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Dissolution of ionic compounds
Where can I find a table that lists the rates of dissolution of common ionic compounds like sodium chloride, potassium permanganate, and ammonium nitrate? Are there tables and equations that relate these rates with the temperature and concentration of the bulk solution?

Not sure that there is such a table, because the rate of dissolution depends so greatly on the form of the substance (massive, large crystals, nanoparticles), the state of hydration (anhydrous copper sulfate generally dissolves more slowly than the pentahydrate, if memory serves me right), agitation, etc.

Thanks for responding. I would assume temperature should have a massive effect though. In a way, I’ve always thought of temperature as a form of agitation. In that at high temperatures the kinetic energy from the motion of the solvent particles perturbs the diffusion layer enough to reduce its concentration and in turn lead to an increase in the rate of dissolution.

There would be too many nonlinear unknowns to have such a table  You would be lucky finding a dissolution study for a particular substance.

No such table exists, because the rate of dissolution depends on the importance of the interface. I just would like to mention a case where this dissolution of $\ce{NaCl}$ is instantaneous. If you inject a dilute salty solution of usual soap in the base of a fuel burner, the flame will produce some clouds of amorphous $\ce{NaCl}$. If this smoke is collected, it is made of uncrystallized sodium chloride. Well, dipped into water, this thin $\ce{NaCl}$ powder is immediately dissolved, as Prof. Wendelin Stark, from ETHZ Zurich, has shown a couple of years ago.


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Finding the percent of light absorbed by a sample
I have an absorbance of $0.266$ and need to find the percent of light absorbed by the sample. Is it found by just turning it back into transmittence% and subtracting 1? So by that, would it be $(1 - 0.542)$ making the percent absorbed $45.8\%$?

Your calculation is correct. The percent absorbed is $45.8$%.

I wonder why have you needed confirmation?

You need Beer-Lambert law
$$\frac{I}{I_0}=10^{-aLC}$$
where $I_0$ is initial intensity $I$ that transmitted by a sample of concentration $C$ with extinction coefficient $a$ in a cell of path length $L$. The ratio $I/I_0$ is called the transmittance and $\log_{10}(I_0/I)$ is called the absorbance or optical density (OD) which is  $aLC$.
As a rule of thumb an OD of $1$ corresponds to $10^{-1} \times 100 = 10\%$ transmittance, and OD of $2$ to $10^{-2} \times 100 = 1\%$ transmittance, etc.
(This law applies at each wavelength used, and assumes that the wavelength spread is tiny compared to the natural shape of a line in the spectrum)
The extinction coefficient is a measure of how strongly a molecule absorbs at a given wavelength and usually has units of $\pu{dm^3 mol-1 cm-1}$ if the cell length is in $\pu{cm}$ and concentration in $\pu{mol dm-3}$.

I made some edits to clear it for OP. That way, the percent transmission is clearly to be $10^{-0.266} \times 100 = 54.2%$.

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Preventing water from freezing (-20 °C) in a vinyl container
What is a safe anti-freezing liquid to use to dilute water, to prevent it from freezing at -20°C, without damaging the vinyl container it's stored in?
This container is a heavy bag, for punching. So the purpose would be to be repeatedly hit with force.
The bag can hold a total of 55 liters of water (by specs).

There is missing the essential context information, what is the purpose of this water in the first place. As suitable freezers depend on that.

I was trying to find a better way to say it, but really "punching". It's a punching bag. Before marking this as off topic, I asked already on the martial art stackexchange about this, but what every person did to help was search in external websites if a certain compound would eat vinyl. This stackexchange got the best (objectively, the ethanol will do exactly what was asked) answer by far.

If vinyl means PVC, there is danger to become brittle at low temperatures (statically or under impact) AFAIK PVC becomes brittle below 0 °C.

If if is polyethylene (perhaps there is a label like these on the bag), than the addition of ethanol suggested and the cold should not be a reason to worry too much.  (I speculate, the small bags for the freezer are made of PE.)

Seems like propylene glycol would be much less expensive and safer than denatured ethanol. Even 60 US proof vodka, as per Maurice’s answer, would be fairly expensive.

@Buttonwood I couldn't find any of the label, so I have to go by what customer support said, which is indeed "vinyl". It's essentially a buoy though, so it must be the same material of the buoys used for boats.

@EdV how much propylene glycol? still 30%?

It is sold as relatively non-toxic anti-freeze, so either the label of the container, or a web search, should reveal the temperature versus concentration information. Maybe straight from the container.

The mixture $30$% w/w ethanol + $70$% w/w water freezes at -$20$°C. It does not damage vinyl containers.

@EdV Would not even pass as a bourbon, either.

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Regioselectivity in the opening of 3-membered heterocycles
(a) Consider the following reactions :

For these two reactions, the nucleophile attacks the less substituted carbon as it is less sterically hindered.
(b) Now consider the following reactions :

For these three reactions, attack occurs on the more substituted carbon. I think the explanation for this is that in the transition state, the carbon undergoing attack has a slight cationic character, and can be somewhat stabilized by hyperconjugation, hence lowering the activation barrier for the reaction.
So my question is this : Among steric hindrance (a) and electronic considerations (b), which is the more dominant factor when deciding which carbon to attack? Is it related to the nucleophile used? Is it related to the nature of the heterocycle? Or is the ring-opening and nucleophilic attack non-concerted in some cases? Is it related to hard soft bases? Thanks for any answers in advance.

As evident by OP's given examples, the more dominant factor here is the strength of the  nucleophile and the reaction conditions. For example, in $\bf{a}$, the  nucleophiles are strong and reactions are under basic conditions (both Grignard reagents and acetylenides are basic). Thus, reaction mechanism is $\mathrm{S_N2}$ type so that  nucleophile is attacking less hindreard carbon (recall the order of the $\mathrm{S_N2}$ is $\ce{CH3} \gt 1^\circ \gt 2^\circ \gt\gt 3^\circ$).
On the other hand, in $\bf{b}$, the  nucleophiles are very weak and reactions are under acidic conditions (reaction should be acid-catalysed with weak nucleophiles; Ref.1 and 2). Thus, reaction mechanism is $\mathrm{S_N1}$ type so that  nucleophile is attacking more hindreard carbon, wgich stabilized the partial positive charge (recall the order of the $\mathrm{S_N1}$ is $3^\circ \gt 2^\circ \gt\gt 1^\circ $ and noreaction with $\ce{CH3}$). The first two examples of $\bf{b}$ are not epoxides, but same principle applies:

References:

David Runciman Boyd and Ernest Robert Marle, "CXCVI.—The velocities of combination of sodium derivatives of phenols with olefine oxides," J. Chem. Soc., Trans. 1914, 105, 2117-2139 (DOI: https://doi.org/10.1039/CT9140502117).
For further information, read this, this, this, and this.


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Why do we clarify water with alum, but beer/cider/wine with bentonite?
My understanding of clarifying (floccing) liquids is that we add a mineral with an electric charge, which attracts suspended particles (like bacteria) which I assume have the opposite charge.  The particles clump together, and then the rate at which it sinks increases due to geometric reasons (the Stokes' drag on a small sphere increases proportional to its radius, but its weight increases as the cube of its radius).
My question is why a drinking-water treatment plant would use alum (aluminum sulfate) while people making alcoholic beverages use bentonite (aluminum phyllosilicate).

I think this is surprisingly broad. Not for the punctual question itself but for the many simplification you took, for the mechanisms of action you don't know, for the fact that flocculation removes many and different thing in the two cases and it can be done with other methods as well. Moreover the properties of water and other drinks are different in themselves and by regulation limits.

Waste water, if not too polluted, is treated in wastewater treatment plants by adding aluminum salts like aluminum sulfate $\ce{Al2(SO4)3}$ in order to remove bacteria. But this solution is acidic, because aluminum sulfate produces aluminum ions that are partly hydrolyzed according to the equation $$\ce{Al^{3+} + H2O -> [Al(OH)]^{2+} + H+}$$ This acidity (plus the sulfate ions) does not disturb the wastewater treatment plant. But it would disturb drinkers if this aluminum sulfate would be added to their beverages.
Bentonite is a special porous clay that adsorbs impurities in its pores. So if you add a sort of mud made of bentonite plus water on top of a barrel of beer, it will not be dissolved. It will not bring anything to the beer. It will slowly fall down the barrel, absorbing bacteria and unwanted stuffs during its way. After one week, it has fallen down the barrel and the beverage can be recovered and drunk safely. Its quality has improved.

Okay, so the alum dissolves into ions, and the bentonite just breaks up into tiny solid particles.  Does the mechanism by which they clump particles together differ, though?  If I google "bentonite wine", I see many links talking about its "negative electrostatic charge" which is the same explanation I see for alum (though I've seen them described as having both positive and negative charges in different references).

I edited the question to specify that I was talking about drinking water, not wastewater.  Alum is commonly used to clarify drinking water.


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How to know the number of valence electrons of f-block elements?
I kinda know how to get the valence electrons for s/p/d- blocks but i don't know about the f-block.
Wikipedia says that valence electrons for f-block is ns + (n-1)d + (n-2)f, so is it true? If so can you give examples?
This is what Wikipedia shows :

And another question : what's the electronic configuration of palladium Pd ions? If i apply what i know i'll get that
$P d^{2+}$ is [Kr] 4d^8 , so is it right?

Take some of a particular lanthanoid and just count.

how? can you give example? should i count the f sub-level with them(like d block transition elements) or just the highest n level?

Example of choosing an element or example of counting? You already have the formula. Keep handy the periodic table.

so the formula is true???

The answers you find yourself are more valuable than those you were told. As these answers teach you the process of finding them. But they will teach you that only if you ask after you have thoroughly tried and failed.

Please, provide context. Lanthanides and actinides are elements for which heuristics build around "valence electrons" break horribly.


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Porous membrane galvanic cell
I'm building a galvanic cell, Zn and Cu as anode and cathode, and ZnSO4 and CuSO4 as the solution respectively.
I want to replace the salt bridge with porous membrane.
Me question is which porous membrane to use? will filter-paper of 2 micrometer work?
thanks

Bridges and membranes are supposed to keep solutions separated while keeping ion conductivity. It is usually trade-off between both conditions.
Does filter paper keep solutions separated?  I do not think so. Difference of pressures or levels would cause flow to the other side.
Somewhat better would be soaked thick paper for drawing/water colour painting. Or even better cardboards  But consider long term desintegrating effects.
Try some ceramics, like porous flower pots(?), long term soaked in some suitable inert salt solution.
Or (not too porous) fritted glass as @Ed V has mentioned.

In the old days, people sometimes used soaked strips of asbestos cloth. A former colleague likes to used folded paper towels. Your answer is much better. Personally, I liked a little glass cell, made by our university glassblower, that had a fritted glass disk between the solution reservoirs. Wish I still had that.

Glassblowers vanished so rapidly from universities.


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Why does calcium emit circularly polarized light?
One page 31 of Quantum Metrology, Imaging, and Communication by Simon, Jaeger and Sergienko, the authors state that:

... calcium has two electrons in its outermost, partially filled shell; these electrons are in $\mathrm{s}$ states, with oppositely aligned spins, so that the total angular momentum is $J = 0$ (a singlet state). Consider then an electron excited to a higher energy singlet state. Intermediate between the ground and excited singlet states, there is a $J = 1$ triplet state. The excited electron can decay in two steps, via this intermediate state. Two photons are emitted in the process. These photons may be emitted in any direction, but if we look only at photons emitted back-to-back, i.e. look at coincidence counts from detectors separated by $180^\circ$, then angular momentum conservation requires the two photons to have the same circular polarization (left-handed or right-handed).

Why should the two photons have circular polarization and not say horizontal or vertical? The angular momentum would still be conserved if the two photons have vertical polarization along say $+z$ and $-z$ directions.

The angular momentum must also be conserved on each step of the two-photon process.

Thanks, @Ivan Neretin. Could you be a little more explicit?

The angular momentum must be conserved upon the emission of the first photon, and then also upon the emission of the second photon.

The two photons are emitted with a delay of a few nanoseconds. Now during this "delay time", the emitted photon has some non-zero angular momentum, whether it is circular polarized or (say) vertically polarized. How is circular polarization special?

I am probably missing something, but if circularly polarized light enters a linear polarizer, the exiting light is linearly polarized and a tiny torque is exerted on the polarizer. Not so for linearly polarized light entering the linear polarizer. I assume the torque was due to the requirement of conservation of angular momentum.

There are various papers like this open access one around: J.-W. Liaw, Y.-S. Chen and M.-K. Kuo, Rotating Au nanorod and nanowire driven by circularly polarized light, Optics Express, Vol. 22, Issue 21, pp. 26005-26015 (2014) https://doi.org/10.1364/OE.22.026005

The transitions are from $m=0$ in the singlet to $m=\pm 1$ in the triplet so will be circularly polarised.  (usually these are observed as $\sigma^+;\sigma^-$ in the Zeeman effect.) Why the polarisation is the same when detected 180 apart is perhaps due to a phase change imposed by the geometry necessary to keep the total angular momentum conserved- but this is just a handwaving guess.

Photons have a spin of magnitude $1$. If we project the corresponding spin state vectors onto the momentum vector of the photon, we get what's called "helicity" with values $\pm1$ which are left and right circular polarization. Helicity 0 is forbidden for massless particles like photons. (You'll have to take it up with the physicists to really explain why. We'll just leave it at that.)
Since the process you describe involves two separate photon emissions, and each one changes the angular momentum of the atom by a magnitude of $1$, the photons that are emitted must also each carry a defined angular momentum to compensate the angular momentum change in the atom. The question is now: Which way are those angular momentum vectors (i.e., spin states) oriented?
Let's say that we place the calcium atom at the center of a Cartesian laboratory coordinate system. Let's further assume that the first photon is emitted along the $+z$ direction with a helicity of $+1$, i.e., the spin vector also points along the $+z$ axis. This would be right circularly polarized light. The atom, meanwhile, gains an angular momentum of opposite sense, described by a vector pointing along the $-z$ direction. Now, the second photon is emitted along the $-z$ axis, carrying away the angular momentum of the atom so that the latter returns to its $J=0$ state. This means that the second photon has its spin vector in the $-z$ direction; again, the helicity (projection of the spin onto its momentum) is $+1$, which is right circular polarization.

Now, let's talk about linear polarization some more, because it gets a little tricky.
A general photon whizzing though free space will have the same energy and momentum no matter whether it's in a $+1$ or $-1$ helicity state. Thus, we can put the photon into any superposition of the helicities and it will still have the same energy and momentum. (In a chemist's mind, think how combinations of complex-valued atomic orbital wavefunctions to the real-valued Cartesian ones still represent stationary solutions.)
This means that, in the general case, describing linear polarization as a superposition of circular polarization states is just as valid as the other way around.1 In the calcium experiment, we are perfectly able to describe the photons as being in a superposition state of horizontal and vertical linear polarization. But if we were to send the through linear polarizers to measure which polarization they have, each one would have a 50% chance for either result.
Note also that a single photon can actually have linear polarization if it is in a equally-weighted superposition state of $+1$ and $-1$ helicity. This does not mean that it has a helicity of $0$; instead, it means that when we measure the helicity, we have a 50% chance to get either result.2

Opposite from the point of view at the atom, but the same from the point of view of each photon in its own direction of travel.

but left circular and right circular are not superimposable by just changing direction of observation ?

@Antimon, when you say "Photons have spin states ±1, which essentially correspond to left and right circular polarization", do you mean the linearly polarized light has no spin?

@Micheal Linearly polarized light has a total spin of zero, but the photons it consists of do have - positive and negative, just cancelling out at the proper phase. And consider it differently: When you say "horizontal/vertical polarization", which plane would that be horizontal/vertical to? There's no plane of preference in e.g. an isolated Ca atom, so photons can only be circular polarized and linear polarization is an emergent feature of multiple photons forming "the" photonic/electromagnetic field

@EdV I guess I should have worded it more carefully. Let's say we are an observer in the fixed laboratory frame. We are looking at the atom so that the first emitted photon comes straight at us. Let's say that it has a circular polarization that is clockwise from our point of view; for the photon itself, that's LCP. Then the next photon flies straight away from us with counterclockwise polarization from our view, which is again LCP for the photon. They key point here is that our perspective is stationary; we're not turning around to look in the travel direction of the photons.

@TobiasKienzler To put a dent in my own explanation, I looked up this exact topic yesterday after posting my answer and unfortunately it seems that your comment is not correct. Single photons can very well be linearly polarized, see this answer on Physics SE or the first sentence in this Wikipedia article. In the end it only matters what you measure. I think I might have to adjust my answer to reflect that as well.

Nicely updated answer that wraps it up well. Thanks for posting the answer!

My pleasure! :) Now I just have to hope that it holds up to scrutiny by those who are more proficient in physics than I am.

@Antimon Good point! Yes, if you "look" exactly, the circular polarization corresponds to the pure two possible spin projections, but like everything in the quantum realm you can of course also have a superposition of both even for a single photon. Sorry for oversimplifying

This is by way of a long comment.
The Ca atoms can take all sorts of orientations in space and presumably they suffer no collisions when in the excited singlet or triplet. The excited singlet must be produced quickly in a time much less that its lifetime. This singlet then decays to a triplet and then this decays to a singlet ground state, in both transitions a photon is emitted. We are told that both are both either right or left circular polarised and detection is at 180 degrees. This means that two identical detection sets are used and at 180 to one another. The fact that detection is at 180 effectively photo-selects the atoms, i.e. only some atoms will have a dipole so aligned to be detected at either position. Once the first photon is detected the second must come from an atom with the same initial dipole direction, provided no collisions or other interaction occurs.
The singlet has $L=1, m=0$,  and the triplet $L=1, m=-1,0,1$. As the initial state is a singlet suppose it generates a photon that is right circular polarised (rcp) with momentum projection labeled  $m_p=+1$, and leaves the atom in the $m=-1$ triplet sub-level and the total change in angular momentum will be zero.  ($L$ changes by 1 singlet to triplet and a photon has 1 unit of angular momentum ($L=1$), plus projection quantum numbers $m_p=\pm 1$ but no $m_p=0$)
The second transition is independent of the first as there is a decay time for both singlet and triplet states, i.e overall not a two photon transition. The triplet decays to the ground state where $L=0, m=0$ from $L=1, m=-1$ so has the same polarisation state as the first transition but can only be detected at 180 (or 0) because of the photo-selection by virtue of where the detectors are placed.
(As an aside presumably detection of both photons at 0 degree needs v faster pd's so it is easier to use two and perform coincidence that way).


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What are compounds where oxygen has oxidation state -1/3 called?
I have heard of oxides, peroxides and superoxides where oxygen oxidation number is $-2, -1, -\frac{1}{2}$ respectively.
But I came across a molecule with the formula $\ce{KO_3}$, oxygen has $-\frac{1}{3}$ oxidation state; what are they called?

These are called ozonides, specifically ionic ozonides. The name clearly comes from the ozone molecule with which the ion is isostructural. Ionic ozonides are known primarily as salts of the alkali metals, with the referenced WP article mentioning $\ce{MO3}$ for $\ce{M}=\ce{Na, K, Rb, Cs}$. The caesium compound has also been used to make tetramethylammonium ozonide in liquid ammonia; the latter has a surprising amount of stability for an organic moiety combined with an oxidizing one:

Tetramethylammonium ozonide, which can be made by a metathesis reaction with caesium ozonide in liquid ammonia, is stable up to 348 K (75 °C):
$\ce{CsO3 + [(CH3)4N][O2] → CsO2 + [(CH3)4N][O3]}$[1]

Alkaline earth metal ozonide compounds have also become known. For instance, magnesium ozonide complexes have been isolated in a low-temperature argon matrix[2].
It is misleading to consider all oxygen atoms in the ozonide ion as having oxidation state $-1/3$ because they are not all equivalent. Like the neutral ozone molecule, ozonide ion has one central oxygen atom uniquely bonded to both others. Specifically identifying an oxidation state for each oxygen atom requires making assumptions (or doing high-powered calculations) on the distribution of the odd electron. For most purposes redox reactions involving ozonide ion may be accounted by treating the $\ce{O3^-}$ ion as a single entity with a total oxidation state of $-1$ for all three atoms.
References
1.
Jansen, Martin; Nuss, Hanne (August 2007). "Ionic Ozonides". Zeitschrift für anorganische und allgemeine Chemie. 633 (9): 1307–1315. https://doi.org/10.1002/zaac.200700023
2.
Wang, Guanjun & Gong, Yu & Zhang, Qingqing & Zhou, Mingfei. "Formation and Characterization of Magnesium Bisozonide and Carbonyl Complexes in Solid Argon". The journal of physical chemistry. A. 114 (2010). 10803-9. https://www.researchgate.net/publication/46392397_Formation_and_Characterization_of_Magnesium_Bisozonide_and_Carbonyl_Complexes_in_Solid_Argon

+1 for a detailed answer. I would had answered by saying that they are ozonides and call it a day. But I knew we have users like you who will go beyond than just one-line answers :)

Just a tip, there was a change in the kind of markup SE uses some(tm) time ago. If you have the > markers separated by a blank line, it will also create separate "quotes". So if they belong together, you need to also quote the blank line to make it appear as one quote.


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α-cleavage of 1-pentene in EI-MS
I am studying the fragmentations in EI-MS, and this is the spectrum of 1-pentene.

The textbook says: The allyl carbocation (m/z=41) is an important fragment in the mass spectra of terminal alkenes and forms via an allylic α-cleavage.
First of all, at the initial ionization event, one electron draws off from π-orbital. In the below image, one can expect the 2nd carbon will bear the positive charge according to the Stevenson rule.

But in this case, I can't imagine the α-cleavage mechanism. Instead, if the first carbon bears a positive charge, I can draw the mechanism like this:

Is this the correct mechanism, although it violates the Stevenson rule, or is this an exception for the Stevenson rule?

Is there space for a McLafferty rearrangement?

@Buttonwood It is written that m/z= 42 is the peak for McLafferty rearrangement.

The main issue is that the structures that are named 1° carbocation and 2° carbocation in the question are not different structures. As a general rule in mass spectrometry, ionization is postulated to occur from the highest occupied molecular orbital, which is usually for alkenes the $\pi$ orbital. Removing an electron from a $\pi$ orbital leads to a singly occupied molecular orbital (SOMO) which is formally a singly-occupied $\pi$ orbital (imagine a double bond but only half occupied), for which there is no sensible Lewis representation. The usual way to represent in the Lewis formalism this $\pi$-type SOMO is to represent it as a resonant structure between the structures that are called 1° carbocation and 2° carbocation in the question, such as below:

So the short answer is that there is no reason, in general, to apply the Stevenson rule to these, as they are not two separate structures but only a single one.
There is nevertheless, for the specific case of 1-pentene an interesting twist. Leif A. Erikson et al. have shown through calculations and ESR experiments in cryo-matrices that the 2-pentene cation follows the general rule given above. On the other hand, the 1-pentene cation has an electronic structure for the SOMO which also involves delocalization of the orbital up to the adjacent sp3 carbon atom. This also translates in a shortening of the C(sp2)-C(sp3) bond upo,n ionization. In this instance, the Lewis formalism cannot be used to correctly represent the SOMO, but it helps in rationalizing the preferential break of the $\alpha$-bond.

The suggested formation of the allyl cation is not a contradiction of the Stevenson rule.

«What is the statement of Stevenson’s rule?
When a fragmentation takes place, the positive charge remains on the fragment with the lowest ionization energy.»
Jürgen H. Gross, public answer key for question 6.1c) on his companion site to Mass spectroscopy, 3rd edition  (here).

As suggested by you, a plausible mechanism for the fragmentation of 1-pentene in a mass spectrometer is the homolysis yielding two radicals, one of them the allyl cation:

The parent for the allyl cation is the neuter allyl radical, $\ce{C3H5^.}$.  In section Gas phase ion energetics data, NIST lists multiple entries for the reaction leading to $\ce{C3H5^{.+}}$.  According to the entry by Kagramanov et al. (1983), this requires $(8.18 \pm \pu{0.07) eV}$ by EI (ref).  This value corresponds to about $\pu{789 kJ/mol}$.
The alternative would is to depart from the neuter ethyl radical, $\ce{C2H5^.}$, to yield the radical cation $\ce{C2H5^{.+}}$.  Here, the most recent entry for EI NIST lists is the by Lossing and Semeluk (1970) with $(8.38 \pm \pu{0.05) eV}$ (ref).  This value corresponds to about $\pu{806 kJ/mol}$, which is greater than the one for the allyl radical.

Energy conversions $\pu{eV} \rightarrow \pu{kJ/mol}$: unitsconverters.com

This answer is misleading : the reaction drawn is a fragmentation of neutral propene, whereas in mass spectrometry one deals with fragmentation of ionized propene, leading to the allyl cation and the neutral radical C2H5°. Second: the IE of the ethyl radical is around 8.2 eV (thus lower than 12.45 quoted in the answer), and the IE of the allyl radical is only slightly lower (around 8.1 eV). Accordingly and following the Stevenson rule, the m/z 29 (C2H5+) is also present, and only slightly less intense then m/z 41.

@PLD «it is more likely that the allyl cation is formed» does not exclude that there is $\ce{C2H5^+}$ is formed. May you put your argument in an answer (well, counting the carbon atoms in my equation [5], it departs from neuter pentene), including an equation about the mechanism and quote of the source of the origin of the numbers you use for the discussion?

@Buttonwodd. I posted an answer to the original question, but it does not contain elements of my comment. The values are from the NIST database as yours. For correct application of the Stevenson rule, one should consider the ioniozation energy between the neutral counterpart and the ion, which means that for cations (and not radical cations as usually stated for Stevenson's rule) one needs to start from the radical (allyl or ethyl) and not from the entire molecule.

@PLD Based on your input, I revised data used and scheme. Does NIST equally list the enthalpy of formation for e.g., the neuter ethyl radical $\ce{C2H5^.}$ since https://webbook.nist.gov/cgi/cbook.cgi?ID=C1981802&Units=SI&Mask=20#Ion-Energetics reads like $\Delta{}_fH (\text{ion})$, too?  These data could be used for a  back on the envelope estimation of $\Delta_rH$ (or, $\Delta_rG$) for the overall reaction.

You have the $\Delta_{f}H$ of the neutral under the "Gas-phase thermochemistry data" section (https://webbook.nist.gov/cgi/cbook.cgi?ID=C1981802&Units=SI&Mask=1#Thermo-Gas). But this value is likely derived from the ionization energy measurements (detailed in the page you refer to) and the formation enthalpy of the ion, so basically we talk of the same data.


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Synthesis of Jet Fuel from Fe-Mn-K Catalyst using Atmospheric CO2
I wanted to do some back of the hand math on how many nuclear reactors it would take to reduce CO2 in the atmosphere to pre industrial levels after reading the article below describing the process.
Regarding this synthesis reaction: https://www.nature.com/articles/s41467-020-20214-z
From the abstract it seems to convert one mole of atmospheric CO2 into a fuel for use either in transportation or as a form of carbon capture (bury it again), it would take (-166kJ + -125kJ + 41kJ) in energy for one mole of CO2 coming to a total of 250kJ of energy required per mole turned into the various fractious products...
Next let's say we want to return the amount of CO2 to pre industrial levels IGNORING output currently, that we were in effect carbon neutral (just for the sake of argument because the numbers are getting outrageous) by 2100. That gives us 78 years to do so.
So. currently, I am reading some random internet sources saying that there is about (1.8 * 10^20) moles of CO2 in the atmosphere currently and that we are at about 400ppm compared to preindustrial levels of 280ppm. So we need a reduction of (120ppm/400ppm) == .3 or 30% reduction in atmospheric carbon from todays levels for a total reduction of (.3 * 1.8 * 10^20) moles of CO2.
Now using nuclear power as it is the most energy rich, but I think it would yield similar cost analysis for other types of power, am I correct in stating that to reduce this in 78 years we would necessarily need at least (.3 * 1.8 * 10^20) moles CO2 * (240kJ per moles CO2) energy to yield that? That would mean we would need to produce 1.29600 × 10^25 joules in 79 years, or in watts 1.29600 × 10^25J / (79*365*24*60*60s) which equals 5.20201144 × 10^15 watts.
I read somewhere the average nuclear power plant yields about 1 gigawatt. If that is true, to reduce atmospheric carbon to preindustrial levels by 2100, we would need ((1.29600 * ((10^25) J)) / (79 * 365 * 24 * 60 * (60 s))) / (1 gigawatt) which yields 5 202 011.44.
Is my off the cuff math correct? In order to return the earth to way it was, minus of course all the nuclear plants, mean that we would need to build roughly 5 million nuclear power stations?

With the same constraints, assuming my crappy math is correct, it would take more land area than the surface of the earth to reduce carbon to pre industrial levels

I didn't check it but wouldn't be surprised. You essentially have to invest almost all the energy used up by humanity up till this point + the inefficiencies. That would immediately put you minimum in the 10s or 100s of thousands of reactors range, so your estimation seems close in order of magnitude. CO2 recycling is a dumb plan from the beginning, but I guess attracts enough governmental and venture money, so people do it anyways. I wish we had something like self-reproducing organic material, that you can just grow from seed to solve this.:)

@Erik Yeah, and it would be the largest industrial project ever, producing enormous waste and use enormous energy, raw materials in the first place.

This type of "science" is just done for funding purposes. The authors will be applauded for publishing in Nature Communications in their respective departments, this will result in one or two conference invitations, the student authors will get an academic job...and the world will move on.

I don't think you can group  "either in transportation or as a form of carbon capture (bury it again)" as a single category with the same energy use. Lots of proposed carbon capture strategies do not involve changing the CO2 to hydrocarbon. For example, pumping the gas into uderground caverns or mineralizing it as CaCO3. THe energy usage for those very different approaches should be calculated separately.

Welcome to Chemistry! On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. I have removed the meta commentary from the title, but leave the (optional) rest to you.


text:
Does the increase or decrease of a pure liquid in an equilibrium system have an effect on the direction of equilibrium?
My textbook reads:

Although pure liquids have no effect on the equilibrium constant, K , pure liquids can affect the direction of equilibrium, according to LeChatelier's principle.

Whereas the Organic Chemistry Tutor, in this video (timestamped),states that pure liquids, along with solids, have no effect on the direction of equilibrium.
So which one is true, if either?

If anything but the amount of pure liquid is constant, the position of equilibrium remains constant.
If a change of the amount of pure liquid changes other relevant parameters like e.g. partial pressures, the position of equilibrium shifts.


text:
What happens to the sea of electrons when a metal melts/boils?
I was taught about metallic solids with the sea of electrons model. My question is how does the sea of electrons function when metal turns into a liquid and a gas. The way I see a liquid is intermolecular forces being constantly formed and broken between particles, but metallic bonds are different in how they are attractions between shared, mobile electrons and nuclei. I can't visualize how particles moving faster have less attraction force with the electrons when they are surrounded by them at all times. From this, it tells me that when a metal melts, the bonds are "loosened" and when it boils, they "break". How does that even work? Is something happening to the sea of electrons that results in some metallic bonds breaking in molten metals and all metallic bonds breaking in gases? In a gas, as the particles get further apart, I doubt the sea of electrons is expanding to fit the entire volume of the gas, because, at that point, it would just be cations and electrons, not atoms, so do the electrons "relocalize" to an ion's valence shell when they can no longer be attracted to multiple atoms?

It seems that you have somehow the picture except for the fact that a metal in gas phase isn't a metal anymore, that is there is not that sea of electron anymore. Until the phase is the melt, you are in presence of a loose sea,as you said. All is a balance between the kinetic energy of the atoms and the attraction. In such a sense is not different than melting and eventually boiling an ionic or covalent compound. I think this comment should suffice.

Liquid metals are (as far as I know) all metallic - they have the 'sea' of electrons, just not in Bloch states any more. Silicon and germanium, semiconductors as solids, become metallic in the liquid phase (and go from 4-fold coordinated in the solid to 8- to 12-fold coordinated in the liquid, the main reason the liquid is denser than the solid).


text:
How can I formally justify the description of an electronic excitation that includes only the relevant orbitals?
I am writing a small program to animate electronic transitions.
To do this, I am animating a Rabi cycle for a single electron as it moves from an occupied into a virtual orbital.
The on-resonance excitation process during the interaction of a molecule with a coherent laser is modelled by the Rabi formula:
$$\Psi = \cos(\Omega_R)\Psi_g + \sin(\Omega_R) \Psi_e e^{-i\omega t},$$
where $\Omega_R$ is the Rabi frequency and $\omega = (E_e - E_g)/\hbar$.
As a minimal example, I will consider the $\ce{H2}$ molecule. The first electronic excitation is $\sigma \rightarrow \sigma^*$.
Formally, the wave function of the ground state is:
$\Psi_g = \det|\sigma_{\alpha}\sigma_{\beta}|$, and the excited state is: $\Psi_e = \det|\sigma^*_{\alpha}\sigma_{\beta}|$.
Intuitively, I think that the description of the excitation process can be understood by considering only the wave function of an isolated electron jumping from a $\sigma_\alpha$ into a $\sigma_{\alpha}^{*}$ orbital, rather than wave function as a whole. In this case, I express the transition state as follows:
$$\Psi = \cos(\Omega_R) \sigma_{\alpha} + \sin(\Omega_R) \sigma^*_{\alpha} e^{-i\omega t}$$
from which an animation of the changing electron density can be constructed by computing isosurfaces of $|\Psi(t)^{2}|$, and assigning the color depending on the phase of the underlying wavefunction at that point:

In a more complicated molecule, the same simplification might correspond to expressing the excitation process as follows:
$$\Psi = \cos(\Omega_R) \psi_{\text{HOMO}(\alpha)} + \sin(\Omega_R) \psi_{\text{LUMO}(\alpha)} e^{-i\omega t}$$
Which, for an s-tetrazine derivative leads to the following animation for the first excited state (side and top view):

My issue is that I am not 100% certain about whether there is a formal procedure that allows me make the claim that these animations provide me with a physically valuable description of what happens during an excitation. It is clear to me that these animations do allow us to understand the symmetry rules and the direction of the transition dipole moment. I am looking for some help in finding the formal way of justifying this approach. Could the "1 particle reduced density matrix" formalism help me here?

Welcome! May I suggest checking out FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. You obviously know the syntax already (most of it is identical to LaTeX); but there are some things which you will find useful, e.g. using \cos{...} instead of just cos ... to correctly typeset cosine and other operators.

Aren't you running (besides details which I might not understand) in the problem to give physical sense to the orbitals, the most usual way to do it being its interpretation as a probability distribution function?

@Alchimista, I should have been a bit more specific about the final step, I will make an edit to clarify. What I animate is the wave function squared, where the wave function is the time-dependent wave function during the Rabi cycle. So, the physical interpretation would indeed be the changing probability distribution of a single electron during the excitation process. I think that this makes intuitive sense. If only one electron changes orbital, then we only have to consider the changing probabilities for that single electron. But I do not know how to justify this assertion formally!

@MaxParadiz I see now. You are looking for something like a energy bases criterion of separation. I am sure specialist will help you. Otherwise try finding done examples, if the final rendering is your main goal.

I'm not sure you can formally justify this approach. Electrons are indistinguishable and therefore it is formally incorrect to localize one electron to one orbital and the other to the other orbital. The full wave function of the system includes contributions from each electron in both orbitals. But there's still no reason why you can't make an animation of only one orbital of the excited state.

https://chemistry.stackexchange.com/questions/159292/how-can-i-formally-justify-the-description-of-an-electronic-excitation-that-includes-only-the-relevant-orbitals?

The answer here depends on what you mean by "formal procedure" and "physically valuable".
The potential flaw in your approach is the assumption that
$$\Psi=\cos(\Omega_R)\sigma_\alpha + \sin(\Omega_R)\sigma^*_\alpha e^{−i\omega t}$$
is a valid description. Formally, the wave function of a two-electron system cannot be split into two independent one-electron wave functions. This is perhaps more clear if you write out the full determinants in the wave functions, which are
$$\Psi_g=\begin{vmatrix}\sigma_\alpha(1)&\sigma_\beta(1)\\\sigma_\alpha(2)&\sigma_\beta(2)\end{vmatrix}$$ and (for a singlet excited state) $$\Psi_e=\begin{vmatrix}\sigma_\alpha(1)&\sigma^*_\beta(1)\\\sigma_\alpha(2)&\sigma^*_\beta(2)\end{vmatrix}$$
where (1) and (2) are the two electrons and $\alpha$ and $\beta$ are spin states.
That is, each wave function necessarily consists of contributions from each possible electron-orbital pairing rather than just assigning one electron to each orbital. (For a triplet excited state, there are actually two additional wave functions to account for the spin possibilities, but that's not necessary to consider for this discussion.)
That said, almost all computational approaches to dealing with such systems apply this invalid approach in order to make the mathematics tractable. The Hartree-Fock method is a very common starting point. By adding various terms to correct for the indistinguishability and interactions of the electrons, the results of these methods can approach the results we would expect to get from a formally correct description of the system (ie solving the Schrodinger equation for a single multi-electron wave function).
Furthermore, treating the total electron density of a molecule (ie the square of the multi-electron wave-function) as the sum of densities of multiple discrete orbitals is a standard pedagogical approach, and predicting the properties and reactions of molecules based on a subset of these discrete orbitals (valence or frontier orbitals usually) works quite well at a qualitative and semi-quantitative level. Whether this pedagaogical value counts as "physically valuable" depends on what you mean by that.
I'll conclude by invoking George Box's maxim that "All models are wrong. Some are useful." While your animations are not defensible on theoretical physics grounds, that does not mean that they are not reasonable and useful representations for pedagogical purposes.


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How to compute a needed volume of solution to adjust to a target pH
I am a software engineer and I would like to solve a chemistry problem. I have revised the subject a bit to understand the basics :)
My goal is to maintain a constant pH of a water tank (tap water). I have measuring devices that can give me an accurate pH adjust according to the temperature. For example a pH of 6 for a 70 L tank.
I have products that can adjust this pH. A pH 2 solution and a pH 11 solution.
My first approach was a naive implementation of adjustment. I defined an optimal pH O and an acceptance threshold K which gives me an acceptable range for the measured pH (O+K and O-K). If the measured value is outside this range, I calculate the difference between the optimal and measured value. The adjustment is done progressively and I tend towards the optimal range. This is a solution that works but it does not optimize the number of adjustments to be made.
My second idea is to directly approach the volume to be distributed. That's why I turn to you. I have read some courses on how to calculate the pH of a mixture. But they always refer to a pure water solution. Moreover, I don't necessarily know the composition of the adjustment product but only its pH. I feel that I cannot simply apply the equations considering pure water and a known acid or base.
How can I approach this problem? I have to admit that some notions escape me. Thanks in advance for your help.

Hi David!  Welcome to Chem SE.  From a practical perspective, it's probably a lot easier to simply monitor the pH and then adjust accordingly. I assume that you will be measuring using testing strips, so the accuracy of the measurement isn't very high.  Measure the pH.  If it's too high, add some of the pH 2 solution.  Mix and then measure again.  Repeat.

Also, to other users... I don't think this question should be down-voted.  The OP is not a chemist or a student.  They have clearly thought about this a lot and are looking for help.  Let's not be snobbish.

Thanks for your answer :) No I use industrial sensors to make my measurements. They are then processed by an embedded system. This system then distributes the adjustment products with peristaltic pumps. Yes, I also think that the first solution is simpler. But the second one will allow me to understand more deeply the evolution of the pH by mixing various liquids.

To clarify my first solution, the error is calculated by making the difference between the optimal pH and the measured pH. This difference becomes the multiplication factor of a constant volume of the adjustment products. For example: optimal pH O = 6, measured pH m = 4.6. Error E = O - m.The volume V is a parameter chosen empirically. The volume I dispense is equal to E * V. Is there a way to calculate this volume V in order to minimize the adjustments? My chemistry notions are very basic (:D)

What is causing the pH to change in your 70L tank? Is it starting with tap water and then something else is being added to it (e.g. waste) that causes the pH to change which you are trying to counteract?

That's the thing. I don't know. This is a prototype and everything is simulated for now. I guess time, surface area of water exposed to the air, and algae must cause the pH to change.

@DavidCruchon
Can your embedded system use a PID controller that can be tuned? That's the outcome for your error = setpoint - process value concept. Calculating a single dispensing increment V requires knowing the chemistry of what's in the tank. If you have a dilute acid inflow vs. a highly basic inflow, the optimal V value will be different in those cases. I'm ignoring other effects like solids, biological organisms which would only further complicate things.

@DavidCruchon
An example of how industry does what you're asking about:
https://www.pumpsandsystems.com/sites/default/files/webinars/2020/1112/BurtProcess-pH-Systems-Design-Slides.pdf

@J. Ari This is exactly what I was missing. The implementation of a PID is probably the way to go. You allow me to widen my research now. Thanks a lot :)


text:
Why do hydrocarbons give out a blue flame on complete combustion?
In complete combustion of hydrocarbons the colour of the flame is blue. But what is the reason behind the flame being blue and not green or black or any other colour?
Edit: I've searched the web for an answer to my question but all of them just equate blue flame to complete combustion and yellow to incomplete combustion. A similar or rather the same question has been asked here on chemistry stack exchange by a user but it just has one relevant answer. The motive behind asking this question again is to get a few more answers or in other words a few more second opinions so as to get a more balanced and clearer understanding of the matter.

@M.Farooq Not exactly. There are three answers in all. Two of them don't answer the question and the third one isn't that easy to comprehend.

@Obina, You can edit/expand your question about what is not clear and show your search effort. Did you search the web?

The answer by Frazky is indeed the answer. The above two are not relevant.

@M.Farooq I have edited the question. Is it clearer now?

Unfortunately, it is closed now. The basic reason for blue flame is that hydrocarbons decompose to smaller simple diatomic molecules like $\ce{C2}$. By the intense heat of oxidations, these molecules are electronically excited. The excited molecules emit blue color.

@M.Farooq Any particular reason why the excited molecules emit blue colour?

It depends on the energy level difference between the ground state and excited state. Blue means that there is a big energy gap.

@M.Farooq Does this mean that if I somehow manage to adjust the energy difference, I can get green fire?

Yes, but molecular energy level in a flame is not in your hand. You can have green colored flame by introducing copper salt in flame. In fact, a lot of colors are possible.


text:
Shielding effect is not constant across a period

Shielding effect stays constant across a period because number of inner electrons stays the same.

Well, that's not true.
It only works for the period =1,2,3
but in period 4, the number of inner electrons of $K$ to $Cu$ is 18,while the number of inner electrons from $Zn$ to $Kr$ is 28.
So the number of inner electrons is not the same, so the shielding effect isn't constant.
Now there is another definition i saw at more than 3 sites which i found it more sensible and works well, is that :

The shielding effect stays constant across a period because number of inner shells stays constant.

And that is true because at period 4, number of inner shells is 3 for all elements in period 4, and this works for all periods.
So what is the right definition?

Neither. Shielding effect is variable across all periods, as verious shielding is provided by inner electrons as well as be the peer electrons from the same orbital group.

dude why the internet is like this loll. Every page i go there is a different answer, i've got to a point where i am so confused if what i learn is right or not smh

Do not look for answers, look for understanding to get your own answers. See Slater rules

I've seen the Slater rules, but we didn't get that far in the course, so i don't think Slater rules will help in my case. About understanding point, i've understood shielding effect but i want to make sure which definition is correct to improve my understanding and help me in my exams, anw big thanks man! :)

It is not about definition, but about the scope of considered factors, which depends on the education level.

Ah okay.. um..I'm first year in college if that helps


text:
How do ketone react with a peroxy acid to yield an ester?
This may be a really dumb question but I can't find any mechanism plot or reasoning how ketone reacts with carboxylic acid and form an ester. I thought R-OH react with RC(O)OOH to form ester..
I tried several iterations to solve the following question. Please let me know how to solve it properly.

That's not a carboxylic acid; a carboxylic acid only has two oxygens. This has one more oxygen. Search in your favourite book, or online, for the Baeyer–Villiger reaction. That should answer most of the question, if not all.

Wikipedia is as a good a place to start as any  https://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation


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Why doesn’t the amount of ligand needed to reach a fraction-bound value of 0.5 increase as the starting protein concentration increases?
In a Kd binding curve of fraction-bound on the y-axis and ligand concentration on the x-axis, it’s known that the Kd is equal to the ligand concentration [L] needed for [PL]/[Ptotal] =0.5. However, I’m confused as to how the ligand concentration needed to result in that fraction-bound value does not increase as you up the starting concentration of protein in solution. Shouldn’t the amount of ligand needed to bind more protein be larger as well? How can you put in the same concentration of ligand and still get an equal fraction-bound value even if you doubled the starting concentration of your protein in solution?

The assumption is that Ptot << Ltot, so that the change in L concentration due to binding to the protein is negligible and [L]free ~ [L]tot. That remains true even if we double the protein concentration. If [P] gets high enough that the assumption does not hold, then we would need to plot using the concentration of free L rather than total L because they would not be approx equal

I see, thank you. That clarified it :).


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Why are almost all fruits acidic?
Almost all the fruits are acidic in nature, but why? At least, fruits like apples should be neutral right? But although apple is sweet, it is acidic. Why?

Oh sorry, I meant neutral. Usually acidic things are sour in taste and bases are bitter in taste. Apple isn't sour nor bitter, so shouldn't it be alkaline? It has a PH level of around 4. Why is it acidic?

Most culinary fruits, including apples, contain organic acids. The most predominant are malic acid and citric acid.    As to why that's the case, you might want to scan through this article:  https://academic.oup.com/jxb/article/64/6/1451/586994

This is a question for Biology.SE, and has already been answered there several times: https://biology.stackexchange.com/questions/38856/why-are-many-fruits-sour; https://biology.stackexchange.com/questions/42673/why-does-citric-acid-occur-in-citrus-fruits

"fruits like apples should be neutral right?"  Why would you think this?

I’m voting to close this question because it belongs on Biology SE.

Fruits have culinary and botanical meaning. For the former, we call fruits those  botanical fruits that are juicy and taste more or less sweet and sour. For the latter, e.g. nuts are non juicy fruits. Tomatoes are fruits botanically, but considered as vegetables for culinary purposes.
Acidic in chemistry context means pH<7, not necessarily tasting acidic/sour. Presence of organic acids ( citric, malic, tartaric acid) causes fruit juice being acidic. Sufficient presence of soluble sugars masks acidic taste by stimulating sensors for sweetness.
Take a juice and dilute it by water. Even if pH raises and objectively it gets less acidic, subjectively it seems more acidic, as sugar concentration gets lower, unmasking sour taste. Acidity decreases slowly by dilution, due higher dissociation of weak acids and due existence of pH buffers. Alkaline fruits would be probably poisonous, full of alkaloids.
The question why there are free acids is rather question for botany. Probably the part of strategy to get seeds distributed by keeping fruits from being attractive until they are ripe and sweet.

In biochemistry, acids are COOH groups, while bases are amines and amides. Nitrogen containing substances are a limited resource in almost any organism. On the other hand, carboxylic groups can almost freely convert to and from sugars.
In regard to fruits:
It looks like a very old kind of co-evolution between the fruit eaters (that spread the seeds) and fruit-producing plants (that pay for the seeds distribution by making edible and tasty fruits).
The high acidic content of the fruit (together with the green color) signals that the fruit has low nutritional value (the seeds are immature and eating the fruit now is not valuable to the plant).
Low acidity, high sugar content, red/orange/yellow color distinct from the green leaves - the fruit is ripe and ready to be eaten (and the seeds are ready to be spread).


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What are the Ionic Properties of Co-ordinate Covalent Bonds?
In my 10th grade book, it says that

A coordinate bond has properties of both covalent and ionic bonds. Therefore, it is also called dative or co-ionic bonds.

I understand that it has properties of covalent bonds, as electrons are still "shared" among the elements(though pair is provided by only one of the atoms), but how does it have ionic properties?

Ionic character means that there is some degree of charge separation, as in standard covalent bonds between atoms of different electronegativity, and even more so due to the mechanism of bond formation.
At an entry level, and I would say even at more advanced level, a formed dative bond is just a covalent bond, albeit a polar - often very polar - one.
You might read this thread
If bond types are in reality intermixed, how come different bonds form completely different structures?
which in part relates to the use of words like ionic character etc. Those terms are perfectly intelligible once one has a clear idea, but it seems that they have some limitations when a clear idea has to be conveyed to beginners.
You can refer to
https://en.m.wikipedia.org/wiki/Coordinate_covalent_bond
to see that in special cases and according to some authors the use of the specific term dative as well as its drawing by an arrow should be preferred.
IUPAC does indeed define a dative bond:
https://goldbook.iupac.org/terms/view/D01523
Note that even without other considerations, IUPAC gives emphasis to the fact that a dative bond tends to break in a heterolytic way, so ionic character it is at least justified when its splitting is considered.
I would finally add a personal note. An ionic bond is per se quite strong. Increasing polarity of a covalent bond also usually increases the same bond strength.
However for a dative bond, according to the IUPAC def. above, peculiarities are a long bond distance and a relative weakness.
So again, the reality must be learned first, than some words just help the communication between initiated. As this community shows, no one use pictorial or even obscure terms just for fun or sake of closure.


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Do degenerate states have exchange energy?
I have read that exchange energy exists between non-degenerate levels $1s$ and $2s$ or between $1s$ and $2p_x$ in the Helium atom in the first excited level:
$$K_s = \left< 1s(1)2s(2)|H'|1s(2)2s(1) \right>
$$
$$K_p = \left< 1s(1)2p_x(2)|H'|1s(2)2p_x(1) \right>
$$
What about the exchange between $2p_x$ and $2p_y$ in the next excited level? Is $$  \left< 2p_x(1)2p_y(2)|H'|2p_x(2)2p_y(1) \right> =0$$? Here, $$H'= \frac {e^2}{r_{12}} $$ is perturbation term. Thanks.

Short answer is yes, there are exchange energies among degenerate orbitals. Most often discussed with respect to d orbitals, probably because of stable transition metals with multiple unpaired electrons in d orbitals


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SMARTS pattern for amines with connected carbons only having bonds to carbon or hydrogen
I'm trying to find a SMARTS pattern that matches primary and secondary amines, where the connected carbon atoms only have connections to other carbon atoms or hydrogens (to exclude amides, guanidines, etc.)
It should match diethylamine, octylamine, t-butylamine or aniline, but not if any other heteroatom is connected to the amine's carbon(s), like in an amide, oxazolidine etc.
I could write a very long SMARTS pattern, where I include all the possible carbon/hydrogen connections that are possible for the carbons connected to the amine, something like this:
[$([#7H2][CX4(C)(C)C, CX3H1(C)C, CX2H2C, CX1H3, CX3(C)C, ...]), $([#7H1][...]]
Is there a shorter/better way to do this?

Search for a nitrogen with three connections NX3
AND has 1 OR 2 hydrogen ;H1,H2
AND is NOT connected to noncarbon ;!$(N[!#6])
AND is NOT concected to a carbon that is NOT connected to carbon ;!$(N#6]~[!#6])
[NX3;H1,H2;!$(N[!#6]);!$(N[#6]~[!#6])] should work.

Thanks for your answer and the detailed explanation! I tried this in rdkit and it almost works, except for aniline C1=CC=C(C=C1)N. I changed the SMARTS to [NX3;H1,H2;!$(N[!#6]);!$(N[#6][!#6])] to explicitly include any carbon, which works for aniline, but unfortunately this now also matches nicotinamide C1=CC(=CN=C1)C(=O)N, where the amide NH2 is matched. Any idea what's the problem?

@snurden The change to #6 was a good idea (I forgot aromatic carbon). The problem with nicotinamide is the double bonded oxygen. I changed the last pattern from single bond to any bond ~. Hope it works now.

Works perfectly, thanks! Totally forgot about the implicit single bond.


text:
What role does ammonium play in the explosion of ammonium nitrate?
I read that ammonium is the "fuel" and nitrate is the "oxidizer". The nitrate oxidizes because it's a nitro group, but what's the mechanism of the ammonium?

Will it help if you know NH3 is flammable ?

I was thinking more along the lines of nitrous oxide being an oxidizer. The ammonium would then be a what combines with the nitrate to produce said nitrous oxide.

@OscarLanzi I see. But N2O is created mainly during peaceful decomposition. It decomposes AFAIK as well during explosion of the pure nitrate, even if it goes better when extra flammable stuff like Al is present.

When heated slowly, ammonium nitrate gets decomposed by the reaction $$\ce{NH4NO3 -> N2O + 2 H2O}$$ This is the industrial way of synthetizing the gas $\ce{N2O}$. The nitrate group is oxidizing and the nitrogen atom in the ion ammonium is oxidized. Both atoms are changing their oxidizing number by the same amount : $4$ units. $\ce{N}$ from $\ce{NO3^-}$ goes down from $+5$ to $+1$. $\ce{N}$ from $\ce{NH4^+}$ goes from $-3$ to $+1$.
Nevertheless in case of overheating, a further reaction takes place : $$\ce{2 N2O -> 2 N2 + O2}$$ and this reaction is exothermic (+$82$ kJ/mol) and exoergic (+$104$ kJ/mol).


text:
Is Co(2+) easily oxidised to Co(3+)?
I think this question lacks some required informations but that's exactly what was given on my question paper. I thought of explaining this in descriptions of crystal field theory but I couldn't solve it at the end.

Why is $\ce{[Co(H2O)6]^2+}$ oxidized to $\ce{[Co(H2O)6]^3+}$ easily?

Can anyone help me get the answer of this question?

The author of the quoted question is obviously mistaken.

The reduced form $\ce{[Co(H2O)6]^2+}$ does not manifest significant reducing abilities and is oxidized with great difficulties only.
The oxidized form $\ce{[Co(H2O)6]^3+}$ is very strong oxidizer.
The respective ox/red pair has the standard redox potential $E^{\circ}(\ce{Co^3+}/\ce{Co^2+})=\pu{+1.82 V}$, what is more than for most oxidizers. (Note that the $\ce{H2O}$ ligands are implicit in this.)
The redox potential can be significantly shifted for other ligands. If they have stronger affinity to $\ce{Co^3+}$ than to $\ce{Co^2+}$, the redox potential gets lower. See the @Maurice answer.

Poutnik should have added that the author of the mentioned sentence is clearly wrong.
But the author would have been right if he would have replaced $\ce{H2O}$ by $\ce{NH3}$. Matter of fact, $\ce{[Co(NH3)6]^{2+}}$ is easily oxidized in $\ce{[Co(NH3)6]^{3+}}$ for example by atmospheric $\ce{O2}$ molecules. The standard electrode potential for the reaction $\ce{[Co(NH3)_6]^{3+} + e^- -> [Co(NH3)6]^{2+}}$ is $0.1$ V. This means that the ion $\ce{[Co(NH3)6]^{2+}}$ can easily be oxidized by most common oxidizing stuffs, like $\ce{O2}$ or $\ce{H2O2}$

I have already exhausted my daily quota of "piece of crap" comments. :-)

Is there a daily quota of everything ? What does it mean ?

No, it was a personal joke. :-)


text:
Is the azo group basic?
I know that amino groups (nitrogen with single bonds with hydrogen and/or carbon) are basic.
I wanted to know if azo groups ($\ce{R-N=N-R}$) are similarly basic.
I searched online but couldn't find any reference whatsoever.
And I'm not sure if being double bonded with another nitrogen might or might not change the basicity of the nitrogen.

Not basic at all, pKa -2.95 https://en.wikipedia.org/wiki/Azobenzene

Nitrogen is strongly electron withdrawing, so... As one can expect situation is similar, if not worse for N2, O2 and CO2 in particular.

@Waylander that is slightly unfair, as the nitrogen is conjugated to aromatic rings which weaken its basicity. Then again, since such conjugation is generally needed to stabilize azo compounds at all against decomposition to radicals, we can argue that the azo function is for all practical purposes not an aqueous B-L base.


text:
Will increasing the temperature of solid reactants increase rate of reaction?
When temperature of aqueous and gaseous reactants increase, the rate of reaction increases since the particles will move more quickly so frequency of effective collision increases. But does this work for solid reactants as well? I know that the particles will vibrate more vigorously, but is it sufficient to result in higher rate of reaction?

yes, in fact a lot of solid state reactions require heating to "push" it along at an appreciable time scale.

For your liquid/gas statement, it depends what counts as a reaction https://chemistry.stackexchange.com/questions/95343/is-negative-activation-energy-possible (Not my DV though.)


text:
Method of cooling a gas without liquid evaporation for Sci-fi novel
sorry for the amateur question here.
I know little about science, but regardless am writing a fictional amateur novel that deals with a lot of chemistry stuff, especially gasses.
My protagonist needs a clever method of cooling a gas.  He has access to several gasses (CO2, O2, H2, nitrogen) in varying temperatures (70c and up).  Unfortunately no water (at least not more than a few bottles).  The atmosphere he is working under (planet atmosphere) is in the range of 150c and at a pressure of roughly 150kpa in the novel.
He also has access to sci-fi style manufacturing machines that can simply 'make' parts, like pipes, valves, pumps, that sort of thing.
I am looking for a clever way or really any way at all, that he can use to cool down some O2 to at least under 40c in such a situation.
My amateur reading online suggests the main way, is via liquid evaporation, but I doubt that would work in the above scenario easily.
Does anybody have any ideas?  Is this even possible?

Let the gas expand into a larger container then the one that is currently holding it. You can calculate that change by using a simple algebraic relationship called Charles law, where (V1T1)/V2=T2 where V1 and T1 are the initial volumes and temp while V2 is the final volume.

Thermal pumps? The same principle as  a fridge or climatization.

@Poutnik definitely more clever and interesting then expansion...

@Poutnik They use some liquid - the point of question is its lack. Or wait, "few bottles" is a lot, so maybe it would work.

@Mithoron It was not clear if any liquid, even in a closed cycles, or one time open evaporation.  // Another option is thermoelectric Peltier effect. Not very effective, but principally suitable.

Just put it on ice

Unfortunately thermoelectric cooling is out.  It would need dedicated device/hardware and I'm keeping that under the 'he doesn't have the right materials to construct) category of things.  Evaporation I looked into and seems too impractical.  As for expansion I actually thought of something like that, but I don't really understand that in terms of the ideal gas law.  If the gas gained access to more volume, wouldn't that just decrease the pressure and not the temperature?  I don't really understand it too well. @dval98

Another option is Joule-thompson effect based cyclical expansion / compression machine, usually used for liquifying gases, this time would be used just for gas cooling.

@Poutnik Trying to understand the Joule-thompson cyclical thing, as that seems to be what I'm after.  If I have a closed loop, maintaining a high pressure in area A and and a low pressure in area B, using at least two pumps (pumping to keep the low pressure flow going and then back into the high pressure area), would that cause area A to be hotter and area B to be cooler?  Or am I being way too naïve here?

@user2765977 - There are several good suggestions here, and I'll make one that really adds nothing but is pretty neat and you could probably get away with including it given artistic license: check out pressure swing adsorption, which is used commercially to separate gas components from air.

@user2765977 A gas is being compressed, warms up by compression and is let  to cool down back to the surrounding temperature. Then it is being expanded through throttling ventil to low pressure (kept by a pump ) what cools it down. A part of cold gas leaves the system for the final usage (or is used as an input for the next unit in case of multi unit system. ), being replaced by input of warmer gas.

Thank you.  I actually already had a friend explain to me what you just commented (he works with 'heat exchange' stuff).  So when the timer is up, will post a variation of that as the answer I'm using and will credit you.

Poutnik's answers led to the result I desired, his explanation was just a bit too technical for a layman like me.  Basically I will use gas expansion for the cooling, that is where the gas is doing the work.
A closed loop, force pumps gas into a high pressure area (low volume) and then uses a sort of valve to let the gas expand into the high volume area.  the pump (intake) and valve (outtake) work in tandem to keep the expansion flow going and causes cooling.
Thanks to all that helped.

This is SCIENCE FICTION you have an opportunity to invent a completely impossible method to cool something that violates all the Laws of Thermodynamics In such a convoluted manner that you probably could apply for a patent on it and you are asking us for help to make it rigorous? Shame!


text:
Synthetic Testosterone Laboratory
I can't seem to find any research articles stating how testosterone is made synthetically in a lab. I am looking for research papers on how anabolic agents (synthetic testosterone) are made in a lab?

Maybe a quick “don’t try this at home” or “don’t use your own product” disclaimer would be nice

Interesting is, the question is not about pharmaceutical production and hormonal treatment, but it is about lab production of anabolic agents. You do not ask about a doping cookbook, do you?

From Wikipedia:

The chemical synthesis of testosterone from cholesterol was achieved in August [1935] by Butenandt and Hanisch. Only a week later, the Ciba group in Zurich, Leopold Ruzicka (1887–1976) and A. Wettstein, published their synthesis of testosterone. These independent partial syntheses of testosterone from a cholesterol base earned both Butenandt and Ruzicka the joint 1939 Nobel Prize in Chemistry.

Both original papers are in German. I don't know if these have industrial relevance nowadays. Might be better to ask on Chemistry SE that.
There's a somewhat fully 2009 article on Pfizer's "2nd/green generation" process. They use androstenedione as the immediate precursor. (The "green" claim seems mainly because of no longer producing heavy metal byproducts.) As they mention both "bioprocess" and the Kalamazoo, Michigan plant in there, which was acquired from Upjohn way back, the androstenedione might be derived from plant sterols (which was Upjohn's big innovation in the 70's), but that's not made explicit in that Pfizer article. You can read about the bioprocess route e.g. in a recent review. It's called a bioprocess because conversion from plant sterols is rather inefficient done by chemical means, so microbes are used. There's another paper/review which talks about that:

For example, the steroid hormone testosterone (TS) is chemically synthesized from the steroidal intermediate 4-androstene-3,17-dione (AD), which is previously obtained from natural sterols by microbial biotransformation (Fernández-Cabezón et al., 2017a).

For details on the chemical conversion from AD, you could probably look at Ercoli and de Ruggieri (1953). More recent papers (like Fernández-Cabezón et al.) propose to use biosynthesis in this step as well. Pfizer's article only speaks of a "proprietary intermediate" being used in this step.
Note that due to extremely short half-life of testosterone in the blood (like 10 mins), medical products use esterified testosterone.

A google search on "synthesis of testosterone" yields this pdf document which describes the steps for making testosterone from cholesterol without experimental conditions (did you want conditions?)  https://www.lkouniv.ac.in/site/writereaddata/siteContent/202003291608409347arun_sethi_steroid.pdf

@Waylander: why are you asking me? I didn't ask the Q. I tried to answer mostly with how it's made in industry nowadays because the Q was posted to Medical Sciences when I answered it. (My answer was migrated together with the Q.)


text:
How to dissolve nickel layer from chromium part
I have a part made of chromium. The part was plated with nickel. How can I dissolve nickel coat from that part?
I knew that nickel could be dissolved in hydrochloric, sulfuric, and dilute nitric acid. But chromium has same properties of dissolution in the same acids. Is there a way to remove the nickel coat without harm to the chromium-made part?

I am afraid chemical ways would have lack of selectivity and the only way is non chemical, e.g. combination of mechanical (peeling off?) and thermal ( different dilation stress?) methods, which may or may not be applicable. Perhaps @blacksmith37 as the respected expert could have some insight here. // Generally, is not it quite unusual, chrome being plated by nickel ? As chrome is like always the top layer.

@Poutnik thank you for fast replay. Yes, it's unusual composition of metals. I have chrome plated with nickel. Usually is nickel plated with chrome.

It seems unlikely that a chemical method will work. Given that pure chromium is very hard and nickel is softer, maybe sandblasting would do the job. Good luck with this!

You do not have a part made of chromium. Maybe Vitalium, a chromium cobalt alloy used for dental protheses, gas turbine blades and other stuff under different names like Stellite. It would be very, very unusual for Vitalium to be nickel plated

For reference, Vitallium is an alloy of 65% cobalt, 30% chromium, 5% molybdenum, and other substances  (but 0% left, so probably some very minor components) // See also Stellite

I meant Stellite 21, but could not remember the number; very similar to Vitallium.

No 21 is mentioned on the Wikipedia page.. See also specialmetals.ir: stellite 21.pdf


text:
Intuitive explanation of why electrode potential of a metal- metal insoluble salt- anion half cell is same as metal-metal ion cell?
My physical chemistry professor in his electrochemistry class derived before us, the formula of the electrode potential for a metal- metal insoluble salt- anion half cell and after a few minutes of putting formulas here and there he finally proved before us that the electrode potential for a metal- metal ion electrode and metal- metal insoluble salt-anion electrode is/comes out to be the same, the formula appeals to me, however, I don't quite get the intuition and the reason behind the same.
Is it because the anion plays a dormant/ insignificant role in the process, if so then why to take it in the first place?

The excess of anion ( like $\ce{Cl-}$ ) determines via the solubility product ( like $K_\mathrm{sp}(\ce{AgCl})  = [\ce{Ag+}][\ce{Cl-}]$ ) the concentration of the metal cation ( like $\ce{Ag+}$ ).
So for the silver chloride reference electrode:
$$\begin{align}
E(\ce{Ag/Ag+})&=E^{\circ}(\ce{Ag/Ag+}) + 0.059 \log {([\ce{Ag+}])} \\
&= E^{\circ}(\ce{Ag/Ag+}) + 0.059 \log {(K_\mathrm{sp}(\ce{AgCl}))}  - 0.059 \log {([\ce{Cl-}])} \\
&= E^{\circ}(\ce{Ag/AgCl/Cl-}) - 0.059 \log {([\ce{Cl-}])}
\end{align}$$


text:
Why are alpha hydroxy acids stable?
We were told by our teacher that 4 hydroxyl groups on same carbon atom results in formation of carbon dioxide by removal of 2 water molecules and he also told that 3 hydroxyl groups on the same Carbon results in formation of carboxylic acid.
So now the carbon with 4 hydroxyl groups can be thought of as a carbon with an acid group and a hydroxy group (which gives us carbon dioxide).
But today he told us that alpha hydroxy acids( comp. with a carbon atom with both acid group and hydroxy group) do not decompose into Carbon dioxide but gave no reason.
Why is this true ?

There's hardly any comparison between ortho acids and alpha hydroxy acids

Still, if you need some logic, then carbocation formed as intermediate in elimination of water in ortho acids is way more stable.

The gist is that 'an acid group' (COOH) implicitly includes a carbon atom. So a different carbon connected to an acid group and a hydroxy group is actually $\ce{R-CHOH-COOH}$ and does not decompose to $\ce{CO2}$.

Simply put, an α-hydroxy acid does not have multiple functional groups on the same carbon atom. It has a carboxyl group centered on one carbon and an hydroxyl group on an adjacent carbon; the latter is the α in "α-hydroxy acid". If the additional hydroxyl group were two carbon atoms away would be a β-hydroxy acid, three carbons away would be a γ-hydroxy acid, and so on. The term ω-hydroxy acid is sometimes used to describe a molecule with a chain of carbon atoms having carboxyl at one end and hydroxyl at the other.
Hydroxy acids are not completely inert. Having both a carboxyl function and an additional hydroxyl function means they can form esters with themselves, which may lead to ringed or polymeric structures. In the case of α-hydroxy acids the result is polymers that may have generated templates for the creation of early organisms.


text:
Nanoscale hydrogen batteries that use water-splitting technology
It was claimed in this article in 2019 that a nano-battery using water splitting technology is developed. It is described as:

The battery gains its charge by interacting with water molecules present in the surrounding air. When a water molecule comes in contact with the reactive, outer metal section of the battery, it is split into its constituent parts — one molecule of oxygen and two of hydrogen. The hydrogen molecules become trapped inside the battery and can be stored until they are ready to be used. In this state, the battery is "charged." To release the charge, the reaction reverses. The hydrogen molecules move back through the reactive metal section of the battery and combine with oxygen in the surrounding air.

There are hardy any details on the article regarding the mechanism of this reaction and I wonder what the material of “the reactive, outer metal section” is. And I have found no paper published with this research. The description makes this battery sounds like a perpetual machine: splitting waters from environment into oxygen and hydrogen, and make hydrogen and oxygen react to create electricity. What would be a good explanation of the mechanisms that makes this possible?  I guess there must be other things going on to keep this described mechanism going, which limits the life of this battery.
It is also claimed that:

The batteries have also demonstrated a power density that is two orders of magnitude greater than most currently used batteries.

But no experimental data is provided (since there’s no paper found).
Also no application of this battery has been recorded. I wonder what is the water-splitting mechanism? And what is the proper material applied in the reactive metal section of this battery to split the water molecule? If it is really this amazing I believe it can be applied inside human bodies to split water and power devices as well, since both nanoscale and larger scale such batteries can be made. But is such technology really practical and applicable now?

You will have look up their original research articles. The essays in your link are written by journalists where the main interest is to create curiousity or present a higly simplied version to public about something complicated. No wonder there is no scientific insight or content there.

@MFarooq Yes and that’s exactly what I did. But as I stated in the question, I found no paper related to this claimed result at all. That why I am asking the question.

I just saw the last lines "The Laboratory team's project will be the first to apply the technique for creating batteries, and at much smaller scales.
The project was funded via the Laboratory's Technology Office Energy Initiative and has entered into phase two of development, which includes optimizing the batteries further and integrating them with sensors."

So this is all imaginary so far. Nothing concrete. They will work on this idea.

Maybe... but it also says “ The batteries have also demonstrated a power density that is two orders of magnitude greater than most currently used batteries.” Sound like they’ve built a prototype already.

Looks and smells like one of those ‘all hat and no cattle’ things. Could be a tight race between these cells and commercial fusion power.


text:
How to make a steady supply of water from a supply of H2 and O2
writing a survival novel where a guy needs to survive and needs to make water to do so.  It is on another planet and he access to sci-fi type manufacturing machines that can make most tings.
He has a supply of O2 and a supply of H2, lets assume at a reasonable temperature.
My understanding is that you mix the two together (preferrably two to one) and ignite it and this produces water in a sort of chain reaction.
Assuming this is done in an inclosed space, does all the H2 and O2 turn into water? Is the chamber left in a vacuum state or close to it after?  There is a lot of heat involved in this process I believe, does that mean the water comes out in the form of steam, or does something about this process end up cooling it into water droplets?
I figure I need to get the answers to some of these questions to be able go about creating a believable scenario.  BTW, please forgive me ahead of time, my chemistry knowledge is basically non-existent, I got what little I know here from googling.

As EdV mentioned in the comments, not premixing the gasses and them suppying them to an intense flame, will get the desired results without any unwanted explosions.  The result will be steam and a lot of heat.

The fuel cell might be the way to go, as he'd get water and electricity in the process...

@ToddMinehardt Thanks, but far too complicated for my reader base.  I need a much simpler technique.

You would burn the two using a ‘glassblower’s’ torch, so the hydrogen and oxygen gases are not pre-mixed and you cannot have an explosion. You get steam and heat and then condense the steam to liquid water.

Recycle urine like the astronauts. Ed V has already given the answer.

In case you are wondering, a glassblower’s torch is a type of total consumption burner. By not pre-mixing, even explosive gas combinations, e.g., hydrogen and fluorine, could be burned. And other exotics, like oxygen and cyanogen. But exotic high temperature flames were made obsolete by argon inductively coupled plasmas, etc.

In case you want to further simplify things:   You don't even need purified oxygen.  Just burn the hydrogen with air (which can also be done with a glassblower's torch), using the right ratio of air to hydrogen needed to get complete combustion (about 2.5:1).  The key differences are that it would burn at a lower temperature, and you might get some nitrogen oxides (from the nitrogen in the air) that (depending on the concentration and toxicity) you might need to allow to vent.

@ToddMinehardt  you deleted my proposed answer, which is fair given you guys are the experts.  But that said, I got enough of what I wanted from this question and am unlikely to come back to it.  So not sure what you want to do here with the question, since apparently I wouldn't know what counted as a legitimate answer if it bit me in the face :)

Conversion of self-answers to selfcriticizing/clarification-requesting comments does not make much sense.

I didn't mean anything by it, was just wondering what should be done with the question and any potential answer.  Given that I might not be unqualified to even no what the correct answer is.  Should I just then ignore this question further and leave it 'unanswered'?  Should I 'transfer ownership' of the question?  (not even sure that's a thing).

@user2765977 It was flagged as such and I almost always defer to the members of the community here. I have undeleted your answer; I encourage you to accept your answer (as you may know, that is a thing) and please don't read anything into it. We welcome you and your presence here so don't be put off.

Hasn't this already be done in fiction? Check The Martian by Andy Weir (or the movie).

As EdV mentioned in the comments, not premixing the gasses and them suppying them to an intense flame, will get the desired results without any unwanted explosions.  The result will be steam and a lot of heat.

Might as well use the steam to run a steam engine or water pump or grain mill, etc. Lots of possibilities.

Might be useful in your current or future writing: https://en.wikipedia.org/wiki/Timeline_of_hydrogen_technologies.


text:
Does Sodium Precipitate with anything to find the level when titrating?
I am trying to find sodium in animal feeds. I already have a procedure to find chloride by using Mercuric Nitrate to form Mercuric Chloride using a titration method but am having a difficult time finding a procedure to precipitate Sodium with another element using a titration method. I know I could reverse calculate to Salt and then Sodium but with animal feeds there can be more than just salt sodium in the sample. Any ideas would be much appreciated!

Is there a reason not to use analytical instrumentation, e.g., atomic absorption spectroscopy or ion selective electrodes? I know titration is well-loved, but sodium compounds are usually soluble.

See https://www.researchgate.net/post/Is-there-any-method-for-determination-of-sulfate-sodium-by-titration for some suggestions on insoluble sodium compounds, though, as mentioned, this is not done routinely because of lack of specificity, expense, etc.

Mercury chloride isn't a good choice either. It's not the healthiest of options and it's quite soluble (in water). Why not use silver nitrate?

@DrMoishePippik Very interesting link to insoluble sodium compounds. Personally, I have always avoided the crown ethers because of the testicular atrophy issue. The sodium zinc uranyl acetate compound is neat: the uranyl ions could be detected subsequently via fluorescence.

Ed V, I did not know crown ethers were so horrible. Thankfully, I don't belong to the era when uranium salts were used to precipitate sodium in all standard labs :-) This test is to be found in most books of 60-70s.

Sodium is the easiest element to be determined by very economical instruments called flame photometers. There are no titrimetric methods for sodium as far as I know. Only gravimetric methods exist and they will require the skill of a highly trained analytical chemist! Consult the classic book called Applied Inorganic Analysis by Hillebrand. It should be available from Internet Archive.


text:
How can Spin-Orbit-Coupling allow interactions between states with different multiplicities?
I can understand how spin-orbit coupling generates the hyperfine structures that we see in textbooks but maybe my view that the L + S momenta interacting to generate a third J momentum is too simplistic. I can not understand how it is possible that SOC effects can allow the mixing of states with different multiplicities.
What is the role of spin-orbit coupling, let's say, when an excited |1/2> spin state in cobalt interacts with the ground |3/2> state or, as another example, on the intersystem-crossing?

In the presence of spin-orbit coupling, spin is not a good quantum number, but only an approximate one. This is because the Hamiltonian contains terms that are off diagonal in $S$ which become larger as the spin-orbit interaction increases, resulting in a mixing of your basis functions of different spin multiplicity. In your example, the excited state gets some character of the same spin multiplicity as the ground state so that radiative decay becomes possible and the excited state lifetime is reduced.

This is almost the same question as https://chemistry.stackexchange.com/q/50436/16683. I'm about to post an answer here, and you'll see that it will turn out to be kind of a carbon copy of my answer there, except that here I'm using braket notation and there I wrote out the integrals explicitly.

Here's a longer sketch of what Paul's comment is describing.
In the absence of spin–orbit coupling (SOC), the available states have well-defined spin and orbital angular momentum components which can be separated. We could, for example, write the total wavefunction as a product of spin and orbital components:
$$|\psi\rangle = |\chi_{\text{spin}}\phi_{\text{orbital}}\rangle \cong |\chi_{\text{spin}}\rangle \otimes |\phi_{\text{orbital}}\rangle. \tag{1}$$
Each spin part $|\chi_{\text{spin}}\rangle$ has a well-defined value of $S$, and each orbital part $|\phi_{\text{orbital}}\rangle$ has a well-defined value of $L$. So the total state $|\psi\rangle$ has well-defined values of both $S$ and $L$.
When a system is placed under some kind of perturbation $H'$, the transition probability per unit time (i.e. how likely the system is to go from state $i$ to state $j$) is proportional to
$$| \langle i | H' | j \rangle |^2  \tag{2}$$
(this is part of Fermi's golden rule). For radiative transitions, the perturbation $H'$ is proportional to the position operator $\vec{r} = (x, y, z)$ (this integral is then called the transition dipole moment). It's not important to actually bother with what this exactly is. The important bit is that this operator depends only on spatial components (it's independent of spin). So, the matrix element in eq. $(2)$ can be rewritten as
$$\begin{align}
\langle i | H' | j \rangle &= \left< \chi_\text{spin}^{(i)}\phi_\text{orbital}^{(i)} \middle| H' \middle| \chi_\text{spin}^{(j)}\phi_\text{orbital}^{(j)} \right> \tag{3} \\[5pt]
&= \left< \chi_\text{spin}^{(i)} \middle| \chi_\text{spin}^{(j)} \right> \left< \phi_\text{orbital}^{(i)} \middle| H' \middle| \phi_\text{orbital}^{(j)} \right> \tag{4} \\
\end{align}$$
Now, notice that the spin part of this integral, $\langle \chi_\text{spin}^{(i)} | \chi_\text{spin}^{(j)} \rangle$, is zero unless the two spin states are the same (because spin states are orthogonal). That means that the matrix element, and hence the transition probability, between states of different spin is zero.

Now, when you turn on SOC, the available states no longer have well-defined $S$ and $L$. In general, they will be linear combinations of different $S$ and $L$:
$$|\psi\rangle = c_1\left|\chi_{\text{spin}}^{(1)}\phi_{\text{orbital}}^{(1)}\right> + c_2\left|\chi_{\text{spin}}^{(2)}\phi_{\text{orbital}}^{(2)}\right> +  \cdots + c_n\left|\chi_{\text{spin}}^{(n)}\phi_{\text{orbital}}^{(n)}\right> + \cdots \tag{5}$$
Crucially, there is no guarantee that the spin components $|\chi_{\text{spin}}^{(n)}\rangle$ all have the same value of $S$. You may mix together singlets, triplets, ... all into the same state.
When you calculate the matrix element $\langle i | H' | j \rangle$ between these states, it's therefore no longer guaranteed that it goes to zero. It's not that the spin selection rule is no longer obeyed; it's more like all the states have a little bit of different-spin-character mixed into them, so that transitions between each pair of states is no longer strictly forbidden.
The slightly confusing part arises because often, the SOC is a fairly weak phenomenon. That means that the extent of mixing is pretty small, or mathematically, one of the coefficients in the linear expansion $(5)$ is much greater than the others, such that
$$|\psi\rangle \approx c_1\left|\chi_{\text{spin}}^{(1)}\phi_{\text{orbital}}^{(1)}\right> + \text{(small contributions from other states)} \tag{6}$$
In this case we can say that the quantum number $S$ is almost well-defined, or that it is an almost good quantum number; and we can label the states as if they did have a well-defined value of $S$. This is the case with atomic term symbols, for example.
However, just because those other contributions are small doesn't mean that they are zero. So, for example, a state that is almost $S = 1$ can have a bit of $S = 0$ character mixed in; that means that the transition between this state, and a second state that is predominantly $S = 0$, is not entirely forbidden.
This mixing is caused by SOC, and hence SOC is precisely the mechanism which allows transitions between states which nominally have different values of $S$. (In actual fact, it is more accurate to say that $S$ is not perfectly well-defined for these states.)

(Many quantum mechanics textbooks cover angular momentum, and angular momentum coupling, in great detail. Normally these are presented using $j_1$ and $j_2$ as the "sources of angular momentum", but in fact SOC is precisely the same thing, just that we use $s$ and $l$ as the labels instead. So if you want a more in-depth treatment of this, you can look in any reputable textbook.)

Thank you so much for such complete answer, it was extremely helpful.


text:
Where does the free electron from from in Ag/AgCl reaction?

For the Ag/AgCl half reaction, where did the free electron that AgCl captured come from? Did it come from the Ag wire? If so, why would Ag wire give an electron to AgCl, both atoms are Ag and have the same electronegativities, right? Also, what's the purpose of the KCl(aq) solution, because I don't see where any of it comes into play.

The $\ce{Ag/AgCl/Cl-}$ electrode is basically ordinary $\ce{Ag/Ag+}$ electrode. The difference is that the concentration of $\ce{Ag+}$ is kept very low by presence of $\ce{AgCl(s)}$ and excess of $\ce{Cl-}$ ( from $\ce{KCl}$ ) due $\ce{AgCl}$ solubility product $K_\mathrm{sp}=\ce{[Ag+][Cl-]}$.
By other words, electrons come from the half reaction $\ce{Ag(s) -> Ag+(aq) + e-}$,
linked to the equilibrium reaction $\ce{Ag+(aq) + Cl-(aq) <=> AgCl(s)}$.

Yosimba2000 asks about the origin of the free electron that AgCl captured. It does not come from some part of this Ag/AgCl electrode. It comes from another electrode. Matter of fact an electrode does not work alone. It has to be coupled with another electrode, in order to produce a cell, whose performance is measured by a potential difference. This other electrode can be any reducing agent like Zinc or Hydrogen dipped in aqueous solution of its ion. The electron is created by an equation like $\ce{Zn-> Zn^{2+} + 2 e-}$ or $\ce{H2 -> 2H^+ + 2 e-}$

I am not so sure he asks about this.

Well, I am pretty sure he is looking for such an explanation. Let's wait his reaction

As I read the OP’s question, I wondered of they were not realizing that a reduction half-reaction was not the same as a properly balanced “full” reaction with no electrons loitering about on either side of the reaction. So I upvoted this answer because I think it addresses the OP’s question. I also upvoted Poutnik’s answer because the two answers wrap it up well.


text:
How to get smiles from pubchem CID
I have downloaded about 10k compounds of interest from pubchem. Im using pubchempy to get smiles for all those structures . I am able to get smiles for each compound one by one but while fetching two many compounds
from pubchempy import Compound
def get_smiles(cid):
smiles = []
mol = Compound.from_cid(cid)
if mol:
    smi = mol.canonical_smiles
    #smiles.append(smi)
    return smi
else:
    return None

i get error
''return opener.open(url, data, timeout)'' and
<urlopen error [Errno 101] Network is unreachable>
how can i get rid of this error

python.stackexchange.com

TL,DR: Assuming you still are connected to NIH, possibly a constrain by the database's rules of access.
I don't understand much why you want to re-define the generation of pubchempy's report of SMILES (which needn't be canonical ones). For example,
import pubchempy as pcp
c = pcp.Compound.from_cid(5090)
print(c.isomeric_smiles)

yields CS(=O)(=O)C1=CC=C(C=C1)C2=C(C(=O)OC2)C3=CC=CC=C3 (condensed from the documentation) works well (just checked with Python 3.9.8).  Of course the above may be extended to query lists of compounds, too.  Not for beauty, e.g.
import pubchempy as pcp
compounds = [5090, 5091, 5092]
for entry in compounds:
    c = pcp.Compound.from_cid(entry)
    print(c.isomeric_smiles)

reports
CS(=O)(=O)C1=CC=C(C=C1)C2=C(C(=O)OC2)C3=CC=CC=C3
CCCC(=O)OC1C(OC(CC1(C)OC(=O)CC)OC2C(OC(C(C2N(C)C)O)OC3C(CC(C(C=CC=CCC(OC(=O)CC(C3OC)O)C)O)C)CC=O)C)C
COC1=C(C=C(C=C1)C2CC(=O)NC2)OC3CCCC3

However, pubchempy has to follow NIH's rules of access like any other mode of access, i.e.

No more than 5 requests per second.
No more than 400 requests per minute.
No longer than 300 second running time per minute.

NIH's documentation, programmatic access

regardless if using NIH's PUG (with time out after $\pu{30 s}$ server time, ref), or an interface like pubchempy which recommends e.g., grouping the requests in smaller portions as strategy.

Thank you for the Clarification it helped me a lot!