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Gaseous metals?
Metals form metallic bonds which explain many of their chemical and physical properties.
This is most familiar in the solid state but metallic properties are still quite recognizable in the liquid state e.g. mercury and molten iron.
However, gaseous metals are much less familiar and it is not obvious whether they would still have a distinct metallic behaviour.  For example, is there a distinction between gas and plasma for a metal?
This article mentions that the metallic bond persists in the liquid state but says nothing about the gas state: Metallic bonding at Chemguide.
So, do metals retain any distinct metallic behaviour in the gas state?
Note that I am not asking only about the electrical properties.  My main question is whether the gas consists of neutral atoms, small molecules, or positive ions in a sea of electrons?

Possible duplicate of Structure of a metal in gas phase

https://chemistry.stackexchange.com/questions/4801/are-there-any-free-electrons-in-metal-vapor

Similar thing as https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions

Similar but not quite identical.  That one seems to concentrate on conductivity whereas my question was more generic and not restricted to their electrical properties.  Also, I have received which I will accept tomorrow unless an even better one comes along.

Also related: https://chemistry.stackexchange.com/questions/104434/is-there-a-metal-for-which-its-gas-is-necessarily-a-plasma/

There's no such thing as "positive ions in a sea of electrons", that's a misconception.

The is confusing of 2 meanings of "metal" in the question:  metal(1) as a substance having metallic properties/behaviour, metal(2) as a chemical element with its solid (or liquid) phase having properties of metal(1). No gas of metal(2) is metal(1).

@Poutnik I meant elements that are metallic in their familiar form at or near STP.  The purpose of the question was whether any metallic properties are retained in the gas phase.  If I already knew then I would not need to ask.  The accepted has assumed the correct interpretation and answered it well.

No, gaseous metals do not retain metallic bonds, nor metallic conductivity, nor  luster, nor any other metallic properties. They are no different from other gases. True, they typically require pretty high temperatures to form, but then again, they are hardly special in this regard, as many non-metallic substances require the same.
See, all metallic properties are in fact collective effects. They are caused by metallic bonding, and not just of two atoms, but of an entire piece of metal (like a thousand, only much more). You don't have bonding in gases. The particles (lone atoms in this case) are basically free to go. Maybe they bump into each other, two or three at a time, but definitely not a thousand.
Plasma is an entirely different thing, unrelated to your initial question. Yes, plasma is pretty different from a gas; to obtain it, you'll have to heat the gas until it ionizes.
So it goes.

There could be molecules/small clusters, but that would not be enough for properties we'd recognize as metallic.

Thanks.  I suspected something like that but wanted confirmation.  So, we can have gaseous iron which is not a plasma?  Just neutral iron atoms (or small clusters) rather ions in a sea of electrons.

Yes we can.$,$

About molecules of metal atoms in the gas phase, I know that Li2 (real dilithium) exists, since it's discussed in my chemistry textbook and plenty of other sources. A look at the citations of this page on Li2: https://cccbdb.nist.gov/exp2x.asp?casno=14452596 leads me this free access paper: https://aip.scitation.org/doi/10.1063/1.2436891 where, in addition to Li2, Na2, NaLi, Mg2, and Si2 are apparently studied experimentally.

Actually, I think that paper just references experimental studies of those molecules, and is not an experimental study itself, but rather a compilation and theoretical analysis based on that data. Still, it clearly is said to be "experimental" data.

All right, Li2 exists, but gaseous Li does not consist of it.

Li2 does exist in the gas phase the amount depending on the temperature. There is much practical information about gaseous metals compositions in the literature about atomic absorption.

I'm not an expert, but I think some metals boil directly into the plasma state, and that other metals (and most other gasses in general) start out as almost entirely unionized gasses when they boil and then slowly become more and more ionized as the temperature goes up. Thus, the difference between gas and plasma is more of a matter of degree than a first-order (sudden) phase change. https://en.wikipedia.org/wiki/Degree_of_ionization
There are two pieces of "evidence" I want to present for what I'm saying.

Certain metals have extremely high boiling points. For example Tungsten and Rhenium both have boiling points that are slightly higher than the surface temperature of the sun. Both elements have lower ionization energies than Iron (https://www.lenntech.com/periodic-chart-elements/ionization-energy.htm), which is mostly ionized at these temperatures even though Hydrogen and Helium aren't (https://www.astro.princeton.edu/~burrows/classes/204/stellar.atmospheres.HR.pdf , pg. 18), so Tungsten and Rhenium should be ionized at these temperatures.

https://www.lenntech.com/periodic-chart-elements/ionization-energy.htm

I've recently been using this USGS paper as a reference. https://pubs.usgs.gov/bul/2131/report.pdf (It does mysteriously repeat several pages with different page numbers at one point, but it's probably mostly trustworthy.) Starting on page 67, this paper contains constant-pressure heat capacities (Cp) at different temperatures for various elements, most of which are metals. This notably includes gaseous data for monatomic calcium, cadmium, cesium, mercury, potassium, lithium, magnesium, sodium, strontium, and zinc.Cp is defined as the minimum amount of energy that needs to be put into a given amount of  a substance in order to raise it by certain temperature. In this paper, it is how many joules are required to raise one mole of a substance by one kelvin. Temperature itself is related in a definite way to the average translational kinetic energy of a particle, which is $\frac{3}{2}$kBT. In a monatomic ideal gas, translational motion is the only significant way that the energy of heat added to a system is stored, so the amount of energy you put in is directly related to the change in average translational kinetic energy, and therefore to the change in temperature. Thus, Cp has the same value for all monatomic ideal gasses (provided it is per mole rather than per unit mass). That value is 20.78615$\frac{J}{mol K}$. However, other types of matter can store some of the energy in other ways, and therefore have higher heat capacities. For example, polyatomic particles (molecules) can store more energy in vibrations and rotations; solids, liquids, real gasses, and supercritical fluids store potential energy in overcoming attractive (and repulsive) forces between their particles that are significant even when not colliding in such cases; and plasmas can store energy by knocking electrons out of their atoms. https://en.wikipedia.org/wiki/Equipartition_theorem https://en.wikipedia.org/wiki/Molar_heat_capacity#Physical_basis_of_molar_heat_capacityReturning to the USGS paper, it can be seen that Cadmium, Mercury, Potassium, Lithium, Magnesium, Sodium, and Zinc all have a Cp of 20.79$\frac{J}{mol K}$ just above their boiling point; Cesium mysteriously has a Cp of 20.78$\frac{J}{mol K}$ just above it's boiling point (possibly an error, though I doubt that); and Calcium and Strontium both have a Cp of 20.84$\frac{J}{mol K}$ just above their boiling point. (Their identicalness is probably mostly a coincidence.) After boiling, Cesium, Potassium, Lithium, and Strontium all show an increase in Cp as temperature increases up to the point where the data stops, which is 1800K in all cases here mentioned. Since all of these substances are described as "ideal monatomic gas, P = 1 bar" by the paper, they shouldn't be able to store energy through rotations, vibrations, or interatomic forces (at least not significantly), so the most likely place for this energy to be going is into ionization, or at least into exciting electrons if not fully removing them from atoms. This would explain why only very electropositive, low-ionization-energy metals show this increase below 1800K. (Other metals probably show the same increase at higher temperatures.) The main hole in this theory is that sodium still has Cp=20.79$\frac{J}{mol K}$ at 1700K and 1800K, where Lithium and Strontium show an increase, even though these latter 2 metals have lower ionization energies. In light of this, let's take a closer look at the relevant data from the paper compared to the ionization energies agreed on by that Lenntech page I linked to, nist.gov, and my CRC (although my CRC seems to have a rounding error on Cadmium, giving 8.993 eV, rather than 8.994):When it is taken into consideration that the paper generally only gives data at 298.15K, 300K, 100K increments above that, and phase-/polymorph-change temperatures, it can be seen that sodium is the only thing about this table that is not consistent with the idea that elements with lower ionization energies show the increase in heat capacity above 20.79$\frac{J}{mol K}$ at lower temperatures, and it is entirely possible that these values were not actually measured for sodium but were just calculated based on the assumption that it was an ideal gas. However, given how little evidence I had for my theory to begin with, this inconsistency actually throws serious doubt onto it. Thus, I tried to find other sources for the heat capacity at constant pressure of gaseous sodium as a function of temperature, but I could only find calculations, no experiments, and the values they gave were so completely different from that USGS paper that I feel I must be missing something. (Moreover, all sources I found seem to be related to cooling systems for nuclear power plants and focus mainly on liquid Na or Na-K alloy.) For example, this paper ( https://www.ne.anl.gov/eda/ANL-RE-95-2.pdf ) gives calculated heat capacities in $\frac{J}{kg K}$ for "sodium vapor" from 400K to 2500K on page 15. Converting 20.78615$\frac{J}{mol K}$ to $\frac{J}{kg K}$, based on the atomic weight of sodium being 22.98976928$\frac{mol}{g}$, gives 0.9041478$\frac{J}{kg K}$, which this paper passes between 400K and 500K for Cp, with Cp continually rising with temperature according to them. (Noting that Cv is heat capacity at constant volume, which should be lower at 12.47169$\frac{J}{mol K}$ = 0.1075126$\frac{J}{kg K}$ for sodium)

So, it looks like I'm still not really sure what heat capacities of gaseous metals say about their ionization, but I still think it's pretty likely that high-boiling-point low-ionization-energy elements like Hafnium, Uranium, Zirconium, and Tungsten are quite ionized as soon as they boil and don't really have any stable unionized gas phase at atmospheric pressure except as vapor in equilibrium with solid or liquid metal. That being said, the type of ionization I've been talking about for most of this post is probably quite low levels of ionization in mostly neutral gasses, so it probably wouldn't be fair to say that strontium, for example, is a plasma as soon as it boils, since it's probably just a mostly neutral gas that has a very small, but non-trivial, amount of ionization. Also remember that it's probably possible that most or much this early increase in heat capacity comes just from exciting electrons into higher energy orbitals within atoms rather than actual ionization.

There is copernicium which supposedly is a gas at room temperature,before it decays radioactively.

I've just realized that the "something" I was missing about sodium vapor might be related to pressure. Also, @Bruh Moments , that does raise the (perhaps somewhat philosophical) question of whether such a thing as Copernicium gas at standard temperature and pressure can exist. I think it probably pretty clearly could (briefly), given that Cn-285 has a half-life of (a whole) 28 seconds, but I imagine it would heat itself up at a pretty significant rate. Hard to say how quickly, though, since most of the radiation would probably escape your gas sample.

@BruhMoments The bigger question is how you could get enough Copernicium to notice (or care) generated AND cool it down and pressurize it to STP before too much of it decayed, and, even if you had quite a bit left, that would affect the purity of your sample, not that that matters too much with ideal-like gases.

I said the "something" I was missing about "sodium vapor" having very different heat capacities from sodium gas (presumably at standard pressure) might be related to the difference in pressure between standard pressure and vapor pressure. However, that really shouldn't affect heat capacity (assuming their not averaging in the heat capacity of the air the sodium vapor is mixed with into it's heat capacity (which would probably make it's heat capacity DROP with increasing temperature)), EXCEPT that it would REDUCE ionization below the SP boiling point. These are not the differences I found.

No metal boils into plasma. The boiling points of even the most refractory metals translate to only a fraction of an electron volt as thermal energy, and ionization energies even of alkali metals are well above that.

@OscarLanzi I thought all that is required is for a substance to evaporate from a liquid into a plasma directly is, for its ionization energy to be significantly smaller than the average kinetic energy of its particles at its "boiling point." So, that's as H. H. said, Hafnium, Uranium, Zirconium, and Tungsten.

@NealConroy That does not work. One electron volt is about 8000 K, assuming an average kinetic energy of (3/2) kT, and everything boils below that. Ionization energies run into multiple eV's.

@Oscar Lanzi I think I was operating the assumption that the "line strength" (brightness of some kind) of spectral lines corresponding to different ionization states of atoms in stars with different "surface" (photosphere) temperatures in that source I mentioned reflected the fraction of that atom in that ionization state. (The graph on page 19 is good, though calcium's ionization is much more clear than iron's at our Sun's photosphere temperature.)

@Oscar Lanzi I don't think more electronegative atoms are stealing electrons, because then you would see anions in the spectra, but now that I think about it, "line strength" likely depends not just on how much of the corresponding species there is in the gas, but also on how often the relevant transitions happen for a that species.

@Oscar Lanzi This source I mean: https://www.astro.princeton.edu/%7Eburrows/classes/204/stellar.atmospheres.HR.pdf

Actually, I suppose more electronegative elements could probably make it a little easier for electrons to enter the electron gas phase, especially in very dense gasses/plasmas (though the photosphere of a star may not be dense enough). That feels like it might be kind of similar to how doping in semiconductors, ?etc., works. It's just that I think the difference in density between a solid and a gas might matter for the nature as ease of formation of an electron gas/sea. In a solid, it might be more like a superposition of orbitals, or a giant covalent bond (a "metallic bond") than like a gas.


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Use of axis of symmetry
Does an axis of symmetry determine chiralty?
Is axis of symmetry considered a true symmetry?
From the above two links, I read that a compound having axis /alternating axis of symmetry is NOT necessarily achiral. For example, trans-1,3-dimethyl cyclopentane has an axis of symmetry but it is still chiral and hence optically active.
My doubt is, then why would we look for alternating axis of symmetry in the first place? Where/How is it useful?

"I call any geometrical figure, or any group of points chiral, and say that it has chirality, if its image in a plane mirror, ideally realized, cannot be brought to coincide with itself." Lord Kelvin, Baltimore Lectures, 1904.

Alternating axes are not axes, much like a Guinea pig is not a pig.
Having axes of symmetry is irrelevant to chirality.
Having alternating axes is a necessary and sufficient condition of non-chirality, that's why we look for them.
So it goes.

Some users may be more familiar with the term "improper axis of rotation" versus "alternating axis."

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Iupac nomenclature problem

I am unable to write the iupac name of the following compound

So what's your guess? It's definitely not from Organic chemistry 101, so I guess you already should have some naming experience.

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Why are the radial wavefunction and radial distribution function different?
I'm having trouble understanding the following two graphs, which are the radial wavefunction of the hydrogen 1s orbital and the corresponding radial distribution function:

Specifically, why is the radial distribution function zero at $r = 0$, but not the radial wavefunction?
Also, why does the radial distribution function have a maximum at the Bohr radius $a_0$?
(Mathematica input for the graphs: Plot[{2*Exp[-x], 4*Exp[-2 x]*x^2}, {x, 0, 6}, AxesLabel -> {r/Subscript[a, 0], Subscript[f, "1s"][r]}, PlotRange -> {{0, 6}, {0, 2}}, PlotLegends -> {"Radial wavefunction", "Radial distribution function"}, BaseStyle -> {FontSize -> 14}, ImageSize -> Medium])

I admit I have a vested interest in this, since I made some nice graphs for it. But... please stop voting to close. Not everything has to be closed as homework. If you foresee a question can have useful answers, consider erring on the side of leniency. And it's already got two, for goodness' sake!

It is actually a quite common question. For beginners it is often not obvious where that $r^2$ term is coming from.

The radial wave function $R(r)$ is simply the value of the wave function at some radius $r$, and its square is the probability of the finding an electron in some infinitesimal volume element around a point at distance $r$ from the nucleus.
But, the infinitesimal volume of space at radius $r$ is $4\pi r^{2} dr$ (it's a spherical shell with thickness $dr$ at radius $r$). That means that probability of finding an electron at radius $r$ is proportional to $R^{2}(r)4\pi r^{2}$. But the behavior of this function is such that the probability of finding the electron at radius 0 is also 0.

The radial distribution has a different form due to integration over the angles:
If we take the absolute square of the wave function $\Psi$ and integrate over the whole volume, we get the Norm of the wave function
\begin{equation}
N = \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ \int \limits _{r=0}^{\infty } |\Psi(r,\theta ,\varphi )|^2r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi
\end{equation}
where $r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi$ comes from transforming the infinitesimal volume element from cartesian coordinates ($\mathrm{d} V = \mathrm {d} x\,\mathrm {d} y\,\mathrm {d} z$) to spherical coordinates.
For the radial probability distribution $P(r)$ we do the same, but omit the integral over $r$
\begin{equation}
P(r) = \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ |\Psi(r,\theta ,\varphi )|^2r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi
\end{equation}
which we can rearrange to
\begin{equation}
P(r) = |R(r)|^2r^{2}\int \limits _{\theta =0}^{\pi }\int \limits _{\varphi =0}^{2\pi }|Y(\theta ,\varphi)|^2\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi
\end{equation}
Where $R(r)$ is the radial part of the wave function $\Psi(r,\theta ,\varphi)=R(r)Y(\theta ,\varphi)$.
For the $1s$ orbital we $Y(\theta ,\varphi)=1$ and straight forward integration over $\theta$ and $\varphi$ yields
\begin{equation}
P(r) = |R(r)|^2r^{2} 4\pi
\end{equation}
So no matter what orbital $\Psi$ represents, $P(r)$ always has a factor $r^2$, which means we always have $P(r=0)=0$.
Your given example is the $1s$ orbital $\Psi_{1s}\propto \exp(-r)$. In contrast to all other orbitals, it does not have a polynomial in $r$. Therefore it has no node at $r=0$.

Why did we omit the integral over $r$ to find $P(r) ? $

@NikolaAlfredi Because we are interested in how the probability changes over $r$. If we integrate over $r$, the resulting function is no longer a "radial distribution".

It is a bit difficult to understand as I am a highschool student.. Please elaborate a little more.

@NikolaAlfredi I am not sure what exactly it is you don't understand. The question is about radial distribution, by definition this excludes integration over $r$.
If we integrate over a variable, the result no longer explicitly depends on that variable. For example if we integrate a simple function $f(x)$ over $x$, the result is a scalar number, not a function of $x$, e.g. $\int_0^2 x \mathrm{d}x = 2$.


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Is there a metal for which its gas is necessarily a plasma?
Related to this question:
Gaseous metals?
The top answer clarifies that even with metals, the gaseous state is distinct from a plasma.
However, my question is, is this actually true in all cases?
Or is there an element and a pressure for which heating it to a gas will necessarily also be hot enough for it to immediately become a plasma?
Or is there a situation in which heating can bring about the transition from solid or liquid straight to plasma while skipping gas?

Well, usually plasma is considered different state of matter then gas, just a nitpick ;) That being said elements with high boiling points under high pressure could go directly from liquid to plasma, I think. There are also more "arcane" ways to make plasma, even very cold one.

The line between neutral gas and plasma is a blurry one. Gases have different amounts of ionization, from almost entirely neutral to almost entirely ionized, and thermal ionization gradually goes up as temperature increases. Also, you may want to look back at "Gaseous metals?" if you haven't since I wrote my answer, since I discussed this a bit and argued (perhaps not very clearly) that heat capacity data probably showed evidence for some ionization or at least electronic excitation in gaseous calcium and strontium at their 1 bar (normal) boiling point.

In principle, all that is required is for a substance to evaporate from a liquid into a plasma directly is for its ionisation energy to be significantly smaller than the average kinetic energy of its particles at its "boiling point". Or in other words, as a very rough estimate this is true when $\mathrm{k_BT_{boiling}≳EI}$.
Going further down in the periodic table, elements tend to have higher boiling points and lower ionisation energies. After reaching the superheavy element region (Z>100) things get much more complicated due to strong relativistic effects, but the elements from Rf to Hs (Z=104 to 108) are expected to have the highest boiling points at ambient pressure of any known chemical substance, possibly well past $\mathrm{6000\ K}$. However, their calculated ionisation energies of $\mathrm{6-8\ eV}$ suggest they would only ionise considerably at temperatures 10 times higher still ($\mathrm{1\ eV/k_B= 11605\ K}$), meaning there is still a long way to go.
As others have mentioned, increasing the pressure is a reliable way to push boiling/sublimation points higher, up to where they can match and surpass the energy regime required for ionisation. However, this would require absurdly high pressures; for example, the boiling point of tungsten at $\mathrm{1000\ bar}$ is still a mere $\mathrm{9000-10000\ K}$ (ref). Thus, a very crude extrapolation would suggest the direct conversion of liquid (or more likely solid) tungsten to a plasma would happen around $\mathrm{90000\ K}$ under a pressure of $\mathrm{20\ Mbar}$ (20 million atmospheres!).

The 90000 K 20 Mbar boiling of Tungsten you imagine is probably impossible, since that is probably far beyond the critical point of Tungsten, based on the less crude extrapolations of Leitner and Pottlacher: https://www.researchgate.net/publication/333306771_Density_of_Liquid_Niobium_and_Tungsten_and_the_Estimation_of_Critical_Point_Data/link/5ce6089ca6fdccc9ddc7019b/download and previous literature they mention. In the supercritical regime, it would just gradually shift from liquid-like to gas-like density.

Yes! If you look at a phase diagram of an element, you will see that with extremely high pressures, extreme heat will keep it a liquid. Eventually, theoretically with some elements, a phase change could occur directly from liquid to plasma, but the pressure would have to be incredibly high. This is caused from the electrons moving too far away from the nucleus.


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What is the real structure of atom showing every ORBITAL
Since the Bohr's atomic  model was rejected and new concept of orbital was introduced so what is the real structure of an atom showing each and every orbital?

The quantum mechanical model of the atom is the best model of the atom so far. The quantum mechanical model can take some time to explain fully, so I will try to type a short summary.
Atoms, in terms of electrons, can be separated into energy levels. Electrons rest in the ground state, and energy can excite these electrons into higher energy levels.
Each energy level has a certain amount of sublevels. Sublevels range from s,p,d,f,g, and so on. The first energy level only has one s sublevel. The second energy level has an s and a p sublevel. The third energy level has an s,p, and d sublevel. This pattern continues theoretically forever.
Each sublevel is broken down into orbitals. An orbital always can only carry 2 electrons. The s sublevel has 1 orbital, so it can carry 2 electrons total. The p sublevel contains 3 orbitals, so it can carry 6 electrons. The d sublevel contains 5 orbitals, so it can carry 10 electrons. This pattern also continues.
To put it simply, an orbital is just a region of space where electrons are to be found in 90% of the time, and the quantum mechanical model explains orbitals in more depth. I suggest you do more research on the quantum mechanical model, so you can understand it.


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Enantiomers and couple of enantiomers
I ask myself a question for some time but it's impossible to me to find the answer!
I can't understand the difference when we say " that molecule has a enantiomer " and when we say " has a couple of enantiomer ?"
*That could by vague, here is an image (the molecule 1)
Then, if I understood well this present molecule has " 1 couple of enantiomer " and " 2 couples of diastereoisomers " because it's a mesomeric compound RS, thus RS = SR.
But let us admit that I shall have taken one element with 2 asymmetric carbons : with 4 different compounds on each of the carbons (as the molecule 2) which is SR, I shall at this moment obtained **"1 enantiomer ": RS, 2 diastereoisomers SS and RR, and " a couple of enantiomer " (= which is in fact the 2 diastereoisomers), it is good that please?
(I join an image also for more understanding.)
Thank you in advance for every person who ventures to help me!

Welcome to chem.SE. Firstly, there's no “énantiomère“ nor “diastéréoisomère” in English. If you mean “enantiomer“ and “diastereomer” (“diastereoisomer”), could you fix it?

You're right sorry ! As I am binational, I make sometimes unintentionally errors like this ...

According to the definitions

enantiomer
One of a pair of molecular entities which are mirror images of each other and non-superposable.

and

diastereoisomerism
Stereoisomerism other than enantiomerism. Diastereoisomers (or diastereomers) are stereoisomers not related as mirror images. (…)

I think it's best to use the terms enantiomers or diastereoisomers only when comparing two structures (or more eventually, for the latter term); as descriptive terms with opposite meaning (but it's more complicated, diastereoisomerism includes also e.g. cis/trans isomerism as well).
E.g. asking “How many enantiomers does this compound have?” can be confusing.
Also “(a) couple” might be confusing as well (in general language it can mean unspecific small number), I'd rather use term enantiomeric pair (even thought it's not an official term).

Thus if I understood well, my reasoning is correct if I use the good terms? Or not at all?
Because my problem is when I've e.g a molecule with 2 asymmetric carbons with 4 different substituants, I don't know if the latter presents "1 enantiomer" or "1 enantiomeric pair" (I don't know what to use as other term beyond that, sorry).
I don't know if I'm rather clear ?...
E.g if I have RS molecule then this one will present a molecule SR that is its "enantiomer" and, "2 diastereoisomers" SS and RR which are in fact one " enantiomeric pair " of the molecule RS ?


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How does the theory of thermodynamics account for warm water freezing before cool water?
According to typical sources on thermodynamics, the warmer a liquid is, the longer it will take to freeze it.
For example, I recently was reading a book on refrigeration and this book specifically said that the hotter the refrigerant is, the longer it will take to condense it, so the refrigeration system has to take that into account when it is designed.
The basic theory is that there are two quantities heat of cooling and then latent heat of melting. So, the refrigeration system must remove both to freeze the liquid. Since the refrigeration system generally has a fixed rate of heat removal, the hotter the gas is, the longer it takes condense it.
However, it is well known that hot water freezes faster than cool water. If two cups containing water are placed in a freezer, one being hot, the other cool, then the hot one will freeze first and this can be proved by anyone. So, how is the anomaly explained theoretically?

Scientifically, if all things are equal, having the hot water freeze faster is impossible. So all things aren't equal. Specifically for example the hot cup melts the frost in the freezer and makes better thermal contact. Another factor is that the hot water can evaporate a substantial amount of liquid.   So the phenomena isn't well known, it just has been well hyped without proper experimentation.

https://physics.stackexchange.com/a/161447/66150

The comment by MaxW is insightful: the phenomenon is not well-known, just well-hyped.
Well, let's assume that the phenomenon is real, that is, can be observed, at least under certain circumstances. If you believe in the phenomenon, even if is is only observable sometimes, if you do enough experiments, you will observe the phenomenon. A good explanation for the inconsistency is that there are factors involved which are energetically much smaller than the energy of freezing, and we don't know how to account for them.
Now, in practice, such a variation in outcomes might cause a process upset: something clogs, or doesn't clog when it is supposed to. In such cases, an overriding effect will be installed (like a stirrer or bubbler) to shake things up so things fall where they are expected.
But there is lots of fun possible designing ever more sensitive experiments to find out why it happens (and why, in the end, the 2nd Law is true). The second law of thermodynamics says that heat flows naturally from an object at a higher temperature to an object at a lower temperature, and heat doesn’t flow in the opposite direction of its own accord. The glitch in the usefulness of the Law here is that thermodynamics does not specify kinetics (the rate of heat flow).

This does not even pretend to answer the question.


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Effect of redefining Avogadro's constant and kilogram on molar mass
Avogadro's constant $N_\mathrm A$ is defined as the number of constituent particles (usually atoms or molecules) contained in the amount of substance given by one mole.
The value of Avogadro's constant $N_\mathrm A$ is going to be set to exactly $6.02214076\times 10^{23}\ \mathrm{mol}^{-1}$ in 2019.
I might add that the kilogram is also going to be redefined in 2019 based on Planck's constant.
Will these changes result in the molar mass of a substance no longer being exactly equal to the molecular mass expressed in grams?

related https://chemistry.stackexchange.com/questions/41273/is-this-definition-of-mole-correct

Atomic weights will no longer be based on carbon 12. But unless you're experimenting to 1 part per billion or better you won't notice the difference.

Will these changes result in the molar mass of a substance no longer being exactly equal to the molecular mass expressed in grams?

Yes. The ratio between the molar mass and the relative molecular mass is called the molar mass constant, and it used to be 1 g/mol exactly. Now that Avogadro's constant is set to a fixed value, it's value has to be determined experimentally.
NIST quotes its value as 0.999 999 999 65(30) x 10-3 kg mol-1 (CODATA 2018).
The relative molecular mass of the carbon-12 isotope is still exactly 12 and the atomic mass unit u (or dalton Da) is still defined as 1/12 of the mass of a carbon-12 atom, so the relationship between relative molecular mass and molecular mass still holds. However, its molar mass is no longer exactly 12 g/mol.
As @MaxW says in a comment, these changes are so minute that they have no practical implications outside of metrology (which was one of the goals of the International Bureau of Weights and Measures when they suggested the change).


text:
Are the molar extinction coefficient and cross-section of an absorption related to the probability of absorption?
I have struggled to find a formal definition of cross-section of absorption; from what I've gathered, its best defined as 'the intensity of absorption'.
Wikipedia's formal definition of the molar extinction coefficient is 'how strongly a species absorbs EMR at a given wavelength, per molar concentration'.
So how are the molar extinction coefficient and cross-section different? I presume that whilst the molar extinction coefficient remains constant for a given wavelength across all concentrations, the cross-section differs for a given wavelength with changing concentration.
Is either the molar extinction coefficient or cross-section of an absorption equal to the probability that a single species absorbs a photon at any one point in time?


text:
Why is ammonia still considered polar despite undergoing rapid inversion?
Here it is stated that a molecule with a rotation axis cannot have dipole moment perpendicular to the axis, as any dipole moment would be changed upon rotation about this axis (which preserves the molecular geometry and hence the dipole).
The ammonia molecule (and other amines) are known to undergo inversion rapidly. When the molecule inverts, its dipole moment would be reversed. So why is ammonia still considered a polar molecule?

Solvents are polar, molecules have dipole moment, and quadrupole, etc. Only for a split second during inversion NH3 is approximately flat and it doesn't do it spontaneously.

possible duplication of https://chemistry.stackexchange.com/questions/62262/chemical-bonding-and-dipole-moment/62316

The ammonia molecule inverts itself roughly 25 billion times per second (https://en.wikipedia.org/wiki/Nitrogen_inversion , http://courses.washington.edu/phys432/NH3/ammonia_inversion.pdf). Thus averaged over a long time the dipole moment of an individual ammonia molecule will be zero, as for every moment there is a dipole pointing in one direction, this will be cancelled at a different time by a dipole in the opposite direction (as the system is charge neutral the dipole takes the same value independent of where you place the origin, so molecular motion won't affect this)
However note the word "long" in the above. Any process that happens on a short time scale will not experience this long time averaging of the property, but rather the instantaneous, non-zero, value of the dipole. So what is short and what is long in this context? Well 25 billion inversions per second means that 1 inversion takes 0.04ns. How far does a molecule move in that time? Well taking an order of magnitude value from https://en.wikipedia.org/wiki/Thermal_velocity of 400m/s a molecule moves 160 Angstroms in the time it takes one inversion of ammonia to take place. This is a HUGE distance for a molecule, thus the motion is dictated by the instantaneous, non-zero, value of the dipole. Thus we talk of Ammonia as a polar species.

At the molecular level, ammonia has a dipole that is parallel to the axis of rotation. At any point in time, the molecule is trigonal pyramidal. Although it can undergo inversion, the trigonal planar structure, with no dipole, is a transition state, with infinitesimal life-time.


text:
What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH?

What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH?
  The $K_a =1.8 × 10^{-5}$ for $\ce{CH_3CO_2H}$. Assume that the volumes of the solutions are additive.

I've done the following:
Write the chemical reaction.
$\ce{CH_3COOH + KOH <=> CH_3COOK + H_2O}$
Then I calculated the number of moles of the reactants.
$\ce n({CH_3COOH})= c \cdot V = 0.001 \text{mol}$
$\ce n({KOH})= c \cdot V = 0.001\text{mol}$
Then I assume the mistake lies here:
I thought because 1 mmol of $\ce{KOH}$ reacts with 1mmol of $\ce{CH_3COOH}$ this will form 1 mmol of $\ce{CH_3COOK}$ which leads to no formation of $\ce{OH^-ions}$. Because there is no excess. Another thing I thought is that $\ce{OH^-}= 1 \text{mmol}$. So I could calculate $\ce{pOH= 3}$ but to no avail because of the answer I later saw that is shown below.
The final outcome is $\ce{pH = 8,72}$.
A hint of where my thought process doesn't make sense would be appreciated? Thanks in advance.

Is there something wrong about my question? I'm ready to change things.

You need to consider pKa of acid in calculation.

Oh, do I need to use the Henderson-Hasselbach equation?

No, for a moment I thought there was more NaOH. Find concentration of resulting salt and plug it into equilibrium - yes you ignored it, acetate is weak base - and calculate pH.

Why $\ce{NaOH}$?  I don't see any $\ce{NaOH-molecule}$. Do you mean $\ce{KOH}$?

Part A:  The conceptually part: Scientific concepts and conception
1-"I thought because 1 mmol of KOH reacts with 1mmol of CH3COOH this will form 1 mmol of CH3COOK"
yes ,the reaction reached at the equivalent point, the weak acetic acid in mixture has been fully neutralized by the stong($\ce{KOH}$) base (no unreacted acid or base present anymore), the solution will only contain $\ce{H2O}$ and the basic salt ($\ce{ CH3COOK}$),so the concentration of the salt :
 $$[\ce{CH3COOK}] =\frac{\text{moles of salt}}{\text{Total volume}} =\frac{1\text{m}mol}{(10+10)\text{ml}}=0.05\text{M}$$
 2-"which leads to no formation of OH−ions. Because there is no excess."
No, the result, at the equivalence point, will be the same as dissolving $(\ce{CH3COOK})$ in water at the same concentration(0.05${M}$) as you have at the equivalence point.
the salt ($\ce{ CH3COOK}$) is  electrically neutral substance formed by cation $\ce{K+}$ (an acid)and  anion$\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COOK   -> K^+_{(aq)}   + CH3COO^−{(aq)}}$$
a)  Since $\ce{K+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{K+}$  actually spectator ion.
b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions$\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$
Part B:: Calculating the  approximate pH of $\pu{0.05 mol L-1} \ce{CH3COOK}$ with neglecting water autoionization:
The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:
$$ K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{a}} =\frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}$$
Assuming :${[\ce{OH-}]=[\ce{CH3COOH}]}$, ${[\ce{CH3COO-}]=[\ce{CH3COOK}]_0}$
Substiute $K_\mathrm{a}\ and\ [\ce{CH3COO^-}]$ in the equilibrium equation and solve for $[\ce{OH-}]$.
$$[\ce{OH-}] = 5.27\times{10^{-6}} , \mathrm{pOH}=5.278 , \mathrm{pH}=8.72$$


text:
Molecular orbitals symmetry - states notation
I'm reading some papers and I'm repeatedly seeing the following notation of system states:

Could you, please, explain to me the meaning of it?
E.g. the state $B^2 \Sigma_u^+$ - I'm aware, that $\Sigma$ is used as a substitute for $A$ in both $C_{\infty h}$ and $D_{\infty h}$. I'm also aware, that $u$ stands for "ungerade", i.e. it tells us, that the state is anti-symmetric with respect to the inversion.
But what does the $B^2$ before $\Sigma$ stand for? And what does $\Sigma^+$ mean?
I'm aware of the Mulliken notation, but it doesn't seem to contain these symbols.

The symbol $B$ is explained in greater detail here: Letter codes in molecular term symbols. In this context, it indicates that it is the second excited state of the same multiplicity as the ground state. The ground state, labelled with $X$, is a doublet; the first doublet excited state is labelled with $A$, and the second doublet excited state is labelled with $B$.
The superscript $2$ denotes the multiplicity, i.e. this is a doublet state, with total spin $S = 1/2$ (technically the spin quantum number) such that $2S + 1 = 2$. The dinitrogen cation $\ce{N2+}$ will have one unpaired electron in its low-lying electronic states, so it's not too surprising that the total spin is $1/2$.
Finally, there is a bit more to the symbol $\Sigma$ than just being a substitute or a generalisation of $\mathrm{A}$. In the linear point groups, this denotes the total orbital angular momentum about the internuclear axis. In this state, there is no orbital angular momentum about this axis, and the relevant quantum number is $\Lambda = 0$, which is denoted by the letter $\Sigma$. ($\Lambda = 1$ is a $\Pi$ state, $\Lambda = 2$ a $\Delta$ state, and so on.)
If you think about the ground state electronic configuration of $\ce{N2+}$, it is essentially the same as $\ce{N2}$, but with one unpaired electron in the $\sigma_\mathrm{g}$ orbital. In order to generate any kind of orbital angular momentum, you need to have an asymmetric population of $\pi$-type MOs. Since this isn't the case, the ground state of $\ce{N2+}$ has no orbital angular momentum, and is consequently a $\Sigma$ state. A similar argument can be applied to the second excited state, which seems to arise from a $\sigma\to\sigma$ excitation.

Just one more question - X, A, B indices provide "indexing" among the states with the same multiplicity, but the same symmetry/angular momentum is not needed? In my example, there are states $\Sigma_g$, $\Pi_u$ and $\Sigma_u$ - are the letter codes really common for all three?

@Eenoku, as far as I know, it is only to do with multiplicity and not the other bits. My sources are in the linked answer.


text:
What is the lining of "Silver-lined" gloves actually made of?
I bought some rubber gloves that claim to be "Silver-lined".  I think this is supposed to make the gloves easier to put on and take off.  Is there an actual silver compound lining the gloves that makes them slippery?
For example these gloves here are only $3.30.  And the description is as follows: "Silver lining enables easy on/off glide and eliminates particle shedding from the glove.".

Are you talking about Silver Sheild gloves?

No, these are fairly regular rubber gloves,  for example, http://www.highgategroup.net.au/silver-lined-rubber-gloves-blue

I don’t see any reason not to assume that they have a very thin silver lining. As in actual metallic silver. Very thin layers cost almost nothing.

True, but it's surprising how hard it is to get any actual information about these things!

its just a different colour, they use a process called chlorination witch makes them glide on and off, otherwise things would stick, alot of latex fashion clothing is chlorinated and even disposable latex gloves but not nitrile,


text:
Are there any descent symmetry tables for character tables of linear structures?
I know about tables containing information about descending in symmetry, i.e. from larger point  groups to their sub-groups.
We can see such tables in this answer, or in this PDF.
My question is - does any similar table exist even for $C_{\infty v}$ and $D_{\infty h}$ point groups?

On page 80 (Chapter 3) of 'Molecular Symmetry and Group Theory' by R. Carter, publ Wiley there are tables for the partial descent from $C_{\infty V} \to C_{2V}$ and $D_{\infty h} \to D_{2h}$ and more importantly with an explanation of how to derive them.

@porphyrin Thank you! Do you mean the 2009 edition?

my edition is 1998 but it should be in newer editions

Here they are, finally I found them in some random presentation for a lecture:
$$\begin{array}{c|c} \hline
D_{\infty \mathrm h} & D_\mathrm{2h} \\ \hline
\mathrm{\Sigma_g^+} & \mathrm{A_g} \\
\mathrm{\Sigma_g^-} & \mathrm{B_{1g}} \\
\mathrm{\Pi_g} & \mathrm{B_{2g} + B_{3g}} \\
\mathrm{\Delta_g} & \mathrm{A_{g} + B_{1g}} \\
\mathrm{\Sigma_u^+} & \mathrm{B_{1u}} \\
\mathrm{\Sigma_u^-} & \mathrm{A_u} \\
\mathrm{\Pi_u} & \mathrm{B_{2u} + B_{3u}} \\
\mathrm{\Delta_u} & \mathrm{A_{u} + B_{1u}} \\ \hline
\end{array}$$
$$\begin{array}{c|c} \hline
C_{\infty \mathrm v} & C_\mathrm{2v} \\ \hline
\mathrm{A_1 = \Sigma^+} & \mathrm{A_1} \\
\mathrm{A_2 = \Sigma^-} & \mathrm{A_2} \\
\mathrm{E_1 = \Pi} & \mathrm{B_1 + B_2}\\
\mathrm{E_2 = \Delta} & \mathrm{A_1 + A_2} \\ \hline
\end{array}$$


text:
What does the volume of a gas depend on?
In my chemistry textbook, it states that the volume of a gas is directly proportional with the amount(number of moles) of that gas. However, it also mentioned before that the volume of a gas depends on its container. So if the volume of a gas depends on its container then wouldn't the volume not change as you add or take away gas molecules? As long as the container doesn't change wouldn't the volume of the gas stay the same?

What your book means is that, at constant temperature and pressure, the volume of gas is directly proportional to the number of moles.  But, in the case of a container, if you add gas molecules, the pressure increases.  So you can't apply the constant pressure rule in this case.

I must clear one thing up. gases always fill their container. If you take away gas particles, it will always fill its container. The volume of gas is proportional to many different factors, though. Based on the ideal gas equation, volume is proportional to moles and temperature. It is inversely proportional to pressure. when the text say that volume is proportional to amount, it is assuming all other variables are constant, and not changed. So yes, in general, as there is an increase in volume, there is an increase in number of moles of gas.


text:
Confusions regarding porous pot experiment
So, I have  studied  about  the porous pot experiment.Although I did understand the concepts of water level falling and rising due to difference in RMMs and thus difference in rate of  diffusion, I  still have two questions left:
Suppose hydrogen is outside the porous pot (in the jar), and carbon dioxide is in the porous pot...

Is the water level in the manometer in general dependent on just the gas in  the porous pot(CO2) or the whole mix of gas in the jar (H2 and CO2)?
Does all of the H2 diffuse into the porous pot or is  some of it left? I thought diffusion is  just the movement of gas particles to make the whole system balanced in concentration, so shouldn't there be an even concentration  of H2 in and out of porous pot?


text:
Why does benzene have 3 π–π* transitions in the UV range, but ethene, butadiene, and hexatriene each have only 1?
Are the π and π* orbitals of benzene much closer together in energy which allows for multiple transitions? Why is this not the case for hexatriene if they both contain the same number of pi bonds?

It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.
If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:

The ground state has overall symmetry $\mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(\mathrm{e_{1g}})^3(\mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:
$$\mathrm{E_{1g} \times E_{2u} = B_{1u} \oplus B_{2u} \oplus E_{1u}}$$
Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state
$$\mathrm{A_{1g} \to B_{1u}, \quad A_{1g} \to B_{2u}, \quad A_{1g} \to E_{1u}}$$
should fall within the UV range.
For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(\pi)^3(\pi^*)^1$ configuration will only correspond to one electronic state.
This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $\langle i | \hat{\mu} | j \rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.


text:
Choose the molecule with the strongest bond
Choose the molecule with the strongest bond: $\ce{F2, Cl2, Br2, I2,}$ or all are equal.
All of them are nonpolar-covalently bonded, so I had thought that the answer would be "all are equal". However, apparently, $\ce{Cl2}$ has the strongest bond, and I couldn't figure out why. Could someone help me understand this? Any help would really be appreciated.

The Bond Dissociation Energies of these molecules are in the order
Cl2 > Br2 > F2 > I2
All these molecules involve a single covalent bond. Covalent bonds rely on sharing of a pair of electrons between two atoms, an electron each from both atoms involved. The better the sharing, the stronger the covalent bond formed. This sharing is reflected in the valence shell orbital overlap that occurs between the two atoms, when a more effective overlap occurs, a stronger covalent bond is formed.
Now, the radii of the halogen atoms increases in the order of writing them down the group, F < Cl < Br < I < At. As the atoms get larger, so do the valence shell orbitals. The increase in size of orbitals carrying the same number of electrons would mean that the electron density would be lower in larger orbitals. So, for the same proportional overlap between two similar orbitals of similar atoms, the sharing of electrons would be poorer and would consequently result in a weaker covalent bond.
This should explain the strength of bonds decreasing as we go down the group.
However, in this orbital overlap consideration gets forgotten another factor that becomes important when atoms are small in size, and that is of electronic repulsions.
Fluorine is extremely small in size and so the electron density is pretty high as well. This results in an unstable arrangement if the atoms come too close, since like charges repel.
The interplay of the above mentioned effects results in Chlorine having the highest bond strength.
I may also add that the same is the case for (1) N-N bond energy as against P-P bond energy and (2) O-O bond energy as against S-S bond energy. But, do keep in mind this stands true only for single bonds.

This is all to do with bond lengths.
F - F bond has the shortest bond length between them all, for its smallest atomic radius (least shielding effect). However, the length is so close, that the lone pairs from the two atoms start to repel each other, therefore the bond would be compromised and slightly weaker than that of Cl - Cl.
For Br - Br and I - I bonds, the bond lengths are longer than Cl - Cl's, resulting in a decrease in bond strength as the attraction between the bonded electron from one atom and the nuclei from the other atom decreases. (further away, smaller the attraction)

Could you tell me which elements other than F have bond lengths so short as to repel each other (and have lower bond energy than it would otherwise)?

also, I'm guessing that in general, the trend is that as you go down a period, you have smaller bond energy — am I right in making this observation? What about across a period?

There are quite a few that have bond lengths so short, for example N-N and O-O if you compare their bond enthalpies with P-P and S-S, if I am not mistaken. But do be careful with multiple bonds. Going across the group, however, to me doesn't seem to have an obvious fixed trend, especially if you compare O-O and S-S with their respected periods. I suggest you to search for some bond length and enthalpies tables. I always recommend the IB data booklet for chemistry.


text:
What component(s) of instant coffee make it deliquescent?
If I put small, separate piles of instant black coffee powder, sugar, powdered creamer, and even something very ionic like table salt out in humid air and come back in a few days, only the instant coffee will have absorbed enough water to change to a sticky tar-like blob while the other three won't change their appearance.
Is it known what substance it is in the  instant coffee powder that makes it hygroscopic and even deliquescent; giving it the ability to almost "make itself into coffee" by removing so much water from the air that it dissolves?

The other three piles serve as everyday-chemistry equivalents of controls.

Interesting. I suspected that instant coffee would start off drier and therefore would be more avid for water, but it seems instant coffee can have as much as 2-3% moisture content, while table sugar is limited to at most 0.04%. Either the surface area to mass ratio plays a large role, or instant coffee really contains more inherently hygroscopic substances.

@NicolauSakerNeto The power is very low density and therefore quite porous, but I don' t know if that's got anything to do with it or not.

How about doing an experiment, for science? Take some sugar and salt and pulverise them as finely as possible in a blender, then spread them out a bit on a plate and check if they now also deliquesce? Another thing to keep in mind is that commercial sugar and salt are intentionally sold impure; among other things, they contain anticaking agents, such as tiny amounts of silica, exactly to avoid deliquescence which is rather inconvenient in storage.

@NicolauSakerNeto 2) I try to avoid doing experiments that are not testing a well formulated hypothesis. My current observations are from everyday spillage, not an experiment. 1) I don't have any of that stuff handy ;-)

The behavior strongly depends on the temperature and humidity in which the coffee was kept and on the specific type of coffee.
From Tea and coffee powders, M. Huang, M. Zhang, in Handbook of Food Powders, 2013:

Being hygroscopic, instant coffee particles are susceptible to the action of moisture – that is, they absorb moisture from the air. If the moisture content increases to 7–8%, the powder or granules can become a paste or solid mass (Clarke, 1987).

Clarke (R.J. Clarke, in Encyclopedia of Food Sciences and Nutrition (Second Edition), 2003) writes:

Instant coffee is relatively hygroscopic, easily picking up moisture from the atmosphere and caking at about 7–8% w/w moisture content. For example, to keep instant coffee below 5% (dry basis) moisture content, the relative humidity of the air with which it is in contact must be below 35–40%, though the precise value depends upon the nature of the instant coffee in question, primarily due to differences in porosity of the particular particles made. It is necessary, therefore, that jars of instant coffee be well sealed prior to sale.

My answer is admittedly incomplete. As pointed out in comments to the post and in the above quotes, porosity (characterized by e.g. the mean particle surface-to-volume ratio) influences condensation, but so do the intrinsic chemical properties of the particle surface.
The chapter by Huang and Zhang referenced above provides the following table (Table 20.1) detailing the composition of instant coffee powders:

Constituent   Content(% dry weight basis)
Minerals                  7.6–14.6
Reducing sugars (glucose) 3.2–13.1
Galactomannan             2.4–10.5
Low molecular organic acids   12
Brown pigments             15–28
Caffeine                   2.5–5.4
Trigonelline               1.56–2.65
Source: Belitz et al. (2009).

Simple sugars are often hygroscopic as may be organic acid compounds, and caffeine. In particular the presence of glucose (part of "invert sugar") could account for much of the absorption of water. In addition, minerals such as various potassium salts (which make up much of the mineral content of processed coffee beans) are known to be deliquescent.
The Belitz reference appears to be: Belitz et al. (2009). Coffee, tea, cocoa. In Food Chemistry (4th Edition), ed. H.-D. Belitz, W. Grosch, and P. Schieberle, 938-970. Germany: Springer. However haven't accessed the original pubs due to paywalls.

It's certainly good to have citations confirming that coffee powder does this, thank you! Do you think it's possible to dig up a full citation for Belitz et al. (2009)?

I'm guessing it's Belitz et al. (2009). Coffee, tea, cocoa. In Food Chemistry (4th Edition), ed. H.-D. Belitz, W. Grosch, and P. Schieberle, 938-970. Germany: Springer.

It's a complicated problem but because of its economic importance I'm sure it's been exhaustively studied. You may need to read a lot to disentangle all of the factors. In fact the presence of minerals is perhaps a challenge as this is ambiguous. The refs I found did not detail the nature of the salts, and perhaps that matters wrt deliquescence. Fun problem...


text:
Is the 9H-fluoren-9-ylium cation aromatic?

In this cation, there are two individual benzene rings which are aromatic. But in the entire π system, there are 12 π-electrons which should make it antiromatic. Which is correct?

Could be called aromatic, but much less than the corresponding anion. (see  https://chemistry.stackexchange.com/questions/88615/acidity-of-cyclopentadiene-when-fused-to-benzene-rings)

And this is why the 4n+2 rule is for single cycles only.  For anything more complex we don't really know much outside of experiments.

According to literature  it is actually anti-aromatic. I guess it depends on when you call something like that aromatic or not. Is the system fully delocalized, are all bonds equal or close in bond lenghts, is there considerable out of plane twisting from the planarity, does the aromatic ring current extend through the whole molecule, ...? Here it seems to be the case and anti-aromatic since both ring-systems are still connected.

@Justanotherchemist The way you have tried to include links doesn't work in comment. You can use [text](link) instead. Anyway, it is understandable this way, too. I think your comment is already halfway to an answer, care to share it as one?

@Justanotherchemist To be precise, it's in quotes: Diphenylmethyl and the related but "antiaromatic" fluorenyl cations have been compared previously…, and it's talking about antiaromatic destabilization.


text:
How does this ring expansion on a steroid system occur?

The correct answer is supposedly B, as shown by Gravestock et al.[1] In this reaction after ozonolysis of the C=C double bond, two ketones are formed. How is a rearrangement or expansion of the ring possible, since carbocations are not formed as intermediates?

Gravestock, M. B.; Johnson, W. S.; Mccarry, B. E.; Parry, R. J.; Ratcliffe, B. E. Acetylenic bond participation in biomimetic polyene cyclizations. Model studies directed toward the synthesis of 20-keto steroids. Synthesis of dl-progesterone and dl-Δ4-androstene-3,17-dione. J. Am. Chem. Soc. 1978, 100 (13), 4274–4282 DOI: 10.1021/ja00481a044.

A pair of ketones is formed by the ozonolysis with reductive work-up, they are then treated with strong base. What is the usually outcome of treating a pair of ketones with base?  Hint - no carbocation is involved at any point.

I took the liberty to tweak some of the stereochemistry, as well as the precise conditions, so that it at least matches a known reaction in the literature. Fundamentally, the question is not changed.

It is clear that the first step is ozonolysis with a slightly reductive workup, yielding two ketones. You seem to have already established that. The second step features a base and an alcohol as a solvent. The hydroxide can react with the ketones in two ways: 1) nucleophilic attack at the electrophilic carbonyl carbon 2) abstraction of the mildly acidic $\alpha$ hydrogen. With the methanol in large excess as the solvent, it could also potentially act as a nucleophile to attack the carbonyl group to eventually form hemiacetals. However, what seems to be the most preferred reaction pathway, based on thermodynamics, here is the aldol condensation to form the $\alpha$,$\beta$-unsaturated ketone. I will now go through the considerations that a problem solver could have made.
Why is the aldol condensation thermodynamically favoured over the other alternative reation pathways?
The key reason is that the formation of the conjugated enone $\pi$ system is quite energetically favoured. On the mechanism of the aldol condensation, Carey & Sundberg (2007) wrote the following, on  p. 684:

In general, the reactions in the addition phase of both the base- and acid-catalyzed
  mechanisms are reversible. The equilibrium constant for addition is usually unfavorable
  for ketones. The equilibrium constant for the dehydration phase is usually favorable
  because of the conjugated $\alpha$,$\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall
  reaction can go to completion, even if the equilibrium constant for the addition step is
  unfavorable.

What about the ketone-hemiacetal equilibrium?
The addition products (i.e. hemiacetals) formed between methanol and the ketones in the basic medium can easily be converted back to the reactants (i.e. reaction is highly reversible). The depletion of the ketone by it reacting in the aldol condensation pathway thus pushes the equilibrium towards the ketone form. However, this consideration is not even important in this case since the equilibrium constant for the ketone-hemiacetal equilibrium is much less than one.
There are numerous $\alpha$ hydrogens but which one would be removed?
To answer this question, usually we would have to know if the conditions provide thermodynamic or kinetic control. Not much information has been provided in this case so I believe it is reasonable to assume thermodynamic control since only one product would clearly dominate while there are various possible kinetic products.
In this case, the hydrogen is first abstracted from the terminal methyl group of the newly-formed ketone, forming the enolate ion. This then acts as a nucleophile attacking the carbonyl carbon of the other newly-formed ketone. Upon protonation, the $\beta$-hydroxyketone is formed. Finally, there is E1cB elimination to form the product, as shown in B.
Other addition products may result either from 1) formation of other enolates from abstracting other acidic $\alpha$ hydrogens or 2) formation of other possible addition products from the enolate attacking other carbonyls (possibly even those on other molecules). Firstly, many of the alternatives would result in ring sizes that would possess too much strain. In this case, a 4-membered ring could form if deprotonation took place at the other $\alpha$ carbon of the same ketone. Intermolecular reaction would not be as favoured here since the proximity of the groups simply make the intramolecular reaction "hard to resist".
Conclusion
I have just gone through some important considerations behind arriving at the structure B, as well as the mechanism of the aldol reaction. Many details provided may seem extraneous though... Hopefully, you are able to understand how the "ring expansion" took place now.
References
Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


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Which has the greater N–O bond length, NO₂⁻ or NO₃⁻?
Which has the greater N–O bond length, $\ce{NO_2^{-}}$ or $\ce{NO^_3^{-}}$?
I don't understand why the N-O bonds should be of different lengths in these two molecules (don't the same bonds have the same length)? I didn't quite know how to determine this and so I've tried googling the answer, but none of the answers already provided helped clear things up for me... Why does $\ce{NO_3^{-}}$ have a greater N-O bond length?

There are two ways to think of it.  First: draw out the (major) contributing Lewis structures for both ions.  Each iin has one pi bond, but it's shared between two linkages in $\ce{NO2^-}$ ion versus three linkages in the $\ce{NO3^-}$ ion.  So the $\ce{NO3^-}$ ion has less pi bonding in each linkage making those bonds weaker overall.
Second: In all of those Lewis structuresthere are nonbinding pairs of electrons on the oxygen atoms, plus a pair on the nitrogen in $\ce{NO2^-}$.  There are more such pairs in $\ce{NO3^-}$ than in $\ce{NO2^-}$.  These pairs tend to repel each other and the bonding pairs, and $\ce{NO3^-}$ requires more room, thus longer bonds, to accommodate the greater number of closely packed nonbonding electron pairs.
These two factors tend to occur together when the molecules have nonnetalluc atoms that fill their votes, as here.

You will need to draw the Lewis structure to determine if the bonds are single, double or triple. Remember that single bonds are the longest, triple bonds are the shortest.
For the nitrite ion (NO2-), there is one single bond and one double bond between the nitrogen and oxygen. This leads to a resonance structure that effectively makes the bond lengths 1.5 each
For the nitrate ion (NO3-), there is one double bond that leads to resonance amongst all three of the nitrogen-oxygen bonds. This effectively gives each bond the characteristic of 1.3 bonds, and slightly longer than in the nitrite ion.


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Molar extinction coefficient a measure of the probability of a photon of the associated wavelength being absorbed?
Obviously the molar extinction coefficient is not equal to the probability of a photon of the associated wavelength being absorbed, but is it a measure of the probability i.e. the higher the value of the molar extinction coefficient, the higher the probability of absorption of the photon per molecule?


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Does conjugation increase λmax of n to π* transitions as well as π to π* transitions?
I would presume yes if the lone pair is on an atom involved in the conjugated system, but no if the lone pair is not on an atom involved in the conjugated system.


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Converting chlorine dioxide concentration to chlorine for the water quality report
I have the data on river water quality analysis indicating chlorine dioxide concentrations of 0.6 mg/l. However, there are no water quality standards for chlorine dioxide, only chlorine. Can I calculate the concentration of chlorine in the water from the concentration of chlorine dioxide?

unless I'm misinterpreting, chlorine concentration = chlorine dioxide concentration * 0.816

Free chlorine and chlorine equivalent are 2 very different things. // There are also 2 ways how to calculate this equivalency. ClO2 -> ClO2- and ClO2 -> Cl- // And yet another thing is functional equivalence of relative effectiveness of water sanitation and of environmental hazard.

You can, in fact, calculate the chlorine concentration from the concentration of chlorine dioxide. Chlorine will normally be in some compound, so when concentration of chlorine is given, it is usually concentration of chlorine compounds.
The concentration of chlorine dioxide is 0.6 mg/L. First convert the mg to g. this results in .0006 g. Divide that value by the molar mass of chlorine dioxide (67.45 grams/mole) to get .000008895 moles of chlorine dioxide, or .000008895 moles of chlorine. Multiply this by the molar mass of chlorine (35.45 grams/mole) to get .0003153 grams of chlorine. Convert this to milligrams to get 0.3 mg (one significant figure), so the concentration of chlorine from chlorine dioxide is 0.3 mg/L.

Be aware: I think you are talking about two different things when you say “chlorine”!
The molecular weight of 35.45 g/mol that Hunter uses refers to a single chloride atom or ion. But usually when talking about “chlorine” in context with water quality (or in processes like bleaching or disinfection), it is actually hypochlorite that is meant.  The formula for Hypochlorite is OCl- and the molecular weight is 51.5 g/mol.
And again, you should be sure if the limit stated is in ppm hypochlorite-ion or in ppm Sodium Hypochlorite (Mw = 74.4 g/mol). Either case, you can use Hunters calculation with the right molecular weight.


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Are most liquids water-based?
Is it that every liquid we see is water, with more or less of x substance?
Example: urine is water with salt and minerals

No, if only because almost anything can exist in the liquid phase.

I'm voting to close this question as off-topic because answer is no, and everyone who knows what oil is can tell you that.

Many liquids that you use on a day to day basis in the kitchen or at home are water. Vinegar is just water with acetic acid dissolved in it. Cool-Aid is just water with flavoring in it. Milk is water with casein proteins and other products dissolved in it. However, many liquids are pure substances. Vegetable oil is just vegetable oil. It has no water. Acetone has no water. Ethanol has no water, yet these are all liquids. Mercury is a liquid.
Water is the universal solvent. That is why water is found in so many liquid mixtures.

Not every liquid is "water" (or better put, contains water); There are a lot of substances that contain no water whatsoever, such as nonpolar solvents (benzene as an example). The reason we hear of so many aqueous solutions in nature is due to the fact that water dissolves most salts and molecular compounds very well. Water also likes to find its way mixed in other polar substances such as alcohol, this is why we hear of 70 or 91% rubbing alcohol (this is a property known as miscibility). Most substances we see will just contain different substances dissolved or mixed with water; however, there are many other substances out there that exist without water. Here is a list of examples:

Elemental Bromine
Anhydrous substances such as anhydrous chloroform
Oils and gasoline
Nonpolar organic solvents such as benzene and diethyl ether


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Why is arsenic toxic to humans?
I don't understand why arsenic is toxic and why it does damage to humans. Could it be that it is so similar to phosphorous? If so, would that make antimony toxic as well?

Arsenic is toxic to humans due to the fact that it imitates phosphorus almost identically in its bonding patterns, as both elements are in the same group.
However, why would this have any effect? The most likely reason for this is due to the fact that when the phosphorus is substituted by arsenic in many of a cell's vital molecules such as DNA and other phosphorus containing groups, the properties of the new substance is slightly different in terms of its polarity, reactivity, etc. Because the proteins and enzymes within the cell interact with these phosphate containing groups in such a specific manner, these slight irregularity in chemical nature can have detrimental effects (you can think of the processes of these proteins and enzymes between the phosphate as clockwork, i.e. if the parts are slightly different, it will not work).
I am not as sure about the chemistry of antimony, as I have not researched that topic greatly, but I would guess it being in the same group adds to its toxicity.

Phosphorous in the human body is primarily in the +5 oxidation state.  A heavier element in Group 15 would have to match that for maximumcsimilarity, and when we get as heavy as antimony in that group relativistic effects start to interfere with that.

Wikipedia has a fairly good article outlining arsenic poisoning. To give a bsic overveiw arsenic imitates phosphorus in numerous enzyme formations, namely acetyl-COA which participates in numerous biochemical reactions including the Kreb's cycle. The Kreb's Cylce is the process used to produce adenosene triphosphate (ATP) which is used for cellular energy. ATP contains 3 linked phosphate groups (triphosphate) and removing a phosphate to form adenosene diphosphate (ADP) is where cell derive energy from. when one of these phosphates is an arsenate instead the cell, cannot use it for productive energy and starts to die.
Antimony is toxic for similar reasons, but antimonate ions are much larger and antimony tends to form in the 3+ state so antimony's oxides do not imitate phosphorus as well as arsenic does. This is reflected in antimony's oral LD50 of $\pu{7000 mg/kg}$ which is 200 times that of arsenic's oral LD50 of $\pu{14.6 mg/kg}$. It should be noted that this decreased toxicity is due to poor absorbtion of antimony and the LD50 for inhalation exposure is substantially lower, though not comparable to arsenic. Also antimony salts tend to even more toxic reflecting that antimony's biggest obstacle to killing you is entering the bloodstream.
Bismuth is even less toxic (non-toxic really) for the same reasons as the properties for bismuth compounds are even more distant from phosphorus than antimony.


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What is the condensed formula of 2-phosphoglyceric acid?
I'm looking for a simple way to visualize the position of the phosphate on 2-phosphoglyceric acid. The condensed formula seems like the best way to do this because it omits the shape of the molecule.

What about $$\ce{HOCH2CH(OPO3H2)COOH}$$
If you really want to visualize the position somewhat more, coincidentally, the C-2 atom is (the only) chiral, so you could draw an asterisk above): $$\ce{HOCH2\overset{\text{*}}{C}H(OPO3H2)COOH}$$
If you think that the phosphate structure is overcondensed, you can write it somewhat more complicated way $$\ce{HOCH2CH(OP(O)(OH)2)COOH}$$
(or with square outer bracket instead).


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Bond lengths in iodine compounds
Why the bond length of $\ce{I-I}$ bond in $\ce{I3-}$ greater than $\ce{I3+}$ . Is that because the former has more p character than s


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Standard for Chemistry Supplies
In searching for chemistry kits it often leads to stuff like this:

I have seen some nicer "boxed sets" for the ~$500 range, but I don't think this is really ideal. This one looks a little nicer. I am wondering two things:

Where the best "complete" home/DIY chemistry set is that you can get online. Including the chemicals themselves, and the equipment like beakers, pipettes, etc.
Where the standard places are that smaller or bigger labs like University student or research labs get their supplies.

I feel like I read somewhere a while back that it is being outlawed because of bad behavior, but I wanted to ask to see what is possible in terms of doing chemistry experiments/analysis in a DIY/bootstrap fashion to learn about things like way in the past. I don't know much about chemistry but it would be interesting to perhaps learn how to do analysis like you might (I imagine) find in biology, geoscience, ecology, or agriculture.

Where the best "complete" home/DIY chemistry set is that you can get online || There is no universal one. Once you decide on specific DIY branch, you should buy a separate set of chemicals and glassware, most of which are reasonably easily accessible. Furthermore, the exact choice of the chemicals and glassware might change with available setup.

This is a bit too opinion based for main site Q&A but perhaps you could ask for input in chat.

Answer on question 2.
As a former chemistry student, I saw that most chemicals were ordered at:

Sigma Aldrich
Biosolve
Fisher Scientific
ThermoFisher Scientific
Local pharmacy (at least in Holland)

Equipment was often ordered directly at the manufacturer (like an HPLC) unless the device is pretty general, like a rotary evaporator.

most of the chemical suppliers refuse to sell to consumers.

@A.K. Depending on country, purchasing chemicals can be highly regulated. The location of OP and whether he has the money or knowledge to run actual lab is not clear.


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Acidity of polysulfanes
In a chemistry olympiad I had this morning, there was a question on the acidity of polysulfanes. Specifically, it asked candidates to explain the increasing acidity of polysulfanes as the number of sulfur atoms in the polysulfane increases. The standard answer would involve usage of the concepts of inductive effects of electronegative elements. However, I thought perhaps, there was more to it. Thus, in addition to writing about the increasing stabilisation of the $\ce {HS_x^-}$ conjugate base via stronger inductive effects as $x$ inceases, I also wrote an additional point about the increasing stabilisation coming from the increased ability of the anionic sulfur atom in $\ce {HS_x^-}$ to accept a hydrogen bond from the other terminal $\ce {H-S}$ as $x$ increases.
I was thinking that as the number of sulfur atoms in the chain increases, the chain becomes more "flexible" and thus, the ease of hydrogen bond formation increases. Of course, this will mean that the acidity would increase up to a certain point, where the ease of hydrogen bond formation no longer increases. After consultation with various general inorganic chemistry textbooks, I was not able to find a reason explaining this trend in acidity.
Firstly, I would like to ask if anyone knows the correct explanation. Secondly, I would like to ask if my proposal is plausible.

Actually H-bonding should go against acidity increase here, IMO. Important would be rather stereoelectronic effects.

https://chemistry.stackexchange.com/questions/7460/is-there-a-general-consensus-on-the-causes-of-the-alpha-effect

@Mithoron But wouldn't that limit it to the addition of 1 sulfur atom? Since it would only be the "alpha" atom that is responsible. And wouldn't the alpha effect suggest basicity of the conjugate base increases with increasing number of S atoms, contradicting the trend?


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Why are there distortions in electron cloud in molecules? How does increase surface area amplify these distortions?
In the physical properties of hydrocarbons,it is said that branched hydrocarbons have relatively less boiling point than it's linear form. From what I have learnt is that it is due to the increase in the surface area of the molecule which increases the intermolecular force. Now, this intermolecular force is the london dispersion force which happens because of the instantaneous distortions in electron cloud. This means the more the surface area of contact the more the ease of these distortions in electron clouds. This explains the strength of intermolecular force of branched and linear form of hydrocarbons and thus the difference in boiling point. But why there are distortions in the electron clouds? Why surface area has to do with these distortions?

The distortions are a quantum effect
Yes, branched chain hydrocarbons tend to have lower boiling points and this depends on their surface area. But the "distortions" in the electron clouds don't depend on the surface area.
The ultimate reason why there are "distortions" (more accurately described as temporary dipoles caused because electrons are not evenly distributed) is because of quantum effects. The uncertainly principle says we can't know the combined location and momentum of quantum things beyond a certain limit. When you work through the mathematics of this, this means that electron clouds always contain random fluctuations that appear as short lived temporary dipoles. The fluctuations in one molecule's cloud can be felt by the electron clouds in other molecules (and can cause second order effects that alter their electron clouds as well as interacting with their random dipoles). When you work out the (fairly complicated) mathematics of this you get a weak, short range, attraction between even non-polar molecules. We call those London Dispersion forces (or Van Der Waals forces).
But the underlying electron fluctuations don't change because of the shape of a molecule. However, the total force of all the fluctuations added up depends on the surface area of the molecule (how can they interact with other molecules except at the surface?) More surface, more opportunity for the fluctuations to interact with other molecules.
So it isn't that the distortions themselves depend on the surface area, it is that in a larger surface there is more opportunity to have them interact and hence the total force caused by the focus is larger.


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How to derive rate law expression for inversion of sucrose with changing pH?
I came across a question worded as follows- "The inversion of cane sugar proceeds with half life of 50 minutes at pH=5 at any concentration of sugar. However, if the pH=6, the half life changes to 500 minutes at any concentration of sugar, the rate law expression for inversion of sugar can be written as?"
"at any concentration of sugar" suggests time period is independent of concentration of sugar and therefore of order 1 with respect to sugar. Reducing H+ concentration by 10 increases half life  by 10, so half life is inversely proportional to [H+], also,as a rule- half life is inversely proportional to [a]^n-1 for any order n and concentration of reactant [a], comparing exponents of a with [H+]- we get n=2, but this is clearly not the correct way to go about it since the rate law is actually given by k(sugar)(H+) while we get k(sugar)(h+)^2; that is order 1 and 2 w.r.t to reactants- So my question is what is the correct method to find rate law expression in such conditions?

Please edit punctuation to make your question more intelligible. As it reads now there seems to be a line ending missing. Also please provide definitions of [a] and n. Overall you have the right rate law it seems though.

I cleaned it up as much as I could. The rate law is not correct as per my textbook though.

d[a]/dt = -k[a][h+], where a=sugar. You seem to confuse [h+] and [a].

can you explain the method behind how you obtained that expression from the information given?

You actually answer this yourself: half life is independent of [a], a =sugar, thus n=1. As for [h+], as you explain half life is inversely proportional to [H+]. Solve the diff eq: separate terms and integrate. Then apply boundary condition sugar = [sugar]0, then solve the resulting equation for the half-life.

It is clear that order of reaction w.r.t. sugar is 1.
Let rate of reaction r=k(sugar)^1(H+)^z
Then you have to define a new half life
            -d(A)÷dt=k(A)(H+)^z                      let (sugar)=(A)
            -d(A)÷d(A)=k(H+)^z dt
             Integrate both side .
  on left put (initial concentration)to ((initial concentration)÷2) and left side from 0 to (half life )
Then ln2=k(H+)^z *(half life)
From here half life =ln2÷k(H+)^z
Then put the values given
500=ln2÷k10^-5z
50=ln2÷k10^-6z
Solve and you will  get z=1


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In a saturated solution, why is there ongoing dissolution and crystallisation?
I was studying equilibrium, and had this question..
In a saturated solution with sugar dissolved, why is there a constant process of dissolving and crystallising? What I mean by this is, why don't the already dissolved sugar just stay that way, and the excess undissolved sugar stay that way? As I thought this would seem easier, and neater (?), wouldn't it??
Is it because the undissolved sugar is still in water, and that means it will want to dissolve, but since the solvent has reached maximum amount of solutes it has to kick some sugar out and hence there is a process of crystalisation??
I am sorry for a rather stupid question... But any explanation is much, much appreciated...

What could cause one of the processes to stop? How would having a lot of dissolved sugar cause sugar to not dissolve, or lots of crystals cause crystallization to stop?

Neater, until you tag the sugar water with radioactive isotope and put untagged sugar crystal in there, and is shocked when you find tagged molecule in the crystal a few minute later.

Because equilibrium is dynamic not static
There are many things in the world of chemistry which could be simpler but are not. Equilibrium is one of them.
There are few chemical processes where equilibrium is equivalent to the situation where everything just stops. Equilibrium normally occurs when the rates of processes going forward are the same as the rates going backward, not when things stop happening. In this case the rate of dissolution is the same as the rate of crystallisation.
On a molecular level things are very dynamic. Molecules in the solution are hitting crystals all the time. Molecules in the crystal are vibrating and being hit by molecules of the solvent: sometimes the molecules in the crystal have enough energy to escape from the crystal and go into solution. Equilibrium is the point where those two reactions have equal rates, not where all the activity stops.
We know this is true in real solutions from several experiments. We can, for example, label the molecules in the crystal with radioactive versions of one of their atoms (eg carbon 14). Even if we start with a saturated non-radioactive solution and add radioactive crystals of the same compound, the radioactivity will spread to the solution (see this question Proof of Dynamic Nature of Equilibrium).
The key thing is that all equilibria in chemistry are dynamic. The major difference in many cases is how dynamic. Some happen very slowly, others very quickly. But the position of the equilibrium only has a weak relationship to the speed of the equilibrium.


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Troubleshooting electrochemical batteries
For my science classes, I have built multiple types of electrochemical batteries utilizing oxidation-reduction reactions, and I have demonstrated that they each produce potential difference across the different solid metals used using a voltmeter.  Unfortunately, I have been unable to harness any of this to light up a simple lightbulb.
For my first demonstration, I used a copper and zinc strip in vinegar solution.  There was a voltage difference across them of about 0.9V.  This was lower than the 1.1V I was expecting, so I hooked up two of these in series to produce 1.8V.  I attempted to light a simple light bulb (which works perfectly with a 1.5V D battery), but there was nothing.  No current flowed.
For my second demonstration, I made a voltaic pile with cardboard in vinegar, zinc discs, and copper discs.  Again, I had a great voltage prepared (about 3V or so), but I could get no current.
For my latest demonstration, I tried a 1.0M CuSO4 solution with a Cu strip in it connected to a 1.0M ZnSO4 solution with a Zn strip in it via a salt bridge of 0.2M KNO3.  The voltage difference was about 1.1V spot-on, but again, I could get no current to flow through the light bulb.
What could I be doing wrong?  Is there something fundamental I am forgetting here, or is there something perhaps wrong with the items I am using?  My original copper and zinc strips were about 1mm thick, if even that, so I replaced them with thicker strips in later demonstrations, but that did not fix anything.

Your experimental procedures seem to be perfect. There is just one minor error in your conclusion: you said that no current flowed. I disagree - current flowed, just not as much as you hoped. Harnessing one of these reactions to light up even the teeniest lightbulb would be similar to harnessing a bumblebee to a model airplane.
Getting a visible change upon connecting the battery is a reasonable goal. In one sense, you have already done it, by observing the needle move in the voltmeter. I know, you would like something more. A mechanical device might work well: say, a compass needle next to a coil of wire. When you connect the battery, some current will flow, some magnetic field will be generated, and the needle will move.
Another possibility: if you take a small battery operated clock (either mechanical/quartz or digital) and remove the battery, you should be able to operate it with one of your batteries (you may have to boost the voltage). These devices use very small amounts of current, in the range that your batteries can provide.

Are there ways to increase the current flowing through this circuit?  Do the thicknesses of the metal samples and/or concentrations of the solutions contribute to the current?  I could not get this setup to run my wall clock, even with 2 cells hooked up in circuit, producing 2.2V.

Current results from the reaction; if more current is needed, the reaction must go faster. But your reaction is limited by the surface area of the electrodes, the concentration of the reagents, and polarization (lack of stirring). When you hook up your battery to a load, its output voltage will drop because of internal resistances. Try measuring current as well as voltage. The power your battery supplies will be its voltage times its current when the load is connected.

You could try considering the Cottrell equation, Randles–Sevcik equation or the mass transfer limiting case of the butler-volmer equation
$$
i = \frac {nFAc_{j}^{0}\sqrt{D_{j}}}{\sqrt{\pi t}}
 $$
i = current, in unit A
n = number of electrons (to reduce/oxidize one molecule of analyte j, for example)
F = Faraday constant, 96485 C/mol
A = area of the (planar) electrode in cm2
cj0 = initial concentration of the reducible analyte j in mol/cm3;
Dj = diffusion coefficient for species j in cm2/s
t = time in s
Also try moving the electrodes closer to one another(thus decreasing the internal resistance) but not so that they are touching
Get the concentrations of the reactants(near each respective electrode) high
Mass transfer losses(concentration loss), Activation loss(Reaction loss) and ohmic losses (resistance) are all responsible for your lowered current and voltage.


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Can aromatic rings be cyanated using cyanide as a nucleophile?
For example, can I convert chlorobenzene into benzonitrile by using potassium cyanide as the reagent?

Replacing Cl in with CN is benzene can be done but requires a catalyst. Uncatalysed nothing with happen.

Can you please tell me the catalyst

There are many sets of conditions in the literature using Pd or Ni catalysis, more info here  https://www.organic-chemistry.org/synthesis/C1C/arenes/cyanations.shtm

If possible, you'd rather start with iodoarenes. One of the well-established method uses copper(I) cyanide. Please see my answer to an earlier question, https://chemistry.stackexchange.com/questions/63892/will-cyanide-substitute-bromine-in-bromobenzene/63905#63905

Chlorobenzene is generally inert to nucleophilic substitution. It was observed by Rosenmund and Braun that using an ethanolic solution of potassium cyanide and cuprous cyanide as catalyst at 200°C, chlorobenzene converts to benzoic acid. But if the ethanol-cyanide solvent is removed and temperature is raised, then the reaction proceeds to form cyanobenzene. Yield was further improved when ionic liquids like DMF was used. This is Rosenmund-Braun reaction.
Note: Working with cyanide is highly dangerous. Do not handle cyanide until and unless you are a professional chemist.
Reference

Wikipedia: Rosenmund-von Braun reaction
Will cyanide substitute bromine in bromobenzene?
Potassium ferrocyanide, $\ce{K4[Fe(CN)6]}$ can be used as a non-toxic source for cyanation.
There are loads of other catalyst for cyanation like $\ce{ZnCl2}$, palladium or nickel salts.[@Waylander]


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Q-chem equivalent solvent settings for Gaussian NMR ab inito methods
I'm attempting to use Q-Chem 5.0 to replicate CHESHIRE-esque  http://cheshirenmr.info/ ab initio calculations of chemical shift values. Unfortunately Gaussian seems to be the predominant tool in this space, and translating from Gaussian16 to Q-Chem is quite a challenge with respect to the solvent model. Currently, just to debug, I am

Performing geometry optimization in Gaussian B3LYP/6-31+G(d,p)
(gas phase)
Using those geometries in both Gaussian mPW1PW91/6-311+G(2d,p)
(giao, scrf) and Q-Chem. For Gaussian, the solvent setting is
scrf=(solvent=chloroform,smd).

For step #2, both agree in the gas phase (to within 3 digits!) but in solvent I get very different answers (Q-Chem is off by ~3ppm in 13C vs Gaussian's 2.1ppm for my dataset.) I think this suggests that I am using the Q-Chem solvent model wrong, which is considerably more complex than what Gaussian seems to expose.
For a start, I have been using:
$rem
BASIS  6-311G**
METHOD  mPW1PW91
JOBTYPE  NMR
SOLVENT_METHOD  smd
$end

$smx
solvent  trichloromethane
$end

in my solvent settings for Q-chem, with the above results. Can anyone recommend a set of Q-chem solvent settings that might yield more accurate chemical shift values and approximate whatever Gaussian's solvent model is doing? I'm also totally open to a different set of best-practices for Q-chem.

There's a Q-Chem specific forum at iOpenShell - either they, or Q-Chem technical support may have better advice. If so, please post an answer so others can find it in the future!


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Why is Buna-S called so?
Styrene Butadiene synthetic rubber is also called Buna-S. It is the abbreviation of Butadiene (Bu) Sodium (Na) Styrene (S).
Why is Sodium included?

Sodium is included in the name because it was essential in the chain polymerization that forms the rubber.
Sodium served as the initiator for the anionic polymerization.


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If the net ionic equation and complete equation balance differently, which one do you use?
Sometimes the full equation of a reaction and the net ionic equation balance out differently. If that is the case, which is the correct one to use?
For instance, in the reaction between a solution of iron(III) chloride and lead(II) nitrate, you get the following for the complete equation:
2FeCl3 + 3Pb(NO3)2 → 3PbCl2 + 2Fe(NO3)2
However, when you use the net ionic equation, you get
6Cl- + 3Pb(2+) → 3PbCl2
However, this can be reduced to
2Cl- + Pb(2+) → PbCl2
This doesn't match with the complete equation.
If the net ionic equation does not match the complete equation, which one should you use?  Or is it ok to use both?

The complete equation is not balanced as written

There is a time and a place for both the complete equation (which is missing 2 Fe(NO3)3, btw) and the NIE.  It sort of depends on what you are discussing /  doing.
As far as the question is it ok to use 2Cl- + Pb(2+) → PbCl2, even though the coefficients are different, yes.  Since the iron and nitrate are acting as spectator ion, they're not important to the NIE.  So, don't sweat it when the NIE looks different from the complete equation.  The chloride ions and the lead (II) ions could have come from (e.g.) NaCl and Pb(ClO3)2.  That's one of the points of NIE's: they focus on only the ions you need to worry about for the (very) specific reaction.


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Where will the positive charge shift in the E1 mechanism of dehydration of an alcohol?

My Attempt:  according to me the double bond should be in position no. 2 but the correct answer has a double bond in position no.1. Why is it so?
Thanks in advance.

Benzene ring in itself is a stable structure. But double bond at position 2 will definitely destruct the stability of benzene ring. While being on position 1, a double bond will be created on adjacent carbon to benzene ring which ultimately distributes the electron cloud more over the structure, leading to more stability.