough has been added, the mixture will have the same density as water.

You can try powdering any fat soluble vitamins from vitamin tablets (i.e. Vitamin E is a fat soluble vitamin) and see if that works.

won't do. First, these vitamin tablets are mostly inert material with small amount of the vitamin and second, none of these vitamins is denser than water.

If the purpose is to increase the viscosity of your liquid (initially water) to let the glitter fall more slowly to the bottom of your test tube, then glycerol may be a non-toxic and affordable chemical to check here.
It is colorless (just as water), but considerably more viscous than water.  If the viscosity is too high (compared to water), you may mix the two liquids in any ratio to obtain a clear solution of the two.  Demixing (like oil and water) are thus unlikely.

It's not, that's what I'm using the glue for (there's also some glycerol actually, but its main purpose is to keep the glitter from clumping). The idea is to create a part that doesn't mix (well) with the rest that's a different color. Thanks for your suggestion nonetheless!


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Advantage of Ir-based vs Alq3 OLED material, is quantum yield more important over the cost?
This Wikipedia page https://en.wikipedia.org/wiki/OLED#Material_technologies mentions a couple of molecules that are used for OLED.
The Ir-based compound has ~ 100% quantum efficiency, while Alq3 may ~30% quantum efficiency
But iridium is way more expensive than aluminum (2,550 USD per troy ounce  ~ 82.00 USD per gram vs 2,846 USD per ton ). (Using cost of metal just a rough measure, e.g., The core compounds at the heart of OLED displays are often made with expensive substances such as iridium, a rare metal that sells for nearly $19 per gram., certainly complete comparison is more meaningful)
Higher quantum efficiency is an advantage. But, does it worth the price difference? Could lower quantum efficiency of Alq3 be compensated by higher energy consumption? Or Ir-based system is in exploration phase, so to speak.
And actually which one or something else is used in the OLED monitor currently in phone/monitor/TV?

Please don't abbreviate Wikipedia as wiki. I personally think this question is too broad and cannot be answered within the scope of this site.

Thanks. Edited. Any suggestions about how to narrow it down? My main question is about Ir vs Al. Sorry that I am not familiar with this site.

@Fancourt3000 To ease comparison between the price per unit of weight, I recommend you use units within the same system, e.g., the price per gram for the more expensive (an example), and the price per (metric S.I.) ton for the more affordable.  It is not wrong if you want to express the price in other units more familiar to you in addition, like temperatures in degree Celsius, in degree Fahrenheit, or in Kelvin. (Even in countries were S.I. units dominate daily life, there are occasions where (troy) ounces remain used.)

I tried to add a paragraph explain my confusing point and narrow the scope of the question, say Alq3 is lower quantum efficiency, but low cost. Could the monitor uses more electricity to make it work as iridium based system? Electricity is a cost, price of metal is also a cost. Please let me know what is needed to reopen the question. Thank you!

Using absolute metal costs is very misleading. What has to be compared is difference of production monitor cost versus difference of lifetime spent energy cost.

The metal cost is somehow a rough measure, I think, e.g., in https://cen.acs.org/articles/94/i28/rise-OLED-displays.html, 'The core compounds at the heart of OLED displays are often made with expensive substances such as iridium, a rare metal that sells for nearly $19 per gram.'

What is the cost of metallic iridium in a single OLED device, compared to the device cost ?

great question, need to look for it


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Why is the transition state the highest energy point?
If partial bonds form during transition state, it means some energy is released, so how can it have the highest energy? Shouldn't the point before transition state be highest energy? (the point where all the bonds are broken and new bonds have not started to form?) Or is it because such a point where all bonds are broken and new bonds have not started to form is non-existent, by which i mean that when the reaction takes place, bonds weaken and partially form at the same time?

Exactly. Also, that’s the definition of transition state, highest energy on the reaction coordinate.

I’m voting to close this question because the OP already knew and stated the correct explanation and merely wanted confirmation.

It would be better to consider the transition state as part of the lowest energy pathway on the change from reactant to product. The transition state itself in 3D has the appearance of a saddle, up and down through the highest point on the reaction path, but is the minimum energy at 90$^o$ to this path. Because of vibrations any particular trajectory will pass close to the minimum energy path but need not be exactly on it. Why is it the highest point is that it is the only point of no return. (btw, we 'know' transition states only from approximate theory, not so far from experiment.)


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When does chain termination occur exactly?
In the reaction of substitution of alkane. There are chain initiation, chain propagation and chain termination. However, when does the chain termination reaction occur? Does it occur simultaneously with the chain propagation reaction and once all the radicals run out, the reaction stop? Or it will occur after the chain propagation reaction finishes?

This is chemistry. If a reaction can occur, it will occur. Really, imagine two radicals which are about to terminate each other. Do you think they will stop at the mere thought that there might be some unreacted starting compounds at the far corner of the vessel?

Reminds me of an answer I wrote a while ago about combustion, which is (for a large part) also a radical reaction. https://chemistry.stackexchange.com/a/101967/17952


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Neutron–proton ratio and isotope stability
I have studied that most of the isotopes (not all of them!) with a neutron–proton ratio of $\ge 1.5$ are unstable; but it is obvious that this is not true in some cases like carbon-14 or technetium-99.
Is it possible for an isotope to have a neutron/proton ratio of more than 1.5 and still be stable? If yes, could you please give an example?

Please refrain from excessive (bold text) and meaningless (> is used strictly for a block quote) emphasis. Note that you are using improper notations for isotopes. For instance, it should be either carbon-14 or $\ce{^{14}C},$ but never C-14. Finally, how do you define "stable", exactly? Have you checked N/Z chart on Wikipedia?

$\ce{^{208}_{82}Pb}$, ratio about 1.54.

U-238 (1.59) is not fully stable, but lasts for billions of years and decsys ultimately to Pb-206 (1.51).

Your two example isotopes have ratios <1.5 and are unstable, but you are asking about the opposite (ratio >1.5 and stable).

See also Valley of stability

In context of this answer, stable means not undergoing a radioactive decay, regardless of the value of measured half-life.(+)

For light nuclei till about 40 nucleons, the neutron/proton ratio of stable isobars(the same nucleon count) is generally around 1.0. ( $\ce{^3He, T}$ are quite exception. )
Nuclei too aside from this ratio ( See Valley of stability ) are unstable, undergoing beta decay ( electron or positron emission, or electron capture ) toward the stability valley.
Extremely proton/neutron rich nuclides may emit a respective nucleon.
For nuclei with more than 40 nucleons, the "stability valley ratios" steadily climb from 1.0 for calcium toward about 1.5 for lead. It is because the short distance strong nuclear force gets harder time at balancing the long distance electrostatic repulsion of protons.
The most stable nuclei are $\ce{^{56}_{26}Fe}$ ( the lowest mass per nucleon ) and $\ce{^{62}_{28}Ni}$ ( the highest binding energy per nucleon ).
The element with the highest number of stable nuclides (10) is $\ce{_{50}Sn}$ - see magic numbers below.
The heaviest stable yet nuclei reach the neutron/proton ratio near 1.5 ( 1.53 for $\ce{^{208}_{82}Pb}$), but they are already on the edge of cohesive ability of the strong nuclear force.
There are nuclides that are observationally stable, but theoretically they could undergo a decay to more stable nuclides.
All known nuclides with Z > 82 ( bismuth and above ) are unstable, adding emission of alpha particle, spontaneous or stimulated fission to the decay repertoir.

There are few other semi-empirical rules for nuclide stability :

If the element has the odd number of protons, it has maximally 2 stable isotopes. Technetium and promethium have bad-luck here, everything taken by even-proton neighbors.
If there are 2 isobars ( nuclei of the same nucleon number) of 2 neighbor elements, (like $\ce{^{40}K, ^{40}Ca}$), maximally one of them is stable ( the one with lower mass ). The unstable one usually decays by some of 3 modes of beta decay. ( $\ce{^{40}K}$ by all of them ).
Generally, nuclei with lowest energy/mass and the highest stability are those with even number of protons as well as of neutrons ( even-even ). Then come odd-even(either way) ones and the least stable are generally odd-odd nuclei. This is related to stabilization of nucleus by nucleon spin coupling.
There are semi-empirically determined so-called nucleon magic numbers
2, 8, 20, 28, 50, 82, 126 ,
related to nucleus quantum structure, predicting relative extra stability of a nucleus. The first 3 magic numbers apply to 3 "double-magic nuclei $\ce{^4_2He, ^{16}_8O, ^{40}_{20}Ca}$. the higher cases for the same numbers are not stable, as they miss the stability ratio, like $\ce{^{56}_{28}Ni}$.(+++)
Some nuclei, often "even-even" ones, like $\ce{^{128}Te,^{82}Se}$ are conditionally stable (not to be confused with meta-stable nuclear isomers), being at the local energy minimum among their isobars. Their "odd-odd" neighbors (like $\ce{^{128}I}$) are even less stable with higher energy, so the regular beta decay is not applicable. But they can undergo very slow double beta decay, simultaneously converting 2 neutrons to protons (or vice versa ): $\ce{^{128}_{52}Te -> ^{128}_{54}Xe + 2 e- + 2 \bar{\nu_e}}$. This can be considered as a kind of quantum tunneling through the state with higher energy than the original nucleus has.
Some of these "change Z by 2" events may be achieved by alpha emission even for medium mass nuclei. The lightest known alpha emitter is accidentally again tellurium:  $\ce{^{104}_{52}Te -> ^{100}_{50}Sn + ^{4}_{2}He}$(++), unless we count the trivial case $\ce{^8_4Be -> 2 ^4_2He}$.

(+) Very strictly speaking,  elements before iron and behind nickel are not thermodynamically  stable, but for most "conventionally stable" nuclei no spontaneous decay/fusion/fission is observed, directly nor indirectly.
(++) This is possibly related to the fact $\ce{^{100}_{50}Sn}$ is "double-magic" nucleus. But in spite of that, it decays anyway by emission of positron or proton, as the ratio 1.0 is too low for tin.
(+++) OTOH, the non-nequal double-magic 20/28 $\ce{^{48}_{20}Ca}$ with neutron/proton ratio 1.4 is unstable, but with exceptionally long half-time $\pu{6.4E19 \mathrm{year}}$, being the lightest nuclide undergoing the double beta decay.

The $N/Z$ numbers you see in general chemistry textbooks are just guiding principles. First of all, ask the question, what is the meaning of a stable isotope? Don't take these numbers and guiding principles too literally. In the comments you have been given so many examples. Let me give another example of $\ce{^{128}Te}$ ($Z=52$, $N/Z= 1.46$). This must be unstable as per the rigid general chemistry rules.
$\ce{^{128}Te}$ undergoes a very rare type of decay and its half life is on the order of $10^{24}$ years. Is this an unstable nucleus? The age of the universe is on the order of $10^{10}$ years. Compare this magnitude. You will have to wait for an infinite time to see it decay. This is stable by all standards. There are so many exceptions.
$\ce{^{124}Xe}$ ($Z=54$, $N/Z =1.29$), also "stable" on an indefinite time-scale. Wait for about $10^{22}$ years to watch it decay.
The examples in the comments deal with U and Pb nuclei with N/Z > 1.5. All isotopes of U are unstable, but $\ce{^{208}Pb}$, is stable.

Surely you notice 10$^{10}$ 10$^{10}$ looks horrible, with two different fonts and two different styles of numbers (oldstyle and lining). The entire thing should be typeset in math mode $10^{10}$ $10^{10}$. Apart from making it look reasonable, this also prevents awkward linebreaks from happening (10 on one line and ^10 on the next line). That depends on the screen width that your reader is using, see https://i.stack.imgur.com/8RX6K.png for an example

Also, I'm no nuclear chemistry expert, but OP asked about stable isotopes with $N/Z > 1.5$. Your examples seem to have $N/Z < 1.5$. So I don't really get the point here.

@orthocresol, Thanks for the typeset. The keypoint is that N/Z >1.5 or N/Z<1.5 is just a guideline.  Of course, the answer is no. Oscar Lanzi gave other examples. You certainly deal with more nuclei than I do :-)

It should be enough to ask for thermodynamic stability.

@Poutnik, I don't think thermo has anything to do with nuclear stability. What was your point regarding thermo. stability?

Time plays no role in TD stability. Either there is  tendency toward  spontaneous nuclear reaction, either not. E.g. diamond or bismuth-209 or tellurium-128 are TD unstable.

Well, being a nitpicker, no very heavy nuclei is TD stable , so let limit ourselves to context of alfa and beta radioactivity.

Electron capture or positron emission is a type beta decay. A colleague who actually has published on tellurium-128 stated that anything having a half life greater than natural U is "stable". I don't know anything about the fundamentals of short range nuclear forces and what causes radioactive decay.

Let us continue this discussion in chat.


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Is an emulsion of dish soap and vegetable oil stable?
If I fill a glass jar with water, vegetable oil, and dish soap, then mix it until the oil is completely emulsified, will the mixture stay emulsified indefinitely?
Assuming the dish soap and oil are chemically stable and do not degrade or otherwise change over time, if I left the jar on a shelf for years, would it remain emulsified? If I heated the jar up (but not enough to boil or decompose anything), or froze it, or shook it, or put it in a centrifuge, would it stay emulsified?

I'm gonna go with yes.

No, it won't remain stable.  You cannot have any arbitrary ratio for a stable emulsion. Recall the types of emulsions oil in water vs. water in oil. If you notice, dishwashing soap has dozens of other additives to stabilize the dishwashing liquid and prevent phase separation. Nobody can make a prediction what will happen after boiling heating.

It would be nice, if you could add some context. What is the background of this query?

Mostly I am curious. There was a comment chain on reddit about washing oily things with soap and someone asked if the oil would become unemulsified. The context seemed to be about pouring grease/oil down the drain being bad vs cleaning up oily dishes being good. That made me wonder, under what conditions if any (applicable to drains and sewers) would emulsified cooking oil become not emulsified.

Ethan, Oil/grease solidification in sewers is a completely different case. Sewage water does not contain enough detergency to keep oil separated and also think of other junk (very complex biology and chemistry there). Also think of precipitation of fatty acids/soaps via calcium and magnesium ions. I lived in a very cold country and it was prohibited to discard cooking oil down the sink.

"Sewage water does not contain enough detergency to keep oil separated" What does this mean? If I am washing a bunch of oily dishes (and dump the wash water down the sink), does the oil become not separated? "Also think of precipitation of fatty acids/soaps via calcium and magnesium ions." TIL

Your dishwashing solution plus oil/fats from dirty dishes dishes sitting in warm water will remain emulsified as long as the oil-soap ratio is appropriate, eventually your sink water will meet the sewage water, which is mostly water and very low detegerent conc. The oil will eventually separate out/oxidize and the detergent will eventually be decomposed by bacteria. UK sewers have tons and tons of solid fat/oil/grease cakes. :-( That is nasty. Youtube has documentaries on it.


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For benzene, why are the first three pi molecular orbitals considered bonding, and the higher three antibonding?

Is it because all the molecular orbitals that are naturally occupied are all considered to be bonding orbitals? So for benzene, since the lowest three orbitals are all occupied (due to benzene having 6 pi electrons), they're all considered bonding, and any higher unoccupied are considered antibonding?
Is this the right way to think about this? If not, how?

No. It is because the lowest three MOs are lower in energy than the corresponding original AOs. In other words, the electrons would rather be there than belong to the separate atoms, as if they wanted the atoms to be together rather than apart. Actually, there is no "as if". These electrons are the force that brings the atoms together. That's exactly what "bonding" means.
The occupancy of orbitals is irrelevant. Sure, the lower-energy ones are occupied first, and they are also more likely to be bonding, but that's about it. Think of the $\ce{O2}$ molecule: it has two electrons sitting on the antibonding $\pi^*$ MO (which also happens to be doubly degenerate, hence the triplet and stuff). On the other hand, boranes are not unlikely to have empty bonding orbitals.
So it goes.

Boranes are actually so arranged that their bonding orbitals are all filled. The use of 3c-2e bonding leads to fewer and more strongly bonding orbitals than would be the case with just pairwise interactions. However, electropositive metal borides commonly have boron structures whose bonding orbitals are filled only by adding electrons, which leads to predominantly ionic metal-boron bnding.


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Can amide form Schiff base?
Can amide ($\ce{RCONH2}$) form Schiff base? When I searched for it, most papers deal with aliphatic amines. Some paper conducted synthesis with amide-containing compound, but iminization occured at β-nitrogen, (e.g. $\ce{RCON^1HN^2H2}$, iminization occured at $\ce{N^2}$) not α-nitrogen. Can α-nitrogen in amide also form Schiff base?

The lone pair on amide $\ce{N}$ is not fully available to make an imine with carbonyl group (it is conjugated with nearby carbonyl; e.g., recall peptide bonds in protein).

The reaction you might expect to give a Schiff base i.e. condensation of an amide RCONH2 and an aldehyde R'CHO gives a dimeric species where 2 amides add to the aldehyde to give (RCONH)2CHR' https://pubs.acs.org/doi/10.1021/ja01335a085


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IUPAC name for compound with various carbonyl groups
I found this question on a resource video for IUPAC Nomenclature for a competitive exam. It would be great if someone provided the approved IUPAC approved name for this compound. Source of the question : https://youtu.be/SpUoxVq-hmI?t=432

As per the IUPAC Nomenclature or what I understood from it, it is given that whenever the carbon atom is not counted we need to use different prefixes. In this case that would mean the substituent with the locant number 4 should be named as methoxycarbonyl and the substituent on the locant number 3, chlorocarbonyl. So as per that, I wrote the answer as 3-chlorocarbonyl-6-formyl-4-methoxycarbonyl-8-oxy nonanoic acid but I don't think that it is right. Any help would be appreciated :)

The carbonyl on carbon 8 is called 8-oxo.

Oh yah i typed it as oxy..... so other than that is the naming of the compound right @MathewMahindaratne ?

@CalistoMathias You might want to check the rules for parentheses. Also note that there is a wrong space in the suggested name.


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Is the reaction stated below a SN1 or SN2 when ethanol is a solvent?
According to me, ethanol has an acidic H (attached to O) so it should support SN1 as a protic solvent.
But in case of the following example, it is taken as an aprotic solvent, and thusly the reaction is stated as SN2

It is not an aprotic solvent. Cyanide anion is a far better nucleophile than EtOH (not ethoxide) so the secondary nitrile is formed by what is probably a mixture of SN1 and SN2

That is  exactly what I’m asking, whether ethanol is aprotic or protic

EtOH is a protic solvent


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Is there a vapor pressure equivalent to freezing point?
Vapor pressure is where liquids are breaking off into the gas phase, below the boiling point. So is there a term where solids break off into the liquid phase, below the freezing point? Vapor pressure is to boiling point as ____ is to freezing point?
There is a equivalent vapor pressure for solids, which can be called sublimation pressure, where solids break off into the gas phase (like with dry ice), but I can't find 1 where solids break off into the liquid phase below the freezing point only when the vapor pressure equivalent surpasses a partial-pressure equivalent.
This is of course, my interpretation of "the other way around." I can't imagine the other way around being certain liquids solidify into solids above the freezing point. So for liquids, when the vapor pressure drops to equal the partial pressure, then condensation occurs - I can't imagine an equivalent where melting occurs above the freezing point (when the vapor pressure-equivalent surpasses a partial pressure-equivalent).

There is known overcooling of vapor and liquid, and overheating of liquid. But, unless I am mistaken, overheating of solid does not happen as it does not need seeding. // At freezing point, saturated vapor pressures over both liquid and solid are equal, with steeper curve for solids.

@NealConroy - Please see https://meta.stackexchange.com/questions/86997/what-can-i-do-when-getting-we-are-no-longer-accepting-questions-answers-from-th and note that you were/are subjected to an automatic ban. The way to lift it is to fix the posts in question.

@ToddMinehardt I been trying so hard to fix the "what are some phenomena that emit the highest % of anti-Stokes shifts" and no 1 has ever answered to me why my question wasn't focused enough. I tried to make it clear as possible, and kept editing from all the hints in the comments. But now that and most of my questions have slowly been deleted, so there isn't anything left to fix.
I don't even know why my "strong bases and oxidizing agents" got somehow deleted too, some of the comments fought to keep from being closed.
Tagging people serves for explicit notification about comments related to the commented question or answer, or at least to adjacent comments. Using it your way is tagging abuse.

@NealConroy - You are still banned from asking questions. See https://chemistry.stackexchange.com/help/question-bans for details on how you can get it lifted. You need to focus on the underlying problem vs. circumventing the ban.

One could say vapor pressure, again. When you have a solid and a liquid phase of a given material, each of these has its own equilibrium with the vapor. The favored phase at a given temperature and total applied pressure is the one with lower vapor pressure, since the other phase would "evaporate" enough to then favor condensation of the lower vapor-pressure phase. The freezing point is then the temperature where the vapor pressures are equal allowing both condensed phases to be in equilibrium with the same vapor.

Some answers I got are:
The broader term would be chemical potential of the solid vs liquid phases.
For solids not much below their freezing point, the 1st few atomic layers on the surface can be disordered and partially mobile, resembling a liquid (but not the same as the actual liquid phase). However, the thermodynamic description of this region is not taken into account with regular thermodynamics, which considers the behavior of only the bulk. If you want to talk about the surface, you have to add extra terms to the potential energy of the molecules, in particular the surface energy.


text:
Why is chemistry based on rules and exceptions when everything can be simulated?
I am a 10+1 student, currently preparing for JEE. It requires me to prepare for Maths, Physics and Chemistry. I particularly enjoy logic related subjects and application, so I find maths and physics enjoyable, however, I am troubled by the way chemistry currently exists (or is taught to an high schooler).
From my current understanding of parts of concepts of Chemical Bonding (Ionic bonds, VSEPR Theory, Molecular Orbital Theory, ...), I understand that bonds are formed on the basis of approach of atoms, achieving lower potential energy, and thus more stability. But from the POV of solving problems, we are expected to learn rules, most of which come with a variety of exceptions. Also, my teachers have told me that all of these are theories, and match the real results to a fairly good extent.
I hope the intro wasn't too long, and here I come to my main question. If I haven't bored you till here, here is my question:-
If chemistry is mainly related to achieving lower potential energy on account of various forces, why can't we simulate all of chemistry?
To elaborate, being a computer sciences enthusiast, I wondered why we can't/don't have perfect knowledge about how things work on an atomic level. My solution/thoughts were that we could have an AI model which is trained and simulates approach of charged entities, achieving stability, then slowly formation of atoms, then molecules by making the model have atoms approach each other, in various conditions; all of this done while scripting the physics laws mainly governing these, i.e., electrostatic attraction and the others involved which I may not be aware of. The thought that made me think of this was that in VSEPR and hybridization theory, we were taught the alignment of hybridized orbitals, and had to memorize the bond angles, when I realized that those were the angles just formed when charged particles were left to revolve in a sphere and would align themselves in a symmetrical position to achieve no net forces acting on each of them. Is it that such an idea has not been thought of(which I highly doubt) or some other more logical reasons?
If you have read my question till here, I am really thankful to you, since I know I can be very annoying and hard to speak to. If you can answer my queries, I am even more grateful and look forward to interacting.
Thanks,
Ravi Arora

"Give me a computer big enough and a memory large enough and I can simulate the world." - misattributed to Archimedes...

What makes chemistry great is its complexity. Chemical reactions are much more complicated than just rolling marbles on a potential energy surface. Typically, higher excited states (difficult to calculate) play a role as well and many particles interact in a complex sequence with short lived intermediate states. Even the simplest reactions are the topic of current research. Machine learning is used in chemistry as well, but this is of course pretty much a black box and does not really provide insight (although it is useful for solving some problems).

By the way, in light of your remark on repelling charges, do you know what VSEPR stands for?

An exhaustive list of why quantum mechanics is hard to model can be made, but in short, one big reason is if we wanted to know everything about the system, we must then know the wave function for that system. This equation contains all the information of the system. Additionally, when we operate on these equations, (e.g. take the derivative or something like that) we cant always find closed form (finite) solutions, leading us to use approximation or numerical methods to find the solution to the wave equation.

Perhaps you might want to just have a look at the Wikipedia page devoted to just H as a quantum system. Just to get a feeling of the complexity. However the complexity of chemistry even exceed the realm of quanta. Think about flow, surface interaction, chaotic systems and so on.

@Paul yes, I am aware of what it stands for. If what I have learnt is correct, it was named VSEPR for Valence Shell Electron Pair repulsion.

@dval98 I am not that well versed with chemistry, considering that I am in 10+1. Any resources where I can understand better your answer? More specifically, what wave functions of a system are? I have learnt about the wave functions of electrons partly in Atomic Structure, but that knowledge is very brief at present

@Alchimista I really don't know chemistry to that extent, I am aware of some parts of the quantum theory, but only the tip of the iceberg. Could you please share resources where I could read about flow, surface interaction and what they are? I have never come across these terms

@RaviArora chemistry goes even to viscosity, turbolence, etc. I wanted to say that practice is even more complicated than solving quantum stuff. Take your time and follow the steps as they are proposed to you from the curricula ;)

@RaviArora, indeed. The fact that electron repulsion is in the name of the model directly implies that the groups arrange to minimize the coulombic repulsion between them.

As an indication of the problem, an accurate calculation is truly vast, for example in benzene with 42 electrons and a minimum of 72 orbitals  there will be $\approx 10^{20}$ integrals to solve. So severe approximation has to be made, even a small fraction of this number say $10^9$ integrals is still an enormous challenge, and benzene is a small molecule.

Related question: https://chemistry.stackexchange.com/questions/50681/is-it-true-that-all-of-the-physical-laws-and-mathematics-needed-to-describe-chem. In particular, note the quote from Dirac in the question.

I have no idea why this has been closed as a question. It has already generated two good answers so doesn't seem to be suffering from a lack of clarity in what is being asked.

We cannot simulate everything because (i) there is no known theory of everything for chemistry and (ii) computers are not powerful enough to numerically solve the equations we already know.

I wondered why we can't/don't have perfect knowledge about how things work on an atomic level

Well... we do. The things you have suggested (like VSEPR theory) are very crude, though, and not suitable for such tasks. Obviously, if it was a matter of just coding in Coulombic forces, somebody would have done it already.
More appropriately, everything should be simulated using quantum mechanics. There are lots of software that will do that.
The question then becomes why can't we just simulate everything using quantum mechanics. Part of the answer is that the Schrodinger equation cannot be analytically solved for anything more complicated than a one-electron system. Thus, there are some approximations involved; and the results are then not guaranteed to be accurate. Going beyond that, the answers here: Why is chemistry unpredictable? will likely explain that; and I'd argue that apart from the misconception that VSEPR is the theory that underpins all of chemistry, this is basically a duplicate question.
That said, quantum chemistry is used a lot in order to predict
(and more often, rationalise) chemical reactions. So it's not impossible to do that. It's just impossible to do it in a total fashion, where literally everything in chemistry is simulated.

I am aware that VSEPR theory is not the theory which truly generalizes chemistry. In fact, our teachers share that at high school level, most of what we are learning is not 100% accurate and, at times, contradicts what someone pursuing sciences would find. And they also instruct us that we have accomplished a lot more in chemistry, but is beyond the scope of school studies.
Coming to my main question, I want to ask, why can't it just be simulated using Coulombic forces and various approaches? What is the need for Quantum Mechanics to be introduced?

You may ask yourself a similar question as to why you can’t simulate the entire universe using classical mechanics. What is the need to bring in relativity and other more advanced theories. Answer is, classical mechanics just isn’t a good enough theory to explain every experimental observation. Same with electrostatics and quantum mechanics in chemistry. In fact you need relativity too in chemistry.

@RaviArora Chemistry is not (only) governed by classical coulombic forces. In fact, chemistry only emerged when the first atoms (helium) became neutral after the big bang. You need quantum mechanics to form chemical bonds and, as pointed out by orthocresol, this is not trivial and requires approximations and numerical approximations. One needs to introduce quantum mechanics because atoms and molecules behave quantum mechanically and not classically. This is just a consequence of many observations.

@orthocresol Paul thanks a bunch for taking the time to write these answers. I just have one last question in mind, which isn't directly related, maybe. That would be:-
Will we ever be able to photograph an atom? I remember from one of my lectures that we can't even shine light on an electron (if that was practically possible) since it would jump to another energy level. So what I am trying to ask more clearly is, do we know if all this is going to be theoretical, or do we not know yet if we can achieve visual view of atoms/molecules?

We can't simulate everything because there isn't enough computing power to do it
The fundamental problem of making chemical predictions is that they are extremely hard to do because of the combinatorial complexity of what needs to be simulated.
Chemists can and do simulate, though. Big computational models can approximate molecular structures and reactions. But it takes a metric shit tonne of computer power even for simple molecules.
Start with the basic equation about the electronic structure of atoms, the Schrodinger equation. We can derive analytical solutions only for systems with a single electron (this is, by the way, also true of the equations of gravity for systems with more than two large objects). So the best possible answers are derived from simulations of the more complex systems.
The big problem is that the complexity of the equations explaining systems grows extremely rapidly with the number of parts and rapidly exceeds all the computing power on the planet.
Consider the supercomputer-level of computer power needed to create AlphaZero, the system that can play extremely good chess. Chess is a very small system: there are only 8 by 8 positions on the board and a small finite number of pieces. The list of rules is short. The system is simpler than all but the most basic molecule. But it still took enormous computational power for a computer to learn good strategies to play the game.
Molecules and their possible reactions are far more complex than that. chemistry is played on an infinite board and the rules about interactions are far more complex than the rules of how chess pieces interact.
Even at a different level of analysis, like finding all the possible proteins with just 52 amino acids in them (never mind determining their structure), is overwhelmingly complex. There are more possible combinations of 20 amino acids than there are atoms in our galaxy.
So, while simulation can sometimes but used in chemistry in limited circumstances, there is not enough computer power in the known universe to turn chemistry into a subject where we simulate everything.

That explains it really well, thanks a lot. Well, I wish there's a way to mark all answers as right since all of them were very explanatory.

What you have in mind sounds like Molecular Mechanics. Atoms are treated there like classical particles and are simulated using classical physics, including Coulombic forces. But we know that this is not sufficient. Effects like protons tunneling can not be explained within that model, just to name one.
Including all known laws of physics into simulations of chemistry would lead to simulations that would require thousands of years( and that is probably a massive underestimation of the time required). That is obviously impractical. So we have to find approximations that are on one hand accurate enough to model the current problem of interest and on the other hand fast enough to be done on a human time scale(days, weeks, months, years at most). Methods that are accurate and fast are tailored to solve particular problems and up to this point, there is no approximate method that is well suited to describe all of chemistry in a fast and accurate manner. Indeed, picking the right method for the chemical problem is by itself a non trivial task and current researchers often have to decide which method to pick for the problem that they are investigating. This is a point where researchers from the theoretical field and experimental field collaborate in many cases.
The development of new methods is also an ongoing process, so its understandable that they aren't taught at school. Why teach a method that might be outdated in a few years ? And learning/teaching all the physics and math required to understand the methods is also beyond school. School only teaches you some basic concepts that allow you to rationalize about chemistry on a shallow level.
VSEPR is an approximate model that allows us to predict equilibrium geometries, and like all approximate methods, there are cases where it works well and cases where it fails. The strongest point of it is its simplicity, we don't even need a computer to do it. Most approximate methods are themselves non trivial and require a lot of computational power. The approximations taught at school are typically methods that are doable by hand, which is probably also more satisfying when you are new to chemistry. Imagine you were taught methods that you could only apply with a large computer cluster, which is most likely not available to the common school student.  I guess that most people wouldn't find that helpful at all and would only be frustrated with chemistry.

I understand your point of the need of a large computer cluster to solve a small problem, but I hope we can reach a stage where this could be done on a more compact system in much lesser time. With the advent of quantum computing, I hope we can get to a point where everything can be simulated. From my current understanding and perspective, that would be very helpful in the fields of medicine, to study adverse reactions of chemicals, and a lot of other fields. I hope we reach such a point someday soon:)


text:
Comparing stability of two carbocations
Which of the following ions(A and B respectively) is more stable? How can we explain this on the basis of resonance?

Here's what the solution said:

Structure (A) is a primary carbocation and has no hyperconjugative stability while (B) is a secondary carbocation having 2 alpha Hydrogens which indicated that (B) must be more stable.  But when I looked at the given solution to the problem, I was a bit confused.
I don't get why there is no resonance possible in (B), aren't they basically the different resonance structures of the same intermediate?
I looked for something like this on the web, and it showed similar results saying (B) isn't planar and hence doesn't have resonance while (A) does, so (A) is more stable.
Obviously, if B wasn't planar then it wouldn't have any resonance but why isn't it planar? Is this about the chair conformation of cyclohexane?
I'm not able to understand these points and would appreciate if someone could help me understand this.

according to JEE level, I agree with your thought process, I think hyper-conjugation is the right way to decide, chair conformation is generally not a part of JEE Syllabus, at least, not till this extent. We generally assume it to be planar.

They: I and II; you: A and B. Where is structure II

Beside hyperconjugation, which is certainly at play here, be careful with the words used. At the root you can't evaluate stability of one limiting form by use of resonance. It is somehow a contradiction in terms. Pedantic: in the above, the limiting form is a hybrid itself,if hyperconjugation is involved. Of course practically no need to sketch it. But I think you get what I mean.

@KarstenTheis (A)->(I) , (B)->(II) they basically named the structures A and B in the question but started referring to them as (I) and (II) in the solution :/

No they did not. Carefully read the question and answer key again.

@PrajwalTiwari about A B I & II: is indeed the pedantic point I've referred to! So I wasn't too pedantic

@KarstenTheis Oh yes I'm really sorry, I mistakenly cropped out that part from the question image, I have edited the question and added the details of A and B in the question statement. Really sorry for the confusion


text:
Fundamentally, what is the difference between redox and acid-base reactions?
For example, consider:
$$\ce{NH3 + H2O -> NH4+ + OH-}$$
Here, the $\ce{NH3}$ is acting as a base (By Lewis theory, it is an electron pair donor).
Why don't we consider this to be a redox (there is a transfer of electrons taking place as $\ce{NH3}$ is acting as an electron donor and $\ce{H2O}$ as an electron acceptor)?
I'm aware that calculation of oxidation states is the best way to figure out if a reaction is redox or not. Then I would like to ask: why exactly do we assign oxidation states according to those rules? Is there an intuitive way to explain why a certain reaction is acid-base and not redox?

You can find useful  Notation basics and How can I format math expressions or chem. equations on Chemistry SE.

"Fundamentally", there's no difference.

In contrary to acid-base reactions(1), a redox reaction is such a reaction, which can be formally divided into two half-reactions. At least one reactant in such a half-reaction explicitly formally acts as a donor or acceptor of one or more electrons.
$$\ce{Cl2(aq) + 2 Fe^2+(aq) -> 2 Cl-(aq) + 2 Fe^3+(aq)}$$
can be formally divided to:
$$\ce{Cl2(aq) + 2 e- -> 2 Cl-(aq)}$$
$$\ce{2 Fe^2+(aq) -> 2 e- + 2 Fe^3+(aq)}$$
In the example, there was no electron transfer between $\ce{NH3}$ and $\ce{H2O}$.
Oxidation states are nothing more but valence electron inventory with defined formal rules of electron distribution. Their sum does not change during a reaction, due the charge conservation law.

(1) A very generalized  Usanovich theory of acids and basis does consider redox reactions as a special case of acid-base reactions. But this theory is not much known nor applied within chemistry community.


text:
Function describing pH vs. volume of base used in titration

Sketch a graph of $\mathrm{pH}$ vs. volume when $\pu{50.0 cm^3}$ of $\pu{1 M}$ $\ce{HCl}$ is titrated with $\pu{1 M}$ $x\,\pu{cm^3}$ $(x \in (0,50))$ of $\ce{NH3}.$

As $\ce{HCl}$ fully dissociates, I know that
$$
\mathrm{pH} = \mathrm{p}[\ce{HCl}]. \tag{1}
$$
When $x\,\pu{cm^3}$ of $\ce{NH3}$ is added, $x\,\pu{cm^3}$ of $\ce{HCl}$ reacts. So, I can write the concentration of $\ce{HCl}$ as the function of $x$ this way:
$$
\mathrm{pH} = -\log\frac{c_0(\ce{HCl})\cdot V(\ce{HCl})}{ V_\mathrm{tot}} = -\log\frac{0.1(50-x)}{50+x}. \tag{2}
$$
Is this function correct? I am doing high school chemistry, hence my knowledge in chemistry is quite limited.

pH=-log ( [H+(aq)])  // Consider pH  buffering by pair NH3/NH4+, as NH3 is a weak base.

Good enough for high school IMO (and nice job with formatting, although I corrected it a bit), but @Poutnik's comment is worth considering. Probably you could tweak your question a bit since currently it appears to one of the am-I-right type questions, which are usually getting closed.

related https://chemistry.stackexchange.com/questions/117984/how-to-derive-the-conductivity-titration-curve-which-accounts-for-salt-formation


text:
Size Exclusion Cromatography - How Does Detector Determine Molecular Mass Distribution in Polymers?
In materials I am using it is written that detectors in SE cromatography determine concentration of polymer molecules of particular mass on exit of column by measuring index of refraction.
How is concentration of particular mass in mobile phase on exit of the column connected with molecular mass distribution of polymers since SE cromatography is used to determine such distribution?

It is explained very nicely here https://en.wikipedia.org/wiki/Size-exclusion_chromatography#Analysis

A refractive index detector cannot determine or find the molecular weight of anything, it just measures refractive index changes of the eluent.

Yes, thank you. Wikipedia states that spectroscopy is used to determine concentration of each molecule in the eluent, but that other techniques are used to determine what these are which in terms of polymers means molecular mass and structure. Spectroscopy does quantitative analysis while other techniques do qualitative.

One should have a standard and elaborate based on results. The connection is due to the fact that known stuff has been measured before. Nothing here is a single reading like measuring T with a thermometer....


text:
Alcohol to Ketone by CrO3 in cyclohexane
I have encountered an exercise question on conformation of cyclohexane and there is a statement saying, 'converting alcohol to ketone by CrO3 is much faster when R-OH is in axial position'. I would like to know why (but I have not learnt this reaction) and found this on the internet:

in first step, it says the chromate ester has a higher 1,3-diaxial
interaction, what I interpret is this axial chromate ester
intermediate is more unstable.

what I can see is the entire chromate ester part is pointing away from the ring, which should  reduce the diaxial interaction eclipsing strain. Can I say this is why the first step (formation of chromate ester) is not the r.d.s. of the entire reaction? Or any other reasons?

in the r.d.s (2nd step), it says 'stabilized T.S. as more strain
released'. What is that 'strain' and how to determine 'more strain is released'
than that of equatorial position? my thought: it should not be 1,3-diaxial strain since I can see the entire chromate ester still
in the same axial position while in T.S.


text:
Does distilled water still have ions in it? Does deionized water lack ions, but still have other "stuff" in it?
This isn't a question about the difference between Deionized and Distilled water, my hope is that for once it's actually a chemistry question related to this topic.
Distilled water is created by repeatedly boiling water then re-condensing it so that all of the "stuff" dissolved in it should be, in theory, dropped out of it.
Deionized water, as is my understanding, is created by running water through an electric current to pull various metal ions out.
So (and forgive me if I'm thinking aloud) does distilled water still have ions? I just did some googling to double-check and while a watermolecule is, in theory, "neutral" google returns that there is still a slight positivity (+) to the H sides of the molecule and a slight negativity (-) to the O which results in Ions(-) still being attracted to the Hydrogens(+).
Part of me wants to say that there is some possibility that distilled water could retain some ions through the distillation process despite the bond being fairly weak. However, my brain wants to lean toward that being unlikely, that Distilled water ought to also be, Deionized.
As for deionized water, my gut is that while it ought not have heavy metal ions and salts and the like in it anymore due to the deionization process, plenty of other "stuff" such as organic material and microbes ought to be present.
So... After having done my best to answer my own question...
Is distilled water also deionized? and does deionized water have "stuff" in it?

Any water contains ions due water self-ionization and usually extra ones due dissolved aerial carbon dioxide ( being a weak acid ).
Distilled water can still contain volatiles(like chloroform from water chlorination), dissolved gases/trace vapours from air (problem for trace analysis if the lab works with solvents) and trace ions from glass where it is stored.
It may also contain traces of salts accidentally travelling in water aerosol formed during distillation. (Neither sea salt stays all in sea in spite of not being volatile.)
Deionized and more generally purified water is produced by number of methods and their combination, like

capacitive deionization, reverse osmosis, carbon filtering, microfiltration, ultrafiltration, ultraviolet oxidation, or electrodeionization.

I would include also ion exchange resins, popular in tap-water filters and among equarists.
At the reverse osmosis, ions partially pass due microscopic membrane ruptures or too big holes, originally present or formed during membrane aging. The osmosis is not particularly effective for organic matter separation. Additionally, there can be ongoing microbial activity on device surfaces behind the membrane.
Electromigration obviously does not address neutral molecules and bigger particles.
These methods are often used combined, like reverse osmosis + electrodeionization.

Distilled water will always contain some ions no matter how pure
There is a simple sense in which even the purest distilled or "deionised" water will always have ions in it. Water self ionises to create small amounts of $\ce{H3O+}$ and $\ce{HO-}$ which is unavoidable whatever the purity.
But the situation is more complicated than that in practice because the purity of different types of distilled water depend on how it is made and what purpose it serves. Not all "distilled" or "deionised" water is the same. And how it is stored can make a difference.
All types of distilled water, for example, will pull some gases from the atmosphere. If they absorb carbon dioxide, for example, there will be some carbonic acid in the water which adds ions.
More importantly, the degree of purity will depend on the chemists need of water that doesn't have certain contaminants in it. Normal lab distilled water just has to be free of gross contamination by many of the things often found it tap water like magnesium and calcium carbonates or sodium chloride in some areas. But storage in glass may lead to trace contamination with ions leached from glass (potassium and sodium, for example).
More specialist "pure" water may need special manufacture and handling to avoid tiny traces of other contaminants when only tiny amounts might interfere with a sensitive reaction or process. The electronics industry, for example, often uses water that has to have contaminant levels of many components below 1 part per billion or better because the processes of making chips is exquisitely sensitive to tiny amounts of some compounds. This water is very expensive.
So most "deionised water" will have small amounts of ionic contaminants plus the ions from self-ionisation. But some specialist types of pure water will have very few ions other than those from self-ionisation. But the number will depend on the type of water and even the purest will still have a very very small amount of non-water ions in it.

Distilled water main impurity is due to the dissolution of atmospheric $\ce{CO2}$ which produces some acidity due to the reaction $\ce{CO2 + H2O -> H3O+ + HCO3^-}$. Usually it is sufficient to produce a $\mathrm{pH}$ not far from $5.5$. This value changes from one place to another one.


text:
When carbon-14 decays into nitrogen-14 during β-decay, carbon-14 emits a electron while also receiving a new one. How does it add up?
As far as I understand, when $\ce{^{14}C}$ decays into $\ce{^{14}N}$, a neutron is turned into a electron and a proton, and a electron is emitted. But here is the part I don't understand.
If an electron is emitted (witch I guess means that one electron is sent away?) while we also get a new electron from the neutron, wouldn't that be +1 electron and −1 electron, witch would result in no change from the initial 6 electrons because they cancel each other out? How can it become a $\ce{^{14}N}$ that has 7 electrons when the resulting electron amount is 6?

I am very new to chemistry and I have searched everywhere to find out how this makes sense but I haven't found anything.

Also, a duplicate on Physics.SE: When Beta decay of C-14 happens, why doesn't the Nitrogen atom produced have a positive charge?

The $\ce{^{14}C}$ atom emits an energetic electron that leaves a $\ce{^{14}N+}$ cation.  Later on, the emitted electron will eject some of its own peripheral electrons and also hit other atoms to ionize them, producing more cations and more electrons. After a while, the $\ce{^{14}N+}$ has become a $\ce{^{14}N^{z+}}$ ion surrounded by a lot of secondary electrons and cations. When these secondary electrons have lost their excess of kinetic energy, $z$ electrons will fall on the $\ce{^{14}N^{z+}}$ ion and discharge it. The other electrons will be redistributed on all the hit atoms around.


text:
What are the major products when naphthalene reacts with ethyl chloride in the presence of AlCl₃?
From Arihant Test Drive, JEE Advanced 2020; Ramchaya builiding, Daryanganj, New Delhi, India; January 2019, p. 342:

Napthalene gives 1-ethylnapthalene on reaction with ethyl choride in the presence of $\ce{AlCl_3}$

My teacher told me the intermediate carbocation is more stable when it's adjacent to the phenyl group, that's why ethyl chloride forms 1-ethylnapthalene. I can't understand why the carbocation will be more stable in the first case.

It appears that your teacher is wrong. Both substitute at the 1-position. See this - https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/22%3A_Arenes_Electrophilic_Aromatic_Substitution/22.08%3A_Substitution_Reactions_of_Polynuclear_Aromatic_Hydrocarbons. In chemistry, if you need to learn facts beyond NCERT, always try to use citable resources. Your own notes, or your teacher's notes, or any coaching textbooks, are not citable resources.

Another problem with these questions is the magnificent ambiguity they leave in terms of reaction conditions like temperature and solvent. As in this case, solvent changes the product. For sulphonation, temperature will change the product.

@trc this is not a coaching textbook, but rather a reference book with questions published by Arihant. Can you specify which books are considered unreliable/not citeable so I'll try not to use them in the future.

See this, in particular ref. 1 and comments: https://chemistry.stackexchange.com/questions/125530/solvent-effects-in-friedel-crafts-reaction

Undergraduate-level introductory textbooks like Clayden are the best to use. Obviously use previous year papers to get an idea of what examiners consider correct. Online sources include MasterOrganicChemistry, Chemistry Libre Texts and even this community for the highly-upvoted answers written by high-rep users (which often contain references within the answers). I think these will suffice. Books made specifically for JEE, almost never cite any primary references (i.e. research papers) so you might run into conflicts, where different JEE books say different things, most common examples being--

-- the order of migratory aptitude, which groups are oxidized or reduced by which reagents and a few other points. I have seen various books say widely different things.

TL;DR: The final product will depend on what solvent you are using, what temperature the reaction is being operated and the concentration of the reactants.
Long answer
From the abstract of this paper1:

The $\ce{AlCl3}$ and $\ce{SnCl4}$- catalyzed Friedel Craft alkylation (methylation, ethylation, isopropylation and tert-butylation) of naphthalenes with alkyl halide was studied in nitromethane, carbon disulfide and benzene solution. Alkylation in nitromethane shows least isomerization giving 75-100% α and 25-0% β substitution. Ready isomerization of αalkylnaphthalene under usual Friedel-Crafts conditions leads to substantially increased β-alkylnaphthalene formation. To ascertain the effect of acid-catalyzed isomerization, the $\ce{AlCl3}$ catalyzed isomerization of α and β-alkyl (methyl, ethyl, isopropyl and tert-buty) naphthalenes were studied, including the determination of equilibrium composition of α/β isomer pairs.

Similar thing happens for FC-acetylation2,3:

The Friedel–Crafts acetylation of naphthalene has been shown to give varying α : β isomer ratios as a function of concentration of the reactants. In consequence the ratio is strongly time- and temperature-dependent. Values of the α : β ratio (determined by an extrapolation procedure, for 1,2-dichloroethane solution) for the limit of zero reaction time, were plotted against reactant concentration, to give a straight line having a near-zero intercept. From this it is inferred that, whereas β-acetylation is essentially first-order with respect to the acylating reagent, α-acetylation is dominantly second-order in acylating reagent.

The Friedel–Crafts acetylation of naphthalene, using acetyl chloride and aluminium chloride in 1,2-dichloroethane solution, has been studied kinetically. The α/β isomer ratio changes as a function of concentration of reactants, and time, e.g. from an initial 4–5 to a final 0.7.

(full image here)
Notes and References

Aromatic substitution. XXXVII. Stannic and aluminum chloride catalyzed Friedel-Crafts alkylation of naphthalene with alkyl halides. Differentiation of kinetically and thermodynamically controlled product compositions, and the isomerization of alkylnaphthalenes, George A. Olah and Judith A. Olah
Journal of the American Chemical Society 1976 98 (7), 1839-1842
DOI: 10.1021/ja00423a032
Kamounah, F. S., Andreou, A. D., Bulbulian, R. V., Gore, P. H., Miri, A. Y., & Waters, D. N. (1981). The Friedel-Crafts acetylation of naphthalene. Evidence for concurrent second- and third-order reactions. Journal of the Chemical Society. Perkin Transactions 2 (2001), 2, 376-378. [55]. https://doi.org/10.1039/P29810000376
The Friedel–Crafts acetylation of naphthalene in 1,2-dichloroethane solution. Kinetics and mechanism, David Dowdy, Peter H. Gore and David N. Waters, J. Chem. Soc., Perkin Trans. 2, 1991, 1149-1159 , DOI: 10.1039/P29910001149
The acylation of naphthalene by the friedel–crafts reaction, J. Chem. Soc., 1949, S99-S103, DOI: 10.1039/JR9490000S99 (acetylation of naphthalene using methylene and ethylene chloride. In absence of solvents, α-substitution occurred and in presence of solvents at higher temperture, β-substitution occurred)
FRIEDEL-CRAFTS ACETYLATION OF 1- AND 2-METHYLNAPHTHALENE By HUBERT EMMANUEL, 1978 (link)


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Why do a stronger acid lose a proton easily when the electromagnetic force in a polar compound should keep it together?
In acid-base theory I've just learned that the stronger the electronegativity, the stronger the acid will be at least it will have an impact on the strength of the acid.
If that so, then the acid should NOT lose a proton due to the electronegativity difference which makes the compound (acid) polar. I mean, shouldn't the electromangentic attraction between the positive side and the negative side ends of the compound have a greater force to keep it together rather than separate it into a conjugate base and a proton?
It does not make sense.

Acids not so much lose protons as donate them - it's the base that makes them act as acids.

I would suggest to please pay attention to correct spelling ('electromagnetic', 'compound', 'separate'), as well as capitalisation, when writing your question. Also, there's no reason to use a bullet point. You can [edit] your question to fix these.

Chemical bonds aren't just electrostatic attractions between some cation $\ce{H+}$ and anion $\ce{X-}$. Well, purely ionic bonds are; but if you don't have a purely ionic bond, then there's some covalent part that also matters, which is the part that you're neglecting. You can't treat a covalent bond as if it were purely ionic—which is what you're implicitly doing, by only talking about electrostatic attractions. By that logic, any bond between two of the same element must cost zero energy to break, since there is no electronegativity difference and hence no ionic component.
Coming back to the issue of acidity: one way of looking at it is that if you have a purely ionic bond and want to break it, then all you need to do is to break the electrostatic attraction. On the other hand, if you have a covalent bond (or something in the middle) and want to break that, then you need to do two steps:

Convert the covalent bond into a purely ionic bond by shifting electron density from the middle of the bond to the part that will become the anion.

Break the ionic bond (i.e. electrostatic attraction).

For a purely ionic bond you only need to do step (2), which is less energy overall, hence easier to lose the proton, hence stronger acid.
You might complain that I've assumed that step 1 must cost energy. Is it possible that step 1 could actually be energetically favourable? Well, almost by definition, no. If it were more stable to form an ionic bond than a covalent bond, then that's what the atoms would do: they would form an ionic bond. Since they form a covalent bond, we must accept that that is a more stable state; and thus it would cost us energy to convert that into an ionic bond.
This analysis might seem simplistic, and it kind of is. But actually, it is not that far off from a sensible thermodynamic cycle, albeit in reverse order: first we break the $\ce{H-X}$ bond homolytically (bond dissociation energy), and then we shift an electron from $\ce{H}$ to $\ce{X}$, i.e. go from $\ce{H + X}$ to $\ce{H+ + X-}$ (ionisation energy + electron affinity).


text:
What would happen if electrons were spin-1?
We were speaking about this in class, but I can't understand it quite well. What would happen if (hypothetically of course) the allowed $m_s$ values were $-1$, $0$ and $1$? What impact does this have on our current Periodic Table? What impact does it have on the other quantum numbers?
I don't see any relation between $m_s$ and the other quantum numbers... which is why I'm asking so I can really understand it.

"I don't see any relation between m_s and the other quantum numbers" You're right; there isn't any. This is an unfortunate choice of spin, though: the Pauli exclusion principle, which guarantees that no two electrons have the same quantum numbers, doesn't work for particles that have integer-valued spins. Google bosons, fermions, and the spin–statistics theorem for more info. Presumably, the question (incorrectly) assumes that the Pauli exclusion principle will still hold, and therefore the 1s orbital can hold up to three electrons, etc., making the Periodic Table look rather different.

...while in fact this would make the 1s orbital capable of holding all them electrons, which in turn would make the Periodic Table look yet more strikingly different, and in particular, not periodic at all.

@Orthocresol so just to reiterate - if the Pauli exclusion principle were to be hypothetically upheld with this change, then the other quantum numbers, subshell letters, etc. and their values would still remain unchanged? The only evident change would be the periodic table being shifted weirdly in order to fit 3, 9, or 13 electrons in each orbital? I'm not sure if the 9 and 13 are correct but I don't understand how I would calculate how many electrons could fit per orbital with this change in m_s values.

@Ivan Neretin I don't understand how the 1s orbital would be able to hold all electrons because of this change... could you explain that?

This is because the Pauli exclusion principle would not hold with bosons.

@diana it has to do with quantum mechanics and is better explained in physics SE than here. Put very briefly, if identical particles have half-integer spins then they come with a symmetry that makes them interfere destructively and cancel out if two of them are in the same state. But integer spin comes with the opposite symmetry and constructive interference, so the identical particles are "attracted" to the same state instead. See also https://en.wikipedia.org/wiki/Pauli_exclusion_principle.

If electrons had the integer spin, there would be nobody asking what if they had spin 1/2. They would all like to sit in 1s orbital, until repulsion would force some to take higher orbitals and grouping would repeat. There would be no bounds as we know, but very different, if any.

Perhaps the question was formulated so that the answer is "there won't be periodic table". There won't be elements.

@diana the Pauli exclusion principle is a direct consequence of electrons having non-integer spin. It is not separable from that idea and you cannot reintroduce it.

The simplest answer is that, if electrons had spin 1, there would be no chemistry and no universe.

There are several issues here which should be carefully disentangled.

Is $m_s$ independent of the other quantum numbers $n, l, m_l$ or do they affect one another?

It is independent. The reason is because $n, l, m_l$ are quantum numbers which deal with the spatial coordinates of the electron, i.e. its position and momentum in 3D space as given by (for example) $(x, y, z)$-coordinates. On the other hand, spin is a separate property, often called an "intrinsic angular momentum" which cannot be visualised in 3D space.*
In nonrelativistic quantum mechanics, the Schrödinger equation actually only describes the spatial part (i.e. $n, l, m_l$); the spin part is just tacked on to that at the end. The spin and spatial parts of a wavefunction are completely separate from one another and don't directly interact. (Indirectly they still do, via things like the Pauli exclusion principle.)

Can spin-1 electrons really support chemistry?

As mentioned in the comments, spin-1 particles (and more generally, particles with integer spin, i.e. bosons) do not obey the Pauli exclusion principle. In the context of an atom, that means that two "spin-1 electrons" can share the same four quantum numbers $(n, l, m_l, m_s)$.†
However, the Periodic Table, and chemistry generally, crucially depends on the Pauli exclusion principle. Consider the case of lithium: we say it has an electron configuration of $\mathrm{1s^2 ~2s^1}$. Why does the third electron go into the $\mathrm{2s}$ orbital? Well, it's because the first two went into the $\mathrm{1s}$ orbital, and thus have quantum numbers $(1, 0, 0, +1/2)$ and $(1, 0, 0, -1/2)$. Since the third electron can't share all four quantum numbers with either of its two predecessors, it's forced into the next-lowest orbital, i.e. $(2, 0, 0, \pm 1/2)$ (the sign is irrelevant).
With spin-1 electrons there is no such constraint, so every spin-1 electron would simply sit in its lowest possible energy state, namely the $\mathrm{1s}$ orbital.‡ The true electronic configuration of the $n$-th element would then simply be $\mathrm{1s}^n$. So, chemistry as we know it would not exist.

If the Pauli exclusion principle still held for bosonic spin-1 electrons, what would happen?

This is the easy part: each orbital would just hold three electrons instead of two. So the 1s orbital would hold up to 3 electrons. The 2p subshell would hold 9 electrons (3 orbitals of 3 electrons each). The 3d subshell would hold 15 electrons.
Consider the 2–8–8–18... pattern that we see in the (normal, spin-1/2) Periodic Table:

2 comes from the 1s orbital (2).
8 comes from 2s + 2p (2 + 6).
8 comes from 3s + 3p (2 + 6).
18 comes from 4s + 3d + 4p (2 + 10 + 6).

In hypothetical spin-1-electron land with hypothetical Pauli exclusion principle, this would be 3–12–12–27:

3 comes from 1s orbital (3).
12 comes from 2s + 2p (3 + 9).
12 comes from 3s + 3p (3 + 9).
27 comes from 4s + 3d + 4p (3 + 15 + 9).

It's not hard to see that these numbers are just the previous numbers multiplied by $3/2$.

Footnotes
* You may have seen oversimplified depictions of spin as the electron rotating about its own axis, like how the Earth rotates to cause the day/night cycle. These depictions are all incorrect and physically unreasonable.
† Note that the Pauli exclusion principle refers to particles that are identical, or indistinguishable. Two electrons on the same atom are indistinguishable, and thus cannot share the same four quantum numbers. Conversely, if we were talking about two separate hydrogen atoms (which are distinguishable), then it is perfectly OK for each of the electrons to have the same four quantum numbers.
‡ In fact, this is not entirely true. If you raise the temperature high enough, then some of the electrons will start to occupy higher-energy states. This concept is similar to a Boltzmann distribution, but is more properly defined by Bose–Einstein statistics. However, the energy gaps between successive orbitals are very large, so at "realistic" temperatures this is not likely to happen.

The problem with theorising about the Pauli exclusion principle in a world with spin 1 electrons is that the principle itself is intimately connected to the spin. In fact it is a direct consequence of non-integer spin and has no independent existence. So hypothesising about it independently makes no sense.

@matt_black Hmm, I don’t disagree, but the question is what it is. You’re welcome to edit my answer to strengthen the wording (if you want to, of course!), or leave it as a comment.

"Realistic temperatures": Strictly speaking, the difference between $n=1$ and $n=2$ in hydrogen divided by Boltzmann's Constant is about 118,000 K.


text:
Identifying the solubility of Polyvinyl Alcohol (PVA) in water versus temperature
Background:
Poly Vinyl Alchohol (PVA) is a plastic used for a sacrificial support material for 3D printing and sewing. This material is used specifically because it is water soluble, and can be dissolved by placing the PVA in water. As dissolving the PVA is basically a processing step, I'm interested in optimizing this step. For me, optimizing this process consists of:

Minimizing the amount of time that the process takes
Minimizing the cost of processing (e.g. water, heating)
Minimizing the waste material (e.g. saturated water)

From my limited experimentation, I know that increasing the temperature of the water increases the speed that the PVA dissolves. I also know that increasing the water volume (either through water changes or just a larger processing tub) also reduces processing time. From my limited understanding, this isn't really that surprising - many chemicals have increased solubility with temperature, and increasing the volume of solvent decreases the concentration of the PVA.
Question:
What is the solubility of PVA in water vs. temperature? Is there a source for a graph that ties the speed of dissolving PVA vs. temperature and PVA concentration? (How could I find this information, if that's better to answer)
(Some) Limits of my knowledge:

I don't know if all PVA is created equal.
I don't really know any organic chemistry.

According to Tager et al.[1]:

All the material described here is thus evidence of the complexity of the PVA-water system, and of the fact that the temperature coefficient of PVA solubility in water depends on the studied temperature and concentration ranges. [...] At medium concen- trations and fairly low temperatures, heating will improve the PVA solubility in water, but it will become poorer again at higher temperatures.

For more information, you can refer to the commentary they made on their Figure 4. Beware though, I read the article quickly and it seems that what they termed "solubility" is not the actual solubility $-$ in $g.L^{-1}$ for instance $-$ but is more like the thermodynamic affinity of the two compounds (water and PVA). I may have misinterpreted something though...
For a clearer view, you may also have a look at Finch[2]:

As you can see, the solubility strongly depends on the degree of polymerization of the PVA. Thus, depending on this parameter in your particular PVA, it may or may not be necessary to heat the water to have a better solubility...

Anyway, in your view of optimizing the process, the final decision should essentially depends on which factor(s) is(are) crucial for you: energy (for heating), water, or time.
Indeed, although the overall solubility ($g.L^{-1}$) of PVA in water can decrease with temperature in certain cases, if the kinetic of the reactions if favoured by an elevated temperature, you may want to heat your water anyway. Please note that solubility $-$ i.e. the amount of substance that you can dissolve in a given volume of solvent (in $g.L^{-1}$ or $mol.L^{-1}$) $-$ and rate of dissolution $-$ i.e. the speed at which a given compound dissolves (in $g.s^{-1}$ or $mol.s^{-1}$) $-$ are two different and possibly unrelated things!
Let say time is crucial for you, you would want to use a lot of heated water and replace it often, so as to "flush" your PVA with hot, 0%PVA water. The dissolution will probably be extremely quick but you will waste much water and energy.
At the opposite, if you are short on water and energy but have plenty of time. You may want to keep the amount of cold water minimal so that at the end of the dissolution, the water is just below the its saturation point with respect to PVA concentration.

Tager et al. The effect of temperature on the water solubility of polyvinyl alcohol, Polymer Science U.S.S.R.
Volume 13, Issue 3 pp751-758, 1971, DOI: 10.1016/0032-3950(71)90042-6
C. A. Finch, Some Properties of Polyvinyl Alcohol and their Possible Applications, Chemistry and Technology of Water-Soluble Polymers pp 287-306, 1983, DOI: 10.1007/978-1-4757-9661-2_17

This is tremendously useful foothold into the subject matter. I couldn't get a copy of [1], but I am reading through [2]. I think you may be right that I'm also interested in the rate of diffusion of PVA.

@W5VO I apologize for I mistakenly wrote "rate of diffusion" instead of "rate of dissolution" (corrected in my answer): https://en.wikipedia.org/wiki/Solubility#Rate_of_dissolution As for [1], I am afraid that the community is against circumventing paywals, so I can absolutely not put you a direct link to Sci-Hub, nor can I strongly advise you to use it so as to quickly find research items using their DOI : https://chemistry.meta.stackexchange.com/questions/3634/what-are-the-site-policies-surrounding-paywall-circumvention

@mrnvick: Very good answer: +1.

@W5VO: If you contact the manufacturer of the PVAl (rather than PVAc), you may be able to get a brochure listing all their products: mw variations, % hydrolyzed variations, and various solubility and mixture effects. This may not give you the exact optimal procedure, but may be a helpful reference and elucidate the molecular interactions going on. It could also lead you to a better PVAl. (BTW, I've also seen PVOH as an abbreviation, to distinguish from PVA = polyvinyl acetate.)

I'm sorry, I didn't mean to imply I wanted you to get me access to that article, only that I don't have an easy way of reading it.  I'm tremendously grateful for any real information on the subject because my efforts turned up nothing. I think in my naivety, I was hoping it was going to be as simple as NaCl in water... and it clearly isn't that.

@JamesGaidis Part of the problem is that I'm purchasing the PVAl as water soluble "fabric" or "filament", and I don't have information about the PVAl source, molecular weight or percent hydrolyzed (wouldn't surprise me if it's a trade secret). Looking for materials information from manufacturers of PVAl is a good tip, I'll see if that turns anything up.


text:
What exactly is a photon gas?
I came upon this question in a JEE Advanced mock test online:

The number of photons of wavelength $\lambda$ required to achieve pressure $P$ in an empty cubical box of edge length $l$ is given by $\displaystyle \frac{Np\lambda l^3}{hc}$. Find $N.$

In the solution, they have assume that the photon particles behave liked a ideal gas and used the equation $\displaystyle PV=\frac{NMV_\mathrm{rms}^2}{3}$. After that they substituted $MV_\mathrm{rms}(V_\mathrm{rms})$ with $\displaystyle\frac{hc}{\lambda}$.
I tried my best and couldn't find the exact source of this question, but I did find this about photon gasses from Franz Schwabl. Chapter 4 Statistical Mechanics, Edition 2; Springer-Verlag Berlin Heidelberg; pp 197-199:

The value of the mean collision time (of photons) is approximately $τ ≈ \pu{10^{31} s}$ at room temperature and $τ ≈ \pu{10^{18} s}$ at the temperature of the Sun’s interior $(\pu{107 K}).$ Even at the temperature in the center of the Sun, the interaction of the photons is negligible. Photons do indeed constitute an ideal quantum gas.
The number of photons is not conserved. Photons are emitted and absorbed by the material of the cavity walls. From the quantum-field description of photons it follows that each wavenumber and polarization direction corresponds to a harmonic oscillator.

Now I have many questions regarding photon gasses. How did the solution assume the number of molecules to be constant if photons can be emitted or absorbed by cavity walls? Why did they substitute $MV_\mathrm{rms}$ with $h/\lambda$ when the de Broglie equation $p=h/\lambda$ is only applicable to massive particles? And is the question and/or the solution even correct in the first place?

Go to Google Scholar, there is a readible article "Teaching the photon gas in introductory physics", by Harvey S.  Leff in the American Journal of Physics. It should be accessible. Direct access is behind paywall.

@M.Farooq https://www.cpp.edu/~hsleff/PhotonGasAJP.pdf I found this free pdf, is it essentially the same thing or do you recommend getting it from American journal of physics

Yes, it is the same article.

@poutnik I think you forgot to add $$ in front of your mathjax equation

De Broglie equation is applicable on particles with momentum, which photons possess. The full form of the famour $E=mc^2$ is $E=\sqrt{{(mc^2)}^2 + {(pc)}^2}$, so for $m=0$, $E=pc \implies p=E/c=hf/c=h/\lambda$.

How did the solution assume the number of molecules to be constant if photons can be emitted or absorbed by cavity walls? At the thermal equilibrium, the rate of photon emission equals the rate of the photon absorption, so the photon count is constant, with existing, but negligible fluctuations.

@Poutnik What does this statement mean then "the number of photons is not conserved". Aren't the two statements contradictory?

If it were conserved, photons could not be either emitted(=created for the photon case) either absorbed(=destroyed). OTOH, the electrons count ( more exactly electron lepton number) is conserved, so electrons cannot be created nor destroyed ( unless a weak force involves electron neutrino to keep the lepton number ).

You might find this answer also helpful: https://astronomy.stackexchange.com/questions/36021/why-do-pair-production-and-photodisintegration-have-such-different-effects-in-co


text:
Hydrogenation of pent-4-en-2-one
If one equivalent of $\ce{H2}/\ce{Pt}$ is made to react with one equivalent of pent-4-en-2-one, what will be the product formed?

Generally the double bond goes, but it depends on the catalyst system used.

Is there a reason to say so?

For your information, actually C=C is stronger than C=O bond with bond strengths 680 & 500 KJ/mol respectively.

@Neon Prince I have edited the question. Thanks

You have not specified the full reaction conditions - temperature and pressure can make a considerable difference - however this quote from Chemistry Libre Texts  here (in turn quoting Basic Principles of Organic Chemistry by Roberts & Caserio) gives us a clear indication of the relative ease of hydrogenation of ketones vs double bonds.

Hydrogenation of aldehyde and ketone carbonyl groups is much slower than of carbon-carbon double bonds so more strenuous conditions are required. This is not surprising, because hydrogenation of carbonyl groups is calculated to be less exothermic than that of carbon-carbon double bonds....It follows that it is generally difficult to reduce a carbonyl group in the presence of a carbon-carbon double bond by hydrogenation without also saturating the double bond. Other reducing agents are more selective:


text:
Is dehydration of cobalt(II) chloride hexahydrate dynamic equilibrium?
I know that the dehydration of cobalt(II) chloride hexahydrate is a reversible reaction, is it a dynamic equilibrium?
$$
\begin{align}
\ce{CoCl2·6H2O &<=> CoCl2·2H2O + 4 H2O}\tag{R1}\\
\ce{CoCl2·2H2O &<=> CoCl2 + 2 H2O}\tag{R2}
\end{align}
$$
Some sources say that it is, but others say that it isn't because the forward and reverse reactions don't occur spontaneously. If it is not a dynamic equilibrium, is it considered a chemical equilibrium? Would be great if someone could please clarify.

I assume you mean dehydration as a state,  not a process. For given T and partial water vapor pressure, there is an equilibrium state of a particular hydration degree. Higher T or lower vapor pressure lead to higher dehydration and vice verse.

Out of curiosity, is there a particular reason you are asking about this very compound? Otherwise I suppose the question could be extended to pretty much all inorganic crystallohydrates.

This was just a common example I saw in my textbooks, but I assume that you should be right to say that this question could apply to other inorganic crystallohydrates.

This compound is used to show a chemical equilibrium in classroom. Dissolving $\ce{CoCl2·6H2O}$ in ethanol produces a blue solution, due to  $$\ce{2 [Co(H2O)6]Cl2 -> [Co(H2O)6]^{2+} + [CoCl4]^{2+} + 6 H2O}$$  The hydrated ion $\ce{Co(H2O)6]^{2+}}$ is pale red. As the complex $\ce{[CoCl4]^{2-}}$ has a dark blue color,  the ethanol solution of $\ce{CoCl2·6H2O}$ is blue. But adding a few drops water, the solution turns red, due to  $$\ce{[CoCl4]^{2-} + 6 H2O <=> [Co(H2O)6]^{2+}  +  4 Cl-}$$. Now heating the solution turns it blue again, by reversing the last equation. And red again by cooling.

And CoCl2 on blotting paper is a classical humidity indicator in air, showing the equilibrium does shift with a change in environment.


text:
DEG monobutyl ether pH change
We are using DEG monobutyl ether (CAS 112-34-5) in our production. Normally it is pH neutral, however suddenly we noticed that it has shifted to being acidic (pH=3.4).
What has changed in the material? I have not been able to identify any decomposition that might have taken place, so the question is if the material is still usable?
pH changes can be caused by minimal changes in the material, however I am out of office and not really able to verify it right now.

That's a pretty large change. Are you confident in your pH measuring system?

Yes, 100%, confirmed via calibrated digital meter several times + pH strips....

Ethers are known to pick up oxygen and become peroxides. Light and access to air accentuate the process. The presence of transition metals can catalyze various oxidations. The unetherified hydroxylic carbon  probably goes to an acid. Whether it's usable depends on your process. Perhaps you could run it thru a bed of limestone to adsorb out the acid.

Thank you for pointing that out @JamesGaidis, we changed the container we store it in recently, the new container is not 100% airtight, so that's the answer....

@Michal: an extensive reference (10 pp) is at http://ccc.chem.pitt.edu/wipf/Web/16340.pdf   But it is scary! But if you keep the air out, you should be OK. Maybe you can add an inhibitor - or maybe your process actually uses the peroxide in some way...


text:
Enthalpy/Temperature change in a chemical reaction
I was wondering how exactly absolute temperature change in a chemical reaction could be calculated.
For example, consider this reaction:
3H₂+N₂→2NH₃, ΔH=-46kJ/mol (298.15K)
Assuming the given reaction goes to 100% completion (which it doesn't, it's the Fritz-Haber reaction, but this is a theroretical question), and is conducted at 298.15K and 1 atm of pressure, so that the enthalpy change is -46kJ/mol, how can we calculate the temperature change?
My thought was that we divide the reaction into two steps of thought; first we let three moles of hydrogen (at 298.15K) and one mole of nitrogen (also at 298.15K) react and form two moles of ammonia gas. The temperature stays at 298.15K, so the two moles of ammonia also are at 298.15K. (Is that correct?)
Then, in a second step (assuming we have a closed container and the released heat is again absorbed by the products) we just add 2 times 46kJ to the products,(correction: 1 times 46kJ) so this problem reduces to a simple heat capacity problem. So Cp(NH3)=35.06kJ/molK roughly, so the temperature of the 2 moles of NH3 should increase from 298.15K to 299.46K.
As I explained, I am absolutely unsure if we can even calculate it like that, so any help would really be appreciated, maybe other examples and sources, too.

This is correct if the reaction is carried out at constant pressure, except that, in step 2, you only add 1*46 kJ to the products.  And also, that's 46 kJ, not 46 J.

Ah thank you. That is because it is 46 kJ per mole of reaction right? Not mole of products

Yes, that is correct.


text:
Can an elementary reaction have zero order?
I am very confused right now about my approach to this question could somebody help me out with this. I reasoned my answer as given below
Let's assume if a complex reaction is of zero order. It implies its R.D.S. has order 0 ,but R.D.S. is an elementary reaction.
So, that particular elementary step of reaction is 0 order.(for the obvious reasons)
So if we take that particular (elementary) step as individual reaction
then it will be elementary and will have 0 order.
R.D.S. here implies Rate determining step
But many people whom I shared my doubt with, explained it on basis of Molecularity = Order Rule for elementary step,but it does not sound logical to me and also bit confusing as if we see pseudo first order reaction there molecularity is 2 but order is 1.

The concept of rate determining step is outdated, but unfortunately still used. However, if one would use its implications and the order of the RDS is in fact zero, all steps must be of order zero since the RDS is the slowest. Physically, this doesn't make sense.

Photolytic reactions $\ce{A ->[light]B}$ have typically molecularity 1 and order 0.

Aren't they first order in photons, the forgotten reagent? What happens at very low pressures? I encountered one zero order reaction that I attributed to product catalysis but never figured out the details.

The combustion of wax in a burning candle is zeroth-order. Has it anything to do with a rate determining step ?

The rate of burning is related to the diffusion of wax in the wick definitely rate determining. It is similar to turning the gas valve on your stove.

As far as I understand, the assumption that the RDS has a zero-order is false. The RDS has its order equal to its respective molecularity, however, in zero-order reactions, the rate-determining step will not have the initial reactants as parts of the elementary step. Thus, when you calculate the rate constant in terms of the initial reactants using the other steps in the reaction, you will get their orders to be zero. Another thing, zero-order reactions are ALWAYS complex reactions. There are no elementary zero-order reactions.

I presume RDS refers to rate-determining step. What does ALWAYS refer to, then?

When a zero order rate is observed it means that the reaction is not a single step reaction and either two mechanisms are involved, or a reactant involved in the rate determining step is being throttled, or possibly there is product catalysis. Such an event invites further investigation. Perhaps someone can tell me what the rate determining step has been replaced by?