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How to obtain the radial probability distribution function of a given orbital from a quantum chemical calculation?
I know that a very similar question has already been asked here and a very competent and useful answer has been provided (see here). However, that previous question was related to the total electron density whereas I am interested in plotting the radial probability distribution function of a given single orbital. Can I also use MultiWFN tool and Gaussian output to do this?

This is actually a lot simpler than I initially thought. I'll be using the same example as previously explained in How to obtain the radial probability distribution function from a quantum chemical calculation? My test setup has slightly changed since then, i.e. I've tried it with Gaussian 16 B.01 and Multiwfn 3.6.0 and it worked the same.
We start the same way as in the linked question, until you have your ar.fchk and load that into Multiwfn, you should see something like this:

 Multiwfn -- A Multifunctional Wavefunction Analyzer
 Version 3.6, release date: 2019-May-21
 Project leader: Tian Lu (Beijing Kein Research Center for Natural Sciences)
 Below paper *MUST BE CITED* if Multiwfn is utilized in your work:
          Tian Lu, Feiwu Chen, J. Comput. Chem., 33, 580-592 (2012)
 Multiwfn official website: http://sobereva.com/multiwfn
 Multiwfn English forum: http://sobereva.com/wfnbbs
 Multiwfn Chinese forum: http://bbs.keinsci.com/wfn

>>

 Loaded ar.fchk successfully!
 Formula: Ar1
 Molecule weight:        39.94800

We need to modify the wave function now:

                    ************ Main function menu ************
 0 Show molecular structure and view orbitals
 1 Output all properties at a point
 2 Topology analysis
 3 Output and plot specific property in a line
 4 Output and plot specific property in a plane
 5 Output and plot specific property within a spatial region (calc. grid data)
 6 Check & modify wavefunction
 7 Population analysis and atomic charges
 8 Orbital composition analysis
 9 Bond order analysis
 10 Plot total DOS, partial DOS, OPDOS, local DOS and photoelectron spectrum
 11 Plot IR/Raman/UV-Vis/ECD/VCD/ROA spectrum
 12 Quantitative analysis of molecular surface
 13 Process grid data (No grid data is presented currently)
 14 Adaptive natural density partitioning (AdNDP) analysis
 15 Fuzzy atomic space analysis
 16 Charge decomposition analysis (CDA) and extended CDA (ECDA)
 17 Basin analysis
 18 Electron excitation analysis
 19 Orbital localization analysis
 20 Visual study of weak interaction
 21 Energy decomposition analysis
 100 Other functions (Part1)        200 Other functions (Part2)

Chose 6:

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   9.000   9.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 34 Set occupation number of inner orbitals to zero
 36 Invert phase of some orbitals

As you can see there is plenty you can do. For this example I will reset the occupation of some orbitals (26). More precisely, I'll transform the wave function so that only the 2s orbital is doubly occupied. Here is how I did it:

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   9.000   9.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 34 Set occupation number of inner orbitals to zero
 36 Invert phase of some orbitals
26

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
0
 Set to which value? e.g. 1.2
 Note:
 You can also input for example "+1.1" "-1.1" "*1.1" "/1.1" to add, minus, multiply and divide the occupation numbers by 1.1
 To recover the initial occupation numbers, input "i"
 To generate occupation state for calculating odd electron density, input "odd"
0
 Done!

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
2
 Set to which value? e.g. 1.2
 Note:
 You can also input for example "+1.1" "-1.1" "*1.1" "/1.1" to add, minus, multiply and divide the occupation numbers by 1.1
 To recover the initial occupation numbers, input "i"
 To generate occupation state for calculating odd electron density, input "odd"
2
 Done!

 Select the orbitals for which the occupation numbers are needed to be changed
 e.g. 2,4,13-16,20 means selecting orbitals 2,4,13,14,15,16,20
 Input 0 can select all orbitals, input q or 00 can return
q

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   1.000   1.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 36 Invert phase of some orbitals
3

 Basic information of all orbitals:
 Orb:     1 Ene(au/eV):  -117.263218   -3190.8944 Occ: 0.000000 Type:A+B
 Orb:     2 Ene(au/eV):   -12.054866    -328.0296 Occ: 2.000000 Type:A+B
 Orb:     3 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     4 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     5 Ene(au/eV):    -9.308431    -253.2953 Occ: 0.000000 Type:A+B
 Orb:     6 Ene(au/eV):    -1.154432     -31.4137 Occ: 0.000000 Type:A+B
 Orb:     7 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B
 Orb:     8 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B
 Orb:     9 Ene(au/eV):    -0.495941     -13.4953 Occ: 0.000000 Type:A+B

           ============ Modify & Check wavefunction ============
 Number of GTFs:    27, Orb:     9, Atoms:    1, A/B elec:   1.000   1.000
 -4 Exclude contribution of some atoms to real space functions
 -3 Only retain contribution of some atoms to real space functions
 -1 Return
 0 Save the modified wavefunction to a new .wfn file
 1 List all primitive function      2 List all basis function
 3 List all orbitals                4 Print detail information of an orbital
 5 Print coefficient matrix in basis function
 6 Print density matrix in basis function
 7 Print various kinds of integral matrix between basis functions
 11 Swap some information of two primitive functions
 21 Set center of a primitive       22 Set type of a primitive
 23 Set exponent of a primitive     24 Set coefficient of a primitive
 25 Set coefficients of GTFs/basis functions that satisfied certain conditions
 26 Set occupation number of some orbitals
 27 Set type of some orbitals       28 Set energy of some orbitals
 31 Translate the system            32 Translate and duplicate the system
 33 Rotate wavefunction, namely X->Y, Y->Z, Z->X
 36 Invert phase of some orbitals

(The values I have typed in all start in the first column.)
Now leave this screen. You may want to safe with 0 the resulting wave function as new.wfn; but it is not necessary. To get back to the main menu simply type -1.
Now that you have modified the wave function, you can again follow the steps from the linked post.
TL;DR:

200 // other functions
5   // RDF
0   // calc
1   // plot
5   // Export to RDF.txt

I'm sure there  is a lot of fancy stuff you can do with that, I used the same gnuplot file to plot the result:

Shameless self-plug:
I have written some startup scripts for Multiwfn, which make some things easier or extend the functionality to run calculations within a queueing system (e.g. QTAIM charges). Find it on GitHub: https://github.com/polyluxus/runMultiwfn.bash


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Ab initio method to calculate C-C bond dissociation energy?
I'm trying to review how to calculate bond dissociation energy for C-C in ethane, a very simple calculation (or should be).
I get the following Hartree energies for SCF 6-31G(p) for methane, ethane, and H2:
ch4_spe            -40.194639920746,
ethane             -79.228125042573,
h2                 -1.126456057886,

I would think that to get the C-C bond dissociation energy, I would simply do (Perl code)
$energies{ethane} - $energies{ch4_spe} * 2 - $energies{h2}

correct?
The reason that I'm asking is that this give an answer of 2.2876 Hartree using NWChem, which converts to 6006 kJ/mol, which is off by about a factor of 17 from the correct value of about 346 kJ/mol.  I know that SCF isn't that accurate, but it shouldn't be this far off. I optimized the bond lengths using Avogadro.
This is the correct method to calculate the C-C bond energy?

Hint:
The balanced "imaginary reaction" you are considering is
H2 + H3C-CH3 -> 2 * CH4   This said, how should you calculate the variation in energy?

I don't see how this reaction equation gives you the carbon-carbon bond dissociation energy; you'll get a mixture of $\ce{H-H}$, $\ce{C-C}$ breaking with $\ce{C-H}$ forming. For the BDE I would consider $\ce{H3C-CH3 -> 2 .CH3}$. (One order of magnitude is not that bad for HF bond energies though. You also might want to consider optimising the structure at the level you're using to calculate the energy.)

I was about to write exacly what Martin wrote. I just tried the calculation, and you can get a very good result by considering the reaction written by Martin. Remember that you should use UHF, with a basis set which includes polarization and diffuse functions when you deal with radicals (I did it with 6-31++G**). For the geometry of the radical, just split the optimized ethane molecule in half and do not optimize it, just run an energy calculation.
See: http://przyrbwn.icm.edu.pl/APP/PDF/127/a127z4p004.pdf

@The_Vinz and Martin, thanks, unrestricted HF on the doublet state gives a dissociation energy of 289 kJ/mol, which is much much more reasonable to the empirical 346 kJ/mol.  If you write your comment down as an answer, I can accept it

You should be able to further refine it: with the method I wrote about I obtained an energy of 351.28 kJ/mol, which is very close to the experimental value :)
Remember that if you are comparing calculations on different molecules, you should apply the same method and basis set to the whole set of molecules

I would like to encourage you to write a self-answer. I further recommend some reading about the UHF/R(O)HF methodology (cc @The_Vinz): 1. U- or RO-method for Singlet-Triplet Gap? 2. Calculating singlet-triplet gap and open-shell singlet character Even though the title suggests something else, they can be applied to open-shell in general.

This "experiment" is about determining bond dissociation energy.
The general way to solve this isn't like calculating enthalpies of reaction, as I initially thought, but rather through calculating the energy of two CH3 radicals, and comparing it with ethane.  As another user pointed out, this should be done with unrestricted Hartree Fock and with diffuse basis sets.
Thus, energy(ethane) - 2*energy(methyl radical) is the solution.

source: Stack Exchange

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

sort: votes


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Predicting colors of aqueous solutions
While going through my AP chem prep book again, I came across questions asking about the color of certain aqueous solutions. For example:

Which of the following substances will produce a colorless aqueous solution?
A) $\ce{Zn2(NO3)2}$
B) $\ce{Cu(SO4)}$
C) $\ce{K2Cr2O7}$
D) $\ce{Co(NO3)2}$

I know that zinc compounds are generally colorless, so I picked that (and it was correct), but I have no idea how to predict the colors of the rest of the compounds.
So, to answer this question (and questions similar to this), would I need to use pure memorization of the common elements that produce color?
If so, where can I find information about these compounds and the color they produce?

The question is probably trying to tell whether you know the difference between a transition metal and other metal ions. Knowing whether there is a colour is the point not knowing what the colour is.

There is no simple way to predict beforehand as far as I know. This is a general knowledge type of chemistry question, you will have to know or memorize the colors of common anions and cations. It will be much easier, if you do so group wise and know the colors of the first row of transition element ions.
Any object which absorbs a certain portion of visible light's spectrum, will appear colored to us. For colored solutions, we will always consider absorption of light in the visible region. So if you see a green solution, it is not reflecting green light, it is absorbing a certain portion of the visible spectrum and the transmitted light [from the solution] appears green to us. Keep this concept in mind because a lot of students have this misconception.
You should know that sulfate ion is colorless but copper (II) ion is blue so most likely copper sulfate solution is blue. In part c, you should know the color of potassium ion and dichromate ion and then guess the color. Do the remaining exercise for others.
It will be very difficult to predict if both ions are colored because it is not like that we are mixing two paints, a yellow paint with blue paint will show a green color. You will have to study the absorption spectrum of that salt. The theory of color of ions and that of solids is quite complex. It was developed by physicists. You can start from crystal field theory if you wish to see how the color of transition metal ions is explained but I will not recommend that you explore it if you are in school.

What about when sometimes its noted that compounds "vary in color". How does this happen if each ion has a set color? Also, it is always an equal distribution between the ions? For example, if you had a blue ion and a red ion in a compound - would it always produce an even purple?

I edited the answer. See the last paragraph.

Cu (II) is only slightly blue in solution, except if you have ammonia in there at pH>8.

https://commons.wikimedia.org/wiki/File:Kupfer(II)-Ionen1.jpg

@Karl, It is funny that you would raise this point to split hairs because we can never define what $slightly$ means spectroscopically. Do you recall dissolving copper nitrate in water in high concentration? It is such a beautiful deep blue solution but I would never call it slightly blue.

The Wiki page on Kupfer(II) shows a really dilute solution of copper (II). I wish the author wrote the concentrations. commons.wikimedia.org/wiki/File:Kupfer(II)-Ionen1.jpg

Its been a long time since I did wet inorganic chemistry. ;) I just remember that the effect of adding a few drops of ammonia solution is quite impressive. Turns really deeply blue.

I also did these experiments almost a decade ago. Yes, one can comparatively say that copper (II) solution is somewhat lighter than copper ammonia complex. That is indeed deeper blue for the same concentration.

I disagree that this is mostly a memorization question. These types of questions are based on the (admittedly simplistic) assumption that color comes from $d \rightarrow d$ transitions in transition metal complexes. That means $d^0$ and $d^{10}$ central metal atoms have colorless complexes, while others are colored. Permanganate and dichromate are the big exceptions ($d^0$ but colored) that you have to memorize, but otherwise, the intent is that students will count $d$ electrons to figure out the answer here.

When a child wants to become a doctor he has to learn the letters and do 2+2 before he can open books on computerized tomography or read surgery texts. This question is posed by apparently a high school going student who was asking about ionic bonds yesterday. There is no point in bringing d-d and teaching exceptions. Even if they do find it out, d-d transitions cannot be applied to dichromate ions.

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


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Murexide or eriochrome black T for EDTA titration with milk
Should I use Murexide or eriochrome black T for EDTA titration with milk? I am heating the milk to different temperatures and trying to find how the amount of calcium differs (or doesn't).

Read this article https://www.sciencedirect.com/science/article/pii/S0022030261898548

Could you give a little background to this? Because obviously the calcium cannot evaporate, there must be another reason why the apparent content might change.

@Karl Calcium in milk is not free, but kept bound in phosphate groups of casseine, encapsulated in protein structure. That may compete with EDTA thermodynamically and kinetically. Additionally, there are protein changes with temperature change.

@Poutnik, This is exactly why pre-treatment of milk is necessary before titration. All info is given that official paper by the dairy association.  sciencedirect.com/science/article/pii/S0022030261898548

Yes, It was clear to me.. :-) But the efficiency of the pretreatment may be temperature dependent.


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Impregnating polyethylene with oil using pressure cooker?
I have some polyethylene spiral wrap that I'm hoping to impregnate with citrus oil as an animal deterrent.
Wikipedia says that polyethylene plastic will pass through "flavourings" fairly easily. It's melting point for the grade I have seems to be around 100C.
Do you think that placing polyethylene and citrus oil into a pressure cooker and keeping the temperature to around 90C for a few hours would infuse the polymer with enough of the flavour compounds from the oils to have any reasonable effect?
Or would it just be as effective to leave the polyethylene in an oil bath and leave it sealed at room temperature for a month or so?

Will sure diffuse in and also come out again, but to what extent, and if it works on whatever animal you want to set it on (and how long), you´ll have to try. Im not sure what exactly "spiral wrap" is, but I´m also not sure you´ll like its mechanical properties after it was at 90°C .

A Google search of "polyethylene spiral wrap" should give you a pretty good idea of the sort of products I'm referring to. If the melting temperature is at least 100C, then, assuming there is no significant mechanical load on it while it is at 90C, and it is allowed to cool down properly, then what kind of mechanic strength can I expect?

The oil likely will; damage the pressure-cooker gasket. Just leave the wrap submerged in a container, and if you wish, warm gently in a hot-water bath.


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Why is iron pentacarbonyl known, but iron(III) hexacarbonyl is not?
Why is the neutral complex $\ce{[Fe(CO)5]}$ known, but the cationic carbonyl complex $\ce{[Fe(CO)6]^3+}$ is not known?

It might be a short question and look simple, but there is more to it than many people might see at first. We have the 18 valence electron rule we also have the problem of when should we use the 18 VE rule and when should we use ligand field theory.

@andselisk why 3+ instead of +3?

@HahaHahaha because it's 'number before sign'

@HahaHahaha +3 implies oxidation number (and it's placed above the symbol of a chemical element). 3+ is used for a charge number. Which one do you think is applicable here?

Ohh, i did not know that nomenclature, i they were interchangeable, so if in an ionic reaction i were to write say $\ce{Fe(III)}$ is what i intend to write, then is 3+ better or +3, because is the os and charge

related https://chemistry.stackexchange.com/questions/7713/why-feco5-exists-but-fenh35-doesnt

There are several things which are needed to be known to be able to deal with this question.
Firstly there is the 18 Valence Electron (18VE) rule (description at libretexts.org). The iron pentacarbonyl is an 18VE complex. It is coordinatively saturated. A low valent transition metal complex with strong ligand field ligands is considered to be coordinatively saturated if the metal has 18 valence electrons (noble gas electron count).
An iron atom has eight valence electrons, each carbon monoxide (carbonyl) ligand adds two electrons to the valence count. Thus we have a total of 18 in total.
Each positive charge which the complex has causes it to have one less valence electron, so the cationic complex will have an iron which will have 17 VE, this alone would make it less stable.
But there is another issue which is more important, the bonding of carbon monoxide to transition metals is dominated by the withdrawal of electron density from the metal atom to the carbon monoxide. So the less electron-rich the metal the less able the carbon monoxides are to bind.
The fact that the iron in the +3 oxidation state will make it less able to bind to $\ce{CO}$ molecules. This is the main reason why it will not be possible to make the complex.
If you consider the series of complexes $\ce{[V(CO)6]-}$, $\ce{[Cr(CO)6]}$, and $\ce{[Mn(CO)6]+}$, the strength of the bonding between the $\ce{CO}$ and the metal decreases along the series $\ce{V, Cr}$ and $\ce{Mn}$ ($\ce{Mn}$ has the weakest bonds).
Secondly, we should only use the 18VE rule when the oxidation state of the metal is low and the ligands are strong ligand field ligands. I do not think that the iron(III) is sufficiently low in oxidation state for the chemistry to be dictated by the 18 VE rule. I would predict that if I was to bubble $\ce{CO}$ through a solution of iron(III) in dilute acid that nothing would happen.
However, if I was to make a solution of an iron(II) complex of a dialkyl glyoxime then it would be able to react with $\ce{CO}$. Here the iron is in a lower oxidation state and has more electron density on it.

In place of 18VE rule, more fancy term is Sidgwick's effective atomic number rule.

I have never heard of Sidgwick's rule, the problem with named reactions is that they give no indication in the name to what they are. I know that every chemist wants a reaction named after them. But things like the aldol reaction and the 18 VE rule are better names as they give an indication in the name of what the rule relates to.

@nuclearchemist it's even worse in astronomy.


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LIF spectrum with laser-coupled levels lines as the strongest
Why, if I excited the OH molecule Q-branch transition, my LIF spectrum (recorded during and after the laser pulse) does not result in only Q line being the most prominent, but also P and R lines with $\Delta J = \pm 1?$
How/why are these coupled?

You give little experimental detail  so is it possible that (a) collisions in the excited state (which can have a lifetime of  several hundred ns) cause population of other levels and (b) your laser bandwidth is wider than  you anticipate and so several rotational levels are populated.


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Ideal characterization technique to study behavior of heated metal nitrates
I have 3-4 metal nitrates as precursors for my coprecipitation reaction. I need to study the behavior of each individual nitrate and also of various combinations at 60 degrees Celsius, which is my experiment temperature. Which characterization technique would be most suitable to study the change in each case if any?
I have the following nitrates-
1)Barium nitrate
2)Aluminum nitrate nonahydrate
3)Iron nitrate nonahydrate
4)Cobalt nitrate hexahydrate
5)Cerium nitrate hexahydrate
The solvent is water for each solution. I need to know whether any nitrate will decompose or undergo any structural changes. I am trying to reach all these nitrates with ammonium carbonate to precipitate carbonates of all.

I find the question unclear. What nitrates, exactly? What change at 60 °C do you expect? What type of characterization do you need? What are you trying to precipitate? What solvent are you working with?

@andselisk I have edited the question.Please take a look

Fe, Co and Ce are variable valence metals: it's not clear from "Iron nitrate" whether it is Iron(II) nitrate or iron(III) nitrate. All of them will undergo structural changes upon dilution, and some may hydrolyse. Whether a metal carbonate will or will not precipitate can be evident from the corresponding $K_\mathrm{sp}$ values or solubility charts all of which are freely available.

@andselisk Its Iron(III).My question is what changes will the metal salt solution will undergo after heating at 60 deg cel, and whether there is any characterization technique to characterize the change


text:
Does the Fermi level correspond to the highest occupied energy level?
I know that, in the band structure of a solid, the Fermi level is the energy level at which there is a 50% probability of finding an electron (at thermodynamic equilibrium), and that it doesn't have to correspond to an actual energy level (ie it's in the band gap in an insulator).
But Atkins and Shriver's Inorganic Chemistry doesn't mention this and instead defines it as the highest occupied energy level at 0 K. (page 103, 5th edition)
I was wondering if these definitions are actually equivalent. If so, can you explain how? Given the fact that in an insulator the Fermi level is in the band gap, it seems like at best the definition in Atkins is oversimplified.

It is not a straightforward concept at all, so I am not surprised that an inorganic textbook simplifies it a bit. You might look at https://aapt.scitation.org/doi/10.1119/1.1629090 for the case of an intrinsic semiconductor.

Be aware of Fermi level versus Fermi energy, as it seems to me the book speaks about the latter.
The quotes below are from both Wikipedia links.

In band structure theory, used in solid state physics to analyze the energy levels in a solid, the Fermi level can be considered to be a hypothetical energy level of an electron, such that at thermodynamic equilibrium this energy level would have a 50% probability of being occupied at any given time.

The Fermi energy is a concept in quantum mechanics usually referring to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature. In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy, whereas in a metal, the lowest occupied state is typically taken to mean the bottom of the conduction band.

The Fermi energy is only defined at absolute zero, while the Fermi level is defined for any temperature.

The Fermi energy is an energy difference (usually corresponding to a kinetic energy), whereas the Fermi level is a total energy level including kinetic energy and potential energy.

The Fermi energy can only be defined for non-interacting fermions (where the potential energy or band edge is a static, well defined quantity), whereas the Fermi level remains well defined even in complex interacting systems, at thermodynamic equilibrium.

Since the Fermi level in a metal at absolute zero is the energy of the highest occupied single particle state, then the Fermi energy in a metal is the energy difference between the Fermi level and lowest unoccupied single-particle state, at zero-temperature.

A collection of non interacting fermions, often called the Fermi gas is a simplified model useful in the solid state physics. There is supposed the considered fermions do not interact, so the energies of these fermion quantum states do not depend on the distribution of  these fermion quantum states.
A typical case would be the classical hydrogen atom model, applied on multielectron atoms. As electrons in the same atom interact quite a lot, it is not a good model.
For solid states, electrons and electron vacancies are often taken as  quasi fermions following Fermi-Dirac statistics for the Fermi gas,
as they are separated enough to be considered as non interacting.

Thank you, that does help. The book calls it the Fermi level but from your answer it seem like they are in some sense mixing the two concepts. When do systems with non-interacting fermions actually arise? One would suppose that electrons in a solid are inevitably going to interact with each other.
Just to be fully clear, is there no particular link between Fermi energy and Fermi level (beyond the fact that presumably Fermi worked on both!)?

Just added the paragraph about the relation of the quantities in a metal context.

Added Fermi gas and non interacting fermions note.


text:
Is the following molecule aromatic or antiaromatic?
This molecule has the same chemical formula $\ce{C20H10}$ as corannulene (which also is non-planar) and corannulene is aromatic, so I'm pretty sure that this molecule should be aromatic too.

I tried to analyse it using the MoCubed app on my phone (on my previous posts, I wrote "WebMo" instead of "MoCubed" because I have both on my phone and sometimes get confused) but it keeps adding unnecessary hydrogen atoms and I can't rotate the molecule to remove all of them so I have to ask here.
P.S. I do know Hückel's rules, but they don't apply to pericyclic rings.

Please don't ever use external storage for keeping essential parts of your question, especially cloud providers such as Google One. Upload the image directly instead, this way the question doesn't loose its meaning once you or the cloud provider you have no control over decides to remove the file. Also, I took a liberty to replace the illustration with the one showing optimized geometry (freeware version of MarvinSketch can do that).

Even though the whole idea of using mobile app for computational chemistry sounds weird to me (to say the least), it would be nice if you would go ahead and correct the previous posts with the correct name of the app instead of posting an explanation here.

Tried to upload directly on SE Android app, failed

Don't use Android app, it hasn't been updated in years and lacks numerous essential features, see e.g. my Meta post. Stick with the web browser instead.

@andselisk using mobile app since I don't need too much info on the exact molecule (not synthesising it even in silico, just as a reference since the molecule I'm designing has an essentially similar part in it, only the aromaticity/antiaromaticity criteria is important)

Essentially similar part = same shaped ring w/same no. Of pi electrons

Isn't it antiaromatic? I mean what's the answer...

@zenix Huckel's simple rule does not necessarily apply to pericyclic hydrocarba.

Yeah, you that in question. I asked for the answer, in case there's one given

Just because two things have the same formula does not mean they are equal in aromaticity.  Styrene is C8H8, cyclooctatetraene is C8H8, and only one is aromatic.

This compound can be rather considered bridged dibenzo azulene. If bridges were saturated, case would be pretty clear, but that's reduced variant, which may somewhat break aromaticity in  five-membered rings.

It turns out this question has some subtleties.  The analysis presented below, based on Huckel MO theory, indicates that:
1.
The compound as presented by the OP is not expected to be aromatic and would be anti-aromatic by the Huckel model -- meaning in practice it would probably distort out of full conjugation to avoid this calamity.
2.
The dication, however, does appear to be aromatic.  Part of the reason is that while we ordinarily suppose that five-membered rings must be negatively charged in an aromatic system, a phenanthrene-like accumulation of three such rings can become aromatic with a positive charge instead.

Same formula, but ...
The OP suggests a comparison with the aromatic compound corannulene, which like the compound in question has the formula $\ce{C20H10}$ and twenty conjugated pi electrons.
Let's compare the Huckel energy levels for these two compounds, keeping in mind that an aromatic system will have (1) a stabilized highest occupied energy level and (2) a wide energy gap between this level and unoccupied states (figure made by the author):

Occupying the lowest available orbitals in corannulene on the left (including two degenerate pairs between $\beta$ and zero that are close together, thus appear as thick bars), we see the desired effect for aromaticity: stabilized highest occupied orbitals, wide gap to the unoccupied ones).  Not so on the right with the OP's compound: here there are nine bonding orbitals so putting in twenty pi electrons means having a mixture of unoccupied and occupied states at the nonbonding level where there is a degeneracy.  This is a classic antiaromatic situation (similar, actually, to triangulene which us invoked in another answer), and thus the answer to the original question is the compound we are examining will not be aromatic.  It will form a diradical or  find a way to get out of antiaromaticity instead.
Thrown for a loop
What happened?  In multicyclic systems aromatic coupling is not just a matter of having the right number of electrons.  Hidden in the energy calculations is a topological difference.  In corannulene five of the conjugated atoms are in an interior closed loop surrounded by the rest of the molecule, while the OP's compound lacks such an internal loop.  This topological change can alter aromatic coupling, which is sensitive to how loops of conjugated atoms are connected to each other.  For this particular combination of molecular formula and pi-electron count, corannulene has a good topology for forming an aromatic system.  The OP's compound does not.
Making lemonade out of lemons
Many Huckel-antiaromatic systems become aromatic if one places the proper charge on them, adjusting the number of pi electrons and thus occupied pi orbitals.  The classic case is cyclootatetraene which goes from an antiaromatic/distorted nonaromatic neutral molecule to an aromatic dianion when we combine it with potassium.  What can be done in this direction for the OP's compound?
According to the Huckel energy levels we ought to either fill the degenerate orbitals as in cyclooctatetraene, or empty them.  We see that in this case the latter is what works.  The dication, as opposed to the dianion, gas a stabilized highest orbital and a wide gap up to what are now unoccupied degenerate states.  In other words, the hallmarks of aromaticity.
But where can these positive charges reside?  We can make the seven-membered ring aromatic with one positive charge, but shouldn't the benzene rings be neutral and the five-membered rings negative like the cyclopentadienyl anion?
We crack open our topology textbook again and we discover that three separate rings are not connected in the same way as the accumulated five-membered rings in the OP's molecule/dication.  We'd better check the Huckel orbitals inherent to a cumulated five-membered ring system like the one in the OP's molecule, having the formula $\ce{C11H7}$ (this figure also by the author):

The seemingly similar phenanthrene system is a fourteen-pi-electron aromatic system.  Here, however, with fewer input atomic orbitals we see the widest gap, thus likely the greatest stabilization, occurs with only ten pi electrons.  This means the eleven-carbon system will have a positive overall charge.  The chain of three "cyclopentadienyl" rings generates an aromatic cation and thus can take on the second positive charge in the dicationic, predicted aromatic form of the OP's compound.

I see. Based on your photo, my molecule seems to be antiaromatic in the 0 oxidation state, but aromatic in the +2, +4 (etc) oxidation states.

To get a good aromatic stabilization you don't just need a HIMO-LUMO gap, you need a wide one.  Only +2 and -4 (which I didn't discuss) really seem to be good candidates.

Corannulene (which is rather planar, the angle between adjacent C6 rings is small) is not a good comparison, as it is just a helicene-type C6 polycycle that forms a ring, basically coronene cut one ring short.
A simple molecule with structural similarity is https://en.wikipedia.org/wiki/Azulene
 and it is indeed aromatic. However I guess the two additional five rings deform the whole thing quite a lot. And they might as well break the aromaticity anyway:
It seems to me there is no single double bond you could leave out of the aromatic system to fix it. That reminds me of triangulene https://en.wikipedia.org/wiki/Triangulene

So my verdict would be no, not (a single) aromatic (system).

I agree.  I've done some calculations based on Huckel theory which indicate the molecule in the question will lack the stabilization present in corannulene.  Hopefully I can find time to formulate an answer, which will be nontrivial.

@OscarLanzi That´d surely be an improvement on my attempt at educated guessing above. ;)

Check out what I came up with.


text:
Is the thermodynamic stability against reduction/oxidation of an electrolyte dependent on the electrode material?
Consider an electrochemical cell with an inert Pt working electrode (WE) and some electrolyte that irreversibly decomposes upon oxidation. A minimum potential $E_{min}$ (vs. some reference) needs to be applied to the WE to allow oxidation to occur (but most likely a higher value is required to actually measure a current). I would call this potential the thermodynamic stability potential limit (SPL).
Does this minimum potential change when exchanging the WE material with another inert material, e.g. glassy carbon? I think it should not, as all that matters is the electrochemical potential/fermi level in the inert electrode.
Could some type of specific adsorption of electrolyte components change (lower) the thermodynamic SPL? To me this sounds like catalysis, which just influences kinetics, not thermodynamics.
What about adding some other substance? Could its reaction products cause degradation of the electrolyte components in some way?

Technically you are correct. The thermodynamic potential of electrolyte decomposition should not be affected by the electrode composition.

which just influences kinetics, not thermodynamics.

The word "just" here is doing way more work than you're giving it credit for. There are situation where the kinetics of electrolyte decomposition are so slow that there is effectively no reaction. The stability window of electrolytes commonly used in Lithium-ion batteries, for example, is smaller than the potentials needed to charge the materials. However, once the reaction begins, initial decomposition forms an inter-phasic layer on the anode known as the solide-electrolyte-interphase (SEI). The kinetics of decomposition on this SEI are slow, which allows the battery to be operated at high potentials[1]. In fact, damage to the SEI has been linked to capacity loss in Li-Ion batteries over extended cycling[2].


text:
How are cells rechargable?
I have learnt that you can test if a reaction is thermodynamically feasible by testing if the cell voltage $E_\mathrm{cell}$ is greater than $0$. Surely, if a reaction is feasible, then the $E_\mathrm{cell}$ of the reverse reaction is negative, and so not thermodynamically feasible; in which case, how can a cell be recharged? Taking the following example:
$$
\begin{align}
\ce{Li+ + CoO2 + e- &<=> LiCoO2} &\quad &\pu{+0.60 V}\tag{1a}\\
\ce{Li+ + e- &<=> Li} &\quad &\pu{-3.00 V}\tag{1b}\\
\end{align}
$$
$$
\begin{align}
\text{In use:} &\quad &\ce{CoO2 + Li &<=> LiCoO2} &&\tag{2a}\\
\text{Charging:} &\quad &\ce{LiCoO2 &<=> CoO2 + Li} &\quad \mathrm{EMF}&= \pu{+3.60 V}\tag{2b}\\
\end{align}
$$
Is it the case that putting an EMF against the EMF makes the reaction feasible? In this case, could any reaction be made thermodynamically feasible by putting a potential difference across it?

I guess you didn't read about electrolysis yet? You should check it out...

Firstly, I would replace the word "feasible" with "favorable". I'm also going to replace $E_{cell}$ with $E_{OCV}$, where the open circuit voltage (OCV) is the potential of the cell without any applied electric potential. So you're right in that if the $E_{OCV}$ is positive, the net reaction is thermodynamically favorable: it will occur spontaneously if allowed to and will produce energy that can be used by the load connected to the battery (i.e. the cell can do work).
You're correct in saying that the reverse reaction is thermodynamically unfavorable, but that doesn't mean the reaction won't happen, just that you need to do work on the battery in order for it to happen. In most cases, this means applying a bias potential ($E_{cell}$) that is more positive than the $E_{OCV}$. When that happens, the current flowing across this bias potential takes energy into the cell and recharges it.
However, not every reaction will proceed in this way. In many cases, if a high potential is applied, another reaction will occur. If you have water in your battery and apply a high potential, you may electrolyze water, splitting into elemental hydrogen and oxygen. This will then reduce the usable capacity and/or cause rupture of the battery.
Lastly, all of this ignores the role of kinetics. Just because a reaction is thermodynamically favorable doesn't mean that it happens at a rate that useful. If you have a cell with $E_{OCV} = 3.60V$, applying $3.601V$ will begin to charge the cell but at such a slow rate than you'll never notice. You generally need to apply at least a few tens of millivolts of over-potential to a lithium ion battery in order to get any appreciable current. So in the case on non-rechargeable (chemically non-reversible) batteries, the over potential needed to reverse the reaction will be big enough that you will instead hit the potential for some other, undesirable reaction.


text:
How to calculate the energy to dissociate a bond into neutral atoms?
I am self studying chemistry through MiT ocw 5.111 . On practice exam 2 problem 2 there is a question which states the following
    element           ionization energy          electron   affinity
    Potassium(K)      418 kJ/mol                 48 kJ/mol
    Fluorine(F)       1680 kJ/mol                328 kJ/mol
    Chlorine(Cl)      1255 kJ/mol                349 kJ/mol

(a) (12 points)   The ionic bond  length for KF is 0.217 nm. Calculate
  the energy (in units  of kJ/mol required  to  dissociate  a single
  molecule of KF into the neutral atoms K and F, using information
  provided above. For this calculation, assume that the potassium and
  fluorine ions are point   charges.

I proceeded to calculate the bond dissociation energy using the following formula $U(r)=\frac{z_1z_2e^2}{4\pi\varepsilon_0r}$  from which I obtained a dissociation energy of $-640$ kJ/mol. Then left with two ions F$^{-1}$ and K$^{+1}$.
 I went on to determine their ionization energies and electron affinities respectively.
To make F$^{-1}\to$ F + $e^{-}$ an electron must be lost thus warranting an ionization energy of $-1680$ kJ/mol and. To make K$^{+1}$+ $e^{-} \to$ K and electron must be gained thus referring to the election affinity of K of $18$ kJ/mol. When equated as (Energy Required - Energy Released) I arrive at
 (Dissociation Energy + Ionization Energy) - Electron Affinity = (-640 kJ/ mol - 1680 kJ/mol) + 48 kJ/mol = -2272 kJ/mol required

However I see in the answer key this is incorrect could someone guide me as to where I am going wrong here thank you.

Your $\pu{640 kJ mol^-1}$ value is correct, so the next step is to neutralize the ions. First add an electron to a potassium ion to get a potassium atom. This releases $\pu{418 kJ mol^-1}$ because it is the reverse of the potassium atom ionization. Then take away an electron from fluoride ion to get a fluorine atom. This requires an input of $\pu{328 kJ mol^-1}$ because it is the reverse of the fluorine electron affinity. So the result is $\pu{550 kJ mol^-1}$, i.e., $\pu{640 kJ mol^-1}$ - $\pu{418 kJ mol^-1}$ + $\pu{328 kJ mol^-1}$.
This figure makes it clear:

Figure copyright information: D.W. Oxtoby, H.P. Gillis, L.J. Butler, Principles of Modern Chemistry, 8th Ed., Cengage Learning, Section 3.8, p. 87, © 2016 Cengage Learning.


text:
Human gastric lipase
Does anyone know if the human gastic lipase has a quaternary structure? If not, why not? I can not seem to find this information anywhere.

I'm voting to close this question as off-topic because it might be better asked in a biochemistry group.

I think the question should be reopened because the OP addressed a real gap in the available information on this protein. I have since edited the images on Wikipedia and Proteopedia - we will see if these edits stick.

There is a crystal structure of human gastric lipase (PDB ID 1hlg). The accompanying paper does not mention any quaternary structure and shows the protein as single chain. Automated analysis of the coordinates support that. The enzyme is glycosylated, so the molar mass does not match the one expected from the sequence. The apparent molar mass determined experimentally is about 50,000 g/mol, corresponding to a monomer.
However, the annotation in the structure (probably erroneously) say that the biological assembly consists of the two chains in the asymmetric unit. Also, there is a report that the size of the molecule changes with pH (90,000 g/mol at pH 6 vs 45000 g/mol at pH 3).
The same group that solved the human lipase structure later solved the dog lipase structure (in an open conformation with a bound ligand) where the annotation in the PDB deposition says it is a monomer. I doubt that there are interspecies differences between dog and human in terms of quaternary structure, so that would be another piece of evidence supporting that the human enzyme is a monomer in its native form.

@Leigh Keep in mind you can also accept answers that were useful to you.


text:
Effect of drinking on internal energy and entropy of the universe
I have just been doing some questions on thermodynamics and have come across a strange question that I can't seem to grasp.
The question asks me to estimate the change in body temperature as well as the entropy change of the universe when a man drinks a litre of 273 K cold beer in an isolated pub. I am given the mass of the man, as well as the specific heat capacity of water, and its density. I also am told to assume that the human body is mainly made up of water, and am given body temperature as 310 K.
I've been thinking that the answer must involve the fact that the pub is isolated and thus that the change in internal energy of the system is constant but I'm not quite sure how to go about it. I have also derived earlier in the question that for a system at constant volume:
$$
\begin{align}
U(T_2) &= U(T_1) + \int_{T_1}^{T_2}C_V(T)\,\mathrm dT,\tag{1}\\
S(T_2) &= S(T_1) + \int_{T_1}^{T_2}\frac{C_V(T)}{T}\,\mathrm dT.\tag{2}
\end{align}
$$
Any hints or help would be much appreciated!

I took a liberty to replace a formulas screenshot with searchable text. Feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, I tried to narrow down the title and removed irrelevant "question about" part; however, since it looks somewhat cheesy, I'd be grateful if someone would go ahead and improve the title:)

I think they are asking you to evaluate the equilibration between a liter of beer at 273 K and a human body at 310 K (focusing exclusively on the thermal equilibration, and neglecting anything else that may be happening)).


text:
Is there a temperature where salt doesn't dissolve in water?
I know that salt dissolving in water is driven mostly by entropy. However, at lower temperatures, entropy is less of a factor in controlling reactions. Is there a temperature where salt doesn't dissolve in water?

Negative-25°C maybe?

Nope! Water freezes before.

That's what I mean.  Then the salt doesn't dissolve :-) .

Solubility is an equilibrium phenomenon, which means that its temperature dependence is a function only of the enthalpy change of the reaction, not the entropy. See the van't Hoff equation for the mathematical relationship.
If by salt you mean sodium chloride, the dissolution in that case is only very slightly endothermic, so the solubility decreases very slowly with temperature.
That means that if you go low enough in temperature, you will reach a point where the solubility theoretically approaches zero. However, the water will freeze long before the temperature gets that low, and solubility will no longer have meaning.


text:
Do all single-electron species have the same orbital energies?
This picture from Ebbing's General Chemistry [1, p. 234] shows the electron orbital energies for a hydrogen atom, where each subshell belonging to the same principal quantum number ($n = 1, 2, \cdots$) are on the same level. Is this true for any particle with only one electron? Would, $\ce{Li+}$, for example, have the same orbital energies?

References

Ebbing, D. D.; Gammon, S. D. General Chemistry, 11th ed.; Cengage Learning: Boston, MA, USA, 2017. ISBN 978-1-305-85914-2.

Please always cite the source the illustration has been taken from.

Since every distinct elementary nucleus has a different charge, why would you expect a single electron system based on that nucleus to have the sameenergy as a hydrogen atom which has only a single positive charge?

@matt_black you're right. They will have different values of energy but the individual subshells will have same energy for a particular value of n for they are degenerate.

It would have to be $\ce{Li^2+}$ or $\ce{He+}$ to have 1 electron only.
And even for 1 electron only ions, energy levels depend on the kernel charge.
For multielectron atoms/ions, energies of different orbitals with the same quantum number $n$ differ, due electron repulsion and kernel shielding. That is the reason why the orbital 4s is occupied before orbitals 3d.
The orbital energies are additionally dependent on the particular electron quantum state distribution.
This affects empirical orbital filling rules, causing exceptions,  and orbital involvement in atom bonding.
So even if the orbital 4s has lower energy than 3d while filling, it can have higher energy when they are filled. Their  relative energy for neutrs atoms also change with the atomic number.


text:
VOC conversion question
I'm trying to find out if a particular paint is considered low VOC. The only piece of data I have available is that the VOC content is $\pu{10 \mu mol / m^3}$. What does that equal in $\pu{g/L}$ of paint?

I just edited, check if actual meaning of your question is preserved..

VOC stands for volatile organic compounds, in this context as mass concentration of vapours in the air.
These 2 mentioned quantities ( concentration in air and the paint ) are very indirectly related via volatility of paint components.
So the same VOC value may belong to huge range of g/L in the paint.


text:
Need help understanding Enthalpy
My chemistry textbook says that:

Enthalpy (Internal Energy) = chemical potential energy + thermal energy

Now, I know that Enthalpy and Internal Energy are different so I really don't understand what my text book is saying.
Also, from searching I understood that $\mathrm{U = q + w}$
since, $\mathrm{w = -p \Delta V}$
And $\mathrm{H = U + \Delta(PV)}$
So for standard pressure :
$\mathrm{\Delta H = q}$ , my text book has stated that

At standard temperature, the difference in enthalpy is equal to the difference in chemical potential energy.

Also I'm not sure I fully understand what it means by potential chemical energy, in general I'm just "super confused". My text book even says that

By potential chemical energy we mean the intermoleculer forces and etc. that exist between atoms and molecules.

But from what I read online, these forces are because of the Internal energy of a system not the chemical potential energy. So if somebody could explain these thing and relate them to my text book somehow I would be grateful :

What is Chemical Potential Energy?

What is Internal Energy?

What is enthalpy? And how do I relate these varying definitions for it to each other?

Note
$\mathrm{U, q, w, H}$ have usual meanings.

It would be helpful if you could provide the exact reference (title, author(s), edition, and publication date), and also provide a picture of the paragraph(s) containing the quotes.  Sometimes textbooks are misquoted, and sometimes the quotes are taken out of context.

It wouldn't be of any help, English is technically my third language and my text book are in another language.

Understood.  It would be hard for me to understand what the textbook is trying to say without actually reading the relevant paragraphs.  For instance, enthalpy and internal energy are not the same but, under certain constraints ($\Delta PV = 0$), $\Delta H$ can have the same value as $\Delta U$.  So I don't know if your textbook is trying to say that U and H are the same (which they're not), or if it is trying to say that, under certain constraints, their changes can be equal (which they can).  Is this a well-established textbook in your country?  And what does your professor say about this?

My text book hasn't mentioned any constraints and has boldly stated said sentence. This is a high school text book and it's the only official text book in the country, but it's famous for being quite terrible compared to western text books but since the college entrance exam is based solely on this book we have no choice but to memorize its nonsense but what I'm trying to do is understand Enthalpy on a deeper level but since it seem my text book is out right flawed in this sence I may have to just accept what it's saying.

Is chemical potential energy a part of Internal energy? If so why do we represent it as U = q + w. What part of this formula accounts for the intermoleculer forces? Also is thermal energy just Q?

I'm curious who is the author of your textbook, and in what country it's used. But I understand if you'd rather not say.  Of course, this is a high school text, so one doesn't expect complete rigor, but it sounds like the problems you're describing are serious.  Maybe you should bring it to the attention of your country's educational board.  And are there any college-level thermo textbooks in your native language? Also, thermal energy is not q. q is the heat flow. Thermal energy is a property of the system. q and w are properties of a process-- of the path through which a system changes state.

The internal energy is defined in (classical) thermodynamics for a finite change as $\Delta U=q+w$ where $U$ is the energy contained in the 'system' called the internal energy, and $q$ is the heat absorbed by the system and $w$ the work done on the system. Often the word 'system' means some ideal gas (often imagined to be in a cylinder with a frictionless piston) that cannot exchange matter with its surroundings but can exchange heat and do work, usually this work is expansion/contraction or $pV$ work although it could also be electrochemical or gravitational work, i.e. used to lift a weight but this is not usually relevant in chemistry. A system that cannot exchange matter but can exchange energy is called 'closed'.
Notice that we do not need to know what the nature the internal energy is to use thermodynamics, i.e. the nature of atoms and molecules is not relevant as thermodynamics was developed before this was understood, and this abstractness is the great strength of thermodynamics, but can make it more difficult to understand. Nevertheless, we do want to understand what the internal energy is.
First, the internal  comprises the kinetic energy of the atoms or molecules, and so defines the thermal energy so an increase in kinetic energy is observed as an increase in temperature. Secondly molecules can rotate in three dimensions thus have rotational kinetic energy and its atoms are bound together by bonds that vibrate thus adding potential and kinetic energy to the internal energy.
The enthalpy is used, as it were, to make a 'level playing field' when comparing experiments because it accounts for the work done $H=U+pV$  by a reaction so the first statement you give is not correct. The work done is small, however,  for liquids and solids as $\Delta V$ is small so that $H \approx U$. (The Gibbs energy $G=U+pV-TS=H-TS$ is similarly used to account for changes in entropy and work done).

(1)  The internal energy chg. is not generally $\Delta U = q + w$.  That only applies to a closed system.  (2)  You'll want to amend your language, changing "isolated" to "closed". What you're describing is a closed system (no exchange of matter with the surroundings) rather than an isolated system (isolated from the surroundings, and thus can exchange neither matter nor energy). I.e., q and w are both necessarily zero for an isolated system. Also, while ideal gases are often used pedagogically, in the actual practice of thermodynamics, the system is typically a real-world reacting system.

Still isolated means no matter no energy exchanges

Thank you for these comments I have corrected my answer.

To understand internal energy*, I prefer to start with this:
Consider a thermodynamic system.  There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy):  flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy).  If the system is closed (no matter can flow in or out), we are left with only the first two, which naturally leads to the following equation:
$$\Delta U= q  + w$$
Now, suppose you are carrying out a reaction under the common condition of constant pressure.  The amount of heat flow associated with this reaction is an important and measurable quantity (measurable using a calorimeter).  So it would be very nice if there were a state function (i.e., one that is independent of path) that happens to give us the heat flow under those conditions.  It turns out that we can construct such a function by adding $pV$ to $U$.  We call that function the enthalpy, $H$:
$$H = U + pV =>  \Delta H = \Delta U + \Delta(pV) =  \Delta U + p \Delta V+ V \Delta p $$
At constant pressure, $\Delta p = 0$, so:
$$\Delta H_p = \Delta U + p \Delta V = q_p + w_p + p\Delta V$$
But at constant $p$,  $p_{sys} = p_{ext}$.  Thus, if we only have $pV$ work:
$$w_p = -p_{ext} \Delta V = - p \Delta V$$
Hence:
$$\Delta H_p = q_p- p \Delta V+ p\Delta V = q_p$$
Let's compare a reaction at constant pressure with one at constant volume.  At constant $V$, with $pV$-work only, $q_V = \Delta U$ (because, with $pV$-work only, $w_V = 0$).  Suppose the reaction is exothermic, and the volume of the products is greater than that of the reactants, i.e., $\Delta V >0$. You can see from the equations that $\Delta H$ will be less negative than $\Delta U$, i.e., less heat is evolved at constant pressure.  We can understand this physically from the fact that, at constant pressure, some of the thermal energy the system evolves goes into the work the system does on the surroundings (because the system expands).  Thus less thermal energy is left over to flow into the surroundings.  I.e., $\Delta H$ is just $\Delta U$ with a "correction factor" to account for $pV$ work.
The other reason that constructing $H$ is useful is that it is an intermediate step in constructing $G$, which is an extremely important function in thermodynamics, as it allows us to determine the direction of spontaneous change (and the equilibrium  state) at constant $T$ and $p$. [For more on this, see my answer here: What is wrong in this argument that dG must always be zero? ]
Finally, a quick (and somewhat simplified) answer on potential vs. kinetic energy (it really merits a separate question): Kinetic energy is the energy bodies have as a result of their motion. This includes translational, vibrational, and rotational motion. Potential energy is the energy bodies have as a result of their position in a potential field.   For example: Two atoms in a strong bond have a lower potential energy than they would when they are separated.  Conversely, if you have particles that repel each other, you increase their potential energy when you compress the system and thus force them to be closer.
*Copying from one of my other answers:
The term "internal" refers to energy internal to the system.  For example, the internal kinetic energy is the energy the system has as a consequence of the thermal motion of its atoms and molecules, as contrasted with the external kinetic energy the system might have as a result of the motion of the system as a whole through space.

The internal energy U is the sum of all energies stored in the chemical bonds of the examined sample of matter. It is a potential energy.
The enthalpy H is the same, but with a correction due to the presence of the atmosphere. The difference between H and U is similar to the difference between weight and apparent weight. Let me go back to this measurement.
If you want to know the mass (or the weight) of an object, you put it an a balance and you read its weight. But this value is not the real mass (or weight) of the object, it is only the apparent weight. If you had done all this job in water, with a balance under water, you would also have obtained a lower value than in air. You would have obtained the apparent weight, namely the weight of the object minus the weight of the same volume of water. This correction is called Archimede's push. In air the correction is about 1000 times smaller than in water, but is is not zero. So usually we ignore this Archimede's correction in air, and we accept the weight in air as if it has been measured in the vacuum.
It is the same for U and H. U is the sum of all energies stored in the chemical bonds of the object, but only in the vacuum. As we are working surrounded by an atmosphere, the object needs to repell this atmosphere simply to exist and to put its atoms and molecules where they are. This work costs a little bit. It costs much more if the object is a gas. If you produce by a chemical reaction a couple of liters of gas, you have to repell air simply to allow your gas to be where it is, and this costs some work, not much, but this work is not zero. That is why H is U plus a small correction, equal to PV. And when you heat an object in our lab, the heat is equal to Delta H and not Delta U.
OK ?

The internal energy is not the potential energy.  The internal energy is the sum of the internal potential and kinetic energy.  For instance, we will increase the internal energy of an ideal gas by heating it, yet an ideal gas has no internal potential energy; it only has internal kinetic energy.  I.e., for an ideal gas, all of its internal energy is kinetic energy.

As theorist said. According to this answer a monoatomic ideal gas would have zero U instead of its kinetic energy. The latter is.

I have never said that the internal energy is due to potential energies.I have always said that it is the sum of all energies of the atoms and molecules. This includes kinetic energies.

"The internal energy U is the sum of all energies stored in the chemical bonds of the examined sample of matter. It is a potential energy.". It is a potential stored in the system sure. It that you're right. But your opening does not says so. There is no space for K in the above. You could amend it as for the rest of the answer is OK and might be useful to starters ( especially the comparison to hydrostatic correction is a valid didactical trick).


text:
What happens and what do we observe during this qualitative analysis of bronze?
What is the observation when we add $\ce{HNO3}$ to bronze followed by addition of dilute HCl and bubbling of $\ce{H2S}$ gas through it.
Please tell the inferences as well.

There are many sorts of bronze. The most widely known is an alloy copper $\ce{Cu}$ and tin $\ce{Sn}$. When dissolved in nitric acid $\ce{HNO3}$, copper is oxidized into a blue solution containing $\ce{Cu^2+}$ ion. Metallic tin $\ce{Sn}$ is not dissolved. It is transformed into an insoluble and colorless oxide $\ce{SnO2}$ which often makes a gel difficult to filtrate. So the gel has often a blue color after filtration, but this is due to the presence of $\ce{Cu^2+}$ ion in the gel. If now you pass $\ce{H2S}$ in the blue $\ce{Cu^2+}$ solution, it will precipitate and make a black and heavy precipitate $\ce{CuS}$.


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Relating Gibbs Free Energy to Equilibrium Constants
I am having some trouble relating these two concepts. I know that if K=1 then delta g is zero, and that we can interpret signs and magnitudes. However, i am having trouble determining required temparature based on given aimed K, and standart Gibbs free energy of a reaction.
Here is the kind of problem that i have trouble with:

Have you considered using the Van't Hoff equation?

I'm voting to close this question as off-topic because you did ask anything, you just posted an exercise. What have you tried, and what do you know about how the Gibbs energy or equilibrium constants change with temperature (the two quantities, btw., are temperature-dependent in distinct ways).

To solve this problem, you suppose that $\Delta$$H°$  and $\Delta$$S°$ do not change with temperature. You calculate $\Delta$$G°$ by two approaches : $1)$ - $RT$ ln $K_p$, and $2)$ $\Delta$$H°$ - $T\Delta$$S°$. Then $T$ can be easily obtained.


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How can a DPPH assay to measure antioxidant activity of beta carotene be conducted?
I am currently doing some research regarding beta-carotene. With reference to other sources and their methodoloy for DPPH assay, I conducted my experiment but failed to find any significant peak in absorbance at 517nm using a UV-spectro.
My sample of DPPH is already dissolved to 0.1mM in methanol.
My methodology was mixing-
0.3cm3 DPPH
0.3cm3 beta-carotene (dissolved in propanone)
2.4cm3 methanol
I increased the dissolved betacarotene in increments of 0.3cm3 ( upto 1.5cm3) while simultaneously decreasing methanol also by 0.3cm3 (upto 1.2 cm3) for a total of 5 readings.
The mixture was shaken and left to sit (in dark conditions) for 30mins before readings were taken using UV-spectrometer. The UV-spectrometer was calibrated with DPPH.
While I achieved the expected colour change from purple to yellow, I did not get a significant absorbance at any reading at 517nm even when using pure beta carotene.
I suspect this may be due to the betacarotene being dissolved in propanone which perhaps may be the wrong solvent, or perhaps i am calibrating the UV-spectro with the wrong substance, however I would like to check if this procedure is incorrect.
Please advice accordingly. Much appreciated.

After taking a glance at https://en.wikipedia.org/wiki/DPPH it seems like you would expect a purple DPPH solution that decolorizes when you add antioxidants or radical scavengers. Maybe try reacting with vitamin C as a positive control.

@Karsten Theis: I think it is actually yellowish at the end (decolorizing is acceptable). :-)

What do you mean by you calibrated UV-spectrometer by DPPH? Is it your control or blank? Also, absorbance of $\beta$-carotene may be interfering too.

I used it as blank.

I used to run this assay and I think 30 min is a lot longer than I let it sit for. Maybe try 5 minutes? Or else leave in the spectrometer and measure the change in the spectrum over time?


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Will dissolving pure linseed oil in ethanol decrease its drying time?
I've looked at some material on the internet and it says that pure linseed oil can take months to dry completely unless a solvent is added, which decreases that time to a few days. I'm wary of industrial solvents due to their toxicity and dangerous fumes, so I was wondering if specifically pure ethanol would be a reasonable and non-toxic alternative to industrial solvents used to speed up the drying time of pure linseed oil.

What do you mean with drying? What are you applying this to? If you dissolve linseed oil in ethanol and let it sit, the ethanol will evaporate (sooner or later) and you still have the remaining linseed oil.

Boiled linseed oil contains mineral spirits to help the linseed oil evaporate. I'm just wondering if ethanol could be used instead of mineral spirits. I'm applying the linseed oil to cloth to waterproof it.

With "dry", you mean solidify, like a resin. The oil does not evaporate. The solvent acts as a plasticiser, until it is finally also evaporated completely. I´ll guess ethanol is gone too quickly. Isopropanol has a slightly higher Bp. If you find nobody who has tried it, you will have do it yourself.

Warning: Cloth with linseed oil, if not stored well ventilated (hanging, not in a heap!) warms up as long as it is not dry and can self-ignite. But it should rather take a week, not months.

Strictly speaking linseed oil doesn't "dry" it slowly polymerises. Since the resulting resin is what you want as an end product (since it protects the surface) you have to allow it to happen. Solvents don't much help speed polymerisation up, except to enhance the penetration into the surface.

@matt_black A plasticiser keeps the resin mobile while polymerising, allowing the reaction to keep going to reach a higher conversion in a given amount of time.

Karl has the most scientifically accurate description.  Linseed oil does not dry, it cures.  Drying is an evaporative action where material is lost, linseed oil cures by taking on oxygen atoms to cross-link and cure (pure linseed oil actually gains 17% WT in oxygen as it "dries" making pure linseed oil 100% solids).  Use a polymerized linseed oil (I make Tried & True) to achieve faster curing - raw linseed oil takes 3-4 weeks to cure, our polymerized linseed oil (danish oil) takes about a day with out any of the toxic drying aids.
Raw linseed oil is a non-polar liquid, alcohols are mostly polar, especially the shorter carbon chains like methanol and ethanol.  This makes them very poorly miscible with most oils.  Once you hit a carbon chain of at least 3+ (propyl, isopropyl, butyl alcohol) then the miscibility of linseed in these longer carbon chains increases as the polarity is diminished but you are still better off using a polar solvent, which has 100% miscibility.


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What is the yellow compound formed in electrolysis of brine?
Last week, I tried to conduct electrolysis of brine in my house using copper electrodes and 6V battery. After sometime, the entire solution turned orange/yellow. The yellow substance formed a kind of cloud at first and later it turned to an orange colloid. Does anyone know what the yellow coloured substance might be?

Given distilled water was used, it's probably copper(I) hydroxide from the electrode. Try to heat it up and see what happens. Yellow-orange $\ce{CuOH}$ should decompose to dark-red $\ce{Cu2O}.$

I think you're right because what I had noticed is that the cathode was getting shorter and used up. Anode was unaffected.

Could be as a result of Na metal in the cathode and the anode is the one used up.


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Which one is the strongest base among the following compounds?

I am confused between option (2) and (4). Both lone pairs of nitrogen are localized, so next, I looked at the hybridization of $\ce{N}$ in each compound. As in (2), the $\ce{N}$ is $\mathrm{sp^2}$ hybridized and therefore is more electronegative, I assumed it would be a weaker base compared to (4).
But the answer is (2). I drew the protonated structures and found that there is resonance in the compound (2).
I was taught that electronegativity is more important in questions like this. Is this a wrong concept? Any help is appreciated!

https://chemistry.stackexchange.com/questions/96513/basicity-comparison-of-ammonia-derivatives-and-guanidine

You are comparing nitrogen with nitrogen; how can electronegativity be an issue?

@Martin-マーチン isn't sp2 nitrogen more electronegative than sp3 nitrogen?

This is also closely related to: https://chemistry.stackexchange.com/q/125449/61578. That link has an answer why guanidine is stronger base.

No that is a myth propagated by some organic chemists. Electronegativity is a property of an element, nothing more, nothing less. Hybridisation is also just a mathematical concept, it doesn't change what the atom or molecule is.

@Martin-マーチン I didn't know that, thank you for the information.

@laksheya: which book is this question from? I'm interested in these questions.

@RahulVerma M.S. Chouhan elementary problems in organic chemistry for JEE. I'm using this for NEET but if you're preparing for JEE, my teachers recommend using the advanced version.

A good way to compare basicity of basic compounds is to draw their corresponding conjugate acids and compare their stability. As you might figured out from drawing resonance forms of compound (2)'s conjugate acid (its protonated form ), the positive charge is going to be shared by the 3 nitrogens in the compound. On the other hand, in compound (4)'s conjugate acid (protonated form) the positive charge is on a single nitrogen. Of course, this gives compound (2)'s conjugate acid higher stability.
Your reasoning about the nitrogen being sp2 makes it less basic in comparison with sp3 nitrogen is correct,however; it only holds true if you were comparing two nitrogens with the only significant difference being the hybridization state, but in some cases, resonance will stabilize or destabilize the conjugate conjugate acid of the base, this is usually have greater impact than hybridization with regard to ranking baisicity of the compounds as in the question presented.
Electronegativity is an important factor in ranking acids/bases,but as I said, it should be used with compounds where the only significant different in the structures is about it (for example, 4 compounds with same structure but only difference is Cl, I, Br, OH at a specific position), in other cases, other factors such as resonance stabilization and aromaticity should be also considered alongside electronegativity.

See the stability of conjugate acid is very important. In case of secondary amine the basicity is appreciable but in guanidine system the conjugate acid is remarkably stable due to stabilizing conjugation that is occurring there.


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Why when we convert from supersaturated to saturated solution we must use the crystal of the same kind of solute?
Why when we convert from supersaturated to saturated solution we must use the crystal of the same kind of solute ?

Could you please add an example or elaborate more? I'm not sure I understand the question.

Says who?$\quad$

Basically crystallization is a problem of kinetics vs. thermodynamics. Basically a supersaturated solution is thermodynamically unstable. However due to kinetics it might take a nucleation center a very long time to form.

Sodium acetate melts at 78°C, If you heat it to 80°C with one or two drops of water, you obtain a saturated solution of sodium acetate. You may cool it down, and you obtain a supersaturated solution at room temperature. The solution should crystallize, but it does not, because no molecule feels more inclined than the next one to start the first crystal. For the crystallization to occur, a small crystal of sodium acetate has to be dropped in the supersaturate solution. It act as a bait. As soon as this crystal touches the supersaturated solution, all the dissolved molecules rearranged themselves around this model and the whole solution is solidified in a couple of seconds. If the small crystal has a crystal structure which is very different from sodium acetate, no crystallization occurs. If the crystal structure is not very different from crystallized sodium acetate, the crystallization starts. Sometimes a dust grain is sufficient to start the crystallization.

The question is,why, if we supersaturate solutions of sodium thiosulphate, sodium sulphate or sodium acetate, why we have to use the crystal of respective salt to start the rapid crystallization ?
We need not to. It is just the most natural and the most reliable way to set the crystallization centre.
Similar effect is achieved in sealed  pocked heaters by bending a chess from some metal alloy, forming crystallization centers at metal lattice irregularities.
Another way is strong shaking or involving other solid particles.


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How can sulfur dioxide and nitric oxide be filtered out from a waste incinerator's smokestacks?
How can sulphur dioxide and nitric oxide be filtered out from a waste incinerator's smokestacks?
I also want to know what it looks like after it is filtered out.

Google is your friend! Search "sulfur dioxide scrubber" and pages like this come up:  https://www.envitechinc.com/sulfur-dioxide-scrubber  or   https://www.yokogawa.com/uk/library/resources/application-notes/sulfur-dioxide-scrubber-ph-control/


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Why can't non terminal carbon in a carbon chain does not form ionic bond?
Why can't non terminal carbons can form like:
                 _---na+
Ch3-ch2-ch-ch3

Nonterminal carbons actually can form ionic bonds or at least bonds with strong ionic character, the latter case being much more common with any sort of carbon.  With magnesium at the other end of the bond, for example, the isopropyl Grignard reagent $\ce{CH3CH(MgCl)CH3}$ is commercially available.


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Kinetics of converting a water + heavy water mixture to HDO
Imagine an initial 50%-50% mixture of normal ($\rm H_2 \rm O$) and heavy water ($\rm D_2 \rm O$).
I think, simply mixing them, the result will exponentially converge to an equilibric state, where about a half of the solution is $\rm{HDO}$.
What is the time constant with what the system is going to the equilibric state?
Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Chemistry SE policy, in contrary to e.g. Physics SE, is to avoid MathJax/LaTeX in titles unless absolutely necessary, for indexing/searching reasons.

I'm pretty sure that proton/deuteron transfer is so fast that the limiting step here is diffusion/mixing. That is, what you're looking for is the time it takes for the protons or deuterons to be randomly distributed throughout the volume. That will be a function of the mixing process used, so it will vary with conditions.

This may help with the kinetics:  https://en.wikipedia.org/wiki/Self-ionization_of_water#Mechanism

As @Poutnik states, it depends on ionization. At room temperature, only ~1 in 10,000,000 molecules are ionized, so the chances of an OD- encountering an H+, or of an OH- meeting up with a D+, must be calculated, and the number of collisions/second, as well. (Actually, the species shown are oversimplified, but you get the drift.)

@DrMoishePippik I think it is not enough, because the cross section of the isotope switch is not known. Actually, the cross section could be calculated by measuring this reaction speed experimentally. But maybe there is some thermodynamical simplification, i.e. calculating probabilities from ionization energies.

Important is also https://en.wikipedia.org/wiki/Grotthuss_mechanism for the proton  interchange. Mobility of H3O+ and OH- gives a hint it must be fast.

https://chemistry.stackexchange.com/questions/15332/normal-and-heavy-water-mixing/15334?r=SearchResults#15334

@NilayGhosh Thanks - it confirms the assumption of the question. But it does not say anything about the reaction speed.

@Poutnik I now incorporated your comment about the Grotthuss mechanism into my answer. I did not see it earlier.

@Karnsten Theis Sure.

[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast.

If you compare the diffusion coefficient of hydroxide ($\pu{5.270e9 m^2/s}$) to that of fluoride ($\pu{1.460e9 m^2/s}$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.
$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$
With one deuterated water, it leads to formation of the desired product:
$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$
If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.
This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.
So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.
The analogous mechanism would work for $\ce{H3O+}$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.

The key word is "mixing". If you take a glass cylinder and fill it halfway with D2O (the heavier water, so put it on the bottom) and fill the cylinder ever so gently with H2O, the mixing will be determined by diffusion (NO mechanical mixing). But the experimental techniques required to analyze and determine the rates would be expensive and time-consuming.
Now if in a practical situation you need to make HOD, you could fill a beaker halfway with one and the rest of the way with the other (doesn't matter which way), and stir with a glass rod or a high speed mixer - or shake the mix in a bottle. The time to make HOD will be as fast as you can mix by shaking or stirring.
However, the idea of watching a diffusion experiment sounds like a fun thing. A cylinder could be filled halfway with dyed water, then covered with pure water, and the color equalization could be observed. It should probably be done with pure water on the bottom and dyed water on top, too, for comparison. And probably with more than one dye (in separate experiments) to see if the dye is important in the diffusion rate, or whether the dye is just carried by larger bunches of water.
The exchange of protons in amines or between H2O and other hydroxylic compounds has been found to be measurably slow (e.g., https://pubs.acs.org/doi/abs/10.1021/ja01652a030, which dates to 1953). A longer article in Wikipedia (https://en.wikipedia.org/wiki/Hydrogen–deuterium_exchange) discusses proteins and other compounds that are separated from the H2O/D2O environment and analyzed separately. What I am suggesting (and may be on shaky ground) is that H2O and D2O in solution together are not as independent as molecules of proteins or alcohols. In H2O, all the hydrogens are equivalent and exchange on a time scale that might be calculable; same for D2O. But when you put bulk H2O and bulk D2O together, the hydrogen bonded clusters will re-form, rearrange, simply by being there, so that if you try to analyze for HOD, you are selecting one molecule (say, by evaporation) from a cluster that is primarily determined by how much mixing you have done. If you add a drop of H2O to a liter of D2O, you may be quick enough to recapture most of that drop. But if you mix the drop (takes time: maybe one or two seconds), those H2O molecules are distributed throughout the liter, forced into new hydrogen bonding by the mixing, and not identifiable as H2O because they are in clusters of molecules. Pull out an oxygen, and it might have an H and a D, or 2D, but not likely 2H, because of the randomness induced by the mixing.

I am not sure in that. The water molecules are very stable, and only about $10^{-7}$ of them is dissociated. I think, a $D_2O+H_2O\rightarrow 2HDO$ requires at least some quantum tunneling of the protons/deuterons (low probability), or some more complex reaction involving $H_3O^+$ / $HO^-$ ions (which are relative rare). Can you please make clear in your answer, why would it be so fast?

A single water molecule is very stable, as you correctly state. If you put a few together, there will be hydrogen bonds and clumps of water which are not just H2O. The few protons are not just H+, not even just H3O+, but more like H5O2+ and H7O3+, and so on until the electrical attraction of the proton is diminished. Perhaps I'm being too practical, not theoretical enough. Measuring the formation of HOD from H2O + D2O requires observing HOD: difficult. Perhaps H2OD+ + D2O --> D2OH+ + HOD could be observed by NMR line broadening. I'm editing my original answer to include more.

Do not forget the fast proton exchanging effect, mentioned in the other comment. H3O+ + H2O <-> H2O + H3O+, resp OH- + H2O <-> H2O + OH-, standing behind very high molar conductivities of both ions.


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As adjectives, not in a math context: 'different' vs 'differential'
I looked up  differential, but still don't savvy why the books below use 'differential' rather than 'different' and vice versa.
Advanced Organic Chemistry: Part A: Structure and Mechanisms by Francis A. Carey, Richard J. Sundberg. p 211.

Electroanalytical Chemistry: A Series of Advances
edited by Allen J. Bard. p 195.

Tietz Textbook of Clinical Chemistry and Molecular Diagnostics - E-Book by Carl A. Burtis, Edward R. Ashwood, David E. Bruns. p 519.

I'm voting to close this question as off-topic because this belongs on an English Language site. The dictionary entry linked has the correct definition.

@JonCuster I disagree.  It is true the dictionary has the correct definition.  But the OP's challenge was, I believe, specifically in how to understand the definition in the context of its application to chemistry.

"Differential" is a compact way of saying "difference between the" (or its equivalent).  For instance, in your first example, where the author writes "Chromatographic separations result from the differential interactions of the enantiomers...", what's meant is  "Chromatographic separations result from the difference between the interactions of the [two] enantiomers...".
I.e., chromatographic separation results from the fact that there is a difference between how the two different enantiomers react with the stationary phase (the solid column packing material).
The first definition given in your link applies here: "Of, showing, or depending on a difference; varying according to circumstances or relevant factors."

Still, I would have written "from the different interactions of....". At least for the first example given the question is more suitable for English SE. The other instances refer to the techniques not the different behavior, so they might have wanted to say that more specifically, like in differential scanning calorimetry etc.  English is not my tongue, though.

I would use differential when there is a quantitative (perhaps subtle) difference, not necessarily a quantitative difference (for the chromatography, if I say different interactions, I would expect different types of interactions, but if I say differential, I would mean different chromatographic behavior because the strength of the interactions are not the same).

@KarstenTheis  I would describe that as a personal choice, since the term can be applied to both quantitative and qualitative differences.  As an example of the latter, consider "differential diagnosis":  https://en.wikipedia.org/wiki/Differential_diagnosis


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Question about haloalkanes
Are haloalkanes considered a homologous series? For example, when assigning chloromethane to a homologous series, do we say that it belongs to the alkanes or haloalkanes?
Edit: I think that it would be a haloalkane. However, I'm unsure if haloalkanes are considered a homologous series.

What do you think? Just include that in your main question. We want to know your views, otherwise it has a full chance to get closed.

I'm voting to close this question as off-topic because you did not give enough context (why are you asking, how did you arrive at your conclusion, did you come across the term homologous series of haloalkanes somewhere).

While homologous series are a very nice thing for teachers to quiz students about, they practically don’t matter to modern lab chemists. Sure, we all know that methanol, ethanol and propan-1-ol form a series and that a $\ce{CH2}$ unit is added when going from one to the next but in terms of actual lab properties the three are different enough to be remembered individually. As for most synthetic applications, linear molecules (the only ones for which a homologous series makes some sense) are not really any target of relevance so they won’t be thought of.
That said, I see nothing against applying a term that works well for many functional groups (e.g. $\ce{OH}$ or $\ce{NH2}$) to another functional group ($\ce{Cl}$) so the only argument against calling them one is ‘my teacher said so and it’s important that I know that in the upcoming exam’.


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Photoelectric effect with 0 Kinetic Energy?
Does the photoelectric effect take place if the frequency of the photon is exactly equal to the threshold frequency of the metal? I know the electron is going to be free, but it will have 0 kinetic energy; it will be stationary. If so, can voltage be still created if the metal surface is connected to a positive terminal side?

In quantum mechanics can anything be truly stationary?

I mean KE of 0.

If you irradiate a metal with the exact amount of energy to produce a free electron with KE = 0, it will just stay free, in the "sea of electrons" of the metal, while the nucleus, with its positive charge, also sits in the same sea - near "other" electrons, each of which has KE due to temperature. Then, after a while, an electron drops in around that nucleus, giving a bit of (thermal) shock. The observable effect will be that irradiation with this wavelength warms the metal as much as possible without exciting any electrons to the "outside" world.
Now if you attach a negative electrode to the metal and a positive electrode close to the surface, you can push/pull that electron (or one other from the sea of electrons) off - the application of a potential gives KE to a free electron. Actually, you don't even have to irradiate: if you apply enough potential, you can pull electrons out of a metal. If you heat the metal, the electrons will come out more easily.
This is the basis of the Edison effect, which makes vacuum diodes possible: a heated cathode "boils off" some electrons which would stay near the wire, but if you have a nearby anode (an electrode made positive by an outside battery), electron current will flow easily from the heated cathode to the (cold) anode. But not from the cold electrode to the hot electrode. So you can rectify AC current to pulsed DC.