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Why does STED microscopy cause photodamage?
STED (stimulated emission depletion) microscopy is a technique which enables sub-diffraction limit of light imaging.
The depletion laser is at 775nm, and about 500 mW, concentrated for a short sequence of pulses on an area about the size of ~0.2 um^2
As far as I can tell very few organic molecules, nor water, absorb in this range of 775nm.
So why does it cause photodamage?
Thermal damage sure, but why bleaching fluorophores so fast?

It is good practice to define acronyms when they are first used. Please define “STED”.

Edited for your comment.

Thanks! I cannot answer your question, but now it has higher likelihood of getting a valid and helpful answer from someone.

@dlight, There are reviews on this topic. There are dyes which excited by red wavelengths and that is what STED uses, right? One of the article suggests "Photobleaching is primarily caused by the depletion light acting upon the fluorophores in the excited states."

See this one too: https://onlinelibrary.wiley.com/doi/full/10.1111/jmi.12698

The depletion laser is at a fluorescence wavelength so should have low absorption by the fluorophore. Normally a pulsed laser is used to get high peak power (to get stimulated emission) but low total energy to prevent heating and damage that way. Any excited state that fluoresces will form triplets and these can react with O2 to produce reactive species that will eventually cause photodamage. Any species in the sample that absorbs (besides fluorophore) can in principle cause photodamage from their excited state. Two photon absorption may be important also.

Ballpark calculation: Wikipedia mentions the sun irradiation received at earth's sea level is up to about $\pu{300 W/m^2}$.  This is after the atmosphere filtered out already much.  You describe an experiment that where $\pu{0.5 W}$ set foot on about $\pu{0.2 µm^2}$ (or, equivalent to to this, $\pu{0.2 × 10^{-6} m^2}$).
This translates to a power density of
$$\frac{\pu{0.5 W}} {\pu{0.2 × 10^{-6} m^2}} = \pu{2.5 × 10^6 W/m^2}$$
which is more than 8000 times larger than above assumption.  This one of the reasons why samples are excited by pulsed laser radiation with repetition frequencies in the range of MHz.  And still degradation by photobleaching may occur, which one already tries to reduce, e.g., by excitation which do not hit the centre, but a tail of an absorption band.
(The mechanisms are different, but the results of exciting a small area with too much energy are similar to microscopic techniques based on second harmonic and complementary third harmonic generation (an example).)


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Is there a known value for the molar extinction (absorption) coefficient for amylose and amylopectin iodine complexes?
I am doing a colorimetry experiment with starch, and have a theoretical model based on Beer's law. To solve it, I need a known value for the molar extinction coefficients of amylose and amylopectin iodine complexes. Are these known values, and where can I find them?
While experimentally finding them would not be too hard, I cannot obtain pure amylose and amylopectin, so I cannot do this myself. I have done some research, but could not find many values(lack of research skills and access to scientific databases).
Any direction to papers or sources would be greatly appreciated.
Alternatively, a reliable and commonly available replacement for a starch standard would also solve my problem, if such a thing exists - this would make things much easier for me. This is a senior school experiment, so a compromise between accuracy and availability is expected.

Do you mean for the starch iodide complex or just plain starch?

Apologies, yes. I meant the starch iodine complex for amylose and amylopectin each.

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Mechanism of hydrolysis of acyl chlorides
The following mechanism for acyl chloride hydrolysis was given by my professor :

However I do not understand why in the second step, oxygen has a partially negative charge, should it not gain a positive charge as it is bonded to three atoms, furthermore shouldn't the $\text{O}-\text{H}$ bond breaking homolytically form a $\text{H}^-$ ion and not a proton?
Should the arrow be reversed in the second step?

Do not confuse real charge with formal charge.

This reaction is mostly ignored in basic organic texts and the mechanism you give is straight out of Wikipedia with some modification. I am sure some serious study exists some where but not on the web. Please bear with me because I can't[or just don't know how to] do drawings.  The first step is probably as written and is most likely slow because it involves charge separations. You are right there is a + charge on the O of water. [btw there is no homolysis involved, that generates radicals, and there is no reasonable path to generate hydride ions] The second step seems to require the participation of a base this may be loosely described as an E2 reaction where a proton is removed from the protonated O atom by another water molecule and the developing negative charge on the other oxygen expels the chloride ion. This mechanism[if plausible] requires enough water for active solvation. This mechanism possibly explains the success of the Schotten-Bauman method where base is added to the reaction as a "catalyst" [tho it is consumed]. If water is limited the base may be the attached chlorine and references do state that dry reactions evolve HCl gas. These are my thoughts on the reaction; library research is in order.

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Multi-step organic synthesis using benzene
I have this project that I've been struggling with. I'm already given the reactant and product, so basically what I have to do is to figure out the steps needed to get the product.

I figured that some of the steps would include the Friedel-Crafts acylation and nitration, but after that, I'm not too sure what other steps I'd have to take. Any help will be very much appreciated!

Welcome to  ChemSE. With some real effort on your part, we will be pleased to help. Get back to us.

How many steps are you considering?

Hint #1: Identify suitable precursors to install any of the three substitutents on a benzene ring. E.g., $\ce{-NO2}$ is introduced by ... . #2: Not all remaining $\ce{C-H}$ e.g., in nitro benzene are equal. Some substitutents direct the next entrant substitutent to ortho / para position, or in meta position when running an electrophilic aromatic substitution. See how these patterns may align with the target of a triple substitution. #3: Some of the groups shown may the result of a substitution and subsequent group transformation  (e.g., redox). #4: Don't limit to electrophilic reaction.

@Waylander has provided a concise route to the nitro compound in question. I am also in accord with his comments regarding the OP's effort and his assessment that benzene, as a starting material, is not a choice by a practicing chemist. Now that he has opened the door to a solution, I offer a more traditional approach.
Classical nitration of ketoacid 7 should be reasonably selective in that the introduction of the nitro group in 8 is both para and meta directed. Keto acid 7 arises from indene 6 via exhaustive oxidation. To access the indene, I would start with benzaldehyde (2), whose preparations from benzene (1) are numerous. The Perkin condensation 2 $\rightarrow$ 3 is known. Hydrogenation of 3 forms carboxylic acid 4, which can be cyclized to indanone 5 with polyphosphoric acid. Friedel-Crafts techniques may also be employed. Reduction of this ketone followed by dehydration leads to the critical indene 6.

The indene cleavage is a nice option. I would be tempted to carbonylate benzene to make benzaldehyde.

@Waylander: I agree with direct carbonylation except that the ones I found on Chem. Absts. require more sophisticated techniques. I opted for something the OP would understand although it is several steps. There is also Cl2CHOCH3/TiCl4 as a CO equivalent but, again, too sophisticated. The whole post is an academic exercise, not something you or I would have executed from benzene.

While I think the OP should make more of an effort at answering this I am going to suggest a route because I think this is an interesting challenge - if a little unreal. In the real world of preparative organic chemistry you would not be starting from benzene. There are many possible approaches to this, but here is mine
Step 1 - carbamoylation of benzene with dimethyl carbamoyl chloride/TMSOTf/Zn(OTf)2 as described here
Step 2 - directed lithiation of the dimethylbenzamide with sBuLi/TMEDA followed by quenching with the commercially available propylene oxide to install the C3 sidechain at the alcohol level.
Step 3 - oxidation with Dess-Martin periodinane to give the benzyl methyl ketone
Step 4 - nitration under standard HNO3/H2SO4 conditions as here should give the desired 1,2,5 substitution pattern as the major component.
Step 5 - amide hydrolysis under strong acid conditions with cHCl. I have chosen this over basic hydrolysis as think the formation of the anion of the starting benzyl methyl ketone activated by 2 EWGs is likely to be an unwanted complication.

The benzene rings in the carbamoylation paper appear to be activated. While toluene is the least activated, benzene is conspicuously absent as an example.

It is absent but think it will work with the addition of Zn(OTf)2


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Why is Ka of weak acid increasing while being diluted?
I am calculating $K_\mathrm{a}$ and percent dissociation of acetic acid being diluted to $\pu{1.00 \times 10^{-1} M}$, $\pu{1.00 \times 10^{-2} M}$, and $\pu{1.00 \times 10^{-4} M}$.
The $\mathrm{pH}$ goes up as the weak acid is diluted, which is to be expected.
The % dissociation is inversely related to the initial concentration of acetic acid.
However, I fail to understand why the $K_\mathrm{a}$ value increases (slightly) as the solution gets more dilute. Isn't $K_\mathrm{a}$ supposed to be a constant (acid dissociation constant) therefore stay constant as long as the temperature stays constant? Isn't according to Le Chatelier's principle, $K_\mathrm{a}$ is a function of temperature?

The simplest solution is that your calculation is wrong or not accurate enough. As you say, it's a constant.

The scale at which you are working, the standard formulas won't work. You will have to solve a quadratic equation to get the results. The standard formulas are just an approximation to make our lives easy.

Have you involved ion activity coefficients ? How exactly have you calculated Ka ? Have you applied any implicit or explict simplifying assumptions ?

E.g. the famous formula pH=0.5(pKa - log c) assumes c >> [H+] >> [OH-].

Can you please edit the question to include the readings you got during the experiment. That might help us understand where the problem is, approximations or just experimental error.

$K_a ≠ x^2/0.100$, rather $K_a ≈ x^2/0.100$. This expression assumes the condition mentioned by @Poutnik. Divide by $0.100-x$ instead. Even after that the results might me screwed because when $x<10^{-6}$,  even self ionization of water causes significant change in $\mathrm{pH}$.

You should rather involve in the question the used algebraic equations explicitly.   A claim not making assumptions may be an assumption itself, not necesserily correct. By other words, if you do not say what you did, we cannot say what you did wrong.

Try to make it complete. And be aware of x^2/(c-x)  not x^2/c-x .Consider also Activity coefficients. Another case is oH meter calibration. How did you do it for low pH values ?

Remember that with increasing ionic strength, activity coeefficients decrease, according to Debye-Hueckel equation. So while activity based Ka is constant, concentration based Ka grows with the acid and therefore ion concentration.


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Alpha-haloketone reaction with H2O in acid medium
I need help with this reaction, I know that in the first step an alpha haloketone is formed, but I can't figure out what reaction is going to happen between the alpha Haloketone and $\ce{H2O/H+}$.

I read your answers and this is some information and assumptions that I can share.
For the reaction between the alpha haloketone and the $\ce{H2O/H+}$ ,I thought that the oxygen of the carbonyl could be protonated so the $\ce{C}$ atom becomes more electrophilic, then the water (the nucleophile) could attack that C atom of the carbonyl group.

I found a paper (Erian, A. W, 2003)  where they report a general mechanism for the reaction of alpha haloketones with nucleophiles

So, considering that, I thought this might be the product of this reaction.

What do you think? I'm not really sure of the answer, but that's what I have so far.
This is the complete reference:
Erian, A. W., Sherif, S. M., Gaber, H. M. (2003), "The chemistry of α-haloketones and their utility in heterocyclic synthesis", Molecules, 8(11), 793–865. (https://doi.org/10.3390/81100793)

The conditions you describe for the first reaction is for a haloform reaction. Normally, acid is the catalyst. What are or your thoughts on the second reaction with a reactive chloroketone?

As usual, identify the nucleophile and identify the electrophile.

I edited the post, hope this can help, what are your thoughts?


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Alkylation of enolates using tertiary halide
I was recently studying the alpha alkylation of enolates using LDA followed by subsequent treatment with primary halides to give the product.I was wondering what happens with teritary halides. My guess is that elimination will take place,and the carbon nucleophile formed by LDA will abstract a proton fron the tertiary halide giving an alkene. Is this chain of reasoning correct?

Sounds good to me!

Also to me. These kinds of alkylation are better done by Lewis acid mediated reaction of the silyl enol ether with the tertiary halide.

@https://chemistry.stackexchange.com/users/55119/user55119 Thanks a lot for your responses people. What happens with secondary halides?


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Substances with high physical expansive property/expansion after reaction
I want to know if there is a substance/mixture of substances that expands more on some chemical or physical change, than what I've already seen.
Elephant toothpaste is a good example, a small amount of two substances expands into something with much more volume.
The list I have created so far is:

Diaper jelly
Orbeez
Elephant toothpaste
Water $\ce{->}$ Ice

A list of multiple substances with most expansive chemical or physical property will be appreciated.
The reason for my question: trying to develop a realistic 'smoke bomb' or some sort for a project photography. Best result would be lets say from small pellet, to huge smoke cloud.

Another example is the pharaoh serpent. But the question could be more defined.

Can you provide parameters for an acceptable final product? For example, I don't think you are including vaporization which increases volume much more than any of the reactions you've cited.

If you want large amounts of smoke, take a bunch of straw and sprinkle some water on it, and then burnt it. Although I am not sure how environmentally safe it is.


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Do high-energy chemical bonds indicate strong chemical bonds?
In my head it seems reasonable that a high-energy bond would indicate that the bond is strong, however a question got me thinking.
Take for example lead azide, a very explosive compound that is often used as a detonator or fuse. It has a decent amount of energy stored in its bonds while being very sensitive to heat, friction, shock, etc.
This indicates to me that the lead azide has a low activation energy and therefore weak bonds, yet still plenty of energy available to be released.

Activation energy and heat of reaction are two seperate quantities.

Energy isn't stored in bonds.  Bonds represent a lower-energy state than the separated atoms. I.e. energetically, bonds are "valleys".  Indeed, it's only because energy decreases with bonding that atoms bond at all, since bonding causes a loss of entropy.  The strongest bonds are the ones that put the bonded atoms in the lowest energy state, and thus take the most energy to break. The most unstable bonds are the bonds that take the least energy to break. Compounds containing such bonds can react to release energy because these bonds can rearrange, through chemical rxns, to form stronger bonds.

There are two factors here to consider.
First, the energy released by a given reaction is (partly) due to the difference between the bond strength in the product vs reactant, not the absolute bond strength, unless the reaction involves just breaking a bond and not forming a new one.
Second, even if your reaction is just a bond breaking, there are multiple ways to break a bond. Reported bond enthalpies are usually defined in terms of the energy required for homolytic bond breaking, ie one electron stays on each component. For example,
$$\ce{HCl -> H. + Cl.}$$
By that measure, the H-Cl bond is quite strong, but we know that in a polar solvent that mitigates the negative effect of charge separation, the H-Cl will readily undergo heterolytic bond cleavage, where both electrons stay on one atom:
$$\ce{HCl -> H+ + Cl-}$$
In that context, we might consider the H-Cl bond to be very weak. (Note, though, that I'm also overlooking the amount of "bonding" between solvent and ions in this case, which means that it's not an example of a pure bond breaking reaction.)
All of which is to say that before you can make a statement about bond energy and bond "strength" you need to define both terms more specifically.


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How to compare stability between 5‐methylcyclohexa‐1,3‐diene and 3‐methylenecyclohexene?
Problem

Compare stability of 1 and 2:

Answer

 1 is more stable than 2.

My approach

Both have the same number of π-bonds.
Both have one resonance structure.
Both are non-aromatic.
In 1 there are three α-hydrogens, whereas in 2 there are four α-hydrogens.

As extent of hyper-conjugation is more in 2 than in 1, making 2 more stable, which contradicts the answer.
Addressing the comment:

A factor you haven't considered is ring strain. I'm not sure which of the two structures would have more ring strain, though.

I think that 2 has more strain because there are three $\mathrm{sp^3}$ carbon atoms which have been forced to be in a 120° angle, while their ideal angle is 109°28′. So, there is more strain in 2 compared to 1 because in 1 only two carbon atoms are forced to do so.

A factor you haven't considered is ring strain.  I'm not sure which of the two structures would have more ring strain, though.

AFAIK, given that short of running an (expensive) numerical simulation, we only got some heuristics to work with, it'd be worth knowing to what extent do those really differ in terms of stability. If it's a small difference (say <1%), then I'd imagine that the heuristics may not be accurate enough to be trustworthy. It'd really help to know whether the authors of the textbook that posed this question have actually made any measurements or used published data, otherwise any answers you get may be misleading (garbage in = garbage out).

If you have access to the two compounds, determine their heats of hydrogenation or combustion. The less exothermic compound is more stable

As far as I have read, in general, exocyclic double bonds (which are outside the ring) are stable for 3 or 4 numbered rings while endocyclic double bonds (which are inside the ring) are stable in 5 or 6 numbered rings.
As you note all the factors, more or less all factors are same in both cases, so it becomes a little handwavy thing.
Additional Reading:

Goumans, T.; Ehlers, A.; Lammertsma, K.; Würthwein, E. endo/exo Preferences for Double Bonds in Three-Membered Rings Including Phosphorus Compounds. Eur. J. Org. Chem. 2003, 2003 (15), 2941–2946. DOI: 10.1002/ejoc.200300120.

Fleck, B. Notes - Stability and Ease of Formation of Endo and Exo Double Bonds in 5- and 6- Ring Systems. J. Org. Chem. 1957, 22 (4), 439–439. DOI: 10.1021/jo01355a023.

Exocyclic vs. endocyclic double bonds in E1 elimination (although this one is not much relevant here).


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Safe chemical reactions to start a campfire
I'm looking for a way to start a small fire electronically with widely available chemicals. It would be great if they are safe to use with metals (not necessarily). But the biggest requirement is that it should be SAFE at medium scales and should not make an explosion.
I have found out $\ce{H2SO4 + KMnO4}$ can start a fire. But is it possible to do it with diluted H2SO4 because concentrated is not safe to handle?
NB: I only want a small ignition. The rest can be continued with some fuel, I believe.
Also, I doesn't possibly need to be a chemical reaction. Anything that can trigger a safe fire is enough

What is wrong with LPG based or petrol based lighters ? They are cheap, safe and used for these purposes all the time. They are not vulnerable as matches are, when get wet.

I want to start it electronically. Like it starts the fire when a button is clicked like so... I can throw some safe chemicals with an electronic valve. Fuel from another valve. Like a little home automation...

That is kind of weird, potentially very unsafe and abusable. If something goes wrong, you can arrive at the burning house.

This is purely experimental and I'll be the only user... But chemicals are altogether unsafe, right? But I'd like it to be as safe as possible... There must be some safer way.. :) And I won't be doing anything inside unless it's VERY SAFE.

Obviously, Extinguishers will be ready to put it off if something goes wrong

I have meant it in the general fire-control safety context, but managing chemicals can bring another factor of their accidental contact. As Gandalf from the LOTR said, Even the wisest cannot see all ends.

@Poutnik I understand... But imagine if no one invented car engines because it work gas or petrol  aren't we scientists crazy enough to do stupid things to learn new stuff?

The mixture made with pure and dry $\ce{H2SO4 + KMnO4}$ produces $\ce{Mn2O7}$, and only with dry substances. This $\ce{Mn2O7}$ is able to start a fire with any organic impurity around due to the oxygen produced by its own decomposition into $\ce{MnO2 + O2}$. But if there is even a small amount of water, $\ce{Mn2O7}$ is not produced, because $\ce{Mn2O7}$ is due to the dehydration of $\ce{HMnO4}$ that is the first intermediate product in the initial mixture. And no fire can be started.

@Maurice So that means I must use highly concentrated $\ce{H2SO4}$ right? I don't that would be safe to be left unattended! Hard to get too!

I think I have found the safest solution with physics... A steel would would create some fire when electricity is conducted... I hope some fuel can take over from there...

@AjithGopi Gas engines do not work with open fire....... With open fire, many things can happen and not all can be easily predicted. It is required to be under direct human control for reason, no matter how it started.

I would not use the mixture $\ce{KMnO4 + H2SO4}$. It is rather corrosive. It may explode without any reason.

I understand that @Poutnik, You know automation doesn't work like that... We study, find errors, and improve the system. It will only be left unattended if everything is completely safe.. If it's not safe, the project will be postponed until a better solution is made... :) But right now, I think using the steel wool is completely safe. It only needs around 9v and a few drop of petrol might be enough to let the fire take over wood...

The problem is that self-igniting binary chemicals are mostly safety risk even if separated. I guess electronically ignited fuel like you mention would be safer.

Most sophisticated approaches to finding a "sexy" chemical solution are likely unsafe, especially compared to long standing solutions we already know work well and are acceptably safe. Like matches.

Small magnesium metal bars are available commercially, packed with a little iron rod for starting fires. You smack the magnesium with the iron rod and get some sparks. If you have some light fluffy kindling, you are supposed to be able to get a fire going. I suppose a little steel wool would help.
If you soak some cotton balls with KNO$_3$, dry them completely and carry them in a sealed plastic bag, they would make excellent kindling.

Pencil shavings also work well as kindling. Though maybe pencils are not used much now.

@Ed V: Actually, that's a very good idea! So, buy a 36 inch x 1/4 inch dowel, cut it into 9 inch pieces (or 6 inch), and get one of those little pencil sharpeners. That should make nice fluffy kindling, without paint or "lead".

I like those old school hand crank pencil sharpeners that used to be in every classroom. Of course, I use mechanical pencils (and pens) now, but the dowel refinement is very good as well! And I think they still sell mechanical pencil sharpeners. By the way, I upvoted: the Mg starter method is most reasonable for most folks.

Apply an electric current to a suitable metal powder mix where, at time of employment, one adds sulfur or potassium nitrate to the open mix (that is, never placed in a confined container). This a possible option and follows the common recipe for flash powder.
However, given the desired applications of an electric current as a fire starting mechanism, particular metals of interest should be good conductors, as well. This could include already cited Fe, as well as Al, Mg and Zn.
Caution, be very careful even remotely electrically igniting such flash powder mixtures as they can result in a quite vigorous thermal reaction. This may potentially result in burns. However, in situations where a common magnesium spark fire-starter, for example, is not working (cold or moisture issues), such a composition may be solely successful.

I have come up with a solution to use steel wool with electric power. Steel wool can generate sparks when electricity is passed through. I hope some fuel can take over from there to start a fire. This might have become off-topic here but I hope someone would find it useful.
Anyways, now I know it's not safe to use $\ce{KMnO4 + H2SO4}$

There would be fine balance between ability to be reproducibly ignited and vapor ability to explode.

+1. A good reference: https://frugalfun4boys.com/burn-steel-wool-science/

Your proposed solution is well known. A simple chemical way is to use a gram or so of potassium permanganate and add a few drops of glycerin (glycerol). But the magnesium fire starter method in the answer by James Gaidis is better than carrying around chemicals: potassium permanganate stains whatever it touches and glycerin is sweet, so it attracts insects, especially ants.


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How is the hydrazone formed in the Wolff-Kishner reduction?
In the Wolff-Kishner reduction the first step is to react a carbonyl with hydrazine to form a hydrazone. The later steps of the reaction requires a strong base to cause loss of nitrogen.
However, the mechanism for the formation of the imine/hydrazone is acid catalyzed:

Wikipedia mentions that the formation of the hydrazone is typically done in situ.  So how does the formation of the hydrazone occur in basic conditions?
References:

https://www.organic-chemistry.org/namedreactions/wolff-kishner-reduction.shtm
https://en.wikipedia.org/wiki/Wolff–Kishner_reduction

I think they prepare the imine in acidic conditions and then switch to base. 1) Hydrazine reaction, 2) KOH/OH- Ethylene glycol.

Thank you @AdityaRoychowdhury, that is what I wanted to know. I edited the title to reflect that answer. I would accept that answer.

Wolff–Kishner reduction is the method for the reduction of aldehydes and ketones to corresponding alkanes. First, the carbonyl compound condenses with hydrazine to form the corresponding hydrazone. The resulting hydrazone is treated with base to induce the reduction of the $\mathrm{sp^2}$-carbon to $\mathrm{sp^3}$-$\ce{CH2}$, and the oxidation of the hydrazine to gaseous nitrogen in the process to yield the corresponding alkane:

Source of the image
The following is the accepted mechanism of the reduction:

Source of the image

How does the formation of the hydrazone occur in basic conditions?

The formation of the hydrazone can be done in either acidic or basic conditions (recall imine formation of aldehydes and ketones with primary amine and 2,4-dinitrophenylhydrazone formation of  them with 2,4-dinitrophenylhydrazine). And, yes, as Wikipedia mentioned, sometimes it is advantageous to use a pre-formed hydrazone as substrate for the reduction. Yet, remember, Wolff–Kishner reduction is better suited for substrates with acid sensitive groups. Therefore, use of acid catalyst to pre-form hydrazone for the reduction is meaningless in that situation.

More on the mechanism of the Wolff-Kishner reduction: https://chemistry.stackexchange.com/questions/85995/on-the-mechanism-of-the-nitrogen-removal-step-of-the-wharton-olefin-synthesis/122597#122597

I can’t find any source for “hydrazine can be formed under either acidic or basic conditions“. On the contrary I find the opposite “ athigh pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O”

Did you look at the sorces given in the answer? For example, look at "imine formation of aldehydes and ketones" in the text.

I did, that’s where the second quote comes from

It'd be nice if the downvoter give some reasoning for their downvote. That's the war to improve the answer.


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At what concentration would copper sulphate crystallise at 25 degrees Celsius?
I am doing an experiment on how increasing concentration of copper sulphate affects the voltage in a galvanic cell. The voltage increased as temperature did until 2M where it started to decrease. I know the reason it does that but I can't find any website saying for 25 degrees Celsius it would start decreasing. I hope that makes sense and thanks for ur help!

The Handbook of Chemistry and Physics has a table giving the dehydration of metallic sulfates. Under copper sulfates, I read that the usual $\ce{CuSO4⋅5H2O}$ starts losing water at $\pu{27°C}$ giving $\ce{CuSO4⋅3H2O}$ which has the same color. Is it of any help ?`In the table of Solubility of inorganic compounds in gram per $\pu{100 mL}$, the solubility of copper sulfate (anhydrous) is reported to be $\pu{20.7}$ at $\pu{20°C}$, $25$  at $\pu{30°C}$, $ 28.5$  at $\pu{40°C}$.

Be aware that for concentrated solutions, many things behave more by empirically observed rules than according suitable theories, that apply on diluted solutions. Like activity coefficients or solution permitivity.

I would also not entirely assume (especially if the experiment is conducted in the presence of oxygen) that one has still a pure CuSO4 aqueous mix with a change in temperature in a galvanic cell setting liberating solvated electrons interacting with H+, SO4(2-), ....


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Why does gas not liquify at a temperature above the critical temperature no matter how much pressure is applied on it? Why?
My textbook says that critical temperature is the temperature above which a gas cannot be liquified no matter how much pressure we apply on it. But why? What is so special about this 'critical temperature'? I mean, if we apply a ridiculous amount of pressure, the gas molecules have to come closer to one another and they have to liquefy. What exactly happens to the gas molecules at critical temperature? Do they become immovable and have infinite inertia or what?

The temperature is too high to allow any inter molecular attractions that'd help them liquefy. Imagine you're running a relay race and have to hand over the baton to the next member, if you're running too fast then the baton cannot be passed. I know it's a funny way to understand it but that's how I think of it.

What is so special about the age of 18 (in many jurisdictions) that makes a person suddenly acquire all those rights? Nothing, except a human convention. Same thing here. Sure, by applying enough pressure you may bring the molecules of a supercritical fluid as close as they are in liquid. It is just that we don't call that "liquid".

Critical temperature means the evaporation enthalpy has become zero. The term "liquify" makes no sense then.

In short, it is not that you cannot push them against each other. It is that they don't stay.

The critical point is a point of convergence of all state properties of the respective liquid and gas. It can be considered as the degeneration point, where there is no difference between gas and liquid and this distinguishing does not make sense any more.
It can be also said the supercritical fluid near the critical point is neither gas neither liquid. It is both at the same time. Farther from critical point, like for permanent gases, either gas-like either liquid-like properties are dominant, depending of the state.
By changing  pressure of supercritical fluid, it gradually transforms between  states, where it behaves more like gas or liquid, without evaporation or condensation. It is similar, as if you reach an elevated level via a big step (liquid/gas or evaporation/condensation) versus  along a slope (supercritical fluid). For the latter, there is no upstairs (gas) nor downstairs(liquid), there is just the slope.
Imagine you figurally walk as the substance on the phase diagram ($T$ grows to the right, $p$ grows upwards) around the critical point. If you do it counter-clockwise, you will evaporate as many times as many rounds you do, without ever condensing. If you do so clockwise, you will condensate as many times as many rounds you do, without ever evaporating.

It really does liquefy.  But it does not do so in exactly the same way as you see below the critical temperature and pressure.
As an example, suppose you heat steam to 400°C and then compress it, isothermally, to 5000 bars pressure*.  When you are done, you find that the water has a density and viscosity more or less similar to ordinary liquid water; what was initially a gas now looks like a liquid.
What's different, however, is you see those physical properties evolve continuously, there is never a specific point where "gas" gives way to liquid droplets that are noticeably denser or more viscous than the gas.  Instead the steam above the critical point becomes liquid-like in a smooth, gradual way during the compression process.  You gradually contracted the gas until it became liquid-like instead of condensing it.
*These conditions are not sufficient to create ionic, superionic or macromolecular phases of water; we may continue to treat the water as a molecular substance.

The smooth, gradual transition can be seen e.g. at 1:11 in this video.

One useful distinction between a liquid and a gas is that while they are in equilibrium with each other, so at the same temperature and pressure but not the same density, molecules of the liquid need to gain energy to leave the liquid phase and enter the gas phase. A surface, with surface energy, exists between the two phases, and allows them to segregate under gravity. The surface energy is due to the mutual attraction of molecules in the liquid being higher than that between those in the gas.
As the temperature is increased, the gas becomes denser, and so the inter-molecular forces within it increase, while the liquid is becoming less dense. At the critical temperature, the energy difference between molecules in the 'liquid' and 'gas' becomes zero. There is no longer any distinction between the two phases, and you have a single fluid.

My textbook says that critical temperature is the temperature above which a gas cannot be liquified no matter how much pressure we apply on it.

Below the critical temperature, you can compress a gas and a second phase (liquid) will gradually form. In places with gravity, the liquid will collect at the bottom of the container. As you decrease the volume, the volume of the liquid will increase and the volume of the gas will decrease, but the pressure will remain the same (at the vapor pressure of a given temperature). If you decrease the volume when the substance is supercritical, no second phase will form, but the particles will of course come closer to each other (just like when you compress a gas without liquifying it).

But why? What is so special about this 'critical temperature'?

In terms of the behavior of the particles, there is nothing special. They have a certain strength of intermolecular interaction when next to each other (like in a liquid), and no intermolecular interactions when further apart (like in an ideal gas). What is special is the behavior of the bulk phase, for example when being compressed.

I mean, if we apply a ridiculous amount of pressure, the gas molecules have to come closer to one another and they have to liquefy.

They don't have to liquify because we can define what we mean by liquid. They do have to come closer to each other on average. In a two phase system, that happens by gas turning into liquid. In a one phase system (gas or supercritical fluid) that happens by gas turning into liquid. In a liquid or solid, compression is hardly possible because of the strength of the repulsive intermolecular interactions at short distances.

What exactly happens to the gas molecules at critical temperature? Do they become immovable and have infinite inertia or what?

Nothing happens to them. They behave exactly as before. This is like at the melting point of water. The individual water molecules don't do anything differently. In bulk, however, you see a different set of properties.


text:
Solubility of Beryllium Compounds in Water
Why is $\ce{BeF2}$ is highly soluble in water but $\ce{Be(OH)2}$ is insoluble?
Both release $\ce{Be^{2+}}$ ion so this means they both have high hydration enthalpies and can easily overcome their lattice enthalpy.
Then, why is $\ce{Be(OH)2}$ insoluble? A simple explanation would be appreciated.

@NilayGhosh I am mostly confused about Be(OH)2

Neither BeF2 neither Be(OH)2 is considered as an alkali halide. The title is misleading.

Also you have to be careful about "Both release Be2+" - the solids are best considered covalent. The species in solution are generally not simple hydrated ions, hydrolysis occurs. This is touched on briefly at https://en.wikipedia.org/wiki/Beryllium

@IanBush Can you elaborate ?

Is the lattice energy of Be(OH)2 more than the hydration energy of Be2+ ?


text:
How can we replace volume in ideal gas equation with V - nb?
We know that ideal gas equation is $PV= nRT$ where P is the pressure of the ideal gas and V is the volume of the ideal gas. Thus, we can write :
$$ P_{\textrm{ideal}}V_{\textrm{ideal}} = nRT$$
Now from volume correction term of van der Waals' equation we know that
$$ V_{\textrm{real}} = V_{\textrm{ideal}} - nb $$ where $ V_{\textrm{real}} $ is the volume of the real gas and $ V_{\textrm{ideal}} $ is the volume of the ideal gas. Clearly, $ V_{\textrm{ideal}}$ and $ V_{\textrm{real}} $ are different.
Then how can we write :
$$P_{\textrm{ideal}} (V_{\textrm{ideal}} - nb) = nRT$$
How can we replace $ V_{\textrm{ideal}}$ with $ V_{\textrm{real}}$?

There is frequent confusion:  The term $V - nb$ is not correction of ideal volume to real volume, but the opposite, it is the correction of the real volume to ideal volume. similarly for  $p + an^2/V^2$ is the correction of the real pressure to the ideal pressure.  The ideal volume would be for the same pressure smaller, not bigger. similarly, the ideal pressure would be at the same volume bigger, not smaller.

Also relevant is that more sophisticated equations come with sets of consistent factors that work together and tearing out a correction term into another set of equations with different assumptions will not go well.

@matt_black Sure, but I have not wanted to go so far. :-)

The actual volume occupied by the gas is less than the measured volume(volume of the container).
In reality gaseous particles are not point objects and don't occupy zero space, which is against one of the postulates of the Kinetic Theory of Gases.
Thus while dealing with real gases we have to account for the volume which has been occupied by the gaseous particles. We do so by assuming the volume of gaseous particles to be $nb$.
Now, the ideal volume(actual volume) occupied by the gas is the measured volume(volume of container) minus the volume occupied by the gaseous particles.
Thus, $V_{ideal} = V_{real} - nb$ and not the other way around.
Note, $V_{ideal} =V_{actual}$ and $V_{real} =V_{measured}$.
The mistake you are making is formula you are using and thus
$P_{ideal}(V_{ideal}-nb) ≠ nRT$, but
$P_{ideal}(V_{real}-nb) = nRT$

In $PV = nRT$, P represents the actual pressure exerted by the gas. Now, we know that the pressure exerted by a ideal gas is greater than the pressure exerted by a real gas since intermolecular forces of attraction come into play in real gas. Thus, $P_{ideal} > P_{real}$. So, $P_{real} = P_{ideal} - something$. From van der Waals, we know that that '$something$' is $an²/V²$. Thus, $P_{real} or P_{actual} = P_{ideal} - an²/V²$. So, $(P_{ideal} - an²/V²)(V) = nRT$. What is wrong here?

@user281837 You did the same mistake again. "real" pressure is not the "actual" pressure. Actual pressure is considered ideal pressure. The measured pressure is the real pressure. Real pressure means the pressure we observe. Now substitute the actual pressure(ideal pressure) in $PV=nRT$ and not the real pressure.

If you understand what I mean than please consider accepting and upvoting the answer.

How $V_{\text {actual}}=V_{\text {ideal}}$? In ideal gas, the molecules have zero volume (point mass).

@Apurvium The actual volume here doesn't mean the volume of the container. It means the volume of the container excluding the volume occupied by the molecules themselves. Thus the next line about real volume = measured volume.

Here I think you are considering a container and labelling it as that of containing a volume $V$ and then considering it to be ideal. Then you go on to derive the Van der waals equation and maybe you think that $V\mathrm{_{real}}$ would thus be equal to the $V\mathrm{_{ideal}} - nb$ as now that $V\mathrm{_{real}}$ shouldn’t have that volume accessible to the particles.(This is what I perceive might be your problem, otherwise you may see the other answers.)
This is actually incorrect.
Consider a container of volume $V$ given below:
`
Here thus we can see that the volume of the container cannot be ideal volume rather it is real volume as it also has the volume of the particles and the inaccessible region around them.
Therefore, $$V\mathrm{_{real}}=V\mathrm{_{ideal}}+nb$$ and thus
$$V\mathrm{_{ideal}}=V\mathrm{_{real}}-nb$$ where $V\mathrm{_{real}}$ is volume $V$ of the container.

Please see Which symbols are written in roman (upright) font and which are italicized? . I have corrected the issue for now

The Volume of a gas filled in a container is defined as the space in which the gas particles can move freely.
In case of Ideal gas where the gas particles have negligible Volume, the whole container is available for ideal gas particles to move and due to this reason $V$of ideal gas=Volume of container.
In the equation $PV=nRT$ the $V$ here is the Space available for the motion of the gas particles freely and not the necessarily the volume of the whole container, while P is the pressure exerted by the particles on the walls of the close container. And I assume that the source of your confusion is assuming that the defination of $V$ in the above equation (i.e. the volume of the gas) is the Volume of the whole container
The actual thing is that the value of the above $V$ comes out to be equal to the volume of container in case of an Ideal gas as explained above
Now in question or whenever, the Volume of gas is measured it is always the volume of the container given, for example,
The measured volume of a real gas = $V^c$ (say)
This means that the volume of the container in which real gas is present is $V^c$ that is $V_{real}=V^c$ thus the actual volume of this real gas ($V$) i.e. the volume available of the motion of its particle, which must be less than $V_{real}$ and the difference must be equal to the the Volume the real gas particles so,
$$V=V_{real}-V'$$
And thus the equation,
$$PV=nRT\Rightarrow (P_{real}+P')(V_{real}-V')=nRT$$
Where $P'$ and $V'$ are Pressure and Volume correction terms respectively.

I realise I am a bit late to answer this, but I found out that no one had given the perfect answer to this. So, here's my two cents.
First off, we should just leave out the words $V_{ideal}, V_{real}$. They're just confusing.
Next, the ideal gas equation taught to you isn't completely correct. The actual equation is:$$P_{ideal} * Free\, Space =nRT$$
where $Free\, space$ means space allowing molecular motion.
Now, when dealing with ideal gases, you must know that their volume can be neglected, and hence the $Free \, Space$ is simply the volume of the container $V$.
However, when dealing with real gases, they are not point objects and don't occupy zero space. So, the value of $Free \, Space$ becomes $V-nb$. Here, I'm assuming you know why and how we took $b$ to be $4*Molar\, Volume$, so I'm not going to show its derivation.
So, the Van der Waals equation then becomes $$(P+\frac{an^2}{V^2})(V-nb)=nRT$$

$V_\mathrm{real}$ is the (measurable) volume occupied by a real gas of given amount, at given p and T. $V_\mathrm{ideal}$ is the volume a real gas would have occupied at the same conditions, if it had been behaving like an ideal gas. For real and ideal pressure analogy, physical chemistry uses for the latter the term fugacity.

Instead of downvoting, maybe y'all could tell me whats wrong with my answer...

V-nb is not Free Space. b is not molar volume of molecules but its quadruple. So abandoning V_ideal and V_real in favour of Free Space is more confusing than keeping them.

Of course b is not molar volume, but I assumed OP knew that. Explaining why we put in $4*molar , volume$ would just unnecessarily lengthen the answer. But V-nb is not free space?? Could you explain why?

Well, it depends what you consider as free space. In sense of space allowing molecular motion it is. In sense of space not being occupied it is not. // Regardless of all that, using V_ideal and V_real, or p_ideal and p_real is clear.


text:
Titration of multi-basic acid

How would a solution made of $\pu{20 mL}$ of $\ce{NaOH}$, $\pu{10 mL}$ of $\ce{HCl}$, $\pu{10 mL}$ $\ce{NaH2PO4}$ and $\pu{5 mL}$ of $\ce{Na3PO4}$, all of them in a concentration of $\pu{10 mmol/L}$, behave? I already tried every way i can think of, but none give the solution of the $\mathrm{pH}$ of this system.

Ways I tried:

After reacting the $\ce{HCl}$ with $\ce{NaOH}$, I use the $\ce{NaOH}$ that i still got to react with $\ce{NaH2PO4}$, and to calculate the $[\ce{H+}]$ I used $[\ce{Na2HPO4}]$ + $\ce{[Na3PO4]}$, but i didnt got any answers.

I used this equation $\ce{NaH2PO4 + 2NaOH -> Na3PO4 + 2H2O}$ to find the concentration and then applied to the amphoteric equation, using $K_{a_2}$ and $K_{a_3}$.

$
\begin{align}
(n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\
(n_{\ce{HCl}})_i= \pu{0.1 mmol}\\
(n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\
(n_{\ce{Na3PO4}})_i= \pu{0.05 mmol}
\end{align}
$
At first the $\ce{NaOH}$ will react with $\ce{HCl}$ as per the following reaction:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} & + & \ce{H2O} \\
\pu{0.2 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\
\pu{0.1 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\
\end{array}
\end{align}
$$
Now, the remaining $\ce{NaOH}$ will react with $\ce{NaH2PO4}$ as per following:
$$
\begin{align}
\begin{array}{ccccc}
\ce{NaOH} & + & \ce{NaH2PO4} & \ce{->} & \ce{Na2HPO4} & + & \ce{H2O} \\
\pu{0.1 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\
\pu{0 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\
\end{array}
\end{align}
$$
Now the resulting solution is  a buffer of $\ce{HPO4^2-}$ and $\ce{PO4^3-}$. So the Henderson equation can be used.
$
[\ce{PO4^3-}]=\frac{0.05}{45} \pu{M}\\
[\ce{HPO4^2-}]=\frac{0.1}{45} \pu{M}\\
$
$$
\begin{align}
\begin{array}{ccc}
\mathrm{pH} & = & \mathrm{pK_{a_3}} + \log(\frac{[\ce{PO4^3-}]}{[\ce{HPO4^2-}]}) \\
\mathrm{pH} & = & 12.35 + \log(\frac{0.05}{0.1})\\
\mathrm{pH} & = & 12.35 + \log(\frac{1}{2})\\
\mathrm{pH} & = & 12.04 \\
\end{array}
\end{align}
$$


text:
Why is the triple phase diagram of water different from the general substances?
The fusion curve of triple phase diagram of water and fusion curve of triple phase diagram of  general substances is different. Why is it so?

Different? To me, they look pretty much the same.

No, they don't. Please look at the images that I have sent (the diagrams were given to me by my teacher).

Your diagrams are exaggerated a great deal. In fact, the curve in both cases is very nearly vertical. The different slope, though, is indeed a fact that requires an explanation. It has to do with the volume change.

There is no such thing as "general substance". There might be typical behaviour and unusual behaviour but "general substance" is bad terminology and a ridiculous overgeneralisation.

The phase diagram of water is peculiar for the negative slope of the line separating the solid, and the liquid for the representation of pressure in function of temperature:

(slightly modified, credit).
As indicated in the comment by @Ivan Neretin, this is due to volume change.  You may recall that $\pu{3.98 ^\circ{}C}$ is special for water because at any temperature above as well as below the density of water is less that at this extreme.  Thus the volume expansion of ice cracks rocks; and while ice skating (i.e., putting you body weight on a pointy small surface area, thus increasing the pressure), you may generate a thin film of molten water.  This applies for the transition between liquid water and hexagonal ice formed under about ambient pressure. (There are multiple other polymorphs of solid water (an entry) in addition to the one frequently seen.)
Much more frequently experienced is a positive slope for the solid-liquid phase boundary, e.g., for $\ce{CO2}$:

(from here).


text:
How can I get big bubbles when I electrolyse water with detergent?
I've made carbon electrodes by putting 12V through a couple of short 15Ω pencils, but when I try to make hydrogen and oxygen bubbles from soapy water, I get tiny bubbles, smaller than table salt.
If I move the electrodes closer together, it's the same tiny bubbles, but a bit faster.
These don't catch fire very convincingly, and my daughter is unimpressed.
How can I get larger bubbles? Different electrodes? Higher voltage? Just shove the AC power in straight from the wall? :)

Kind of related (if you're looking for an at-home demo), and a hit for folks of all ages: https://en.wikipedia.org/wiki/Glowing_pickle_demonstration

@ToddMinehardt I did the electric pickle demo a few times over the years. Fun, but the electrocuted pickle smell is unexpectedly foul. ;-|

This might be tricky. The demos I've seen mostly rely on collecting the hydrogen and oxygen, separately or together, and then bubbling them through a soap solution.
If you've got the soap in the electrolyte, you might have a hard time getting anything more than two piles of foam, one of which is slightly flammable. I doubt the hydrogen foam would even rise in air; the volume percentage of gas is too low.

I know I did this demo when I was a trainee teacher (before I came to my senses) and I don't recall doing anything more than shoving the electrodes into a thimble-sized plastic beaker. Maybe I've blanked out collecting tubes...

I found what presumably actually happened: https://edu.rsc.org/experiments/exploding-bubbles-of-hydrogen-and-oxygen/752.article
One very full beaker of water and electrolyte (bicarbonate of soda is apparently better than salt, lest you get chlorine instead of oxygen), a cork with two electrodes and a thin plastic tube, one end of which is level with the base of the cork to exclude as much air as possible, the other passing into a smaller plastic beaker of water with detergent or some other means of creating explosive form.
The combined gas is pre-mixed, and satisfactory loud.
I still have no memory of the primary beaker though: presumably the Thargons have been altering reality again, just to mess with me.


text:
Molecular Docking physiological effects
If I wanted to perform a Virtual Screening for one molecule and several ligands, take for example, the docking between TDO protein and several antidepressant drugs:
How would I be able to correlate the docking affinity with a physiological effect?
Is it necessary to perform clinical studies also?
I think this question could be extrapolated to many cases of MD, but let's take the molecules discussed as an example. TDO catalyzes the oxidation of L-tryptophan (L-Trp) to N-formyl-L-kynurenine, as the first and rate-limiting step of the kynurenine pathway. It is an enzyme, and as such a molecule with a high docking affinity would probably inhibit it´s action.
But what if the molecule was a membrane transporter? How could we determine if the binding to a ligand prevents the binding of the receptor, or on the contrary favors it?
PS: Not sure if this question belongs to this forum, if it´s more suitable in bioinformatics or another let me know.

You need to know of course if your molecular simulations are accurate (whether they reflect what happens in the organism), but you would also need a biochemical model of the target's function and how processes at the cellular/tissue/organ level relate to a response at the organism level.

Regarding your second example (about transporters) you might want to be more specific regarding what receptor you are referring to. Can you provide a more concrete example?

@BuckThorn thx a lot for your answer. I guess I could know if my simulations are accurate based on the scoring function?  But how do I connect this with the physiological effect, how can I get a biochemical model of the target´s function? is there a way of doing this in silico? I want to perform a MD for TrkB, this receptor interacts with BDNF, so performing MD doesn´t really achieve anything if I can´t explain whether this docking would inhibit the interaction between this two molecules, or, in the contrary, enhance it

You would need to compare experimental (in vitro presumably) data to simulations to see if your scoring functions are accurate. You might use relative scoring functions, either way experimental reference data would be important. I would not know how to set up a model to predict binding effects (I can imagine how it might work but it is not my specialty).

This article deals with the interaction network of TrkB: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3889250/

Regarding your question on whether binding of a small molecule interferes or enhances binding of TrkB to BDNF, you might start from a model with or without the small molecule and compare docking of BDNF to this.

@BuckThorn thx again for your answer, really appreciated. I think I´ll find a reasonable way to establish a connection between previous experimental data and my MD results. Regarding the TrkB molecular interaction network, however, I could not find a way to link this information and my study. Isn't that a little out of the reach of a virtual screening?

I know too little about your specific system to answer on how to link these but a true expert on this would be able to tell you (I'm guessing since you are running such sims you'd be able to).

@BuckThorn I think the interactome may influence the criteria to consider in order to choose an appropiate crystalline structure. That is, it´s conformation, maybe even pH.. I will take all that into consideration. Really appreciated, it was helpful


text:
Why is only mole fraction used in Henry's law and not some other measure of concentration?
Why is some other quantity like molality or molarity used?
I understand why mass percentage cannot be used as it is difficult to measure mass of gas...
Now is mole fraction used because solubility is directly proportional to the pressure and moles are directly proportional to the pressure?
Also, how do they measure the number of moles of a gas?

There are multiple ways to express Henry's law (which originate different but equivalent Henry's law constants), some of which involve mole fraction, others use concentrations in mol/L.

Okay thank you but then will the unit of Henry's law constant change?

Yes, it changes accordingly.


text:
Flavonoid Xanthine Oxydase Inhibitors vs Hydroxychloroquine
Regarding COVID-19 and hydroxychloroquine (HCQ), HCQ is known to increase pH in lysosomes by 1-2 orders of magnitude, and also results in lethal intracellular heme overload in parasites responsible for hemolytic diseases.  Question is, if circulating levels of xanthine oxidase (XO) are also known to be increased in hemolytic diseases, if one considers the natural flavonoids Kaempferol, quercetin, or luteolin, which are known xanthine oxidase (XO) inhibitors, antivirals, and anti-oxidants, what would be different about pharmacology of natural XO inhibitors and HCQ?
Is HCQ an XO inhibitor because it results in heme overload in parasites?  What about anti-oxidant properties of HCQ in terms of radical scavenging?  There is no information supporting the notion that HCQ is an antiviral.
Sticking with hemolytic diseases, for which the goal is overload of intracellular heme in parasites, what would happen if a human took high levels of an XO inhibitor that wasn't HCQ during malaria?
XO inhibitors usually result in an overall decrease in uric acid, essentially rendering them as a natural supplement for gout -- i.e., replacement to allopurinol, which has horrendous side effects.

HCQ does show antiviral properties in vitro, but, in vivo (in humans), when taken after an initial dose of 800mg for 4 days at 600 mg/day, it was not found to prevent COVID-19 in high-risk exposure settings.  Thus, it didn't stop people from getting SARS-CoV-2 infection followed by onset of symptoms/severity.
I suspect this is exactly what would happen with flavonoids, many of which have shown antiviral properties in vitro (laboratory settings).  The information on flavonoids and SARS-CoV (2003) is very sparse, but several flavonoids did inhibit SARS-CoV in vitro (see last paragraph below).  I do know there was a clinical trial approved in 2020 by NIH in which the investigators received an IND (From FDA)  to use diosmin as a supplement to treat patients with COVID-19 (i.e. hospitalized).  However, I don't know if was completed.  Diosmin is available OTC as a supplement, and used via prescription under the brand names Dafflon® and Vasculera® for chronic venous insufficiency (CVI), which is caused by venous hypertension (VH). VH, in turn, aggravates metabolic imbalances leading to a self perpetuating cycle of further metabolic changes, including venous acidosis. These changes promote further inflammation in vascular tissue leading to edema, skin damage and possible ulceration and deep vein thrombosis (DVT). Varicose veins and hemorrhoids are also part of the spectrum of CVI disorders.
Interestingly, it took 16 years before a group published the most comprehensive paper on in vitro inhibition if SARS-CoV (2003) by many flavonoids that were considered.


text:
Why did we introduce the concept of oxidation number in chemistry?
Isn't oxidation number a weird concept? It assumes complete transfer of an electron from a more EN to a less EN atom per covalent bond formed. This assumption isn't even true. We know that partial $δ^+$ and $δ^-$ charges develop on the atom. Why did we introduce such a weird concept? And what is it good for?

If assuming something weird helps us predict something better, we assume it. For example we also assume hybridization which in reality doesn't happen. But it helps us predict structures. Oxidation state likewise helps us predicting the oxidizing or reducing properties of a substance.

In short: if something looks weird, but works, then it isn't all that weird.

" It assumes complete transfer of an electron" - it doesn't assume anything.

https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number

Oxidation number is a deliberate approximation to make keeping track of electrons in molecules easy
You are quite right to say that oxidation number isn't "true". It is clearly a simplified idea and not a perfect representation of "reality" whatever that is.
But there is a point to it. That point is to enable chemists to keep track of the electron count in molecules. Electron distributions may be infinitely variable in bonds but electrons are discrete and keeping count of them is important. In transition metal complexes, for example, it is important to make sure all the bonding electrons are correctly accounted for. Many transition metals can have variable oxidation states with very different properties and the (crude) approximation of where those electrons are helps us classify those differences even when the metal-ligand bonds are covalent and the electron distributions are not clear-cut or ionic.
So there is no intent to correctly represent the distribution of charge in bonds. But pretending the oxidation number is an integer helps keep track of things and is often very useful at explaining properties.

Another example is from biochemistry: It would be very difficult to count all valence electrons in a protein just to figure out what is going on when a disulfide forms. Instead, we say that the oxidation number of the sulfur atom in cysteine changes, and all other oxidation numbers are untouched. That tells us that there has to be an oxidizing agent that gets reduced to accept the electrons released during the disulfide formation.


text:
Effect of concentration on voltage from the perspective of energy
As I understand it, the voltage in an electrochemical cell is due to the electrons transferred having a decrease in electrical energy as they go from one species to another.
I assume that this change in potential is due to the electrostatic attractions and repulsions of the protons and other electrons of the species reacting with each other. However, I fail to see how the concentration of species would change the potential difference.
For example, I feel like the change in electric potential for an electron moving from $\ce{Zn}$ to $\ce{Cu^2+}$ in the reaction given below should not depend on the amounts of $\ce{Zn^2+}$ and $\ce{Cu^2+}$ present, just on the atomic level.
$$\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$$
Could someone please explain why the voltage is not constant, from a conceptual standpoint? I considered that it may be that $\ce{Zn^2+}$ attracts electrons as well, but the solution remains electrically neutral due to the salt bridge, so I am not sure if that is the reason.

Your conceptual notion on this is simply incorrect. For an extremely thorough discussion of the findamentals, see this paper: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell, which is what you specifically ask about, is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior.

Thank you, this was exactly what I was looking for.


text:
Is ionisation an equilibrium process?
I read in a page that ionisation is an irreversible process. But, as equilibrium is always reversible would that mean equilibrium does not involve ionisation process.
And than dissociation is a reversible process therefore only dissociation process is seen in ionic equilibrium.
Now if the above statement is correct would that mean only covalent compounds take part in ionic equilibrium?

A ionic compound like NaCl(s) performs dissociatiating/associating dissolution equilibrium. A covalent compound like acetic acid performs ionizating/recombining acido-basic equilibrium. Both reversible in common and thermodynamic sense.

An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas.
Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (while temperature is constant) until eventually the boiling point is encountered. At this point the pressure is equal to the vapor pressure (at that temperature) and the gas and liquid can coexist (all of this by the way is simpler to illustrate with a T-p phase diagram). Further compression condenses more gas until all of the substance is liquid. Below the boiling point pressure none of the substance is liquid, and above that pressure none is gas.
Next consider solubilization of a substance in a fixed volume of liquid (with T,p constant). One could traverse the solubility diagram by changing the temperature, but it is for the present discussion simpler to consider changing the concentration. At very low concentrations (small amounts of substance) all of the substance solubilizes. Exactly  none (yes, none) is present as the solid. But as you increase the amount of added substance a concentration is reached at which solid remains. That exact point is the solubility limit of the substance at that temperature. It can be thought of as equivalent to the boiling point pressure (coexistence point) of a gas and liquid at a given temperature. If you increase the amount of added substance at that point, no more substance will solubilize (exactly none).
So when a salt is added below its solubility concentration limit no solid remains, all of the substance is in the liquid, and there is no equilibrium only in the sense that there cannot be solid below the solubility limit.

This is also why the equilibrium equation for solubilization of a substance is written as
$$K_{\mathrm{sp}} = \frac{a(\textrm{soln})}{a(\textrm{s})}= a(\textrm{soln})$$
where a's are activities (for small concentrations, the activity of the solubilized substance is equal to concentration ). We ignore the activity of the solid (it is constant and assumed equal to 1) and above the solubility concentration limit the activity of the solubilized substance is also constant (equal to the solubility product).

For most salts the equilibrium constant for ionization is very large and thus the reaction practically proceeds only in the forward direction and so we generally say it is irreversible.
But in reality no chemical reaction is irreversible and some backward reaction always takes place.
So, in a strict sense ionization is a reversible equilibrium process but for practical  we assume it to be irreversible.

Note that in context of thermodynamics, reversibility versus irreversibility is not both way versus one way processes. The process occuring in system is reversible, if returning the system  surrounding to its initial state returns the system to the initial state as well. This happens, if during all the procedure we can consider everything being all the time in equilibrium. In reality, no process is fully reversible in this sense, as reaching equilibrium takes infinity time.

@Poutnik I understand that but I didn't mean reversible in thermodynamic terms but in the common terms. As per the OP's question it seems like he is not referring to reversible and irreversible from thermodynamic terms but with regard to one way or two way process. Or in simple words I believe OP is asking about the extent of reaction when referring to reversible and irreversible.

well, equilibrium involves thermodynamics.

@Poutnik I do accept that the terms I have used are a little ambiguous. But I wanted to convey the answer to the OP in a layman's manner. This is a very standard question from high school or secondary school where they still don't understand the thermodynamic meaning of reversible and irreversible but use the term for the extent of reaction.

Ok. keep it. :-)

Thank you guys...so let me just sum it up..all reactions are reversible(In a strict sense as stated) and therefore can be considered in equilibrium.(and yes I am a high school student)


text:
Hydrolysis of nitriles: Amide vs Carboxylic acid
I was recently studying the preparation of amides,and one of the methods listed was acid hydrolysis of a nitrile. Up until now I thought acid hydrolysis of nitriles yielded carboxylic acids. How are the methods to get amides and acids different?
I can view the amide as sort of an intermediary stage while getting to the acid,but how do you stop the reaction at amide? Do you use stoichiometric quantities of the acid? This is puzzling me.

I believe to prepare amides from nitrile you need to do an alkaline hydrolysis and not an acidic one.

@NisargBhavsar I believe alkaline hydrolysis of nitriles leads to a carboxylate ion and ammonia. Could you check? Thanks for answering!

Boron trifluoride acetic acid complex is a reliable reagent for the conversion of nitrile to primary amide.

Hydrolysis of nitrile is pH dependent. It will lead to formation of different products based on pH.
Theoretically, nitrile hydrolyze to form amide and then ammonium salt.
$$\ce{RCN ->[H2O] RCONH2 ->[H2O] RCOONH4}$$
But in practice, the reaction between nitriles and water is so slow that it is considered negligible. The nitrile is instead heated with either a dilute acid such as dilute hydrochloric acid, or with an alkali such as sodium hydroxide solution to accelerate the hydrolysis reaction. But this pH dependent reaction yields different products.

Acid hydrolysis: yields carboxylic acid

$$\ce{RCN + 2H2O + HCl -> RCOOH + NH4Cl}$$

Alkaline hydrolysis: yields sodium carboxylate salt

$$\ce{RCN + H2O + NaOH-> RCOONa + NH3}$$
To obtain amide, hydrolysis is done selectively and in controlled manner. It can be done in both acidic and alkaline medium. In acidic medium, nitriles are converted into amide in a single step process using TFA-$\ce{H2SO4}$ mixture as reagent.

In alkaline medium, it can done using alkaline solution of hydrogen peroxide (see @Light Yagami's answer).
Reference

Chemguide
Facile and Highly Selective Conversion of Nitriles to Amides via Indirect Acid-Catalyzed Hydration Using TFA or AcOH−H2SO4, Jarugu Narasimha Moorthy and Nidhi Singhal, The Journal of Organic Chemistry 2005 70 (5), 1926-1929
DOI: 10.1021/jo048240a

So,how do you get the amide?

@VenkaatBalaje see updated answer.

@VenkaatBalaje if this answer helps, feel free to accept the answer by clicking the checkbox.

Yes, you are correct. For converting a nitrile to amide, we do controlled hydrolysis.
So, first amide is obtained and then ammonium salt of the carboxylic acid is obtained.
So, the reaction would be $$\ce{RCN}+2 \ce{H_2O_2}+\ce{OH^{-1}} \ce{->} \ce{RCONH_2} +\ce{O_2}+\ce{H_2O}$$
Which is done at $\mathrm{pH}$ range of 7-8.
If you increase or decrease the $\mathrm{pH}$ more than this, you will get the acid.
Additional Reading:

Mcisaac, J. E.; Ball, R. E.; Behrman, E. J. Mechanism of the base-catalyzed conversion of nitriles to amides by hydrogen peroxide. J. Org. Chem. 1971, 36 (20), 3048–3050. DOI: 10.1021/jo00819a034.

The Hydrolysis of Nitriles: Chem Libretexts

I understand the hydrolysis of nitriles using alkaline H2O2,the question was about the acid hydrolysis of nitriles.Thanks for answering! BTW,are you the same Light Yagami on AoPS?


text:
How to convert cell diagram to a chemical equation?
How would one convert on the right side of the cell diagram:
$$\ce{Zn(s) | Zn^2+(aq) || MnO4-(aq), Mn^2+(aq), H+(aq) | Pt(s)}$$
$$
\begin{align}
\ce{Zn(s) &-> Zn^2+(aq)} \tag{ox}\\
\ce{MnO4-(aq) &-> Mn^2+(aq)}\tag{red}
\end{align}
$$
Obviously, one would then balance and combine the half-equations. I just need to know if the unbalanced half-equations are correct.

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.
See also upright vs italic and Math SE MathJax tutorial.

The reduction «or something like that» does not make sense.  Most often classes introducing redox chemistry  refer to the reduction of permanganate $\ce{MnO^-_4}$ into $\ce{Mn^{2+}}$ if the reaction is conducted in acidic media.  Reductions in neuter, and basic media yield other products.

Assuming you meant $\ce{MnO^-4}$ and not $\ce{MnO^{4-}}$, your reduction half reaction is incorrect. Because according to your equation $\ce{Mn^{2+}}$ is being oxidised to $\ce{MnO^-4}$ rather than being reduced.
The correct half cell reactions would be:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{MnO^-4 + 8H+ + 5e- -> Mn^{2+} + 4H2O}\\
$$
And so the overall balanced reaction would come out to be:
$$\ce{5Zn + 2MnO^-4 + 16H+ -> 5Zn^{2+} + 2Mn^{2+} + 8H2O}$$

It was typed wrong on the tutorial sheet and I did not spot the error before submitting the question. The corrected cell diagram makes far more sense; before it was conflicting against cell diagram convention.

To clarify: The H+ on the RHS of the cell diagram simply means the reaction is being performed under acidic conditions?

@jseissmann Yes, it shows that the medium is acidic.

@jseissmann For a reaction equation to be balanced, the number of atoms on the right, and on the left hand side must be ... equal. Simultaneously, the sum of the charges on the left must equate the sum of the charges on the right hand side of the arrow, too. The reduction, as written by you: $\ce{ MnO^-4 -> Mn^{2+} }$. Left hand side: four atoms of oxygen, right hand side: none; thus not balanced.

@NisargBhavsar This is true, I forgot to set the @ sign; the comment was to the OP.

@Buttonwood I am aware they are unbalanced - I simply wanted to check that my translation of the cell diagram to the unbalance half-equations was correct as the H+ on the RHS threw me. I can balance half-equations just fine and therefore didn't need full answers on that part


text:
Why are Andrews' isotherms divided into three fragments at low temperature and low pressure?
In college, I studied that for an ideal gas $PV = nRT$ , that is, at constant temperature and constant number of moles, $P \propto \frac {1}{V}$. Thus, graph between $P$ and $V$ is a rectangular hyperbola when $dT =0$ and $dn=0$. But in my textbook there is this topic called 'Andrews' isotherms' wherein they have given some $P$ vs $V$ curves for gases and at low temperatures and low pressures the graphs seem to be divided into three fragments. How is this possible? Shouldn't the curve be a rectangular hyperbola? How is it possible? Shouldn't ideal gases obey $PV = constant $ at all temperatures and pressure. That is literally the definition of ideal gas : A gas that obeys ideal gas equation at all temperatures and pressure.
I also studied that real gases show deviation from ideal gases. But when we draw $P$ vs $V$ curves of a real gas, they deviate from the $P$ vs $V$ curves of ideal gases slightly at high pressures and low temperatures. They almost remain rectagular hyperbolas. But the $P$ vs $V$ curves of real gases don't literally break into three fragements like Andrews' isotherms. The curves in Andrews' isotherms don't even come close to rectangular hyperbolas. What type of curve is being shown in Andrews' isotherms? How are these curves disobeying Boyle's law to such a great extent, not even coming close to a rectangular hyperbola?

Is the figure in this Physics SE question what you mean? If so, the answer is that these fragmented curves are cases where the the substance is transitioning from a liquid to a gas.

It's been a long time since I studied this.  But Andrews' isotherms are not referring to an ideal gas the curves a real system that, at its critical point, the homogeneous substance is exactly balanced with both the liquid and gas form are at the same temperature, pressure and density.
You also might want to read Boltzmann's explanation of the Joule-Thomson effect along with Gibbs' and Maxell's explanation of the {U, S, V} minima where Gibbs developed his famous Gibbs free energy which was the missing potential function in real thermodynamic systems, not ideal analysis of those system.  The U,S,V curve is much more important than analyzing Andrews' isotherm by the ideal gas law.
In fact, Maxwell, the most famous physicist in the world realized Gibbs was correct that the potential function minima that determines the direction of a spontaneous reaction is Gibbs free energy, but it was already after Maxwell published his own book erroneously stating that it must be a minimum in the potential function of internal energy, U that determines the spontaneous direction of a reaction.


text:
Meaning of the term "Activation of alkene"
While reading organic texts, I have come across authors referring to "Activation of alkene" what does that mean !? Does it mean to include the alkene in resonance or what else exactly ?
I saw this notation, while reading about Heck reaction:

The palladium-catalyzed C-C coupling between aryl halides or vinyl halides and activated alkenes in the presence of a base is referred as the "Heck Reaction"

I also observed this while reading about nucleophilic conjugate addition:

Ordinary nucleophilic additions or 1,2-nucleophilic additions deal mostly with additions to carbonyl compounds. Simple alkene compounds do not show 1,2- reactivity due to lack of polarity, unless the alkene is activated with special substituents.

Activation of an alkene just means that the double bond has a higher electron density than that of a normal isolated double bond. That is, the electron density in the double bond is greater than the one observed in ethene $\ce{CH2=CH2}$.
Activation, in organic chemistry, generally means the compound displays a greater nucleophilic nature than it normally should due to increased electron density. For example, $\ce{-OCH3}$ group activates benzene when it forms toluidine. This makes the benzene ring more electron rich and so make it easier to react in nucleophilic reactions.

"Activation of an alkene just means that the double bond has a higher electron density" Does it really? One C atom could have partial negative and the other positive - still more reactive; or two positive - enables it to act as electrophile - why exactly nucleophilic character would be only important?

The term was used to refer to an $\alpha, \beta$ unsaturated carbonyl in my notes for the Michael reaction. Doesn't that mean it's an electron deficient bond?

Also, the wiki on the Michael reaction has the phrase 'The R" substituent on the activated alkene'. Seems to imply it's electron-deficiency that's meant.

@SafdarFaisal: please respond if you're available.


text:
Stabilization of multiple bonds through inductive effects
In the conversion of m-chloroanisole to m-anisidine using $\ce{NaNH2}$ and liquid $\ce{NH3}$, apparently the reason the $\ce{NH2-}$ goes for the meta- position instead of the para position is that, in the reaction mechanism for the conversion to m-anisidine, the benzyne intermediate involved has the triple bond closer to the $\ce{-OCH3}$ group than in the reaction to the para product, thus stabilizing the intermediate due to the stronger inductive effects of $\ce{-OCH3}$.
I know the triple bond in benzyne isn't exactly your typical triple bond, but is there always some(if negligible) stabilization of all multiple bonds through electron-withdrawing inductive effects?

Use $\ce{}$ for chemical compounds, italicised formulas are typographically incorrect.

https://chemistry.stackexchange.com/questions/150830/which-sides-hydrogen-is-removed-during-benzyne-formation-in-substituted-haloare

Since electron-withdrawing groups (EWGs) lower orbital energy (we can rationalise this in terms of reduced electron density leading to reduced shielding), they will stabilise multiple bonds - you see this in the Diels-Alder reaction, where EWGs on the dienophile reduce the energy of its LUMO and so accelerate the reaction by reducing the gap between the dienophile LUMO and diene HOMO (conversely, electron-donating groups on the diene destabilise its HOMO by the same logic and so also speed up the reaction).

I don't know if I understand your reasoning here. EWG's lower the orbital energy of the multiple bond LUMO, but what does that say about the total energy of the molecule? OP is asking why one isomer is more stable than the other, and I don't think you can invoke LUMO lowering as an explanation.

I should have been clearer - EWGs lower the orbital energy of all orbitals (including the LUMO), so by lowering energy of the occupied orbitals they will have a stabilising effect (the stabilisation of the LUMO is simply a specific example relevant to the Diels-Alder reaction). OP's question was specifically whether electron-withdrawing effects stabilise multiple bonds.


text:
Effect of stirring and shaking on chemical kinetics
Why don't we take the effects of stirring and shaking on the rates of chemical reactions? While studying chemical kinetics, I've seen that the rates depend on the concentrations of various species, temperature, pressure, etc. But I've never seen an equation taking into account the mechanical stirring, shaking and geometry of the beaker, presumably because these things are hard to experimentally measure and form a physical model which incorporates these things. For example, if the iodine clock reaction changes its rates due to stirring, every time I do the reaction, I shall get different times for colour changes. Then why do chemical kinetics equations make sense at all?

Stirring and shaking provide some kinetic energy to the system. It increases slightly its temperature and the reaction rate. Very slightly.

@Maurice I don't think that slight increase in temperature changes anything. It's probably that stirring mixes the reactants well, and prevents localized concentration deficits.

@Maurice we can keep the reaction setup on a thermostat in keep the temperature constant. We would still see a rate change.

For a homogeneous reaction, as the mixing is increased, the rate will approach the theoretical rate. For aqueous reactions of fairly dilute reactants that happen slowly enough to measure kinetics, a “normal” amount of stirring is sufficient to be very close to that theoretical value and differences in rates between mildly different experimental setups are negligible

For fast reactions there are special setups to achieve mixing that is rapid enough to eliminate mixing as a kinetic factor

Apparently, shaken or stirred ethanolic solutions can be differentiated by bond. https://en.wikipedia.org/wiki/Shaken,_not_stirred


text:
How many stereoisomers are present in octahedral complexes of type MA3B2C?
Does $\ce{MA3B2C}$ type have optical isomerism or just geometric isomerism? I was told that it has three isomers, two cis and one trans. Are the two cis isomers optically active?

Welcome to Chemistry! What have you attempted to do with regards to the answer to this question? We need to make sure we aren't doing your homework for you.

well i worked out that for the two cis isomers, 'b' could be trans to 'a' or 'c' , and they would be different because of the difference in arrangement, but what I cant figure out is if they're optical isomers of each other, because when I take one of the cis isomers which is 'b' being trans to 'a'  and draw its mirror image and rotate it I'm not able to bring 'b' trans to 'c' , but its mirror image does seem to be non-superimposable

Don't draw mirror images, better look for a symmetry plane in the molecule, because it is there.


text:
Do acetals and hemiacetals give 2,4 DNP test?
Do acetals and hemiacetals give positive 2,4 DNP test?
The reagent for the reaction is described as:

An aqueous solution of 2,4-dinitrophenyl hydrazine (DNP) is known as Brady's reagent

I recently learnt that acetal don't give positive tollens test whereas hemiacetals did.
Following from the same conclusion I believe that since its an aqueous solution and seems alkaline aswell so according to me acetals should give negative 2,4 DNP Test whereas hemiacetals should give positive.
However I fail to find any literature on this, are my conclusions correct or are there any twists to it!?

The standard mixture for a 2,4-DNP test contains 2,4-DNP and conc. Sulfuric Acid in EtOH/water here. This is sufficient to hydrolyse most acetals and hemi-acetals so they will give a positive test. Cyclic acetals such as dioxolanes are more stable and may not.


text:
Performing XPS on Pt–Ru/C nanoparticles
From what I understand, to get clear peaks, $\ce{Pt}$ is analyzed at $\ce{Pt}$ 4f; and $\ce{Ru}$ at $\ce{Pt}$ 3d.
However, if i'm not mistaken, charge correction is needed.
But $\ce{Ru}$ 3d(280/284 eV) overlaps with $\ce{C}$ 1s(284 eV), therefore I cannot use $\ce{C}$ 1s for charge correction. Hence, I mix in some $\ce{Au}$ powder and use $\ce{Au}$ 4f(84 eV) for charge correction and then analyze $\ce{Ru}$ at $\ce{Ru}$ 3p(461eV).
But now, $\ce{Au}$ 5p(74 eV) overlaps with $\ce{Au}$ 4f(74 eV) so I cannot analyze at $\ce{Pt}$ 4f
and I try to analyze at $\ce{Pt}$ 4p(519 eV).
However, my problem is that I'm not getting clear enough peaks. The obvious thing I can think to do is increase the number of scans or lengthen the dwell time.
But I was wondering if there are other solutions other people usually do that I might be missing.
I do see some papers that try to deconvolute $\ce{Pt}$ and $\ce{C}$ in the same region, but my professor says that this is an incorrect method because it is impossible to consider the correct charge correction that way.
Does anyone have any suggestion on how I should proceed?

Despite resources like this, especially the numbers in parentheses (e.g., in Ru3d(280/284)) are difficult to understand. You may ease the reader's access to your problem if the relevant parts of the language of your special question is explained briefly.  Possibly an image / a diagram may illustrate your problem further.

However, my problem is that I'm not getting clear enough peaks. The obvious thing I can think of to do is increase the number of scans or lengthen the dwell time.
But I was wondering if there are other solutions other people usually
do that I might be missing. I do see some papers that try to
deconvolute Ru and C in the same region but my professor says that
this is an incorrect method because it is impossible to consider the
correct charge correction that way.

Do you have an Atlas of XPS spectrum of elements? Check the instrument's manual. I do recall CASA used to have element by element spectrum.
Thermo Fisher has one periodic XPS spectrum here. If gold is causing problems, choose another metal whose peaks do not overlap with your sample.
The only solution for weak signals is that you collect the data for a longer time (ensemble averaging).
Don't try deconvolution in XPS especially when it is peak fitting. Your mentor is right. It is not reliable. You can fit as many peaks as you like. There must be a rationale for fitting peaks in an analytical signal (you exactly know peak shapes, peak model, baseline etc.)


text:
Dependance of heat capacity at constant volume on temperature for solids and liquids
In thermodynamics we have an equation for the change in internal energy:
$$\mathrm dU = nC_V\,\mathrm dT\tag{1}$$
$$\int{\mathrm dU} = \int{nC_V\,\mathrm dT}\tag{2}$$
Since $C_V$ is independent of temperature, for an ideal gas
$$ΔU = nC_VΔT\tag{3}$$
Why is $C_V$ independent of temperature for an ideal gas? On the other hand, in case of solids and liquids, $C_V$ may or may not depend on temperature. Why?

Ideal gases are made of moving points, without any volume, and without any internal degree of freedom like vibration and rotation. So $C_v$ is only due to the three degree of freedom called translations, and this number does not depend on temperature.

In engineering, we consider the heat capacity of a real gas in the ideal gas limit to be a function of temperature, and we call this the ideal gas heat capacity.

An ideal gas is defined as one of non-interacting point particles and its internal energy depends on temperature alone, thus the slope of a plot of $U$ vs $T$ is a straight line with slope $C_V$.
A real gas (or liquid or solid) is made up of molecules and these have internal energy levels due to a molecule's ability to rotate and vibrate. At low temperatures these energy levels are mostly not excited (or occupied) and so the internal energy is low, however, as the temperature is increased more levels become excited and the internal energy increases. These levels are not equally spaced from one another (due to the nature of the molecule) and so the internal energy initially rises slowly with temperature then more rapidly, i.e. initially the slope of $U$ vs $T$ is small but gradually increases until at v high temperature the slope becomes effectively constant. Thus as the heat capacity is the slope of $U$ vs $T$ it is initially small and increases to a constant value at high temperature.
(At very, very low temperatures all molecules are in their lowest energy levels and a slight increase in temperature may not be enough to excite any other levels so the heat capacity remains at zero (from zero K up to a few degrees K) then increases as more levels become excited as the temperature increases)

The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy.
The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, vibrational and rotational energy contribution depends on ratio of mean energy per degree of freedom $E=\frac{1}{2} \cdot k_\mathrm{B} \cdot T$ versus the energy of quantization steps.
The capacity raises as rotation energy contribution kicks in, and later vibration energy as well, and finally electronic energy, as their quantization steps grows in this order.
There is a similar temperature dependent effect for liquids and solids, where quantized vibration is essentional, especially for solids.

The classical model of ideal gas does not take into account quantum effects. By warming up of an ideal gas, the mechanical energy gets evenly distributed between energy of linear motion, bond vibration and molecular rotation, proportionally to absolute temperature.
Quantum effects, taking part in behaviour of real gases, liquids and solids affects energy distribution. Vibration, rotation or electron energy are allowed by laws of quantum mechanics to have just some energy levels.
Imagine you throw small stones upwards to a steep hill. Stronger you throw it, the higher it would stay. This is analogous to an ideal gas.
Now you try to throw such stones to open windows of a tall building. If you make weak throwing, it will not flow even to the nearest window and will fall back. Stronger and stronger throwing ( = higher temperature ) would eventually cause some stones would reach at least the 2nd floor. Later even 3rd one etc.  This is like quantization of possible energies for real gases and condensed phases.

Can you please explain in simple language? I am just a Grade 12 student and I do not have higher knowledge about science.

Well, asking about temperature dependency of heat capacity does not match your statement much. :-)  I will try...


text:
Reason for enthalpy being non-zero at constant temperature

Suppose a reaction is occuring at constant temperature. Then $\mathrm dH=nC_{p}\mathrm dT = nC_{p} × 0=0$.

Please explain what the mistake is that I have done. Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change. But according to the equation $\mathrm dH=nC_{p}\mathrm dT$, if temperature $T$ is constant then $\mathrm dT =0$ and $\mathrm dH = 0$. But this is clearly not true. There are reactions with constant temperature and $\mathrm dH≠0$. Please explain what is wrong?

dH = nCpdT is true only at constant pressure.. dH = dU - dW

Please refrain from using MathJax in titles, we generally discourage it here.

@SafdarFaisal $\mathrm dH=nC_{p}\mathrm dT$ is valid for all cases, and not just at constant pressure. Of course, this is with the assumption that Cp is calculated for an ideal gas. check here: https://physics.stackexchange.com/questions/203605/delta-h-c-p-delta-t

You are aware that enthalpy is a function not only of temperature and pressure, but also, for a multicomponent system, the amounts of the various components present (which change during a chemical reaction), right?

The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats up by an amount dT and then is returned to the set T. If you were able to measure these infinitely small changes and also to know the instantaneous Cp (which changes as the reaction composition changes), you could sum them up and get a non-zero value for the enthalpy change. Obviously, that's not a practical approach.
Instead, to determine the enthalpy change experimentally, it is easiest to perform the reaction within an insulated apparatus such that the temperature is not constant and the change in temperature can be used to determine the enthalpy change.

For a multicomponent reactive system you would rather write the enthalpy differential as
$$
d(nH) = \left(\frac{\partial (nH)}{\partial T}\right)_{P,n_i}dT + \left(\frac{\partial (nH)}{\partial P}\right)_{T,n_i} dP + \sum_{i = 1}^N \left(\frac{\partial (nH)}{\partial n_i}\right)_{P,T,n_{j\neq i}}dn_i
$$
Where $H$ is the molar enthalpy. At constant temperature and pressure, $dP = 0$ and $dT = 0$, therefore
$$
d(nH) = \sum_{i = 1}^N \left(\frac{\partial (nH)}{\partial n_i}\right)_{P,T,n_{j\neq i}}dn_i
$$
Note that for a chemical reaction $dn_i$ for each component can be related to the others by stoichiometry. For example, for a simple $\ce{A -> B}$ reaction, $-dn_A = dn_B = d\xi$, where $\xi$ is the extent of reaction. Thus, for this example,
$$
d(nH) = \left(\frac{\partial (nH)}{\partial n_A}\right)_{P,T,n_B}dn_A + \left(\frac{\partial (nH)}{\partial n_B}\right)_{P,T,n_A}dn_B = \left[-\left(\frac{\partial (nH)}{\partial n_A}\right)_{P,T,n_B}+\left(\frac{\partial (nH)}{\partial n_B}\right)_{P,T,n_A}\right]d\xi
$$
Expressing this as a total derivative,
$$
\frac{d(nH)}{d\xi} = -\left(\frac{\partial (nH)}{\partial n_A}\right)_{P,T,n_B}+\left(\frac{\partial (nH)}{\partial n_B}\right)_{P,T,n_A} = \text{heat of reaction}
$$
Note that the RHS is the difference between the partial molar enthalpy of each component. In ideal solutions, these correspond to the molar enthalpies of the components individually. In other words, we get the expected formula that the heat of reaction is the difference between enthalpy of products and enthalpy of reactants (weighed by stoichiometric coefficients).
So your heat of reaction is still there, you have a mixture with variable composition (a chemical reaction) therefore the enthalpy differential has an extra term related to the chemical composition.

[OP] But according to the equation $dH=nC_{p}dT$, if temperature $T$ is constant then $dT =0$ and $dH =0$.

Even if this equation would describe a system completely, a constant temperature does not imply that $dH$ is zero. Instead, you can have a temperature bath of infinite size (n approaches infinity) so that the average temperature does not increase. Of course, there would be temperature fluctuations while the reaction goes on, even if it is a glacially slow reaction.

[OP] Clearly, there are many reactions which take place at constant temperature and have a definite enthalpy change.

From the discussion above, it might be clear that it is not so easy to run an exothermic reaction at constant temperature. Instead of having a huge water bath, you can also buy an instrument for isothermal calorimetry (ITC) that monitors the temperature and cools or heats the reaction mixture to keep it at constant temperature. In this instrument, the amount of heat tranfered is logged, and that is how you measure the enthalpy change. This is distinct from measuring with a traditional calorimeter, where you monitor the temperature change and estimate the heat capacity of the calorimeter. Estimating the heat capacity is difficult because it is affected by the reactants and products you are measuring.
What this means is that there are multiple sources and sinks of thermal energy. If you are successful in balancing them perfectly, it will be an isothermal process. Often in thermodynamics, you are making arguments that rely on an ideal system. The real systems might approach that ideal, so the observation will also approach that of the ideal system.

OP might have meant in a theoretical sense, not practically measuring.

@SafdarFaisal Yes, they might have. Maybe they will be able to choose their favorite answer (or most useful one) once there are some more. Often, theory and measurement are related to each other, though.

This really doesn't address the main question of why the described application of the equation is invalid.

@Andrew I edited my answer, trying to address your concern while attempting to keep the approach different from your answer, which is a good one.


text:
Ought I add salt to water bottles, if I'm putting the bottles in front of a fan to improvise an air conditioner?

Is this advice correct? Are they alluding to Freezing Point Depression? Adding salt to tap water lowers the Crystalloid's freezing point.

Pour 3 tbsp (51 g) of salt into each of your 3 plastic water bottles. Use disposable plastic bottles for the easiest set-up and clean up. Pour 3 tablespoons (51 g) of table salt per bottle. Put the caps back on and shake the bottles to thoroughly mix the salt.

I live in a studio flat. I know of the Law of Thermodynamics, also here  – but my plan is different. Where I live, night time is way cooler than day time. I'll freeze the bottles over night, because even if my freezer heats up more than usual, my night room temperature will still be cool enough. During day time when it's hot, I'll place the frozen bottles  in front of my fan to air condition my studio flat. Does this work? I read Quora.

It makes sense physically, but the question is if you would have enough bottles for that for sufficient effect.
Twice water  heat capacity than ice about compensate lower amount of water, so it can be considered about the same.
The major effect is increased heat transfer keeping it colder for most of the time, as melting, absorbing the most of heat, would occur at lower temperature. Also, liquid provides better heat transfer due convection.
The strongest effect would be using as much salt for solution to still get frozen in the freezer.
About 330 g / 1 L water makes solution with freezing point near $\pu{-20^{\circ}C}$.
50 g / 1.5L PET would be like $\pu{-2^{\circ}C}$.
If we take the heat for ice melting as 100%, water warming up by $\pu{12^{\circ}C}$ takes  15% and ice warming up by $\pu{16^{\circ}C}$ takes just 10%. Shifting melting point by $\pu{2^{\circ}C}$ causes increasing total water heat capacity by 1.25%.