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Derivation and visualization of cell potential dependency on temperature and entropy
Working from the thermodynamic identity
$$\Delta{S} = -\left(\frac{\partial{\Delta{G}}}{\partial{T}}\right)_p,$$
is it allowed to integrate both sides to obtain an expression derived from the Nernst equation
$$E_\mathrm{cell} = \frac{T\,\Delta S}{zF}?$$
Can one plot the results and obtain $\Delta S$ from the gradient, assuming $\Delta S$ is constant for the reaction and $E_\mathrm{cell}$ values at some temperatures are given?

You need to decide how to integrate both sides of $\displaystyle \Delta S=\left( \frac{\partial E}{\partial T}\right)_pzF$.

What you are suggesting in your equation is that $\Delta G_m=-T\Delta S_m$ but you know that in fact $\Delta G_m=\Delta H_m-T\Delta S_m$. Your math is incorrect because you are missing a constant of integration.
The expression $\Delta G_m=\Delta H_m-T\Delta S_m$ can be rewritten using the Nernst equation as
$$-nFE = \Delta H_m -nFT\left(\frac{\partial E}{\partial T} \right)_p$$
by using the definition you provide for $\Delta S_m$ as the temperature-derivative of $\Delta G_m$, and the Nernst equation $\Delta G_m = -nFE$. In fact
$$\Delta S_m = nF\left(\frac{\partial E}{\partial T} \right)_p$$
Therefore if you plot $nFE$ as a function of T the slope of the curve at a given temperature will provide the value of $\Delta S_m$ at that T.


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Alternatives to Bayer process for extracting aluminum from lunar regolith
I am exploring ways to produce metallic aluminum from lunar regolith. According to most published results of lunar soil, $\ce{ Al2O3}$ comprises roughly 14% of mare samples by mass with $\ce{SiO2}$ being main impurity (40%). From what I been able to understand so far, both Bayer and Serpeck’s processes rely on reagents that can't be produced in-situ. Hence, the challenge is to use the least amount of chemistry to refine lunar material to the point when electrolysis can be used. Maybe magnetic separation?  Is Al2O3 (aluminum oxide) diamagnetic or paramagnetic?

I maybe have missed something so let tell me what you are asking.

If the Moon's crust is anything like Earth's, most of the 14% Al2O3 is not a physical mixture with the 40% silica, plus whatever else is in the regolith.  Most of the alumina is chemically combined with silica, so we will need to be able to extract the alumina or some other aluminum compound chemically.  Sounds like fun :-( .

source: Stack Exchange

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What do ε₀ and V mean in a Pourbaix diagram?
From Wikipedia:

What does this graph mean?
What is $\varepsilon$?  It looks like $\varepsilon_0$, like the vacuum permittivity, but what does this have to do with the polymorph and/or isomers of ferrite or hematite?
And what is the V or 'five' in square brackets after the $\varepsilon_0$?

This diagram, called E-pH diagram or Pourbaix diagram, makes it possible to know the domains of existence or of predominance of redox species as a function of pH.
The abscissa is the pH, the ordinate the reduction potential (with respect to the standard hydrogen electrode): thus, it is not a question of the size epsilon but of the potential $E$ which has been a little "stylized" as $\mathcal{E}$, which is expressed in volts (hence the symbol V).

This is a Pourbaix Diagram (I assume this is where you got this figure). While it's not a common notation as far as I'm aware, $\epsilon_0$ is being used to represent the reduction potential in volts (V). As mentioned at the top of the Wikipedia page, this is basically a phase diagram, that tells you the ranges of voltage/pH for which a particular species of (in this case) iron is most thermodynamically favorable. The dividing lines coincide with an even mixture of the surrounding species.

It's probably meant to be $\mathcal{E}_0$ $\mathcal{E}_0$, as in electromotive force.

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Depicting a mechanism of peptide bond formation with protonated and deprotonated amino group
I've noticed that some textbooks and video lectures use an amino acid wherein the amino group has two hydrogens, and others use an amino acid with three hydrogens in the amino group (in the formation of a peptide bond).
How come there are two variations? And also, if the starting amino acid has three hydrogens in its amino group, what is the mechanism for the part that the nitrogen bonded to the carbonyl ends up having one hydrogen attached to it?
I worked up the mechanism on the former condition, but starting with three hydrogens, I don't quite get it. I hope you someone can enlighten me.

I think it might be helpful to link to some of those video lectures/textbooks. They could be publishing errors fixed in another edition or acknowledged in an errata.

I'm reading Solomons' Org chem book and then I also watched a video from Khan.

General advice for written communication: don't use ALL CAPS and make sure the title allows the reader to understand what exactly the question is about without opening it.

Also the word is "protonated" not "with three hydrogens".

In neutral aqueous solution, the amino group is protonated (3 bound hydrogen atoms ) and the carboxilic acid group deprotonated (no bound hydrogen atom). In other solvents, you might have both groups in the neutral form the name amino acid suggests.
The mechanism of peptide bond formation is complex and involves intermediates with tRNA in the cell and starting materials distinct from amino acids for synthesis in the lab.


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Ice sublimes on the moon?
I read that the ice sublimes on the moon as the vapour pressure will be really low, but won’t a low vapour pressure increase the boiling point?
(Might be wrong, kindly don’t judge.)

Instead of asking not to judge your question, you could fortify your question by adding conditions at the Moon's surface (dark side?) you are asking about and provide your thoughts after looking at the water phase diagram.

It is on the process of becoming a good question. You need to add more details in order to get good answers.

Boiling is not a part of the picture at all.

There is no equilibrium vapor pressure on the moon. That's because the moon does not have a significant atmosphere to speak of, not even on the dark side. Any water that becomes gas is likely to drift off into space. Note even Earth loses water to space (although mostly as other molecules or atomic hydrogen and oxygen).
In a closed container there is at each temperature below the triple point of water a value of the vapor pressure at which solid and vapor can coexist such that without additional input of energy the amounts of both remain constant over time. This is what is meant by equilibrium: their relative amounts, as well as T and P, remaining unchanged. The equilibrium (coexistence) temperature at a given vapor pressure below the triple point is the sublimation temperature $T_{\textrm{sub}}$ at that pressure.
Since the moon's atmosphere is certainly not enclosed and not likely very stable, with pressure negligible, none of this holds, that is, there are no equilibria. In addition there is no feedback mechanism, at least at the low pressures under consideration here: the chemical potential of water in ice is independent of that in the vapor
(except at high pressures).

Lower vapor pressure lowers the boiling point. Don't you use a pressure cooker to increase the boiling point of water so that food can cook rapidly?
Why go to the moon? Ice sublimes in Canadian weather, where temperatures can go as low as -40 to -50 Celsius in some cities. People claim that ice sublimes on the roads and never melts. I never did this experiment but wet clothes are claimed to become dry in those extremely cold weather conditions.
From NASA's website Ice on the Moon

The Moon has no atmosphere, any substance on the lunar surface is exposed directly to vacuum. For water ice, this means it will rapidly sublime directly into water vapor and escape into space, as the Moon's low gravity cannot hold gas for any appreciable time. Over the course of a lunar day (~29 Earth days), all regions of the Moon are exposed to sunlight, and the temperature on the Moon in direct sunlight reaches about 395 K (395 Kelvin, which is equal to about 250 degrees above zero F). So any ice exposed to sunlight for even a short time would be lost. The only possible way for ice to exist on the Moon would be in a permanently shadowed area.

I really appreciate your reply, but sources say that a lower vapour pressure increases the boiling point, what am I missing? Anyway, thanks for your help

"Lower vapor pressure...", this is incorrect. Lower atmospheric or ambient pressure reduces the boiling point of a liquid.


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leaving a spoon in a cup of tap water for about 12 hours makes a pattern of bubbles, why?
I left the spoon in the water after showing the cup from the tea that I was drinking last night makes a spoon like the pattern of bubbles

Maybe it is just the gases naturally contained in the tap water, or maybe the spoon reacts with water. The former option seems more likely, but we need more experiments.

If the tap water is relatively rich in gaseous $\ce{CO2}$ at the source, which is easier at low temperature, this gas will slowly get out of the water when stored in a warmer place where the air contains few $\ce{CO2}$. The spoon may here serve as catalyzer of this reaction.

I agree with @Ivanneretin it looks like escaping of dissolved gases, as they have lower solubility in warmed up water. The spoon acts as a condensation center for gas bubbles  and temporal heat conductor during water warming up. The spoon edges are the most exposed to the water, so the most effective in both ways.

What you're asking isn't just why do you get bubbles, but why do the bubbles form a pattern at the surface that corresponds to the outline of a spoon.  I'm guessing what's happening is the spoon-like pattern on the surface is caused because bubbles selectively form at the edges of the spoon.  Then, when they rise straight up to the surface, you get a pattern that replicates the edge of the spoon. This naturally leads to the question:  Why do the bubbles selectively form at the edges or, to put it another way, why do they selectively form on extremely high-curvature surfaces?


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Why is there no salt bridge for hydrogen–silver bromide cell?
What is the reason there is no salt bridge for this cell?
$$\ce{Pt(s) | H2(g) | HBr(aq) | AgBr(s) | Ag(s)}$$

The function of a salt bridge or a diaphragm is to avoid mixing of 2 different electrolytes or electrolysis products from the respective half-cells, while allowing ion electro-migration between them.
This cell does not need a salt bridge as it uses the only electrolyte $\ce{HBr(aq)}$ and the is no need to separate the $\mathrm{AgBr}$ electrode from gaseous hydrogen.


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Why do bubbles form and stay when I shake a bottle of tea?
I have some bottled tea and I noticed that shaking it creates some foamy bubbles that stay for quite a while (at least half an hour if not more). I originally thought it has something to do with my saliva, but then I realized that a brand new bottle of tea also does the same thing. The listed ingredients are water, oolong tea and vitamin C. What could be the reason?

I'm inclined to close this as a duplicate of https://chemistry.stackexchange.com/questions/6957/how-does-shock-trigger-nucleation-of-gases-dissolved-in-a-liquid

@andselisk I think the OP does not ask why bubbles are created – what is the nature of that linked question – but why they persist as a minor foamy section. That makes it a different question. Additionally, the tea is not saturated by $\ce{CO2}$ as the other drink is.

Poutnik is right. The persistance of these bubbles must be due to some surfactant effect. Maybe oolong tea contains an ingredient which acts as detergent and decreases the surface tension of water.

@Poutnik I think the underlying concept is the same, the only difference is the dissolved gas (air vs carbon dioxide). Whether we need a separate question for each and every gas, is an opened question for the community.

@andselisk I think it is not about a dissolved gas in this case at all, but about dispersion of the already present gaseous phase and about foam stabilization by liquid composition. Similarly, the foam on the poured-in Czech beer is more about beer liquid than about $\ce{CO2}$.

@andselisk I agree with Poutnik and Maurice. The OP is mainly asking about the persistance of the bubbles/foam layer and what, if anything, in the chemical make-up of the bottled tea may be stabilising the bubbles / acting as a surfactant. And the quick answer is : probably the Vitamin C. Ascorbic acid is known to have surfactant properties in a number of solitions. Not published as an answer because I don't have time to go searching for references right now ...

What other liquids are you comparing it to? Are these other liquids also in a closed bottle. One determinant of bubble lifetime is the humidity of the air. The other is the composition of the bubble.

Tea is may be considered as aqueous solution of many different substances extracted from solid tea leaves.  (It is not so much about «tea» as in tea, or a tisane.)
In this solution may be ingredients altering the surface tension of water sufficiently enough that air bubbles are trapped for a prolonged time.  It could be a component added to act as a preservative, which equally happens to act as a surfactant, or an ingredient intentionally added to ease formation and preserve these bubbles; like the tapioca pearls in  bubble tea.
Compounds acting as surfactants may be artificial (like CTAB) made for cleaning, or processes from natural sources, e.g., polysaccharides (like starch), or plain natural.  Washing clothes with leaves of climbing ivy, for example  works because they contain a small amount of saponins.  That a compound acts as a surfactant does not say if it is safe in food, or not.

This article has a nice introduction about what determines foam stability, and describes how to increase it: https://www.pnas.org/content/early/2017/09/06/1705181114.full

Reminds me about the coffee-experiment NYTimes reported in 2017. On the more scientific side of the same body of work, Nature communications and its sequel by a Dutch group (for once, its not about waves.


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Can cis-trans isomers interconvert?
Can cis-trans isomers interconvert via tunneling ?

Cis trans tunnelling in a search renders quite some papers. Samples were specially prepared, tunnelling involves hydrogen/deuterium positional exchange rather than a "rotation".

I have just searched for "geometrical isomers tunnelling". Found things involving D as a label so I imagine it is easy for hydrogen atoms to tunnel.

geometric isomerism [obsolete] Obsolete synonym for cis-trans isomerism. (Usage strongly discouraged). I emphasise: strongly. Otherwise the question is good.

@MilanPaul that was a pictorial way to say that the tunneling involves somehow one nucleus tunneling to the other side (but I am not even sure such a decomposition makes sense, I just want to suggest that a long alkyl chain won't tunnel. Or, it makes sense but it is implicit in the concept of tunnelling).

If you have, say, HDC=CHD then if both H and D on the same carbon could exchange via tunnelling a close to 1:1 cis/trans mixture would exist: it might be possible  to measure this ratio precisely as one species may be slightly more stable than the other. Do you know if this type experiment has been done, I would be surprised it it hadn't.

@porphyrin If we started with pure cis/trans and the barrier is high, the probability will be very low right? So how can we expect close to 1:1 mixture ?

The barrier is often too high for thermal interchange (absorbing a photon will work though) but protons and deuterons can tunnel so the height of the barrier is less important. I'm assuming that H/D have similar tunnelling rates when saying 1:1 but in fact its more complicated and is $\sim e^{-c\sqrt{m}}$ where $c$ is the area of potential energy barrier above any given energy $E$.

This conversion is possible photochemically, after absorption of a sufficiently energetic UV photon.

Dear <@porphyrin> Does e stand for electric charge or Euler's number?  M is mass?  And c is area of potential energy, not potential itself?

In my comment $e$ is the exponential i.e.  $\exp(-c\sqrt{m})$

Electron tunneling
Tunneling refers to elementary particles moving "through" a potential barrier in the absence of sufficient energy to go "over the barrier". This is roughly equivalent to saying that a particle can occupy two locations and traverse between them even though all paths between those locations have a zero probability of the particle being there.
Some orbitals such as the p-orbital describe a probability density with areas of non-zero probability ("lobes") separated by space with zero probability ("nodes").
When electrons are transferred from one electron carrier to another, the transfer can occur through bonds or through space; the through-space transfer is either based on orbital overlap or on tunneling.
Proton tunneling
Proton tunneling is much slower because the proton is about 2000-times heavier than the electron. Tunneling has been used to explain reactions involving protons that are faster than expected from inspecting the potential energy landscape. This proton tunneling also explains proton transfer in hydrogen bonds.
For formic acid and acetic acid, proton tunneling explains cis-trans isomerism at ultra-low temperatures. Acetic acid, at room temperature, has a cis and a trans conformation.

There is a rotational barrier (which one may explain via resonance involving the carbonyl double bond) which is lower than thermal energy at room temperature [Ref 1]. For related systems like the peptide bond, cis and trans forms are isomers because the rotation barrier is higher.
At ultra-low temperature in a nobel gas matrix, the trans form is dominant and does not convert into the cis form unless you excited the molecule at the right wavelength. This means that the cis and the trans form are isomers with individual properties. Under these conditions, it is possible to enrich the cis form, and to observe the conversion to the trans form in the dark [Ref 2]. The temperature-dependence of the kinetics (and kinetic isotope effects) suggest that this cis-to-trans reaction occurs via tunneling.
Hydride tunneling in enzymes
Many redox reactions involve hydride transfer, e.g. from alcohol to NAD+ in the alcohol dehydrogenase catalyzed reaction. For enzymes, it is typically not possible to measure reactions at ultra-low temperatures; instead, a key experiment is to study the primary and secondary kinetic isotope effect [Ref 3]. From studies of multiple enzymes that catalyze hydride transfer it is evident that tunneling does occur, and the extent of tunneling is modulated by mutations in the active site. The review makes the argument that tunneling is easier to study for hydrogen (i.e. proton, hydrogen radical and hydride) transfer than for electron transfer because in the former case, isotopes (with a different deBroglie wavelength) are available as a probe. While the examples in the review are all redox reactions rather than E/Z isomerization reactions, one could envision the latter based on hydride tunneling as well.
"Double" tunneling
If there are two atoms connected to a double bond (such as in the molecule $\ce{H2C=CDH}$ mentioned in the comments), isomerization through tunneling would be much harder because the hydrogen and the deuterium have to somehow switch position. To avoid a clash, they would have to tunnel simultaneously, and could not both take a path along the saddle point of the potential energy landscape. Therefor, a cis-trans isomerization of $\ce{H2C=CDH}$ seems much less feasible, and there does not seem to be any experimental evidence for it.
References

Barriers to rotation adjacent to double bonds. 3. The carbon-oxygen barrier in formic acid, methyl formate, acetic acid, and methyl acetate. The origin of ester and amide resonance. Kenneth B. Wiberg and Keith E. Laidig. Journal of the American Chemical Society 1987 109 (20), 5935-5943. DOI: 10.1021/ja00254a006
Rotational isomerism of acetic acid isolated in rare-gas matrices: Effect of medium and isotopic substitution on IR-induced isomerization quantum yield and cis→trans tunneling rate. E. M. S. Macoas, L. Khriachtchev, M. Pettersson, R. Fausto, M. Rasanen
J. Chem. Phys. 121, 1331 (2004); DOI:10.1063/1.1760733
Tunneling and Dynamics in Enzymatic Hydride Transfer.
Zachary D. Nagel and Judith P. Klinman, Chem. Rev. 2006, 106, 3095-3118, DOI:10.1021/cr050301x

Tunneling between atoms may be difficult according to the other answer, but in some environments a scrambling mechanism may work.  In the presence of superacid, a bridgehead hydrogen in bicyclic compounds can be removed as a hydride ion, forming a carbocation.
Upon re-attachment the hydride ion may return to either of two positions mixing the cis and trans isomers, and eventually equilibrating them.  This process is documented for two such hydrocarbons by Olah et al. [1].

From Ref. [1]
Reference

George A. Olah, G. K. Surya Prakash, Jean Sommer, Arpad Molnar, Superacid Chemistry (John Wiley & Sons, Mar 26, 2009), p. 533.

Do the authors claim this is due to tunneling? The question asks about tunneling in the main body but not in the title. I'm not sure a hydride ion (i.e. a multiple particle system with electrons attached to protons) can tunnel.

This is entirely non-QM in nature.  I specifically point out that this is an alternative to tunneling, which appears not to work for atoms based on another answer.  The hydride ion scrambles (by switching between substrare and graphene layer) rather than tunneling.

@karsten theis I’m not familiar with this specific example, but there are many other examples of claims of hydride tunneling. See for example Judith klinman’s work on alcohol dehydrogenase reactions

@Andrew Thanks for the pointer, I included a reference to and discussion of  DOI:10.1021/cr050301x in my answer.

Have a look at the literature report by Helen Leung et al. (2013):

We synthesized cis-1,2-difluoroethylene from the dechlorination of 1,2-dichloro-1,2-difluoroethane. The products contain a cis/trans mixture of 1,2-difluoroethylene (...).

Here's the experimental evidence of tunneling using microwave spectroscopy (Leung, 2013, fig. 4a):

Edit
I think we should strive to use E/Z instead of cis/trans.

Reference
Leung, H.O., Marshall, M.W., Mueller, J.L., and Amberger, B.K. (2013) "The molecular structure of and interconversion tunneling in the argon-cis-1,2-difluoroethylene complex" J. Chem. Phys. 139:134303. DOI: 10.1063/1.4823494

We get a mixture of isomers, but do they interconvert?  The mixed product could come from mixed confirmations of the reactant.


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Would adding a water soluble salt (e.g. NaCl) affect a water sensitive reaction?
Let's take for example a reaction that would benefit from having less water in it, ester formation from alcohol and acid. We know that the equilibrium can be shifted to the right (ester formation) if water is constantly removed through distillation. My question is: would adding an excess of NaCl affect the reaction and push the reaction to the right?
The water would solvate the NaCl and theoretically be "preoccupied." I would like to think that water adding back to the ester to form an alcohol and acid would be more energetically favourable, but I don't really have enough knowledge to answer that on my own. Any thoughts?

Esterification
$\ce{R_1-COOH + HO-R_2 <=> R1-COO-R2 + H2O}$
has the equilibrium constant, expressed in compound activities:
$$K = \frac{a_\mathrm{ester} \cdot a_\mathrm{\ce{H2O}} }{ a_\mathrm{acid} \cdot a_\mathrm{alkohol}}$$
The water activity is decreased by dissolved salts by 2 ways:

decreasing the molar fraction of water by dissolved salt
decreasing its activity coefficient by bounding it by ion hydration.

As the result, dissolved salts decrease the numerator of the equilibrium expression and shift the equilibrium toward production of ester.
Present dissolved salts will partially affect activities of the acid and the alcohol as well, but in lesser extend.

A limited search yielded a work by Huang et al. (2016RSCAdv2106) cites chlorides of Cr, Sn, Zn, Al as previously used «used in the conversion of carbohydrates (e.g. cellulose, glucose and fructose) into platform molecules (e.g. HMF) through their Lewis acidity and Brønsted acidity.» then sticks (1 mmol scale, microwave heating; 400 W, 110 C and 10 min heating) to 20mol% of $\ce{Al2(SO4)3}$ ».  Question @Poutnik Are you aware of examples (no microwave heating) of greater scale than, e.g. 0.2 mol you could include as reference?

@Buttonwood I am not aware of any particular references ( nor have I currently time resources to find any ), applying just general principles on high level.


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Would this be a correct resonance structure?

In the (B) option, second resonance structure is supposed to be valid.

But according to my textbook, many adjacent charges would make the resonance structure too unstable and hence it would not be accepted as a valid resonance structure.

At the end is matter of where to put the discriminating limit. Your question makes sense but the meaning of the exercise is another one. Another example of exercise that shouldn't be given. Consider that some are certainly not limiting forms of the same molecule. This is the key to the exercise. If you knew it already than you should have specifically asked a slightly different question.

Yes, second B structure is important contributor.

Resonance formulae differ by localizing the electrons with dashes, plus and minus signs.  Thus correct, B is the sole line matching the criterion.
This contrasts, e.g. with line A depicts what is known as keto-enol isomers, which is: a change of how atoms are connected with each other.  Here, keto-O once binds (or not) to an hydrogen.  See guides, for example here.
Some of the forms may appear «high energy» for, e.g. writing charges in one form, but not the other.  As in the example of diazomethane, line B, or with DMSO:

The mere presence of atoms with a charge in one resonance formula does not render the formula wrong, but -- compared to the form without charges -- may represent a lesser or minor contribution to the overall structure represented, than the other.  In decreasing priority, there is a set of rules to consider.  To quote from Wikipedia, formulae must

obey as much as possible the octet rule [...]
have a maximum number of covalent bonds;
carry a minimum of formally charged atoms, with the separation for unlike and like charges minimized and maximized, respectively;
place negative charge, if any, on the most electronegative atoms and positive charge, if any, on the most electropositive;
do not deviate substantially from idealized bond lengths and angles [...]
maintain aromatic substructures locally while avoiding anti-aromatic ones [...]

You will encounter nitro group's Lewis formula as a potential pitfall searching in databases, too.


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Can I measure the molarity of Acetic Acid titrating Vineger with Sodium Hydroxide?
Imagine I have 5 g of vineger which I titrated with 33 mL of 0.1M Sodium Hydroxide (NaOH).
I tried V₁S₁=V₂S₂ formula but It needs the volume of the acid.
I have tried to calculate the volume with W=SVM/1000 formula which gives me 250/3S mL (where S is the molarity of Acetic Acid), but at the end S is cancelled out.
Is there any way I can calculate the molarity of Acetic Acid?

You were able to weight 5 g in a container.  Maybe you still have some of the vinegar, may record the weight of an empty graduated cylinder (as $m_0$).  Then fill this cylinder again until it contains again 5 g (as $m_1 = m_0 + \pu{5 g}$) and read the volume of the vinegar at the wall of the graduated cylinder.  The density should be somewhere between 1.00 and 1.30 g/cm³ (reference, acetic acid).

According to the Handbook of Physics and Chemistry, the density of vinegar ($6$% $\ce{CH3COOH}$) is $1.0069 g/mL$ at $20°C$. So you can solve your problem  yourself.

@Maurice with the precision shown on the numeric values, I believe we can just write $1.0 g/mL$

@Nicolas Yes we can !

Is there a good reason for deviating from standard variables' symbols for molarity $(c→S)$ and volume $(V→W)$? $S$ is entropy and $W$ is work, so one would have to bend their mind trying to read your formulas. Kinda reminds me of this episode of Andrew Dotson — Physics Professors Be Like (YouTube).

Typically in such a situation the liquid wouldn't be weighed but rather dispensed with a volumetric pipette. Conversely if the liquid was weighed, then you might calculate the mass percentage of acetic acid. It is really an odd technique to measure a mass and then convert that to a volume.

To use the formula that you indicate ($V_1 * C_1 = V_2 * C_2$), it is therefore necessary to know the volume of acetic acid $V_1$ or you only have the mass $m_1$.
To switch from one to the other, you need to know / determine the density of the acetic acid used : you just need to take a volume $V$ of acetic acid and weigh it, the ratio $\frac{m}{V}$ corresponds to the density $\rho_1$ -> by transferring into the dosage equation you thus arrive at $\frac {m_1}{\rho_1} C_1 = V_2 * C_2$, you know everything except $C_1$ so you can determine the concentration

Actually it should be $\frac {m_1} {\rho_1}$ to get volume.

Of course, I didn't read it well. So I modify with the correct formula. Thank you @MaxW for the vigilance


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E1 selection rules for transitions between two Σ states
For E1 selection rules (embodied in the electronic transition moment), to have an allowed transition the direct product of irreducible representations (IRs) must contain the totally symmetric IR. So, for transitions between two $\Sigma$ states, the following transitions are symmetry allowed: $\Sigma^- \leftrightarrow \Sigma^-$ and $\Sigma^+ \leftrightarrow \Sigma^+$.
But I read in Herzberg's book "Molecular Spectra and Molecular Structure I: Spectra of Diatomic Molecules" that in Hund's case b when we have a rotational distortion of the electronic motion or spin-orbit interaction, the $\Sigma^- \leftrightarrow \Sigma^+$  transition can appear.
Although for rotational distortion I understood that this violation is due to the $- \leftrightarrow +$ rule. I am wondering whether the direct product rule does not apply anymore?
Could you please tell me if there are other cases to consider where the $\Sigma^- \leftrightarrow \Sigma^+$ transition is allowed? And did I misunderstood the selection rule governing the transitions between two $\Sigma$ states?

The z transition dipole belongs to $A_1\equiv \sum^+$ symmetry species so product $\sum^+\times  \sum^+\times \sum^+\equiv A_1$ and similarly $\sum^-\times  \sum^+\times \sum^-\equiv A_1$. If there is spin orbit 'distortion' which has symmetry $A_2\equiv \sum^-$ then the case is $(\sum^+\times \sum^- )\times  \sum^+\times \sum^-\equiv A_1$ so is allowed

Thank you very much porphyrin for your answer, it is clearer now.


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Virtual screening lib
I had a quick question on virtual libraries. I'm hoping to perform a virtual screen using a library that contains very small, soluble compounds - think glycerol and smaller (<100MW). As far as I can tell, the majority of virtual libraries contain mostly larger, drug-like compounds. Is anyone aware of any libraries that might contain ideally 10,000+ or so of these tiny soluble compounds?
If not, any idea how difficult it might be to construct a custom virtual library enriched for these compounds?
Thank you!!

Just a guess: COSMO-RS based calculations are pretty good for most small molecules, as far as I know.


text:
How to name this organic isomer compound?
The chemical formula is C10H22O. I don't know how to assign which group gets priority when it is formulated this way. I think there is an iso- or sec- involved, but I'm not sure. I've looked at other questions for guidance but nothing is similar to this. I don't know where to go from here and where to start with naming this. Please lend help as to where to start.

Please consider punctuating the question nicely. It will surely improve it.

Welcome to Chemistry SE! Have a look at some of the IUPAC recommendations.

"i believe the chemical formula is C10 H22 O, i think it is related or similar to 2-propylheptanol" – This looks like you just counted the atoms to get the molecular formula and then googled "C10H22O". This is not a useful approach in organic chemistry as your example nicely illustrates.

"i don't know how to assign which group gets priority" – That might be because there are zero groups in this compound that could get such kind of priority.

"i think there is an iso- or sec- involved" – No. And preferably, you don’t use “iso” and “sec” at all.

You are right on this point: it is not an alcohol because there is no $\ce{-OH}$ group ; it is an ether oxide.
To name it, we are looking for the largest carbon chain: here it is $ \ce {CH_3 -CH_2 -CH_2-CH_2-CH_2-CH_2-CH_2-CH_3} $ with 8 carbon atoms so the suffix of the compound is octane.
We then add as a prefix the smallest carbon chain (here 2 carbon atoms so ethoxy) to which we add the number to indicate the position where the side chain is attached on the main chain (3 here) -> the prefix is therefore 3 -ethoxy
So ultimately your chemical is 3-ethoxyoctane

"octane" is not a suffix.


text:
Seeking safe, water-soluble chemicals for an IR spectroscopy experiment
I am working with a custom-built device which is intended to perform some mid-infrared chemical measurements.  It consists of four thermopile sensors with optical windows.  In front of each window is a band-pass filter.  Each filter admits a distinct range of wavelengths.  We performed an initial experiment which suggests that the manufacturer may have installed the filters in the wrong positions.  I am trying to design a more definitive follow-up experiment.  I can't tell the filters apart by looking at them.
These kinds of sensors are typically used in the automobile industry to characterize engine exhaust gases.  We happen to be measuring aqueous solutions instead, so I am trying to identify useful reference solutions we can prepare.  Ideally, the solutions would absorb well at one of the three wavelengths, and transmit well at the other two.  I don't care whether the reagents are organic or inorganic.  Safe, water soluble, inexpensive, and already present in my lab are all desirable qualities though.
The center wavelengths of the three optical filters are 6.7 μm, 8.3 μm, and 9.6 μm.  In wavenumbers, these are around 1490, 1200, and 1040 cm⁻¹.  The bandpass range on each of the filters is about 0.2 μm wide.  I also have a fourth, broad-band filter which I will identify by process of elimination.
Using my limited knowledge of IR fingerprints and the NIST Webbook database of infrared spectra, I have identified two candidate compounds.  Glycine absorbs preferentially in the 6.7 μm band, and trimethyl phosphate absorbs preferentially in the 9.6 μm band.
I still need a good candidate compound for the middle band: 8.3 μm / 1200 cm⁻¹.  The best I have found so far is n-decanoic acid, but it's not really that selective in absorbing at 8.3 μm.  It also isn't particularly water-soluble.
I would greatly appreciate it if anyone has any suggestions.  Thanks!

You might say why you suspect that the filters are in the wrong position. You might ask for the help of the near lab if the filter can be easily transported. It will be very quick to check what is what. But I agree with Farooq's answer. IR isn't something fitting your requirement of characterising aqueous solutions. At least not in standard analysis. One can rarely say never. But it would always be the wrong technique.

Thanks for your reply, Alchimista.  This is the thermopile sensor: https://dexterresearch.com/product-finder/st60-quad-silicon-based-thermopile-detector-4-channels/  Those windows are just under 1 mm across and the optical filters are sealed inside the can.  I need to test the device intact.

Disclaimer: I assume you refer to compounds which may be «soluble in water», not necesarily to constrained to «recording the spectra in aqueous solution» for limitations of IR transmission spectroscopy which are a bit lesser present with IR ATR spectroscopy, and basically not present for Raman (M. Farooq's answer). I assume you mention water soluble because you want to clean your setup after recording the data simply with water.  Note: often, the recordings mentioned below are literally about thin films of theses substances, i.e., an optical pathlength across the sample much less than $\pu{1 mm}$.
Complementary to NIST webbook chemistry offering a search by the vibrational energy here is the Japanese Spectral Database for Organic Compounds, SDBS. (Querries may be either in Japanese, or English.)  It equally offers a «reverse search» starting with spectral information you already have in hand:

(credit)
I'm not aware about the spectral width of the sensors you mention, if these record e.g., at $\pu{(1200 \pm 10) cm^{-1}}$, or say $\pu{(1200 \pm 50) cm^{-1}}$.  Obviously, the larger the allowance, the list of compounds indexed in the database matching this criterion will increase. In a query with the former threshold, browsing across the list of hits however were entries like formic acid (attention, potential danger because it is corrosive), or acetone  (attention, potential danger because it is flammable).
By back-and-forth between the listing (to rule out compounds not water soluble, requiring some intuition / knowledge) and the spectra, you may identify if there are other bands of (dis)interest to you, too, e.g.

(example acetone, loc. cit.)

We happen to be measuring aqueous solutions instead, so I am trying to identify useful reference solutions we can prepare.

This experiment is really problematic. mid-IR is not meant for aqueous solutions. Water is a no-no in infrared spectroscopy. Look at the low transmittance of water in mid IR range. Liquid water is going to swamp everything. What will be your container (cuvet) material? This is why Raman spectroscopy is the way to go with aqueous solutions.

Thanks for your reply.  We know that water is a pain to work with.  Notice that all three of our optical filters are targeted to the 1400 - 1000 cm⁻¹ range where water transmission is relatively high.  For this quick experiment, I don't even plan to use a transmission cuvette.  I will point the sensor at an open container of the test solution.  I don't need quantitative or sensitive results.


text:
Why IUPAC name this amine is written as 2-methylpropan-2-amine and not as 2,2-dimethylethanamine?
Why IUPAC Nomenclature of following amine is written as 2-methylpropan-2-amine and not as as 2,2-dimethylethanamine?

"2,2-dimethylethanamine" is not only a wrong name, it also describes a different compound.

This question is tagged with "amine"; however, the numbering of 2-methylpropan-2-amine is not limited to amines. You can see the analogous numbering for example in 2-methylpropan-2-ol and 2-chloro-2-methylpropane. Can you spot your mistake now?

The basic principle in nomenclature is to look for the longest carbon chain: here it is the chain with 3 carbon atoms so the suffix prop; since it is an amine connecting to the carbon 2 it gives the suffix propan-2-amine.
As you have a $ \ce {CH_3} $ group grafted in position 2 of the main carbon chain, you must add the prefix 2-methyl
So you have 2-methylpropan-2-amine.

It is should be propan-2-amine right?

Of course, i edit my post

It's so because while doing IUPAC naming we have to always choose longest branch where the number of functional group (here Amine) should be at least number. That's why we here call it 2-methylpropan-2-amine and not 2,2-dimethylethanamine. You can clearly see in the actual name it's propane which mean 3 carbon atom while as the name you want to give it has name ethan which means 2 carbon. Also if you will write 2,2dimethylethananime it means you are considering 2 methyl group which means only 1 carbon is left which will be called methane not ethane.
Hope it helps

"Also if you will write 2,2dimethylethananime it means you are considering 2 methyl group which means only 1 carbon is left which will be called methane not ethane." : There will be 2 carbons left if one considers 2 methyl group substitutions to Central Carbon, wouldn't there? H3C-C(CH3)2-NH2


text:
Is it correct to look at plane of symmetry in substituted cyclohexane for determining a meso compound?
I read this question and it's answer by @orthocresol. Now, he has written that the compound in question is meso not because there is a plane of symmetry but because of the rapid flipping of the cyclohexane ring, which makes sense to me. In reality the ring is not planar which seems to be in 2D. But, if we don't consider that ring is 3D and that it actually isn't planar, and we assume that it is planar, answer still comes out to be same, that is, for an actual meso compound, we are able to find a plane of symmetry in the 2D drawing and for a non meso compound, there is no plane of symmetry in the 2D drawing.

So, my question is why from that view also answer still comes out to be same? Is it a coincidence?
In this question, @Chakravarthy Kalyan writes "stereochemical analyses such as the preceding one can be done with drawings using planar rings", but with no explanation why. If we consider the compound 1-(chloromethyl)-3,5-dimethylcyclohexane, this is a meso compound with no plane of symmetry in the 2D drawing. So isn't it a counter example to his claim? If that is true, then why at all  people use the method of plane of symmetry for determining meso compounds in cyclic rings?

As Alchimista points out the chloromethyl can rotate, probably with a period far less than a nanosecond and so, in effect, there is a plane of symmetry involving the Cl atom, but not as you happen to have drawn it. On average the Cl atom can be considered to lie on the  'north-south' line of the figure.

Operatively:
A plane of symmetry in a 2-D projection of a cyclohexane is a sufficient condition.
Alternatively, and lengthy, you conduct an analysis involving conformers but not limited to the flipping of the ring, that means you must let the chloromethyl rotate, in your example.

In more details:
The answer to your title question is YES providing that the molecule isn't blocked in one conformer.
The latter situation requires low T, or the presence of more than one bulky substituent. At room temperature and within the context of common practice and exercises we can be sure that the condition is met.
I suspect that cases in which the situation can rapidly escalates towards complexity exist, as for heavy substitutions alters the conformational distribution, too.
The second question is basically
"How is that/why a plane of symmetry in a 2-D projection of conformers which themselves have no symmetry but are enantiomers does effectively work as a criterion?".
Well, a laconic but solid answer is that there is no other possible outcome.
There must be a formal mathematical way to express this fact; I am thinking of symmetry and projection operations.
Yet, I can't do better than visualise the situation, but this is again following the reasoning and depicting the sketch within the answer you have linked, although mentally.
It is the flipping between enantiomeric conformers that makes the two simultaneously equalise each other, and this necessarily requires the two stereocentres having opposite absolute configuration and being at opposite sides of a symmetry plane of the molecule's projection, otherwise they would forcedly be different isomers and there won't even be this discussion.
You can also think of the structure as having the axial and equatorial substituents in an averaged  intermediate position, as for the two limits can't be resolved.
Indeed, a 2-D projection is in a way a real representation of the molecule - not brutal alteration is operated. It just says what it's supposed to tell, i.e. how one would see the molecule as watched from above the projection plane.
During the flipping, one would see a rapid on-plane oscillation of the ax and eq substituents in an alternate fashion, and this require a plane of symmetry in order for the two to compensate their effects.
To make this visualisation clearer, I've sketched a rudimental figure showing the motion of ax / eq substituents and their average position as ×

Considering that the motion of the substituent can be taken as a handle for the entire conformational flipping, one can see the relevance of a symmetry plane of the projection. In such a picture there is no reason to expect optical activity, as if the molecule would be flat.
However, while the picture just above is a nice visual, the optical inactivity originates from the simultaneous presence of both enantiomers, and this should be right the reason for deprecating the old term internal raceme in favour of meso compound.

So, there is a plane of symmetry in the 2D projection of 1-(chloromethyl)-3,5-dimethylcyclohexane?

No. I have said that symmetry in one of the conformers is sufficient, not that must be there for every conformer. In other words the conformer you have drawn isn't meso. The compound is.

I see. But you didn't answer why is it alright to consider only the 2-D projection when in reality it isn't 2-D?

@LightYagami then I don't understand the question or better it seems to me that no other outcome is possible. It is OK only if a plane is there. In your example isn't possible.    If you would have drawn a conformer in which Cl point above or under the projection plane you would have conclude that the compound is meso, again without thinking of flipping.

I know that it is me who lack conceptual clarity. My understanding is, that a compound is meso if there exists a plane of symmetry in it. I would have been satisfied if the chair conformer had a plane of symmetry. Ok, so you said that if a particular conformer has a plane of symmetry, then it would be a meso compound. Does any conformer of the chair form also has a plane of symmetry in it...maybe by some rotations... if there is, then I am satisfied. If still my question doesn't make sense, would it be OK to you if I delete this post?

I do not think that it does not make sense. You are asking why a on plane projection with a symmetry plane does the job of taking the 3D structure and analysing its various conformers. If the question is really so plz confirm it. For me it is visually clear, as for the two centers do each the opposite of the other. But perhaps I OR ANOTHER USER may think of a more formal answer.

Yes, this is exactly the question.

Thank you! that cleared my confusion now.

Overall I agree with Alchimista's answer, though I think it might be quite complex to grasp, seeing the comments that followed it.
If it helps, here's my down-to-earth version, which in fact covers chirality in general, not just cyclohexanes.
The 'executive summary' of my answer would be: not only it is correct to use an 'unnatural' or 'high energy' conformation of a molecule to look for its symmetry elements, but it's a fundamental part of the abstraction / rigidification that inevitably accompanies symmetry considerations.
Side note: the chloromethyl group in your example is not 'asymmetric' just because the Cl atom points to one side: this is just a side effect of the limitation of representing 3D objects in 2D. You could just as well write it with the Cl pointing toward you, and you would have 2 H atoms pointing symmetrically to the back of the group.
Symmetry is more fundamentally related to imagining the object in 3D (which I understand is not everyone's strongest suit, but nowadays you can find free 3D simple molecular modelling software by which you can explore these concepts).
Lengthier explanation below.

First, it should be clear that a molecule is usually an extremely dynamic entity, it never stands still, it keeps vibrating, and its bonds and the groups connected by them rotate in space in very complicated ways (just think of all the docking studies).
By such motions, the molecule will assume different conformations, some of which have higher energy and some lower energy.
However, between the extremes of these energy curves there are a multitude of conformations, and the molecule can spend some time in each of them [except of course those that are not physically allowed (typical case: atropisomers)].
I'm stressing this point just to clarify that there is nothing scandalous or outlandish in imagining the molecule in a conformation that is not the lowest energy one or a relative minimum: it is still one possible conformation, we are not breaking or forming any bonds to make it.
Even more fundamentally, the act of looking for 'planes of symmetry', 'axes of symmetry', etc., of a molecule, implies by necessity that we imagine a rigid abstraction of it: we are looking at it frozen in a state or conformation that is definitely unnatural for it, no matter how well we choose such state, even if it's a relative minimum in the conformational energy curve.
So it is totally acceptable to choose any conformation that is 'convenient' for studying the stereochemistry of the molecule.
Why? Very simply, because in any case if a molecule is chiral, you cannot possibly find any (physically allowed) conformation with a plane or centre (or more generally, rotary-reflection axis) of symmetry (i.e. 'symmetrical' in the common sense).
It follows that, on the contrary, finding even just one (physically allowed) symmetrical conformation guarantees that the molecule is achiral (so, meso if some stereoisomers of the same molecule in other configurations are chiral).
If you want to see this in a mathematical sense, you can think that the existence of one symmetrical conformation implies symmetry in the whole set of conformations the molecule can assume.
For each conformation A there is for sure a conformation A' (of the same molecule, of course) that is a mirror image of A.
In a more physical sense, once you have achieved symmetry in one conformation, for each molecular motion that breaks the symmetry, there is a corresponding, opposite motion that cannot be 'distinguished' (energetically) from the former.
On the other hand, if a molecule has no such symmetry in any of its conformations, this is necessarily never the case.
Here's a concrete example:

I'm sure you would have no doubt that this molecule is a meso stereoisomer.
Even if for some electronic reason the most stable conformer of this molecule were this:

which is not symmetrical, it would not make the molecule chiral, would it?
For any N conformations of the molecule you can imagine, there are for sure N corresponding mirror image conformations of it that are still the same molecule.
If you take this one instead:

it is sure that you will not find even a single conformation whose mirror image is still a conformer of the original molecule.

Thank you (+1), this cleared a lot of confusion, but you have explained "why it can be assumed a molecule is achiral even if it has a plane of symmetry in only one conformation?" , my question was "why a 2D projection of the molecule takes into account the effect of a 3D molecule?" which has now been answered by alchimista. :)


text:
Why does chemisorption decrease with increasing temperatures?
Why is it that chemisorption decreases with higher temperatures?
The explanations I found online were quite vague and one of them cited Le Chatelier's principle, but I thought chemisorption is an irreversible process (because covalent bonds are formed between the adsorbate and adsorbent) so there is no reversible equilibrium.
Explanations I found online:

"Further increase will increase the energy of the molecules adsorbed and will also increase the rate of desorption. Therefore, the extent of adsorption decreases." - why does the rate of desoprtion increase?

"The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterward is due to the exothermic nature of adsorption equilibrium."

The chemisorption barrier depend on the electronic states involved which form a barrier to dissociation, i..e an activation energy.  Adding energy , i.e increasing the temperature, brings the molecules energy closer to the top of the barrier and so it can dissociate more easily. This is true of chemical reactions in general.

@porphyrin Oh, I see. Thank you for your answer! Also, is it that dissociation occurs with higher temperatures because the increase in kinetic energy breaks the covalent bonds? Just wanted to clarify because I'm a high school biology student, but I need to explain some chemistry concepts in my report.

Not just kinetic, molecules have vibrational and rotational energy also but eventually a molecule will find that it has enough energy to surmount the energy barrier and break a bond. The fraction of energy at any given temperature is given by the Boltzmann distribution.

Let's answer this question with the help of some thermodynamics and kinetics. Chemisorption is a chemical process with $\Delta S<0$ and $\Delta H<0$. Thus, in simpler words it is an enthalpy driven exothermic equilibrium
Every chemical process is an equilibrium with varying extent of completion. So your idea of chemisorption not being an equilibrium is wrong. Every chemical reaction is theoretically reversible in nature. What we call irreversible reactions in general are such reactions which are heavily favoured on one side of the equilibrium.
Now,
$\ln K = -\frac{\Delta H}{RT}  + \frac{\Delta S}{R}$
It is evident from this above expression that the $K$ will decrease when we increase $T$ when $\Delta H<0$ and thus the aforementioned conclusion.
But this is only half the story. Like all other chemical reactions, chemisorption also needs some activation energy to overcome the transition state. At lower temperatures there are very few molecules which have the kinetic energy enough to overcome the activation energy. The fraction of these molecules increases with increase in temperature and thus increasing the extent of chemisorption.
Conclusion
Due to the combination of thermodynamic and kinetic effects the rate of chemisorption first increases with the increase in temperature but after a certain saturation point the thermodynamic effects start dominating and thus the extent of chemisorption starts to decrease.
The graph below might help you understand how the extent of chemisorption changes with change in temperature.

Here,
$x$ = Mass of adsorbate
$m$ = Mass of adsorbent


text:
Related to liquid-liquid extraction
I was reading about purification of organic compounds and I came across this.
I quote below, a line from the excerpt that caused my confusion.

The organic solvent is later removed by distillation or by evaporation to get back the compound.

My question:
Why can't we just seperate the organic compound from the water by distillation/evaporation directly?

My approach:
The aim of the procedure is to - seperate out from water, the required organic compound that is presented to us in an aqueous solution form.
If the organic compound has relatively higher boiling point, use evaporation.
If the organic compound has relatively lower boiling point, use distillation.
If more than two compounds are to be recovered, use fractional distillation.

I felt like it was 'unnecessary' to mix it with an organic solvent and then proceed with evaporation/distillation of the new organic solution without water content. We could just proceed with evaporation/distillation of the aqueous solution directly and recover the organic compound - I've stated how in the 'My approach' section.

Self-evaluation:

I might not be aware of the various reasons because of my lack of knowledge and hence consider it unnecessary.
Distillation might cause reactions/decomposition because it is being heated in an aqueous medium while it might not react in an inert organic solvent.

Typically there are other components in the mixture that are left behind in the water. Hence the “purification” aspect

Water can, of course, be removed by evaporation (or freeze-drying), but if you used mains water then be aware there is a mineral content that will be left in your organic product.

That's highly dependant on context this question lacks.


text:
ELSD detection principle
While documenting myself on the detectors used in HPLC, I came across one that I did not know: ELSD.
I understood that the sample with the mobile phase was nebulized at the column outlet but I do not understand what is measured next.
Could you clarify this for me?

Was this link not successful in explaining ? Evaporative light scattering detector

ELSD is an acronym for Evaporative Light Scattering Detector. It is used as a quasi-universal detector for HPLC.
The first step is the nebulization of the column effluents with a dry gas as you have mentioned. If the column effluents contain any non-volatile analyte, then those will form minute particles. The mintue particles become suspended in the gas stream and form an aerosol. The gas stream is made to pass through a focused light beam (usually a diode laser) which is perpendicular to the flow direction.
If aersol is present, the light will be scattered, and can be detected/analysed by a detector photodiode which is placed perpendicular to both the light beam and flow direction.
There is a nice picture that I got from this site:

Reference- Analytical Chemistry, G. D. Christian, P. K. Dasgupta, K. A. Schug, 7th ed., Wiley, 2013

Nice explanation, the key point is that the analyte has to relatively non-volatile with respect to the mobile phase.


text:
Compound with pressure sensitive melting point at moderate temperature and pressure
I'm in need for a compound with ideally all of the following properties:

A melting point of 5 °C or less at 1 atmosphere (or slightly lower pressure)
A melting point of 12 °C or more at less than 20 atmospheres.
A significant enthalpy of fusion

(long term stability, a boiling point > 50 °C, and less than a 3 digit \$ cost per kg would also be nice).
Seeking such a compound, a melting point range this wide appears to be unusual at such small pressure differences. So far, the best I have is cyclohexane, but the melting point is a bit too high at 6.5 °C at normal pressure, and it needs over 100 atmospheres before the melting point gets above 12 °C, which is impractical.

I am afraid the required parameters are not realistic.

@Poutnik A negative would also be helpful as an answer, as I can then conclude that I would have to look for other solutions.

It would help ( unless it is a research secret ), if you reveal the background, what is the intended application of this material. As it is possible others could suggest alternative ways to reach the goal, which is unknown for now.

@Poutnik It's as mundane as a cooling requirement for an application where providing electricity is troublesome. If this compound existed, it would be convenient. If not, there are already many slightly less convenient solutions. (gas-liquid phase transitions being the most on-topic). It's a solved problem, I was just hoping an even simpler solution existed.

I'm not going to do your research for you (at least not for now—perhaps if the mood takes me I'll do the Mathematica programming at some later point), but I will give you guidance that will help you find the answer:

It's hard to find data on melting points at different pressures for a wide range of compounds (and you want to search through a wide range of compounds).  What is much more comprehensively available are data that can be used to calculate such pressure dependency, namely the specific densities of substances in their solid and liquid phases, and their enthalpies of fusion.  So:

According the Clapeyron equation,
$$\frac{dT}{dp}=\frac{\Delta V_m}{\Delta S_m}$$
where $T$ is the phase transition temperature, $\Delta S_m$ is the molar entropy change for the phase transition, and $\Delta V_m$ is the molar volume change for the phase transition.
For a phase transition, $\Delta S_m$ = $\frac{\Delta H_m}{T}$, where $T$ is in Kelvin.  And you can determine $V_m$ (in $\pu{\frac{{cm}^3}{mol}}$) from the molecular mass, $M_r$ (in $\pu{\frac{g}{mol}}$), divided by the specific density, $\rho$ (in $\pu{\frac{g}{cm^3}}$):
$$V_m = \frac{M_r}{\rho}\equiv \frac{\pu{\frac{g}{mol}}}{\pu{\frac{g}{cm^3}}} = \pu{\frac{{cm}^3}{mol}}$$
Thus:
$$\Delta V_{m, fus} = \frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}$$
The subscript "fus" means "fusion", indicating this is a solid-liquid phase transition.
Substituting, we obtain:
$$\frac{\Delta T_{fus}}{\Delta p} \approx \frac{dT_{fus}}{dp}=\frac{\Delta V_{m, fus}}{\Delta S_{m, fus}} =\frac{\frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}}{\Delta S_{m, fus}}=T_{fus}\frac{\frac{M_{r, liq}}{\rho_{liq}} - \frac{M_{r, solid}}{\rho_{solid}}}{\Delta H_{m,fus}} $$
What you want is to identify the substances that have the largest values for the above expression, and then filter according to your other criteria.
Note that this expression only gives an exact value for $\frac{\Delta T_{fus}}{\Delta p}$ if you integrate with respect to $T_{fus}$ using $\Delta H_{fus} (T), \rho_{liq}(T) \text{ and } \rho_{solid}(T)$—i.e., if you account for the temperature-dependencies of each of these. However, since you are determining $\frac{\Delta T_{fus}}{\Delta p}$ over a range not far from $T_{fus}$ at 1 atm, and since the densities for the phases that are stable at room temperature are probably at $25 ^\circ C = 298 K$, which is close to your desired $T_{fus}$, and since they will not change much over your pressure range, this should give you a very good approximation. Once you narrow it down to the actual substance, you can apply corrections to get a more precise estimate.

You could do this search manually by plugging in values into an Excel spreadsheet, or semi-manually by finding downloadable data tables with these values and importing them into Excel, but the best approach would be to use a program like Mathematica*, which has access to large chemical databases, and ask it to search for all substances for which it has density data in the solid and liquid phases, and enthalpies of fusion, calcualate the values for the above expression, and sort by magnitude.  You can then apply additional filters (e.g., melting point and enthalpy of fusion), as desired.

[*This of course requires knowing how to use Mathematica, or learning to use it.  It's possible someone on MathematicaSE could help you write such a query. Alternately, it's possible the online CRC is also searchable in this way (certainly seems it should be), though I don't know myself. One way to get help might be to check the reference desk of your local library.  They might have the capability to query chemical databases in this way.]
Once you have this, you can then identify the best candidate compounds, and further filter by cost and stability.

Note: It may not be possible to find a compound whose melting point changes enough (a comprehensive chemical database search would help you determine this).  Finding a large pressure-induced change in melting point is challenging, because substances don't change volume that much in going from a solid to a liquid. Where you do get a large pressure-induced change in a phase transition temperature is when going from liquid to gas because, under normal conditions (room temp and 1 atm), gases have about 1000x the molar volume of liquids. That's why water boils at a notably lower temperature when you go into the mountains, even though the change in pressure there is a mere fraction of an atmosphere. [By "notably lower" I mean enough of a change to affect cooking times.]

Plus you make your search even more challenging by requiring a high $\Delta H_{fus}$, since a high $\Delta H_{fus}$ lowers $\frac{\Delta T_{fus}}{\Delta p}$ (everything else being equal).

Will report back if this yields results.

@SE-stopfiringthegoodguys  When I gave my initial answer I forgot to include the entropy term!  I've amended my answer accordingly.

After trawling through large tables, there's indeed nothing of the sort to be found.


text:
what is the concentration of EDTA after titration with Zinc Metal?
I take $\pu{1g}$ of zinc metal and after dissolving, diluting it to $\pu{500 ml}$ and adjust the $\mathrm{pH}$ of $\pu{50 ml}$ of it to make titration with an unknown concentration of EDTA.
I found $\pu{1g}$ of zinc took $\pu{25 ml}$ of EDTA that means $\pu{0.004 g}$ of zinc equivalent to $\pu{1ml}$ of unknown concentration of EDTA.
my question is how can I calculate the concentration of EDTA from the above information?

From the mass of zinc, you can calculate the quantity of titrated material (using its molar mass).
You write the relation between the quantities of matter at equivalence (according to the equation of the dosage reaction): the only unknown is the concentration of EDTA

please more clarify what do you mean?

what is the equation of the dosage équation ?
you know the mass of zinc $m_{zinc}$ —> what is the relation beetween $m_{zinc}$ and the quantity of matter $n_{zinc}$ ?

Zn2+ (aq) + EDTA4- (aq) ---- Zn(EDTA)2- (aq)

With this equation, what is the relation beetween the quantities of matter ?

(mass of zinc / molar mass of zinc )X (1000)= ( molarity of EDTA x Volume Of EDTA )

it's correct using the right units (be careful with the factor 1000)

what is the mass that should be compensated in equation 1g or 0.004?... 2)(be careful with the factor 1000) ....what do you mean?

« that means 0.004 g of zinc equivalent to 1ml of unknown concentration of EDTA. » : you have the response

you mean the concentration of EDTA become 0.004 mol?! ....really I'm very confused

No, you say that 0,004g of zinc is équivalent to 1mL from the unknown EDTA : use the equation beetween the quantities of matter

(0.004/65.38)X1000=(molarity of EDTA x 1) ...am i right ?

Yes that’s correct

Also, OP's statement of "$\pu{0.004 g}$ of zinc equivalent to $\pu{1ml}$ of unknown concentration of EDTA" is incorrect as well.

Since OP has productive conversation with Nicolas, I think the information extracted from chain of comments is good enough to convince OP has made fair enough effort to solve this homework. The following steps might be helpful as well:
OP didn't indicate what $\ce{Zn/EDTA}$ molar ratio would this reaction take. As rule of thumb, it is generally $1:1$. Thus, $\pu{1 g}$ of $\ce{Zn}$ took $\pu{25 ml}$ of $\ce{EDTA}$ solution at the end point, that means the amount of $\ce{EDTA}$ in $\pu{25 ml}$ of the solution is $\frac{\pu{1 g}}{\pu{65.38 g mol-1}} = \pu{0.015 mol}$ (where $\pu{65.38 g mol-1}$ is the atomic mass of $\ce{Zn}$). Therefore, molarity of $\ce{EDTA}$ solution is $\frac{\pu{0.015 mol}}{\pu{0.025 L}} = \pu{0.6 mol L-1}$.
Note: According to the given data, the answer should be in one significant figure.

the very helpful answer, thanks a lot


text:
How to relate the conductance of an aqueous electrolyte to its conductivity
In this answer on Physics Stack Exchange, the author states that conductance $G$ is related to conductivity $\sigma$ by $$G = \sigma\frac{A}{l}$$
where $A$ is the cross-sectional area perpendicular to the direction of electric current and $l$ is the length of the conductor.
How does this relate to the conductivity of solutions such as aqueous electrolytes?
Papers in the literature often give the conductivity of aqueous electrolytes in units of $\pu{S/m}$ or $\pu{mS/cm}$.  Here is an example from Wu et al., J. Electrochem. Soc., 2015 where the authors plot conductivity in $\pu{mS/cm}$ versus molar concentration in $\pu{M}$:

But, in looking for conductivity values for sulfuric acid as a function of molar concentration, I found this report.  It states the conductance, rather than conductivity, of aqueous electrolytes (Or so it seems).
On page 7 of that pdf, for example, the authors provide a plot of $\pu{micromhos}$ versus percentage by weight:

I think that $\pu{1 micromho} = \pu{1 S}$.  What information would I need to determine the conductivity (in units of, for example, $\pu{S/m}$ or $\pu{S/cm}$) of an aqueous solution of sulfuric acid from the above plot?

A “mho” was an old unit that was the reciprocal of an ohm. The name is simply “ohm” spelled backwards and the unit symbol was the inverted upper case greek letter omega. So 1 mho equals 1 S.

@Ed V: So, OP's given relationship of $\mu \pu{mho} = \pu{S}$ is incorrect. It should be  $\mu \pu{mho} = \pu{\mu S}$?

@MathewMahindaratne, Good catch, it must be micromhos =microSiemens. It reminds me of another concentration unit which was used with conductivity. Fortunately, it vanished. It was called demal.

@MathewMahindaratne You are correct, From the wiki article, which I only saw after my comment, it seems that “mho” was suggested by Lord Kelvin back in 1883!

The answer is quite simple, if you look at the equation carefully. Conductance and conductivity are related by cell constant, $\displaystyle\frac{\text{Area}}{\text{Distance}},$ where the area is the area of the electrodes and the distance is the distance between the electrodes.
$$G = \sigma\frac{A}{l}$$
In conductivity measurements, the cell constant was either given by the manufacturer or it was determined experimentally by making a certain concentration of $\ce{KCl}$, the conductivity of which was well established and standardized.
In short, you cannot convert micromho to conductivity because you do not know the cell constant. And weight by weight conversion to molarity is not that trivial. You need to know exact density of each solution.
Did you see the end pages of the manual from you posted the old graph, there is a useful chart (for future readers).

(+1) Good answer and useful for future reference.

In the "Index of Electrolytes" on page 3 of the document at the source of the problem (Ref. 1)

note that the third column is titled "Maximum conductance and point of inflection at 25°C [$\pu{μmhos/cm}$/% by wt.]". However, the entries are given in units of conductivity ($\pu{μmhos/cm}$ or $\pu{μS/cm}$)! The first number for each entry in that column corresponds to the maximum conductivity in units of $\pu{\mu S/cm}$.
For sulfuric acid the entry is

If you look at the plot on page 7, sure enough, the maximum conductance occurs at 30%/wt, when the conductance is 825 mS (unspecified uncertainty but ~$\pu{\pm 3 mS}$) and (according to the table in the Index of Electrolytes) conductivity is 825 mS/cm.
The cell constant therefore has a value
$$C=\frac{\kappa}{G} = \pu{1 cm^{-1}}$$
If you compare values in the Index with those in other plots you'll identify a similar pattern. Therefore, although reported as conductances, the values in these plots can be read as conductivities, the cell constant for conversion between the two values being  $ \pu{1 cm^{-1}}$.
References
Note that the source of the data seems legitimate (Ref. 1) and compares well to other (official and unofficial) sources found during a quick web search (e.g. Ref. 3). The plots represent a compilation from various sources specified in the bibliography of Ref 1 and includes International Critical Tables (Ref. 2).

Conductance Data For Commonly Used Chemicals.
Rosemount Analytical - Emerson Process Management,
44-6039/rev. B December 2010

International Critical Tables, Vol. Vl, pp. 230-258; McGraw Hill, 1929.

G.W. Vinal, D.N. Craig. Resistivity of Sulphuric Acid Solutions and its Relation to Viscosity and Temperature. Journal of Research of the National Bureau of Standards 13 689-697 (1934).

Looks like I am still the only upvote for your answer. I also upvoted M. Farooq’s answer because the two answers give a good overall answer.


text:
Why is my dried out copper sulphate green when wetted with water?
I dried out some blue copper sulphate in an oven to see if it would turn white which it did.
I can't remember exactly but I left it overnight in the oven at around 50 C. It was white the following day.
When I put some water drop by drop on it it did not return to the original bright blue. Instead it took on a more green colour.
Why did this happen ?
The copper sulphate was bought from an chemistry chemicals supplier and the water was normal tap water which was treated for domestic use in a city. Water is drinkable.
There was definitely no ammonia in the vicinity.

Two questions : 1) What was the final temperature of your oven ? 2) Are you sure that there was no ammonia in the vicinity ? Rather small amounts of ammonia may react with copper ions to produce complex ions whose color is a different sort of blue.

I dried it at around 50 C. There was definitely no ammonia in the vicinity.

@Maurice Sorry, it turned green not purple, I don't know why I typed purple. Major typo.

The hydrate is blue , anhydrous is whitish.

@Kantura. And maybe tomorrow you will suddenly remember that the temperature was not 50°C and that it was not copper sulfate, but maybe who knows ...

@Maurice Tomorrow is here , your “maybe” didn’t happen :) Current info still stands.

Municipal water treatment often includes both chlorine and ammonia, creating chloramines for a stronger effect.
The color from the reaction with ammonium hydroxide is here - I don't think this matches what you describe.
However, the production of copper chloride (this video shows a different method) seems to be consistent with a green color.  You can search more images online to see if that matches what you saw.
There's a reason why chemistry labs like distilled water. :)


text:
Is it possible to do Cobalt catalyzed directing group asisted meta C-H functionalization of arenes
Directing Group assisted Pd catalyzed (sp2)meta-C-H functionalization has a growing attention in recent years.
In addition, few examples can be found for meta-C-H functionalization via Rhodium. I am wondering whether it is possible to do meta-C-H functionalization using Cobalt as its a first-row transtion metal. Since Cobalt chemistry is somewhat different what sort challenges we are going to face? If someone could share any reported articles based on this chemistry, it is much appreciated.


text:
Can the total amount of solution be found as a ratio between molar mass of a component and total mass of solution?
I wonder whether the following relation is true:
$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}},$$
where $M$ is the molar mass of the component, $n$ is the amount of substance and $m$ is the mass.
It was derived assuming $n = m/M,$ $n = n_\mathrm{solvent} + n_\mathrm{solute}$ and $m = m_\mathrm{solvent} + m_\mathrm{solute}.$
I don't think this is true, but I wanted to be sure before doing anything weird on a test.

No, I'm afraid it's completely wrong. Try re-doing it by substituting n_solvent=m_solvent/M_solvent, and n_solute=m_solute/M_solute. Also, include the units as a check. Note that your expression doesn't have the right dimensions.

Molar mass of solute and solvent is different, therefore, based on that alone, your relationship is incorrect.

Even if M for both solvent and solute was accidentally equal, it would be still wrong. As molar amount is proportional to mass, not reciprocal to mass.

To sum up the comments, only the following relation for the total amount of solution $n_\mathrm{tot}$ is universally true:
$$n_\mathrm{tot} = n_\mathrm{solvent} + n_\mathrm{solute} = \frac{m_\mathrm{solvent}}{M_\mathrm{solvent}} + \frac{m_\mathrm{solute}}{M_\mathrm{solute}}\tag{1}$$
The best you can do is to assume that $n_\mathrm{tot}\approx n_\mathrm{solvent}$ for the diluted solutions of small molecules. Also, if the molar masses are similar $(M_\mathrm{solvent}\approx M_\mathrm{solute}\approx \bar{M}),$ the expression can be lead to a common denominator:
$$n_\mathrm{tot} \approx \frac{m_\mathrm{solvent} + m_\mathrm{solute}}{\bar{M}}\tag{2}$$
This can be the case, for example, for the solution of ammonium nitrate $(M(\ce{NH4NO3}) = \pu{80.043 g mol^-1})$ in dimethyl sulfoxide $(M(\text{DMSO}) = \pu{78.13 g mol^-1}).$
Algebra aside, your mistake was also neglecting $M_\mathrm{solvent}.$
Keep in mind you can always check dimensions.
For your proposed formula

$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}}$$

it doesn't work out well:
$$\mathrm{dim}~n_\mathrm{tot} = \mathsf{M}\cdot\mathsf{N}^{-1}\cdot\mathsf{M}^{-1} = \mathsf{N}^{-1} \neq \mathsf{N}\tag{3}$$
Illustrating with common units used in chemistry:
$$[n_\mathrm{tot}] = \frac{[M]}{[m_\mathrm{solvent}] + [m_\mathrm{solute}]} = \frac{\pu{g mol^-1}}{\pu{g}} = \pu{mol^-1} \neq \pu{mol}\tag{4}$$

I made a typo in the latex, I was referring to if this was true $n_{solvent}+n_{solute}=\frac{m_{solvent}+m_{solute}}{M}$ and not $n_{solvent}+n_{solute}=\frac{M}{m_{solvent}+m_{solute}}$ still not correct, right?

@Xetrez In general, no, the expression is still not correct with a minor exception I addressed in my edit.

and what about it is wrong? is it that the $m_{total}\neq m_{solvent}+m_{solute}$?


text:
Amides and Alcohol acidity
Are amides more acidic than alcohols? I thought this regarding a secondary amide and a tertiary alcohol functional group since amides have both electronegative oxygen and nitrogen, compared to alcohol which just has an oxygen

Their acidities are similar.

This may assist you  https://owl.oit.umass.edu/departments/OrganicChemistry/appendix/pKaTable.html

Are amides more acidic than alcohols?

This is very broad question. The very short and general answer would be both have similar acidities (as Mithoron pointed out in his comment elsewhere). To support this statement, there is some evident in the Table provided by UMass website that mentioned by  Waylander in his comment. For example, the $\mathrm{p}K_\mathrm{a}$ value for the $\ce{N-H}$ in ehtanamide $(\ce{CH3C(O)NH2})$ is about $17$ $(\ce{CH3C(O)NH2}/\ce{CH3C(O)NH-})$ while that for the $\ce{O-H}$ in ethanol $(\ce{CH3CH2OH})$ is about $16$ $(\ce{CH3CH2OH}/\ce{CH3CH2O-})$. In that account, ethanol is more acidic than ethanamide (about by 10 times). However, the acidity would be all most identical if you add one extra $\ce{CH3}$ group to the carbon bearing $\ce{O-H}$ group on ethanol, making it $2^\circ$-alcohol. For instance, $\mathrm{p}K_\mathrm{a}$ of 2-propanol $(\ce{(CH3)2CHOH})$ is about $17.1$. Further, the acidity would be reversed if you add another $\ce{CH3}$ group again to the carbon bearing $\ce{O-H}$ group (to make it $3^\circ$-alcohol. Best example is tert-butanol $(\ce{(CH3)3COH})$, $\mathrm{p}K_\mathrm{a}$ of which is about $18.0$ (about 10 times less acidic than ethanamide). Based on these examples alone, can you say any alcohol is more acidic than particular amide? The answer is no. That's why I suggest that this question is broader than it seems. Thus, I advise OP to narrow down their scope so can give reliable answer.


text:
Finding the percent volume of each gas in the gas mixture
A gas mixture of $\ce{H2}$ and $\ce{N2}$ weighs $\pu{2.00 g}$ and have a volume of $\pu{10.0 L}$ at $\pu{700 mmHg}$ and $\pu{63.0 °C}$. Calculate the vol.% and the partial pressures of the two gases in the mixture.
I have only found the total number of moles using the formula:$PV=nRT$.
But I cant get any further! Any ideas?

I suggest you to start with what is being asked, not a general formula that seems to fit the scope of question, and then unfold the solution in terms of symbolic algebra. Can you write the formulas for the volume fractions $\varphi_1,$ $\varphi_2$ and partial pressures $p_1$ and $p_2$ (indices 1 and 2 refer to hydrogen and nitrogen, respectively) and try to deduce where the missing variables can be found?

When they say volume percent, what they really mean is mole fraction times 100%.

Let's call $x$ = number of mole $\ce{H2}$, and $y$ = number of moles $\ce{N2}$.
The mass of the gas mixture is : $$x·2 + y·28 = 2 g $$ The total number of moles is $$n = x+y = \frac{pV}{RT} = \frac{700}{760}101325 Pa·\frac{0.01 m^3}{8.314 J K^{-1}mol^{-1}·336 K}= 0.3341 mol$$Here you get two equations with two unknowns. This could be solved by choosing : $x = 0.3341 - y$. The first equation yields : $$2 = 2x + 28 y = 0.6682 - 2y + 28 y = 0.6682 + 26 y$$ $$y = \frac{2 - 0.6682}{26} = 0.0512 mol$$ $$x = 0.3341 mol - y = 0.3341mol - 0.0512mol = 0.2829 mol$$ $$x/n = \frac{0.2829mol}{0.3341mol} = 0.8467 $$ So the molar fraction of $\ce{H2}$ in the mixture is $84.67$% $\ce{H2}$. If you want the volume percent of $\ce{H2}$, you must calculate the volume of $\ce{H2}$ alone, which is $$V(\ce{H2}) = xRT/P = 0.2829 RT/P$$ As the total volume is $V = nRT/P= 0.3341RT/P$, the volume percent of $\ce{H2}$ is $\frac{0.2829mol}{0.3341mol} =  84.67$%. And by the same reasoning, the partial pressure of $\ce{H2}$ is $84.67$% of $700 mm$ Hg, which is $592.7 mm$ Hg.

Please do not forget to utilize \pu macro available on Chemistry.SE for units. Unit symbols must be upright, otherwise they can be interpreted as variables, and they also have to be separated with a space from the numerical values they attached to. $1mm$ Hg is wrong, but $\pu{1 mmHg}$ is correct. See the difference? The proper formatting is achieved with $\pu{1 mmHg}$. For more info, check out Units formatting on Meta.

It's also a good practice to avoid the buzzword "number of moles". Despite a widespread usage, it's an illiterate term. See Meta post on this issue. It's advisable to use a proper term amount of substance (e.g. "amount of nitrogen" in place of "number of moles N2"), especially when you are supposedly teaching younger students so that the improper terminology doesn't stick with them.

In addition to the answer given in elsewhere, I just want to point out one easy way to calculation. In this way you can avoid too many unit conversions to achieve your answer. Be aware that the Gas constant $R$ can be given in few different values based on its units.
Accordingly, you can use $R = \pu{62.36 L Torr K-1 mol-1}$ since given data $(p = \pu{700 mmHg} = \pu{700 Torr}$; and $V = \pu{10.0 L})$ are consistent with that value. Thus, from gas law for ideal gases, $pV = nRT$:
$$n = \frac{pV}{RT} =  \frac{\pu{700 Torr} \times \pu{10.0 L}}{\pu{62.36 L Torr K-1 mol-1} \times \pu{336 K}} = \pu{0.334 mol}$$
Since, $n = n_\ce{H2} + n_\ce{N2}$, and $2 \times n_\ce{H2} + 25 \times n_\ce{N2} = \pu{2 g}$, You have following two equations with two unknowns:
$$n_\ce{H2} + n_\ce{N2} = 0.334 \tag{1}$$
$$2n_\ce{H2} + 28n_\ce{N2} = 2 \tag{2}$$
Take $(2) - 2 \times (1)$, and you get:
$$26n_\ce{N2} = 2 - 0.668 = 1.332 \; \Rightarrow \; n_\ce{N2} = \frac{1.332}{26} = \pu{0.0512 mol}$$
Thus, from $(1)$:
$$n_\ce{H2} = 0.334 - n_\ce{N2} = 0.334 - 0.0512 = \pu{0.2828 mol}$$
Once you know $n_\ce{H2}$ and $n_\ce{N2}$, you can find their volumes under given conditions:
$$V_\ce{H2} = \frac{n_\ce{H2}RT}{p} = \frac{\pu{0.2828 mol} \times \pu{62.36 L Torr K-1 mol-1} \times \pu{336 K}}{\pu{700 Torr}} = \pu{8.46 L}$$
$$ \therefore \  \% V_\ce{H2} = \frac{\pu{8.46 L}}{\pu{10.0 L}} \times 100 = 84.6\%$$
Therefore, using simple mathematics, you can get $ \% V_\ce{N2} = 100 - 84.6\% = 15.4\%$.


text:
Can freezing temperatures cause chalk to fragment?
I am a physics student, with little experience in chemistry. I am researching for a project which involves freezing and its capabilities on humans and different minerals like chalk.
I have been researching chalk for the last couple of months as part of my project. Can the cracks generated cause the mineral to break into fragments under freezing conditions?
It would be appreciated if anybody can explain this to me or even send/link me some papers that I can read further about this effect.

I am intrigued in how freezing temperatures can make chalk crack or "explode" as some would call it. I will try to edit my question to make it more clear, sorry in advance if my question is not clear in the first place.

Sure, do that. As stated before, I am completely new to Stachexchange, so I am still learning to post questions in here. Thank you very much for understanding. Do I have to give you permission anywhere to edit, or can anybody do that?

Weathering of rocks through freeze/thaw cycles is a normal thing.

Chalk cannot at given conditions explode chemically, spontaneously turning into other compounds.
It could locally "explode" mechanically as consequence of mechanical stress caused by thermal dilation due changes to temperature. But calling it explosion would be very much exaggerated. It would be rather just cracking.
There is a way such cracking would be more probable, if the chalk is porous, soaked by water. The internally forming expanding ice could crack the chalk easily  But it would not happen at the first freezing event. It would need probably repeated freezing and melting, similarly like long term freeze weathering of rocks. The speed of such process depends on porosity..
Note that such cracking would be rather progressive than sudden, with cracks or crevices developing progressively, with occasional small pieces or blocks  falling away. Sudden break of the whole block into small pieces is highly non-probable

That's exactly what I have been reading and understanding. Should I edit my question to "crack" instead?
Furthermore, could chalk - let's say a chalk vase - in theory, "explode" in any way, if we put that vase outside in freezing conditions?

See the A update.

@Poutnik Just wanted to point out a typo "changes to temperature" in the 2nd para.

@Poutnik where do I see the "A update"?

A as the answer, Q as the question. There was added another paragraph to the answer. BTW StackExchange sites are primarily Q/A sites.


text:
How do point mutations affect complementary base-pairing and widths of the nucleotide pairs?
Point mutations are divided into two categories: transitions and transversions. Since point mutations only occur during DNA replication (an integral process of both mitosis and meiosis), the mutated nitrogenous bases only appear on the newly synthesized DNA strands, rather than on the parental strands that serve as templates during DNA replication.
A transition occurs when a would-be purine is replaced by another purine, or a would-be pyrimidine is replaced by another pyrimidine. In contrast, a transversion occurs when a would-be purine is replaced by a pyrimidine, or vice versa. Because the two strands of a DNA molecule always follow the strict rule of complementary base-pairing, there must be some impact on chemical structures of the DNA molecule when either a transition or a transversion occurs.
Specifically, under normal circumstances, an adenine forms two hydrogen bonds with its complementary thymine, and a cytosine forms three hydrogen bonds with its complementary guanine. When a transition occurs, such as when a would-be thymine is replaced by a cytosine, would hydrogen bonds still form between the two bases? And how many hydrogen bonds will there be? Same questions for when a transversion occurs, such as when a would-be thymine is replaced by a guanine.
Another relevant point to ask is the distance between the two DNA strands that constitute a DNA molecule. The rule of complementary base-pairing helps to maintain equal widths for each complementary base pair. Thus, an A-T pair and a C-G pair are of the same width. But when a transition occurs, would the new A-C combination have a different width? Same question for when a tranversion occurs.
Thanks to anyone who answers my questions.

Although this is certainly in the realm of chemistry I think that it can better fit in Biology SE. They can already know the answer without much drawing and thinking of molecular models.

Here is a paper https://doi.org/10.1093/nar/gkv254

When Point mutations occur in one strand of DNA, irregular base-pair emerge. One of the main features of this irregular state is that the width of the double-strand differs from the standard form(usually larger width as the new, irregular base pairings are weak) in that specific area. In other words, the portion of DNA which has the defective base-pair bulges out or protrudes leading to the asymmetry of DNA.(analogous but not the same as Thymine dimers) Most of the times the DNA repair system finds this and deals with it in the correct way. I do not know about the exact number of hydrogen bonds that appear in these situations but perhaps it obeys either Watson-Crick, Hoogsteen, or Wobble rules.


text:
Origin of electrode potential and its connection with electric double layer
Recently, I was learning about electrode processes and I came across the topic of "electrode potentials".
I understood that at equilibrium, a potential difference was generated at the interface between the electrode and the electrolyte because of the formation of an electric double layer.
I was reading
this
article which says that on dipping metals into their respective salt solutions, positive ions readily form and move into the solution leaving behind electrons in the electrode and a dynamic equilibrium is achieved soon.
Across different metals, only the position of the equilibrium (magnitude of negative charges on the metal electrode) differs due to their differing tendencies to oxidize (or reduce).
(I understood this part)

On the other hand,
I’ve also seen in some other texts that at equilibrium, the charge acquired by the metal might be either positive or negative depending on its tendency to oxidize (or reduce)
(unlike the $1$st source where they say the charge will be always negative).
Examples of
$\ce{Zn}$ ($\ce{Zn}$ has greater tendency to form $\ce{Zn^2+}$ ions hence metal acquires negative charge)
$$\ce{Zn<=>>Zn^2+ + 2e^-}$$
and
$\ce{Cu}$ ($\ce{Cu^2+}$ ions have greater tendency to get reduced hence $\ce{Cu}$ metal acquires positive charge)
$$\ce{Cu^2+ + 2e^-<=>>Cu}$$
are usually given.
(I also understood this explanation)

But what is confusing me is which of the above explanations is correct (or more accurate)?
Because if we go by the $2$nd view,
we have that $\ce{Cu}$ ions are more likely to get reduced so let’s say the $\ce{Cu}$ electrode becomes positively charged when ions from the solution stick to the metal, but at the same time few other $\ce{Cu}$ atoms in the metal will also get oxidized (because there are limited electrons in the $\ce{Cu}$ electrode) and dissolve so there’s no net oxidation or reduction. If this indeed happens then at equilibrium the $\ce{Cu}$ electrode will have no net charge (which isn't possible).

Note-I've seen a few other answers here on this site like this, this and this but none of these seem to address the confusion I'm having.
Any suggestion will be of great importance.

Poutnik, We should not trivialize such questions. I do not think anyone can adequately address the origins of electrode potentials in detail. Do we microscopically understand the origins of electrode potentials? I don't think so. It is beyond StackExchange Q&A and volunteer written webpages. The phenomenon had eluded Nobel laureates.

@M.Farooq Could you please tell me why this is so difficult? Moreover, as you said none of the resources (I found) talk in detail about this. Most of the texts I found on the web including the book I have mention the 2nd view in my question. And lastly, is it hard to even tell if the 1st or the 2nd view makes more sense? I'd almost believed the 2nd approach but just because of the libretexts article, I got confused. The first view is also talked about in Jim Clark's page.

The reason it is difficult is because it a very fundamental question. The question is why do metals when dipped in solutions of their ions develop a certain charged state. Therefore, there is a potential difference between the solution and the metal itself. Yet we cannot measure this potential. Nobel laureates like Ostwald, Nernst investigated these phenomena but they could not succeed, because nobody can experimentally measure the absolute electrode potential of a single electrode.

@M.Farooq Thanks a lot for helping me out!

I revoke my prior comment, due confusion of the meaning of the electrode potential – in context of this question is meant the potential difference wrt electrolyte.

Let's see if we can tease out the main points here.  The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook.
Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight positive charge in the region of the solution close to the metal surface owing to the sudden generation (and dissolution) of positively charged metal ions.

The amount of negative charge generated is metal specific and as of now, unknowable in an absolute sense.  Therefore, we are required to compare the extent of this process at one metal electrode to that of another.  If we compare the Zn electrode above to a Pb electrode, we would find that the Pb electrode has less negative charge on its surface relative to Zn.  Both of the electrodes, however, have a slightly negative charge at their surfaces.  Because the charge is greater at the Zn electrode, we say Zn is easier to oxidize than Pb.
To answer your question which of the above explanations is correct, the answer is both (although the clarity of the explanations is up for debate).  A negative charge is generated at both electrodes, but the the amount relative to one another differs, and that gives rise to the notions of easier/harder to oxidize/reduce.
(1) In my experience, microscopic theory of electrochemical processes typically assume the presence of the metal and metal ion in part to avoid issues which identifying what in the system is being oxidized and reduced.

Theoretically, we could do any reaction.
Practically, some are only possible in one direction; others are under special conditions.
For the case of copper: the standard potential of the pair $ \ce {Cu^{2 +} / Cu} $ being positive, this means that naturally the reaction is done in the direction $ \ce {Cu^{2+} -> Cu} $.
But we can still force the reaction in the other direction by imposing a potential.
For the case of zinc ($ \ce {Zn^{2+} / Zn} $), the natural reaction is that of oxidation so the reaction is $ \ce {Zn} -> \ce{Zn^{2+}} $.
So if we put copper ions $ \ce {Cu ^{2 +}} $ and zinc together, a reaction takes place between these species; if, on the other hand, we place $\ce{Cu}$ and $ \ce {Zn^{2 +}} $ ions, nothing happens naturally.
To know if a reaction is possible or not, we must look at the standard potentials of the couples (for similar concentrations otherwise we must calculate the potential using the Nernst relation): the reaction is naturally possible then between the oxidant of the highest potential torque and the lower potential torque reducer.

Thanks, but I was asking about what happens in the half cells at equilibrium (no ext. potential applied or connection between two half cells)


text:
Can a ferrite magnet be impregnated with polymer to reduce its fragility?
Ferrite magnet is a ceramic material and consists of agglomerated crystalline domains with spinel structure. Like other ceramic materials, ferrite magnets are brittle. For example, here is a photo of newly manufactured $\ce{Sr_xBa_{1-x}FeO2}$ tablet which already has chips and sharp fragments stuck on a surface:

This is one of the primary reasons as to why ferrite magnets for the consumer market are sold wrapped in plastic shells or coated with resin. There are also flexible magnets, a composite material consisting of a polymer matrix impregnated with ferrite magnet powder. Flexible magnets do not crumble and are more  impact-resistant, but they are also significantly weaker due to lower density of magnetic domains.
Would it be possible to do the opposite, i.e. impregnate the ferrite ceramic body with polymer to create a less brittle composite material still retaining magnetic properties comparable with those of a bulk permanent magnet?
I suppose the main difficulty stems from the fact that the manufacturing technology requires sintering at the temperatures around 1000 °C which limits the variety of binders and polymers to choose from. If there is no alternatives to currently used inorganic salts, can the impregnation be done afterwards? One suggestion could be creation of a slight increase in porosity of ferrite and filling the voids with polymer in vacuo at lower temperatures, but I didn't find any sources as to how legitimate or cost-efficient this approach can be.

Organic polymers like the ones to create the VHS K7 will have difficulties to sustain temperatures above 300 C.  Do you need to retain their mechanical flexibility?  What magnetic strength should these provide, still like the bulk ferrite ones, or would something as week as the tapes suffice?  Blending the concept of magnets with a head resistant matrix ... the ceramic tiles of Buran spacecraft, or lightweight aeorogels an early example ... just some thoughts.

@Buttonwood Thank you for additional examples with the tape and space shuttle. Basically, what I have in mind is a solid bulk ferrite reinforced with polymer to reduce brittleness. Sounds quite simple on paper IMO, but I didn't find anything related despite an existing demand for a material like that.

By their nature, highly magnentic materials are brittle, and inducing porosity decreases magnetic strength. "Rare-earth" magnets are usually plated with a malleable metal to reduce shattering, somewhat. Iron pole-pieces may be cemented to ferrite magnets for the same purpose.

@andselisk your username didn't trigger autocomplete in The Periodic Table so I'll leave a temporary delete-able comment here: I looked up that term and wow, I appreciate your swift action! (also, now I can't stop thinking about a somehow-microfabricated zillion tiny cubes or close-packed sphere rare-earth micro-magnets within a flexible matrix)