text:
Blue Tinge for White Paint
I am trying to make an aqueous white paint with Mica and TiO2 as mainstay that gives a slight tinge of blue when it dries but should remain white while in solution.
So far I have tried optical brighteners but they havent been successful. What can I add in my paint for this?

But that would be expensive + paint will appear blue? I need to keep the paint looking white while it is in liquid state. Blue tinge only when it is dry.

Mica does not dissolve in water to yield a solution.  Neither rutile, nor anatase (both $\ce{TiO2}$) enter an aqueous solution either.  Perhaps, finely ground, you may get them stabilized in an emulsion (like, e.g., white tooth paste).  Dyes (soluble color agents) and pigments (color agents not soluble in water) of blue color you find in nature and which resist extended exposure to sun light are rare (e.g., lapis lazuli from Afghanistan).

Buttonwood pointed out to lapis lazuli in the comments, but I want give an interesting perspective. There is a very beautiful blie pigment used in South Asia as for a  post-laundry washing. It is literally dirt cheap and but pretty stable as a dye. It is an inorganic dye used to make white look bluish white. "Ultramarine blue is an inorganic and complex compound formerly obtained from lapis lazuli but now prepared artificially. This famous blue pigment dates back to the early 19th century. Chemically, these are the most complex out of all inorganic class pigments.

It consists of sodium, aluminum, silicate lattices and polysulphide (-S-S-S) chains. It is produced by heating kaolin, sodium carbonate, sulfur and other inexpensive ingredients together. The blue color of the pigment is due to the S3-radical anion, which contains an unpaired electron. "

Quoted from here: sciencedirect.com/science/article/pii/S2090536X15000027

@M.Farooq I posted the first comment, suggesting lapis lazuli, but the OP rejected lapis as too expensive and they want the liquid to be white until it dries, at which point they want a blue tint. So I deleted that comment. Apparently neither Han blue nor Egyptian blue would be acceptable either.

The OP can check "robin blue" on Amazon pigment which is nothing but dirt cheap from of lapis lazuli. A synthetic and safe pigment.  There is no such paint which is white in solution and bluish when dry.

@M.Farooq    The only thing I can think of is having a brightener plus a volatile quenching agent in the liquid. Then as the paint dries, the quencher evaporates and the brightener does its brightening. But the list of safe and volatile quenchers is short, e.g., nitromethane is highly toxic.

@EdV, Don't the brighteners bleach after a while?

@M.Farooq Yes they do, as far as I know. Then you are stuck with dingy, aka “ecru”. That is why I suggested lapis lazuli in the first place, but it requires dropping the white while liquid condition.

Yes, the white suspension condition has to be dropped.


text:
Best bet for separating fused glassware
I have a sintered glass funnel stuck in a flask with a 24/40 ground glass joint.
I regularly used this glassware to process very dilute solutions of uncured photopolymer resin dissolved in ethanol. Unfortunately, I believe I accidentally hit the joint with some stray UV laser light while it was wet, thus curing some of the resin, and binding the glassware together:

Because it was so dilute, combined with the flask being under vacuum, I guess it was able to work its way well into the space in the joint.
I already consider this glassware to be lost; it's been in storage for a while and now I'm cleaning house. Still, I figure I'll go for one more Hail Mary attempt at separation before trashing it.
What's my best chance for separating these? Is there something I can do with heat, or a solvent besides ethanol, or some other trick that might help?

Notes:

I've got all the necessary PPE to apply brute force if needed. However, I do not have a fume hood or a lot of lab experience, so things like e.g. hot piranha solution are aren't an option for me. My setup is wood / metal shop, not so much chemistry lab.

The only thing I tried when it originally got stuck was soaking the joint in ethanol for a while at room temperature, hoping to dissolve any resin; but it didn't work; or at least, it didn't separate with the max amount of force I was comfortable exerting (with gloves on) before I felt it would break.

If all else fails, then I'm going to cut the top of the flask off and grind everything down mechanically to try and recover the funnel at least. I want to try to save both, first, though.

Maybe using acetone and an ultrasonic bath. For a day, and with plenty of ventilation. Good luck!

@EdV That is a great idea and, frustratingly, my ultrasonic cleaner is barely too small to accomodate the glass. I'll think about this a little though, maybe I'll get some inspiration. Or maybe I know somebody with a larger capacity cleaner. Thanks.

I would recommend putting it in a soap bath for a few days and then heating the joint with a hot air gun or blow dryer. I've done this a couple of times, and it seems to work fairly well -- although it might take a while!

After you get these unstuck, your metal shop comes in very handy: just make a PTFE sleeve for future use between the filter and flask. Or buy the inert adaptor, maybe! ;-)

Seek a professional glass blower. While these become more and more rare on campus of a research university for specialized glass ware you can't (yet) purchase from a catalogue for chemistry and physics, these have large baths and the experience e.g., to flame heat just the outer neck of the flask to disentangle the connection. Or contact a manufacturer (e.g., Chemglass). With standard lab equipment, however, curation by UV across glass is less likely than getting stuck by aged grease, reagent alkalinity, or cycles of heat and (ambient) cold.

source: Stack Exchange

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

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text:
Why are some colorings called "Fast", i.e. Fast Yellow AB (E105)?
Every time I look into things like this, I keep seeing talk about "fast food logos", so I've hit a dead end.  What makes a specific color "fast"?
I see Fast Yellow and Fast Green on this Wikipedia page, but no mention of why it's called "fast".  Is it just a name, like how some colors are called "Brillant"?  Is it a color that looks like something frequently used on race cars or fast food, so they just call it "Fast XYZ"?

Perhaps they can be made color-fast?

Usually when dipping a fabric or a cloth into a dyeing solution, it takes a long time to have it colored. It is necessary to wait long enough for the dye to enter the inner parts of the fibres. It is not the case for the fast dyes. As soon as the fabric or the cloth is dipped into the fast coloring solution, it is perfectly colored.

Colour fastness

http://www.chm.bris.ac.uk/webprojects2002/price/fastness.htm

Use quotes in your google search: $$\mathrm{"fast\ blue" meaning}$$

During the search, I came across acid-fast bacteria, those that don't lose a stain when washing with acid: https://pubmed.ncbi.nlm.nih.gov/30725806/

It's the same as in "Fasten your seatbelt", i.e. to make the seatbelt "fast".

@Loong that article is an example of useless Wikipedia articles: it doesn't tell whether fastness of a "fast to fade" dye is higher or lower, while it's crucial due to being counterintuitive.

A good dictionary is your best friend. Fast as an adjective for a color

Of a colour or dye: that will not readily fade or wash out; permanent. Now also in fast to: not susceptible to fading from. The usage dates to 1658!
1658   W. Sanderson Graphice 80   Fast and firm colours, as Umber,
Oke.
[Unabridged Oxford English Dictionary]

This word has a different root than your "fast" food. It is cognate with German fest which means firmly established.
https://www.collinsdictionary.com/us/dictionary/german-english/fest
Recall that the dye industry fluorished in Germany from coal tar.

Also compare "fasten", "fastener", expression "fast asleep". Most Germanics actually adopt a version of "fast" with this meaning as well.

Given that the term "Lake" seems to also be derived from other historical terms, I had a feeling it was some kind of reference to that, and I take it that in that case, Fast Yellow would be more permanent than other dyes that aren't Fast?  Also I completely forgot to address color fastness in my original question, as I was wondering if that's what it was about.  Thank you!

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


text:
When uranium or thorium is broken down in molten salt reactors, will it produce reactive chlorine or fluorine gas?
In molten salt reactors, uranium or thorium exist as $\ce{UCl4}$ and $\ce{ThF4}$ in a 1:4 stoichiometric ratio. When a uranium is split into a barium and a krypton, the barium becomes $\ce{BaCl2}$ while the krypton is inert. As a result, two chloride ions are forced to give off their electrons and become elemental chlorine that can react with the structure. Does that mean as the chain reaction proceeds, the molten salt will become more and more oxidative? How can we maintain the stoichiometric balance in molten salt reactors?

Fission creates many different pairs, not just Ba and Kr. Each pair will have different salts. Plus there are buffers included to reduce any mismatch.

And nuclear fission will produce so much energy that puny electrostatically driven chemical bonds with diffuse electron clouds are blown to the tenth dimension. You'll get a lot of ions (nuclei even?) and electrons until things cool down and recombine in some way.

@OscarLanzi - Only 10 dimensions??? Ha! My string theory goes to 11...

Tenth dimension is more poetic. Besides, I have heard of theories with 26 dimensions.

Ah, “It goes to 11.” ;-)

molten salt reactors do produce gases and often include system to remove the gas from the molten salt. Corrosion is probably not the biggest concern as some of the products interfere with the nuclear reaction and have to be removed to ensure the reactor keeps working optimally. Hence this is not a fundamental problem, but an engineering one.

As mentioned in the comments including mine (now deleted), fission of uranium produces not only barium and krypton but various other pairs and its salts. You can find a detail explanation of all the fission product chlorides in the referenced paper. The free chlorine that you mention is also present in abundance when the fission product chlorides decomposes at high temperatures.
Reference

Bradley, D. The preparation and properties of the chlorides of uranium, plutonium, and thorium and of the fission product chlorides. United Kingdom: N. p., 1957. Web. (link)


text:
Biocompatible positively charged functional groups on the pH range 5–9
In drugs and biomolecules, there are quite a few groups that are negatively charged around neutral pH (pH 5 to 9): carboxylates, sulfates, sulfonates, phosphates, phosphonates, tetrazoles, certain polycyclic phenolates.
There seem to be fewer options for positively charged groups: protonated amines (primary, secondary, tertiary), quaternary ammonium, and guanidinium. Notably, all these involve nitrogen. Imidazole doesn't stay positive in this entire pH range (pKa = 6.9). Do medicinal chemists have any other options for positively charged groups available?

Proteins binds to cations like magnesium, calcium or zinc when "they need a positive charge". A crown ether with potassium would also work. However, why not just use a nitrogen compound?

@KarstenTheis Crown ethers are an interesting idea. So is binding divalent metal ions. I’ve tried amines and guanidinium but they don’t quite have the right geometry and charge distribution for my application. Just trying to find some new ideas.


text:
How to remove crystal water from potassium ferrocyanide trihydrate?
For potassium ferrocyanide/yellow prussiate/potassium hexacyanoferrate(II), I identified only commercial sources of the trihydrate, $\ce{K4[Fe(CN)6] * 3H2O}$.
What are the conditions (temperature, because I guess heating in an oven might suffice) to completely remove the crystal water in this chemical?

Anhydrous trihydrate is a glaring example of an oxymoron, much like "Little Big Horn" or  "honest politician".

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

One suitable approach is to look up the information provided by the chemical suppliers about the trihydrate (example Millipore/Sigma) and to note the CAS registry number.  Use this number in chemistry relevant databases (e.g., Reaxys by Elsevier, Scifinder by the American Chemical Society) to look up primary references describing this (if you really need this, check with your TA) in a research library.  As the librarian of your uni for assistance how to use these sources.

Just heat it.

Alternatively refer to this paper.


text:
Fluorescence: what mechanism regulates the rate of discharge of energy?
The other day, I noticed a strip of fluorescent tape when the lights in the room were turned off.
The energy was discharged as a continued, faint glow, rather than one burst or flash of light.
How so? What prevents the molecules from de-exciting all at once?

Appears to be phosphorescence rather than fluorescence. The former can last minutes or longer, e.g., glow in the dark products, while the latter typically lasts less than milliseconds. Look up phosphorescence in wikipedia.

related https://chemistry.stackexchange.com/questions/31286/fluorescence-versus-phosphorescence

Phosphorescence is driven by the slow loss of energy from excited electron states
Firstly, the phenomenon you describe is phosphorescence not fluorescence: they have different mechanisms.
Fluorescence happens (to simplify a little) when light kicks electrons from their normal orbitals in some molecules to a higher excited state orbital (often via a multi-step pathway involving multiple orbitals). Those electrons drop back to their low state very quickly, emitting specific colours of light.
Phosphorescence is more complex. The basic idea is the same: light excites some electrons to higher energy orbitals. But, in this case, there is an internal mechanism that rapidly allows the electron to drop into an orbital state where there is a major barrier to the transition back to the ground state (specifically one that involves changes in the spin). These transitions are "forbidden" in some cases for symmetry and quantum reasons. But even "forbidden" transitions can happen but only when the exact molecular shape and circumstances are right. In some cases this happens because of a rare combination of molecular vibrations and these happen far more slowly than normal non-forbidden electronic transitions.
The overall effect is that, in some molecules, electrons can get "stuck" in higher energy orbitals until some much slower internal process breaks the rules that disallow the transition back to the ground state. This means that the decay process is billions or trillions of times slower than fluorescence and can take minutes or hours. So you can often see the light emission from the transition decaying slowly over minutes or hours.

The after-glow is called phosphorescence. Fluorescence stops immediately the moment light is turned off because the process is ultrafast!
Phosphorescence on the other hand requires a change in the spin of the electron in a molecule, such transitions are forbidden, this is why it is slow. You can check the Jablonski diagram to understand this better, see here https://micro.magnet.fsu.edu/optics/timeline/people/jablonski.html
Can you see how many rates and competing processes are involved?

It is not (strictly) true that fluorescence is instantaneous; many types of molecules fluoresce with lifetimes of hundreds of nanoseconds (although most lifetimes are shorter than this) and phosphorescence in some types molecules is shorter than hundreds of nanosec, but in a given type of molecule fluorescence is always faster than phosphorescence, due to the spin forbidden nature of the latter process.

Yes, I agree. One can see in the figure that there is "delayed" fluorescence too, along with their lifetimes but the OP has not responded at all.

Yes, but delayed fluorescence is rather rare, eosin etc,  as triplets are usually too low in energy and internal conversion between them  too fast for this to be significant. However, there has been more research on this latterly to improve  fluorescence yields when in devices. There is also T-T annihilation producing fluorescence as well as excimers and so forth.

Although the previous answers correctly explain the different mechanisms of fluorescence and phosphorescence, they do not really touch upon why the light is not emitted all at once. Even in the very fast fluorescence, not all light is emitted in one burst (although one needs fast detectors to see this).
The reason for the slow de-excitation of the molecules is that each of the excited states has roughly the same and independent chance to de-excite under emission of a photon at a certain time. Compare this with tossing a hand of coins, removing all the coins that show head and tossing the coins again that show tail. After the first toss, roughly half of the coins is removed, after two tosses, 75% of the coins is removed etc. Replace the coins with excited molecules and the tosses with increasing time, and you see why not all molecules emit light at the same time: the probability for de-excitation is less than unity and you get an exponential decay of the excited states. The difference between fluorescence and phosphorence rates is that states that can decay by fluorescence have a much higher chance to do so than states that have to decay via phosphorence.

To answer why the energy is not emitted as a burst we have consider the chance of a photon being emitted.
To match experimental data we have to assume that each event (photon emission) is randomly occurring and is independent of all others and that each occurs in a non-overlapping time interval (i.e. its own). This means that in a time interval $\delta t$ (a) the chance of only one event being recorded is $\lambda \delta t$ where $\lambda$ is a positive number, (b) The probability that zero events happen is then $1-\lambda\delta t$ and (c) the probability that two or more events occur is zero, which means that $ \delta t$ has to be very small. The constant $\lambda$  has units of 1/time and is usually called the rate constant.
The chance of no event occurring at time $t\to t+\delta t$ (where $t$ is fixed) is
$$\displaystyle p(t+\delta t)=(1-\lambda)\delta tp(t)$$
which is the product of the chance of no events up to time $t$ and of no events in time $t\to\delta t$. Rearranged this is
$$\displaystyle \frac{p(t+\delta t)-p(t)}{\delta t}=-\lambda p(t)$$
and in the limit of small time increment is the first order differential equation;
$$\displaystyle \frac{dp(t)}{dt}=-\lambda p(t)$$
Integrating and multiplying by $N_0$ the number of molecules initially present that can react, fluoresce or decay radioactively give the familiar first order decay, $N=N_0e^{-\lambda t}$ where $\lambda$ is the rate constant for the process and its reciprocal the lifetime.
In your example the emission is long lived because the transition is 'forbidden'. Usually this is because the spin of the ground state is different to that of the excited state in which the molecule is now in. The spin forbidden process can be overcome by an interaction that changes the angular momentum such as spin-orbit coupling and if this is weak the excited state decays slowly. Paramagnetic species or heavy atoms will induce spin-orbit coupling.
Initial absorption is usually a spin allowed process but in the excited state interactions can cause a fraction of molecules to cross to a triplet state (spin 1) so a transition is forbidden to the ground state (spin 0) in an aromatic molecule for example. In molecules containing transition metals the spin state can be complicated by the spin of the metal and in general then we would refer to emission as luminescence, and keep fluorescence for spin allowed process and phosphorescence for triplet to singlet transitions.


text:
Maintaining our supersaturated oceans
Supersaturates solutions can be unstable and decay beautifully once a seed crystal is introduced.
Our oceans near the surface are supersaturated with respect to calcium carbonate. This map shows the solubility product of calcium and carbonate ions, scaled such that a value of one indicates saturation with respect to aragonite. The thermodynamically stable form of calcium carbonate is actually calcite which means the supersaturation level is even higher. Falling levels of supersaturation are a climate change concern for the welfare of calcium-shelled organisms.
What prevents the seed crystals (of which there are plenty) from removing this super-saturation?

Your question is addressed in detail in a 1984 paper titled "Apparent calcite supersaturation at the ocean surface: Why the present solubility product of pure calcite in seawater does not predict the correct solubility of the salt in nature?" by Cooke, Robert C., and Paul E. Kepkay. Check Google Scholar for this title.

If there was oversaturation, sea creatures forming aragonite shells would not have recently observed troubles to form them.

Hint: water evaporates at the surface. What happens if crystals do form in deeper layers?


text:
How to highlight the substructure of a molecule with thick red lines in RDKit as SVG (high res)
I have the following code:
from rdkit import Chem
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem import rdDepictor
from rdkit.Chem.Draw import rdMolDraw2D
from IPython.display import SVG

m = Chem.MolFromSmiles('c1cc(C(=O)O)c(OC(=O)C)cc1')
substructure = Chem.MolFromSmarts('C(=O)O')
print(m.GetSubstructMatches(substructure))
m

Which produces the following plot.

However the code above doesn't produce the high resolution image.
I'd like to have the SVG.
I tried this:
drawer = rdMolDraw2D.MolDraw2DSVG(400,200)
drawer.DrawMolecule(m,highlightAtoms=m.GetSubstructMatch(Chem.MolFromSmarts('C(=O)O')))
drawer.FinishDrawing()
svg = drawer.GetDrawingText().replace('svg:','')
SVG(svg)

But I get:

What's the right way to do it?
The code can be tested in my Google Colab.

In your code for SVG you use GetSubstructMatch instead of GetSubstructMatches so only one match is found.
To get all matches you have  to use GetSubstructMatches and then transform the matches in one single tuple for the highlights.
from rdkit import Chem
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem.Draw import rdMolDraw2D
from IPython.display import SVG
from itertools import chain

m = Chem.MolFromSmiles('c1cc(C(=O)O)c(OC(=O)C)cc1')

sub = m.GetSubstructMatches(Chem.MolFromSmarts('C(=O)O'))
print(sub)
((3, 4, 5), (8, 9, 7))

allsubs = tuple(chain.from_iterable(sub))
print(allsubs)
(3, 4, 5, 8, 9, 7)

Now use allsubs and you get the your image.
drawer = rdMolDraw2D.MolDraw2DSVG(400,200)
drawer.DrawMolecule(m,highlightAtoms = allsubs)
drawer.FinishDrawing()
svg = drawer.GetDrawingText().replace('svg:','')
SVG(svg)

It may be useful to work RDKit's cookbook from top to bottom.  I.e., to start with drawing a molecule
# source: https://www.rdkit.org/docs/Cookbook.html
from rdkit import Chem
from rdkit.Chem.Draw import IPythonConsole
from rdkit.Chem import Draw
IPythonConsole.ipython_useSVG=True

def mol_with_atom_index(mol):
    for atom in mol.GetAtoms():
        atom.SetAtomMapNum(atom.GetIdx())
    return mol

# Test in a kinase inhibitor
mol = Chem.MolFromSmiles("C1CC2=C3C(=CC=C2)C(=CN3C1)[C@H]4[C@@H](C(=O)NC4=O)C5=CNC6=CC=CC=C65")
# Default
mol

For me, it works good enough (though in a Jupyter notebook I prefer the output as .png):

Then, running your example as a subsequent cell
# your example
m = Chem.MolFromSmiles('c1cc(C(=O)O)c(OC(=O)C)cc1')
substructure = Chem.MolFromSmarts('C(=O)O')
print(m.GetSubstructMatches(substructure))
m

both yields
((3, 4, 5), (8, 9, 7))

as well as the captured illustration below:

Note, it might be that RDKit packaged for your OS lags a little behind the one packaged by Miniconda (cf. tangents in this answer).  On occasion, this may affect some of the functionality/syntax at your disposition.


text:
Formation of α,β-unsaturated ketone from tertiary propargyl halide

My attempt: I went ahead with an $\mathrm{S_N1}$ type reaction followed by the attack of $\ce{OH-}$ on the 3 degree carbocation formed.
This does not match with the answer given (see image below), in which they have first formed a cumulated diene with the $\ce{OH}$ on the benzylic position. This enol tautomerizes to give an α,β-unsaturated ketone.

The aryl-vinyl carbocation should be more unstable than a 3 degree carbocation with 6 α-hydrogens, so I don't understand why the major product is not formed by the attack of $\ce{OH-}$ on the 3 degree carbocation. What's the proper mechanism here?

Consider an SN2 process with OH attack at the benzylic position. Is the benzylic position less sterically crowded than the tertiary centre?

But why can't we consider SN1 mechanism? Br- is a moderate leaving group and we are getting a 3° carbocation. Also we are getting a unstable cummulated Diene in the SN2 one which is unstable though it does get converted keto form but isn't the first reasoning better? Am I ignoring something important?

I don't think it is possible to assign one mechanism over the other. Both are possible pathways and both probably occur. What you cannot dispute is the identity of the product. Remember, all the mechanisms we draw for these reactions are only models.

Related: https://en.wikipedia.org/wiki/Meyer%E2%80%93Schuster_rearrangement

@Shreyans if you got your answer, feel free to self answer it. Self answer are always appreciated and it also help future readers.


text:
How do I balance a hydrogen peroxide redox equation?
I have this equation:
$$\ce{H2O2 + MnO4- + H+ -> O2 + Mn^2+ + H2O}.$$
How do I balance it, since O has two different oxidation states on one side?

Check out Karsten’s answer here: https://chemistry.stackexchange.com/a/128423/79678.

You separate the reduction and oxidation half-equations, and then there is no issue with the oxidation state of oxygen on either side.
For the oxidation half-equation you start with (in acid solution)
$\ce{H2O2 -> O2}$
Clearly the hydrogen on the left is given off as $\ce{H^+(aq)}$ and the two oxygen atoms, being oxidized from -1 to 0 each, lose two electrons total. So the balanced oxidation half-reaction is
$\ce{H2O2 -> O2 + 2H^+ + 2 e^-}$
The reduction half-reaction should be no issue as it is just Mn(VII) being reduced to Mn(II). Then, as usual for redox reactions, identify multiples of the two half-reactions that balance the electrons, combine and cancel the balanced electrons to get your final result.


text:
How much salt in a cup of water would it take to neither hydrate nor dehydrate the average person?
I don't know a lot about chemistry, and this question came up during lunch with my friend. I was interested in this question so I thought I could ask here. Sorry if I didn't use the right tags.

Isotonic saline, which is 9.0 g salt per liter (= 2.1 g/ 8-oz. cup) is used to maintain fluid balance. But I don't know if "maintaining fluid balance" means neither hydrating nor dehydrating--e.g., it could be hydrating since, to maintain flluid balance, you need to make up for fluid loss.  That might be a question better posed on Biology SE.

Is the saline for drinking, intravenous injection or taking a bath?

This is more of a human physiology question. As theorist said, theoretically, an isotonic saline solution (0.9% NaCl) wouldn't hydrate or de-hydrate a person, and it prepared at this concentration because it matches the osmolarity of blood plasma. But it may be administered intravenously to patients in a hypotonic dehydration state, for example, so the amount of salt in a cup of water--or in a solution administered to a patient, for that matter--would have to vary, depending on the hydration state of the person. Also, from an osmosis ("hydration") point of view, many other substances can have the same effect on the body as salt (e.g.: sugar).


text:
How does part of frozen liquid ( like soup) boil and still be frozen at same time if temperature stays same while melting?
I was reading the following basic question about temperature change during phase change (During phase change in matter, why doesn't the temperature change?), and the answers provided clearly explained the concept. However, in everyday life, for instant a pot of frozen soup, this does not always prove true. Part of the soup in the pot can be boiling, while there are parts that are still frozen. They cannot possibly all be the same or nearly the same temperature. How can this be explained thermodynamically?

You are not at equilibrium so cannot use normal thermodynamics. Thermal diffusion is  slow compared to the rate of heating  is giving rise to what you observe, usually there is melting round outsides of the pan and solid in the middle.

Without even mentioning thermodynamics, doesn't this look even obvious to you?

I was reading the following basic question about temperature change during phase change (During phase change in matter, why doesn't the temperature change?), and the answers provided clearly explained the concept.

One of the answers in the link spells out the condition for the temperature not changing: "Mix water and ice in a container. Keep stirring the mix while the ice is melting." Ideally, the heat transfer is slow compared to the mixing so that the temperature is identical throughout the sample. When you try to find the melting or boiling point of a substance, that is your goal.

However, in everyday life, for instant a pot of frozen soup, this does not always prove true. Part of the soup in the pot can be boiling, while there are parts that are still frozen.

Yes, your observation is correct. This is not an ideal experiment but an everyday observation. When you want to eat warm or hot soup, you are not trying to determine the melting point of water (it is not pure water anyway). So you just crank up the heat to get your soup hot as fast as possible without burning any of it.

They cannot possibly all be the same or nearly the same temperature. How can this be explained thermodynamically?

Different parts of a sample can have different temperatures. In those cases, the system will never be at equilibrium. Most of the well-defined thermodynamics are about equilibrium states, so there is no contradiction of theory and observation here. Even when you melt pure ice very slowly, it is not an equilibrium process; however, you can describe it partially through equilibrium thermodynamics because you are very close to equilibrium in that case. When you boil soup while there are still chunks of ice in it, you are not.

The system you describe is not at thermodynamic equilibrium.  The system must be spatially uniform only if it is at thermodynamic equilibrium.


text:
How does change in temperature affect experimental calculation of molar mass
This is coming from AP Chemistry 2009 q2 f. To sum it up, a student undercalculates the molar mass of a gas that is flushed into a sealed flask of dry air using DRT/P, and you have to justify if the following occurrence is responsible for the error. "The temperature of the air was 23.0°C, but the temperature of the CO2(g) was lower than the reported 23.0°C." The explanation provided says that it would not have been responsible because the result would be greater density and then larger mass therefore higher calculated molar mass. My question is if the temperature increases the volume and therefore the density, would it not also increase the pressure, balancing the increase in density out, and create the same calculated molar mass?

If you know the temperature, it works at any temperature. If you plug in the correct density but the wrong temperature, the resulting molar mass will be off.

My question is if the temperature increases the volume and therefore the density

Let's stop here for a second. Volume is inversely proportional to density according to the equation:
$$D = \frac{m}{V}$$
Therefore, if the temperature increases the volume as you had said, the density would actually decrease. You are right about how increasing the temperature will also increase the pressure. Now, looking at the equation:
$$M=\frac{DRT}{P}$$
You can see that a decrease in density and increase in pressure would actually decrease the molar mass. I hope this would explain it.

I'm not aware how you get the gas into a sealed container. It seems that a weighed vessel of air of known volume at a specified measured T was flushed with a gas and resealed at atmospheric pressure: constant P constant V but the gas was colder so that means there were more molecules in the volume more mass and a higher MW.  If the results were too low the more reasonable source of error would be impure gas mix in the sample, air or water.


text:
How can semicarbazide occur during meat processing other than from nitrofuran treatment/feeds?
Since nitrofurans are banned e.g. in the EU from use in food-producing animals, but not entirely as medications for humans, there have been some stories/reports of possible diversion of human-intended medications for animal use, e.g. in Portugal in the previous decade.
In 2018 the USDA ran an investigation apparently prompted by the import ban of a shipment of US chicken on such grounds, but concluded that the semicarbazide (SEM) byproducts were likely not from the animals being treated with nitrofurans but "by-products formed during food processing".

While the detection of SEM in pre-intervention samples could have been indicative of nitrofurazone use, its absence in pre-intervention samples in this study suggests that the subsequent detection of SEM in the sampled products is not indicative of nitrofurazone use and may be a result of by-products formed during food processing.

However the paper is merely discussing sampling and testing, but offers no more detailed mechanism of how semicarbazide could have resulted during food processing. So, how could that have happened?
(I see that azodicarbonamide (E927) is a flour treatment and this breaks down to SEM as well, but I haven't read of this being ever added to meat. E927 is banned in the EU too, but not in the US. Slightly more usefully, there's a paper that discusses SEM occurence in some processed dairy products (whey & milk protein concentrates) and crustacean products (shrimp and prawn powders). Although not mentioned in the abstract, inside the paper, chicken powder and egg powder are discussed/measured as well. The [chemical] gist of that seems to be "The reaction of urea with hydrazine, favoured at
higher temperatures > 80°C, can lead to the formation of
SEM." Still I'm not sure how/if that has any relevance to raw chicken meat, which is what the USDA tested.)

Amusingly perhaps, this seems to be related to the "chlorinated chicken" controversy. Towards the very end of the powders paper (that I mentioned in my question), a 2004 paper is cited which found that compounds used in cleaning equipment and/or the meat itself may cause SEM as a byproduct.

A significant formation of SEM was observed in samples treated with hypochlorite commonly used in food processing for disinfection or bleaching. SEM was formed in different kinds of nitrogen compound-containing samples (0.3–20 μg kg−1) after treatment with 1% active chlorine. It was detected in the mg kg−1 range after hypochlorite treatment (0.015% active chlorine) of creatine. Lower levels were also formed from creatinine, arginine and urea.


text:
How to know the group number from the ionization energy?

The successive ionization energy of the first four electrons of a representative element are $\pu{738.1 kJ/mol}$, $\pu{1450 kJ/mol}$, $\pu{7730 kJ/mol}$ and $\pu{10500 kJ/mol}$. Characterize the element according to the periodic group.

From this data page, I predicted that the element is $\ce{Mg}$ which belongs to Group 2. But I don't have a systematic way to predict the group number or the electron configuration from the question. Is there any approach for this kind of question ?

(1) Note that units like kJ/mol should be written in upright, not italics. It's not correct to simply place them in math mode $...$ without anything else; either use $1000~\mathrm{kJ/mol}$, or perhaps more conveniently, $\pu{1000 kJ/mol}$. See https://chemistry.meta.stackexchange.com/q/86/16683, especially section 3.3, for more info. (2) When copy-pasting links from Wikipedia, please try to edit out the links to the mobile version, this is not good for people using it on a desktop. You can do this by changing en.m.wikipedia.org to en.wikipedia.org.

Another way is using the Wikipedia Android application, (guessing iPhones have one too), which provides desktop links.

The particular values are not that important. Their pattern is. Notice the big relative step between the 2nd and the 3rd ionization energies.
The first two values are relatively small as these electrons are quite shielded by more inner electrons.
The other two need much more energy to be taken away, being shielded much less, as the  "peer electrons" from the same orbital group do not shield the nucleus charge much for each other.
That means the first two electrons have started another element period with another quantum number $n$.
So the element group is the second group.


text:
What Chemical Database Extensions for Structure and Similarity Search are there?
I am looking for extensions/add-ons for relational databases (RDBMS) that enable storing and handling chemical structures including the possibility for substructure searches and similarity search.
I am currently dealing with a custom application that is built on top of an Oracle 11gR2 database with the OrChem extension. In the words of the accompanying scientific publication, OrChem is

[..] an extension for the Oracle 11G database that adds registration and indexing of chemical structures to support fast substructure and similarity searching.

This is what the application needs.
The problem is that OrChem does not seem to be maintained anymore, the  last release happened almost 10 years ago. From what little I could find in my research, I got the impression that OrChem also only seems to play nice with Oracle 11G, which will soon reach the end of its extended support lifecycle.
So, since I have the need to upgrade the underlying database, I am looking for information on alternatives (or, if someone can offer, experiences with running OrChem with versions >= Oracle 12).
For possible alternatives, projects that I could find are

MyChem, an extension similar to OrChem for the MySQL database (versions >= 4.0) and MariaDB. Of the projects I could find, this one is the least outdated, but even that means that the last release happened 6 years ago.

Pgchem::tigress, a cheminformatics extension for the PostgreSQL DBMS (version > 8.x). Similarly to OrChem, the last commit happened almost 10 years ago.

chemsearch, for which again, the last release happened about 10 years ago. This one seems to be relatively mature (version number is 3.2) and it supports at least the Oracle and PostgreSQL RDBMS, the source code zip also has folders named mssql and mysql. However, since I could not find a real documentation, I am unsure which versions of the different RDMBS are actually supported.

Can anyone offer deeper insights concerning the compatibility of the listed cheminformatics extensions with concrete RDBMS versions and/or more current alternatives that are not listed here?

You do not specify the type of molecules.  If these are small and deposit as SMILES strings, you may consider to compute fingerprints for a subsequent similarity search.  See, e.g., this entry from the OpenBabel documentation, or Greg Landrum's presentation at RDKit user group meeting 2012 here.

It seems that my search was missing the correct keyword. The database extension modules I was looking for are usually called cartridges.
A search for rdbms chem cartridge yielded a few more recent options:

Bingo (open source, for Oracle, PostgreSQL and MSSQL Server)
RDKit PostgreSQL cartridge (open source, for PostgreSQL)
JChem Cartridge for Oracle and JChem PostgreSQL Cartridge (commercial, for Oracle and PostgreSQL)
MolCart (commercial, for MySQL)

This list is not meant to be exhaustive, just thought I'd share what I could find already with little effort.

Just ran into NextMove's Arthor: https://www.nextmovesoftware.com/arthor


text:
How much nitrogen in fertilizer
I’m trying a little exercise for fun in chemistry but I don’t find where I’m wrong.
Here it is:
In $\pu{2 kg}$ of a fertilizer, $\pu{39.7 mass\%}$ is $\ce{(NH4)2SO4}$, and $\pu{15.2 mass\%}$ is $\ce{NH4NO3}$.
How much nitrogen (in $\pu{kg}$) contains this sample?
So I calculate the molar mass of both, which is $M(\ce{(NH4)2SO4}) = \pu{132 g/mol}$ and $M(\ce{NH4NO3}) = \pu{80 g/mol}$.
And then I calculate the number of mol of both so there is 39.7/132 = 0.3 mol and 15.2/80=0.19 mol of each substance
And then I figure out how much mol of nitrogen there is, so in 1 mol of $\ce{(NH4)2SO4}$ there is 2 mol of N and same for the other
So finally we have $2 \cdot 0.3 + 2 \cdot 0.19 = 0.98$ mol of N in the whole fertilizer and finally the mass of N is $\pu{0.98 mol}/\pu{14 g mol-1} = \pu{0.07 g}$.
And multiply by $\pu{2 kg}$ of fertiliser, we have $\pu{0.14 kg}$ of nitrogen.
But it seems the true answer is $\approx \pu{270 g}$.
So where is it wrong ?

First thing wrong was inconsistency of unit usage. Other things just followed as playing with numbers and wrong units seemed right.

(39.7 g/100 g) / ( 132 g/mol ) = (0.3 mol/ 100 g)

To get a mass from the amount of substance, you have to multiply by the molar mass. If you carry the units, you would see that dividing by the molar mass is incorrect for that step.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

There are two mistakes in your calculation. Everything is correct to the point where you get $\pu{0.98 mol}$ of $\ce{N}$. You do not mention it, but this value is valid for a total mass of fertilizer of $\pu{100 g}$, because you started from $\pu{39.7 g}$ of fertilizer. And these $\pu{39.7 g}$ correspond to  $\pu{39.7 wt\%}$ of $\pu{100 g}$.
Now first mistake is the calculation of the mass of $\pu{0.98 mol}$ of $\ce{N}$. As $\pu{1 mol}$ of $\ce{N}$ weighs $\pu{14 g}$, $\pu{0.98 mol}$ weighs $\pu{0.98 mol} \cdot \pu{14 g/mol} = \pu{13.72 g}$. This is the amount of nitrogen in $\pu{100 g}$ of fertilizer.
Then, $\pu{2 kg}$ contains $20$ times $\pu{100 g}$. Here is your second mistake. So the total amount of nitrogen is $20 \cdot \pu{13.72 g} = \pu{274.4 g}$.

Note, $\pu{wt\%}$ describes percentage by mass equally known as mass fraction.  There equally are $\pu{mol\%}$ about mole fraction and $\pu{vol\%}$ about volume fraction, too.

I suggest to carry the units whenever possible within reason.  A subsequent dimensional analysis is one of the tools available to check the results for plausibility and chemistry.se may be a vector to disseminate this technique to its readers.  On chemistry.se, the \pu{} syntax works reasonably well within equations, too; the occasional escaping of % is a small investment in addition.


text:
Charge polarisation in acetamide molecule due to different lithium salts (varying anions)
I am trying to understand how will acetamide molecules behave in the presence of lithium salts with different anions (nitrate, bromide and perchlorate).
Among the three, it is evident that bromide is the most electronegative, therefore should induce maximum polarisation on acetamide, am I right?
Secondly, what will happen to the atomic partial charge distribution of the acetamide molecule in the presence of these anions (and of course cations)? Will the amide group become more positive and the carbonyl group become more negative?
PS: I am trying to get a crude partial charge distribution (without making quantum calculations) to run molecular dynamics simulations of mixtures of acetamide with these salts.

You mean what happens if you dissolve lithium salts in acetamide? Thinking about some vague polarisation does no good here. Li coordinates to carbonyls, and original ligands, amide groups hydrogen bond with these ligands and carbonyls, and that's still a crude description.

Yes, in this case dissolving lithium salts at a particular mole ratio in acetamide makes it a eutectic mixture. I am trying to simulate them. You're right about the fact that lithium coordinates with carbonyl and the anions H-bond with the amide group, but apart from there is also significant charge transfer, which can ideally be obtained from QM calculations, but I am just wondering there is some hand-waving arguments to figure this out.

Anions can be compared based on how weakly coordinating they are. Less coordinating anion makes coordination with carbonyls more preferable.

I am sorry, don't the anions coordinate with the amide group in this scenario?

I meant with lithium - amide displaces anions, particularly weakly coordinating, like perchlorate. Anions also hydrogen bond with N-H, but interaction is pretty weak.


text:
Is this a valid interpretation of the Mean relative speed between two (Maxwell-Boltzmann) species?
In textbooks, the collision frequency $Z_{ab}$ is always given as something like:
$$Z_{ab} = n_an_b\sigma_{ab} \Bigg(\frac{8 k T}{\pi \mu}\Bigg)^{1/2}= n_an_b\sigma_{ab} \bar{C}_{rel}$$
where the reduced mass $\mu = m_am_b/(m_a+m_b)$. Given that the mean speed of a species "i" is given by $\bar{C}_i = \Big(\frac{8 k T}{\pi m_i}\Big)^{1/2}$, then, algebraically,  the first equation is equivalent to:
$$Z_{ab} = n_an_b\sigma_{ab} \sqrt{\bar{C}_a^2 + \bar{C}_b^2}$$
Is there a reason that this is not a valid interpretation of the collision frequency? I have not found a single source which presents either the collision rate $Z_{ab}$, or the mean relative speed $\bar{C}_{rel}$ in this way.

Mathematically it is the same with you definition of $C_i$.

I posted the same question on the physics forum to see if it would get any more traction. https://physics.stackexchange.com/questions/674124/is-this-a-valid-interpretation-of-the-mean-relative-speed-collision-rate-of-part


text:
Why do we need Equilibrium Constant as well as Acid Disocciation Constant?
I did the experiment to find out the dissociation constant of a weak acid using Henderson equation. However, after going through the theoretical part of the experiment, I am wondering why we need two separate equations for finding out the Acid Dissociation constant and the Equilibrium constant.
Simply put, the expression for dissociation constant and equilibrium constant is identical. As such, when an acid dissociates, we could have explained it using the equilibrium constant equation as well; then, why do we need a new term called Acid Dissociation constant for explaining the same?
One explanation for this question that I can think of right now is - we can use the equation for equilibrium constant only when an equilibrium is achieved, and the concentration of acid, i.e., [Acid] must be present in the denominator of the equilibrium constant equation. However, we can use the equation for acid dissociation constant irrespective of the fact that the reaction reaches an equilibrium or not. So, the presence of the [Acid] term in the denominator is of little significance in case of strong acid since they get dissociated completely, however, it will still have a great significance for dissociation of a weak acid.
I have got an explanation in terms of activity from this answer, but is there a simpler explanation? A more intuitive one will help me understand the concept better.

Acid dissociation constant = equilibrium constant of acid dissociation.

@Poutnik, Okay. But, why do we need the acid dissociation constant equation to calculate the dissociation constant of the weak acid? Can't it be done using equilibrium constant equation?

See the first comment. They are synonyms. The same equation using different synonyms is still the same equation. // Provide eventual equations you refer to.

The answer by Lautman has provided the required equations. I guess he has made your point clearer.

It was your job to make your question clear.

Does this answer your question? Acid dissociation constant and equilibrium constant

It does, but in terms of activity. I wanted an intuitive one

For any system at equilibrium we can define an equilibrium constant as follows:
$$\ce{A(aq) + B(aq) <=> C(aq) + D(aq) }$$
$$K_\text{eq} = \frac{[\ce{C}][\ce{D}]} {[\ce{A}][\ce{B}]}$$
When the solutions are aqueous, we might use the subscript $c$ for the $K$, to indicate concentration.  When we are using gas pressure instead of concentration, we use the subscript $p$.
In the case of acid dissociation, it's just another reaction at equilibrium with $\ce{H3O+}$ in the products.
$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A^-(aq)}$$
$$K_\text{a} = \frac{[\ce{H3O+}][\ce{A^-}]} {[\ce{HA}]}$$
The term "acid dissociation constant" is used instead of "equilibrium constant" for added clarity.  In this situation they mean the same thing.

So, having a hydronium is essential in case of acid dissociation constant equation? We can identify an acid dissociation reaction explicitly by the presence of hydronium ion?

Yes.  The presence of hydronium as a product is what makes it an acid dissociation.


text:
Are there any binary ionic compounds with carbon in it?
It may be organic, may be inorganic but is ionic and binary in nature and contains carbon. Is there any such compound?

Metal carbides, even of alkali metals, likely have a good deal of covalent character. But see this answer.

Also related: https://chemistry.stackexchange.com/questions/40598/has-a-carbon-compound-ever-been-found-having-an-ionic-bond/40600#40600

I think that they are probably asking about simpler ionic compounds with carbide anions.  Like sodium carbide.

Does this answer your question? Has a carbon compound ever been found having an ionic bond?

The well known compound calcium carbide is an ionic compound containing carbon (as $\ce{C2 ^2-}$ units).
The compound was once very common as its reaction with water produces acetylene which was used in lamps and even early car headlights. It is also used in some chemical processes.
The invention of an easy way to make it in the late 19th century was a major part of the industrial revolution in the chemical industry.
It is probably the best known ionic compound containing carbon.

Aluminum carbide, $\ce{Al4C3}$, is ionic and reacts strongly with water according to : $\ce{Al4C3 + 12 H2O -> 4 Al(OH)3 + 3 CH4}$


text:
How is formula for mean ionic activity derived from its constituent activities?
I spent hours in order to understand the following - but without success:
At many places the mean ionic activity coefficient  for the reaction
$$\ldots  \rightarrow \nu_A A^{z_A+} + \nu_B B^{z_B-}$$
is defined as
$$\gamma_{\pm} = \left(\gamma_A^{\nu_A}\gamma_B^{\nu_B}\right)^\frac{1}{\nu_A+\nu_B}$$
The single activity coefficients for A and B are, using Debye-Hückel's law and the ion strength I:
$$\log \gamma_A = -A \cdot z_A^2 \cdot \sqrt I$$
$$\log \gamma_B = -A \cdot z_B^2 \cdot \sqrt I$$
This means
$$\gamma_A = 10^{-A \cdot z_A^2 \cdot \sqrt I}$$
$$\gamma_B = 10^{-A \cdot z_B^2 \cdot \sqrt I}$$
and
$$\gamma_A^{\nu_A} = 10^{-A \cdot z_A^2 \cdot \sqrt I \cdot \nu_A}$$
$$\gamma_B^{\nu_B} = 10^{-A \cdot z_B^2 \cdot \sqrt I \cdot \nu_B}$$
Finally, this gives
$$\gamma^{\pm} = \left(10^{-A \cdot z_A^2 \cdot \sqrt I \cdot \nu_A} \cdot 10^{-A \cdot z_B^2 \cdot \sqrt I \cdot \nu_B}\right)^\frac{1}{\nu_A+\nu_B}$$
$$\gamma^{\pm} = 10^{-\frac{A \sqrt I \cdot (z_A^2 \nu_A+z_B^2 \nu_B)}{\nu_A+\nu_B}}$$
$$\log \gamma^{\pm} = -\frac{A \sqrt I \cdot (z_A^2 \nu_A+z_B^2 \nu_B)}{\nu_A+\nu_B}$$
But I read always
$$\log \gamma^{\pm} = -A \cdot |z_A \cdot z_B| \cdot \sqrt I$$
So what's going on here?
EDIT:
According to the hint below (see comment)
For charge balance I have to set
$$z_A=\nu_B$$
$$z_B=\nu_A$$
and so I get for the factor on the right
$$\frac{z_A^2 \nu_A+z_B^2 \nu_B}{\nu_A+\nu_B} = \frac{z_A^2 z_A+z_B^2 z_A}{z_A+z_B}= z_A z_B$$
So it is solved.

Don’t you have charge neutrality on the right hand side of your first expression? Seems like that provides the missing piece.

@Ed V: You mean what I added to my question at the end (its easier to write there): how embarrassing it is to me - I've fiddled for hours and then it's so easy...

No biggie: glad to have helped a bit!

What a shame...sometimes I feel like an idiot afterwards ;-)

Happens to us all!


text:
Modified Arrhenius equation in the presence of a heterogeneous catalyst
Suppose we are doing heterogeneous catalysis, for example, between nitrogen and hydrogen to create ammonia in the presence of a solid catalyst, like rhuthenium. It's obvious that the gas molecules don't "collide" directly and, therefore, the Arrhenius equation does not completely apply. Does a modified version of the equation exist that takes this into account?
My guess would be that since catalysts generally decrease the activation energy of a reaction, the activation energy value would possibly have a fractional component to it. Furthermore, because the reaction is occurring on a surface, the decreased direct collisions may affect the rate. I would greatly appreciate it if someone could point me in the right direction.

Being catalyzed is not the major modification factor. The major factor is becoming a reaction on solid surface, instead in free volume.

I would start with the 2D model of the multistep reaction mechanism, each step having its own activation energy. ( so does not catalyzed reaction if free space )


text:
Reaction of Grignard reagents with phosphoric acid derivatives
The use of Grignard reagents that people are most familiar with is the production of tertiary alcohols by reaction with a suitable carbonyl compound, whether it be an ester, ketone, or aldehyde. As I understand it, Grignard reagents are strong nucleophiles and will readily attack any electrophilic centers. If so, how would a reagent having the formula RMgX react with a phosphoric acid derivative $\ce{(R^1)(R^2)(R^3)P=O}$, where $\ce{R^1}$, $\ce{R^2}$, and $\ce{R^3}$ can be either alkyl, alkoxy, $\ce{-OH}$, or $\ce{-H}$? Does it result in simply $\ce{(R3)P=O}$, a mixture of products where one or more of the $\ce{R}$ groups in the phosphoric acid derivative is replaced by that of the Grignard reagent, or something else? Is the result controlled by stoichiometry?

According to this JOC paper by Hays,[1] diethyl phosphonate $\ce{(EtO)2P(O)H}$ has been reacted with alkyl or aryl Grignards to give secondary phosphines oxides and unsymmetrical tertiary phosphine oxides.
According to this JOC paper by Berlin and Peterson,[2] trimethyl phosphate $\ce{(MeO)3PO}$ has been reacted with phenyl Grignard to give methyl diphenylphosphinate $\ce{Ph2P(O)OMe}$ and triphenylphosphine oxide.
It is worthy of note that this patent[3] describes the use of triethyl phosphate $\ce{(EtO)3PO}$ as a solvent for the preparation of Grignard reagents.
References:

Hays, H. R. Reaction of diethyl phosphonate with methyl and ethyl Grignard reagents. J. Org. Chem. 1968, 33 (10), 3690–3694. DOI: 10.1021/jo01274a003.
Berlin, K. D.; Peterson, M. E. Nucleophilic displacement reaction on organophosphorus esters by Grignard reagents. II. Evidence for competitive displacement on carbon and phosphorus in the reaction of Grignard reagents with simple phosphates. J. Org. Chem. 1967, 32 (1), 125–129. DOI: 10.1021/jo01277a032.
Matsui, M.; Kitahara, T.; Yabuta, G.; Yamada, T. Triethyl phosphate as the solvent for the preparation of Grignard reagents. Proceedings of the Japan Academy. Ser. B: Physical and Biological Sciences 1999, 75 (2), 35–36. DOI: 10.2183/pjab.75.35.


text:
Nitro as a leaving group in an aromatic ring
I came across a chemistry problem posted on Discord. The objective is to propose a reagent that will enable the following reaction to occur:

One proposed answer for the mysterious Reagent D is sodium methanethiolate.The claim is made on the basis of the fact that patent US6777575B2 describes this exact sort of synthesis and calls for using MeSNa as a reagent in this situation. A counterargument is that the nitro group is stabilized through conjugation, and that methoxy or trifluoromethyl would be a better leaving group in this situation. Which claim is correct? In the reactant, can nitro act as a leaving group and allow for substitution with methanethiolate? Why or why not?

It sure is usual leaving group in plenty of SNAr reactions. CF3 isn't better and exchanging methoxy wouldn't be thermodynamically favorable afaict.

The nitro group (or nitrite anion) can indeed act as a leaving group in nucleophilic aromatic substitution, especially if the ring is electron-deficient (as it is in this problem). Wikipedia offers another example, also using an electron-deficient ring:

In the compound methyl 3-nitropyridine-4-carboxylate, the meta nitro group is actually displaced by fluorine with cesium fluoride in DMSO at 120 °C.[1]

What enables displacement of the nitro group is its electronegativity, combined with the fact that it actually does not conjugate all that well with the ring. Conjugation of the nitro group to the ring would require increased formal charge separation with a positive charge being imparted to the ring, which becomes more unfavorable if the ring is electron-deficient.
Cited Reference

Tjosaas F, Fiksdahl A (February 2006). "A simple synthetic route to methyl 3-fluoropyridine-4-carboxylate by nucleophilic aromatic substitution". Molecules (Basel, Switzerland). 11 (2): 130–3. https://doi.org/10.3390/11020130. PMC 6148553. PMID 17962783.


text:
Dehydrating hydrated Magnesium chloride
I came to know that hydrated Magnesium chloride $\ce{MgCl2.6H2O}$ is dehydrated not directly by heating as it gets hydrolysed to $\ce{MgO}$ . But instead it is dehydrated by passing through $\ce{HCl}$ gas at $\pu{650 K}$. But on doing so how is formation of $\ce{MgO}$ prevented?

It is prevented by shifting the equilibrium
$$\ce{MgCl2(s) + H2O(s,g) <=> MgO(s) + 2 HCl(g)}$$
to the left, by removal of $\ce{H2O}$ and providing enough $\ce{HCl}$.

Basically I agree, but do we get MgO? Could form the hydroxide or even Mg(OH)Cl.

At 650  K and in HCl stream? I doubt that. Mg(OH)2 reportedly decomposes at 350°C. I bet Mg(OH)Cl would not held the stand much better, even if I do not reject possibility of temporary a/o minor occurrence.


text:
NMR spectrum with impure solvents
How to find the proton signals of CHCl3 impurities in NMR spectrum recorded in CDCl3? What are their intensities?

You can look up the chemical shift. The relative intensity of signals from different molecules depends on their relative concentrations.

Nobody would really consider CHCl3 to be an "impurity" in this context - it is almost universally known, and accepted, that deuterated solvents are not 100% deuterated. "Residual protonated solvent" is a more generally appropriate term (notice that the references in Buttonwood's answer also use this phrase). "Impurity" would usually refer to some other species.

A classic and still handy reference to identify signals of impurities in NMR spectra is this compilation: Gottlieb, H. E.; Kotlyar, V.; Nudelman A. NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities in J. Org. Chem. 1997, 62, 7512–7515 (doi 10.1021/jo971176v).  On occasion, you find a selection of this compilation in tables by vendors of NMR solvents (e.g., Sigma Millipore).
Both an update as well as an extension of the former is, e.g., Babij et al. NMR Chemical Shifts of Trace Impurities: Industrially Preferred Solvents Used in Process and Green Chemistry in Org. Process Res. Dev. 2016, 20, 661–667 (doi 10.1021/acs.oprd.5b00417, open access).
Both consider $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectroscopy in separate tables and a range of the more frequently used solvents such as, e.g., $\ce{CDCl3}$, $\ce{d_6}$-acetone, $\ce{D2O}$, etc.


text:
What are the intermolecular forces between the following compounds in a mixture?
I need to figure out the intermolecular force between the following compounds in a mixture:

acetone-aniline,
acetone-chloroform,
ethanol-acetic acid,
acetone-ethanol,
ethanol-water

Can someone please share the type of 'intermolecular' bonding between these compounds and the diagrammatic representation as well?

These images are ignoring the bond angles. Do this look right? What about interactions other then hydrogen bonds and dipole-dipole interactions?

Become familiar with the different types of intermolecular forces (H-bonding, van der waals, etc...). Look at and maybe draw each structure of the listed molecules. Identify the functional groups and try and reason your way to some answer. Then come back and post what you have.

@dval98 I already know those concepts, I am just not sure how these would work for the specified chemical compounds. I have questions such as, where 'dimers' form and would I only show hydrogen bonding or other dipole-dipole interactions as well, will there be other dipole-dipole interactions, etc. Anyway, I have added my attempt to the question by making edits, please take a look at that, and give me some feedback, thank you.

Well, ignoring bond angles, the pictures are right, so what do you wanna know? Other interactions? Well, I imagine there's some halogen bonding between acetone and chloroform, for example.

See also https://en.wikipedia.org/wiki/Van_der_Waals_force

You can do much beyond that. Specialists are still studying how H+ is solvated in water based solutions, just to give you an example of the complexity of the subject.


text:
Maxwell's Relations and Differentiating With Respect to Nᵢ
I was wondering if anyone could write out Maxwell's relations for partial derivatives with respect to particle count $N_i$. Starting from the fundamental thermodynamic relation,
\begin{align}
  \mathrm{d}U(S, V, N_i) &=
    T\mathrm{d}S - P\mathrm{d}V + \sum_{i}\mu _i\mathrm{d}N_i\\
  \mathrm{d}U(S, V, N_i) &=
    \left(\frac{\partial U}{\partial S}
      \right)_{V, \{N_i\}}\mathrm{d}S
    + \left(\frac{\partial U}{\partial V}
      \right)_{S, \{N_i\}}\mathrm{d}V
    + \sum_{i}\left(\frac{\partial U}{\partial N_i}
      \right)_{S, V, \{ N_{j\neq i}\}}\mathrm{d}N_i
\end{align}
I tried to write a relation by differentiating with respect to particle number, but I want to make sure all of my subscripts are correct. I wrote, for a first Maxwell relation,
$$
  \left(\frac{\partial T}{\partial N_i}
    \right)_{S, V, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial S}
    \right)_{V, \{N_{j\neq i}\}} =
  \left(\frac{\partial ^2U}{\partial N_i \partial S}
    \right)_{V, \{N_{j\neq i}\}}
$$
Are these subscripts correct? I just want to make sure this is accurate. If someone could write out the remaining Maxwell relations, that would be great. There should be seven more relations left.

It seems to me what you did is correct.

@ChetMiller Should the middle partial derivative have all $N_i$ held constant or just $N_{j\neq i}$?

You're right.  I missed that.  It should be all of them.

@CalebWilliamsUC, if you got your answer, feel free to self answer it. Self answer are always appreciated and it also help future readers.

@NilayGhosh Alright! I will self-answer a bit later.

Just now getting back around to this.
\begin{align}
  \mathrm{d}U(S, V,
\left \{ N_i \right \}) &=
    T\mathrm{d}S - P\mathrm{d}V + \sum_{i}\mu _i\mathrm{d}N_i\\
  \mathrm{d}U(S, V,
\left \{ N_i \right \}) &=
    \left(\frac{\partial U}{\partial S}
      \right)_{V, \{N_i\}}\mathrm{d}S
    + \left(\frac{\partial U}{\partial V}
      \right)_{S, \{N_i\}}\mathrm{d}V
    + \sum_{i}\left(\frac{\partial U}{\partial N_i}
      \right)_{S, V, \{ N_{j\neq i}\}}\mathrm{d}N_i
\end{align}
$$
  \left(\frac{\partial T}{\partial V}
    \right)_{S, \{N_{i}\}} =
  -\left(\frac{\partial P}{\partial S}
    \right)_{V, \{N_{i}\}} =
  \left(\frac{\partial ^2U}{\partial S \partial V}
    \right)_{ \{N_{ i}\}}
$$
$$
  \left(\frac{\partial T}{\partial N_i}
    \right)_{S, V, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial S}
    \right)_{V, \{N_{j}\}} =
  \left(\frac{\partial ^2U}{\partial N_i \partial S}
    \right)_{V, \{N_{j\neq i}\}}
$$
$$
  -\left(\frac{\partial P}{\partial N_i}
    \right)_{S, V, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial V}
    \right)_{S, \{N_{i}\}} =
  \left(\frac{\partial ^2U}{\partial N_i \partial V}
    \right)_{S, \{N_{j\neq i}\}}
$$
\begin{align}
  \mathrm{d}H(S, P,
\left \{ N_i \right \}) &=
    T\mathrm{d}S + V\mathrm{d}P + \sum_{i}\mu _i\mathrm{d}N_i\\
  \mathrm{d}H(S, P,
\left \{ N_i \right \}) &=
    \left(\frac{\partial H}{\partial S}
      \right)_{P, \{N_i\}}\mathrm{d}S
    + \left(\frac{\partial H}{\partial P}
      \right)_{S, \{N_i\}}\mathrm{d}P
    + \sum_{i}\left(\frac{\partial H}{\partial N_i}
      \right)_{S, P, \{ N_{j\neq i}\}}\mathrm{d}N_i
\end{align}
$$
  \left(\frac{\partial T}{\partial P}
    \right)_{S, \{N_{i}\}} =
  -\left(\frac{\partial V}{\partial S}
    \right)_{P, \{N_{i}\}} =
  \left(\frac{\partial ^2H}{\partial P \partial S}
    \right)_{ \{N_{ i}\}}
$$
$$
  \left(\frac{\partial T}{\partial N_i}
    \right)_{S, P, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial S}
    \right)_{P, \{N_{i}\}} =
  \left(\frac{\partial ^2H}{\partial N_i \partial S}
    \right)_{P, \{N_{j\neq i}\}}
$$
$$
  \left(\frac{\partial V}{\partial N_i}
    \right)_{S, P, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial P}
    \right)_{S, \{N_{i}\}} =
  \left(\frac{\partial ^2H}{\partial N_i \partial P}
    \right)_{S, \{N_{j\neq i}\}}
$$
\begin{align}
  \mathrm{d}F(T, V,
\left \{ N_i \right \}) &=
    -S\mathrm{d}T - P\mathrm{d}V + \sum_{i}\mu _i\mathrm{d}N_i\\
  \mathrm{d}F(T, V,
\left \{ N_i \right \}) &=
    \left(\frac{\partial F}{\partial T}
      \right)_{V, \{N_i\}}\mathrm{d}T
    + \left(\frac{\partial F}{\partial V}
      \right)_{T, \{N_i\}}\mathrm{d}V
    + \sum_{i}\left(\frac{\partial F}{\partial N_i}
      \right)_{T, V, \{ N_{j\neq i}\}}\mathrm{d}N_i
\end{align}
$$
  \left(\frac{\partial S}{\partial V}
    \right)_{T, \{N_{i}\}} =
  -\left(\frac{\partial P}{\partial T}
    \right)_{V, \{N_{i}\}} =
  \left(\frac{\partial ^2F}{\partial V \partial T}
    \right)_{ \{N_{ i}\}}
$$
$$
  -\left(\frac{\partial S}{\partial N_i}
    \right)_{T, V, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial T}
    \right)_{V, \{N_{i}\}} =
  \left(\frac{\partial ^2F}{\partial N_i \partial T}
    \right)_{V, \{N_{j\neq i}\}}
$$
$$
  -\left(\frac{\partial P}{\partial N_i}
    \right)_{T, V, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial V}
    \right)_{T, \{N_{i}\}} =
  \left(\frac{\partial ^2F}{\partial N_i \partial V}
    \right)_{T, \{N_{j\neq i}\}}
$$
\begin{align}
  \mathrm{d}G(T, P,
\left \{ N_i \right \}) &=
    -S\mathrm{d}T + V\mathrm{d}P + \sum_{i}\mu _i\mathrm{d}N_i\\
  \mathrm{d}G(T, P,
\left \{ N_i \right \}) &=
    \left(\frac{\partial G}{\partial T}
      \right)_{P, \{N_i\}}\mathrm{d}T
    + \left(\frac{\partial G}{\partial P}
      \right)_{T, \{N_i\}}\mathrm{d}P
    + \sum_{i}\left(\frac{\partial G}{\partial N_i}
      \right)_{T, P, \{ N_{j\neq i}\}}\mathrm{d}N_i
\end{align}
$$
  -\left(\frac{\partial S}{\partial P}
    \right)_{T, \{N_{i}\}} =
  \left(\frac{\partial P}{\partial T}
    \right)_{P, \{N_{i}\}} =
  \left(\frac{\partial ^2G}{\partial T \partial P}
    \right)_{ \{N_{ i}\}}
$$
$$
  -\left(\frac{\partial S}{\partial N_i}
    \right)_{T, P, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial T}
    \right)_{P, \{N_{i}\}} =
  \left(\frac{\partial ^2G}{\partial N_i \partial T}
    \right)_{P, \{N_{j\neq i}\}}
$$
$$
  \left(\frac{\partial V}{\partial N_i}
    \right)_{T, P, \{N_{j\neq i}\}} =
  \left(\frac{\partial \mu _i}{\partial P}
    \right)_{T, \{N_{i}\}} =
  \left(\frac{\partial ^2G}{\partial N_i \partial P}
    \right)_{T, \{N_{j\neq i}\}}
$$


text:
Is it possible to electrochemically remove the aluminum fouling a steel file?
I have a number of metal files that, to my regret, I have used on aluminum which gets stuck to the teeth.  As suggested on several mechanics forums,  I have tried soaking the files in concentrated NaOH solution without result.
Would connecting a DC source to a file anode and a copper cathode in an acidic CuSO4 solution preferentially remove the aluminum from the file?  If so, what pH and voltage would work best?
If not, is there another method that would work?

You will be at the edge of a problem. Files are very hard steel ( eg Rockwell C 64), Steels hydrogen stress crack when the hardness is above about Rockwell C24.  Acids generally put hydrogen into steel , cathodic charging really puts hydrogen into steel. I would try very hard to remove the aluminum with wire brushing. If you try acid and the files do not appear cracked , bake them at 300 F for an hour to remove hydrogen. That temperature is unlikely to soften/temper the files . ( 400 F is more typical bake but that may cause softening ).

It requires time ( minutes to days) for the hydrogen to enter . diffuse to some location and concentrate enough to cause a crack. So it could be possible to clean the files then immediately bake them . Unfortunately , I don't think  anyone could say how much time.

Have you tried mechanical removal by using a file card?

@the last line of the question: how about ultrasonic cleaning? Some powerful units can induce cavitation that might knock off aluminum particles.

How about phosphoric acid, it will dissolve the aluminum and put a phosphate coating on the iron at the same time.  Not sure what best conc would be, it could be looked up.

Commercial cleaning agents containing nitric acid will passivate non-stainless steel, but nitric acid doesn't dissolve aluminum readily, so plain nitric acid (at concentrations ~27% or more) won't attack the steel or the aluminum.
However, the addition of a little fluoride will allow a small degree of corrosion of the steel. In the case of stainless steel, the steel will remain passive after extraction (and rinsing) from the solution, even tho a small amount of steel may have been removed.
On the other hand, aluminum can be cleaned with a relatively low concentration of nitric acid if the solution contains a significant amount of fluoride.
Granted, fluoride (as HF solution or NH$_4$HF$_2$) is dangerous, but can be handled safely if sufficient care is used.
It would seem that some balance of water, fluoride and nitric acid could result in rapid corrosion of the aluminum while at the same time barely touching the steel. I'm going to make a stab at 27% HNO$_3$ + 2% HF (or 3% NH$_4$HF$_2$) + 70% H$_2$O for a starter attempt. This is guaranteed to fizz up, dissolving the finely divided aluminum. If the steel is affected too much, increase the nitric acid concentration (yeah, counterintuitive, but higher HNO$_3$ passivates steel better).
I could name a company that sells these materials.

Try a variation of the classic bleach battery which has an anode of aluminum in aqueous NaOCl/NaCl. The latter is commonly sold as disinfecting chlorine bleach prepared from the action of chlorine on aqueous NaOH.
However, in place of the usual noble cathode of copper, I would try instead a freshly prepared inexpensive high surface area carbon cathode. For example, I have experienced good results from burnt food, to quote a source:

In cooking utensils, burnt food may produce carbon deposits with resulting rapid galvanic corrosion from the aluminum-carbon couple [474, 475].

Consider burnt bread with which I have had some success. Studies on advanced oxidation processes (AOP) interestingly cite, per my recollection, also the use of burnt coconut shells (likely reflecting local access).
Also, applying heat should promote the rate of the anodic corrosion of the aluminum (alloy?), which normally involves an inception period (which is characteristic of an electrochemical reaction).
Rationalization: Your Al contamination on the steel file may consist of an aluminum alloy, making it less anodic (so that, its tendency to corrode is reduced). Also, the relative surface area of the Al-steel anode to the cathode may not be small, which also reduces its rate of corrosion. Hence, the recommended use of a high surface area carbon cathode. Now, one could purchase activated carbon but I would try my home recipe if your problem is expected to be ongoing and you are looking for an efficient low-cost solution.
Note: In the current situation, the iron/steel file is the cathode which is coated with aluminum. As such, the cathode has little surface area relative to the anodic Al, which is the reverse of what one should have for fast electrochemical dissolution.

@AJOER Thanks for your answer. I will give your recommendations a try.   It seems less perilous than concentrated HCl, HNO3, etc.  A few questions:  Would a charcoal briquette work as the cathode in place of the burnt toast or coconuts?   Could I speed up the reaction by connecting the electrodes to an external power supply?  (Fe/Al +, C-?)


text:
Natural Sources for S-(-)-Limonene?
I'm interested in developing an undergrad organic lab that delves deeper into distillation/extraction, optical rotation and enantiomers. Our lab currently does a steam distillation extracting R-(+)-Limonene from orange peels but I am wanting to explore if there is a natural source that is rich in the S enantiomer. I have checked various places on the internet to little avail. As yet, I haven't been able to find a reliable source describing any natural occurrence (Wikipedia mentioned something about it being present in some mint essential oils but left it vague and I couldn't find any other sources to back that up). I did see some is available from Sigma Aldrich for purchase but I think it would be much more illustrative for the students to obtain this via another extraction (Or divide up into teams). Could anyone provide some insight or direction for me in this? Or else confirm that this is not richly available in natural products (Which I suspect might be the case from my superficial understanding of biology's chiral preferences)?
I have also read that the extract from oranges is actually just enriched in the R-(+) enantiomer but I am not particularly equipped to try and separate enantiomeric mixtures like that. Perhaps if someone has suggestions for making this a simpler option (enantiomeric separation) for undergrads, I would love to hear that too.
Thanks in advance!

The (S)-form of limonene is naturally found in pine needle oils and mentha oils. However, the concentration is lower (by way of comparison, the (R)-form is present in citrus peel oils at a concentration of over 90 %). In case of pine needle oil, you would have to separate it from terpineol and other similar components.
It is also present in ginger, nutmeg, pepper, mace, coriander, and other herbs and spices.


text:
Non-RSPO MB glycerin vs RSPO-MB glycerin
Does non-RSPO-MB glycerin function differently than RSPO-MB glycerin? For example in a cosmetic serum /lotion?

A good practice is to explicitly explain less common acronyms. Even if readers do not know them and are not experts in the field, they may bring some insights from their own domain of expertise.

No. The RSPO-MB (Roundtable on Sustainable Palm Oil- Mass Balance) certification has nothing to do with the purity or quality of the glycerin. It is a certification of the supply chain for the palm oil which is used to make the glycerin.
Ref: RSPO Supply Chains


text:
What is the cheapest way to give someone 1 mol of gold?
Let's say I've lost a bet to a friend who is a molecular biologist and I have to give her 1 mol of gold. What is the cheapest way to accomplish this?
It was a fun bet so I don't need to give her the gold, but I want to do it anyway. Just in a fun way.
Are there scientific tricks I can pull on her? Here are some ideas that might help:

Is there a different definition of a mole?
Is there a cheap and useless chemical compound containing gold?
Is there a kind of wordplay I can pull?

(1) No. (2) No. (3) Maybe, but that's not a question on chemistry.

Using 1E+10 m³ of seawater would not be practical, but it would give about 1 mol of gold.

Golden mole

Send it to me immediately, and I'll forward it to your friend.

@IvanNeretin About the word play, I meant something in the language of chemistry. Since I don't know much about chemistry, I thought chemists might have some jokes and stuff between themselves.

Czech language calls "mol" a small flying insects, eating either wool (clothing moth), either stored food. (mill moth). So take a dead moth and make it golden by golden dust/glitters or whatever it is called. // Otherwise 1 mol of gold is about 197 g, so think twice.

@Poutnik What if I give her 197g of another element and call it a DIY gold pack? "Just open the pack, perform fusion and get your gold*!" *results may vary

Use Fool’s gold. Very cheap. Too bad this was not posted six days ago, on Mole Day.

The only stable gold isotope 197Au is possible to make from the stable platinum isotope  196Pt by neutron absorbtion  and following beta decay. :-) But  it would be very expensive :-D

@EdV If the she-friend realizes it is fool's gold, she would be unhappy for several reasons. :-D Hopefully in good joking mood only. :-D

@Poutnik Well it doesn't have to be practical, but physically possible.

Another option is lead iodide $\ce{PbI2}$ which was called "mad gold". Made of yellow, shiny platelets looking like gold. it is easily prepared by mixing a lead nitrate solution ($\ce{Pb(NO3)2}$) and a potassium iodide solution ($\ce{KI}$). The reaction produces a yellow precipitate of $\ce{PbI2}$ which is not crystalline. But by diluting the mixture and heating it to $100$°C, the precipitate is soluble. And by cooling it down slowly, it produces wonderful shiny platelets of "mad gold". It is wonderful ! It can be filtered. My students are always filled with wonder when doing it in the lab.

There is also a critter which is named "mole." So a gold plated statue of the critter would be a "mole of gold."

@MaxW See Loong's comment :)

If you really want to do it, go to used jewelry or coin shops. They often sell gold at a multiple of the spot price in forms that do not have additional value, for use by jewelry makers and other artists. Purity is not assured, but these sources are much cheaper than technical materials suppliers.

According to  Lamb, A. E., Anderson, C. W. N. & Haverkamp, R. G. (2001). The induced accumulation of gold in the plants Brassica juncea, Berkheya coddii and chicory. Chem New Zealand. 2011, 34-36., the leaves, stems, and roots of the mustard plant contain $\approx \pu{100 mg}$ of gold per kg of dried biomass in certain soils.  So if a fresh mustard plant weighs, say, $ \pu{200 g}$, and has a water content of $85\%$, then its dried weight is $\pu{30 g = 0.030 kg}$, and it would thus contain $\pu{ {100 mg gold}//kg dried biomass}\, \times \, \pu{0.030 kg dried biomass = 3.0 mg gold = 0.0030 g gold}$.
To get a mole of gold ($\pu{197 g)}$, you would need $\pu{197 g}$/$\pu{0.0030 g = 66,000}$ mustard plants.
Mustard seeds are very small: $\pu{1 to 2 mm}$ in diameter.  Let's assume they're spherical, and $\pu{1.5 mm}$ in dia.  That gives a volume equal to:
$\frac{4}{3} \pi \Big( \frac{1.5\pu{mm}}{2}\Big) ^3 =  \pu{1.8 mm^3} =\pu{ 0.0018 cm^3}$.  If we assume their specific gravity is about the same as water's, each seed would weigh about $\pu{0.0018 g}$.
To get $\pu{66,000}$ mustard plants, we'd need many more than $\pu{66,000}$ seeds, since some would not be viable.  I don't know what the viability of mustard seeds is, but let's say $\pu{100,000}$ would do it.
$\pu{100,000}$ mustard seeds would weigh $\pu{180 g}$.
So you can give your friend about $\pu{200 g}$ of mustard seeds, and tell her all she has to do is plant these in appropriate soil conditions, and she will have herself $\pu{1 mole}$ of gold.

Maybe this calculation is valid in New Zealand. But I doubt that this mustard can contain gold when growing in a soil which has no gold.

But it would make better seasoning than gold, no?

(+1) For bringing in a gold hyperaccumulator and doing the math! My favorite hyperaccumulator is the New Caledonian tree that has blue-green latex (“rubber tree-like”) sap that is more than 1 M in nickel citrate.

@Maurice:  I already already included qualifiers about soil conditions—twice— in my answer:  "in certain soils" "in appropriate soil conditions".

The nice thing about theorists is that they can come up with ideas that are so impractical that some chemist will now wonder if this is challenging enough.

@JamesGaidis  Yes, we can be trouble-makers, but sometimes the experimentalists come up with these schemes entirely on their own:  "Fritz Haber did research on the extraction of gold from sea water in an effort to help pay Germany's reparations following World War I.  Based on the published values of 2 to 64 ppb of gold in seawater a commercially successful extraction seemed possible. After analysis of 4,000 water samples yielding an average of 0.004 ppb it became clear that extraction would not be possible and he ended the project."  https://en.wikipedia.org/wiki/Gold#Seawater

Thanks @EdV!   It's ironic that this happened on a closed question, particularly since I often find myself complaining that questions have been incorrectly closed :).  So now that I have access to moderator tools (no idea what that means yet), how much trouble can I cause?

@EdV  To quote Napoleon, it seems "Glory is fleeting"  :D


text:
What are the physicochemical properties related to medical drugs in the PubChem database?
I am not a chemist, in fact, I come from a computer science background. However, I am involved in a project related to artificial intelligence-based drug discovery. For this, I am trying to make a dataset that contains smiles notation of some chemical compounds along with their chemical properties.
In the PubChem database, there are lots of available properties, which are:
MolecularFormula, MolecularWeight, CanonicalSMILES, IsomericSMILES, InChI, InChIKey, IUPACName, XLogP, ExactMass, MonoisotopicMass, TPSA, Complexity, Charge, HBondDonorCount, HBondAcceptorCount, RotatableBondCount, HeavyAtomCount, IsotopeAtomCount, AtomStereoCount, DefinedAtomStereoCount, UndefinedAtomStereoCount, BondStereoCount, DefinedBondStereoCount, UndefinedBondStereoCount, CovalentUnitCount, Volume3D, XStericQuadrupole3D, YStericQuadrupole3D, ZStericQuadrupole3D, FeatureCount3D, FeatureAcceptorCount3D, FeatureDonorCount3D, FeatureAnionCount3D, FeatureCationCount3D, FeatureRingCount3D, FeatureHydrophobeCount3D, ConformerModelRMSD3D, EffectiveRotorCount3D, ConformerCount3D.
Surely, I don't want to select all of them as it will result in an overly high-dimensional dataset.
So my question is:
What are the physicochemical properties related to "medical" drugs I should keep?

There have been many attempts at defining the properties that make a molecule ‘drug-like’, often based on retrospective analysis of approved/marketed drugs.
The most famous of these is the Lipinski rule of 5 which define a set of parameters based on hydrogen bond donor count, hydrogen bond acceptor count, molecular mass, and LogP. Although many medicinal chemists use this to guide decisions, it’s worth noting that less than half of new drugs actually obey these guidelines.
The properties you use also depend on what kind of drug-like molecule you’re tying to find - if your target is in the brain, you want a very different set of properties to a drug that acts in the lungs.
LogD, Molecular weight, aromatic ring count, rotatable bond count, tPSA, hydrogen bond donors/acceptors are a fairly standard set of properties considered routinely (and all able to be easily computed).
The reality is that despite huge amounts of data, we’re really very bad at predicting whether any given molecule will end up being ‘drug-like’, let alone whether it has the potential to be taken forward to an actual clinical candidate.
(As an aside the issue is confounded by our inability to predict in silico some properties: xLogP is a calculation, but rarely correlates to an actual measured LogP value so if you use xLogP to make a prediction about drug-likeness, you’re starting off with a fairly large error)

Thank you @NotEvans for the amazing answer. I would like to ask about LogD and aromatic ring count, how can I compute them as they are not available in the provided properties?

You’re very welcome.    LogD, like LogP is an experimental value - but there are various calculators (the good ones are commercial). Aromatic ring count could be worked out from the structure if you wanted, but there are libraries available to do so - RDKIT is highly useful for calculating these kinds of parameters.    There is also an open source GUI called DaraWarrior that can calculate simple properties based on structure (inchi/smiles)

@AmineChadi Have look in the chat room here.


text:
Calculating the electric potential of a fruit battery
In a fruit battery, if I am using a reactive metal electrode for the cathode but the cathode reaction is the reduction of the hydrogen ions (in the acid of the fruit), how can the type of the cathode metal affect the voltage. In my experiment I used the same metal for the anode and different metals for the cathode and the resulting voltages are different. I am confused because the resulting voltage should be the difference between the anode reaction and the cathode reaction which I expected to be the same.

Give us information about the metal used both for the cathode and the anode. The effect could be entirely different if you choose copper or zinc, or any other metal.

Narrowing «fruit battery» to «lemon battery», there seems to be a correlation between concentration of lemon acid (which depends e.g., by the variety of the lemon tested) and the voltage recorded (figs 3 and 4 of Construction and evaluation of electrical properties of a lemon battery by Mazumder et al. in Journal of Chemical, Biological and Physical Sciences, 2018, 92-101, e.g. found on ResearchGate.)

(+1) For a lemon cell (aka “battery”), with zinc anode and copper cathode, the open circuit voltage is different than if you use zinc anode and silver cathode. I did this myself and was puzzled because I thought, as apparently you did as well, that the cathode reduction only involved hydrogen ions being reduced to hydrogen gas. But I think there may be an over-voltage effect that accounts for the difference in open circuit voltages. In any event, good observation!

I think this Wikipedia article is quite useful here https://en.m.wikipedia.org/wiki/Lemon_battery


text:
Substitution in aryl halides
I need to propose a route for the synthesis of 2-(bromomethyl)benzonitrile from 1-bromo-2-(bromomethyl)benzene. In the attached figure, the structure of the starting compound and the species to be obtained are indicated.

I thought it could be done by treatment with NaCN in acetone. But, under those conditions, my teacher has told me that you replace the bromine attached to the methyl, since aryl halides hardly undergo substitution reactions.
So I really don't know very well how it could be done. I hope you can help me.

This question is likely to be closed because you have not made any effort to solve the problem. Therefore, I will not offer a solution but make you work for it. Instead of worrying about how to make the less reactive site react selectively, ask yourself what if both of them react with cyanide. This product can lead to a "vinylogous malonic acid". What might homophthalic acid do on thermolysis? So how do you convert this aromatic carboxylic acid to the product? Let's see what you come up with. @Waylander

@user55119  If we select conditions to give the bis-nitrile then a straightforward acid hydrolysis will give homophthalic acid. I cannot find a reference to it, but I presume from your comment that it decarboxylates to give o-toluic acid. Formation of the primary amide followed by dehydration with POCl3 or SOCl2 gives the 2-cyanotoluene. A straightforward benzylic bromination (NBS/AIBN) gives the required. Steppy but reliable.

I'm afraid it has to be steppy. I found a photochemical decarboxylation with low quantum yield isolated in 7% yield [DOI:10.1248/cpb.40.2188]. The problem with thermolysis may be anhydride formation. Perhaps prolonged heating in acid. The bromination of o-toluic acid is known. A simpler route:? Oxidize homophthalic and form phthalic anhydride. NaBH4 reduction to the lactone and proceed from there. In reality, who would do this?

Related: https://en.wikipedia.org/wiki/Letts_nitrile_synthesis

See the Addendum to my answer subsequent to your accept.

As a complement to @Waylander's solution to this problem and after some exchange of Comments on other possible solutions, other methods for arylcyanation provide an alternative methodology. The benzylic bromide in dibromide 1 is problematic because it is the more reactive of the two bromides. Accordingly, dibromide 1 is temporarily converted into the methyl ether 2. Traditional formation of internally solvated Grignard reagent 3 followed by implementation of one of two possible transnitrilation reagents. 2-Pyridyl cyanate 4 has been shown [1] to transfer cyano groups upon reaction with aryl Grignard reagents. Alternatively, a number of dialkylated malononitriles have been utilized as cyano group transfer reagents with Grignard and organolithium reagents. Typically, dimethyl malononitrile 5 [2] functions admirably in this regard. The success of each of these reagents depends upon the lose of of a stabilized leaving group --- the anion of 2-pyridone in the former case and the anion of isobutyronitrile in the latter. With arylnitrile 6 in hand, reincorporation of the benzyl bromide moiety to form target nitrile 7 is accomplished through the agency of hydrogen bromide with the concurrent lose of methyl bromide and water.

Addendum (21 Dec 2021; user55119): The transformation 6 $\rightarrow$ 7 needs clarification. If $\ce{CH3Br}$ is formed by protonation of the ether followed by cleavage to form $\ce{CH3Br}$ then the resulting benzylic alcohol could add to the nitrile faster than the alcohol is converted into the desired bromide. If cyclization occurs, a pathway to 7 may be lost. Ideally, protonated 6 must preferentially cleave at the benzylic site to form 7 and methanol, which is then converted to $\ce{CH3Br}$.@Waylander's suggestion of $\ce{BBr3}$ is a possible, viable alternative.

J. S. Koo, J. I. Lee, Synthetic Communications, 1996, 26, 3709.

J. T. Reeves, C. A. Malapit, F. G. Buono, K. P. Sidhu, M. A. Marsini, C. A. Sader, K. R. Fandrick, C. A. Bussacca, C. H. Senanayake, J. Am. Chem. Soc., 2015, 137, 9481.

is the nitrile group going to survive the HBr treatment?

@Waylander--I didn't mean to imply 48% HBr. I had HBr in inert solvent -- benzene, perhaps.  If Me is cleaved first rather than the benzylic site, then cyclization could be a problem. But the intermediate iminolactone could cleave to the product.

@Waylander: The Grignard reagent with orthocresol in place of methoxide is known. This should remove the issue of any cyclization via a benzylic alcohol.

BBr3 may be a better option for the demethylation than HBr as it does not react with nitriles - there are examples of its use in MeCN

@Waylander--Certainly BBr3 is an option. Its use in MeCN may not be relevant if nothing comes of complexation. In nitrile 6, N-complexation may lead to oxygen addition to the nitrile with Me and/or benzylic site reacting with bromide. It could be complicated as you mentioned earlier. Nothing ventured; nothing gained.

This is a non-trivial problem. As your teacher has pointed out the bromomethyl group is very reactive to nucleophiles, so the issue is to find conditions for the reaction of the aryl nitrile that do not affect it.
The standard way of transforming an aryl bromide into an aryl nitrile is Pd-catalysed cyanation with a source of CN such as Zn(CN)2. This Organic Letters paper offers  a set of mild conditions for the transformation, but I cannot say for sure if the bromomethyl group would survive. If it does not then you are faced with transforming it into another group that would survive and can be taken back to the benzyl bromide afterwards; -OAc might be a good choice if this is required.


text:
Are there any cold temperature reactions in liquid where ice forms as a precipitate?
The title asks the entire question, but I'll elaborate:
Precipitates form in aqueous reactions when some product is insoluble in water. Is there a reaction in some other solvent where solid water forms as an insoluble product -- solid/insoluble rather than liquid/immiscible because it happens below water's freezing point?

Nothing particular in mind, but guessing any reaction in cold nonpolar solvents where water is eliminated.

That being said, most reactions where water is a product either wouldn't go at low temperature, or would raise the temperature as they go. Finding a reaction which evades both these obstacles is somewhat of a challenge.

One idea is to use a cryptand to dissolve sodium or potassium hydroxide and tgen add a carboxylic acid. The main obstacle with this approach is getting a solvent where this works and the freezing point is low enough. Benzene, for instance, disappointingly freezes at +5.5°C.

Closest I can think of are the Icicles of Death: http://thescienceexplorer.com/nature/mysterious-phenomena-known-icicles-death.

I have a feeling Oscar's suggestion is closest to the mark. Suitable solvents would be toluene (m.p. -95 °C) or possibly better, chloroform (m.p. -63 °C), both of which dissolve water to less than 0.1% by weight at r.t. (and even lower at sub-zero temperatures). I'd suggest tetrabutylammonium hydroxide as a base instead, but preparation of anhydrous TBAOH solutions is rather difficult, so not all of the water that "crashes out" would come from the acid-base reaction itself.

How about ammonium hydroxide, $\ce{NH4OH}$, and hydrogen chloride, $\ce{HCl}$?
At a concentration of 33% $\ce{NH3/H2O}$ by weight, the melting point (m.p.) is 173 K (-100 °C)! A 40% hydrochloric acid solution has m.p. of 247 K (-26 °C). The reaction yields $\ce{NH4Cl}$ and $\ce{H2O}$, and both would certainly be solids at those temperatures... though since ammonium chloride is exceedingly soluble, the water might be tied up as "water of hydration".
BTW, Harry Clement Stubbs, AKA Hal Clement, wrote some excellent science fiction such as Mission of Gravity and Star Light, drawing on the physics and chemistry of ammonia and water, among other things.

Would they, really? Soluble salts tend to keep water liquid well below 0°C. Also, the neutralization reaction is highly exothermic.

Ammonium hydroxide do not exist. It is a solution of $\ce{NH3}$ in water

We could precipitate a predominantly water-ice phase from ... (sea)water. Methane clathrate-hydrates are commonly made on sea-floors by organic reduction of carbon dioxide[1]. One factor favoring clathrate hydrate formation is the relatively low solubility of the methane in liquid water, which enables the clathrate to form with methane while the carbon dioxide from which it is derived remains dissolved in the liquid. Other common gases such as carbon dioxide itself and hydrogen sulfide are more soluble and clathrate hydrates involving these are much rarer on Earth.
See also this answer from Earth Science SE.
Reference

W. P. Dillon, in Robert A. Meyers, Encyclopedia of Physical Science and Technology, Third Edition (2001).


text:
Why do electron-withdrawing groups deshield outer ring protons?
I understand that ring protons experience deshielding because the induced magnetic field (outside the ring) has the same direction as the external applied field, but shouldn't electron withdrawing groups that delocalise electrons throughout an extended aromatic system also lead to a smaller induced magnetic field outside the ring (thereby lowering the effective field experienced by the protons and diminishing their chemical shift)?
With electrons spread out in a larger conjugated system (and thus, with potentially smaller ring currents?), I expected to see lower values for the protons; but, by studying multiple spectra, I came to notice that electron-withdrawing groups actually lead to increased chemical shift values (/heightened effective magnetic fields).
I suspect the reasoning behind electron donating groups decreasing the chemical shift is similar, but I can't seem to grasp the logic, since I am correlating electron density with increased ring currents (and increased induced magnetic field). Any help would be appreciated.

I agree that without data one can reason in both ways. However even if you can assume smaller ring current and therefore less deshielding, the electron density is moved toward the electron withdrawing group, and this deshield the protons as well - especially in an aromatic ring. Reality tells that is the second effect that is prevalent.
Alternatively and shorter, you can see the electron withdrawal as from the hydrogens, not from the ring current. As you know, in delocalised electronic system the inductive effects persist over few bonds.
I think this can rationalise the data, at least at posteriori.

Thank you - I didn't know exactly which of the two effects - deshielding versus smaller ring current - had the biggest impact, thereby being predominant.

@mateis the point is that, let me use  the following expressions for sake of brevity, as you suck electron density from the ring, the ring sucks in electron density from the C-H. Hope it helps.