text:
Do strong acids actually dissociate completely?
Do strong acids actually dissociate completely (every single molecule dissociates), or are they just assumed to do so for the sake of simplicity? That seems odd, considering weak acids, many of which are very similar to strong acids, don't fully dissociate.

I doubt that an answer can be given to this question, because there is no easy way of measuring the concentration of ions in concentrated solutions of strong acids. The glass electrode gives only the activity and not  the concentration of $\ce{H3O+}$.

@Maurice Are you kidding? Answer is simple - yes, it's an approximation for sake of simplicity. While there's lots of problems with estimating pKa lower then 0, in highly concentrated solutions strong acid simply can't be fully dissociated, because there's more acid then water. Levelling effect disappears and acid shows its real strength (not having much to do with pKa).

@Mithoron. Do you have any idea how the pKa of strong acid like HClO4 have been determined ?

@Maurice I already linked last time this topic was up http://www.nrcresearchpress.com/doi/pdf/10.1139/v78-385

Yes, for sufficiently strong acids, in sufficiently dilute conditions, in sufficiently basic solvents. However, things are hazier than you might expect, and depending on your definition, "clear-cut" examples of complete dissociation only become possible when you start swatting flies with nuclear bombs.
The problem is that chemistry works on equilibria, and at some point it becomes difficult to measure tiny concentrations - in a haystack of one billion straws, how can you be sure there isn't at least one needle? And in one trillion? And what about in a septillion?

First, let us perform a qualitative investigation by considering the generic acid dissociation equilibrium in water, as follows:
$$\ce{HA(aq)->H+(aq) + A-(aq)} \quad \quad \quad \mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}}=10^{-pK_a}}$$
To simplify the following discussion, we will make the very crude approximation where we equate the thermodynamic activity of a species with its molar concentration. In other words:
$$\mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}} \approx \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}}$$
Our objective now is to virtually eliminate the presence of undissociated $\ce{HA}$ molecules. For clarity, we rearrange the above equation:
$$\mathrm{[HA(aq)] \approx \frac{[H^+(aq)][A^-(aq)]}{K_a}}$$
Clearly, to make $\ce{[HA]}$ as small as possible, the numerator must be very small, and the denominator must be very large. Let us set a goal of making $\mathrm{[HA]<10^{-27}\ mol\ L^{-1}}$, which implies less than $\mathrm{0.001}$ molecules of undissociated $\ce{HA}$ in a litre of solution.
At this point, I should mention that aqueous solutions of acids in ambient conditions must always have $\mathrm{[H^+(aq)]>10^{-7}\ mol\ L^{-1}}$, thanks to the self-ionization of water. We take the best-case scenario (which physically corresponds to very dilute solutions):
$$\mathrm{10^{-27} \gtrsim \frac{10^{-7}\ [A^-(aq)]}{K_a} \quad \longrightarrow \quad 10^{-20} \gtrsim \frac{[A^-(aq)]}{K_a}}$$
We now take a typical strong acid, hydrogen chloride ($\ce{HCl}$) which has an estimated pKa of -5.9 in water. In other words:
$$\mathrm{10^{-20} \gtrsim \frac{[Cl^-(aq)]}{10^{5.9}}=\frac{[Cl^-(aq)]}{790\ 000} \quad \longrightarrow \quad [Cl^-(aq)] \lesssim 8\times 10^{-15}\ mol\ L^{-1}}$$
We see therefore that meeting the criterion of full dissociation requires preparing an aqueous $\ce{HCl}$ solution with femtomolar concentration. Essentially every solution used in a chemistry laboratory will have a higher concentration, so every $\ce{HCl}$ solution anyone has ever touched will have undissociated $\ce{HCl}$ molecules. Strictly speaking, you could say the dissociation is never complete. However, if you have a mind to make a fully dissociated solution of $\ce{HCl}$, it is physically possible.

Just diluting acid solutions into oblivion until every molecule dissociates is kind of boring. The more exciting prospect is, of course, to increase the strength of the acid. The acidity scale goes far, far beyond aqueous solutions of $\ce{HCl}$.
Interestingly, for sufficiently strong acids (typically $\mathrm{pK_a<-10}$, meaning acids at least 10 000 times stronger than aqueous $\ce{HCl}$), it is possible to isolate hydronium salts - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, where the cation is $\ce{H3O+}$ and the anion is supplied by the conjugate base of the acid. The stoichiometry can be made exact. Surely this represents full dissociation of the acid?
Well, again, it depends on your strictness. If you actually look at the crystal structures of many of these salts, it's clear that there is a strong, close  interaction between the $\ce{H3O+}$ cation and the anion, and you could argue this still does not count as complete dissociation.
That said, you can go crazy and consider some of the strongest acids we can put in a bottle, such as carborane superacids. The corresponding hydronium salt of these superacids can have extremely weak interactions between the hydronium cation and the anion. For example, when the hydronium salt $\ce{(H3O+)(CHB11Cl11^-)}$ is prepared in a benzene solution, the solid that forms actually contains three molecules of benzene surrounding the hydronium cation, which is completely detached from the anion. If this doesn't imply "complete dissociation" of the carborane superacid in water, I don't know what would.
The acidity scale goes further still than even carborane superacids. There are a few known acids which will practically always fully dissociate, no matter what. The representatives of this class are rather exotic things such as $\ce{H_3^+}$, $\ce{HN2^+}$, $\ce{HNe^+}$, $\ce{HHe^+}$, and so forth. These acids can only be detected in the gas phase, as they fully dissociate by protonating virtually anything they come into contact with (including any anion, which is why they cannot be isolated as a neutral compound).

The last way to ensure full dissociation of an acid in a solution is to change the solvent in which it is dissolved. All things considered, water is a pretty weak base. If, for example, ammonia (am) were used as the solvent, the relevant equilibrium for $\ce{HCl}$ would be:
$$\ce{HCl(am)->H+(am) + Cl-(am)} \quad \quad \quad \mathrm{K_a(am)=\frac{a_{H^+(am)}a_{Cl^-(am)}}{a_{HCl(am)}}=10^{-pK_a(am)}}$$
If I were to give a crude guess, I'd imagine that the $\mathrm{pK_a}$ for $\ce{HCl}$ in liquid ammonia would very roughly be somewhere around -15; that is, dissociation of $\ce{HCl}$ in liquid ammonia is about a billion times easier compared to water. This is because ammonia is a stronger base than water.
You know how we were talking about hydronium salts as if they were some unusual thing? Well, the equivalent concept in ammonia would be the formation of ammonium salts. And indeed, $\ce{HCl}$ does easily form a salt with ammonia in a exact 1:1 molecular ratio, namely ammonium chloride, $\ce{NH4Cl}$. Not exactly an unusual compound.
There's nothing stopping you from using an extremely strong base as a solvent. For example, in pure liquid phosphazene superbases, even very weak acids may be fully dissociated in ambient conditions, even in the most rigorous sense.

Fluoroantimonic acid is so strongly disassociating that molecules of $\ce{HSbF6}$ apparently have never been isolated.  Only compounds in which a proton has been transferred to form ionic species are seen.  (Replacing the fluorine with chlorine or bromine also gives an isolable compound, which may or may not be molecular depending on if water of hydration accepts a proton.)  In its usual application as a solution in hydrofluoric acid solvent (the latter, by the way, is itself close to being a superacid, even without the fluoroantimonic acid solute), fluoroantimonic acid consists overwhelmingly of solvated $\ce{H_{n+1}F_n^+}$ cations and $\ce{Sb_nF_{5n+1}^-}$ anions, which equilibrate with $\ce{HF + SbF5}$ rather than with a molecular $\ce{HSbF6}$ species.  When a "pure" fluoroantimonic acid compound is obtained from the solution, we still do not get association into neutral molecules:

Two related products have been crystallized from HF-SbF5 mixtures, and both have been analyzed by single crystal X-ray crystallography. These salts have the formulas $\ce{[H2F+][Sb2F_{11}−]}$ and $\ce{[H3F2+][Sb2F_{11}−]}$. In both salts, the anion is $\ce{Sb2F_{11}−}$.[1]  As mentioned above, $\ce{SbF_6−}$ is weakly basic; the larger anion $\ce{Sb2F_{11}−}$ is expected to be still weaker.

Cited reference:
1.
Mootz, Dietrich; Bartmann, Klemens (March 1988). "The Fluoronium Ions $\ce{H2F+}$ and $\ce{H3F2+}$: Characterization by Crystal Structure Analysis". Angewandte Chemie International Edition. 27 (3): 391–392.  Link


text:
Hydrogen bonding in acetylene and other terminal alkynes
My textbook says that hydrogen bonding is possible only when there is acidic hydrogen connected to an electronegative atom like $\ce{N}$, $\ce{O}$ or $\ce{F}$ and a basic lone pair of electrons on another electronegative atom.
Recently, when I was going through a material on organic chemistry online, I read that $\ce{sp}$ hybridised carbon atom is more electronegative than $\ce{sp}^{3}$ hybridised nitrogen atom. Assuming that is true, we can infer that $\ce{sp}$ hybridised C-H bond is more acidic than $\ce{sp}^{3}$ hybridised N-H bond.
I already know that ammonia has strong hydrogen bonding and the nitrogen atom in ammonia is $\ce{sp}^{3}$ hybridised. So I wanted to know if acetylene has strong hydrogen bonding.
I asked my teacher about this but he simply replied by saying 'no'. I tried googling my doubt but couldn't find any resource which gave me a satisfactory answer. I also wanted to know that, if acetylene shows hydrogen bonding, can other terminal alkynes show hydrogen bonding?

Have you looked up the pKa values of acetylene and ammonia?

"Assuming that is true, we can infer that sp hybridised C-H bond is more acidic than sp3 hybridised N-H bond." That logic is not compelling.

Water and HF have strong hydrogen bonding. Ammonia is a gas. Bp. 240K compared to 293 and 373. Hydrogen bonds between water and ammonia are strong.

@Karl Yeah I just checked the pKa values... Saw a drastic difference in those values between acetylene and ammonia

Still it is true that a C sp is more electronegative than a C sp3. Just think as the situation seen by a charge test approaching it from the two side, or what is seen by the terminal H of acetylene as compared to an H atom in ethane.

https://en.wikipedia.org/wiki/C%E2%80%93H%C2%B7%C2%B7%C2%B7O_interaction https://en.wikipedia.org/wiki/Pi-Stacking_(chemistry)

https://pubs.acs.org/doi/10.1021/ct300100h https://www.sciencedirect.com/science/article/abs/pii/S0022286016304938 https://pubmed.ncbi.nlm.nih.gov/23247410/

Thank you people for the help. I have better clarity now.

source: Stack Exchange

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text:
Reaction analogous to burning that causes cooling rather than releasing heat
Is it physically and chemically possible to design both a gas and a solid substance, such that there would be a technologically feasible initiation of a chain reaction analogous to the spark that starts a fire, with the unusual feature that the reaction cools the region?
We are familiar with a spark being applied to the surface of a solid surrounded by gas, so that the spark is applied where there are both solid and gas.  Would it be possible to instead start a chain reaction in the gas, with the gas then changing form and affecting the solid?
I acknowledge that I haven't provided any information about the process that initiates the chain reaction, except via the vague analogy with a spark.  For example, perhaps  spraying a liquid catalyst into the gas is what initiates the chain reaction.

Endothermic reactions are a real things. Look at Wikipedia for them. This is because is quite difficult to get what you mean by analogous here. Opposite, if the reaction must be a chain reaction, I am not aware of such processes. The point is that every step would require activation from a cooled down microscopical surrounding.

In order for a reaction to be spontaneous and endothermic, it must produce a great increase in entropy. A simple way to increase the entropy is to produce a gas. For example, the reaction $\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$ is strongly endothermic, but produces a large increase of entropy due to the production of a gas, $\ce{CO2}$. Now it seems dubious that a reaction consuming a gas could be endothermic. It should produce more gas than it consumes. The only case I know is the reaction $\ce{C + CO2 -> 2 CO}$ which happens at around $1000$°C. But it is not a chain reaction.

I think one would be hard pressed to find an example. The whole point of the "burning" is that the exothermicity of the reaction helps sustain the reaction. A "reverse" process would inhibit the process and thus, be unlikely to be self-sustaining.

By burning do you mean combustion in an oxygen atmosphere?

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text:
Finkelstein reaction of vicinal dihalides
I recently came across the following in my OChem class:
Ethylene dichloride on reaction with  NaI in acetone reacts via E2 mechanism to give ethene.
I have three questions:

Since it is an E2 mechanism, a strong base should be present, but the iodide ion is a weak conjugate base of its corresponding acid. So why is the mechanism E2?

Bases are thought to be as proton acceptors. According to the above reaction, however it seems that the iodide ion attacks the chlorine on one of the carbons. Why is that so?

Chloride ion is less stable than iodide anion as more charge dispersion takes place in the latter. Then why is the chlorine a leaving group here, i.e. why does the reaction even occur?

I have just started OChem, so please correct me if my question has mistakes.

NaCl is insoluble in acetone, this is what drives Finkelstein

Ohh right.. completely overlooked that part about Le Chatlier's ...that does answer the third part. What about the E2 part, however how does the reaction proceed via E2 when iodide ion is not a strong base?

@Waylander can you please share some insights

I would if I could find a reference to this reaction. 1,2-Diodoethane is a stable compound see https://chemistry.stackexchange.com/questions/123561/why-is-1-2-diiodoethane-unstable


text:
How does X-Ray Crystallography determine the components (as opposed to the structure) of a molecule?
Thanks for some online resources (i.e. here) I have an understanding of what X-Ray Crystallography does.
By blasting X-Rays at a crystal and recording the scattering patterns on a plate behind the crystal, it becomes possible to determine the shape of the molecule in the crystal.

Where I'm confused though is in diagrams of protein molecules mapped out by x-ray crystallography, scientists are, incredibly, able to map out extremely complex structures. But what I don't understand is, how are they able to determine which atoms, or other molecules, they are seeing in the structure?
How do they know which part of the structure is which amino acid? Which part is hydrogen? Which part is nitrogen? Etc. I don't see any way of determining this from a diffraction pattern on a plate.

Why, that's simple. Carbon is black, nitrogen is blue, and hydrogen is white.

Seriously, we don't see atoms at all, let alone molecules. We just see some clouds of electron density. We know how many electrons are in each region. Also, we know how many electrons are in each atom. See where this is going?

More electron density is more scattering. Combine with horrendously complicated computation, a supercomputer, and some good initial guesses and you get a solution.

I'd suggest to read about x-ray crystallography of small molecules. You might as well grasp basics enough to answer answer your current question. Literature on protein crystallography focuses on more specific problems assuming you already know the basics (which you likely don't judging from the question).

https://chemistry.stackexchange.com/questions/124245/what-do-x-ray-diffraction-and-nmr-tell-about-the-proteins

That is because the solution to the structure isn't done just via the plate. The signal you see on the plate is what is called "structure factors" of the experiment. They are associated with the fourier transform of the electron density. There are computational ways of obtaining the electron densiy once you have the structure factors. When you do that, you arrive at something like this:

Then, once you have something like that, at least for proteins, you need to find the best possible geometry that fits the density field. So if you succeed you might end up with something like this:

The trick here is that the molecular biologists and physicists who work with proteins already know something of what they are looking for. They don't try to fit a haphazard collection of atoms to the electron density map, they start with a polypeptide chain, often one for which the sequence has already been experimentally determined so they know a lot about the composition of the molecule well before any structure resolution takes place.
This is the tl;dr version of the thing. In reality there are a number of things I have omitted. For instance, it isn't always necessary to obtain the electron density in order to resolve the structure. There are ways to theoretically compute the structure factors from a test geometry such that a structure is considered resolved if the theoretical and the actual form factors coincide up to a certain tolerance.
The structure is also resolved in steps. Often when you crystallize something, a lot of elluents and additives are added to the mix in order to make the crystallization possible, particularly for proteins, which have very fragile and sensitive crystallization behaviors. So researchers first try to resolve the geometry of the main protein chain, after that is done there will be residuals which can be adjusted by trying to find molecules of the additives and elluents in the unit cell. That is how we end up with sulfates, waters and other small molecules and analytes floating amidst the protein structure for the majority of the PDB structures resolved via X-Ray crystallography.
That said, for very small molecules and an electron density with a very high resolution, it is possible to distinguish one atom from the other by the shell structure around them. But, again, composition is often an information that the researcher has a priori, not something the diffraction is expected to provide.
P.S. If you have time, you can check the Yale Course of Professor McBride Freshman Organic Chemistry I. It is totally free an in the internet. In it, he does a very pedagogical and awe-inspiring introduction to the ins and outs of X-Ray diffraction (and other experimental techniques) that go well beyond the scope of this answer. He does not focus on proteins, but on the basics, small organic molecules, but it does wonders to clear the fog. The strictly relevant classes are classes 5 and 6, but I desperately urge you to go through everything that came before instead of just jumping in.

That is a very good and comprehensive answer!

Thank you so much for the incredible answer and for the linked course!


text:
Water Gas Shift Reaction
Considering the water-gas shift reaction
$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)},$$
I would like to understand how evolves $K^\circ$ the equilibrium constant of this reaction when I add water with a fixed temperature (and pression).
I know that
$$K^\circ = \frac{x(\ce{CO2})\cdot x(\ce{H2})}{x(\ce{CO})\cdot x(\ce{H2O})}.$$
I am wondering how is the equilibrium affected when we add some water.

If you add water without changing the temperature and the pressure, the equilibrium constant $K°$  is not modified. The only effect would be that the amount of $\ce{CO2}$ and $\ce{H2}$ goes up.

Thank you Maurice, how can we prove that?

@Poutnik that's right, I was rather unclear. I am asking how does change the latter compared to the former.

@Poutnik I get that. If I am posting this question it is because I don't know where precisely to search this information and I was looking for someone to get me on the right path. I don't need people to do my own work, but I would love some help for me to see it through.

@Poutnik Thank you. I must have made a confusion. Sorry for bothering.

No bothering, you are welcome. This site prefers teaching fishing instead of giving fish.

You may confuse equilibrium constant and reaction quotient.

In chemical thermodynamics, the reaction quotient is a quantity that provides a measurement of the relative quantities of products and reactants present in a reaction mixture for a reaction with well-defined overall stoichiometry, at a particular point in time. Mathematically, it is defined as the ratio of the activities of the product species over those of the reactant species involved in the chemical reaction, taking stoichiometric coefficients of the reaction into account as exponents of the concentrations. In equilibrium, the reaction quotient is constant over time and is equal to the equilibrium constant.

By adding water, you change the reaction quotient, but not the equilibrium constant.
If you add some reagens to a systém in equilibrium, you shift the quotient value away from the equilibrium constant. The reaction is then ongoing in the direction which causes convergence of the reaction quotient toward the equilibrium constant.
The equilibrium constant depends only on temperature.


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Why does osmolarity take into account both penetrating and non-penetrating solutes?
According to Wikipedia:

… osmolarity takes into account the total concentration of penetrating solutes and non-penetrating solutes, whereas tonicity takes into account the total concentration of non-freely penetrating solutes only.

But if osmolarity has units of $\pu{Osm L^-1},$ shouldn't it include only non-penetrating solutes, because the penetrating solutes won't contribute to the osmoles?

Why do you think they do not contribute ?

Because they can move freely across the membrane and are thus not osmotically active @Poutnik .

But if the solutes can pass freely through the membrane, this decreased activity will be equal at both sides of the membrane @Poutnik .

What if they flow before the stream flows @Poutnik ?

Let us continue this discussion in chat.

Collected and elaborated info from the chat with the OP:
To summarize the osmolality and tonicity difference, elaborated in the below link:

Osmolality takes into account all solutes.
Tonicity excludes solutes able to pass the semipermeable membrane.

So isotonic solutions need not to have the same osmolality.
The osmotic pressure is caused by decreased water activity, what does not depend on if solute is or is not able to pass the membrane. By other words, osmotic pressure and related osmolality  is property of solvent, not solutes, similarly as kryoscopic and ebulioscopic effects, all belonging to the class of colligative properties( depending on solute molality, but not on solute nature )
Primary is osmolality and osmotic pressure difference it causes. The difference then causes water osmotic flow.
When the osmolality is getting about balanced, then there is progressively coming into the play the difference in tonicity.
The same osmolality but higher tonicity means there is lower osmolality and therefore indirectly concentration of permeable solutes. This means there is concentration gradient powering the diffusion flow toward more tonic side. It is taking water with it, keeping osmolality balanced.
As result, volume of more tonic side increases even with the same osmolality.
At certain circumstances, the residual osmotic pressure may be compensated by the permeable solute concentration gradient, keeping status quo.


text:
What would oxygen or nitrogen snow look like?
I was watching a video of what would happen if the earth got kicked out of the solar system and at one point it was mentioned that oxygen and nitrogen snow would form. So would such snow be any different physically to good ol' water snow? Would it be more fine or less dense? I'm asking this because I'm thinking of writing a story or something that uses that. Curiously, there seem to be almost no related results on the internet.

Welcome to chemistry stack exchange SomeMan. This is a beautiful question. I had not really considered it until you asked it here, thank you for that. Now let us hope some atmospheric chemist show us what is up (Ozone is the answer. Ozone is up).

You may be interested in the short story "A Pail of Air", which has a very similar setting.

I guess our ordinary water snow may vary widely in that how fine or dense it is. Then again, a shape of individual snowflake is an interesting question in and by itself.

Oxygen snow would be pale (or deeper) blue.

Have you ever discharged a CO2 fire extinguisher? How "different " would you say is that from regular snow?

It is all physics - there aren't any chemical reactions going on here. Why is this question on this site?

Neptune's moon Triton has a thin atmosphere that includes crystals of nitrogen ice: https://en.wikipedia.org/wiki/Atmosphere_of_Triton

@PeterMortensen Chemistry does bother with phase changes where there is no chemical reaction either. Similarly for the all the thermodynamics of ideal gas. Domains of interest are not sharp-cut, separated, boxed and labeled, but there are fuzzy boundaries, covered by interdisciplinary science.

I don't know if there would actually be oxygen and nitrogen snow. I would think under those circumstances, there might be carbon dioxide  snow, but the rest of the atmosphere would probably liquify first, there may be oxygen and nitrogen rain with some hail or sleet, but snow? Not sure it would happen.
If it did, oxygen and nitrogen snow would be different from our normal water based snow because of the absence of hydrogen bonds.
Individual snowflakes hexagonal in shape would not form, as this is a function of hydrogen bonds and this would not happen with oxygen or nitrogen. Water ice is also less dense than liquid water because of the way hydrogen bonds cause "voids" in the solid structure. This would not be the case with oxygen and nitrogen. So you'd probably get something more like needles than snowflakes.
Also, solid oxygen is blue as noted by Ed V below, so blue snow if it happens at all.
Qualifier:
If the earth is out there (beyond the orbit of Pluto) long enough, enough of the atmosphere would eventually freeze and the pressure may drop below the triple point of nitrogen. If that happens nitrogen would sublimate, then fall as snow as temperatures fluctuate.
But even Pluto's surface temperature and pressure is above the triple point of oxygen most of the time (temperature does sometimes drop below though) so any oxygen in the atmosphere would probably fall as rain (blue rain) or possibly hail/sleet rather than snow.
Another thing to remember is that if the atmospheric pressure is that low, there won't be much drag/resistance either, so no "softly drifting" snow. Those needles will reach terminal velocity pretty damn fast and shatter on the ground.

I also thought most of the atmosphere would come down as rain rather than snow. However, depending on how the atmospheric pressure changes as it cools, it may be possible to go below the triple point temperature/presssure of nitrogen and oxygen, at which point any further cooling would trigger precipitation as snow.

“Solid, liquid, and compressed gaseous oxygen is blue.” Reference: First sentence of S.C Tsai, G.W. Robinson, “Why is Condensed Oxygen Blue?”, J. Chem. Phy., 51(8), 1969, 3559-3569.

For both oxygen and nitrogen the triple point pressure is practically a vacuum, so that won't really come into effect. The triple point temps are both under -200°C but not absolute zero so I suppose it's not entirely impossible, but rather unlikely

Nitrogen's triple point is 63 K, 13 kPa (on the order of 0.1 bar). That is a bit more than vacuum. I don't think there is any liquid nitrogen on Pluto - it either sublimates or snows.

@Peter Mortensen Sure. But compared to cutrently? Earth would need to lose a large chunk of its atmosphere to get to that. And again, sure, Pluto's surface temperature is below the triple point temp of nitrogen anyway, even if the high atmosphere isn't, and its atmospheric pressure maxes at only 1-2Pa so yeah, nitrogen snow could be a thing there. Not so much oxygen, since it doesn't really feature in Pluto's atmosphere. That doen't mean the same holds for earth.

Also, from the surface pics of Pluto, it looks like most of the nitrogen ended up liquid first - ocean of nitrogen with water ice floating on top - before it froze


text:
Why does the C−H bond dissociation energy vary in a homologous series of primary alcohols?
Specifically, for the primary carbon atom in the alcohol. Here is bond dissociation energy (BDE) data from chapter three of Luo's Comprehensive handbook of chemical bond energies [1] (boldface refers to dissociated atom):
$$
\begin{array}{llc}
\hline
\text{Compound} & \text{The broken bonds} & \mathrm{BDE}/\pu{kJ mol-1} \\
\hline
\text{Methanol} & \ce{\textbf{H}-CH2OH} & 401.92±0.63 \\
\textit{n}\text{-Ethanol} & \ce{\textbf{H}-CH2CH2OH} & 410.0±8.4 \\
\textit{n}\text{-Propanol} & \ce{\textbf{H}-CH2CH2CH2OH} & 406.3±8.4 \\
\hline
\end{array}
$$
Many websites like Chemistry LibreTexts describe how the average C−H BDE is different from the BDEs in specific compounds. Some explain how the hybridisation of the C−H bond has an effect or whether you're considering the primary, secondary, or tertiary carbon makes a difference.
In the case of the homologous series of primary alcohols though, the hybridisation for the C−H bond is always the same. Also, I'm only using values from the primary carbons. I checked the bond lengths for the C−H bond on the primary carbon for all the alcohols on Molview too. They're all the same.
What other factors can influence the variation in the BDEs?
References

Luo, Y.-R. Comprehensive Handbook of Chemical Bond Energies; CRC press, 2007. ISBN 978-0-8493-7366-4.

An idea my chemistry teacher described:
Perhaps, the intermolecular forces could influence the bond enthalpy between C-H bonds?
For example, as the homologous series of alcohols gains more methylene groups, the London
Dispersion Forces (LDF) between each alcohol molecule increase. So let's say you're heating up the alcohols to break the C-H bonds. In the process, some of the heat energy will go towards overcoming the LDF's attraction between molecules. Then, additional energy would be needed to break the C-H bonds after that.
At the same time though, you would have to consider the effect of decreasing polarity of the alcohol molecules and decreasing dipole-dipole forces. Does anyone know what kind of data could be used to quantify the strength of these different effects?

Think about the inductive effect

There is a big error in measurement for ethanol and propanol.

I wouldn't trust Molview data on equal bond length though. Molview only roughly predicts the values. Experimental data collected by using, say, microwave spectrometry or x-ray diffraction (or even better, neutron diffraction since we are looking for hydrogen atoms), will show some deviation.

Better think about products - what radical you get.


text:
How to extract limonene from the orange peels?
What is a reasonable procedure to extract the orange essential oil from the rinds?  I thought this would be a common experiment.
Here's what Wikipedia had to say, they  identify the essential oil as mostly D-limonene (it's an isomer):

Orange oil is an essential oil produced by cells within the rind of an orange fruit (Citrus sinensis fruit). In contrast to most essential oils, it is extracted as a by-product of orange juice production by centrifugation, producing a cold-pressed oil. It is composed of mostly (greater than 90%) d-limonene, and is often used in place of pure d-limonene. D-limonene can be extracted from the oil by distillation.

A centrifuge seems to be more involved than I'd like, since these are just household objects.  I considered for example boiling the water with orange rinds in it (as would be the recipe) except maybe something evaporates into the air.  Should I consider using a different solvent? Wikihow suggested I use vodka or olive oil. There's also a video on YouTube for extracting lemon oil.  It doesn't look so elementary.

Can you handle steam distillation?  https://edu.rsc.org/download?ac=12160


text:
What explains the trend of ionization potential in coinage metals?
The expected pattern for ionization potential is
$$\ce{Cu} > \ce{Ag} > \ce{Au},$$
but the observed patter or experimental value is
$$\ce{Au} > \ce{Cu} > \ce{Ag}.$$
What could be a justification for this?
The simple way to look upon is a general increase in size which fails here. So, I suspect introduction of f-orbital in gold has to do something with it which is absent in copper and silver.

Think about and search for possible answers thoroughly before posting a question to prevent writing up redundant information.
Include what you already  know/thought about, what you found in textbook or online resources, what you have understood or not/tried/failed. Written effort attracts elaborated answers. Questions without elaboration effort are not very welcome, and may be closed.

Poor nucleus shielding by f electrons, lanthanoid contraction and relativistic 6s orbital contraction may play the role.  Compare the trend in in neighbor groups.

@Mithoron my question talks about ag also


text:
Is selenium disulfide polar or non-polar?
So I saw in some internet site that $\ce{SeS2}$ is a polar molecule. When I drew the Lewis structure of the molecule, it showed up as a linear molecule like this:
$$\ce{S=Se=S}$$
The electronegativity difference of $\ce{Se-S}$ is just $0.01$ and the linear form shows that vectors cancel each other. Yet, some websites state that this molecule is polar, when everything to me indicates is a non-polar molecule.
Are they wrong or am I missing something here?

Selenium disulfide has a composition that approximates to SeS2 and is sometimes called selenium sulfide. However, as used in proprietary formulations, it is not a pure chemical compound but a mixture where the overall Se:S ratio is 1:2. The compounds are Se–S rings containing a variable number of S and Se atoms, SenS8−n.  https://en.wikipedia.org/wiki/Selenium_disulfide

I understand your assumption of selenium disulfide being a simple molecular like that. It is similar to the mistake of silicon dioxide, which at first glance we might assign it to be a linear molecule like carbon dioxide. However the crystal structure of silicon dioxide is far more complex.
In selenium disulfide, the selenium and sulfur atoms form a 8-membered ring, similar to the 8-member rings you would find in certain allotropes of either elemental selenium or elemental sulfur. The ratio of $\ce{Se:S}$ atoms is not exactly $\mathrm{1:2}$ as the chemical formula implies (that would give $\mathrm{\frac{8}{3}}$ atoms of $\ce{Se}$ and $\mathrm{\frac{16}{3}}$ atoms of $\ce{S}$ per ring). Indeed the ratio is closer to $\mathrm{3:5}$ ($\mathrm{3:6}$ is $\mathrm{1:2}$ anyways). A better representation of the compound is $\ce{Se_{n}S_{8-n}}$.

This compound is non polar because it is symmetrical in shape which makes the vector sum equal to zero. I think the electronegativity will not play any major role as opposite vectors are cancelling each other..

This would be true if that were the actual structure of the compound. But it isn't.


text:
Can nitrous dioxide develop when copper and dilute nitric acid react?
When doing an experiment in the laboratory with copper, dilute nitric acid and finally hydrochloric acid, the first reaction between copper and dilute nitric acid should end in nitric monoxide, cupric nitrate and water:
$$\ce{3 Cu(s) + 8 HNO3(aq) -> 3 Cu(NO3)2(aq) + 4 NO(g) + 2 H2O(l)},\tag{R1}$$
but somehow nitric dioxide, a reddish-brownish gas appears with a poignant smell, develops from this type of reaction:
$$\ce{Cu(s) + 4 HNO3(aq) -> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}.\tag{R2}$$
There was 5 mL of nitric acid and as we added nitric acid we quickly covered it with a watch glass. Then we put it in a sand bath instead of heating it normally with a Bunsen burner. Unfortunately, I cannot as it was one of the experiments that we just wanted to know what was going to happen and write down our observations and conclusion.

Strongly related questions: https://chemistry.stackexchange.com/questions/17067/how-does-copper-reduce-dilute-nitric-acid-to-nitric-oxide-and-concentrated-nitri ... https://chemistry.stackexchange.com/questions/43168/why-does-copper-react-only-with-nitric-acid

There are two ways to interpret an observation nitrogen dioxide.

Reactions between a metal and nitric acid rarely proceed by a single path to generate a specific nitric acid reduction product; in all likelihood there will be a combination of multiple reactions and thus a mixture of products.  With copper you could get both $\ce{NO}$ and $\ce{NO2}$.  And with a more powerful reducing metal like magnesium even more products could get into the mix, including ammonium nitrate.

You may have conducted the reaction in the presence of air.  $\ce{NO}$ is readily oxidized by the air to make $\ce{NO2}$ "after the fact".  This is in fact part of the natural production of nitric acid in thunderstorms.  Further complicating this is the fact that nitrogen dioxide in turn reacts with water, forming nitric oxide and regenerating some of the nitric acid (this is also indicated in the above link).  To get the true product mix out of the nitric acid you needed to use an environment free of air and unintended water.

The cited reaction of Copper metal and dilute Nitric acid has long known to produce ${NO}$ gas, although the precise reaction mechanism remains complex.
The mechanics, in my opinion, likely involves a REDOX pathway with, for example, ${NO2}$ as a radical product in the case of concentrated Nitric acid proceeding, in this case, somewhat simply as follows:
$\ce{Cu + 2 HNO3 -> Cu++ + 2 OH- + 2 .NO2}$
and, with more acid, Copper nitrate, also a cited product, with the net reaction now being completely in agreement with noted Reaction R2 above.
Perhaps, less complex is to quote Wikipedia on the mechanics of the interaction between Nitric oxide and oxygen:

When exposed to oxygen, nitric oxide converts into nitrogen dioxide:

$\ce{2 .NO + O2 → 2 .NO2}$

This conversion has been speculated as occurring via the ONOONO intermediate.

Now, I actually agree with the speculation that ${N2O4}$ is likely of assistance here, but how?
To answer, I refer to another educational source, which again repeats, albeit, noting the improbability of the termolecular pathway, quoting:

An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps:

$\ce{2 .NO + O2 → 2 .NO2}$

However, of import on the speculated pathway, to continue quoting:

An elementary reaction is at equilibrium when it proceeds in both the forward and reverse directions at equal rates. Consider the dimerization of ${NO}$ to termolecular, with ${k_1}$ used to represent the rate constant of the forward reaction and ${k_{-1}}$ used to represent the rate constant of the reverse reaction:

$\ce{.NO + .NO ⇋ N2O2}$

So, my opinion as to the speculated pathway that avoids the unlikely, but widely promulgated termolecular reaction (which may instead involve ${N2O4}$) is as follows:
$\ce{.NO + .NO ⇋ N2O2}$ (fast equilibrium)
$\ce{N2O2 + O2 -> N2O4 (slow)}$ Source here
$\ce{N2O4 ⇋ .NO2 + .NO2}$
where the net reaction is in accord with the more widely claimed single step reaction:
$\ce{2 .NO + O2 → 2 .NO2}$
implying, I would argue, only apparently speculatively (as there is a confirming source) that there is no direct action of oxygen on the ${NO}$ radical absence water (in complete contravention to the widely promulgated reaction above), but relatedly per a cited reference, instead on its possible self-reaction non-radical product ${N2O2}$, which then forms the non-radical intermediate ${N2O4}$, being probabilistically more likely the precise path.
Those doubting this comment are welcomed to review the associated biochemistry of the ${NO}$ radical in the human body as, for example, presented in this 2015 article Breathing new life into nitric oxide signaling: A brief overview of the interplay between oxygen and nitric oxide☆.

It is absolutely impossible to produce any $\ce{OH-}$ ion out of a reaction consuming an acid like $\ce{HNO3}$. In case this would happen, the $\ce{OH-}$ ion would be immediately destroyed according to : $$\ce{OH- + HNO3 -> H2O + NO3^-}$$

@Maurice: Actually what it implies, is that an excess of HNO3 is favorable in moving the cited net reaction to the right!

Supporting my comment to Maurice, interesting readings here https://chemistry.stackexchange.com/questions/17067/how-does-copper-reduce-dilute-nitric-acid-to-nitric-oxide-and-concentrated-nitri which was already a cited reference by Nilay Ghosh above.

Continuing, reading a prior question's answer link just provided an interesting reply by Oscar Lanzi, which accounts for NO in dilute conditions via a secondary reaction involving NO2 with excess water.


text:
Which are the Geometrical isomers of the complex [Cr(NH3)2Cl2(en)]+?Is it one or two?
I was recently studying coordination compounds and i came across optical isomerism shown by complexes.I think that the complex i gave above have 4 geometrical isomers(I think this is where I am wrong, that it only has three geometrical isomers),but among those four geometrical isomers,how many of them are optically active?.According to what i studied only two of the three geometrical isomers should be optically active,other two having plane of symmetry(I am not too clear about the plane of symmetry concept),but all  the solutions i looked up for the question said the complex only have 3 Geometrical isomers.Where am i wrong?.Is it that among the four geometrical isomers, two geometrical isomers are same?,if yes then how?

I suppose the easiest approach is to draw them all and then visually decide. The rules for being optical isomers are clear. They are mutually the mirror image for each other and at the same time they are not identical molecules.

You have an octahedral complex where two of the adjacent positions are occupied by the ethylenediamine ligand. Let's make these equatorial for a clearer picture.

There are only two positions in the remaining four positions that are trans (C and D). You can choose to put two chlorides there, two ammines there, or one of each. Those are the 3 geometric isomers.
If the trans ligands are the same, then there is a plane of symmetry that goes through C-M-D and bisects the ethylene diamine ligand, so the complex will not be chiral. But if the two ligands are different in positions C and D, then there will not be a plane of symmetry and the complex will be chiral. You can draw them out to see what they are.
To summarize, there are three geometric isomers, one of which is chiral. If you have trouble with them types of problems, sometimes, it's helpful to just consider the cis/trans or mer/fac relationships between groups of ligands.


text:
Order of solubility of sodium halides and potassium iodide

Order the following ionic compounds in decreasing order according to their ease to dissolve in water.
a) NaBr; b) KI; c) NaF; d) NaCl.

d > a > b > c
a > b > c > d
b > c > d > a
c > d > a > b

So, according to what I researched the solubility $s$ for each one of these compounds in water (at $\pu{20 °C})$ is:
$$
\begin{array}{lc}
\hline
\text{Compound} & s/\pu{g L-1}\\
\hline
\ce{NaBr} & 905 \\
\ce{KI} & 1400 \\
\ce{NaF} & 40.4 \\
\ce{NaCl} & 359 \\
\hline
\end{array}
$$
Therefore, my answer would be, from more soluble to least soluble:
b) KI > a) NaBr > d) NaCl > c) NaF
But this answer is not among the list. I think option 4 might be "correct", but it's kind of backwards.
Are the given answers wrong or am I doing something wrong?

You know you're supposed to be able to answer this question without looking up the numbers, right? And it doesn't matter if it's molar or mass solubility. You only need to know one little fact.

Teach me please, or tell me where to go. I'd like to know because these online classes are sooo bad.

NaF is not very well soluble. The halogens Cl, Br, I are rather similar in behaviour, flourine and the flourides (nearly) always stick out. You cannot be expected to know those numbers by heart, you know that NaCl is very soluble, so the outlier must be NaF, on the lower end.

The order b-a-d-c is correct, if the solubility is in gramm per liter. If it is in mole per liter, the order is a>b>c>d, as the saturated solutions of NaBr is the highest ($8.78~ mol L^{-1}$), then KI ($8.48~ mol L^{-1}$), then NaCl ($6.19~ mol L^{-1}$), and NaF ($0.95~ mol L^{-1}$)


text:
Removing ashes of 3d printing resin
I'm not a chemist so excuse me if my question will bite your eye or ear.
A resin 3d printing uses polymers that cure under exposure to UV light. There is a special sort of 3d printing resins made for casting metals. Basically these resins produce smaller amount of ash during casting. I'm looking for a way to clean-out a metallic part made from aluminium (or other metals in a list below) from 3d printing-resin-ash that occurs in place of contact with metal.
I'm unable to apply mechanical force to clean out the model due to it's geometry so naturally I would like dissolve the remnants of polymer (supposedly carbon) with some sort of acid or anything else that would eat-out ash while preserving metal as much as feasible.
The list of metals can be any of the following:
Aluminum, chrome, copper, nickel, steel, silver (ordered from most preferable to least preferable)
Is there a chemical compound capable of achieving the goal.
Thank you in advance!

If the polymer ash is substantially carbon, nothing can dissolve carbon that would not destroy the mold.

One of my friends suggested burning it in pure oxygen, which should gradually oxidise C into CO and CO2? Is it true?

Can your mold withstand the temperature of burning carbon??? Use an oxyacetylene torch, then.


text:
How one calls temporal changes of chemical content due to nuclear reactions?
So, I am trying to find the term for subject question: temporal change of chemical content due to nuclear reactions. Specifically, I am interested in time evolution of the number of atoms of each isotope in, let say, spontaneous decay chain but without any interest in potential chemical compounds. I was googling for "stoichiometry dynamics" and "nuclear chemistry kinetics" but these seem to have specific meaning clearly distinct from what I had in mind.

Do you know the Szilard-Chalmers effect ? It happens when a stable isotope is bombarded with neutrons to produce a radioactive isotope emitting gamma rays. In the gamma-ray emission, the recoil is sufficient to break a chemical bond existing with this isotope. For example, if  $\ce{NaClO3}$ is hit by neutrons, the stable isotope $\ce{Cl-35}$ is transformed into $\ce{Cl-36}$, which emits gamma rays. This atom is pushed out of the chlorate ion and can be recognize by adding $\ce{AgNO3}$ in the solution : this produces a precipitate of $\ce{AgCl}$.  The fate of the 3 oxygen atoms is not known.

Maybe “radioactive nuclei decay kinetics”?

@Maurice - I wasn't aware of this specific effect, but something along these lines. This is more of a "singular" transformation, I had in mind for an example decay of U-235 where you have rich dynamics, cascade of decays described by a Batemans equation.  I just didn't want to reinvent the wheel if there is nomenclature for this.

@Andrew - yeah that could work. Basically, difference between dynamics and kinetics is that dynamics goes into specifics of the cause of the temporal change, but kinetics just describes time evolution. Just wanted to check if there is an established nomenclature.

Does "transmutation" not serve your purposes?

Well, depends how you see it - transmutation is, by my understanding, the very nuclear reaction that changes Z and A in the nucleus. Such events take place in very short period and are stochastic in its quantum nature. Like I (unsuccessfully?!) explained, I was more interested in meso-temporal regime where such individual events are not observed but the whole process can be described with smooth, derivable functions.

Isn't decay what are you looking for???


text:
Find the electrode potential of a solution
I'm stuck with the next exercise.
What  is the electrode potential (against NHE) in the solution prepared by adding 90 mL of $\ce{ KI }$ 5.0 M to 10 mL of $H2O2$ 0.10 M, with a buffer at pH 2.0?
First I wrote the reduction half-reactions
$$ \ce {I2(s) +2e- ->2I-} $$
$$ \ce {H2O2 +2H+ +2e- ->2H2O} $$
then I wrote the complete reaction where $ \ce {H2O2} $ reduces and $ \ce {I-}$ oxides with the number of initial and final moles
$$ \ce {H2O2 +2H+ +2I- ->2H2O + I2(s)} $$
$ \ce {Initial:   0.001 n H2O2 , 0.045n 2I-  } $
$ \ce {Final:   x  H2O2 , 0.043+x  2I-   } $
and as the products are solid or liquid they have standard concentrations
Since it is not easy to know x, I proceeded to use Nernst's equation to find the potential of the electrode with the pair of iodes and knowing that the $E$ for this is $0.5345 V$
$$\ce {E = 0.5345 - \frac{0.05916}{2} log({\frac{0.043n}{0.1L}}})^2 = 0.5305V$$
but when I see the correct answer it appears as $0.419V$ and I don't know where is my mistake :c thanks for read. <3

The initial solution contains $ 0.45 mol$ $\ce{KI}$ in $100 mL$, with a concentration $4.5 M$. It contains also $0.001 mol$ $\ce{ H2O2}$, with a concentration $0.01 M$. The final concentration of $\ce{KI}$ is $4.48 M$. This is sufficient to allow you to calculate the electrode potential of the $\ce{ KI}$ solution


text:
Why does equilibrium for the addition of the Claisen enolate to an ester not favor the product?
I am currently reading chapter 26 of Clayden's Organic Chemistry, and a diagram for the elementary steps of the Claisen condensation with ethyl acetate in ethoxide is provided. It is obvious why the net reaction is favorable. The 1,3-dicarbonyl compound has a pKa of ~10 while the pKa of ethanol is 16 (so the more acidic 1,3-dicarbonyl compound will be the product in its deprotonated form). Even though the first few elementary steps may not favor the product, the last step is essentially irreversible.
I am having trouble understanding why the second equilibrium arrow claims to favor the formation of the reactants and not the products.
The first equilibrium arrow is obvious, the pKa of ethanol is 16, and the pKa of the ester enolate is ~25 (when protonated), so the less acidic ester will be favored in equilibrium. However, in the second equilibrium arrow, the pKa of the starting enolate is ~25 (when protonated) while the pKa (when protonated) of the product alkoxide is likely somewhere around 16. Wouldn't this mean the product for the second equilibrium arrow would be favored thermodynamically? I have considered that this step does result in a decrease in entropy, but I wouldn't expect this to have a drastic impact on the direction of the equillibrium.

Eli: I think you are reading too much into the magnitude of the equilibrium arrows. If the reaction is conducted in the presence of ethanol, then the arrows are likely pointing in the right directions.  Notice that the three sets of arrows are all the same and are probably pre-set as a symbol. Realize that the reaction is not catalytic in base.

@user55119 Is the direction of the equilibrium arrow due to the fact that this is a bimolecular process occurring in dilute solvent (i.e. more molecules are dissociating than coming together due to the dilution)? I know it is not that important to the net reaction, but I am just curious!

I think that there is little thought to the magnitude of the arrows. Based on pKa's they are probably correct.

The second step is the nucleophilic attack of enolate on the ester, it is not a deprotonation, so pKa values would not affect the equilibrium of that step.


text:
Is there an actual specific heat for aluminum in its gaseous state?
I've scourged the internet and I can't seem to find it anywhere although I've found the specific heat values for aluminum in both solid and liquid states. I'm trying to construct a heating curve and I need the specific heat capacity of aluminum in its gaseous state in order to calculate the amount of energy that would be necessary to bring aluminum from its liquid state to a gaseous state.

By "bring aluminum from its liquid state to a gaseous state", I assume you mean to a gaseous state at a temperature above aluminum's boiling point—since to merely convert it from a liquid to a gas, you only need the enthalpy of vaporization.
According to https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node18.html,
most metallic vapors have molar heat capacities very close to those of ideal gases, i.e., $C_v = 3/2 R = 12.47$ J/(mol K) and $C_p = 5/2 R = 20.79$ J/(mol K).  Thus you can obtain aluminum's specific heat capacity by dividing by the atomic mass, giving $C_v = 0.46$ J/(g K)  and $C_p = 0.77$ J/(g K)
The reason for this is that metals don't form metallic bonds in the gaseous state (metallic bonding is a collective phenomenon that requires bulk metal), and are thus monatomic*.  And the heat capacities of monatomic gases are simple, since they have only translational degrees of freedom.
[*As real gases, they may not be exactly monatomic, since they can still transiently associate via van der Waals forces.  But treating them as monatomic is a good approximation.]
The exception to the above would occur when the temperature is sufficiently high to cause the metal to ionize, in which case you would have a higher heat capacity because you would now have two (or more) particles for every ionized aluminum atom. But since aluminum's first ionization energy is 5.985 eV/atom = $577$ kJ/mol, and $RT$ at $3000$ K is $25$ kJ/mol, I don't expect there to be significant ionization close to its $2792$ K boiling point. [Also, some might argue that, once it starts ionizing, it is no longer a gas, but a gas-plasma mixture.]

I have read somewhere gaseous mercury is an exception among metals not forming diatomic molecules, being called for that pseudonoble gas. But it may be true that those diatomic molecules are just negligible parts for most metals and conditions. I remember posting recently some link about a study of temperature and pressure dependency of Rb2 occurrence. It was not negligible.

https://en.wikipedia.org/wiki/Dirubidium  Dirubidium is produced when rubidium vapour is chilled. The enthalpy of formation (ΔfH°) in the gas phase is 113.29 kJ/mol.[2] In practice, an oven heated to 600 to 800K with a nozzle can squirt out vapour that condenses into dimers.[3] The proportion of Rb2 in rubidium vapour varies with its density, which depends on the temperature. At 200° the partial pressure of Rb2 is only 0.4%, at 400 °C it constitutes 1.6% of the pressure, and at 677 °C the dimer has 7.4% of the vapour pressure (13.8% by mass).

Practically no data there, but I guess the page would not exist if dialuminium was not a thing.

@Poutnik Mercury does seem to be an exception. But as far as dialuminum being "a thing", that page is thin evidence. It could have been auto-generated purely as a placeholder by a computer algorithm after doing a webcrawl and finding articles like this, which specify "dialuminum trioxide", which is not Al2(g). https://www.spiedigitallibrary.org/conference-proceedings-of-spie/10035/100352X/Modeling-diagnostics-of-trioxide-dialuminum-content-in-gas-aerosol-medium/10.1117/12.2247833.full?casa_token=W_OZWXEyVQgAAAAA%3aMjfO5UeRQacrPmv1iDgJhbqaN61p31D_FMIx4nqXBcAgeH6bAm_oyhE6DvQSYRQ5ydgErBm3&SSO=1

I have not done detailed research, so I was not going to make any hard claim about Al2(g). But I guess for majority, but not all, of metals and conditions, temperature is too high and/or vapour density too low to form considerable amount of dimers, affecting the heat capacity values. BTW there is no need to form metallic bonds that are a bulk property. The old good covalent bond is enough. I suppose noticable dimerization may occur mainly for metals with low boiling point like alkali metals, or just for sensitive spectroscopic methods, e.g. measuring vibration-rotation IR spectra.


text:
Is Porphyrin Planar?
I have been looking at this diagram from quite some time, I know the 18 pi electron cycle that forms porphin(e) is definitely planar but what about the whole molecule?... The Nitrogen's act as a polydentate ligand with a metal complex so that may be a reason that the whole molecule may not be planar..
Can someone elaborate?

Our respected member @porphyrin definitely is not planar.  But I guess he would know for sure for the compound. I do not see any reason why it should not be. AFAIK, the H-N are not neighbors, but are placed across the hole.

only 18 electrons are in the same plane, the rest are not, hence prophyrin is not planar. am i correct? @Poutnik

I do not think so. Those other ones are just not part of the conjugated ring. See also https://en.wikipedia.org/wiki/Porphyrin#/media/File:Porphin-18e.png  But the H in H-N may make it nonplanar.

Porhyrins readily undergo distortions, and speaking of porphyrin metal complexes, they are as planar as metal centers allow them to be.


text:
Do changing opt=modredundant to opt in Gaussian makes geometry optimization not to take into account frozen angles?
I am a newbie to Gaussian and just generated an input for the geometry optimization for some molecules with multi ring system. However, in the article that was a reference for those calculations, some of the dihedral angles were freezed, so I did the same with adding the needed dihedrals to constrain with lines D X X X X F.
However, someone superior told me that I shouldn't use the opt=modredundant, but opt instead. Does it mean that I shouldn't constrain angles then? Or should I just manually change it to opt and the program will still understand that the angles are constrained?

Have you looked at the Gaussian manual (which includes a description of the keyword)? Did this person say why you should not use modredundant?

"Does it mean that I shouldn't constrain angles then?" Of course not, if you wan to recreate the result in the original article. However you may then want to try again without freezing coordinates. Why were they frozen in the original article in the first place?

@BuckThorn, thank you so much for the comments. Angles were frozen due to the fact that this molecule is part of a protein (i. e. nonplanar distortions that are being caused by it), so to ensure the overall structure of these molecules and their alignment, wouldn't change so much, I believe. However, we are trying to make this work with very similar system, so it's not exactly about recreation and more about getting new results.

Well, as I mentioned before, if you want to recreate the literature results impose the constraints. If you want to explore more of the energy surface, don't. This goes beyond technical aspects and is a question of opinion which might be impossible to answer without knowing more about your project.

@BuckThorn, Yes, of course, I have looked into the manual. It is written like this: "Except for any case when it is combined with the GIC option (see below), the ModRedundant option will add, delete, or modify redundant internal coordinate definitions (including scan and constraint information) before performing the calculation. This option requires a separate input section following the geometry specification". My guess would be that deleting the modredundant keyword would mean that the angles wouldn't be constrained, but I am not sure, so that's why I am asking this question here.

So you wondering how modredundant works? Yes, you can use it to impose constraints, as your line shows. Try it with something very simple first (H2 say) then progressively more complex.

@BuckThorn, thank you for the answer! However, the part is that not clear to me is that if you delete the keyword modredundant, but leave all of these constrains D X X X X F, does the program still understands them as frozen coordinates or just not reads them at all then?

The following illustrates the use of modredundant in a simple optimization (water angle at fixed bond lengths).
Input:
# B3LYP/6-31G(d) Opt=ModRedundant

test to see how to freeze coordinates

0 1
H     0.0 0.0 0.0
O     1 1.0
H     2 1.2 1 105.0

1 2 F
2 3 F

Optimized internal coordinates:
                           ----------------------------
                           !   Optimized Parameters   !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name  Definition              Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! R1    R(1,2)                  1.0            -DE/DX =   -0.0225              !
 ! R2    R(2,3)                  1.2            -DE/DX =   -0.1043              !
 ! A1    A(1,2,3)               99.0431         -DE/DX =    0.0                 !
 -------------------------------------------------------------------------------

Note the angle was not constrained so the final value differs from the input.
Now change the route to
# B3LYP/6-31G(d) Opt

and the output changes to
                       ----------------------------
                           !   Optimized Parameters   !
                           ! (Angstroms and Degrees)  !
 --------------------------                            --------------------------
 ! Name  Definition              Value          Derivative Info.                !
 --------------------------------------------------------------------------------
 ! R1    R(1,2)                  0.9687         -DE/DX =    0.0                 !
 ! R2    R(2,3)                  0.9688         -DE/DX =    0.0                 !
 ! A1    A(1,2,3)              103.6489         -DE/DX =    0.0                 !
 -------------------------------------------------------------------------------

So sure enough, removing modredundant loosens restraints under the optimization.

Thank you very much for the help! Just tried it myself with $H_2O_2$, so ensured that keyword is a must for constraining, despite the additional input of constrains (just was thinking there is somehow more complicated methodology to this). Sorry for bothering with relatively trivial matters.

@aerospace No trouble, and it's not entirely trivial. I just happened to have done this before. But there might be other solutions. I don't claim modredundant is the only way to constrain, someone else might have another answer.


text:
What are the applications of tetraxenonogold(II)?
I recently stumbled upon a pretty interesting chemical: tetraxenonogold(II) $\ce{AuXe4^2+}.$ Basically, an atom of gold with four xenons around. It is a product of a superacid reaction and it is interesting because both gold and xenon are quite inert elements.
What could be the application of such a molecule? It reminds me of some cancerostatic compounds.

That's not exactly how it works. You look at your application, whatever it might be, and then think of the properties a molecule should have in order to fit.

As far as I can tell it's most important application is to act as an example of a comparatively stable chemical species containing bonding between two inert elements

It only exists in very special conditions, so that rather strongly limits its applications.

That's like asking for the practical use in sending a probe to Neptune or Pluto. There is none whatsoever. We still do it, because it has a lot of benefical economical and cultural side effects.

Related: https://chemistry.stackexchange.com/questions/87830/were-goldii-complexes-synthesized-for-the-first-time-in-2017?r=SearchResults

I am currently writing an article (fun science stuff) about superacids. When it comes to carborane superacids, they are used to create that, so I wondered whether it's just a cool toy molecule, or if it is actually used for something.


text:
Tetrahedral complex and vibronic coupling or different rule for d-d transition?
I'm currently preparing a few slides for an upcoming talk in our group. I would like to mention some of the very basics regarding crystal field and ligand field theory as well, though this will not be the main topic of the talk.
I started from allowed transitions and how the (transition) dipole moment operator causes the transition between two states to be allowed or not (from just pointing out the area under an odd and even function and from group theory basics).
Then I'd like to continue to octahedral complexes. But here you've got two problems, one is the d-d transition that is forbidden and the other one is the Laporte forbidden transititon. When I need to compare this to the next slide with tetrahedral ligand fields I can always say that the Laporte rule doesn't work here since there is no inversion center.
So how is the Laporte rule still achieved for octahedral complexes? Due to something we call vibronic coupling. I always assumed this meant that a vibration, prior to the exitation deforms the octahedron to something that has less symmetry (which is probably happening, too) until I recently heard about transforming your set of p-orbital originating $T_\mathrm{1u}$ along the vibrational mode T1u that is also the dipole moment operator in this case, as x,y,z transforms as $T_\mathrm{1u}$ in this point group. Transforming $T_\mathrm{1u}$ along $T_\mathrm{1u}$ in an octahedral complex gives you, among others also a $T_\mathrm{2g}$, so the $T_\mathrm{2g}$ for the transition can also be described in terms of p-Orbital and thus you get p-orbital contribution. (Please correct me if I've understood anything wrong here).
Now this is a good explanation, what it doesn't explain however is how actual d-d transitions are allowed now? This means there is either another thing happening here or that there is also a vibronic coupling for the tetrahedral case.
Now, before I read into those group theory explanations I assumed that the d-d transitions happens because we are not talking about actual d-orbitals but we are talking about electrons in bonds. As the $T_\mathrm{2g}$ requires a π-type bonding and the $E_\mathrm{g}$ uses σ-bonding the actual transition is from a π-orbital into a σ-orbital, so p-s or if you take both sides pd-ps.
The other alternative would be that also the tetrahedral complex uses vibronic coupling. I can't find much on this though. There are calculations in combination with Jahn-Teller effects but no clear answer as with the octahedral case.
Hence my question: When talking about d-d forbidden transitions without taking the Laporte rule for odd even parity into account, is the often cited mixing of d-states with p- and s-states from something coming from the bonding situation or is the prohibition lifted due to the same vibronic coupling mechanism as used for partiy forbidden transitions?

For  any transition to be allowed the transition moment integral $\int\psi_i\mu\psi_f \ne 0$ . In symmetry terms this means that $\Gamma_i\Gamma_\mu\Gamma_f$ contains the totally symmetric representation. Thus a vibration can distort, say $\psi_i$ to make this allowed, i.e. $\Gamma_i\Gamma_\mu\Gamma_f\Gamma_{vib}$ contains $A_1,;A_{1g}$ etc, where $\Gamma$ means symmetry species.

First of all thank you, I do understand that principle but I was curious if this was also the explanation for all d-d transitions in general or if, leaving group theory aside, the bonding situation in, say an octahedral complex, also adds this effect of what we call p-d and p-s mixing for the transition. Because we can still see that octahedral complexes show weaker absorptions due to parity.

I'm not sure I quite understand, but if you mix p-s or dp-ps or whatever you will change the shape of the wavefunction and this may make the integral non zero.In this case a vibration may not be needed. The bigger the overlap $\psi_i\mu\psi_f$ the more intense the transition will be, obviously.

Yes, your mathematical and group theoretical approach proves it of course and I do understand the point of the irred. repr. in the end. My problem is I need something to destinguish the absorbance in tetrahedral vs. in octahedral complexes and something to lift the d-d problem as well. The former one is as you say vibronic coupling and can easily be shown using group theory. But I'm struggling a bit for the latter one. Like does this pd-ps mixing really happen through bonding situation or do we also consider mixing due to vibronic coupling when we refer to tet. complexes in this case?

Like literature will tell you d-d transitions are forbidden (ignoring parity here now). But, they happen due to p and s orbital mixing. Now if you search for this mixing you will find examples of vibronic coupling and parity included. So is there ever any other mechanism besides vibr. coupl. or Jahn-Teller involved here or are dd and parity rules lifted by the same mechanism?

spin orbit coupling and Herzberg-Teller effect (or intensity stealing) are the only other effect that immediately come to mind.

Oh my god I just saw my logical mistake, the direct product of the dipole T2 with the two states T2 and E already contains A1 (at least to the table I found), in contrary to an octahedron where the dipole T1u adds an odd symmetry that requires a vibrational mode to make it even again. Okay, now it makes sense to me. Thank you for the explanations though.


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Why can't nitrate ion in aluminium nitrate solution oxidise copper when a copper spoon is used to stir the solution?
I have been recently studying electrochemistry and I came across the electrochemical series. A question was asked saying what will happen if a copper spoon is used to stir an aluminium nitrate solution,the answer was no reaction will take place as Aluminium is more stable in its ionic form and copper cannot reduce aluminium ion, but can't nitrate ion which is above copper in the spectrochemical series oxidize copper and be transformed into NO which is more stable than Cu. Why doesn't it happen?

Minor correction: It is not the spectrochemical series but rather the electrode potential tables. Those tabulated values are given under standard conditions
You are indeed correct in thinking that nitrate ion should oxidize copper. It is a game of pH. If you have very low pH and nitrate ions, the following reaction takes place (in the absence of other ions)...
$$\ce{Cu(s) + 4 HNO3(aq) -> Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}$$

I used Mathpix Snip to convert an equation image into LaTeX. It is very useful tool for converting scanned equations/ mathematical papers into editable format. See https://mathpix.com/ if interested.

Well, Mathpix Snip does well for plain math, but fails horribly at chemical expressions. Please check the output for sanity before copy-pasting.

Mathpix Snip is an excellent tool for scanned math equations. I do not know of any better freeware. It is not meant for chemical structures. Anyway the sky did not fall apart if the state symbol for liquid appeared as 1 by my oversight.

I just would like the post by a 20k user to look professionally. Don't blindly accept what some black box tools spoon-feed you with. As for Mathpix, it's not the first and not the greatest OCR tool for LaTeX documents. Just as many others, it fails at recognizing anything complex and compiled with extra packages. What's even worse, it's not a tool, but a service which doesn't let you do anything before you sign up for it, which makes it useless for me.

There was an important question in medieval times "How many angels can stand on the point of a pin?" https://en.wikipedia.org/wiki/How_many_angels_can_dance_on_the_head_of_a_pin%3FWhether Mathpix Snip is a tool or service, after 100 years, if SE survives, people would view this $excessive$ obsession with LaTeX, fonts sizes of symbols or trivial terminologies exactly like the above question. We should focus on the concept.

I don't know about that, but disregard for standards and lazy formatting will lead to the future not far from the movie Idiocracy. I think you still don't get the point and think that LaTeX markup is an unnecessary tool invented by nerds to make life of ordinary people harder. BTW, a real tool and not not a service like Mathpix allowing you to use it max. 50 times a month. Also, SE doesn't use LaTeX. It's a different technology called MathJax that pretty much borrowed macro names, syntax and typeface from LaTeX for the use in web space.

What you call "excessive obsession" is in fact literacy. Currently, LaTeX is the only free and open-source tool available to every person on the planet that allows for literate written communication in all areas of expertise from linguistics to physics, also complying with notations and formatting from every known time period of a human history. Since you also mentioned the life cycle, LaTeX is already 36 years old, TeX syntax is over 40 years old, and still there is no good alternative, imagine that. Sure, focus on the concept, but do it in a correct and a sustainable way.

First, an observation that Aluminum nitrate is the product of a weak base and a strong acid. As such, expect the aqueous $\ce{Al(NO3)3}$ salt solution to be but mildly acidic.
Now, if the copper metal is fully immersed in a mildly acidic salt solution with no other factors (like dissolved oxygen exposure), I suspect this pH level alone is not likely sufficient to result in any noticeable chemical reaction including gaseous $\ce{NO}$ generation.
However, exposure to an oxygen source introducing some dissolve ${O_2}$, could in time result in an electrochemical based reaction resulting in the observed formation of an insoluble basic salt (see relatedly, for example, Equation (7) in this Wikipedia discussion on the industrial production of basic copper chloride).
So possible corrosion of the copper metal, especially if aged with a protective cuprous oxide coating. This is based on the electrochemical half-cell reactions:
$\ce{ Cu -> Cu+ + e- }$
$\ce{ Cu+ -> Cu++ + e- }$
$\ce{4 H+ + O2 + 4 e- <-> 2 H2O }$ (Source, see for example, p. 21-2 here)
resulting, in the current case, the following net reaction, for example:
$\ce{4 Cu+ + O2 + 2 H+ -> 4 Cu++ + 2 OH-}$
with potentially some insoluble basic nitrate formation observed.
As such, an aerated version of this acidic ionic mix may still not be a completely inert (non-corrosive) situation for the copper metal with time. As added support, I reference an educational source on the history of The Statue of Liberty, to quote:

The Statue of Liberty is a landmark every American recognizes. The Statue of Liberty is easily identified by its height, stance, and unique blue-green color (Figure 1). When this statue was first delivered from France, its appearance was not green. It was brown, the color of its copper “skin.” So how did the Statue of Liberty change colors? The change in appearance was a direct result of corrosion. The copper that is the primary component of the statue slowly underwent oxidation from the air. The oxidation-reduction reactions of copper metal in the environment occur in several steps...

A more technical reference Atmospheric corrosion prediction: a review makes several important observations, to quote in part:

For another example, when the nitric acid content in the atmosphere is high, the corrosion rate of copper is also independent on temperature (Samie et al. 2007)... When the temperature increases, the solubility of oxygen decreases and the corrosion product is more compact, the corrosion reaction slowed down as a consequence. However, temperature increase also leads to the increase of diffusion rate of oxygen and chloride ions, which accelerates the corrosion reaction.

Also:

Acidified aerosol is another important pollutant which has the potential to make a significant contribution to the atmospheric corrosion process (Cole et al. 2009)...While gas absorption may occur relatively close to the source, acidified aerosols can be transported over some distance from the original gaseous source. The aerosols are characterized by low pH, fine size (typically 1–100 mm) and high dissolved ionic salt content...Take zinc as an example, the presence of acidified aerosols may lead to enhanced corrosion by disrupting any protective oxide films and subsequently establishing electrochemical cells, relatively uniform oxide layers are formed, which may be dissolved...

The OP is asking " but can't nitrate ion which is above copper in the spectrochemical series oxidize copper" not whether copper will corrode or not. Copper even starts to dissolve slowly in vinegar in the presence of oxygen.

@Farooq: Yes, and answered, to quote: "if the copper metal is fully immersed in a mildly acidic salt solution with no other factors (like dissolved oxygen exposure), I suspect this pH level alone is not likely sufficient to result in any noticeable chemical reaction including gaseous NO generation." However, is it clear that the copper is completely insulated from air? And if not, the claim of no reaction is, in time, not likely accurate, as claimed per my sources and presented chemistry.

I should also add per the question itself, to quote: "when a copper spoon is used to stir the solution", which implies a partial exposure of the copper spoon to both air and the acidic ionic salt, which would likely display corrosive effects at the interface of the copper metal spoon and the solution with time, or upon removal of the spoon and left to dry (especially a wet-dry cycle).


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Nomenclature of aldehydes with double bonds and stereoisomerism
I'm struggling to name this compound shown below:

I'll post my chain of thought here so it can be corrected!

longest chain is marked as below:

this is because the functional group with the highest priority is the aldehyde (HCO) functional group, so the chain should contain this functional group
the chain continues down to to bottom-right to obtain the maximum length with the greatest number of substituents (although actually looking at it now, could the carbon chain go left and skip the double bond?)
there's a butyl side chain on carbon 3
there's a double bond at carbon 3
there's an ethyl side chain on carbon 4
there are two methyl groups on carbon 5
following CIP rules, the groups with highest priority are arranged opposite each other, so the molecule is E

Therefore, would the molecule be (E)-3-butyl-4-ethyl-5,5-dimethylhex-3-enal?
Or have I got myself completely mixed up...
Thanks for any help!

Kipf: What you have suggested is reasonable but apparently not the latest version of the rules. (E)-3-(2,2-dimethylpentan-3-ylidene)heptanal.

That’s not the longest chain

IUPAC has changed some of the rules of nomenclature as of 2013. In aldehyde 1 both the aldehyde group and double bond are included in the base name (noun) and the methyl group is the substituent (adjective) because both chains are $\ce{C4}$. In structure 2 the methyl group of 1 has been changed to ethyl. Now the longest chain is $\ce{C5}$ containing the aldehyde group and the double bond, as a substituent is the modifier methylene. In structure 3 the branch at $\ce{C3}$ has two $\ce{C3}$ units. The one containing the double bond takes priority as the "noun". In structure 4 the chain containing the double bond is shorter, which becomes the "adjective". In your compound 5, the noun is heptanal (in black) and the adjective, which contains the double bond, is in red. The two entities are numbered independently.


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Resources to learn second quantization formalism and how to implement it in code for quantum/computational chemistry research
I want to know what are some useful and practical resources to learn the formalism of second quantization as it pertains to fermions to be used in solving Quantum Chemistry problems. I am particularly asking for resources that go beyond the symbolic formalism and have guides/examples on how to implement the second quantization ideas in code (regardless of comp. language).
The pink book: Molecular Electronic-Structure Theory from Helgaker is a good start on the topic with all the formalism but I am looking for other resources with coding examples or practical implemmentations?

You could check out any code set that implements configuration interaction wave function methods.  The Journal of Chemical Physics published an open-access article reviewing the state of ab-initio software, and it lists code sets that use CI.
 https://aip.scitation.org/doi/pdf/10.1063/5.0023185
Notably, NWChem uses the second quantization approach and is described in this open resource paper:  https://arxiv.org/pdf/1904.01131.pdf
Finding algorithms to rapidly create valid CI expansions was a big research area in the late 1970's, and the Graphical Unitary Group Approach was created.  There are several open source packages that use the GUGA, including GAMESS. https://www.msg.chem.iastate.edu/gamess/download.html   You can get GAMESS by requesting access and registering.  You could then review the implementation assuming a basic knowledge of FORTRAN.


text:
How to turn oil into in a gel?
I've seen some EPR experiments where they add gelatin to water with spin-probes in it, as it "hardens" it immobilizes the spin-probes so that you can check effect of tumbling rate on the EPR. Cool
I would like to do the same with asphaltene in crude oil.... So I'm looking for a chemical that I can add to oil that will turn it into a more-or-less solid black gooey mess. It shouldn't be something too reactive though, because I don't want it to change the chemical aspects of the asphaltene (e.g. not make it self-aggregate). Just lock it in place.
Thoughts?

Asphaltene is too large and anisotropic to tumble effectively on an EPR timescale, isn't it? Anyway, can't you just vary the temperature? Crude oil should turn into a solid black mess at -20°C.

Why single out asphaltenes ( which may or may not be in a crude oil ) of the myriad of components ?  Crude may be anything from a water-like liquid , to something you can walk on ( at 20 C ).

Asphaltenes are usually the only paramagnetic component within crude oil. An individual molecule is about 1kDa but will get bigger if they self-aggregate, and they usually do!! For a recent oil that I was looking at, I got a good EPR match against EasySpin by using a tumbling rate of 45 nano-seconds. Hmmm. Only -20degC?? I had assumed much lower. Yes, I should check that.

Heavy crudes are diluted with thinner oils or heated to be pumpable.

Put your asphaltene source on a silica gel column and elute with hexane and then hexane/toluene mixture to remove paraffins (saturated compounds), aromatics, and resins (small polar compounds). At last, remove the asphaltenes (high molecular weight polar compounds) with pure toluene, evaporate to dryness (a brittle black solid), then with heat you can disolve/disperse the asphaltenes in a simple high molecular weight neutral material like petroleum jelly and let it cool. This will give a high and/or adjustable concentration of what you want to look at. https://en.wikipedia.org/wiki/Asphaltene

Thanks for the feedback James. I worry that the spectrum of the asphaltene would change quite a bit if it was taken through that the extraction process. Indeed, there are papers which describe how that process creates spectra that depend on the solvent polarities used to create the standalone asphaltene. So my hope is to leave it somehow "unmolested"  in the oil and then monitor the change in spectrum as the tumbling rate is frozen, and in such a way that I neither trigger self-aggragation nor destruction.

In that case, you could add a wax, heat to dissolve, mix well, then cool. Waxes come in a wide range of melting points from petroleum jelly to paraffins (up to 60C mp) and microcrystalline waxes (up to 90C mp). Your comment suggests that some self-aggregation may be a natural condition.

Hi James, we had indeed arrived to the same conclusion, we are going to try candle wax this afternoon. Yes, some self-aggregation maybe unavoidable, indeed, I'm not exactly sure if I would know. But I'm excited about the idea of getting an EPR "powder" spectrum of the crude. It might tell me a lot. Mostly likely it will just confuse me though :-)


text:
Carbon properties
Can carbon in the form of graphite melt or even boil? Under what experimental conditions? I ask this question because I found in a very serious book (Tables CRM Commission romande de chimie) information that I do not understand:

sublimation at 3650 ° C
boiling at 4827 ° C

But if the carbon sublimates at 3650 ° C, how does it boil? Is the work already done?
How valid are these values. thank you in advance

That book more pretends to be serious. You are right, carbon cannot boil at 4827°C if it already sublimes at 3650. The fact is that carbon has it's triple point somewhere around 11 MPa and 4600°C. (wikipedia gives +-300 on that temperature. It's extremely  difficult to maintain such a pressure and temperature at the same time, let alone measure anything.)

Don't forget, different behaviours are possible under different conditions (eg pressure). What conditions did the source specify?

Mithoron's reference displays a diagram showing that no liquid carbon can exist at less than 100 atm, whatever the temperature. Carbon sublimes at ordinary pressures.


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Is there a better coordinate framework and symmetry operator for constructing molecular orbitals of molecules with higher-order geometry?
The following excerpt is taken from [1] (with a few rewording, emphasizes are mine).

In an attempt to construct molecular orbital (MO) of any molecule, one needs to determine the symmetries of possible symmetry-adapted linear combinations (SALCs) that are formed from pendant-atom atomic orbitals. This prequisite can be proceeded with following steps:

Use directional properties (i.e. vectors) of orbitals on the outer atoms as a basis for a representation of the SALCs in the point group of the molecule.
Generate a reducible representation for all possible SALCs by noting whether vectors are shifted or not by each class of operations of the group.
Decompose the representation into its component (irreducible representations) to determine the symmetry species of the SALCs.

So far, I have practiced of constructing MOs of several molecules having simple geometry (low-order molecules) e.g. $\ce{H2O}$, $\ce{NH3}$, and others but I am worry with higher-order molecules, particularly adamantane, $\ce{Re2Cl8}$ cluster, arachno-$\ce{B4H10}$, and others.
I want to accomplish those tasks with hands and pens. I know there are several softwares where one can draw the molecules, assign the point group, and draw the molecular orbital but here, I want to know the process behind it and I want to figure it without using any software, so I am alright if the tasks will be very exhaustive and tedious. I want to know if there exists any method to simplify the process. At there, I want to ask several questions as below.
On the first step above from the excerpt, to generate group orbitals based on the representation of basic symmetry requirements, I am getting difficult with the natural way to place several atoms on Cartesian coordinate and I am wondering if there is any applicable projection or cartography that can ease up the process of generating the coordinates because most molecules are not planar. There are some books that present $\ce{CH4}$ molecule in a cubic framework which can ease the assignment of the coordinates for the atoms but I can not find any evidence of placing those higher-order molecules on polyhedral framework (perhaps like Bravais lattices that I can define the unit cells and so on). I got several papers of how these molecules had processed on the softwares but it turned out that the molecules were having coordinates that I worry I will get sloppy when I need to symmetrize it again. Is there any literature that can deal with this problem?
For the second step from the excerpt, is there a generator or a (matrix) operator that can ease up the process? Most matrices I found are already worked in backward process, that after visual inspection of symmetry operation, then the matrices were generated, which is totally very useless to me in this case. If this exists, it would be very relieving because so far I can not find any digestable mathematical method (not visual inspection) or algorithm that can perform such symmetry operations on those higher-order molecules.
Thank you.
P.S.: If possible, please illustrate your answers like a student trying to understand and answer the problems so I can work on those molecules later. Use molecules that are not as simple as water but not fullerene (small cluster molecules that are on par with molecules that I will work on) because most papers give examples of the latter one that I feel too strenuous to draw it with hands and pens.
References

Carter, R. L. Molecular Symmetry and Group Theory (Wiley, 1998).


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How many possible nucleotide types can be created?
In the RNA world hypothesis, it has been argued that RNA is made of 4 nucleotides and it can act as a template to create another copy of itself.
And it's been claimed that we only have 5 nucleotides, 4 of which is used in RNA.
And I also searched and realized that a Nucleotide is made of a phosphate groul, a sugar, and a nitrogenous base.
My question is, in the world of possibilities, how many types of nucleotids can be created? And if it's more than 5, then why nature has only created 5 of them?

https://en.wikipedia.org/wiki/Base_pair#Unnatural_base_pair_(UBP)

How many? This question simply makes no sense. It's only a matter of what one would still classify as nucleotide. Why the 5 ones? We obviously do not know. Oh, I'm pretty sure one could write a book about what are various guesses about that, but it only makes this question fall into like all close reason this site has.

How many is a pointless question. "Why those five exactly" makes a lot more sense, although it would still invite for a lot of guessing. "Why not more than five" is easy: five seems sufficient. Perhaps careful engineering could lead to a working system with four, but that's again guesswork.

I don't know why RNAs with 5 nucleotides became so prominent in our biosphere but you can bet that it is probably because it was better (even slightly so) than other RNA models it may eventually have competed with. So in order to arrive at a nice hypothesis about why this thing exists, we need to understand what was the competition like and how would our 5-symbol code benefited against rivals.
Consider the setting we are discussing. We have self-replicating RNA strands competing for the building blocks in the primordial soup (and maybe in each other)  in order to use those to make more of itself. The single most important resource that these replicators will have is time and speed. RNA strands aren't terribly stable and will hydrolyze after a certain window of time, meaning that our critter have a lifespan. If an RNA strand is to work as a self-replicator it must be able to assemble copies of itself within its lifespan otherwise its code will die with it.
We know that the supply of nutrients must have been quite limited because if you put mindless replicators that can eat each other in a soup of free dinner they reproduce at an exponential rate until the supply and demands are met and the population reaches and equilibrium appropriate to the free energy available. A population in this level leaves very little free nutrients around.
So in any given pool teeming with RNA critters, life must have been hard and dinner difficult to come by, because the time of bounty, when life was rare and food was abundant was followed by "that time when everyone ate like nutjobs until all the food was in their bodies rather than out there".
We also have reason to accept that there must be an upper limit to the length of the self-replicating strands. Longer strands are more prone to hydrolysis, oxidation and radiation. Environmental hazards that may lead to mutations and the rupture of the strand, leading to the corruption of the code and potentially the loss of its self-replicating quality, effectively resulting in its death. Longer chains have a shorter lifespans and take longer to copy than longer chains.
These two things generate selective pressures that might help us explain why the amount of nucleotides isn't larger. On one side, individuals with too few symbols (shallow code) will require longer chains to hold their code. This will shorten their lifespans and make their code harder to copy, since it is longer. So going too shallow will hamper an individual's ability to reproduce by increasing the time required to make copies and shortening the lifespan. Eventually, there must be a specific chain size above which the average copy/lifespan is less than 1. With a shallow code only the simplest individuals can reproduce reliably and that leads to low genetic diversity and less adaptability.
On the other hand, individuals with too many symbols (deep code) can have very short chains which are, in theory, quicker to copy but must compete for a more varied assortment of resources and must search harder and longer to meet their nutritional needs. If the code gets too deep, the time needed to procure the neccessary building blocks will exceed the critter's lifespan, meaning that, on avarage, an individual using this code will die before it can make a single copy of itself. With deep codes we can have rich, dense complexity but its maintenance cost can be prohibitive for replicators of this type.
So here is a possible explanation. Maybe 5 nucleotides is the best bang for our buck. The longest code that would have granted our RNA ancestors enough genetic diversity to survive crises that may have wiped shallower rivals while keeping their average copies/lifespan greater than their deeper rivals.
While this might look like an appealing explanation, I urge you not to take this at face value. I have zero evidence to corroborate with any of this. Since your question was so broad, I do believe that some speculation is in order. However the example here may help you learn how to think these questions through and arrive a satisfactory hypothesis to test in a suitable laboratory. Send the link up here when you publish it, cheers.


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Determine mechanism: E1 vs. E2
I am asked to indicate what is the product of this reaction, and the mechanism by which it is formed:
3-Bromo-3-ethylpentane + sodium hydroxide in methanol, 50 ºC
I identify it as an elimination reaction. I would say that it is a bimolecular elimination, E2, because a tertiary alkyl halide and a strong base (NaOH) are used. However, my teacher told me that it was unimolecular, E1, and I don't understand why.

If there is strong base present, then E2. If only methanol and no base, then its E1.

I would argue that you can posit a guess, but without experience, you cannot be certain unless you have experimental evidence. Methoxide/hydroxide is strong-ish, but it would be a better base if it weren't in a protic solvent. Heating is frequently indicative of E1. I would lean that direction, given that the electrophile is somewhat bulky, but again, you can't be 100% sure a priori.

@user55119 but it is that in the reaction are the two conditions

3-Bromo-3-ethylpentane is so bulky that an E2 elimination seems to be unlikely as the $\ce{OH}^{-}$, despite being a strong base, has a very low chance of directly attacking the beta $\ce{H}$. Moreover, proceeding by the E1 mechanism, we have a tertiary carbocation which is quite stable compared to primary and secondary carbocations.
However, as @Zhe has pointed out, looking at the experimental data gives the best answer. A general rule of thumb (although exceptions may exist) is that tertiary alkyl halides tend to follow E1 mechanism rather than E2 even under strong bases.

Well, I thought that the unimolecular elimination reaction, E1, takes place when a tertiary alkyl halide or a secondary alkyl halide reacts in the presence of a weak base (iodide anion, water...)

The unimolecular elimination reaction, E1, takes place when a tertiary alkyl halide is treated with a base (both weak or strong) or when a secondary alkyl halide reacts in the presence of a weak base.

@Adiboy it’s not true that if we have strong basic conditions then E1 would be favoured for tertiary alkyl halides. E2 dominates in that case as in the transition state, you form a more substituted alkene for a E2 and also E2 requires strong base. The only thing that would be required to see is if the solvent is protic or aprotic.

The only catch I find in this question is that, it’s given that you are using methanol as your solvent and that’s a protic  solvent . It will be the case that the base $\ce{OH^{-}}$ would be solvated by methanol and so it cannot efficiently grab a proton at the literal time of the leaving group from the $\beta$ carbon (we also know $\ce{Br^{-}}$ is a very good leaving group, so it leaves out fast). So the reaction favours E1 mechanism.


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What is the definition of energy-independent basis?
I am reading this article by Zatsarinny and Fischer [1].
At some point they span a wave function $\Psi_{E}$ in terms of energy independent basis
$$\Psi_{E}=\sum_{k} A_{E k} \Psi_{k}$$
For example, The Structures of Binary Compounds mentions that LCAO is an energy-independent basis [2] (Google Books):

The non-linearity of the one-electron eigenvalue problem represents a
considerable complication of this scheme, which does not occur if an
energy-independent basis set, such as the LCAO, is used.

What is the energy-independent basis?
Reference

Zatsarinny, O.; Fischer, C. F. The Use of Basis Splines and Non-Orthogonal Orbitals in $R$-Matrix Calculations: Application to Li Photoionization. J. Phys. B: At. Mol. Opt. Phys. 2000, 33 (3), 313–341. DOI: 10.1088/0953-4075/33/3/303.
The Structures of Binary Compounds; Boer, F. R. de, Pettifor, D. G., Hafner, J., Eds.; Cohesion and structure; North-Holland: Amsterdam, 1989. ISBN 978-0-444-87478-8.