text:
Does hot water make more foam?
A message has been floating around saying that it is better to wash your hands with warm water, as it helps the soap create more foam.
Is it true? If it is true, why is it so? Is there a temperature dependant factor in the soap's cleansing mechanism?

Foam is hardly relevant at all. It might be true that warm water cleans better, though.

All chemical processes have a temperature dependance, higher temperatures increase reation rates. The breakdown of fats (including virus particles) by soap is a chemical process.

@Waylander A virus particle is hardly a fat, and soap does not chemically break down fats either. ???

The lipid bilayer is similar enough for someone who is not a chemist, as is the dispersal of fats by detergent

At high temperature, the dissolution rate of any substance in water is greater than at low temperature. This has nothing to do with the nature of the solute, soap or not soap.

https://chemistry.stackexchange.com/questions/113103/does-the-main-washing-effect-of-soap-come-from-foam

Is there a temperature dependant factor in the soap's cleansing mechanism?

Yes, because hydrophilic-hydrophobic interactions are always driven by entropy. Entropy contributes to Gibbs Energy which is the final quantity determining whether a reaction takes place or not (or in which direction it takes place for that matter). But your question implies other questions that should be answered.
Does soap create more foam with warm water?
I don't know. In general, mixtures exhibit phase separation at a lower temperature bound of miscibility and a upper temperature bound of miscibility. Some mixtures do not have a temperature range where no phase separation occurs, some are miscible in all circumstances, you get evaporation/freezing before you get phase separation, think water and acetone.
Does more foam in my soap help clean my hands?
No. Foam/bubbles are small sheets of water with surfactant towards the gaseous phase on the in- and outside. Hence the oily look and texture. What you cleans your hands is the surfactant, because it also helps solubilize hydrophobic stuff like certain proteins. The cleaning is done by hydrophilic stuff being dissolved in the water, and hydrophobic stuff being covered inside micelles of the surfactant, which are soluble in the water.
Does more soap help against viruses?
If your hands are really dirty, maybe. Surfactants are known to denature proteins. You wish to denature the virus proteins so that they can no longer carry out their functions. This would also rob your skin of its protective fatty outside layer. You need not worry to denature your own proteins unless you are trying to drink soap water. Also, aggressive surfactants like SDS work better than your standard fatty acid soap, being even more damaging to your skin as well.
How can I optimally protect myself against viruses on my skin?
Use a desinfectant like ethanol or iso-propanol with at least 70% concentration. It denatures proteins quite effectively. Usage for at least 30 seconds is the official recommendation to kill most common viruses and bacteria. I personally don't wash off, but wait for it to evaporate, which is a good heuristic for the 30 seconds. Make sure to use some sort of lipid replenishing cream every now and then to recover your skin.

The T-dependence of a reaction (whether it shifts to the left or right, and by how much, as the temperature is changed) is determined by the T-dependence of the equilibrium constant (Keq), and the T-dependence of Keq is determined by the sign and magnitude of the enthalpy, not the entropy.  Yes, the T-dependence of the Gibbs free energy is determined by the sign and magnitude of the entropy, but Keq  is related to the Gibbs free energy through a factor of T, which shifts the T-dependence from the entropy (in the Gibbs free energy) to the the enthalpy (in Keq).

@theorist higher temperature means higher entropic contribution to reaction directionality, no matter how you put it. Either it increases the factor in front of dS or it acts as a divisor for dH.

Nope, that's incorrect.  The entropy has no effect upon the T-dependence of the equilibrium constant (i.e., upon the first derivative of the equilibrium constant with respect to T).   The T-dependence of the equilibirum constant is given by the van't Hoff equation:  $\frac{d ln K_{eq}}{dT} = \frac{\Delta H}{RT^2}$  Note that $\Delta S$ is not present in that equation. To understand why $\Delta S$ disappears, you should go through the van't Hoff equation's derivation.  This can be found in any good pchem textbook.

If something is not present in the first derivative of something, it could as well have a constant contribution to an quantity, right? The van't Hoff equation shows that with rising temperature, the contribution of enthalpy to the equilibrium constant approaches zero.

Now you're confusing two different things, the T-dependence of Keq, and the value of Keq.  At the beginning of your answer, you were saying there is a T-dependence in Keq because of the entropy, which is wrong, as shown by the van't Hoff eqn.  Now you've switched to talking about the value of Keq itself, which does depend on the entropy change, but misapplying the van't Hoff equation to the value of Keq, when in fact the van't Hoff equation is for the T-dependence of Keq.


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How to compare the acidity of this multicyclic hydrocarbon to the acidity of cyclopentadiene?
This is the question:

Which one is more acidic?

According to me, cyclopentadiene should have been the answer, as losing a proton would make it aromatic; whereas the first option doesn't need deprotonation to be aromatic, it already is. However the first option is given as the correct one.
I have come up with the following reasonings as to why the first option should be more acidic:

Deprotonation might impart aromaticity to the pentagonal rings too which would act as the driving force to lose the proton.
The negative charge on central carbon (after deprotonation) would be in conjugation with the $\pi$ electrons of the three rings which would make the entire molecule planar and hence reduce the strain on the three bonds that are directly linking the benzene rings. This could be another driving force for deprotonation.

The second reasoning seems much convincing to me but I fail to understand how can I compare that factor with the aromaticity factor involved in cyclopentadiene.

related https://chemistry.stackexchange.com/questions/124356/what-is-the-most-acidic-neutral-hydrocarbon

"...as losing a proton would make it aromatic..." - cyclopentadiene would become anti-aromatic instead (note the 4n electrons, and planar molecule)

@RahulVerma It's 6 electrons in conjugation (4n+2).

@AumkaarPranav: Oh yeah, you're right! Late night, brain farts.

The method presented by Yusuf is a simplified model, which works best for well-behaved molecules. Unfortunately in this case it presents the 'correct' result based on incorrect assumptions. (Even though I did a calculation, I am not 100% convinced this is actually the correct result in the first place.)
In order for the resonance to stabilise a negative charge, the π-systems must be aligned, which is the case for the deprotonation of cyclopentadiene, but it certainly is not true for the other molecule F (Loong said it's called 3a2H-benzo[3,4]pentaleno[2,1,6,5-jklm]fluorene). Here is the calculated structure at the ri-BP86/def2-SVP level of theory:

As you can see, it is domed, and there are three (almost) independent aromatic systems. Through deprotonation this will not change much.
A reason for its relatively high acidity can be found in applying Coulson's Theorem, where the bond angle between two $\mathrm{sp}^x$ hybrid orbitals can be calculated from their angle $\omega_{ij}$ and their hybridisation indexes $\lambda_x$:
$$\cos(\omega_{ij}) = -\frac{1}{\sqrt{\lambda_i\lambda_j}}$$
Since the molecule is C3v symmetric, therefore $\lambda_i = \lambda_j \mathop{:=} x$, this simplifies to:
$$ x = -\sec(x)$$
The inner angle is $\approx\pu{102^\circ}$, therefore $x \approx 5$. The term for hybrid orbitals is actually just an abbreviation:
$$\mathrm{sp}^{x} = \mathrm{s}^{\frac{1}{x+1}}\mathrm{p}^{\frac{x}{x+1}}$$
Given that we just form linear combinations, we can see:
\begin{align}
 1\times\mathrm{s}, 3\times\mathrm{p}
   &\leadsto 3\times\mathrm{sp}^5, 1\times\mathrm{sp}
\end{align}
Given the rounding I have done, the remaining orbital is very roughly speaking an $\mathrm{sp}$ orbital; in the same ballpark as acetylene, from which we know that it has a very acidic proton.
The reason for the high acidity is therefore not the stabilisation of the conjugate base, but the destabilisation of the acid. This is the completely reverse reasoning as cor cyclopentadiene, in which the corresponding base is massively stabilised by resonance.

Given the above, how are you supposed to know, and how are you supposed to judge these compounds?
In my opinion, the exercise is something more on the level of quantum chemistry course and belongs into the canon of a masters student. While simply models will accidentally produce the correct result, it is important to see, that they completely fail for F. Structure-property relationships cannot be inferred from a 2D-graph of the molecule. To be honest, the reasoning above was also rather post-hoc, after I have calculated the molecule and analysed its structure.
I believe, (as long as you are not a human calculator), that there is no simple way to tell which of those molecules is more acidic; there is no easy way to weigh the stabilising effect of resonance against the destabilising effect of pyramidalization.
On the ri-BP86/def2-SVP level of theory (which is rather crude, but I don't have a supercomputer at home) it really comes out that F is more acidic. The following isodesmic reaction yields a difference in electronic energy of about $\Delta E_\mathrm{el} = \pu{-92.8 kJ mol-1}$.

Important: Even though the above sound reasonable, it still neglects plenty of things and has other shortcomings; to name the most important: thermal corrections, solvent effects, level of theory.

Addendum:
Upon deprotonation the structure does not change very much (see larger image), it remains very domed, see the overlay of the structures below.

Judging from the three highest occupied molecular orbitals (HOMO, HOMO -1, HOMO -2) we can still presume some delocalisation of the negative charge into the adjacent aromatic rings.

I'm talking more about hybridisation, Bent's rule, and Coulson's theorem in the following: Are the lone pairs in water equivalent?

Interesting take.  One thing that comes to mind:  Does the polycyclic structure become more planar when we remove the proton?  If so, would that mean we can still invoke the aromaticity/resonance effect?

@OscarLanzi not significantly, it remains domed. If I don't forget, I can post a picture of the anion, and I could have a look at the orbitals.

Thanks.  Something else:  You render three central orbitals as $sp^5$ meaning each has 1/6 of the $s$ orbital.  If so, the orbital with the proton has the remaining half of the $s$ orbital and so is all the way to $sp$.  May want to check this.

@Oscar I've added the images. Thanks again for catching my calculation mishap.

You can take both factors (1) and (2) combined as the reasoning for the greater acidity of the former.
When you are comparing stability (in thermodynamic terms) of two conjugate bases, you are basically trying to position them with respect to each other on the energy axis of the energy profile (or reaction coordinate diagram).
In this case, the reaction coordinate (RC) for both the intermediates remains fixed (as a similar reaction, namely, deprotonation has occurred for both the reactants), while you try to use qualitative arguments to position them relative to each other on the energy axis.
In that spirit, one basic idea which we can use to check is that by seeing that:

More is the energy released by the system in form of resonance, more will be the stability of the negative charge.

Note that if you look at the reactants themselves, the first species is releasing roughly 3 times the resonance energy of benzene more as compared to the second one due to the presence of three benzene rings. Hence, we can estimate that the former should have had a lower energy positioning than the latter to start with.
After proton abstraction from both, like you said, 3 more five-membered aromatic rings will be generated in the former system, plus the negative charge can be delocalised over the three benzene rings. So now, if you see, the latter species is stabilising by an extent equivalent to the resonance energy of one cyclopentadienyl anion.
For the former, the system is releasing energies equivalent to (resonance energy of 3 cyclopentadienyl anions) + (resonance energy of 3 benzyl anions).
So, by a qualitative estimate, the (1) species should be energetically more stable than the (2).

Thank you for the answer. Can you also please explain (or maybe provide a link to do so) the nature of aromaticity in the first case: do all the three cyclopentadienyl rings show aromaticity along with the benzene rings at the same time, or is there some kind of partial aromaticity? I am unable to wrap my head around aromaticity of the conjugate base of the first molecule (and aslo, do you know what it's called?).

I'm sorry, but I have to disagree with this simplified reasoning.

@Martin - マーチン♦Well,I am not a master's student(undergrad presently) so I had no knowledge of the calculations you have presented in your answer,I simply used what I have been taught by my instructors till now. However,the 3-D structure was quite enlightening about the how the overall aromaticity of the system was functioning,I didn't think about looking at that. Thank you for your answer :)

Yes, you are at no fault at all. I believe many of these exercises are hastily given out, without the proper reasoning beforehand. I have noticed that manytimes such exercises heavily rely on these inaccurate models. That is mainly a problem within the educational system, but I do not wish to diverge in politics. I hope that 3D-structure considerations will play more important roles in your future career ;)

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is bleach residue (after evaporation) safe?
I have been preparing bleach disinfectant lately in a spray bottle and has been spraying on my couch, door mat, door knots, etc...
I don't rinse it... So I assume there is bleach residue left after evaporation?
Is it safe to leave the residue?
I felt some very mild burn like feeling on my arms when putting my arms on my couch lately; I don't know if this is some kind of placebo effect or if it is real.

Liquid bleaches are stable in highly alkaline solution. You should not be spraying it on couches. If you are feeling irritation, wipe the couches with a wet towel (in warm water) several times. Let them air dry. There is no need to be afraid or panic out of the current situation. Caution and patience is the way to go.

Bleach, especially employed in laundry applications, may have, in addition to the active ingredient sodium hypochlorite (NaOCl), an equal concentration of NaCl (as it is prepared from the action of Cl2 on NaOH) and at times, an alkaline, like lye (NaOH) or Washing Soda (sodium carbonate, Na2CO3). The latter can cause skin rashes and irritation (and will be present on drying).
The rationale for the latter is to keep the pH elevated in aqueous sodium hypochlorite as upon exposure to CO2, there is the possible introduction of hypochlorous acid. While HOCl is more powerful (and reportedly nearly a hundred times more potent disinfectant than the associated hypochlorite) is far less stable (a short shelf-life). Further, the decomposition product is O2 and also HCl (and not NaCl).
The created hydrochloric acid can further react with the sodium hypochlorite or hypochlorous acid as follows:
$\ce{HCl + NaOCl -> NaCl + HOCl}$
$\ce{HCl + HOCl <=> Cl2 + H2O}$
leading to a chain decomposition of the bleach, hence the advantage of adding NaOH or Na2CO3. To quote the cited source:

The pH has a significant effect on the stability of sodium hypochlorite solutions. Below pH 11 the decomposition of sodium hypochlorite is significant due to the shift in the equilibrium in favor of the more reactive hypochlorous acid. A pH between 12 and 13 gives the most stable solution. This equates to 0.4 to 4.0 grams per liter (gpl) excess NaOH. Greater concentrations will not improve the stability. Excessively high improve the stability. Excessively high alkalinity will damage textiles and retard the bleaching and disinfecting actions of the hypochlorite.

I would suggest that you can even dilute your current NaOCl dosing and add some lemon juice, just prior to use, to create (per the source reference noted above) a much more powerful mix (and a more potent bleaching agent, so be careful where you select to employ it) albeit, with a much shorter shelf-life.
To rapidly remove bleach smell (likely Cl2O from the dehydration of HOCl), one can apply a spray of dilute H2O2, which rapidly converts NaOCl to NaCl and oxygen. Caution: this further promotes bleaching action via intermediate radical formation (see 'Hydroxyl radical from the reaction between hypochlorite and hydrogen peroxide'), so again be careful where you apply it.

@ Ajkoer. All you say is exact, but these rections happens in a solution of bleach. If the laundry is dry, these reactions will simply not happen, because NaClO is decomposed when dry. So there is no risk of having HOCl, Cl2 or Cl2O in Yeoman's laundry.

I did also say that NaOH (using in the prep of bleach) and perhaps Na2CO3 could be leftover upon evaporation. NaCl is also found and I explained how from its prep. Also explained why one would want to make the bleach more alkaline. And yes, I added some side comments on how to remove bleach smell with H2O2.

I also went off-topic and discussed the conversion of the NaOCl with weak lemon juice to the 80 to 100 times more powerful (including in disinfecting per the provided reference) hypochlorous acid, HOCl, which is albeit more unstable, so this implies a time-of-use decision.

Bleach is usually a mixed solution of sodium hypochlorite NaClO and a lot of sodium chloride NaCl. NaClO is decomposed when dry. So your couch is probably soaked in salt (NaCl). It may produce an uncomfortable feeling.

so it is not dangerous right? with what's going atm I am kinda paranoid and trying to disinfecting everything

It is not dangerous. But it may be disagreeable.

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Which type of deviation from Rault's law is shown by mixture of water and acetone?
There are dispersion forces in acetone amd hydrogen bond in water so what they should show negative deviation but just want it to be confirmed

Hydrogen bond will be formed between acetone and water.

Water and acetone form a polar/non-polar mix and due to strong bonds, the mix is less volatile than would be predicted by the Ideal Version of Rault's Law (which is a linear parametric weighing of the respective partial vapor pressures).
This results in what is referred to as a negative deviation (as the curve is beneath the straight line associated with the Ideal Version of Rault's Law).
Here is a good educational video on the topic.


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A question about potential difference across a membrane
Membrane potential of living cells arising from the difference of concentration of ions on both sides of the membrane, but why didn't someone use selective polymers as AEM (Anion Exchange Membrane) and apply pressure to cause accumulation of anions on one side of the membrane forming a potential difference?

I think it has been done...

@Martin-マーチン i didn't see anyone doing it

@mohamed, search Donnan potential and papers like these https://pubs.acs.org/doi/10.1021/jp0342273


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Colligative properties of colloids
The following lines are from my textbook:

Colloidal particles being bigger aggregates, the number of particles in a colloidal solution is smaller than a true solution. Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of smaller order compared to true solutions.

What is the explanation supporting the above statement?

Roughly (in the ideal world), colligative properties depend on the number concentration of particles, not on what those particles are. A particle can be a micelle or vesicle or other colloidal particle such as a polymer, or a nonassociated solute molecule. For instance, a protein is a polymer usually consisting of a linear chain of aminoacids. It is considered a colloidal particle. If you compute the colligative properties of a protein solution based on the total aminoacid concentration you obtain a different result than if you compute them based on the protein concentration, which may be orders of magnitude lower than the aminoacid concentration. The same applies to other polymers or to association colloids such as micelles. The concentration that determines colligative properties is that of the micelle or polymer, not that of the (associated or linked) monomers which combined make up the polymer or association colloid.
By the way, the statement "the number of particles in a colloidal solution is smaller than a true solution" is clearly potentially confusing. Presumably what is meant is that a "true" solution does not contain polymer or associated molecules (forming say micelles).

Suppose a solution $0.01$ M of sugar or any moderately big organic compound. If $N$ is the Avogadro number, $1$ Liter of this solution contains $0.01·N$ molecules, and this is about $6·10^{21}$ molecules. The osmotic pressure will be  $ p = cRT = 0.01 RT$, because $c$ is the concentration of individual particules in the solution, here molecules.
Now suppose the molecules in this solution are not individually separated, but they are grouped in micelles containing $1000$ molecules. Each micelle is an independent particule in the solution. The number of these particules (here micelles) is $6·10^{21}/1000 = 6·10^{18}$. In moles, it is  $n/N = 6·10^{18}/(6·10^{23})$ = $10^{-5}$ mol. This number determines the osmotic pressure. The osmotic pressure will be : $p = cRT = 10^{-5} RT$.  This is $1000$ times smaller than without micelles.

Here is a simple example that may help from application of the Rault's Law (see this video), where colligative properties of a solution mixes apply (linearly or with deviations) to the relative molar concentration of solute compositions and not the identity of the solute.
First, take a known amount of CaCl2 and place it into a known volume of water, but do not stir (so there are large undissolved pieces of the CaCl2, or more formally, nonassociated solute molecules). Now, by Rault's law, the partial vapor pressure of the water is lowered, but only by the ratio of the moles of water present to the sum of both moles of water plus moles of dissolved CaCl2. Clearly, this is not as effective in reducing the partial vapor pressure as if you have totally dissolved the CaCl2.
Similarly, the freezing point lowering advantage of CaCl2 (employed commercially to 'melt' ice) is not achieved if pieces of the CaCl2 are too big to be effectively dissolved.
Finally, if one lets your anhydrous CaCl2 get exposed to air, there is created some insoluble masses, which do not dissolve even on boiling. However, the boiling now occurs at a low temperature (at least theoretically) than as if the particles had dissolved.
The above directly provides an explanation of the statement: "Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of smaller order compared to true solutions"

"by Raoult's law, the partial vapor pressure ... sum of both mole s of water plus moles of dissolved CaCl2. " Applying the same logic to the 'freezing point situation' I believe we can conclude the same for F.P. as well But then you say,"The freezing.. is not achieved if pieces of the CaCl2 are too big to be effectively dissolved'--- here the mole fraction theoretically becomes 1. So doesn't this mean that the freezing point gets reduced even more? I'm sorry if this sounds crazy but I'm new to high school chemistry and Raoult's Law!


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Desilylation Mechanism
I want to understand the mechanism of desilylation of trimethylsilyl under fluoride conditions in methanol, because I can't find any useful info that makes me understand the mechanism that takes place.
Thank you in advance!

https://chemistry.stackexchange.com/questions/47698/relative-basicity-of-fluorine-and-oxygen-in-silyl-ether-deprotection

It does help a lot, but my case is a little bit different because a terminal alkyne is bonded to the silyl group instead of oxygen. Why would that bond be broken and not any other carbon bond?

Think about the exceptional strength of an Si-F bond and the pKa of terminal alkynes vs other types, sp2 and sp3, C-H bonds.


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Is this approximate equivalence table between Pople's basis sets and Jensen's DFT attuned polarization consistent basis sets (pcseg-n) correct?
I have some calculations results I runned previously using Pople's basis sets, mostly 6-311+G(d), under Gaussian09. These days I read some texts on Frank Jensen's family of polarization consistent, segmented basis sets, optimized for DFT (pcseg-0, pcseg-1, pcseg-2, pcseg-3, pcseg-4 and the respective augmented versions). Now I'm thinking about trying to reproduce my results using Jensen's basis set family and Gamess-US.
Said that, I'm not sure about the correspondence between the two sets. I understand in Pople's basis sets, for light elements (first 3 periods of the periodic table) it's common to see people using double zeta(DZ) 3-21G for rough calculations, and either double zeta 6-31G or triple zeta(TZ) 6-311G for more precise work. Both 6-31G and 6-311G can have up to two polarization functions (or none) and up to two diffuse functions (or none) associated, resulting in 2x3x3 = 12 combinations between them (not counting heavier elements, that would require f polarization orbitals). In the table 5 of Nagy, Balazs, and Frank Jensen. “Basis Sets in Quantum Chemistry.” Reviews in Computational Chemistry (2017): 93–150. Print., they group together 3-21G and pcseg-0; 6-31G(d), cc-pVDZ and pcseg-1; and 6-311G(2df), cc-pVTZ and pcseg-2. So I assume each basis set inside these 3 groups to be equivalent (but not sure). As pcseg-2 is the only triple zeta option, despite already falling in the range of f-polarized basis, I suppose that, to map these 12 Pople's basis into the respective Jensen's basis, I need some combination between pcseg-0, pcseg-1, pcseg-2, aug-pcseg-0, aug-pcseg-1 and aug-pcseg-2 over H and heavier atoms (6x6 = 36 possibilities). For me it's not trivial to choose which of the 36 Jensen's possibilities in this range best match the 12 Pople's ones not explicitly cited on the paper. My guess at the closest mapping between the 2 sets is as follow:
Type    POPLE's         DUNNING     JENSEN'S CLOSEST
DZ      3-21G                       pcseg-0 on all atoms?
DZ      6-31G                       ?
DZP     6-31G(d)        cc-PVDZ     pcseg-0 on H, pcseg-1 on heavier?
DZP     6-31+G(d)                   pcseg-0 on H, aug-pcseg-1 on heavier?
DZP     6-31G(d,p)                  pcseg-1 on H, pcseg-1 on heavier?
DZP     6-31+G(d,p)                 pcseg-1 on H, aug-pcseg-1 on heavier?
DZP     6-31++G(d,p)                aug-pcseg-1 on H, aug-pcseg-1 on heavier?
TZ      6-311G                      ?
TZ      6-311+G                     ?
TZP     6-311+G(d)                  pcseg-0 on H, aug-pcseg-2 on heavier?
TZP     6-311G(d,p)                 pcseg-1 on H, pcseg-2 on heavier?
TZP     6-311+G(d,p)                pcseg-1 on H, aug-pcseg-2 on heavier?
TZP     6-311++G(d,p)               aug-pcseg-1 on H, aug-pcseg-2 on heavier?
TZP     6-311G(2df)     cc-PVTZ     pcseg-2 on all

Is my reasoning sound and the proposed equivalence table correct, or did I get it all wrong? If wrong, could somebody please give the correct mapping from Pople's traditional basis sets to Jensen's optimized ones?

My personal opinion is that Pople-basis sets should have left serious calculations ten years ago. That being said, I would ignore the previously used basis set and simply pick the basis based on the standard criterion: what is the best basis set that I can afford for the calculation that I want to do? If this does not meet publication-level accuracy, obtain more computational resources. I also would not mix basis sets, the risk for errors due to internal BSSE is too large.

@ksousa, since this question went 1 month with no upvotes and no answers, yet within 24 hours on Matter Modeling SE you got an answer from Frank Jensen himself (the inventor of the Jensen basis set you're trying to come up with), and this seems to have happened to you 3 times now, could you help us by writing an answer describing your experience, here? https://materials.meta.stackexchange.com/q/1/5

Thank you @user1271772, I'll write it. If somebody else happens to stumble in the same problem and find this question, please don't use my attemped table. I got it mostly wrong. The correct table was given here.


text:
Which compound will undergo dehydration more readily?

Which will undergo dehydration more readily under acidic conditions, alpha-hydroxy ketone or beta-hydroxy ketone?

My guess would be the second one will undergo dehydration with ease relative to the first one although I am not sure. One reason that comes to my mind is that the carbocation formed in the first one will be destabilized by the ketone. Is my guess correct? If yes, am I correct in my reasoning too? Or are there any other dimensions that I am missing to look into.
Any help would be appreciated.

Sounds good, Tony!

@Tony also alpha-hydroxyketo will be relatively  more stable than beta due to a Hydrogen bonding, which will be better in the former compound since OH and O of keto will be closer. This should hinder the enolisation of alpha compund further.


text:
Identifying Copper Purity At Home
Are there any ways to figure out if the copper foil and mesh I have are pure, as-close-to 100% copper as possible (and as advertised)?

How pure do you need it? 99% (2 9's)? 99.9999999% (9 9's)? Question without context is meaningless.

@DrMoishePippik I am a layman– I don't have a good sense of what the difference is, but I just want to know if what I have is an alloy or is mostly pure copper.

There is no easy or simple home test for copper purity.

The colour is a good hint. It changes remarkably when you go towards really pure.

@Karl It's a conductive copper foil used for emf shielding and 'slug protection'. It does almost have a pinkish hue to it.

I don´t remember numbers and details, you´ll have to find sth online. But "pinkish" sounds good,  purer than your ordinary rain gutter material.

I doubt color could be really useful. First it does not give you any numerical data. And metallic copper sheet are usually superficially oxidized, and covered by black copper oxide. I remember having deoxidized a small sheet of copper by heating it in a Hydrogen atmosphere. The copper so obtained is nearly pink or pinkish ! So the typical brownish color of commercial copper is due to a partial oxidation process.

As a home tool, but not necessarily a definitive solution, I would suggest the dissolution of your copper metal/alloy in a galvanic cell where the Cu is the anode and as a noble cathode, use a source of graphite.
This approach for the dissolution has been favorably argued in commercial patents in a 1999 U.S. Patent 6,692,632. To quote in parts:

The method is preferably used in synthesis, but can also be useful in the destruction of pollutants etc. In certain embodiments of this nature, the use of external current applied to the system is not necessary, and a simple galvanic cell can be set up...

On copper particularly, for the home chemist, I have demonstrated a so-called Bleach Battery approach as an electrosynthesis path discussed in 'Subject: Experimental Variants of Bleach Battery to Obtain Metal Oxides (like CuO). However, I would replace the cathode, as in the current case as Nickel could be present as an impurity also in your copper metal sample, with very noble graphite in its path to CuO and other metal oxides that may become evident (or be chemically identified).
As the copper can be alloyed with metals like Zinc, Tin, Aluminum, Nickel, Silicon,..., with the first three creating respective oxides from the dissolving a small piece of the copper metal/alloy, potentially revealing distinct metal oxides as, for example, CuO is black, ZnO is white, SnO is blue-black (as produced by heating the tin(II) oxide hydrate, SnO·xH2O
where x<1), and Al2O3 varies from white to a clear transparent gel. The metals Nickel and Silicon are expected in their metallic state.
I say, convert a bit of it into metal oxides and proceed from there!

How can we estimate the purity of copper from obtaining other metal oxides?

+If you are in the US ( and many other places)you have copper 101 or 102 ( CDA = Copper Development Association designations), with 99.95 % minimum Cu. CDA says this Cu is what is supplied under the following ASTM Material Specifications- B48, B132 ,B152 ,B187 , B272 , B432 ,and F68 ; all flat Cu products. The same Cu is supplied in other shapes in other ASTM specs. Do you know the ASTM designation of your material ? What is your application where 99.95% + Cu content may not be sufficient ?

Review of a old ASTM indicates your material is likely B152. The 1981 revision lists several UNS copper grades , all range from 99.90 to 99.99 % Cu. I doubt the chemical requirements changed much since then.

Excellent answer! I upvoted because, unless one has access to proper spectroscopic analysis instrumentation, home analysis is completely futile. I happen to have a fair bit of copper on hand, including two pieces of OFHC copper sheet: 24"x8"x3/16" and 15"x8"x3/16". Fortunately, copper has to be highly pure for most of its intended industrial purposes.


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Determine the pH and pOH of pure water at 50 °C

Determine the pH and pOH of pure water at 50 °C.

I’m not sure what formulas to use on this problem.

Welcome to Chemistry.SE! We have a site policy for homework questions which can also apply to self-study questions. Please edit your question to include your thoughts, where you got stuck, and let us know where you're finding difficulty so we may best help you.

Use the same formula as for 25°C. The key is not the formula, but information you have to search for.

Think of Delta G vs. T

According to the “Self-ionization of water” article in wiki, the pKw of water at 50 degrees Celcius is 13.26. There is a nice table for other temperatures.


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Reactivity of different alcohols with NaNH2 and CH3I

Identify the final product (B) in the following reaction:

I'm not sure how $\ce{NaNH2}$ and $\ce{CH3I}$ will react with the different types of alcohol present here. According to my solution,ether was formed in place of $\ce{CH2OH}$ in the end, whereas the answer says that $\ce{CH2OH}$ group will remain unaffected. What am I doing wrong? Can anyone explain to me the full mechanism?

Tertiary alcohols are usually more reactive than primary alcools.

But the pKa of primary alcohols is lower  see http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf.    In real life I would not expect much selectivity in this reaction.

t-BuOH is more acidic than primary alcohols in the gas phase but not in solution.

Per my answer, a more likely explanation relates to an issue with reagents and/or procedure.

I had since edited my answer, and in particular, suggest checking the quality of the NaNH2 given its sensitivity to moisture/water/air.

Sodium amide is more than strong enough of a base to generate the primary alkoxide, which is stabilized best by solvation, so your proposed methylation at this position is perfectly viable. Cold $\ce{KMnO4}$ will leave the tertiary alcohol untouched and oxidizes the secondary one to the ketone.

I suspect, possibly just to a minor extent, that you have also managed to produce an expected photolysis product here (see below).
Interestingly, the photo sensitive methyl iodide, in the presence of a suitable catalyst ($\ce{CO2, Cu,}$..) with lab light (rich in UV) may have commenced in a radical reaction chain in the presence of ethanol:
$$\ce{CH3I ->[$h \nu$] ^.CH3 + I^.}$$
$$\ce{C2H5OH + I^. -> HI + C2H5O^.}$$
$$\ce{C2H5O^. + C2H5O^. -> (C2H5)2O2}$$
$$\ce{(C2H5)2O2 + ^.CH3 -> (C2H5)2O + CH3O^.}$$
....(the reaction chain continues)
where, I have stopped at the formation of Ether, albeit possibly created in very small amounts but detectable by its smell, nevertheless.
Repeat the experiment controlling for light/catalyst and see if you can still smell the Ether.
My advice, any time working with a photoreactive iodides, be mindful of light exposure.
[EDIT] The implication of my analysis, check your product yields, reagents and procedure, as a minor photo-induced side product may not (but still could) be connected to your results. In particular, I would start with the $\ce{NaNH2}$ as per Wikipedia:

$\ce{NaNH2 + H2O → NH3 + NaOH}$
$\ce{4 NaNH2 + 7 O2 → 2 Na2O + 4 NO2 + 4 H2O}$

which suggests that either exposure to moisture/water/air can degrade the compound.

I can't see how it answers OP question?

I have edited my comments. My answer clearly presents an understandable analysis and further notes it may NOT be even related to the poor results. So, look elsewhere and do remove the light factor.

No step is photochemically-induced free radical reaction. First there will be acid-base reaction, followed by SN reaction and finally oxidation.. These type of questions aren't to know possible issues with reactants or procedure, rather what final product wil be and why?

Zenix: My edited answer points to NaNH2 (which readily degrades). Your comment is definitely valid if it is even apparent that the reaction is at all proceeding. But with only a minor photolysis side product being created, this suggests to me that the reaction may, indeed, not be proceeding!

On further thought, the ethanol employed (the source of the Ether per a radical mechanism) may not be absolute, the source of water.

Zenix, thanks for discounting the photo radical path leading to the report of Ether. I readily await your claimed path and, if you can't, I even more readily await that admission, so I could postulate other routes.

The issue is that the reaction hasn't actually been done in reality and is just a hypothetical paper-pen reaction.. So, it is ideal conditions outside the reaction..


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Acidity order of organic compouds
Here is the question:

We have to find the order of their acidity.
Here is my thought process:
We check the acidity order by finding which conjugate base is more stable. Thus we remove $\ce{H+}$ from all of them. One question I have here is that in (IV), from which position should I remove $\ce{H+}$? Will it be the position from where resonance can take place or from the middlemost carbon?
In (I), (II), and (III), resonance can take place in all of them after removing $\ce{H+}$. Resonance will be more in (II) and (III) than in (I).
Conclusion:
(II) and (III) are more stable than (I). I am uncertain about the stability between (II) and (III) and about (IV).
Please tell me if I have been wrong somewhere and how I could proceed further...

Consider the pKas of dinitromethane, malonitrile and malonate. Dintiromethane is by a long way the most acidic pKa 3.6, malonitrile is a bit more acidic than malonate 11 v 12.9.

Okay, so if we talk about the acidic strength of a compound, consider adding a base to a solution of all the above. Practically, a reaction is bound to take place-if possible-due to thermodynamic reasons. Hence, reaction in IV will take place since the base will try to find the most acidic Hydrogen, which is why one should consider the removal of H atom from which resonance can take place.
Now for the order of acidic strength, I, II and III are individually more acidic than IV, due to extensive conjugation in all three but the last. This consequently makes the conjugate bases stable and acids stronger. Now for the order among these, notice that Cross-conjugation takes place in III, as we have COO- group. Hence, III will consequently be less stable and a weaker acid.
The order and reason stated by you for II and I are correct, and hence the order for acidic strength is:
II>I>III>IV
i.e. option A.

After we remove H+, resonance can also take place in (IV). Moreover, more stable the compound is, less acidic it will be. Cross conjugation can also take place in NO2.

Why does cross conjugation reduce stability?

@aitchessbee I would suggest you check out the instability caused by cross conjugation yourself https://en.wikipedia.org/wiki/Cross-conjugation here. No, the more stable the compound after removing H atom, the less basic the conjugate base and hence the more acidic the compound. This is the basic principle which you yourself have used to solve some part of the question. And also, the bond in NO2 is a 'co-ordinate bond' and does not allow cross-conjugation.


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Why dosen't NO₂ form according to the octet rule?
Supposedly every atom that is in the second period of the periodic table has to obey the octet rule; this means that they have to be surrounded by 8 electrons (or so I was taught). Studying resonance in an exercise, they showed how nitrogen as NO₂ can form 5 bonds, which would mean that NO₂ does not behave according to the rule previously mentioned. I am at a loss; any help would be great.

Does this answer your question

https://chemistry.stackexchange.com/questions/9448/structure-of-no2-compound

Does this answer your question? Why don't unstable odd electron species dimerize so that they become stable?

Capitalization can be important - your question title is asking about the diatomic Nobelium molecule, and I have no clue if that is even a thing...

Because not everything has to.

No atom has to obey the octet rule. It is a rule. It has no theoretical justification. It works often. But that's all that could be stated. And a lot of compounds are known that do not follow this rule like $O_2, NO, B_2H_6$, etc. And Nitrogen does not form 5 bonds. It cannot. $4$ bonds is the maximum for atoms of the second period (Li, Be, B, C, N, O, F).


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Number of nodes in Hartree-Fock solution
The Hartree-Fock equation for atoms  is of the form
$\left(\frac{d}{dr^2}+f(r)-\epsilon\right)P(r)=g(r) \tag1$
Usually algorithms to solve this equation assumes that the number of nodes of $P(r)$, $n_r$, is of the form
$$n_r=n-l-1 \tag2$$
where $l$ is the orbital quantum number and $n$ the principal quantum number.
For closed shell $(1)$ is spherical symmetric so $(2)$ is justified.
What about non closed shell, what is the theoretical justification for $(2)$?

This may be of value:' On the Solution of the Hartree‐Fock Equation in Terms of Localized Orbitals', J. Chem. Phys. 34, 89 (1961); https://doi.org/10.1063/1.1731622 .


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Boiling Soda Reaction - Does It work?
I've come across a series of reactions, where $$\ce{2NaHCO3 <=> Na2CO3 + H2O + CO2}$$ at temperatures above 50 degrees Celsius, and $$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$$ I Have 3 questions about these equations:

Will the above reactions proceed in a boiling Baking Soda solution set to reflux until virtually all of the baking soda is converted to lye and CO2?
If so, would the reaction follow a half-life pattern, over a certain period of time?
How would I calculate the amount of time required to convert a saturated baking soda solution into a solution with a certain concentration of lye?

I appreciate any advice provided.

The decomposition of heated NaHCO3 is rapid. It is the fundamental reaction for preparing Na2CO3 by the Solvay process, but it is not carried out in solution. It is done with dry powder. It is not necessary to make any rate calculation.

Actually, if one adds your two reactions:
$\ce{2 NaHCO3 <=> Na2CO3 + H2O + CO2}$
$\ce{Na2CO3 + H2O <=> NaHCO3 + NaOH}$
Upon cancelling, to derive the reaction:
$\ce{ NaHCO3 <=> NaOH + CO2}$
which implies that heated Sodium bicarbonate is a chemically active source of Lye, if one allows the CO2 gas to escape (which answers (a) of your question).
To answer (b) and (c), add Aluminum to the heated aqueous mix and it will dissolve, and you could collect the generated hydrogen gas over water and record volume increase over time. Reaction:
$\ce{2 Al + 6 NaOH -> 2 Al(OH)3 + 3 H2 (g)}$
Now, per this Aluminum shiny finish destroyed in dishwasher, you have an understanding as to why this occurs with Aluminum ware in the presence of NaHCO3, which is a popular spotless-rinse agent employed in dishwasher's washing powder.

Did you read Maurice's comment? It's quite correct this time, and your post isn't.

Note, I stated upon 'canceling' meaning removing compounds appearing on both sides of the reaction. This implies that the liberated H2O (as water vapor from heating the dry powder, for example) remains in a closed system where it could possibly interact with Na2CO3. This is not likely to occur upon decomposing NaHCO3 by heating in open air, which is the context of Maurice comment.

Thanks Mithoron, for your point which is certainly worthy of clarification.

The point is while bicarbonate easily decomposes when pure is not a proof it will in solution.

Your aluminum pan being disfigured in a dishwasher in the presence of NaHCO3 should be all the proof you need. By the way, I have performed the reaction of attacking preheated Al foil (to remove any protective annealing, oils,..) in aqueous NaHCO3 in a vessel with microwave heating which turns out to be vigorous, give it a try,

Here is a link to an external forum with my comments and reference on this topic https://www.sciencemadness.org/whisper/viewthread.php?tid=71493#pid479202 .

2NaHCO3 --> Na2CO3+H2O+CO2 at temperatures above 50 degrees Celsius: TRUE.
However, Na2CO3+H2O --> NaHCO3+NaOH is a hydrolysis reaction: it only goes to a small extent, maybe 2-5%, and on evaporating the water, it goes back to Na2CO3. Na2CO3 melts at 851C without decomposition.
So, if you take a solution of NaHCO3 (pH ~8.5) and boil it, you will generate a solution of Na2CO3 (pH ~12) by loss of CO2 and if you keep evaporating the water, you will get Na2CO3 as a crystalline powder (and when it cools, it will begin absorbing water from the air to form one or more of its many hydrates).

I never planned to evaporate the water: "...a boiling Baking Soda solution set to reflux"


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Effect of pH on sulphur isotope variation on a landscape
I recently ran a forest regression model, this model showed me that increasing soil pH increases sulphur isotope variation across the landscape. I am not how this is possible and if it makes sense

Sulfur has a natural variation in the proportions of its isotopes (consider volcanoes, for example, which act as natural isotope fractionators). So the variation due to biochemical action is expected.
The United States Geological Survey (USGS) discusses the use of sulfur (and oxygen) isotope ratio to determine the source of sulfur, temperature and pH in forests.
The National Center for Biotechnology Information (NCBI) discuses the fractionation of sulfur isotopes in beech forests: "The investigation of the fractionation of S compounds in forest soils is a powerful tool for interpreting S dynamics and S biogeochemistry in forest ecosystems. Beech stands on high pH (nutrient-rich) sites on Flysch and on low pH (nutrient-poor) sites on Molasse were selected for testing the influence of stemflow, which represents a high input of water and dissolved elements to the soil, on spatial patterns of sulfur (S) fractions." This seems to me to explain the results of your tests.


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Importance of concentration of gold in gold parting
I read following lines about gold parting:

Gold is not attacked by $\ce{H2SO4}$ or $\ce{HNO3}$, but copper and silver dissolve in them, when concentration of gold is less than $30\%$. If however, the gold content in sample is more than $30\%$, then copper and silver are also not attacked by acid of any strength. Hence, before acid treatment, impure sample is melted to add necessary amount of silver to reduce gold content to quartation.

My question is why aren't silver and copper attacked by acids of any strength when concentration of gold exceeds $30\%$?
Also at the end it's written that:

Gold is separated and fused with borax and nitre and $100\%$ gold is obtained.

Why do we fuse with borax and nitre $(\ce{KNO3})$ at the end? I don't see any need to do as gold won't react with either of them.

The fusion is to oxidize impurities and separate them into the easily discarded borax slag. It would be helpful to have a reference for your first statement (the 30% thing).

@Ed V my teacher shared notes... Since, our schools are closed due to COVID-19 lockdown. Won't HNO3 oxidise the impurities, since Cu and Ag are our main impurities in gold parting.

Between getting the gold alloy dissolved and the fusion step, you have to precipitate the gold, which will be powder that is not completely pure. Then you mix that with the fusion mix and fuse it.


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How do you manage to liquefy a gas by applying only pressure?
How can you liquefy a gas by applying only pressure? To increase the pressure of a gas you need a compressor, and when the compressor does work on the gas , it will increase his pressure but at the same time it will increase his temperature, and it will still be a gas. To become a liquid it will need to have his temperature decreased bellow his boiling point at the correspondent pressure. So how can u liquefy a gas by just incresing his pressure?
I read that in butane lighters, the butane is liquid inside the lighter because it is at a pressure of around 2 bar and at ambient temperature the butane is liquid if it has a pressure of 2 bar. But isnt that 2 bar the vapour pressure inside the lighter? If the temperature increases the pressure/vapour pressure inside will increase and if the temperature decresses the pressure/ vapour pressure will decrease and the butane will still stay at liquid state. So i dont undersand why people say that the butane is liquid because of the increase in pressure, because at the ambient temperature to become a liquid needs to be at 2 bar. And the butane that is put inside the lighter is liquid but is it already at 2 bar or is at any pressure?

First, not all gases can be liquefied at room temperature by increasing pressure. If the gas is above the critical temperature, it cannot be liquefied by any increase in pressure; it becomes a supercritical fluid. Supercritical fluids have some of the properties of a gas (e.g. diffusing through fine openings), ans some of liquids (e.g. dissolving solids and liquids). For that reason, supercritical $\ce{CO2}$ makes an excellent dry-cleaning agent.
Second, the supercritical point for butane is 152 °C at a pressure of ~3,800 kPa. At that temperature, a bit above boiling water, the pressure in the lighter would be about 38 times atmospheric pressure... if it hadn't yet vented its contents.
As for the manufacture of the lighter, butane could easily be liquefied by cooling to it's boiling point at standard pressure, ~0 °C, easily reached in a freezer (which is why butane is not the best fuel for camping in the snow). However, since the fumes must be contained anyway, a pressurized system is needed, and two atmospheres pressure is easily handled. The same for liquefied propane, which is pumped at room temperature into steel tanks.

But what do you mean by "gases can be liquefied at room temperature by increasing pressure"? if i increase the pressure of a gas at room temperature i will also increase his temperature and it will remain a gas, only if i decrease his temperature after increasing his pressure it will liquefy. Am I thinking correctly?

It depends on the phase diagram. Some gases can be liquefied by increasing pressure alone, without removing heat of compression, others require cooling, and yet others, such as hydrogen, are so far above critical temperature that they must be chilled considerably before they can be liquefied at all. Consider liquefied butane in a sealed container: heating it to a temperature below 152 °C will increase the pressure, but does not cause the butane to completely vaporize.

As one increases the pressure, the volume occupied by the gas decreases. Like you said, the temperature should increase with increasing pressure. Equivalently you could say that the thermal energy of the molecules of the gas has increased which manifests itself in increased collisions and vibrations of the constituent molecules. The decrement in volume serves as a constraint as this means that we are physically bringing those molecules closer. All this leads to increased interactions between those molecules and you get a liquid.
P.S. : I didnt read the supporting question as I am sleepy and therefore less motivated to read that.
P.P.S : This is my first answer here, so I am sorry for the construction of this answer. If you find any mistakes, kindly notify me (I'm not sure how it works).
Enjoy!

Hi, thanks for answering. What do people mean by  "gases can be liquefied at room temperature by increasing pressure"? if i increase the pressure of a gas at room temperature i will also increase his temperature and it will remain a gas, only if i decrease his temperature after increasing his pressure it will liquefy. Am I thinking correctly?


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extract cellulose from CMC using cuoxam reagent (schweizer's reagent)
Good day, I am new in the field of chemestry. My question is, if it is possible to reverse or extract (I am not very sure how to say it), cellulose from carboxymethyl cellulose (CMC) by the use of schweizer's reagent. I read that this reagent is capable to "regenerate" cellulose from a different sources, but unfortunately I havent found any paper that can support this idea.
I want to thank in advance for your time and help.

Nobody thinks of regenerating cellulose from carboxymethyl cellulose. it is not worth while. Cellulose is a cheap product. Carboxymethyl cellulose is a rather expensive product. So why spend much money from an expensive substance to be converted in a cheap derivate ?

The reason, is because in the laboratory where I am working, we have a huge amount of CMC, that nobody use and we have the materials to do the cuoxan reagent. That is the main reason


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Discrepancy between the apparent volume of the solution and the volume of the solute arising from the definition of solubility
I have solubility $s$ given in $\pu{g/mL}.$ I understand this as mass of solute $m$ divided by volume of solvent $V.$
But how about the volume of the resulting solution? I read online I should pour the solute into a measuring cylinder/cup first, and then add solvent until the volume fits the calculated value $(V = m/s),$ but this seems to be at odds with the definition of solubility.
What is the correct way to understand solubility? I can't know the volume of a solution based on solubility.

In a typical scenario, I would start by adding some solvent to a volumetric flask, maybe 25 to 50% full. Then quantitatively add the weighed mass of the solute. Rinse any solute down into the flask, swirl to dissolve the solute, and then add solvent to the mark on the volumetric flask. A stirring bar, if used, has a small volume that has to be compensated and volumetric flasks are calibrated for a specific temperature. That is the nutshell version.

I'll add that in general volumes are not additive when using mass/volume units. // There is a measure % v/v where a solution could be made by mixing volumes. Say 50% water and 50% ethanol by volume. However if you mix 50 mL of water with 50 mL of ethanol you won't get exactly 100 mL of solution.

I am just wondering whether what i am doing is right. Calculating volume of solvent from solubility, then adding the solute. What's bothering me is, that of course, the volume of the solution is not the same as that of the solvent. For example, i mixed a kilo of solute into 650 mL of solvent (water) and the volume of the solution jumped to about 1000 mL. I didn't expect this, though I should have and I am wondering whether I am interpreting solubility correctly. How am I to know what volume of solution I will obtain? This is important so i can select the right container for the solution.

@user1095108 - See EdV's comment. You need to put the kg of solute in a 1 L volumetric flask, then add enough water to fill the flask about 2/3 full. Dissolve the solute, then add enough water to get to the 1 L mark on the flask.

Shouldn't I add exactly as much volume of water, as the solubility equation says I should (650 mL) and then end up with whatever volume I end up?

I have solubility s given in g/mL. I understand this as mass of solute m divided by volume of solvent V.

Your confusion is arising from mixing the concept of solubility and concentration. Unfortunately, both are expressed the same way. You have to see the context. Most authors (like good scientists) try to write their experiment explicitly so that others can understand and repeat their experiment.
See the examples in this link: Solubility Concepts
Let us say, we find from the literature that the solubility of NaCl in water is 360 g / 1L. It means that if we take 1 L of pure water, we should be able to dissolve 360 g of salt in it. Note by adding 360 g of NaCl, the volume of water may not be exactly what 1 liter anymore. I cannot calculate concentration out of this information.
Now some authors may refer to concentration when they talk about solubility. For very dilute solutions it does not make a difference. So it is always good to see the context.

I read online I should pour the solute into a measuring cylinder/cup first, and then add solvent until the volume fits the calculated value (V=m/s), but this seems to be at odds with the definition of solubility.

Don't trust the webpages made by amateurs or pages from non-reliable sources. With time you will develop the sense which sources are more reliable. There is a saying that paper never refuses ink. It means one can write whatever on paper, and the paper will never say no, even if it is wrong. The same is true for the web. If we search the web for corona virus, you will find a lot of non-sense. Only reliable sources are to be trusted.


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How can I measure the amount of quinine in a home-made tonic water?
I'm into mixology, and lately I've been looking into experimenting with home-made tonic water, by extracting quinine (quinine salts) from cinchona barks in an acidic aqueous solution. I've read a bit about the potential dangers of quinine overdose, so I want to make sure I know how much the amount of quinine is in my extract.
I know it is fluorescent under UV; I thought maybe there might be a minimum quinine-to-water ratio for fluorescence. If this is the case, I can dilute the solution gradually until the solution is no longer fluorescent. However my knowledge on this subject is fairly limited.
Is there a home-accessible way of measuring the amount of quinine in an aqueous solution?

Without standard solutions and a fluorimeter, you can do a side-by-side comparison using a commercial tonic water and your solution. I just prevailed in getting a poor photograph of the beautiful blue fluorescence from a commercial tonic water (begins with an S), using 405 nm excitation (BluRay laser diode). The photo is below:

The two cuvettes contain tonic water (undiluted on the right) and diluted about 5 times (on the left). The laser entered from the right. The bubbles must be removed because they cause significant light scatter that gives red artifacts in the camera. By naked eye, it actually looks lots better: no red artifacts.
Removing the dissolved gas, to avoid bubble formation, is often done done by using an ultrasonic bath, sparging or by simply shaking the bottle repeatedly and waiting. Adding NaCl is a no-go because it quenches quinine fluorescence (a nice Stern-Volmer experiment).
Fluorescence is very sensitive, but the standard joke is that only 10% of all substances fluoresce ... and they are the wrong 10%. But this is a good case!
So, you would simply A-B the solutions: make dilutions of the commercial tonic water and see which one most closely matches your home-made tonic water. Cheers!
Note: you can use near-UV LEDs, i.e., 395 or so nm peak emission wavelength. The absorption max for quinine sulfate in dilute sulfuric acid is around 350 nm, and the peak emission wavelength is around 460 nm. But the photo shows that 405 nm works fine: it is not always necessary to be at the optimum excitation wavelength. Driving an LED is trivially simple: just a DC voltage supply, such as a battery or 'wall wart', and a current limiting resistor. Or use a constant current source that can output 20 or so mA.
One more thing: if you dilute the tonic water sufficiently, you will not be able to see the fluorescence with the naked eye. But a fluorimeter can detect it at even lower concentrations. Eventually, with sufficient dilution, you reach the limit of detection. But that is way below whatever you might want in usable tonic water.

Years ago, I bought 100 UV LEDs, with 397 nm nominal peak emission wavelength. I only used one, for a paper I published, so I have 99 left. I am thinking it would be fun to use two more to illustrate how to do the A-B comparison. Stay tuned!

Amazing images! I am always amazed by fluorescence. If you can find a test for checking olive oils with your UV, that would be great! They fluoresce in red color (the good ones). I wrote something for The Physics Teacher on olive oil fluorescence demos but could not investigate more.

The OP can do ImageJ analysis and construct a calibration curve with intensity after taking pictures like yours. ImageJ is free of cost. This is certainly home doable :-). He just needs one standard of quinine.

Wow, I had no idea the good olive oils fluoresced red! The laser I used is built into a “Space Rays” toy (long story) that I used in lectures to pop balloons when doing the VSEPR lectures. I can even burn holes in inflated balloons without poping the balloons! I will do some olive oil testing tomorrow morning!

Yes, the red glow of virgin and high quality olive oils is really beautiful (from chlorophyll). "I can even burn holes in inflated balloons without popping the balloons!" Never heard of that!

Ah, chlorophyll! Should have guessed. The hole burning was entirely by surprise: red and black balloons simply pop, but blue balloons get drilled and the edges of the tiny hole are essentially ‘cauterized’, so the balloons ‘bleed out’, as it were.

You've also mentioned the fluorescence quenching property of NaCl, can't that be also be exploited to measure the amount?
Great answer btw.

Yes it can! This is sometimes called “quenchometric analysis”, a rather ugly phrase, but accurate. Chloride, bromide and iodide ions all quench quinine’s fluorescence and the Stern-Volmer equation is easy to use.

I think a very useful home measurement, can be devised if we happen to know such specific amount of Chloride ions that can quench some other amount of quinine’s fluorescence until it is no longer visible to the naked eye. This way given a sample, one can take a specific amount of the "home-made" tonic water and dissolve table salt in it until it's no longer fluorescent. How can I calculate such amounts?

All I can suggest is empirical, because it depends on the conditions: whose eyes are involved, room lighting or dark room, how much liquid volume is being observed, etc. So this is analogous to the 'Scoville units' for grading the hotness values of various peppers. It works, but is not highly quantitative. Other than that, the Stern-Volmer method works, but then you need a light detector (or smart phone with that ImageJ program mentioned by M. Farooq), etc.

@Ed V: Is there dilute sulfuric acid in the pictures that you posted.  Do I really need the 0.05M or so sulfuric acid if I want to conduct quenching experiments?

@rdemyan No, just tonic water right out of the bottle. If I had an ultrasonic bath at home, I would have sonicated the tonic water to get rid of the dissolved carbon dioxide to avoid the pesky bubbles. Standard quinine solutions are typically made in the sulfuric acid solution, but that is to ensure accuracy if highly accurate quantum efficiencies are important. I do not think you need the sulfuric acid at all since you are really doing relative intensities.

@rdemyan By the way, you can easily find LED UV flashlights for sale, both 365 nm and 395 nm. They are inexpensive and will work very well. So no need to mess around with single LEDs.


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How to estimate hydrogen peroxide concentration in an easy way?
I have bought 20 Liter of Hydrogen Peroxide about 2 years ago which I was told that it was pure at that time. I'm aware of instability of H2O2, which easily decomposes into water and hydrogen.
The H2O2 at my premises had probably been decomposed partially within this period. According to its current practical effectiveness, I guess the concentration is still above 15%, but I can not be sure.
How can I estimate its current concentration with using equipments at home?

Note that I can find a small batch of known 3% H2O2 solution for some kind of comparison, if needed.

Pure hydrogen peroxide is very dangerous! Even 30% is nasty stuff. Are there any environmental authorities you can contact to safely take it away? It is a bad idea to try to guess how concentrated it might be and a do it yourself test could have very bad consequences.

This H2O2 is actually a component I'm using for etching PCB along with HCl, so I'm using it for some other purpose (I don't want to give it away). I have doubts about its purity in the first place, it's only what I'm told by the seller but it probably isn't the case. Anyhow, I'm aware of its dangers and always careful about the security measures, so you can be sure that I'll follow a safe path.

Glad to hear it! So one simple test would be to see how much oxygen gas could be evolved for a given volume of hydrogen peroxide solution catalytically decomposed. Somewhat awkward to do at home, but possible to kluge together a setup as per high school chemistry gas experiments.

How can I be sure that the sample is fully decomposed? What material should I use for that purpose? Is shaking or gently heating (placing the H2O2 cup inside a boiled water cup) sufficient for proper decomposition? I can precisely measure the H2 emission by using syringe as a measuring flask.

A simple catalyst, e.g., manganese dioxide or KI solution, will do the job and you don’t need total decomposition just to get an estimate.

I think I can extract manganese dioxide from a regular single-use battery. I suppose I should place the catalyst at the bottom of the measuring flask because I guess water will be on top of H2O2, like oil on water. Is that accurate?

Let us continue this discussion in chat.

One path to safely determine the strength of H2O2 is:

First dilute it to 1/3 strength (this is for safety when dealing with higher concentrations of hydrogen peroxide).
Slowly titrate against a known concentration of NaOCl until the vigorous oxygen liberation ceases.
One can also collect the volume of O2 evolved.

Reaction equation:
$\ce{NaOCl + H2O2 -> NaCl + H2O + O2 (g)}$
So, moles of oxygen liberated equates to moles of hydrogen peroxide. Also, moles of sodium hypochlorite employed should also equates to the moles of H2O2.

I used to do this as a lecture demo: put 30% hydrogen peroxide in a fishbowl, kill the lights in the lecture hall, and squirt in Chlorox bleach (5.25% by weight NaOCl solution). Beautiful red, but weak, forbidden singlet delta oxygen emission! Students loved this chemiluminescence demo.

The reaction you referenced is apparently a function of concentration and the pH difference between solutions. Another safety advantage of working with dilute solutions (per Step 1) is that this very interesting reaction no longer proceeds to any extent of being at all visible, so no Singlet Oxygen from dilute ingredients!

I do not know why your answer got a downvote, because it is fine, especially if the OP only needs a reasonable estimate of the hydrogen peroxide concentration. So I am upvoting.


text:
Why does cinnamaldehyde pass the ferric chloride test?
I was given some data to determine what the unknown compound was and figured out that it was cinnamaldehyde. After reviewing the classification test given to me by my professor, I was a bit confused as to why it would pass the ferric chloride test as well as pass a Lucas test but yet show no reaction.
Can someone explain to me why that is possible?
Also, in terms of solubility, why would cinnamaldehyde be soluble in sulfuric acid only and not both sulfuric acid and $\ce{NaHCO3}?$

The ferric chloride test is generally for phenols and enols, i. e., acetoacetic ester (purple color. $\ce{FeCl3}$ is also an oxidant which may cause a change in color.

I think the cinnamaldehyde gives $\ce{FeCl3}$ test due the similar reason why acetic acid can give $\ce{FeCl3}$ test as $\ce{Fe^3+}$ can act as oxidizing agent.
As in case of acetic acid $\ce{Fe(CH3COO)3}$ (red colour) is formed, a similar complex may be produced.

The formation of ferric acetate is not an oxidation.

Ferric ion may oxidize the aldehyde group to carboxylic acid group

True enough but that doesn't account for a necessary, definitive color change.


text:
Why are the acidities of methane and water so different?
I was comparing the acidity of methane $\ce{CH4}$ and water $\ce{H2O}$, and a lot of things weren't making sense. Generally, the process of turning $\ce{HX}$ into $\ce{H+}$ and $\ce{X^-}$ can be broken into 2 hypothetical steps:

Homolytic cleavage of the $\ce{H-X}$ bond, followed by
Electron donation from $\ce{H}$ to $\ce{X}$.

The energy required for step 1 is correlated with bond length, while the energy released by step 2 is correlated with electronegativity of $\ce{X}$.
The way I see it, electronegativity comes as a direct consequence of atomic size. Adding an electron to smaller element means the electron is added to an orbital that's relatively close to the nucleus, which is more stable. Thus, smaller elements are generally more electronegative, particularly across a row in the periodic table.
Now we consider $\ce{CH4}$ and $\ce{H2O}$. Oxygen, being smaller than carbon, will be more electronegative; thus, $\ce{OH-}$ is more stable than $\ce{CH3-}$ and step 2 will be more favorable for $\ce{H2O}$. But oxygen being smaller than carbon also means the $\ce{O-H}$ bond length is smaller than the $\ce{C-H}$ bond length, and thus harder to homolytically break. Overall, I'd expect these two steps to roughly cancel, and their acidities to be at least in the same ballpark. Yet the $\mathrm{p}K_\mathrm{a}$ for $\ce{H2O}$ is 15.7, while the $\mathrm{p}K_\mathrm{a}$ for $\ce{CH4}$ is approximately 55!
How can I make sense of this? What are other factors that impact electronegativity and homolysis that I am not considering? Or are there other aspects to this problem that I am not considering entirely?

Your assumption that the process can be broken into two steps might be theoretically helpful, but it isn't what happens in the real world for hydrogen-based acids. And your assumptions about the drivers of bond length and electronegativity are also wrong. You should start with actual observations not crude theory.

Generally, the process of turning $\ce{HX}$ into $\ce{H+}$ and $\ce{X^-}$ can be broken into 2 hypothetical steps [...]

Indeed, this process is slightly simplified because acidities are typically measured in solution, whereas your thermodynamic cycle only deals with gas-phase energetics.
However, in the present case, it's actually enough to look at gas-phase energetics (as will be shown below). The reason for this is because the difference between water and methane is so stark, that even ignoring solvation effects, we can arrive at the same result.

I'd expect these two steps to roughly cancel, and their acidities to be at least in the same ballpark.

Well, considering that water and methane have quite different acidities, it's almost certainly going to be the case that these two factors don't cancel out. The analysis below is probably not going to be very thorough. But again, because the difference is so huge, even some inaccuracies are going to be OK.
Let's look at factor 1, the bond dissociation enthalpy. Blanksby, S. J.; Ellison, G. B. Acc. Chem. Res. 2003, 36 (4), 255–263 quotes values of $\pu{104.99 kcal/mol}$ for methane, and $\pu{118.82 kcal/mol}$ for water. Indeed, the bond in methane is easier to break, which is in favour of methane being a stronger acid.
For factor 2, the electron gain, the NIST WebBook has (computationally obtained) values of $\pu{0.12983 eV}$ for the methyl radical and $\pu{1.80321 eV}$ for the hydroxyl radical. As far as I can tell, these are electron affinities (i.e. $\Delta U$, not $\Delta H$). So, one must note that a more positive value actually means that adding an electron is more favourable, as the Wikipedia page on electron affinities points out (as well as several textbooks). With this in mind, it's clear that this step favours the hydroxyl radical, suggesting that water should be a stronger acid.
The remaining question, of course, is which of these two factors is more important. We need to do some unit conversion, but it is not too difficult, especially if you use this handy online calculator. It turns out that
$$\begin{align}
\overbrace{\pu{(118.82 - 104.99) kcal/mol}}^{\large \text{Factor 1}} &= \pu{13.83 kcal/mol} \\
&\approxeq \pu{0.600 eV} \\
&< \pu{1.67338 eV} = \underbrace{\pu{(1.80321 - 0.12983) eV}}_{\large \text{Factor 2}}
\end{align}$$
Factor 2 is almost three times more important than factor 1. These values are quite rough, but certainly enough to draw the right conclusion.

Here is a bigger chunk of the periodic table with acid dissociation constants of the hydrides:

Source: Libretexts
The authors claim that the two factors are electronegativity (how much of a positive partial charge the hydrogen has in the covalent bond) and bond strength (expressed as estimated homolytic cleavage energy).
There is no strong anomaly for the three elements (N, O, F) that are implicated in strong hydrogen bonding.

The energy required for step 1 is correlated with bond length

The correlation is not that strong if you compare methane, ammonia, water and hydrogen fluoride. The correlation with electronegativity is better.

Source: Libretexts

As far as the acidity of water goes, it is a very special case that observes the process of homoassociation. This is when an acid forms stabilizing hydrogen-bonds with it's conjugate base This allows water to be the more stable acid, because hydrogen bonds are observed between $\ce{OH-}$ and $\ce{H2O}$, while none are observed between $\ce{CH^-_3}$ and $\ce{CH4}$
This factor might not be significant enough to provide explanation for all of the difference that you are seeing, so there is still another culprit out there.

I mean, if we're talking about aqueous pKa's, then yes, OH- is stabilised by hydrogen bonding; but the difference in pKa is still very large even in aprotic solvents, so there's still an additional factor. I also don't get how CH4 has anything to do with stability of CH3-. Your CH3- is going to be either in the gas phase (where it's surrounded by effectively nothing) or dissolved in something (where it's surrounded by solvent molecules), and that solvent isn't gonna be methane.

I am not referring to hydrogen bonding with the solvent, because in this case water is not necessarily the solvent, just the acid. So, assuming you have more than one molecule of water, this would still apply, even if it is in an aprotic solvent. Also, the fact that CH4 has nothing to do with stability of CH3- actually is part of what my argument is based on.

Uhm - let me get this right. You're saying that if we were to dissolve two molecules of water in DMSO and one dissociated, then the OH- would stick to the remaining undissociated H2O? What evidence do you have for this?

It probably wouldn't work like that if you had too much DMSO because the H2O and the OH- would never come into contact, but yeah, that's the basic principle. Also, like I said, it's called "homoassociation." look it up.

The examples provided on Wikipedia, such as HF, involve concentrated solutions, where you have lots of undissociated molecules – I definitely agree with you that these can influence observed acidities. But I'm really not sure they are relevant to the pKa values we're talking about. These are usually measured at rather dilute concentrations – for a discussion of water in DMSO see e.g. https://pubs.acs.org/doi/pdf/10.1021/ed086p864. No mention of association is made. But, I will let other people judge.

Okay. I'll review my sources again. But unless you have an alternative to my answer, we shouldn't be arguing on the comment thread anymore.

That's a very good point about solvents - comparing acidities in water isn't a good idea. In DMSO, water's pKa is 32, while methane's is roughly 56. It's certainly closer, but still a very large gap, considering that the electron affinities (and atomic sizes) of carbon and oxygen are pretty close. As far as I can tell, homoassociation is only relevant at large concentrations, so I'm still left with my original question.

Well, I guess my answer wasn't as good as I thought. If you want me to, I can delete my answer, but for now I'll just revise it.

No need to delete!

Okay then. I'll leave it as-is.


text:
Why do atoms stay nearer to each other even when delocalized electron moves away
When the delocalized electron moves from between two atoms to other then why yet the measurement of distance between those older two atoms is less than normal single bond.

Under what circumstances? Crystal bonding? Molecular bonding? Unclear.

Molecular bonding

There is a significant quantum probability a delocalized electron is there, what in average attracts atoms together, compared to the state there was no such an electron.
E.g. for benzene, the distance is shorter than for a single bond, but longer than for a double bond, being effectively 1.5 bond.

Poutnik can you elaborate

Well, elaboration is expected for the question in the first place in this site.. The question could be closed for low effort of the OP.

I meant that there are delocalized electrons in benzene ring. Those electrons travel throughout the ring. So when one such electron moves away from a position which was in between of two atoms then after that electron going away why yet the atoms are close to each other?  They have nothing to share now until another delocalized electron comes in then why they still stay nearer even though no electron between them for pi bonding

If you were just 1 minute thinking about any analogy of classical mechanics, you would not ask.


text:
Which is the most reactive metal: Caesium or Francium?
I searched on Wikipedia for reactivity series but it showed me Caesium as the most reactive metal. I know and have studied that reactivity increases as we go down the first group. Then shouldn’t it be Francium more reactive than Caesium than the other way around.
PS: I study in class 10 so please explain in that way only.

Oscar, if they are mere predictions then can’t we predict that it will be more reactive than Caesium. If you want to you can explain in more detail.

Francium is less reactive due to relativistic effects. This earlier Q&A entitled "Why do the trends in reactivity not apply for francium?" may be helpful.

For chemical purposes, francium pretty much does not exist.

The reactivity series consists of elements whose properties are FULLY known and understood. It is on the basis of whether an element can displace another from its salt solution (this is what you've been told at grade 10 Indian schooling level I guess?).
Francium, on the other hand is a radioactive element, with a half life of only 22 minutes for its most stable isotope
(Francium-223). Therefore, we can only make predictions, and not observations, about the nature of its salts and whether they would displace other elements from their salt solution or not.

Sorry for the bad English, it's my second language.

We do not really know.  Francium is known only as highly radioactive, short-lived isotopes and we have very little experimentally based chemical information about it.  Most of its properties are mere predictions.

Caesium is more reactive. Generally as you go down the periodic table, electro-positivity increases. However, for the really heavy elements, the presence of so many positively charged protons in the nucleus has the affect of causing the electrons to move round at incredibly fast speeds approaching the speed of light. As Einstein realised at such speeds strange things begin to happen. The electrons become a little closer to the nucleus than expected and they also become slightly harder to remove than expected.
Thus Caesium is more reactive than Francium.
For More Information https://www.quora.com/Which-is-most-reactive-element-francium-or-cesium


text:
does Hydrogen and nickel reduce c=c
In my course, I've come across H2 and Ni catalyst as a reducing agent for alkenes and LiAlH4/NaBH4 as a reducing agent for carbonyl groups. I know LiAlH4/NaBH4 can't reduce c=c as it is nucleophilic but can H2 and Ni reduce the carbonyl group? I've found uncertain and conflicting answers over the internet. My text book says it is possible and even gives the conditions for the reaction but past examination papers say otherwise.
question(compound Q):

answer:

IF Ni/H2 reduced  -CO- group, it would not be used for fat hydrogenation for food purposes, as fats would not be fats any more.

you do have a point but Is there any specific reason why?Im uncertain about what I'm supposed to write on the paper as the CIE examination board has implied that it does in fact reduce the c=c bond in another place.


text:
Why benzene takes in more delocalized electrons from its substituents even though it has six of them
Benzene has six delocalized electrons in its structure then where does it places the new electrons which it gets from other substituents like oh, etc. Why does it accepts those new ones?  Please give an elaborated answer.

That's not the best way to think about it; the benzene ring in phenol doesn't exist in isolation, so you have to consider the system as a whole. In benzene itself, there are no substituents and thus there's no problem having six π-electrons into six spaces. In phenol, there are eight π-electrons in total and eight spaces for them. The difference is that the spaces are distributed differently: there are >6 spaces on the benzene ring and <2 spaces on the oxygen atom for them. It still adds up to 8; there's no overcrowding. A proper answer would illustrate this using the molecular orbitals.

Please explain using molecular orbital

I'm sorry, but that will have to be somebody else's job; I've got other stuff to do. Why not try looking it up in a book?

@orthocresol slight language issue.  When you say eight places for the delocalized electrons in phenol I assume you mean four bonding ot nonbinding orbitals, but people might think it's eight conjugated atoms when there are really only seven.  I think we should be more precise.

@OscarLanzi I appreciate the clarification. Indeed - four orbitals is what I meant, which collectively can hold eight electrons.

Which book should I use?

Using atomic orbitals to construct molecular orbitals involves some shifting and redrawing because of overlaps. Benzene is symmetrical, so whether you draw six carbon atomic p-orbitals or squash and spread them out, you get the idea. Now, when you look at phenol, the C-O bond is unsymmetrical, so if you want to be more precise, you say that oxygen draws some electron density from the carbon thru the sigma system (because oxygen is more electronegative).
The oxygen becomes richly negative at carbon's expense, but this isn't fair(!), so it gives back a little electron density thru the pi system of molecular orbitals that it doesn't care about so much (because the pi lobes are farther from the oxygen nucleus). If you draw molecular orbitals, draw the sigma bond with more electron density toward oxygen, and draw the pi system with electron density a wee bit toward carbon.
The energy diagram of the molecular orbitals of benzene will be shifted by putting a hydroxyl on the ring. One one hand you can consider the six orbitals of the aromatic system merely being shifted in the space above the carbon ring, or, on the other hand, you could consider the three filled pi orbitals to remain essentially the same while oxygen donates some electron density to the unfilled (anti-bonding) orbitals (ignoring the sigma system, which just gets shoved around a little).
So the bottom line is that you are not putting whole electrons into someplace new, but swelling up some electron balloons while squashing some others.

James gaidlis I just don't understand one thing that now the electrons flow in a joint p orbital then how does the charge develop on para and ortho

The molecular orbitals define electron density for symmetrical molecules as a symmetrical smooth function. When you perturb one end of a nice symmetrical orbital, it adjusts in such a way that more electron density builds up on certain atoms. Without going thru the mathematics, use resonance diagrams to view the charge repositioning in phenolate (PhO-) at ortho and para positions. These resonance diagrams are partial contributors to the whole electron density positioning, so the electron density on the oxygen is slightly less than 1.00, and slightly more than 0.00 on o and p carbons.

thanks for this answer, Got the intuition, Will learn rigorous theory in Quantum Mechanical Model Soon


text:
White hot hydrogen gas, like that during a big bang
Difficult to know for certain which field this is. According to what I can discover about it, the big bang explodes into being as white hot hydrogen gas, millions of degrees hot.
I'm okay with that my question is what colour would it fade to as it cools down? Is there a spectrum I can refer that would illustrate that. I've seen such diagrams for metals as they cool down.
Difficult times people, thank you for your time and input, stay safe. Cheers!

I don't know how the spectra of gases at millions of degrees look like, but at cooler temperatures, say, 10000 K, the spectrum of the sun is a good example.  What did you read about the metal spectra ?

Hey thnx. I have a chart somewhere showing the colours white hot metal fades to as it cools down but that of course is metal. It first fades to a deepr white then yellow and then shades of red colours. Was wondering about white hot gas. It's a tricky one! Cheers

Is this incandescence can anyone tell me?

I guess black body radiation with given temperature will make a good approximation.

Yes, that is incadescence which is very low temperature phenomenon (below 3000 K). The spectra of gases at high temperatures is not continuous like a black body.

How soon after the big bang? See e.g. https://en.wikipedia.org/wiki/Recombination_(cosmology)

Hydrogen gas cannot be millions of degrees: the bond dissociation energy is 436 kJ/mol, so it will dissociate to atoms as you increase the temperature. Then the ionization energy of a hydrogen atom is 13.6 eV, so all you have is protons and electrons. The recombination link by @BuckThorn is worth reading.

(This might get more/better answers on Physics Stack Exchange)
At the high temperatures of the early Big Bang, you have individual protons and electrons, not hydrogen atoms and molecules.  This is because the $kT$ thermal energy is much, much more than the binding energy of the hydrogen atom.
In that case, the moving individual charges create a continuous black body spectrum:

(from Sun.org)
The light will look like a 10,000K source: a bit on the blue side.  But most of the energy will be at wavelengths much shorter than visible light.  The Wien displacement law shows that:
$$\lambda_{\rm{peak}} = {{b}\over{T}} = {{3\times 10^{-3}m \cdot K}\over{1\times 10^6 K}} = 3\times 10^{-9}m$$
A 3nm wavelength corresponds to soft X-rays (or perhaps the very hardest ultraviolet light).
You'll note that the vertical axis below is logarithmic.  At 100X temperature to the 10000K like, the peak will be at the shorter (left) wavelengths of X rays, but also much higher.  The energy flux goes as $T^4$, so a 100X temperature is a factor of $10^8$ in intensity:  There will be a lot of X rays at these early temperatures.


text:
Determining direction of equilibrium for metal complex

Determine if $K$ is greater or less than $1$.

At first I thought the equilibrium would favor the right side because this could have to do with ligand binding where the negative charge would probably cause binding to occur tighter than on the left side.
However, I was told that the left side could be favorable due to resonance stabilization. How to properly estimate the value of equilibrium constant?

I got the answer, equilibrium favors the right side because (Cd) is a soft metal and (S) is soft too, so it binds better to (Cd) than (O), in accordance with the HSAB theory.

Is there a third choice, where the cadmium is hexacoordinated (to both O and S), or is the question just whether cadmium prefers sulfur or anionic oxygen?

I think it is hard to predict the value for the equilibrium constant, there is the problem that it will be likely to be depend on the environment. I strongly suspect that for the complex dissolved in water the value will be different to if it is dissolved in dichloromethane.
I did a quick search of it in the literature and I can not find the cadmium complex. But there is a closely related compound. It is a copper complex, please see
An electrochiroptical molecular switch: Mechanistic and kinetic studies
Barcena, Homar S.; Liu, Biao; Mirkin, Michael V.; Canary, James W, Inorganic Chemistry, 2005, vol. 44, # 21, p. 7652 - 7660


text:
Question on chemical equilibrium
8 moles of so2 and 4 moles of O2 are mixed in a closed vessel.the reaction proceeds at constant temperature. By the moment when equilibrium sets in,80% of the initial amount of so2 enters into the reaction. The equilibrium pressure would be?    My attempt: reaction will be 2so2+O2---->2so3. There will be 1.6mole of so2,0.8 mole of O2 remaining and 6.4 mole of so3 will be formed. Now how to find the equilibrium pressure?

Is 4 bigger than 3.2 ?

Oh sorry I misread, I have edited.

As the volume is not given you cannot, unless you a priori know the K expressed in equilibrium pressures, as the reaction Quotient is pressure dependent.

This question given to me had an option 2.17 and an option bearing 'can not be calculated' and clearly answer given to me was 2.17 atm. so I feel it is solvable.

So let have 1 mol of oxygen, at unknown temperature and volume. What is the pressure ? It is as asking for p from p=nRT/V, where you know neither T neither V.

That's why I am asking this question. It must be solvable somehow

Another option is accidentally missing data for the task. Or, you may express the relative pressure, you need not T nor V for it.

May be I don't know they meant standard condition?¿

Can you elaborate how to do by relative pressure?

Not that you may. You should now, unless you are very underprepared.

I know the method which we write by partial pressure raised to their stoichiometric coefficient which require total pressure as delta n which is (2-(2+1)) not zero . So we would require total pressure. Can you further explain?

HINT - Start with "8 moles of so2 and 4 moles of O2 are mixed in a closed vessel." The problem doesn't specify the volume of the container, so let's just assume that the total pressure of the gas is $p$. What is the partial pressure of $\ce{SO2}$ and $\ce{O2}$ in the unreacted gases?

@MaxW I have tried. Why don't you write a answer.

@SMSheikh -- Sorry, I misread the problem.  I was thinking of calculating the equilibrium constant. User Poutnik's comment is correct. You need more information. Did you write the problem exactly as given?

@MaxW  you can check the book "Modern approach to chemical calculation: an introduction to the mole concept" by RC mukerjee,page number 835 ,question number 164, chapter : miscellaneous objective questions have this question and answer is clearly 2.17 atm.

@MaxW yes I am 100 % sure. I have written the ditto question.


text:
Projection Operator Method on Cl4+
I am not quite sure how to start on 1b.We are given 2 diagrams but I am not sure how to use the second one in the question, or if we have to. I got the point group for Cl4+ to be (D2h) however, but after that I do not know how to proceed.

To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors, e.g.  $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought as for vibrational normal modes..
The equation to make a linear combination of vector names $L_M$ is
$$\displaystyle L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$
where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized $d/h$ can be ignored. One further point is important. In the $C_{3V}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120^ \text{o}$ to the right, say, will turn it into a different vector than turning $120^\text{o}$ to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(v)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$.
The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.
The first step is therefore to choose vectors to represent, say $\pi$ orbitals and operate on them to form the reduced representation and so find the only those symmetry species that can be present then use the formula above and record the names of the total result  produced for each symmetry species in turn, e.g.  $2\pi_1-\pi_2$ etc. Using the nodes produced you can guestimate the order of energy levels.
(You can find details of forming a reduced representation here Understanding group theory easily and quickly)


text:
Why we need the average atomic mass?
I know , Average Atomic Mass of an element is - the sum of the Masses of its isotopes,each multiplied by its Relative Abundance percentage (in decimal notation).
But what confusing me is, what is the concept of Average Atomic Mass. In other words,why we want to take average atomic mass of every element on the periodic table,why not Atomic mass ? Why we need it and when to use it?

It depends on the branch of chemistry you are operating in, but for organic chemists calculating the molecular weight of molecules and molar quantities it is essential.

Why should we need average atomic mass ? It is about the same question as : Why should we go to the Moon ? Why should we try and prove that the biggest prime number does not exist ? Why should we determine the number of beetles  in the world ? Why should we study how Eratosthenes measured the radius of the Earth ? Why read Shakespeare ?

If you want to separate the two peaks in mass spectrometry having the mass 28, you must know the precise atomic masses of C, N and O isotopes, because $^{12}C^{16}O$ seems to have the same mass as $^{14}N_2$.

Suppose you have a bottle of NaBr and you need to use an amount equal to 1 mole. Sodium is all 23Na isotope, but bromine is 79Br and 81Br, with isotopic abundances 50.69% and 49.31%, respectively. So the NaBr in the bottle has both bromine isotopes and you cannot use the molar mass of NaBr with 79Br or the molar mass of NaBr with 81Br. You must use the average molar mass of NaBr if you want an accurate weighing of the NaBr. Only if you have isotopically pure chemical could you use the relevant non-average atomic isotopic weight.

The average atomic mass is useful because its numerical value is equal to the molar mass of the element. This in turn is useful to know how much of a solid to take when you wish to react it with some known quantity of another reagent, because weighing is typically the easiest way to quantify a substance.

If you know precisely which isotope of the element you sre dealing with then go right ahead and use the molecular mass but generally we dont know which isotope is available to us and moreover there are mamy impurities a single gram might have a mixture of kany isotopes.
So for reducing the error to the maximum limit possible statistically we multiply the mass of that particular isotope with its relative percentage abundance, you can imagine this as if a particular molecular mass is more available in nature than its more probable that is the isotope we have or that isotope is more in abundance in our mixture and hence the value of the average molecular mass we are using as simplification should tend more to that isotope