text:
Why is it important to know the spin of bioinorganic complexes?
During a bioinorganic chemistry course, my teacher gives a lot of importance to spin states of metal ions and how they changes (or not) after the formation of a complex.
For example, when hemoglobin's Fe(II) ion couples with oxygen molecule, it's spin changes from high spin to low. My teacher says that is very important but he doesn't give a reason.

In short, understanding chemistry -> understand structure -> understand how it carries out its function. In this case, the structure around Fe is very important in understanding the cooperative binding of O2.


text:
How appropriate/adaptable is HPLC for in situ sample analysis on Titan?
I have a limited background in chemistry (nothing beyond general Chem 2), so I am asking for help with a group project. We have two chemists on our team, but communication is limited at the time and I am trying to answer some specific questions for myself so that we can move forward. We're designing a probe to study the chemical composition of Titan's lakes and sediment deposits in the lake beds, among other functions. We would like to adapt HPLC for the project using UV-Vis spectroscopy as the detection method. However, from what I understand, a UV-Vis spectrophotometer must be calibrated using all known solutes, which we can't know in this situation. I read that when the detector is showing a peak, some of what is passing through the detector can be diverted to a mass spectrometer and the fragmentation pattern can be used to identify unknown compounds, which is what we would like to do. However, I'm not sure how this would work or whether it would be wise given the constraints of our project. Our motivation for adopting HPLC is that we can keep all substances in their original phase for analysis. But if I'm not mistaken, mass spectrometry will require vaporization. My question then is, is there a way to use HPLC to identify unknown compounds suspended in the lake without disturbing the state of the sample? Any advice would be appreciated.

It is a very good question because this is an active area of research. Having heard talks from the actual people who are making analytical instruments for space analysis, I can mention the actual issues before coming to your actual question so that you are aware of the latest problems. As far as I know, HPLCs have not been sent, but GC systems have been successfully used.
Key problems
(i) Weight. It costs an enormous amount (in millions) of money per kg to send anything in space. Routine HPLCs are very heavy. We need two heavy duty persons to lift the stack of standard equipment.
(ii) Latest HPLCs can now be reduced to the size of a small brief case. Check capillary chromatography. Also check HPLC on a chip. It is the size of a hand palm.
(iii) Elecric power consumption: Think how much power is consumed by the motors in the pump (huge) and other analytical equipment including the mass spec. You don't have lot electrical power in space. This the biggest problem besides weight issues.
(iv) Think of excessive radiation. Can the plastics and equipment tolerate huge amount of radiation from the Sun and elsewhere.
(v) Temperatures: Think of temperature extremes with liquid as solvents.
(vi) There is no HPLC column (or GC column) in the world which can separate everything. May be you would attach at least 7-8 columns at once and see which one separates the sample (having a system with multiple columns at once is commercially available. This is done by pharma people)

Our motivation for adopting HPLC is that we can keep all substances in
  their original phase for analysis.

Yes, HPLC is a non-destructive technique. Mass spectrometer is not. However, UV-detection is useless for unknowns because you have no clue about their absorption spectra. Even if you have an absorption spectrum, it provides very little information about the full structure. Only mass spectrometer can be used in case of complete unknowns. Keep in mind that a typical HPLC C-18 column can only separate hydrophobic compounds under ideal conditions. The sole job HPLC or GC is to simplify the mass spectrum of a sample. Imagine you inject a sample containing 50 analytes into a MS. Imagine the complicated mess. With an HPLC, only compound will reach the detector one at a time. Mass spectrum is very simple then.

My question then is, is there a way to use HPLC to identify unknown compounds suspended in the lake without disturbing the state of the sample? Any advice would be appreciated.

Short answer no. HPLC cannot detect or separate suspended matter. If you meant substances dissolved in the lake, yes, then it is possible. Every sample has to be carefully filtered to remove any particles.
Since you have two chemists, why don't you just do a real experiment on lake water.

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text:
Density and concentration of hydroiodic acid
I weighted my hydroiodic acid sample: 25 ml weight 34.1 g, so the density is about 1.364 g/ml.
I want to know its concentration based on density. Does anyone have the corresponding density chart?

You should just titrate it.

Your right, but i just run out of indicator, have to resupply soon.

In 1981, Nishikata et al. published Viscosities of Aqueous Hydrochloric Acid Solutions, and Densities
and Viscosities of Aqueous Hydroiodic Acid Solutions in the Journal of Chemical Engineering Data.
Table III from this paper shows what you are after.

I don't know what temperature you used to make your density measurement, but assuming it was 20 °C, then your $\ce{HI}$ looks to have a concentration of about 39 weight percent.

Here is the density/concentration chart for aqueous $\ce{HI}$ at $\pu{20 ^\circ C}$:
$$
\begin{array}{ccc}
\hline
\text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) & \text{Concentration, %}(w/w) &  \text{Concentration, }(\pu{mol/L})  \\ \hline
1.0342 & 5.2 & 0.4204 \\
1.0812 & 10.8 & 0.9129\\
1.1226 & 16.4 & 1.4393 \\
1.1765 & 22.4 & 2.0603 \\
1.2333 & 27.2 & 2.6226 \\
1.2918 & 33.1 & 3.3429 \\
1.3605 & 38.7 & 4.1163\\
1.4208 & 42.9 & 4.7653 \\
1.5072 & 48.7 & 5.7385 \\
1.5913 & 53.0 & 6.5936 \\
1.6933 & 57.0 & 7.5458 \\
\hline
\end{array}
$$

Evidently, the density and molarity of aqueous solutions of $\ce{HI}$ is directly proportional at $\pu{20 ^\circ C}$ (LHS Chart) with the equation of straight line is $y = 10.965x - 10.87$ where $x$ is density in $\pu{kg/L}$ and $y$ is the molarity ($\pu{mol/L}$). Thus,
Molarity of your $\ce{HI}$ solution is: $10.965\times 1.364 - 10.87 = \pu{4.086 mol/L}$.
I also make a plot to see how percentage of the solution behave with density. Evidently (and also agree with literature finding), the density and $w/w\%$ of aqueous solutions of $\ce{HI}$ have second order relationship at $\pu{20 ^\circ C}$ (RHS Chart) with the equation of $y = -72.044x^2 + 274.62x - 201.55$ where $x$ is density in $\pu{kg/L}$ and $y$ is the percent concentration ($w/w$). Thus,
Percent concentration of your $\ce{HI}$ solution is: $-72.044\times 1.364^2 + 274.62\times 1.364 - 201.55 = 38.99\% (w/w)$.
Reference for densities and percentages:
E. Nishikata, T. Ishii, T. Ohta, "Viscosities of aqueous hydrochloric acid solutions, and densities and viscosities of aqueous hydroiodic acid solutions," J. Chem. Eng. Data 1981, 26(3), 254-256 (https://doi.org/10.1021/je00025a008).

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text:
What chemical can be used for getting rid of hard water spots on cars?
Most water spot removers available on Amazon or automotive stores are just fine abrasives which remove a small amount of your car's clearcoat every time you use them.
I'm looking for a chemical which can dissolve the water spots made out of limestone, chalk and gypsum.
Any acids suggested should not be strong enough to damage clearcoat (diluted), should not be abrasive.
Obviously I will not leave it on my car for very long, just apply, maybe rub around and rinse off with water.
I've not had any success with vinegar.

see https://chemistry.stackexchange.com/questions/117570/will-using-vinegar-harm-car-paint , who has tried an experiment.

Maybe the water spots are not limestone, chalk or gypsum, all of which are dissolvable in acid. Maybe they are mud spots, which could contain clays which adhere to the car paint.
If a toilet bowl cleaner based on citric acid (stronger acid than acetic, also a chelator) or Coke (acidic, contains phosphate) do not work quickly (without damage), perhaps an alkaline dispersing agent (like a liquid laundry soap) plus some elbow grease could diminish the visibility of the spots. Dispersing agents for clays include phosphates and anionic polymers like polysilicates (but they are typically quite alkaline - need to check damage potential).
Then, a good waterproofing (waxy) polish could make it easier to resist and remove the spots. Discovering where the spots come from (rain, splash from puddles, lawn sprayer, etc.) could help suggest what materials are present in the spots.


text:
pH of an acetate buffer after dilution
100 mL of a 0.1 mM buffer solution made from acetic acid and sodium acetate with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution?
The pKa value for acetic acid is given as 4.75.
The answer provided in my textbook is pH = 5.53. The closest that I can get to that is pH = 5.507. I know that's an almost insignificant difference but I'm just confused how they got to their answer, so if anyone can give me some suggestions that would be awesome. I'll outline my approach below:
Using the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] in the 100ml:
$$10^{5-4.75} = \frac{[A^-]}{[HA]}$$
$$[A^-] = \left(\frac{10^{0.25}}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 6.4{\times}10^{-5}M$$
$$[HA] = \left(\frac{1}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 3.6{\times}10^{-5}M$$
The addition of 900ml of water to 100ml of buffer is a 10x dilution:
$$[A^-] = 6.4{\times}10^{-5} \times 0.1 = 6.4{\times}10^{-6}M$$
$$[HA] = 3.6{\times}10^{-5} \times 0.1 = 3.6{\times}10^{-6}M$$
$$\ce{[H+]} = 10^{-5} \times 0.1 = 10^{-6}M$$
ICE table to determine equilibrium concentrations after dilution:
$$
\begin{array}{c|c@{hello}c@{}c@{}c@{}c@{}c@{}c}
   &   [CH_3 COOH] & + &[H_2 O] & {}\rightleftharpoons{} & [H_3 O^+] & + &[CH_3 COO^-] \\
\hline
I & 3.6 \times 10^{-6} && - && 10^{-6} && 6.4 \times 10^{-6} \\
C & -x && - && +x && +x \\
E & 3.6 \times 10^{-6} -x && - && 10^{-6} +x && 6.4 \times 10^{-6} +x \\
\hline
\end{array}
$$
$$K_a = \frac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}}$$
Then substituting in Ka and the equilibrium values from the ICE table: $$(10^{-4.75}) = \frac{(10^{-6}+x)(6.4 \times 10^{-6} + x)}{(3.6 \times 10^{-6} -x)}$$
$$x^2 + 2.518 \times 10^{-5}x -5.762 \times 10^{-11} = 0$$
$$x = 2.111 \times 10^{-6}$$ $$ [H^+] = 10^{-6} + 2.111 \times 10^{-6} = 3.111 \times 10^{-6}$$
$$ pH = -log_{10}(3.111 \times 10^{-6}) = 5.507 $$

Your intermediate results have two significant figures. You answer and the textbook answer are the same at two significant figures. If you want to get closer, keep some extra digits in the intermediate results.

Calling a solution where the total concentration of acid and base combined is 0.1 mM (or 0.01 mM) a buffer is unconventional, to say the least. As you can see from your calculation, the conjugate acid is no longer a major species. The buffer capacity will be very low, and adding an indicator at concentrations where you can see it will change the pH by a lot.

That was one of my first thoughts, but even using a value for x calculated using unsimplified values for [HA] and [A-] throught the calculation results in an almost identical answer: pH = 5.50716.
I also agree that the solution isn't exactly a great buffer - but then that depends on what it has to buffer. My problem isn't with the peculiarity of the question, but rather with the approach to finding the pH of a highly diluted weak acid/buffer system.

You might have spent more time on the question than the person how gave the textbook answer...

I fear that might be the case...

If the initial pH was 5.0 you know that only has one significant figure, right? I hate these sort of stupid problem where they state the problem with few significant figures. If they want 3 significant figures in the answer then the problem should be setup in such a way to support that. // As it is the book's problem solver probably rounded somewhere and as you now know you'll go crazy trying to figure out where. // In the good old days I use a slide rule for about 10 years. 2 significant figures was typical, three if you were careful.

Depending on your first calculations before dilution:
\begin{align}
[\ce{CH3COO-}]&=[\ce{CH3COONa}]=\pu{6.4\times{10^{-5} M}}=[\ce{Na+}]\\
[\ce{Na+}]_\text{(After dilution)}&=\pu{6.4\times{10^{-6} M}}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-4} M}}~\text{ (Before dilution)}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&=\pu{1\times{10^{-5} M}}~\text {(After dilution)}
\end{align}
So,you have the the following $5$ independent equations in order to specify the unknown concentrations of $5$ species as present in an aqueous solution consisting of $\pu{1\times{10^{-5} M}}$  buffer solution made from acetic acid and sodium acetate :
\begin{align}
6.4 \times 10^{−6} &= [\ce{Na+}] \tag{1}\\
K_\mathrm{a} &= \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}
     = 1 \times 10^{−4.75} \tag{2}\\
K_\mathrm{w}
    &= [\ce{H+}][\ce{OH-}]
     = 1 \times 10^{−14} \tag{3}\\
1\times{10^{-5}}
    &=[\ce{CH3COOH}] + [\ce{CH3COO-}]\tag{4}\\
[\ce{H+}]+[\ce{Na+}]
    &= [\ce{OH-}] + [\ce{CH3COO-}] \tag{5}
\end{align}

I appreciate any fruitful comment


text:
coffee concentrate and water separated after 2 hours. How/why?
I'm asking because this makes no sense to me and making sure I'm not stupid.
I brewed 10 gallons of cold brew concentrate(TDS 12.2) and mixed it with 10 gallons of water.
I moved 10 gallons of the mixture into kegs and they ended up at TDS 6.4. Two hours later I filled the rest of the kegs. One keg was at TDS 4.8 while the rest were at TDS 0.4.
Will coffee and water separate? I've never seen this happen before and previous kegs have never tasted differently until just now.

Then you most probably didn't mix properly.

I guess my main question is: how are the water and coffee concentrate not homogenous? and how would I ensure proper mixing?
Again, we've never seen this happen before.


text:
How can I find the smaller symmetric structure from big crystal unit cell?
I have Pyrope (Mg3Al2(SiO4)3) crystal structure, downloaded from materialsproject:
https://materialsproject.org/materials/mp-6073/#
It appeared as triclinic cell, but if you download CIF file and open it from softwares like Avogadro or VESTA, it appeared as a orthogonal simple cubic (a=b=c, alpha = beta = gamma = 90 degree). I also checked that this geometry is the same with what I downloaded from American Mineralogy Material Database.
Now, I wish to run DFT or quantum calculations for them. So, I excluded atoms in periodic boundary, unit cell contains total 160 atoms. I'm afraid this might be too big, might takes too long time to calculate via DFT or any other quantum simulations. All I need to do is just geometry optimizations, but still, 160 atoms seems too big.
Using VESTA or Avogadro or any other software, is this possible to find the symmetric structure inside this or other "big" unit cell, and make a smaller / simpler version of unit cell for faster calculation?
Or, can I manually cut this unit cell in 1/4 size, 40 atom system with 1/2 size for x, y, and z direction, and run the minimization? It seems that the Pyrope unit cell have some sort of repeating structure inside the big unit cell. But I'm not sure if it is "safe" to find symmetry and divide the unit cell manually.
Thank you

A fully ab initio geometry optimisation of pyrope has already been done: https://pubs.acs.org/doi/pdf/10.1021/jp050316z

@IanBush Thank you, but I need to do some other things, like volumetric expansion and compression and etc.... which requires good amount of computational resources.

You don't cut unit cell in half out of the blue. It would be about as sensible as cutting your car in half, in the hope that it will still be able to carry half the load, say, yourself and one passenger. (Besides, cutting a cell in half along x, y, and z would give you 1/8, not 1/4 the volume.) Also, pyrope is definitely neither triclinic not orthogonal nor simple cubic. As for the symmetry, it is written right there in the CIF file, as plain as it gets, and is pretty high (think of 1/48, not 1/8). Whether your software of choice can read and properly account for it is another question.

@exsonic01 Note the paper referenced calculated the full phonon spectrum of pyrope as well as the structural optimisation, and is14 years old. 160 atoms should be reasonably routine nowadays on a small cluster, people are doing well over 1000 atoms in optimisations. I've just submitted a paper with around 2800 atoms in the structure, studying the structure and properties as a function of the metal at the heart. That required a fairly chunky machine, but only ~2000 cores which is not outrageous.

@IanBush Well, not all research centers have affluent amount of computational resources, and I have other guys in my lap to run some simulations... Oh well, I will see what I can do. Thanks anyway.

@IvanNeretin After checked CIF files, Now I know it is Isometric - Hexoctahedral. Then, when I run DFTs, can I use the cell dimensions (a,b,c,alpha,beta,gamma) written in CIF file? Thanks again

No, the unit cell is the smallest translational repeating unit, and any truncation of that would mean that you're calculating a different solid which may or may not exist.
160 atoms is a decent sized unit cell, but is still routinely doable with the right resources. (The materials project page you link to is in fact an automated DFT geometry optimization of that solid, so you could use that)
If you still want to do the calculation yourself then you have to use the whole cell; however, if the unit cell has some additional symmetry (e.g. center of inversion), then a good DFT program will take advantage of that to speed up the calculation.

Thanks. But I have a question, the unitcell CIF file I downloaded from materials project (and american mineralogy) actually have Mg24 Al16 Si24 O96. But we know the original formula of structure is Mg3 Al2 Si3 O12. Material database one actually 8 times bigger than formula. That was why I was thinking it might be possible to devide the unit cell.

Speaking of DFT, I'm using Quantum Espresso 6.4.1, and as far as I know default option consider the symmetry...

The ‘actual’ formula you’ve mentioned above is the empirical formula and not the structural formula, so you cannot do what you are proposing. (http://webmineral.com/data/Pyrope.shtml)

OK, thanks to let me know, I'm checking CIF file and now I know I was totally wrong. Purpose of my questions was, I just wished if such things are possible, just in case to check.

If you want to impose symmetry restraints (i.e. keep the symmetry), there are only 4 unique atoms, and most of their coordinates are fixed. The asymmetric unit (in this case 1/48th of the unit cell) is the unique volume, not the unit cell.

The structure contains 4 independent atoms, Mg, Al, Si and O. The space group is a cubic one containing 96 asymmetric units. If you place an atom on a general position, it will appear 96 times in the unit cell. Mg, Al, and Si are located in various special positions, i.e. lie on symmetry elements. Placing the Mg and the Si atom will result in 24 atoms in unit cell, while placing the Al atom in its position (on the origin) will result in 16 atoms in the unit cell (on a three-fold and on the center of inversion).
So you could say that the asymmetric unit contains one oxygen atom, one quarter Mg and Si atom respectively (24/96), and one sixth Al atom (16/96). One of the sites you cited in a previous question gives the fractional coordinates as:
Si: 0.3750 0.0000 0.2500
Al: Origin
Mg: 0.1250 0.0000 0.2500
O: 0.0341 0.0490 0.6530
If you look at the space group, http://img.chem.ucl.ac.uk/sgp/large/230az1.gif,

you can see that Al, Mg, and Si are on special positions (you can either find the locations on the diagram or apply the 96 operations (48 combined with the I-centering) listed on the right to the coordinates and see how many unique locations you get), while oxygen is not.

can I manually cut this unit cell in 1/4 size, 40 atom system with 1/2 size for x, y, and z direction, and run the minimization?

If you software is aware of crystallographic symmetry, it is a very simple system with 4 degrees of freedom (the oxygen position and the lattice parameter a). Everything else is fixed if you impose the symmetry of garnet.

I have Pyrope (Mg3Al2(SiO4)3) crystal structure, downloaded from materialsproject: https://materialsproject.org/materials/mp-6073/#. It appeared as triclinic cell...

There are some inconsistencies in the way the site shows the data. The space group is (correctly) given as Ia3d. The 3D image does not show a body-centered cubic cell, but instead the primitive rhombohedral cell (the smallest unit that recreates the entire crystal through three translations). In one iteration ("primitive"), the coordinate file lists the symmetry as 1, applied the Ia3d operators to the atomic positions (maybe to "not to confuse" users unaware of crystal symmetry), and switched to the rhombohedral cell.


text:
Will using vinegar harm car paint?
After having parked my car in a newly constructed airport parking garage, I came back to find something had dripped through a crack in the ceiling of the structure and left considerably thick film on my car.  On a glass window surface, I tested soap and water, mineral spirits, two types of latex paint remover (i.e. goof off), and household chlorine bleach.  None were effective.  But plain white 5% vinegar dissolved the film if I let it soak or scrubbed on it with a cloth.

My question: is an acid of the strength of vinegar a risk to automotive "paint" (the top layer is probably "clear coat").

I'd try using a paper towel wet with vinegar and just stick it to the surface of the car. Then use a spay gun to give it a good squirt. I think the worse thing you can with such a coating is rub. I'd be worried about scratching the clear coat and/or paint. Take the paper towel off and rinse with water while you rub gently with your fingers.

Vinegar is not a strong enough or corrosive enough substance to damage paint and you can minimise the risk by using a cloth or sponge wet with vinegar on the glass and not spraying the whole car or leaving it in the surface for an extended period. Also, it is a good agent for removing what is almost certainly a limescale-like layer (basically a carbonate that will react with acids).

That is a valid question, because acid rain has been known to damage paints, particularly high-gloss, i.e. clear coat. Best would be to test the effect of vinegar on a hidden area, and double-check under bright light from different angles. Your idea to test on glass is good, because it shows how long a soak is needed.

If it appears unaffected, rinse afterwards with a weak solution of baking soda or ammonia, and then rinse again with plain water to remove any remaining acid.
If damage is slight, can car wax effect a good repair?
If vinegar does damage in the test spot, you might look at various car polishes or even cutting compounds, and then wax (or reapply a gloss coat, which may mean refinishing a large section of bodywork to get it to match).

Vinegar Test: No Visible Change After Two Hours
As advised above, I decided to test vinegar on an inconspicuous spot.
First I created a pattern of exposed paint using masking tape.  The shape was just something random that I thought I could most easily recognize, even if just a slight difference in the car's finish:

Then I placed vinegar-soaked cloth on the test area of the car:

After two hours (which was longer than I expected the cleaning of the other parts of the car to take), I removed the cloth, rinsed the area, and finally just washed the are normally:

And finally took the car into various lighting conditions.  There was no detectable difference in the finish that I could see:


text:
Can you use a photon's torque to change a molecule?
Can you use the torque of a photon to change bond angles, etc.? 11-Cis-Retinal to the all trans version, in  the eye, is a good illustration [1].
Note: This phenomenon can be observed by closing blinds, allowing the sun to shine through, you get light beams. If you blow a smoky material past it, you can observe this phenomenon of light.
References

Rego, L.; Dorney, K. M.; Brooks, N. J.; Nguyen, Q. L.; Liao, C.-T.; Román, J. S.; Couch, D. E.; Liu, A.; Pisanty, E.; Lewenstein, M.; et al. Generation of Extreme-Ultraviolet Beams with Time-Varying Orbital Angular Momentum. Science 2019, 364 (6447), eaaw9486. https://doi.org/10/gf4f6h.

Molecules are solid, they don't change. Except the rare cases when they do. Can you change a piece of metal with your bare hand?  Normally no, but sometimes yes. Same thing here.

Probably not, normally the total energy of a photon is needed to produce an excited state, (by being absorbed by the molecule), and from which isomerisation/bond breaking  can occur. This is the case in your example where 2 to 3 eV are needed, and the torque is not likely to be of sufficient energy. You should ask a question in Physics StackExchange to better understand what this torque actually is, as opposed to circular polarisation, for example.

Hey Ivan, thanks for the information!

Porphyrin- I am not trying to get the photon to be absorbed by an electron. I am wondering if bombarding a specific bond with photons with torque could change the orientation of a bond. i only provided the study to give more information. Thank you! Just trying to figure things out.

What makes you think that the reference you've provided is relevant to 11-Cis-Retinal?

The torque could be used to align whole molecules, see the info at CFEL in Hamburg https://www.controlled-molecule-imaging.org/, and corkscrew -laser technique

@porphyrin At technique uses optical spin angular momentum (which is tied to the polarization), not the orbital angular momentum (tied to the spatial dependence) which is at play in the referenced publication, i.e. the two have very little to do with each other. They're similar, but they operate at vastly different length scales - OAM will only couple to objects at the size of the wavelength.

@Emilio Pisanty Thanks for clarification.

Hey Guys, i only added the study to show photons can have a torque associated with them. That is really all. The same answer basically for the question about 11-cis-retinal. I only included it to give an example. Thanks guys!

What relevance does this have to spin?? "This phenomenon can be observed by closing blinds, allowing the sun to shine through, you get light beams. If you blow a smoky material past it, you can observe this phenomenon of light."

I think it is cool to observe natural phenomena that surrounds us. I would say with 95% confidence that the beam of light created by the method i described has photons with a torque. I think it confirms the experiment i linked to since you can actually see the phenomena as the study i linked originally mentioned states when interacting with matter. I can literally see the movement.
https://www.nature.com/news/people-can-sense-single-photons-1.20282

@kevindelahoussaye If your core question is whether optical forces are sufficient to directly twist a molecule, then your reference acts in a way to make it directly harder for your question to be answered. (This is because the reference you provided is highly suboptimal to addressing molecular structures compared to e.g. optical centrifuges.) If you're asking about the applicability of that specific paper to twisting molecules, then ask explicitly. (Full disclosure - I'm an author in the reference you cited.)

Hey Emilio, Thanks for the response! As mentioned above the only reason i included your paper was to show photons can have a torque. I believe your paper does that. I am not asking about the applicability of your paper to twisting molecules; only to show they have a torque. Which i then wondered if you could use the torque on photons, that exist in plenty in the world, to change a bond. I do not know of any techniques to produce photons with a certain torque. That is also another reason. I think it is probable that photons can change orientation given it seems to happen in our bodies. Thanks!

Sorry if my reference skills are not up to par. I was not even aware of optical centrifuges. Thank you for your contribution! Please pass along my warmest regards.


text:
Using concentrated sulfuric or nitric acid during titration of permanganate with oxalic acid
During titration of $\ce{KMnO4}$ with $\ce{H2C2O4}$, can we use concentrated $\ce{H2SO4}$ or $\ce{HNO3}$ instead of bench sulfuric acid?

Concentrated acids are never used for titration due to fluctuations in density/concentration and, most importantly, you are always going to greatly overshoot the equivalence point.

@AdityaJain We prefer to not use MathJax in the title field due to issues it gives rise to; see here for details. It is also preferable to avoid abbreviations and spell out chemical names in title. As for your answer, plagiarisms are not tolerated on Chemistry.SE. If you are using third-party sources (text/graphics/data), always make sure to distinctly separate quoted material (e.g. with > markup) from your work and put a proper reference.

Also, conc. $\ce{H2SO4}$ and $\ce{HNO3]$ are oxidisers themselves and would, thus, confound the result.

And you really don't want to pour concentrated sulfuric acid into anything, due to heating, and both acids pose too much of a danger for casual use in titration.


text:
How can one synthesise a conjugated alkyne chain?
Let's take the conjugated alkyne chain (e) below, taken from Wikipedia:

How would one typically go about synthesising such a conjugated chain with a series of alkyne functions?

A carbenoid Fritsch-Buttenberg-Wiechell (FBW) rearrangement of a substituted dibromoolefinic precursor is used to generate a lithium acetylide, and subsequent trapping with carbon-based electrophiles provides a wide range of di- and triynes. The lithium acetylide formed from the FBW reaction can also undergo transmetalation to provide zinc, copper, tin, or platinum acetylides [1]. The paper described the preparation of poly-ynes of up to 6 triple bond units.

Image source: organic-chemistry.org
References

Luu, T.; Morisaki, Y.; Cunningham, N.; Tykwinski, R. R. One-Pot Formation and Derivatization of Di- and Triynes Based on the Fritsch−Buttenberg−Wiechell Rearrangement. J. Org. Chem. 2007, 72 (25), 9622–9629. https://doi.org/10.1021/jo701810g .

Limiting the target to diynes I mention the earlier still valuable oxidative coupling of terminal alkynes, copper catalysed, in air.


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Which transparent plastic to use for electrolysis of dilute sulphuric acid
I plan to do an experiment which involves the electrolysis of dilute sulfuric acid (1 M or 1 mol/L). I would like to carry it out in a long transparent plastic container (easier to handle and clean). What would be the suitable plastic for this?

Just curious, why not use glass since it’s already transparent and inert with sulfuric acid? And probably also easier to clean than plastic?

The problem is that I need a pretty big 'tub' or container for this experiment, and I imagine that this will take too much weight. And also I'm not sure were I would find a deep rectangular glass container, at least where I live.

At 1 M, most plastic containers, such as polyethylene (polythene, in the Old World), polypropylene (better at high temps and a bit more transparent), acrylic or polycarbonate should be usable for at least a few days' use.


text:
How to explain Gibbs free energy is a pressure-dependent state function?
I'm aware that through deriving Gibbs free energy to infinitesimal changes, we could get the formula: $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$, giving that Gibbs free energy is pressure-dependent.
However, while dealing with the definition of Gibbs free energy: $ΔG = ΔH - TΔS$ (and $ΔH$ must be held at constant temperature, am I wrong?) most textbooks stated that it is held under constant pressure.
I still couldn't understand the relationship between Gibbs free energy and pressure and why: $\mathrm dG = V\,\mathrm dp$.

$G$ is a function of both pressure and temperature. Only at constant temperature is $dG=Vdp$. At constant pressure $dG=SdT$.

$G$ is defined as $H - TS$, and $H$ is $U + PV$.
Since in closed system you have: $\mathrm{d}U = T \mathrm{d}S - P \mathrm{d}V$, it follows that:

$\mathrm{d}H = T \mathrm{d}S + V \mathrm{d}P$
$\mathrm{d}G = -S \mathrm{d}T + V \mathrm{d}P$


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AB5E type molecule
In $\ce{AB5E}$-type molecules, why are the lone pairs in axial bonds?

If the lone pair is present in axial bonds it repels four other bonds. On the other hand if the lone pair is in equatorial bonds, it repels only two other bonds.

So what is the reason behind this?

There are no axial bonds here.

Why? Because there is no other choice. Starting from the AB6 octahedral configuration, all six vertices of the octahedron are symmetric, so it doesn't matter whichever one you “choose” to replace by the lone pair. All will yield the same final configuration.


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What were the tests to determine that a substance is an element or not?
What tests were conducted in the ancient days to determine whether a given substance was an element, say, how could one say that mercury, gold, silver, sulphur, phosphorous were elements?

I think you are projecting a modern concept (element) onto a not-yet-scientific period.

Not necessarily, but he should not use misleading term ancient for 17th-18th century.

1669 is the year of the discovery of Phosphorous attributed to Brand as the last discovery of the elements mentioned by you.  Which is prior to modern chemistry, yet not ancient.

by ancient is meant when modern tools were not available, say 1669 as mentioned by Buttonwood concerning Phosphorous. Regret use of 'ancient'


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What chemical could dissolve low-density polyethylene?
What chemical could dissolve plastic aside from acetone? Preferably, low-cost and environmentally friendly. Any suggestions would be helpful.
I'm targeting for thin plastic (low-density polyethylene). Environmentally friendly for me is that it follows the EPA standards from the USA. I'm targeting for around 150 USD. I'm okay with solvents that make the plastic swell, but a chemical that could dissolve it is preferable.

150 USD for... what? Per liter, per tonne, per day? Can you say anything at all about what you're trying to do?

The safest solvent might be supercritical carbon dioxide, $\ce{CO2}$. Supercritical carbon dioxide is also used in "green" dry-cleaning. Use for dissolving polymers requires a pressure vessel for ~5 to 50 MPa, able to work at ~40 to 100°C. There are (paywalled) articles at ACS and NCBI, if you want to investigate further.
According to Agilent, 1,2,4-Trichlorobenzene (TCB) is effective.
Some other solvents mentioned at Research Gate include toluene and xylene.

N.B. The aromatic solvents all are toxic, some quite so, and are volatile and flammable. All of them require temperatures near the melting point of polyethylene, ~100°C, to dissolve much of the polymer, so a reflux or pressurized system is required.
Supercritical $\ce{CO2}$, while not very toxic, can cause asphyxiation above ~5%, and the container is, effectively, a bomb containing a warm or hot pressurized fluid.

Thanks!! I'll look into that chemical!!


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Equivalents Triethylamine in Swern-oxidation

Hey folks,
so I've been wondering why the use of 5 equivalents of Et3N in this case of a Swern-oxidation is necessary. I have done this reaction succesfully quite a few times now, but all the literature uses 5 equivalents as well. I am fully aware of the mechanism (shown below). I can explain one eq. for the HCl that emerges in step one:
And in the second step we need the base to form the sulfurylide:

but this should not be a equilibrium as the Ylide has no net charge and in my opinion is more stable.
If it were to just enhance the reactivity a bit, I'd think that 2 more eq. (so 3 eq. in total) are sufficient.
I hope you can help me with this.

Have you tried the reaction with less equivs?  Do you need extra to raise the pH especially in the presence of carbon dioxide?  Is some trapped as a hemi-aminal?  Just guesses ...

3 equivalents of Et3N is stoichiometric. If the ylid is reversibly protonated by Et3N-HCl, why not use excess Et3N? It certainly isn't detrimental. Note that each successive reagent is in excess.

If isn't broke don't fix it! Et3N is cheap.


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How can I calculate the equilibrium between several reactions?
If I mix xNaOH + yCdCl2 + zZnCl2 in an aqueous medium, how can I calculate the final equilibrium quantities of the resulting salts (which I suppose may be the following):
Q1 NaOH
Q2 CdCl2
Q3 ZnCl2
Q4 NaCl
Q5 Cd(OH)2
Q6 Zn(OH)2
Also (only if it will take more than an hour for >99.999% of the reaction to occur), how can I calculate the respective quantities of each vs time?
Suppose the reaction occurs at ~25C, unless a better prescribed temperature is suggested.

Don't think in terms of salts (which, BTW, half of these aren't), think in terms of individual ions and their concentrations. Write down all conservation laws, solubility equations, etc. See where this gets you.

Is this in the solid state or in aqueous solution? Also, the x, y, z are meant as initial amounts or concentrations, not as stoichiometric coefficients, right?

@KarstenTheis This is in aqueous solution (I didn't think of specifying but now corrected thanks to your comment). Your assumption is correct, x, y and z are the initial amounts (I probably could have written it more clearly as Q1(t=0), and Q1(t= n), Q1 (final) or something the like).


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Why are carbonyl groups so important

The most important are π orbitals as they are lower in energy than σ orbitals and the carbonyl group (C=O) is the most important of these—indeed it is the most important functional group of all. ~Clayden, Chapter 5

Why is it that Carbonyl Groups in organic chemistry are given so much importance? I am still in highschool, and at this level every book (like Peter Sykes, Solomon, Clayden) or my teachers, all talk about how Carbonyl compounds are the most important. But why is it? Why are they given so much importance?

Carbonyl compounds are the bridging for the synthesis of almost all functional groups in organic chemistry via one or the other organic reaction

I don't think there's a definitive answer to this, apart from @YUSUF's comment. Why don't you complete your high school organic course and self-realise its due (or undue) importance?

@DivMit — Actually, all chemical moieties are important in organic chemistry (OC), and that is the base of OC. The carbonyl group (found in ketones or aldehydes) is especially reactive because of the difference in electronegativity between C and O. But to be honest, your point of view will change when you learn more about phosphorus or silicium!

Carbonyl groups participate in many (the majority?) of biochemical reactions

Many ancient cities rose to prominence as important crossroad points. Same thing here.


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Dangers of using activated carbon past its lifespan
Activated carbon (AC) is one of the most widely used media for water filtration. It's effective at removing chlorine taste and odor, and it's recognized as being safe.
However, most activated carbon filters have an indicated lifespan of about 4-6 months, after which they need to be replaced. But is there any risk to using an AC filter past its lifespan other than the fact that it won't effectively filter contaminants anymore?
I've read in passing that an AC filter past its lifespan can actually release contaminants in water, because some contaminants in the influent water may be preferentially adsorbed by the AC, replacing existing contaminants previously trapped in the AC and causing them to be released in the effluent water. Is this something that has been explored in research?
In particular, has it been shown that using a saturated AC filter can cause the levels of certain contaminants — e.g. a trihalomethane such as bromoform, or perhaps fluoride or lead — to rise to a level that is well above the EPA maximum contaminant level?

The only danger I see is that the activated carbon isn't doing its job right, so you've basically answered your own question in the last paragraph. As for research: you will find plenty of articles on preferrential adsorption at the expense of desorption.

You may also get fine carbon particulates working their way through the filter medium with time.

Every adsorbent has a saturation capacity depending on its active surface area. The idea to replace carbon filter is not only due to the saturation capacity but imagine fungi, bacteria may be adsorbed as well. It is worth replacing.

Thanks. I understand that it's well worth replacing, but what I'm wondering is how real the risk is of contaminating the water with previously trapped compounds such as trihalomethanes, or perhaps even heavy metals like lead which may have a small affinity to AC.

This is a difficult question, because you have to see what is the equilibrium constant of adsorption of trihalomethanes on AC and in water. Assume an activated carbon sample which is saturated with trihalomethanes. When you pass ultrapure water through a column of AC, depending on the equilibrium constant, you might be able to detect trihalomethane in that ultrapure water.

Is this something that has been explored in research?

Definitely yes, and the keywords you're looking for are: adsorption, desorption, and fluid transport in porous media. Activated carbon is a nanoporous material with high specific surface area, and the kinetics of adsorption, desorption and transport in liquid phase in such media are well understood from a fundamental point of view.
You can look at papers, reviews and book chapters like this one to look more in depth into the questions. At the macroscopic scale, the behaviour of the complete filter can be predicted by equations of fluid dynamics, local thermodynamic equilibrium, and first-order kinetics.

Upvote. I'm not sure why someone downvoted your answer... I understand the mechanics of adsorption and desorption, but is it realistic that a saturated AC filter could cause the effluent water to have unsafe levels of lead, for example?


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Will cyclopenta-2,4-dienecarbaldehyde undergo aldol condensation or cannizzaro reaction?
The following question was given in an exam.Which of the following undergo aldol condensation ?
Given answer :  a,b,c

My answer: b,c
My Logic :
Cyclopenta-2,4-dienecarbaldehyde(1) on reaction with a base gives carbanion whose resonance structures are 1a to 1e as shown below.This anion is an aromatic anion .Being stable ,tis aromatic anion now in oresence of base would undergo Cannizaro Reaction to give 3 and 4.

My question is, will cyclopenta-2,4-dienecarbaldehyde  undergo aldol condensation?

Well the CH group between 2 double bonds and a CHO group seems very "deprotonable" to me. But this is just to make your imagination "bubbling" !!

Hi, please don't make new tags - although this is something the system allows you to do with a handful of reputation, on Chemistry.SE we don't usually do this without some sort of consensus on Meta. In this case, there were already appropriate tags to describe your question well. If you're unsure you can always ask somebody what is a good tag. Also, please avoid the all caps.

Somewhat ambiguous in the wording of the question. Presumably, we're talking about self aldol?

Anyway, the only legitimate reference I can find to the starting material in the literature is this https://onlinelibrary.wiley.com/doi/abs/10.1002/jlac.19646780107, where it's made as a mixture of isomers/tautomers, so it seems that (as has been the case before) this is a made-up question which cannot be seriously answered. (Reaxys turns up a few more hits for the compound, which upon closer inspection are revealed to be typos.)

@orthocresol ,this question was given to students preparing for IIT engineering exam in my city during a weekend test.My point is ,since carbanion is involved in aromaticity, it would remain stable,and therfore may not participate as enolate. And there may no be an aldol.

I think the formation of an aromatic anion does increase the favourability of forming it, lowering activation energy of the 1st step of the reaction, allowing it to take place more quickly. But of course, it should not end there because there is additional stability to be gained from the aldol condensation. However, perhaps, the aldol may take take place at the alpha carbon and it may occur at another carbon in one of the other resonance structures, to allow for the elimination of water in the condensation.

@orthocresol ,I didnot know I created  a tag .I have no intention. Is there any  mechanism to untag, please let me know.

@Zhe , I edited the question.Which path will the molecule take,an aldol path or cannizzaro path , given ,the intermediate is aromatic. Would it loose aromaticity to again form energitically higher non aromatic species to give aldol or remain aromatic ,in which case lead to cannizzaro reaction  ?

@Tan Yong Boon , great idea.However , my point is , aromatic carbanion is stable , and therefore may not undergo self aldol.If carbanion occurs at different point ,then, this carbanion would loose aromaticity, in which , itwould be energitically be more abd aldol self condensation may not take place.


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What is the major product of the reaction of 2-methylbuta-1,3-diene with HBr?
What is the product of the reaction of 2-methylbuta-1,3-diene (isoprene) with HBr?

What I understand is that carbon (2) will get a positive charge after protonation.
But after this, which should be the major product?

3-bromo-3-methylbut-1-ene – because in this case the carbocation would be stable because of being tertiary, but the double bond would have less number of hyperconjugated structures and will be less stable

1-bromo-3-methylbut-2-ene –  as the carbocation would be less stable because it would be primary but the double bond would be much more stable because of 8 $\alpha$ hydrogen available


text:
Doubt on process to calculate pKa
I have to solve an exercise that I've been trying for a long time without success.
Information that I have:

pH @ certain temperature of an aqueous dissolution of an acid
molar concentration of the acid

Unknown:

approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

So far I tried to express the chemical equation and its two steps to finally get $\ce{H3O+}$ and an oxid.
I've calculated backwards the Hydronium concentration $[\ce{H3O+}]$ with the $\mathrm{pH}$ formula. $$ \mathrm{pH} = -\log([\ce{H3O+}])$$.
I've written down the formula to calculate $K_\mathrm{a}$ from where I expected to be able to calculate $\mathrm{p}K_\mathrm{a}$.
I don't know how to extract now the concentrations of every component on the $K_\mathrm{a}$ formula in order to be able to calculate that. Moreover, what am I suppose to do with the temperature?
I hope you can give me guidance in the process to solve such exercises.

approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

The first approximation: the equilibrium concentrations of $\ce{H2SO3}$ and of $\ce{HSO3−}$ are controlled solely by the
equilibrium
$$\ce{H2SO3 <=> H+ + HSO3−}$$
Because the first $K_\mathrm{a}(K_\mathrm{a1} = 0.017$) is much larger than the second$(K_\mathrm{a2}=1\times{10^{−7.19}})$
The second approximation:$\ce[\ce{H+}] \approx [\ce{HSO3-} ]$

Say your acid is monobasic: suppose its formula is $\ce{HA}$.
So, it dissociates according to following equation:
$$\ce{HA + H2O <=> H3O+ + A-}$$
Suppose its molar concentration (given) is $c$, and it dissociate $\alpha$ amount in water (assume total volume is $\pu{1.0 L}$) at given temperature at equilibrium. Thus, equilibrium concentrations of $\ce{HA, H3O+}$, and $\ce{A-}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively.
$$\therefore \; K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{\alpha \cdot \alpha}{c-\alpha}$$
Since $\alpha = [\ce{H3O+}]$, $-\log \alpha = \mathrm{pH}$ (which is given) and you know $c$, which is also given, you can estimate the $\mathrm{p}K_\mathrm{a}$ (and hence $K_\mathrm{a}$ if you needed).

Late edition:
Take (-$\log$) on both side of above equation. Thus:
$$\mathrm{p}K_\mathrm{a} = -2\log \alpha + \log{(c-\alpha)}=2(\mathrm{pH})+ \log{(c-\alpha)}$$
Most of the time (in cases of $c\gt\gt\alpha$),  $\mathrm{p}K_\mathrm{a} \approx 2(\mathrm{pH})$

Thank you very much. Now I am gonna insert the values and show you what I get. I have no clue if it makes sense or not.
I start with an aquous dissolution of $\ce{H2SO3}$. I know it will dissociate in two steps, however I read in a book that the first dissociation is normally the strongest and almost every time you can easily ignore the second one, so I proceeded like it follows:
$$p\mathrm{H} = -\log([\ce{H3O+}]) = 1.1$$
This value was given. Then we know that the dissolution has a concentration of $0.5\,\mathrm{M}$. Thus:
$$\ce{H2SO3 + H2O -> H3O+ + HSO3-}$$
$$\ce{c - x -> x   +x}$$
$$K_a = \frac{[\ce{H3O+}][\ce{HSO3-}]}{[\ce{H2SO3}]}=\frac{x·x}{c-x}=\frac{x^2}{c-x}$$
I don't know how to determine this without a calculator. Is there an  approximation for this?
$$K_a = \frac{10^{-2.2}} {0.5-10^{-1.1}} = 0.015$$
From this, and using the expression by the last equation:
$$pK_a = -2 \log(x) + \log(c-x) = 2.2 + \log(0.5-10^{-1.1}) = 1.82$$
Does that make any sense to you?

Welcome to ChemSE.  While retaining the numeric data provided by you, I worked in markup formatting your answer.  Be invited to learn some of the techniques accessible thanks to mathJax and mhchem (mhchem thankfully is available as assisting usepackage in LaTeX, too) for which an outline is provided here: https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here -- some of it (e.g. $\ce{2 H2O <<=> H3O+ + OH-}$) equally works well in comments here on ChemSE and is quite easier to use.


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Difference in major product of sulphonation of phenols
When phenol is sulphonated by fuming sulfuric acid at a lower temperature, the ortho product is the major whereas at higher temperature the para product is formed as the major.
Why does this happen, because looking at the reaction mechanism, it appears to me that para should always be the major.
Is it the kinetic control vs thermodynamic control effect?? If yes, then why is para thermodynamically more stable and ortho kinetically faster??

Check this

It's because at lower temperature H-bonding exert extra stability to ortho product while at higher temperature H bond get break so stability of o-product become less than para product


text:
Chlorine gas and stainless steel
After dealing with a leak from a 73 kg compressed chlorine gas cylinder at 6 bar, there was green powder/rust/discoloration on the stainless steel tubing similar to copper chloride or copper sulphate. Any thoughts of what this might be?

It is likely to be compounds of the main elements in stainless steel: Ni, Cr and Fe. Special alloys are often used when corrosive gas exposure is expected, e.g., Monel. In your case, the exposure was accidental.

Thanks Ed V ,  but would any of these have the chractrustic blue/green of Copper Chloride/sulphate ?

All three of the metals have some green compounds. There are numerous varieties of stainless steels, e.g., 316 and 304, but I don’t know what stainless steel you are dealing with. I will check to see if I can find the colors of the chlorides of Ni, Cr, and Fe. But I am guessing someone here will post an actual helpful answer. If I find anything useful, I will post as a comment. If an actual helpful answer is posted by someone, I will delete my comments: I think (I am still new here) that comments are supposed to go away sooner rather than later.

BTW, some stainless steels that are resistant to dry chlorine are attacked by moist Cl2, so check for any water in the environment.

From wiki, the hydrated chlorides are various greens. Also, wiki says HCl attacks all varieties of stainless steel and contact should be avoided. I know this is correct from my lab experience. When chlorine gas encounters water vapor in the air, some HCl gas is formed, so I think you have some hydrated chlorides. Hope this helps!

@DrMoishe Pippik Excellent point! Sadly, though, my lab environments were never dry! When it rained hard, I even had rain in one of my hoods!

I will post some images tomorrow to illustrate - but thanks for the I input so far


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What materials reversibly swell up slightly when in a moist environment?
I am aware that some "Super Absorbent Polymers" can swell up enormously when they absorb water. However, I am looking for a material which swells up (maybe about 30% to 100%) when in a moist environment, and which will contract again when moved to a less moist environment.
I.E. by measuring the expansion of the material, one could determine the moisture content of the environment the material was in. It would act like a moisture measurement device.
Are there any such materials? Perhaps there are some "Fairly Absorbent Polymers"

Superabsorbers are perfect for that. They swell by an order of magnitude or more in (distilled) water, but in moist air, you surely stay below a factor two.

@Karl - Thanks. And is the effect reversible? Moved back to les moist air, will they reduce in size again?

Quite reversible, if you keep the swollen material at low stress.

A classic hygrometer consists of bundle of a dozen or so human hairs (blonde are supposedly best, because they're typically thinnest, i.e. react fastest to humidity changes), stretched between a metal spring with a pointing device and a fixed point.
A uniaxial effect is much easier to record precisely, and hair is thin (=fast) and far more mechanically reliable than a piece of swollen hydrogel.

Nice answer +1!  That's the Hair tension hygrometer one of the possible classical hygrometer, right?

@GM I think that's the classic version. What others are there?

I know there is also  the metal-paper coil type which is one of the classical cheap hygrometer http://hygrometer.net/types-hygrometers-multiple-uses/

Many hygroscopic materials have this behaviour but sometimes they might have some "memory" of the previous swellings cycles, or the swelling might vary depending on the arrangments of the polymers chains, making them often not so reliable for such a device. You might want to do some testing on Nylon which is hygroscopic (more detail in the link). Also, Polycarbonate and Polyethylene Teraphthalate are hygroscopic and swells due to RH fluctuation. I personally worked on natural polymers used as artworks materials these are some results I collected on parchment and paper samples.

Monitoring microclimate-induced deformations on hygroscopic materials
  using conoscopic holography sensors , Paper 11058-37, SPIE2019, Munich
  to be published


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Is “mass-action ratio” identical to “reaction quotient?”

Mass-action ratio
Reaction quotient

According to their definitions expressed in the above Wikipedia articles, these two concepts seem to be totally identical. Is that true? If not, what is the difference?

The concepts are extremely closely related.  The difference is that the mass action ratio is a formally a ratio of concentrations, or, for gases, of partial pressures.  The mass action ratio, and the term "mass action" in general, refers to situations where the chemical potential of the reactants and products is directly proportional to the concentration, or pressure, or mass, of the species involved.
The reaction quotient is is a tiny bit more general; the implicit assumption of mass action is not necessarily made.  The reaction quotient is the ratio of chemical potentials of reactants and products.  It works even in systems that don't follow the law of mass action, i.e. where the activity coefficients vary during the equilibration.
For all intents and purposes, the two concepts are the same.  If you want to split hairs, you'd probably do it the way I did here.


text:
How are phase transition diagrams calculated?

How are diagrams such as above plotted? Since empirical studies are slightly off from the 'ideal', what is the source for the phase plots?

That particular diagram is from Thermo-Calc (https://www.thermocalc.com/ - the triangle symbol in the lower left indicates the program). Thermo-Calc implements thermodynamic models based on Calphad (Calculation of Phase Diagram) principles. One selects appropriate models for the Gibbs free energy of the phases, takes the known experimental data, and fits parameters of the model such that the calculations 'best fit' the data. There is a journal named Calphad which has a lot of literature (and other journals) and several books and conferences on the topic.

I see. Are models on it citable or is it more of a toy-model deal?

Searching Web of Science for topic "Calphad" results in 3202 results. It is a (growing) field of materials science, with many applications. There is a lot of good thermodynamics, crystal structure, and more applications of first-principle (e.g. DFT) to get at relevant thermodynamic parameters. So, no, it is not a toy model thing (or hasn't been for the past 30 year or so). I have several answers on Chemistry that use Calphad results to explain what is going on in binary phase diagrams.

https://chemistry.stackexchange.com/questions/19454/boundary-lines-in-phase-diagrams-and-the-lever-rule/19468#19468 might help visualize how the set of Gibbs free energy curves of the various phases ends up in a binary phase diagram.

For starters see: Clausius Claperon


text:
How to calculate the amount charge transfer from a natural population analysis?
I am trying to reproduce the quantity referred to as the degree of charge transfer reported in this publication by Zhu et al. [Ref. 1] for the F4-TCNQ molecule. I am interested in section 3.2 in the main text about using the bond length as an estimate for the charge transfer amount.

In figure 3b, the authors show the correlation between the degree of charge transfer, CT$_{pop}$ from their natural population analysis compared to the approximation of the bond lengths of F4-TCNQ from equation 1, shown below.
$$\rho = \frac{\alpha_{CT}-\alpha_{0}}{\alpha_{-1}-\alpha_{0}}$$
with $\alpha_i = l_3 (l_2+l_4)$ where $l$ is the bond length, and subscript $i=0, -1, CT$ denote the netural molecule, the anion, and the compelx, respectively.
It's not clear to me, how the authors determined the values on the y-axis in Figure 3b, and there are no further clarification or references to a methodology for the values within the text or supplement information. Both axes in Figure 3b use the F4-TCNQ anion as an arbitrary reference.
I can calculate $\rho$ from the above equation, but I want to compare $\rho$ to the CT$_{pop}$ at the level of theory I am employing, verifying the accuracy of my calculations.
Through reading and googling, I believe the CT$_{pop}$ refers to a population analysis such as a Mulliken or natural population (AKA Lowdin population), referenced in Mendez et al. [Ref. 2] (bottom of page 9 in the Theory footnote). I do not know how to obtain the degree of charge transfer from the resulting population analysis.
I have one reference from Joo et al. [Ref. 3] who gives more details on a similar method, writing on page 2, left column:

The calculated atomic charges were then used to obtain the degree
of charge transfer between the donor and acceptor molecules,
which may be defined as the sum of all atomic charges on the
donor part of the complex.

The above text leaves me to believe the method of determining the degree of charge transfer is as follows:

Calculate a population analysis of the DA (donor and acceptor) complex to determine the atomic charge
Determine the atomic charge of each of the fragments D and A by summing the atomic charge for the D and A fragments, respectively
Report the atomic charge of fragments as the degree of charge transfer.

The last aspect is Joo et al. [Ref. 3] reports the value for the electron donor while Zhu et al. [Ref. 1] indicates the value for the electron acceptor, which is F4-TCNQ.
Does this procedure sound reasonable? Any thoughts or comments would be appreciated.

TL;DR
How can I reproduce the values shown in Figure 3b on the y-axis, the degree of charge transfer from a natural population analysis?
References

Zhu L; Kim E.-G; Yi Y; Bredas J.-L. Charge Transfer in Molecular Complexes with 2,3,5,6-Tetrafluoro-7,7,8,8- Tetracyanoquinodimethane (f4-Tcnq): A Density Functional Theory Study. Chemistry of Materials 2011, 23 (23), 5149–5159 DOI: 10.1021/cm201798x.
Méndez Henry; Heimel, G.; Winkler, S.; Frisch, J.; Opitz, A.; Sauer, K.; Wegner, B.; Oehzelt, M.; Röthel Christian; Duhm, S.; et al. Charge-Transfer Crystallites As Molecular Electrical Dopants. Nature Communications 2015, 6 (1) DOI: 10.1038/ncomms9560.
Joo B; Kim EG. Model-Independent Determination of the Degree of Charge Transfer in Molecular and Metal Complexes. Chemical Communications (Cambridge, England) 2015, 51 (81), 15071–15074 DOI: 10.1039/c5cc05274b.

I will evade answering the main question because I am not feeling expert enough, but I want to point out that Natural population analysis (NPA) is very different (and better) from the Loewdin one. Loewdin is computed from orbitals of a special basis set (although it comes from the density, it is still free-atom-derived basis set which is used to obtain the density matrix). NPA (part of NBO) is from density matrix which is adjusted to the particular molecular environment, hence better representing the "real" atomic population. NPA is available in the commercial NBO program, old version of which is also included with Gaussian package (maybe also in other packages, I don't know). You can also try AIM charges which are absolutely basis-independent (if Your density is good enough).
Personally I would think the analysis You proposed is fine with NPA or AIM charges, but I state again that I am not an expert in the topic.
See https://www.q-chem.com/qchem-website/manual/qchem43_manual/sect0061.html and also http://www.huntresearchgroup.org.uk/teaching/teaching_comp_chem_year4/L7_bonding.pdf .

Related: How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory? Please note that the NBO version included in Gaussian is long deprecated; worst of all, results are not consistent with newer developments, and there are studies that point out why. I strongly recommend not using anything earlier than NBO6.

Thank you @Igor Mihailovs, I didn't realize there was a difference between NPA and Loewdin. Does AIM stand for Atoms In Molecules by Bader? I don't have much experience with either an NPA or AIM analysis. I have been playing around in Multiwfn link to the site, and I did recognize AIM analysis implemented.

Yes, AIM is for "Atoms-In-Molecules". Actually, Gaussian seems to also have some AIM-related optionas computable (look at IOp(6/35) options). But I have, too, never used it, so I cannot comment further. Just running IOp(6/35=2) for AIM charges produces no result.

@Martin, I think that this essentially means we should use AIM? In addition, NBO7 is not for free…

@igor I think there is a free implementation of NPA, try looking up janpa. Multiwfn can do aim charges, it's actually easy and well explained in the manual. NBO is - in my experience - well worth the money, more so than Gaussian for example.

I am not 100% certain, but I read on the CCL archive that AIM in Gaussian was unstable and therefore deprecated/never fixed. Also on the matter at hand, I am not familiar with the literature on the subject. While AIM has a rigorous condition for charges, it is a more or less arbitrary one. It doesn't hurt to look at other decomposition schemes to find whether they agree.

@Martin-マーチン, I see now, thank You! My comment related to the price of NBO was from the viewpoint that if one needs just one of the many functionalities of the program (the charges), it is hardly worth it buying the whole suite. On the other hand, maybe these many features will turn out needed when they are available :)

As a little off-topic: do You think Gaussian is worse for SCF-based procedures than some other commercial code? Prices are usually in the same range...

Already pointed out by @IgorsMihailovs, NBO isn't free, and my university doesn't own a copy. I am going to, work worth the AIM in Multiwfn and see if I can obtain similar correlations as shown in the main post.


text:
How do fuel/oxidizer combustion mechanics change when the constituents are supercritical?
The problem to solve is this: I want to create a heat generator [or heat absorber] that functions at atmospheric pressure to the pressure at the bottom of the sea. It will be used to drive a Stirling engine.
As such, I'm worried that a simple burner with adjustable flow rates won't act predictably at extreme pressures.
Givens:

1 atm - 1000 atm ambient pressure
I can maintain a constant differential pressure between fuel/oxidizer bottles and the reactor
1 kW output power (small in scale)
Nitrous oxide/hydrocarbon (probably propane) constituents [edit: not set in stone]

Will the combustion chemistry work out? Is the reaction so fast that stable combustion is impossible at low flow rates?
I am also reading up on supercritical hydrothermal combustion as an alternate solution. I have contemplated using liquid nitrogen as a heat absorber, but it is appx 1/100 mass-efficient in terms of energy input/output. Alternate solutions that avoid the problem entirely are also welcome. <3

Pu238 would work well at any pressure (short of being at the center of shaped charges), and would not need to be refueled frequently, but it would be difficult to get... not even NASA has a steady supply.

Haha! That may get some eyes on my pet project that I'd rather not attract... That would be an excellent solution, however ;)


text:
Gauche conformations for 14-electron AH2BH2 systems
The hydrazine molecule, and the similar sulfonium/Oxonium ylides, all have a gauche conformation that is more stable than the anti conformation.
Is there any particular reason for this pattern? Orbital Interactions in Chemistry tries to use overlap between the 2 “lone pair” orbitals on the carbons as a justification (since it’s a 2 center 4-electron interaction).

They don't try to use that, they do it, and it does explain the structure. I think the ordinary organic chemist (and probably most of the other chemists, too) would call that negative hyperconjugation.

https://doi.org/10.1063/1.1749518 https://doi.org/10.1021/ja00762a030


text:
How electric field work inside a battery?
Suppose Beaker A contains Zn rod submerged in zinc sulphate solution.At the electrode/electrolyte interface ,since the force felt by the solvent molecules and ions are anisotropic because of the electrode phase, the electrical neutrality breaks down on the solution side,orientation of solvent dipole take place and a net charge develops in any volume element in the solution phase near to the electrode.This creates a net electric field across the phase boundary which result in the motion of electrons on the electrode according to the field. Suppose the net charge in the solution phase is positive and induces a equal amount of negative charge on the electrode.This induction of negative charge on electrode results in some of the Zn2+ ions who have shed the electrons to leave the electrode and enter the solution resulting in a excess charge in the bulk of the solution.To maintain electrostatic condition this excess charge reside on the surface of electrolyte.The same process happens in Beaker B containing Cu and Copper sulphate solution,but the induced charge on Cu electrode is less negative compared to Zn and the excess charge residing on the electrolyte surface is also less compared to Zn.Since Zn electrode is more negative ,the direction of field lines is from Cu to Zn.On the electrolyte Since Zn electrolyte is more positive the direction of field is from Zn to Cu.Since the both field is equal and opposite they nullify each other .Thus net field is zero.If this is the case then how come electrons moves when we hook up a wire across this electrode.Can I solve this problem just by connecting a salt bridge.Or is there any flaw in the explanation


text:
Investigating the effect of solvents (protic vs aprotic) on the alkaline hydrolysis of ethyl acetate using a conductivity meter
What different areas about solvents can be explored with the conductivities collected of the alkaline hydrolysis of ethyl acetate in propanone and ethanol?
$$\ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH}$$
Currently, I am looking at the effect on activation energy, rates, and rate constants and the overall conversion of ethyl acetate to sodium acetate at the end of the ten minutes of the reaction in the two solvents (to see which is more favorable).
Will I be able to integrate transition state theory and Gibbs equations in any way possible?


text:
Bohr's model of an atom
This question is not very mighty. But I'm having a problem wrapping my head around the Bohr's atomic structure. The textbook says electrons revolve around the nucleus in fixed energy shells, with each shell having its different value for energy. What I want to know is, what does it mean for a shell to have an energy level? Where does the electron get the energy to revolve around the nucleus from? And when the electron jumps from a ground state to other shell, after gaining energy, why does it have to return to its original shell? I mean it has satisfied the criteria of going into the next shell, so why does it return.
And my main concern is the energy part. Where does electron get the energy to revolve in a shell.
If possible, please refer me a source to study this in depth.

Forget quantum stuff for a moment. Forget electricity altogether, it's complicated. Say I'd lift a stone and quickly release it, so for a brief moment there will be just a stone in the air. Being at some height above the ground, it must have some potential energy ($mgh$, they say). Where did it get it from? And more importantly, why does it have to return to the ground, if it has the energy?

Atoms were formed whatever mechanisms starting from more energetic states. The issue "fixed" by the model is that of discrete abs/emiss lines and finally of stable atoms existence. For the " why electrons return", it is the same as throwing a stone up.....

@IvanNeretin If the nucleus is attracting the electron, just as your analogy, the earth attracts the stone, there must be a constant supply of energy to the electron to keep it from falling into the nucleus.

@IvanNeretin Where does that energy come from?

Congratulations, you are precisely at the point where the whole world's physics stood just before the Bohr's model, and I'm not mocking or joking. It is a genuine problem, and a huge one at that. Bohr solved it (sort of) by postulating the existence of certain energy states. The electron can only exist at one of these states, and consequently, can't fall "below" the lowest state, much like a stone can't fall through the ground. (That's why it is called ground state, BTW.) Admittedly, the explanation was rather laconic and pretty much boiled down to "Because I said so, that's why".

@user231094 Even if you accept the Bohr model that electrons rotate around the nucleus (and it is a very simplistic model that was rapidly rejected by science), they don't need a constant input of energy to do it. Nor do planets evolving round the sun. They revolve because they have momentum. They keep revolving until something changes that momentum by adding or subtracting energy from the system. Same in Bohr's model. Electrons stay in the same orbit with constant momentum until energy is added and they move to a different orbit.

1. what does it mean for a shell to have an energy level?                                                2. where does the electron get the energy to revolve around the nucleus from?            3. when the electron jumps from a ground state to other shell, after gaining energy, why does it have to return to its original shell?

Think of the planetary system which inspired Bohr to think about his atomic model. The Earth is rotating around the Sun in a fixed orbit. What keeps the Earth rotating around the Sun? The mechanics of the planetary motion, and other electrical phenomena were very well understood in Bohr's time, so much so that by late 1880 to early 1900 a physics professor advised his student, who would become a Nobel laureate later, not to study physics because there is nothing left to do in physics
...here might still be a little dust or bubble at one or the other angle to check and classify, but the system as a whole is quite secure, and theoretical physics is noticeably closer to that degree of perfection that geometry has had for centuries..
"Max Planck: Wege zur Physikalischen Erkenntnis."
So classical electrodynamics (motion of charges) predicted in Bohr's time that a moving negatively charged body should spiral into the positively charged proton. Obviously this was not happening. Secondly, a moving charge should emit electromagnetic radiation because the great Maxwell had already established that.
Bohr had a moment of enlightenment. He just said let us say that the electron does not spiral into the nucleus but stays in a fixed orbit nor it emits radiation. Lo and behold, that assumption helped him explain most the hydrogen atomic spectrum in the UV, Visible, and in the near infrared. This was a big big feat! That is why when you read Bohr's theory, his assumptions are called Bohr's postulates.
What is a postulate in logic as per a dictionary's definition? "A fundamental principle, presupposition, or condition, esp. one assumed as the basis of a discipline or theory; (also) a proposition that is (or is claimed should be) taken as granted; esp. one (to be) used as a basis for reasoning or discussion, a premise."
"Taken for granted" is the keyword, which means when you are reading my theory, accept my assumptions that they are true-don't question them.

... my main concern is the energy part. Where does electron get the energy to revolve in a shell?

M. Farooq's answer makes the analogy with planetary motion. Continuing the analogy, where did our planet earth get the energy to revolve around the sun? We don't really know, but we know that there is potential energy (gravitational field) and kinetic energy (the movement of the planet). The gravitational pull is just strong enough to keep the planet in its orbit (accelerate it towards the sun so that it stays on its curved orbit), so we could say kinetic and potential energy are matched. The classical picture of the electron bound to an atom is exactly the same, with an electrostatic field instead of a gravitational field.

what does it mean for a shell to have an energy level?

Every planet in the solar system is on a separate orbit, with a different potential energy. It would take a different amount of energy to remove each planet from the solar system (even if they all had the same mass). The new thing about Bohr's model (mentioned by Ivan Neretin in a comment and by M. Farooq in an answer) is that the energy is quantized, i.e. only certain energies are allowed. This was known in Bohr's time from the hydrogen spectrum, but there was no theory yet to explain it.

when the electron jumps from a ground state to other shell, after gaining energy, why does it have to return to its original shell?

See this question at Stackexchange Physics: https://physics.stackexchange.com/questions/110431/why-do-electrons-in-an-atom-fall-back-to-the-ground-state. It does not have to return (at very high temperature, there is a mixture of ground state and excited states), but usually it does, and it is a combination of quantum mechanics and thermodynamics.

The trick is a very basic physical intuition error - one that one should not be blamed for, because it took the likes of Galileo and Newton to realize what was wrong with it, and as an idea it held as orthodoxy for ages and across the world prior to the innovations of their genius. It's much like the Flat Earth - there's actually no blame in entertaining the idea, it only gets silly today when you insist on refusing to capitulate to the evidence and reasoning behind why it cannot be so under the guise of finding the belief in some giant "conspiracy" more appealing on an emotional level. And the physical intuition error here is the idea that in order for something to be moving and stay moving, it needs to have a "power source" - something continually inputting energy into it, all the time, to keep it going, and when that power source runs out of energy, the thing will stop moving.
This, however, is not how our universe actually works to the best of our knowledge, though there's no reason a universe couldn't work this way. Instead, motion, in fact, while it can be and is associated with an energy, called the kinetic energy, but this energy, once imparted to an object, will stay with it until something comes along to take it away, meaning that an object left to its own devices and set into motion, will always keep moving, as far as we can tell.
The reason why it was thought for a long time that motion requires a continuous input of energy is from naive observations of motion in the environment on the surface of the Earth. Here, if you throw a rock, say, it eventually stops. It hits the ground and skids to a stop. You can push a wheelbarrow, get it going, then let go, and it soon stops, if not stopping immediately. Indeed, you can't very easily find something on Earth that won't just move and keep moving of its own accord. But today, we understand this to be the result, not of motion being inherently unsustainable, but rather because there are in fact agencies in the environment which are acting to rob energy from the moving object, like air drag and frictional forces (both the rock and barrow are subject to these). Thanks to these, we have to continually resupply energy to the moving object with some sort of engine to maintain the motion, hence why your car is always burning fuel as you are driving.
In lieu of such forces, though, these items would continue to move, and we can demonstrate this by considering identical objects given identical initial impetuses in contact with differing kinds of surfaces. When we do that, we find that the time it takes each to stop depends on the type of surface, strongly suggesting the stoppage is actually due to the interaction with the surface, and not due to some "inherent" unsustainability of motion.
Of course, the ultimate demonstrative proof of this has only been done very recently with the space program: when we send a craft to, say, Mars, the rockets need only burn to lift it from the Earth and then set it on its way. Once that is done, there is no further rocket burning required (and very thankfully for us, otherwise it'd be FAR harder to do space travel than it is now: think about how much fuel a rocket like that burns in just a few seconds, now remember it takes about 20 MEGAseconds to travel to Mars from Earth, and imagine if you had to burn fuel at that rate for all that time, literally millions of times longer! You'd need a rocket with a fuel tank a million times bigger.).
So, finally, we get to Bohr's model. Bohr's model imagines the electrons as free particles in the vacuum of space ("inner space", not outer space, in a way) following Newton's laws of motion, and not subject to any dissipative forces like with pucks and wheelbarrows grinding on hard terrestrial surfaces. Thus, their motion requires no ongoing input of energy to maintain itself.

Add: I see you also inquire about some other features of Bohr's model, like what it means to "have an energy level". What this refers to is the quantization condition. The reason for this requires us to once more think about what I just said regarding dissipative forces. You see, the model that Bohr's was based on, the so-called "Rutherford model", imagined the atom as essentially identical to a miniature solar system, with the electrons being like "planets" and the nucleus serving as the "Sun", and the gravitational force has been replaced by the electric force as the appropriate central force confining things to orbits. However, the theory of electromagnetism, as finalized by James Clerk Maxwell, does predict the existence of a kind of dissipative force in this situation: namely, the emission of electromagnetic radiation, like a radio transmitter. The atom should "broadcast" away all its energy until there's none left, the electrons' orbits decaying like a satellite experiencing atmospheric drag, and finally crashing and burning in the "nuclear Sun" at the center.
Yet clearly, we don't have that, we have stable, interactive and lively atoms, not shriveled, neutral and inert, worthless nuggets of electron-dapped nucleum! Based on this, and some other experimental data, what Bohr suggested was that the orbits must be constrained somehow: and the way he did this was to suggest that the electrons are only "allowed", for some reason that he did not know, certain levels of orbital mechanical energy (not quite, but to keep things simple), and these levels are isolated, fixed numbers, not a continuum of possible values. Moreover, there was an absolute minimum, nonzero orbital energy for each electron, which would ensure that the electrons would never collapse to the bottom. Note that this also requires a violation of EM theory because effectively you have a contradiction in that even with this constraint, the electron still must be orbiting, so what is preventing it from radiating? Hence it's not a particularly satisfactory model, but this is how it works.
The phenomenon of discrete energy levels is, however, an actual empirical fact, which comes from spectroscopy - that's why Bohr used it, so given the above, it must have a different explanation, and this is what led to the development of the fully modern theory of quantum mechanics. In this model, considerably more radical changes to physical theory are required - in effect, we introduce a parameter $\hbar$, which acts as a sort of (in a very rough sense) cosmic "resolution limit" - think about a computer which stores numbers with finite precision, but done more eloquently and subtly - that limits the amount of information allocated to describe the parameters of particles such as their positions, velocities, rotational speeds, and otherwise, and done in such a way that the permitted energy values end up also as being discrete, yet from a powerful, general set of principles that can be used to accurately describe the behavior of any physics at the microscopic scale, not just atoms.

The first two paragraphs are useful in remediating naive concepts about force and Newtons first law


text:
Why doesn't tantalum and tungsten form amalgam?
User @Poutnik has recently answered a question in which he quoted this statement from Wikipedia:

Almost all metals can form amalgams with mercury, the notable
exceptions being iron, platinum, tungsten, and tantalum.

I am not sure about the validity of this statement as I found contradicting results regarding iron and platinum amalgams. There has been discussion around iron amalgams on internet. I found this reddit post and on further research, I came to know that iron indeed form amalgam albeit unstable (See here). This has also been discussed in an another question. User @DrMoishePippik has pointed out that iron can form mercury-based alloys(amalgams?) at specific conditions. So, there is a contradiction regarding formation of iron amalgam.
Some searching also proved that platinum can also form amalgams:

Platinum Amalgam is obtained by trituration of platinum sponge with
mercury in a warm mortar; it cannot be obtained by direct union of
platinum foil and mercury.
The amalgam has a silvery appearance, and with 12 per cent, of
platinum is soft and greasy to the touch, but higher percentages of
platinum increase its stiffness. When heated strongly the mercury is
volatilised and platinum remains as a grey residue.[...]

I searched the same site for tantalum alloys and gave me following information:

Sodium, potassium, mercury and silver do not alloy with tantalum even
at high temperatures; attempts to prepare alloys with arsenic,
antimony, lead, zinc and tellurium have also failed, but the formation
of an alloy with silver, copper and tin for making a dental amalgam
with mercury has recently been claimed.

No information about tungsten amalgam were given on that site. Other sources is giving me irrelevant information although the only other thing I found out is that since they doesn't form amalgams, they can be separated from a mixture of other metals.
So, the questions are:

Does tantalum and tungsten really doesn't form amalgams?
If yes, what is the reason?
If no, is there any evidence/scientific papers to prove it?

I just saw a comment User @Mithoron has left in the question "Why does mercury not form amalgam with iron?"

Pt, W and Ta also resist dissolution with Hg

Can anyone elaborate on this statement?

Interesting question. The ASM Alloy Database does not even have an entry for Hg-Ta, strongly suggesting that it has not been studied much if at all. It does have Hg-Ti, and the phases there are quite different from, say, Ag-Hg which does suggest there may be difficulties forming an amalgam.

It's also interesting that the metals concerned melt at rather high temperatures: 3290 K for Ta, 3695 K for W. Whatever is holding those metallic crystals together isn't letting much Hg sneak in.

I guess that original mentioning of Fe, Pt, Ta and W on Wikipedia could be meant implicitly at usual conditions. As general validity of statements about existence of compound or particular matter form largely depends how hard we try and what criteria we use. // The whole question could be also reverted: Why almost all metals form amalgams, considering many metals do not form mutual alloys ?

Does tantalum or tungsten really form amalgam(s) (with mercury)?

By X-ray analysis, G. Jangg and E. Lugscheider in 1973 found that tantalum $(\ce{_{73}Ta})$ or tungsten $(\ce{_{74}W})$ does not form amalgams with mercury $(\ce{_{80}Hg})$ based on their analytical measures. The authors describe that these metals dissolve only negligible amounts of mercury. They also found out that not only $(\ce{_{73}Ta})$ and $(\ce{_{74}W})$, but also other metals such as aluminum $(\ce{_{13}Al})$, titanium $(\ce{_{22}Ti})$, vanadium $(\ce{_{23}V})$, chromium $(\ce{_{24}Cr})$, iron $(\ce{_{26}Fe})$, cobalt $(\ce{_{27}Co})$, nickel $(\ce{_{28}Ni})$, zirconium $(\ce{_{40}Zr})$, niobium $(\ce{_{41}Nb})$, molybdenum $(\ce{_{42}Mo})$, ruthenium $(\ce{_{44}Ru})$, rhodium $(\ce{_{45}Rh})$, rhenium $(\ce{_{75}Re})$, osmium $(\ce{_{76}Os})$, iridium $(\ce{_{77}Ir})$, thorium $(\ce{_{90}Th})$, and uranium $(\ce{_{92}U})$ dissolve only negligible amounts of mercury under the conditions used. The abstract states following:

Investigations by X-ray analysis proved that $\ce{V, Nb, Ta, Ti, Zr, Cr, Mo, W, Fe, Ni, Co, Re,}$ $\ce{Ru, Rh, Os, Ir, Al, Th,}$ and $\ce{U}$ dissolve only negligible amounts of mercury. The solubility of mercury in platinum up to $\pu{250 ^\circ C}$ is small; at $\pu{250 ^\circ C}$ the solubility increases abruptly to ca. 15.5% and alters little with further temperature increases. The jump at $\pu{250 ^\circ C}$ corresponds to a peritectic reaction in the system $\ce{Pt−Hg}$. The data obtained are in good agreement with data in the literature. Manganese dissolves at $\pu{500 ^\circ C}$ ca. 0.3–0.5% $\ce{Hg}$; the temperature dependence of the solubility was not examined. Rhenium forms no compounds with Hg, at least at $100$–$\pu{500 ^\circ C}$. The metals of the Va and VIa-group of the periodic table do not dissolve any measurable amounts of $\ce{Zn}$.

Following are the conditions used:
$$
\begin{array}{lcr}
\text{Atomic number} &\text{Metal} & \text{Heat Treatment}/\pu{^\circ C} \\\hline
13 & \text{Aluminum} & 260 \\
22 & \text{Titanium} & 280, 400-500 \\
23 & \text{Vanadium} & 800 \\
24 & \text{Chromium} & 400 \\
26 & \text{Iron} & 800 \\
27 & \text{Cobalt} & 800 \\
28 & \text{Nickel} & 260 \\
40 & \text{Zirconium} & 280, 400-500 \\
41 & \text{Niobium} & 800 \\
42 & \text{Molybdenum} & 800 \\
44 & \text{Ruthenium} & 800 \\
45 & \text{Rhodium} & 800 \\
73 & \text{Tantalum} & 800 \\
74 & \text{Tungsten} & 800 \\
75 & \text{Rhenium} & 100-500^a \\
76 & \text{Osmium} & 800 \\
77 & \text{Iridium} & 800 \\
90 & \text{Thorium} & 800 \\
92 & \text{Uranium} & 800 \\
\hline\\
\end{array}\\
^a\text{No evidence of compound formation with mercury over the temperature
interval indicated}$$
It is interesting to see, according to the table (except for $\ce{Th}$ and $\ce{U}$) that most of $\mathrm{d}$-block elements does not make amalgams with mercury. Seemingly, only $\ce{_{25}Mn}$ $(\chi_\ce{Hg} = 0.004)$, and $\ce{_{30}Zn}$ $(\chi_\ce{Hg} = 0.020)$, amongst 3d-elements significantly makes an amalgam with $\ce{Hg}$ at $\pu{773 K}$ and $\pu{293 K}$, respectively $(\chi_\ce{Hg} \text{ represent the mole fraction of mercury in the mixture})$.
The solubility of $\mathrm{3d}$-block elements in mercury has also been checked in 1932 with similar results (Ref.2). The percentage solubility determined in this work is $0.0010\%$ $\ce{Mn}$ and $0.0020\%$ $\ce{Cu}$ amongst $\mathrm{3d}$-block elements tested. The other percentages obtained were: titanium, $ \lt 1 \times 10^{-5}$; vanadium, $ \lt 5 \times 10^{-5}$; chromium, $ \lt 5 \times 10^{-5}$; iron, $ \lt 1 \times 10^{-5}$; cobalt, $ \lt 8 \times 10^{-5}$; and nickel, $ \lt 2 \times 10^{-5}$. The authors have also checked a $\mathrm{4d}$-block element, molybdenum $( \lt 2 \times 10^{-5})$ and a $\mathrm{5d}$-block element, tungsten $ (\lt 1 \times 10^{-5})$ with similar results.
Similar to $\ce{Cu}$, $\ce{Ag}$ has shown best solubility $(\chi_\ce{Hg} = 0.364)$ amongst $\mathrm{4d}$-block elements at $\pu{373 K}$ and $\ce{Au}$ has shown best solubility $(\chi_\ce{Hg} = 0.092)$ amongst $\mathrm{5d}$-block elements at $\pu{663 K}$ (Ref.3).
References:

G. Jangg and E. Lugscheider, "Die Löslichkeit von Quecksilber in verschiedenen Metallen (Solubility of mercury in various metals)," Monatshefte für Chemie 1973, 104, 1269-1275 (DOI: https://doi.org/10.1007/BF00910041).
Nevill Maxsted Irvin and Alexander Smith Russell, "117. The solubilities of copper, manganese, and some sparingly soluble metals in mercury," J. Chem. Soc. 1932, 891-898 (DOI: https://doi.org/10.1039/JR9320000891).
H. M. Day and C. H. Mathewson, "Solid Solubility of Mercury in Silver and in Gold," Transactions of the American Institute of Mining and Metallurgical Engineers 1938, 128, 261-281 (DOI: https://doi.org/10.1039/JR9320000891).

It was hard to decide, Mathew, I would have liked to grant the bounty to both. Thank you for the answer, I have upvoted it.

Thank you @Poutnik for bountying my question. I definitely need an metallurgical expert to clear the myth. It produce two beautiful answer and I had a hard time choosing which one to accept. But I chose this answer because I got one specific information I needed. Anyways, I will go through the answer over the weekend. Cheers :)

@NilayGhosh It is more than fair here if  one answer gets the bounty and the other is accepted. :-)

Imagine a lump of a metallic element in close contact with an equal amount of liquid mercury. Amalgamation might proceed to some degree if the metallic element has an even higher surface tension than mercury. A surface layer of adsorbed mercury will be the first step in amalgamation.
This first step, wetting, but not involving extensive solution of, or into, the metal, is well illustrated by this image of a mercury switch (Ref 1). The electrodes are well wetted, but must remain integral for many years thru electric currents, temperature changes and sparking.

Mercury can penetrate some metals, like aluminum, probably faster thru grain boundaries, thereby causing the metal to disintegrate. There is an added complexity that aluminum has an impermeable oxide layer that must be breached first.
Atoms may disperse thru mercury until a solubility limit, or more accurately, a viscosity increase is reached that effectively immobilizes the mercury. The term amalgam commonly means a fluid or semisolid material like a dental amalgam, which, although it finally becomes hard enough to bite on, is usually encountered in a semisolid state (see the image below, Ref 2). The time required for complete dispersal of the mercury in the liquid into the solid particles of the silver-tin alloy determines the alloy’s usefulness.

The phase diagrams of other mercury alloys shows the effect of melting point of the non-mercury element. The thallium (mp = 304 C) alloy is liquid at room temperature to about 45% thallium (Ref 3).

Uranium, melting at 1132 C), can stiffen a 1% alloy at room temperature, and 3% uranium will solidify the alloy as the mercury boils away at 357 C! (Ref 4)

Another way amalgams can form is by addition of the other metallic element, atom by atom, one at a time. Such is the usual method of preparing sodium amalgam electrolytically. A pool of mercury in contact with a solution of a sodium salt is made negative enough to attract sodium ions, where they will discharge and enter the mercury mass before reacting with water. This can be continued until the mercury is stiffened, which amounts to reaching a solubility limit, and at that point, any more sodium formed will react with water.
Aluminum and sodium have relatively low melting points, suggesting that they have low cohesive strength. “Low melting point” also includes silver and gold, with melting points around 1000 C. These metals can be expected to alloy, or amalgamate, easily with mercury.
But high melting point (for Pt, 1768 C) is not a guarantee of complete non-amalgamatability. The OP mentioned a way of making what appears to be a soft solid at 12% Pt from Pt sponge, which can be stiffened with more Pt, but this cannot be done with Pt foil. Pt foil can be obtained from Aldrich as thin as 0.025 mm; Pt sponge is composed of particles less than 10 microns thick - i.e., with 2500 times the surface area (Wikipedia). A surface reaction product Pt(Hg)4 has been found, but no penetration into the Pt (Ref 5) until the temperature is raised to 250 C (Ref 6), when the solubility reaches 15.5%, but forms another solid, not liquid, phase.
X-ray analysis showed that V, Nb, Ta, Ti, Zr, Cr, Mo, W, Fe, Ni, Co, Re, Ru, Rh, Os, Ir, Al, Th, and U dissolve only negligible amounts of mercury (Ref 6). So melting point can be a good indicator of the ability of a metal to alloy with mercury. Higher melting point suggests greater difficulty in atomizing the metal. There could be a correlation with surface tension of the metal. And one more quantitative measurement should be the heat of vaporization (not fusion, since the heats of fusion are quite a bit smaller). The heats of vaporization of tungsten and tantalum are higher than all the other elements I have checked by a factor of about two! (WebElements)
Temperature of preparation is important because some mercury-metal alloys may form only at higher temperatures. Whether the amalgam is liquid or solid will depend on the concentration of the metal in the mercury and the observation temperature. And time may affect properties like fluidity and stability (e.g., note the instability of ammonium amalgam, although it is easily prepared).
Ref 1. https://en.wikipedia.org/wiki/Mercury_(element)
Ref 2. https://pocketdentistry.com/14-silver-amalgam/
Ref 3. https://www.researchgate.net/figure/Phase-diagram-of-binary-system-Hg-Tl_fig3_47393888
Ref 4. https://www.researchgate.net/publication/238894897_The_HgU_Mercury-Uranium_System
Ref 5. S. K. Lahiri and D. Gupta, Journal of Applied Physics 51, 5555 (1980); https://doi.org/10.1063/1.327440
Ref 6. G. Jangg & E. Lugscheider, Monatshefte für Chemie / Chemical Monthly 104, 1269–1275 (1973)

Gallium is an interesting non-amalgam former in which you get two immiscible liquids from 30C to fairly high temperatures.