text:
Racemization at stereogenic centre
I am getting a bit confused about why the configuration at the optical centre changes from d to l after enolization?

Any stereogenic centre next to a carbonyl group will be destroyed by enolization. Such compounds racemize quickly.

[OP in comments] In the figure, it is mentioned that optically pure l isomer is formed, but instead it will be more like a racemic mixture?

For equilibrium reactions, the chemical equation does not tell you what the net direction of reactions will be. When you reach equilibrium, you will have some of each of the two enantiomers, and some of the enol. If there is not special circumstance (chiral cosolvent, etc.), the two enantiomers will be present in equal amounts.

I am getting a bit confused about why the configuration at the optical centre changes from d to l after enolization?

What is confusing is that the molecules are labeled "optically pure isomer". This does not make much sense in this context. It is true that no matter whether you start with optically pure d or l enantiomer, you will end up with a racemic mixture.
Molecular formulas do not indicate pure or mixture, they just denote a given molecule, so labeling them "pure" is a bit strange.
Taking a snapshot at equilibrium, some centers will have changed from d to l, and some will not have changed. Also, some centers are in the process changing from planar to d or l, and some d or l centers are in the process of changing to non-chiral. This is what we call dynamical equilibrium.

Forming an enol causes the carbon attached to the methyl group to essentially form a trigonal planar structure. This means that in the eventual conversion back to a ketone, there are two sides the hydrogen can attach to, often referred to as the re and si faces (see prochirality).
Therefore, the 'optically pure' d-isomer would actually exist in equilibrium with the l-isomer as the acetoacetic ester, which forms especially stable enols, will basically force prochirality on the carbon attached to the Me group.

In the figure, it is mentioned that optically pure l isomer is formed, but instead it will be more like a racemic mixture?

@HiyaJandial The picture doesn't say a pure optical isomer if formed: it shows a reaction path where the two pure isomers are in equilibrium.


text:
How to convert molecule structure to 3D PyTorch tensors for CNN?
I want to try convolutional neural networks for drugs classification. I use PyTorch for 3D CNN implementation. Is there a way to obtain 3D tensors from SMILES or SDF/PDB structures?

PDB is already a 3D representation, but SMILES is not.

This seems like a much better fit for Matter Modeling

I've seen the conversion of SMILES into 1D and 2D representations.  Is there any reason you specifically wish to use 3D tensors? I haven't seen 3D tensors in literature, but that isn't to say that it doesn't exist.
torchdrug converts SMILES molecules graphs based on NetworkX. You can install it using:
pip install torchdrug
Be aware that it only works for python 3.6-3.9.
pysmilesutils converts SMILES into a vector. You can install it using the following commands:
python -m pip install git+https://github.com/MolecularAI/pysmilesutils.git
PyTorch Geometric converts SMILES into a matrix constructed of property vectors of each atom in the molecule.

Thank you! The three-dimensional structures that can be obtained from molecular docking provide useful information about the geometry of the molecule within the binding pocket. But PyTorch CNN cannot work with SDF or PDB files, it requires tensors. In the case of three-dimensional structures - three-dimensional tensors. This link can give you more information about this approach -

I know a computational chemist who specialises in ML for drug-target interaction, I'll ask them about this and get back to you!

source: Stack Exchange

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

sort: votes


text:
Correlations between the pK values of acids or based in polar and non-polar solvents
I try to predict pKa values of acids (in water), computing their energy of deprotonation with quantum chemistry; and I found that the slope of the linear correlation "pK vs Energy" is more close to the theoretical value, if the computation uses an advanced method of computing the solvent effects. I try to explain this obsevation theoretically. Please comment, whether the following argumentation is correct:
Strong acids contain more polar OH fragments (in the neutral form), so strong acids in the neutral form are more strongly stabilized in polar solvents than in non-polar ones. As a result, in polar solvents, compared to non-polar solvents, strong acids become slightly weaker, and weak acids become slightly stronger.
Is this statement true? If we take a number of e.g. acids and measure their pKa in water and non-polar solvents, will it turn out that a polar solvent slightly reduces the dependence of pKa on the properties of the molecules themselves, i.e. in polar solvents, strong acids become a little weaker, and weak acids a little stronger due to the solvent?
In other words, I want to find a data on pK values of acids in non-polar solvents (or at least less polar than water), and check the correlation between the pKa values in water and in these solvents; the most important value is the slope of this correlation (whether it is not 1.0).

That's not true - afaik strong acids tend to be weaker in lower polarity solvents. More precisely, the effect you consider is present for some acids, but should be rather negligible.

And HCl solutions in toluene are not conducting electricity.

For analysing strength of acid, it's rather good to consider situation of ions after dissociation. Here comes solvent's dielectric constant, showing how well ions are "shielded" from each other

@Maurice Sure, but it only says it can't protonate toluene, if you added some methanol, this would change, I think.

@Mithoron can you provide some links for articles?

Articles about experimental data, or theoretical predictions? As far as values are concerned, you can just check pKa table with water vs DMSO values.

@Mithoron I thought I will not to have to build this correlation myself, or at least I will not have to google the pKa values for each compound manually. Maybe at least there are papers where these values are shown together at a table?

Well, I just told you about pKa tables, for example https://organicchemistrydata.org/hansreich/resources/pka/

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


text:
M- 15 Peak of 1-pentene in EI-MS
This is the spectrum of 1-pentene.

And my professor says that the peak M-15 (55) came from this process:

But how this is possible, when I tried to prove through mechanism, no one situation match with it.
I tried these:

How can this CH3 peak exist with my professor process?

You know you can just cut the bond on the other end? I could see issues with other peaks, but this one...

Yes, I know, however, how does my professor find this peak (M-15) by his mechanics?

You don't need any mechanism for such elementary thing in MS. If you really need it for some reason, you could write there's a half bond (1 electron instead of 2) instead of any you want to break and make a fishing hook from it to the atom that you want to wind up as a radical.

That's for single bond, why would you try breaking double?

In an ionization, the electron in a pi bond is more accessible to leave than the electron in a sigma bond. The second picture is schema that my professor showed us. In this picture, the CH3 leave from sp2 carbon and not from other side.

Maybe if it was chemical ionisation then it would be an issue, but in MS you get cation radicals that are so excited that they break apart from surplus energy, and then break again! You're prof. seems to be confusing for no good reason.

Note that according to the second picture, the unique group that can be leave when the bond is break, is the CH2..

In the third picture, I put possibility, and no one they show a CH3 group leaving when the bond between carbons that before had pi electron is broken by ionization.

If the product is written exactly as stated on your answer, it cannot be true, as the ion depicted has m/z 57 and not 55. m/z 55 is a methyl-allyl cation, and thus has a double bond.


text:
Can a non-rate determining step control what products form in a reaction?
In Organic Chemistry by Clayden, the stereoselectivity and regioselectivity of the E1 mechanism is explained in terms of the differing transition state energies (and therefore rates) of the possible deprotonation steps.

However, the rate-determining step of the E1 mechanism is the formation of the carbocation, not the deprotonation. How is it that the deprotonation is able to control the stereoselectivity and regioselectivity of the reaction despite not controlling the overall rate of reaction?
I have a potential explanation for this, though I am not sure if it is correct. I think that the rate-determining step only limits the rate of the overall elimination, and has no impact on the stereoselectivity. Although carbocation deprotonation has to occur much faster than the carbocation formation, when comparing the two deprotonation routes, one route is still much faster than the other (even though both are quite fast compared to the carbocation formation), which means that the deprotonation is stereo and regioselective, because the majority of the carbocations will react via the faster route. Is this thinking correct?

The rate-determining step determines the rate of the overall reaction. However, that is not to say that there is only one transition state that determines the regio- and stereo-selectivity of an entire reaction.
With these mechanisms, many transition states can form, and in chemistry, reactions tend to follow pathways that flow with lower energy demands.
For example, take a waterslide with large bumps, which represent the different transition states you have to pass. Climbing to the top of the waterfall is your first step, and that requires the most energy, just like the formation of the carbocation which has the highest activation barrier. On the top deck, there are several options for slides, and again, on each slide, there are many bumps. On some rides, they are much larger, corresponding to less favoured pathways. Although you will get past them since you are at the top, your ride would be much slower, which would not as fun. The ride with the smallest bumps will get you down the fastest, and so you'd choose that one.
For all the pathways, forming E/Z-product or more the more/less substituted product, you will encounter the formation of a carbocation. From there, it is simply that the pathways that result in E and more substituted products go through lower-in-energy activation barriers. Thus, those pathways are favoured (kinetically and thermodynamically) and the observed regio and stereo-selectivity become present.


text:
How to obtain reliable ligand binding energy in QM manner?
I wish to compute the binding energy of the ligand to the known cavity of the receptor.
My idea was to:

Perform docking to the recognized binding site using AutoDock Vina
Export bound ligands and interacting aminoacids using Chimera
Run optimization on the ligand-aminoacids system (ONIOM; high layer ligand: B3LYP/6-31+G(d,p); low layer aminoacids: PM6), ligand alone (B3LYP/6-31+G(d,p), and aminoacids alone (PM6). These along with empirical dispersion GD3 and PCM water solvent.
Recompute single point energies using larger basis set, like B3LYP/def2TZVP
Compute the binding energy according to the equation: ΔE=[E(system)-E(ligand)+E(aminoacids)]
Assess binding affinity (K) as: K=e^(-ΔE/RT)

Could someone confirm whether this is a valid pathway? If no, what shall be changed?


text:
Lower melting points of zinc, cadmium and mercury
I am aware of the fact that d-block elements like zinc, cadmium and mercury have lower melting points than other d-block elements. Also I am aware that these three metals have a fully filled d shell.
How does the fully-filled d orbital help in explaining the reduced melting point? What effect explains mercury being a liquid and zinc and cadmium being solids at room temperatures?

Unanswered duplicate: https://physics.stackexchange.com/questions/421835. As for the last question, https://chemistry.stackexchange.com/questions/13667.

The two questions in the second para are separate and needs to be separate posts.

Mercury is a rather special case, which cannot be compared to Zn and Cd. Mercury (Z = $80$) belongs to the heaviest elements of the periodic table. Pekka Pyykkö's calculations have shown that nuclei having more than $70$ or $75$ protons are surrounded by electrons whose orbital velocity is not far from the velocity of light, in the Bohr's model. Of course the Bohr model is wrong, but similar calculations have been done without Bohr's model, which show that outer electrons have a relativistic behavior in the heaviest atoms. When a particle velocity is not far from the velocity of light, its dimensions become smaller. So when electrons become relativistic, they become smaller and tend to stay nearer the nucleus. Seen from outside, relativistic electrons "disappear" inside the "outer shell" of non-relativistic electrons. This effect is dependent on the total of the quantum numbers n+l.
Mercury atoms have the electronic configuration $\ce{(Xe) 4f^{14} 5d^{10} 6s^1}$. The corresponding n+l values are rather high:  $4+3=7, 5+2=7, 6+0=6$. So these electrons are "absorbed" under and inside the corresponding xenon orbitals. For approaching atoms, mercury looks like a sort of "heavy xenon". And xenon is a gas with a boiling point at $161$ K. This is why mercury is "nearly" a gas: it is a liquid with a relatively low boiling point, compared to other metals.
Ref. Pekka Pyykkö, Jean-Paul Desclaux, Relativity and the Periodic System of Elements, Accounts of Chemical Research, Vol. 12, No. 8, August 1979, 276 - 281.

Why does fully filled d explain reduced boiling point?

Fully filled $d$ orbitals have nothing to do with boiling point

I am sorry, melting point!


text:
Confused why delta ∆Q and dQ don't make sense for heat Q
In my chemistry teacher's notes, some notations concerning the heat $Q$ are marked as inappropriate.

$Q$: yes
d$Q$: no
$\delta Q$: yes
$\Delta Q$: no

In the second bullet in the screenshot below, why does ∆Q have an "x"" beside it and others have a check mark? I had tried to understand but I failed to do

I'd think twice before using notes where a simple and correct $$Q<0$$ is replaced with  invented abomination like $$Q=\ominus\mathrm{ve}$$

Related: https://chemistry.stackexchange.com/q/22171

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

Heat is not a state function. For any state function $X$ (like enthalpy, entropy, Gibbs energy, etc.), it makes sense to use d$X$ (total or exact differentials) and $\Delta X$. For quantities like heat or work that are path-dependent, you use the $\delta$ notation (inexact differential), e.g. when integrating along a path.

$Q$ and $W$ already include implicit delta ($\Delta$), being specific forms of energy (ex)change.
In contrary to the operator $\mathrm{d}$, operator delta $\Delta$ is not  limited to functions with exact differentials. So $\Delta Q$ (similarly $\Delta W$ )could still be used for cases like $\Delta Q = Q_2 - Q_1$.

Intuitive approach: If you have amount of money $M_1$ in your bank account at some time, then you receive some income $I$ and give out a loan $L$, what equation can you write relating this quantities and your new amount of money $M_2$?
Hopefully you'd come up with some equivalent form of $M_2 = M_1 + I - L$, which we can write as $\Delta M = I - L$. This is a conservation law (we made the implicit assumption that money cannot be created or destroyed, only transferred and accumulated or depleted).
What sense would the terms $\Delta I$ or $\Delta L$ make? I'm sure in some context it could make sense to talk about your "change in income" or "change in amount loaned", but that would require comparing two scenarios, here we just want to perform a simple balance on a single system over a well-behaved time period. The income and loan are amounts of money that are transferred, they are ways in which the actual quantity changes. That is, only $M$ changes, and $I$ and $L$ are contributions to that change.
Maybe it would be more rigorous to think of the first law this way (standard scientific sign convention):
$$\Delta E = \Delta E_{work} + \Delta E_{heat}$$
[Change in internal energy] = [energy change due to energy transfered as work] + [energy change due to energy transfered as heat]
Then give a simpler symbol for amount of energy transferred as work: $\Delta E_{work} \equiv W$ and likewise for heat.


text:
Whats a good solvent to analyze amygdalin in FTIR?
What solvents would be good to analyze amygdalin using FTIR spectroscopy? Will carbon tetrachloride react with the amygdalin if used as a solvent?

KBr disc is the standard method. Why complicate it with solvents/liquid phase spectra? It is a sugar, will it dissolve in carbon tetrachloride? Think about it.

If you aim for a quick substance identification, why not opt-in for ATR-FTIR to record the data of the neat substance? Then, there is no need to identify a suitable solvent. Then, the solvent does not influence the position of the absorption bands and there aren't absorption bands of the solvent to account and correct for. But you should apply (and note in the report) a correction for the wavelength dependent dispersion/penetration depth between your window (Si, Ge, diamond) and sample -- check the spectrometer software for this.


text:
Why are electrons paired up in hexaaquacobalt(III)?
I know water is in the middle of the spectrochemical series generally being the differentiator between the weak field and strong field ligands.
I have generally seen water being a weak field ligand, but in the case of hexaaquacobalt(II), it pairs up the electrons of $\ce{Co^3+}$ in the complex $\ce{[Co(H2O)6]^3+}.$ Why is water acting as a strong field ligand here?

Cobalt(II) or (III)? The question seems to mix them up.

If it's Co(III) https://chemistry.stackexchange.com/questions/112675/is-conh%e2%82%83%e2%82%84cl%e2%82%82cl-paramagnetic-or-diamagnetic/115285#115285 gives the answer, and explain why almost all Co3+, and in fact the vast majority of d6, complexes are low spin

Whether a complex is high or low spin does not solely depend on the ligands attached. There are other factors such as the size of the metal, the charge on the metal, the coordination of ligands, and so on, that can determine whether a complex exhibits high or low spin in Crystal Field Theory.
In this case, while $\ce{H2O}$ is lower on the spectrochemical series, that doesn't imply that the complex should have a low splitting and thus be high spin.
For example, $\ce{[Co(H2O)6]^{2+}}$ tends to adopt a high spin configuration. In fact, $\ce{[Co(H2O)_{6-n}(NH3)_n]^2+}$; $n=0-6$ was shown to favour high-spin configurations through DFT calculations even though $\ce{NH3}$ is considered a stronger ligand than water [1].
Therefore, $\ce{[Co(H2O)6]^3+}$ being high spin and diamagnetic doesn't mean that the ligand $\ce{H2O}$ is a strong field ligand. It just implies that all factors considered, the complex is more stable in a low-spin configuration.
Sources:
[1] https://pubs.acs.org/doi/abs/10.1021/ic0257930


text:
Fermi Contact Interactions and resultant coupling contants
I am currently learning about spin-spin coupling mechanisms and have become very confused about the mechanisms by which spin-spin coupling happens in NMR.
I understand how we can draw diagrams where the Fermi contact, Pauli interaction and Hund interactions can give us an image of favourable spin states in a spin system. However, I find it challenging to use these diagrams to predict the energies of spin states.
From such diagrams, how can we estimate whether the scalar coupling is positive or negative?
For example, for a 1H-12C-12C-1H system or a 19F-12C-1H system.
Example diagram:


text:
Apparent circularity of the definition of mass
I am not a chemistry student, so forgive me for the naïvety of the question. The atomic mass of an atom is defined relatively to the mass of 1/12 of the mass of an unbound neutral atom of carbon 12 in its nuclear ground state and at rest.
But here it seems that the mass of an atom is defined in terms of the mass of carbon itself! How then, is the mass of carbon 12 calculated? What even is mass? It seems obviously circular! If one says that the mass of carbon is calculated by adding the mass of a proton or some other quantity, how is that mass then defined?
A simple layman answer would be appreciated.

Yes, you are right. The atomic masses are somewhat circular with reference to carbon-12. It has a 200 year long history. We always need a reference to assign a number, right? These atomic masses are relative atomic masses. Mass itself is another story; but first read the Wikipedia entry on mass for starting. Mass, like charge is a property of matter.

Note that what you're describing is the atomic mass, not the mass. And the definition of atomic mass isn't circular, because the atomic mass of carbon-12 is defined to be exactly 12. Relating the atomic mass to an actual mass (in kg) would be a different thing altogether—this has to be physically measured, see e.g. https://en.wikipedia.org/wiki/Dalton_(unit)

…and if you are asking about mass as physical quantity on general, note that kilogram is defined via fixed numerical value of the Planck constant and is not bound to carbon-12.

This is not so different from the historic definition of mass of macroscopic objects.  You would have a set of weights kept by some town official, and these would define certain masses. During market in that town, those weights are the official ones. Merchants might have their own set of weights to sell their potatoes, but if there is conflict, you use the official weights.
Of course, a different town might have a different set of weights, making things a little bit fuzzy.
The good thing about using an atom as official standard is that the mass of a given isotope, different from a potato or a man-made weighing standard, is the same in every village.

The mass of the carbon atom is not calculated. It is measured. Let me explain the principle of this operation.
To measure the mass of the carbon atom, a carbon sample has to be heated in a small box (as big as a match box) to such a high temperature that carbon is in a gaseous state. The whole is in a vacuum. Now there are two needles going one through the ceiling and the second through the floor of the box. The needles do not touch one another. An electric spark is then created between the needles. The spark is made of electrons getting out of one needle and jumping to the second needle. During their flight, these electrons are hitting some carbon atoms inside the box. This collision removes one electron from the hitted carbon atom, producing carbon ions $\ce{C^+}$.
Now the opposite (vertical) walls of the "matchbox" are made of metallic plates, one being charged positively and one negatively. So the positively charged carbon ions are moving in the direction of the negative plate. Most fall on this plate. But the plate is punched by a circular hole in its middle. So a small proportion of the $\ce{C^+}$ ions are just directed to the middle of the whole hole. They don't know to which part of the hole they have to go. Instead they just cross the center of the hole, specially if there is another plate parallel to the punched one, but being charged at a higher negative potential than the punched plate. This second plate attracts the flying ions stronger than the punched one. So the ions go through.
You may guess that the second plate is also punched in its middle just in front of the first hole. This produces a narrow jet of $\ce{C^+}$ ions, which are attracted by a third plate. And between the 2nd and the 3rd plates, an electric or a magnetic field is created that deflects the jet. But light ions are more deflected than heavier ones. So measuring the deflection allows you to calculate the mass of the ion, and of course the mass of the Carbon atom, which is defined as $12.000$.
The nature of the gas included in the original "matchbox" can be changed, and produces other ions. If by chance, another ion is deviated twice as much as the $\ce{C-12}$ atom, its mass will be defined as $6.000$. As a consequence, the mass of any present ion can be determined by this apparatus.

Are you describing a mass spectrometer?

Yes. it is a mass spectrometer. I have not told it, because the author of the question said that he is not a chemistry student.


text:
Trouble understanding the physics of a voltaic cell
I recently watched a few videos on the electrochemistry of voltaic cells on youtube and then after watching a video of a physical demonstration of a voltaic cell in action, I realized the videos were insufficient in describing all the processes involved in a voltaic cell.  Specifically, I noticed in the demonstration video that no current was being registered on the ammeter until the salt-bridge was connected.  This led me to reformulate a model of how the voltaic cell works.
My current model of how I understand a voltaic cell:

A solid mass of zinc is placed into a glass filled with a liquid solution of zinc(2+) ions and sulfate SO4(2-) ions.  Additionally, a solid mass of copper is placed into a separate glass filled with a liquid solution of copper(2+) ions and sulfate SO4(2-) ions.
Since there is physical connection between the two masses, no electrons can pass between them.
Before connecting a wire, the mobile electrons are moving randomly within the masses.  After connecting a wire solely to the zinc mass, the electrons will begin to move into the wire and a potential gradient will form along the length of the wire and begin causing electrons to drift down the wire.
The transfer of electrons from the zinc mass to the wire will result in the creation of zinc ions.  The zinc ions will detach from the main zinc mass and move into the zinc sulfate solution increasing the positive charge of the liquid.
After connecting the other end of the wire to the copper, the gradient will begin moving the electrons at the end of the wire into the copper.  Here, the electrons bond to the copper(2+) ions in the solution and the ions attach themselves to the zinc mass.  This results in an increase in negativity of the copper sulfate solution.
If there is no salt bridge, then the zinc sulfate solution will become positive and attract the electrons of the zinc mass which eliminates the electric potential gradient within the wire which halts the net flow of electrons from the zinc mass through the wire into the copper mass.
By adding a salt bridge, the net positive charge in the zinc sulfate solution becomes reduced due to the inclusion of the negative chlorine ions and the net negative charge in the copper sulfate solution becomes reduced due to the inclusion of positive sodium ions.
The salt bridge therefore maintains the electrical neutrality of the two solutions which allows the zinc-wire-copper gradient to exist uninterrupted.

My question is:

Is this model correct?
What exactly is the timescale on this?  If the salt bridge is removed, are the zinc ions pretty rapidly being removed from the zinc mass and overwhelming the zinc sulfate solution with positive charges?  It must be a very quick process and the amount of zinc ions necessary to disturb the neutral net charge of the zinc sulfate solution must be invisible to the eye if this is the case.

Have you tried other sources than videos?  As this is very frequent topic,explained in textbooks and online resources in many ways and places.

See also CH SE META post about policy for "Am I right?" questions

Propagation of electrostatic force due charge disbalance is extremely fast and the force proportionality factor to the net charge is enormous. This is not specific to electrochemistry.

Your model is nearly correct. I would modify it this way.
When a zinc plate is dipped into a $\ce{ZnSO4}$ solution, one or two $\ce{Zn}$ atoms are giving their electrons into the metal, and these atoms become $\ce{Zn^{2+}}$ ions which pass into solution. But this phenomena lasts less than one nanosecond. It stops because the $\ce{ZnSO4}$ solution becomes then positively charged, and this positive charge prevents new positive $\ce{Zn^{2+}}$ ions from entering the solution.
A similar phenomena occurs on the copper plate when dipped in a $\ce{CuSO4}$ solution. As soon as the copper plate is dipped into such a solution, one or two $\ce{Cu^{2+}}$ ions are taking electrons from the copper plate, to become $\ce{Cu}$ atoms. But this phenomena lasts less than one nanosecond. It stops because the corresponding $\ce{SO4^{2-}}$ ions have not been modified (discharged) and remain in solution: As a consequence the solution becomes negatively charged. Ions from such a negative solution cannot pick up new electrons from the copper plate.
So if now the two electrodes are connected with a wire connecting both plates, nothing new will happen.
But, if now a $\ce{NaCl}$ salt bridge is installed connecting both solutions, this creates a voltaic cell which will work for two reasons : First the positively charged solution around the zinc plate attracts the negative ions from the bridge and around the copper plate. Second, the negatively charged solution around the copper plate attacks the positive ions from the bridge and from around the zinc solution. The electric neutrality of all parts of the voltaic cell is maintained. So it works.


text:
Conductance during titration
Why does the conductance of a solution changes during titration. And what is the relation of conductance of a solution with titration?(for both acid base and redox titration)

Here is some information for acid/base titrations. You can improve your question by editing it, incorporating what you know and what you need help with. This paper also discusses redox titrations, but it is behind a paywall. However, once you understand the acid/base titrations, you should be able to figure out which redox titrations might work well.

I believe this question is missing context about what really is the misunderstanding. What are you confused about?  Do you understand why solutions are conductive in the first place?  Or is it applying this knowledge to titrations?  I think some clarity about what you know and where you’re finding trouble connecting ideas would help us write a good answer.

Let's take an example. A solution of hydrochloric acid contains two ions : $\ce{H3O+}$ and $\ce{Cl-}$. The ion $\ce{H3O+}$ has a specially high conductivity at infinite dilution ($\ce{349.8 cm^2ohm^{-1}mol^{-1}}$), about $5$ times as high as the conductivity of $\ce{Cl- (76.4cm^2ohm^{-1}mol^{-1}}$). In a titration with $\ce{NaOH}$, the ion $\ce{H3O+}$ is replaced by the ion $\ce{Na+}$ whose conductivity is low, even lower than $\ce{Cl-}$ : the conductivity of $\ce{Na+}$ is $\ce{50.1 cm^2ohm^{-1}mol^{-1}}$.
So the $\ce{HCl}$ solution has a high conductance, which is mainly due to its "quick" ion $\ce{H3O+}$. If this ion is replaced by the "slow" $\ce{Na+}$, the conductance of the solution decreases.

Before you post and ask you need to look information on internet because there is a lot of information about this kind of titration.
You must think about what produce the conductivity and what are the changes during redox reactions. In this case theres a change in the free ions and their charges that form the analyte. When you add the titrant you are reacting it with the analyte and makes a stable compund with low conductivity. You assume that the coplete reaction as been done when you reach the equivalent point of conductance. If you still adding titrant you will see an increase of conductivity because of the free ions of this compound.
So when you are looking for the conductance you are looking for the point that make the system stable as you add the reactant that it will be minimum at the equivalent point that means that the complete reaction has been done.
https://www.vedantu.com/chemistry/conductometric-titration
https://www.britannica.com/science/titration


text:
alternative reaction mechanism for acetal formation from hemiacetal
Most textbook would depict the reaction as following:

(from Organic Chemistry by Craig B. Fryhle, Scott A. Snyder, T. W. Graham Solomons)
But why wouldn't the second step be like this:

$\ce{^-OH}$ is a very strong base and therefore a bad leaving group.

the leaving group here is the protonated form of OH, which is a good leaving group?

Sorry, I misrepresented your drawing. Carbocations are generally less stable than oxonium, especially if the they have a low number of substituents bonded to them. In your proposed mechanism, the carbocation is only mono-substituted, which is very unstable and would lead to a higher activation energy.

Well yes, it could be. But I think it is oversimplistic to suggest that the electron rich oxygen right beside it plays no part.

$\ce{O-R}$ groups are electron-withdrawing, so it would just make the carbocation even less stable by making it even "more positive". I don't see a point in bringing up resonance structures considering they're non-equivalent in this case, and the major contributor (oxonium) is significantly more stable than the minor contributor (carbocation) due to octet completion and other factors mentioned.

In fact, the mechanism you've drawn is entirely the same as the textbook mechanism (apart from the fact that your final product is missing an OR group, which I assume is an unintentional error).
The reason is because $\ce{C+-O}$ is entirely equivalent to $\ce{C=O+}$; they are two resonance forms of the same thing:

much like the resonance structures of benzene:

and you certainly wouldn't say that one resonance form of benzene was different from the other.
However, there is one difference: the $\ce{C=O+}$ resonance form is more accurate than the $\ce{C+-O}$ resonance form. That's because it is a larger contributor to the resonance hybrid (which in turn can be rationalised using many rules—in this case, the octet rule is relevant).

I think octet explanation would be better for OP to understand why oxonium ion ion is governing.


text:
Detecting epoxy in a polyester powder coat paint
I have 2 steel parts which were both powder coated in a polyester powder coat, however one of them does not perform exactly as I would expect it in my unusual use case which depends on temperature. I suspect this is because the well-performing one contains epoxy - but I am not sure. I would like to sort of narrow down my options for ordering new parts.
Both parts look virtually identical, and unfortunately I only have 1 of the well performing part, which I also need to use constantly.
Is there a reasonably accessible (common household/industrial chemicals or tools) way to detect if either of the powder coats contains epoxy, without it being triggered by polyester, either chemical or otherwise?
From what I gathered, acetone would dissolve both of them, so that likely won't yield any results without some detailed analysis.

Can you scrap a little the one that is working better? The idea is to take a sample and run some analysis. The most simple one will be an FTIR. Here is an example.

You should mix both in some acidified ethanol mixture. Acid can be muriatic acid. Water and dilute acid will hydrolyze epoxy to a diol but won't be able to break ester.
This diol when reacted with periodic acid will release a ketone which you can then apply a tollen's test for.


text:
Which heat pad content materials will keep me warm the longest?
I am trying to find a good heating pad/bag for myself that keeps me warm for as long as possible (nonelectrical, heated once, then used until it gets cold). The classic is a hot water bottle, and according to wikipedia

Liquid water has one of the highest specific heat capacities among common substances, about 4184 J⋅kg−1⋅K−1 at 20 °C

this is already good compared to other common materials.
Throught the internet, other materials for heating bags/pad are sold. Among them:

watery gel (medical heat/cold pads)
different types of grains/seeds (rice, wheat, buckwheat, peas, cherry stones)
sand
moor mud

And I often find opinions about them, like

heat pads are better than water bottles
grain/seed pillows are better than water bottles
moor mud pillows keep warm the longest
and so on. But these are bare statments, I could not find explanations of any of these claims. So from the little I know about physics I thought it might have to do with specific heat capacity of these materials. But those cannot be found anywhere. Of course they will not have ONE value; all listed materials are material mixes so they would have a range, but I could not find any range either.

Can anyone provide such values/ranges?
Or are there other things to consider beside heat capacity?

Not all the warming is based on heat capacity, some is based on active release of heat, activated on demand. // For the former, not only capacity, but also shape of contact surface, initial surface temperature and rate of heat provision(heat conductivity) are important.

You state, "nonelectrical, heated once, then used until it gets cold;" does that include phase change heat sources (\e.g., sodium acetate)? Does it exclude chemical reactions, such as rusting?

@DrMoishePippik I am not trying to create my own heat pad, I merely want to buy one that one can buy on the market. I am not aware of any type of reusable heating pad that includes phase changes or chemical reactions.

@Kjara, standard operating procedure for winter camping: put a hand warmer in the sleeping bag before retiring, e.g., https://www.amazon.com/Hand-Warmers-90-Count-Activated/dp/B09P4M8DC2/, which relies on rusting, or https://www.amazon.com/Instant-Reusable-Heat-Camping-Wave/dp/B000E4C5DC/ , using phase change, (nor recommendations, just examples)


text:
Is the periodic table ordered by bonded atoms and does that change the electron configurations and ordering?
In my book (Mortimer, The Basic Knowledge of Chemistry) the electron structure of the elements is introduced, there the relative energies of the atomic orbitals are shown:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s... (Tab. 6.22)
Then the aufbau principle is explained and on a whole page each period with its exceptions is gone through with the electron configurations.
Up to that point, everything was understandable, but the last paragraph now says the following:

The electron configurations mentioned in the preceding paragraphs and in Table 6.6 are based on measurements of unbound single atoms in the gas phase. For the periodic table, however, the classification according to chemical properties is decisive, i.e. of bonded atoms. For bonded atoms of the d- and f-block (not the s-block), the energetic sequence of the orbitals is somewhat different from that shown in Fig. 6.22, so that the d-levels come just below the following s-levels (see marginal columns). This does not affect the order of the elements in the periodic table, but it does affect the chemical properties.

Is the periodic system ordered with the properties as if they were bound atoms? Do these bound atoms also have a different electron configuration or still the same as unbound, but the energy sequences are now shifted (do the electrons change their orbital as soon as the energy sequence changes) and why doesn't that affect the order of the elements in the periodic table?
And why can d-block elements still be classified as d-block when energetic sequence changes?

Have a look at some other versions of the periodic table: last I check, there are more than 100! This tells you that ‘one size doesn’t fit all’. Put another way: there are horses for courses! My favorite is this wonderful 1949 periodic table by John Drury Clark: https://chemistry.stackexchange.com/a/128563/79678.

Is the periodic system ordered with the properties as if they were
bound atoms?

I think it will help you understand the context better if you read a little about the history of the periodic table. The repeating pattern of the elemental properties was recognized long before electrons were discovered. People had  a very little understanding of the structure of atom then (electron, neutron and protons) and we still do not know enough. It is called a periodic table because the elemental and chemical properties in the bulk repeat regularly after certain periods (similar to periodic mathematical functions). Thus, the periodic table has nothing to do with electron configuration at this moment.
Much later, when people started to learn about how hydrogen, helium, and other lighter atoms, such as Li/Na, emitted light in the gas phase, they realized that the electrons in a given atom must have different bound state energies with the nucleus. This is what Mortimer is discussing unbound single atoms in the gas phase. You simply heat or electrically excite a given element such that the given element is vaporized into gaseous atoms. For example, $\ce{H2}$ will dissociate into hydrogen atoms and emit light in an electric discharge tube.
All electron configurations were based on extensive calculations of the hydrogen atom. A theory that closely matched the experimental results of the hydrogen spectrum was retained, and the previous theory was discarded. Based on the possible electron configuration of hydrogen and helium, researchers extended these ideas to other heavier elements, too much of an extrapolation. Therefore, for heavier elements, consider the concept of electron configurations with a grain of salt, i.e., do not worry about it too much until and unless you wish to seriously pursue theoretical chemistry or physics. Otherwise, there is a lot of hand-waving esp. in general chemistry textbooks.

Do these bound atoms also have a different electron configuration

Yes, in a sense they are called molecular orbitals instead of atomic orbitals.

The periodic table has been and still is not directly related to the electronic structure of the elements. It is obtained by alining the elements by order of increasing atomic masses, then cutting this long series into groups just after each noble gas, and superimposing the obtained groups of elements. This is all. As the groups do not have the same number of elements, some further breaks have been done to take care of this fact. But the electronic structure does not intervene in the definition of the table.

So is the block division more of an approximation? The division into S, P, D, F blocks only works for unbound atoms (because if the energy sequence of bound elements changes and, for example, 3d goes below 4s, some elements would also have to change into another block because the last added electron would go in the other subshell)?


text:
Is there a formal definition of “identical” molecules?
I was reading about atropisomers and their impacts on drug design, and I started wondering whether we have a formal definition for identical molecules. Atropisomers are not the same molecules because they can’t easily interconvert, so I was thinking a definition of “identical” might involve conformations, as it seems like conformers are chemically different (for example, only certain (cyclo)alkyl bromide conformers can undergo elimination via an E2 pathway).
One definition that I think could work would be that two molecules are identical if they can quickly interconvert between the same set of conformers (though I imagine the timescale of the interconversion necessary for the molecules to be identical depends on the purpose of the molecule). Two conformers would be considered the same if they are geometrically identical and could be superimposed upon each other. Would a definition like this differentiate all isomers, or would it classify some isomers as identical? Is there an accepted definition of identical? Is it something that is defined based on the context of the situation?

related https://chemistry.stackexchange.com/questions/89909/atropisomerism-of-naphthyl-alcohol

Typical cut-off for atropoisomers is 1000 seconds, afaik. https://en.wikipedia.org/wiki/Atropisomer

Following up on Mithoron's comment, atropisomers are "identical" in that you don't have to worry about the conformational energy barrier when you interconvert them on paper or in your mind. However, in practice, a sample containing atropisomers will contain molecules with different physical properties relevant for instance when interacting with a biomolecule binding site. It's all about the timescale.


text:
How to calculate the solubility for the hydrogen concentration based on the pressure and temperature in the Metal Hydride Hydrogen compressors?
My name is Ketan Karkare and I am a student in the University. I am currently working on the metal hydride compression and have gone through the following paper: Mathematical modeling for the PCT curve and have DOI uploaded below:
DOI: https://doi.org/10.1016/0360-3199(94)90097-3
I was trying to work on the derivation for the hydrogen composition based on pressure and temperature which is the equation (1) and also given below:
$C_{H}=AP^{\gamma/2}\exp{\frac{-\gamma V_{H} P}{RT}}\exp{\frac{\Delta H_{s}}{RT}}$...… (1)
Can someone please explain the derivation or share the document which has the derivation for this formula

The paper cites "Z. Q. Zhou, F. Feng, J. L. Zhang and J. S. Ge, Proc. Syrup. on Hydrogen Storage Materials and their Applications, p. 139. Tianjin, China" as the source for the equation. I'm afraid you might need to find the original text. As for the question, I'd suggest to include definitions for all non-standard variables (just don't post screenshot from a paper, please).

I could be wrong, but it seems the exponential functions might have been derived from the Van't Hoff integrated equation: $ln(P)=\frac{\Delta H}{RT}-\frac{\Delta S}{R}$

Hello Everyone! Thank you for your reply , @andselik I am trying to find the following book as you stated but not able to get the original book as well.


text:
Are the physical and chemical properties of a molecule an average of the properties of its conformers?
Conformers sometimes differ in chemical reactivity. For example, only certain alkyl bromide conformers have the appropriate geometry to undergo elimination via an E2 mechanism. I imagine that conformers differ in physical properties as well, for example I think they might have different dipole moments, differing abilities to form hydrogen bonds, and differing abilities to pack together (if any of these are not true please let me know), leading to different melting and boiling points.
If a molecule can rapidly convert between different conformers, are the properties of the molecule that we observe a weighted average of the properties of the conformers, with the most stable conformers contributing the most to the properties? Could there be some kind of “synergistic” effect, where, for example, two different conformers are able to hydrogen bond well with each other, leading to a lower overall boiling point than the boiling point of any individual conformer? If this synergy is present, could it lead to conformers that we would expect to be high in energy based on steric arguments actually being lower in energy than we expect due to interactions with other conformers, and due to this lowered energy, the molecules might become this conformer more frequently?

That would be yes, yes and yes and... I dunno who many questions you managed to pack in there, but your analysis seems correct. Read a bit about proteins if you wanna go down this way. For example prions, or perhaps binding of intrinsically disordered proteins.

How different do the properties of individual conformers tend to be?

It depends? And what properties? Some things can be hydrophobic in non-polar solvent and vice versa, for example.

I am thinking of properties like dipole moment, melting point, and boiling point.

See, the thing you seem to not get is difference between stuff you can get in different forms and what you can't. If you can it's not called conformers. That's pretty much why atropoisomerism was "invented".

That's why proteins can get chemists' pants in a twist ;) They defy what I said in earlier comment

Is that because a folded and unfolded protein can be viewed as different conformers of the same molecule?

Let us continue this discussion in chat.

Conformers sometimes differ in chemical reactivity. For example, only certain alkyl bromide conformers have the appropriate geometry to undergo elimination via an E2 mechanism.

Studying a reaction might suggest something about the distribution of conformers, as you point out. By investigating the reaction rate under different conditions and proposing an appropriate mechanism you can attempt to deduce the fraction of reactive molecules in the total ensemble at any given moment, and how quickly they interconvert to generate more reactive molecules.

If a molecule can rapidly convert between different conformers, are the properties of the molecule that we observe a weighted average of the properties of the conformers, with the most stable conformers contributing the most to the properties?

The main tool chemists use to study conformers, their properties and their distribution is spectroscopy, typically the interaction of molecules with different types of electromagnetic radiation including scattering.  Separation techniques can also be useful. What you observe in your sample depends on the timescale of the experiment. If the conformers interconvert more rapidly than that timescale you observe a single average. Otherwise you might observe a superposition of the properties of different conformers, as if you were performing separate but simultaneous experiments on these conformers. Intermediate situations occur too.

Could there be some kind of “synergistic” effect, where, for example, two different conformers are able to hydrogen bond well with each other, leading to a lower overall boiling point than the boiling point of any individual conformer?

A (well-defined) boiling point is by definition an ensemble property of a large homogeneous sample of molecules in a liquid state. So the question does not make sense.

If this synergy is present, could it lead to conformers that we would expect to be high in energy based on steric arguments actually being lower in energy than we expect due to interactions with other conformers, and due to this lowered energy, the molecules might become this conformer more frequently?

Intermolecular interactions can play a very important role in determining preferred conformations. This can be the case whether those interactions are with other molecules or the same type.


text:
What happens when iron(III) chloride is dissolved in ethanol?
I dissolved $\pu{25 mmol}$ $\ce{FeCl3*6H2O}$ in $\pu{25 ml}$ ethanol and got a clear ethanol solution of $\ce{FeCl3}$ 4 days ago. And I keep it in a drawer from light these four days.
However, when I opened my drawer today, the solution is not clear anymore.
And it does not recover to clear after ultrasonic treatment.
It seems that there are some yellow suspended drops/particles in the solution. I wonder what they are and what happened to my solution of ferric chloride.

The insoluble stuff in suspension is probably a precipitate of $\ce{Fe(OH)3}$ produced by the hydrolysis of $\ce{FeCl3}$ according to $$\ce{FeCl3 + 3 H2O -> Fe(OH)3 + 3 HCl}$$ For some unclear reason, this reaction is more important in ethanol than in water.

Well, it's important in water, too. The solution should have been acidified to prevent that.

You also dissolved FeCl3 in a mix of ethanol and water [the water of hydration plus any water in the ethanol]. What ions are being formed, however slowly, and what is not soluble involving those ions? It probably is not a clean reaction.

See https://www.degruyter.com/document/doi/10.1515/zna-1983-0614/pdf on disproportionation.


text:
Notation for spectral density in experimental community
In an open quantum system setup where the system is coupled to infinite harmonic oscillators as bath (as in the Caldeira leggett model, for example), through the Hamiltonian,
$$H = \frac{\hat{p}^2}{2m} + V(\hat{x}) + \sum_n \frac{\hat{p}^2_n}{2m_n}+ \frac{1}{2}m_n \omega_n ^2 \hat{x}^2 - \hat{x} \sum_n g_n \hat{x}_n$$
Here, $\hat{p}, \hat{x}, \hat{p_n}, \hat{x_n}$ are, respectively, system momentum and position operator and nth bath oscillator momentum and position operator and $g_n$ is the coupling strength between the system and the nth bath oscillator.
One then defines the spectral density as following,
$$j(w)=\frac{1}{m}\sum_i \frac{g_i ^2}{\omega_i ^2}\delta(\omega-\omega_i)$$
The dimension of $j(\omega)$ can be calculated to be $\frac{mass}{time}$.
But in the same paper as linked above (in the supplementary info), the spectral density is later expressed in terms of ${time}^{-1}$. (please refer to the following image). What happened to the mass dimension? Are they now calling $\frac{j(\omega)}{m}$ spectral density?

On the other hand, the following paper expresses the weighted spectral density $\rho(\omega)\omega^2$ as $cm^{-1}$. (please refer to the following image).It appears that in the notation of this paper, $\rho(\omega)$ is the spectral density. If so, then the correct dimension of weighted spectral density should be $\frac{mass}{{time}^3}$. How is it $cm^{-1}$ instead?

I know that sometimes, in experimental literature, frequency is expressed as wavenumber and one needs to multiply the term with speed of light expressed in cms ($c=3 \times 10^{10}$)in order to get the frequency. But in the second paper I have linked, one would need to multiply the expression with $c^3$ instead, because the original expression is in terms of $\frac{mass}{{time}^3}$.

I believe the main problem is the pigeon hole effect with this terminology. Too many people use the term "spectral density" and it is a different version for each in each paper. It is a loosely defined term in my opinion.

You will literally have to follow each equation in a given paper and try to recreate their figure if possible.


text:
Is the formation of starch from glucose in plants an endothermic or exothermic reaction?
If I understand it correctly photosynthesis is an endothermic reaction.
Energy from the sun is used to form glucose.
Next the glucose is turned into starch.
Would this be a combination reaction?
If so, is that then an exothermic reaction?

Consider also energy outcome of the whole biochemical enzymatic process of starch synthesis, that may have different energy outcome than the formal direct starch synthesis from glucose.

Related: Why are condensation reactions endergonic?

If you want to see what drives this low energy reaction, you have to consider that e.g. the released water is likely to partially evaporate. And the starch crystallises, i.e. removes itself from the reaction equillibrium.

The formation of 1,4-alpha glycosidic bond between two glucose monomers in starch, and more generally of glycosidic bonds between monosaccharide monomers to form polysaccharides, involves a dehydration reaction, a type of condensation reaction, but is not a combination reaction since two products form.
A condensation step in the polymerization may be written schematically as
$$ \ce{-(C5H10O4)C(H)-\color{red}{OH} + \color{red}{H}-OC(H)(C5H10O5) -> -(C5H10O4)C(H)-O-C(H)(C5H10O5) + \color{red}{H2O}}$$
Note what happens:

one C-O bond is cleaved (dehydration)
one OH bond is cleaved (dehydration)
one C-O bond forms (the bridging glycosidic ether bond)
one OH bond forms (forming water)

So no net bonds are formed.
You can estimate the enthalpy from the difference in bond dissociation energies $^1$ for OH in water and in methanol, 463.5 kJ/mol versus 437.6 kJ/mol, respectively, and assuming the CO bonds have similar energies. The difference is -25.9 kJ/mol, so slightly exothermic.
You can also estimate the enthalpy change from heats of combustion$^2$ for glucose and maltose (or starch of course), ie the following reactions:
$ \ce{2 C6H12O6(s) + 12 O2(g) -> 12 CO2(g) + 12 H2O(l)} ~~~ \Delta_\textrm{c} H = -2 \times 3138.0 \pu{kJ/mol}\\ \ce{C12H22O11(s) + 12 O2(g) -> 12 CO2(g) + 11 H2O(l)} ~~~ \Delta_\textrm{c} H = -5655.7 \pu{kJ/mol}$
The heat of formation of maltose from glucose is equal to the difference between the heats of combustion in the previous reactions, or -620.2 kJ/mol. You can estimate a similar value for starch. However this number is computed for formation of one solid from the other, not for addition of a monomer to starch in solution (the initial and final states matter).
The two estimates are very different but then again, the real hydrolysis or formation of such a glycosidic bond in starch is bound to be more complicated$^3$.
Note the drawn reaction is the net process and that as is typical of biochemical processes, the actual reaction involves various enzymes, here that activate the glucose monomer by converting it into glucose-ADP by reaction with ATP, and then facilitate encounter of the activated monomer with the growing starch polymer. You can read more about the definition of individual terms elsewhere, such as in the Wikipedia.
References

CRC handbook of Chemistry and Physics, 85th ed., D. R. Lide, ed., CRC press, Boca Raton, FL, 2004, p 9-65.

Sánchez-Peña, M. J.; Martínez-Navarro, A. G.; Márquez-Sandoval, F.; Gutiérrez-Pulido, H.; Pacheco-Moisés, F. P.; González-Ortiz, L. J. Heats of Combustion of the Main Carbohydrates Contained in Plant-Source Foods. Nutrition Reviews 2020, 78 (5), 382–393. https://doi.org/10.1093/nutrit/nuz063.

(1) Sørensen, T. H.; Cruys-Bagger, N.; Borch, K.; Westh, P. Free Energy Diagram for the Heterogeneous Enzymatic Hydrolysis of Glycosidic Bonds in Cellulose. Journal of Biological Chemistry 2015, 290 (36), 22203–22211. https://doi.org/10.1074/jbc.M115.659656.

Only one new O-H bond is formed, since the OH comes off the substrate intact so it's just HO + H, so I think your calculation is not valid

There is no difference in number of bonds. There's only slight difference between bond strengths and an entropic effect. Both aren't all that obvious. Whatever they are, polymers like this tend to hydrolyse in acids, so overall the polymerisation isn't particularly favourable.

@Andrew  Yes, what a miss. I try again.

@Mithoron Are you saying the equations are unbalanced (they are not), or that the heat of reaction glucose->maltose is not equal simply to the difference in heats of combustion of those two reactions?

This -620.2 kJ/mol way to high, and I was wondering why. Difference between solid and aq is not enough, but I don't think I found the reason yet.


text:
Calculate the concentration of HF in aqueous solution of NH4F and NH4HF2 mixture
There is a similar question on this site, but here i have a mixture.
I have a mixture of NH4F (35% weight) and NH4HF2 (8% weight) in water. How do I calculate the weight % expressed in HF? or the molar concentration of HF?
I arrived to calculate the mol/L of each component of the solution, being 10.49M of NH4F and 1.64M of NH4HF2. Now how do I calculate the HF conc?
Thank you

As usually in all but standard scenarios with ready to use expressions:  Formulate all equilibrium equations, amount inventory equations and charge balance equation. Use substitution method to solve the equation set and  to express desired concentration as the function of the known. Use simplifying approximations based on strong inequalities wherever applicable.

Not that at such high concentrations, all calculations are off by large extent. And I do mean large.

No problem about the accuracy of the calculation, i just need an idea of the weith or volume ratio expressend in NH4F:HF:H2O

@Ale Are the concentrations given initial or final?

Concentrations given are initial ones

The initial equilibrium systems that lead to formation of $\ce{HF}$ as an intermediate are:
$$
\require{cancel}
\begin{align}
\ce{NH4HF2 (aq)&<=>\cancel{\ce{NH4+(aq)}}+\cancel{\ce{F- (aq)}}+HF (aq)} \tag{$K_1$} \\
\ce{\cancel{\ce{NH4+(aq)}}+\cancel{\ce{F- (aq)}} &<=> NH4F(aq)} \tag{$K_2$} \\
\hline
\ce{NH4HF2(aq) &<=> NH4F(aq) + HF(aq)} \tag{$K_3$}
\end{align}
$$
Let:
$A$ represent $\ce{NH4HF2}$
$B$ represent $\ce{NH4F}$
$C$ represent $\ce{HF}$
$D$ represent $\ce{H+}$
$E$ represent $\ce{F-}$
$W$ represent $\ce{H2O}$
At equilibrium, the concentrations of $A,B,C$ are:
$$C_{A_1}=C_{A_o}-x$$
$$C_{B_1}=C_{B_o}+x$$
$$C_{C_1}=x$$
Using the relationship between the overall equilibrium constant $K_3$ and the individual ones, $K_1$ and $K_2$:
$$K_3=K_1\;K_2=\frac{C_{B_1}\;C_{C_1}}{C_{A_1}}=\frac{(C_{B_o}+x)\;x}{C_{A_o}-x}$$
$\ce{HF}$ dissociates partially in aqueous solution:
$$\ce{HF (aq)<=>H+(aq) + F-(aq)}$$
The concentrations of $C,D,E$ after dissociation are:
$$C_{C_2}=C_{C_1}-y=x-y$$
$$C_{D_2}=y$$
$$C_{E_2}=y$$
Using relationship between equilibrium concentrations and $K_a$:
$$K_{a}=\frac{C_{D_2}\;C_{E_2}}{C_{C_2}}=\frac{y^2}{x-y}$$
We can calculate $K_1$ and $K_2$ using the solubilities of $A$ and $B$ at 25°C:
$$K_1=\left(\frac{S_A}{M_A}\right)^2=\left(\frac{\pu{630g/L}}{\pu{57g/mol}}\right)^2=122.2$$
$$K_2=\left(\frac{S_B}{M_B}\right)^{-2}=\left(\frac{\pu{835g/L}}{\pu{37g/mol}}\right)^{-2}=\pu{1.97e-3}$$
Then, we calculate $K_3$:
$$K_3=K_1\;K_2=(122.2)\;(\pu{1.97e-3})=0.241$$
Then, we calculate the initial density of the mixture (i.e. when only $A,B,W$ are present):
$$\rho_o=X_{Ao}\;\rho_A+X_{Bo}\;\rho_B+X_{Wo}\;\rho_W=(0.08)(\pu{1500g/L})+(0.35)(\pu{1010g/L})+(0.57)(\pu{1000g/L})$$
$$\rho_o=\pu{1043.5g/L}$$
Next, we calculate $C_{Ao}$ and $C_{Bo}$:
$$C_{Ao}=x_{Ao}\;\frac{\rho_o}{M_A}=0.08\;\frac{\pu{1043.5g/L}}{\pu{57g/mol}}=\pu{1.46mol/L}$$
$$C_{Bo}=x_{Bo}\;\frac{\rho_o}{M_B}=0.35\;\frac{\pu{1043.5g/L}}{\pu{37g/mol}}=\pu{9.87mol/L}$$
Solving for $x$:
$$\frac{(9.87+x)\;x}{1.46-x}=0.241$$
$$x=\pu{0.0347mol/L}$$
The $K_a$ of $\ce{HF}$ at 25°C is approximately:
$$K_a=\pu{6.76e-4}$$
So we can solve for $y$ now:
$$\frac{y^2}{0.0347-y}=\pu{6.76e-4}$$
$$y=\pu{0.00452mol/L}$$
Finally, we can calculate the concentration of $\ce{HF}$ in the final solution:
$$C_{C_2}=x-y=\pu{(0.0347-0.00452)mol/L}=\pu{0.0302mol/L}$$

@Poutnik $K_{sp}$ (in this case, $K_1$ and $K_2$) can be estimated using solubility of $\ce{NH4HF2}$ and $\ce{NH4F}$ in water.

$\ce{NH4HF2(aq)}$ means $\ce{NH4+(aq) + HF2-(aq)}$ //  $\ce{NH4F(aq)}$ means  $\ce{NH4+(aq) + F-(aq)}$

Some sources show $\ce{NH4HF2}$ as $\ce{NH4^+ + HF + F-}$. Check ionic structure.

If there is not enough initial F- to deny dissociation of HF2-, according to the equilibrium. But the solution has plenty of F-. Reaction F- + HF <=> HF2- has K near 4. // Note that very concentrated HF very raises in acidity and pure liquid HF has acidity comparable to sulphuric acid, due 3 HF <=> H2F+ + HF2-. ( +  forming $\ce{H_{n-1}F_{n}-}$  )

I misunderstood one of the moderators. I thought that syntax was for units only. Thanks for the correction.

OK, thank you. What about the weight ratio of HF, NH4F and H2O?

There are these possible equilibrii, but the first 2 do not apply, as there are no solid salts. (5) can be neglected due low enough $\mathrm{pH}$.
\begin{align}
\ce{NH4HF2(s) &<=> NH4+(aq) + HF2-(aq)} \tag{1}\\
\ce{NH4F(s) &<=> NH4+(aq) + F-(aq)}\tag{2}\\
\\
\ce{F-(aq) + HF(aq)  &<=> HF2-(aq)}\tag{3}\\
\ce{HF(aq) &<=> H+(aq) + F-(aq)}\tag{4}\\
\\
\ce{NH4+(aq) &<=> NH3(aq) + H+(aq)}\tag{5}
\end{align}
For the reaction (3), $\log{K} = 0.6$, i.e. $K \approx 3.98=\frac{[\ce{HF2-}]}{[\ce{HF}][\ce{F-}]}$ so very rough estimation of [HA] can be done.
$$[\ce{HF}]=\frac{[\ce{HF2-}]}{K[\ce{F-}]}=\pu{\frac{1.64}{3.98 \times 10.49} mol L-1 HF} \approx \pu{0.04 mol L-1 HF}$$
For the reaction (4), $\mathrm{p}K_\mathrm{a} = 3.17$
$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[F-]}}{\ce{[HF]}}}=3.17 + \log{\frac{10.49}{0.04}} \approx 5.6$
For the molar percentage of $\ce{HF}$, there is needed the solution density. You know it, or can measure it. Then it is trivial calculation, reversed to the ones you used to calculate molar concentrations of the salts.
$$w\%=\frac{c[\pu{mol L-1}]M[\pu{g mol-1}]}{\rho[\pu{g L-1}]} \cdot 100$$

An alternative approach, using Excel solver with Least Squares approach to solve the set of nonlinear equations:

Conc of
variables
values
units
equations

F-
x
10.98
mol/L
Ka = xt/z

HF2-
y
1.32
mol/L
K=y/xz

HF
z
0.030
mol/L
x+2y+z=a+2b

H+
t
0.0000018
mol/L
x+y=a+b

5.7
pH

NH4F
a
10.49
mol/L

NH4HF2
b
1.64
mol/L

Ka
0.000676083

K
3.98

OK, thank you. What about the weight ratio of HF, NH4F and H2O?


text:
Synthetic resin method to remove permanent hardness of water
Why is $\ce{R-SO3H}$ first treated with NaCl and then the RNa form i.e., the sodium form of resin is used for treating hard water? Why cannot we directly treat hard water with $\ce{R-SO3H}$? What is the purpose of NaCl?

It depends if you want to have as counterparts of anions of strong acids Na+ or H+.  For bicarbonates, the latter leads to release of CO2.

The short answer is do you want to drink HCl or NaCl rather?

@AChem can you explain elabrately with reaction please, I'm not getting??

The ionex in H cycle converts dissolved salts to respective acids. The ionex in Na cycle converts salts to respective sodium salts.

@DAYA'SSIMULATION, Can you please read Wikipedia article on ion-exchange, https://en.wikipedia.org/wiki/Ion-exchange_resin. Also, can you write ion-exchange equation in the main question? I want to see if you have understood ion-exchange itself or not?

@Poutnik so we are delebrately preventing HCl formation ,got it thank you!

@AChem actually I don't know the reaction and it is not in my book,only the above paragraph is present, that's why I'm bit confused ,some people say NaCl is treated with R-SO3H to make R-SO3H soluble in water as it's is insoluble in water.

Hard water is a water containing the ions $\ce{Ca^{2+}, HCO3^-, SO4^{2-}}$. When this water crosses a layer made of beads having the formula $\ce{RSO3H}$, the following reaction happens : $$\ce{Ca^{2+} + 2 RSO3H -> (RSO3)2Ca + 2 H+}$$ $\ce{(RSO3)2Ca}$ remains fixed on the resin beads. But the ion $\ce{H+}$ replaces $\ce{Ca^{2+}}$ in solution. So the solution becomes acidic. If the negative ion remaining in solution is $\ce{HCO3^-}$, this is not too serious, because both ions $\ce{H+}$ and $\ce{HCO3^-}$ react and produce $\ce{H2CO3}$ which is quickly decomposed into $\ce{H2O}$ and $\ce{CO2}$. But if the negative ion is $\ce{SO4^{2-}}$, it produces a solution containing sulfuric acid $\ce{H2SO4}$ which is rather corrosive.
This unfortunate result can be avoided if the resin has been previously treated by a solution of $\ce{NaCl}$. Such a solution contains the ions $\ce{Na+}$ which react with the resin according to the reaction $$\ce{RSO3H + Na+ ->RSO3Na  +  H+}$$ Here too, the resin is modified, becoming $\ce{RSO3Na}$, and the $\ce{H+}$ ions are released into the solution. But as this treatment has been done in advance, before adding hard water, these acidic ions can be eliminated from the resin beads by washing. With this new resin beads containing now $\ce{RSO3Na}$, hard water may be used to cross the modified resin layer. In this case, the $\ce{Ca^{2+}}$ ions of the hard water will react with the modified resin according to : $$\ce{Ca^{2+} + 2 RSO3Na -> (RSO3)2Ca + 2 Na+}$$ After this treatment, the resin is once more modified, becoming $\ce{(RSO3)2Ca}$ but the water coming out of the resin layer contains $\ce{Na+}$ ions as only positive ions. The negative ions of the solution have not been changed. They are $\ce{HCO3^-}$ and $\ce{SO4^{2-}}$. The solution getting out of this treatment with $\ce{RSO3Na}$ corresponds to a solution made by dissolving $\ce{NaHCO3}$ and $\ce{Na2SO4}$ which are both not corrosive at all.
This is why it is better to use the modified resin $\ce{RSO3Na}$ instead of the original acidic resin $\ce{RSO3H}$, when treating hard water.

Better yet treat with a mixed bed column or a subsequent anion exchange resin adding an equivalent amount of OH-.  Neutral water!


text:
Concentration dependence of electron transfer
Regarding a single electron transfer:
$$O +e^- \rightarrow R$$
We find the current dependent on the overpotential as specified by the Butler Volmer equation:
$$i = i_0 (exp(\frac{\eta}{b})-exp(\frac{-\eta}{c}))$$
With $\eta =E^{applied}-E^{eq}$ the overpotential, $i_0$ the exchange current density, b and c Tafel slopes for the anodic and cathodic reaction. $E^{eq}$ is defined by the Nernst equation:
$$E^{eq}=E^0-\frac{RT}{F}ln(Q)$$
We find thus that the overpotential at the electrode surface is dependent on the concentration at the surface. The exchange current density itself however is said to be:
$$i_0 = Fk_0C_O^{\alpha}C_R^{\beta}$$
In some books, pages online, and review articles I've seen these thrown around so much that it is not clear anymore how the current density is affected by the concentration. If you rework the first equation with the Nernst equation one can get the Q out of the exponential, but is this already taken into account in the exchange current density factor or not? Is the current density "doubly" dependent on the concentration as in the overpotential is dependent on the current density and there is some dependence of the exchange current density?

Yes, the current density is "doubly" dependent on the concentration, in the following sense that you have written:

$ C_O(0,t) $ and $ C_R(0,t) $ appear in $E^{eq}$ within the overpotential defind as $ \eta = E - E^{eq} $ where $ E $ is the potential difference between the electrode surface and the bulk solution, i.e., $ E = \phi_M - \phi_S $.
$ C_O(0,t) $ and $ C_R(0,t) $ appear also in the exchange current density, which formally, has the following form

$$ j_0 = F k^\Theta \bigg(\dfrac{C_O(0,t)}{C_O^\Theta}\bigg)^{1 - \alpha} \bigg(\dfrac{C_R(0,t)}{C_R^\Theta}\bigg)^\alpha  $$
where $ \alpha $ is the symmetry factor, and $C_O^\Theta$ with $C_R^\Theta$ are the reference concentrations associated to the rate constant $ k^\Theta$.
You can see this "double" dependence in textbooks that derive B-V kinetics by using the Transition State Theory, or Eyring Theory, applied to charge transfer reactions.
Disclaimer: If you always talk about equilibrium, the concentrations at the electrode surface and the bulk are the same, so there is only one concentration to talk about. However, the kinetics are evaluated at the electrode surface, so it is a good practice to put $ (0,t) $, when it applies. When you write "$C_O$" or "$C_R$" in electrochemistry, my first question is, where is the $O$ and $R$ you are talking to me about? (my explanation does not take into account double-layer effects, where the evaluation is more subtle)

I do understand the need to specify it is the concentration at the electrode surface when talking about heterogeneous electron transfer. (In reality of course there is no specific point where charge transfer is taking place, it is spread out over space). Thank you for your answer, it cleared up some doubts I had. If only modern electrochemistry articles were better...


text:
Finding the pH of a solution from neutralizing hydrochloric acid and potassium hydroxide

What is the pH of the solution if $\pu{30.58 cm3}$ of hydrochloric acid ($\ce{HCl}$) with a concentration of $\pu{0.2855 mol dm-3}$ and $\pu{29.41 cm3}$ of potassium hydroxide with a concentration of $\pu{0.2971 mol dm-3}$ are mixed into a $\pu{250 cm3}$ flask ($\ce{KOH}$) solution, and then fill the flask after the temperature has returned? $K_\mathrm{w}=1.00 \times 10^{-14}$

The acceptable pH given is 8.85 but I got 9.45. Could someone share the right way to do this?
My methodology:
n(HCl) = 0,00873059 mol  < n(KOH) = 0,000873711
remaining n(KOH) = 0, 000 006 821 mol
c(KOH) = n/ V  = 0,000027284 mol/dm3
pOH = -log(0,...27284)
pH = 14 - 4,45 = 9,45

So how do you get 9.45? Did you note that the flask is filled so you dilute the mixture after combining the stocks?

I calculated the remaining amount of substance (KOH) and then used 250cm3 as the volume of the substance to get the concentration ( c = n/V ).

Several issues are there, but the big one is the book is wrong. They apparently forgot to divide by the $250$ mL flask volume.
First the potassium hydroxide addition was not properly transcribed, it should read $0.008737711$ mol with two consecutive 7's. This gives a net potassium hydroxide addition after tge acid of $0.000007121$ mol. Dividing this by the filled flask volume of 250 mL then gives $0.000028484$ mol/L.
When I was in graduate school I remember having to take an entrance exam for the Ph.D. program. The exam was of course directed more to critical thinking than computing power, and one of its questions asked for the pH of various hydrochloric acid solutions at ambient temperature -- including $10^{-8}$ molar, where if you weren't careful you found a basic solution with pH 8! I did catch that; ever since, I learned to include water autoionization, which gives a correct answer. So we render the net potassium hydroxide concentration as actually the difference between hydroxide and hydrogen ion concentrations:
$\ce{[OH^-]}-\ce{[H^+]}=0.000028484$
$\ce{[OH^-]}×\ce{[H^+]}=1.00×10^{-14}$
Solving the difference equation for the hydroxide ion concentration and plugging that into the product equation leads to the quadratic equation
$\ce{[H^+]}^2+0.000028484\ce{[H^+]}-1.00×10^{-14}=0$
with the positive root
$\ce{[H^+]}=3.51×10^{-10}\text{ mol/L}$
and with the logarithm, $pH=9.45$.
So apart from a minor transcribing error, you were correct. Clearly it is the book that was wrong.
The difference in pH between the answers is $0.60$ unit, whose powers of $10$ is close to $4$. If you (or in this case, the book) forgot to divide by the volume of $250$ mL above, or equivalently assumed a one-liter flask, you would have gotten the equations
$\ce{[OH^-]}-\ce{[H^+]}=0.000007121$
$\ce{[OH^-]}×\ce{[H^+]}=1.00×10^{-14}$
from which the reader can then calculate the book's incorrect $pH=8.85$. It looks like this question (unwittingly) tested critical thinking, too!


text:
Making a super-eutectic
A eutectic system is a mixture of substances which has a lower melting point than any of it's components. For example salt to melt ice. Also, a eutectic has the right mixture of components so that it freezes all at once (the liquidus and solidus temperatures are the same).
It seems that the eutectic effect generally grows stronger when we mix three components instead of two. For example in a magma chamber.
If we blend more and more components, how much further can we take this effect? Could we have room temperature inorganic halide salts? An alloy with over a dozen metals liquid below -100C?

Well, it can get pretty far, but I think that results of adding more components are somewhat diminishing.

@Mithron: Has anyone tried alloying all the alkali metals and alkaline earth metals together? Start with CsNaK and slowly add in other metals to chip away at the solidus temperature?

https://en.wikipedia.org/wiki/Sodium%E2%80%93potassium_alloy#Further_alloys_with_low_melting_points only goes as far as CsNaK

@Mithron: Why hasn't anyone tried adding lithium or rubidium to CsNaK to see if they can get any lower?

IIRC alloys of more than 3 elements are still pretty fresh research topic, go for it if you dare ;D

@Mithoron: I may ask CodysLab. Since partial freezing of a non-eutectic tends to pull things toward eutectic it makes the experiments less picky about exact proportions, which is a must when there are like 6 metals at once.

related https://en.wikipedia.org/wiki/Wood%27s_metal#Related_alloys

https://link.springer.com/article/10.3103/S1062873807050139

See Wikipedia for molten salts: https://en.wikipedia.org/wiki/Molten_salt , but even ethylammonium nitrate (q.v.) will only get you to 10°C m.p. BTW, triple-eutectic galinstan, https://en.wikipedia.org/wiki/Galinstan, is useful, since it's much safer than NaK or Hg. (I remember when aircraft and race car valves used NaK for cooling...)

The ASM ( American Society for  Metals) has a phase diagram handbook with many three component systems .

This is a quite broad question, and I will try to keep at a reasonably high level. Undoubtedly there are specific systems (more below) that will have unusual features that differ from this outline.
Lets look solely at the liquid phases of mixtures. To zeroth order, one treats a random (unordered) phase as an ideal gas - no atom-to-atom interaction (attractive or repulsive) contributing to the enthalpy, and the entropy comes from random mixing of the A and B atoms. This entropy term ends up being
-RT($x_{A}\ln x_{A} + x_{B} \ln x_{B}$)
with $x_{A} + x_{B} = 1$ and is a maximum when $x_{A} = x_{B} = 1/2$.
The first order treatment of solutions adds an enthalpy of mixing term $\Omega$, the heat of mixing. This comes from considering the energetics of A-A, B-B, and A-B interactions. So, the adjustment to the enthalpy takes on the form
$\Delta H_{mix} = x_{A}x_{B}\Omega$
If the A-B interactions are stronger than the A-A and B-B interactions, the heat of mixing is negative and eutectic may form. A positive heat of mixing leads to miscibility gaps. Again, the enthalpy of mixing takes on a maximum at $x_{A} = x_{B} = 1/2$ for a regular solution, being $0.25\Omega$.
Now, real alloys tend to be non-ideal and non-regular, but lets stay up at 30,000 feet for now.
An extension of the entropy of mixing to ternary solutions is straightforward, looking like
-RT($x_{A}\ln x_{A} + x_{B} \ln x_{B} + x_{C} \ln x_{C}$)
with $x_{A} + x_{B} + x_{C} = 1$, and is a maximum when all compositions are $1/3$.
The enthalpy term gets a bit trickier, since you have to consider A-A, B-B, C-C, A-B, A-C, B-C, as well as A-B-C interactions. So one can write:
$\Delta H_{mix} = x_{A}x_{B}\Omega_{AB} + x_{A}x_{C}\Omega_{AC} + x_{B}x_{C}\Omega_{BC} + x_{A}x_{B}x_{C}\Omega_{ABC}$
As you add more components to the mixture, you need to add more interaction terms.
OK, where are we now? First, the entropy term increases (is increasing negative) as the number of components in the solution increases. For 2 components, the natural log terms come to $-0.69$, for a ternary it is $-1.09$, and for 10 components in equal proportions it is $-2.30$. Not too surprising, the more components the higher the possible entropy.
On the enthalpy side, things look pretty good as well initially. Lets ignore the ternary interaction terms like $\Omega_{ABC}$. Still one is left with more interaction terms that could add up to something. If $\Omega_{AB} = \Omega_{AC} = \Omega_{BC}$ (and all are negative) then the enthalpy of mixing is ${1 \over 3} \Omega$, which is larger than the enthalpy of mixing of the binary solution. As the number of components increases, one goes from $0.25$ for a binary to $0.33$ for a ternary to $0.45$ for a 10-component system. Wow, we win on both counts!
Well, except that it would be highly unlikely that $\Omega_{AB} = \Omega_{AC} = \Omega_{BC}$ (and even less likely as the number of components increases). If the A-B term is more negative than A-C or B-C, then the lowest enthalpy of mixing is found away from the equal 1/3 compositions, so the entropy term is no longer minimized. If the D-E term in a 5-component system is highly positive, that skews things even worse.
As one example, lets look a bit at the Galinstan alloy mentioned in comments. In the Ga-In system, modeling (Ch. Li et al, J. Phase Equilibria 21(4) 357 (2000)) indicates that the enthalpy of mixing term is actually positive - the Ga-In (small) eutectic is driven mainly by the entropy term. For Ga-Sn (Bhupendra Kumar et al., Metals 2021 (11) 1363), the mixing term is also positive. Again, the small eutectic is from the entropy. For In-Sn (X.J. Liu et al., J. Electronic Materials 30(9) 1093 (2001)), the enthalpy term is indeed negative, but fairly small. Why does Galinstan melt "so" low? Well, it is mostly gallium so melts pretty low already, the Ga-In and Ga-Sn eutectics are way over by Ga to begin with, so a bit of both helps add more entropy, and the In-Sn negative enthalpy of mixing adds to that effect.
Could there be better systems? Perhaps, but all the stars have to align with respect to the enthalpy of mixing terms.
Added mostly as a comment, but a paper hot off the press (Journal of Phase Equilibria and Diffusion) might be interesting. The point is to try and use the Calphad method to rapidly find possible eutectic points in a quaternary system. To quote the abstract:

In
multicomponent systems, finding eutectic points can be
challenging due to the complexity of the systems...

There is debate whether Galinstan melts at 11C (only 19C below Ga) or -19C (49C reduction).


text:
Heating effect of alkaline earth metal chloride
Why does $\ce{MgCl2.8H2O}$ undergoes internal hydrolysis?
$$\ce{MgCl2.8H2O(s) ->[\Delta]MgO(s) + H2O + HCl}$$
while Calcium, Strontium, and barium undergoes dehydration and form anhydrous chloride.
$$\ce{MCl2.xH2O ->[\Delta] MCl2 + xH2O (M = Ca/Sr/Ba)}$$
Even though unlike $\ce{BeCl2}$, all of them have same ionic nature.
I didn't find much data on it and what I did find went over my head. Any help would be appreciated. Thanks!

Closely related: https://chemistry.stackexchange.com/questions/124974/when-mgcl%e2%82%82-nh%e2%82%82o-is-heated-to-the-boiling-point-of-h%e2%82%82o

The short answer is that the magnesium compound does not directly form the oxide, but forms a basic salt that decomposes further with additional heating. Moreover, calcium chloride can also be converted to the oxide by this method. See answers to the related question noted above.

So that means if for given appropriate conditions, same test is done all alkaline earth metal. They will all show similar behaviour. And there won't be any anomaly.

The answers to this related question address the thermohydrolysis of not only magnesium chloride but also calcium chloride. Details are in andeselisk's and my answers to that question; here I give a summary.

Magnesium chloride is more prone to thermohydrolysis than heavier AEM chloride because of the small size of its cation, but calcium chloride can be made to react also.

Magnesium chloride gives the normal salt upon dehydration in an HCl atmosphere, otherwise it gives a basic salt. The basic salt then requires further heating to decompose giving the oxide.

Calcium chloride gives the normal salt upon dehydration alone, but above the melting point it reacts with water or oxygen to give the oxide. Since the reaction occurs with the chloride molten, any intermediate basic salts would likely be dissolved in the melt rather than be observed as a separate product; in this case only the more refractory calcium oxide is so observed.

What you say is correct. But it does not answer the question  : Why ? Why is it so ?

Refer to the related question for details. And I do refer here to the small size of the magnesium ion.

"thermohydrolysis" fancy ;)

@Oscar Lanzi In your answer to the related question you wrote "The driving force for the hydrolysis is the stronger bonding of alkaline earth cations with oxide or hydroxide ions than with chloride ions". Does the same reasoning goes for carbonate ions also.  As magnesium bicarbonate decomposes to give magnesium hydroxide unlike calcium bicarbonate which decomposes simply to calcium carbonate?

@AyushAggarwall this is not a good comparison. No alkaline earth bicarbonate exists as a solid. Solubility trends in carbonates versus hydrixides appear to correlate with magnesium bicarbonate solution forming the hydroxide.


text:
Threshold of steric hindrance for the axial position in cyclohexanes
My professor mentioned that isopropyl is able to go into the axial position on the chair conformation of a cyclohexane, while tert-butyl cannot. This made me wonder then, at what point is a substituent on a cyclohexane too sterically hindered to ring flip to the axial position?
More specifically, is there a specific A value threshold for which a substituent is too sterically hindered to be in the axial position on a cyclohexane?


text:
Multi compartmental ODE's in chemical engineering
I am alien to the field of chemical engineering. And I am confused about equations, written by chemical engineers, including several variables belonging to different compartments.
Assume that,

$[X]$ is in concentration units and belongs to the $V_X$ compartment.
$[Y]$ is in concentration units and belongs to the $V_Y$ compartment.
$X$ has a negative feedback on itself and $Y$ increases $X$.

I would write my equation as following,
$\frac{d [X]}{dt} \times V_X = [X]\times V_X - [Y] \times V_Y$.
I would write this way, because these should be amounts not concentrations (right?).
However the equations are,
$\frac{d [X]}{dt} = [X] - [Y]$.
I don't understand how concentrations can be used that way. I would be very happy if anyone could help me to solve this issue.
Thanx!

I can be written in 2 ways. Using intensive quantities (and their derivatives) like concentrations, or using extensive quantities like molar amounts, but then volumes must be used to express relations between both.

but both are accepted as correct in chemical engineering, I see. Thank you!


text:
Why do chlorophyll need to be dissolved in organic solvents to observe fluorescence?
First, chlorophyll extraction was done with ethanol. I used my phone's flash light to see the fluorescence which was a little pinkish red. Then, I added an ether solvent to the extracted chlorophylls(which contain ethanol), and saw the fluorescence again which was more of a red color than before.
I've learned that chlorophyll dissolve well in ethers than in ethanol, so the fluorescence is higher in an ether solvent. What is the reason for this? There are same amount of chlorophyll but just a different amount of 'how much they are dissolved'. Do chlorophyll need to be 'dissolved' in a solution(organic solvent) to make fluorescence?

You may be mixing up some concepts. Solubility and fluorescence intensity would be expected to depend on each other (since a higher concentration of the fluorescing substance will initially produce a more intense signal), but the color detected might be shifted by the solvent in a way that has nothing to do with concentration. Is your question why chlorophyll is soluble in organic solvents?

It is most likely concentration being greater in the ether solution making the fluorescence brighter. Of course if it does not dissolve there is none of it there to fluoresce. Chlorophyll in leaves fluoresces so it does not need an organic solvent.  If it does not dissolve well in a solvent it may aggregate and this quenches any fluorescence.

Fluorescence color and intensity can be affected by the solvent and dilution of the fluorescent species for a few reasons, including:

The state of aggregation, e.g., from dissolved to amorphous to crystalline, affects the molecular environment.
As the concentration changes, new species may form.
The polarity of the solvent affects the vibrational relaxation time of the fluorescent solute. Note that ethyl alcohol is far more polar than diethyl ether.


text:
What kind of isomers are ethyl butanoate and propyl propanoate?
Are ethyl butanoate (1) and propyl propanoate (2) metamers or functional group isomers? My book claims they are metamers, but here one group is $\ce{-CO-OR}$ and another is $\ce{-CO-OR'}$ where $\ce{R}$ is $2$ carbon and $\ce{R'}$ is $3$ carbon. Though both of them are esters, aren't the functional groups different?

https://chemistry.stackexchange.com/questions/78783/what-is-metamerism?rq=1

Metamerism occurs when isomers have different alkyl chains on the sides of the functional groups. In the case of an ester, the relevant functional group is the ester (..O-C=O..) group. The length of R and R' has no bearing on the functional group in this case. An example of a functional isomer would be something like ethyl butanoate and caproic acid.


text:
Why are the electron affinity values of alkali metals still negative, despite the process being endothermic?
I understand that electron affinity is generally exothermic(groups 16 and 17). But, alkali metals are reluctant to gain an electron because losing electrons is beneficial for them. So, the electron gain enthalpy is termed endothermic (source: Chem libretexts) for alkali metals because it is hard to add an electron to their atoms and you need to invest energy. If it is endothermic, the electron affinity values must be positive, but they are still negative. Why? Why do the values still represent the release of energy?
In the image, the EA values of alkali metals are still negative, not positive. But in the description, it says that the process of adding an electron to alkali metals associates with the absorption of energy (endothermic- positive)
Source https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Electron_Affinity

That means you should stick to the source with electron affinity values not some "Chem libretexts". https://chemistry.stackexchange.com/questions/51400/why-do-alkali-metals-have-an-exothermic-electron-affinity?rq=1

Wow that's terrible. It completely contradicts the rest of the content and even the values listed below the text, in the example.  The electron affinities are not endothermic in those cases. It might be a matter of emphasis however. Maybe it is meant that among metals there is a decreased affinity. May be worth contacting the authors.


text:
Find the minimum concentration so that it doesn't precipitate

Calculate the solubility of silver chloride.
a) in pure water
b) Calculate the minimum concentration of ammonia that prevents precipitation in a solution that
contains $0.1$ mol of AgNO3 and $0.01$ mol of NaCl per litre. Data: Kf [Ag(NH3)2+] = $1.6×10^7$ Ksp(AgCl)= $1.8×10^{−10}$

For part a) I'm totally sure on how do it: $$AgCl\rightarrow Ag^+ +Cl^-$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ s \ \ \ \ \ \ s$$ $K_s=s^2\implies s=\sqrt{K_s}$
But for part b) I'm really stuck, I've tried $$\ce{Ag^+ +2NH3 \rightarrow Ag(NH3)2}$$ and $$\ce{AgCl\rightarrow Ag+ +Cl-}$$ but not sure where to go from here.

Hint: $\ce{[Cl-]} = \pu{0.01 mol}$, when you only consider one liter of the solution

yes so then we can find the $[Ag^+]$ with the constant of solubility which gives $1.8*10^{-8}$ but then I don't know what to do

You also know Kf, can you do something with that?

@SafdarFaisal something like this maybe? $K_f=\frac{1.8*10^{-8}x}{x}$ but I'm unsure since we didn't know the moles of $\ce{Ag(NH3)2}$ either

In pure water, you have probably seen that $s \ce{= 1.34·10^{-5} M}$.
In a solution containing $\ce{Cl-}$ ions in concentration $0.01$ M, the concentration in $\ce{Ag+}$ is then :
$$[\ce{Ag+}] = \frac{K_\mathrm{s}}{[\ce{Cl^-}]} = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8}$$
The definition of $K_\mathrm{f}\ce{(Ag(NH3)2^+)}$ is
$$K_f = \frac{[\ce{Ag(NH3)2^+}]}{[\ce{Ag^+}][\ce{NH3}]^2} = 1.6\times 10^7$$
so that the critical concentration of $\ce{NH3}$ is given by:
$$[\ce{NH3}]^2 = \frac{[\ce{Ag(NH3)2^+}]}{[\ce{Ag+}]\cdot K_\mathrm{f}} = \frac{0.1}{1.8\times 10^{-8}\times 1.6\times10^7} = 0.347$$
The final $\ce{NH3}$ concentration is $\sqrt{0.347} = \pu{0.59 M}$

The value of $K_\mathrm{s}$ given was $1.8 \times 10^{-10}$, I took the liberty to change that

@Maurice amazing! But why is $\ce{[Ag(NH3)2+]}=0.1$?

@ Aley20. Silver ions are present under two forms : $\ce{Ag^+}$ and $\ce{Ag(NH3)2^+}$. The total of these two ions is $0.1$ . But, as $\ce{[Ag^+] =  1.8 · 10^{-8}}$, this value is negligible with respect to $0.1$, so that $\ce{[Ag(NH3)2^+]}$ is practically equal to $0.1$

@Maurice oh okay! So does it always add up to 0.1 or is it only because we had 0.1M of $\ce{AgNO3}$? I'm guessing the latter but I want to be sure.

@ Aley20. You are right. The given value ($0.1 M$), is the concentration in $\ce{AgNO3}$.


text:
Why and how are mullites (and in extension solid solutions) formed and how are their formulas conceived as opposed to regular crystalline solids?
From doing more Phase ID for XRD on aluminosilicate based coal fly ash samples, I've seen mullite phases appear a few times which have the non-stoichiometric chemical formula $Al_{4+2_x}Si_{2-2x}O_{10-x}$ where $x \approx 0.14 - 0.49$ according to this reference which also details in a good Phase ID methodology for mullites.
My questions are the following:

Why are some solids crystalline instead of being a solid solution; how is the latter formed and why?
How are solid solution chemical formulas identified as opposed to easier-to-understand stoichiometric chemical formulas as seen in crystal structures (e.g. SiO2 in quartz).
What common methods exist to identify the $x$ parameter in solid solution formulas (apart from identifying common ratios between characteristic peaks as mentioned in the reference above).

Those solutions are crystals.

Wait, are "solid solutions" synonymous with polycrystalline solids?

No, not at all. In a bit I’ll get an answer out…

Being a solid solution is separate from being a crystal. Some alloys occur with one single stoichiometry (plus/minus defects). Many can accommodate a range of compositions.
Lets start easy: The Au-Cu binary system shows complete solubility across the entire composition range. The crystal structure is fcc, and Au and Cu atoms sit randomly. The lattice parameter will change a bit going across the composition range and that is about it. Similarly the Si-Ge system also shows complete solubility in the diamond cubic crystal structure.
The NaCl-KCl pseudo binary system has a slightly different response across the composition range:

Below 800K the fcc-based NaCl (or KCl) crystal structure has a miscibility gap - only so much K in a primarily NaCl (or Na in a primarily KCl) crystal fits. Above 800K you see complete solid solubility, with any amount of Na or K fitting in between the Cl sublattice.
The GaAs-AlAs pseudo binary has complete solid solubility, again with any amount of Al/Ga in the As sublattice. The lattice parameter and the band gap changes across the composition range.
The B2 crystal structure is based on the bcc unit cell, with the corner and center atoms being different sites. The exemplar is CsCl. However, there are many binary metallic systems where the B2 structure shows up, but often with a reasonable composition range. One example is in the Al-Ti system:

Here one sees a variety of phases, some line compounds (Ti$_{3}$Al$_{5}$ for example), but many with wide composition ranges. These include, notably, quite reasonable solubilities of Al in both hexagonal and bcc Titanium as a random solution with the Al sitting on hcp or bcc lattice sites. In addition, the light blue B2 $\beta_{0}$ phase is an ordered bcc phase, with the Al preferentially occurring on one of the B2 sites. This additional symmetry of the crystal will be clear in diffraction patterns.
The sublattices do not all need to be completely full. One example is in the Ti-N system:

Again one sees that the hcp and bcc Ti phases can accommodate some amount of nitrogen in them - however these N atoms are on interstitial sites not in place of Ti. There are also three stoichiometric compounds (line compounds). Finally there is the TiN phase, which is an fcc Ti lattice with varying amounts of nitrogen on tetrahedral sites. This is modeled as a two-lattice system with either N or a vacancy on the one one site, and fcc Ti on the other. You can only get up to a 1:1 stoichiometry range, then you run out of the space on the (N,Va) sublattice.
In the BaO-CaO system, there is limited solid solubility in the halite (BaO) or periclase (CaO) phases. BaO-SrO behaves much like NaCl-KCl. Al$_{2}$O$_{3}$-BaO only shows line compounds.
For many mineral oxides (including the ones you are looking at, but basically about a million discovered minerals in nature), one has an oxygen sublattice with octohedral and tetrahedral sites (or more complex oxygen lattices but the point remains), and various metals will prefer to sit on one site or the other. In some cases this is a strong preference, in other cases some will sit on either site without caring much. Sometimes this requires vacancies on one sublattice or the other (well, really multiple sublattices), charged defects, or whatnot. Once you get oxygen and Al, Fe, Si, and maybe a few other impurities, well, the sky is the limit.
Your particular case of mullites (see [Journal of the European Ceramic Society][4] for example) shows one set of AlO$_{6}$ octahedra, with "cross-linking" of tetrahedral chains of AlO$_{4}$ or SiO$_{4}$ connecting them. The composition is Al$_{4+2x}$Si$_{2-2x}$O$_{10-x}$ showing you can put some Al in place of the Si, but you also have to get rid of some of the oxygens for charge compensation. A quite complex structure all in all.
[4]: https://www.sciencedirect.com/science/article/pii/S0955221907002634#:~:text=The%20crystal%20structure%20of%20mullite,%2CSi)O4%20tetrahedra.

Thanks for the very diligent answer! Before I can fully understand if your answer solves my question I think I need to spend more time understanding some background here in your answer. Could you please advise what types of diagrams you refer to so I can do some reading along with some sources? Thanks so much, I'll return here and mark it as officially answered (although I've +1'd) once I'm sure of the answers :)

Those are binary phase diagrams (some are pseudo-binary). Porter and Easterling have a classic book called Phase Transformations in Metals and Alloys that gives the basic background to them in the first few chapters. Sorry, I assumed (bad of me) that someone doing phase identification in ceramics would have done a course somewhere on phase diagrams.

All good, it's a reasonable assumption to make, I'm still a beginner in some ways which is why I hadn't heard of them (although I've seen them before quite a bit).