text:
Finding out the validity of the electrolysis of CaS to Ca(OH)2 to S in alkaline electrolyte
I would like to ask for the advice of chemists on how to go about validating a reaction that I have hypothesized as one stage in a chemical plant I am trying to improve. I want to find out whether a solution of CalciumSulfide (CaS) in an aqueous solution of high pH electrolyzes into solid CalciumHydroxide ( Ca(OH)2 ) and solid (monoclinic) sulfur ( S ). From datapages (wikipedia and chem-libretexts) on electrode potentials, I have determined that the most probable outcome of subjecting this electrolyte to a (absolute) voltage of about 0.4 V yields the production of hydroxide ions (along with hydrogen gas) at the kathode and the oxidation of sulfur anions at the anode (under the assumption that [OH-]~[S-2]~1 M). My hypothesis (and the necessary condition for my proposal to work) is that the hydroxide from the kathode reacts with the Ca(OH)+ in the electrolyte to form calcium hydroxide precipitate settling at the bottom of the electrolyte vessel, and that the sulfur (upon oxidation) leaches to the anode as a (monoclinic) crystalline solid. This would mean that the two products can be extracted separately in somewhat pure form, and I would be a happy man.
Half-reactions:
2H2O(l) + 2e − → H2(g) + 2OH−(aq) (−0.829 V)
S −2 (aq) → S(s) + 2e − (+0.407 V)
Total reaction:
2 H2O(l) + S −2 (aq) + Ca+2(aq) → H2(g) + S(s) + Ca(OH)2(s) (−0.422 V)
I have not studied chemistry (physics rather), but I still want to find out whether this reaction takes place under the above voltaic potential difference (so as to confirm my belief of the power consumption to be ∆G = n · F · epsilon_0 = −2 · 96, 485 · 0.422 kJ/mol = ~80 kJ/mol(S, Ca(OH)2, H2) )
I have two ways of finding out:
Does any of you chemists reading this know (a) if this reaction takes place under steady-state pH~12-14 (b) if this the only reaction to take place under steady-state pH~12-14, and (c) if this reaction takes place under the above-mentioned potential difference of epsilon_0 = 0.4 V?
In case I have to do the experiment myself in order to confirm my hypothesis would this cause me any problems? Namely, I will either (a) buy pure CaS from the internet, or (b) buy CaO and S from the internet, then produce H2 from hydrolysis myself and combine H2+S+CaO -> CaS + H2O under high temperature. Having obtained CaS through either option, I will dissolve CaS under constant stirring and under electrolysis at various voltages in order to document the molar quantities of the products from some energy investment. My problem (especially with option b) is that it is not legal for me to buy these chemicals. Remember, I do not have any degree or license in chemistry so this might raise eyebrows with law enforcement, which I do not want.
Thank you for reading. I hope this explained my situation well enough.

The trouble is that calcium sulfide $\ce{CaS}$ is not soluble in water. It is contained in the mineral called oldhamite. It cannot be electrolyzed in water.


text:
Why are acidic solutions often neutralised with carbonates rather than hydroxides?
In many syntheses stuff like calcium carbonate is used to neutralise acids, but this seems annoying, as you have to add it very slowly to avoid too much fizzing.
Could someone explain why anyone would voluntarily use carbonates over something like $\ce{NaOH}$ or $\ce{Ca(OH)2}$?

They buffer near neutral if you add a little bit excess. With NaOH, the pH goes up really high if you overshoot even a little bit if no other buffer is present

Related to the previous comment: excess hydroxide is a reasonably good nucleophike and may trigger side reactions. Carbonate would be weaker.

And there are expense and safety issues, too.

It depends, because the neutralization with $\ce{Na2CO3}$ offers the byproduct of $\ce{CO2}$, while the neutralization with lye does not.
If you perform the neutralization in an open system, you may monitor the evolution of the gas typically leaving the reaction mixture (you have to account for some $\ce{CO2}$ being soluted in water, like in a seltzer, though).  Thus, provided there was enough time for the neutralization reaction, the cease of gas evolution is a cheap indicator that $\ce{HCl}$ was neutralized and any additional addition of aqueous $\ce{Na2CO3}$ only increases the pH value of the aqueous phase much above 7.  Contrasting to this, if you use lye, the determination of the endpoint of titration requires either a) the addition of an indicator to an aliquot of your reaction/titration, b) adding an aqueous droplet of your reaction mixture on strip of pH paper, or c) an electrochemical determination of the pH value.
In organic chemistry, the use of $\ce{Na2CO3}$/$\ce{K2CO3}$ e.g. in a malonic ester condensation equally has the advantage that $\ce{CO2}$ escaping the open reaction mixture 1) drives the deprotonation to completion and 2) inhibits the back reaction, which otherwise would yield an equilibrium between acid (and corresponding conjugate base) and added base (and its corresponding conjugate acid).
For a neutralization in a closed system, the generation of $\ce{CO2}$ and generated pressure is a potential safety issue and may be a no-go.  Balanced with other parameters (e.g., price of the reagent per equivalent to quench one acidic function, process parameters and scale of reaction, OSHA/safety for the employees), lye then may be an alternative.

It's simple. Hydroxides are often strong bases, strong enough to react rapidly with e.g. human skin, should any be spilled on an operator's hands. In contrast, carbonates are weak bases- they don't react readily with human skin, so the operator can just wash the carbonate out with water (attempting to do that for hydroxide would result in an even faster reaction) without any noticeable harm.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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text:
What are skew positions in the conformation of cyclohexane?
I encountered a question which mentioned the skew position during discussion of the conformers of cyclohexane (chair/boat).
Here is the question (grammatical errors have been emended):

Which of the following statements is/are wrong about the greater stability of the chair form over the boat form?
(A) In the chair conformation, all $\ce{C-H}$ bonds in adjacent carbons are in the skew position
(B) In the boat conformations, there are four skew interactions and two eclipsed interactions
(C) The boat form has angle strain
(D) In the boat conformation, there are three skew interactions and three eclipsed interactions

It talks about skew positions. What does that mean? Is a skew position a staggered form? Or does it mean an axial position, or an equitorial position?

Please do not forget that the boat conformation is a transition state, it is not a minimum on the potential energy surface. Some people might sell this fact as angle strain.

It also appeared that the term skew position (or symmetrical skew) was used to refer to the chair conformation in some early research papers. I found two papers published in the early 1960s where the term was used in that way.

...with the growing interest in the family of cyclohexane conformations which include the boat and the symmetrical skew as extreme cases...

The symmetrical skew cyclohexane with neither bond length nor bond angle distortion, alternatively referred to as the stretched conformation or the "halfrotated" form, is shown by the numerical analysis to have dihedral angles of 33.1, 33.1, and 70.67", and not the 30, 30, 60" values quoted, for instance, by Klyne and Prelog (1960).

The skew conformation has the minimum energy of the flexible form...

References
Angyal, S. J., & Hoskinson, R. M. (1962). 578. Cyclitols. Part XIII. The conformation of cyclic diketals: the skew conformation. Journal of the Chemical Society (Resumed), 2991-2995.
Bottomley, G. A., & Jefferies, P. R. (1961). The Dihedral Angles of Cyclohexane. Australian Journal of Chemistry, 14(4), 657-659.

I googled "conformation skew staggered" and found:

The infinite number of possible intermediate conformations between the two extreme conformations are referred to as skew conformations. In these conformations, hydrogen atoms are closer than in staggered but away than in eclipsed conformation.

Source: https://knowledgebin.org/entry/stereochemistry-7/
Then I googled "torsion angles boat chair conformation" and found:

Source: Libretexts

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why do ketoses undergo dehydration more rapidly than aldoses?
I was reading about Seliwanoff's test and Wikipedia mentions the principle behind this test

This test relies on the principle that, when heated, ketoses are more
rapidly dehydrated than aldoses.

Well, as to what might be the reason behind the rapid reaction of ketoses, my conjecture is that as 5HMF is the final product of the dehydration of both aldoses and ketoses is a 5 membered compound, so there will be a extra step of ring contraction in dehydration of  aldoses which increases the activation energy and thus decreases the rate of the reaction.

Seliwanoff's Test - The Favouritism for Ketoses

TL;DR

The Seliwanoff's Test is a method to distinguish ketoses and aldoses.
The test uses resorcinol + dil. $\boldsymbol{\ce{HCl}}$ as the reagent.
The test involves saccharide undergoes triple dehydration, forming
furfural (5-hydroxymethylfurfural), and the furan derivative reacts
with resorcinol to form the red colouration.
Francisco Sánchez-Viesca and Reina Gómez's paper "Reactivities Involved in the Seliwanoff Reaction" explains the difference in reaction times between glucose and fructose.
They found evidence for a C-2 to C-1 intramolecular hydrogen-transfer during the acid-catalized (sic) isomerization of D-glucose to D-fructose.
The dehydration of fructose to 5-hydroxymethylfurfural occurs through the fructofuranose form, which can be originated either by glucose isomerization to fructose or more directly from the fructopyranose form.
Fructose presents common reactivities in the synthesis of hydroxymethylfurfural, not being the case with glucose or glucose isomerization to fructose.
Slower reactions resulted in accord with less common reaction steps, i.e., with especial reactivities.

Quoting the NCERT Lab Manual:

Ketohexoses give red colour and ketopentoses give blue-green colour. Aldoses take longer time to produce colour because under the same conditions, aldoses form furfural slowly, probably because β
-elimination is required before dehydration to furfural. Therefore prolonged heating should be avoided.

Introduction
This is a test to distinguish ketoses and aldoses. The reagent used here is resorcinol + dil. $\boldsymbol{\ce{HCl}}$. The ketoses show a positive test (red colouration of furfural derivative).
Mechanism is as given below:

The saccharide undergoes triple dehydration and forms furfural (5-hydroxymethylfurfural as OP has said) and the furan derivative finally reacts with the resorcinol to form the red colouration.

The answer
Francisco Sánchez-Viesca and Reina Gómez had written a paper titled "Reactivities Involved in the Seliwanoff Reaction" (source: Google Scholar) that had explained everything in concern.
Quoting:

Since glucose reacts much slower than fructose as observed in the Seliwanoff’s Test, we thought that this delay could be attributed to a previous isomerization of glucose to fructose. Thus, we looked for the current trends on this isomerization. We found that there is evidence for a C-2 to C-1 intramolecular hydrogen-transfer during the acid-catalized (sic) isomerization of D-glucose to D-fructose. The reaction from β-D-glucopyranose to α-D-fructofuranose is depicted in the below figure (sic), from a novel method of direct degradation of cellulose in to 5-hydroxymethylfurfural with sodium or potassium salts of acidic salts (sulphates and phosphates), a green HMF production.

The dehydration of fructose to 5-hydroxymethylfurfural occurs through the fructofuranose form. This can be originated either by glucose isomerization to fructose or more directly from the fructopyranose form. Dehydration starts at the hemiketal, and the intermediate enol rearranges to the aldehyde. The second dehydration gives the α, β-unsaturated derivative. A third water loss affords hydroxymethylfurfural.

These mechanism hold high significance with respect to ketoses and aldoses, and the hydride shift here makes the huge difference.

What are the consequences?
The article says a lot of things, but here's a table summarising everything, comparing and contrasting glucose and fructose.

Reactant
Key Steps
Distinctive Features

Fructose
Protonation of anomeric OH
Initiated by protonation of more reactive OH

-do-
Tautomerism
Resonance Stabilisation

-do-
Dehydration of tetrahydrofuran
Allylic OH dehydration prior to tautomerism

Glucose
Protonation at O-2
Not most basic OH

-do-
Cationic rearrangement contraction

-do-
Two dehydrations to HMF
Distinctive ring contraction step

In a nutshell: Quoting the article once again:

Fructose presents common reactivities in the synthesis of hydroxymethylfurfural, this not being the case with glucose or glucose isomerization to fructose. This explains the different reaction times observed in the test and it’s utility for ketose identification. Slower reactions resulted in accord with less common reaction steps, i.e., with especial reactivities.

In common man language: compare it to climbing a ramp and a staircase. The ramp is easy to climb, needs less physical stress, while the staircase needs more physical exertion. Strike an analogy for this as: Ketoses just need to climb a ramp, while glucose has to climb a steep staircase of a mechanism.

Sources:

Sánchez-Viesca, Francisco, and Reina Gómez. "Reactivities involved in the Seliwanoff reaction." Modern Chemistry 6.1 (2018): 1-5.

Ketoses are more rapidly dehydrated than aldoses due to their molecular structure. In ketoses, the carbonyl group is located at the second carbon atom, making it more reactive and prone to dehydration reactions. When heated with a strong acid, ketoses can quickly form furfural derivatives through dehydration.
Aldoses, on the other hand, have the carbonyl group at the end of the molecule, which makes them less reactive in comparison. They require a β-elimination step before dehydration can occur, which increases the activation energy and slows down the reaction.

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Could Nernst Equation fail at high Ion concentration diffrence
When I was learning about neuroscience, I came across the Nernst equation and found out that it is only dependent on the quotient of the ion concentration outside compared to the inside. So, now I thought, how is that possible? But, then found out that electrostatic force is much stronger so the effect on the concentration is negligible. But, now my question is: What about if we had a huge difference for example if $$\mathrm{Ion}_\mathrm{out} = \pu{150 mM}$$ and $$\mathrm{Ion}_\mathrm{in} = \pu{0.000015 mM}?$$
So, $$E=\frac{RT}{ZF}\ln(Q),$$ where $Q$ is $$Q = \frac{150}{0.000015}.$$
Would the Nernst equation still work, or would diffusion affect it too much. Or is diffusion already counted in, than how and why?

Be aware that N.E. describes equilibrium. There is no equilibrium when there is (net) diffusion.

Nernst equation works correctly at high dilution. At concentrations higher than about $0.1$ M, the concentration must be replaced by the activity. The activity may be considered as a sort of concentration calculated by dividing the number of moles of solute by the volume of the "free" water (water not electrically attached around the ions). This volume may be much smaller than the volume of the solution. And it is also necessary to take into account the probability of recombination of the separated ions in solution.


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Electrode potential of Br2/HBr half cell
I am wondering if it is possible to find the electrode potential of such a half cell:
$$\ce{Br2 (l) + 2 e- <=> 2Br-(g)}$$
The standard electrode potential is for $\ce{Br-(aq)}.$ However, is it possible to measure such a half cell? And if yes, will I be able to find this value from a reputable source?

Not all redox potentials are measurable, but they can be formally calculated from reaction Gibbs energy with involved SHE half cell.

Half cell potential of $\ce{Br2/Br^-}$ has a standard redox potential equal to +$1.06$ V. It can be measured if this half-cell has a platinum electrode which is coupled with any other half-cell.

@Maurice Have you noticed (g)?

@Poutnik. I have not seen the (g). But it is impossible to have a lonely ion in the gaseous state after such a half-reaction. So this question does not have any meaning


text:
Peak in beer freezing temperature plot
I left an open beer in the freezer while monitoring its temperature. The first minutes works as expected. But after the plateau, at about −8 °C, a peak appeared:

I thought it was an artefact because the can was open and the sensor was inside.
So I repeated the experiment with an intact, closed can.

The same spike occurs slightly below −10 °C. Well into the superfreezing phase, which is much wider this time.
Since beer is a liquid solution, I think this must be some component that changes phase before the others. Or maybe something related with solubility of $\ce{CO2}$ or ethanol as temperature changes.
Do you know what it is due to?
Update: reproducibility
After defrosting the can and putting it back in the freezer, the peak appeared again. Same result.

In order to discard a sensor issue, this is the freezing curve of the sensor itself:

Curve of full freezing process:

What disturbs me is the peak sharpness and symmetry, giving a strong hint it may be an artefact. Try repeated measurement to test the peak location reproducibility, together with measurement of the open space T.

Thank you @Poutnik. I reproduced the result and updated the question. I cannot measure the open space at the same time since I only have one sensor. But I don't find any reason why the freezer should suddenly increase 5ºC at this exact moment.

It  could be interesting to replicate experiment with plain and  carbonated (open and closed) water.

@Poutnik it might be an artefact after all. I dissasembled my setup and found condensation water in my temperature probe. That spike may be caused by the latent heat of that small amount of water phase-transitioning. I am repeating the experiment with a new probe.

The strangest thing is the fact that the thin exothermic peak happens always at nearly the same temperature, -$9$°C or -$10$°C, independently from the end of the supercooled water, which happens sometimes at -$5$°C and sometimes at -$15$°C, as if it was the crystallization of a very small amount of an impurity in the beer. This crystallization is very different from the crystallization of supercooled water, which is responsible of the sudden and lasting increase of temperature observed in all curves at various temperatures (from -$5$°C to -$15$°C).

There is no process in the bulk liquid volume that could cause so rapid cooling.

So it finally turned out to be an artefact, sorry.
The probe was protected by a small metallic tube. Somehow water condensed inside. Since this water is isolated from the main liquid, it performed its own phase-transition releasing its latent heat right next to the sensor.
The effect was not appreciated when freezing the probe alone. I don't know why, maybe because of a faster cooling o due to its physical position. But it was very noticeable when measuring the beer can.
I removed the tube and cleaned the sensor. I also ensured the thermosensitive part is physically in contact with the can; and I insulated it better from the environment. Now the measurement is more accurate and no weird peaks.
Thank you all for your time.

A nicely done experiment!
Consider hand-warmers containing sodium acetate trihydrate, $\ce{NaCH3COO.3H2O}$. When heated above the melting point and then cooled, the compound does not quickly solidify, but can be greatly supercooled. Given an impetus to begin  crystallization, such as the shock-wave produced by a "clicker" (or nucleation by a speck of dust), the molecules reorganize from liquid to solid, giving off heat to warm your digits. The increased order of the solid frees up some energy.
In a similar way, when supercooled water freezes, it gives off heat. Your graph seems to show that effect.
To see it even more vividly, put distilled water in a polypropylene food container with lid on loosely. Boil the water in a microwave oven for just a few seconds to remove most dissolved air, which otherwise would nucleate freezing too early. Do not boil a second time, or the air-free water can superheat, and then flash into steam all at once, blowing open the microwave and blowing an internal fuse. [Lest you ask, yes, this happened to me - no burns, but I had to disassemble the oven to replace the fuse.]
Put your thermocouple on the outside of the container (or it, too, could nucleate freezing), and place all in a fairly cold freezer. Likely, the freezer will not be cold enough to freeze that water, and your cooling graph should be smooth. Now, take out the container and give it a sharp rap on a counter... watch the water, and observe the temperature change.
Of course, freezing beer drives out the bubbles, so I hope you don't mind flat beer. ;-)

Hmm, if it was freezing of supercooled beer, there would not be fast temperature return to previously achieved temperature. There would be reaching the freezing point, forming a plateau until bear is frozen. With some drift, as freezing point would not be constant with freezing progress.

P.S.:The plateau like on the second picture of the original post.

Thanks for your explanation. But I am not sure to understand the reason of a sharp peak in the middle of the process. Would you mind to clarify, please?

@Poutnik you will not see a clear plateau in the "closed can" case because the liquid is contained in a constrained space. The ice formation raises the pressure and lowers the freezing point. Eventually the can will size-up or crack (or the freezer's temperature is reached). I uploaded a picture of the long term so you can appreciate it better.

@EyC What this means then? : So I repeated the experiment with an intact, closed can....  with the clear plateau in the picture. OTOH for the open case,the freezing plateau occurs before the peak. So in neither of cases us the peak related to the freezing up period.

@Poutnik you're right. I just wanted to support your first comment and emphasize that the plateau will not be at constant temperature, since the freezing point drifts like you said.


text:
Does lithium form ionic hydride?
It is stated in my chemistry textbook that lithium indeed forms lithium hydrides. However, significant covalent characters could be found in lithium hydrides (like least reactivity).
But in some other sources, it is stated that Li, along with Be and Mg, form covalent hydrides, not ionic, and are stating reasons like Fajans' rules, ionization energies and electronegativities for the same. I'm totally confused about whether lithium forms ionic hydrides.
Does Li form partially covalent hydrides or partially ionic hydrides? (Since I know, no compounds can be 100% ionic or covalent.)
Or is the % of covalency greater than electrovalency?

There is no "or". It is partially covalent and partially ionic.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

As described in the comments, there really is no sharp boundary between "ionic" and "covalent" bonding. With metal hydrides the situation is muddled further because the usual electronegativity-difference rules don't really work. Not only lithium but also magnesium forms a hydride whose ionic character far exceeds what would seem to be expected from electronegativity differential.
Magnesium hydride, sometimes called "covalent" or "borderline covalent" (I've even seen "polymeric") with its electronegativity difference less than unity, nonetheless has a crystal structure (rutile) generally associated with compounds having a higher degree of ionic character; the magnesium is octahedrally coordinated just as in the oxide or fluoride (the latter also has a rutile structure). Moreover, it serves readily as a hydride-ion source to water, if the passivating hydroxide layer is broken by either physical milling or chemical doping [1].
Lithium hydride shows similarly high ionic character even though here too the electronegativity differential would seem too small. It has a rock-salt structure like the fluoride or chloride, and reaction with water is again limited by saturation of the solution [2] (a more forgiving limit than with magnesium hydroxide) rather than lack of hydride-ion availability.
What happens in terms of atomic structure is the hydrogen 1s valence orbital is simply too small to offer good covalent overlap with the relatively diffuse valence orbitals offered by most alkali and alkaline-earth metals. The orbitals of lithium among alkali metals and magnesium among alkaline-earth metals are just a little too diffuse to optimize covalent bonding to hydrogen, and so the hydrides tend more to ionic bonding than would otherwise be the case (beryllium, with a smaller cation radius, does offer better overlap and its hydride does show more covalency). Heavier metals in both groups serve only to make the overlap situation worse and the bonding forced even more towards ionic.
The role played by the relatively small size of the hydrogen is consistent with Fajans' Rules after all, as these rules favor ionic bonding with small anions. Hydride ion is very much "puffed-up" compared with the parent atom but still is comparable in size with fluoride ion. This similarity has even prompted the use of flouride-hydride ionic solutions as hydrogen storage media, as with the magnesium counterion [3].
References

Liuzhang Ouyang, Miaolian Ma, Minghong Huang, Ruoming Duan, Hui Wang, and Min Zhu. "Enhanced Hydrogen Generation Properties of MgH2-Based
Hydrides by Breaking the Magnesium Hydroxide
Passivation Layer", Energies 2015, 8, 4237-4252. https://doi.org/10.3390/en8054237.

J. H. Leckey, L. E. Nulf, and J. R. Kirkpatrick. "Reaction of Lithium Hydride with Water", Langmuir 1996, 12, 26, 6361–6367. https://doi.org/10.1021/la950843h.

Terry D. Humphries, Jack Yang, Richard A. Mole, Mark Paskevicius, Julianne E. Bird, Matthew R. Rowles Mariana S. Tortoza M. Veronica Sofianos Dehong Yu, and Craig E. Buckley (2020). "Fluorine Substitution in Magnesium Hydride as a Tool for Thermodynamic Control".
J. Phys. Chem. C 124, 17, 9109–9117.
https://doi.org/10.1021/acs.jpcc.9b11211

According to this reference Oxford Reference quoting A Dictionary of Chemistry here

The bonding in lithium hydride is believed to be largely ionic; i.e. $\ce{Li+H-}$ as supported by the fact that hydrogen is released from the anode on electrolysis of the molten salt.


text:
Balancing disproportionation reaction of hypoiodous acid
In aqueous solution, the acid $\ce{HIO}$ disproportionates according to the following equation where m, n, p and q are simple whole numbers in their lowest ratios.
$$\ce{mHIO → nI2 + pHIO3 + qH2O}$$
This equation can be balanced using oxidation numbers.
In $\ce{HIO}$ and $\ce{HIO3}$, the oxidation state of iodine is +1 and +5 respectively. Therefore in $\ce{HIO}$, iodine lost 4 electrons to iodine atoms, therefore n is 2 and finally p is 1. However, when the electrons are received by the iodine atoms, they should form iodide ions instead of molecules. How would you tackle this question?

They are not and they should not. // Do not you think HIO is strong enough oxidizer to prevent creation of easily oxidized I- ?

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

The disproportionation rection of $\ce{HIO}$ cannot produce $\ce{HIO3}$ and $\ce{HI}$ (or $\ce{I^-}$ ions) in acidic solution. The reason is that $\ce{I^-}$ and $\ce{IO3^-}$ ions do react with one another, if $\ce{H^+}$ ions are present, to give : $$\ce{IO3^- + 5 I^- + 6 H^+ -> 3 I2 + 3 H2O}$$ If no ion $\ce{H^+}$ is present, the previous reaction does not occur, and the solution contains simultaneously the ions $\ce{I-}$ and $\ce{IO3^-}$


text:
How to derive the relation between entropy and heat described by the equation dS=dq/T?
I know that entropy is measure of disorder, and a book (Intro to Thermal Physics by David J. Schroeder) states that $\mathrm dS=\mathrm dq/T$, where $S$ is entropy, $q$ is heat flow, and $T$ is temperature.
What I don't understand is the relation between these two interpretations.
Taking the definition of entropy as measure of disorder, I am looking for an intuitive proof of the equation $\mathrm dS=\mathrm dq/T$, or an explanation as to why this equation  represents measure of disorder, or both.
I would very much like to know why entropy is not a path function, as it is dependent on heat (which is a path function).

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

While 'measure of disorder' is correct it does not help much especially as 'measure' tends to get dropped. It is better to think of it as the number of ways energy levels can be filled (as in Boltzmann definition $S=k\ln(\Omega)) $. In this sense ,heat being transfer of energy  in a in a random manner, it is easy to follow why entropy increases.

Entropy is not a measure of disorder. This statement has been corrected in several textbooks. Order/disorder is related to information entropy.

The post is probably about relation of information entropy and thermodynamic entropy.

I just try to answer the second question as I am just a university student that took an Introductory Physical Chemistry course a few months ago.
$\ce{\mathrm{d}S=\frac{\mathrm{d}q}{T}}$ is the definition for entropy in the classical thermodynamic view. A measure of energy dispersal at a specific temperature. Although it does not have "S" sitting on one side like many other equations (e.g. $E=mc^2$), it is how we define entropy. It only has a derivative form. This is because it has to associate with the change of heat and the equation for heat and temperature is different under different cases. Thus, we write this as the general form. When we apply this definition under certain condition, we may simplify it further by doing integration.
Actually, more specially, the definition is:
$\mathrm{d}S = \frac{\delta q_{rev}}{T}$
The heat in this equation refers to the reversible heat pathway only, (side note: "δ" sign means path function differential, "d" sign means state function differential). Entropy is a state function because it only refers to one and only one path for the heat, the reversible pathway. There is no other pathway, so it becomes a state function.
I am not sure how scientists in the past came up with this equation for the definition. Why divided δq by temperature? Why it need reversible heat? Maybe some else will answer it.

It seems like there are two questions you care about:

Given that $dS = dQ/T$ and that $Q$ is a path function, why is $S$ a state function?
How do you relate the interpretation of $S$ as 'measure of disorder' and as 'quantity related to heat production'?

tl;dr: Entropy is related to heat production via $dQ = TdS$, and heat characterizes disordered microscopic motion, so entropy is thus linked to disorder.

The two questions are kind of muddled up in terms of terminology, so let me hide some preconceptions by defining a new function $X$ such that $dX = dQ/T$. This is a useful definition because it turns out that $X$ is a state function. (Mathematically, $dQ$ is an inexact differential, which corresponds to a path function. Inexact differentials can be made into exact differentials by multiplying them by integrating factors. In this case, on physical grounds, the integrating factor for $dQ$ is $1/T$.)
The identification of $X$ as a state function follows from the analysis of Carnot cycles by the eponymous Sadi Carnot, for which it was found that $$\frac{Q_H}{T_H} + \frac{Q_C}{T_C} = 0$$ for the heat exchanged ($Q$) and temperature ($T$) of a hot ($H$) and a cold ($C$) reservoir. Because all thermodynamic cycles can be constructed as a limit of infinitely many infinitesimal Carnot cycles, the identity above establishes the exactness of the differential $dX$, but a satisfactory interpretation of this curious quantity $X$ was never established in macroscopic thermodynamics. (A common incorrect explanation for $S$ being a state function despite $Q$ being a path function is the imposition of a reversible path for the process. This is certainly true, but does not explain why $S$ is a state function. After all, there are still infinitely many reversible paths between two states.)

Thus we turn to statistical mechanics. The most basic form of statistical mechanics deals with a microcanonical ensemble with some number of microstates $\Omega$. In order to relate this to macroscopic thermodynamics, let us assume the existence of an extensive thermodynamic variable $S$, which we will call the entropy, that relates the number of microstates to thermodynamic properties. Mathematically, we take $S = f(\Omega)$ for some as yet undetermined function $f$. By extensivity, scaling up this system by a factor of $n$ will give us an entropy $nS$. We should also expect to have $\Omega^n$ microstates, assuming each subsystem is approximately independent. Thus, on physical grounds, we identify the functional equation $$f(\Omega^n) = nf(\Omega).$$ This can be transformed into Cauchy's functional equation, and has as its only continuous solution $$S = f(\Omega) = k\ln\Omega$$ for some constant $k$. This is the expression for the Boltzmann entropy.
The Gibbs entropy is a generalization of the Boltzmann entropy to canonical, rather than microcanonical, ensembles: take the total entropy (which you cannot measure in practice, not being aware of the number of available microstates) and subtract the entropy associated with the microstates to end up with the entropy associated with the macrostates, $$S = -k \sum_i p_i\ln p_i,$$ where $i$ enumerates your macrostates. For the precise derivation, see Blundell and Blundell's Concepts in Thermal Physics, Ex. 14.5. (There is a bogus derivation that writes the microcanonical entropy as $S = - k\ln p$ and then purports to take the average of this entropy to get $\langle S\rangle = -k\sum p\ln p$, but this does not stand up to scrutiny.)
Now that we have an expression for the statistical-mechanical entropy, we can see what thermodynamic variable it relates to: not surprisingly, the conclusion is that $$X = S,$$ so the mysterious state function of macroscopic thermodynamics is actually the entropy, which is concerned with the number of microstates of the system! The derivation involves writing the first law of thermodynamics in macroscopic and microscopic form, matching terms, and following a similar functional-analytic treatment as above. See Hill's An Introduction to Statistical Thermodynamics, Ch. 1.4. (I believe this particular example of functional analysis was first advanced by Schrodinger, him of quantum-mechanical fame.)

Finally, after having derived that $X$ was a state function just from thermodynamics, and having interpreted $X$ using statistical mechanics, we revisit thermodynamics from a more enlightened standpoint. The amazing simplicity of thermodynamics is a reduction from $6N$ microscopic degrees of freedom to just three (in the simplest case) macroscopic variables: $S$, $V$, and $N$. What is so special about these variables that they would survive this coarse-graining procedure?
It turns out that $V$ and $N$ are intimately linked to conservation laws and hence preserved by symmetry via Noether's theorem, whereas $S$ basically lumps together all the other microscopic degrees of freedom that were coarse-grained away. This distinction underlies the treatment of $pdV$ and $\mu dN$ in the first law as work terms associated with ordered microscopic motion, whereas $TdS$ is a heat term associated with disordered microscopic motion. See Callen's Thermodynamics and an Introduction to Thermostatistics, Ch. 21.


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Molecular orbitals in an ionic diatomic compound
The wavefunction of a heteronuclear diatomic molecule, after the orbital approximation and the Born-Oppenheimer approximation, is
$$\psi = c_A \chi_A \pm c_B \chi_B $$
where $\chi$ are the starting AO wave functions. Calling
$$\gamma = \dfrac{c_A}{c_B}$$
the coefficient value can be calculated by
$$ c_A = \sqrt{\dfrac{\gamma^2}{\gamma^2 + 1 + 2\gamma S}} $$
where $S$ is the overlap integral
$$ c_B = \dfrac{c_A}{\gamma} $$
In an ionic diatomic compound, $c_A = 0$ and $c_B = 1$: how can $c_B = 1$ if
$$c_B = \dfrac{0}{\gamma}$$

Hint: CB can be 1 if gamma=0. It might be easier if you start by writing it as CA=gamma*CB rather than as a ratio, so avoiding confusing division by zero problems.

So, your advice is: $c_A = 0 \cdot c_b$ ?

Close. CA=0 implies gamma=0 from your first relation as CB can not be zero under such conditions, then calculate CB by the normalisation condition. Sorted.

As
$$\gamma = \dfrac{c_A}{c_B}$$
when $c_A=0$, $\gamma$ must also be zero. This follows as $c_B$ can not be also be zero as the wavefunction is normalised to unity. Thus under such conditions it is invalid to divide through by $\gamma$. Instead simply use the normalisation condition and the fact that $c_A$ is zero to show that $|c_B|=1$ .

Using the normalisation condition
$$c^2_A + c_B^2 + 2c_A c_B S = 1$$
while $c_A = 0$
$$0 + c^2_B + 2 \cdot 0 \cdot c_B S = 1$$
$$ c^2_B = 1 $$
$$ c_B = \sqrt{1} = \pm 1$$


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How can I dissolve carbon using generally available substances?
I am asking this question out of despair: my wife burned to death crèpes dough into our induction stove.

I tried all usual products (the ones to clean the stove, vinegar, then sodium bicarbonate left for some time, gentle mechanical prying, ...) but it is stuck.
Is there a way to dissolve carbon (burned organic substance) using a reasonably available substance? (say, HClO3 would be doable,  H2SbF6 or C2N14 - not that much). For the sake of the exercise, let's assume that it would be handled by a physicist.

Unfortunately, this may require suitable abrasives, a high speed rotary tool and lots of elbow grease and time. Good luck!

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

@Community: well, I am not sire what is not clear: I have burned organic stuff on an induction stove and I would like to use a substance that is generally available to the public to dissolve it. What exact extra information are you looking for, exactly? (the answer from Maurice is the kind I was looking for)

Please let us know if something actually works and also please edit to post the “after” photo!

From the brown color, your stain is still not pure carbon (perhaps polymerized fats/oils and charred sugar). There are chemical ways to oxidize organic matter from glass but not applicable domestic objects. The lye method (NaOH) is perhaps still your savior as suggested in the answer.

The best way to dissolve a black deposit on glass is to use a concentrated solution of $\ce{NaOH}$, about half water - half $\ce{NaOH}$. But take care ! This dissolution is highly exothermic. Try to prepare not more than a couple of milliliters of it in a test glass tube. The test tube will get hot during the mixing and dissolution operation. And take care ! This solution should not touch your skin, as it will soon destroy it. Be ready to intervene with sodium bicarbonate : this powder  destroys $\ce{NaOH}$.
A couple of milliliters of this $\ce{NsOH}$ solution is enough for your cleaning purposes. Have one or two drops of it on the burned organic spot. Rub with some blotting paper or toilet paper to remove the spot. Don't wait ! The dark stuff will quickly be absorbed by the paper. But take care, not to touch the stuff with your fingers. Rub with new clean paper, and then rinse several times with water each time with a new piece of toilet paper. I have used this process many times on the window of my oven, and it has always worked perfectly.
Last advice. Don't wait cleaning the clean glass surface. If you forget to clean the glass surface quickly, the $\ce{NaOH}$ solution will slowly corrode the glas itself. And take care to avoid contact with aluminum pieces. NaOH destroys aluminum rather quickly.
In case of an accidental contact with $\ce{NaOH}$, go for a shower and wash a long time, until no soap feeling are felt any more on the $\ce{NaOH}$ spot.

I would suggest that the OP use some PPE, like eye protection and rubber gloves, and use a scrap piece of cloth or paper towel: toilet paper shreds easily, by its design, when wet.

I agree with everything except the "Don't wait" part. You'd probably have to wait for some time, and even then I wouldn't be too optimistic.

There is a commercially available solution of KOH in polyethylene glycol/water gel. In the UK this is marketed as Ovenmate, this may be useful

@Ed V and Ivan Neretin. I repeat. I have done it several times at home and also in a chalet in the mountain. I always worked properly and quickly. Of course, gloves are an efficient protection. I remember that once I thought it was useful to wait half an hour or so before cleaning it. Now the window is frosted.

@ Waylander. Thank you for the information about the commercial gel. The trouble is that you are obliged to buy a whole bottle. And you only need one or two milliliters.

@EdV: yes of course. This is what I meant by the fact that this will be done by a physicist - I did a lot of experiments and safety is important and a given. But it is always good to remind that, thanks.

@Maurice It keeps. Good for cleaning pyrex, stainless steel and cast iron pans too.

Unfortunately, the stains did not bulge. I bought yesterday NaOH, it came as a powder (small crystals) and I tried a 1:1 dilution (indeed, it was noticeably exothermic) and then a stringer one but there was not a single black particle on the towel. Thank you for the idea, though, it may be useful in another case.


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Is there really any difference between Resonance and Mesomeric effect?
I have been in the classes of two different teachers of organic chemistry and they both agree on the fact that there is a difference between the mesomeric effect and resonance effect (they didn't agree mutually, I just compared their answers) but both of them tell the differences which are unique or simply don't match. So I decided to find the difference via research. I searched online and in the books but answers there also vary. Here are some findings of my search

Answer by @NotEvans (here) mentions that in the gold book of IUPAC, they have been taken as synonym terms.
Answer by @Raju (here) mentions the difference as "Resonance refers to delocalization of electrons in a given system. The mesomeric effect is the electron-donating or withdrawing nature of a substituent due to resonance." The answer matches @KshitizSharma answered on the link where @NotEvans answered.
A page on the website Pediaa mentions the difference to be "Resonance is the effect that describes the polarity of a molecule that is induced by the interaction between lone electron pairs and bond electron pairs. The mesomeric effect is the effect of substituents or functional groups on chemical compounds."
There are several other places but as expected, all their answers differ

My one teacher let's say A (don't want to disclose the name) says that the Resonance effect is a broader umbrella term that includes the mesomeric effect. In resonance, we have delocalization of electrons and the delocalization can be of several types such as pi-electron delocalization, sigma electron delocalization, dancing resonance, etc (these phenomena are shown in images below) but in mesomeric effect, we can only have pi-electron delocalization.
Dancing Resonance

Sigma Bond Resonance

Pi Electron Resonance

Now my other teacher, let's say teacher B sort of agrees with the definition and difference of Resonance and Mesomeric Effect given on the website Pediaa which is mentioned above in point 3.
My opinion (not conclusion) on all of this is that technically my teacher A seems to be correct to me. But I can't decide and the definition on the IUPAC book confuses me even more so I came here to ask for help. Thanks.

Notice: The website called Pediaa might have its maintenance going on because of which the server seems down.So please believe my mentioned words, they have been exactly copied.

Have you read the poem "Five Blind Men and the Elephant", where each person tries to explore an elephant and comes with their own version. Someone touches elephant's trunk and calls it a snake, someone touches the legs and calls it a tree trunk and so on. The world of molecules and sub-atomic world is indeed that elephant, and chemists are unfortunately unable to see them- to see what is happening. Hence each one comes up with an intelligent story only to be improved or discarded later. So the lesson is that don't take everything literally (incuding the statements made by high school or college teachers). Moreover, do not rely on the web-answers including this one ultimate facts.
In the same way the concept of resonance was developed really a long time ago. Keep in mind that none of the structure you draw with resonance reflects a reality. It is  a subtle and an esoteric way of human expression that we do not know how it looks like in reality. X-rays allow you see the arrangement of solids in space but it does not let you see electrons, there is no physical arrow pushing going on and there no electron dance going on. It is all human imagination.
Think of the classic benzene example, (image from Wikipedia). None of the structures reflect a reality, hence the term resonance hybrid, because benzene does not chemically behave as if its double bonds were "static". The C-C bond distances in the ring have to be different, otherwise.

There is a very nice article by Robert C. Kerber titled If It’s Resonance, What Is Resonating? in the Journal of Chemical Education (It is freely available on Google Scholar). Have you wondered why the name "Resonance"?

History The use of multiple structures to represent compounds with
(what we now call) delocalized bonding (4) was pioneered in Germany by
Arndt (5) and in Britain by Ingold (6). Their ideas were based
essentially on chemical intuition. In the United States, Pauling
$(7,8)$ in his pioneering applications of the principles of quantum
mechanics to chemistry, came to a similar description from a different
starting point. Whereas the concepts of describing a single substance
in terms of multiple structures may not have differed significantly,
the terminology did. Arndt coined the term $Zwischenstufe$ for the
hybrid structure, while Ingold opted for the equivalent term (derived
from Greek) mesomer. He proposed for the concept the terms mesomeric
effect or mesomerism. The cognate term mesomerie came into use in
French and German. Pauling preferred the term resonance, derived from
valence bond theory (see below). Pauling and Ingold were well aware of
each other's terminology, and each offered reasons for rejecting the
alternative proposal: Because the resonating system does not have a
structure intermediate between those involved in the resonance, but
instead a structure which is further changed by the resonance
stabilization, I prefer not to use the word 'mesomerism,' suggested by
Ingold, for the resonance phenomenon.

L. Pauling (9) Pauling describes the phenomenon under the name 'resonance,' which, as is well known, is based on the mathematical
analogy between mechanical resonance and the behavior of wave
functions in quantum mechanical exchange phenomena. There appears,
however, to be some possibility that this method of description may
suggest an analogy which has never been intended.
C. K. Ingold (6) Pauling's argument conflates the concepts of structure and energy. Moreover, his chosen term resonance is derived
from an analogy to coupled oscillators, which also lacks energetic
implications: "There is no close classical analogue of resonance
energy" (10). So neither term includes explicit reference to energy
considerations, weakening Pauling's argument. For use in discussing
the structure (i.e., geometry) of the hybrid, Ingold's argument seems
the more persuasive. His recognition of the gratuitous oscillatory
aspect of the analogy is prophetic of the subsequent experience of
generations of students.*

Your next query on mesomerism and IUPAC. In terms of terminology, most chemists follow the IUPAC recommendation's on terminologies. You can call mesomerism as a type of resonance. See this reference "Introductory Organic Chemistry and Hydrocarbons: A Physical Chemistry Approach" by Caio Lima Firme on pg 111-112.

As Ingold himself defined: "Mesomerism is an extension of valency
theory (that is the classical valence bond theory) and, like all
valency theory, is founded in the quantum theory (...). The
fundamental wave-property in the theory of valency is resonance $-$
the resonance of connected standing waves; their mutual perturbation
replaces these waves by new standing waves" (Ingold 1938). Ingold
exemplified the resonance theory by using the two resonance
structures: $\mathrm{R}_{2} \mathrm{~N}-\mathrm{CH}=\mathrm{N}^{+}
> \mathrm{R}_{3}$ and $\mathrm{R}_{3}^{+}
> \mathrm{N}=\mathrm{C}(\mathrm{H})-\mathrm{NR}_{2}$ with their own
standing waves which resonate to form the real molecule. Ingold stated
that the term mesomerism was given to account for the special
importance of resonance in organic chemistry. He also said: "When
mesomerism was first recognized as a general phenomenon in organic
chemistry, it was appreciated as an electron displacement than as an
energy disappearance" (Ingold 1938). From this point on, we can say
that mesomerim is equivalent to resonance type 3 (resonance involving
only charged and/or neutral covalent structures), although IUPAC
recognizes mesomerism as essentially synonymous with resonance (which
could include all types of resonance). Nonetheless, IUPAC adds that
mesomerism is "particularly associated with the picture of
$\pi$-electrons as less localized in an actual molecule than in a
Lewis formula” (IUPAC Gold Book). Then, mesomeric state is the same as
resonance hybrid and energy of mesomerism is synonymous to resonance
energy (Ingold 1938).

Thank you, seems like it's all the result of conflict between Ingold and Pauling.

Again, just recall the blind men and the elephant (molecule)! Both were highly respected chemists.

Yes, I understand


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Copper powder from aqueous copper sulfate solution help
I’ve been wanting to make copper powder for some homemade pyrotechnics. When placing an iron object in a copper sulfate solution, copper powder is produced. However, sooner or later, the reaction slows down and the solution begins to turn a teal color instead of its previous blue color. From what I’ve learned, the teal color is from iron sulfate. Therefore, the copper sulfate eventually gets used up.
I know another way to produce copper powder. Place two COPPER electrodes in the copper sulfate solution and the anode dissolves while copper powder builds up on the cathode. My question is, do I need to replace the copper sulfate solution if I do it this way? It seems I would not have to because the copper ions are being replaced by the dissolving anode.

Though the copper "sacrifial" electrode replaces that lost from solution, you may get solid copper plated out, rather than a fine powder.

I tested it today, and with my experimenting, a lower voltage(3-5) will result in plating and a higher voltage(12-24) will result in nice fluffy copper powder.

The net result of the 2 copper electrode scenario is moving copper from one electrode to another. If original copper is pure enough, you need not to replace the solution for  long time. Note that the key is not the applied voltage, but resulting current density.

In your first experiment, the copper sulfate eventually gets used up. But in your second experiment, the copper anode also gets used up. There is no miracle. Whatever your choice, there is an end to these experiments.

Right, but I’ve got loads of copper and not much sulfate. Granted, there are probably more efficient ways of producing copper powder.


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Mass spectrometer Ions with same mass/charge ratio but different charges
The mass spectrometer identifies ions based on mass/charge ratio. What happens when two ions have the same mass/charge ratio but different charges or masses?

This situation is not uncommon. A mass spectrometer has no way of knowing the absolute mass or absolute charge of ions. Suppose compound A has a peak that appears at $m/z$. You suggest another compound B, which has $2m/2z$ or $nm/nz$. This is the only condition where you will see a peak overlap in MS. Note, $m$, is the nominal mass. Exact masses, measurable by ultrahigh-resolution mass spectrometers, will not masses of two compounds match precisely. Only enantiomers and structural isomers will have identical exact and nominal masses.
However, there are tricks to discern doubly charged ions. For example, you can look at the isotopic patterns. In order to avoid the problems, mass spectrometers are coupled with a chromatography system. With that, such conditions rarely arise.


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What does it mean by spatial neighbor?
What does it mean by a spatial neighbor (SN) in the case of a protein residue or protein chain?
Why is it different or special as opposed to other neighbors?

A neighbor in the space? // Amino acids that are not neighbors in protein primary structure, can be spatial neighbors in secondary/tertiary/quarternary structure. // Just guess, due lack of  context.

There are chain neighbors and then there are spatial neighbors.

Proteins are long chains of amino acids joined together. Atoms or amino acid residues that are a few bonds away from each other can be considered "neighbours".
But the long chains of atoms/amino acids usually show complicated folding in 3D space. Indeed, the specific shape of that folding is critical for the protein to function. The implication of the folding in 3D is that residues or atoms that are far apart if we count bonds can be very close in 3D space. The last amino acid in the chain might be hundreds of bonds away from the residue at the other end of the chain, but might sit right next to it in 3D space. That would be a spatial neighbour.
The point is that, in complex 3D structures, we can count "neighbours" by either the number of bonds between them or by how close they are in space when the protein is folded.


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How to remove starch from silicone ("Oogoo")?
Oogoo is an easy and cheap way of creating silicone rubber objects from a hand-formable putty. Basically: just mix silicone caulk with starch (I'm using acetate moisture absorbing silicone caulking and corn starch), the silicone in the mixture sets in about 10 minutes. Links (there are also some relevant videos on Youtube on the TKOR channel):
http://www.instructables.com/id/How-To-Make-Your-Own-Sugru-Substitute/
http://www.instructables.com/id/7-Things-To-Fix-Or-Do-With-Oogoo/
http://www.instructables.com/id/Oogoo-mixer-on-the-cheap-avoid-messy-hands-/
I'm assuming that the starch doesn't chemically react with anything in the Oogoo mixture, so I was thinking that it may be interesting to try and remove it from the silicone rubber. I imagine this could create a light-weight, porous silicone rubber.
What method could I use for removing the starch without degrading the silicone?
I was thinking maybe hydrochloric acid could hydrolyse the starch? I think it wouldn't harm the rubber? Assuming that could work, what concentration of acid should I use at what temperature? If I used 33% hydrochloric acid, should I dilute it first, and what would be the temperature for the hydrolysis of starch?

33% HCl is pretty aggressive stuff. How about filling the caulk with something more soluble to begin with, like superfine sugar or NaCl? Maybe even grind the sugar or salt up finer yourself with a mortar and pestle?

I'll try using sugar or salt instead of starch and see how it goes. Possibly I should mix either sugar or salt with a small amount of water, to make the caulking cure fast. However I don't know how will this work.

Magnesium sulfate (Epsom Salt) is available as a heptahydrate. It can be powdered easily, and might provide the H2O. Or MgCl2.6H2O (winter deicing salt).

@JamesGaidis very interesting, thanks.


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Natural preservatives of ink
Would pure compounds like thymol, carvacrol, and cinnamaldehyde work as preservatives of fountain pen ink? I'm specifically interested in three properties -

Prevention of growth (mostly fungal) in the bottle of water-based ink.
Not be caustic to fountain pen materials - metals, plastics, and rubber.
Does not cause an allergy when handling or if some ink ends up on the user's skin.

Would the compounds I mentioned be suitable for this purpose? Thank you very much in advance.

Natural also means more organisms may already have a way to deal with it.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

That I would be willing to accept, but what about the caustic or noxious nature of the chemicals to humans and pens?

The most noxious chemicals are natural ones. The boundary is not synthetic=bad and natural=good.

Are thymol, carvacrol and cinnamaldehyde corrosive to metals/plastics/rubber?


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Adsorption on solids - homogeneous/heterogeneous surface definition
How could we define homogeneous and heterogeneous surfaces (rigorously)?
It is among my exam questions (but I could not find any definition in my textbook or on the internet).
I would say the homogeneous surface is a surface made of one type of molecule (opposite of the heterogeneous surface). But my knowledge about surfaces is very surface (and I am surprised it is listed as an exam question) so are there any other conditions or is it enough?

Homogeneous surface is not a thing at all.

Why? Not even in an ideal case?

We talk of homogeneous mixtures and reactions when there is one phase and no surfaces at all. Otherwise we use the word heterogeneous. That's all.

Yet it is literally one of the questions which my professor gave me

@IvanNeretin All depends on context.  We can talk about homogenous/heterogenous surface in context of the surface itself, not the whole system.  Like an object made of solid pure compound can have homogenous surface, but if made of a metallic alloy, it can have heterogenous surface, with properties location dependent.

Yeah, I guess we can.

Homogeneous surface is the surface where we can (in the context of the scenario, like the purpose and  required/available resolution and sensitivity) neglect any surface variability. Otherwise, the surface is heterogeneous.
It is important to say it is relative evaluation. For a set of surfaces, it can reach both extremes like "everything is homogeneous" to "everything is heterogeneous".
A visually homogeneous surface can be considered heterogeneous when evaluated by the enough zooming-in optical or electron microscope.
A visually heterogeneous surface can be considered homogeneous when evaluated from far enough or with low resolution or sensitivity for differences.


text:
Reducing sugars and redox potential
I am wondering if reducing sugars are those that can act as reducing agents, and they act as such with Tollens and Fehling reaction, then what's the redox potential here? And is it a reversible reaction like with for example redox cofactors, or once the sugar is reduced it cannot go back?

It is known that the reducing cofactor of dopamine-$\beta$-monooxigenase is a sugar, L-ascorbic acid. That means, the oxidized dehydroascorbate has to be reduced back to ascorbate for this system to work. Thus, it is safe to say oxidation of reducing sugar is reversible process depending on the conditions used.
Nonetheless, although all redox half-reaction systems are shown reversible, the reversibility is based on the conditions. For example, it is shown that the redox reduction half-reaction of silver is reversible:
$$\ce{Ag+ + e <=> Ag }  \qquad E^\circ = \pu{0.7996 V} \tag1$$
It doesn't mean all the condition this reaction is reversible. For instance, suppose you used a weaker reducing agent (stronger oxidizing agent) than silver:
$$\ce{Au+ + e <=> Au }  \qquad E^\circ = \pu{1.692 V} \tag2$$
Then, the dominant reaction will be the backward reaction as shown in the equation $(4)$:
$$\ce{Au+ + e <=> Au }  \qquad E^\circ = \pu{1.692 V} \tag2$$
$$\ce{Ag <=> Ag+ + e }  \qquad E^\circ = \pu{-0.7996 V} \tag3$$
$$\ce{Ag + Au+ <=> Ag+ + Au }  \qquad E^\circ_\mathrm{rxn} = \pu{0.8924 V} \tag4$$
On the other hand, suppose you have used a stronger reducing agent (weaker oxidizing agent) than silver such as glucose:
$$\ce{glucose + H2O <=> gluconic acid + 2H+ + 2e}  \qquad E^\circ = \pu{-0.050 V} \tag5$$
Then, the dominant reaction would be the forward reaction as shown in the equation $(6)$:
$$\ce{2Ag+ + glucose + H2O <=> 2Ag + gluconic acid + 2H+}  \qquad E^\circ_\mathrm{rxn} = \pu{0.7496 V} \tag6$$
As shown for the silver, gluconic acid can be reduced to glucose (Ref.1) similar to glucose been oxidized to gluconic acid.
References:

J. W. E. Glattfeld and Edna H. Shaver, "The catalytic reduction of d-gluconic acid to d-glucose," J. Am. Chem. Soc. 1927, 49(9), 2305–2308 (DOI: https://doi.org/10.1021/ja01408a026).

Downvoters, please let us know what's wrong with this answer to be fair.


text:
Does ozone have polar bonds?
Does $\ce{O3}$ have polar or non-polar bonds? My teacher says that $\ce{O3}$ has non-polar bonds.
But chem-libre texts and many other sources state that a non-polar bond has equal distribution of electrons, which doesn't seem to be the case in $\ce{O3}$.
(Not asking whether the molecule $\ce{O3}$ is polar but rather if the bonds in $\ce{O3}$ are polar.)

Nonpolar bonds have insignificant partial charge on respective atoms. Ask yourself if it is the case of ozone

@poutnik yes it does have the partial charge. Is the bond non polar?  A bond is said to be polar when there is even distribution of electrons between atoms. But the distribution (from what I have seen in the electron density graphs that isn't the case.)

A bond is said to be non polar when there is (about) even distribution of (bond) electrons between atoms. When this distribution is shifted toward one of atoms, forming partial atom charges, such a bond is polar.

To complement my old comment, electronegativity-wise non-polar bond can be polarized by already existing opposite charges on the bound atoms, as is in the ozone case.

You could say that the bonds are non-polar and the partial charge is due to the imbalance in non-bonding electrons. However, once there is a dipole due to those, you would induce a shift in the bonding electrons as well.
This view is mostly an artifact of a single electron wave function view.  The reality is a bit more holistic with a wave function that describes multiple indistinguishable electrons, so the important thing is the actual dipole moment, not assigning “blame” for it.

As we know resonance operates in Ozone molecule, so bond order between O-O bind is 1.5
As in the resonance hybrid, there is evrn distribution of electrons between O-O bonds, I think so that it is non-polar


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Reaction Mechanism for a radical ring opening
I have a radical reaction which I got from an organic chemistry lecture and was drawing the possible mechanism for it.
Fig 2. shows my thought process. The I is removed making a radical and ISnBu$_3$. The radical opens the ring and the new radical is formed which then reacts with the hydrogen radical. However Fig 1. shows the reactants and products from the lecture which don't match with what I thought.
I don't see how the carbon on the "top" of the propane ring becomes part of the chain.
Could the lecture slides have been wrong or am I missing something?

Without consulting the literature, there is the possibility that fig. 2 is a product of kinetic control whereas fig. 1 is the product of thermodynamic control. The second radical in fig. 2 would recyclize under dilute Bu3SnH concentration. This is certainly the case for the bromo epoxide analog of the cyclopropyl iodide.

@user55119 is there be a reaction mechanism for the thermodynamically controlled product? I just don't see how the carbon can become part of the chain

After you form the radical, break the "bottom" bond in the cyclopropane ring. That produces your terminal double bond and a resonance stabilized benzylic radical. Breaking the other cyclopropane bonds does not produce a stabilized radical.

@Ron just beat me to it.

On OP's comment:

I just don't see how the carbon can become part of the chain!

shows confusion on the mechanism. However, the comments by user55119 and Ron correctly suggested the path of the mechanism. I thought it would be helpful to put the matter as a solution. Following scheme shows the possible mechanism to achieve the sought product:

Note that, there are two bonds that can open the cyclopropyl ring. The bond the OP has shown opening gives the primary radical, which is the less stable. The bond opening shown in my scheme gives the benzyl radical, which is stabilized by resonance. Thus, the given path is the thermodynamically stable path.


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The existence of sodium carbide
(Here, I quote a question from an Indonesian undergraduate Chemistry competition: "sodium carbide reacts with water to form ethyne gas." This is the problem that caused me to ask this question on the site.)
Please provide references of the existence or inexistence of sodium carbide.
Thank you in advance for you who have answered my question.
Edit 2: The type of answers I look for are the ones which includes  analysis of probable sodium carbide structures, but deeper explanation will also be highly appreciated.
Thank you to "Community" who suggested an edit to this question.
Edit 3: I (Galen) have changed the title of this question and rephrased my question. Thanks to Oscar Lanzi and others who have attempted to answer my question.

This patent describes the production of sodium carbide from sodium and carbon monoxide  https://patents.google.com/patent/US2642347A/en

@Waylander , thank you! Through your comment, I also found about Google Patents for the first time.

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

Housecroft and Sharpe give the acid-base reaction of sodium amide with acetylene in liquid ammonia to produce sodium carbide and ammonia

As the comments imply, sodium carbide certainly does exist. However, it is difficult to get because it is only metastable. Therefore it cannot be made from direct combination of the elements in thermodynamically stable form, a procedure often used (in some cases indirectly) with stable binary compounds.
Sodium carbide can be made by using the sodium/CO reaction commented by Waylander (https://patents.google.com/patent/US2642347A/en). This patent gives a brief reference of some other methods:

It has been proposed to produce sodium carbide by reacting sodium vapor with carbon in an electric arc (German Patent 526,627), by reacting calcium carbide with sodium monoxide or sodium hydroxide (Vaughn U. S. P. 2,156,365) or by reacting metallic sodium with acetylene (British Patent 336,516 of 1930).

Note that in no case is the carbon in a thermodynamically stable form when the carbide is formed; for example without the sodium component the carbon monoxide in the sodium/CO reaction would spontaneously decompose on cooling to the actual carbide formation temperature. In the case of the electric arc the arc takes the carbon (and the sodium) into a higher-energy, metastable state.
Sodium carbide is not alone. Magnesium carbide presents a similar issue.
Wikipedia gives a curious explanation of the lack of full stability (and thus the impact on possible synthesis techniques) in terms of the related process of forming graphite intercalation compounds:

Different from other alkali metals, the amount of Na intercalation is very small. Quantum-mechanical calculations show that this originate from a quite general phenomenon: among the alkali and alkaline earth metals, Na and Mg generally have the weakest chemical binding to a given substrate, compared with the other elements in the same group of the periodic table.[1] The phenomenon arises from the competition between trends in the ionization energy and the ion–substrate coupling, down the columns of the periodic table.[1]

Cited Reference
1.
Liu, Yuanyue; Merinov, Boris V.; Goddard, William A. (5 April 2016). "Origin of low sodium capacity in graphite and generally weak substrate binding of Na and Mg among alkali and alkaline earth metals". Proceedings of the National Academy of Sciences 113 (14): 3735–3739. arXiv:1604.03602. Bibcode:2016PNAS..113.3735L. https://doi.org/10.1073/pnas.1602473113. PMC 4833228. PMID 27001855.

Sodium carbide isn't hypothetical. It does exist and can be synthesized. In @Oscar's answer, there is a patent which shows the synthesis reaction of sodium carbonate and carbon at temperature 1050 °C-1200 °C and contacting the resulting gaseous mixture of carbon monoxide and sodium vapor on steel surface.
There is another patent which shows a different synthesis reaction which uses sodium hydroxide. The temperature required is also less:

The method of forming sodium carbide which comprises reacting a mixture containing free sodium hydroxide and free carbon at temperature of around 700 °C-900 °C and free carbon being in excess of the stoichiometric requirement for the formation of said carbide from said sodium hydroxide.

Moreover, there is one paper (M S Swapna et al 2017) which discusses a synthetic route involving a natural precursor, Pandanus.


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Databases for Enthalpy of Solution and Ionic Radii for Ionic Salts
I am looking to investigate the relation between experimental enthalpies of solution and theoretical enthalpies of solution calculated using theoretical equations for the lattice enthalpy and hydration enthalpy for ionic salts. I read that because of the covalent nature of ionic salts with a lower difference in electronegativity, the experimental lattice enthalpy of these salts is actually greater than would be expected theoretically. Would this be worth investigating?
To investigate this, I am looking for public databases with data for the solution enthalpy and ionic radii of ionic salts however I cannot find any other than the CRC handbook for physics and chemistry (and only the 85th version; would it be better to get a more recent version or should these data be the same in more recent versions?). I would like to use more than one database so could someone please let me know of any other public databases with these data?

You can refer to the NIST Chemistry WebBook for thermodynamic data by species or reactions. I also found a database for lattice enthalpy here. For ionic radii, you can refer to this database.

I am trying to find data for the enthalpy of solution specifically but I can't find it anywhere other than the CRC handbook. The NIST chemistry webbook does not seem to include it either. Do you know where I could find data for the enthalpy of solution?

I made a brief search and found similar results...Maybe you should look for research papers on specific compounds instead.


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Cleaning used frying oil with salt water : what is the point of using salt?
In the process of making soap from used frying oil, I need to purify the oil.
I filtered the cooking oil but I also want to remove the polar compounds that are formed during frying : free fatty acids, monoglycerides, diglycerides.
To do so, I've seen that I can separate the polar compounds from the oil using a salt water solution (sources : 1, 2).
The process is slightly different in the 2 sources but it always consist of mixing salt water in the oil and letting it rest until water and oil separates, and that impurties will be in the salt water.
I understand we use water so that it can "catch" the polar compounds from the oil, but what could be the role of the salt in this process ?

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Salt can increase solubility of some proteins: https://fac.ksu.edu.sa/sites/default/files/bch_332_lectures_3_and_4.pdf

Search the concepts of salting-in and salting-out.

So we have here a salting-in process ?
In the meantime, I found this article that says that adding salt allows to reduce the interfacial tension beetween water and oil. This also seems good for mixing the 2 more easily..

Additionally, the two phases can layer back easily


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What gives natural rubber a permanent shiny finishing?
Natural rubber is made by vulcanizing latex mixed with sulphur. Commonly, the resulting natural rubber is dull instead of shiny. However, there are also products made out of natural rubber that are shiny, such as the wellington boots in the picture below.
My question is:
In addition to vulcanization, what additional step is done in production to make this rubber shiny? What chemical process is at play?
Possible ideas

As the picture shows boots which are shiny, my first thought was that it might just be polish. However, it appears to me that this is wrong, because a polish is a hydrophobe wax or cream that has to be applied regularly to keep shoes waterproof and shiny. To the boots shown below no cream or wax needs to be applied to maintain the shine and there is no cream rubbing off. So I would assume that there is something else going on.

In semiconductor manufacturing, silicon can be polished by applying chemicals. Maybe this could also be used for rubber? If so, what chemicals could be used to create the shiny finishing of the boots pictured below?

Background
Originally, I posed this question to my chemistry teacher, but she couldn't provide me with an answer. So I would be very happy if anyone of you could help me!
Any hint would be welcome! Thank you very much for your time!

This is called polishing. It has nothing to do with chemistry.

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

Thank you for your feedback! I've edited my question to make it more understandable.

Permanent shiny surface can be achieved also by simultaneous application of high temperature and sufficient pressure of shiny surface of mold(cast, form). Future rubber matter is originally liquid and later plastic, before finally gets elastic. // It may be also possible the form surface has applied polishing agent, which is by pressure incorporated into surface layer of rubber, so it does not go away easily.

@Poutnik Thank you for your answer! Two follow-ups:

When using high temperature and sufficient pressure, do these conditions need to persists during vulcanization, or is the surface ‘cured’ to a shiny finishing that persist without temperature/pressure such that the material can be vulcanized later?

What kind of polishing agent is used for incorporation into the surface layer of rubber?

I would say filling the form and vulcanisation comes in the same step. The rubber us not taken away to get vulcanised.// It was a speculation. The main purpose would be prevention rubber getting glued to the form.


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confusion regarding meta-cresol and phenol
when I googled the answer, i found out that m-cresol has a PKa value of 10.1 whereas phenol has 10.0.
I am simply unable to use any logic to compare their strengths.
We have been taught only the resonance effects and inductive effects (also exceptionally the ortho effect).
Here, none of these apply ! Steric hindrance doesn't come into the picture, too !
Based on what am I supposed to compare their basic strengths ?

Well the difference is quite subtle. However the presence of methyl is right what you need.....

I upvoted the question because the idea that logic can explain tiny differences requires an amount of discussion that is inversely proportional to the magnitude of the difference. Sometimes we have to act philosophically.

Why would such poorly written question not be downvoted? Punctuation, capitalisation, formatting, buzzwords...

The pKa of meta-cresol suggests that it is slightly less acidic than phenol. As commented, the difference is quite subtle. You could invoke the greater electronegativity of the carbon in methyl as compared to hydrogen, you can neglect any hyperconjugation, and you can simply dismiss it. What, really, is the significance of that tiny difference in pKa? You just keep invoking and/or dismissing the effects you have been taught until you come up with the "right" answer, the one that appears to explain the effect.
Or, to phrase the question differently, is there some difference between phenol and meta-cresol that might explain, or contribute to the slight difference in pKa? My first question is How significant are the numbers in the measurement? If you Google the pKa of phenol, on the first page of hits, you can find 9.88, 9.95, 9.98, 9.99, 10 (no decimal), and ~10 (with roughly sign). If you do the same for meta-cresol, you get 10.09, 10.1 and 10.13. So the various numbers (and there may be more) indicate a range of uncertainty.
My second question is What other differences are there? Melting point, solubility... Phenol is a solid at room temperature, mp = 40.5$^o$C. But with 11% H2O, it is liquid. Meta-cresol is liquid, mp = 11$^o$C.  It is difficult to turn these facts into an explanation, but there is something interesting about the solubilities. Meta-cresol is soluble in water 23.5 g/L at 20$^o$C, 58g/L at 100$^o$C. Even tho it is liquid at room temperature, it is not completely miscible in water even at 100$^o$C. On the other hand, phenol is more soluble in water at room temperature (83 g/L at 20$^o$C), liquid with addition of 11% water, and miscible above its melting point.
These little bits of information suggest that phenol is more highly hydrated in water. Except for that extra lattice energy that makes it a solid, phenol solubility in water would appear dramatically different from meta-cresol. Imagine that room temperature is about 42$^o$C (a bit warm). Phenol is liquid, miscible with water; meta-cresol is liquid, not very soluble in water (~3-4%).
So now we ought to consider hydration effects in addition to the standard inductive and resonance effects. Among the three, one is likely to be able to explain the tiny difference.
Perhaps a fair answer would be that resonance effects are nil because the methyl is in the meta position. Although the carbon is electron withdrawing relative to hydrogen, the methyl group (sp$^3$) is less electronegative than the ring carbon (sp$^2$), and has more electrons to donate, so the negative charge on oxygen has less room to expand in. And the fact that phenol is better hydrated suggests the H2O molecules are helping phenolate spread the charge more than they help meta-cresolate.
And your teacher will appreciate such a long-winded discussion of all the effects you can possibly think of.


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Why is "Carbon can form many compounds with long carbon chains" not correctly explained by "Carbon atoms can share electrons with one another"?

Why is "Carbon can form many compounds with long carbon chains" not correctly explained by "Carbon atoms can share electrons with one another"?

I am a high school student just started learning organic chemistry and the above was a problem from a past public exam in my city. It is obvious that both statements are true but the answer key says that the later is not a correct explanation to the first. However, what I found on the Internet told me that the fact that carbon can form many compounds with long carbon chains is indeed due to the fact that they can form very covalent bonds with themselves. So what is wrong about the explanation?

Nearly all atoms can share electrons with many other atoms. Carbon is not special in this regard.

@Ivan what will be the real explanation then?

These statements are only tangentially related. Do you know what "explanation" means?

If you're saying that A: "Carbon can share electrons" and B: "Carbon can form long chains" and that B is the explanation for A, then it definitely isn't. You can say A explains B, but not B explains A. Your question seems to indicate that you think B should explain A.

@TRC I meant the opposite.

Somehow it is "that it can share electron with more others, up to four".

You can take the prompt and turn it into a single sentence:

Carbon atoms can share electrons with one another, therefore carbon can form many compounds with long carbon chains.

Now you can try this for other elements to see whether there is a cause and effect.

Fluorine atoms can share electrons with one another, therefore fluorine can form many compounds with long fluorine chains (?)

The first part is true, e.g. $\ce{F2}$, but the second part is not true. You might say that fluorine is a bad example because it usually just makes one bond when forming molecules.

Oxygen atoms can share electrons with one another, therefore Oxygen can form many compounds with long oxygen chains (?)

The first part is true, e.g. $\ce{O2}$ or ozone, but the second part is not true. Oxygen in molecules often makes two single bonds, so if that were the only requirement, it should make long oxygen chains but it does not.

Nitrogen atoms can share electrons with one another, therefore nirogen can form many compounds with long nitrogen chains (?)

The first part is true, e.g. $\ce{N#N}$ or $\ce{H2N-NH2}$, but the second part is not true. Nitrogen in molecules often makes three single bonds, so if that were the only requirement, it should make long nitrogen chains (or even networks) but it does not.
We can go from carbon down the group to silicon, and still no long chains:

The silanes ($\ce{Si_nH_{2n+2}}$) are less thermally stable than alkanes ($\ce{C_nH_{2n+2}}$). They tend to undergo dehydrogenation, yielding hydrogen and polysilanes. For this reason, the isolation of silanes higher than heptasilane has proved difficult.

Source: https://en.wikipedia.org/wiki/Binary_silicon-hydrogen_compounds
So what is wrong with the cause-effect relationship is that there are counter-examples where statement 2 ("long chains") does not follow from statement 1 ("share electrons"). There must be some other requirements for long chains of the same element, and it must be quite restrictive because carbon seems to be the only element making those chains, at least under our current terrestial conditions.


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How switching values of $R$, $K_p$ and $K_c$ does not alter the value of change in standard Gibbs energy?
$\Delta_\text r G=RT\ln \frac{Q}{K}$
If we switch $K_p$ from $K_c$, accordingly $Q_p$ will change to $Q_c$. Therefore, value of $\Delta_\text r G$ remains the same.
$ΔG^\circ=−RT\ln K=ΔH^\circ−TΔS^\circ$
If we switch $K_p$ from $K_c$, accordingly $\Delta H^\circ$ and $\Delta S^\circ$, which were defined at 1 atm will change their values according to $\pu{1 mol/L}$. Therefore, value of $\Delta G^\circ$ remains the same.
Now consider, $K_p=K_c(RT)^{\Delta n}$
$\implies K_p=K_c(0.082\times T)^{\Delta n}$ $\pu{(atm)}^{\Delta n}$
$\implies K_p=K_c(8.314\times T)^{\Delta n}$ $\pu{(pascal)}^{\Delta n}$
Which is fine, but consider:
$ΔG^\circ=−RT\ln K_p$
$\implies ΔG^\circ=−0.082\times T\ln (K_c\times0.082\times T)^{\Delta n}$ $\pu{L.atm}$ $=−8.314\times T\ln (K_c\times0.082\times T)^{\Delta n}$ $\pu{Joule}$
and also as $\ln K_p$ is dimension-less,
$ΔG^\circ=−8.314\times T\ln (K_c\times8.314\times T)^{\Delta n}$ $\pu{Joule}$
Why are there two different values of change in standard Gibbs energy? What did I miss?

The relationship between $\Delta G^\circ$ and $K$ is only defined for the dimensionless value $K_{eq}$.

@Andrew $K_{eq}$ will be one of these $K_p$, $K_c$, $K_x$, etc.?

short version: if all of your reactants and products are dissolved species, $K_{eq}=K_c$, and if all are gases, $K_{eq}=K_p$, but if you have any gases in your reaction, $K_{eq}\neq K_c$! Look up the defintion of $K_{eq}$ and "activity" in any general chem text for a full explanation.

@Andrew Suppose the reaction is a heterogeneous solid-gas phase and the value of $K_c$ is given. Then we will convert $K_c$ to $K_p$ by taking the value of $R$ as $\pu{0.082 L atm mol-1 K-1}$ and then we can use, $\Delta G^\circ =-RT \ln K_p$ $\pu{L atm}$?

There is no reason to change the units and value of R that you are using, since the standard units for $\Delta G^\circ$ are still J/mol, and the logarithm is dimensionless. So R should always be in J/(mol K).

@Andrew What I meant is: In the expression, $K_p=K_c(RT)^{\Delta n}$, $R$ must be in $\pu{L atm mol-1 K-1}$ for converting $K_c$ to $K_p$ in $(atm)^{\Delta n}$? But if $K_c$ was dimensionless, $K_p$ will not remain dimensionless as $RT$ has dimension of $\pu{L atm mol-1}$?

The conversion between $K_p$ and $K_c$ needs to change units, because both of those values have units. In contrast, $K_{eq}$, which you use in the relationship with $\Delta G^\circ$, is dimensionless. It happens to have the same magnitude as $K_p$ or $K_c$ under some circumstances, but it's not the same thing.

Recall that $\Delta G^\circ$ represents the difference in Gibbs free energy between an equilibrium state and a "standard state." This standard state is completely arbitrary, and the definition has changed slightly over time. Currently, IUPAC defines it as having all dissolved substances at a concentration of 1 mol/L and all gases at a partial pressure of 1 bar.
When we calculate the magnitude of $K_{eq}$, we use dimensionless quantities known as "activities", which are approximately equal to the amount of substance relative to the reference state. The deviation of the magnitude of the activity from the magnitude of the concentration is beyond the scope of this answer. For now, just assume they are always equal if the units are correctly used as described below.
Thus, the activity of a dissolved substance is the concentration in mol/L divided by the reference state of 1 M, which simply removes the units. But it requires that the units are M. For gases, the same process applies, except that the reference state is 1 bar, so units must be bar (or you could convert the reference state of 1 bar to whatever units you are using, but that's a bit silly).
For solids and bulk liquids, the activity is always 1, because they cannot differ in concentration, so they are always the same as the reference state.
In contrast to $K_{eq}$, $K_c$ and $K_p$ are not defined with respect to a reference state. They are simply ratios of either concentrations or pressures of products and reactants at equilibrium. If all of our reactants and products are dissolved species and we express their concentrations in mol/L, then the magnitude of $K_c$ is equal to that of $K_{eq}$. Likewise, if reactants and products are all gases, the magnitude of $K_p$ is equal to that of $K_{eq}$. But if we have gases in the reaction, the magnitude of $K_c$ is not equal to that of $K_{eq}$ because we haven't accounted for the reference state of 1 bar.

I am a bit confused about your first sentence: At equilibrium, to define $\Delta G^\circ$, all the reactants and products must be in standard state, say 1 mol/L. That means in any other instance they were not in that "standard state".


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Product of reaction between cyclohexene and bromine in methanol at 273 K
What product could we expect when we let cyclohexene react with an equivalent of bromine in methanol at 0 °C? Is it just the trans-addition of bromine (rac.) or also the SN1-susbtitution on a carbon due to MeO? Maybe twice? See below:

The first product (bromoether) is correct but the mechanism is wrong. Consequently,  the second and third structures do not form. Have you tried googling "bromohydrin formation"?

Thank you. I just looked it up, so the bromoaddition gives us an intermediate where meoh can attack, but I wasn’t sure if it’ll really deprotonate

The oxygen of MeOH attacks the intermediate bromonium ion from the rear (SN2) then the proton is lost. With no strong base, no prior deprotonation of MeOH occurs.

Note that at 273K MeOH is not a strong enough nucleophile to attack a secondary alkyl bromide so the third product, the di-ether, will not be formed.

It starts with the formation of brominium ion, as in direct bromination of alkenes. If you do not know the mechanism, you can refer to it here, at MasterOrganicChemistry.

But here, the $\ce{MeOH}$ can itself act as a potential nucleophile. It is similar to if $\ce{H2O}$ is used as a solvent - refer the formation of halohydrin here.
Since $\ce{MeOH}$ is available in larger concentrations, it more readily attacks the carbon. Markovnikov rule is followed for determining which carbon is attacked, but in this case, both are symmetrical.
The attack proceeds through $\mathrm{S_N2}$ mechanism, forming two enantiomers. After the $\ce{C-O}$ bond is formed, the $\ce{MeOH}$ loses $\ce{H+}$ to form ether and $\ce{HBr}$.


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Reasons for scratching the wall of a flask when precipitate crystals
I heard that scratching the wall of the flask helps precipitate crystals when obtaining crystals.
By scratching the wall of the flask, the crystals from it fall in the center and act as seeds to help produce crystals?
So, can it be said that the process of scratching the walls of the flask induces heterogeneous nucleation and helps crystallization?
To understand this, would it be helpful to study 'crystallization (creation of crystalline nuclei and growth of crystals)?
Will this process of scratching the wall of the flask ultimately help increase the yield?

Because nucleation is a very rare event (after all, you need a single nucleus for making crystalls with millions of particles in them), and very hard to study. It is possible that all nucleation events are heterogeneous (you need to have crystalls forming in a containerless droplet without any impurities such as dust to have a valid counterexample that could be homogeneous nucleation - and it still could happen on the surface).

How do you define yield? Size of crystalls, or total mass of crystals. For large crystalls, you need a single nucleus - too effective nucleation would be counterproductive.

If your product is most stable as a solid (at ambient conditions) and you want to harvest/recuperate this product by filtration from a liquid/solution, the more crystalline precipitate staying to top of your Buchner filter, the higher the yield based on isolated material.  This obviously does not affect the yield based on conversion of the reagents.


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How to troubleshoot absorbance analysis when readings fluctuate
I am trying to measure creatinine in fluids by using an enzymatic (creatinine) test. I try to determine the concentration by the difference in absorbance before and after 5 minutes after a reagent is added. Have a look:
diazyme test
My problem is that the measurements ‘jump’ all over the place, but seem centered at the right value. My absorbance is on the order of 0.0010 (the photometer provides 4 decimal places resolution) for my sample material for a concentration of say 60 umol/l. That means a change in the least significant digit in the absorbance gives an error of minimum 10%. I think this could be causing my fluctuating results, because simply rotating the cuvette changes the absorbance by at least 0.0001. I also see that the temperature in the room change 2 degrees during the day. Perhaps it could influence the speed of the reaction.
What is the right way to troubleshoot a situation like this to find out why my measurements are unstable?

Instead of thinking about the instrumental (spectrophotometer) problems, re-think your experiment. You are testing the instrument at its typical performance limit. The absorbance of the method is too low! Let us convert the absorbance 0.0010 to percent transmittance. It is 99.77%
What does that mean physically? You are "requesting" the instrument to differentiate between 100 % transmittance and 99.77% transmittance. No, this difference will be buried in noise (and hence the fluctuations).
The only solution to your analytical problem is to re-make the samples in such a way that the absorbance >0.1 and it should not exceed 1.5.
In short, prepare a 100-fold more concentrated sample. All these suggestion assume that you are not making any mistake in solution or reagent preparation. Re-check your solution preparation as a sanity check.


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What does one mean by saying "in conjugation"
In my chemistry class in high school we just started with resonance and my teacher said that its the phenomenon of moving around of negative charges, lone pairs or pi electrons with no change in position of atoms and in conjugation with pi electrons. Could someone explain what the statement in bold is supposed to say?

Charges do not actually move around in resonance. They are not structures in equilibrium. They are simply ways of representing a hybrid structure, when no diagram may be used individually to do so.

You should ask your teacher to better illustrate what he/she means. Anyhow, you can read the Wikipedia page on "Conjugated system" to get more info on what he/she is trying to say. It basically has to do with the definition of the word "conjugated".

"Conjugated double bonds in a molecule mean that the single and double bonds alternate". If you can write a Lewis structure like that, every atom in a chain will have an atomic orbital that can interact with its two neighbors (typically, a p-orbital), giving rise to the molecular orbitals that span multiple atoms in a chain or ring.

Did you look up conjugation in the dictionary? It implies joining, but in chemistry it is joining the atoms in a special way.
The chemical meaning is straightforward, see https://goldbook.iupac.org/terms/view/C01267

In the original meaning a conjugated system is a molecular entity
whose structure may be represented as a system of alternating single
and multiple bonds

More explanation here Chemistry LibreTexts

This answer illustrated that conjugation has in principle a purely topological meaning. It is worth to note that the word conjugation used to mean delocalisation it is a somewhat a contradiction in term, in the sense that a perfect organic metal would have no alternating bonds. Still, the existence of alternated ie conjugated limiting structures is a requisite for resonance / delocalisation, and this makes the two terms synonyms in practice.


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(How) can a reaction be both stereoselective and stereospecific?
From what I have learnt, stereospecific and stereoselective reactions are defined as follows -
Stereospecific - The stereochemistry of the reactant dictates the stereochemistry of the product. Thus, an optically pure compound yields only one product.
Examples - SN2 mechanism which leads to complete inversion of stereochemistry, bromination of alkene which always proceeds via anti-addition (hence, a cis-alkene will yield two enantiomers with the $\ce{Br}$ groups on opposite sides).
Stereoselective - The reaction yields two (or more) products, out of which one is major and others are minor.
Examples - elimination reactions yield a mixture of cis and trans products, but the trans product is generally formed in a greater amount.
It seems to me that a reaction can either be stereospecific (yielding only one product) OR stereoselective (yielding >1 products with one in greater yield) and hence can't be both. However, my teacher recently told that some reactions are both stereospecific and stereoselective (and things being online, it'll be some time before she answers my query regarding her statement). I've also seen the same thing appear in some test papers that I solve. Answers to this related question also seem to indicate that stereospecific reactions are a subset of stereoselective. I cannot understand how that is possible. Any insight would be appreciated.

@Jay Thank you. I've seen the Wikipedia page, but it only mentions the definition and examples of stereospecificity itself, not about how it is related to stereoselectivity.

Buttonwood answer is correct as per IUPAC so follow it. But no one will ask such in exam as per my experience.

Stereoselectivity and stereospecificity are two unrelated parameters, one describing product configuration depending reaction parameters, the other on educt structure.

@Karl I'm sorry, I thought both were used to describe reactions based on what product they formed. Is that not the case?