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does table salt dissolve in glycerin
Tell me, does table salt dissolve in glycerin?  And is it possible to increase its solubility?
I have vegetable glycerin with nicotine that I use for vape.  So that nicotine would not be bitter, on the forum I was advised to dissolve salt in it.  But salt dissolves very poorly.  How can I speed up this process?

Adding water will certainly help with this issue, probably at the cost of introducing other issues.

This is probably a bad idea - the salt will deposit on the heating element, possibly risking corrosion and lowering the surface area of the element

Can you use something other than sodium chloride?  Sodium bromide seems to be much more soluble in methanol:
Sodium chloride = 14.9 g/kg
Sodium bromide = 168 g/kg
A similar trend would be expected with other alcoholic solvents such as glycerin.
You may also consider lithium chloride.  Lithium salts are more soluble in many polar solvents than heavier Group I salts due to ion pair or covalent molecule formation.  Lithium chloride has high solubility in many polar solvents, including over 400 g/kg in methanol and over 240 g/kg in ethanol.  This again suggests that the salt will work well in other alcoholic solvents like glycerin.


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Electron shells and subshells energies
According to what i learned in university, the energy of orbitals increases when the principal quantum number n increases. Also, in a given shell, s orbitals have lower energy than p orbitals which have lower energy than d orbitals, which in turn have lower energy than f orbitals.
And according to Aufbau principle: "Electrons fill lowest energy orbitals first."
But when we write the electron configuration of atoms, it goes as follows:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p
For example, why is 4s filled before 3d?
I think i have a big misconception here.
Another question circulating around the same topic is the order of electron removing when we want to form cations. To take as an example, let's take the electronic configuration of the atom Mn and its cation Mn2+
Mn: [Ar] 4s2 3d5
Mn2+ :[Ar] 3d5
Why did we remove electrons from the 4s before the 3d? Shouldn't the order of removing electrons be exactly the opposite pathway of Aufbau principle? And what is exactly the method used to remove electrons from an atom to form its cation?

"s orbitals have lower energy than p orbitals which have lower energy than d orbitals" This statement is only true when there is more than one electron.

https://chemistry.stackexchange.com/questions/73325/while-filling-electrons-we-follow-aufbau-principle-but-not-while-removing-them

https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium

That's a duplicate of these^ posts.

source: Stack Exchange

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Ce₂(SO₄)₃ solubility in water
I have mostly come across examples of solids whose solubility increases in liquids with an increase in temperature. However, there seem to be exceptions to generalized rule this such as the solubility of $\ce{Ce2(SO4)3}$ in water. What makes this substance so special?

No need to go after anything this exotic. There are great many exceptions, to the point that there is hardly any rule at all.

@IvanNeretin While exceptions are not exactly rare, the general rule (or trend) is that solubility of salts increases with temperature. http://www.chemgapedia.de/vsengine/media/vsc/de/ch/11/aac/vorlesung/kap_7/kap7_5/kap7_52/grafik/t_abhaengigkeit.gif Here's another one, CaCrO4 http://docplayer.org/docs-images/80/80977997/images/2-0.jpg And the well know NaCl, whose solubility is nearly temperature independent.

See for reference, sodium sulfate: https://chemistry.stackexchange.com/questions/10457/why-does-sodium-sulfate-have-an-unusual-solubility-temperature-curve?r=SearchResults

https://chemistry.stackexchange.com/questions/51862/why-does-the-solubility-of-some-salts-decrease-with-temperature

It depends on just what part of the phase diagram you are on. Coming up from freezing. Solubility increases. Past the eutectic solubility decreases up to the boiling point if pure water.

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Why is the binding angle of HOF lesser than that of H2O, despite the lone pairs and size of the flourine?
Why is the binding angle of HOF(101°) lesser than that of H2O?(104.5°) Since fluorine has unpaired electrons wouldn't it repel hydrogen more strongly than the corresponding hydrogen of H2O? Also, the size of fluorine is bigger than hydrogen, which also should have made the repulsion greater.

https://chemistry.stackexchange.com/questions/57665/which-has-the-largest-bond-angle-between-water-oxygen-difluoride-and-dichlorine

In $X_2O$ molecules, such as $H_2O$, $F_2O$, $Cl_2O$, the oxygen is at the center of a tetrahedron and therefore theoretically the X-O-X bond angle should be 109.5°. However, this is never the case and it's always a balance between the repulsion of the two lone pairs of the oxygen(they repels more strongly than the bond pairs) and the steric repulsion of the X atoms.

Among the series H, F, Cl, hydrogen is the least electronegative and fluorine the most.
F is very electronegative and thus the electron density of the O-F bond near O is less than that in O-H.
If the electron density is farther away from the oxygen, then the 2 lone pairs of O squeeze the two bonding pairs closer together and the F-O-F bond angle will be less than the H-O-H angle.
Based on the electronegativity, you should expect that the Cl-O-Cl bond angle is less than H-O-H but more than F-O-F. This is not the case and the reason is that chlorine is big enough so that the steric repulsion prevails.
In your case you are considering the molecule HOF:
According to the logic above, you may expect that its bond angle is less than $H_2O$ but more than $F_2O$. Instead, H-O-F bond angle is 97° so even less than $F_2O$!

The reason for this unexpected small angle is that the bond length O-F and O-H are very different and their electron density(bond pairs) don't interact much and therefore they repel even less than in the F-O-F case(see picture above, where electron density is represented in blue).


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Rechargeable Battery Based on Neutralization Reaction of Acid and Base
The neutralization reaction, such as $\ce{HCl + NaOH → NaCl + H2O}$, generates heat, which is around $\pu{57.30 kJ/mol}$.
Can we use this energy to design a battery?
This idea occurs in my mind for a long time.

---------- Selection for Acid and Base -------------
For acid, $\ce{HCl}$ is a good choice, $\pu{36.46g}$ for $\pu{1 mol}$ of $\ce{H^+}$.
For base, $\ce{NaOH}$ is Okay, but not the best.
Problems for both $\ce{NaOH}$ and $\ce{LiOH}$ is that they both absorb water when forming crystal from water.
$\ce{LiOH·H2O → LiOH + H2O}$, ($\pu{100-110 °C}$)
In my idea, I consider hydrogen chloride ($\ce{HCl}$) as the acid and ammonia ($\ce{NH3}$) as the base. Surprisingly, it turns out they have a very high energy density. Also, they both can be stored in gas phase, which is hard to solve for sulfuric acid ($\ce{H2SO4}$) with high concentration.
$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}, \quad ΔH = \pu{-176.0 kJ/mol}$
Energy density $= \pu{176.0 kJ}/\pu{53.49g} = \pu{3.29 MJ/kg} = \pu{913.9 W·h/kg}$, which is better than lithium-ion battery ($\pu{265 W·h/kg}$).
As a comparison, for Hydrogen Combustion,
$\ce{H2(g) + 1/2 O2(g) → H2O(l)} + \pu{572 kJ}$ ($\pu{286 kJ/mol}$), it gives energy density of $\pu{286kJ}/\pu{18g} = \pu{15.89MJ/kg}$.
To be noted, $\pu{1eV} = \pu{96.485 kJ/mol}$, which means the battery should give $176.0/96.485 = \pu{1.82 V}$.
In water, higher than $\pu{1V}$ might break $\ce{H2O}$ into $\ce{H2}$ and $\ce{O2}$. In my design, the energy in each proton is released in two steps, where each gives $1.82/2 = \pu{0.91V}$.

---------- Properties for Acid and Base -------------
$\{1\} \; \ce{HCl(g) + H2O(l) -> H3O+(aq) + Cl-(aq)}, \quad ΔH = -92.31 - (-167.16) = \pu{-74.85 kJ/mol}$
$\{2\} \; \ce{NH3(g) + H2O(l) -> NH_4+(aq) + OH-(aq)}, \quad ΔH = -46.11 - (-80.29) = \pu{-34.18kJ/mol}$
$\{3\} \; \ce{H3O+(aq) + OH-(aq) -> 2H2O(l)},  \quad ΔH = \pu{−57.30 kJ/mol}$
$\{4\} \; \ce{NH4+(aq) + Cl-(aq) -> NH4Cl(s)}, \quad ΔH = \pu{–20 kJ/mol}$
$\{4.1 ~Lattice ~energy\}$:
$\ce{NH4+(g) + Cl-(g) -> NH4Cl(s)}, \quad ΔH = \pu{–705 kJ/mol}$
$\{4.2 ~Hydration ~enthalpy\}$:
$\ce{NH4+(aq) -> NH4+(g)}, \quad ΔH = \pu{307 kJ/mol}$
$\ce{Cl-(aq)  -> Cl-(g)}, \quad ΔH = \pu{378 kJ/mol}$

---------- Battery Design -------------
With the aid of a small electrolytic capacitor, we can release the energy in neutralization reaction into electricity. The reactions highly rely on proton transmission.
Solvent: all reactions are in aqueous solution (water).
Capacitor: anode with $\ce{HCl(aq)}$, cathode with $\ce{NH3(aq)}$.

$\ce{HCl}$ gas is connected to anode, and $\ce{NH3}$ gas to cathode, both connections are controlled by valve (switch).
Moreover, anode and cathode are connected by two channels (with switch): PEM membrane channel, pipe channel.
--------- energy release -----------
The battery energy is released by repeating the following two steps.
Step 1: Capacitor charge with reactions {1}, {2}, and {3}:
$\ce{anode + Cl- - e-  -> anode^+(Cl-)}$
$\ce{cathode + NH4+ + e^- -> cathode^-(NH4+)}$
Switch on PEM membrane channel, and switch off pipe channel. $\ce{H3O+}$ flow through PEM membrane and react with $\ce{OH-}$ {3}, then $\ce{Cl-}$ flow to anode, and $\ce{NH4+}$ to cathode, causing potential ($<\pu{1V}$) in capacitor, and electrons flow from the anode to the cathode.
Step 2: Capacitor discharge with reaction {4}:
$\ce{anode^+(Cl-) + e^-   -> anode + Cl-}$
$\ce{cathode^-(NH4+) - e^- -> cathode + NH4+}$
Switch off PEM membrane channel, and switch on pipe channel. $\ce{Cl-}$ and $\ce{NH4+}$ flow through pipe channel and produce $\ce{NH4Cl(s)}$, causing potential ($<\pu{1V}$) in capacitor, then electrons flow from the cathode to the anode.
------------battery recharge---------------------
To charge the battery (input energy), reverse the above two process.

Possible Issues
i) Not environmental friendly, $\ce{HCl}$ and $\ce{NH3}$ gas are toxic.
ii) Too much pressure in $\ce{HCl}$ gas and $\ce{NH3}$ gas.
iii) Electron exchange rate limits the battery power density.
iv) Hard to design capacitor for $\ce{NH4Cl}$.
v) switches of the membrane or pipe are hard to design.
vi) energy efficiency might not be as good as expected.
Welcome to discuss the above issues and other practical problems.
----------------------------update on 14.08.2019--------------
This is respond to the comments by @Oscar Lanzi, about the issue of low potential.
Actually, the concept of storing energy in acid and base for battery is already studied in hybrid battery. Check this research paper "Three electrolyte high voltage acid–alkaline hybrid rechargeable battery", they use electrolyte configuration of 1 M H2SO4|0.2 M K2SO4|2 M KOH to increase the voltage.
(https://www.sciencedirect.com/science/article/pii/S001346861101228X)
Go back to the design of HCl,NH3 battery, the low potential is indeed a problem as mentioned by @Oscar Lanzi, and it also brings many practice issues such as the membrane. But I still want to discuss about the possibility or physical/chemical principles behind those problems, regardless of the limitation of materials.
For the voltage potential, 0.8V is the best for HCl + NH3 reaction in water solution, the reason is that water is so easily to break down into H+ ions, self-ionization of water is only $10^{-14}$ at room temperature.
However pure NH3 itself does not produce H+ easily,
$2NH_3 ⇌ NH_2^− + NH_4^+, k = 1.9 \times 10^{-33} ~( - 50 °C)$
This effect is called Leveling Effect.
Therefore, can we just select another approriate solvent and achieve  $1.82V$ potential?
One possible solution might be to mix water with other solvent so that the ionic product of water (or equivalently pKa) is reduced from $10^{-14}$ to $10^{-32}$, one example is to mix water with Dimethyl Sulfoxide (DMSO).
As is shown here, pKa value for $H_2O$ in liquid water and DMSO are 15.7 and 32 respectively, that means $[H3O+][OH-]/[H2O] = 10^{-32}$ in the mixture solution of water and DMSO (if I understand correctly).
(http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf)
Well, I am not sure about whether nernst equation still works in this situation since the solvent inside and outside the membrane is not the same, i.e.  (water) / (water DMSO mixture).
The following calculation is based on the assumption that nernst equation still works.

Illustration of new design
~==========================================~
~   $<A>$   ||     $<B>$     ||     $<C>$
~        ||        ||
~  HCl+H2O  ||   Cl-   NH4+  ||   NH3  +  H2O
~        || ↗      ↖ ||
~- - - - - - - - - - - - - ||xxxxxxxxxxxxxxxx||  - - - - - - - - - - - - -~
~    ↓↓   ||xxxxxxxxxxxxxxxx||    ↓↓
~   [H+]   ||xxxxxxxxxxxxxxxx||   [OH-]    ~
~    ↓↓   ||xxxxxxxxxxxxxxxx||    ↓↓
~- - - - - - - - - - - - - ||xxxxxxxxxxxxxxxx||  - - - - - - - - - - - - -~
   $<D>$   H2O   +    DMSO
~==========================================~

Then, assume we have the following ideal battery design where
1)  $H^+$ penetrates from HCl(aq) to DMSO(aq),
2) $OH^-$ penetrates from NH3(aq) to DMSO(aq),
3) $Cl^-$ → $<B>$
4) $NH_4^+$ → $<B>$
Here is the calculation of potential induced by the ion transmission, based on Nernst Equation.
For ammonia (NH3), 1.0 M aqueous solution (1 mol/L) has a pH of 11.6, which gives $[OH-]_C = [NH4+]_C = 10^{11.6-14} = 10 ^{-2.4} mol/L$.
For hydrogen chloride (HCl), the aqueous solution can achieve pH of -1 (i.e. 10  mol/L), which gives $$[H^+]_A$ = 10 mol/L$.
For ammonium chloride NH4Cl(aq),  the saturated aqueous solution gives 39.3g per 100ml at 25°C, which gives $[NH_4^+]_B ≈ [Cl^-]_B = 393/53.49 = 7.35 = 10^{0.866} mol/L$.
For water and DMSO mixture, $[OH^-]_D = [H_3O^+]_D = 10^{-16}$.
Based on Nernst equation, pH sensor will change its output by 59.16mV for every change in one pH unit, i.e. ten fold difference in concentration of $H^+$.
(https://www.vernier.com/news/2018/04/06/the-theory-behind-ph-measurements/)
As a result,
For 1) $E_1 = + 59mV * ( (16) - (-1))  = 17 * 59mV :[H^+]_A / [H_3O^+]_D$
For 2) $E_2 = + 59mV * ( (16) - (2.4)) = 13.6 * 59mV :[OH^-]_C / [OH^-]_D$
For 3) $E_3 = + 59mV * ( (-0.866) - (-1))  = 0.134 * 59mV :[Cl^-]_A / [Cl^-]_B$
For 4) $E_4 = + 59mV * ( (-0.866) - (2.4))  = -3.266 * 59mV :[NH_4^+]_C / [NH_4^+]_B$
Note that in 4) the $NH_4^+$ ion needs energy to pump from $<C>$ into $<B>$.
In total, we have $ 27.468 * 59mV = 1.625 V$, still a little far from the best possible potential $1.82V$.

My poor eyes, also whole idea isn't particularly practical.

@Mithoron Yes, many practical issues. Here, I didn't put much effort in the materials such as PEM membrane and capacitor.

One problem is there is not enough voltage for this thing to sell.  Say you have  molar acid and 1 molar base, both strong and both functioning as ideal solutions.  Thermodynamically you have one electron transfer associated with the net reaction
$\ce{H^+}_{aq}+\ce{OH^-}_{aq}\to\ce{H2O}$
and the equilibrium constant is $10^{14}$ at room temperature.  Put that into the Nernst Equation and you get about 0.8 volt.  Not enough to compete with what is commoly used.  You need a bigger free energy punch, and generally redox reactions, which involve a greater amount of electron transfer, provide more than acid-base reactions.

Thanks for your comments, however, I think it just proves that proton-based battery works as good as electron battery. 0.8 volt indeed comes from the H+, while the other ions Cl- and NH4+ will also contribute to the same amount of voltage. In other words, 0.8 V is the potential voltage for half cell in Nernst Equation.
This corresponds to the two steps of energy release in the capacitor where each give 0.8 V.
The voltage problem might be solved by increasing the gas pressure since the concentration of HCl and NH3 in water will be increased under high pressure.

How do you get 0.8 V from the alt ions?  Show your reactions and potentials maybe?

Maybe you are right. But this idea is not new, I just found another paper published 3 years ago, based on the same idea (neutralization reaction), "An Acid-Base Electrochemical Flow Battery as energy storage system" (https://www.sciencedirect.com/science/article/abs/pii/S0360319916325289?via%3Dihub),  they claim an energy efficiency of 55%, i.e. only achieve half of the ideal potential.


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Raman Spectrum of Water
This is the Raman spectrum of water. There is $\pu{1635 cm-1}$ Raman peak corresponding to $\ce{HOH}$ and $\pu{3410 cm-1}$ Raman peak corresponding to $\ce{OH}$. But why is there general nonzero background in intensity? What caused it?
This is taken using 532nm wavelength raman system:

This is taken using 785nm wavelength raman system up to 2750 cm-1 only (versus 4000 cm-1 of the first spectrum):

In the above picture. Why is the background dimishing to zero towards 2750 cm-1. My guess is that as wavelength gets higher, water absorbs more. But can it absorb the noise too?

All detectors have noise. All spectrometers aren’t perfect. There is always background in every measurement.

You mean for very expensive advanced raman spectrometers, the noise would be less and less? You mean the above background is simply noise and not any characteristics of water such as fluorescence?

You need to keep the spectrometer and the detector separate in your mind. It is a quite broad question to determine all the possible noise sources in the whole system. But, consider a high resolution spectrometer- you only pass a very narrow frequency range to the detector. So, not many photons. But the dark counts are whatever they are, so your signal to noise is low.

The above photo was sample shown to me by manufacturer taken using 532nm laser and visible wavelength detector. Are you saying if 785nm is used, the noise would even be more??

See (added) 2nd picture using 785nm spectrum of water. Why is the background dimishing to zero towards 2750 cm-1. My guess is that as wavelength gets higher, water absorbs more. But can it absorb the noise too? What do you think cause it to go down to zero?

You just put side by side pictures taken with different source and detector. Ergo the absolute values at y do not tell info. Fluorescence is present in Raman spectra and is difficult to remove as for it requires the sample to be in, it has no physical blank to run a background.

porphyrin, water becomes more absorbent as wavelength is increased. So water can absorb IR more than visible. So in the 2nd picture.. As wavelength increased, it's absorbing the fluorescence? why do you think this theory is wrong? The 2nd device used the First Defender RMX https://www.thermofisher.com/order/catalog/product/FIRSTDEFENDERRMX  the seller gave me sample picture of water spectra taken direct 3mm from water without passing through any container.

535nm principally scatters a factor 4.7 more than 785, not only Raman scattering but also elastic Raleigh scattering, which is one of your major noise sources. Also the second raman with the NIR source is perhaps a modern FT Raman. And the NIR source makes less flourescence, which is the worst noise source of all in most cases.

@porphyrin: 3000 1/cm from 785 nm is about 1 micron (1026.8 nm).

@Samzun That 2nd device is a handheld detector which is not designed to output high-quality spectra. It isn't a general-purpose spectrometer. It's used to detect specific compounds, probably by pattern matching the data to a reference library. So it just has to be good enough for that purpose. You have to make a lot of compromises to shrink a spectrometer to that size, and the loss of intensity towards 3000 cm-1 may well be one of those compromises.

It's been a while since I have run any Raman spectra but the constant noise level is bothersome. I suspect less than optimal instrument conditions and too much stray light in the spectrum.

This answer is based on the educated guess that the green excitation Raman spectrum was acquired with a dispersive Raman instrument*.

But why is there general nonzero background in intensity? What caused it?

For any real-life instrumentation, you should never assume to measure 0 signal for 0 analyte concentration (see also Does a calibration curve of absorbance vs concentration need to pass exactly through the origin?) nor expect the baseline to be at zero unless corrected for, even if there really were no Raman scattering outside those two bands, the instrument and/or sample preparation procedures (electronics, optics, masking, cross-sensitivity, ...) will often lead to some non-zero signal.
The CCD in the spectrograph has a so-called dark current. So the raw signal you observe is dark current plus photon signal. Dark current is influenced by settings of the CCD (you trade off sensitivity and dark current) and by temperature.
The constant part of the dark current shouldn't trouble you too much: you can measure and then subtract it (which is in fact what spectrometer software does where you see the baseline at zero). However, as subtracting spectra sums their noise** (error propagation). As Raman measurements typcially suffer from low signal-to-noise-ratio, you may want to avoid this, so Raman spectrometer software usually either outputs raw spectra or offers you the choice whether you want to subtract the dark spectra.
For SNR of Raman spectra, you'll have to take into account shot noise (Photon noise): the noise is directly proportional to the signal intensity, and that includes optical background such as fluorescence or elastically scattered or reflected light as well as your Raman signal. (And then come the other sources of noise such as flicker noise and the various sources of noise related to your application).
For a thorough discussion of the noise properties of Raman spectrometers, there's a whole chaper of McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5 on the topic
(for FT instruments, see Zhao & McCreery: Multichannel FT-Raman Spectroscopy: Noise Analysis and
Performance Assessment, ApplSpec, 1687 - 1697, 1997.)

About the 785 measurement in comparison:
The green excitation y-axis looks like raw counts (the data itself would reveal more: raw counts are usually integer numbers). In contrast the red spectrum's y-axis goes from 0 to slightly above 1.0. Which means that some transformations have been done (which is impossible to say from the data alone - you'll have to talk to the manufacturer what they do).

 water absorption

Absorbance of in the range below 1100 nm is < 0.2/cm. So yes, you loose about 1/3 of the intensity through 5 mm water and back - but you can avoid that by focusing not as deep into the water.
What hurts your signal more is that the quantum efficiency down there is pretty awful for many CCDs (around 10 %, there's a reason why their spectral range ends in the mid of the -CH stretching vibrations).

* A rather safe guess, as FT instruments for such short wavelength AFAIK exist at the moment only as "home-built" instruments developed in some research facilities - whoever uses one of those is going to be fully aware that they are using something special.
** more precisely: the variances of independent random variables are additive

Do you think I should get this 785nm model https://www.shopstellarnet.com/785nm-preconfigured-raman-spectrometer-system/ or this 532nm model http://enspectr.com/portfolio/enspectr-r532/ ? I just want to buy one for learning molecular structure.. which would you buy if you were just a beginner?

btw.. when you said "absorbance of in the range below 1100 nm is < 0.2/cm". Did you mean by below is like 900nm or below is like 1200nm? I was asking water absorbance between 800nm to 1050nm. how many percentage is the absorbance through 5mm of water using 785nm laser? I want to study water hence can't get 532nm off my mind. Thanks so much.

cbeleites, the Stellar model has 499mW laser but at 785nm. The Enspecr has 30mW laser only at 532nm. I know 532nm has 5 times intensity than 785nm. But it's like equal to 30mWx5= 150mW laser only. What substance that you need the full 499mW on the 785nm.. which still has 3 times the laser power compared to 150mW (30mW x 5). I guess this will be my deciding factor which of them to get. Also need to know absorbance of water between 785 to 1015nm. Is it like 15% only or 1/3? Thanks!

900 nm is below 1100 nm (so all your Raman range).

You cannot conclude too much from the 500 mW vs. 15 mW: λ⁴ dependency is one thing, quantum efficiency and the losses in the optical path are another thing. You'll have to measure to find out how well the instruments do.

But for visible light 532nm laser raman spectrometer, there is no losses (or almost insignificant absorbance) via the water since it's not IR, right? I want to study effect of 1% solution in water which the 532nm system claims to do.

Samzun: From what you have written so far, the question which Raman spectrometer is best for you cannot be answered. I also have the impression that you need a better understanding of the instrumentation basics and tradeoffs here. I feel that thisis leaving the region of what is generally useful to the public. As I do advise (and teach) professionally on these topics, you can contact me (see profile) if you are interested in getting professional help with the decisions for purchasing an instrument. Otherwise, reading & understanding the literature I've linked and working through a list of the

...questions that need to be answered in order to get to a good decision (both in my answers and in the comments of others!) should get you a good understanding of what to look for.

A background on Raman:
Samzun (OP), buried in your query might be the taught belief that a Raman scatter spectrum is independent of the excitation wavelength, and this is mostly true, except for a scatter intensity that scales as (1/wavelength)^4 of the excitation.  This wavelength-independence makes Raman useful for chemical species identification, and being able to chose the excitation wavelength makes it amenable for various carrier media, including water, for the analyte under observation.
So while the ideal (theoretical) Raman spectrum is fixed, various pieces of your optics will multiply their signature onto the spectrum.  As noted in cbeleites' response, a dispersive grating can add its signature.  Unlike a simple prism, a grating's signature can be non-monotonic and is often oscillatory, especially in highly efficient holographic blazed gratings.
Another significant contribution is the quantum efficiency of your detector.  I suspect you are using a silicon detector for both your 532nm and 785nm excitation.  Silicon's quantum efficiency drops off slowly and as it nears 900nm drops off precipitously to zero at 1050nm.  At an excitation of 785nm a lot of the scatter will be at those longer wavelengths.
Other pieces of the optics will also add signature to your spectra, e.g. anti-reflection coatings, filters, polarization dependence, etc.  A lot of care must be taken to remove these various contributions.  As you note, the carrier media (water in your case) can also be an absorber.  But in your case for 785nm excitation, you will find that water absorption is not the main contributor, and the detector quantum efficiency is.
If you know the signatures of the various contributors, you can divide them out to get the ideal (wavelength-independent) Raman spectrum of your analyte.
Water spectrum:
An additional question that you ask is about the background around and between the HOH and OH vibrational modes.  I do not believe this has been properly addressed in the other replies, and unfortunately the answers are buried in academic literature.  This additional background of water is also bothersome for the shot noise is adds across the whole spectrum.  Polarization dependent Raman will be the easiest way to convince yourself this broad background in water's spectrum is real.
The likely answer for this broad background is hydrogen-bonding between the water molecules.  It loosely bonds multiple water molecules and you then have the vibrational modes of a larger amorphous superstructure.  This is very complicated, and so only anecdotal and handwavey discussions are found in literature.  This may be unsatisfying, especially to a theoretician, but in your practice it may not matter, e.g. if water's background stays constant and consistent and you know its input variables.
Good luck observing!


text:
What makes ammonium nitrate soluble?
Ammonium nitrate is one of the classic examples of an endothermic dissolution.  People will usually explain why it's soluble by stating that the spontaneous dissolution process of ammonium nitrate is what drives the reaction, meaning that the obvious answer is it's soluble because ΔG of dissolving it is negative.  Here is an example.
However, after thinking a bit more about it, this explanation seems very vague and doesn't explain it well.  Enthalpy usually has a stronger role in determining the solubility of salts, since their ΔH values are usually much larger than the TΔS values they would have, especially at room temperature.
Both calcium and aluminum fluoride are also salts that have an endothermic dissolution, but by the number of ions dissociated, one would think that ΔS of dissolving both of these salts would be much higher than that of ammonium nitrate.  Ammonium nitrate is still several times more soluble though.
My question is, what mechanisms allow ammonium nitrate to have such a high ΔS when dissolving in solution that allow it to still be very soluble, despite having an endothermic dissolution?

At a complete guess, hydrogen bonding between ammonium and water partially destroys the order that hydrogen bonding gives to liquid water. But ultimately it's all hand-waving - the energetics are the only thing that really matters.

Here is some data on other well-soluble salts:

Comparing the values, both the anion and the cation contribute to magnitude of enthalpy and entropy. The larger cations seem to contribute to a larger entropy of dissolution, as do the larger anions. The entropy is concentration-dependent, so I assume the data is for a standard state of about 1 mol/L dissolved salt.
There is a nice discussion about the dissolution of ammonium chloride here. The enthalpy of dissolution (14.4 kJ/mol) and the entropy of dissolution (75.3 J/(K mol), source) are close to that of KCl, so there is nothing super-special about the ammonium ion.

[...] since their ΔH values are usually much larger than the TΔS values they would have, especially at room temperature.

At the beginning of the dissolution process, the entropy dominates. At the solubility limit, entropy and enthalpy contributions are equal and add up to zero.

What makes ammonium nitrate soluble?

I did not find a data table that includes entropy and enthalpy of dissolution for ammonium nitrate, but you could easily calculate these from entropy and enthalpy of formation of the reactants and products. On second view, I am having trouble finding the data for aqueous nitrate, but this textbook has some values. There is a report that nitrate comes in two forms in aqueous solution, perhaps complicating things.
Some people seem to think that the dissolution of nitrates is special in terms of enthalpy (compare to e.g. carbonate) and others argue that the entropic contribution is special. It would be easier to answer the question, for example, "what makes ammonium nitrate more soluble than ammonium carbonate".


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What is the safest way to hook up car battery jump cables and why?
I know that when both connecting and disconnecting battery jumper cables between two car batteries there is a prescribed safe way to do it. I can never remember the correct way because I can’t understand the reasoning behind the prescribed sequence. Is there an easy explanation / (chemical?) principle that can make this procedure (the sequence of both connecting & disconnecting the four posts) more logical and therefore easier to remember?
My confusion is that at some point you must make a connection that causes sparks (which risks battery explosion) because it completes the circuit. Completing the circuit can’t be avoided, so why does it matter which post you are connecting to when you complete the circuit?

I'm voting to close this question as off-topic because if that's about chemistry then every experienced driver should get a diploma.

Karl has the right answer. This is not about chemistry, it would be a better fit on mechanics.stackexchange where it has already been asked many times. https://mechanics.stackexchange.com/questions/22296/when-reconnecting-a-battery-why-connect-the-positive-first

I think this is a valid chemistry question, wondering wether there can be a chemical reason behind the recommended procedure.

My fault. I’m a new user and was attempting to answer a chem question that was protected and required 10 reputation. So I entered this question, but got stuck in chemistry and very confused about how to navigate this site. I’d be happy to move it if only I could figure out how to do that. Let me know. Thanks.

Many recommended procedures such as this one call for connecting (+) to (+), then (-) on the "good" battery to a metal component in the "dead" car engine.  The negative terminal of the battery is grounded to the metal components, so we may think of the last connection as (-) on the dead car.  If a spark were to form when the circuit is established on the last connection, it would occur at the metal component and not at the terminal where it could interact with electrolytically generated hydrogen or oxygen gas.

Ive heard the same reasoning, but what hydrogen concentration could there be? Theres a good reason why lead batteries must be located in a well ventilated position. Also the dead battery specifically, being dead, is most unlikely to have recently produced any hydrogen.

The "dead" battery could still have some voltage left, just not enough to start the car.  Hydrogen could come from a small amount of water decomposition, which may not be complrtely avoided in an aqueous electrolytic cell.

Surely, below ten volts, it cant start a car. But I thought hydrogen generation is (or was, rather, with modern batteries) usually occuring during recharging, isnt it?

Recharging is basically what we're doing.  And I agree the risk is less now but safety procedures should always be redundant.

But it only starts when you have already connected the cables. One legitimate reason could be that an old, oldstyle battery with low electrolyte  might contain quite an amount of H2+O2 mixture in its housing. That would be a bomb.

You connect to the chassis away from the battery on the dead car to give low resistance path to the starter motor and not too low resistance path to the dead battery, but that's not in OP. You connect negative last, because if you connect positive last you could slip, short circuit and cause huge current, fire (and if H2 present, explosion.) This answer is getting upvotes on ChemistrySE because it contains the word Hydrogen but it's not very relevant. Short circuit risk is ALWAYS there. Chance of sufficient H2 to explode is small, as noted in links to MechanicsSE in my comment on the question.

Done that dozens of times, have never seens sparks fly. The current flowing is not exactly huge, you are connecting (first) + to +, and (then) - to -, after all. The voltage difference is perhaps 3V, not 12.
I don't think there is an electrochemical reason to do it in this order. Afaik it's just so you dont short-circuit the good battery if you slip with the second cable and touch the chassis. That would give a huge spark, probably weld the connector to the metal, and definitely leave a hole.
From such a discharge, a spark might even find it's way into the battery, which, at least in the old days, might contain some highly explosive electrolysed $\ce{2 H2 : 1 O2}$ mixture. Especially if the battery was old, hadn't been refilled with dest. water for a while, and was therefore weak, which is why your car didn't start in the first place ...
Modern car batteries are usually sealed.

Perhaps its also safer to first contact + on the empty battery. That way any mishap with the other end of that cable is less devastating.

This is the right answer. The negative terminals are connected to the car bodies. So if you connect the negative lead first, then it is possible to accidentally connect the positive lead to one battery and some part of one of the cars' bodies, leading to a massive electrical discharge and fire. Get the positive lead securely fastened first, then you can connect the negative lead safely.

There is a chemical reason in that the battery might contain hydrogen gas which can cause the battery to explode. I'd guess that this is less of a problem now with "sealed" batteries, but in the olden days when the batteries weren't sealed and you had to monitor the fluid level in the batteries it was a very real problem. No just from the shrapnel, but also from the sulfuric acid that got spayed on everything.

@MaxW Not just hydrogen gas, but a nice stochiometric 2:1 mixture or H2 and O2. But it is inside. I guess together, these two arguments make sense: A mishap with the cables, a strong electric discharge, and a spark flying from there finding it's way into the battery. I'll update.


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What is the meaning of the phase state label (cr,l) appearing in ATcT thermochemical data tables?
I was examining data associated with this post on the heat of solution of ammonium nitrate when I came upon the NIST ATcT data page which lists the heat of formation of $\ce{(NH4)NO3(cr,l)}$. In this formula, what states does "(cr,l)" refer to?

I think it is cr = crystal and l = liquid.

The nearest I came to an explanation is an article (Ref. 1) by authors of the database explaining that, in the case of $\ce{H2O (cr,l)}$, the notation (cr,l) refers to the condensed state. Also, a link from the above data page explains that "cr=crystal" and "l=liquid" (thanks MaxW for hinting at the solution).
A further update: it finally dawned on me that the website sometimes provides enthalpy data at two different temperatures ($\pu{0 K}$ and $\pu{298.15 K}$) in one line, so evidently the lower T refers to the crystalline state, the higher to the liquid state. But it remains somewhat confusing notation. In the case of an aqueous solution of $\ce{(NH4)NO3}$, it would seem to make sense to use the notation "(aq)" for the liquid state. Evidently "(l)" is synonymous with "(aq)" in the case of solutes in aqueous solution, and "(cr)" refers to a frozen dilute aqueous solution at $\ce{0 K}$. But this remains confusing, since in this table, the last entry describes the following reaction: $$\ce{(NH4)NO3 (cr,l) → [NH4]+ (aq) + [NO3]- (aq) }$$ Here a distinction is made between the hydrated (or aqueous) ions and the species $\ce{(NH4)NO3 (cr,l)}$, whatever it may be. It seems unlikely that $\ce{(NH4)NO3}$ can exist as a molten salt. It seems (cr,l) is used to signify a condensed phase of the pure substance, whichever makes sense under the given conditions. In the above reaction this is presumably the crystalline solid, and the reaction represents solubilization.
References

Ruscic et al., J. Phys. Chem. A, Vol. 108, No. 45, 2004. See the caption to Table 1.


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Negative hyperconjugation

Compare the acidity of the following compounds: $\ce{CF3OH}$ and $\ce{CCl3OH}$.

My attempt: according to me $\ce{CCl3}$ group will show a stronger Negative hyperconjugation effect than $\ce{CF3}$, and thus would stabilize the anion more effectively.
But the answer is $\ce{CF3OH}$ is more acidic. What is wrong in my approach?

What is negative hyperconjugation?

Fluorine is the boss of all electronegative things.

@electronpusher typically, lone pair to $\sigma^*$ donation

$\ce{CF3OH}$ is more acidic because of the inductive effect of the $\ce{-CF3}$ group: Fluorine is very electronegative and therefore it pulls the electron-density toward itself weakening the $\ce{O-H}$ bond in $\ce{CF3OH}$.
Chlorine is less electronegative than fluorine, thus $\ce{CCl3OH}$ is less acidic than $\ce{CF3OH}$.
Because $\ce{H}$ is even less electronegative, then $\ce{CH3OH}$ is less acidic than all its halogenated derivatives.
A related question can be found here:
Why is CF3COOH exceptionally acidic?
The negative hyperconjugation is a type of resonance effect where there is a donation of electron density from a filled π- or p-orbital to a neighboring σ*-orbital. It does not occur in your molecules.

According to me, while comparing acidic strength, we should remove $\ce{H+}$ from both the compounds, hence we get $\ce{CF3-O-}$ and $\ce{CCl3-O-}$.
Now, we will compare stability using priority order:
Aromaticity > Resonance > Hyper-conjugation > Inductive effect
Now, we can see that, in $\ce{CF3O-}$ only inductive effect is applicable whereas in $\ce{CCl3O-}$ both inductive effect and d-orbital resonance can be seen.
So, as we know acidic strength is directly proportional to the stability of the conjugate base. In other words, it is only dependant on the stability of the negative charge on the compound, and as we can see the negative charge is more stable on (b) than in (a).
Therefore, the acidic strength order would be $\ce{CCl3O-}$>$\ce{CF3O-}$.


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Why aerosol forms when you take a hot shower in a very cold day?
When temperature is about $\pu{17 ^\circ C}$ and I open the hot shower in maximum temperature (for a common electric shower) the air gets almost immediately full of droplets and this lasts the entire bath.
I guess there is water vapor in the air and it gets cooled fast and condenses. If this is true, I don't get where the vapor comes from.

I guess you heard about mist? Or that water vaporises?

@Mithoron Yes, I did. As you can see in my question. I understand the vaporization of water, and its condensation. My questions is not about those generic phenomena but about the specific situation.

As air temperature increases(in the vicinity of the hot shower) air can hold more water molecules, or with more technical terms: the saturation water pressure increases with the temperature.
Near the water, the air temperature is high and since there's plenty of hot water that evaporates then the vapor pressure increases. Then because of convective motion, new cold air replaces the hot one and it becomes full of water vapor as well. The previously warm air is now far away from the hot water and it is starting to cool down. When its temperature decreases it can't hold so many water molecules and therefore you observe the small droplets of condensed water.

Ref. picture https://www.chemguide.co.uk/physical/phaseeqia/vapourpress.html


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Is scandium considered a transition metal?
I have seen so many conflicting answers to this question in various places so I wanted to ask it again.
IUPAC defines a transition element as

an element whose atom has a partially filled d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

By this definition, scandium would be a transition element as its atom has an incomplete d shell.
However, I have seen many other definitions which require an incomplete d shell as a stable ion. This also adds confusion as the common scandium ion is $\ce{Sc^3+}$ which has no d electrons, so wouldn’t be a transition element, yet it’s my understanding scandium can display a +2 oxidation state also (albeit less commonly).
What’s the correct definition? Is scandium considered a transition metal or not?

https://chemistry.stackexchange.com/questions/91487/can-we-classify-all-the-d-block-elements-as-transition-metals

Whenever this question turns up, i feel a strong urge to ask "what else should it be".

It may depend on who you ask, and maybe even when you ask.  The tl, dr of what is below is that we are moving towards counting scandium as a transition metal.
Wikipedia follows the IUPAC definition and accepts the label for scandium because it forms atoms with partially filled $d$ orbitals, and also copper because it forms ions with partially filled $d$ orbitals in common settings.  Either possibility is admitted by IUPAC and thus by Wikipedia.
On the other hand, Britannica does not use the label "transition metal" for Group 3 elements, because they do not retain $d$ electrons in their ions.  Instead they render a separate classification, the "rare earth metals", in this group.
Britannica is based in the UK, of course, and we may be seeing a difference based on different countries.  Let's look at a couple other sources:
https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/trans.php, based in the United States, says yes, again following the IUPAC definition.  https://www.chemguide.co.uk/inorganic/transition/features.html, based in the UK, reports that most UK syllabuses use a definition that excludes scandium from the transition metals.  The latter source does state, however, that "the most recent IUPAC definition" would include scandium because of its neutral atom configuration.
The development of organometallic chemistry may have something to do with this evolution.  Many elements, including magnesium and more recently calcium, show lower oxidation states in some organometallic environments.  In the case of scandium this shows up as an atom retaining its $d$ electron while using the $s$ electrons and/or vacant $d$ orbitals for bonding to organic ligands.  Going back to the Wikipedia article on scandium, this gives references to several organometallic examples in which scandium has reduced oxidation states, from +2 all the way down to 0.  For example, Cloke et al. [1], show a single scandium atom in a low oxidation state in a setting similar to how a later transition metal might appear:

Cocondensation of scandium vapour with 1,3,5-tri-tert-butylbenzene affords the first example of a scandium(0) complex, the sandwich compound bis(η-1,3,5-tri-tert-butylbenzene)scandium; ESR and UV studies also the reveal the presence of a second compound from this reaction, formulated as a similarly unique example of scandium(II) arising from activation of one of the methyl groups of a tert-butyl ring substituent.

Such developments favor, and may have motivated, the broader, scandium (and yttrium) inclusive definition of a transition metal now used by IUPAC.
Reference
1.
F. Geoffrey N. Cloke,   Karl Khan  and  Robin N. Perutz, "η-Arene complexes of scandium(0) and scandium(II)", J. Chem. Soc., Chem. Commun. 1991, 1372-1373.


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Different shapes of fluorescence spectra in Raman
Note: I haven't owned any of these spectrometers. I'm asking so I can decide whether to buy 532nm or 785nm wavelength raman to probe molecular structure (especially solutions) and I'm concerned about fluorescence versus signal intensity. The first picture is from 16:00 minute time of this video https://www.youtube.com/watch?v=VHjmlQdzpW4&t=1099s
Why is the shape of fluorescence spectrum different between systems using 532nm and 785nm wavelength laser in Raman Spectroscopy? In the following using the 785nm RMX, the fluorescence is maximum at the left (starting at 100 cm-1) and minimum at the right (ending at 2750 cm-1).

While in the following where the sample is the coke itself using 532nm raman spectrometer, the fluorescence has the peak at center with each side at minimum (also started a 100 cm-)?

This is with the coke removed and spectra saved. The peak is at 2400 cm-1.

My question is that both started a 100 cm-1. But in the case of the 785nm raman, the fluorescence spectrum started at maximum already compared to the 532nm raman which has the peak at higher cm-1. Why?

This question is ill-formed and very incomplete, as it stands. Please re-do it, ditch the photo with the soda pop bottle, tell what the sample under test happens to be, and tell the location of the maximum in each spectrum. Try to facilitate having people help if they are of a mind to do so and have the time to spare.

The sample of the 532nm raman is the coke itself.

please see edited post. The peak is at 2400 cm-1 and the vertical at about 5000 counts.

This is my last advice: use toluene as the sample and see if you get the same highly detailed Raman spectrum as reported here: https://chemistry.stackexchange.com/q/46626/79678 . The spectrum was obtained using 532 nm excitation. Then use the "First Defender" RMX, that uses 785 nm excitation, to get a comparison Raman spectrum. Good luck!

Well. The first picture is from a video whose link I added in the edit. I haven't bought or owned the 785nm RMX or 532nm yet or any raman spectrometer. That's why I can't test them. I'm asking questions so I can decide whether to buy the 785nm RMX or 532nm model.

You need to do some serious thinking about just what you want (or need) to accomplish with a Raman spectrometer, even if just a (less expensive) handheld one. There are lots of factors to consider and homework to do. What you do not want is to pay tuition to the school of hard knocks and get stuck with an instrument that is woefully inadequate. If at all possible, try before you buy and pick a test substance with some nice sharp peaks, as given in a publication in the refereed literature: you need  to check resolution, sensitivity, S/N and so on. Good luck!

I live in a country where we don't have a single raman spectrometer. I just want to know for now why the fluorescence spectra varies between 532nm and 785nm wavelength laser. I spend many days reading in the internet about raman.

Might be well the detectors or whatever. Your 5000 counts or whatever might be very little in absolute terms. Raman spectrophotometers aren't fluorimeters at the end. Peaks might relate be harmonics and relate to the exc wavelength,  though I didn't check the values.

There is no reason why fluorescence emission background should look similar at all with green and red excitation.
Different substances

You say the green excitation spectrum is of Coke.
According to the video, the red excitation spectrum is of the peel of a tangerine.

Those samples consist of totally different substances so they will have different fluorescence spectra.
Different emission wavelength ranges
Assume this black line is the fluorescence emission spectrum of one sample:

the 532 nm excitation (dark green line) Raman spectrum is in the wavelength range 532 to 781 nm (green shaded area)
the 785 nm excitation (red line) Raman spectrum is in the wavelength range 801 to 1014 nm (red shaded area)

There isn't even any overlap between the spectral ranges you look at.
Different Excitation wavelengths
In first approximation (slight) changes in the excitation wavelength won't change the shape of the fluorescence emission, only the intensity.
However, if the excitiation wavelength changes sufficiently to excite a different fluorophore, also the shape of the fluorescence emission spectrum can (will) change.

About buying a Raman system
Before going deeper into the buying decision I suspect you'll need to learn more about the basics of Raman spectroscopy (and maybe about fluorescence). I recommend starting maybe with
some textbook chapters about spectroscopy for chemical analysis in general e.g. Skoog et al.: Principles of Instrumental Analysis if you directly like to have a focus on instrumentation (possibly also get some introductory text about spectroscopy for structure determination), and then to get deeper into Raman McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5.
Any serious advise about which system is better for you will need a whole lot of further background information: e.g.
what samples will you measure (phase, analytes, matrices), what applications (qualitative, quantitative, spatially resolved, restrictions on measurement time, lab use vs. field use, need for polarized measurements/depolarization ratio, ...),
will these specification change in future (other research projects).
As you comment that there is no Raman instrument in your country,

maybe you could visit a reputable lab that does Raman on similar systems as the one you want to study?
Raman system manufacturers are usually quite willing to demonstrate the ability of their instrument to measure what you need with a few of your samples, either by you sending them a couple of samples or even by you visiting them with your samples so you can try out the handling of the instrument yourself.
Do this at least with the few instuments are in your shortlist so you do have a basis for comparing actual performance for your samples.

A few comments about the Raman experiments in the linked youtube-video

Vodka-in-bottle spectrum: measuring through or on glass usually doesn't disturbn Raman spectra for green (e.g. 532 or 514 nm excitation) and also 633 nm red excitation is OK. With 785 nm excitation you'll get very bad background.
The reason why the handheld measurement works is its long working distance: the focus is deep inside the bottle, thus hardly any scattered light from the bottle gets to the detector.
Sugar Raman measurement background: not all background in Raman spectroscopy is fluorescence. White powders (that are actually clear substances with lots of refraction and reflexion due to the small particle size) will give a lot of elastically scattered light (i.e. still excitation wavelength). Depending on how good the spectrograph is, such scattered light may end up in the detector at the wrong pixel and thus be interpreted to have another wavelength.
Such scattering background will usually decrease the further you get from the Rayleigh line. A conclusive experiment to show that it is scattering and not fluorescence is: dissolve (or melt) the powder and deposit the substance as a clear film (after drying). Scattering will then (mostly) be gone.
Sugar vs. Sugar in fruit and the all-same-food spectra: solid sugar vs. sugar dissolved in water won't have the same vibrational spectrum (the spectra of some substances do not change much when dissolving in certain solvents, but sugar has tons of interaction with the surrounding water molecules due to its -OH groups - and that does affect the spectrum.)
As for the "almost all food has the same NIR spectrum": right in that wavelength range of 940 - 99 nm are overtones of -OH (water, carbohydrates, alcohols, H-bridges to oxygen). As all those fruits and veggies (including the peel) are in first approximation a mix of water and carbohydrate, there shouldn't be any surprise in those spectra being very similar.
With a good spectrometer and good data analysis (which includes good design of the calibration experiments) you can approach questions like "is this apple ripe for harvesting yet", but this requires the spectra to be available with a noise level (reproducibility) that allows measuring small shoulders of the absorption band.
Besides what the video discusses, there are a some other important things Tellspec claimed but cannot measure for physico-chemical reasons: mono-atomic ions such as Na⁺. They don't have molecular vibrations (again with super-good spectroscopic equipement you may be able to measure the effect of the ion on the water, but that's not going to work in practice). Which means, if you scan a soup and then add salt, it won't see the difference. Yet they claim to tell you sodium intake...

Hi how come the 785nm excitation starts from 801nm and not 785nm?

The data sheet of the handheld instrument gives the low wavelength end of the spectrum as 250 1/cm Raman shift. (For the green I couldn't read the axes properly, but my guesstimate was "close to 0")

In your experience. Are there many things you can't scan using 532nm that you can with 785nm? I read whole week and many references. That 532nm can't scan organic substance. Usually are most materials that is raman-active organic or not? I just want to learn raman spectroscopy and want to buy one for hands on...


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Resonance and mesomeric effect

Can you please explain to me why are A and E not relevant resonance structures of the molecule F?
And is B aromatic?

For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.
For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.

The relevant resonance structures of the given molecule are as follows.

Structure A is shown below. NItrogen with 2 sigma bonds is linear as shown below. However , nitrogen in the ring is strained with an angle of $\ce{120^o}$ (figure 1).

In figure 2 ,an unstrained Nitrogen with $\ce{120^o}$ bond angle is shown for your reference. Therefore , this structure $\ce{A}$ is irrelevant resonance structure .

Structure $\ce{E}$ is not a resonance structure of $\ce{F}$ or $\ce{1}$ in my structure. The carbanion in $\ce{2}$ is in conjugation with imine bond ($\ce{ C=N }$). This gives $\ce{3}$.

Structure $\ce{B}$ in not aromatic .


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What are Mulliken atomic charges, and how are they computed?
I was reading about the self-consistent extended Hückel method (SC-EHT), and stumbled upon various formulae for the diagonal elements of the Hamiltonian matrix. I noticed that many formulae depend on either the gross atomic population or Mulliken atomic charges. I know the formula for the Mulliken population analysis, but what is the Mulliken atomic charge and how can I evaluate it?
Thanks!

This is a helpful resource- http://cmt.dur.ac.uk/sjc/thesis_prt/node113.html

The charge results from adding up all orbital populations of an atom and subtract it from the nuclear charge.

Mulliken atomic charges can be defined as[1]:
$$q_A = Z_A-\sum_{\mu\in A}\left( \mathbf{P\cdot S} \right)_{\mu\mu} \tag{Szabo 3.196}$$
Here I have used the same notation as in Szabo[1], with $\mathbf{P}$ being the density matrix and $\mathbf{S}$ beign the overlap matrix. $Z$ is the nuclear charge, and using the greek letter $\mu$ as indicies indicates that we are working in atomic orbital basis (not in in molecular orbital basis).
The sum runs over $\mu\in A$ meaning, that we only consider atomic orbitals that are centered on the $A$th atom.
we can therefore note that Mulliken charges are only defined when we used atomic-centered basis functions (which we most often do).
[1] : Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Attila Szabo and Neil S. Ostlund


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Number of sterioisomers of 2,3,5,6-tetrachloroheptane

What is the total number of stereoisomers for the compound 2,3,5,6-tetrachloroheptane?

I figured the answer to be 16, but the answer given is 10.
Here is a similar question

Apparently, some of your 16 are the same thing spelled backwards.

but why 10 ..pls explain for better understanding

Your link is not similar. It is a tetrachlorooctane, not heptane. Write the 16 possibilities as XXXX using R and S. Now cast out the duplicates.

@Ivan Neretin, in his comment soon after your post, essentially solved the problem for you. In a comment I subsequently tried to set you in the right direction. Now that the ten structures have been posted in an answer and you have had time to think about a solution, here is the methodology to get the correct answer if you haven't solved it on your own.All of the 16 permutations are drawn as shown in the chart below. The 10 isomers are in black and the 6 duplicates are in red. Within each group, a red assignment has an identical black one. As an example, in the second group of four, RRSR = RSRR owing to the symmetry in the compound. That is, 2-3-5-6 = 6-5-3-2. The central group of six has "four meso structures", two of which are duplicates.
Note that the number of possible permutations in each group are the coefficients of each term in the expansion of (R + S)4.

I wrote only those that are optically active and have omitted meso and duplicates in my answer. However , I am well aware of these.I corrected my mistake.

Can you please give the reference of this answer (for example, name of source from which the concept of eliminating R-S left right similarity has been taken)?

@S Das: I have no source, just many years of experience. Imagine a model of the RSSS molecule mounted on the teacher's desk viewed by the class of students reading left to right. The teacher is on  the other side of the model. What does s/he see reading left to right? Has the molecule changed?

The given compound 2,3,5,6-tetrachloroheptane has 10 isomers. These have been shown below.

The compound 2,3,5,6-tetrachlorooctane in the link does not have a meso compound. The terminal groups, methyl and ethyl are different and therefore meso form is not possible.
The possible combinations have been given below, which are 16.


text:
Does an electronegative atom which carries a negative formal charge withdraw electrons by induction?
Electronegative atoms withdraw electrons by their inductuve effect. Does this fact remain true even when they are heavily loaded with electrons through negative formal charge? Are these two ideas related?

No, deprotonation changes things.


text:
Identify the type of structural isomerism
Question:

My Approach:
Since both the compounds have the same principal functional group, they can't be functional isomers. Obviously they are different compounds. Now options A and C are ruled out. Further the substituent ketone in first compound and aldehyde in the second compound have the same locant, so the given compounds cannot be positional isomers. So the compounds must be Chain Isomers. Further, we can just say, as there is difference in the chain length, the compounds are chain isomers. But the answer is incorrect.
The solution in my book, for this problem states, since compounds have different functional groups ketone in first compound and aldehyde in the second compound, they are functional isomers.
Now, I am totally confused. Please explain how to attend this question. Is the solution incorrect?

The solution is correct: if you have the same molecular formula but different functional groups, then you have functional isomers

@blupotatos, But here we have the same principal functional group - carboxylic acid, but different substituents. There is also difference in chain length. Can it be both chain and functional isomers?

Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. Constitutional or structural isomers are compounds with the same molecular formula but their structural formulas are different. They also called chain isomers because of the possibility of branching in carbon chains (e.g., n-butane and 2-methylpropane). Note that just the twisted versions of the original molecule can be mistaken by novices as structural or chain isomers (e.g., conformers).
Positional isomers: Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups. They differ from each other in the location of the functional groups on or in the carbon chain. In other words, in position isomerism, the basic carbon skeleton of isomers remains unchanged, but important groups on the skeleton are moved around on it (e.g., 1-bromopentane, 2-bromopentane, and 3-bromopentane).
Functional isomers: Functional isomers are constitutional isomers bearing different functional groups (e.g., propanoic acid, methyl ethanoate, and 3-hydroxypropanal).
Accordingly, the answer given is correct: These are functional isomers, since each have at least one different functional group changed (keto vs aldehyde).

Thank you for your answer. Here both the compounds have the same principal functional group, carboxylic acid. Aldehyde and ketone groups are just substituents on the basis of seniority order. Further I see difference in the principal chain length. So should I call it more like a chain isomer and less like a functional isomer.

See, you must use the concept of principal functional group only for nomenclature purposes. Here the compounds you mentioned in your question have two different groups in addition to carboxylic group, i.e., ketone and aldehyde. These compounds are chemically different. They show the traits of ketone as well as aldehyde in the respective compounds. So it is better to call them functional isomers. The answer in your book is correct. No need to worry about that.

If you could choose multiple correct answers then yes. Else you must go with functional isomers only. Otherwise both are correct.

The two compounds aren't identical by simple observation.
Now as far as chain isomers are concerned it cannot be the case as the functional groups are different. Chain isomers have same functional groups at same positions
They can be functional isomers as they do have different functional groups.
They can't be position isomers as well as the nature of functional group shouldn't change.


text:
How does the energy released during a bond formation typically manifest itself on atomic level?
I've been reading about how energy is released when new chemical bonds are formed but it's hard to find info on how that typically manifests itself (besides that it's via heat). I read somewhere that the valence electrons will move into lower energy orbits during a bond, but I thought that always created a photon.  How and where is the heat actually "created?". When two H atoms form H2, does the new molecule start moving faster (discounted for increased mass)? What happens if there are no surrounding molecules?

Electronically excited states transition into vibrationally excited states, which desactivate nonradiative by transfering energy to colliding neighbouring atoms, creating heat.

https://chemistry.stackexchange.com/questions/111936/how-is-energy-released-when-a-chemical-bond-is-formed

https://chemistry.stackexchange.com/questions/55914/how-is-heat-physically-released-in-an-exothermic-process

https://chemistry.stackexchange.com/questions/63597/how-does-a-diatomic-molecule-vibrate-in-response-to-heat

Second link is for duplicate imo.

Possible duplicate of How is heat physically released in an exothermic process?

As pointed out by @Paul in case of the reaction between 2 hydrogens atoms you need a third body to take away the excess energy. This third atom(M) can be another hydrogen molecule or other inert gases(eg. Ar, He, N2).

Or you can have the presence of a wall(eg. gas-phase hydrogens atoms in a container):

Another possibility is adding a reactive gas(eg.oxygen) and in this case, the mechanism is much more complicated with many steps.
All these reactions are discussed in ref.1.
Comment: when two hydrogens atoms combine 4.5eV is released. In the case of an inert third body, I haven't found any quantitative information about how much energy is transferred to the third inert gas. If all of it is transferred to the third body(eg.H2 molecule) as kinetic energy you end up with an H2 molecule moving at 20'800m/s that corresponds to a kinetic temperature of 34'800K!
Another example is the CsBr formation in the presence of a third inert body(R):

From ref.2:

Analysis of the detailed dynamics of recombination at small impact parameters showed that the stabilization of an ion pair occurs through the collision of the third body with the two ions and that the dynamics of energy transfer in this region is almost
  completely determined by the repulsive interactions of the Xe atom with both the ions.
In the range of impact, the main mechanism of stabilization is energy transfer energy through repulsion between the third body and both the ions or through consecutive collisions of the third body with the ions approaching each other.

Regarding how much energy is transferred to the third inert body:

The amount of energy transferred to the third body is apparently determined by the collision configuration and by the phase of the collision of the ions at the moment of energy transfer.

References

Bennett, J. E., and D. R. Blackmore. 1971. “Rates of Gas-Phase Hydrogen-Atom Recombination at Room Temperature in the Presence of Added Gases.” Symposium on Combustion 13 (1): 51–59.
Kabanov, D. B., and L. Yu Rusin. 2012. “Mechanism of the Direct Three-Body Recombination of Atomic Ions in a Central Collision.” Russian Journal of Physical Chemistry B 6 (4): 475–85.
Stepukhovich, Aleksandr Davidovich, and Viktor Markovich Umanskii. 1969. “Kinetics and Mechanism of Three-Body Recombination of Atoms and Radicals.” Russian Chemical Reviews 38 (8): 590.

What if there isn't a third atom?  How does the energy increase manifest itself in the H2 atom?  increased kinetic energy?  Thanks!


text:
Is the equation dG = Vdp − SdT valid only for a reversible process? Can it be applied for an irreversible one too?
In the derivation, we write:
$$
\begin{align}
G &= H - TS &\quad&\to&\quad \mathrm dG &= \mathrm dH - T\,\mathrm dS - S\,\mathrm dT \\
H &= U + pV &\quad&\to&\quad \mathrm dH &= \mathrm dU + p\,\mathrm dV + V\,\mathrm dp \\
U &= Q + W &\quad&\to&\quad \mathrm dU &= \mathrm dQ - p\,\mathrm dV \\
  & & & & &= T\,\mathrm dS - p\,\mathrm dV
\end{align}
$$
So solving these gives
$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$$
But what we actually write is
$$\mathrm dH = \mathrm dU + p_\mathrm{int}\,\mathrm dV + V\,\mathrm dp$$
And in
$$\mathrm dU = \mathrm dQ - p\,\mathrm dV = T\,\mathrm dS - p_\mathrm{ext}\,\mathrm dV$$
so on solving, it should be
$$\mathrm dH = T\,\mathrm dS - (p_\mathrm{ext} - p_\mathrm{int})\mathrm dV + V\,\mathrm dp$$
And only in reversible process, should it be valid to write
$$p_\mathrm{ext} - p_\mathrm{int} = 0$$
But my teacher argued that because this is a state function, it should be valid for all types of processes. I'm a bit confused about how his claim is valid. Can someone please explain what he means or even if he is correct or not?

Since you've already posted several (good) questions, I allow myself to add a couple of notes on formatting: please do not add hyphens after colons and stick with uniform notations, e.g. don't mix "$p$" and "$P$" for pressure (I personally prefer lowercase $p$, while $P$ is ambiguous and is also used for power).

The easiest way to avoid confusion is to start from the expression for the $pV$ work associated with a reversible process:
$$\mathrm dw_\mathrm{rev} = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV$$
where $p$ is the internal pressure of the system, and here the system is subjected to $pV$ work only. Yes, it is absolutely true that for reversible $pV$ work the condition $p_\mathrm{ext} = p$ holds. While you came to that conclusion, you did so in a flawed way because your expressions for $\mathrm dH$ are not correct.
The above definition of reversible $pV$ work leads to the expression for $U$ in terms of $pV$ work and entropy $(\mathrm dU = -p\,\mathrm dV + T\,\mathrm dS),$ and ultimately, through the definition $H = U + pV,$ to the expression $\mathrm dH = V\,\mathrm dp + T\,\mathrm dS.$

But my teacher argued that because this is a state function, it should be valid for all types of processes.

The laws of thermodynamics hold irrespective of path. Provided the end points are the same, two different paths will result in the same changes in state variables.
In practice, when attempting to determine the change in a state variable by evaluating a path integral, you want to choose a convenient path. This is usually a reversible one because then the work can be related to equation of state variables, and the entropy change can be related to the heat via the second law.

As far as I can understand is , you solve for reversible. How is my expression for dH wrong ?? . Because for work we are talking about external pressure. But for U, H its internal pressure . How can you just cancel that out if process is irreversible??

The expressions following where you write "but what we actually write..." are "incorrect" ($dH=dU+p_{int}dV +Vdp$ and the expression for dU). Well, depending on what you assume. As soon as you insert $dQ = TdS$ you implicitly assume reversibility, so $p_{int}=p_{ext}$. As for the expression for H, you are right, it involves the internal pressure, but again, you fall back to this implicit assumption $p_{int}=p_{ext}$ as soon as you insert an equilibrium expression implying $dQ = TdS$.  Try using the first law in its "raw" form instead and see that it also works.

dU = dQ - P(ext)dV : This is true for all Thermodynamic process
For reversible process
dQ = TdS
P(ext) = P(int)
So the equation becomes :
dU = TdS - P(int)dV
This is a equation in State functions only , so it will hold for all processes , both irreversible and reversible.
This is where you made the mistake .


text:
Coordinate bonds between molecules
For the following question

Which molecules react to form a dative/coordinate bond?
A. $\ce{CH4}$ and $\ce{NH3}$
B. $\ce{C2H2}$ and $\ce{Cl2}$
C. $\ce{NH3}$ and $\ce{HF}$
D. $\ce{Cl2}$ and $\ce{HF}$

the answer scheme says C is correct, but I do not see how two already complete molecules can form coordinate bonds. $\ce{NH3}$ has a lone pair of electrons, $\ce{HF}$ having none. How can a coordinate bond between this lone pair and $\ce{HF}$ happen? I am confused.
The others are also equally confusing.

I think the question means the H-bonding between HF (H donor) and NH3 (H acceptor)

@szentsas I wouldn't call it H-bonding. Wouldn't ammonia, being a lewis base, abstract a proton from HF?  In some low-level chemistry courses this happens to be a fundamental example for dative bonds.

I don't think you ever consider this a dative bond. If you transfer the hydrogen, it's definitely not dative. That is a new covalent bond. @WilliamR.Ebenezer

@Zhe and that's why it's called coordinate covalent bond. It is virtually indistinguishable from a regular coordinate bond. Kindly read: https://www.chemguide.co.uk/atoms/bonding/dative.html, and then pass your judgement.

@WilliamR.Ebenezer no need to be passive aggressive. The point that was being made was that, despite what chemguide says, the majority of chemists would consider something like $\ce{OC:}\rightarrow\ce{BH3}$ to be a dative bond, but not the N–H bond in $\ce{NH4+}$.

The fact that the two are indistinguishable is likely where I derive the strongest opposition. In a traditional dative complex like $\ce{BF3\cdot OEt2}$, there is no such ambiguity. Otherwise, you're starting to model some sort of partial dative bond (with resonance?), and it's no longer clear what that means. Moreover, the dative bond is a dative bond irrespective of how it was created (where your citation's argument falls short). When you look at the ammonium ion and you can't decide which bond is wholly belonging to the nitrogen, is any one of them really dative?

@orthocresol I apologise if I sounded blunt. No agression intended howsoever. I just wanted to point out that we have to sometimes recall some basic (and perhaps obsolete) perspectives of the dative bond that we tend to exculde when we attain knowledge of relatively sophisicated topics in chemistry. Only then can this question can answered. And Zhe, I agree with your perspective but also believe that this question might not benefit greatly from it. Excuse me if I have been naive.

Option C is correct.
You could start by looking up the definition of a coordinate bond.
Here's a tiny excerpt from the above link:

A coordinate bond is a kind of 2-center, 2-electron covalent bond in which the two electrons derive from the same atom.

In your case, as you have correctly noted yourself, the $\ce{NH3}$ molecule will have an "lone pair" of electrons. However, your judgement that $\ce{HF}$ has none is quite incorrect– the fluorine atom has $3$ lone pairs. (not that it's very relevant to the question, but you can look up the structure of $\ce{HF}$ here)
Since the $\ce{NH3}$ is a lewis base, its lone pair of electrons would abstract the $\ce{H}$ from $\ce{HF}$.

$\ce{NH4+}$ has four equivalent bonds. You could call one of them the "dative" bond, but there's really no way to distinguish them from each others in general cases.
The concept of dative bond comes here just because the bond between nitrogen and the hydrogen from $\ce{HF}$ involves 2 electrons (like all single bonds), both of which are contributed by the nitrogen atom (unlike most single bonds).
(Image source)


text:
What is the correct definition for the activation Gibbs energy?
i found several definitions of the activation energy in electrochemistry, but i dont know why they are different. For example in Atkins "Physical Chemistry" the activation gibbs energy is defined as:
$\Delta^‡G_c=\Delta^‡G_c(0)+\alpha F\Delta\phi$
and he assumed that $\Delta\phi$ is equal to $E^{eq}$ when no current flows through the battery.
In the Book "Electrochemical Methods" of Bard and Faulkner another definition ist used:
$\Delta G_c^‡=\Delta G_{0c}^‡+\alpha F(E-E^{0'})$
$E^{0'}$ is the formal potential.
I have also found a possible explanation in the Book "Electrochemical Dictionary"
, but I do not quite understand. The following sentence is in the book:
"If a reference electrode is used the Galvani potential difference can be replaced by $(E-E^{0'})$, i.e., with the difference of the electrode potential related to a reference electrode and the formal potential"
. But Why we need the reference to the formal potential?
I would be glad if someone can explain me why the used different definitions?

Related: https://goldbook.iupac.org/terms/view/G02631


text:
Can five membered ring form through Friedel-Crafts alkylation?
Our teacher told us that five membered ring can’t be formed through Friedel-Crafts alkylation but acylation is OK. Why is that? Is it because alkylation is an unfavored 5-endo-trig reaction but acylation is 5-endo-dig, thus favored?

I don't understand the premise of this question. Why do you consider either ring closure an endo cyclization?

Maybe your teacher means to say that because further reactions would happen due to the further activation of the ring due to the alkyl group as opposed to the deactivation due to the acyl group.
Also further catalyst induced polymerization reaction can occur apart from possible polyalkylation
In general alkylation on rings are not preferred because of the above reasons as well


text:
Convert lactose in milk to Galactooligosaccharides
There are few articles describing the conversion of lactose to GOS on Wiley Online Library with different enzymes. Yet none of them really talks about what % of the lactose in milk can be converted to GOS.
Is it possible to convert all lactose to GOS, and still be able to consume the milk/dairy product?

It may help to link the articles you have already read through, just so answers don't have to search through the same articles.


text:
Finding mathematically the ground state density in DFT
To find the ground state density in DFT, you set the following Lagrangian:
$$L = E[\rho(\vec r)] - \mu\left(\int \rho(\vec r) \mathrm{d}\vec r - N\right)$$
While minimising with respect to the density, we get:
$$\frac{\partial L}{\partial \rho} = \frac{\partial E[\rho(\vec r)]}{\partial \rho} - \mu = 0$$
My problem is with the explicit derivation of the energy functional. Let's consider the Thomas-Fermi-Weizsacker kinetic energy functional:
$$T_s(\rho) = C_F \int \rho^{5/3}(\vec r)\mathrm{d}\vec r + C\int \frac{| \nabla \rho (\vec r)|^2}{\rho (\vec r)}\mathrm{d}\vec r$$
For example, how do you derive this with respect to the density?
$$\frac{\partial T_s(\rho)}{\partial \rho} = C_F \frac{\partial}{\partial \rho}\int \rho^{5/3}(\vec r)\mathrm{d}\vec r + C\frac{\partial}{\partial \rho}\int \frac{| \nabla \rho (\vec r)|^2}{\rho (\vec r)}\mathrm{d}\vec r$$
For the first term, you could use Leibniz integral rule, since the density very far away from the nucleus is zero. However, I really have no idea how to tackle the second term. Moreover, what is the algorithm afterwards to find the ground state density? How does one figure out this density from the equation?

Pople's paper on KS-DFT in a finite basis set may be a starting point for you: https://doi.org/10.1016/0009-2614(92)85009-Y

What we have to do here is a functional derivative.
Lets consider the following method of finding functional derivartives:
$$F[\rho+\delta \rho]-F[\rho] = \int_\Omega \delta\rho \frac{\delta F[\rho]}{\delta\rho}\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{1}\label{FuncDer}$$
Lets now consider the Thomas-Fermi kinetic energy functional using the above equation.
The Thomas-Fermi functional is given as:
$$T_\mathrm{TF}[\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}\tag{2}$$
Lets now try to evaluate $T_\mathrm{TF}[\rho+\delta\rho]$:
$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\left(\rho+\delta\rho\right)^{5/3}\mathrm{d}\mathbf{r}\tag{3}$$
Now lets do Taylor expansion of the integrand:
$$T_\mathrm{TF}[\rho+\delta\rho] = C_\mathrm{TF}\int_\Omega\rho^{5/3}+\frac{5}{3}\rho^{2/3}\delta\rho+\mathscr{O}(\delta \rho^2)\mathrm{d}\mathbf{r}\tag{4.1}$$
$$T_\mathrm{TF}[\rho+\delta\rho]= C_\mathrm{TF}\int_\Omega\rho^{5/3}\mathrm{d}\mathbf{r}+\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{4.2}$$
Now subtracting $T_\mathrm{TF}[\rho]$:
$$T_\mathrm{TF}[\rho+\delta\rho] - T_\mathrm{TF}[\rho]=\int_\Omega C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta \rho^2)\tag{5}$$
Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:
$$\frac{\delta T_\mathrm{TF}[\rho]}{\delta\rho}=C_\mathrm{TF}\frac{5}{3}\rho^{2/3}\tag{6}$$
Now for the second part of the kinetic energy functional in your question, i.e. the Weizsacker kinetic energy functional:
$$T_\mathrm{W}[\rho]=C\int_\Omega\frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r}\tag{7}$$
Again lets evaluate $T_\mathrm{W}[\rho+\delta\rho]$:
$$T_\mathrm{W}[\rho+\delta\rho]=C\int_\Omega\frac{\nabla(\rho+\delta\rho)\cdot\nabla(\rho+\delta\rho)}{\rho+\delta\rho}\mathrm{d}\mathbf{r}\tag{8}$$
Now doing a Taylor expansion again:
$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho} + \frac{-2\nabla^2\rho+\frac{\nabla\rho\cdot\nabla\rho}{\rho}}{\rho}\delta\rho+\mathscr{O}(\delta\rho^2)\mathrm{d}\mathbf{r}\tag{9.1}$$
$$T_\mathrm{W}[\rho+\delta\rho] = C\int_\Omega \frac{\nabla\rho\cdot\nabla\rho}{\rho}\mathrm{d}\mathbf{r} + \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{9.2}$$
Now subtracting $T_\mathrm{W}[\rho]$:
$$T_\mathrm{W}[\rho+\delta\rho] - T_\mathrm{W}[\rho] = \int_\Omega C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\delta\rho\mathrm{d}\mathbf{r}+\mathscr{O}(\delta\rho^2)\tag{10}$$
Thus by comparing with Eq. (\ref{FuncDer}) it can be seen that:
$$\frac{\delta T_\mathrm{W}[\rho]}{\delta\rho} = C\left(-\frac{2\nabla^2\rho}{\rho}+\frac{\nabla\rho\cdot\nabla\rho}{\rho^2}\right)\tag{11}$$
Now for the second part of your question.
For molecular calculations involving kinetic energy functionals you have to look into orbital-free density functional theory (OF-DFT).
The kinetic energy functionals are not used in Kohn-Sham density functional theory (KS-DFT).
In KS-DFT the kinetic energy are calculated from the Kohn-Sham orbitals instead of the density.
Often when working with OF-DFT the equations can be solved numerically using finte element method (FEM) for a nice introduction to this take a look at Joel Davidsson master thesis [1].
Keep in mind that doing OF-DFT calculations is not a trivial task!
[1] : https://www.diva-portal.org/smash/get/diva2:864857/FULLTEXT01.pdf

Do you know of any other theses or relevant material in the subject? This thesis was a very very interesting read!

I am afraid not, I have only look at OF-DFT very briefly were I had to implement in Python using Fenics during a partiel differential equation course. @Raphaël
I mainly just uses the thesis I linked to as a reference for what I needed to do.


text:
Reacting ZnSO4 with KCl
I'm considering the following reaction:
$\ce{ZnSO4} + 2 \ce{KCl} \ce{->} \ce{ZnCl2} + \ce{K2SO4}$
The high solubility of $\ce{ZnCl2}$ would favour the $\ce{ZnSO4}$ & $\ce{KCl}$ direction of the reaction, but I'm wondering if I could turn this around by making the reaction occur in a flow of boiling hot water.
The idea would be to subject (initially dry) $\ce{ZnSO4}$ & $\ce{KCl}$ stationary phase to a flow of water in the 80-100°C range. The comparative solubility of $\ce{ZnCl2}$ at this temperature being high, I'm wondering if it being comparatively flushed away more than the other species could help drive the reaction forwards in the opposite direction?
I wouldn't get pure $\ce{ZnCl2}$ of course but I'd be happy if I could get a 90% $\ce{ZnCl2}$ solution from dry $\ce{ZnSO4}$ & $\ce{KCl}$.

There is no reaction, nor is there a need for one. A solution of ZnSO4 and KCl is the same as solution of ZnCl2 and K2SO4.

Is it not called a double-replacement reaction?

It would be, if some of the participating salts were insoluble. That's not the case, though.

If one of the salts can be otherwise partly isolated from the reaction medium, is it not somewhat equivalent to one being insoluble? Could the "reaction" proceed by one of the products being comparatively carried away many times more than the reagents combined?

It is lower, not higher solubility, that might separate one of the products from the reaction mixture and thus drive the reaction that way. With all four salts rather well-soluble, we're out of luck.

Yes, I understand that (if the solvent is stationary). But would it not change if the solvent was limited and moving across a stationary, mostly undissolved mix of the salts?

That would be a setting too exotic to tell for sure.

Would you even get pure zinc sulfate and potassium chloride again if you dissolve them both in water and then boil them off?  I'm guessing since zinc chloride is more soluble, there would be more zinc sulfate that forms though...

@chemN00b Of course, but this can only happen to the limit of existing sulphate ions, which as you correctly write are less soluble than chloride ions and thus would only be a minor part of the "washed-out" solution. And they can actually be somewhat further separated by the drying process you refer to (initial precipitate would be richer in sulphate ions and later precipitate richer in chloride and potassium ions).


text:
Are there any forbidden collisional electronic transitions?
There are spectroscopy selection rules marking some radiative electronic transitions as forbidden. For example, it is stated that single electron of H atom can not move from S2 to S1, or from P2 to S2 (and vice versa) with photon emission/absorption.
But is this true in the case of collisional (or, more generally, radiationless) transitions? Are there any selection rules and what they are if any?
I found the Wigner spin conservation rule. Are there any more rules?

I think what you are asking about is pressure broadening: https://en.wikipedia.org/wiki/Spectral_line#Pressure_broadening . From my experience in NMR, all transitions are allowed during relaxation induced by weak interactions with the "lattice." Not sure how this translates when looking at electronic spectroscopies.

The photon carries a unit of intrinsic angular momentum which, with spin conservation, leads to selection rules of a change of 1 in angular momentum in optical transitions in the atom, so $s\to p,;p\to d$ allowed but not $s\to s;; p\to p;,s\to d$ etc. In a collision, as generally the partner has no intrinsic angular momentum (excludes rotation) then there should be no other restrictions other than that to conserve spin.

@porphyrin It makes sense. And can the partner bring/take enough rotational angular momentum to cause changes like s→p etc?

I think that that is unlikely to influence electronic transitions, however, at high pressure 10 to 100 atm otherwise 'strictly forbidden' bands are observed in what have been called 'pressure induced' bands. This may be due to the mutual distortion of the charge distribution of the molecules because of interaction at close range, and  electric dipole selection rules are broken.


text:
How to estimate the maximum chlorine hazard from 1 g of Palladium(II) chloride?
I plan to work with an electrolyte that contains 1 g of Palladium(II) chloride. I am trying to estimate how much of a hazard this system could possibly pose (and provide ventilation, chlorine monitoring and other safety measures accordingly).
How would you go about determining what max. ppm concentration of chlorine might one need to expect in this case?

The question cannot be answered as is because determining ppm concentration would require knowledge of the approximate volume of your work space. With that information, it's basically a stoichiometry problem.

What is of interest is the total maximum amount of chlorine released (a), the rate of it being released with an uncertainty range (b) and the volume which it is released into. One could create a simple formula from (a), (b), and (c) and learn about their relationships even if (b) and (c) are unknown or uncertain. In other words, just treat the volume as unknown variable x; or assume an enclosure of volume 50 liter if that is easier.

Just assume all of the chlorine is released at once and you should have a upper bound.

What's the point? It's not like magically whole thing would instantly turn into the gas and you'd breath in whole. In labs there are much more serious risks.

The point is exactly to estimate that risk. If you think there is no serious risk then that's great; please articulate to people less experienced in this area why that is and how to think about.

There is no hazard whatsoever. There is no need for any safety precaution.

Consider the molecular mass, and the mass of chlorine in 1 g of $\ce{PdCl2}$. Then consider the volume of the work space, and how quickly the $\ce{Cl2}$ disperses to fill that space.
Until it disperses, the concentration at the source might be 1E6 ppm (i.e. 100%), but is that meaningful? If it disperses to fill a flask, what is that volume? Are you doing this under a fume hood?
BTW, if no hood is available, you could enclose the reaction vessel(s) in a container with some sodium thiosulfate, $\ce{Na2S2O3}$, solution, which was used to absorb chlorine in gas mask canisters.

The sodium thiosulfate sounds like a great idea. It sounds like it would also help with any smell issues. Do you think it may be sufficient to just have an open dish with some  sodium thiosulfate near the electrolysis cell inside the same enclosure (which is a 50 x 50 x 50 cm enclosure)?

Is it a true enclosure (e.g. glove box)? In that case, a large pan (lots of surface area) or a piece of cloth soaked in the solution might suffice.  If it's a fume hood, there should be no need for the thiosulfate.

The enclosure is basically a temperature-controlled fridge. It's not 100% airtight but close. I could add some basic air exchange but it would complicate things. I'll probably go with the large pan of sodium thiosulfate and accept that small amounts of Cl2 will gradually leak out into the ambient (which is reasonably well ventilated).


text:
Which compound has resonance structures?

Which compound has resonance structures?
A. $\ce{C6H12}$
B. $\ce{CH3CHO}$
C. $\ce{NaBr}$
D. $\ce{Na2CO3}$

The answer key says D is correct, but I do not understand why $\ce{Na2CO3},$ which is an ionic compound between the polyatomic ion carbonate and sodium, can have resonance structures.
The other choices are equally confusing.

Draw the structure of the carbonate ion, and you'll see.

Arguably, A and B both have resonance structures, although those structures are not as good compared to what you could draw for D.

@Zhe what are the resonance structures in A and B? I can only see possible hyper conjugation in B but nothing in A

Technically speaking, you can just break a carbon-hydrogen or carbon-carbon bond and create an ion pair. You might do this to describe hyperconjugative stabilization of a cation or a 1,2-group shift. Like I said, it's not as good of a resonance structure as for D. Note that I didn't make this suggestion for C because that's almost purely an electrostatic interaction. But if you wanted to go nuts, draw a covalent bond between those two atoms. It is a resonance structure for the standard charge separated ion pair.

@Zhe ok I get your point

As you can see in this structure 3 equivalent resonance structures are possible by simple conjugation between negative charge and the π bond. Hence being covalent is not a condition for possible resonance even compounds with some ionic character can show resonance besides all compounds have some ionic and covalent nature both present.
In the other options there is no scope for resonance as none of the resonance conditions are formed
Source- https://commons.m.wikimedia.org/wiki/File:Carbonate-ion-resonance-2D.png


text:
Do external electronic shells influence internal ones?
There is a well-known electron shielding effect: negatively-charged internal electronic shell screens external shells from positive nuclei, thus increasing their radii.
Thus, 2S orbital radius of Li atom (ground state) should be greater than radius of excited Li2+ cation with single 2S electron.
But is there a reverse influence: from external electronic shells to internal ones? For example, are radii of 1S orbital in Li and Li+ equal, or not?
I am sure there is not a big difference anyway, but this is theoretically interesting. Probably it can be proved by comparision of internal shell ionisation energies of Li and Li+ but I am not sure where to look for appropriate experimental data.
I have some intuitive considerations, while they did not lead me to certain answer:

Concept of Faraday cage looks to be irrelevant, because electron
shells are not literally material conductive shells, but they are
filled space regions.
Shells interpenetrate, thus external shell
can a bit force out internal shell from nucleus by increasing
electron density in space  adjacent to nucleus.
Classically,
internal one of two concentric  elastic charged spheres should
shrink. If we imagine big spheres, their small areas should interact
repulsively as parallel plates. But considered spheres are actually
not "big".

If you like quantum mechanics, go to the source: Quantum Mechanics of One and Two Electron Atoms by Bethe and Salpeter. It will tell you a lot about electron-electron interactions.

Interesting question... I'd expect your third bullet point to correspond to a Faraday cage, which might be right to the first order. More subtle effects may be due to magnetic interactions, which are not shielded (a compass will work inside a Faraday cage).

Not only do "outer" electrons impart some shielding to "inner" ones, but an analytical technique is based on that fact.
In X-ray photoelectron spectroscopy, binding energies of electrons just below the outer shell are measured.  These binding energies are typically hundreds of electron volts, roughly 280-300 eV for carbon as a typical example.  An important feature is that this binding energy is measurably not fixed for a given element, rather it varies according to how the atom is chemically bound.  As an example, the 1s electrons in carbon lie at about 285 eV binding energy in hydrocarbons, but shifted by as much as 8 eV by bonding to fluorine (see here), which is more electronegative and draws electrons away from the carbon 2s and 2p  shells.  Similarly, the 1s electrons in oxygen range from about 529 to 534 eV in various oxides, with the higher values tending to occur in nonmetal oxides (see here).


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Chirality in carbocations
The following compound is a carbocation.

Clearly, the given compound does not have any element of symmetry. The compound is non planar due to the presence of ethyl group. Thus it must be chiral. But my book says the compound is achiral. How is this possible? Please explain.

It is planar.$,$

@IvanNeretin, The carbocation carbon is $sp^2$ hybridised but the ethyl carbon is $sp^3$ hybridised, so the leftmost methyl group may be out of plane right?

The entire compound is not planar, but the bond geometry around the positive carbon is (trigonal) planar. All orientations of the ethyl group are equivalent.

@electronpusher, So the compound has plane of symmetry(plane along the screen) and thus is achiral right?

Yes, that's right.

We must be checking the planarity only at the carbocation carbon and treat every other substituents as simply a spherical atom. I thought different conformers in the ethyl group leads to non-planarity, which can cause the system to exist in left and right handedness, thus chiral. But it seems too different. We must be considering only one carbon.

@GuruVishnu Essentially correct. You may want to look into the difference between Rotational Isomers and Optical Isomers.


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Lattice Energy of MX2 type
While finding the solution to the question that which among BaO and BaCl2 has hi that is energy I found that it was BaO whose lattice energy is high and there were explaining it just with charges. So I am bit confused that why did they not consider the the fact that there were 2 Cl atoms in BaCl2


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How to remove the foggy discoloration in polycarbonate tubes caused by acetone?
Unknowing that the tubes used in the centrifuge are made of polycarbonate, I cleaned them with acetone. Following, a white smoky discolouration appeared and the tubes turned opaque. Is there any way to remove this?

I think it's hopeless.

About the only way I can think of is throw out the old ones and use new ones.

I agree with the comments, but maybe it does not matter much? If the tubes are completely opaque, then you need new tubes.