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Why can't there be a gas-dispersed-in-gas colloid?
Colloids, a special type of heterogeneous mixtures, typically have two components. One is the dispersed phase (DP), which is the "solute-like" part and another one is the dispersion medium (DM), which is the "solvent-like" part. The different interactions of the DP and DM define the properties of a colloid.
My textbook has a few examples based on the different types of DP and DM, but it doesn't have any reference to a colloid where both DP and DM are gaseous. I asked my teacher about it and he told me that it was because gases can flow into each other and hence no heterogeneous mixture, and hence no colloid, of them can ever be possible.
That makes sense, but doesn't it also rules out the existence of colloids with DP and DM both being liquids? Liquids can also flow fairly easily right? Why is the science behind both of them not the same? And, WHAT exactly is the science behind both of them?

https://chemistry.stackexchange.com/questions/43320/why-do-we-consider-all-gasses-to-be-in-one-phase https://chemistry.stackexchange.com/questions/43118/can-gases-be-immiscible

https://chemistry.stackexchange.com/questions/58420/why-does-a-distinction-between-homogenous-and-heterogenous-mixtures-exist

The nature of gases is their molecules are separated while the nature of heterogenous dispersed mixtures is they are not separated. Therefore a heterogenous ( in equilibrium ) mixture of gases is an oxymoron.
Liquids are different case, forming droplets, kept by the surface tension. In contrary to gases, mutual dissolving of liquids depends on intermolecular interactions. such droplets may even have micelar nature, like n-amylalcohol in water.
Once I have tried to replace unavailable i-amylalcohol by available n-amylalcohol for extraction from water and deeply regretted it, seeing the stable emulsion. i-amylalcohol does not have such behavior, as the $\ce{(CH3)2CH^{}-}$ group is an obstacle for building such structures.

If I am getting it correct, what you want to say is that liquids don't exist seperately as molecules, but as droplet like entities and that is the reason why it isn't possible for them to behave like gases while forming solutions and mixtures?

Non miscible liquids form droplets. Miscible ones dissolve each other into true solutions with separated molecules, interacting all mutually by van der Waals forces and eventually forming hydrogen bonds. // Why not to read what gas and liquid mean on molecular level? All this is direct conclusion of that.

Let's take gas 1 as the dispersed phase (which is already at molecular level) which disperses evenly throughout the dispersing medium which is  gas 2 (also at molecular level). Thus, the system is already homogeneous at the molecular level. So, we can consider it "solution". For e.g. The atmosphere is a solution. Take the major two gases namely nitrogen and oxygen and we can consider the atmosphere to be "oxygen dissolved into nitrogen".
For a detailed discussion, see: Name of particles in a suspension


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What is the most stable cycloalkene?
I'm looking specifically at cyclopentene and cyclohexene. Supposedly, greater instability would lead to a more exothermic enthalpy of combustion. However, based on my data, cyclopentene is more unstable than cyclohexane and yet has a less exothermic enthalpy of combustion. Is my data wrong? What is the most stable cycloalkene?

Welcome to Chemistry SE!
Be aware that laconic, not elaborated questions are usually closed on the StackExchange network.    How do I ask a good question?.

Molar or specific enthalpy of combustion ?
For the former, larger molecules have obviously more negative molar combustion enthalpy.
For the latter, hydrocarbons with higher H/C ratio ( 1.6  cyclopentene vs 1.67 cyclohexene) have  more negative specific combustion enthalpy.
Instabilities play rather minor role in slight shifting the values. The best is comparing isomers, if applicable, or to make corrections for different number of "ordinary" $\ce{C-C}$ and $\ce{C-H}$ bonds and different combustion enthalpy contribution by their conversion to $\ce{H2O}$ and $\ce{CO2}$.

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Could you explain why the iron (ii) carbonate re-dissolves and how this can be described as an acid/base reaction , by mixing FeSO4 and Fe2+ salts
Iron(II) carbonate is insoluble but in class you might not have seen any solid. The precipitate often redissolves because the iron(II) sulphate has acid added to it. Can you explain why the carbonate redissolves and how this can be described as an acid/base reaction
I know little to nothing about the topic and have done quite a bit of research but I can't find a simple definition that I am able to understand

Welcome to the chemistry site of the StackExchange (SE) network.  Be aware that laconic, not elaborated questions are usually closed on the SE.
The site expects that you include the compact summary of your related current knowledge, involving your conclusion of searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself. Effort not shown may be considered as effort not done, possibly leading to the question closure. How do I ask a good question?.

Ferrons carbonate $\ce{FeCO3(s)}$ is like calcium carbonate $\ce{CaCO3}$ : it is soluble in carbonic acid with formation of the hydrogenocarbonate ion $\ce{HCO3^-}$in a reaction : $$\ce{FeCO3(s) + H2CO3 -> Fe^{2+} + 2 HCO3^- \tag{1}}$$ This means that $\ce{Fe(HCO3)2}$ is soluble in water.
$\ce{FeCO3(s)}$ can also be dissolved by the weak acidity produced by the hydrolysis of the  $\ce{Fe^{2+}}$ ions coming from $\ce{FeSO4}$ in water, according to the two equations ($2$) and ($3$)  : $$\ce{Fe^{2+} + H2O <=>  [Fe(OH)]^+ + H^+ \tag{2}}$$
$$\ce{FeCO3 + H^+ -> Fe^{2+} + HCO3^- \tag{3}}$$

$\ce{FeCO3}$ is an insoluble substance that does exist in water. $\ce{FeSO4}$ does not exist in solution, as it is soluble, and produces $\ce{Fe^{2+}}$ ions and $\ce{SO4^{2-}}$ ions. So the two reactions (1) and (3) of $\ce{FeCO3(s)}$ with acids do not occur with $\ce{Fe^{2+}}$ as it would mean that $\ce{Fe^{2+}}$ produces $\ce{Fe^{2+}}$ which is absurd.

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Determining chemical state and oxidation state from XPS analysis
This is an call for help! I have performed XPS analysis of my samples to get chemical state in which each element is present. I had given the samples to a third party lab. The operator has provided me with an excel sheet and I am attaching a picture of the peak tablehere. I have no prior experience in analysing XPS data and am having a hard time in deciphering the chemical state of each element.

The operator has given me the fitted data for each element and binding energy range for each element. I was under the assumption that I just need to compare the peak binding energy of each element to the reference databases available. But for my peak binding energy/energies there is not reference chemical state for that element.
For eg. For Pt 4F5/2 orbital, peak BE is 74.76 but there is no reference chemical state for this Pt with this orbital. I tried searching in open sources databases and in literature. In this case, according to papers there can be a shift depending on the chemical environment surronding the atoms but how to decipher this information i.e what shift is happening and due to what?
Any help or any link to any good literature which illustrates the process of finding chemical state of elements from fitted data will be highly appreciated.

XPS manufacturers also have reference tables for each element. I recall that very well. Please ask the operator to provide the name of reference handbook and then visit a good library. He/she must be using them regularly. Don't waste your time on web searching and open databases. Not everything is free. There are thick reference books that tabulate these but their name escapes now.


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Can halogens show -3 or -5 or -7 oxidation state. Why?
I mean can we think it like if the electrons get excited into the d orbitals and because there are now more than one unpaired electron, why does it only loose them and why don't they accept more than one electrons?

Welcome to the chemistry site of the StackExchange (SE) network.  Be aware that laconic, not elaborated questions are usually closed on the SE.
The site expects that you include the compact summary of your related current knowledge, involving your conclusion of searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself. Effort not shown may be considered as effort not done, possibly leading to the question closure. How do I ask a good question?.

Multiple negatively charged ions are possible when the final ion respects the octet rule. This is the case for $\ce{O^{2-}, N^{3-}, C^{4-}}$. But halogens needs only one electron to follow the octet rule. They cannot hold more than 1 electron.

https://chemistry.stackexchange.com/questions/59146/is-it-possible-for-halogens-to-show-oxidation-states-less-than-1

@Poutnik - just a comment, but 'laconic' may not be a common vocabulary word for many folks who don't have English as a first language (and indeed, even for those who do).

@JonCuster I do not have EN a a primary language. It is rather a part of background cultural knowledge, it is not an English word, coming from Laconia/Lacedaemonia = ancient Sparta. They sent to king Philip of Macedonia a short answer: 'If...." Since then comes the adjective laconic as extremely short and no descriptive.

@Poutnik - I understand that, but again a simpler, more common word might be easier for many folks to understand.

@JonCuster I admire Poutnik's canned comment and I think it's been thought out pretty well. Shall "laconic" be unknown to the reader, the follow-up phrase "not elaborated questions" will do just fine. Besides, I find it to be a beautiful word (next to cellar door level (YouTube — Donnie Darko Cellar Door Scene)), and, a foreigner from the Eastern Bloc, I also have no issues with it as there is also a cognate word in my first language (cf. Russian «лаконичный»). Finally, dumbing down the language won't do any good for anyone.

@andselisk - personally, I've always preferred 'pithy' to 'laconic'. I love the richness of the English language, but if the goal of SE is to communicate broadly then one should ask if their word choice is helping of hindering such communication. It isn't dumbing it down, it is making a choice to use widely understood vocabulary.

@JonCuster That's a fair point, can't argue with that. Although, since we cannot censor each other's vocabulary (well, at least as long as it doesn't become vulgar), let me suggest the following to resolve this little conflict: post a comment using  preferred words if you can get ahead of Poutnik — this way you deliver the message to OP the way you like it, and his exquisite choice of words will be a stimulus to do it faster and more often:)

@JonCuster I admit I do not know the word "pithy" at all, but searching for it, it looks it has quite different meaning to laconic. The latter, translated to Czech, is just "lakonický",  in German, "lakonische", in French "laconique", Italian "laconico", Spanish "lacónico", Swedish "lakonisk", Hungarian laconicus, Finnish lakoninen, so it seems to me like an international word.

@Poutnik - 'international' in the sense of deriving from classical Greek and Latin and thus spreading across Europe readily (pithy is more localized, coming from Old English). There are many languages less influenced by Western Europe classics education. I don't mean to censor, quite the opposite - I'm trying to make it easier for people to figure out what comments mean.

@JonCuster I am aware that influence on e.g. SE Asia languages is lower, but in spite of that, I think there is bigger chance they would know laconic than pithy. Its usage escaped from classic education.

Different atoms/molecules/ions combine to gain more stability always. Any specie will not ever accept or gain electrons naturally that will lead to increase in their potential energy (i.e., less stability). Thus, halogens only need one electron to complete their octet and attain noble gas configuration. So, they always prefer to accept one electron only. If anyhow they accepts more than one electron, then their Potential energy will increase leading to the loss of electron, and thus they will again have accepted one electron in overall process. Thus, we always say that halogens can only accept one electron naturally.


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How to calculate ionization energy of an atom provided energy of the electron at ground state?
It is given that in ground state the energy of electron in hydrogen atom is $-2.18\times10^{-18}\mathrm{J}$. I'm required to calculate the Ionization Energy in kJ/mol
Question for reference:

At ground state the eng of electron in hydrogen atom is $-2.18\times10^{-18}\mathrm{J}$. Calculate the I.E in KJ/mol

I could not find it in my book and also tried googling. None of these worked.
It is not a homework question. It is from a question booklet which have high chance of being asked in the upcoming exams.

As Homework is considered literal homework, exams, self-study questions, puzzles, worked examples etc. // There is no need to find solution in books, it is based on elementary knowledge that IS in the books and on many internet places. You do know how to multiply one number by another.

I was confused because 1312.0 is the Ionisation Energy of Hydrogen atom. But it is for one atom. I was asked in terms of kJ/mol which i evaluted to 1312.8 which is pretty close

I guess you do know the relation of 1 atom and 1 mol of atoms. BTW the ionization energy 13.6 eV is quite well known value.

Yes. 6.022 10 to the 23 atoms per mole. Some books say 6.023

1312.0 is for 1 mole. Sorry for the mistake

That  I understand. :-).  13.6 eV = 13.6 . 1.602e-19 J.

Ionisation energy : It is the minimum amount of energy required to remove an electron from the neutral gaseous atom.

If the electron is bound to the nucleus then its total energy will be negative. To separate the electron from the nucleus we should supply energy until the total energy of electron just becomes zero.  So by definition the minimum amount of energy required to remove an electron from ground state will be $2.18 \times 10^{-18}J $, which is nothing but the Ionisation energy.


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Acidity and Basicity of titrate and titrant in Acid-Base titration
I want to know that do we always use monoprotic acid and bases in the Acid-Base Titration. And, if the answer is No. Then, I am confused in the definition of Equivalence Point, because I had seen on internet (on Khan academy site) that at Equivalence point moles of acid and bases are equal. (You can also see it below in the image):

So, in the lase line they had written that line for which I'm talking about. This line can we wrong, if we will use Diprotic acid with Monoprotic base or vice-a-versa. Because then equal number of moles of acid and bases, will have different number of H+ and OH- ions.

Please correct me if I'm going conceptually wrong.

It makes sense to always use a monoprotic strong acid or base as the titrant. The analyte might be monoprotic or polyprotic (you don't have a choice if that is your sample).
In general (any titration, not only acid/base), the equivalence point is reached when the compound of interest in the analyte is used up. You would want a reaction that goes to completion, and be aware of all other reaction apart from the one you are using to quantitate the analyte. For a polyprotic acid or base, you have to know the protonation state before you start. Then, you can titrate either to the first or the second equivalence point, or (let's say you are monitoring with a pH meter) jot down both points and use them in your quantitation.

I am confused in the definition of Equivalence Point, because I had seen on internet (on Khan academy site) that at Equivalence point moles of acid and bases are equal.

In general, you have to check the stoichiometry of the reaction. In acid/base titrations, the relevant stoichiometry is usually 1:1 because most acids give of protons one at a time, and most bases pick up protons one at a time. However, if you use barium hydroxide, you would have to consider that it contains two hydroxide ions per formula unit, so there would be a factor 2 in your calculations. I know of no reason to use barium hydroxide, though, given that sodium hydroxide is much more soluble.

There are polyprotic acids that will generate multiple equivalence points during titration. One example is $\ce{H3PO4}$ being titrated by $\ce{NaOH}$.
Initially, you would react $\ce{H3PO4}$ with $\ce{NaOH}$ to form essentially $\ce{H2O}$ and $\ce{NaH2PO4}$ at the first equivalence point. Now, technically, $\ce{NaH2PO4}$ is a salt, and the moles of $\ce{NaOH}$ added should be the same as the moles of the acid $\ce{H3PO4}$. So the last statement of that definition is not necessarily wrong. However, the difference is that $\ce{H2PO4-}$ will serve as the acid to react further upon adding more $\ce{NaOH}$.


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How long can I keep HDPE plastic in an oven for before it starts to burn?
I am recycling HDPE plastic (chips) in an oven. I left it in the oven for 10 mins and it melted the outside layer of the plastic fine. When I cut into the plastic with a saw after it cooled, there was a layer that hadn't melted.
I think 180 degrees Celcius is about the right temperature (it's round about the minimum melt temperature for HDPE), but I would like how long I can keep it in the oven for before it combusts?
My thinking is if I leave it in the oven longer than 10 mins, it will also melt the inside. But I especially don't want to leave it in the oven long enough to make nasty combustion fumes.
Any help with time I should leave it in the oven would be appreciated ('practical' answers preferred).

Your spouse or significant other is very accommodating, that's all I can say...

I hope you're not using the same oven you use for regular cooking.  The vapors from the plastic could deposit themselves on the interior of the oven, and then deposit themselves on your food the next time you cook.  Also, this is best done in a garage with the door open (you want lots of ventilation).

Practical answer is: "Never do anything like that at home".

Clean, colourless HDPE is not going to emit toxic fumes, but recycled material ?! Anyway, in what way is this "recycling"? You obtain a solid blob of plastic, probably with a lot of air bubbles in it.

Was going to cut shapes out of it using a bandsaw. When I cut into it previously, there weren't any bubbles, just an unmelted section. Also the plastic in the photo has been folded ... Might get a better sheet if it's cooked for longer.

The plastic chips were shredded milk jugs.

Good news: it is not going to combust at 180 °C. Bad news: some toxic fumes will probably be released anyway.

I highly recomend going to a trift store and picking up a cheap panini press, or a toaster oven. On the panini press, theres usually no ability to change temperatures, but I find its extremely hard to melt it unless you walk away for like 20 minutes. On the toaster oven, 160° should be perfect and i've never had it burn at that temperature. I hope this helps :)


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Can a plain sheet of stainless steel adhering to the 316 grade be regarded as food safe at every finish level?
My question is not purely about chemistry but it definitely involves it.
I was looking for a while to buy a food-grade stainless steel plate that would fit my home pizza gas oven (to have the dough placed on), and from my research the best possible solution was to find someone who can specially cut 316 stainless steel to the befitting proportions.
I only found one place that agreed to consider such a small order, and it should be done by tomorrow. But now I'm not certain I can automatically count on its food-safety just because the steel is 316 grade. I read at least regarding regular carbon steel sheets they are often hot rolled, which leaves "mill scale" that might chip off, and that they also may be treated with acids that might still be present if the finish level is crude.
I wasn't told the specific finish type for the sheet used, but they said "it's fairly smooth, not mirror-polished". Can that be a food-safety concern even when 316 stainless steel is the material? Is it advised I'll process it in some way to reduce the chance of hazardous chemicals leaching into the food?
EDIT: Here are photos of the plate I got:

Personally I would wire-brush the whole shebang and then wash like a normal dish. That should bring it down to a "true" 316 surface and remove whatever industrial residues that possibly remain.

@AngerDensity As I already had pizza dough made in advance and the people there claimed as it's 316 it should be safe already, I washed it with soap and dish scrubber, heated it up in the oven for 20 or so minutes and later washed it again. I'd rather have it polished first but let's hope it's indeed fine as is.

@AngerDensity I will update that while the sheet didn't seem to become warped during warm up, when I placed the pizza dough parts of the sides went up 1 to 2 centimeters. I suppose this is because the upper surface became colder due to the dough absorbing heat and that made it contract towards to middle? Anyway, it seems to correct itself after removing to pizza and letting the temperatures even up, but I wonder if that repeating warping–unwarping could result in the metal breaking at some point?

There are no special "food grades" although advertisements often claim it. There are grades most commonly used for food equipment; food equipment is primarily 316 and 304. The 316 has some advantage for corrosive foods. A home oven would be such mild service that either grade would be fine. That is, if food has been processed it has very , very likely been contacted by 316 and/or 304 so they are "food safe". Stainless sheet is cleaned ; there are about 4 common commercial finishes according to a net search. Some combination of pickling, heat-treatment, and rolling, and possible sanding. I expect washing with soap and water will produce a safe surface after any of these processes.  I can't imagine you would encounter scale as may be on thick plate.

Well, the gas oven is rather high temperature, it can reach about 500 Celsius near the burners, but to my understanding the 316 should still withstand this. The sheet will be 2mm thick (I actually planned on 3mm originally but the price became close to what I paid for the pizza oven, which seemed excessive) if that gives any indication. But I still read on some sources stainless steel is also often "pickled" with acids to remove surface roughness: is that a concern?

I picked up the plate today, it looks a bit like it's made of scrap metal (photos are added to the question). While they told me that as it's 316 (regarding which, the most I could verify is that it's non-magnetic, but frankly what I have is their word) it is fine like that and can go into the oven with food, I wonder if all of this surface junk is something that should be cleaned/polished out somehow or not. The sheet manufacturer's logo for example, how do you think it's imprinted?

Washing with soap and water should be fine; scouring may be needed to remove stencil ink. 316 may be anywhere from non-magnetic to very magnetic depending on the amount of cold work.  Annealed is non-magnetic which is apparently what you have. Specifications such as ASTM generally require metals to be marked with certain information. ; usually an ink jet stencil. US government generally requires marking for country of origin.

Soap and water removed some of the visible dirt, although not everything. The serial number visible on the piece returns nothing on Google, although there was probably more written which was cut off from this piece. My cousin does some work with metals and said the imprint is usually laser carved? The one thing that is a bit strange is the weight being 1420 grams rather than the expected 1460 grams (calculating cylindrical volumes against about 8,000KG/M² density of stainless 316), but it MIGHT be that the thickness is just 1.9 and a bit, it's hard to tell with a mechanical caliper.


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Is this formula for number of optical isomers correct?
For a symmetric molecule with an even number of chiral centres (for acyclic molecules with chiral centres only, not considering $\pi$ bonds or rings), the formulae (also mentioned in this question) are:

No. of meso isomers $=2^{\frac n 2 -1}$
No. of optical isomers $=2^{n-1}$

However, when I went to test it out on $\ce{CH3-CHD-CHD-CHD-CHD-CH3}$ ($n=4$), my results were not matching with the formula.
I got $2$ meso isomers:

And I got $6$ enantiomer pairs ($12$ optical isomers):

The formula for meso isomers seems to match; $2^{{4 \over 2} -1} = 2^1 = 2$
But the formula for optical isomers doesn't match; $2^{4-1} = 2^3 = 8 \neq 12$
Where did I go wrong? Or is the formula wrong? If so, what is the correct formula?

No simple general formula exists, unless you introduce more restrictions than given by the literal interpretation of the sentence "for a symmetric molecule with an even number of chiral centres". For one, you can have optical isomers with no chiral centers.

@NicolauSakerNeto Sorry for not mentioning that, this formula is for acyclic molecules with chiral centres only, not considering $\pi$ bonds or rings (basically the simple case)

I feel that the formulae should satisfy the condition:
$($No. of meso isomers$)\ \times\ 2\ +\ ($No. of optical isomers$)\ =\ 2^n$.
$$$$
This would give us: No. of optical isomers $= 2^n - 2^{\frac n 2}$,
which matches with the example I took. Is this correct?

I think that the enantiomeric pairs 1 and 4; 2 and 3 are identical. Just rotate the molecule by 180 degrees about an axis perpendicular and passing through the centre of CH3 - CH3 line. This counting problem arises when symmetrical molecule can be numbered from two ends. Hope this helps..

So just reduce the number of enantiomers you counted by 4 i.e. there are 8 enantiomers and the formula is correct.
Also when there are odd number of chiral centres (similar to above problem), the total number of stereoisomers = 2^(n-1).

Why don't you add your comment to own answer?

You disproved the formula for the number of optical isomers. You are better off to count off $2^n$ different combinations of $R/S$ configurations, then note that the mesos are counted twice (because the intended mirror-image pairs are just one isomer for meso compounds).
So the true number of optical isomers is $2^n-2^{(n/2)-1}$, out of which $2^{(n/2)-1}$ are the meso compounds and the rest consist of enatiomeric pairs.
Thus for $n=4$ you have $2^4-2^1=14$ total optical isomers, which matches your two meso compounds and 12 others that are six pairs of enantiomers.

I detect a down vote, now if I could detect suggestions from the voter for improvement that would be awesome!

Try to use these formulas for finding number of stereoisomers for a given compound:

Geometrical Isomers:

If n is odd: $2^{n-1}+2^{\frac {n+1}{2}-1}$
If n is even: $2^{n-1}+2^{\frac {n}{2}-1}$

Optical Isomers:

(2.1) If there is no meso compound:
Optically Active Isomers(O.A): $2^{n}$
Racemic Mixtures obtained: $\frac {O.A}{2}$
(2.2) If there is a meso compound:
(2.2.1) (For even values of n)
Optically Active Isomers: $2^{n-1}$
Meso Isomers: $2^{\frac {n}{2}-1}$
(2.2.2) (For odd values of n)
Optically Active Isomers: $2^{n-1}$
Meso Isomers: $2^{\frac {n-1}{2}}$


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Definition of Molecularity of a reaction?
If nucleophilic attack occurs simultaneously at two seperate electrophilic sites of the molecule following SN2 at each site, will the moleculularity still be 2? Does molecularity depend on the total number of reacting entities, or the number of reacting entities at a particular site?
I think molecularity will be defined wrt a specific site, in a molecule.
Please elaborate.

It's the molecularity not some "sitality" or whatever. It's not like simultaneous attack is likely - and truly trimolecular reactions common.

@Mithoron But if we have a very large molecule, like some really complex organic molecule, having different active sites, then can't multiple attacks take place in different parts of the molecule simultaneously?

Being very large hardly changes anything. Important issue is that kinetics deal with practical result - even if your contrived scenario happened once in a billion times, all others would be simply two step and effect on reaction would be none.

If the reaction requires simultaneous nucleophilic attack of both sites to proceed through an elementary reaction step, then we would describe that step as having molecularity of three.
If it just so happened that a particular molecule in a reaction had two simultaneous nucleophilic attacks at different reaction sites that could each have proceeded at different times, then the two would be considered different elementary steps each with molecularity of two.


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How can we explain solubility of hydrogen sulfide in water?
On my most recent high-school chemistry exam I had a longform question where I was supposed to compare the solubilities in room-temperature water under normal pressure of a few molecules. According to the question $\ce{H2S}$ is very weakly soluble in water, and I needed to explain why.
I suggested that the difference in electronegativity between $\ce{H}$ and $\ce{S}$ is around $0.38$ (on the Pauling scale), which is very close to the threshold we use to categorize polar and non-polar bonds $(0.4).$ So,  although the $\ce{H-S}$ bond is technically non-polar, making the whole molecule non-polar, it is very close to being polar, which might make the molecule a very slight dipole, favorizing solubility in water, a polar solvent
The teacher accepted this explanation, although they took away points because I did not mention the hydrogen bonds that supposedly form between the hydrogen sulfide and water molecules.
This confused me because I thought that hydrogen bonds can only form between hydrogen atoms in a $\ce{H-O},$ $\ce{H-N}$ or $\ce{H-F}$ bond and a very electronegative atom in a covalent bond such as $\ce{N},$ $\ce{O},$ or $\ce{F},$ which is what I find in most high-school level resources.
What am I missing out on? Can $\ce{H2S}$ form hydrogen bonds (even very weak) with water when dissolved?

Then your teacher would seem pretty good, but solubility of H2S is by no means very weak. Was this his idea or yours?

@Mithoron Well, the data I’m given is 2.5 mol/L for H2S, which I’m to compare to 0.0013 mol/L for diiodine and 50 mol/L for NH3. She did not say that my remark that H2S is weakly soluble is wrong, but just said that the part about the hydrogen bond is missing.

This source says H2S does not form hydrogen bonds, and given that is a publication of the American Chemical Society, I am inclined to believe it  https://www.acs.org/content/dam/acsorg/education/publications/ch8.pdf

@Waylander yes but this is "pure" H2S. My question was not whether H2S could form hydrogen bonds with itself, but rather if it can with H2O... although the two might be related? I’m not sure. Thanks for this resource.

The important thing is that H2S is an acid - that's what makes it way more soluble then CH4. Then again NH3 has much more pronounced acid-base reaction with water. Another issue is boiling point, or rather equilibrium vapor pressure.

$\ce{H2S}$ does make hydrogen bonds with $\ce{H2O}$, but these bonds are very weak. On the other hand, it should be known that solubility is one of the chemical properties which is not yet completely understood. Nobody is able to explain why $\ce{CaCl2}$ is extremely soluble in water ($600$ g /L water), and why $\ce{CaF2}$ is nearly totally insoluble in water, or why $\ce{NaClO4}$ is extremely soluble in water ($2090$ g/L water), and potassium perchlorate $\ce{KClO4}$ is nearly insoluble in water ($7.5$ g /L).

Related: https://chemistry.stackexchange.com/questions/84698/is-dissolution-of-h2s-in-water-exothermic

Hydrogen sulfide is actually relatively strongly soluble in water compared with other low-polarity gases. In Earth Science SE, this answer explains that water-ice hydrates are most likely to be found with gases having low solubility in liquid water such as methane, and in particular methane hydrates are more common than hydrogen-sulfide hydrates because (on a molar basis) hydrogen sulfide is 80 times as soluble as methane.
Hydrogen sulfide is too weak as a hydrogen-bonding agent and an acid for either of these attributes to play much of a role in its relatively large solubility. Rather, the relatively large sulfur atom is more easily polarized than the hydrogen and carbon of methane, so hydrogen sulfide can interact more strongly with water through a dipole-induced dipole interaction. The induced polarity imparted to the hydrogen sulfide also contributes to its being more acidic than methane despite carbon and sulfur having nearly identical electronegativities.

Very helpful, Thank you so much!!


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Interpreting Ternary Phase Diagram Stoichiometry
I am trying to locate the lowest-melting mixture in a system of three nitrate salts: sodium, potassium, and calcium nitrate. I have the relevant ternary phase diagram, but I am unsure exactly about interpreting it because of the "doubled" stoichiometry of the sodium and potassium salts on the plot.

I posit that along the drawn-in red line, there is a 50/50 molar mixture of Ca(NO3)2 and (KNO3)2. At the zero NaNO3 limit, this would give 0.5 mol Ca(NO3)2 and 1 mol KNO3, so I understand the composition to be 33.3 mole percent Ca(NO3)2 and 66.7 mole percent KNO3. Is my interpretation correct?

An interesting way to portray the ternary, but really no weirder than some others. Looks like they wanted to focus/normalize on the (NO3). Given that, yes, I think your interpretation is correct. You might wish to look at the KNO3-Ca(NO3)2 binary and confirm your supposition.

Ha, unfortunately the same source (Phase Diagrams for Ceramists) gives the same double stoichiometry for each of the applicable binary diagrams! I suppose I will go back and follow the original references in that compilation to see how they're plotted there, but that requires me a physical trip back to the library.


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Preparation of tertiary Grignard reagents
Has anyone a good and detailed experimental procedure to prepare tertiary Grignard reagents from tertiary alkyl halides on the lab scale (1-5 mmol)? The actual procedures in the literature are very scarce and limited to patents (large amounts). Older procedures exist only for tBuMgCl and not after 1932.

Hi, can you give some more detail about what you are trying to do?

"hindered" gringards are hard to make. Most require extreme conditions (well extreme by the standards of making gringards from primary alkyl halides). The least you will need to do is to reflux the solution in THF. But many will require extra steps to activate the Mg (like adding ground glass to Mg powder. Or even starting with something like Mg-anthracene. A few experiments like that are well worth attempting, but literature techniques often leave subtle bits out.

The likely problem with tertiary Grignard reagents is that they may, owing to their basicity, be prone to effect elimination of the HX from halide in solution to form an alkane and alkene. For this reason, vinyl magnesium bromide cannot be prepared in ether but it may be prepared in the better coordinating solvent tetrahydrofuran (THF).  t-BuMgCl is sold in THF. To activate magnesium, try the classic crystal of iodine or 1,2-dibromoethane.

A few mls of Dibal-H soln are also good for activating magenesium.

Sterically hindered GRs are hard to make as @matt_black mentioned and although its synthesis looks single step and straight forward, most literature leaves bits out. I did find a recent publication where they mentioned the steps:

tert-butylmagnesium chloride
$$\ce{t-BuCl + Mg ->[I2,anhyd. Et2O]  t-BuMgCl}$$
In a 3L, three-necked, round-bottom flask, fitted with a mechanical stirrer and a Hg seal, a 500-mL separatory funnel, and an efficient reflux condenser, were placed Mg turnings or powder (144g, 6.0 mol). The Mg was then covered with anhyd. $\ce{Et2O}$ (400 mL). Pure $\ce{t-BuCl}$ (~5 mL, with a crystal of $\ce{I2}$) was added to start the reaction. Stirring was begun and a solution of pure $\ce{t-BuCl}$ (555g, 6.0 mol) in anhyd. $\ce{Et2O}$ (1.5 L) was dropped slowly onto the Mg during 6-8 h. The reaction was practically complete when all of the halide had been added, but stirring was continued for further 15 min.
Science of Synthesis: Houben-Weyl Methods of Molecular Transformations Vol. 7: Compounds of Groups 13 and 2 (Al, Ga, In, Tl, Be...Ba), Georg Thieme Verlag, 2014

Alternatively, you can look at the orgsyn article or this 2001 WOL article but the steps/reagents are more or less same.

All these procedures shares with the patents I found on the subject, the large scale (usually from 500 mmol to 1 mol). On the smallest scale (2-8 mmol), I had no success. Magnesium is very active (washed with diluted HCl, H2O, EtOH, Et2O and dried) and it starts normal Grignard in seconds. Here I tried to further activate it with I2 or dibromoethane, but again no success. An old JACS ("The Yields of Some Aliphatic Tertiary Grignard Reagents and the Limits of their Usefulness as Synthetic Reagents", JACS 1933, 1559) discusses these problems, but again reactions are made on 1 mol scale...

@saiwa What exactly are you trying to convert? Is it just a haloalkane like tert-butyl chloride, or are there other functional groups? Nitro groups in particular can kill the Grignard reaction by oxidizing both the intended product and the magnesium metal.


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How to guess the Chemical state in which particular element will be present from XPS data?
Since my last post yesterday, I found some leads and started to analyse my data. But then I started encountering more doubts, questions and confusion. Any help will be highly appreciated and will help me in speeding my research.
For analysis, I am matching the peak binding energy for each orbital for each element with the NIST Reference Database. During the analysis process I was confused how to guess in what form is the element/element's present. I am not sure what I am doing is correct or not.
I am analysing XPS results of Pt impregnated on gamma alumina. I have 4 samples each subjected to different conditions. I am trying to figure out in what oxidation state/in what form the Pt will be present.
So I'll give an example -
For Pt, I got a peak binding energy of 74.76 eV for the 4f7/2 orbital. When I searched this peak in the reference database, I got the following possible candidates with the following reference binding energies
a) PtO - 74.6 eV
b) PtO2 - 74.6 eV
c) Pt - 74.5 eV
This particular sample was fired in an engine and the temperatures were around 300 deg cel, and we don't know exactly in what form the Pt will be present. How do I decipher in what state the Pt will present from the XPS data ?

Where do you get these data? I get from https://www.jp.xpssimplified.com/elements/platinum.php: 71.0, 72.4, 74.9 for the same compounds in the given order, in which case your compound is more definitely PtO2.

Recently, there a lot of XPS questions here which are being closed or not receiving an answer. The problem is that you are not sharing enough experimental details so nobody has an answer and nobody knows if your 74.76 eV peak is at the true position or not (see the reasons below). You also need get a good XPS textbook written by the very inventor of XPS. Check Wikipedia about the inventor of XPS.
When people wish to assign XPS to a certain oxidation state or assign a certain compound, here are some of the steps that help people in interpretation.

One should be aware of the fact that there are two types of X-ray photoelectron spectra. Survey scans which are acquired rapidly, or high resolution scans which take several hours or even a full day. Peak assignments should be done on high resolution scans. A survey scan is a quick and dirty analysis and this is most likely done in your case. High resolution scans are expensive.

Peak positions are very sensitive to charged state of the surface. By chance, if you have seen scanning electron microscopy of particles, one can see particles flying from the surface due to charging. Similar charging effects exist in XPS. If the XPS substrate to be analyzed is being electrically charged by impinging X-rays, the peak positions will be shifted.

Usually adventitious carbon (always present on all samples) is used to correct the peak positions. Peak assignments should not be done on uncorrected spectra.

Now the 74.76 eV value can be a useful number OR it can be completely useless if steps or conditions (1) to (3) were not fulfilled.

The fourth step is to use chemical intuition about the nature of the surface. Do you expect such compounds on the surface? Was Pt present in an oxidative environment? If you think there is negatively charged platinum species, the charge must be balanced by some cations. Is there an indication of cations present on the surface. If you assign PtO formation, is there a oxygen peak in similar proportion to Pt atom %?

Look at peak widths of the element in question. A wide peak indicates multiple environments or oxidation states. This is done by peak fitting (wrongly called deconvolution in common parlance) in the software. CASA XPS manuals are online.

In short do not look at just at a single number from the NIST table, look at the conditions in which that NIST data was acquired and also see how good quality XPS data is summarized. They write all the details. Ask your XPS data provider to show all the conditions too (peak position correction, baseline correction method etc).
Your 74.76 eV can only be assigned to a certain oxidation or compound if and only if your AND their experimental conditions match closely. Otherwise, everything will be handwaving. A lot of XPS data in analytical literature is based on handwaving because XPS peak assignment is not a trivial task. This is from the NIST website. Note they do not mention uncertainity, which is also important.

XPS for Pt
.

Thank you so much for providing this detailed answer. This certainly helped me a lot and cleared a lot of my doubts. I also went through a basic XPS textbook, which also helped in grasping the basic textbooks. Is it okay if I dm you? ,I had a few more detailed doubts after following your answer

Wulf, I am not an expert in XPS at all. I was just a user for my research work. You can try posting crystal clear questions so that everyone can benefit from the discussion. I don't know how dm works.


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Is there oxygen isotope exchange between dissolved CO2 and H2O?
If you had a sample containing an elevated concentration of $\ce{H2^{18}O}$, and bubbled $\ce{C^{16}O2}$ through it, would some of the oxygen-18 isotope be transferred from water to carbon dioxide?
I am aware that this occurs with hydrogen isotopes and am curious if it works for larger and heavier atoms such as oxygen.

Yes, absolutely. In (at least) aqueous geochemistry, the phenomena is used to determine source rock and age for various parts of the hydrologic system. See this publication for more information.

It is quite obvious it does, via its equilibrium reactions with water,  forming carbonic acid and bicarbonate/carbonate ions.

@Poutnik, I realized that as I read the paper Todd recommended. I wonder if it would still occur if the molecule did not dissociate. Like if the gas were ethanol instead.

It would need some organic chemist to tell. I can imagine some nucleophilic substitution replacing OH by other OH. But I am afraid bubbling ethanol vapor through water is funny idea, as you would prepare ethanol solution.

@ericnutsch It doesn't have to dissociate, but a reaction is obligatory. With alcohol, if pH was high enough, substitution would be possible.

Ethanol does not exchange with water in weakly acidic conditions; see the article by J. Dunbar, https://doi.org/10.1524/zpch.1982.130.2.247.

As Todd Minehardt points out, not only can the oxygen be exchanged, but this exchange is applied in aqueous geochemistry.
The exchange occurs through the formation of carbonic acid, given in blue below:
$\ce{CO2 + H2O <=> \color{blue}{H2CO3}}$
Once the carbonic acid molecule is formed, its oxygen atoms effectively become equivalent because they rapidly exchange hydrogen ions with the water solvent. So when the above (dynamically equilibrated) reaction is reversed, any one of the carbonic acid oxygens might end up as a water molecule, including atoms that were originally in the carbon dioxide.

Now I have to wonder if, during a chemical reaction, certain atoms are more likely to end up in certain places. E.g. if (hypothetically) H2CO3 did not exchange hydrogen ions, would the "central" O atom be from the CO2 molecule that went into it? (It "looks like" that arrangement should require less bonds to be broken and re-formed during the reaction)

That depends on if you allow tautomerization from the direct adduct $\ce{H2O - CO2}$ to $\ce{HO - C(O)OH}$. With the tautomerization, the "central" oxygen atom would come from the carbon dioxide, but the hydroxyl oxygens would come from both reactants and thus mixing would occur. But without the tautomerizaton, the oxygen atoms cannot mix because the water and carbon dioxide oxygens never become equivalent.

Hmmmm... just guessing here: $\ce{CO2 ->[H2O] CO2 (aq)}$, $\ce{CO2 (aq) + H2O <=> H+ (aq) + HCO3+ (aq)}$, $\ce{HCO3+ (aq) <=> H+ (aq) + CO2^2+ (aq)}$. The Grotthuss mechanism accounts for the mobility of protons, and carbonic acid will most likely be in the smallest concentration of all involved species. Like actually lower that $\ce{2 H2O (aq) <=> OH- (aq) + H3O+ (aq)}$.

@user253751 the particular reaction is not a good example, but otherwise, marking atoms by their (stable or radioactive) isotope and looking where they go in a reaction is a powerful tool for studying reaction mechanisms. Stable isotopes (e.g. Hydrogen-2) are traced by NMR and radioactive isotopes (this is the older approach) by their radiation.


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How to find ion concentrations at various points in a solution that will precipitate multiple insoluble salts?

The solubility product constant of lead (II) iodide is $1.2 \times 10^{-5}$ at 273 K.
The solubility product constant for lead (II) carbonate is $1.5 \times 10^{-13}$ at the same temperature.
50 milliliters of a solution at this temperature contains an unknown concentration of carbonate ions and 0.050 M iodine ion concentration.
A secondary solution of 0.10 M lead (II) nitrate is added. After 32 milliliters of the lead (II) nitrate solution is added, the lead (II) iodide begins to precipitate.
Calculate the original concentration of the carbonate ions in addition to the concentration of carbonate ions in the solution when the lead (II) iodide begins to precipitate.

I attempted to find the concentration of carbonate ions when the lead iodide started to precipitate by dividing the lead carbonate's solubility product constant by the concentration of lead ions at this time, since the lead carbonate precipitates first.
So I concluded that it will be at equilibrium at this point ($1.5 \times 10^{-13} / (0.10 \times 0.032 / 0.082))$ and then multiplying this by the current volume (0.082 L) and dividing by the original volume (0.05 L) to get the original concentration of carbonate ions, at $6.3 \times 10^{-12}$ M.
To get the concentration at the time when the lead iodide started dissolving, I took my answer from the first part of the earlier calculation ($1.5 \times 10^{-13} / (0.10 \times 0.032 / 0.082))$, for a final carbonate ion concentration of $3.8 \times 10^{-12}$. However, I am unsure if the process I used was correct.
Any help is appreciated. Thank you.

When writing a text, the writer must  remember others will have to read it. Your text is a challenge for reading and orientation.  // Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps focusing on principles and spotting mistakes. It also improves orientation, Q/A reusability and the permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

Make yourself sure the process is correct before trying to apply it.

Hint: Define an algebraic variable for the unknown carbonate concentration and the function calculating the PbI2 precipitation point from this unknown concentration. The inverse function will then provide you the result.

Your reasoning is too much complicated. If all the data are calculated in moles, the result is immediate. Let's try !
$32$ mL of $0.1 M \ce{Pb^{2+}}$ contains $0.0032$ mol $\ce{Pb}$. As a consequence, the initial carbonate solution contains the same amount of $\ce{CO3^{2-}}$. As these $0.0032$ mole are dissolved in $50$ mL. It means that the initial concentration $\ce{[CO3^{2-}]}$ is $\ce{0.0032 mol/0.05 L = 0.064 M}$


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How are antioxidants different from other compounds that contain hydroxyl group?
From my book:

An antioxidant is a substance that prevents oxidation by reacting with an oxidizing agent

If that's the case, why do we distinguish between antioxidants? Are all compounds containing hydroxyl group not capable of donating protons to oxygen-containing free radicals?

What book? Please name it.

Your question is a precise equivalent of "How are red-haired men different from tall men". They are different in that they have red hair; also, not all of them are tall. That's about it.
Same thing here. Donating a proton has nothing to do with antioxidant properties. Indeed, among the volumes of forgotten lore of inorganic chemistry one can find the notion of acids; these are distinguished by their ability to donate protons. Are they antioxidants? No, and often quite the contrary.
Antioxidants are compounds that oxidize easily, and oxidation has to do with transfer of electrons, not protons. That they all seem to contain -OH groups is just a feature of our biology, which is a consequence of our chemistry, which in turn is centered around water. Look at other compounds in our bodies, most of them (except fats) have -OH groups too. There is no fundamental chemical reason behind that.
So it goes.

Yes sorry, I'm trying to get the hang of it. Is there a particular video or book that helped you understand the concept of oxidation and reduction?
And I'm sorry if it was a stupid question, what I wanted to ask was what separates antioxidants from other compounds that can oxidize. I believe you also asked about which book I got the quote from and it was from [Fundamentals of General, Organic, and Biological Chemistry with MasteringChemistry, SI Edition] by McMurry.

No, it wasn't me who asked about the book, for I don't believe it is important. For the same reason, I don't remember where I got the concept of redox reactions. Why on Earth would anyone remember that? Do you remember where you learned that 2+2=4? Oxidant gains electrons, reducing agent loses them - it is as simple as that.

Reducing compounds are like weapons - some can be as weak as a table  knife, some as strong as a nuclear bomb (figurativelly).
As antioxidants

Antioxidants are compounds that inhibit oxidation, a chemical reaction that can produce free radicals and chain reactions that may damage the cells of organisms. Antioxidants such as thiols or ascorbic acid (vitamin C) may act to inhibit these reactions. To balance oxidative stress, plants and animals maintain complex systems of overlapping antioxidants, such as glutathione.

are usually considered such reducing compounds that are neither too strong neither too weak, reducing just too strong oxidants, including oxidizing radicals and dissolved oxygen. There is additional requirement to be chemically and/or biologically compatible with the material protected against oxidation.
Water does contain $\ce{-OH}$ bond and is not considered as antioxidant, reducing just extremely strong oxidants like e.g. fluorine. Similarly ethanol is very weak antioxidant.
Reducing abilities of $\ce{-OH}$ group is very different across compounds. Typical antioxidants are polyphenols like epigallocatechin gallate, one of green tea infusion components. Other ones are e.g. pyrogallol(used in gas analysis for oxygen absoption) or ascorbic acid, also known as vitamin C, or tocoferol - vitamin E.

Why are antioxidants usually considered reducing compounds that are neither too strong nor weak? Is it because once they donate electrons they, in turn, become radicalized, but since they're not strong or weak, they won't be able to further oxidize other reducing compounds?

@Patrick Too strong = reduced  is even stuff which should not be. Too weak = ineffective to act as antioxidant, as the stuff to be protected could be oxidized with higher priority.

@Patrick Not all OH containing substances have antioxidative properties ( cellulose, starch, table sugar), not all antioxidants must contain OH groups (even if very most does, especially natural/natural-identical).


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Can a thin stainless steel plate warping and unwarping in a gas oven due to temperature gradients break at some point?
This is in continuation of a previous question:
I have a specially cut 316 stainless steel plate, 345mm diameter * 2mm thicknesses, used in a butane pizza oven that can reach 500 degrees Celsius as a base for the pizza dough.
The plate doesn't seem to warp at all during the warm up period, but when I place dough onto it the plate's sides warp upwards a centimeter or two. I suppose this is because the dough cools the upper side of the plate too quickly and that causes it to contract asymmetrically.
When the pizza is removed the plate seem to flatten out pretty quickly. Will such cyclic deformation pattern damage the integrity of the material in the long run?

No, the temperature differences are relatively small in a home oven. It will distort but not crack. Thermal fatigue can be caused near temperatures like 1600 F over thousands of cycles. Heat-treatment containers are typically nickel alloys which have physical properties very similar to 316 ( austenitic structure). They can be cycled to very high temperatures ( 1800 F) every day for years before possibly cracking.

I'd say 500 degrees Celsius (932 Fahrenheit) is above what's usually defined as home oven temperatures, but I suppose it's still within safe range according to your answer. The fact that it flattens back to normal tells me the distortion doesn't really affect the molecular structure of the plate. Although I did read something called carbide precipitation can start occurring above 430 Celsius or so, but I'm not sure what that means.

And the warping means that the thermal stresses are being relieved. Certainly flexing from a pizza a day for the rest of the OP's life will have no fatigue impact at all.

@JonCuster But here the warping is caused by the thermal differentiation when one side of the plate is cooled rapidly by the pizza dough, which seems more stressful than a slower and more symmetrical cooling such as when the plate is left to cool off after the oven is turn off. And I don't make pizza every day, but when I do it's about 4 pizzas one after the other, so there could be 4 warp-unwarp cycles within 20 or so minutes.

The number of cycles is still so low as not to be an issue. For real fatigue failures you need fairly high stresses, and your plate readily flexes to relieve them.

@JonCuster To extrapolate on the situation, the effect as is happens with the dough being dropped in using a floured pizza peel. The issue is that the heat transfer is not that great actually and the bottom crust is a bit lacking, so I think of trying to actually oil the peel (and thus the dough's underside). Oil would increase heat transfer, but perhaps that would also cause more serious warping that could prove detrimental? I do think I may lower the oven heat in that situation as the oil may burn too much which would be unhealthy and unsavory.


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Is the shortness of boron trifluoride's bonds better understood and explained now than it was "way back" in the 20th century?
Wikipedia's Boron trifluoride; Structure and bonding says:

In the boron trihalides, BX3, the length of the B–X bonds (1.30 Å) is shorter than would be expected for single bonds,7 and this shortness may indicate stronger B–X π-bonding in the fluoride. A facile explanation invokes the symmetry-allowed overlap of a p orbital on the boron atom with the in-phase combination of the three similarly oriented p orbitals on fluorine atoms.7 Others point to the ionic nature of the bonds in BF3.8
7 Chemistry of the Elements N. N. Greenwood, A. Earnshaw, Elsevier (1997)
8Gillespie, Ronald J. (1998). "Covalent and Ionic Molecules: Why Are BeF2 and AlF3 High Melting Point Solids whereas BF3 and SiF4 Are Gases?". Journal of Chemical Education. 75 (7): 923.

Google's English dictionary uses Oxford languages and gives the following definition and synonym for "facile":

adj ignoring the true complexities of an issue; superficial. "facile generalizations"
Similar: simplistic, superficial, oversimple, oversimplified, schematic

I also noticed that the references cited are from 1997 and 1998 and in the subsequent near-quarter-century quantum-mechanical computational techniques have improved substantially. So I'd like to ask:
Question: Is the shortness of boron trifluoride's bonds better understood and explained now than it was "way back" in the 20th century?

I guess the disappointing answer to your question is: No. Look up charge-shift bond and see for yourself if that makes understanding and explaining easier. || @Tyberius You've linked to the same post twice; I can only guess that this was not intended.

Semi-related post about BCl3 dimerization that touches on the bond lengths. Also, a much earlier study linked in the comments about simulating the BF3 with ab initio methods.


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If Graham's law doesn't apply to the ammonia and hydrogen chloride diffusion in a glass tube demonstration, is there a law that does?
A comment below What exactly are "white fumes" and why does holding a bottle of ammonia (conc) next to a bottle of HCl (conc) make them? links to The
Royal Society Of Chemistry video Diffusion in action | The reaction of ammonia and hydrogen chloride demonstration

Declan Fleming revisits the classic diffusion demo of the reaction of ammonia with hydrogen chloride  – providing some great tips for doing it even better.
Download the instructions: https://rsc.li/3xsjpxS

Fleming says:

As Graham's law doesn't apply to this situation I tend to leave things here...

and

Although the presence of air molecules makes Graham's law inapplicable...

and the instructions linked in the video's description includes the following:

However, this method neglects the fact that the assumptions behind Graham’s law cannot apply. The different sizes of the molecules and their collision cross sections with other molecules come into play. In other words, the rate of diffusion depends not only on the molecules of HCl and NH3, but also on the properties of the (mainly) nitrogen and oxygen molecules into which they are diffusing. As such, the observed ratio may be slightly less than that predicted in the simplistic model above – the demo is perhaps best kept as a qualitative one at this level.

The ratio of the distances from the boundary where smoke particles are produced and the two sources is 15.5 cm / 8 cm ~ 1.94 and the square root of the mass ratio (if it did apply, which it does not) is (36.46 / 17.03)1/2 ~ 1.46.
Both gasses are diffusing in air, which is not at all the same as effusing from an aperture smaller than the mean free path (of order 10 microns). So I understand why Graham's law does not apply here. So I'd like to ask:
Question: If Graham's law doesn't apply to the ammonia and hydrogen chloride diffusion in a glass tube demonstration, is there a law that does, or at least gets close?

Annotated screenshot from The Royal Society Of Chemistry video "Diffusion in action | The reaction of ammonia and hydrogen chloride demonstration" https://youtu.be/SWByFMo32Qg

Almost certainly Fick's laws, which take into account the mixture density explicitly.

@ToddMinehardt and yet while taking the time to negate Graham's law in several places, there's no mention of Fick's in the video.

Well, just a suggestion from me. I didn't bother with the video. There will likely be something that fits, and it will be along the lines of describing kinetics in viscous media.

@ToddMinehardt Also, the moelcules first have to get from aqueous solution to the gas phase, and this process might have distinct kinetics for the two molecules.

Here is a treatment of diffusion of oxygen in different other gases: http://compost.css.cornell.edu/oxygen/oxygen.diff.air.html

The experiment and modeling of the position using differential equations is described in a paper behind a paywall (doi:10.1063/1.5083927).
Here is the abstract:

Vapor-phase ammonia, NH3(g), and hydrochloric acid, HCl(g), undergo a series of complex reactions, including nucleation and growth, to form solid ammonium chloride, NH4Cl(s). The counterdiffusional experiment, whereby HCl(g) and NH3(g) diffuse from opposite ends of a tube and react to form spatiotemporally complex patterns, has a rich history of study. In this paper, we combine experimental data, molecular simulations, and analysis and simulations of a partial differential equation model to address the questions of where the first unobserved vapor product NH4Cl(g) and visually observable precipitate NH4Cl(s) form and how these positions depend on experimental parameters. These analyses yield a consistent picture which involves a moving reaction front as well as previously unobserved heterogeneous nucleation, wall nucleation, and homogeneous nucleation. The experiments combined with modeling allow for an estimate of the heterogeneous and homogeneous nucleation thresholds for the vapor-to-solid phase transition. The results, synthesized with the literature on this vapor-to-particle reaction, inform a discussion of the details of the reaction mechanism, including the role of water, which concludes the paper.

Another paper has diffusion coefficients for hydrogen chloride and for ammonia in nitrogen (or air). The diffusion coefficient for ammonia is about 1.5 larger than that for hydrogen chloride, explaining while ammonia travels further before the reactants meet.
In a random walk, the average distance from the source increases with the square of the time. It is not clear to me what happens in the cylindrical geometry, but the rate of transport probably is linear after a steady state has been reached.
There are several complications that would explain why the distance traveled might not be inversely proportional to the gas diffusion coefficients. First, the substances have to undergo a phase transition from dissolved to gas. Second, there might be reactions in the gas phase with water in air. Third, you don't see the ammonium chloride in the gas phase; rather, it is the crystalline ammonium chloride you see, first as smoke and then as solid on the wall of the glass tube.
The simplest empirical law would be "8 cm away from the source of HCl". Once you start varying parameters (temperature, pressure, distance, geometry, humidity, isotopes, different reactants, gas different than air) you would have to include these parameters in the law.
Or you go to the local university library and check out the theoretical model described in the literature cited above and how it compares to experimental results.


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effect of solution contact angle with KCl and NaCl ore on dissolution rate
I am working on a potash solution mining project (NaCl solution injection) and it is said that in the mine cavern, the contact angle of the solution with cavern wall can have impact on the dissolution rates, i.e. on the bottom of the cavern, the dissolution rate is much lower than at the top of the cavern (due to a variety of factors), but we are just trying to capture it with a multiplier on the default rate (the kind of rate measured in lab on vertical surface).
Here is a very rough diagram showing inside the cavern, solutions are injected into it to dissolve ore on all faces. On all surfaces, solution and brine will enter and leave and there will be some flow along those angled faces, which will result in some differences in the dissolving rates.

I did a lot of search online and in various libraries, journal articles, lab reports, I am surprised that there is very scant information on this (i.e. contact face's orientation impact on dissolution rate). Any pointers would be greatly appreciated.

Can you elaborate little more what you mean by the contact angle? Possibly with illustration? Does it mean the normal vector orientation of the contact surface solid-liquid? In such a case, density gradient would definitely affect disolution rate by supported or suppressed convection. Hard to quantify as diffussion coefficients dramatically change with salt concentrations and also there is diifficulty with defining properties of transient layers.

@Poutnik I've updated the question with a cartoon drawing of the cavern. Again, I can find no definitive lab experiment or qualitative description of the differential dissolution rates based on the contact angle inside the cavern.

Hard to quantify but qualitatively, is not it obvious? From upper surfaces, enriched dense solution flows down, replaced by fresh liquid. At bottoms surfaces, it tends to stay sitting down, slowing dissolution. Even with forced mixing, the effect partly remains. But I guess this you know and ask for something else. // Do you mean really potash (K2CO3) or rather potassium(K) in form of KCl?

It is KCl solution mining. Yes, qualitatively it is clear that the rate would be faster at the top than at the bottom. I am looking for quantitative measure of the magnitude of such differences based on contact angle, most of the lab measurements I saw were done on vertical surface, I wish there are documents/labs done to quantify such effect, any pointers?


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Are 2-pyridone and cycloheptatrienide antiaromatic?

Determine which one of the following compounds is not aromatic:

I think 2-pyridone (preferred name: pyridin-2(1H)-one) (1) is non-aromatic. It already contains three double bonds. If I take the lone pairs on the nitrogen atom and consider them π-electrons, this would result in eight π-electrons which results in antiaromaticity.
I am guessing maybe the $\ce{C=O}$ double bond doesn't count as part of the ring and therefore if the lone pair on nitrogen is $\mathrm{sp^2}$-hybridized, it will result in an aromatic compound.
Similarly, cycloheptatrienide (2) seems to also be non-aromatic as the $\mathrm{sp^3}$-hybridized carbanion causes a non-planar structure. And if $\mathrm{sp^2}$-hybridized, creating eight π-electrons and therefore becoming antiaromatic.

This can be confusing, especially when you are given a molecule that has a pendant carbonyl oxygen like your first molecule.
Strictly speaking, the $4n+2$ and $4n$ electron rules apply only when the conjugated electrons form exactly a single ring, meaning there can be no pendant carbonyl oxygen atoms (or other double-bonded goodies like sulfur atoms or imino groups).
Therefore your first molecule may indeed be aromatic, as I discuss below, but the electron-counting rule you're supposed to use does not tell you that unless you have been taught the appropriate modification for this type of situation. Absent that condition, the problem is not well-posed.
There is empirical evidence for such a modification in the case of pendant carbonyl oxygens, which conveniently have both a high electronegativity and a high rate of occurrence compared with other double-bonded ligands. If polarizing the carbonyl group to $\ce{\overset{+}{C}-\overset{-}{O}}$ leaves the ring fully conjugated with an appropriate $4n+2$ count of remaining π-electrons, then the compound shows properties consistent with this “aromatic zwitterion” model (such as an enhanced dipole moment and a more strongly basic oxygen atom than would be expected of a ketone). Trying this modification with your first molecule does indeed leave a conjugated ring with six π-electrons, so the aromatic-zwitterion structure is favored. The nitrogen atom being able to donate an electron pair helps this along, but even without such an atom cyclopropenone and tropone show similar aromatic-zwitterion character.
The second molecule, actually the $\ce{C7H7^-}$ ion, is also not quite cut and dried. Of course it cannot be aromatic, as the OP correctly points out, but is it nonaromatic with a nonconjugated, $\mathrm{sp^3}$-hybridized carb-anion center, or antiaromatic with all carbons $\mathrm{sp^2}$-hybridized and the charge delocalized? For larger rings the energy penalty for antiaromaticity drops off, so it's not as clear-cut as say cyclobutadiene would be. Experimental investigation or sophisticated computation is needed to prove that $\ce{C7H7^-}$ is in fact nonplanar and nonaromatic.


text:
Calculating the Hopping distance in K+ F- bonds

This is a graph showing the formation of K+ F- ionic bonds. To my understanding, when the atoms are far apart, their neutral state is more stable. But when they get closer, they form ionic bonds to become more stabilized.
My question is: how do we calculate the distance when K+ and F- start to form ionic bonds? ie the hopping distance?

Good question
For the neutral state, I think, you can consider only the LJ  potential. On the other hand, you can calculate the stabilisation of the LJ potential considering the two ions. So LJ potential and Coulombian effect.
Intrinsicly one ion in the void (or gas phase, as you prefer thinking) is less stable than the neutral atom, here is why the $\Delta E$ is positive when the two are apart.
When the two start approach, the Coulombian interaction starts to take over and the total energy of the system starts to decrease.
Here you can calculate the zeros of the function and you can get the "hopping distance"


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Comparison between Atomic Spectrum of Hydrogen and Alkali Metals
I posted a question on Chemistry Stack Exchange a while back. It was related to the naming of atomic orbitals. One of the answers to it mentioned a research experiment. The link to the question is given below.
Size of Orbitals, Making Intuitive Sense of Quantum Model, Nomenclature of Subshells in the Quantum Model
As mentioned in the answer, alkali metals show various different bands of spectrum, on the basis of which we named the S, P, D and F orbitals.
But I didn't get why we don't have these seperate spectra in Hydrogen. (I've studied Hydrogen Spectrum in my classes, and it doesn't talk about anything like this. It only talks about Bohr using them to justify the quantisation of energy.) So what makes Hydrogen different?
Also, do elements other than alkali metals show a similar spectrum pattern or is it a single spectrum like Hydrogen? Which ones do? What's the defining criteria for a single spectrum and multiple spectrum?

The fact a hydrogen atom has just a single electron is a deal breaker. Similarly as for the classical gravitational analogy, the general analytical solution of the system state exists only for systems with 2 (point-like or spherically symmetric) objects.

The basis for orbital naming was the hydrogen spectrum not the alkali metal spectrum. And I have no idea what you mean by "we don't have these separate spectra in hydrogen". The biggest difference between the hydrogen spectrum and others is that hydrogen has only one electron making it far simpler (and analytically soluble).

The H and alkali atomic spectra are very similar in structure. The main difference being that the upper levels of the alkalis are different in energy between the s, p, d orbitals because the inner electrons shield the nucleus. For H the levels with the same n are of equal energies. Herzberg, Atomic Spectra and Atomic Structure, Explains this nicely and in excruciating detail

As mentioned in my previous answer to your first query, sharp, principal, diffuse, and fundamental notation for modern orbital "labels" comes from alkali and alkaline earth metal spectra. This is very well established in historical spectroscopy. It did not come from analyzing hydrogen spectrum. Why? The reason is that these visual "labels" for spectrum lines come from the time when the electron was not discovered and it was very easy to study alkali and alkaline earth metals by electric arcs or sparks. It was just pure mathematical analysis (hence the name series spectra...the series is the empirical mathematical series).
Hydrogen atom is the simplest atom with only one electron. How did they find out why hydrogen has only one electron is another story. This was the first atom that was tackled with sound and rigorous physics principles known to man at that time. Those wonderful years were the mid-1920s for physics but perhaps not for mankind. In fact, hydrogen atom spectrum lines also showed "mathematical series" behavior, their names were Lyman, Balmer, Paschen, Pfund series etc, named after European spectroscopists. They did not call them sharp, principal, diffuse, fundamental series.
Today these s,p,d,f labels just have a notational meaning related to orbital angular momentum $l$ values in Schrodinger's equation. If $l$ =0, you label it as $s$, $l$ =1, it is $p$ etc. This is pure algebraic use. Do not associate any other historical meaning to them now.

Okay right, being a multielectron system makes a lot of difference over here. That is exactly why the Bohr model failed and we had to look up to Quantum Menchanics right, because it couldn't explain the spectra of multielectronic species. Thanks a lot, @M. Farooq!

Yes, multielectron atoms are still a challenge.


text:
Expressing ΔH and ΔS using heat capacities and then using these to express ΔG
Essentially, I think I derived the correct expressions for ΔH and ΔS, but for some reason I can't combine them into the correct form of ΔG.
Workings below:

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$$\Delta G(T_1) = \Delta H(T_1) - T_1 \Delta S(T_1)$$
$$\Delta G(T_2) = \Delta H(T_2) - T_2 \Delta S(T_2)$$
$$\Delta G(T_2) - \Delta G(T_1) = \Delta H(T_2) - \Delta H(T_1) + T_1 \Delta S(T_1) - T_2 \Delta S(T_2) = \\
\int_{T_1}^{T_2}{\Delta C_\mathrm{p}(T)\mathrm{d}T} +  T_1 \Delta S(T_1) - T_2 \left(\Delta S(T_1)+\int_{T_1}^{T_2}{\frac{\Delta C_\mathrm{p}(T)}{T}\mathrm{d}T}\right) = \\ \int_{T_1}^{T_2}{\Delta C_\mathrm{p}(T)\mathrm{d}T} + (T_1 - T_2)\Delta S(T_1) - T_2 \int_{T_1}^{T_2}{\frac{\Delta C_\mathrm{p}(T)}{T}\mathrm{d}T}$$
For $\Delta C_\mathrm{p}$ constant with $T$:
$$\Delta G(T_2) - \Delta G(T_1) = \Delta C_\mathrm{p}(T_2 - T_1 ) + (T_1 - T_2)\Delta S(T_1) - T_2 \Delta C_\mathrm{p} \ln{\frac{T_2}{T_1}} = \\ (\Delta C_\mathrm{p} - \Delta S(T_1))(T_2 - T_1 )  - T_2 \Delta C_\mathrm{p} \ln{\frac{T_2}{T_1}}$$


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DNA staining agent with good absorbance at 532 nm
I need to stain some double-stranded DNA with an intercalating fluorescent dye. For imaging, I am using a microscope setup equiped with a 532 nm green laser.
What is the staining agent of choice to use with a 532 nm excitation source?
Sytox Orange may work but it is a bit pricey.

Did you try  propidium iodide? It is less fluorescent than Sytox Orange, but cheap.

@MathewMahindaratne It is certainly cheap enough and absorbs at the right wavelength, I will try that

Check this as well: https://www.fishersci.com/shop/products/live-or-dye-594-614-50as/501964599


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How is it possible for Enzymes to get absorbed undigested
We as medical practitioners frequently prescribe enzymatic preparations like Trypsin-Chymotrypsin, which actually are proteins. Often I wonder how come a protein gets absorbed undigested, through our intestine, and reaches the site of its action (inflamed tissues)? After all we have been taught in biochemistry & physiology that protein, before getting absorbed, must be broken down into its constituent amino acids.

Is it actually absorbed, or does it act locally within the digestive tract?

@DrMoishePippik. I think you are referring to enzymes like PANCREATIN which are used solely as digestive enzyme medicine.Yes pancreatin is intended and in fact does act locally inside the intestine. But TRYPSIN-CHYMOTRYPSIN type of preparations which , although may be used as digestive enzyme to act locally, are more commonly used with an intent to reduce inflammation at post surgical or traumatic wounds. I wonder how on earth an enzyme , taken orally , can bypass the digestive degradation to amino acids and reach the  distant sites of inflammation, say in our limbs?

This question is about physiology but with a healthy dose of biochemistry, so probably ok for the site although better suited for biology SE. However, it would be much improved with citations to sources.

I’m closing this question because it’s not about chemistry.

There is no biochemistry involved in this question. It’s physiology and medicine.

Too long for a comment:
In order for a protein to be absorbed, two things are required: (1) It needs to reach the absorption sites of the gut intact; and (2) It must be able to pass through the intestinal lining.
Gerhard Lorkowski [1] claims "Due to enteric coating, degradation of the protein structure to single amino acids or peptides in the acidic environment of the stomach is reduced." and "Research has confirmed that peptides and larger protein molecules pass through the mucosal barrier of the gastrointestinal tract."
The first claim is plausible to me.  I don't know about the second.  It's intriguing, but Lorkowski didn't provide any references to support it.
I'd recommend asking this question on Biology SE. You can relate what Lorkowski wrote as a starting point.
[1] Lorkowski, Gerhard. “Gastrointestinal absorption and biological activities of serine and cysteine proteases of animal and plant origin: review on absorption of serine and cysteine proteases.” International Journal of Physiology, Pathophysiology and Pharmacology, vol. 4,1 (2012): 10-27.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3312459/


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Vapor-Liquid Equilibrium and partial pressure of ammonia
I tried searching the forums for a similar question but I do not think it was entirely answered:
Suppose I have a closed system in a container with solvent and a certain concentration of gas, say ammonia ($C_{NH3}$), in the liquid phase at a fixed temperature which is at equilibrium (assuming all ideal gases for now). The gas phase will have a pressure based on the individual partial pressures of the components in the headspace, where each individual partial pressure can be expressed as the mol fraction multiplied by the total pressure (for ideal gases):
$P_t$ = $P_1$ + $P_2$ + $P_3$...
$P_i$=$x_i$*$P_t$
Question 1: If the desire is to increase the concentration of ammonia (${NH}_3$) in the liquid phase at a fixed temperature, would adding more nitrogen gas ($N_2$) into the gas phase help with this to increase the overall pressure of the vapor space above the liquid ? My initial thoughts here are to apply Henry's law where the solubility of a gas in a liquid is proportional to its partial pressure (at a certain temperature): $S=k*P$, Henry's Law link, but I am not clear as to what happens to the partial pressure of ammonia in this example (total pressure increases, but mol fraction of ${NH}_3$ decreases)
Question 2: Will the liquid concentration of ammonia, $C_{NH3}$, be affected by changes in the gas-phase volume at a fixed liquid volume and temperature (i.e. keep the same liquid volume in larger and larger closed containers). My initial thought here is, as the gas-phase volume (or headspace volume) increases, more moles of the components will move from the liquid to the gas phase to achieve the same equilibrium pressure at a fixed temperature, but not entirely clear here.
Thanks for the help and let me know if I can provide more clarity to the questions. Relatively new to this forum as well

Ammonia is very bad example, unless it is explicitly your case. It's extreme solibility significantly and not very linearly  modifies properties of initial water solvent.

Increased total pressure would  cause slight increase of chemical potential of dissolved gas, what would lead to slightly higher gas partial pressure with the same dissolved gas concentration.

@Poutnik, so you are saying then increasing total pressure would not significantly change the partial pressure of ammonia in the gas phase


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How does protein denaturation speed change with temperature?
So this is something that I had been wondering since middle school. We "roughly" know that protein denaturation occurs above a temperature of 72 degrees Celsius, and is the reason why it is usually not recommended to boil milk that has already been pasteurized. My question is, what is the relation between the speed of protein denaturing and the temperature at which it is observed? For example, does denaturing at 72 degrees occur twice as slowly as denaturing at 144 degrees assuming all other conditions remain unchanged? Also, what is the standard metric for measuring the speed of protein denaturing? Any links to published research are highly appreciated. Please note that I'm not looking for a precise answer, I'm looking at a rough idea on a high-level that can be explained to people that have no experience towards the field of biochemistry.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5409303/

"We know that protein denaturation occurs above a temperature of 72 degrees Celsius" Why does everything after "We know" is usually questionable, or like here completely misleading? Every single protein has different denaturation kinetics and you could write a few books about this. And then a new book about new discoveries every few years,

Well ofcourse the question needs to have some sort of generalization, since I don't have a background in Chemistry beyond high-school level. I understand the chemical world is very vast and to expect a precise answer is unreasonable. Hence I'm asking about general pointers.

Check out https://en.wikipedia.org/wiki/Arrhenius_equation for starters

There are thermophilic organisme that live at temperatures well above 100 °C, without having their proteins coagulated  by the heat. See https://en.wikipedia.org/wiki/Hyperthermophile

Thinking about rates of chemical processes more broadly, since that underlies your question, a reaction will not be twice as fast at 144 degrees C as at 72 - it will commonly be much faster. Rate of reaction depends exponentially, not linearly, on temperature - there’s a good explanation at https://en.m.wikipedia.org/wiki/Reaction_rate . It mentions an oft-quoted rule of thumb that for many reactions, for every 10 degree Celsius rise in temperature, reaction rate doubles. If denaturation of proteins obey this rough approximation, then the rate at 82 degrees would be double that at 72.
There’s another common misconception in your question: 144 degrees Celsius is not double 72 degrees. The choice of the freezing point of water as zero degrees Celsius is arbitrary, and doesn’t tell us anything about the laws of nature. If you more appropriately use Kelvins as your unit of temperature (where 0 Kelvin is -273 degrees Celsius - absolute zero, the lowest possible temperature), then 72 degrees Celsius=345 Kelvin so doubling the temperature would give 690 Kelvin = 417 degrees Celsius!


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Chelation of iron: is it removed or just contained?
I have a question that came up in my research (neurobiology), maybe someone can help me out! I read that neuromelanin, which is found in certain brain structures, can chelate iron and protect cells from the toxicity of iron. I also read that chelation means "incorporation of a mineral ion or cation into a complex ring structure" and I know chelation is used to remove iron e.g. in case of overload. However what I don't understand is if chelation will "degrade" the iron molecule (I'm not familiar with the right terminology, sorry!) or if it will just "contain" it but the iron is still intact. The reason why I'm asking is because there is this notion of iron influencing MRI signal in certain regions that contain neuromelanin which binds/chelates iron, but at the same time chelation seems to mean removal. So I'm wondering if chelated iron still has the same properties as "free" iron. Thank you in advance!
"Neuromelanin pigment is able to accumulate different metals, mainly iron. Neuromelanin seems to be the most effective system for scavenging iron, which results in a long-term immobilisation of iron inside neurons." (Ward et al., 2014, The role of iron in brain ageing and neurodegenerative disorders)
"Low Iron in LC Neurons Is Chelated by NM. NM is a strong chelator of heavy metals because of the presence of catechol groups in its structure (31). NM has strong chelating ability for iron, which provides another important mechanism of protection from iron mobilization and the consequent toxicity." (Zecca et al., 2004, The role of iron and copper molecules in the neuronal vulnerability of locus coeruleus and substantia nigra during aging)

It would be helpful to readers if you put those citations (or references) in your question.

Thank you for the tip, I added a few! @MathewMahindaratne

Iron is an element. It can be bound in a molecule, e.g., to neuromelanin, or to another chemical, e.g., to ethylenediaminetetraacetic acid (EDTA) or to "NM" (which needs to be defined). What "NM' (or EDTA) do to living organisms is another question, and whether these chelate other vital metals, such as calcium, yet another. BTW, the description seems to border on pseudo-medicine. Chelation is useful for heavy metal poisoning, and has been used to treat those who ingested plutonium, with mixed results.

Well, if you used a chelator like EDTA, it would chelate metal ions and later be removed with urine, I guess. This neuromelanin stays where it is along with whatever it chelates, afaict.

if chelation will "degrade" the iron molecule … or if it will just "contain" it but the iron is still intact.

Iron is an element.  That means that short of nuclear reactions, it is indestructible and will always remain as iron. It can't degrade.
In chelation, other elements react and combine with the iron to produce a larger, more complicated molecule that contains the iron atom in its interior, where it can no longer be seen (chemically).
As far as the body is concerned, it no longer exists.
With an appropriate choice of chelator though, the resulting molecule will be one that the body will recognize as unwanted and will dispose of it, typically by filtering it out of the blood through the kidneys and mixing it into the urine.

Chelation is an operation where a chemical substance is surrounding entirely a metal atom like $\ce{Fe}$. This means that this atom cannot interfere any more with any other components of the liquid in which it is dissolved. These components (proteins, enzymes, etc.) will not be puzzled or modified by the metal atom which is chelated. It looks as if the metallic atom has been removed from the solution.

@Maurice, thanks.  I've updated accordingly.


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Finding the molar solubility of silver carbonate in a solution of 0.125 M sodium carbonate
In calculating the molar solubility of silver carbonate in a solution of $\pu{0.125 M}$ sodium carbonate (knowing the $K_\mathrm{sp} = 8.46 \times 10^{-12}$) I have a question with the solution provided which is as follows:
$K_\mathrm{sp} = [\ce{Ag+}]^2 [\ce{CO3^2-}]$, the initial concentration of $\ce{CO3^2-}$ is $\pu{0.125 M}$. Let $x$ denote the change in concentration for $\ce{Ag+}$.
Therefore $K_\mathrm{sp} = x^2 \left(\frac{1}{2}x+0.125\right)$. Assuming the change for $\ce{CO3^2-}$ is negligible, that gives $K_\mathrm{sp} = x^2\cdot(0.125)$, which substituting the $K_\mathrm{sp}$ value gives $x \approx 8.227 \times 10^{-6}$, which is the molar solubility.
What I'm confused about is shouldn't the molar solubility be half that number since $\ce{Ag2CO3 <=> 2 Ag+ + CO_3^2-}$, meaning that for each mole that $\ce{Ag+}$ increases, $\ce{Ag2CO3}$ only decreases by half that amount? Could someone just explain whether my reasoning is correct or not?

There is justified simplification all carbonate ions come from sodium carbonate. You can verify it at the final check.

Hi, you might find it interesting to briefly scroll through https://mhchem.github.io/MathJax-mhchem/ -- many of the things you wrote can be done in an easier way using mhchem (the available syntax goes beyond just enclosing normal maths in \ce{...}).

Also, yes, I'd expect the answer to be divided by a factor of 2.

@Poutnik I don't think that is what I was asking for, I understand why the assumption was made, but I don't understand why the $x$ in the given solution actually represents the molar solubility.

X represents c(Ag+)=2c(Ag2C3)

Suppose in equilibrium, the soubility of $\ce{Ag2CO3}$ is $s \ \pu{mol L-1}$ in an aqueous $\pu{0.125 mol L-1}$ $\ce{Na2CO3}$ solution. That means, amount of $s \ \pu{mol}$ of solid $\ce{Ag2CO3}$ would be dissolved in one leter of $\pu{0.125 mol L-1}$ $\ce{Na2CO3}$ solution and in equilibrium as:
$$\ce{Ag2CO3 (s) <=> 2Ag+ (aq) + CO3^2- (aq)}$$
Therefore, by definiion, for $\ce{Ag2CO3}$, $K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CO3^2+}]$ (all concentrations are in equilibrium).
Since, solid $\ce{Ag2CO3}$ is dissolved in an aqueous solution of $\pu{0.125 mol L-1}$ $\ce{Na2CO3}$, you may have to consider $\ce{Na2CO3}$ concentration as well. However, as Poutnik pointed out in his comment, it is justified simplification that all carbonate ions comes from sodium carbonate solution alone since it is in higher concentration compared to carbonate ion concentration from dissoled $\ce{CO2}$ in the water. Let's look at the corresponding ICE chart (ignore solid $\ce{Ag2CO3}$ in the solution):
$$\begin{array}{l|cccc}
& [\ce{Ag+}]  & [\ce{CO3^2-}]   \\
\hline
\text{Initial concentration}      & 0  & 0.125       \\
 \text{Change in concentration}  & +2s & +s  \\
 \text{Equilibrium concentration}   & +2s & 0.125+s   \\
\hline
\end{array}$$
Thus at equlibrium,
$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CO3^2+}] = (2s)^2 \cdot (0.125+s) = (2s)^2 \cdot 0.125 = 0.5s^2$$
Since $K_\mathrm{sp} = 8.46 \times 10^{-12}$ is given,
$$0.5s^2 = K_\mathrm{sp} = 8.46 \times 10^{-12} \ \Rightarrow \ s = \left(\frac{ 8.46 \times 10^{-12}}{0.5}\right)^\frac{1}{2} \approx \pu{4.11 \times 10^{-6} mol L-1}$$
Thus, as as orthocresol pointed out in his comment, the given answer should be divided by a factor of 2.


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How can the reaction heat be ignored in an energy balance?
I have received from my professor the following (summarized) problem:
A first-order exothermic reaction $\ce{A->B}$ took place in a typical tubular plug flow reactor (PFR). The tube was cooled to a constant wall temperature $T_\mathrm{w}$ and the total amount of heat removed from the reactor $Q$ was given in Watts. All needed constants and geometrical parameters and $T_\text{inlet}$ and $T_\text{outlet}$ were provided.
The goal was to find the (constant) value of the wall temperature $T_\mathrm{w}$ that ensures heat $Q$ is being removed from the reactor.
My question is not about the final value but rather the logical aspects. In my professor's solution, they simply used the following formula from a heat exchanger:
$$
Q = U \times A \times LMTD
$$
where $LMTD$ is the logarithmic mean temperature difference based on $T_\text{inlet}$, $T_\text{outlet}$ and $T_w$ (to be determined). $Q$, $U$ and $A$ are given.
This approach makes sense for a tubular heat exchanger but as soon as a reaction is happening inside the inner duct, the energy balance doesn't result in the $LMTD$ anymore. The PFR heat balance in this scenario is:
$$\frac{\mathrm{d}T}{\mathrm{d}z} = \frac{k_{W}}{u\rho c_P}\frac{4}{D}\left(T_\mathrm{w}-T\right)+\frac{r(-\Delta_R H)}{u\rho c_P}$$
If the reaction were non-existing ($r=0$), one could derive the $LMTD$ approach to solving this from the balance above. But that's not the case.
What do you think?
Given $Q$ (along the entire reactor), $T_\text{inlet}$, $T_\text{outlet}$ and all other constants/parameter, can you calculate $T_\mathrm{w}$? Is my professor's approach correct? If so, how can one derive this formula from the energy balance as one does for a simple heat exchanger?

What are the definitions of $\pu{k_w, u, \rho, c_P, D}$ ?

Kw is the heat transfer coefficient (also known as U), u is the flow velocity, rho is density of the fluid, cp is heat capacity and D the diameter of the reactor

I agree with your assessment.  What does it mean to say that all the heat is removed?  Does that mean that the inlet and exit temperatures are the same?  Is the reaction rate supposed to be constant, or is it varying with z?

Update: Comments reveal I missed a major point below, which invalidates this answer, which is that the reaction rate $r$ depends at least on $z$ and probably also on $T$.  For now I'm keeping the answer up in case it's useful to anyone else; I (or someone else) might delete it soon.
WRONG ANSWER BELOW
The energy balance you have is:
$$\frac{dT}{dz}  =  \frac{k_W}{u \rho c_P} \frac{4}{D}(T_W - T) + \frac{r\left(-\Delta_RH \right)}{u \rho c_P}$$
This is a non-homogenous differential equation in $T$.  "Non-homogenous" means the equation is like $\frac{dT}{dz} = f(T) + K$, where $K$ is a term that does not contain any dependence on $T$.  (In the general case, $K$ may be a function of the dependent variable $z$, but you enjoy a simpler case where $K$ is just a constant.)
To be specific, for your energy balance, the term not dependent on $T$ is:
$$K = \frac{k_W}{u \rho c_P} \frac{4}{D}T_W + \frac{r\left(-\Delta_RH \right)}{u \rho c_P}$$
It's instructive to compare to the case with no reaction.  The non-homogeneous part of the equation gets simpler:
$$K_{\text{no reaction}} = \frac{k_W}{u \rho c_P} \frac{4}{D}T_W$$
However, in neither case does this non-homogeneous (some call it hetereogenous) part depend on $T$, or on $z$.
But, the "homogeneous" part of the equation is the same, that is, in either case,
$$f(T) = - \frac{k_W}{u \rho c_P} \frac{4}{D} T$$.
What I would recommend doing is to just substitute the symbol $K$ for the non-homogeneous part and solve the differential equation
$$\frac{dT}{dz} = -\frac{k_W}{u \rho c_P} \frac{4}{D} T + K$$
After you solve it, and generate the log-mean temperature difference equation, you can then substitute whatever symbols you'd like for $K$ into your solution; that is, you can generate solutions for both the no-reaction heat exchanger case as well as the case with the reaction.
Hopefully this will convince you that your professor was right.  If you know the inlet and outlet temperatures, you can just use the simple equation.  It's obviously true, however, that if you compared two PFRs, one acting as a simple heat exchanger with no reaction, and the other with a reaction, and both PFRs had the same flow rate, heat capacities, and inlet temperatures, the outlet temperature would be very different in the reacting PFR vs. the inert heat exchanger one.

Hi Curt, thanks for the reply. However, I realize I should have explicitly mention that a chemical reaction rate is dependent on the concentration of the species, which decreases throughout the coordinate z. Meaning, K is not a constant as you assumed it to be, but rather a negative exponential function of z because of the term "r" (reaction rate).

Oh, you're right.  I should have realized that!  Well, in that case, my answer doesn't work and I think you, not your professor, are probably right!

I wonder if I should delete the answer.  It's wrong, but I wonder if it's wrong in a useful way?

I think it's useful, and to be honest I had already settled the discussion with other PhD students who agreed the proposed solution was flawed. I should have closed the question but I forgot I posted it here haha


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My particles won’t separate without neutralizing the solution. I want to understand why?
I etched active carbon with a concentrated KOH solution and added HCl to neutralize the KOH. Separeted the particles with centrifuge at 4000 rpm. They should be micron sized.
My boss asked me to stop using HCl and just separated the KOH with centrifugation. But the separation didn’t work. What influenced this?

Your active carbon particles may have been partially coated by carboxyl groups $\ce{-COOH}$, which have acidic properties. In contact with $\ce{KOH}$, these acidic groups are transformed into $\ce{-COO^-}$. So all particles are negatively charged and will repell one another. These repulsions are such that the particules will stay in emulsion and will not be separated by centrifugation.

@Maurice That makes sense

As a consequence, the only way of separating your charcoal particles by centrifugation is to first neutralize them by adding enough $\ce{HCl}$.

Also, what is the density of the KOH solution? If it approaches that of the activated carbon (which has air bubbles), centrifugation might be difficult.

I added diluted HCl until the pH of ~1 and that helped to separate the particles