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How can a system that has phases of different compositions be in equilibrium?
I understand determining conditions of equilibrium for a unary system; each of the separate phases can be treated as a one-phase system with its own set of extensive and intensive properties. However, I am having difficulty understanding multi-component systems. Despite the addition of more components, would it be the similar to a unary system, in that for the system to come to equilibrium, the temperature, the pressure, the chemical potential of each component must be the same in all of the phases?

See: Gibbs Phase Rule

Yes, that is the requirement for thermodynamic. equilibrium of a multicomponent multiphase system.


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Why does Raoult's Law have a weighted sum instead of a direct sum?
I'm fully aware of the various questions asked about Raoult's Law, and I'm asking this after having gone through all of them and not finding a suitable answer to my question.
I have studied, that at a certain temperature, the equilibrium vapour pressure of a liquid remains constant, irrespective of the amount of liquid taken. From this, I conclude that if I take a mixture of two liquids, the equilibrium vapour pressure should just be the sum of their individual equilibrium vapour pressures, since both liquids attain equilibrium when the partial pressures of their vapours are equal to their equilibrium vapour pressures. However, Raoult's Law states that the vapour pressure depends on the amount of each liquid in the solution.
This question almost answered my question, but I do not understand why the concentration of the liquid matters when it's in a solution. I understand that this is not due to any intermolecular forces, since the mixture in consideration is assumed to be ideal.
We had a long discussion in class about this, and in the end we concluded that the equilibrium vapour pressure would depend on the surface area of the liquid, and in a mixture the surface is not completely available to a single component of the mixture. I'm not sure if this explanation is correct, and I would like your opinions about this.
This question talks about a different explanation, but I believe it talks about solutions of solids in liquids. Is a similar explanation valid for mixtures of liquids? I can't seem to get my head around the explanation that the liquid molecules occupy a fraction of the surface which is equal to their mole fraction, which reduces the equilibrium vapour pressure of each component of the mixture.

When a liquid is in solution, the rate of its escaping molecules per unit surface area is decreased. But the rate of its condensing molecules, for the given vapour pressure and the unit surface area, is not decreased. The result is that at equilibrium, the vapour pressure is lower.
Therefore, the total equilibrium vapour pressure is weighted sum of equilibrium vapour pressures of pure liquids, where the weights are molar fractions.
The direct sum of partial pressures of pure liquids is applicable, if they are kept separated or if they are not miscible (like water and essential plant oils).

Why does the rate of escaping molecules decrease?

Imagine students running out of school. Does the rate of boys running out depend on percentage of boys? Is this rate the same if it is boys only school as well as if there is just 10% of boys? Molecules do not escape 10 times faster after realization there is 10 times less of them.

Of course, the area argument is only a starting point for understanding—it's highly simplistic. Vapor pressure lowering is, as you wrote, dependent on mole fraction.  Suppose you have two solvents, A and B, where the surface area of A is twice that of B.  Then in a 1:1 mixtue of A and B, the fractions of surface taken up by A and B are 2/3 and 1/3, respectively, rather than 1/2 each.  Thus, by the area argument, the vapor pressures of A and B would be reduced to 2/3 and 1/3 of their pure vapor pressures, respectively.  But that's incorrect.  Ignoring non-ideality, they're each reduced to 1/2.

@theorist by surface area of A being twice that of B, do you mean the surface area of the molecule? In that case, even if the molecules of A were to occupy twice the surface occupied by the molecules of B, the number of molecules of A and B at the surface would be equal, right?

@UmeshKonduru  That's right, I was referring to the relative surface areas of the individual molecules.  [More exactly, to the relative surface areas the molecules take up at the gas-liquid interface, which could actually be different from their relative surface areas, e.g., if they are elongated and oriented vertically—but that additional complication is not needed to demonstrate my point.]  And yes, assuming ideality, the relative numbers of A and B molecules at the surface would equal their relative numbers in the bulk solution which, in my example, was 1:1.

@UmeshKonduru Ultimately what drives this is the entropy of mixing of the two liquids, which lowers their chemical potentials and thus their vapor pressures.  Poutnik knows this, but it's difficult to give a mechanistic explanation of a statistical, emergent property like entropy, so the area argument is often given to students before they are ready to study this at a more advanced level, which is why Poutnik provided it (plus you were asking about area arguments).  So it's fine, but it's also nice to know that it is a simplistic picture.

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Aromaticity of 1,4-Dioxin
Structure of 1,4-dioxin is:

Wikipedia page says that it is non-aromatic.
Pubchem page shows its structure is planar.
In a planar structure, obviously there will be 8 (= $4n$) pi electrons in conjugation, making it antiaromatic, contrary to what Wikipedia says.
Also, I feel the molecule has a decent chance to break antiaromaticity by becoming non-planar i.e. if the left half of the molecule bends at an angle to the right half, like a semi-open book. That is a structure which would preserve the planarity of individual $\mathrm{sp^2}$ carbons, while allowing oxygen atoms to have a bond angle corresponding to $\mathrm{sp^3}$ hybridization.
Where is my reasoning wrong? What is the actual nature and structure of this molecule? References would also be appreciated.

https://chemistry.stackexchange.com/questions/44743/aromaticity-of-1-1-bicycloprop-2-en-1-ylidene-and-1-4-dioxine

Aromaticity is a fuzzy subject and there are no strict mathematical rules. There is no recognised unilateral theory that does explain it. This becomes a real issue when we are dealing with compounds that are generally hard to describe. One such compound is 1,4-dioxin.
This makes it a very unfortunate battle ground for dogma, which the answer provided here and comments following are proof of. You cannot draw a line and multiple textbooks have published and teached a romanticised version of this topic. It is way more complex than drawing a molecule and counting π-electrons (which strictly do not exist but are rather electrons in π-orbitals).
The Gold Book defines aromaticity (https://doi.org/10.1351/goldbook.A00442) this way:

The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.

Please note that there is no mentioning of the Hückel (4n + 2) rule and no mentioning of antiaromaticity.
The IUPAC Gold Book gives us another definition under the term aromatic (https://doi.org/10.1351/goldbook.A00441):

In the traditional sense, 'having a chemistry typified by benzene'.
A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1HNMR spectrum.
See also: Hückel (4n + 2) rule, Möbius aromaticity
The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition states. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Finally, we need to look at the definition of the Hückel (4n + 2) rule (https://doi.org/10.1351/goldbook.H02867):

Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system. [...]
Systems containing 4n π-electrons (such as cyclobutadiene and the cyclopentadienyl cation) are 'antiaromatic'.

Unfortunately this definition also has some issues (especially the chosen terminology which I consider wrong), but I do not want to go deeper into that.
However, it must be noted that from a purely technical point of view the last sentence is misleading. Cyclobutadiene as it exists is not an antiaromatic molecule, because it avoids this destabilisation by distorting. So there is a conundrum.
For that reason, it is safer to ignore this sentence.

On the DF-B97D3/def2-SVP level of theory (and all others I have checked), 1,4-dioxin is planar. Also for all intents and purposes it does not matter whether it is perfectly planar or slightly distorted. Therefore, for the sake of this argument it is planar.

In this special case, it has D2h symmetry.
The molecule is a monocycle.
Let's have a look at the π-orbitals. In the following the blue/orange ones are doubly occupied, the red/yellow are virtual orbitals.

We can see that (in this framework) all atoms are trigonally coordinated (or bent). We can count the occupied orbitals, 4, and hence have 8 π-electrons.
From this alone, applying only the definitions, the molecule must be antiaromatic.
It isn't though.
The problem with antiaromaticity arises that two (or more) degenerate orbitals are singly occupied in such hypothetical structures and that there is at least another structure that is lower in energy that doesn't have this problem.
Let's take cyclobutadiene. If all carbon carbon bonds were the same, it would have D4h symmetry and there are degenerate orbitals half occupied. A structure with different bond lengths doesn't have resonance stabilisation, but it also doesn't suffer from the degeneracy issue. This rectangle-like structure is lower in energy than the hypothetical one that is 'antiaromatic'.
For 1,4-dioxin we do not have this problem. The orbitals are not degenerate. The two oxygen break symmetry enough so that fürther distortion is not necessary.
Here are the energies of the orbitals shown above:
25              1.135267952     Unoccupied      B2G
23              -1.097443828    Unoccupied      AU
22              -4.154667088    Occupied        B3U
21              -6.911474284    Occupied        B1G
19              -9.230446836    Occupied        B2G
16              -10.846271644   Occupied        B3U

Let's circle back to Hückel. This rule is best understood as a guideline to explain aromaticity after the fact. It is not a good predictive tool. Heteroatoms were not included when this rule was conceived.
Most aromatic molecules do not follow Hückel's (4n + 2) rule, but are still considered aromatic.
If an experiment says a compound lacks the characteristics of an aromatic molecule, it isn't. I do not have access to the source Wikipedia cites, but it appears to be a study on exactly that issue.

Thus is a nice answer cleaning the point that the application of the Huckel rule shall be applied when possible, unless extension which are not available by a simple glance at the chemical structure sketch are implemented. The only criticism I have, is that you always insist on the fact that antiaromatuc compounds do not exist. I am aware that you (and sometimes I myself) do so to make clear a point often misunderstood. But when we say so, likely we propagate the confusion. An antiaromatic compound exists, if it does, by being not aromatic. Which is what the term ultimately means.

I have to add that at the end that the message isn't that different from what I was conveying ultimately conveying in my DELETED by Mods, answer (I can't read the comment that was placed concomitantly). No simple Huckel rule shall be applied to the molecule in case. And it does not clear the misconception in which OP imagined a change in hybridisation as a mechanism to avoid aromaticity when needed. That never happens. If instead of attacking my word "hybridised atom" you would have start with something so sensible as the above ans all the discussion won't have even started.

Hello! Martin, do you have access to these papers https://doi.org/10.1002/hlca.19780610422  https://doi.org/10.1016/0368-2048(80)80007-7 ? If so, could you please reference to experimental evidence as well ?

@permeakra Unfortunately I don't have access to these. Otherwise I'd assume they would make a good addition here. If you have access, you can judge if they would stand alone as an answer. Usually experimental results do. Thanks for these recommendations nevertheless.

Assuming the structure is indeed planar*, the presence of two oxygen rather than carbon atoms as one opposing pair breaks the symmetry and therefore breaks the orbital degeneracy on which the 4n+2/4n rule is based.
[EDIT: If it is planar, the symmetry group is D2h, if non-planar it is C2v. Neither group has any degenerate irreducible representations.]
Thus if you had had all carbon atoms in the conjugated ring, with eight pi electrons you would have a pair of half-occupied, degenerate orbitals leading to antiaromaticity. With the oxygen atoms there one of these degenerate orbitals drops off in energy and then takes both electrons into a more stable state, so you avoid the antiaromaticity.
*The JSMol structure given in the WP Infobox shows the oxygen atoms very slightly out of the plane of the carbon atoms. Below is an eclipsed view taken from the Interactive Image:

What is a WP Infobox? And why would this data be any more authoritative than the planar structure of pubchem? | In any case, the degeneracy (of benzene) is broken, so even the angle doesn't matter. I'd guess the algorithms tries to make it planar but fails. My old xtb installation makes it reasonably planar.

It may be the the pubchem structure is the one I showed, actually. You may want to see if you can rotate it to generate my view.

Can you provide an explanation/reference where I can read up in detail as to why the presence of oxygen breaks the MO degeneracy and leads to double occupancy in one orbital in place of single in two?

I also checked the pubchem again, unfortunately it is exactly planar and not even slightly distorted.

@Martin-マーチン «WP Infobox» likely refers to the interactive structure representation in English edition of Wikipedia. See, e.g., about furan, second entry in the section «identifiers» with «3D model» pointing to https://chemapps.stolaf.edu/jmol/jmol.php?model=c1ccoc1. Possibly, this is being used for quite some time (and runs on a computationally constrained/affordable side of theory).  Robert Hanson opened a dedicated thread about this in Jmol's public user list a couple of days ago (entry: https://sourceforge.net/p/jmol/mailman/message/37393900/).

@OscarLanzi What is the origin of the interactive representation? Is the interactive model based on findings like X-ray diffraction, spectroscopy .or. by assembling atoms and subsequent running an optimization like e.g., UFF, or (since this entry is an organic molecule) MMFF94s in Jmol? The reason for is this uncertainty in my perception is this: orient the model so that $\ce{C-O-C}$ faces you (like above). Right-click by mouse should open a pull-down menu (at least in Linux Debian 12 running Firefox 91.4.0esr (64-bit)). Then select computation (22nd entry) -> optimize structure.

@OscarLanzi I'm on my way to illustrate this bent being ironed out in the chemistry.se chatroom, link: https://chat.stackexchange.com/transcript/message/59921065#59921065

Can you explain to me for which reason the oxygen electrons shall be considered in p orbitals to start? There is nothing to avoid in 1,4-dioxine, being it really planar or slightly out of plane. Why everyone in this thread seems to have difficulty seeing it as an ether?

@TRC Though not 1,4-dioxin but 2,3,7,8-tetrachloro-dibenzo-p-dioxin as an entry from the COD, this one appears pretty flat without distortion.  And this model is based on a X-ray diffraction study of the crystalline solid (https://pubs.acs.org/doi/abs/10.1021/ba-1973-0120.ch003).

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Is tetraoxidocobaltate(V) low-spin
After hearing that hexaamminecobalt(III) and hexafluoridonickelate(IV) are both low-spin due to the central metals' high oxidation states (even though both bear pi-donor ligands), I began to wonder if the percobaltate anion, i.e. tetraoxidocobaltate(V), is also low-spin. It is already known that even tetranorbonyliron(IV) is low-spin (compare: iron(III) is usually high spin even in octahedral environments), so maybe the high oxidation state of cobalt in tetraoxidocobaltate(V) can make an 'exception' to the rule that certain types (i.e. with pi-donors or tetrahedral coordination) of complexes are always high-spin.
Is my reasoning correct, and is there any general piece of data in the literature in this manner?

The answer, at least for potassium tetraoxidocolbate(V) in dilute tripotassium phosphate solid solution, appears to be high-spin despite the high oxidation state of the cobalt.
Brendel and Klemm[1] oxidized a 3:1 atomic mixture of potassium and cobalt (the metals being in the form of oxides) with $\ce{O2}$ and obtained a product solid-phase soluble in $\ce{K3PO4}$, which is rendered as the isostoichiometric and isostructural $\ce{K3CoO4}$. To assess the electronic spin state, the magnetic moment per cobalt atom was obtained as a function of composition of the solid solution:
$3.2$ a/o* $\ce{Co}\implies 5.1$ Bohr Magnetons
$8.1$ a/o $\ce{Co}\implies 4.3$ Bohr Magnetons
$56$ a/o $\ce{Co}\implies 2.6$ Bohr Magnetons
*a/o = atomic percent, $\ce{Co}/\ce{Co + P}$
The theoretical magnetic moment for four unpaired electrons, meaning high-spin in the tetrahedral $d^4$-core complex, is $\sqrt{24}\approx4.9$ Bohr Magnetons, and the most dilute solution is taken to agree with that figure. More concentrated cobaltate solutions appear sysceptible to antiferromagnetic coupling of cobalt atoms from neighboring $\ce{CoO4^{3-}}$ ions.
The high-spin outcome is unusual for such a highly oxidized metal core, but favored by several factors:

The complex is tetrahedral rather than octahedral, which cuts the cryatal field splitting by more than half. It also renders the low-spin configuration for a $d^4$ core with zero rather than two unpaired electrons, which feeds into the next factor.

Exchange energy is strongly more favorable for four rather than zero unpaired electrons. This effect is magnified by the quadratic dependence of exchange energy on unpaired electron count.

The oxide ion ligands are stronger pi donors than fluoride ligands, due to the greater negative ligand charge and higher orbital energies of oxide versus fluoride. As described here this tends to favor oxo complexes of high-oxidation-state transition elements with few $d$ electrins in their core, and cobalt(V) is a suitable case of this effect.

Reference

Brendel, C. and Klemm, W. (1963), Weitere Versuche zur Darstellung von Kaliumcobaltat (V). Z. anorg. allg. Chem., 320: 59-63. https://doi.org/10.1002/zaac.19633200109

In tetraoxidocobaltate(V), coordination no. of cobalt is 4, has d4 configuration, strong field ligand as Δt is higher than pairing energy, hence (e)2 (t2)2 splitting configuration so the hybridisation is dsp2 with square planar geometry.
The no. of unpaired electrons is four so the magnetic moment

μ = √24B.M.
Therefore it is a high spin complex.

Unfortunatly this does not seem very coherent. The answer begins with arguments that the complex would be low-spin (strong field ligand, high oxidation stare) but then claims the complex is high spin.

Please see Why it is wrong to use the concept of hybridisation for transition metal complexes

I understood that valence bond theory and crystal field theory cannot always completely explain some complexes but I still have to learn more about molecular orbital theory.

Valence bond theory can definitely explain any complex correctly, but it might not be convenient. Crystal field theory on the other hand is outdated and incomplete. Please don't lump the two together.

But CFT is a damned sight better than hybridisation


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P2O5 concentration % based on a 75% Phosphoric Acid Solution
I have a very poor chemical background, so please be indulgent in the case my question seems silly.
I need this clarification for a practical application:
If I want to obtain a P2O5 concentration of 4,7 % in my final solution, what % of Phoshoric Acid Solution  at 75% I need to put in?
My guess is about 8,7 % of 75% Phoshoric Acid Solution, is it that correct?

Why guess and not simple calculation? 1% P2O5 is equivalent 1.3807% H2PO4. The needed mass of H3PO4 solution is inversely proportional to its percentage. Needed volume of H3PO4 solution is equal to its mass and density ratio.

Congratulations ! Your calculation is correct. Here is the proof.
$1$ mole $\ce{P2O5}$ weighs $142$ g, and produces $2$ moles $\ce{H3PO4}$, which weighs $2·98 $ g = $196$ g $\ce{H3PO4}$. So pure phosphoric acid can be made with $142/196 = 72.4 $% $\ce{P2O5}$.
As a consequence, a solution containing $75$% phosphoric acid can be made with $0.75·0.72.4$ % = $54.3$ % $\ce{P2O5}$.
If now you want to dilute it to get a solution containing $4.7$% $\ce{P2O5}$, you have to take $4.7/54.3 = 8.65$% of your initial solution.
All these values are supposed to be made in mass, and not in volume. Let's transform these results into volumes.
In tables, $75$% $\ce{H3PO4}$ has a density of $1.579$. $\ce{H3PO4}$ coming from $4.7$ % $\ce{P2O5}$ contains $6.5$ % $\ce{H3PO4}$ and has a density of $1.034$. As a consequence, if you want to make $1$ liter containing $6.5$% $\ce{H3PO4}$, this liter weighs $1034$ g. And it should be done with $0.0865·1034 = 89.3$ g $\ce{H3PO4}$. And this mass of $\ce{H3PO4}$ is contained in a mass $m$ and a volume $V$ of phosphoric acid (density $1.579$). $m = 89.3$ g/$0.75 = 119$ g solution; and $V = 119 $g$/1.576$ g/mL = $75.5$ mL
Conclusion. To make exactly $1.000$ liter of the final solution, you have to measure $75.5$ mL of the initial $75$% solution, put it in a $1$ liter flask, then dilute with enough water so as to obtain exactly $1$ liter at the end.

OP beware of using a scale

@Alchimista. A scale ? What do you mean ?

I mean something to measure mass. This might be the case in which a perfectly correct answer can be wrong in practice. Unless OP use a scale, in fact.

Your calculations are for mass context. For volumes, you have to include density.


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Specifically how does the presence of moisture and oxygen affect the yield of ferrocene?
In a lab this week at uni we made ferrocene (I'm a second year Chemistry student). This was done by reacting nBuLi with cyclopentadiene (kept in freezer) to create the Cp anion in-situ.
To this mixture I mistakenly added iron(+3) chloride hexahydrate instead of anhydrous iron(+2) chloride, which was supposed to have been used later on to observe the oxidation of ferrocene to ferrocenium. I was told to continue with my synthesis, and I did make some ferrocene but only in mg amounts. I ask what was it in the mixture that reduced the iron?
In terms of my observations - as the hydrated salt was added a sludgy black precipitate began to form (solvent was THF) I cleaned the Schlenk tube at the end with washes of 6M HCl and small amounts of this solid dissolved with each wash and the solution turned green each time. Could someone tell me what this solid was?
Also could you suggest some of the effects of water and oxygen on the mixture? Why does the reaction need to be performed under inert atmosphere considering ferrocene is stable to the air? What are some of the unwanted reactions that would occur?

Butyllithium is an immensely powerful base. Any water present will instantly react with it so it will not be available to react with the cyclopentadiene. There is water in the air which is why you keep the system under dry nitrogen.

Okay, thanks! What effect would adding the hydrated salt in the second step have had, instead of something anhydrous - assuming the LiCp had already formed?

@Tilly you were supposed to add Iron (Ii), by the way...

The effect of adding the hydrated salt would be to regenerate the cyclopentadiene as the anion of cyclopentadiene will steal a proton from the water

The fact that the final product is stable to air and water doesn't matter. All the key ingredients are highly sensitive to water and air and will disappear in side reactions if you don't exclude the air and water.

@matt_back - Could you give some specifics about the side reactions that would occur? Thanks for your reply

@Alchimista - haha yeah I am aware. I realised about 3 minutes after adding it


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Polarity vs Acidic Strength (Alcohols and Inorganic acids)
If $\ce{HI}$ is more acidic than $\ce{HCl}$ because less polarity in $\ce{H-I}$ makes it easier to donate $\ce{H+}$ ion thus increasing its acidic strength, how does more polarity in $\ce{O-H}$ bond in primary alcohols (and also water) due to less electron releasing alkyl group translate to more acidic strength??

Less polarity alone does not mean more acidity. If it did, hydrocarbons would be very acidic.

If I recall correctly, the reason that H-I is a strong acid is that the valence orbitals of iodine are simply too large to overlap well with the small valence orbital of hydrogen.

I hold the view that the X-H bond needs to be polarized to some degree to make the hydrogen have some positive charge. Consider for a moment the hydrogen atoms bonded to ruthenium in RuH2(CO)(PCy3)3. These are very basic hydrides as they have partial negative charge chartacter and will act as Bronsted bases. While the hydrogen atom in HCo(CO)5 has a strong partial positive charge and acts as a Bronsted acid with great ease.


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What is the origin of the word salt as used in chemistry?
I have done a number of searches to determine why compounds of acids and bases are known as salts but came up empty. The fact that common salt dominates the word's usage means all my searches bring up the history of sodium chloride.
Is there a known origin of the word salt as applies to general compounds of acids and bases? Just like there is a well known origin of the word alkali.

This question better suits to SE site History of science and mathematics

https://www.etymonline.com/word/halo-

The origin of the word ${\it sal}$ in chemistry is Latin and is associated in general with things from the earth. Usage of the term dates to the pseudo-Aristotle's ${\it De\ Perfecto\ Magisterio}$ of mediaeval time; and is also apparent in Arabic alchemy, where one sees ${\it sal\ ammoniac}$ appearing in reference to "the eagle."
There are 3 individuals named as contributing significantly to chemistry: Libavius, Agricola, and Rudolph Glauber. Glauber, for example, contributed his interpretation of the story of the Golden Fleece to include mention of "Sal Mirabile," which is actually "Glauber's salt."
This is the foundation of the answer to your question,

Is there a known origin of the word salt as applies to general compounds of acids and bases?

Keep in mind that the use of the word 'salt' (in any language, with respect to chemistry) was put forward by those who kept using it. So to a large extent, one can say that it's always been done that way. On the other hand, the systematic naming of salts as a chemical class - and not a name - comes later.
In this answer, I will be taking material from:

M.P. Crosland. ${\it Historical\ Studies\ in\ the\ Language\ of\ Chemistry}$. Dover Publications, New York. 1978.

I can provide the references he uses if requested.
The problem was introducing a systematic nomenclature for describing substances because there was an ever-increasing number of them being discovered.
The general historical evolution starts with the fields of botany and chemistry at the beginning of the 18th century, where Linnaeus' botanical binomial nomenclature was used by Bergman toward the end of the century. Guyton de Morveau extended and refined Bergman's scheme and collaborated with Lavoisier to establish what we work with today.
In the 16th century, Paracelsus referred to potassium bisulphate (which is more soluble than potassium sulphate) as ${\it sal\ enixum}$ in the alchemical literature. Paracelsus was famous (in part) for putting forth the theory that metals are composed of "Mercury, Sulphur and Salt," arising in alchemy.
Later, in 1664, the French chemist Le Fubure published work in which he differentiated salts of nitre by crystal habit. Duhamel, in 1736, showed that the salts of potassium and sodium were distinguishable by their bases.
In any case, there was a pressing need to formalize a nomenclature. Torbern Bergman (1735-84) is credited with taking on the task of reform in the first orderly manner.  He introduced (via publication) his ideas towards a binomial system for naming, which was limited although a good start.
Guyton de Morveau expanded on Bergman's ideas and was able to systematically name 500 salts from 18 acids and 24 bases (earths, alkalis, and metals).
The big arrival was ${\it Méthode\ de\ nomenclature\ chimique}$ in 1787 and was authored by Guyton de Morveau, Lavoisier, Berthollet, and Fourcroy. It is at the point that the system for naming chemicals was in place. From here, the contributions of British, German, and other chemists was incorporated.
By the early 19th century, additions and revisions to the above were made and the systems of naming was largely in place. The evolution of naming things from the earth as 'salts' to naming things in a certain class as 'salts' largely took place during the last quarter of the 18th century.

You write  "Usage of the term dates to the pseudo-Aristotle's De Perfecto Magisterio of mediaeval time, and then write immediately after "There are 3 individuals named as contributing significantly to chemistry: Libavius, Agricola, and Rudolph Glauber."   Named where?  In De Perfecto Magisterio?  It couldn't have been there, because the latter was probably written in the late 1200's (see https://www.britannica.com/topic/The-Sum-of-Perfection-or-the-Perfect-Magistery), and Glauber was born in 1604.

@theorist Crosland, page 6 (pseudo-Aristotle's work referencing (among other things) sal ammoniac); Crosland references Libavius, Agricola, and Glauber on p. 9 in the context "Having mentioned the names of Libavius and Agricola in connection with the legend of the Golden Fleece, we might conclude this section by referring to an interpretation of this legend by a third great figure in the history of chemistry, Rudolph Glauber."

The word salt had had different meanings in alchemy and ancient chemistry (> 1000 years ago). Some of the meanings are somewhat unexpected as you will see below. When new researchers say all their searches came up empty, it often means they are searching the free web. The fact is that most of the professional and high quality science is still behind paywalls.
Salt meant the following, selected from the unabridged Oxford dictionary (behind paywall):

Alchemy and Old Chemistry. One of the supposed ultimate elements of
all substances. Obsolete.
Example: 1723   J. Clarke tr. Rohault's Syst. Nat. Philos. I. i. xx.
109   Hence they conclude, that these five Substances, viz. Mercury,
Phlegm, Sulphur, Salt and Caput mortuum..are the only and the true
Elements of all..mixed Bodies.

Old Chemistry. A solid soluble non-inflammable substance having a taste. Obsolete.

Finally, the modern meaning appears from 1779...

Modern Chemistry. A compound formed by the union of an acid radical
with a basic radical; an acid having the whole or part of its hydrogen
replaced by a metal. (In wider theoretical use the term ‘salt’
includes acids as salts of hydrogen.) Also, ethereal salt, an ester.

The first marked step towards the modern conception of a chemical salt
was Rouelle's definition (a1770) of a neutral salt as a compound
formed by the union of an acid with any substance serving as a base
for it and giving to it a concrete or solid form. Various
modifications of this or earlier views were put forward until the
publication of Lavoisier's definition of a salt as the union of an
acid with an oxide; this definition, however, was found to be
inadequate when the composition of the halogen compounds, sulphides,
etc. came to be accurately known. A further revolution in the theory
of salts was made by Berzelius, who divided them into two classes;
viz. haloid adj. and n.   salts, formed of an electropositive element
and a halogen, and amphide adj.   salts, resulting from the union of
an acid and a base; the latter he subdivided into oxy-salts,
sulpho-salts, selenio-salts, and telluri-salts. According to more
recent conceptions (Arrhenius 1888) salts, including acids, are
regarded as composed of positive ions or cations (hydrogen and metals)
and negative ions or anions (halogens and acid radicals).

Any chance you could include bibliographic details of the included references, eg Rouelle, Berzelius and Arrhenius?

Andrew, this will require severals hours of research. These works are in German and French. Interested researchers should start from A Source Book in Chemistry, 1400-1900
By Henry Marshall Leicester, Herbert S. Klickstein · 1952.

got it. I thought they were included in the work you quoted. No need to go to great lengths. Great answer even without them.

@Andrew, Here is the original paper that defind salts https://gallica.bnf.fr/ark:/12148/bpt6k3542h/f470.vertical

As a complement to M. Farooq explanations, I would add that in the beginning of the $19$th century a salt was a substance produced by the reaction of an acid on a hydroxyde, like the sea salt which can be obtained by the reaction $\ce{NaOH + HCl -> NaCl + H2O}$.
In general, acid and hydroxyde were always reacting according to the general equation : $$\ce{Acid + Hydroxyde -> Salt + Water}$$


text:
Can nickel(IV) be stabilised in water using appropriate ligands?
Platinum(IV) is perfectly stable in water as its hexachlorido complex, and so is palladium(IV). However, nickel(IV) is not stable in water even in its hexafluorido complex- the exchange-inertness of hexafluoridonickelate(IV) (which is low-spin despite the strong pi-donor property of fluoride; nickel(IV) is "just too high in oxidation state" to admit a high-spin configuration) does not reduce the strong oxidising ability of the central nickel(IV).
My initial guess was that hexaamminenickel(IV) and/or hexacyanidonickelate(IV) would be stable in water, since the otherwise hydrolytically unstable cobalt(III) can be stabilised that way, but I could not find any information about the hydrolytic stability of either ion. So I'm asking it here, just in case stronger sigma-donors than ammonia/cyanide are necessary.

P.S. Hexafluoridocobaltate(III) is high-spin, although that fact would likely not be relevant here.

related https://en.wikipedia.org/wiki/Ni(IV)_organometallic_complex

https://www.sciencedirect.com/science/article/abs/pii/S0277538711005195

Seems dimethylglyoxime can stabilise it enough for strongly basic solution https://www.sciencedirect.com/science/article/abs/pii/0022072864800826

Nickel may be oxidized from nickel(II) to nickel(IV) with diacetylpyridine dioxime ligands[1] and also in polyniobate complexes[2], in the latter case using hypobromite as the oxidizing agent. The latter reference also notes a Ni(IV) molybdate complex.
References
1.
Everett I. Baucom and Russell S. Drago. "Nickel(II) and nickel(IV) complexes of 2,6-diacetylpyridine dioxime". J. Am. Chem. Soc. 1971, 93, 24, 6469–6475.
https://doi.org/10.1021/ja00753a022
2.
Charles M. Flynn Jr. and Galen D. Stucky."Heteropolyniobate complexes of manganese(IV) and nickel(IV)".
Inorg. Chem. 1969, 8, 2, 332–334.
https://doi.org/10.1021/ic50072a029


text:
Values of Hammett Acidity function for differing concentrations of H2SO4?
I am styling the effect of the concentration of different catalysts on the hydrolysis of sucrose to study it's reaction mechanism. I will be using the Hammett-Zucker hypothesis to find the molecularity of the reaction but I am unable to find a H0 value for H2SO4 anywhere. Is there some source I am missing or can I calculate it?

More like a chart for H0 value for H2SO4 solutions. For pure it's about -12, but you need diluted - it's -1 for 20% for example.

How did you know it is -1 for 20%? Do you have some table that I can refer to?

Only got few example values, but I'm pretty sure more data could be found on internet.


text:
Which steps of the lithium ion battery chemistry are the main contributors to the energy transfer?
A couple of days ago Jordan Giesige, who runs the informational youtube channel 'The limiting factor', released an ambitious video about Lithium-ion battery chemistry. The target audience: interested people with no background in chemistry. That video goes as far in depth as is possible within the length of that video.
My background is physics, I am active on physics.stackexchange, answering questions.
Eager to learn more I started looking for additional information. It appears there is little to no information at the level I am looking for. That is not surprising, of course, it is a rather specialized subject.
The following is an overview of how I currently understand the chemistry and electrochemistry of Lithium-ion batteries. My request is to correct me if you notice a misconception.
Rather than using the words 'anode' and 'cathode' I will simplify and I will use the names 'the graphite' and 'the cobalt' generically. That is, I'm aware there are other chemistries, such as iron phosphate chemistry, and that the cobalt is used in the form of a cobalt oxide; I'm using the name 'cobalt' generically to refer that aspect/side of the chemistry, whatever the actual implementation.

In the uncharged state there is a large population of lithium at the cobalt side, and the cobalt is in a particular oxidation state.
The electrolyte is an organic fluid, with an additive that readily forms a complex with lithium ions. That complex readily dissolves in the organic fluid. The manufacturing process introduces a population of lithium ions in the electrolyte.
(The first charging cycle (which happens at the factory), triggers an initial chemical process; that initial process is outside the scope of this question.)

Charging
When a standard charging cycle starts the following happens:
Electrons are being withdrawn from the cobalt side and are being delivered to the graphite side.
The electron depletion at the cobalt side facilitates a change in oxidation state of the Cobalt. This change in oxidation state drives release of lithium ions into the electrolyte. At the graphite side lithium ions enter the graphite structure, and obtain an electron from the graphite, from there on residing in the graphite in an intercalated state.
Maintaining charge
The electrolyte is designed to present a high energy barrier to dissolving of negative ions. Because of that energy barrier the probability of an electron migrating from the cobalt side to the graphite side is low.
Discharging
In the charged state the graphite is slightly electron enriched, the cobalt side slightly electron depleted.
As current start flowing the available electrons are attracted by the cobalt, and the cobalt changes oxidation state. The now negatively charged cobalt oxide causes lithium ions to go out of solution, joining the cobalt.
At the graphite side electrons are withdrawn, and as a consequence the graphite is withdrawing electrons from the intercalated lithium atoms. Lithium ions go out of intercalation, and dissolve in the electrolyte.

What I don't know is whether it is the case that on both sides the process of electron transfer contributes to the overall voltage of the battery cell.
It could also be, I don't know, that the maintenance of the discharge current is all due to one side, with the other side actually presenting a bit of a negative contribution.
If it is the case that one side presents a negative contribution: I don't know which side that might be.
It could be, I don't know enough about electrochemistry to guess, that during discharge the chemical process of the graphite robbing the Lithium's outer electron releases the energy that is driving the overall process.
Hence the title of this question:
The overall chemistry is like a bucket brigade, on uneven terrain. I'm interested to know which stages in the process are driving force, and which stages are being driven.

This new edit is a comment, I write it here because comment space is (for good reasons) small. This is not a request to expand the answer further. This is feedback on how the answer has helped me.
Independent of the details of the chemistry at the graphite side and at the cobalt side: if at one side more Lithium ions would be entering the electrolyte than at the other end would be leaving the electrolyte the electrolyte would become positively charged. Becoming positively charged would come at an energy cost, so that is not going to happen; something else will happen, whatever it takes that keeps the net charge of the electrolyte close to neutral.
I am aware that the direction in which a chemical reaction proceeds is also sensitive to the concentration of the reactants, and the smaller the difference in electrochemical potential the larger the contribution of sensitivity to concentration of reactants.
To my understanding: when the net reaction consists of a cascade of sub-steps then it isn't necessarily meaningful to try and attribute a property of 'being the driver of the overall reaction'.
To my understanding: since the amount of electrolyte is quite limited the rates of Lithium ions entering/leaving the electrolyte are stronly coupled, and as a consequence the rates of the reactions at the graphite side and at the cobalt side are strongly coupled.

You probably mean this video How a Lithium Ion Battery Actually Works

While charge bilance of both electrodes must match, part of the charge exchange can be spent by parasitic reactions, like deploying metallic lithium on the "graphite" surface or forming oxygen on "cobalt". Lithium then further slowly reacts with solvent.

Czech language has a saying ( and the movie too), literally translated as: A single hand does not clap.  It fits well to fossil fuels (gas without oxygen nor vice versa  does not work) and to galvanic cells too (reductant and oxidant).
Both "hands" are the source of energy, not just the fuel nor just one of reductant/oxidant. For the fuel case, carbon and hydrogen atoms of fuel molecules release energy by forming stronger bonds with oxygen.  The other "hand", oxygen atoms of oxygen molecules release energy by forming stronger bonds with carbon and hydrogen.
The rechargable cells are based on general reaction
$$\ce{Ox1 + Red2 <=> Red1 + Ox2}$$
This reaction has its value of the reaction Gibbs energy $\Delta G_\mathrm{r} [\pu{kJ/mol}]$ as the maximal theoretical value of non volume (e.g. electrical) work the reaction can provide.
There is the well known equation from electronics, relating energy, charge and potential/voltage: $ \mathrm{d}E = U \cdot\mathrm{d}q$, directly applicable to galvanic cell, being a bridge between chemistry and electronics. This cell voltage is related to above by equation
$$\Delta G_\mathrm{r}=-nFU$$
where n is number of exchanged electrons (here n=1) and F is the Faraday constant.
In water based electrolytes, there is convention to define the standard electrochemical potential related to standard hydrogen electrode $\ce{H+(aq,1M)/H2(g,1 atm)/Pt}$, for the reaction
$$\ce{Ox1 + n/2 H2 -> Red1 + n H+(aq) + n e-}$$
which is not applicable for waterless systems.
In galvanic cells, the system of the overall net reaction is spatially separated to regions at the cathode and the anode.
$$\ce{Ox1 + n e- <=>[fw, charging, cathode][bw, discharging, anode] Red1 }$$
$$\ce{Red2  <=>[fw, charging, anode][bw, discharging, cathode] Ox2 + n e- }$$
For Li-ion cobalt cells:
$$\ce{CoO2(s) + e- + Li+(solv)<=>[fw, charging, cathode][bw, discharging, anode] LiCoO2(s) }$$
$$\ce{C-Li(s)  <=>[fw, dicharging, anode][bw, charging, cathode] C(s) + e- + Li+(solv) }$$
Reactions are simplified. There are no distinct red/ox phases, but there is  continuously changed non-stoichiometric composition of active electrode materials.
Electrolyte is then $\ce{Li+(solv) + PF6-(solv)}$ dissolved in (usually mixture of ) dialkylcarbonate solvent $\ce{R1-O-CO-O-R2}$, where R is usually methyl, ethyl or bidental ethylene.
During charging, there is transfer of lithium ions in direction $\ce{Li_xCoO2 -> LiPF6/(RO)2CO (electrolyte) -> Li_yC_z}$ and vice versa during discharging.
Each electrode material acts like a charge pump, trying to keep the particular potential of the respective electrode, at which both reduction and oxidation rates have zero net effect.
The "graphite" electrode keeps low potential, as it is eager to provide electrons to an external circuit and release lithium ions to electrolyte. The "cobalt" electrode on the other hand keeps high potential, eager to accept electrons from an external circuit and absorb lithium ions from the electrolyte.
And all vice versa at discharging.

The cobalt electrode is at the beginning made of $\ce{Li2Co2O3 = Li2O·2CoO}$, and not $\ce{Li_xCoO2}$ as you state.  It is transformed in the charging process into $\ce{Li^+ + e- + Co2O3}$. The $\ce{Li+}$ ion passes into solution and the electron is sent to the carbon electrode. This graphite electrode is made of pure carbon before charging process. At the end of this charging process, it contains some lithium coming from the $\ce{Li+} $ moving in solution, plus the electron brought from the electric connection outside.

@Maurice I have not described the initial state at all. At initial charging graphite electrode gets also covered by products from reaction of lithium with dialkyl carbonate..Within the operational life there is LixCoO2, where X must not be too close to 0, as it would tend to release oxygen, nor to 1, as both oxide then tend to irreversibly separate.

During charging, the cobalt is partially oxidized to the +4 state, with some lithium ions moving to the electrolyte, resulting in a range of compounds Li
xCoO
2 with 0 < x < 1. lithium cobalt oxide

@Poutnik I'm not sure, but I get the impression that you read only the title of the question, and that you rather skimmed the body of the question. Anyway: as Jordan Giesige remarked: the complexity of a rechargeable battery is such that it is kind of like an organism. The chemistry is almost metabolism-like. I now feel that asking about rechargeable batteries is not a good fit for stackexchange. I feel the question needs a threaded forum, so that there is room for follow-up questions. I expect to submit this enquiry in a chemistry forum in the near future

@Cleonis You jump to conclusions too fast. Search also for prior Q/A on the site, as it was discussed few times  You you are free to ask anywhere, but aside of that, elaborate  your question more or post another question, if it is better idea. // Comparing rechargable cells to organisms is overkill  Li-ion cells have just more complicated chemistry, mainly due side reactions of metallic lithium with dialkyl carbonates.

@Cleonis The question is like if asking which electrodes of a capacitor contributes to the stored energy more: the positive one or negative one? The cell voltage is in direct relation to the Gibbs energy change $\Delta G_r = -nFE$ of the net reaction(simplified) $$\ce{C-Li  + CoO2 -> C + LiCoO2}$$

@Poutnik I've been meaning to get back to this. Thank you for expanding the answer. Comment space is (for good reasons) small, so I will comment in the form of an edit to the original question.

@Cleonis already done :-)


text:
Phosphate transfer in hexokinase

In glycolysis, hexokinase transfers a phosphate group from ATP to glucose. This is an example of the coupling of an exergonic reaction with an endergonic reaction, such that the endergonic reaction is  forced  to  occur.
Please explain how  the  enzyme  ensures  that  the  phosphate  group  ends  up  on the glucose and does not simply get transferred to a water molecule.

In trying to work out this problem, I came up with the following explanation:

The hexokinase has two different states: open and closed form.
In the open state, only glucose can bound thus the enzyme will change its shape
When the enzyme change its shape, the enzyme will close.
Once the enzyme is in the close state, the enzyme can bind the ATP (the phosphate to the glucose).

This sequence will make the phosphorylation possible without getting the phosphate transferred to a water molecule.
So, is the explanation correct or am I wrong?

In crystals, hexokinase is able to bind to ATP (or ADP or AMP-PNP) in the absence off glucose. It has a miniscule ATPase activity (40,000 times slower than the hexokinase activity.

Source: https://febs.onlinelibrary.wiley.com/doi/full/10.1111/j.1432-1033.1970.tb01075.x
The two pieces of evidence combined show that ATP does not "have to wait" for glucose to bind and the enzyme to close. Comparing the Km of the hexokinase to the ATPase activity, it does seem the ATP binds a bit tighter when glucose is bound.

[quoted by OP] This is an example of the coupling of an exergonic reaction with an endergonic reaction, such that the endergonic reaction is forced to occur.

This is not the best example of coupling because the reaction is a phosphate transfer from ATP to glucose, an exergonic reaction. All the enzyme has to do is to catalyze that reaction without letting water in the position that the glucose hydroxyl group occupies when bound. Better examples are e.g. ion pumps where ATP does get hydrolized to form ADP and phosphate. Here, the enzyme has to be a little more "sophisticated" to couple the two reactions.


text:
Can I use the Winkler method as a way to measure the rate of photosynthesis?
I want to make an investigation where I measure how the color of light affects the rate of photosynthesis.
The idea is to put an Elodea plant in a test tube and shine with different color of light during time. I want to measure the dissolved oxygen in the beaker every minute for 5 minutes for different colors. As I have no access to a dissolved oxygen sensor I thought I could take out different samples of water every minute and separate them in bottles to conduct the Winkler method on each one. Hence, calculating the mg/L of dissolved oxygen and make a line of best fit for each color.
Is this a functioning idea? Could I get feedback?

Good thought experiment. Winkler's method is applicable to dissolved oxygen. A concern, if I were doing this experiment is the formation of oxygen bubbles. Wilker's titration would not be able to detect the oxygen which escaped as bubbles.
What you can try is stuffing a large number of aquatic plants in a one or two liter beaker and invert a funnel and a testtube over it as shown below : This is a standard experiment over centuries. Now you can illuminate the beaker with various colors and measure the volume of oxygen. Now oxygen has some solubility in water, when you see bubbling, water is already saturated with oxygen.
In order to make your experiment reliable try Winkler's method and this gas-volumetric method.


text:
The products of the reaction between cupric oxide and nitric acid
I am a third year student in junior high chool and I am having some difficulty finding information about the reaction of metal oxides and acids, and I cannot find a more accurate one online.
What exactly is produced by the reaction between curpic oxide and nitric acid?
Why does the concentration of nitric acid cause the products to be different?
Here're something I've found:
$\text{CuO}+2\text{HNO}_3\text{(dilute)}=\text{Cu(NO}_3\text{)}_2+\text{H}_2\text{O}$
$\text{CuO}+6\text{HNO}_3\text{(concentrated)}=\text{Cu(NO}_3\text{)}_2+4\text{NO}_2\uparrow+\text{O}_2\uparrow+3\text{H}_2\text{O}$
$2\text{CuO}+8\text{HNO}_3\text{(concentrated)}=2\text{Cu(NO}_3\text{)}_2+4\text{NO}_2\uparrow+\text{O}_2\uparrow+4\text{H}_2\text{O}$

See these questions: https://chemistry.stackexchange.com/questions/97090/which-is-a-better-oxidising-agent-concentrated-nitric-acid-or-dilute-nitric-aci ... https://chemistry.stackexchange.com/questions/17067/how-does-copper-reduce-dilute-nitric-acid-to-nitric-oxide-and-concentrated-nitri?rq=1 ... https://chemistry.stackexchange.com/questions/158763/why-does-mercuryii-nitrate-form-only-when-using-concentrated-nitric-acid

All your equations are the sums of the two following equations $$\ce{CuO + 2 HNO3 -> Cu(NO3)2 + H2O}$$ and $$\ce{4 HNO3 -> 4 NO2 + O2 + 2 H2O}$$. They can be added with any numerical parameter before being added.

Thanks! Btw, do similar ecomposition reactions occur on $\text{HCl}$ and $\text{H}_2\text{SO}_4$?

No ! $\ce{H2SO4}$ does not react with $\ce{HCl}$, because in the $19$th century $\ce{H2SO4}$ was used for synthesizing $\ce{HCl}$, at $300$°C, by the reaction : $\ce{H2SO4(l) + NaCl(s) -> Na2SO4(s) + 2 HCl(g)}$

I think you may have misunderstood me a little (Pardon me, I am not a native speaker of English). I meant that do similar reactions occur when curpic oxide reacts with  $HCl$ and $H_2SO_4$ under different concentrations?

@DrThomasPeng. I am like you. I am not a native speaker of English. I will try to answer your question. Cupric oxide reacts with all acids, and in particular with $\ce{HCl}$ and $\ce{H2SO4}$, according to : $$\ce{CuO + 2 HCl -> CuCl2 + H2O}$$   $$\ce{CuO + H2SO4 -> CuSO4 + H2O}$$ and it does not depend on the concentration. The reaction is the same at all concentrations of the acids

You have my gratitude. That's a very clear explanation.


text:
Why is supersaturated sodium acetate so (meta)stable?
Sodium acetate is a common example used in general chemistry lectures to illustrate the idea of supersaturation, and that precipitation of a supersaturated solution can be initiated by a seed crystal. It makes a great demo, and there are many excellent videos online that illustrate this phenomenon.
My question is: What is it about sodium acetate that makes it so well-suited to this particular demonstration? Is there an explanation from experiment, theory, or simulation that explains why sodium acetate is better than, say, sodium chloride? (Or perhaps potassium acetate? though that may be a cost issue.)

Beside NaOAc (used in hand warmers), super cooling the melt of benzoic acid is an other attractive example, too. It is not obvious if the presence of a second type of (solvent) molecules eventually helps or hampers the crystallization (e.g., viscosity, kinetics vs. thermodynamics). On top, conditions favourable to create (microscopic) germs of crystallization may differ from those of crystal growth (e.g., here.

Good question! As @Buttonwood states, viscosity plays a major role. Consider that silicate glasses (e.g., soda-lime or "window" glass) devitrify (recrystallize) if heated for a long time  and allowed  to cool slowly -- and at these temperatures, glass is very viscous. See http://franklin.chem.colostate.edu/glassguy/devit.html

I think it would be interesting to consider the nature of the activation barrier to crystallization. Is the activation barrier mostly entropic or enthalpic? If I had to guess, I would think it is an entropic bottleneck, where it is difficult to induce the necessary molecular rearrangements to generate critical nuclei of the crystal.


text:
Why are hydrated lithium ions' radii larger than hydrated sodium ions' radii?
Why are hydrated lithium ions' radii larger than hydrated sodium ions' radii i.e. $r_\ce{Li+(aq)}>r_\ce{Na+(aq)}$?
If ionic radii increase down the group i.e. $r_\ce{Li+}<r_\ce{Na+}<r_\ce{K+}$ then the radii of hydrated ions also increase down the group i.e. $r_\ce{Li+(aq)}<r_\ce{Na+(aq)}<r_\ce{K+(aq)}$. But according to my class notes, The radii of hydrated ions are larger than those of gaseous ions. The radii of hydrated ions decrease as the group. I'm not sure I understand. But, according to the class notes and Quora answers, how can the size of an ion alter the hydration radii?
According to a Quora answers,

Hydration energy is the amount of energy liberated when an ion gets hydrated that is , it accepts water molecule. e.g. lithium ion is small in size so it accepts large number of water molecule so it's hydration energy is more whereas caesium ion is bigger in size so it accepts less number of water molecule and has less hydration energy.

Another answer says,

As the size of ion increases its polasiation tendency decreases therefore the tendency to polarise nearby water molecules decreases hence decreasing its hydration radii. This effect can be seen by observing the trend of hydartion radii in alkali metal ions.

Look around the site, it was discussed recently.

Please check you greater/smaller than signs. Are they like you intended?

This current profile is unregistered. Is this your another profile? https://chemistry.stackexchange.com/users/100905/huzaifa-abedeen

Down a group, ionic radii increases. But at the same time, due to decrease in charge density, the ions attract less number of water molecules. The smallest cation (with the same charges) having highest charge/size ratio attracts maximum number of water molecules.

Image credit: https://rwu.pressbooks.pub/bio103/chapter/the-chemical-context-of-life/
Compare the relative increase in ionic radii down a group to effective radii of hydrated ion, i.e., with attached whole water molecules. Clearly, there will be a decrease in radii of hydrated ions.


text:
How substituted/contaminated can a hydrocarbon be to still be considered a hydrocarbon?
I have a doubt about any type of hydrocarbon with a Carbon-Heteroatom bond, instead of a C-H bond like oxygenated, nitrogenated, brominated hydrocarbons. My question is: how polluted can they be to still be called a hydrocarbon?
Kind regards and thanks for reading!

Generic names such as oxygenated, nitrogenated, brominated hydrocarbons are just layman's terms. They are not acceptable in formal chemistry writings.

There's no objective answer to this. The term one uses to describe a molecule will depend heavily on the context. How many fillings do we need to put on a piece of bread before we call it a sandwich?

Hydrocarbons(1) cannot be substituted at all, as they contain just carbon and hydrogen by definition.
Hydrocarbons(2) as the base of substituent nomenclature can be formally as substituted as you wish.
1-chloropropane is not a hydrocarbon(1).
Hypothetical hexaazabenzene is still hydrocarbon(2), even if there is no carbon nor hydrogen, all formally replaced by nitrogen.

It is a matter of convention, which may evolve over time and context, including who you are communicate to.
A strict application of the definition of «hydrocarbons» requires that there is at least one atom of carbon plus an other of hydrogen per molecule, and that any atom of the molecule is either of carbon, or hydrogen.  However, hydrocarbons like methane (in natural gaz), propane/butane (for camping and lighters), and isomers of octane (for combustion driven engines) are only a small subset of organic molecules.  Much of (one might argue, more) interesting chemistry actually only is possible if hydrogen and carbon atoms are replaced (substituted)* by other atoms.  The term paraffin you may encounter for this hydrocarbons actually derives from Latin parum affinis which may be translated as of little relationship (only) and (in chemical terms) of low reactivity.
The terms of «organic» and «inorganic» in chemistry are a bit younger, and still inherit a convened discern that the former traditionally requires a molecule of at least two atoms binding to each other: one of carbon, the other not a carbon atom. (In databases like Beilstein, a few classes like carbonates [e.g., $\ce{CaCO3}$] equally are considered as inorganic, too.) These definitions are old, and coined when (because of lack of e.g., spectroscopic methods) knowledge about chemical constitution and structure if accessible was derived by comparison: how your unknown compound would react in reactions already investigated for known compounds?
Over time, means of structure elucidation (e.g., spectroscopy) and synthetic methods evolved; now, the separation of «organic» and «inorganic» is not as sharp as it was at time of e.g., Liebig and Wöhler (terms like e.g., «organometallic»). As an illustration, in heterocyclic compounds, where atoms other than carbon co-constitute the rings, nitrogen is one of the more frequently seen replacements for carbon:

(Sometimes, molecules are stable enough to be isolated as such [e.g., pentazole], sometimes only as motif within a molecule [e.g., 1,2,4,5-tetrazine].)
Similar to the above line leading up to pentazole, given the context and convention an other example which would not fit the above traditional definition of organic is the motif of the phosphazene ring. However, in modern terms, it is an example of organophosphorus compounds with many examples known (a listing by PubChem):

(edited illustration of an abstract).
*) This replacement of atoms is not a pollution, or contamination.  If you use the term «pollution», you assume the simultaneous presence of at least two chemical compounds in a sample.

I guess OP may have used "pollution" as a figure of speech, like if hydrocarbon structure were "polluted" by atoms of other element.

Indeed it may be understood as a figure of speech.


text:
Can two noble gases attract each other?

If two hydrogen atoms are far apart, they have no effect on one another. But as they are bought closer together, they begin to excerpt an effect. The two nuclei, having the same positive charge, repel one another, and the two electron clouds also repel one another. However, most important of all is the attraction between the nucleus of one atom and the electron cloud of the other atom. As the atoms approach, the electron clouds are pulled toward the region between the nuclei (d). - Chemical Principles, Dickerson, Fourth Edition

First of all, how is it that the electron clouds do not repel each other (as they approach and do not yet have a binding molecular orbital)? Shouldn't (d) look more like this if the first attractive force is from London dispersion force (is it?):

Is the attraction between the nucleus and the cloud only so much stronger because the nucleus is stationary and therefore the electrons repel each other much less than the nucleus attracts them due to the constant change in their residence probability?
In principle, two Xe atoms, for example, should first attract each other in this way, shouldn't they? The only difference is that they do not form a bond. I know that this behaviour is described by the Lennard Jones potential but I would just like an explanation for all this.

Consider that at low temperatures, noble gases do liquefy, i.e., condense because there is attraction between atoms. All, except for He, even form solids at low enough temperature, with atoms regularly aligned. (Willem Hendrik Keesom did solidify He, only under pressure of 2500 kPa.)

See: https://www.journal.csj.jp/doi/10.1246/bcsj.57.3062

You make this far more complex than you need to. The issue is that there are several different forces involved, some attractive, some repulsive and the overall result is a combination of all of them.

See https://physics.stackexchange.com/q/684215/313612.

I've written a lengthy answer to a similar question here. But in essence, I think you have the right idea with this part of your question:

First of all, how is it that the electron clouds do not repel each other (as they approach and do not yet have a binding molecular orbital)? (...) Is the attraction between the nucleus and the cloud only so much stronger because the nucleus is stationary and therefore the electrons repel each other much less than the nucleus attracts them due to the constant change in their residence probability?

The electrons on one atom feel more of the stationary nuclear charge of the other atom than its dispersed electron cloud. That's why you see the overall "inwards" polarization of the electron densities.
However, the text book you cite gets one detail slightly wrong:

However, most important of all is the attraction between the nucleus of one atom and the electron cloud of the other atom.

It's not the electron cloud of the other atom that a nucleus is attracted to, but its own shifted electron cloud.


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Reasoning for why anions have larger radius
I understand one thread of logic that adding electrons to create an anion creates more electron repulsion hence increasing the ionic radius.
What's wrong with the reasoning that there is now a larger negative charge and hence the positive nucleus and negative electron cloud have a stronger attraction so the ionic radius decreases. One thought was that the number of electrons increases and hence the nucleus has to pull more electrons toward it so the pull becomes weaker, but I'm not sure if a combination of charge and number of objects is a relevant metric.
Can someone explain the best way to think about this and how all of these concepts (electron repulsion, charge of electron cloud as a whole, and number of electrons) relate with each other and contribute to ionic radius? Thanks!

I think it's rather sad that this gets downvoted. Obviously OP's thought process is completely wrong, and I've pointed out as much in my answer; but if they were completely right, then they would not need to ask a question here. And having been here and having seen questions over many years, I must say that this confusion over electrostatic forces is actually quite common.

What's wrong with the reasoning that there is now a larger negative charge and hence the positive nucleus and negative electron cloud have a stronger attraction so the ionic radius decreases.

The problem is that you're treating it as if all the electrons can be collectively reasoned about as one single negative charge. This isn't true. The nucleus is considered as one single positive charge, because all of those positive charges are crammed within such a small space that we may as well consider them as being at the same point in space. On the other hand, this is not true of electrons. They don't glue themselves together like the nucleus does. Each electron must be considered individiually when describing the attractive/repulsive forces.
So, in other words, the negative charge has not gotten larger. There is no "the" negative charge. What has happened is that there are now more negative charges.

One thought was that the number of electrons increases and hence the nucleus has to pull more electrons toward it so the pull becomes weaker

This is not correct either; it is not even based on anything scientific. It sounds as if it's a human trying to lug a weight along, and a heavier weight is harder to move than a lighter one.
Electrostatic attraction simply doesn't work like that, though. The force between two charged objects is $F = kq_1q_2/r^2$, where $k$ is some constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. Clearly, if $q_1$ or $q_2$ become larger, the attractive force between them increases, not decreases as this line of argument would have one believe.
Having said that, note that this equation is not even applicable to the current situation. As mentioned previously, the addition of an electron does not increase the size of $q_1$ (or $q_2$). It is not "adding" to some existing basket of negative charge thereby making it more negative. The electron is a negative charge of its own. So there are at least two reasons why this line of logic is wrong.

Can someone explain the best way to think about this and how all of these concepts (electron repulsion, charge of electron cloud as a whole, and number of electrons) relate with each other

Of these three factors, only the first (electron repulsion) is actually useful. The "charge of electron cloud as a whole" has no useful meaning, as explained above, because the electrons are not to be considered as a collective entity.
Circling back to the original issue about why anions have a larger radius than the neutral atom: imagine that you have a neutral oxygen atom with 8 protons and 8 electrons, and this atom has found its equilibrium, most stable arrangement with an atomic radius of $r$. This means that all the electrons are within a sphere of radius $r$.*
What forces are at work here? Well, we have 8 different electrons, each of which are attracted to the nucleus and would like to be closer to the nucleus. That gives us 8 electron–nucleus attraction terms. Attraction is good, so we want to try to keep the electrons as close to the nucleus as possible. On the other hand, every electron is repelling every other electron. The number of pairs of electrons is (8 × 7 / 2) = 28, so we have 28 electron–electron repulsion terms. Repulsion is bad, so we want to try to keep the electrons as far apart as possible. Note that not all of these terms have the same size, so just because there are 28 versus 8 doesn't mean that the 28 is more important than the 8. I'm just listing this to drive home the point that electrons must be considered individually, and that it's not just some "nucleus versus all the electrons" tug-of-war.
Before adding in one more electron, let's consider why this radius $r$ is chosen. Why don't the existing 8 electrons, for example, cram themselves into a slightly smaller sphere, of radius $r - \delta$? Well, if they did, then the attractive forces would increase, and that's a good thing. On the other hand, because the electrons are now contained within an even smaller space, the repulsive forces would also increase, and that's a bad thing.
It turns out that $r$ is the special radius which perfectly balances these two factors. That is, if you take a slightly smaller radius $r - \delta$, the repulsions get too big; and if you take a slightly larger radius $r + \delta$, the attractions start to diminish.
Now consider adding one more electron to the mix. Where would this electron go? Of course, the new electron itself will want to be as close to the nucleus as possible. However, cramming it within the sphere of radius $r$ will also inevitably increase the electron repulsions inside; and recall that $r$ is already that radius which happens to nicely, snugly, balance the attractive and repulsive forces for 8 electrons. By adding this 9th electron within this sphere, we're totally destroying this balance as it introduces far more repulsions, while not changing the 8 original attractions. If the 8 other electrons had their own way, the 9th electron would simply be repelled; it wouldn't even be able to enter this sphere.
The expansion of the atomic/ionic radius is thus best viewed as a compromise between these two factors. The nucleus wants the 9th electron to be as close as possible, but the 8 existing electrons would rather not admit it into that same sphere. What happens in the end is something in-between: the sphere of electrons expands a little bit to fit in the 9th one, allowing it to stay somewhat close to the nucleus, but without increasing repulsions too much.

What about removing electrons?

Simple: when you remove an electron, there are fewer remaining repulsions, so the electrons happily move into a smaller sphere in order to maximise their attraction to the nucleus.
It would be wrong to think of this as being a "larger overall attraction per electron". The attraction between the nucleus and electron is not an averaged quantity. There is simply just one attractive force between the nucleus and every electron.

But books talk about "number of electrons"...

Yes, they do. The reason for this is because the number of electrons is an indirect proxy for the number of electron–electron repulsions. More electrons = more repulsions. If all else is equal, the most stable radius will be pushed outwards, in order to reduce the repulsions.
The "shielding" effect is a different way of expressing the same thing. Effectively, what it means is that the electron–nucleus attraction (i.e., the desire for the electrons to be as close to the nucleus as possible) is "countered" or "weakened" or "watered down" by the electron–electron repulsions (which make it very hard to cram electrons within that small space). This can be pictured as the nucleus not being "as positively charged" as it really is, which gives rise to the "effective nuclear charge" $Z_\mathrm{eff}$.
At the end of the day, fundamentally, everything boils down to electron–nucleus attractions and electron–electron repulsions.

Can I have an analogy

Yes, you may. If you want an analogy that is more true to the underlying physics, then think of this: the nucleus is a piece of tasty cake at the centre of a large room. The electrons are humans trying to get to the cake to have a nibble, but all while trying to maintain social distancing.

Footnote
* From a quantum mechanics point of view, this answer cuts a lot of corners, like in this phrase that electrons may be found within a sphere of radius $r$. But that's way, way beyond what you need to know at this point. I only mention this so that nobody complains about it in the comments.

Thank you so much for the thorough response! This clarifies a lot of disparate concepts together. Appreciate it.


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Can reverse micelle's exist outside solution?
As in reverse micelle's, the polar liquid remains inside the micelle while the exterior is one of lower dielectric constant; if the liquid were to be carefully removed, would the micelle structure remain?
I reason so as the air or medium outside the micelle would still be of a lower dielectric constant, so it is still favourable to remain with that structure. Is this sensical?

Putting the polar groups (e.g., CTAB) close together in the centre of a reverse micelle likely is more costly than letting the alkyl chains mingle in normal one exposed to polar media. In your question, do you permit the inverse micelle to be covered by a «normal» one?  This then leads to liposomes which are used as carrier for dyes (-> spectroscopy) and medication (phototherapy, carrier of mRNA/DNA vaccines, etc.) ...


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What are the hydrodynamics behind a Soxhlet extractor?
So I'm trying to wrap my head around the physics of how a Soxhlet extractor works, not the chemical, solvent and chemical components part, but rather the physical, hydrodynamic, siphoning part : why exactly does the solvent flow back through the siphon tube into the flask ?
I imagine Bernoulli's principle, comes into play, but, unlike a usual siphon tube, there's nothing or no one physically initiating the siphoning by sucking air out, or lowering the pressure in the flask.
I've tried to find any theoretical study explaining the physics behind a Soxhlet extractor, but to no avail.
Any help would be appreciated, thanks.

Soxhlet extractor, source: https://commons.wikimedia.org/wiki/File:Soxhlet_mechanism.gif

Vapours of the solvent continuously rise, condense on the Dimroth condenser, and drop as liquid into the thimble.  The level of the liquid in the thimble rises, where it yields a hot solution (liquid/solid extraction).  The thimble is penetrated by the solution, passing into the outer volume.  The level of the liquid in the Soxhlet rises up to the point where the level in the inner is greater than the one in the equally rising arm of the siphon closer to the inner side of the Soxhlet.
By cohesion (of the solution) and gravity (acting on the solution), the potential energy of the solution now is lowered as it passes across the more distant arm, falling (again) from the higher reservoir to the lower one, the still pot.  When the solution returns to the lower flask, there is little reason to think there would be a vacuum in the lower flask because vapours still ascend to the Dimroth condenser on top (which, for obvious safety reasons, stays open to atmosphere/is connected to the Schlenk line for pressure equalization).
As for the Bernoulli principle: Indeed it is helpful that the open diameter of the siphon is much smaller than the one containing the thimble.  Thus, the local pressure difference generating by the solution returning falling down is large, causing suction next to the thimble during the then swift transfer.

+1 I think I doubt that Bernuilli's principle has any significant effect here however ...

@Karl As long as liquid flows in the siphon, some suction may occur. With fixed length, filling a tube of small section requires less volume of liquid than a tube of large section when length. I agree that a capillary would not be efficient since a small section equally limits flow (as in volume per unit of time) with liquids of fixed density and impractical for cleaning.

Not sure I follow. Bernuilli says that a flowing substance reduces the static pressure. The pressure difference between the just emptied vessel and the end of the tube is however simply proportional to the filled length of the downward part of the tube minus the filled length of the upward part.

Not sure I meant this one, intention was «greater velocity for smaller sections if flux is fixed». But ok, Bernoulli dynasty in mathematics, as the Couperin in music.


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When people say "The energy of an orbital depends on its ℓ value due to screening effect" , what is screening what?
My book says:

Both the attractive and repulsive
interactions depend upon the shell and shape
of the orbital in which the electron is present.
For example electrons present in spherical
shaped, s orbital shields the outer electrons
from the nucleus more effectively as compared
to electrons present in p orbital. Similarly
electrons present in p orbitals shield the outer
electrons from the nucleus more than the
electrons present in d orbitals, even though all
these orbitals are present in the same shell.

Here, they've mentioned outer electrons. What exactly do they mean by those?
Because when I tried to reason out how shielding affects the value of ℓ, I thought that 4s had more orbitals between it and the nucleus (everything between the nucleus and 3d, and then 3d too), and thus 4s should be more shielded than 3d.

Note that orbitals are theoretical constructs and not physical objects. The term orbital is primarily the wave function describing the particular electron quantum state. But it can be also understood as this quantum state itself or the 3D geometrical representation of the  spatial presence probability of such an electron.
An outer electron is mostly more distant from the nucleus than an inner electron. Therefore the mean radial repulsion of the outer one by the inner one aims outwards.
There is said mostly above, as there is significant probability that e.g. 2s or 2p electrons are closer to the nucleus than 1s electrons, even if vice versa is (much) more probable. That is because of orbital geometrical overlapping.
The resulting effect is like if the outer electron were attracted by a nucleus of less than actual charge. This effect is called nucleus shielding/screening.
The effect intensity for 2 considered orbitals depends on their mutual probabilistic overlapping (more overlapping = less screening) and inner orbital shape (closer to  spherical symmetry = better screening)
Electrons in s orbitals have significant occurance probability near nucleus and in inner regions. Therefore, they are less shielded than electrons in other orbitals with the same $n$ and have the lowest energy of them.
OTOH, due their spherical symmetry, they are the best in shielding, unless overlap level overrules that.

What exactly is shielding what? Is the the orbital in which the electron is in shielding the electron from the nucleus? Thanks for answering!

Orbitals are not a physical thing. Shielding is effect of classical electrostatics. Electrons shield and are shielded by other electrons, depending on how much their mean mutual repulsion  subtracts from nucleus attraction.

See also https://en.wikipedia.org/wiki/Slater%27s_rules

"OTOH, due their spherical symmetry, they are the best in shielding, unless overlap level overrules that." - Aren't all l subshells overall spherically symmetric?

@Antimon I was speaking about orbitals, more exactly about probability density of their present electrons, not about subshells. For the latter, they are not if they are just partially occupied.

@Poutnik "ns orbitals have significant n-1 local inner radial probability maxima and therefore are generally shielded less than p or d orbitals, therefore have lower energy" could you please explain the first part of the sentence?

The paragraph has been rephrased.

If I understand your question correctly, you are asking how the value the orbital angular momentum $\ell$ affects the shielding -- or in other words the energy. To answer this question, let's first look at the hydrogen atom. In the hydrogen atom, a single electron is bound to a single proton. Because of the simplicity of the system, the potential energy is purely Coulombic, that means that $V\propto r^{-1}$. As a consequence, the different $\ell$ levels in hydrogen for the same principal quantum number $n$ are degenerate (ignoring relativistic effects). This probably makes sense, because if you have only one electron there is no shielding.
Now, let's look at a multielectron atom. When we look at atoms with more than one electron, we still often use the solutions of the Schrödinger equation of the hydrogen atom, that is, we use one-electron wavefunctions and ignore the interaction between the different electrons. Although this might sound like a very rough assumption, it is not too bad in practice. If you consider one of the electrons of a multielectron atom, the other $Z-1$ electrons together with the nucleus of charge $+Z$ give a nett positive charge close to $1e$.
The situation is thus very similar to the case of the hydrogen atom. A single electron bounded to a core with a positive charge. The description will get closer to that of hydrogen if the electron under consideration is in a highly excited state corresponding to a large clasical radius so that the electron effectively sees a single core. For lower electronic states, the electron can penetrate the "cloud" of other electrons, resulting in a slightly different potential energy.
But why does this energy then depends on $\ell$? Well, if you look at the radial part of the Schrödinger equation with partial wave expansion (known as the Coulomb equation), there is a potential energy term in the equation of the form
$$
\frac{\ell(\ell+1)}{r^2}.
$$
This centrifugal barrier makes it more difficult for higher $\ell$ states to get close to the nucleus and experience the effect of the other electrons. As a consequence, orbitals with $\ell\ge 3$ behave almomst like those of hydrogen (meaning that they are degenerate and have a so-called quantum defect close to zero).


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Can a salt bridge run out and become a regular bridge?
The goal of a salt bridge is to maintain electroneutrality in both of the half-cells.
At the anode half-cell, the anode is oxidized. The resulting electrons travel up the wire to the load while the electrolyte gains a positive charge from the extra solvated cations. At the cathode half-cell, the cathode is reduced. The incoming electrons allow the solvated cations in the electrolyte to solidify and deposit onto the cathode. These two redox reactions lead to the formation of a positive charge in the anode half-cell and a negative charge in the cathode half-cell.
Of course, to keep the current flowing through the load, electroneutrality must be maintained: the two half-cell electrolytes must remain neutral. To do so, a crude solution would be to just use a second wire, or a water bridge. Several answers already point out that doing so just means that the current flowing through the load will be severely reduced as the half-cell electrolyte ions would just flow through the water bridge instead of the load.
Using a salt bridge avoids this, as the chosen salt is usually inert and does not react with the electrolytes. However, the salt bridge eventually runs out of the initial salt. The anion of the salt keeps flowing into the anode while the cation of the salt keeps flowing into the cathode and they eventually end up completely depleted.
If I understand correctly (according to this discussion), the salt bridge does not, in fact, run out, since the ions of the electrolytes actually take over as charge carriers. However, in this case, are we not just back to square one, which is to say, just a water bridge? Or is there some benefit to the reaction being already underway?

A water bridge works poorly until the ions from both solutions flow into it. A salt bridge works well from the start.

Ions flow to the bridge too, not only from it. // The second wire would create 2 cells from 1.

@Mithoron it does confirm my understanding that the original charge carriers of the salt bridge can get depleted, in which case the electrolyte ions of the half-cells take over as charge carriers in the bridge. However, it doesn't confirm whether the performance of the salt bridge drops once the initial charge carriers are depleted.

As a toy example, just imagine a salt bridge with NaCl, except it has a single molecule of NaCl. That single molecule would get very quickly depleted, as the Na+ would go to the cathode and the Cl- would go to the anode. Was there even any point in using a salt bridge with so little salt?

Negative electrode at the anode  and a positive electrode at the cathode ?.

@DavidCian The point is, salt bridges use intentionally quite high salt concentrations.

Apparently your problem is not chemical. It is related to electric charges being created and/or destroyed in solution.
Let's examine the "fate" of electric charges with a simple example : Zinc metal as anode and a copper plate dipped into $\ce{Cu^{2+}}$ as cathode (plus the same amount of corresponding negative ions like sulfate $\ce{SO4^{2-}}$ ions).
When the cell works, new $\ce{Zn^{2+}}$ ions are created at the anode, and $\ce{Cu^{2+}}$ ions are disappearing at the cathode. So positive charges are created around the anode, and the same charges are disappearing from the cathode, also creating an excess of negative ($\ce{SO4^{2-}}$) ions around the cathode. Without a bridge, the anode will become more and more positively charged, and the cathode more and more negatively charged. After a while, no more positive ions could be created in such a highly positive anodic region.
The cell can only work if these charge excesses disappear. This is done by allowing the new anodic cations $\ce{Zn^{2+}}$ to join the excess of negative ions from the cathodic region. Of course both ions $\ce{Zn^{2+}}$ and $\ce{SO4^{2-}}$ are attracting each other and will move to join in the salt bridge. But the bridge is filled with a salt solution containing positive and negative charges like $\ce{K+}$ and $\ce{NO3^-}$. At one entrance of the bridge, the positive $\ce{Zn^{2+}}$ ions attract the negative $\ce{NO3^-}$ ions and repell the positive $\ce{K+}$ ions. On the other side of the bridge, the negative $\ce{SO4^{2-}}$ ions produce exactly the same chemical migration. And the cell may work on. If the migration rate are forgotten, the local charge concentrations do not change, even though the chemical concentrations are changing all the time.  The solutions are always electrically neutral.
In the long run, $\ce{Zn^{2+}}$ will slowly replace $\ce{K^+}$ ions in the bridge, and $\ce{SO4^{2-}}$ will slowly replace $\ce{NO3^-}$ in  the same bridge.
So from the beginning to the end of this long "cell story", the electrical neutrality of the solutions is maintained, even though the composition of the solutions do change a lot.


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Evidence that electrons don't exist between orbitals?
I don't understand the logical jump that was made early on in the atomic age, to conclude that electrons jump in and out of existence when moving between orbitals? Similarly, why do electrons need to occupy everywhere in the orbital at once with a probability, rather than simply moving around it quickly? This is a problem I have with the overall logic of  quantum interpetations of physics and the uncertainty principle.
For instance, if in the smallest amount of time we could measure, an electron travelled around its orbit, say, 400 million times, and it took 1/800 millionth of that time to pass between orbitals, obviously we would be unable to observe it. But that doesn't mean it isn't happening.
Even the uncertainty principle seems to me to apply similarly to our scale of the world as to electrons: if I am walking, you can't technically measure my position and my path or speed at the same time, only one or the other. That doesn't mean I don't exist when you don't know my location, it just means you don't know where I am currently existing. I don't see why this has to be different for subatomic particles? For instance, why can't it just be a particle moving along a wave path, why does it need to be a particle and a wave at the same time? I don't see where this logic came from or what supports it. For instance, in a double-slit experiment, what logic  excludes the possibility of a particle that is simply moving in a wave-like path? If only one particle passes through only one slit, it doesn't cause an interference pattern, not because it was detected and therefore ceased to be a wave, but simply that there was not another particle to cause interference. Particles travelling in a wave-path might of course be expected to travel through either one or the other slit, but surely each individual particle only goes through one slit?
I have a similar problem with describing orbitals as probability clouds. A probability isn't something real, it is merely the quantification of an expectation based in some data. For instance, a coin being flipped may be expected to have approximately 0.5 probability of landing on heads. But it either happens or it doesn't, it doesn't 0.5 happen. So the probability was never a real event, just an expectation. So why can't an electron be physically travelling around its orbital really fast? Of course there will be a higher probability of it being in a particularly well-teavelled part of the orbital at any one time, but in reality it isn't everywhere, it is just in one place, or indeed travelling along it's path, but at a scale of time so many times faster than we can measure that it seems to be everywhere at once. But it, in fact, isn't. Can anyone explain to me why the less sensible logic is widely accepted, but the more intuitive explanation doesn't fit the data?

To be honest, I'm not certain whether "is quantum mechanics correct?" is a relevant question for Chemistry.SE. I guess it might be ok, but Physics would definitely have been better. One obvious objection to "things moving very fast" is that there's a speed cap on things moving quickly. [Of course, we could then move on to asking what evidence there is for relativity, but well...] Another objection is Bell's inequality.

Note, also, that when writing Markdown you need to enter two line breaks to create a new paragraph.

Evidence that something does not exist is not possible. There is possible evidence (or lack of it) that something exists. // Like the famous request for proof of the unicorn non-existence.

You're hardly the first or last person who thinks quantum mechanics is counter-intuitive. But with a century of experimentation covering ins and outs which not many can even conceive, Nature operates to an incredible precision as if it were quantum-mechanical, so you might as well say that it is. Rather than trying to seek flaws with pure argumentation, you should find the humility to fix your intuition and definitions (which may require throwing them in the bin entirely and rebuilding from scratch) and spend the effort to understand some of the actual mathematical framework.

There is no between. The very notion is simply missing; the electrons can't be there, much like they can't be blue, or stubborn, or upside down.

Welcome to Chemistry.SE! Your question clearly shows an effort to put your own reasoning into words, and probably represents a line of thought that a number of students of chemistry struggle with. There are many subquestions in your post, which makes it somewhat difficult to provide a single answer, however. I'd recommend reading about the history of quantum mechanics to dig deeper.

"you can't technically measure my position and my path or speed at the same time, only one or the other." Yes I can! I measure your speed continuously via  the Doppler effect of light reflected at your surface while you pass through a light switch.

I'm upvoting, not because this is an advanced question on a controversial matter (it is not), but because it is an important question that nearly every student who encounters quantum mechanics will grapple with.

Many scientists over the past century have struggled and continue to struggle with the representation of events at the atomic scale, as exemplified by the Wikipedia article on interpretations of quantum mechanics or recent headlines like "even physicists don't understand quantum mechanics". Schrödinger himself is known to have been disillusioned with the interpretations others ascribed to his wavefunction. So the questions above are not easily answered.
We can, however, point out a fundamental issue with any whizzing particle model: accelerated charges should radiate out energy according to the Larmor formula as observed in particle accelerators.
If you conceive of the electron as a point-like particle with a defined trajectory over time, then logically you should also be able to define its speed and acceleration. Any nonzero acceleration should lead to an emission of radiation and hence an energy loss over time, with the electron inevitably crashing into the nucleus within a fraction of a second. This clearly does not describe atoms as we know them.
It is the inability to fit atomic observations like non-radiation into purely classical models that led to the development and adoption of the abstract description of the atom in quantum mechanics. Not very intuitive, but we'll have to put up with it until something better comes along...

Any nonzero acceleration should lead to an emission of radiation
This is only partially true, as constant linear acceleration does not lead to radiation what is used in linear accelerators. As this radiation is proportional to the time derivative of acceleration, which is zero for constant acceleration vector.

P.S.: I have thought the same in past, but was corrected by physicists it is not about proportionality to (x'')^2, but about x' . x'''.

Thanks for the comment - proportionality to x'' is in indeed the widely reported condition, but I know there are a few subtleties involved as highlighted in e.g. https://en.wikipedia.org/wiki/Nonradiation_condition or
https://skullsinthestars.com/2008/04/19/invisibility-physics-acceleration-without-radiation-part-i/. If you have any reference to the x''' dependency, I'd be interested.