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If two atoms join, do they lose their original spectral lines?
Each element and even isotope of element has its own set of unique spectral absorption lines. What happens to these lines after two or more atoms join together into molecule?
Are all of the lines that pure element has in monoatomic form lost? Or is it that some or all lines are preserved?
Also, does it depend on the type of bond (ionic, covalent)?

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Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.

This is not homework question. All my research was googling it and not finding anything relevant. I decided to not even mention my research to shorten my question as I found people have short attention span and are more likely to read and respond to shorter text.

On CH SE, the shorter the question, the higher risk to be closed, instead of being responded.

It depends on which part of the spectrum you're looking at, for different parts of the spectrum involve different electrons.
Visible and UV spectra involve transitions of the valence electrons, which certainly change their states as atoms are joined chemically. The colors we see in chemical substances, such as iodine vapor being purple or nitrogen dioxide being brown, arise from such spectral shifts.
When we move to the far UV and X-ray portions of the spectrum, which involve transitions of more tightly bound inner electrons, this changes. These inner electrons remain in essentially the same quantum states. The spectral lines associated with them remain largely intact with only minor variations. X-ray methods for identifying elemental composition, such as energy-dispersive spectroscopy, rely on this constancy.

O2 is not blue in the gas state the blue color is from molecular association in the liquid and solid.

There is a 2-paged article in the Journal of Chemical Education on the blue color of liquid oxygen. Here at https://pubs.acs.org/doi/10.1021/ed042p647 and they have a very different explanation of this beautiful color. They say it is a phenomenon in which a single photon excites two electrons in two different O2 molecules which have collided.

This is getting long due to comments that no longer apply because of answer edits. May want to remove.

Thank you very much Oscar Lanzi , this was excellent answer, just what I was trying to find.

The lines of the atomic spectra are not maintained in the spectrum of the molecule. For example, the hydrogen atom lines are defined by the so-called Lyman, Balmer, Paschen, etc. series. The hydrogen molecule do not show these lines. It has its own series of lines.

Is it always like that in all molecules? I mean, did they never find even one spectral line being preserved in all molecules that were ever examined?

Slightly amplifying what Maurice said, isolated atoms have atomic emission lines in the IR, visible and UV spectral regions. But atoms in molecules are not isolated, so they do not have those atomic emission lines. Instead, molecules have a richer set of opportunities to absorb or emit light.

Just to emphasize what Maurice says in his answer, that I upvoted a year ago, here I show my two dimensional spectrum of the light emitted by my hydrogen discharge tube: https://physics.stackexchange.com/a/768678/313612. The high voltage discharge results in some hydrogen molecules dissociating to atoms. Then light can be emitted from excited hydrogen atoms, e.g., the Balmer series shown, plus light can be emitted from excited hydrogen molecules, e.g., the Fulcher alpha bands.


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Confusion regarding 1st and 2nd electron gain enthalpy
$\ce{O}$  has the 1st electron gain enthalpy $\pu{-141 kJ mol-1}$. $-ve$ value implies that energy is released when electron is added to an isolated atom. This also means that if $\pu{141 kJ}$ energy is supplied to $\pu{1 mol}\ \ce{O-}$ it forms $\pu{1 mol}\ \ce{O}$.
The second electron gain enthalpy has value $\pu{+780 kJ mol-1}$. This implies that energy is to be supplied to form $\ce{O^2-}$.  But if we supply $\pu{780 kJ}$ energy to $\ce{O-}$, would not the 1st added electron be removed; since it takes $\pu{141 kJ}$ of energy to be removed?
What am I missing?
Edit $1$- You guys are very rude. Seriously, $3$ downvotes but $0$ comments for a relatively well written question? Get over your attitude people.

@Poutnik Are Ionization enthalpy and Electron Gain Enthalpy not different?

@Poutnik Yes, I know they are different! My question was about electron affinities, but your comment referenced ionization energies.

@Poutnik Very sorry for the misunderstanding. Please see the edit.

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$  to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$.

If downvotes are not commented, one can imply the default downvote reasons that appear at the downvote triangle mouseover. I have not downvoted, but element symbols in italics is a typographic sin. // Complains should be rather in  comments than in question.

The correct answer is that the oxide dianion is thermodynamically unstable in vacuo
(Warning: PDF autodownload) with +780kJ/mol of excess energy, i.e. it does take +780kJ/mol of energy to add an electron to an "oxygen monoanion", AND that the oxide ions we see in, say, magnesium oxide crystals, are sufficiently lowered in energy, by the magnesium cation's positive charges' attracting the oxides' electrons, that they do not have any excess energy and are therefore thermodymically stable.

Thanks for the answer. I understand that $O^{2-}$ is unstable, but how can it be formed, since addition of the second electron requires an energy more than that required to remove the 1st added electron?

Let one say that the neutral oxygen atom has energy 0kJ/mol. The oxygen monoanion has energy -141kJ/mol and the oxygen dianion has energy (-141)+(+780)kJ/mol. NOT (-141)-(+780) kJ/mol.

To say it in different words; you are trying to subtract the "+780" your currenrmost result -141, when in fact you have to add it.

But is it not true that if we add the first electron affinity to the $O^-$  it becomes $O$?

O-: -141kJ/mol and O: 0kJ/mol

(-141) + 141 = 0

cf. O2-: 639kJ/mol and O-: -141kJ/mol

(639) + (-780) = -141

Sorry, but can you please elaborate?

Let us continue this discussion in chat.

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What non-toxic non-corrosive liquids are used for heat transfer?
I have a steel vessel placed concentrically within a larger steel vessel, resulting in air-filled annulus (air gap) between the two vessel walls (figure below). Due to the existing vessel construction, the only way to heat the inner vessel is to externally heat the larger outer vessel. The air gap makes this extremely inefficient. I would like to fill the annulus with a liquid to help with the heat transfer. I am needing to heat the inner vessel to 400 degrees Fahrenheit. Since the liquid will be open to atmospheric pressure, a liquid that wont boil at 400degF and 1atm pressure is needed. A non-toxic fluid is needed (we don't want to be breathing any harmful fumes). Is it possible to dissolve a particular kind of salt into water to achieve this goal? Is there any other type of liquid to consider?

Mineral/silicone oil typically used for oil heated baths is rated for approx. 200 to 250 °C (Sigma).

Note that for many mineral oils, the flash point is below the boiling point (as an example, one SDS I checked had boiling point of 260-340C, flash point of 146C, or under 300F). Not good for an open container.

Low Mw polypropylene glycol might work?

If a solid is allowed, maybe powered iron. Sand is not as good, but no significant safety issues compared to any liquid.

Why dom't you fill the gap with metal beads, or metallic powder ? Iron or aluminum ! Heat capacity of metals is much smaller than water, and heat conductivity is higher than air.

@Maurice and Ed V (+1), those are excellent ideas! I guess there will still be some air gaps via the porosity. I’ll have to consider their thermal expansions and melting points (don’t want them to become stuck in the annulus).

@Maurice Ha ha! So just visit a machine shop and offer to take their swarf! Win-win for everybody!

That is why I suggested iron powder: it packs great. I used to have two 5 lb. bottles of it. Really cheap, too.

@EdV how about aluminum? Any experience with it as well? It seems like it has a high thermal conductivity.

Aluminum is better than iron for this, but finely powdered aluminum is problematic and relatively expensive. You could always use a mix of aluminum turnings and chips, i.e., machine shop swarf, and powdered iron to avoid air gaps.

With aluminum powder you have taken the first step to a thermite bomb.

@blacksmith37 what about aluminum beads/bbs?

You can try with hot air circulation

I remember glycerol (b.p. 290 °C) is used sometimes, but not sure what temperature is safe wrt eventual pyrolysis and creation of traces of acrolein, in case of long term exposure.

Is this an XY problem? Why do you need to use this particular vessel, which is clearly not good for heating?

@Poutnik is great, but please do not consider glycerol: it is sweet, attracting ants and other insects. It is also very sticky, hygroscopic, always gets spilled somehow, and it will make your life miserable!

@EdV You should write that up as an answer. I know a liquid was specified, but that seems to be an assumption on the part of the OP, rather than a hard requirement.

@EdV  Good is to use monolithic diamond, diamond crystals, sand or dust, as diamonds have typically 10 times greater heat conductivity than the best conducting metals (  diamond conductivity grows dramatically with purity). Unfortunately, such heating bed is not exactly cheap.

Mineral oil should be good (aka  white oil). Vary similar to motor oil , flashpoints not easy to find but around 450 F. Or, for a much higher flash point- phosphate ester. I don't know what it is but we used it for lube oil in very high temperature applications.

A good starting suggestion is probably a mineral oil. These are often used in small scale labs for heating baths that need to reach temperatures much higher than boiling water. So they should be easily available and their properties are well known.
A possible alternative would be a molten salt mixture. Some of these are available even for low temperature situations but many are fine at higher temperatures (see this, for some examples, though it isn't a review for this sort of application, just an indication that such salts are available). Many molten salts are non-volatile and non-flammable but whether they are better than mineral oil will need some investigation.
Another alternative (though possibly more appropriate for lab-scale not large scale) is to use Armor metal beads. These are small metal beads that can be used as an alternative to liquids in heating baths in laboratories (either to avoid liquid contamination or to reach higher temperatures). They are quite expensive, but (obviously) are not flammable, can reach higher temperatures than many liquids and can be reused indefinitely.

Aren't molten salts highly corrosive?

@toolforger not universally. There is enough knowledge in the field to choose one that will work with the materials you need.

You can use a heat transfer oil like Mobiltherm 610.  They also have some other oils that you can use, if you are able to shield the surface of the oil from the air. If you use a lower grade oil, it can still work, but the oil may slowly decompose.
Dont just buy some generic mineral oil, without checking if it will decompose or catch fire under the conditions you are expecting.
Heat transfer oils can be poured at room temperature, unlike some molten salt baths.
Phillips66Lubricants makes some heat transfer oils, but afaik none that can handle your temperatures in an open system.
Chevron has this line in the data sheet for HTOs: The oil surface in contact with air in open systems should not exceed 107°C (225°F)

This seems a very bad idea to me. Please tell us how you intend to heat the outer tank and some idea of actual dimensions; then consult a professional chemical engineer with experience in heating. There are industrial grade heating pads or sheets that reach 250C that can fit in very small spaces. Investigate also the possibility of inductive or infrared heating. This will involve some expense but if you can't use good, appropriate, equipment you should reconsider doing this.
One suggestion would be to fill the space with aluminized Mylar helium balloons; that will give more heat conductivity than air and they should not melt [mp~240C].

"Please tell us how you intend to heat the outer tank and some idea of actual dimensions" — this should be a comment. Please ask OP prior to posting an answer if something is unclear to you. Note that unit symbols should be separated from the numbers with the non-breaking space, and C is the symbol for coulombs, not degrees Celsius.

I purposely put the question in an answer because I was getting perturbed by suggestions that could possibly lead to a large scale industrial accident.

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Synthesizing octopine from arginine and 2-bromopropionic acid
I am a biologist, and organic chemistry lessons are a distant memory now. I am interested in synthesizing a small amount of octopine. Octopine is a natural molecule resulting from an enzyme-catalyzed reaction between pyruvate and arginine, as shown below:

Sato et al. [1] synthesize octopine  as follows:

Octopine was prepared by synthesis from ʟ-arginine and ʟ-α-bromopropionic acid according to the method of Abderhalden. Purification of octopine was done by ion-exchange chromatography using Dowex 50W-x8 (H+ type) and recrystallization from aqueous ethanol. The purity (>99%) of the synthesized octopine was checked by high performance liquid chromatography.

I could not find the original Abderhalden paper (E. Abderhalden and R. Haase, Z. Physiol. Chem., 202, 49-55 1931). From what I understand based on other papers mentioning this name, the "method of Abderhalden" is simply a nucleophilic substitution between the α-amine and the halogenoalkane.
However, my understanding is that in this case it should lead to two issues:

Multiple reactions of the alkyl halide on the target α-amine, leading to e.g. tertiary amine products.
Reactions between the alkyl halide and the guanidino group of the amino-acid.

Are there reasons why these would not be a problem? Should one use special reaction conditions to obtain the correct product? Are there alternative easy ways to synthesize this molecule, other than purifying the enzyme and using the biosynthetic reaction?
Reference

Sato, M.; Nakano, T.; Takeuchi, M.; Kanno, N.; Nagahisa, E.; Sato, Y.; Kobatake, Y. Effects of Octopine on the Serum Cholesterol Level in Rats. Biosci. Biotechnol. Bioch. 1996, 60 (1), 154–156. DOI: 10.1271/bbb.60.154.
Abderhalden E.; Hase E. Gewinnung von Iminodicarbonsäuren aus Aminosäuren und Halogenfettsäuren. Z. Physiol. Chem., 1931, 202 (1), 49-55. DOI:10.1515/bchm2.1931.202.1-3.49

Both issues are realistic concerns. I can only presume that the purification steps referred to removed the side products which are likely to form.

I have a vague recollection that sodium cyanide borohydride will selectively reduce imines over ketones, as long as the imine is protonated. You might be able to use that to do a biomimetic synthesis

@Andrew  you are correct, though it will give diastereomers. You probably don't want to do it with pyruvic acid as it may be strong enough to liberate HCN from cyanoborohydride

Arginine methyl ester with a nitro group on the guaninidine is commercially available. A reductive amination with the commercially available methyl pyruvate (ethanol or methanol with sodium cyanoborohydride or sodium triacetoxyborohydride at room temperature) will give you the dimethyl ester of nitrooctopine. This should be a good reaction and it is easy to purify by chromatography at this point. Mild basic hydrolysis of the esters followed by hydrogenation over Pd/C to remove the nitro group will give octopine.
edit: As noted by @user55119 this will give a mixture of diastereomers.

The reductive amination is likely to give diastereomers, one of which is the undesired epimer of the dimethyl ester of nitrooctopine. This could be a serious problem if the epi-diastereomer dominates. How confident are you about the hydrogenation?

@user55119  Hydrogenation of the nitro to deprotect the guanidine of arginine is the first method mentioned in Greene's "Protection Groups in Organic Synthesis", see J. Org. Chem. vol 44 (1979) 3442. You are correct about the possible diasteromers,

I have several questions: (1) There are 2 C=O's on (methyl)pyruvate; is there a reason to expect only one of them to react? (2) What is the purpose of the esters? Solubility in the solvents used for the reaction, maybe? (3) Wouldn't the hydrogenation of the nitro group give an aminoguanidine instead of regenerating the desired guanidine? (4) Wouldn't the secondary amine is the guanidine (the guanidine nitrogen closest to the carboxyl) also react and create secondary products?

One of the C=O's is in the ester group, its presence activates the ketone C=O to attack by the -NH2 2) The reductive amination can be done on pyruvic acid but the conditions are more complex - this is an easy reaction with the esters. 3) The hydrogenation cleaves the N-N bond (see the ref. in my comment above). 4) The presence of the nitro group withdraws electron density from the whole guanidine group making it less nucleophilic.


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Relationship between d group and $\Delta_O$ value found in Tanabe-Sugano diagram
So this question came up in class recently and got me thinking, is there any relationship between the metal d group and the $\Delta_O$ found for some allowed transition in Tanabe-Sugano diagram?
I mean, for a $d^3$ complex the value found for $\Delta_O$ is the same of the highest energy transition, is that the same for all other $d^n$? And of not, is there any other relationship between $\Delta_O$ and the highest energy transition?

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AlCl3, Al2O3, Al(OH)3. What is more ionic?
I tried with fajans' rules.
First I got AlCl3 and Al2O3. There, because O2- has high negative charge I decided that Al2O3 is less ionic. But my teacher told us that we can get the same thing by comparing the electronegativity difference. But When I applied it to this pair, an error occurred. Al-O has higher electronegativity difference, but it is less ionic. Is that supposed to happen ?
When I take all those 3 compounds I have no idea how to compare AlCl3 and Al(OH)3. Their charges are the same. I think (OH)- ion is much larger than Cl3. But not sure. Therefore I thought AlCl3 is more ionic. But it turns out to be that Al(OH)3 is the answer. A lot of places suggest that AlCl3 and Al2O3 are covalent. Can't I use fajans' rules to more covalent compounds?

OH group is smaller than Cl and Al(OH)3 isn't molecular, unlike the chloride.

In the pure state, $\ce{|AlCl3}$ dimerizes as a molecule $\ce{Al2Cl6}$. The structure is made of two squares having one edge in common (like an open book). Each aluminum atom is in the center of each square, and the six $\ce{Cl}$ atoms are at the six apexes. Covalent bonds $\ce{Al-Cl}$ are along the diagonals. The two aluminum atoms are negatively charged. The two central chlorine atom are positively charged. The whole is not ionic.


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Is hydroxylamine monodentate or ambidentate ligand?

What is the total number of geometric isomers of the complex $\ce{[Pt(Cl)(py)(NH3)(NH2OH)]+}$?

If $\ce{NH2OH}$ acts as an ambidentate ligand, then there should be six geometric isomers. Or is it strictly a monodentate ligand?

What did the book think?

I think I've only seen $\ce{NH2OH}$ as monodentate ligand (M−N bonding). Deprotonated hydroxylamines are a totally different story, there are plenty of both bi- and ambidentate complexes with those.

The title of this question doesn't match with the question in the text body. For example, hydroxyl amine acts differently with different metals. See this and this articles for examples.

@MathewMahindaratne That's mostly editor's fault (mine too), not OP's. We tried to make the question less of am-I-right type and more conceptual.

I asked this question because it's a exam question, and 6 and 3 both option were present, book only has answer not explanation.


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Do the electronegativity of atoms affect the molecular geometry of molecules?
I would like to know whether the electronegativity of atoms will affect the molecular geometry of a molecule.
For example, the molecule $\ce{CHFCl2}$. Originally I thought that as fluorine was the most electronegative, it would have a stronger partial negative charge, so the shape of the molecule would not be tetrahedral. I also thought that as carbon has a higher electronegativity, there would be a partial positive charge at the carbon atom. Can anyone explain whether the electronegativity of atoms in a molecule can affect the shape of the molecule? In $\ce{CHFCl2}$ why can $\ce{CHFCl2}$ even have a tetrahedral shape although I think the charges cannot cancel out?

It can, but so can the atom radius.

Could you give some examples? Are there certain conditions for that to happen?

I will wait a day, and if you will not have been able to find any and will have justified it, I will provide some. // Before all that, check if your assumptions are correct

Hint: look up "VSEPR"

Under special circumstances, yes. For example, the NIST Web Book entry for tris(trifluoromethyl)amine (not the preferred IUPAC name, written here in an acceptable name for the sake of intuitivity) clearly gives a trigonal planar coordination geometry at nitrogen, while e.g. the valence isoelectronic trimethylamine has the VSEPR-expected (3BD1LP) triangular pyramidal geometry at nitrogen- this is due to negative hyperconjugation from the nitrogen lone pair to the C-F antibonding orbital, which is promoted by the so high electronegativity of fluorine. This effect is quite general, and is called the anomeric effect(Warning: PDF auto-download)

I suspect OP meant slight deviations from perfect tetrahedron not entirely different classes of structures


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Cant figure out whether Lithium or Strontium is in my substance?
So for uni I got a substance and I have to basically figure out which anions and cations are in it. I get two chances to guess otherwise I get a new substance and have to start from scratch.
Anyway yesterday I made my first guess:
Cations: Kalium, Ammonium and Lithium
Anions: Chloride
My anion was correct however they said either CO3, NO3, or SO3 is in there as well. I tested negative for all 3 of them, so not sure how to continue here.
My main issue though is the cations:
-Ammonium, this one Im sure about. I did the test with those two circle glasses where you put ph-paper on one side and NaOH with my substance on the other and then put them together. Did that multiple times, the paper turned blue in seconds every time.
-Kalium, semi sure on this one. I mixed it with HOAc and Na3(Co(NO2)6 and it turned yellow-orange. I put in the centrifuge after and its precipitation was yellow as well. Any opinions on the thrustworthiness of this test?
-Lithium, now this one I feel is the wrong one, because I did the flame colour test and as you all know its dark red looks the same as Sr. Since I have absolutely no idea how to differentiate them I just guessed Lithium. Please if you have any idea how test for either of them, help me out because I dont know how continue. I might just guess Sr for my second try.

You're asking us to do what your TA / instructor should be doing in the seminar belonging the to he lab course: Teaching qualitative wet inorganic analysis. You're welcome to ask about a specific issue with one of the tests, but we can't give a lecture that easily takes half a semester here.

Anyway welcome to the ion lottery! We've all been there ... ;)

Lol thank you! Yes Im aware its just that the instructors most of the time are no help at all so I appreciate any other help I can get. Like yesterday I wanted to test for Kalium Cations and the instructions said to use Kaliumchromate. But we only have Kaliumdichromate in the lab. So I asked my instructor if I could use that as well. Instead of answering he told me that is a good question, find out by yourself and then explain it to me. Any chance you could help me with that one? I cant find the answer googling; the instructor wont help and I dont know where else to look.

Keep in mind the recommended chemical name is potassium,instead of kalium.

Lithium is more magenta-colored, while strontium is bright red.

I failed lol but thank you anyway. Turns out it wasnt either strontium or Lithium, it was both which I hadnt even considered and explains why I wasnt able to tell them apart. They really said f^ck you with this substance lol

Lithium produces a red purple flame with only one big line in the spectrum. Strontium produces an orange flame with plenty of lines in the red, orange, yellow and green region of the spectrum. Lithium has no lines in the green. The color of the flames are not the same. Sr is orange red. Lithium is more red purple.
You may also test the solutions of your substances. Solutions of lithium salts do not react when adding a sulfate solution (like sodium sulfate $\ce{Na2SO4}$). Solutions containing strontium ions produced a white precipitate of insoluble strontium sulfate $\ce{SrSO4}$.

So if I solve my substance in water and add Na2SO4 or any other Sulfate and white precipitate is produced I can safely assume that it is Strontium and not Lithium?

Yes. Adding some sulfate ions to an aqueous solution is a proof that the solution contains strontium, baryum or lead (and calcium in concentrated solutions)


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How do I measure the percentage of CO2 in the air?
What tools can, in a very simple home environment, be used to easily measure CO2 levels in the air, for use in a home school science experiment?

Googling “home co2 sensor” brings up a variety of options

Those measure overall air quality (the ones I found, anyway). I'm trying to measure plant CO2 production in a closed system, I don't think they'll do the trick, sadly...

Ok, ‘greenhouse co2 sensor’ pulls up some others. Including ones for Arduino.

Look up Orsat gas analyzer in  Wikipedia. It has limited accuracy , but when I used one over 60 years ago it worked easily.

I bought one[PTH-9 Serie] works fine, very responsive, battery life is poor tho. There must be continuous air flow thru the meter for accurate results. If you are measuring CO2 uptake or emission over time a continuous flow system seems needed.

Please update your question. What accuracy do you need, what CO2 range do you expect?

I think it has actually been answered. How do I close the question when the best answer is in a comment?

You may determine the amount of $\ce{CO2}$ in air by using a saturated solution of calcium hydroxide $\ce{Ca(OH)2}$ ($\pu{0.0234 M}$ at $\pu{20°C}$). If you take $\pu{25 mL}$ of this solution and titrate it with $\ce{HCl 0.05 M}$ and thymolphtalein as indicator, you will need $\ce{23.38 mL HCl}$.
Now if you put $\pu{50 mL}$ of the same calcium hydroxide in a $5$ liters flask, close it and stir for half an hour, part of the calcium hydroxide will react accordingly to $$\ce{Ca(OH)2 + CO2 -> CaCO3(s) + H2O}$$ So the solution gets turbid. Titration of $\pu{25 mL}$ of the obtained solution with the same $\ce{HCl}$ solution will give you less than $\ce{23.38 mL}$, as $\ce{CaCO3}$ will not be destroyed by $\ce{HCl}$ if the indicator is thymolphtalein ($p\pu{Ka = 10}$). So the difference between the two volumes of $\ce{HCl}$ gives you the amount of $\ce{CO2}$ in the air of the big bottle.


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Calculation of the Seperation factor of Ethanol / Water rectification based on two saturated vapor temperatures
I want to calculate the efficacy of a Water / Ethanol rectification based on two temperatures with help of a vapor-liquid equilibration diagram.
The first temperature is measured right above the liquid.
The second temperature is measured right before condensation.
The rectification factor doesn't has to be extremely exact.
The calculation is carried out by a computer. The curves of equilibrium diagram is calculated by regression based on a few data points.

Is the "rectification factor" in mole/mole or mass/mass ? What is ${ T_1}$ red, ${T_1}$ blue, ${T_2}$ red, and ${T_2}$ blue ?

Does it make any difference if it is in mol/mol or mass/mass? T1 and T2 are variables. They will be measured during the process. The values from the image (blue) are just for example.

You are not very serious. The mixture 1 mol ethanol (46 g) plus 1 mole water (18 g) corresponds to a mixture 50% water + 50% ethanol in mole. But it is also 46/64 = 72% ethanol in mass, plus 28% water in mass. 50% is different from 72% !!

I know that the molecular weight of different molecules is different. But why is that important in this case? I don't want to know the concentration of Ethanol. Im pretty sure, if I would use a diagram that uses mass instead of mol, the result would be the same. However, if it really has an influence, let's say I would use mol, like in the diagram.


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IAPWS-IF97 empirical equations for water and steam
IAPWS-IF97 provides empirical equation the dynamic viscosity and the thermal conductivity for water, are there any formulas by IAPWS-IF97 for the density and the specific heat capacity of water?
I found "some" empirical formulas for both, the density and the specific heat capacity of water by going through some papers, but I wonder about recommendations by the IAPWS-IF97 if there is any.
Do you have any recommendation for similar equation for steam?

How accurate do you need to be? Have you considered using "steam tables" or an equation of state?

@electronpusher I would like to avoid using tables just to make everything $T$ or $P$ or both dependent; in terms of the accuracy, I would like to get "close" to the IAPWS-IF97 values in the region 0 - 150 °C and up to 1 MPa, where the standard EoS may already fail (still even up to 0.4, 0.5 MPa it won't be that accurate). Recently, I found some formula for $C_P$, but I cannot find any for the isochoric heat capacity, and there is no direct (constant) relation to obtain the other from knowing the $C_P$, e.g. Mayer.

I'm guessing the generalization of Meyer's equation is no good without an equation of state. I'm not sure what your options are if theoretical models (EoS) aren't sufficient and you don't want to use tabulated empirical data.

I was also not sure and that's why I asked :), but it's fine, alternatively I can just search though the tables.


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Kolbe's electrolysis reaction for alkenes' production
I came across a question in which the major product was asked for Kolbe's electrolysis of potassium salt of 2,3-dimethylbutane-1,4-dioic acid. The answer was given as trans-(but-2-ene). I couldn't understand how does the trans isomer form as the specific major product from this reaction, following free radical mechanism. Would someone please help me with this. Thanks!

As to your concern about the lack of references to the Kolbe process, try: J. Meyers, et al., Intramolecular Biradical Recombination of Dicarboxylic Acids to Unsaturated Compounds: A New Approach to an Old Kolbe Reaction, ChemElectroChem, 2020,  7 (24), 4873-4878.

Which question, exactly?

Generally during electrolysis of dicarboxylic acid we will get a few different products but the main product is unsaturated hydrocarbon. Maybe at this point, based on ChemElectroChem, 2020, 7, 4874 - down of page, the mechanism reaction is like in drawing:

Thanks for the answer. My concern with the mechanism shown is that according to this article, the syn-conformer is more stable than the anti-conformer that you have drawn here.

@ananta: Just because the syn-clinal conformation may be more stable in solution, does not imply what goes on when radicals are formed.

@user55119 under kinetic conditions (for example, high volatage), the syn-conformation may not to turn into the anti-conformation in time. We need to know how much of the cis- and trans-isomers are formed vs the voltage, temperature, pressure graphs. Do some metallic surfaces promote the formation of cis-conformer? Most studies are carried over a pristine platinum surface. Surely, there are unanswered questions here, and an incompletely understood mechanism.

It has little to do with mechanism and more to do with the fact that stable products are, in general, the major products in most of the chemical reactions.
In this case,  trans-(but-2-ene) is more stable than cis-(but-2-ene), due to steric reasons.


text:
Does zinc oxide lose its UV blocking properties when mixed with oil?
Zinc oxide provides strong protection against UVA and UVB radiation. Does $\ce{ZnO}$ retain this protection when it is dissolved in oil?
When $\ce{NaCl}$ is dissolved in water, the $\ce{Na+}$ and $\ce{Cl-}$ ions are pulled apart by the water molecules. I would assume that a similar thing happens when $\ce{ZnO}$ is dissolved in oil, where the $\ce{Zn^2+}$ is separated from the $\ce{O^2-}$ ions.
Also, according to Wikipedia — Zinc oxide,

ZnO reacts slowly with fatty acids in oils to produce the corresponding carboxylates, such as oleate or stearate.

But that still doesn't tell me whether that would lead to a loss of the UV protective properties.

$\ce{ZnO}$ is not soluble in oils. But it can form dispersion in oils. $\ce{ZnO}$ may react with traces of fatty acids in oils, forming respective zinc salts. But for UV screening context, it is negligible negative effect. $\ce{ZnO}$ does not lose in such a case its properties, there is just little less of it there.
The major factor in UV screening context is difference in refraction indexes of $\ce{ZnO}$ ( or $\ce{TiO2}$ ) particles and the fluid (liquid or gas). The bigger difference means stronger light scattering and longer average UV path through the given layer.
With a longer optical path comes also better UV absorption by the used liquid, more exactly by UV absorption additives.


text:
Alkynes product with bromine water
Has anyone analyzed the products of the reaction of any simple alkyne (eg.: acetylene) with bromine water? Most online references state that a dihalo-addition or tetrahalo-addition happens, but the only experimentally-based reports I found are more than 90 years old and state that dihalooxy-addition happens (although with hypobromous acid), leading to α,α-dibromoaldehydes and α,α-dibromocarboxylic acids (I expect α,α-dibromoketones possible with internal alkynes). Any new data on that?

Well, why would they do more research on that?

Please reference those experimental reports

Internal alkynes afford $\alpha$, $\alpha$-dibromoketones with tribromocyanuric acid in acetic acid. DOI: 10.2174/1570178615666180803152951

There are 4 kinds of reactions which occur for sure:

addition of dihalogen to alkene and alkyne in carbon tetrachloride gives vicinal dihalide(alkane and alkene respectively).
addition of hydrohalic acid to alkene and alkyne gives monohalogenated alkane and alkene respectively.
addition of dilute hydrobromic acid to alkene gives alpha-bromo alkanol
addition of hydrohalous acid to alkene also gives the same product

If such kind of reactions are possible then I think that the reaction about you are talking should also be possible via tautomerism of alkenol, but I can't see how can carboxylic acids form.
Addition of HOBr across C=C
I think this could be helpful


text:
Why don't we take activities in the rate law?
I am slightly confused by the fact that concentrations and not activities are taken in the rate law which is different to equilibrium and seems to lead to a discrepancy. For example, for the reaction $aA(g)+bB(aq)+cC(g)⇌eE(g)+fF(aq)+gG(aq)$, the value of $K_{eq}$ is:- $$K_{eq} = \frac{(P_{\ce{E}})^e[\ce{F}]^f[\ce{G}]^g}{(P_{\ce{A}})^a[\ce{B}]^b(P_{\ce{C}})^c}$$
where all quantities are dimensionless.The above expression follows from the definitions of activities of gases and solutes.
Now, the rate law for the forward reaction will be $v_f= K_{fwd} [A]^a[B]^b [C]^c$ and for the backward reaction will be $v_b= K_{bwd} [E]^e [F]^f[G]^g$ where [X]=$ \frac{P_x}{RT}$. Since, at equilibrium, $v_f=v_b$, so  $K_{fwd} [A]^a[B]^b [C]^c=K_{bwd} [E]^e [F]^f[G]^g$. Since $K_{eq}$=$\frac{K_{fwd}}{K_{bwd}}$ so we have $$K_{eq} = \frac{[{\ce{E}}]^e[\ce{F}]^f[\ce{G}]^g}{[\ce{A}]^a[\ce{B}]^b[{\ce{C}}]^c}$$
which is incorrect. So, do we take activities in the rate law(and concentration is a common misquotation) or is there something else I'm missing?

Try to use activities instead of concentration.

@Poutnik That was what I suspected but many texts say to take concentration. Is it a common misquotation?

It is rather empirical pragmatism. It is much easier to use concentrations.

Expressions with activities are exact, but mostly unusable. Concentrations are often acceptable approximations.

@Poutnik Why is $v_f= K_{fwd} (P_A)^a[B]^b (P_C)^c$ unusable?

Substitute literal values for activities for solutes, or fugacities for gases. As partial pressure is similar approximation as concentration.

Let us continue this discussion in chat.


text:
stable conformation of 2,3-dimethylsuccinic acid
I was researching for answering this question on Kolbe's electrolysis of 2,3-butanedicarboxylic acid (succinic acid). Turns out, this compound is not very well researched in particular, and shows few results on search engines. I did find some results for 2,3-dimethylbutanedioic acid (2,3-dimethylsuccinic acid). PubChem shows the anti-conformer as the standard structure.

Now, the compound has two stereogenic carbons and 3 isomers:

R-R
R-S
S-S

Do any of these isomers under any conditions show higher stability for the syn-conformer? I ask because of the possibility of $\ce{H}$-bonding:

Would the stability due to $\ce{H}$-bonding be enough to overcome the steric repulsion?

As told in comments, the above schematic has a 7-membered ring, which are generally less stable than 6-membered rings. This prompted me to draw another diagram which has two 7-membered rings instead:

I see that the rings are somewhat strained, but that doesn't mean that, at least, one of the two hydrogen bonds does not form.

As to your concern about the lack of references to the Kolbe process, try: J. Meyers, et al., Intramolecular Biradical Recombination of Dicarboxylic Acids to Unsaturated Compounds: A New Approach to an Old Kolbe Reaction, ChemElectroChem, 2020, 7 (24), 4873-4878.

There is no cis-isomer. If you are addressing the favored hydrogen-bonded conformation, the least one can say is that what is true for the R,R-stereoisomer is true for the S,S. Remember, you are dealing with a 7-membered ring, not a 6-membered ring.. Perhaps a computational guru can help with an answer.

@user55119 You say one 7-membered ring. I draw two 7-membered rings. What do you say now?

I believe your original question was about which diastereomer is more strongly, intramolecularly hydrogen-bonded. That issue is still in abeyance. As to your new doubly hydrogen-bonded structure, it does not add to the resolution of the question at hand.

Yes indeed, the question does deal with the diastereomers as well. I did find some evidence suggesting that the syn-conformer is more stable than the anti-conformer.


text:
Difference between number of hydrogen bonds formed and number of hydrogen bonds in a molecule
I would like to know how to count the number of hydrogen bonds formed in a molecule and the number of hydrogen bonds that a molecule has. For example, water forms two hydrogen bonds with its two hydrogen atoms, while accepting two hydrogen bonds from other molecules' hydrogen atoms, which makes a water molecule have four hydrogen bonds.But some molecules such as ammmonia have only one lone pair at the electronegative atom, while the number of hydrogen atoms in the molecule is higher than the number of receptors in the molecule, while in some cases, the number of lone pairs in the electronegative atom is higher than the number of hydrogen atoms? How do I find the number of hydrogen bonds formed? How do I find the total number of hydrogen bonds in a molecule?

The number of hydrogen bonds formed by an atom depend on several factos:

Presence of small, highly electronegative elements ($\ce{O}$,$\ce{F}$,$\ce{N}$, and sometimes $\ce{Cl}$)
Number of hydrogens covalently bonded to such elements
Spatial considerations
State-dependent factors (Temperature, Pressure, etc.)

You are confusing covalent bonds present in water with hydrogen bonds. In the stucture ice (phase?), each water molecule does form four hydrogen bonds in a tetrahedral geometry (wiki).
You may also find this article useful.

So each water molecule forms how many hydrogen bonds?

four hydrogen bonds in ice, fewer in liquid state, and close to zero in vapor phase

Then what about the lone pairs of oxygen that also take part in hydrogen bonds in liquid water? Do they count as 'forming' hydrogen bonds other than the two hydrogen atoms?

Yes! as long as there is a strong interaction between a small electronegative element (in this case oxygen atom) and an electropositive hydrogen, it will count as a hydrogen bond. Although, in case of liquid water, because of the thermal energy, each water molecule may not be able to form four hydrogen bonds.


text:
Relationship between molar flux and density of current
I am studying the mass transport for a charged species in a stirred solution. So, I have obtained the Nernst-Planck equation:
\begin{equation}
\mathbf{J} = -D\nabla C - \frac{zFDC}{RT}\nabla \phi + C\mathbf{v}
\end{equation}
where $\mathbf{J}$ is the molar flux, $C$ is the molar concentration of the charged species, , $z$ is the ion valence, $R$ is the gas constant, $F$ is the Faraday constant, $D$ is the diffusion coefficient, and $\mathbf{v}$ is the stirring rate of the solution.
Now, I want to relate $\mathbf{J}$ with the current $\mathbf{I}$. By definition the current sign is given by the movement of the positive carriers. I have found this relation in many electrochemistry books:
\begin{equation}
\mathbf{J} = -\frac{\mathbf{i}}{zFA}
\end{equation}
where $A$ is the area. I have nearly understood this relation, but I can't understand the sign minus. Why? If $z>0$, the carriers are cathions, positive charges. Therefore their movements should be in the same direction of the current. Where is my mistake?


text:
Dissolved Oxygen Meter

I have an Extech Heavy Duty Dissolved Oxygen Meter (Model 407510). Tech support for this company is terrible, so I'm asking questions here.
The manual says to expect a reading of 20.9 for calibrating to ambient air.
I'm getting 21.1. Is this acceptable variation? I am at 2133m/7000 ft ASL.
What is the clear membrane (if that's the right word) end of a DO meter made of?
Presumably it is some sort of gas-permeable polymer (?).
Photos here

That is not an unusual reading for an oxygen monitor at 5500 ft ASL in a well ventilated place. What is the quoted calibration?

The manual says to expect a reading of 20.9 for calibrating to ambient air. And yes, 0.2 doesn't sound too far off, but then again 19.5, a difference of 1.4, is not considered safe. Just asking.

Listed accuracy is +/-0.7% so seems OK. Are you intending to use that in life safety applications? (“Safe” is a imprecise word, if you don’t understand impacts of oxygen deficient environments please talk to somebody who does.)

Not an oxygen deficient environment in the sense of someone entering an enclosed space, but thanks for pointing that out.


text:
Manganate and Permanganate
In my chemistry course, the addition of the “per” prefix generally means that the oxoanion of relevance will have one more oxygen than the “base” ion.

e.g. chlorate = $\ce{ClO3-}$, perchlorate = $\ce{ClO4-}$

However, permanganate is $\ce{MnO4-}$ and manganate is $\ce{MnO4^2-}$.  The charges are different but there are still four oxygen atoms in each, so it seems to violate the convention.  Any thoughts?

Does this answer your question? What is the meaning of the prefix "per-"?

It is a little unclear.  It mentions “oxygenated states” but does not cover examples where the number of oxygen atoms is the same.

Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial.

Because it has nothing to do with number of atoms?

https://chemistry.stackexchange.com/questions/59302/nomenclature-of-polyatomic-ions https://chemistry.stackexchange.com/questions/40506/what-do-the-prefixes-meta-ortho-pyro-mean-in-inorganic-chemistry

Does this answer your question? Nomenclature of polyatomic ions

It explains the per and other prefixes using the number of oxygen atoms, which, according to the information I have received in this question, I believe is a bit imprecise.

A usual exercise is to take polyatomic ions of the form $\ce{MO_{x}^{y-}}$ and compute the oxidation state of $\ce{M}$. Then look at the names of these ions.

They ran out of names, maybe?
Permanganate does have more oxygen in the sense that it could formally be considered a possible product of adding oxygen to manganate:
$\ce{4MnO4^{2-} + O2 + 2H2O -> 4 MnO4^- + 4 OH^-}$
This theoretical reaction goes along with manganese having oxidation state +7 in permanganate versus only +6 in manganate. We may say that "per-" simply means higher oxidation state.

The general convention does not follow the number of oxygen atoms, but the order of oxidation states of the central atom.
For the increasing order of oxidation states, there are these prefixes/suffixes:

hypo-  + -ite
-ite  ( -ous for acids )
-ate  ( -ic for acids )
per- + -ate

The naming starts with -ate (-ic), Then -ite ( -ous ) is added if there are 2 oxidation states. Which of the other two is added as the third one depends on the oxidation state of the least common one.


text:
At constant solvent activity, do solution properties depend on identity of the solute?
When a substance is dissolved or mixed with water in high concentrations, the activity of water drops from 1 to a value that depends on the substance and its concentration. Is it correct to say that water solutions with the same activity have the same colligative properties irrespective of what the added substance is?
To give an example, let us say that dissolving 100 mg/ml of NaCl in a water sample decreases its activity to 0.9. and that dissolving 50 mg/ml of CaCl2 in water will decrease the water activity to 0.9. Will the two solutions behave similarly? Examples of what I mean by behavior are physical properties such as the freezing point, viscosity, and solubility of certain gases? Is there a rule of thumb to say what properties will stay the same and what properties will be different?
Please note that the used numbers in the question are arbitrary, and are used to make the question easier to understand.

If by "behave" you mean "have the same water vapor pressure" yes. You need to clarify what you mean by this. It depends on the property. The conductivity would not be the same. And the term "colligative property" is typically introduced in the context of ideal solutions. At high concentrations you observe non-ideal behavior which might include temperature-dependent changes in water-solute interactions and in the self-association of the solute.

@BuckThorn Thanks, I updated the question. From your comment, I think the simple answer is no. Is there a rule of thumb to say what properties will stay the same and what properties will be different?


text:
Rate law for thermolysis at temperatures well above the decomposition temperature
If the temperature is higher than the decomposition temperature, are thermal decomposition reactions reversible? For example, which of the following schemes would be more appropriate for temperatures higher than the decomposition temperature of calcium carbonate $\ce{CaCO3}$?
\begin{align}
\ce{CaCO3(s) &-> CaO(s) + CO2(g)} \tag{R1}\\
\ce{CaCO3(s) &<=> CaO(s) + CO2(g)} \tag{R2}
\end{align}
Also, which of the following equations would be correct for the reaction rate $v,$ where $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ are the rates of the forward and reverse reactions, respectively?
\begin{align}
v &= k_\mathrm{fwd} \tag{1} \\
v &= k_\mathrm{fwd} - k_\mathrm{rev} \cdot p(\ce{CO2}),\tag{2}
\end{align}

related https://chemistry.stackexchange.com/questions/88780/equilibrium-between-barium-carbonate-and-barium-oxide-and-carbon-dioxide

In principle, the pyrolysis of $\ce{CaCO3}$ is an equilibrium, which should be described with a double arrow. But usually this equation is considered as not reversible, because $\ce{CO2}$ does not have the opportunity to stay in contact with the solid $\ce{CaO}$. In practice, $\ce{CaCO3}$ is heated in an open system. $\ce{CO2}$ is eliminated as an unwanted gas, because $\ce{CaO}$ is the wanted goal of the operation. If this reaction would have been done in a closed flask, without any possibility for $\ce{CO2}$ to escape, the reaction would have been reversed by cooling the flask. And $\ce{CaCO3}$ would have been "back-synthesized" at room temperature. But, I repeat, this back-reaction is not wanted, as the whole reaction is carried out in order to produce the biggest amount of $\ce{CaO}$ as possible. Anyway it is not easy to do this reaction in a closed flask, as the pressure will quickly increase to a rather huge value, which is not simple to maintain at a high temperature.

Hi, thank you for the answer. So if I understand correctly, in a closed flask at temperatures higher than the decomposition temperature, the backward reaction will take place simultaneously with the forward reaction, right? Given the forward reaction is much faster at temperatures higher than the decomposition temperature?

In theory, in closed flask, the backward reaction will take place simultaneously with the forward reaction. But this is impossible to make in the real world, because the pressure in the flask will soon increase to more than $100$ bars, and even maybe more than $1000$ bars. And no material (glass, ceramics or metal) will resist to such a huge pressure at the temperature of decomposition (about $800°$C).


text:
Chlorine Cleaner for Car Interior
I am interested in removing cigarette odor from one car and a vinyl/plasticy smell from another.  I have found a product that is supposed to work for this:  NuVinAir  (www.nuvinair.com).  It works by releasing chlorine into the car's interior, which is supposed to permanently neutralize the odors.  A concern I have is any harmful reactions with the chlorine and previous-used vinegar and/or other cleaners on the car's interior.  The SDS of the NuVinAir cleaner is found here:  https://www.nuvinair.com/wp-content/uploads/2020/10/ReStore-ReFresh_SDS_092920_English-6045980bd30c9f400c01222df27ca08b.pdf
Thanks!

It certainly does not seem a good idea to have chlorine gas released inside a car. Why even consider such a product? If enough gas were released to reach all surfaces and to react with all traces of odor, it likely would be deleterious to both the car and to you. See J.D.Power for ideas: https://www.jdpower.com/cars/shopping-guides/how-to-get-smoke-smell-out-of-a-car

Upon further review it is chlorine dioxide gas, not chlorine.  Does this make it safe to use?

Chlorine and chlorine dioxide are both toxic gases. If there is a slight excess of any of these gases in the car, it will be deleterious to you.

The cigarette odor and the vinyl odor come into your nose, but their reservoir is not in the atmosphere in the car, but rather on the surfaces where high-boiling compounds have adsorbed and are slowly evaporating.
Fabric especially will hold lots of foul-smelling compounds on the fibers.
I doubt that there is much cigarette odor held inside fibers or plastic parts, but the vinyl odor could well be plasticizer slowly coming to the surface.
You might think that leaving the car in the sun with the windows open would slowly evaporate the odorous compounds, but while this might be somewhat true for the cigarette odors, if plasticizer is the source of the vinyl odor, the heat will just bring more up to the surface.
Probably the best way to reduce the odors would be a serious cleaning of the surfaces with a mild soap/detergent, with lots of paper towels - as close as possible to what a professional auto detailer does - followed by placing a car deodorizer somewhere in the car to mask the odors with something more pleasant, like pine.


text:
How specific is the exception of negative charge location (orbital vs atom) for determining stability of conjugate base?
Chapter 3 Acid-base reactions in Klein's Organic Chemistry As a Second Language explains an exception to the typical order of important factors for determining how stable a negative charge on a conjugate base is [1, p. 61]:

This example illustrates that a negative charge on an $\mathrm{sp}$  hybridized carbon atom is more stable than a negative charge on an $\mathrm{sp^3}$ hybridized nitrogen atom. For this reason, $\ce{NH2-}$ can be used as a base to deprotonate a triple bond.

How specific is this exception? Does it only hold true for nitrogen? Or could an oxygen atom be $\mathrm{sp^3}$ or $\mathrm{sp^2}$ hybridized and still be less stable than the structure on the left despite its higher electronegativity?
Reference

Klein, D. R. Organic Chemistry As a Second Language: First Semester Topics, 4th ed.; Wiley: Hoboken, 2016. ISBN 978-1-119-11066-8.

Exception to typical order is kind of ambiguous. Why isn't this just included in the typical order? In fact, what is the typical order that is listed in this book?

The enhanced electronegativity that comes from having carbon with $\mathrm{sp}$-hybridized orbitals is enough to allow terminal acetylenes to protonate nitrogen-anion bases like $\ce{NH2^-}$, but (in the absence of additional enhancements such as conjugation) not an oxyanion base like $\ce{OH^-}$. In effect carbon with $\mathrm{sp}$-hybridized orbitals lies between ordinary nitrogen and oxygen atoms.
We can get the same effect with $\mathrm{sp^3}$-hybridized orbitals on carbon in a cyclopentadiene ring, as with the parent compound $\ce{C5H6}$ which is an even stronger acid than acetylene. The acidity comes from the ring becoming aromatic when it is deprotonated and negatively charged; as a result cyclopentadiene is weakly deprotonated even by highly concentrated sodium hydroxide solutions, as well as being strongly deprotonated by amide ions. When you study aromaticity you should see this effect, as well as the opposite effect of cyclopropenyl compounds acting as Lewis bases to make that ring aromatic.


text:
Are there any monocations with negative electron affinities known?
It is known that most dianions are unbound per se, i.e. that the corresponding monoanions have negative electron affinities, and the "dianions" we see in e.g. metal oxides can exist as such only due to the additional positive charges (on top of the "weaker" positive charges on the e.g. oxygen cores) on the countercations' stabilising effect on the dianion's electrons, rendering them bound. It is also known that most neutral molecules are electronically bound; i.e. that the corresponding monocations have positive electron affinities.
However, some monoanions do have positive electron affinities- for example, the dodecahydrido-closo-dodecaborate dianion is bound, rendering its (radical) monoanion's electron affinity being positive.
My question now follows- are there any monocations with negative electron affinities; i.e. whose neutral counterparts are electronically unbound?

I am not aware of any atom or molecule with negative ionization energy. Probably because their bound electrons have lower energy then free electrons.

Are you asking whether there are any electronically unstable atoms prone to spontaneously lose an electron and form stable cations? In the absence of a solvent?

@BuckThorn yes.

Atoms OR neutral molecules.

Better question would be if it's possible at all, because it actually may be, even if the whole idea is way more outlandish than it may seem.

A monocation with an endothermic (negative) electron affinity is the same as a neutral species with an exothermic (negative) ionization energy. Reframing it this way may turn up more information.
The neutral substance with the lowest ionization energy I know is tetrakis(hexahydropyrimidinopyrimidine) ditungsten(II), also called ditungsten tetra(hpp). With a positive ionization energy of +3.51 eV, it is still very, very far from the goal. It does not seem possible that a reasonable chemical system would manage to bring the ionization energy all the way down below 0 eV. The problem is that ionization energy/electron affinity are defined and measured in a vacuum, but opposite net charges in a vacuum are completely unscreened and attract far too strongly.
But what if you do screen the charges? In a much weaker sense, you might consider that electrides kind of fit the requirement - you can think of these as neutral compounds which, in a condensed phase (typically the solid state), spontaneously eject an electron as if it were a separate anion. The condensed phase surrounds the charges and does the job of screening them enough so that the electrons and the cations can exist as "separate" entities.

Meh, I never liked such view on electrides.

You are talking about superalkalis, which are defined as groups of atoms that have a lower ionization energy than any alkali metal atom -- that limit being about 3.9 eV for caesium. ("Pseudo-alkali" is used for iniization energies below the 5.4 eV value for lithium.)
As noted by Nicolau Saker Neto, no superalkali is known with ionization energy less than 3.5 eV, so we do not have anything that spontaneously gives off an electron to form a gas-phase  superalkali electride. However, lattice-energy stabilization effects do stabilize some condensed electrides, such as the cryptand-alkali metal electrides given in Ref. [1].
Some superalkalis have a lower ionization energy than the electron affinity of atomic chlorine (which actually beats that of fluorine). Trilithium oxide, $\ce{Li3O}$, is one example; the species identified by Nicolau is another. For such species $\ce{M}$, the combination $\ce{M^+, Cl^-}$ would be more stable, without ion-pairing or condensation, than the neutral $\ce{M}$ group plus a neutral chlorine atom. This comparison does not, however, account for the bonding of chlorine in its normal state into diatomic molecules.
References

James L. Dye (19900). "Electrides: Ionic Salts with Electrons as the Anions". Science 247, 4993, 663-668.


text:
Performing a reaction on one side of piperazine ring
I want to perform a reaction on one nitrogen of piperazine ring while avoiding the reaction on both nitrogens, I am familiar with Boc piperazine however it's expensive and would take 2-3 months to import to my country, does anyone know an alternative method ?

Making your own BOC piperazine is not hard, the general principle is to do the reaction with the monohydrochloride - there is a good procedure here from piperazine, piperazine di-HCl, t-BuOH and carbonyl di-imidazole.
The same document also has a procedure for the monoacylation of piperazine.
A different procedure, which is general for di-amines, using Me3SiCl/SOCl2 to create the mono-hydrochloride and then reacting with BOC2O is here


text:
Predicting whether NaOH will react or not
I recently came across a problem which made use of the fact that NaOH doesn't react with benzyl alcohol
The problem was of a past year paper of an engineering exam in my country which I had attempted incorrectly.
Although now that i know that the pka value of benzyl alcohol,i know that the reaction won't take place to a large extent,but as an examinee who wasn't aware of the pka value,is there any method to decode whether the reaction takes place or not(or any such reaction mechanism for the same)??, as the acidic hydrogen tempts us to treat it as an acid.

Are you asking, "how could I know from the structure of benzyl alcohol that it will form an alkoxide in presence of NaOH?"? Because as it stands it's not really clear what you are asking. I can flip this around: if you did know that it forms an alkoxide readily, would you have predicted that it reacts?

It is good that you understand the role of the $\mathrm{pK_a}$ to decide whether an acid/base reaction can occur or not.
I am not aware of an absolute way to determine, just by looking at the structure, what the $\mathrm{pK_a}$ value would be. There's two things, however, that might be of help:

If you learn the approximate $\mathrm{pK_a}$ value of some compounds (or just whether they react with NaOH I guess), then if you are faced with a similar compound, you might be able to tell whether it is a stronger or a weaker acid than the ones you know, and that might help you place the molecule on your mental scale.

If not in an exam situation, then you should be aware that the $\mathrm{pK_a}$ values are very often found in organic chemistry tables (example). Using this tables then, you will be able to at least get a $\mathrm{pK_a}$ value for an unsubstitued compound, and you only need to consider the electronic (or sometimes steric) modification of the substituents.

As for how you could've known the $\mathrm{pK_a}$ value of benzyl alcohol, consider the following way of thinking: the methyl group connecting the phenyl and the hydroxyl group sort of acts as an insulator between the two functional groups, making benzyl alcohol more similar to ethanol than to phenol - indeed the $\mathrm{pK_a}$ of ethanol and benzyl alcohol are quite close. Then you need to know that ethanol does not react with NaOH (this was probably covered in your class, so  you might know this, see point 1. above), and then you can conclude than benzyl alcohol won't react either

Technically sodium hydroxide does react with ethanol, although sodium metal offers better conversion to form the sodium ethoxide product.

This is a bit surprising - it might be that you can make them react under extreme conditions, but I'm pretty sure they won't react if you mix them together. In fact, you can even buy sodium hydroxide dissolved in ethanol.

Maybe the the extreme condition is the ethanol must be anhydrous. Likely the commercial product contains some sodium ethoxide, which would not really affect its acid-base reactivity.

Hydroxide, water and alcohols establish equilibria among the various alkoxides etc. depending on solvent and original oxide concentrations.


text:
Best material selection for a component containing hot saturated-level saline water with high chloride content (Polymer vs coated metal)
I have a component (Which can be seen as the attached file)
and inside the component, there would be hot saturated-level saline water with high chloride content (Not shown)
Considerations:

The operating pressure inside the component is 5% to 10% higher than the atmospheric pressure.
The saline water temperature would be about 100 C
The chloride content of water is at least +230 ppt
Sealing of the components is important
There are some millimetric holes on the surface

Our preferences:

Not to use expensive metals such as Titanium or Super duplex
Polymer is preferred as the material of the component
Cheaper metals with a non-expensive coating are also preferred
The expansion coefficient of the material is preferred to be as low as possible to prevent distortion and shape change (such as many polymers) otherwise the sealing and maintenance would be difficult

Based on the above explanations, What do you think?
What would be the best cheapest available material with the cheapest production process? Do you recommend polymer materials (which one and why) or coated metals?

I don't understand your comment well, You say that we should have more chloride in water for saturated-level saline water? The TDS of the water is +350,000 ppm and the chloride content is more than 80,000 ppm. Actually the relation between saturated-level saline water salt and chloride is not important what is important is material selection

Saturated saline could have, just roughly from the top of my head, some 350 g NaCl/L, what is some 350*35.5/(23+35.5)=212 g Cl-/L. That definitely is not 80 ppt as parts per trillion, like 80 ng/kg. ( Errorneously taken before as ppb, 1000 times more ) // What exactly do you mean by +80 ppt and what is its relation to saturated saline.?

Yes, you are completely right, the condition at least has to be 80 g Cl-/L and could reach 212 g Cl-/L. In addition to Na+ we have Ca2+, Mg2+, SO4 2- (the condition is not steady). I have corrected and edited my question. Our concern is which material would be the cheapest and the most efficient material for our application in water with super high chloride content.

For strange reasons you insist on errorneous usage of parts per trillion.

Does it make difference using ppt, ppm or g/L, or other units? if using ppt is not professional please let me know

Using ppm, ppb, ppt is professional, but not recommended, as there must be clearly stated if you mean mass, volume or molar amount. Also, it is sometimes incorrectly used with mixed quantities. E.g. ppm as mg/L is not fully correct, as it implies 1 L has mass 1 kg.

In the olden days in the USA, ppt meant parts per thousand. Sad, but true. For decades now, ppt means parts per trillion.

One more reason why not to use ppx concept. Using ppt as part per thousand is very strongly discouraged, leading surely to confusion with part per trillion. %, promile or as the best option explicit units like g/L or g/kg are suggested.

Depending on size, fiberglass would likely be cheapest; one would need to research epoxy or polyester for binder. More costly would be rubber lined steel. Rubber lined steel rail cars and storage tanks have been used many decades for super phosphoric  acid  at about 300 F. I do not recall the specific rubber.

Could FRP coating also resist under such conditions? for example ST37 steel+frp coating?

Rubber is what is used; and any steel. Some coatings would likely work but they would need test history or new tests.


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With respect to nomenclature, when is parent chain length a bigger priority than having functional groups before double bonds?
I came across this problem in Organic Chemistry as a Second Language. In the book, it is taught that one must prioritize functional groups over double bonds, while also trying to keep both as low in number position as possible. My numbering (top) does this, but disobeys the hierarchy in that a substituent comes before a double bond. The book's numbering, however, disobeys this hierarchy by putting a double bond before a functional group. The book's solution also makes the parent chain longer, which I know is better.
Both ways defy the hierarchy in some way(s). I want to know why the book's solution is considered better.

I think, you misunderstand the concept of the priority. If the group has the priority over another group, it does not mean, that this group's locant has to be smaller. It just means, that the group with the highest priority needs the lowest locant possible. Therefore you should do this:

What is the group with the highest priority? -OH. (Therefore the name of the compound is something-ol)
How many of them are there in the molecule? Only 1.
What is the principal hydrocarbon – the longest one with the maximum (in this case only one) groups with the highest priority? The whole hydrocarbon! (Since the -OH group does not divide this hydrocarbon in two.)
How long is the principal hydrocarbon? 8 carbons. (Therefore the name of this compound is octan-?-ol with some modifications we add later.)
And now the part with the priority: Should I number the principal hydrocarbon from the left or from the right? From the right, because the -OH group is "closer" to the right. Therefore it is octan-4-ol. If numbered from the left, it would be octan-5-ol.
And that is it. Now just add the other groups. In this case, it is just a double bond. Therefore oct-2-en-4-ol.

One last thing is stereochemistry, if you need to do that. In many cases you do not need to. As far as I know, the cis-/trans- nomenclature is obsolete now, if you speak about double bonds. Nowdays you use (E) for trans and (Z) for cis. If you understand German: E means entgegen (on the oposite side) and Z is zusammen (together). Nowdays cis/trans is used only if you speak about cycle molecules.
Therefore (E)-oct-2-en-4-ol.

Here is a summary of what you should do about this example:
First, find the parent hydrocarbon. This is based on the longest continuous carbon chain.
Second, find a functional group with the highest priority and give numbers to the carbons from the nearest position of the functional group to the end of the chain. Here, while the hydroxyl has higher priority than the double bond, both are numbered from the same side of the chain.
Using this rule, you can find the correct answer as the below name (Trans or (E)-oct-2-en-4-ol).
https://www.chem.ucalgary.ca/courses/350/Carey5th/useful/nomen.html


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What are good ways to anticipate co-crystalization of different metal ions?
What rationale (purely theoretical/assumptive, or known by experience / backed by evidence) would you use in an attempt to effectively anticipate which other metal or elemental ions are most likely to co-crystalize with a given metal / elemental ion?
In other words, to generally anticipate which elements are likely to easily substitute for a certain element during crystalization (which, as we know,  generally promotes the exclusion of anions/cations different from that constituting the bulk of the crystal)?
Please feel free to reword my question or its description if you have a better way to put it right.

Major factors are shared valency, similar ionic radius, shared relevant chemistry regarding counter-ions. Like Al, Cr and Fe(III) in alums. // The universal, but demanding  good way is obtaining good chemistry background knowledge. There are usually no simple rules to know all in chemistry. (except this one)

Would you like to copy/paste your comment into a formal answer?

Apart from alums, there are not too many examples of co-crystalization. In usual alum $\ce{KAl(SO4)2·12H2O}$, $\ce{K}$ may be replaced by $\ce{Rb, Cs, Tl, NH4}$, and $\ce{Al}$ may be replaced by $\ce{Cr, Ga, In, Co(III), Mn(III)}$.  But there are some strange situations. Iron may replace $\ce{Al}$ only with ammonium and $\ce{Tl}$. And thallium makes alums only with $\ce{Al, Cr, Fe}$.

@Maurice: Can K not also be replaced by Au(I) which has a radius way closer to that of K than Cs does? Or is it because it has a different chemistry?

@Hans Au(I) is not stable in water solutions and disproportionates, unless in stable complexes like [Au(CN)2]-.

Major factors for coprecipitation are shared valency, similar ionic radius, shared relevant chemistry regarding counter-ions. Like Al, Cr and Fe(III) in alums.
Note the various forms of co-precipitation (Quote form the above Wikipedia link):

There are three main mechanisms of coprecipitation: inclusion, occlusion, and adsorption.

An inclusion (incorporation in the crystal lattice) occurs when the impurity occupies a lattice site in the crystal structure of the carrier, resulting in a crystallographic defect; this can happen when the ionic radius and charge of the impurity are similar to those of the carrier.
An adsorbate is an impurity that is weakly, or strongly, bound (adsorbed) to the surface of the precipitate.
An occlusion occurs when an adsorbed impurity gets physically trapped inside the crystal as it grows.

Each of the above factors applies in various extent on them:

Inclusion is the most sensitive to the ion radius and valency.
Occlusion has almost no prerequisities to occur.

The universal, but demanding and unpopular good way to obtain predicting knowledge is obtaining good chemistry background knowledge. There are usually no simple rules to know all in chemistry. (except this one)

I'm looking at atomic radio tables on wikipedia. They have two different ones: (1) Crystal ionic radii and (2) Effective ionic radii in pm of elements as a function of ionic charge and spin (ls = low spin, hs = high spin). Am I correct to assume that crystal ionic radii is what is relevent for the present purpose?

Is it the low spin or the high spin radius of an element that should be considered?

@Hans The ion spin and spin dependable radius would depend on the charge and geometry of anions and or the lattice geometry. But I do not know enough here to give an authoritative answer.


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Why is rapid expansion/compression reversible?
I am looking over the Otto Cycle on this MIT website and it says at one point "the processes from 1 to 2 and from 3 to 4 are isentropic" in reference to the expansion and compression of the piston. I understand however that the compression and expansion in an internal combustion engine is quite violent and quick. And according to wikipedia "Throughout an entire reversible process, the system is in thermodynamic equilibrium, both physical and chemical, and nearly in pressure and temperature equilibrium with its surroundings." However this isn't the case for our expansion and compression is it? the pressure and temperature of our expansion/compression is not in equilibrium? It also says "reversible processes are extremely slow (quasistatic)". But once again, our expansion/compression isn't slow at all?
I've noticed, there are many processes that seem to be described as isentropic that from what I can tell are not in equilibrium with their environment throughout and are not quasistatic (my book for example says pumps, turbines, nozzles, and diffusers perform isentropic operations)? So through what justifications are these approximations made?
The main reason I ask this question is because I am attempting to calculate the pressures in a gun barrel, and I have seen the process described as isentropic (adiabatic and reversible) but for the same reasons as the Otto cycle, cannot understand how this would apply. The gun barrel acts similar to a rapidly expanding piston. Ignition, rapid expansion of gases against bullet (piston), then bullet (piston) leaves and the gases diffuse rapidly into the atmosphere. How is this reversible or quasistatic in any way? If anyone has an idea regarding the gun barrel and what process it is that would be appreciated as well.

The process described in the theoretical Otto cycle is idealized, it is performed so quickly that there is no opportunity for heat to be exchanged (and is therefore adiabatic). In essence any process that is too fast to allow significant heat exchange with the surrounding can be modeled as approximately adiabatic. In addition the work during the expansion/compression is assumed to be performed against an external pressure at equilibrium at all times (without loses). This satisfies the condition that the process be described as reversible.
The heat exchanged in the process can be related to the change in entropy through the second law of thermodynamics, which can be stated as
$$dS = \frac{dq_{\textrm{rev}}}{T}$$
For a reversible adiabatic process $dq_{\textrm{rev}}$, the quantity of heat exchanged during an infinitesimal step of the process, is zero, and it follows that the total entropy change is zero:
$$\Delta S = \int_1^2 dS = \int_1^2 \frac{dq_{\textrm{rev}}}{T} = 0$$
This explains the conclusion that the process is isentropic. The diagrams in the MIT course lecture notes show that this is an idealization. The true Otto cycle is not isentropic. The ideal and real cycles are arguably similar, however.
When a bullet is fired there is a chemical reaction that generates very high pressure (as during the ignition of fuel in an internal combustion engine), in this case within the casing between the bottom of the barrel and the base of the bullet. The exit of the bullet can be argued to be analogous to the rapid expansion of a piston, and occurs quickly on the scale of heat exchange with the surroundings, so is arguably adiabatic. Here the expansion is against a constant atmospheric pressure, so the reversibility condition seems more difficult to justify.

Back pressure is greater than atmospheric bullet velocity is mach 2 or 3 . Adiabatic is a coarse approximation. You have probably not fired a machine gun or BAR the barrels  get much too hot

@jimchmst I thought of the compression front as the bullet exits. Presumably it leads to a closer match between internal and external pressures, but this does not strike me as clearly rendering the process reversible. Physics of firearms is tbh not something I know much about, but the OP wrote this info was "optional" so that last paragraph is my 2 cents, otherwise I would have withheld posting an answer. If you can clarify this part and answer the more general question the OP posted you are welcome to do so.

Further reading on isentropic compression: https://www.grc.nasa.gov/www/k-12/airplane/normal.html


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What does it mean when we say an atom has infinite energy levels/shells?
My chemistry teacher told that an atom has infinite energy shells according to Bohr model , where electron reside according to its angular momentum and energy.
But in lower classes/ grades  I have learned that when we give excess energy to atom, the atom loses electron.
But now they are saying that when electron gain energy it move on to higher energy level.
So in this way, an electron should never 'jumps out'(sorry for that word) of atom? And keep going to higher energy levels as atom has infinite energy levels?
And should always be bound with the atom and never lost.
However electrons  do actually leave atoms when given excess heat.
So how does atom lose electron and not always being bind with atom when given excess heat in spite of infinite energy levels?
What does my teacher actually mean by saying it?

The energy levels are not evenly spaced in energy: they get closer and closer. See the little figure here, for an example: https://chemistry.stackexchange.com/a/119468/79678.

How many integer numbers n > 0  is there for -a/n^2  < 0 where a is positive real number?

Please check your spelling before posting.

Simply put, an electron in a hydrogen atom in a molecule of water in your fingertip could be detected in your elbow, or even on the moon. It does not have a location until detected, only a probability function describing where it can be found.
Raising the energy level changes the probability of where it can be detected, and, as your teacher stated, there are an infinite number of levels. Notice in the energy level diagram at HyperPhysics, or from physicsforums.com, below, that for higher n, levels are closer together.

This explains how there can be an infinite number of levels in a finite range of levels. This is a converging infinite series. For example, in Zeno's paradox of Achilles and the tortoise, 1/2 + 1/4 + 1/8... an infinite number of terms never exceeds one.
So, in a sense, even if the hydrogen is in an ionized state, the electron is never completely lost... though it might be indistinguishable from other electrons.
[Irrelevant thought for the day: can atoms be unionized, if they want better working conditions?]

"an electron in a hydrogen atom in a molecule of water in your fingertip could be detected in your elbow" Strictly speaking, this would require excitation (I'm not sure tunneling to forbidden regions of a potential are possible, that would violate energy conservation?), so that electron is presumably constrained never to be detected in your elbow.

@BuckThorn If the potential is infinite, then the wavefunction is 0, but if it's finite then there is some probability. So it depends on how "forbidden" we're talking.

@orthocresol good point. But how do you circumvent the issue with conservation of energy? "Quantum tunneling allows electrons and other small objects to have a small probability of being in regions where conservation of energy prohibits them." (from https://web.phys.ksu.edu/vqm/vqmnextgen/qmbasics/exploringquantumtunneling.pdf) Hmm, it might be my turn to post a question.

@BuckThorn The act of measuring the particle is what changes its state and thereby its energy. I don't know enough to answer this, but I strongly suspect the question becomes one of how the measurement is done. Measuring the position / momentum / etc. of a particle necessitates its interaction with the measurement apparatus in some way, so you only need the energy of the entire system (particle + measurement apparatus) to be conserved. This is veering into more philosophical, interpretations-of-quantum-mechanics grounds -- https://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics

@orthocresol yeah, it gets messy. A few posts in physics SE try to answer this, some along the lines of what you propose: https://physics.stackexchange.com/questions/111697/extra-energy-in-quantum-tunneling

@BuckThorn, there are some details in quantum physics it's hard to get one's mind around; such as zero-point energy and the Casimir effect, or Hawking radiation, or photon scattering by transient vacuum charge fluctuation. In the specific, they seem to violate conservation, yet on the average, Maxwell's demon can't win -- TANSTAAFL.


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how to make a PEM (proton exchange membrane) out of Nafion Dispersions?
I would like to use a Nafion membrane for a fuel cell project, but the Nafion sheets are hard to integrate. Therefore, I would like a gel type of Nafion that can take the shape of the container I put it in e.g. 5 cm^3 Cube). When researching, I found out about Nafion™ Polymer Dispersions which are a liquid form (thick).
Is there a guide on how to use it to fabricate a proton exchange membrane? Do I require other materials (chemicals) to mix with the Dispersions to make the PEM?
My ideas:
1- Dry the Nafion solution under 120C until all H2O is evaporated hoping for the dry polymer to be adhesive.
2- Mix the Nafion solution with either PVA or agar to give it a gel-like structure. I don't know if that will affect protons' pathways.

It'd be helpful if you put your thought in the text so that we can suggest whether it'd work or not and other proposals, which would work.


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Why is the relationship between a compound in its elements expressed in moles or particles? Is there a law dedicated to this?
I understand that a ratio can be made for the number of elements in a compound and the actual compound. Depending on the need, different units can be used. It can be both moles. It can also be entities where the entity is an ion, molecule or atom.
For example, in $NO_3$ a ratio can be made where 3 mol of O is to 1 mol of $NO_3$. Another ratio is 3 atoms of O is to 1 formula unit of $NO_3$. What is the law that says this, and why can different units be used for this formula? What are the other possible units aside from mole and particles? Why are moles or particles the basis?
I know the topic is fairly simple but most discussions I've seen just focus on other things. For example, there is more emphasis Law of Definite Proportions and Multiple Proportions. Then when the part where this ratio is used in other lessons such as percentage composition or conversions between mass, mole, and atom they just briefly explain that you can make a ratio because that's how it is presented in chemical formula. I mean I can accept it but I'd like a little more explanation that really focuses on ratio between a compound and an element it contains. It's not that I don't completely get it but I'd like a more intuitive grasp.

Do you got this bit about how proportion works? Unit is irrelevant.

" a ratio can be made for the number of elements in a compound and the actual compound" Sorry but this is unclear. What do you mean with "number of elements in a compound"? For instance, in methane there are two different elements. Is that what you mean?

It is like a common sense law, not related to chemistry, applied since ancient times, e.g. for scaleable cooking.

One mole of one substance contains the same number of particles as one mole of another substance. This follows from the definition of the quantity “amount of substance” and its unit, the mole. Thus, mole ratios are always equal to particle ratios, and you don’t need a separate rule for distinct contexts.


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Tips on understanding one-bond J couplings in NMR
I have just started a spectroscopy course in nuclear magnetic resonance, which I have loved thus far, but have been extremely puzzled regarding 1J coupling constants.
I was wondering if any of you had tips to further understand 1J coupling constants as a whole. For example, I'm confused about bonding between two of the same atoms (119Sn-119Sn) and between those with different spins (14N-15N), and how it would relate to the intensity on a spectrum.
I have talked to my professor about it; however, he is unclear about these concepts specifically, I would appreciate any help on this subject :)

Do you have a textbook, or a more specific question? You might look at the further reading list on this site: https://chemistry.meta.stackexchange.com/questions/2889/resources-for-learning-chemistry

Usually with couplings it is the splitting pattern that is a concern. You are asking about peak intensities, and those depend on the nucleus too of course, and can depend on the extent of splitting for various reasons. Perhaps you can be more clear about what you are wondering about (splitting patterns and intensity of couplings versus peak intensities).

One-bond couplings are not any different from any other couplings. There’s nothing special about how many bonds the coupling occurs over (except for the magnitude of the coupling, but that’s just a number which kind of falls from the sky anyway). Your question about same/different nuclides suggests to me that the confusion is not about one-bond couplings per se, but rather about couplings between equivalent or inequivalent nuclei. But you need to explain your question better, preferably with one or more example compounds.

(… In particular, if you are puzzled by some specific question or problem that was given to you, then just put that question here.)


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Hydrogen evolution on gold
I am working on intercalating an organic compound ( an amine) in another material. This has to be done in acidic medium and thus I am using HCl with gold as the cathode. I am wondering at what voltage the hydrogen may start to come out because of the reduction of H+ ions since I will have to keep my operating voltage under that one.
Thanks for your help in advance.

Consider voltammetric curves for both electrodes. It then gives you the voltage for the given current and vice versa. And also look at overpotential of hydrogen on gold, possibly also for oxygen on anode metal, if applies.   https://en.wikipedia.org/wiki/Overpotential

Besides, it is also convenient to subsitute HCl with H$_2$SO$_4$, you may inadvertingly  oxidize the chlorine anion.


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Which ionic radius is most relevant for anticipating co-crystalization?
In the context of anticipating the likeliness that elemental impurities co-crystalize into the crystals containing an element of similar radius (see this question), which is the relevant radii to consider?

Is it the crystal ionic radii or the  effective ionic radii in pm of elements as a function of ionic charge and spin?

AND

Is it the low spin or the high spin radius of an element that should be considered?


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How is Magnesium Ethoxide appropriately prepared, stored and used to dry ethanol?
I found many procedures vaguely describing the preparation of magnesium ethoxide from elemental magnesium in excess dry ethanol in the presence of I2 as activator (in absence of air), I also know that detailed procedures exist in publications I cannot access.
What is the smallest I2 to Mg ratio I can use without significantly affecting the speed of the reaction (assuming I'm fine to let it sit a few hours at max 40C)?
Is it appropriate to store it under 5x excess ethanol, itself under argon, in a glass bottle at 5-25C?
When used to dry 90-95% ethanol, is it appropriate to add only say 110% its molar counterpart in water content, and just stirr it an hour at ambient temperature? Given that after filtration traces will be left dissolved in the dried ethanol, can it still be safely distilled at atmospheric pressure knowing there will be a little bit of air in contact with the ethanol? (Magnesium Ethanoate, alone, having a flash point of 43C)

Generally you would use magnesium metal rather than magnesium ethoxide to dry ethanol. See this video from NurdRage on methanol drying, though it should be closely analogous for ethanol. Ultimately, molecular sieves are better, though 90-95% ethanol is a rather wet starting point.

Thank you for your response and for the video. In the video this man does dry the alcohol with magnesium methoxide (formed in situ). He mentions himself dropping iodine to accelerate the process, which he carries out under heating with methanol (which is much more reactive than ethanol). All my questions stand.

This video on solvent drying includes EtOH at about 7.45  https://www.youtube.com/watch?v=RawVXnMSyXc&t=103s

@Waylander: Thank you for the video. I had read the paper that he took the table from but I would like to use magnesium ethoxide because it's fairly fast acting, efficient, and doesn't require me to buy anything I don't already have at hand.

@Hans  I have never made and stored Mg ethoxide so cannot offer any advice.

To dry absolute ethanol [not 95 -99.5%] magnesium ethoxide is generated in situ by adding a small amount of dry Mg turnings, the same quality used to make a Grignard reaction, to the ethanol in a flask with stirrer and condenser. The reaction usually starts by the addition of a tiny iodine crystal that is allowed to settle on a chip, a second crystal is allowable. I recommend that this process be practiced on small amounts of alcohol and Mg to make sure the alcohol is dry enough to start and the magnesium is not too heavily coated with oxide. In general take all the precautions one takes in running a Grignard reaction.  This procedure also works for drying iso-propanol and, if I remember correctly, for tertiary butanol. I never tried it for methanol.  If the starting ethanol is the 95% azeotrope preliminary drying with calcium oxide is necessary. I found that the high quality commercial grade of absolute alcohol gave superior results and uv analysis showed no detectable benzene. Today solvents are of higher quality.

What is the smallest I2 to Mg ratio I can use without significantly affecting the speed of the reaction (assuming I'm fine to let it sit a few hours at max 40C)? Is it ok to use just 1% as much I2 as Mg if using 99.0% Ethanol?

Given that after filtration traces of magnesium ethanoate will be left dissolved in the dried ethanol, can it still be safely distilled at atmospheric pressure knowing there will be a little bit of air in contact with the ethanol? (Magnesium Ethanoate, alone, having a flash point of 43C)


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Degree of dissociation and conductance
My book says ratio of molar conductance at a given concentration  of solute and molar concentration at infinite dilution is equal to degree of dissociation for weak electrolytes.
I understand it intuitively but I feel like I need a proof to believe it which i cannot find anywhere on internet.
Can you prove the relation indicated by the equation below?
$$\alpha = \dfrac{\Lambda_\mathrm{M}}{\Lambda^{\circ}_\mathrm{M}}$$

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

I have written what it says exactly in questions and I am a newbie here so i can't seem to get around formating yet, sorry

No problem, that is why those comments are posted, to be handy for you for future cases.

Note that terms "proof" and "science" do not like much each other, unless you mean the formal mathematical proof of consistence of theoretical model. Considering matching theory and experiment, there are used terms experimental confirmation or refutation of theoretical claims of models, hypothesis or theories.

The expression follows (1) from the principle that ions forming due to dissociation of electrolytes are the charge carriers responsible for the conductance of electrolyte solutions, (2) from an empirical observation (Kohlrausch's law of independent migration of ions), and (3) from the definition of the limiting value of the molar conductivity as the concentration of an electrolyte as the value of the molar conductivity as the electrolyte concentration goes to zero.
The limiting conductivity of electrolytes is more than just a definition, of course, it has the important property that it is a constant, which is consistent with the independent behavior of individual ions as the electrolyte concentration goes to zero, and is therefore consistent with Kohlrausch's law.
The introduction of the degree of dissociation reflects the fact that ions are charge carriers, whereas undissociated neutral species are not. Therefore only the fraction of electrolyte that has dissociated contributes to the molar conductivity.