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How would I find R/S configuration in this case?
I know about the rules to assign R/S configurations when the orientation of each bond to the central Carbon atom is given- the whole, give them a priority order based on their Atomic number/masses, move the group with the least priority as far as possible from the eye, and then draw an arrow joining the other groups in order of their priority. If the arrow goes clockwise, it's an R configuration, otherwise it's an S configuration. But how would I do the same when the orientation of none of the groups are given? The answer key says that compounds I and II have R orientation- but when I simply apply the steps above to the  compound I, I get S. How do I do this?
https://i.stack.imgur.com/2yPaI.jpg is the question

While assigning priority in I l, CHO is second priority group instead of $CH_2OH$ as in CHO there is double bond between C and O so we can consider the duplicate form,i.e, C is bonded by two O atoms and O is bonded by two C atoms. So, the rotation is counterclockwise but H is in front of observer, so we consider the opposite rotation, i.e, clockwise. Thus it is R configuration.

Instead of using the rule that we have to consider the opposite rotation (while move the fingers from more priority group to less priority), when the least priority is in front of the observer. You can visualize it this way, when you see the the compound I, drawn on paper, then H and OH is towards you and CHO and $CH_2OH$ is away from you and the rotation is counterclockwise. But when you see the same compound below the paper, the H and OH is away from you and CHO & $CH_2OH$ is towards you but now the rotation is clockwise. So now we consider this rotation to say that it is in R configuration.

@Manu Thank you! I understand, but why are we allowed to assume that the H is in front of us? Why couldn't it already be behind the compound?

In Fisher projection, the groups on horizontal line are toward the observer and on vertical line are away from observer(when observer observe the molecule from on the top of paper).

While assigning priority in I, $\ce{-CHO}$ is second priority group instead of $\ce{CH2OH}$ as in $\ce{CHO}$ there is a double bond between $\ce{C}$ and $\ce{O}$, so we can consider the duplicate form, i.e. $\ce{C}$ is bonded to two $\ce{O}$ atoms and $\ce{O}$ is bonded to two $\ce{C}$ atoms. So, the rotation is counterclockwise, but $\ce{H}$ is in front of the observer, so we consider the opposite rotation, i.e. clockwise. Thus it is R configuration.
Instead of using the rule that we have to consider the opposite rotation (while moving the fingers from the more priority group to the less priority), when the least priority is in front of the observer. You can visualize it this way, when you see the the compound I, drawn on paper, then $\ce{H}$ and $\ce{OH}$ is towards you and $\ce{CHO}$ and $\ce{CH2OH}$ is away from you and the rotation is counterclockwise. But when you see the same compound below the paper, the $\ce{H}$ and $\ce{OH}$ is away from you and $\ce{CHO}$ and $\ce{CH2OH}$ is towards you, but now the rotation is clockwise. So now we consider this rotation to say that it is in R configuration.
The figures are basically the Fischer projections of the corresponding molecules. The actual representation of Fischer projections is below.

Hence the 'pushing $\ce{H}$ as far as possible' thing is done when $\ce{H}$ is on dashed bond (into the screen). Just use the rules of Fischer projections and change the molecule to the form which you feel most comfortable to determine R/S configurations.

I think you have a problem with assigning priority order in these groups. Actually, we assign priority according to the atomic number. Let's say you have two $\ce{CH3Br}$ and $\ce{CH3Cl}$. Now look upon the first atom of both the group. If both the group have the same atomic number then move to next. If there is a double-bonded atom, like $\ce{O}$ so if it is like $\ce{=O}$, then we would treat it like $\ce{-O-O}$. And then same priority rule according to the atomic number will continue, and clockwise rotation will give R and anti-clockwise rotation will S. If you jump hydrogen while moving clockwise or anti-clockwise, then R is converted to S and S is converted to R. There are some more rules for some typical example. Since you have asked for this question so I haven't mentioned them. You can get the answer.


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Reaction most dominant product for a low concentration nucleophile
If i got this reaction, and I determine that it will react in a Sn1 mechanism, will the dominant product will be in the results of the EtOH/H2O or CN-?

I think, the main product will be with the CN- after the hydrid shift in the Sn1 mechanism, and we will get a non chiral product, but im really not sure cause it says low concentration of the CN- nucleophile.

What are your thoughts on this?

I think, the main product will be with the CN- after the hydrid shift in the Sn1 mechanism, and we will get a non chiral product, but im really not sure cause it says low concentration of the CN- nucleophile

Edit your question to include your thoughts. The question is poorly phased in that it does not indicate whether an equivalent of CN- is present even if it is in low concentration. Secondary alkyl halides undergo substitution by a mix of SN1 and SN2 and the conditions are set up to favour SN1

As SN1, this is definitely a solvolysis.

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Contradiction in variation of conductance and conductivity on dilution
I do conceptually understand that on dilution conductivity decreases and conductance increases but if we look at the formula of conductivity, a contradiction seems to arise.
$$\kappa=GG^*$$
where $\kappa$ is conductivity, $G$ is conductance and $G^*$ is cell constant.
It appears that on dilution if conductivity decreases then conductance should also decrease which is obviously wrong. Where is the problem?

Very interesting question. I don't know the exact answer,but the reasoning may help. The $G^$ also changes with dilution. Suppose you have a cube of volume V and of side area A and edge length L. So $G^$ = L/A and electrolyte is filled in the cube. So for dilution cube volume should increase suppose it to be 2V after dilution. Then for constant separation between plates(L) area of plates become 2A. Thus new $G^{'}$ become $G^/2$. The new $G'$ is somewhat less than $2G$, so overall $\kappa$ decreases.

So here I assume that for dilution, plate separation remain same and only the plate area increases which causes lowering of cell constant. I don't know the reasoning behind this assumption. Perhaps there could be some different reasoning behind the observation of lowering of $\kappa$, if this assumption is not correct!

You get it wrong.
With the dilution, both conductance and conductivity ( specific conductance ) proportionally decreases, bound by the conductometer cell constant $G^* = \dfrac{\kappa }{ G}$.
What increases is molar conductivity $\Lambda_\mathrm{m}= \dfrac{\kappa }{ c}$ toward its infinite dilution value by the Kohlrausch equation $\Lambda_\text{m} =\Lambda_\text{m}^\circ - K\sqrt{c} = \alpha f_\lambda \Lambda_\text{m}^\circ$.

Note that conductance is NOT the ease with which ions move. You confuse it with electrical ion mobility, that is closely linked to the molar conductivity.
In fact conductance and conductivity have nothing to do with ions at all all.
Ionic solutions are just one special case how to manage these general current related quantities.
See conductance   and Conductivity on Wikipedia.

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why disproportionation of free radicals does not happen in kolbe's electrolysis if we use secondary or tertiary carboxylic salt as substrate?
In notes given by chemistry teacher, for secondary carboxylic salts the final products are alkane and ester.

but i guess, as the reaction intermediates involve secondary/tertiary free radicals, some radicals should also go under disproportionation reaction and produce alkene.

why this doesn't happen?

Because of the free-radical-scavenging activity of the secondary/tertiary  carboxylic salts?
Hence, radicals that could possibly undergo disproportionation reactions resulting in alkene are removed.
As a possible example, see 'Evaluation of the Free-Radical-Scavenging Activity of Diclofenac Acid on the Free-Radical-Induced Haemolysis of Human Erythrocytes'.


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Heterolytic vs homolytic cleavage of the C−H bond in cyclopentadiene
Consider the cleavage of the indicated hydrogen in cyclopentadiene:

The hydrogen indicated will be easily removed as $\ce{H^.}$ or $\ce{H^+}$. Assume a nonpolar medium.

My question was that would aromaticity give enough stabilization for hydrogen to be more easily removed as $\ce{H+}$?

Well, cleavage of a bond depends on the nature of the process. For instance, when the given molecule is reacted with a base, you'll get a heterolytic cleavage whereas exposure to light might pose a different issue.

Forming aromatic systems is generally highly favourable energetically

In addition to the process, the environment will also make a difference: gas-phase or solution? In the gas-phase, you won't get heterolytic cleavage.

The cleavage of the C-H bond could be envisioned to take place in two steps:

Homolytic cleavage to hydrogen atom plus cyclopentadienyl radical
Ionization of the hydrogen atom to $\ce{H+}$ and addition of the electron to the cyclopentadienyl radical

Ionization of the hydrogen atom requires 313.8 kcal/mol. Aromatic stabilization, calculated from unexpected differences between calculated and experimental heats of hydrogenation, yield an "extraordinary 36 kcal/mol" greater stability than expected. Well, 36 kcal/mol could be extraordinary under certain circumstances, but compared to an ionization energy, it's not very large.
The molecular orbital situation for the cyclopentadienyl radical isn't exactly non-aromatic, it's just not as fully aromatic as it could be. The five pi electrons of the radical are all bonding, but there's room for one more bonding electron. The question is whether the extra stabilization of that electron is sufficient to cause the ionization of the hydrogen atom. A typical carbon-carbon single bond has a dissociation energy of about 100 kcal/mol and a pi bond would be a bit less. A bond would involve two electrons, so one electron would be responsible for about half that, or less than 50 kcal/mol.

So between the "extraordinary" stabilization of benzene (36 kcal/mol) and the extra bond energy of the sixth electron (50 kcal/mol), I don't think there is enough energy available to ionize the hydrogen atom. Homolytic cleavage would be easier than heterolytic. However, if you provide an ionic solvent that stabilizes the proton and cyclopentadienyl anion, circumstances could change completely.

Difficulty of heterolytic cleavage can be seen by its low acidity: the pKa of cyclopentadiene is 16 (https://en.wikipedia.org/wiki/Cyclopentadiene). But that's still a lot more acidic than methane:  Its pKa in DMSO is estimated to be 56. (https://en.wikipedia.org/wiki/Methane)


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Why is the lone pair of electrons of nitrogen delocalised in quinoline?
I am aware that the lone pair of electrons on nitrogen in pyridine is located in an orbital which is not parallel to the orbitals of carbon and hence it is not delocalised. How do they then delocalise in quinoline?

Exactly the same

Your question: I am aware that the lone pair of electrons on nitrogen in pyridine is located in an orbital which is not parallel to the orbitals of carbon and hence it is not delocalised. How do they then delocalise in quinoline?

I am a little confused about your question. Your title definitely indicate lone pair of nitrogen in quinoline is delocalized. To me, your question indicate it may of may not delocalize. Nonetheless, my answer is it do not delocalize (behave same as in pyridine).
The following is what most recent book about chemistry of heterocycles (Ref.1) has to say:

Physical Properties: Quinoline is a colorless, hygroscopic liquid with a smell similar to pyridine. Quinoline is only sparingly soluble in water but miscible with nearly all organic solvents. Quinoline when exposed to light for a longer duration turns yellow and subsequently brown. The bp of quinoline at $\pu{760 mm}$ was found to be $\pu{237.1°C}$. Its freezing point was reported to be $\pu{−14.94°C}$, whereas the calculated value was $\pu{−14.85°C}$. The density of quinoline at $\pu{25°C}$ has been reported to be $1.08981$. The dipole moment of quinoline has been determined both in the vapor phase and in the liquid phase. In the vapor phase the dipole moment was $\pu{2.29 D}$, whereas in carbon tetrachloride it was found to be $\pu{2.27 D}$.
Quinoline is a base because the lone pair of electrons on the nitrogen atom of pyridine is not involved in the formation of a delocalized $\pi$-molecular orbital. Quinoline is aromatic with a resonance energy of 47.3 and is considered to be a resonance hybrid of the following contributing structures (Figure A):

Structures I, II, and III are of lower energy; however, additional charged structures IV–VIII are also possible because of the presence of the electronegative nitrogen atom. The dipole moment of quinoline is $\pu{2.10 D}$, which confirms the presence of charge separation within the ring.

As in the case of pyridine, quaternization of the $\ce{N}$-atom of quinoline is also common because of the availability of its lone pair to the suitable electrophile. This markedly enhances the susceptibility of the hetero-ring to nucleophilic attack. If quinoline is quaternized with benzoyl chloride and the product treated with potassium cyanide, the dihydroquinoline derivative (IX, a Reissert compound) is formed (Figure B). Compounds of this type are of interest in that they undergo a complex hydrolytic cleavage on treatment with concentrated hydrochloric acid, giving eventually quinoline-2-carboxylic acid (X) and an aldehyde (Ref.2).
References:

Vishnu Ji Ram, Arun Sethi, Mahendra Nath, Ramendra Pratap, In The Chemistry of Heterocycles: Chemistry of Six-to-Eight-Membered N, O, S, P and Se Heterocycles; Elsevier Ltd.: Amsterdam, The Netherlands, 2019, “Chapter 2.3: Six-Membered Isolated and Benzo-Fused Heterocycles with One Nitrogen Atom,” p. 77 (ISBN: 978-0-12-819210-8; and DOI: https://doi.org/10.1016/C2018-0-04877-X).
P. W. G. Smith, A. R. Tatchell, In Organic Chemistry for General Degree Students, Vol.2: Aromatic Chemistry; First Edition, Pergamon Press Ltd.: Oxford, United Kingdom, 1969  (ISBN: 978-0-08-012948-8 and DOI: https://doi.org/10.1016/C2013-0-05520-X).


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Orange flakes forming after salt water electrolysis with rusty iron electrodes
I am using two old iron bars, rusty from the years, into a solution of tap water saturated with NaCl.
After a few minutes at 32.6V and between 1 and 2 Ampere, I get a clear looking yellowish solution, containing HOCl (which I am after) along other products.
Invariably, letting the solution sit for a while, either in the electrolysis vat itself, or after having immediately transferred it to another container (via a spigot I installed above the electrodes in the vat, which lay on the bottom of it, so any leakage from the eletrodes themselves should stay on the bottom and I should extract only the "clean" solution above), I notice many orange flakes floating around, that end up sedimenting at the bottom.
After this happens, the yellowish solution becomes mostly transparent.
These are the flakes that form in the external container, where I transfer the electrolysed solution of tap water, NaCl and a touch of vinegar, I seal the top, and put it away from sunlight:

These are the flakes forming in the electrolysis vat, where I left sitting the electrolysed solution of tap water and NaCl (no vinegar added) near the window and with open top (you can see the electrodes submerged to the bottom through a sealed hole on the side)

The flakes do not appear ferromagnetic, they are impalpable (as soon as I try catching them with tweezers they disperse or just float away). I suspect they are iron oxides, but not sure.
I would like to know if such solution is "safe" to be used in a pressure spray painter or if I risk clogging the nozzle.
Also, I would like to know if investing in graphite electrodes will prevent such by-products.

It is Fe(OH)3. No idea if it will clog the nozzle. It is even difficult to be filtered out, so perhaps not.

@Alchimista how comes it flakes up only after a while, when the solution has been sitting in the container? Before that, the liquid is yellowish in colour, but totally clear

The formation of a precipitate is a reaction like any other, it may take time to complete. You may try to somewhat reverse it by vigorous stirring, perhaps in an old blender.

@TAR86 thank you for explaining it's a precipitate; my aim would be to completely avoid having that in the solution, so I will need to go with graphite electrodes.


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Mass Transport and Ion Transport - Li-ion Batteries
I am new to the field and trying to grasp the theories/ physical phenomena behind the Li-ion batteries. I came across two phenomenons 'Mass transport' and 'Ion transport'.
As I understood Mass transport is the Diffusion process (Fick's First law) which describe the transport mechanism how Li-ions are transported from one electrode to the other through the electrolyte.
Now, I wonder what is covered by the Ion transport equations which are based on Fick's Second law and Stokes-Einstein equation.
As I understood, they both describe the same thing [probabl two parts of the same phenomena which is Mass (which is Li-ions) transport through the electrolyte]. But I don't quite get it, Could someone please shed some light on this matter? Any explanation including explanation relating to physical phenomenon is much of my liking.
Many thanks in advance.

The Nernst-Planck equation is predominately used to describe this

many thanks....any suggestions for books to learn electrochemical basics related to Li-ion batteries if you know some?

Honestly i just used Wikipedia. Learn the Nernst Equation, Butler Volmer Equation, Overpotential losses including concentration/reaction/ohmic losses, randles sevick equation & cortrell equation, Electrochemical impedance spectroscopy, Galvanic cells, electrolytic cells. For physical modeling you will need the essential Nernst-Planck eqn


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Doubt about definition of boiling point
The definition of boiling point says boiling point is the temperature at which the vapour pressure of liquid becomes equal to vapour pressure of atmosphere. But how does the vapour pressure of the liquid form above the liquid when the surrounding (the portion just above the liquid) is having the atmospheric pressure. Where does the vapour pressure develop?

Note often on the stove you get bubbles...

Equilibrium vapor pressure! 2) The point is that bubbles form inside the liquid. 3) Partial pressure above is largely irrelevant.

Obviously if you are boiling chemical X in an open beaker then you can't create a vapor pressure of 1 atm of chemical X over the beaker. However as Jon Custer noted in his comment you can see bubbles forming in the beaker which indicate that the liquid is boiling. A thermometer in boiling liquid would give you the boiling point, at whatever the local atmospheric pressure is.
A way to actually have 1 atmosphere of vapor pressure over a boiling liquid would be to use a reflux setup. The vapor evaporating from the liquid pushes its way up the condenser column, where it cools, forms a liquid and drips back down into the pot. Thus the vapor over the boiling flask is the pure compound which is boiling. The thermometer shown would measure the temperature of the vapor which would be the same as the temperature of the liquid. (You could use two thermometers, one in the vapor and one in the liquid to check this...)


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Distance requirement for hydrogen bonding to occur
I am trying to understand the distance requirement for hydrogen bonding to occur (more specifically in the context of protein folding). There needs to be a donor and acceptor atom for bonding to occur. What is the range of distances between these atoms where one can say that a hydrogen bonding is likely to occur when the polypeptide is in an aqueous environment? Is it element pair specific?

Yes. there is element and orientation dependence. There is no simple answer to this, lengthy papers and even lengthier discussions are being had over this issue.

You will probably find all this information in biochemistry textbook. To give you an idea in proteins, typically; alpha helices H..O distance is 2.06 Angstrom, and N..O, 2.99. In beta sheets (parallel) H--O 1.97, N--O 2.29 and in anti parallel sheets, H--O 1.96 and N--OI 2.91. These H bonds are not linear, e.g C=O..H-N typically 150 to 160 degree for COH or NHO angle.
In Watson-Crick base pairing in DNA, (C-G pairs) the linear H-bond distances  are  $\mathrm{N_4-(H)-O_6}$ 2.91, $\mathrm{N_3-(H)-N_1}$ 2.95, and $\mathrm{O_2-(H)-N_2}$ 2.86.

Can you point me to a specific textbook or website where I can get this information? I am looking for some rough rules of thumb in the form of: If the distance is below a certain threshold and the other conditions are met, the two atoms could form a hydrogen bond.

I found this link right after the answer was posted. http://www.bio.brandeis.edu/classes/biochem104/hydrogen_bonds.pdf

@Chris Here is another source, citing a book by Jeffrey on hydrogen bonds: https://proteopedia.org/wiki/index.php/Hydrogen_bonds#Distances_and_Energies


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How can I have iron powder without a fast reaction with oxygen?
I've seen this video in which fine iron powder reacts immediately with oxygen when exposed to the air, turning into rust.
Now, will this reaction always occur with iron powder? Or is there something special with this specific sample?
I'm writing a steampunk story where people are liberating iron powder in the air. Now, I know that iron powder exists without creating sparks, but I don't want to make some huge mistake, so I'd like to understand a little more about this reaction.
Also, the presence of coal combustion smoke or water steam would have some influence in the reaction?

Iron will react with the oxygen in air, just at a very slow speed. When you make the particles really fine, however, their surface to volume ratio becomes bigger, the energy that is set free by the reaction is enough to heat up the small iron particles and the reaction will increase in speed due to the temperature increase, resulting in a thermal runaway. Finally, you'll have glowing particles which even may start a fire.
For more insights, do some research on  "pyrophoric iron", which is the technical term for it.

All is matter of how fine the powder is ( surface per mass ratio ) and what is the level of prior surface reaction with oxygen.
If the powder is very fine ( I mean very ) and the surface is "naked iron", it would burn in a small firework of sparks.
If the powder is course or the surface is already passivated by oxygen because of the powder history of preparation, it would not provide noticeable reaction.
I remember some old book of chemical experiments, preparing pyrophoric powder by pyrolysis of ferrous oxalate,
but it was not iron powder, but pyrophoric ferrous oxide $\ce{FeO}$:

Preparation
$\ce{FeO}$ can be prepared by the thermal decomposition of iron(II) oxalate.
$$\ce{FeC2O4  -> FeO  +  CO2  +  CO}$$
$\ce{FeO}$ is thermodynamically unstable below $\pu{575 °C}$, tending to disproportionate to metal and $\ce{Fe3O4}$:
$$\ce{4 FeO(s) -> Fe3O4(s) + Fe(s)}$$

With the following oxidation by air;
$$\ce{3 Fe(s) + 2 O2(g)-> Fe3O4(s)}$$
respectively;
$$\ce{6 FeO(s) + O2(g)-> 2 Fe3O4(s)}$$

Are you sure that the substance obtained by pyrolysis of ferrous oxalate is FeO ? I remember having read that it was pure iron.

I remember reading it in that book, and independently it is also listed in the ferrous oxide Wikipedia page, so I would take it as OK.

It may be, that in the end there is pyrophoric $\ce{Fe}$ carried by $\ce{Fe3O4}$


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Probabilistic molecular graph models
Let a collection of molecular graphs have at most $N$ nodes, $d$ node types (atom type), and $t$ edge types (bond types). A graph from this collection is normally represented by the tuple $(F, E)$ with
\begin{equation}
\text { node-label matrix } F \in \mathbb{R}^{N \times(1+d)}\quad \text { and }\quad \text{edge-label tensor } E \in \mathbb{R}^{N \times N \times(1+t)}
\end{equation}
In a statistical model, $i$-th row in matrix $F$ represents the probability of different node types for a node $v_i$. Similarly, filament $(i,j,:)$ of tensor $E$ represents the probabilities of edge $\epsilon_{i,j}$ in the graph.
In these models, usually, probabilities of nodes and edges are assumed to be independent. This, however, is not a realistic assumption. Type of atoms and bonds are obviously not independent from each other. For a node of certain type, the incident edges collectively must have total type values not exceeding a threshold, where the “type value” is some numeric property of an edge type.
Question:
I was wondering if the type of the node is only dependent on its immediate neighboring edges (incident edges) or we can argue that it also is affected by the edges farther from the node.

Assuming that "atom type" already accounts for the actual nuclei and only refers to the different chemical character, I can think of examples of influence from a distance.

The one with the smallest distance is the acidity of hydrogen attached to carbon. When we compare the hydrogens compared to the central carbon in Propane and Propionaldehyde, we observe a large difference.
Comparing any carbon of Benzene with one of the inner carbons of Butadiene, we observe a large difference in reactivity towards electrophiles.
The same comparison for a carbon of Benzene and the carbon in the 4 position of Pyridine yields a drop in nucleophilicity for the latter.


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What are chemical bonds made of, and how do they form?
All atoms are linked together by chemical bonds, but what are the bonds themselves made of? I'm sorry that's a stupid question, but I'm only 14, and haven't entered high school yet, and my science class isn't really teaching me much I haven't learned. So please try to simplify your answer too.

In a nutshell chemical bonds are how atoms share electrons.

@MaxW I know that, but what are the connections made of? Electrons? And if so, how are they attracted to each other?

A bond is an abstract concept meant to capture how atoms stay locked together. Behind the scenes, it's just quantum mechanics applied to electrostatic interactions.

@Zhe So really...there is no answer?

Of course there is an answer, but it contains a whole first year of school chemistry. We cannot condense it to sth you can inhale in half an hour. ;-)

@Karl So I need more patience...

Yea, sorry. :-) I hated it when I wanted to learn sth and had to wait for it. I´m curious however, how does a "science class" work?.How many hours is that per week? When I was 14-15, I had two hours per week of each biology (4th yr), physics (2nd yr) and chemstry. (1st yr)

@Karl Well, in my school, it's about 50 minutes of science. I'm in 8th grade, but did homeschooling for 7th grade, and studied a textbook by myself. I learned a lot already before 8th grade, so many of the things taught in class I had already learned. And we barely spent time on chemistry.

In chemistry there aren't any "particles" in an atom other than protons, neutrons and electrons. The protons and neurons are in the nucleus. A nucleus is always positively charged so two bare nuclei would repel each other. To make molecules atoms must share electrons in some manner. To start learning about how atoms share electrons read Wikipedia's simple article  Chemical bond, then the more detailed article Chemical bond.

I'll assume that you know that charges of opposite sign attract each other and charges of like sign repel each other. Every atom consists of a tiny, positively charged nucleus and one or more negatively charged electrons (such that the overall charge is balanced to neutral).
Let us talk through what happens when two hydrogen atoms come close to each other. The (single) electron of the first atom will be attracted to its "own" nucleus, but it will also be attracted to the second nucleus. The same happens vice versa. However, there is also repulsion between the electrons. It turns out, due to fundamental properties of the electrons (they are smeared out over space, you may have heard that in some experiments, they are like a wave), the repulsion between the electrons is a little less than their attraction to the nuclei. There is also nucleus-nucleus repulsion, but they are not that close.
Overall, it is a balance. When the distance between the nuclei becomes too short (the bond becomes too short), the repulsion between the nuclei will increase and the electrons will be pushed into too small a space, also leading to more repulsion between them. When the nucleus-nucleus distance becomes too large, the nuclei-electron attraction will become a little less, but the repulsive forces far more so, leading to a net attraction, which pulls the bond back together.
Of course, it is more complicated than what I have written here. In reality, electrons are indistinguishable from each other and do not really belong to a particular nucleus. In order to understand why hydrogen and hydrogen form a chemical bond, but helium and helium do not, some quantum physics (or at least results thereof) must become involved.

Thank you for the simple but well-explained answer.

I think you'd do better to talk about two main types of bonds. Use NaCl for ionic bond and H2 for covalent bond.

@MaxW I really wanted to do without the octet rule, which I would have to grab out of thin air.

I don't mean to go that far down the rabbit hole. Just that sodium cations and chloride anions attract because of electrostatic force + to -, whereas in H2 the atoms share the electrons equally.

You may also consider an image I read once in a book that makes science accessible to the general public. It starts from the idea that electrons have a property called spin that corresponds to male or female behavior : electrons try to make pairs if they are alone. If half of the electrons are male and half female, there are two sorts of Hydrogen atoms, one half with male and one half with female properties. So if you send one million H atoms in a close container, they will collide randomly and repel each other. When two male or two female electrons collide, nothing happens. But if a male and a female electron collide, they may repel one another because of their charge, but they tend to stay together. Each electron accepts the presence of the other electron in the same region of space. They will never get completely separated in the future. This makes a bond between the two H atoms : they will stay together, and form a molecule. At the end, all $\ce{H}$ atoms are engaged in molecules $\ce{H2}$. Of course it is an image.
I perfectly know that this image does not correspond to the reality. Electrons have no sex. But they have a spin which may be either up or down. In a chemical bond, there are two electrons, and the two electrons have opposite spins. So if you replace the notion of sex by the notion of spin, it can help you understand and see how a chemical bond is made. Images are often useful when newcomers are approaching new concepts, even if everybody knows that this image is a little bit ridiculous. It may help. I even would be pleased to know our young boy's opinion.


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Dissolving hydrated vs non-hydrated salts
in chemical catalogs, salts are often sold in hydrated and non-hydrated versions.  Like sodium sulfate anhydrous and sodium sulfate decahydrate.  I tried dissolving the anhydrous and it didn't dissolve very easily, instead it clumped.  I'm wondering if I would have had easier dissolving with the hydrated form, but I don't have it to check.  Is it often the case that easier dissolving occurs with the hydrated salts with water complexed to the ion?

Usually hydrated salts are quite easily dissolved in water. I have often observed this phenomena, for example with $\ce{Na2CO3}$. But I don't know whether it is systematic.

Thanks @Maurice, that is helpful. I was setting up an electrolysis experiment and figured sodium sulfate was the best salt to use, but purchased the anhydrous and the clumps had to be mechanically broken up by tapping with a glass stirring rod.

On the other hand, there are some salts of supernal solubility. Read "The Endochronic Properties of Resublimated Thiotimoline" by Asimov, https://archive.org/details/AstoundingScienceFictionv41n1/page/n120/mode/2up


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Bioinformatics tools for predicting synthetic reaction for identified drugs?
We are aware of many bioinformatics open source tools like AUTODOCK, Smina for drug discovery through structure based virtual screening.
I was wondering if there are validated bioinformatics tools for predicting the synthetic reaction for the identified compound.
On browsing literature, I came across ReactPRED(1)
But again, I have not heard about this tool a lot. What are the tools available for predicting synthetic reaction for identified drugs in both open source and paid?
Edit:
By synthetic reaction, I am referring to the synthesis reaction of the compound. The tool should be suggesting the synthesis reaction for synthesizing the identified compound.

Tadi Venkata Sivakumar, Varun Giri, Jin Hwan Park, Tae Yong Kim, Anirban Bhaduri, ReactPRED: a tool to predict and analyze biochemical reactions, Bioinformatics, Volume 32, Issue 22, 15 November 2016, Pages 3522–3524, https://doi.org/10.1093/bioinformatics/btw491

Also I wanted to highlight that a tool for such an important requirement may not actually exist


text:
Raman scattering and "electron and proton center" of molecule
I am currently studying the textbook Infrared and Raman Spectroscopy, 2nd edition, by Peter Larkin. In a section entitled The Raman Scattering Process, the author says the following:

Light scattering phenomena may be classically described in terms of EM radiation produced by oscillating dipoles induced in the molecule by the EM fields of the incident radiation. The light-scattered photons include mostly the dominant Rayleigh along with the very minor amount of Raman scattered light. The induced dipole moment occurs as a result of the molecular polarizability $\alpha$ where the polarizability is the deformability of the electron cloud about the molecule by an external electric field. Fig. 2.9 shows the response of a nonpolar diatomic placed in an oscillating electric field.

Here we represent the static electric field by the plates of a charged capacitor. The negatively charged plate attracts the nuclei, while the positively charged plate attracts the least tightly bound outer electrons resulting in an induced dipole moment. This induced dipole moment is an off-resonance interaction mediated by an oscillating electric field.

Can someone please better explain what the "electron and proton center" in figure 2.9 is supposed to be, and how it relates to Raman scattering? Is the "electron and proton center" of the diatomic molecule only at that single point that is pointed to by the arrow in figure 2.9? If so, then why? Or is each of the three points with the identical symbols an "electron and proton center"? Again, if so, then why?
I would greatly appreciate it if someone would please take the time to clarify this.
EDIT:
What does it even mean for a diatomic molecule to be "homogeneous"? I understand the concept of "homogeneous" mixtures, but not a "homogeneous" molecule.

In the water molecule, the electron center is at the center of the Oxygen atom, on its nucleus. The proton center is in the center of the triangle ABC where A is the nucleus of the Oxygen atom, and B and C are the nuclei of the 2 H atoms. The proton center is not far, but not exactly on the Oxygen nucleus.

@Maurice Are you saying that the textbook author is incorrect? Is water even a "homogeneous diatomic molecule"? What does it even mean for a diatomic molecule to be "homogeneous"? I understand the concept of "homogeneous" mixtures, but not a "homogeneous" molecule.

@porphyrin Ahh, ok, I see. So, in that case, all three points with the same symbol are "electron and proton centres"?

@porphyrin Thanks for taking the time to explain this to me. Feel free to post an answer, so that I can accept it and close this question.

I guess for homogeneous read homo-nuclear. He is trying to explain how the electric field of the radiation induces a dipole by shoving the electrons around, hence reference to polarisability: this relates to the standard description. This induced oscillating field then radiates, hence scattering. The electron and proton centre is when the field is zero, i.e. whatever that distribution is in the molecule in no external field


text:
Octahedral Crystal Field Splitting Orbital Energy Levels

(image source)
In the above diagram, where do the 3/5 and 2/5 come from? Also, how are the $\ce{e_g}$ orbitals degenerate with each other?

This isn't a homework question. After the semester ended (I don't go to MIT), I ended up on MIT open courseware to watch some videos about areas of chemistry I haven't covered yet or haven't covered well. I am asking the question here because I have no other avenue in which to ask questions to other people. I'm just trying to use my time during the pandemic to build my knowledge of chemistry

The total energy of the d orbitals before and after the application of the octahedral (cubic) field has to be the same. As three levels fall and two rise, the ratio is as shown. The $\Delta_0$ arises from the solution of the equations. (In older literature the total gap is $10Dq$ with the levels $6Dq$ and $-4Dq$.)
If $V_0$ is the interaction energy $3\Delta_0/5=\int\psi(e_g)^*V_0\psi(e_g)d\tau$ and $-2\Delta_0/5=\int\psi(t_{2g})^*V_0\psi(t_{2g})d\tau$. In practice $\Delta_0$ is found empirically.
The order of levels is reversed in a tetrahedral field and the size of $\Delta_0$ is also changed to $-9/4\Delta_0$.

"The total energy of the d orbitals before and after the application of the octahedral field has to be the same". Why?

You are not creating nor destroying energy, you are just moving electron density into metal centered orbitals that do not lie in the path of the approaching ligands. To do this those orbitals that are in the way have to be higher in energy so the other ones are filled first. But if you lift one you have to lower another one to conserve energy, that's the lever principle.


text:
Kolbe Schmitt followed by methylation

The first step in this synthesis is a Kolbe Schmitt process, but why is two equivalents of base needed? Also, what is the purpose of the two equivalents of methyl iodide? I assumed that there will be the protection of the phenol (to create a methyl ether), which would justify the two equivalents of base while converting the carboxylate into a methyl ester. However, the base is shown in separate steps, which makes me doubt if base would be available in the second step. Also, another problem is that if you follow the sequence (2 Grignard reagents used to convert the ester to a tertiary alcohol, followed by dehydration, followed by hydrogenation of the double bond just formed, followed by a Friedel Crafts acylation), you get to compound R, which only has 13 carbons, when if you had methylated the phenol, you would have 14. Does this mean that the hydrogenation somehow deprotected the methyl ether? Even more confusing, there is methylation occurring from R to S, which tells me that the phenol was protected at some point. How is this possible?
EDIT: Here is the structure of the final product. The problem was about derivatives of natural products from Salvia plants, but I could not find a total synthesis of this compound anywhere. Also, there is no easy name for it so I couldn't search it well. If you can find the source, please let us know. Note that there are other steps in the synthesis that I didn't include in this question.

I wonder why these textbook syntheses tend to be terribly impractical...

You have a sharp eye. There is no need for 2 eq. of CH3I if you are not forming o-methoxy methyl benzoate. I feel that R is C14H18O4. If not it is the phenolic cmpd C13H16O4.

The main thing that I don't like is there is a clear methylation step from R to S and you are definitely not methylation the carboxylic acid, meaning there much be a phenol in R. But how could of the initial ether have been converted back to the phenol?

Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol O undergoes thermal dehydration catalyzed by the acidic phenolic hydroxyl group. Steps P --> Q are straightforward. Step R --> S methylates the phenol hydroxyl and excess KOH insures that any ester that is formed is saponified. Clemmensen reduction of S and remethylation afford T.
Addendum: The OP's comment reveals that T is a carboxylic acid, not a methyl ester. Presumably the Clemmensen reduction effected some demethylation which is the reason for the last methylation step.

Thank you for your help. The only problem i see is from S to T. The formula says there are 14 carbons but the methyl ester would give 15 carbons. I'm not sure what the methylation after the Clemmensen is supposed to be but you don't add any carbons. Also, the next step in the sequence after T is a SOCl2 step then AlCl3, meaning that a carboxylic acid would be converted to an acid chloride then intramolecular FC acylation, pointing to there being an acid in T. Also, the problem stated that T had acidic character. Maybe methylation is to replace any demethylated methyl ether from Clemmensen?

I think your right about S --> T. Too many carbons in T. Of course if I had known that T was a carboxylic acid as you did, I would have come up with your solution. I'll change the structure for T. BTW: I have done Clemmensen reductions on anisole type cmpds without demethylation.

What was the source of this problem?

There are two unexpected demethylations in this sequence. I would like to see the original literature it is based on and read the rationale for them.

@Waylander: Agreed. The Grignard demethylation is reasonable given the chelation and the presence of iodide. The Clemmensen less so. I'll do a Chem Abst search.

@Waylander: I could not find the Grignard demethylation. I found S --> T by NaBH4 reduction and hydrogenolysis. Tet. Lett. 30, 3469 (1989).  BTW you will be pleased to know that T was cyclized with PPA. ;)

I couldn't find a source in the literature but I included the final product in an update to my question. Maybe this question is just wrong but I doubt it.

The question is how did you get the synthetic scheme? I did a Chem Abst search for diketone 5 with no luck. I did find an S --> T reduction with triethylsilane. I suspect that the Clemmensen step may have been contrived.

@user55119 Apparently, it's from the 2020 International Chemistry Olympiad preparatory questions. https://icho2020.tubitak.gov.tr/icho-2020-hazirlik-sorular%C4%B1

@orthocresol: Yes and page 46 has this one: https://chemistry.stackexchange.com/questions/132027/elimination-of-bromine-with-sodium-iodide-and-electrocyclization/132090#132090


text:
Do amines or alcohols have stronger intermolecular hydrogen bonds?

Which has stronger hydrogen bonding, $\ce{CH3OH}$ or $\ce{CH3NH2}$

I think it comes down to which has more dominance; number of hydrogens, number of lone pairs, or electronegativity.

Alcohols form stronger H-bonds compared to amines.

The boiling point of non-ionic compounds are highly depend on their H-bonding abilities. For example, boiling point of water (molar mass: $\pu{18.02 g/mol}$) is $\pu{100 ^\circ C}$ at $\pu{1 atm}$ while that of ethanol (molar mass: $\pu{46.07 g/mol}$) is $\pu{78.4 ^\circ C}$ at $\pu{1 atm}$, even though ethanol is heavier and have more other intermolecular attractions, excluding H-bonding, such as van der Waals interactions than water. This difference in boiling points displays clear dominamce of H-bonding (HB) on boiling point over other forces on these non-ionic compounds.
Therefore, since molar masses of methanol and methylamine is very close ($\pu{32.04 g/mol}$ for $\ce{CH3OH}$ versus $\pu{31.05 g/mol}$ for $\ce{CH3NH2}$), comparition of their boiling point would be excellent indicator of the comparison of their H-bonding strength: Boiling point of $\ce{CH3OH}$ is $\pu{64.7 ^\circ C}$ at $\pu{1 atm}$ while that of $\ce{CH3NH2}$  is $\pu{-6 ^\circ C}$ at $\pu{1 atm}$. Therefore, it is safe one to say that $\ce{CH3OH}$  has stronger H-bonding capabilities than that of $\ce{CH3NH2}$ even though $\ce{CH3NH2}$ has extra $\ce{H}$ compared to only one in $\ce{CH3OH}$. The best comparison measures of these two types of hydrogen bonding (HB) are the $\Delta H_\mathrm{HB}$ values in Wikipedia (Hydrogen Bonding):

$\Delta H_\mathrm{HB} = \pu{21 kJ/mol}$, listed for $\ce{O−H···:O}$ in water-water.
$\Delta H_\mathrm{HB} = \pu{13 kJ/mol}$, listed for $\ce{N−H···:N}$ in ammonia-ammonia.

Mathew Mahindaratne has provided analysis based on experimental values of the boiling points of the two compounds. I would like to offer a different view using bonding analysis.
Before I begin tackling the question, we shall first clarify the concept of the hydrogen bond. While the popular view of the hydrogen bond is as a particularly strong type of "dipole-dipole" interaction (more of a charge-dominated interaction), the more chemically accurate view manifests itself as a type of "fractional chemical bond" (more of a covalent interaction) between the two participating molecules (Weinhold & Klein, 2014). In fact, in their paper, authors define the hydrogen bond as:

a fractional chemical bond due to partial intermolecular $\ce {A-H...:B <-> A:^{-}...H-B^{+}}$ resonance delocalisation (partial 3-centre 4-electron sharing between Lewis bases), arising commonly from quantum mechanical $n_\ce{B} \rightarrow \sigma^{*}_\ce{A-H}$ donor-acceptor interaction.

Based on the above definition for the hydrogen bond, we can easily explain why $\ce {O-H...:O}$ would be a stronger hydrogen bond than $\ce {N-H...:N}$. To adopt this type of resonance, there is a need for heterolytic cleavage of the single bond. As the more polar of the two, the $\ce {O-H}$ bond undergoes heterolytic cleavage more easily and hence, the hydrogen bond formed between the alcohols would be certainly stronger than that formed between the amines.
To explore this further, we can look at the whole range of hydrogen bond enthalpies provided on the Wikipedia page:
\begin{array}{|c|c|c|c|} \hline
\text{Hydrogen bond} & \text{Enthalpy, } \pu{kJ/mol}\\ \hline
\ce{F-H...F} &  161.5\\ \hline
\ce{O-H...N} & 29\\ \hline
\ce{O-H...O} & 21\\ \hline
\ce{N-H...N} & 13\\ \hline
\ce{N-H...O} & 8\\ \hline
\end{array}
Notice that as the polarity of the $\ce {X-H}$ bond increases, the hydrogen bond strength increases. Comparing between $\ce {O-H...O}$ and $\ce {O-H...N}$, we also observe that as the the availability of the lone pair on the hydrogen bond acceptor increases, the strength of the $\ce {H}$ bond also increases. In terms of molecular orbitals, we can say that the $\ce {\sigma^ {*}_\ce{O-H}}$ is lower in energy and more energetically accessible to the lone pair orbital of the acceptor, compared to $\ce {\sigma^ {*}_\ce{N-H}}$ while the $\ce {n_N}$ nonbonding lone pair orbital is higher in energy and is able to better access the $\sigma^{*}$ orbital of the hydrogen bond donor, compared to $n_\ce{O}$.
Reference:
Frank Weinhold, Roger A. Klein, "What is a hydrogen bond? Resonance covalency in the supramolecular domain," Chem. Educ. Res. Prac. 2014, 15, 276-285 (DOI: 10.1039/C4RP00030G).


text:
Acidic strength of diethyl malonate vs ethyl acetoacetate

Which would be more acidic, diethyl malonate $\left(\ce{CH2(CO2Et)2}\right)$ or ethyl acetoacetate $(\ce{CH3COCH2OOC2H5})$

In the above compounds, I have marked the hydrogen which I think will be the most acidic. Now removing that hydrogen would yield a carbocation. Both the carbocations would be equally stabilized due to resonance effect. In fact diethyl malonate acid would be more stabilized due to the extra inductive effect of the extra oxygen. Yet Wikipedia shows that ethyl acetoacetate has a p$K_\text{a}$ less than that of diethylmalonate acid.
What is making it more acidic?

That's diethyl malonate, not malonic acid... If you're looking at malonic acid on Wikipedia, that's going to be a completely different pKa value.

good catch.. edited it

https://chemistry.stackexchange.com/q/18987/16683

Also carbanions not cations...

The hydrogens which you have chosen are correct but both the carbanions are NOT eqaully stablised via the resonance. In case of malonic ester, the elctrophilicity of the carbonyl(c=o) is reduced due to the resonance from the oxygen beside it. So the carbanion will not be pulled very effectively into resonance as the carbonyl becomes electron rich due to the oxygen of the ester.
However, in the ethyl acetoacetae only one of the carbonyl groups is affected by this reasonace. The other one is still highly elctrophillic. So it stablises the carbanion even more. So it has a lower pka(more acidic) than the malonic ester.


text:
How to predict which type of optical isomerism
When I learnt about optical activity in school, my teacher told me that there is no way to theoritically predict whether a compound is dextrorotatory or leuvorotatory. How is that possible?

Hope you make it to OCSC!

Because it may happen that some compound is dextrorotatory for one color (one wavelength) and levorotatory for another color (another wavelength)

You can estimate via computational chemistry.

In terms of why it's hard, consider the amino acids for the biologically prevalent enantiomer. If I told you the specific rotation for one of them, can you tell me the sign of the specific rotation of any other?

@WilliamR.Ebenezer I thought OCSC term is used in my country only, is it used all across the world? Also have you attended OCSC?

@Zenix, Nah, I couldn't make it. Didn't score enough. And yes, I believe the term is used worldwide.

Thanks to everyone who spent time to reply to my question. Thank you William for your wishes. Maurice, I understood your point. I didn't quite understand what Zhe meant in his second comment.

A more accurate statement might be "not by simple inspection". And there might be a significant margin of error even in the best theory.

Related: https://chemistry.stackexchange.com/questions/83990/what-is-occuring-on-the-quantum-level-when-a-molecule-rotates-plane-polarized-li/84136#84136


text:
Type of structural isomerism
What is the type of structural isomerism in which isotopes move around?
Examples

$\ce{CH2D-NH2}$ and $\ce{CH3-NHD}$
$\ce{CH2D-NH-CH3}$ and $\ce{CH3-ND-CH3}$
$\ce{CH2D-NH-CH2CH3}$ and $\ce{CH3-NH-CHD-CH3}$ and $\ce{CH3-NH-CH2-CH2D}$

here $\ce{D}$ refers to deutrium.
My progress: So I know that there are various types of isomerism, like

Tautomerism - Differ in the position of protons and electrons. Usually I see that this occurs with changes in bond orders. So I don't think it will be this.
Ring-chain isomerism - When there is a difference in the number of carbons in a ring
Functional isomerism - When the functional group changes. I don't know if $\ce{-NH2}$ is a different functional group from $\ce{-NHD}$.
Metamerism - This is when there is a change in carbon chain on both sides of a bridging group. I don't know if a change in isotopes would count as a change in carbon chain.
Chain isomerism - When the carbon skeleton changes. Here the skeleton carbon is the same.
Positional isomerism  - When a functional group attaches itself on different carbon of the same chain. Not applicable here.

This is not a homework question. This question was to fill the gap in my knowledge of isomerism with isotopes.

According to Wikipedia, this type of isomerism is known as Isotopomerism

Isotopomers or isotopic isomers are isomers with isotopic atoms, having the same number of each isotope of each element but differing in their positions. The result is that the molecules are either constitutional isomers or stereoisomers solely based on isotopic location. For example, $\ce{CH3CHDCH3}$ and $\ce{CH3CH2CH2D}$ are a pair of constitutional isotopomers of propane

For a different source, you can also check out the IUPAC Gold book's definition of an isotopomer.


text:
Rate law of reactions, what about the proportions?
Consider the reaction
$$\ce{A + B + C -> D + E}$$
and $\mathrm{rate} = k[\ce{A}][\ce{B}]^2.$ My book said if we double the amount of $\ce{A},$ then the reaction rate will double.
I understand this, but what about the law of definite proportion?

"…but what about the law of definite proportion?" — what about it? Could you please elaborate more?

we cant simply put more of a substance to get more of the product, the reactants must have a fixed ratio.

Ehhm, and what is has to do with the law of definite proportions?

i dont understand how you can put more of just one reactant and end up influencing the others. is this thinking correct?
some of the extra stuff will be remaining, something like a catalyst

for example, if you double the amount of A say it become 2x, then the reaction happens twice as fast. why is this? also, is only x amount of A is used?

Either you interpreted your book incorrectly or your book in wrong. If you double the concentration of $\ce{A}$, then the rate of reaction will increase $\ce{A}$. This has nothing to do with the Law of Definite Proportions because the reaction will proceed to the make the product in the correct ratio and will stop when we run out of the limiting reactant.

oh ok, so if we double the amount of only one substance, that substance will get left?

No ! let's take a numerical example. If you start with all concentrations  equal to $0.1$ mol/L, and the rate is $\ce{10^{-4}}$ mol/L/s, it means that after $10$ s, the concentration of A, B and C has decreased by $0.001$ mol/L, and becomes [A] = [B] = [C] = $0.1 - 0.001 = 0.099$ mol/L. Now if you double the amount of A, the rate will become equal to $\ce{2·10^{-4}}$ mol/L/s. It means that after the same delay ($10$ s), the amount of A, B  and C will be respectively : [A] = $0.2 - 0.002 = 0.198$ mol/L, [B] =  [C] = $0.1 - 0.002 = 0.098$ mol/L.

ohhhh ok I get it! but wont there end up with 0.100 mol/L of [A] unused (limiting reactant)?

I don't understand how you arrived at that number without any of the initial concentrations specified. The rate law is a general description of how any mixture will proceed. Are you assuming a certain ratio?

[OP, in a comment] oh ok, so if we double the amount of only one substance, that substance will get left?

Yes. The rate law tells you how fast product is being made. The change in the different species is still given by the chemical reaction equation (which tells you in which proportion the amount of species will change, not in which proportion they are present at the beginning or at the end of the reaction).
You can see that when you look at the definition of rate:
$$\mathrm{rate} = \frac{d[X_i]}{\nu_i dt}$$
where $\nu_x$ is the (signed, i.e. negative for reactants) stoichiometric factor of the $i$-th species. Dividing by the stoichiometric factor ensures that the law of definite proportions is followed.

[OP in another comment] we cant simply put more of a substance to get more of the product, the reactants must have a fixed ratio.

The law of definite proportions is not about the amounts present, but about the change in those amounts. I can set up a reaction with 1000 times more elemental oxygen than hydrogen, but they will still react in a 1:2 ratio if the reaction yields water.

thank for clearing it out! my book never said anything about there being residue of the limiting reactant.


text:
Why are chelating ligands often used to treat poisoning by heavy metals (e.g. mercury, lead)?

Why are chelating ligands such as EDTA often used to treat poisoning by heavy metals (e.g., mercury, lead)?

I have a vibe that it has something to do with the lone pairs of electrons that are able to bond to the heavy metals to form a stable coordination complex but I'm not sure about the in-depth process or explanation behind this.
Also, does it have to do with it being a polydentate ligand and not a monodentate ligand?

Yep. It's all about complex stability. (And low toxicity of the chelator, if certain constraints, like calcium and magnesium balance, are observed).
Polydentate complexes (usually) are more stable.

Lead, mercury and other heavy metals form complexes with both monodentate and polydentate ligands. But the complexes made with polydentate ligands like EDTA are much more stable than with monodentates. Usually metallic complexes become toxic if the central metallic ion (non-complexed) is released or not entirely surrounded by ligands. Monodentate complexes do not surround completely the central metallic ion, and they may be decomposed by the acidity of the stomach. Polydentate complexes resist to this acidity, and often they surround the metallic ions much better than monodentate. So if a person has ingested one of these heavy metal ions, its toxicity will be hidden by a complexing agent like EDTA.

Chelation therapy, which is a very effective way to treat heavy-metal poisoning, is a chemical process in which a synthetic solution is injected into the bloodstream to remove heavy metals and/or minerals from the body. British Anti-Lewisite (BAL) was one of the first chelating agents to be developed as an anti-dote for war gas, dichlorovinyl arsine (Lewsite) during the Second World War. The breakthrough in the development in chelating therapy came after the introduction of ethylenediaminetetraacetic acid (EDTA), initially to combat lead intoxication (Ref.1).$^*$

$^*$During Second World War, dimercaprol (BAL), an organic dithiol compound, was developed as an experimental antidote against the arsenic-based poison gas Lewisite. After World War II, mass lead poisoning was observed in large number of navy personnel, later identified as a result of their jobs repainting the hulls of ships. This introduced the medical use of EDTA as a lead chelating agent. (Ref.2)

A specific salt of EDTA used to bind metal ions in the practice of chelation therapy is calcium disodium ethylenediaminetetraacetate (or sodium calcium edentate; $\ce{CaNa2EDTA}$), specifically treating mercury and lead poisoning. When this EDTA solution is injected into the veins, which is a slow process (administration must be intravenous because it has very law intestinal uptake), it chelates with (or "grabs") heavy metals and minerals such as lead, mercury, copper, zinc, iron, arsenic, aluminum, and calcium. The heavy metal-EDTA complex will be then removed from the body through urinary system.
For example, lead poisoning is generally treated by using the chelating salt, $\ce{CaNa2EDTA}$, which is the calcium chelate of EDTA. This chelating agent has a great affinity to heavy metals, the removing agents such as $\ce{Pb^2+}$. The point is this chelating agent has a greater affinity to lead than to calcium (see large $\log K_f$ values of heavy metals compared to that of $\ce{Ca^2+}$ in following table) and so the lead chelate is formed by exchange. Then, the lead chelate would be excreted in urine, leaving behind harmless calcium. The disadvantage of this therapy is though it removes only extracellular heavy metals.
$$
\begin{array}{c|ccc}
\text{metal} & \log K_f \text{ of EDTA} & \log K_f \text{ of DTPA}  \\
\hline
\ce{Al(III)} & 16.13 & 18.40 \\
\ce{Ca(II)} & 10.96 & 10.74 \\
\ce{Cd(II)} & 16.46 & 19.31 \\
\ce{\color{red}{Cu(II)}} & \color{red}{18.80} & \color{red}{21.53} \\
\ce{Fe(III)} & 25.10 & 28.60 \\
\ce{Hg(II)} & 21.80 & 27.00 \\
\ce{\color{blue}{Li(III)}} & \color{blue}{2.79} & - \\
\ce{\color{blue}{Na(I)}} & \color{blue}{1.66} & - \\
\ce{\color{red}{Mn(II)}} & \color{red}{14.04} & \color{red}{15.60} \\
\ce{\color{red}{Mn(III)}} & \color{red}{24.80} & - \\
\ce{Pb(II)} & 18.04 & 18.80 \\
\ce{\color{red}{Zn(II)}} & \color{red}{16.50} & \color{red}{18.75} \\
\hline
\end{array}
$$
Further, $\mathrm{pH}$ also is an important factor influencing complex formation and stability. Most chelating agents are unstable at low $\mathrm{pH}$, whereas at high $\mathrm{pH}$ metals tend to form insoluble hydroxides which are less accessible to chelating agents. This feature becomes significant in pathological conditions leading to acidosis or alkalosis (Ref.2).
Note that the value of EDTA as clinical chelating agent was reduced by the need for slow intravenous administration, law intestinal uptake, exclusive extracellular action, and high stability constants with essential metals such as Zn, Cu, and Mn (Ref.1, Ref.3).
References:

Kiran Kalia, Swaran  J. S. Flora, “Strategies for Safe and Effective Treatment for Chronic Arsenic and Lead Poisoning,” Journal of Occupational Health 2005, 47(1), 1–21 (https://doi.org/10.1539/joh.47.1).
Swaran  J. S. Flora, Vidhu Pachauri, “Chelation in Metal Intoxication,” Int. J. Environ. Res. Public Health 2010, 7(7),  2745–2788 (doi: 10.3390/ijerph7072745).
Sobrasua E.M. Ibim, Jason Trotman, Paul I. Musey, Wasswa E.B. Semafuko, “Depletion of essential elements by calcium disodium EDTA treatment in the dog,” Toxicology 1992, 73(2), 229-237 (https://doi.org/10.1016/0300-483X(92)90105-N).


text:
316 stainless steel and high concentration, high temperature Hydrogen Peroxide
I'm looking at doing some experiments with small scale rocketry and I need to find the right materials for this project, where I will be burning through lots of HTC (peroxide above 90% concentration). I've found this datasheet here that says grade 316 stainless steel is suitable to temperatures up to 48*C. Can anyone tell me whether its suitable for temperatures much higher than that, or at least point me in the right direction to find any research on the subject? I'm having trouble finding the data I need.
This is more for experimental purposes than any practical applications so the total mass of the materials used is a non-issue.

You better change your approach, as starting with high-test peroxide strikes me as unreasonable, to put it mildly.

Umm... Stainless steel in direct contact with H2O2? You might want to read John D. Clark's Ignition!, particularly Ch. 4: "And  a  disconcerting  number  of  things  could  start  the  decomposition  in  the  first  place:  most  of  the  transition  metals  (Fe, Cu,  Ag,  Co, etc.)  and  their  compounds;  many  organic  compounds  (a  splash  of peroxide  on  a  wool  suit  can  turn  the  wearer  into  a  flaming  torch, suitable  for  decorating  Nero's  gardens)... "


text:
Where do the electrons in lead–acid battery come from?
$$
\begin{align}
\ce{PbO2 + H2SO4 &-> PbSO4 + H2O + O} &\quad &\text{(anode)}\\
\ce{Pb + H2SO4 &-> PbSO4 + H2} &\quad &\text{(cathode)}
\end{align}
$$
The cathode’s $\ce{H2}$ joins the anode’s $\ce{O}$ making $\ce{H2O}$ in solution.
The book says the cathode gets four electrons, but from where? Two liberated hydrogens have one each, that’s two. Where’d the other two come from?

That's lead storage battery.

I adjusted the formatting of the reactions, but the first one looks wrong. It would be nice if you could add the source you've taken them from and cite the textbook.

These aren't equations for battery, but just for reactions with conc. H2SO4. @andselisk  Question could use your closehammer, I think ;)

@Mithoron I'm not sure about single-handedly closing this one. I don't think this is necessarily a bad or a homework question; it's just somewhat hard to trace the OP's reasoning in the absence of the source they took the reactions from.

@andselisk Well, that would be like "trihandedly" now ;) and it seems real unclear.

These equations look as if they had been taken from 19th century books, when ions and electrons were not yet known. In order for these former scientists to describe how Pb and PbO2 get converted into PbSO4 (without electrons), the only possibility is to admit that Pb "produces" H2 from H2SO4, and that PbO2 looses one O atom, so that both H2 and O are allowed to "combine" into H2O at the end. This is of course hard to believe today. But it was the only possibility of explaining the lead battery in the middle of the 19th century.

The half-equations are not correctly written. There is no Oxygen atom released, and no H atom emitted, as the author proposes. And the cathode does not get 4 electrons, as he or she states. The correct half-equations should be, first at the anode : $$\ce{Pb  +  SO_4^{2-} -> PbSO4 + 2 e^-}$$ And at the cathode it is : $$\ce{PbO_2 +  4H+ + SO_4^{2-} + 2 e^- ->  PbSO_4 + 2 H2O}$$

Can we consider $\ce{PbO2}$ as a positive electrode

Yes ! $\ce{PbO_2}$  is a positive electrode.

To be fair, there is generally hydrogen release in a lead-acid battery, especially during formation, but that's not the desired reaction.


text:
Variations in solubility of total dissolved salts in water with temperature
I want to know the solubility of "total dissolved salts" or mainly Calcium Chloride and Magnesium Chloride causing hardness of water.
Is this data published anywhere at different temperatures?

Google is your friend. Using "Calcium chloride solubility", you find plenty of references where this solubility is given at different temperatures. Expressed in grams $\ce{CaCl2}$ in $100$ mL water, it is $59.5$ à $0°$C, $64.7$ at $10°$C, $100$ at $20°$C, $128$ at $30°$C, $137$ at $50°$C, $147$ at $70°$C, and $159$ at $90°$C. Do the same with Magnesium chloride.


text:
Any large databases of IR Spectra that you are able to download many spectra in bulk
I am currently working on a machine learning project, in which I need access to many IR spectra, the more the better, in order to build a classifier that will used some unsupervised learning in order to work out the functional groups present in the IR Spectrum. The only databases I have found so far do not have such a feature. Any help would be greatly appreciated.

This might fit better on the new Materials Modeling SE. They already have a good couple questions about database requests/machine learning

I’m voting to close this question because this is not really a site to give outside resource recommendations outside of our master list.


text:
Why is dinitrogen tetroxide a planar molecule?

I was wondering why $\ce{N2O4}$ is a planar species in spite of having a σ-bond in between the two $\ce{N}$ atoms for free rotation. Does it not form a conformer?

Because pi conjugation.

Why there would be delocalisation of pi electron cloud over N-N bond   as the RS formed would be highly unstable as it would have + charge on oxygen atom with  high electron deficiency

No, it wouldn't have + on oxygen.

The oxygen with the -charge is  in conjugation only with the oxygen present in the same N atom .It can't be in conjugation with the =O present in the other N atom . How there can be partial bond between 2 N atoms

Well, just like that. Each O with minus. Each N with plus. Double bond between N.

I have added the diagrams of RS

Good. Now see, there is one more resonance structure.

I think it is a very good question (why the down vote?) - by comparison the geometry that the isoelectronic oxalate ion adopts in crystals depends upon the counter cation (https://en.wikipedia.org/wiki/Oxalate) indicating that in that case at least the energy well keeping the ion planar is not that deep - so why is N2O4 planar? pi conjugation would suggest double bond character in the N-N bond, but as is well known that bond is weak and long, so I don't find that a very satisfying explanation.

I think that one would expect that a stable form of the molecule might be planar and have D2h symmetry.
Some support for the claim:

Early crystal structures suggested that the ONO bonds on either side
of the N-N bond are easily modeled by low level DFT
(doi.org/10.1039/C19660000082).
More recently, simulations of the trans and cis N2O4 structure show
planarity when optimized at B3LYP/ aug-cc-pVTZ (aka, pretty good not
amazing) See here.

The simulations also suggest that the barrier to isomerization is low, meaning that while the stable structures are planar, there are low barrier (low temperature) out of plane rotations possible.
Summary:

Just because something has sigma bonds doesn't imply that it will immediately be non-planar
yes the molecule can form a conformer


text:
can i use an isobutane stove (like jetboil) or butane blowtorch with borosilicate glass?
I don't have a bunsen burner, but have a camping stove (similar to https://www.amazon.com/Stansport-183-B-Isobutane-Backpack-Stove/dp/B004EPWM5S) and a blow torch.  I'd like to do some distillation with things like salt water and juice, and am looking at a distillation set (https://www.amazon.com/Distillation-Apparatus-Equipment-Chemistry-Glassware/dp/B07R8JFK97/). Is it safe to use this with a blowtorch or camping stove?

You should look into a water or sand bath and temperature controller. According to your link, that glassware is good to 800 Celsius. But using a blowtorch is a point source of high heat, and a camping stove lacks temperature control.

While this would have been considered normal in the past (I used such a burner when I was a kid), today bunsen burners are considered outdated as heat sources.
Of course it will still work fine fine, but our view on risk has changed.
Normally one wants to avoid having any source of ignition in the lab, and burners are also difficult to control.
That being said, a burner like this one  will make for a very nice burner with a focused flame. This type of burner works really well when you need an open flame, like for melting salts onto magnesia rods, sterilizing inoculation spatulas, and showing how metals change flame colors.
You can even buy a portable bunsen burner that is pretty much the same:

One big downside is that the burner will transform into a flame thrower if is falls over.
Also, you will use a lot of gas. If you like working with an open flame, I would recommend to buy a small propane tank with a pressure regulator and a regular bunsen burner, a setup like this:

Just get the smalles propane tank feasible, like a 2 or 5kg tank, just to keep the total amount of flammable material as low as possible.
Any store that sells gas supplies will be able to set you up for little money, just buy a bunsen burner for propane.
Doing a distillation of salt water will work, but it will be difficult to control and you can easily shatter the flask by uneven heating. Maybe that is a good lerning experience!
Distilling anything flammable, such as alcohol, on an open flame would be reckless.
Some general advice:

Never leave a burner unattended, not for the shortest moment
Remove anything flammable nearby. Work outdoors if possible.
In any distillation setup, always be able to remove the source of heat very quickly. Don't use the burner as a support, so that you can take out the burner without your setup falling apart. That is especially important for a burner with an attached gas tank.
Make sure the gas tank does not get hot.
Always be prepared for any flask to break at any time. What would happen if the boiling liquid would be released. Would you get injured?
Wear safety glasses
Wear non flammable clothing that is closed at the neck. A traditional lab coat is not a good protection for your chest. Buy something like this
Learn about superheating and use boiling stones

Thanks. Would a hot plate be better than a burner? Should I just save up for one of those hot sleeve things?

depends on what you want to and how much you plan to distill. Chemists used open flames for a long time and it sounds like a useful and fun learning experience. Please don't distill anything flammable on an open flame. A hot plate doesn't work well because the thermal contact is bad, but you can use an oil bath (messy), just a metal pot filled with oil on top of the plate. you need to put the hot plate on a little scissor jack so you can remove the heat and don't overheat your residue. Plenty of heating mantles on ebay starting around $50,  that will be most convenient and safest way.

A cool thing to do is steam distillation of essential oils. https://www.engineering.iastate.edu/brl/files/2011/10/brl_essentialoils.pdf You can mix directly the plant material with the water and don't need to buy the special receiver. The oils will float on top.


text:
Why 4‐methylidenespiro[2.4]heptane cannot be converted to 1‐chlorobicyclo[3.2.0]heptane?
According to the solution, the following reaction is not correct:

But why? I have tried solving in the following steps:

Forming a carbocation by breaking the double bond.
Expanding the ring.
Adding a $\ce{Cl-}$ anion.

This should thereby form the compound that is shown. But the solution is given that this reaction is not correct. Can you please explain to me why?

Count the carbons!

It has lost a methyl group.

Yes got it... sorry for such a silly question

@ManuShukla No need to apologize, I find the question to be fine. Plus you added your thought process which is always a good thing to do.

Here the concept of angular strain comes into the picture. The ideal tetrahedral and is around 109° 29'. As the angle between the C−C bond increases, it is more away from the ideal $\mathrm{sp^3}$ hybridization and hence will become more unstable.
For a cyclohexane molecule, the angle between the C−C bonds is 120°, therefore the deviation from the ideal angle is 10° 31', whereas for a cyclopentane the angle between C−C bonds is 108°, therefore the deviation from the ideal tetrahedral angle is around 1.5°. This is a very low deviation.
Since the transition is from cyclopentane which has a relatively lower deviation from the ideal angle to cyclohexane which has a relatively higher deviation from the ideal angle, the above transition does not take place.
Plus the number of carbons are not balanced in your reaction.

Only your last line is relevant. (1) There's no cyclohexane in the question(!!). (2) Angle strain is only one part of a strain in a molecule. (3) The difference in angle strain is not as large as you think it is; certainly not enough to prevent a cyclopentane -> cyclohexane ring expansion from occurring. These types of reactions occur all the time. Although it feels kind of moot discussing this in detail here given point (1).

what i ment was that the angel strain for the cyclopentane is lesser than that of cyclohexane and hence the transition between to cyclohexane from cyclopentane does not take place


text:
Bonding and coordination of oxygen in a Ga2O3 crystal structure
I'm trying to find the crystal structure of gallium oxide ($\ce{Ga2O3}$). However, I find the images of the crystal structure in the peer-reviewed journals problematic.
First, I hop to the wikipedia article of Gallium oxide and find this image:
There was no labeling of the color, but I think Cyan is Gallium, and Red is Oxygen. The trouble is Ga typically has 3+ and Oxygen has 2- valence. Yet in this image we see Ga have 4 and 6 bonds! How is this possible?
The source is from a paper:  J. Åhman, G. Svensson and J. Albertsson: A Reinvestigation of β-Gallium Oxide. In: Acta Cryst. (1996). C52, 1336-1338. (The image was made by a wiki admin on chemistry structure, who made many structures for wikipedia.)
So I check out the article and it has this figure:

That's basically unreadable, so I go to Acta Crystallographica Section C and it has an interactive 3D chem model viewer and a model for Gallium Oxide:

Again I see Gallium with 6 bonds and oxygens with 3 bond! Is this correct? please tell me how?
In Galazka's β-Ga2O3 for wide-bandgap electronics and optoelectronics,I find this description:

Crystal structure of $\ce{β-Ga2O3}$

$\ce{β-Ga2O3}$ crystallizes in the base-centred monoclinic system in the
  space group C2/m. The unit cell (figure 1) contains 20 atoms
  consisting of crystallographically inequivalent $\ce{Ga^3+}$ and $\ce{O^2−}$ atoms. In this low symmetry structure Ga atoms are coordinated tetrahedrally and
  octahedrally (Ga1 and Ga2, respectively), while O atoms are
  coordinated three and fourfold (O1/O2 and O3, respectively). O1 shares
  two bonds with Ga2 and one bond with Ga1, O2 shares three bonds with
  Ga2 and one bond with Ga1, while O3 shares two bonds with Ga1 and one
  bond with Ga2. The lattice parameters of $\ce{β-Ga2O3}$ are listed in table
  1. There are two easy cleavage planes: the (100) plane formed by O3 atoms and (001) plane formed by O1 atoms.

and this figure:

Can someone please explain the crystal structure of $\ce{Ga2O3}$, beta-gallium oxide?

Umm, for an OP with such skill in writing posts and searching, your query is rather disappointing. There is nothing strange about this. Much like there isn't about FeF6 anion or H3O+ cation.

https://chemistry.stackexchange.com/questions/33964/why-can-oxygen-have-a-coordination-number-of-3-in-a-complex-between-silver-and-s

@Mithoron, The first link does not clarify my issue, since in that link it specifically say the oxygen atom has a positive charge, but in Ga3O2 structure there is no charge.

@Mithoron, coordinate number is not the same as number of bonds. Are you saying in the ball-and-stick figures above, the sticks are not covalent bonds, but something else?

It is a low symmetry crystal. None of the Ga or O are perfectly happy, but altogether they are happier in that structure than in another. Actually not that weird of a structure compared with many other oxides. It does seem strange coming from, say, Al2O3, but one of the reasons you see that first is because it is simple and straightforward.

Well, it would have plenty of formal charges, if they were written down there. Trivalent Ga is Lewis acid, divalent oxygen - Lewis base. That would  be very strange if coordination num. would be that low. It's exactly the same with H3O+

Does it help if you think of this as an ionic compound? In ionic compounds, there are no directed bonds. The coordination number depends on the size of the ions, coupled with the need for electroneutrality.
For example, in calcium oxide, both calcium and oxygen (in red) have octahedral coordination, with 6 neighbors each:

Granted, the electronegativity difference in calcium oxide is 2.44, while in gallium oxide it is only 1.65, and not different from that of silicon oxide, which has a network covalent structure.
Also, consider structure like that of hexagonal ice, where each oxygen is surrounded by four hydrogen atoms, two connected by covalent bonds, and two connected by hydrogen bonds. In this case, the atomic distances are clearly different, and you would be correct to say that oxygen makes two bonds, with a tetrahedral coordination if you consider the hydrogen atoms from neighboring atoms.
To distinguish these three extremes (ionic bonds in one case, covalent bonds plus contacts with neighboring atoms, network covalent structure) and to categorize what is going on with gallium oxide, you would have to look at the gallium oxygen bond lengths, which should be available in the paper you cite.

Thank you Karsten! So if I understand correctly, I should not assume the stick in the above ball-and-stick models are always representing covalent bonds. Instead the can represent any bond, and I should look to bond length for guidance. Yes, there are definitely different bond lengths of gallium oxide crystal. Thank you for pointing me in the right direction!


text:
Why can SN2 reactions occur when the reagents are a primary substrate and a weak nucleophile/base?
The reagents are ethyl bromide and water. In this Khan Academy video at 5:20, the narrator says because ethyl bromide is a primary substrate, the reaction is neither SN1 or E1. This makes sense because SN1 and E1 favor tertiary carbocation.
But between SN2 and E2 reaction, the narrator says SN2 is more favorable. However, water is a weak nucleophile/base, while SN2 or E2 reactions require a strong nucleophile/base.
Why are SN2/E2 able to happen in this case? Is it because water can only act as a weak base but not a weak nucleophile so that SN2 is more favorable?


text:
Does SN2 favor strong nucleophile and SN1 favor weak nucleophile? Does E2 favor a strong base, and E1 favor a weak base?
I understand if you have a protic solvent, it will stabilize the strong base (to form weak acid) or the strong nucleophile. The protons would react with them. To have an SN2 or an E2, so you need an aprotic solvent. I also know that in general, a tertiary substrate will favor SN1/E1 reactions, because a tertiary carbocation is more stable. A primary substrate will SN2/E2 reactions, because of less stereo-hinderance.
But why does SN2 favor strong nucleophile and SN1 favor weak nucleophile from the first place? Besides, does E2 favor a strong base, and E1 favor a weak base? I also don't get the reason.

That's not the right view.
The $S_{N}2$ mechanism does not favor strong nucleophiles. That statement doesn't make sense at a high level. Mechanisms can't favor anything.
The key here is to look at what controls reactivity in $S_{N}2$ and $S_{N}1$ reactions.
The former is concerted where the bond breaking and bond forming take place at the same time. Therefore, the strength of the nucleophile is important as is the propensity of the leaving group to leave.
In the latter mechanism, there are two steps, bond breaking and bond forming, where bond breaking is the slower and more difficult step. Since the nucleophile is not involved in that first step, the strength of the nucleophile simply doesn't matter.
Similar considerations apply when looking at elimination reactions.


text:
Is silicon dioxide decomposition possible?
I was thinking about this, I made Gibbs free energy calculations and I can't figure out if
$$\ce{SiO2(s) -> Si(s) + O2(g)}$$
is actually possible, even at a ridiculous high temperature.

I suggest having a look at Si-O phase diagram, e.g. https://doi.org/10.1007/s11669-007-9062-5 (or search images). By the way, you shared neither your Gibbs free energy calculations nor the temperature you arrived at; however, the $\ce{SiO2}$ decomposition temperature definitely lies above 3000 °C at 1 bar. Not sure it this counts as "ridiculous high" though.

Per @andselisk's comment the melting point of $\ce{Si}$ is 1414 °C, and  $\ce{SiO2}$ is 1,713 °C so the decomposition reaction would have to be: $$\ce{SiO2(l) -> Si(l) + O2(g)}$$

@MaxW I think it would be rather gas phase, as andselisk mentioned temp. needs to be higher then 3000 °C and SiO2 has boiling point (i.e. it doesn't decompose before boiling)

Well, gaseous SiO2 has to decompose at high enough temperature, but reaction may be rather be: SiO2 (g)⟶SiO (g) + O (g)

@Mithoron - The decomposition could be a gas phase reaction. The B.P of SiO2 is  2,950 °C, so just below 3000 °C andselisk mentioned. // Thanks for making this point.

related https://chemistry.stackexchange.com/questions/37562/difference-between-dissociation-energy-and-dissociation-temperature-for-water

Anything decomposes at high enough temperatures, though at that point you may end up with a gas of free atoms or a plasma. There is no silicon dioxide in the sun (though silicon monoxide has been detected in stars )

@andselisk your link paywalls.  I would think a figure is available without one.  Thanks.

@OscarLanzi True, the article is paywalled, just like many other scientific articles are; that's why I added direct link to Google Images. Besides, there are means to overcome this limitation that exist for nearly a decade.

Well, SiO2 seems to decompose to large extent just during vaporisation https://arxiv.org/ftp/arxiv/papers/1312/1312.7592.pdf

Silicon dioxide is difficult to decompose directly, but it can be broken down in the presence of other reagents at more moderate temperatures.  In particular, the reaction of silica with elemental silicon has been known since 1905:

In 1890, the German chemist Clemens Winkler (the discoverer of germanium) was the first to attempt to synthesize $\ce{SiO}$ by heating silicon dioxide with silicon in a combustion furnace.[1]
$$\ce{SiO2 + Si ⇌ 2 SiO}$$
However, Winkler was not able to produce the monoxide since the temperature of the mixture was only around 1000° C. The experiment was repeated in 1905 by Henry Noel Potter (1869–1942), a Westinghouse engineer. Using an electric furnace, Potter was able to attain a temperature of 1700° C and observe the generation of $\ce{SiO}$.[2] Potter also investigated the properties and applications of the solid form of $\ce{SiO}$.[3,4]

The reaction may be used to remove silica from ores.
Cited references:

C. Winkler Ber. 23, (1890) p. 2652.

J. W. Mellor "A Comprehensive Treatise on Inorganic and Theoretical Chemistry" Vol VI, Longmans, Green and Co. (1947) p. 235.

U.S. Patent 182,082, July 26, 1905.

E. F. Roeber H. C. Parmelee (Eds.) Electrochemical and Metallurgical Industry, Vol. 5 (1907) p. 442.  Link

Cool, but that doesn't answer the question. Reduction is sure more reasonable, but very different thing.

Did I not make that qualification at the beginning?

You sure did, but why talk about entirely different reaction?

I rashly assumed that this approach could best be explored by giving an example of a possible reaction.

Well, this does suggest that SiO2 may decompose rather into SiO, so it's still can be valid point...


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Possibility of SET mechanism of Wurtz reaction
I was studying the Wurtz reaction, and I wanted to figure out the mechanism. When I searched the web, I came across the single electron transfer (SET) mechanism explanation. How can the alkyl group and the halogen form free radicals when the halogen is clearly much more electronegative then the alkyl group?

How can the alkyl group and the halogen form free radicals when the halogen is clearly much more electronegative then the alkyl group?

To answer your first question, it is important to note that the halogen does not form free radicals in this mechanism. Instead, the driving force is the net formation of sodium halides, as illustrated in the mechanism I drew below. In the preceding sections, I will try to compare the Wurtz mechanism to a perhaps more familiar one, the Grignard mechanism. This should help elucidate some general aspects of organometallic mechanisms.
If you are familiar with the Grignard reaction, the following comparison may help clear up the involvement of radical species in the mechanism. The Wurtz reaction is in fact very similar to the Grignard reaction. Both involve the 'insertion' of a metal species in between the carbon-halogen bond. In the case of the Grignard reaction, this metal is magnesium. This allows for the formation of a simple alkyl magnesium halide; however, with the Wurtz reaction, the addition of a sodium atom in between the carbon-halogen bond results in the formation of neutral sodium halide and a carbon radical. This carbon radical will rapidly react with another equivalent of sodium metal to form a carbanionic species, which will in turn displace another halogen in an Sn2 reaction.
The Wurtz reaction is not a super useful reaction due to the formation of this reactive radical. Most chemists would opt to use the much gentler Grignard reagent. Below, I have drawn out the mechanism of the Wertz reaction of 1-bromo-3-chlorocyclobutane to form bicyclo[1.1.0]butane. In this, you can see that a carbon radical is formed. This radical must come in contact with another equivalent of sodium before it is neutralized to form a carbanionic species. For this reason, the radical will also have plenty of time to react with other parts of the molecule in unwanted ways. The Grignard reaction, on the other hand, results in the formation of a radical in the 'reduction' step; however, this is neutralized very quickly in the recombination step and will thus not react unnecessarily with other parts of the molecule. I have included a diagram of the Grignard reagent formation below.
Mechanism of the General Wertz Reaction

Mechanism for the Formation of a Grignard Reagent


text:
Why is HCl the acid of choice for swimming pools?
Why is muriatic acid the go to acid to put into pools? Does it work better than nitric acid, sulfuric acid, or another strong acid? I mean they all will add the same amount of H+ right? Note, we use some cyanuric acid as well, but mainly HCl for adjusting pH.

The HCl not only increases the acidity of the pool but also increases the concentration of "free chlorine", or as chemists like to call it, hypochlorous acid. This is why most pools smell strongly of bleach. Also, HCl is much cheaper and less dangerous than nitric and sulfuric acid.

"muriatic acid" term is not used for many decades, some young chemists may not even know what it is. Hydrochloric acid is the term to use. I see this is not the first post combining the archaic term with pools and the author (you?) was already informed.

@Poutnik well I took a few chemistry classes so I know it as HCL. But when marketed as pool supplies, they call it muriatic acid. Even says so on the box

Yes, I have already guessed it would be the case.

There are several reasons not to consider these acids to be interchangeable for use in swimming pools. A total answer covers aspects of chemistry, biology, safety, availability, and cost. As hydrochloric, nitric, and sulfuric acids are all strong acids and can provide the desired acidification, they are very different when you consider their respective counter ions. Do you want chloride, nitrate, or sulfate in your pool water?
Safety First:

Hydrochloric, nitric, and sulfuric acid are all strong acids that are capable of causing chemical burns and present inhalation hazard from their fumes.
Additional safety information can be obtained from PubChem.:
Hydrochloric acid
Sulfuric acid
Nitric acid

Chemistry:

Nitric and sulfuric acids are more corrosive to copper and stainless steel. Copper (I) chloride is significantly less soluble in water than copper salts of nitric and sulfuric acids. Swimming pool heaters are likely to contain copper pipe.
As an added benefit for salt pools, which use an electrochemical chlorine generator, the Cl- ions are converted to chlorine for disinfecting the pool water.

Biology:

As pointed out in another response, nitrates can promote algae growth.

Availability and Cost:

Hydrochloric acid is commonly available as muriatic acid (31.5% HCl, ~10 molar) from home centers, building and construction suppliers, and hardware stores, because it is used in many construction-related activities, with cost ranging from \$5 - 9 (USD) per US gallon. A gallon will contain approximately 37 mole equivalents of H+.
Sulfuric acid is available from auto parts retailers as battery acid (37% H2SO4) and costs approximately \$10 (USD) per US quart ($40 per gallon). A gallon of battery acid will contain approximately 36 mole equivalents of H+ (adjusting for normality). Sulfuric acid costs 4-8 times more than hydrochloric acid for equivalent pH reducing capacity.
Nitric acid is less readily available to the consumer market. One on-line supplier offered reagent grade (70%) for \$125 (USD) per liter.

Added Note:
Cyanuric acid is a weak acid and serves as a stabilizer. Quoting from PubChem:

Cyanuric acid is used in swimming pools to lower the rate of photochemical reduction of chlorine, hypochlorous acid and hypochlorite ion.

About the price of H2SO4 versus HCl, I have always thought the latter is generally more expensive with H2SO4 being the cheapest mineral acid. Is it possible that the "accumulator grade" H2SO4 is extra expensive, because like any product aimed to cars is often much more expensive than similar product not aimed on cars ? At least I have such experience.

@Poutnik Price and availability may depend upon country. I'm speaking for US markets and don't know what the situation is in Europe or other locations, which may be very different. For the US, sulfuric acid is also found in some drain cleaners, but those products contain additives, like gels and "metal inhibitors," that seem undesirable for a swimming pool. Muriatic acid remains the most cost effective. Acid concentration and additives are also treated as trade secrets for some products.

I had rather meant acid itself may be much cheaper than final products aimed to particular customers.

Comparing prices from Sigma-Aldrich (now Millipore), reagent grade H2SO4 (95-98%) costs $85 (USD) for 2.5 L versus reagent grade HCl (37%) which costs $108 (USD) for 2.5 L. Sulfuric acid is cheaper than hydrochloric acid, but readily available muriatic acid (31.5% HCl) is still less expensive for consumer markets.

Reagent grade is too good for pools, I guess.

I would agree with that.

The equivalent amounts of hydrochloric, sulphuric or nitric acid are interchangeable, acidity-wise. But there are other properties that may lead to prefer one over the others:
Nitric acid would provide nitrates as food to algae or other microflora.
Sulphuric acid is unpleasant for manipulation, as even tiny droplets  create during the time holes in clothes.
Additionally, but not sure if it happens, calcium rich water, when mixed with the acid, may locally and temporarily create a haze of calcium sulphate due its limited solubility, what could raise questions if anything is wrong.
Cyanuric acid is a weak acid ( weaker than acetic acid ), not well suitable for pH adjustment. It is rather used as the chlorine stabilizer.

Cyanuric acid is used as a chlorine stabilizer in swimming pools. It binds to free chlorine and releases it slowly, extending the time needed to deplete each dose of sanitizer.


text:
Why does pouring carbonated beverages onto fruit cause them to fizz?
I have noticed that making mimosas, the champagne fizzes more than just pouring it into a flute. Likewise, pouring beer onto citrus fruit or strawberries fizzes a lot. My theory is that the pH drop associated with the acidic fruit causes the solubility of CO2 to decrease and hence, the fizz. Can anyone verify this?

https://chemistry.stackexchange.com/questions/98739/why-is-there-only-a-slight-change-in-ph-when-soda-loses-its-fizz https://chemistry.stackexchange.com/questions/330/why-does-ice-cream-make-soda-fizz https://chemistry.stackexchange.com/questions/47071/what-causes-soda-to-fizz-and-how-can-it-be-stopped

All carbonated drinks are oversaturated solutions of $\ce{CO2(g)}$. Pouring over fruits lead to contact with many residual air bubbles, which serve as  fizzing places for $\ce{CO2(aq) -> CO2(g)}$.
It is more or less the same as the ( previously already opened ) drink shaking, where violent fizzing occurs as well.
It is a similar process activation, like if you bend the activation chaff in a pocket heater with molten $\ce{CH3COONa \cdot 3 H2O}$, or if you pour overcooled water from a freezer ( see YouTube ) and watch fast ice growing.


text:
Is enthalpy affected by isotopes?
Enthalpy is a sum of $U$ and $PV$, and I thought that for different isotopes, internal energy values may differ. Is it true? There are three stable isotopes of $\textrm{Mg}$, mass number 24, 25, and 26. If enthalpy is different, which one is a standard for enthalpy $0$? I read somewhere on the web that enthalpy of a covalent bond is affected by isotopes, but I'm confused when there are no covalent bonds.

Enthalpy is slightly affected by isotope composition, but mostly below enthalpy measurement errors.
Generally, the standard enthalpy is considered for natural mixture of stable isotopes, unless considered for particular isotope explicitly.