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Can water be generated artificially
As we all know water that in many of the countries or states we find a lot of water scarcity. I wonder if water could be produced by artificial technology. I searched about this in google but I find other techniques like recycling of water,and also about bringing out the penetrated water etc. But I could not find such machine or technology that is able to produce or generate water. If the idea of generating water artificially successful then we could be able to overcome the water scarcity . I wanted to know whether this kind of technology is already invented or not.

Creating water artificially is super easy; that's not the question. The question is "where is the money".

Burn natural gas, maybe?

Not really generated, but in effect the same: There is an ongoing project with already built prototypes of capacity hundreds L of / day, working with solar energy, based on condensation of air humidity on cold condensers. They are aimed on hot arid countries, where is plenty of solar energy, but water is scarce.

P.S.: Solar Air Water Earth Resource (S.A.W.E.R.)

Does this answer your question? Can you artificially create molecules with a machine?

2/3 of the world is covered with water. water isn't scarce. Potable water is scarce. So why "create" it from scratch when you can desalinate it to make it potable?

@matt_black globality versus locality. Where is no water it cannot be desalinated.

@Poutnik How many countries with notable populations are entirely desert with no nearby access to water? And, if there are any, what is the relative cost of building a pipeline to the sea versus water vapour capture or (as the question seems to want) the cost of shipping hydrogen and oxygen to the location so water can be created?

@matt_black Producing water from transported hydrogen or building a pipeline across the Sahara desert are not enterprises I would like to invent in.

@Poutnik Precisely. But there are already pipelines across the sahara, though they transmit oil and gas, not water. On the other hand almost nobody lives in the locations where the oil is produced so that is pretty irrelevant. If people wanted to live in the middle of a desert with no local access to water (and they never do), then by far the cheapest solution would be a pipeline plus desalination.

@matt_black There are many living locations across the world where water sources exist, but are limited, and where is not enough people to put a pipeline from the very far, or money for it. The pipeline end water quality could be very questionable. Additionally, there may be not an external source of electricity. A supplementary and autonomous water source, e.g. for agriculture activities, or even drinking, can be handy.

Let us continue this discussion in chat.

I see others go to details and variations. The answer is easy. Yes one can generate water by any combustion. So you can se that is not viable. Except if one burn hydrogen. But the latter must be made. This requires energy. The only way would be to get that energy from the sun. This is ongoing research, indeed. However the day it will be available, in the form of hydrogen, it will be firstly used to replace fossil. That can be also used for desalination, rather than direct water production.

@Alchimista Sure, when sea water is available, desalination is preferred.

OT but not much. Just not at country scale. Niche applications do exist. For instance fuel cells are a real thing. They provide a controlled oxidisation of oxygen leading to both electricity and, after addition of salts, drinkable water. Except for the scale this is what most answers your question. Used in space, for instance.

Put some $O_{2}$ and $H_{2}$ in a chamber, give it some activation energy and boom, you got some water.
However, mixing a flammable gas with a strong oxidizer can make things a little interesting.
Plus we would be manufacturing large amounts, making the process very dangerous (i.e. expensive equipment is needed)

the process can be done safely on any scale. that isn't the problem. How many countries with no access to, say sea water to desalinate, have ready access to supplies of oxygen and hydrogen?

@matt_black I would assume that a country who has trouble accessing H2 and O2 supplies would also have trouble accessing the equipment required to safely carry out the reaction.

@matt_black the reactions would require large reaction vessels, condensers, etc. So yea access to the necessary reactants may be limited, but so is the access to a lot of other things. Just two perspectives on the same problem.

Water for steam oxidation of silicon in semiconductor processing is made by burning hydrogen and oxygen. Primarily this is done to meet the purity requirements.

@JonCuster I never suggested that the reaction between hydrogen and oxygen isn’t used. Simply, to produce the quantities needed to supply a city or developing country would required large equipment that can withstand the reaction. Therefore the process is expensive and dangerous, but (obviously) not impossible

@dval98 - understood. Actually the comment was more to show how rare it would be to do water generation at an industrial scale. Ultra high purity is the only reason to do it for semiconductor processing.

@JonCuster my apologies then. Disregard my comment. Sorry for the misunderstanding

@dval98 - I'm not sure I phrased my comment clearly enough, so no worries...


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How do i get acetic acid from vinegar?
I'm trying to improve my rocket project, but, since the competition only allows the use of vinegar and sodium bicarbonate as its fuel, I had an idea of using the vinegar in a different way. My plan was to get the acetic acid from the vinegar and substitute it for the acetic acid, which is probably more efficient and powerful in terms of acceleration. So, for that, I have some doubts about how should I do it

You can concentrate it quite a bit with fractional freezing.

Wouldn't that be a cheating as "the competition only allows the use of vinegar and sodium bicarbonate" ?

Azeotropic distillation.
Example from this patent: here

Acetic acid is difficult to separate from water by conventional distillation or rectification because of the close proximity of their boiling points. Acetic acid can be readily separated from water by using azeotropic distillation. Typical examples of effective agents are ethyl n-valerate and 4-methyl-2-pentanone.

source: Stack Exchange

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

sort: votes


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After neutralizing a free radical, why don't anti-oxidants become free radicals themselves?
This is my understanding: A free radical is a molecule with a single unpaired electron. This molecule is highly reactive because it wants to steal an electron from another molecule (oxidation). Anti-oxidants are strong reducing agents that can donate their electrons to free radicals to neutralize them.
In order for this mechanism to be effective, the anti-oxidant should itself remain stable after the reaction. However, at this point the anti-oxidant would be missing a single electron. What general mechanisms keep them stable, so we don't get an endless chain of free radicals? Are there general mechanisms or do they each work differently (I know there are a huge amount of different anti-oxidants)?

They do. But these radicals are way less aggressive.

There are oxidating and reducing radicals. Then there are oxidating and reducing agents that are not radicals. Then there are also radicals which recombine with other radicals. Then there are agens able to be reduced or oxidized, but not eagerly. Then there are antioxidants not forming radicals when oxidized.

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Why is entropy change in an irreversible adiabatic process positive?
I've read this in a book

During irreversible compression, large quantity of heat is produced due to friction, while the decrease in volume is less. Thus, increase in entropy due to rise in temperature exceeds the decrease in entropy due to decrease in volume.

I don't get this at all. Is the friction they're talking about the one between pistion and container? What if we consider an ideal condition with no dissipation?
I also want to know the physical interpretation of increase in entropy in adiabatic (irreversible) process. Shouldn't it be zero? This book tried to explain it but it was confusing and I'd love to get further insight on this.

Difficult question because you can't really use thermodynamic equations to explain non equilibrium processes

I agree that the paragraph is highly confusing. For one thing speaking of "heat" in the context of an adiabatic process requires an explanation as to where the heat is going. Clearly for an ideal gas in an adiabatic container the final temperature of the gas depends only on the total work done on gas, and irreversibility must be tied somehow to a difference in the applied external pressure and the internal pressure of the gas. For a reversible process the two would match each other during the compression. In the irreversible case there is a mismatch and some of the work done by the surroundings goes to waste (the applied pressure exceeds the internal pressure), so in the end less work is done (on the gas) than in the reversible case.
The proof that an irreversible adiabatic process has an associated nonzero entropy change starts from the observation that the terminal states for reversible and irreversible adiabatic processes are not the same (in this case for instance the volume is larger in the irreversible compression). To compute the entropy change for the irreversible adiabatic compression you need to identify a reversible process (of which there are many) between the same terminal states as the irreversible process. Evaluating the reversible process that will take the system between the end points of the irreversible process you find that an entropy change is required, even though no heat is exchanged in the actual irreversible adiabatic process and despite reversible adiabatic processes not generating a change in entropy.

The average internal pressure may exceed the external pressure (the pressure exerted by the inside piston face), but the internal gas pressure at the piston face must match the pressure exerted by the inside piston face, because the two constitute an action-reaction pair.

The friction they are talking about is viscous friction of the gas itself.  This comes into play when the gas is deformed rapidly, and is roughly proportional to the square of the rate of gas deformation.
Regarding entropy, the change is entropy is equal to the integral of dq/T only for a reversible process.  In an irreversible process, in addition to this exchange of entropy with the surroundings, there is also entropy generated within the system itself.

Alright. What if I consider a reversible adiabatic process with same initial and final state variables as the irreversible adiabatic process into consideration. Since Entropy is a state function change in entropy will be same for both the processes. The heat flow in this reversible adiabatic process is undeniably zero, then so is the entropy change. Being a state function, shouldn't it be zero for the irreversible adiabatic process, at least on paper?

It is impossible to define a reversible process between those same two states that is adiabatic.

If you want to get a better understanding of all of this in detail, see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11, Problem 11D.1, Equation of Entropy Change.  There, they provide explicit expressions for the local rate of entropy generation during an irreversible process.

A point I would like to add is that you can't devise a reversible path for a irreversible process for a system with same surroundings which irreversible process occured. You would need to find to put the system in different surroundings for such processes.

That, of course, is correct.

Here is the reasoning behind why friction produces heat.
The frictional forces arise from the interactions with the gaseous molecules and the surface of the container along with some resistance to motion caused from the interactions of the gas molecules. However, the reasons for which they arise are a little different (i.e. surfaces cause friction cause they are imperfect/not smooth; friction in liquids and gases arise from the interactions of the individual atoms).
Entropy is the amount of micro-states available for the molecules. So, even though the compression decreases the amount of micro-states (negative entropy), the heat introduced from friction causes a positive change in entropy, which has a greater magnitude than the decrease caused by compression. Therefore, the gas molecules have more available micro-states.
Entropy in reversible adiabatic processes is always zero. For irreversible adiabatic expansion, this is not the case as the gas does work and expands, causing the amount of micro-states available to increase. It is irreversible as the gas loses the ability to do work when the process is over and the gas won't spontaneously return to its compressed state.


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Inputting liquid species in chemkin pro
I have been trying to model various species decomposition/combustion in Chemkin Pro.
Chemkin only lets you to allow to input gas phase and surface site solid species in your input files, but I have various fuels/propellants in liquid state whose decomposition/combustion I need to model.
Is there any way to input liquid species reaction mechanism's in Chemkin by inputting them as gas species but somehow manipulating the Arrhenius coefficients, so that the software will ultimately model the liquid species in question?
Also if there's some other way to accomplish this instead of manipulating Arrhenius coefficients, I would be grateful for the input


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Despite being less reactive,alkanes are highly flammable
I read in my book that alkanes are saturated compunds and the valencies of carbon are satisfied by single bonds. I knew that propane and butane are highly flammable and could cause explosion.
My question
If alkanes are less reactive then why they are highly flammable and could cause unfortunate explosion?

Because "less reactive" is still pretty reactive.

They do not self-ignite nor rapidly oxidize themselves during contact with air, do they ? 10^30 is smaller than 10^40. Is it small ?

In chemistry, reactive and flammable are two different term. For example, Diethyl ether is a rather non-reactive material but it is highly flammable.
In your case, propane or butane are mostly unreactive to common chemicals, but it does react with oxygen at an elevated temperature- that is they are flammable. Under similar condition, propene or butene are more flammable.


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How to correct the pH and composition of bottled water?
For some time I've been buying a brand 'A' of (non-sparkling) bottled water.
It wasn't easy to find one that I liked, because I make tea with it, and most other brands I tried made murky, smelly, dark tea, with a disgusting film on top, whereas this one was perfect.
Unfortunately, I just found that the supermarket doesn't stock brand 'A' anymore (I know, it's not a big problem, by any standards; just discussing a bit of science, for fun).
So I picked the closest one I could find (brand 'B'), naively thinking that the main 'factor' to look at was pH.
Brand 'A' had a pH pf 6.7; brand 'B' was 6.9. I thought: close enough.
Actually: wrong.
I made tea with brand 'B' water, and while it wasn't as horrible as with higher-pH ones I had tried before, it was nowhere as good, it made that slimy film, and smelled a bit off.
So I looked at the composition of the two brands side by side, converting the concentration from mg/L to mmol/L:
$$
\begin{array}{lcc}
\hline
\text{Component} & C_\mathrm{A}/\pu{mmol L^-1} & C_\mathrm{B}/\pu{mmol L^-1} \\
\hline
\ce{HCO3-} & 0.38 & 4.72 \\
\ce{SiO2}  & 0.32 & 0.00 \\
\ce{Na+}   & 0.13 & 0.18 \\
\ce{K+}    & 0.02 & 0.16 \\
\ce{Mg^2+} & 0.04 & 1.60 \\
\ce{Ca^2+} & 0.10 & 1.07 \\
\ce{F-}    & 0.01 & 0.00 \\
\ce{Cl-}   & 0.02 & 0.06 \\
\ce{NO3-}  & 0.03 & 0.00 \\
\hline
\end{array}
$$
Clearly there is much more 'stuff' in water 'B' than in 'A', and in fact the residue is 289 mg/L in the former, 43.6 mg/L in the latter.
One thing that caught my eye was the charge balance, which seems a bit off for 'B' (+0.91 mmol/L), whereas it's a bit closer to 0 for 'A' (-0.02 mmol/L).
But OK, my main question at this point is: can I add something to water 'B' so that it 'behaves' a bit more like 'A', for the use that I mentioned?
My hypotheses so far are that:

the pH should be lowered
$\ce{Ca^2+}$ and $\ce{Mg^2+}$ are responsible for the formation of that 'film', and I might prevent that by adding something that forms soluble coordination compounds with these ions (and more stable than the ones they form with whatever it is in tea that reacts with them)

Obviously I am not going to use stuff that is not edible, like EDTA; and for the pH correction, I'll have to rely on calculations, not going to stick a pH-meter into it.
So my best bet at the moment is: citric acid.
Though I am not sure how I should calculate the correct quantity to add to the 5 L bottle I buy (surely I can't aim to convert most $\ce{HCO3-}$ to $\ce{H2CO3}$, or the final pH would have to be too low).
What do you think?
Any ideas/comments/suggestions?

Where do you live that you cannot take tap water to make tea? Most bottled water is high on minerals.

Search for a brand C,closer to A in the bicarbonate content. The bicarbonate content
of B is nowhere close to make a good tea from it. You may try a Brita ionex based filter or similar.

@Poutnik : thanks, I will look for that!

@Karl : obviously if I could use tap water I would. It's even worse in my case, full of calcium, very high pH, and actually all the people who live here advise not to drink it. And no, wrong, not true that 'most' bottled water is high on minerals; there are several  brands with rather low residue, like brand 'A' that I described.

BTW, to the 4 people who voted to close my question without actually commenting or explaining what they don't like about it: well done! We're never going to get any shortage of sad, cowardly backstabbers in this world!

Remember bicarbonate content, aside of heavy metals like iron and manganese, is probably the most critical parameter. Boiling converts it to rather alkaline CaCO3 or Mg(OH)2, what leads to scale deposites and adsorption/oxidation/condensation/darkening of various tea flavonoids and unpleasant sensoric perception, together with eventual precipitation of hydrated oxides of Fe or Mn.

@user6376297 You shouldn’t jump to assumptions about the close votes. I can see them; three are on needs more focus, the fourth is on we do not give personal medical advice. Personally, I disagree with both but neither would immediately suggest adding a comment. (Indeed, unless a custom off-topic reason is used, most close votes here come without comment and are considered self-explanatory.)

@Jan : OK I see your point; still, several people, including yourself, managed to comment usefully and informatively about the situation I described and the question I asked, so it does not seem as off-topic or lacking focus as that. I don't know, for me anyone who's so dead set on just shutting other people up for no reason embodies the exact, antipodal opposite of the curious, open mindset that is required to do science (and actually to be a decent human being). So I do deeply and personally resent this insufferable, obtuse way of saying 'no' to everything a priori.

@user6376297 $\ce{MgSO4}$ contributes to the permanent hardness of water (which can't be removed by boiling) and adds a bitter taste to the water, too (literally «bitter salt»); doesn't water B contain equally quite some of $\ce{SO4^{2-}}$, too?  I'm not sure if a journal like JChemEduc has published a procedure easy and safe to replicate «DIY» to soften water (here especially to remove Ca/Mg) which would skip both distillation of the water (e.g., sun oven), and percolating tap water across exchange resins.  It might be a question for geology folks (identification of natural zeolithes).

@Buttonwood : thanks, sounds like an idea to try out!

If you are prepared to do chemistry to "fix" bottled water, I'd say it might be cheaper if you tried the same with your tap water. Just saying. Most of the time people avoid tap water because it is contaminated with bacteria, or terribly chlorinated because of the same. Both of which go away with boiling it.

@Karl : tap water in the building where I live is very hard, but that's not the main issue. The building is quite old, so the water pipes are not in a very good state. So, whenever you open the tap after it's been closed for a while, the first 10 L or so of water that comes out has a weird, unpleasant smell, which I would describe as 'moldy' for want of a better word. Even when I use tap water to boil food I have to let it run for 1 min or more before it's 'OK'. Not to mention when I go away for a while. Then it even smells 'rusty', and it takes days to clear out. A no go for tea, as I said.

The ones you want to look out for are calcium and magnesium. Either try finding a water that has lower content of those two ions or try finding a water filter that can soften the water.
I grew up in an area with very hard (but potable) tap water and such a water filter was constantly used before boiling our tea water. It had the nice added side-effect that there was significantly less limestone deposition on our kettles.
I would not recommend trying to fudge around the pH by yourself. The pH values of both bottled waters are very close to neutral meaning that additions of minute quantities of acid will likely have a large effect. But that notwithstanding, both the process of brewing tea and potentially adding milk (or lemon, if you are into that) to a black tea will alter it anyway.

Long time ago, I supervised an undergraduate on a summer tea project (for fun). In South Asia there is a natural strawberry colored tea made from slightly matured green tea. You boil the leaves in water with a small amount of sodium bicarbonate. Ice cool it, and boil again and aerate it with a ladle. The tea becomes reddish on aeration and after adding milk, it becomes pinkish. So I can recall your experiences because I tried this recipe in another country. No matter what I did, the right color would not develop. It remained murky brown. It turned out that the chlorinated water was ruining the natural colors, it was particularly destroying the reddish pigment. Passing the tap through a carbon filter+ion-exchanger did not work as calcium also seemed important for the right color.
Finally a mineral water bottle worked and the undergrad saved a project for his poster presentation.
Anyway, it seems your tap water is of bad quality or it is hard. This tea "scum" are Ca/Mg salts. You guessed it right. My suggestion is do not add any acid such as citric acid. Tea is not meant to be made in the presence of citrate salts especially if you add milk to it. I do not know what the taste be like. Most likely horrible!
I do not know how easy it is find potable distilled water in stores. Ask the manufacturer if it is fit for human consumption. Distilled water is not ultrapure water as it may have organic impurities or volatile impurities. It is only free from water soluble salts. All these tea scum problems should vanish.

Thanks; I think distilled water is positively NOT fit for human consumption, at least that's what I am told. Unfortunately before I saw your comment I already tried adding citric acid. I assumed 1 mol citric acid per mol of bivalent ions would do, so 2.67 mmol/L, which worked out to about 2.8 g citric acid monohydrate to add to the 5 L bottle. So... hmm... let's say that I now have 5 L of rather sour water to go through :( Not going to back-correct it with bicarbonate, I messed it up enough as it is :)

both waters are bottled, according to the question. 2) Traditional Earl Grey Tea drinkers may heavily object to not adding citric acid to their tea but they would also not add milk. Some Germans took that idea too far and tend to add a slice of lemon to whatever black tea they make until I request they stop …

Adding a few drops of lemon juice to green tea or black tea (without milk) is very common throughout the world. This is done after boiling the tea. If we boil the tea with a few drops of lemon juice, it will be a horrible taste. But lemon juice or slice is not equivalent to citric acid alone-hence my suggestion.

@user6376297, Drinking distilled indeed not fit for consumption because of the lack of ions, however when you boil tea it is no longer distilled water and it will have ions.

Your brand B mineral water is pretty hard water. Boil it for 5 minutes, cool, decant the water. Do your tea with that.

Distilled water unfit for human consumption should be identified on the label with giant red letters; otherwise it should be mineral-free and quite pure. Getting an accurate pH will be difficult since there will be such low ionic conductivity. Possible contaminants like low vapor pressure organic compounds? Like ethanol, methanol, benzene? Highly unlikely.
Distilled water is bland - perhaps too bland for drinking straight. And by the time you boil out the little bit of $CO_2$ present, it will be blander than bland! Therefore perfect for your tea. If by chance, the distilled water is of low quality, try another brand. I suspect that distilled water is made in truckload quantities and sold to bottlers who do a good job and to bottlers who are sloppy.


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Calculating the amount of moles from given number of atoms
Complete the sentences below by doing the mole calculations:
i)In $\pu{20 mol}$ of copper(II) phosphate, there are ______$\pu{mol}$ of copper ions and ______$\pu{mol}$ of oxygen atoms.
ii)A sample that contains $1.8 \times 10^{23}$ atoms of lead has a mass of ______$\pu{g}$.
iii)A gold crown that contains $7.95 \times 10^{22}$ atoms of gold has______$\pu{mol}$.
These questions are from TopHat, if anyone can help me that would be awesome.

I’m voting to close this question because, like your professor, we feel you should learn the material by doing it yourself.

In a crammed class of 100-300 how can professors manage to ensure that each and every student is following them? Worse, if the classes are online. The attention span of younger students is hardly 10-20 min. Too many distractions and no habit of taking notes. Above it, many lazy teachers provide them with PowerPoint print "notes".

Actually, you should learn these in high school. Google "mole concept" and you will realize these are pretty straight forward math question once you realize what is Copper(II) phosphate [you can google that too].

This is not a title to a question. A title should reflect the body of the question (in this case, there is none).

You only need to know how to write chemical formulae and know two facts:
I) 1 mole of a substance equals it atomic or molecular weight.
II) 1 mole of a substance has $\ce{6.02×10^23}$ particles or units.
Think of moles like a dozen. Have you eaten a double-yolked eggs by chance?
If we buy 1 dozen double yolked eggs, how many eggs yolks will be there: 12 x 2 =24.
If you have 20 mole of $\ce{CuCl2}$ unit, there are 20 moles of Cu atoms and 2x20 moles of Cl atoms. In order to solve copper phosphate, you need to know how to write the formula of copper phosphate.
Also, develop a habit of opening textbooks. If you only rely the "web" and on the professor, they will not take you anywhere. Textbook's solved examples are a great resource which modern students rarely bother to read.

I always like to have two or three books for the same topic. Reading the slightly different worded explanations from different authors really helps understanding.

Certainly, this should be the habit of any serious student of science whether he is a teenager or 80 year old retired scientist.


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13C-NMR of aniline: Why are carbons that are further from the amine group more downfield than carbons that are closer?

A $\ce{^{13}C}$-NMR of aniline shows that the carbons further from the $\ce{-NH2}$ group are more downfield than the ones near the top of the benzene ring. If electronegative atoms like nitrogen have a deshielding effect on nearby Carbons, why is this the case? Is it because of dipole vectors?

I strongly suggest using deshielded/shielded instead of downfield/upfield, as the latter terms are historical relics from an old way of doing NMR (continuous wave). Deshielded/shielded is really more intuitive and easily relatable to other chemistry concepts.

Although the nitrogen atom is more electronegative than carbon, nitrogen will readily share its electron pair, allowing one to draw a resonance structure where the nitrogen is donating electron density into the ring. Therefore, the donating of the lone pair of electrons overcomes the inductive effects arising from the difference in electronegativity.


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Why would Auger electron detection have potential for higher spatial resolution than EDX
I understand that Auger electrons are emitted when an electron falls from a higher energy level to a open inner shell and EDX uses electrons that fall from outer shells when an inner shell is kicked off. I also have read that AED has a higher spatial resolution but I don't understand WHY this spatial resolution is higher. What is different about the Auger electrons that increases resolution?

This reads like a standard homework question. What do you know about the differences of the two techniques?

I have never used the techniques my self so all i know is the theory. In principal, when a beam of electrons is aimed at my material, a core electron will be kicked out leaving a hole. A higher energy electron will fall to fill that hole. When this happens it can either release an x-ray or it can dislodge another electron which would be an Auger electron.

I realize that auger electrons have a small mean free path so they are less likely to reach the detector but how would this give better spatial resolution? I would have predicted this would give worse spatial resolution since your signal to noise ratio would be so low

this is not a standard homework question but even when i post questions that have nothing to do with homework you all still say they are homework questions. So i don't even know what a "non-homework" question would look like at this point

Correct. Now, assuming the struck atom is not on the surface, either the x-ray or the (low) energy electron have to get to the surface and then to the detector. So compare mean free paths.

i get that the mean free path of auger is smaller what i dont get is why this gives it a better spatial resolution

As for the non-homework, well, having taken a class many moons ago on analytical techniques, yes, this is a question to test if you understand what is going on.

Either the Auger electron or the x-ray is emitted in a random direction. The x-ray can emerge from a larger volume of material and still hit the detector.

this was a discussion we had with 6 other people. this is called discussion, maybe things have changed since you took classes but nowadays sometimes people have discussions in seminars and we walk away with questions that we want to understand

what do you mean the auger or x-ray is emmitted in a random direction? are you saying one or the other is?

@HarleyMcFarlen - An excited atom is typically thought to emit either an Auger electron or an x-ray. Whichever is emitted is emitted in a random direction.

I think this is a good question. The problem is more fundamental. In both cases, you would be using electron beams on the sample.
In EDX, you are shining a beam of electrons and detecting the X-rays. Here comes the main problem. It is hard to focus the X-rays. Mea culpa here. As Jon commented it is the depth of penetration of the electron beam and the corresponding X-ray emission that main affects spatial resolution. This picture roughly tells the story:

Auger electron electron spectroscopy, you are focusing a beam of electrons and you detect electrons. It is relatively easy to focus electrons using "lenses". Hence you can point point the exact origin where the electrons are being ejected.
Also, think about the depth profiling. EDX is not a surface technique unlike Auger. I used to have this misconception until I attended a workshop on surface analysis. The electrons can easily penetrate the sample and similarly originate from the some depth as well.

The 10's of kV incident electrons penetrate into the sample. The 100's of eV Auger electrons only escape from a shallow depth. In contrast the x-rays generated in EDX escape from a greater depth. It is this difference that leads to the resolution difference, nothing to do with 'focusing' or not of the x-rays. They come from a greater volume of material.

Thanks for the pointer. Do surface scientists distinguish surface (x-y plane) vs. depth resolution (x,y, z) resolutions?

The problem is that the incident beam blooms out laterally as it goes through the sample, so x-rays generated below the surface come from a larger lateral dimension than the incident beam. This means that the x-ray 'source' is bigger than the beam in x-y and can come from a greater depth range. In contrast, the Auger electrons only get out from the near-surface region, and there has not been much beam blooming by then. So, for Auger vs EDX the two (lateral and depth) go hand in hand.

Thank you, I really appreciate you taking the time to answer.

Also, note that one is not getting position information through 'imaging' the electrons of x-rays. The incident electron beam (perhaps 5-10 nm in diameter) is scanned in x and y, and counts at the detector are attributed to the current beam position on the sample.


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Modeling activation energy function
I'm quite new to the field, and actually am a mathematician. I was wondering if there is a way to model (find the equation of) the reaction progress (more specifically, the activation energy functions) using some equation?

Optimally I'd like to use some graphing calculator like Desmos to write the equation and get a similar function.

The reactions you show are actually v complex. Start with something such as atom+diatom reactions, H+OH, F+H2 etc. based in LEPS potential energy surfaces.  see Polanyi & Woodall, J. Chem. Phys. 57, 1574, (1972); Polanyi & Schreiber, Faraday Disc. Chem. Soc. 62, 267, (1977) and textbook by Steinfeld, Francisco & Hase, Chapter 9,' Chemical Dynamics & Dynamics'(Prentice Hall 1999);Levine & Bernstein 'Molecular Reaction Dynamics and Chemical Reactivity' (OUP 1987).

Yes, you definitely can, but it's complicated. Because you are a mathematician it might be easier to understand.
Introduction
On your picture is one-dimensional representation of the energy function of multiple variables. Variables are coordinates of all atoms in the molecular system. This function define the potential energy surface or PES. Sometimes you can simplified it and represent as a three-dimensional surface using a linear combination of coordinates. Your picture is the energy profile of the PES. Where minimums are equilibrium states of the system and saddle points (minimax) are transitional states.
The reaction progress on the picture is the intrinsic reaction coordinate (IRC) which is a one-dimensional representation of the a linear combination of vectors of your molecular system vibration.
Real energy profile
Consider, for example, the reaction $N_2O_2$ with $H_2$. Atoms in both molecules can oscillate in a specific well-defined way which gives you the molecular vibration. You need to solve quantum differential equations for the molecular system to get vibrational frequencies and vibrational vectors. To solve it analytically is incredibly difficult, but you can solve it numerically using quantum chemistry software. It will compute energy over intrinsic reaction coordinate (IRC) as a set of points with a certain displacement or step. Using simple spline function you can get a real energy profile.

Approximation
It is possible just to use a square function. According to the harmonic approximation it is precise at the vicinity of equilibrium states, but useless with transitional states. Sometimes to avoid this problem people use energy profile diagrams. Usually people are not interested in the form of function but the energy differences (activation energies, Gibbs energies and etc.)

If precision is not the case you just need to know energies of initial separated molecules, the transitional state energy (at the saddle point) and the energy of the product (which you can compute with quantum chemistry software) and join square functions at the found energies. You also need to remember that the energy profile is made for the elementary reaction.
Calculation
There are a lot of tutorials how to compute energies of initial systems, transitional states and products with quantum chemistry software. It should be mentioned that you will need to sum equilibrium energy of computed separately molecules to get your initial energy. The most difficult part is to compute the transitional state energy. You need to find a molecular geometry which would be very close to your potential transitional state. You need to calculate Hessian (second derivatives) and start minimax optimization (saddle point optimization).


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Resonance involving carbanion
Suppose a carbanion is involved in resonance and thus its electrons conjugate with unhybridised p orbital
of the nearby sp2 hybridised carbon and a pi bond is formed. Since carbanions are sp3 hybridised, how would it form a pi bond or will the hybridisation of the carbon change?

Yes, when bonds are formed, the hybridization can change in a carbon atom.
I think it’s important to remember something. Hybridization is not a real phenomena that occurs in reality. It is a model we use to simplify the complex bonding that occurs between atoms.
So, yes hybridization of the carbon will change, as it is not concrete and can change as bonds are made.


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How to make anhydrous or monohydrate FeSO4 from heptahydrate form?
I need the anhydrous form of $\ce{FeSO4}$ (7720-78-7). I have $\ce{FeSO4.7H2O}$, is there anyway it can be converted to anhydrous form, or at least to monohydrate?
Do you know where I can purchase it? Most vendors only have the heptahydrate form, even vendors that listed the anhydrous as available, but when I asked, they only have the heptahydrate.

Have you tried heating it?

@Alchimista - 175 °C? // Wikipedia lists 60–64 °C as decomposition for heptahydrate, 300 °C as decomposition for
monohydrate, and as 680 °C decomposition for the unhydrated salt.

Sorry folks, I was speaking about Fe(III). Comments deleted to avoid confusion for future readers

It'd be nice to know what you need it for, and why you would also accept the monohydrate. Also makes people more keen to answer.

Wikipedia renders multiple stages of decomposition as the common heptahydrate is heated.  Based on the data summary at the beginning of the article and the quoted passage below, the anhydrous salt is obtainable between 300 and 680°C.  A lower temperature, starting at 60-64°C, gives the monohydrate.  An oxygen-free atmosphere is necessary to prevent iron oxidation to Fe(III).

On heating, iron(II) sulfate first loses its water of crystallization and the original green crystals are converted into a white colored anhydrous solid. When further heated, the anhydrous material releases sulfur dioxide and white fumes of sulfur trioxide, leaving a reddish-brown iron(III) oxide. Decomposition of iron(II) sulfate begins at about 680 °C (1,256 °F).

$\ce{2 FeSO4 → Fe2O3 + SO2 + SO3}$

To use a temperature as hot as 300 °C I wonder if oxygen in the air could oxidize Fe(II) to Fe(III). You'd need a special vacuum oven to get to 300 °C.

Yes air can oxidize it even at room temperature, slowly.  I'll add the atmosphere needs to be oxygen-free.


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entropy calculation in an irreversible process
I found out that if we want to find the entropy change in an irreversible process let's say $\text{state A} \to \text{state B}$ , we find or make an reversible path for the same process and find the entropy change for that irreversible process. The thing I don't get is that, even if entropy is a state function, does it mean that the change in entropy for the irreversible process we needed to find is the same as the change in entropy for the reversible path we constructed? Or am I being mistaken? If its the latter, please provide me a solution for it. Thanks.

That's exactly right.

@BuckThorn I actually can't make sense of it, like how can it be the same? even though its a state function, I can't get why is it true. Thanks for your reply, I hope that you can elaborate a bit on it.

You make a reversible path between the same two end states as the irreversible process.  The entropy change for the irreversible process will be the same as for the reversible path.

I really can't get as to why it's true, I know it's a state function, but I really cannot make sense of why the entropy would be same for both the irreversible and reversible process.

Quantities such as heat and work can vary depending on the path (how you get from an initial to a final state). In contrast a defining aspect of a state function is that it is independent of the path.
Entropy is a state function, as a corollary therefore a path-independent property. The experimental determination of entropy involves measuring the heat transferred as a function of temperature during a reversible process and computing $\int \frac{dq}{T}$. That is another defining property of the entropy. It is not trivial to wrap your head around why it should be so, but entropy can be thought of as helping quantify a limiting or ideal property, in particular the maximum work that might be obtained from a process, or the least amount of work required to bring about a non-spontaneous change. It defines processes in which there are no losses (losses being consistent with irreversibility).
As an example, at constant T and p the Gibbs free energy change (which is related to the limiting work) is related to the entropy as follows:
$$-\frac{\Delta G}{T}=\Delta S_{system}+\Delta S_{surrounding}$$

oh alright, I was having this idea, but the thing was that I was just wrapping my head to see why is it really true. Thank you for your answer, appreciate it!

but why do you relate entropy as being thought of as “maximum potential to do work” ?

Hopefully my edit clarifies one way.

The answer above is perfectly fine. There is another way to see it. It is very important to realize what is a "system" and a "surrounding", what changes when we talk about Entropy and other state functions? when we talk about Entropy, we say "The total change in Entropy of Universe is never negative, it can be either zero or positive". We usually define a system and a surrounding but take surrounding in discussion only when we talk about entropy.
when we are saying any quantity is state function, we mean it with respect to the system. So, both of them are not same.
Take for example, any process which is performed by two ways:

PV work (irreversibly), say, total change in entropy is "x".
Electrochemically (reversibly), total change in entropy of system is "y" (since the process is reversible, realize, that total change will be zero)
if someone asks you to find the change in entropy of surrounding in an irreversible process.
Then, the way around this problem is that you simply use third law.
whether the process is done irreversibly or reversibly, the del S system is same. So, "y" is del S (system), x has to be positive. x-y is the answer!
You can find such problems as well as detailed concept in Klotz and Rosenberg's book on chemical thermodynamics.

"since the process is reversible, realize, that total change will be zero" this is wrong.. it's only true for the full cycle for reversible process

Entropy change for the system between two points can be determined simply from plugging the final and initial states into the state equation of entropy. The fundamental idea is that entropy change only depends on the state of the system and the state of a system is described by the state variables.
However, the total entropy change (sys + surr) depends on how the irreversible process was. Suppose the total entropy change in an irreversible process for the whole universe was some quantity $ \Delta S_{net}$, and entropy change for the system along any reversible path is $ \Delta S_{sys}$ the entropy change in surroundings is:
$$ N = \Delta S_{net} - \Delta S_{sys}$$
Note: A reversible process is characterized by $N=0$, that is there is no  entropy production and $N>0$ if irreversible

More on this extra entropy concept
For an example of this using physical situation, see the answer by Chet Miller here
For details on the terms in the formula, see the answer by Chet Miller here

Your answer and Shivani's are admittedly very useful (perhaps more than mine). However for a reversible process we do say that N=0.

Good point I"ll add it in


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Should equiatomic iron–cobalt alloy be written as FeCo or CoFe?
I have an alloy made of 50% iron and 50% cobalt. I see occasionally in the literature this is written as FeCo alloy.
But shouldn't it be CoFe alloy since the chemical symbols are supposed to be arranged in alphabetical order?
Is there somewhere where these rules are written down?

There is a lot of history and accustomed usage wrapped up in alloy names. Just the way it is (and no, having looked at many many phase diagrams I don't think I can come up with many rules. Well, hydrides always have the metal first, but then there is water, so it just goes to show...

I will add that, for your particular alloy, it is shown as FeCo on the phase diagrams at the ASM Alloy Phase Diagram Database. But I don't think there is a signed treaty of understanding between metallurgists and chemists.

If you want to become known as a IUPAC evangelist ... you will find a few guys on each side who are prepared to go for each other's throats on such issue, and the rest will be glad when they've strangled each other. ;-)

@Karl - yes, we need a new SALT treaty (Similar ALloy Terminology) to bring peace to the world...

@JonCuster https://en.wikipedia.org/wiki/Contradictio_in_terminis :-))

Does this answer your question? What is the IUPAC nomenclature for Intermetallic Compounds?

@NilayGhosh I think it would be better to close the other way round.

Current IUPAC recommendations suggest following electronegativity ordering. The symbol for iron (a less electronegative element among the two) should be written first: $\ce{FeCo}.$
From the section IR-4.4 Sequence of citation of symbols in formulae [1, pp. 58–59]:

IR-4.4.2 Ordering principles
IR-4.4.2.1 Electronegativity
If electronegativity is taken as the ordering principle in a formula or a part of a formula, the atomic symbols are cited according to relative electronegativities, the least electronegative element being cited first. For this purpose, Table VI* is used as a guide. By convention, the later an element occurs when the table is traversed following the arrows, the more electropositive is the element.
[…]
IR-4.4.3. Formulae for specific classes of compounds
IR-4.4.3.1 Binary species
In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1) is most often used in binary species.2
[…]

For intermetallic compounds, earlier recommendations prescribed alphabetical ordering
rather than by electronegativity (see Section I-4.6.6 of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990, ed. G.J. Leigh, Blackwell Scientific Publications, Oxford, 1990).

Examples:
[…]

$\ce{Rb15Hg16}$
$\ce{Cu5Zn8}$ and $\ce{Cu5Cd8}$

Table VI as seen in [1, p. 260]:

Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


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How would I correct this Hess cycle?

Given
$$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \quad \Delta_\mathrm{r}H = \pu{-196.0 kJ mol^-1}$$
and $\Delta_\mathrm{f}H(\ce{SO2}) = \pu{-296.0 kJ mol^-1},$ find $\Delta_\mathrm{f}H(\ce{SO3}).$

I have chosen to solve this by constructing a Hess cycle and have come up with the following:

This method has lead me to the correct procedure of solving the equation: $$\Delta_\mathrm{r}H = -2\Delta_\mathrm{f}H(\ce{SO2}) + 2\Delta_\mathrm{f}H(\ce{SO3})$$
However, it is not entirely complete in my opinion and I am unsure on how to proceed in completing it, specifically on what I should place below the two arrows.
I am aware that it should be some combination of $\ce{S + O2}$ but am unaware on how exactly the coefficients would work.

Sounds like you are overthinking the problem: there is no need to go back to the elements. Also, please note that enthalpy is not a dimensionless quantity (see the edit).

@andselisk but if I am showing the $\ce{\Delta H_f}$ shouldn't I also show the reaction for it aswell?

You did the right thing. However the method you used sounds confusing sometimes, it gave me a hard time too. Hess's law says that the resultant enthalpy change in a reaction is same whether it occurs in one or several steps .
We can use that to our advantage by assuming a hypothetical situation where the given reaction proceed in a number of steps for our convinience.
In your example, we disassociated $\ce{SO_2}$ molecule to corresponding standard state constituents and the enthalpy change associated with it will be $-\Delta H_f(\ce{SO_2})$.
$$\ce{2SO_2(g) + O_2(g) ->[-2\Delta H_f(SO_2)] 2S(s) + 2O_2(g) + O_2(g)->[2\Delta H_f(SO_3)] 2SO_3(g)}$$
I took the following reaction through above mentioned 2 steps $$\ce{2SO_2(g) + O_2(g)->[\Delta_r H] 2SO_3(g)}$$
According to Hess law, the Enthalpy of reaction $\Delta_r H$ will be the summation of the all enthalpy changes occuring in the hypothetical intermediate steps.
$$\Delta_r H = -2\Delta H_f(\ce{SO_2}) + 2\Delta H_f(\ce{SO_3})$$

In the first reaction where the $SO_2$ is converted into free state elements, you didn't consider the chemical changes of the molecular oxygen, why?

Heat of formation is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its standard state. $\ce O_2$ is already in it's standard state and by convention, the enthalpy of substances in their standard state is taken as zero.

Oh nice, that makes sense. Thank you!


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Why do neutral metal atoms have affinity towards electrons?
Since first electron affinity is almost always negative, neutral atoms of most elements (including metals) must have an affinity towards electron. But shouldn't metals want to lose electron(s) rather than having an affinity towards it?
Take lithium for example. It can attain noble gas configuration as it loses an electron. But first electron affinity of lithium is -60 KJ/mol. So, it has an affinity towards electron although it wants to lose one. But wasn't it supposed to have a positive electron affinity?

The negative ions of the noble gases are not stable. He- is kind of metastable with a lifetime of around 10 microseconds, the rest get rid of the extra electron much quicker.

Now, body of your question is asking quite different thing then title - so either if they are atoms with positive affinity (yes there are) or why ionisation energies of neutral atoms and molecules are always positive, no atom "wants to" lose electrons - pick one.

BTW why am I getting ideas like going back in time and pre-emptively killing the dude (ahem, great scientist) who came with this dumb idea of atoms losing electrons?

@Mithoron is it ok now?

Somewhat better, it's still mixing ionisation and affinity, which are quite different, but if your misunderstanding stems from this...

Nobody wants to lose electrons.

Related question with a similar answer to yours: if atoms are neutral, why do they form bonds and interact with each other?


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Why isn't sodium hydrogen phthalate used instead of KHP?
Why is potassium hydrogen phthalate (KHP) ubiquitously synthesized and used as a primary standard, while sodium hydrogen phthalate isn't even available to buy?

My guess is that NaHP is hygroscopic, but KHP, which is a primary standard, is not hygroscopic. This is helpful when it comes time to weigh out the standard. A quick search shows 5 suppliers of NaHP, none of which I have ever heard of previously.

For classical analytical problems you have to check at least 100 year old literature. It is easy in Google Scholar. Set the date limits to 1920s or 1950s whatever you wish to try.  In older times, the conventional wisdom, which still persists, was that one should use a higher molecular weight standard. The rationale was that the relative error will be small on a balance. KHP has a higher molecular weight than NaHP. This is one advantage. Second, advantage is that it available in extremely high purity. It can be dried as well.
Both KHP and NaHP work perfectly. There is a caveat with NaHP, that it crystallizes as a hemihydrate i.e., for two NaHP units, there is a water molecule. KHP crystallizes as a normal anhydrous salt. Analytical chemists run away from hydrated salts because you would never know what is the hydration level, what was the humidity level at which they were stored. We would never know the "true" molecular weight of the substance which is being weighed out. There are salts such as ferrous ammonium sulfate with water crystallization, but it is very stable with respect to storage. NaHP, who knows how it loses its water of hydration.
Search "Acid potassium and acid sodium phthalates as standards in acidimetry and alkalimetry" on Google Scholar. The reference is Journal of the American Chemical Society 37, 10 (1915): 2352-2359.


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Does it make sense to differentiate the Arrhenius equation with respect to temperature?
I am trying to answer the question: "between concentration and temperature, which has a greater significance on the rate of reaction"
Hence, I'm trying to find a method to systematically compare the impact/significance of concentration and temperature on rate. This is where I ended up trying to differentiate the Arrhenius equation:
$$k = A\mathrm e^{-E_\mathrm a/(RT)}$$
I thought that deriving rate constant $k$ with respect to temperature is finding the rate of change of $k$ with respect to temperature. Is it conventionally and mathematically correct to do this?
If it is, then continuing the thought process:
I differentiated the equation assuming that the Arrhenius constant...is...constant... But is that true? Does $A$ change when temperature change (since my variable here is temperature)?
If everything was not non-sense until this point, I ended up with the derivative:
$$\frac{\mathrm dk}{\mathrm dT}=A\mathrm e^{-E_\mathrm a/(RT)}\times E_\mathrm a/(RT^2)$$
Is this correct chemistry/maths?
EDIT
Is there a better way to compare the significance of concentration with that of temperature on the rate of reaction? Maybe using percentage increase of each factor with the percentage increase of rate?

One problem is that dk/d(conc) and dk/dT will have different units, so you can't easily compare them quantitatively to each other.

@Andrew yes that makes total sense I did not think this through...I was also trying to ask how should I acutally go about comparing these two factors or if it's too complicated to compare them at all.

Does it mention anything about the order of the reaction?

@dval98 It's my vague research question so I can choose what order of reaction I want to look at.

The question lacks details, and is thus effectively asking for a coarse-grained answer.
So:
The answer depends strongly on how the activation energy compares with RT.  Let's compare doubling the (absolute) temperature with doubling the concentration.
Concentration: If the reaction is first-order in the reactant in question, then we get a $2$-fold increase in rate; $2$nd-order => $4$-fold increase.   If we have two reactants, and the reaction is first-order in one, and second-order in the other, and we double the concentrations of both, we get an $8$-fold increase. And so on.
Temperature:  If we ignore the T-dependence of A and Ea, doubling the absolute temperature increases the rate by:
$$\frac{e^{\left(-\frac{\text{Ea}}{2 \text{RT}}\right)}}{e^{\left(-\frac{\text{Ea}}{\text{RT}}\right)}} = e^{\frac{\text{Ea}}{2 \text{RT}}}$$
If $\text{Ea} \approx \text{RT}$, then doubling T increases the rate by a factor of $e^{(1/2)} \approx 1.6 $.  And if $\text{Ea} < \text{RT}$, doubling T increases the rate by a factor less than $1.6$. Thus, in this regime ($\text{Ea} \lessapprox \text{RT}$), one could say concentration is more important (except in the trivial case of a zeroth-order reaction, or in the unusual case where the order is between $0$ and $1$).
On the other hand, if Ea is much larger than RT (which is common for chemical reactions, if T is about room temperature*), then one could say temperature is more important.  For instance, if  $\text{Ea} \approx \text{10 RT}$, then doubling T would increase the rate by a factor of $e^{5} \approx 148$.
*At $298$ K, RT is only $\approx 2.5$ kJ/mol.
Indeed, if we consider a "typical" activation energy of  $\approx 80$ kJ/mol**, then even a modest increase in T (from $298$ K to $340$ K) will increase k by about 50-fold.
**How to relate a reaction barrier to the time the reaction needs to proceed?
Another way to think of this is that, if RT is close to (or greater than) the activation energy, then there is abundant thermal energy relative to activation energy.  Consequently, the availability of thermal energy isn't strongly limiting the rate of this reaction, so increasing the temperature doesn't have a strong effect on the rate.
However, if RT is much less than the activation energy, then the lack of available thermal energy is strongly limiting the rate of reaction.  Consequently, increasing the temperature will have a large effect on the rate.

I remember van't Hoff ( I guess it was him ) empirical rule "Temperature increase of 10 degrees leads typically to chemical reaction rate increase 2-4 times". It would be 16-256 times per 40 degrees ( 340-298 = about 40 )

@Poutnik  Yes, that rule is nicely consistent with what I indentified as a typical Ea, since if Ea = 80 kJ/mol, and T_initial = 298 K, then increasing T by 10 K would increase k by 2.8 times, which is right in the middle of that rule's range.  More broadly, if that rule assumes T_initial = 298 K, then its 2 - 4 x rate increase is tantamount to assuming an Ea range of 53 – 107 kJ/mol (298 K ->308 K with Ea = 53 kJ/mol => 2-fold rate increase; 298 K->308 K with Ea = 107 kJ/mol =>4-fold rate increase).

Note that $A$ is a kind of a collision frequency factor, which is temperature dependent via the mean molecule speed, even if far less then the Boltzmann exponential factor.
$$\mathrm{d}k/\mathrm{d}T = \frac {\mathrm{d}A}{\mathrm{d}T} \cdot \exp{(\frac {-E_\mathrm{a}}{RT})} + A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1}$$
$$A = A_0 \cdot {({\frac {T}{T_0})}}^{1/2} \tag{2}$$
$$\frac {\mathrm{d}A}{\mathrm{d}T} = A_0 \cdot \frac 12 \cdot {({\frac {T}{T_0})}}^{-1/2} \cdot \frac {1}{T_0}=\frac {A_0}2 \cdot {({\frac {T_0}{T})}}^{1/2} \cdot \frac {1}{T_0} =\frac {A_0}2 \cdot \frac 1{\sqrt{T \cdot T_0}}  \tag{3}$$
By subtitution in (1), using  (2) and (3):
$$\mathrm{d}k/\mathrm{d}T = \left( \frac 12 \cdot \frac 1{\sqrt{T \cdot T_0}}   +  {({\frac {T}{T_0})}}^{1/2}  \cdot \frac {E_\mathrm{a}}{RT^2} \right) \cdot A_0 \cdot \exp{(\frac{-E_\mathrm{a}}{RT})}\tag{4}$$
As AJKOER properly noted, temperature dependance  $A=A(T)$ can be usually neglected and we can use the simplified (1), as $\frac {\mathrm{d}A}{\mathrm{d}T} = 0$:
$$\mathrm{d}k/\mathrm{d}T = A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1a}$$
unless the activation energy is extremely low and reaction rate is driven by diffusion ( typical one is $\ce{H3O+(aq) + OH-(aq) -> 2 H2O(l)}$  ).
In solvents, we cannot usually ignore intermolecular interactions, so temperature dependence of $A$ may get quite complicated.

With respect to your question on the math, I start with the Arrhenius equation:
$$k = A\mathrm e^{-E_\mathrm{a}/(RT)}$$
and would recommend simplifying the math by first proceeding with the introduction of a natural log transformation (this can also be used to linearly estimate $E_\mathrm{a},$ see discussion here):
$$\ln k = \ln A - \frac{E_\mathrm{a}}{RT}$$
Then, take the derivative of $\ln k$ with respect to $T$ we have:
$$\frac{\mathrm d \ln k}{\mathrm dT} = \frac{E_\mathrm{a}}{RT^2}$$
Note the positive impact (on the natural $\log$ of $k)$ with respect to the activation energy scaled by the inverse square of the absolute temperature.
To answer your second question, "Does 'A' change when temperature change (since my variable here is temperature)?", here are some pertinent comments per Wikipedia, to quote:

Given the small temperature range of kinetic studies, it is reasonable to approximate the activation energy as being independent of the temperature. Similarly, under a wide range of practical conditions, the weak temperature dependence of the pre-exponential factor is negligible compared to the temperature dependence of the $\mathrm e^{-E_\mathrm{a}/(RT)}$ factor; except in the case of "barrierless" diffusion-limited reactions, in which case the pre-exponential factor is dominant and is directly observable.

So, 'A' apparently only displays a weak temperature dependence, to answer your question.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Keep in mind \ce{…} macro is strictly for chemical equations and chemical formulae and should never be used for anything else. Symbols for math operators, mathematical constants and textual labels must be upright to avoid confusion with variables which are written in italics.

You should start from an idea of the variation in T and concentration that might be encountered, call these $\delta T$ and $\delta c$. If you have a model of the rate $r(T,c)$ then the associated variation in $r$ with respect to the individual variables can be determined from the partial derivatives as
$$\delta r_i = \left(\frac{\partial r}{\partial x_i}\right)_j\delta x_i$$
The different $\delta r_i$ can be compared directly to see to changes in which variable $r$ is most sensitive.
Remember that whatever equation you end up using to describe $r$ is just a model, it may be regarded as correct insofar as it accurately predicts data. Since you do not discuss any particular data that you are attempting to model, presumably you may assume $A$ and $E_a$ are constant.

You may want to double-check the indices in your chain rule.

Yes it does make sense to differentiate wrt temperature but in the Arrhenius model $A$ is a constant. However, if the pre-exponential $A$ is replaced with terms representing the bond that dissociates and other partition functions, i.e we replace $A$ with $A(T)$ a function that depends on temperature. The value of this term depends on the type or reaction, unimolecular, hard-sphere collision, atom + diatom etc. For two hard spheres $A(T) \sim T^{1/2}$ for example. This means, in effect, that the Arrhenius model is abandoned and transition-state theory is used instead.


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Keeping a catalyst suspension suspended
I am currently working on a heterogeneous photocatalyst project.  We are having significant problems with the catalyst not staying suspended in an aqueous solution.  What can I do to prevent aggregation of an oxide catalyst.  I suspect there are lots of possible solutions.
Some extra details about my specific system:

Aqueous solvent (pH: ~7)
Nanoparticles around 100-200 nm
Isoelectric point around pH 3 - 4
Roughly 0.2 mg of catalyst / mL

Nobody can suggest anything to a generic query without knowing any details. Mention the particle size of the oxide.

I will add some details

It would depend on the reactivity of the oxides, and the tolerable viscosity. A first thought would be some organic dispersant, perhaps siloxanes. See https://www.sciencedirect.com/science/article/abs/pii/S0927775714001691

How about you do your literature study?

To float for how long? An initial thought was add either sodium dodecylsulfate (SDS) (example and or polyethylene glycol (PEG). Vendors' lists may be an inspiration (example), «NP ink»  and «quantum dot» (both used e.g., for printing organic electronics) equally may yield hits.


text:
Polymerization of propene chemical equation
What is the equation to represent the polymerization of propene?
It has to be in the form of general formula:
$$\ce{H-R-OH + H-R-OH -> H-R-R-OH + H2O}$$
I have this so far:
$$\ce{CH3-CH=CH2-OH + CH3-CH=CH2-OH -> ...  + H2O}$$
I'm confused as to how this reaction would work.

What do either of your equations have to do with propene?

4 bonds of a carbon atom should be enough, don't you think ? BTW, others are confused as well, as propene polymerization does not work this way. It is not polycondensation.

The enol form of propaldehyde .... no, you gone up to a very wrong start. Grab your textbook, and try again after reading it once more!

Propene condenses to give Polypropylene, but your question body asks something very different from the title...

@AniruddhaDeb   Propene polymerizes to PP, (poly)condensation assumes elimination of small molecules like water.

Propene performs radical ( denoted by $\cdot$ ) driven polymerization:
$$\ce{R. + CH2=CH(CH3) -> R-CH2-CH(CH3).}$$
where the radical end is eventually terminated by recombination,stopping the polymerization. Polymerizations typically convert a double bond of a monomer to a single bond, freeing extra electrons to create extra bonds between monomers.
OTOH, polycondensation eliminates small molecules like water from atoms of bonding functional groups, like :
Polyethylene terephthalate ( PET ) production eliminates water molecules :
$$\ce{R-COOH + HO-CH2-CH2-OH + HOOC-C6H4-COOH \\-> R-CO-O-CH2-CH2-O-CO-C6H4-COOH + 2 H2O}$$
Similarly polycarbonate eliminate hydrogen chloride ( that is immediately neutralized by hydroxide):
$$\ce{R-OH + COCl2 + HO-dian-OH ->[NaOH] R-O-CO-O-dian-OH + 2 HCl}$$


text:
Average bond energy of H−F and H−O hydrogen bonds

Which of the following hydrogen bonds (shown by the dotted line) has the highest average bond energy?
A) $\ce{F-H\bond{...}F}$
B) $\ce{F-H\bond{...}O}$

I am thinking the answer is $A$ since the covalently bonded F atom, being highly electronegative, would induce a high partially positive polarity on the H atom. But on the other side, the more electronegative atom F will form a stronger Hydrogen bond than the comparatively less electronegative O atom.
Any flaw in my reasoning?

So fluorine does have the ability to hydrogen bond. However, they are usually transient and depend significantly on the species that is donating the H-bond along with some other factors (i.e. what phase the molecules are in, temperature, etc...)
Fluorine is so electronegative that it holds its electrons much "tighter" then other atoms, making it a poor H-bond acceptor. Oxygen, being less electronegative, is a better H-bond acceptor since the electrons aren't held onto as tightly, allowing them to interact more with other atoms.
Since bond energy is "the average value of the gas-phase bond-dissociation energy for all bonds of the same type within the same chemical species. The larger the average bond energy, per electron-pair bond, of a molecule, the more stable and lower-energy the molecule" (IUPAC WEBSITE), I would suggest that the H-bond with oxygen has a higher bond energy, as it is the stronger interaction of the two.
EDIT
The OP requested some sources, in which I failed to do initially, so I apologize for that but here they are.

J. Am. Chem. Soc. 1983, 105, 10, 3206–3214

Tetrahedron, Vol. 52, No. 38, pp. 12613-12622, 1996

J. Am. Chem. Soc. 1962, 84, 16, 3221–3222

Here is a quote from reference 2.

Theoretical calculations variously estimate 5 the strength of a F...H bond to be between 2 to 3.2 kcal mo1-1.
This can be compared 6 to an O...H hydrogen bond which is typically between 5 - 10 kcal mo1-1. Consistent
with this the electrostatic influence of fluorine is approximately half that of oxygen 7. Thus the greater
electronegativity and lower polarisability of fluorine over oxygen, suppresses its electrostatic influence and
renders it a poorer hydrogen bond acceptor.

The last reference discusses hydrogen bonding in halides, not too sure how much it compares them to oxygen bonding but it is certainty worth taking a quick look.

I've searched on the internet and at many reliable sites, I found that the strength of the hydrogen bond depends on the difference of electronegativity between Hydrogen and the Hydrogen bond acceptor atom, "H...F>H...O>H...N" specifically written.

@AryanVerma Please reference the websites you are referring too. In the meantime I will edit some references into my answer. Just so we are clear, you are suggesting that fluorine is a stronger H-Bond acceptor then oxygen?

https://nimysart.com/hydrogen-bond/
https://courses.lumenlearning.com/introchem/chapter/hydrogen-bonding/#:~:text=Hydrogen%20bonds%20are%20strong%20intermolecular,increase%20in%20hydrogen%2Dbond%20strength  ('Key Point' number 2)
https://chemistryguru.com.sg/factors-affecting-hydrogen-bond   (Under the heading "Polarity")
I personally think that larger the electronegativity of the Hydrogen acceptor atom, larger will be the attractive force exerted by it on the electron cloud of the Hydrogen atom.. leading to greater bond strength. Therefore, I guess H...F Hydrogen bond is stronger

@AryanVerma In your reference, it is talking about the polarity of the molecule, not its ability to form strong hydrogen bonds. Because fluorine is SO electronegative, its electron shell is not very polarizable and it cannot form interactions as strong as oxygen. However, I can see how that can be misinterpreted, as the author does not make it that clear.  Additionally, sometimes supporting ones claims requires primary sources, rather than online tutoring services. That is just some advice from someone who has done the same thing.

In gas phase rotational spectroscopic experiments, the H-bond force constant has been related to bond strength (JACS, v109,356,1987.) via 'electrophilicity' of donor and 'nucleophilicity' of acceptor. This shows that FH-OH2 has approx twice the strength of FH-FH hydrogen bonds.


text:
Naming and structure of PtCl4.2HCl (complex/coordination compound)
Given that $\ce{PtCl4.2HCl}$ when reacted with excess of $\ce{AgNO3}$ produced zero moles of $\ce{AgCl}$, I understand that all the chlorine atoms are inside the coordination sphere. But what will the formula and thus name of this complex be? And what could the coordination polyhedron look like? Would the name be
$$\ce{H2[Pt(Cl)6]}$$ or  $$\ce{[Pt(Cl)4(HCl)2]}$$
If someone could explain why the name is one and not the other, I'd be very grateful. Thank you in advance!

Platinum(IV) chloride adopt an octahedral coordination geometry. This geometry is achieved by forming a polymer wherein half of the chloride ligands bridge between the platinum centers. So, whenever it is dissolved in some acids, say $\ce{HCl}$, the chloride bridging ligands breaks and the molecules gets added giving $\ce{H2PtCl6}$, also called chloroplatinic acid.

But the story doesn't end here. Chloroplatinic acid is highly hygroscopic. It pulls water from air and forms an ionic complex consisting of  hydronium ($\ce{H3O+}$) cation and hexachloroplatinate ($\ce{PtCl6^{2-}}$) anion. Its actual formula is $\ce{[H3O]2[PtCl6](H2O)_x}$ where $\pu{0 ≤ x ≤ 6}$. But for simplicity, we write the former formula.

but which one out of $$\ce{H2[Pt(Cl)6]}$$ and  $$\ce{[Pt(Cl)4(HCl)2]}$$ is the right formula for the coordination complex i mentioned? besides i dont this is platinum chloride, or is it?

@sam the first one is the correct one. Its name is chloroplatinic acid.

Neither $\ce{H2PtCl6}$ nor $\ce{PtCl4•2HCl}$ is strictly correct. As explained by Nilay Ghosh, only $\ce{(H3O)2PtCl6}$ (typically with additiinal water of hydration) is known in practice, for the methods used to produce the acid all involve some form of aqueous processing. Any proposed hydrogen ions or hydrogen chloride will thereby be leveled to $\ce{H3O^+}$. Even after that levelling the compound is still hygroscopic tending to draw in additional water of hydration from air or liquid water. The formula $\ce{H2PtCl6}$ is only a shorthand version of the true formula.


text:
Stability of Gauche and Anti form in conformational Isomerism
I came through this question: Which form is stable Gauche or Anti in F-CH2-CH2-F?
My answer: Due to lone pair repulsion Anti form should be more stable but the answer is given Gauche form. Please tell me how is gauche form more stable in this case?

in gauche form there is intramolecular hydrogen bonding between H and F which makes it stable...

This problem should be referred to concept called stereoelectronic effect. It is stabilized by the donation of $\sigma_{C-H}$ bond to $\sigma_{C-F}$ bond, leading to decrease energy of the whole system.


text:
How is pumping "Air" into a bicycle make dW (delta work )negative , when in fact there is a increase in pressure and a slight increase in volume?
Considering that it's an adiabatic process hence $\mathrm dQ=0$, and $\mathrm dU=\mathrm dW$. And since the work is done on the system then $\mathrm dW$ must be positive.
(The answer which is in the textbook is that $\mathrm dU$ is positive, while $\mathrm dW$ is negative.)

The book must be using the sign convention that dW is the work done by the sysetem on the surroundings, and -dW is the work done by the surroundings on the system.

Depending on context and domain dependent convention, the work done is assumed in 2 different, opposite ways, with opposite signs.
Work done on a system $\Delta U = Q + W$, preferred by chemistry
versus
work done by a system $\Delta U = Q - W$, historically preferred by physics. It is used by the former way as well recently.

Speculatively said, physics way was defined during creation of theoretical background for thermal machines. Physicists may had been focusing on work done by machines.
Later chemists came, who invented their way, focusing on energy of systems.

As a prevention, it is always good to explicitly mention which way is used, to prevent confusion.
If applied, your pump is doing positive "chemical" work on the system=tyre, but the system=tyre does negative "physical" work on the pump. The textbook takes the convention of physics.

In an important sign convention, preferred in chemistry, work that adds to the internal energy of the system is counted as positive. On the other hand, for historical reasons, an oft-encountered sign convention, preferred in physics, is to consider work done by the system on its surroundings as positive.

Wikipedia - Work_(thermodynamics)

It's a simple matter of energy conservation. Here no heat can flow between surrounding and system. If work is done by the system, where do you think that energy comes from?
When a gas does expansion work, a part of kinetic energy is used up in doing work and average kinetic energy of the gas decrease. I ask you, what is the measure of overall kinetic energy of a gas? It is the temperature. Since kinetic energy decreases, the temperature also decreases. Internal energy for an ideal has depends only on temperature of the gas, so the internal energy decreases when work is done by an ideal gas in adiabatic process
The equation for first law of thermodynamics might look different for physics and chemistry but it is just a matterof sign convention.
For convenience, this equation will always be true:
$$Q = \Delta U + \int p dV$$
For adiabatic processes this becomes
$$-\Delta U = \int p dV$$
On expansion $dV$ increases so $U$ decreases. $dV$ and $\Delta U$ will have opposite signs

I think the most common error in thermodynamic chemistry is the convention of the equation. Some supposed that the first law equation should be $U = Q + W$, the others supposed $U = Q - W$. Because of the sign in equation W with respect to $pdV$. But all these things do not change the system intrinsic.

There are two categories of scientists on this planet. Some, like you, Bagya Aluthge, consider that all energies (heat $Q$, work $W$) entering a system are positive, and contribute to increase the internal energy $U$, so that $\Delta U = \Delta Q + \Delta W = \Delta Q - p\Delta V$. But some engineers consider that the system is a machine which, when heated ($\Delta Q > 0$), must produce some work $\Delta W$, which is then positive : $\Delta W = + p\Delta V$. For those engineers, the internal energy $U$ is the part of the heat that is not transformed into work : $\Delta U = \Delta Q - \Delta W = \Delta Q - p\Delta V$.
So, both both categories of scientists, the result is the same for $U$ : $\Delta U = \Delta Q - p\Delta V$. Only the sign of the work $p\Delta V$ is different.

It is rather that physicists, creating background for thermal machines and focusing on work done by machines, came before chemists, who invented their way, focusing on energy of systems.


text:
Eudiometer Problem?
In lab, students collect 42.6mL of N2 gas over water 24 C and 751 torr. The water in the eudiometer is 16.5 cm higher than the water outside.
Calculate the dry standard volume of N2?
I'm given the volume with water vapour i assume and i need to calculate the dry volume
"  The water in the eudiometer is 16.5 cm higher than the water outside. "
This sentence brings all the confusion, does someone has a picture that explains it? does it mean that the atmospheric pressure is higher by 16.5cm water, then i convert that to mmHg.
hence
PN2=Patm-P(difference)-Pwater
and then
PV=nRT
V=nRT/P
but still i have no moles, any hints?

@Poutnik so my sentence in the original post makes sense that atm pressure is higher by 16.5 cm H2O?

Ok 16.5cm H2O is 12.13mmHg=12.13torr
that means that the pressure of the gases is 751-12.13=738.87torr.
how from there i can get the volume of dry N2?
@Poutnik

I know im given the partial pressure of water vapour at a specific temperature so i can find the partial pressure of N2 alone, but still i dont have the dry volume or the moles, if i had the moles i can calculate the dry volume but i dont.

I guess you need not us to check if you can resolve simple arithmetic. Involve independent investigation and express it in the question, otherwise it could get closed for insufficient explicitly written effort of the OP. Hint - search for saturated vapour pressure table, in case the task expects the correction. Note that n=pV/(RT)

Yes but what is V? is it 42.6ml? because that is the volume of both gases (Water vapour and nitrogen) can i still use it? @Poutnik

Really ? Imagine it is not a wet gas, but air with O2 partial pressure 21 kPa. How would you calculate n for O2 ?

I truly apologize if i seem so dense, i dont know why this is confusing me, if i put 42.6 as volume i can get the number of moles of gas that exist in the volume which is N2 and H2O.

Surely you can. the question remains, if they expect water vapour correction or not, but I cannot decide that.

Let us continue this discussion in chat.

HINT - The picture below shows the situation.


text:
Can we relate approximate Hartree-Fock orbitals to true solutions in the basis-set limit?
Assume that we've "solved" the RHF Hartree-Fock equations in some finite basis, that is, we have arrived at a self-consistent set of coefficients $c_{\mu{}i}$ such that each spatial molecular orbital is expanded as $\chi_i = \sum_{\mu=1}^K c_{\mu{}i}\varphi_i$. By construction, this set of orbitals is orthonormal, and forms a collection of eigenfunctions for the Fock operators in the given basis, so $f_i |\chi_i\rangle = \varepsilon_i |\chi_i\rangle$ where $\varepsilon_i$ is the $i$th orbital energy.
Question: what, if anything, can we say about these orbitals in terms of their relationship to the "true" orbitals in the basis set limit, denoted $\chi_i^\mathrm{BS}$? Here, I mean those orbitals that would be obtained if the calculation were to be extended to an (infinite) complete basis set. (I realise that the assumed existence of a single unique set of true orbitals may not be justified...)
For example, is it possible to estimate, bound, or otherwise describe the overlap $\langle \chi_i | \chi_i^\mathrm{BS} \rangle$ between an approximate solution and the true solution? How about matrix elements such as $\langle \chi_i | \mathcal{H} | \chi_i^\mathrm{BS} \rangle$, where $\mathcal{H}$ is the electronic Hamiltonian, or indeed any self-adjoint operator?
I suspect/hope that some kind of a statement towards the first part might be possible, given that the Roothan-Hall equations are effectively a collection of Galerkin equations. One of the standard hand-waving statements about solutions obtained using Galerkin methods is that they are in some sense projections of the true solution onto finite-dimensional subspaces, and have residual vectors that are "orthogonal" to that subspace(for a potentially complicated definition of "orthogonal").
(I'm also interested in answers that would apply in any of the HF settings, i.e. UHF, ROHF, etc.)

This is a very interesting question. The answer also has many possible practical uses that have not been explored. For instance, extrapolation from the finite basis to the complete basis is often done to estimate the CBS energy, but as far as I'm aware this has rarely, if ever, been attempted for the orbitals themselves. I suppose this is because one generally cares less about the actual HF orbitals than observables associated with them. I'll be interested to see any answers.

In quantum chemistry numerical experimentation is the norm and rigorous mathematical proof the rare exception. As a result, not many rigorous results are know, not even for the good-old Hartree-Fock method. For a review of mathematical results I recommend you have a look at these references:

Claude Le Bris, Computational chemistry from the perspective of numerical analysis, Acta Numerica (2005), pp. 363–444
Pablo Echenique and J. L. Alonso, A mathematical and computational review of Hartree–Fock SCF methods in quantum chemistry, Molecular Physics, Vol. 105, Nos. 23–24, 10 December–20 December 2007, 3057–3098

In particular, section 3.3 of the first reference is relevant, but don't expect to find mathematically strong results. For example, here's a quote:

The choice of an AO basis for solving
a given problem mostly relies upon some practical know-how. The lack of
rigorous understanding is a pity, because the output of the calculations
(typically some molecular property) might be very sensitive to the choice
of the basis set. The only available measures of the quality of the basis
set are obtained, in the chemistry literature, by choosing test cases, i.e.,
reference systems, where the solution of the exact Schrödinger equation may
be computed, mostly through numerical computations and, when possible,
with the help of an analytic calculation.

From a practical point of view I don't see why occupied Hartree-Fock orbitals expanded in the usual Gaussian basis sets should not converge (in the integral norm sense, ie leaving aside the lack of cusp at the nucleus) to the exact solutions of the Hartree-Fock equations. I have little idea as to what happens to the virtual orbitals.


text:
Acidity of carboxylic acids
is there a reliable database with the pKa of halogenated carboxylic acids? I am also looking for the pKa of ethanoic acid substituted with other electronegative elements (like sulfur and oxygen).

You might want to try “Dissociation Constants of Organic Acids and Bases” in CRC Handbook of Chemistry and Physics.

With some effort, you may retrieve some data from one of the now preserved public Reich collections, e.g., pages 4  and following in William's compilation. Note, water needn't to be the only solvent to record these data, this signpost for example hints to many additional primary references about pKa based on experiment or/and computation, e.g. in acetonitrile / MeCN.


text:
How to extract pure potassium carbonate from ash?
In our village, lots of ash is produced which is thrown off. I want to extract pure potassium carbonate from that ash. I am able to extract the potassium compound and a soluble compound by recrystallizing and evaporating but I found impurities. The first test I performed is electrolysis. This test confirmed the presence of chlorine gas which is able to change the litmus paper to white. The second test I performed is flame test. During the flame test, I got a yellowish orange color while my expectation was bluish violet. So, I assume there is sodium. How can I extract the pure compound?
To conclude, there is a small amount of potassium chloride, sodium chloride, sodium carbonate which are the impurities.

Please try to structure your question. Literally.

In scientific texts, it is recommended for readability reason to keep average word count per a sentence 20-25 words with avoidance to use more than 35 words. Your first sentence has 87 words.

@Ananiya I corrected the grammar and punctuation in your question. If I alter your question in any way, feel free to roll back the edit.

Historically speaking, potassium derives from pot ash and now potash when -- if there were no geologic sources of $\ce{K2CO3}$ -- woods were burnt in a pot.  The addition of water will dissolve $\ce{K2CO3}$ and other high soluble matter, but not the ash.  According to wikipedia, the addition of amine may lower the amount of KCl, but basically then, as by Poutnik's numbers, it is fractional crystallization.
Additions:

You may find temperature dependent solubility of $\ce{K2CO3}$ as sole solute in water, e.g. by Schultz et al. in Ullman's Encyclopedia of Industrial Chemistry (doi 10.1002/14356007.a22_039):

\begin{array}{c|c}
\mathbf{Temperature (^\circ C)} & \mathbf{Solubility(g~K_2CO_3/ 100~g~H_2O)} \\\hline
\text{0} & \text{105.5}\\
\text{10} & \text{108.0} \\
\text{20} & \text{110.5}\\
\text{30} & \text{113.7}\\
\text{40} & \text{116.9}\\
\text{50} & \text{121.9}\\
\hline
\text{60} & \text{126.8}\\
\text{70} & \text{133.1}\\
\text{80} & \text{139.8}\\
\text{90} & \text{147.5}\\
\text{100} & \text{155.7}\\\end{array}
The caveat however is that the precipitate obtained from these solutions in this temperature range isn't $\ce{K2CO3}$, but the sesquihydrate, $\ce{K2CO3 * 1.5 H2O}$.  Like $\ce{K2CO3}$, it has to be stored in a dry environment or otherwise, deliquesces.  It looses water if heated to $\pu{130-160 ^\circ{}C}$.

Consider the economic and ecologic perspectives: Synthetic $\ce{K2CO3}$ may be much cheaper than collecting and washing the ashes, and subsequent steps of purification.  It may be purer, with less salination problems for farmers, too.  E.g., how do you will treat waste waters containing other salts than $\ce{K2CO3}$?  How do you generate the heat to recrystallize the material?  Don't get into the rut by yourself or / and your friends to collect wood to heat the dishes, which generates new ashes, which again will be leached ... leading quickly into a disaster.  If put into execution, see how solar energy (e.g., evaporation under the sun to preconcentrate, solar ovens to boil) may be used instead.

The composition of the wood ashes varies with the nature and the origin of the wood. In the average, it is a mixture of $40$ % - $70$ % $\ce{CaCO3}$, $5$% - $10$% $\ce{MgCO3}$, $5$% - $10$% $\ce{K2CO3}$, $5$% - $10$% $\ce{Na2CO3}$, $2$% - $5$% $\ce{SiO2}$, $2$% - $5$% $\ce{Ca3(PO4)2}$, $0.5$% -$2$% $\ce{Al2O3}$, <$1$% $\ce{NaCl}$, <$1$% $\ce{KCl}$, plus traces of manganese and iron. So it is difficult to purify, and/or to extract any of these compounds, as a lot of these compounds are insoluble in water: $\ce{CaCO3, MgCO3}$, $\ce{SiO2, Ca3(PO4)2, Al2O3}$. The only soluble compounds are $\ce{K2CO3}$ and $\ce{Na2CO3}$, plus traces of $\ce{NaCl}$ and $\ce{KCl}$, and they form less than $20$% of the total mass of the ashes. The solution obtained from washing the wood ashes is a mixture of sodium and potassium carbonate. The problem is how to separate those two carbonates. Fractional crystallization does not seem easy. I would rather transform them into nitrates by adding nitric acid : at $0°$C, $\ce{KNO3}$ is  not very soluble ($5$ g/$100$ mL), which is much less than $\ce{NaNO3}$ ($73$ g/$100$ mL). So obtaining $\ce{KNO3}$ from the mixture by cooling a hot solution should be relatively easy.
Ref.: J. Hartwig, Ann. der Chem. u. Pharm. XLVI, p. 97. A. Wurtz, Dictionnaire de chimie pure et appliquée, Vol. 1, p. 641 (1891).

It seems that by partial crystalization, it can be somewhat purified as the most soluble salt, but cannot be easily purified to high purity degree. It may be andvantageous to crystalized near water freezing point.
Solubility in $\pu{g/100 mL}$ for $\pu{0^{\circ}C, 20^{\circ}C, 100 ^{\circ}C}$
$\ce{KCl}$:28, 34.1, 56.3
$\ce{K2CO3}$: 105, 111, 156
$\ce{NaCl}$:35.7, 35.9, 39
$\ce{Na2CO3}$: 7, 21.5, 45.5
Potassium salts and Sodium salts

The trick is to convert the soluble K2CO3 to the more insoluble KHCO3.  Extract the ash with warm water; cool, filter to remove CaCO3 and other insolubles. Get together with the beer and wine makers and bubble their CO2 in and precipitate KHCO3, Filter, dry and heat to get K2CO3. You must refine the process to get good yields. It might be practical to recover Ca and Mg carbonates by treating a slurry of the filtered solids with CO2 since their bicarbonates are reasonably soluble. Boil the solution to precipitate the carbonates. BTW your tests for purity are specious. There are very low levels of sodium, chlorine in fresh water vegetation; that is why herbivores look for salt licks.

Why the downvote?  This is the only answer that makes chemical sense and is not simply  a bunch of hand waving.


text:
Regarding reversible phase transition of a real gas
According to change in entropy for a reversible phase transition, say from liquid to gas at constant pressure $p$ and temperature $T$
: $\Delta S =\frac{\int dq_p}{T} $  which comes out to be $\Delta S =\frac{\Delta H_{vap}}{T_b}$. This means that this process is spontaneous since Entropy change is positive. But the Gibb's Energy change for this process is $0$. Why is that?
Change in entropy for this process suggests that this is a spontaneous process but Gibb's Energy change suggests it is not. Why does irregularity occur?

A positive change in entropy does not necessarily mean that a process is spontaneous.  Where did you get that idea?  That is only true for an isolated system.

See the answer I have written here, I have derived spontaneity condition of Gibb's from scratch

$$\Delta S_\mathrm{sys} =\frac{\Delta H_\mathrm{vap}}{T_b}$$
is the entropy change of the system.
$$\frac {\Delta G}{T_\mathrm{b}} = \Delta S_\mathrm{tot} = \Delta S_\mathrm{ext} + \Delta S_\mathrm{sys} = \Delta S_\mathrm{ext} + \frac{\Delta H_\mathrm{vap}}{T_b} = 0$$
is proportional to the total entropy change.

There are 2 unreasonable points your question raised.
Firstly, you made a mistake when you transformed $\triangle S = \frac{\int dq_p}{T}$, because $dq$ should equals $-dH_{sys}$. The above answer replicated that mistake too. Because $\triangle S_{sys}$ is state function, it can not be transformed like that, actually $\triangle S_{ext} = \frac{\int -dH_{sys}}{T}$, so $\triangle S_{tot} = \triangle S_{sys} - \frac{\triangle H_{sys}}{T}$. (those equations can be found in any physical chemistry books).

Secondly, you can not distinguish between standard-state Gibbs energy and Gibbs energy. We have that: $\triangle G = \triangle G^o + RTlnQ$ with Q is the proportion of component. In order to make one reaction spontaneous, Gibbs energy $(\triangle G)$ should equal 0, even though whatever $\triangle G^o$ is.

It's unclear what you're trying to say. According to https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node41.html the change in entropy of the system for a reversible process is in fact $\int \frac{dq}{T}$. It was just that in this case the pressure and temperature were constant because of the phase transition. So the heat flow is equal to the change in enthalpy $\Delta H_{sys}$. And for $\Delta G$ I applied the equation $\Delta G= \Delta H - T\Delta S$.

Q is defined as an amount of heat transferred to out of the system. However, change of enthalpy is defined as change of heat of the system. So, $\triangle H_{system}$ must have different sign regarding to $Q$ or $\triangle H_{system} = - Q$.

Additionally, we assume that: $\triangle S_{tot} = \triangle S_{sys} + \triangle S_{ext} = \triangle S_{sys}+ \frac{-\triangle H_{sys}}{T}$So: $\triangle H_{sys}-T\times \triangle S_{sys} = -T\times \triangle S_{tot} =-\triangle G^o_{sys}$ (let's remember $\triangle G^o$, not $\triangle G$)

NO!! change of enthalpy IS NOT defined as change of heat of the system at all. This statement is only true for isobaric process. $\Delta H=q_p$, $\Delta H$ will not be equal to actual heat transfer in the system. Q is not defined as the amount of heat transferred out of the system. It is simply the transfer of heat in or out of the system. It doesn't matter if it is coming in or going out.

$Q_p$ and $Q$ are different, my friend. These make physchem so hard.

I'm not saying they're the same. I'm saying $\Delta H\neq Q$ except for isobaric processes.

Yeah, but different sign. Sign of any quantity depends on how we define it.

No, not just different sign but they are not even numerically equal.


text:
A problem regarding the first law of thermodynamics

2 moles of an ideal gas are in a closed system with a temperature of $25\ \mathrm{^\circ C}$.
The gas is cooled under constant pressure and afterwards heated gas under constant volume.
During the cooling process the gas has lost $1000\ \mathrm J$ of heat to the environment.
It is known that the molar heat capacity of the gas is $C=25\ \mathrm{J/(^\circ C\ mol)}$
What is the work done by the system during the cooling and heating combined?

Official answer: $333\ \mathrm J$.
The problem: I understand how to get to this solution, however I don't understand why an alternative method yields an incorrect answer. Here's what I tried:
First, during the heating process there is no work done because the volume is constant.
$Q=-1000\ \mathrm J$
We know that:
$\Delta E=\frac32\cdot n\cdot R\cdot\Delta T$
Now, $Q=n\cdot C\cdot \Delta T$, therefore:
$-1000=25\times2\times\Delta T$ and therefore $\Delta T=-20$
So $\Delta E=\frac32\cdot n\cdot R\cdot\Delta T=\frac32\times2\times8.314\times(-20)=-498.84\ \mathrm J$
By the first law of thermodynamics: $\Delta E=Q+W$, therefore $-498.84=-1000+W$ and $W=501.16\ \mathrm J$
Could anyone please tell me where is my mistake?

When a problem requires calculations using values, always write the values with the correct units and carry the units through the calculation. Do not omit the units while performing intermediate steps and do not just reintroduce units at the end of the calculation.

The question didn't specify which process the given molar heat capacity was specified for. So we will assume that it is molar heat capacity for the first process. The first process is isobaric so $$q=nC_p\,\Delta T$$
You calculated $\Delta T$ correctly. The only mistake you did was assume that the gas was monoatomic. The expression for $\Delta U$ you wrote was $\Delta U=\frac32nR\,\Delta T$. This expression would be correct if this was a monoatomic gas. But we don't know that do we?
The correct expression to use here is $$\Delta U=nC_V\,\Delta T$$
It is given that molar heat capacity for first process (constant pressure) is $C_p=25\ \mathrm{J/(^\circ C\ mol)}$ and  we know that $$C_p=C_V+R$$
So you can calculate the value of $C_V$ and use it to get the value of $\Delta U$ which I calculated to be $$\Delta U=667.2\ \mathrm J$$
Plug this into first law of thermodynamics and you will get the correct answer.
Edit: I used the sign $\Delta U$ for Change in Internal Energy intead of $\Delta E$

Firstly, I do not know what coefficient $3/2$ of your $\Delta E$ means, it makes none sense.
From data of the problem above, we can consider this process as an irreversible one. Additionally, because the latter process is isochoric process, the work done by that also equals $0$. So the work done by all process is the work done by isobaric process.
In the other hand, similarly to your calculation of $\Delta T$, now we can calculate the first work by equation $\Delta W=-p\,\mathrm dV=-n\times R\times \Delta T=- 2\ \mathrm{mol}\times8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}\times 20\ \mathrm{^\circ C}\approx-333\ \mathrm J$.
So, the work done is $333\ \mathrm J$.

What is it about the data of the problem that leads you to conclude that this process is irreversible???

I managed to get the answer using this method but can't figure out what's wrong with the other method. For an ideal gas we were taught that the energy is 3/2nR*T

That's only for a monoatomic ideal gas.

The Equation you used is the internal energy involving in kinetic molecular energy, not the change in the internal energy.

@ChetMiller This was never mentioned in my class and was used extensively even when it wasn't known that the gas was monatomic.

You should distinguish some basic concepts. They can be found in any physical chemistry textbooks.

I'm shocked.  What kind of Thermo course is this?  Please see the Answer given by @Eyy boss (which solves this the correct way).

The problem is the question requires to determine the work done by all, so that $/Delta U = Q + W = -1000 + W = 667.2 J$ => W = . You see that point!!!  actually, we do not need to calculate by a sophisticated way like that.

You said the the process is irreversible, so $p_{ext} \neq p_{int}$ . Work done for irreversible process is $W = -p_{ext} \Delta V$. You can't use ideal gas equation here. According to you $p_{ext} \Delta V = nR\Delta T$, but that's wrong. The correct equation will be $p_{int} \Delta V = nR\Delta T$. That's why thermodynamics focuses on reversible processes because for reversible process $p_{ext} \approx p_{int}$ and we can then use ideal gas equation.