text:
Could a positron be made to stably orbit an atom?
Suppose you have a single atom of neon, argon or some other noble gas of your choice. While the entire atom is electrically neutral a nearby positron would probably be more strongly attracted to the electron cloud than it would be repelled by the nucleus [since the nucleus is arguably further away]
I wonder if this means, given sufficient energy (but not too much), if it is possible for the positron to form a positron orbital around the atom.
Has this been done? Could it be feasible?

It would probably mutually circling with a particular electron before their fast annihilation. See also positronium.

Besides positronium, I guess there might be some short-lived bound states of (positron + normal atom).

Neon is a very complex atom. Forget about it for the time being. What do you think will happen if you had an anti-proton and positron just like normal hydrogen has a proton and an electron? What will happen to its spectra? If you were another world, you would not be able distinguish the normal hydrogen spectrum and the anti-hydrogen atom!

I wonder if some insight could be gained from the analytical solution of the Schrödinger equation for the $\ce{H2^+}$ cation (or maybe even the $\ce{HD^+}$ cation), if the mass for one of the hydrogen nuclei can somehow be allowed to vary all the way down to the mass of a positron. There's even a chance that, should a bound $\ce{Hp+}$ system exist, it may have a longer half-life than positronium, because the hydrogen nucleus will try to spatially separate the electron and positron, due to their respective attraction and repulsion.

A positron cannot orbit an atom. As soon as it will approach an ordinary electron, both would be annihilated, and transformed into two gamma-rays.

@NicolauSakerNeto Nice idea, in principle the Hamiltonian is the same so, after adapting the mass, you could take the bound states of that solution and answer "yes" to the question. The problem is H2+ is normally solved within the Born-Oppenheimer approximation, which will not be applicable here at least when considering the behaviour of the electron and positron.

Inspired somewhat by @NicolauSakerNeto one idea to get at least an answer is to solve separately for an isolated hydrogen atom and an isolated positron, and then use the coupling terms between them as a perturbation. All you have to do then is prove that there is at least one state of the system with E<0 to answer the problem. Much too tired to solve that now, but it might not even be too bad, the perturbation will be much weaker than in the comparable case where perturbation theory is used in the helium atom.

@NicolauSakerNeto, your idea seems to be worked out in this per by Matyús etc al.

As far as exotic atoms go, positron is essentially a nucleus not a electron. Binding it to an atom makes a molecule, not a bigger atom. For example there's positronium hydride.

Without going into the mathematics, as others have done, consider that an "orbit" describes the probability of finding a particle. So, if there is even a very small, but non-zero chance that the antiparticle gets within a certain distance to a normal electron (cloud)... Poof! you have two 0.5 Mev photons, and two fewer particles. This applies to any metastable quantum system... eventually, they decay.

Positron chemistry is a thing, though experimental lifetimes are in the nanosecond range due to annihilation.
The noble gases seem to have a relatively favourable electron configuration, reserving a separate shell for any positron that might care to join $-$ thus minimising the risk of premature encounters with their negative brethren. However, in order to bind an electron an atom must either have net electrical charge or an unfilled electron shell. The noble gases have neither.
Halide ions are more attractive, and have been observed to bind positrons to form $\ce{[X-;e+]}$. Nevertheless, the positron appears not to form an atomic orbital in these cases (possibly impeded by the large size of $\ce{X-}$ ions), but rather acts as a second nucleus in a molecule, as with positronium hydride.
Positrons can of course form stable orbitals around antiatoms, as evidenced by the synthesis of antihydrogen at CERN, but stable orbitals appear out of reach for normal atoms.


text:
Is each bag of plastic a polymer/macromolecule? If yes, why does it tear so easily?
For example, polyethene is a macromolecule. So each bag of polyethene should be one molecule. Doesn't tearing it therefore involve breaking strong covalent bonds? Shouldn't the energy needed to do this be very high? But it tears pretty easily, wondering why this is so.
Any insights welcome!

https://chemistry.stackexchange.com/questions/47522/why-does-plastic-tear-easier-when-you-pull-it-quickly https://chemistry.stackexchange.com/questions/31133/how-many-molecules-are-there-typically-in-items-made-of-plastics https://chemistry.stackexchange.com/questions/31299/ripping-apart-plastic-is-this-a-chemical-or-physical-change

PE/PP bags are not formed by a single macromolecule.
Their macromolecules are linear with very occasional branching. The mass of each such a macromolecule is many orders of magnitude smaller than the mass of the bag.
So called giant macromolecules aka network covalent bonding have mass/size equal or at least comparable to the mass/size of the solid object. Their atoms need bonding to 3-4 neighbor atoms, forming 3D structure of chemical bonds (diamond, quartz), or 2D structure (graphene, independent or as a part of graphite)
The tear beyond elasticity likely starts with breaking these intermolecular bonds keeping the individual molecules together; these are weaker than the covalent $\ce{C-C}$ and $\ce{C=C}$ bonds within a polymer chain.

@buttonwood I have taken the tearing question as conditional one, with the condition not met. :-) Thanks for improving my English and Android mistyping.

No worry. Both words probably are not used frequently enough to be considered in the default spell check, except if they offer to add words (like the ones for Firefox, or Emacs, etc.) because their user encounters them with intent. In a case of changing semantics (like calender/calendar/colander), the addition for subsequent automatic correction (flyspell like) might even be considered as the road not to be taken.

source: Stack Exchange

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

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text:
How many characters does the SMILES notation uses?
I am not a chemist, in fact, I come from a computer science background, however, I am working on a project that involves using machine learning for SMILES generation.
My question is, how many characters are used in the SMILES notation and what are they?
I looked it up with no success!

For more than two decades, Wikipedia dedicates an entry about SMILES, including references to primary references (including Weiniger's seminal paper in 1988).  There are multiple variations of SMILES (e.g., unique, canonical, inchified; see e.g. here), and other string notations (like the more recent InChI), too.

It's also worth mentioning that for generative ML applications, DeepSMILES or SELFIES are consistently better.

Thank you for this information. However, I still cannot find any source listing all the characters used in the SMILES notation.

A brief glance on the documentation by Daylight suggests for the typical organic molecules:

you need all characters to represent a chemical element by the symbol in the periodic table of elements.  Likely, this condenses to [A-Za-z].  Note, though, while upper case C and lower case c both refers to carbon, the second one is part of an aromatic system; e.g. C1CCCCC1 about cyclohexane, but c1ccccn1 about pyridine.

coverage of all [0-9], because you may refer explicitly to isotopes if necessary, e.g. [13C] about $\ce{^{13}C}$.

= and # marks double and triple bonds.  Single bonds may be noted by - (but often are implicit); on occasion you meet :.

stereochemical configuartion may be indicated by / and \ around double bonds (used pairwise) about (cis/trans) relationships; or more general (E/Z).  For sterogenic centres (mainly carbon C, but not limited to this [e.g., sulfoxides]) you need the individual @ or @@ to indicate them as either (S), or (R), respectively.  Depending on the intended use, these four indicators may be absent or partially missing.

Pairwise, parentheses ( / ) and square brackets [ / ] are used.  E.g., C(O)C is about ethanol, while COC describes dimethyl ether.  With the brackets you mark start and end of the elements with more than a single letter; e.g., [Sn] about tin, while S were about sulfur and n were about an aromatic nitrogen.
Brackets are in place to mark charges, like [Fe+2] or [Fe++] (functional equivalent) about $\ce{Fe^{2+}}$, or [OH-] about $\ce{OH^-}$.

there are complex ring closures (e.g., in buckyball fullerenes) described with the % sign.  (Credit to @Geoff Hutchison to pointing to this detail.)

You need the full stop / period . to join e.g., multiple molecules or ions into one SMILES string, like for sodium acetate: [Na+].[O-]C(=O)C

If you stop here, the set of A-Za-z0-9=#:+-[]()/\@.% is a key space of 76 characters.
The lower part of the document documents some keywords reserved to describe the spatial constitution of complexes of transition metal complexes, and the use of > to encode reactions.
Noel M O'Boyle wrote a paper about the methods to identify a canonical SMILES representation of (organic) structures which provides insight here, too.
Towards a Universal SMILES representation - A standard method to generate canonical SMILES based on the InChI in J. Cheminf. 2012; doi 10.1186/1758-2946-4-22.

Great summary. You forgot % for extended ring closure, e.g. C2%13%24 because some systems need more than just one digit for ring closures (e.g., fullerenes, other complex ring systems)

@GeoffHutchison Thank you for pointing the %.  So far, I didn't use SMILES about fullerenes and it was a new for me.

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text:
acetic acid 50% safety and storage
I need <5% acetic acid in some solutions (for microscopy) and if I am not wrong I need around 15-20% solution to be able to make these (other components like alcohols or glycerol I can find and use near 100%). I can find 50% acetic acid easily to purchase, and I am planning to dilute this to 20% or 25% immediately for storage and use later (all at home). Other than having straightforward PPE (goggles, gloves, lab coat, respirator if needed) and following the basic acid dilution procedure (adding acid to water etc.), is there any safety issue I should be aware of? I.e. I do not have a fume extractor.
I see different information in a few places but usually I read organic acids are compatible with alcohols for storage, so I assume I can store it together with for example IPA 99% ? I have a safety cabinet for flammable liquids which I use to keep alcohol based things.

As you are diluting a solution which is not concentrated, you can simply add your acidic solution in enough water, and stir. Nothing will happen. There are no safety issue that you should be aware of.

OK, so can I derive a take home message that there is a major difference, in terms of handling, between 50% and glacial forms ?

No ! Dilutimg glacial acid is weakly exothermic when mixing it with water. But you should remember that acid must always be dropped into water, and not the contrary.

OK. The reason I said this is reading/watching things about glacial acetic acid worried me a bit, so I want to know if/to what extent 50% is safer. I also see glacial has a bit different categorizations (transport numbers etc.) than solutions. Now from your answer I understand although glacial acid is more concentrated, it is (being weak) not very dangerous and 50% is not very different than that in terms of handling (naturally still with enough PPE, but proper/full lab environment not needed). Is that right ?

Acetic acid is called glacial when it is 100% pure. This name comes from the fact that, if a bottle of this acid is dipped into a bath made of melting ice, it freezes. This does not happen with significantly diluted solutions of acetic acid.


text:
water to hydrogen to water
I'm trying to find out if there is any loss of water when converting water to hydrogen and then back into water and can't find an answer anywhere I've searched. I understand that the water produced is a gas, vaporized, but I assume it can be contained and cooled back down into water. So, if you hydrogenize a quart of water and then provide the correct proportions of oxygen to turn it back into water in an exothermic reaction do you get a quart of water back? If there is no loss, then, using solar-energy, can a closed-system engine be created?

How far do you need to go in searching  to stumble upon the law of mass and energy conservation ?
Theoretically, there is no loss. Practically, it is question  of design and material science.
There can be losses in incomplete capture of hydrogen or water vapor. There can be leaks and secondary reactions like material oxidation or corrosion, consuming oxygen. Hydrogen  is vulnerable to diffusion through tiniest holes and even into or through some materials, or forming metal hydrides.
But in large extent, losses are negligible.


text:
Is there any way to make ferrofluids float on water?
I am trying to do an oil spill clean up experiment, and I was thinking of using magnetism to clean up oil. I could make a ferrofluid/ferripaste using some type of oil, such as motor oil or mineral oil, and iron oxide or magnetite powder. I don't want it to sink in the water because if we used this in real life, you don't want the oil to sink in the ocean. Here is the video on how to make ferrofluid: https://www.youtube.com/watch?v=CWH9C__jn2w.
Thank you!

Perhaps using a froth of bubbles? You could try adding a bit of detergent to the ferrofluid and use an aquarium bubbler to see:

Does this create froth?
Will that froth (if it is created) float?
Will it draw in added oil?
Can magnets remove this from the water?
Can the excess oil be removed to regenerate the original ferrofluid?

Without step 5, your idea would be economically unfeasible.


text:
What volume does CO2 converted from O2 have?
Let's say a human inhales air that contains exactly 1 Liter of pure O2 (about 4.76L regular air) and no CO2. All at normal pressure.
They manage to hold their breath until all the O2 has been converted into CO2 and diffused back into the air (of course unrealistic).
What volume will the CO2 of the exhaled air have?

My attempt:
A very trivial idea with the tiny amount of highscool Chemistry knowledge I thought of is to calculate through the number of Mole because we know that every O2 molecule will result in one CO2.
For O2: 1.331Kg/m^3 density and one Mol weighs 31.9988g.
Therefore the liter weighs 1.331g and has 1.331/31.9988mol = 0,0415953098241184 mol of O2
For CO2: 1.842Kg/m^3 density and one Mol weighs 44.0095g.
Therefore the resulting mol for the O2 will translate to 0,0415953098241184 * 44.0095g = 1.8305g
Which due to its density, takes a volume of... about 1L too?
Can this be correct?
It seems nonintuitive since we have many more atoms in the CO2 after all.

Huge thanks in advance :)

p.s. before someone claims this is for homework, I'm a software dev working on a game that shall contain a somewhat realistic simulation of the Pulmonary system :P

At the same temperature and pressure O2 converted to CO2 would not change volume. If warmer outlet temperature is considered, then volume is proportional to absolute temperature.

You treat both gases as ideal, and every ideal gas has same standard molar volume. More to the point, yes, it's (approximately) correct, as long as molecules have negligible volume. BTW even if it's for work or self study, things like that still fall under homework-type.

At the same temperature and pressure, oxygen converted to carbon dioxide by glucose oxidation would not change volume. If warmer outlet temperature is considered, then volume is proportional to absolute temperature.
$$\ce{C6H12O6(aq) + 6 O2(g) -> 6 H2O(g) + 6 CO2(g)}$$


text:
How long do temporary dipoles last?
Theoretically, if you put 2 neutral atoms/compounds together in a controlled space with no external electrical fields, would a temporary dipole last forever, or would the jumping of electrons break the induced dipole?
Generally speaking, how long do temporary dipoles last? Milliseconds? Less? More? In what scenarios would a temporary dipole last for a "long" time and in what scenarios would it last for a short time?

How long would last disturbance of spherical symmetry of electron probability density of a H atom, if the random electrostatic field disappears ?

You're not getting the concept here. While resulting vdW molecule could last indefinitely, these instantaneous dipoles arise and all the time - are basically fluctuations of electron density that happen on extremely short timescale.


text:
How to apply a certain voltage vs SHE
For one of my projects, I must apply a voltage at the electrodes of -1V and +2V vs SHE. But my device potentiostat picks a voltages of 3V. How am i certain that it doesn't apply the voltage between -2V and +1V (and if that happens can I change this somehow). Or should I just use a reference electrode?

Sorry ! I don't understand how you can imagine having an inversion of polarity. If you install a potential of + $2V$ vs. SHE, it will not suddenly change sign and become $-2 V$. Please explain !

@Maurice I do not think OP had in mind polarity inversion, but potential shift maintaining the same difference, that just accidentally  looks like polarity switch A: +2V, B: -1V ->A:  +1V, B: -2V.

Maybe have a look at something like this and other potentiostat discussions. Searching here is helpful, though potentiostats can be a bit hard to understand.

You have to use 3-electrode system, with 2 power electrodes and the SHE as the reference electrode.
The problem is that if you apply the external voltage 3 V on both power electrodes, their potential wrt SHE depends on the particular electrochemical system. The only value that is a priori given is the difference of their ECH potentials.
You need to have such electrochemical redox systems which potentials are close to the target ones. But I am afraid both potentials are extreme and there will be rapid formation of gases on the electrodes
Electrodes will establish such potentials, where there is the same (but the sign) passing electric current.
You may will need to tune the electrode potentials for the given or chosen current and/or tune current to tweak both potentials at the same time.

Increasing current decreases the potential of the cathode and increases the potential of anode (non-linearly and independently) and vice versa.

Increasing concentration of oxidized form at an electrode increases its potential for the given current and vice versa.

Increasing concentration of reduced form at an electrode decreases its potential for the given current and vice versa.

Note that 3 V potential difference is quite a lot. Either the cell resistance is high, either there is high current.


text:
Why do the splitting pattern on my NMR differ from those stated in literature?
I attempt the synthesis of p-toluic acid using the Grignard reagent (prepared from 4-bromotoluene and Mg) and dry ice. The NMR of the raw product was recorded using $\ce{CDCl3}$ and a peak showing water is present.

According to literature, the peaks at around $\pu{8 ppm}$ (hydrogens in ortho positions with respect to the $\ce{-COOH}$ group) should have a splitting pattern of a doublet, but I got a multiplet instead. Does anyone know why?

Those aren’t actually supposed to be doublets; see eg https://chemistry.stackexchange.com/questions/84390/how-can-multiplets-in-para-disubstituted-benzene-rings-be-described and links therein. Your spectrum is entirely correct.

If you don't have a matching splitting somewhere else, then it can't be a doublet.

Beside the point these signals are AA'XX' as mentioned by @orthocresol (tangential, drawings in this question show why stating these signals as doublets lacks support by theory), be glad the experiment reveals the more complex pattern (e.g., by sample concentration, solvent, spectrometer resolution, field strength).  The «water peak» in 1H NMR recorded $\ce{CDCl3}$ equally may indicate the bottle of $\ce{CDCl3}$ wasn't kept as closed as it should, and $\ce{CDCl3}$  isn't dry (anymore), too.

For a simulation of these pattern (even more useful for partially fluorinated compounds), see e.g. MNova's resources.


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Converse of ΔU = 0 for isothermal processes: is it true?
For an ideal gas undergoing an isothermal process $\Delta U = 0.$
But is the converse true? Let's say, $\Delta U$ for an ideal gas is zero. Is its temperature constant then? Can we say that it's undergoing an isothermal process?

If  there is ongoing reaction between ideal gases at constant V in isolated container, delta U = 0 but T can change very much.

We can't say that the ideal gas has undergone an isothermal process.  All we can say is that, at the two end points of the process, the temperature is the same.

@ChetMiller That's a valuable point, and I agree with you. But for state functions, I believe it doesn't actually matter.

@electronpusher I agree also with poitnik’s comment

Isothermal process means that the temperature of the system (here, ideal gas) is constant throughout the process.
For an ideal gas, internal energy is a function of temperature only and its change can be expressed by:
$$\Delta U=nC_V\,\Delta T$$
Therefore,
$\Delta U=0$ implies that $\Delta T=0$.
However, this only implies that initial temperature and final temperature are same.
The temperature may/may not be constant throughout the process.
Therefore, $\Delta U=0$ is not a sufficient condition to say that ideal gas is undergoing isothermal process.

U is independent of path. the Second condition to define an ideal gas involves V or P: [partialU/partialV][T]  equals zero. This strongly implies that a constant internal energy for an ideal gas means isothermal. Of course if there is a change in chemical potential E or obviously a latent heat change at constant T this isn't true. This is like saying Boyle's Law takes place at constant T and U instead of just constant T.


text:
Effect of H+ concentration on equilibria
Considering the two equilibriums below, when the $[\ce{H+}]$ decreases, my book says that equilibrium \eqref{rxn:2} will shift to the left by a greater extent compared to equilibrium \eqref{rxn:1} because equilibrium \eqref{rxn:2} has a greater amount of $\ce{H+}.$ (4 mol compared to 2 mol)
$$
\begin{align}
\ce{VO2+ + 2 H+ + e- &<=> VO^2+ + H2O}\label{rxn:1}\tag{R1}\\
\ce{VO3- + 4 H+ + e- &<=> VO^2+ + 2 H2O}\label{rxn:2}\tag{R2}
\end{align}
$$
However, I do not understand the link between the amount of $\ce{H+}$ and the extent of shift in position of equilibrium when faced with falling $[\ce{H+}].$ Why will equilibrium 2 shift by a greater extent?

You do not distinguish $\ce{VO2+}$, written as $\ce{VO2+}$ and $\ce{VO^2+}$ written as $\ce{VO^2+}$ // For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ).

Try to enumerate equations properly, this skill is based on counting to 20 and on conservation of atom and charge counts.

You can apply \ce{} to the whole equation, not for each formula, using -> as the reaction arrow.

If you write the equilibrium expressions, the second one has hydrogen ion concentration to the 4th power in the denominator while the first one has hydrogen ion concentration only to the second power. So changing hydrogen ion concentration has more effect on the second equilibrium. So decreasing hydrogen ion concentration by a factor of 2 results in reducing product concentration by a factor of 4 for R1 and by a factor of 16 for R2.

Since you lost 50 rep with the bounty, but got good answers, I will give you 10 back.

If we take the half reactions as given (so ignoring Maurice's insight about the stability of the species at different pH), the answers are in the comments.

[Ed V] If you write the equilibrium expressions, the second one has hydrogen ion concentration to the 4th power in the denominator while the first one has hydrogen ion concentration only to the second power. So changing hydrogen ion concentration has more effect on the second equilibrium.

and

[Poutnik] If you formulate Nernst equation for a particular half-reaction, you can easily derive the rate of potential change with changing pH.

Intuitive answer?
For each turnover in the forward direction of R1, two $\ce{H+}$ ions have to come together. Increasing the $\ce{H+}$ concentration by a factor 10 (lowering the pH by one unit) will make that 100-times more likely.  For R2, you need four $\ce{H+}$ ions. Increasing the $\ce{H+}$ concentration by a factor 10 will make that 10,000-times more likely.
The pH dependence will be the same no matter whether you formulate an acidic or basic half reaction. You can write R2 as a basic reaction,
$$\ce{VO3^- +  2H2O + e- <=> VO^2+ + 4OH- \tag{R5}}$$
and instead of four $[\ce{H+}]$ on the left side you have four $[\ce{OH-}]$ on the right side. Lowering the pH by one unit would in turn lower $[\ce{OH-}]$ by a factor 10, making it 10,000 times less likely that four $\ce{OH-}$ ions would come together. So written as R2 or as R5, this reaction is more sensitive to changes in pH.
Reaction vs half reaction
Both R1 and R2 are half-reactions that would not attain equilibrium unless they are combined with a second half reaction. For a half-reaction, you could discuss the reduction potential and its dependence on pH. Or you could combine both half reactions with the same other half-reaction (which should be independent of pH to keep things simple).

The last sentence in my comment was sloppy: I should not have said 4 and 16, as though no reactant was being produced by the equilibrium shifts to the left. So your answer cleaned up my error nicely!

@Karsten Theis Thank for the answer! Can i clarify one thing: Does the reaction quotient tell you how likely the species in a reaction are going to come together?

@john No, comparing the reaction quotient Q to K tells you how far from equilibrium you are, and also the Gibbs energy of reaction as $$\Delta G = R T \ln \frac{Q}{K}$$
This is related to the entropy of mixing, which most folks don't consider intuitive. However, you can use a kinetic argument for changing parameters of a reaction that has reached equilibrium like I did above.

Before studying your two redox equations (R$1$) and (R$2$), it is necessary to simply compare the ratio of the two ions dioxovanadium(V) $\ce{VO2^+}$ and vanadate $\ce{VO3^-}$  which both exist in solutions containing Vanadium at the oxidation number +$5$. Both ions occur in the left-hand side of your two equations (R$1$) and (R$2$).
Let's proceed like this. These two Vanadium containing ions are both interrelated by an equilibrium like $$\ce{VO2^+ + 2 OH- <=> VO3^- +  H2O \tag{R3}}$$  which is equivalent to $$\ce{VO2^+ + H2O <=> VO3^- + 2 H+\tag{R4}}$$If sodium vanadate $\ce{NaVO3}$ solution is acidified by adding some $\ce{HCl}$ solution, the reaction (R$4$) gets inversed and produces the dioxovanadium(V) ion $\ce{VO2^+}$. This reaction is reversible. By adding some solution of $\ce{NaOH}$, the dioxovanadium(V) ion $\ce{VO2^+}$ is transformed into vanadate $\ce{VO3^-}$ like in (R$3$).
The system dioxovanadium(V)/vanadate is working like an indicator. Dioxovanadium(V) ion exists mainly in acidic solution. Vanadate ion exists mainly in basic solutions.
Now, to go back to the original questions, you should realize that the equation (R$1$) is the only one that does make sense, because you may imagine a half-cell containing both $\ce{H+}$ and $\ce{VO2^+}$ simultaneously in solution. In contrary, equation (R$2$) is purely theoretical, because the ion vanadate $\ce{VO3^-}$ does not exist significantly in acidic solution. Before taking part to the redox reaction, it is immediately transformed by $\ce{H+}$ into dioxovanadium(V) ion. This is why the reaction (R$2$) needs more $\ce{H+}$ ions than (R$1$).

You don't need to spam dots like you do with .........(3), just tag the reactions with \tag{R<number>} (I use prefix R to differentiate chemistry from math). If you refer to the reaction in text, label the reaction and use \eqref{<label>} to create a dynamic link. For the ease of reading and parsing, aligning reactions about arrows is a good idea, too (similarly as one does with multiline formulas/cases/series of equations). See my edit to the main question for the example how it can be done. Finally, chemical names should not be capitalized unless the first word in sentence.

@Putnik. You are right about the possible confusion between $\ce{VO2^+}$ and $\ce{VO^{2+}}$., I have not made this mistake.

@Maurice Hi, I don't quite understand how this allows us to compare which equilibrium has a larger shift in POE when concentration of H+ decreases?

@john If you formulate Nernst equation for a particular half-reaction, you can easily derive the rate of potential change with changing pH.

@Poutnik. You are right. Sorry !  I should have used the correct name, which is dioxovanadium(V) for the ion $\ce{VO2^+}$. I will edit this change.

@Andselisk. Thank you for the information about the use of \tag{ } for labelling a reaction.


text:
Which form does Casein take in milk (before and after extraction)?
In this video, NileRed said

casein is exists in milk as a salt Calcium caseinate. And that structure of the salt is composed of 3 forms of casein. Alpha, beta, kappa. Together they form a stable structure called a micelle.

Further on he said that the kappa's polar part is bonding with the water and the non polar is bonding with the alpha and beta parts.
Then he showed that the sub-micelles are tied together with calcium phosphate.
My question is :
1- does casein actually exists in milk as a salt Calcium caseinate?
Because in this site they said :

Calcium caseinate is produce by changing the pH of milk to neutral or acid. In this state, casein becomes insoluble in water... . After separation, manufacturers combine casein with calcium hydroxide at high alkaline levels and dry the protein.

Which shows that Calcium caseinate is a product not the input. or they just re-produced Calcium caseinate again?

The second confusion is
2- Is the calcium that is tying the sub-micelle together the same that is mentioned in the name Calcium caseinate?

If YES, then by breaking the bond, do we just end up with pure casein? And why isn't the phosphate present in the name?
If NO, then which one did we break from?

Is "Nile Red", a good replacement of authentic books or published references? Many dot.com type websites without an author are dubious. Why should we trust "Live Strong"? Try searching your question in Google Scholar. I upvote because you made an effort to search elsewhere besides "Nile Red".

Please check the title question. Its meaning seems unreadable, to me.

@M.Farooq , I totally agree with your point. it's not like "Nile Red" is better than authentic books but he is better than me. that's why I'm asking other people who are also better than me. that being said, I'll check google scholar. thanks!

@Alchimista , oh my bad dude. i've no idea how I messed up the title this way. it should be fixed now.

In this article they said :

Caseins form complex particles or micelles, which are usually complexes of calcium caseinate and calcium phosphate. When milk clots or curdles as a result of heat, pH changes, or enzymes, the casein is transformed into an insoluble caseinate-calcium phosphate complex.

Which means :

Yes, in milk, it exists in calcium caseinate form
No, There are 2 different calcium in the formula
We broke the Calcium caseinate NOT calcium phosphate


text:
why Tin/Zinc alloys can be used to solder aluminum and Tin alone does not "wet" the aluminum?
It is impossible to solder aluminum with only tin, tin does not "wet" aluminum surface because of the instantaneous reaction aluminum/oxygen on the surface of the base metal (aluminum). I suppose the problem is that the tin will be in contact with aluminum oxide and not with the metal.
It is known (in this industry field) that an alloy of Tin/Zinc can solder aluminum metal, this alloy will "stick" to the aluminum surface, so what about the role (interaction/reaction)of Zinc in the "wetting" of the passivated aluminum ?
EDIT:
aluminum can be soldered by tin/zinc alloys
https://www.belmontmetals.com/popular-uses-for-tin-zinc-solders/
here a specific product
https://www.belmontmetals.com/product/60-aluminum-solder-tinzinc/
I can say from my experience that you can solder aluminum even with a homemade alloy of tin/zinc 70/30 or 80/20 using a normal gas torch in free air.

You should be more careful with capitalisation and spelling. BTW "It is known" was OK for Dothraki in Game of Thrones, but we like sources here ;)

How do you know that an alloy zinc/tin, aluminum metal can be soldered ? Have you got a reference ? I thought aluminum can be soldered under argon and only this way.

ok I edited the question, hope that now it is more clear. I insert sources about factory made products (tin/zinc) that are used to solder aluminum in open air.

@Maurice, you use the right term "miracle", I want add that even the flux does nothing if you don't add zinc, i don't have found nothing on the net, but I suppose that Zinc can "capture" the oxide leaving the Tin a free aluminium surface to wet, in absence of air (because under the melted solder drop). If you perform a initial wetting with Tin/Zinc you can than solder with Tin only on that area.

Thank you for the added information.I don't have the slightest idea about how to explain this "miracle", which implies that the protecting alumina layer is removed. There may be some similarity with the action of mercury. If you dip an aluminum piece in a $\ce{HgCl2}$ solution, the following reaction occurs $$\ce{2 Al + 3 Hg^{2+} -> 2 Al^{3+} + 3 Hg}$$. The metallic mercury atoms make an alloy with the aluminum surface. And this alloy is no more protected by the alumina layer, So it reacts quickly with water producing $$\ce{2Al + 3 H2O ->  Al2O3 + 3 H2}$$

@Mithoron It is known was OK  for Dothraki in Game of Thrones, but we like sources here  :  Generally true, but for general truth like 6x7=42 is often hard to find references (but Hitchhiker's Guide to Galaxy) and similarly for lifetime experience. :-)

I think the alloy is used to give a melting range ( 200 to 300 C) rather than a melting point ( 231 C) for pure tin. The zinc might reduce oxygen in the melt but I doubt it can breakdown aluminum oxide to clean the surface.

Just speculatively, zinc being in the same group as mercury, could melted zinc have similar destroying effect to aluminium oxide as mercury?

Interesting because zinc can cause liquid metal embrittlement of steels and stainless steels.


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How does double bond placement, in an alkene, affect the boiling point of the alkene?
Will having the double bond be closer to the middle of the molecule result in a higher or lower boiling point than that of an isomer with the double bond closer to one of the ends? Also, does the same relationship apply to alkynes?

Prediction of boiling and melting points are amongst the most difficult things to predict before hand. Don't worry too much about it. Most software, like ChemDraw, churn out ridiculous values. I don't how it makes such predictions.

Perhaps it could be done, even by feeling, if you focus on a pair of such isomers. For instance, I suspect that something more symmetric would have a higher BP. Check it out. For the rest see M. Farooq comment.

Perhaps chances for an educated guess being correct (say, with some $\Delta{}T \approx \pu{10 K}$) if one narrows the scope to simpler classes for some prior training.  E.g., otherwise non-branched alcohols, carboxylic acids with either odd or even number of carbon atoms.


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(IUPAC) How do you number an alkane that has two equally small starting points?

For example, do you number this from left to right or right to left? How do you know?
Would you get 2,4-dibromo-6,6-dichloro-5-methylheptane or 4,6-dibromo-2,2-dichloro-3-methylheptane?

There are always two methyl groups in a linear alkane :) Though I get what you mean. It is the presence of other substituents that breaks the symmetry. Your teacher or IUPAC gives the rules.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; (…)

Therefore, the example is named as 4,6-dibromo-2,2-dichloro-3-methylheptane rather than 2,4-dibromo-6,6-dichloro-5-methylheptane since the locant set ‘2,2,3,4,6’ is lower than ‘2,4,5,6,6’.


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Why is my ethanol–water mixture showing negative change in enthalpy of mixing?
I am conducting an experiment where I mix two solutions $\text{A}$ and $\text{B}$ and then I am measuring the temperature change (for real solutions, i.e., $\Delta_\mathrm{mix}H\neq 0).$ I took the following mixtures which showed the following changes in temperature:
$$
\begin{array}{lllcc}
\hline
\# & \text{A} & \text{B} & \Delta T_\mathrm{surr} & \Delta_\mathrm{mix}H \\
\hline
1 & \text{Acetone} & \text{Aniline} & > 0 & < 0 \\
2 & \text{Acetone} & \text{Chloroform} & > 0 & < 0 \\
3 & \text{Methanol} & \text{Acetic acid} & > 0 & < 0 \\
4 & \text{Acetone} & \text{Ethanol} & < 0 & > 0 \\
5 & \text{Ethanol} & \text{Water} & > 0 & \color{red}{< 0} \\
\hline
\end{array}
$$
The above mentioned are the results of my experiment which I conducted in my school chemistry lab at temperatures around $\pu{30 °C}$ with approximately no change in atmospheric pressure.
As seen, the first four cases show results as expected. However, in the 5th case, it's written all over the internet that an ethanol-water mix shows positive deviation, however, as seen through my experimental results, $\Delta_\mathrm{mix}H < 0$, i.e., it shows negative deviation. Can someone please tell me why this is happening?
Note: I repetitively cleaned all my equipment thoroughly with water before using the apparatus. Also, I conducted 3 trials in total for each mix and they showed similar results, so there is no chance of random errors leading to outliers.

They must mean positive deviation from Raoult law, with mixture having higher vapour tension than expected? As it is well known ethanol-water mixing enthalpy is negative.

@Poutnik Yes, it is a positive deviation from Raoult's law, but wouldn't that also mean that the temperature would fall for the surrounding since the enthalpy of intramolecular bond formation after the solutions are mixed is greater than the enthalpy of intramolecular bond formation for pure solutions?

The enthalpy is lower, ergo the mixing is exothermic


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By how much does borax reduce the melting temperature of gold?
I'm researching ways in how to salvage gold from old electronics. I think I got the process down but I have one question. When you have salvaged the gold from the electronics you are left with gold flakes.
This you want to melt down into a single piece of gold. I have been told that the addition of borax powder reduces the melting temperature of gold, but I'm unsure of how much. My blowtorch can go to 1000 °C and I want to know if this would be adequate to melt the gold/borax mix.

I don't believe it does. That's not what flux is for.

Is it really 24 karat gold - only then the mentioned temp. is important, otherwise m.p. can be much lower.

The melting point of gold is 1064ºC, so your blowtorch is inadequate. You have to melt the gold in something, and the borax acts to cover it, allowing the dense gold to settle, dissolving various metal oxides that might contaminate your flakes, and floating above the gold.
You would really need to get a hotter flame, with a fairly large zone of high temperature, because you need to heat all this stuff (container, gold, impurities and flux to somewhat above 1064ºC (1948ºF). There are gold melting electrical furnaces that would be more convenient.

The one shown is among the least expensive - you can pay a lot more, but it is advertised as a gold melter, with about 150ºF extra, so it might work for you. A lot more convenient than aiming a blasting flame for 15 or 20 minutes, trying not to burn the house down.


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Do solids in a solution also apply partial pressure? If yes, can this be used as an intuitive explanation for osmosis?
I know that in a mixture of different gases, we can assign every type of gas molecules a partial pressure so that the total pressure is the sum of all partial pressures. This can be extended to gas molecules dissolved in a liquid when there is a liquid phase next to a gaseous phase.
Can we also assign partial pressure to a solid that is dissolved in a liquid? For example, can we define partial pressures for ions in a $\ce{NaCl}$ solution so that
$$p_\mathrm{tot} = p(\ce{Na+}) + p(\ce{Cl-}) + p(\ce{H2O})?$$
This question came up to me when I tried to come up with an intuitive explanation for osmosis. If such a partial pressure phenomenon would exist, a solution with high concentration of solids would have a lower partial water pressure than a solution with low concentration of solids, assuming both solution have the same total pressure. This would result in a net flow of water from low concentration to high concentration if both solutions were divided by a selectively permeable membrane.
The usual explanations for osmosis I found on the internet using the chemical potential were quite unintuitive to me since chemical potential is quite an abstract concept to me. It would be nice to have an explanation on a basic level.

No. That is not intuitive at all. If you remove water, the dissolved stuff would simply fall at the bottom of the container,. Even with water, the only P exerted is hydrostatic. I do not see how this kind of forced analogy helps as compared to the right description of the phenomenon.

Such an analogy would help me. I want to understand osmosis more on a molecular level. Simply accepting the mathematics does not make me feel like I truly understand whats going on. But I see your point that the analogy to the gas does not make sense. Still, there has to be some explanation whats happening on a molecular level, without refering to abstract observables from statistical mechanics, or no?

You can explain the osmosis with pressure exerted by molecules to each other. But that is far to be an analogy to partial pressure of ideal gases. Perhaps try reading first the article Chemical potential and then Osmosis at Wikipedia. - I see that while writing this comment the answer has been added.

But unfortunately it is not quite correct except for the part in the link.

The vapor pressure of ions like $\ce{Na+}$ and $\ce{Cl-}$ is effectively zero compared to the vapor pressure of water, so your equation is essentially correct though $p_{\mathrm{total}} = p_{\ce{H2O}}$ (assuming the system has no air). The ions do lower $p_{\ce{H2O}}$ compared to a system with pure water, and the effect (the change in pressure) is proportional to the mole fraction of ions.
As for assigning a pressure to a solution in the context of osmosis, there exists the concept of osmotic pressure, which is defined as
$$\Pi = cRT$$
Where $\Pi$ is the osmotic pressure, $c$ is the solute concentration (often multiplied by $i$, the Van't Hoff factor), $R$ is the universal gas constant and $T$ is the absolute temperature.

Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.

In the absence of such external pressure in the semipermeable membrane system, solvent will move from the region of lower to higher osmotic pressure, i.e. the presence of solute generates an inward flux of solvent. The osmotic pressure is not related to the total pressure (like in your equation).

The second part is acceptable but the first section has the defect of mixing different things and ideas, including a reference to vapour pressure. And in the second part, it is rather a P acting on the liquid and not on the pressure of ions or solute. Which is ultimately affected by the presence of the latter.

OP asked about partial pressures and gas pressures and included an equation that adds them up to the total pressure, so I decided to mention the effects of ions on vapor pressure (which are indeed related to the total pressure) since this is where I think OP's confusion might be. As for the second part, I see that osmotic pressure is a property of the solution, not of individual solutes (I'll try to improve the answer).

Yes I think that the way the question is posed makes everything more difficult. Just saying that it is almost nonsensical won't be polite enough. I even suspect that playing enough one can create connections between the various phenomena we have already mentioned in terms of (if would be there) mechanical pressure. All this won't help OP neither. And myself I cannot write a better answer than those commonly found, as in the Osmosis page on Wikipedia etc. Perhaps you can stress the last line. In particular adding solute suck liquid inwards. The picture of OP can't even work as for it would....

.... require the liquid to move in order to increase its pressure. And there is no such a principle.


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Do the phosphoester and glycosidic bonds of dNTPs increase thermal stability?
A while ago, I provided an answer to a question on Biology.SE concerning the half-life of dNTPs under PCR conditions.  At the beginning of my answer, I give a qualification:

Caveat: ignoring the bonds between the different parts of dNTPs likely misrepresents the stability of the intact molecule, given that the covalent bonds between the nucleobase, sugar, and phosphate pieces each have their own relative stabilities.

How wrong is my approach to model the stability of the whole dNTP by the stability of its pieces? To what extent do the covalent bonds between the constituent molecules (indicated with red arrows below) contribute to the thermal stability of the dNTP as a whole? What is the organic chemistry explanation (resonance?) for why the intact molecule would be more (or less) stable than its least stable constituent in a neutral-buffered aqueous solution at 95 °C?

Edit, re: Karsten Theis's comment

You don't suggest a mechanism of degradation, or a degradation product.

Per the linked Biology.SE question, I'm interested in any degradation mechanism that would lead to the dNTP being no longer be polymerizable by DNA polymerase. As I understand it, the incoming dNTP base-pairs (hydrogen bonds) with the template, allowing the ribose 3' hydroxyl group to attack the triphosphate group on the incoming nucleotide, yielding a phosphodiester bond and a pyrophosphate leaving group.
So, any mechanism that results in a product that (1) interferes with hydrogen bonding associations of the dNTP with its template or (2) prevents subsequent phosphodiester bond formation is fair game. Even though deamination of dCTP may inhibit specific polymerases, my understanding is that dUTP can still be incorporated by the polymerases used in PCR, so that specific reaction is not within the scope of my question.

You don't suggest a mechanism of degradation, or a degradation product. Here is one possibility described in the literature, deamination of dCTP to dUTP, with inhibitory effect on archeal polymerases and an enzyme that helps to lift inhibition: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC117351/

When the glycosidic bond hydrolyzes, the sugar will have a richer chemistry (because the ring can open up). If that translates into lower stability under PCR conditions is another question.

@KarstenTheis thank you for catching that. Fixed.

https://chemistry.stackexchange.com/questions/159043/do-the-phosphoester-and-glycosidic-bonds-of-dntps-increase-thermal-stability

How wrong is my approach to model the stability of the whole dNTP by the stability of its pieces?

There are two classes of errors that such an approach introduces. First, it is possible that the pieces have different reactivity in the context of the complete molecule. Second, this approach does not consider breaking the molecule at the connections.
Do the connecting bonds stabilize the pieces?
The glycosidic bond prevents the ribose ring from opening via the usual linear/hemiacetal pathway. In the linear form, the aldehyde could be oxidized to a carboxylic acid or reduced to an alcohol. In the ring form, this is not possible. The triphosphate group on its own could lose the alpha or the gamma phosphate in the same way (it is a symmetric species with anhydride linkages on both ends). Once it forms an ester, this would change the reactivity of the anhydride linkage on the alpha carbon side. Once the nucleobase is connected to the sugar, the linking nitrogen no longer has a protonated and deprotonated form. This might also change some degradation reactions.
Are the connecting bonds part of degradation mechanism?
The glycosidic bond is subject to hydrolysis (see e.g. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC523769/pdf/nar00047-0143.pdf). Once the glycosidic bond is gone, the sugar will potentially also be degraded. The phosphoester bond is cleaved in vivo by enzymes; non-enzymatic hydrolysis is possible under harsh conditions (see e.g. http://www.tud.ttu.ee/im/Tonu.Reintamm/shabarova/4.5.html).

So, any mechanism that results in a product that (1) interferes with hydrogen bonding associations of the dNTP with its template or (2) prevents subsequent phosphodiester bond formation is fair game.

There are two types of degradation products that should be considered; those that the polymerase no longer incorporates into the growing strand (having those accumulate would make the syntheses a bit slower) and those that inhibit polymerization (even small concentrations of the latter could inhibit polymerization strongly).

What is the organic chemistry explanation (resonance?) for why the intact molecule would be more (or less) stable than its least stable constituent in a neutral-buffered aqueous solution at 95 °C?

The explanation would depend on the specifics (conditions during the PCR - pH, presence or absence of divalent cations, other additives, which nucleotide are we talking about). Resonance is not the core explanation of the effects because the linkages are pretty much single bonds. If we are just substituting -OH with -OR or -NH with -N-C..., there is no drastic effect on the aromatic systems and double bonds in the constituents.


text:
How can Phthalocyanine be aromatic when it is not planar?
Even as Hückel's rule is not applicable in every scenario, especially in metallic clusters, it seems that organic structure need to be planar to show aromaticity. That being the case, Pubchem's representation of Phthalocyanine shows a non-planar structure. Part of it is completely out of phase and perpendicular to the whole structure, just as in Biphenyl, with this making the latter experimentally confirmed as non-aromatic (the whole molecule, not each phenyl ring):
Phthalocyanine Pubchem
Is this representation mistaken or is there something more to this?

Related: https://chemistry.stackexchange.com/questions/51484/does-aromaticity-require-a-co-planar-pi-system/51485#51485

As mentioned in that link, they can be twisted to a degree and be aromatic. Yet in Phthalocyanine part of the side is completely out of phase with the rest just as in Biphenyl, which should render it non-aromatic.

Is it really? Why did you look at a lame molecule viewer and not crystal structure? It's flat all right and even if it wasn't flat as a whole then a single aromatic ring would make it aromatic.

Aromaticity, especially if you consider systems with more than a single cyclic structure (e.g., benzene, pyridine, pyrole, thiophene, furane, cyclopentadienyl anion), isn't on/off like like a light switch.  See, for example [18]annulene (wikipedia entry here).  For phthalocyanines, it actually equally matters if these are empty or complex a transition metal.  Thus, related.

In addition, many of PubChem's 3D structure entries are the result of putting the atoms together (in a computer) and optimizing the structure in a molecular force field at low computational cost.  This may, but need not be the most favourable conformation in solution, and even less in the solid/crystalline state experimentally recorded.

Mithoron/Buttonwood, I was not aware of such compromises and limitations of 3D singular molecular representation. Makes more sense now on it planarity. I believe this question can be closed, unless someone wants to add anything else. Thank you!

Just to add a little on the Huckel's rule: it CAN be applicable to aromatic all-metal clusters as well. ;-)

Just to be precise: this is the protonated form, and I think you are interested in the deprotonated form coordinating to some metal ion, aren't you.

You read my mind Greg. Any suggestions on good modelling software for this that is free and accessible?

@C-Consciousness I'd recommend using xtb which is a fast approximate quantum method available on Linux, Mac, etc. and Avogadro but I admit to being biased as an Avogadro developer.

Phthalocyanine is flat, for example in the crystal structure (linked on PubChem) or PQR or Wikipedia

As noted in the comments, PubChem uses an efficient force field (MMFF94) which does not consider aromaticity. Since the record you linked is protonated (though strangely in a cis-oid configuration, not the typical trans-oid depiction) the steric considerations between the nitrogen atoms and the protons break planarity.
In reality, due to delocalization around the macrocycle, breaking planarity and aromaticity is energetically disfavored.
While there are a few force fields which consider aromaticity effects (e.g., MM3 / MM4) generally one needs at least an approximate quantum chemical method. PQR (which I used for the image) uses OpenMOPAC and XTB / GFN2.


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Why does energy of non-axial d-orbitals decrease in an octahedral complex?
In an octahedral complex, if we assume the metal ion to be inside a cube then ligands will approach from its faces. So  the axial d orbitals will experience more electronic repulsion as compared to non-axial d orbitals. That's why they are higher in energy. But what I am finding difficult to understand is why non-axial d orbitals decrease their energy below the excited degenerate d orbitals level. Some answers that I found on other sites were mentioning about conservation of energy and that total decrease in t2g level and eg level are equal but still I cannot understand it.

The crystal field splitting depicts the energy levels in an octahedral complex, relative to one where the ligand charge density is spherically distributed. That is to say, the repulsion is already there, it's just being redistributed (hence the 'conservation of energy'). It's not being compared against a free ion in which there is no external repulsion.

In an octahedral complex dx2-y2 and dz2 will experience more repulsion by ligands and their energy will be raised so the extra energy gain is due to repulsion but why the non axial orbitals decrease their energy? In spherical field if they were excited, then when the octahedral complex is formed they will be experiencing more repulsion(not more than that faced by axial orbitals),then going by that logic they should have higher energy than in spherical field but their energy decrease further. Is it so that their energy is being used for exciting dx2-y2 and dz2 orbitals.

I don’t follow your train of thought. The xy, yz, and xz orbitals point away from the ligands in an octahedral field, whereas in a spherical field they can’t help but point towards some charge density (since it is uniformly distributed over all directions).So I don’t see why there should be more repulsion. It seems to me that there should be less, which corresponds to the “decrease” in energy.

Of course, this is all quite handwavy, and I wouldn’t blame you at all if you didn’t want to take my comment at face value and instead wanted a mathematical proof of the above. But that’s an entirely different question, IMO.

In spherical field the whole set of d orbitals gets excited. When the ligands gets closer and closer and they form octahedral complex then there comes a difference in repulsion faced by axial and non axial orbitals. So I get it that the axial orbitals will face more repulsion and non axial will face some less but the magnitude of repulsion faced by dxy, dyz and dxz won't be less than that when they were experiencing a spherically symmetrical field(of course because in spherical field the ligands were far away)


text:
What causes water to condense when air pressure is reduced?
In this video: https://youtu.be/IyBRGhXBuGw
The cabin pressure is reduced and subsequently a mist is formed.
What exactly is the cause?

Consider what happens to the temperature of a gas as pressure is reduced. PV=nRT

Note that requests for video comments  are not very welcome. It is not critical here, as video is not even needed. But generally, the least you should do is providing the relevant video timestamp. Community readers are not really fans of watching videos to provide answers.

Please describe the content of the video and refer to it with creator and title. Links may die as videos get deleted, renamed, set private again, etc., and this question losses all context. Please [edit] accordingly. In its current state it should probably be closed.

See Gay-Lussac's Law.      https://en.m.wikipedia.org/wiki/Gay-Lussac's_law

I wrote "In this video [..] the cabin pressure is dropped and a mist is formed". How is that not a description? I'm not going to change it for no reason, so just close it, if that makes you happy.

honestly you lot need a day off lol

Reducing pressure causes temperature drop, as part of thermal energy of air is spent on mechanical work during air expansion.
That is why there is cold air at high altitudes. Air cools down by 1 °C when air  climbs 100 metres upwards until first condensation.
When temperature drops below the dew point, mist starts forming.


text:
What gas particles are formed when diamond turns to gas?
If you heat up diamond away from oxygen (air completely) until the solid diamond is now gaseous, what are the gas particles made of? Just the element carbon without a charge?

The same as for graphite.

Diamond will cease to be diamond long before sublimation.

Yes. Since diamond is a covalent network solid, it is made up of nothing but pure carbon atoms. Being a network solid, it is extremely stable (only at high temperatures and pressures in this case). If you have heated it enough to cause it to sublime, then the carbon atoms are in such an excited state to make it impossible for any bonds to form between them. Therefore, you would be left with gaseous monatomic carbon.

It just occurred to me that I should mention that you would have to have it at very specific conditions to prevent it from turning into graphite - a lower-energy structure. Even then, its melting point would be well over 4000 °C, about 3/4 the temperature of the surface of the sun, which would be an insanely difficult temperature to maintain for long enough to cause the entire diamond sample to sublime. And you wouldn't have to do it in just an non-oxygenated environment, the diamond, being pure carbon, would try to react with a huge variety of other elements if present.

I'm not entirely confident that the result would be monatomic carbon in all circumstances. I'm sure a range of small molecules could occur depending on the exact conditions. But this is an empirical question that might actually have been addressed.


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Spin-only magnetic moment of [Fe(en)(bpy)(NH₃)₂]²⁺

The one that can exhibit highest paramagnetic behaiour among the following is ($\ce{gly}$ is glycinato; $\ce{bpy}$ is 2,2ʹ-bipyridine):
(a) $\ce{[Pd(gly)2]}$
(b) $\ce{[Fe(en)(bpy)(NH3)2]^2+}$
(c) $\ce{[Co(ox)2(OH)2]-}~(\Delta_0 > P)$
(d) $\ce{[Ti(NH3)6]^3+}$

In option (c) $\ce{[Co(ox)2(OH)2]-}~(\Delta_0 > P)$ cobalt has valence shell configuration $\mathrm d^4.$ Since the pairing energy $P$ is less than $\Delta_0$, there should be pairing. So, the number of unpaired electrons here should be $2.$
According to me, iron in (b) $\ce{[Fe(en)(bpy)(NH3)2]^2+}$ also has valence shell configuration $\mathrm d^6.$ Since all the ligands have nitrogens as donor sites, there shouldn't be any pairing.
I was told when the element is of $\mathrm{3d}$ series and in $+2$ oxidation state, ligands having nitrogen as donor sites are considered as weak field ligands and only ligands like carbonyl and cyano are considered as strong field ligands, and thus the number of unpaired electrons should be $4$ in this case, resulting in the highest paramagnetic behavior for (b).
However, the answer says the opposite. Is the aforementioned fact wrong?

Now, you wrote, "since all the ligands have nitrogens as donor sites, there shouldn't be any pairing." Actually, bpy is usually considered a π-acceptor ligand, which makes it more likely that the complex is low-spin. However, it's IMO quite unfair to expect a student to guess whether a given Fe(II)-bpy complex is high- or low-spin, because they're often "on the edge", i.e. need to be experimentally determined.

@orthocresol Okay, my teacher asked me to learn it and thus this confusion

When taken as a 'fact', it is wrong that iron(II) complexes hexacoordinated by nitrogen atoms will always be high-spin (no pairing). This is exemplified by the complexes of 1-propyl-1H-tetrazol-κN4 (ptz): both the high-spin and low-spin complexes of $\ce{[Fe(ptz)6]^2+}$ are known, isolated and characterised; I wrote about this in a previous rather long answer. In the complexes, six nitrogens are donating to an iron(II) centre.
It is also obvious that the question and its solution hinges on whether the specific iron complex is high or low spin, as both the palladium and titanium complexes can be easily ruled out. High-spin iron(II) complexes have four unpaired electrons as you correctly noticed while low-spin iron(II) complexes have two.
Unfortunately, as the correct solution goes against your teacher's rule of thumb and as examples for both spin types exist, I cannot really help you here. At best, you might be able to look at the bpy ligand, notice that it has a multitude of π-interaction capabilities and take that into account – although I myself wouldn't know whether the π interactions would be enough to change the spin state.
I hope you aren't presented with such a question in your exam.


text:
Way of determining based on atomic characteristics what to expect during reaction
So I believe the question was clear, but here goes.
I would like to know the rules for determining the following properties of a molecule, and/or of a reaction it produces with another molecule or atom.
Of the molecule/atom itself

Explosiveness
Toxicity
Combustibility

Of the reaction

Releases toxic byproducts
Releases heat
Releases light
Explodes
Combusts

I want to be able to do this by the atomic structure of the molecule and/or the number of subatomic particles in the atom, to be able to predict per mole certain aspects of an interaction.
I want to be able to do this to isolate things that are hazardous or produce neato effects like flashes of light based off a rule set by running a recursive algorithm that can determine extended and simple reactions given subsets of data.
I'd also like to know based off physical properties what the necessary catalyst is to make the reaction occur, application of heat, cold, some form of solvent etc.
I just need pointed out to the proper set of principles and reading.
In physics you can plot out things based on physical properties to get a guess. That's really all I'm hoping for here.  That and perhaps work that has been done to match moles of reactants to physics units such as lumens,  joules (for heat) etc.

It is too broad and deep question for Q/A site format. It calls for several books, which would cover just surface of the topic.

@Poutnik one thing at a time then. how about toxicity.

I will correct myself: Instead of many books, just few ones. Your questions are kind of Chemistry counterpart of the Ultimate Question of Life, the Universe and Everything from the Hitchhiker's Guide to Galaxy.

Chemistry is semi-empirical science. There are few laws, many semi-empirical and empirical rules and many exceptions. There is no shortcut to learn it.
First you have to memorize or search for the facts to know them. Later, when knowledge and experience comes, you will learn to see patterns in the facts, related to underlaying known or unknown rules. It will allow you to reduce the data you need to remember or search for. But there will be always a lot of behaviour that cannot be predicted from a simple rule set.
Many facts can be predicted, but it may be inaccurate and very complicated.

so how about a table of hazardous, volatile reactions and substances ? is there any such thing along with applications and measurements like lumen release during reaction etc ?

I am no expert on chemistry, or computer programming, but it seems like it might be possible to program in the bonding behavior of all the elements, polyatomic ions, and even compounds, and then compile an array (or set of arrays) of what reactions would be physically possible. The problem with this is you would have to eventually hard code in a cap for how many atoms can be in a single reaction, or else your computer would be processing for eternity and never give you a result. The other problem is it would have to be one crazy supercomputer to be able to process all that information and give you any useful answers within your lifetime.
The biggest problem however comes from the nature of chemical reactions themselves. You can take two of some of the most dangerous substances access to the general public, combine them, and get harmless salt water. (NaOH + HCl => NaCl + H2O).
In other words, while it would be theoretically possible to calculate what chemical reactions can occur and their products, it is impossible to predict, with any degree of certainty, the properties of those products.

but wouldn't you be able to say that the dangerous chemicals combining themselves produced dangerous byproducts ? and also the key there is that THEY are the dangerous components that would be redflagged. the question is HOW is it physically determined that they are dangerous ?

No, in my example, you combine two dangerous chemicals and get, not only a harmless, but essential chemical for life. And how dangerous a chemical is can be determined by the number and severity of medical issues it can cause. For example, HCl, can cause chemical burns, scarring, permanent eye damage, lung damage, suffocation, stomach damage, and death... NaOH can cause very similar problems. Yet when combined, it produces standard table salt and water. My point is the properties of the reactants cannot be used to predict the properties of the products of a reaction.


text:
What might cause overestimated measured pH value of sodium hydroxide solution pumped with carbon dioxide?
I tried an experiment at home where I had a $\ce{CO2}$ tank pumping $\ce{CO2}$ into a solution of $\pu{28 g}$ sodium hydroxide in $\pu{200 ml}$ water. The initial $\mathrm{pH}$ was $11.24.$
I expected it to form sodium carbonate and bicarbonate, but instead, the $\mathrm{pH}$ went up to $11.5.$ I should point out that my $\mathrm{pH}$ sensor needed a few minutes of cleaning in distilled water to prevent inaccuracy because the sodium hydroxide particles stuck to it.
Is this supposed to happen, or is my $\mathrm{pH}$ sensor inaccurate? Is there another way to do it? I also found some videos online and in those videos, the $\mathrm{pH}$ went acidic instead of the $\mathrm{pH}$ of sodium carbonate which is between $8$ and $10.$

The initial pH should be >14, and after adding CO2, it should drop. NaOH is very soluble in water, so it is strange that you talk of sodium hydroxide particles.

pH measurements at extreme pHs are not accurate. It is not a surprise that you find unpredicted changes in pH after bubbling carbon dioxide because the sensor is not meant for such a high concentration of NaOH. There are well known sodium errors associated with glass electrodes. The sensor is not faulty, it is not designed to be used in concentrated alkalies! The electrode starts to respond to Na ions instead.
If you are talking of "particles", sodium carbonate can precipitate in strong caustic solutions, so you can indeed see particulates of sodium carbonate depending on the concentration.

Thanks for your help but do you recommend I use a better pH sensor or would that not work as well?

No pH sensor will work at such a high concentration. What are you trying to do? What is your experiment about?

I am trying to add co2 to sodium hydroxide to form sodium carbonate for a science experiment. If a pH meter can't measure that high, should I just make carbonic acid, use the pH sensor,  and then add sodium hydroxide and not measure it?

You cannot make carbonic acid. It does not exist. If you explain more what you are trying to demonstrate in your experiment, one can think of a solution. Is this a student project or you are teaching ?

Just for a student projet, I am trying to demonstrate the ansrobsion of co2 through a sodium hydroxide solution and turning it into sodium carbonate. A few months back I did the same thing with water instead and the ph dropped from 7.4 to 4.7 in the co2/water experiment. But I don't understand what you mean when you said carbonic acid doesn't exist.

Instead of pH sensor use a universal indicator to show the changes in pH and use dilute 0.01 M NaOH. Why did you choose such a high conc. to begin with?

I added a high concentration because I thought the pH of the sodium hydroxide solution should be between 12-14

Your concentration of NaOH is way more than 1 M.

Thank you for your time, but I have one more question, will adding co2 to a sodium hydroxide solution have the same effect as adding sodium hydroxide POWDER to carbonic acid?

There is no carbonic acid. If you are interesting in doing a demo, I told you the universal indicator way with a dilute NaOH solution (< 1 M) to show how CO2 is decreasing the pH of the caustic solution.

Would pH strips also work?

Yes, they should.

Thanks for the help! I redid it with ph strips and the ph went from 14 to 11!

Well, if you start with a pH of 11.24 and it jumps to 11.5, that would be more basic, not more acidic. Using some pH calculators, the pH of your initial NaOH solution should have been 14.54, NaOH being very basic. After pumping excess CO2 through it, the pH of your Na2CO3 solution should have turned out to be 10.25. For this situation, I would say you either have a faulty pH sensor, or there is some other source of error that would change the results so drastically. Most likely the former.
This is completely up to you, but if you are planning on getting a new pH sensor, I recommend Sensorex. They have a few options, including one for about \$50, or a more sensitive one for around $100.
https://sensorex.com/laboratory-ph-electrodes/?hsCtaTracking=6c3ce343-9554-4fba-8b74-8d232f941f6f%7Cf9b4d806-dea3-4aff-a8e9-7df730ead108


text:
Why is acidic strength of A>B?
a) H-(C=O)-OH
b) H-(C=O)-O-OH
The acidic strength of the two structures given below  was a>b .
I think the explanation to that is in structure A, the O atom of carboxyllic acid is in resonance with O anion therefore more stable thereby more acidic.
While in b the O anyone is not in resonance.
However why can't the O anyone br in lone pair lone par resonance with the previous oxygen atom which makes the chain complete there by making b more acidic ?

In order to make the resonance in the conjugate anion, the deprotonated oxygen atom has to share an electron pair with its neighbor and that extra shared pair then overlaps an antibonding orbitals to the rest of the molecule. So the resonance/delocalization has to weaken a bond in order to make the new one. With the anion of (A) the weakened bond is a $\pi$ bond, while in in the (B) anion we have to weaken a $\sigma$ bond which is a greater energy cost. While there are molecules where resonance/delocalization impacts $\sigma$ bonds, they are less common than those where weaker $\pi$ bonds are affected in this way.
WP indicates that the $pK_a$ for the peroxy acid B is about 7.1, versus about 3.7 for the "normal" acid A.
This effect is seen with methyl groups on the catboxyl carbon too. WP articles for the corresponding acetic acid species indicate that the $pK_a$ for the two acids is about 8.2 for the proxy acid versus about 4.8 for the "normal" acid.


text:
Why do two programs assign different SMILES to one amino acid sequence?
I have an amino acids string "MKGK" and I would like to convert it into SMILES format.
I tried two approaches.
First with PepSMI which gives me:
N[C@@]([H])(CCSC)C(=O)N[C@@]([H])(CCCCN)C(=O)NCC(=O)N[C@@]([H])(CCCCN)C(=O)O

Second, using ChemAxon Marvin's molconvert command
/Applications/MarvinSuite/bin/molconvert --peptide MKGK smiles

I get:
[H]N[C@@H](CCSC)C(=O)N[C@@H](CCCCN)C(=O)NCC(=O)N[C@@H](CCCCN)C(O)=O

Both of them give different result. Are they chemically the same?
If not, which result should I trust?
Finally is the conversion from amino-acid to SMILES is a lossless process?
Is there anything we should be careful about?

TL,DR:  Despite differences on the string level, they all may be correct.
Very likely, the string MKGK you state is the compact representation of the primary structure of a structure (as in sequence of amino acids) by the FASTA format.  For this purpose, FASTA is a more efficient string notation than SMILES.  Note, their scope is complementary to each other; FASTA has large (bio)molecules in mind (proteins, RNA/DNA), SMILES' focus in comparison are smaller molecules, though there is some overlap where both may be practical.  If choosing, you equally have to consider ease of storage, processing, already existing data and programs in the field you want to use when choosing one, or the other (e.g., BioPerl, BioPython).
FASTA and SMILES are both a simplified description of a structure.  Say, a tripeptide is described initially in a .mol2 file.  Then, the description in a SMILES string will remove specific atomic coordinates in a description like «atoms $\ce{A1}$, $\ce{A2}$, $\ce{A3}$, $\ce{A4}$ are all around atom $\ce{A5}$ to form a (R) configurated stereogenic center» and only states that these form a (R) configuration. In addition, a typical SMILES string will remove all hydrogen atoms except those around a stereogenic centre. (You may retain these as explicit ones, though).  FASTA's description of amino acids in sequence equally is loosing some information, too.
What do you want to do once you have the SMILES string in hand?  At present, your question does not state this clearly enough.
The potential obstacle using SMILES is that it is one name of multiple definitions.  Often, programs do not explicitly state if the output is e.g., canonical, universal, or inchified to mention some of the options.  Other programs let you choose, e.g., OpenBabel.  Thus, once you pass a certain threshold of structural complexity of a molecular structure, the very same molecule may be described with multiple SMILES strings -- each of them may be a correct description as such, yet the at the level of the SMILES string, they differ. See, for example, O'Boyle's paper mentioned in this recent answer to resolve this potential confusion.  So it is up to you to check what form of SMILES the version of your program used provides and if this understood (input and output) by any other program you use while processing these strings.  (And turn off the autocorrect in the spread sheets, if you use them; it is an avoidable pitfall (see e.g., Nature).
What do you want to do with the SMILES strings, once in hand?  Do you want to use them (eventually) to create atomic coordinates?  These won't be the originally recorded ones.  Do you want to engage in protein folding (possibly assisted by machine learning like AlphaFold [then see e.g., this potentially interesting thread here about the now open submission of candidates])?  For small molecules InChI is a more recent string standardized notation, and its hashed form InChIKey provides a computationally affordable approach to check them for structure identity, e.g., in a database.
References implicitly mentioned:
Noel M O'Boyle Towards a Universal SMILES representation - A standard method to generate canonical SMILES based on the InChI in J. Cheminf. 2012; doi 10.1186/1758-2946-4-22
Dyani Lewis Autocorrect errors in Excel still creating genomics headache in Nature, 13th August 2021; doi 10.1038/d41586-021-02211-4

Thank you so much for your clear and elaborate answer.
I'm using SMILES to extract features (e.g. number of amines, benzene, etc) and do machine learning predictions.
Another usage of that SMILES is to do docking on a receptor using Vina Autodock (for example).
I need it because the peptide I have is so short (4 amino acids). It's hard to create a reasonable PDB structure using Alphafold etc.
Do you think my intention and approach make sense?
Your expert advice is much appreciated.

Four amino acids per molecule, these still may be processed to generate a conformer (e.g., OpenBabel) or multiple ones (e.g. obabel) or to count fragments (e.g., RDKit). If you prefer a more GUI centered approach for e.g., a substructure count, DataWarrior offers this as well.

Contrasting to OpenBabel and RDKit, DataWarrior permits to initialize the tested conformations on data inferred from the COD, or -- like the other two -- from a force field.  Then it may be worth to convert FASTA (or SMILES) into .sdf (which OpenBabel offers on the fly, too) with atomic coordinates at least as an intermediate format.  However, I do not know, if .mol2 or .sdf (the later may contain multiple .mol2) is an input format Vina Autodock accepts .or. if a conversion of .sdf into .pdb by, e.g. OpenBabel suits your needs.

thanks so much again. May I know what is a conformer. What's the command line for OpenBabel to generate that. Also for RDKIt how can we know which substructure to extract? Is there a way to extract all?

@scamander Since this requires multiple lines, it is better to continue this in the chemistry.se chatroom here.

My answer may be too late, but there is very simple way to answer question initiating this topic.
SMILES representations should be converted into InChIKey. If they present the same molecule, they should result in the same InChIKey.
In contrast to SMILES, InChIkey for the same molecule is always the same. It is easy to be compared manually due to fact, that it contains only 27 characters.
I have translated above SMILES strings into InChIKey using Marvin JS molecule editor.
Both of them have given InChIKey=FGIPJUHCFCYWLW-KKUMJFAQSA-N.
For further reading:
Annotation of Peptide Structures Using SMILES and Other Chemical Codes–Practical Solutions. Minkiewicz p. et al., Molecules 2017, 22(12), 2075;
https://doi.org/10.3390/molecules22122075

You wouldn't actually need InChIKey, the standard InChI would have sufficed. So would have cononicalising the SMILES. Anyway, still a valid point.


text:
Can PV vs P plot be called an isotherm?
Normally we call plots drawn at constant temperature as isotherms, e.g. $P$ vs $V$ plot, Langmuir isotherm etc.
Using that logic, shouldn't the $PV$ vs $P$ plots also be called isotherms?
Personally, I haven't read in any text about such plots being called isotherms.

If the T is kept constant why not? Perhaps it is not very useful to plot something that is expected to be constant for a closed system?

@Alchimista   ... Only if ideal gas is involved.

@Poutnik yes of course but it might be related to a entry level. Otherwise why the question?

@Alchimista Yes it's an entry level question by one of my student. I also thought on your lines but just wanted to confirm. :) Sometimes there are IUPAC definition or similar standard things.

All curves at constant T are or can be called isotherms. But they are often not called that way explicitly. Explicit term is usually used in scenarios where other variables could be kept constant instead, having isobars, isochores, isotropes,...


text:
In Kohn Sham DFT, what do we know theoretically about the exchange correlation energy?
As I understand Kohn Sham theory, it is proposed that the ground state energy for a molecule can be derived from the ground state of a fictitious set of non-interacting electrons moving in a fictitious potential. This fictitious potential is conventionally written as $E_T+E_V+E_J+E_{XC}$ where we have known formulae for the kinetic energy $E_T$, the potential energy $E_V$ and the Coulomb energy $E_J$ but the "exchange correlation" energy $E_{XC}$ is a priori unknown.
I have two questions:
(1) what theoretical reason do we have to believe that the ground state energy can be derived in this way? Clearly we have practical reasons to know that this approach can work pretty well, but is it possible that there are some molecules where the Kohn Sham approach simply won't work?
(2) is there any theoretical reason to believe that the $E_{XC}$ term has anything in common between different molecules? I have seen it suggested that with heuristic approaches to finding $E_{XC}$ "different models work better in different scenarios" which makes me suspicious that we have simply turned a question of heuristically finding the ground state energy for an individual molecule into a question of heuristically finding the exchange correlation energy for that molecule?

If you reduce this to just one question instead of two, it could be worthwhile to ask on Matter Modeling Stack Exchange.


text:
Why is the the temperature change due to the combustion of paraffin wax so high (10000 Kelvin)?
I'm trying to work out the temperature change (I got something like 10k Kelvin,hotter than the Sun's core...) caused by the combustion of paraffin wax which has a  molar enthalpy of combustion $E^{wax}_{m}=15MJmol^{-1}$.
$$38O_2+C_{25}H_{52} \to 25CO_2 +26 H_2O \tag{1}$$
so for a given number of moles of oxygen $n_{O_2}$, the reaction requires $(1/38)n_{O_2} $ moles of wax ($C_{25}H_{52}$) and produces $(25/38)n_{O_2}$ moles of $CO_2$ and $(26/38)n_{O_2}$ moles of $H_2O$.
Consider a thermal-dynamically isolated system composed of 1 Litre of pure $O_2$ under STP (plus the required amount of $C_{25}H_{25}$ for complete combustion). Calculate the temperature change due to the combustion. The molar heat capacity of $CO_2$ and $H_2O$ are both approximately $30Jmol^{-1}K^{-1}$
From $Q=mc\Delta T$:
$$n_{wax}E_m^{wax}=n_{CO_2}c_{m,CO_2}\Delta T+n_{water}c_{m,water}\Delta T $$
$$ \left(\frac{1}{38}n_{O_2} \right)E_m^{wax}=\left(\frac{25}{38}n_{O_2} \right)c_{m,CO_2}\Delta T+\left (\frac{26}{38}n_{O_2} \right)c_{m,H_2O}\Delta T$$
$$\Delta T=\frac{(1/38)E_m^{wax}}{((25/38)c_{m,CO_2}+(26/38)c_{m,H_2O}}$$
$$\Delta T=\frac{E_m^{wax}}{25c_{m,CO_2}+26c_{m,H_2O}}=\frac{15MJmol^{-1}}{(25+26)\times 30Jmol^{-1}K^{-1}} \approx 10000 K$$
What have I done wrong?

Heat capacity does not really stay constant at temperatures that high. Also, much of the heat is irradiated outside or passed to the surroundings. If we neglect both these effects (even though they are not really negligible), then that's what we would have.

@Ivan Neretin
Thanks.  So my working is not wrong then I guess? Another thing that intrigued me was that $\Delta T $ is independent of the number of moles of $O_2$ initially present?

Looks right to me. As for the moles of O2, there are (presumably) exactly as much as needed for the full combustion. Take more O2, and you'll have an excess of it; take less, and you'll have an excess of paraffin. Both would appear in the denominator and thus change the ΔT.

Cross posted: https://physics.stackexchange.com/q/673720/313612.

As a curiosity exercise try to paper-see what thermometer (range) you would need using a calorimeter taking some data from your exercise or the Web in addition to those you already have. Perhaps it would look less spectacular.


text:
Can gases float on each other?
So I want to know do gases have the same concept as water and oil ? Like if I put some hydrogen and helium in a room with 0% air will they float on each other like the hydrogen go down and the helium goes up or will it mix?

They mix, via diffusion and convection. Think perfume and air: perfume would be useless if if was only located near the floor.

Why not?

While there is no anything similar to layering of liquids in terms of long lasting stability I've decided to add this comment as for important SAFETY implications. Release of gases in confined environment such as cabinets but even rooms is dangerous. Do not expect anything to leave and mix so fast. Think of the many accidents with CO2 involved. And even a leaking pressure bottle can pose dangers. Asphyxia is particularly sneaky because we feel lack of oxygen indirectly by means of CO2 in the blood stream. Gradually reducing oxygen does not prompt us to breath, just makes us confused/ sleepy.

There is difference between liquids and gases, as all gasses are mutually mixable.
Hydrogen and helium would do both mixing up and layering, with hydrogen going up and helium down.
Layering is just a temporary thing until all is mixed up. Similarly as ethanol and water can form 2 layers, if brought together carefully,as mixing would make a single phase from them.
Note that gas mixing up happens much faster and easier than for liquids. There is much smaller density difference and more intensive diffusion and convection.
The bigger the density difference is, the more easier is forming layers and the slower is their mixing.
The equilibrium condition in static scenario builds very slight concentration gradient where gravity induced sedimentation is balanced by gradient net diffusion. Real life scenarios in air context bring strong vertical and horizontal convective mixing.

And that SF6 leak will find the low spots and make them potentially hazardous, but not forever. Confined spaces are nasty things.

maybe worth noting that in everyday life, density differences due to temp differences are much more common than density differences due to molecular composition, and layering of (less dense) warm air above (more dense) cold can be stable for quite a while

@Andrew  Surely the differences because of temperature are much more important in the air context, forming labile, conditionally stable and stable temperature  gradients ( I have background in air force field meteorology). But we do not speak in the context of air here.

It depends on how much force the particles experience. Proteins in solution don't settle in a beaker on the benchtop, but they do in an ultracentrifuge, and isotopes in the gas phase may be separated by centrifugation as well. Hydrogen can escape from the upper atmosphere (Jeans escape, https://en.wikipedia.org/wiki/Atmospheric_escape#Jeans_escape). Maybe the latter does not count as settling.

@KarstenTheis I am well aware about centrifuging of proteins and uranium. But it is not like forming 235UF6 and 238UF6 layers. It means upper layer is at equilibrium just slightly enriched by 235U. As at some concentration gradienct, settling is compensated by diffusion. Hydrogen escape is about escape velocity. BTW, I have said atmospheric. composition very slightly change with altitude, what implied increased hydrogen few hundreds km above sea.

I see, no layers of pure substances like water and oil.

The scenario might change in very big gas columns. Miscibility does not prevent gravity to be at work. So there would be a composition gradient in ideally static atmosphere of planetary scale. Not really floating  nor layers, though.

@Alchimista The lower 80 km of atmosphere is consider ( but vapor ) as having about constant composition. As I said, slight gradient  keeping diffusion and settling in equilibrium, for rather unreal static scenario. For air with dT/dh < -0.0098 K/m, it gets mixed by adiabatic convection, in spite of being stable by density gradient, not speaking about wind.

@Poutnik I often upvote comment by mistake. All right. But I am pointing to a principle. Planetary was the scale of a gas colum only, in my comment. I used "ideally" in fact. It is clear that Sun and Earth rotation overrule a simple gravity compositional gradient. And even the P gradient is somewhat an approximation. Actually if we take just the question in itself, which has no reference to long lasting situations and even mentioned a carefully prepared room, the answer is even yes. See the link by Nilay Ghosh for a nice entertaining display.

Yes, gases do mix but if you leave them overtime they will slowly settle down less dense one on the top and more dense one at the bottom (Of course if there not chemically reacting.).The example you have selected is so hard-the density difference of hydrogen gas STP is $\ce{0.09 kg}$ $\ce{m^{−3}}$ and helium gas is about $\ce{    0.178 kg m^{−3}}$ which gives a density difference of $\ce{ 0.088  kg m^{−3}}$ which is very low might take so much time to settle down. Gaseous have very low inter molecular interactions that they mix up fast but separate out very slow due to the mobility of denser gas is less in closed environment.
You could try few examples,

Electrolysis of salt water and taking Hydrogen and Chlorine gas to the same container Chlorine will sink to the bottom and Hydrogen will go to the top. (Hint you might need to fix two pipes to the top and bottom to take gaseous out and test).

Try putting bit of dry ice (Frozen $\ce{CO_2}$) in to a warm water jar slowly poor it to a candle near by the gas will extinguish the fire. I am sure there is lot of YouTube videos about that

To see the effect in more depth as others have explained try to get the difference of the densities of the gaseous higher. like $Chlorine$ gas and $\ce{CO_2}$ is a good example for every day use. Even the fun experiment $\ce{SF_6}$ gas (the Aluminum boat floating in air experiment).
You should check this video for what happens to gaseous when there is no gravity. The denser gas sticks to where ever it is created in zero gravity. Layers will be observed if there is gravity due to buoyancy force.

You use the adjective gaseous as a noun. // Rate of gas diffusion and convection is many orders higher than the rate of settling down. At equilibrium, composition changes with altitude very slightly.,// What changes with altitude is air pressure and therefore oxygen partial pressure, not air composition.

Yeah probably right atmosphere is not a closed system after all. I will delete that example

No, that first sentence is wrong. You can create temporary layers of gases with very different densities but, if left over time, they will mix, they won't settle out.

Then what about the buoyancy force acting upon the less dense gas from more dense gas ? Yes Hydrogen and Helium gas isn't a good example to explain those gaseous will mix.

To add a point not covered; Although they generally mix , hydrogen has such a low density it tends to rise or float and concentrate. So, when inside, flammable levels of hydrogen can collect at a ceiling. Industrial sites will use explosion proof light fixtures and other precautions when hydrogen is present. I was impressed by this situation at my first job ( about 1958 so most precautions were not yet developed ) when a hydrogen explosion blew out a brick wall. The room contained about 50 conventional steel bottles of hydrogen that were connected by manifolds ; my job was to go into the room each day and open and close valves as needed to maintain the hydrogen pressure/ flow for a process. When I heard fire engines , I went to see what was happening ; my boss was very glad to see me as they went going through the broken brick looking for me.( I don't get to tell that story much).


text:
DI water vs. Tap water in ethanol-water and salt phase change reaction
We did a very simple lab like one you would do in highschool showing that you can seperate the phases of water and ethanol using salt. As simple as it sounds, we could not get it to work no matter what we tried—until finally a student assistant used tap water instead of DI water and we saw the phase transition.
Why did the tap water work and not the DI water??

It should have worked. Was you DI water contaminated or you didn't add enough salt in DI to cause separation. Tap water has divalent cations like calcium, magnesium which can affect the salting out effect.

My hypothesis is tap water salts form fine precipitate when mixed with ethanol, forming centers for creation of water phase.

You really need tons of salt and saltparticulates therein should be sufficient to cause phase separation. I call such experiments, one time flukes. DI water works, you just need enough salt.

@M.Farooq It would still be interesting to compare it side by side. I would not exclude possible oversaturation.


text:
Preference for most acidic hydrogen

I understand that deprotonation from the carbon circled in red is feasible because deprotonation results in a resonance-stabilised structure. But according to it, deprotonation is also feasible in blue-circled carbon. Then why is that the red carbon hydrogens are more acidic?

Aldol reactions like this are thermodynamically controlled, so which hydrogen is most acidic isn't the issue here, it's which product is the most stable.

I can see here not two but five acidic centers. All of them would be deprotonated to some extent, but what comes later is important, as ortho said.


text:
atomic structure - why only hydrogen be separately written in periodic table

The other
exception is hydrogen. It has only one
s-electron and hence can be placed in group 1
(alkali metals). It can also gain an electron to
achieve a noble gas arrangement and hence it
can behave similar to a group 17 (halogen
family) elements. Because it is a special case,
we shall place hydrogen separately at the top
of the Periodic Table

lithium can also loose one electron and behave like a noble element so should we also place all ns1 and ns2 element separately at top ?

You neglected question preparation. How many electrons does H atom need to gain electron configuration of the next noble gas? And Li atom?

While most alkali metals do form anions, these do not have a noble gas configuration. Only hydride ion from this group does. Also hydride ions exist in a radically different environment from natride and heavier congeners.

In short, H and He are the two having 1s electrons configuration. It is quite natural to place them in the same period.

That's not the complete explanation why Hydrogen is placed separately. Hydrogen is highly electropositive like alkali metals and forms bonds with halogens. It also has high ionization enthalpy like halogens while alkali metals easily loose electrons. Similarly various other factors contribute special placement of hydrogen and not just it's electronic configuration.


text:
Why did these magnetic ball bearings corrode?
Before the pandemic, I bought a set of magnetic ball bearings (a few mm in diameter) and attached them to a side panel of my desk (presumably, the panel is iron coated in grey paint).
On returning to the office, I found that the ball bearings are badly corroded (more specifically, it looks like the interior of the bearings have corroded, with an outside coating flaking off).
What happened? Was there some kind of sacrificial oxidation of the ball interiors (even though the paint presumably insulates the bearings from the desk)?
It's not impossible that some kind of cleaning fluid was sprayed on the bearings at some point, though other ball bearings nearby (on a non-magnetic part of the desk) are still pristine.

They appear to be plated, maybe with nickel, and made of some inexpensive magnetized metal or ceramic. Perhaps put on safety glasses, and crush one of the bad ones with pliers, to see if it is metal or ceramic.

These magnets are usually NdFeB ceramics coated with something like nickel. So what you are seeing is not so much the corrosion of the magnet but the breakup or corrosion of the surface coating, revealing the ceramic inside. The coatings are fragile and easily corroded or disrupted. So the effect is common as plenty of things can disrupt the thin coating on the outside.

Yes, I agree some electrochemical corrosion occurring.
Per the picture, there is especially evidence of corrosion in particular spots, for example, at the end.
My guess is that one of the contributing factors is your touching the ball resulting in the deposit of NaCl. With a change in temperature and humidity, you likely fostered the creation of an electrolyte. The only missing ingredient is a weak acid, perhaps CO2 or a vinegar mist as a result of, yes, someone cleaning your desktop.
Now, ScienceDirect notes, to quote:

The rate of the galvanic corrosion depends on the relative surface areas and the corrosion potentials of the anode and cathode.

which suggests oxygen-rich air was the cathode, with your low surface area metal undergoing anodic corrosion with added salt and an acid source.
Interestingly, per a source, to quote:

To put it simply, the total cost of corrosion across the globe is astronomical. It adds up to an estimated $2.5 trillion (USD), and that number doesn’t even include individual safety or environmental consequences, according to the report.

So, you apparently made but a small contribution to the annual global cost of galvanic corrosion.