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Will bubbling nitric oxide through a solution of sodium percarbonate neutralize the NO?
The sciencemadness wiki page on Nitric Oxide (https://www.sciencemadness.org/smwiki/index.php/Nitric_oxide) states that NO can be neutralized by "bubbling it in a sodium percarbonate solution". Unfortunately, no further source or information on this technique is provided.
My inexperienced guess is that the H2O2 from the dissolved sodium percarbonate oxidizes the NO to NO2, which reacts with water to form HNO3 and is finally neutralized by the carbonate ions in solution. Does this sound anywhere close to correct? However, since NO is almost insoluble in water, will bubbling with an inverted funnel trap really react with it quick enough before the bubbles reach the surface?
Finally, is there any danger to attempting this scrubbing method other than unreacted NO reaching the surface and turning into NO2 gas?
Any help is appreciated, thank you.

I would not say danger, but percarbonate should to be kept in excess to stop neutralisation at bicarbonate level. Otherwise evolved CO2 could be stripping out part of NO.


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Methods for immobilization of DNA on gold surfaces?
I've been looking for methods for immobilizing 5'-Thiol modified dsDNA on gold slides. Most of what I've come across suggests using DTT followed by a desalting step or using TCEP. However, I haven't had any luck reducing and then conjugating the DNA onto the gold slides. I've seen some methods use DTT, and some use TCEP, and some include an incubation step with MCH. It's been difficult finding a complete and satisfactorily detailed method online.
The DNA I am using is a 1 kb long double-stranded molecule with a 5'-Thio modification at one end and a 5'6-FAM modification at the other end. I need to synthesize the strands myself via PCR using two modified primers and a template strand rather than ordering the fully modified 1 kb strand, so my working concentrations are usually fairly low (~70-100 ng/uL after PCR and spin-column purification).
Can anyone provide a detailed method that has worked for them for reducing thiol modified DNA and immobilizing it onto gold surfaces or gold monolayers?

The MCH method is described for ssDNA. Would it still work for dsDNA is up in air.

The most of the procedures you have described are for binding single-stranded DNA (ss-DNA) to gold surfaces via thiolate bonds. To my knowledge, I'm not sure (note that I'm not an expert on this field) these methods would work for double-stranded DNA (ds-DNA) without some modifications. However, the authors of Ref.1 have claimed, to their knowledge, the first procedure for binding DNA (supposedly ds-DNA) via thiolate bonds to gold surfaces.

Accordingly, thiol-substituted deoxy-oligonucleotides were used as primers in PCR, which were obtained by esterification of a 6-mercapto-hexanol-group to the 5'-phosphate of oligonucleotides (20-mers) using a  $\beta$-cyanoethyl-phosphoramidite C6-thiol-modifier. The fragments were amplified by PCR using either thiol-modified oligonucleotides or, as a control, unmodified oligonucleotides. To avoid oligomerization of the oligonucleotides during the PCR reaction, dithiothreitol (DTT) with final concentration $\pu{2 mM}$ has been used in the polymerization mixture. After PCR, the DNA fragments were extracted with phenol, precipitated with ethanol, and purified by agarose gel electrophoresis. Since residual agarose, even in minute amounts, might be mistakenly interpreted as DNA threads, the thiolated DNA fragments were further purified by chromatography on Elu Tip d according to a standard protocol given in Ref.2. Purified DNA was dissolved in either C Tris-$\ce{HCl}$, $\mathrm{pH}$ 7.5, or $\pu{10 mM}$ ammonium acetate.

Prior to use, a  200-fold excess of DTT should be added to $\pu{0.1 \mu M}$ solution of DNA fragments, and incubated for $30$-$\pu{60 min}$. The template-stripped gold surfaces were prepared as described in Ref.3. Following the stripping with tetrahydrofurane the mica sheet was lifted gently using a  slight vacuum suction. The gold-platelets were dried in a  desiccator for $\pu{15 min}$ prior to use.

Routinely, a  droplet of $\pu{1 \mu L}$ ($0.3$-$\pu{0.5 fmol}$) of DNA in DTT and either $\pu{10 mM}$ ammonium acetate or Tris-$\ce{HCl}$ buffer was transferred onto the template-stripped gold surface, and left for $45$-$\pu{60 min}$ at room temperature at 100% relative humidity; then dried for $10$-$\pu{30 min}$ in a  desiccator, and rehydrated in either ammonium acetate or Tris-buffer. This cycle of drying and rehydration - which should not irreversibly denature the DNAs - was found to lead to a  fair amount of DNA being anchored to the gold surface. After mounting in the instrument to be used, the fluid cell was flushed with either ammonium acetate or Tris-buffer, which removed non-thiolate-anchored DNA. In principle, thiolated DNA fragments can be anchored to gold directly from the droplet solution without drying; however, the amount bound was then found to be quite small.

Also read Ref.4 for most recent work in the subject.
References:

Martin Hegner, Peter Wagner, and Giorgio Semenza, "Immobilizing DNA on gold via thiol modification for atomic force microscopy imaging in buffer solutions," FEBS Lett. 1993, 336(3), 452-456(DOI: https://doi.org/10.1016/0014-5793(93)80854-N).
Seldon, R.F. and Chory, J. (1993) in: Current Protocols in Molecular Biology (Ausubel, F.M., Brent, R., Kingston, R.E., Moore, D.D., Seidman, J.G., Smith, J.A. and Struhl, K. eds.) Greene Publishing Associates and Wiley-Interscience: New York, NY, chap. 2.6, Suppl. 20.
Martin Hegner, Peter Wagner, and Giorgio Semenza, "Ultralarge atomically flat template-stripped Au surfaces for scanning probe microscopy," Surface Sci. 1993, 291(1-2), 39-46 (DOI: https://doi.org/10.1016/0039-6028(93)91474-4).
Janese C. O'Brien, Vivian W. Jones, Marc D. Porter, Curtis L. Mosher, and Eric Henderson, "Immunosensing Platforms Using Spontaneously Adsorbed Antibody Fragments on Gold," Anal. Chem. 2000, 72(4), 703–710 (DOI: https://doi.org/10.1021/ac990581e).

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Electrolytic Cell and Electrochemical Cell
Why is the Anode positively charged in Electrolytic Cell and negatively charged in Electrochemical Cell when all it does in both cells is oxidation, i.e. lose electrons.
Similarly, for the Cathode as well.

Review the guides Asking and How to ask. Not following the guidance may lead to lack of satisfying answers, objections, question down-voting or even question closure.

Imagine the galvanic cell Zn|ZnSO4||CuSO4|Cu. Imagine applying external voltage on it at such a value the current is zero. Now imagine what would happen if you 1/ increase the external voltage a little 2/ decrease the external voltage a little.  // At more positive electrode (Cu) would occur oxidation(anode) in the (electrolytic) case 1/, reduction(cathode) at (galvanic) case 2.

Nomenclature The anode is not positively charged, neither the cathode is negatively charged. There is not an accumulation of charge in the electrodes, rather we have electron transfer from one to another. If there was an accumulation of electrons in the electrodes, batteries would not exist, because electrons would not be able to flow externally and try to use them to produce work.
The correct propositions are:

In a galvanic cell, the potential of the cathode is positive with respect to the anode.
In an electrolytic cell, the potential of the anode is positive with respect to the cathode.

This is due to the fact that it is impossible to measure the potential of a phase, or try to know its sign. Nevertheless, we still refer to the cathode as "positive" and the anode as "negative" in the galvanic cell...
Question Respectively:

The galvanic cell is the spontaneous operation of an electrochemical system, chemical energy (stored in the reactants) is transformed into electric energy. Thus, upon connection of these two electrodes, the electrons or negative charges will flow externally from a region of lower potential to a higher potential in this spontaneous process. This is, from the anode to the cathode, and thus the anode is said to be "negative".
The electrolytic cell is the inverse operation, the surroundings (you as electric energy) try to bring back the chemical energy that was stored in the reactants. Thus, you "give" energy to the cell so that electrons non-spontaneously go from a region of a higher potential to a lower potential. More precisely, you give energy so that the inverse reactions take place. This is, from the anode to the cathode, and thus the anode is said to be "positive".

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Decomposition of Al2S3 in water
Al2S3(s) + 6H2O(l) → 2Al(OH)3(s) + 3H2S(g)
I found this reaction on the web.I am not able to understand why this reaction proceeds from a weaker acid(H2O) to a stronger acid(H2S);contradictory to the usual behaviour.

It is H2S(g), you said so yourself; do you know what that (g) means?

Salts of weak bases and weak acids are regularly hydrolyzed in large extent, especially if hydrolysis is supported by leaving gaseous products or by precipitation of solid ones. Additionally, Al has stronger chemical affinity to O than to S.

Note that $\ce{HS-}$ is very weak acid.

As Ivan says, hydrogen sulfide departs as a gas, thus upsetting the proposed equilibrium between a weak acid and a stronger one. Precipitation of the aluminum hydroxide also pulls the reaction away from the equilibrium that would have been expected based on $\pu{pK_a}$ values alone.
Aluminum sulfide is not the only sulfide that does this. Most sulfide salts that are otherwise soluble in water react similarly and emit hydrogen sulfide gas when exposed to moisture, including sodium sulfide where the hydroxide would be water-soluble.


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Thermodynamics of borax dissolution
I'm currently experimenting with the dissolution of borax $\ce{Na2B4O7.10H2O}$. The procedure was quite simple, samples of saturated borax at different temperatures are titrated, then the value of $\ce{K_{sp}}$ can be determined. From different values of $\ce{lnK_{sp}}$ and $\ce{ln{1/T}}$, some thermodynamics parameters such as $\Delta H$, $\Delta S$  can be found.
However, the main problem is that the solubility of borax we found by titration is extremely low compared to the information we can find on the Internet. We suspect the source of error lies in the preparation of a saturated borax solution. So, my question is, How can we prepare such a saturated solution at different temperatures?
Here are some of my observations after doing this experiment over and over:

We looked up the solubility of borax (about 30g/100 mL of water at $\ce{60^{o}C}$), so we weighed 30g of borax and put it in 100mL of water and heat it to $\ce{60^{o}C}$. However, according to my observations, there is still a lot of undissolved borax. Does It mean that this is way too wrong? Like, should I do another method that I've read that I should put the solute in gradually and when all of them dissolved, just add a little more until they are no longer soluble? Did I make a supersaturated solution or something like that?

Some prelab guides I found on the Internet say that I should continuously stir while gently heating the solution. I tried but it didn't help at all. The substance that should be crystallized at the bottom of the container is now dispersed throughout the solution, which is a potential source of error.

Some say I should let the solution at much rest as possible for about 10 min to ensure that equilibrium is reached. With all equipment that I have, I can just let it rest in the hot bath that was used for heating. However, what if the temperature rises over the point that borax got dehydrated (just a little bit over $\ce{60^{o}C}$) or decreases more quickly by interacting with the surrounding? I tried but this seems too hard.

Finally, my idea for changing the temperature of the solution is to let that solution at $\ce{60^{o}C}$ cool down gradually. Several editions of lab reports I found on the Internet show that I should prepare, like, 5 solutions for 5 different temperatures. And also, is it true that using ice can make the temperature change so quickly that the equilibrium between borax (solid) and borax (dissolved in the solution) cannot be established? So if I cannot use ice, how can I prepare a saturated solution of borax at some low temperatures like $\ce{20^{o}C}$?

I've tried several times and the results are still quite bad. Hope I can find reasons for it. Thanks for reading such a looong question of mine and I would be really really appreciate your help.

In the 2nd paragraph you state that the solubility of borax is extremely low compared to the data found in Internet. Extremely low ? In the 3rd paragraph, the theoretical solubility of borax at 60°C is 30 g/100 mL water. In the Handbook of  Chemistry and Physics, the solubility of borax is 14.2 g/100 mL at 55°C, and 201 g/100 mL at 100°C. Its solubility shows a huge variation between 50°C and 100°C. It is 15 times more soluble at 100°C than at 55°C.  Maybe your temperature is not the same in all parts of your solutions ? This may explain your results.

Some substances dissolve slowly, so giving sufficient time may be an issue. Also, the directions to produce a truly saturated solution at 60°C would expect some undissolved solids at bottom -- the solution above is saturated, and can be separated by decantation or filtering.


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Can you infer toxicity of an ionic compound based on the toxicity of its ions in other configurations?
Suppose I have two ionic compounds: "NaY" and "XCl". If these were both 'safe' to eat, would it be true that the compound XY would also be 'safe' to eat, because in solution, these would dissociate into  X+, Y-, (which are also produced by our two original 'safe' compounds?).
Since safety is dose dependent, perhaps another way to ask would be whether the behaviour of the individual ions would be the same, regardless of what compound they were in prior to dissolution.

No, it doesn't work like that. Particularly because of varying solubility and/or coordination of counterion.

It is far from an additive effect; you have to take the specific properties of the salt as a whole into account.
To cite one example: barium salts are toxic when dissolved, but relatively insoluble barium sulfate is benign enough to be ingested for examination of the digestive system. See https://medlineplus.gov/druginfo/meds/a606010.html.

Or HCl and NaOH, which produce relatively benign NaCl. Unless you're on a low salt diet...

Generally the biological effects or functions of ions are relatively independent of the source however the total effect depends on the overall ion concentrations. This is accommodated in blood analysis by the anion gap. Your body needs certain concentrations of ions such as Na+, K+, Ca++, Mg++, etc. and anions such as Cl- and HCO3-. The proper concentrations can be arrived at by various methods within reason. It would be foolish to add K+ and Na+ by eating only NaHCO3 and KHCO3 and not eating salt. Ca And Mg are usually ingested as carbonates or organic salts. All that sodium and potassium benzoate used as preservatives contributes Na+ and K+ while the benzoate is excreted as hippuric acid. The combinations will probably not be toxic but the balance must be maintained.


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Can Acetone act as a ligand?
I have never seen any Carbonyl Oxygen acting as a ligand. But a naive intution says that since carbonyl oxygen has two lone pairs, it should act as a donor atom, donating one of its lone pairs as shown below:

In general can aldehydes and ketones act as a monodentate ligand? If no, then what's the reason?

Why, transition metals are routinely coordinated with acetone, there are hundreds if not thousands of known unique complexes with acetone alone. Even tetrakis(acetone)lithium and hexakis(acetone)sodium are known.

@andselisk can you provide some references and images? I can't find any. Also can you give an example of transition metal complexes?

Li: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=MEHBEG; Na: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=TOPYUS; Ca: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=TITQOF; Fe: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=HOLPUW; Pt: https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=OJECOX.

Like everything can coordinate if there's right Lewis acid present.

Examples are known. Coordination can take place through the oxygen atom, as shown in the question, or through the carbonyl pi bond so as to form a metal-carbon-oxygen triangle. See, among others, Huang and Gladysz 1. A couple sample figures from the reference are given below.

Reference

Yo Hsin Huang and J. A. Gladysz (1988). "Aldehyde and ketone ligands in organometallic complexes and catalysis". _J. Chem. Educ. 65, 4, 298. https://doi.org/10.1021/ed065p298

Nice downvote there. Would be better if you could give ideas on how to improve this answer.

Yes, there are numerous examples. Check out some of them:

Chemistry in Acetone Complexes of Metal Dications:  A Remarkable Ethylene Production Pathway
Jianhua Wu, Dan Liu, Jian-Ge Zhou, Frank Hagelberg, Sung Soo Park, and Alexandre A. Shvartsburg
The Journal of Physical Chemistry A 2007 111 (22), 4748-4758
DOI: 10.1021/jp068574z
$\ce{[M(Ac)6^2+][A−]2}$ where M is Mg(II), Ca(II), Sr(II), Mn(II), Fe(II), Co(II), Ni(II), and Zn(II); and $\ce{A−}$ is $\ce{FeCl4−}$ and $\ce{InCl4−}$:  Driessen, W.L. and Groeneveld, W.L. (1969), Complexes with ligands containing the carbonyl group. Part I: Complexes with acetone of some divalent metals containing tetrachloro-ferrate(III) and -indate(III) anions. Recl. Trav. Chim. Pays-Bas, 88: 977-988.DOI: 10.1002/recl.19690880811
Troepol'skaya, T.V., Sitdikov, R.A., Titova, Z.S. et al. Synthesis and structure of complexes of acetone acylhydrazones with some transition metals. Russ Chem Bull 29, 903–907 (1980). DOI: 10.1007/BF00958804
$\ce{[Fe(C19H25N5)(C3H6O)(H2O)2](BF4)2·2C3H6O}$: Kilner, C. A. & Halcrow, M. A. (2006). An unusual example of a linearly coordinated acetone ligand in a six-coordinate iron(II) complex. Acta Cryst. C62, m437–m439. (Link)


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Why is molality measured in terms of the solvent instead of the whole solution?
The molality $b$ of a solute dissolved in a solvent is given by:
$$b_\mathrm{solute} = \dfrac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$
where $n_\mathrm{solute}$ is the amount of solute (e.g. in mol) and $m_\mathrm{solvent}$ is the mass of solvent (e.g. in kg).
Why is molality measured in terms of the mass of the solvent instead of the whole solution? Is there any specific advantage to taking the ratio this way?

We already have a unit to measure with respect to the full solution, molarity (because for a solution, the volume matters and atoms/molecules/ions per litre), as you may have seen in problems of ionic equilibrium. And we do know the density and other properties of the solution to convert the volume into mass of solution. But with respect to mass of solute, it is used in industrial areas, where volume is not used, because the per gram of solute intake matters for the body (food stuff) and makeup cream and other.
I don't know of any other significance for the two : molarity, and molality, units of concentration.

The choice of molality in cryoscopie measurements is due to practical reasons. It is coming from the question : Which temperature do I obtain when adding $m$ grams (or $n$ moles) of a salt to $1$ kg ice ? The final volume of the obtained solution and its final mass are of no importance. The final temperature depends on the ratio $n/m$, i.e. on the molality of the future solution.

The choice of molality is due to practical reasons. It is coming from the question : Which temperature do I obtain when adding  grams (or  moles) of a substance to $m_o = 1$ kg ice at $0°C$, if the substance becomes a solute later on ? The final volume of the obtained solution and its final mass are of no importance to the scientist. The final temperature $T' < 0°C$  depends on the ratio $n/m_o$, i.e. on the molality of the future solution. The law of cryoscopy states that the temperature of fusion of ice + salt is $T'$ < $0°C$ and that this value does not depend on the nature of the dissolved substance : $$T' = - 1.86 n/m_o$$ This result is valid for all sorts of solutes, provided that they get dissolved in water without dissociation. If they get dissociated, the parameter $1.86$ must be multiplied by the number of ions produced by the dissociation. As a consequence, the fall of temperature $T'$ depends only on the molality $n/m_o$ of the substance to be added to the ice. Here is the origin of the use of the term "molality" in this sort of experiment.

You have already answered this question over three years ago: https://chemistry.stackexchange.com/a/125105.

You are right. But I was and still am unable to find the "old" report I wrote three years ago.


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Non Aromaticty due to Cross Conjugated Systems
I was going through this site looking for some answer on aromatic character, definition and other ways of judging an aromatic compound, when is tumbled upon this question:
Are p‐xylylene and fulvene aromatic?
There, an answer stated that because there is cross conjugation in the structure due to which it is non aromatic. I doubt if that is true (may even be the reason for the 2 downvotes) since we have squaric acid:

which is cross conjugated and aromatic. But I am never sure about organic chemistry, because I don't actually guarantee to know all rules for aromaticity, so i would just like to confirm. Does cross conjugation lead to non-aromaticity?

Aromaticity, and its antonym too, are the properties of simple cyclic systems of alternating single and double bonds. The use of the term elsewhere is an abuse of language.
That said, in practice people tend to apply the word "aromatic" to all kinds of systems. One just has to remember, though, that it is not a clearcut black-or-white characteristic and not a precise measure of anything. King Richard I of England was said to have the heart of a lion. That's a metaphor. When someone calls, say, naphthalene aromatic, it is the same thing.
Now to the point: is that non-rigorous use of "aromatic" justified when applied to cycles with side chains? The answer is: yes or no, depending on the particular system. You may look up the bond length, or compare the stability via enthalpies, or (if neither of these is available) calculate the orbital energies and see how it all turns out. Indeed, squaric acid is not really aromatic in that vague sense, but its dianion definitely is. So are the dianions of its triangular, pentagonal, and hexagonal siblings, known as deltic, croconic, and rhodizonic acids. On the other hand, 1,4-benzoquinone does not look aromatic at all.
Nature is under no obligation to give us a rule of thumb for deciding aromaticity just by looking at the formula. That such rule exists for simple cycles is a rare and lucky coincidence.
So it goes.

The aromatic character of the carbonyl ring anions is one factor that makes the neutral compounds relatively strong acids. Croconic acid, in particular, has a second deprotonation constant approaching that of sulfuric acid, without the use of halogen or cyano functions.

Yes, that's pretty impressive for an organic acid that does not even have a -COOH group.

Ok, so i get what you are trying to say, but just another thing to be sure of, is it possible for a monocyclic planar molecule with 4n+2 π electrons to be non aromatic just because it does not have continuous conjugation (and rather has cross conjugation)?

Why, sure, the example from your very question will do: fulvene.


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pH of a solution from mixed solution
I am not a chemistry student. I have a problem with my preserved food. I am afraid that the pH is not below 4.6 for it to safe to eat. I want to check the pH of the food but it has small amount of liquid. So I took a sample of 6 mL of it and then adding 100 mL of water. I don't have distilled water so I measure the pH of my water and it has pH of 7.1. Now I combine them together and check for the pH with my pH meter and the result is the mixture has pH of 3.6.
From my research I found this formula and I reverse the step to find the pH.

Find $[H^+] = 10^{-pH}$
$[H^+]_{mix} = \frac{[H^+]_1 \cdot V_1 + [H^+]_2 \cdot V_2}{V_1 + V_2}$ where the volume is in Liters
$pH_{mix} = -\log{[H^+]_{mix}}$

From there I solve for $[H^+]_2$ which I defined as for the fermented liquid. From my calculation I get that the $pH_2 = 2.35$.
Am I correct? Or is this formula wrong?
My formula is based from here

TL;DR: Measuring pH of diluted solution can generally provide just rough estimation of possible pH range of the original solution.

You have based your mathematical analysis on the false premise of conservation of $\ce{H+}$ ion amount.
It is not like for other ions like $\ce{Na+}$ or $\mathrm{Cl-}$. Food frequently contains (natural, formed or added) weak acids (citric, acetic, tartaric, maleic or others) and their respective salts. With solution dilution or mixing, further $\ce{H+}$ can be formed by acid dissociation or the opposite, consumed by recombination, to keep respective chemical equilibrium.
Therefore, $\mathrm{pH}$ calculations are not that easy like for strong acids like $\ce{HCl}$. Generally, if there is a solution of known $\mathrm{pH}$ and if you dilute it in some ratio, there is no way to tell the final $\mathrm{pH}$, unless you know the solution composition or unless you measure the pH. E.g. if you diluted vinegar 10 times, $\mathrm{pH}$ would not increase by 1 like it would for $\ce{HCl}$, but just about by 0.5. If there was also a salt of the acetic acid present, it would increase much less, being a pH buffer.
So, if some solution has $\mathrm{pH}=3.0$ and if you dilute it 10 times by distilled/deionized water, $\mathrm{pH}$ of the diluted solution can be in the range $3.0 - 4.0$.
Additionally, tap water can also affect the resulting $\mathrm{pH}$, e.g. due dissolved carbon dioxide and/or bicarbonate.


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Polarity and solubility
I have a doubt about solubility: a molecule is polar when it has a dipole moment that is ≠ 0. Very often it is said that like dissolves like, that is, that polar substances dissolve in polar solvents, and apolar substances dissolve in apolar solvents. But this is not always true: for example, 1 - hexanol is polar (its dipole moment is ≠ 0), but it is not soluble in water.
This means that polarity does not imply water solubility.
That said, what do you rely on to roughly predict solubility?

Also organic substances are better dissolved in organic solvents and water definitely isn't  a organic solvent.

Being polar in context of whole molecules or chemical bonds means rather significant than nonzero dipole moment, while nonpolar means rather small than zero dipole moment.
Comparing dipole moments makes sense for molecules of similar sizes. Consider there are polar=hydrophilic atomic groups and nonpolar=lipophilic atomic groups. Dominating group characters determines overall hydro/lipophility of molecules.For n-alkanecarboxylic acids, hydrophilic group is the same, but there are growing lipophilic n-alkyl chains, progressively turning character from hydrophilic (formic acid) to lipophilic (stearic acid).

A dipole moment does have the strong chemical meaning, but it cannot be taken mechanically as the only decisive parameter. One has to consider the molecular context. You may think about nonpolar parts of molecules to be effectively attracted to nonpolar solvents while polar parts to polar solvents. With the polar part attraction significantly stronger, so nonpolar lipophilic parts must be bigger/longer to get to fuzzy boundary between lipophilic and hydrofilic overall behaviour. CO2 has zero dipole moment, but has symmetrical polar bonds, so it has quite good water solubility.

so, generally, it is correct to evaluate the ratio between polar and non-polar fractions in a molecule. With this, It Is possible to evaluate if a molecule is effectively polar or non polar and where It Will be soluble. This is regardless of the presence or absence of a dipole moment, which is probably more important for estimating the polarity/solubility of small molecules, such as chloroform, water, carbon dioxide or CCl4, etc..

@Luckenberg It can have a polar end and a nonpolar end, like classical soap, or most of surfactants.

What are you referring to? (In relation to my comment).

@Luckenberg "...It Is possible to evaluate if a molecule is effectively polar or non polar..."

oh sure, we have to evaluate the balance between polar and non-polar Groups in a molecule. So, does what i wrote in the comment above make sense (more or less)?

The longer the chain of carbon atoms, the less influence on polarity the alcohol moiety has. Methanol, ethanol, and 1-propanol are all miscible. 1-butanol is soluble ~70g/L, 1-hexanol: ~6g/L and 1-heptanol: ~0.1 g/L.
Each additional carbon link in the chain reduces solubility by a factor of more than 10.
However, add some more alcohol groups, and solubility increases. For example, 1,3-butanediol is miscible with water, and 1,6-hexanediol is soluble at 500g/L.

So polarity does not depend (exclusively) on the dipole moment of a molecule, but more on the number of polar groups that appear in it and, therefore, more on the ability of the same to form hydrogen bonds?
So, a molecule can have a dipole moment = 0, but still be water soluble (and polar). Probably the dipole moment has a more "physical" than "chemical" meaning (?)

The diols with alcohol groups at both ends have negligible polarity, but are more soluble than those with single alcohol group... Your conclusion seems reasonable.


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Is there a property that indicates the presence of superacid?
Superacids are often characterized as acids that are stronger than 100% sulfuric acid. However, this definition is somewhat arbitrary, and accurately determining the acid strength in concentrated solutions, non-aqueous solvents, or other phases such as gas and liquid can be quite challenging.
Furthermore, it is not clear, at least to me, how one would compare the acid strength of a solid or gas with that of a solution containing 100% sulfuric acid.
So I was wondering if there is one property that is common to all superacids. I know that superacids can stabilize carbocations, although it is unclear whether this is a property common to all superacids.

Related: https://en.wikipedia.org/wiki/Hammett_acidity_function

Check out, for example, this analysis: https://en.wikipedia.org/wiki/Carborane_acid#Acidity

I don't believe there is a 100% correct answer for what you are looking for. You could try using the Hammett acidity function as suggested by @ivan-neretin, where an acid with $\ce{H0≤−12}$, which is the value of 100% sulfuric acid, would indicate superacidity. However, obtaining H0 for non-liquid acids is challenging at best.
Another approach you could try is to compute the deprotonation energy (DPE) of your acid, which is the energy required to remove the H+ ion. DPE can be calculated as:
$$\ce{DPE = E_{H+} + E_{A-} + E_{HA}}$$
asuming the species $\ce{H+}$, $\ce{A-}$ and $\ce{HA}$ from the following chemical equilibrium:
$$\ce{HA <=> H+ + A-}$$
However, this method might also be time-consuming and making meaningful comparisons, especially when comparing acids in different phases such as solid, liquid, and gas, could be difficult.
You have made a valid observation that the definition of superacid is somewhat arbitrary, and this is supported by George Olah in his book "Superacid Chemistry"

Gillespie proposed an arbitrary but widely accepted definition of superacids, defining them as any acid system that is stronger than 100% sulfuric acid, that is, H0 ≤ - 12.

While it is true that one common property of superacids is the ability to prepare and stabilize carbocations, which earned George Olah a Nobel Prize, I'm not entirely sure if this is a distinctive feature of superacids, nor I'm sure if all superacids meet the criteria of being stronger than 100% sulfuric acid can make long-lived carbocations.

Not all carbocations require superacids to stabilize. Cyclopropenyl cation was precipitated from a nonpolar solvent. Carbocations generally require superacids to stabilize them if the positive charge is not incorporated into or conjugated with aromatic rings.

Is the block quote exact? The missing negative sign in $H_0\le-12$ is a major error.

@OscarLanzi you are absolutely correct. I will change it. Thanks

@OscarLanzi I'm not stating that carbocations are only stabilized by superacids, just that often superacids can stabilize carbocations. But according to the example you gave where $\ce{C3H3+}$ is stabilized after reacting with $\ce{SbCl5}$ which is not a superacid, means that this is clearly not a property that should be used to identify a superacid

I don't believe there is a 100% correct answer for what you are looking for. // This may actually be the 100% correct answer.  :-)

Since superacids are defined as more acidic than 100% $\ce{H2SO4}$, one possible characteristic is the ability to form salts of the conjugate acid $\ce{H3SO4^+}$.
Such a synthesis has been carried out on $\ce{HF +SbF5}$ by Minkwitz and Seelbinder[1]. Instead of $\ce{H2SO4}$ itself, the substrate used was bis(trimethylsilyl) sulfate, $\ce{((CH3)3Si)2SO4}$, yielding the salt $\ce{[H3SO4^+][SbF6^-]}$.
Reference

Minkwitz, Rolf & Seelbinder, Raphael & Schöbel, René. (2002). "Protonated Sulfuric Acid: Preparation of Trihydroxyoxosulfonium Hexafluoroantimonate H3SO4+SbF6−". Angewandte Chemie International Edition. 41. 111 - 114. https:// doi.orf/10.1002/1521-3773(20020104)41:1<111::AID-ANIE111>3.0.CO;2-6.


text:
Does "brine rejection" happen for dissolved gases as well?
"Brine rejection" is the effect of sea ice pushing out dissolved salts, forming sweetwater ice and brine. I understand "freeze distillation", a method to concentrate alcohol in fermented brews, is effectively the same thing.
Do liquids, if they form crystals, "reject" dissolved gases the same way they reject other dissolved substances? Is the brine from arctic ice enriched in $\ce{CO2}$, for example?

One difference is that the salt has nowhere to go, other than to remain in the brine or form a crystalline solid, whereas the gas can equilibrate with the atmosphere if the latter is in contact with the ice/brine, at least before ice completely covers the brine. After that the gas concentration under the ice can increase. This might be important in shallow areas.

While there may well be counterexamples, a paper in Cryobiology (G. Lipp, Ch. Körber, S. Englich, U. Hartmann, and G. Rau in Cryobiology Volume 24, Issue 6, December 1987, Pages 489-503) titled "Investigation of the Behavior of Dissolved Gases during Freezing" includes the statement:

In a manner similar to that of dissolved
salts, gases are enriched in the liquid
during freezing.

In particular, for oxygen the segregation (distribution) coefficient was measured to be 0.048. In other words, in steady state segregation one expects the oxygen concentration immediately in front of the solid-liquid interface to have about 20x the oxygen in the liquid as in the solid.
While perhaps overly broad, the general concept that gases that do not strongly interact with water would prefer to remain in the liquid vs the solid is straightforward as an entropy effect.
Bottom line, yes, ice has less dissolved oxygen than the liquid.


text:
Explaining Vapour Pressure and Evaporation by Chemical Potential
According to the coexistence curves, both gas and liquid phases can only exist when the chemical potential of both phases are equal.
Does this mean above and below the coexistence curve of liquid and gas, there is only pure liquid or pure gas phase present?

For example, if we consider a closed system in a rigid container having only water present initially at, say 20°C, with remaining volume empty (vacuum), will the water evaporate and create vapour pressure? How can we explain this using chemical potential?
In another case, if we have a closed container with a piston to ensure a constant external 1 atm pressure on the container filled with water at 20°C, will this mean there will be no vapour pressure/evaporation in the system, as the pressure and temperature conditions correspond to pure liquid on the phase diagram of water.
In general, how can we explain the spontaneity of evaporation of any liquid (in a closed container) at a given temperature and pressure using its chemical potential and phase diagram?
Further, if a liquid in closed container is in equilibrium with its vapour (rate of condensation = rate of evaporation), is it true that the point corresponding to its (P, T) will necessarily lie on the coexistence curve (this does seem true by definition, but I might be wrong).
And can we say that a liquid will evaporate if and only if the conditions of pressure and temperature lie on the coexistence curve, and otherwise it will exist either as a pure liquid or as a pure gas? If yes, isn't it a contradiction to evaporation under standard conditions?
My concepts might be really flawed but I could not find any sources explaining my dilemma.
Any help is greatly appreciated!

Make a thought experiment with system containing water + vapour at isothermic conditions, with a piston very slowly changing its pressure ( to allow heat exchange and keep dT=0)  along the isotherm. What would happen on the path there and back ?

@Poutnik While compression, the pressure increases, so the equilibrium of water + vapour shifts towards the water and the vapour condenses at a higher rate? I can explain this using equilibrium but not through chemical potential though...

Does this mean above and below the coexistence curve of liquid and gas, there is only pure liquid or pure gas phase present?

Yes. For example, water in vapor-liquid equilibrium at $\ce{1 atm}$ can only exist at $\ce{100 °C}$. If you change the $P$ while keeping $T$, or you change $T$ while keeping $P$, you will have water in the gas or liquid state.

In another case, if we have a closed container with a piston to ensure a constant external 1 atm pressure on the container filled with water at 20°C, will this mean there will be no vapour pressure/evaporation in the system, as the pressure and temperature conditions correspond to pure liquid on the phase diagram of water...

Yes. If you keep water at $\ce{1 atm}$ and $\ce{20 °C}$, as the phase diagram points out, water can only exist as a liquid.

Further, if a liquid in closed container is in equilibrium with its vapour (rate of condensation = rate of evaporation), is it true that the point corresponding to its (P, T) will necessarily lie on the coexistence curve (this does seem true by definition, but I might be wrong).

Yes. Phase transitions for pure fluids take place at constant pressure and temperature. For example, if you heat saturated liquid at $T = 100 °C$, saturated gas will form in presence of (now in a less quantity) saturated liquid. If more heat is added, then you will end up only with saturated gas. All this changes take place at constant temperature. However, heat a bit more, and you will have a superheated gas but now $T \neq 100 °C$.
The process must be done sufficiently slow so that thermodynamics still apply and our statements hold. Also, this evolution cannot be traced in the P-T diagram, but you can clearly describe it in the P-V diagram.

And can we say that a liquid will evaporate if and only if the conditions of pressure and temperature lie on the coexistence curve, and otherwise it will exist either as a pure liquid or as a pure gas? If yes, isn't it a contradiction to evaporation under standard conditions?

Yes. Remember that a pure fluid at VLE has $1$ degree of freedom. So, as long as you can mantain the pressure, more precisely the saturation pressure, it will evaporate at constant $P$ and $T$. If the pressure is changed, you move out of the V-L coexistence curve and another process will take place.

Some Mathematics of Evaporation
I see that you are interested in the process of evaporation. Thermodynamics will tell you nothing about rates, but we can make up a simple situation that you may like. Lets consider the following figure:

Say we heat an initial number of moles $n_{L0}$ of saturated liquid from the bottom at a heat rate $\dot{Q}$. As a consequence, we have an intlet molar flow $\dot{n}$ from the liquid phase to the gas phase. As the liquid phase evaporates, the energy that is losing has an enthalpy of $h_L^\pu{vap}$. We will apply the laws for both separate systems, and thus, we have to apply the laws for open systems. We disregard energy such as potential energy and no work is being done.
The mole balances for both phases yield
$$
  \frac{dn_L}{dt} = -\dot{n} \hspace{2 cm}
  \frac{dn_G}{dt} = \dot{n} \tag{1,2}
$$
The 1st law yields
$$
  \frac{d(n_Lu_L)}{dt} = \dot{Q} -\dot{n}h_L^\pu{vap} \hspace{2 cm}
  \frac{d(n_Gu_G)}{dt} = \dot{n}h_L^\pu{vap} \tag{3,4}
$$
Principal hypothesis (the one that interests you): the pressure of both systems (and thus the temperature by Gibbs phase rule) is mantained. This is, the chemical potential of both phases are equal. Therefore, all enthalpies and internal energies are constant and may be taken out from the time derivatives. With Eq. (4) we can say the following
\begin{align}
  u_G^\pu{vap}\frac{dn_G}{dt} &= \dot{n}h_L^\pu{vap} \hspace{1 cm} \text{(use Eq. (2))} \\
  u_G^\pu{vap}(\dot{n}) &= \dot{n}h_L^\pu{vap} \rightarrow
  h_L^\pu{vap} = u_G^\pu{vap} \tag{5}
\end{align}
And Eq. (3) tells us the following
\begin{align}
  u_L^\pu{vap}\frac{dn_L}{dt} &= \dot{Q} -\dot{n}h_L^\pu{vap} \hspace{1 cm} \text{(use Eqs. (1) and (5))} \\
  u_L^\pu{vap}(-\dot{n}) &= \dot{Q} -\dot{n}u_G^\pu{vap} \\
  \dot{Q} &= \dot{n}(u_G^\pu{vap} - u_L^\pu{vap}) \rightarrow
  \boxed{\dot{n} = \frac{\dot{Q}}{\Delta_\pu{vap} u}} \tag{6}
\end{align}
Eq.(6) may seem kind of obvious, but we have to prove things. Consequently, if the heat rate $\dot{Q}$ is constant, so will be the molar flow rate $\dot{n}$.
Of interest in thermodynamics is the quality, i.e., the mole fraction of gas in our system. Eqs. (1) and (2) integrated yield
$$ n_L(t) = n_{L0} - \dot{n}t \hspace{2 cm} n_G(t) = \dot{n}t \tag{7,8} $$
Eq. (7) states that we will run out of liquid when $t = n_{L0}/\dot{n}$. Finally, combining Eqs. (6-8)
\begin{align}
  x(t) &= \frac{n_G(t)}{n_L(t) + n_G(t)} \\
  x(t) &= \frac{\dot{n}t}{n_{L0}} \rightarrow
  \boxed{x(t) = \bigg(\frac{\dot{Q}}{n_{L0}\Delta_\pu{vap}u}\bigg)t \hspace{0.5 cm}
  0 \leq t \leq \frac{\Delta_\pu{vap}u}{n_{L0}\dot{Q}}} \tag{9}
\end{align}
Eq. (9) is the most simple but also beautiful result that we can get from an evaporation process if thermodynamics holds. Everything of it makes sense:

The quality will be an increasing linear function.
Higher the internal energy of evaporation, the harder it will be to evaporate the liquid. We may regard it, with a cautious distance from kinetic processes, as a kind of "thermal resistance".
Higher the heat rate, more rapidly the quality will increase.

That clarifies so many things, thanks a lot!! If I have understood correctly, during evaporation, initially, when only liquid is present in a closed container with vacuum, the chemical potential of gas is lower so the liquid evaporates spontaneously (increasing pressure) until the chemical potential of gas and liquid is equal and they are in equilibrium at a (P, T) corresponding to the coexistence curve, and heating it further (isothermally and isobarically) will alter the amount of gas and liquid.

I haven't addressed the vacuum case, I didn't notice. That chamber has water liquid and air on top, so we don't have a single component anymore, but lets simplify by stating that the air is our pressure. If sufficiently low, then the liquid water will evaporate at the temperature $T$ it is. E.g., in a hot day of $\pu{40 °C}$ that would happen at $\pu{7375 Pa}$. So if $P >7375 ; \text{Pa}$ we have liquid and nothing happens. Exactly at $P =7375 ; \text{Pa}$ it will boil at constant $P$ until everything evaporates. However, you must mantain that $P$ from the outside.

And what about for $P<7375 \  \text{Pa}$? Will it evaporate spontaneously? And I assume everything will evaporate at constant (7375 Pa, 40 ∘C) if external energy is provided.

@Omega No, remember, pure fluids evaporate at constant pressure. So, once everything is evaporated, if you continue lowering the pressure, your answer is in the phase diagram: it is a superheated gas. Imagine a vertical arrow going down at $T=40 °\pu{C}$

So, assuming an isothermal process, a system of only liquid cannot go from being a pure liquid to reaching the coexistence curve unless pressure is changed by external means?

@Omega Yes. At sea level, you can only have in your hands water at $T=25 ;°\ce{C}$ and at $P=1 ; \ce{atm}$, approximately. So, to lower the pressure, you have to do something. How we do it in real life , you have done it already: you use a mix of gases in your kitchen, have a combustion reaction, and then the water is boiled in a pan. In this case, the pressure is the same, but you increase $T$.

Alright that clears it, thanks!


text:
How can Joule-Thomson inversion coefficient of ideal gas be zero? Does temperature of ideal gas not change on adiabatic expansion?
Does Joule-Thomson coefficient being zero means there will be no change in temperature? If that is the case then according to first law of thermodynamics, E = Q+W (don't know how to add delta sign). So, if expansion takes place in adiabatic conditions, the gas must cooldown.
I am confused, any explanations of the Joule-Thomson effect would be appreciated.

J-T E is not an ordinary AB expansion. The gas does not do any work  expanding against external pressure there. At ordinary adiabatic expansion, all gases cool down.

$\Delta U = Q + W$  ->  $\Delta U = Q + W$

Expression formatting reference:
Basics/CH+Math expressions/Upright vs italic/Math tutorial/Advanced

@Poutnik Thanks, I think I understand now. The textbook I was studying from began the discussion on J-T-E using adiabatic expansion, that got stuck in my head.

@Poutnik Just to be clear, when an ideal gas is taken from a region of very high pressure to a region of very low pressure suddenly, it will not cool down or heat up because the gas doesn't do any work against external pressure. Is that right?

It depends on omitted details in the mentioned scenario. If the ideal gas was in an isolated balloon, lifted 1 km in the atmosphere it would cool down. If it expanded against vacuum between 2 chambers in isolated, constant volume system, it's temperature would remain.

@Poutnik I get it. Thanks.


text:
Total ion current normalisation of intensities derived from mass spec - does this produce compositional data?
A definition of total ion current normalisation from https://www.scirp.org/journal/paperinformation.aspx?paperid=86606, i.e. for each feature in the sample, divide its intensity value by the total for the sample.
If I have my features in columns and samples in rows, and for each sample/row I divide the intensity of each feature by the sum of all intensities for that sample (e.g. R code would be ...
  TIC <- apply(data, 1, function(i){
   out <- i/sum(i)
  }
    )

When I then check the sum of the values for each sample, this equates to one, much like the calculation of relative abundance from OTU data, where the relative abundance sums to 1 or 100% for each sample. Is the calculation of TIC in this format correct and does this now result in compositional data? Most of the R packages that would have automated the calculation of the TIC are no longer available, so any input on calculation would be appreciated. Thank you


text:
Single or multiple calibration curves for interpreting datasets?
I ran three batches of samples through the HPLC. Each batch run contained standard solutions and a set of experimental samples. The standard solutions included in each batch run were exactly the same, but the experimental samples in each batch run represented different times the experiments were run. The composition of all of the experimental samples and the standards are exactly the same. I'm not sure what is the best practice for interpreting the the unknown concentrations in the experimental samples.
(1) Individual calibration curves: Should I interpret the experimental samples associated with a batch run with a calibration curve I make from the standards of that run?
OR
(2) Consolidated calibration curve: Could I make a single calibration curve using the responses from the standards across all three batch runs to interpret the experimental samples?
I can provide my data as an example. Thank you!

For serial analysis, there is often used calibration curve and regulation diagram based on control samples. The calibration curve is kept until regulation diagrams breaks the thresholds according given rules. Not sure if seriality applies in your case.

If there are isolated 3 sets, I would go rather for individual calibration curves. As involving standard solutions from all batches into a single curve could bring inter-batch bias errors.


text:
Query regarding covalent bonding
Why don't covalent bonds result in the atoms involved in the bond being charged?
Let's take the Chlorine/Cl2 molecule as an example.
So each Chlorine atom in the Cl2 molecule has 7 valence electrons. Each Chlorine atom contributes 1 electron to the covalent bond; thus, 2 electrons are shared between the two chlorine atoms in the molecule and each Chlorine atom has a complete octet. The shared pair of electrons belong to both Chlorine atoms, meaning that each Chlorine atom has 18 electrons and 17 protons which is 1 more electron than the total number of protons. This implies that each chlorine atom has a net -1 charge.
However, the Cl2 molecule has 34 electrons and 34 protons, implying that it doesn't have a net charge.
Why has such a contradiction arisen?
Can someone help explain this?

The charges do arise in covalent bonds due to different electronegativities in case of heteronuclear molecules and oscillating electron cloud in homonuclear molecule but those charges are very slight and always complimentary for the two atoms meaning that partial charges created are equal and opposite due to which overall molecule is neutral.

@Pscionexam2003 It's not a question about partial charges, but about electron counting.

There is no contradiction.
From both shared electrons, each chlorine atom "owns" one half.
$$17 = 16 + \frac 12 + \frac 12$$
If a husband and his wife share their family car, they do not have two cars, do they? Each of them has one half, together one car.

The theory of molecular orbitals considers forming 2 molecular orbitals (bonding and antibonding) as the linear combination of original atomic orbitals. The one with lower energy (bonding orbital) is filled by the electron pair, each electron from the respective atom.
This bonding orbital is shared between both chlorine atoms.

For your case, there is need to distinguish "having" as inventory and "having" as quantum feature/functionality.
In context of inventory, each atoms has 17 electrons.
In context of functionality, each atom has 18 electrons, as 2 electrons are servants of 2 lords. They occupy the bonding molecular orbital, stretched over both chlorine atom.

Thanks for your answer. However, my query is a bit different...After the formation of the bond between the 2 chlorine atoms,2 electrons are shared between the two. This means that each chlorine atom has an extra electron as the shared pair of electrons are not "owned" by any one atom, they belong to both. If each chlorine atom were to "own" after bond formation one-half of the shared electrons then each Chlorine atom would have an incomplete octet. Am I missing something?

If 1 electron belongs to 2 atoms, each has effectively 1/2. It is like in CsCl ionic lattice, where each Cs+ is surrounded by 8 Cl- ions. But each Cl- ion is shared between 8 such crystallographic cells, so Cs+ ion "owns" 1/8 of it. And 8 times 1/8 is 1. / See also the answer update.

If the bonding orbital is filled by 2 electrons donated by each Chlorine atom in the Cl2 molecule and is shared between both chlorine atoms then doesn't each chlorine atom effectively get only 1 electron as a result of which both the atoms have an incomplete octet which in reality I suppose isn't the case.

Thanks for the answer update...I get your point. What I don't understand is why the covalently bonded chlorine atoms which should have a complete octet that is 8 electrons in the valence shell possess a net charge of -1. I know that the number of protons and electrons in the Chlorine molecule are equal and therefore it should be electrically neutral and therefore should not possess a net charge.

They do not possess a net charge. From 2 x 17 electrons, 2 X 16 are around their Cl atom and remaining 2 are around both atoms. Thinking there is net negative change (each atom claiming bot for itself) has no bigger reason than net positive charge (each atom giving it up for the other one.// By other words, 2 electrons extended their spheres of interest to both atoms.


text:
Can electrical conductivity be measured by colorimetric or photospectrometrical method?
I would like to measure electrolyte leakage from plant material after exposure to stress. Stress damages cell membrane integrity, resulting in efflux of intracellular components to the surrounding solution. This can be measured by means of an Electrical Conductivity (EC) meter which measures the potential of an electrical current to be transported through water.
Plants have a wide range of electrolytes which leak out of the cell, such as K+, Na+, Cl–, Ca2+, Mg2+, O2-, NO3-, PO43− the list is near endless. When a cell membrane is damaged due to (a)biotic stress, electrolytes leak towards the outside of the cell. A standard protocol to measure such electrolyte leakage can be found here.
I want to make my protocol high-throughput and optimized, and measuring every sample separately with an EC meter is very time demanding. Therefore I wonder if there are specific color reagents or chemical reactions which change their color in response to electrolyte presence. This is much easier to translate to for example a plate format. I find literature on this topic very scarce, I was for instance able to find some on free ion concentration by means of ion increment method, or colorimetric assays for specific ions but nothing as an indicator for EC. Would this theoretically be possible?

I see as the only chance and only in specific cases some random correlation without causality.

What is the specific electrolyte you want to measure? Is there any reason to believe there are more elecrtolytes in the plant cell than outside?

A chromogenic polymer-based material with properties very similar to what you are seeking has been developed by the authors of Ref. 1. The color of the partially sulfonated polystyrene particles changes with ionic strength.
From the abstract:

The color of the resulting hydrogel-like ordered structures can be continuously shifted over the entire visible range (405–760 nm) by changing the content of ions over an extremely wide range of concentration (from about 70 μM to 4 M). The optical response is completely independent from pH and temperature, and the initial color can be fully recovered by washing the sulfonated opals with pure water. These new smart photonic materials could find important applications as ionic strength sensors for environmental monitoring as well as for healthcare screening.

References

Luca Nucara, Vincenzo Piazza, Francesco Greco, Valentina Robbiano, Valentina Cappello, Mauro Gemmi, Franco Cacialli, and Virgilio Mattoli. ACS Applied Materials & Interfaces 2017 9 (5), 4818-4827. DOI: 10.1021/acsami.6b14455

It is a good step toward the goal. Correlation between ionic strength and conductivity is strong for the (near) same electrolyte diluted to various degree. It is weaker for similar electrolytes and weak generally. But for specific plant scenarios it should be usable.

@Poutnik Yes, the path to a solution often involves many small steps. Hopefully this answer can be of some assistance to the OP in indicating that there might be solutions closely related but perhaps not identical to the one outlined.

Thank you, this solution seems very feasible. I will need to review the ion concentration range, and the synthesis possibility of Stimuli-responsive photonic crystals (PCs).


text:
How do I go about calculating final temperature of tank draining at saturation?
I am working on calculating pressure in a tank where the fluid is sitting at its vapor pressure. For example, imagine a 12 in diameter 100 L tank of nitrous oxide at room temperature (745 psi vapor pressure). This tank would be drained at 1 L/s from the liquid side, and some of the nitrous oxide would boil off in order to maintain vapor pressure, but would come at the cost of reducing the temperature and vapor pressure in the fluid.
How do I find what the final fluid temperature at drain would be (assuming no heat being input into the system)?
I was thinking there would be some combination of Antoine's, Clausius-Clapeyron, ideal gas law, and some derivation of an energy balance equation, but I am not too sure.
The experimental situation would literally be a tank (assume perfect insulation) that is filled with nitrous oxide, say a 100 L tank because it is arbitrary. This tank would then be drained through a choked orifice into the atmosphere. The flowrate at the start would be 1 L/s, but as pressure drops, would decrease in flowrate. I want to find out the ending pressure and temperature when the liquid would run out. Experimental data points to the tank draining from liquid before running out, similar to how a propane barbeque runs out of liquid and becomes cold with lower pressures.

If you can truly assume an isolated system, then the final temperature would be about -88.5 Deg C (nitrous oxide boiling point), varying with atmospheric pressure. Assume it would be high T and high p scenario for water. Final temp would be 100 Deg C. Similarly for LPG with complication of being a mixture. But in all cases it would be asymptotical final state.

Unless you mean the lowest temperature still able to supply 1 L/s. Final T would be then the T with pressure able to provide the given flow with the fully open valve. There is not enough data for that.

I guess my question is what would be the temperature where the fluid runs out? The drain rate would be out of a choked orifice, so upstream pressure would affect the downstream flowrate. We have empirical data that shows ending pressure is around 75% of the starting pressure but want to be able to actually calculate/model this.

How well insulated the real system is? As with total insulation it would not run out, being in mechanical equilibrium with atmosphere. // I advice to properly describe scenario in the question to prevent task misinterpretation by readers.

The situation would literally be a tank (assume perfect insulation) that is filled with nitrous oxide, say a 100L tank because it is arbitrary. This tank would then be drained through a choked orifice into the atmosphere. The flowrate at the start would be 1 L/sec, but as pressure drops, would decrease in flowrate. I want to find out the ending pressure and temperature when the liquid would run out. Experimental data points to the tank draining from liquid before running out, similar to how a propane barbeque runs out of liquid and becomes cold with lower pressures.

If a LPG canister had been isolated, the liquid phase would not have run out as there would have been atmospheric pressure inside. This is observed in similar scenario of limited usability of LPG at low T, even if initially warm.// The deviation from being isolated leads to running out.  // For all cases, the key is to know ratio of the specific or molar liquid heat capacity times T difference and evaporation enthalpy. if low enough, the isolated running out is possible.

@AlexanderPatrus I was about to answer this question, but something is strange. You state that your tank is in vapor-liquid equilibrium all the time? So, while the pressure of the tank goes down, $ T = T^\pu{sat} $ at the corresponding pressure?

@MetalStorm Yes that would be the condition in the tank. I am not sure how to calculate the final temperature in order to give me the final pressure. Basically I don't know how to calculate the temperature drop

@AlexanderPatrus My main concern is that phase transitions for pure fluids (like here with $\ce{N2O}$) take place at constant temperature (or pressure). So, if you have saturated liquid $\ce{N2O}$ and it starts boling at $T^\pu{sat}$, it will keep evaporating until all you have is saturated gas at the same $T^\pu{sat}$ (at least, thermodynamically). Do you have some data?

@MetalStorm As the tank is isolated, the liquid N2O had to cool down to compensate evaporation enthalpy. It is similar as when one CH SE user was surprised that his chlorine containers were covered by ice. He was assuming it is due adiabatic cooling. He was not aware chlorine is provided liquefied and it was due evaporation enthalpy.

@Poutnik Hello Poutnik, thanks for the comment. But the tank is not isolated, there is draning. Do you think a process is possible such that, as time passes by, there is always a VLE equilibrium between phases? The mental image would be that of a guy walking the P-T curve, from right to left, along the gas-liquid equilibrium curve.

@MetalStorm By isolated, I have meant insulated  (assume perfect insulation) I do agree by an isolated system is meant an insulated closed system. Sorry for confusion. (Czech language uses the same term for both, easy to confuse it.)

Since the tank would be drained from the liquid portion, the ending pressure would certainly not be atmospheric, it would be something higher. With ending defined as liquid running out.

For calculations near critical point, evaporation enthalpy, specific liquid and gas heat capacities, gas and liquid densities and adiabatic expansion quite strongly depend on p and T. I am not sure data are available. I am afraid the easier way may be empirical evaluation if applicable.

Let m be the mass of N2O in the tank at any time, x be the mass fraction liquid N2O,V be the tank volume, v be saturated volume per unit mass, u be saturated internal energy per unit mass, and h be saturated enthalpy per unit mass.  At any time, the internal energy of the tank contents will be:  $$U=m[xu_L(T)+(1-x)u_V(T)]$$If we apply the open system version of the 1st law of thermodynamics to the contents of this adiabatically insulated tank, we obtain:  $$dU=h_L(T)dm$$or$$[xu_L(T)+(1-x)u_V(T)]dm-m(u_V-u_L)dx+m\left[x\frac{du_L}{dT}+(1-x)\frac{du_V}{dT}\right]dT=(u_L+Pv_L)dm$$or
$$-(u_V-u_L)dx+\left[x\frac{du_L}{dT}+(1-x)\frac{du_V}{dT}\right]dT=-(1-x)(u_V-u_L)d\ln{m}+Pv_Ld\ln{m}$$This equation is subject to the constraint that the total volume is constant:  $$m[xv_L+(1-x)v_V]=V$$or$$x=\frac{v_V-\frac{V}{m}}{v_V-v_L}$$These equations provide the basis for determining x(m) and T(m).


text:
What is the actual complex formed in Fehling's solution?
What is the actual complex formed in Fehling's solution when Fehling’s solution A and Fehling’s solution B are mixed in equal amount?
Wikipedia suggests: $\ce{[Cu(C4H4O6)2]^2-}$
But the book I follow suggests: $\ce{[Cu2(C4H4O6)2]}$ .
It states the structure of the complex as:

Another source suggests the structure as:

So, the complex a/c to the above source is $\ce{[Cu(C4H4O6)2]^4-}$
So, altogether my questions are:

What is the actual structure of the complex?
How many copper ions are actually coordinated?
To neutralize the charge, do Cu(II) ions act as counter ions(e.g.: considering of Wikipedia : $\ce{Cu[Cu(C_4H_4O_6)2]}$ )?

H in bound OH frequently goes away as H+, e.g. in citrate complexes. // ACS has some paywalled article https://pubs.acs.org/doi/10.1021/ja01501a013

You didn't read Wikipedia carefully "The tartrate serves as a ligand. However, the coordination chemistry is complex and various species with different metal to ligand ratio have been determined."

I had read that @Mithoron and there are several citations too. But I was in a search for a "commonly adopted structure" or such like which has major proportion in the solution(due to their different stabilities). Their stabilities comparisons(although, yes they are unstable but we can state their 'relative' stabilities). Moreover besides that statement , why Wikipedia stressed over $\ce{[Cu(C4H4O6)2]^2-}$ ? Altogether, I was in a search for a more rigorous answer. At least it could bring some good answers(and some knowledge and clarity to me).

As far as formula is concerned [Cu(C4H4O6)2]2− is most reasonable one, but it obviously depends on pH (forget this monster with two Cu ions).

Tetraanion is unlikely and if it could be made, reagent wouldn't decompose to hydroxide, I think.


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What's the priority order of primary, secondary and tertiary amino groups in IUPAC nomenclature
It's a really simple question but I didn't get answer to this anywhere. What's the priority order of primary, secondary and tertiary amino groups in IUPAC nomenclature.
For example what would be the name of the following compound:

It's possible to answer the question 'how is this named?', but it doesn't involve comparing between primary, secondary, and tertiary amino groups at all (see e.g. https://iupac.qmul.ac.uk/BlueBook/P6.html#6202040102), so let's take a step back first. Presumably you have identified the 'parent hydride' or the 'parent chain' as a hexane. Amines are suffixes (unless there's another more important group, which isn't the case here), so the name is going to look like (...)hexane-x,y,z-triamine; and the part which you are stuck at is how to number the chain to determine x,y,z, is that correct?

@orthocresol Yes exactly.

An example that fits better with the title would be a case where you can only choose one of the amino groups as a suffix like 2-{3-(2-aminoethyl)-5-[2-(methylamino)ethyl]phenyl}-N,N-dimethylethan-1-amine.

It's possible to answer the question 'how is this named?', but it doesn't involve comparing between primary, secondary, and tertiary amino groups at all.
The most relevant part of the IUPAC Blue Book is P-62.2.4; even if you don't want to read the fine print, there are a few examples there which might let you see a pattern.
You have identified the parent chain as a hexane. Now, the idea of 'priority' between functional groups is used when deciding which of them becomes the suffix. For example, carboxylic acids have quite high priority, so where possible, you want to name acids as ...-oic acid; see Table 4.4 in P-43. If you scroll down to entry 51 in that table, you'll see that amines are also suffixes, but there isn't a distinction between primary, secondary, and tertiary amines.
In fact, in your compound, all of the amines can be suffixes, so you don't need to choose between them. The name is therefore going to be something like (...)hexane-x,y,z-triamine, where (...) are some substituents, and x,y,z are the locants for the three amines. Note that the extra alkyl groups on the nitrogens also become substituents.
How do we choose those values x,y,z? Well, there are two options:

The left gives us hexane-1,3,4-triamine, and right gives us hexane-3,4,6-triamine. P-14.4(e) says that the lowest locants should be chosen for the suffixes. Because the locant set 1,3,4 is lower than 3,4,6, we choose the numbering on the left.*
To finish up, we just need to add in the extra substituents. These are:

a methyl group on N attached to C–1
an ethyl group on N attached to C–1
a methyl group on N attached to C–4

To name these, we use P-62.2.4 again:

The locant of the parent hydride to which the nitrogen atom is attached is cited as a superscript to the letter locant N, for example, N2, N5, etc..

giving us the full name
N1-ethyl-N1,N4-dimethylhexane-1,3,4-triamine
(where ethyl comes before dimethyl due to the usual alphabetical ordering).

* This means a point-by-point comparison of the two locant sets (see P-14.3.5). So, you compare the lowest member of 1,3,4 with the lowest member of 3,4,6, i.e. 1 vs 3; and since 1 is smaller than 3, 1,3,4 is lower than 3,4,6. Some places still suggest that you should sum up all the locants and compare them, which is not correct at all.


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Size of Lanthanum
I was going through the values of atomic and covalent radii of the d-block elements.
The atomic radii of $\ce {Sc},\ce {Y}$ and $\ce {La}$ are $164, 184$ and $187$ pm respectively (Source: Wikipedia). My question is, why is the radius of Lanthanum not much bigger than Yttrium, when Yttrium and Scandium have a difference of $20$ pm?
The reason cannot be Lanthanoid contraction, since Lanthanum is before the Lanthanides. I thought it might be due to the differential shielding effect of $3$d and $4$d electrons, but I can't find any reliable sources stating this.

Empirical atomic radius column from Wikipedia data page says for Sc/Y/La 160/180/195 pm, for metallic radius 162/180/187 pm

@Poutnik I referred to the Wikipedia page of each element, where the radii (empirical) are in the box to the right of the page. But even with the metallic radius values you provided, why is there a 7pm increase versus an 18 pm increase?

It is similar for Ca / Sr / Ba (met radius 197 / 215 / 222.)


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Chemical potentials of mixture of two kinds of ideal gases vs. ideal gas separated by partition
Consider the below physical situations:
(i) Closed rigid container of volume $V$ that is isolated from the rest of the universe with uniform temperature $T$ and pressure $P$ filled and mixed with two species of ideal gases 1 and 2 with particle counts $N_1$,$N_2$. The total number of particles $N=N_1+N_2$ is fixed.
(ii) Closed rigid container of volume $V$ that is isolated from the rest of the universe with uniform temperature $T$ and pressure $P$ divided into two components of equal volume 1 and 2, separated by a partition that allows particle exchange. Both components are filled with ideal gas of the same kind with particle counts in both components $N_1$,$N_2$. The total number of particles $N=N_1+N_2$ is fixed.
In both physical situations (i) and (ii), it seems like the chemical potentials of 1 and 2 are
$$\mu_1=\mu_1^\circ + k_B T\ln \left(\frac{N_1}{N}\right)$$
$$\mu_2=\mu_2^\circ + k_B T\ln \left(\frac{N_2}{N}\right)$$
Is that true?

What do you mean by "allows particle exchange" in the case 2? Are they separated or not?

@Poutnik In case ii, components 1 and 2 are separated but are allowed to exchange particles with each other.

Then state functions are functions of time.

@Poutnik Sorry, I don't understand what do you mean.

I may have misunderstood the scenario. Is composition of compartments identical?

Let's assume that compartments have the equal volume..

This is. a potentially deep if very poorly thought-out question. An ideal gas does not consider differences in the identity of molecules since they have no size, forces etc. Removing the partition between two volumes of an ideal gas will have no effect on the energy etc. since the gas laws follow the First Law and more or less ignore the kinetic theory of gases.  It becomes interesting when the Second Law and kinetic theory are involved. Maybe rethink the question. A mix of He and Xe or of N2 and CO would be interesting.

Different diffusion rates due different molecular mass and speed would lead to slight pressure difference to balance it, until composition and pressure reach equilibrium.

If I read correctly in (i) you say that particles 1 and 2 are both in volume V, and in (ii)  there are two regions of volume V/2 one with just particle 1 and the other with particle 2, then you allow them to mix so that you have the situation in part (i).  If so you know your answer.


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By what mechanism might the decarboxylation of sodium benzoate in the presence of heat and sodium lye occur?
I am trying to get into chemistry and I saw a video by Nile Red in which powdered sodium benzoate and sodium hydroxide are mixed and heated and a reaction proceeds as follows:
$\ce{NaOH (s) + C6H5COONa (s) ->[200^\circ C] C6H6 (g) + Na2CO3}$
I don't understand how the bond between the benzene ring and the carboxyl group is being broken because a benzene anion seems very unstable, and I also don't get how the hydroxide is being ripped apart considering the OH bond is very strong. Can someone explain a plausible mechanism for how this reaction might take place?

Probably two step substitution + proton exchange (kinda like haloform reaction, with benzene instead). It happens so looking unlikely is irrelevant.

Ok, so the carboxylate carbon undergoes nucleophilic attack by the hydroxide and forms a bicarbonate ion whose proton is then transferred to the newly formed aromatic anion? Is that a reasonable possible mechanism? I want to understand this stuff so I can predict reaction products and invent syntheses.

Probably something like that. Proton transfer could be concerted, so there may be no separate phenyl anion involved.

@EatingTechnobladesRemainsAt3am it would not be bicarbonate. With the carbon in the $\ce{CO3}$ moiety also bound to the hydrocarbon group it would be an orthocarboxylate.

I have viewed the presumed Nile Red video to which the OP refers on the thermal decomposition of sodium benzoate in the presence of sodium hydroxide. An intimate, powdered mixture of sodium benzoate (0.69 mole) and sodium hydroxide (1.5 moles) contained in a metal paint can and fitted with a distillation apparatus was heated over a camp burner.[1] Although the OP indicates that the temperature of the reaction was 200oC, there is no evidence that the temperature of the reaction was determined. At these elevated temperatures, it is clear that the entropy (TΔS) outweighs enthalpy (ΔH).
Two previous Answers to this question offered variations on an ionic fragmentation of the carbon-carbon bond in question of sodium benzoate to afford benzene and carbon dioxide. The prospect of a radical fragmentation may also be considered. One source of hydrogen atoms is surreptitious water with a bond dissociation energy (BDE) of 119 kcal/mol. Alternatively, hydroxide has a BDE of 110 kcal/mol. There is the possibility that, under the vigorous conditions of the reaction, addition of hydroxide to the carbonyl group of the benzoate anion 1 occurs to form adduct 2. Bond dissociation of 2 leads to phenyl radical 4 and radical 3. continued...

The BDE of the O-H bond in 3 is expected to be much weaker than that in the other candidates. The BDE of the C-H bond in formic acid 7, although not formate, is 96 kcal/mol. However, the BDE for the same bond in the radical 8 is only 12 kcal/mol. Presumably, a radical cage effect would be operable prior to diffusion.
BDE's are taken from S. J. Blanksby and G. B. Ellison, Acct. Chem. Res., 2003, 36, 255.

My advice, don't try this at home.

Thanks for your answer but I am going to try this at home. If I get cancer in 20 years we will know why, it will be from benzene inhalation :D

https://bard.cm.utexas.edu/resources/Bard-Reprint/247.pdf suggests that both mechanisms are possible.

An alternative possibility is proposed here, which does not require formation of a free phenyl anion. Instead the phenyl-anion moiety is incorporated into a delocalized bonding in an intermediate state.
In this alternative, the hydroxide ion first adds to the carboxyl carbon to from the orthocarboxylate salt on the left. Ordinarily, we would not expect a carboxylate ion to act as an electrophile, but here the addition of heat energy and the strong basicity of the hydroxide ion may enable such an addition. The orthocarboxylate then undergoes an electronic rearrangement to separate the carbonate portion and leave the decarboxylated hydrocarbon behind.

Did you intend for your mechanism arrows to move electrons as shown?

Yes. One pair moves into the C-H bond region and the other becomes the second bond in the carbonyl portion of the sodium carbonate.

I meant have the arrows go in the opposite direction.

I don't have the nice apps to edit easily, could you show me? Thanks.

OL: If this diagram is unsuitable, feel free to go back to the original.

OK, thanks to you

To be clear, you mechanism also proceeds through a phenyl carbanion, though you've just shown it as concerted. You mechanism involves the nucleophilic attack of hydroxide to a carboxylate (which is odd) but has the advantage of being intramolecular. The phenyl carbanion doesn't distinguish our proposed mechanisms however, I've just explicitly addressed it

Note my addition to the first paragraph.

@OscarLanzi your minor edit looks reasonable to me and tbh your answer "feels" to me more correct (maybe because it's more parsimonious/intramolecular), especially given the unusual reaction conditions

Decarboxylations tend to follow the mechanistic path I've drawn below. You're right in that this is a bold transformation--partly this is driven by the high heating. Decarboxylations are also driven by entropic and enthalpic forces.

Entropically, you are starting with one compounded (sodium benzoate) and producing two (phenyl carbanion and CO2 gas). Entropy is frequently described as "disorder" which you can quantify as how many microstates are possible. If you have a loaf of bread and a four-shelf book case, you can put in on the first, second, third or fourth shelf. If you tear the loaf of bread in half, you can for example, put one loaf half in the third shelf and the other loaf in the first shelf—you can see there is many more combinations of options or "microstates". Replace "bread" with molecules and "book shelf" with energy level and you have entropy. Breaking the molecule in two increases entropy, especially because the CO2 is a gas, which has more entropy than a solid.
Enthalpic means, put simply, how strong the bonds are. While you do raise legitimate enthalpic concerns, this will be offset in part by the strong double bond you form making CO2 gas (a double bond is stronger than a single bond). If you make more strong bonds than you break, this is enthalpically favoured.
CO2 becomes NaCO3H and Na2CO3 readily. In fact this happens in your blood and is why people used to recommend breathing into a brown bag—increase the CO2 concentration which pushes towards NaCO3H. On that same point, decarboxylations are also an extremely common reaction in the body, albeit not with sodium benzoate.
Finally, the proposed reaction proceeds through phenyl carbanion (all other proposed mechanisms involve the same), though the mechanism may be concerted. This is a bit strange, but I'll just reference one or two previous posts on this species and point out that the aromaticity is not disturbed, as shown in the second figure. Typically you may see phenyl carbanions stabilised as lithiated species (i.e., using extremely strong organolithium bases to generate), which I think points to why such extreme heating was required by NileRed. Also, knowing his channel is sort of "garage chemistry", heating is much accessible than unusual reagents.

Looks like 2 people have suggested 2 possible mechanisms. The one you suggested involved the dissociation of sodium benzoate into CO2 and a benzene anion as the rate limiting step, and the other commenter suggested that nucleophillic attack of hydroxide on the carbon in the carboxyl group would be the rate limiting step. Could we test these 2 hypotheses by measuring the effect that the lye concentration has on the reaction rate?

Sodium hydroxide is introduced as a solid, not a solution species. So we can't define a variable sodium hydroxide concentration to follow this approach. Quenching with an alkyl halide and seeing if an orthoester is recovered might work, or it might not if the sodium hydroxide consumes the alkyl halide first.

That's a good question regarding distinguishing mechanisms. One possibility would be to do computational studies, which may distinguish which is more energetically reasonable--attacking a carboxylate anion with a hydroxide anion, or the presumably difficult generation of the phenyl anion. You can potentially monitor for CO2 gas evolution; "capture" the phenyl anion intermediate; or try a second reaction with a strong but nucleophilic base. Phenyl anions are generated all the time (though unusual looking) and partially stabilised so I'm not sure proposing an alternative mechanism is motivated


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Temperature Dependence of Crystal Formation in Storm Glass
The storm glass is an interesting historical device which claims to be able to predict the weather (for example, clear liquid implies fair weather, murky liquid implies rainy weather and large flaky crystals correspond to cloudy skies and snow in winter).
The liquid in the glass is a mixture of distilled water, ethanol, potassium nitrate, ammonium chloride, and camphor and it has been shown that the crystals which form are camphor.  To be more specific, it is a camphor-ethanol solution with aqueous ammonium chloride and potassium nitrate solution. Studies have found that as expected the ability of the device to predict the weather is questionable and that the appearance of crystals is affected only by temperature.
If fair weather is considered ''warm'' (no crystals) and rainy weather is considered ''cold'' (murky water), it seems like decreasing the ambient temperature generally helps crystals to form.
Roughly speaking, what is the reason that one might expect this temperature dependence to occur based on the above ingredients of the storm glass?

Specify "this", as the context it may point to within the post is too broad. T based or weather based behaviour? The former has easy explanation by solubility curve, where the camphor solution is at ambient temperature close to saturated one.

Temperature dependence.  I don't think there is any actual evidence for weather-based behaviour.

Similar effect could be achieved with water solution of salt with strong temperature dependency of solubility, like potassium nitrate, if there is prepared saturated solution at threshold temperature. Camphor may manifest at given solution extraordinary effect, as there is competing solubility versus being salted out, so possibly standing on an edge, being tilted by T to either side.Possibly kind of a bistable system, analogy to bistable electronic circuits.

@Poutnik I am not sure how this works, why does one need both the camphor-ethanol solution and the aqueous ammonium chloride and potassium nitrate solution, is it important that potassium nitrate's solubility in water increases with temperature?

Not sure, but it may have amplifying effect.

Yes I looked into it a bit more and I think the inorganic salts don't seem to be essential.  Small variations in the amounts of the salts has not been found to make visible changes to the storm glass.

The form, number and size of crystals forming depends not just on the temperature, but also very much on the rate of temperature change.

The (perhaps implicit) assumption of the device' design could be that every thunderstorm is a cold front underpassing wedge like a layer of ground level warm and humid air.  Then, while surface / ground air temperature drops, the temperature of the exposed device equally decreases.  Often, lower temperature of a solvent corresponds to lower solvent capacity for the solute (keyword solubility product) and precipitation to establish (again) a new equilibrium between material as solid and dissolved material.  This could lead to a less transparent solution in the tube recognized by the trained eye if the device is constantly in an instrument shelter.
Further on, while the humid warm air is pushed upwards by the cold front, these masses' temperature decreases because they reach higher heights.  Then,  their capacity for gaseous water is reduced (keyword humidity).  This leads to  condensation to form little droplets of water (related: Mollier steam diagrams); depending on currents of rising air (updraft) the potential for rain, and eventually thunderstorms.  For such a mechanism leading to a thunderstorm, the concept could work, possibly  complemented with visual observation (slow formation of characteristic shapes of clouds, etc).  In comparison to an electronic device, it wouldn't be easy to be used while hiking in alpine regions, where swifter changes of weather merit a likely faster response to changes (time constant in signal processing).
But recording the temperature (variation) as correlation to and sufficient predictor of thunderstorms isn't sufficient.  Assume you placed the device in your living room (like some do for an aneroid barometer) for convenience of reading.  It is plausible that if you air your room in winter, the mere drop of temperature the device is exposed to now leads to a reading "storm ahead".  A better approach to predict the likelihood of a storm includes monitoring air pressure (and its variation) both over time, as well as in space.  Literally, there are ground (or surface / sea level) highs and lows of air pressure, and in greater heights altitude highs and lows.  Currents of air / storms as a release of pressure gradients then may be vertical, horizontal (e.g. updrift as used in gliding), attenuated by woods, and in  more complex trajectories because of obstacles like mountains in the terrain. It is not obvious how this setup is sensible to these variations.


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Boiling points of hydro- and fluoro-carbons
I am teaching organic chemistry to a high school student. While discussing Teflon, we referred to the corresponding section in the 10$^\text{th}$ edition of "Organic Chemistry" by Solomons and Fryhle, where it is mentioned that "fluorocarbons have a lower boiling point when compared to hydrocarbons of the same molecular weight:"

Now, we know that carbon—fluorine bonds are more polar and higher polarity should result in a higher melting/boiling point due to higher interatomic forces. Moreover, higher molecular weights usually translate to higher boiling points. How do I explain it to my student.
Furthermore, the example taken here does not validate the statement made in the book because the molecular weights of $C_5F_{12}$ and $C_5H_{12}$ are about 288 and 72 g/mol, respectively, and not the same. I collected the data from here and here. On combining the two, the said relation does seems noticeable. What is the author trying to say here? Do we compare the molecular weights or the molecular formula?

I am also having trouble understanding the underlying principle as well, and if someone could elucidate it more, it would be really helpful.

Plot
For easy comparisons, I have plotted the boiling points of hydro- and fluoro-carbons for up to 8 carbon atoms:

Except for the case of 1 carbon (methane and tetrafluoromethane), the trend seems to hold true.

You accidentally copied the boiling point of ethane instead of propane in your table. Fixing it grealy reduces the apparent discrepancy between propane and octafluoropropane.

https://chemistry.stackexchange.com/questions/124744/large-difference-in-boiling-points-of-tetrafluoromethane-and-tetrachloromethane

@NicolauSakerNeto thank you for noticing the discrepancy, I have corrected the same.

@Mithoron do you mean to say that hydrogen is more polarizable than fluorine?

Molecules of halogenated hydrocarbons (check also CCl4 ) of the same/similar molar mass as respective hydrocarbons are much more compact with much weaker intermolecular van der Waals forces.

Yes, polarizability is lower. Also related https://chemistry.stackexchange.com/questions/171723/why-is-the-boiling-point-of-a-iodoalkane-greater-than-that-of-the-corresponding

Four factors (that have been pointed out) are at play here: molecular size, molecular weight, polarity and polarizability. How do we account for all four, and more (if any), in explaining the plot I have added in the question?

Eventual slight partial charges of F atoms would lead rather to decrease of the overall attractive intermolecular forces by mutual repulsion.

The following explanation is given by the Ref.1 for the vast difference in physical properties between perfluoro-alkanes and their hydrocarbon counterparts:

The physical properties of fluorinated organic compounds are in many cases very different to those observed for nonfluorinated molecules. In a very simple way, the majority of these changes can be rationalized by the atomic properties of the fluorine atom. For example, it has a high ionization potential and a low polarizability, which consequently leads to very weak interactions between perfluorinated moieties. Due to the electronegativity of fluorine, $\ce{C-F}$ bonds are always strongly polarized; nevertheless, perfluorocarbons belong to the most nonpolar compounds known. This is explained by the fact, that the dipole moments in a perfluorocarbon molecule cancel each other. Consequently, partly fluorinated compounds can have a significant polar character. Overall, the physical properties of hydrofluorocarbons are often very different from those of both, perfluorocarbons and hydrocarbons. Finally, the properties of fluoroorganic compounds are also determined by the excellent match of the corresponding $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of carbon and fluorine.

Since fluorine is a larger substituent than hydrogen, the structural behavior of perfluorocarbons differs from that of their nonfluorinated analogues. Typically, linear hydrocarbons adapt a linear zig-zag conformation, which is not the case for the corresponding perfluorocarbons. The steric repulsion of fluorine substituents in relative 1,3-positions renders a zig-zag conformation very unfavorable, so that nonbranched perfluorocarbons have a helical structure. Expectedly, helical chains are less flexible than zig-zag conformations, so that perfluorocarbon molecules occur as rigid rods.

I'm not sure this would answer completely your question, nevertheless, it is worth mentioning. Interesting point of view (also note that perfluoro-alkane compounds have significantly lower dielectric constants than expected, representing their low polarizability).
If you are interested in learning boiling point comparisons of perfluoro-ethers and -amines to that of hydrocarbons, it is worth reading Reference 2.
References:

Axel Haupt, "Chapter 11: Fluoroorganic compounds – unusual properties and versatile applications," In Organic and Inorganic Fluorine Chemistry: Methods and Applications; Walter de Gruyter GmbH: Berlin, Germany, 2021, pp. 283-300 (DOI: https://doi.org/10.1515/9783110659337-014)(ISBN: 978-3-11-065933-7).
A. A. Woolf, "Relative boiling points of fluoro-ethers, fluoroamines and other fluorocarbon derivatives to fluorocarbons," Journal of Fluorine Chemistry 1999, 94(1), 47-50 (DOI: https://doi.org/10.1016/S0022-1139(98)00336-4).