text:
Counting total number of stereoisomers of 1,2-dibromo-3,4-dimethylcyclobutane

I came across this question in one of my tests.
The answer assumes the major product (the anti markovnikov one) being formed. That is, I wish to determine number of stereoisomers of 1,2-dibromo-3,4-dimethylcyclobutane. I had a hard time figuring this out.
I have found a total of 10 stereoisomers - 2 are meso-compounds and 4 pairs of enantiomers, just by drawing and visualizing the possible 3-d structures.
However, I am quite unsure of this answer. Does anyone have a systematic way of finding the number of stereoisomers of this compound? Can someone confirm that there are 10 isomers only?
Any help would be appreciated. Thanks in advance!

Wow, do you got an even more ill-conceived question? Read anything about cyclobutadiene and ditch whatever is the source of this nonsense you're citing.

I agree with Mithoron, the exercise is terrible. From the technical point of asking this question here, that's better. It can still be improved by actually including drawings of what you found. However, given that the exercise is a lost cause, I'm not sure whether this would do any good...

Yes, I understand that this is an anti aromatic- highly unstable reactant to begin with, but I am just interested only in the stereochemical aspect.

Rishi Shekher: I am in agreement with Mithoron and Martin that the cyclobutadiene shown is a poor chemical example. Nonetheless, it does not preclude addressing the stereochemical question. The choice of 2,3-dimethylcyclohexa-1,3-diene as a substrate.satisfies both problems. At issue here is the recognition that the peroxide effect (Kharasch reaction) is operative and no other regioisomers are to be considered.. All 16 possible permutations for drawing 1,4-dibromo-2,3-dimethylcyclohexane are shown below. Filtering through these structures is tedious. However, there is a more convenient approach to addressing the question at hand. continued

A systematic presentation of the 16 possible descriptors are shown below that follow the coefficients of a binomial expansion for $\pu{2^4}$ (1,4,6,4,1). Owing to the potential symmetry in the molecule, there is ample opportunity for duplication. There are two meso compounds: meso 1 (RSRS = SRSR; red) and meso 2 (RRSS = SSRR; blue). There are four pairs of enantiomers 1 through 4. There is a redundancy of structures for enantiomers 3 (green) and enantiomers 4 (magenta). In total there are 10 stereoisomers for 1,4-dibromo-2,3-dimethylcyclohexane.


text:
Does decreasing pH decrease solubility of adipic acid?
I added 3 grams of adipic acid to 150ml deionized water with heating + mixing until it was completely dissolved. This was then boiled and cooled to allow crystals to form. The maximum amount of adipic acid recovered was 2.62g. Is there a way to recover more of the acid without completely boiling off the solution? Does decreasing the pH decrease the solubility of adipic acid?

The solubility of a substance in pure water depends only on the temperature. But adding some strong acid like hydrochloric acid decreases the solubility of weak acids like adipic acid (Le Chatelier's principle)

Add salt - sodium chloride. See https://en.wikipedia.org/wiki/Salting_out

Why dissolve it in the first place? And how do you know how much you had before you dissolved it?

Yes it will.  Protonated adipic acid is less soluble in water than sodium adipate.  Colder temperatures also make it less soluble.
It can be extracted into a non-polar solvent that is more easily evaporated.  Drying under excessive heat may degrade adipic acid.  Vacuum may be better.
Some adipic acid will remain in solution if water is still present, but with multiple batches, losses can be reduced by using the same "mother liquor".
Try putting your re-crystallization solution in the refrigerator.

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text:
How to separate the salts within homemade sea salt
Goals
There are a few things that I'm looking to do:

Separate salt from seawater via some evaporation technique.
Separate the each of the major salts from the resulting sea water to get each into their purest possible form.
In particular--and if possible--I want to try and separate any iodine-containing salts from the salt water.

General Information
As I've found, salt water is composed of much more than just $\ce{NaCl}$. According to Britannica, 99% of all sea salts are composed of the following ions by weight: $\ce{Ca^2+, Mg^2+, K+}$, and $\ce{Na+}$ cations, as well as $\ce{Cl-}$ and $\ce{SO4^2-}$ anions (they have a table showing concentrations of all of these ions). There are many more molecules and ions to find to find in the ocean, but these are the major ones in terms of weight. With respect to iodine (you'll see why this is relevant later), I've found other sources that say that iodine is one of the more abundant micronutrients in seawater.
These dissolved ions, without the presence of water, can combine with themselves and other substances found within seawater to create salts including calcium carbonate, gypsum (hydrate of calcium carbonate), sodium chloride (halite), potassium chloride and magnesium chloride (source). Note the way that the salts are separated by solubility here.

My Plan (WIP)
First, I'm going to get a boat out into the middle of the ocean and get a few liters of sea water--ideally, as much as I can take.
Then, I'll reduce the solution until only particulate matter is left, either by evaporating the sea water outside with the sun and/or boiling the seawater in batches over a stove. Once the particulate matter is isolated, I could dry it further in an oven. I'm not sure about any of the advantages or disadvantages with these methods aside from power consumption, so I'll make more than one batch and subject each of them to different procedures so as to see what kinds of yields I get for each batch.
Finally, I'll have to separate the salts somehow, and this is where I'm lost. I have a couple of ideas for doing this:

Using a microscope, I could separate crystals with visually different shapes from others. I would like to avoid this, since it seems both inefficient, imprecise, and time-consuming. Plus, I don't own a microscope.
In one of the sources above, they say to separate the salts by solubility. I have never separated soluble solids before by differing degrees of solubility, so this would be new for me, since I wouldn't be able to just dissolve a completely insoluble solution in water and filter the concentrated particulate from the water. However, this seems like the most reliable method I could find for separating sea salts, so I think this would be the best method for doing this.

Tl;dr, I plan to make my own sea salt by getting salt water from the ocean and evaporating the water out (either via boiling or leaving it out in the sun) until only salt is left. Although, instead of using the sea salt as is, I want to see if I can separate it to see what particular kinds of salts are left behind. In particular--and if possible--I'm looking for iodine salt, to see if I can make my own iodized salt from just seawater. However, I don't really know how to separate water-soluble solid substances from each other, and I don't know if I'll even be able to obtain any salt with iodine in it, so I'm looking for any tips, guides or procedures for doing this.

Not all sea water is the same. If you want iodine salts, you might need to start with some that is high in iodine. OR, perhaps better, start your experiments by acquiring some sea salt (ie flakes produced by evaporating sea water which saves a step in your experiment).

Sounds like you're on the right track -- but rather than separating crystals by hand, use the difference in solubility of salts to make the work easier.
In commercial salt evaporation ponds, with a concentration gradient, the least soluble salts will precipitate first. This is perhaps the least work-intensive way to separate the various salts. So you might try letting the brine slowly trickle down an incline in the sun, and see what precipitates out at various places. This Wikipedia solubility table might give you an idea of what will precipitate, knowing the concentrations of species in sea water, for example. BTW, this may be used to recover the now-valuable lithium from brines.

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In which cases do we use the the reduced mass formula?
Edit: Karsten added a description of reduced mass:

"It is a quantity which allows the two-body problem to be solved as if it were a one-body problem."

I do not really have a strong Physics background (my highest level is undergraduate level General Physics).
The solution of a problem from Chapter 1 of McQuarrie's Quantum Chemistry, 2nd edition, involves the concept of reduced mass, supposedly because of the phrase singly ionized.

1-34. Using the Bohr theory, calculate the ionization energy (in electron volts and in kJ·mol^(- 1))
of singly ionized helium.

I struggle to understand why sometimes the concept of reduced mass is used, especially after a quick read of a part from Chapter 8A from Atkins' Physical Chemistry, 11th edition page 305:

In all except the most precise work, the reduced mass can be replaced by me.

Can somebody give me an explanation on this matter? Also, in which cases do we not use the reduced mass concept?

"It is a quantity which allows the two-body problem to be solved as if it were a one-body problem". If the smaller mass is much smaller than the larger, you can estimate the reduced mass by the smaller mass. This is not the case when considering rotational energy of diatomic molecules (masses are similar), but it is when looking at planets and the sun, or electrons bound to nuclei in Bohr's model. There is no connection to the ionization state.

See e.g. phys.libretexts.org   -  A_The_Bohr_Model_of_Hydrogen-like_Atoms  // Notice that the electron reduced mass μ=0.99946 . m_e

The time-independent Schrodinger equation is a linear ordinary differential equation.
In QM the wavefunction $\Psi(x)$ of a system isnt some linear combination of the wavefunctions of the individual elements which compose the system it is $\Psi(x_{0},x_{1})$ so immediately for 2-body systems the wavefunction of such system is converted into a system of differential equations , which only some forms are solvable.So we use the reduced mass to find the center of mass of a 2-body system and by finding the center of mass we convert it to a 1-body system so it can become solvable.


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Are there any examples of tricoordinate hydrogen?
There are cases where hydrogen forms two covalent bonds (of order 0.5 I assume) in certain molecules and ions such as $\ce{H3^+}$, $\ce{CH5^+}$, $\ce{B2H6}$, $\ce{HF2^-}$, and hydrogen-bridged cations (below is an example from wikipedia about hydrogen-bridged cations), but are there cases where a coordination number of 3 or higher is achieved, even theoretically? Aside from maybe the more salt-like compounds, e.g. $\ce{NaH}$.

If the illustration reproduced is from a publication, or a book: Don't forget to state the source of the illustration used.  For one, credit to whom credit is due.  For two, the publication where this illustration was used may provide context, and serve as source for inspiration among future readers of your question.

I would not include $\ce{CH5+}$ or rather $\ce{[CH3(H2)]+}$ in the mix. https://en.wikipedia.org/wiki/Methanium Also, trivalent is the wrong word. You are looking for tricoordinate. Valence is a fixed value for each element; and that value for hydrgen is one. Valency is also a very ambiguous word, many people use it for different things, and in combination with other terms (e.g. hypervalent) it has become quite muddy. Since you obviously look for a coordination number, use that word to avoid ambiguity.

@Martin-マーチン, I agree I probably shouldn't have used the term "trivalent", but I don't see how methanium is unique from my other examples with 2e3c and 4e3c bonds.

Well, there certainly is room for debate. And yes, looking back at all of those, they all seem rather different and unique. I guess I cannot come up with my thought process (again) from my last comment.

There are  hundreds of metal complex structures with tricoordinated hydrogen, it's nothing special or exotic these days (e.g. search for μ3-H).

Magnesium hydride, with the rutile structure, has each hydrogen atom bonded to three magnesium atoms, but this is generally considered predominantly ionic (see here for a discussion of the structure and ionicity of alkali and alkaline earth hydrides).
A better candidate for covalently bound, three-coordinate hydrogen is the iridium complexes given by Chodosh et al.[1]. The formulae of these complexes are described as follows:
$\ce{(IrH2LL')3(\mu_3H)(BF4)2}$
$\ce{L: PPh3,P}\text{(cyclohexyl)}_3$
$\ce{L':}\text{ pyridine, acetonitrile}$
In these complexes the three iridium centers are each bonded to two outer hydride ligands and rwo neutral organic ligands, with a central hydride ligand joined to all three iridium centers holding the complex together.
The introductory section of the article (available in front of the paywall) mentions several other, previously reported examples.
Three is actually not the maximum. Cousins et al. [2] have obtained four-coordinate hydrogen incorporated into alkali metal phenolate oligomers, the hydrogen taking the place of a lithium or sodium atom and thus showing alkali-metal-like characteristics.
Reference2

Daniel F. Chodosh, Robert H. Crabtree, Hugh Felkin, Sheila Morehouse, and George E. Morris (1982). "Trinuclear iridium cluster containing a tricoordinate bridging hydrogen ligand: structural and chemical studies". Inorg. Chem. 21, 4, 1307–1311.
Publication Date:April 1, 1982
https://doi.org/10.1021/ic00134a005

David M. Cousins, Matthew G. Davidson and  Daniel García-Vivó (2013). "Unprecedented participation of a four-coordinate hydrogen atom in the cubane core of lithium and sodium phenolates". Chem. Commun. 49, 11809-11811.
https://doi.org/10.1039/C3CC47393G.


text:
Determining volume of hydrogen gas evolved in water splitting reaction
I want to carry out a photocatalytic water spliting reaction. This reaction will generate Hydrogen and Oxygen gas in 2:1 ratio according to the following reaction -
2H2O + energy (light) → 2H2 + O2
So, if I collect the produced gas over water in a test tube, should I be able to determine the volume of hydrogen gas in that mixture? From literature I came to know gas chromatography is used to determine the volume and evolution rate of hydrogen. Is that necessary for a small lab experiment?

It depends on your desing. You could use a gas burette or even a graduated test tube uf you are working with a small set-up and small volumes of gas.

Collecting any amount of the combined gasses would be an explosion hazard.

If you collect oxygen and hydrogen in a single tube, the volume reflects both. If you know that the molar ratio is 1:2, you can calculate the volume of hydrogen gas by multiplying the total volume by 2/3. If you want to verify the molar ratio without expensive instrumentation, you might consider a chemical reaction that only reacts with oxygen or only reacts with hydrogen.
There is a standard lab to determine the oxygen content of air. It uses steel wool to react with the oxygen (but not the nitrogen) in air, turning it into "rust". The volume change observed corresponds to the oxygen that reacts. A mixture of oxygen and hydrogen is explosive (more so at a ratio of 1:2), so I would keep it at a small scale and avoid sparks.
I'm sure there are also convenient reaction to use up the hydrogen while leaving the oxygen, but the most common search result on the internet is the reaction with oxygen, which would remove both oxygen and hydrogen from the gas phase rather than isolating one of them.

With electrolysis, H2 bubbles up from the cathode.  O2 from the anode.  The apparatus is H shaped glass tubing with Pt electrodes on the bottom and measuring burettes on top.
Around 1.23 V is required to split water.  This can easily be provided from a solar panel.  Salt, such as KOH, improves the conductivity of water.  It will help improve yields of both gasses.
Should you wish to continue on to a scale-up, it is possible to pressurize your H2 and O2 yields by simply having the water intake from a height above the apparatus.  Every 20 feet above gives 8 psi.  No need for a compressor after the gasses are made.

Electrolysis powered by a solar panel could be called "photocatalytic", but it is far from what plants (like biology, not like chemical industry) do to split water. To be fair, conventional photosynthesis also does not yield hydrogen but rather reduced cofactors like NADPH.

@Karsten I'd set mine up right next to the desalination plant, somewhere like in San Diego, CA.  Plenty of sunshine.  Pour water in, out comes pressurized H2! Scalable.  Burn H2 for energy, back comes H2O for irrigation.

@RobertDiGiovanni Thank you for your response. But I'm not using any electrodes actually. I'll place my photocatalyst inside water and irradiate it with light of proper frequency. This should result in splitting of water.


text:
Why can i not use glue (which contains cyklohexane, petrol and ethyl acetate) for gluing PVC materials?
The manufacturer do not recommend using glue to gluing PVC. But Why? Thank you for your help.

Always write clear questions with included relevant details, purpose, context or background info to receive useful answers. Otherwise the answers may be just guesses. // With comparable lengths of the question title and body, the former is too long and the latter too short.

Which sort of glue ? There are a lot.

Generally, PVC is supposed be glued only by glues explicitly intended for PVC glueing.
Otherwise, the glue solvent may not prepare PVC surface for glueing sufficiently and the glue residue after solvent evaporation may not be PVC compatible to form reliable binding.


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Using Hydrogen as a carrier gas for a GC; consideration of reactivity with analytes
In many cases people use H2 as a carrier gas for a GC. But does this exclude analysis of (say) systems with unsaturation or aromatic compounds etc. If a analysis uses a temperature program say up to 300 C (quite typical) doesn't the carrier gas H2 react with your analytes as this temperature?
Just wanting to know if there are specific guidelines about what sort of analytes H2 can be used as a carrier gas with. Or are the Temperatures / partial pressures too low to worry about internal reactions. As compared to other carrier gases e.g. N2 or Argon H2 would seem unusually reactive and hence I wonder.
Notwithstanding all the other advantages of H2 in terms of cost, viscosity etc. what about sheer reactivity?

Hydrogen, even at elevated temperatures, generally needs a catalytic surface to react with unsaturated systems


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Why are carbon atoms of glucose considered chiral?
A chiral carbon atom is defined as

"a carbon atom with four  different groups attached to it."

I am aware that the open-chain form of glucose has four chiral carbon atoms.
However, I do not undertstand why that is, since the alleged chiral atoms do not have four different substituents: For example, the 3rd carbon atom is bound to one hydrogen atom, one hydroxylgroup and two neighbouring carbon atoms.

2 same atoms does not mean 2 same atomic groups.

Okay, but why is the last carbon atom at the bottom NOT considered chiral then?

Exception: 2 same atoms means 2 same groups if  the atom is the whole group.

Groups R-COOH and R-CH2OH are not the same, even if both are attached to R via C atom. Or, do you think they are the same?

@Johanna Similar to an onion, starting from the central atom to check if it is a chiral one / a stereogenic centre (and if so, on can assign R, or S configuration), one works toward more remote "shells".  User @user55119 compiled instructive answers; just two of them here, and here.


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Heat transfer calculation in open tank equalization
Thanks for your replies guys, sorry for unclear and undefined question. Let me try to reparaphrase my question.
Updated info:
This is the way i try to solve this problem.
I calculate the pipe length required to cool fluid inside open tank. With the data i had shown yesterday. I assume that no radiation, tank wall is insulated because the outer surface doesn't contact with ambient air, heat from ambient air is negligible (no longer using natural convection).
You can see my calculation and the pipe sketch inside the tank in a file I’ve attached below
With those data i did my calculation based on internal forced convection of fluid inside pipe. I also assume the pipe diameter with 2 simulations scenario using 10 inch and 1 inch pipe. I notice that the result doesn’t make sense. The calculation using 1 inch pipe has smaller pipe length compare with 10 inch pipe. The way i calculate is from the heat required from the tank then I assign it as heat for force calculation inside pipe (because Qin = Qout). I assume that outlet temperature of pipe is equal with desired tank (30 C). My reference is based on textbook Heat and Mass transfer fundamental and application by Yunus.A Cengel, Afshin J. Ghajar, in chapter 8.
Based on the Nusselt number (Nu) of each simulation, the smaller pipe has the greater value than the larger pipe. That’s because the flowrate in smaller pipe is higher, so that Re (Reynold Number) will be increased as well. With the higher Re can cause hl (convective heat transfer coefficient) higher too which will result to lower heat transfer surface contact. This is also applied on larger pipe with opposite result.
The focus here is to cool the tank. According to my understanding, to cool the tank with those data, the smaller pipe should has longer pipe length than the larger pipe. Based on my calculation, the pipe length for 1 inch is around 966 m, while for 10 inch is around 6810 m, so from this it’s likely the pipe diameter is proportional with the pipe length. I think that’s impossible, imagine if using pipe diameter closer to 0 it's likely that the pipe length is also very short. With short pipe length and smaller diameter, it’s not enough to cool the tank.
But that’s what happened to my calculation. My guess is because the way I calculate is (maybe unconsciously) focusing on heating the pipe from 28 to 30. But I don’t know, I don’t have any idea why this can be wrong.
[enter link description here][1]
So what do you guys think?
*I try to simulate the calculation without natural convection, when the force convection one is already clear I will try to consider this too.
*Let me know if there is something still unclear for you, I will try my best to make it more understandable
Click the link below to see my calculation
[1]: https://drive.google.com/folderview?id=1TlV9CxFvpx_A6q1GA0PLx6JI4_u5TXQC

Why not to involve calculation attempts you have done until now?

Its hard to know what you are doing from your description, could you add a picture of the set up? For example is cooling via a coil inserted into the tank or just by adding water . Also, you could do a quick estimate by working out the energy you need to loose to cool to a given temp and so volume of water needed.

Do you have to cool down the tank with one batch of 740 m³ of waste water or is there a continuous waste water stream going in and out?

Why is the tank insulated if it's at 60 °C and you want to cool it down?

You have omitted the heat transfer resistance on the outside of the pipe which is going to be greater than the resistance inside the pipe.  What kind of agitation is in the tank?

Let's see your heat balance equation on the fluid in the tank and your heat balance equation on the fluid in the pipe.


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Does a porous medium have to be charged in order for the fluid to establish the streaming potential?
In Wikipedia, I read:

Adjacent to the channel walls, the charge-neutrality of the liquid is
violated due to the presence of the electrical double layer: a thin
layer of counterions attracted by the charged surface.    The transport of counterions along with the pressure-driven fluid flow gives rise to a net charge transport: the streaming current.

But when I asked the chat.openai, it said

No, a porous medium does not necessarily have to be charged to
establish streaming potential. Streaming potential can arise in a
porous medium when a fluid flows through it, and there is a pressure
gradient across the medium. This can create a charge separation in the
pore fluid due to the movement of ions, resulting in an electric
potential difference across the medium. The zeta potential, which is
the potential difference between the surface of the porous medium and
the bulk fluid, can influence the magnitude and direction of the
streaming potential. However, even in the absence of a zeta potential,
a pressure-driven flow can induce a streaming potential in the pore
fluid.

So, now I'm a bit confused.

Please be more precise ! You start by speaking of "Adjacent to the channel walls" . Well ! Which channel ? Which walls ? What is adjacent ?

https://en.wikipedia.org/wiki/Double_layer_(surface_science)  In surface science, a double layer (DL, also called an electrical double layer, EDL) is a structure that appears on the surface of an object when it is exposed to a fluid. The object might be a solid particle, a gas bubble, a liquid droplet, or a porous body.

in the specific case of my question it is the walls of the porous body that is saturated by the fluid

Is chat.openai a brother of chatGPT?

You have asked everyone, what do the library and the books there say?

@AChem they told me to post my question here


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reaction of PCl5 with alcohol
I am able to think of two possible mechanisms for alcohol+PCl5 here, SN1 and SN2. In SN1, Inversion won't occur whereas, in SN2 inversion will occur. Which of the two is correct?

Does this answer your question? Reaction of alcohols with PCl5 and PCl3

@NilayGhosh thanks for suggestion but that post does not talk about inversion that's why I have posted new question

Reaction of PCl5 with alcohols leads to inversion. This was demonstrated with Malic Acid by Walden - so the process is sometimes referred to as Walden inversion.
See discussion here or here


text:
Insoluble ammonium salts?
Most solubility rules I can find state that "most ammonium salts are highly soluble". I wonder if there are actually any insoluble salts containing the ammonium ion. An answer here states that ammonium diuranate is insoluble but it is downvoted and I couldn't verify that being true.
 If none of ammonium salts is soluble, can there be other ways of getting rid of ammonium ions in water? That is, precipitation or liberation as NH₃?

If uranyl salts are not available, ammonium salts are recognized by heating it at high temperature. Mixed with some $\ce{Na2CO3}$, all ammonium salts produce some $\ce{NH3}$ which is a gas recognized by a pH band. It turns out to a basic color in presence of $\ce{NH3}$

Cation exchange resin in ammonium cycle is quite insoluble.  :-)

According to the CRC Handbook, of Chemistry and Physics (a great place to start looking), ammonium phosphomolybdate monohydrate is soluble to 0.02 g/100 ml water (presumably at ~25 °C). That's not insoluble, just slightly soluble, as far as I'm concerned ;-). Regrettably, though, no data on uranium salts' solubility (classified??),

@DrMoishePippik Wouldn't adding to water affect hydration level and thus affect solubility? Also, can it be made by precipitation? It is stated here that it can be made by heating ammonium orthomolybdate combined with phosphoric acid and nitric acid, yielding ammonium nitrate, water, and a yellow precipitate of ammonium phosphomolybdate.

There are at least 3 analitical reactions in wich ammonium ion becomes a part of insuluble compound:
$\ce{2NH4+ + Na+ + [Co(NO2)6]^3- -> (NH4)2Na[Co(NO2)6](s)}$
$\ce{NH4+ + 2[HgI4]^2- +4OH- -> [O=Hg2=NH2]I(s) + 7I- + 3H2O}$
$\ce{NH4+ + Mg^2+ + HPO4^2- -> MgNH4PO4(s) + H+}$


text:
How does an electrons's wave function change when it moves between energy levels?
I'm taking a class on QM and we're simulating the wave function of an electron in a box at the lowest energy level and I'm supposed to change the simulation to show the wave function for the next energy level.
The problem is that I don't quite understand the relationship between position and energy level.  I'm fine with the formula for the energy levels, but I don't see how to relate it to $\psi(x)$, as it doesn't even involve $x$.
And yet, I know there is a relationship -- if I understand correctly, electron orbitals are just PDFs of position obtained from normalizing and squaring the position wave functions of electrons with different quantum numbers, one of which is energy level, and these orbitals are very obviously different, as can be seem from the many diagrams of them.  So, clearly energy level does effect the position function, and, for that matter, so do the other quantum numbers.
But what's the specific mathematical relationship, either for a theoretical electron in a box or for an electron in an atom?  Like, say I already have an expression for $\psi(x)$ at an arbitrary energy level.  How would I modify it to model an electron at a different energy level?
I did see these questions:
Relationship between Quantum Numbers and the Wave-function
Relationship between Energy Level and electron position
https://physics.stackexchange.com/questions/355461/energy-eigenfunction-completeness
https://physics.stackexchange.com/questions/345244/in-qm-we-have-position-and-momentum-space-what-about-energy-space
but none of them answer the question of what the specific mathematical relationship is in a useful form.
I'm not really sure if this is a physics or chemistry question, but I'm picturing electron orbitals in atoms so I'm leaning towards it being more chemistry.

Electron in a box has a totally different wavefunction from an electron in an atom because the conditions are totally different. But I'm sure they taught you how to figure out the lowest energy level (or at least prove that it is one and gave you some idea why it works - not how to discover it from scratch). It has to be a sine wave with zeroes at the edges of the box, yes?

@user233751 Isn't the particle in a box a simple model of a one electron atom though?  And then there's the Kronig-penney model for electrons in a semiconductor, which IIRC, is basically a sequence of interconnected finite energy wells.  I took semiconductor physics last semester and we spent a while with that model and the approach to finding wavefunctions for electrons with this model was very similar to how we did it for the particle in a box problem.

@user253751, right but my question is, if I have the formula for the position wavefunction at, say, n=1, how do I find the formula for the wavefunction at some other value of n?

I don't think there is a way to do that.

Often, the one-electron functions with higher energies have more nodes. Often, the functions are of the same type, just with different parameters. So knowing one solution of the problem might help to find others by analogy.

You need to go back to the very start. Here, you're kind of asking: I have a solution $\psi_0$, how do I get the next solution $\psi_1$? The answer is to look at how $\psi_0$ was obtained, and it turns out that that same process which yielded $\psi_0$ will give you all the $\psi_i$'s along with their associated energies $E_i$.
Physical systems tend to admit a series of wavefunctions*
$$\{\psi_0(x), \psi_1(x), \cdots\},$$
which satisfy the time-independent Schrödinger equation
$$\hat{H}\psi_i(x) = E_i \psi_i(x)$$
for all $i$. The way to obtain the wavefunction is to solve the Schrödinger equation. For a particle in a box, you have that
$$\hat{H} = \frac{p^2}{2m} = -\frac{\hbar^2}{2m}\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\right)$$
(for $0 \leq x \leq L$, $L$ being the length of the box) and so to obtain the wavefunctions you need to solve the differential equation
$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = E\psi, \label{eq:de}\tag{1}$$
which also respect the boundary conditions $\psi(0) = \psi(L) = 0$. It turns out the functions $\psi$ have a general formula of
$$\psi_n = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),$$
with the energies
$$E_n = \frac{n^2\hbar^2\pi^2}{2mL^2}.$$
So, generally speaking, you don't generate one wavefunction from another one: you obtain all of them at the same time by solving the differential equation (eq. $\ref{eq:de}$).
The same is true of pretty much any other system, including the hydrogen atom (which gives you the orbitals). It's just that the Hamiltonian is different, so you get a different differential equation, and ultimately different solutions.
Now... there are some cases where you can go from one stationary state to another. A notable example is the quantum harmonic oscillator, where you can use 'raising' and 'lowering' operators to go between stationary states; however, this isn't something you can conclude by looking at the final wavefunctions, it's something you have to figure out by studying the form of the Schrödinger equation. So again, you have to go from the start.
TLDR: The "specific mathematical relationship" you're looking for is that they all satisfy the Schrödinger equation.

* To be technically correct, these are stationary states; they are 'special' wavefunctions which do not change over time (in that the expectation values of observable quantities like position and momentum are independent of time). Non-stationary states are perfectly permissible too, and can be constructed as linear combinations of stationary states.

The Schrödinger equation is just a normal type of differential equation. We interpret it as describing how the operator for the sum of kinetic and potential energy (the Hamiltonian) produces the total energy. It has the form $H\psi_i=E_i\psi_i$ with lots of solutions $E_i$. The Hamiltonian has the form $\displaystyle H=-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)$  where $V(x)$ is the potential energy, zero in the case of a particle in a box, but $kx^2$ for a harmonic potential as, for example, in a vibration; $k$ is a constant.
However, our description is not complete, because the 'boundary conditions' have not been specified. To solve this differential equation to find $\psi$, integration is needed and is done using normal mathematical methods. We know that in integrating, limits are always involved in one form or another. These limits are called 'boundary conditions' when solving differential equations and you always have to use them to get to any solution. In the case of a particle in a box, these are that $\psi$ is zero at each side of the box, because here the potential energy is infinite by definition. This condition forces integer solutions onto the problem; the wave functions $\psi$ are sine waves that are always zero at the ends of the box. Using the words 'integer solutions' is just another way of saying that quantum numbers are involved. What boundary conditions are used depends on the particular problem you are trying to solve, and changing the boundary conditions will change the solutions.
Finally, notice that we find all values for $E_i$ and $\psi_i$ at the same time, but how we interpret the wave function is terms of probability of being at a certain position $x$, i.e., $\psi(x)^*\psi(x)$ has nothing to do with the maths, but how we choose to interpret these via quantum theory. To illustrate the ordinariness of the maths involved in the Schrödinger equation, a similar equation and solutions are found when a taught wire is plucked, as in a guitar string.

@PeterMortensen The spellchecker marks wavefunction as bad spelling but this is common usage. A matter of taste I guess: https://www.reddit.com/r/Physics/comments/3vvjsm/wave_function_or_wavefunction/

We do write wavelength and wavenumber, eigenvalue and eigenfunction, perhaps if wavefunction is commonly used it will appear in dictionaries in the future.

A quick intuition to understand how energy affects the position distribution can be obtained by looking at the s orbitals.
For orbitals $1s, 2s, 3s, ..., ns$, the increasing principal quantum number means that you have increased energy. Correspondingly, the average distance between the electron and the nucleus also increases. Thus, the higher energy allows the electron to stray further from the nucleus.
Mathematically, all these orbitals must be solutions to the Schrödinger equation. As pointed out in previous answers here, when you solve these equations you don't obtain one solution, but instead, a class of solutions that, once you impose boundary conditions, will result in wave functions for various different energy levels.


text:
Can I separate paint thinner into acetone and toluene by simply boiling it?
I have a paint thinner that contains 60% toluene and 40% acetone. I would like to obtain some more-or-less pure toluene. Considering the big difference in boiling points (110°C toluene vs 56°C acetone), and because I only need the toluene and not the acetone, I just tried to pour the paint thinner in a pot and put it on the outside stove.
What I expected to happen, based on my small understanding of chemistry, was that the temperature would rise until 56°C where it would stay for a while and bubble off all the acetone, and then to stop bubbling and the temperature to increase until 110°C where it would start to bubble again.
However what really happened was that both the temperature never stopped rising and the liquid never really stopped bubbling once it had started (although it slowed down to some degree after the first 30 seconds or so). When I only had like 20% of the original amount left, I accepted I have no idea what I'm doing and took the pot off the stove.
...and now I'm here, asking for advice from people who know better than me. Does this whole procedure even make sense? And if yes, how should I approach it? Maybe try to keep the temperature precisely on a certain level?

Hopefully you had lots of fresh air flowing into the room…

@JonCuster Very attentive, that's why I wrote "outside" stove.

Why not just to get pure toluene?

This is beyond a little dangerous.  Acetone has a very low flash point. Vapors of both are heavier than air.  You might be better off as suggested, buying pure toluene.

@Poutnik It doesn't seem to be THAT easy to find, not in stores at least. I found some online but it's more expensive than the paint thinner and I would have to pay for shipping. (Also, I love to try and learn new things.)

I assume xylenes may be present too. They would be harder to separate than acetone.

The paint thinner may contain some additives other than the two solvents, but a possible separation could be made by adding water.
Acetone is completely miscible with water at all ratios; toluene is hardly soluble in water at all. If you add water to your paint thinner, it will separate into two layers, probably in a nasty-looking way, as the first water-acetone mixture is less dense than toluene (sp.gr toluene = 0.866; sp. gr. acetone = 0.750; sp.gr. H2O = 1) and will float, but will sink as more water is added.
The ternary phase diagram has been published (Ref 1) but the images may be copyrighted. The image can be found on a google search for the terms: "phase diagram toluene water acetone"; the image comes up on google, but not at the actual url of the publication because of a paywall. The phase diagram is very similar to the one of water-ethanol-toluene however (Ref 2).

Imagine that "ethanol" in the diagram = acetone. A separation technique could be as simple as adding enough water to the paint thinner until the mixture separates cleanly with the toluene on top, then siphon off the toluene from the two-phase mixture, leaving behind what will hopefully be a small interfacial foamy layer. (The hashed area in the figure is the two-phase region.) That is the best scenario. But a formulated paint thinner may well contain surfactants or other additives that will complicate the separation by making the interfacial layer quite thick and/or stable; but it might settle down on standing overnight.
Ref 1. https://pubs.acs.org/doi/10.1021/je7000182   Surface Tension of the Ternary System Water + Acetone + Toluene; Sabine Enders, Heike Kahl, and Jochen Winkelmann; J. Chem. Eng. Data 2007, 52, 3, 1072–1079
Ref 2. https://www.researchgate.net/figure/VLLE-mol-diagram-for-the-water-ethanol-toluene-ternary-system-at-1013-kPa_fig1_39441801

A good idea.  Toluene recovered may have a bit of water in it.  A follow-up with dryers may be needed.

Thanks, this seems to be the way to go, especially because it doesn't require any fancy equipment I don't have. If I mix water and paint thinner 50:50 in a measuring cup and stir it, the layers shift to like 60:40 with practically no "interfacial foamy layer". The toulene (top layer if I'm not mistaken) however does look very cloudy, so I assume that's where @RobertDiGiovanni s comment comes in? Are "dryers" some other substances I need to buy? Or some kind of device?

@MaxD if it's worth your time, wash repeatedly with water.  There are "drier" compounds people use to make 100% ethanol.  But, for all your trouble, you might just "cut" your thinner with more toluene.  Toluene may be less soluble in water at lower temperatures too.  Curious, what will  be your application with the toluene?

@RobertDiGiovanni I'm experimenting with making lava lamps, and toluene should dissolve the solvent dye I have bought, mix with mineral oil and not mix with water (because the solvent dye doesn't dissolve in mineral oil directly, and I think I need some kind of intermediate substance).

@MaxD you might be better off with acetone/toluene than water/toluene.  You may try other solvents like alcohol, turpentine, or mineral spirits (or more toluene).  Unless there is a specific reason to eliminate acetone, you might just keep it.  Toluene is only slightly less flammable.  Acetone should dissolve in mineral oil.

@RobertDiGiovanni Yes but acetone and alcohol also mixes with water, and the dye doesn't dissolve in white spirit (I tried). Thought is I want to avoid the colors bleeding from the lava into the water. (Maybe that wouldn't happen either way, I haven't tried yet, but I assume it would.)

@MaxD so the "lava" is dyed solvent that does not mix with water.  Worried about toluene, it's flammable, toxic, and does mix with water somewhat.  You may want to try another dye.  A little research (so what other people do) and some experimentation might solve this.  Good luck.

@RobertDiGiovanni The lava is like 60% mineral oil and 40% tetrachloroethylene to get the density right, and just enough toluene to get the solvent dye to mix. Anyway, if you have a suggestion for another dye (NOT a pigment) that would work in this scenario, I would love to hear it. I have been digging a ton into this subject and solvent dyes were the only stuff that sounded somewhat useful. (I will try pigment colorings with paraffin wax in another lamp, but for this lava which is permanently liquid, pigments are gonna be a problem as they sediment out over time.)

@MaxD Klean Strip Toluene.  Ace Hardware.  100% Toluene.

@RobertDiGiovanni A recipe doesn't really exist for this kind of lamp, I'm just figuring things out on my own. I also assume that since I'm using just a little bit of toluene, if in that little bit there are a few little impurities, it wouldn't be too much of a problem. Also, thanks for the suggestion, but I'm from Europe and my local hardware stores don't have it so I would have to order online and pay shipping. Maybe I'll give the water cleaned toluene a chance before I do that. Thanks for the advice!

Your logic of separating the acetone from toluene by distillation is reasonable, and would work well for separating many substances, such as propane from water (e.g., at a gas well).
However, the primary issue is that those two boiling points are close enough to require fractional distillation. Consider that the starting mixture is more toluene than acetone, and that the kinetic energy of each molecule differs, so that even at 56 °C, there are enough toluene molecules with sufficient energy to leave the mixture (along with the acetone). This is another way of saying the vapor pressure of toluene at that temperature is significant, about 0.2 atm (~20,000 Pa). As the concentration of toluene rises, there is increased probability that a toluene molecule evaporates, rather than an acetone. So one would need a fractionation column, providing a temperature gradient, to get a better separation.
Another possibility is that acetone and toluene might form an azeotropic mixture (though I could not confirm that; see this list, for example) . From Wikipedia: "An azeotrope... is a mixture of two or more liquids whose proportions cannot be altered or changed by simple distillation... because... the vapour has the same proportions of constituents as the unboiled mixture."
For example, separating water from ethanol is difficult, since at ~95% alcohol/water, the condensate contains the same proportion of water as the starting mix. Other means are needed to remove the remaining moisture, so Absolut® не правда; it's not absolute alcohol.

As DrMoishe Pippik pointed out in his answer, fractional distillation would separate certain amount of toluene from the mixture (Ref.1) without complicating things like adding water (Water make azeotrope with both acetone and toluene).
$$
\begin{array}{c|ccr}
n \ (\text{Frac. #}) & \text{% toluene in the frac.} & \text{mole frac. acetone (calc.)} & \pu{T ^\circ C} \\
\hline
1 & 20 & 0.85 & 57-60 \\
2 & 30 & 0.77 & 60-68 \\
3 & 50 & 0.59 & 68-73\\
4 & 60 & 0.49 & 73-78 \\
5 & 70 & 0.40 & 78-87 \\
6 & 80 & 0.27 & 87-97\\
7 & 90 & 0.14 & 97-103 \\
8 & 98 & 0.03 & 103-106 \\
9 & 100 & 0.00 & 106-109\\
\hline
\end{array}
$$
Note that, in this separation, initial mixture has contained 30% acetone by volume.
References:

H. M. Coleman, "Laboratory demonstration of fractional distillation," J. Chem. Educ. 1967, 44(8), 476-476 (DOI:  https://doi-org.srv-proxy2.library.tamu.edu/10.1021/ed044p476).

If the OP could manage some even primitive column, like a wide inert glass/metal tube, filled by suitable glass/ceramic/inert pieces, not packing themselves too tightly to have enough air around, he could manage distilling out most of acetone before significant amount of toluene evaporates too. But I admit it may be challenging to manage.

The column used in this research is just a air condenser with glass wool in it (I believe).  I agree, it'd be challenging and get complicated if decide to add water to azeotrope one or both.

Glass wool IMHO could act as a little strange multiple floor distillation column. It could work for acetone/toluene well. (The link seems to be behind paywall or requires registration and perhaps student/academy status)

I think glass wool would plug up too easily.


text:
Combustion of Sodium (Na) Metal in presence of Excess Air
There was a question which I was recently solving regarding the products formed after combustion of Sodium Metal in presence of Excess Air.
The given answers were Na2O2 (sodium peroxide) and Na2O (sodium oxide).
But on the contrary, a textbook I refer to gave the products to the reaction as Na2O2 and NaO2 (superoxide)

I was wondering if someone could clarify what the correct reaction and final products would be to the reaction ?
Reference:

Ncert textbook (grade 11) S-Block elements

@NilayGhosh although the answer explains how Na2O2 is formed, my primary doubt is whether NaO2 or Na2O is formed in the reaction along with Na2O2.

@Hydro_7, All three can be formed as a mixture. The exact proportion will depend on the exact conditions. So you can follow your textbook.


text:
Does increasing gas pressure in a galvanic cell decrease or increase the voltage?
This is for the case of a hydrogen fuel cell.
My reasoning was that, similar to rates of reactions, increasing gas pressure increases the 'concentration' of the molecules, thus there's more reactions. I guess that that would more likely affect the current though.

When working with galvanic cells always start by writing the  redox reactions as they occur.  This will help answer your question.

Theoretically, as in the answer below, there's little effect. Practically, though, pressure on a physical cell can force the electrodes into closer proximity, reducing cell resistance and increasing current. This can show as increased voltage, if the meter or circuit has finite resistance. Of course, sufficient pressure could destroy the cell, too.

Pressure has significant influence on the cell voltage only if some active components of redox systems are gaseous.
Otherwise, the effect is very small, depending on slight volume and/or compressibility changes. These can be both ways and higher pressure would support the state with smaller volume, with (near) negligible effect.
There cannot be therefore said in general if higher pressure would increase or decrease the cell voltage.

In later added context of hydrogen fuel cells, the cell  voltage would increase with pressure of hydrogen and oxygen gases for two reason:

Thermodynamically, electrode potential of the oxygen electrode would increase due increased activity of oxidant(oxygen). And vice versa for hydrogen as reductant. Both effects would increase the open voltage of the hydrogen fuel cell.
Kinetically, increased pressures would increase rates of redox reactions, decreasing the cell voltage drop under the load, so it would for the given load provide the dynamic part of voltage increase.

There will be also possible minor effects caused by pressure affecting geometry of the cell design.

Sorry, I realised now that I never gave context. I just editted the question; I was referring to the hydrogen fuel cell.

I have updated the answer. // Always include enough relevant details of the question purpose, context or background to make the question clear,  preventing  misunderstanding and need of  clarification.


text:
How would you find the Q value in the Nernst equation if you were to change the temperature?
I'm studying how changing the temperature of the anode would affect the voltage produced within a galvanic cell. I think that the Nernst equation plays a role here but I am unsure how the reaction within the anode would be affected due to temperature. Would the reaction quotient change as temperature changes? Is it possible for the value of Q to be calculated during the reaction by using the voltage?

You have to consider also changing of the standard redox potential for different temperature that is linked to temperature change of the Gibbs energy of the redox reaction.

I'm unsure how to find the change to standard redox potential as it isn't feasible to find the change in entropy and change in enthalpy during the reaction. Can you elaborate further? @Poutnik

I have not said you can find it, but for simple cases. For many cases, it is matter of experimental work.


text:
Nomenclature of monocyclic molecule with multiple insaturation
The PIN (Preferred IUPAC Name) of this molecule

is (Z)-cycloocta-3-en-1,5-diyne, but at first I thought it was (Z)-cycloocta-1-en-3,7-diyne, following the rules of the Blue Book, P.31.1.3
Can someone help me to decipher the logic and the application of the naming rules to find the correct name of this type of structures?

The preferred IUPAC name is (3Z)-cyclooct-3-ene-1,5-diyne; note the three small differences.  Subsection P-31.1.3 contains multiple rules. Which rule did you follow so that the numbering is changed?

Is "PIN" an IUPAC abbreviation? If so, please don't use it.  If not, just don't use it. ;)

If by P.31.1.3, you meant P.31.1.3.1, then perhaps you misread it (emphasis mine):

In monocyclic homogeneous unsaturated compounds, one double or triple bond is always allocated the locant ‘1’. When alone, the locant ‘1’ is omitted in names.

Both of your given names are in accordance with this rule.

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (...) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. (...)

Since the set of locants [1,3,5] is lower than [1,5,7], the name (Z)-cycloocta-3-en-1,5-diyne is preferred.


text:
What is the chemical mechanism at work during the carboxylation of graphene oxides?
The literature describes carboxylation as an efficient technique to increase the ability of graphene oxide to adsorb metals, as it increases the content of carboxylic acids in the material, while decreasing the content of C-O bonds of epoxide and hydroxyl groups [1,2,3]. Basically, a chemical treatment is carried out where acetic acid is used as a source of carboxylic acids, in a basic environment produced by the addition of NaOH (3M). Can anyone please tell me what is the mechanism involved in this carboxylation reaction? Specifically, why is a basic environment necessary for carboxylation and what are the chemical reactions involved?
Thanks
[1] HE, H.; KLINOWSKI, J.; FORSTER, M.; LERF, A. A new structural model for graphite oxide. Chemical Physics Letters, v. 287, n. 1-2, p. 53-56, abr. 1998. Elsevier BV. http://dx.doi.org/10.1016/s0009-2614(98)00144-4.
[2] WHITE, R. L.; WHITE, C. M.; TURGUT, H.; MASSOUD, A.; TIAN, Z. R. Comparative studies on copper adsorption by graphene oxide and functionalized graphene oxide nanoparticles. Journal of the Taiwan Institute of Chemical Engineers, v. 85, p. 18-28, abr. 2018. Elsevier BV. http://dx.doi.org/10.1016/j.jtice.2018.01.036.
[3] ZHAO, L.; CHEN, J.; XIONG, N.; BAI, Y.; YILIHAMU, A.; MA, Q.; YANG, S.; WU, D.; YANG, S. Carboxylation as an effective approach to improve the adsorption performance of graphene materials for Cu2+ removal. Science of the Total Environment, v. 682, p. 591-600, set. 2019. Elsevier BV. http://dx.doi.org/10.1016/j.scitotenv.2019.05.190.

:/ Even this "carboxylation" term sounds like they don't know what they exactly do.

@Mithoron, why do you say so?

Because it's a made up term? Meaning they are introducing carboxyls "somehow". You can acetylate, esterify, substitute... but they use a vague term because of not knowing the answer to your question.

@Mithoron, yes, I get it...


text:
Markonikov's Rule (Cyclohexene reacting with H-Br)
Based on Markonikov's rule, H-X hydrogen will be added onto the carbon (double bond) with more hydrogens. Therefore, the LHS and the middle makes sense. But why is the RHS product possible ? Hydrogen is adding onto a carbon which is not a double bond

The reason that that RHS product is possible is because a hydride shift can take place. As a result, the positive charge is in a more stable position (tertiary).
To summarize, the electrons from the pi bond will attack HBr and will form a carbocation. A hydride shift will occur which places the positive charge at the methyl group. Then the bromide ion will attack which yields the product on the RHS.


text:
What is the correct way to specify the weight fraction from bulk compositional data with a listed total % (e.g. from XRF, ICP)?
TL;DR = With bulk composition results, such as from X-Ray Fluorescence (XRF) testing, is the mass fraction of a given analyte that analyte given as a % or the analyte divided by the given total mass %?
Perhaps a silly and very elementary question, but I want to be certain as I've been wondering for a while now and never 100% sure if I'm right or not. But I'd like to be to minimise any potential compounding errors.
I've recently received some X-Ray Fluorescence (XRF) bulk composition analysis of some coal fly ash material I'm going to analyse. The full table of results is as below where the units are listed as %.
(note that I did try to put everything in Chem SE table format but it was too difficult to do, format and the help page didn't provide too much clarity from my POV):

Note that the total % provided is 100.21%. I presume they're not always 100% due to errors with instrumentation detection limits (etc.). However, when I do the sum (i.e. Each analyte % + LOI %) I get 100.06 % =/= 100.21 %.
I understand that the mass (or weight) fraction of a given analyte is given by the formula below where $w_i$ and $m_i$ is the mass/weight fraction and mass of the given analyte respectively and $m_{tot}$ is the mass of the total mixture:

I have been told by some to just assume that the % given for each analyte is the mass fraction but I feel that may not be right and I'd rather less compounding errors than more.
So with this in mind, my questions are:

Is the mass fraction of SiO2 from the above tables 54.19% as listed or, for a given mass of 100.21 g for the parent material, $\frac{54.19 g}{100.21 g} \times 100\%$ = 54.08%?

Can the moisture and LOI analytics be added as part of the total? I.e. should I assume from this table of results that 0.01% of the total mass is H2O and 0.01% is carbon respectively?

It is not a silly or an elementary question, but perhaps you can add more details about  analytical details. XRF and ICP-OES are apples and oranges, for moreso for a heterogeneous mixture like fly ash.

The only other details I could think of maybe adding from the report is detection limits. What else could I look for to add?


text:
Crystal structures exhibiting dimeric arrangement of tetrahedrally coordinated cations
I am looking for inorganic crystal structures that exhibit tetrahedrally coordinated cations arranged in a vertex-sharing dimer arrangement, where the cation - shared vertex - cation angle is close to 180°:

Are there any known instances of such crystal structures? If there are, then any number of examples would be welcome. If there are examples of compounds with just similar features (dimers with different coordination, for instance), then this would also be appreciated.

A starting point would be to search for molecules with the same symmetry  as the structure you show.

I did. I checked every database i have access to, and found nothing. I have repeatedly expanded the search range, and found nothing. Further searching would entail just checking every compound with a cation that might assume tetrahedral arrangement. Which would take a very long time.

Why aren't the silicates known to yield all kinds of "tetrahedra" polyhedra connected via corners, ridges, and surfaces suitable here?

@Buttonwood I am not just looking for compounds with tetrahedral coordinations in general. If you know of silicate compounds that exhibits isolated, vertex-sharing dimers, then please inform me.

Given the unit must be isolated and should not be the subunit of an ionic network, the following graph model for CCDC ConQuest's (ver. 2022.3.0) query with restraints based on connectivity $(\mathrm T_n)$ and fixed the central bond angle within 170° to 180° range gave 119 hits.

The unique found moieties are as follows with exemplary references from CSD database:

Formula
Name
Bridging angle
Conf.
CSD ID

$\ce{[Fe2OCl6]^2-}$
(μ-Oxo)bis[trichloroferrate(III)]
$\angle\ce{Fe-O-Fe} = 180°$
Staggered
ADETIM

$\ce{[Fe2OBr6]^2-}$
(μ-Oxo)bis[tribromoferrate(III)]
$\angle\ce{Fe-O-Fe} = 180°$
Staggered
AWIDEQ

$\ce{[Be2F7]^3-}$
(μ-Fluoro)bis[trifluoroberyllate(II)]
$\angle\ce{Be-O-Be} = 180°$
Staggered
SEYHOT

$\ce{[Hg2I7]^3-}$
(μ-Iodo)bis[triiodomercurate(II)]
$\angle\ce{Hg-O-Hg} = 180°$
Staggered
GODDUY

$\ce{Zn(CH3)2}$
Dimethylzinc
$\angle\ce{C-Zn-C} = 180°$
Staggered
SAJDEO

$\ce{Cd(CH3)2}$
Dimethylcadmium
$\angle\ce{C-Cd-C} = 180°$
Eclipsed
AROPAA

$\ce{[Tl(CH3)2]^+}$
Dimethylthallium(III)
$\angle\ce{C-Tl-C} = 179°$
Eclipsed
DOMHAO

$\ce{[Cu(CH3)2]^-}$
Dimethylcuprate(I)
$\angle\ce{C-Cu-C} = 180°$
Staggered
DAZWIK

$\ce{[Cu(CF3)2]^-}$
Bis(trifluoromethyl)cuprate(I)
$\angle\ce{C-Cu-C} = 180°$
Staggered
NONVIW

$\ce{Hg(CF3)2}$
Bis(trifluoromethyl)mercury
$\angle\ce{C-Hg-C} = 180°$
Staggered
DTFMHG

$\ce{Hg(CHCl)2}$
Bis(dichloromethyl)mercury
$\angle\ce{C-Hg-C} = 180°$
Staggered
PIPFEZ

$\ce{[Mg(NH3)2]^2+}$
Diamminemagnesium(II)
$\angle\ce{N-Mg-N} = 180°$
Eclipsed
QUQQEB

$\ce{[Ag(CF3)2]^-}$
Bis(trifluoromethyl)argentate(III)
$\angle\ce{C-Ag-C} = 180°$
Staggered
ATABAB

$\ce{[Ag(NH3)2]^+}$
Diamminesilver(I)
$\angle\ce{N-Ag-N} = 178°$
Staggered
AWEGOZ

$\ce{[Au(CF3)2]^-}$
Bis(trifluoromethyl)aurate(I)
$\angle\ce{N-Au-N} = 178°$
Staggered
QOCPAC

$\ce{[Au(CH3)2]^-}$
Dimethylaurate(I)
$\angle\ce{C-Au-C} = 180°$
Eclipsed
WEMGOK

$\ce{[Au(NH3)2]^+}$
Diamminegold(I)
$\angle\ce{N-Au-N} = 180°$
Staggered
DIWZOB

$\ce{[Au(GeCl3)2]^-}$
Bis(trichlorogermyl)aurate(I)
$\angle\ce{N-Au-N} = 180°$
Staggered
RAGGEL

Notes

There are different conformations as well as several entries that do not follow your cation-anion assignment exactly.
There is a bias for protons placement for the structures refined with Shelx due to the limitations of the riding model, so the confirmations for H-containing (methyl, ammonia) ligands might be incorrect.
The database contains about four structures with linear diethyl ether moieties, which I omitted as outliers.

Unrelated, but I'd be grateful if someone could suggest the way for preventing automatic line breaking for MathJax in Markdown tables so that chemical formulas don't split across the lines.

This is very close to exactly what i was looking for, but I appear to have made my phrasing of the question insufficiently clear. You made your search on the assumption it "should not be the subunit of an ionic network". This is the opposite of my intention. It should be a subunit of a crystalline structure, with the An positions bonding into a wider structure, but dimeric. Not sharing futher vertecies/edges/faces with other TME coords. As an example, the structure of La3FeO6, but with the isolated Fe tetrahedra arranged as isolated dimer instead, in the configuration i showed in the question.

@AChemicalEntity The moieties here are the units of the structure. I stressed out the fact they are not arbitrarily selected from the infinite chains or sheets, i.e. they are not covalently bound to other atoms, i.e. fulfilling "not sharing further vertices/edges/faces with other TMEs" criteria.

Ok, in that case it is exactly what i was looking for. Thank you.

@AChemicalEntity No prob. You can select either "Unit cell" or "3x3x3 packing" on the page for each structure (below the 3D viewport) to visualize the moieties in the context of the whole structure.

@andselisk The answer by @pawel provided here with multiple instances of   apparently does «the trick».  Indeed a bit ugly, I didn't identify a way to state something in line of "do this x times" to keep the underlying code easier to read and maintain. Eh bien ... it is an attempt.  The edit affected both the chemical, as well as the mathematical formulae (first and third column).

@Buttonwood Thank you very much for the edit and the link! Hacky, but as long as it works… I also tried to make the table more fit by removing less relevant formatting details such as braces for bond angles to eliminate horizontal scroll bar.

@andselisk While «filling up» one column, I noticed «the other» required some of these special characters; an iteration over multiple steps. And this was at the rim of «surprise», and «annoying» (which is a biased comparison between LaTeX and markdown / JavaScript) because there is trailing space in columns 1, 2, 3 as if there were trailing(!) padding. Or, markdown simply assumes non-proportional characters all on the same base line (no horizontal shrinkage for the sub- and superscripts, either) and hence has a particular difficult task. Maybe something for the site's engineers?

Sodium sulfide has an anti-fluoritecstructure, in which the sodium ions are tetrahedrally coordinated and the sulfide ions are cubically coordinated. If you look along any body diagonal of the cube of sodium ions coordinating the sulfur, you see your dimeric structure. Each sulfide ion is shared by four such dimeric arrangements oriented in different directions.
Magnesium silicide has a similar structure, but with less ionic bonding and more electron delocalization leading to a narrow indirect gap for semiconduction [2]. This electronic property has led to potential thermoelectric applications.
Reference

Dr. Hiroshi Mizoguchi, Dr. Yoshinori Muraba, Prof. Daniel C. Fredrickson, Dr. Satoru Matsuishi, Prof. Toshio Kamiya, Prof. Hideo Hosono (2017). "The Unique Electronic Structure of Mg2Si: Shaping the Conduction Bands of Semiconductors with Multicenter Bonding". Angewandte Chemie 129(34), 10269-10273.
https://doi.org/10.1002/ange.201701681

Hmm... This isn't really what I was after. I am not looking for a three-dimensional structure where you could rotate and choose the axis to pigeonhole the arrangement, but rather a structure where the dimer is a distinct structural feature within a separate matrix.

@AChemicalEntity How would the matrix be attached to the dimer? Via ionic interactions? The dimer shown in the question has 7 anions and 2 cations, so it is likely that there is a net charge.

The dimeric units share a common center. So your count multiple-counts an anion.

@Karsten The dimer is part of the unit cell of a larger inorganic crystal structure. Essentially, just a feature in that full crystal structure. Yes, it does have a net negative charge.

@Karsten To add on the previous comment, the entire crystal structure is ionic.


text:
Potassium permanganate changes from pink to clear on reacting with urine
KMnO4 was used by the cleaners to clean the water tank in our building. For a day or two all the water in our and taps cisterns was pink. On using the toilet before flushing I noticed that the colour of the water changed from pink to colourless because of urine. I couldn't understand why this happened, kindly help out.

Permanganate just did its work, being outnumbered and all spent.

Call the cleaning crew for more, maybe?

The chemical reaction is the sum of the two following half-equations, one due to urea $\ce{CO(NH2)2}$ and one to the pink permanganate ion $\ce{MnO4-}$ :
$$\ce{CO(NH2)2 + H2O -> CO2 + N2 + 6 H+ + 6 e-}$$
$$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}$$
The total of these equations needs to engage $30$ electrons, which gives :
$$\ce{5 CO(NH2)2 + 6 MnO4- + 18 H+ -> \\ -> 6 Mn^2+ + 5 CO2 + 5 N2 + 19 H2O}$$
As you see, urea destroys permanganate ions in acidic solutions.
The final solution is colorless, because it contains $\ce{CO2}$ and $\ce{N2}$ which are colorless, and $\ce{Mn^2+}$ which is nearly colorless.

Not sure about acidic enough environment. Rather expecting some organic complexes of Mn II/III, otherwise MnO2(s) could be formed.  Urine is not just urea. // You can use \ as wrapping the MJ output.


text:
Will adding sodium carbonate to a low pH solution of sodium hydrosulfide have an adverse reaction?
I am managing an HSE project to dispose of an expired NaHS solution. There was approximately 500 gallons of liquid in two tanks. Both tanks have been pumped down, and there is little residual liquid with ~ 100lbs of solid in the bottom of the tanks. The H2S readings in the tank are approaching IDLH conditions. I am assuming that over time, the pH of the NaHS solution decreased due to exposure to air which is causing the high concentration of H2S in the tank.
We want to suppress the H2S vapors so that when the solids are placed in drums for disposal, there won't be any off gassing in the headspace of the drum. Would adding solid sodium carbonate to the solid in the bottom have any adverse reaction? I am assuming that some heat will be generated due to it neutralizing the H2S, but this is taking place outdoors and we would add the carbonate slowly to minimize heat generation.

There won't be much heat, but there won't be much neutralization either. H2S is a weaker acid than H2CO3.

@IvanNeretin if it is actually $\ce{Na2CO3}$, it could work because carbonate ion is a stronger base than hydrosulfude ion. Once you have bicarbonate ($\ce{NaHCO3}$), however, you're dead.

@OscarLanzi True. But then again, what is it that we are trying to avoid? Is it the possibility of pressure buildup? Then carbonates are no better than sulfides.

Pressure buildup yes, toxicity they're different. The OP needs to clarify his objective.

For disposal of solids, from  https://www.cpc-us.com/images/2017SDS/sulfursds/SDS_Sodium_Hydrosulfide_Flake_February_14_2019.pdf : "Use a Ferric Chloride solution to precipitate sulfur and ferrous sulfide, then filter and send the cake to a landfill for industrial waste." BTW, if the pH is too low (e.g., excess NaOH added), then waste is classified as corrosive, as well as toxic.

@OscarLanzi we are trying to increase the pH of the solid/sludge mixture to decrease the rate of hydrogen sulfide evolving from the contents of the tank, so that when it is placed in closed containers in can be handled safely. Why would bicarbonate forming be a problem? It’s a weak base with a pH of 9 in solution

Bicarbonate forming would be no problem, it's just ineffective. You therefore need the normal carbonate salt and then you get only 1:1 molar neylutralization, not 2:1.


text:
How does the electron get back to the ground state at the energy level?
I have a doubt about how this emission process is carried out in the atom where we have the orbiting electron and we want the electron to return to the fundamental value.
I know that in absorption we inject a light on top of the atom the electron at a certain frequency electron becomes energized it starts to jump from the fundamental level to the excited level. I don't know what method is used when it is emitted, I would like an explanation, please, on the subject, the little I understood is by temperature, but it was not clear to me.

There are a number of ways an electron can return to ground state from a higher metastable state. Here are two:

Spontaneous transition, as in fluorescence and phosphorescence, where the electron emits radiation, removing the energy.
Radiationless decay, AKA  external conversion, in which collision with other particles transfers energy as heat.

BTW, a metastable state can be quite long-lasting. Thermoluminescence is used to date ceramic, igneous rock and other materials thousands or millions years old. A different meaning to dating for the long term...


text:
Are there real chemistry bottles with this kind of label?
I watched a movie recently which had a close-up of this weird bottle:

The way it reads is very strange to me. Is this just complete movie nonsense, or are there really bottles in real life sold with such strange labels full of exclamation marks and whatnot?

Hehe haha … no.

It is not fluoroantimonic acid (CAS Number 16,950-06-4). And IUPAC names are not registered trademarks.

Is that a glass  bottle? "[fluoroantimonic acid] cannot be contained directly in glass carboys, as it attacks glass." https://en.wikipedia.org/wiki/Fluoroantimonic_acid Also note spelling, flouroantimonic [sic].

@DrMoishePippik It could some special ingredient for making antimony-containing bread. ;-)

Something is not right. Not chemical, at least. Fluoroantimonic has u before o; the label on the questionable bottle has o before u. And Salvadeo Company Inc. does not come up on a google search.
There is some reality to the image, however. A search for fluoroantimonic acid brings up an image for Magic Acid (a blend of fluorosulfuric acid and antimony pentafluoride: Ref 1) which has some common elements: the number 17,508-0 (top center) seems to be a catalog number for 100g of the material. The OP's image and this one also contain the words spectrograde, Freeze!, Corrosive!, Hygroscopic!, and Store under nitrogen in similar positions. And with a red border. So the OP's image could be a label for Magic Acid that was promoted in acidic strength to Fluoroantimonic Acid and assigned to Salvadeo Company, but misspelled.

The bottle in the OP's image does seem to be glass, so it would not be appropriate for Fluoroantimonic Acid. The YouTube video (Ref 2) says that fluoroantimonic acid must be contained in a Teflon bottle (see image below for 50g).

Magic Acid (being only the second most acidic superacid) might be contained in glass. But the ingredient listed on the OP's label is incompatible with the material of the container.
Ref 1.https://en.wikipedia.org/wiki/Magic_acid
Ref 2. https://www.youtube.com/watch?v=UWBNcMyfiGQ

So, ultimately, the answer to the title question is an unqualified Yes. Seems like the prop department knows their stuff…

@JonCuster, not unless they learn to spell: it's flourantimonic acid on the label.

@DrMoishePippik - pretty harsh on some (almost certainly) non-chemists to get that close...

Well, I'd be concerned whether malic or maleic acid is used in a a flavored apple tart ;-)


text:
How does the Szilard-Einstein refrigerator design evaporate its refrigerant?
In their 1927 patent, Leo Szilard and Albert Einstein use butane as a refrigerant in their non-compressive refrigerator design. In the patent the authors begin the description of the principle of operation by stating:

A suitable refrigerant, for instance butane, in liquid form is contained within evaporator 1. An inert gas, for instance ammonia, is in troduced into evaporator 1 through conduit
30 and distributor head 31. The refrigerant evaporates in the evaporator in the presence of the inert gas due to the fact that the partial pressure of the refrigerant is reduced thereby and the resulting gaseous mixture passes through conduit 5 to within condenser 6. (Emphasis added)

Why would adding gaseous ammonia to the butane-ammonia mixture result in additional butane evaporating?
The quote refers to the parts of the diagram shown below. A full color-coded diagram of the refrigerator design can be found here.

What is the driving force for evaporation?

@ViníciusGodim evaporation is driven by the molar concentration (i.e. partial pressure) of gas molecules. Equilibrium occurs when the molar concentration is high enough that an equal number of gas molecules return to the liquid phase as leave the liquid phase. An change in average kinetic energy (i.e. temperature) will change the equilibrium point.

I would say the driving force is the difference between the partial pressure at the interface, which is given by equilibrium considerations (in fact it's the saturation/vapor pressure), and the partial pressure of the bulk (distant from the interface). It's all about this difference. Diluting the gas phase will increase the diference (driving force), so more evaporates.


text:
Does boric acid really rearrange phenols?

This reaction was on prepchem.com and I want to know whether it could work. Most of their reactions are well sourced but this is rather scant (and can't find the patent referenced).
It is possible that it is actually the ether that is relocated, but I can't deduce a mechanism for either reaction despite a lot of searching through papers on all sorts of boron reactions.
There are all sorts of ways that boric acid can complex with oxygen, so I feel like it might just be real, so if there is some boronic dark magic I would love to know.
Someone did claim they could explain it and said:

The OH of the phenol adds into the BOH3,
You make BOH4,
Which is an ionic,
That anion is unstable so it’ll release a OH-,
Which then nucleophilic attacks para,
Electrons move and you eliminate BOH3,
Then you rearomatize

They also drew an image that I linked below but it looks like the boric acid just gains an OH- from somewhere to make tetrahydroxyborate, and the phenolic oxygen somehow loses an electron? IDK it just looks like a + charge symbol next to it? Really confusing exchange.
https://i.stack.imgur.com/AqzcQ.png

I can see the phenol- boric acid cationic complex forming, similar to phenol-BF3, and losing water to give a neutral species, but I cannot see it losing hydroxide ion which in any case would be mopped up by the boric acid present in the reaction mixture.

Drop the first 0 in the reference and search for US patent 4013727 (seven digits, not eight). You should be able to see the original patent.

Prepchem.com apparently misinterpreted what was in the patent.
The abstract of the patent reads:

A process for preparing hydroxyphenyl ethers by oxidizing phenyl ethers with hydrogen peroxide in the presence of a ketone or with a ketone peroxide.

This has nothing to do with rearrangement. Instead we start with a phenyl ether, in this case anisole, and then the oxidation using a peroxide compound (as referenced in the abstract) plus boric acid (mentioned in the body of the patent) yields a mixture of hydroxyl-bearing products. Since the reaction with the peroxide is essentially an electrophilic aromatic substitution and the ether function is activating and ortho/para- directing, we would expect to get a mixture of ortho and para products. Neither product would be expected to rearrange to the other; they both come from the parent ether.

Thanks for that! They must have started using an automated system to scrape these, not even close. Certainly an interesting reaction, but several other reactions produce a similar mix of products and I was looking for a way to convert the unused ortho product into the para. Glad to learn about the inaccurate patent number too since I think I have come across similar erroneous numbers before and might be able to use them after all.


text:
What is ionic reaction equation for reaction between carbon dioxide and limewater?

Write the ionic reaction equation for the reaction
\[\ce{Ca(OH)2(aq) + CO2(g) <=> CaCO3(s) + H2O(l)}.\]

I first wrote the complete ionic equation as such:
\[\ce{Ca^2+ + 2 OH^- + CO2 <=> Ca^2+ + CO3^2- + H2O}.\tag{R1}\]
Then I crossed out the common $\ce{Ca^2+}$, and got
\[\ce{2OH^- + CO2 <=> CO3^2- + H2O}.\tag{R2}\]
How is this wrong?

Did you consider solubility rules?

Also, some CO2 might be considered carbonic acid, https://en.wikipedia.org/wiki/Carbonic_acid

It's a trick question. There aren't any spectator ions you could cancel out. In fact, unless you write it in steps like in Maurice's answer, there are only two ions (from the soluble calcium hydroxide). I will edit your question to add the physical states, that might clear things up. Carbon dioxide could also be written as in aqueous solution, that is a matter of preference or how the reaction is set up.

$\ce{CaCO3}$ is a precipitate. It is not made of separated $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ ions. So the equation is : $$\ce{Ca^{2+} + 2 OH^- + CO2 ->  CaCO3 + H2O}$$ This equation may be considered as two successive equations, namely :$$\ce{CO2 + 2 OH^- -> CO3^{2-} + H2O}$$ and $$\ce{Ca^{2+} + CO3^{2-} -> CaCO3}$$


text:
Is it easier to produce hydrodynamic cavitation in a liquid close to boiling point?

Cavitation is defined as the process of formation of the vapor phase of a liquid when it is subjected to reduced pressures at constant ambient temperature. Thus, it is the process of boiling in a liquid as a result of pressure reduction rather than heat addition. Source

Hence, if the liquid is already boiling, does it then reduce the energy needed to produce cavitation, of the sort which can then damage solid materials through the collapse of the bubbles?
E.g. If a pump's impeller in a water pipe is able to operate in 25°C water at just below the speed at which cavitation starts to occur and damage it, if one then raises the temperature of the water to 90°C, will cavitation then occur and start damaging it, (all else held constant)? (Obviously heating the liquid may produce/exacerbate other harmful effects also, but just interested in the cavitation effect)

Already boiling or a bit below b.p.?

@Mithoron Yea in the example a bit below bp - but its the same q essentially - does a higher temperature of the liquid mean damaging cavitation (must be damaging) is easier to produce

Your presumption is absolutely correct! This was answered on Physics Stack Exchange, too.
However, the relation between temperature and cavitation is far from linear, as discussed in Water cavitation from ambient to high temperatures by Magaletti1, Gallo and Casciola (kudos to the authors for making this open access, not behind a paywall!).
BTW, consider boiling plain water in a microwave. After boiling a while to remove dissolved air, and then cooling and reheating, a container filled with superheated liquid is ready to flash into vapor, in effect cavitating explosively. When I did that accidentally, reheating water for tea, on touching the microwave door button to remove the cup before the timer expired, the water suddenly boiled, slamming open the door from the pressure, blowing the interlock fuse. Luckily, I was protected by that door from serious scalding.

Great answer thanks. However, my q wasn't v clear have edited - is the cavitation produced in the hotter liquid also just as damaging as that in the cooler liquid? If the cavities (bubbles) have been produced because the bulk fluid pressure is closer to the vapor pressure, then is the collapse of the cavities consequently less violent and therefore damaging also? i.e. if I want to damage something with cavitation is it easier to do it with hot water or cold?

If you read at the above reference, https://www.nature.com/articles/s41598-021-99863-z , cavitation pressure increases with temp!


text:
Have we ever managed to "add" a proton to a nucleus?
Have chemists or physicists ever managed to "add" a proton to an atom in order to create a new element? For instance, a hydrogen atom has one proton; has anyone managed to give a proton to a Hydrogen atom to create a helium atom (as well as an extra neutron, for stability)?
I would imagine that this would take an immense amount of energy to do, so I would understand if this were not possible. However, even if it isn't currently feasible, is there any theoretical way to do this, possibly with technology that we don't currently have access to?

Yes, this is the basis of fusion reactors. Search on fusion and you'll find lots of info. A common reaction is combining tritium with deuterium to make helium and a free neutron, as described here:https://en.wikipedia.org/wiki/Deuterium%E2%80%93tritium_fusion

(p,n) reactions are quite common in nuclear physics research. See, for example, https://iopscience.iop.org/article/10.1088/0034-4885/50/6/001/meta - "The (p,n) reaction and the nucleon-nucleon force" in G F Bertsch and H Esbensen 1987 Rep. Prog. Phys. 50 607.

You would also have to add an electron if you want to turn a neutral atom into a neutral heavier atom, for charge neutrality.

Fusion of protium (protons) to helium is extremely slow proces even in Solar core, as conversion of extremely unstable diproton to deuteron is extremely rare. Massive and luminant stars with hot enough core usually fuse hydrogen via catalytic CNO process (Wikipedia). .

You do not need the extreme conditions of accelerators or nuclear fusion. Nuclear reactions with protons can be observed even in the pure water of light-water-moderated fission reactors. By collisions of fast neutrons with hydrogen atoms in the coolant $\ce{H2O}$ (i.e. the desired moderator effect), recoil protons are produced which can react with $\ce{^16O}$ to form $\ce{^13N}$ or with $\ce{^18O}$ to form $\ce{^18F}$.
$$\ce{^16O (p, \alpha) ^13N}$$
$$\ce{^18O (p, n) ^18F}$$


text:
Can benzene diazonium chloride always participate in a coupling reaction with any compound that contains a phenyl group?
I know that benzene diazonium chloride couples with phenol and aniline. One N atom of the diazonium ion forms a bond with the carbon at the para postion of the ring. That made me think, can a coupling reaction occur between benzene diazonium chloride and say, benzene or chlorobenzene or any compound that has a phenyl group?Ann if not, then what kind of compounds will it react with?

Only strongly activated rings - primarily what you mentioned. If it could react with benzene, you'd probably see such example.


text:
Retrosynthesis of 1-(3-chloro-5-nitrophenyl)-2,2-dimethylpropan-1-one
I was tasked with figuring out the order of attaching the functional groups  to a benzene ring for the synthesis of 1-(3-chloro-5-nitrophenyl)-2,2-dimethylpropan-1-one:

I propose the following order:

the ketone from an acid chloride;
the chlorine with $\ce{Cl2}$ and $\ce{FeCl3};$
the nitro group with $\ce{HNO3}$ and $\ce{H2SO4}.$

My logic is that after adding the carbonyl group, the chlorine group should be added because adding the nitro group second would make the ring too deactivated for the chlorine later on. However, I am not sure about the relationship between the strength of the deactivator and its ability to meta-direct.
$\ce{Cl},$ being a ortho/para director, would want the nitro group to be added ortho or para to itself, but the carbonyl would prefer meta substitution. Because the carbonyl is a moderate meta director and halogens are weak ortho/para directors, does this mean that the carbonyl’s influence would “win out”?

Your approach is reasonable. Acetophenone forms 3,5-dinitro-acetophenone upon nitration. This product undergoes selective Zinin reduction to 3-amino-5-nitro-acetophenone. A Sandmeyer reaction here would be expected to substitute chlorine for the amino group. Your ketone would likely be amenable to this synthetic route.

A follow up. You plan to start with benzoyl chloride. What is your nucleophile, t-butyllithium? I would be concerned about over addition in spite of the bulkiness of the ketone that is formed. A safer route is benzaldehyde and  t-butyllithium followed by oxidation of the secondary alcohol to the desired ketone.


text:
Stereochemistry of reaction of thionyl chloride and phosphorous pentachloride with alcohol
When a secondary alcohol reacts with SOCl2 (or even PCl5), does the reaction proceed via SN2 or SNi?

see https://chemistry.stackexchange.com/questions/171573/reaction-of-pcl5-with-alcohol/171575#171575  for PCl5

Also see: https://chemistry.stackexchange.com/questions/69063/if-socl2-reacts-with-alcohols-via-sni-why-doesnt-pocl3

For SOCl2 it depends on the reaction conditions- if a base such as pyridine is present the reaction goes with inversion, in the absence of base it goes with retention
Full explanation  here


text:
Why will changing the solvent crack the column for column chromatography?
When we perform column chromatography with silica to purify the reaction crude product, we sometimes need to change the solvent system to better extract our target product.
For example, changing from pure hexane to 1:4 hexane:DCM. However, if the polarity is not change gradually, it often leads to column cracking, which means there are air bubbles (or is it solvent bubbles?) in the column, and it will affect elution the pattern.
Why might changing solvent polarity crack the column? And where is the air coming from? I do not see how the air is getting into the column when there are solvents sitting above it.

Based on both answers, when the run is finished it may be best to run a gradual gradient back to non-polar solvent (hexane) as well.

@PeterMortensen Why a silica gel and why not a link for that one? This is some English that I, a non-native, don't understand. Can you provide more context for that, please. Are the links to such common chemicals as hexane and DCM really necessary?

The fundamental reason for cracks in packed beds is that all chromatography columns (used in organic labs or even HPLC) do not have particles packed in a regular crystalline manner. This means that the bed density is neither radially nor axially uniform. At best we have random closed packing in HPLC columns, which are typically packed at 10,000 to 30,000 psi (!!) but the "home-made" organic lab column has a loose-packed structure. This means that there are large gaps in this structure. As an analogy consider a heap of quasi-stable grape bunches (randomly arranged spheres). Grape columns can collapse when a few bunches of grapes are moved.
Now there are two types of solvents (i) aggregating and (ii) dispersive if we classify them on the basis of their suspension properties. You can readily check a suspension of silica in those solvents under an optical microscope. When you change solvents, you actually change how those loose particles stick to each other. If a bed is loosely packed, and suddenly a solvent is added which is strongly aggregating, cracks may develop in the bed.
P.S.: The solubility of gases may cause air bubbles to form at the interface of two different solvents. Helium sparging has become almost obsolete because it is becoming expensive and there is a predicted shortage of this element in the future. If you are concerned, vacuum degassing under an ultrasonic bath is a routine procedure these days.
Regardless, the punch line is that an ordinary silica bed is loosely packed. If the particles were arranged in a perfectly crystalline manner, cracks would never appear regardless of the solvent used. We can only dream about discovering a magical way to arrange silica particles in crystalline form.

Helium sparging is obsolete?  I think the "punch line" is not to change the ratio of polar and non-polar so quickly.  Any movement of column material will be a result of  permeating and swelling of silica by the polar solvent.

In this case, hexane is the "aggregating solvent".  But this is "normal phase" chromatography.  That's what you start with.  The more polar mobile phase is used to elute the components that are more attracted to the silica.

It's the polar mobile phase that's going to change the column if it is added in too abruptly.  Check the packing/storage solvent.  I would not give up on a "cracked" column without washing it for a few hours.

"When you change solvents (mobile phases), you actually change how those loose particles stick together".  This can "crack" even a highly compressed silica mass.  The polar solvent will swell a mass of silica.  Then add hexane, what happens?  But I would agree a more consistant particle size, shape, and method of packing would be contributing factors.

The model of liquid chromatography with a stationary phase (silica, alumina, ...) and a mobile phase (your eluent / eluent mixture) is a simplification. If your liquid chromatography is not based on size exclusion, then recurrent adsorption of molecules to, and desorption of molecules from the stationary phase are detrimental to the success of your intended purification.
Note, this back and forth isn't only about the molecules you want to separate.  It equally happens to the molecules of your eluent (which itself can be a mixture), which are of much higher concentration.  Hence, the dissipation of heat of adsorption can become an issue, especially for a sudden substitution of one eluent (mixture) by one other, or if the gradient to change the eluent's composition is too steep: The temperature can rise this much that outer walls of the (preparative) column feels warmer, and that solvents of lower boiling point pass this threshold.  A preparative column filled by gravity / settlement of a suspension then can break.  Perhaps someone else can back my impression with experimental data, the larger the ratio of diameter to enclosed volume of a preparative column, the more likely this is to happen.
(A rise of temperature equally can be observed preparing the silica suspension if solvent is added to still dry silica.)
VWR / Ace glass offers jacketed chromatography columns with dimensions suitable for preparative scale; in principle, their connection with a thermostat could attenuate the problem. So far, I didn't use them; their threads suggests they might anyway be designed for automated use with a fraction collector, or a MPLC:

(picture credit VWR)

Silica gel is polar.  Hexane is non-polar.  DCM (dichloromethane) is polar.
Adding polar solvent to the column too quickly could cause uneven permeation into the column packing.
But the main concern is solubility of gasses in DCM vs hexane.
When the DCM mixes in with hexane in the column, outgassing may occur.
Sparging mobile phases with helium is known to reduce outgassing and might help if running a strong gradient.
If possible, ease off on the rate of gradient change as well, and let the best column storage solvent elute through for a while at the end of each run.


text:
Potentiometric conundrum
I'm trying to develop a simple method of analysis based on the half-reaction:
$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}$$
Making up the cell is a $\ce{Ag/AgCl}$ reference electrode. The cell is only $\pu{5 ml}$.
Dress rehearsal:

I measured the EMF of $4$ zinc sulfate stock solutions from $\pu{0.5 M}$ to $\pu{1.6E-3 M}$. Each point was replicated once. I obtained the following correlation:

Using $\log_{10}$ I got a slope of:
$$\pu{8.2\pm 2.2 mV}$$
But using Nernst's Equation at $\pu{298 K}$ the value should be about $\pu{28 mV}$, so the discrepancy is enormous! And this means reduced sensitivity, of course.
Additional problems which may or may not be related to the first problem are:

reproducibility wasn't great
long equilibrium times (over $5\text{min}$)
$\ce{ZnSO4}$ turned out of poor quality

Does anyone know what causes the low slope problem?

Are you using a voltmeter designed for potentiometric measurements, i.e., an electrometer? An ordinary inexpensive DMM, in voltmeter mode, has too low input impedance, by a factor of about 100,000. And reduction of hydrogen ions to hydrogen gas is also a competitor.

No, I'm using a general purpose multimeter. What instrument would you recommend instead?

Basically the kind of voltmeter used with pH electrodes is needed. They have input impedances sufficiently high to reduce current loading to around a pA. There are various high impedance voltage follower front ends that can be built relatively inexpensively and such a buffer stage would be used between the electrode and the input to a standard 10 M Ohm input impedance DMM.

OK. Where you write And reduction of hydrogen ions to hydrogen gas is also a competitor wouldn't that invalidate the Daniell cell?

Breakfast done, snow is only an inch, so not worth removing. The zinc metal gets oxidized, so no competition with hydrogen ion reduction. That leaves the electrometer issue. One possibility, aside from making a FET-input op amp front end for the DMM is just make a concentration cell: two solution reservoirs, a salt bridge, two zinc electrodes, zinc sulfate solution in the salt bridge and reference half-cell reservoir. That solution would be relatively concentrated and effectively constant concentration.

Yes, a concentration cell. But how would that solve the problems of low slope and long equilibration times? How does it circumvent the need for an electrometer?

You get decent currents, unlike the pA with the Ag/AgCl reference electrode. So an ordinary DMM is fine. And the Nernst equation essentially then boils down to just the indicator side, so it should work well.

Okay, thanks for your input. Will definitely try the CC.

The potential you measure at a metal electrode is not necessarily only the metal couple, but a mixed potential of all the redox reactions taking place at your electrode. If there is a significant rate of any other reactions happening at your electrode you will get a different potential. You can have for example Zn oxidation whilst having proton reduction and/or oxygen reduction.

At first glance the cell potential seem close to theory, except for the slope.

In my shed's rubble I found the remains of an old pH-meter, with console ('electrometer'). I'll see if I can breathe sole life into that...


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Valence Electrons and Ionisation
For bromine within it’s electron configuration the valence electrons are 4s2 and 4p5 and not 3d10 and 4p5, my chemistry teacher explained this as 4s2 being less shielded than the 3d orbital.
Valence electrons are the electrons that get ionised correct? If the 4s2 orbital experiences more Zeff then wouldn’t it be harder to ionise as there’s a greater degree of electrostatic attraction? And the 3d orbital, since it experiences more repulsive forces, would be easier to ionise?

I really don't understand where this question is going. Are you talking about the first ionisation energy of bromine? Why would s/d orbitals be involved with that?

The first ionisation energy depends on the highest energy electrons which would be the valence electrons, is that correct? If so the valence electrons 4s has a lesser degree of electron shielding than 3d, if this is the case then the 4s electrons should experience more Zeff and thus should require a higher ionisation energy as they are more attracted. But 3d orbitals have more repulsion so therefore why aren’t they ionised first as it would require less energy?

The highest occupied orbitals are the 4p orbitals, so I really don't understand your question.

What about the second ionisation process, the electron in the 4s orbital would become ionised no?

Most likely your teacher misspoke. The $4s$ electrons clearly suffer less nuclear charge and are more reactive; in fact $4s$ and $4p$ the only reactive electrons in bromine.
In this answer the change in electronic energy levels as we add more nuclear charge is discussed: it causes the $3d$ orbitals to ultimately fall below the $4s$ in energy; $3d$ meets up with its $3s$ and $3p$ shell-mates as the electrons sort out according to their proncipal shells. With chromium and iron, the elements discussed in that answer, we need to form a positive ion to make this switch happen; but by the time we get to bromine there is enough unshielded nuclear charge even in the neutral atom to sink the $3d$ level below tge $4s$. Thus in bromine the $3d$ electrons are sunk into the core along with tge rest of the $n=3$ shell, leaving only the $4s$ and $40$ electrons to act in ordinary chemical processes.