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Looking for gas or liquid for story research
I am working on a story for adaptation into animation. Main lead is a chemist in a traditional fantasy setting. The chemist uses her own compound to battle using fire or explosions. I love the chemistry to be grounded. I am looking for a very combustible gas or liquid that has a color. Preferably green-blueish. Preferably lighter than air. Does this exist? Any other suggestions about other cool looking compounds are very welcome.

This is not going to be well received here; maybe it would work at worldbuilding. For chemistry it is too vague and if you get the chemistry right, it wouldn't be exciting for your viewers.

The only gases which are combustible and lighter than air are Hydrogen $\ce{H2}$ and methane $\ce{CH4}$ ; unfortunately these gases are colorless.

Please look at the table here. I would use a mixture of trifluoronitrosomethane (for color) plus hydrogen and oxygen. And, if that answer is useful to you, please consider upvoting it when you get enough rep!

Or have your heroine chemist modify this stuff to remove the ‘pesky drawback’. ;-)

@Maurice Acetylene is a little lighter than air and CO sneaks in just barely. And diborane.

Ammonia and chlorine maybe? $\ce{NCl3}$ gives a good bang. Off-yellow color probably not a winner though....

Chlorine dioxide ($\ce{ClO2}$) is explosive at conc. above 10%. But the color is somewhat yellowish-green not quite you are looking for. Meanwhile, nitrosyl cyanide is blue-green but I'm not sure if it is flammable.

@EdV. Acetylene and $\ce{CO}$ have densities which are respectively $26/29 = 0.90$ and $28/29 = 0.96$ with respect to air. Instead of going upwards in the atmosphere, these gases will soon mix with air. And included in ballons, they will not float, due to the weight of the balloon. Anyway they are colorless.

@Maurice Sure, but the OP said “Preferably lighter than air.” So lighter than air is not an absolute requirement. If it was, they would be out of luck. But they are doing fiction, so they can take liberties by using mixtures, etc.

Since this is for a story, how about going with an aerogel containing explosive nano-encapsulated diborane and trifluoronitrosomethane and maybe some of the other chemical substances mentioned in the comments? Otherwise, there simply is no low molar mass gas that has color in the visible spectrum, i.e., 400 nm to 700 nm. Good luck with the story!

@Todd You do not make NCl3, as it is a totally impractical high explosive no less. Dust, light, any organic, ...will cause it, for no apparent reason, to produce a huge explosion. Like playing with drops of nitroglycerine. As for ClO2, a light sensitive explosive and also toxic. Any idea of mixing chlorine and organics could easily produce a very toxic (immediate or slow) brew.

Here is one possible embodiment for your story that is grounded in actual chemistry.
Warning dangerous explosive mixture is possible in reality, so as to mitigate potential danger, I recommend the use of only small amounts (say, a test tube) of hydrogen and chlorine gas in a well-ventilated setting.
The first step is to mix hydrogen and chlorine gas in a vessel. I would recommend for safety a deflated clear plastic bag devoid of any air/oxygen contamination.
Upon intended or accidental detonation, that can occur apparently via several paths, for example, per a reference, to quote:

The reaction of chlorine and hydrogen can be initiated by direct sunlight, other sources of ultraviolet light, static electricity, or sharp impact.

And produce primarily a shock wave with associated high temperatures and possible container shrapnel.
With all appropriate safety measures in place, expose the H2/Cl2 gas mix to a LED light consisting of UV light, or a spark. This results in a chain reaction cumulating in a very vigorous powerful explosion. Also, sunlight exposure may also result in unintended detonation in situations where there is a higher amount of reflective UV rays (as, for example, from a lake, ocean, snow,..).
Chlorine gas has some coloration against a white background. However, due to hydrogen dilution, likely less intense here, so close-up photography is likely required. Try increasing chlorine concentration to alleviate the coloration issue, as per a source, to quote:

mixture of hydrogen and chlorine is exploded by almost any form of energy (heat, sunlight, sparks etc). Explosive range: 5-95%.

Also, try dissolving the Cl2 in CCl4 (this may also help with the O2 contamination which stops the chain reaction).
See associated available videos here.
A novel last embodiment, yet to be verified (albeit, the associated chain reaction chemistry appears supportive of) is the reaction between chlorine (absence of any oxygen), hydrogen with added select metals imbued with hydrogen.
To construct the latter, for example, the insertion of Aluminum or Magnesium metal in HCl will, at least temporarily, possess an active surface presence of absorbed hydrogen (see this commentary by ScienceDirect and more extensive from De Gruyter).
As per a source already noted, various dry metal hydrides can ignite in chlorine, so this likely results in a detonation of the mixture. Additional caution should also be noted, as select metal hydrides, as powders, can even augment explosive powers.


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Why is there preferential adsorption of a common ion during preparation of colloids by peptization?
There was a similar question asked here on Chem SE, but the answer was not very satisfactory to me, so I am asking a fresh question.
Peptization is the process of creating a colloid from a precipitate by shaking it with a suitable electrolyte. In this process, the rule of preferential adsorption is followed which states (source) that:

The Particle Constituting the Dispersed Phase adsorb only those ion preferentially which are more related chemically to it or which are common with the Lattice ion of Dispersed phase

So what is the reason behind this selectivity? I think it might have to do with the lattice of the precipitate, because the precipitate is in a solid-state and even after being dispersed it will also remain in the solid-state but as small particles.
So, do the lattice ions try to extend their lattice by adsorption and in this process get separated into small particles? Lattice ion adsorption puts little strain on the lattice structure, because having a similar dimension they simply fit into the gaps in the edges of the crystal, whereas ions other than lattice ions will create some distortion because of its dissimilar dimension?
I found a paper related to this and according to this paper size and deformation of lattice can play an important role in adsorbtion of ions

Fajans and Erdey-Grdz (7) pointed out that in addition to solubility
other factors, such as deformation, size of the adsorbed ions, and
dissociability of the adsorption complex formed have to be taken into
ADSORPTION ON IONIC LATTICES 1029 account. Moreover, the charge
(valence) of the adsorbed ions and the deformability of the lattice
ions will play a part. Considering the many factors that influence the
adsorbability, it is quite plausible that the socalled
Paneth-Fajans-Otto Hahn rule can be only of a highly approximate
character.

Reference
I.M. Kolthoff, "Adsorption on Ionic Lattices", J. Phys. Chem., 1936, 40(8), 1027–1040, (Link to the article)

Mention the reason for downvote , please

Numerous reasons: (1) Improper use of boldface emphasis; (2) improper capitalization, both upper and lower; (3) linking a Quora question as a "source"; (4) improper punctuation, including comma, period, and question mark abuse; (5) almost completely incoherent word salad of a post; and (6) a very clear indication that you expect to get something without investing a modicum of effort here. Welcome so chem.SE. Read the rules. Tour the site.

@Samardeepsingh Please note the edits in the reference section. There is no need to put the whole reference in bold font, and it's good to follow common referencing styles.

@Samardeepsingh, What does the Kolthoff's article say? I don't know the exact reason, but must be related to fitting of the ion in the crystal structure. Don't worry too much about downvotes. It is a disease.

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What is the reaction steps taken when producing auroral forcing?
As described in here:
https://www.nasa.gov/feature/goddard/2018/sounding-rocket-mission-will-trace-auroral-winds
A mixture of Trimethylaluminium, Barium and Strontium reacts and ionize in the upper atmosphere to produce the spectral emission seen. What is the exact reaction pathway taking place?

There is some information on strontium and barium here: an earlier version of the publication below, freely available on the web, appeared in Journal of Pyrotechnics, No. 18, 2003. Color Values and Spectra of the Principal Emitters in Colored Flames, W. Meyerriecks* and K. L. Kosanke.

The chemical reaction is close to irrelevant -- high-energy ions and electrons in the aurora knock apart the chemical compounds, leaving free ions of Al, Ba and Sr, if the chemicals are sufficiently dispersed.

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Can a metal be forced to form an anion theoretically?
I know that metals have the capability to lose electrons and form cations, but is it also theoretically possible to supply an electron to a metal so that it forms an anion?
If so, has it ever been done?
I referred this question (Can two metals combine to form a compound?) but could not get a satisfactory solution from that.

See Electron_affinity_data_page

Absolutely! You will find these mostly in electride systems and off these, mostly in alkali metals.
Here is an example research paper:
"Superakali-Alkalide Interactions and Ion Pairing in Low-Polarity Solvents, J. Am. Chem. Soc., 2021, 143(10), 3934–3943 (https://pubs.acs.org/doi/10.1021/jacs.1c00115)
Remember, metals have a positive charged when ionized because it is energetically more favorable to lose electrons than to gain them, this being of course an oversimplified version of electron orbitals and shells. If you have a situation in which this is reverse or not possible, you will get a negative metal ion

Also, to your original question. Your question in if metals can form compounds between themselves and your follow up asking if they can form ions makes it seem as if you have completely rejected the idea of metals forming covalent bonds between themselves and could only be plausible on the basis of ionic bonding. Think again! Look at this absolute monstrosity containing a quintuple - yes, quintuple

covalent bond between two chromium atoms: [CrC6H3-2,6-(C6H3-2,6-(CHMe2)2)2]2

Your answer would be more complete if it included metal carbonyl anions like $\ce{[V(CO)6]-}$ and also things like $\ce{CsAu}$.

https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.200352314 is yet another example, containing Pt2-

Also it is not energetically favorable to lose electrons from an atom. It is energetically favorable to form covalent bonds, ionic bonds or solvate ions.

The electron affinity, defined as the energy released in the reaction
$$\ce{X(g) + e^- -> X^- (g)}$$
is positive for all but the noble elements and a few elements with half-filled shells (see a data table here), which is consistent with the anion being more stable than the neutral atom. This makes it difficult to understand the motivation for the first half of your question, since the answer is a resounding yes, there is no spontaneous driving force that would drive (most) isolated metal anions to dissociate giving up an electron. There is at first glance no theoretical basis to suppose such anions cannot be formed.
On the other hand, the ionization energy of all (gas phase) atoms is positive, that is, loss of an electron is an unfavorable process. So it is not a good idea to generalize that "metals have a positive charge when ionized because it is energetically more favorable to lose electrons than to gain them". Just the opposite. Most metals have favourable electron affinities.
Making anions that are stable is another matter, precisely because most other elements also attract additional electrons.
A useful measure of the relative attraction of elements for electrons is the electronegativity, of which the Mulliken variant is defined as an average of the first electron affinity and ionization energies. And since metals have smaller electronegativities than non-metals, they tend to lose electrons rather than gain them. While most metals have positive electron affinities, their ionization energy is often lower than the electron affinity of other atoms.

Yes, there are a large number of negatively charged metal ions, many of them containing several covalently bound atoms. The positively charged counterion is typically a group 1 metal.
Wikipedia defines a zintl phase as the product of a reaction between a group I or group II metal, and a post transition metal or metalloid, but most of the examples are based on elements that are not unambiguously metals (such as silicon.)
https://en.wikipedia.org/wiki/Zintl_phase
There are however separate pages for plumbides and stannides. The stannide page claims Sn(4-) in Mg[2]Sn which is a surprising amount of charge on one atom. Polymeric (Sn(2-))[n] has more reasonable double negative charge.
https://en.wikipedia.org/wiki/Plumbide
https://en.wikipedia.org/wiki/Stannide

To make a capacitor out of metal, you have to have negatively charged metal atoms on one of the plates (although I guess one could argue that the negative charge is not tied to a particular atom, but shared among all the atoms).


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Can two different definitions of acidity and basicity be used together?
I wanted to predict the pH of $\ce{PH3}$, it has basic nature as it has a lone pair of electrons. But it also has acidic nature as its $\ce{P-H}$ bonds are weak. Now according to the  Lewis definition $\ce{PH3}$ is a base and according to Bronsted-Lowry definition it is a acid. Can we say from this that $\ce{PH3}$ is neutral?

Nope, you definitely can't say that PH3 is neutral. It depends on the case- PH3 will react like a Bronsted Lowry acid with Bronsted Lowry base, and it will react like a Lewis base with a Lewis acid.

A rather weak base and an awfully weak acid, I must say.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.  And by convention, after the full stop/period at the end of a phrase, we use a space.

Put simple: these are two mutually independent definitions.

The Bronsted concept is based on the possibility to donate a proton (i.e., an acid), or to accept a proton (i.e., a base).  But it isn't only the presence/absence of a proton in a compound, it equally depends on the reaction partner if a molecule may act as an acid, or as a base.

The Lewis concept is based on the presence of an non-bonded electron pair (Lewis base), or empty molecule orbital (Lewis acid).

Since the two concepts take a different perspective, which is independent of each other, it is possible (but not a requirement) that the two concepts independently assign a compound to be an acid, or a base.  It happens e.g., the hydroxide $\ce{OH-}$ anion indeed simultaneously is both a Bronsted base as well as a Lewis base.  But since the two concepts are set up independently of each other, you can't «neutralize» acidity assigned by one theory with basicity assigned by the other theory.


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Why are there only 14 types of Bravais lattices and not 28 when there are 7 types of unit cells and each can have four variations?
As the title suggests, I can't understand why certain kinds of variations (like Face-centred or Body-centred) are restricted to certain types of unit cells. An orthorhombic unit cell has Primitive, Body-centred, Face-centred and End-centred variations, why can't the same be applicable to other crystal systems too?
Has this something to do with symmetry, but in that case a cubic unit cell should have four variations (it only has three). There are some patterns, like all crystal systems which have angles $=90^\circ$ have Body-centred variations, but beyond that it doesn't make sense.

Yes it has to do with symmetry. Say, a body-centered cubic cell is impossible due to the lack of cubic symmetry.

Related: https://chemistry.stackexchange.com/questions/47834/why-can-a-face-centered-cubic-lattice-not-be-redrawn-as-a-body-centered-tetragon/47868#47868

@Ivan "a body-centered cubic cell is impossible". So iron at ambient temperature does not exist (it's bcc). Typo maybe?

@Oscar Of course. I meant to say "base-centered".

Essentially, certain combinations of the possible point-group symmetries (cubic, tetragonal, hexagonal, trigonal, orthorhombic, monoclinic, triclinic) and possible translational symmetries (simple, base-centered, face-centered, body-centered) end up having identical overall lattice symmetries and thus you don't get $7×4$ unique lattices.
For example, suppose you propose a base-centered cubic lattice. Base-centered means you get the same lattice back if you translate the corners of each unit cell to the centers of a pair of specified opposing faces (the "bases"). But "cubic" means you get the same unit cell back when you rotate each face to match an adjacent one (rotating around a body diagonal of the cube). So to have both the base-centered translational symmetry and the cubic point-group symmetry, you have to allow translation of the unit cell corners to the centers of all the faces, not just one opposing pair, and your intended base-centered cubic lattice is really face-centered cubic.
Let's try a different example. Suppose you try to construct a base-centered tetragonal lattice by allowing translations of the corners onto the centers of the opposing square faces of the prism. Tetragonal point-group symmetry does not include a roration around a body diagonal or any other operation that would shift the square faces onto another face, so you avoid the trap of turning "base-centered" into "face-centered" like what happened with cubic symmetry. You really do have a specific pair of "bases". But now there is a different trap: you can draw a smaller unit cell, with smaller square faces, that is a simple tetragonal lattice. So again your intended base-centered lattice is not unique; in this case it is just another simple tetragonal lattice.
When we work through all the constraints with each of the seven point-group symmetries, we find that a unique base-centered lattice exists only for the orthorhombic point-group symmetry, a unique body-centered lattice exists only for cubic, tetragonal and orthorhombic point-group symmetries, and so on. Thus only 14 out of the apparent 28 point-group/translational symmetry combinations actually form different Bravais lattices.

Why can't a triclinic structure have a base centered lattice?

You can draw a smaller cell that is still triclinic.

Any reason an orthorhombic crystal can be face centred but not a tetragonal one, since both can be redrawn into smaller orthorhombic/tetragonal lattices, and the second one has "more" symmetry

I assume you know how to make a smaller body-centered tetragonal cell out of a larger face-centered one. If you try this with orthorhombic symmetry, however, the intended body-centered cell does not have rectangular bases like the actual standard body-centered orthorhombic cell. Instead the bases are rhombi. Equivalence occurs only when the rectagle/rhombus is a square, in which case the point group upgrades to tetragonal.

Whether to use a centered cell or the smaller primitive cell is a question of convention. The conventions are guided by making life easy. You could take a primitive triclinic cell and redefine the cell axes and lengths to make it body-centered but that would be kind of silly. There is no gain in simplifying symmetry operations, and you just made your cell volume twice as large and added a symmetry operation you didn’t have for the primitive case.
On the other hand, a body centered orthorhombic cell is easier than the corresponding smaller cell with non right angles and symmetry elements going along diagonals.
For cubic space groups, we don’t expect an A-centered cell because the presence of a 3-fold along the body diagonal would result in centering of the other faces as well.


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Obtaining orbitals from a many-body wave function
I have seen how sometimes people talk of orbitals obtained from a CAS or DMRG calculation. I haven't found any explanation of how this is possible. Given the many body wave function (obtained from exact diagonalization, Lanczos, DMRG, whatever), $\Psi(x_1,\ldots,x_n)$, how could one possible define a set of orbitals, which are one body functions?   I think (not sure) that this is what computational chemists refer to as Natural Bond Orbitals, but I have only found conceptual explanations, not the mathematical details. Probably is related with diagonalizing the one-particle density matrix?  Can someone provide mathematical details/references?

For any wave function, you can define the one particle density matrix $\int \psi(x_1, \dots, x_n)\psi^\ast(x_1^\prime, x_2, \dots, x_n) dx_2\cdots dx_n$. The eigenfunctions of this operator are single particle functions and are called natural orbitals. For a treatment of reduced density matrices see for instance [this book] (https://www.springer.com/gp/book/9783540671480).


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I mix table salt and baking soda with water to make mouth rinse, and leave them for several days. It started to form precipitation, and a bad odor
I mixed 2 spoons of baking soda with 1 spoon of salt and mixed them with water (but not mixing totally) and left them for several days. At first I saw some bubble but the water was clear, but then it got this layer of precipitation on top. I still use it to rinse my mouth, but then I notice it has this really bad odor that grows worse day by day. I threw it all away, but the mixture/precipitation stick to the side of the glass too and it's very hard to brush off.
What happened? Was there any toxicity? I googled everywhere, but they say there couldn't be reaction between $\ce{NaCl}$ and $\ce{NaHCO3}$. Could it be the air and or something in the water that added to it? Or the glass?

Fungus and mold growth. The layer of "precipitation" on the top, as you said, is fungal/bacterial growth. Common sense dictates that you should discard such solutions, let alone use them for rinsing your mouth.

I have used that mouthwash for several years with no problem.  I mix  a cup of each dry powder together then use a small amount to make fresh  water solution a few times per day.

It is not much different than seawater, and many things grow on that.

All aqueous solutions when maintained several days in the open air, get dust, batteries  and pollen from the wind. This produces mold growth and bad oder. Nothing unusual. Discard these solutions !

First, per a reference on personal care products, to quote:

Many personal care and pharmacy products contain a mixture of citric acid and sodium bicarbonate.

So chemically, you may actually have a mix of citric acid and NaHCO3, which per an educational  source can react as follows, to quote:

In the presence of water, citric acid and sodium bicarbonate (aka baking soda) react to form sodium citrate, water, and carbon dioxide. Students investigate this endothermic reaction...

So one possible explanation upon mixing NaCl, citric acid and NaHCO3 with a limited amount of water could form a citrate as discussed in this Wikipedia article on trisodium citrate.
Interestingly, its reported appearance is a white crystalline powder. Further, quoting the Wikipedia article:

Sodium citrate is employed as a flavoring agent in certain varieties of club soda. It is common as an ingredient in bratwurst, and is also used in commercial ready-to-drink beverages and drink mixes, contributing a tart flavor.

So your tasting experience may have help in the identification of this distinct unexpected salt.
I would also note as a possible explanation that adding CO2 and NaCl to tap water lowers dissolved oxygen content. This may foster, in time and exposure to dust, a more biological-based growth that would, however, not have a crystalline salt appearance.
I hope this helps.

How have you come to conclusion citric acid is involved in any way? Batches of baking soda or table salt do not contain citric acid.

First, I cite the citric acid presence as a 'possibility'. I also vaguely recall reading the contents on bicarbonate mixtures in the grocery store. Per a source: "baking powder (a mixture of sodium bicarbonate and at least one other compound). Baking powder differs from baking soda in that it includes an acidic compound that reacts with sodium bicarbonate to produce carbon dioxide. One of the most common compounds mixed with sodium bicarbonate in baking powder is tartaric acid..." at https://www.encyclopedia.com/science-and-technology/chemistry/compounds-and-elements/sodium-bicarbonate


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How do you make an SDS for a milligram sample synthesized for the first time?
There is this impurity detected by GCMS that we got custom synthesized at a 15 mg level. There's this absolutely tiny vial that needs to be couriered to the lab for further analysis.
Fedex insists on an SDS. Other than identity and a few basic details nothing more is known at this point about the sample (e.g. flash point, LD50 etc.)
But of course, the quantity is 15 mg so one would not reasonably anticipate a hazard. I am not aware of an existing SDS nor do we have enough information on the hazards. For all we know this substance has no commercial uses and may not have been manufactured at scale ever before. That is why we had to do a custom synthesis (Sigma / Merck etc.did not offer it in their catalogs)
How would the chemists deal with a situation like this one?  It seems to transcend the reasonableness standard.

Of course it depends on dose and pathway of application when LD50 is reached (e.g., botolinum toxin).  If your compound is an analogue/homologue in structure of already known compounds, may you infer from their SDS suitable precautions and mention that these are preliminary?  If you have their CAS registry numbers and know some Python, find_sds is one helpful tool to collect such data (you might need to remove the explicit version request for entry idna in file requirements.txt to get it installed).

I have sent - and received - many samples and I've always filled in terms as non flammable non explosive etc. Somewhere, but I forgot if it was in an accompanying letter, I've made clear that it was a new compound for academical research ONLY. I've never run into troubles of sort.

@Alchimista And that's how it should be! Sadly sometimes the people involved follow scripted processes: Chemicals = SDS. Maybe this is because it's an international air shipment? Who knows!

@curious_cat Perhaps in this case. But what I mean is that the SDS is replaced by an accompanying letter (in addition to the shopping formular where you keep low profile in all entries and describe your sample as for research only. The letters can be a copy of your research correspondence to the recipient, this I was doing to prevent problems at the custom, especially in certain times (anthrax crisis, for example). But it is now quite some time I am not sending/receiving by couriers...


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Solubility and covalent character
We have heard many times that this XYZ solute is soluble in this ABC solvent. But what does being soluble in mean in terms of bonding? What makes a compound soluble in some solvent and not in some other. In my text it is written that:

Lithium chloride is considered to be an ionic compound, but it is soluble in alcohol, which suggests that
it also possesses a small amount of covalent character.

What does covalent character have to do with solubility? On what factors does solubility depend? Please clear this confusion.
Edit:

Inter-ionic interaction is strong and solvent has to have strong interaction with ions to dissolve them, and that means it has to be very polar solvent.

This sentence helps the most!

Well, if something is soluble in ethanol one could suspect it's somewhat similar to this solvent (like in the like), which is covalent, isn't it. It's pretty lame reasoning, but apparently the one used there.

Solubility is one of the numerous parameters that have not yet been understood in chemistry. There are plenty of empirical rules that work perfectly for a group of substances. But not for all substances. There are too many exceptions. Just consider this example ! In principle, $\ce{F, Cl, Br}$ and $\ce{I}$ should have similar properties. Matter of fact, $\ce{CaCl2, CaBr2, CaI2}$ are similar. They are all extremely soluble in water. Unfortunately $\ce{CaF2}$ is absolutely insoluble in water. It is the main mineral source of fluor in the world. Why is it insoluble ? Nobody knows.

@Maurice, Thanks for emphasizing this point "Solubility is one of the numerous parameters that have not yet been understood in chemistry." Students should understand that.

@MFarooq I would change that to "easy to understand" or "fully understood in all cases".

The key for solubility are comparable intermolecular forces solute-solute, solute-solvent, solvent-solvent.
Inter-ionic interaction is strong and solvent has to have strong interaction with ions to dissolve them, and that means it has to be very polar solvent.
If there is covalent character(true covalent molecules, or tight ion pairs), intermolecular solute-solute forces are much weaker. If solvent-solvent forces in polar solvents are much stronger, replacing them by weak forces solvent-solute is not favourable. OTOH non-polar solvent have weak mutual forces, so there is not energy problem and dissolving is generally preferred by mixing entropy raise.
Therefore, ionic salts are generally much less soluble in ethanol, compared to water. Covalent character brings compounds closer to ethanol in involved forces and solubility increases.
---- Response to comments
The strength of intermolecular forces solute-solvent is usually somewhere in the middle between solute-solute and solvent-solvent, so forces have to be evaluated in their mutual relation. If forces solute-solvent are strong, but solvent-solvent forces are stronger, solubility can be lower than for case all are rather weak.

So basically, can I conclude ( up to high school level) that the more a solute forms bonds or has greater attraction force with a solvent, more soluble it will be?

See the answer update.

lithium chloride is considered to be an ionic compound, but it is soluble in alcohol, which suggests that it also possesses a small amount of covalent character.

This is a somewhat elaborate case, where the underlying point is that what solubilizes in alcohol is not the same as what solubilizes in water.
Water is a very small molecule with a very large dipole moment (intramolecular charge separation). The water molecule can easily accommodate charged species (ions), which it surrounds with its dipoles, forming stabilizing ion-dipole interactions. In addition water has a large dielectric constant, which results in screened interactions between ions that reduces their mutual attraction at longer distances. A third factor is that water has structure due to hydrogen bonding, which can change in character when a solute is incorporated, with associated changes in entropy and enthalpy. When the solvent becomes more structured, as happens when a highly charged ion is dissolved, the entropy of surrounding water is lowered, which discourages solubilization. However, as mentioned before, there are strong ion-water interactions which result in a large magnitude of the enthalpy of solvation (which is negative). Importantly, the devil is in the details of the relative magnitudes of the entropy and enthalpy changes.
Alcohols have neither as large a dipole moment nor dielectric constant (they are also not as structured by hydrogen bond networks, but that counts as a plus when it comes to solubilizing power). The net result is that ions are generally less soluble in alcohols. The strong mutual attraction of oppositely charged ions is not easily overcome by attractive interactions with the solvent. The result is a very low solubility of the ions.
But there is more (the catch). $\ce{LiCl}$ can solubilize as dimers composed of one $\ce{Li+}$ and one $\ce{Cl-}$ ion. The author interprets this as suggesting that there must be a covalent interaction that holds these atoms together (although why ionic interactions are not enough is not explained). The most important point here is that the energetics of dissolution of a covalent solid (for instance sugar) versus an ionic solid (table salt) are very different, because the strength of solute-solvent interactions are very different, as are solute-solute (or ion-ion) interactions. Ions interact much more strongly with the solvent and with each other, and in order to solubilize an ionic solid you need strong ion-solvent interactions that can compensate for the strong ion-ion interactions, if ion-ion interactions are indeed broken! In the case of $\ce{LiCl}$ breaking those interactions is not entirely necessary. In a covalent solid with small solutes, solute-solute interactions are weaker and solute-solvent interactions do not need to be as strong to compensate for disrupting the solid lattice.
Overall this is not entirely satisfying. If all this can seem confusing, I agree, there are a lot of details to unravel and some of the terminology sounds very similar (solvation versus dissolution, for instance). In any case, it is important to keep in mind that what happens in the solvent is only half of the picture. The other half is how the ions are organized and interact in the solid lattice. When dissolution of an ionic solid is discussed, it is often in terms of two contributions: the lattice enthalpy (which describes how strongly ions interact in the lattice, and the enthalpy of solvation (which describes how strongly the solvent interacts with the ions)*. If you ignore how strongly the ions attract in the lattice you will not fully understand why an ionic solid solubilizes.
*entropic factors can also be important as not all solutes dissolve exothermically (driven by an enthalpy change).


text:
Boiling of ammonia water
L̶e̶t̶'̶s̶ ̶m̶a̶k̶e̶ ̶i̶t̶ ̶s̶i̶m̶p̶l̶e̶.̶ ̶I̶ ̶h̶a̶v̶e̶ ̶a̶ ̶p̶o̶o̶l̶ ̶o̶f̶ ̶a̶m̶m̶o̶n̶i̶a̶ ̶w̶a̶t̶e̶r̶,̶ ̶a̶n̶d̶ ̶w̶h̶e̶n̶ ̶I̶ ̶h̶e̶a̶t̶ ̶t̶h̶e̶ ̶a̶m̶m̶o̶n̶i̶a̶ ̶w̶a̶t̶e̶r̶,̶ ̶a̶m̶m̶o̶n̶i̶a̶ ̶w̶i̶l̶l̶ ̶d̶e̶s̶o̶r̶b̶ ̶a̶n̶d̶ ̶v̶a̶p̶o̶r̶i̶z̶e̶.̶ ̶I̶n̶ ̶t̶h̶i̶s̶ ̶c̶a̶s̶e̶,̶ ̶(̶1̶)̶ ̶w̶i̶l̶l̶ ̶t̶h̶e̶ ̶w̶a̶t̶e̶r̶ ̶b̶e̶ ̶v̶a̶p̶o̶r̶i̶z̶e̶d̶ ̶a̶s̶ ̶w̶e̶l̶l̶?̶ ̶(̶2̶)̶H̶o̶w̶ ̶d̶o̶ ̶I̶ ̶d̶e̶t̶e̶r̶m̶i̶n̶e̶ ̶h̶o̶w̶ ̶m̶u̶c̶h̶ ̶o̶f̶ ̶t̶h̶e̶ ̶w̶a̶t̶e̶r̶ ̶w̶i̶l̶l̶ ̶b̶e̶ ̶v̶a̶p̶o̶r̶i̶z̶e̶d̶?̶
̶
̶T̶h̶a̶n̶k̶s̶ ̶i̶n̶ ̶a̶d̶v̶a̶n̶c̶e̶.̶
Thanks for the replies. Since now I have known that no matter how I boil the ammonia water to desorb ammonia, water will always present. And since there is people that wanted me to elaborate my question, here it goes.
I am doing a Fluent simulation of boiling 20% ammonia water under 4 atm using multiphase model. The issue is I need to define the mass transfer from liquid phase to vapor phase, and the mass transfer model I plan to use is evaporating-condensation. However, to use this evaporation-condensation model, I will need the saturation temperature. As far as I know, the saturation temperature is used to define the further dissolving of ammonia. However, since a small fraction of water will be vaporized as well, how do I determine the temperature for the water to start vaporizing (can I use the boiling point of water under 4 atm)?
Thanks in advance.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

If you heat the commercial concentrated ammonia solution ($25$% $\ce{NH3}$) at usual pressure, it will boil at $32$°C and the vapor contains $3$% $\ce{H2O}$ (and of course $97$%  $\ce{NH3}$). So the liquid looses much ammonia and nearly no water. Its total volume decreases a bit but the concentration of ammonia decreases more, so that it is necessary to heat to a higher temperature to continue distillation.
When the concentration of ammonia is down to $23$%, the liquid has to be heated at $35$°C, and the vapor contains $4$% water.
When $\ce{[NH3]}$ = $21$%, the liquid boils at $40$°C, and the vapor contains $5$% water.
When $\ce{[NH3]}$ = $18$%, the liquid boils at $50$°C, and the vapor contains $9$% water.
When $\ce{[NH3]}$ = $15$%, the liquid boils at $55$°C, and the vapor contains $12.5$% water.
When $\ce{[NH3]}$ = $10$%, the liquid boils at $70$°C, and the vapor contains $30$% water
If you want other numerical values, look for "Diagram ammonia - water" over Google. And you will find all the data you want, even at higher pressures.


text:
What role does carbon play in the calcination of strontium carbonate in step 1 of the strontian process?
Wikipedia attempts to discuss the chemistry here, with the concern,

Where does the second C go?

Clearly, someone has subsequently attempted to balance $\ce{SrCO_3}+\ce{C}+\ce{H_2O}\stackrel{?}{\to}\ce{Sr(OH)_2}+\ce{CO_2}$ by doubling $\ce{CO_2}$ on the right-hand side, but this doesn't balance oxygen. Indeed, it's not hard to see this equation cannot be balanced without removing elemental carbon. If for every mole of the consumed carbonate we consume $a$ moles of $\ce{C}$, we must produce $a+1$ moles of $\ce{CO_2}$, and hence to balance hydrogen (oxygen) requires $1$ ($2a+1$) moles of $\ce{H_2O}$, so $a=0$.
This is all an attempt to make sense of the following paraphrase of Das Mineralreich:

Strontium carbonate is calcined with carbon in the presence of steam to form strontium hydroxide. The strontium and carbon dioxide formed are rejoined later in the process, forming strontium carbonate once again.

Does "calcined with carbon" imply carbon isn't consumed, but is important to have present (e.g. as a catalyst)? Or should another species be present in this equation? I assume the former, because if so the entire process has no net chemical effect; it simply relocates sugars in molasses, which is the goal of the process. If I'm right, carbon's role in the calcination will be in affecting the reaction mechanism for $\ce{SrCO_3}+\ce{H_2O}\to\ce{Sr(OH)_2}+\ce{CO_2}$, hence my tag choice.

IMHO, there is error in the article, as 2 CO are formed, not 2 CO2. Reduction CO2 + C -> 2 CO supports calcination by removing CO2.

@Poutnik Are you saying the step is $\ce{SrCO_3}+\ce{C}+\ce{H_2O}\to\ce{Sr(OH)_2}+2\ce{CO}$ followed by $2\ce{CO}+\ce{O_2}\to2\ce{CO_2}$, so net $\ce{SrCO_3}+\ce{C}+\ce{H_2O}+\ce{O_2}\to\ce{Sr(OH)_2}+2\ce{CO_2}$?

I think Wikipedia's equation is not a correct representation. In the Mineralreich  (Wiki link is based on Scribd, whereas the same is available on Internet Archive). With a little bit of effort one can locate the relevant paragraph

Verwendung. Strontianit ist das am leichtesten zu verarbeitende
Rohmaterial für Strontiumpräparate und wird daher in Westfalen, wo er
in genügender Mächtigkeit vorkommt, durch Bergbau gewonnen. Für seine
Verwendung ist bemerkenswert, dass die Kohlensäure nicht wie bei
Kalkspat durch einfaches Erhitzen ausgetrieben werden kann, er muss
hierzu entweder in Wasserdampf oder mit Kohle, am besten mit beiden
erhitzt werden. Das so oder aus dem Nitrat erzeugte Strontiumoxyd und
-hydroxyd wird in der Zuckerinduslrie verwendet, um aus der Melasse noch den  letzten Rest Zucker zu gewinnen. Zucker verbindet sich mit
Strontian zu schwer löslichem Saccharat, einer Strontianzucker
verbindung, und kann als solches von der Melasse abgeschieden werden.
Usage. Strontianite is the easiest raw material for strontium
preparations to process and is therefore obtained by mining in
Westphalia, where it occurs in sufficient abundance. It is noteworthy
for its use that the carbonic acid cannot be expelled by simple
heating, as is the case with calcite; it must be heated for this
purpose either in steam or with coal, preferably with both. The
strontium oxide and hydroxide produced in this way or from the nitrate
is used in the sugar industry to extract the last bit of sugar from
molasses. Sugar combines with strontian to form sparingly soluble
saccharate, a strontian sugar compound, and as such can be separated
from the molasses.
Translated with www.DeepL.com/Translator (free version)

I think the above paragraph is self-explanatory. $\ce{SrCO3}$ cannot be decomposed like calcium carbonate and we need a powerful reducing environment. It must be a two step reaction. One would get $\ce{SrO}$ first by strong heating in the presence of carbon and steam would convert the $\ce{SrO}$ to  $\ce{Sr(OH)2}$. Wiki's equation attempts to combine both, as if it were a simultaneous reaction.

Thank you. I think @Poutnik's comment implies the step is $\ce{SrCO_3}+\ce{C}+\ce{H_2O}\to\ce{Sr(OH)_2}+2\ce{CO}$ followed by $2\ce{CO}+\ce{O_2}\to2\ce{CO_2}$, so net $\ce{SrCO_3}+\ce{C}+\ce{H_2O}+\ce{O_2}\to\ce{Sr(OH)_2}+2\ce{CO_2}$. Do you think that's right?

@J.G. I suppose you are right. //  it must be heated for this purpose either in steam or with coal, preferably with both.  Both ways shifts the equilibrium by removing the calcination products. Steam removes $\ce{SrO}$,  coal removes $\ce{CO2}$.

@Poutnik Ah, so it's three reactions, $\ce{SrCO_3}+\ce{C}\to\ce{SrO}+2\ce{CO}$ followed by $\ce{SrO}+\ce{H_2O}\to\ce{Sr(OH)_2}$ and $2\ce{CO}+\ce{O_2}\to2\ce{CO_2}$?

@J.G.  It is rather a chain of reactions: $\ce{SrCO3(s) <<=> SrO(s) + CO2(g)}$ // $\ce{SrO(s)  + H2O(g) <=>>Sr(OH)2(s)}$ /AND-OR/ $\ce{CO2(g) + C(s) <=>> 2 CO(g)}$ //  $\ce{2 CO(g) + O2(g) <=>> 2 CO2(g)}$ // It is rather on the chemist, if to focus on particular reaction steps or on the partial or total net reactions.


text:
Sodium metal and alcohol reduction of imines. Is it possible?
It is known that sodium metal in ethanol with toluene under reflux can reduce nitriles to amines1 and can also reduce oximes and schiff bases to primary amines and secondary amines respectively2. Is it possible to reduce primary imines where $\ce{R1R2C=N-H}$ or $\ce{R1R2C=NMgBr}$(imine anion) to amines? I have of yet not yet been able to find literature confirming this.
References

Walter, L. A.; McElvain, S. M. (1934). The Reduction of Cyanides. Journal of the American Chemical Society, 56(7), 1614–1616. doi:10.1021/ja01322a051
The Reduction of Aliphatic Cyanides and Oximes with Sodium and n-Butyl Alcohol, C. M. Suter and Eugene W. Moffett, Journal of the American Chemical Society 1934 56 (2), 487-487
DOI: 10.1021/ja01317a502

I couldn't find a paper which directly discusses imine reduction with the help of sodium metal and alcohol but I did found its mention in a different paper which discusses the reduction using calcium and ethanol. I believe sodium metal follows similar reaction pathway.

The reduction of imines is one of the most significant and useful methods for preparation of the corresponding amines. Many reduction processes were therefore widely investigated, which include catalytic reduction as well as the use of metal hydrides ($\ce{LiAlH4, NaBH4}$ or others) transition metals complexes (such as iridium, rhodium, molybdenum, nickel, tin, titanium, ruthenium, samarium), different boron compounds or complexes or metal reduction systems- metal (Na, Zn, Al or Mg) in alcohols, alkaline or acidic aq. solution. Most of the general methods of imine reduction present several disadvantages in the view of cost, safety and operations. As a solution to these problems, we paid much attention to the reduction of imines in water or alcohol as both hydrogen source and solvent.  Now, we wish to report the reduction of aromatic and aliphatic imines using metallic calcium in ethanol.

Ref.: RAPID AND CONVENIENT APPROACH TO AMINES: REDUCTION
OF IMINES USING METALLIC CALCIUM IN ETHYLALCOHOL, Yoshiharu MITOMA, Naoyoshi EGASHIRA, Cristian SIMION and Masashi TASHIRO, Revue Roumaine de Chimie, 2006, 51(7-8), 839–842 (link)

magnesium has also been used as the reducing agent for reductive amination so I would expect sodium would work.  https://pubs.rsc.org/-/content/articlelanding/1996/p1/p19960000265/unauth


text:
The total number of stereoisomers of 1,2-dibromo-3,4-dichlorocyclobutane
I tried writing all the geometrical isomers for 1,2-dibromo-3,4-dichlorocyclobutane, and then saw if they were optical or not. And I am getting the answer as 10:

However, the given answer is 8. I can't really figure out where I went wrong. I am not sure whether the last one is a stereoisomer or not.

The last two are both optically inactive ( they have axis of symmetry..).Hope that helps.

@Abhinav Can you  please once say where the axis will be?

The two Cl atoms are not symmetric right? One is above the plane and one is below...

Huh? It won't be even 8 as at least two of them are identical.

I think if the question is asking for number of stereoisomers, OP is correct about 10. If it is about optically active isomers, then answer is 8 as shown in following image:

Since the given molecule contains four chiral centers, theoretically it should have maximum of $2^4 = 16$ stereoisomers. The best way to find the correct number of isomers is to assign the $R/S$ (Cahn-Ingold-Prelog) configuration for each chiral center. Since the molecule has at least one plane of symmetry based on the stereoisomer as shown for the structure 1, the structure can name two different ways: For example, for the structure 1, it could be $(1S,2R,3R,4S)$ going counter-clockwise (red-numbering) or $(1R,2S,3S,4R)$ by going clockwise. Thus, the corresponding mirror images would have $(1R,2S,3S,4R)$ going counter-clockwise (red-numbering) or $(1S,2R,3R,4S)$ by going clockwise configurations. As shown in the image, these for configurations are identical. Thus, structure 1 is not optically active (meaning its mirror image is superimposable because it has plane of symmetry as indicated) and called meso-isomer. Yet, it is a stereoisomer.
The same is true for the structure 4 as well and hence optically inactive (the second meso-isomer). However, structures 2, 3, 5, and 6 have non-imposable mirror images as indicated in above image (show the assigned $R/S$ configurations). As a consequence, these four and their mirror images are optically active (total of eight structures). Thus, 1,2-dibromo-3,4-dichlorocyclobutane has 10 stereoisomers: 8 optically active isomers and 2 meso-isomers.
OP's doubt about last structure can be solved by assigning corresponding $R/S$-configurations as well: For the given structure it is $(1S,2S,3R,4R)$ (clockwise) or $(1S,2S,3R,4R)$ (counter-clockwise; same as clockwise). The assigned mirror image is $(1R,2R,3S,4S)$ (clockwise) or $(1R,2R,3S,4S)$ (counter-clockwise; same as clockwise). There are no coinciding names. Keep in mind that there is no plane of symmetry because 1,2-dibromo versus 1,2-dichloro here. Thus, structure 6 is optically active.

@Mather Mahindaratne Thnx for the solution. In the question they have mentioned to find the number of stereoisomers and not optically active stereoisomers, but seeing as you too got 10 for stereoisomers, the answer must be wrong... Again thnx for clarifying my doubt in this much detail!

@MathewMahindaratne downvoted because you assign two different names to each structure, although only one is correct

@Loong: Not necessarily. The two names ($R/S$-configurations) are based on numbering. Rest of the name is the same.

Exactly. One numbering is correct and the other one is wrong.

@Loong  I dont think  he numbered/labeled the whole compound as either R or S, he just numbered all the chiral carbons individually.


text:
Metal oxide formation minimum temperature?
Is there such thing as minimum temperature for creation of metal oxide?
Let's say a highly reactive metal, lithium, comes into contact with solid oxygen at a cryogenic temperature of 20 Kelvin, would oxide form, or will lithium act more like a precious metal at these super cold temperatures and remain oxide free?

Since you are having reaction between solids, diffusion is the limiting factor affecting kinetics. I highly doubt at these temperatures it's substantial enough to make an observable oxide layer. Keep in mind that alkali metals also visibly react with ice only when they are pre-heated. If they are not causing ice melting, there is no reaction.

Then again, oxygen is a biradical and might well react with zero activation energy, and then...

@IvanNeretin. Will oxygen react with lithium or with iron at $20$ Kelvin ?

With iron, it doesn't react much even at room temperature. With lithium, I don't know what to expect.

@andelisk Ok, then I change my question to liquid oxygen at 60 Kelvin. Will it make the oxide then?

It may be rather quantitative than qualitative question: What rate of oxide creation would be enough to observe oxide creation ? //  Remember that individual atoms or molecules do not have temperature. Different temperature just means different statistical energy distribution. Lithium atoms and oxygen molecules with enough energy may react at any temperature.

@Poutnik Since velocity of atoms at given temperature has large random distribution, I think I made mistake by asking this question in terms of temperature. What I really want to know is how much electron volt quantity of energy is required to create the oxide, if there even is any energy barrier, I want to know if this reaction theoretically requires zero energy that means it can happen at 0.001 Kelvin.

Since there is need to break O2 molecule to form 2 oxide ions, it is not IMHO justified to expect near zero activation energy.

@Poutnik I thought that since lithium is so electropositive, it might tear the oxygen atom away. Or alternatively, 4 lithium atoms might connect with dioxygen without any oxygen splitting needed.  I was surprised before at quantum effects affecting chemistry so despite being aware of the o2 bond energy, I thought this question is worth asking.

@poutnik you may not need to break the oxygen-oxygen bond completely. Lithium can form a peroxide, even if this would be only metastable. Even so, formation of the peroxide ion still requires a reduction of bond order, so the activation energy is merely less instead of zero.

@OscarLanzi I am aware of peroxide,  I used context of oxide. Another possible factor for nonzero activation energy is electron work function within metallic structure.

If there was near zero energy barrier, oxide formation reaction rate constant would be (very)  roughly $k \approx a \cdot \sqrt{T}$, driven by thermal collision rate. I very doubt this is the case. Usually, T change by 10K causes reaction rate change by 2-4times. The above case would need 4 times higher T for 2 times faster rate.


text:
Hydrogen atom orbital projections resemble to some Chladni figures! Mathematical reasoning?
Chladni figures are formed by sprinkling salt on a vertically vibrating membrane, f.e. . Some of them are purported to be projections of hydrogen atom orbitals f.e. 1s, 2s, 2p, 3d, e.t.c., f.e.
One could bring it down to that the vibration patterns of Chladni figures are, as known, two-dimensional standing waves (this part is well established) and also to that -according to the Copenhagen (Born) interpretation- the hydrogen electron distribution also follows a (three-dimensional) standing wave pattern.
So far, so good, but what about the mathematical reasoning lying behind?
Could we claim that the two-dimensional Schrodinger equation for the hydrogen atom (i.e. the radial associated Laguerre polynomial, Ln) and the Chladni figures equation (i.e. the Bessel function of the first kind, Ja) are linear to each other?  Any thoughts on the mathematical reasoning would be highly welcome..

It would be helpful to cite a source for the resemblance and show some examples of ones that actually match orbitals.

Indeed. Examples added.

Modes of a 2D membrane and modes of 3D spherical harmonics will have a number of similarities just because of symmetry requirements. That is all.

Also suitable for Mathematics if you are looking for mathematical "finesse".And I would say what are the mathematical similarities or something like that. There shouldn't be any real reasoning, perhaps just reasons.

I was hoping -rather in a naive way- to find some linear relation between associated Laguerre polynomial (L) and Bessel function of first kind (J)-the main part of the Chladni figure equation. I have found a number of relations between but none which shows a straightforward relation, at least in my eyes..


text:
Lithium Air vs Li Fluoride battery capacity
Lithium air aka lithium oxygen batteries are described as highest theoretical capacity battery possible. In wikipedia, it says lithium air has energy density 5210 Watt hour per kg, including weight of the oxygen.
What if we used  fluorine instead of oxygen? Obviously, voltage would go up since fluorine is more electronegative than oxygen but when it comes to energy to weight ratio, I am not so sure if it will be better.
Fluorine is element number number 9 and oxygen 8, that means its heavier by one proton and one neutron, furthermore it can take only one electron as opposed to the oxygen which can accept two so you need twice as many atoms to make the reaction.
With all that said, which one is the true ultimate theoretical combination. What is the theoretical energy density of lithium fluorine battery in watt hours per kg?

What is your supposed theoretical arrangement of such a battery, considering fluorine reactivity?

@Poutnik There is no arrangement, theoretical maximum capacity does not take arrangement into consideration. Its like 100% carnot efficiency, or energy density of fuel material, its just a value that cannot be ever exceeded. Go into wikipedia "metal air battery" page and you will find bunch of these values, hydrocarbons have similiar values too, only difference is that they use mega joule per kilogram instead of kWh/kg.

Even theoretical analysis must take into account the theoretical considerations. The is no point in considering theoretical potentials when everything would blow up or if the whole cell burns in contact with fluorine. To consider specific energy, you have to evaluate the particular electrochemical system.

@Poutnik No, you are wrong. This isnt even theoretical analysis, its maximum theoretical energy. Whether it blow up instantly is irrelevant in same way as you will never extract 120 mega joules out of kilogram of hydrogen in heat engine. Yet 120 MJ/kg is still number that is usefull, despite being unrealistic to attain in any system of any possible arrangement.

I insist that to consider specific energy, you have to evaluate the particular electrochemical system, as it is not only about the storage capacity but also voltage. And regarding specific storage capacity, the molar masses of involved active substances must be known as well.

@Poutnik I think you misunderstood my question, I again suggest you look at the "metal air battery" wikipedia page. Voltage is always lower in real battery, especially as it depletes than theoretical electrode redox voltage. The value I am asking about, the maximum theoretical Joule/kg or watt hour/kg, does NOT take into consideration REAL world, realistic, attainable voltage in system of given arrangement. You are thinking in realistic values, like those we get from real batteries, but this is completly different value that is calculated in completly different way than what I am asking about.

Let us continue this discussion in chat.

@Soliton. It is no use reasoning like you do. Fluorine is a gas that reacts with everything. Water is destroyed in presence o fluorine. Plastics are destroyed. Metals are destroyed. Sometimes violently, with flames. The only substance that does not react with fluorine are the fluorides, which can be considered as the ashes produced by the action of fluorine. You may imagine cells with fluorine. They will never be working ! Never !

@Maurice And despite all of that, you can still calculate theoretical energy density of Lithium - Fluorine battery, isnt that amazing? The calculation doesnt care if its possible in real world, only how much energy it has.

@Soliton. You could also calculate the theoretical pressure of a gas contained in a $1$ liter flask containing $10^{100}$ kg at a temperature of $10^{100}$ K. I am sure this calculation is possible.


text:
Reaction between sodium hydroxide and silicon dioxide
I was trying to make sodium silicate according to the following reaction:
$\ce{SiO2 + 2NaOH -> Na2SiO3 + H2O}$
I used colloidal silica available as an intestinal adsorbent since reaction with crystalline SiO2 would require a lot of heat. I mixed a solution containing 20 g of NaOH with a suspension of 15 g of SiO2, however, nothing seemed to happen. I checked the pH and it was still highly basic. Is there any way I could get the chemicals to react? Should I try bringing them to a boil?

I would use silica powder from a reputable source, if the silica from the intestinal adsorbent has had surface treatment (which it might) that would clobber things up. Always use laboratory chemicals.

$\ce{SiO2}$ reacts with $\ce{NaOH}$ only at temperatures higher than 350°C. And then the obtained silicate is hard to dissolve in water. It lasts hours of dealing it with hot  water under pressure.

@StianYttervik Should I look at silica cat litter?

@kwangmyonguser No, you should not.

UPD: The next day the solution turned perfectly clear, and I could see no traces of silicon dioxide in it. I decided to boil the water off, and at the point when there was little to no water, the solution looked rather gooey and glue-like. In an aqueous solution it releases white precipitate upon acidifying, thus I think that I succeeded and that the product definitely contains some sodium silicate.

Definitely, you should boil the solution of silica and sodium hydroxide few hours. I did this experiment many years ago. And, as I remember this, it is extremely dangerous and very boring experiment. So boring that I still remember it approximately after 40 years.


text:
Can endergonic reactions be exothermic?
Pretty much the title. I've heard it said that no (see, e.g here, answer by Lukas Schaedler).  To quote his argumentation:

No. Exergonic reactions may be both exothermic or endothermic. Endergonic reactions are endothermic only.
Endergonic and exergonic relate to changes in free energy (delta G), while endothermic and exothermic are related to changes in enthalpy (delta H).
Gibb’s equation written in form of free energy (G), enthalpy (H) and entropy (S) changes:
delta G = delta H - T . delta S
Therefore, it’s possible for an exergonic reaction (delta G < 0; decrease in free energy) to be both exothermic (delta H < 0; heat is released) or endothermic (delta H > 0; heat is consumed). The second scenario may happen if :
T . delta S > delta H
On the other hand, endergonic reactions are endothermic only; delta H needs to be positive and greater than T . delta S in order to delta G > 0

But it's not really clear to me why. In particular, in the example given, why is $\Delta S \ge 0$ (which in my understanding would be necessary to force $\Delta G = \Delta H - T\Delta S$ to be negative)?

Reactions with ΔG negative are simply non existant.

@Maurice - Huh? The definition of an endergonic reaction is one where $\Delta G^\circ > 0$. There are plenty of reactions where this is not the case.

@Damian Birchler:  Posts outside chemistry.se may move (or become defunct).  Thus, it is better to repost (within reason) the argumentation clearly labeled as a quotation (the > sign at the beginning of the line yields the vertical bar).

@Todd Minehardt. Challenge : Which reaction can you give as an example of ΔG° < 0 ?

@Maurice - Any spontaneous reaction? I mean, the answer to the OP's query is that "endergonic reactions are defined as those where $\Delta G^\circ > 0$." That's it. It's a definition.

@Todd Minehardt. Of course, all spontaneous reaction are those where ΔG° < 0. But the question was : Does it exist equation where ΔG° > 0 ? The answer is  NO ! No reaction are possible with ΔG° > 0, because these reactions would not be spontaneous. Or give me an example !

Equilibrium reactions go both ways at the same time and only one way can be spontaneous. // KNO3 dissolution is exergonic and endothermic. KNO3 crystallization is exothermic and endergonic.

The person who posted that answer on Quora was probably confused about the significance of $\Delta S$ as it relates to spontaneity. For isolated systems in general the sign of  $\Delta S$ (the change in the entropy of the system) is the relevant criterion determining whether a process is spontaneous, since the entropy of the surroundings remains constant, so changes in the entropy of the universe are limited to the system.
$\Delta G$ is of course the relevant criterion for constant temperature and pressure processes. When the change is negative (exergonic) the process is spontaneous. This places no constraints on $\Delta H$ other than that it should satisfy
$$\Delta H \le T\Delta S$$
which can be interpreted as meaning that changes in the entropy of the surroundings must more than compensate for any potential decrease in the entropy of the system: if the entropy of the system decreases (is negative) then the entropy change of the surroundings ($-\Delta H/T$) must be positive and greater than this in magnitude:
$$\Delta S_\textrm{surr} \ge -\Delta S$$
Freezing of any pure substance above its melting point is an example of a process that is not spontaneous but exothermic. By extension, the reverse is also true: melting of a substance above its freezing point is exergonic (spontaneous) yet requires an input of heat (endothermic).
Note also that $\Delta G^\circ$ determines the position of equilibrium since $\Delta G^\circ = -RT \log K_\textrm{eq}$, but does not suffice to say whether the reaction is spontaneous. For that you need to compute $\Delta G = \Delta G^\circ + RT \log Q$, that is you need information about the starting concentrations.

"Freezing above the melting point", can this be achieved? In the lab?

Should it have been "below its melting point"?

@DamianBirchler No, I imagine not. That was the point of the (endergonic) example, since the question was whether an exothermic process could be endergonic.

Yeah, I'm starting to get it now. "Endergonic/not spontaneous" means doesn't happen under the given conditions (temperature, pressure, concentrations, etc.), period. I think my confusion stemmed from the fact that I've always considered "the reaction" to be, e.g., A + B -> C. But it doesn't really make sense to say "A + B -> C" is (not) spontaneous, one would have to say "A + B -> C is not spontaneous at temperature T, pressure p and so forth". To summarize, any reaction is both exo- and endergonic - it depnds on the conditions (temperature, pressure, voltage, ...).

I've also learned from Wikipedia (can't recall the page anymore, though) that the plain "A + B -> C" would be called a "process", which only together with a range of conditions makes for a full "reaction". I.e. "process" and "conditions" = "reaction". I find this terminology helpful. I don't know, however, if it's widely shared.

A chemical reaction involves making/breaking bonds. A process is a more general term and includes phase transitions and just about anything else, for instance a piston working could be called a process. To be fair I have not looked up the authoritative definition of process but that is my past experience. It even has its own Wikipedia page: https://en.wikipedia.org/wiki/Thermodynamic_process // Defining spontaneity of a reaction should include composition (starting and ending) and T, p. Yes, conditions can modify the extent of reaction.


text:
Meaning of $^2[\,1/2]$ terms in atomic spectroscopy
I was trying to analyze some atomic emission spectroscopy data and I encountered unfamiliar notation of the terms for me.
Commonly, the electronic states of the atoms can be described by quantum numbers using the term notation ("L-S coupling"). The terms are given as:  $^{2S+1}L_{J}$, where $S$ is the total spin quantum number, $J$ is the total angular momentum quantum number and $L$ is the total orbital quantum number in spectroscopic notation.
However, in this table there are also terms having notations without letters ($S$,$P$,...), like $^{2}[\,1/2]^\mathrm{o}$ (black box). What does that mean and why are there separate cases for $J = 0$ and $J = 1$?
[Edit] The data has been taken from NIST Atomic Spectral Database for Ne I (neutral neon in spectroscopic notation).
NIST database
The notation, asked in this posted has already been explained here Notation key by NIST

The notation of coupling schemes are explained here https://www.nist.gov/pml/atomic-spectroscopy-compendium-basic-ideas-notation-data-and-formulas/atomic-spectroscopy-2

I am not familiar with this type of coupling but it is called LK couping. The number in square brackets is K. It is always good to a give a reference when you paste an image.

I dont understand that notation. But my lecturer told me to study up L-S coupling so this kind of notation confused me.

But the transition looks like this? For example: I have a wavelength 352 nm so the transition is $^2[1/2]$ to $^2[1/2]^0$?

This notation is certainly not LS coupling. Ask your professor if he expects you to learn LK coupling.

It is probably, just not written with letter because L is not in the form of integer because the databese uses classic LS form for other elements so I guess the notation must remain the same

In addition to previous comments, you may find helpful to have a look at Herzberg's Atomic Spectra and Atomic Structure,

May also be helpful to read this original paper by Cowan and Andrew in 1965 "Coupling Considerations in Two-Electron Spectra" https://opg.optica.org/DirectPDFAccess/42879AAE-F5CE-452A-82787AE00DCEA0A0_52651/josa-55-5-502.pdf?da=1&id=52651&shib=578648&seq=0&mobile=yes


text:
Dextrose equivalent values (DE) of the most common sugars

What are the dextrose equivalent values (DE) of lactose, and other common sugars?

Some data like the values for sucrose (0), maltodextrin (x<20), maltose (52), and glucose (100) can be found at different places on the net, but others such as lactose are not as easy to come by.
An answer containing the DE values of common sugars, including lactose and fructose would be appreciated.

Contradictory findings! https://academic.oup.com/clinchem/article-abstract/38/7/1301/5649980?redirectedFrom=fulltext

Dextrose equivalent $(DE)$ is a measure of the amount of reducing sugars present in a sugar product, expressed as a percentage on a dry basis relative to dextrose (glucose). Commercial sugar products are generally characterized by their dextrose equivalent $(DE)$ value (Ref.1). The theoretical definition of $DE$ is given by:
$$DE = \frac{\text{Reducing power of Syrup Solids}}{\text{Reducing power of D-glucose}} = \frac{\text{# of Reducing-End Groups in Syrup Solids}}{\text{# of Reducing Groups in D-glucose}} $$
However, number of reducing Groups are inversely proportional to the molecular weight of solid products (Ref.1 and Ref.2).
$$ \therefore \ DE(\%) =  \frac{M_\mathrm{glucose}}{M_n} \times 100 \tag1$$
where $M_\mathrm{glucose}$ is the molecular weight of D-glucose and $M_n$ is the "molecular weight" of syrup solids. Keep in mind that this is the theoretical definition of the $DE$.  The standard method of determining the $DE$ is the Lane-Eynon titration, based on the reduction of copper(II) sulfate in an alkaline tartrate solution, an application of Fehling's test. To see the deviation between theoretical and  Lane-Eynon determination, Ref.1 has given following examples:
$$\begin{array}{l c r}
\text{degree of polymerization} & \text{theoretical }\%DE  & \text{*Lane-Eynon* }\%DE \\
\hline
1\text{ (D-glucose)} & 100 \text{ (by definition)}  & 100 \text{ (by definition)} \\
2\text{ (maltose)} & 52.6   & 58  \\
3& 35.7  & 39  \\
4  & 27.0   & 30  \\
5 & 21.7   & 24.2  \\
6 & 18.2   & 20.8  \\
\hline
\end{array}$$
Following are three common sugars (lactose and maltose, dimers of glucose; and sucrose, a dimer of fructose and glucose):

As indicated by green circles, only lactose and maltose have at least one reducing end remaining while sucrose have none. Thus, $\%DE$ of sucrose is zero. Since, both lactose and maltose are dimers of glucose and they have identical molecular weights. Hence, according to the equation $(1)$, both of them have the identical theoretical $\%DE$, which is $52.6\%$ (see the chart from Ref.1). To determine Lane-Eynon values of them, you shoud do appropriate titration of analysis (e.g., Ref.1 and Ref.2).

Late edit:
As requested by OP, I have searched for $\%DE$ value for fructose. Evidently, fructose is also a reducing sugar (Ref.3) since all monosaccharides are reducing sugars. Thus, since fructose has the same molecular weight as glucose, its theoretical $\%DE$ should be $100\%$.

References:

M. G. Fitton, "Rapid Determination of Dextrose Equivalent by Cryoscopy," Starch 1979, 31(11), 381-384 (DOI: https://doi.org/10.1002/star.19790311107).
Y. Rong, M. Sillick, and C. M. Gregson, "Determination of Dextrose Equivalent Value and Number Average Molecular Weight of Maltodextrin by Osmometry," J. Food Sci. 2009, 74(1), C33-C40 (DOI: https://doi.org/10.1111/j.1750-3841.2008.00993.x).
Although D-fructose would appear to be nonreducing, it readily undergoes keto-enol tautomerism at high $\mathrm{pH}$ to form a mixture of D-glucose and D-mannose: M. H. El-Rafie, Hanan B. Ahmed, and M. K. Zahran, "Facile Precursor for Synthesis of Silver Nanoparticles Using Alkali Treated Maize Starch," International Scholarly Research Notices 2014, Article ID 702396 (DOI: https://doi.org/10.1155/2014/702396).

+1. Thx for your answer. I added a query about “fructose” at the end of the question. Would you be able to also give a DE figure for that?

@Constantthin: Fructose is not a reducing sugar so $%DE$ should be zero like sucrose.

It may be so, but do you have a reference to back up that? This website:    https://www.bionity.com/en/encyclopedia/Reducing_sugar.html     says it is. Is the author(s) wrong?

@Constantthin: I have added value for fructose. based on the reference sited.

@Constantthin: It's possible. The $\mathrm{pH}$ of blood is 7.4, a little basic. With enzymes capable of doing so, fructose can isomerize. In Lane-Eynon  $DE$ determination, the solutions are basic so it has non-zero value.


text:
Can Chlorhexidine Gluconate (CHG) be used for disinfection of medical device (e.g. of polyethylene material)?
Chlorhexidine is commonly used as a skin antiseptic, hand disinfectant, endodontic irrigant etc. It possess antibacterial and antifungal properties, and is considered low toxicity when used in low concentration.
Some articles mentioned that Chlorhexidine is also used in disinfection of medical device/ surfaces in their description.
However, most commercial solutions are formulated for skin disinfection, and there are almost none of the studies that evaluates the antimicrobial effect of Chlorhexidine on medical device/instrument.
Is Chlorhexidine suitable to be used as disinfectant for medical instruments (non-invasive) or hard surfaces? Since the antimicrobial effect is comparable to another common agent, povidone-iodine.


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Oxygen-releasing material at ≤ 200 °C
I have to anneal a ceramic in high oxygen pressure, in an autoclave. For this purpose I need a chemical which can release oxygen at temperature less than or at 200 °C. Would appreciate any suggestion.
I am aware of $\ce{KClO3},$ but it seems that it releases $\ce{O2}$ at 300 °C.

Steel can do what you ask. Specifically a steel container with liquid oxygen inside it.

KClO3 decomposition can be catalyzed by MnO2, but not sure how it goes down with T. You may be already aware of it.

Hydrogen peroxide solution, with any convenient catalyst, e.g., manganese dioxide.

With MnO2, can KclO3 decompose below 200 C? Any reference for it. Thanks.

if yoy could provide a little more context in would be nice. does it have to be solid? does it have to have a lot of oxygen per weight?

I have to anneal a ceramic in high oxygen pressure, in an autoclave. Need a source of oxygen for this purpose. Oxygen/weight is not so critical. Thanks

@user49535 It would be better if you can add all the information into the question body. Comments are temporary and it can be deleted.

Some ideas, looking around at some potassium salts:

Potassium permanganate, $\ce{KMnO4}$: decomposes at 240°C, probably the most readily available as I commonly see it in chemical laboratories.

Potassium manganate, $\ce{K2MnO4}$: decomposes at 190°C. Tends to disproportionate with water, but can be made from the permanganate plus concentrated potassium hydroxide solution.

Potassium ferrate, $\ce{K2FeO4}$: decomposes at 198°C. Breaks down readily in the presence of water. Listed as "Danger" is the GHS safety classification.

All such compounds should be handled with care and especially avoiding contact with organic material which they may cause to combust.

Thanks. I am wondering, when KMnO4 decomposes at 240 C, K2MnO4 is a byproduct. Does it also decompose further to liberate more oxygen? I have not found yet the chemical reaction of decomposition of K2MnO4 at 190 C.

The K2MnO4 would decompose further according to its WP article, but it gives no more details. A combination of K2O, MnO2, and lower potassium-manganese oxysalts plus evolved O2 would be a plausible outcone. One advantage of using transition metal oxysalts like these is they have no potential to form volatile compounds apart from O2.

Wikipedia reports the existence of lower manganese oxyanions, so likely these are formed with additional evolution of O2.

@user49535 You can check some papers: https://pubs.rsc.org/en/content/articlelanding/1971/j1/j19710001821 ... https://link.springer.com/article/10.1007/BF02549317. In short, $\ce{KMnO4}$ decomposition at 250-300°C is noted ($\ce{10 KMnO4→2.65 K2MnO4 + [2.35 K2O_{7.35},MnO_{2.05}] + 6O2}$, the bracketed part being $\ce{δ-MnO2}$). At higher temperature, $\ce{K2MnO4}$ decompose to form lower potassium-manganese oxysalts: $\ce{K4Mn7O16,  K3MnO4, K3(MnO4)2}$. You can also check the kinetics or the reaction: https://www.sciencedirect.com/science/article/abs/pii/0040603185854794


text:
Why does boiling occur when vapor pressure equals ambient pressure?
Here is my understanding of the boiling process: pressure of the atmosphere is pressing down on the liquid and this pressure is propagated throughout the liquid. That is, pressure everywhere inside the liquid is equal to the ambient pressure (I'm ignoring the weight of the liquid columns for simplicity). In order for bubbles to form inside a liquid, the gas molecules must push outwards with a greater pressure than the existing ambient pressure to "make room" for the bubble. The pressure that the bubbles are able to exert on its liquid surroundings is the vapor pressure and hence we need to be at a temperature where vapor pressure at least equals the ambient pressure.
But my source of confusion is that ambient pressure is not the only pressure pressing down on the liquid. There is also the vapor pressure of the liquid in addition to the ambient pressure. Then according to my understanding of how boiling works, this means that the pressure inside the bubble (which equals the vapor pressure of the liquid) must be greater than the ambient pressure plus the vapor pressure of the liquid which makes no sense. So I'm assuming I have an incorrect understanding of how boiling works and was wondering if someone could help me correct it.

see  site:stackexchange.com+Why+does+boiling+occur+when+vapor+pressure+equals+ambient+pressure

@Poutnik Thanks for the reference. Is the answer in this post basically saying that the vapor pressure is higher at the bottom of a container than at the top of the surface (due to higher pressure at the bottom), allowing the bubbles to form at the bottom?

Which answer in which post ? (the link was a hint for you that you are supposed to do basic search before posting a question) //  Saturated vapor pressure depends on T, not p (1). But when p is higher, higher T is needed to p_vap = p, therefore the boiling point slightly depends on the depth. As convection is involved, there may occur streams of temporarily overheated metastable liquid.// (1) In your Q context, otherwise things are more complicated in tiny details.

The ambient pressure is just one, there is no such a thing you describe. Or you have a pressure cooker, and indeed boiling stops.

https://chemistry.stackexchange.com/questions/74651/real-boiling-temperature https://chemistry.stackexchange.com/questions/107362/how-does-the-process-of-nucleation-work-for-boiling-liquids

Not to be super critical, but, in line with@Alchimista comment, consider "hot smoker" hydrothermal vents. No container, but the pressure of sea water prevents boiling. https://www.whoi.edu/feature/history-hydrothermal-vents/discovery/1979-2.html .

@DrMoishePippik Are you saying supercritical or super critical?

@KarstenTheis, sorry, that was a bad pun on supercritical -- above critical temperature and pressure, there is no boiling.

But my source of confusion is that ambient pressure is not the only pressure pressing down on the liquid.

Pressure does not press down. When something is under pressure, it exerts it in all directions. Maybe it would help imagining this problem in the absence of gravity (you would have to put the sample in a stretchy balloon to get some pressure while being able to change the volume).
So the balloon would exert some pressure on the liquid. If the liquid is hot enough that the vapor pressure is higher than the pressure, and if you can get some tiny bubbles going, they will expand as more and more liquid goes into the vapor phase, increasing the volume of the bubbles while maintaining sufficient pressure not to collapse.

Thanks for the reply. So to summarize: the reason why the boiling occurs when vapor pressure equals ambient pressure is bc bubbles would otherwise collapse if the vapor pressure inside the bubble was less than the ambient pressure it feels from the liquid. Is that correct?

@rofimfao Yes. Getting it to make tiny bubbles is a question of kinetics and nuclei forming, more complicated... when you listen to a pot about to boil you can witness some of that

@rofimfao Here a link to the question exploring the sounds: https://chemistry.stackexchange.com/q/125368/72973


text:
Comparing covalent characters of two halides of group 15
Which is more covalent : NCl3 or PCl5 and why?
This question was asked on a test, and the given answer was NCl3. Shouldn’t PCl5 be more covalent since pentahalides are more covalent than trihalides?

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Pentahalides are more covalent than trihalides of the same element with the same halogen. For instance $\ce{TaCl5}$ is more covalent than $\ce{TaCl3}$ (the respective WP articles give their melting points as 216°C for the pentachloride and 440°C with decomposition for the trichloride); but we cannot make such a simple comparison of $\ce{TaCl5}$ with either $\ce{TaBr3}$ or $\ce{PCl3}$. Similarly the pentahalide/trihalide rule could be overriden by the difference between nitrogen and phosphorus in your case.


text:
Why is this compound more basic than the other?
So, the original question was which compound is more basic.

Compound A and Compound B respectively.
My reasoning is that on protonation, Compound B would be more stable than Compound A as well, Compound B is an allylic nitrogen cation. Thus, compound B should be more basic.

Why don't nitrogen cations act similar to carbo cations in this case?
But the given answer is that compound A is more basic, because of being connected to a sp2 hybridized carbon. Another given reason was that the first cation was stabilized by water while the second one wasn't?
Where am I wrong?

Being connected to a sp2 hybridised carbon makes it weaker not stronger. Still, enamines can be even weaker because of mesomeric effects.

@Mithoron, I know right! I was like, that sp2 reasoning doesn't make much sense.

The lone pair in B is conjugated with the C=C bond. Similar to the nitrogen lone pair in an amide, which is also not very basic

What may be happening is that the A cation has a resonance structure where the $\pi$ bond polarizes to the nitrogen and thus turns the $sp^2$ carbon into a carbocation center (leaving the nitrogen formally uncharged). This carbocation contribution is enhanced by the $sp^2$ carbon also hyperconjugating with the carbon atom next to it in the ring. Also the B cation cannot form an allylic structure because protonation of the nitrogen atom in that compound saturates it.

The A ammonium ion is sp2 hybridized so less basic[really the ion is more acidic] than an sp3 nitrogen. However, the enamine structure in B delocalizes the electron pair over the nitrogen and the double bound reducing basicity even more. There is no resonance of an ammonium ion with a double bond; there are no orbitals available on the nitrogen to accept electrons to reduce the charge[except antibonding].


text:
What is the maximum length of peptide which considered to be hard for synthesis by the current technology?
I am a molecular biologist and work with recombinant proteins. I am wondering what the cut point, in terms of length, is at which I should produce the peptide using heterologous expression rather than ordering a peptide synthesis service? What is the maximum length which makes it hard to chemically synthesis the peptide? Any ideas are welcome.

The protein lysozyme (160 amino acids)  has been synthesized using a flow reactor according to https://www.chemistryworld.com/news/protein-synthesis-revolution-on-way-as-large-peptides-made-in-hours-not-days/4011891.article . The same article suggest 50 amino acids as a cut-off where the traditional solid state synthesis would be infeasible.


text:
Relation of partial charge with wavefunction
Suppose we have an one electron system. The probability $P$ that electron's position is in some interval $x + dx$ is given by:
$$P = \int_x^{x+dx}|Ψ(x)|^2 dx$$
So if we measure $N$ times the electron's position the number of times that is in the interval $x+dx$ is $n$ and it holds that:
$$P = \frac{n}{N}$$
Lets move now into a polar molecule e.g. $\ce{HCl}$. In order to find the partial charges we must know the wavefunction of the bonding molecular orbital. Based on electronegativity differences we should expect $\ce{Cl}$ to have a negative partial charge and $\ce{H}$ to have a positive charge. (I will work with one dimension for sake of simplicity).
The value of $|Ψ(x)|^2$ should be higher in the region of $\ce{Cl}$ and lower in $\ce{H}$. I was thinking that this partial charge could be explained by "cutting" the distance between the two nuclei and then integrating $-e|Ψ(x)|^2$ at the new seperated spaces. So the partial charge at $\ce{Cl}$ would be:
$$δ^- = \int_{x_0}^{\infty}-e|Ψ(x)|^2dx$$
where $x_0$ is defined as the point of seperation of two nuclei. In the same way we can find the partial charge of $\ce{H}$. The problem is that I can't interpret it. I mean $-e|Ψ(x)|^2$ gives the probability of finding a charge of size $-e$ at a given region (in analogy of finding electron at a given position).
Edit
First I would like to apologize for being unclear. Introducing time was a bad idea. What comes to my mind is "Shut up and calculate". In QM all physical quantities are calculated with expectation values so the same must holds for partial charges (dipole moment more correctly). Now why that must be the case (expectation values) may be related to Ergodic hypothesis as Ian pointed out in the comments.
I don't know if that is correct so I can write it as an answer. The question is about how assymetric wafevunction translates into partial charges.

I've read this over a few times, and maybe it's just because it's late, but I'm afraid I don't understand what the problem is. "But this doesn't make sense because we know that the molecule partial charges doesn't change over time." - what is the contradiction exactly? Up until now you haven't discussed anything that changes over time, so where does the "time" consideration come in - are you saying that the value $n/N$ changes as you make more and more measurements?

@orthocresol Almost exactly my thoughts last night. On re-reading this morning I'm still not much better, maybe I need more tea. I have a sort of feeling the view of the world being expressed is something akin to the Ergodic Hypothesis but I'm struggling to quite make the connection.

The $P$ is really $P_{x+dx}$ so changes as $x$ varies and in this way you have  a set of values $P_{x_1+dx}, P_{x_2+dx} $ etc. that when compared will give you a value proportional to the charge distribution. Forget about measuring $N$ times, $\int \psi^*\psi$ is already an average.


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How to add custom shading to a PDB file or viewer
I have a PDB file and viewing it through molstar, I have a numpy array of confidence intervals. Essentially an output from DeepMinds alphafold. Does anyone know how I can add a custom shade so that I can color the atoms/sub-structures according to their confidence interval?
I have not tried anything because I am not sure where to start.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Since when janitors do bioinformatics? ;) This question might be better off on Bioinformatics.SE

Here is an example where they colored by structure quality, but I don't know where those values are uploaded: https://molstar.org/viewer/?snapshot-url=https%3A%2F%2Fmolstar.org%2Fdemos%2Fstates%2Fcytochromes.molx&snapshot-url-type=molx

The easiest approach would be to utilize the b-factor column in the pdb-file as suggested in the pymol wiki:
https://pymolwiki.org/index.php/Color#Reassigning_B-Factors_and_Coloring

I can only use molstar, how do I accomplish this on molstar?


text:
How is this Clemmensen Reduction taking place?
How did that isotopically labelled Carbon migrate? I know that an carbanion is formed, but I am not able to conclude the final product. Please Help

The carbanion formed at one aldehyde attacks the other aldehyde by performing nucleophilic addition to form a cyclopropane ring. This ring opens up in such a way to give a more stable carbanion, which here is going to be on the C14 carbon (because of the electronic effects of oxygen).

Can you kindly show mechanism like written sort of or text picturised solution

And how can u compare carbanion stability it is identical

See the mechanism for clemensen reduction.


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What is the error in the number of aromatic isomers of C7H8O?
I am a student studying basic organic chemistry (isomerism) and was asked a question to find the number of aromatic isomers of $\ce{C7H8O}$.In fact this was a question in IITJEE exam. So I tried to draw all and I found all but the problem is that I found more than all. The answer to the question is 5 which may be wrong(I don't know) but several websites state this. Here are some links to those:

https://www.toppr.com/ask/question/what-is-the-number-of-aromatic-isomers-for-c7h8o/
https://www.vedantu.com/question-answer/number-of-aromatic-isomers-for-c7h8o-class-11-chemistry-cbse-60153d93925e1033660d420c
https://ask.learncbse.in/t/in-c7h8o-how-many-aromatic-structure-isomers-are-possible/69466/2

Here is the list of isomers I found:

So what is wrong with the list of my isomers?

Obviously, by the author's intention, you were not supposed to think of furanes.

@IvanNeretin but should the actual answer also include furanes?

"Actual answer" is a social construct.

@IvanNeretin What does 'social construct' mean here?

I mean, some people say something, and others say otherwise, and it all boils down to some subtlety in the definitions, and then someone brings up the three-membered rings, and chaos ensues.

@IvanNeretin Chemistry is a systematic scientific study, should it not have a clear answer to a particular "finite" and "inside the domain'  question? If it should have then this question surely holds an appropriate and factually correct answer which is unique. By that logic I meant what should be the answer of this question?

True, but you were not supposed to know all of chemistry while answering this. Setting the educational issues aside, your answer is more correct than the author's, but then again, three-membered things actually do exist, don't they?

@IvanNeretin Yes, I agree. Thanks for your help, I was having doubts about my answer and learnt from you about the flexibility of answers.

@Ritanshu  You are missing at least 2 more

@Waylander Thanks for your help with extra information, I will definitely add this to the actual answer.

@Ritanshu  3-cyclopropylfuran & 2-cyclopropylfuran

Piling on: 2-prop-1-enylfuran https://pubchem.ncbi.nlm.nih.gov/compound/53915682 and there are plenty more down this avenue

I'm not good with names, but take a molecule of cyclopropenone and replace the hydrogen atoms with a tetramethylene bridge.

You are completely correct - the furan derivatives you've drawn display unquestionable aromatic character, though the "strength" of their aromaticity is somewhat lower than various other prototypical organic aromatic compounds.
By aromatic, the examiners really were assuming you would think benzenoid. While the latter term is more precise for the intended purpose (evaluation of fundamental organic chemistry structural knowledge), it is also rather technical and somewhat obscure. However, most people with a high-school understanding of chemistry are not likely to consider more complex aromatic systems in a testing environment with time constraints, and so these terms are often used as very crude synonyms.
In a more general sense, consider that university entrance exams require thousands of variations of simple questions to be churned out every year, and there's only so many mistakes/oversights that quality control can pick up on. These questions are also designed to be read and answered in a handful of minutes. This leads to a lot of "lossy compression" and corruption of meaning, where subtleties in content are overlooked, intentionally or unintentionally. If you're looking at exam answer sheets as the ultimate dictators of truth, then you're doing it wrong, as Ivan pointed out.
But ultimately, good job for looking at the finer print!

I can draw a dozen more isoners with a furan ring, and I haven't even counted what can be made with a cyclopropenone ring. Please do not attempt to formulate problem statements when drinking the real Budweiser beer from the Czech Republic!

I just wanted this question not to be unanswered when the answer was itself subtly discussed in the comment section. Hence here's the part of discussion which provides some answer to this question. Of course the knowledge was borrowed and this is from @IvanNeretin .
Ivan Neretin emphasizes the flexibility of answers and the social structure of education. Hence for the needs of basic education, the question only includes 5 isomers which include benzene as the core but the "more" correct answer should be the structures I drew. And according to @Waylander, there are at least two more isomers available for the formula which are 3-cyclopropylfuran & 2-cyclopropylfuran. @Martin also states that that there are several other isomers missing of which he gives one example  2-prop-1-enylfuran .
Thanks to Ivan Neretin and all others for their help and time.


text:
What happens when you dilute a surfactant solution at the critical micelle concentration?
Say if instance I have a 100 ml solution with micelles formed using some surfactant. Now if I pour that 100 ml solution into 500 ml of water, will those micelles break up into free surfactant again?
Also, is there a way to calculate the binding energy of a micelles and see if the new solution with micelles (immediately after mixing) within the 500 ml of water will overpower that binding energy, to force the micelles to break up?

The answer to the first question is yes. The micelles will disappear.

Diluting a detergent solution at the CMC will result in a transition from a micellized into a disaggregated state. The definition of the CMC is that it is the concentration of onset of micellization. Below this concentration micelles do not exist.
For the diluted solution, surfactant in micelles will have a higher free energy than free detergent. In general, if you can determine (or compute, say) the free energy of the detergent in the free state versus in the micelle then you can make a prediction as to which state is preferred. Such a computation might not be simple, depending on the degree of detail with which you seek to describe the interactions in the micelles versus the freely solubilized surfactant.


text:
Why we are dividing rate of reaction with respect to a species by its stoichiometry coefficient to find rate of that particular reaction
If there is a reaction like this
$$\ce{ H2 + Cl2 -> 2HCl}$$
I can understand that each individual species (reactant or product) can have different rates and we can compare them like this,
$$-\frac{d[\ce{H2}]}{dt} =- \frac{d[\ce{Cl2}]}{dt} = \frac{d[\ce{HCl}]}{2dt}$$
But I'm having problem with defining the rate of reaction (for the chemical reaction shown above) as the above expressions.
Can someone please explain me why it is like this?

You are missing negative signs, but think about this: the rate of decrease, in either reactant concentration, is half the rate at which product concentration increases. Simply, when one hydrogen molecule bites the dust, two HCl molecules are produced.

Yeah but why rate of reaction of this reaction is [HCl] /2dt why it can't be - 2[Cl2]/dt.why we have to go for simplest ratio(dividing by stoicheometry) .???

Well, so long as you are consistent and others know what you are doing, no problem. The 2 has to go somewhere.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

It is simple: The reaction rate must be the same regardless of the component rate you use to describe it. The reaction does not go faster nor slower if we describe it by the rate of HCl or Cl2.

If rate of reaction with respect to HCl is 4 and with respect to cl2 is 2.  Then if we divide them by stoicheometry we will be getting 2 as the rate of reaction for the above reaction. Why can't i multiply the rate of reaction with respect to cl2 by 2 where it will be2*2=4 and say rate of reaction of the sytem is 4, Instead of dividing HCl formation rate by 2.

What if there were several different st. coefs? What number would you choose for such multiplication?

Will dividing everything by stoicheometry give the true rate for that particular reaction? Can we find the real rate of a paticuler reaction by dividing it by stoicheometry.

So is that mean we are dividing by stoicheometry to have a constant value for rate. Its not actually the real rate of that reaction??

Reaction rate is convention, there is no true reaction rate. Important thing is to keep this convention clear, not deliberate. If there was 2A + 3B -> C + 4D,  which (absolute value ) of component rates would you choose as reaction rate?

This is what my problem is. I don't know, there are 4 components where all 4 has different reaction rate. But according to what  i have learnt, i have to choose C component as it have stoicheometry 1.

I think i got it. Its a rule we have to divide the reaction rate with related to a component by its stoicheometry. Thank you soo muchhh