text:
Stereoisomerism
The C-C bond in alkanes can freely rotate , then how can you call 2 alkanes (having atleast one chiral carbon) as diastereomers if they have different arrangements of groups (atoms)  in space & are Non super imposable on thier mirror images because they can rotate around C-C bond and become same as the other molecule
For example
 2-bromo 2-methyl 1-phenyl ethanol
Here there are 2 chiral carbon , one with phenyl (let us call this First chiral carbon) and another with bromine (second chiral carbon). In first chiral carbon,  if we interchange the position of phenyl and OH group, we get 2 structures which are  not mirror images of each other and hence my sir told me that they are diastereomers. But I think the C-C bond between the two chiral carbon can freely  rotate ,then by rotating the C-C bond keeping the 2nd Chiral carbon fixed we get the same structure . So I got confused here please help.

Welcome to ChemSE. Please clarify your question. Perhaps provide an example. One center of chirality does not a diastereomer make.

Also please split your question in two or three sentences, with some punctuation? Its barely understandable.

Sorry I have changed the question you can see

You can't change the stereochemistry of a chiral centre by rotation of single bonds. So, you cannot make the two diastereomers "the same", no matter how much you twist and turn it. Make a model and try it out.

Please explain how otherwise send any 3d animation of it

Here is a "3D animation" per your request. Changing the configuration at C1 in structure 1 produces diastereomer 2. They are diastereomers (constitutional or structural isomers) because they have the same atom connectivity but different arrangements in space. The same is true of the enantiomeric pair ent-1 and ent-2. The diagonal relationships , 1 <--> ent-2 and 2 <--> ent-1, are diastereomeric as well.


text:
How to charge titanium dioxide particles for a DIY e-paper capsule?
I'm trying to make a single e-paper capsule (1 cm × 1 cm) at home. At first, I tried to charge the $\ce{TiO2}$ particles by putting them in $\ce{H2O}$ with $\ce{NaOH}$ and then putting that whole solution in baby oil, but it turns out I shouldn't have any water in the whole process. I also tried adding a surfactant (liquid soap) but it also has water. Water is a problem because it forms micelles around the $\ce{TiO2}$ particles and prevents proper spreading. How do I charge the $\ce{TiO2}$ particles without water (and $\ce{NaOH}$). What oil-soluble surfactant I can use that doesn't contain water? Does anyone know any details about how the chemicals of the e-paper display work?
I researched a lot and it turns out the big companies use complicated processes and mixing different chemicals with expensive lab machines to produce the desired charged $\ce{TiO2}$ solution. Is there any way I can do it at home?

From the abstract of an article describing a TiO2 preparation protocol$^1$:

It was found that highly charged TiO2 particles could be prepared by milling 1.0 g of TiO2 particles, 0.5 g of silane additive as a charge controlling agent and 0.5 g of dispersant in 100 mL of cyclohexanone as a suspension fluid for 8–10 h using pot mill, followed by treatment with 0.4 g of acetic acid.

What is a common silane additive used as a charge controlling agent? And what would be an appropriate dispersant here?
Can anyone here recommend additional preparation protocols, perhaps something that might be considered a "standard recipe"?
References

Park, Lee, Park, Jin, Choi, Hae, Han, Yoon Soo, Kwon, Younghwan, Choi, Hyung. Fabrication of charged particles for electrophoretic display.
Current Applied Physics 6(4):644-648.

Why would "big companies use complicated processes and mixing different chemicals with expensive lab machines to produce the desired charged TiO2 solution" if there was an easy and cheap way to do it?

True, but I spent my whole summer vacation researching and experiment and it's hard to let go.

At the beginning, it was as easy as mixing H2O, NaOH and TiO2, that was the first step. But then I learned that I shouldn't have any H2O anywhere, and it turns out, charging TiO2 particles without water is hard.

Start by figuring out the composition of the commercial stuff. Then reverse engineer...

The engineering is already known, https://www.researchgate.net/publication/232414208_Fabrication_of_charged_particles_for_electrophoretic_display it's just too complicated for a 17 year old to do at home without the budget and proper equipment. My last option is to find charged TiO2 online to buy, or very cheap e-paper display.

It is extremely cool that you are pursuing this, and you may yet find some way to make something similar work. You might want to look inside that article for more details, unfortunately I can't help you with that.

The following product is listed as a "charge control agent for toner cartridges": https://www.gelest.com/product/SID3395.6/

You have regular and uniform $\ce{TiO2}$ nano-particles? Those aren't too easy to come by either...

@StianYttervik I asked people at ClearInk and E-Ink and they said their process of making the fluid is a trade secret and they don't share it just like that. There are a few documents leaked though and it's pretty complicated. It requires special machines and so on...

Heh, yeah I should think so. But it isn't that complicated to just make the nano particles (the regular and uniform bit can be, though) - it could be out of reach of regular household items. There are pyrolysis routes that I am vaguely aware of. If there is a university nearby, I'd try to ask one of the chemistry professors. The method you describe in the post is... interesting. Silane ($\ce{SiH_4(g)}$) is a quite explosive gas, I wouldn't try milling it without having the atmosphere in order.

@StianYttervik I think that recipe refers to a substituted silane, dispersed in solvent, and not silane gas.

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text:
How to predict the decomposition products of for example barium carbonate?
I am a high school student studying basic chemistry and I am confused about the process of writing decomposition reactions.
The question I have encountered asks me to decompose $\ce{BaCO3}$.
Online, I have found that this decomposes to $\ce{BaO + CO2}$.
My question is, essentially, is there a method to determine which elements go where. I'm having trouble finding answers.

A hint is usually in the name. Carbonates usually turn into $\ce{CO_2}$, hydrates become $\ce{H_2O}$ - nitrates become $\ce{NO_x}$ - nitrides becomes $\ce{N_2}$... and so forth and so on. These substances were named before we had spectroscopy - many of them were named after the gas that they decompose to.

I disagree with the question being labelled as too broad. The answer winds up being too broad, in that there aren't a set of rules that could be condensed into a stackexchange post, but the OP's question is clear and M. Farooq's answer gives a clear explanation of why the question is not as simple as it might seem.

Chemistry, despite being several hundred year old science, it is still very much empirical. In the beginning you will have to learn and accept some facts. Later you will start to see patterns (hence the periodic table). If you learn that the carbonate of barium decomposes into carbon dioxide and barium oxide, it will be safe to say that Mg, Ca, Ba, Sr, will do the same. Here you have learnt four more chemical reactions. If you recall the periodic functions in mathematics, which repeat themselves periodically, in the same way, elements repeat their properties in the periodic table.
When you strongly heat inorganic salts, usually you get smaller products (e.g., carbon dioxide and barium oxide) which are energetically more stable with respect to the starting material.

The recipe isnt really complicated:

(I guess you already know that) Barium metal and oxygen are very reactive with each other. So you would have to go to very high temperatures to reverse that reaction.
Carbonates contain $\ce{CO3^{2-}}$, which is $\ce{CO2}$ with an extra Oxygen anion $\ce{O^{2-}}$. Such decomposition reactions usually make as little rearrangements as necessary, so it is likely that the $\ce{O=C=O}$ motive is retained.
You might also know that $\ce{CO2}$ is very stable, and only decomposes to $\ce{CO2 -> CO + C}$ at rather high temperatures (Bouduard equillibrium).
More complicated structures (molecules) are usually unstable at higher temperatures, so we can just forget about them.
You probably already know other carbonates give off carbon dioxide upon heating.
Gas evolution is usually favoured, because gas has high entropy.

Voilà, thats it. The thing was more or less clear after the second of my bullet points, but the rest fits in as well. If the solution was sth else, that would be quite unusual and you would either have heard about it or your teacher wouldnt expect you to know. ;)

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text:
What's the biggest organic molecule that could have a smell?
For a molecule to have a smell it's necessary that the molecule be volatile enough to be in the air. So I think that excludes molecules which are solid at room temperature and atmospheric pressure. Maybe the question then is equivalent to: what is the highest molecular weight organic compound which is liquid at room temperature and atmospheric pressure?

It can't be a simple as solid vs liquid. Some solids, for example, have distinct odours. Some of these are not, strictly, the odour or the material but of reaction products of the material (osmium has a sharp and dangerous smell because of the formation of osmium tetroxide). And menthol is a solid with a high vapour pressure and a very distinct odour.

A smell is not a property of a chemical, but of a nose ability to detect it  Without knowledge of threshold partial pressure sensitivity for all compounds, the question cannot be answered.

Wax, as in a candle, has a pretty distinctive smell, and it's pretty big.

Being solid or liquid is not relevant. Is there a vapor pressure and do the molecules affect the sensors in our [or the dog, cat, butterfly, hyena etc.] sense organs. Even metals have an odor. The question shows a profound lack of observational power and lack of chemical knowledge.

I'll quote from $\ce{[1]}$:

The general requirements for an odorant are that it should be
volatile, hydrophobic and have a molecular weight less than
approximately 300 daltons. Ohloff (1994) has stated that the largest
known odorant is a labdane with a molecular weight of 296. The first
two requirements make physical sense, for the molecule has to reach
the nose and may need to cross membranes. The size requirement appears
to be a biological constraint. To be sure, vapor pressure (volatility)
falls rapidly with molecular size, but that cannot be the reason why
larger molecules have no smell, since some of the strongest odorants
(e.g. some steroids) are large molecules. In addition, the cut-off is
very sharp indeed: for example, substitution of the slightly larger
silicon atom for a carbon in a benzenoid musk causes it to become
odorless (Wrobel and Wannagat, 1982d).
A further indication that the size limit has something to do with the
chemoreception mechanism comes from the fact that specific anosmias
become more frequent as molecular size increases. At the “ragged edge”
of the size limit, subjects become anosmic to large numbers of
molecules. An informal poll among perfumers, for example has elicited
the fact that most of them are completely anosmic to one or more musks
(e.g. Galaxolide® mw 244.38) or, less commonly, ambergris odorants
such as Ambrox®, or the larger esters of salicylic acid.
One can probably infer from this that the receptors cannot accommodate
molecules larger than a certain size, and that this size is
genetically determined (Whissel-Buechy and Amoore, 1973) and varies
from individual to individual.

N.B.: Labdane's molecular formula is $\ce{C20H38}$, which gives a molecular weight (MW) of $\pu{278.5 Da}$ (Da). $\ce{[5]}$  Thus either the $\pu{296 Da}$ value is a typo, or the authors were quoting the MW of a labdane derivative.
Note added in response to answer posted by John Cuthbert (which was a nice find!):
While iodoform, at $\pu{394 Da}$, does indeed exceed the $\pu{>300 Da}$ "general requirement" provided above by Turin & Yoshii, a comparison of its estimated density to that of, e.g., labdane, indicates it's a much smaller molecule (iodoform's three iodine atoms add a lot of mass without a lot of size, at least relative to carbon, hydrogen, and oxygen):
I couldn't find labdane's density, but I found the density of one of its diols (i.e., labdane with an $\text{–OH}$ substituted for $\text{–H}$ in two places). So if we use its density, along with labane's molecular weight, we obtain:
$\pu{MW = 278.5 Da}$, $\pu{\rho = 0.9 g/cm^3}$ $\ce{[6]}$
=> estimated molecular volume ≈ $\pu{510 Å^3}$
Iodoform:
$\pu{MW = 393.732 Da}$, $\pu{\rho = 4.008 g/cm^3}$ $\ce{[7]}$
=> estimated molecular volume ≈ $\pu{160 Å^3}$
Even if the density of labdane were, say, 20% higher than that of the diol, we'd get a molecular volume of ≈ $\pu{430 Å^3}$, which is still far above that of iodoform.
This makes it clear that the limiting attribute is physical size rather than molecular weight, and that Turin & Yoshii were using molecular weight as a shorthand surrogate for size. This works reasonably well when comparing oxygenated hydrocarbons, but obviously breaks down when the compounds contain significantly heavier nuclei.  As Turin & Yoshii write more precisely at the end of the quoted passage:  "One can probably infer from this that the receptors cannot accommodate molecules larger than a certain size." [Emphasis mine.]
References
$\ce{[1]}$: "Structure-odor relationships: a modern perspective", by Luca Turin (Dept of Physiology, University College London, UK) and Fumiko Yoshii (Graduate School of Science and Technology, Niigata University, Japan), which appears as chapter 13 of: Handbook of Olfaction and Gustation. Richard L. Doty (ed.). 2nd ed., Marcel Dekker, 2003.
$\ce{[2]}$: Ohloff, G. Scent and fragrances: the fascination of odors and their chemical perspectives. Berlin, Springer, 1994.
$\ce{[3]}$: Wrobel D, Wannagat U. SILA PERFUMES. 2. SILALINALOOL. Chemischer Informationsdienst. 13(30), Jul 27, 1982.
$\ce{[4]}$: Whissell-Buechy D, Amoore JE. Letter: Odour-blindness to musk: simple recessive inheritance. Nature, 245(5421):157-8, Sep 21, 1973.
$\ce{[5]}$: https://pubchem.ncbi.nlm.nih.gov/compound/Labdane
$\ce{[6]}$: https://www.chemsrc.com/en/cas/10267-22-8_1640218.html
$\ce{[7]}$: https://en.wikipedia.org/wiki/Iodoform

Highly enlightening. As per Google, steroids cause body odor but do they have their own distinct smell?

The chapter from which I quoted indicates they do ["some of the strongest odorants (e.g. some steroids) are large molecules"], though  don't know what they smell like.  And I found this:  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2822864/

@M. Farooq some have urine like odour some are musk and animalic with sandalwood reminiscences. Also dry egg fishes, see " bottarga", might have a musk and mushroom/forest scent that I, perhaps erroneously, link to estradiol or triple C-C containing hormones.

I'm looking at the table of contents of "Handbook of Olfaction and Gustation. Richard L. Doty (ed.). John Wiley & Sons, 2015" and I don't see this listed... are you quoting from a physical copy of the book or something?

@pentane Sorry, the source from which I'd obtained the chapter incorrectly cited it as coming from the 3rd edition. It's actually from the 2nd edition (see: https://trove.nla.gov.au/work/29206984?q&versionId=46501355 ). I've corrected this in my answer.  A draft of the entire chapter (titled "Structure-odor relations...", rather than "Structure-odor relationships...."; the latter is how it appears in the book's TOC), uploaded by Fumiko Yoshii (the 2nd author), can be found at:  https://www.researchgate.net/publication/241197841_Structure-odor_relations_A_modern_perspective

Why does it need to be hydrophobic? Don't odorants solve in water when they are in the nose, so a receptor can detect them?

@cheeesus It's not the water solubility that's important, it's that hydrophobicity implies weak electrostatic interaction and therefore greater volatility. Hydrophilic molecules are polar (and often include hydrogen bond donors or acceptors) and therefore have strong electrostatic interactions among themselves in addition to strong interactions with water. For instance, ethanol, which is polar and acts a both a H-bond acceptor and donor, has a vapor pressure of 5.95 kPa, while propane, a nonpolar molecule with a similar molecular mass and geometry has a vapor pressure of 853.16 kPa.

@WaterMolecule except ethanol has an odor and propane does not...

@pentane Yeah, it's probably not a great example, since they are both very volatile. How about N-methylacetamide and methyl acetate. N-methylacetamide has a very hydrophilic amide group, a very low vapor pressure (~20 Pa), and is a solid under standard conditions. I've used it in the lab and never noticed any smell. Methyl acetate, although similar in geometry, is less hydrophilic, has a much higher vapor pressure (~20 kPa), and is a liquid under standard conditions. It has a pleasant smell.

(on behalf of the old-school chemists)
Iodoform (393.732 g/mole) has a strong characteristic odour.

Nice find--thanks!  See my edit above.


text:
Why not create a latent heat battery?
Since latent heat is not sensible, it seems like it would be useful in creating an efficient battery. Sensible heat interferes with electron flow in standard lead-acid and other batteries. The latent heat-electron interface might occur separately from the energy storage. Has this idea been explored?
Of course, I may be incorrectly assuming that latent heat does not itself add resistance to an electric circuit...

Maybe something like this? https://www.sciencedirect.com/science/article/pii/0022072894037025

Maybe! The discussion is inaccessible to me but it appears there may be a state change involved. The part about the interaction of H2O w/ an ion is intriguing.


text:
Equivalent mass in case of disproportionation reaction
I came across a question where the equivalent mass of hypophosphorus acid is to be found in the given reaction:
$$\ce{2H3\overset{+1}{P}O2 -> H3\overset{+5}{P}O4 + \overset{-3}{P}H3}$$
If $M$ is the molecular mass of $\ce{H3PO2}$ then $E_\text{oxidation}=M/4$ (for change in oxidation state from $+1$ to $+5$) and $E_\text{reduction}=M/4$ (for change in oxidation state from $+1$ to $-3$)
Equivalent mass of $\ce{H3PO2}$ in this disproportionation reaction
$$= E_\text{oxidation}+E_\text{reduction}= M/4+M/4= M/2$$
Now, I am not able to comprehend that, why these two types of equivalent mass has been added to arrive the equivalent mass of $\ce{H3PO2}$.
My take: The concept of equivalent mass collapses in this situation!

in my understanding, there is no sense in the eq. mass addition. Consider formally both original molecules as different compounds.

I like and dislike the question. I assume this is from your Indian university entrance exams. The good part is that it is tricky, the sad part is that the concept of equivalents is so outdated that any North American student will barely recognize it. They should abolish teaching equivalents exactly because this problem you are encountering.
Fact number no. 1, unlike molar equivalents of a given molecule equivalent weights are reaction dependent. Hence addition of equivalent weights is irrelevant. Equivalent weights are useful in analytical chemistry because they simplify calculations. The case of disproportionation reactions and their equivalents is rarely discussed because nobody uses disproportionation reactions in titrations, gravimetry etc.
In your case, you will have two equivalent weights one with respect to oxidation, and one respect to reduction. Solve the disporptionation reaction below and see what you get as an equivalent weight for $\ce{Br2}$ in each case.
$\ce{3Br2 + 6OH^- -> 5Br^- + BrO3^- + 3H2O}$


text:
Radiofrequency transmitter in an NMR experiment: Is there an involvement of (electromagnetic) radio wave?
A student in Chemistry StackExchange asked the effect of radio waves on matter, which led to an interesting set of arguments by some users. It was pointed out that in a proton nuclear magnetic resonance (or any other nuclei for that matter), there is no absorption of the photon corresponding to the radio waves. The following reference was cited NMR signal reception: Virtual photons and coherent spontaneous emission which says (in the abstract)

In portions of the magnetic resonance community, there is a misunderstanding of the process of nuclear magnetic resonance (NMR) signal generation and reception, and even in accepted texts, it is frequently described in terms of absorption and emission of radio waves, or radiation, by a two‐level quantum system. While this explanation can be refuted, for those who do understand that the NMR free induction decay signal is easily explained by Faraday's law of induction, reconciling the presence of an induced electromotive force with an apparent absence of transitions between nuclear energy levels causes conceptual problems.

I had never heard of emission of (electromagnetic radio waves) by the precessing nuclei in an NMR experiment, but a Google search of NMR + "absorption of radio waves" shows the concept of radio wave absorption is very common.
This leads to my main question: What does the RF transmitter do in the NMR experiment and what is practically happening in the RF coil?
(a) My understanding was that the RF transmitter, emits (electromagnetic) radio waves. If the RF coil is being energized with AC current of MHz frequencies, isn't the coil emitting electromagnetic radiation in MHz range? The magnetic field component of the radio waves interacts with the nuclei magnetic dipole.
(b) Another user stated "The "transmitter" powers the pulsed radio frequency B1 field. It never decouples from the probe coil to become electromagnetic radiation, i.e. actual photons. " Although this word "decouple" is more of a play of words, can an isolated oscillating B exist without a corresponding oscillating electric field?
I thought to learn from physical chemists and engineers who will have a better picture of the RF transmitters used in the NMR than us analytical chemists. My key question is related the Rf transmitter, not FID.
Thanks.

Note that it is closely related to how caesium atomic clocks work. In NMR case, radio photons flips proton spins, as there is spin energy difference in magnetic field. In the clocks case, there is Cs kernel spin energy difference due kernel electron spin coupling leading to hyperfine spectrum.  Effectively, the unpaired electron acts as atomic embedded magnet, creating magnetic field, instead of external magnets.

I think this is $the$ key point, is there an absorption of radio waves (as you call it radio photon) in an NMR experiment? I disagree with point (b).

I agree with (a) energy is absorbed by the sample and spins are moved from one energy levels to another. (b) implies that the coil and sample are coupled, it is just a matter of how you look at this, (as you write a play on words). Of course they are coupled, energy is absorbed.  The quote you give mixes up quantum and classical ideas. NMR is a funny mix of individual spin changes studied in a quantum way and then objects such as the magnetisation which are more easily understood in a semi-classical way.

See https://physics.stackexchange.com/questions/334826/potential-energy-in-nmr-and-energy-level-transition/335683#335683 or other similar NMR questions on Physics SE.

NMR requires a QM description to make sense of more advanced concepts, even if some of the basics can be explained using a classical description (Bloch equations). I have to admit that I don't really understand the quoted paragraph. Perhaps I belong to the portion of the NMR community that misunderstands NMR signal generation/reception ;-)

I finally asked an NMR spectroscopist who was the editor of the Journal of Magnetic Resonance and knows the inside out of NMR spectrometers. He sent a scanned copy of Hoult's paper from 1989. The title explains it all "The magnetic resonance myth of radio waves
." Surprisingly very low citations (15). It is amazing how much nonsense we have been fed with radio waves absorption. Sometimes innocent student questions on these forums can lead to introspection.
Here is the article Radio Waves Myth in NMR. Karl will like it.
The conclusion is that there is no involvement of radio waves, neither in absorption nor in emission (of course). The emission process is easier to understand but the transmission process seems to have confused even the experienced spectroscopists and more MRI people.
This is from another paper of Hoult. "The Origins and Present Status of the Radio Wave Controversy in NMR"

The origins, history, and present status of the controversy
  surrounding a quantum description of the NMR signal as being due to
  radio waves are traced. With the Principle of Relativity and Coulomb’s
  Law as formal starting points and the minimum of mathematics needed
  for understanding, the derivation of a classical electromagnetic
  theory of signal reception is first given. The agreement between that
  classical theory and a recent NMR experiment is then presented,
  leading to proof that, except for the highest field imaging
  experiments, there is no significant contribution of radio waves to
  the signal.

An energy exchange (i.e. absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.
Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.
An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.
(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)
The energy transfer during the $B_1$ pulse (your original question) happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.
The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.

I might also add that the matter is somewhat more complicated in high field MRI with humans, where the sample dimensions reach a similar dimension than the wave lenght of photons with that energy level.

That last paragraph is incredibly cryptic.

@BuckThorn Ja. ;) It d be great if anyone could improve on that. The last sentence reads like a "stop bugging me you idiots, I havent understood that myself".

@BuckThorn On second thought, I think the statement is still true and to the point. Imagine putting a powerful dipole emitter inside the probe. Photons (EM radiation) coming out of it would not be absorbed by the probe coil, at least not strongly. Ergo, the probe also cannot produce much photons on its inside.

Long time ago, a physicist had posed a question that a famous physicist had asked him: How will you convince an 11 year old about the size of the molecules in the air? He said nobody could provide a decent answer and he was curious to know our opinions. I suggested, what if we ask the child to look at the dust particles in the air near a sunny window. Since they are floating and constantly moving, chances are whatever is hitting those dust particles is invisible to us and must be smaller than the dust particles. The physicist said that it is a good start and the theme is correct... contd

contd....So the same goes for our friend Karl. A good rule of thumb in chemical education is to avoid complicated jargon because then it reduces to a mere play of words or as others said becomes cryptic. Often, at a deeper level it doesn't mean anything. I wasn't interested in the $acquisition$ of FID signal. And a preparatory $\ce{B1}$ pulse $alone$ doesn't mean anything either. I am more interested in knowing what happens in the RF transmitter coil and does it emit radio waves or not. Apparently there is an emission of radio waves and after all why it is called a RF transmitter?

@Karl Ok, thanks for freeing me of the shame of continuing to be confused :-)

@M.Farooq I think I understand better what your question is about. I would say, of course there is an RF field in the probe. The question seems more to be, how does that field interact with the nuclei? I think this can but certainly need not be treated semi-classically (it could be treated in a fully QM way, but why would you), ie using a classical Hamiltonian to describe the field but QM to describe the nuclei.

@M.Farooq The word "transmitter" is just jargon. Its a hf amplifier that could also be used to power a radio station. The actual probe circuit (coil+capacitors) is not part of the box labelled as transmitter. In between, there is usually a T junction with crossed diodes, where the "receiver" (the preamp, and mixer, ADC etc. behind it) are connected.


text:
Why does coumarin predominantly exist in keto form?
By applying the logic that phenol exists in enol form rather than keto form as it attains aromatic character, why is it not the case in coumarin?

Which enol form would you suggest? Try drawing it out! That may give you your answer.

If it was amide, the question would make sense...

Does 2-Hydroxyquinoline come in the enol or the keto form?

I have drawn the form that I think should be the predominant form

Oxygen is pretty electronegative, and so it doesn't really "like" to be positively charged.  Also in general it takes energy to separate charges, so zwitterionic forms aren't favored in general, especially without any proton transfers.

@CurtF. I think your comment is a complete answer.

@CurtF. I agree with your explanation.

There simply is no tautomer possible. There is no keto group in that compound to begin with, and there are no enolisable protons either. What you have drawn are two resonance structures. Both of them are valid configurations, but neither actually exist; see What is resonance, and are resonance structures real? On that note, I don't think the answer by @CurtF. has anything to do with it.

Good point Martin, I think you are right and thanks for reminding me why I made my response a comment rather than an answer!

The key point (I think) is that if atom nuclei (including hydrogen/protons) don't move between two structures, they aren't tautomers, but are different resonance structures.  Then the question is which resonance structure is "closer" to the real structure.  Perhaps my comment explains why the left structure is closer to reality, but the key point is that the analogy to phenol tautomers is invalid since here we are dealing with resonance structures instead of tautomers.

The answer is as simple as it will be unhelpful to your original question: There is no tautomer possible.
Coumarin does not have a keto group, and there are no enolisable protons either. Coumarin consists of a benzene ring, connected to a heterocycle with the functional group lactone. The molecule is flat and the conjugation extends throughout it.
What you have drawn are two resonance structures. Both of them are valid configurations (there are more), but neither actually exist; see What is resonance, and are resonance structures real? In both of your drawn structures the nuclei positions (in the clamped nuclei approximation) do not move.
The concept which applies here is completely different from keto-enol-tautomrie, where there are two different, observable molecules.
The question which you are probably asking is: Which of the drawn structures is the major contributor to the (so-called) resonance hybrid.
I have run a quick calculation of the DF-B97D3(BJ)/def2-SVP//GFN2-xTB level of theory, which suggests that the conjugation is skewed towards the lactone group. More investigations would be necessary, but the π orbitals should give you a general idea.

You have correctly pointed out what my question should be ! So, can the explanation for skewness towards the lactone form would be Curt's explanation in the earlier comment ?

@Abhigyan Essentially yes. Have a look at the Becke dipol-corrected partial charges of coumarin (DF-B97D3(BJ)/def2-SVP//GFN2-xTB); they clearly indicate that both oxygen atoms will accumulate electron density. Conjugation in this regard rather withdraws electron density from the adjacent ring. Note that this are quite crude simplifications, further study should be undertaken.


text:
Methanol mixed with varsol
If I have varsol in a tank and methanol is put into the same tank will methanol rise to the top or will it separate? Or will it mix together ?

please give a citation for Varsol, this is not a well-known material

Note that I have never heard about Varsol until now, living in the Central Europe. By searching Wikipedia ( what you could do as well before asking ), I realized the below:
Varsol alias White spirit alias many other synonyms, used as a painting solvent, is based on C7-C12 oil hydrocarbons, with density 0.765(type T1), 0.780(type T2), 0.795(type T3)
( divided by volatility, T1 the most volatile )
Methanol has density 0.792. So it would rather sink under T1, T2, or being indecisive with T3 ( this may be temperature dependent )
Note that methanol has very limited solubility in non-aromatic hydrocarbons.
Note: with reported partial aromaticity, methanol will be somewhat soluble.

Poutnik has given an excellent answer in general for your question. Since you asked about Varsol™ in particular (not white spirit in general), I'd like to give an additional insights for your benefit. Yet you didn't mention the paticular Varsol™ brand so I'll give physical data for all 3 Varsol™ brands avilable.
Varsol™ is an ExxonMobil Chemical brand for a line of conventional non-dearomatized aliphatic fluids boiling in the mineral spirit or white spirits range. ExxonMobil offers several Varsol™ brands, namely, Varsol™ 1 (distillation range: $\pu{156-204 ^\circ C}$), Varsol™ 18 (distillation range: $\pu{159-203 ^\circ C}$), and Varsol™ 110 (distillation range: $\pu{249-294 ^\circ C}$). Varsol™ is an aliphatic fluids, but is non-dearomatized. Their aromatic content ($wt \%$) could go as high as 29% depending on the brand. I have no experience with Varsol™, thus, I can only speculate that any Varsol™ brand's solubility level in methanol can be depend on their aromatic content values. For example, toluene and xylenes are missible in methanol but hexanes and heptane are not. However, since aromatic content is less than 50%, I can assure that there would be two immersible layers, assuming you are adding equal amount of methanol (based on the amount of methanol you are adding). In that case, which layer be on top can depend on which brand of Varsol™ you are using: Densities of Varsol™ 1, 18, and 110 are $0.789, 0.790$, and $\pu{0.825 gcm-3}$, respectively at $\pu{15.6 ^\circ C}$. According to the values given by Poutnik, Varsol™ 1 and 18 would be the upper layer while Varsol™ 110 would sink to the bottom when you add enough methanol to the brand of Varsol™ in your tank.
Source: Fluids at a glance - ExxonMobil Chemical


text:
Reaction of ether with HCl
Most of the reactions undergo ring expansion whenever favourable.
But this reaction [1, p. 514] doesn't seems to do so:

Practice Problem 11.18 Provide a mechanism for the following reaction.

Could anybody explain why?
References

Solomons, T. W. G.; Fryhle, C. B.; Snyder, S. A. Organic Chemistry, 12e ed.; John Wiley & Sons, Inc: Hoboken, NJ, 2016. ISBN 978-1-118-87576-6.

Presence of a good nucleophile Cl- that captures the stable cation faster than any ring expansion. Also 5 ring is pretty stable so expansion to 6 is not particularly favoured.

But it does rearrange to a tertiary cation.....wouldn't it be better if it's a tertiary cation in a 6 membered ring?

No, 6 ring is not better than 5

Oh, I had thought otherwise. What's the reason for it?

Ring strain slightly lower with 5 than 6

In the future, please add complete reference to the textbook you are quoting from (page number(s), edition, publisher etc.: see my edit as an example). Also, this question somewhat looks like a "covert" homework question: probably you can share with us some of your own thoughts. Other than that, it's a nice question IMO.

Yea will see to it that happens from the next time.

How would you get a tertiary cation if ring expansion occurs?

There are 2 things to consider: What would be the energy difference between before and after rearrangement from initial $2^\circ$-carbocation (I) and each of rearranged carbocations II and III:

It is obvious that the rearrangement of carbocation I to give carbocation II by ring expansion have minimal energy change since both initial and final carbocations are $2^\circ$ and ring expansion does not relieve too much energy because it is cyclopentyl to cyclohexyl. However, the rearrangement of carbocation I to give carbocation III by 1,2-hydride transfer relieves good amount of energy since initial carbocation is $2^\circ$ and final rearranged carbocation is $3^\circ$. Although the carbocation II would undergo further stabilization by 1,2-hydride transfer to relieve extra energy, I to III happens much faster than I to II transfer to give 3 as the major product.


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Did the finding of the electron from cathode rays/ oil drop experiment at least temporarily throw Dalton's Atomic theory "out the window"?
Because chemists learned that the atom was divisible, did Dalton's theory get any criticism or did they just adjust the theory to fit the new idea?

I'm voting to close this question as off-topic because it belongs on another site, probably https://hsm.stackexchange.com/

I'm not sure a 'chemist' at the time could be likened to a chemist today. While textbooks give an account that seems clear, logical, and direct, the reality is that figuring all this stuff out was pretty darn messy.

Dalton's atom was the most primitive idea. Keep in mind many physicists were against atomistic theory and cathode rays were not immediately recognized as electrons or a subatomic particles. The beauty of science is that it is self-correcting. This happens on daily basis still today. Nobody actually throws any theory out of the window or badly criticizes another until and unless they are bitter enemies. There are successive improvements in models which explain the experimental observations better and better.
A German company Miele has a nice logo "immer besser- always better". This is true for science. On the other hand, science is still in infancy. For example, can you define what is charge? Or What is mass? If someone comes up with an explanations, would we throw away all the previous work of physics or chemistry or hold the previous scientists in high esteem? Same thing happens with Dalton. His name will always have the due respect for proposing the clear cut idea of atoms.

Dalton's theory was about how atoms behaved: the observation of Cathode rays provided part of the explanation of why they behaved like this
The key elements of Dalton's theory were observational. He explained how the idea of atoms could explain chemical reactions and many observational facts about them. This was an important breakthrough and explained far more than previous ideas. But it didn't provide much of an explanation of why they behaved that way.
The observation of cathode rays was one of the first steps in providing the why not a fundamental challenge to Dalton's actual theory. Chemists didn't get a (mostly) complete explanation of what atoms behave the way they do or of some of the characteristic features of atoms until the discover of the neutron in the 1930s. Why, for example, do atomic weights not always follow the atomic numbers (or the pattern of chemical behaviour observed across the periodic table)?
We now understand that chemical properties are determined by the electrons in atoms and the number of electrons depends on the number of nuclear protons. But the mass also includes neutrons which occasionally alters the apparent periodic relationship between atomic mass and chemical properties.
None of this is a fundamental challenge to Dalton's ideas. All of it is a deeper explanation of why the observed properties and behaviours of atoms in Dalton's theory work as they do.
So we didn't throw Dalton's ideas out. We refined them with a more fundamental theory that explained why they behave like that.
This contrasts with other changes in our understanding of chemistry. For example, we discarded the old phlogiston theory of the explanation for combustion because observations didn't refine it but completely refuted it.

Atoms are not divisible in  chemical sense.
They are smallest amounts of chemical elements, similarly as molecules are  smallest amounts of  chemical compounds.


text:
Mass percentage of water in hydrate?
1.034 g of a metal sulfate hydrate was heated to a high enough temperature to remove all the water of hydration. The mass of the anhydrous salt left behind was found to be 0.956 g. What is the mass percentage of water in the hydrate?
I did $1.034-0.956=0.078$ grams of water
Then I did $\frac{0.078}{1.034}$= about $7.5$%
Is this correct?

Was not calculation of percentage taught in a basic/elementary school ? It is a real fundamental knowledge for anyvody who deals with chemistry. Rather a shame to ask.....

@Poutnik but is my interpretation of the question correct. Im not asking if the algebra is correct.

Lack of effort..

Chemistry calculation problems often disappear, if stripped down from the "dark chemistry magic".
It is really just percentage calculation, especially if everything has been commented explicitly like here.
Yes, it is correct.


text:
Why is the product of the reaction "Bi + O2" Bi2O3 instead of Bi2O5?
I am stuck on a homework question that states:

A piece of solid bismuth is heated strongly in oxygen.

The answer key states the answer as $\ce{Bi + O2 → Bi2O3}$. However, it is to my understanding that the metal Bismuth has a charge of +5 because of its location on the periodic table. Therefore, $\ce{(+5)[charge of Bi]*2 + (-2)[charge of oxygen]*5 = 0}$ giving me $\ce{Bi2O5}$. Please advise.

Many elements have multiple valencies. The most common oxidation state of Bi is +3. Bismuth (V) is also possible. Read more here on Bismuth oxides.

It is true that bismuth burns with a bluish flame in the presence of air to form bismuth trioxide($\ce{Bi2O3}$) as @M.Farooq mentioned. Bismuth pentoxide is not a pure compound but a mixture of various compounds like water, bismuth trioxide and bismuth tetroxide. It is also not stable at strong heat, decomposing at 393 K. Also, there is no direct preparation for this compound. You have to first make bismuth trioxide and then react with a bunch of other chemicals to form bismuth pentoxide.
So, the balanced reaction would be:
$$\ce{4Bi + 3O2 ->[\Delta] 2Bi2O3}$$


text:
Why can't O form OF6?
At first I thought that the reason is what I was taught, the absence of vacant d-orbitals but then I did some research. Most of the answers stated what I initially thought but one particular answer told that it's a common misconception (d-orbital theory). According to it, there can be two possible reasons-

Not much electronegativity difference between O and F.
The atomic size of Oxygen is not sufficient for it to hold 6 atoms.

So which one is likely to be more appropriate, the d-orbital theory or the electronegativity and atomic size theory?

See this question and the links therein.

Specifically this answer: https://chemistry.stackexchange.com/a/100938

While "OF6" looks pretty absurd, there's possibility of existence of OF3+ cation which would have IV+ ox. state for oxygen!

Whose to say it can't?  Maybe we just need super low temperatures and gigapascals or terapascals of pressure....

@mithoron wow!  Inquiring minds want to know your references!

@OscarLanzi https://www.sciencedirect.com/science/article/abs/pii/S0022113999001396

https://chemistry.stackexchange.com/questions/23496/how-does-absence-of-nd-orbital-in-oxygen-affect-its-valency

One possible explanation involves atomic size.  You need a large enough atom in the middle to stabilize all the occupied orbitals in the octahedral structure; sulfur makes it but oxygen does not.
This reference gives the molecular orbitals for sulfur hexafluoride.  A key feature of these molecular orbitals is a doubly degenerate pair, labeled $\mathrm{2e_g}$ in the above source, which effectively contains the "extra" valence electrons in fluorine-based orbitals.
The picture below, taken from a different source due to limited downloading options from the free one, includes one of these combinations (here labeled $\mathrm{3e_g}$, as different nomenclature is used) among others.

If we look carefully we see that the ostensibly nonbonding orbital has antibonding overlap between adjacent lobes and so is technically antibonding.  The same is true of the other component of this degenerate pair.  So to make the orbital effectively nonbonding and keep the octahedral structure from blowing apart, we need a large central atom to act as a "spacer" between orbital lobes.  Apparently sulfur is large enough, oxygen is not.

Nice answer, but what is "cathedral arructure"? I assume a typo, but not sure what you meant!

Corected, I think.  Tiny keys ơn my device +  fat fingers + overeager autocorrect = huge struggles for me to type.  Is there a stylus for Galaxy A50 phones?

@OscarLanzi Sorry for the off-topic comment, but I see you complaining about problems with typing and formatting quite often. The problem is, pretty much every smartphone is a device intended for consumption, not creation. All the external wireless or on-board keyboards I tried were horrible for anything remotely productive. If you would ask me, you were better off buying a refurbished Thinkpad (X series if you need something compact) for about $100, use smartphone as a hotspot and have a superior typing experience.


text:
Temperature measurement
Consider some liquid in a beaker. I am measuring its temperature using a thermometer. When i put my thermometer's bulb in the liquid, the particles of the bulb gain the same kinetic energy as of the liquid particles. Now, my question is that
Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid?
I have read somewhere that there is no effect of increase of rotational energy or vibration of molecules (as a whole) on temperature readings. Why so? Do the vibrations of the molecules don't change kinetic energy of the thermometer's bulb's particles?

Near the room temperature, many of the vibrational degrees of freedom in common molecules are frozen. They don't get any energy. The rest get $kT/2$ per each.

i am not specifying the temperature of the liquid.It is at any temperature. So, is rotational energy of the molecules or intramolecule vibrations get transferred to the bulb??

It surely does.

Let me suppose that the molecules of substance under measurement has 0 translational energy. So, the molecules of the substance which are very very near to the bulb having only intramolecular vibrations hit the particles of the bulb, wouldn't there be increase in the translational kinetic energy of those particles? In this way, if the substance under measurement has no translational kinetic energy of its molecules, but intramolecular vibrations, then too there will be some temp reading on the thermometer. Theoretically, its temp was to be 0( as translational kinetic energy is 0)

That's right (except the last sentence). Also, when the atoms of those molecules hit the atoms of other such molecules, they give them some translational kinetic energy, so in a very short order all energy in the sample will be distributed equally between all possible degrees of freedom.

it means that it is impossible for me to have 0 translational kinetic energy for molecules, if they have non-0 intramolecular vibrations.Then what about rotational kinetic energy. If i suppose that the molecules have no translational energy, but non-0 rotational energy. Is it going to affect the translational kinetic energy of the particles of the bulb (& other atoms)?

Same thing here. A rotating molecule will hit another and send it flying away.

Then what happens in the case of water at 100°C? When more heat is provided, why intramolecular vibrations increase, not the translational kinetic energy? Also, if once intramolecular vibrations increase, won't they hit with other molecules in the system increasing their translational kinetic energy and hence increasing the temp? But it is observed that temp of water doesn't increase after 100°C while it is getting converted to steam.

Water is complicated. Think of H2. It has some translational kinetic energy, and also some rotations and some vibrations. You add more heat, and all of a sudden the bond snaps. The molecule is no more; there are just two lone atoms. Where is the rotational energy? It is gone; the very concept does not apply anymore.

Please be specific to H2O. I suppose that in some way the heat provided increased only the rotational and intramolecular vibrations of the H2O. Then, when steam comes in contact with any object, let it be our skin, it causes more burns, compared to water at the same temp i.e. both have same translational K.E. of particles

1) You are correct that the thermodynamic temperature is a measure only of the translational kinetic energy. Intramolecular vibrations do not contribute to the temperature.
2) You are also correct that rotational and vibrational KE can be converted to translational KE in a collision (as long as total momentum and kinetic energy of the system are conserved).
To reconcile these two statements, we need to remember that temperature is a bulk property of the population, not a property of individual molecules. If we say that a mole of gas is as 273 K, it does not mean that every molecule of gas has the same amount of translational KE. It only tells us the population average.
We also need to remember that the average distribution of energy between translation and other movements (eg vibration and rotation) in the ground state is a fixed property of a substance in a given phase, represented empirically by the heat capacity.
If we look at a single collision between two molecules, we might see a change in the net translational KE if some energy is transferred to a rotational mode (resulting in a loss of translational KE). However, that means that the molecule in question is now in an excited rotational state. Across the entire population, that excitation will be offset by a different molecule dropping from an excited rotational state to the ground state by transferring some rotational KE to translational KE in a collision. The number of molecules in the excited state will on average remain at the value we calculate using stat mech.
So just as there are fluctuations in the distribution of translational KE among molecules within a bulk population due to changes in speed, there are also fluctuations due to transfers of energy between different types of motions, but the population average remains essentially constant if the population is large enough that stat mech approaches are relevant.
In your example, the local translational KE at one point in the thermometer bulb might increase, but it will be offset by a decrease somewhere else, such that the measured temp is an accurate reflection of the translational KE of the combined thermometer+material system.
In the specific case of solids, the idea of translational KE is confusing, since the molecules seem stationary. However, there are small movements of the centers of mass of the molecules relative to each other even if the molecules are confined. This is distinct from vibrations, in which the center of mass remains fixed as the atoms move relative to each other.
Consider, for example, $\ce{CO2}$. If the C atom remains in a fixed position and the C=O bonds stretch asymmetrically, there is a net movement of the center of mass. Likewise if the O's remain stationary during the same asymmetric stretch movement. Likewise for a bending mode. In reality, the motion of the atoms at any given moment is a complicated mix of the possible vibrational, rotational and translational movements.

Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid?

No, but they will have the same temperature. Solids don't have any translational energy, the atoms time-averaged positions are constant. If you want to link the temperature to the energy of the particle, it is easiest to start out with a monoatomic ideal gas because here, there is a linear relationship.
Relationship between change in temperature and energy
Although under some ideal circumstances, particles have an energy of $\frac{1}{2} k T$ per degree of freedom, how much energy is needed to raise the temperature of a sample depends on the sample (and the temperature). The heat capacity is a measure of change in energy per change in temperature. So the heat capacity tells you the total energy going into all degrees of freedom when the temperature increases. The splitting up into translational, rotational and vibrational energy depends on the nature of the sample (for example, the physical state) and the temperature.

if the particles of solids have 0 translational energy, then why do they have non-zero temp? isn't the temp of a substance the average translational energy of particles?

@user36956 No. The vibrational energy of the particles in a solid also matters. Heat isn't just translational energy.

OP is correct that for any state of matter temperature is a measure of translational KE only. Constant time-averaged position does not mean no translational KE. The center of mass can move back and forth around a fixed average position.

@Andrew This Q&A agrees with you: https://physics.stackexchange.com/questions/210685/do-solids-have-translational-energy. Why call the vibration in a ionic or metallic solid translational energy, but the vibration in a molecular liquid or gas not?

Vibration is movement of atoms within a molecule relative to each other such that the center of mass does not move. If the center of mass moves, it’s translation.  When you’re dealing with a metal instead of a covalent solid, the independent unit is an atom instead of a molecule. When a metallic solid is melted, the atoms move freely from each other, so it makes sense.

@Andrew So the substance has to "know" what we consider the independent unit so it can adjust its translational energy to match its temperature? I am thinking of diamond or graphite (for the latter, movement perpendicular to the plane of the layers counts but in plane does not?), or ionic compounds with polyatomic anions (movement of atoms in anion does not count, but that of monoatomic cations does?).

@KarstenTheis - It doesn't have to "know" anything. The termperature is a measure of the ability to transfer heat through collisions (ignoring for this discussion radiative transfers). At the bulk population level, that's a measure of the translational kinetic energy only. I'll try to explain that in a longer answer.

@Andrew Here is my new question: https://chemistry.stackexchange.com/questions/121313/what-are-the-degrees-of-freedom-in-an-ionic-solid-such-as-sodium-azide


text:
Why does bubbling carbon dioxide through calcium hydroxide result in a precipitate?
Why does the addition of carbon dioxide to calcium hydroxide create the precipitate calcium carbonate? What happens chemically for such a molecule to form? How exactly do their interactions come into play?
Now CaOH isn’t terribly soluble in water but the parts that do dissolve, disassociate into their respective OH- and Ca2+ ions. So when we bubble CO2 through the CaOH solution the calcium ion is attracted to the partial negative charge on the oxygen atoms of carbon dioxide. So that forms CaCO2 so my question is where does the other oxygen of calcium carbonate come from? Is it snatched from a hydroxide ion?
Calcium carbonate also has a resonant structure, I don’t know what that is exactly but it’s like electrons are being shared uniformly throughout a given structure so that means carbons original double bonds get disbanded too I guess? If anybody can explain what happens here as well, it would be great.
Here are my drawings, they may be simplified but it’s all I know:

There are millions of chemical reactions that just happen all the time. Why this one?

What exactly do you mean? I want to understand the interactions between the molecules and how they form the end products of calcium carbonate and water. Like the oxygens of the hydroxide groups and co2 are partially negative etc, I just want to understand how their interactions come into play

You put your comment and more rather to the question itself, for the below:

We have a policy which states that you should show your thoughts, effort and attempts to answer your question yourself. It'll make us certain that we aren't doing your homework for you, and that the Q/A is beneficial for broad audience.
As "homework class questions" are considered literal homeworks, self-study questions, puzzles, worked examples etc.
Please [edit] in your full reasoning or thoughts on this. See Homework
Otherwise, the question may get closed.

Oh alright, it’s not a homework question I’m just curious. Will edit.

The important thing to consider is the chemistry of carbon dioxide in water, forming carbonic acid. Note also that calcium hydroxide is at edge between being a strong or a weak base.
$$\begin{align}\
\ce{CO2 ^ + H2O &<=> CO2 . H2O} \tag{1} \\
\ce{CO2 . H2O &<=> H2CO3} \tag{2} \\
\ce{H2CO3 &<=> H+ + HCO3^-} \tag{3} \\
\ce{HCO3^- &<=>  H+  + CO3^2- } \tag{4} \\
\ce{ Ca(OH)2 &<=> CaOH+ + OH- } \tag{5} \\
\ce{CaOH+ &<=> Ca^2+ + OH-} \tag{6} \\
\ce{Ca^2+ + CO3^2-  &<=> CaCO3 v} \tag{7}
\end{align}$$
Yes, bonds in $\ce{CO3^2-}$ are equivalent, with strength and length somewhere between a single and a double bond. It is similar as  for $\ce{SO4^2-}$, $\ce{SO3^2-}$, $\ce{NO3^-}$, $\ce{ClO4^-}$, with sharing the "doublebondness" within the former two, and sharing the charge in all of them.


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The fifth shell of an atom
According to the $2n^2$ rule, the fifth shell should accommodate 50 electrons.
However, my chemistry book says that the $2n^2$ rule does not apply to the fifth shell.
Could you explain why this happens?
P.S. I have studied quantum numbers, Heisenberg uncertainty principle, Pauli exclusion principle, and the quantum mechanical model of an atom; please answer accordingly.
Thank You

It means very little.  We begin to occupy the fourth shell in neutral atoms before the third one is even half done, and by the time we would get to filling the fifth shell we are in a spaghetti bowl of shell and angular momentum combinations.

The $2n^2$ rule is nothing but a mathematical consequence of the mathematical description of electrons around an atom – what chemists call quantum theory and is currently the most accepted description of atomic matter by a mile. Thus, until quantum theory is significantly amended, the fifth shell does, in theory, have the capacity to hold $2\times 5^2=50$ electrons.
I say in theory, because shells aren’t filled all at once bottom to top. In fact, the $n^\text{th}$ shell has $n$ subshells denoted by the letters s, p, d, f, g, h, … which in neutral atoms are (usually; there are exceptions) filled in a specific order known as the aufbau principle. For the first two shells this doesn’t have any effect, but the 4s shell is filled before the 3d shell and the 6s shell before the 4f shell, to name just two examples.
To completely fill the fifth shell (i.e. fill the 5g shell), we would need to also fill the 6s, 6p, 6d, 7s, 7p and 8s shells. The heaviest known (i.e. synthesised according to the standards laid out by IUPAC and IUPAP) element as of 2019 is oganesson (Og) which also happens to be the last 7p element. Two heavier elements (the 8s elements) must be discovered before the third in line will most likely be the first 5g element. The g subshell can hold 18 electrons, so until we have one with a completely filled 5g shell some time will pass.


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Why are the phase diagrams showing $G_s$ and $G_l$ situated so close on one end and then widens up at the other extremes?

With reference to the phase diagrams drawn in b and d why are the Gibbs phase lines coming closer from being so further apart?

Those lines are just examples. As the temperature changes, the shape and relative position of the liquid and solid phase free energy curves will change. That is all.

After some thought, it is probably worthwhile to actually write an answer rather than a short, fairly useless comment.
Lets look at the Cu-Ni phase diagram, as calculated using the CALPHAD parameters given in the paper referenced in many of the pictures below (X.J. Liu et al., J. Electronic Materials 37(2) 210-217 (2008)).
Overall:

This looks a lot like your pictures above. Remember, red is liquid, blue is the fcc solid. The temperature (1800K) is above the melting point of both elements. The liquid and solid Gibbs free energy curves (top plot) are pretty close together on the Ni side, which makes sense since the Ni just melted. Cu melted some 400K earlier, so it should be expected that the solidus is well above the liquidus line there.
In the middle, the curves look like:

You can see the tie line around about 33 percent Ni. Pure copper has melted, so the liquidus is below the solidus. Pure nickel is still solid, so it is the other way round on the right side of the phase diagram.
Below the melting point of Cu, it looks like:

Now the solid free energy curve is below the liquid free energy curve everywhere - both pure Cu and pure Ni are solids. The liquidus on the copper side is closer to the solidus - the copper is closer to the melt temperature.


text:
What exactly is "oxidation state"?
the textbooks say that it is the amount of electrons lost or gained by an atom in a bond but in a covalent molecule the atoms never technically lose electrons ,instead share them to in such a way to complete their octet.for eg:- In OF2 molecule, it is said that the oxidation state of oxygen is +2 which implies it has lost 2 electrons. However it also must be gaining electrons from the other flourine atoms, right? please help me out with this.

The oxidation state is indeed the amount of electrons lost or gained by an atom in case of some highly contrived and artificial process which never really happens.

Whatever we say it is, maybe?

I don't understand at all

For calculation all those bond are treated as if there were ionic.


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How to calculate size of a specific cavity in a protein for every frame in trajectory file?
Here is the situation, I have trajecory file of a protein of interest. That protein has 2 main cavities that are participating in interraction with ligands or other proteins and also other cavities. Also residue numbers around this cavity are more or less known. I'm looking for a method or software to get volume of this cavities for every frame in trajectory.
I want to specify a cavity around a residue and  get output for this cavity. What I am looking for is something like this:
Frame Cavity volume
1 1500 A^3
2 1499 A^3
3 1499 A^3
4 1499 A^3
5 1498 A^3

I tried CASTp, it was okey for singe frame if wrote it as pdb but not promising for a 100ns trajectory.
Also tried CAVER. CAVER was able to open a trajectory file and calculate cavity volumes for the frame that I chose but it was not possible to calculate cavity volumes for all of the frames and get output from it.
If anyone has any suggestion I'll be greatful.

If you can get one, then you can get many - it's just a matter of making right scripts for sequential initiation of program an gathering the output.

@Mithoron Well I gladly would do that but It has graphical user interface so I dont know how can I do that.  Also that program has problematic naming for cavities, for different frames same cavity has different names. If someone confirms it is possible with this software,then I'll dig for a solution.

@OnurOzcan CASTp seems to operate as a standalone webserver at which you have to submit your data, so it doesn't sound like a good solution. On the other hand the CAVER program seems to allow for analysis of a series of frames, see the examples that come with the program

They both provide pymol plugins, but the CASTp plugin is for visualization only. It may be possible to interact with the CAVER plugin using python.

In short, POVME exactly does this.


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Changes in conductivity of CuSO4 solution after NH3 solution is added
In my high school chemistry class we are writing a lab. Originally, I wanted to measure the rate of a ligand exchange reaction at different temperatures by measuring changes in conductivity. After some discussion, I came to the conclusion that there would be no change in conductivity and therefore I would not be able to use this method. Nonetheless, I decided to do the test the method; the results can be seen below:

This was quite a surprising result and so I decided to look at the reaction that occurs more closely. The full reaction is as follows:
Initially the solution consists of $\ce{[Cu(H2O)6]^{2+}_{(aq)}}$.
Once ammonia is added, the following reactions occur:
$$\ce{NH3_{(aq)} + H2O_{(l)} <=> OH- _{(aq)} + NH4+_{(aq)}}$$
$$\ce{Cu^{2+}_{(aq)}+ 2OH- _{(aq)} <=> Cu(OH)2 _{(s)}}$$
A continued addition of ammonia results in:
$$\ce{[Cu(H2O)6]^{2+}_{(aq)} + 4NH3 _{(aq)} <=> [Cu(NH3)4(H2O)2]^{2+} + 4H2O_{(l)}}$$
Looking at the reactions above which I obtained from here, I would assume that the conductivity increases due to the dissociation of water in the presence of ammonia to form ammonium and hydroxide ions. Is this assumption valid? Also, my experiment would like to use conductivity to measure how long it takes for the ligand exchange reaction to occur between the water ligands and ammonia ligands of copper. Would this be feasible, or would I only be measuring the rate at which ammonium and hydroxide is formed?
Note:
- The graph was obtained by using a conductivity sensor
- 25 ml of 35% ammonia solution was added to 25 ml of 0.5 mol $\ce{CuSO4}$ solution in this reaction.

I took a liberty of numbering your reactions so that they could be referred to. Also, we decided against using MathJax in titles. Now to the point. Did you add the NH3 solution gradually or at once?

Thank you for informing me, I was not aware of this. The NH3 was added all at once, i.e. I poured it in using a beaker as fast as possible.

Equation 3 is wrong. Please correct.

Well, now it is clear. See, the dissociation of water is pretty much instantaneous; it does not take 6 seconds, nor does it take 1. Heterophase reactions, on the other hand, are much slower. As for your reaction (3), it doesn't happen exactly as written, because after the reaction (2) is complete, you don't really have any $\ce{Cu(H2O)6^2+}$ around.

@IvanNeretin That makes sense, but what would cause the increase in conductivity over 6 seconds?

As solid Cu(OH)2 turns into the soluble ammonia complex, we're getting more ions in the solution.


text:
Converting AlCl3 to Al2O3
Since oxygen is more electronegative than chlorine, if I introduce $\ce{O2}$ gas to $\ce{AlCl3}$ and apply some heat, will the chlorine be replaced by oxygen, resulting in $\ce{Al2O3}$ or $\ce{Al(ClO2)3}$, then evolving the remaining $\ce{Cl2}$ into the air? I suspect not, but why?
Is there a standard way of converting $\ce{AlCl3}$ to $\ce{Al2O3}$?

You probably can, but it would make little sense; industrially, it's the other way around as aluminium(III) oxide is easily available in large quantities. You can hydrolyze $\ce{AlCl3}$ to $\ce{Al(OH)3}$ and calcine it above 600 °C to eliminate water and what's left would be quite pure $\ce{Al2O3}.$

Thanks @andselisk. Though it makes little sense in the current world we live in, my motivation behind this question was imagining a how to create a chlorite in a world without electricity (no electrolysis)

Yes, you can make aluminium oxide from aluminium chloride pretty much in your way but then again as @andselisk said, it is a pretty inefficient reaction and wastes a lot of fuel and energy and also produces many byproducts. Hence, the reaction is the other way round if considered on an industrial scale. The main preparation method is already described by @andselisk in the comment section. Only to verify its authenticity, here is a patent claim:

A process for preparing alumina which comprises heating aluminum
  chloride hexahydrate at a temperature of from about 225 to 600 F. For
  a period of time such as to liberate only a portion of its chlorine
  content, commingling a basic reagent with the resultant
  chlorine-containing residue to form aluminum hydroxide, washing and
  filtering the latter and then calcining to produce the oxide.


text:
Differentiating between Elementary and Non Elementary reaction
Is it possible to tell just by looking(reaction's equation) that whether a given is reaction is elementary or non elementary ?

No, nobody on Earth can say that just by looking at the reaction equation.

There are ways to experimental assess elemetary vs. non elementary reactions. See here https://www.quora.com/How-do-I-determine-whether-a-reaction-is-elementary-or-not

Sometimes an educated guess can be made, though.

You can't even necessarily decide for sure with an experiment.

I think the wikipedia gives a pretty clear exposition on the subject: https://en.wikipedia.org/wiki/Elementary_reaction

Reactions that are probably not elementary
Whenever there are more than three reactants, it is unlikely that the reaction is elementary. Tenary elementary reactions are already quite rare. If you have a reaction like
$$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$
it is impossible for 27 particles to come together at once, so you would expect some intermediates. Even for a reaction with just 3 reactant molecules such as
$$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$
it is highly unlikely that a single collision would be sufficient to break 6 bonds and make 6 new ones.
Conversely, a reaction with a single product such as
$$\ce{O + O2 -> O3}$$
is probably not bimolecular because there is no way of getting rid of the excess energy. Instead, the reaction proceeds like this:
$$\ce{O + O2 + X -> O3 + X}$$
where X is "a third molecule" that leaves the collision chemically unchanged but with higher kinetic energy (source: https://cimss.ssec.wisc.edu/wxwise/ozone/OZONE2.html).
Reactions where you can't tell
Bimolecular simple reactions could well be elementary, but often are not. The reaction
$$\ce{H2 + I2 -> 2 HI}$$
shows two distinct mechanisms, one with one elementary step as written (the iodine has to be vibrationally excited, favored by high temperature) and the other one with iodine first dissociating into iodine atoms (source: https://link.springer.com/chapter/10.1007/978-3-642-80299-7_12).
Is it possible to tell just by looking whether a given is reaction is elementary or non elementary?
No, but sometimes it is obvious that the reaction has to be multi-step. In cases where a mechanism is dependent not only on the net reaction, but also on reaction conditions, it becomes obvious that just knowing the net reaction is not sufficient.

I am not sure about $\ce{N2O4}$.According to the definition of an elementary reaction, I would argue that in a steady-state all that matters there is what fraction of molecules are activated, not how they become activated. The temperature would define the Boltzmann fraction of activated molecules.

@BuckThorn Good point. For atmospheric reactions (ozone...) the mechanism sometimes includes a second unspecified molecule that does not react, but is involved in a collision.

So the idea is that a collision with an inert gas will transfer more energy than would a collision with another molecule of the same type? Is that what makes the reaction not "unimolecular"? But don't all gas molecules - assuming ideality - carry on average $3kT/2$ of kinetic energy? Another way of putting it is that the reaction is unimolecular in the activated species, and the fraction of activated species depends on T, not on the gas composition. So my point remains, what makes the dissociation of $\ce{N2O4}$ not elementary?

Cannot consider myself an expert on the subject, but this suggests the dissociation is unimolecular: https://www.researchgate.net/publication/257992714_Transition_state_and_dynamics_of_unimolecular_no-barrier_fragmentation_Thermal_dissociation_of_N2O4

@BuckThorn Yes, you are right. I will fix it.


text:
Chirality as atom feature
I have been reading some literature on molecular energy prediction using machine learning techniques and I have noticed that one of the input parameters in many of the available models is "chirality". In these works "chirality" is defined as an atom feature. Meaning that for every atom in the molecule, a one-hot or null vector is used to assign R/S to the atom. As far as I understand, chirality is a molecular property and only the atom which is the chiral center can be assigned R/S.
I was wondering if my understanding is correct and if not, what am I missing. Thanks!
PS: One such a paper is this.

I don't think you are missing anything. An atom is assigned as R/S or nonchiral (null vector). While atoms can be chiral centers, chirality is inherently a molecular property, with the whole molecule not being superimposable on its mirror image.

Thanks @Tyberius. I just found an exact description in an implementation of such algorithm, stating "chirality`: (categorical) R, S, or not a Chiral center (one-hot encoded)." which makes it clear that it only assigns R/S to chiral center.

Just to clarify: a molecule may be achiral and have an even number of stereogenic ("chiral") centers greater than 0 (meso compound). A chiral molecule may also have more than one stereogenic center.


text:
How do elemental halogens get their colour?
I learnt that in halogens the colour is due to transfer of electrons from HOMO to LUMO during excitation but I have trouble understanding what that means. Also I don't understand how iodine is blackish in solid state while violet in gaseous state.

Related: https://chemistry.stackexchange.com/questions/33861/why-does-color-of-halogen-deepens-and-the-boiling-point-decreases-down-the-group?r=SearchResults

To the 2nd part, notice that dye based (= not pigment )  black colour of markers or inkjet printers is usually dark violet colour.


text:
C-C bond lengths in hydrocarbons
We had a question in a rest comparing the bond lengths of hydrocarbons fluorides, and the answer key says CF3- CF3's supposed to have a longer C-C bond than H3C- CH2--F.
Why is this so? Wouldn't the C-F bond have less s-character, leading to increase in s-character in the other bonds? That way, CF3-CF3 should have the shortest bond length.
Thanks!

I would argue mainly that fluorine is bigger than hydrogen (more electrons) increasing repulsion. You might also argue that the electron withdrawing tendency of the fluorines leaves a reduced amount of bonding charge between the carbons.

Alternatively said write two delta+ on both carbon atoms. This immediately aids.


text:
CNDO/2 clarification
I wrote a piece of code on RHF using Python a while back, and figured I'd extend it to also implement CNDO/2 (yes, I know it's old but I figured it'd be easier for me.)
The approximated Fock matrix elements, taken from the original paper, are as follows:
$$F_{\mu\mu} = U_{\mu\mu} + [(P_{AA} - Z_A) - 0.5(P_{\mu\mu}-1)]\gamma_{AA} + \sum_{B ≠ A}(P_{BB} - Z_B)\gamma_{AB} $$
$$F_{\mu\nu} = \beta_{AB}^0S_{\mu\nu} - 0.5P_{\mu\nu}\gamma_{AB}$$
where $U_{\mu\mu}$ is an empirical parameter measuring "orbital electronegativity", $P_{AA}$ measures the total electron population on atom $A$, and $P_{\mu\mu}$ measures the electron population of basis function $\mu$.
Unfortunately, I got quite unusual results (read: the lowest orbital energy being about $-50$ hartrees) when I tested this approximation on molecules like $\ce{CO2}$ or $\ce{H2O}$. (I used a STO-3G basis set from Basis Set Exchange).
For the population terms, I used the formula $$P_{\mu\mu} = 2\sum_{i, \mathrm{occ.}}c_{\mu i}^2$$
and
$$P_{\mu\nu} = 4\sum_{i, \mathrm{occ.}}c_{\mu i}c_{\nu i}S_{\mu\nu}$$
and accordingly defined $$P_{AA} = \sum_{\mu ∈ A}P_{\mu\mu}$$
Is there something wrong with my equations?

It may also help to include your python code if it isn't too long. Your error could just as easily be in the implementation rather than the underlying equations.


text:
What happens when a super-critical fluid is compressed?
Suppose that I have some substance with a critical temperature of 20°C. That means above that temperature, the substance exists neither as a gas nor a liquid, but instead as a super-critical fluid. Does it mean that no matter how high a pressure I apply on the fluid, it is not going to turn into a liquid above 20°C? Why is that so?
A substance is a liquid if the particles are close enough to interact, i.e. the space between particles is similar across different liquids. It seems that if we increase the pressure (i.e. decrease the available volume), particles should get close enough to interact.
I know that the higher the temperature of a gas, the higher the speed of the particles. How does this affect the phase transition?
What is the role of intermolecular forces? They are part of the explanation why at lower temperatures, gases condense into liquids.  It seems that we are ignoring them in the case of super-critical fluids. Are we now saying I just need to apply pressure to decrease the distance between particles? If it weren't for intermolecular forces, I would have to apply much higher pressure to turn a gas into a liquid (at temperatures below the critical temperature).
So what happens when a super-critical fluid is compressed, and how is the interplay of particle speed, intermolecular forces and distance of particles different below and above the critical temperature?

If you are worried about density alone, then of course you can put your particles as close as you wish by applying enough pressure. There is nothing special about 20°C in this regard.

Then when the density of a given gas becomes too high, won't it turn into a liquid??

so, is it only the closeness of particles of a substance that tells whether it is a liquid or a gas, or it is affected by the speed of these particles too?

It is complicated. First of all, here is you compound compressed to such-and-such density; how do you tell if it is a liquid or a gas?

As per me, anything is a liquid if its particles have got particular distance between them, say=x. Is it true?

Not quite. What people typically care about is the phase transition, with surface and bubbles. Then you can tell for sure that the thing below is a liquid, and the thing in the bubbles is a gas. Now that's what ceases to be true above the critical temperature.

So, for the people, what is the criterion for phase transition gas-->liquid? If particles are coming too close to each other, aren't they going to show us as if they are liquid like something?

Two phases separated by a surface, that's the criterion.

I am not able to get why it is so?

Because we decided so.

Let me suppose that all of a sudden i reduced the temp of a liquid and increased the pressure. In no time, it changed to a solid. Now, isn't it a solid?? We didn't observe this slow phase change.

Okay, so please form a answer and tell me that if i keep on increasing the density of a gas above its critical point, what happens to its state and what happens to its state below the critical point if density is increased? If there is no state change in the first case, then why?? If yes, then why do we say gases don't liquify above critical temp?

There is no such thing as phase transition happening too fast to be observed. We have pretty advanced cameras these days. Now to the point: we say that gases don't liquefy above critical T precisely because there is no phase transition in that case. There is no so surface, no two phases. There is just one fluid gradually becoming more and more dense.

There is no discrete phase order transition in the critical region. That is not exactly the same as saying you can't distinguish between liquid and gas. The question is about the way the transition between the two states occurs. There is no discrete transition, it is impossible to define a boundary "beyond" the CP.

So, what is the change observed when some gas is highly compressed which is above its critical temperature? There is no liquid i 'm going to see. But why??

You can see it as you see both the gas and the liquid at the same time, as they have identical properties, so there is no difference between them. Below critical temperature, their properties converge toward the critical point. Imagine a beach (liquid), the cliff(gas) and the cliff edge(phase transition). You go along the beach, which climbs and becomes more like cliff, the cliff descends and become more like a beach. At some point the cliff edge disappear and you cannot say what is what, all is the same, with no sudden change with altitude. Single phase with gradual change with pressure.

Note that along the visual line your eye  - air - glass barrier - supercritical liquid,  it may look like liquid, due its refraction index.

When you heat up a liquid at constant volume (leaving sufficient space for the gas phase), the density of the liquid will decrease and the intermolecular interactions will weaken. Some of the liquid will transition to vapor, so the vapor above the liquid will get denser and the frequency of intermolecular interactions will increase (it behaves less and less as an ideal gas as the number of collisions increases with increasing density). At the point where the density of the liquid and the gas are the same, there is no more phase boundary (with gravity, we picture liquid on the bottom and gas on the top, but if the density is identical, that is no longer the case).
This is all pretty outlandish for most of us because we are used to constant pressure, not constant volume. Here are two videos showing the process:
Liquid chlorine: Here, it looks like the liquid level stays constant - the expansion of the liquid and the evaporation balance each other out. If you look carefully, though, you can see how the vapor phase color gets more intense, indicating more molecules in the vapor phase. At the same time, the color of the liquid gets less intense, indicating a decrease in density.
https://chem.libretexts.org/Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/10%3A_Solids%2C_Liquids_and_Solutions/10.13%3A_Critical_Temperature_and_Pressure
Carbon dioxide: When approaching the critical point, the system gets turbid because the difference in density between liquid and gas is no longer sufficient to keep the liquid down and the vapor up. The system is a heterogeneous mixture of liquid and vapor for a while, and then a single pure phase.
https://www.youtube.com/watch?v=GEr3NxsPTOA
Making of video: https://www.youtube.com/watch?v=-gCTKteN5Y4

I understand that this is meant for beginners, but " the intermolecular interactions will get stronger" is not true. The pair potential need not change, but yes, with increasing density the frequency of interactions increases.

@BuckThorn I replaced my incorrect statement with your correct description - it is not more difficult to understand for beginners. (I also edited the question in hopes to get it re-opened because I think the underlying conceptual question is valid and insightful).

There is still room for confusion. For instance water density increases with heating in a small range of T (although this can be considered anomalous) :-). This is a difficult topic to describe in layman's terms. For instance, near the CP the liquid-like phase is sometimes referred to as a "compressible liquid". Imho the substance is more like a liquid than a gas near the CP. Our liquid/gas binary terminology sort of just fails in this region. But your description of the behavior as you move along the coexistence line is reasonable. I was not the downvoter btw.

@BuckThorn I appreciate your input! I also appreciate the votes, in either direction.


text:
Surface modelization of an electrode
I am trying to model the interaction between an electrode and a saline solution through an RC circuit.

A reference electrode is placed in the saline solution to measure the voltage between the electrode used to inject or retrieve charges and the reference electrode.
I then perform a 3 phases cycle using current square waves.
Phase 1: Charge injection through the electrode
During this phase, $e^{-1}$ are injected in the saline solution. At $t = 0$, $C$ is not charged, thus the voltage measured between the electrode and the reference electrode is equal to the votlage drop on $R1$.
This phase uses a current $i$.
Phase 2: Wait a bit
During this phase, charges are not injected or retrieved through the electrode. Thus, $C$ is discharging its charges in $R2$.
This phase is very short, and $C$ will not be fully discharge at the end.
Phase 3: Retrieve charges through the electrode
During this phase, the current flows in the other direction: $e^{-1}$ are retrieve from the saline. At $t = 0$, since $C$ was not fully discharge, the voltage measured between the electrode and the reference electrode is equal to the votlage drop on $R1$ plus the votlage drop on $R2$.
This phase is 3 times longer than the first one and uses a current $-i/3$. Thus, the same amount of charges is injected and retrieved.
The measured voltage looks like:

The general equation for the voltage measured given by the circuit above is:
$$V_{measured} = K*e^{-\frac{t}{R2*C}} + R1*i + R2*i$$
with $i$ the current through the system.
As you can see, the time constant only depends on $C$ and $R2$, however, when fitting an exponential on the data, I get 2 completely different time-constant for the first phase ($e^{-1}$ injection phase) and the third phase ($e^{-1}$ retrieve phase). For the third phase, the time-constant is about 5 times larger...
My current guess is that either $C$, or $R2$, or both are not linear and do not behave the same way depending on the current flow direction. Especially since I think phase 3 is influenced by phase 1. What could explain this behavior? What chemicals reaction happens when a current is injected and retrieved? How could I model this phenomenon?
Thanks for the help to understand this behavior :)

This question seems to be perfect for Physics.SE. There is not much chemistry going on.

@andselisk I will post it as well on that forum, ty. I thougt it was a good match with chemistry because I feel like the problem lies in the ions movement and the chemical reaction between the electrode and the saline.

I'd probably wait for a while, and if there won't be any answers on Chemistry.SE, it would make sense to move your question to Physics.SE. Cross-posting isn't always a desirable thing to do.

Typically, reference electrodes are designed to conduct extremely low currents, e.g., a few pA at most. I suggest looking up three and four wire potentiostats (and how they work): there should be information available about solution impedances. Another posting option is at EE.SE, if posting here or at Physics.SE does not get results.

@EdV Well it is not really an electrical problem as the Capacitance/Resistance equation part is clear. The only problem is that the values of this capacitance are not the same depending on the current flow direction... And we are using mA currents and checked the reference electrode already.


text:
Obtaining conformational energy from Newman projections in first place
In this source, the conformational energy from Newman projections is obtained. Is there a formula or quantum chemistry approach to obtain the values in first place (pretend that I don't know the source)?

This has 3 votes for homework and 1 vote for too broad. I don't particularly agree with either, but I do think it's very unclear what you're asking. If you know the "energy cost" of each gauche interaction or eclipsing interaction, then it is a simple matter of adding these up to get what can be loosely described as the total energy of the conformer. Is the question about how to add them up, or is it about how to obtain those values in the first place?

@orthocresol The question has been modified. I want to know how to obtain the values in first place.

@Mithoron Unfortunately, it is not a homework question.

@orthocresol well, I guess I should have voted it "unclear as you are asking" but my first interpretation was that it was about the origin of the energy values, and that it therefore was "too broad". I recommend either contacting the author of that document or looking for associated documents, or basic literature including textbooks. Relative energies of different conformers can in some cases be obtained experimentally (by measuring populations spectroscopically) and estimated theoretically using QM.


text:
QM9 dataset with chirality
I am trying to find chiral centers of molecules in QM9 dataset. Browsing their SMILES representation, I noticed SMILES yield using
datasets.get_qm9(GGNNPreprocessor(), return_smiles=True)
don't have @ or @@, but it seems that some of these molecules have alternative representations. For instance, both 'Cc1ccoc1C(C)O' and 'CC@Hc1c(C)cco1' represent the same molecule, but of course the latter one outputs the chiral center:
[(1, 'S')]
while only the former is available in the dataset. I found this at https://github.com/chemspacelab/qm9-C7O2H10. Does anyone know if this can be generalized to all the molecules in the QM9 dataset? Thanks!

I'd be very, very careful using random versions of QM9. The original files are available albeit a bit non-standard - but your problem is based on incorrect stereo-stripped SMILES - however you got them.

@GeoffHutchison Cc1ccoc1C(C)O is from chainer_chemistry package. CC@Hc1c(C)cco1 is from github account above, which I realized that belongs to one of QM9 authors (von Lilienfeld).

I don't know about that particular GitHub archive, although I'll point out that the smiles.csv contains several kinds of SMILES - the final column containing the correct SMILES.
IMHO, this is an unfortunate result of people mis-using SMILES. As you note, it's perfectly possible to take a chiral molecular structure and output a SMILES with no stereochemical information.
Take alanine - to a chemist, if you simply say 'alanine,' they immediately think 'amino acid' and translate to L-alanine, or C[C@@H](C(=O)O)N as the so-called "isomeric SMILES":

Unfortunately, if you go to PubChem for L-Alanine, you'll see a "canonical SMILES" which sounds official and is CC(C(=O)O)N - no stereochemistry!
The official version of QM9 can be found on FigShare based on the corresponding publication.
While the official files are in a somewhat non-standard XYZ format, they do have the stereochemically-correct SMILES at the end of the file, e.g. C1N2[C@H]3[C@@H]4[C@@H]5O[C@]13[C@H]2[C@H]45 Just watch that some numbers are formatted like -8.0386*^-6 (i.e., $-8 \times 10^{-6}$) and can sometimes be misinterpreted by programs like Open Babel or RDKit .. practically everything I've seen.

Thanks for sharing the original source. Although I was familiar with the publication, I never checked their original dataset for some reason, maybe never thought it would be any different. I'll try see if the results improve with this.


text:
Enantiomeric excess catalysed
Enantiomeric excess is defined as
$$ee = \frac{[R]-[S]}{[R]+[S]}$$
I found this problem in an IChO paper:
"When using the enantiomerically pure (BINOL)Al(OiPr) as a catalyst for reduction of αbromoacetophenone, the ee of the product equals 81%. What is the ee of the product if the catalyst ee equals 50%?"
The problem as stated is not that difficult: if given the ee of the product mixture ($ee_{p1}$) when the catalyst is enantiomerically pure ($ee_{c1}=1$) then (after noting that mole fraction of [R] and [S] are simply $f_R=\frac{1}{2}(1+ee)$ and $f_S=\frac{1}{2}(1-ee)$ respectively) we say that $f_{Rp2} = f_{Rc2}f_{Rp1} + f_{Sc2}f_{Sp1}$ from which we can calculate $ee_{p2}$, which after a little algebra reduces to $ee_{p2} = ee_{c2} ee_{p1}$. In the above case, then, we'd get $ee_{p2} = 0.5*0.81$.
It's quickly possible based on this formula ($ee_{p2} = ee_{c2} ee_{p1}$) to show that, if given one value $ee_{p1}$ for the product mixture's enantiomeric excess when the catalyst is of a certain enantiomeric excess $ee_{c1}$, we can calculate the product's predicted enantiomeric excess $ee_{p2}$ for any other catalyst enantiomeric excess we try $ee_{c2}$ through the equation
$$ee_{p2} = ee_{p1} \frac{ee_{c2}}{ee_{c1}}$$
What I am wondering is, is there a way to derive this latter formula from the beginning without going through the step of first considering an enantiomerically pure catalyst case? I would not have thought of that case had the contest writers not specifically mentioned it. But not being able to derive the formula from first principles makes me feel I don't really understand the way the various mole fractions would contribute etc.


text:
To move one electron in one H atom from the ground state to the second excited state, 12.084 eV are needed

How much energy is needed to cause 1 mole of H atoms to undergo this transition

Assume Bohr's model of quantisation
The energy for transition is
$$1312\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
Where the electron moves from $n_1$ to $n_2$.
$$1312\left(1-\frac19\right)$$
$$1166.22\ \mathrm{kJ/mol}$$
The answer given is 1164. I know we can just account for this by approximation, but I wanna know why is this arising in the first place. Why do I have to approximate? I feel I am doing something wrong, so please help me with my problem.

Your 1312, whatever units it has, is approximate.

Yes it does. But I assumed it was conventionally used. However, after your comment, my concept of reality has completely shattered.

@Aditya Please try using proper quantity equations and check your units.

Also, notice that the answer you obtained is off by less than 0.2%, so is it really that wrong?

For $1 \mathrm{H}$ atom $=12.084 \mathrm{eV}$
Now, For 1 mole,
$$
\mathrm{E}=12.084 \times 6.023 \times 10^{23} \mathrm{eV}
$$
$$
1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}
$$
So $\mathrm{E}=12.084 \times 6.023 \times 1.6 \times 10^{-19} \times 10^{23} \mathrm{~J}$
$$
=12.084 \times 6.023 \times 1.6 \times 10^{4} \mathrm{~J}
$$
$$
=116.45 \times 10^{4} \mathrm{~J}
$$
$$
=116.45 \times 10 \times 10^{3} \mathrm{~J}
$$
$$
=116.45 \times 10 \mathrm{~kJ}
$$
$$
\mathrm{E}=1164.5 \mathrm{~kJ}
$$

Slight differences of calculated and expected numerical results are matter of different input data, different rounding and sometimes different computation procedure or even an error.
Eventual acceptation of slight deviations or insisting on the exact match is rather matter of the local education policy, regardless of which is correct.
Chemistry itself has little to do with that.


text:
Using MP-AES, can I accurately say I've found a metal concentration on one wavelength and not another?
I've been given the opportunity to use an MP-AES for metal detection. I'm testing traces of metals in plants. The plant matter is digested so I have a liquid solution, I expect there to be a lot of matrix affects within this.
I have tried to use a couple of wavelengths for each metal cadmium, arsenic and lead.
Arsenic has a lot of carry over, and may be sticking to the plasma?
I'm ignoring it for now.
What I want to understand is the different ppm readings for the wavelengths, when looking at cadmium and lead. The 'u' symbol means that the concentration is below the limit of detection, but on another wavelength there is some concentrations of metals that are detected.
Is it biased to only be using the data from the wavelengths where metal has been detected? Even then, can I say that this concentration would be the correct amount in the solution?
Should I be trialling other wavelengths?
I'm struggling with what to search for when looking into the literature for something like this.

There are many details, but it is certainly true that, for a given element, some emission wavelengths are “more sensitive” than others. More accurately, some wavelengths (other things equal) can yield lower detection limits. With commercial systems, the manufacturer usually provides some general guidance regarding wavelength selection and how to deal with matrix effects, etc.

A fancy instrument isn't a panacea. You still have to consider background and matrix effects. Also the software which is being used to reduce the raw data can be problematic.