text:
How do you achieve phase separation of a methanol / chloroform solution?
If I'm washing a methanol solution with chloroform, can I just add water in a 1:1 mixture with the methanol and expect it to break? Do I need to do anything else to cause the aqueous phase to separate (centrifugation, etc..)? Alternatively, are there any better methods, like salting-out the chloroform?

Late edit:
My intention:

I'm just using the chloroform to get rid of other compounds, so once I separate it out, I expect my molecule to be in the aqueous phase, and then to vacuum distill it out.

This likely was a misstep. Look up a solubility table (like this) and you see that chances are darn low. The best for now is to distill the solvents off (with reduced pressure) which still should be easy (no H2O). For future reference, next time, take a test tube, fill it with some of solvent A and solvent B, shake it, and watch if there is a phase separation within a reasonable time. It doesn't cost much (say 5 mL of each), but guards you against troubles for extraction and TLC.

that table says they are miscible. At least in basic testing, they don't seem to separate within a reasonable time (2h @ 25 C -- Room Temperature).

I just aaded your comment for jimchmst's answer. I think it's important to be in the question. If you don't agree, just roll over.

The OP's comment: I'm just using the chloroform to get rid of other compounds, so once I separate it out, I expect my molecule to be in the aqueous phase, and then to vacuum distill it out.

This indicate OP's in the process of isolating a water soluble compound from a (bio)metrix. Probably a natural product. In old days, there was a procedure we have follwed to isolate low polar to high polar compounds. The procedure as follows:

The extract was first dissolved in 90% methanol/water mixture (90:10 methanol:water). This solution is subjected to solvent-solvent extraction with petrolium ether (b.p. $40$-$\pu{60 ^\circ C}$) or even hexanes. Two or three time extraction removes most of non-polar compounds from the initial matrix.
The initial methanol solution is then made upto 75% methanol/water mixture (75:25 methanol:water) by adding enough water. This new solution is then subjected to solvent-solvent extraction with chloroform or dichloromethane. Again, two or three time extraction would remove most of mildly-polar compounds from the initial matrix. The main point here is even in 75% methanol/water, the two layers separate with chloroform. It is true that some amounts of high-polar compounds are also transform to organic layer (here it is chloroform or dichloromethane), yet separation is imminent.
Then again, the initial methanol solution is made upto 60% methanol/water mixture (60:40 methanol:water) by adding enough water. The new solution is then subjected to solvent-solvent extraction with ethyl acetate. Again, two or three time extraction would remove most of fairly-polar compounds from the initial matrix.
Lastly, methanol in the aqueous layer is removed in vacuo and water was removed by freeze-dry method to obtain highly polar water soluble materials.

I think, OP can follow this procedure if OP need only water solubles in the original matrix.

In step #1 why do you add the water first? I would think adding the chloroform first would be beneficial, since you get a solution, as opposed to having to force mix the two phases?

My compound if soluble in methanol (and slightly soluble in water). I think I'd just vacuum distill both of them off.

@Sargun Dhillon: To my understanding, I think you have a methanol extract and you want to remove unnecessary organic solubles to get water solubles. I think I was wrong about water solubility. Yet, this method gives you judt water/methanol solubles at the end.

Here is the phase diagram, good luck!  Addition of water will eventually give two phases but your extract will most likely distribute between them. Depending on what you were extracting the advice to carefully vacuum distill at very low temperature the solution is the best idea; then I would carefully progressively extract the residue with solvents of increasing polarity to see what extracted.  My suggestions are: pentane, methylene chloride, possibly ethyl acetate or chloroform, methanol, [if you are after fluorinated compounds perfluoro-2-propanol] water. Be observant. If done properly this can be the poor man's liquid chromatography.

Please cite illustrations appropriately without links to the search engine queries with irrelevant tracking junk.

I'm just using the chloroform to get rid of other compounds, so once I separate it out, I expect my molecule to be in the aqueous phase, and then to vacuum distill it out.


text:
What is the cathode in NiMH batteries
What is the cathode in $\ce{NiMH}$ batteries?
I’m not sure if it is $\ce{Ni(OH)2}$ or $\ce{NiOOH}$. I’m confused because I thought that these batteries are like $\ce{NiCd}$ batteries except better (more energy density, less toxic, etc.) and use a metal hydride instead of $\ce{Cd}$ as the anode (this is all info from my textbook). So I thought the cathode in both kinds of batteries was $\ce{NiOOH}$ because that was in my textbook (it said the cathode in a $\ce{NiCd}$ battery is $\ce{NiOOH}$ and mentioned nothing about the cathode for a $\ce{NiMH}$ battery).
Here are sources I've found:
$\ce{Ni(OH)2}$ as cathode:

“The components of $\ce{NiMH}$ batteries include an anode of hydrogen absorbing alloys $\ce{(MH)}$, a cathode of nickel hydroxide $\ce{(Ni(OH)2)}$ and a potassium hydroxide $\ce{(KOH)}$ electrolyte.”
https://www.sciencedirect.com/topics/materials-science/nickel-metal-hydride-battery

"Cathode: Nickel oxyhydroxide $\ce{Ni(OH)2}$.” https://www.powerstream.com/BatteryFAQ-nickel-cadmium.html

$\ce{NiOOH}$ as the cathode:

“The chemical reaction at the positive electrode is similar to that of the nickel–cadmium cell $\ce{(NiCd)}$, with both using nickel oxide hydroxide $\ce{(NiOOH)}$.”
https://en.m.wikipedia.org/wiki/Nickel–metal_hydride_battery

The cathode is made of $\ce{NiOOH}$ when the cell is new. This can also be written $\ce{NiO(OH)}$,where the oxidation degree of $\ce{Ni(III)}$ is more evident.  The cathode is made of $\ce{Ni(OH)2}$ when the cell is  discharged. Whatever the nature of the anode (cadmium, hydrogen, or any other material), the cathode is working according to the equation $$\ce{NiOOH + H2O + e^- -> Ni(OH)2  +  OH^-}$$
Usually the batery is built up with $\ce{Ni(OH)2}$ and it must first be charged by electrolysis according to the reverse reaction, which is often mentioned in the publications.

Got it. Thank you so much. I forgot to consider the charged vs discharged state of these batteries

According to Nickel Metal Hydride (NiMH): Handbook and Application Manual of Energizer Nical Metal Hydride:

The nickel-metal hydride battery chemistry is a hybrid of the proven positive electrode chemistry of the sealed nickel-cadmium battery with the energy storage features of metal alloys developed for advanced hydrogen energy storage concepts. This heritage in a positive-limited battery design results in batteries providing enhanced capacities while retaining the well-characterized electrical and physical design features of the sealed nickel-cadmium battery design.

The only difference between two kinds of batteries are nickel-metal hydride battery does not use cadmium electrode. Instead, it uses am alloy-metal hydride. The eelectrochemistry as follows:
At charge state at the negative electrode:
$$\ce{Alloy + H2O + e- <=> Alloy(H) + OH-}$$
The water in the electrolyte is decomposed into hydroxyl ion and hydrogen atom, which is absorbed into the alloy.
At positive electrode:
$$\ce{Ni(OH)2 + OH- <=> NiOOH + H2O + e-}$$
As evident, the charge reaction is based on the oxidation of nickel hydroxide just as it is in the nickel-cadmium couple.
At discharge state at the negative electrode:
$$\ce{Alloy(H) + OH- <=> Alloy + H2O + e-}$$
The hydrogen atom is desorbed and combines with a hydroxyl ion to form water while an electron is released, which is contributing an electron to the circuit.
At positive electrode:
$$\ce{NiOOH + H2O + e- <=> Ni(OH)2 + OH- }$$
As evident, nickel oxyhydroxide is reduced to its lower valence state, nickel hydroxide.
For complete review of these batteries, read here and here.

Thanks for your answer. I didn’t have the equations in my book and I suspected they would look like this but it’s nice to be sure

source: Stack Exchange

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text:
Can there be a gas-phase electrolyte?
Reading the first paragraph in Wikipedia's  I found one sentence particularly interesting:

Some gases, such as hydrogen chloride ($\ce{HCl}$), under conditions of high temperature or low pressure can also function as electrolytes.[clarification needed]  - https://en.wikipedia.org/wiki/Electrolyte

note: as mentioned in comments the "clarification needed" has since been removed.
Question: Can there be a gas-phase electrolyte? If so, is there any more in depth or academic resource that:

describes the nature and behavior of a gas-phase electrolyte
descrbies gas-phase HCl as one?

They don't. That's why "clarification..."  was added, but it should be just deleted. Probably it was a hint about plasma, but it's considered a state of matter distinct from gas.

I suspect that this is abuse of language. HCl can be electrolysed when fed to a reactor as a gas phase reagent, see eg https://www.sciencedirect.com/science/article/pii/S0013468619316512 That would not make HCl a "gas phase electrolyte". The Wikipedia page definitely needs clarification.

BTW I did not see a "clarification needed" label.

Buck Thorn, seems that the clarification label has been removed, albeit not the original claim.

@BuckThorn I made some edits to the question; maybe "No. Electrolysis can happen at a gas-solid interface but that doesn't make the process electrolysis proper. is all that's needed as an answer?

e.g. Gas Phase Electrolysis of Carbon Dioxide to Carbon Monoxide Using Nickel Nitride as the Carbon Enrichment Catalyst

I think, it is a result of the word having multiple definitions in different contexts. Specifically, the original definition in context of chemistry

Electrolyte is a solid, causing electrical conductivity when dissolved
in a liquid.

Later works extended it to non-solids, of course, but the root "lyte" (AFAIK, Greek for "earth") remained.
However, in context of technical application it is clearly different, since, for example, it is common to hear a term 'solid electrolyte'.  In addition to that, electrolytes are solutions that can be electrolized, but there are many things that can be electrolized, such as molten salts. "molten salt electrolyte", "solid electrolyte", "glass electrolyte" (sic!) are terms you can find. In this context a different definition is in effect

Electrolyte is a medium with ionic conductance.

This includes:

solutions of eletrolytically dissociating compounds
many molten salts
some auto-dissociating liquids
solids with ionic conductance

Assuming the second definition, if there is a gas medium with significant concentration of ions, it could fit. Normally, however, gases are molecules, and if the gas has high conductivity, it is usually mostly due to free electrons and falls under domain of plasma physics.
The possible borderline case is supercritical fluids. Well above pressure and temperature of critical point you can't really talk about 'gas' and 'liquid' since there is no phase transition between them. The density of the medium, however, might be high enough to stabilize positive and negative ions. Indeed, "supercritical electrolytes" are a thing in literature, though usually term "supercritical electrolyte solutions" is used. I didn't find anything about individual $\ce{HCl}$ specifically, but, for example, $\ce{H2O-HCl}$ mixtures at high temperature and pressure might qualify. Similarly, there are works on electrochemistry in supercritical $\ce{CO2}$

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text:
How can a hydrogen atom ever emit an X-ray photon?
From Scientific American, February 2014:
The Proton Radius Puzzle:

...we had to tune the laser so that it came in with exactly the right amount of energy. The atom would make the jump to the higher state only if the energy of the laser perfectly equaled the energy difference between the 2S and the 2P state. If the wavelength were a bit off, nothing would happen. How did we know if the atoms were making the jump? Any atom bumped up to the 2P state would quickly release a low-energy x-ray photon. If we found these photons, we knew we had the right energy...

But articles about the hydrogen spectral series say that even the highest-energy photons released, the so-called Lyman series, max out at about 91 nanons, which is UV territory, not X-ray....
Thinking further: when measuring the 'Lamb shift' of hydrogen, they also measured the Lamb shift of muonic hydrogen.  Perhaps muonic hydrogen has much higher energy levels and level differences than regular hydrogen? Actually, the whole point of using muonic-hydrogen to measure Lamb-shifts was because its energy shifts are greater than regular hydrogen.

Low energy x-ray, high energy UV… Lots of overlap in various definitions.

different but related: What is a “hydrogen-like” or “hydrogenic” atom?

Similar as overlapping X ray and gamma ray regions, based on the way of creation, not on energy.

I believe this is about the muonic hydrogen experiment at PSI. The electron is replaced by a muon which is 200 times heavier and shifts the energies accordingly.

In an approximate calculation the hydrogen Rydberg constant $R_y$ is multiplied by $m_{\mu}/m_e=200$ and using the Balmer type equation the energies are 200 times larger as Paul writes, so the 2p-1s energy is $200\cdot3R_y/4$

Combine the comments by Jon Custer and Poutnik and that is.

X ray photon energies are usually rendered in keV, so let's say that by the time we get to 1 keV energy we may reasonably call a photon "X-ray".
Normal hydrogen has an ionization energy from the ground state of only 13.6 eV, not enough really. It's less than five times the max for visible light, which peters out around 3 eV, and most people would allow more than that factor for the "ultraviolet" energy range.
But a muon is about 200 times as massive as an electron and, since this leaves the muon still lighter than even a hydrogen-1 nucleus, that means the ionization energy quoted above gets multiplied almost by that factor of 200. Now we're over 2 keV, with a 1s -> 2p transition taking three-quarters of that amount, and that may be reasonably considered as reaching "X-ray" territory.


text:
Why is oil solubility improved in solvents at higher temperatures?
This is in the case of extracting plant oil using a solvent extraction method for the purposes of biodiesel production.
In a paper, it is stated that increased oil yield may be due to the 'increase in oil solubility at a higher temperature' when cottonseed oil was extracted using hexane and ethanol (separately) as solvents. But it does not elaborate on why this occurs.
I had a search online, and I understand that the solubility of solids improve due to increased kinetic energy which allows the solute particles to gain more movement and break apart in the solvent. Gases seem to be the opposite. But what happens for liquids?

It can have thermodynamic, kinetic and structural reasons.
Higher temperature helps to neglect energy difference between intermolecular bonds, what increase solubility. Entropy raise due mixing helps either. But during extractions, solubility is seldom the limiting factor.
The extraction rate depends on temperature. Regardless of if it is more soluble at higher temperature or not, it dissolves and gets extracted faster.  With the same effort, one gets higher yield at higher temperature, unless there is a product deterioration at higher temperature.
Another factor may be stronger desintegration of plant structure to have better access to embedded oil.

Gases are all infinitely miscible. Gaseous molecules can occupy the whole available volume, whatever their interactions. Liquids should have the same behavior. But they don't behave like gases. Sometimes they are not miscible. Why ? The reason is the existence of H-bonds ( or other van der Wals forces).
When water is to be mixed with another substance (solid or liquid), a lot of H-bonds have to be broken. And this operation requires a lot of energy. This may be accepted if enough new H-bonds are made between water and the new solute. This happens with substances like methanol and ethanol for example. Furthermore, entropy increases. But if the number of new H-bonds is too small, as with octanol or higher alcanols, Nature is reluctant to separate water molecules from one another. Not enough new H-bonds are created in the mixture. As a consequence, the two liquids stay separated, even though it is against entropy gain.
But at high temperature, H-bonds are not so strong in pure water. At boiling point, they nearly disappear. So it is easier to mix polar substances in boiling water than at room temperature.


text:
Does lithium aluminium hydride react with conjugated dienes?
I have searched many websites, but was unable to find if $\ce{LiAlH4}$ is able to reduce conjugated dienes, such as styrene to ethylbenzene or 1,3-butadiene to 1-butene.

It doesn't, if it could you should have been able to find an example of that.

I agree, it does not reduce alkenes or poly-alkenes


text:
Why does van der Waals' equation of state give only one P, V pair for a particular temperature here?
I came across this question which I initially ignored believing that the data was insufficient:

Calculate the pressure exerted by one mole of $\ce{CO2}$ gas at $\pu{273 K}$ if the van der Waals' constant $a = \pu{3.592 dm^6 atm mol^{-2}}$. Assume that the volume occupied by $\ce{CO2}$ molecules is negligible.

I saw that I had 2 variables $P$ and $V$ and logically from what I'd dealt with with real gases I concluded that there must not be a unique solution, but the answer (which can be incorrect) mentioned a unique value of $P$.
I tried solving by writing the equation of state:
$$\left(P+\frac{a} {V_\mathrm{m}^2}\right)V_\mathrm{m} = RT$$
By writing it as a quadratic in V we get
$$PV_\mathrm{m}^2-V_\mathrm{m}RT+a=0$$
The discriminant of this quadratic is $D = R^2T^2 - 4aP$.
This implies that not all pressures are feasible since $D<0$ for some values of $P$. But for most other values of $P$, $D>0$ which implies that there are 2 values of $V_\mathrm{m}$ for a certain $P$. But looking at the compressibility factor ($Z$) vs $P$ graphs for various gases I found that for a certain $P$ there was only one $Z$ at a temperature, which means there should be only one solution? So the only possibility is that $D=0$ which yields a fixed value of $P$ and hence a fixed value of $V_\mathrm{m}$. That gave me the right answer.
But that seems really wrong since temperature alone shouldn't be able to dictate a unique state and that also implies that volume or a real gas is independent of the volume of container which sounds terribly wrong.
I strongly feel $D=0$ caused this but I don't see why not.
I would like to observe only the gaseous state, i.e, set the temperature above the Critical Temperature.

Why do you think that having multiple possible values for V is illogical?

@IvanNeretin I have looked at the compressibility factor (z) vs pressure for various curves and have found that there was only one z for a given p

Hence doesn't that mean there is only one volume for a certain (p,z) pair?

[The remaining contents of my original comment have been removed, since the edits have rendered them obsolete.] Please see: FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?.

You must have heard about the phase transitions. Scenario 1: you have 9 moles of water and 1 mole of saturated water vapor. Scenario 2: you have 1 mole of water and 9 moles of vapor at the same pressure. Is volume the same in both scenarios?

@IvanNeretin Okay makes sense if we consider phase transitions. But what if we confine our discussion to only gases, i.e, set the temperature above the critical temperature?

I couldn't comment because I have less reputation to do so, but I wanted to suggest looking at this text here https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/16%3A_The_Properties_of_Gases/16.03%3A_A_Cubic_Equation_of_State Although it worked in your case, all roots may not be identical and at subcritical temperatures, 2 of the 3 roots (if real), do have a physical significance

@Anmoldeep Thanks! That link was extremely helpful.

Fundamentally, the problem is in the problem statement. The proper form for the Van der Waals equation is
$[P+(a/V_m^2)](V_m\color{blue}{-b})=RT$
which would give a cubic equation when we try to isolate the molar volume. It is this cubic equation that usually has one or three roots that can be correlated with the possible phases, as described in the other answers. Apparently the bookmakers wanted to simplify the equation to a quadratic one, but in so doing they give it spurious mathematical properties which do not match the physics and thus confuse you.
The quadratic equation is accurate in the limit of high temperature and low pressure, where the liquid volume can indeed be ignored ... for the larger root only. The smaller root in this situation will be smaller than the liquid volume which the full VdW equation models as $b$, so properly putting in that $b$ term would blow that small root out of the water. The small root is not really a root, even approximately.
So pick the larger of the two roots which is the only one that really matches with the proper VdW equation.

The van der Wals expression can be shown graphically at constant temperature by the curve ABCDEFG in following diagramm, if the temperature is lower thant the critical temperature.

Try to follow the line. Start at A with a large volume of gas and a small pressure. Increase the pressure. The volume decreases from A to B and C. But after C, along CDE, the volume and the pressure decrease simultaneously : this is absurd and ridiculous. Something should be done to explain this stupid CDE fragment. After E, the line EFG makes sense : the pressure must increase to get a smaller volume.
The reason of the nonsense is that when going from A to B, something new happened to the compressed gas. At B, the first drop of liquid appears. The system has to be cooled, but when the volume decreases to V < V(B), the pressure and the temperature do not change. The amount of liquid gets more and more important. When it arrives in F, the gas is totally transformed into a liquid. So the behavior of the system is described by the horizontal line BDF. If on wants to still decrease the volume , the pressure becomes quickly very high, as everybody knows that liquids are nearly incompressible.
So the compression of a gas A is described by the line ABDFG, where the surface BCD is equal to the surface DEF. The line AB corresponds to a gas. The horizontal line BDF corresponds to two phases (gas and liquid). The line FG corresponds to a liquid.
The two solutions of van der Wals equation correspond to the points F and B.


text:
Why don't the MO energies sum up to Hartree-Fock SCF value?
I would like to calculate Hartree-Fock energy of $\ce{H3^+}$ ion from orbital energies. So I performed the following calculation in Gaussian 16:
# hf/sto-3g
1 1
H                     0.        0.        0.
H                     0.        0.        0.85261
H                     0.        0.       -0.85261

Then I took these two energy values from the log:
 nuclear repulsion energy         1.5516354770 Hartrees
...
Alpha  occ. eigenvalues --   -1.07569

Unexpectedly, the sum 1.5516354770 - 2 * 1.07569 = -0.599765 Hartree is not even close to "SCF Done:" value computed by Gaussian:
SCF Done:  E(RHF) =  -1.20555771776

Since I didn't explicitly require any corrections or solvation model, I wonder what contribution to SCF energy did I miss?
Input file and log are uploaded on https://github.com/nkrivoshchapov/h3plus_calculation

Bit off-topic, but I personally prefer using https://gist.github.com for this kind of thing instead of having to keep a repo around forever.

This is a standard result in HF theory, see e.g. section 3.3.1 in Szabo and Ostlund. The issue is that summing the single particle energies double counts the electron-electron interactions.

For the problem at hand, the log file of the calculation would fit in its entirety into this post. However, it is completely unnecessary as in this question Gaussian is merely an example. You'd still have the same problem if you'd use Orca, Psi4, Gamess, … , Pen and Paper. In the g16 case, the input file (and version) should be enough to reproduce the calculation with miniscule differences; scientific rigor should still stand. But, since you have posted it: Do not use NBO 3.1! That is now deprecated for at least a decade, we're at 7, now. It really has issues with theory. Don't use it. Please.

The key misconception you have is that the sum of the orbital energies is equal to the electronic energy. Compare the formulas for the two (taken from Chapter 3 of Szabo and Ostlund's Modern Quantum Chemistry)
$$E_0=\sum_i^Nh_{ii}+\frac{1}{2}\sum_{ij}^N\langle ij||ij\rangle$$
$$\sum_i^N\epsilon_i=\sum_i^Nh_{ii}+\sum_{ij}^N\langle ij||ij\rangle$$
If you simply sum up the orbital energies, you are double counting the electron-electron repulsion interactions.
For a closed-shell system, we can write the electronic energy in terms of the orbital energies, but we need additional terms:
$$E_0=\sum_i^{N/2}(h_{ii}+\epsilon_i)$$
To get Gaussian to print the core Hamiltonian (along with its components), you need to include the keyword iop(3/33=1) (or iop(3/33=5) if you also wanted to see the two electron integrals for some reason). You will also need pop=full to print out the MO coefficients so you can transform the core Hamiltonian to MO basis.
For convenience, I'll just copy the results here:
MO coefficients
0.58017   0.00000   1.42841
0.31436   0.78557  -0.94873
0.31436  -0.78557  -0.94873

Core Hamiltonian
-0.163989E+01 -0.107667E+01 -0.107667E+01
-0.107667E+01 -0.136306E+01 -0.377849E+00
-0.107667E+01 -0.377849E+00 -0.136306E+01

When this is done, we can now evaluate the Hartree-Fock Energy, which I did using Python:
>>> import numpy as np
>>> C = np.loadtxt("C.txt")
>>> H = np.loadtxt("H.txt")
>>> H_MO = C.T @ H @ C
>>> H_MO[0,0]
-1.6815253284170897
>>> e_MO = -1.07569
>>> E_NR = 1.5516354770
>>> E_HF = H_MO[0,0] + e_MO + E_NR
>>> E_HF
-1.2055798514170895

This is in much better agreement and the only reason for the deviation is that the MO energies/coefficients and core Hamiltonian aren't printed to sufficient precision.
You can also get the Hartree-Fock energy by summing the individual components of the energy printed after the SCF, along with nuclear repulsion energy (requires #p in the input line)
 SCF Done:  E(RHF) =  -1.20555771776     A.U. after    4 cycles
            NFock=  4  Conv=0.11D-09     -V/T= 2.1301
 KE= 1.066737227463D+00 PE=-4.429751734382D+00 EE= 6.058213121791D-01

Where KE/PE/EE is the kinetic/potential/electronic energy.

As well as the maths you might consider a physical picture - the orbital energy of electron 1 is from electron 1 moving in the potential of all the other electrons, so it includes the interaction of electron 1 and electron 2. Similarly electron 2's orbital energy includes the interaction with all other electrons, including electron 1. Thus if you sum the orbital energies you include each e-e electron twice. Thus you need to half it to get the right energy.


text:
Dehydration reaction of an alcohol, 2,3-dimethyl-2-phenylbutan-1-ol, with conc. sulfuric acid (at high temperature)

What's the major product of the dehydration reaction of 2,3-dimethyl-2-phenylbutan-1-ol with conc. $\ce{H2SO4}$ (at high temperature)?

I am aware that in this reaction $\mathrm{E}$1 mechanism will be followed, and I'm getting 2-methyl-3-phenylpent-2-ene as the answer whereas the answer given in my book is 4-methyl-2-phenylpent-2-ene!
The procedure which I followed is depicted in following scheme:

Help me to find out my mistake here.

Please edit your post for grammar, punctuation, and capitalization; and use MathJax for typesetting.

What's to explain? My answer is identical with my answer. How am I to explain why your answer (which I don't know) is different from the answer in your book (which I don't know either)?

Post your work and how you got your answer. Otherwise, this would be closed as a homework question in few minutes.

Does your book gives 2,3-dimethyl-1-phenylbut-1-ene as the answer, for curiocity?

Please see the edited question

@MathewMahindaratne I'm getting the same product - 2,3-dimethyl-1-phenylbut-1-ene.  That being due to migratory aptitude of phenyl group being more than the alkyl groups. Is that your reasoning too?

@TRC: Yes, I'd say so. In addition, phenyl is known for great neighboring group participation, which would be great possibility here to reduce high energy of primary carbocation.

This might be the mechanism based on the answer in your book.
The isopropyl group might shift because of the resonance stabilization due to phenyl group and inductive effect of the methyl group in the resulting carbocation. This carbocation might be the most stable one compared to the other carbocations' possible.

The issue appears to boil down to whether the isopropyl group migrates, or whether the methyl group migrates. Why should iPr migrate over Me? "Because it gives the book's answer" is not really satisfying enough.

@orthocresol True - I always knew phenyl group to have greater migratory aptitude than alkyl groups...

@cva You are correct! I checked other possibilities as well and found that this carbocation is the most stable among all (having not only resonance stabilization and inductive effect but also the maximum number of alpha-Hydrogen atoms available for hyperconjugation) and that's why it gives the major product. Thanks

Agree with @orthocresol .Further , what is the book & author ?


text:
Why does adding excess NaOH prove amphoteric nature?
When 3+ hexaaqua ions react with hydroxide ions, the following set of reactions occurs:

When we get to stage 3, a neutral ion has formed, and a white precipitate is created. Now, we know that for amphoteric species, adding $\ce{H+}$ ions will reverse this process, hence removing the precipitate. Surely this is an effective test for amphoteric nature?
However, according to my exam board, if you add an excess of hydroxide ions (hence forming a negative ion) after initially adding a small volume of hydroxide ions, and then the precipitate dissolves, this proves that the metal is amphoteric. This does not make sense to me because the definition of an amphoteric species is one that reacts with both acids and bases; yet, in this test, the species has only reacted with hydroxide ions; no acid - so there is no proof of amphoteric nature.
So, why is a species amphoteric if it forms a white precipitate after reacting with hydroxide ions, then dissolves after adding an excess?
An example of a problem is: "how do you prove lead (II) hydroxide is amphoteric, whereas magnesium hydroxide is not?" The answer is: add an excess of $\ce{NaOH}$ - magnesium forms a white precipitate, whereas lead(II) hydroxide forms a white precipitate, then dissolves. There is no mention of acid, so how does this prove amphoteric nature?
If anybody can clear my doubts or explain where I am wrong in thinking, I would be very grateful.

:/ And you know any hydroxide that doesn't react with acids? No? I guess not - that's a given, just reaction with the base needs to be checked.

The definition of an amphoteric species that you are using is akin to the definition of a dog as a furry creature that runs and bites. It is not wrong per se, but relies on a good deal of prerequisites to be understood correctly.

@Mithoron I don't see how the comment is helpful, it just potentially adds to the confusion of the OP.

@IvanNeretin I think the OP has a right to be confused. A more helpful comment would provide some guidance or ask for clarification.

According to IUPAC Goldbook:

A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.

Most common example for amphoteric compound is water, which can act either as an acid or base depending on the solute present:
$$\ce{HCl + H2O -> H3O+ + Cl-} \tag{acting as an base}$$
$$\ce{NH3 + H2O -> NH4+ + OH-} \tag{acting as an acid}$$

OP describes the problem as: "This does not make sense to me because the definition of an amphoteric species is one that reacts with both acids and bases; yet, in this test, the species has only reacted with hydroxide ions; no acid - so there is no proof of amphoteric nature".

The problem is the question does not clarify which compound is the amphoteric. The starting compound, $\ce{[M(H2O)6]^3+}$, is not amphoteric for sure. I choose the most suitable amphoteric compound in this senario as the neutral precipitate, $\ce{M(H2O)3(OH)3}$ because it can shows the rest of amphoteric compounds in the series. For instance, according to the IUPAC description, this precipitate is an amphoteric compound because it will react with both acids and bases (as  Mithoron pointed out in the comment) as follows:
$$\ce{M(H2O)3(OH)3 ->[H3O+] [M(H2O)4(OH)2]+ ->[H3O+] [M(H2O)5(OH)]^2+ ->[H3O+] [M(H2O)6]^3+}$$
$$\ce{M(H2O)3(OH)3 ->[OH-] [M(H2O)2(OH)4]- ->[OH-] [M(H2O)(OH)5]^2- ->[OH-] [M(OH)6]^3-}$$
Note: Strictly speaking, according to the IUPAC description, all of these metal hydroxides except for $\ce{[M(H2O)6]^3+}$ and $\ce{[M(OH)6]^3-}$ are amphoteric compounds since each can go either way with an acid or a base.
For more examples of amphoteric compounds including lead(II) hydroxide, read here and here. For example, lead(II) hydroxide is an amphoteric compound, which shows following reactions with both acids and bases:
$$\ce{Pb(OH)2 (s) + 2HCl (aq) -> PbCl2 (aq) + 2H2O}$$
$$\ce{Pb(OH)2 (s) + 2NaOH (aq) -> Na2Pb(OH)4 (aq)}$$

Thank you for the response.
I understand why the species is amphoteric. I just don't understand why if a species reacts with an excess of NaOH and dissolves, it is amphoteric. Could anyone help to clarify this?

An example from my text book is:
https://paste.pics/D2MUR
Apparently, the second reaction proves that lead hydroxide is amphoteric...?

@Matthew: I edited according to your concern. Hope you understand it better now.

@MathewMahindaratne Thank you, this helps a bit more; I now understand that M(H2O)3(OH)3 is the amphoteric species.
However, I am still unsure as to why only adding an excess of NaOH (and seeing if M(H2O)3(OH)3 dissolves) proves amphoteric nature. Surely, like you described, you would also need to add acid to prove that M(H2O)3(OH)3 reacts with both acids and bases and is thus amphoteric?

Here is a reference question from my exam board which explains why I have concerns over this:
https://paste.pics/D2NVH


text:
Would a solution with a hydronium concentration of 1e-10 be basic?
So obviously the pH of the solution would be 10, which is basic on the scale. But isn't adding hydronium ions to a neutral solution that is water fundamentally acidic.
Or is it that water itself has a concentration of 1e-7 hydronium ions at any given point making any addition of hydronium ions tip the scale to acidic?

Does this answer your question? Calculating the pH of a highly dilute solution of HCl

Are you asking about a solution which actually has $\ce{[H+]} = \pu{10^{-10} mol dm^{-3}}$, or are you asking about one where $\pu{10^{-10} mol dm^{-3}}$ of $\ce{H+}$ has been added to water?

Obviously, we have to take account of the concentration of hydronium ion released from water.
Water releases 10-7 ions and the other substance used in the given problem produces 10-10 H+ ions. So total concentration is 10-7 + 10-10.
Taking negative log of the stuffs, we get the result pH = 6.999956
The result shows that since the substance is very dilute, it hardly could make a major change in the pH. But never the less, keeping true to its nature, it made the solution only a slightly acidic (<7).
Remember, it's never possible to create a basic solution by just adding H+ ions to water..!


text:
Miscibility gap of heptane-methanol mixture
To explain my question a bit better: I added heptane and methanol together in equal ratios (1:1) and observed a separation in phase where there appeared to be an oily layer sitting on top of the other liquid.
But when I increased the amount of methanol in a ratio of 1:2 and 1:3 (heptane to methanol), this appearance disappeared.
I understand that heptane and methanol are immiscible, but what led to the change? I apologize if this is something really obvious, but I would greatly appreciate any help.

Methanol is a polar solvent; heptane is very non-polar. If you start with methanol as solvent, you can dissolve a little heptane into it; if you keep adding heptane, you reach the solubility limit of the solute heptane in methanol, and the excess heptane floats.
If you start with heptane as solvent, you can dissolve a little bit of solute methanol into it; at its solubility limit, excess methanol will sink.
Now, when you take methanol and heptane in a 1:1 ratio, both solutes have exceeded their solubility in both solvents, and you should wind up with a heptane layer on top, with a little bit of methanol dissolved in it, and a methanol layer below, with a little heptane dissolved in it. In some cases, a metastable intermediate layer forms, but will separate into the upper and lower layers. (Those molecules just can't get it all figured out instantly!)
The phenomenon is called a miscibility gap, where it seems that two liquids go together but only so far. Data on methanol-heptane were not available, but a phase diagram of methanol-cyclohexane is shown below:

The x-axis goes from zero methanol (i.e., all cyclohexane) to all methanol (no cyclohexane). The large central area is the two-phase region. At each extremity, whether methanol-rich or cyclohexane-rich, there is some solubility, which increases if you raise the temperature, until at about 320K (47$^o$C), the mutual solubility equals complete miscibility, and at this temperature and higher, there is only one phase.
A somewhat different plot (of enthalpies of hexane and methanol mixtures) shows the same kind of miscibility gap. Hexane and methanol are completely miscible above about 45$^o$C, but as the temperature is lowered, mixtures with ratio of 1:1 will separate into two phases. Mixtures with only about 20% of the other solvent are stable at room temperature. But from the chart, it would seem to be a good bet that reducing the temperature will make the miscibility gap even wider.

If I were to give a simple definition of the miscibility gap, would this work: "when two liquids with partial miscibility are mixed and become completely miscible at a specific temperature or higher". I'm having difficulty comprehending the concept, so I'm wondering whether this is an appropriate way to describe it.

@alwaysconfused_kat: Sure, but two liquids with partial miscibility (i.e., limited solubility) are mixed, there is no absolute requirement that there be some temperature above which both are completely miscible and still liquid (i.e., below the boiling point). Of course, above the boiling points of both, the vapors (gases) will be freely mixable.

There is another two-phase vs temperature novelty in practical chemistry: ethoxylated surfactants frequently have cloud points, which are temperatures above which a solution will separate into two phases. The polyoxyethylene group become less soluble in water as the temp rises, and the surfactant, which is homogeneously dispersed in micelles at lower temps, de-solves, rises and become a distinctly separate phase. Dishwasher detergents use this phenomenon to limit foaming inside the dishwasher; not a property of dishwashing (sink) detergents, which favor foam.

The disappearance of boundary is an optical illusion. Whenever we see two boundaries in a liquid, it means that their refractive index is different. If the composition of two phases is such that their refractive indices are very close or match, the phase boundary disappears. It does not mean that separate phases do not exist.
Heptane and methanol are partially miscible though, in heptane methanol system, the mole fraction of methanol is stated to be 0.270 at room temperature.*
*Ref: Savini, C. G., D. R. Winterhalter, and H. C. Van Ness. "Heats of Mixing for Partially Miscible Systems: Methanoi-n-Hexane and Methanol-n-Heptane." Journal of Chemical and Engineering Data 10.2 (1965): 171-172.
Also note that you did not mention the purity of heptane. Low grade heptane from cheaper sources is always contaminated. If you really care with this phenomenon, you have to use the highest purity solvents like LC-MS grade solvents.

Mr. down voter would you elighten us with your wisdom please rather than pressing buttons and walking away?

I agree with you that downvoters should make an explanatory comment. (I didn't downvote your answer.)

Down voting is not a bad thing if done properly with a constructive tone. The problem is this button is available in every hand and we ave no clue if grade 5 student did it or an Einstein was not satisfied.


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How Sodium Nitrite Interacts with Air and Humidity
What happens to integrity of sodium nitrite (NaNo², CAS: 7632-00-0) of AR grade when it gets exposed to air?
If container has been opened and chemical has been exposed to air for a few minutes and then closed again, it's integrity keeps getting affected? Is purity affected as well?
If air happens to contain humidity, does it potentiate the effect on chemical's integrity?
Does this type of chemical expire and how long can it be stored unopened? What happens after chemical has reached its expiration?
How long can it be stored after being exposed to air and then closed in it's container for later projects? Additionally, as far as integrity and purity are concerned, is it reliable for repeated use for important project?
Thanks

"Sodium nitrite comes in the form of white or off-white powdery crystals that gradually oxidize to sodium nitrate. It is hygroscopic and easily soluble in water (making the water solution alkaline)" according to this source https://www.ube.com/contents/en/chemical/industrial_chemicals/sodium_nitrite.html. Clearly it should be stored in a dry cabinet and exposed to the air as little as possible.

What is $\ce{NaNO^2}$?


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Predicting the solubility of sulfate salts
I came across a problem that asks to predict the solubility of alkaline earth metal sulfates: Magnesium, calcium, strontium and barium sulfates. I know that the spontaneity of dissolving can be roughly understood, in terms of lattice energy and the hydration enthalpies of the resulting cation and the anion. In this case, the anions are the same, so the enthalpy of hydration of an anion is the same for every salt. As for the lattice energy and the enthalpy of cation hydration, these two get smaller as the cation gets bigger. They have an opposing effects on solubility. so, how is it possible to predict the relative solubility at all in this case?

The question is not identical but there is an answer suitable for both questions.

From that question, I found this website "http://www.chemguide.co.uk/inorganic/group2/problems.html", the explanation to me is more convincing than the ones in the question itself. I found the answer.


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Reactions of P2O5 and P4O10 with sulfuric acid
I have  seen two reactions of $\ce{H2SO4}$, in one it reacts with $\ce{P2O5}$ and in the other $\ce{P4O10}$.
The Reactions are as follows:
$$\ce{H2SO4 + P2O5 -> SO3 + 2HPO3}$$
$$\ce{6H2SO4 + P4O10 -> 6SO3 + 4H3PO4 }$$
Since it is a dehydration reaction the oxidation states of sulphur and phosphorous would be the same in reactant and product, this is true for both the reactions.
Are the both Products correct $\ce{HPO3}$ and $\ce{H3PO4}$  correct?
If yes what I am missing?

P2O5 and P4O10 is the same thing.

I agree with Ivan Neretin. Basically, $\ce{P4O_{10}}$ is the dimeric form of  $\ce{P2O_{5}}$.

To clarify, read here.

Yes @IvanNeretin ... I agree with it but why two different products . Then may be one of them is incorrect then which one ?

@MathewMahindaratne I object. P4O10 is the only form of P2O5 (not counting the polymeric forms, that is). There is no monomer of which it is a dimer.

As for the reactions, both are correct.

@Ivan Neretin: Well received. I was just fugaratively speaking. See my answer for confirmation.

Note that $\ce{P4O10}$ is one of four polymorphs of phosphorus pentoxide and is meta stable, prepared by its rapid vapor condensation. The other three are polymeric. It makes an analogy to elemental phosphorus. // Therefore,not every $\ce{P2O5}$ is $\ce{P4O10}$.

@AyushShankaram A more specific question would be: "Which product formation is more preferred, H3PO4 or HPO3?". Since the anhydride(P4O10) is removing water, it will always want to form the highest hydrate possible which is H3PO4  ($\ce{P4O10 + 6 H2O → 4 H3PO4}$). HPO3 is the lowest hydrate of the anhydride and it forms when phoshoric acid loses water through heat ($\ce{H3PO4 ->[\Delta] HPO3 + H2O}$). Since the reaction doesn't involves heat, it proceeds to form H3PO4.

@Nilayghosh The question is, if water activity in anhydrous H2SO4 is high enough for forming H3PO4. It may quite well be equivalent to heating of H3PO4.

Phosphorus pentoxide is the common name of a chemical with molecular formula $\ce{P4O10}$. The common name is derived from its empirical formula, $\ce{P2O5}$. This white crystalline solid is the anhydride of phosphoric acid:

It is a potent dehydrating agent and used as a powerful desiccant (as indicated by the exothermic nature of its hydrolysis) in limited capacity:
$$\ce{P4O10 + 6H2O -> 4H3PO4} \qquad \Delta H = \pu{-177 kJ}$$
However, the desiccating power of $\ce{P4O10}$ is strong enough to convert many mineral acids to their anhydrides. For examples, $\ce{H2SO4}$ is converted to $\ce{SO3}$:
$$\ce{P4O10 + 6H2SO4 -> 6SO3 + 4H3PO4} $$
Therefore, it can be concluded that the second reaction is the correct one (although it may or may not be balanced). The first one is incorrect, solely based on the fact that $\ce{P2O5}$ does not exist.

Hm, I have never seen P4O10 during describing dyhydration reactions to produce H3PO4. Especially when dehydrating other dehydrating agens. I usually see written the product as polymetaphosphoric acid (HPO3)n


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Grignard Reagent Facilitates Epoxide Opening
I have read from my textbook, that the opening of an epoxide ring with a strong nucleophile like a Grignard reagent, takes place from a less substituted side, with a higher SN2 character.
The acidic workup of the product obtained above gives alcohol.
Also, if ring-opening takes place in an acidic medium, it takes place with a higher SN1 character (even though carbocation formation doesn't really take place) and the epoxide ring opens from the more substituted side
My question is if one can protonate the epoxide in an acidic medium, then facilitate the reaction with Grignard Reagent (RMgX), would the regioselectivity be according to SN2 type or SN1 type, i.e. from the less substituted side, or more substituted side
I propose that since the epoxide ring is now protonated, it can easily leave than it's unprotonated form, and the Grignard Reagent can now attack from the more substituted side as well.
Edit:
The question I was talking of in the comments.
According to my teacher, the H+ is the reminscent from the epoxidation reaction as benzoic acid is formed as a product. I can see the the epoxide gets protonated, but the subsequent attack of Grignard Reagent is what confuses me

Answer given is A part, which corresponds to ring-opening from more substituted side.

Yeah, no, you can’t do that. Grignard plus acid do not go well together.

There are some nice examples of using Lewis acids + Grignards to control/change selectivity of ring opening. But as above, protic acids are a no go

I agree with you guy 1000%, even I understand acids turn Grignard reagents back to alkanes, the thing is, I have a question in my textbook which I have added in the edit above.

@Anmoldeep That H+ is most probably signifies acidic hydrolysis.

@NisargBhavsar But acidic hydrolysis of epoxides also requires a nucleophile, which is the Grignard reagent in the next step, which loops back onto my question, with what regioselectivity will the ring open

This looks like an error in the textbook. If the epoxide is protonated then the MeMgBr reacts with the proton to generate methane. As noted above lewis acids such as BF3 will bind to epoxides and enhamce their reactivity.

@Waylander, ok so there is no possibility of reaction here, MeMgBr even if in excess cannot open the epoxide?

If you use excess Grignard to neutralise all the H+ then eventually you will get Grignard attack on the epoxide, but why do it this way? You will get opening to give B and are using MeMgBr as a base, using up a costly reagent for no gain. If you want the epoxide opening to give A use BF3 as the complexing agent on the epoxide

@Waylander, thanks a tonne, I get it now


text:
Nitration of haloarene
Question: Convert benzene to 4-bromonitrobenzene.
My attempt:

Answer given:

Is my solution correct? Can aryl halodes react directly with $\ce{AgNO2}$ to form aryl nitrate?

AgNO2 may react with the two bromine atoms.

Or, in this case, with neither - this is a method for aliphatic nitro compounds not aromatic.

(1) The bromine workup in your answer can be quite messy. If you increase the concentration of bromine, it can lead multiple isomer(both o- and p-isomer). It can also lead to formation of the tribromo- compound. (2) Why do you think AgNO2 will react with one bromine atom? (3) AgNO2 does react more with alkyl halides than aromatic halides


text:
Why don't carbon nanotubes behave like quantum dots?
If carbon nanotubes (CNTs) have the size required to have an bound electron energy state, why don't CNTs behave like quantum dots, such as having optical tunability, amongst other properties?

What is optical tenability? What other properties you are looking for? Also, have you considered that nanobutes are 1D while quantum dots are 0D structures?

But somehow they do. It is like asking why a molecule is not a quantum dot. It is, and a very small one. What is more quantum dot than a polyene?

Carbon nanotubes are 1-d materials which are confined in 2 dimensions. You cannot compare it with quantum dots which show confinement in 3 dimensions.The difference in optical properties maybe be primarily due to the very small size and 3-D confinement of quantum dots.

If anything, QDs are 0D, not 3D. Sorry, I don't find this answer to be convincing or comprehensive.

@Isaac said that the confinement is 3D (ie the structure is a 0D)


text:
Does the Atomic Mass listed for each element on the periodic table reflect the relative abundances of their isotopes on Earth? Or in the Universe?
Somehow, I cannot find a site or book or paper explaining exactly how the average atomic masses for the elements on the periodic table are weighted....
I posted a question either here or on Physics S.E. about the possible inclusion of atomic mass defects and such on the calculations of average atomic mass, but apparently forgot to mention WHERE the isotopes are coming from...
I thought I had posted this question before on S.E. somewhere, or someone else had, but I cannot find it now....
If I had to guess, the isotope abundance ratios usually listed, and their weighted average that makes up an element's 'Atomic Mass', is based only on relative isotope abundances found on Earth (not including mantle or cores), and, maybe, the rest of the easily-visible Solar System...

Related: https://chemistry.stackexchange.com/questions/87929/is-atomic-weight-universal/87930#87930

https://chemistry.stackexchange.com/questions/33056/is-the-relative-natural-abundance-of-isotopes-of-an-element-the-same-everywhere

Most of values are are for earth, or at least the portion of earth we can get our hands on.

I think is even more practical. Meaning that it reflects the isotopic composition of whatever (not enriched) chemical that can arrive at our bench / reaction vessels.


text:
Does Ei or pyrolytic elimination reaction undergo carbocation rearrangement?
I've read about E1 reaction (unimolecular elimination reaction) which forms the most stable carbocation first and then undergoes the elimination.
Ei is also mentioned to be "unimolecular" in my book, where its rate α [substrate].
Question:
Can anyone give an example showing whether or not it processes the formation of the most stable carbonation?

Read here for better description.

The answer to your question is no! Ei is an elimination reaction which doesn't doesn't undergo any carbocation rearrangements and also it does not depend on carbocation stability, the carbocation rearrangement generally occurs in acid catalysed dehydration. pyrolytic elimination reactions proceed through a cyclic transition state, this is a beta elimination with syn stereoselectivity.
This kind of elimination occurs in:
1)esters when they are heated to 500 degrees also called pyrolysis of esters.
2)tertiary amine oxides when heated till 100 degrees also called cope elimination.
3)xanthates when heated to 200 degrees also called pyrolysis of xanthates.
All of the above reactions yield alkenes through elimination mechanism via a cyclic transition state.
Here's a quick mechanism for these reactions:
1)cope elimination

2)pyrolysis of ester

3)pyrolysis of xanthates(Chugaev reaction)

these are the mechanisms and its important remember that the alkene is the major product and alkenes are formed because of internal elimination reaction, this internal elimination is favored in above cases because it forms a cyclic transition state which is very stable.
These cope reaction in particular is very useful as it proceeds through 5 membered ring transition state which needs relatively low temperature.
but for Chugaev reaction and pyrolysis of esters the transition state is a 6 membered ring which needs more temperature than cope reaction.
I hope this helps clarify your doubt!!

Can you change the images? It is hard to see. It is better to use a transparent background. It is more better to use chemical structure drawing softwares. See: https://chemistry.meta.stackexchange.com/questions/3736/how-do-i-draw-chemical-compounds-and-chemical-reaction-mechanisms-and-add-it-to/3737

Why are there so many dislikes? :(


text:
How to derive the number of spectral lines?
Recently, I found that when electrons in an atomic sample de-excite from a higher energy level ($n_2$) to a lower energy level ($n_1$), the number of spectral lines observed in the spectrum is
$$\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$$
Can anyone please tell me how to derive this?

When you move from level $n_1$ to level $n_2$, the total number of energy levels are $n_2-n_1+1$ (including $n_1$ and $n_2$).
Now, you want to find the total number of possible spectral lines in the atomic sample.
Note that if you chose any two energy levels (say $n_i$ and $n_j$), you will get a unique spectral line corresponding to those energy levels.
So, the total number of spectral lines possible are $$\binom{n_2-n_1+1}{2}=\dfrac{(n_2-n_1+1)(n_2-n_1)}{2}$$

Great answer and is the number of combinations of $n_2-n_1+1$ objects taken 2 at a time. Just for completeness, the brackets mean $\displaystyle \binom{a}{b}=\frac{a!}{b!(a-b)!}$

This expression is also equivalent to $\Sigma(n_2-n_1)=\Sigma{\Delta n}=n+(n-1)+(n-2)+...+1$

@Apurvium That’s because that is another way of showing that through rearranging the possible terms and doing summation! :)


text:
What does water saturated oxygen mean?
What does water saturated oxygen mean in this question:

The vapour pressure of water at 80º C is 355 torr. A 100 ml vessel contained water saturated oxygen at 80º C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50.0 ml, vessel at the same temperature. What were the partial pressures of oxygen and of water vapour, what was the total pressure in the final equilibrated state ? Neglect the volume of any water which might condense

Oxygen saturated by water vapor, similarly as air saturated by water vapor, i.e. having 100% relative humidity.

Does that mean the water(l) and water(g) are in equilibrium?

Yes, if water(l) is present. It need not to be.

Please, Prabodh ! Try to start some beginning of calculation. The Forum will never do all the job, and you nothing. We are here to help those who try to solve their homework and are stopped by a difficulty. We will help them. So go ahead, start some trials. Tell us where you are stopped. We will help you.

@Maurice In my understanding, the OP have not asked us to solve the task, but to explain the expression, providing the task as the expression context.

@Poutnik Indeed I am not asking to solve the task, I was just confused by the terminology.

Consider the partial pressures separately. If you answer these questions you have your own solution.
Oxygen does not condense, may be deemed an ideal gas. What does halving the volume do to its partial pressure? (We assume that exactly half the volume, 50 ml vs 100 ml, is available because we neglect the liquid volume in this problem.)
Water can condense, adjusting its partial pressure back down to equilibrium. What would that restored equilibrium pressure be versus the vapor pressure given in the problem?
Finally, what does not condense is still essentially an ideal gas, add the partial pressures to get the total pressure.
Good luck!


text:
Best way to electrochemically etch metal
I've got some plates of cast iron that I'd like to etch electrically  in a pattern by masking certain portions, as is commonly done for chemical etch.  I found several sources all indicating that the metal to be etched should be positive with respect to the counterelectrode.  So I hooked up about 3VDC to the work piece to be etched and negative terminal to the counterelectrode, all submerged in a saltwater bath. This voltage sufficed for about 10 amps for the ~ 5cm^2 pieces I am testing. When I do this  I get bubbling off the counterelectrode, and none off the work piece.  Is this to be expected? I was imagining that the active bubbling was the etching occurring in which case I'd want to put the counterelectrode positive w.r.t the work piece.
In any case will adding some acid eg. 10% HCl speed things up? I need to get thru 3mm iron.

drive-by downvotes are so annoying - at least state what you think is missing

I have down-voted, because 1. I don't fully understand what you are trying to do, i.e. do you just want to dissolve the iron 2. there is not enough detail about the actual conditions to count as a chemical experiment, 3. "best" implies that you are asking for an opinion as no rigid criteria are given 4. the question is better suited somewhere else, maybe [diy.se] 5. There is little to no actual evidence that you did do some research on what you are trying to attempt.

Flake type ( grey) or other types of cast iron ? The graphite flakes may have a effect ; there are some ordinary corrosion situations where the iron is removed leaving a surface of mostly flake graphite, ( "graphitic corrosion").

The question is really in need of an answer. Chemical expectations are that bubbling = a reaction, therefore, no bubbling = no reaction. This especially affects non-chemists, who add NaHCO3 to vinegar to clean sinks and such. We need to explain how the absence of bubbling can still cover a rapid reaction or useful chemical effects - or how bubbling is a positive indication but no bubbling is not proof of inactivity.

Bubbling off the (negative) electrode is hydrogen being evolved as sodium cation is drawn to that electrode. Your electrolyte also has a negative chloride ion (anion) which is being drawn to the (positive) cast iron piece, but instead of evolving as Cl$_2$ or even O$_2$, it attacks the cast iron and dissolves it as Fe$^{+2}$. Instead of extracting electrons from the anion, the process is extracting electrons from the iron atoms in the cast iron and putting these iron cations into solution - you should see a slight color change.
So, consider that the bubbling at the cathode represents a certain amount of reaction which is happening invisibly also at the anode. If you hook an ammeter in series with your circuit, you can determine the rate of iron dissolution.
In my experience, electrolysis is good for electropolishing or for small amounts of reaction, but 3 mm of iron is going to take a long time, depending on the area involved. I would suggest using hydrochloric acid to start, perhaps tempered with urea. Cleaning and etching solutions are commercially available, or you can make your own from urea and muriatic acid (https://images.homedepot-static.com/catalog/pdfImages/b3/b3dfe4ac-18ae-40c1-be18-81d192a75709.pdf). After you get about 2.5 mm etched off, you can swith to electrolysis if you like.

Please don't use MathJax this way: Cl$_2$. It has the potential to completely break on other screens than yours and after updates and changes. Either use \ce{Cl2} (preferred) or Cl2.


text:
Non - Polarizable Electrode/Interface
Non-polarizable electrode or interface is an electrode without charge separation on electrode/electrolyte interface or without double layer at which electrochemical reaction happens infinitely/arbitrarily fast.
Platinum electrode used in SHE redox system is an example of such electrode.
However, if there is no charge separation how does SHE develop its equilibrium potential? Since developing electrode potential requires development of electrical double layer?

Developing electrode potential different to equilibrium potential requires development of electrical double layer, acting as the capacitor dielectricum.

Equilibrium potential also requires double layer formation, whenever there is excess charge on either phase or polar molecules it will form.


text:
Using samples in titrations
I've never done an experiment before. So, the methodology provided in this pdf is quite confusing.
Link to the pdf: https://www.canterbury.ac.nz/media/documents/science-outreach/vitaminc_iodine.pdf
So, this pdf here details the methodology for conducting an iodometric titration to find ascorbic acid content in fruits/vegetables. On page 2 under "Method" section, its written how to make a 100 ml of sample solution with 100 g of the vegetable.
But in the titration experiment, we use only 20 ml of this sample solution. If we're taking only 20 ml for titration, wouldn't the tested ascorbic acid content be low? I mean, if you tested only 20 ml of the sample solution, then how is it representative of the amount of ascorbic acid in 100 g of that vegetable?

If you have prepared 100 mL of the solution to be analyzed, and if you only use 20 mL of it, it means that you are supposed to repeat this titration up to four times, in case of difficulty.

Obviously also a 20 ml aliquot from 100ml of solution contains 1/5 of the total analyte. You typically use multiple samples to get a reasonable mean and a standard deviation for the analysis which will indicated good technique.

There are (at least) two reasons to consider: accuracy and precision.

Accuracy as in preparing a stem solution of $\pu{100 mL}$ with a balance right to say $\pu{0.1 g}$ or a graduated cylinders right to the full $\pu{1 mL}$ is easier, than for $\pu{20 mL}$.  Depending on the class, volumetric flasks may be better in terms of absolute / relative error than graduated cylinders; look this up.

Precision as eventually drawing an aliquot of $\pu{20 mL}$ of $\pu{100 mL}$ of a stem solution per individual measurement allows you to perform the analysis four times, to determine at least arithmetical mean value, standard deviation, interval of confidence and identify outliers. (Well, with four recordings this is not as solid as with e.g., 20 measurements, yet you may look up a $t$-value table and engage a Dixon, or a Grubbs test).

I understand the reasons about precision...but still, if I'm trying to find the amount of ascorbic acid in 100g of a particular vegetable, I don't get how using just 20ml of the solution is gonna give me that answer...also, if I'm taking only 20ml, then I would not even know how many grams (out of 100) have I tested out of the vegetable. I may get a precise answer with 4 repeats, but I wouldn't be able to say in "x" amount of vegetable, I found a particular amount of ascorbic acid, because I don't know the "x" in the first place i.e. the tested amount of vegetable.

@ZetaFox  Say you start with $\pu{100 g}$ of lemons to prepare $\pu{100 mL}$ solution. You assume all of the vitamin of the fruits now is in this stem solution. From this, you draw an aliquot of $\pu{20 mL}$, and determine this fifth of the $\pu{100 mL}$ stem solution contains $\pu{0.5 g}$ ascorbic acid. Because this reading is only about a fifth of the sample, you infer the total in $\pu{100 g}$ of the lemons was five times as much, i.e. $\pu{2.5 g}$.  Ok?


text:
Calculating molarity of iodine solution after reaction
I made an iodine solution referring to the methodology provided in this link:
https://www.thoughtco.com/vitamin-c-determination-by-iodine-titration-606322
$$\ce{KIO3 + 5KI + 3H2SO4->3I2 + 3H2O + 3K2SO4}$$
I want to calculate the molarity of the iodine solution produced from reacting $\pu{0.268g}$ of potassium iodate and $\pu{5g}$ of potassium iodide. I'm not sure how to calculate that.
I tried finding the limiting reactant and calculating the number of moles based off of that, but I'm not sure if that's right. Also, in order to calculate molarity you need volume. Based on the procedure outlined in the link, first I dissolved $\ce{KI}$ and $\ce{KIO_3}$ in $\pu{200ml}$ of water, but then I'd to add $\pu{30ml}$ of sulfuric acid and make it up to $\pu{500ml}$. So, I wonder what volume should I take into account when calculating molarity, $\pu{200ml}$ or $\pu{500ml}$?

You must calculate the molarity in the final $\pu{500 mL}$ solution. If you do it in the $\pu{200 mL}$ solution, it is no use, because there is no reaction, and so no iodine $\ce{I2}$ in the solution containing just $\ce{KI + KIO3}$, and no acid.
Now, if you want to solve this problem, the first thing to do is to calculate the amount of $\ce{KI}$ and $\ce{KIO3}$ in moles.
Well ! $\ce{ 0.268 g KIO3}$ contains $0.268 g/214 = 1.252·10^{-3}$ mol $\ce{KIO3}$. Do the same calculation for $\ce{KI}$. You'll see that this product is present in rather large excess (about $25$ times too much). The only important product is the potassium iodate.
Mixed with $\ce{KI}$ and $\ce{H2SO4}$, $\ce{KIO3}$ reacts according to the equation given in your reference : $\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$. As a final result, the amount of iodine $\ce{I2}$ is $3$ times the amount of iodate. So you will obtain $3·1.252·10^{-3}$ mol $\ce{I2}$. We will not do all the calculation. Do the calculation yourself ! You will obtain $a$ mol. This iodine reacts with the iodide ion in excess according to $\ce{I2 + I^- -> I3^-}$, which can also be written $\ce{I2 + KI -> KI3}$.  As a consequence, you obtain a solution containing the same amount $a$ of  $\ce{I3^-}$ (in moles) dissolved in the final volume $\pu{500 mL}$. Do the calculation of the concentration $\ce{[I3^-]}$ which is numerically equal to the concentration $\ce{[I2]}$. Some teachers wants the student calculate the amount of $\ce{I2}$ produced before reaction with $\ce{KI}$.

Hello, I did the calculations and found the answer to be 0.0075M. Now, I wanna ask the possibility of calculating molarity of this iodine solution through titration. I've already done the experiment where I titrated it with standard ascorbic acid solution. I did some calculation and found the answer to be somewhere around 0.005M. Why is this so? I expected the answer to be the same (0.0075M) since it was the same iodine solution...do you think its just a calculation mistake or is it wrong to find molarity through titration for this problem?

Of course the molarity of the iodine solution can be calculated by titration, provided you are sure of the concentration of the ascorbic acid. And you should get the same result.

I see, so there must be a mistake in my calculations. I'll check that. Btw, I'd like some clarification on your answer. You said that mixed with KI and H2SO4,  KIO3 produces I2. And then this reacts again with KI to produce I3− (triiodide). First question, does KIO3 react with only H2SO4 to produce I2 or do all three KIO3, KI and H2SO4 react to produce I2? Secondly, why does I2 react again with KI to produce I3, is it because KI is in excess? If the final product is I3-, then why is I2 included in the equation and not I3-?

In solution $\ce{KIO3}$ does not react with $\ce{H2SO4}$ or with $\ce{KI}$ alone. In solution $\ce{KI}$ does not react with $\ce{H2SO4}$ alone. But, if mixed together, the three substances $\ce{KIO3, KI, H2SO4}$ do react and produce some $\ce{I2}$. Now $\ce{KI}$ reacts with $\ce{I2}$ because it is in excess.. In fact, if you know that $\ce{KI}$ is in excess, you can add the two equations. But if $\ce{KI}$ is not in great excess, the reaction produces $\ce{I2}$ that makes a precipitate, as $\ce{I2}$ is poorly soluble in water. So you choose the equation if you know of the excess of $\ce{KI}$.

"add the two equations", do you mean I could substitute I2 with I3- in the first equation?

NO ! Add the equations, as if they were algebraic equations. Add $\ce{3 I2 + 3 I- ->3 I3^{-}}$ to your first equation. Add the two left-hand sides together. Add the two right-hand sides together. Suppress the $\ce{I2}$ in excess. It gives $$\ce{KIO3 + 8 KI + 3 H2SO4 ->  3 KI3 + 3 K2SO4 + 3 H2O}$$


text:
Do baking soda or vinegar reduce toxicity?
Both baking soda and vinegar can be used to remove cigarette odor from cloth and surfaces, at least temporarily. Do they reduce toxicity of the third hand cigarette smoke too or just 'hide' the smell?

The toxicity of cigarette is due to nicotine. Nicotine is not deeply modified by adding vinegar or baking soda. Such an addition changes nicotine into an ammonium derivate that suppresses its odor and smell, but it does not modify its toxicity. Nicotine and its ammonium derivate have the same toxicity.


text:
How are protons transferred from an acid to a base in aqueous solution?
I have read in my textbook that an acidic proton is transferred to a base through direct contact. Since acid is ionized in water, wouldn't it be more likely that the surrounding water molecules become protonated to a greater extent than the base (which has less contact with acid)?

Yes, but then $\ce{H3O+}$ comes into contact with the base, and you end up with the same outcome (the base being protonated), just indirectly.

[OP] Since acid is ionized in water ...

Weak acids are largely protonated, so they still can act as a acid. Strong acids, however, are fully ionized (deprotonated) in water, and can't act as an acid anymore. The protons of strong acids are transfered to water, yielding hydronium ions, which in turn can react as a acid.

[OP] ... wouldn't it be more likely that the surrounding water molecules become protonated to a greater extent than the base (which has less contact with acid)?

The water molecules are able to act as bases, so you can have an acid:base reaction with water. It is true that water is usually present in higher concentration than everything else (otherwise, we would not consider water to be the solvent). For that reason, the most likely encounter is between two water molecules, followed by water with a different species (hydronium ion, hydroxide ion, other acid or base), followed by encounters by non-water species.

[OP] I have read in my textbook that an acidic proton is transferred to a base through direct contact.

This is correct in the sense that an isolated proton does not exist in aqueous solution. So the proton will only dissociate from an acid if it almost immediately associates with a base (this could be water, hydroxide, or another base). Where the proton ends up (or where is predominantly is attached) depends on the $\mathrm{p}K_\mathrm{a}$ values of the conjugate acids of the available bases.

Thank you for your response! Regarding your last statement, why would the location of the proton depend on the pKa values of the conjugate acids of the available bases? Would the strongest base not get protonated to the fullest extent (as described by its Kb value) then the next strongest base and so on?

@Justin It is an equilibrium. So if the strengths are similar, it is near 50:50, and if they are very different, it is almost 100:0, but not quite. It also depends on the concentration of the species.

The proton transfer is like this:
$$\ce{HA + B <=> A- + BH+}$$
Where

$\ce{HA}$ is any acid, including $\ce{H3O+}$ and $\ce{H2O}$
$\ce{B}$ is any base, including $\ce{H2O}$ and $\ce{OH-}$

Note that formally neutral $\ce{HA}$ and $\ce{B}$ may be either neutral molecules or ions.
Concentrations of respective acidic and basic forms are mutually linked by their own equilibrium constants for their water interaction, called acidity resp. basicity  constant:
$$K_\mathrm{a}=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}$$
$$K_\mathrm{b}=\frac{\ce{[BH+][OH-]}}{\ce{[B]}}$$
and by water auto-dissociation constant:
$$K_\mathrm{w}=\ce{[H3O+][OH-]}$$


text:
What is the difference between structure assignment and structure prediction?
I am absolutely new to structural bioinformatics (only started last week). I am working on the secondary structure assignment/prediction (actually I am not sure) of proteins using machine learning.
When it comes to protein analysis, most of the books or research papers talk about protein structure prediction. Only a handful of books or articles talk about protein structure assignments.
What is the difference between structure assignment and structure prediction?

I suppose these resources do not speak about these terms out of context. Is this context the same or different for both terms ?

My comment was a hint for you to finish your homework of elaboration of the question. // Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own effort.

https://rostlab.org/papers/2008_rev_assignment2/paper.html#DSSP

https://pubmed.ncbi.nlm.nih.gov/8749853/

Assignment refers to a statement that a particular amino acid residue
(1) has a given backbone conformation described by dihedral angles falling within certain limits
(2) based on the local conformation of a series that includes adjacent amino acids, the residue can be said to be part of a secondary structure such as one of various types of helix, sheet or turn (or simply lacking specific structure, ie intrinsically disordered).
The basis for the determination of dihedral angles in NMR is to pool together various spectroscopic variables and use these as constraints during molecular dynamics simulations of the protein, in order to generate an ensemble of possible structures, then to evaluate the dihedral angles in those structures. The data usually includes NOEs, which are signal enhancements observed when two nuclei are close and provides a sensitive constraint on the magnitude of their separation. NOEs are labelled based on the atoms involved, including the distance between the residues (in the primary sequence) to which each atom belongs. Certain patterns of NOEs are consistent with particular secondary structures and can provide qualitative evidence of such structure before applying the more elaborate analysis previously described. Other valuable data include heteronuclear chemical shifts and various J-couplings. A high-quality set of NOEs and complementary data can in principle constrain the conformation of the protein down to a non-redundant choice.
It is possible for (1) and (2) to be inconsistent, that is, to assign an amino acid to particular secondary structure, but for that particular amino acid to reveal, on the basis of spectroscopic data and/or simulations, conformations inconsistent with that secondary structure. In that case the assignment is said to violate the data (or vice-versa), making the assignment suspect if not falsifying it, and calling details of the analysis into question. The quality of an ensemble is assessed on the basis of a statistical analysis of the agreement between the data and the assigned secondary structures.
Secondary structure prediction is a term used to refer to the use of known correlations between secondary structures and observables, and can be said to include using spectroscopic or crystallographic data to estimate what the structure of a protein is like, including how flexible. Assigning residues to particular secondary structure is a form of prediction, but when the input is experimental data and the degree of confidence is sufficiently high the process is not usually called prediction. Prediction involves an input of data, possibly sparse and usually including parts or all of the primary sequence, obtained for a protein of interest, into a prediction algorithm which takes such inputs and returns a list of possible secondary structures ranked by likelihood of being accurate. The algorithm is developed (a step in some cases called training) by optimizing the agreement of the output obtained for proteins of known structure.

Is "Assignment" a part of the "Prediction"-process?

Structure assignment is relating e.g., a measured absorbance frequency (like a triplet in $\pu{^1H NMR}$), or a structural feature (like $\Phi$ and $\Psi$  in a Ramachandran plot) to a structure already known.  It equally may be based on a putative / postulated structure; then, you compare e.g., how a simulated NMR spectrum is similar to the experimentally recorded one.  If the two spectra match 1:1, the putative structure likely is a correct one for this NMR experiment.
Structure prediction departs knowing e.g., only a sequence of $\alpha$-amino acids, e.g. Ala-Leu-Cys, and offering a guess about their spatial arrangement (e.g., folding).  Its foundation may be more or less educated / reasonable, which is why one typically infers from already known relationships of sequences of amino acids, and their 3D structure already established, e.g., by NMR, or X-ray crystal diffraction.  Machine learning (ML) calls this stage of establishing a model as training.  Aside from comparing with other algorithms, you eventually benchmark your  ML algorithm in multiple runs with different training sets (e.g., sequences) to watch the outcome of the predicted structures; for one, comparison with experimentally determined data intentionally excluded from the training set now offer a check about the correctness of the results.  For two, this equally offers to estimate the likelihood a folding  predicted in silico to equally occur in vitro e.g., reporting a range for a particular $\Psi$ between two adjacent amino acids.

I am working on the secondary structure assignment/prediction (actually I am not sure) of proteins using machine learning.

Secondary structure assignment is an automated method that defines the secondary structure (helix, sheet, loop) based on a known three-dimensional structure. Secondary structure prediction refers to taking the primary structure (the sequence of amino acid residues in a protein) and predicting the secondary structure.
There are also some experimental methods that help to characterize secondary structure in the absence of a 3D structure. Circular dichroism spectroscopy gives an estimate of the secondary structure content. Assigned chemical shifts of C-alpha and C-beta atoms from NMR experiments give a residue-by-residue prediction (or characterization) of secondary structure.

What is the difference between structure assignment and structure prediction?

Without the qualifier "secondary", I am unfamiliar with structure assignment. If you google "protein structure assignment", most hits will talk about secondary structure assignment. NMR resonances are assigned to protein primary structure (i.e. which residue gives rise to which signal).
Structure prediction most often refers to tertiary structure prediction. However, you can also predict secondary structure, protein-ligand structures (docking studies) and quaternary structure.

I need a layman's explanation of "protein secondary structure assignment". Adding a diagram or picture would be much appreciated.

Secondary structure refers to the conformation of the main chain in a protein, and the hydrogen bonds between the carbonyl oxygen and the amide hydrogen that are observed. The example below shows that the two secondary structure elements called beta sheet and alpha helix differ in the conformation of the main chain and in their hydrogen bonding pattern.

Source: https://commons.wikimedia.org/wiki/File:Alpha_beta_structure_(full).png
To assign a secondary structure based on an atomic model, you would measure the main chain torsion angle and deduce hydrogen bonds from distances of hydrogen bond donors and acceptors. Often, the secondary structure assignment is shown juxtaposed with the primary sequence and sometimes with other annotations. Here is an example screen shot. The highlighted Cys30 is part of a helix (red shade) and makes a disulfide bridge with Cys115 (see additional annotation).

Source: https://www.rcsb.org/sequence/1LYS
Different software packages usually agree on the presence of helices and sheets, but sometime have small differences in the exact start and end of the secondary structure elements (see e.g. here). It is also possible to assign secondary structure from electron density or cryo-EM maps, see e.g. here.

Is "Assignment" a part of the "Prediction"-process?


text:
Does amide ion prefers to act as a base than a nucleophile?

In this question, I think that amide ion, being a good nucleophile should attack on the terminal carbon through  $\mathrm{S_N2}$ mechanism to form compound (c), but the answer given is (b), and the solution:

But I am not able to understand the kinetic factors that favour the amide ion to act as a base rather than a nucleophile. I think it is much easier for a strongly nucleophilic amide ion to act as a nucleophile through a one step $\mathrm{S_N2}$ reaction (polar aprotic solvent will also favour it), rather than first acting as a base and then the carbanion acting as a nuceophile. The $\ce{C-Cl}$ bond is polar, which also increases the chances of an $\mathrm{S_N2}$, but the $\ce{C-H}$ bond is not that polar.
So, my question is, will (b) be the major product, and if yes, than what are the kinetic factor that favour it in comparison to (c)?
(Note: The reaction is assumed to be under kinetic control.)

Your mistake here is to think of amide anion as strongly nucleophilic. It is a very strong base and the protons next to the sulfonyl group have a pKa ~30. It is also poorly soluble in DME

Please note that it's written $\mathrm{S_N2}$ $\mathrm{S_N2}, not $\ce{SN^2}$ $\ce{SN^2} (the latter means sulfur and nitrogen squared).

In general, acid base reactions are very fast. See https://www.masterorganicchemistry.com/2012/05/21/acid-base-reactions-are-fast/

I second what @mcole said. In general, acid-base reactions are faster than nucleophilic substitution. Which is also the reason why Grignard reagent cannot be used for nucleophilic addition on aldehydes/ketones which have acidic hydrogen like $\ce{-OH}$, $\ce{-COOH}$ and so on.

After acid base reaction has occured, the substitution by carbanion will be faster than amide ion since intramolecular substitution is faster because of greater chances of attack. Am I right?


text:
Is it possible to create hydrophobic water?
I recently saw "hydrophobic water" in my school science fair. I have no idea on the procedure of how to make it, so, can I make it? I did google it and read some articles (zero helpful) and see some images (which seem to match with the one I saw). So, could you tell me how to make hydrophobic water if I even can? Thanks in advance.
Articles that didn't help:
How to make stuff super hydrophobic
How to make stuff hydrophobic or hydrophilic

Like an autoimmune reaction where an organism attacks the very organism itself, this seems not sustainable (from perspective of thermodynamics).  How if possible water could yield droplets (fog, rain, rivers, snow, ice) if the molecules would constantly repel each other?

Hydrophobic water would make an exemplary oxymoron.

Please do explain us what you really have seen in your school science fair.

Sodium-ion-phobic sodium-ions do exist. @IvanNeretin

@KarstenTheis Point taken. Then again, as I see now, hydrophobic water exists as well, but it still makes an exemplary oxymoron.

Boiled-water is hydrophobic.

I found an interesting yt short which might be the experiment you are talking about. The yt short

You can't make hydrophobic water molecules. You can, however, make hydrophobic droplets containing mostly water. They are not made of pure water, but are coated with a substance that remains on the surface of the water droplet and changes the properties of the surface.
Here is an example of making such a hydrophobic drop: https://www.youtube.com/watch?v=H0spGzO2FSo.
And here is a demonstration of the properties (this drop also contains hydrophilic dye which remains mixed with the water in the bulk of the droplet: https://www.youtube.com/watch?v=MkLbVLGcn-A.
Here is another demonstration with a drop of pure water (top left) encountering a coated drop (bottom right, pushed with a finger):

Source: https://www.youtube.com/watch?v=Z4bVP7hEcKI

Well done....   :-)

I think writing "hydrophobic water droplets" is misleading. The droplets are not hydrophobic, they repel each other, but not because they contain water. All the same,  I guess it might not be worth arguing about (but still rankles!).

@BuckThorn I changed it to hydrophobic droplets containing mostly water.

I'm not sure that edit addresses the point I was trying to make. My point is that whatever is inside the droplets is of secondary importance, as it is what is stuck on the surface that is responsible for repulsion. So "hydro" in hydrophobic seems a misnomer in this context. Still, that's presumably how people refer to the phenomenon when they see it, so pressing the subject seems pointless.

Here's an analogy: if I throw a water balloon at you, you will not regard yourself as hydrophobic or hydrophilic based on whether the balloon ruptures on impact.

@BuckThorn The pure droplet of water is hydrophilic. If you touch it with your finger, it will wet it. The other drop is hydrophobic. If you touch it with your finger, it rolls away. In your analogy, the balloon is hydrophobic, not me. To see whether I am hydrophilic or hydrophobic or Telfon, you have to try water and oil droplets.

Ok, yes, you are right of course. The coated droplet repels an uncoated water droplet. Does that make water hydrophobic? I don't know anymore.

@Buckthorn Is hydrophobicity property of a substance or a phase ? If the latter then the phase originally based on water can be made hydrophobic without making water hydrophobic and the question would be rather Can we make water phase hydrophobic?

Water cannot be made hydrophobic. Water drops can be set to act like if water were hydrophobic, but it would not be pure water anymore.
The drop surface can be made hydrophobic by hydrophobic liquid, film or dust. See the links in the Karsten's answer.

A dirty trick could be electrostatic charging of separated water drops, e.g. in the Kelvin_water_dropper, eventually modified for drop repulsion demonstration.

I think it is perhaps the first explanation. I don't think it would be the second as their grades depend on it. I don't completely understand the last explanation but it might be the last one as well.

I vote for the first one as well, I have added the rest for alternatives.

tl;dr–  Hydrophobic-water would be water that doesn't significantly cohere to itself.  So, water's hydrophobic when it wouldn't form a condensed-phase (e.g., if it boils and preferentially exists as a gas at equilibrium).  Can't have hydrophobic-water in an equilibrium-preferred condensed-phase, as a condensed-phase means that the material's sticking to itself and thus isn't phobic of itself.

Two species are phobic within some medium under some conditions when they don't here together (cohere if they're the same species; adhere otherwise).
To be clear: phobic species don't really repel so much as they're not very attracted to each other.
Species can become self-phobic.  Of course, it's important to remember that members of the species aren't repulsing others of their own kind so much as just aren't very attracted to them.  And then since they're not really attracted, they do the natural thing and just tend to disperse, filling the volume containing them.  In other words, they vaporize.
Water can be vaporized, e.g. through heating.

Note: Can't have hydrophobic-water as a condensed-phase.
A condensed-phase (like liquid or solid) is defined, in part, by the material sticking to itself.
You can have hydrophobic-water physically inside of a condensed-phase, e.g. you can have molecular-water as a minority-component of a larger material-phase in a scenario in which the molecular-water wouldn't be so cohesive, but you can't have hydrophobic-water as a condensed-phase, because it wouldn't cohere (and thus wouldn't be condensed).

aren't the water molecules still attracted to each other even in gaseous state and their intermolecular forces are just too weak?

@ShirsakMajumder: Yup!  If you check out this data page for the Van der Waals equation-of-state, the values for $a$ (which corresponds to intermolecular attraction) are positive.

then doesn't that make them hydrophilic?

@ShirsakMajumder: Nope!  Hydrophobic stuff attracts water too, just relatively weakly.  When you see water bead up on a hydrophobic surface, it's not that the surface is repelling the water, but rather water's self-attracting itself away (often in a quasi-spherical shape, like a droplet) and the surface's own attraction is too mild to counteract it.


text:
Is there a parameter for degree of ‘mixedness’
So I was making coffee this morning and when mixing some milk into the black coffee, I wondered if there was some parameter or value in chemistry which describes how well mixed two liquids (or gases) are.
Imagine water and oil in a beaker (I choose these because they can be completely separate). This parameter that I am thinking of could have a value of 0 where the water and oil are in completely separate layers. It would then have another value, say 1, when all molecules are perfectly mixed amongst each other (after a hypothetical, infinitely efficient shaking of the beaker).
Would this just be entropy or something derived from entropy? Or is there no such thing?

Just intuitively, it could be based on degree of sample homogenity and the sample/system mass or volume ratio.

The parameter you are looking for is the entropy, which is steadily increasing in mixing milk into the black coffee.

I have wondered about this as well, also in the context of miscible liquids, and for flow reactors.

I don’t see a way to measure the entropy change as mixing progresses, or to compare how it increases over time for different mixing methods. It would be interesting to have something measurable, so that different mixing methods could be compared. The measurable quantity might then be related to entropy through theoretical models.

Entropy of mixing is well defined, but when you mix there is always diffusion and convection and possibly mechanical stirring to consider so some volumes in the total will mix well before others so it will be a mess basically. If you put a dye into one liquid then measured the total absorption change throughout the whole spatial extent of the sample this might give a sort of average value vs time.

Don’t forget the enthalpy of mixing as well. The resulting G of the system vs composition determines if there will be phase separation or not.

Chemistry is messy, but the closest I can think of to what you have in mind is solubility: https://en.wikipedia.org/wiki/Solubility Keep in mind that neither coffee nor milk would be considered homogeneous nor thermodynamically stable, so in that case it would be both difficult to reduce liquid-liquid miscibility to a single parameter.

Look up "solubles concentration" in the context of coffee preparation, eg https://en.wikipedia.org/wiki/Coffee_extraction

From the perspective of engineering / physics, there is research about mixing coffee and milk to yield layered latte (e.g., this open access publication in 2017 Nature Communications echoed in NYTimes and Smithsonian).


text:
What is the n-factor for reduction of chromate(VI) to chromium?
Here a reaction is given:
$$\ce{CrO4^2- -> Cr}$$
I don't know the full reaction, but I made some estimates. One estimate was that on balancing charges we get two electrons on right-hand side.
I will consider two different methods that I used for finding $n$-factor:

$n$-factor is the number of electrons participating in a reaction;
$n$-factor in the total change in oxidation state per molecule.

But when I solved by the first method using charge balancing, I got $n$-factor as $1,$ but when I solved by the second method I got $6.$
I am confused in this about how to get the correct $n$-factor. Please check my work and help me to reach out the answer.

The corresponding half-reaction is :$$\ce{CrO4^{2-} + 8 H^+ + 6 e- -> Cr + 4 H2O}$$ This reaction is used to deposit a bright layer of chromium metal on steel (bumpers) by electrolysis.

$n$-Factor is an obsolete concept that is essentially useless for over a century.

@andselisk But it is extremely important for solving problems in physical chemistry for the exams which I am preparing for...

@MridulKumar In the context of electrochemistry and Nernst equation in particular, you need the the number of transferred electrons, which is not the same as a more broad and ephemeral term "$n$-factor". If you'll try to use the concept of $n$-factors for, say, multistep electrode process, you'll fail big time. My advice is to avoid erroneous terminology and obsolete concepts, as well as the textbooks that use them.

@andselisk  Do you mean that the $n$ in the nernst equation is not n-factor... As my teacher told me that it is n factor.

The $n$ in Nernst equation is numerically equal to the n-factor. Your teacher is right but this is bad way to connect Nernst equation with equivalents.

Please tell your teachers that the concept of chemical equivalents is obsolete. It was taught 50 years ago. Hope your teachers can communicate to the public exam authorities!
$$\ce{CrO4^2- -> Cr}$$
If it were simple molarity, you would take the experimental weight of dichromate (Salt) and convert that into moles by dividing the mass by formula weight of dichromate salt.
Now you are interested in converting the molar concentration into normality $N$.
In the case of redox reactions, molarity can be converted into normality by dividing it by an integer $n$ and this is your n-factor.
Check how many electrons are transferred: Cr (VI) to Cr (0). So $n$=6. So whatever molarity you have, divide it by 6 to obtain the corresponding normality.
Just follow your second method. Your first point

$n$-factor is the number of electrons participating in a reaction;

does not mean anything.