text:
What the link between radical(root) and its meaning in chemistry inside etymology
All i know radical mean root and when i heard root all what i think about is the root in plant so what the link?
Why they chose that word?

Can you explain what you mean when you say a group of atoms is linked to a radical?

@SafdarFaisal Im sorry about my (not clear question-i edit it) i was mean root is thing i can found in plant so why they use it in chemistry? And what the link?

The history is explained very well, https://en.wikipedia.org/wiki/Radical_theory. Read it and modify the query if you still do not understand.

@AChem That what i found : "A radical was identified as the root base of certain acids (The Latin word "radix" meaning "root"). The combination of a radical with oxygen would result in an acid. " so... Is that the answer? If yes so that all the radicals is just can make acid with oxygen or..?(maybe i still dont understand)

Words in any language can have one etymology but different senses and meanings.  Yes, radix means plant roots but radical in chemistry and mathematics have nothing to do with plant roots. They have completely different meanings. Now one can ask why mathematicians call something radical? The square root sign is basically an "r" of the mathematical radical. Plant roots have only a figurative connotation with the radical as something which is the "basis". A radical person sticks to closely to the original roots of his/her idealogy (often used in a negative sense).
In a chemical usage, one can add a little bit more to the above answer, radical or radicle (in even  older books) simply meant:

An atom or group of atoms regarded as a (or the) primary constituent
of a compound or class of compounds, remaining unaltered during
chemical reactions in which other constituents are added or removed.
Now more widely: an atom or group of atoms which behaves as a unit and
is regarded as a distinct entity (OED entry)

Suppose a chemist analyzes sulfuric acid $\ce{H2SO4}$ in the 17-18th century. He finds that sulfur and oxygen ratio is 1:4 by mass. Next he add a base, sodium hydroxide to neutralize the acid, and after evaporation he isolates the salt. The salt is analyzed and it is also found the contain sulfur and oxygen ratio of 1:4. For that ancient chemist, $\ce{SO4}$ is a sulfate radical. It is a group of atom that is behaving as a unit in a chemical reaction. Forget about charges on sulfate, that did not exist when "radicals" were discussed
Looks at a snippet from 1858 book, that shows organic radicals= group of atoms that behave as a unit.

The above answer by Nilay shows more modern meaning of radicals.

Below is the text from the Wikipedia article of Radicals:

The term radical was already in use when the now obsolete radical theory was developed. Louis-Bernard Guyton de Morveau introduced the phrase "radical" in 1785 and the phrase was employed by Antoine Lavoisier in 1789 in his Traité Élémentaire de Chimie. A radical was then identified as the root base of certain acids (the Latin word "radix" meaning "root").

At that time, chemists considered "radical" as the "free element" of the acid entity. For e.g. sulfur for sulfuric acid etc. He even considered that the word before acid is the radical like "acetic" for acetic acid, "muriatic" for muriatic acid (hydrochloric acid).

When Lavoiser dealt with the nature of acids, he assumed them to be oxygenated substances, i.e. oxygen combined with an entity which he chose to name a radical. According to his concept, the radical might be a single element in the case of the inorganic acids (sulphur in sulphuric and sulphurous acids, phosphorus in phosphoric and phosphoraus acids, carbon in carbonic acid, the "radical muriatique" in muriatic acid, the "radical boracique" in boric acid), or it might be some stable combination of carbon and hydrogen in the various organic acids. The designation of an element as a radical quickly passed out of favour, particularly after Davy demonstrated that oxygen is not an essential part of an acid.

Then as both organic and inorganic chemistry grew, chemists were identifying more and more "free" species. Like the first organic "radical" identified was the triphenylmethyl radical, $\ce{(C6H5)3C•}$ by Moses Gomberg in 1900 which abolished the idea that "radicals" shouldn't be the elemental species but rather the "free" species originating (rooting) from an established/stable species.
The meaning of "radical" has evolved through phases and now in modern chemistry, radicals are more than just "roots".
References:

THE HISTORY OF FREE RADICALS AND MOSES GOMBERG'S CONTRIBUTIONS, AARON J. IHDE, Chemistry Department, University of Wisconsin,
Madison, Wisconsin 53706, U.S.A.
Walton, J. C., Nonhebel, D. C. (1974). Free-Radical Chemistry: Structure and Mechanism. Kiribati: Cambridge University Press.

@NilayGhosh thank you for your answer. Can you help me if i understand well? First- what i understand is that they use radical which mean root from latin word radix as something essential like in plant the root is something essential. Second - the chemist considered the radical is free element because its can react fast or its have the ability to react.  Third- i couldnt understand that: *by Moses Gomberg in 1900 which abolished the idea that "radicals" shouldn't be the elemental species but rather the "free" species originating (rooting) from an established/stable species. * - Plz help

@Pcp115 1. Yes. But it essentially doesn't have to be a plant. Root of anything is radical. 2. No. They initially thought that the acids came from free element like sulfuric acid came from sulfur. So, they considered sulfur as the radical. Even a "stable" species containing hydrogen and carbon reacting with oxygen to give acid were considered as radical. These assumptions were later proved wrong. 3. When Gomberg identified "triphenylmethyl radical", it abolished the idea that radicals are not free elements nor stable species rather free species from a stable species- triphenylmethane


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Why is some first electron affinity endothermic
Elements like N and period 2 elements have endothermic EA. What's the explanation for that as isn't it usually exothermic?
Does this also imply that some elements subsequent EA is exothermic instead of endothermic?(this sounds impossible but I don't know)

Answer to the second part here. https://chemistry.stackexchange.com/questions/170991/why-is-the-first-electron-affinity-exothermic-and-successive-ones-endothermic

The total orbital angular momentum contribution is zero for these elements. Leading to the symmetrical effect described in the answer below.

The 2nd electron affinity is endothermic for all elements.
Only $\ce{Be}$, $\ce{N}$, $\ce{Ne}$ from the 2nd period  have negative electron affinity.
Notice that all elements with the negative electron affinity(*) have kind of "pseudo noble gas" configuration, in sense of having about spherical symmetry of electron density distribution in valence orbitals. It means the whole group of valence orbitals is half or fully filled.
See Wikipedia: Electron affinity data_page

(*) Or, at least their electron affinity is shifted toward zero compared to their neighbours.

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Removing 9-Aminoacridine from tissue?
There's this compound 9-Aminoacridine, that I need removed from the tissue slides it was applied to. Thing is everytime I use ethanol, Acetone, water, methanol, all the usual solvents; our pathologist says there are still remnants of it in tissue. It's key that we can remove it but it remains there nonetheless. I'm not sure what other solvents exist that can be used, it does happen to have a 2 aromatic rings in the structure. Let me know if there are solvents or a way to find them. In the SDS it says ethanol but that is a lie for my case I guess. Here's the structure:

Maybe add some acid?

If you use the acridine as a stain / as an indicator and it hampers subsequent analysis / work, is there an established protocol in your field (maybe histochemistry) which uses a different compound instead which does not adsorb or (chemically) bind this much? Perhaps indication of the tissue (dermis, epidermis, subcutis, etc), protocol applied, intended analysis after the rinse can provide a direction. E.g., what concentration of remaining acridine would be below the limit of detection of the next analysis, or (compared to your signal of interest) weak enough (with a signal/noise ratio of 3)?

The issue with removing biological stains is that they interact with cellular components and adhere to them. 9-aminoacridine intercalates with DNA, so it would take a large amount of solvent to remove it, by the law of mass action.
Since it is freely soluble in ethanol, the best course might be many repeated washings in ethanol, perhaps with gentle agitation (e.g., magnetic stirrer) at elevated temperature. However, that treatment is also likely to destroy the structure of the specimen.
It might be best to measure the effectiveness and harm of washing on a few specimens.

Using a UV lamp and UV-blocking filter on a light meter or camera (by looking at exposure info), measure fluorescence of the 9-aminoacridine to establish a baseline on how much stain is present.
Wash a set number of times.
Measure the amount of dye left, as above.
Examine under a microscope for damage.

You'll likely see a logarithmic drop in fluorescence, and then have to decide at what point the game is worth the candle, i.e., if dye removal is worth the damage to the slide, the time and the effort.

Thank you so much this is monuemntal help! 9-aminoacridine  has a natrual fluoresence correct?

@Sciguyyy, "9-Aminoacridine is a fluorescent weak base which distributes across the plasma membrane according to the pH gradient, and can be used to monitor changes in intracellular pH in eukaryotic cells." https://www.sciencedirect.com/science/article/pii/0014579384811400

So do you think under a hood UV light, I should see some fluoresence? And your comments are super appreciated, this could beyond help my work, thank you!

@Sciguyyy, sorry, but it's doubtful I can be of much practical help beyond the above suggestion. Try it and see.

Thanks!!! I'll play around with some tools.

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will adding metal nano particles to a CNT-polymer matrix increase the conductivity of the polymer?
I was wondering would just physical mixing of metal nano powders with MWCNT in a polymer matrix improve the conductivity of the said polymer? Would the metal particles interact with the CNT network. I see some papers where they chemically reduce the nanoparticles while mixing them with a CNT dispersion improving conductivity. But I was wondering if would just physical mixing of powders result in a similar increase?


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How can titanium be separated from magnesium by washing?
I saw this question in a GCSE past paper:

After this reaction, there is a mixture of the solids magnesium, titanium and magnesium chloride. Titanium does not react with dilute hydrochloric acid. Suggest a simple method to separate titanium from the mixture.

And in the mark scheme, one of the ways you could get the full 2 marks was with this answer:

filter the mixture (to remove titanium) / filter off the
titanium (1)
wash the titanium (1)"

I was wondering how this would separate the titanium from the magnesium (the other way to get 2 marks was to use $\ce{HCl}$ to react with the magnesium before filtering)?

The mark scheme for the other explanation makes no mention of HCl - only "washing" the titanium after it's been filtered.  It's an Edexcel paper btw.

Yes. If your mixture contains titanium $\ce{Ti}$, magnesium $\ce{Mg}$, and magnesium chloride $\ce{MgCl2}$, you may first get rid of magnesium chloride by washing the mixture with water, which dissolves $\ce{MgCl2}$
alone. Then, you add diluted hydrochloric acid $\ce{HCl}$. The metal magnesium reacts easily with diluted solutions of $\ce{HCl}$ (maybe $0.1$ M), producing gaseous hydrogen $\ce{H2}$ and magnesium chloride which remains in solution. At the end, when no bubbling appears on the metallic powder, the remaining solid contains pure titanium.


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Why are p orbitals most important for stability?
Is there nice way to see/show/derive why p orbitals are so important for stability? Or asked differently: Why are elements with filled p orbitals on the right side of the periodic table? Why aren't elements with fully filled s or d shell like Beryllium or Zinc noble?

What should I change about my question? It has been closed for being "to broad", but my question is what it is and I assume many people have the same question. Where else should I ask it?

This question is legitimate, I do not know if this will be give a complete answer, but here is the basic principle.
We have to define first the orbital at the outermost shell.
When you have to solve the hydrogen atom, there is a term $+\frac{l(l+1)}{r^2}$ from the angular momentum $L^2$ which gives a positive energy when angular quantum number $l>0$. This means that the $s$-orbitals ($l=0$),  are closer to the nucleus as compared to the other orbitals which have a kind of "centrifugal force" and are farther from the nucleus, $s$-orbital cannot form the outermost shell (farther away from the nucleus).
Now in the case of $d$-orbitals ($l=2$), $f$-orbitals ($l=3$) and $g$-orbitals ($l=4$) :
They are farther away from the nucleus compared to the $p$-orbitals, however their radial extent is not larger than the $p$-orbitals because their filling starts from $n \ge 3$  whereas the $p$-orbitals and $s$-orbitals already have a large spatial extent (more number of nodes). Therefore, $d,f,g$ orbitals are more localised as compared to $s,p$ orbitals.
To conclude, a $p$-orbital is always the most delocalized orbital because of its orbital angular momentum and its spatial extent, so it defines the outermost shell.
This outermost shell $p$-like defines the stability when it is filled because another added electron will be so far from the nucleus that the compound will be very unstable (low electron affinity $A$). $p$-orbitals are filled when the potential of the nucleus is larger as compared to the innermost orbitals, $p$ block elements are often more electronegative (ionisation potential $I$ is high). When $p$-orbitals are filled, the compound is very stable $I-A>>$ : Noble gases. The filled $s,d,f$-orbitals have lower $I-A$ values than that of noble gases.


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Atomic terms and perturbation theory
I have a question about the derivation of the possible atomic terms of an atom. Consider an atom with $N$ electrons in the central-field approximation, with hamiltonian
$$
H_{0} = \sum_{i=1}^{N} \left[ - \frac{1}{2} \nabla ^{2}_{i} + V _{c}(ri)\right]
$$
We can write its states as the product of the states of one electron, which are characterized by the quantum numbers $n,\ l,\ m_{l}$ and $m_{s}$. We know that the configurations we get like this are degenerate: all configurations with same values of $n$ and $l$ but differents $m_{l}$ and $m_{s}$ have the same energy. These configurations also have well-defined values of total  orbital angular momentum and spin $L$ and $S$, and we can classify them in atomic terms $^{2S + 1}L$, that have the same energy. However, if we consider corrections for the central field approximation, the degeneracy of these terms is broken, and we can determine which ones have higher or lower energy by Hund's rules.
My doubt is the following: thinking from the perspective of perturbation theory, I am considering the various states of the same electronic configuration as unperturbed states:
$$
| \psi^{(0)} \rangle = |n_{1},l_{1},m_{l1},m_{s1}, ..., n_{N},l_{N},m_{lN},m_{sN} \rangle
$$
and taking as perturbation a potential equal to $V_{ee} - V_{c}$, where $V_{ee}$ is the Coulomb interaction between the electrons. The use of degenerate perturbation theory generates $|\psi \rangle$ eigenstates that are also functions with well-defined $L,\ S,\ M_{L}$ and $M_{S}$, since the perturbed hamiltonian commutes with the spin and orbital angular momentum operators (I'm ignoring spin-orbit interaction for now). But why do the possible $L,\ S,\ M_{L},\ M_{S}$ values of the disturbed states have to be the same as the values of the undisturbed states? Couldn't the perturbation generate states with well-defined angular moments but with different eigenvalues?

Afraid in general you must include the spin-orbit coupling term in the Hamiltonian for that porpouse. However, you can distinguish between states of different total spin $\mathrm{S}$, eg. triplet and singlet , since the wavefunction are built using a different set of spin-orbitals.

As you know very well, this angular degeneracy is due to the fact that the potential $V_c=V_c(r)$ is purely radial and the angular part do not affect the energy. The angular solution are degenerate if we do not include interactions like spin-orbit interaction.
The interaction $V_{ee} \propto \frac{e^2}{|\vec{r}'-\vec{r}|}=V_{ee}(r,\phi)$ has an angular dependency. If you solve the equation $H|\psi \rangle = E |\psi \rangle $ with $H = H_0 + V_{ee}$ the angular degeneracy is broken.
If you consider $V_{ee}$ as a perturbation, acting on the state $|\psi^0\rangle$, the correction $\langle \psi^0 |V_{ee}|\psi^0\rangle$ will not break the degeneracy, the eigenvalues will be only shifted compared to the previous equation.

thanks for the answer, but i still have a doubt. For example, consider the carbon atom in the ground state. In this configuration, we know that there are only 3 possible terms, $^{1}S,\ ^{1}D$ and $^{3}P$. To deduce this, I need to use a central field approximation and the Pauli exclusion principle. However, how do I know that if I consider the electron-electron interaction and solve the Schrodinger equation, the only possible terms will be these three as well? Does it make sense to talk about the Pauli exclusion principle when not using an independent particle model?

I understand your concern, your central field is only radial and leads to levels with only a certain correction and the Pauli principle (PP) the electronic distribution. The most important part of $V_{ee}$ remains radial, the angular part of this potential breaks the angular degeneracy but this splitting is often weak that in a mean field approximation a degeneracy can still noticeable. Therefore approximately the configuration remains  and the huge part of the work is to correct the energies. Even when not using an independent model, PP gives the e- distribution in the levels.


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To check if something is prefixed as a Pyro
How to figure out if something's a Pyro by looking at its formula?
The thing I know about Pyro compounds is that

Ortho acid + Ortho acid - $\ce{H2O ->}$ Pyro compound

I have no issues if I end up with the ortho acid I am quite familiar with. For example, everyone pretty knows what $\ce{H3PO4}$ is and thus if I am asked if $\ce{H4P2O7}$ is prefixed with a Pyro, I wouldn't have any trouble.
The problem comes when I deal with compounds of $\ce{As, Sb, Si}$ and I have to check if they are prefixed with a Pyro or not....
For example, take $\ce{As2O3}$. How do I know if that's a Pyro? Now take $\ce{H4As2O7}$. How do I know if that's a Pyro?
One way I think would be to check if their oxidation states comply to their occurrence states. But that is definitely not an ideal approach. So how do I go with this? Do I need to remember these ortho acids or can I go just remembering a few?
Please someone help me in this
PS: I have already checked this webpage but I couldn't find the solution to my problem in this- What do the prefixes meta, ortho, pyro mean in inorganic chemistry?

The rule of thumb is the compound has to be known long enough and widely enough to deserve the pyro- prefix before the prefix was stopped being used for such condensed anions. By other words, the prefix is used for historical reasons.

Oxides do not use pyro prefix in their names. // For acids/ anions of acids, there is not a simple general rule, you need to know historical circumstances. Memorizing those few where pyro- is used may be easier. And even then, it is not the preferred naming. // Absence of a simple rule, based on basic principles, is rather a rule than an exception in chemistry.

@Poutnik I didn't know this..... it's weird they still ask questions based on Pyro acids and yes my bad, an oxide is not prefixed with a Pyro

Pyro compounds contain $2$ central atoms (different from oxygen and hydrogen). Exemple : $\ce{H4As2O7}$ contain $2$ arsenic atoms. It can be considered as a pyro- acid.

okay, I did not see this either, thanks @Maurice


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Esterification of Serine

I came across this question above, and I'm unsure about the reaction with H+(aq). The answer is that the amine group is protonated as usual, but I am not sure why H+(aq) doesn't catalyse esterification of serine with itself?
As you'll be able to tell by the question, I don't have much experience and couldn't find anything that explained this in a way I can understand (so I apologise if the question is simple!). If someone could explain what happens and why, in simple terms, I would be grateful.
Many thanks

It does not catalyse the esterification because it is not anhydrous acid conditions, any ester that forms is hydrolysed back to free acid because of the water present. You need anhydrous conditions - such as the iPrOH/H2SO4 condtions shown.

Great, that makes sense - thank you!


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Do we know the asymptotic behavior of density functional in DFT?
I was reading a book on DFT the other day, and the author included asymptotic behavior for $E_{xc}[\rho(r)]$.
Given that, do we know the asymptotic behavior for kinetic energy and/or total energy in DFT?

It would help tremendously for context of you'd include a citation and maybe a quote.

What do you mean with "asymptotic behavior" ? At long distance ? At short distance ? At infinite time ?

Either one is good. I'd to know either (or maybe both)

A very important advice will be to read again the principles of density functional theory to better understand the approximations of $E_{xc}[\rho(r)]$ from LDA to GGA climbing gradually the Jacobs ladder.
There is no expression of the kinetic energy (KE) depending on the density $T[\rho(r)]$ in DFT, however there are tricks to deduce KE after solving the  Kohn-Sham's equation.
DFT has generally no problem at small distances (interatomic) where the density tends to be homogeneous, even a crude approximation like LDA can handle this situation.
At large distances, the density is more difficult to handle for DFT (inhomogenious, or atomic-like) even by using corrections like GGA. A non-local approximation like Hartree-Fock (HF) is more suitable for the situation. If you plot the dissociation energy or the interatomic potential (Total energy vs interatomic distance) by using a simple DFT functional compared to HF you will notice this huge difference for $r>>$, the asymptotic behaviour of DFT to the dissociation energy is pathological.
The main drawback of DFT is the fact that eigenvalues are upshifted, called self-interaction. This pathological asymptotic behavior is due to this self-interaction (separating DFT and HF). The intuitive correction is to move towards HF. Therefore, there are hybrid functionals integrating HF exact exchange. Until now, major corrections are range-separated functionals.
Obviously, all the corrections affect only  $E_{xc}[\rho(r)]$ not the kinetic energy consequently the total energy.

Nitpick - there is no KE functional in DFT as most commonly used, that doesn't mean such functionals don't exist.  Orbital Free DFT is an example of a method that does use a KE functional, https://journals.aps.org/prresearch/abstract/10.1103/PhysRevResearch.3.033198 is a recent article and a review is at https://link.springer.com/chapter/10.1007/0-306-46949-9_5

@IanBush  Yes why not Thomas-Fermi directly ? There is no reliable one.

That's kind of the problem with HK theorem, it's non constructive existence theorem. The functional exist, but we don't really know the formula for that functional.


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Chemical Test for Hydrogen
I know that to perform a chemical test for $\ce{H2}$ we have to take a burning candle flame near to the mouth of the test tube and a blue flame with a characteristic pop sound confirms the presence of hydrogen. But my question is that, is there any other chemical test for hydrogen except this one with a pop sound?

No common chemicals tests exist for hydrogen. Certain metals absorb hydrogen but their prices are so high that it is impossible to include them in teaching schools. In fact this test is also non-specific. Methane would do the same. However physical tests like spectroscopy from a hydrogen discharge tube is extremely reliable.

As per the spectroscopy suggestion by @AChem, here is the visible region spectrum produced by electrical excitation of hydrogen gas in a gas discharge tube: https://physics.stackexchange.com/a/768678/313612. Nothing else looks like that, even deuterium gas.

Anyway if there is a $CO_2$ dectector and the reaction do not produce $CO_2$, the gas is likely to be $H_2$.

There are metals that promote the catalytic oxidation of hydrogen in air, such as platinum and palladium. While these are fairly expensive in bulk, there are reasonably inexpensive sensors which use micrograms of these catalytic metals to detect hydrogen, and other gases.
However, as you can see from the specifications for the MQ-8 sensor, it is not hydrogen-specific. That sensor also detects "alcohol" vapor, and even propane, though it is less sensitive to those gases than hydrogen.
BTW, the "POP" test is certainly not specific, either, and you'd get a bigger bang (and soot) from acetylene.
So, this test might not be exactly what you want:

It relies on electrochemistry.
It does not differentiate $\ce{H2}$ from $\ce{CH3OH}$, $\ce{C2H5OH}$, $\ce{C3H8}$, etc.

However, such a sensor is very useful where batteries are charged, or hydrogen gas is used, to warn of leaks.

Possibly hydrogen is distinguished by the voltage at which it oxidizes, versus other gases the sensor detects.

@OscarLanzi, actually, the sensor is based on electrothermal, rather than electrochemical measurements. It is not potentiometric electroanalytical chemistry. https://www.figaro.co.jp/en/technicalinfo/principle/catalytic-type.html

I would be inclined to create two half-cells, the one a standard hydrogen electrode, and the other, the gas under test, connected with a narrow tube as recommended with the Standard Hydrogen Electrode to avoid mixing.
The source for the hydrogen of the reference tube can be from purchased Hydrogen gas, or also from electrolysis with water formed hydrogen.
When the half-cells are connected with the narrow tube, with one half-cell having the gas under test and the other half cell having $\ce{H2}$, the potential difference can be measured with a digital volt-meter as the potential difference between the cathode and the anode under very small current flow conditions.
The picture link for one half of this setup is here
Author attribution:

The scheme of the standard hydrogen electrode:

platinized platinum electrode
hydrogen gas
solution of the acid with activity of $\ce{H+} = \pu{1 mol dm−3}$
hydroseal for preventing oxygen interference
reservoir through which the second half-element of the galvanic cell should be attached. The connection can be direct, through a narrow tube to reduce mixing, or through a salt bridge, depending on the other electrode and solution. This creates an ionically conductive path to the working electrode of interest.

If the voltage difference is $0$ $\pu{Volts}$ then the gas is diatomic hydrogen.  Just to mention that below this result is in combination with the flame test which eliminates inert gases as producing the zero voltage.  Other reference gases (like Oxygen) can be also used in the reference cell verifying the Standard Electrode Potential tables values given there -- to eliminate gases that are producing zero voltage because of their lack of reactivity at one of the electrode sites.
It can (and should in my opinion) also be further tested with the flame test above.  The test setup can be verified as previously stated using diatomic oxygen using the standard electrode potential for diatomic oxygen from the standard electrode potential table.
The details about calculating the potential difference between two standard potential cells is at this link here.
Together with the flame test above with its pop as kind of a generic test, the two-electrode cell test using two separate reference gases like $\ce{H2}$ and $\ce{O2}$ should provide the additional confidence that the gas is indeed hydrogen.

Small amounts of hydrogen can be tested by methods using electrochemistry, as illustrated in two answers above.
Larger amounts, on the order of 50 - 100 mL, can be tested and identified by the unique property of hydrogen gas: its low density.
This is the procedure: weigh a small rubber balloon. Blow it up with air to about 50 - 100 mL and weigh again. The weight should be marginally higher to reflect the contents - and higher pressure - (on the order of 50 mmHg, or 6.5% of normal air pressure) inside the balloon. For highest accuracy, it's best to pump up the balloon with a tire pump rather than with your mouth, in order to eliminate the variability of humidity/saliva and CO2. The increase in weight for 100 mL of air will be about 0.129 gram, easily observable. This also gives you some practice and can be done repeatedly to ascertain error limits.
It may be more difficult to figure out an easy way to pump the suspicious gas into the balloon, but connecting the balloon to the gas volume and flooding that container with water can be arranged to drive the gas into the balloon. Corrections for the water (humidity) concentration can be made if necessary. Measure the volume of the inflated balloon; small balloons can be purchased that blow up almost spherically, and the calculations can be approximations. H2 at atmospheric pressure, neglecting the water vapor contribution, will weigh 0.00899 grams, giving a buoyancy of 0.120 grams, which is easily measured on a flat-top balance (see picture). Even a mechanical balance can be used (second picture). The open top allows balloons of various, even large, sizes.

The buoyancy for 100 mL helium will be 0.102 grams, and to be realistic, this will probably be within your error of measurement. However, helium won't burn; it will even snuff a (small) flame.
I've done this to distinguish between hydrogen and methane, and confirmed it by the soot spot left by methane on a cool watch glass.
The methane was also confirmed by gas IR. Gas IR can be used as a partial identification of hydrogen by virtue of no peaks in the usual IR region, compared to, say, a hydrocarbon gas. Helium would also not absorb, but would be eliminated by its nonflammability. A gas IR cell can be made with a section of copper tube (about 7 cm long, or as long as can be placed in the IR analyzer) with salt plates attached to either end. Suitable holes drilled into the copper tube with attached copper tubing can be provided with stopcocks and the cell can be filled with the aid of a water pump arrangement as discussed above.

There is a recent article in J Chem Edu (https://doi.org/10.1021/acs.jchemed.3c00129
) that describes an assay based on Pd-catalyzed reduction of organic dyes (methylene blue or thionine). The palladium is loaded on the surface of ZnO-particles to decrease cost and toxicity, allowing the experiment to be used in a middle-school setting. The assay is reusable as you can reoxidize the dyes with molecular oxygen.
The article mentions a commercially available kit (e.g. https://www.h2bluetest.com/) to test hydrogen concentration in water. Apparently, this caters to a new fad of ingesting elemental hydrogen as an antioxidant (doi: 10.18632/oncotarget.21130).


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Why does color of substances not disappear after some time?
Some compounds are colored due to HOMO-LUMO transition. We see the complimentary color to the frequency which the electron absorbs to get the high molecular orbital. But after some the electron should return to the ground state emitting the same frequency again, and, hence, after some time the compound should start appearing white since all frequencies will be present in the light reflected by it.
Or, if the electron does not return to the ground state, there must be a point where saturation occurs, that is, all the possible electrons have gone to the excited state, in which case the compound should again appear white. So why are compounds colored?

Most absorbing substances do not re-emit, but convert the electron energies to other forms. Those that do usually emit longer wavelengths as fluorescence.

Most of the electronic transitions happening after absorption of a photon happen very quickly. So there is little scope in most molecules for a "buildup" in the excited state. And there are multiple pathways for the electrons to get back where they started many of which don't involve visible light.

Looks like poorly worded duplicate of https://chemistry.stackexchange.com/questions/88767/why-objects-usually-dont-exibit-fluorescence-phosphorescence

If you are seriously interested in understanding why things are colored, there is a book called The Physics and Chemistry of Color: The Fifteen Causes of Color,  by Kurt Nassau. It should be available in any good library, and Internet Archive has it as well. The main theme is that there are 15 separate color causes. In your case, there are a couple of very common misconceptions due to web searches.

The complementary color is a very crude estimate of the reason behind the apparent color of solutions and even gases. In my opinion, this should never be applied to proper chemistry.

Once a molecular species absorbs light of a certain frequency, the electrons relax to the ground state but without emitting light. Perhaps a physicist can explain the exact mechanism of those radiationless transitions. Most of the absorbed energy of light is released as heat. Certain molecules do not do that, instead they emit light of a longer wavelength. This is called fluorescence (like your highlighter marker ink).

What you are describing happens only in extremely rare cases for example in hot gas phase atoms. For example, you have vapor of sodium atoms in a flame, and you shine 589 nm at the sodium vapor, the atoms absorb 589 nm and re-emit 589 nm. The is called resonance fluorescence.

Finally,

where saturation occurs, that is, all the possible electrons have gone
to the excited state,

Now you are talking about lasing phenomenon (of lasers), which is another world to explore.
One can now see how complex the scenario is. This is why the suggested book is > 500 pages.

Nice answer I already upvoted! I never got around to getting the book by Kurt Nassau, so I should fix that. There seems to be an uptick in causes of color questions: I just answered one here.

Absorption and transmission do play a role in the color of the compound and absorption of a certain region of the visible light results in transmission of the remainder of the region, often resulting in complementary colors.
Excitation and relaxation very often occur via different mechanism due to selection rules. Thus, the same frequency of photon that was absorbed is not emitted, which would have resulted in a white colored compound.
Example of Ruby

Ruby has a brilliant red color. It occurs because of several factors:

Excitation from the $^4A_2$ ground state of chromium ion (the impurity responsible for color in ruby) to the $^4T_2$ or $^4T_1$ states (broadened by vibrational and rotational energy levels) corresponds to violet and green-yellow parts of white light passing through ruby resulting in transmission of strong red which is partially responsible for the red color of ruby.

The relaxation, however, does not occur directly from $^4T_2$ or $^4T_1$ to $^4A_2$ due to selection rules and instead passes through an intermediate $^2E$ state. Part of this causes slight warming, and the other part result in bright red fluorescence.

Could you please add a reference for the source of the image? Thanks!

@EdV I reconstructed the image from the source hyperlinked in the heading Example of Ruby (Britannica).

Thanks! Generally we like to have sources listed for all images, to avoid copyright issues, unless the images are made by the poster. Links, as you know, sometimes go bad.


text:
Boiling point of 2-methylpyridine is less than that of 3-methylpyridine?
As I was reading about physical properties of pyridines and substituted pyridines, I came across this:

2-methylpyridine boils at 129°C while 3-methylpyridine boils at 15° higher than 2-methylpyridine

And the reason that is given is:

This lowering of boiling point is explained in terms of hindrance to association offered by the groups in the vicinity of the nitrogen atom

Assuming that the book is talking about hydrogen bonding as "association", wouldn't 2-methylpyridine boil at higher temperature than 3-methylpyridine as the methyl group is closer to nitrogen atom in 2-methylpyridine and hence can form better hydrogen bonds than 3-methylpyridine ?

Pray, tell me why my question have been downvoted ? At least leave a reason for downvotes!

Disregarding the question itself (which is fine IMO), there were a typo in the title ("32-methylpyridine"), so do not hesitate to proofread your own posts. Also, every quote needs to have a clearly stated source, i.e. here a reference is missing, which I would guess is the primary concern of downvoters. A brief guide how to cite a textbook is available, and make sure to include edition and page numbers. Otherwise it may be considered as plagiarism.

Two things:

The nitrogen in pyridine doesn't hydrogen bond with saturated carbons. It's more likely that you'd get pi-pi stacking, or interactions between the lone pair and the aromatic ring of a second molecule. Having a methyl group near to the lone pair disrupts the latter interactions.

If the nitrogen in pyridine did form interactions with the methyl group, 2-methylpyridine would still boil at a lower temperature.  Internal interactions don't need to be broken for a compound to pass into the gas phase.


text:
Why does increasing the surface area of the cathode submerged in the electrolyte increase the current in a galvanic cell?
I can understand why increasing the surface area of the anode would increase the current, because the rate of oxidation will be increased resulting in a greater rate of flow of electrons through the wire to the reduction half-cell. However, increasing the surface area of the cathode also seems to increase the current.
Even though the rate of reduction is increased by increased submersion of the cathode, what bearing does this have on the current?

Thanks for the suggestion! Unfortunately I'm looking at galvanic not electrolytic

In an electrolysis, the current increases if the surface of the electrodes increases. And it also increases if the distance between  the electrodes decreases.

I think you are referring to cell constant $G^* = \frac{l}{A}$

Yes. I am referring to the cell constant l/A.


text:
What is the effect of dissolving ammonia in water on hydrogen bonding?
Hydrogen bonding exists between $\ce{NH3}$ molecules. When $\ce{NH3}$ is dissolved in water, I know that hydrogen bonds are formed between $\ce{NH3}$ and $\ce{H2O}$ molecules. However, do the $\ce{NH3}$-$\ce{NH3}$ hydrogen bonds still exist?

It depends on the concentration. For an ammonia concentration of 0.06 mol/L, you would have a thousand water molecules for every ammonia molecule, and the hydrogen bonds between ammonia (if you started with liquid ammonia) will be replaced by ones with water. Also, you will have more hydrogen bonds (ammonia has only one donor, so you can have one hydrogen bond on average per ammonia molecule in pure ammonia). Up to four water molecules can have a direct hydrogen bond with ammonia.
At very high concentrations of ammonia, you would expect occasional direct  hydrogen bonds between two ammonia molecules. Because in a liquid, hydrogen bonds are temporary and dynamic, this would eventually be replaced by a hydrogen bond with water, and other pairs of ammonia molecules might form a hydrogen bond.

Also diluted NH3 is protonated to greater degree.


text:
Combining two overlapping wavefunctions
Let's assume that we solved the many-body Schrödinger equation for two molecules and we can now write their wave functions as
$$
|\Psi_1\rangle = O_1|\operatorname{vac}\rangle\,,\\
|\Psi_2\rangle = O_2|\operatorname{vac}\rangle\,.
$$
where $O_{1,2}$, according to the definition above, are the operators creating states $|\Psi_{1,2}\rangle$ from vacuum.
Moreover, let's assume that both equations are written in a basis wherein the two molecules are spatially separated yet their wave functions do have some noticeable overlap. Now we would like to construct a "composite state" of two molecules.
It seems natural to define such a composite state as
$$
|\Psi_1,\Psi_2\rangle = O_1 O_2|\operatorname{vac}\rangle\,.\\
$$
However, I think that this definition would only make sense if the two operators commuted or anti-commuted. (Otherwise, should we use $O_1 O_2|\operatorname{vac}\rangle$ or $O_2 O_1|\operatorname{vac}\rangle$?..) Note that for non-overlapping sets of orbitals the question does not arise since the two operators clearly commute with each other.
Now, consider two ways of defining such operators.

In the coupled cluster ($U=e^{T}$) or unitary coupled cluster ($U=e^{T-T^\dagger}$) approaches, operator $O$ can be expressed in terms of an exponential of a polynomial in creation and annihilation operators applied to an operator creating some reference state from vacuum:
$$
|\Psi_{1,2}^{\vphantom{\dagger}}\rangle = U_{1,2}^{\vphantom{\dagger}} |\rm ref\rangle = U_{1,2}^{\vphantom{\dagger}} a^\dagger_{\rm ref} |vac\rangle\,,
$$
Therefore one option is to postulate $O_{1,2}^{\vphantom{\dagger}}=U_{1,2}^{\vphantom{\dagger}} a^\dagger_{\rm ref}$ for CC or $O_{1,2}^{\vphantom{\dagger}}=U_{1,2}^{\vphantom{\dagger}} a^\dagger_{\rm ref} U_{1,2}^\dagger$ for UCC. However, it is not clear to me how two such operators $O_1$ and $O_2$ commute with each other. Moreover, operators $U_{1,2}$ are typically constructed in a way that their action is only sensibly defined when they are applied to the vacuum state. The latter, however, could be fixed if $U$ was defined adiabatically or through similarity transformation (basically, $U$ should be the wave operator[1]), in which case its action would be sensible on all possible states.

Once the states $|\Psi_{1,2}\rangle$ are found, one can write them down as polynomials $P_{1,2}$ of creation operators only acting on the vacuum state, and postulate $O_{1,2}=P_{1,2}$. In this case, it is clear that two such operators commute.

It is tempting to choose the second approach to defining $O_{1,2}$, but I still don't understand the physical motivation besides using this particular definition and not something else.
The motivation for my question stems from the fact that I would like to be able to construct $O_{1,2}$ in some way resembling approach 1. (even in an approximate way)
UPDATE
I've just noticed that on Physics SE there already exists precisely the question I ask here, and nobody has got close to answering it in 10 years.
References:

Kato, T. Wave operators and similarity for some non-selfadjoint operators. Math. Ann. 1966, 162 (2), 258–279. DOI: 10.1007/BF01360915.

How can we create a state from vacuum ?

Apply a creation operator

A question for my own understanding, although it might also help your own... but aren't all of these electronic structure calculations conducted 'in a vacuum'? My understanding was that they are, which is why you either need to include explicit solvation molecules, or apply solvation models to calculations you want to compare to solution-based experiments...

@isolatedmatrix The word 'vacuum' in your comment (as in 'calculation in a vacuum') and in the original question (as in 'from vacuum') means two entirely different things; the latter implies the level of generality to which we chemists rarely have to ascend. Maybe this is a physics question, after all.

@isolatedmatrix The question is mostly in the language of second quantisation, a very powerful formalism that is not usually met until graduate level or later, and then usually only on the more theoretical side. It has creation operators, which increase the number of particles in the system by 1, and annihilation operators, that reduce the number of particles by one. "Vacuum" is simply a state of the system with no particles.

@IvanNeretin and IanBush: sorry, my mistake

Thanks for comments. I think this question would better fit the Physics SE indeed.


text:
Graph of kinetic energy of ejected electron vs intensity
While studying photoelectric effect I got a doubt in the graph of kinetic energy of ejected electron vs the intensity of electromagnetic radiation.
Everywhere this graph is given as

But here for 0 intensity, a certain value of kinetic energy is given but 0 intensity means no light is falling so how can kinetic energy of ejected electron can have a value other than zero. As per me the graph should start from origin and then suddenly or drastically rise with infinite slope(with increase in intensity) to a point which is kinetic energy for that particular frequency. Or is it like that zero intensity does not mean no light instead it means negligible light. Correct me if I am wrong.

@Poutnik Does that mean that at 0 KE point, we are talking about the electron's that energy that it has being bound within the atom instead of being as photoelectron?

But if we have 0 persons we can't take any value of height I think. We would simply take for 0 persons height is 0 as it cannot be measured. Similarly for 0 ejected electrons kinetic energy of ejected electron should be zero as it cannot be measured.

I suspect that, mathematically speaking, you are correct. But most people reproducing the chart leave out the subtlety that it is valid only for any positive value of intensity and meaningless at zero. But a careful mathematician would not show a line from zero to the ke for that frequency but a single point at zero and a flat line for any positive value of intensity.

Clearly if there are no photons then nothing happens, so the graph starts at an intensity equivalent to 1 photon.

You use zero with 2 different meanings and confuse them together. Zero as value of physical quantity and zero as arbitrary qualitative symbol for missing measurement. If you could not measure temperature, it does not mean $T = \pu{0 K}$. Missing measurement would be marked by a dash or similar, not by $\pu{0 K}$.

Zero energy has no special meaning for emitted electrons that are not here.
$$\lim_{n \to 0+}{(\frac {E_\mathrm{em}}n)} \ne 0$$
$$\lim_{n \to 0+}{(\frac {E_\mathrm{em}}n)} = E_\mathrm{1e}$$
where

$E_\mathrm{em}$ is the total energy of emitted electrons
$n$ is the number of emitted electrons
$E_\mathrm{1e}$ id the energy of 1 emitted electron

If the total energy of emitted electrons is zero, it does not mean the energy per electron is zero. Because you divide zero by zero, what has not particular general value, what you learn when learning mathematical limits. Some ratios with both terms converging to zero converge to zero, others converge to either infinity and yet other converge to finite numbers.

I think applying the uncertainty principle, we can say there always will be at least one photoelectron and hence a non-zero KE. Is it correct to say?

It has nothing to do with the uncertainty principle. Rather, the intensity is never zero. but even if is,  still  $\frac 00 \ne  0$


text:
Why is more than one product formed in Swarts Reaction?
I have got a very clear idea about Swarts reaction for monohalo compound. But I found the following reaction in Chrome

Now, my question is why are 2 products formed? Secondly why is there no trifluoro compound? What is the possible mechanism for this reaction?
Source: https://protonstalk.com/wp-content/uploads/2021/02/image1-e1612596758363.png
Source of the image

Do you have a reference for just  mono-fluorination of multi-chlorinated compounds? I'm finding otherwise. Just see here.

Since $\ce{SbCl5}$ is a Lewis acid, we can consider a carbocation mechanism.
In the first stage, a chlorine atom is preferentially taken off the $\ce{CCl2}$ carbon because then the carbocation is stabilized by pi back-donation from the other chlorine. The antimony in $\ce{SbF3}$, being in a lower oxidation state and less acidic than its counterpart in $\ce{SbCl5}$, gives off a fluoride ion to restore neutrality to the organic compound.
In the second stage, the carbocation is stabilized by pi back-donaation from the fluorine, which is very electronegative but has available electrons and good overlap with the carbon. The $\ce{R}$ group lacks lone pairs and is less effective at stabilizing the carbocation, so difluorination at what was originally the $\ce{CCl2}$ carbon is a likely end point.


text:
Mechanism for synthesis of neopentyl iodide using triphenylphosphite/methyl iodide
I was wondering how the reaction between neopentyl alcohol and triphenylphosphite/methyl iodide looks like from a mechanistic point of view.
It should involve the formation of a phosphonium iodide salt, which can then use the phosphorus' oxophilicity; however, for that to work, the alcohol needs to act as a nucleophile, which in my opinion is not feasable, as alcohols in general are weak nucleophiles.

The ability of alcohols to act as nucleophiles on phosphonium ions is central to reactions like the Appel reaction (https://en.wikipedia.org/wiki/Appel_reaction) and the Mitsunobu reaction (https://en.wikipedia.org/wiki/Mitsunobu_reaction).

As evident by the Ref.1, the first step is the reaction of triphenylphosphite and methyl iodide to give methyltriphenoxyphosphonium iodide (See path B of Ref.1). This reaction is conditional and some details of conditions are described in Ref.2. Then, one of the phenolate ion in methyltriphenoxyphosphonium iodide is replaced by neo-pentyloxy group giving away a phenol molecule. The resulting compound, suggested by the original authors of the method (Ref.2), undergoes $\mathrm{S_N}2$ mechanism to give the resulting neo-pentyl iodide in good yield:

However, the suggested $\mathrm{S_N}2$ mechanism leans heavily on the claim that rearrangement does not occur, but in light of the finding of Ref.3 (the presence of more than 6% of t-amyl iodide in the reaction mixture that can only achieved by the rearrangement of neo-pentyl group, loses much of its credibility. Accordingly, to explain the rearrangement, Ref.3 suggests following two paths for the mechanism:

According to Ref.2, $\bf{A}$ and $\bf{B}$ are in equilibrium. The covalent form $\bf{A}$ would undergo “cis elimination” thus giving rise to a cyclic transition state, $\bf{a}$ leading to give pure neo-pentyl iodide (See Path 1). Alternatively, the ionic form $\bf{B}$ would break up to give an ion pair $\bf{b}$, which collapses to neo-pentyl iodide before rearrangement occurs (Path 2). However, when combination of neo-pentyl and iodide ions is delayed, the rearrangement would occur and t-amyl iodide would be formed (Also, according to Ref.3, the material balance has never exceeded 80% and the deficiency may well be due to olefin formation).
References:

H. N. Rydon (Checked by W. Fuhrer, R. Keese, and A. Eschenmoser), "Alkyl iodides: neopentyl iodide and iodocyclohexane (Propane, 1-iodo-2,2-dimethyl- and Cyclohexane, iodo-)," Organic Syntheses 1971, 51, p.44 (DOI: 10.15227/orgsyn.051.0044); Coll. Vol. 6 (1988), p.830 (4 pages).
S. R. Landauer  and  H. N. Rydon, "458. The organic chemistry of phosphorus. Part I. Some new methods for the preparation of alkyl halides," J. Ckem. Soc. 1953, 2224-2234 (DOI: https://doi.org/10.1039/JR9530002224).
Nathan Kornblum and Don C. Iffland, "Neopentyl Iodide," J. Am. Chem. Soc. 1955, 77(24), 6653–6654 (DOI: https://doi.org/10.1021/ja01629a083).


text:
Can H2CO3 and NaHCO3 be in the same solution at the same time?
In my opinion they can't. Because the product of reaction will have $\ce{H2CO3}$, and end up with $\ce{H2O}$ and $\ce{CO2(g)}$. And gas can not be part of a solution.
$$\ce{H2CO3 +2NaHCO3 -> Na2CO3 + 2H2O + 2CO2(g)}$$

Well, if you boiled it out and then heated the rest, you would end with Na2CO3(s), H2O(g) and CO2(g). Otherwise, turning drinking water with carbon dioxide/carbonic acid/bicarbonate pH buffer to washing soda solution  is not observed.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

There are ongoing four linked equilibrium reactions with their respective equilibrium constants:
$$\ce{CO3^2-(aq) <=>[+H+][-H+]HCO3-(aq)<=>[+H+][-H+]H2CO3(aq)<=>[-H2O][+H2O]CO2(aq)<=>CO2(g)}$$
For drinking water with natural bicarbonate $\mathrm{pH}$ buffer, the concentrations of $\ce{HCO3-(aq)}$ and $\ce{CO2(aq)}$ are comparable, with minor concentration of $\ce{H2CO3(aq)}$. See also carbonic acid on Wikipedia.


text:
Why is halohydrin not formed In bromination of stilbene?
Stilbene can be dibrominated using $\ce{HBr}$ and hydrogen peroxide in ethanol.  Common sense would suggest that the major product should be the bromo-ethoxy product due to reaction with bromonium ion followed by the solvent since it outnumbers the strong nucleophile.  But, in practice this is not seen, or only seen in very small amounts.  Why is this?

What is the mechanism for trans-stilbene to stilbene dibromide?

As stated in reference 2, both anti- and syn-products are formed, so your intuition may be correct, and you may, partly, be proposing the correct mechanism for the dibromination of stilbene.

Reagents produce $\ce{Br2}$ in-situ:
$$
\ce{2HBr + H2O2 -> Br2 + 2H2O}
$$
This reaction mechanism may be favorable at lower concentrations of $\ce{HBr}$. However, according to several references (e.g., 1, 2), an appreciable concentration of $\ce{HBr}$ is required for dibromination of stilbene.
Moreover, bromide ion, being negatively charged, is a better nucleophile than ethanol, being neutral. Thus, you might expect your product to react again to form syn-dibromo product, due to anti ethanol addition followed by $\mathrm{S_N}2$:

Both anti- and syn-products are formed, so your intuition may be correct; however, anti-product is more favorable at the commonly used concentrations.

Typically, 70% yield is obtained in the experiment.$^\text{1}$
References

McKenzie, L. C., Huffman, L. M., and Hutchison, J. E. (2005). The Evolution of a Green Chemistry Laboratory Experiment: Greener Brominations of Stilbene. J. Chem. Educ., 82(2), 306-310 (DOI: https://doi.org/10.1021/ed082p306).
Kilway, K. V., Horn, J. D. V.(2006), X2 Addition to Alkenes: Bromination of trans-Stilbene. Available as PDF.

In both of first two schemes, the ethoxy function has extra carbon. Would you correct it as soon as possible?

@MathewMahindaratne thank you for the suggestion, have made corrections; I made some changes in the second mechanism as well.

Again, 3rd structure of 1st scheme has peroxide addition (shouln't oxygen have plus charge) instead of ethanol addition. Also, if $\ce{HBr}$ leaves, 4th structure of 1st scheme should not have hydrogen on oxygen.

@MathewMahindaratne thank you again, must be that I am not very familiar with the software.

Still, 3rd structure of 1st scheme has peroxide addition instead of ethanol addition. It can easily fix by putting Et for OH.


text:
How Can the First Ionization Energy of Helium be Accurately Calculated Using Basic Quantum Mechanics?
How Can the First Ionization Energy of Helium be Accurately Calculated Using Basic Quantum Mechanics?
There is a book which summarizes the fairly accurate calculation of the First Ionization Energy of Helium, but it would be nice to see if there is a more direct answer than the one calculated, that might have better accuracy even.
The experimental ionization energies for Hydrogen and Helium are given at this link: Experimental Values of Ionization Energies and in Table 3 in the linked Chemistry book here:
Table 3: Ionization Energies (kJ/mol)

Element
1
2
3
4
5
6
7
8

H
1312

He
2372
5250

Li
520
7297
11810

Be
899
1757
14845
21000

B
800
2426
3659
25020
32820

C
1086
2352
4619
6221
37820
47260

N
1402
2855
4576
7473
9442
53250
64340

O
1314
3388
5296
7467
10987
13320
71320
84070

F
1680
3375
6045
8408
11020
15160
17860
92010

Ne
2080
3963
6130
9361
12180
15240

Na
496
4563
6913
9541
13350
16600
20113
25666

Mg
737
1450
7731
10545
13627
17995
21700
25662

Are these the most currently-accepted experimentally-confirmed values?
Now how can these energies be calculated using radially-symmetric basic Quantum Mechanics Theory accurately?
I have made fairly accurate Helium ionization calculations at this link.  What are other accurate approaches that use just the radially-symmetric Schrödinger wave equation, with a few small simplifications?

You should define what you mean by "accurate" in context of your post. As any quantum calculations of a three-body system are inaccurate, unless we put the inaccuracy threshold above their inaccuracies.

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
// Optionally, here are formatting guides for texts and formulas/equations/expressions.

@Poutnik I try to reference the picture as a link and a picture only in the case where the copy link option **produces an error and does not work **.  If the link to the picture works, I always use the link, whenever possible.  In these cases where I got an error message with the link option, I took a screenshot of the picture and posted a separate html link reference.  Also, the accuracy calculation is given in the online book link reference.

The point is, the question content policy focuses on questions being self-contained, with quoted or reproduced essential content. Links are provided rather just as the content origin reference, or for further reading. By other words, the question should make perfect sense without following any link. // Instead of table images, there are preferred tables created by SE Markdown or by MathJAx/LaTex  code. See e.g.  SE meta -  tables // I think this your case is acceptable, just for future.

Is this an ad for your book?

@Buck Thorn : I do not make any money from this online version of the book.  At this point, I am trying to figure out how to apply its principles to explain the Periodic table beyond Hydrogen and Helium, call it Volume 2 versus the referenced Volume 1 of the problem (and solution).  I feel that the screening calculations need to be accurately made in order to understand the current problem more clearly.  So, as to be short, it is not an advertisement.  (I have not even bought the hard copy myself!)

Hello @Buck Thorn  :I thought about your feedback - the question being self-contained.   The calculations that I used were tedious - relying on College Calculus Integrals.   To not detract from the core of the answer (getting buried in the integrals), I was thinking that a separate reference was best.  The current version of the book that I downloaded, my calculated ionization energy for Helium was 22.1433 eV (2136.6417 $ \pu{kJ/Mol} $ ) conversion at https://www.unitsconverters.com/en/Ev/Particle-To-Kj/Mol/Utu-6180-6179 .  Then the accuracy is $ {(2372-2136.6417)}/2372*100 = $ +9.9% accurate.

Also, I asked, "Are these most now-accepted experimental values?" because I remember writing a separate reference (not published here for reference) where the comparison between the CRC value for the first ionization of Helium and the calculated version of the fist ionization of Helium was within some 0.3% of the 24.5874 eV experimental (Helium) from the [CRC link][ https://is.muni.cz/do/rect/el/estud/prif/js11/fyz_chem/web/podpora/ELECTRON_CONFIGURATION.pdf ].  Now, there is an ionization theory error of ${(24.5874-22.1433)}/{(24.5874)}*{(100 {%})} = 9.94% $ accuracy.  I need to re-calculate.

There is a long history of calculating two-electron atoms starting with Bethe and Salpeter's classic book. https://journals.aps.org/pra/pdf/10.1103/PhysRevA.98.012510 claims calculations of non-relativistic energy levels in He to 28 or more significant digits.

You also might be interested in https://doi.org/10.1103/PhysRevLett.127.093001 and references therein.

my calculated ionization energy for Helium was 22.1433 eV ((2136.6417 kJ/mol)  -  I would turn down the valid digits for it to be taken seriously. I do not believe the model is so accurate.

There is no need to reach the experimental results by basic methods, the wave function corresponding to $He$ atom has no analytical expression. The numerical methods used do not often bare a significant scientific meaning. Trying to reach this experimental result is more or less a computational challenge. The expansion of the ground-state function in a minimal basis-set leads to an accurate description depending on a right choice of the basis functions for this expansion. We are more interested in the suitable basis functions for reducing the computational cost.

The current version of the book that I downloaded, my calculated ionization energy for Helium was 22.1433 eV (2136.6417 kJ/Mol ) conversion at unitsconverters.com/en/Ev/Particle-To-Kj/Mol/Utu-6180-6179 . Then the accuracy is (2372−2136.6417)/2372∗100= +9.9% accurate

Assuming that your second ionization energy is given by the analytical formula of hydrogenic atoms $E_2 = \frac{Z^2}{2n^2}$ a.u. , your total energy, related to the first ionization energy $E_1=22.14$ eV will be $E_{tot} = -(E_1+E_2) = -76.54$ eV. This ground state energy is not accurate at all for a numerical method. A variational method applied considering $V_{ee}$ as a perturbation gives an analytical solution of $E_{tot} = - 77.7$ eV more accurate that your numerical result.


text:
Which gamma radiation had more high-energy?
I am working on a short story centered around the Fukushima Daiichi nuclear disaster and I was wondering which $\gamma$ radiation had more high-energy: the radiation from the cooling water flowing through the core caused by the decay of $\rm{^{16}N}$, or the $\gamma$ emission caused by the decay of $\rm{^{137}Cs}$?
I understand that $\rm{^{137}Cs}$ has a much longer half life than$\rm{^{16}N}$ but which is the more powerful emitter of $\gamma$ rays during their respective half-lives? (Reference Link)

Your reference link leads to nowhere.

Why do you decide to study Cs-137 alone ? In the nuclear fission, isotopes of maybe half the periodic table are produced. Cs-137 is only one of these fission fragments. Why choose Cs-137 ?

I think I correctly fixed the link

Also $^{16}$N emits gamma irradiation with energy 6.13 MeV. $^{137}$Cs gamma rays have energy 0.662 MeV.

Maurice, Cs137 is an important fission product. It has a specal mobility. The fission process forms Xe137 (half life 3.8 minutes) which has time to diffuse and otherwise travel before it decays into the medium lived Cs137. As a result in nuclear fuel pins and reactor systems you can often find Cs137 in places remote from the uranium / plutonium fuel. Also for fuel which has cooled for more than a few years the Cs137 is the main gamma emitter, it tends to dominate the gamma dose to workers in reprocessing plants.

There is no simple relationship between the half life of a radionuclide and the decay energy. Radioactivity only occurs when it is thermodynamically possible for a nuclear reaction to occur in the nucleous. But the nucleous must overcome a kinetic barrier for the decay to occur.
A good example of this is the alpha decay of the more heavy elements, the decay energies of Pu239 and Pu240 are almost the same but the half life of Pu240 is much shorter than that of Pu239.
Gamma emission is when nuclei which are in excited states relax into lower energy states. Normally this happens very quickly but there are examples of long lived excited states such as Ir-192m2 which has a half life of 241 years. The ground state of Ir-192 has a half life of 74 days. Thermodynamically the excited states of iridium should be able to undergo the beta decay into Pt-192. But instead the excited state of Ir-192 undergoes a internal conversion into the ground state which then does the beta decay.

This doesn't answer the actual question: which of the specific isotopes has the most energetic gamma emissions.


text:
Which reaction occurs in calcium vanadate when it comes in contact with water?
I am a physics student currently studying the electrical properties of $\ce{CaVO3}$ and I have noticed that the surface of my thin film samples, which were originally smooth and transparent, now have blue and orange spots all over. These spots vary in size, some being only a few $\pu{\mu m}$ or smaller and the largest being approximately $\pu{200 \mu m}$ across but consisting of many spots combined. Visually it appears like it might have been water condensation droplets that caused this after having been inside of a helium flow cryostat at temperatures between $\pu{4 K}$ and room temperature. I have also noticed a change in the electrical properties of my samples, which I might be able to understand if I know what sort of compound these spots are. I do not know for certain that these spots were caused by a reaction with water, I am purely basing this on the visual appearance of droplets.
The change in the electrical properties has happened over time. I first measured the properties when the samples were new and a year later I repeated these measurements and noticed an increase in the resistivity for the entire temperature range as well as a change in the charge carrier density, accompanied by these colorful spots, which makes me believe that they may be responsible for it. The spots are too small to be measured individually with my methods.

Are your film samples made of pure calcium vanadate ? Why study calcium vanadate ? Why study calcium vanadate at 4K? Which electrical properties have changed ?  Is this change due to the temperature ? to the humidity ? Does it happen only on your spots ?

@Maurice Yes, my samples are pure CaVO_3 and I study them to better understand the effects of electron correlation in correlated metals. Temperature-dependent measurements allow me to better understand what is happening in the system and 4K is the lowest temperature we can achieve at our lab and therefore the closest we can get to the ground state of the system.

https://www.sciencedirect.com/science/article/pii/S0304885309004399 might be of interest

If you mean $\ce{Ca}\left( \ce{V}\ce{O_{3}} \right)_{2}$ aka $\ce{Ca} \ce{V_{2}}\ce{O_{6}}$ then $\ce{Ca} \ce{V_{2}}\ce{O_{6}} + 2 \cdot \ce{H_{2}}\ce{O} \to \ce{Ca^{2+}} + 2 \cdot \ce{H_{2}}\ce{V}\ce{O_{4}}^{-}$ should apply.
You can also find the reaction here in equation $81$: $$
\ce{Ca_{n}}\ce{V_{2}}\ce{O_{5 + n}} + \left( 3 - n \right) \cdot \ce{H_{2}}\ce{O} \to n \cdot \ce{Ca^{2+}} + 2 \cdot \ce{H_{3 - n}}\ce{V}\ce{O_{4}}^{n-}
$$ where $n = 1$.
So your calcium vanadate ($\ce{Ca}\left( \ce{V}\ce{O_{3}} \right)_{2}$) reacts with water ($\ce{H_{2}}\ce{O}$) to calcium ions ($\ce{Ca^{2+}}$) and dihydroxy-oxido-oxovanadium ($\ce{H_{2}}\ce{V}\ce{O_{4}}^{-}$ aka $\ce{V}\ce{O_{2}}\left( \ce{O}\ce{H} \right)_{2}^{-}$).
The whole thing could theoretically continue to react at high enough temperatures: see here (page $8$)):
$$
4 \cdot \ce{V}\ce{O_{2}}\left( \ce{O}\ce{H} \right)_{2}^{-} \to \ce{V_{4}}\ce{O_{12}}^{-4} + 4 \cdot \ce{H_{2}}\ce{O}
$$
But predicting the color of the whole thing is not that easy. On the one hand, there can be many other reactions, especially if your substance contains other particles. On the other hand, it's really difficult to find references for this. Also, more information would be useful, like its new properties.

My material is actually $\ce{Ca}\ce{V}\ce{O_{3}}$

I would assume that $\ce{Ca}\ce{V}\ce{O_{3}}$ dissociates first in water:
$$
n \cdot \ce{Ca}\ce{V}\ce{O_{3}} + m \cdot \ce{H_{2}}\ce{O} \rightleftharpoons n \cdot \left( \ce{Ca^{2+}} + \ce{V}\ce{O_{3}}^{2-} \right) + m \cdot \ce{H_{2}}\ce{O}
$$
Because of the $\ce{V}\ce{O_{3}}^{2-}$ ion's peculiar electron configuration, I would assume that it would "steal" the hydrogen ($\ce{H}$) from the water ($\ce{H_{2}}\ce{O}$). The resulting oxygen in turn wants to react and could also react with the ion. This would result in something of the form:
$$
n \cdot \ce{Ca^{2+}} + m \cdot \ce{H_{2}}\ce{O} + n \cdot \ce{V}\ce{O_{3}}^{2-} \to n \cdot \ce{Ca^{2+}} + \left( n - w \right) \cdot \ce{V_{x}}\ce{O_{y}}^{z-} + \ce{R_{1}}
$$
If we let it dry again, we should just get a new calcium vanadate again (and some remaining part $\ce{R_{1}}$). Some vanadium-oxides are kinda yellow-orange or dark blue like vanadium($\rm{IV}$)-oxide. Here $\left\{ w,\, x,\, y,\, z \right\} \in \mathbb{N_{0}}$, $n - w$ is the number of this special kind of vanadate (the vanadate $\ce{V_{x}}\ce{O_{y}}$), $x$ and $y$ are forming something which is supposed to imply the type of vanadate, $z$ is the electric charge of this vanadate / vanadate-ion and $\ce{R_{1}}$ is a remaining part resulting from the reaction, which in turn can be another vanadate or it can be a completely different compound.

Thank you for your answer. My material is actually CaVO$_3$ and sadly I have also not found references on this subject, especially for this form of calcium vanadate. I have included the change of electrical properties as an edit in my original post, I hope this helps.

The famous book by Cotton and Dickinson, Advanced Inorganic Chemistry, states that the blue oxide $\ce{VO2}$ can be fused with chalk and produce calcium vanadate $\ce{CaVO3}$ and ... "Little is known of its structure"...

@ Kevin Dietrich. Your last reaction is wrong. Calcium vanadate cannot produce $\ce{CaOH}$ in water, as this substance $\ce{CaOH}$ does not exist. Calcium hydroxyde is $\ce{Ca(OH)2}$

@Maurice Then I'll delete it.

@ Kevin Dietrich. Your new last reaction is wrong. It produces a strange stuff called $\ce{R1}$ which corresponds to nothing on the left hand side. This is the same for the parameters $w, x, y, z$, whose meaning is unknown.


text:
Why Does Milk Seal Sidewalk Chalk?
I work at a public library and we recently had a sidewalk chalk hour for families. Someone gave us a tip that if we brush milk with a paintbrush over the finished chalk drawings, it helps seal them and makes them last longer. It has been several weeks, two heavy rainstorms, and a hail storm, and all of the drawings we brushed with milk are still there! The ones we didn't brush with milk are long gone. What are the chemical properties of milk that make it do this to sidewalk chalk?

I suspect the protein in the milk acts as sealant: https://www.ehow.com/how_12343389_create-beautiful-pastel-art-milk-chalk.html and  https://richesonart.com/products/paints/richesoncasein/richcaseinfaq.html#:~:text=Gouache%20is%20similar%20to%20Casein,that%20gouache%20does%20not%20have.

Casein has been used in paint and glue since time immemorial

Did you use whole milk (4% fat) or skim milk (~0% fat, normal casein)? Next time, do a "chemical" experiment.

Proteins coagulate, i.e., harden. Egg, as well milk, is used in tempera, too.
Ah! A teachable moment -- have the children's section display cardboard box sarcophagi, painted with tempera made by grinding colored chalk and milk or egg. This would actually have some authenticity, from the citation above, as an alternative to making canopic jars with the remains of dinner.


text:
The electron structure in transition metallic bonding
Copper has an electron structure of $\ce{[Ar] 3d^10 4s^1}$. In salts it may form two stable ions, the $\ce{Cu^+}$ and $\ce{Cu^2+}$. (It's actually not very clear to me why the 2+ ion is common, why does it readily lose one of its d orbital electrons, but I'm willing to wave my hands and say "something quantum").
In pure copper metal, in the usual high school picture of metallic bonding, the conduction band electrons delocalise leaving a structure with ions "floating in a sea of delocalised elections".  In copper, how many of its electrons delocalise?  Does it make sense to describe copper metal as $\ce{Cu^+}$ or $\ce{Cu^2+}$ ions surrounded by an electron ocean?

Image of metallic bonding from BBC bitesize
The linked page says that in group 2 metals, both valence electrons are delocalised, but I've also read that in copper, the whole $\ce{3d^10 4s^1}$ electrons are all valence electrons.

https://chemistry.stackexchange.com/questions/24873/why-do-copper-ii-complexes-contain-so-many-valence-electrons

Use the word "delocalized" carefully, this sea has a bonding contribution, $11$ $e^-$ will not have a collective motion when $\Delta V$ is applied, however only the states near to the Fermi surface will have such a behavior, from the density of states about $n_e = 1.4 $ in agreement with the result of the linked answer.

Can you exapand on that comment, if necessary in an answer to the duplicate.

I guess he means there's plenty of covalent component to the bonding - electrons that don't delocalize.


text:
Mixture of two ideal gases
Why is $\Delta H_{\mathrm{mix}}=0$ in a mixture of ideal gases? If $\Delta H=\Delta U + \Delta PV $ and I know that in ideal gases there are no interactions between molecules so $\Delta Q=0$ and I consider $\Delta V=0$
How do I prove this? Do the gases need to have constant pressure so $\Delta H_{\mathrm{mix}}=Q$? If not, I really don't understand how could it be $0$ otherwise.
Can someone please tell me if this correct?
So if $\Delta Q=0$ and $\Delta V=0$ then from the first law of thermodynamics I can say that $\Delta U=0$ so the temperature $T$ is constant. That would also tell me that pressure $P$ is constant.
From that I can say that $\Delta H_{\mathrm{mix}}=Q$, so $\Delta H_{\mathrm{mix}}=0$.
Is this correct?

Of course it is implied that you mix them at constant pressure, otherwise you might expect a lot of unpleasant things, like work.

@IvanNeretin The thing that I'm really struggling with is: this constant pressure is the pressure of the gases or the pressure exerted on the gases.( I know its probably a stupid question). Because I think that when the gases ended their mixing process there would be a constant pressure even the if their initial pressures are different.

"Constant" here means "the same before and after".

$\Delta H_\mathrm{mix}=0$ at p, T constant is the implicit part of definition of ideal gases.

I will give the general proof as to why $\Delta_\mathrm{mix} H=0$ for a mixture of ideal gases. You will not be able to reach a conclusion with the equations you use, we need a bit of theory about the thermodynamics of solution. I will give some definitions, and then see what this definitions tell when we apply the ideal-gas model.
1. Properties of a solution There are three guys, all different, that represent different thermodynamic properties in a solution. For a general property $M$ we have:

$M(p, T, y_1, ..., y_N)$ is the thermodynamic property of a solution as a whole. It is a function of pressure $p$, temperature $T$, and all molar fractions ($N-1$ becase one is linearly dependent).
$M_i(p, T)$ is the thermodynamic property of species $i$ as pure. It is a function of pressure $p$, temperature $T$.
$\bar{M}_i(p, T, y_1, ..., y_N)$ is the partial molar property of species $i$ in a solution. It is a function of presure $p$, temperature $T$, and all molar fractions ($N-1$ also). This is the new guy in solution thermodynamics and is defined as

$$ \bar{M}_\mathrm{i} := \left[\frac{\partial (nM)}{\partial n_\mathrm{i}}\right]_
   {p, T, n_\mathrm{j\neq i}} \tag{1} $$
where the symbol $j \neq i $ says that the differentiation must be done keeping all the amounts constant, with the exception of $n_i$. Eq. (1) seems obscure but is simple to understand. Say the property $M$ is the molar volume, so $nV$ will be the total volume. $\bar{V}_i$ will say how much the total volume $nV$ changes when we add a differential amount of species $n_i$ in the solution. Partial molar properties were introduced by Gilbert N. Lewis, who also introduced fugacity and the model of ideal solutions, so I am just repeating what he said.
2. Summability relation Now, in solution thermodynamics, every thermodynamic property in a solution will not be just a function of two intensive variables. It will be a function of the different amounts of the species present in a solution. You may imagine that a mixture of hexane and chlorobenzene will have different properties whether the molar fraction of hexane is $0.1$ or $0.9$. Hence, the total differential of the property $M$ is now
\begin{align}
  \mathrm{d}(nM) &=
  \left[\frac{\partial (nM)}{\partial p}\right]_{T, n} \mathrm{d}p +
  \left[\frac{\partial (nM)}{\partial T}\right]_{p, n} \mathrm{d}T +
  \sum_\mathrm{i} \left[\frac{\partial (nM)}{\partial n_\mathrm{i}}\right]_
  {p, T, n_\mathrm{j\neq i}} \mathrm{d}n_\mathrm{i} \\
  &=
  \left[\frac{\partial (nM)}{\partial p}\right]_{T, n} \mathrm{d}p +
  \left[\frac{\partial (nM)}{\partial T}\right]_{p, n} \mathrm{d}T +
  \sum_\mathrm{i} \bar{M}_\mathrm{i} \mathrm{d}n_\mathrm{i} \tag{2}
\end{align}
where we used Eq. (1) to replace the partial derivative. The new term here is the final sum. If we: (1) open up $\mathrm{d}(nM)$, and (2) use that $n_\mathrm{i} = x_\mathrm{i} n$ or in a total differential form $\mathrm{d}n_\mathrm{i} = n\mathrm{d}x_\mathrm{i} + x_\mathrm{i}\mathrm{d}n$ we get two equations when we group the terms with $n$ and with $\mathrm{d}n$
\begin{align}
  & n\mathrm{d}M + M\mathrm{d}n =
  n\left(\frac{\partial M}{\partial p}\right)_{T} \mathrm{d}p +
  n\left(\frac{\partial M}{\partial T}\right)_{p} \mathrm{d}T +
  \sum_\mathrm{i} \bar{M}_\mathrm{i} (n\mathrm{d}x_\mathrm{i} + x_\mathrm{i}\mathrm{d}n) \\
  & n\mathrm{d}M + M\mathrm{d}n -
  n\left(\frac{\partial M}{\partial p}\right)_{T} \mathrm{d}p -
  n\left(\frac{\partial M}{\partial T}\right)_{p} \mathrm{d}T -
  n\sum_\mathrm{i} \bar{M}_\mathrm{i} \mathrm{d}x_\mathrm{i} - \sum_\mathrm{i} \bar{M}_\mathrm{i} x_\mathrm{i}\mathrm{d}n = 0 \\
  & \left[M - \sum_\mathrm{i} x_\mathrm{i}\bar{M}_\mathrm{i} \right] \color{blue}{\mathrm{d}n} +
  \left[\mathrm{d}M -
  \left(\frac{\partial M}{\partial p}\right)_{T} \mathrm{d}p -
  \left(\frac{\partial M}{\partial T}\right)_{p} \mathrm{d}T -
  \sum_\mathrm{i} \bar{M}_\mathrm{i} \mathrm{d}x_\mathrm{i} \right]\color{blue}{n} = 0 \tag{3}
\end{align}
For a system of a total amount $n$ and a total variation of $\mathrm{d}n$, two equalities rise in Eq. (3) in order to be zero. The second one, with a bit more of manipulation, gives the famous Gibbs-Duhem equation. The first one is the summability relation
$$ M = \sum_\mathrm{i} x_\mathrm{i}\bar{M}_\mathrm{i} \tag{4} $$
This equation is also simple to understand. It is a connection between the property of the solution $M$, and all the partial molar properties. For example, if we have access in a mixture of hexane$(1)$+chlorobenzene$(2)$ to $\bar{V}_1$ and $\bar{V}_2$, then applying Eq. (4) will give us the molar volume of the solution in the whole range of composition, by simple algebra.
3. Gibbs theorem It would be nice if we could get values for the partial molar properties. There are two approaches to this. The first one is going to the laboratory, and the second one is to impose a model behavior for the fluid. For the ideal gas model, we can get all the molar properties with the addition of Gibbs theorem:

A partial molar property (with the exception of molar volume) of a species in a mixture of ideal gases at a pressure $p$ and $T$, is equal to the molar property as a pure ideal gas at temperature $T$ but at the partial pressure in the mixture.

Mathematically, this is
$$ \bar{M}_\mathrm{i}^\mathrm{ig}(p,T) = M_\mathrm{i}^\mathrm{ig}(p_\mathrm{i},T) \hspace{1 cm}
    M \neq V \tag{5} $$
4. Enthalpy of mixing Now we apply Eq. (5) for the enthalpy. But this case is really easy, because for ideal gases, the enthalpy of a pure species is independent of pressure. It is only a function of temperature, so we can evaluate the pressure at any point we want and the result will not change
$$ \bar{H}_\mathrm{i}^\mathrm{ig}(p,T) = H_\mathrm{i}^\mathrm{ig}(\color{blue}{p_i},T) =
    H_\mathrm{i}^\mathrm{ig}(\color{blue}{p},T)  \tag{6} $$
Combining Eqs. (4) and (6) when $M=H$
\begin{align}
  H^\mathrm{ig} &= \sum_\mathrm{i} x_\mathrm{i} \bar{H}_i^\mathrm{ig} \\
  H^\mathrm{ig} &= \sum_\mathrm{i} x_\mathrm{i} H_\mathrm{i}^\mathrm{ig}(p,T) \\
  H^\mathrm{ig} - \sum_\mathrm{i} x_\mathrm{i}  H_\mathrm{i}^\mathrm{ig}(p,T) &= 0
  \rightarrow
  \boxed{\Delta_\mathrm{mix}H := H^\mathrm{ig} - \sum_\mathrm{i} x_\mathrm{i} H_\mathrm{i}^\mathrm{ig}(p,T) = 0}
  \tag{7}
\end{align}
The left side in Eq. (7) is the definition of the enthalpy of mixing, and for ideal gases it is equal to zero. In fact, for any property $M$, its value of mixing is defined as
$$ \boxed{\Delta_\mathrm{mix} M = M -
   \sum_\mathrm{i} x_\mathrm{i} M_\mathrm{i}(p,T)} \tag{8} $$
As you see, for a mixing process, the change in the property is not just some typical subtraction like "$H_2 - H_1$".
You had some doubrs about the pressure in a process of mixing. I think Eq. (8) will clear them. For example, if you could have an enthalpy-meter, you would have to:

Measure each enthalpy of the pure species to get the values of $H_\mathrm{i}(p,T)$. This measurement must be done with every species at $p$ and $T$.
Measure the quantities being mixed, so you know also the values of the $x_\mathrm{i}$'s.
Mix the species, so you get the solution, and measure the value $H(p,T, x_1, x_2, ..., x_N)$. This measurement, also, needs to be done with the solution at $p$ and $T$.

Thus, a mixing process is just made by changing the different amounts of the species. The pressure, regarding the enthalpy of mixing, refers to the same value before and after. You can imagine a process of mixing where the pressure may not be constant. However, if you can guarantee experimentally that the final state of the solution is pressure $p$, your results will still be accurate because the enthalpy is a state function.
Typically, heat experiments are done to get enthalpy data of solution, and volumetric experiments are done to get molar volume data of solution. These are the easiest ways to calculate partial molar properties.
You may ask why we arbitrarily relate partial molar properties with pressure $p$ and $T$. We could have chosen molar volume $V$ and entropy $S$ to define the partial derivative in Eq. (1). This is because of practical convenience, since $p$ and $T$ can easily be manipulated experimentally.
As an exercise, you can do the same with the molar volume $V$, and reach that it is also zero. For the Gibbs energy $G$ and entropy $S$ you will not get a zero value.
References

J. M. Smith, H. C. Van Ness, M. M. Abbott, "Introduction to Chemical Engineering Thermodynamics", 7th ed., Mc-Graw Hill, 2005.
J. M. Prausnitz, R. N. Lichtenhaler, E. G. de Azevedo, "Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd ed., Prentice Hall PTR, 1999.


text:
Doubt regarding isothermal enthalpy change of ideal gases
In the book, I have seen that it is written that in an isothermal change, $$\Delta H =\Delta U + \Delta n_\mathrm{g}RT$$
I can clearly understand the derivation where $\Delta H = \Delta U +\Delta (PV)$
And as $PV=nRT$, $\Delta (PV)=\Delta(n_\mathrm{g}RT)$
And as $T=\text{const.}$
$\Delta (PV)=\Delta n_\mathrm{g}RT$
We plunge this value in the above equation. But what my question is, if the process is isothermal, internal energy must be constant i.e. $\Delta U=0$.
Then the equation must be
$$\Delta H = \Delta n_\mathrm{g}RT$$
I want to know where is my concept wrong?

CH SE site prefers not to use MathJax in titles.

Wikipedia say ∆U=0 for ideal gas in isothermal process.

∆U=0 reflect the balance between heat (coming/going) and work (done by system/done on system).

But what my question is, if the process is isothermal, internal energy must be constant.

Thermodynamics in this contexts evaluates changes of state variables. The "state" means their change depends only on the initial and final state, not on the path how the final state has been reached.
Therefore, being isothermal means here that the initial and final temperature are the same, while the temperature profile between these states does not matter.
It says nothing about the $\Delta U$ value. The eventual ongoing process may be exothermic, endothermic or thermally neutral. If you take natural gas and air at $\pu{25 ^{\circ}C}$, burn the gas and cool the products back to $\pu{25 ^{\circ}C}$, it is an isothermal thermodynamic process.
For $\Delta n_\mathrm{g} \ne 0$, a chemical reactions must occur. In such a case, $\Delta U \ne 0$ even if $\mathrm{d}T=0$.
Without any reaction, $\Delta n_\mathrm{g} = \Delta U = \Delta H = 0$ unless the system is open.
With a reaction, generally, $\Delta H = \Delta U + \Delta(pV) \ne 0$ as $\Delta U \ne 0$ and possibly $\Delta (pV) \ne 0$.

If $\Delta n_g$ = 0, $p\Delta V =0$ from Avogadro's law of gases ($ V \propto n$) which further means $\Delta H = \Delta U = 0$. I think this is what you mean or not?

Yes, it is what I mean by other words. For $\Delta n=0$ is by consequence of the ideal gas definition $\Delta (pV)_T = 0$

Your equation is based on the condition that a chemical reaction is occurring within a closed system.  Even for an ideal gas chemical reaction, $\Delta U$ and $\Delta H$ are not zero even if the change in temperature is zero.  Only if the process is adiabatic and in a constant volume system is $\Delta U=0$, but the temperature is not constant in such a case.


text:
Confusion about the number of microstates for orbitals
I am very confused about the microstates for a specific orbital.
Let's suppose I have an electronic configuration
$$
\ce{[\dots] 2p^5}
$$
Considering the symmetry of the orbitals and the "Electron-Hole" idea, one would get for this state the same Term Symbols for the possible states, i.e.:
\begin{align}
L &\in \{1\}\\
S &\in \{1/2\}\\
J &\in \{1/2,3/2\}
\end{align}
With that, one gets the following terms:
$$^2P_{1/2} \text{ and } ^2P_{3/2}$$
My confusion starts here. If I simply draw every possible state with the electron-box diagram, I would get 6 possible states:
\begin{array}{|l|c|c|c|}\hline
1:& \uparrow\downarrow & \uparrow\downarrow & \uparrow\\\hline
2:& \uparrow\downarrow & \uparrow\downarrow & \downarrow\\\hline
3:& \uparrow & \uparrow\downarrow & \uparrow\downarrow\\\hline
4:& \downarrow & \uparrow\downarrow & \uparrow\downarrow\\\hline
5:& \uparrow\downarrow & \uparrow & \uparrow\downarrow\\\hline
6:& \uparrow\downarrow & \downarrow & \uparrow\downarrow\\\hline
\end{array}
What am I thinking or doing wrong?

Do you know Hund's rules ? Most of your possible states are inconsistent

Everything looks correct so far. You can calculate the M_S and M_L values of the configurations, then draw a table with M_S and M_L as row and column and then extract two tables that correspond to your term symbols. One table should be 2x2 (S=1/2, L=1, J=1/2) and the other table 4x4 (S=1/2, L=1, J=3/2). What exactly are you unsure about?

Makes sense. But if the energy shift of the $L \cdot S$ coupling for fixed L and S only depends on the values for J : $\Delta E_{FS}=c/2\cdot(j(j+1)-l(l+1)-s(s+1)$, can we say that the ground state, according to the Hund's rules, can actually assume any one of the 2 states of j=1/2, as stated by you?

I will explain in details why I stated (in the comments) that some states are inconsistent. The spin-orbit (S.O) coupling correlates the orbital angular momentum $\vec{L}$ and the spin angular momentum $\vec{L}$ in a way that the eigenvalues of these operators are not good quantum numbers. This means naively that the ground-state is not an eigenvector of these operators anymore, but an eigenvector of the total angular momentum $\vec{J}$. This is a non surprising result as the orbit of a charged particle independent of its spin around a heavy charged particle creates has an angular momentum, a magnetic moment and a magnetic field which acts on the spin of the particle : in a more intuitive picture, one has to consider a relative motion of the heavy charge around the light particle, giving in the frame  of the latter the superposition of $\vec{L}$ and $\vec{S}$. When someone is used to deal with dipole interactions, there is an intuitive deduction that the orbital magnetic moment derived from $\vec{L}$ and the spin magnetic moment proportional to $\vec{S}$ should be antiparallel in the ground-state : therefore the classical direction of the rotational motion should be chosen depending on the orientation of the spin to the particle. Considering a weak interaction ($^2P$), the S.O. coupling gives the correction :
$$ \Delta H_{S.O.} = \lambda \vec{L}\cdot \vec{S} = \lambda L.S.\cos(\theta)$$
There is no need to use this expression $ \Delta E_{S.O.} = \beta \left[j(j+1)-l(l+1)-s(s+1)\right]$  to  derive the third rule :  if $\lambda > 0 $, the only stable configuration is $\cos(\theta) < 0$, this means $ \frac{\pi}{2} < \theta < \frac{3\pi}{2}$. Because, $\vec{L}$ is quantized along $z$, $L_z |l,m\rangle = \hbar m_l|l,m\rangle$ with  $m_l$ integers,  and $S_z |l,m\rangle = \hbar m_s|s,m\rangle$ with  $m_s$ half integers, the angle  $\theta \approx \frac{9}{10} \pi \neq \pi$ for an antiparallel  configuration of $\vec{L}$ and $\vec{S}$ (should be checked). With S.O. coupling, $m_j$ eigenvalues of  $\vec{J}$ are good quantum numbers. The previous results shows that $\vec{S}$ and  $\vec{L}$ should be antiparallel for a stable configuration. This means that if you consider $\vec{J} = \vec{L} + \vec{S}$
as the total angular momentum, $m_l<0$ should match $m_s>0$ and inversely, the filling is given step-by-step  in the table below :
\begin{array}{|l|c|c|c|c|}\hline
m_l & +1 & 0 & -1\\\hline
    &  \downarrow & & \\\hline
    &  \downarrow & \downarrow & \\\hline
    &  \downarrow & \downarrow & \downarrow \\\hline
    &  \downarrow & \downarrow & \uparrow\downarrow \\\hline
 (4)   &  \downarrow & \uparrow\downarrow & \uparrow\downarrow \\\hline
\end{array}
The final states gives $J=|-1+\frac{1}{2} |= 3/2$ and is the ground state, some of the other states do not satisfy $\Delta E_{SO} <0$ especially if $J=1/2$. The state (3) is also correct only if one starts by $m_l=-1$ :
\begin{array}{|l|c|c|c|c|}\hline
m_l    & -1 & 0 & +1\\\hline
 (3)   &  \uparrow & \uparrow\downarrow & \uparrow\downarrow \\\hline
\end{array}
However, the state (1) represent the conventional filling picture, from left to right, where the empty states are on the right, this is the reason why the third rule has the term $J=|L-S|$ for less than half filled and $J=L+S$ for more than half filled shell. Obviously, if you fill the shell from left to right until an half filled shell an then right to left,  $J\equiv |L+S |$.

But your assuption for a stable state is for the ground state. My main question,initially, didn't have to do only with the ground state, but rather with all allowed states, even if they aren't stable. But at the same time, your answer clarified my comment about the possible ground states, i.e. two of them are equally possible, if I understood correctly.

The possibles states are (permutation of 2 spins in 3 levels) and  (2 spins -> 3 levels) gives $3!\times 3!$ = 36 possibilities without Pauli exclusion and reduced to 6 by applying the exclusion principle, so you was right. Three possible values of $L$, therefore two states with $L=0$ ($J=1/2$), two states $L=-1$, one with $S=1/2$ ($|J|=1/2$) and  two states $L=1$, one ($S=-1/2$) ($|J|=1/2$). To summarize four states $|J|=1/2$ and your are right two of them are the ground state ($|J|=3/2$). My answer should not have been so long and so far from the main question.


text:
How to achieve highest CO2 pressure from acetic acid + sodium bicarbonate reaction?
I'm planning to make a high-end vinegar + baking soda rocket and started doing some researches for the best ratio of the reagents's concentrations to achieve maximum pressure. Furthermore, I know that the temperature of the vinegar affects the reaction, so, what's the best temperature, the hotter the better?
If you know any other ways to make the reaction faster, I'd like to know. For context, I have a 2 liter bottle and would like to achieve 250+ psi. Is it possible to achieve this pressure with the reaction happening inside this bottle, or a separate vessel would be needed?

Reaction can hardly be "better" - results can be, and in your case it's speed, or more precisely higher rate. To achieve that, not only temperature and concentrations are relevant, but primarily the speed of mixing. Also don't add "thanks", you can do it later in comment.

https://chemistry.stackexchange.com/questions/16758/comparing-rate-of-reaction-of-acetic-acid-and-sodium-bicarbonate-at-different-te

250 PSI seems impossible, see here.