text:
Why does N(OH)3 not exist?
The hydrolysis of Group-15 tetrahalides gives us their hydroxides via a $\ce{S_{N^2}}$ mechanism:
$$\ce{PCl3 + 3H2O -> H3PO3 + 3HCl}$$
Nitrogen Trichloride, however, yields a different set of products, not forming the trihydroxide as expected.
$$\ce{NCl3 + 3H2O -> NH3 + 3HOCl}$$
A little bit of research led me to this exact same question (with 2 unsatisfactory answers) and this article, which claims that it exists, but I cannot read it. Why does $\ce{N(OH)3}$ not exist?
Another thought along similar lines is the existence of gem-diols in general, which are quite unstable, as well as $\ce{C(OH)4}$, which does not seem to exist as well. Is this an abnormal property of period 2 elements? All the other elements in group 14 and 15 form stable hydroxides whereas the first members do not seem to do so.

You are confusing two entirely different and unrelated questions. The direction of hydrolysis is determined by the difference in electronegativity. Indeed, why the hydrolysis of PCl3 does not end up in PH3 and HOCl? This is not because PH3 or HOCl doesn't exist; in fact they both do. Then why?

When you say that a molecule like $\ce{N(OH)3}$ does not exist, you must specify under what conditions too. $\ce{N(OH)3}$ certainly doesn't exist in aqueous solution, but in free space it probably does.

What you render as $\ce{H3PO3}$ is not predominantly $\ce{P(OH)3}$ either.  In Group 15 you have to go to arsenic to get predominantly the hydroxide under normal laboratory conditions.

Even if you could make it, wouldn't it just decompose into nitrous acid and water? I haven't done the math, but I'd expect the decomposition to be both energetically and entropically favorable, and I doubt the kinetic barrier is high enough to prevent it at normal laboratory temperatures.

Saying sth doesn't exist is a big statement. I guess a H3NO3 acid (which is full of single bonds) could theoretically exist, just not as stable as something like HNO3 (with stabilized resonance structure) and hence is not favored under normal condition. P and O is less favorable to form double bond because of longer bond length.

even after reading all those great answers I still don't understand why Na3PO3 does not/can not exist.

@undefined I assume you meant Na3NO3. Anyways, there is no such $\ce{{NO_3}^{3-}}$ ion. The Na3NO3 we are seeing is an oxide-nitrate salt $\ce{Na3[NO2-]O}$ containing nitrite ion. See this article: https://onlinelibrary.wiley.com/doi/epdf/10.1002/anie.197603762

@NilayGhosh I'm sorry, yes that what I meant. Thank you for your answer. I understand what you mean but what I do not fully understand is, why is there no such ion? your linked article is sadly behind an expensive paywall, so I can not read it

@undefined the crystal structure on Na3NO3 is given here.  Basicaly it's a double salt containing ordinary nitrite and oxide ions.

Well let's not focus on your text but on the actual question here about $\ce{N(OH)3}$.
I can only give you some ideas here but not a definite answer.
Whenever we have tests for our first semester students there are some who forgot what a nitrate was, or nitric acid. But as we ask for a nitrate as trivial name an ortho-nitrate is an accepted answer as well. And many people will just try to balance things with $\ce{X(OH)_n}$ until it fits. It's not wrong in a test but we have to think about these compounds.
Let's take silicates. There are ortho-silicates, so salts that derive from the ortho-silicic acid $\ce{H4SiO4}$ or $\ce{Si(OH)4}$. If you try to make this acid however it will start to polymerize while water is being produced. The reason for this is that you have a positively charged $\ce{H^+}$ that can be deprotonated easily close to an $\ce{Si^4+}$ center in your $\ce{Si-OH}$-bond. Hence two of these centers will react to something like $\ce{Si-O-Si + H2O}$ until something like a meta-silicic acid forms.
Now going back to our nitric acid first with $\ce{N^{+5}}$. We know the acid is called $\ce{HNO3}$. If we add a water to this we will get the ortho-nitric acid $\ce{NO(OH)3}$ or $\ce{H3NO4}$. Much as with the ortho-silicates we can find examples like the compound $\ce{Na3NO4}$ which could formally be the salt of the ortho-nitric acid.
Now if we want the same for nitrous acid $\ce{HNO2}$, well let's add water $\ce{H3NO3}$ or $\ce{N(OH)3}$. Does it exist? Well there is a salt $\ce{Na3NO3}$. The problem is, this is no ortho-nitrite but an oxide-nitrite $\ce{Na3[NO2]O}$.
I can't tell you however why there hasn't been an ortho-nitrite formed, yet. These ortho-acids are just not very stable.

Although the acid and its salts are unknown, orthonitrite esters have been found, and from a natural source at that [1].  The orthonitrite function, which is trivalent rather than monovalent as nitrate would be, is incorporated into a macrocycle that hinders its decomposition.
The picture below is from Ref. 1.
Reference:
1.
Leanne Murray,* Tang K. Lim,* Graeme CurrieB and Robert J., "Aplidites (A-G): Macrocyclic Orthonitrites from
an Australian Tunicate, Aplidium sp.", Aust. J. Chem., 1995, 48, 1253-1266.Link

There are several dozens of structurally characterized μ₃-nitrato metal complexes, so probably there is no need to go that exotic:)

Maybe the nitrato complexes are the more exotic ones depending on your point of view.  The reference says the orthonitrite function "has not been previously described", perhaps meaning "...in organic chemistry".

@andselisk can you provide amy reference(s)?  Google searching of the phrase "mu3-nitrato complexes" does not work for me.  Thanks.

Here you go: https://doi.org/10.1016/j.poly.2005.03.169; https://doi.org/10.1002/chem.201504799; https://doi.org/10.1016/j.cplett.2016.03.020; https://doi.org/10.1021/ja509214d; https://doi.org/10.1039/b903467f.

Thank you @andselisk.  When I apportion charges with normal or article-specified metal oxidation stares I find each $\ce{NO3}$ moiety has only $-1$ charge instead of the hoped for $-3$.  It's as if one covalent bond and two ionic ones were delocalized over the three linkages (like the $\ce{S-F}$ bonds in $\ce{SF6}$), versus three separare covalent linkages as in the orthonitrite esters/alkaloids of my Ref. [1].

The question presupposes that $\ce{N(OH)3}$ does not exist.
However, Ab initio molecular dynamics evidence of a new stable symmetric $\text{C}_s$ structure for N(OH)3 Chemical Physics Letters volume 435, pages 34-38 says:

We point out that very few studies have addressed N(OH)3 and this study is relevant in the context of its possible detection in the gas phase.

See also A new non-symmetric N(OH)3 species: Comparison with the C3 species and thermochemistry at the HF, DFT, MP2, MP4 and CCSD(T) levels of theory Journal of Molecular Structure: THEOCHEM, volume 802, pages 111-115.

Did yhey consider tautomerization like phosphorous/phosphonic acid?

@OscarLanzi no, I read the full text of both articles and they don't mention anything where an H is bonded to the N.  Only three slightly different N(OH)3 species, one with $C_3$ symmetry, one  $C_1$,and one  $C_s$.  The J. Molecular Structure article simulates IR spectra, saying "...can perhaps be used as reference data to compare with experimental IR results obtained in the gas phase in the near future".

It isn't just nitrogen that fails to form an orthoacid with the formula $\ce{X(OH)3}$.  The same thing happens with phosphorus.  The phosphorous acid $\ce{H3PO3}$ is not $\ce{P(OH)3}$, with three equivalent $\ce{OH}$ groups  as you might believe. This phosphorous acid has one H atom that is impossible to neutralize. Its structure is better described by $\ce{HPO(OH)2}$, with a central $\ce{P}$ atom surrounded by one $\ce{H}$, one $\ce{O}$ and two $\ce{OH}$ groups. Neutralized by a hydroxide it can produce two series of salts, and not three. With $\ce{NaOH}$, it gives: $\ce{NaH2PO3}$, $\ce{Na2HPO3}$. $\ce{Na3PO3}$ does not exist.
Arsenic, the next element in Group 15, does form the acid $\ce{H3AsO3}$ in water solution, and normal salts thereof.

Going across from left to right on period 2 and 3, the likelihood of forming stable hydroxides seems to go down. Lithium hydroxide, boron hydroxide, indium hydroxide, they all exhibit pretty good stability. As atomic radius decreases, the electronegative oxygens can be expected to drawn closer. Even nitric acid utilizes resonance to maintain stability.
This question has not been answered formally by modern research.


text:
How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock?
As a non chemist I am most often charmed when visiting Wikipedia articles of chemical elements and see images of very pure specimens of element after element, proton by proton, and often also metal cube specimen made from smithing similar pure deposits.
The wiki article Periodic table allows me to do so easily; here are some elements I found looking almost the same and don't think I personally could distinguish between them without some instrument:

molybdenum and manganase
titanium and chromium
rutenium and cadmium
sodium and aluminium
silicone and germanium

My problem
Generally all examples are of synthetic specimens of about 99.9% purity but that me to wonder what test can a chemist make in a natural environment, say inside a natural cavemine, to determine which natural deposit contains what after the mining and positioning a seemingly multimineral natural deposit;
Of course, I understand that with natural deposits of mainly carbon or copper, or iron (water exposed and rusty) natural deposits, let alone with very pure specimens of these elements, it will be easy because of color, but I believe that generally or mostly color by itself won't be enough due to various color biases and I assume that today there are in-place very-accurate tests that can be done, especially for different parts of a multimineral deposits.
My question
How can chemists distinguish pure chemical element specimens that look almost "the same" as well as what deposit is what in a multimineral mined rock?

Most elements are not mined pure. Those which are (say, gold and silver) are relatively easy to tell apart.

As you enjoy reading Wikipedia this entry may interest you https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy

I am quite sad this question got a downvote ; in my opinion, images of very pure deposits of elements might be one of the best ways to do public understanding of science ; I don't even call myself a chemistry amateur although I really like learning new things about chemistry in accessible terms for general audience of chemistry knowledge and here I tried to help others and myself get one of the most fundamental understandings about chemistry.

I’ll give you an upvote for making the point about facilitating public understanding of science! Maybe this link will be of interest: https://theodoregray.com/PeriodicTableDisplay/ . One minor thing: what you call deposits are usually called specimens. Also, manganese, ruthenium and silicon are mispelled.

@EdV thanks ! By specimen did you mean to any "a whole, mined, multimineral deposit" or just to a particular "sub deposit cutted out of a mined multimineral deposit"?

No. I mean element specimens such as the machined rods, arc-melted pellets, etc. shown at, e.g., this link: https://elementsales.com/ . Go there and look at cobalt or hafnium or whatever element you want. These are not mineral specimens and, like Ivan said, are not mined as pure elements.

Thanks @EdV I have just made the question more helpful for future readers (hopefully) by accessing the important distinction you presented, in the simplest words I could.

What you're describing is essentially the field of "qualitative inorganic analysis". That's a very broad subject, but a short description is that prior to the advent of advanced instrumentation, chemists developed extensive series of tests specifically to determine the elemental composition of samples.
For example, one would start with basic physical properties like color, density, and hardness, then look at reactions with heat, acids, bases, oxidants, etc. If the compound could be solubilized in any way, then one can look at how the solubility/color/etc is affected by addition of known cations and anions. If it was a geological specimen, then the location and the types of surrounding minerals would help inform the analysis as well.
A well trained inorganic chemist would have standard flow-charts of tests and could fairly confidently identify the elemental composition of common minerals and metals.

The qualitative inorganic analysis is a science in itself. To know what sort of elements you have in your sample, you have to first make it react with an acid like concentrated nitric acid $HNO3$. At the end of the reaction, you obtain a solution and an insoluble substance. You filtrate the residue. And now you get a solution S1 and a deposit D1.
You add some other solution, say $HCl$, to the solution S1. This will produce a precipitate D2 and a solution S2, that you separate by filtration.
Simultaneously you treat the deposit D1 by a new reactant, like $NaOH$. Part of the deposit is dissolved. You obtain a new solution S3 and an insoluble part D3, that you separate by filtration.
Afterwards, you treat each solution and deposit by a new reagent, to obtain a new solution and a new deposit. It is a long job. And I cannot describe to you how to separate all different metals of the periodic table. It takes months of lab study to master this technique. But the principle is the same overal. At the end you obtain a precipitate which contains only one element. By weighing it, or by titration, you can determine its weight. You may need a lot of different reagents to carry out the nth separation.

source: Stack Exchange

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

sort: votes


text:
How is red oxide of copper converted into black oxide & vice versa?
I can't really find any information about it.
Any help would be appreciated.

Red Copper oxide is Copper(I) oxide Cu2O. Black Copper oxide is Copper(II) oxide CuO. Copper(I) oxide is oxidised to Copper(II) oxide in moist air.
Copper(I) oxide is oxidised to Cu(II) by dissolving in dilute nitric acid which can be thermally decomposed to Copper (II) oxide by heating at 180C.
Copper(II) oxide can be reduced back to Copper metal by Hydrogen, Carbon Monoxide or Carbon. This may then be heated with oxygen to make Copper(I) oxide. Copper(I) oxide may also be made by reaction of solutions of Copper(II) salts with Sulfur Dioxide.
Source - Wikipedia entries for Copper(I) oxide Cu2O and Copper(II) oxide CuO

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


text:
Molarity of two solutions

Question: Two solutions of ethanol marked as X and Y are labeled as 25% ethanol by mass and 25% ethanol by volume respectively. If the density of solution Y is 0.789 g/mL and that of solution X is 0.968 g/mL, identify the solution with higher molarity.

Solution :
Let total volume = $\pu{100 mL}$
Volume of ethanol = $\pu{25 mL}$
Mass of ethanol = density×volume = $0.789×25 = \pu{19.725 g}$
Molarity of solution Y = Mass of ethanol/Molecular mass of ethanol×1000/Volume of solution (mL)
Molarity = $\frac{19.725}{46}\times\frac{1000}{100}=\pu{4.28 M}$
Hence, molarity of solution Y = $\pu{4.28 M}$
For solution X: Let total mass of solution = $\pu{100 g}$
Mass of ethanol = $\pu{25 g}$
Volume of solution=Mass/density = $100/0.968=\pu{103.30 mL}$
Molarity = Mass of ethanol/Molecular mass of ethanol×1000/volume of solution (mL)
Molarity of solution X=$\frac{25}{46}\times\frac{1000}{103.30}=\pu{5.26 M}$
So, the molarity of solution X is more than that of Y. In the calculation of molarity of Y,  density of ethanol is taken as $\pu{0.789 g/mL}$. But according to me, that is the density of the solution and  not of ethanol. Please clarify my doubt.

Question: Two solutions of ethanol marked as X and Y are labeled as 25% ethanol by mass and 25% ethanol by volume respectively. If the density of solution Y is 0.789 g/mL and that of solution X is 0.968 g/mL, identify the solution with higher molarity.

The question is a mish-mash of data. It is a bad problem. I hate stupid problems!!
Mixture X is supposed to be 25% ethanol by mass and has a density of 0.968 g/ml.
Mixture Y is supposed to be 25% ethanol by volume and has a density of 0.789 g/ml.
According to tabular data a solution of Z% by mass should be slightly less dense than a solution of Z% by volume.
The solution with a density of 0.789 g/ml is either pure ethanol, or nearly so. (The temperature of the density data wasn't specified in the problem.) Thus mixture Y has the higher molarity of ethanol.

Now forgetting any real chemistry, let's solve the problem as given. The molecular weight of ethanol is 46.07 grams/mole.
Mixture X
25% ethanol by mass and has a density of 0.968 g/ml.
Thus 1000 ml of solution X has a mass of 968 grams. 75% of the mass is water and 25% of the mass is ethanol.
Thus the molarity is:
$\mathrm{M}_\mathrm{X} = \dfrac{968*0.25}{46.07} = 5.25\ \mathrm{molar}$
Mixture Y
25% ethanol by volume of solution Y has a density of 0.789 g/ml.
Thus 1000 ml has a mass of 789 grams. 75% of the volume is water and 25% of the volume is ethanol.
Now the problem can't be solved without extra data and assumptions. Perhaps the simplest is to assume that:

Pure water has a density of 1.000 grams/ml
The volumes are additive.

The density of water isn't too bad of an assumption, but assuming that the volumes are additive is a very poor assumption.

Both the concentrations of Solution X and Solution Y are correct. On the other hand, the value of the density that you used on the calculation of the mass of ethanol is also correct.
The direct relationship between the volume of ethanol on the solution and the density of the solution demonstrate the mass of the ethanol on the solution.
For example:
Mass of ethanol = (density of the solution X)(volume of ethanol)
Mass of ethanol = ($0.968 g/mL$)($25 mL$)
Mass of ethanol = $24.20 grams$
However, if you use the density of ethanol, it will result on the mass of the ethanol as ethanol itself and not as ethanol on the solution stated on the problem.
For example:
Mass of ethanol = (density of ethanol)(volume of ethanol)
Mass of ethanol = ($0.789 g/mL$)($25 mL$)
Mass of ethanol = $19.73 grams$
I use the Solution X to provide a clear comparison between the mass of ethanol using the density of the solution and the mass of ethanol using density of ethanol itself.
I think, you get confused with the calculation of the mass of ethanol using the density of Solution Y, because both of the density of ethanol and Solution Y are the same. The problem itself is not correct, it states that Solution Y is only $25$$%$ ethanol yet its density corresponds to the density of an ethanol.

The question is bad enough to put on-hold, but I'd give my take since OP tried hard to solve it.
It is given that mixture $X$ is 25% ethanol by mass ($w/w$) and has a density of $\pu{0.968 g/mL}$.
It is also given that mixture $Y$ is 25% ethanol by volume ($v/v$) and has a density of $\pu{0.789 g/mL}$.
Let's take mixture $X$ first:
Since it it has a density of $\pu{0.968 g/mL}$, $\pu{1000 mL}$ of this solution weighs $\pu{968 g}$. Yet, only 25% of this mass is ethanol.
Therefore, mass of ethanol in $\pu{1000 mL}$ of solution $X = \pu{968 g} \times 0.25 = \pu{242 g}$. Or,  $\pu{242 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} = \pu{5.249 mol}$, since molar mass of ethanol is $\pu{46.1 g/mol}$.
Thus, molarity of solution $X$ is $\pu{5.25 mol/L}$.
Now, we can look at mixture $Y$:
Since it has a density of $\pu{0.789 g/mL}$ (given), we can calculate that weight of $\pu{1000 mL}$ of solution $Y$ is $\pu{789 g}$. However we cannot go further without knowing additional date from here as  MaxW instate elsewhere. As MaxW's conclusion, we have no choice, but assume that (1) density of water is $\pu{1.00 g/mL}$ at operating temperature; and (2) water and ethanol are additives (meaning no contraction or expansion when mixed together).
If we can assume these two facts, then we can conclude that 25% of this $\pu{1000 mL}$ of solution $Y$ is ethanol ($\pu{250 mL}$) and other 75% ($\pu{750 mL}$) is water.
Therefore, mass of water in $\pu{1000 mL}$ of solution $Y = \pu{750 mL} \times \frac{\pu{1.00 g}}{\pu{1.0 mL}} = \pu{750 g}$. That means, mass of ethanol in $\pu{1000 mL}$ of solution $Y = \pu{789 g} - \pu{750 g} = \pu{39 g}$. Or, $\pu{39 g} \times \frac{\pu{1 mol}}{\pu{46.1 g}} = \pu{0.846 mol}$.
Thus, molarity of solution $Y$ is $\pu{0.846 mol/L}$.

Note: This calculation is similar to that of MaxW. Yet, just for pure curiosity, I put this answer since none of the answers have given the molarity of solution $Y$.


text:
Ammonia Vapor Absorption Refrigerator
I am working on fabricating the icyball system. I am planning to charge it with the help of convex mirrors. Whether it is necessary to create a vaccum inside the refrigeration closed cylinders before filling ammonia?

These refrigerators are not very efficient anyway, but for demonstration purposes, it will work, albeit even less efficiently, with residual air.


text:
Spectrum Of Metals
Recently, I've read that when metals are exposed to a high voltage source, metals produce a spectrum. I am curious on what is the minimum value of that 'high voltage'? Is the spectrum due to the high voltage depends on the atomic properties of the metal? Any help is highly appreciated. Thank You.

Do a search for DC arc spectroscopy or arc spectroscopy or similar terms. There is a vast amount of published research in the area!

Your question will probably be closed, I think it is too broad and contains parts that could be on-topic in Physics SE but not here. A very important aspect is the atmosphere where the electrical discharge is happening. Air molecules slow the electrons considerably, so even if you have 1,000 volts in your discharge, the average kinetic energy of the electrons could be tens of volts or less, so you need to consider the power (voltage and current) and how it is heating the metal and removing atoms and their chemical and excited state processes as vapor. It's a big hot mess!

Only slightly related: In the recent Con-Ed transformer “fire”, what exactly produced the color of the huge blue glow over New York City?

Is the spectrum due to the high voltage depends on the atomic properties of the metal?

Yes, the atomic emission spectrum is unique for each element. The electrical voltage plays a secondary role in my opinion, it is rather the temperature which determines which line would be visible in the emission spectrum. These days DC arc spectroscopy is not that popular for spectral purposes as better and higher temperature sources such as inductively couple plasma are available. The minimum voltage has to be above break-down voltage of air.


text:
Would Zirconium (IV) Silicate be stable in a hot oxygen atmosphere?
As per the title, really. I potentially need to protect a bit of pipework from hot oxygen (800-1000 K), and I've seen this refractory paint for sale. It says it's good up to 1750 C, but hot oxygen is another kettle of fish. I'd have thought since it's formed from oxides at high temperature and it doesn't lose any oxygen in the process that it'd be fine? Just thought I'd check with the experts here before burning a hole in my kit!

In addition to asking here, you should also contact the manufacturer.  They should be the best experts on their own product.

Zirconium only goes to +4 oxidation state so it can't get oxidised any further which is a plus. Zirconium silicate is used as an enamel and glaze so clearly high temps are not a problem. All sounds promising but I'm not an expert.

Yeah, I've contacted them too. Just thought I'd ask here as well to see if anyone had a technical reason why it wouldn't work, rather than the typical manufacturer answer of "we haven't tested that scenario".


text:
What are common solvent removal methods?
In a laboratory setting, the most common solvent removal method is rotary vacuum distillation. This is something I have performed many times, starting in organic chemistry labs. However, I am interested in learning of other solvent removal techniques.
Specifically, I am having a lot of trouble removing trace levels of solvent (chloroform, heptane, toluene, cyclohexane) from nicotine, which is an oil in its pure form. I have been using a rotary vacuum system to remove the solvent, though traces of solvent always remain. Nicotine is heat sensitive and degrades easily in air. Are there any other methods I should consider?

Freeze-drying is usually an excellent option, but nicotine has a very low melting point, so that's probably not feasible. Stirring the oil under high vacuum (<1 mbar)  for a few hours could work, and you might be able to afford heating it mildly (60-80 °C) in the absence of oxygen.


text:
Using osmotic pressure to find the molar mass of a compound

Working on-board a research vessel somewhere at sea, you have (carefully) isolated 12.5 micrograms $(\pu{12.5E-6 g})$ of what you hope is pure saxitoxin (a non-electrolyte) from a poisonous (and quite cross) puffer fish. You dissolve this sample in $\pu{3.10 mL}$ of water and determine that the osmotic pressure of
  the resulting solution is $\pu{0.236 Torr}$ at $\pu{19 °C}$ $(\pu{760 Torr} = \pu{1.00 atm}).$ What is the molar mass of the compound?

I am having trouble reaching the correct answer of $\pu{310.8 g/mol}$ for this question. I first found the pressure to be $\pu{0.0003105 atm}$ and plugged numbers in for the formula $\Pi = MRT.$ I got $\pu{2.04E-5 mol L-1}.$ I used this to find the amount of substance and got $\pu{6.3E-8 mol}.$
Lastly, I divided $\pu{12.5E-6 g}$ by $\pu{6.3E-8 mol}$ and got $\pu{1.98E-14 g/mol}$ which is not the right answer.
I feel like I have the right process so if you can help show me where I messed up, I would appreciate it.

Note that calling the physical quantities by the units is incorrect, e.g. saying "I found atm" or "I found the number of moles" is illiterate and should be avoided. Instead, use the exact terms such as "pressure" and "amount of substance" (see my edit).

Also, it is generally advised not to perform intermediate calculations (solve algebraically first, plug in the numbers at the end) unless you are explicitly asked to do so, and if you decided to go against the recommended approach, make sure you aren't using textual description for the calculations; post proper math formulas and notations instead.

I couldn't resist. The osmotic pressure equation is
$$Π = cRT,$$
where $Π$ is osmotic pressure in atmospheres, $c$ is molar concentration, $R = \pu{0.082057366080960 L atm mol-1 K-1}$ (exactly) and $T$ is temperature in Kelvin.
In this problem, $Π = \pu{0.236 torr}$ times exactly $\pu{1 atm}/\pu{760 torr}$ and $T = \pu{(273.15 + 19.0) K}.$ So $c = \pu{1.2953E-5 M}$. But $\pu{12.5E-6 g}$ of saxitoxin was dissolved in $\pu{3.10 mL}$ of water, which is $\pu{4.032E-3 g/L}$. Hence $\pu{1 mol}$ of saxitoxin has a mass of $\pu{311 g}.$

Comments are not for extended discussion; this conversation has been moved to chat.


text:
Gold refining using HCl
I am trying to dissolve base metals off of gold plated electronic scrap, and I am using HCl to dissolve some of the base metals.
I pored in the HCl into a flask containing the scrap ( the contents of the flask where later moved into the 2 liter jar as seen in the photo ) Once I dissolved the base metals I tried to neutralize the acid with baking soda. However this neutralization was taking too long and about 400 ml of baking soda later I decided to try to react the rest with sodium sulfate to bring the ph down.
 I figured this would produce sodium chloride witch wouldn’t be a big deal. This did help by reacting the rest of the HCl and it brought the ph down to about 5, but there was also baking soda along with the sodium sulfate so I don’t know witch brought the ph down. The sodium sulfate was reacting though because before I had a odorless gas ( CO2) and now I was getting chlorine gas.
However the day after the gold off the scrap had disappeared.  The piece of scrap in the photo was a piece I observed during the reaction. At the beginning it had about 2mm ( my measurements are not off it was very heavily plated ) of gold plated onto the surface. And at the end the gold had dissolved away and the copper base metal was coated in a black durable substance. This deposit was not on the gold it was on the copper the gold was once plated on. I even cut it in half to be sure that it wasn’t just deposited onto the gold. The only other chemical that was possibly in it was about 25-50 ml of acetone wich I used to dissolve the plastics with. At first the solution was a brownish yellow then the brown particles settled out and the solution was yellow.
 My question is where did the gold go and what gold compound produced.
 And for those who say the gold didn’t dissolve it did. I had over a quarter of a pound of gold plated scrap disappear and the solution turned yellow. The gold doesn’t just evaporate. !

I do not see reason for adding sodium sulphate. It is pH indifferent salt. Baking soda solution is not the best choice, as its solubility is limited and produces a lot of gas. HCl wil leave a lot of metal not dissolved, like copper..HNO3 would dissolve copper and silver too. When Metals are dissolved, why not to get rid of extra HCl first, and neutralize it separately ?

I agree with Poutnik. Sodium sulfate is not going to help.

I do not have access to nitric or sulfuric acid

If you do not have access to nitric or sulfuric acid, you will never be able to dissolve gold. Gold is only soluble in a mixture of HCl and HNO3.

Adding baking soda will never bring the pH down. And sodium sulfate does not react with gold, copper or any other metal. And finally you do not tell us what you have really done. You say that you have smelted the odor of chlorine. It is impossible to obtain chlorine gas Cl2 with the metals and chemicals you have used.

Chemical reactions are not used to participate in element transmutations, otherwise alchemists would already use it. How can $\ce{HCl}$ and $ce{Na2SO4}$ produce carbonic acid, if neither of them contains carbon ? In fact, they do not react, if they are what you say they are. If mixed, they form a solution of 4 respective ions. What sulphate may do in concentrated solution is to restrict solubility of hydrogenchloride.

I suggest you to break your questions into paragraphs. It looks like a block of text.

Per a 2017 article from the Journal of Molecular Liquids some interesting comments.
Yes, you can use HCl albeit in conjunction with a chlorate (or bromate or iodate..) and added chloride, to quote:

It has been found that mixtures between sodium chlorate and HCl (NaClO3 + HCl ⇆ HClO3 + NaCl) in 20 mL are effective media for the dissolution of pure gold-wire...In HClO3 solution containing abundant halide ions (X- = Cl-, Br-), the Au metal should be oxidized mainly by X2 to be AuX4-, where X2 is produced by the oxidizing power from HClO3, just as dilute HNO3 (< 2 mol dm-3) containing abundant NaCl.

Also interestingly, hypochlorous acid with added NaCl may also work, to quote from the same source:

The gold dissolution in hypochlorous acid (HClO) containing halide ions has been also examined...

The last comment may imply that treating a concentrated solution of calcium hypochlorite with CO2 (creating, in situ, HOCl and a precipitate of CaCO3, see as a reference, and also here ), then removing the calcium carbonate, and further adding more CaCl2, may actually dissolve gold!

Seven mistakes in seven lines!!
Caleb is of course allowed to write down any sorts of equations. It is not forbidden by law. The trouble is that it is no use creating equations for reactions that simply do not happen in the reality. For example,
1) $\ce{Pb}$ does not react with $\ce{HCl}$, as he imagines. It does not react at all.
2) A mixture of $\ce{HCl}$ and $\ce{Na2SO4}$ will not produce sulfuric acid, as he says. In solution it simply produces a mixture of $\ce{Na^+, H3O^+. Cl^-}$, and $\ce{SO4^{2-}}$ ions.
3) Sulfuric acid does not react with metallic gold $\ce{Au}$, despite his assertion.
4) Sulfuric may react with some metals to produce sulfates, but will never produce  perchlorates, despite his view.
5) Chlorine cannot be produced by a mixture of $\ce{HCl}$ and sulfuric acid, with or without some other compound like $\ce{PbSO4}$ or $\ce{PbCl2}$.
This being said, Caleb should absolutely learn that :
1) Sulfuric acid is not $\ce{HSO4}$. No! It is $\ce{H2SO4}$, with 2 $\ce{H}$ atoms
2) Lead chloride is not $\ce{PbCl}$ but $\ce{PbCl2}$. $\ce{PbCl}$ does not exist!

"A mixture of HCl and Na2SO4 will not produce sulfuric acid, as he says."  It could produce bisulfate, though.  Bisulfate dominates over sulfate if pH<2, as that number is the pKa for bisulfate.

I don't get it! Why such a high rep user don't use \ce{^} at all..


text:
Estimating maximum size of magnetite particles in this Nile Red ferrofluid video?
In the Nile Red video Making ferrofluid from scratch a suspension of small magnetite ($\ce{Fe^{2+}Fe_2^{3+}O_4^{2−}}$) particles is produced from a combination of ferrous chloride and ferric chloride.
The nanoparticles are then given a coating of oleic acid ($\ce{CH3(CH2)7CH=CH(CH2)7COOH}$) to provide steric stabilization to prevent aggregation, then dissolved in kerosene.
After about 23:20:

…the magnetite isn’t truly dissolving, it’s going into suspension. This is because the nanoparticles aren’t soluble in the solvent, but they’re small enough that they can still easily get dispersed.
The oleic acid coating that helps with this, by preventing two particles from coming too close together, and by also allowing them to interact with the non-polar solvent. The result of this is a stable suspension known as a colloid, where it’s almost like the magnetite being dissolved.

As an exercise I'd like to estimate the maximum size of the oleic acid-coated magnetite particles that can be stable in this solution. Some considerations are:

While the magnetite particles themselves have a high density (over 5 g/cm3), the oleic acid molecules might be arranged radially making the volume of the complete particle much larger than the magnetite core.

While large magnetite particles would sink quickly to the bottom due to gravity, the thermal agitation (Brownian motion) at room temperature is sufficient to keep these in colloidal suspension if they are small enough. This suggest that there might be some ratio of $k_B T$ with some expression of gravitational potential energy.

Question: How would I go about estimating the maximum size of the magnetite nanoparticles that could remain in stable suspension in this case?

Assuming you could make a more dilute colloid (not the black blob) without changing anything essential, i.e., particle sizes, how about Mie scattering? Per wiki, this is used for particle sizing when the particles are below 50 microns. The reference is ISO 13321:2009. Or maybe the Tyndall effect could give an estimate of the range of particle sizes.

@EdV by estimate I mean using pencil and paper or thinking or a white board, not a real measurement. I've asked for a way to estimate the maximum size that won't settle out of suspension.

I hope you get an answer: every paper I find is way too complicated. Based on the fact that Brownian motion is modeled  using the diffusion equation (in part), I would assume spherical particles should have diameters small enough so that the diffusion model is valid. But there are so many factors: particle density, viscosity, fluid density, and so on. My wild guess, just a hunch, is that particle diameter must be less than 10% of the diffusion length in the fluid. I really hope you get an answer so I can see how bad I guessed!

@EdV I don't know about the oleic acid "wrapping" and the effective density of the coated particles, but I have a hunch that the maximum density will come from the same effect that keeps the argon from falling to the bottom of a closed box of air. Thus my comment about"some ratio of $k_B T$ with some expression of gravitational potential energy".


text:
Options to keep someone awake against their will?
I already asked this on Worldbuilding Stack, but they told me to ask here. It is for a RPG campaign.
Are there any options that an evil person could use to keep someone awake against their will? Like a gas, an injection, or any substance or simple practice that wouldn't allow them to sleep even if they wanted to?
If any costs are involved, I would like to know the most cheap and viable ways to achieve this.
Furthermore, if there is a way for the victim to be be unaware of why they can't sleep, I would like to know that too.
The technology context is 80's decade and the purpose is a that a organization want to make some experiments about the consequences of sleep privation, especially if there is a point when allucinations begin.

Well there is hydrogen sulfide whose smell of rotten eggs would likely to keep one awake.

I understand this is a role-playing game (not rocket-propelled grenade) question, and I think it really could be modified to fit in World Building SE. If I were you I'd try again to ask it there, and spend some time formulating the question to fit the site, adding specific details, etc. It's a bit of a weird site and some users are less cordial than others, but I'd still give it a try.

Why do you want to know this? Are you one of these evil persons ?

@Maurice I think me asking this is suspicious, so here is my topic on Worldbuilding: https://worldbuilding.stackexchange.com/questions/167459/options-to-keep-someone-awake-against-their-will
As I said, I'm writing a RPG campaing.

A bit of a primer first: The body's sleep cycle is regulated by Melatonin, which is a hormone produced by the pineal gland. You can read more about this hormone here but in summary, for a biochemical solution to keeping someone awake, you would want to either catalyse the breakdown of melatonin in the body or inhibit melatonin secretion by the pineal gland.
The latter option has been tried: tests have been conducted and show that Serotonin N-acetyltransferase is the enzyme responsible for the diurnal rhythm of melatonin production in the pineal gland of animals and humans. The compound N-bromoacetyltryptamine was shown to be a potent inhibitor of this enzyme in vitro and in a pineal cell culture assay. Note that the first paper regarding this was published in 1998 and may not fit into your timeframe.
As for a cheaper solution, one way is to just keep the person in a room where the light is on 24x7. This is a common torture trick used by several nations as it offsets the circadian rhythm and tricks the body into believing that it is morning all the time.
References:

Khalil EM, De Angelis J, Ishii M, Cole PA. (1998). Mechanism-based inhibition of the melatonin rhythm enzyme: pharmacologic exploitation of active site functional plasticity. Proc Natl Acad Sci USA.
Ferry, G., Ubeaud, C., Mozo, J., Péan, C., Hennig, P., Rodriguez, M., Scoul, C., Bonnaud, A., Nosjean, O., Galizzi, J.‐P., Delagrange, P., Renard, P., Volland, J.‐P., Yous, S., Lesieur, D. and Boutin, J.A. (2004), New substrate analogues of human serotonin N‐acetyltransferase produce in situ specific and potent inhibitors. European Journal of Biochemistry
De Angelis J, Gastel J, Klein DC, Cole PA. Kinetic analysis of the catalytic mechanism of serotonin N-acetyltransferase. Journal of Biological Chemistry

P.S: I believe any worldbuilding user with a bit of background in chemistry/biology would have been able to answer your query on that SE.

Warning! Not for migraine patients, you'd necessarily cause severe headache and eventually make them fall asleep by light exposure...Mission failed!

Jokes apart, this was a well researched answer.Thanks, got to learn a lot of new things :)


text:
What is Rinman's green used for?
I can't find any info about it's uses neither in book nor in internet.
Any help would be appreciated.

Well my friend Google says it is Cobalt green. https://en.wikipedia.org/wiki/Cobalt_green

And it also says
"It is rarely used because it is a weak chromophore and relatively expensive compared to chromium(III) oxide"
So, aren't they used at all?

Due to it being a weak chromophore and relatively expensive than chromium(III) oxide, its use as a pigment has become obsolete. Since it is kind of malleable and it is resistant to oxidation, it is hence used for sheeting, coating galvanized iron, etc. It is used in making brass and other alloys, and is also largely consumed in electric batteries. Also, it magnetic at room temperature, so it is used in spintropic devices.
References

Wikipedia
Findwords

I can't really imagine that it's used as pigment. First of all it contains cobalt in comparison to the less harmful Cr(III). Another thing that bothers me about the compound is its composition. While Thenard's Blue is what we call a spinel-structure, Rinman's Green isn't one. Although textbooks sometimes tell you this and quite often students try to convince me it's just not the case. At this point I will then ask my students for the color and one simple concept that I often use is that we start from ZnO in a wurzite structure. In there some zinc ions are replaced by cobalt. This gives us cobalt in a $\ce{[CoO4]^6-}$ tetrahedral environment. While octahedrally coordinated cobalt(II) is often pink, tetrahedrally coordinated cobalt(II) is usually blue. We see this in the example of Thenard's Blue $\ce{Co[Al2O4]}$. So we expect the cobalt part in Rinnman's green to introduce a blue color but what we get is green.
And the easiest way to explain that is that $\ce{ZnO}$ tends to become yellow at elevated temperatures. Usually this is reversible but let's just say a yellow color introduced by high-temperatures solid state phases containing $\ce{ZnO}$ is at least plausible. Then those two components will mix to form the color green.
When we do this as test for zinc in the lab I get a lot of different colors from blue to green. If you heat it too much or too long or you use too much cobalt you will even end up with black. And the cobalt(II) composition in the literature varies quite much.
So I can imagine, and that's really just my idea here, that it might be hard to reproduce the exact same green colors every time which makes it not very suitable for a commercial product. But I could as well just be mistaking here and there's an industrial way. If we make it with students however you will always find impurities or not fully mixed phases.

The main use of Rinmann's green is analytical. It may be used to identify small amounts of zinc in a mixture. If a precipitate containing metal hydroxides and sulfides is burnt and calcined on a porcelain lid in a Bunsen flame, one may usually obtain a white or grayish mass. If now one drop of a diluted solution of $Co(NO_3)_2$ is added to the mass, and then calcined again, a green and non fusible mass is obtained if there is $Zn$ in the mass. This reaction is characteristic for zinc $Zn$. It is easy to do in the lab.

A test for Zn would be a qualitative analysis, not an analytical analysis.


text:
Is terylene a copolymer?
Terylene is a polyester made from terephthallic acid and ethylene glycol. Even though we use two reagents, they undergo esterification to give bis(2-hydroxyethyl) terephthalate (monomer) which then undergo polymerization.
So, is it correct to consider terylene as a copolymer? My reasoning tells that it should be a homopolymer. Same line of reasoning gives nylon, phenol-formaldehyde resin, bakelite — all as homopolymer.
But majority of questions which I go through considered the above polymers as copolymers. What is correct? Are they homopolymers or copolymers?

Whatever we decide to be correct is OK. This is a human convention, not a fact of nature.

It is not a requirement to be monomer ester to prolong the polymer chain. The EG can attach itself to the polymer carboxy end, the acid to alkoholic end. But better then copolymer is copolycondensate, as water is eliminated.

It's not a copolymer and common name is PET not some "terylene".

Is terylene brand name for PET

PET = PolyEthyleneTerephthalate = polyester of terephthalic acid and ethylenglykol. So yes, If terylene is what you say it is, then it is the brand name for PET.

No, PET it is definitely not a copolymer. Neither is e.g. Nylon 6,11 (or any other polyamide with two numbers). The monomers alone would not polymerise, or at least only by a very different reaction (e.g. ethylene gylcol gives PEG).
What you can have are alternating copolymers, but there the sequence is a statistical property. It comes when a reactive chain end B* has a higher probability to add an A (to give BA*), and a chain end A* to add B (to give AB*). That probability is sensitive to monomer composition, the temperature, pressure, catalyst, etc. If you want to read more about that, "copolymerisation parameters" is the topic to look up. The monomers you see there are afaik invariably olefines, i.e. they have a double bond that undergoes polymerisation.


text:
Fehling's solution turning yellow and then black?
While testing for formaldehyde (an aldehyde), I added Fehling's solution to the sample and let the test tube simmer in a water bath for some time. I returned ~15 minutes later to see a black/dark grey precipitate awaiting me, which looked very metallic (Almost like Tollen's test, but darker)
I try again and this time heat the tube for only five minutes: I obtain a buff yellow coloured precipitate which, on standing, gives me the same metallic precipitate I observed earlier.
I expected that Fehling's solution will give a red precipitate of $\ce{Cu2O}$ but it looks like the formaldehyde seems to be oxidizing it all the way down to metallic copper. Also, what is the reason for a yellow precipitate to form rather than a red one?
I could not take any pictures: We're not allowed to carry phones/cameras into the lab.

Yeah sometimes you even end up in a blue-green solution, with no precipitate, something you'd confidently conclude to be a negative result. But later I was told by my teacher that it actually is an aldehyde and that turning green is a sign of a reaction undergoing. Net result: I scored poorly in that test -_-

In the reaction with aldehydes, Fehling's solution produces $\ce{Cu2O}$ as an orange or yellow precipitate. I don't know why the color is not always the same. The oxidation of an aldehyde produces an acid. And an acid usually cannot be further oxidized.
There is one exception: methanal $\ce{HCHO}$ that can be oxidized to formic acid $\ce{HCOOH}$, and then into $\ce{CO2}$. That is why Fehling's reagent produces metallic copper $\ce{Cu}$, $\ce{CO2}$ and no acid when reacting with $\ce{HCHO}$.

Alright that does make sense. I did not consider the further oxidation. However, I don’t think I saw any bubbles of CO2 emerging. Could you please also provide a reference/reaction mechanism?

By all means, precipitated copper will not appear black. Sorry, the explanation is not correct. Please Google Oxidation of formaldehyde by copper(II) in aqueous solution byJ. J. BYERLEAYN D W. K. TEO. It specifically says that no gas evolution seen and they rule out the oxidation of formaldehyde to CO2.

Who said there was a black precipitate ? Nobody !


text:
Why is H⁺ ion considered the source of acidity, and OH⁻ considered the source of basicity?
As the title implies, what caused the association between $\ce{H+}$ and acidity, and between $\ce{OH-}$ and basicity?

What causes the association between the red color and the word "red"?

Well, that's just a definition, are acidity and basicity also just definitions?

What else are they?

Positron12 is right. It is a definition of acidity and basicity.

But the question still holds. What makes us generalize hydrogen ions for acidity, that we consider every solution that ionizes and produces it to be acidic?

That is true just for Arrhenius' theory, which is rather outdated, pretty much made obsolete by Bronstedt.

Yours is a historical question. Note that in modern chemistry, it is not required that acids have H+ or bases have OH-. The words acidity or basicity referred to certain common behaviors such as acids will liberate hydrogen when come in contact with metals, or they will decompose carbonates to carbon dioxide. Similarly, bases had other common characters. Indicators behaved the same way with acids or bases. No one knows why bases were called bases. Acids we understand, that they tasted sour (=Latin). The term alkali (Group I bases) makes more sense (from Arabic for soda ash or potash).
Definitions are made by humans and I am glad you are not blindly accepting the definitions and at least thinking about them. Arrhenius was among the first ones to think why solutions conduct electricity. When Arrhenius proposed this idea in his PhD, but he knew that his mediocre committee members will not like it. How can anything ionize in water? It was a stupid idea to them. It turned out that most common acids and bases had very high conductivity better than anything else known, even better than salt solutions. Say 0.1 M HCl had way higher electrical conductivity than 0.1 M NaCl? Why? The "proton" in aqueous solution has the highest conductivity known today. What comes next? The hydroxide ion. Also in Arrhenius' time electrical decomposition was known. Faraday did a lot of electrochemical experiments then.
When you electrolyze an acid solution, what do you get at the electrode which has a negative electrostatic charge. Hydrogen gas! What appears at the positive electrode, chlorine gas. How can "H" and "Cl" reach the electrode, this is possible when they have a charge on them as well. So, "H" in acids must exist as "H+"; as a positively charged entity. I have not done justice to Arrhenius history or the history of acid bases.
If you are really (or better word genuinely) interested then search the Theory of Ionization+ Arrhenius History in Google Scholar. Partington's History of Chemistry will have it too but it is meant for serious scholars of chem. history (multiple volumes). It is available online from the Internet Archive.

Why are bases called bases?  Read the Etymology section here.

Interesting, story! The word base did not have English origins and even the German use the word base. It sad that Jensen stopped writing for J.Chem.Ed after the change of editors. He published a couple of my history questions too.

Aqueous solutions
In aqueous solution, $\ce{H+(aq)}$ and $\ce{OH-(aq)}$ are always present because of the autoionization of water:
$$\ce{H2O(l) <=> H+(aq) + OH-(aq)}$$
So in aqueous solution, I would be comfortable saying there is an "association between $\ce{H+}$ and acidity, and between $\ce{OH-}$ and basicity". When the two concentration are equal, the solution is neutral. When the concentration of $\ce{H+}$ is larger, the solution is acidic. Finally, when the concentration of $\ce{OH-}$ is larger, the solution is basic.
This is just the definition of acidic and basic aqueous solutions, and can be used to define acids and bases in an empirical way (i.e. not knowing the underlying details of atomic structure and chemical bonding). The chemistry of aqueous solutions often is vastly different depending on whether they are acidic, neutral and basic, so it makes sense to introduce these concepts (and even have a quantitative measure for them, the pH of the solution).
Arrhenius vs. Brønsted-Lowry definition
The hydrogen ion plays a role in defining an acid both according to Arrhenius and Brønsted-Lowry. It seems that in defining a base, however, that the hydroxide ion is required according to Arrhenius and not mentioned according to Brønsted-Lowry. If you combined water autoionization with the Brønsted-Lowry definition, you get the hydroxide ion back. So why does the definition of acid stay the same and the definition of base changes or suddenly apply to different substance. In other words, is sodium hydroxide just an Arrhenius base and ammonia just a Brønsted-Lowry base (see an answer here: Is Sodium Hydroxide a Bronsted-Lowry Base?)?
Ionization vs. dissociation
Some substances, such as $\ce{HCl(g)}$ or $\ce{NH3(g)}$ ionize when they react with water, either giving off a hydrogen ion or picking up a hydrogen ion. Ionization is a good term for that process because neutral species react to form ionic species.
When bound to these species, the hydrogen atom is connected via a covalent bond. There is no ionic substance I know of that contains $\ce{H+}$-ions. When we write $\ce{H+(aq)}$, we imply that the hydrogen ion is covalently bound to a water molecule ($\ce{H3O+}$) or a cluster of water molecules.
On the other hand, there are ionic species that contain $\ce{OH-}$ ions, for example $\ce{NaOH(s)}$ or $\ce{Ca(OH)2(s)}$. When these are added to water, they dissolve into solvated ions, e.g.
$$\ce{NaOH(s) -> Na+(aq) + OH-(aq)}$$
This process could be described as dissociation, and does not look like it involves hydrogen ions at all. If you isotopically labeled the sodium hydrozide, however, you would realize that they mostly react to form water, as in:
$$\ce{Na^{17}OH-(s) + H2O(l) -> H2^{17}O(aq) + OH-(aq) + Na+(aq)}$$

Why is H⁺ ion considered the source of acidity, and OH⁻ considered the source of basicity?

It is not. The source of acidity is a substance that has an ionizable covalent bond to hydrogen. The source of basicity is a substance that can form a covalent bond with hydrogen by reacting with a hydrogen ion. This includes hydroxide ions. In aqueous solution, the consequence of adding acids or bases to aqueous solutions is a increase of the concentration of hydrogen ions or hydroxide ions (with a decrease in hydroxide or hydrogen ions, respectively). In non-aqueous systems, where there might not be any source of hydroxide ions or they would not be solvated, the acid/base concept is limited to hydrogen ion transfer, or expanded to consider a different anion (such as $\ce{NH2-}$ in liquid ammonia).
Can we make this easier to learn?
It is a conceptual hurdle to introduce $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$ as the first examples of acids and bases because they are so different from each other. It would be easier to start with acetic acid and sodium acetate, or ammonia and ammonium chloride. A nice pair of strong acid and strong base would be $\ce{HNO3}$ and $\ce{Na2S}$, or $\ce{HCl}$ and $\ce{CH3CH2ONa}$.
Once acid/base concepts have settled in, the somewhat weird case of $\ce{NaOH(s)}$ could be introduced, along with mentioning that while this is an example of an ionic substance that already contains hydroxide as an anion, no comparable ionic substance containing hydrogen ions as a cation exist.

For a wacky take on the role of hydroxide in defining how acidic or basic a solution is, see https://chemistry.meta.stackexchange.com/q/4431

Why don't you submit this idea for the Journal of Chemical Education after expansion, especially isotopic labeling plus your "wacky" take. Might be worth the attention. They have a new editor now, hopefully the quality will improve further.

"There is no ionic substance I know of that contains H+-ions." That is still true, but there are hydronium salts such as hydronium perchlorate or tosylate, see solid salts of hydronium ion. So sodium hydroxide and hydronium perchlorate could be introduced at the same time.


text:
How are CH3COOH molecules ionisied into H+ and CH3OO-
It's written in my chemistry book ,under weak acids, that the molecules of organic acids  are partially ionised in water into ions. But I don't understand how molecules are ionised.
I'm an ol student.

See this link: https://chemistry.stackexchange.com/a/17343/79678 . I did not downvote the question or answer, by the way.

"Note that in the above text I have used the word "ionize" to describe molecular acids because by definition when these acids react, they ionize. Or form ions. On the other hand salts of acids and bases such as sodium hydroxide are said to "dissociate," because ionic compounds are made of ions, and it wouldn't make much sense to say that ions ionize or become ions." Oh god that makes so much sense now!

The $\ce{O}$ atom in the $\ce{O-H}$ bond is more electronegative than $\ce{H}$ atom leading to a slight shift of the electron density towards the $\ce{O}$ atom and the development of a small negative charge, $\delta-$ on oxygen atom and $\delta+$ on hydrogen atom.
Now, a water molecule also has an oxygen atom which attracts some electron density towards itself from the two hydrogen atoms on it's sides causing $2\delta-$ charge on itself and a $\delta+$ each on both the hydrogen atoms.
This $2\delta-$ charge having $\ce{O}$ atoms of the water molecules surround and attract the $\ce{H}$ atoms of $\ce{O-H}$ bond of $\ce{CH3COOH}$ molecules due to the $\delta+$ charge on them. The hydrogen atoms of water in turn surround and attract the $\ce{O}$ atom of the $\ce{O-H}$ bond of $\ce{CH3COOH}$ molecules.
This attraction leads to separation of the $\ce{O-H}$ bond releasing $\ce{H+}$ and $\ce{CH3COO-}$ ions with a shell of water molecules surrounding both of these ions and stabilising them.

In the molecule $\ce{CH3COOH}$ there is a final bond $\ce{-O-H}$ on the right hand-side of the molecule. This bond is made of a doublet of electrons and the bond is not very strong. Apparently the $\ce{H}$ atom, or better the $\ce{H}$ atom without its electron, i.e the ion $\ce{H+}$, is attracted by the next molecule of the solvent $\ce{H2O}$, and may quit the molecule $\ce{CH3COOH}$ to produce an ion $\ce{H3O+}$, and leaving an ion $\ce{CH_3COO-}$. For some reason, only a small fraction of the $\ce{CH3COOH}$ molecules are behaving this way. This phenomena is called ionisation of the acid.


text:
Formating a chemical equation - Physical states
I study chemistry in an University and I have two professors that says opposing things. One says that the physical state must be subscribed (1 at the picture) and other that it must be written to the left side (2). Since both of them are posdocs and have way more experience in chemistry than me, I really don't know to whom I should list. i've tried to look for a recommendation in the IUPAC site, but I couldn't find anything about that. Someone could help me with a link from some organization (preferably an international one) that recommends about this topic?

The ACS Style Guide suggests using your second option, with the physical state in parenthesis, directly to the right of the compound, and not subscripted.

This is indeed a duplicate of the question asked on Meta: What is the standard way to denote physical states in a chemical reaction?

Overall science is not affected by whether you write subscript or a superscript or in line. Afterall all notation was invented by humans. When professor quibble over these things my suggestion for them is to read the problem of "How many angels can stand on the point of a pin?" Apparently when Constantinople was falling, allegedly scholars were discussing these problems. Anyway, the lesson is that science is more important, notation comes later.
The International Union of Pure and Applied Chemistry (IUPAC), makes recommendations to systematize notation. Show the book "Quantities, Units and Symbols in Physical Chemistry" Third
Edition, https://doi.org/10.1039/9781847557889. It is available for free.
There was a similar question What is the standard way to denote physical states in a chemical reaction?
but the reference is old. The above reference is recent but there is no difference. Just for your interest, to see how old that reference is, it looks like type-written text!

They suggest to write the state of aggregation as they call it, just next to your symbol like $\ce{Na(s)}$, not as a subscript.

I would not downplay the importance of "notation", it is extremely important. The question is, is the notation self-evident, is there room for misunderstanding or confusion? I have personally never seen version 1 in the OP and would suggest sticking to 2 which is standard convention. I would be interested to know who is the source and propagates use of 1.


text:
When exactly does the first drop of distillate appear in a distillation?
The below image (source: here) is from a fractional distillation of a mixture of two liquids, one with a boiling point of 69 degrees celsius and the other with a boiling point of 98 degrees celsius. In this diagram, it appears that the first drop of distillate occurs at the boiling point of the component with the lower component.

However, I was wondering why the first drop of distillate appears at the boiling point of the species with the lower boiling point, instead of being in between the boiling points of the two species.
My reasoning is here: If you start with a mixture of A and B, represented by point $a_1$, when you heat it, the first vapor will appear at point $a_2$ which corresponds to temperature $T_2$, which is in between the boiling points of either component.
Why does the first drop of distillate appear at a temperature equal to the boiling point of the lower boiling component, instead of in between the boiling points of the two components?

Because the T is being read where the vapour condenses.

Can you elaborate please?

T in a distillation apparatus is that of the condensing phase. It is the T of the first drop of distillate. That is why fractional distillation works.

Or in other words the T in the graph has nothing to do with the temperature of the liquid phase. Would the distillation have failed there will be no flat portion of the curve. Look for fractional distillation principle. Having no column would results in basically the situation you have described, in which you consider the T at which the liquid starts boiling. That is not the distilled phase.


text:
Is there any kind of reaction with two types of reactants (gases) and one product (also gas) such that the total volume after the reaction increases?
I would like to think of the following equation:
$$\ce{aA(g) + bB(g) -> cC(g)}$$
where $a + b < c$ and $\ce{A},$ $\ce{B}$ and $\ce{C}$ are different gases.
Is there such chemical reaction?

I think extractions and ire synthesized isobutane from 2 separate gases but he couldn’t correctly synthesize it. You might want to check his channel out on YouTube if your interested. He might of made it correctly because he made multiple videos on it and I didn’t see all of the videos.

There are plenty of reactions like this, especially if you go beyond STP, so it would be nice if you also specify the conditions.

Burning of nearly any hydrocarbon would do, provided that it returns water in vapor form (which of course it does).

@IvanNeretin Burning hydrocarbons won't satisfy one of his conditions, which is that the reaction produces a single product.

@theorist Oops, I missed that one. Well, then all of a sudden there are not that many reactions, and those which are still there look highly unnatural and contrived, like O3+C3O2.

@IvanNeretin Does Ozone react with Carbon Suboxide? If so, thank you very very much for providing your answer! I've been pondering on a seemingly flawed highschool chemistry problem and was thinking of a counterexample to show that it is indeed flawed. Anyway, thanks for the answer!

Well, ozone oxidizes pretty much everything, and C3O2 reacts with pretty much everything, so what are their options?

What about the photochemical reaction $\ce{2O_3 \to 3O_2}$ if you let A and B be the same.

This is a fun question. The essence of what makes it challenging is that you're doing a synthesis -- combining two different species into one -- yet ending up with more particles, not (as is typical) less.  This limits the number of possible reactions significantly.  Here, however, is one that meets the conditions:
Octasulfur ($\ce {S_8}$) boils at 444.6  $^\circ \text{C}$ at standard pressure.  So, above that temperature, one could have:
$$\ce{ S_{8(g)} + 4O_{2(g)}-> 8SO_{(g)}},$$
where $1+4 < 8$.
Of course, if octasulfur were combusted with oxygen, it's likely that many other species of sulfur oxides (i.e., compounds of the form  $\ce {S_xO_y}$) would be produced as well.

@Theorist : In the vapor phase, sulfur is not made of S8 molecules, but S2.

@Maurice Do you have a reference to support that?   According to https://aip.scitation.org/doi/abs/10.1063/1.1734241 [J. Berkowitz, Equilibrium Composition of Sulfur Vapor, J. Chem. Phys. 39, 275 (1963)]: "Sulfur vapor, as well as...have been studied mass‐spectrometrically. The results indicate the presence of all possible S$_n$ molecules between S$_2$ and S$_8$ in measurable concentration between room temperature and the boiling point of sulfur."  Given this, one should also find S$_8$ just above the bp.  And, of course, the proposed rxn could also take place below the bp.

@Maurice This is not true, see https://doi.org/10.1007/b12405. Reading the abstract should be enough to find out that S₈ is one of the dominating forms in gaseous phase at the temperatures up to 1000 °C. Later, in the chapter it is mentioned that "the ratio of the partial pressures calculated by Detry et al. for saturated sulfur vapor at a temperature of 300 °C is as follows: S₈:S₇:S₆:S₅:S₄:S₃:S₂ = 1:0.35:0.44:0.026:0.001:0.001:0.007." Also, see this table for 500 °C. The theorist's answer looks perfectly fine to me.

Decomposition of ozone:
$\ce{O3 + O3 -> 3 O2}$
Technically, not 3 different gases, but the introduction of isotopes may be judged as in compliant (and yes, contrived).


text:
Why is the HOMO-LUMO gap in N2 and CO of different sizes?
Why is the HOMO-LUMO gap in $\ce{N2}$ and $\ce{CO}$ different even though they have the same number of valence electrons?

So what if they have the same number of valence electrons? See, CH4 and H2O have the same number of valence electrons; does that mean they are similar in any other way?

@IvanNeretin Maybe the OP wanted to say that N2 and CO are isoelectronic, have the same symmetry, and their atomic orbitals are at similar energy because they are neighbors in the periodic table.

As correctly noted in the question, $\ce{CO}$ and $\ce{N2}$ are iso-electronic, so they should be comparable. But a difference remains: $\ce{CO}$ is hetero-nuclear, $\ce{N2}$ is homo-nuclear. The latter case typically means that there is better overlap between the orbitals (because they are of similar extent/energy). This translates into a greater gap between the bonding and the anti-bonding combination of the orbitals. This will definitely change the HOMO-LUMO gap. However, without performing a detailed quantum chemical calculation, it is not clear to me whether the HOMO-LUMO gap should be bigger or smaller for the homo-nuclear case, because one has to take the entire molecular orbital diagram into account.


text:
Computing the bond spring constant from first principles
Given a set of atoms and their positions in space, is there a way to compute the bond spring constant for each pair of atoms that are considered to be bonded (I can already determine the bonds with OpenBabel)? I don't want to rely on empirical data, and it's okay if the spring constant is wrong by up to about 20%. I already know the bond length and bond energy, if that helps.
For background: my ultimate goal is to compute the normal modes of a molecule, by finding the generalized eigenvalues of an undamped mass-spring system. I already know the mass matrix, but I don't know the stiffness matrix.

I guess one thing I could do is use OpenBabel to find the bond energy, then perturb the location of one atom in a bond, find the bond energy again, and divide the change in bond energy by the distance the atom was perturbed.

tldr, you want the force constant from the Hessian matrix .. ideally in internal coordinates
First off, you're saying 'first principles,' which most computational chemists would take to be some level of quantum chemical method, whether density functional or wavefunction-based. None of that is available in Open Babel - it only implements a very small number of classical force fields.
Even if you have optimal bond lengths and energies, that doesn't directly give you the force constants. For that, you want the second derivative.

In principle, you could start perturbing atoms and evaluating the energy. In that way you could build up a stochastic approximation of the PES and evaluate numerical second derivatives. I don't recommend that.
A more accurate way would be at least to get the forces / gradients at each of those points, which you can use to build up the Hessian matrix. There are some notes about that as part of the psi4numpy notebooks which can be used to drive the Psi4 quantum program.
The problem is that for any molecule over a few atoms, the Cartesian Hessian matrix won't directly give you a bond force constant. Most vibrations include multiple atoms, and are rarely along any particular Cartesian axis.
The psi4numpy notebooks give some indication of how you can evaluate the Hessian in internal coordinates, which should be closer to what you want.

yes, I just put "first principals" in the title to distinguish from the myriad of questions about computing the bond spring constant from IR spectroscopy data and the like

So if I understand correctly there is no transformation that will allow you to convert a (diagonal) matrix of force constants for a vector of normal modes into a (diagonal) matrix representing displacements along connected atoms. So the answer to the actual question should be "no" and then the second question (requesting accuracy bounds) would also be "no"?

@BuckThorn - it depends on the coordinate system for your Hessian. As I said above, if you evaluate the Hessian in internal coordinates, then you'll have force constants for the bonds in your z-matrix.

@GeoffHutchison Maybe I am mixing topics, but my understanding is that the weighted Hessian has cross-terms, it is not usually diagonal in any but normal coordinates (except maybe for a simple molecule where they coincide). Question becomes to what extent you can ignore those cross-terms if you transform into internal coordinates (a transformation which you can do starting from Cartesian if you assume small enough displacement ie linearize). I realize force fields often ignore cross terms, but the question is how big the error is. Perhaps this is totally unimportant.

The only way to know would be to do it for a molecule of interest. Indeed, for larger molecules, the normal modes typically involve multiple atoms and thus the internal coordinates won't help much - but my point is that the Cartesian coordinates aren't useful.


text:
What is electrolytic protection?
My highschool textbook tells me that sacrificial protection (more technically-cathodic protection) and electrolytic protection are different. I have searched high and low for an explanation of electrolytic protection, but I got nothing.
It describes the electrolytic protection, stating that the steel/iron structure is hooked up to an electrolytic (not electrochemical!) cell as the negative terminal (cathode) and the anode is an inert material (titanium, in the example).
Why does this work, if it does at all?

Electrolytic protection and sacrificial protection are synonymous, in my opinion. It is obtained by applying a layer of zinc metal on a piece of iron exposed to rain. The iron piece will not get oxidized, because zinc will be oxidized before iron.

The world would be rather different without CP .Nat gas piped to homes, etc, would be rare. Many would not have water piped to homes ; many municipalites do not apply CP and hope for the best with their cast iron pipe , just replace as needed , the taxpayers will pay for it. Gasoline and plastics would be much more expensive because of transportation costs . I hoped a high school science text would not be so incompetent . But not surprised after looking at the local university chemistry. text.

I understand why a HS text might call it "electrolytic" , but in the real world it is called "cathodic protection" or CP. That is what to look for on the net. It is a multibillion dollar industry. Essentially all pipelines ( oil, gas, products, water , etc.) use it . A large majority off offshore steel structures ( oil  platforms, ships, etc) use it. And many steel products are galvanized . There are two basic types ; sacrificial coatings and impressed current. Sacrificial coatings are represented by galvanized steel , also aluminum and cadmium coatings. Impressed current, CP is also  two categories : Sacrificial anodes , usually aluminum ( which do not use rectifiers) , and inert anodes , usually titanium or graphite. These are buried in soils made more conductive with graphite powder (typically) and electrically connected to rectifiers  which are then connected to the steel to be protected . The voltage is adjusted and monitored along the protected steel , eg. pipeline. Pipelines are coated with an electrically insulating coating to reduce the amount of current required to maintain a protective voltage , normally lower than - 0.65 V ( Cu - Cu SO4 half cell) Too low a voltage can cause hydrogen gas bubbles under the coating and hydrogen embrittlement of certain steels.There is also anodic protection but it has very specific limited applications. So, there is such a thing as "electrolytic" protection.

How do those impressed current systems actually prevent corrosion? As long as there is any oxygen close to the metal (especially in conductive soils/waters), you can have a "side reaction" of surface corrosion/rust formation, completely separate from whatever flow of electricity from the CP-system...

An over simplification is the CP raises the voltage of the steel so the oxygen emf is not strong enough to take any electrons.

But voltage relative to what? The oxygen that exists everywhere in the conductive soil/water environment  anyway (which actually causes the corrosion), or just to some random distant anode installed somewhere else? Having a cathodic voltage merely relative to the latter seems pointless -- it's the oxygen omnipresent everywhere in the conductive soil/water environment that causes the corrosion. How would having a voltage relative to some distant anode in a specific location protect against that?

Electrolysis anodes can only protect metal cathodes from oxidants present nearby the anode surfaces. In other words, for example, establishing a voltage difference between a steel cathode and some distant graphite (or whatever) anode (this is what impressed current (ICCP) diagrams often seem to look like anyway!) does not automatically also establish a significant voltage difference between that same steel cathode, and other oxygen molecules that instead exist next to the surface of the cathode, as well.

CP is such a large industry you can get all the info you want on the net under cathodic protection.

Interesting, it seems I was quite overestimating the rate of rust formation / oxidation on the metal parts in those conductive soil/water environments anyway. If ICCP is able to deter / slow down corrosion and/or somewhat lessen its dangerous effects even somewhat, it could still be quite a useful investment...

As far as I know sacrificial protection refers to the protection of iron specifically, while electrolytic protection is for all metals.
The concept of electrolytic protection is that a valuable metal is connected with an invaluable and more reactive one to create an electrolytic cell. The invaluable is the anode as it is more reactive and it gets oxidized by the environment, while the less reactive valuable metal forms the cathod and is protected from oxidization.
A note : Electrochemical cells are cells that produce electricity and cells that electricity is applied to them. Electrolytic cells are cells that electricity is applied to them. So they are not exactly the same thing.


text:
How does the rate constant change with the change of the temperature?
In class, we discussed the following multiple-choice question.

The rate expression for a reaction is:
$$rate = k[X][Y]$$
Which statement is correct?
A. As the temperature increases, the rate constant decreases.
B. The rate constant increases with increased temperature, but eventually reaches a constant value.
C. As the temperature increases, the rate constant increases.
D. The rate constant is not affected by a change in temperature.

Considering the Arrhenius equation given below, the rate constant is dependent on the absolute temperature of the reactants and activation energy of the reaction.
$$k = Ae^{-\frac{E_\mathrm{A}}{RT}}$$
Taking the following limit as the absolute temperature approaches infinity reveals that the rate constant converges to the Arrhenius constant at very high temperatures. This is also evident when graphing the Arrhenius equation. Therefore, we conclude that option B is correct.
$$\lim_{T \to \infty} Ae^{-\frac{E_\mathrm{A}}{RT}} = A\lim_{T \to \infty} e^{-\frac{E_\mathrm{A}}{RT}} = A$$
However, when checking the mark-scheme, option C is listed as the correct answer. Is there a mistake in my reasoning?

The Arrhenius "constant" is actually only a pre-exponential factor which is in itself dependent on the temperature. The activation energy is also dependent on the temperature.

@Martin-マーチン"The activation energy is also dependent on the temperature."How?

If the rate constant is not affected by a change of temperature, it means that the activation energy is zero.

If this is a test question on Arrhenius' Law, at more elementary levels, in my opinion, it maybe an acceptable question. Otherwise per this source, for example, a possible issue from this reference, to quote::

Defying Arrhenius’ law
In a theoretical paper, Ian Smith and colleagues at the University of Birmingham, UK, proposed several mechanisms by which neutral species reacting under the near-vacuum pressure and extremely low temperature conditions of interstellar clouds could defy Arrhenius’ law and actually react more rapidly at lower temperatures [2].

Also:

Arrhenius’ law doesn’t apply here, said Smith, because ’there is no well-defined barrier on the minimum energy path leading from reactants to products in these reactions.’

And:

Chemical reactions at extremely low temperatures, for instance in interstellar clouds, can run at surprisingly fast rates, report astrochemists who have used a combination of theoretical and experimental methods to work out why.

Answer B is technically correct although C could also be accepted. Assuming $E_A$ to be constant, at low temperatures because of the inverse $T$ in the exponential $\exp(-E_A/(R\cdot \text{small number}) \equiv \exp(-\text{big})$ which is a small number. At high temperatures $\exp(-E_A/(R\cdot \text{big number})\equiv \exp(-\text{small})$ is a big number, so the rate constant increases with temperature.
At temperature such that the exponential  $\to 1$ the rate constant is $A$ for all practical purposes. Usually this limit is not reached unless $E_A$ is very small and then as $A$ is also a function of temperature this ($A$) term becomes now important. Generally, however, for the vast majority of reactions the exponential term is the most important.


text:
Why does Larmor precession occur in NMR?
Larmor precession is the phenomenon that in NMR the spin (and by default also the magnetic dipole moment) of protons does not line up with the applied magnetic field but rather precesses around this direction at a certain angle.

The concept of Larmor precession confuses me for several reasons:

The analogy is made to a children's toll which precesses in the gravitational field of the earth. And then it is claimed that since these subatomic particles spin around their axis something similar happens. BUT subatomic particles don't really spin around their axis, rather the spin is a quantum property, so how does that add up?
Second of all it is stated that because of their quantum nature protons can only display 2 states, ie parallel or antiparallel to the magnetic field. But if Larmor precession takes place they are no longer parallel or antiparallel?
The calculation of the energy of the proton in the external magnetic field comes down to:

energy= spin magnetic moment x magnetic field strength
Nowhere is there a cosine to indicate the effect of the spin magnetic moment being of axis. Surely the angle between both must matter?

Look for the 'vector model of the atom', this is used describes the behaviour of 'spin', electrons usually, but they have the same spin properties as protons so the physics applies to them also.  To answer (1) I agree that idea of fundamental particles spinning is rather odd and misleading,  but the equations that describe their behaviour are similar to those for macroscopic spin, and the name stuck. (2 and 3) are answered by the vector model.

If you have a lot of protons, the component of the magnetic dipole along the axis adds up (to the bulk magnetization) and the components perpendicular will cancel out. To understand the basics of pulsed NMR, you only need to consider the bulk magnetization, see e.g. http://www-keeler.ch.cam.ac.uk/lectures/understanding/chapter_3.pdf

@Karsten, does this mean, for the sake of argument, that when you would only have one proton in your sample there would be no Larmor precession?

@Stikke, I would suggest that you forget about about nuclei for a moment and see why a gyroscope does precession. Here is a nice illustration. The idea is derived from mechanics. http://mriquestions.com/why-precession.html

@Stikke No, it is the opposite. With a lot of protons, we can ignore the precession (for some purposes) and focus on the bulk magnetization. With just one, we can't ignore the component of the vector perpendicular to the magnetic field.

The quantum way to view precession that also connects to the idea of only having two quantum mechanical states is to say that when the spin is measured, you will always get aligned or anti-aligned with the field. However, the particle can actually exist in any linear combination of these two states. There will then be two time-dependent phases which correspond to states of different energies. Therefore, the average (classical) direction of the spin is neither up nor down, but at some angle to the field, where that angle is related to the energy difference of the two states.


text:
Chemical Reaction for making Acetic Acid from Glycerol inside Fermenter
I have been trying to find a chemical reaction for the formation of acetic acid from glycerol. I have been searching different literature, but apparently the reaction can not be found. I was hoping maybe anyone here would know the reaction.
$$\ce{C3H8O3 + X -> C2H4O2 + X}$$
Some Background
I am working on producing Succinic Acid (SA) from Glycerol (GLA) inside a fermenter. I know that inside the fermenter, the GLA is converted into succinic acid, water and acetic acid. I have been able to find a reaction for the formation of succinic acid, which is given by the following,
$$\ce{C3H8O3 + CO2 -> C4H6O4 + H2O}$$
But I can not seem to find a reaction for the acetic acid. I need an equation because I want to know the stoichiometric relation between the GLA and AA.
Additional Question
Another thing is the buffer solution reaction required to keep the $\mathrm{pH}$ constant inside the fermenter. I have been studying that either ammonia or $\ce{Ca(OH)2}$ can be used. I know the chemical reaction,
Calcium Hydroxide + Succinate -> Calcium Succinate + Water.
The calcium succinate can be reacted with conc. sulfuric acid in order to crystallize the SA crystals, since sulfuric acid can release the free succinic acids. I understand the theory behind these reactions, but I can not seem to find chemical reactions, from where I can see the stoichiometric coefficients.

What organism are you using? The specific metabolic pathways can change the stoichiometry. Also, aerobic or anaerobic?

@Andrew I am using E. Coli and it is anaerobic. Hence, I am using CO2 with E. coli.

Short answer: under anaerobic conditions, it is impossible to produce acetate as the sole product from glycerol with wild-type E. coli.
Explanation: Under anaerobic conditions (and assuming no substrates for anaerobic respiration have been added), a significant concern is redox balancing. Since electrons are not transferred to oxygen (as they are under aerobic conditions), the number of electrons in the substrates must equal the number of electrons in the products.
Glycerol as a substrate presents problems in this regard, because it is slightly more reduced than typical hexose substrates (eg glucose). Furthermore, the desired product acetic acid is more oxidized than glycerol (or hexoses for that matter).
Looking at the core metabolic pathways of E. coli, we immediately see that 1 molecule of glycerol is converted to 1 molecule of acetate. E coli is unable to utilize all three carbon atoms of glycerol make 3 molecules of acetate from 2 molecules of glycerol. The third carbon atom of glycerol is lost as either formic acid or $\ce{CO2}$ (with concomitant production of hydrogen gas). So we can write an initial stoichiometry of
glycerol ($\ce{C3H8O3}$) + $\ce{H2O ->}$ acetic acid ($\ce{C2H4O2}$) + formic acid ($\ce{CH2O2}$) + 4 electrons + 4 $\ce{H+}$.
[Note that the formic acid can be converted to $\ce{H2}$ and $\ce{CO2}$. The exact ratio of hydrogen to formate will depend on conditions.]
Unfortunately, there is no pathway for disposing of the excess electrons. You would need to add a respiration substrate such as fumarate (converted to succinate) or nitrate or nitrite (both can be reduced to ammonia).
One alternative approach would be to ferment the glycerol to a mixture of ethanol and formic acid or hydrogen gas and then chemically oxidize the ethanol to acetic acid. The balanced stoichiometry of that fermentation is
glycerol ($\ce{C3H8O3}$) $\ce{->}$ ethanol ($\ce{C2H6O}$) + formic acid ($\ce{CH2O2}$)
or
glycerol ($\ce{C3H8O3}$) $\ce{->}$ ethanol ($\ce{C2H6O}$) + $\ce{H2 + CO2}$


text:
Help to understand the fundamental equation of thermodynamics
The fundamental equation of thermodynamics states that $\mathrm dG = V\,\mathrm dp-S\,\mathrm dT$. I am having difficulty understanding this equation. The Wikipedia page for Gibbs free energy says:

… is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric).

Isn't this in direct contradiction to the fundamental equation of thermodynamics? If you have constant pressure and constant temperature, then $\mathrm dp$ and $\mathrm dT $ are both equal to $0$, so $\mathrm dG = 0$?
What am I not understanding?

You don't have any issue with thermodynamics, just with article...

What do you mean?

The form of the equation you wrote is for the case of pure substance or mixture of constant composition.  Otherwise, there are additional terms involving the chemical potentials of the various species times the change in the number of moles of each species.


text:
Salinity difference in KCl and NaCl
If I calibrate my conductivity probe using KCl calibration solution, then test my aquarium water which is NaCl, will I get a false reading? When I use a digital refractometer calibrated to zero using distilled water and test my saltwater aquarium, it reads 35 PPT but when I use a conductivity probe calibrated using KCl 53,000 uSiemens it gives me a slightly different reading every time. Is this because I am calibrating to KCl but trying to measure NaCl?
Thanks for any clarification with this.

$\ce{KCl}$  is used to calibrate conductometers via determining the device proportionality constant between the measured conductance and the known solution conductivity.
The relation between conductivity and molar concentration / mass concentration / salinity is for obvious reasons for different salts different.
It is possible the device has a built-in calculation of the $\ce{NaCl}$ salinity from conductivity, or a table or a formula is available.
If the conductometer is calibrated directly in $\ce{NaCl}$ salinity, then it must be calibrated by $\ce{NaCl}$ solution of the known salinity.
If other salts are present in sea water, they affect the measurement.
Note that conductivity raises with temperature due increased ion mobility. It is much more sensible then refraction index, so it's results change with temperature more then refractometric results.
Warmer water would lead to higher reported salinity due higher conductivity, unless there are made automatic or manual corrections.


text:
How to predict disproportionation products from Frost diagram?
I read that concave points in a Frost diagram disproportionate. But what are the products of the disproportionation?
For example, in the Frost diagram for manganese at pH = 0, the species $\ce{HMnO4-}$ is concave with respect to $\ce{MnO4-}$ and $\ce{H3MnO4}$. But at the same time, it is also concave with respect to $\ce{MnO4-}$ and $\ce{MnO2}$. How do I know which set of products $\ce{HMnO4-}$ disproportionates to in acidic solution at pH = 0?

Interesting question. First of all understand what is on the x-axis and on the y-axis. You y-axis is Gibbs free energy/Faraday constant. Choose three consecutive oxidation states, the middle oxidation state, if it lies above the two, is likely to disproportionate.Read more here on Frost and Latimer digram

In your case look at the example of $Mn^{3+}$, if you connect $Mn^{2+}$ and $MnO_2$, $Mn^{3+}$ lies above the two, so 3+ oxidation of Mn is unstable with to disproportionation.
In short, just by using Frost diagram you cannot predict disproportionation of Mn (VII) because if it were to disproportionate, you will need Mn(VIII), which is not shown, so Frost diagram analysis only works for three points and you cannot say anything about the boundary values.

I am asking about the products of the disproportionation of the +7 oxidation state.

It will $not$ disproportionate, as I said, it will work with three points not an isolated boundary value.


text:
Calculating the energy of a molecule using probability distributions
I wish to calculate the energy for a protein where the position of all the C-alpha atoms is known.
One way is to calculate the pairwise distances between the atoms and then look up a probability distribution to determine the likelihood that the distance for that particular pair is the given value, take the negative logarithm of that value, sum this over all pairs and this is a proxy for the molecular energy. This is the approach taken by the machine learning community. (Physics-Based Design of Protein-Ligand Binding)
Can anyone point me to a sample calculation or publication where this is elaborated in some detail?
I have very little background in chemistry beyond the introductory college courses. The question that I am trying to research is this: Given two spatial configurations of a molecule, how does one determine that one configuration has a lower energy than the other? What is the benchmark that is used for measurement? When Ramachandran generated the plots of torsion angles, how were the optimal angles calculated?

This isn't actual energy, but a scoring function that can be called like that (good to remember that). This one seems kinda simplistic one, not taking actual sequence into account?

What is your source for "this is the approach taken by the machine learning community"? Maybe that would be a starting point for finding a related publication.

There are a wide variety of ways of calculating energy or scoring - but none that I know of would be compatible with this type of descriptor.

You ask about the torsion angles - the first plots were done from experimental crystal structures. So 'ideal' torsions come from finding mean/median torsions across a lot of proteins and domains.

Thank you to all for your assistance.


text:
Liquid neon freezer?
Liquid nitrogen is approximately -330 degrees Fahrenheit. Liquid neon is approximately -413 degrees Fahrenheit. I want to know about if this type of freezer is easily possible in practice and if such a freezer actually exists. The idea is to keep the matter as cold as possible and combine it with a noble element that doesn’t react with anything that we know of due to electron configuration. The idea of liquid neon is to stop effectively all biological activity and chemical reactions, because liquid nitrogen is both hotter and not a noble gas, though liquid neon is not a noble gas though neon usually is.

What's the point? Liquid nitrogen is quite enough to stop all biological activity and nearly all chemical reactions, and itself is quite inert at these temperatures. In the rare cases where you genuinely need to go lower, helium is your choice.

We should note that some chemical species have only been observed in matrix isolation. However, one typically uses argon (because it is commonly available) or helium (because the lower temperature is needed). One could suppose the use of neon for reactions at temperatures higher than liquid/solid helium that also necessitate a matrix more inert than argon.

Neon has the advantage that it has about $20\times$ the molar specific heat of helium, so that even if it doesn't get things as cold it can cool them off quicker. The huge disadvantage is that neon is much more expensive than helium or nitrogen.

Liquid nitrogen is probably sufficient, but neon is both more inert and colder/less hot.

I'd suspect that liquid neon would be significantly more expensive than liquid nitrogen with scant reduction in chemical reactivity and scant further reduction in biological activity.

Just because it is slight doesn’t mean it doesn’t happen

You're really not talking about a self-contained "freezer" but a cryostat where a liquid coolant will need frequent replenishment.

Sorry I should have been more clear. By "scant reduction" I mean yes there is a theoretical reduction but that the reduction is in reality so small as to be immeasurable. // Excess isn't always better. Ever try to park a semi-truck in a typical parking garage for cars?

Makes sense, so nitrogen is probably sufficient


text:
How are reaction enthalpy and entropy affected by temperature
The equation for Gibbs Free Energy demonstrates how the standard change in Gibbs Free Energy for a reaction is affected by temperature.
$$\Delta G^\circ =   \Delta H^\circ - T\,\Delta S^\circ$$
However, to my understanding, the equation makes the inherent assumption that the standard change and enthalpy and the change in entropy for the reaction is unaffected by temperature. I have heard that this is not true and that reaction enthalpy and entropy are indeed dependent on temperature. However, I do not know how exactly temperature affects the two and why.
Would a reaction occurring at a higher temperature be more or less exothermic? Or does it perhaps depend on the reaction itself?
I know that the entropy of a substance increases at higher temperatures but I do not know how this correlates to the situation of a reaction since I'd assume that both the reactants and the products would be at a higher entropy.

There is nothing to know here. $\Delta H$ may change either way, depending on the reaction. Ditto for $\Delta S$.

The standard Gibbs free energy, enthalpy, and entropy changes, $\Delta G^\circ, \Delta H^\circ,\text{and }\Delta S^\circ$, are all formally temperature-dependent.  Indeed, one could write them as: $\Delta G^\circ (T), \Delta H^\circ (T),\text{and }\Delta S^\circ (T)$.
"Standard" simply means that the pressure is 1 bar.  See "standard thermodynamic quantities",  https://goldbook.iupac.org/terms/view/S05927
Most commonly, standard thermodynamic quantities are tabulated at $298.15 \text{K}$, but they can be given for any temperature.  Indeed, the reason why one needs to specify the temperature is that they are temperature-dependent.
However, while the following equation does not assume the temperature-independence of any thermodynamic quantity, it does assume that the temperature is constant for the change in state:
$$\Delta G^\circ =   \Delta H^\circ - T\,\Delta S^\circ$$
This is because, by defintion:
$$G = H - TS \Rightarrow dG = dH - TdS - SdT$$
Then, under the assumption of constant T,
$$dG = dH - TdS \Rightarrow \Delta G = \Delta H - T \Delta S$$
As to the specific temperature-dependence of each of these thermodynamic quantities, we have:
$\require{begingroup} \begingroup \newcommand{\pd}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}}$
$$\pd {\Delta G}Tp = -\Delta S$$
$$\pd {\Delta H}Tp = \Delta C_p$$
$$\pd {\Delta S}Tp = \dfrac{\Delta C_p}{T}$$
Note the above hold for any change in state in which the pressure is constant, which of course includes when the pressure is the standard pressure.
In case you are confused by $\Delta C_p$, note that it is just the difference between the constant-pressure heat capacitity of the products and reactants.  I.e., just as $\Delta G$ is $G_{prod} - G_{react}$, $\Delta C_p$ is $C_{p, prod} - C_{p, react}$.
Hence, as the temperature increases, if (say) $C_{p, prod} > C_{p, react}$, then:
$$\pd {\Delta H}Tp > 0,$$
which means the reaction will become less exothermic/more endothermic as the temperature increases.  Note, however, that $C_p$ is itself also temperature-dependent.
$\endgroup$

For each species you could use Kirchoff's equation to calculate the change in enthalpy with temperature, in its integrated form (at constant pressure) this is $\displaystyle \Delta H_2-\Delta H_1=\int_{T_1}^{T_2}\Delta C_p dT$. The heat capacity $C_p$ changes with temperature also, but for small temperature changes it may be considered to be effectively constant, then $\displaystyle \Delta H_2-\Delta H_1=\Delta C_p(T_2-T_1)$, if not, the heat capacity can be approximated by $C_p=a+bT+cT^2 \cdots$ and integrated but the constants should be determined experimentally.
The entropy is zero at zero Kelvin for a pure substance so must increase with temperature and as $\displaystyle S_T=\int_0^TC_p/T dT$ you can calculate the change in entropy between two temperature.
However, there are phase changes as the temperature increases (solid, liquid, gas) then there are discontinuities in the heat capacity and the entropy becomes $\displaystyle S_T=\int_0^TC_p/T dT+\Delta H_{fus}/T_{fus} + \Delta H_{vap}/T_{vap}$. The terms with phase changes cancel out in $\Delta S$ if your species are in one phase, i.e. all reactants are in the gas phase.
(If you are familiar with partition functions in Statistical Mechanics then thermodynamic properties of molecules can, in principle, be calculated directly.)