text:
Prigogine vs. Bronsted and the minimum entropy production principle
I apologize for the length of this question that was asked here prigogine-bronsted but got no reply; anyhow, Bronsted's name is much better known among chemists than among physicists...

Prigogine's minimum entropy production principle fails in the simplest heat conduction case, because the temperature distribution that minimizes the entropy production rate formula $\sum_k J_k X_k$ contradicts Fourier's law. In fact, the resulting temperature has an exponential distribution for a homogeneous rod between two fixed temperatures, see [4]. The minimization gives the correct linear temperature dependence $T(x)=c_0+c_1T$ only if the conductivity $\kappa = \kappa(T)$ has a quadratic temperature dependence $\kappa(T) \propto T^2$, a non-physical result.
This failure is well-known and Prigogine & Glansdorff explained it away by saying that the principle should only be used for small deviations relative to steady-state and the minimizing functional is to depend on two temperature distributions simultaneously, the unknown dynamical and the stable steady state distribution in which the latter temperature distribution is the thermodynamic force (ie., the gradient $X$) while the former is used in the definition of heat transported (ie.,current $J$), see eq 2.8 in [5].
I am wondering if it were possible to save Prigogine's minimum entropy production principle for heat conduction by combining it with Bronsted's "Energetics". I know Bronsted is out of fashion but would you hear me out for a couple of minutes.
According to Bronsted we can calculate the dissipated work (he calls it "lost work") for each interaction as the drop of the extensive quantity $da_0$ from a high potential (intensive) $Y_0$ to a lower one $Y_1$:
$\delta w(a)=(Y_0-Y_1)da_0 \ge 0$
He also makes the point that when entropy $dS_0$ is transferred the dissipation is $\delta w(s) = (T_0-T_1) dS_0$ where $\delta q = T_0 dS_0$  is the thermal energy, "heat", transferred into the system from $T_0$ to $T_1$.
For Bronsted the sum of these dissipated terms and then their volume integral over the system represents the total dissipation. If that happens at temperature $T_1$ (system temperature) then the entropy produced within the system is $T_1\sigma = \sum \delta w$, and in the special case when only heat transfer is important then $\sigma = \frac{1}{T_1}(T_0-T_1)dS_0$.
Now Prigogine would rewrite this as $\sigma = \frac{1}{T_1}dS_0=\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\delta q$
and say that the thermodynamic force $X$ is the term $\frac{1}{T_1}-\frac{1}{T_0}$in the big parentheses as it moves "heat" $\delta q$ across the gradient the result of which is the wrong temperature distribution when minimizing entropy production.
But following Bronsted who says the thermodynamic force is really the drop (gradient) of the intensive parameter (pressure, gravitational field potential, temperature, electric/magnetic field potential, etc.) for that is what moves the extensive "charge" (volume, mass, entropy, electric charge, dipole moment, etc.,) and results in "lost work", ie., dissipation.
If we do that then instead of a linear relationship between heat and temperature gradient we should look for a linear law between the transferred entropy and temperature gradient that moves it. In other words we should take as linear constitutive relationship $\delta S = L_s \delta T$ where $L_s=L_s(T)$ is the "new" Onsager conductivity coefficient but now of entropy and not of "heat". With infinitesimal deviation (force) the entropy production is then $\delta \sigma = \frac{L_s}{T}(\delta T)^2$ and its variational integral is
$$ \sigma = \int_{T_\ell}^{T_h} \frac {L_s}{T} \left(\frac{dT}{dx}\right)^2 dx \tag{1}\label{1}$$
When this $\eqref{1}$ relationship is used to minimize the entropy production formula we do get is $L_s=\kappa T$ where $\kappa$ (heat conductivity coefficient) is a constant and the minimizing temperature is a linear function of the coordinate $T(x)=T_{\ell}+(T_h-T_{\ell})x$, as was to be expected from Fourier's law, and of course we also have $q=-\kappa \frac{dT}{dx}$.
If this line of thought is correct then the problem is not with the minimum entropy production as such but rather with our choice of thermodynamic force and "charge". Instead of transferring "heat" in our calculation of entropy production we should consider only the transfer of entropy from a given temperature over a given temperature gradient, and then minimize entropy production.
Bronsted emphasizes that his way of calculating total dissipation (he calls it "loss of work") from the product of a transferred "charge" (mas, chemical species, electric charge, dipole moment, entropy, etc.,) falling down a potential drop is correct even when the potential drop $Y_0-Y_1$ or $T_0-T_1$ is not infinitesimal but finite, so may be this is the right thing to do...
I would appreciate any comment you may have as I am trying to understand both Bronsted and Prigogine.

La Mer, Foss, Reiss: Some New Procedures in Thermodynamic Theory         Inspired by the Recent work of J. N. Bronsted, ANNALS NEW YORK ACADEMY OF SCIENCES May 1949; https://doi.org/10.1111/j.1749-6632.1949.tb27295.x
Leaf: The Principles of Thermodynamics, J. Chem. Phys. 12, 89 1944    https://doi.org/10.1063/1.1723919
Bronsted: The Fundamental Principles of Energetics, https://doi.org/10.1080/14786444008520665
Palffy-Muhoray: Comment on ‘‘A check of Prigogine’s theorem of minimum entropy production in a rod in a nonequilibrium stationary state’’, Am. J. Phys. 69(7), July 2001; https://doi.org/10.1119/1.1371916
GLANSDORFF, PRIGOGINE: ON A GENERAL EVOLUTION CRITERION IN MACROSCOPIC,  Physica 30, 351-374 PHYSICS https://doi.org/10.1016/0031-8914(64)90009-6


text:
What is the product of the following reaction(different sources give different answers)
$$\ce{FeS +dil HNO3}$$
$$\ce{FeS +conc HNO3}$$
I got different answers to these reactions in different sources
Source1 Jd lee:
Source 2 Op Tandon:

Can anyone please explain which is the correct one and the reason for the same

Please don't use images in place of text, add conditions, cite the sources.

What do you mean you "don't have idea about the sources"? Was it someone else who wrote this question for you? Source 1 looks like a handout from a lecture/presentation, and source 2 looks like a scan from a textbook. Add corresponding references for both, it's that simple. All third-party sources must be cited, the preferred citation style for references is ACS.

As for markup, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Please spare colloquialisms and slang for informal chatting, this isn't the right place for either.

Whatever the experimental conditions, $\ce{FeS}$ does react with any acid $\ce{HX}$ like $\ce{HNO3}$ according to $$\ce{FeS + 2 HX -> FeX2 + H2S}$$ This reaction proceeds like this if the acidic solution is rather diluted. If  $\ce{HNO3}$ is more concentrated, both $\ce{H2S}$ and $\ce{Fe^{2+}}$ may be oxidized according to one of the following equations : $$\ce{H2S + 2 HNO3 -> S + 2 H2O + 2 NO2}$$ $$\ce{H2S + 8 HNO3 -> H2SO4 + 8 NO2 + 4 H2O}$$ $$\ce{Fe^{2+} + HNO3 + H+ ->  Fe^{3+} + NO2 + H2O}$$
It is hard to know which ones of these equations is favored at a given concentration of nitric acid.

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text:
Interaction energy of an hydroxide ion with a Pd surface
I need to estimate the interaction energy of a hydroxide ion with a palladium surface of $-0.6$V in water.
I know that the electric field of a charge $e$ is
$$
E = \frac{1}{4\pi\epsilon}\frac{e}{r^2}
$$
where $r$ is the distance and $\epsilon$ is the dielectric constant.
Also, I know that for an infinity charged plane the electric field is
$$
E=\frac{\sigma}{2\epsilon}
$$
where $\sigma$ is the surface charge density. I have the area of the surface, $A=0.125$ cm$^2$. But I wonder how can I calculate this interaction energy.

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text:
Why is Molar Mass of a gas twice the vapour density?
I've seen this relation at a number of places, and I can't get why. A simple Google search doesn't give me a simple answer either.
Also, are there any exceptions/conditions for this relation to be true?

Vapour density $(\rho_\mathrm{v})$ of a gas is not the absolute density, such as mass density $m/V$ or number density $N/V$. It is in fact relative density, compared to the density of some other, reference gas at the same temperature and pressure:
$$\rho_\mathrm{v} = \frac{\rho}{\rho_\text{ref}}.\tag{1}$$
It is a dimensionless quantity (just a number, without a unit). The choice of the reference gas is arbitrary and should be clearly stated. Usually, either hydrogen $(\ce{H2}),$ oxygen $(\ce{O2}),$ or air is chosen.
If the reference gas is hydrogen $(M(\ce{H2})\approx \pu{2 g mol^-1}),$ and we use the ideal gas approximation $(\rho \propto M),$ we can derive the following relation:
$$\rho_\mathrm{v} = \frac{\rho}{\rho(\ce{H2})} = \frac{M}{M(\ce{H2})} \approx \frac{M_\mathrm r}{2} \implies \boxed{M_\mathrm r \approx 2\,\rho_\mathrm{v}},\tag{2}$$
where $M_\mathrm r$ is the relative molecular mass of gas.
If the reference gas is oxygen, then the relation is $M_\mathrm r \approx 16\,\rho_\mathrm{v}$.
The relationship holds only approximately, and it becomes more wrong as we deviate from the ideal gas conditions (e.g. low pressure, high temperature).

Congratulations on your first answer, very well written. A couple of nitpicks regarding typesetting and notations. First, using complex abbreviations in math mode is a bad idea: $VD$ reads as $V$ times $D.$ A single letter with an index is preferred, like $\rho_\mathrm{v}$. Second, don't leave orphaned braces around math mode, enclose them too with $(…)$. This way they have appropriate height and won't be left hanging at the EoL (see my edit for examples).

@andselisk, thank you for your comments and your edit! I do not like abbreviated symbol names either, but I used $VD$ as I couldn't find any other more standard symbol for vapour density, and $VD$ appeared several times online.


text:
Why can I extract all metabolites from blood plasma with one solvent, but to extract only fatty acids I must use many solvents?
I want to do a metabolite study using GC- without derivatization and UPLC-MS (RP and NP) on samples of plasma and urine. My hypothesis is agnostic about which metabolite will correlate with the disease picture that I study. So I want to separate as many compounds as possible to see what correlates with the disease. There is a step called ‘metabolite extraction’ in metabolite identification literature. I have questions about it.
It seems in plasma for example, some authors say there is exactly one step in metabolite extraction, often involving a so-called ‘extraction solvent’. For example this one:
https://youtu.be/MSZXsNBpWf8 [Video summary: (1) The scientist says that ‘the extraction protocol for all sample types are the same’. (2) The scientist simply adds a so-called extraction solvent to sample aliquots and collected the supernatant after cooled centrifugation]. Here is a reference on so-called extraction solvents in the context of so-called metabolomics:

Zeki, Ö. C., Eylem, C. C., Reçber, T., Kır, S., & Nemutlu, E. (2020). Integration of GC–MS and LC–MS for untargeted metabolomics profiling. Journal of Pharmaceutical and Biomedical Analysis, 190, 113509.

In contrast, I can see some authors want to study a single compound type, for example phospholipids,
organic acids or sterols, and they all use different protocols to
do extraction (and many solvents). Even within the same type of compound, the authors use very different protocols. For example the sample preparation-sections of:

McDonald, J. G., Smith, D. D., Stiles, A. R., & Russell, D. W. (2012). A comprehensive method for extraction and quantitative analysis of sterols and secosteroids from human plasma. Journal of lipid research, 53(7), 1399-1409.
Amaral, C., Gallardo, E., Rodrigues, R., Leite, R. P., Quelhas, D., Tomaz, C., & Cardoso, M. L. (2010). Quantitative analysis of five sterols in amniotic fluid by GC–MS: Application to the diagnosis of cholesterol biosynthesis defects. Journal of Chromatography B, 878(23), 2130-2136.
Castillo-Peinado, L. S., López-Bascón, M. A., Mena-Bravo, A., de Castro, M. L., & Priego-Capote, F. (2019). Determination of primary fatty acid amides in different biological fluids by LC–MS/MS in MRM mode with synthetic deuterated standards: Influence of biofluid matrix on sample preparation. Talanta, 193, 29-36.

Questions:

I take it the theory is to use LLE. Are all metabolites methanol-soluble? What isn’t methanol-soluble in plasma?
If it is within your time budget to explain from a high level of abstraction why some of these protocols are elaborate, it would be appreciated. I worry that I won’t be extracting the maximal number of metabolites. If it's too elaborate, can you suggest targeted literature?

Can you provide a summary of the relevant content of the youtube video?  Or provide a reference that describes the extraction solvent for metabolites? The main question is probably one of solubility. The metabolites might have a rather high aqueous solubility compared to some lipids. Some sterols are very insoluble and present in aggregates (LDL fraction etc), therefore the need for more nonpolar solvents.

It is not generally true to say that one solvent can extract all metabolites from a fluid such as blood or urine, this would be an over-generalization. In a urine sample, there are thousands of known compounds, each with varying polarities and solubilities, same with blood. Extraction with a solvent, say 50% acetonitrile (ACN) in water, may capture a large number of metabolites, but it won't capture all of them. When you spin blood down you get a number of layers which is evidence of the broad diversity of matrix components in blood. Another challenge may be solvent miscibility. For instance, methanol is quite miscible in water so LLE is difficult because you cannot separate the methanol from water, therefore you'd have to use a less miscible solvent such as hexane or chloroform.
Metabolomics sample preparation is often overlooked as a means to an end, rather than an important experimental parameter that can provide different insights into the same sample with differential extraction. Sometimes an easy sample preparation such as chloroform is sufficient for metabolomics, but other times the matrix is too complex following extraction, or there is a loss of metabolites or low extraction yield, so it is necessary to optimize the sample preparation, often adding complexity.
I'll also add that GC has a number of sample introduction techniques. I assume you want to do a liquid injection, but approaches such as headspace and SPME can be very insightful.


text:
How can I determine the purity of an oligonucleotide solution?
After desalting an oligonucleotide via gel chromatography, I am left with (presumably) an oligonucleotide substance suspended in a buffer. What is the simplest way to verify the oligonucleotide's purity to make sure the desalting process was successful?


text:
How to dissolve metallic lead fast in nitric acid?
We were trying to dissolve a piece of metallic lead in concentrated nitric acid until we found out that concentrated nitric acid is not a suitable media, as long as it produces a layer of lead oxide which is harder to dissolve.
We tried then with nitric acid 1:4 in water and could notice a little advance in the digestion but it still looks like it is going to take a long time...
Does anyone have struggled with this before?
Edit: It dissolved overnight. Sorry for being that impatient and thank you for all the help.

Powder it, maybe?

Thank you for the suggestion. I think it would be problematic since it is a relatively expensive certified reference material for isotopic composition (produced in 1991, the certificate does not contain a section for handling instructions), and would like to manipulate it the minimum possible

How about hot nitric acid?

Use pure nitric acid maybe?

Using subdestilled nitric acid was our first option but it forms a layer of PbO which does not allow the reaction with the acid to proceed anymore :(

Do you have an ultrasonic bath?

@Waylander I do have and will try it today. Thanks!

For your future reference look up https://en.wikipedia.org/wiki/Lead(II)_nitrate. In the Chemical Properties and Reactions section they show the solubility of lead nitrate in nitric acid solutions, which decreases strongly with acid concentration. It's due to the common ion effect, which is significant because the concentration of nitrate ions from your 1:4 diluted nitric acid solution is comparable with that of the saturated water solution of the lead salt.

You are right, thank you. I should have done some search first before just attempting to get it into solution by brute force.

Some ideas are given by Bell[1]. The author recommends:

Experiments in this laboratory have shown that mixtures of fluoroboric acid, hydrogen peroxide, and ethylenediaminetetraacetic acid (EDTA) rapidly dissolve lead alloys at room temperature. The addition of EDTA is necessary to prevent the formation of insoluble salts during the dissolution step.

Understanding, and optimizing, nitric acid
Nitric acid performs relatively poorly because the lead nitrate (not any oxide) produced by the reaction  has more limited solubility in nitric acid solution than in plain water. This is due to the common ion effect, and also at high concentration to nitric acid being a poorer solvating agent for ions than plain water (as distinct from its power as a reactant).
It follows that to maximize the solubility the acid should indeed be diluted with water. Using lead nitrate solubility data from Ferris[2] and the stoichiometry of the reaction between lead and nitric acid, the salt solubility is limited by the acid if the initial acid (before it reacts) is 21-23 weight% $\ce{HNO3}$ or more, depending slightly on whether the acid is reduced to $\ce{NO}$ or $\ce{NO2}$. With a more dilute acid, including the 1:4 diluted mixture that was actually used, saturation is avoided and the reaction proceeds until either the metal or the acid is exhausted.
References
1.
Harry F. Bell (1973). "Rapid dissolution technique for lead alloys". Anal. Chem., 45, 13, 2296–2297. https://pubs.acs.org/doi/10.1021/ac60335a027.
2.
L. M. Ferris (1960). "Lead nitrate-nitric acid-water system". J. Chem. Eng. Data, 5(3), 242. https://pubs.acs.org/doi/10.1021/ac60335a027.


text:
Can I take H = 0 at absolute zero
I am a 12th grade students making a reaction engine(includes state change, solvation also) on java, I was wondering can I take enthalpy(heat not change in heat) of a substance 0 at P(ext.) = 1, Kinetic energy = 0 and temperature = 0k(-273.15°C). from there I can take H approximately at temp. = t(at any temperature from there on). Note: I want to calculate H not ΔH. Or You may have any suggestions then please tell me.

You may just as well take H=0 at any other arbitrarily chosen conditions. The result will be equally valid.

@IvanNeretin thanks for your response sir, I was thinking, should I take H = heat of formation at NTP as these values are already given in books, or defined by experimentally  as standards at NTP. What do you say?

Of course. That's what the books are for.

First of all, I deeply appreciate your work. At absolute zero, only the entropy of a perfect crystal is zero. It is the third law of thermodynamics. A system has energy, even if the temperature is $0$ K.
Consider for example a monodimensional quantum oscillator. Solving the associated Schrödinger equation, you obtain the energy
$$
E =\biggl(n + \frac{1}{2}\biggr)h\nu
$$
where $n$ is the vibrational quantum number, $h$ is the Planck constant, and $\nu$ is the frequency associated with the oscillator.
At $0$ K, you are in the lowest possible state. Therefore $n=0$, but the energy is not zero
$$
E=\frac{1}{2}h\nu
$$
This energy is known as Zero Point Energy, then the energy is not zero, otherwise, it will violate the Heisenberg principle.
I want to remark on the physical chemistry of the problem. You can do this approximation, saying that the heat at $0$ Kelvin is zero. You consider this heat as a constant and impose it to zero.


text:
Do I need to 'neutralise' iron/steel/metal with a base after removing rust with acid?
I've never asked a question here before but these forums seem to have some extremely knowledgeable people so I thought I'd give it a try. I know it's a 1st grade question but I honestly don't have the first idea about this. So, I have a motorcycle, the fuel tank was completely full of rust, extremely bad. I've finally managed to get the tank clean by using a combination of things, mainly white vinegar (at least I know this is acetic acid 5%). I got this idea from watching many Youtube videos and reading online articles. I'm now at the stage where I need to get the vinegar out and clean and dry the tank (preventing as much flash rusting as I can).
My question is... without exception, all of the resources I've seen/read have said "after removing the vinegar, it's important to neutralise the fuel tank with baking soda/washing liquid" - I was planning to just follow this as it sounds good! This is apparently to 'stop the vinegar reaction/ return to a safe ph level'....I've seen how quickly flash rust takes hold and I'm wondering if using baking soda is really necessary. Surely it would need to be in the tank for a few minutes at least for it to do anything? - What exactly is this doing? Looking into this I've also read that baking soda and vinegar will create salt, this is concerning as I'll then need to make sure any salt is thoroughly removed. I could understand the idea to say, neutralise the vinegar itself before disposal (maybe?) but does the vinegar have any lasting effect on the steel after it's been removed?
As a side question, is there any recommendation on the best possible ways to prevent flash rust? I've seen/read many ideas which I think probably work to varying degrees of success, mainly to get the tank dry as quickly as possible then coat it with an oil/wd40/fogging oil/kerosene. I've just discovered water based corrosion inhibitors and was thinking these might be the ultimate best option, so after going through whatever process I need to, as a final step I would rinse with a corrosion inhibitor?

You might want to migrate this question to Motor Vehicle Maintenance and Repair SE for a more practical and less theoretical range of answers.
That being said, plain water is sufficient for rinsing out any remaining acetic acid. (It's a gas tank, not lab equipment.) However, unless you take further steps to stop the tank from rusting while in service, your efforts will all be wasted. The tank rusted because of water in fuel, most likely due to condensation of water from humid air as the tank "breathes". You now have bare, clean steel with no protection from future rust.
Forget the oil / kerosene / what have you. Its protective effect will last only until the first fuel-up, when the gasoline will dissolve it.
I suggest that you coat the interior of the tank with a product that is designed for the purpose, such as POR-15 Fuel Tank Sealer. I've used it. It works. It will completely coat the inside of the tank with a hard polymer coating that is impervious to fuel and water, and it will prevent any future rust.

Your suggestion of the interior sealer is a good one. A tank that was "completely full of rust, extremely bad" is a leak waiting to happen.

In case of future link rot, the "this one" link above refers to POR-15 Fuel Tank Sealer.

Thank you MTA, very useful and practical advice. That's exactly what I was thinking,  The main reason for the question was due to thinking about how to avoid as must flash rusting as possible, if I have to rinse it again after using the sodium bicarbonate then it's an extra step maybe unecessary, delaying the drying process and inviting flash rust. I think a small amount of flash rust won't really be a problem. I was planning to use a tank sealer however I've heard a lot of horror stories about them pealing and clogging up the fuel filter, that's what getting the rust out is supposed to avoid

@Darren That is correct, flash rust or any light, sound rust is not a problem for the product I mentioned. Stories of coatings that peel off are most likely due to improper surface preparation: either trying to coat flaking rust or coating over an oily surface. When the surface is properly prepared, this stuff can only be removed by grinding it off.

@MTA I do suspect that poor preparation and pre existing heavy rust may be the reason some people have had poor results from using a tank sealing product. I think you've allayed my fears and I'm going to give it a try. If they really bond to the tank that well, then that's a perfect solution

Proper surface preparation may include a primer before apply the actual sealant, depending on the actual product used.

Apparently you mean the interior? The reaction of sodium bicarbonate with acetic acid is very rapid, slosh it around then pour it out. Being damp it will tend to get a very thin rust on it quickly unless coated with an oil as you suggest.  I would not try to used corrosion inhibitor, that is a whole different science. Gasoline is not very corrosive to steel, it is stored in steel tanks in refineries and shipped in steel pipelines with no problem (Refinery tankage does rust on the bottom because water settles out of the gasoline causing corrosion.). I would think any rust would not be a problem because it should be caught by the filter. I have not looked for years but steel gas tanks were normally lead plated (terne plate) or galvanized. If you do this again, I would suggest "conversion coating",  phosphate compound such as Naval Jelly. That leaves a bit of a protective coating on rusted steel.

Hi blacksmith and thank you very much for your answer. Good point, I do indeed mean the inside of the tank. That's useful to know that I can just slosh it around and then get it straight back out. I'm still wondering what this is actually doing? I know the sodium bicarbonate will react with the vinegar but there will be minimal vinegar remaining in the tank. Is the intention to make sure the vinegar doesn't mix with the fuel that will be going in or is the purpose to restore the steel somehow? I'll definitely think about other solutions and look into Navel Jelly if I ever do this again.

It's called Naval Jelly, not Navel Jelly. :D BTW, a lot of carbonated drinks such as Coca-Cola also contain phosphoric acid and can be used as makeshift rust removers in a pinch. Just make sure to use the sugar-free kind if possible, to avoid the risk of gunking up your fuel tank with leftover sugar residue.

Rust is a spot phenomena. It does not proceed on the whole surface of iron and steel plates. Rust starts at and around surface microcrystals of "impurities". When started, it goes on around the same microcrystal, digging deeper and deeper at the same point. The metallic surfaces more than about $1$ millimeter away form the spot are nearly not oxidized. As these cracks and clefts may be thin and deep, but contains acidic solutions, it is difficult to remove these acidic residues by simple washing with water. It is much more efficient to use bicarbonate powder that may enter the cracks. Corrosion will be stopped. And if too much has been used, it won't bother the metallic piece.

Hi Maurice and thanks for your very in depth answer. So to summarise in layman's terms, the purpose of the flush with sodium bicarbonate is to remove/neutralise any remaining trace of acid, and is NOT to counter any effect the acid had on the steel?

@Darren. You are right. Bicarbonate neutralizes any remaining trace of acid.

I'd say the "trick" with the bicarbonate for removing the acidity is that it doesn't need to enter the cracks - it will remove acid equivalents in the bulk solution, and protons are much faster transferred from anywhere (including the cracks) compared to ions actually moving by diffusion. However, I'd think OP will want to get rid of the remaining  electrolytes as well. Meaning they anyways need to rinse thoroughly and wait for the stuff to diffuse out of the pores.

As @MTA already pointed out, you will get moisture again into the interior of the tank, and therefore need to apply an appropriate coating.
The conclusion from that wrt. to your question of how to prepare/rinse the tank now is: you need to follow the instructions of the actual sealing product you use. My guess would be that it likely requires the tank interior to be dry and free of any oil/fat/grease, and may suggest to use certain primers first. And, even if not specified, you'll also not want to have residual acid nor salt.
Some considerations:

Bicarbonate ($\ce{NaHCO_3}$) removes residual acidity by reacting with the acetic acid producing sodium acetate and $\ce{CO_2}$. The $\ce{CO_2}$ "contains" the acidity and as gas can quite easily be gotten out of the tank.
This reaction is very fast, since it does not require diffusion of the (large) acetate ions but only proton transfer.

However, you'll have salt left (both excess sodium bicarbonate and sodium acetate) which you want to rinse out: in case of future moisture they'd give you a more corrosive mixture than only the pure water.
Rinsing is most efficient if you do it often (smallish amounts of water each time are sufficient, but you want to be sure to reach all the inner surface area). This process requires the salt ions to travel (diffusion), so is somewhat slower than the acid-base reaction above.

(Personally, I'm not sure I'd have used acetic acid in the first place)

Also, it may be worth while considering to use the high vapor pressure of the excess acetic acid (dry in a warm place, ideally with a ventilator) rather than introducing more electrolytes via the bicarbonate. You'd still need to rinse afterwards to remove acetate, though.

Thanks cbeleites, future rusting is the biggest concern and there doesn't seem to be any 'perfect' solution to that. It seems most people just keep the tank fully topped up with fuel but when the bike is in use, it's hardly practical to keep refueling after every journey. The tank sealer would be ideal but I've heard many people say they wish they hadn't used one as the ethanol eventually breaks through and causes the seal to peel and clog up the fuel filter, so I'm in two minds about that. I probably wouldn't use acetic acid again. I just wanted the rust out and I know nothing about chemistry

@Darren, rust will also clog up the filter... make sure the product is suitable for gasoline tanks, if it doesn't clearly say so, ask whether that also applies to high ethanol content. However, if you seal in any moisture, the tank will go on rusting. If that happens, there is no way to prevent the "peeling", because you could also say that the tank peels away from the sealer... So make really sure that the interior is entirely dry and properly primed.

Yes exactly, but it would be rather annoying if the product that was intended to stop the rust forming, and then clogging the filter gave you the same or a worse problem. MTA has mentioned the tank sealers are solid if used properly with proper prep as you say. I think the complaints about them may source from improper preparation.  I think I'm going to give it a try.

@Darren: IDK about the relative risk of clogging the filter, but a sealant should have a much lower risk of fuel leaks, as another answer pointed out.  You already know that a lot of the metal in the tank wall is rusted away, so further rust could go all the way through.

As a side question, is there any recommendation on the best possible ways to prevent flash rust?

I own a company that builds parts washing equipment. The machines use water and alkaline cleaning solutions.  They are bare steel on the inside and they don't rust even when they are dry so long as the solution pH is above 10/10.5.  We have machines that are over 30 years old in the field with no rust. We have stored machines for months without rust so long as the wash solution dried on the bare steel and isn't washed/rinsed off.  I suspect the Bicarb rinse is meant to raise the pH above neutral which will protect the steel somewhat better than leaving it acidic when it dries until a protective coating is applied. The pH of sodium bicarbonate is not very high - around 8.5. I would use NaOH as the neutralizer and make sure the final rinse ends up with a pH above 10. Once your tank dries it won't flash rust or show any signs of rusting for days.
Regarding water based corrosion inhibitors: In our shop we use water and 1000ppm sodium nitrate as the rust inhibitor for all our testing.  It is effective at preventing rusting of the  machines before shipping.  The machines use hundreds of or even thousands of gallons of water. The steel sheets get wet and dry many times during our tests. Usually the shop people don't accurately measure the nitrate as it only takes a very small amount to prevent rust.  If they see rust they add another couple of pounds to the test solution.  The canning industry sodium nitrate to prevent rust in canning before the inside of cans are coated. It is also used in hotdogs and other meats as a preservative. It works best with solution pH of 9. Our shop is in the Midwest and only during the most humid months of July and Aug do we sometimes see light surface rust on our steel.

Unfortunately I do not see how this answers the question.

@Martin-マーチン Unfortunately the OP left the Q a little open by slipping this in: "As a side question, is there any recommendation on the best possible ways to prevent flash rust?"


text:
How do I keep my homemade liquid fabric softener from separating?
I have a recipe for a homemade liquid fabric softener that smells good and works great, but I have to shake it every time before I use it. It gets old fast.
The recipe calls for:
12 cups hot water, almost to boil
6 cups distilled white vinegar
4 cups of favorite hair conditioner
1 cup baking soda
Mix hot water and conditioner completely
Mix in vinegar completely
Mix in baking soda slowly till it no longer rises then add all the rest and mix completely. Set aside for a few hours and then mix it back together and done.
Well, no not done cause it won't stay together....please help

Any other methods you've tried?

This is not a good idea to use fabric softener like this. You final rinse will have all these unwanted chemicals on your clothes, towels, bedsheets. Hair conditioner is meant for a temporary exposure.

Don't the vinegar and baking soda mostly destroy each other?

Maybe adding some soap might help. Taking chunk out of the soap in your bathroom and grinding it, or purchasing some sodium hydroxide and mixing with leftover oil.

Though the recipe seems pointless, to keep material from separating, you might try ultrasonic homogenization.


text:
How to measure organics in liquid solution
We have a solution that has been put through a filter, it has gotten down to 20ppm. We want to know what the remaining organics are so we can try to filter those out. Is there any machine we can buy to determine this? What would this method of figuring out this problem called? Any direction to help me with this would be much help thank you.

The instruments are very expensive. You cannot buy it for your school project. For this type of work, a system called gas chromatograph with a mass detector (GC-MS) is used. See Wikipedia for more details.

"Filter them out"? What filter will remove dissolved organic chemicals? Activated carbon? How will it work?


text:
Reaction Workup and Quenching
In the supporting information of synthetic paper by Buchwald's group, They use 2-MeTHF as solvent (See B. General Procedures for Amination part, page 8). After the reaction ends and cools the mixture to room temperature, they dilute it using dichloromethane.
Question 1: Why do they use dichloromethane, instead of 2-MeTHF which is the solvent of the reaction? Any particular reason? Also I have seen another paper that Buchwald's group also used dichloromethane in prior to filtration, and in that paper they used toluene as solvent.
Question 2: What is 'kinetic' quench? I searched for it literally in Google, but I couldn't find anything about it.

I have never heard the phrase "kinetic quench". As for the procedure there may be not stronger reason than DCM is the preferred solvent within the group - it works so don't fix it.

@Waylander What I wonder is why they used other solvent instead of original solvent, 2-MeTHF. If the purpose is quenching, I don't think that addition of just another solvent may quit the reaction.

Or, perhaps they mean 'kinetic' quench that by diluting the mixture so lowering the reaction rate? If so, but still I don't understand why they did not use original solvent.

2-Me-THF is a lot more expensive than DCM so it may simply be a cost issue


text:
Dominance of electronic factors in cupric ion
I know that Cu+ ( cuprous ) ion is less stable than Cu²+ ( cupric ) ion . There are some reasons explaining this by high hydration enthalpies and low ionization enthalpies . Even though , these reasons explain so much , I wanted to know that how do these dominate over electronic configuration. How can we define more stability in terms of energy; like I can easily see the energy released by hydration and absorbed by ionization, but how can I compare it with the potential energy of Cu+ ion which is lowered by it's stable configuration ?
I searched on internet , but was unable to find data of the potential energy of the ions . Is there any theoretical reason which can explain that why hydration enthalpy and other factors must dominate the stable configuration?

To say that $\ce{Cu^{2+}}$ (or more accurately, $\ce{Cu^{2+}}$ plus an equimolar amount of metallic copper, which conserves mass and charge) is more stable than $\ce{Cu^+}$ is something of a misstatement. Among the fourth period metals every $2+$ ion plus an equimolar amount of metal is more stable than the $1+$ ion when solvated by water, except obviously potassium. As described here, the extra electrostatic energy from water solvation of an ion with one more charge balances about $3000$ kJ/mol ionization energy, and among the $3d$ transition elements the second ionization energy always comes in under that. (So does the third in some of these elements.)
The real question is actually the reverse: why is $\ce{Cu^+}$ relatively stable so that we can ever see it at all in aqueous chemistry, unlike all other fourth-period metals besides potassium? The second ionization energy, which is the main energy cost to forming the $\ce{Cu^{2+}}$ ion, is raised somewhat by ionizing from a full $3d$ subshell, but beyond that the $\ce{Cu^+}$ ion is relatively good at forming bonds having covalent character with soft, polarizable bases.


text:
Why does Samarium have such a large c-axis lattice constant? (>26 Å) Are there significantly larger lattice constants of the elements?
Figure 1.2 on page 12 of The Structure of Rare-earth Metal Surfaces linked below shows a repeating stack of hexagonal layers arranged as A B A B C B C A C...  and several sources give the c-axis lattice constant for Samarium of over 26 angstroms!
Question(s):

Why does Samarium have such a large c-axis lattice constant?
Are any significantly larger lattice constants seen in pure crystal forms of the elements?

I understand that the short answer to #1 may simply be that this just so happens to be the lowest energy configuration, but something about this "long" range order in a crystal of a single type of atom somehow feels unusual.
As the image shows, for many elements we see A B and A B C stacking of hexagonal layers in the form of fcc and bcc structures (cf. this answer).
Has the reason for this been studied? And explained?

The Structure of Rare-earth Metal Surfaces By Stephen David Barrett, Sarnjeet S. Dhesi (2001)
The Crystal Structure of Samarium A. H. Daane, R. E. Rundle, H. G. Smith AND F. H. Spedding, Acta Cryst. (1954). 7, 532

Fig 1.2 Conventional unit cells for the crystal structures based on the hexagonal lattice.

While the case of samarium is indeed interesting, I'd probably omit the third question "...of any inorganic compound?". The answer is obviously yes, and, having little to do with samarium the element, is going to attract numerous answers with huge molecular oxides, inorganic polymers and quasicrystalline structures making it too broad.

The questions regarding unit cell in general could be addressed or fully answered by a couple of queries to WebCSD — Unit Cell Search (provided you have the means to access it).

Alpha Mn has 58 atoms in its unit cell, but is cubic rather than elongated so it misses out …

@JonCuster Interesting! A question about largest unit cell volume or atom count for a single element seems equally interesting to this one. I'll let answers to this one unfold for a while, then think about asking that. On the other hand I'd be delighted if someone went ahead asked it first.

This is the answer to the question

Are any significantly larger lattice constants seen in pure crystal forms of the elements?

and provides arguably the largest single linear parameter of a unit cell for an element, $\pu{45.46 Å}.$
Loa et al. [1] reported the Ba-IVc host-guest phase stable in the range from 18 to 21 GPa and possessing a supercell (ICSD #186822) with the following parameters:
\begin{align}
a &= \pu{33.99 Å} &\quad \alpha &= 90.00^\circ \\
b &= \pu{45.46 Å} &\quad \beta &= 90.00^\circ \\
c &= \pu{13.41 Å} &\quad \gamma &= 90.00^\circ \\
V &= \pu{20720.94 Å^3} \\
Z &= 768
\end{align}

Figure 1 | Basic host–guest motif of the Ba-IV crystal structures.
The red lines indicate the basic host unit cell.

This is far from “ordinary” crystal structures, as authors explain:

[…] the crystal structure lacks translational symmetry along one direction (along $b$) so that neither a crystallographic unit cell nor a space group can be assigned.
Nonetheless, the supercell not only indexes all the reflections, both sharp and diffuse, but also has the size of the smallest possible unit cell of fully ordered variants of this structure (Supplementary Information).
It is therefore appropriate to discuss the Ba-IVc structure in terms of a representative unit of that size, containing 768 atoms.
[…]
Our final model of the full Ba-IVc host–guest crystal structure is illustrated in Fig. 3, with the guest substructure being of particular interest.
All of the chains of guest atoms take one of only two different positions along the chain direction (the $c$ direction), which we will call the up and the down positions hereafter.
In Fig. 3, the up and down chains are coloured yellow and blue, respectively, and
from the relative arrangement of the two types of chain an intricate
two-dimensional pattern emerges (Fig. 3c).
The pattern consists of intertwined S-shaped units, each comprising 12 adjacent chains of the same type, up or down.
At first glance, the pattern in Fig. 3c looks well ordered, but it contains two different configurations of these S-shaped units that are mirror images of each other, and the tiling includes an element of randomness.
It is this tiling of relatively large building blocks with a certain degree of randomness that leads to the unusual diffraction patterns with sharp and diffuse
reflections.
We find it rather striking to observe such structural patterning in an elemental crystal, reminiscent in some ways of Escher’s drawings of plane-filling patterns.

Figure 3 | Schematic view of the crystal structure of Ba-IVc.
a,b, Views of a 768-atom representative unit along the crystallographic $c$ and $b$ directions, respectively.
Host atoms in alternate layers perpendicular to $c$ are shown dark and pale grey.
Guest atoms in the up and down positions are coloured yellow and blue, respectively.
Light blue and light yellow backgrounds highlight the pattern formed by the up and down guest chains.
The red square in a indicates the basic Ba-IV unit cell.
c, Pattern representing the up and down guest chains and decomposition into zigzag and square-wave rows.
d,e, Side views along the $b$ direction of the zigzag and square-wave rows.
f, Displacements of the guest atoms perpendicular to the chain direction.
The reﬁned atomic displacements are shown ×5 magniﬁed.
The chains at the corners of the S-shaped building blocks, marked ×, show a clear zigzag displacement pattern.

Reference

Loa, I.; Nelmes, R. J.; Lundegaard, L. F.; McMahon, M. I. Extraordinarily Complex Crystal Structure with Mesoscopic Patterning in Barium at High Pressure. Nature Mater 2012, 11 (7), 627–632. DOI: 10.1038/nmat3342.

Well this is going to keep me up at night for sure! I think it is an answer rather than the answer to question 2, as what they have produced is not quite a rigorous 3D crystal, though definitely not a quasicrystal. But I'll go off and read this (and the paper) more carefully this weekend.


text:
Can water synthesis and decomposition be in a dynamic equilibrium?
When heated at extreme temperatures, water can spontaneously decompose.
According to Wikipedia (https://en.wikipedia.org/wiki/Water_splitting):

In thermolysis, water molecules split into their atomic components hydrogen and oxygen. For example, at 2200 °C about three percent of all H2O are dissociated into various combinations of hydrogen and oxygen atoms, mostly H, H2, O, O2, and OH. Other reaction products like H2O2 or HO2 remain minor. At the very high temperature of 3000 °C more than half of the water molecules are decomposed, but at ambient temperatures only one molecule in 100 trillion dissociates by the effect of heat.[15] The high temperatures and material constraints have limited the applications of this approach.

At those temperatures, will the hydrogen and oxygen (and combination of those) still react to form water? If so, could this be considered a dynamic equilibrium?

I'm intrigued by the statement "at ambient temperatures only one molecule in 100 trillion dissociates by the effect of heat.". It seems the author might be conflating the self-ionization of water and dissociation into hydrogen and oxygen molecules, even though these are quite different processes. At room temperature, the latter process is vastly more suppressed, but in principle still happens.

At room temperature, $\pu{\Delta G = -228.6 kJ/mol}$. then $\pu{\frac{\Delta G}{RT} = -92.267}$. And $\pu{K =  e^{-\frac{\Delta G}{RT}} = 6.48 · 10^{-41}}$ As  consequence there is not even $1$ molecule $\ce{H2O}$ out of $1$ mole water which is decomposed into $\ce{H2}$ (or $\ce{H}$) and $\ce{O2}$ (or $\ce{O}$) at room temperature.

The reaction you are looking at is
$\ce{H2 (g) + 1/2 O2 (g) ->  H2O (g)}$
The equilibrium constant can be characterised by the reaction's standard Gibbs free energy, which can be calculated from the enthalpy and entropy.
We have, at standard conditions, $\Delta H° =\pu{-241.83 kJ mol-1}$ and $\Delta S°= \pu{-44.16 J K-1 mol-1}$. At 25 °C, this gives a large negative value of the Gibbs free energy, so at equilibrium, we almost exclusively have product (we say that the reaction is irreversible).
However, we can calculate the equilibrium constant from $K=e^{-\frac {\Delta G°} {RT}}$ and as the entropy is negative, we will get a positive $\Delta G$ at some (very high) temperature.
The exact calculation is complicated by the fact that we can't just plug the numbers above into the formula for the Gibbs energy, because the enthalpy and entropy changes significantly over the given temperature range. But if you found the heat capacity data for the three gases, you could calculate $\Delta G°$ at 2200°C.
And to answer your question, the reaction can be considered as a dynamical equilibrium, just like any other chemical reaction. And at a sufficiently high temperature, there will actually be a 50-50 equilibrium between the species.


text:
Mechanism of terpene alcohol rearrangement

I thought about all kinds of hydride shifts that could be possible here, but I still could not figure out how the mechanism would work.

Please edit the title to reflect more specifically what you are asking, perhaps including the type of reaction.

Welcome to ChemSE! Your question will likely be considered as homework and be voted down. You need to make some effort to answer the question yourself. Hydride shifts are a secondary consideration when formation of a C-C bond is the primary issue. Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange

Redraw the starting molecule so that the double next to the hydroxy group is pointing downwards.
Protonate the hydroxy group so that it forms water leaving a tertiary cation.
See how the double bonds are lined up to do a cascade forming the [6] ring and giving a tertiary cation at the opposite end of the newly formed-ring.
Capture that cation with water to give the product.

...and if there is still any doubt, invert the product vertically. ;)


text:
Calculate the volume of acid needed to change the pH of an unbuffered or buffered solution using a strong acid, weak acid, and diprotic acid
It's been a while since I've been in a chemistry class and I'm having a hard time figuring this out. I need to reduce the pH of a solution from 5 to 4. There are 3 acids I can use to do this: 20% v/v sulfuric acid, concentrated hydrochloric acid (assume 36% w/w), or glacial acetic acid. I calculated the molarity of each in the table below using density and molecular weight. For the calculations I'll assume the solution I need to acidify is 1 liter, but I'll actually be scaling this up to a large tank.

Value
20% v/v H2SO4
Concetrated HCl (36% w/w)
Glacial Acetic Acid

mol/L
3.73
11.75
17.42

pKa1
-10
-7
4.76

pKa2
1.99

The starting solution has $[H^+]_1 = 10^{-pH_1} = 10^{-5} \frac{mol}{L}$, and I need to increase it to $[H^+]_2 = 10^{-pH_2} = 10^{-4} \frac{mol}{L}$. For a 1 liter solution, that's $(10^{-4} \frac{mol}{L} - 10^{-5}\frac{mol}{L})*1L = 9.0x10^{-5} mol\,H^+$.
I think I got the HCl alright because it's a strong acid, so 1 mol HCl $\rightarrow$ 1 mol H+. Therefore
$9.0x10^{-5} mol\;H^+*\frac{1L}{11.75mol}=7.66x10^{-6}L$ HCl solution needed.

For the H2SO4, if I ignore the weak acid HSO4-, then I can do the same calculation as for HCl and get
$9.0x10^{-5} mol\;H^+*\frac{1L}{3.73mol\;H_2SO_4}=2.41x10^{-5}L$.
But I'm not sure if I can ignore the second proton since the pKa2 is still pretty low at 1.99. Calculating [H+] from HSO4- than I get
$K_{a2}= \frac{[HSO_4^{2-}][H^+]}{[HSO_4^-]} \rightarrow 10^{-1.99}=\frac{x^2}{3.73M} \rightarrow x=0.195M=[H^+]$. Does that mean that 1 L of 3.73 molar H2SO4 will contribute 3.73 + 0.195 = 3.925 mol H+? In that case I'd need
$9.0x10^{-5} mol\;H^+*\frac{1L}{3.925mol}=2.29x10^{-5}L$ of the H2SO4 solution.
That's 5% difference, which will be a lot when I scale this up.

For the glacial acetic acid, I can do the same calculation I did for the HSO4-.
$K_a= \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \rightarrow 10^{-4.76}=\frac{x^2}{17.42M} \rightarrow x=0.0174M=[H^+]$.
$9.0x10^{-5} mol\;H^+*\frac{1L}{0.0174mol}=0.0517L$ of the acetic acid solution.

So to summarize, to decrease the pH of a 1 liter solution from 5 to 4, I need to add 9.0x10^-5 mol [H+], or
7.66 microliters of 11.75M HCl,
24.1 microliters of 3,73M H2SO4, or
51.7 milliliters of 17.42M CH3COOH (glacial acetic acid).
Is this correct, or am I missing something? Now the kicker - how does this change if the solution I'm trying to acidify is buffered? What do I need to know about the buffer in order to calculate the pH change with these 3 acids?

All your calculations only make sense if you know the nature of the original solution. You have to know the formula or the pKa of the initial acid.  If the initial solution with pH 5 contains a strong acid, the calculation is rather different from the calculation made with a solution of a weak acid.

Are my calculations correct in the case of a strong acid solution, ie not buffered? I think I can determine the pKa of the pH 5 solution experimentally via titration, but how would the calculations change if it is a weak acid?

What is the background of your question? If you are away of education, you are unlikely to do such tasks, that seem to be made up for sake of education.

I work for a small manufacturer that mostly relies on pre-mixed proprietary chemical mixes, but does some R&D here and there. Some chemical suppliers are willing to help with these sorts of inquiries, but others can't be bothered.

If the solution is unbuffered, you could ignore that it has a pH of 5 and start with pure water instead for a rough estimate.
The calculation for $\ce{HCl}$ makes sense. The calculation for $\ce{H2SO4}$ does not; you know that the final pH will be 4. At pH = 4, sulfuric acid will have lost 2 protons (you should have used 0.0001 for the $\ce{H+}$ concentration rather than "x"). So you need half the amount of sulfuric acid compared to hydrochloric acid.
In practical terms, you would hardly be able to add exactly 7.66 $\mu$L HCl to your solution. Typically, you would get a rough estimate of how much you need to add and then add an appropriately concentrated solution dropwise until you reach the desired pH.
If the solution is buffered, you need to know the concentration of the buffer substances along with their pKa values. Or you would need to know the buffer capacity between pH 5 and 4. In this scenario, you might take a small sample and adjust the pH, giving you a sense how much you need to add to the entire solution without zipping past the desired pH.

Thanks for your explanation. Dealing in μL is a bit ridiculous, but this is just a feasibility calculation for a large chemical tank which is 100s of gallons. I'll probably end up trying to adjust a small sample like you suggested, but I also want to understand what's going on chemically, not just operationally.


text:
Schrodinger operator with matrix potential and Green's function
Clearly Schrodinger operators with matrix potentials appear very naturally in molecular dynamics/quantum chemistry, particularly when considering a crude adiabatic basis or diabatic basis for an ansatz solution to the time independent Schrodinger equation of a molecule (see for example Sections 12.1/12.2 of Tannor's "Intro to Quantum Mechanics, a Time Dependent Perspective").
Would knowledge of the existence of a Green's function/fundamental solution for such Schrodinger operators with a matrix potential be of any scientific or computational use?
More precisely, I'm interested in proving the existence of Green's functions/the fundamental solution to  Schrodinger operators of the form $-\Delta \otimes I_{n \times n}  + V$ on an open set $\Omega \subseteq \mathbb{R}^d$.  Here, $I_{n \times n}$ is the $n \times n$ identity matrix, $V = V_+ + V_\epsilon$ where $V_+$ is a Hermitian, positive semidefinite (a.e.) $n \times n$ matrix function and $V_\epsilon$ is some "small perturbation" of $V$ (for example perhaps some Kato type class? I'm not quite sure what would work yet.)


text:
Why is the brown ring complex coloured?
In my textbook it is stated that the brown ring complex formed during the test of nitrate ion is coloured due to charge transfer phenomenon but I think it is due to d-d orbital electron transition. Which is correct and why?

See: https://chemistry.stackexchange.com/questions/22368/are-these-reaction-equations-for-the-formation-of-the-brown-ring-complex-correct?rq=1

@NisargBhavsar there is only one iron atom in the complex

There is an excellent discussion of this compound in this paper:
Monsch, G. and Klufers, P. $\ce{[Fe(H2O)5(NO)]^2+}$, the "Brown-Ring" Chromophore. Angew. Chem. Int. Ed. 2019, 58, 8566-8571. DOI: 10.1002/anie.201902374.
The authors note that this complex is difficult to interpret in the context of the IUPAC definition of formal oxidation state, and that reasonable arguments can be made for Fe(I), Fe(II) or Fe(III) as the proper oxidation state of the central atom, with the NO oxidation state varying accordingly so that the total charge of +2 is retained.
As for the origin of the brown color, how one describes that is also somewhat dependent on what the oxidation state assignment is. There are two electronic transitions associated with the color - one being from the Fe-NO $\pi$ bond orbitals to completely metal centered d orbitals and the other being from the same starting orbital but going into the Fe-NO $\pi^*$ orbital. Whether you describe these as $d \rightarrow d$ transitions on the metal or as ligand-metal charge transfer depends whether you assign the $\pi$ bond electrons to the metal or the ligand in your determination of the oxidation state.

The reactions involved here are:
$$\ce{2 HNO3 + 3H2SO4 + 6FeSO4 -> 3Fe2(SO4)3 + 2NO + 4H2O}$$
$$\ce{[Fe(H2O)6]SO4 + NO -> [Fe(H2O)5NO]SO4 + H2O}$$
In this case, the brown complex is $\ce{[Fe(H2O)5NO]SO4}$ and the colour is determined by the excitation of electrons in d orbitals of the $\ce{Fe+}$. It's also a Laport allowed transition, because the $\ce{NO}$ breaks the central symmetry of the previous complex.


text:
Is N-bromosuccinamide more selective than N-Chlorosuccinamide or TCCA when conducting radical halogenations?
Radical bromination in UV light for instance, is known to strongly favor bromination at the most substituted positions compared to chlorination if elemental bromine and chlorine are the halogenating reagents being compared. Does the same hold for NBS vs NCS or similar chemicals?

NBS is not the overall brominating agent. Bromine is. HBr liberated in the radical bromination catalyzes the slow release of bromin from the NBS, which is only slightly soluble in CCl4. I suppose the same applies to NCS.

The reaction tends always to generate the more stable intermediate (carbocation, carbanion or radical).
With reagents like NBS or NCS the difference is that you generate in situ the halogen radical you need, and is much more selective. Usually, these types of reagents are used to create an allylic halide or benzylic ones.


text:
Reaction between sodium hydroxide and phosphorus trioxide
My chemistry textbook [1], chapter Periodic Properties of Elements, section p-Block elements — formation of oxides suggests that the reaction is
$$\ce{P2O3 + 2 NaOH -> 2 NaHPO3},\tag{R1}$$
but surfing over the internet I couldn't find that compound anywhere (at least in any verified places) which got me thinking whether it was a typo which should be corrected as $\ce{Na2HPO3}$. In fact, the overall reaction I think should be
$$\ce{P2O3 + 2NaOH -> 2Na2HPO3 + H2O},\tag{R2}$$
but that's just my attempt of making a balanced equation. So, which one is correct and why?
Reference

Sanjit Kumar Guha. Chemistry Class XI–XII, New and Revised Ed.; Lecture Publications Ltd.: Bangladesh, 2019; Vol. 1. ISBN 978-984-90642-1-3. (in Bengali)

Yes, that's a typo. Disodium hydrogen phosphite contains the anion $\ce{HPO3^{2-}}$: https://en.wikipedia.org/wiki/Disodium_hydrogen_phosphite

Balanced? Count the number of sodium atoms on each side of the equation.

Phosphorus(III) oxide $\ce{P4O6}$ (molecular formula for “real” cluster of $T_\mathrm d$ symmetry is preferred over historically inaccurate $\ce{P2O3})$ is hydrophobic and undergoes hydrolysis over time when dispersed in water in solid or liquid form (m.p. ~ 24 °C).
Addition of an inorganic base is an nucleophilic attack on $\ce{P4O6}$ that promotes hydrolysis to form corresponding inorganic phosphites.
There is a summary of phosphorus(III) oxide chemistry by D. Heinz [1].
Under ambient conditions $\ce{P4O6}$ hydrolyses over the course of weeks to phosphorous acid $\ce{H3PO3}$, which is contaminated with phosphorus(V) acids and phosphorus suboxides.
However, it has been shown that under mild basic conditions, such as sodium bicarbonate aqueous solution, and lower temperatures (0 to 10 °C) about 70 % of $\ce{P4O6}$ is hydrolysed to diphosphite $\ce{H2P2O5^2-}$:
$$\ce{P4O6(s) + 4 H2O(l) -> 2 H2[H2P2O5](aq)}$$
Heinz further points out that depending on temperature, ionic strength, pH and other factors there are numerous reaction paths leading to different hydrolysis rates and condensation grades.
The most common products, however, are still phosphites and diphosphites.
Products with condensation grade greater than 2 are only intermediate products, and pretty much nonexistent under temperatures over 50 °C due to instability of chain-linked bridged $\ce{P-O-P}$ bonds.
To sum it up, the higher the temperature, the greater the hydrolysis rate and the smaller the diphosphites content among the products.
Phosphorus chemistry is not all that primitive, and there is no way to provide an exact answer without knowing the conditions the author of your textbook implies.
The only fact that is certain is that hydrolysis with $\ce{NaOH}$ will pretty much always yield disodium hydrophosphite $\ce{Na2HPO3},$ but your textbook made a typo and left incorrect stoichiometry (as other answers pointed out).
If you are still looking for the referenced reaction, there is one listed in Lidin's  Reactivity of Inorganic Substances [2, p. 167] for concentrated solution and, as I would speculate, room temperature (albeit the reaction is going to be exothermic):
$$\ce{P4O6 + 6 NaOH (conc)-> 2 Na2HPO3 + Na2H2P2O5 + H2O}$$
References

Heinz, D. Zur Chemie von Phosphor(III)-Oxid. Pure and Applied Chemistry 1975, 44 (2), 141–172. DOI: 10.1351/pac197544020141. (Open Access, in German)
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Your reaction scheme is correct, but the equation isn't correctly balanced and the stoichiometry written in your book is written with a typo. It should be $\ce{Na2HPO3}$. While I haven't found the exact equation in Inorganic Chemistry by Housecroft and Sharpe (my go-to inorganic chemistry reference), they say that $\ce{P4O6}$ reacts with water to form $\ce{H3PO3}$. And I assume that the reaction is carried out with aqueous $\ce{NaOH}$, not the molten version (I don't know if that would work too).
The $\mathrm{p}K_\mathrm{a}$ values of phosphonic acid are $2.00$ and $6.59$, so both will be deprotonated by a hydroxide solution, leading to $\ce{Na2HPO3}$.
It is interesting to note that the third hydrogen is not attached to oxygen, but to phosphorus, so the $\ce{PO3^3-}$ ion does not form.


text:
Why would 1,2,3,4-tetramethyl-1,2-dihydro-1,2-diazete be aromatic?

If the lone pair of the nitrogen on the bottom were to make the double bond with the nitrogen at the top, then the nitrogen on the top would have 10 electrons, meaning that the lone pair on the top nitrogen needs to shift to the left carbon.
Which will cause the π-bond electrons to shift and form a double bond with the bottom nitrogen.
In the end, each nitrogen will have a trigonal planar shape with two double bonds (one with carbon on the left and one with each other) and one single bond(?).

I do not understand how this structure would make sense.
This seems very unstable to be aromatic.

There's no such structure and the compound isn't aromatic.


text:
Atomic Radius of Manganese and Chromium
It is clearly a general trend that the atomic radii of the 3d elements in ascending order of atomic weight, leads to the following trend as in the graph below.

But, it was pointed out to me that, Mn has a higher metallic/ionic $[M^+]$ radius than chromium in reality, as given the radius values from literature.
How can this anomaly be explained?
The only thing different in Mn from Chromium is one proton more in the nucleas and the presence of 5 unpaired electrons(due to filled 4s orbital) in Mn, than in Cr(which has 6 unpaired electrons in 4s and 3d orbitals).
But again the 4s orbital is outside to the 3d orbital and therefore it is not contributing to the shielding, at least to an appreciable extent.
Please explain.

The reason for your confusion may be the difference between atomic radius and ionic radius, as well as the specific comparison between ions that your source was making. Comparing manganese and chromium ions with the same charge, spin and coordination numbers, chromium ions are larger about half of the time (see http://abulafia.mt.ic.ac.uk/shannon/). However, since they didn’t seem to specify which ionic radius they were talking about, I suspect that whoever told you that manganese ions were larger was comparing the two elements in their most stable and common oxidation states, and $Mn^{2+}$ ions are indeed larger than $Cr^{3+}$.


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Relevance of equilibrium constant Kp for reaction when reactant is solely a solid and is absent
Consider a reaction where a solid dissociates to give gases. There must a equilibrium between them, define the equilibrium constant in terms of pressure as $K_p$. Suppose I take the gases alone in a container without the solid. The amount is such that the $Q_p$ comes out to be less than $K_p$. Now would they show back reaction?
While being introduced to concept of $K_c$ we are told concentration of solids is constant so it isn't considered in equilibrium constants. But here the concentration is altogether 0, so why won't/how will we take that into consideration.

The concentration of the solid is constant but not that of the gases. If you mix gases that can react to form a solid they will do so until the $K_p$ is satisfied.

Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?

Think of ammonium chloride $\ce{NH4Cl}$. It gets dissociated into two gases : $\ce{NH3}$ and $\ce{HCl}$ at high enough temperature.

@andselisk the answer well explains why concentration of solids is constant, but that too is when the some solid is indeed present, here it is completely absent initially, so wouldn't we observe back reaction?


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Cannizzaro reaction via electrolysis
As cannizzaro reaction reaction is a redox reaction carried by adding base to an aldehyde, is it possible to carry out the reaction using electrolysis where graphite anode and copper cathode is used. Is it possible to carry out electrolysis of benzaldehyde in water solution to get benzoic acid and benzyl alcohol?

Let's start with reduction processes. In his book, Alfred J. Fry states in § 6.1.3. (page 217) that :

The electrochemical reduction of carbonyl compounds to alcohols is of limited synthetic interest, in view of the wide variety of chemical reagents that can also effect this conversion.
But the electrochemical route is one of the best methods for preparation of pinacols derived from aryl aldehydes and diaryl ketones. They may be reduced to pinacols in alkaline or aprotid solvents : yields are usually high ($\pu{85\%}$)
The reduction of carbonyl compounds to saturated hydrocarbons in strong acids at cathodes of lead compares favorably with procedures like Clemmensen and Wolff-Kishner reactions in both yield and simplicity.

Let's now discuss electrochemical oxidations. In page 273, the same author points out that :

At present, electrochemical oxidation of organic compounds is not nearly as valuable a technique as is electrochemical reduction. This is due to the fact that the mechanism of anodic processes had been studied far less than cathodic processes. There is an exception : the Kolbe reaction, where salts of carboxylic acids result in dimeric products plus $\ce{CO2}$

Ref.: Albert J. Fry, Synthetic Organic Electrochemistry, Harper & Row Publishers, New York 1972, 336 p. Library of Congress Number 72-82902

So is it possible to carry out cannizzaro reaction electrochemically? considering the electrolysis if water produces at cathode:2 H2O (l) + 2 e - → H2 (g) + 2 OH- (aq) and at anode:2 H2O (l) → O2(g) + 4 H+ (aq) + 4 e -  so can the OH- ion formed at cathode somehow do the same reaction as in basic solution? Can the reduction and oxidation take place within solution via electrolysis and form the cannizzaro products?

I took the liberty to edit your answer by blockquoting the text. Feel free to revert back if you don't  like the change. Is there any reason to put page numbers inside dollar sign?


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How to prepare standard sample of benzene for gas sampling bag?
I need to make a 1 ppm gaseous standard of benzene in nitrogen gas within a Tedlar® gas sampling bag. Is this as simple as injecting 1 μL of neat benzene liquid into 1 L of nitrogen gas within the bag?
Or do I need to weigh out the benzene liquid and use a calculation including the molecular mass of benzene?

You need 1 microlitre of benzene vapor per 1 L of N2 at the same conditions as the used N2, if ppm(v/v) is assumed. Or recalculated V for different p or T. I suggest to use a kind of concentrated  benzene vapor in N2 mixture and use this for final standard.

Here we go with ppm again…

Consider obtaining a crash course about gas laws and metrology basics  or delegation of  the task.


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Can metal displacement reactions happen with insoluable salts?
I know how normal metal displacement reactions work. If one metal is more reactive than the other, they will displace.
For example, normally for this reaction:
$$\ce{Fe_{(s)} + 2AgCl_{(aq)} -> Fe^{+2}_{(aq)} + 2Ag_{(s)} + 2Cl^{-}_{(aq)}}$$
I would say that because the iron is more reactive than silver, the iron pushes it's electrons onto the silver.
But in this case, can this reaction depicted actually occur? Because $\ce{AgCl}$ is an insoluble salt. This means that the $\ce{Ag+}$ ions are not floating around to touch the iron metal. This means the reaction cannot occur.
I think in reality since $\ce{AgCl}$ can dissolve a little bit due to equilibrium, the reaction would go ahead, but if we pretend equilibrium dissolution doesn't occur, can this reaction proceed?
I think it can occur if the iron is touching the AgCl directly (allowing for conduction). I also think it cannot proceed if the iron was not touching the AgCl. Is this correct? For the purposes of answering my question, please ignore the solubility of AgCl.

If one metal is more reactive than the other, they will displace.  // While it is generally true, even less reactive  metal ( slower reaction kinetics) can displace the more reactive metal from its salt, if it is preferred by the reaction thermodynamics by the sign of the Gibbs free reaction energy. // Insolubility is rather a vaque name convention for salts with very limited solubility. Reaction would be very slow, unless it is supported by dissolving of iron in acidic solution and by reduction with transient atomic hydrogen.

It doesn't matter if the reactants are salts or not: consider thermite, where aluminum displaces iron from its oxide. What would happen if AgCl, in contact with Fe, were warmed to 455 °C?

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

In fact, at least on the basis of thermodynamics, a small amount of reaction between iron metal and "solid" silver chloride can be expected in water solvent.
This is because silver chloride is not completely insoluble in water. So if the thermodynamics is favorable, at least a small amount of reaction can occur.
First off, Ref. [1] quotes a standard reduction potential of -0.44 V for iron(II)/iron metal, versus +0.80 V for silver(I)/silver metal (all data and calculations are for 25°C and one atmosphere). Applying the Nernst Equation then gives  the following equilibrium constant:
$\ce{2 Ag^+ + Fe <=> 2Ag + Fe^{2+}}, K = [\ce{Fe^{2+}}]/[\ce{Ag^+}]^2=8.5×10^{41}$
Taking silver chloride as having a solubility product of $1.77×10^{-10}$, we add in the dissolution reaction and render
$\ce{2 AgCl + Fe <=> 2Ag + Fe^{2+} + 2Cl^-}, K = [\ce{Fe^{2+}}][\ce{Cl^-}]^2=2.6×10^{22}$
This equilibrium condition is far beyond the range of concentrations/activities in a real aqueous ferrous chloride solution. So, what small amount of silver chloride dissolves into the water can undergo displacement favored by the large equilibrium constants above, producing ferrous chloride and silver metal, whereupon the removal of silver ion enables more silver chloride to dissolve. The reaction will eventually slow down due to accumulating chloride ions decreasing the solubility of the silver chloride, but the thermodynamics and mechanism are there for at least a little reaction to occur.
References
1.
Veleva L., “Soils and Corrosion” (Chapter 32), in Corrosion Tests and Standards: Application and Interpretation, 2nd Edition, R. Baboian Ed., ISBN: 0-8031-2058-3, ASTM International, OH, pp.387-404, 2005.

Thank you - but you haven't answered my question. I know that AgCl is not fully insoluable. Indeed I have referenced this in my original question, and how the slight dissolution would continually shift the equilibrium. My question was actually, what if we assume no Ag+ ions dissolved into solution? Would it still cause the deposition of silver onto the nail?


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Why aren't electrolytic cells usually separated?
Galvanic cells need to be separated so that electrons will be forced to move through the external circuit, but I was wondering why electrolytic cells are usually not kept separated but rather as a single cell? Is it just convenience or do the reactions take place differently if the two electrodes of an electrolytic cell were to be separated?

Galvanic cells need to be separated so that electrons will be forced to move through the external circuit [...]

In a Galvanic cell, the reaction has a negative Gibbs energy of reaction. If the reactants are allowed to come in contact, the reaction proceeds without generating electricity (generating more heat instead). In some cases (like a lead acid battery), the reactants can't come into contact because they are both in the solid state, one on the anode and the other on the cathode. In other cases, they would come into contact, so there has to be a barrier (and separate half-cells).

[...] but I was wondering why electrolytic cells are usually not kept separated but rather as a single cell?

In an electrolytic cell, the reaction has a positive Gibbs energy. If the products are allowed to come into contact, they would react (unless there is a need for a catalyst) in the reverse direction, leading to a futile cycle. In many cases (where the products are gases or solids or liquids that don't mix with the electrolyte), the products are unable to come into contact, like in molten NaCl electrolysis. In other cases, like the electrolysis of water, the products could potentially mix in the gas phase but are not sufficiently reactive to support the reverse reaction (for water electrolysis, you would have to collect oxygen and hydrogen in a common vessel and then ignite them with a spark, either on purpose or by accident).

Is it just convenience or do the reactions take place differently if the two electrodes of an electrolytic cell were to be separated?

You can't make a generalization. It depends on the specific reaction.

@planckton. Galvanic cells need not to be "separated" as you state. What has to be separated is not the cells, but the half-cells. In the Daniell cell, Zinc is the anode and emits electrons which move through the external circuit. But this phenomena also produces $\ce{Zn^{2+}}$ ions in solution. This is only possible if some negative ions are present in solution. These negative ions are coming from the cathode region, which is filled with $\ce{CuSO4}$ solution. As the $\ce{Cu^{2+}}$ ions are discharged by the incoming electrons, the sulfate ions are repelled, and must leave : they cross the membrane or the bridge between the two half-cells, to arrive in the anodic solution, where the $\ce{Zn^{2+}}$ ions are attracting them.


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Just how accurate is this XKCD comic about building a real-life periodic table with cubes made up of each element?
I was reading the book “What If?” by Randall Munroe and this part was very intriguing to me. He answered the question in the title and I was curious if his answer was realistic, and if not, which parts he got wrong.
This is the comic I'm talking about. I'm especially curious if Astatine is really that dangerous.

Yes, pretty much. This guy is generally very good with his estimates. Astatine does not exist; if it would, then yes it would be that dangerous.

Astatine exists, just not for very long…

https://chemistry.stackexchange.com/questions/141585/could-there-ever-be-a-way-to-safely-handle-visible-amounts-of-elements-like-asta

More reliable sources reckon there is a total amount of 45 milligrams in the earth's crust. The difference to "doesn't exist" is imo negligible. ;) https://ataridogdaze.com/science/As-Earth-crust.html Although if you had some magic to collect those 45 mg in your hand in one moment, you'de be indeed dead  in the next.

While his general point about the lower-down radioactive elements is completely right, there are several uncharacteristic errors in the chapter (though it is broadly correct despite them.) He calls ammonia an element; he confuses two different stories about russian defectors being killed (the one killed with polonium was poisoned via tea, the umbrella incident used ricin).

Is your question about the entire chapter (which is likely too broad as it would require a substantial amount of fact-checking), or just the astatine part? If it's the latter, could I encourage you to [edit] your question title & text to more clearly reflect that? Thanks.

BTW, I'm reminded of a science fiction story about an intelligent, ant-like race, found on a planetoid far from it's sun. The workers were separating and collecting, in a circular amphitheater, each element, starting with an outer ring of frozen hydrogen, then liquid He, and so forth. A human explorer contacts and befriends the queen, in the amphitheater's center. He realizes she should stop before uranium, but she continues until the explosion destroys her... and lofts her eggs to colonize other planetoids.  Anyone know name or author?

@DrMoishePippik I don't. Maybe Someone one Scifi.SE...

Randall does say (p.37 bottom) "the other half of the data is from Look Around You, which was a mock-"educational" show presenting utter nonsense (in serious voice)...  Also keep in mind his disclaimer "I like when things catch fire and explode, which means I do not have your best interests in mind." elsewhere in the book


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Is there a good citation for the impact of inelastic neutron scattering on the measurement of the intramolecular geometry of H2O?
I have tried to find a good citation for the experimental measurement of the intramolecular geometry of H2O in the liquid phase and come across a problem:
It seems that the method of choice for such a problem is the use of elastic neutron scattering/diffraction and that this exact method cannot be applied to liquid water due to the properties of protons, specifically their nearly identical mass compared to neutrons.
I can find a bunch of Wikipedia references that explain this issue, but, as per usual, there is no usable citation attached to the relevant paragraph. The Wiki article on neutron diffraction states:
"It is undesirable to work with the relatively high concentration of H in a sample. The scattering intensity by H-nuclei has a large inelastic component, which creates a large continuous background that is more or less independent of scattering angle."
https://en.wikipedia.org/wiki/Neutron_diffraction#Hydrogen,_null-scattering_and_contrast_variation
I remembered that I've heard something similar in regards to the necessity to use heavy water (D2O) instead of regular water in nuclear reactors due to the effect of efficient transfer of momentum from neutrons to hydrogens. Apparently the efficiency of this kinetic energy transfer increases the smaller the differences in mass between two colliding particles/atoms are (neutrons and hydrogens obviously have almost identical masses). So in order to achieve high enough criticality in a nuclear reactor (sustained fission with an average emission of 1 neutron per fission reaction), D2O needs to be used as a mediator because the hydrogens of regular water would absorb the kinetic energy of all emitted neutrons and cause the nuclear reaction to cease.
"Water makes an excellent moderator; the ordinary hydrogen or protium atoms in the water molecules are very close in mass to a single neutron, and so their collisions result in a very efficient transfer of momentum, similar conceptually to the collision of two billiard balls. However, as well as being a good moderator, ordinary water is also quite effective at absorbing neutrons. And so using ordinary water as a moderator will easily absorb so many neutrons that too few are left to sustain a chain reaction with the small isolated 235U nuclei in the fuel, thus precluding criticality in natural uranium."
https://en.wikipedia.org/wiki/Pressurized_heavy-water_reactor
So I know that elastic neutron scattering cannot be done to a high level of precision in liquid water, which is why studies that measure the intramolecular geometry of water using neutron scattering use D2O instead of H2O.
https://www.tandfonline.com/doi/abs/10.1080/00268977700103011
And I am pretty sure I've understood why that is. Now: why can't I find a good article or book excerpt to cite for my report to outline this issue as to why comparisons between force field calculations of the water geometry and experimental data are an issue?
Can anyone find what I can't?


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If there are two possible reactions, will the one with a more negative Gibbs Free Energy be favored?
I was thinking of the reaction between methanol (CH3OH) and sodium amide (NaNH2) and wondering if I could get methylamine (CH3NH2) from it. One reaction forms methylamine and sodium hydroxide, while the other produces sodium methoxide and ammonia. When calculating the Gibbs Free Energy for both reaction at STP, the Gibbs Free Energy for the methylamine and sodium hydroxide reaction is about x1.5 more negative compared to the other reaction. However, many websites say that the sodium methoxide reaction is the one that happens. Is this true? If so, why?

Favoured thermodynamically or kinetically?

Thermodynamically. I don't know the kinetics of it, but from what I can conclude they should be about the same.

Hmm, I would say the acidobasic reaction would be much faster than substitution.

I understand that substitution reactions are generally slower than acid-base reactions, but is there a way to make it more favorable so that the reaction will proceed?


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What factors influence the upper size limit of micelle's?
Is there an inherent size at which micelle's cannot grow larger in volume?
I have been trying to find a quantifiable method of measuring a size limit with all the variables that affect micelle formation and growth (solvent polarity, temperature, etcetera), but I've failed at finding a good resolute answer. Was wondering how one could compute a result for this.

Micelles form from individual surfactant molecules , beginning at the CMC (critical micelle concentration), which varies for different surfactants, but is in the range of about 100 ppm (Ref 1).

At one end of the scale, sodium octyl sulfate (mw = 232) requires 0.13M to form micelles because it is soluble enough to exist surrounded by water molecules until there are enough other octyl sulfate anions nearby that the octyl groups associate. The surfactants with lower CMC associate more at lower concentrations.
The first micelles will be small, but grow larger as the concentration of surfactant increases. As long as they stay dispersed (Brownian motion), they might as well be soluble. However, if the concentration of the surfactant increases, it will become a separate phase, a separate layer (usually floating). Some dispersible surfactants (like polyethylene oxide surfactants) have a cloud point, where raising the temperature decreases the solubility to the point where micelles prefer agglomeration rather than dispersion in water. Just below the cloud point is where the micelles would be largest. Above that temperature, if two of these very large micelles collide, they begin forming the separate (floating) phase.
Stirring a concentrated surfactant solution with high shear would generate some very large (but unstable) globs which resemble micelles - not unlike making an oil-in-water emulsion. But this is probably not what you want.
Even with surfactants without a cloud point, increasing the concentration to just below the solubility limit would create the largest possible micelles.
Surfactant solutions (even below the cloud point) show a Tyndall effect, scattering of light. At the concentrations producing large micelles, the micelle size would be significantly larger than the wavelength of light, and a drop of the solution between glass slides could be examined with a good microscope. If you insure the micelles are not squashed, a good estimate of diameter, and thus volume, could likely be obtained.
An off-the-top-of-my-head estimate for the largest possible micelle volume would be the volume of a particle with the density of the surfactant that can be suspended indefinitely by Brownian motion against settling by gravity. Smaller particles would be more stable; larger particles would coalesce. Thickening the water with gums like xanthan could stabilize much larger micelles. A surfactant with density very close to H2O would also confer greater stability (in H2O) to large micelles. And lower gravity (space station?) might work too.
Ref 1. https://en.wikipedia.org/wiki/Critical_micelle_concentration

I'm thinking of soap bubbles blown in air. It possible to make them quite large. Surface tension is probably a factor that limits the size of a micelle in a particular system.


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Experimental Verification of Van der Waals parameters?
I am currently trying to find an academic paper that experimentally verifies the a and b values of Van der Waals forces for carbon dioxide. I can't seem to find this, and am wondering if I am missing something. Could somebody either give me a reason this is so hard to find, or link me a paper I have proven myself incapable of finding?
If this is impossible, is there a paper that empirically computes the answer?

It is the best tested by verification of critical parameters, which are directly linked to the vdW a,b parameters. Measuring V=f(p,n,T) or similar would give different parameters,  depending on the used p,T  region and deviation of the real gas from vdW model.

There are a dozen of far better gas equations. Nobody uses vdW, because if you actually need the improved accuracy over the ideal gas equation, you'd have to either experimentally determine good parameters for the p,V range you are interested in, or derive them from one of the other equations (which have up to ten parameters!). The first is a lot of work, the second is pointless, because you can directly use the more accurate equation.


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Can entropy of universe remain constant in a process where P and T is constant?
We know $$\Delta S(\text{univ})=\Delta S(\text{sys})+\Delta S(\text{surr})$$ at constant $P$, $T$.
$$\Delta S(\text{sys})=\Delta H(\text{sys})/T$$
$$\Delta S(\text{surr})=\Delta H(\text{surr})/T$$
As $\Delta H(\text{univ})=0$, $\Delta S(\text{sys})+\Delta S(\text{surr})=0$
Or $\Delta S(\text{univ})=0$
But we have $\Delta S(\text{univ})>0$ is a must for spontaneous process. That would mean no process is spontaneous at constant $P$, $T$.
What is the fallacy here?

What are the temperature and pressure of universe to keep them constant?

@poutnik i don't get what you mean. I have taken P, T for system only, not for universe

It is not about constant pressure or temperature but on reversibility. What ever may be the process the change in entropy for universe is positive. But zero for reversible process. This is popularly called second law of thermodynamics.

@Infinite can you elaborate where reversibility comes into play here ? And what is wrong with my proof ?

The formula that you used ( ΔS = ΔH/T ) is applicable for reversible process but not for irreversible process. And in reality no process is reversible.

@Infinite ok, i missed that

An irreversible process at constant temperature and pressure means that the temperature of the surroundings (assumed ideal) is held at the same temperature T as the initial temperature of the system throughout the process.  So, for the surroundings, $$\Delta S_{surr}=\frac{-\Delta H_{syst}}{T}$$.  For the system, according to the Clausius inequality, $$\Delta S_{syst}=\frac{\Delta H_{syst}}{T}+\sigma$$where $\sigma$ is the amount of generated entropy due to irreversibility (always > 0).  If we add these two equations together, we obtain:  $$\Delta S_{universe}=\Delta S_{syst}+\Delta S_{surr}=\sigma>0$$If the process is carried out reversibly, then $\sigma=0$.


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Size of Orbitals, Making Intuitive Sense of Quantum Model, Nomenclature of Subshells in the Quantum Model
Alright, so I am doing the Quantum Mechanical (or what some people call the Wave Mechanical model) of an atom. There's this part where we have different zones of the probablity of finding electrons, the orbitals. The orbitals are a part of our imaginary concept of shells and subshells. (I know that it might not be completely true at the higher academic levels, but that is the vocabulary I know for now.) I have a few different questions about this atomic model.
A. What is the size of an orbital, or a subshell, or a shell in this model? Is it somewhere related to the Heisenberg postion uncertainty? Like, the size of an orbital being equal to how much uncertain we are about the postion of an electron?
B. My teacher keeps saying that all these shapes of orbitals and even their existence is purely mathematical. Is there really no intuitive sense we can make of the reason of their presence as such without involving any differential equations?
C. Why are the names of subshells based on completely random letters (s, p, d and f)? I tried searching for some source on the internet but it simply dwells deeper into their shapes and planes and graphs rather than explaining their names.

Hey. I posted the question and would just want to add a few more things. First thing, I know my understanding of the Quantum Model might be flawed, simply because of how little we are taught about it and how difficult a concept it is. Please give any suggestions you might have about that. Second, please answer my questions based on the academic level of a typical high school student. Thanks a lot.

The format of this Qustions/Answers site is intended rather for filling the gaps, not for laying the foundations. Consider studying available textbook and online materials and rasing follow up questions. // Search valuable sites with the term  like site:libretexts.org or  site:hyperphysics.phy-astr.gsu.edu or site:en.wikipedia.org.  See also Chemistry SE: resources-for-learning-chemistry

Thanks @Poutnik for the suggestion, I will definitely look up these resources. But I am afraid that most books don't answer these questions about size of orbitals and other stuff. I looked up a few books before posting the question myself, but in vain. It would be helpful if you could lead me to some specific text. Thanks, again.

Especially the libretexts site is specifically focused of education at level you can understand. Hyperphysics format is then done by very usable form of a "cheat cards", very easy to digest. See googled topic orbitals on the respective sites Libretexts and Hyperphysics

You may remember the order of the subshells $spdf$ by the following reasoning : $s$ is the first letter of "spherical", because the $s$ orbitals have a spherical symmetry. $p$ is the first letter of  "plane", meaning that the $p$ orbital has a planar symmetry. $d$ is the first letter of "different", because these orbitals have different symmetry surfaces. Other orbitals are defined by alphabetic order after "$d$", avoiding the letter $e$ which is used for exponential. Historically speaking, the letters $s, p, d, f$ were named after other criterias, which are harder to explain.

Addressing the last part of your post

Why are the names of subshells based on completely random letters (s,
p, d and f)? I tried searching for some source on the internet but it
simply dwells deeper into their shapes and planes and graphs rather
than explaining their names.

These letters are not random and as you can see all over the internet (e.g., Wikipedia) they stand for sharp, principal, diffuse and fundamental. After that the story ends without telling what is sharp, principal, etc. What are those adjective for and what do they describe? These letters described the nature of the lines in an atomic spectrum and this system was not universal (Taken from "A Report on Series Spectra by A Fowler).

Basically, long before the concept of orbitals originated, people were fascinated by the spectrum of alkali metals. They have a rich spectrum if you excite them under an electric arc. When you view the arc through a spectroscope (=a device that separates wavelengths), you see a number of lines these letters described those lines as they appeared to the viewer! Pure visual labels.

Now you don't see fundamental there because the fundamental series was discovered in the infrared region. So no visual label, it is just a label. Nothing fundamental about it.
Now, spectroscopists, mathematicians, physicists wanted to find out a mathemtical relation among the wavelengths of those lines. They came up with mathematical series that described the wavelengths as terms of a series. When quantum mechanics was developed this nomenclature was retained since they were able to assign the electronic transitions in orbitals corresponding to those wavelengths. I will have check original papers from 1920s to see who first assigned them.
P.S. Saw an interesting comment below the original post. I want to correct it since some student later on start to spread this reasoning in future on other websites.

You may remember the order of the subshells spdf by the following
reasoning : s is the first letter of "spherical", because the s
orbitals have a spherical symmetry. p is the first letter of "plane",
meaning that the p orbital has a planar symmetry. d is the first
letter of "different", because these orbitals have different symmetry
surfaces. Other orbitals are defined by alphabetic order after "d",
avoiding the letter e which is used for exponential. Historically
speaking, the letters s,p,d,f were named after other criterias, which
are harder to explain.

I am afraid this is completely imaginary and there is no historical work which justifies the choice of the names orbitals using the above reasoning. The letters do not stand for spherical, plane, and different. The letter f was not chosen to avoid "exponential".

Thanks a lot @M. Farooq! This was exactly the kind of answer I was looking forward to my question. I have a few counter questions, and will be really glad if you answer them too. First, isn't the spectrum you mention the same thing we do for hydrogen and what Bohr used for justifying the quantisation of energy? Why don't we have different bands like these there as well? Second, can you please tell what source are you referring to, since that will be really helpful to me for further information. Thanks. (:

Ayush, Welcome, the source is already mentioned, "A Report on Series Spectra" by A Fowler. Go to Internet archive and locate the book. And post a separate question for Bohr.

The orbital sizes and shapes represent the probability of finding an electron in that space with 95% confidence. You are correct that it is related to the uncertainty principle - we can only describe electron positions as probabilities.
s, p, d, f are short terms for the funny names physicists give to things when they first observe them but don't know what they are. In the case of orbitals, these are "sharp," "principal," "diffuse," and "fine" (or "fundamental").
Based on your question and curiosity abut these things, I think you may enjoy reading Richard Rhodes' book The Making of the Atomic Bomb. The first half of the book is a fascinating and accessible history of quantum physics. That's where I learned what s, p, d, and f stand for.

The probabilistic interpretation of QM (see Born rule) has not all that much to do with the Heisenberg uncertainty principle at all. IMO, the HUP is often invoked in places where it shouldn't be, or needn't be. Basically: orbitals have a size because they represent a probability distribution -- that's true. The HUP doesn't state that, though; the HUP only enforces a constraint on that probability distribution (which the orbitals satisfy).

@orthocresol, that is exactly I want to know. If it isn't the uncertainty principle which predicts the size of an orbital, then what does? All I have read about quantum numbers only tells about "how many" electrons are located in a specific orbital and not about where they are located.

@AyushRoy That depends on how you define the "size" of an orbital. Orbitals don't have a "volume"; they actually extend out to infinity (the conventional pictorial form essentially truncates it), so if you wanted to go by where their furthest extent is, then all orbitals are the same (a very lame conclusion). A more sensible metric might be the mean distance from the nucleus, which you can obtain through integration once you know the mathematical forms of the orbitals (e.g., google '1s wavefunction'). I don't want to go into the details here, but none of this has to do with the HUP, really.

Here is the connection between Born, Heisenberg, and the hertofore not mentioned Jordan. It's a bit easier to comprehend than the previous link. https://aapt.scitation.org/doi/10.1119/1.3009634

@orthocresol, I meant orbital to the region of the maximum probability, something like 90-95 percent. Anyways, thanks a lot orthocresol and LynneKLR. I will definitely look up these sources.