text:
Find ratio of quantities of water and of argon in airtight container
I am working on my EJU exam past papers (2013) and got stuck at this question. Knowledge of high school or first year university chemistry is assumed. I honestly do not know where I should start.

When argon and water were placed in an airtight container with a
  constant volume and the gas mixture was kept at 373 K, the pressure of
  the gas mixture became P1 with a portion of water remained as liquid.
  When the whole was heated to 403 K, all the water was vaporized and
  the pressure became P2. From (1) to (6) below choose the correct
  formula representing the ratio, water/argon, of the quantities (in
  mol) of water and of argon in the container. Assume that the vapor
  pressure of water at 373 K is Pv, that argon is an ideal gas not
  dissolving in water, and that the volume of water before heating to
  403 K can be neglected.

Please note that answer key provided is (1). I tried to think that we could form an equation from total moles of gases equate to the sum of total moles of water and argon at 403 K since all are vaporized. Then use the ideal gas law to find the total amount of gases at 403 K. However, I cannot connect these pieces and information given at 373 K together. Could anyone give me a solution to this question or at least a hint so I can continue working on it?

You should use the ideal gas law to convert between $n_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write
$$\begin{align}n_{g,1}=\frac{p_1V}{RT_1}\\n_{g,2}=\frac{p_2V}{RT_2}\end{align}$$
where
$$\begin{align}n_{g,1}&=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&=n_{H_2O}+n_{Ar}\end{align}$$
You also know that for each individual gas the partial pressure is proportional to the number of moles (Dalton's law):
$$n_i=\frac{p_iV}{RT}$$
 The "trick" is that at $T_1=\pu{373K}$
$$p_{Ar} = p_1 - p_V$$
where $p_V$ is given as the partial pressure of water at that temperature.
You should be able to show that
$$\frac{n_{H_2O}}{n_{Ar}}=\frac{n_{g,2}}{n_{Ar}}-1$$
and making appropriate substitutions it should not be difficult to arrive at the answer.


text:
How will the periodic table change due to redefinition of the kilogram (May 2019)?
The definition of a kilogram has recently been changed. What effect, if any, will this change have on data on the periodic table?

https://chemistry.stackexchange.com/questions/110189/why-has-the-mole-been-redefined-for-2019

Possible duplicate of Is the kilogram still consistent with the old definition?

@Mithoron I think the question and answer linked don't work as duplicates as the question refers to an earlier change of the definition of the kilogram (mass of a cubic centimeter of water changing to a reference cylinder of platinum), while the answer discusses a proposal for a new definition that wasn't ultimately used (the number of atoms in a sphere of silicon-28).

Check out this video by the chanel veritasium which explains about the various implications of redefining the kg in detail

The number of significant figures of carbon-12’s molar mass changed on May 20th, but for a different reason. Otherwise, no changes.

Molar masses
The molar mass of the carbon-12 isotope used to be 12 g/mol exactly (by definition). Now, it is experimentally determined, with the current value $\pu{11.9999999958(36) g mol-1}$ (CODATA 2018). So the molar mass of carbon-12 changed. The standard atomic weights (relative atomic masses) shown in the periodic table are not affected by this. However, atomic weights (dimension one) are no longer exactly equal to molar mass divided by 1 g/mol. Instead, they differ by a tiny bit (you would need 9 or 10 significant digits to see the difference).
No visible changes
Here is an animated gif of a portion of the periodic table at ptable.com over the past four years (Jan 23rd 2016, Jan 10th 2017,Jan 6th 2018, May 2nd 2019). Nothing much has changed near carbon:

The standard atomic weights change a bit each year, as better estimates of the isotope compositions of elements with multiple stable isotopes are made. Also, the publishers of this specific periodic table decided to truncate atomic weights to 5 significant figures in 2018. The biggest change occurred when elements in period 7 were discovered and named.
The last image is from May 2nd 2019, before the change of the kilogram (and mole) definition. On May 29th 2019, the data shown at ptable.com is identical.

I have not checked for every element, but I don't expect any difference to be significant as the goal of any international convention on that subject was to keep "backward compatibility" (as we would say for computers) with existing measures.
Well, of course, the 7th or 8th decimal can change, but it won't change the face of the world. At least, we won't suddenly grow up 4cm :-)

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text:
Calculate the pH of buffer solution
PROBLEM: A solution was prepared by mixing $\pu{25.00 mL}$ of $\ce{NaOH}$ $\pu{(c = 1mol/L)}$ and $\pu{10.00 mL}$ of acetic acid $\pu{(c = 2.5 mol/L)}$ in water to give $\pu{250 mL}$ of solution.
What is the $\mathrm{pH}$?
$K_\mathrm{a}=\pu{1.76\times 10^{-5} mol/L}$
Ok, so I know acetic acid will react with sodium hydroxide do form sodium acetate:
$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$
initial quantities of $\ce{CH3COOH}$ and sodium hydroxide are $\pu{ 0.025 mol}$  (c*V)each
after the equilibrium is reached all base and acid have reacted, so we have $\pu{0.025 mol }$ of $\ce{CH3COONa}$
I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because concentration of acid is 0.
I got a response that I should solve it like this (after I do ICE table like I did above):
$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$
$[\ce{OH-}]=\sqrt{  K_\mathrm{b}\cdot c_\mathrm{salt}}$
and after I calculate $\mathrm{pOH}=5.12$ I can calculate the $\mathrm{pH}=8.88$
I dont understand the last part, why is this so? what is
$$\ce{CH3COO- + H2O <=> CH3COOH + OH- }$$
representing?
Thanks for answering! :)

There are some calculations in this answer https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566

Note that sodium acetate solution cannot be considered as buffer. It is very sensitive to excess of both sodium hydroxide and acetic  acid, forming with the latter the acetic acid/sodium acetate  buffer.

I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because the concentration of acid is 0.

Yes, as commented above, sodium acetate solution cannot be considered as buffer;the result at the equilibrium , will be the same as dissolving  $\pu{0.025 mol }$ of $\ce{CH3COONa}$ in $\pu{250 ml}$ water,so you have the salt solution of $\pu{0.1M}\ce{[CH3COONa]}(c_\mathrm{\text{salt}}=\ce{[CH3COONa]_\mathrm{I}}=\pu{0.1M})$

I don't understand the last part, why is this so? what is...

The salt ($\ce{ CH3COONa}$) is  electrically neutral substance formed by cation $\ce{Na+}$ (an acid) and  anion $\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COONa   -> Na^+_\mathrm{(aq)}   + CH3COO^−_\mathrm{(aq)}}$$
a)  Since $\ce{Na+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{Na+}$  actually spectator ion.
b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions $\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$ $$(K_\mathrm{b(\ce{CH3COO-})}=\frac{K_\mathrm{w}}{K_\mathrm{a(\ce{CH3COOH})}}=\frac{10^{-14}}{1.76\times 10^{-5} }=5.68\times{10^{-10}})$$
c)The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}[\ce{CH3COOH}]_\text{equilibrium}}{[\ce{CH3COO-}]_\text{equilibrium}}$$
Neglecting water autoionization and assuming :${\ce{[OH-]_\text{equilibrium}}=\ce{[CH3COOH]_\text{equilibrium}}}$, ${[\ce{CH3COO-}]_\text{equilibrium}=c_\mathrm{\text{salt}}=[\ce{CH3COONa}]_\mathrm{I}}=\pu{0.1 M}$,so:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}^2}{c_\mathrm{\text{salt}}}$$
$$[\ce{OH-}]_\text{equilibrium}=\sqrt{  K_\mathrm{b}\cdot c_\mathrm{salt}}$$
You  can calculate $\ce{[H+]}$ -without neglecting the water autoionization  and without approximation- by substituting in the following cubic equation and solving bywolframalpha
$$\
 [\ce{H+}]^3
+( K_\mathrm{a}+[\ce{CH3COONa}]_0) [\ce{H+}]^2
  -K_\mathrm{w}[\ce{H+}] -    K_\mathrm{w} K_\mathrm{a}
  $$
$\ce{[H+]} =1.32658 ,\pu{pH}=8.877$

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text:
Book reference for electronic occupation of TM
Is there a book I can reference that has a table like this? I can't reference an online source on my thesis and all the pictures I have found online like this don't make their reference explicit.
Thanks!

Common set of oxidation states of transition metals can be found in virtually every inorganic chemistry textbook. The rest is from the periodic table of elements. Here is one by NIST. IMO this is generic stuff you don't need to cite at all, but consulting with your supervisor is the way to go.

Any general chemistry textbook, e.g. https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chemistry_(OpenSTAX)/06%3A_Electronic_Structure_and_Periodic_Properties_of_Elements/6.4%3A_Electronic_Structure_of_Atoms_(Electron_Configurations)

I need to cite something since I am not a chemist (and neither are my professors), I need to show something to corroborate what I am saying. Thank you for the links provided, but I don't see a way to cite the first one, or a list of books in the second one I could get.

You can cite the first two or three inorganic texts here https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry/37305#37305 and maybe this as well https://www.amazon.co.uk/Chemistry-Elements-N-Greenwood/dp/0750633654 if you want to go full overkill.

Element Name and Symbol + Atomic Number + Electron Configuration
NIST Standard Reference Database 111 provides an up-to date reference for the electron configuration of neutral atoms in the ground state [1] in both tabulated and periodic table formats.
Common Oxidation States
An overview of elements' oxidation numbers is provided in practically every inorganic chemistry textbook. Arguably, the most extensive one is given throughout Greenwood and Earnshaw's Chemistry of the Elements [2].
More compressed information is provided in CRC handbook of chemistry and physics (Section 4. Properties of the Elements and Inorganic Compounds) [3, pp. 4-1 — 4-92].
There is also a periodic table at the end which includes oxidation numbers of the elements in the upper right corner of each cell:

References

Kramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2018). NIST Atomic Spectra Database (ver. 5.6.1), [Online]. Available: https://physics.nist.gov/asd. National Institute of Standards and Technology, Gaithersburg, MD. DOI: https://doi.org/10.18434/T4W30F
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3


text:
Safely mixing methanol and chloroform
Sorry for what is likely a naive question. I haven't worked with chemicals at all and have only had some brief lab safety instruction.
I need to create a mixture of chloroform and methanol 2:1 for a lipid extraction from bone.
I have read the safety data sheets for both chemicals but can't find any other information about how to safely create this mixture other than doing it in a fume hood. I need about 200 ml total.
If anyone has any information or can direct me to where to look that would be MUCH appreciated.
Thank you!!!

What kind of work environment are you in?  What kind of equipment do you have access to?

Mixing chloroform and methanol presents no hazards, except for those presented by handling chloroform by itself and methanol by itself.  That is, mixing these reagents does not create any dangerous reactions.  However, chloroform is a toxin and methanol is both a toxin and is highly flammable.

It will be in a lab at a university, although one that is not used frequently. I have access to everything I imagine I'll need safety wise (gloves, lab coat, safety glasses, etc). There is a fume hood, glassware, that kind of thing. Anything specific I will need?

I guess I just wanted to make sure there weren't any special precautions to be taken. So I can literally just pour the right amounts into a beaker or something like that? I wasn't sure if it had to be done in a special way.

@Heather, that's right.  Just mix.  Nothing will react.

Just be sure the hood draws air.

And dispose of residues into a chlorinated solvents labelled vessel

And mix well because their densities are very different

Don't over think.  At this level every simple manipulation would take weeks ;) I assume you don't that near flame or red hot surface nor you indulge in smelling atop.

I am sorry to give my opinion here but I feel compelled to do so!
Security officers or the people writing the safety data sheet are motivated by only one thing: they don't want to be held liable! For them, everything is deadly and they don't even try to mitigate that (true) fact according to their audience.
As a result, all the risks are presented at the same level and nobody can really adjust his/her practices according to that.
I vividly remember one of my students at Oxford University. The rule there is(was?) that the student should copy the whole safety sheet on a form submitted to the Safety Officer before he/she can even get the chemical. My student had to do that for a ridiculous amount of cyanide and he was so stressed out that he ended up shaking and putting cyanide everywhere! I mean on the balance, around the balance, on the floor etc... I had to be besides him "It's OK, it's no big deal" while making signs to the others to not come around because there was cyanide everywhere.
So @Heather, I really advise you to ask your co-workers about the true risks in your field. You won't die because you had a whiff of chloroform or methanol, even though you should always work under a hood so you don't have a whiff of nasty materials. Also the safety sheets must be adapted to your own risks.


text:
What am I doing wrong with the fuel and packing I am using for a cannon I made?
Recently I have been using flash powder in a cannon that I made, for a chemistry project, however, it is not causing the rubber ball I put in it to propel out of the tube. First I put in 30g of flash powder, then pack a small wad of paper on top, and lastly insert the ball in an aluminum foil bowl. The fuse is then inserted into the bottom and lit on fire, but the only thing that has been happening is fire flaming from the end of the tube and melting parts of the rubber ball. Do you have any small suggestions, as I am running out of time to figure this out? I attached a picture of the cannon. (Also I do not have time to make black powder).

Just curious about the "running out of time" aspect for figuring out how to get your cannon operable...Aside from that, flash powder doesn't have the bang-for-the-buck that black powder does. You might try compressing and shaping the charge, thereby increasing the energy density and subsequent yield on ignition.

I only have a few class periods left to complete this project and get raw data, so it would be difficult for me to make black powder since it can take a while. How would you suggest I do that?

What is your "flash powder" made of?

Maybe rather than finishing the cannon project by building a working cannon, instead get some raw data showing why it is not working so well. You could try to measure the volume change from the flash powder, or test whether pumping air into the tube would propel the ball (I don't know how good the fit has to be). Anyway, most projects involve troubleshooting, and most troubleshooting involves testing individual aspects of the problem, and running some positive and negative controls.

Flash powder and gun powder are two different mixtures, designed to do two different things. Gun powder is designed to produce a lot of gas very quickly (Carbon dioxide, Sulfur dioxide, Nitrogen) which propels the projectile out of the barrel.
Flash powder - typically potassium chlorate/perchlorate and aluminium powder, examples here and here, produce a lot of heat and light very quickly but not much gas - so no propulsion.


text:
mechanism in bi cyclic compounds

In both cases ring expansion is happening. What is Carbon tri chloride purpose?

I would suggest running a Google search on "carbene addition to alkene". $\ce{CHCl3}$ in alkali medium forms dichlorocarbene $\ce{:CCl2}$.

Your question b has already been answered  https://chemistry.stackexchange.com/questions/41500/reaction-of-furan-with-chloroform-in-base

" What is Carbon tri chloride purpose?" is not answered in the link provided but the answer starts with a carbene attack on furan .
Trichloromethane in presence of a base undergoes 1,1-elimination (or an alpha elimination) because both groups are removed from the same carbon.The electron-withdrawing effect of the halogens makes the hydrogen acidic enough to be removed by a strong base such as hydroxide or alkoxide ion.
A halide ion then acts as a leaving group from the conjugate base, producing the
 carbene.1

Carbenes find a pair of electrons  to complete their valence shell of electrons.carbenes attack a lone pair, a C=C double bond (electron-rich or -poor), or even a C–H bond.In its reactions ,

It is an electrophilic reagent and is known to give ring expansion of aromatic rings.
Electrophillic attack takes place on pyrrole where the intermediate is most stable.

This leeds to normal Reimer–Tiemann reaction product pyrrole 2-aldehyde.
Carbene attack of a C=C double bond leads to an ring expansion leading to 3-chloropyridine.

The reaction of bicyclic compounf acts on similar lines of dichloro carbene attack.2

References
1 ORGANIC CHEMISTRY ,JOSEPH M. HORNBACK ,UNIVERSITY OF DENVER
2 MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE ,SIXTH EDITION

The second diagram is quite loose with the hydrogen on the nitrogen atom.

I missed this the first time, but the arrow pushing is backwards in the first step of the pyrrole example.

@Zhe yes it is pyrrole that donates to dichloro carbene and not other way around.I will corect shortly.


text:
Why are my microgels not centrifuging (sticking to tube)?
I have created a hydrogel water in oil emulsion using a microfluidic flow focusing device. I have been able to collect the microgels/particles from the device and then pipette/manipulate them. When I try to centrifuge them to wash them they seem to aggregate and stick to the tube walls instead of collect at the bottom of the tube. What could be causing this and how do I fix it? Below are some relevant details.
Water phase: 2 wt% Norbornene-modified hyaluronic acid, DTT (crosslinker), .1% Irgacure I2959 (photoinitiator)
Oil phase: Light mineral oil with 2% span 80 (surfactant), .1% Irgacure I2959 (photoinitiator)

Have you tried to change the material of the centrifuge tubes? It looks like you are using PP or PET; try PTFE instead.

Thanks so much for your suggestion, I really appreciate it. I am using polypropylene tubes. I tried looking for PTFE but I don't see any in a clear/conical form. I ended up adding acetone and that seemed to solve the problem. I'm not sure exactly why though. I read about it being able to pull the surfactant out of the oil phase due to polarity, does that seem correct?


text:
Hydrogen + Fluorine Adiabatic Flame Temperature
The theoretical adiabatic flame temperature of Hydrogen and Fluorine in their stoichiometric mixture is often quoted at around $\pu{4300K}$, such as in this presentation on page 2, or this paper which is shown entirely in the pubs preview. However, this thermodynamics table from JANAF makes it seem as though the temperature should be much higher. I suspect I'm missing something, but after quite a lot of searching I can't find it! (So it's probably something very basic)
My work: The Gibb's energy for the reaction is quite negative even out to $\pu{6000K}$, so the combustion should happen effectively to completion regardless of temperature ($K = 307 = \exp\{--285.7\times1000/(8.314\times6000)\}$. The formation enthalpy is listed $\pu{-273kJ/mol}$, and we could either do the temperature integral of $C_p$, or scan down the H-H(Tr) column to find this puts us way beyond $\pu{6000K}$, let alone $\pu{4300K}$.
Anhydrous $\ce{HF}$ gas apparently does form dimers, trimers, etc. but these should only increase the adiabatic flame temp if present, and they (understandably) don't appear to be stable beyond a couple thousand K.
Thanks in advance for the help!

Unless I'm missing something, your assessment looks correct.

Glad it's not just me, thanks for giving it a look! There's a mystery afoot.


text:
Thermodynamics: ΔU = Q + W or U = Q?
Heat is the total kinetic energy of all atoms of the system. When work is done on the system it means that a part of system kinetic energy is used to do the work, and this work makes the surrounding warmer.
So "$\Delta U$" of the system is equal to "$Q$".
And now, why we use the work of the system in: $\Delta U = Q + W$?

"Heat is the total kinetic energy of all atoms of the system." That's incorrect. Heat is the energy exchanged with the surroundings in the absence of work. When volume is constant and no other work is done, $\Delta U = Q_v$. More generally $\Delta U = Q + W$

Heat is energy transferred because of a difference in temperature. You know from your own experience that if something is colder or warmer than your  hand (say)  you can feel the difference due to energy leaving your hand (if what you touch is colder) or gaining energy if warmer.  If the temperature is the same there is no energy transferred, i.e. no heat.

The questions is a good one, and answering it winds up being central to a lot of key concepts in thermodynamics. A brief correction before starting, though I don't think it's central to the question,  "When work is done on by the system it means that a part of system kinetic energy is used to do the work..."
The disconnect between your understanding and that used in thermodynamics is that in thermodynamics, the system and its surroundings are separate. So, when our system does work on the surroundings that energy won't come back to heat our system. If we do as you say and use our system to do work on the surroundings, turn that work into heat, and then apply that heat back into our system, the result is just as you say! The kinetic energy in our system isn't lost. Or put in another way when we're allowing heat to flow in and out of our system, all of that heat ultimately winds up as kinetic energy and $ΔU = Q.$
Let's look at a couple examples of putting $\pu{1 J}$ of heat into a system with different surroundings to see how this works out. Our system will be an insulated piston filled with an ideal gas (say helium), which can expand and push on one of 3 things in its surroundings: a mass, a vacuum, or a generator.

Mass: This follows $ΔU = Q - W$.  Of the $\pu{1 J}$ of heat we apply, the portion that goes into warming up our system and increasing internal kinetic energy is $C_v$ (the molar heat capacity is $1.5R),$ whereas the portion of that instead goes into doing work on the surroundings is $R$ (the universal gas constant). So our system gains
$$[\pu{1 J}]\cdot \frac{C_v}{C_v + R} = \pu{\frac{3}{5} J}$$
of thermal energy, and does
$$[\pu{1 J}]\cdot \frac{R}{C_v + R} = \pu{\frac{2}{5} J}$$
of work on the surroundings by expanding and pushing up the mass. This $\pu{\frac{2}{5} J}$ of work energy won't return. These portions $C_v/(C_v + R)$ and $R/(C_v + R)$ show up a lot later on in thermodynamics as their inverses, $k$ and $k/(k - 1)$, respectively.
Vacuum: This follows $ΔU = Q$. There is no surroundings to do work on, so all of the expansion energy goes into slamming the piston against its backstop, and after everything settles this ultimately results in the full $\pu{1 J}$ of heat becoming thermal energy inside our piston.
Generator: This also follows $ΔU = Q$. It's the same situation as the mass, with $\pu{\frac{3}{5} J}$ of thermal energy and $\pu{\frac{2}{5} J}$ of work done on the surroundings, but that $\pu{\frac{2}{5} J}$ of work can be turned into electricity and then be used to heat our piston.
Just like in the vacuum case, all $\pu{1 J}$ of heating ultimately winds up as thermal energy in the system, though it spent some time as work on the surroundings during the process. So, $ΔU = Q - W + W.$ I think this final case is the closest to matching your question.

Tech detail: please don't use \pu{…} everywhere in math mode. Its sole purpose is to handle physical units (hence the shortcut for the macros: \pu).

Sorry, I don't quite understand what you're asking for. I can't seem to find a markup how-to page, but looking at your edits it looks like its best to handle equations by putting them within dollar signs to display them with TeX? Also, thanks for editing, it looks very much nicer!

It's not excessive use of $$, it's using a macro for physical units on the entire expression. If you are familiar with LaTeX, the \SI{}{} command is similar to \pu of MathJax. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also: MathJax basic tutorial and quick reference.


text:
Is vapour pressure a surface phenomenon?
As a fact, I know that vapour pressure doesn't depend on surface area.
Does this imply it is not a surface phenomenon also?
It shouldn't be as then it would depend on surface area of molecules and the mole fraction of liquid in a solution violating the definition of colligative properties.
If not, then why doesn't it depend on surface area exposed?
vapour pressure doesn't depend on area

Note that it does depend on the shape of the surface as $$p=p_{\mathrm{0}} + \frac{2\gamma}{r}$$ where $p$ is saturated vapour pressure over a spherical liquid particle of radius $r$ and surface tension $\gamma$ , p0 is the ordinary saturated vapour pressure over flat surface.

This is why is a Pressure, isn't?

Vapour pressure is normally defined as an equilibrium phenomenon. In statistical mechanics terms it is the point where an equal number of molecules are leaving the liquid and entering the vapour and leaving the vapour and entering the liquid.
This means that, however small the exposed surface, enough molecules have gone into the vapour phase to create the equilibrium. But most systems are not observed at equilibrium. The kinetics of the process leading to equilibrium clearly depend on the surface area of the liquid as there is far more opportunity for molecules to escape into the vapour phase if the exposed area is larger. But, if you are prepared to wait for the equilibrium to be established, the surface area doesn't matter.
Hence the slight intuitive confusion: what we mostly observe depends on surface area because what we observe is mostly systems that have not yet reached equilibrium so surface area matters. But on a strict equilibrium definition surface area doesn't matter but you might have to wait a long time to see the eventual equilibrium position.

What processes do we define as "surface phenomenon"?

When dealing with particles of extremely small size (such as dispersed water droplets), the surface area may relate to the bonding phenomenon at the surface. I would say that, at this scale, the surface area does matter. The comment by @Poutnik shows an inverse relation with $r$, suggesting the dependence on surface area diminishes for larger scales.

Well, the other answer frankly seems too complicated using statistical mechanics and all, let me give you a simplified version
Imagine a beaker, when you zoom in on the surface you see all the atoms and the molecules jiggle faster with more energy than the other atoms just below it. The vapour pressure is "almost" due to these atoms, by "almost" I mean that it is these atoms which get yanked off of the liquid and create a "commotion" at the top of the beaker.But there is a limit as to how many of the molecules get yanked of the surface which is dependent on the amount of heat you provide. Now take the beaker and empty its contents to another beaker which has more surface area.
If we zoom in on the surface of the new beaker we can see that there are more atoms at the surface which now "snap" out and create an even greater "commotion" at the top of the beaker, thus creating more vapour pressure.
Now according to the definition of pressure, increasing the area should decrease the vapour pressure but it doesn't, thus we conclude and say that it is not a surface phenomenon.

Can yall give me the reason why you are downvoting me? please it would really help either in changing my mindset on answering question or yours


text:
How to find the concentration of barium ions in a mixture of barium sulfate and barium thiosulfate?
I have a hard time with selective precipitation problem.

A mixture of $\ce{BaSO4}$ and $\ce{BaS2O3}$ is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is $[\ce{Ba^{2+}}]$ in the saturated solution? $K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$ and $K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$.

My solution:
Because $K_\mathrm{sp}$ of $\ce{BaSO4} < \ce{BaS2O3}$, $\ce{BaS2O3}$ will dissolved first. When $\ce{BaS_2O_3}$ dissolved, the concentration of $[\ce{Ba^{2+}}] = [\ce{S2O3^2-}] = \sqrt{K_\mathrm{sp}(\ce{BaS2O3})} = 2 \times 10^{-5}$.
That means, in the solution, there are:
\begin{align}
[\ce{Ba^2+}] &= 2 \times 10^{-5}\\
[\ce{S2O3^2-}] &= 2 \times 10^{-5}\\
[\ce{SO4^2-}] &= \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} = \frac{9 \times 10^{-11}}{2 \times 10^{-5}} = 4.5 \times 10^{-6}
\end{align}
But the solution is $\ce{[Ba^2+]} = \sqrt{4.9 \times 10^{-10}}$.
What have I missed?
EDIT 1 :
\begin{align}
\ce{BaSO4 &<=> Ba^2+ + SO4^2- }\\
\ce{BaS2O3 &<=> Ba^2+ + S2O3^2- }
\end{align}
The Solubility Product Equation :
\begin{align}
K_\mathrm{sp}(\ce{BaSO4}) &= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
K_\mathrm{sp}(\ce{BaS2O3}) &= [\ce{Ba^2+}] \times [\ce{S2O3^2-}]
\end{align}
When the solids dissolves, it will produce $\ce{Ba^2+}$ ion, $\ce{SO4^2-}$ ion and $\ce{S2O3^2-}$ ion.
$\ce{BaSO4}$ will stop dissolving when $\ce{Ba^2+}$ ion concentration in solution when multiply with $\ce{SO4^2-}$ ion concentration is equal to solubility product of $\ce{BaSO4}$, which is $K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$
And at the same time,
$\ce{BaS2O3}$ will stop dissolving when $\ce{Ba^2+}$ ion concentration in solution when multiply with $\ce{S2O3^2-}$ ion concentration is equal to solubility product of $\ce{BaS2O3}$, which is $K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$
Solubility of $\ce{BaSO4}$ in water is
\begin{align}
 s &= \sqrt{K_\mathrm{sp}(\ce{BaSO4})}\\
 s &= \sqrt{9 \times 10^{-11}} \\
 s &= 3 \times 10^{-5.5}
\end{align}
Solubility of $\ce{BaS2O3}$ in water is
\begin{align}
 s &= \sqrt{K_\mathrm{sp}(\ce{BaS2O3})}\\
 s &= \sqrt{4 \times 10^{-10}}\\
 s &= 2 \times 10^{-5}
\end{align}
Now, when the solids is mixed in water, There should be $[\ce{SO4^2-}] =3 \times 10^{-5.5} $ and $[\ce{S2O3^2-}] =2 \times 10^{-5} $.
Now I confuse what ion concentration should I use to determine $[\ce{Ba^2+}]$
EDIT 2:
\begin{align}
K_\mathrm{sp}(\ce{BaSO4}) &= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
\frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} &= [\ce{SO4^2-}] \\
\end{align}
and
\begin{align}
K_\mathrm{sp}(\ce{BaS2O3}) &= [\ce{Ba^2+}] \times [\ce{S2O3^2-}] \\
\frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} &= [\ce{S2O3^2-}] \\
\end{align}
From @santimirandarp answer, I got that :
\begin{align}
[\ce{Ba^2+}] &= [\ce{S2O3^2-}] + [\ce{SO4^2-}]\\
[\ce{Ba^2+}] &= \frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} + \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &= \frac{4 \times 10^{-10}}{[\ce{Ba^2+}]} + \frac{9 \times 10^{-11}}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &= \frac{4.9 \times 10^{-10}}{[\ce{Ba^2+}]} \\
[\ce{Ba^2+}]^2 &= 4.9 \times 10^{-10} \\
[\ce{Ba^2+}] &= \sqrt{4.9 \times 10^{-10}} \\
\end{align}
My problem solved. Thank You

There is no "dissolved first". The salts don't have the central command to tell them when to start dissolving, nor the intelligence to find out what other salts are sitting nearby. They all start dissolving at once.

@IvanNeretin okay, what do you think happened ? What is the $[\ce{Ba^{2+}}]$ ?

I think that both salts have dissolved a little, and you have all three ions in the solution.

@IvanNeretin Yes, Both solid will dissolved and make a equilibrium with each other and their concentration will be related with their solubility product. Since the solubility product of \ce{BaS2O3} is bigger, It means the solubility of \ce{Ba^2+} from \ce{BaS2O3} is bigger than the solubility of \ce{Ba^2+} from \ce{BaSO4} and I started from that to count other ion concentration. What do you think I should start to work on first ?

Salts don't have military insignia either. Any individual $\ce{Ba^2+}$ does not know whether it is from BaSO4 or from BaS2O3. Write down all unknowns, then all equations, then see what you can do.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

$[\ce{SO4^2-}] =3 \times 10^{-5.5} $ is a very odd way to write a concentration. Integer powers of 10 are expected.

in my opinion, the answer you gave is better than the one I post. You should post it and accept it, and I can remove mine. But I'll let you decide !

@santimirandarp I got that idea after watching your answer. I think your answer still better. As you say in the first line of answer, Let user have another perspective to approach a problem in many different ways. btw, Thanks

@valkedin nice! good luck!

This is just one possible approach:
First of all, the constants doesn't seem different enough to argue that $[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]$. So for now consider:
$$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}] \hspace{2em} (1)$$
and dividing both $K$s we can find: $$\frac{K_{\ce{BaS2O3}}}{K_{\ce{BaSO4}}}=\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}=4.44 \hspace{2em} (2)$$
Now we can solve for sulphate: $$K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}=[\ce{SO_4}^{2-}]^2(1+\frac{[\ce{S_2O_3}^{2-}]}{[\ce{SO_4}^{2-}]}) \hspace{2em} \mathrm{using}\hspace{1em}(1)$$
Using equation $(2)$ we get $[\ce{SO_4}^{2-}]=4.03\times 10^{-6}$
And for thiosulphate: $$K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}=[\ce{S2O_3}^{2-}]^{2}(1+\frac{[\ce{SO4}^{2-}]}{[\ce{S2O_3}^{2-}]}) \hspace{2em} \mathrm{using}\hspace{1em}(1)$$
Using inverse of equation $(2)$ leads to $[\ce{S_2O_3}^{2-}]=1.8\times 10^{-5}$
The sum yields
$[\ce{Ba^{2+}}]=[\ce{S_2O_3}^{2-}]+[\ce{SO_4}^{2-}]=0.00002215015=\sqrt(4.906295\times 10^{-10})$


text:
Which has a high melting point, cocoa powder or cocoa butter?
I have been doing some research into chocolate, and after looking at the chemical compositions of cocoa butter and cocoa powder, struggled to determine which had a formula/structure that would have a higher melting point.
Which has a higher melting point, natural cocoa powder or cocoa butter?
(For those unfamiliar with the differences between cocoa powder and cocoa butter, here is some helpful information:
Cocoa powder:
https://en.wikipedia.org/wiki/Cocoa_solids
Cocoa butter:
https://en.wikipedia.org/wiki/Cocoa_butter
Composition of both:
https://www.icco.org/faq/61-physical-and-chemical-information-on-cocoa/106-physical-and-chemical-information-on-cocoa-beans-butter-mass-and-powder.html
Thanks!)

If you hold a piece of chocolate, containing cocoa butter, in your hand, does it melt ? Yes, it does.  If you do the same with cocoa powder, does it melt ? No, it does not.

I rather doubt cocoa solids actually melt on heating, they are rather more likely to decompose

None of them have melting point  For their behaviour see comments above.


text:
Le Chatelier's Principle Having No Effect
Consider the following equilibrium system:
$$\ce{2 IBr(g) <=> I2(g) + Br2(g)}$$
Would a decrease in the volume of the system effect the equilibrium?
I think it would not affect the equilibrium of the system.
This is because a decrease in volume would increase the pressure of the system. Le Chatelier's principle states that a decrease in volume would favour the side with less moles. Since both sides have an equal amount of moles, there would be no shifting of equilibrium.
A decrease in volume would also increase the overall concentration of all substances within the system. Considering that it affects the overall concentration and not just one side, an increase in concentration would merely increase the rate of reaction. Not the equilibrium of the system.

I see it as right evaluation.

This is assuming you are operating in the case where there are no deviations from ideal gas behavior. Given the size of these molecules, that is highly unlikely.

Are you responding to Poutnik or me? Is it highly unlikely that it would not affect equilibrium?


text:
Which of the following is the most basic compound?

Which of the following is the most basic compound?

$\ce{HS-}$
$\ce{H3Si-}$
$\ce{H2P-}$
$\ce{Br-}$

I can't figure out which of these is the strongest base. Would it be $\ce{H3Si-}$, since $\ce{Si}$ is the least electronegative of all these central elements?

The size of the atom carrying the charge is generally more important than the electronegativity.

I agree with Doctor Insult's argument that "the stronger the base, weaker is the conjugate acid." In that sense, higher the $\mathrm{p}K_\mathrm{a}$ of acid, weaker is the acid, and hence, stronger is the conjugate base. The $\mathrm{p}K_\mathrm{a1}$ values of $\ce{HBr}$, $\ce{H2S}$, $\ce{H3P}$, and $\ce{H4Si}$ are -8.7, 6.89, 27, and ~35, respectively (Ref.1). Therefore, the order of the strength of their conjugate bases should be:
$$\ce{H3Si- \gt H2P- \gt HS- \gt Br-}$$
Hence, strongest base is $\ce{H3Si-}$.
Note: Ironically, this is the same decreasing order of electronegativity of center atoms $\ce{Si, P, S, Br}$: $1.90 \lt 2.19 \lt 2.58 \lt 2.96$, respectively.
References:

William L. Jolly, In Modern Inorganic Chemistry; 1st Edn.; McGraw-Hill, Inc.: New York, NY, 1984, p 177.

Basic strength of a base is inversely proportional to the electronic negative charge density on the charged atom, because it will more readily to try to stabilize it charge by abstracting a proton. Size varies as Br>Si>S>P, so $\ce{H2P-}$ will be the strongest base. Even from the ${K_a}$ data, I concluded that $\ce{H2P-}$ will be the strongest base. Look up the ionisation constants of their conjugate acids. The stronger the base, weaker is the conjugate acid. Hope that helps.

In this case, the charged atoms are all similar in size, so the electronegativity is the dominant effect.


text:
The maximum no. of isomers (including stereoisomers) that are possible on mono-chlorination of the compound: 3,3-dimethylpentane is?
The maximum no. of isomers (including stereoisomers) that are possible on mono-chlorination of the compound: 3,3-dimethylpentane is?
 The answer given is 5.
My working: The fourth carbon is chiral when chlorine is added to it, so it'd have an enantiomer, and there are two other carbons where chlorine can be substituted, so how is the answer 5?

$\ce{CH3-CH2-C(CH3)2-CH2-CH2Cl}$,
$\ce{CH3-CHCl-C(CH3)2-CH2-CH3}$ (this is chiral and would have an enantiomer),
$\ce{CH3-CH2-C(CH2Cl)(CH3)-CH2CH3}$

are the compounds I could make out.

Can you show us the structures you have come up with?

2 is same as 4 so one less.

@MichaelLautman I've edited the question. Please see if it answers your question.

Who said the answer was five? You've got it all right.

only 4 products are are formed.


text:
How to remove sugar from juice
My son has autism and we noticed that blueberries make him feel much better. The problem is blueberries contain sugars which interfere with diet he has (low carbohydrate).
How do I filter out the sugars (fructose?) without interfering with other compounds present in the juice? We can't add yeast and wait. Perfect option would be to add something that interacts only with the sugars and produces something neutral to the human body or is easily removed.

You could culture the juice under aerobic conditions. Tart blueberry juice actually sounds delicious (to me anyway).

You could try unripe blueberries.

What in the blueberries makes him feel much better? I wonder (as a contrarian) if the low carbohydrate diet makes him feel worse, so the sugars make him feel better. A test would be to give him some honey (~40% glucose, 40% fructose) diluted in water (maybe add some blue food color?) to see if that makes him feel better. Fructose is sweet; blue is pretty; glucose is a brain food...

A blueberry contains ~85% water, 5% glucose, and 5% fructose (and it is generally considered a low glycemic index food, see e.g. this site). The solubility of fructose in water is extremely high (wikipedia says ~4000 g/L (25 °C), pubchem says "freely soluble in water"), and that of glucose is pretty high too, so there is essentially zero chance of removing these sugars through crystallization or other method without significantly altering the fruit or its juice.
If you are still interested in generating an extract of the fruit, this article might be useful. It suggests two different approaches, juicing and drying. The first will generate juice and a press cake, and since the sugars are highly water-soluble the cake should be the reduced-sugar component of interest to you. You could also freeze dry the juice, assuming this contains the important components, then partially rehydrate the dry product and attempt to crystalize out sugars. Or you could dry the whole fruit and then attempt a separation with ethanol, with sugar extracted into the alcohol to obtain a reduced-sugar extract which can then be dried. Further separations with more apolar solvents might extract hydrophobic "phytonutrients" from the juice. But you are likely to remain somewhat in the dark with respect to what you've removed from the berry without further analytical tools to help you characterize the products.
There are alternatives to extraction of the sugar that involve biochemical conversion of the monosaccharides. One alternative is to treat crushed blueberry with xylose isomerase, which would convert the glucose to fructose. Fructose has a low GI ranking, which might be a benefit. Another alternative is to treat the blueberry with a mixture of enzymes (perhaps including sucrose-sucrose 1-fructosyltransferase (1-SST), see e.g. this example) that converts the glucose/fructose mixture into largely indigestible polysaccharides (soluble dietary fiber).
Otherwise the best solution might remain to eat the fruit raw or to seek other foods with similar composition and without sugar (perhaps you should consult a dietitian).

Offering the blueberries in moderation while reducing carbohydrates in the remainder of the diet seems a good approach.

Is there a compound that reacts only with glucose/fructose producing something that's easier to separate?

@blueb I don't know off the bat of one but there would surely be some way, although perhaps not currently practical. Also, I added a similar idea to the answer.


text:
Multiplicity letter codes for large amount of states
Well, let's say we have a large amount of states computed (e.g. over 50), all with the same multiplicity.
So, our ground state can be $X^2\Sigma^+_g$ and we can continue with $B^2, C^2$ etc. as described in Molecular orbitals symmetry - states notation and Letter codes in molecular term symbols.
But what is the convention of the number of states is larger, than the number of letters? Does any convention like $B^2_2$ exist?

My guess is that this system of letter codes loses its convenience if you are interested in more than ~10 states. It would likely be more intuitive to switch to a numerical system like $S_0$ for ground, $S_1$ for first excited singlet, etc. One issue I have hear of with this letter code is that the labels often go unchanged if a state is latter discovered to be in the wrong order. http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.700.1780&rep=rep1&type=pdf

@Tyberius Thank you for the answer! The new notation seems quite nice, but, unfortunately, it omits the information about symmetry which I consider to be really important. Have you ever seen some sort of "joined usage" like $D_0\Sigma^+_g, D_1\Sigma^+_u...$?

I haven't seen it before, but I think it would be understood correctly.


text:
Bromohydration regioselectivity
In which of these cases would the addition of $\ce{Br2/H2O}$ be highly regioselective?
$$\ce{CH3CH=CHCH2CH3}$$ or $$\ce{(CH3)2C=CH2}$$
since both are asymmetric alkenes, how do I answer the question?

Which one will give the most stable cation?

@Waylander Comparing the stability of bromonium intermediate, I think the second compound will be more stable due to the presence of a  3-degree carbon which has a stronger inductive effect than those with degree 2 as in the first compound. But does the carbon number matter (because the first compound has a greater number of carbons on both sides of the double bond)?

Tertiary cations are always more favoured over secondary

@Waylander, okay, so, how is stability related to regioselectivity?

With molecule (2), you have two possible cations from addition of Br+; one is tertiary, one is secondary. Tertiary is much more stable so you will get very little product resulting from water reacting the secondary cation = high selectivity observed.
With molecule (1), Br+ addition gives 2 possible secondary cations; nothing much to choose between them = poor selectivity, ~1:1 mixture observed.


text:
Aromaticity of conjugated cyclic ketones
Consider the following compounds:

Cyclopenta-2,4-dien-1-one
Cyclopent-2-en-1-one

In both compounds, I can visualize a resonating structure in which the ring follows Huckel's rules by:

The $\ce{C=O}$ $\pi$-electrons transfer to $\ce{C}$.
The $\ce{C=O}$ $\pi$-electrons transfer to $\ce{O}$.

A test that I gave, gave 2 as aromatic but not 1, why?
Also can you recommend a lucid source with an explanation of aromaticity that has lots of examples?

You know Huckel's rules? There's no reason to think any of the two would be aromatic.

Possibly the second compound is misprinted.  Cycleprop-2-en-1-one would have an aromatic contribution.

Cyclopent-2,4-dien-1-one is not aromatic as you suggested by the resonance structure (see A on the following diagram). That resonance is forbidden and does not exist, ever (I marked it with a red cross). Also, Note that the conjugated $\mathrm{sp^2}$ carbons of the original structure are not cyclic. Nonetheless, its allowed resonance is given in B, and therefore, cyclopent-2,4-dien-1-one is considered to be antiaromatic (Ref.1) due to a significant polar resonance structure (the carbocation is formed at the carbonyl carbon; see B on the diagram):

It is a very unstable compound and reacts with itself in a Diels-Alder reaction to give a more stable dimer, resemble cyclopentadiene. It is noteworthy that all attempts to synthesize cyclopenta-2,4-dien-1-one have been failed and, yielded only a dimerization product (see C on the diagram).
Cyclopent-2-en-1-one is, on the other hand, not aromatic or antiaromatic, but a $\alpha,\beta$-unsaturated cyclic ketone, which is also very stable.
Note: You may read https://nptel.ac.in/courses/104103071/module7/lec17/2.html, https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Aromaticity, and https://www.masterorganicchemistry.com/2017/02/23/rules-for-aromaticity/ to learn about aromaticity.
References:

R. Pal, S. Mukherjee, S. Chandrasekhar, T. N. Guru Row, “Exploring Cyclopentadienone Antiaromaticity: Charge Density Studies of Various Tetracyclones,” J. Phys. Chem. A 2014, 118(19), 3479–3489 (https://doi.org/10.1021/jp5010924).

But this article says it's partly anti aromatic https://pubs.acs.org/doi/abs/10.1021/jo034401j

I mentioned that in my answern(Ref.1). The question is not about antiaromaticity, any way.

Ok I misunderstood the answer. I the paper where they referred to dimer as something else than the diel's alder reaction.

its allowed resonance is given in B, and therefore, cyclopent-2,4-dien-1-one is considered to be antiaromatic but wouldn't B resonance structure be extremely unstable and therefore contribute very less to hybrid and therefore hybrid will be more close to A which will be non-aroamtic?


text:
Can we use potassium iodide as pH meter electrode storage solution.? If not then why?
If KCl is not available can KI be used to store pH probe meters? Both are potassium salts.

If I remember correctly, $\ce{KCl}$ is used due to the diffusion coefficients of both ions being similar, but I might be wrong though.

@FelipeS.S.Schneider You are correctly pointing to Hittorf's ion transfer numbers (e.g., https://en.wikipedia.org/wiki/Ion_transport_number).


text:
Strongest overlap among s–s, p–p, s–p
So, I was wondering, when overlap of orbitals takes place, among s–s, s–p, p–p which type of overlapping is stronger?
Because in s–s more volume is being used in overlap, but the electron density is less; whereas in p–p electron density is more, but overlap is less; or s–p because it has partially both the effects?

Related: https://chemistry.stackexchange.com/q/81456/44877

Looks like a repeat Q to me. But there may be more to add to the answer to the question linked by @Tan Yong Boon

I'm not talking about hybridised orbitals , more like in H-H and F-F type

@user232243 sigma bonds are not reliant on hybridization. The assignment as sigma, pi, delta solely relies on the symmetry, not the constituent orbitals. The latter can be used to find the symmetry involved.


text:
Why is there no effective alternative to metal salts in antiperspirants?
Metal salts, especially aluminum compounds are used in antiperspirants to prevent sweating. I found out aluminum chloride hexahydrate, which is found in most antiperspirants, undergoes an acid-base reaction with water which causes the proteins to denaturate, thus these proteins are precipitated out as a gel blocking the sweat glands. Additionally, aluminum is able to bind water molecules.
However, I would like to understand the biological and chemical processes in more detail as I am not quite sure why there is no effective alternative to metal salts to prevent sweating. There are many compounds that could undergo an acid-base reaction with water. There are also different metal salts which are able to bind several layers of water.
My specific questions are:

Which biological and chemical processes underlie the effect of antiperspirants that make metal salts so effective?
Why are aluminum salts used most commonly?

Zirconium  is also used. See https://en.wikipedia.org/wiki/Aluminium_zirconium_tetrachlorohydrex_gly

Is there any reason to believe that aluminum is the 'most effective' at preventing sweating, as opposed to 'effective enough' given cost, toxicity, manufacturability, and a dozen other dimensions?

Aluminum ions are also used for medical purposes (hyperhidrosis).
If there was a better option, wouldn't it be used for those applications?

Great question! I've often thought this myself!

Which biological and chemical processes underlie the effect of    antiperspirants that make metal salts so effective?

Aluminium-based complexes present in the antiperspirants react with the electrolytes in the sweat to form a gel plug in the duct of the sweat gland. The plugs prevent the gland from excreting liquid and are removed over time by the natural sloughing of the skin.  From a biological point of view, it interact with the keratin fibrils in the sweat ducts and form a physical plug that prevents sweat from reaching the skin’s surface.

Why are aluminium salts used most commonly?

Aluminium salts have a  astringent effect on the pores causing them to contract, further preventing sweat from reaching the surface of the skin. Aluminium salts are also natural antimicrobial agents, so they control bacteria on the skin, reducing unpleasant odors.
Some of the aluminium salts used in antiperspirants:

Aluminium chloride
Aluminium chlorohydrate
Aluminium zirconium tetrachlorohydrex gly (an aluminium-zirconium compound)


text:
Concept of Adjusted Retention Time in Chromatography
Why is the relation between the adjusted retention time, retention time and dead time as follows :
tR' = tR - tM ?
To find the time that the sample spends in stationary phase (adjusted retention time), we just need to subtract the time, in which it is in the column but not in stationary phase. Right ?
But dead time is different from this idea. How does it help in finding the time spent in stationary phase ?

To find the time that the sample spends in stationary phase (adjusted retention time), we just need to subtract the time, in which it is in the column but not in stationary phase.

This is a misconception (a subtle point here). The adjusted retention time is the time an analyte spends in the column not the stationary phase. Remember that a chromatography column is made up of a stationary phase plus the mobile phase. The sample continuously partitions between the mobile phase and the stationary phase.
Dead time simply refers to the time it takes for a sample to travel from the injector to the detector without having any interaction with the stationary phase. Imagine a garden hose, when you open the tap, it take a while for the water to reach the other end. Dead time is exactly like that.
Calculating the fraction of the time spent by an analyte in the stationary phase is slightly more involved. Check this link on "Migration rate of solutes in a chromatographic separation"
https://web.njit.edu/~kebbekus/analysis/chromatography%20overheads.htm

But then, what is the difference between retentiin time and adjusted retention time?

Retention time includes the dead time of the sample, adjusted retention excludes the dead time of the sample.

@M.Farooq How we measure dead time if the column is always filled with mobile phase? The detector would always get signal from the mobile phase.

@adosar, You can post this as a separate question.


text:
Create a buffer using CH3COOH and NaCH3COO that has a pH of exactly 3.75.?
a 50 mL sample of your buffered solution will have to be able to withstand the addition of 25.0 mL of 0.100 M NaOH solution.
the buffered solution will break after the addition of no more than 35.0 mL of the 0.10 M NaOH.
Can someone give me an overview of how to approach this? Thanks


text:
Why are d(xy),d(yz) and d(xz) orbitals involved in d³s hybridisation?
Shouldn't d(x²-y²) dxy and dz² orbitals participate? Since this corressponds to the best case overlap?
What factors decide the participating orbitals in hybridisation?
P.S I am well versed with the concepts of CFT and VBT but know little to nothing about other theories lik MOT and Group theory.

Learn about MOT and group theory.  Hybridization is an inferior concept that should be thrown into the ash heap of history.  Start here.

@OscarLanzi so this can't be explained by simple hybrididsation?

Please, forget hybridisation especially for transition metals, read https://chemistry.stackexchange.com/questions/76726/why-is-it-wrong-to-use-the-concept-of-hybridization-for-transition-metal-complex . Oh and all atomic d orbitals have the same amount of nett overlap with an s function - zero, as they are eigenfunctions of the same Hermitian operator and are hence orthogonal.

@user226375 I am not saying it can't be explained bybhybridizatiin.  I am saying it shouldn't.  See also the preceding comment.


text:
What is jellium aromaticity?
As I understand, if a molecule is cyclic and planar, then every atom in the ring must be $\ce{sp^2}$-hybridized and the molecule must have [4n+2] π electrons, where n∈N (Hückel’s rule).
An article in Chemistry World by Sam Howell ("New rule predicts stabilising aromaticity in open-shell metal clusters", 21 MAY 2019), talks about jellium aromaticity.

image source : https://www.chemistryworld.com/news/new-rule-predicts-stabilising-aromaticity-in-open-shell-metal-clusters/3010515.article
My  question is what is jellium aromaticity?
Does it have anything to do with Hückel's rule?

Helpful: https://www.ncbi.nlm.nih.gov/pubmed/30997449

@user79161 this link leads to abstract and a paywall.Further it talks about spherical aromaticity. Now that

@user79161 Sperical aromaticity ??? That will wartent a seperate question.

I can't help with the paywall in any legal way, I'm afraid. Secondly, it pertains to jellium aromaticity as analogy to spherical one, as you noted. Thirdly, I regret you didn't find it helpful.

I didn't mean they're analogous. The paper deals with them in that way. I suppose we can get to an answer if we try to understand what we are typing here, for starters. Besides, how much information do you expect out of a 3-page paper. The link says helpful, not answer.

As a comment I can say that aromaticity is used to mean an extra stability. For starting refer to Hirsch aromaticity in Wikipedia

Please note that atoms cannot be hybridised in the sloppy way it is said "sp²-hybridised atom"; this usually implies that an atom has sp² orbitals, or is trigonally coordinated. In any case, the continuous loop of orbitals is more important. Also, while all Hückel aromats are aromatic, not all aromats obey Hückel's rules.


text:
How to capture the gas produced by combusting alcohols
I am burning various alcohols and am attempting to capture the carbon dioxide produced. I could burn the alcohols in a conical flask with a cork on the top, with a tube directing the gas into a gas jar. However, due to the small volume of a conical flask, there would be almost no oxygen to burn and the flame would extinguish quickly. Is there any experimental setup I might have access to with high-school lab equipment that allows me to burn alcohols in an environment with plenty of oxygen, while capturing the gas produced? Thanks

If you look at the overall reaction, there is actually no gas produced, but there is a consumption of gas:
$\ce{CH3(CH2)_n-OH + \frac{3}{2} (n + 1)O2 -> (n + 1)CO2 +(n + 2)H2O}$
So 1.5 mol of oxygen are consumed, while only 1 mol of carbon dioxide is formed, so the volume will be less if the temperature is maintained.
To perform this experiment, you must collect the carbon dioxide in a solution of alkali and then titrate the solution.
The carbon dioxide will be converted to carbonate:
$\ce{2NaOH + CO2 -> Na2CO3 +H2O}$
A constant flow of air must be put into the system and the gas that comes out must be bubbled with a diffuser through the alkaline solution.
The air flow must be slow, so that all the carbon dioxide is washed out of the gas stream.
In the titration there will be two equivalence points, the first one of the excess of alkali and the second one of the carbon dioxide.
This one has a much smaller pH change.
Be sure to make a careful blank of the alkali, because it normally has some carbonate that will add to the value of carbon dioxide.

Hi, thanks for the help, just a few questions: 1. Why does the solution that the CO2 is bubbled through need to be alkaline? 2. How could I pump a constant flow of air into the system, realistically, in a lab setting? Thank you again.

Look at the second reaction. The carbon dioxide is converted to carbonate which is not a gas and will stay into the solution. In our lab we have a line of compressed air. There are also small pumps wich give low flows of air. They are simple and cheap

Is there a reason you can think of that could justify this experiment to determine the CO2 emissions of alcohols over just using their mole ratio in the reaction you provided to find out? Possibly to find the environmental emissions per litre of fuel used as opposed to per mol? Thanks

If you have an unknown alcohol, you could determine the chain length by doing this experiment, and if an alcohol is mixed with water you could, in principle, determine the content. However, the precision of the results will probably be very low, so it will be kind of orientative only. To analyze water in alcohol it is better to use Karl-Fisher


text:
Ozone cross-section in the near-infrared
I am searching for ozone absorption cross-section in the wavelengths range 1.1 μm to 10 μm. So far I looked in the following databases:

HITRAN
Max-Planck Institute for Chemistry Mainz

and of course, googled it.
Any idea where I can find such information?

By searching for ozone ir absorption I found this http://vpl.astro.washington.edu/spectra/o3.htm which gives hitran spectra  over several spectral ranges.


text:
K in Gibbs free energy
When we write for an equilibrium that $$\Delta G^\circ = -RT \ln(K)$$  Which type of  $K$ , ie is it $K_\text{concentration}, K_\text{pressure}$ or $K_{\chi}$ because all these have different numeric values which will give different values of $\Delta G^\circ$ which is obviously not possible.
Edit: In the question marked as being duplicate of, they still don't tell which $K$ it is. Even if we divide by the appropriate standard value, and make it unitless, $G$ will still give different values based on $K_\text{concentration}$, $K_\text{pressure}$ or $K_{\chi}$.
Also can $K_{\chi}$ also be used as it is dimensionless less too?

You may want to see https://chemistry.stackexchange.com/questions/43531/why-are-kc-and-kp-interchangeable-in-the-free-energy-equilibrium-equation?rq=1 and links therein to additional posts.

The answer / question there does not include the mole fraction part. Because mole fraction is dimensionless is it used to for activity? Also they both answers don't tell what to do ,because in highschool atleast , we have to deal with this only. Even if we take activity , then too ,eg in formation of NH³ , we will get different answers as such for concentration and pressure

$\Delta G^\circ$ is defined for a certain standard state. No matter whether you use mole fraction, concentration or pressure  for a given species, as long as you divide these by the standard mole fraction, concentration or pressure of that species, you will obtain the correct (dimensionless) value for K. Another way of saying this: as long as Q = 1 for the standard state, you are good. This means the numerical value of K will change only if you switch to a different standard state (in this case, the value of  $\Delta G^\circ$ will also change).

@KarstenTheis But how is that possible, the Gibbs free energy has a definite value.How can it change based upon the different values of K we take.

@NightWriter This question has been marked as duplicate but the other answers don't fully answer it. How to reopen this question.?

Concerning reopening: You edit your question to mention the other Q&A and how it does not fully answer your question.
Concerning values of K: The Gibbs free energy does not have a definite value. It depends on the temperature and whether you use the "chemical" or the "biochemical" standard state.

@KarstenTheis At a given temperature it will have a definite value , which should not change based upon the calculation we do on pen and paper , but using different values of K , will give different values of G at the same temperature. Also can $K_{chi} $ be used as it is dimensionless (mole fraction)

$K$, the standard thermodynamic equilibrium constant, is computed from $\Delta G^\circ$, using
$$\Delta G^\circ = -RT\log K \tag1$$
Generally speaking, $K$ in equation (1) is unitless. Its value depends on the specified reference standard states and $T$ (and obviously on the equilibrium activities of reactants and products).  Both $K$ and $\Delta G^\circ$ change (and are certainly allowed to) if you change the choice of standard states for any species.
You can also compute an equilibrium constant from a value of $\Delta G^\circ$. To do this reverse operation you need to know the standard state ($p^\circ=\pu{1 bar}$, $c^\circ=\pu{1 M}$, or $m^\circ=\pu{1 molal}$, $\chi^\circ=\pu{1}$) associated with each species in the equilibrium equation. At equilibrium each product/reactant is at the same temperature and in the same phase as in its associated reference standard state (if not at the same partial pressure, concentration, or mole fraction).
When you use either $K$ or $\Delta G^\circ$ in practice, you need to know the standard state of each species (this is often apparent from additional information that is provided with these parameters). If you are working with a gas-phase reaction then you can refer to $K$ as $K_p$. If you are using standard states then all reference states are $\pu{1 bar}$ $^\dagger$, and partial pressures computed directly from $K_p$ are in $\pu{bar}$ units (these can obviously be converted later).
To compute $K_c$ from say $K_p$ you have to convert units with an appropriate conversion factor. This is described in the linked post. Same applies to problems where $K_\chi$ is provided, where concentrations are described in mole fraction units. Note that if you change units, for instance to compute $K_c$, then you have (perhaps implicitly) changed the reference state (perhaps to a nonstandard state), and in that case $\Delta G^\circ$ also changes. This is also explained in an answer to the linked post.
$\dagger$ Components whose activity is constant can usually be ignored


text:
What easily available chemicals do I use to turn oak light grey?

I have 10 square meters of dry and freshly sanded oak that I would like to artificially age/weather to the colour of the above photo. Once treated the oak is to be used for a bedroom interior. I was wondering if there are any chemists out there who may know how to do this.

Graying is symptomatic of bleaching (oxidation) caused by air and sunshine. You can use chemical bleach for a similar effect.
See for instance: https://www.wikihow.com/Bleach-Wood.
There are a lot of guides online on the subject. This one seems pretty useful, discussing different bleaches, their safe handling, and treatment steps. Even the US Department of Agriculture has a guide on the subject.
The appropriate bleach for your application appears to be the "two-part (A/B) wood bleach" in which a mix of peroxide and sodium hydroxide reacts with the wood. Commercial kits are available. The USDA guide has the following to say:

One of the most powerful, satisfactory, and widely used bleaches in recent years is based on  a  strong  solution  of  hydrogen  peroxide  (30  or  35  percent  strength).  The  wood  is  first coated evenly with a solution of sodium hydroxide (caustic soda, or ordinary lye) containing pounds of the solid per 10 gallons of solution.  After this has dried for about 30 minutes, an even  coat  of  the  hydrogen  peroxide  is  applied.  The  sodium  hydroxide  applied  as  the  first coat  renders  the  hydrogen  peroxide  unstable,  whereupon  it  decomposes  and  liberates  a  large amount of oxygen in a chemically active condition, which does the bleaching. The liquid on the  surface  should  be  allowed  sufficient  time  to  dry,  during  which  time  the  bleaching  occurs; then  the  wood  should  be  rinsed  thoroughly  with  cold  water.  To  neutralize  any  remaining traces  of  sodium  hydroxide,  the  surface  should  then  be  washed  or  sprayed  with  a  solution containing 1 pound of acetic acid (or 1 pound of oxalic acid) per 10 gallons of water, allowed to stand for 15 minutes, then rinsed thoroughly with water and allowed to dry. After the wood is sanded lightly to remove raised grain the new finish may be applied.

See also the section labeled "precautions".
Regarding the chemistry of the bleaching process, Wikipedia provides the following summary:

Hydrogen peroxide
Using hydrogen peroxide to delignify chemical pulp requires more vigorous conditions than for brightening mechanical pulp. Both pH and temperature are higher when treating chemical pulp. The chemistry is very similar to that involved in oxygen delignification, in terms of the radical species involved and the products produced.
Oxygen
Oxygen exists as a ground state triplet state, which is relatively unreactive and needs free radicals or very electron-rich substrates such as deprotonated lignin phenolic groups. The production of these phenoxide groups requires that delignification with oxygen be carried out under very basic conditions (pH >12). The reactions involved are primarily single electron (radical) reactions.[14][15] Oxygen opens rings and cleaves sidechains giving a complex mixture of small oxygenated molecules. [...] While the radical reactions are largely responsible for delignification, they are detrimental to cellulose. Oxygen-based radicals, especially hydroxyl radicals, HO•, can oxidize hydroxyl groups in the cellulose chains to ketones, and under the strongly basic conditions used in oxygen delignification, these compounds undergo reverse aldol reactions leading to cleavage of cellulose chains.

Lignin is a polyphenolic compound that lends wood much of its color, so it is the main target of bleaching. The cellulose is best left alone, so a relatively mild bleaching protocol is best for wood.
Another important factor affecting the aged appearance, evident in the posted image, is more difficult to replicate with a chemical treatment. That is water loss, which is responsible for cracking. A quick surface treatment might not provide the highly aged appearance in the picture$^\dagger$.
Update (and somewhat off-topic)
$\dagger$ Oak is apparently among the most difficult woods to dry without it developing cracks. If the wood is already dry then it is unlikely to crack, but if it is fresh and is allowed to dry quickly (for instance by setting in the sun for a few weeks) than it is almost certain to crack.

Thanks for that excellent, informative and concise answer! +1

Thanks Night Writer, I have access to a pretty good wood yard from whom I can source some low grade oak hence already containing the required faults at a somewhat reduced price.


text:
When will acid-base reactions occur before substitution reaction
For example, given the reaction below:

The substrate is secondary, the solvent is aprotic and it is to react with a strong bulky base. From this, I assumed that the reaction would be $\ce{E2}$. However, it instead begins with an acid-base reaction where the tert-butoxide removes the proton of hydroxide and then proceeds with a $\ce{Sn1}$ reaction where the oxygen attacks the carbon of the leaving group to form an epoxide.
How can you determine if an acid-base reaction will take place before proceeding with a substitution reaction?

Acid-base reactions (like the one shown, involving proton transfer between electonegative atoms) are almost always extremely fast - for all practical purposes instantaneous (diffusion-controlled), and therefore much faster than substitution reactions. In the example given, the bromocyclohexanol is of comparable acidity to tert-butanol (maybe even more acidic), so we would expect instant proton transfer, and the observed result.
The initial proton transfer might not be so favorable if you used a much weaker base, so you could get a different result - or no reaction.


text:
Why is nitroglycerin so unstable?
I know that nitroglycerin is an incredibly powerful explosive due to its three nitro groups, but why is it so unstable?
What in its chemical structure makes it so sensitive to shock and temperature? Why is TNT, in contrast, so stable?

Welcome to Chemistry.SE! Please do spare some time to go through the tour and the help center. We're glad you could join us!

partial reason is, nitrate esters ( C-O-NO2 ) are less stable than nitrocompounds ( C-NO2 ). Another partial and  speculative reason is being liquid.

Maybe it is explosive combination and orientation of the atoms in nitroglycerin that even slightest impact or friction cause it to explode releasing large amount of energy, atoms being rearranged to form stable molecules like nitrogen, oxygen etc.

An explosion is a type of spontaneous chemical reaction that, once initiated, is driven by both a large exothermic change (great release of heat) and a large positive entropy change (great quantities of gases are released)   $\ce{^1}$ from  reactants to products (a thermodynamically favorable process).
The energetic stability of the gaseous products and hence their generation comes from the formation of strongly bonded species like carbon monoxide, carbon dioxide, and (di)nitrogen, which contain strong double and triple bonds having bond strengths of nearly 1 MJ/mole.
Consequently, most commercial explosives are organic compounds containing $\ce{-NO2, -ONO2}$ and $\ce{-NHNO2}$ groups that, when detonated, release gases like the aforementioned (e.g., nitroglycerin, TNT, HMX, PETN, nitrocellulose).
In general, Detonation is used to describe an explosive phenomenon whereby the decomposition is propagated by an explosive shock wave traversing the explosive material at speeds greater than the speed of sound within the substance.
The following factors affect the stability of an explosive: (1) Chemical constitution.
(2) Temperature of storage. (3) Exposure to sunlight (4) Electrical discharge.
Certain groups like nitro  (–NO2), nitrate (–ONO2), and azide (–N3), are intrinsically labile or Kinetically there exists a low activation barrier to the decomposition reaction which makes compounds exhibit high sensitivity to flame or mechanical shock.
Nitroglycerin $\ce{^2}$ is a primary explosive  that is extremely sensitive to stimuli such as impact, friction, heat, static electricity, or electromagnetic radiation ultimately leading to detonation.

You can see that 4 moles of nitroglycerin produces 35 moles of hot gases.The concentrated gases are under very high pressure, so they expand rapidly. The heat speeds up the individual gas particles, boosting the pressure even higher. The gas expands faster than the speed of sound, it generates a powerful shock wave.
As a rule of thumb, explosives with an activation energy of decomposition
higher than 170 kJ/mol are stable for thousands of years at room temperature, whereas for values below 155 kJ/mol, chemical stability is limited.$\ce{^3}$
As can be seen from Table below $\ce{^4}$, the commonly used explosives from the classes of aromatic and aliphatic nitro compounds, secondary nitramines, and organic azides are very stable, whereas aliphatic nitrate esters suffer from much lower chemical stability.

TNT is a secondary explosive less sensitive than a primary explosive and requires substantially more energy to be initiated.
2,4,6-trinitrotoluene (TNT) is a molecular explosive that exhibits chemical stability due to its high performance, comparatively low shock sensitivity, low melting temperature of TM = 353–354 K (80–81 °C)$\ce{^5}$ $\ce{^6}$
References

https://en.wikipedia.org/wiki/Explosive#Stability
http://www.ch.ic.ac.uk/rzepa/mim/environmental/html/nitroglyc_text.htm
Chimia 58 (2004) 401 -408
https://www.ingentaconnect.com/content/scs/chimia/2004/00000058/00000006/art00010?crawler=true
M. L. Hobbs and M. J. Kaneshige, in Proceedings of the Society for Experimental Mechanics Annual Conference, Albuquerque, NM, 2009
APPLIED PHYSICS LETTERS 104, 021911 (2014) ,
" Chemical stability of molten 2,4,6-trinitrotoluene at high pressure "
Dana M. Dattelbaum, Raja S. Chellappa, Patrick R. Bowden,Joshua D. Coe,
and Madeline A. Margevicius


text:
What polymer can be dissolved in ethanol and is insoluble in water?
I need a polymer with parameters noted above. How to determine the solubility theoretically? Hansen/Hildebrand solubility parameters? Wypych is not the ultimate solution...

As a preface, I'm not sure what the exact solubility levels are, so this answer is qualitative.  Hopefully someone else can improve upon it or suggest something better, but for now:

Poly(1,2butylene glycol) (PBG) of moderate molar mass (Mn ~1000) is soluble in ethanol but sparingly soluble in water. Most solids will wind up being soluble in water if they're soluble in ethanol, owing to the two solvents' similar intermollecular forces but water's smaller size and greater overall solvent strength. This said, materials with the ethyleneoxy- repeating unit like Polyethyleneglycol (PEG), Poly(1,2butylene glycol) (PBG), or crown ethers have enhanced solubility in ethanol due to ethanol's ability to line up along the backbone of the polymer. Meanwhile, the ethyl pendant group hanging off the backbone of Poly(1,2butylene glycol) strongly decreases water solubility. The end of the ethyl group is far enough away from the oxygen that it's completely non-polar and the shape winds up causing something called clathrate formation in the water.
This polymer is mostly used as a Solgel or a coating. It could also be used for chelating some metal salts in solution. However, it would make a rather terrible structural polymer.


text:
what's the mechanism of formation of succinimide ester?
I am forming succinimide ester:

Below is the experimental protocol I have adapted from (ACS Chem. Biol. 2012, 7(2), 316–321) pg 7 in supporting information:

Formation of Activated Ester
N-Succinimidyl Trifluoroacetate (15.0 mmol, 5 equiv) was placed in a flask. The flask was
backfilled with Nitrogen, capped with a rubber septum, and maintained under an argon atmosphere. To the flask were
added THF (15 mL, 0.2 M) and 2,2-dimethoxyacetic acid (3.00 mmol, 1 equiv). Pyridine (1.69 mL, 21.0 mmol, 7 equiv) was added by syringe dropwise (over 3
min, slightly exothermic). The mixture was stirred in the dark (3 h, producing a homogeneous solution), quenched with water (5 mL, 5 min), and diluted with DCM (45 mL) and water (50 mL). The organic phase was isolated, the aqueous phase was further extracted with a mixture of DCM and THF (3:1, 25 mL), and the combined organic phases were washed with sodium bisulfate (0.1 M aqueous, 40 mL) and dried with sodium sulfate. Volatiles were removed under reduced pressure

Could somebody provide the mechanism of this reaction? My guess is that pyridine abstracts hydrogen in the carboxylic acid, and the lone pair on oxygen attacks the carbonyl centre in N-Succinimidyl Trifluoroacetate.. Is this correct? I could not rationalise why 5 equiv. would be needed.

The reaction did not quite work out and I think it may be attributed to the fact that I did not have pure N-Succinimidyl Trifluoroacetate to begin with. When forming N-Succinimidyl Trifluoroacetate, I saw two peaks in 19F NMR at $\delta \: \pu{-72.14 ppm}$ and $\delta \: \pu{-75.62 ppm}$. I believe a peak at $\delta \: \pu{-75.62 ppm}$ was unreacted Trifluoroacetic anhydride. The 1H NMR showed that there was some unreacted N-OH succinimide. I want to figure out what kinds of side products these unreacted materials have led, but to do that, I believe I must understand the mechanism of this reaction first.

Would appreciate any help or guidance

Your suggested mechanism is only the first step which would generate the mixed anhydride of TFA and dimethoxy acetic acid. What has to happen then is the O- of the N-hydroxysuccinimide has to attack the mixed anhydride. Remember the pyridine that is present can also act as a nucleophile displacing the hydroxy succinimide from the starting material. I think the mixed anhydride is the key intermediate.

@Waylander: Agreed! The pyridine cleaves the TFA (exotherm) group and the resultant salt forms the mixed anhydride. This step should be reversible back to the starting material because the TFA half of the mixed anhydride is more reactive. Ultimately, the dimethoxy acetic acid half of the mixed anhydride reacts irreversibly with N-hydroxysuccinimide.

@Waylander thank you. But by the sound of it, I still should have got some product out but I don't see anything. Do you think there is any chance that sodium bisulfate has hydrolysed the acetal group?

The pKa of sodium bisulfate is 1.99 so yes it is quite possible you hydrolysed the acetal

@Waylander Would using brine, NaHCO3 and NH4Cl solve the issue?

The reason for using bisulfate is to remove pyridine. Do you need to remove this for the next step? Hard vac down, redissolve in DCM and citric acid wash should remove most of it. A further wash with aq Cu(II)  should remove it all.

@Waylander Thanks for your insight. I may skip using citric acid as it can again destroy my acetal but will use Cu (II) solution

Pyridine boils at 115 so dilution with toluene followed by hard vac down should remove most of it.

I would also pay attention to the water content of all substrates and reagents used. THF is also notorious for picking up water.

@Waylander so using N-Succinimidyl Trifluoroacetate has failed and thinking of using DIC with N-OH Suc... (want to avoid using DCC as I have been told that removing DCU is extremely difficult) would DIC (diisopropylcarbodiimide) work in this case?

It would, but there are also diimides with a remote tertiary amine that are easier to remove than DIC residues

Have a look here:  http://documents.bachem.com/coupling_reagents.pdf

@Waylander Thank you. They suggested EDAC · HCl etc but I am scared that the HCl might hydrolyse my acetal groups.. maybe I should try with DIC first.

The HCl is tied up with the tertiary base so should not be able to protonate your acetal. If in doubt have a bit of extra tertiary base present.

@Waylander I have done it. After stirring for some time, I have filtered using celite, and dried of solvent (THF). I have 3 major singlets: $\delta : \pu{4.840 ppm}$ (1H), $\delta : \pu{3.432 ppm}$ (7.26H), and $\delta : \pu{2.731 ppm}$ (6.36H) + many other impurity peaks. As you can see, the integration ratios are all off (while chemical shifts kinda make sense) think my crude is just too impure to draw out any conclusion. The issue is.. I don't think my product will stay stable in the silica. Do you have any suggestions? Would appreciate any advice& thoughts

What is your ultimate goal with this product? Are you reacting it with a nucleophile?

@Waylander Bioconjugation. It will be reacting with the amine group on my amino acid

How precious is the amino acid? I would be inclined to run the reaction with the crude N-hydroxysuccinimde material and try to purify the product. Alternatively you could try to make the mixed anhydride of the dimethoxy acetic acid with BOC-anhydride using the conditions described here https://www.nrcresearchpress.com/doi/pdf/10.1139/v87-106


text:
Why is the trimethylammonium group a stronger electron withdrawing group than the nitro group?
I was working on a question that basically boils down to deciding whether the nitro group ($\ce{-NO2}$) or the trimethylammonium group ($\ce{-N+(CH3)3}$) on a benzene ring, is a stronger electron withdrawing group (EWG).
The solution says that the trimethylammonium group is a stronger EWG since it has an overall positive charge, while the nitro group is overall neutral.
But I am getting confused here. Why should the nitro group be neutral? The $\ce{N}$ atom has 4 bonds (a single bond to an oxygen atom, a double bond to another oxygen atom, and another bond to the carbon of the ring), so the $\ce{N}$ atom in $\ce{NO2}$ should also have a positive charge?
Further I analysed that since $\ce{O}$ is more electronegative than $\ce{N}$, it tries to shift the $\pi$ electrons of the double bond towards itself, hence increasing the electrophilicity of the nitrogen of $\ce{NO2}$, while in the trimethyl amino group, the $\ce{N}$ being the more electronegative atom gets electrons donated to it due to the +I effect of the three $\ce{CH3}$ groups, thus reducing the electrophilicity of $\ce{N}$. So $\ce{-NO2}$ should be a stronger EWG than $\ce{-N+(CH3)3}$.
Where am I going wrong?

You are confusing the way the nitro group is drawn with a partial +ve charge on the N for an actual charge. It is not charged, this is just a convenient way of modelling it. The NMe3 group has an actual overall charge.

@Waylander, Sorry for the late reply, but does it mean that the $NMe_3$ group exists with the ring as an ion? Like $Na^+Cl^-$?

Yes, there will be an anion associated with this such as Cl-


text:
What quantity of dessicant agent
It is known that dessicant agents such as magnesium sulfate can be used in order to trap water. But I was not able to find what quantity has to be added ideally compared to the amount of water that would be trapped. I know you can put it in excess but it could be a waste sometimes.

This is not something you typically measure with accuracy. A shovelful per bucket will do.

Mag sulfate is cheap; wasting some is insignificant compared to the hassle of having to redry your solution

Too much of it might adsorb a product for wich one worked months long.

You add dissicant until it suspends easily in the solvent when you swirl the flask. Some people say: "Until it snows".
When you add a bit of magnesium sulfate or other fine powdered dissicant, it forms a lump and goes to the bottom.
Many times it just sticks to the bottom when you swirl the flask.
You add more solid and swirl until it suspends and then slowly falls to the bottom, like it is snowing. At that point it is taking no more water and it has no sense to add more.


text:
Reaction quotient and Gibbs free energy at the start of a reaction

According to the equation 33 above, shouldn't be all the reactions be spontaneous initially? Because the reaction quotient Q, is zero at the start of the reaction and logarithm of Q makes it a very large negative value and ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

Yes, basically this is stated in an answer to this post: https://chemistry.stackexchange.com/questions/115541/if-change-in-free-energy-g-is-positive-how-do-those-reactions-still-occur/115544#115544

Have you taken the image from which textbook?

Your interpretation of the equation is correct. $\Delta G^o$ gives you thermodynamic favorability of a reaction under standard conditions (Q=1) and even reactions with positive values of $\Delta G^o$ (unfavored under standard conditions) can be driven to proceed if the concentrations of the reactants and products are made extreme enough.
The best way to think about the process is via a curve of G vs reaction extent $\xi$, this is related to Q and ranges from 0 (pure reactants, Q = 0, #1 on the graph) to 1 (pure products, Q = ∞, # 2 on the graph):

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/23.4%3A_Free_Energy_and_the_Gibbs_Function
Any reaction will have an equilibrium point on this curve, corresponding to a most stable composition (# 3 on the graph). The value of the equilibrium constant (K) tells us the x coordinate of the most stable position (i.e. how much reactant vs product we have in the equilibrium mixture).
No matter what the value of K, it will always be in the range $0<K<∞$. If we start the process with pure reactants, $Q = 0$ and we are to the left of the equilibrium point. The system will shift to the right and some of the reactants will convert into products.
This is what you mean by

ultimately we'll have ∆G to be a large, negative value which makes the reaction spontaneous?

you can calculate a negative $\Delta G{_r}$ for the reaction when you have no (or infinitesimal amounts of) products.
Many texts are moving away from the term "spontaneous" for this situation, because the system may not shift to the right (reactants convert to products) if there is a kinetic barrier to prevent this, so the forward reaction may not actually be observed. The term "thermodynamically favored" is often used instead, meaning having a negative $\Delta G{_r}$ under these conditions.
You can find lots more about these concepts at the following posts:
If change in free energy (G) is positive, how do those reactions still occur?
Calculating ΔG at the extremes of reaction extent

What you write is correct about all reactions proceeding if starting with reactants only, these will decrease in concentration and products increase until equilibrium is reached. How long this takes is a matter for chemical kinetics not thermodynamics. Strictly speaking thermodynamics has nothing to say about this process as it deals only with equilibrium situations.  Think of the word 'spontaneous' in thermodynamics as not having the same meaning as the word used in general language, but describes only that $K_p > 1$ so that $\Delta G^{\mathrm{0}}<0 $.
The graph below shows $\Delta G=\Delta G^\mathrm{0} +RT\ln(Q)$ where for a reaction $\ce{A <=>B}$ at equilibrium $\displaystyle Q\equiv K_p=\frac{p_B}{p_A}$ where $p$'s are the partial pressures. The $\Delta G$  is the gradient of $G$ with extent of reaction $\xi$ which is zero when only reactants are present and is $1$ when $1$ mole of reactants have been converted to products. Thus $\Delta G$ is the slope of the curve shown in the plot in the previous answer.
An alternative way of looking at this is shown in the figure below. You can see that as $Q$ becomes smaller $\Delta G$ becomes larger and negative, and vice versa. The slope of the plot is $RT$.


text:
Mechanism of Fritsch–Buttenberg–Wiechell rearrangement
The following reaction was given, which uses potassium t-butoxide.

The solution to this problem was given as follows:

Step 1 is the attack of strong base, t-butoxide, on the vinyl proton.
Step 2 seems to have the chlorophenyl group shift to the neighbouring carbon, forcing the cleavage of the $\ce{C–Br}$ bond.
Here, the bromine is conjugated to the alkene, since the lone pair on bromine is in resonance with the alkene, leading to partial double bond character between $\ce{C}$ and $\ce{Br}$. In such a case, the $\ce{C–Br}$ bond usually cannot be cleaved.
My question is

Is this carbanion intermediate responsible for the product?
What is the mechanism to form 1-chloro-4-(phenylethynyl)benzene?

As your heading correctly states, this reaction take the path of Fritsch–Buttenberg–Wiechell rearrangement (Wikipedia).   The reaction mechanism of Fritsch–Buttenberg–Wiechell rearrangement (FBW) is described as:

The strong base deprotonates the vinylic hydrogen, which after alpha-elimination forms a vinyl carbene. A 1,2-aryl migration forms the 1,2-diaryl-alkyne product. The mechanism of the FBW rearrangement was a subject of on-surface studies where the vinyl radical was visualized with sub-atomic resolution (Ref.1).

However, if you carefully read this reference, you'd realize they use special method to generate diradical from gem-dibromo-alkene stepwise. Thus, the proposed mechanism for every FBW-type reactions. Earlier, A. A. Bothner-By (Ref.2) from  Harvard reported that in a diaryl-substituted system of a bromoalkene (e.g., see structures 1 and 4 in the scheme 1), the major migrating group is the aryl moiety located anti to the halogen (the fact, which has been proven by $\ce{^{14}C}$-labeling experiments). Particularly, this migration of an aryl group between multiply-bonded carbons is one of the several unusual features of the rearrangement (Ref.2):

Yet, there are plethora of data to support the 1,2-phenyl group migration through single bonds (e.g., the benzilic acid rearrangement), and at least one example for migration of an aryl group between multiply-bonded carbons (e.g., the Beckmann rearrangement). Thus, based on these facts, I suggests following arrow-pushing mechanism, which would be very reasonable:

References:

N. Pavliček, P. Gawel, D. R. Kohn, Z. Majzik, Y. Xiong, G. Meyer, H. L. Anderson, L. Gross, “Polyyne formation via skeletal rearrangement induced by atomic manipulation,” Nature Chemistry 2018, 10, 853–858 (https://www.nature.com/articles/s41557-018-0067-y).
Aksel A. Bothner-By, “Stereochemistry of the rearrangement involving migration between multiply-bonded carbons,” J. Am. Chem. Soc. 1955, 77(12), 3293–3296 (https://doi.org/10.1021/ja01617a042).

The link for ref. 1 is the same as ref. 2. Ref 1 should be https://www.nature.com/articles/s41557-018-0067-y.pdf......@Mathew


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Calculating bond length in VMD (Visual Molecular Dynamics)
Is there a way to calculate bond length in VMD (Visual Molecular Dynamics)?
I have a lot of trajectory files of water molecule, and I need to calculate the bond length of $\ce{O-H}$ bond, and find the average.
I know that there is a GUI based method to calculate bond lengths in VMD, but can I do the same using a script?

The answer is "of course". The thing is there are a lot of ways to do this. If you are looking for a "canned" way to do this, you can wait for someone to answer. Otherwise I recommend searching the VMD site including sample scripts. You can also write your own script in your own favorite programming language. If you don't have one it might be a good opportunity to learn one.

@NightWriter I would love to write my own script, perhaps in Python or Fortran. But I haven't really been able to, because I don't know how to extract data from .dat files efficiently. Do you have resource recommendations for that?

That sounds like a chicken and egg problem. If you know how to program then writing a program to inspect such a file should not be too difficult. You may want to look at this: https://www.ks.uiuc.edu/Training/Tutorials/vmd/tutorial-html/node7.html  I can't elaborate on this at present, but the native VMD TCL scripting language is not necessarily the easiest to wrap you head around, and scripts may not work straight from the webpage (you might have to do some tinkering with the code).

Also,  what do you mean efficiently? If you hope someone will post an answer here you may want to include more information into the question by editing it. What does your .dat file look like?


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Which is more stable: isobutylene or cis-2‐butene?
How do you compare which is more stable without using the heat of hydrogenation or combustion or any other numeric factual value:
$\ce{(CH3)2-C=CH2}$ or $\ce{CH3-CH=CH-CH3}$ (cis)?

Wouldn't we expect the most substituted alkene to be most (thermodynamically) stable?

@electronpusher both the compounds have two methyl groups each.

@Karl I was taught "fill in the correct term for two substituents at the same carbon">trans>cis

Numbers are better than intuition. Both alkenes have the same molecular formula. Heats of Formation (kcal/mol) are helpful: isobutylene (-4.3), (Z)-2-butene (-1.7).

@Groverkss You're right http://www.chemgapedia.de/vsengine/vlu/vsc/de/ch/12/oc/vlu_organik/alkene/alkene_kiel.vlu/Page/vsc/de/ch/2/oc/stoffklassen/systematik_struktur/acyclische_verbindungen/ungesaettigte_kohlenwasserstoffe/monoene_monoine/alkene/stabilitaet.vscml.html

@user5519, I need logic; the reason behind this.


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Limitations of extremes for the exchange-correlation functional - DFT
In DFT the exchange-correlation function is critical for the correct description of the system. I have a question about the applications and shortcoming of some functionals.
LDA delocalizes electrons so are good for conductors. And pure HF localizes electrons so it's good for molecules/atoms.
So, is it safe to say that if one doesn't include correlation and only uses HF exchange energy that crystals, or any material with long range ordering, couldn't be properly described?
And the same around, if one only uses correlation, conducting crystals are properly described but no molecules/atoms could be studied correctly?
I want to understand the applications and limitations of both extremes and I am having a hard time finding that kind of analysis online.
Thanks!

One way to begin to gain some insight in this is to consider the work done on benchmarking density functionals. I shamelessly plug the work done by my old group, which can be found here. The GMTKN30 data include results for Hartree-Fock and LDA functionals with large basis sets. The description of the several benchmark sets involved and the analysis in the actual paper will illustrate many of the shortcomings and strengths of these approaches (and why better, more pragmatical DFT approaches are more widely used). While I have no access to the GMTKN55 paper, I trust that it contains references to other benchmark papers that may actually deal with crystals.


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Is barium sulfate soluble in dilute mineral acid?
I have read that $\ce{BaSO4}$ is soluble in $\ce{HCl}$ , but I am not sure whether this works for dilute acid as well.
The reaction is -
$\ce{BaSO4 + 2 HCl ->BaCl2 + H2SO4}$

H2SO4 does not form, $\ce{HSO4-}$ does. And you have to add the physical states as you are documenting how a substance goes from solid to aqueous. Another hint: chloride is a spectator ion.

The key point is to have an acid with such a high $\ce{H+}$ activity that in large extent shifts the below reaction to the left :
$$\ce{  HSO4-  + H2O <=> SO4^2- + H3O+ } \\ K_\mathrm{a2}=1.0 \cdot  10^{-2}$$
Then activity of sulphate ions from dissolution of $\ce{BaSO4}$ is decreased by the equilibrium reaction and the solubility increases.
$$\begin{align}
K_\mathrm{s}&=a_{\ce{Ba^2+}}\cdot a_{\ce{SO4^2-}}=1.1\times 10^{-10}\\
\frac{K_\mathrm{a2}}{a_{\ce{H+}}}&=\frac{a_{\ce{SO4^2-}}}{a_{\ce{HSO4^-}}}\\
\end{align}$$
The  effective solubility product
$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{a_{\ce{H+}}}{K_\mathrm{a2}}$$
Note that despite activity coefficients {gamma} decrease with increasing ionic strength, for concentrated acids this tendency reverses and raises very rapidly
[HCl]  Gamma
0.0005 0.975
0.01 0.904
0.1 0.796
1 0.809
2 1.01
5 2.38
10 10.44
12 17.25

Therefore, $a_{\ce{H+}}$ of 10 M HCl = 104.4.
$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{104.4}{K_\mathrm{a2}}\\=1.1\cdot 10^{-10}\cdot 1.044\cdot 10^4=1.148\cdot 10^{-6} \mathrm{~ mol^2 L^{-2}}$$
Solubility would be then $ M \cdot \sqrt{K_\mathrm{s,ef}}=233.4 \cdot 0.001072=0.25\mathrm{~g/L}$
Conclusion: The $\ce{BaSO4}$ solubility in concentrated $\ce{HCl}$ is limited, in diluted one is very low. The solubility in concentrated sulphuric acid would be much higher, as $a_{\ce{H+}}$ would be very high.