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what is an organic compound that can meet these conditions?
im looking for a gaseous substance (or airborne solid like dust) to use in my story that is organic, has a sweet odor, or at least pleasant enough to not cause any suspicion from main characters, is nontoxic to inhale, has some coloration (whether that be changing the color of a substance it dissolves in, being visible in the air, or having colored condensation), and is combustive enough to explode from some reaction (preferably with oxygen and high pressure/heat) while not active enough to react with other gasses in the atmosphere. one substance i've found that seems close to what i'm looking for is benzene but due to having no coloration and being harmful from prolonged exposure doesn't seem to be quite what i need. thank you! (edit: from what ive been told having all of the features desired but not required isnt possible so please stop saying what i want is impossible and instead just try to have as many as possible, im not looking for all of them im looking for most of them)

Benzene does not meet the safety requirements.,too. It is a powerful carcinogen.

What do you mean with "my story" ? Are you a writer ? Substances that are both gaseous and colored are not numerous : $\ce{Br2, NO2, Cl2, ClO2}$. They are all toxic and corrosive. Furthermore they cannot burn. So I am afraid the substance you are looking for does not exist.

@Maurice i am actually but i felt this was not fitting for the other stack exchange branches, also i did allow solid airborne particles as well as gasses if that makes it easier.

@zackit. I repeat : what do you mean with "my story" ? Are you writing a novel ? a science fiction ? Has it to make sense ? Because you will never find a non-toxic substance which has a color, a pleasant odor, and may explode when mixed in air.

@Maurice its not science fiction, and i said "preferably with oxygen" it doesn't have to be oxygen. also, it doesn't need to color the air, just have some sort of coloration. please reread and pay attention to what is required and what is preferred.

If we share the same notion of perfume, then the likelihood of a finely dispersed aerosol behind a person, an animal, an object to meet all of your criteria is slim.  If the characters of your plot posses a better protected respiration system, than the typical human being, they may find the fumes of nitric acid beautiful (e. g., Derek Lowe's How Not to Do It).

Your best bet is probably some finely divided dust. Many finely divided dust particles (eg flour, coal, wood) can explode given a suitable trigger. So there may be one which is also coloured (and there are few volatile substances that have much colour so dust is the way to go).

Custard powder? https://www.youtube.com/watch?v=aGBT5pwxThU

The way you treat language, I do not believe you are a writer; I do, however, believe that the bounty you have offered prevented closure of this question. To be honest, I would treat this the same way as I would treat homework. You have not offered any own research. Additionally, you have failed to present your query well.

Benzene, which has already been mentioned in the comments is colourless. The frequency of light absorbed corresponds to the complementary colour of the colour we see in azo dyes. Benzene only contains a small area of delocalisation, which is not sufficient to lower the energy gap to the visible part of the spectrum, meaning benzene absorbs UV light and is colourless. On the other hand, It has a sweet odour and is highly flammable. Benzene evaporates into the air very quickly. Its vapour is heavier than air and may sink into low-lying areas.
NOTE: benzene may possess a pale yellow colour at room temperature.
Also, though benzene is deadly in case of long-term exposure (Long-term exposure means exposure of a year or more) and causes harmful effects on the bone marrow and can cause a decrease in red blood cells, leading to anaemia, Acute (short-term) inhalation exposure of humans to benzene causes no more than a bit of drowsiness, dizziness or headaches, in relatively lesser concentrations

with this extra information about benzene, i feel the substance i'm looking for may be some solution composed of benzene and a few other substances. though it does not provide a complete answer for what it i will mark this correct anyway as it is seemingly what i need.

Given that I am a new member and don't have much reputation, I can't comment on your post, so I am answering here...ozone has a pale bluish colour and also a metallic odour and if it is present in minimal concentration it is not deadly toxic, just might result in chest pain and laboured breathing...In the stratosphere, ozone is formed when a molecule of oxygen absorbs a photon of light with wavelengths shorter than 200nm, is split into atomic oxygen and the atomic oxygen reacts with an oxygen molecule to form ozone, but i am not sure whether it is volatile or not..however, ground-level ozone is formed by reaction betweeen oxides of nitrogen and VOCs (volatile organic compounds)

Why not edit your first answer and add this to it?

Because they're different solutions or something

Actually you have enough rep to comment now, so perhaps you could delete the answer and comment instead (with 86 characters removed if you can: just remove the unnecessary dots, definition of VOC, extra 'u' in odor, maybe call nitrogen "N" and oxygen "O", etc.).

Yes, now I have enough rep, earlier I had 1 rep when I posted this,,, but what is the advantage of doing that - deleting this and commenting?


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Does KCN prefer SN2 or nucleophilic addition?
If we have a 1° halide that also contains a carbonyl group (like 5‐chloropentan‐2‐one), and we add 1 mole of KCN in DMSO to it, what will happen?

Will the SN2 product by substitution of chlorine or nucleophilic addition at the carbonyl carbon be the major product, and why? And does using HCN instead of KCN have a different effect?

Cyanide addition to carbonyl gives cyanohydrins but this is reversible and usually under acid catalysed conditions. Nucleophilic displacement of the chloride will predominate.

@Waylander thanks for the answer. So will the SN2 product be dominant even under acid catalysed conditions? And does using HCN instead of KCN change anything at all?

If you introduce acid cyanohydrin formation is going to be favoured https://en.wikipedia.org/wiki/Cyanohydrin_reaction  To get the chloride displacement you want no source of hydrogen to protonate the cyanohydrin alkoxide, so polar aprotic conditions with KCN

source: Stack Exchange

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

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What is the role of the Magnesium Ion in Chlorophyll
Chlorophyll as the main photoreceptor molecule in photosynthesis seems to have three unique features 1) a lot of conjugated double bonds, 2) a ring like structure and 3) a central metal ion (Mg).  The conjugated double bonds make sense, as this creates delocalized electrons, resulting in excitation energies in the optical spectrum, which allows the molecule to absorb light.  The ring-like structure seems to be designed to chelate the central Mg ion, but what is the role of the Mg ion? Does it add electrons to the delocalized molecular orbital.  Its interesting that pheophytin, the next step in the redox chain,  seems to be identical to chlorophyll except it doesn't have the Mg ion.  Is this something like electron doping in a semiconductor PN junction biasing the flow of electrons towards the pheophytin?

I am no expert in the field, I'm not sure what you are referring to as the "redox chain". Is it photosynthesis? (I can be dumb sometimes, hope I can excused :)

Besides CSE article, you may find some info here as well.

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


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What happens when tran-2-butene reacts with aqueous bromine in the presence of chloride?
$\ce{Br2-H2O}$, containing some $\ce{NaCl,AlCl3}$ ?

I initially though that there would be anti addition of Br and OH to give 3-Bromobutan-2-ol  but then I also got the intuition that NaCl might ionize in water and the $\ce{Cl-}$ might be involved here as well (obviously, with different stereochemical products)

But when I referred my book, it showed the following products only:

Why does this happen?

First $\ce{Br2}$ is forming first a positive ion with one $\ce{Br}$ atom fixed on the double bond. It makes a positively charged triangular structure with the two central carbon atoms and one bromine atom. Then a negative ion may react with this positive ion. And here there are two possibilities. Either the bromide ion remaining after this first reaction, or a chloride ion from the solution. The probability of having chloride to react is greater. So $\ce{Cl-}$ attacks the triangular cation, and gets fixed from the opposite side of the triangle.

@Maurice Yes, I know that the bromonium ion is formed as a transition state. What I'm asking is: Why does $\ce{OH-}$ also participate and attack the cation?

@PrajwalTiwari OH- is present in very low concentrations in water. Usually, its H2O which attacks the bromonium intermediate, however, in this case there are lots of Cl- around, and Cl- is a better nucleophile than water, so Cl- is the one that attacks (mostly)


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How does NaOH modify pyranodioxins?

I am given the molecule in black which is treated with $\ce{NaOH}$. The product is in blue.
Could someone explain the mechanism? All I have so far is that the $\ce{-OH}$ group on the bottom right carbon has been deprotonated to $\ce{-O-},$ but am not sure how the rest of the reaction should proceed.

In similar tune @MathewMahindaratne's comment, because many of the reactions with carbonyl groups are ruled by equilibria (e.g., aldol reaction vs. retro-aldol reaction), see if you draw the mechanism starting from the structure depicted in blue toward the one one depicted in black.  It might be the opposite direction of what you want, but may help you to identify key intermediates.

The molecule in black consists of a cyclic ketal and a hemiacetal. The hemiacetal is known to be in equilibrium with its aldehyde form, which is the the molecule in blue, in aqueous solutions even without $\ce{NaOH}$. Recall the reaction of glucose (a cyclic hemiacetal) with dilute sodium hydroxide, which is known as Lobry de Bruyn-van Ekentein rearrangement (Ref.1).
Nonetheless, a hemiacetals can be decomposed to the relevant carbonyl and the alcohol under basic conditions as explained here (a reliable mechanism is given). In your case both carbonyl and the alcohol groups are within the same molecule (the molecule in blue):

All acetals and ketals are stable to alkaline conditions since they don't carry free hydroxyl group.

References:

Roy L. Whistler, J. N. BeMiller, "Alkaline Degradation of Polysaccharides," Advances in Carbohydrate Chemistry 1958, 13, 289-329 (DOI: https://doi.org/10.1016/S0096-5332(08)60359-8).


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Why is Vapor Pressure of Liquid Equal to Ambient Pressure at Boiling?
What does phase change from liquid to gas have to do with equality of vapor pressure and ambient pressure? I understand that at higher pressures more kinetic energy is needed for molecules in liquid to spread out to form bubbles of vapor since vapor has lower density than liquid which means that higher temperature is needed. But why should equality of vapor pressure and ambient pressure determine when phase change takes place?

Related: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

When the vapor pressure equals the applied ambient pressure the liquid can form vapor in the bulk liquid hence "boiling".  To accurately measure this care must be taken to mix the liquid well and to prevent overheating. At lower vapor pressures evaporation is only from the exposed surfaces.


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Why is magnesium ascorbate, which is a conjugate base of a weak acid, acidic?
Why are magnesium ascorbate and potassium acetate, which are salts of strong bases with weak acids, acidic?
I discovered with litmus paper that magnesium ascorbate is acidic and found out from https://foodb.ca/compounds/FDB015417 that potassium acetate should also be acidic.
How can this be? Shouldn’t conjugate bases of weak acids, ascorbate and acetate in this case, be basic and the strong dissociated spectator ions, $\ce{K+}$ and $\ce{Mg^2+}$, neutral?

The site you mentioned has given $\mathrm{p}K_\mathrm{a}  = 4.54$ for $\ce{CH3CO2K}$. That is actually $\mathrm{p}K_\mathrm{a}$ of acetic acid. As we all know, $\ce{CH3CO2K}$ is a slightly basic salt.

If your potassium acetate were acidic by actual measurement, I would guess that it was not pure. If your source for magnesium ascorbate is a food supplement, I would question its purity. Perhaps you could try a recrystallization.

PubChem Magnesium ascorbate $(\ce{Mg(C6H7O6)2})$ shows a monobasic salt. Wikipedia says pKa's are 4.10 (first), 11.6 (second)
In addition to Poutnik's answer, the crystal ionic radii in pm according to wikipedia of
$$\ce{Mg^2+ is 86pm and Ni^2+ is 83pm}$$
This following coordination complex formation reaction is known
$$\ce{Ni^2+(aq) + 2(dmgH2)->Ni(dmgH)2(s) + 2 H+(aq)}$$

Nickel(II) is square planar.[2] It is surrounded by two equivalents of
the conjugate base (dmgH−) of dimethylglyoxime (dmgH2).
(https://en.wikipedia.org/wiki/Nickel_bis(dimethylglyoximate))

Four five member rings center at the Nickel(II) ions.$$\phantom{1}$$ Similarly, after this reaction
$$\ce{Mg(HA)2(s) ->[H2O] [MgA2]^2-(aq) + 2 H+(aq)}$$
Rings could center at the magnesium $$\ce{[(C4H6O4)-O^--C=CO]Mg[OC=CO^--(C4H6O4)](aq)}$$
Therefore, this complex formation reaction could be possible
$$\ce{Mg(HA)2(s) ->[H2O] [MgA2]^2-(aq) + 2 H+(aq)}$$
due to

similar ionic radii
existing coordination complex

$\mathrm{p}K_\mathrm{a}$ relation in the answer of Kav

Wikipedia says pKa's are 4.10 (first), 11.6 (second)

gives the hint a monobasic salt with alkali metals should be very mildly alkalic, as pKa2 is farther from pH 7.
$pH \approx \frac{\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2}}{2} = \frac{4.10+11.6}{2}=7.85$
OTOH, it will probably form a complex with Mg, affecting acidity of the 2nd hydrogen. Like
$$\ce{2 H2A(aq) + Mg^2+(aq) -> [MgA2]^2-(aq) + 4 H+(aq)}$$
respectively
$$\ce{Mg(HA)2(s) ->[H2O] [MgA2]^2-(aq) + 2 H+(aq)}$$
See also citrate or EDTA complexes for this effect. AFAIK, before suitable metalochrome indicators came, metal content with excess of Na2H2-EDTA had been sometimes determined by back titration using NaOH.


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Will OTs ever act as a nucleophile and substitute with an alcohol?
If you reacted TsOH with a secondary or tertiary alcohol, will you ever see a racemic mixture of OTs substituted products after the water group leaves? Is OTs just too weak of a nucleophile to ever substitute?

It is really a very poor nucleophile

You are reacting a very weak base with very weak acid like a tertiary alcohol, so no.


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Why is Ag+ not hydrolysed?
Usually, strong acid anions and strong base cations don't get hydrolysed. But I've noticed through some questions that $\ce{Ag^+}$ doesn't seem to get hydrolysed, why is that? Is there any concept that I'm not aware of that predicts which ions get hydrolysed and which don't, or is it just the ions from strong acids/bases don't get hydrolysed?

Does Ag2O precipitate when AgNO3 solution neutralised by NaOH ?

Why would it get hydrolyzed? Big cation, low charge.


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Question regarding the colour of Benzoic acid and Benzyl alchohol
This question is primarily based on the Cannizzaro reaction, where a base induces a disproportionation reaction of non-enolizable aldehydes (benzaldehyde, formaldehyde, etc) to form an alcohol and a carboxylic acid.
In the video by Nile Red exploring the Cannizzaro reaction (https://www.youtube.com/watch?v=qEBC204WTKs), the product mixture contained both Benzoic acid and Benzyl alcohol and had an orange color. I am aware that the benzoic acid was in the form of sodium benzoate due to sodium hydroxide also reacting with the acid produced.
My question is, where does the colour come from?
I am guessing it is either from the benzyl alcohol or sodium benzoate, but I have learned that most organic compounds are colorless.
Moreover, if the aldehyde used was something like formaldehyde, will the orange color also form?
Thanks in advance for the clarifications.

none of the components you mentioned (Benozic acid, Benzyl alcohol, sodium benzoate) has a yellow colour.
The colour in organic compouns is closely tied to ther delocalized sytems of pi-electrons and depends on the gap between the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO).
When the energy of this gap corresponds to a wavelength that is visible, the compound absorbs this wavelength and the colour of the compound is the complementary colour.
A good example are condensated aromats. For benzene or naphthalene the gap between HOMO and LUMO is too big therefore, the corresponding wavelength is in the area of UV light.
For anthracene, the absorbed wavelenght is violett and therefore the colour of the compound is yellow.
Back to your question. Because the reaction happends under strong basic conditions, high temperature and with many aromatic compounds, I think sidereactions will occure where multiple aromats condensate together and therefore the solution turns yellow to orange.

Thanks for the answer. Could you suggest what sort of side products could produce that yellow/orange color?

No, I am sorry. The reaction goes up to 90% yield so 10% of the molecules could undergo sidereactions. If I had to guess, I would not got condensated aromats but for some kind of structure similar to benzophenone were the C=O double bond can expand the conjugated pi electron system. benzophenone itself is still colourless but something similiar might become yellow or orange.

As it happens pure anthracene crystals are white and a solution colourless. The first absorption band is below 400nm. if its coloured there is most likely some oxidation product.

Ah ok. I knew that is often the case with naphthalene, I produced white naphthalene in the lab but yellow anthracene. Thanks for the comment


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Octahedral, Tetrahedral or Square planar?
In cases where a metal ion like, Fe(III), Cr(III), Pt(II) is made to react with a ligand species like oxalate, CN-, CO etc and it forms a complex then, How do we determine whether 6 ligands will be coordinated to the central atom(i.e. in case of octahedral) or 4 ligands will be coordinated (as in tetrahedral or square planar arrangement) ?

https://chemistry.stackexchange.com/questions/4649/tetrahedral-or-square-planar https://chemistry.stackexchange.com/questions/112856/why-do-some-ions-form-tetrahedral-complexes-more-readily-than-others

From your question, I suppose you do not know about crystal field theory. To properly understand how and why certain complexes are octahedral and other ones are tetrahedral you need to fully understand this theory.
You can find information in Greenwood and Earnshaw, Chemistry of the Elements. A more detailed info about the theory is in Theory of Groups in Chemical Applications of Group Theory of Cotton.
I will explain the basis concept here. The isolated ion has a spherical symmetry, then all the microstates for a specific electronic configuration have the same energy. The ligands can be imagined as a source of electric potential. This field alters the symmetry, than the microstates have not all the same energy. In particular, the 5d orbitals will not have the same energy in presence of the ligands because they are shifted up and down. A degeneracy remotion, as always in QM, will alter the symmetry of the system, that is the structure will change. Then the geometry of a complex depends upon:

the nature of the metallic center
the oxidation state of the metallic center
the nature of the ligands ($\pi$-basic, $\pi$-acid, $\sigma$-donor and so on).

Thanks a lot, I do know the basics of cft. One of the questions I came across asked the reaction of Fe3+ with oxalate ligand in aq. sol containing KOH. Although octahedral geometry felt right to me I wasn't able to rule out the reason why it is not tetrahedral for instance. Is it this something based on experimentation ? Or is there any logic using which If given a pair of metal and ligand I can figure out what the geometry will be.

Fe(III) has a d$^5$ electronic configuration and oxalate is a $\sigma$-donor ligand, then it will be an high spin complex, no net gain in energy. Anyway, remember for other cases that the octahedral field is always stronger than the tetrahedral one. In this specific case it is more convenient to form six ligands because there is not so much steric hinderance. But, for other metals other factors can play a role like the Jahn-Teller distorsions in Cu(II) ammonia complexes


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how to prepare acid washed sand for disruption of cell wall and the amount required for disruption
the sample is a mixed sample of maize(yellow and white) undergoing fermentation. while conducting an analysis for  extraction of crude enzyme.Acid washed sand was required for the liberation of enzyme trapped in the cell

Easiest way is with a sintered glass filter funnel. Place the sand in the funnel, add the acid (1M HCl or H2SO4 will do for this), stir it around then apply the suction, Turn off the suction, add clean water, stir it around then apply the suction again. Repeat the wash a couple of times then dry under suction, pour the sand out of funnel and oven dry.


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Carcinogenicity of home coated cast iron pans
Cast iron pans are a traditional cooking utensil and are becoming more and more popular in recent  years.
I myself switched from Teflon/ PFC pans to cast iron pans.
To use the pan they are normally coated with oil, which is baked in. For this mostly line seed oil is used, as it polymerizes well.
The normal procedure is explained in this video
Now my question is how likely is it that this process produces carcinogenic substances like PAH (Polycyclic Aromatic Hydrocarbons) which are often associated with heating oil.
EDIT:
As matt_black pointed out, I am interested of the amount of these substances. And if food that is cooked in a freshly coated pan contains significant more, than a pan with other coatings. The next question would be of course if so, how long this increase can be detected after multiple uses.

This is the wrong question. The right question is whether there is any evidence that seasoned cast iron cookware contains enough nasties for anyone to notice (over and above what cooking itself produces). The dose makes the poison.

"Line seed" oil? Do you mean linseed oil? Note that food-grade linseed oil, AKA flax seed oil, is not considered toxic, but "boiled" linseed oil may be, particularly if it contains plasticizers and polymerization catalysts.


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How can you calculate ring strain in beta lactams for different penicillin drugs?
I am currently working on a lab, where I am trying to calculate whether the R group in penicillin type drugs has an affect on the ring strain of the beta lactam. I know how to find ring strain in cycloalkanes through the molar heat of combustion, however I can't find any information about the combustion of beta lactams as I'm sure it doesn't happen often, so is there another way I can calculate the ring strain of the beta lactams in different penicillin type drugs?
Thanks,

Are you looking for an experiment to confirm the ring strain or is a computer model also an option?

I cannot currently do an experiment in a lab so I am looking for a way to calculate ring strain with data available online. Computer modelling is definitely an option. I have also tried using Webmo to calculate the ring strain, however the strain energy calculations were completely different each time even if I was using the exact same penicillin drug.

Would be the heat of formation  instead / in addition to the heat of combustion you an alternative suitable for you? This then wider scope may increase the likelihood to identify literature.  A first entry found was «AM1 study of the electronic structure of benzylpenicillin» by Venugopal et al. in IJOBC here, e.g. about the influence to protonate the lactam-N.

For the calculation of ring strain for ring-opening metathesis polymerization, I used DFT calculation.
In generell you open your molecule at the position you want to and saturate the bonds. Then you compare the enthalpy of products and educts.
As an example you can look at cyclopentene.If it is opened during Metathesis and you saturate the doublebonds at the end of the chain with two $\ce{CH2}$ or one ethene molecule.

This has also been reported in literature for example:
https://onlinelibrary.wiley.com/doi/full/10.1002/pola.28695
I have also done a quick search and in this paper they have used the same technique (they just used ethane instead of ethene because they saturated single bonds instead of double bonds). They used post Hatree Fock B3LYP to calculate the enthalpy. I think you are fine to use either one.
https://chemistry-europe.onlinelibrary.wiley.com/doi/10.1002/cphc.201900173


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Confusion in comparing the contribution of resonating structures to resonance hybrid
So I was comparing the $+M$ effect of $\ce{-CH^-_2}$, $\ce{-NH^-}$ and $\ce{-O^-}$. Answer was given as:
$$ \ce{-CH^-2 > -NH^- > -O^-}$$
So, I attach them with ethene and made their respective resonating structures as:
$\require{enclose}$

$\ce{\underset{(a)}{CH2=CH-\overset{\ominus}{C}H2} <-> \underset{(b)}{\overset{\ominus}{C}H2-CH=CH2}}$

$\ce{\underset{(a')}{CH2=CH-\overset{\ominus}{N}H} <-> \underset{(b')}{\overset{\ominus}{C}H2-CH=NH}}$

$\ce{\underset{(a'')}{CH2=CH-\overset{\ominus}{O}} <-> \underset{(b'')}{\overset{\ominus}{C}H2-CH=O}}$

So my doubt was whether $b$, $b'$ and $b''$ have same contribution to their respective resonance hybrid or it's different ?
Or it's that they have same relative contribution but order of $+M$ is determined by inductive effect in resonating structures or inductive effect determines resonating structure contribution ?

For future reference: bodies of questions, answers, and comments on ChemSE (on occasion, equally sibling sites on stackexchange.com, too) equally understand mathjax and mhchem.  Especially mhchem offers an easier input and better rendering of chemistry-related content (e.g. simple formulae in reaction equations) and is presented e.g. here.  Note: because browsers do  not universally understand this addition, do not use them in the title of a question / answer.

Both the structures a' and b' are incorrect. Nitrogen does not have d orbitals and cannot expand its octet.

The $\ce{-NH2}$ group with a negative charge indicates pentavalent nitrogen. So I corrected it.

For $b$, you have equivalent resonating structures as the structures are identical.
The structures $a'$ and $b'$ are not correct as nitrogen cannot expand it's octet.
For $a''$ and $b''$, the first structure $(a'')$ is more stable as there is negative charge localised on electronegative atom as compared to $b''$ where it is localised on carbon.
For the second question, I think inductive effect is completely different than the mesomeric effect. For determining the order of the strength of mesomeric effect, you can simply compare their electronegativity orders, more is the electronegativity, lesser is the strength to donate electrons.

Thanks ! I got what you said but can you comment on whether we can use inductive effect for comparing contributions of resonating structures or just the rules like "full octet,more pi bond ,-ve charge on more EN atom,...." are valid ?

Why should inductive effect be considered for resonating structures? It has to do with stability, the one which is more stable will have more contribution in resonating structures. Inductive effect is an entirely different thing than stability.

But in cases when we compare stability of carbocation and carbanion, We also use inductive effect to determine stability order of them , So, I was applying the same logic in resonating structures. Is it wrong to do so ?

While comparing the stability among carbocations and carbanions, we use inductive effect because it gives the idea of electron density on that species, electron deficient species would get stabilsed if it is attached to ERG, similarly carbanions will get stabilised if they are attached to EWG. When comparing "resonating structures", we have to chose which among them is more stable, for which we have different parameters like seeing if their octet is complete, no. of $\pi$ bonds, charge separation, negative charge on EN atom and so on... inductive effect is not relevant here.

So how we'll be comparing -M effect of $\ce{-COH}$ and $\ce{-COR}$ ?

Here, you will use inductive effect, as the alkyl group has more electron releasing tendency, so it will have less -M effect as compared to $\ce{-CHO}$. Wikipedia is a good place to read about it.

Earlier you were saying inductive effect isn't relevant for resonating structures but won't the order of $-M$ effect will be determined by resonating structures itself ?

What I mean to say is, in the original question that you have asked, there is no role of inductive effect. If situation arises, that we also need to see the inductive effect also, then we will see it. I think this would be helpful.

Both the structures a' and b' are incorrect. Nitrogen does not have d orbitals and cannot expand its octet.

@AdityaRoychowdhury I agree. I didn't notice that part, thanks for pointing it out.


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What is the difference between sublimation and instantaneous fusion+vaporisation?
Sublimation is defined by the IUPAC as The direct transition of a solid to a vapour without passing through a liquid phase.
A sublimation equilibrium can occur in two sets of conditions:

at a temperature-pressure set given by the phase diagram of the pure substance. This is only possible if the pressure is lower than that of the triple point.
E.g. sublimation of $\ce{CO2}$ at -78.5 °C and 1 bar:

at any other temperature-pressure set, provided the temperature is lower than the melting point and the pressure is higher than the vapour pressure of the solid. This is possible even at a pressure higher than that of the triple point, if the gas phase in the system is not the pure substance (such as a system with air). E.g. sublimation of $\ce{I2}$ at 100 °C and 1 bar:

However, what happens in a case far from equilibrium, such as when applying a red-hot metal rod on an ice cube? (i.e. an object with a very high thermal conductivity and at a temperature much higher than the boiling point of the substance it is applied to). Some will consider this a sublimation, others will describe it as an instantaneous fusion+vaporisation (my issue is with the use of the term "instantaneous").
For this situation, I have coined several propositions, both on the physical process and on the terminology:

The temperature gradient is so great, the conditions are so far from equilibrium, that the substance will not go through a liquid phase: it is sublimation.

Since the atmospheric pressure is higher than the pressure of the triple point, the system will go through a liquid phase, however short-lived it is: it is fusion followed by vaporisation.

From a molecular point of view, the molecules do received enough energy to melt, then to vaporise; however, the state of a phase is not defined at the molecular level: considering a liquid phase in these conditions does not make sense. Considering a "fusion+vaporisation" does not make more sense than talking about sublimation. From a thermochemical point of view, the presence of a liquid phase cannot be evidenced experimentally since $\Delta H_{sublimation} = \Delta H_{fusion} + \Delta H_{vaporisation}$, enthalpy being a state variable.

Even though the presence of a liquid phase cannot be evidenced (or if considering a liquid phase does not make sense), the temperature at which the process takes place is higher than the melting point: it is, by convention, not considered sublimation.

Even though, from a thermodynamics point of view, the system should go through a liquid phase, the two processes ("sublimation" and "fusion+vaporisation") are in competition. The two phenomena could occur simultaneously. The idea, here, is similar to phenomena such as supercooling and superheating of liquids. The solid-to-liquid transformation, while thermodynamically favoured, could be kinetically unfavoured in such extreme conditions. I would consider this a similar process to MALDI: using a laser to "desorb" (as per the acronym, but is the process really different from sublimation?) a solid which would, upon slower heating, decompose instead. If sublimation does not occur in the case of the metal rod on ice, would a laser (e.g. IR) be able to do it? What would be the difference between the two?

Which of these propositions would you consider the most appropriate? Do you have another view of the process? Regarding the terminology: when would you use "sublimation" in such cases?

Close to the metal rod you will have a very large temperature and pressure gradient that might result in rupture of the solid or even an explosion, depending on conditions. Furthest from the rod and adjacent the solid you will have a layer of water. You will not see sublimation.

@BuckThorn "or even an explosion": because of the rapid formation of a gas. Does that not correspond to the definition of sublimation?

Sublimation is strictly speaking a first-order phase transition involving direct conversion from solid to gas. In your experiment you go through the liquid, if for no other reason that the pressure is so high. The phase diagram wins. But it is "almost" sublimation. The dangling molecules at the interface however won't care what you call the process.

However this does make me think of laser-induced cavitation: https://www.sciencedirect.com/science/article/abs/pii/S0301932220305425

I agree that the molecules do not care what we call the process. I'm also interested in the terminology. So... Would you agree with my proposition #3? Or are for #2?

I suspect either case might be valid depending on experimental circumstances. Sorry about the slippery answer. Surface energy would become important if there is partial fragmentation ("explosion"?) of the solid. Imagine small chunks of solid flying off, temporarily in a low pressure setting. You are trying to describe a highly non-equilibrium system so time is a variable and the appropriate description will obviously change with time.


text:
Can I use silica gel to perform column chromatography at home?
While this question has been asked and answered, it may indicate that a column chromatography qualitatively is possible. At home, is it possible after collecting silica gels from small bags of them from snack packages? Secondly, if qualitative analysis is possible, how about quantitative?

The type of silica that's used for flash chromatography is a very specific one, in particular the particle size is very important (see this one for instance). I would be very surprised if the one from snack packages were the same grade. And BTW it's really not healthy to inhale it, so I'd definitely not use it at home / without proper equipment like a fume hood.

Silica gel from snack packages does not even look like what you need for chromatography, not even remotely.

Silica gel from snack packages are designed to absorb moisture from atmosphere. Silica gel used in laboratories have designed to separate compounds according to their polarities, etc. using a mobile phase. This same thing  Ivan Neretin has said in one sentence. :-)


text:
How do you distinguish ferrocene, acetylferrocene and diacetylferrocene in a NMR spectra?
I made an experiment where I made a Friedel-Crafts acetylation of ferrocene. Now I would like to know if/how I can distinguish ferrocene, acetylferrocene and diacetylferrocene in a NMR spectra? For my reaction I used ferrocene, acetic anhydride and phosphoric acid.

The presence of the methyl singlet from the acetyl group at 2.4 is pretty diagnostic and its intergration against the ferrocene protons will tell you whether there are 1 or 2 acetyls present

Can you show the NMR spectrum, and also how much you have been able to analyze the peaks?

$\mathrm{^1H \ NMR}$ spectra of pure ferrocene, acetylferrocene, and diacetylferrocene are very distinguishable:

$\mathrm{^1H \ NMR}$ spectrum of ferrocene consists of only a singlet at $\approx \pu{4.2 ppm}$.
$\mathrm{^1H \ NMR}$ spectrum of acetylferrocene consists of 4 signals: a triplet or multiplet around $\pu{4.8 ppm} \ \ce{(2H)},$ a triplet or multiplet around $\pu{4.5 ppm} \ \ce{(2H)},$ a singlet around $\pu{4.2 ppm} \ \ce{(5H)},$ and a singlet around $\pu{2.4 ppm} \ \ce{(3H)}$.
$\mathrm{^1H \ NMR}$ spectrum of diacetylferrocene consists of 3 signals: a triplet or multiplet around $\pu{4.8 ppm} \ \ce{(4H)},$ a triplet or multiplet around $\pu{4.5 ppm} \ \ce{(4H)},$ and a singlet around $\pu{2.4 ppm} \ \ce{(6H)}$.

Accordingly, the number of signals in each spectrum determine which compound it is. Since this is a Friedel-Crafts reaction, I assumed diacetylferrocene means acetylation on each cyclopentadiene ring (I also assumed you have separated each compound).


text:
Gibbs free energy for the evaporation of water
Problem

Find the value of $\Delta G$ for the evaporation of water at $\pu{100 °C}$ and $\pu{1 atm}.$

Answer

 $\pu{0.3 kJ mol^-1}$

Question
I have the following data:
$$
\begin{array}{lcc}
\hline
\text{Compound} & \Delta H/\pu{kJ mol^-1} & \Delta S/\pu{J mol^-1 K^-1} \\
\hline
\ce{H2O(l)} & –285.8 & 69.96 \\
\ce{H2O(g)} & –241.8 & 188.7 \\
\hline
\end{array}
$$
$$T = \pu{100 °C} + 273.15 = \pu{373.15 K}$$
I calculated $\Delta G$ of the product (gas) as
$$
\begin{align}
\Delta G &= \Delta H + T\Delta S \\
  &= \pu{-285.8 kJ mol^-1} - (\pu{69.97 J mol^-1 K^-1})(\pu{373 K}) \\
  &= \pu{-26395}
\end{align}
$$
Then I calculated $\Delta G$ of the reactant (liquid) as
$$
\begin{align}
\Delta G &= \Delta H + T\Delta S \\
  &= \pu{-241.8 kJ mol^-1} - (\pu{188.7 J mol^-1 K^-1})(\pu{373 K}) \\
  &= \pu{-70625}
\end{align}
$$
Finally,
$$
\begin{align}
\Delta_\mathrm{r}G &= \Delta G(\ce{H2O(g)}) - \Delta G(\ce{H2O(l)}) \\
  &= -26395 + 70625 \\
  &= -44230
\end{align}
$$

Well, to begin with, you are probably expected to know that $\Delta G$ is somehow related to $\Delta H$ and $\Delta S$. Rings a bell?

I get -0.3 kJ/mol. Are you sure about the answer?

@IvanNeretin Yes, I'm aware of the formula ▲G = ▲H - T *▲S but I'm still getting it wrong ahaha

@BuckThorn Yes, I checked and it's positive. May I ask you how you got to that solution though?

Those values you gave are for the standard state at 25 C and 1 atm. Of course, for 100 C and 1 atm (an equilibrium change) $$\Delta S=\frac{\Delta H}{T}$$or $$\Delta G = 0$$

$((-241.8--285.8)1000-373.15(188.7-69.96))/1000$

@BuckThorn Ohh got it. I forgot to multiply the enthalpy by 1000. Hmm, I'm also getting a negative solution though; could be my book's mistake...

Did you read my comment?

@neavys It's Gibbs, not gibb's. And the main issue with your calculation is that you omitted units and allowed yourself to sum up numerical values for J and kJ, which you should've never done. I tried to brush up notations and math, but I cannot finish corrections since it would break initial intend. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

There is wrong sign (+), should be $\Delta G =  \Delta H - T \cdot \Delta S$.

See my edit to your original post, starting with bold Here is the correct way to solve this.

@ChetMiller This should the answer, not edit.

My previous answer was hidden.  What was that about?

@ChetMiller It wasn't hidden, it was converted to a comment as it wasn't a complete answer, rather a hint. Hints should never be posted as answers.

@andselisk I contend that my answer was a complete answer.

The average heat capacity $C_p$ of liquid water and water vapor over the range between $\pu{25^\circ C}$and $\pu{100^\circ C}$ are, respectively $4.18\ \pu{kJmol^{-1}K^{-1}}$ and $1.72\ \pu{kJmol^{-1}K^{-1}}$.  So the enthalpies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$H_l(\pu{100^\circ C})=-285.8+(0.018)(4.18)(75)=-280.2\ \pu{kJmol^{-1}}$
$H_g(\pu{100^\circ C})=-241.8+(0.018)(1.72)(75)=-239.5\ \pu{kJmol^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta H=40.7\ \pu{kJmol^{-1}}$
The entropies of liquid water and water vapor at $\pu{100^\circ C}$ are:
$S_l(\pu{100^\circ C})=0.06996+(0.018)(4.18)\ln{(373.15/298.15)}=0.08684\  \pu{kJmol^{-1}K^{-1}}$
$S_g(\pu{100^\circ C})=0.1887+(0.018)(1.72)\ln{(373.15/298.15)}=0.1956\  \pu{kJmol^{-1}K^{-1}}$
So, at $\pu{100^\circ C}$, $\Delta S = 0.1088\ \pu{kJmol^{-1}}$ and $T\Delta S=40.6\ \pu{kJmol^{-1}}$ So the change in Gibbs free energy between saturated liquid and saturated vapor at $\pu{100^\circ C}$ and 1 atm is $$\Delta G=40.7-40.6=0.1\ \pu{kJmol^{-1}}$$ which, to within roundoff error is zero (as expected).
The Spoiler answer is obviously incorrect.


text:
How the plate height is assumed to be the standard deviation?
Fundamentals of Analytical Chemistry book states that the  plate  height  can  be  thought  of  as  the  length  of  column  that  contains  a  fraction  of  the  analyte  that  lies  between  $L$  and  $L - \sigma$ [1, p. 870].
But what I know is that $H = \sigma^2 / L$ (At the retention time), so, how can plate height be equal to $L - (L - \sigma)$?
$H$ is the plate height. $L$ is the column length. $\sigma$ is the standard deviation.
Reference

Skoog, D. A.; West, D. M.; Holler, F. J.; Crouch, S. R. Fundamentals of Analytical Chemistry, 9th ed.; Brooks Cole: Belmont, CA, 2013. ISBN 978-0-495-55828-6.

I feel that you misunderstood what the authors were trying to say in that textbook. Forget about everything for the time being.
Imagine if you injected a single analyte band into an HPLC column of length L at a constant flow rate. You waited for some time, and now the band is exiting the column. For simplicity, assume that the band is rectangular.
Ask yourself, how much average distance did the analyte band travel? Indeed, it must have traveled a distance L if it has to exit the column. Most modern HPLC columns are 10 cm long; assume analyte traveled an average of  L=10 cm.
Now a band cannot be infinitely thin inside a chromatographic column. The analyte spreads in space, i.e., its width increases as it passes through the column due to diffusion and other processes.
So let us say the band is now a 0.05 cm wide rectangle when it reached the end of the column. Half of it is inside the column, and half of it has exited the column. What width of the band is inside the column, and what width is outside the column? You would say 0.05/2 cm of the band is inside the column, and 0.05/2 cm of the band is outside the column.

Total distance traveled by the portion of the band which has exited = 10 + 0.05/2 cm
Total distance traveled by the portion of the band which is inside = 10 - 0.05/2 cm

Now in chromatography, as you have seen, peaks are not rectangles, but they are Gaussians. Furthermore, Gaussians are characterized by their "standard deviation," which is a measure of their spread. If you apply the sample logic

The mean (=center of the Gaussian) has traveled a distance of 10 cm or L
The portion of the Gaussian which is still inside traveled a distance L-1*standard deviation
The portion of the Gaussian which exited the column traveled a distance L+1*standard deviation.

Note that conventionally a full Gaussian band at the base is considered to have a width of 4xstd. dev.

Thanks for the answer. I got this part, but what has this to do with the plate height? Why the book assumes that the plate height is equal to the distance from the middle of the Gaussian shape to the point the tangent line intersects which is equal to $\sigma$, though the equation stats that $H = \sigma^2 / L$?

This is property of a Gaussian function. Again before you read Skoog section, see Figure 1 and Figure 2 here https://www.chromatographyonline.com/view/chromatography-fundamentals-part-vi-gaussian-distribution-and-moment-analysis

You are right that the description in SWH&C is a bit nebulous and potentially misleading.  What the book attempts to do is to explain the meaning of $\sigma$ as it relates to the peak shape and peak area. The catch is that you can't report a single value of $\sigma$ as representative of the efficiency of a column, since it would not be constant under all circumstances. But you can report one value of H (for a given flow rate and solute/mobile phase) and this will be representative. Then if H and L are known the standard deviation of the Gaussian peak shape in distance units can be computed as $\sigma=\sqrt{LH}$.
One might wonder why $\sigma$ is not constant, or why plate height is not proportional to $\sigma/L$, as the textbook might appear to suggest, so I will explain why this is.
The reason is that one would like to have a property that depends on the composition of the column but not on its dimensions (or the elution time). If you double the length of a column, then a solute will take twice as long to elute (if the mobile phase flows at the same rate), but the width of the peak will increase  $\propto\sigma \propto t^{1/2}\propto L^{1/2}$, meaning $\sigma/L\propto L^{-1/2}$, so $\sigma/L$ is not a property that is independent of the length of the column. On the other hand, since $\sigma^2 \propto t \propto L$,  $\sigma^2/L$ is independent of L and of the elution time (provided the flow rate is constant).
Additional relations that allow you to draw these conclusions are  $L\propto\mu t$ and $\sigma=\sqrt{2Dt}$, where $\mu$ is the mobility of the solute and D its diffusion coefficient. D is a constant property for a given solute and column phases, and $\mu$ is proportional to the flow rate and phases, but not dependent on column dimensions.  Therefore $D/\mu$ is independent of column dimensions or of the elution time, and by extension so is  $\sigma^2/L$.
Note this only covers the effect of diffusion of the solute in the mobile phase (the efficiency is also a function of the flow rate for other reasons).
For more information on how H is determined in practice you might want to look at a GE Healthcare application note (Ref 1) or the reference therein (Ref 2).
References

GE Healthcare Application note 28-9372-07 AA

Hagel, L. et al. Handbook of process chromatography 2nd ed., John Wiley and Sons, Inc., New York (2008).

In isocratic chromatography, the peak width measured as sigma = t_r/ sqrt(N), where N is the column efficiency. Therefore, the standard deviation increases linearly with time. I feel your proportionalities are not correct. Am I missing something? So in a rather strict sense, sigma not sigma square is proportional to t_r, where, t_r is the retention time. Since, H has been normalized with respect to length, it is indeed a length independent measure. I agree Skoog's version is twisted.

@M.Farooq There are techniques like isoelectric focusing that will change some of the relationships. My point is that if one looks at simple diffusion as the source of broadening, then it makes sense to compute H as an invariant property for the column as the ratio of the peak variance (in distance units) to the column length. I suspect some of the trouble in Skoog's book has to do with units (converting between time and distance) and an attempt to be concise.

H cannot be an invariant property of a column because it is very much kinetics and flow rate dependent and hence analyte dependent. In short, Skoog made the matter worse by trying to assign a physical meaning to H. We should just treat it as a number.

@M.Farooq You're right, H is a function of the flow rate, so I'll have to modify my answer to explain what I mean. You're right that the solute identity matters, and that this is all approximate. The simple answer to the question is that the OP is basically right, the book convolutes the definition of H with that of $\sigma$.


text:
critical temperature and critical pressure condition data of Starch (Corn)
does anyone know where can I find the critical temperature and critical pressure condition data of Starch (Corn)?

Can't find non-existent things.

@Mithoron The OP confidence that it exists is worthy of my admiration. I suspect he may have become a victim of a joke of "go and find it."

Hypothetical temperature of starch melting or even boiling is much higher than temperature of starch pyrolysis, not speaking about critical values.
Additionally, starch is mixture of molecules of various lengths and branching. Therefore, melting, boiling nor critical parameters would not have been defined, even if starch had been thermally stable.


text:
Dielectric polarization: how to identify what part of the total polarization (dipolar+electronic+ionic) I need (conceptual question, I use DFPT, VASP)
I am doing polarization calculations to get data for use in other calculations.  To start with, I want to calculate the static dielectric constant. However, I'm a newbie to polarization and it is a struggle to relate my calculation output to the static dielectric constant. If someone could point me in the right direction, I would be grateful.  I'm using VASP software for the first time.
The total polarization is mainly the sum of dipolar, ionic, and electronic terms.  VASP implements a method (DFPT; density functional perturbation theory) for polarization calculations that seems to calculate only the ionic and electronic terms.  This is a concern because my fluid may have a significant dipolar component.  If anyone thinks I'm wrong on this, a heads-up would be great.
The reason I mentioned that DFPT may only calculate the ionic and electronic contributions to total polarization is that I have managed to get VASP output for the electronic term.  In the output file (.xml, for any VASP aficionados), it is in the form of a dielectric function that has six elements.  Three of the elements belong to a "real" part and three belong to an "electronic" part of the dielectric function. From my reading, it looks like the "real" part should somehow correspond to the static dielectric constant. My question is, which part of the total polarization is this likely to represent?
In other words, does each of the three terms in the total polarization contribute to the static dielectric constant, where this one is likely to be the electronic contribution?  Or is the output I have likely to be the total (dipolar too) static dielectric constant?  I ask because I read that only the dipolar and ionic terms contribute significantly to the static dielectric constant.

@GeoffHutchison, thank you.  Yes, there are even VASP and DFPT tags there.  I tried but one of the moderators deleted the bottom two paragraphs and it didn't make sense or address what I needed, so I thanked him, deleted it, and moved it here.


text:
Does anyone know what program was used to make this molecular structure picture?
I am a quantum chemistry beginer, and this picture color scheme is classic. I wonder how to make such pictures by myself.

"… this molecular structure picture": you have two pictures, which one you are interested in? The first one is pretty straightforward and can be obtained with Avogadro/Jmol/Mercury/MolView/Olex2/… The second one looks like an ORTEP diagram for atomic displacement parameters refined isotropically. Either way, you have to attribute all third-party illustrations.

Here's a web-based molecular 3d diagram maker: http://wwwtyro.github.io/speck/

The first type of image is very easy to make. I would suggest Avogadro with POV-ray. Avogadro can be used to draw the structures (or open a file containing the information, like .mol2, .sdf etc.) and then high-quality images can be generated using POV-ray.
To do this, install Avogadro, then draw a structure, and then File> Export > POV-ray. A new subwindow will open, where you can specify the location of the POV-ray executable.

In my case, the executable is pvengine64.exe, it might be different in different platforms, or in different versions.
You can also change the resolution here. Finally, click "Render", and a new windows with POV-ray should open, and it should display the image, after a few seconds.

This is the image of the molecule that you showed in your question. The background looks like that because the "Alpha transparency" option is turned on in the Avogadro subwindow, which makes POV-ray render only the molecule, and there is no background.
As for the second picture, it looks like something that would be used in crystallography to represent probability ellipsoids of atoms. I don't know much about this, so I can't help you there.

OK. btw, i found there's a bit difference between Avogadro and  Avogadro 2 version.

@kamisamaintherain Yes, Avogadro 2 is still under development afaik. Does it work with POV-ray? I have never used it.

After File> Export, a Windows file window popped out directly.

@kamisamaintherain Ah, maybe it saves the .pov file directly? If it saves the file, then you can right click and press "render with pov-ray" (on Windows at least)

I downloaded the first version, and it works fine. once again, thanks for your patience!!!


text:
Multiple stir bars have lost their magnetism
I'm attempting to produce sodium metal chemically. It is a proven process. It requires 15 hours of heating with continuous stirring. I've run the process 4 times now. Every time the stir bars get demagnetized and of course don't work anymore. I've replaced the stir bars with different sizes, types, and from different manufacturers. Regardless the stir bars inevitably quit working. The temperatures used are 120 C for two hours, and then 200 C for the remainder of the 15 hours. The process calls for quantities of Mg, NaOH, Menthol Crystals, and Mineral Oil. So, that's what the stir bar is stirring. The rate of spin is not high since the stir bar would be thrown to the side of the flask, so the rate is very carefully set. The rate, however, doesn't seem like it would be the issue here.  Does anyone have any information or suggestions. please.

Heating the bars  causes spontaneous demagnetisation and by reaching the Curie temperature even gaining paramagnetism. Try experimenting with different magnetic materials where these phenomena happen at higher temperature. Consider alternative ways of mixing, if applicable.

I would thought about the apparatus + heater/heating bath bearing placed on a horizontal linear or circular  shaker. These are used e.g. for large volume extractions or  in pharmaceutical stability studies, simulating tablet dissolution.

@Poutnik If the Curie temperature is the factor here, then OP must have some poor-quality bars. Usually magnet stirring bars should be good for heating up to at least 350 °C.

I suppose eventual demagnetization comes before paramagnetic transition, I have mentioned that for completeness. I admit am I not an expert in magnetic properties. @HSChem  Are the bars usable after cooling down again, or is the change permanent ?

I really appreciate each of the responses. Thank you! For Poutnik's question about bar usability after they've cooled down, my response is they are not usable. They remain demagnetized. It seems illogical to me because the temperature doesn't seem high enough to consistently cause problems. I've used bars from three different manufacturers if I recall correctly. Is there a brand of stir bars that has worked well for anyone.

It is possible that the stir bars are being chemically damaged by the harsh reaction (teflon stir bars, for example, are certainly not compatible with liquid sodium and possibly not with fused alkali at high temperatures). So, if the coating is damaged, the contents might suffer reactions that destroy the magnet.

Thank you matt_black for your response! It certainly is a consideration and may well be the cause. The mixture of magnesium and sodium hydroxide as a heated solution is destructive. Without making sure the NaOH is very dry the bottom of the flask can be damaged. The stir bars always come out coated with a relatively hard black substance that must be scraped off. A mixture of sulfuric acid and hydrogen peroxide won't touch it after hours of soaking. Do you know if there is a different type of stir bar that might not be affected in this way? Thanks again for everyone's responses!

The reaction involves "Mg, NaOH, Menthol Crystals, and Mineral Oil." Somehow, I doubt that recipe will make any Na metal! And, as has been mentioned, heat destroys magnets.

That's what I thought Dr. Moishe Pippik, when I first heard about this process. Then I watched it happening. Here's a link if you're interested: https://www.youtube.com/watch?v=BsNoiFj3wlw&t=138s                                                   If you do watch the video I'd be interested in your thoughts. Regardless, please accept my appreciation for your response concerning this issue. It means a lot to me for someone to take the time to share their thoughts.


text:
Triple Point of a compound and critical point
Is it necessary for all compounds to have a triple point? Also, how is the triple point determined?
Suppose a substance does not have a triple point, so how do researchers agree on the impossibility of existence of a triple point in such a case? Also, can there be multiple triple points for a compound?
Also, since the vapour equilibrium curve ends at point E, what is the phase of water beyond that point?

All compounds except helium have a triple point, unless they decompose before reaching it. As for the second question, the phase beyond E is called supercritical fluid.

So what is the phase of water in supercritical fluid?

The phase is supercritical fluid. To call it a liquid would be wrong. To call it a gas would be wrong too.

Imagine analogy: There is a beach(liquid) and a steep cliff plateau above it (gas). when you walk along the shore ( heating up), the beach  climbs up and the cliff plateau goes down, each becoming more similar to the other. At some point ( critical point ) they meet and become the same.  Is it a cliff or a beach ?

@Poutnik Yes,I get you, each deriving properties of the other with also remaining itself... it could be anything depending on P,T and other values.

It is evidence you have not got it yet. Their ALL property values converge to each other. Imagine a fog and a foam, each under intensive mixing, where you gradually add water or air respectively, at some point, there is no difference between fog and foam.

Do you mean to say there is no distinction between both and they become same eventually ,meaning there is no vapour nor liquid  ?

Exactly. If you mix 1 kg of substance and 1 kg of identical substance in identical state, what you get ?

Something which is not either of them but all over same (homogeneous sort of)

Let us continue this discussion in chat.

@IvanNeretin - helium does have a triple point where the liquid (He-I), superfluid (He-II), and gas phases meet. It is at 2.17 K and about 0.03 bar. Just not the 'normal' triple point, but 3 distinct phases coexist there.

@JonCuster That's right, point taken.

@Poutnik "Is it a cliff or a beach ? "        depends:are you at sea level? ;)

@buckthorn Not, you are not. :-)  It was said the beach slowly climbing ( or is prolonged ) and cliff does the otherwise. As some point, there is no cliff in the sense of an edge.

@buckthorn Another analogy could be a skipping gramophone groove, where the skip is the ordinary phase transition, while the plate rotation is going around the triple  point to do the same phase transition again and again. Like only evaporation and no condensation. Or vice versa.

The essential condition is the compound thermal stability.
If it decomposes below its melting point, it does not have a triple point.
If it decomposes before properties of gaseous and liquid phases converge to each other,  it has just estimated, extrapolated triple point.
Beyond CP we talk about supercritical fluid, it kind of shares many properties of both liquid and gaseous worlds.
It is usually determined experimentally, or as an extrapolation.
See the picture of liquid/gas phase system below, at and above the critical point on Wikipedia: Critical_point

You must first understand what happens when water is heated along the boiling curve.
Start from the point $\pu{100°C}$and $\pu{1 atm}$. At this point there are two phases, one liquid whose density is nearly $\pu{1 g/cm^3}$ and exactly $\pu{0.96 g/cm3}$. And the vapor has a density of $\ce{0.0006 g/cm^3}$. If now you heat this system in a closed volume, the temperature and the pressure will increase like in a steamer. But the densities of the two phases vary in opposite directions. The liquid expands, and the vapor contracts.
At $\pu{180°C}$ , the liquid has a density of $\pu{0.87 g/cm3}$ and the vapor $\pu{0.0079 g/cm3}$
At $\pu{312°C}$, the liquid has a density of $\pu{0.71 g/cm3}$ and the vapor $\pu{0.046 g/cm3}$
At $\pu{374.8°C}$, both the liquid and the vapor have a density of $\pu{0.32 g/cm3}$
If you look at the surface of the liquid just before $\pu{374.8°C}$, the surface of the liquid gets less and less visible. It is transformed into a sort of haze, and slowly disappears. Of course the pressure is extremely high ($\ce{217 atm}$) and the curve pressure vs temperature stops at $\pu{374.8°C}$.


text:
Is glycinate ligand a monoanion or a zwitterion?
Glycinate as far as I can guess from the structure is a monoanion with negative charge on oxygen atom (though that negative charge is resonating). But when I used this logic in my exam it costed me 4 marks. I don't understand why.
Is it because it is derived from glycine in strongly basic medium which is not a monoanion but a zwitterion? Still, it doesn't seem to be making sense.

Solid glycine is in zwitterionic form. That's why it is hard to dissolve. That's probably what the question asked. Can you elaborate a bit?

In solutions more basic than pH 9.6 glycine mainly exists as the carboxylate anion you describe. Between 9.6 and 2.34 it exists mainly in the zwitterion form.

titration curve from here

So, what exactly do I consider glycinate ligand to be-mono anion or zwitterion,  from an exam point of view ?

It entirely depends on the pH of the solution it is in. What exactly does the question say?

You may find more info here.

,Assuming glycinate is in its anionic form ($\ce{H2N-CH2-CO2^-}$), one thing to consider is it may not act just as a simple anionic ligand like chloride or hydroxide.  You can count off four atoms from the nitrogen to either carboxyl-ate  oxygen and those atoms both have available electron pairs.  So you add the metal atom and you get a five-membered ring similar to that formed with ethylenediamine.  Thus, glycinate anion may act as a chelating agent.

An example of this is ferrous glycinate  http://ve.scielo.org/scielo.php?script=sci_arttext&pid=S0004-06222001000500003


text:
Calculating nanocluster's Lewis acidity
In general, fluoride ion affinity, ammonia binding energy, photoacoustic IR spectra of adsorbed species (pyridine, CO) are used to measure the Lewis acidity of a material.
But in the computer simulation method, such methods are not suitable to calculate nanoclusters Lewis acidity because of the susceptible electronic environment of the system
[if extra species placed near a nanocluster, nanoclusters change their configuration and
form chemisorption as nanoclusters are unstable (tend to form bigger species: nucleation.
Furthermore, it might be a global minimum (GM)].

Thus, I concluded that calculating electron affinity (EA) might be the most suitable method to calculate the Lewis acidity of nanoclusters (overall, not for each cation/active site).
What do you think about the method to calculate the Lewis acidity for the nanoclusters system?
or is there a method to calculate the Lewis acidity of each atom site for the nanoclusters system?


text:
In sourdough bread baking what do the wild yeast and bacteria not eat during dough fermentation and why?
I bake bread using a sourdough starter. The starter is a culture of lactobacillus and some kind of yeast. They apparently eat the sugars or carbohydrates in the dough during fermentation but when I look at the nutritional info for fully baked sourdough bread I see 56gm of carbohydrates per 100gm and my question is, why do these organisms stop eating and what is it exactly that they leave behind? They seem to consume something in the dough then they stop as if they have run out of food at which point you need to bake the bread else it becomes "over-proved". Why do they stop eating and what is actually left over when the fully fermented dough goes in the oven?

I guess they eat up the monosaccharides, which are an easy food. The rest (which is the bulk of the dough) they are either unable to consume, or it takes much longer.

Plus, but I am inviting, they can be regulated. I.e when they are full the reaction slow down.

@Ivan They are fine with all kinds of carbs. Sourdough "is fed" with flour, not sugar. Bacteria don't get full and then rest, instead they divide when they have sufficient nutrients.

@Ivan does flour contain a lot of monosaccharides? Are these essentially sugars?

@Karsten, but there comes a point where they dont eat any more and you need to eaither feed them more flour or they will die apparently, that is why you keep a starter in the fridge to slow the consumption speed but I am still unclear on what is left over? if they eat carbs why don't they eat the rest of the 56gm of carbs left over per 100gm?

The purpose of "feeding" is to dilute the acids and replenish the carbs. It also dilutes the concentration of bacteria. The combined effects result in better conditions for cell life and reproduction. Every time the "eat" some carbs, they also produce some acid and ethanol. It is the waste products that keep them from reproducing more as the sourdough matures.

perfect, that completely answers my question, thank you.. shoul I wait or vote now?

@KarstenTheis you said they don't get full and rest and one line below affirm the opposite. The possible regulation I mentioned.

Flours contain a diverse array of carbohydrates which vary depending on the specific type of flour (wheat, rye, teff, etc). There will usually be a lot of starch (itself having variable amounts of different types of branching), some small mono and oligosaccharides (again of varying types) and "fiber", which is a broad term for polysaccharides that are more resistant to digestion.
To add to the variables, the specific types of bacteria and yeast in a sourdough fermentation can vary in their ability to digest the different types of carbohydrates.
So one part of the answer to your question is that yes, there are carbohydrates that cannot be digested, but the specific molecules that are not digested will depend on the flour and the microbes. The cause of the undigestibility is either 1) the monosaccharides are linked in a way that they cannot be cut apart or 2) the microbes lack the metabolic capability to digest the monosaccharide.
But generally in a bread dough, the digestible carbohydrates have not been completely consumed, so some of the remaining carbohydrate is digestible. The reason it is not used up is that the baker stops the fermentation before it reaches that point so that the dough still has a good consistency.
It is also true that the fermentation activity slows down considerably before this point, even though it does not stop. There are two reasons for this. 1) the pH drops, which slows the metabolism of the bacteria and 2) the easily digestible carbs get used up first, so the later fermentation rate can be affected by the slow rate of breakdown of the more difficult to use carbohydrates. Ultimately, the dynamics of the fermentation are determined by the choice of flour, the microbial population and the conditions (temp, water content, whether and how often the baker works the dough, etc).

Is there a difference between 1) and 2) reasons for indigestibility? If a microbe lacks the capability to digest a sugar, isn't that because it doesn't have enzymes capable of breaking that glycosidic bond?

(1) means they could use the monosaccharides if the glycosidic bond is broken, eg cellulose. (2) means they can’t use the free monosaccharide even after the glycosidic bond is cleaved. For example, some organisms cannot transport some monosaccharides. Others cannot phosphorylate or epimerize them into metabolic intermediates. Mannose and galactose are examples of mono saccharides that some organisms cannot utilize.

From the introduction in a paper on the different lactic acid bacteria (LAB) cultures in sourdough:

Sourdough LAB give rise to the characteristic qualities of several kinds of foods (flavor, texture, taste, and shelf life) by producing metabolites such as lactic acid, acetic acid, and ethanol. However, it is not easy to control sourdough LAB fermentation, due to the spontaneous occurrence of complex phase successions of LAB communities during subculture. Previous studies have characterized three phases, each dominated by the genera Weissella, Pediococcus, and Lactobacillus, whose succession determines the balance of metabolites in sourdough.

Over time, the pH of the dough changes (hence sourdough).

One of the major characteristics of sourdough fermentation is a drop in pH proportional to the maturation of the LAB community that produces lactic and acetic acid, eventually reaching a pH of approximately 4.0. Since pH changes must induce stress on the LAB community, assessment of pH conditions is necessary to understand and control the evolution of the LAB community in sourdough.

Follow-up question

[OP, in comments] but there comes a point where they don't eat any more and you need to either feed them more flour or they will die apparently, that is why you keep a starter in the fridge to slow the consumption speed but I am still unclear on what is left over? if they eat carbs why don't they eat the rest of the 56gm of carbs left over per 100gm?

The purpose of "feeding" is to dilute the acids and replenish the carbs. It also dilutes the concentration of bacteria. The combined effects result in better conditions for cell life and reproduction. Every time they metabolize ("eat") some carbs, they also produce some acid and ethanol. It is the waste products that keep them from reproducing more as the sourdough matures.

So you are saying that once the PH reaches a certain level the bacteria and yeast can no longer eat the food anymore?

Yes, they make their own habitat inhabitable. Brewer's yeast does the same with alcohol. Maybe a cautionary tale for humanity...

ok so it is not that they run out of food, it is that their habitat becomes untolerable to keep eating and reproducing... hahah yes that is a good point. Maybe coronavirus will come in the sourdough too and start infecting them :)

There are indeed viruses that attack bacteria. They are called bacteriophages (or phages in lab jargon), and phage-infections are a plague on labs that study or utilize bacteria. Mixed metaphors... plague is caused by another bacterium.

But even at low pH, the yeast continue to ferment if there is food. In most sourdoughs, the fermentation is ended by the baker, not by actual cessation of microbial activity. Also, a lot of the carbs are in fact undigestible because neither the flour nor the microbes have the right enzymes to break them down. So it's a combination of multiple factors

Like humans, unicellular organisms don't do well in their own poop metabolites.

Your question is based on a false premise. They don't stop when they run out of food. When you overproof the dough, there is still tons of edible carbohydrates left in the dough, mostly starch.
If you wait for the fermentation to stop on its own, this will happen not due to running out of food, but due to the accumulation of too much waste material. The usual story taught is that it is the ethanol - this is why we didn't have strong alcohol before destillation was discovered, you can only ferment up to 15-ish percent alcohol before your fermenting organisms die (and you usually get this percentage in wine, beers are below that, even with tricks for producing strong beer). I could imagine that other stuff combines with the ethanol to kill them off, the whole ammonia, thioles etc. produced by vigorous fermentation are not very friendly to microorganisms.
But you are not even waiting for the fermentation to stop. Overproofed dough is still very actively fermenting. It just goes into a state where its gluten can no longer trap gases - I don't know if this is because you have too much gas, or because the microstructure of the gluten mesh changes, probably both. Anyway, at that point, you cannot bake good bread, but this has nothing to do with stopped fermentation.
To answer literally your question of "why do they stop eating"

if you leave them alone until they stop on their own, they stop because they die out due to pollution (poisoned by their own waste)
if you throw your dough because it overproofed, they don't stop eating
if you use your dough for baking before it is overproofed, they stop eating because they die from the heat in the oven

To sum it up, fermentation processes in the kitchen (not only dough, but also yogurt or sauerkraut) do not mean that the microorganisms use up their food. On the contrary, they use up a tiny fraction of the food available, and most of it is left unchanged in the finished product.

"they use up a tiny fraction of the food available" - That's good because we want to get most of the nutrients when we eat the food. (Unless we're talking diet beer.)

What yeast does:
glucose => etanol + carbon dioxide.
What lactobacilus does:
glucose => lactic acid
They usually don't find glucose around them, so they break down the starch first (starch is abundant in the dough).
They both also do other things, like making new cells from the ingredients they find around them, but these other things are of less importance in bakery.
Why they stop at some point? Well, in the usual bread-making, they stop only when you bake the bread and the temperature kills them.
They can be significantly slowed down (like any other organism that cannot control its own temperature) by cooling down the dough. That's how ready-made dough is sold.
If you leave the dough alone, they will live and reproduce up to the point where etanol, carbon dioxide and lactic acid are too much for them. Depending on the water content of the dough, the availability of oxygen, the ability of the carbon dioxide to escape and the initial mixture of bacteria and yeast, you may get some kind of beer. In general, no more fermentation happens in the beer, as long as there is neither oxygen nor dilution. The microorganisms will just slowly die off because they don't tolerate that much of their own products around.
Or you may get something less pleasant than beer, because the dough is usually not air-tight and because other microorganisms are always present. Given time, air and if the mass is not too dry, the end result will be carbon dioxide, nitrogen and some inorganic salts.


text:
What is the voltage of an electrochemical cell at equilibrium?
I am having a hard time understanding what happens at equilibrium in an electrochemical cell (a full cell with two electrodes). At equilibrium, according to Nernst equation, the cell potential is $E_{cell}=E^0_{cell}-\frac{RT}{zF}lnQ_r$. Is this cell potential equal to the voltage of the cell ? At equilibrium, the voltage is zero, so does it mean that $E_{cell}=0$ and that both half cells have the same potential ? I am confused about this, as I also understood that $E_{cell}$ at equilibrium is the electromotive force of the cell, which means that it isn't zero.
Thank you for your help !

There is a cell equilibrium and 2 independent half-cell equilibrii.
There are 2 half-cell equilibrium potentials and a cell open voltage as EMF as the difference of 2 half-cell equilibrium potentials, because the cell in the whole is not in equlibrium.
When a cell gets into equlibrium as the whole system, the cell voltage is zero.


text:
Why is vicinal diiodopropane unstable, leading to deiodination?
I came across this reaction where simple electrophilic addition mechanism will result in a vic-diiodide.
$$\ce{H2C=CH-CH2-I ->[HI, excess][CCl4]}$$
But the solution says that vicinal iodine groups are unstable and the following reaction occurs:
$$\ce{CH3-\underset{(unstable)}{\underset{\underset{\huge I}{|}}{C}H-\underset{\underset{\huge I}{|}}{C}H2} ->[][$-$I2] CH3-CH=CH2}$$
I am unable to figure out how the dihaloalkane is unstable. It might be due to steric reasons, but still the two iodine can arrange in a way that reduces the repulsions.
But even if it is unstable, how did $\ce{I2}$ leave the haloalkane to yield an alkene? This dehalogenation does happen when a metal is used though (Zn dust, Mg in acetone, I guess) but what would be the mechanism here?

As for the question, I doubt one can really comment on the stability of the compound without knowing what else is present besides the compound. Apparently, the source also omitted the presence of any reagent that could have been used to cause the dehalogenation, only showing the $\ce{(-I2)}$.

That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.

Is this in clayden @user55119 ? And can this happen in other like vicinal chlorides / bromides ?

Hey i found an interesting observation, C-C single bond length is 134 pm. While van der wall radius of Iodine is 198 pm and that of chlorine is 175 pm. That means even if we count opposite orientation in Iodine, van der wall repulsions are too significant unlike in case of Chlorine.

Could you share a reference for the mechanism?

Ashish Ahuja: Try this. https://pubs.acs.org/doi/pdf/10.1021/ja01607a037


text:
Does the anti-rust properties of bluing carbon steel disappear at a certain temperature?
Yesterday I blued my new carbon steel wok over a very high (200K BTU) burner which is capable of getting the wok to 1000+ degrees with ease.
I understand that bluing adds a rust-resistant layer which is triggered by a chemical reaction at around around 575 F.
In this case I decided to blue the wok all the way up to 1100-1200 F given that this will be used on a very high temperature stove, and while I won't cook at those temps, the pan itself will get that hot from time to time (through accident, burning off stuck on oil, etc).
As I was doing this it brought some questions to mind. I tried researching but I have no knowledge of chemistry so I came here instead.

Does the rust-resistant property of tempered carbon steel vanish after you leave the blue temperature zone? In this case, I got it to the "very slight red" or "slight red" zone.

Are other properties affected as well at these temperature zones, in addition to rust resistance? It seemed from Googling that hardness is likely plummeting, which I suppose isn't important for a wok. It got me to thinking what else is changing in this pan.

Thanks!
In process:

Completed:

Those are temper colors caused by different thickness of iron oxides. Little or no rust protection. The time at temperature can be estimated by the colors. Gold > pink > blue > purple, nominally. They thicken with longer time and higher temperature. Stainless steels go through the same color changes (in air) at higher temperatures.


text:
Intermediates in aqueous bromination of phenol
A question in my book (an archive for IIT JEE exam questions) asks:
In the reaction:

The intermediate(s) is/are?
The correct answers are given as

However, as per my knowledge, compound (b) does not appear in the reaction mechanism. Is this an error in the book or do I know the wrong mechanism?
Correct mechanism according to me:

Can you give the reference for the book?

Also, (b) and (c) are not intermediates, they are products.

@Mathew If (IV) is the product then (b) and (c) could be intermediates.

@Shoubhik R Maiti: In that way, (I) and (III) are also intermediates. But they were not included in the answer. That's why my comment. In my experience, (II)-(IV) are in the product mixture, in which (IV) predominates.

@MathewMahindaratne its an archive for IIT JEE (an exam I'm preparing for) questions. The questions considers (IV) to be the "final" product and so anything in the mechanism before that as intermediates, and it was a multiple choice question, so not all the intermediates are given as answers. However is the mechanism that I gave correct?

@Hridai Khurana: I understand now. Yes, your mechanism is acceptable. It's believed to be step by step addition of $\ce{Br2}$ to the ring. However, I doubt ipso addition of $\ce{Br2}$ (structure (b)) exists because of inductive effect of $\ce{Br}$ (it is not in your mechanism).

In a 1922 paper, the intermediate of panel B is shown:

More than a 100 years later, this looks a bit funky.
A more recent paper from 1989 shows a dienone intermediate for the reaction in neutral or acidic aqueous solution. However, there is only one and not two bromine atoms in the para-position:

These are both screen-grabs from the freely available first page, thus the low quality.
It is possible that the mechanism varies with solvent and pH. Panel C shows the product of the first bromination step in its deprotonated form. Not sure why this would be called an intermediate, other than that the subsequent bromination steps are faster. This means that in the presence of an excess of bromine, you probably could not isolate the singly-substituted product.

+1. This question was in my mind for a long time. Glad you answered it. I have linked the other question which I answered through comments but since this post didn't have a decent answer, I didn't refer this post in my own answer.

In NaOAc/HOAc phenol forms 2,4,4,6-tetrabromocyclohexa-2,5-dien-1-one. An Organic Synthesis Preparation

@user55119 I see. The reaction posted by the OP is basic, not acidic, though. The 1922 paper does not mention pH on the first page. I wonder what they mean with "tribromophenol bromide", and whether it is related to the dienone in your comment.

@Karsten: I suspect that the tetrabromide in strong  base may form the tribromophenoxide or at least be in equilibrium.

The correct answer per my checks is B, however I stand for correction. Since the solvent is a polar solvent, the OH of the solvent will abstract the H of the OH on the phenol. Making the phenol a phenolic oxide PhO. Which will activate or the electrons on the O of the phenoxide will be pushed to the Ortho and the Para position on the ring, making them more activated hence the formation of a trisubstituted ring.

Is this answer wrong?


text:
Why is the Ag+ concentration in a beaker fixed in the reaction between excess AgCl (s) and KI (aq) from the start?
This arises as a sub-question for the following question: Calculate the concentration of I- in a solution obtained by mixing 0.10 M KI with an excess of AgCl(s).
I tried to look at How does addition of AgCl affect concentration of iodide in KI solution?, but it didn't answer my question which I detailed below.
My textbook says to solve it in the following way:
1) Ignore the 0.10 M KI since it doesn't matter.
2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its sp.
sp=1.77×10−10=^2
=1.3×10^−5
This is the Ag+ concentration
3) Substitute calculated Ag+ concentration into equilibrium for sp of AgI.
sp=8.51×10^−17=[Ag+][I−]
8.51×10^−17=[1.3×10^−5][I−]
[I−]=6.4×10^−12
My confusion arises due to step 3. Why can you assume that the Ag+ concentration is fixed and substitute the concentration into the sp for AgI? Why would you need not create an ice table for the sp of AgI where the first calculated Ag+ concentration is the "initial" condition? For example, this is what I mean: I inserted a picture of an ICE Table that I would think to use, but the textbook thinks otherwise. I'm confused why I'm wrong. Thank you!!

The book is very wrong in step 2. The $\ce{Cl-}$ and $\ce{I-}$ quantitatively exchange. So in the end the solution is 0.1 molar in $\ce{Cl-}$ with just a trace of $\ce{I-}$ .  I'll try and edit my answer at https://chemistry.stackexchange.com/a/147936/22102 to clarify the answer.

I added a new answer to the original question explaining why the book answer is wrong. If that answer doesn't explain the conundrum please ask there for further clarification.


text:
Are compounds which are used to seed clouds like silver(I) iodide hygroscopic?
I have read a teensy bit of the subject matter and the explanation they provided for silver iodide's ability to being a good site for ice to form is because of its size resembling an ice nucleus.
How exactly does silver iodide adsorb water onto its surface? Is it hygroscopic?
Ice crystals growing larger due to vapour deposition or rapid freezing of supercooled water upon contact with ice makes sense. But how does silver iodide do it?
If the hygroscopic properties of silver iodide are what enables it to seed clouds, can the common salt also be used to seed clouds? I do apologize as the question wasn't aptly structured. Please do help in understanding this!
P.S. I found an article which I find is relevant to this question [1], but unfortunately, I'm not experienced enough to go about reading scientific articles.
Reference

Pena, R. G. D.; Caimi, E. A. Hygroscopicity and Chemical Composition of Silver Iodide Smoke Used in Cloud Seeding Experiments. Journal of the Atmospheric Sciences 1967, 24 (4), 383–386. DOI: 10.1175/1520-0469(1967)024<0383:HACCOS>2.0.CO;2.

No, it's rather the opposite.

Silver iodide is not hygroscopic. It is used to seed clouds, because when the atmosphere is saturated with water vapor, liquid water should start condensing. But to do this, it needs a seed, a place when the first molecules will start to condense. No water molecule feels that it should be the first one to show other molecules where to be condensed. But, as silver iodide has the same hexagonal shape and same dimensions as snow, it may make believe it is an ice crystal. And the neighboring water molecules will be mistaken, follow the example and deposit on it. The common salt cannot do this. It does not look like ice.


text:
Milliliters of NaBr when changing molarity
I was very confused by this problem. I was able to get the molarity, but I wasn’t certain how to go from there:
If $\pu{2.60 g}$ of $\ce{NaBr}$ is dissolved in enough water to make $\pu{160.0 mL}$ of solution, what is the molar concentration of $\ce{NaBr}$? How many milliliters of $\pu{0.10 M}$ $\ce{NaBr}$ would you need to supply $\pu{2.60 g}$ of $\ce{NaBr}$?
$\text{Molarity} = \pu{0.158 moles/L}$
I wasn’t certain how to answer the second part of the problem. I kept getting $\pu{159 mL}$, but the answer is $\pu{250 mL}$. What am I doing wrong?

Edit:
I am not entirely used to the M symbol for molarity, so I was mixing myself up. Instead, I equated $\pu{0.10 mol/L} = \frac{(\pu{2.60 g} \times \text{molar mass NaBr})}{x \ \pu{L}}$ — I solved from there for $x$ and got $\pu{250 mL}$.

My crystal ball is experiencing inference from a current lightning storm. Please explain how you calculated 159 ml and we can tell you what you did wrong.

I understand what I was doing wrong. I am not entirely used to the M symbol for molarity, so I was mixing myself up. Instead, I equated 0.10 moles/L = (2.60 g X molar mass NaBr)/X L — I solved from there and got 250 mL.

It should be mass/molar maa to get mol. Not mass x molar mass.

@MathewMahindaratne I got the right answer, but thank you for pointing out that arithmetic slight when posting. As you note, I meant to write mass X 1 / molar mass = g X (mole/g).

I decided to give a help here since OP has given enough effort to solve this problem. First, OP needs to know the molarity of a solution is defined as the amount of solute (in this case $\ce{NaBr}$) in $\pu{mol}$ per $\pu{1.0 L}$ of solution. The first part of problem states:

If $\pu{2.60 g}$ of $\ce{NaBr}$ is dissolved in enough water to make $\pu{160.0 mL}$ of solution, what is the molar concentration of $\ce{NaBr}$?

To solve this problem you may need to know the molar mass of $\ce{NaBr}$, which is not given. But you can find it by googling it. The molar mass of $\ce{NaBr}$ is found as $\pu{102.8 g mol-1}$. Now you can find the amount of $\ce{NaBr}$ in $\pu{160.0 mL}$ or $\pu{0.160 L}$ of solution (volume has to be in $\pu{L}$), which is:
$$\frac{\pu{2.60 g}}{\pu{102.8 g mol-1}} = \pu{0.0253  mol}$$
Thus,
$$\text{Molarity of the solution} = \frac{\text{amount of NaBr}}{\text{volume of solution}} = \frac{\pu{0.0253  mol}}{\pu{0.160 L}} = \pu{0.158 mol L-1}\\ = \pu{0.158 M}$$
The second part of the question is:

How many milliliters of $\pu{0.10 M}$ $\ce{NaBr}$ would you need to supply $\pu{2.60 g}$ of $\ce{NaBr}$?

This supposed to be solved using the first part of the reaction. So, by first part, you know now that $\pu{0.160 L}$ of $\pu{0.158 M}$ $\ce{NaBr}$ contains exactly $\pu{2.60 g}$ of $\ce{NaBr}$. so using the dilution principle, $M_1V_1 = M_2V_2$, you can find your unknown:
Here, $M_1 = \pu{0.158 M}$, $V_1 = \pu{0.160 L}$, and $M_2 = \pu{0.10 M}$. Your unknown is $V_2$, you can find using $M_1V_1 = M_2V_2$:
$$V_1 = \frac{M_1V_1}{M_2} = \frac{\pu{0.158 M} \times \pu{0.160 L}}{\pu{0.10 M}} = \pu{0.253 L} = \pu{253 mL} $$
Since olarity of the solution is given as $\pu{0.10 M}$, the answer should be $\pu{250 mL} $ with correct significant figures.
Note: @Entropy has shown the correct calculation as you have done. However, it is a different method to find the volume of solution. I'm sure the one wrote the question intended to use the first part of the question to solve the second part.

Great alternative approach! Thank you :)

Whoever down voted this answer is a gutless coward because he/she didn't give a good reason for their action. I'd like mediator to look at this matter.

It comes to be $\pu{252 mL}$ precisely:
$$V = \frac{\pu{2.6 g}}{\pu{103 g}} \times \frac{1}{0.1} \times \pu{1000 mL} = \pu{252.4 mL}$$
$\text{Molar Mass of NaBr} = \pu{103 g mol-1}$.

That is correct, but using significant figures, the answer is rounded down.


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Chemically producing sodium metal from magnesium
Anyone with experience making sodium metal ($\ce{Na}$) with the following process? Reactants: $\pu{14 g}$ magnesium powder, $\pu{20 g}$ sodium hydroxide; Catalysts: $1$ to $\pu{2 g}$ menthol crystals; Reagent: $\pu{125 mL}$ mineral oil with no additives. Can use $3$ to $\pu{4 g}$ sodium metal to remove any $\ce{H2O}$ in the $\ce{NaOH}$ during the first two hours of stirring at $\pu{120 ^\circ C}$. Then $\pu{13 h}$ stirring at $\pu{200 ^\circ C}$. My understanding is this process was identified by NurdRage on YouTube and is a proven process. I've run it 4 times and the sodium never solidified even though it is present in the resulting paste like product. I'm looking for someone who's run this for advice. I hope this is an appropriate question to ask. I'm new and don't know the protocols followed by everyone. If not an appropriate question please let me know so I can avoid such questions in the future. Thank you!

Theoretically it is not possible: $$\ce{Na+ + e- <=> Na} ,, E^\circ = \pu{-2.71 V}$$ and  $$\ce{Mg^2+ + 2e- <=> Mg} ,,  E^\circ = \pu{-2.372 V}$$ Probably, in this case it is: $$\ce{Mg(OH)2 + 2e- <=> Mg + 2OH-} ,,  E^\circ = \pu{-2.690 V}$$  May be heating assists here to work.

https://chemistry.stackexchange.com/questions/81805/reaction-between-magnesium-metal-and-sodium-hydroxide-does-this-even-happen

@MathewMahindaratne Std. potentials hardly say about what's theoretically possible.

https://chemistry.stackexchange.com/questions/146991/can-be-elektron-alloy-used-to-make-sodium

I greatly appreciate your responses and the links provided to related issues. This has all been helpful as further thinking to consider as I continue this process. If I get it to work perhaps I'll be able to provide some information about it that others will find interesting. Thanks again!

@HSChem I guess, you may use a fuse and dry ingredients as shown in another nurdrage video.

Hey Aditya Roychowdhury...Thank you very much for your response. I'll double check but I believe I've viewed and duplicated the dry ingredients thermal process. It does produce sodium, though it is highly contaminated with Mg. I did use magnesium power, however, and I've got some Mg ribbon to cut up into small pieces to use and see if the results are different. I will double check the link you've taken the time to send to me. Thanks again!


text:
How to make spectra smooth by using gaussian or lorentzian function
I want to use some origin spectrum to fit to my experimental spectra. However, my experimental spectra is not so smooth because of finite resolution of spectrometer. I used to use 'origin' to make it smooth, but it seems not reasonable because I don't know how 'origin' makes it. Then I was suggested using Gaussian or Lorentzian function.
A lot of nonsense, in one word, please teach how to use gaussian or lorentzian function.

you could fit by hand by adding gaussians or lorentzians to reproduce the spectra. but I think the more numerically justified way of doing it with chi squared fitting or some other type of numerical fitting method.

It would help us to answer your question if you posted an example spectrum.

considering that this question was closed, can you consider removing it and posting it on https://mattermodeling.stackexchange.com? (generally software based questions are better suited there)

please teach how to use gaussian or lorentzian function.

It will take several hours of discussion or 4-5 lectures. You are mixing several things.
a) Briefly, your collected spectrum is noisy. You want to remove the noise. What type of spectra is this?
b) The finite resolution of your spectrometer has nothing to do with noise. Increasing the resolution mathematically or so-called deconvolution is another story.
c) The easiest solution is to "fit" Lorentzian or Gaussians to your experimental spectrum after denoising it (step a). Experimental spectra are neither pure Gaussians nor pure Lorentzians.
Here are the steps. I am mentioning the keywords so that you can search the right terms.

Can you get hold of PeakFit software? It has a better function library than Origin.
Denoise your spectrum, using Savitsky-Golay filter.
Once excessive noise has been filtered, use a trial function such as Gaussian or Lorentzian and see which one fits better. This is checked by residuals.
In an experimental spectrum, not all peaks have the same width so you will have to use multiple Gaussians or Lorentzians to fit a segment of a spectrum.

Search "Pragmatic Introduction to Signal Processing Applications in scientific measurement 2019 edition" by Prof. Tom O'Haver. It is a free book and contains all the practical stuff which would be very useful to you.

Thank you for your answer. I'm grateful for your detailed answer.

My present condition is that I have gotten the spectrum of several metal solution by using a program which will be used to fit my experimental spectra. I'm sorry I didn't make it clear because I am not familiar with this subject. Anyway, I will follow your advise first. Thanks a lot.

Sorry, your experiment is still not clear. What is type of spectrum and what is the instrument?


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Lattice Energy and Stability in s block compounds
Lattice energy decreases down the group but stability of the compound still increases why?
For the purpose of this post I'll take alkali earth metal carbonate as an example.
As we go down the group size of cation increases and hence the lattice energy decreases, however my book stated that

The thermal stability (of carbonates) increases with increasing cationic size

I've heard that big cations are able to form better stable compounds with big anions. However we say that lattice energy has decreased. Then what is the factor that is providing this stability in an ionic compound ?

In this case the smaller cation will polarize the lager cation and thus decrease stability.

So it is basically that the development of covalent character is the factor responsible for decreasing stability ? and so can I conclude that as we are going down although lattice energy is decreasing but ionic character is becoming greater which is making it more stable !??

Yes, that is right.