text:
Electronic vibrational rotational transition of electrons -- for a transition from $\Sigma$ to $\Pi$ how are P and R branches possible?
Suppose we have a diatomic molecule in a $^1\Sigma$ state, and it transitions to an excited $^1\Pi$ state.  Note that the total spin of the electrons remains unchanged ($\Delta S=0$) as we can assume spin-orbit coupling is negligible and hence the dipole operator involved in the transition cannot act on the spin.  The electrons have gained some orbital angular momentum ($\Delta\Lambda=1$) which is fine, this comes from the incoming photon, which has 1 unit of angular momentum.  Then if we look at the rotational fine structure, we see that a $\Delta J=0$ transition is allowed, since we've already changed angular momentum elsewhere.  A "Q-branch" in my fine structure is allowed, angular momentum is conserved, everyone is happy.
My question is what about the "P" and "R" branches, which are where $\Delta J=\pm1$?  These are usually introduced in spectroscopy courses first, before the Q-branch, because you consider a simple transition from one rotational state to another, ignoring possible electronic changes.  You argue that because angular momentum must be conserved, when the molecule absorbed the photon it has to take its angular momentum, hence $\Delta J=\pm1$.
In my reading, I've noticed when people talk about electronic spectroscopy, where the orbital angular momentum of electrons could change (e.g. $\Sigma\rightarrow\Pi$), they say that one might expect P, Q and R (i.e. $\Delta J=0$ and $\pm1$) fine structure.  I just don't see how the P and R branches are allowed though, if we've already used up the angular momentum to change the orbital.

It is essentially because in a $\Pi$ state there are two angular momenta, end over end and along the internuclear axis (in a diatomic) and it is involved. Best to look for a full explanation as in book by M.J. Hollas Chapter 7, p257 4th edition of 'Modern Spectroscopy' publ. Wiley


text:
Why is there a difference in crystal shapes of sodium chloride (NaCl) and potassium chloride (KCl)?
According to Sodium chloride and Potassium chloride:
Sodium chloride (NaCl) crystallizes, among various other shapes, into octahedron and tetragonal pyramid shapes which are absent in potassium chloride (KCl). I am a quite confused about this because both KCl and NaCl have a crystal structure. What is the explanation?
And what are some books where I can find the shapes that salt crystals form?

Crystal shapes are a complicated story.

There are some papers on that topic. The crystal habit/shape or in broader terms morphology depend a lot on concentration, temperature, impurities, etc. According to some papers that I found an octahedral shape can be achieved for KCl as well. So you might have to search for 'KCl octahedral morphology' and follow the cited papers. But it's a complicated story.

I feel the urge to reiterate once again that it is a really, really complicated story. There is no simple rule.

Maybe you want to remove the last question as it is too broad.

Crystal habits from solutions containing additives differ because the interactions with these additives differ. Here is one example for sodium chloride: https://pubs.acs.org/doi/10.1021/acs.cgd.7b01170

To compare between $\ce{KCl}$ and $\ce{NaCl}$ structure, you need to know what crystallographic planes are and their "distance between the atoms ratio".
$\ce{KCl}$ adopts a simple cubic system in which each atom lies at the corner of the cube.

The first set of planes ABFE, CDHG, ADHE, BCGF, ABCD, EFGH are alike. They are called (100) planes. Let the distance between the atoms be $\mathrm{d_1}$.
Parallel planes ADGF and ABGH are called (110) planes and are inclined at 45° to the cubic faces. Distance between atoms $\mathrm{d_2 = \frac{d_1}{\sqrt{2}}}$
Third set of planes like AFH are (111) planes. Distance $\mathrm{d_3 = \frac{d_1}{\sqrt{3}}}$.
Thus, for simple cubic system, $\mathrm{d_1 : d_2 : d_3 = 1 : \frac{1}{\sqrt{2}} : \frac{1}{\sqrt{3}}}$.
This ratio is in agreement with practical values. However, this calculation doesn't answer if the $\ce{K}$ and $\ce{Cl}$ atoms are in alternative fashion. This can be answered when compared to $\ce{NaCl}$.
The values obtained for the planes of $\ce{NaCl}$ are almost similar to that of $\ce{KCl}$ except a weak reflection at the (111) planes at about 5°. This makes the ratio,  $\mathrm{d_1 : d_2 : d_3 = 1 : \frac{1}{\sqrt{2}} : \frac{2}{\sqrt{3}}}$ which agrees that $\ce{NaCl}$ adopts a face-centred cubic system.

This difference can be explained by the fact that the number of electrons (atomic number) determines the amount by which an atom scatters X-rays. Because the atomic number of potassium and chlorine are not that apart, $\ce{KCl}$ had adopted the simple cubic lattice but in case of $\ce{NaCl}$, difference between the atomic number between sodium and chlorine is more, hence they adopted the face-centred cubic lattice. But, we can say that both $\ce{NaCl}$ and $\ce{KCl}$ have same structure but with different arrangement.
Pictures taken from Modern Physics by Kiruthiga Sivaprasath, S. Chand Publishing, 2008

I thought NaCl and KCl crystallized in the same spacegroup, with anions fcc and cations in "holes" with octahedral coordination. For KCl, your picture does not show a complete unit cell but just an eighth of it (KCl is not simple cubic).

At STP NaCl and KCl do crystallize in the same structure, see e.g. https://en.wikipedia.org/wiki/Category:Rock_salt_crystal_structure . It is the CsCl structure that can be viewed as interpenetrating primitive lattices. I really don't understand what "both NaCl and KCl have same structure but with different arrangement" is trying to say.

source: Stack Exchange

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

sort: votes


text:
O-C-O bond angles in carboxylates vs carboxylic acids?
I am wondering how different the O-C-O angle is in a carboxylate vs its acid form.
Does it (noticeably) increase because of the electron-electron repulsion caused by an additional lone pair?
Bond lengths are given here, but angles are mentioned only when discussing neutral acid (giving 122°).

Not a proper answer, but the O-C-O bond angle in a given carboxylate depends also on their counterion. In aqueous solution you may have the free carboxylate or some form of a metal carboxylate complex. In the solid dry state of a metal carboxylate, it will be  different than in solution. I also would assume that the O-C-O bond angle in dissolved, neutral carboxylic acids depends on the solvent as well as their concentration, as they tend to form dimers. Could you specify a little more what exactly you are interested in?

This answer might help you: https://chemistry.stackexchange.com/a/60645/17368

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


text:
How to make a paste out of vinegar?
I am trying to make a smooth vinegar cream for short term use of up to about 3 months (i.e. no preservatives are needed if stored in a cool place).
Vinegars such as apple cider vinegar or pure acetic acid tend to be sold as liquids.
These liquids are diluted like water and sometimes it's hard to administer them on a tissue; for example, on mushrooms or inside a candy, or as part of a cleaning material or cosmetic appliances.

Starches gave a thick, hard paste and sank
Pectin or Gum Arabic or Xanthan gum made clusters of solid material paste
Cream of tartar worked nice but if I recall correctly the outcome still wasn't "creamy"
Aloe Vera gel worked nice, but if I recall correctly it was often too dilute in my opinion

How to make a paste out of vinegar?
Perhaps Borax? Perhaps some vegan butter such as Shea butter, Cocoa butter, Mango butter, etc.?

Could you use cream of tartar or citric acid instead?

@KarstenTheis I tried cream of tartar but I don't recall smoothness of industrial creams.

@KarstenTheis I believe that citric acid would make the paste significantly more acidic than I need it but I should try it.

Such thickeners may need conservants, unless for short-time applications.

@Poutnik thanks, I have edited the question to clarify that it's a short term product without the need of conservants.

Maybe better answered at https://cooking.stackexchange.com/?

Hello @Snijderfrey, I am sure that not, because I am also asking about cosmetics. If there was a "cosmetics" tag here, I'd add it.

Use some methyl cellulose or hydroxyethyl cellulose. Both are use to make personal care products "creamy"

Downvote? A useful question with much prior research.

Borax contains "large amount" of Bromide Boron   , not bromide. // BTW you discuss thickeners only qualitatively, not quantitatively.  It is hard to believe none provides suitable result at any ratio. // Have you tried pectin , used widely as additional gelling agens in homemade jams? It may be comfortable with acidity of acetic acid.

@Poutnik thanks a lot for you remarks; I did try pectin; it was hard for me the stir and the outcome had clusters (lumps/chunks).

Thickeners are sometimes very eccentric. They just need to be coddled.
For instance, xanthan will certainly work in your application, if you treat it nicely - or get the right xanthan gum. There are various grades of xanthan gum, even various grades of the food grade gum. One grade is called readily dispersible - but it is made by using formaldehyde, and restrictions on formaldehyde are currently making it difficult to find.
The problem with xanthan gum is that it disperses readily in alkaline solutions (pH ~10), but begins swelling in lower pH solutions so fast that the particles become sticky on the surface and clump together so quickly that the inner portions of the particle are then removed from easy access to water which would swell the particle.
So, one solution would be to get food grade (is that necessary?) readily dispersible xanthan gum. Another, slightly more complicated, would be to disperse the not-so-readily dispersible xanthan gum in water with a slightly alkaline pH (pH 8 will have 100 times more OH$^-$ ions than H$^+$ ions), until the mixture is smooth (and way too thick!), then dilute with your vinegar and mix thoroughly. I've done this on a large scale (1000 gallons). A little NaHCO3 or even Na2CO3 will get you the pH =8, then the vinegar will totally overcome the alkalinity, and the xanthan gum will be already swollen and thickened.
Some thickeners may work better at the low pH of vinegar, like the suggested modified celluloses such as methyl cellulose or hydroxyethyl cellulose, but making a pre-thickened mix and then diluting with vinegar might work better in all cases.
In a similar thickening situation, egg albumen coagulates in vinegar. "...by adding vinegar, we get ...increased acidity to help the egg white coagulate and form a solid white". (Ref) So don't bother trying egg white to thicken your vinegar!
Ref: https://www.wgtn.ac.nz/science/ask-a-researcher/why-do-we-add-vinegar-when-we-are-poaching-an-egg

The best solution is polyvinyl alcohol $5$% in water. The trouble is that it takes many hours to make it dissolved, even if you heat the solution. The solution is viscous like glycerin. But if you mix it 1:1 with another solution made with borax $5$% in water, the resulting mixture is much more viscous, nearly solid. It looks like the silly matter called slime. Borax is hydrated sodium diborate $\ce{Na2B4O7·10H2O}$
But be cautious ! Don't let the mixture dry on your skin : when dry, it sticks firmly to your hair ! Not to the skin, but to the hair : Impossible to remove !


text:
How does Calanolide A react with NaOH solution?
Calanolide A is an anti-HIV drug

I assume the ester group, phenolic hydroxyl group and carboxyl group formed by hydrolysis in the molecule can react with NaOH. Is that correct?
How many moles of NaOH will be consumed when reacting with 1 mole of Calanolide A?

Remove the last para as it is a HW question unless you have done some effort.

Coumarins are hydrolysed by hydroxide as described in this paper
Coumaric acid contains a phenol and a carboxylic acid group and can consume 2 eqs of hydroxide, though hydrolysis can occur with only 1 eq giving the salt of the carboxylic acid and the phenol un-ionised. It is important to not overcook the reaction as the coumarin/cinnamic acid group is able to degrade under strong base conditions.


text:
How do I calculate the empirical formula when ash is present in the percentage composition?
I am trying to calculate a chemical formula. How do I calculate the empirical formula when ash is present in the percentage composition?
Example-

C- 47.7%
H- 7.2%
N- 1.8%
S- 0.9%
O- 38.0%
Ash- 4.4%

Since the ash is present at 4.4%, which could not be added into a chemical formula, should I exclude it?
If yes, should I just use the current percentage composition and neglect ash OR find the new percentage composition based on the sum of 95.6 g of other compositions?
For context, above is the composition of wastewater, where it is almost impossible to accurately obtain the composition of ash due to multiple components and mixtures available in it.
How do I solve this issue?

Our university microanalysis lab used to see this problem. It was due to students submitting samples that were contaminated with scraped off filter paper. ;-)

You cannot, unless you know the ash composition.

Does the ash come from an external source? If so, ignore it and just compute on what's left. That makes some pretty strong assumptions though...


text:
Using neutral or basic alumina in column chromatography for purification of amines
I am conducting a synthesis involving amines, and it is found that some spots still hardly moves from the starting point in spite of the dichloromethane 9:methanol 1 eluent condition. Addition of triethylamine doesn't seem to help the situation because I'm guessing there are two different compounds in the streak (that is they are similar in Rf and hardly move from the starting point). So I am considering using alumina TLC plate for identification and alumia column chromatography for purification. Here are the questions with it.

There are two types (except acidic) of alumina, neutral and basic. Is it sufficient to use neutral one for amines? Or should I use basic one?
Are all procedures same to the column chromatography using silica?

Either alumina will potentially work. The procedures are similar to silica chromatography.  I have extensively used ammonia in MeOH with DCM for silica chromatography of amines.


text:
Does heat increase the hydrolysis of urea hydrochloride?
The preparation of ammonium chloride by reacting urea with HCl produce urea hydrochloride the reaction take too much time to get ammonium chloride depending on ratios.

So can heat increase the process?

Thermodynamically or kinetically? For the latter, it speeds up all processes. Some more than others, so it can shift equilibrium both ways.

I have cropped your image to highlight the relevant text but it's better to write the text as pictures are not searchable.

You may look here to understand the reaction.


text:
Performing the mean and standard deviation for a extraction based on averaged content determination

Is it okay to perform the mean and standard deviation for a triplicate extraction based on averaged values of content determination?

I performed an extraction in triplicate. Each one of the replicates was injected into the HPLC system three times and the values were averaged. Is It okay to express the results as the mean of the three averaged values (and standard deviation of them) to show the error of the replicates of extractions? My procedure is based on the literature (see the fragment of a paper attached below):

But I don't know if it is okay to use the averaged HPLC determinations. To propagate the errors of replicates and injections could be an alternative but the literature says that it is not always feasible and the authors that make the procedure similar to mine do not look to measure the pooled variance. So, is It okay to use the averaged values as mentioned?

There are no hard and fast rules in HPLC determinations but rather good practices, until and unless you are doing legal analysis where the analytical requirements can be very stringent.
For routine academic work, some people make three or more injections from the same sample and express the standard deviation. This error indicates the precision of the instrument only, mainly the injection precision.
If you make three different extractions of the same plant sample and run HPLC of each in triplicate, this is also fine. You will have nine readings in total (say HPLC peak areas or heights). If you calculate the mean and the standard deviation of those nine readings, this time the standard deviation will reflect the error from the HPLC injection and your own handling of the three extractions. This approach is better and one must clearly write all the details of sample handling and calculations for the readers, though the first one is the bare minimum.
Error propagation formulae can become a nightmare for real experiments. Avoid them.

Comments are not for extended discussion; this conversation has been moved to chat.


text:
What chemical mechanisms or parameters altered during the usage of buffered extractant for leachability of a soil or sediment?
Toxicity characteristic leaching procedure (TCLP) is an extraction method to simulate a soil or sediment leaching. The method employed two extraction fluids - one is a buffered
solution of acetic acid and sodium hydroxide while the another one is an unbuffered acetic acid solution. [1] Although this protocol is widely used by many government instances, I still can not find any meaningful justification on whether first solution should and should not be able to thoroughly leach a sodiment or soil. I also can not find any theoretical, computational, or spectroscopic basis that can explain the mechanism behind the dissolution of such soils by unbuffered and buffered solutions, let it be a common ferrihydrite or goethite soil. Could there be any evidence that it might transform into an unstable hydroxide-hydrate-acetate ferrihydrite species?
Any insight from research article is appreciated and please do not put simple textbook explanations. Thank you very much.
[1] https://www.epa.gov/sites/default/files/2015-12/documents/1311.pdf


text:
Why does 1-ethylidene-2,3-dimethylcyclopropane exhibit cis-trans isomerism?
I think the carbons 2 and 3 have different configuration according to the CIP rules which makes the groups different:

Am I correct? Is there any official criteria regarding these cases ?

This example should answer your question: https://chemistry.stackexchange.com/questions/134700/can-geometrical-isomers-be-enantiomers/134722#134722. Cis/trans doesn't mean much here.

This is a related question and have accepted answer.


text:
Why is the 'side' of a double bond defined across the double bond horizontally rather than vertically?
When we are determining cis-trans isomerism, we seem to refer to a molecule's 'side' in relation to the double bond in a horizontal manner, as if we had drawn a mirror line through it horizontally. This lead us to say that the cis configuration of any molecule has atoms on the 'same side' of the double bond and a trans isomer on the 'opposite side' of the double bond.
If we had drew the line vertically and there was two species of the same type on one of the carbons in a double bond, this would then not exhibit isomerism as it would not have different groups attached, for example the arrangement in figure (2). But is there a geometric reason why we refer to the 'sides' in this way, or is it just convention as it necessarily must be this way?

In case of (2), the two options are structural isomers (in the example, either a linear chain of four carbons or a chain of three carbons with a branch). The definition is such that if you could rotate around the double bond, trans and cis (or E and Z) would swap. On a different note, why does a mirror switch left and right and not top and bottom?

You can still get isomers from "the other line"; it's just that they aren't cis/trans [or (E)-/(Z)-] isomers.
The reason why cis/trans isomers are "special" is that the molecular connectivity is the same. If you took the two isomers and looked solely at how each atom was bonded to which, while ignoring the 3D layout (the geometry) of the molecule, then the two isomers would be completely equal. This type of isomerism is called stereoisomerism, and cis/trans alkene isomers are one example of it.
On the other hand, if you were to swap a hydrogen and methyl on your alternative diagram, you would get a completely different connectivity with a linear instead of a branched carbon chain. The carbon on the left, instead of being doubly bonded to one carbon only, would now have a double bond to carbon and a single bond to carbon. These are still isomers, but they are structural isomers, not stereoisomers.


text:
What does it mean by pruning and enrichment in the case of Rosenbluth method?
As far as I understand:

Pruning means deleting something.
Enrich means to enhance/increase weight.

Now my question is, in the case of PERM algorithm,

Are we deleting some polymers chains? Or, are we deleting some beads in all polymer chains?
Are we increasing the weights of some polymers? Or, are we increasing the weights of some beads/atoms of all polymers?

When we are modifying the pruned weights, what are we replacing them with?

You posted a duplicate of this in mattermodelling? If so, could you please link each question to the other so that once it is answered on one site, the answer can be accessed from the other site as well.

https://link.springer.com/article/10.1007/s10955-011-0268-x


text:
Relationship between cell potential and applied potential in electrolysis
My question is about this problem, and I got the correct answer (which is 495g), but I am confused about the concept behind the correct approach:
"Calcium metal is obtained by the direct electrolysis of molten CaCl2. If a metallurgical electrolysis apparatus operates at 27.6 A and 1.2 V, what mass of calcium metal will it produce in 24 hours of operation?"
Basically, I did current * time to find the charge, then divided by (96485*2) since Ca has a 2+ charge. I got the correct answer. However, what I don't get is why I shouldn't take the standard cell potential into account. Specifically, why is it not correct to find the cell potential, then use the applied potential to find dG, and then divide that by the cell potential and the charge to find out the amount of Ca(s) that can be formed? Why is voltage not used in the problem at all? While the question is only asking about the formation of Ca(s), the reaction involves both the reduction of Ca2+ as well as the oxidation of Cl-, and I thought the oxidation step would need energy too, since these two half-reactions complete one redox reaction.

Relation of used energy and cell potential difference is much more complicated than relation of passed charge and molar amount. As analogy, you can solve linear algebraic equation by general formula for equation of the 3rd order, but why??

BTW standard redox potentials in molten salts are seldom known. Tabelated ones are for water solutions.

Thank you for your answer. I guess a follow-up question would then be, you said it's much more complicated than that and I think I understand your comment, so why is it conceptually wrong to do energy=current * applied voltage = -nFE0? (specifically, I found a number when I did I*V, then divided by -nFE0 to calculate the "number" of this redox reaction that has happened in 24 hours. I think there definitely is something wrong with this approach because it gave me ~130g yield, but I don't know where I did wrong.) Thank you again.

Rather Applied energy is integral of U.dq over time. But part of it is spent on Ohmic loses and part on transition potential differences. Such a big current means being miles away from an equilibrium state.

The current alone determines how much metal is deposited. Each charge carried replaces one valence electron, so, as you state, Ca++ requires two electrons to be transferred.
How much metal should you get?  ~6.25 * 1018 electrons = 1 Coulomb, or 1 Ampere-second. 96,485 coulombs = 1 Faraday, equivalent to 1 mole of a univalent metal.
Over-voltage, beyond cell potential, is changed to heat, but it does not electrolyze additional metal.


text:
Considering the odd mass number rule, how does $\ce{^{14}{N}}$ NMR work?
In my high school chemistry class, we learned that NMR only works for nuclei with odd mass number. Hence, we often use the $\ce{^1{H}}$ and $\ce{^{13}{C}}$ spectra. However, on this page about NMR I read, it says that $\ce{^{14}{N}}$ is also used, but with a mass number of 14, this evidently violates what we learned in class.
What am I missing here? Is the "odd mass number" thing just a simplification? If so, what's the actual criteria for whether or not something can be used for NMR?

A better way of saying this is that a nuclei with a non-zero spin is NMR active. This means that the nuclei must have an odd number of neutrons and/or protons. N-14 has a spin of 1, so it is NMR active.

See this link. This answers your question in detail. https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/nmr/nmr2.htm

An odd mass/nucleon number guarantees one of proton or neutron number is odd as well (odd + even = odd). Therefore the mandatory  NMR condition of  the nonzero nucleus spin is met. nitrogen-14 as the double odd nucleus (odd + odd = even) meets this condition by 2 ways.  BTW nitrogen-14 is it the heaviest (2H, 6Li, 10B, 14N) stable odd-odd nucleus (with one exception of metastable isomer tantalum-180m1)

For an even simpler example, $^2\ce{H}$ NMR is also possible

Your high school chemistry class was wrong. It's not NMR active if it has an even number of protons and an even number of neutrons. Otherwise, active. Having an even mass number does not imply this condition, but having an odd mass number immediately implies the opposite.


text:
What theory accurately explains metallic bonding in beryllium?
Since beryllium is an alkaline earth metal, the bonds between beryllium atoms could be considered metallic and we can use molecular orbital theory (MOT) to explain metallic bonds in metals.
Consider metallic bonding in lithium metal. Say, $n$ lithium atoms combine to form $\ce{Li_n}$. Here $n$ atomic orbitals (AOs) combine to from $n$ molecular orbitals (MOs). Since there is only one valence electron per atom in lithium, and a MO can hold two electrons, only half the MOs are filled. As the result, all bonding molecular orbitals (BMOs) are occupied, and antibonding molecular orbitals (ABMOs) left unfilled.
I tried to explain metallic bonding in beryllium metal in a similar way. But the results are weird. This is because all the $n$ MOs (formed after combining $n$ beryllium atoms) get completely filled with electrons. As there are equal number of electrons in BMOs and ABMOs, this type of bonding is not possible in beryllium metal. A similar explanation is used to describe the instability of the $\ce{He2}$ molecule.
According to MOT, beryllium metal $\ce{Be_n}$ shouldn't exist. But we know that beryllium is metal exists with strong metallic bonds. Isn't a contradiction to MOT?
How are beryllium atoms in a metal are linked? Is there any new theory to explain this or is there any concept I missed in MOT?

The usual MOT explanation for why the group IIA elements are metals is that the band caused by the overlap of the np orbitals is sufficiently wide to overlap that due to the ns orbitals (the one you describe above). As the bands overlap there is zero band gap, and hence the elements are metals.

Possible duplicate of How does metallic bonding in alkaline earth metals work?

@andselisk the question seems to look like a duplicate but it isn't. Because both the contents in the question and the answer are different. I was searching for the missing concepts in MOT that I hadn't learned.

The fundamental problem is that solid state band structures are not molecular orbitals, period. The band structure of beryllium is described in https://journals.aps.org/pr/abstract/10.1103/PhysRev.133.A819 and looks nothing like what you think (particularly the 'cigar and cornet' Fermi surface).

The usual MOT explanation for why the group IIA elements are metals is that the band caused by the overlap of the np orbitals is sufficiently wide to overlap that due to the ns orbitals (the one you describe above). As the bands overlap there is zero gap, and hence the elements are metals.
To test this proposal I have done a quick calculation for Be with CRYSTAL, an ab initio code that uses a local basis set (i.e. something like atomic orbitals) for calculations on periodic systems such as crystals. It's not the most wonderful calculation ever, it's only LDA for instance, but it should be enough to show what is going on.
Here is the Density of States in the region of the Fermi Energy (the energy scale is shifted so the Fermi Energy is at zero), along with the projected density of states onto the s type functions in the calculation, and a similar projection onto the p type functions:

You can clearly see that the lower band is predominantly of s type character, and the upper is mostly p, with a little mixing. There is also clearly a dip at the Fermi energy.
Zooming into the region of the Fermi energy shows us the below:

It is clear that the Total DoS is not zero at the Fermi energy because the "p band" just overlaps with the "s band". Thus there are unoccupied states at and infinitesimally higher energy than occupied ones. Thus Be is a metal because the unoccupied p band is wide enough to overlap with the fully occupied s band.
The input file for the SCF calculation that I used is below. The basis set is Be_pob_TZVP_rev2
BERYLLIUM
CRYSTAL
0 0 1
194
2.29  3.59
1
4   0.333333333333 0.666666666667 0.25
END
4 5
0 0 6 2.0 1.0
 4700.2365626        0.00023584389316
 704.82845622        0.00182437910190
 160.43110478        0.00939661482240
 45.425347336        0.03690892415900
 14.798334125        0.10897561281000
 5.3512452537        0.21694284551000
0 0 2 2.0 1.0
 2.1542044819        0.44695408857000
 0.9336374440        0.20866985771000
0 0 1 0.0 1.0
 0.3017450800        1.00000000000000
0 0 1 0.0 1.0
 0.1411145200        1.00000000000000
0 2 1 0.0 1.0
 0.6526215700        1.00000000000000
 99 0
END
DFT
XLGRID
EXCHANGE
BECKE
CORRELAT
PZ
END
SHRINK
12 24
TOLINTEG
12 12 12 12 16
MAXCYCLE
150
NODIIS
FMIXING
50
TOLDEE
8
END

And for the properties calculation
BASISSET
0
NEWK
12 24
1 0
DOSS
2 200 -1 -1 1 12 0
-0.5 0.75
8 1 2 3 4 8 9 10 11
6 5 6 7 12 13 14
END

I didn't understand what these graphs are. But okay I got the main point i.e., p band overlaps with s band. So what actually happens after overlapping ?

@Infinite I have a feeling VBT might be more accessible for you. TL:DR is because of s-p mixing, p orbitals are also used for bonding, while for He energy difference is too big for that. You could imagine Be wants to have octet - is this simplified enough, or too much?

Observation: even though the overlap at the Fermi level looks small, beryllium has a higher electrical conductivity than any other alkaline earth metal; actually the heavier congeners strontium and barium are the ones where conductivity drops significantly.

@Mithoron VBT could give a rough idea about bonding. But MOT has several advantages over VBT in explaining the observed facts. I would like to understand the concept in detail so MOT could be useful.

@Infinite Thing is VBT is actually equivalent to MOT in practice - both are bases for computational chemistry. Now that I think of it, you should probably rather read about electronic band structure

@OscarLanzi I would hesitate a good while before I read anything quantitative into the above calculation, it really is at the level of "Oh, look, I have a geometry and there's a basis set that google turns up, let's do the quickest and dirtiest thing I can to show the principle". In fact I've just done a slightly better calculation including a better level of theory and a geometry optimization - and the latter shrinks the cell a bit so noticeably increasing the band overlap. But to do something near respectable we need a basis set I trust, and I haven't time to look into that.

@IanBush - the classic paper would be https://journals.aps.org/pr/abstract/10.1103/PhysRev.133.A819


text:
Can chloride donate an electron pair to silver(I)?
According to the Lewis dot structure of $\ce{Cl-}$, $\ce{Cl-}$ has octet fulfilled and should be able to donate a pair of electrons as it is a Lewis base. The Lewis dot structure of $\ce{Ag+}$ seems weird. As it is a Lewis acid, I should assume that it can accept a pair of electrons.
However, now the problem arises. If $\ce{Cl-}$ donates a pair of electrons to $\ce{Ag+}$, there should be a coordinate covalent bond between them. In reality, $\ce{AgCl}$ is an ionic compound, and there is no such coordinate covalent bond, so I'm a bit confused.
In simpler words, $\ce{Ag+}$ is a lewis acid.$\ce{Cl-}$ is a lewis base. Can $\ce{Cl-}$ donate an electron pair to $\ce{Ag+}$?
Could someone please point me in the right direction?

Cl- ion is capable of donating electron pairs. See the question (...)

(...), and then see the answer.

If $\mathrm{Cl^-}$ donates a pair of electrons to $\mathrm{Ag^+}$, there should be a coordinate covalent bond between them. In reality, $\mathrm{AgCl}$ is an ionic compound, and there is no such coordinate covalent bond, so I'm a bit confused.

Yes, there is no dative bond between them. $\mathrm{AgCl}$ is an ionic compound with partially covalent character. So, the bond between $\mathrm{Ag^+}$ and $\mathrm{Cl^-}$ will be a simple ionic bond.

In simpler words, $\mathrm{Ag^+}$ is a lewis acid. $\mathrm{Cl^-}$ is a lewis base. Can $\mathrm{Cl^-}$ donate an electron pair to $\mathrm{Ag^+}$ ?

Yes, $\mathrm{Ag^+}$ can act as Lewis acid and $\mathrm{Cl^-}$ can act as Lewis base. But actually $\mathrm{Cl^-}$ can't donate an electron pair to $\mathrm{Ag^+}$. This is because both are weak Lewis acids and bases.

If $\mathrm{Ag^+}$ can't accept an electron pair, how can it be a Lewis acid?

It may not accept an electron pair in this case. It doesn't mean that it isn't a Lewis acid. Lewis acid doens't mean that it can accept an electron pair from any type of ligands. For example, $\mathrm{Na^+}$ is also a Lewis acid. Eventhough it is a Lewis acid it frequently forms ionic compounds rather than forming covalent compounds.
I can give an example where $\mathrm{Ag^+}$ can act as a Lewis acid. $\mathrm{AgCl}$ is insoluble in water but it is soluble in ammonia. This is because $\mathrm{AgCl}$ forms complex with $\mathrm{NH_3}$ which accounts for its solubility. During formation of complex $\mathrm{NH_3}$ will donate an electron pair to $\mathrm{Ag^+}$. So here it acts as a Lewis acid. I can give its equation -
$$ \ce{ Ag+ + 2NH3 -> [Ag(NH3)2]+}$$

According to the Wikipedia definition, a Lewis acid is one that can accept an electron pair. If Ag+ can't accept an electron pair, how can it be a Lewis acid?

Sorry, it is not actually a question posted here. It just for your clarification.

Okay I will include answers to this comments in my post.

@Infinite is it always true that weak lewis bases don't form dative bonds with lewis acids?

@Arpan no, it is not true. If cation has high charge density for example Cr(III) ion can act as LA in presence of Cl- ion.

@Mithoron I wonna know where I was incorrect ?

Easier to tell where you were correct, just about last paragraph. Cl does coordinate Ag all the time.

Correct? You said it was thoroughly incorrect, right?

Yeah, Cl will coordinate to Ag all the time. If it doesn't coordinate, then formation of the compound would be senseless.

@Mithoron Is there a coordinate covalent bond between Ag+ and Cl- in AgCl?

Oh, why you even try to answer such a lame question. AgCl is more covalent then ionic. Coordinate bonds are no different then "non-coordinate". Cl− can donate an electron pair to Ag+. And they aren't "weak" Lewis acid and base. There's no single scale of strength of them anyways.

Chlorine $\ce{(Cl)}$ needs to gain one electron in order to complete its octet (forming the $\ce{Cl-}$ ion), while silver $\ce{(Ag)}$ needs to lose one electron (forming the $\ce{Ag+}$ ion). This is why it is an ionic reaction. Unlike in a covalent bond, the electrons aren't shared; they are donated, forming dissociated ions in solution:
$$\ce{Ag^+(aq) + Cl^-(aq) <=>> AgCl (v)}$$
Note that $\ce{AgCl}$ generally forms as a precipitate. The reaction is technically reversible, but only proceeds to a very limited extent in solution.
Indeed, as you noted, $\ce{Ag+}$ is a Lewis acid, and $\ce{Cl-}$ is a Lewis base. Therefore, a Lewis acid-base adduct can form between the two. In fact, this is precisely why $\ce{AgCl}$ precipitates in solution! The $\ce{Ag+}$ and $\ce{Cl-}$ ions hold tightly together, forming a bond that has an extremely high covalent character. The bond between them is very difficult to break apart, which is why $\ce{AgCl}$ forms as a precipitate rather than being solubilized in water.

According to Wikipedia, "A Lewis acid (named for the American physical chemist Gilbert N. Lewis) is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct." If Ag+ can't accept an electron pair, how is it a Lewis acid?

Thank you for responding, sir! However, I'm genuinely confused as to why Ag+ is a Lewis acid if it can't accept an electron pair.


text:
Ion Exchange chromatography

Suppose that Anion exchanger is used in the stationary phase
normally -ve charge A.A will be eluted first as it will not attach
but my question is does the zwitter ion(have net charge 0 will be the same as -ve charge A.A?
or it will take a longer time than -ve charge A.A?

It depends on the charge distribution on the surface of the protein, and usually the best way to tackle this is to do experiments with different pH values. There are too many parameter (charge distribution, pH stability, interaction with other proteins) to get more than a conceptual overview. Everything else is empirical.


text:
Fragmentation behavior of spots in TLC
I tested multiple development using eluent condition hexane + 1% of ethyl acetate. Then I observed kind of fragmentation of spot.

I'm not sure, but I guess that successive spots are from the skyblue spot in the bottom. What is the cause of this phenomenon? I know there can be a tailing when the concentration is too high, but this seems somewhat different from the tailing.

Can you explain what do you mean by fragramentation of a spot? This is not classical tailing. Are you thinking that those blue spots are all of the same compound? This is quite unlikely. How many compounds were expected?

@AChem I expect that there is about 6 compounds, but I can't explain those if the blue spots are all different compounds. As I continue multiple development, blue spots successively locates at the same positions. Are there chances that this behavior is observed when dealing with amine compounds? Even, if I use more polar eluent system then another color of 'fragmentation' is observed.

I think, they are all different but closely related compounds and that is the same fluorophore is present in each (beautiful blue!). One compound will not show this "spotted" behaviour.

You may try using another solvent system as well to develop the TLC. One solvent system cannot tell us anything. Another more advanced trick is to use two-dimensional TLC.

@AChem Guess I'd better try 2D TLC... Thank you.

2D TLC might be very telling. Your compound could be decomposing (silica is a Lewis Acid, and wet silica is a Bronsted acid).

@BenNorris Oh, I missed the possibility of decomposition. Thank you. I would try it.


text:
Can I use baking soda to precipitate soluble copper out of a citric acid bath into a non-soluble form for disposal
I may be at the wrong forum with such a basic question but trying to ferret out the answer on Google has not been productive.  Apologies in advance.  I do art metal, working almost exclusively in copper.  As copper hardens quickly when worked I frequently anneal it by heating the workpiece to 1200° F and quenching.  Following  annealing the copper is placed in a citric acid bath of approximate PH 2.0 which removes all of the oxides on the surface.  Over time the acid bath becomes heavily laden with the dissolved copper oxides, turning the solution blue. As I've seen on the internet, this dissolved copper is harmful to fish and other aquatic life so removing it is necessary before disposal. One technique I've found states that by neutralizing the acid/copper solution with baking soda the soluble copper will precipitate out in a non-soluble form which can then be disposed of. So my question is: Is this true?  Is this all that is necessary to both neutralize the spent acid bath and change the dissolved soluble copper to a non-soluble form which will render the remaining liquid safe for the drain and the precipitated copper safe for disposal in the trash?  Thank you.

Yes, basic copper carbonate is fairly insoluble, so it should work.
Citric acid itself should be environmentally safe, if diluted with enough water (don't let it sit full strength in the P- or S-Trap of a sink, though, or it might eventually corrode a copper waste pipe). The reason it must be neutralized with baking soda is to precipitate the copper.
BTW, though insoluble in water, the salt produced in this reaction is soluble in ammonia solution, the basis for making cuprammonium rayon (an experiment you can try with the waste solution). Make rayon at home!


text:
What is it about nuclei with low gyromagnetic ratios that make them unreceptive in NMR?
I'm having some trouble rationalizing why nuclei with low gyromagnetic ratios are considered unreceptive in NMR. I know it's because it will take a higher magnetic field to "hit" its Larmor Frequency (putting aside more advanced techniques like cross-polarization or DNP), but what physical significance does the value have in NMR beyond that? I know it's a ratio of magnetic moment to angular momentum, so could it have to do with speed of Larmor precession that makes it unreceptive, or maybe a smaller population difference due to small spacing between energy levels?
For instance, comparing 103Rh to 1H, both are almost 100% abundant (103Rh is slightly more abundant, in fact), but 1H is enormously more receptive than 103Rh, which I can only attribute to the difference in gyromagnetic ratios since they're both spin-1/2 and don't have a quadrupole moment.
I'd appreciate any direction you could offer!

The SNR varies as $\gamma^{11/4}$, which is explained further in this question & its answers: https://chemistry.stackexchange.com/questions/131689

[For 103Rh specifically, there is another complicating factor in that you need to make sure that you have a probe that can go to such low frequencies... many probes out there only go down to 15N which is still ~3x higher frequency than 103Rh.]


text:
Blue copper acetate solution turning brown
I used white vinegar,hydrogen peroxide and some coins coated in copper and some copper wire. My solution started to have a blue color at the bottom but there were some white particles floating around. After 2 days my solution turned brown. No more blue. If accidentally some of the wires contained iron,could it be ferous acetate that gives this brown? My solution was all the time exposed to light,but I don't believe that this is the problem. I want to use the solution for shifting colors in natural dyeing.

Copper wires dipped in a $\ce{Cu^{2+}}$ solution produces a slow chemical reaction $\ce{Cu + Cu^{2+} -> 2 Cu^+}$.  This cuprous ion $\ce{Cu^+}$ makes dark mixtures with cupric ions, according to Qualitative Analysis, by F. P. Treadwell, 1924.

One source of your problem could be said coins. The coins are "coated in copper" but would have probably more reactive metals beneath, which reduce the copper back to metal as the base gets exposed to the solution. The displaced metallic copper, formed as fine particles in the solution, can then appear brown. If you used pennies where the base is zinc, that would give colorless ions when oxidized, so the only color you would see is that brown.
Copper wire is much more pure and so less prone to this side reaction. Using the wire alone as the copper source is more expensive, but recommended here.


text:
Calculating ionic strength at different concentrations

Zn$^{2+}$ reacts with a molecule, B, in a solution at several different concentrations of NaCl.
Determine the charge of the biomolecule and k$_0$ (second order) (assume that Zn$^{2+}$ does not change the state of charge)
The pH is adjusted with HCl
[ZnCl$_2$]$_0$ = $1$ mM
pH = $3.0$
[B]$_0$ = $\pu{1.40e-6}$ M
The concentrations of B at different concentrations of NaCl are:
$$
\begin{array}{lr}
\hline
\text{time}/\pu{min} & \ce{NaCl}/ \pu{50mM} & \ce{NaCl}/ \pu{75mM} & \ce{NaCl}/ \pu{100mM}\\
\hline
1 & \pu{1.39e-6}& \pu{1.38e-6} & \pu{1.37e-6}\\
2 & \pu{1.37e-6} & \pu{1.35e-6}& \pu{1.33e-6}\\
3 & \pu{1.36e-6} & \pu{1.33e-6}& \pu{1.30e-6}\\
4 & \pu{1.34e-6} & \pu{1.31e-6}& \pu{1.27e-6}\\
5 & \pu{1.33e-6} & \pu{1.29e-6}& \pu{1.24e-6}\\
\hline
\end{array}
$$

I began by calculating the rate constants for each concentration of NaCl (converted in M$^{-1}$s$^{-1}$) and got:
k$_{50}$ = $135$ M$^{-1}$s$^{-1}$
k$_{75}$ = $206$ M$^{-1}$s$^{-1}$
k$_{100}$ = $314$ M$^{-1}$s$^{-1}$
I then know I should calculate the ionic strength of each concentration in order to then be able to plot $\log (k)$ against I$^{1/2}$. However, I am confused as to how to calculate the ionic strength given B$_0$. If B$_0$ was not given then I would have calculated the ionic strength as:
$I = \frac12 \sum z^2(b_i/b_0) = \frac12 \left( 2^2 \cdot \ce{[Zn^2^+]}^2]+ (-1)^2 \cdot \ce{[Cl-]} + 1^2 \cdot \ce{[Na+]} + (-1)^2 \cdot \ce{[Cl-]} \right)$
Which would give me the answers:
$I_{50} = \frac12 ( 2^2 \cdot (\pu{1e-3})+ (-1)^2 \cdot (\pu{2e-3}) + 1^2 \cdot 0.05 + (-1)^2 \cdot 0.05) = 0.053$
$I_{75} = \frac12 ( 2^2 \cdot (\pu{1e-3})+ (-1)^2 \cdot (\pu{2e-3}) + 1^2 \cdot 0.075 + (-1)^2 \cdot 0.075) = 0.078$
$I_{100} = \frac12( 2^2 \cdot (\pu{1e-3})+ (-1)^2 \cdot (\pu{2e-3}) + 1^2 \cdot 0.1 + (-1)^2 \cdot 0.1) = 0.13$
These would then be plotted as $\log(k)$ against I$^{1/2}$. But what is the B$_0$ for then? I would assume that it should be incorporate into the calculation of the ionic strength somehow but I don't understand how.
All help is appreciated!

What is the equation of the reaction ? What is the unit of the given numerical values in the table ? Is the ion Cl- included in the reaction ? Why do you calculate the ionic strength ? Is the concentration not sufficient ? What is the meaning of NaCl/50 mM ? Does it mean that the concentration of NaCl is 50 mM ? What is the difference between Bo and bo ?

@Maurice The equation of the reaction is not given. The values are the concentration of B thus I would assume they are in the unit M (it is not specified). I want to calculate the ionic strength so that I can determine k$_0$ from the equation logk = logk$_0$ + $1.02$z$_A$z$_B$$^{1/2}$. Yes, sorry it means that the concentration of NaCl is $50$,$75$ and $100$. B$_0$ is the initial concentration of B while b$_i$ is the molality of the ion and b$_0$ is the standard state $1$ mol kg$^{-1}$.

It's difficult to understand your question. I would assume that since the concentration of the biomolecule is so low it does not affect the ionic strength. It should be easy to test whether this affects the result. The $B_0$ provide additional data points for the fits (at time 0, presumably in the determination of k).

Are you sure of your formula $\pu{log k = log k_o + 1.02 z_A√z_B }$ ? I suppose that $z$ is the charge of the ion. This formula shows that $ k$ does not change with the concentration and the ionic strength...

@Maurice Sorry, an "I" is missing in the equation: logk=logk$_0$+$1.02$z$_A$z$_B$√I


text:
Which would be the undefined article in singular for Ag0/AgCl reference electrodes?
The Google exact match searches for "an Ag0/AgCl reference electrode" and "a Ag0/AgCl reference electrode" yields the same few results. Which usage would be more appropriate, for example, in the next sentence?

Electric potential differences were measured using a lead ion-selective electrode and a/an Ag0/AgCl reference electrode.

Would it be more consistent to say a silver/silver chloride reference electrode?

I'd use "a" because I would pronounce "Ag" as "silver".

It all depends how one would read Ag/AgCl. If you have silver in mind then spell it out, if you have Ag in mind then "an" should go with it. Google scholar shows about 98,000 results with "an Ag/AgCl", and about 62000 with "a Ag/AgCl".


text:
Can general purpose compressors be used for a specific gas without modification?
Air compressors are commonly used for all sorts of purposes related to compressing air above 1 atm. My question is, could an air compressor be used for a specific gas without modification?
For example, could I use an air compressor to compress Hydrogen, Oxygen, or Nitrogen, abet at a lower pressure?

Specific gas properties have to be considered. E.g. compressed oxygen cannot come into contact with organic substances like oil or grease to avoid explosion.

And hydrogen can leak through fine openings, past compressor seals, and is very easy to ignite over a wide concentration ratio. Using it for H2 compression involves the Hindenburg certainty principal.

Having seen the result of using a non-oxygen-rated fitting in an oxygen line (potentially mislabeled by the supplier!), extreme caution and care is needed. Do not use any equipment outside of its instructions and labeling. Pressurized hydrogen has potential excitement written over it as well.

Yes, without further specifics. Assuming you mean a modest size reciprocal type. As commented, there are special consideration for primarily hydrogen or oxygen, and of course if there are specific corrodents. Large recips typically have cage/disc/flutter valves. I have seen many valves because they are a problem. I remember only one type- hard 410 stainless , so they are not altered for various gases ( refineries, chem plants and pipelines pump a variety of stuff). Get information for anything extreme. One story; Motion compensator for off shore drilling; They use 2000 psi air. They operate 24 hours/day. Some lube oil gets in the piping. Rarely it can cause a detonation of the air piping ( as a detonation ,it ruptures the pipe in several places  and throws shrapnel around).

So you are saying "yes, but really no"?

Tentative "yes", depending on specifics.

Can't help, but notice you should be more careful with punctuation and capitalisation. You should edit posts to fix such errors, otherwise it's like you scribbled it in a hurry.


text:
Triethylamine is detected by ninhydrin during TLC
Continued from this question. I purchased new triethylamine from Sigma Aldrich, spotted freshly opened triethylamine to the TLC plate, dipped into ninhydrin solution, and heated it. Again the purple spot showed. The ninhydrin solution seems to have no problem because there were no purple spots when I treated it with empty TLC plates. I have used ninhydrin staining using eluent containing the triethylamine before (TLC monitoring of Buchwald-Hartwig coupling reaction using ninhydrin stain), but there was no problem. I don't know why. Also in one site, it says that staining with ninhydrin shortly after eluting with 1% triethylamine is one of rookie mistakes in TLC. Is staining with ninhydrin 'shortly' after eluting with triethylamine the problem? Should I wait a minute after developing the plates with the eluent containing triethylamine? How can ninhydrin detect the triethylamine which is a tertiary amine?

When you have a lot number you can access Certificate of Analysis on Sigma. How pure is TEA? Someone answered it earlier stating that TEA might be impure. Just out of curiousity, do you have access to an HPLC that can be used for reaction monitoring? If your compounds somewhat are hydrophobic and soluble in ACN or MeOH, a C18 column should be able to separate most of the organic compounds

@Achem I thought that he or she is saying that the TEA is corrupted during storage. So I thought that there would be no problem with fresh TEA. Unfortunately our lab doesn't have HPLC.

Okay, still access the CoA from Sigma by entering the lot number (written on the bottle) and see how pure is TEA. The person who responded earlier is saying that TEA is not pure as received. It does not degrade during storage.

@AChem The CoA says that the purity of triethylamine is ≥99%.

Can you use ammonia in MeOH as your eluent instead? Otherwise leave the plate in the fume hood (or blow it with a heat gun) for a while to allow the residual TEA to evaporate.

@Waylander I would try, although I am a little afraid of affecting some compounds in sample. And a few postings say that TEA is detected by ninhydrin, probably it seems true..

@KrangLee, have you not tried to let the TLC plate rest for a while after the previous question? You should have done so, I think, before asking this question. In case you can get a certificate of analysis from your supplier, it might state something like "ninhydrin positives". The corresponding number should be low for you, obviously. Triethylamine itself is not reactive towards ninhydrin.

Does this answer your question? Does ninhydrin staining detect triethylamine?

@Snijderfrey Although there is little information, I found some paper or posting which says that triethylamine is detected by ninhydrin: https://pubmed.ncbi.nlm.nih.gov/12793457/ and https://chemtips.wordpress.com/2013/01/31/pick-your-poison-solvents-gradients-and-adulterants-in-flash-chromatography/.

@Snijderfrey Resting for a while after development is... I would test it.


text:
In calculating the radius ratio in a tetrahedral site in an FCC lattice, why do we take tetrahedral void in the middle of the body diagonal line?
When calculating the ratio of cation / the ratio of anion in a Tetrahedral system in an FCC lattice, why do we take the void to be in the center instead of it being √3a/4 from the corner.

Where did you get the value of $\sqrt(3)a/4$ from ? What do you mean by ratio of cation/ratio of anion ? Do you mean that "G" is an atom? As it stands there is a lack of clarity in the question.

That is not the center. The image shows 1/8 of the unit cell. It is unfortunate that they labelled the edge with $a$, conventionally used for the unit cell length. It does not matter for the result, though, because they express it in terms of ratio of radii.


text:
In acids why does polarising the O-H bond weaken rather than strengthen it?
In hexaaquaions it is generally true that aqueous solutions of complex ions with a higher positive charge on the central metal ion will have a lower pH. Explanations typically note that the greater charge density of the central metal ion has a stronger polarising effect on the OH bonds in the aqua ligands, which weakens the O-H bonds, increasing the dissociation of H+ ions.
How I'm struggling with this is that it's also common to predict that bond strength increases with polarity, to the point that Bond dissociation energy values are used to calculate Pauling values.
So we end up with apparently contradictory statements such as:
"The bonds to H also follow the expected trend of increasing bond strength with increasing electronegativity difference." (1.4: Bond Polarity and Bond Strength, Chemistry Libre Texts, https://bit.ly/3FbzexP)
and
"For a series of mononuclear oxoacids of an element E, the strength of the acids increases with increasing number of oxygen atoms. This trend can be explained qualitatively by considering the electron-withdrawing properies of oxygen. The oxygen atoms withdraw electrons making the O-H bond weaker." (Inorganic chemistry, Shriver and Atkins, 4th, p.120)
Taken in tandem, my reading of all this is that bond strength generally increases with polarity, yet when a nearby electron-withdrawing moeity polarises the O-H bond in an acid, the bond is weakened, not strengthened.
I'd be grateful for any explanation anyone can offer. Thanks.
EDIT: I may have answered my own question. Viewed in the sense of repulsion between the positive central metal ion and the delta positive hydrogen atom(s) in the aqua ligands, a more polar O-H bond will produce greater repulsion between the metal ion and the H atom. So a bond is not intrinsically stronger for being polar, it's context specific.

"Bond strength" as measured by dissociation energies is a measure of the energy required for homolytic cleavage of the bond, where one electron goes with each atom. Acidity is a measure of heterolytic cleavage, where both electrons stay with the base and the H+ ion separates.

Your edited conclusion is not correct. The apparent contradiction results from the definition of bond dissociation energy which refers to homolytic cleavage of a bond, where one electron from the bond goes with each atom:
$$\ce{A-H -> A. + H.}$$
When an acid dissociates, the bond is broken heterolytically, with both electrons staying on the base:
$$\ce{A-H -> A- + H+}$$.
A highly polarized bond tends to be more difficult to break homolytically, but easy to break heterolytically, and a simple explanation for this is that the second electron is at a much lower energy if it stays with the electronegative atom rather than going with the electropositive hydrogen.

Thanks for that. Just to clarify, the incorrect part of my conclusion is that the effect of polarity on bond strength is context specific, right? It's not incorrect to reference the greater repulsion between the higher charged central metal ion in an aqua complex and the H atom of the more polarised O-H bond in the aqua ligands is it?

@thetada Right.


text:
Why urea crystallizes in aqua regia?
Recently I stripped gold from old cpu's with ferrous chloride after that I rinsed the cpu's and dissolved them in the aqua regia to get the gold inside when all done, I added urea solution to the filtered aqua regia to precipitate gold if there any. But all I did get is a white crystals.

Introduced into concentrated nitric acid, urea produced a white precipitate of urea nitrate $\ce{CO(NH)2·HNO3}$, which can be filtrated and dried. But it must be kept dry. It is decomposed by water into urea and nitric acid in the presence of small amount of water. This compound is known to be a dangerous explosive, which is simple to synthesize. It has been used in Afghanistan war by amateurs who did not get chemical laboratories to synthesize usual explosives. It is extremely difficult to detect after explosion because it is destroyed by water.

@Marice Plus presence of copper and tin and other element. So how to neutralize it, I don't want that in my home.

The best thing to do is to destroy it by mixing with enough water. I don't know how much. An equal volume of water should be sufficient, as Wikipedia mentions that this urea nitrate is difficult to maintain in a pure state. The smallest amount of water destroys it into urea and HNO3. But apparently the urea nitrate is soluble in water, but it is quickly decomposed when in solution. And the solution contains only nitric acid and urea. Nitric acid can be neutralized by sodium carbonate, or sodium hydrogen carbonate.

@Tartard It's better to have a look at the MSDS of urea nitrate: https://www.chemblink.com/MSDS/MSDSFiles/124-47-0_Pfaltz-Bauer.pdf


text:
Can I synthesize 4-aminophenol and/or acetic anhydride at home?
I am going to try to synthesize acetaminophen as a possible science fair project, or just for fun, but doing it with precursors to the precursors of it might be more fun for myself then just using the easy to find things, but I couldn’t find any easy ways to synthesize either 4-aminophenol or acetic anhydride. So if anybody knows of methods of doing so that I could do at home, I would greatly appreciate it.

For your product: https://prepchem.com/synthesis-of-4-aminophenol/ For its precursor: https://www.prepchem.com/synthesis-of-phenylhydroxylamine/ Two-step synthesis starting from nitrobenzene. You will need zinc, ammonium chloride, sulfuric acid and sodium bicarbonate.

Chemistry should be done in an appropriate laboratory setting, with a fume hood and proper protective equipment.

I just don't think you could do it at home or maybe even school lab.

Commenters already provided the main points to your question. So, I am providing some more info. to make it into an answer.
@TheRelentlessNucleophile already provided the prepchem article for synthesizing 4-aminophenol [phenylhydroxylamine + sulfuric acid (for phenylhydroxylamine precursor, you need nitrobenzene, zinc and ammonium chloride. If you don't have nitrobenzene, you can make it by nitrating benzene)]. For acetic anhydride, you need acetyl chloride and sodium acetate.. Once you have these two reagents, you can make acetaminophen (paracetamol). I recommend to watch this NileRed's video (taken from 6-part series of converting aspirin pills to paracetamol).
Again I am stressing what @Zhe said. Please do your experiments in a proper laboratory settings, with fume hoods or proper protective equipments. Do not do any kind of such experiments if you are not experienced in handling chemicals or don't know what you are doing. If you are doing for your science fair project, I recommend to take help from a person who is experienced in handling chemicals.

Not to beat a dead horse, but the days of swashbuckling, cavalier chemists are over. We're professionals now, with professional standards for safety and environmental awareness. Gone are the days of doing chemistry without goggles or characterizing your product as "having the characteristic taste of brucine."


text:
Why does heavy water have two different solvent peaks with DSS and TSP?
I was going through the NMR solvent peaks chart of Cambridge isotope laboratories (PDF) and I noticed that $\ce{D2O}$ had two solvent peaks: one at 4.8 with sodium trimethylsilylpropanesulfonate (DSS) and one at 4.81 with trimethylsilylpropanoic acid (TSP).
Why do we not see two peaks for other solvents as well? Is it necessary to use an internal reference with $\ce{D2O}$ and why can we not just reference the $\ce{HOD}$ peak?

As you know "D2O" will not show a signal in proton NMR. Something else must be generating this signal. In the footnote of the solvent chart it reads "When the exchange rate between H2O and HDO is slow on the NMR timescale the water peak appears as two peaks"

I don't think there are two peaks being observed, I think it's more like the chemical shift is changed a bit depending on what internal standard is used

@orthocresol I agree with you that we get peaks at different positions based on the internal standard that we use. I was curious as to why we don't see this phenomena with other solvents. Suppose we were to add a slight amount of TSP or DSS to other solvents would they show different chemical shifts?

@AChem that was my second question. If we take $CDCl_3$ for example it will also not show a peak but because of proton deuterium exchange some of the $CDCl_3$ becomes $CHCl_3$. The same should occur for deuterium as well. We should see a HOD peak or a $H_2O$ peak. Why can't we just reference that peak and not use an internal reference? However when I look at literature they seem to use an internal reference only when performing $D_2O$ nmr.

@carbon12, Yes you are right that the NMR signal can arise only after isotope exchange but not from CDCl3. D2O also exchanges protons very fast with labile protons in alcohols or carboxylic acids... recall the term D2O shake. H2O should always be present in all solvents until and unless they are bone dry. I don't know the signal strength ratio of D2O vs. H2O (adventitious water). For  peak referencing question, orthocresol is the NMR expert.


text:
Will iron(III) sodium EDTA release its nitrogen into a water medium and become bioavailable to aquatic plants?
I am wanting to create an N-free Hoagland's solution and have discovered a recipe that includes sodium ferric ethylenediaminetetraacetate salt $(\ce{C10H12FeN2NaO8}).$ Will the nitrogen in EDTA release into the water and become bioavailable to aquatic plants?

In what form would you consider the nitrogen to be available to plants - nitrates, nitrites, free ammonia, glycine?

I will be measuring both nitrate and nitrite in the water, so those, I guess.

Then I think you are in the clear. There seems little chance that the nitrogens from EDTA will end up as nitrates or nitrites - see the reference in my answer below.

This paper which discusses microbial degradation of EDTA is the most relevant I could find. It identifies the major products of degradation:

Included among these compounds were both isomers of ethylene-diaminediacetic acid (NN-EDDA, and NN'-EDDA), EDMA, NTA (nitrilotriacetic acid), and glycine.

No mention is made of nitrates, nitrites or free ammonia if those are the forms of nitrogen you are principally concerned about.


text:
What is the reason for extracting copper from copper pyrites and not iron?
Why can copper be extracted from copper pyrites ($\ce{CuFeS2}$) but not iron? Is there some intuitive logic behind it?

Without more context, the question is in danger to be closed for lack of details. // The site expects that you write explicit compact summary results of your prior effort, based on your knowledge and searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself.
Effort not shown can be considered as effort not done and such a question may be  closed.  How do I ask a good question?.

Economics. Copper is expensive compared to iron. There are far better, cheaper sources of iron. But sources of copper are rarer.

Quoting from this source

Copper pyrites contains only 2-3% of copper. The rest of ore contains iron as sulphide, silica, silicious materials, sulphur, arsenic etc as impurities. Froth floatation is carried out by xanthate and pine oil. The froth is collected and dried when concentrated ore is obtained which contains 25-30% of Cu. Then smelting of the roasted ore in blast furnace is done.
The minor reactions that occured during roasting continue here. Ferrous oxide combines with sand to form a fusible slag. Cuprous oxide formed combines with ferrous sulphide to give ferrous oxide and cuprous sulphide. This is because iron has more affinity for oxygen than copper.
FeO+SiO2→FeSiO3
During roasting, copper pyrite converts into a mixture of
FeS+Cu2S→FeO+Cu2O

Copper has high value . Iron has low value and if contaminates with sulfur even a negative value, so recovery of this iron would be an economic mistake.

Chalcopyrite contains 34.5% Cu, 30.5% Fe, and 35.0% S. The mineral is the principal source of copper metal, and one of the major source of sulfur. Since copper is in majority, it is called "copper ore" from where copper is extracted. The iron portion is converted into iron(II) silicate and removed as slag.
$$\ce{2CuFeS2 + 2SiO2 + 4O2 → Cu2S + 2FeSiO3 (slag) + 3SO2}$$
If you want to extract iron, you need iron ore which are rich in iron content.
For more detailed explanation, see my answer: Extraction of copper from chalcopyrite


text:
Validating Resultant Structures of Computational Quantum Chemistry Optimisations
I have tried finding sources (publications, websites, blogs) that outline the logic of accepting the resultant structure of optimisation. I understand using a Hessian to determine if the structure is a local minimum. I am more interested in understanding what steps I should take to make sure that my functional and basis set are accurately determining the bond lengths and angles. I would need to know for benchmark testing so I can choose a functional and basis set that balances accuracy and cost.
Are there any resources that explain this type of process in-depth?
EDIT:
I found these answers on this forum from LordStryker and Martin that both give pretty thorough explanations for how to validate structures and choose a good functional for a project
Dipole moment of cis-2-butene
DFT Functional Selection Criteria

Do you have questions that go beyond the ones you have already found? If so, please be more specific. Currently it's hard to add to that,


text:
Can two S=1/2 centers have J and D parameters simultaneously?
I am new to EPR/ESR and need some help in connecting theory with a published experimental data. Following is my query.
What I understood from EPR theory is that two active EPR centers with S=1/2 interact with each other through zero field splitting (D/E) if they lie on the same ion/atom. On the other hand, when they exist on two neighboring ions/atoms, the interaction is through exchange interaction (J).
On the other hand, I came across a paper recently which suggests that zero field splitting parameter D is large if both the interacting S=1/2 centers are on the same ion/atom and small if they are on different neighboring ions/atoms. Which means in either case the concept of zero field splitting is valid! Also, the same paper talks about exchange interaction between same S=1/2 states.
Now, my question is which of these two things are correct. Or they both are correct and I am  misunderstanding them.
Will appreciate any help.


text:
Does 3-hydroxy-penta-2-one give positive Tollen's test?
I know that alpha hydroxy ketones that can tautomerise to an aldehyde give positive Tollen's test because an aldehyde gives a positive Tollen's test.
This molecule however doesn't tautomerise (or double tautomerise) to an aldehyde, yet it gives a positive test why?
EDIT: the question linked in comments shows the mechanism of the reduction of Tollens reagent only for an $\alpha$-hydroxy ketone  that has the hydroxy group on the terminal carbon.

Alpha hydroxy ketones always give a positive Tollen's Test. Refer this: Why do α-hydroxy ketones give Tollens' test?

@NisargBhavsar Yes they do, I just want a mechanism in which the hydroxy group is not on the terminal carbon but somewhere in between, because if the hydroxy group is on the terminal carbon it easily forms an aldehyde. However, this is not the case with the compound above.

https://youtu.be/xlaKvKfNidA this answers everything


text:
What will happen if the electrolysis of Brine takes place for a long time?
This is the reaction:
$$\ce{2NaCl(aq) + 2H2O(l) -> 2NaOH(aq) + H2(g) + Cl2(g)}$$
If the above reaction goes on for long enough, we will eventually run out of Cl- ions. There will only remain an aqueous solution of NaOH in the pot as the hydrogen and chlorine gases have floated away.
Then the electrolysis of NaOH solution will start, and H2 gas will be produced at the cathode and O2 gas will be produced at the anode. Now, if this electrolysis also goes on, at some point, only solid NaOH will remain in the pot.
Is my analysis correct?

Related post

At not very low Cl- concentration, oxidation of H2O is preferred to oxidation of Cl-.

https://www.reddit.com/r/chemhelp/comments/ut3ybn/comment/i987eoi/?utm_source=share&utm_medium=web2x&context=3

If the electrolysis is done without a separation membrane, the chlorine gas will partly be dissolved in the solution and react with NaOH according to : $\ce{2 NaOH + Cl2 -> NaClO + NaCl + H2O}$. So the composition of the solution changes gradually. The solution becomes bleach.
If a membrane separates both electrode zones, you may be nearly right. After a very, very long time, the water will be nearly totally consumed. But you will not obtain NaOH solid, because the contact between both electrodes will be interrupted when the NaOH is highly concentrated. The remaining  $\ce{NaOH}$ drops will be more and more separated, and the contact between them become poor. The inner resistance of the system will increase a lot, so that the current will become smaller and smaller, to come to zero before total consumption of water.

I've put more thought into this, and have realized that this is a complicated matter. Let me run the whole process, as I understand it, by you. Let us assume that we have a 1L 1M NaCl aqueous solution. So, we have 1 mole Na+ and 1 mole Cl- ions roaming around in the water, and we have 55.55 moles of H2O. Now, 10^-7 mole of that 55.55 mole will dissociate into H+ and OH- ions. So, number of mole of H+ ions = number of mole of OH- ions=10^-7 mole. Now let us dip two graphite electrodes in the solution. We have two positive ions: H+ and Na+, and we have two negative ions: Cl- and OH-. (...)

(...) Now, the electrolysis will begin. H+ and Cl- will be discharged at their respective electrodes as they have higher reduction and oxidation potential than Na+ and OH- respectively (calculated using the Nernst equation). Now, the critical matters start. After the electrolysis process goes on for some time, the concentration of Cl- will decrease, and the oxidation potential will also decrease. The difference between the oxidation potential of Cl- and the oxidation potential of OH- will get smaller and smaller. At one point, the two oxidation potentials will become equal. (...)

(...) Which will get oxidized then: OH- or Cl-. I do not know. Let us shift our focus to the cations. Unlike Cl-, there is a large reserve of H+ ions in the H2O molecules. Whenever the concentration of H+ ions starts to dwindle, H2O molecules dissociate to form H+ and OH- ions through the following reaction:
$$\ce{H2O(l)<=>H+(aq) + OH-(aq)}\tag{K_c=$10^{-14}$}$$
The H+ ions liberated as H2 gas will be replenished by the H2O molecules in the above reaction. However, over time, the concentration of OH- ions should start to get greater than the concentration of H+ ions. (...)

(...) It should take longer for the concentration of H+ ions to get low than the concentration of Cl- ions. The product of the concentration of H+ ions and the concentration of OH- ions should always be 10^{-14}. In other words,
$$\ce{[H+][Cl-]=10^{-14}}$$
At one point the concentration of H+ ions should get really low (and the concentration of OH- ions will get really high). Probably, the discharge of Na+ ions will start then. I don't know for sure. (...)

@tryingtobeastoic  There is the important note that cathodic/anodic reactions are not limited to cations/anions. Water can be reduced at cathodes to hydrogen or oxidized at anodes to oxygen as well. Electrolysis of concentrated KOH solution can hardly rely on negligible content of H+(aq). Similarly, for H2SO4 solutions, oxidation of water is much easier than of sulphate.

@Poutnik When you say that H2O is oxidized, do you mean that this happens:
$$\ce{4OH-(aq) -> 2H2O(l) + O2(g) + 4e-}$$
Similarly, when you say that H2O is reduced, do you mean that this happens:
$$\ce{2H+(aq) + 2e- -> H2(g)}$$
?

@tryingtobeastoic  OH- does not seem to me like water, neither does H+(aq). // Consider rather including you later thoughts and analysis into the question elaboration, (or own answer if it answers your question), instead of in extensive comments. It is not good if sum of your comments is much longer than the question itself.

@tryingtobeastoic $\ce{2 H2O + 2e- -> H2 + 2 OH-}$ // $\ce{2 H2O -> O2 + 4 H+ + 4 e-}$

@Poutnik I'm reluctant to ask new questions because if I do so, I might get banned.

@tryingtobeastoic I have not advised posting a new question. Focus on improving quality of your existing questions.


text:
Is H2 gas produced when a mercury electrode is used as cathode in the electrolysis of Brine?
If platinum electrodes are used as electrodes, H2 gas and Cl2 gas are produced at the anode and cathode respectively.
However, if a mercury electrode is used as the cathode, an exception is seen. Instead of H+ ions, Na+ ions are discharged first. This is because Na+ ions become mixed with mercury and form mercury amalgam (Hg.Na). As the mercury amalgam is formed, the tendency of Na+ ions to be reduced becomes greater than that of H+ ions. So, Na+ ions are discharged first.
My question is that after all the Na+ ions have been discharged, through the formation of Hg.Na, will H+ ions start to get discharged?

There is not meantioned the additional factor of high overpotertial for H+ reduction on Hg cathode.

@Poutnik Sir, I don't understand what overpotential is. Is the understanding of "overpotential" absolutely necessary?

But you do understood Google search, don't you ? Overpotential

And please don't say that the $\ce{Na+}$ ions become mixed in mercury. They don't. Only non-charged mercury atoms are soluble in mercury. Sodium ions are discharged at the mercury surface before being dissolved and form an amalgam.

@Maurice Oh, I see. So, Na atoms are soluble in mercury, not Na+ ions, right?

Yes. Metallic sodium is soluble in mercury, and not its ions.


text:
Effect of Pressure on Equilibrium
This might be a little bit of a basic question, but I've been trying to get my head around pressure and Le Chatelier's principle and I can't quite understand it.
Currently, my understanding of Le Chatelier's principle is that: if a change to the position of the equilibrium's position is made, the system will shift to restore the equilibrium position to be consistent with the equilibrium constant.
My current understanding of equilibrium position is essentially the ratio between concentrations of reactants and products.
So here's my problem: I understand that adding, say, more of a reactant will change the equilibrium position directly by increasing the concentration of reactants, resulting in more products being produced to restore the equilibrium position back to consistency with equilibrium constant. However, I do not understand how altering the pressure of a system has any impact on the equilibrium position? It doesn't appear to increase the concentration of either the reactants or the products? If the volume of a reaction vessel is decreased, wouldn't the concentration of the reactants and products increase proportionately?
For example, if I have 4 reactants and 2 products in a reaction vessel of 10 units cubed. Let's say the concentration of reactants is 4/10 which is 0.4, while the concentration of products is 2/10 which is 0.2. The ratio of reactants to products is 0.4:0.2 which is 2:1.
If I increased the pressure by decreasing the volume of the reaction vessel to 5 units cubed, the concentration of reactants would be 4/5 which 0.8 while the concentration of products would be 2/5 which is 0.4. The ratio of reactants to products is 0.8:0.4 which is still 2:1.
I hope I have phrased my confusion clearly. Please, if you can help explain this to me or point out a conceptual misunderstanding, I would highly appreciate it.
Thank you!

Speaking about pressure impact on equilibrium, it is much more common to use rather partial pressures, as there is usually gaseous context. For concentrations in condensed phases without involved gases, pressure has minimal impact.

@Poutnik so, does this partial pressure have an effect on the concentration of the gases and if so how?

Considering it as your rhetorical question which you answer yourself. Relation of gas partial pressure and and its formal concentration is determined for for ideal gas approximation by the state equation pV=nRT and concentration definition c=n/V // Consider $\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$ and how the change of total pressure affects the reaction quotient.

Have a look at this answer https://chemistry.stackexchange.com/questions/153545/how-does-the-value-of-the-equilibrium-constant-give-information-about-the-extent/153551#153551  which gives an example showing that the extent of reaction changes as pressure changes.

Equilibrium problems involve chemical species in (g) and (aq) states (gas and aqueous solution, respectively).
For an ideal gas:
PV = nRT
Concentration is defined as:
C = n/V
If we algebraically manipulate the first equation, we get:
n/V = P/RT
Therefore:
C = P/RT
In other words, the concentration of an ideal gas is a function of both temperature (T) and Pressure (P):
C = f(T,P)
However, at isothermal conditions (constant temperature), the concentration of an ideal gas is only dependent on Pressure:
C = f(P)
For a reversible reaction of the form:
aA(g) + bB (g) ↽−−⇀ cC(g) + dD(g)
The ratio of products to reactants you're describing is a constant (K) which can be expressed in terms of concentration (Kc), partial pressures (Kp), or even molar fractions (Kx):
Kc = (CC^c)(CD^d)/(CA^a)(CB^b)
Kp = (PC^c)(PD^d)/(PA^a)(PB^b)
Kx = (XC^c)(XD^d)/(XA^a)(XB^b)
And the three of them are related in the following way:
Kp  =  Kx(P)^Δn  =  Kc(RT)^Δn
(where Δn = c+d-a-b)
From here, we can observe that at constant temperature (T), all K's are a function of pressure (P), unless Δn = 0.
When Δn = 0, all K's become independent with respect to "P", and they all become equal to each other:
Kp = Kx = Kc
In conclusion, we have two cases:

If Δn = 0, then the equilibrium constant will not depend on pressure:

K ≠ f(P)

If Δn ≠ 0, then the equilibrium constant will change if pressure changes:

K = f(P)

First examine the equilibrium constant expression. The numerator is the forward rate, the denominator the reverse rate. If all are the same phase and the number of product and reactant moles in the equation the same the effect of pressure is minimal.
Example: gas phase H2 + I2 = 2HI Same moles no major effect of pressure there will be a minor effect from nonideality.
below the iodine triple point:  I2(s) + H2 = 2HI Now the [I2] in the vapor is almost constant and pressure change affects [H2] and [HI] and will change the equilibrium position in this case a pressure increase removes HI.
Reactions involving phase changes must consider the volume changes in the phase change and the latent energy change. There is also a small change in vapor pressure with inert gas pressure.
In a gas phase reaction when the equation moles of products and reactants differ increased pressure favors the side with the fewer moles.
Solids and liquids have the same considerations fewer moles are favored by higher pressures altho the behaviors of some high-pressure metallic phases are hard to explain. The pressures involved are high for condensed phases. In a phase change for a pure substance increased pressure favors the denser phase. Increased pressure by an inert gas raises the vapor pressure of solids and liquids.


text:
Can sodium sulfate be used in the salt bridge?
My book mentions sodium sulfate ($\ce{Na2SO4}$) as an electrolyte that can be used in a salt bridge. I'm asking this question because ChemLibreTexts says that

It is also important that the charge balance in each of the half cells facilitated by the ions in the salt bridge occurs at the same rates. That means that the halide anions moving from the salt bridge into the anode to balance out the excess $\ce{Zn^{2+}}$ ions do so at the same rate as the alkali cations moving from the salt bridge into the cathode to balance out the depletion of $\ce{Cu^{2+}}$ ions. Ions have a property known as mobility and the mobility of an ion depends on its size. Smaller ions have a higher mobility than larger ions. That means that the ideal species for a salt bridge should have a cation and anion of the same size and charge.

In $\ce{Na2SO4}$, the sizes of $\ce{Na+}$ and $\ce{SO_4^{2-}}$ are approximately equal (0.242nm and 0.227nm). The problem is that the magnitudes of the charges on them are not equal ($+1$ and $-2$). Also, there are twice as many $\ce{Na+}$ ions than $\ce{SO_4^{2-}}$ ions. So, will $\ce{Na+}$ ions and $\ce{SO_4^{2-}}$ ions enter the solution at the same rates?
My thoughts:
Probably the positive and negative charges are flowing in the two half-cells at the same rate. It's true that the charge on $\ce{SO_4^{2-}}$ is $-2$ (twice that of $\ce{Na+}$'s charge) and $\ce{Na+}$ has a $+1$ charge, but the number of $\ce{Na+}$ ions is twice that of $\ce{SO_4^{2-}}$ ions. So, probably the rate at which the ions are entering the cells is the same. Am I correct?

If the mobility of cation and anion differ, won't the salt bridge develop a net voltage? It may not much affect the tiny current draw of the cell, but how about voltage measurements of each half-cell?

Mobility depends on both size and charge. See this image.

It is of course better to use cations and anions of the same size and charge. But it is not a requirement. The cell and the bridge are also working if these parameters are different. The only effect to be observed at the end of the life of the cell is a depletion of ions in the copper zone. This will increase the inner resistance of the cell, and decrease the available voltage observed between the electrodes.
On the other hand, $\ce{Na2SO4}$ can be used to fill a bridge in the Daniell cell ($\ce{Zn/Cu}$). It must absolutely be avoided if one of the electrode is made of lead, because $\ce{Na2SO4}$ will react with $\ce{Pb^{2+}}$ and produce a white insoluble precipitate of $\ce{PbSO4}$, which will destroy the cell.


text:
Stable conformer of protonated ethane-1,2-diamine
I had been studying on the stable conformers of ethylene glycol and ethane-1,2-diamine.
Due to hydrogen bonding in the vicinal $\ce{-OH}$ groups, ethylene glycol prefers to show gauche conformer. The same can be said for ethane-1,2-diamine in neutral medium.
My query arises for ethane-1,2-diamine in pH<7, where I expect the $\ce{-NH2}$ group(s) to be protonated. Can we predict the most stable conformer now for these cases? Does protonation affect the extent of hydrogen bonding?

Wait, you really ask about diprotonate en? Well, yeah at such pH it would be, but it's not like any intramolecular hydrogen bond would be possible.

After @Geoff Hutchison 's answer I now asked for both cases without mentioning a specific pH, hope that's all right

@KimDokja Generally, it is discouraged to expand the scope of a question after having received an answer.

@Tyberius, Will take care of that in the future . Thank you for pointing it out.

I'm not 100% convinced that both amines will protonate at the same time, since protonation of a charged molecule is different than the neutral. At the moment, I don't have the time to calculate the $\mathrm{p}K_\mathrm{a}$ of the first and second protonation.
But let's go with your assumption - that you have two $\ce{NH3+}$ groups. Interestingly, there's much less known about conformational preferences of charged molecules.. which is a project currently underway in my group.
Here's my initial guess .. that the $\ce{NH3+}$ groups would function like $\ce{CH3}$ sterically, and the positively charged groups would want to move as far apart as possible due to electrostatics:

We can test this using quantum chemical methods. A tool called CREST performs driven molecular dynamics on the charged species, and returns anything within $\pu{X kcal mol-1}$ (The default is $\pu{6 kcal mol-1}$, which is what I ran).
And .. that's the lowest energy conformer. Nothing else is within $\pu{6 kcal mol-1}$ .. not only by removing the possible hydrogen bonding, but also because the two charged $\ce{NH3+}$ groups repel.

Thank you very much, this was very helpful! As it is interesting, I have now edited my question regarding the number of protonated groups. I appreciate that this is an ongoing project, but what would you predict for the monoprotonated compound?

The monoprotonated compound exhibits a strong hydrogen bond between the $\ce{NH3+}$ and the lone pair on the other $\ce{NH2}$ group.


text:
Is it possible to modify how energy is released in the combustion of a fuel like methane?
From what I understand (please correct me if I'm wrong), combustion is an exothermic oxidation reaction where, for example, a fuel like methane is combined with oxygen, i.e. CH4 + 2O2 → CO2 + 2H2O. My question is in this reaction, in what forms is the energy released? I know it releases light and heat, but how is the heat created? Is all the energy released as photons, which then increase the kinetic energy of the surrounding molecules? Does the reaction intrinsically cause the kinetic energy of the molecules to increase somehow?
The reason I'm interested is I'm curious about what would be the most efficient way of capturing all of the energy released from this reaction. E.g. with some combination of heat exchange and radiation capture.

https://chemistry.stackexchange.com/questions/55914/how-is-heat-physically-released-in-an-exothermic-process https://chemistry.stackexchange.com/questions/119093/how-does-the-energy-released-during-a-bond-formation-typically-manifest-itself-o

But if you really want practical side then https://en.wikipedia.org/wiki/Fuel_cell

If the real question here is the one in the last paragraph, I suspect editing the title would make this a much better question. The title is generic; the last paragraph is interesting.

My back-of-the envelope calculation says ~10^4 times  more thermal energy than light energy is released by burning methane. The thermal output of a stove-top burner on high is ~2 kW. I'd estimate the light energy is ~< 0.1 W.

@theorist do you consider "light energy" to be all of the EM radiation or just the visible portion of the spectrum?

@JBaczuk  Ah, good point. I was considering only visible light, not other portions of the EM spectrum.  And IR emissions could be substantial.  Though now I'm curious how much of the thermal energy released by the combustion of methane is in the form of IR.   I suppose that would depend on the combustion conditions.

There are alternatives to combustion as a way of releasing energy but only in specific reactions
In a typical combustion reaction (or any exothermic reaction) it is hard to control how the energy gets released. The reacting molecules bump into each other, react with each other and the resulting excess energy is distributed rapidly in a variety of ways. Some goes into vibrational energy, some to kinetic energy, some into electronic transitions (hence light). But however the very specific reaction starts off, the energy is rapidly redistributed across many other molecules (other molecular interactions happen frequently even in the gas phase and result in vibrational or kinetic energy being spread around other, neighbouring, molecules in ways that end up as "heat").
This makes it hard to modify how the excess energy is extracted from the system (but, having said that, many boilers or engines are very efficient at extracting heat energy).
Occasionally, though, it is possible to drive a reaction down a path where how the energy is extracted is not primarily heat. Trivially, this is what glow-sticks do. But the amount of energy there is small and it isn't even obvious that the majority appears as light. But it is also what Fuel Cells do.
Fuel cells use moderately complex systems where a catalyst on a surface can encourage the energy to emerge as electricity. The simplest are based on the reaction between hydrogen and oxygen. We could burn those gases to get mostly heat. But a fuel cell encourages the gases to combine to give water but instead of producing most of the excess energy as heat, capture it as electricity which can be directly used to drive machinery without needing a turbine or generator to transform heat into electricity. And some, more complex, fuel cells can use methane.
Most fuel cells rely on complex setups and precious metal catalysts to encourage the reaction to happen on a surface where instead of heat, (and simplifying a lot) electrons can be captured and used to drive an electrical circuit.
This can be more "efficient", but the cost of the devices and their inflexibility compared to heat-driven generators means they are not usually a cheaper way to generate electricity. Yet.

The king of this sort of process would seem to be cellular respiration, which has an efficiency of 80%-90% even after taking the elecrochemical potential it generates and converting it to chemical energy.  For comparison: Steam-electric power plant: ~30% - 50%. Combined cycle gas turbine electric power plant: ~50% -60%.  Fuel cell: ~50%-60%.  https://www.nature.com/articles/s42003-020-01192-w

All the chemical energy comes from the relative positions of electrons being attracted to nuclei. Efficiency depends on the path that the energy is transferred. Efficiency is an engineering concern; Making everything work right electrically and mechanically. Thermodynamics just does its best to determine what is possible. Thermodynamics tells us that to get the maximum work out of a heat engine the temperature difference must be a large as possible and the final T as low as possible. [The lowest practical T is ambient.] [By the Way that is one reason Global warming is giving us problems; We are raising the ambient T.] The maximum work possible from a process is given by the Free Energy Function: G = H-TS. This means that energy must be lowered and entropy increased to do work. If one looks at it closely the only way we can survive as a functioning entity is to maximize the increase of entropy of the Universe that we cause. We must use LOW Entropy energy as efficiently as possible and Expel to the Universe High entropy Energy maintaining a reasonable temperature to function properly.
Back to the first sentence! An electrochemical potential difference is the maximum work that can be gleaned from a chemical process. Delta G = EnF. So a chemical cell gives the most hope of using chemical energy efficiently. The battery people are working on this. The only other method of high efficiency is a simple photochemical process the emission or capturing or of a photon or the transfer of an electron between energy levels by emission or capture of a photon. Even that interacts with the motion of molecules. That gives us a clue [I must admit this was in the back of my mind, another Commenter stated this recently and I am giving my incomplete thoughts here If I can find the comment I will quote it]. Transfer of energy relies on conservation of momentum and total energy in every collision. Collision between molecules involves distortion of the electron orbitals. The idea that the distortion resulted in differential infrared emission and absorption between the different bonds makes sense both as a means of energy transfer by conduction and as the cause of black body radiation. Energy is transferred by transfer of electrons to lower lower energy orbitals because the nuclear attractions in some chemicals are greater than others. To do work this energy must be converted to electricity thru an electrical cell or to heat by intermolecular collisions that lower the quantum state energy into thermal energy, heat and the difference in heat content exploited [less efficient], any energy loss by direct emission of photons is sending possible low entropy energy to the universe and should be contained.


text:
What metals are safe to use as electrodes when doing electrolysis of water (I am trying to produce hydrogen gas)
I am going to be a DC power converter (with fairly high voltage and amps) to turn water into hydrogen (and oxygen) gases. However I have a few questions about the safety of this process.

What electrolyte should I use in the water? I was thinking of using Magnesium sulfate (epsom salt) or Sodium bicarbonate (baking soda).

What metal should I use for the electrodes? I am not sure if graphite is conductive enough to be efficient. I want to use stainless steel, but have heard this may produce hexavalent chromium. Is this true? What about iron or copper? Are these safe? Do they produce any other toxic byproducts? (I would use platinum or platinum-coated titanium, but these are expensive).

Is it safe to use these electrolytes and electrolytes together?
Thanks!

Efficiency of graphite depends on its resistance. Dimensions are important.

Platinum is excellent .

Magnesium sulfate is not the best choice when carrying out an electrolysis. Because, near the cathode, the solution becomes basic, and magnesium hydroxide $\ce{Mg(OH)2}$ will be produced, and it is poorly soluble. So the solution will soon become more and more turbid.
Sodium bicarbonate is not a better choice, because first it is not very soluble. Solutions have to be as concentrated as possible, to get a good yield. Second in the anode region, some $\ce{CO2}$ will soon be mixed with oxygen, due to the solution becoming acidic.
Graphite is a good conductor, and is efficient in electrolysis. But it has a tendency to crumbling. Tiny grains of black stuff are continuously produced at the surface of the anode, and fall back on the bottom of the container. This is disagreeable. Iron, copper and stainless steel are all slowly corroded by oxidation at the anode.
The best practical method for producing hydrogen by electrolysis is using platinum electrodes in a $\ce{H2SO4}$ or $\ce{Na2SO4}$ solution

Also, Faraday's experiments with electrolysis for the production of hydrogen from water resulted in the discovery that it takes between 0.5v and 1.24 volts to split a water molecule, anything more exponentially increases molecular movement, i.e. raises temperature, like an electric coffee pot. Amperage requirements depend on surface area of the electrodes. Lye in water solution and stainless steel electrodes, graded for corrosion prevention like 316 or 405 stainless, are sufficient and economical, butter knives work wonderfully for a quick demo, platinum definitely best, bare loose computer hard drive disks are plated lightly and work well, but if you aren't doing the electrolysis for profit, then stainless will work. In this cell, as the water is electrolyzed into lighter than air H and O2, the naturally rise out but the lye stays in solution and a single cell voltage of 1.5v contributes very little if any heat, so the experiment can be maintained by simply adding water to maintain coverage of the electrodes... Under a venthood if indoors, hydrogen will gather at the ceiling and any small spark will ignite it explosively especially with the increased oxygen, or in a closed cell the "dihydroxy" gas can be piped into a combustion engine or separated by electromagnet in a clear pipe divided, and fed into an acetylene oxy torch setup with the H in substituted for acetylene.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.