text:
Why are steric effects repulsive?
Normally, I think about orbitals interacting in the context of bonding. When two atomic orbitals overlap, they can do so in-phase or out-of-phase. The in-phase overlap results in a bonding molecular orbital with lower energy than the original AOs; conversely, the out-of-phase overlap results in an antibonding MO with higher energy. Valence electrons fill the bonding MOs first, resulting in a bond.
The explanation of steric effects that I often hear is that they are caused by the overlap of two electron clouds, resulting in repulsion. Since electron clouds are just composed of occupied orbitals, how is this "overlap of electron clouds" different from the picture of bonding I described above? Does the steric repulsion just come from occupied antibonding MOs? I imagine that inner shell electrons and Coulomb forces must play a role as well, right?
@matt_black answered a related question about steric vs eletronic effects by saying:

The normal distinction between "steric" and "electronic" is based on whether the effect is transmitted through space or through bonds

I guess I'm essentially asking a follow-up question on how effects are transmitted through space.

The point is right that combining all atomic orbitals you already have the possible solutions. What is left out is electrostatic repulsion. I see the exclusion principle already included in the orbital itself.

In particular thinking of conformers might help, eg rotation about a bond.

Exact details are largely irrelevant - you just can't put one object into another.

It's all about the Pauli exclusion principle. Assume the available MOs in each of two approaching molecules are fixed, and that each molecule is in a closed shell configuration (all electrons are paired). The electrons in closed shell systems cannot accept additional electrons that share principal, azimuthal and magnetic quantum numbers, since all the spin QMs are taken up. In fact, other than filling higher energy (LUMO) orbitals there is no way to accept additional electrons while retaining the current structure. The solution is to deform the structure of the molecule so as to generate new orbitals that allow formation of novel MOs. The deformation is associated with a potential energy barrier, since it is assumed that the colliding molecules are in low energy electron configurations. Therefore the instantaneous response to the collision is repulsive.

Cool, that mostly makes sense to me as a simple chemical biologist! Can you elaborate a little on why the structure has to deform to allow formation of new MOs?

@Andrew My point is that two molecules can in fact react to form a new one. That can happen in different ways, but if both reactants are initially stable closed shell molecules then it requires substantially modifying the electronic arrangement, which amounts to creating new MOs. It's not very deep. The devil is in the details. The key point is that these are closed shell stable molecules. Closed shell means no room for additional electrons within the current configuration (a bit circular of an argument perhaps).


text:
Why is bleaching with chlorine permanent but with sulphur dioxide temporary?
I was reading the properties of chlorine when I came across this statement:

Chlorine is a powerful bleaching agent; bleaching action is due to oxidation. Chlorine bleaches vegetable or organic matter in the presence of moisture. Bleaching effect of chlorine is permanent. $$ \ce { Cl_2 +H_2O -> 2HCl + O} $$ $$ \ce { Colored Substance + O -> Colorless Substance } $$

I searched it up and found out that bleaching with $\ce{Cl2}$ is an oxidation and is made powerful with the help of the nascent oxygen, hence it is permanent, but bleaching with $\ce{SO2}$ is a reduction and hence it is temporary.
Could someone please help me understand why the oxidation is permanent and the reduction is not? Is it powerful to an extent of the reaction not being able to get reversed? If yes/no, is there any quantitative/qualitative measure for the same?

Because our air is (re)oxidizing, maybe?

Are you sure you refer to $\ce{SO2}$, sulfur dioxide, and not $\ce{S2O^{2-}4}$ like in sodium dithionite, which may, for example, be used to reduce temporarily blue indigo into yellowish / colorless leucoindigo (eventually oxidizing on air again) as used for blue jeans (see, e.g. here)?.

@Buttonwood - I am referring to $SO_2$ only. I am not sure about how to relate the leuco dye mechanism with this, but it seems to me that atmospheric oxygen aids in the replacement of the oxygen that is lost in the bleaching action of $SO_2$ , which gives back it's original color. What about the $Cl_2$ bleaching mechanism though, do you have any clue if it is possible to re-obtain nascent oxygen? Thank you.

The bleaching action of Chlorine is permanent because it involves the process of oxidation. $\ce{Cl2}$ reacts with water to produce nascent oxygen which combines with the material in question (by oxidising it, as pointed out by @Mathew Mahindaratne
) and makes it colourless.
The bleaching action of $\ce{SO2}$ is temporary because it involves the process of reduction. $\ce{SO2}$ removes oxygen from the coloured substance to make it colourless. Atmospheric oxygen slowly takes place of the removed oxygen, because of which the material regains it's colour.

Hey, I'm sorry, I don't seem to get "why" this happens. I get it that bleaching with $Cl_2$ produces nascent oxygen, which makes the material in question colorless, but why? Is it because nascent oxygen cannot be re-obtained? In case of $SO_2$ it makes sense that there is an actual replacement of the original oxygen, so thank you for that, but for the $Cl_2$ part, I don't get it yet. Could you please elaborate on that if you know it please? Thank you

Well, since the material gets oxidized with $\ce{Cl2}$ , and our atmosphere is an oxidizing atmosphere at present, it makes sense that the material would remain oxidized (and thus bleached). If our atmosphere was a reducing atmosphere (the Oparin-Haldane Theory states that early Earth had a reducing atmosphere), then intuitively, it would make sense that  $\ce{SO2}$ would be a permanent bleaching agent and  $\ce{Cl2}$ a temporary one.

Woah that explains it well! Thank you. I read about the Oparin Haldane theory once you mentioned it. It seems very interesting.

@Meta xylene: Bleaching action is not removing oxygen or giving oxygen. It is about the breaking chromophore (see my answer).

@MathewMahindaratne Not oxygen, nascent oxygen represents oxidation, right? Isn't that what you mean?

@Meta xylene: Regardless of oxidation is by $\ce{O2}$ or  $\ce{O^.}$, it breaks the conjugation of chromophore, which is responsible for the color. Reduction  do the same, but relevant chromophore would able to get its conjugation with help of  atmospheric $\ce{O2}$ (or  $\ce{O^.}$).

Yes, which was my point. The oxidising atmosphere means that reducing bleaching agents only have temporary effects. I do agree that chromophore breakdown is responsible for it (wasn't aware of the same, thanks), but the question was about why one was permanent and the other temporary. But I'll edit my response to make it clear that it's oxidation and not nascent oxygen that does the bleaching :)

Everybody knows chlorine ($\ce{Cl2}$) is a bleaching agent acting on oxidation. However, what majority of us do not know is sulfur dioxide ($\ce{SO2}$) is also a bleaching agent, which in contrast to $\ce{Cl2}$, acting on reduction (though we know about other reducing sulfur agents such as sodium dithionite and thiosulfate). The well-known Encyclopaedia Britannica simply explain it as:

Wood and animal fibres are bleached by acidic reducing agents such as sulfur dioxide. In the pulp and paper industry chlorine dioxide, hydrogen peroxide, sodium peroxide, sulfur dioxide, sodium bisulfite, and sodium hydrosulfite are commonly used.

Bleaching agents, regardless of oxidizing or reducing, are formulations that whiten or lighten a substrate by solubilizing color-producing substances or by altering their light-absorbing properties. The decolorizing reaction generally involves the removal of chromophoric sites in which electron delocalization over conjugated double bonds has made the substrate capable of absorbing visible light. The bleaching agent will typically react by irreversibly cleaving or adding across these double bonds (Ref.1). If the bleaching is done by oxidation, the conjugation is broken irreversibly (e.g., converting double bonds to diols), and permanently because there is no reducing agents in air. Even sunlight helps the oxidation, which is the chief bleaching method of textiles (a process known as crofting) used before the discovery of the element chlorine in 1774.
If the bleaching is done by reduction, it is a different story. According to Ref.1:

A few bleaching compounds act by chemical reduction; these include sulfur dioxide ($\ce{SO2}$), sulfurous acid ($\ce{H2SO3}$), hydrogensulfite ($\ce{HSO3−}$), sulfite ($\ce{SO3^2−}$), and dithionite ($\ce{S2O4^2−}$), as well as sodium tetrahydroborate (borohydride) ($\ce{NaBH4}$). Their application is primarily in pulp and textile manufacturing, where the bleaching action is thought to occur by reduction of a chromophoric carbonyl group. Other applications include the bleaching of glues, gelatin, soap, and food products.

These reduced chromophores would be regenerated slowly by the atmospheric oxygen (recall oxidation and reduction of indigo dye as Buttonwood explained in comment section). Thus, the bleaching by reduction is not permanent in most cases.

Late addition: After I submitted my answer, I have seen the conversation between OP and other commenters. I felt OP have not so correct idea of bleaching. Bleaching action is not removing oxygen or obtaining nascent oxygen. It is about the breaking the chromophore responsible for the color. I think I have explained the principle quite clearly in my answer above. Basically, regardless of bleaching of oxidation or reduction, it breaks the conjugation of the chromophore, which is responsible for the color (you want to get rid of). If it is done by oxidation, the color can't get back because no reducing agents in our atmosphere. Yet, if the bleaching is done reduction, the relevant chromophore would able to get its conjugation back with help of atmospheric $\ce{O2}$ (or $\ce{O^.}$).
Also, I like to point to a nice demonstration of bleaching by $\ce{SO2}$ here.

References:

Marianna A. Busch, Kenneth W. Busch, in Encyclopedia of Analytical Science (Third Edition);  Alan Townshend, Editor-in-Chief; Elsevier Science Ltd: Amsterdam, Netherlands, 2019 (ISBN: 9780081019832).

Hey, thank you for your answer. I would just like to include here that I have not misunderstood bleaching per se, I was trying to correlate oxidation and reduction with its permanence. I saw the visual in your link in the Late Addition and I have gotten the $SO_2$ bleaching part of it clear now, thank you.

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text:
Microcopy Fluorescence filter set usefulness
I have a brief doubt and maybe someone around here can give as some help. We are doing a pilot study on microplastics using a fluorescence method (red nile). The dye is solvachromic, so it's emission spectrum changes by the polarity of the media. We do have the spectra for the excitation (mostly blue light) and emission (for the solvent used).
We have an old zeiss microscope equiped with epifluorescence. The filter has emission in 460-490, FT510 and LP at 520.

My question is if the emission spectrum of red nile includes 510-520 but maxima is achievied at 560, the filter set we have will work and we will see green fluorescence?
In others types of microplastics the spectrum of emission is shifted towards the red but it includes a small extitation in green. Is this filter useful?
Hope someone can help we have applied the protocols but we doubt if there is a problem in the red nile solution or in the fluorescence filter

The  nile red fluorescence ranges from approx 530 to 700nm depend on solvent polarity so a fixed filter will be of v limited use. You really need to see the spectrum so it would be best to try and get a ccd attached to the microscope. There are many now that have fibre couplings and are relative cheap, I've used ones from Ocean Optics in the past. They probably have couplers to fit microscopes.

Thanks for responding,  we do have a camera in a modern microscope, but we do not have the fluorescence accesory. The FT and LT in the filter set will block the emitted fluorescence? We try using blue leds and arduino as suggested in some papers but we see red fluorescence or yellow and it is supposed to fluorescence in green. How can we easily test if the red nile solution is working fine?

Im not sure what you mean by the filter has emission in 460-490, and what exactly are the FT510 and LP520 filters i assume that hey transmit at these wavelengths. Are they narrow band or edge interference filters or something else?

You first need to look at the transmission characteristics of each filter before choosing any filter.

Hi. I read more and more and as far as I understand, the filter set we have emits blue light (460-490), has dichroic beamspitter FT510 and a longpass filter of 520. So as much as I understand any emission above 520 will be visible. In fact we see red and yellow fibers after red nile dying. I suspect that the dye may not be working well since the stock solution was prepared long ago in DSMO and were kept at dark but not refrigerated. We are new at this, have followed the protocols but we cannot clearly see the classic green fluorescence of most microplastic.

@AnastD, You are using the wrong word for filters. A filter does not emit. It transmits a given range of light.

As @M.Farooq says, your filters are characterized by their transmittances versus wavelength. It seems like you want to illuminate your plastic specimens with blue light, then the dichroic beamsplitter does something unclear to me and then you want to use the 560 nm long pass filter to observe the green fluorescence. Is this anywhere near accurate? It would help us if you edited to include a simple diagram. By the way, it is Nile red, not red nile.

From the wikipedia article on Nile red, you will get different emission spectra depending on the solvent. So look at the spectra there, see what it shows for DMSO (if it does) and what solvents give green emission spectra, and then try exciting with a cheap UV flashlight or your blue LED.

I also think you should sketch what you have as setup or in mind. I imagine that you have white light or so light, and want to narrow it in a way to observe fluorescence with higher contrast. But it is not clear..

Thanks for your comments, and apologize for some incorrect expressions due to my English language. We now have the spectra of different types of plastics dyed with Nile Red with different solvents to explore. I did not clearly understood the LP specifications of the old epifluorescence microscope and the question was not properly formulated. We will try with a UV or a blue led lamp. Thanks for your time reading this!

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text:
Retention Time Change in Reversed Phase Chromatography (revised)
I previously learned in books that in reversed phase chromatography, retention time changes depending on the content in mobile phase, which means as the polarity of mobile phase increases, retention time decreases. For example, let's say there are two kinds of combination of water and methanol(5:5 and 6:4 ratio). The latter one's polarity is stronger. So, if we run an experiment with the latter one, retention time decreases than the former one. Can somebody explain why?

Please clarify "content of "what" in the mobile phase"? I can provide some explanation but clarify your question with a concrete example.

You are thinking just the opposite. RPLC is usually done for hydrophobic compounds. C18 columns do not retain hydrophilic compounds; say uracil or cytosine will not show much retention.
Suppose you have a hydrophobic compound such as naphthalene and do RPLC using 50:50 Water: Methanol and 60:40 Water: Methanol. Which mobile phase would show higher retention for naphthalene? It will be 60:40 Water: Methanol system? Guess why? C18 column consists of long C-18 alkyl chains. What would naphthalene prefer? A highly hydrophobic C-18 alkyl chain or a water-rich mobile phase?
This is the typical surface chemistry of a RPLC column. Hydrophobic compounds prefer to stay sticking to the stationary phase. We need a higher methanol or acetonitrile concentration to elute them. Adding more water to the mobile phase increases the retention time.


text:
What is the term for a collection of chiral molecules of unequal amounts of left- and right-handed enantiomers?
This is what I found on wiki:

In chemistry, a racemic mixture, is one that has equal amounts of left- and right-handed enantiomers of a chiral molecule.

Homochirality can also refer to enantiopure substances in which all the constituents are the same enantiomer ...

But what is the term for a collection of chiral molecules of not exactly the equal amounts of left- and right-handed enantiomers?

‘Scalemic’ mixture was used in older literature to describe a non 1:1 mix of enantiomers. Not seen so much now days, presumably as much easier to quantify the exact e.e/e.r

There is no specific word, but we can describe something called as enantiomeric excess. Search this term.

I would say simply "optically active".

"Rotatory", but without specifying the direction if unknown...


text:
Is it true that the better the trial function, the better approximation for the energy in variational principle?
While studying the variational principle in McQuarrie's Quantum Chemistry, I came across the following problem: to relate the difference between an approximation $\phi$ and the exact ground-state wave function $\psi_0$ with the difference between the approximate energy $E=\int\phi^* \hat H \phi \, \text{d} \tau$ and the exact energy $E_0=\int\psi_0^* \hat H \psi_0 \, \text{d} \tau$. For instance, if I know that $\phi$ and $\psi_0$ differ by at most, say, $O(\epsilon)$ over all space, can I say that $E-E_0 \approx O(\epsilon)$?
It is intuitive to think that the better approximation $\phi$ is, the better the energy approximation gets, but I cannot prove this.
Without loss of generality, suppose $\phi$ and $\psi_0$ normalized. By writing $\phi$ as a linear combination of $\psi_n$, we get
$$
\phi=\sum_{n=0}^{\infty} c_n\psi_n
$$
such that $\sum_{n=0}^{\infty} c_n^* c_n=1$.
Therefore
$$
\hat H \phi=\sum_{n=0}^{\infty} c_n\hat H\psi_n = \sum_{n=0}^{\infty} c_nE_n\psi_n
$$
Hence $E=\int\phi^* \hat H \phi \, \text{d} \tau=\sum_{n=0}^{\infty} c_n^* c_nE_n$.
From here, the author shows that $E \geq E_0$, but how can I estimate the difference between the approximate energy and the exact energy?

Probably better suited for Matter Modelling SE: https://mattermodeling.stackexchange.com/

For me these two concepts are synonymous. "A trial function is better" is another way of saying that "A trial function produces a better approximation of the exact energy"

I guess even Maths SE might have a good answer. I was told that on point it might be not true and is a bit alike to the situation around local minima... But I am not sure. But it is a sort of minimization, to answer your very last question.

TL;DR
No this is not true. Although a "better approximation" might yield a lower energy, this is not always the case.
Full answer
There is a major issue in the formulation of your problem: You want

to relate the difference between an approximation $\phi$ and the exact
ground-state wave function ψ0 with the difference between the
approximate energy $E=\int\phi^* H\phi \mathrm{d}\tau$ and the exact energy

The second part of that statement includes a clear mathematical definition of the target property (energy $E=\int\phi^* H\phi \mathrm{d}\tau$). However, there is no clear definition of what a better approximation of the wave function actually is. How do you want to define or measure it, so it can be expressed in a mathematical way?
There a various options to define such a measure, which would allow comparing two trial wave functions and judge their quality. The most obvious and commonly used one is the total energy. But this is the property you want to compare with, so we need something else. Examples for further options are:

overlap with the exact wave function (if available) $\langle \phi|\psi_0\rangle$
expectation values other than total energy, e.g. $\langle r\rangle$ or $\langle r^2\rangle$
any molecular properties of the chosen model system (e.g. dipole moment, excitation energies, etc.)

The next problem is: the choice of the measure for a good approximation might influence the result. For example the overlap $S_i=\langle \phi_i|\psi_0\rangle$ for two different trial wave functions $\phi_1$ and $\phi_2$ (possibly each with a different variational ansatz) might yield $S_1 > S_2$ (meaning $\phi_1$ would be the better approximation), while the total energy of the second trial wave function might be lower (and therefore close to the true ground state): $E_2 < E_1$.
The different measures on how good of an approximation a given trial wave function is do not correlate, therefore there is no general answer which one is actually better. It depends on the property you want to calculate.
There is a even paper available covering this issue and including a more detailed example. Sadly, it is paywalled although it is primarily intended for educational purposes: https://pubs.acs.org/doi/10.1021/acs.jchemed.8b00959

how can I estimate the difference between the approximate energy and
the exact energy?

In essence, this is the central question quantum chemistry is concerned about. And the field developed a whole lot of different methods to achieve reasonable approximations. Thus, I am afraid there is no simple answer to this question.

Good point on that a higher cosine similarity (overlap) does not necessarily imply a lowering of the expectation value.

Disclaimer: As Feodoran has correctly pointed out in his answer, a trail function $\phi$ with a bigger overlap $\langle \phi|\psi_0\rangle$, does not necessarily yield a better approximation for the ground state energy $E_0$. I will leave my answer nonetheless, since I believe it can help to understand the variational principle.
Let us assume a wave function $\phi$ describes a state for our system of interest. For simplification, we assume that $\phi$ is normalized. In order to determine the properties our system will have when it is in the state $\phi$, we apply the corresponding operator. For the total energy $E$, it is the Hamiltonian $\hat{H}$. The problem in QM is that in contrast to classical mechanics we don't necessarily get a single value for the state, but a discrete distribution of values. The possible values are the eigenvalues of the operator $E_n$. The square of the overlap (dot product) between $\phi$ and the corresponding eigenfunction $\psi_n$ gives the probability $p_n$, which is the probability for $E_n$ to be observed:
$$p_n = c_n^* c_n =\langle\phi|\psi_n\rangle\langle\psi_n|\phi\rangle$$
The last relation essentially already answers your question, since it directly connects the probability of observing $E_n$ with how close $\phi$ is to $\psi_n$. The dot product serves as the measure of similarity. In the special case where the state $\phi$ coincides with $\psi_n$, the two functions fully overlap and, due to orthogonality of the eigenfunctions, has no overlap with the other eigenfunctions $\psi_{m \neq n}$. Hence, $E_n$ is 100% observed.
We cannot calculate the $p_n$'s without knowing the eigenfunctions, but we can calculate the expectation value $\langle E \rangle$:
$$\langle E \rangle = \langle\phi|\hat{H}|\phi\rangle = \sum_n p_nE_n$$
and we have of course $\langle E \rangle \geq E_0$. The better $\phi$ is an approximation for $\psi_0$, the larger the overlap is and the more probable $E_0$ becomes. Hence, minimizing $\langle E \rangle$ by varying $\phi$ will gives us the best possible approximation to $E_0$, which is the entire premise of variational theory.

"The dot product serves as the measure of similarity". My doubt is precisely to find an upper bound for the difference between the exact and the approximate energy, given an upper bound for the difference between $\phi$ and $\psi_0$. How can I translate this "measure of similarity" into a mathematical estimate?

I'm not aware of a way to estimate the error in variational theory. The dot product is useless, since you have to know the eigenfuctions beforehand. Maybe a similar approach to a basis set extrapolation could be done.

This statement is strictly speaking not correct: "The problem in QM is that in contrast to classical mechanics we don't necessarily get a single value for the state, but a discrete distribution of values. The possible values are the eigenvalues of the operator En."
Yes, we get a discrete distributions of these values. But each of these energy values represents a single state, meaning: each state has a single energy eigenvalue. What you are probably referring to is the probability distribution, which is not the eigenvalue of the Hamiltonian, but the absolute square of the eigenfunction.

"But each of these energy values represents a single state, meaning: each state has a single energy eigenvalue". What you are referring to are the eigenstates that the state of interest will "collapse" to during the measuring process. The state of interest itself obviously does not have a single energy value, except if it coincides with an eigenfunction.


text:
EDTA titrations of singly charged cations
It is well known that EDTA can be used for titrations involving complexations of highly charged metal cations, with the most well known being those of the alkali earth metals and transition metals. However, would it be possible to analyze metal cations such as the alkali metals or singly charged transition metals?

EDTA titrations can be done for a large number of metals except the alkali metals.

If EDTA wants to make a complex with a metal, it has to be able to include it between two of its "fingers" or "claws". One finger is not enough.  And -COOH is one of these fingers. Now EDTA  has four fingers, all terminated by a - $\ce{COOH}$ group. And it has to loose at least two $\ce{H}$ atoms for gripping one metallic atom.  With one finger, you cannot clip anything. This can only be done with doubly charged ions, (or more heavily charged ions). The reaction of EDTA with a doubly charged ion $\ce{M^{2+}}$ is if the EDTA is partially ionized : $$\ce{H2Y^{2-} + M^{2+} -> MY^{2-} + 2 H+}$$

Titrations involving chelating of metal ions with ethylenediaminetetraacetic acid (EDTA) is an extremely useful analytical tool and a very well studied system in analytical chemistry. It can be bind to virtually any metal ion, evidently even to alkali metals. For instance, sodium ion complexes with EDTA in certain situations (Ref.1). The abstract of the reference states that:

The crystal structures of tetrasodium (1,2-ethanediyl-dinitrilo)tetraacetate pentahydrate and disodium dihydrogen (1,2-ethanediyldinitrilo)tetraacetate dihydrate have been determined. $\ce{Na}$ ions prefer to act as bridges between different $\mathrm{edta}$ ligands and only in the tetrasodium salt is the $\mathrm{edta}$ ligand hexadentate to one $\ce{Na}$ ion. $\ce{Na}$ ions display seven-coordination, when coordinated to one $\mathrm{edta}$ ligand (distorted pentagonal bipyramid), and six-coordination, when linked to several $\mathrm{edta}$ ligands (from distorted octahedral to trapezoidal bipyramidal geometry). An $\ce{Na}$ ion in the tetrasodium salt shows five-coordination, because the sixth site is blocked by the $\mathrm{edta}$ ligand. $\ce{H}$ atoms in the disodium salt are linked to amine groups. Only these two salts can be obtained on varying the $\mathrm{pH}$ of an aqueous solution containing $\ce{Na+}$ and ethylenediaminetetraacetic acid. The first is obtained around $\mathrm{pH}$ 2-5 and the second around $\mathrm{pH}$ 12.

Nonetheless, there are no evident in the literature that EDTA titrations can be done to determine alkali metal concentration (agreeing with M. Farooq in his comment). As a matter of fact, disodium ethylenediaminetetraacetate (dihydrate) salt is the EDTA source it has been almost always used in most of EDTA titrations to determine  a large number of metal cations including alkaline earth and transition metals. Keep in mind that $K_f$ values of alkali metal complexation with EDTA are very law compared to other metals (Ref.2):
$$
\begin{array}{|c|c|c|c|}
\hline
\text{Cation} & K_f & \text{Cation} & K_f & \text{Cation} & K_f & \text{Cation} & K_f\\
\hline
\ce{Li+} & 6.17 \times 10^2 & \ce{Be^2+} & 1.58 \times 10^9 & \ce{Cr^2+} & 3.98 \times 10^{13} & \ce{Sc^3+} & 1.26 \times 10^{23}\\
\ce{Na+} & 4.57 \times 10^1 & \ce{Mg^2+} & 6.17 \times 10^8 & \ce{Mn^2+} & 7.41 \times 10^{13} & \ce{Y^3+} & 1.23 \times 10^{18}\\
\ce{K+} & 6.31 \times 10^0 & \ce{Ca^2+} & 4.90 \times 10^{10} &\ce{Fe^2+} & 2.09 \times 10^{14} & \ce{La^3+} & 3.16 \times 10^{15}\\
\ce{Ag+} & 2.09 \times 10^7 & \ce{Sr^2+} & 5.37 \times 10^8 & \ce{Co^2+} & 2.04 \times 10^{16} & \ce{In^3+} & 1.00 \times 10^{25} \\
\ce{Tl+} & 3.47 \times 10^6 & \ce{Ba^2+} &  7.24 \times 10^7 & \ce{Ni^2+} & 4.17 \times 10^{18} & \ce{Bi^3+} & 6.31 \times 10^{27} \\
\hline
\end{array}
$$
This lack of complexation of alkali metals with even strong chelator such as EDTA can be attributed to their lower Lewis acidity (EDTA is a strong Lewis base because it contains both amino and carboxylate groups).
Apart from alkali metal cations, singly charged transition metal cations form  complexes with chelates and can be determine by relevant titrations. For example, concentration of silver(I) ions ($\ce{Ag+}$) can be determined by cyanide titration (a monodentate):
$$\ce{Ag+ + 2CN- -> [:\!N#C-Ag-C#N\!:]-}$$
The fairly high $K_f$ values for those singly charge cations (e.g., see the values for $\ce{Ag+}$ and $\ce{Tl+}$ in the table above) other than alkali metals suggest that they were able to involve in complexation titrations (They are compatible with alkaline earth metals).
References:

M. Font-Bardia, X. Solans, M. Font-Altaba, “Sodium ion complexes with ethylenediaminetetraacetic acid,” Acta Crystallographica Section C 1993, C49, 1452-1456 (https://doi.org/10.1107/S0108270193001350).
Arthur Earl Martell, Robert M Smith, Editors, In Critical Stability Constants, Volume 1; Plenum Press: New York, NY, 1974, pp. 204-211 (ISBN-13: 978-0306352119).


text:
Quantum numbers associated with electron
I wanted to get your opinion on something. I was given the electron 5p^3 and was told to find the quantum numbers associated with it. I got:
n=5
l=1
Ml=1
Ms= +1/2
However, my teacher has told me that I am incorrect and that my magnetic quantum number (Ml) should be -1. Could anyone possibly explain why this may be the case or if my teacher is incorrect. Thank you in advance!

Do you mean you were given the electron configuration $\mathrm{5p^3}$ ? I don't understand which electron's quantum number you are trying to find.

@ShoubhikRMaiti Yes! Sorry for the notation. I am still trying to look into how to do it on here.

$\mathrm{5p^3}$ means there are three electrons. So if it is the ground state atom (i.e. follows Hund's rule) then you will have $\mathrm{n=5}$, $\mathrm{l=1}$ for all electrons. Then for three electrons—($\mathrm{m_l=1,m_s=+1/2}$), ($\mathrm{m_l=0,m_s=+1/2}$), ($\mathrm{m_l=-1,m_s=+1/2}$). All electrons have parallel spin. Are you trying to calculate the total orbital angular momentum quantum number($\mathrm{L}$) for the configuration, by any chance? Because each electron has a different value of $\mathrm{m_l}$.

As for the notation on this site, there is a handy guide here: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

@ShoubhikRMaiti No I am not. Yes that is why I am a little confused on the question itself knowing that there are different values for my magnetic quantum number. However, the way I tried to envision it was if I was trying to draw an orbital diagram of just the electron given and seeing which box is the last arrow drawn present within. I believe that would have been the the third box that would have given me the value of 1 for my magnetic quantum number.

If you draw three boxes, then usually from left to right, they are +1, 0, -1. So, the third box can have $\mathrm{m_l=-1}$. However, there is no agreed convention on the order of drawing orbitals, so this is not an answer that I can justify scientifically. Someone else may draw the orbitals in a different order.


text:
Hydrogen peroxide and potassium iodide reactions
What is the explanation for the two possible reactions of $\ce{H2O2}$ with $\ce{KI}$ in acid media (iodide-catalyzed decomposition of hydrogen peroxide or iodide oxidation by hydrogen peroxide) and what determines the prevalence of one over the other?
It is known that if the hydrogen peroxide is concentrated, the pathway which results in the decomposition of hydrogen peroxide is favored. Otherwise, the iodide oxidation is predominant.
Two well-known chemistry demonstrations (elephant's toothpaste and the iodine clock) rely on these reactions.
The first one involves the decomposition of hydrogen peroxide
$$
\ce{2H2O2 ->[\ce{KI}] 2H2O + O2}
$$
which is believed to occur through the following mechanism
$$
\ce{H2O2 +I- + H+ -> HOI + H2O}
$$
$$
\ce{HOI + H2O2 -> H2O + O2  +I- + H+} \text{(rate-determining step)}
$$
On the other hand, the reaction that happens in the iodine clock is
$$
\ce{H2O2 + 3 I- + 2 H+ -> I3- + 2H2O}
$$
I would like to understand this phenomenon based on thermodynamic or kinetic data, since I could not find an explanation for it.

related https://chemistry.stackexchange.com/questions/107884/can-an-iodine-clock-reaction-work-without-using-a-strong-acid

What is the mechanism of the 'clock' reaction? Does it also form $HOI$ initially? What is its (likely) rate-determining step? Assuming that both types of reactions can happen in parallel, what decides which one is prevalent must be kinetics, as far as I can tell. So if you can write the differential expression of the rds of each, you should be able to answer your question.

See: https://chemistry.stackexchange.com/questions/8857/redox-reaction-with-hydrogen-peroxide-with-potassium-iodide ... https://chemistry.stackexchange.com/questions/61260/what-happens-in-the-reaction-of-hydrogen-peroxide-with-potassium-iodide-in-acidi


text:
p-Nitroaniline reaction with sulfuric acid
An alternative version of the well-known demonstration carbon snake (the dehydration reaction of sugar by concentrated sulfuric acid forming a column of graphite) consists in mixing  paranitroaniline and sulfuric acid.
This reaction is very vigorous and occurs suddenly, forming a similar column of black solid. However, except from the article "Studies on nitroaniline–sulfuric acid compositions: Aphrogenic pyrostats" (https://onlinelibrary.wiley.com/doi/abs/10.1002/app.1970.070140813), I could not find any relevant information about it. I would like to know if the products of this reaction or its mechanism have already been studied.

related https://en.wikipedia.org/wiki/Polyaniline

I think the abstract of the reference did not give away what's going on the reaction. However, I assume that the full paper should have described what's happening to at least certain extend. Since I don't have access to this paper at the moment, only I can do is speculate. Yet, I'd direct you to reliable source for certain information. Accordingly, when you heat the mixture of p-notroaniline and conc. sulfuric acid:

The only gaseous product you would obtain is $\ce{H2O}$ (with some nitroaniline subliming at higher mole ratios) up to the temperature of $\pu{230 ^\circ C}$. At $\pu{230 ^\circ C}$, it is called pre-expansion stage. Sulfur dioxide has been absent from gaseous products (confirmed by iodine test). It was noted that dehydration and sulfonation are the main chemical processes taken place at this stage.

There are two more stages to follow before you see the carbon snake, after stage of which called oxidative decomposition of the residue. This UCSB article also provides some other interesting excerpts from the original article (Ref.1), and listed below for readers' benefit:

The mole ratio of nitroaniline to sulfuric acid, however, is not critical; voluminous foams have been obtained from mixtures over a mole ratio range of 0.3 to 2.0 if a proper heating schedule is maintained. Sulfur dioxide and water are the principal gaseous products evolved during these expansions.

The foams were good thermal insulators inert to chemicals, e.g., concentrated mineral acids, alkalies, organic and inorganic solvents, and oxidants.

The exact structural nature of the insoluble residue constituting the foam is unknown. All the sulfur in the residue is present as sulfonic acid groups attached to the benzene nuclei. Secondary amino or azinium groups resulting from deammoniative coupling must be present in order to account for ammonium ions and Nitrogen content less than 2 per benzene nucleus. A high nitrogen-to-oxygen ratio indicates that few if any nitro groups remain. Thus, both nitro arid sulfonic acid groups are involved in oxidative processes.

Also, followings are three relevant demonstrations of the reaction: here, here, and here. Also, here is a demonstration of the dehydration reaction of sugar by concentrated sulfuric acid forming a column of graphite for comparison.

It is worth noting that the carbon form created by the p-nitroaniline reaction with sulfuric acid is found to be useful in $\ce{CO2}$ adsorption (Ref.2).

References:

A. C. Poshkus, J. A. Parker, "Studies on nitroaniline–sulfuric acid compositions: Aphrogenic pyrostats," J. Appl. Polym. Sci. 1970, 14(8), 2049–2064 (doi: https://doi.org/10.1002/app.1970.070140813).
Enrico Andreoli, Andrew R. Barron, "$\ce{CO2}$ Adsorption by para-Nitroaniline Sulfuric Acid-Derived Porous Carbon Foam," J. Carbon Res. 2016, 2(4), 25 (9 pages) (doi: https://doi.org/10.3390/c2040025).


text:
Why does Cody use Litharge?
In many of the precious metal refining videos Cody (Cody's Lab) takes the sample with the precious metal and puts Litharge into it. After heating it in the furnace, while pouring it out the molten lead comes out and Cody says that the precious metal is all inside the molten lead(Later extracting it using cupellation). How does this happen?
https://www.youtube.com/watch?v=vltBwXncjCI&t=369s
From 3:01
Edit: My doubt is specifically the processes involved during Litharge Smelting.(that result in the precious metal being alloyed with lead)Why is Cody confident that all of the silver has been alloyed with the lead? Why cant the silver form a separate mass? My speculation is that all the impurities except the precious metal melts, the litharge is reduced into lead, the molten lead being denser than other molten impurities sinks to the bottom ,the silver being the densest goes to the bottom to meet lead and they alloy. Is this possible?

If you want help, I would recommend posting a link to a specific video.  You should then note the time points, in the video, at which the particular steps about which you have questions take place.

Cody explicitly says he uses a process of cupellation. What's unclear about that?

Also, the time stamp in the link is 6:09, the time stamp in the text is 3:01. What footage should we look at, after all?

Litharge is $\ce{PbO}$. It is probably introduced to save silver from compounds containing oxidized silver like $\ce{AgCl}$. If metallic silver powder is mixed with $\ce{AgCl}$, this compound will react with $\ce{PbO}$ according to $$\ce{2 AgCl + PbO -> PbCl2 + Ag2O}$$ And $\ce{Ag2O}$ is an instable compound being easily transformed into $\ce{Ag + O2}$ at high temperature. So all metallic silver and silver from silver compounds are finally dissolved into lead, because lead dissolves metallic silver and not silver compounds. Without litharge, silver compounds like $\ce{AgCl}$ are lost.

So all metallic silver and silver from silver compounds are finally dissolved into lead This is what my doubt is about.

I have edited my previous message, to take your remark into account.


text:
Differences in AChE inhibition kinetics between paraoxon and methyl paraoxon
According to literature, dimethyl paraoxon has a reaction rate constant for the aging (spontaneous dealkylation) of acetylcholinesterase of $\pu{0.186 h-1}$, a spontaneous reactivation rate constant of $\pu{1.01 h-1}$, and an inhibition reaction rate constant of $\pu{1.2 \times 10^6 M-1 min-1}$, compared to $\pu{0.022 h-1}$, $\pu{0.022 h-1}$, and $\pu{2.2 \times 10^6 M-1 min-1}$, respectively for paraoxon.[1] It has been shown that the process of aging, at least for the nerve agent tabun, proceeds through cleavage of the $\ce{O-C}$ bond [2], which would presumably mean that a carbocation is generated as an intermediate, and thus, ethyl breakage, rather than methyl, should be more favorable. However, as the referenced paper shows, this is not the case, and methyl paraoxon, rather than paraoxon (with ethyl groups), ages more rapidly, and is also more susceptible to spontaneous reactivation. Furthermore, paraoxon (with ethyl groups) also has a higher inhibition rate constant. Why is this so?
References:
[1] Franz Worek, Horst Thiermann, Ladislaus Szinicz, Peter Eyer, "Kinetic analysis of interactions between human acetylcholinesterase, structurally different organophosphorus compounds and oximes," Biochemical Pharmacology 2004, 68(11), 2237–2248 (doi: https://doi.org/10.1016/j.bcp.2004.07.038).
[2] Eugénie Carletti, He Li, Bin Li, Fredrik Ekström, Yvain Nicolet, Mélanie Loiodice, Emilie Gillon, Marie T. Froment, Oksana Lockridge, Lawrence M. Schopfer, Patrick Masson, Florian Nachon, "Aging of Cholinesterases Phosphylated by Tabun Proceeds through O-Dealkylation," Journal of the American Chemical Society 2008, 130(47), 16011–16020 (doi: https://doi.org/10.1021/ja804941z)

Aging is an unfortunately complex mechanism that was more well studied for branched phosphorus esters like sarin (R=isopropyl) or soman (R=neohexyl) that when attached to a cholinesterase, looks like Ser-OP(O)(CH3)-O-R.  Protonation of the ester oxygen allows delocalization of the cation to a branched carbon that can undergo elimination forming the aged product.  It is unclear although unlikely that methyl paraoxon and paraoxon undergo traditional 'aging' as defined by cation rearrangement but could undergo a hydrolytic mechanism (attack of OH- at the phosphorus) that might explain why methyl ester undergo this subtype of 'aging' more rapidly than the ethyl ester.  The relative rates, are in fact, similar to what one might find when comparing the hydrolysis of methyl and ethyl carboxyesters.


text:
Why does water weaken ion ion attractions?
My lecturer has told me that water, having a high dielectric value, will "shield" ions and reduce ion-ion attractions.

I really don't see why. Having water molecules surround these two ions doesn't mean that Coloumb's force doesn't exist anymore - nothing in $F=\frac{kQq}{r^2}$ has changed just because water molecules are surrounding them. I recognise that the water molecules will arrange to face the ions in a polarised way (oxygen to the $+$ ion and hydrogens to the $-$ ion). However, I don't see how having a shell of water surrounding you will reduce the attraction. I guess it would prevent them from coming closer due to the sheer physical size of the water shell, but wouldn't there still be attraction?
Can someone explain to me why the ion-ion attraction decreases when water is there?

Even following your reasoning there would possibly be a bigger r between them. So the force is reduced even with no consideration of the field.

@Alchimista IMHO, this is obvious to OP. He has specifically asked about the effect of permittivity.

Ok than also K changes. I am afraid that I don't understand the question. Important is that OP understand my comments and your answer.

Electrostatic field caused by ions or any other external source causes partial orientation of molecular electric dipoles along the field vector. Their orientation causes displacements of center of positive and negative charges. This displacement causes induced electrostatic field, that decreases the primary field intensity by the value of relative permittivity.
By other words, you replace the vacuum permittivity in the Coulomb law by the water permittivity.
$$F = \frac {q_1 \cdot q_2}{4 \cdot \pi \cdot \epsilon  \cdot r^2}$$

As a complement of Poutnik's answer, the distance $r$ between the center of positive and negative charges increases. So the force between them decreases

So from what I'm understanding, the water can orientate itself (as it is polar) between the ions. This reduces the electric field strength as the water's electric field cancels it out. Hence force decreases?

Yes. But it is not limited to water. It happens in various degree for any material. The direct application is e.g. for capacitors. The same charge causes different voltage, depending of the permittivity of the dielectric.

In addition to the answer by Poutnik, here is another perspective:
Although in the force between the ion pair nothing changes, this is not the only force at play here. Water is a highly polar molecule and its dipole (permanent as well as induced by the ions) interacts with the ions. Thus, we have to consider another Coulomb force term for each water-ion pair in the system. These Coulomb force will be smaller individually, but there are many of them. Due to the alignment of the dipole, each water molecule will provide some negative charge to the cation and some positive charge to the anions. Effectively, this will a) shield the ions charges from each other and b) somewhat smear out the charge over a larger area/volume reducing the charge density.
In essence, this is captured in the water permittivity. But a more detailed (and more cumbersome) description would consider explicite Coulomb terms between every water-ion pair.

As another perspective consider the interaction energy between two ions, this is
$$\displaystyle E=\frac{q_1q_2}{4\pi\epsilon_0}\frac{1}{\epsilon r}$$
where $\epsilon$ is the dielectric constant of the solvent, approx 2 for hexane/benzene and 80 for water and $r$ is the ionic separation. ($\epsilon_0$ is the permittivity for free space)
At $1$ nm separation the energy is
$$\displaystyle \frac{-(1.602\cdot 10^{-19})^2}{4\pi \cdot 8.854\cdot 10^{-8)} }\frac{1}{\epsilon 10^{-9}}\approx 2.3\cdot 10^{-19}\frac{1}{\epsilon} \;\mathrm{J}$$
and so for a non polar solvent the energy is $\approx 10^{-19}$ joules which is a lot, thermal energy at room temperature is $k_BT=4.1\cdot 10^{-21}$ joules so thermal energy is no where near enough to separate the ions, it is $28 k_BT$. In a polar solvent such as water at the same distance the energy is $2.8\cdot 10^{-21}$ so comparable to thermal energy so the ions can drift away from one another.


text:
What exactly is hydrolysis? What are the products of hydrolysis of aluminium?
I came across three reactions while studying p-block compounds in inorganic chemistry.
$$\ce{2Al + 2NaOH + 6H2O -> 2 Na[Al(OH)4] + 3H2} \label{eq:1} \tag{1}$$
$$\ce{Al2O3 + 2NaOH + 3H2O -> 2 Na[Al(OH)4]}  \label{eq:2} \tag{2}$$
$$\ce{Al2O3 + 6NaOH + 3H2O -> 2 Na3[Al(OH)6]} \label{eq:3} \tag{3}$$
Look at the above reactions. $\eqref{eq:1}$ and $\eqref{eq:2}$ have different reactants(aluminium and aluminium oxide) but they give the same product.
On the other hand, $\eqref{eq:2}$ and $\eqref{eq:3}$ have same reactants but give different products.
What is exactly going on in these reactions. How do I predict what product is going to be formed in the major amount?
The writers of the book haven't specified the reaction conditions.
Thanks

What is strange about different reactants giving the same product or the same reactants giving different products ? It is quite common in chemistry. Chemistry is not mathematics. There are few laws , there are many more or less empirical rules and rest is to be remembered or found if needed. Unless experience teaches you to see unseen behaviour patterns.

What is exactly the question? R1 and R2 differ in the productS. R1 involves a redox R2 & R3 don't. Also they're are unbalanced.

For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use  MathJax

Hydrolysis ... is any chemical reaction in which a molecule of water breaks one or more chemical bonds. The term is used broadly for substitution, elimination, and solvation reactions in which water is the nucleophile. Wikipedia

Thank you so much Poutnik for formatting and answering to my question. Will use MathJax in future as I have learnt how to use it.

See: https://chemistry.stackexchange.com/questions/80446/why-does-sodium-hydroxide-react-with-aluminum-oxide ... https://chemistry.stackexchange.com/questions/60452/reaction-between-alumina-and-aqueous-sodium-hydroxide-which-ones-right

I will answer in a more general way to your question, because you have a specific one. Aluminium is absolutely a metal, but it can manifest peculiar features of non-metals. Its oxide, $\text{Al}_2\text{O}_3$ is very inert, while the hydroxide $\text{Al}(\text{OH})_3$ is a light-blue jelly solid with an amphoteric behavior, that is it can react both with acids and bases. I suggest you to read the "bible" Chemistry of the Elements of Greenwood and Earnshaw.

I was aware of this passive nature of its oxide and amphoteric nature of aluminium metal. However, I was not able to grasp the particular equation. It has now been brought to understanding that it just depends upon the availability of NaOH. In case, the concentrations changes, the products change. Thanks for your advice too.

The difference between $(2)$ and $(3)$ is the number of $\ce{NaOH}$ that has been used. If few $\ce{NaOH}$ is available, $\ce{Al2O3}$ reacts according to $(2)$. If much $\ce{NaOH}$ is available, it reacts according to $(3)$. So equation $(3)$ is equal to $(2)$ plus twice the following equation $(4)$ $$\ce{Na[Al(OH)4] + 2 NaOH -> Na3[Al(OH)6]\tag{4}}$$
This is the same for the reaction of metallic aluminum. If only a few $\ce{NaOH}$ is available, the reaction will occur according to $(1)$. If enough $\ce{NaOH}$ is available, it will react according to $(1) + 2·(4)$ giving an equation $(5)$, which is : $$\ce{2 Al + 6 NaOH + 6 H2O -> 2 Na3[Al(OH)6] + 3 H2 \tag{5}}$$

So, can I assume that metallic aluminium and aluminium oxide react with NaOH the same way? That's a great explanation BTW. Thanks for explaining.

Both $\ce{Al}$ and $\ce{Al2O3}$ react with $\ce{NaOH}$ to produce first $\ce{Na[Al(OH)4]}$ (and of course something else, $\ce{H2}$ or $\ce{H2O}$). Now if enough $\ce{NaOH}$ is available, $\ce{Na[Al(OH)4}$ reacts with it to produce at the end $\ce{Na3[Al(OH)6]}$


text:
Possible mechanism by which a methyl glycinate hydrochloride salt in toluene (insoluble) reacts with an acid chloride through the nitrogen atom?
I'm trying to work out a possible mechanism for the reaction of a methyl glycinate hydrochloride salt in toluene (it is insoluble) with an acid chloride, reacting through the N-terminus. As the amine is protonated (hydrochloride salt), it is non-nucleophilic and insoluble in the toluene. The question states that upon addition of the acid chloride and heating under reflux for 4 hrs the insoluble salt slowly goes into solution.
Given the salt is non-nucleophilic, I am struggling to see a way that the reaction could take place. My current thoughts are that a small amount of the salt must exist in the non-polarised form (an amine + HCl) and this is what reacts as the amine is a nucleophile and so able to react with the acid chloride, forming a soluble product hence the gradual dissolving over time.
Would this be correct, or am I missing something? Any suggestions much appreciated!
Edit: There is no base such as Et3N in the solution, although I imagine that could solve the problem quite nicely!

Your logic seems sound to me. The heating should support conversion to the more thermodynamically stable product. It would help if the HCl byproduct was being removed from the system, but that doesn't seem to be the case.

@electronpusher thank you - that would seem to make sense about the heating promoting formation of the more thermodynamically stable product which then can react.


text:
why we can use the equipartition theorem for translational motion of molecules at room temperature and above because quantization is unimportant
From the book Chemical Principles The Quest for Insight, 5th Edition by Peter Atkins, Loretta Jones

The equipartition theorem is a result from classical mechanics; so we can use it for
translational and rotational motion of molecules at room temperature and above, where
quantization is unimportant.

I know quantization describes that photons, separate "packets,"are what energy is thought to happen in a subatomic level. However, I don't how quantization is related here: because quantization is unimportant, we can use it for translational and rotational motion of molecules at room temperature and above? and also why quantization is unimportant here? I believe translational and rotational motion of molecules happen in a subatomic level. Moreover, why can't we use the equipartition theorem for vibrational motion?

What they mean is that for translational motion the quanta are so small that using classical mechanics produces no error, i.e $k_BT$ is far far bigger than the energy gaps between translational quanta, ($k_B$ is the Boltzmann constant) the same may be true of rotational motion where quanta may be only be separated by a few wavenumbers vs $k_BT\approx 210$ wavenumbers at room temperature. It is generally true for rotational motion but not all e.g. H$_2$. Vibrations have quanta of at least hundreds of wavenumbers  and $v=1 , v=2$ etc. are hardly populated at room temperature.
We can only use the equipartition theorem if $k_BT \gg \Delta E$ where $\Delta E$ s the gap between energy levels, i.e the fact that there are discrete levels is no longer important. This can occur if the energy gaps are v small or the temperature very high or both.
(Technically the 'equipartition theorem' of classical statistical mechanics states that the mean (average) value of each independent quadratic term in the energy is equal to $k_BT/2$)


text:
Why don't beryllium and magnesium dissolve in ammonia?
From my textbook:

The [alkali and] alkaline earth metals dissolve in liquid ammonia to give deep blue solutions forming ammoniated ions. $$\text{M}+(x+y)\text{NH}_3 \rightarrow [\text{M}(\text{NH}_3)_x]^{2+} + 2[e(\text{NH}_3)_y]^-$$

However, my teacher has told me beryllium and magnesium are anomalous in that they don't dissolve in ammonia solution. I couldn't find a reference for this claim in JD Lee's Concise Inorganic Chemistry, Fifth Edition.
I would like to know two things:

What determines a metal's solubility in liquid ammonia?
Why don't beryllium and magnesium dissolve in ammonia?

They do not dissolve in water either. Is it really surprising they do not dissolve in ammonia ? Beryllium does not dissolve even in most acids.

Ray Bradbury's problem is : Why does metallic calcium dissolve in liquid ammonia, and not magnesium ?

I remembered beryllium has a pretty high ionization energy and now its low solubility isn't surprising anymore. I kept thinking about how its small ionic size should make it soluble.

@poutnik a cording to the WP article on the element, "Beryllium dissolves readily in non-oxidizing acids, such as HCl and diluted H2SO4, but not in nitric acid or water as this forms the oxide.".

@Oscarlanzi Hmm, you are right..

Strongly related: https://chemistry.stackexchange.com/questions/82874/solutions-of-group-1-and-group-2-metals-in-ammonia

Magnesium and beryllium are not really anomalous.  Their poor solubility or nonsolubility in ammonia can be correlated with the fact that their ions are more densely charged than those of the alkali and heavier alkaline-earth metals that do readily give solvated electrons in liquid ammonia.  However, at least for magnesium solvated-electron solutions can be formed in some other solvents or, in ammonia, with an electrolytic process.
Let's look more closely at magnesium, for which much more is known experimentally about solubility and solvated-electron generation than for beryllium.  Compared with lithium, calcium and the elements below these, the greater charge density of the magnesium ion (and by extension, the beryllium ion) inhibits solvated electron formation in at least two ways.

The metal gets passivated.  Solvated-electron solutions of metals in ammonia are not completely pure metal-ammonia solutions.  The highly electropositive metals will slowly react to form the amide and hydrogen, but the amide ion in solution stabilizes the solvated electrons and thus also slows the reduction reaction.  But with the doubly charged and compact magnesium ion, the hard-base amide ion form a salt with little solubility.  This inhibits metal dissolution rather than stabilizing solvated electrons.

What does dissolve, reacts faster.  With its compact size and double charge, magnesium ion polarizes the ammonia molecules and tends to displace protons.  These then react (as solvated ammonium ions) with the solvated electrons.

IUPAC has published a review by Lagowski [1] which includes a discussion of metal solubilities in ammonia solvent.  Magnesium is considered "soluble" in their table, but the text indicates that this is a borderline case and other solutes, such as ammono-bases, are needed to get a stable solvated-electron solution.  Magnesium is shown with a higher ion charge density than any of the clearly soluble metals, including the lanthanides europium and ytterbium.
Thus compared with metals that form those blue solutions, magnesium dissolves more slowly and reacts faster, and so few solvated electrons can accumulate in a proposed magnesium/ammonia solution.
Breaking through
Magnesium is certainly electropositive enough to generate solvated electrons.  Xu and Lerner [2] report that magnesium along with its heavier congeners calcium, strontium and barium generate solvated electrons in ethylenediamine, these electrons being captured by graphite to form blue or green intercalation compounds.  (Curiously, magnesium and barium appear to adopt a +1 oxidation state while the intervening metals go to +2.)  Significantly, magnesium requires more aggressive conditions of higher temperature and longer time to react, a difference the authors attribute to the effects of smaller ion size (which would be correlated with passivation and low solubility) and compare with lithium versus other alkali metals.
More relevant to dissolution in ammonia is the electrolytic process described by Combellas et al. [3].  In this process, the magnesium is dissolved electrolytically into a magnesium salt solution, generating reasonably stable solvated-electron solutions.  Such solutions find use in reductive activation of PTFE surfaces without undesirable carbonization that alkali metals tend to produce; and since the process is electrolytic it can be controlled by way of the electric current.  The electrolytic process breaks both barriers cited above.  With the polarization the magnesium dissolves much faster than it would as an isolated metal, and observations of solvated electron lifetimes made by the authors indicate that the dissolved salt is stabilizing the blue solution against hydrogen evolution.  (The authors do not give a solution-chemistry explanation for this, but the observations are consistent with the magnesium being incorporated into singly-charged ion pairs rather than remaining as independent +2 ions.)
References
1.
J.J.Lagowski, "Solution Phenomena in Liquid Ammonia", http://publications.iupac.org/pac/25/2/0429/pdf/index.html.
2.
W. Xu and M. M. Lerner, "A New and Facile Route Using Electride Solutions To Intercalate Alkaline Earth Ions into Graphite", Chemistry of Materials 2018 30 (19), 6930-6935. https://doi.com/10.1021/acs.chemmater.8b03421
3.
C. Combellas, F. Kanoufi, A. Thiébault, "Solutions of solvated electrons in liquid ammonia: Part 1. Chemical properties of magnesium solutions", Journal of Electroanalytical Chemistry ,
2001 499, 1, Pages 144-151.   ISSN 1572-6657, https://doi.com/10.1016/S0022-0728(00)00504-0


text:
How is phenolic resin made?
The following chemical structure is phenolic resin.  How is this chemical made?

Here is an experiment that all my students have carried out in the lab.
Make a solution with about $3$ mL $40$% solution of methanal and an equal volume of phenol. Transfer the solution into a small paper cup. Add a equal volume of concentrated $\ce{HCl}$, when stirring with a glas rod. This produces a strong exothermic reaction, and the result is the described phenolic resin, whose common name is bakelite.

The phenolic resin called bakelite is rather a
phenolformaldehyde resin.

My synthesis is producing a phenol formaldehyde resin, as methanal is also called formaldehyde,

Yes, that is correct. You have initially written methanol, what is understandable as easily done typo. More exactly, it would be rather 40%(probably) solution of methanal, which is itself gaseous.


text:
Temperature changes in exo/endothermic reactions
In exothermic reactions, heat is released, whereas heat is absorbed in endothermic reactions. But exothermic reactions are often described as being associated with an increase in temperature, and the opposite is the case for endothermic reactions.
How does heat release result in a higher temperature of the system, and heat absorption result in lower temperature? Shouldn't it be the other way around?

Endothermic reaction is cooling the mixture, exothermic is heating - clear now? BTW this thing was asked here like a dozen times already and isn't any better because of it.

Exothermic means that, to hold the temperature constant during the reaction, heat has to be removed.    If the heat is not removed, the temperature must rise.

Your confusion is understandable.  I looked at the answers linked in the comment on your OP, and I'd like to explain it differently from how they do.  Hopefully this explanation will get to the heart of your confusion.
You are thinking (incorrectly) that an exothermic reaction works something like this:
The system has a certain amount of thermal energy.  In an exothermic reaction, some of the system's thermal energy is relased to the environment, and thus the system must lose thermal energy and thus decrease in temperature.  I.e., you're imagining something like what happens when a warm block of metal is placed in a cool cup of water.  The metal releases some of its thermal energy to the water, and thus cools.
But that's not what happens in an exothermic reaction.  In an exothermic reaction, the system internally generates extra thermal energy. This is generated through a chemical reaction in which chemical energy (essentially, the energy in chemical bonds) is converted to thermal energy.  Now, if the system is internally generating extra thermal energy, any of a number of things can be done with this energy.  The two most common pictures are as follows:

The system is in a large (relative to the system) constant-temperature heat bath, and is surrounded by diathermal (heat-conducting) walls.  In this case, because the temperature of the system is kept constant, the thermal energy flows out (i.e., is released) into the bath.

The system is adiabatic, i.e., is surrounded by a perfect thermal insulator.  In this case, the extra thermal energy stays within the system, causing its temperature to increase.

The reverse of what I wrote above applies to an endothermic reaction.


text:
How much energy is required to operate a Sabatier reactor?
I am trying to model some industrial processes and having a hard time finding information about the Sabatier reaction. The Sabatier reaction combines $\ce{CO2}$ (and possibly $\ce{CO}$) with $\ce{H2}$ to produce $\ce{H2O}$ and $\ce{CH4}$.
On the one hand, this reaction is exothermic. On the other hand, it requires high heat and pressure. Does the reactor need initial heating and pressurizing of the gases, and then is it self-sustaining (or rather requiring cooling)?
What is clear is that some energy is be needed to compress gases and flow them over the catalysts.
Are there any estimates of how much energy a reactor would use? I understand that scale would also be an important parameter, but any preliminary estimates are useful to me.

It is a chemical kinetics issue.  Raising the temperature increases the rate of the reaction.

Useful paper on this topic regarding the need to initially heat and then remove heat from reactor. https://pubs.rsc.org/en/content/articlelanding/2018/se/c8se00073e#!divAbstract

It is rather chemical engineering question, comparing rate of heat production and rate of heat sink at given operational temperature.

I'm a chemical engineer. What is the heat of reaction for this reaction? I mean the place I work at is doing an even stupider reaction so this one deserves a bit more looking into

@Maurice your answer has zero value. Do you know what "modeling" means. Maybe you should climb down from your high horse so you can read the question better?

@dlight. Please explain ! I have not written anything about this question.

@Maurice I guess your comment was deleted by admins yesterday, although it should have been deleted the same day you posted it.


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Is there a generalize-able relationship between Ecell and T for a galvanic cell?
Is there a generalize-able relationship between Ecell and T for a galvanic cell (assuming constant concentrations of all components)? (ie: higher temp = bigger/smaller Voltage)
Or is it entirely dependant on the underpinning thermodynamics - ie: if a given galvanic cell is -∆Srxn, then it will become LESS spontaneous at higher T, thus lower Ecell (&vice versa)?
And does this have anything to do with the two other temperature & electricity relationships I know of - as T increases, R increases; & storing batteries in the fridge makes them last longer? Or are both of those totally unrelated to Ecell (/is the fridge one an old wives tale?)

You apply $ \pu{\Delta G = - zE_{cell}F = \Delta H - T \Delta S}$. If the chosen temperature is not too far away from $25°C$, it makes sense to consider that $\pu{\Delta H}$ and $\pu{\Delta S}$ are constant. As $\pu{z}$ and $\pu{F}$ are constant, applying this formula gives the cell potential $\pu{E_{cell}}$ at any temperature.

Awesome. That is exactly the connection my brain was missing. Thank you so much! Much easier to see the relationship in this formulation than in Nernst! :)

The above stays for the equilibrium. If current is involved, not only thermodynamics, but kinetics must be involved as well. Like charge transfer rates of electrode reactions, ion mobility, molar conductivity, diffusion coefficients....  The topic is rather a scope for a long article or a small book.


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What does onset of oxidation mean in Cyclic volammetry
I've read a few posts on here on how to calculate onset of oxidations in Cyclic voltammetry. However, I dont know what it actually means when comparing the onsets of oxidations of two different molecules, for example, two conjugated semiconducting polymers.

I am not an expert on this subject, but my guess is that the onset potential means the voltage at which the activation barrier for the electrochemical reaction is overcome.

Not really about the onset itself but with conjugated polymers you can have peaks broader than expected due to a distribution of effective conjugation length whatever reason.

The concept is explained very nicely here Cyclic voltammetry. I would say the term onset of oxidation is slightly vague but one should talk about $E_{onset}$ as shown in the figure. Qualitatively, the current (on the y-axis) begins to rise very fast, that corresponding potential $E_{onset}$ tells us that after this potential, the kinetics of oxidation will become appreciably fast.


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MEA triazine and formaldehyde reaction
I have been making a formulation of roughly 50/50 MEA triazine (CAS no. 4719-04-4) and 37% formalin, with a <1% of a benzalkonium chloride.
The formulation has been pretty stable, but it has occurred that after being subjected to strong agitation it suddenly starts reacting increasing the temperature to more than 100ºC and giving off fumes that smell very strongly like amine.
What I mean by strong agitation is the sloshing of transporting it in a tanker truck filled up to 70% of its capacity and/or filling a tank from the top letting a stream of the product fall the height of the tank.
Thanks!

Your formulation is MEA-triazine and aqueous formaldehyde or MEA + formaldehyde to get MEA-triazine? Please elaborate your formulation. Also, what chemical is transported by the truck before you put your formulation?

"Free formaldehyde can boost the capacity and breakthrough efficiency. However, it makes triazine degradation for shorter shelf life as well."  Quoted from this reference https://prism.ucalgary.ca/handle/1880/112374

Thank you for your answers!
The formulation is actually aqueous MEA triazine + aqueous formaldehyde. It has no free MEA.


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octahedral interstitial sites located entirely within a full hexagonal HCP unit cell at a height of $c/4$ up from the bottom of the unit cell
I have been trying to visualize and find the number of them for a while, through the image attached.
I can't tell whether the c/4 part includes the octahedral site or not...how should I go about approaching this problem?

Those white dots are the octahedral/tetrahedral sites (respectively). Those white dots on the left are at c/4 and 3c/4...

so would you say there are 3 full sites in the c/4 region or 0?


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Why is HF much more covalent than KI despite having a greater electronegativity difference?
Generally, the percent of ionic character in a two-element compound correlates quite well with the difference in the electronegativities of the two elements making up the compound, as can be seen in this graph:

(Image originally by the University of Florida, via user4076 here at ChemSE.)
Even here, though, an anomaly is visible.

HF has an electronegativity difference of 1.893 (4.193 - 2.300).
LiBr has an electronegativity difference of 1.773 (2.685 - 0.912).
KI has an electronegativity difference of 1.625 (2.359 - 0.734).
LiI has an electronegativity difference of 1.447 (2.359 - 0.912).

Based on this, HF should be mostly ionic - certainly more so than LiBr, KI, or especially LiI.  Instead, judging from the graph, HF is slightly more covalent than ionic, with ~45% ionic character, whereas LiBr is ~60% ionic, KI is ~75% ionic, and even LiI is ~52-53% ionic despite having a considerably-lower electronegativity difference than the ~45%-ionic HF.
Going along the sequence KI - LiBr - HF, the ionic character of the compound actually declines steeply as the electronegativity difference gets somewhat greater.
What is going on here?

Hints: Look up hydrogen fluoride and behold its extravagant chemical behavior. Also note hydrogen bonding is involved. Very involved!

This calculation of values is funny - precision to one thousandth, really? You're treating stupid electronegativity way too seriously, and you're only proving that you shouldn't.

https://chemistry.stackexchange.com/questions/17064/is-kf-the-most-ionic-compound

Do you consider H+ would be just a proton/deuteron  with the extremely strong electrostatic field ?

@Poutnik Well, in the optics of OP an ionic bond to something would fix that.

Does this answer your question? Is KF the most ionic compound?

@Alchimista: No, it doesn't.

@Mithoron: All the electronegativities in question are available with that degree of precision, so why not?

@EdV: That still doesn't explain KI v. LiBr.

In my judgement, no need: the table is pretty much a waste. Chemistry has lots of trends, patterns and exceptions. Electronegativity is only one factor that, unfortunately, gets over-emphasized in early chemistry courses. And there are several ways to define electronegativity, leading to arguments such as “Is nitrogen more electronegative than chlorine or vice versa?”

@EdV: Except that difference-in-electronegativity is directly linked to bond polarity and ionic character.

So why is the plot not a straight line if the difference in EN is the sole factor? I will stay with what I said.

"difference-in-electronegativity is directly linked to bond polarity and ionic character" in which electronegativity? Allen electronegativity you used has no direct connection to bonds at all.

@Vikki-formerlySean that is a bit surprising. In my opinion it should. If you stick to the fact that bigger the delta El Neg more polar the bond is, you won't obviously get out of this.

https://en.wikipedia.org/wiki/Fajans%27_rules


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Why does this acid-base reaction occur?
My teacher told me about the reaction of sodium peroxide with water:
$$\text{Na}_2\text{O}_2 + 2\text{H}_2\text{O} \ce{->} \text{H}_2\text{O}_2 + 2\text{NaOH}$$
Now, the concentration of products may be comparable to that of the reactants (warranting a $\ce{<=>}$ sign instead of $\ce{->}$), but why does the forward reaction occur at all?
Hydrogen peroxide is a stronger acid than water, and sodium peroxide isn't a stronger base than sodium hydroxide.
In an acid-base reaction, a strong acid converts to a weaker acid and a strong base to a weaker one. The opposite seems to be happening here.

$\ce{HO2-}$ is a weaker base than $\ce{NaOH}$. but $\ce{O2^2-}$ is much stronger base than $\ce{HO2-}$ and does not occur in water solutions in significant amount. But its salts ( sometimes in the form of hydrates ) can be precipitated at highly alkaline solutions of hydrogen peroxide.
Additionally, lack of product presence supports the respective equilibrium reactions to be pushed forward. There would be mix of hydrated $\ce{Na+}$, $\ce{OH-}$, $\ce{H2O2}$ and $\ce{HO2-}$.

Aside of that, a weaker acid convert a salt of a stronger acid to the respective acid, even if not quantitatively. As all is about the equilibrium and initial concetrations of involves acids and salts.

The equation should be rather written like this:
$$\ce{Na2O2(s) + H2O -> 2 Na+(aq) + HO2-(aq) + OH-(aq)} \label{1}\tag{1}$$
$$\ce{HO2-(aq) + H2O <=> H2O2(aq) + OH-(aq)}\label{2}\tag{2}$$
Note that after the $\eqref{1}$ is performed, there is abut the same concentration of $\ce{OH-}$ and $\ce{HO2-}$. But while there is $\pu{55.x mol/L}\  \ce{H2O}$, there is still none $\ce{H2O2}$. So even if $\ce{H2O2}$ is stronger acid than $\ce{H2O}$, and even if $\ce{OH-}$ is stronger base than $\ce{HO2-}$, the reaction $\eqref{2}$ goes forward until the equilibrium is reached.
BTW, $\ce{H2O2}$ does not convert to $\ce{H2O}$, it is not a redox reaction.

I don't think I understand. Basically, my problem is: if the forward reaction occurs to near completion, the acid on the product side should be weaker than the acid at the reactant side, and the base at the product side should be weaker than the base at the reactant side (i.e., it should dissociate less).
In this reaction, the acid on the product side is H2O2. The acid on the reactant side is H2O. Wouldn't H2O2 convert to H2O? And then, wouldn't the concentration of H2O be greater than that of H2O2? But here, the concentration of H2O2 is higher. Ditto for bases.

I think your teacher's version of hydrolysis of sodium peroxide is straight out from Wikipedia article, which follows the reference 1. Wikipedia states that:

Sodium peroxide hydrolyzes to give sodium hydroxide and hydrogen peroxide according to the reaction: $$\ce{Na2O2 + 2H2O -> 2NaOH + H2O2}$$

However, even though it can seemingly be considered to go to the completion at once as above equation showed (because of the extremely high reactivity of $\ce{Na2O2}$), actual total reaction should establish as combination of two following reactions, one in equilibrium, as Poutnik showed in the answer elsewhere:
$$\ce{Na2O2 + H2O -> NaOH + HO-O^-Na^+} \tag1$$
$$\ce{HO-O^-Na^+ + H2O <=> HO-OH + OH- + Na+} \tag2$$
Keep in mind that $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O2 + H2O <=> HO-O^- + H3O^+}$ reaction is $11.75$ (Wikipedia) and hence $\mathrm{p}K_\mathrm{b}$ of the reaction $(2)$ should be $14 - 11.75 = 2.25$ (or $K_\mathrm{b} = 5.62 \times 10^{-3}$). Therefore, this equilibrium is feasible with $[\ce{HO-O^-}] \gt [\ce{H2O2}]$.

Because of the high reactivity (e.g., Ref.2) of $\ce{Na2O2}$ seemingly misguided the final outcome of sought reaction. It is note worthy that anhydrous $\ce{Na2O2}$ cannot be prepared by $\ce{H2O2 + 2 NaOH}$, which gives $\ce{Na2O2.8H2O}$ instead even at high temperature. However, anhydrous $\ce{Li2O2}$ has been prepared by this method (Lithium and Sodium Peroxides in pages 979-980 of Ref.1).
References:

E. Dönges, “Section 18: Alkali Metals,” In Handbook of Preparative Inorganic Chemistry, Volume 1; 2nd Edition; Georg Brauer, Editor (Reed F. Riley, trans. Ed.); Academic Press, Inc.: New York, NY, 1963, pp. 950-992 (ISBN-13 : 978-0121266011).
Hector Russell Carveth of Roessler and Hasslacher Chemical Co. (New York, NY), “Process of producing sodium peroxide,” US Patent 1931, 1,796,241.


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What difference between pi bond and bent bond? Is it possible to replace sigma-pi bond model and use only bent bond instead?
Between sigma-pi bond and 2 bent bonds, is it actually equiavalence? Or there is specific molecule that require to be bent bonds and cannot exist as pi bond and vice versa? I felt like we actually don't need pi bond model at all if we consider it all to be bent bond
If it not, then how can we consider which chemical result will be bent bond, and which will form pi bond?

@Poutnik - I assume OP is referring to the bent-bond or tau-bond model that Pauling and others favored over the sigma/pi description of double bonds.

A good reference is Wintner J Chem Educ 1987, 64, 587: https://pubs.acs.org/doi/pdf/10.1021/ed064p587

Does this answer your question? What utility does the Tau bond model of orbital overlap have?

@Mithoron That's giving me more information but still it not so clear yet. Did it means Tau bond only apply to alkenes and alkynes, and every other molecule still being sigma and pi strictly?

This earlier answer may be helpful.

In the famous words of George Box, "All models are wrong. Some are useful."
The bent bond and sigma/pi descriptions are two apparently distinct models for representing the underlying structure of electron density of molecules. Both are oversimplifications, but the key question (as with any model) is which is better for accounting for all of the known molecular structures and predicting the results of observations we have not yet made.
There are examples that support both. Bent-bonds have been cited famously as better predicting the conformation of propene, but the sigma/pi gives more intuitive results for reactions like the Diels-Alder. As in the case of the VB vs MO theory debate, both models can be elaborated with greater complexity to the point that they give indistinguishable results, but the sigma/pi approach has largely taken over because in its basic form it is more intuitive and gives accurate predictions in most cases.
The general concept of a bent bond remains useful for highly strained systems such as cyclopropanes, where the electron density of a single bond does not appear to align with the axis connecting the atoms.
To answer your final question,

If it not, then how can we consider which chemical result will
be bent bond, and which will form pi bond?

it is rare that the sigma/pi model will fail you in modeling a molecule with a double bond, and since it is the prevailing model pedagogically, you would do best to use it preferentially and only invoke bent bonds for strained single-bond systems.


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Is it possible to store moist air at high pressure?
Is there any way to store moist air inside a high pressure tank?
Let's suppose that I have humid air (30%) at room temperature and I want to store it at 300Bar in a tank. when I extract the fluid from the tank (isothermal expansion) to bring it back to ambient pressure will I keep the same percentage of humidity?
In most cases where it is necessary to store high pressure air to be used in a system that requires humid air, a cylinder is used where ambient air is stored which then pass into a humidifier. I'd like to know if it is possible, conceptually, to remove the humidifier and directly store moist air in some way

There is one importing principle to consider.
Air does not store vapor. Space does. Less space means less vapor capacity, no matter how much of air is there. (*)
If air is being compressed, the partial vapor pressure increases. It it reaches the saturated vapor pressure at given temperature, water vapor starts to condensate. When air is then expanded, its  absolute and relative humidity decreases, according to the expansion ratio.
E.g. Lets have 1 cubic metre of compressed air with 100% relative humidity. If it expands to 2 cubic metres and temperature raises back to the initial temperature, its relative humidity will be 50 %.
So compression provides liquid water and expansion needs liquid water to restore humidity.
If there is compressed air with cold water and if pressure drop e.g. 20 to 1 atm) then its relative humidity suddenly drops from 100% to 5% (when air warms up back after adiabatic cooling down.). Water exposed to low humidity air will start slow evaporation.
It will not help much in humidity restoration, unless some active evaporation technique is involved a/o the air release is slow, passing evaporation unit.

(*) Exactly speaking, saturated vapor pressure and therefore vapor capacity of space very slighly increases with total pressure. But it is negligible in context of your question.

Note that absolute humidity is in g/m3 - it is intensive, not extensive variable. Rather, the vapour content  = absolute humidity * volume.

let's see if I understand correctly, the absolute maximum humidity depends only on the temperature and the volume, if I compress the humid air, I reduce the volume and therefore the absolute maximum humidity. therefore as I compress the water it condenses inside the tank.
I have another question, if I have two fluids in the high pressure tank (air and condensed water) if I abruptly reduce the pressure a certain percentage of liquid water should turn into vapor?

Sudden decompression would also typically be accompanied by a change in temperature, which would also reduce the rate of evaporation.

@michaelseifert Sure, I have implied it in the note about temperation of the final air volume.

@PatrickDanzi It will turn into vapor but only slowly.  No matter how fast or slow you draw the first air that comes off will be at 1.1% RH.  Draw it slow enough and it will eventually rise to 100% RH--but slow enough is measured in days.  Probably lots of days.


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Decomposition of Hydrogen Peroxide Catalysed by Potassium Iodide in Basic Solution
As part of my high school project, I had to measure the initial rate of decomposition of hydrogen peroxide catalysed by potassium iodide at different pHs. At higher pHs (alkaline conditions), I found the initial rate of reaction to be slower, and I have repeated my experiment multiple times to confirm this. Any possible explanation for these results? I seem to have some trouble finding any answers online. Would appreciate some help on this, thank you!
FYI, I measured initial rate through the volume of gas evolved. There was still gas evolved for high pH, except that the rate was slower

My blind guess is that in alkaline condition, some of the $\ce{H2O2}$ exists in deprotonated form: $\ce{HO2-}$, and the deprotonated form does not react with iodide($\ce{I-}$). (two anions are unlikely to come close enough for any reaction to occur).


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Why is the shape of sulphates and perchlorates such that lattice enthalpy doesn't decrease down the group?
The solubility of carbonates, bicarbonates and hydroxides of s-block elements increases down the group.
But, the solubility of their sulphates and perchlorates decreases down the group.
The reason for this is: down the groups, ions get larger. Thus, hydration enthalpy decreases, and solubility should take a hit. However, for hydroxides and (bi)carbonates, solubility increases. Lattice enthalpy is related to the Coulombic force between the charge centres, which decreases as the square of internuclear distance. Thus, lattice enthalpy goes down much faster than hydration enthalpy does, so on the whole, solubility does increase.
However, according to this answer, sulphates and perchlorates have a shape that ensures that lattice enthalpy decreases much slower than hydration enthalpy. Of course, this is only pushing the question one rung deeper. What is it about the shape of sulphates and perchlorates that takes care of the square factor?

They are tetrahedral and rather bulky, maybe?

https://chemistry.stackexchange.com/questions/96822/which-has-a-greater-dominance-hydration-enthalpy-or-lattice-energy

@OscarLanzi I thought their tetrahedral shape might have something to do with it, but what?

@Mithoron Thanks, the second answer helps.

Maybe at least delete this "special shape" from title, my nonsense detector is screaming ;) I don't think this old post is nearly as useful as its score would suggest.

@mithoron you're evil!  The score is 0!

@Mithoron Uh, which post is less useful than its score suggests -- the one you linked or mine?

The one you linked. With "special shape".


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How to calculate pH of an CH3COOH solution?
please, does anyone know, how to properly (with calculation procedure) calcutate $\mathrm{pH}$ of an $\ce{CH3COOH}$, when you know only:

-$\ce{CH3COOH}$ is 8% (water solution)
-density of $\ce{CH3COOH}$ - $\pu{1.01 g cm-3}$
-$K_\mathrm{a}$ - $1.74 \times 10^{-3}$

I know, how to calculate pH and I tried to do it myself. But...
I have done this:
 - V is multiplied by 0,08 for 8% solution.
and then calculated pH from concentration.
pH = - log [H3O+]
But still, I am getting wrong numbers...

EDIT:
I found this:

and calculated $\mathrm{pH}$ 2,71, but this is still way too far from 2,9...

It wasn't clear that what percentage of the solution. It could be $w/w$ or $w/v$ or etc. The density of the solution is given for a reason. Did you use it?

$\pu{K_a}$ is not $1.74·10^{-3}$ as you state. It is $1.74·10^{-5}$

I have asked OP to verify the solution concentration but didn't get the answer. Thus, I assume it is $8\% \ (w/w)$. Thus, if you assume $[\ce{HA}] = c$ then:
$$ c = 8\% \ (w/w) = \frac{\pu{8 g}\text{ of HA}}{\pu{100 g}\text{ of sol}} \times \frac{\pu{1.0 mol}\text{ of HA}}{\pu{60.05 g}\text{ of HA}} \times \frac{\pu{1.01 g}\text{ of sol}}{\pu{1.0 mL}\text{ of sol}} \times \frac{\pu{10^3 mL}\text{ of sol}}{\pu{1.0 L}\text{ of sol}}\\ = \pu{1.35 mol L-1}$$
Acetic acid ionization according to:
$$\ce{HA + H2O <=> H3O+ + A-}$$
If $\alpha$ amount of $[\ce{HA}]$ is ionized at equilibrium, concentrations at equilibrium would be  $[\ce{HA}] = c - \alpha$, and $[\ce{A-}] = [\ce{H3O+}]  =  \alpha$. Thus:
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} = \frac{\alpha^2}{c- \alpha} \tag1$$
Since $c = \pu{1.35 mol L-1}$ and $K_\mathrm{a} = 1.77 \times 10^{-5}$, $c - \alpha \approx c$, the equation $(1)$ can be simplified here to get $\mathrm{pH}$. Take $\log$ on each side of the equation:
$$\alpha^2 = K_\mathrm{a} \times c \ \Rightarrow \ 2 \log \alpha = \log K_\mathrm{a}  + \log c$$
Since $\alpha = [\ce{H3O+}]$:
$$2\times \mathrm{pH} = \mathrm{p}K_\mathrm{a} - \log c  = 4.75 - 0.13$$
Thus, $\mathrm{pH} = 2.31$.

Late addition:
OP insists that his given answer for $\mathrm{pH}$ of the solution is $2.9$. Since I suspected OP's given data for the problem, I assumed the concentration of the acetic acid solution must be $0.8\% \ (w/w)$ instead of $8\% \ (w/w)$. When calculating with that value, you get $c = \pu{0.135 mol L-1}$. Hence,
$$\mathrm{pH} = \frac{1}{2}\left(\mathrm{p}K_\mathrm{a} - \log c \right) = \frac{1}{2}\left(4.75 + 0.87\right) = 2.81$$

Sorry @Mathew Mahindarante this cannot be the solution. Thus pH of the 8% CH3COOH is 2,9, how I mentioned in question...

And 4,75 - 0,13 is 4,62

@HASHTAG: You forgot to divide by 2.

@HASHTAG: Also, you forgot to correct your data. For example, percentage of HAc. Its $K_\mathrm{a}$ is not correct either. Hence, I assume your suggested answer might not be correct. My answer is based on $8% \ (w/w)$ concentrated HAc solution with known $\mathrm{p}K_\mathrm{a}$ of 4.75 (I just assume the given density is correct).

@MathewMahindaratne Unit symbols should never be appended with "of something". If you want to show what physical quantity does a particular value refer to, use an equation in symbolic form, and then plug in the numbers, say, next line for the ease of comparison.

IMHO, the task is errorneous and the input data do not match the expected result, if calculation is done correctly. What you, @HASHTAG , ask for is what incorrect calculation fits the incorrect, but expected result.


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Is Polyethylene Glycol always derived from Ethylene Glycol? If not, how does it differ?
I have heard Polyethylene Glycol is derived from Ethylene Glycol, so was doing some research to see how they differ and found this.
Now this states that:

Polyethylene glycol is produced from the reaction between ethylene oxide and water, ethylene glycol or ethylene glycol oligomers.

As I understand it ethylene glycol is made from ethylene oxide and thus I assume that means PEG can be derived from either, is that what the above statement is saying? Though I'm not sure how ethylene glycol oligomers come into play here.
What I'm trying to work out, is there any difference in the PEG depending on what it is derived from or does it come out exactly the same?
If so, what is the purpose of deriving it from one or the other? Different uses? Safety elements? Human consumption?
I have also read that the molecular weight in turn determines the characteristics of each type or category of PEG.
With that in mind and the difference in what PEG is derived from, does the PEG differ at all depending on what it is used in? (Food, Medicine, Cosmetics, Vaccines etc)

The main reason why alternative routes are used to produce polymers is that they give different results (eg the polymer chain length). Sometimes alternative routes can be used to give the same polymer and the choice is purely economics or convenience. Whatever the source, the polymer will be graded to meet the requirements of  specific use by a variety of techniques.


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How to draw chair-form for trans-decalin (with slight variation)?
I understand how to draw chair form for trans-decalin (figure 1) for the usual structure, but I am having difficulty if taken as shown in figure 2 wherein the hydrogens at bridge-head carbons are in opposite orientation. Please help me with relevant chair form diagram.

See a model kit (example, for this type of question a bit too few atoms, though) if 1 and 2 are really different from each other.

I know that they are identical, since on rotation about an axis perpendicular to the plane of molecular, they are inter-convertible. But if given figure 2, translating it directly to a chair form gives the highly strained trans-decalin as shown in:
https://www.masterorganicchemistry.com/2014/08/05/fused-rings/#seven

In figure 1, when we translate directly to chair form, the top bridgehead carbon can easily have the hydrogen take up the position axially pointing upwards and similarly for bottom bridgehhead hydrogen, thereby making the chair form stable.
Whereas in figure 2, doing the same yields the improbable highly strained conformation. How do I resolve this issue?

@Ritwik_Kumar Since 1 and 2 represent the same molecule (including atomic spatial arrangement in 3D-space) I suspect the question is essentially how to make the illustration for the second viewpoint look better. Is something like this is what you are looking for?

Yes, I was getting the same. Is it a problem that it doesn't look like chair form? I am unable to resolve this mental issue when solving or teaching this type of representation...

@Ritwik_Kumar Well, one representation is better than the other. In 2D both structures look fine, in 3D not so, and higher dimensions will likely cause more confusion.

There is no short way to avoid building a 3D model of the two alternatives. Get a kit and build both. And this should clarify that cis-decalin is not "improbable or highly strained". It is less stable than the trans molecule but not by a huge amount mainly because the rings are not locked and can therefore interact a bit spatially in some conformations.

I did not understand. What is wrong with drawing just the bottom left for both the structures at the top?

I need to emphasize that I am asking for a solution to paper chemistry (i.e., chemistry done on paper using 2d projections of 3d structures). I am able to visualize that the two are identical and superimposable on rotation. Using kits and models is a solution but our first instinct is to use 2d projections for simple molecules. I want to know if there is a solution to resolve the 2d projection issue I am facing.

@Alchimista, When we try to convert figure 2 directly (without rotating to figure 1) to chair form projection representation, it will come out as linked by andselisk in a previous comment as a .png

Use of 3D models
3D version of Figure 1:

Viewed upside down (rotate about z by 180 degrees):

Now it looks like we are viewing from the bottom, so we will turn the molecule a bit around the x axis:

Use of 2D projections
If you want to use the 2D projections, consider whether hydrogens are pointing up or down on the bridge head. First, we will interpret the chair as viewing the top of the ring from above (left panel) rather than viewing the bottom of the ring from below (right panel) to avoid ambiguity.

No matter whether you use two fused chairs as drawn by the OP (Fig 1 bottom panel) or its mirror image, the front hydrogen will point down and the back hydrogen will point up. This works well for Fig 1.
In order to use the same chair building block but show the front hydrogen pointing up, you could rotate the flat model, and connect two chairs accordingly, as shown below. In effect, you choose the location of the two bridging position to match the stereochemistry you want.

I appreciate the 3d models Karsten but I was looking more for a solution in the realm of 2d projections as shown in question. We can use the final 3d structure that you have provided to draw the 2d projection but it looks a bit complicated to teach.

@Ritwik_Kumar I added a second part to my answer.

Because the 2D images in the original post are identical, the 3D image shown fits either one. The reason this image is usually drawn is that the methine hydrogens are not visually obscured by other bonds. See Karsten Theis's final two images with methine hydrogens in place.

A common notation in crystallography used for denoting atoms above and below the plane of the paper for the structures with quasi-flat overall shape is the use of filled (⚫) and open (⚪) circles, respectively.
I think you can use this technique for decalin as well.
For trans-decalin with both rings in chair conformation there is even no need to change your drawing for pseudo-3D (red plane refers to the plane of drawing, not symmetry) (scheme A).
Also, note that in crystallography there are also plus (+) and minus (−) signs used for the same purpose, e.g. denoting an object (including atoms) that lies above and below the plane containing the reference axes, respectively.
I personally find this notation less applicable here because of the possible confusion arising with the clash of notation for charge distribution (scheme B).