text:
Is there a list of must-have columns for LC-MS/MS used in biochemistry/clinical lab setting?
Considering a wide variety of chromatographic columns used for (U)HPLC based on

phase type (normal or reverse);
column dimensions (geometry, length, inner diameter);
particle material $(\ce{SiO2},$ $\ce{ZrO2},$ $\ce{TiO2},\ldots),$ size, shape and porosity;
dopants/modifiers etc.,

as well as proprietary manufacturing techniques involved, is there a narrowed-down list of the bare-minimum column types that every analytical lab doing specifically liquid chromatography tandem mass spectrometry must have to cover the majority of the analytical routines (qualitative and quantitative analysis of short-chain fatty acids, primary metabolites, amino acids etc.)?
Background: a new LC-MS/MS lab is being created within a university's biochemistry division from the ground up both for the fundamental research and providing analytical services, so listing the most versatile and recent columns would be appreciated.
I do realize the question is borderline off-topic being somewhat opinion-based, so it would be nice if the answer were based on a trustworthy source.

Would it suffice to provide an academic review article on the topic, or maybe even a vendor-sponsored review: https://www.americanlaboratory.com/163469-Review-of-HPLC-2014-Advances-in-HPLC-and-So-Much-More/ ?

You really might get a lot of mileage from talking to a vendor (even if they might want to sell you more than you need): https://www.thermofisher.com/se/en/home/industrial/chromatography/liquid-chromatography-lc/hplc-uhplc-columns.html

@BuckThorn Call up all vendors. Their salespeople will cue up at your door, and each of them knows best about the shortcomings of their competitors products. ;)

I think it matters if the new lab will be primarily a university or college facility for professors and their research students or if the lab is primarily intended to provide wide ranging analytical services to various paying clients. I assume the latter, in which case it would be helpful to know what service needs are presently not being met or are lacking in the prospective service area. In any event, congratulations for having such a pleasant problem to solve!

This answer will change fairly frequently because technologies are always improving.  These days, I think a bare minimum list might be something like:

a 2.1 mm × 50 mm C18 column for rapid reversed-phase methods.
a 2.1 mm × 150 mm C18 column for slower, broader-coverage reversed phase methods.
a 2.1 mm × 50 mm HILIC column for rapid methods to detect polar metabolites.
a 2.1 mm × 150 mm HILIC column for slower, broader-coverage methods to detect polar metabolites.

Some further notes:

If your lab is going to use ion-pairing based methods, you will probably want separate columns for ion-pairing methods and "clean" methods without ion-pairing reagent.  Ion-pairing agents like tributylamine and/or perfluoroheptanoic acid are very difficult/impossible to remove from columns once they have been used for this purpose.
Fairly soon the best answer might be to use 1.0 mm diameter columns instead of the "standard" 2.1 mm diameter that is most common nowadays.
The most widely used HILIC column seems to the "BEH Amide" column chemistry from  Waters.  Following close behind may be ZIC-HILIC from SeQuant and/or Agilent's HILIC-Z chemistry.
For C18 chromatography, performance differences between different vendors are comparatively minor (although real).  A wide variety of chemistries are used.


text:
How to obtain curve energy and non-bonded interaction parameters (e.g. Lennard Jones parameters) with Gaussian?
I am parameterizing a force field and would need to obtain non-bonded interaction parameters of atoms (e.g., Lennard-Jones parameters).
My primary idea is to obtain a potential energy curve using the Gaussian 09 software and fit the data on that curve to an equation that has non-bonded interaction parameters of atoms (e.g., Lennard-Jones curve).
What I thought was to put two identical molecules and vary the distance between two atoms of these molecules (e.g., using the scan function on redundant coordinator editor) and, therefore, get the energy curve. What I was trying to do was a simple example to get interaction parameters for the oxygen atoms of two water molecules.
I was taking a test with Gaussian at the theory level B3LYP/6-311G++(d,p). Here is the input file:
%nprocshared=4
# opt=modredundant b3lyp/6-311++g(d,p) geom=connectivity

Title Card Required

0 1
 O                 -3.09468841   -0.10392610    0.00000000
 H                 -2.13468841   -0.10392610    0.00000000
 H                 -3.41514300    0.80100974    0.00000000
 O                  3.39491938    0.10392610    0.00000000
 H                  4.35491938    0.10392610    0.00000000
 H                  3.07446479    1.00886193    0.00000000

 1 2 1.0 3 1.0
 2
 3
 4 5 1.0 6 1.0
 5
 6

 B 1 4 S 10 0.200000

Something I noticed is that the level of theory used (obviously) influences the calculations results. In addition, the initial distance of the water molecules also influences whether the simulation will fail or not (in this example, it works).
Could you kindly tell me if this approach makes sense? Or should I use another software, another type of calculation? Do you have any suggestions on how I could get an energy curve by varying the distance of two non-bonded atoms in the Gaussian? Examples would be appreciated.
With Gaussian and a similar approach I was able to obtain parameters such as bond stretching, angle deformation and dihedral torsion. However, it would be lacking to obtain the non-bonded interaction parameters of atoms.
Things I should consider:

Do a rigid scan instead a relaxed scan (like the code presented).
I need a dispersion-corrected density functional methods.

No worries about that, I'm just telling you to help. Before we do a deep dive into the level, I suggest reading DFT Functional Selection Criteria, where I and others have gathered quite a bit on the matter. Tl;dr: It's complicated and depends on what you are looking for. What you most definitely will need is some dispersion correction. But I guess we'll burn that bridge once we've crossed it. In its core, the question is good, so let's hope  someone can provide an answer.

Also (loosely) related: How to calculate Lennard-Jones potential with quantum mechanical methods, How do water models fit into force field methods? I was pretty sure there are more related questions, but I just can't find them. I recommend browsing around with the search.

Thanks. I saw this reference, but my question, at first, would be how to build a coherent simulation in Gaussian to obtain the energy curve!
"what should i take into account in this kind of simulation?" I mean. As you said, two things are important: level of theory and dispersion correction.

Is it for fun or for production? You are aware that real dispersion forces are multi-body interactions and DFT does not calculate them, right?

To start off, you definitely need to use dispersion-corrected density functional methods if you want to obtain reasonable non-bonded interaction parameters. I've seen many cases where B3LYP (with no dispersion) yields a fully-repulsive potential energy scan.
I also agree with your assertion about doing a rigid scan. In principal, a relaxed scan is a good idea, but consider that most force fields separate out the non-bonded interactions. Thus, for parameterizing, you don't really want to change the bond lengths, angles, etc. from a relaxed scan.
There are a few ways to get the parameters. As mentioned above, there's already a similar question with tips on curve fitting.
Multiple papers have also been published on fitting force field parameters from quantum chemical results, e.g:

QubeKit J. Chem. Inf. Model. 2019, 59, 4, 1366-1381 - code at GitHub
QuickFF J. Comput. Chem. 2015, 36, 1015– 1027 - code at GitHub
ForceBalance J. Chem. Theory Compute. 2013, 9, 1, 452-460 and J. Phys. Chem. Lett. 2014, 5, 11, 1885-1891 - code at GitHub
ForceFit J. Comput. Chem., 2010 31: 2307-2316
Parfit J. Chem. Inf. Model. 2017, 57, 3, 391-396 - code at GitHub

Update there's a paper on this very specific problem:
J. Chem. Theory Comput. 2020, 16, 2, 1115-1127

Here, we utilize a previously described Minimal Basis Iterative Stockholder (MBIS) method to carry out an atoms-in-molecules partitioning of electron densities. Information from these atomic densities is then mapped to Lennard-Jones parameters using a set of mapping parameters much smaller than the typical number of atom types in a force field. This approach is advantageous in two ways: it eliminates atom types by allowing each atom to have unique Lennard-Jones parameters, and it greatly reduces the number of parameters to be optimized.

I've probably missed some of the various "fit force field from QM" papers since it's a bit outside my expertise. Suggestions welcome.

Even if you do your own fitting, I suggest at least reading these papers, since they've all spent time considering requirements.

Thanks a lot Geoff Hutchison! I will certainly read the articles you recommended. Your comments are an excellent starting point.

Dear Geoff Hutchison, I performed several tests with silica-based compounds and the best method I obtained to represent bond lengths and angles was HF 6-31G (2d, p), in this case (particularly) I would not need dispersion-correction to obtain non-bonded parameters, right?

@EmersonPL - Because HF doesn't fully include electron correlation, it does not properly handle dispersion either. There are empirical dispersion corrections for HF as well, e.g. HF-3c, HFD, .. depending on the program you use.

If you have the time / computational resources, I'd recommend an RI-MP2 method, eg with cc-VDZ or def2-SVP since it would also include better treatment of dispersion.

Thanks a lot Geoff Hutchison! I'm using Gaussian, btw.

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text:
Where do the percents in empirical formula questions come from?
A typical empirical formula problem reads something like this:

Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur. source

How would those percentages be determined in real life? Spectroscopy?
A percent composition problem would obviously yield percentages, but those start with compound formulas--there'd be no need to determine the empirical formula from the percentages.

I believe its through chemical processes like combustion(the sulphur would form $\ce{SO2}$, which you can then capture and quantitatively estimate), reactions with acids, alkalis, halogens, titrations(if it's an acid) etc.

It is impossible to make any general statements about how the atomic percentages would be determined. There are a zillion different kinds of samples and a zillion different analytical techniques.

How would those percentages be determined in real life? Spectroscopy?

You are right, this analysis comes in the domain of analytical chemistry. The fundamental requirement for formula determination is that your sample must be extremely pure.
There is a whole field. So there are two broad classes of approaches. One is called classical analysis where one would dissolve the sample in an "appropriate" acid or base mixture, and this is followed by target element specific titration or precipitation. This technique results in very reliable data but it is very time consuming
Today most people would use an instrument called the inductively coupled plasma optical emission spectroscopy (ICP-OES), you "srpay" the dissolve sample into a state of matter (plasma) which is as hot as the surface of the Sun. Under those those conditions, compounds break down into their constituent atoms and emit characteristic wavelengths.
For example, if we are interested in knowing how much aluminium is there, we would construct a calibration curve and interpolate the amount of Al in your unknown sample.
The same thing (although less effectively) can be done for sulfur.
Sometimes you need both classical or instrumental analysis.

These percentage are experimental. Suppose you mix 36 g aluminium powder and 64 g sulfur powder, and you heat the mixture. A exothermic reaction will occur. As a final result, a pure compound is produced, which is aluminium sulfide. With any other proportion, the same aluminium sulfide is formed, but some aluminium or some sulfur remains unchanged at the end of the reaction.

Thank you for the attempt, but this answer does not address the method of analysis.

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text:
When and how do mineral chelates separate in the body?
If one consumes an amino acid chelate such as iron bisglycinate, it is generally absorbed intact as far as I am aware. At some point, presumably the iron could be used for say hemoglobin synthesis and the glycine for collagen synthesis. How exactly does the dissociation of this chelate occur in the body to support this?

I have no idea of the specifics here, but every ion and its ligand have an equillibrium.

Better in Biology SE perhaps


text:
Are there many examples of stable, well characterised organometallic compounds of saturated hydrocarbons with beta hydrogens?
One of the first things you tend to learn in organometallic chemistry is that transition metal compounds containing beta-hydrogens are unstable (there are plenty of stable main groups ones; eg grignards seem pretty stable for most saturated aliphatics). There is a tendency for the beta hydrogen on the alkane to be abstracted by the transition metal leaving a metal hydride and a coordinated alkene, possibly followed by other reactions.
Aryl and alkene ligands are mostly immune to this reaction for various reasons. Similar hydrocarbon ligand with no beta hydrogens are usually much more stable.
But are there any well characterised examples of stable transition metal organometallics with beta hydrogens?

Are you asking only about  transition metal compounds? Then you should highlight is as there's plenty of main group organometallics where there's no such tendency.

@Mithoron Good point. I've clarified.

Zinc compounds are pretty stable. Isopropyl zinc bromide is commercially available.

@Waylander Zinc isn't considered a transition metal https://en.wikipedia.org/wiki/Group_12_element#Classification


text:
How to extract force when the center of mass of molecules is put in the box in GROMACS?
To extract the force as in the title I used the gmx trjconv command like below.
gmx trjconv -f alad.trr -s alad.tpr -n alad.ndx -pbc mol -novel -force -o alad.gro

However, the force was not extracted in alad.gro file.
Of course, I got the coordinates and the force at equal intervals when getting the alad.trr file.
Another way, I used gmx traj command to get force.
gmx traj -f alad.trr -s alad.tpr -n alad.ndx -pbc -mol -ox coordi.xvg -of force.xvg

However, the force obtained from force.xvg didn't look like the value I wanted.
So, how do I get the force of molecules put in a single periodic box?

Could you include a figure showing force.xvg, and a better description of what it is exactly that you would like to measure, and why you think the output is not correct.

However, I'm still wondering why coordinates of alad gro and coori.xvg are not same.

I solved the problem. As gmx trjconv -pbc mol option only make molecules translate, the cooridinates obtained from alad.gro can correspond to forces obtained from gmx traj -f alad.trr -s alad.tpr -n alad.ndx -of force-dump.xvg(same results using gmx dump command). However, coordinates of coordi.xvg doesn't match with alad.gro . Also I couldn't trust force.xvg. whether the data is correct is tested by my own work ouput. So it's hard to say simple..

There is a proposal for a Stack Exchange site for Materials Modeling which is currently in the commitment phase. There are a lot of experts currently there and questions like these would get more activity there. It would be great if you could help building the site by committing to the proposal.

I solved the problem. As gmx trjconv -pbc mol option only make molecules translate, the cooridinates obtained from alad.gro can correspond to forces obtained from gmx traj -f alad.trr -s alad.tpr -n alad.ndx -of force-dump.xvg.


text:
Is there any way to chemically check sweetness level in any drink or water or tea etc?
I was wondering if there is a test or way to find out how sweet any drink is? For example, if I mix sugar in water, how can I test its level or number of sweetness?

I mean you can check for the presence of a sugar, such as how much sucrose is present, and base the level of sweetness off that.
Check out this other thread as the answer also can pertain to this question
https://chemistry.stackexchange.com/questions/51107/measuring-sweetness

No. Sweetness is a highly specific characteristic of compounds. Some sweet compounds have nothing in common and some carbohydrates (sugars) are not sweet.

There must be a scale of the type Scoville I guess. But if I knew it I forgot. I just guess this based on the fact that each sugar and each sweetener has a number expressing its sweetening power. So there should be a kind of reference albeit subjective. If it is so someone here should know for sure. But you can't check chemically for sweetness, strictly speaking.

Post by the GodlyBeast

@TheGodlyBeast if so how do you explain sweetness of artifical sweetners?

It could be in theory possible if the human GPCR proteins which sense the sweetness were available. For instance, they could be bound to an electrode, or somehow monitored via UV-Vis, and their structural variation could be related to the sweetness level. But, without receptor, there is no "sweetness". Sweetness is not an universal thing: some animals perceive sweetness in different ways, for instance gorillas don't perceive brazzein as a sweet compound, whereas humans do: different proteins give different answers.

@Alchimista what you are mentioning is the perceived sweetness level compared to sucrose :)

@permeakra, I was referring to identifying specific sugars, such as sucrose, not sugars in general.

@The_Vinz yes but I meant there should be a systematic way to do it. Though op is propably asking for something in which perception is not involved which seems to be hard to find.

As asked by the OP, I think the answer to the question has to be no. If it was known that a drink contained only glucose as a sweetener, then it would presumably be possible to measure accurately the glucose concentration in the drink. But maybe other substances in the drink would be interferences in the measurement. Following on, as @Alchimista commented, perhaps a Scoville-type sweetness test could be used. But the possible interferences problem complicates this as well. In addition, perceptions of sweetness may also be affected by seemingly innocuous substances in the drink, e.g., I find that a dash of table salt makes watermelon taste sweeter and I have no clue how that might arise.
Then there is the matter of inorganic substances. The old fashioned name for lead acetate was ‘sugar of lead’. Poisonous though it is, it earned its deadly sweet name. As well, some salts of beryllium ('glucine'; all berylium salts are highly poisonous) and yttrium are sweet tasting: see wiki. So it is not just various organic substances that can taste sweet and who knows what else is sweet tasting: no sane chemist (nowadays) tastes chemical substances.
My guess is that there are quite a few chemicals that would taste sweet, but we will probably never know what they are. Then how would sweetness levels of two lead acetate aqueous solutions be compared? I don’t see how it can be assumed, a priori, that sweetness is directly proportional to lead ion concentration and no one should do such an human subjects experiment anyway. And what about a drink containing glucose and lead acetate? Even if the lead acetate did not interfere with the quantification of the glucose concentration, the drink would be sweeter than expected from just the glucose measurement result.

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

if I mix sugar in water, how can I test its level or number of sweetness?

Number of sweetness is perhaps subjective. However, sugar content can be measured by refractive index changes (if you are just talking about sugar and water and no colors). Refractive index linearly increases with sugar content in water.
Keep in mind it is not a highly selective test but it quite routinely used as well. You need a refractometer for that purposes.
Another simple test is density measurement for sugar content, and there is a special unit. Read more about Brix unit elsewhere.


text:
Why can zwitterions be difficult detect by HPLC-MS?
I understand that to separate molecules by HPLC (or SFC) you commonly adjust the pH of the mobile phase to get preferable charges (i.e. negative charges) on the molecules - so that they interact with the column and are therefore well separated and thus give a large distinct peak.
Im not completely sure I understand what happens to zwitterions like amino acids when you try to separate them. Is it correct to think that they randomly interact with the column and thus "spread out"resulting in a chromatogram with a lot of noise? Why is it that the connected MS also has problems detecting them (should it not continuously detect the molecule)?
Example of amino acids that could be difficult to detect:

Same as with other molecules, you adjust the pH of the mobile phase to get the anion or cation depending on what you are detecting. At neutral pH zwitter ions carry no net charge but are polar species so do not interact with the column like other neutral molecules.

That is not the right structure for aspartate.  You've drawn the side chain with an aldehyde group instead of a carboxylic acid.  Also, the amino acid on the left, 2,4-diaminobutyric acid, is a bit unusual.  It's perfectly similar to the "usual" proteinogenic amino acids, and the same methods would work OK with it, but you won't see it in most published methods.

Zwitterionic amino acids are routinely detected by HPLC and by MS.  The only challenge is to find an appropriate stationary phase.  The most common stationary phases for this type of separation would include octadecyl-derivatized silica (ODS, also known as C18) and a HILIC stationary phase.  Many column vendors make columns explicitly marketed for separation of amino acids.
Examples:

Sciex markets an LC-MS method using a C18 column.
Agilent markets an LC-MS method using a HILIC column.
A decently-cited paper from 2016 describes a 3-minute LC-MS method for amino acid separation using HILIC columns.

What needs to happen is that analytes need to reversibly equilibrate between the stationary and the mobile phase via a single "mode".  If there are multiple modes by which analytes can interact with the column, then peak shapes will be highly distorted and/or flattened into noise.  An example of this would be if the analyte can both interact hydrophobically with the C18 moiety of ODS and via Lewis acid / Lewis base interactions with bare silanols (Si~OH) groups on the surface of exposed, underivatized silica.  These days, column manufacturers go through great lengths to insure that very few if any exposed silanols remain exposed on derivatized silica resin used for analytical LC columns.  Zwitterions are really no different than other analytes in this regard.
At the MS, zwitterions will need to be transferred to the gas phase as charged ions for detection.  But this isn't really a problem either, at least for e.g. electrospray mass spectrometry.


text:
Does PEO particle size decrease when dissolved in water?
I am looking to make metallic ink for inkjet printing and using PEO 1 wt%  (MW ~600,000) as a binder. The print head I am hoping to use has a nozzle diameter of 80 microns. The metal powder dispersed in the ink is ~100nm. If I dissolve the PEO into deionized water, will the PEO particles be too big to pass through the nozzle?  The PEO purchased is from sigma aldrich (https://www.sigmaaldrich.com/catalog/product/aldrich/182028?lang=en&region=IE)
(Physicist asking a chemistry question!)
Thanks :)

PEO is very soluble in water. At 1 wt% concentration, you should expect it to be individual molecules throughout the solution. A PEO molecule with MW of 600,000 has ~13,600 repeating units. You can imagine them to be roughly ~200 nm blobs calculated based on the radius of gyration of a linear polymer in a good solvent.


text:
hydrodynamic thickness of polymer vs. concentration of electrolyte?
I've measured the hydrodynamic thickness of a NaPSS layer on polystyrene latex and plotted it against the concentration of the supporting electrolyte, NaCl.
My graph shows a general decrease in thickness with some fluctuations at earlier concentrations. I thought that thickness would increase with concentration due to more charge screening. Can anyone help explain these results?

Osmotic pressure?


text:
Confusion in half filled or full filled electronic configuration
At the end of electronic configuration, we were taught that, electron orbitals are most stable when they are either fully filled or half filled.
E.g., the final valence configuration of chromium is $\ce{(4s)^1 (3d)^5}$ and not $\ce{(4s)^2 (3d)^4}$.
But the final electronic configuration of chlorine is $\ce{(3s)^2 (3p)^5}$ and not $\ce{(3s)^1 (3p)^6}$.
Why doesn't chlorine follow the half filled or full filled theory?

Having a half-filled or filled subshell is stabilising, as you say. But for Cl, the difference in energy between 3s and 3p is greater than the additional stabilisation from filling the 3p subshell. 3d and 4s are closer in energy, so for Cr it is favourable to promote a 4s electron and have a 4s1 configuration.

Great answer @atbm! I would like to invite you to participate as one of the first group of users in the Materials Modeling Stack Exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MGY4NjYyM2ZlY2NkNDI0MWRmYjhjMGI1N2RmM2U3ZDk1N2VjOWQ1MGU1MmVhNDZkZjI5MzNiNzY1MmE2ZjRhOYrAzULyFn6juGAbZ7wko5LvlWXZ0iQfr620nQ70gVNl0 We will soon be in the private beta phase, where only a select group of users can participate. These users help to determine the style of questions for the future. Can you help us by committing?

In the case of 3$d$ and 4$s$ energies, there is an "anomaly" in the ordering, being the 3$d$ orbital energy higher than that of 4$s$. However, you are comparing the configuration of chromium (with 3$d$ and 4$s$ orbitals) with chlorine. The valence shell of the latter contains 3$s$ and 3$p$ orbitals, and the energy of 3$s$ orbitals are lower than that of 3$p$.

The reason we see these Aufbau's principle exceptions in transition metals is because the $4s$ and $3d$ orbitals are very similar in energy.
In chromium, having a $4s^2$ $3d^4$ configuration results in electron-electron repulsion due to the two electrons in the $4s$ orbital. For this reason, chromium adopts a $4s^1$ $3d^5$ configuration, in which each electron occupies its own orbital. Recall that Hund's rule essentially states that you fill each orbital once before going back with the second electron in regards to orbitals of the same energy (in the same subshell). In the case of chromium, we are dealing with orbitals that are almost in the energy ($3d$ and $4s$) so you can essentially view this as a special case of Hund's rule that extends to orbitals of nearly the same energy.
For what it's worth, the other notable exception is copper, which adopts a $4s^1$ $3d^9$ configuration over $4s^2$ $3d^8$. For reasons that go beyond the scope of this answer, before you hit the transition metal elements ($Z \le 21$), $3d$ is higher in energy than $4s$, so we fill $4s$ before $3d$. But beyond this point, $3d$ becomes lower in energy than $4s$, so you lose $4s$ electrons first when we're talking about ionization. Knowing this, it is energetically more favorable to have $4s^1$ $3d^9$ because the higher energy orbital only has one unpaired electrons and the lower energy orbitals have paired ones.
The important thing to note from these two examples is that there is no special added stability form merely having a half full electron shell (i.e., something is not magically more stable because of having a half filled subshell). Adopting this form of configuration has consequences that are perfectly explainable to students of intro chemistry—do not brush it off as mere "magic."


text:
Can I determine the rate law order from stoichiometric coefficients?
I've read in two books, and seen in three separate videos courses, that you cannot determine the rate law order from stoichiometric coefficients.
Then, I have seen homework and "practice" problems that ask you to do exactly that.
Needless to say, I am confused.
This is how I am presently making sense of the issue: Given a balanced equation, and absent any other information, you can (and must) use the stoichiometric coefficients to determine the rate law. So, for example, given the equation $2A+B \rightarrow C + 3D$, the rate law would be $rate = k[A]^2[B]$.
But, this would only hold true if this were an "elementary reaction" (i.e., no equilibrium).
So, I wonder why the books and videos emphasize that you can only determine order by experiment and not from the stoichiometric coefficients. I am guessing that they do so because, in practice, you do not know if the reaction is an elementary reaction or not and therefore experiment is the only way to make that determination. And, because you don't really know if it is an elementary reaction, you can only determine the rate law experimentally.
The bottom line:

as a practitioner, determine rate law experimentally;
as a student, given no information other than the balanced equation, determine the rate law from the stoichiometric coefficients;
as a student, given rate and concentration data, determine the order (and rate law) from the data.

That is a pragmatic, but unsatisfying, solution to the problem but I would prefer a clearer explanation of what is going on.

In general, three body collisions are rare, so 2A + B will not be reacting via a three body collision and, therefore, it is not an elementary step reaction. Coincidences aside, and there are plenty, rate law orders are determined experimentally unless the reaction is an elementary step reaction. And this cannot just be assumed.

The stoichiometry indicates only how many moles of A and B etc react to give products and nothing about the mechanism of the reaction, even for apparently simple reactions. The mechanism can only be determined by experiment.

Thanks for the comment. I understand that "multi-body" collisions become rarer as the required number increases, but I had not yet "put two and two together" that the third order reaction [A]^2[B] would not be very probable as a practical matter. That is important to understand, obviously. But, it really does not change my position that the student is in no position to do anything but use the stoichiometric coefficients if no other information is given (assuming the student is not supposed to break the reaction into its intermediate states) (see my comment below to @Zhe's answer).

@RandallBlake if that is really the only information given, then the question is underspecified.

There are multiple issues here:

You should not assume that there is a simple rate law. Chemistry is complicated, and even simple reactions may proceed through multiple pathways in parallel.
More generally, you should not assume that you are given enough information to determine the solution. Just because the stoichiometric coefficients are there, does not mean that you can use them to determine the answer. So, if you just have a balanced reaction, you know nothing. If you want to become a scientist, you should learn to be comfortable with not having the answer.

So, I would say that your bottom line is incorrect.
The experimental rate law should be determined experimentally.
Given a putative mechanism with elementary steps and knowledge of rate constants for those steps, you can propose a rate law that for that mechanism. Do not assume it has any bearing on reality, since your putative mechanism may be wrong.
Frequently, the approach is to take the experimental rate law and see if the derived rate law for a proposed mechanism is consistent with the experimental result.

thanks for the explanation. I think "I get it." But, I have a small "quibble." You say that my "bottom line is incorrect" but I disagree. I agree that the rate law MUST be determined experimentally. Everybody says that and I have no problem with that. BUT, as a student answering a homework question, if I am ONLY given the balanced equation and asked to determine the order, then the ONLY thing I can do is use the stoichiometric coefficients (unless I'm being quizzed on multi-step reaction mechanisms). A practitioner, on the other hand, would simply go with the experimental measurement.

@RandallBlake "the ONLY thing I can do"... no, you could say there is insufficient information to arrive at a specific answer.

Oh, Karsten, if it were only so simple! Homework nowadays is served by heartless computers which demand specific answers of specific type. Non-conforming answers are judged by these inanimate tormentors as simply wrong. The student victim is left to "suffer the slings and arrows of outrageous fortune" wherein one MUST give the expected answer to the question that cannot be answered. So be it. There is some consolation in the knowledge that the experts here seem to agree that such questions are ill crafted and ill conceived (and that my computer oppressor will never know what it means to love).

@RandallBlake To be fair, it was always that way, except that that machines of old were just heartless people. On occasion, one was lucky to have a good teacher who understand that chemistry was a discipline filled with subtlety and nuance. I specifically remember getting 30% or the like on a quiz in eighth grade because I knew about orbitals, but I did not provide the "correct" answer.

Nobody can determine the order of a reaction with just the stoichiometric coefficients.
For example look at the reaction $I_2 + Acetone -> ICH_2COCH_3 + HI$. It has a rate corresponding to the law : -$d([I_2])/dt = k[H^+][Acetone]$. The concentration $[I_2]$ does not appear in the rate law. It may seem incredible, unbelievable. Nobody can guess it. It is an experimental result.
On the other hand, if a reaction A + B --> .....   has a rate law like : - d[A]/dt = k[A][B], in this case one can say that the reaction is an elementary reaction.

Just because the rate law matches the stoichiometry does not mean it is an elementary reaction. Lots of multi-step mechanisms would result in the same rate law. On the other hand, if you specify that a reaction is an elementary one, you can infer the rate law from the coefficients of the reaction as written.

"[i]f you specify that a reaction is an elementary one, you can infer the rate law from the coefficients of the reaction as written."   Karsten, if I had read your simple, concise statement in my chemistry book (or the many other sources I consulted), I would never have had to ask this question!! In hindsight, those sources suggested as much, but none, in my opinion, stated the matter so clearly and plainly. It would have saved me much heartache.


text:
Solubility Of Benzoic Acid
Why is benzoic acid well soluble in benzene (and other unpolar solvents, see e.g. wikipedia). What causes the hydrogen bonds in the solid to break? Something (i.e. polar and ionic substances) is soluble in water because of the hydration enthalpy with water, but i can see no similar cause for a solubility in benzene!

Are you saying benzoic acid dissociates in benzene?

@EdV So says wikipedia. https://en.wikipedia.org/wiki/Benzoic_acid

@Karl The wiki article gives pKa values for benzoic acid in water and DMSO, but benzoic acid also dissolves in hexane and carbon tetrachloride. No pKa values are given for those.

@EdV Ah, sorry. My eye read "dissociates", but my brain got "dissolves". ;) I meant the solubility lists in the box on the right. Nothing in benzene can take the proton, so no.

https://chemistry.stackexchange.com/questions/1289/how-do-non-polar-substances-dissolve-in-non-polar-solvents

What do you mean by "benzoate bond"? That is not a chemical terminus.

Solubility is one of the most complicated phenomena in chemistry. The "why" part of science is extremely difficult to answer in general. As a result one can ask one millions questions as to why iodine is soluble is benzene? What is so unique about benzoic acid? So many organic acids dissolve in benzene including acetic acid. Even a trace of water is soluble in benzene. Everything dissolves everything, to a very very small extent (depends on your limit of detection).
So the point to keep in mind is that although benzoic acid has a carboxylate group, it is a very hydrophobic acid and it literally floats on water. Similarly, you can sublime benzoic acid easily, so the intermolecular bond in benzoic is apparently not that strong.

So does the benzoic acid dissociates like that in water or does the benzene break the intermolecular force( van der waal) force of the benzoic acid molecules to distribute it i.e to dissolve it.

benzoic acid cannot dissociate in benzene. If it did, it the solution would have some electrical conductivity (that is how you can check if a molecule is dissociating in a solvent, by measuring their electrical conductivity). So benzene must be overcoming the intermolecular attraction among benzoic acid "molecule".

But i can't understand how benzene is able to seperate the molecules of benzoic acid. What's the mechanism behind this?

At a molecular level do not try to think in a literal way, as shown in animations that molecules are coming close, they pulling apart benzoic acid. Most of these animations on the false until and unless someone has done a real quantum calculation. The rest are cartoons. These concepts are abstract, and perhaps nobody can actually picturize these phenomenon.

I must say I strongly disagree to the sentiment in this answer. If I cannot think of a good reason why some substance A is well soluble in some other substance B, then either A is indeed not very soluble in B, or I haven´t been thinking hard enough.

I would guess that benzoic acid forms dimers or trimers connected by hydrogen bonds in nonpolar solvents. To the outside, these only show the phenyl rings.
The colligative properties of those solutions would clearly show how many individual particles they contain. I´ll leave the literature research for you. ;)

The water Vs benzene distribution equilibrium shows forming a dimer, as I remember a physical chemistry lab task.


text:
Is Gibbs energy minimized for processes at constant temperature are pressure only?
I've had this doubt for quite a while, This link
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.08%3A_Dependence_of_Gibbs_Energy_on_Temperature_and_Pressure
Clearly states that Gibbs energy can be used to predict the reactions that are happening at constant temperature and pressure.
Still I've faced many questions like "Effect on equilibrium on a system of ideal gases (in a closed container) upon addition of inert gas (while maintaining constant volume and temperature of the container). The answers of such questions involve the concept of equilibrium and they say - Since the partial pressures of the species involved in the reactions isn't affected (upon adding inert ideal gas) so the equilibrium is not affected. Thus the reaction doesn't move in any direction, it remains in equilibrium. (The equilibrium constant involve partial pressures of gaseous species)
But my question is why can we even use the equilibrium constant and reaction quotient to predict the direction of reaction as the total pressure of the system clearly rises. (Its not constant).
The usual expression of equilibrium constant relating activities of reactants and products is derived from taking Gibbs energy change for a reaction and then using dG = 0 at equilibrium. But the latter is only true for systems under constant pressure and temperature.
Example of the link to question -
Addition of inert gas at "constant volume"?
Please tell me where I'm wrong as I'm really confused right now!

Why isn't someone replying? Is the question not clear?


text:
Reason for double acylation of ferrocene only occurring with certain reagent combinations
Can anyone explain why the acylation of ferrocene produces the following results:
Ferrocene + acetic anhydride + AlCl3 ==> mono-acylated ferrocene
but:
Ferrocene + acetyl chloride + AlCl3 ==> di-acylated ferrocene (an acyl group on each ring).
I have seen various statements confirming this, but not a single explanation for it! My current thoughts are that it must be something to do with the carboxylic acid produced from the reaction with the anhydride somehow hydrogen bonding/blocking the second site, but not really sure how it would work.
Any explanations greatly appreciated!

Could it also maybe be that the HCl produced by the first acetyl chloride-catalyzed acylation is a stronger acid than the HOAc produced by the acetic anhydride acylation, and that the resulting acid could promote the 2nd acylation?  (Honestly not sure, just a thought...)

One factor:  since ferrocene is covalently bonded, the rings "communicate" with each other via the molecular orbitals passing through the iron atom.  Thus the acetyl group placed on one ring not only deactivates that ring, but to some extent deactivates the second ring as well.  The second ring would still be attacked preferentially in the second stage but may require a stronger electrophile, which acetyl chloride is compared with acetic anhydride.

Thank you - this seems to make sense!


text:
Is the speed of sound the fastest rate that a chemical reaction can progress through a substance?
I read that a martensitic transformation can travel at the speed of sound in that material. Is that confirmed as being the fastest speed of a chemical reaction (transformation)?

If the reaction front travels faster, you have an explosive.

@Karl Posted links in his comments here: https://chemistry.stackexchange.com/q/54282/79678 . Great link, by the way, and I will delete my other comment, since Karl has this!

@EdV On second thought, i think it is not clearly impossible that a solid solid transition spreads at a higher velocity. But of course ot cannot be much faster.

@Karl Based on wiki, stishovite is a polymorph of quartz that is produced in hypervelocity meteorite impacts. There is also a wiki article on “shocked quartz”. I have always assumed that melting was not involved, just shock.

Does a reaction or a phase transition A - - - > B need to propagate in general and in principle? Besides of seeds, this is obscure to me.

Certainly, chemical reactions can propagate far faster than the speed of sound. If the reaction emits electromagnetic radiation, e.g. light, and is itself initiated by light, then the speed of propagation is the speed of light in that medium. See this article from Sandia National Labs, or from the Royal Society.

"If the reaction emits electromagnetic radiation, e.g. light, and is itself initiated by light, then the speed of propagation is the speed of light in that medium"  In such a case the propagation speed wouldn't be the speed of light, because you need to account for the time for the reaction itself. I.e., a particle emits a photon, it travels to another particle, a reaction (which takes times) takes place in which that particle emits a photon, that photon then travels to the next particle, and so on.

@theorist, no, that is only the case where the medium is so closely packed that light cannot propagate within it, and then the maximum speed would be far less. Consider a large lump of nitrogen triodide on one counter, and another a meter away. Theoretically (pardon the expression), light from one lump detonating should reach the other in 1/3*10^8 seconds. BTW, the Sandia lab is based on the idea that the detonation is propagated by light.

The propagation speed of a reaction is generally taken to mean the speed with which a reaction propagates through the reacting medium. That doesn't describe your system, in which you have two lumps of reactant separated by a meter in a darkened room and shine a light on the first lump, leading to the second lump reacting 1/3 *10^8 seconds later. That's simply a trivial consequence of any reaction that both produces, and is initated by, light. In addition, it makes the term "propagation speed" not chemically meaningful, since you now have a propagation speed that is independent...

.... of the specific physical chemistry of the reaction itself (except that that it produces, and is initiated by, light).  It is telling that the Sandia article you reference doesn't contain a single use of the terms "propagate" or "propagation"; it only uses "initiate" and "initiation".  The Royal Society article is in German, but the English abstract, again, only uses "initiation", not "propagation".  Perhaps a better phrasing to describe your set-up is that is allows separated reactants to communicate at the speed of light.

A better example is UV initiated epoxy resin curing. I just don't know if they are that transparent to UV.

@user2617804  I don't see how that would count as light-speed propagation.  With UV-initated curing, you're simply initating the reaction with something  that travels at c (photons).  Once the UV hits, and initates the reaction, that reaction doesn't then propagate the next reaction.  That requires another photon.  So the best way to describe this would be that the reaction initiator moves through the reacting medium at c.


text:
Sodium hydride 60% dispersion in mineral oil
I've got a bottle of sodium hydride 60% dispersion in mineral oil. It's not "in mineral oil" it's just powder with an oil coating. I'm using it in an inert reaction, is there a way to remove the oil?

Place your weighed sodium hydride dispersion in the reaction flask. Swirl with hexane/heptane, let it settle, pipette off the clear solvent. Repeat then add your reaction solvent. The mineral oil is inert and will not interfere with your reaction and runs at the top of a tlc/column. Most of the time it is not necessary to remove it.


text:
Memory foam chemicals from pillow spoiled bed and rendered it useless
I have recently purchased a memory foam pillow and in two days i ve experienced breathing problems and stomach problems and eyes irritation due to the harsh toxins in it and I returned the memory foam. But, The memory foam has touched the walls next to the bed and the plywood on the bed. Because of which i m unable to use the plywood which is still causing the harsh effects of the memory foam pillow whenever i m spending time on it. I recently bought another plywood it picked the same toxins ( may be from the wall) and i m again facing the same problem. Need suggestion for a way to clean these chemicals away from plywood and wall ?

Write the manufacturer... or better, have your solicitor drop them a note. Start politely, and if you do not get satisfactory response, you can also go to social media.

it is not smells. I m guessing it is fumes of chemicals which are persisting.

I would suggest that perhaps it wasn't the pillow but another issue altogether.
If in two days your pillow outgassed enough to saturate the wall and the plywood, a couple days should also be sufficient for those to also release any trapped gasses. As I doubt they have any properties better at trapping gases than the foam in your former pillow.
Just a couple thoughts:
Do you have a carbon monoxide monitor in your home? A faulty furnace or water heater could be causing some of your problems.
Have you opened a window to provide better airflow?
Have you febreezed the area? The active ingredient in febreeze, cyclodextrin, does trap unwanted chemicals and gasses.
Or another chemical trap, is activated charcoal.

I have given airflow for many days. a month after the first plywood is outgassing and i got rid of it i purchased a new plywood and placed it. The new plywood started outgassing in 2 days. There is a cardboard box also next to the bed it was there when i got the pillow. I even use a quality air purifier ( which is supposed to absord harmful chemicals too ) all night that is also not helpful.

Actually the issue was caused just in 2 days after i bought a memory foam pillow. and when i removed the plywood from the room the air is normal but when i bring it back into room in 2 days the room is full of harsh chemicals in air. When i brought new plywood it also got the same chemicals on it.

Thanks for this question, as poor quality memory foam mattresses likely pose a significant health risk in certain locale.
Per Wikipedia on associated hazards of memory foam, to quote:

Emissions from memory foam mattresses may directly cause more respiratory irritation than other mattresses. Memory foam, like other polyurethane products, can be combustible.[9] Laws in several jurisdictions have been enacted to require that all bedding, including memory foam items, be resistant to ignition from an open flame such as a candle or cigarette lighter. US bedding laws that went into effect in 2010 change the Cal-117 Bulletin for FR testing.[10]
There is concern that high levels of the fire retardant PBDE, commonly used in memory foam, could cause health problems for users.[11] PBDEs are no longer used in most bedding foams, especially in the European Union.
Manufacturers caution about leaving babies and small children unattended on memory foam mattresses, as they may find it difficult to turn over, and may suffocate.[7]
The United States Environmental Protection Agency published two documents proposing National Emissions Standards for Hazardous Air Pollutants (HAP) concerning hazardous emissions produced during the making of flexible polyurethane foam products.[12] The HAP emissions associated with polyurethane foam production include methylene chloride, toluene diisocyanate, methyl chloroform, methylene diphenyl diisocyanate, propylene oxide, diethanolamine, methyl ethyl ketone, methanol, and toluene. However, not all chemical emissions associated with the production of these material have been classified. Methylene chloride makes up over 98 percent of the total HAP emissions from this industry. Short-term exposure to high concentrations of methylene chloride also irritates the nose and throat. The effects of chronic (long-term) exposure to methylene chloride in humans involve the central nervous system, and include headaches, dizziness, nausea, and memory loss. Animal studies indicate that inhalation of methylene chloride affects the liver, kidney, and cardiovascular system. Developmental or reproductive effects of methylene chloride have not been reported in humans, but limited animal studies have reported lowered fetal body weights in rats exposed.[13]

Here also is a rather condemning depiction of possible issues per an organic competing company (which provides source links), to quote in part:

Toxic chemicals in memory foam
  So what are those “additional chemicals”? What do we know about memory foam?

There have been legal consequences for memory foam companies which claimed their products are free of volatile organic compounds (VOCs).
Some memory foam mattresses contain toxic chemicals such as formaldehyde, benzene, and naphthalene.
Memory foam may contain isocyanates, which, according to the Occupational Safety and Health Administration, can cause irritation of the eyes, nose, throat, and skin. Exposure to isocyanates can also lead to chest tightness and asthma.

Polyurethane: flame retardants, off-gassing, and safety
  Memory foam mattresses often contain chemical flame retardants which have been linked to a litany of health problems, including developmental brain disorders, cancer, and obesity. Organic wool, however, is a soft, natural material that not only can allow a mattress to pass the mandated flame test but adds a layer of airy cushioning to the top of your mattress.
It’s not uncommon to notice a chemical off-gassing smell upon unpacking a memory foam mattress. The best way to get rid of the odor is to allow the mattress to air out by moving air through the room. This can be done by opening doors and windows and running a fan.
Memory Foam Health Risks
  So what’s the risk? The risk is the unknown. We can’t positively say a memory foam mattress will cause “this effect.” Every person is different. Every mattress is different. Some memory foam mattress users complain of waking with headaches. Others develop respiratory issues such as asthma. Still, others have concerns about the long-term health effects of years of nightly exposure to this material.

If one decides to employ a cheap version of a memory foam product, against likely good advice, I would advise allowing for a lengthy period of aeration on say a porch.
To clean up contaminated areas, I would recommend first creating some aqueous zinc acetate from the action of H2O2 (now sold in stores in large bottles for laundry purposes) plus vinegar plus NaCl plus a piece of graphite on a piece zinc metal. Jumpstart the electrochemical reaction in a microwave for a minute. Let stand for a few hours and add water to dilute.
Next step, apply a mix of H2O2 plus Washing Soda (Na2CO3) to the target areas.
Finally, while wet, spray on a mix of vinegar plus the aqueous zinc acetate.
The basis for this recipe, see "Development of Bicarbonate-Activated Peroxide as a Chemical and Biological Warfare Agent Decontaminant".

is there an option to buy aqueous zinc acetate online readily ?


text:
Strong exothermic mixing of two 'safe' liquids
I have read (thanks to Stack Exchange) that mixing water and sulfuric acid is exothermic, as is NaOH and HCl.
Are there two 'safe' liquids that can be mixed and produce anywhere near the same amount of heat? By 'safe' I mean they don't come with the same heavy warnings as sulfuric acid or lye. I would like them to be safe enough to be stored in a home and that a spill would probably cause some damage but clean up could be performed by the home owner safely.
I understand that the actual mixing might not be considered safe if they do release a lot of energy.
I'm only considering liquids because they can be kept separate and pumped together.

Being the two individual liquids safe means that they are both relatively inert. So the heat will be merely that of mixing. I don't expect any couple to release too much of heat by just that. Different is the case in which you mix an acid with a base, or when one liquid can be easily ignite. For instance interaction between flammable solvents and very active silica gels can cause fires.

@Alchimista: Inert? Not necessarily. The reaction may require a catalyst that's present in only one of the two liquids. The question doesn't appear to require that both liquids are pure.

Acetone could well be found in the home, but I think that dichloromethane might be rarer. Either is likely to cause damage when spilt as they are both good solvents for a range of materials. Does "highly flammable" count as a "heavy warning" for the purpose of this question?

@MSalters I said relatively inert. I think is pretty clear what I meant.

If you need a quick source of moderate heat, why not use a single liquid? Sodium acetate trihydrate ($\ce{CH3COONa·3H2O}$), for example, releases heat energy on crystallization... and it takes very little to trigger that crystallization! Commercial heat packs contain a "clicker", which makes a sharp vibration. You can make your own heat pack fairly safely and inexpensively: sodium acetate can be found for ~US$8/450 g, and a Safety Data Sheet states, "it's  not classified for physical or health hazards under GHS."
Not only can you get a noticeable temperature rise quickly, it's reusable.
It would be more difficult to find liquids that quickly release energy that do not have physical hazards (e.g. ignition) or health hazards. Hypergolic fuels, for example, tend to be rather toxic, as well as explosive. You might get a bang out of reading John Drury Clark's Ignition!, though.


text:
PdPS interlayer distance
PdPS is being considered as a layered material but when I look at its crystal structure it does not look like that. Fo example I could not figure out its interlayer distance. Does anyone familiar with this structure?

Does this help or this or this?

Could you post the information about the structure that you have trouble interpreting?

Here is a visualisation that shows the layered structure of the material.

From the side:

This is very helpful, and it becomes possible to deduce the layer distance from the length of the unit cell vector a. Could you post a view of a single layer in a rotated view as well? S is yellow, P orange, and Pd bluish, right?

Yes, that's is the colour scheme.


text:
How to obtain the linear form of Redlich Peterson isotherm model by algebraic transformation or any other ways?
The Redlich Peterson isotherm model is known as:
$$
\begin{aligned}
q_e=\frac{AC_e}{1+BC^g_e}
\end{aligned}
$$
Is there any way to obtain the linear form of this equation by algebraic transformation or any ohter ways(like using software origin)?In other words,given that $q_e$ and $C_e$ as variables and A,B,g as parameters,how can I make linear fitting based on this equation?

What about taking logs on both sides?

@M. Farooq I am not sure of it.Could you give details?

a= c/d, log (a)= log (c/d), which is log (a)= log (c)= log (D). Also you can invert the equation 1/qe= 1/AC+ AC/BC^g? Does it make same sense?

Is this its linear form?

If you are familiar with Origin (which you suggest) you can fit your data directly. To get initial values for the fitting, for A try the value of q when C is small, and A/B when C is large.

a= c/d, log (a)= log (c/d), Sorry a typo, log (a)= log (c)- log (D). This is also a linear form 1/qe= 1/AC+ AC/BC^g? If it still does not make sense, watch youtube for linear equations.

If you want to use a linear form rearrange to  $\displaystyle  \frac{C}{q}=\frac{1}{A}+\frac{B}{A}C^g$ and a plot $C^g$ vs $C/q$ which has intercept $1/a$ and slope $B/A$.

Directly fitting the equation, as per the first comment by @porphyrin, is the best way to proceed: it does not alter the noise structure and programs like Origin make this curve fitting trivially simple. The linearization idea is from the days before curve fitting software. Personally, I wish it would fade away.


text:
Fractional index in a chemical formula of metal oxide
I have just read an article in which they used $\ce{WO3}$ and $\ce{WO_{2.9}}$ as precursors, whose objective was to form $\ce{WS2}$, in an atmosphere of argon by CVD. In one boat there is the powder of $\ce{S}$ and in another boat there is $\ce{WO3}$ or $\ce{WO_{2.9}}$, in an oven with a double temperature zone. In the article they conclude that $\ce{WO_{2.9}}$ forms larger layers of $\ce{WS2}$ in the substrate giving the following justification:

They found that $\ce{W^6+}$ cannot be directly sulfated by $\ce{S},$ unless some intermediates are formed due to the high energy of the $\ce{W-O}$ bond. The reduction from $\ce{W^6+}$ to $\ce{W^5+}$ is mandatory for the incorporation of sulfur in the $\ce{WO3}$ network. For $\ce{WO_{2.9}}$ in our case, its partial $\ce{W^6+}$ ions have been reduced to $\ce{W^5+}$ or $\ce{W^4+}$ ions. Therefore, we find that replacing $\ce{W^5+}$ or $\ce{W^4+}$ with $\ce{W^6+}$ in the initial stage facilitates the growth of the single crystal $\ce{WS2}$ film.

I didn't realize what the difference is between $\ce{WO3}$ and $\ce{WO_{2.9}}$ and where these ions come from that they talk about in this part of the article. I also don't understand what a fractional index means in a chemical formula. Could you help me to interpret that part of the article and explain these doubts that I have please?

Similar question for W3O8: https://chemistry.stackexchange.com/questions/65862/oxidation-state-of-tungsten-in-w3o8

There is nothing special about fractional indices in chemistry, it is just a pointer signifying we are dealing with a non-stoichiometric compound.
Group 6 metal are known not only for their regular oxides existing in numerous polymorphic modifications, but also for forming homologous series of nonstoichiometric oxides $\ce{\ce{M^{VI}_nO_{3n-1}}}$ and $\ce{\ce{M^{VI}_nO_{3n-2}}}$ with the general formula $\ce{M^{VI}O_{3-x}}$ $(\ce{M} = \ce{Mo}, \ce{W};$ $x \in (0; 1)).$
Now, to make things clear, notation "$\ce{WO_{2.9}}$" does not refer to a molecule; $\ce{WO_{2.9}}$ is a formula unit showing composition of a network solid.
To further illustrate the origin of the fractional coefficients here, let's refer to the crystal structures of the stoichiometric and non-stoichiometric tungsten(VI) oxides $(\ce{WO3}$ and $\ce{WO_{2.9}},$ respectively).
Depending on the temperature, $\ce{WO3}$ adopts several polymorphs with various crystal structures; at the room temperature the most common structure it adopts is the one of distorted $\ce{ReO3}$ type, being made of tilted $\ce{[WO6]}$ octahedra cross-linked via the corners (for the illustration, see e.g. Wikipedia).
On the other hand, $\ce{WO_{2.9}}$ is an oxygen-deficient compound in comparison.
To account for that and eliminate the presence of point defects, crystal structure undergoes a transformation resulting in increased number of edge-sharing groups of $[\ce{WO6}]$ polyhedra and reduced number of the corner-sharing ones, e.g. the structure gets more "packed", and, more importantly, a crystallographic shear is introduced (which is not an impurity).
Let's have a look at the asymmetric unit of $\ce{WO_{2.9}}$ of the crystal structure determined by Magnéli [1]:

Figure 1. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing an asymmetric unit $\ce{W10O29}.$
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.
Due to the asymmetric unit $\ce{W10O29},$ the chemical formula (formula unit) is reduced to $\ce{WO_{2.9}}.$
One can also fill the entire unit cell knowing the number of formula units $(Z)$ per unit cell; here, $Z = 2$ and this results in equal formula $\ce{W20O58},$ often seen in the literature:

Figure 2. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing packed unit cell.
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.
Finally, to illustrate the location of the aforementioned shear plane, let's complete coordination shells on all tungsten atoms and view the structure along the $b$ axis:

Figure 3. Crystal structure of $\ce{WO_{2.9}}$ [1, ICSD-24736] showing polyhedral representation ($[\ce{WO6}]$ octahedra) with the shear plane along $a$ axis denoted with red arrow.
Color code: $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#2194D6}{\Large\bullet}~\ce{W}$.
As for the assignment of the formal oxidation numbers for the non-stoichiometric compounds like this, this is typically considered a non-trivial task of a questionable value.
Also, see the answer of mine to the related question Oxidation state of tungsten in $\ce{W3O8}$.
References

Magnéli, A. Structure of β-Tungsten Oxide. Nature 1950, 165 (4192), 356–357. DOI: 10.1038/165356b0.

There are also references to a $\ce{W_{18}O_{49}}$ such as this one.  Would this be a combination of $\ce{W_nO_{3n-1}}$ and $\ce{W_nO_{3n-2}}$ with $n=6$?

@OscarLanzi Judging from the structure (ICSD 202488 by Lamire et al. 1987), $\ce{W18O49}$ is a different story: there is a tungsten with C.N. 7 (pentagonal bipyramid, like in many uranium(VI) componds) forming pentagonal assemblies with other $[\ce{WO6}]$ octahedra, which is even more efficient packing (than shear plane) that can afford to account for further O-deficiency.

Thanks.  One minor point:  "Further O deficiency" seems to be a judgment call.  $\ce{W_{18}O_{49}}$ is less deficient than $\ce{W3O8}$; the latter, on an equal tungsten basis, would be $\ce{W_{18}O_{48}}$.

@OscarLanzi True, let's just agree on oxygen deficiency.


text:
Why isn't ΔH = TΔS in ΔG = ΔH - TΔS?
Applying a Legendre transformation on $U = q - w$ we get the familiar $G = H - TS$.
Making an innocent approximation delivers $\Delta G = \Delta H - T\,\Delta S$.
When one wants to predict the spontaneity of a chemical reaction, one gets the $\Delta H$ and the $\Delta S$ for the reaction in question from tables, and both refer to the system undergoing the transformation. Therefore we can write:
$\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - T\,\Delta S_\mathrm{sys}$
In which:
$\Delta G$: Variation of the system's Gibbs energy
$\Delta H$: Variation of the system's enthalpy
$\Delta S$: Variation of the system's entropy
However: $\Delta S_\mathrm{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).
So: $\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - q$. If we are making predictions based on Gibbs energy, we are under constant pressure and so $\Delta H_\mathrm{sys}=q$.
Therefore: $\Delta G_\mathrm{sys} = q - q$, $\Delta G_\mathrm{sys} = 0$ and we are always on equilibria.
Where am I mistaken? I've already seen some say that some stuff refers to the surroundings, but that makes no sense to me. If both are at thermal equilibrium, $q$ is going to have the same effect on the entropy of the system and that of the surroundings, from $\Delta S = \frac{q}{T}$.
I'm a chemistry undergratuate and have seen no mention to anything related to this on common p-chem textbooks (Atkins, McQuarrie, Ball).

You suppose that the number of moles is constant. It is not the case in chemistry. In a chemical reaction ΔG° is measured by ΔG° = -zEF or by ΔG° = -nRTlnK. And ΔH is measured by  ΔH = mCΔT. ΔG changes a lot with temperature. ΔH does not change so much. Both curves (ΔG° vs. T and ΔH vs T) intersects at O K. The difference is TΔS.

I believe my answer here directly explains why it's not the case that "$\Delta G_{sys} = q - q$, $\Delta G_{sys} = 0$ and we are always on equilibria.", while at the same time addressing your question about the entropy of the system vs. the entropy of the surroundings:  https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/124470#124470

The following thread in another forum specifically addresses your question is great detail: https://www.physicsforums.com/threads/thermochemistry-challenge-problem-chets-paradox.913567/ DrDu in post #2 does a masterful job of answering.

If the process is reversible, $\Delta S_{universe}=0$. If in addition the temperature is constant (the "innocent" approximation), $\Delta S_{sys}=\frac{q}{T}$ and $\Delta S_{surr}=-\frac{q}{T}$. If the pressure is also constant and there is only pV work, then $q=\Delta H$ and $$\Delta S=\frac{\Delta H}{T} \tag{system}$$ Therefore $\Delta G=0$ for a reversible process at constant T and p.
If the process is irreversible, $\Delta S_{universe}>0$. If in addition the temperature is constant, $\Delta S_{surr}=-\frac{q}{T}$, but now $\Delta S_{sys} \neq \frac{q}{T}$. If the initial and final states of the system are the same as in the reversible process, then $\Delta S_{system}$ is equal to the change for the reversible process (one reason the 2nd law of thermodynamics is useful), so that $$\begin{align}\Delta S_{univ,irrev}>&\Delta S_{univ,rev}\end{align}\\ \Delta S_{sys,irrev}+\Delta S_{surr,irrev} > \Delta S_{sys,rev}+\Delta S_{surr,rev} \\ \Delta S_{surr,irrev} > \Delta S_{surr,rev} \\ q_{surr,irrev} > q_{surr,rev} \\ q_{irrev} < q_{rev} $$
(the convention used here is that, unless noted otherwise, q is the heat exchanged by the system and is positive for an endothermic process).
So the answer is that there is no error. A reversible process is one in which the system passes through an infinite series of equilibrium states, an admittedly odd conclusion.

However: $\Delta S_{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

No. $\Delta S_{env} = \frac{q}{T}$, assuming the process is reversible (as usual). So you get $\Delta S_{universe} = -\frac{\Delta G}{T}$ for a process where the heat exchange with the environment is reversible (and there is no non-PV work).
For more background, see https://chemistry.stackexchange.com/a/114323

Why tho? Why is it that the entropy change of the system is not equal to its enthalpy change over temperature but the surrounding's is?
If I were to insert some heat into the system wouldn't its entropy change be given by heat/temperature? It seems that there is some not clear distinction here. What I got is that the heat of one can be used to calculate the change in entropy of the other and vice versa but not the change in itself entropy.

@BananaAsker If your argument would hold then there could be no change of entropy in an insulated system. q would always be zero, so no change in entropy. However, all you have to do is to mix two substances, and the entropy increases. So there is a flaw in the argument.


text:
Raoult and Henry's law
According to my textbook, for Raoult's law, we have:
$$μ_i(\mathrm{vap}) = μ_i^* (p_i^*,T) + RT \ln x_i$$
where $μ_i^* (p_i^*,T)$ is the chemical potential at $T$ and $p_i^*$, the pressure of pure $i$. So, when $x_i=1$, we have $μ_i (vap)=μ_i^* (p_i^*,T)$. That's okay and makes sense.
For Henry's law, we have:
$$μ_i(\mathrm{vap}) = μ_i^∞ (p_j^*,T) + RT \ln x_i$$
where $μ_i^∞ (p_j^*,T)$ is the chemical potential at $T$ and $p_j^*$, the pressure of pure $j$ (the other component of the mixture). So, when $x_i = 1$ we have $μ_i^∞ (p_j^*,T) = μ_i^∞ (p_j^*,T)$ and that doesn't make any sense, since we have $i$ pure and the equation says that the chemical potential is the chemical potential of pure $j$.
Am I interpreting something wrongly?

Your last equality has two identical symbols on both sides of the = sign ! You simply repeat the same symbols on both sides. These formula cannot be different. It is like writing an equation :  a = a.

In addition to Maurice's comment, your last statement "we have i pure and the equation says that de chemical potential is the chemical potencial of pure j" is inconsistent with the equations you have written, which describe the chemical potential of i (not j).

The first equation is for the chemical potential of species i in an ideal liquid solution, neglecting the Poynting correction.  The second equation is an approximation to the chemical potential of species i in a real solution in the region of low values of the mole fraction of i, and apparently includes both the Poynting correction, hence evaluation of the chemical potential at the total pressure p (not just pj) as well as a term for RT times the natural log of the infinite dilution activity coefficient of species i. So the equation only applies at low values of the mole fraction, and the first term in the equation represents the extrapolation of the low mole fraction limiting behavior to a mole fraction of 1.


text:
Oxidation states the elements in nitrosyl chloride
I have read that $\ce{NOCl}$ dissociates to $\ce{NO+}$ and $\ce{Cl-},$ and the various reasons stated as such seem fine.
But what I don't get is this: if we follow the actual rules of assigning oxidation numbers, we find that nitrogen has got an unit positive charge with an unit positive charge on chlorine too (if we follow the structure of $\ce{NOCl}$ closely).
So, am I wrong in judging the oxidation number of nitrogen in $\ce{NOCl}$ to be $+1$ instead of $+3,$ as the dissociation of $\ce{NOCl}$ seems to go against my interpretation? What should be the oxidation number of nitrogen in $\ce{NOCl}$ in that case?

Related: Why does NOCl dissociate as +NO and Cl-?

If i assume the reverse case then we would have cl+ which is much much less stable than a cl- ion.It is better to have a positive charge on NO as it can be resonance stabilised.I'm just a 11th grader and i've been taught to believe that whenever i get a positively chared halide radical,it's a red flag for me(positive charge is really unstable on a halide radical but not practically impossible)


text:
Activation energy vs temperature
We know that activation energy is independent of temperature. But during the dehydration of alcohol or any reaction, activation reaction can be increased by following the thermodynamically controlled product, which is favourable at high temperature. So how is it possible?

The thermodynamically controlled product is irrelevant to activation energy.

There are two misconceptions in the OP's question

We know that activation energy is independent of temperature.

This is not true in general. Granted, over fairly small temperature intervals (for example, going from zero to hundred degrees Celsius is only about 25% change in absolute temperature terms) Arrhenius plots mostly show a fairly straight line (i.e. constant activation energy). However, there are examples of significant non-Arrhenius behavior, e.g. https://pubs.acs.org/doi/pdf/10.1021/j100310a026.
See also: Is activation energy temperature-independent?

[...] activation reaction can be increased by following the thermodynamically controlled product, which is favourable at high temperature

As pointed out by Ivan in the comments, once you have a thermodnamically controlled product, the kinetics no longer determine the outcome. So the explanation for the different product ratios is not that the activation energies of the two competing reactions have different temperature dependence; instead, at the high temperatures the reverse reactions also take place, and the product of lower Gibbs energy forms (i.e. the product ratio is determined by the equilibrium constants of the two processes).

Activation energy is an experimental result, obtained from the slope of the log of the rate constant vs $1/T$. It is usually determined over a rather small domain of temperature, much less than $100\ \mathrm{^\circ C}$. Who knows whether this slope is dependent on $T$, if it could be measured between, say, $10\ \mathrm K$ and $1000\ \mathrm K$?


text:
Why is it not practical for insoluble salts to be prepared from other insoluble salts?
My textbook says that it is not practically feasible to make insoluble salts (say PbSO4) from other insoluble salts (say PbO), without explicitly mentioning the reason for the same.
My guess is: if the product is also an insoluble salt, it may deposit on the insoluble reactant, and stop further reaction. However, isn't this problem solved by simply stirring? (Is the constant stirring required the reason why such a method would not be "practically feasible"? But isn't stirring rather common for chemical reactions?)
So, what is the actual reason?

How would the counterions that constitute the product ever encounter each other if not dispersed as intermediates in a solvent? You could have interfacial reactions occur in solids but those would proceed slowly.

@BuckThorn I had guesses the same initially, but then I wondered : wouldn't this reasoning imply that you cannot make any salt from an insoluble salt (which isn't the case, as you can make lead nitrate from lead oxide by adding dilute HNO3)?

@BuckThorn Another guess I have is: if the product is also insoluble, then it will form a layer on the surface of the insoluble reactant, ceasing further reaction. Only caveat: this must be easily solvable by just stirring the mixture? (maybe constant stirring is why it's not 'practically feasible'? But, then again, stirring is rather common for chemical reactions, isn't it?)

I think the key point is that there are kinetic factors at play that might interfere with progress to thermodynamic equilibrium. As you yourself suggest, there are a lot of possible scenarios one can envision, some approximating what occurs in reality. The generalization is that insoluble means here also that you don't form reactive intermediates. Anything that encourages formation of reactive intermediates (here adding a particular acid to some solid) will encourage formation of the thermodynamically favored solid.

In few words,from thermodynamic point of view, it does not matter they are insoluble. From kinetic point of view, it is the major thing that matters. It would be very slow in the best .

@Arjun Constant stirring sounds very innocent, but when you have solid-liquid suspensions, it's quite power-intensive. It's quite a hassle even when the product is soluble (see the phosphoric acid production process where rock phosphate gets a passivating sulphate layer.)

The comment "My textbook says that it is not practically feasible to make insoluble salts (say PbSO4) from other insoluble salts" is not precisely accurate. Limited solubility in water does not necessarily imply a lack of solubility in an acidic medium.
For example, I have prepared Silver acetate (moderately water-soluble, see Table) from the action of dilute acetic acid and 3% H2O2 in the presence of an electrolyte (a small amount of KNO3). Jumpstart the reaction in a microwave. The created Silver acetate is somewhat soluble in acetic acid (see this), hence there is no issue in the creation reaction.
The addition of moderately soluble PbCl2 to Silver acetate/acetic acid likely creates insoluble AgCl and Lead acetate (where the Silver acetate is also moderately water-soluble with respect to water). An explanation of solubility properties, per reference above, relates to a possible creation of an acetic acid salt.


text:
Do metals react with ammonia gas (not liquid phase)?
In my chemistry notes I have two chemical equations as follows:
$$
\begin{align}
\ce{2 NH3(g) + 6 Li(s) &-> 2 Li3N(s) + 3 H2(g)}\tag{R1} \\
\ce{2 NH3(g) + 2 Na(s) &-> 2 NaNH2 + H2(g)}\tag{R2}
\end{align}
$$
Before digging into the reasons for different products in two equations, when I searched in the internet to find the clear picture of what are the metals (like alkali metals, alkaline earth metals, transition metals) that can react with ammonia in gaseous form, I could only find metals dissolving in liquid ammonia.
I would be grateful if someone could tell that metals don't react with gaseous ammonia, or, if it reacts, what are the metals those react with gaseous ammonia? And are there any conditions needed for the reaction (like heat)?

@andselisk , thanks for editing the question to give it a clean view, How can I input chemical equations in the format that u have edited? And thanks for the comment for the question that I can cling to the fact that Alkali metals react with ammonia gas with elevated temps.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Thanks @Nilay Ghosh the content in the link is exactly what my second question in the mind.

That is how nitride layers are made on steel - react in ammonia.

From Chemiday.com, it is stated that magnesium reacts with ammonia to form magnesium nitride and hydrogen gas. The temperature is somewhere between 600 and 900 degrees Celcius. The give several references.

@ed can you offer some of the references?  WP does not give any (see my answer) and Chemiday by itself is not trusted in this forum.  Thanks!

I will see what I can find and thanks for the Chemiday input: I know nothing about it and will not reference it again. One thing, though. If magnesium (not actually burning) reacts with ammonia gas at 700 deg C, as you have in your answer, then I find this quite reasonable. Certainly Mg burning in air will continue to burn if the air is replaced by nitrogen. Then if the nitrogen was replaced by ammonia, I think the Mg would keep burning: the ammonia would thermally decompose to nitrogen and hydrogen gases.

Wikipedia reports that magnesium nitride can be prepared:

By passing dry nitrogen over heated magnesium:
${\displaystyle{\begin{matrix}{}\\{\ce {{3Mg} + N2->[{\ce{800^{\circ }C}}]Mg3N2}}\\{}\end{matrix}}}$
or ammonia:
${\displaystyle{\begin{matrix}{}\\{\ce{{3Mg} + 2NH3->[{\ce{700^{\circ }C}}]{Mg3N2} + 3H2}}\\{}\end{matrix}}}$

No references are cited.

Per this source https://www.chem.tamu.edu/rgroup/goodman/pdf%20files/cat-ltr-72-197-01.pdf even at 500 C with varying metal catalyst, per Figure 4, there is apparently significant decomposition of ammonia. As such, it is not likely that the active agent reacting with Mg, at 700 C, is, in effect, NH3. In other words, the cited reaction is not likely accurate. Check out my answer.

As to the question does gaseous NH3 directly react with metals, other than complex formations, a more accurate short answer is likely no. A better answer is a reaction may proceed from heat (or other energy sources, like light) induced breakdown products of ammonia interacting with the metal.
A review of the literature suggests that if ammonia is heated to high temperature or subject to photodissociation, then a reported radicalization may occur per a source, Journal of Chemical Physics: 'Dissociation of NH3 to NH2 + H ':
$\ce{NH3(g) + Heat/UV light/ArF laser -> •NH2 + •H}$
The laser reference is provided here. A thermal reference relating to the heating of ammonia on a Tungsten metal surface states:

When NH2 forms by the dissociation of NH3 with the release of H, the bond becomes covalent...

Thus, it is possible in the presence of say a metal M:
$\ce{M -> M+ + e-}$
$\ce{•NH2 + e- -> NH2-}$
For a net reaction:
$\ce{M + •NH2 -> MNH2}$
Note, the so-called amino radical can be created at near room temperatures as well, as per Wikipedia, to quote a reaction between ammonia and the hydroxyl radical:
$\ce{NH3 + •OH → •NH2 + H2O}$
And, may I suggest a related, but slower radical reaction, as a possibility with say surface absorbed •H on Aluminum (discussed on a prior thread), placed in a NONE aqueous medium containing ammonia (in relative excess to the hydrogen radical):
$\ce{NH3 + •H → •NH2 + H2}$
Source: See Reaction [10] in this 2008 article:'Photochemical Behavior of Ammonia in Aqueous Suspension of TiO2' by Ki-Min Bark, available online.
$\ce{•NH2 + •H → •NH + H2}$
Added source: See ebook. And also possible an unwanted reverse reaction like:
$\ce{•NH2 + •H → NH3}$
Nevertheless, in the presence of select metals some possible creation (per reaction scheme outlined above) of corresponding salts of interest with ammonia. For example, per this 2000 source with magnesium:

MgNH2 could be produced in interstellar/circumstellar gas from the association reaction of Mg+ + NH3, as predicted by theory.

It should be further noted commencing with the formation of $\ce{NH4+}$ salt from ammonia, the following reaction pathway can lead to other salts:
$\ce{NH4+ ⇆ H+ + NH3}$
$\ce{H+ + e- ⇆ •H}$
$\ce{NH3 + •H → •NH2 + H2}$
An example includes $\ce{ZnNH2}$ whose surface formation on ZnS in the presence of ammonia is noted. From a thesis by Peng Huang (available online), 'CHITOSAN IN DIFFERENTIAL FLOTATION OF BASE METAL', to quote:

Due to the cationization effect, ZnNH3+ represents ZnNH2, indicating that strong interaction occurred between –NH2 and zinc on sphalerite surface. This hypothesis was further supported by the XPS results (i.e., amine group NH2 was the dominant species...

My final example is a bit explosive, I am referring to Silver nitride (Ag3N) whose formation is reportedly accompanied with secondary products of Silver amide (AgNH2) and the imide (Ag2NH). As a clue to the reaction mechanics, I quote a source noting the conditions favoring Ag+ creation:

Elemental silver ionises rapidly when exposed to an aqueous environment such as wound exudate.[66]

So, assuming the silver is an alloy (and not 100% Ag) in the presence of aqueous ammonia, to cite a source:
"Ammonia precipitates Ag+ as brown/black oxide.
$\ce{2 Ag+(aq) + 2 NH3(aq) + H2O(l) -> Ag2O(s) [brown/black] + 2 NH4+(aq)}$
In excess, ammonia dissolves the precipitate, forming a diamine silver complex.
$\ce{Ag2O(s) + 4 NH3(aq) + H2O(l) -> 2 [Ag(NH3)2]+(aq) + 2 OH−(aq)}$ "
Unfortunately, on standing with water evaporation and, I would argue, likely exposure to oxygen and light, a reverse reaction reforming Silver oxide and NH4+. More recently, a 2011 article noting that the action of light on Ag2O forms an Ag/Ag2O pair that is powerful photocatalst. The introduction of electron holes and e- likely attacks the ammonium ion as I detailed above leading to AgNH2 and beyond.
For the general interaction of the amino radical with organics, a good review can be found here.


text:
Description of a van't Hoff apparatus
A "van't Hoff apparatus" is mentioned as a means of performing a reaction involving gases in a reversible manner, for instance here.
Unfortunately I have been unable to find a description of such a device using a web search (including on this site). It may be that the device has other names. I intuit from the post above that the apparatus is similar to the device used to measure the Joule-Thompson coefficient of a gas.
Could someone provide a description of the nature and operation of such an apparatus (even if it is only a theoretical device)?

Van't Hoff Equilibrium Box

I got here through Enrico Fermi Circa 1936
Title: thermodynamics, pp101-109, Chapter VI Gaseous Reactions, section 22. Per Fermi this is a gedunken featuring hypothetical ‘perfect semipermeable membranes’ that only allow a certain gas to pass through. Maybe helpful?

Thanks for the reference. I think you mean "gedanken experiment"

The Van't Hoff equilibrium box is a chamber containing a mixture of reactants and products at equilibrium.  It is connected to an array of cylinders, each containing one of the pure reactants or products.  The chamber itself is operated at constant total pressure and temperature.  The cylinders are each connected to the equilibrium box through a semipermeable membrane that only passes that species into the chamber.  Pure species are introduced in stoichiometric proportions to the chamber through these membranes at the same partial pressures as in the reaction mixture, and pure products are removed in stoichiometric proportions at the same partial pressures as in the reaction mixture.  So the operation of the equilibrium box is reversible, and the delta G in going from pure reactants in to pure products out is zero.

Ok, thanks! A search for Van't Hoff equilibrium box did turn up something substantial, such as this: http://www.eoht.info/page/Van%27t+Hoff+equilibrium+box . I suppose that, in order to separate the products, the pistons must completely evacuate the reaction chamber.

No.  Please see my answer.  The reaction mixture in the chamber does not change.  The products are removed at the same rate that the reactants are introduced.


text:
Do isotopic groups cause optical activity?
I understood the simplified reason for optical activity of a compound based on the difference in the angular velocities of counter-rotating circularly polarized light. The difference is due to the spatial arrangement of different groups with different electronic configurations. Even though it wasn't specified in my textbook, I guess the difference in the electronic configurations cause some electrical constraints to the oscillating electric field of the plane polarized light, thereby rotating the plane of polarization.
Do isotopic groups cause optical activity?
For example, is $\ce{CHDTF}$ optically active due to the presence of different isotopes of hydrogen?
I understood that $\ce{CHTDF}$ is chiral or it exists as a pair of enantiomers. But I don't know whether different isotopes of hydrogen account for any optical activity as I think optical activity is something to do with the electronic configurations rather than the atomic masses.

related https://chemistry.stackexchange.com/questions/93737/does-cht-cht-have-a-geometric-isomerism-because-of-tritium-atoms https://chemistry.stackexchange.com/questions/44004/are-elemental-compounds-composed-of-different-isotopes-polarised

That's an example of cryptochirality - https://chemistry.stackexchange.com/questions/103048/are-there-chiral-compounds-that-dont-rotate-plane-polarized-light

The book Organic Conformational Analysis and Stereochemistry from Circular Dichroism Spectroscopy has a chapter on the CD of ketones which are chiral only through deuterium substitution.

@Tyberius: Thanks for the comment. However, I don't have a copy of that book. Is there any mention of optical activity (specific rotation) and explanation based on the difference in the counter rotating circularly polarized light components?

I will add an answer in a bit, but most of that chapter is available as a preview from Google books (though it doesn't offer a lot of explanation of why this occurs, only that isotopically chiral molecules have been synthesized and shown to be optically active)

Thank you Tyberius. Reference for those who don't have a copy of the book mentioned : Instances of deuterium

Optical rotation (OR) emerges from a difference in the index of refraction of a compound towards left/right circularly polarized light. All chiral have some difference in these indices of refraction. Your question specifically relates to isotopically chiral molecules where the effect is more subtle.
You are correct that (OR) is fundamentally an electronic effect and that isotopic substitution shouldn't directly affect that. However, vibrations can cause the molecule to oscillate around the the equilibrium structure to geometries that have a different OR and isotopic substitution alters the potential energy surface. The transient chirality of the molecule from its zero point vibrations is enough to produce OR. The effect of these vibrations can actually be fairly significant, accounting for as much as 20-30% of the observed optical rotation.
For isotopically chiral molecules, the effect is small, but can still be measured. As an example, 2-deuteriocyclohexanone has an optical rotation of $2.5^\circ$ and a number of other substituted cycloalkanones have been synthesized since.

From a 2007 paper::

Here we show that instrumental advances in Raman optical activity, combined with quantum chemical computations, make it possible to determine the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane. This saturated hydrocarbon represents the archetype of all molecules that are chiral as a result of a dissymmetric mass distribution. It is chemically inert and cannot be derivatized to yield molecules that would reveal the absolute configuration of the parent compound. Diastereomeric interactions with other molecules, optical rotation, and electronic circular dichroism are, in contrast to the well-known case of bromochlorofluoromethane, not expected to be measurable. Vibronic effects in the vacuum ultraviolet circular dichroism might reveal that the molecule is chiral, but the presence of nine rotamers would make it extremely difficult to interpret the spectra, because the spatial arrangement of the rotamers’ nuclei resembles that of enantiomers. The unequivocal spectroscopic determination of the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane therefore presented a major challenge, one that was at the very limit of what is possible.

Reading between the lines:

The described molecule looks like this:

Optical rotation is not expected to be measurable.  Probably some fantastically minute but nonzero specific rotation could be computed with advanced quantum mechanical methods, but it would be no fun to try to measure it.
The difference in the electronic configuration of the isotopically substituted enantiomers in the ground state is negligible.
But the electronic configuration of rotationally or vibrationally excited states is not negligible.  Therefore, some technique that can measure the optical properties of rovibrational excited states might be able to discern the difference.  (I'll leave it to experts in this area to explain why Raman optical activity worked better than vUV-CD.)


text:
The hydration energy of NaCl is −774.3 kJ. Then why are we not able to harness that energy and run machines just by mixing salt in water?
The hydration energy of NaCl is −774.3 kJ. Then why are we not able to harness that energy and run machines just by mixing salt in water?
Where does all the energy go which is released when we mix salt with water?

The dissolution of NaCl in water is slightly endothermic, so the one number you have does not tell the whole story.

Do you actually have any salt at home? Try to mix it with water. Go on, do it. Chemistry is an experimental science, after all. Got any energy?

@IvanNeretin With all due respect, this isn't helpful.  It sounds like you're giving the questioner a hard time for asking the question—something I've noticed you often do to first-time questioners. This seems like a violation of the SOC to be nice to new contributors.  In addition, your comment implies one can't get energy from dissolving salt in water, which is incorrect.

That energy is in kJ just from one mole NaCl. And where does all the energy go?

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

EDIT: I just did the experiment that Ivan Neretin suggested. I used 250 mL of water in a 400 mL PTFE beaker. I measured the water temperature with a Hg thermometer and got 20.5 degrees Celcius. I then added 14.51 g NaCl, stirred gently to full dissolution, and the new measured temperature was 19.5 degrees Celcius. So the salt water was about 1 M and the temperature dropped by 1 degree.
From the wiki article on "Enthalpy change of solution", we have the following quote:

The first two steps are endothermic and only the third is exothermic. According to the OP, that last step has hydration (i.e., aqueous solvation) enthalpy of -774.3 kJ/mol. But the sum of the first two steps, via Born-Haber cycle, is +778.2 kJ/mol, so the sum is +3.9 kJ/mol. Thus, we see that dissolving salt in water is slightly endothermic, as shown in the table from the wiki article:

So the hydration enthalpy is not for the entire dissolution process, i.e., solid NaCl dissolving in water. It is only for the last of the three steps.
An alternative to the three conceptual steps is to combine the second and third steps. As before, the first step simply involves putting in sufficient energy to convert the salt crystal to ions that are far enough apart so that coulombic interactions can be ignored. Then the combined second and third steps conceptually involve putting all those ions into water. This scheme is illustrated, for KF, in the following figure (© Brooks/Cole, Cengage Learning):

From the figure, the enthalpy of solution is exothermic, but only -16 kJ/mol, while the enthalpy of hydration is -837 kJ/mol.
Bottom line: the OP's confusion was in assuming enthalpy of hydration was equal to enthalpy of solution.

The amount energy available to do work is the free energy (not the enthalpy). If you dissolve NaCl in water at 25 °C until you reach saturation, the Gibbs free energy change is about −8 kJ/mol. You can indeed harness this energy by using it to generate osmotic pressure, which can in turn be used to, say, raise a weight in the surroundings.  [Strictly speaking, since this is not a constant-pressure process, the Gibbs free energy doesn't exactly give the energy available to do work; but since there is little volume change, the discrepancy is small.]
BUT:
The osmotic pressure apparatus required to harness energy in significant quantities would be complex, and the amount of energy produced is quite small (relative to that obtained, say, from burning natural gas).  World salt production is 259, 000,0 00 metric tons/year.  Even at 100 % energy conversion of the dissolution of all of that salt, the plant would only generate 1 MW!  That's likely significantly less than the energy cost of doubling current world salt production (even if that were possible) (needed to feed the plant at that level), as well as other energy costs associated with plant operation, maintenance, and construction.
In sum, you'd be losing energy (and a lot of money).
Finally, generating energy from turning fresh water into salt water is the opposite of what humanity is currently working to do: instead, we are using energy to turn salt water into fresh, with desalination plants.

Do you have any estimates for the energy input required for desalination? I know it is something of an apples and oranges comparison, given non-standard states, but it would be interesting to have a ballpark estimate.

@EdV Accoring to https://en.wikipedia.org/wiki/Desalination#Energy_consumption, the most efficient desalination plants use 3 kWh/m^3 to remove salt from seawater.  However, with reverse osmosis, it is possible to achieve desalination of seawater using < 2 kWh/m^3.  Given that the molarity of salt in seawater is ~0.6, that works out to 18 kJ/mol and < 12 kJ/mol, respectively (where "mol" refers to  moles of NaCl removed).


text:
URGENT Concentration of Vinegar | Titration?
Example of acid base titration

The moles of acid and base are equal in the end point, so $n_A=n_B$ where $A$ and $B$ are acid and base, respectively. The number of moles can be written as $n=C*V$ where $C$ is the molar concentration and $V$ is the volume. So
$n_A=n_B$
$C_A*V_A=C_B*V_B$
$C_A*10mL=15.7 mL*0.05 M$
Threfore
$C_A= 0.0785M$
Now, this is moles of acetic acid per liter. The mass per mole of acetic acid is $M_r=60,052 g/mol$. So in grams per liter this is
$0.0785M*60.052 g/mol= 4.71 g/L$
So now you have 4.71 g in 1000 mL (1L). The only thing left to do is a simple rule of three because %m/V es mass of acetic acid in 100 mL, not 1000 mL. Then
$100*4.71/1000=0.471$

Welcome to Chem SE! Thank you for your contribution. But this site works on a few principles, we don't answer a question where the OP shows no effort from his side. We don't do the entire problem for them, we only clear conceptual doubts in their workings. We guide them through constructive comments. So I would encourage you to do the same :)


text:
Suspension liquid for ferrofluid that makes it float
I am currently devising a project to create a kinetic sculpture, similar to Sachiko Kodama. Does anyone know any suspension liquids that when applied, the ferrofluid floats on top?

Given that ferrofluid lava lamps now exist, this suggests that what you want to do is possible. I hope you succeed!