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What is the relation between solubility and thermal stability?
The thermal stability of alkaline earth metals increases down the group for hydroxides i-e., Be(OH)2 is less stable than Ba(OH)2.
The solubility also increases down the group for these compounds i-e., Be(OH)2 is less soluble in water as compared to Ba(OH)2. Hence for Hydroxides of group 2 elements the solubility and thermal stability trends are same i-e. Both increases down the group. But the solubility and thermal stability trends become inverse of each other for the "sulphates of group 2 elements". Does this mean that there is no relation between solubility and thermal stability?

There's no reason to think there would be such relation, imo.

Nobody is able to foresee the solubility of a product. There are some experimental rules, but they all have exceptions, that nobody is able to explain.
Just have a look on the Calcium salts made with the halogens (F, Cl, Br, I). There is a nice analogy among Cl, Br and I, but not F. Look ! The Calcium chloride $\ce{CaCl2}$, bromide $\ce{CaBr2}$ and iodide $\ce{CaI2}$ are all extremely soluble in water. They are all soluble in less than their weight of water. But calcium fluoride $\ce{CaF2}$ is among the least soluble product on Earth. The principal mineral for Fluoride is $\ce{CaF2}$, and it can be found everywhere at the surface of the Earth. If this mineral would have been at least a little soluble in water, the rains would have washed away this mineral in the geological times. Why is there such a huge difference between calcium fluoride and the other halogenides ? Nobody knows !
Another example. Potassium perchlorate is the only potassium salt which is very weakly soluble in water. By comparison, Sodium perchlorate is extremely soluble in water. Why? Another example: the number of Silver compounds which are soluble in water is limited. In organic solvents, it is even worse. But the Handbook of Chemistry and Physics says that Silver perchlorate is soluble in toluene. Why? Nobody knows.
From time to time there are articles published in journals like the Journal of Chemical Education. The author is proud of displaying a theory filled with new parameters, for explaining the solubility of quite a lot of chemicals. But there are always exceptions, that he regrets not to be able to explain.

While solubility, thermal stability and density are pretty hard to explain as @maurice said. But, there is good logic for the solubility of group 2 hydroxides. The lattice energy of a solid is inversely proportional to the radius ratio (as lower the radius ratio greater the packing efficiency). So, hydroxide being a small ion, makes a more stable aqueous solution with beryllium than barium.
Similarly for the sulphates, sulphate is a large ion, hence, beryllium has more stable sulphate.
This radius ratio logic can explain solubility of nearly all group 1 and group 2 salts
(I say nearly because of carbonates and hydrogen-carbonates, which follow a reverse trend)
For a more detailed explaination see this (1)
(1) https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/1_s-Block_Elements/Group__2_Elements%3A_The_Alkaline_Earth_Metals/1Group_2%3A_Chemical_Reactions_of_Alkali_Earth_Metals/The_Solubility_of_the_Hydroxides%2C_Sulfates_and_Carbonates


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MOPAC: can we run THERMO immediately after the geometry optimzation?
With MOPAC, I am trying to set up single run with geometry optimization followed by calculation of THERMOdynamic parameters.
Is there viable way for that?

It will surely be possible by bash scripting, but you should check the optimized structures are sensible, converged and actually the conformer you are interested in.

OK, in between I found it, it is the GEO_DAT keyword, for example, reading in previously converged PM7 geometry of methane:
 AUX LARGE CHARGE=0 SINGLET  GEO_DAT="metan_geomopt_pm7.arc" THERMO(298,298)  PM7

source: Stack Exchange

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

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Explanation of the third ortho effect
The third ortho effect is, according to Wikipedia, the phenomenon where in a ring with a meta and an ortho-para director both meta to each other has the incoming nucleophile go ortho to the meta director rather than para. Why is this? I would think that the ortho-to-meta position would be destabilized due to a stronger inductive effect.

There was no good explanation of this effect known as of 2001, according to this textbook: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6236433/
I don't know if one has been found since.

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


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Extraction Techniques for Organic Compounds from Aqueous Media
I am looking to attempt an extraction of an organic compound that had just been prepared using a Friedel-Crafts acylation. The compound was prepared in MDC (~20 mL in total). After the reaction, the organic compound was poured into ~20 mL of ice-cold distilled water. The protocol I am referring to states

"The reaction mass was cooled and water was added slowly under stirring condition. Product was extracted in MDC (12 X 2) and washed with water followed by treating with saturated sodium chloride solution (20 mL) ... "

I don't want to make any inferences as to what this means since I am new to synthetic lab work, but my guess is that it means 2 portions of 12 mL of MDC were used to separate the compound from the aqueous phase. I am not sure why they follow this by washing with water unless I am interpreting what "Extracted in MDC (12 X 2)" means. Similar sections in the paper include a little more detail that confirms my original speculations regarding what is meant by (12 X 2), but it still does not make sense to me why this would be followed by washing with water and not skip immediately to washing with a saturated NaCl solution. Here is the DOI to the paper: https://doi.org/10.1016/j.bmc.2018.05.016. The section I am looking at is 2.1.4.

Sorry for this seemingly trivial question. I am just looking to gain a more complete understanding of reading experimental sections, as I have not had much experience with this.

You are reasoning as if DCM and MDC are the same substance ? I am not sure. DCM may be dichloromethane. But MDC ?

MDC and DCM are the same (methylene dichloride and dichloromethane)

Sorry for the confusion, I did not realize that I used two different acronyms until I had posted the question. I will change it to be more clear.

The backwash with water is to remove any inorganics that may have been carried through into the DCM product solution. It is a good idea to keep all your aq wash fractions until you have isolated your product. If your yield is low that is where the rest of your product is.

Is the amount of DCM used in the reaction a big deal? The protocol just states that the reactant was dissolved in DCM (doesn't specify how much). I went with ~20 mL to dissolve the product in the reaction flask since this is about how much is used in the 2 consecutive 12 mL DCM washing steps.

The amount of DCM doesn't really matter as it will be removed under reduced pressure. I rather doubt they precisely added 12ml each time. 20ml washes would be perfectly OK


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Electron Ionization and the Franck-Condon Diagram: vibrationally excited and vibrationally ground states
I am currently studying Mass Spectrometry: A Textbook, third edition, by Jürgen H. Gross. On page 41, the author says the following:

As explained by the Franck-Condon diagram, hardly any molecular ions will be generated in their vibrational ground state. Instead, the majority of the ions created by EI is vibrationally excited and many of them are well above the dissociation energy level, the source of this energy being the 70 eV electrons.

On page 37, the author presented the following Franck-Condon diagram:

In this diagram, the energy is plotted on the ordinate, and the bond length is plotted on the abscissa.
My understanding is that the vertical transitions represent the electron ionization of the neutral $M$.
Following this diagram, on page 38 the author said the following:

The counterpart of the vertical ionization is a process where ionization of the neutral in its vibrational ground state would yield the radical ion also in its vibrational ground state, i.e., the $(0 \leftarrow 0)$ transition. This is termed adiabatic ionization and should be represented by a diagonal line in the diagram. The difference $IE_{\text{vert}} - IE_{\text{as}}$ can lead to errors in ionization energies in the order of 0.1 - 0.7 eV.

So, If i'm correctly interpreting the description below the Franck-Condon diagram, the only way for the molecular ions to be generated in their vibrationally ground state is if the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?
If my interpretation so far is correct, then two related questions immediately follow:

Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?
How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)

I'm studying this from a non-chemistry background, so I'd appreciate it if people would please take the time to explain this in a more novice-friendly way.

When the textbook says "hardly any molecular ions will be generated in their vibrational ground state", I think that implicitly assumes the electron ionised is not non-bonding (the details are too long to fit into a comment, but loss of a non-bonding electron does not significantly change the wavefunction, so the Franck-Condon principle means that the most probable transition is to the same vibrational state, as that has greatest overlap with the original wavefunction). I'd be interested as to others' thoughts.

@atbm Hmm, what does the electron ionised being non-bonding have to do with this?

The FC principle means that the probability of transition is dependent on the overlap between the initial and final vibrational wavefunctions. If the electron removed is non-bonding, then there is little change in the wavefunction so overlap between the initial and final vibrational ground states is almost 1 (and, because the eigenfunctions of the Hamiltonian are orthonormal, overlap with the other wavefunctions is almost 0). So the ion would be most likely to be generated in the ground state.

@atbm Ahh, interesting. There seems to be something implicit in this that I'm not sure I'm understanding correctly: Are you saying that the neutral molecule M (the original molecule) is in a vibrationally excited state? Because, based on what you just said, it seems that this is implied, since, as the author stated, most of the molecular ions generated will end up in their vibrationally excited state?

I'm assuming the neutral molecule is in the ground state - if it's in, say, the 1st excited state the molecule would be generated in the 1st excited state. This assumes the electron removed is non-bonding - which would mean the internuclear distance doesn't change, hence the link to your question. If a bonding electron is lost, the bond length will increase as in your example.

@atbm But does this not contradict what you just said? If the FC principle means that the probability of transition is dependent on the overlap between the initial and final vibrational wavefunctions, and the molecule M is originally in the ground state, then would we not expect that most of the molecular ions will be generated in their vibrational ground state?

Let us continue this discussion in chat.

Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?

It's not so much about whether the bond length in $\ce{M^.+}$ is larger or smaller than that in $\ce{M}$; it's more about whether it's different. If you think pictorially about the vibration of a diatomic molecule or a spring, it makes sense that as you increase the energy (i.e. go to higher vibrational states), the amplitude of the vibration increases. The only way for a molecule to access bond lengths that are substantially different from its equilibrium bond length is therefore to be in a vibrational excited state.
Ionisation of $\ce{M}$ effectively always takes place from the vibrational ground state, so the initial bond length will always be very close to the equilibrium bond length of $\ce{M}$. If ionisation is almost instantaneous, such that the bond length cannot change during the ionisation, then $\ce{M^.+}$ is going to be generated with a bond length that is far away from its own equilibrium. This corresponds to an excited vibrational state. Whether it's larger or smaller than equilibrium doesn't matter; it just matters that it's different.
Of course, to say that this is the case presupposes that the equilibrium bond lengths of $\ce{M}$ and $\ce{M^.+}$ are substantially different. But this appears to be the assumption that the author makes. For more insight into this, @atbm's comments are spot on: the change in the equilibrium bond length will depend on the bonding character of the electron being removed. If a bonding electron is removed, then the bonds holding atoms together are weakened, and consequently $\ce{M^.+}$ will have a larger equilibrium bond length than $\ce{M}$; similar considerations hold for nonbonding and antibonding electrons.

How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)

I suspect that the FC principle is no longer relevant to the adiabatic ionisation process. Adiabatic processes are associated with "slow" transitions: see e.g. https://en.wikipedia.org/wiki/Adiabatic_theorem, whereas the FC principle is to do with "instantaneous" processes where there is no possibility of the bond length changing during the ionisation.
This is not my area of expertise, but I will go out on a limb and say that the adiabatic ionisation process is largely hypothetical (at least in the present context). Wikipedia writes that

However, due to experimental limitations, the adiabatic ionization energy is often difficult to determine, whereas the vertical detachment energy is easily identifiable and measurable.

I am not sure what "error" the author is talking about, but my interpretation is that the adiabatic ionisation energy is the "true" or "actual" ionisation energy of the molecule; however, because physical ionisation does not take place in an adiabatic fashion, the measured ionisation energy differs from the "true" ionisation energy by ca. 0.1–0.7 eV.
So I would not interpret this as being two different physical processes that actually occur in the mass spectrometer; rather, it seems to be more of a thought experiment to me. Of course, I am open to being corrected on this.

Thanks for the excellent answer! I’m going to make some comments as I read through your answer.

This agrees with what you said regarding how quickly ionization occurs: "According to the Born-Oppenheimer approximation, electronic motions and nuclear motions can be separated due to the large mass difference between nuclei and electrons. Furthermore, the Franck-Condon principle states that electronic transitions will occur on a much faster timescale than it takes the nuclei to move to their new equilibrium positions. [...]

[...] Applied to the interaction of an energetic electron with a gaseous molecule this means that the positions of the atoms and thus bond lengths remain unaltered during ionization. In diagrams where energy is plotted on the ordinate and bond length on the abscissa, such transitions are represented as vertical transitions."

The author then seems to emphasise the bond lengths becoming longer: "No matter where the electron has formally been taken from, ionization tends to cause weakening of the bonding within the ion as compared to the precursor neutral. Weaker bonding means longer bond lengths on the average and this goes with a higher tendency toward dissociation of a bond." Do you think that this is true most of the time?

This might initially seem contradictory, but I don't think it is. The first quote says that the bond lengths remain unaltered during ionization. The second quote then, if I'm interpreting it correctly, seems to emphasize that the bond lengths usually become longer after ionization.

The author provides the following metaphor, which I think does well in helping us conceptualize this phenomenon: "To understand the situation of the molecule imagine an apple through which a bullet is being shot: the impacting bullet passes through the apple, transfers an amount of energy, tears some of the fruit out and has long left it when the perforated apple finally drops or breaks into pieces."

The one question I have is regarding your response to my second question. It seems to me that you answered my second question in your response to my first question; specifically, you said the following: "The change in the equilibrium bond length will depend on the bonding character of the electron being removed. If a bonding electron is removed, then the bonds holding atoms together are weakened, and consequently $\ce{M^.+}$ will have a larger equilibrium bond length than $\ce{M}$; similar considerations hold for nonbonding and antibonding electrons." [...]

[...] Does this not account for the situation where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? I ask because it seems that you answered my second question by remarking upon how you think that the FC principle is no longer relevant to the adiabatic ionisation process, but it seems to me that your aforementioned response already answers my second question?

Regarding the seemingly contradictory quote, your reading of it is fairly accurate. The equilibrium bond length of M•+ is (typically) larger than the equilibrium bond length of M. However, because of the FC principle, during the physical process of ionisation, the bond length does not change: so immediately following ionisation, the bond length of M•+ is far away from its equilibrium, which corresponds to an excited vibrational state. If M•+ were to relax back to its ground state, then the bond would indeed lengthen. However, sometimes it does other stuff too, like fragment.

Regarding your question: it is more complicated than that, I believe. Even bearing the FC principle in mind, ionisation will typically proceed to give a mixture of vibrational states. So, it doesn’t actually mean that every molecule of M•+ will be formed in an excited state; it only means that it’s most likely to be formed in an excited state. More formally, it depends on the Franck–Condon factors which atbm has mentioned. The probability of forming state x is related to the FC factor for state x. These in turn depend on the overlap between the vibrational wavefunctions of interest.

So, with an instantaneous ionisation process, you get what is essentially a mixture of vibrational states of M•+. This mixture is, by virtue of the FC principle, biased towards excited states, but it does contain some ground state. On the other hand, an adiabatic process usually indicates one which is carefully controlled such that the wavefunction does not change during the course of the process. In this case, that refers to ground state (gs) to gs. And this will occur (almost) without fail, such that (nearly) all of the molecules go from gs to gs: no longer a mixture.

Wow, such excellent explanations! Thank you so much for taking the time to explain this!

The reasons for the change in internuclear separation and the imporance of  the Franck-Condon factors, as has been clearly pointed out in answers and comments. The FC factors determine the strengths of transitions from $M$ to $M^{+.}$ and to clarify this figure below shows a simple calculation based on harmonic oscillator wavefunctions of the effect of displacement on the spectrum produced.

The FC factors with vibrational wavefunctions $\varphi$ are defined as $\displaystyle F=\left |\int \varphi_n^*(R-R_n)\varphi_m(R-R_m) \right|^2$ where $R$ is bond extension and $R_n,R_m$ equilibrium values. In the case of harmonic oscillator transitions from the ground state vibrational level on the lower (ground) electronic state to a higher electronic state is $\displaystyle F_{0,n}=\frac{x^n}{2^nn!}e^{-X/2}$ where the displacement parameter $X=\mu\omega(R_n-R_m)^2/\hbar$ with reduced mass $\mu$ and frequency $\omega$.
The log10 of the FC factors for different $X$ are shown in the figure. Small displacements lead to low quantum numbers in the excited state and large displacements to high quantum numbers.

The use of the word 'adiabatic' is strange here as the 0-0 transition will occur with a small or large probability, as determined by the FC factor; if small it might be 1000 times less intense than the most intense transition. Usually the term refers to situations when there is a interaction between potential energy surfaces that causes them to move apart (avoid one another or repel one another) in the adiabatic picture. This is the usual case that we draw for an energy profile leading to a transition state 'at the top of the hill'. The wavefunction then changes smoothly from that of reactant to product. This is not usually what happens in an optical transition as the interaction between states is minute in the absence of the photon. Thus to cause a adiabatic change, as shown in your picture, the two state have to be caused to mix at the position of the minimum of the $M^{+.}$ state which presumably means starting with energy high up in the ground state and using a photon to reach the minimum in the higher state but this is a lot like a normal transition.  Alternatively a short pulse could be used, say 10 femtosec long, to couple the two states , and while they are in superposition ( about 10 fs before they de-phase) somehow to cause the transition to occur with a second pulse.


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Nomenclature of a Benzopyran (Chromene) as a Substituent of an Aldehyde

I am currently writing a paper on the benzopyran above.
I am attempting to name the compound using IUPAC conventions. Because there is an aldehyde moiety, I suspect I will need to name the compound as a substituent of methanal (so 1-(substituent name)methanal). I could alternatively name the aldehyde as a formyl moiety, but this would not be following IUPAC priorities. Benzopyrans have the following locants (with the 2H indicating an extra hydrogen at the 2 position). So, because the aldehyde is at the 6th locant, the compound would be named 1-(XXXXX-6-yl)methanal. The substituents on the benzopyran are as follows

2,2-dimethyl
2,7-dimethoxy
8-(1-oxo-butyl)

By assembling the everything in order, I get:
1-(5,7-dimethoxy-2,2-dimethyl-8-(1-oxo-butyl)-2H-1-benzopyran-6-yl)methanal
I am not sure whether or not I should have named this compound based on the aldehyde moiety since it would be so trivial to name it as a formyl substituent instead?

Based on this  http://www.chemspider.com/Chemical-Structure.9053158.html?rid=bd0a0081-98c9-4f63-9096-4c7722547f57  I would say your compound is 2,2-Dimethyl-5,7-dimethoxy-8-(1-oxo-butyl)-2H- chromene-6-carbaldhehyde

I had forgotten to account for the fact that the aldehyde was attached to a cyclic moiety, so I see why my name seemed so unnecessarily complex!

@Waylander The preferred prefix for “1-oxobutyl” is “butanoyl”. And the systematic benzo name (here: “2H-1-benzopyran”) is the preferred IUPAC name for “chromene”.


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How can you have a reduction potential for a non redox reaction?
The standard reduction potential for the half cell $$\ce{FeCO3_{(s)} + 2e- <=> Fe_{(s)} + CO3^2-}\quad\quad(E^\circ=-0.756\ \mathrm V)$$ and the standard reduction potential for the half cell $$\ce{Fe^2+ + 2e- <=> Fe_{(s)}}\quad\quad\quad\quad\quad\quad(E^\circ=-0.44\ \mathrm V)$$ are as given.
From these, I calculated the $E^\circ$ for the reaction $$\ce{FeCO3_{(s)} <=> Fe^2+ + CO3^2-}\quad\quad(E^\circ=-0.31\ \mathrm V)$$ How do I interpret this number because the dissolution is not a redox reaction? How do I calculate $\Delta G^\circ$ for the reaction since $n=0$ in the dissolution reaction ($\Delta G^\circ=-R\cdot T\cdot\ln K$ will not work)?

What? Where is Fe (0)?

@Alchimista In the dissolution reaction, Fe stays in the +2 oxidation state on both sides of the equation so n = 0.

I don't get it then... It is like wanting the circumference of a square or even worse.

The standard reduction potential for the half cell $$\ce{FeCO3_{(s)} + 2e- <=> Fe_{(s)} + CO3^2-}\quad\quad(E^\circ=-0.756\ \mathrm V)$$ and the standard reduction potential for the half cell $$\ce{Fe^2+ + 2e- <=> Fe_{(s)}}\quad\quad\quad\quad\quad\quad(E^\circ=-0.44\ \mathrm V)$$ are as given.

If these are half cells, you have elemental iron forming at the cathode and being used up at the anode. To express this, you could write the redox reaction as:
$$\ce{FeCO3(s)_{cat} + Fe(s)_{an} <=> Fe(s)_{cat} + Fe^2+(aq)_{an} + CO3^2-(aq)_{cat}}\tag{1}$$
with subscripts "cat" and "an" designating cathode and anode species, respectively.
$n$ will be equal to 2, and now calculating the Gibbs energy will make perfect sense.

From these, I calculated the $E^\circ$ for the reaction $$\ce{FeCO3_{(s)} <=> Fe^2+ + CO3^2-}\quad\quad(E^\circ=-0.31\ \mathrm V)$$

That does not make any sense. If you have the iron carbonate just dissociating in a one-pot reaction, there is no electron transport (where would you attach your voltmeter?) and assigning a reduction potential is wrong.
However, the Gibbs energy of reaction of the one-pot reaction will be the same as that of the electrochemical cell described in reaction (1).

One has to take the last reaction as the net reaction of the cell, based on the half-cell reactions above, not as a single place reaction.


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Carbon dioxide electronic geometry
It is known that allene (propadiene) $\ce{H2C=C=CH2}$ geometry is such that its hydrogen atoms are not in the same plane – do not form rectangle, but “extended tetrahedron”

I wonder if structurally analogous carbon dioxide $\ce{O=C=O}$ exhibit similar geometry, i.e. do its lone electron pair orbitals form extended tetrahedron as well (IUPAC officers probably would not like following structure diagram)

or are they in the same plane? And why?

AFAIK, neither. Oxygens are probably sp hybridised - this stuff with el. pairs is tricky.

@Mithoron Well, what's then your proposal of its Lewis formula? I bet they are sp² hybridized, like in oxo group in organics.

It has nothing  to do with Lewis formula. You assume lone pairs have to be identical, but it's not true. Martin could elaborate more about this stuff.

https://chemistry.stackexchange.com/questions/50906/are-the-lone-pairs-in-water-equivalent

https://chemistry.stackexchange.com/questions/76908/what-is-the-hybridization-of-chlorine-in-vinyl-chloride https://chemistry.stackexchange.com/questions/80962/what-is-the-hybridization-of-terminal-fluorine-atoms-in-molecules-like-boron-tri


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Interpretation of Carbon-13 NMR Spectra for Chemically Similar Carbon Atoms
I am analyzing the Carbon-13 NMR data from a reaction I will be modifying. There is not NMR data available for the specific compound I am attempting to synthesize; however, the compound I am synthesizeing only differs by the number of carbon atoms on the acyl group (from a Friedel-Crafts acylation). The compounds I will be synthesizing and analysing are 1-(5,7-dimethoxy-2,2-dimethyl-2H-1-benzopyran-8-yl)butanone (1) and 1-(5,7-dimethoxy-2,2-dimethyl-2H-1-benzopyran-6-yl)butanone (2) respectively. The compounds synthesized in the protocol I am following were 1-(5,7-dimethoxy-2,2-dimethyl-2H-1-benzopyran-8-yl)ethanone (4a) and 1-(5,7-dimethoxy-2,2-dimethyl-2H-1-benzopyran-6-yl)ethanone (4b) respectively. I will include images below. 14 peaks are expected from each of the carbon-13 NMR spectra since each carbon is chemically different; however, compound 4 only showed 13 NMR peaks. My hypothesis was that this was due to the fact the the methoxy carbons are so chemically similar that they appear as a single peak; however, I am not sure if this is the case. I expect that I can use this to my advantage in distinguishing compounds 1 and 2 when I synthesize them. The doi to the article is: 10.1016/j.bmc.2018.05.016. I will also include the NMR data below:

I apologize for the terrible quality image. I cannot figure out how to increase the resolution. The carbon-13 NMR data is also included in the paper if you cannot make this out.

I did not expect it just by looking at the molecules, but apparently (via software simulation, both by nmrdb.org and mnova) the methoxyl carbons have very close shift in 4b (they are 55.8 and 63.05 ppm in a4 and they collapse in 56.2 ppm in 4b)
Probably higher resolution 13C NMR, and for sure a HETCOR experiment (as the proton shifts are clearly non-degenerate), will distinguish them

I did not know there was a website that allowed you to calculate the theoretical NMR spectrum. Thanks for pointing that out! Unfortunately, I only have access to a 300 MHz spectrometer, so I will probably just use the fact that these peaks collapse on each other to distinguish my two compounds, as the proton NMR spectra of both of these compounds are highly similar.

Yes, generally all the NMR prediction methods are not very accurate in every case, so use them carefully, but they can give some insights, and on sets of "typical" and lucky molecules like this, they can give satisfactory results (mnova gave a very good prediction).


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when Al2O3 melts which bonds are broken?
Previously, I was trying to understand why $\ce{Al2O3}$ has a lower melting point than MgO. An answer on quora said

Do not confuse the character of the bond between the metal and oxygen with the bonds between molecules in the compound. The polar character of the bonds between aluminum and oxygen is indeed less than between magnesium and oxygen but this describes the energy needed to separate the metal from oxygen. When a solid melts it doesn’t break down the bonds between oxygen and aluminum or magnesium. The intermolecular bonds (that is bonds between one $\ce{Al2O3}$ and another $\ce{Al2O3}$) are in fact weaker than the bonds between two magnesium oxide molecules. This results in the greater melting point of magnesium oxide in spite of the less covalent character of the bonds within the molecule.

But if $\ce{Al2O3}$ is ionic and forms a giant lattice, which bonds that are "between one $\ce{Al2O3}$ and another $\ce{Al2O3}$" are broken?

Right, the text you are mentioning is not clear enough. It is not correct  to say that the molecule $\ce{Al_2O_3}$ exists and is separated from the next $\ce{Al_2O_3}$. Such a molecule does not exist. A sample of $\ce{Al_2O_3}$ is made of a superposition of $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ ions. It is not made of a superposition of molecules $\ce{Al_2O_3}$. Now the fact that MgO melts at a higher temperature than $\ce{Al_2O_3}$ is due to the stronger attraction between $\ce{Mg^{2+}}$ and $\ce{O^{2-}}$ ions. For some reason, the attraction between $\ce{Al^{3+}}$ ions and $\ce{O^{2-}}$ is weaker.

Actually, it is well known that in the liquid phase Al is predominantly 4-fold coordinated with O. Oxide melts are not simple, generally speaking. Now, this does not imply that there are specific, identifiable, fixed molecules floating around - each atom prefers a surrounding amenable to it, which constrains the available liquid configuration.

A substantial part of this answer is beating around the bush and can be replaced by a single mention that $\ce{Al2O3}$ is a formula unit for the network solid and not a molecular formula. I doubt you can directly compare magnesium and aluminium oxides in terms of $\ce{M-O}$ bonding like that. Both have different crystal structures (trigonal vs cubic) and coordination environments, and it's not clear how you draw a conclusion about the "attraction" (whatever this means) between metal and oxygen atoms. Sorry, but to me this looks like an answer for the sake of answering.

The quote you give  clearly summarises the situation. In a solid (or liquid) there has to be some significant interaction between individual molecules otherwise there would only be gasses. This inter-molecular interaction is not that forming chemical bonds, i.e. a solid is not a super big molecule, but is additional to these and is due to the nature of the molecules and has several different forms.
The general term is inter-molecular forces, sometimes also call van-der-Waals forces and these are generally much weaker than a chemical bond. These forces arise from charge-charge (Coulomb), charge-dipole, dipole-dipole, dipole--induced-dipole and induced-dipole--induced dipole (also called dispersion or London forces) interactions between molecules.
You can find a summary in many phys. chem. textbooks but for all the gory details see 'Intermolecular and Surface Forces' by J. Israelachvili, publ Academic Press.


text:
NMR magnetically equivalent protons for a 1,4-disubstituted benzene ring
Below I have drawn a molecule with random substituents $\ce{X}$ and $\ce{Y}$:

If the distances from $\ce{H_b}$ and $\ce{H_a}$ to $\ce{H_d}$ are considered, they are different - $\ce{H_b}$ is 5 bonds away and $\ce{H_a}$ is 3 bonds away, so $\ce{H_b}$ sees $\ce{H_d}$ differently compared to $\ce{H_c}$. The same argument can be made vice versa and also between the same pair of protons towards $\ce{H_c}$. This was what the lecturer told me regarding this compound below.
This led me to believe then that all the protons in the molecule are magnetically inequivalent and using up to $\ce{^4J}$ coupling, all protons show a doublet of doublet.
However, when I was in school, I was told that if symmetry was a property of a molecule, as above, $\ce{H_a}$ and $\ce{H_b}$ were equivalent, as are $\ce{H_c}$ and $\ce{H_d}$, and each pair of protons gives a doublet. Researching this matter on Google still held true in my undergraduate studies.
What am I getting wrong here?
I read a quote on another question highlighted here

For two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.

So as $\ce{H_a}$ and $\ce{H_b}$ are chemically equivalent but $\ce{H_b}$ couples differently to $\ce{H_c}$ than $\ce{H_a}$ does, that therefore means that surely $\ce{H_a}$ and $\ce{H_b}$ are magnetically inequivalent?

when I was in school, I was told that [...] each pair of [equivalent] protons gives a doublet

This is only strictly true for magnetically equivalent protons. If there is magnetic inequivalence, such as in the p-disubstituted benzene ring, then it is no longer true.
It's not as simple as a dd either; when you have magnetically inequivalent protons that see each other, the first-order rules (the so-called $n+1$ rule) doesn't work so well anymore. You can see some real-life examples of how these molecules look like here: https://organicchemistrydata.org/hansreich/resources/nmr/?page=05-hmr-15-aabb%2F For the p-disubstituted benzene it sort of looks like a doublet if you zoom out far enough, but if you look closely, there are additional peaks flanking the major peaks.

I say "see each other" because if you think about it, the two protons labelled HA above are also magnetically inequivalent; but since they are so far apart, the two benzene rings are essentially independent of each other, and give rise to the same NMR spectrum.

Ah right that makes sense then. So if I considered $\ce{^4J}$ coupling as I mentioned under first-order rules (because this is being examined soon) I would be correct given the information I know of at this moment in time in my studies?

I'm not sure what answer to give you: on one hand, I could tell you that technically, given what you know at this stage, you could claim that it's a dd. It is certainly a good guess, because you can't realistically be expected to produce the second-order spectrum shown above. But that's not in line with reality, and I don't think it makes sense to pretend that it's right and then unlearn it later...

I understand - I think I am covering the disadvantages of first-order interpretations later on so I will tick your answer anyway but thanks so much for the help! I have read your content but I will wait for the lecturer to offer his clarification before proceeding onwards!

No problem. Don't be too overwhelmed by the second-order spectra when you get to them; there isn't all that much rhyme or reason to them. It suffices to know that they are, well, weird. (There are equations that describe them, and the website I linked explains it very well, but in practice it's... not very useful.)


text:
Which is more stable? CF2 (singlet) or CF2 (triplet)
Generally, we know that triplet carbene is more stable than singlet. But in my book, it is written that singlet difluorocarbene is more stable than triplet difluorocarbene.
What's the reason for it?

Wikipedia article provides some explanation for the same. If it helps...

you might be interested in this related question and answer https://chemistry.stackexchange.com/questions/50277/how-does-spin-flipping-of-triplet-carbenes-occur

So generally singlet is less stable as triplet has half-filled orbitals which leads to stability.
However whenever a group that can give  electrons by back donation (examples being lone pair carriers like $\ce{-OH},$ $\ce{-NH2},$ etc. are attached, they stabilise the singlet by back donation of their lone pair into the vacant orbital of the singlet carbene.
This is similar to back bonding in $\ce{BF3}.$

i think you can add back bonding?

Singlet carbene has a vacant p orbital. Whereas triplet carbene doesn't. Same with the case of singlet and triplet difluorocarbene. Due to the availability of vacant orbital back bond is established between F and C singlet, hence more stable.


text:
Synthesis of the tetrachloroaluminate(III) anion
Looking for random chemicals, I found a reference on the anion and thought that it could be synthesised from common laboratory reagents:
$$
\begin{align}
\ce{2 Al + 2 NaOH + 2 H2O &-> 2 NaAlO2 + 3 H2}\\
\ce{NaAlO2 + 4 HCl &-> NaAlCl4 + 2 H2O}
\end{align}
$$
The first reaction obviously occurs at high yield at STP. Does the second reaction occur at STP or not? If not, how can the $\ce{AlCl4^-}$ ion be synthesised?

You do not synthesize ions, you synthesize compounds.

Note that the first reaction would require substantial amount of heat to assure the designated product $\ce{NaAlO2}$ is formed (nowhere near STP):
$$\ce{2 Al(s) + 2 NaOH(l) + 2 H2O(g) ->[\pu{\approx 450 °C}] 2 NaAlO2(s) + 3 H2(g)}$$
In aqueous solution the product would be tetrahydroxoaluminate $\ce{[Al(OH)4]-}:$
$$\ce{2 Al(s) + 2 NaOH(aq, conc) + 6 H2O (l) ->[Δ] 2 Na[Al(OH)4](aq) + 3 H2(g)}$$
Synthetic procedure from "common laboratory reagents" is described in Brauer's Handbook of Preparative Inorganic Chemistry [1, p. 816]:

Sodium Tetrachloroaluminate
$\ce{NaAlCl4}$
$$\ce{\underset{58.4}{NaCl} + \underset{133.4}{AlCl3} = \underset{191.8}{NaAlCl4}}$$
The reaction is carried out in a Pyrex vessel (Fig. 245), which should be as compact as possible.
  The stoichiometric amounts of C.P. $\ce{NaCl}$ and freshly sublimed $\ce{AlCl3}$ are introduced into the pear-shaped part of the reaction vessel.

  Fig. 245. Preparation of sodium tetrachloroaluminate.
The filling is done under nitrogen and the vessel should be very clean and dry.
  A melting point capillary is then affixed underneath the ground joint; the apparatus is evacuated to a high vacuum and torch-sealed.
  It is then immersed as deeply as possible in an oil bath at 200–240 °C.
  The $\ce{AlCl3},$ which tends to sublime onto the cool parts of the wall, is driven back by passing a luminous flame over the condensate spots.
  When the flask contents have become a nearly clear melt, they are poured hot into the side flask of the vessel.
  The product solidifies on cooling.

References

Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1963; Vol. 1.


text:
A compound that reacts with ethyl magnesium bromide and also decolorizes bromine water
Which of the following compounds reacts with ethyl magnesium bromide and also
decolorizes bromine water solution:

Is the answer only option (c)?

Can you explain what made you think so?

@RahulVerma The double bond present outside the ring will decolorize bromine water and the -OH becomes -OMgBr and $C_2H_6$ is formed. However, in option (d), the nitrile also reacts with Grignard to give an imine and eventually a ketone and the ring (which isn't benzene) has double bonds but I'm not sure if those double bonds can decolorize bromine water.

Good analysis. I bet that d) is missing a double bond.

Oh, my bad. I didn't noticed (d) wasn't benzene deriv.


text:
Up to which dilution can one assume that 1 mg/l = 1 ppm?
For aqueous solutions it is often assumed that 1 milligram per liter equals 1 ppm. Are there any guidelines up to which concentration this shortcut is admissible?
Are there any formulas which calculate the error when using this shortcut?

One can always calculate the correct concentration and see. It much depend on the application. Worse case is the often "wrong" use of % in which one does not even know what denominator has been used.

"Parts per …" are dimensionless entities: to derive $\pu{ppm}$ from mass concentration $γ$ (e.g. $\pu{1 mg L-1})$ one uses density of water $\rho(\ce{H2O}) = \pu{1E6 mg L-1}$ for normalization:
$$\frac{γ}{ρ(\ce{H2O})} = \frac{\pu{1 mg L-1}}{\pu{1E6 mg L-1}} = \pu{1E-6} = \pu{1 ppm}$$
Once the solution is too concentrated, you obviously are not allowed to use $ρ(\ce{H2O})$ and have to use the real concentration of the solution.
However, this occurs at the concentration levels way higher than those to be expressed with $\pu{ppm}$ anyway: that's why "parts per …" are so ubiquitously used in biochemistry. Higher molar masses, lower molar concentrations, primarily aqueous media: no need to account for the error stemming from the difference in the densities of pure water and the solution.
Note that "parts per something" units are deprecated and are not recommended for use, see e.g. Is 1 ppb equal to 1 μg/kg?.


text:
Water-stable organoaluminum compounds in which aluminum is part of the aromatic ring
The more common organoaluminum compounds, such as trialkylaluminum, decompose on contact with oxygen-containing molecules (water, ethanol etc). However, my guess is that organoaluminum compounds with an aluminum atom being part of the aromatic system might be more stable towards oxygen compounds, since the high affinity for oxygen of aluminum could be (at least partially) offset by the high energy barrier to destroying the aromatic ring.
My questions are- is my guess correct, and is there any specific aromatic organoaluminum compound that is not destroyed (or acts reversibly at the worst) with water, ethanol, THF (etc)?

You're probably not going to get what you want putting aluminum into an aromatic ring.  Aluminum and carbon have relatively poor pi overlap, so your ring will not have the full delocalization necessary for string aromatic stabilization.  And even strong aromatic stabilization may not be all that strong compared with aluminum preferring to bond with oxygen or nitrogen rather than carbon.
Instead of trying to design a compound that somehow tames the reactivity of aluminum-carbon bonds, choose the right polar solvent.  Protic solvents are bad.  According to Wikipedia, organoaluminum compounds "readily form adducts with bases such as pyridine, THF and tertiary amines. These adducts are tetrahedral at Al."  The named examples are aprotic and coordinate through nitrogen or oxygen.


text:
Do tetrachloroaluminate salts hydrolyse?
Tetrafluoroborate salts, such as lithium tetrafluoroborate, are reasonably stable in aqueous solutions. Does the same hold for sodium tetrachloroaluminate as well or does it end up as $$\ce{NaAlCl4 + 4H2O -> NaAl(OH)4 + 4HCl}?$$

NaAlCl4 will hydrolyze according to the following reaction $$\ce{AlCl4^- + \ 2H2O ->  Al(OH)2^+  +  2 H+ +4 Cl-}$$and if there is enough water it could goo further away like $$\ce{AlCl4^- + \ 3H2O ->  Al(OH)3  +  3 H+ +4 Cl-}$$ But the reaction will never produce $\ce{NaAl(OH)4}$ as you think. The substance $\ce{NaAl(OH)4}$ can only be produced in very basic solution, and here the solution is acidic.

Does it happen at STP or does the mixture have to be heated or something?

Then, can NaAl(OH)4 be generated if the HCl is boiled away?

This hydrolysis occurs at room temperature by simple dilution. And even if HCl is boiled away, it will not generate NaAl(OH)4. Al(OH)3 is the end point of this hydrolysis. To reach NaAl(OH)4, a large excess of OH- is necessary.

The sodium simply ends as NaCl.


text:
Compounds with aluminum and nitrogen atoms directly bonded to each other as part of an aromatic ring?
Just hours ago, I asked a question about whether aromatic organoaluminum compounds with aluminum and carbon bonded together as part of an aromatic ring would be stable in water. The answer seemed to be that "it wouldn't even work at all since aluminum and carbon have relatively poor pi-overlap, and even then the resulting compound would still end up complexing with nitrogen- or oxygen-based protic solvents due to the high affinity for nitrogen and oxygen of aluminum", implying that aluminum-nitrogen bonds are reasonably stable, at least when they're alone by themselves.
My questions are- do aluminum and nitrogen have sufficient p-orbital overlap to be directly bonded to each other as part of an aromatic ring, and, if they do, how stable are they to different kinds of solvents (protic, aprotic, polar, non-polar etc), and is there any specific compound reported in the literature that is water-stable and in which there is a direct aluminum-nitrogen bond as part of an aromatic ring? (Note that these need not be organoaluminum compounds- that is to say, the aluminum atom in question need not be bonded to any carbon atoms)

Again no.  The aluminum-nitrogen bond is much stronger as a sigma bond than as a pi bond.
Borazine, with boron rather than aluminum in the six-membered ring, is known as an aromatic ring and even called "inorganic benzene".  Like its benzene counterpart, the borazine ring can be fused to give polymeric structures ultimately leading to hexagonal boron nitride.  But with aluminum nitride, where pi overlap is weaker and the bonding more polar, such a layered phase is not known, this nitride most commonly having a wurtzite structure.  In effect, the Al-N pi bonds are broken up by cross-linking with other nitrogen atoms even where the pi bonded structure is ostensibly aromatic.
Take a cue from this answer.  You are on firmest ground with aromaticity when the elements involved are a subset of boron, carbon, nitrogen, oxygen, sulfur.

I see. Then why does Wikipedia say that "silabenzene has been determined to be aromatic, although in a different way that benzene" (reworded from the article on silabenzene)?

Those are the best elements.  I would expect less stabilization as we move to other elements.

Aha. Still, I'm pretty sure that 1-ala-2, 6-diazabenzene would be as aromatic as silabenzene, since Al and Si are only one electron and one proton apart.

@JustAYoungArtist What you're imagining is an analogue of acac complex. Such complexes aren't aromatic.

https://en.wikipedia.org/wiki/NacNac


text:
Separation and qualitative identification of cations
A liquid problem may contain the following cations: $\ce{Pb^{2+}}$, $\ce{Cr^{3+}}$, $\ce{Fe^{2+}}$, $\ce{Ca^{2+}}$, $\ce{Fe^{3+}}$, $\ce{Ag^{+}}$, $\ce{Au^{3+}}$, $\ce{Co^{2+}}$. I need to devise a scheme to separate these cations individually.
I have managed to prepare the following scheme:

But, I don't know how to separate the following cations: $\ce{Cr^{3+}}$, $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$, $\ce{Co^{2+}}$.
I thought I could use ammonia or sodium hydroxide as a reagent. If I use this reagent, all the species would be precipitated, if I am not mistaken, and therefore I could not separate them. Perhaps they could be separated by using solvents and then using identification reactions (for example, I know that Co(II) precipitates with 1-Nitrous-2-naphthol). But, at first sight, I see it as very complex.

Why don't you do systematic tests of cations from group 0 to group 6?

I assume you're referring to the use of an analytical march.

Idk what analytical march is... Is it chemistry journal? I was talking about doing systematic wet tests (or you can also do flame test for $\ce{Ca^2+}$) for basic radicals. You can do flame tests for $\ce{Co^2+}$ or $\ce{Cr^3+}$ but you won't be able to differentiate between them. So we generally use it for tests of group 4 cations

An analytical march is a technical and systematic process (a series of unitary operations), of identification of inorganic ions in a solution by means of chemical reactions in which the formation of complexes or salts of unique and characteristic color takes place. For example, the analytical march of the $\ce{H2S}$

You mean a thing like that?  https://bit.ly/2MNGINs

Yes, you should ideally do so, @Zenix is correct

But, for example, if a mixture can contain As(III), Bi(III) and Cu(II). When hydrogen sulphide is added, I get a mixture of precipitates (I don't know which cations it belongs to, whether it belongs to all of them or only to one).

There are subsequent test... If you don't know about qualitative analysis method for finding out the salt, where did you got this question?

I got this question from a sample subject test.

and then how would they come up with the scheme? @Zenix +ç

Well there is a pretty simple way, that you can do in any laboratory by standard reagents too.
React the unknown mixture with $\ce{NH4OH + NH4Cl}$
(ammonium chloride is put as a standard to prevent the precipitation of Zn, Mn, Cu and Ni hydroxides, however, you don't have those in your sample, so you can proceed with only ammonium hydroxide too)
now you will observe precipitate as
Green- confirms $\ce{Cr^{+3}}$
Reddish-brown- confirms $\ce{Fe^{+3}}$
White- confirms $\ce{Al^{+3}}$
(These are colours of the corresponding hydroxides)
Now if you have multiple salts, you carry out the following conformatory tests
For $\ce{Cr^{+3}}$, add a few dorps of strong oxidising agent in an acidified medium ($\ce{KMnO4}$/$\ce{H+}$ would do), this oxidises $\ce{Cr^{+3}}$ to $\ce{Cr2O7^{-2}}$ which gives the super famous chromyl chloride test.
For $\ce{Fe^{+3}}$, add a few drops of $\ce{CH3COONa}$ and you will initially observe a dark-red solution of $\ce{Fe3(OH)2(CH3COO)6}$, which on heating precipitates as blood red $\ce{Fe(OH)2CH3COO}$
For $\ce{Fe^{+2}}$ simply add KI, very distinct black $\ce{FeI2}$ precipitates out
For $\ce{Al^{+3}}$ add ammonium phosphate, you will see a cloudy white precipitate
You can read more here, Vogel is pretty much the most solid and trustworthy book for practical inorganic chemistry
You are welcome!

I had thought of using sodium hydroxide. But, if the liquid sample is made up of all or most of those cations, a mixture of precipitates will be obtained. And, in that case, I won't be able to distinguish which species it is.
Obviously, if when I add the hydroxide, I get only one green precipitate, I know that the mixture only contains chromium and the other cations are not part of it.

well, standard tests get hard for multiple salt compounds. You would have to carry out specific confirmatory tests,  and you cannot do this easily in a small laboratory. So it'll be a pretty tough task for say an undergrad level student.

Should I send you link, where you can give a deep study to the topic? There are confirmatory tests for each radical, you can accurately find if any particular cation is there or not?

Yes Zenix that would be nice, thanks a ton!

You could re-edit and add it in my answer too, that would be helpful for the entire community!

Also, how do you get that fancy font doe the elements with the charge in? Can't seem to find a way using a standard dell keyboard :( and I see the entire community do it (sorry I'm a bit of a noob in tech)

There you go, the answer is fully complete now.

If you add NaF, it will make a complex with Fe(III) that will not react any more with any reagent later on.
Cobalt may be recognized by adding solid NH4CNS and some amyl alcohol. If there is Cobalt in the solution, the alcohol phase will get blue. Iron must be previously eliminated by adding a little solid potassium or sodium fluoride.
Fe(III) may be recognized by adding a few drops of a dilute NH4CNS solution in the solution. A red color proves the presence of Fe(III).
Cr(III) may be recognized by bleach (NaOCl solution). if heating the solution with bleach produces a yellow solution, it proves that Cr was present initially, as this yellow color is due to the chromate ion. Of course, it will be necessary to filtrate the solution, as all the other metallic ions will make a dark precipitate of mixed hydroxides after adding bleach.


text:
Clarification on usage of dU =nCvdT
In many places I have come across the following derivation
Work done in isothermal expansion

By first law of thermodynamics

$\mathrm dU=Q+W$

As $\mathrm dU=nC_V\,\mathrm dT$, and in isothermal process $\mathrm dT = 0$

$\therefore\mathrm dU=0$
$W=-Q$
i.e heat absorbed is completely converted into work

The second step doesn't make any sense at all. As far as my understanding $\mathrm dU=nC_V\,\mathrm dT$ is true only when volume is constant, but it's clearly mentioned expansion in the statement.
So how to justify the use of $\mathrm dU=nC_V\,\mathrm dT$  expression even when volume is not constant?
Please consider the fact that I'm a school student so please answer in a simple manner.

The reason why statement #2 doesn't make sense to you is because it is not generally true!
For a closed system of constant composition, with P-V work only,
$$dU = \left(\frac{\partial U}{\partial T}\right)_V dT +  \left(\frac{\partial U}{\partial V}\right)_T dV = C_v dT + \left(\frac{\partial U}{\partial V}\right)_T dV$$
The difference between that statement and your statement is the extra term at the end, which is the amount by which U changes with V at constant T, times the change in V.
[Minor note: I've dropped the "n" you use, as I'm using total heat capacities, not molar heat capacities.]
However, if we have an ideal gas, U is independent of V, which means:
$$\left(\frac{\partial U}{\partial V}\right)_T=0,$$
Hence, with an ideal gas, the term on the right goes away, giving us your statement #2:
$$dU = C_v dT$$
I.e., your statement #2 doesn't follow directly from #1.  Instead, it's specific to ideal gases.
So:  If a process invoving an ideal gas is carried out at constant pressure instead of constant volume, it is true that $dV$ may not be zero.  But since that non-zero $dV$ doesn't matter when it comes to calculating $dU$ for ideal gases, we can use $C_V dT$ even when $dV \neq 0$.

May I know how we got this relation $$\left(\frac{\partial U}{\partial T}\right)_V dT = C_v dT$$ , I’m aware that it’s a general fact that heat  added at constant volume is Equal to increase in internal energy. Is it possible to prove this relation mathematically

@Chemist  Any standard physical chemistry textbook will cover this. For instance, go to this URL at Google Books for Castellan, Physical Chemistry, 3rd Ed., and click on the link for Page 117: https://books.google.com/books?id=YI9aDwAAQBAJ&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q=117&f=false

I am a 12th grader, so I'll explain in the most colloquial way possible.

Internal energy is a state function.
A state function is a property whose value does not depend on the path taken to reach that specific function or value. ... Path functions are functions that depend on the path taken to reach that specific value. (1)
This implies a derivation for internal energy will stand true, even if if you change the process, as far as you keep the initial and final conditions same, as it is a state function as defined above, so it will depend only on the initial and final conditions and not the path taken to reach so  (By conditions I mean the conditions that internal energy depends on, which is merely temperature)
The significance of this is that $C_v$ is merely appearing because you derived it using an isochoric process. If you would have used an Isobaric process, you would see ($C_p$-R), now would you wonder why $C_p$ is appearing when the pressure is not constant?

So basically $C_v$ appears as a consequence of the first law of thermodynamics. Not as a consequence of constant volume

Thank you for answering , Moreover as mentioned in 2nd statement why is it essential to keep the initial and final conditions same what are its implications, kindly explain (12th graders are undergrads)

Let us continue this discussion in chat.

I have added it in my answer check it out

This is related to the difference between how Cv is measured experimentally and how it is applied in solving practical problems.  To measure Cv experimentally, we hold the volume constant and determine the change in internal energy U by measuring the amount of heat added.  However, for practical purposes, we know that, for an ideal gas, U and Cv are independent of volume (and depend only on temperature).  So, once Cv is known from a constant volume experiment, we can use the measured value to calculate the change in internal energy for any situation involving a change in temperature (and volume).  For a situation where the temperature does not change, the change in U is zero, irrespective of what the volume is doing.

This has to do with the simplifications we make regarding the energy of molecules in an ideal gas (they've been largely discussed here if you want to have a look).
But essentially, in an ideal monoatomic gas, we can state that all the energy is translational kinetic energy, and so we can write that the mean energy is: $$U=n \frac{3}{2}RT=n c_vT$$
So, the expression $\delta U = nc_v \delta T$ is only valid for ideal gases.
In all other cases: $\delta U = T\delta S - p\delta V$,
where $S$ is the entropy of the system.

You may calculate dU and/or dH in any transformation, at constant volume or constant pressure, because U and H are state functions. These variations of U and H may not be easy to obtain, but they can always be calculated.


text:
How is NaBH4 able to reduce tertiary alkyl halides to alkane but LiAlH4 create an alkene?
When attacking tertiary alkyl halides, aren't both of them just supposed to follow $\mathrm{S_N2}$ and substitute the leaving group with Hydride ion? Apparently only $\ce{NaBH4}$ does it and $\ce{LiAlH4}$ causes an $\mathrm{E1}$ elimination resulting in an alkene. Why does that happen? $\ce{LiAlH4}$ does secondary and primary alkyl halides to alkanes, so why not tertiary as well?

The correct terminology is "primary/secondary/tertiary", not "1/2/3 degree".

That "degree" terminology has been regarded as arcane and outdated since at least the middle of the last century. Whoever still teaches that should not be approached during a full moon, or without a handful of garlic.

Do you think that the solvent in the NaBH4 reaction may play a role in the mechanism of the reduction? Do you  have a reference for the data you have presented?

As far as i know LiAlH4 also does E1 here

That "degree" terminology is widely used in the Indian senior secondary classes.

This is a classic case of $\ce{S_N2}$ nucleophilic substitution versus elimination. Sodium borohydride gives the former reaction, using a transferred hydride ion as the nuceophile, while lithium aluminum hydride produces elimination.
Two factors explain the different courses of the two reactions.
Getting down to basics. Generally, stronger proton bases favor elimination because they cleave the carbon-hydrogen bond more readily. Lithium aluminum hydride (which will deprotonate water and other protic solvents even under strongly basic conditions), is clearly a stronger proton base than sodium borohyidide (which is reasonably stable in water at sufficiently alkaline pH).
Elbow room. $\ce{S_N2}$ on tertiary substrates is made difficult by steric hindrance. A more compact nucleophile then becomes more favorable to overcome this barrier. Borohydride ion with its boron atom fits that compared with aluminohydride having the larger aluminum atom. The larger size of the aluminum atom also leads to weaker, more nearly ionic bonding to the (hydride-like) hydrogen and thus stronger proton basicity, which goes back to the basicity factor.


text:
Reaction of diamines with carboxylic acids to form imidazole
I've recently came across this reaction and my guess is that firstly an amide forms that later reacts with the second amine group. Is this a general reaction? Does it matter if the compounds used are aromatic?

This is a general reaction of diamines, aromatic or otherwise, and carboxylate groups - whether acid, esters or amides. Uusally needs high temperatures and long reaction times to give a decent yield. Good candidate for being done in a microwave reactor.


text:
E1 in 6 membered rings
According to my lecturer, if the geometry prevents planarization of the C-centre than an E1 is impossible. As chair and boat are the only stable conformations of a 6 membered ring of carbons, does this mean that such a ring can never undergo an E1?
Furthermore, why must the C-centre be planar for E1? I guess it allows for maximal stabilization by hyperconjugation?

The chair and boat are conformations of cyclohexane. But cyclohexane isn't the only compount with a six-membered ring of carbons. For example, cyclohexene exists...

Well, a ring can twist into an unstable conformation..

Bridgehead atoms can't. You need to think about product - cyclohexene is perfectly reasonable. Double bond at bridgehead, not really.

Well, the answer to your question, without any complications, is pretty simple
Whatever can be forced into planarity in order to bear the $sp^2$ carbocation will undergo $E1$, this can be easily achieved in cyclohexane, as it can easily have a twisted conformation which is relatively less stable, but yet can bear the carbocation. Same is not the case with bridgehead carbons.


text:
GAMESS(US) Polarizable Continuum Model frequency calculations are giving me what seems to be garbage
I'm trying to run Polarizable Continuum Model (PCM) calculations using the GAMESS(US) software. I'm not that well versed in what exactly these calculations are doing (beyond the whole, 'gas within a solvent cavity' explanation). So I will preface and say the calculations might be working and I am simply not understanding the results.
I am running a simple calculation just to start understanding PCM calculations, so I'm using this input:
! AutoGAMESS Version 1.1.8
!  by Brian C. Ferrari
!
 $CONTRL SCFTYP=RHF MULT=1 NPRINT=0 COORD=UNIQUE
 RUNTYP=HESSIAN ICUT=12 ITOL=25 DFTTYP=B3LYP
 MAXIT=200 QMTTOL=1E-7 ICHARG=0 ISPHER=1 $END
 $SYSTEM MWORDS=800 MEMDDI=800 $END
 $STATPT OPTTOL=1E-6 NSTEP=200 $END
 $PCM SOLVNT=CH3OH $END
 $FORCE METHOD=SEMINUM NVIB=2 PROJCT=.TRUE. $END
 $SCF DIRSCF=.TRUE. FDIFF=.FALSE. CONV=1d-7 $END
 $DFT JANS=2 $END
 $BASIS GBASIS=CCT $END
 $DATA
AutoGAMESS COx Energetics
Cnv 4,

 C           6.0   0.0000000000   0.0000000000   0.5267400687
 O           8.0   0.0000000000   0.0000000000  -0.5996000687
$END

The calculation terminates normally but the results it prints out are the following:
  MODE FREQ(CM**-1)  SYMMETRY  RED. MASS  IR INTENS.
     1       0.000    A       12.000000    0.001636
     2       0.000    A       12.000000    0.001636
     3       0.000    A       12.000000    1.444876
     4       0.000    A       15.994910    0.001228
     5       0.000    A       15.994910    0.001227
     6       0.000    A       15.994910    1.084002

This seems like garbage to me, since the frequencies are all 0 but there are some IR intensities. Which leaves me wondering what is going wrong to cause this error in the calculation? What should I change in the input file?

For the matter at hand: These low lying modes correspond to rotation and translation; I was always under the impression that gamess does project them out. The intensities may well be numerical noise, after all you are using a lot of approximations. I don't think we have many gamess experts here, so it might be a while before you get a proper answer.

I believe Martin is on the right track. With projct=.True., it should project all the translations and rotations to have zero frequency. However, that should only account for the first five in this case, so one of the options used must be making GAMESS project the bottom 6, which is the typical case for nonlinear molecules, rather than correctly zeroing out the first 5

I have found the following on the force keyword: At stationary points, the projection simply eliminates rotational and translational contaminants.  At points with non-zero gradients, the projection also ensures that one of the vibrational modes will point along the gradient, so that there are a total of 7 zero frequencies. Could you please check and post the gradient, too. Did you get a message that you have reached a stationary point? For comparison, you could run the calculation without solvent and with analytical derivatives.

@Martin-マーチン With the PCM calculation it says it has not reached a stationary point however for the non-PCM calculation with the same parameters it does reach a stationary point. I'll try putting stricter convergence criteria on it and see if that helps.

Well, that at least explains the extra mode. Using tighter convergence is often a good idea, but check carefully if it oscillates. Tighter scf convergence might also help, and a larger grid will almost always improve the calculation (at a cost of course). As long as there is no proper answer, you might also want to update the post. To avoid lengthy discussions here, you can also find me in [chat]. Good luck!


text:
How does the vapor pressure in a vessel affects the total pressure after a combustion reaction?
The problem is as follows:

A sealed container of $3\,L$ in capactity is filled with hydrogen and
  oxygen at $314\,torr$ each one at a temperature of $27^{\circ}C$. A
  spark produces the reaction between oxygen and hydrogen. Find the
  total pressure after the reaction supposing that the temperature does
  not change. (You may use the vapor pressure of water at $27^{\circ}C$
  is $27\,torr$)

The alternatives given are:
$\begin{array}{ll}
1.&157\,torr\\
2.&184\,torr\\
3.&312\,torr\\
4.&468\,torr\\
5.&494\,torr\\
\end{array}$
What I've attempted to do to solve this problem was to establish the initial conditions for the reaction:
$$O_2+2H_2\rightarrow 2H_{2}O$$
The number of moles for each gas can be found as follows: (for the sake of brevity I'm omitting units)
$n_{O_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503$
$n_{H_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503$
Both gases have the same number of moles.
From the reaction I could spot that the limiting reagent in this case is the hydrogen thus it will the one to be used for the calculation of the amount of water produced.
Since both hydrogen and water are in the same proportion the moles of water will be the same as what are the hydrogen.
$n_{H_{2}O}=0.0503$
While for the amount of oxygen which will remain will be half of the value since:
$0.0503\,mol H_{2}\times\frac{1\,mol\,O_{2}}{2\,mol\,H_{2}}=0.0252\,mol\,O_{2}$
Therefore this will be established as follows:
$\begin{array}{llll}
O_2&2H_2&\rightarrow&2H_{2}O\\
1\,mol&2\,mol&&2\,mol\\
0.0503&0.0503&&0\\
-0.0252&-0.0503&&+0.0503\\\hline
-0.0251&-0.0503&&+0.0503\\
\end{array}$
From this I went on to calculate the pressure for each:
For the water:
$P=\frac{nRT}{V}=\frac{0.0503\times 62.4 \times 300}{3}= 313.82$
For oxygen:
$P=\frac{nRT}{V}=\frac{0.0251\times 62.4 \times 300}{3}= 156.62$
But since it was given that the vapor pressure of water is $27\,torr$ made me confused on how should I use this given information. Does it mean that should I subtract from the oxygen?.
If I do that I obtain:
$156.62-27=129.62$
Then I assumed that this pressure can be added to the pressure of water:
$313.82+129.62=443.44\,torr$
But using this is way off from any of the alternatives in the given answers:
But If I add all the numbers:
$313.82+156.62+27=497.44\,torr$
Which is seems close to one of the alternatives but the number swapped. $468\,torr$
I'm not certain if this is due significant figures or an error in the problem. The source of my major confusion is exactly how should it be assesed the information regarding vapor pressure in a combustion reaction?. I'm assuming that after the combustion the water vapor formed will excert a pressure in the container so does this must be added to the pressure of the water formed resulting from the moles produced?.
Can somebody help me with this and clear out my doubts?.

First, kudos for actually showing work on a problem!
Second, I'd solve this by inspection, rather than using the Ideal Gas Law, where time is critical on a test.
As you correctly calculated, the limiting reactant is $\ce{H2}$. Therefore the pressure of $\ce{H2}$ is nil; all the $\ce{H2}$ is in the water.
Half of the $\ce{O2}$ is used, reacting with the $\ce{H2}$ to make $\ce{H2O}$, so the remaining partial pressure of $\ce{O2}$ is ~157 torr.
The volume taken up by the liquid water is ~1/2,000 that of the original gases, so the volume left in the flask is still ~3 l, though slightly less. Since the answer choices are not close together, and the decrease in volume is negligible, I'd ignore it.
The vapor pressure of water at 25° C is ~24 torr, and at 30° C it's ~32 torr, so at 27° C, it would be about halfway, by linear interpolation, say ~28 torr.
The sum of those partial pressures is close to an answer choice.
Trivia: This question seems to be citing pi with the initial value, 3[.]14 torr. If the temperature had been -8° C (ice at that temperature has a v.p. ~2 torr), the answer would have been 159 torr, so the first six digits of pi could have been used; i.e. 314 ->159.

Interesting trivia. I totally overlooked that. However I'm still confused about the answer. Should it be $314+157+27=498$? (although it doesn't match any of those answers with 494 the only closest one). Btw the partial pressure of water at that temperature was given in the problem. I'm not sure if the water vapor pressure should be subtracted or added and the justification for that. Can you help me with this part?.

@ChrisSteinbeckBell, from where did you get a partial pressure of 314 torr at the end? At the beginning, there was H2 and O2, but at the end, you state all the H2 was used up, as well as half the O2.

I didn't noticed. So the water formed at that temperature became condensed the only gas remaining in the flask is oxygen and vapor pressure due water, hence $157+27=184$?. Is this alright now?.

It looks good to me. Again, when taking multiple-choice tests, there are usually a few ridiculous answers, a close-but-not-quite-right choice, and the correct one. When pressed for time, see which of the two best answers make the most sense.


text:
Why is high molar molar mass a requirement of good primary standards for volumetric analysis?
I have been taught that using primary standards with a relatively high molar mass will reduce the error between the equivalence point and the end point during titration. However, I am unsure as to why this is the case.
It would be great if someone could point out to me what the purpose of using primary standards with a high molecular mass is.

The larger the sample, the less chance for an introduced impurity to affect it.

@DrMoishePippik I see it other way around: for the same mass of the probe, the higher the molar mass, the smaller is the amount, so impurities are actually going to interfere more IMO.

@andselisk You can likely afford to buy a higher quality standard (or spend a bit more time purifying it), but not to buy a more precise scale.

@Karl A decent lab balance is a one-time investment, whereas a set of HQ-standards for the lab performing a wide range of analyses might cost a fortune. But that's not at all what I was commenting about.

@andselisk Can`t argue much with that: I just checked the prices, and a good scale with a hundred gram scale and ten microgram resolution goes for just 2k€ today, including taxes. In the end you just have to do the math and compare the Mw and purity (and prices) of different standards in combination with the scale you have and/or want to buy.

This practice is done in analytical chemistry in order to minimize the relative weighing error on the balance. Preferring a larger formula weight for a primary standard has nothing to do with impurity levels. We have to start with the highest purity standard.
For example, you wish to prepare a 0.010 M solution of oxalic acid dihydrate in 1 L flask. Its formula weight is  126.07 g/mol. You need to weigh out 1.2607 grams of the acid on the balance. Imagine the balance has a small constant error, the relative error in weighing is smaller for a larger formula weight compound as compared to a smaller formula weight standard for the same concentration. Imagine if the molecular weight of oxalic acid were 60 g/mol, how much do you need to weigh out?


text:
Optical activity checking
In the following cyclopentane deratives, why can't I say that a,b,e have a center of symmetry so they are optically inactive and c as it has a plane of symmetry as shown is also optically inactive? The book tells that only c is optically inactive.

Your "centre of symmetry" works only in one direction, that is to say, the left-right direction. Try inverting the front two carbons through your centre of symmetry; you'll find that it actually isn't a centre of symmetry. You can't pick and choose which atoms are inverted and which aren't. Also, it's helpful if you can rotate your image 90° before posting it, and I don't mean that you should post it upside down.

Alright, from what I know, I am certain that any cycloalkane with an odd number of carbon atoms will never have a center of symmetry. So your argument of $a$, $b$, $e$ having centre of symmetry is outright wrong, since upon inverting the front most carbon in cyclopentane derivatives you would see that cyclopentanes do not have a center of symmetry. Cycloalkanes with odd number of carbons can however have plane of symmetry or axis of symmetry.


text:
Neutralizing a sulfuric acid spill
From my AP Chem Prep Book:

What should you do if you spill sulfuric acid on the countertop?
A) Neutralize the acid with vinegar.
B) Sprinkle solid NaOH on the spill.
C) Neutralize the acid with NaHCO3.
D) Neutralize the acid with an Epsom salt $(\ce{MgSO4})$ solution.

The answer given was C — stating that it should be $\ce{NaHCO3}$ because it is a weak base solution.
However, I do not understand why the strength of the base is of relevance (given the situation) and how to determine the strength of the base — at least relative to other bases.

A strong base would later be a problem itself (NaOH is corrosive, you can feel the saponification of your fingers if you touch it). A weak base like bicarbonate (which you can even drink with no relevant problems) is a much wiser choice!

@The_Vinz If the slippery feel is saponification, why does baking soda also feel slippery? Is it reacting with free fatty acids on the skin, or is there something potentially wrong with the "slippery" explanation?

https://chemistry.stackexchange.com/questions/35016/acid-accident-treatment https://chemistry.stackexchange.com/questions/71486/realistic-chemical-spill-accident-in-high-school-chemistry-class

First, why other options are not really the options:
A: vinegar, being a weak acid, doesn't neutralize sulfuric acid and only dilutes it;
B: solid sodium hydroxide, a strong base, does neutralize sulfuric acid, but it does so vigorously releasing substantial amount of heat per unit of time:
$$\ce{2 NaOH(s) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l)}$$
Using solid NaOH it is also tricky to guarantee it will cover the spill and there won't be any unreacted acid or excessive hydroxide (which is equally unwanted) left behind as there is little to no visual clue whether the neutralization is complete, unless you test various spots with, let's say, pH paper.
D: Magnesium sulfate solution doesn't react with sulfuric acid and only dilutes it.
Second, sodium bicarbonate solution not only neutralizes the acid
$$\ce{2 NaHCO3(aq) + H2SO4(aq) -> Na2SO4(aq) + 2 H2O(l) + 2 CO2(g)},$$
but also can be (and should be!) used in excess to assure complete acid neutralization.
Once the acid is neutralized, you are left with solution of sodium bicarbonate and sodium sulfate which is only slightly basic due to cation hydrolysis.
Another important advantage of using $\ce{NaHCO3}$ is visual control: once the process is complete, gas evolution stops.

However, the effervescence might also be a disadvantage if it's distributing fine droplets of the liquid.

Practically, I thought that. I've used (solid) bicarb on real spills of battery acid, but you really want lots to get rid of the still acidic foamy mess. But still best option compared to NaOH which leaves you either a hazardous acid or a hazardous alkali).

Avoiding the quick release of heat is also a point of emphasis IMO - you sometimes do not want to burn through the floor

The other thing you want to be careful of is creating an acid mist.  Many reactions with sulfuric acid are very exothermic and result in such an energetic reaction that a rather stable acid mist is formed.  As you might imagine, you do not want to inhale sulfuric acid mist. That acid is very nasty.


text:
Can Ca and K exist as an ionic compound?
Recently in the Chemistry.SE chatroom a user posted this to celebrate the New Year:

Can Ca and K exist as an ionic compound? If not, why is it impossible?

Ca with atomic number 20, K 19
  so 2019 ->Happy New Year->2020

No.$\mathstrut$

Guys, seriously? I came up with this in five minutes in the middle of the night on Jan 1st and I already had to explain myself in chat on Chemistry.SE and then in chat on English.SE. It's illiterate and makes no chemical sense and was only made for the sake of creating a non-standard congratulation keeping a LaTeX compiler busy at the new year's eve:)

Where is this Chemistry.SE chatroom ?

@Maurice There is a general-purpose The Periodic Table chat; also, you can see all chats by clicking the chat link in the community menu in the upper right corner (screenshot).

I'm voting to close this question as off-topic because it is not chemistry, it is about a joke in chemistry which is about arranging symbols to make a word or slogan

@NuclearChemist I agree. A while ago, I came across a similar question of wordmaking using chemical formulae. I don't know if these are on-topic but a meta post discussing about its on-topicness will do justice.

Some years ago in the UK the number plate on a car had the following format, Letter, a number (1,2 or 3 digits) and then a three letter code. There was a debate in chemistry world's letters pages about the number of formulas or molecules you could see on number plates. HCN was one which came up. You could also have nitrosyl flouride (ONF). You can also have calcium oxide (CaO), KCN, Carbonyl sulfide COS or SCO or OCS. You can have the organometallic ligand COD (cyclooctadiene). The Solvents DMF and THF can appear also.

More numberplate chemcials include cyanogen halides (NCI and NCF), radium oxide RaO, Barium oxide (BaO), Plutonium nitride (PuN), lithium flouride (LiF), Aluminium nitride (AlN), With the pre 1980s numberplate formate which was three letter code, number then single letter (year letter). You could have SiO2 (silica), PuO2 (plutonium dioxide). With a 1980s car you could have C60 at the start of the number plate

As a creator of this memetic illustration, I feel obliged to post a brief answer.
Potassium is known for forming several metal alloys, such as a widely used Sodium-potassium alloy NaK.
However, CaK doesn't exist [1]:

The Ca-K system was found by Klemm and Kunze [67Kle] to exhibit virtually complete immiscibility in both solid and liquid states. No compounds were found by X-ray diffraction.
Cited Reference
*67Kle: W. Klemm and D. Kunze, "Systems of Alkali and Alkaline Earth Metals," Proc. Int'l. Symp. on Alkali Metals, London Chem. Soc., Special Publ. No. 22, 3-22 (1967). (Equi Diagram; Experimental; #)

References

Pelton, A. D. The Ca−K (Calcium-Potassium) System. Bulletin of Alloy Phase Diagrams 1985, 6 (1), 34–34. DOI: 10.1007/BF02871178.

As the creator you deserve the "accepted answer," thank you.

No for a couple of reasons:

Ionic bonds are between a metal and a non-metal - Ca and K are both metals.
K and Ca have a difference of .18 in the Pauling Scale - and definitions of ionic bonds are often stated as having above a 1.7 difference.
Both atoms, Ca and K, aim to lose their electrons to form a full valence(thus forming cations) and have negative oxidation numbers.

Caesium and gold are both metals, but they easily react to form form an ionic compound, caesium auride.


text:
Why are there two structures of L-Glucose in the Fischer projection?
We are told that L-Glucose is an enantiomer of D-Glucose, which would mean that they are mirror images of each other. But then we're also told that the L and D forms is determined by the spacial relationship to glyceraldehyde.

Which of the structure is correct? Are they the same?

Strictly on the left there is not a projection. It is a projection like every other structural formula

It looks to me like the left one is just wrong.
It looks like who made the image took D-glucose and just flipped the 5-OH.
The result? The molecule at left is not glucose, but L-Idose!


text:
Surely distilled water is the answer to a polluted planet
I have been researching a long time on best water to drink and find that there are so many sources of pollution from industry, farming, agriculture and so forth creating run off into our fresh water sources, even the treatment plants seem to add to the problem by adding chemicals.
Distillation separates the many compounds from the water through evaporation then condensing therefore leaving behind what we don’t need in water.
All our minerals are really found in food.
I feel it’s the safest way to keep us healthy as don’t want these dangerous compounds in my body.

Do you think your food is magically free from the same sources of pollution?

What is your question?

Distilled water can be very dangerous if  large quantities are taken ; causing electrolyte depletion. A friend ( Phd Chem Engineer) once drank distilled water  exclusively for a couple days to get a drunk-like "high".

I am even thinking that getting a drunk like feeling by distilled water is stressing the body more than ethanol. It is basically ionic imbalance and we, at least roughly, know that among other things ions are essential for moving muscles, the first of which we need is heart.

The water which is distributed in the public is not polluted. It has been first filtrated through a layer of fine grains of activated wood charcoal that fixes most unwanted impurities. Then it is treated by gases like chlorine or ozone that eliminates unwanted substances or toxic bacteria that would have passed through the charcoal filter. Don't forget that bacteria in water are much more dangerous than chemical impurities.
Nobody wants dangerous compounds in the drinkable water. That is even the reason why Health officers are checking the quality of the water continuously, all over the world. In case of a contamination, they take steps to eliminate this drawback. They are paid for doing that.

It is probably worth noting that water purification to render it potable, as described in Maurice's answer, is much less costly per gallon than distillation.


text:
Can separated and purified substances be obtained from a chromatogram?
I am studying chemistry, and I am learning about chromatography.
But I am confused about how chromatography is used to ‘separate and purify’ substances.
I have learnt about chromatography paper, which is used to check if a substance if pure or not. Spots of the substances to be tested are place on the chromatography paper, which is then dipped in solvent. If there are several components, the spot separates into several spots (1 per component) as they travel up the paper.

Is it possible for separated and purified components obtained from the separate spots on the chromatogram, as usable substances, or do chromatographs only show the purity of substances?

Have you tried a basic searching before asking ?  See e.g   https://en.wikipedia.org/wiki/Column_chromatography

If HPLC is available, then: 1. You first take small aliquots of the sample, inject them and try out different methods. This allows choosing the best method and finding at which Retention Time you can expect the analyte to elute 2. Now inject the rest of the sample and direct the chromatography output to different vials depending on previously known Retention Time 3. Now you can dry down that vial which gives you the purified substance 4. You may want to dilute it back to the right concentration and run another LCMS for Quality Control.

Generally, all chromatography is done on silica gel. It's the exact same procedure for TLC (thin layer chromatography), except that instead of paper you have a thin layer of glass covered with silica. What can be done in that case is scraping the spots, then extracting the product with organic solvents and filtration. There exists plates made for this, which are thicker than just for analysis.
Of course this is not viable for large amounts of products. In that case, we use a column of silica (see the image below, taken from https://bitesizebio.com/29947/basics-chromatography-column/).

This time, gravity makes the solvent go through the column and come out at the other end. Compressed air is often used to make this process go much faster.
You will however think "how do you reveal the products IN the column?" As the products aren't always colored, you can't really tell when what you want is coming out. Therefore, everything is collected in tubes of the appropriate size and revealed afterwards. If everything goes right, there is no overlap between your products and you get everything out of the column. If there is an overlap, you have to either do the column again for the mixed fractions or just discard those and keep the tubes containing only the pure product you want.
This technique is used pretty much all the time, as it can separate anything (as long as it doesn't degrade on silica) and can be adjusted by changing which solvent or mix of solvents you use.

In reality the outflow is monitored in almost real time by quickly sampling the portion and checking if it contains eluate(s), and eventually running a quick TLC of it.

“All chromatography is done on silica gel” is not remotely true.

There are many different ways of using chromatography. You may use it as an analytical tool, like in your figure. But you may also use it for preparatory goals. For example if you have a liquid containing some impurity that you know is adsorbed on silica or alumina, you may use a long and broad glass tube, put it in a vertical position and fill it with alumina or silica powder. Then you pour your impure liquid into the tube. The liquid will slowly get through the powder and get out in a pure state. The impurity will be fixed in the alumina or silica powder


text:
Is an amorphous solid a single phase?
Amorphous solid possess short-range order but no long-range order (e.g. amorphous silica, the short-range order is originated from the tetrahedral bonding between Si and O). Is it possible to have multiple domains with different short-range orders in one amorphous solid?
If so, should we treat this solid as a single phase or multiple phases? Can we detect the existence of multiple domains?
Relative concept: polyamorphism
(It seems that polyamorphism implies that amorphous phases can be distinct from each other.)

Amorphous silica, https://en.wikipedia.org/wiki/Structure_of_liquids_and_glasses

When you say "amorphous solid" it means 100% amorphous. I don't know that there is a single word that means "partially amorphous  and partially crystalline." // So yes some sample could be partially amorphous and partially crystalline.

Yes. But what we are discussing is domains with different short-range orders that are still amorphous.

Short-range orders is essentially what amorphous means. There is also a quasicrystal.

According to the Wikipedia article you linked polyamorphism requires that amorphous phases be distinct from each other, i.e. "polyamorphism requires two distinct amorphous states with a clear, discontinuous (first-order) phase transition between them." So yes, the two amorphous states would be considered different phases. At the coexistence boundary in the phase diagram you may observe both phases coexisting.

It is difficult to give a general answer. A glass is an amorphous solid that may change its composition in a continuous way from one point to another. When melting brown bottles and colorless bottles in the oven of a glass factory, you may obtain a non homogeneous mixture, with regions more or less colored. It is obviously one and the same phase all over the mixture. But as the mixture is not homogeneous, its composition is not the same everywhere.

"A phase is a form of matter that is uniform throughout in chemical composition and physical state (Atkins, 2010)." Do you mean "no boundary between domains" with "the same phase all over the mixture"?


text:
Diastereomers, enantiomers, and meso compounds of molecules with lots of chiral centres
For a molecule with 2 chiral centres, there are four possible combinations, (R,R), (R,S), (S,R) and (S,S). For example, 3-bromo-2-butanol has four stereoisomers: (2R,3R)-3-bromo-2-butanol, (2R,3S)-3-bromo-2-butanol,
(2S,3R)-3-bromo-2-butanol and (2S,3S)-3-bromo-2-butanol.
Of these combinations, the following pairs are diastereomers:
S,R and S,S
S,R and R,R
R,S and S,S
R,S and R,R
I notice that only one chiral centre has been flipped to form these pairs.

The following pairs are either enantiomers or meso compounds (e.g. for (2R,3S)-tartaric acid):
S,R and R,S
S,S and R,R
I notice that both chiral centres have been flipped to form these pairs.

Could we  generalise that flipping an even number of chiral centres always gives either an enantiomer or the same molecule if it's a meso compound, and flipping an odd number of chiral centres always gives a diastereomer?

Could we generalise that flipping an even number of chiral centres always gives either an enantiomer or the same molecule if it's a meso compound, and flipping an odd number of chiral centres always gives a diastereomer?

No. The generalization is slightly different: Comparing chiral centers in two stereoisomers with more than two chiral centres, if

all chiral centres are the same, it is the same molecule.
some but not all are switched, it is a diastereomer.
all are switched, it is an enantiomer (or meso compound).

Open-chain carbohydrates are a good example (compare D-glucose, L-glucose, D-galactose, D-talose etc.). For this class of molecules, there is one more term, epimer (when exactly one chiral centre is switched).
This generalization excludes the more exotic cases of stereoisomers, see e.g. Why are allenes chiral? and Why atropisomers are called conformers?


text:
Question On Balancing
I recently worked on a lab which involved the following reaction:
$$\ce{HCl(aq) + CuO(s) -> CuCl2(aq) + H2O(l)}$$
In my description of the reaction I wrote the following
"The equation shown above represents a multi–step reaction between copper (II) oxide and hydrochloric acid in aqueous solution. To predict the products, we must recall that a metal oxide turns into a metal hydroxide when placed in water. The initial copper ion had a +2 charge, so the hydroxides must have a net charge of –2. This is accomplished through two hydroxyl groups. Once copper (II) hydroxide has formed, it may react with hydrochloric acid. These reactants undergo a special type of double replacement reaction known as a neutralization reaction. In essence, the metals combine ($\ce{CuCl2}$) and water is formed"
From here, I was wondering how exactly to balance the equation. Specifically, do I have

$\ce{CuO(s) + 2 HCl(aq) -> Cu(OH)2(aq) + 2 HCl(aq) -> CuCl_2(aq) + 2 H2O(l)}$
$\ce{CuO(s) + 2 HCl(aq) -> Cu(OH)2(aq) + 2 HCl(aq) -> CuCl2(aq)  + H2O(l)}$

It seems like I can "cancel out" a water molecule when going from middle reactants to final reactants. Is this true?

CuO is very poorly soluble in water and there is no intermediate copper hydroxide formation in HCl solution. The proposed description looks like a Rube Goldberg machine to me. Do you have any experimental evidences to support suggested "mechanism"?

Your equations are not balanced (count the atoms!). First balance, then cancel out. Then you'll get a single result. Besides that, as @andselisk said, there's no reason to assume intermediate copper hydroxide.

As andselisk points out, CuO is poorly soluble, but the hydroxide is listed in the CRC Handbook (blue, insoluble, no numbers), so it can be assumed to exist under some conditions. The Reference Book of Inorganic Chemistry, Latimer and Hildebrand, 3rd ed, 1951, p110, mentions Cu(OH)2 being formed in solution by addition of hydroxide, cold, to Cu++ as green or bluish-green (in hot solution, black CuO is formed). Cu(OH)2 is a very weak base; K = 5.6 e-20 (this must be Ksp).
Now, your experiment: you start with black CuO and wind up with green CuCl2. Because the blue/bluish-green Cu(OH)2 color will be so faint because of its low solubility, I'm going to suggest that under ordinary concentrations of HCl (~0.1 - 1 M), you will not be able to detect the very fast appearance and disappearance of Cu(OH)2 in solution. However, with a large excess of CuO and very dilute HCl, you might, under some conditions, be able to get some Cu(OH)2 into solution if you did it delicately. But I wouldn't think this would be easy. If you wanted to see Cu(OH)2, the way to go would be to add cold NaOH solution to a cold copper salt solution.
Now, for your equations: #1 is not balanced: you have two total hydrogens on the left and four on the right.  #2 is balanced right and left, but the middle has too many hydrogens.
I think you overthought the reaction. It might be two-step, but that intermediate step is very difficult to investigate and can be swept under the carpet. But its better to overthink and walk back than to underthink and miss the boat.