text:
If sulphuric acid barely evaporates, why we can smell it?
If sulfuric acid barely evaporates, why we can smell it? Or does it give off sulfur dioxide gas?

Are you talking about laboratory-grade 98% sulfuric acid? As one of the largest volume production chemicals in the world, it comes in all kinds of grades and mixtures, so that whatever else is in it may be responsible for the odor.

You can smell salt at the sea shore even if salt does not evaporate at all.

There are at least two plausible explanations for why an odorless compound with very low vapor pressure could nevertheless generate an odor:
Sulfuric acid is highly reactive.  If you are pouring or dispensing sulfuric acid, you are probably generating some mist. This is a concern in industrial settings where sulfuric acid is used:
https://www.cdc.gov/niosh/docs/81-123/pdfs/0577.pdf
Thus, even if the sulfuric acid itself has neither a significant vapor pressure nor an odor, as the mist enters your nose it could react with substances in your nasal passages to create compounds that have an odor.  Some studies have reported the perception of odor from sulfuric acid mists:
https://www.epa.gov/sites/production/files/2014-09/documents/sulfuric_acid_interim_dec_2008_v1.pdf
And even without misting, particulates in the air, as they fall into an open container of sulfuric acid, could react with sulfuric acid to generate compounds that create a smell.  This effect is seen with metals -- the metallic "smell" is not from the metals themselves (which have an extremely small vapor pressure at room temperature), but from reaction products of the metals:
Why can we smell copper?

According to wikipedia:

Sulfuric acid (alternative spelling sulphuric acid), also known as vitriol, is a mineral acid composed of the elements sulfur, oxygen and hydrogen, with molecular formula $\ce{H2SO4}$. It is a colorless, odorless, and viscous liquid that is soluble in water and is synthesized in reactions that are highly exothermic.

Hence, you can't smell it.

But then why it is said: If you are exposed to concentrated sulfuric acid in air, your nose will be irritated and it may seem like sulfuric acid has a pungent odor (https://www.atsdr.cdc.gov/phs/phs.asp?id=254&tid=47).

As conc sulphuric acid is a strong dehydrating agent, I think it would absorb the water or the mucus in your nose and you will feel irritated

But the issue is how the sulphuric acid molecule gets into the air and into my nose?

Welcome to the site! I would urge you to go through this and this. Please try to familiarize yourself with this MathJax and $\LaTeX$ tutorial for proper formatting in posts. Additionally, plagiarism is a serious offense: when copy-pasting from Wikipedia, make sure you add a reference. And also enclose quoted text in block-quotes using >.

Sulfuric acid can get into your eyes and nose as fine droplets, or in a water-based  mist depending on how a spillage occurs.

The only subjects that might care for plagiarism or copyright in this context are Wikipedia and SE themselves. This is for clarity. Though proper citations make the answer more precise and informative.

To OP. Actually the impression of smelling can be triggered by phisical parameters, too. Whatever reaction / interaction between the nose sensors or even the mucose might trigger that. It is not really a smell but a discomfort (pungent...). For how it reaches our nose Iron is even more astonishing. I suppose that a bit of molecules, droplets, or fine dust are always in the air.

I see. I guess Alchimista is right. Thanks!

Most of the sulfuric acid in the air is formed from the sulfur dioxide released when coal, oil and natural gas are burned. The sulfurous anhydride that is released forms sulfuric anhydride slowly, and it reacts with water in the air forming sulfuric acid. Sulfuric acid dissolves in water present in the air and can remain suspended for varied periods of time; It is removed from the air in the rain. Sulfuric acid in the rain contributes to the formation of acid rain. Sulfuric acid in water separates resulting in hydrogen and sulfate ions. The ability of sulfuric acid to change the acidity (pH) of water depends on the amount of sulfuric acid and the ability of other substances in the water to neutralize hydrogen ions (buffer capacity).

It may reach our nose due to a varietyof reasons. The sensation of smelling is it probably because sulfuric acid absorbs water from the nose and irritates it. Similar to how eyes are irritated while cutting onions due to formation of sulphuric acid. Also, it isn't necessary for it to vaporize. It can be just fine droplets that reach the nose and cause irritation.


text:
How does pH of soda drink vary with time
Basically I have done an experiment where I measure the pH of my soda drink at a specific temperature over $\pu{45 min}$. At every temperature, the $\mathrm{pH}$ of my soda drink increases, but much faster for higher temperatures.
For instance, I measured at $\pu{25 ^\circ C}$ how fast the pH increases of my soda drink over $\pu{45 min}$. Next I measured at $\pu{35 ^\circ C}$ how fast the $\mathrm{pH}$ increases of my soda drink over $\pu{45 min}$. Continued for the next 4 temperatures.
I plotted the graph of how $\mathrm{pH}$ increases with time for 6 temperatures (getting me 6 graphs), but I have difficulty finding out the relationship I have is linear or exponential. The current trend I have only seems to be linear from the $\pu{45 min}$ time period I performed my experiment for, but I was wondering, if it just looks linear because my experiment was faulty, or it is actually supposed to be linear in relationship. I basically do not know if there is a theory which states that the relationship I am looking for is or is not linear.
I will appreciate the assistance.

I will make few simplifications:

I will neglect presence of other acids like citric or phosphoric acid, considering just water and carbon dioxide.
I will use the simplified exquation of $\mathrm{pH}$ of a weak acid (  :

$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a} - \log {[\ce{CO2}]}) \tag{1}$$

I will  consider $\ce{CO2}$ escaping as the exponential process of the 1st order kinetics.

$$[\ce{CO2}]=A + B \cdot \exp{(-Ct)} \tag{1a}$$

I will  neglect the final equilibrium $\ce{CO2}$ concentration, compared to initial one:

$$[\ce{CO2}]=B \cdot \exp{(-Ct)} \tag{1b}$$
We can realize that the exponential function for $\ce{CO2}$ escaping kind of nullifies the logarithm function of $\mathrm{pH}$ definition:
$$\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {\left(B \cdot \exp{(-Ct)}\right)}\right) \tag{2}$$
$$\mathrm{pH}
\simeq \frac 12 \left(\mathrm{p}K_\mathrm{a} - \log {B} + \frac1{2.303}Ct\right) \tag{3}$$
Therefore, $\mathrm{pH}=f(t)$ is approximately linear.
It is not linear if we consider all factors, especially if other acids are present.

Additional requested clarification:
The equation (1) is quite notoriously known simplified equation for $\mathrm{pH}$ of weak acid. It can be directly derived from definition of a dissociation constant of an acid:
$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{4}$$
involving 2 simplifications:

solution is acidic enough to ignore water auto-dissociation:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{\ce{[HA]}} \tag{5}$$

solution contains enough of a weak acid, so it's majority is not dissociated:

$$K_\mathrm{a} \simeq \frac{\ce{[H+]}^2}{c} \tag{6}$$
where
$$c=\ce{[HA]}+\ce{[A-]} \tag{7}$$
Then
$$\ce{[H+]}=\sqrt{ K_\mathrm{a} \cdot c} \tag{8}$$
After logaritmization of (8), we get (1).

May I know how that equation you gave regarding the 1/2 (expression) came about in the very first line of the post? I would appreciate it.

See the answer addition.

Thank you, additionally i do not understand how equation (2) became (3), where did 2.303 come from?

As an undergraduate majoring in mathematics, you should know that in chemistry context, log means log10,  log(exp(x))= ln(exp(x))/ln(10)~=x/2.303. Additionally, the omission of A relates to omission of the final CO2 equilibrium content.

Oh yea thats true, my bad

Pure water has a pH that goes form 7.00 at 25°C, to 6.92 at 30°C  and to 6.13 at 100°C. As you obtain an effect which is contrary to this one, it means that your solution is loosing dissolved $\ce{CO2}$, so that the solution becomes more and more alcaline (or basic). After some time, you may obtain a precipitate of $\ce{CaCO3}$, depending on the composition of your soda.

source: Stack Exchange

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

sort: votes


text:
Why do boiling points of alkanes and alcohols become closer with increased chain length
I came across this graph, which clearly shows a large difference between the boiling points of alkanes and their equivalent alcohols.
However, it is clear that as chain length increases, the difference in boiling point between alkanes and alcohols decreases.
What is the main reason for this?
I have read that it is due to the dispersion forces increasing in both alcohols and alkanes, but wouldn't this not really have an effect on the hydrogen bonding of the alcohol? (meaning that boiling point should be consistently higher)

It is simple. Longer the chains higher the similarity between them. In other words the OH group are getting "diluted".
It is more difficult for the donor to find the accepting counterpart, and the number density of the hydrogen bonds, both for volume and mass, gets lower.
The interaction between molecules becomes dominated by VdW forces.
To visualize this think of methanol and of a very very long primary alcohol. In the second case it is somehow "more difficult" to spot the difference versus the parent alkane.

Thanks, could you explain a bit more about how the OH group is diluted?
Is it simply that it is more difficult for hydrogen bonds to form since the donor and acceptor are less likely to be close together?

Yes. And also on number per mole vs mass or volume. You still have one group for each chain, but simply the chain gets longer. Number of OH / number of C or #OH / mass etc. You have many way to express or visualize this...

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


text:
How electron configuration is related to the oxidation state of an element such as sulphur and can this be extended for elements in other blocks?
The question arises from trying to get where does the values of $-2$, $0$, $+2$, $+4$ and $+6$ from the electron configuration.
The electron configuration for sulphur $(Z=16)$ is $[Ne]3s^23p^4$.
It makes sense that to complete the octet sulphur needs to gain two electrons hence it will have $-2$, but it can lose all the electrons $2+4=+6$. But I have no idea what sort of explanation can be made for obtaining the other values?. Does it exist a rule or something that am I unware of?.

It is pretty rare that Sulfur looses two electrons. It may "loose" the four electrons 3p in substances like sulfites and SO2. Same thing for the six electrons 3p and 3s in substances like sulfates and SO3. Anyway the term "loose" is not ideal, because molecules have molecular orbitals, which are not always simply related to s and p atomic orbitals. It is not correct to say that Sulfur "looses" electrons. it would have been a real loss only in the case of an ionic bond.

:/ You couldn't make it any broader? This question may count as a duplicate of https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number

Does this answer your question? An introduction to "oxidation state"/"oxidation number"

Wikipedia has a nice article on the topic of oxidation state. It quotes the IUPAC definition:

IUPAC: Oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds [...]

For sulfur in particular, if there are no unpaired electrons, no bonds to another sulfur (i.e. all heteronuclear bonds) and sulfur has an octet, the oxidation state of sulfur will come out as an even number. The range is -2 (all electrons in octet counted towards sulfur, i.e. $\ce{H2S}$) to +6 (no electrons in octet counted towards sulfur, i.e. $\ce{H2SO4}$). For each heteronuclear bond, the electrons associated with it are either all counted toward the sulfur atom, or none of them are.

[OP] The question arises from trying to get where does the values of −2, 0, +2, +4 and +6 from the electron configuration.

The oxidation states are listed by wikipedia as the major ones (the ones in bold, see below), but others such as +1 (e.g. $\ce{H2S2}$, dihydrogen disulfide) are also possible.


text:
Is ethane or carbon dioxide easier to liquify?
Let's say $\ce{CO2}$ and $\ce{C2H6}$ are both in gas form, not vapour. Now which one will be easier to liquify? My take is that the molecule of $\ce{C2H6}$ should be much larger than $\ce{CO2}$ and hence have greater van der Waals forces. And also the critical temperature of $\ce{C2H6}$ is higher than that of $\ce{CO2}$, so it should be relatively easy to make $\ce{C2H6}$ vapour. So I think the answer should be $\ce{C2H6}$.

Note that in context of chemistry, vapour is gaseous phase of substances considered liquid or solid at normal conditions. So gas  vapour does not make sense for CO2 or ethane.

You can liquify ethane at standard pressure (1 bar) simply by lowering the temperature (cooling), as indicated by its phase diagram (see for instance here). On the other hand, if you look at the phase diagram of $\ce{CO2}$ you find that starting at standard pressure (1 bar) and room temperature,  it cannot be cooled into the liquid state. Rather it will solidify upon encountering the transition temperature (-78.464 °C, see e.g. the wikipedia) and become dry ice. In order to liquify carbon dioxide you need to compress the gas above its triple point pressure (518 kPa) before cooling. This makes the process of liquifying $\ce{CO2}$ somewhat more complicated.


text:
Lowering water hardness using Ion exchange method
I want to remove water hardness using Ion exchange resin filter. Does anyone tried this method I want to know if this method really lowers the hardness. What is the best method of regenerating resin filter? How much water it uses for regenerating process?

That is a standard water softener in the US :The resin exchanges Na+ for Ca++ , typically.

Yes - the process works
If you are just looking to remove hard water salts (divalent, trivalent) the sodium chloride regenerable resin ("softener" resin) works fine and you can regenerate and put the waste to sewer.
If you want to remove all metals, then you'll need to acid regenerate and neutralise the wash before discharge or treat as hazwaste.
If you want to remove both cations and anions then you'll need either mixed bed or separate cation and anion resin. Don't try to regen mixed bed resin at home - it's a PITA.
Or you can step up to the plate to spend some real money :

Carbon filter - removes organics and breaks down hypochlorite found in most town water
Reverse osmosis - concentrates and dumps 90+% of dissolved ions -> take permeate and follow by
UV sterislisation (technically unneccessary - but, hey, belts and braces)
follwed by
PFOS specific resin followed by
separate cat/an resin beds followed by
mixed bed nuclear grade resin into sealed, nitrogen blanketed tank

Tada - 18MOhm sterile analytical grade water.
(if you REALLY want to go all out - package it and then gamma irradiate - take no prisoners!)

The ion exchange resin works pretty well. It removes the hardness. This is the standard method for producing demineralized water in practice. If you want to regenerate the used resin, you treat it with concentrated hydrochloric acid, and then wash it with plenty of water until the water is not acidic any more when getting out of the resin. You may also replace HCl by a concentrated solution of sodium chloride. But in this case it is difficult to know when the regenerating process is finished and the refreshed resin ready for use.

Or you buy a new cartridge, avoiding the hazardous waste if you don't run a chemical laboratory with a hazardous waste management plan.

@KarstenTheis I wonder if the kitchen you or / and your friends use has a dishwasher.  All the models I encountered so far regenerate(d) their water softening resin by adding just sodium chloride to the correct opening (https://en.wikipedia.org/wiki/Dishwasher_salt).  Hydrochloric acid is something for larger scale application, but even the dishwasher in the lab regenerates multiple times with salt before its hip-high cartrige then is regenerated at the plant.


text:
Why does the binding of DMSO to metal via the S atom (k-S) strengthen the S=O bond?
So, DMSO (dimethyl sulfoxide) can form metal complexes in both $\ce{\kappa-O}$ and $\ce{\kappa-S}$ mode i.e. binding with the oxygen or the sulfur respectively. The general explanation given is that softer metal ions bind to the soft centre $\ce{S}$ and hard metal ions bind to the hard centre $\ce{O}$.
From the IR data it can be inferred that the binding of DMSO to a metal ion via $\ce{O}$ causes the $\ce{S=O}$ bond stretching frequency to go down. For example the $\ce{S=O}$ stretch in free DMSO (liquid) is $\pu{1055 cm-1}$, whereas the stretch in $\ce{[Co(dmso)6]+}$ appears at $\pu{950 cm-1}$. This makes sense to me because the oxygen can only act as $\pi$ donor (apart from the usual $\sigma$ donation), and donate electrons to the metal $\mathrm{t_{2g}}$ from bonding $\pi$ orbitals, which will weaken the bond between $\ce{S}$ and $\ce{O}$.
However, when DMSO binds to a metal ion via $\ce{S}$, the $\ce{S=O}$ bond stretching frequency goes up. For example, in $\ce{PdCl2(dmso)2}$ shows the $\ce{S=O}$ stretch at $\pu{1116 cm-1}$.
But how is this possible? The sulfur is already electron deficient due to the strongly electronegative $\ce{O}$, and should be a $\pi$-acceptor, and accept electrons from metal into the $\pi^*$ orbitals. This means that the $\ce{S=O}$ bond should also be weakened in this case. But this does not happen. Why? Does it have something to do with the molecular orbitals of DMSO?
[Note: According to this post, the $\ce{O}$ is mainly $\pi$-donor and the $\ce{S}$ $\pi$-acceptor for DMSO, and I agree with the argument in that answer. But it does not say anything about the strengthening of the $\ce{S=O}$ bond in $\ce{S}$ bonded complexes.]
The pictures of HOMO and LUMO of DMSO respectively:

Just guessing here, but I think you're confusing yourself by thinking of the S-O bond as having a lot of pi character. It's essentially just a sigma bond strengthened by the ionic interaction of S+ and O-. That's why the geometry of DMSO is more like trigonal pyramid than trigonal planar. So in kS, as long as S donates more density to the metal via a sigma interaction than it gets back in a pi interaction, it becomes more charged and the ionic component of the S-O bond strengthens. With kO binding, the charge on O decreases, weakening the ionic component of the S-O bond.

@Andrew I am not sure that there can be full +1 and -1 charges on S and O though. The electronegativity of S is 2.58 which is nearly around C (2.55). If C=O bonds have a lot of pi-character then S=O bond can too. Is it possible to estimate how much of S=O bond is pi and how much is ionic?

-1 For seemingly so well researched question, to think of S-O bond as double, not dative? Seriously?

The C=O bond is not a dative bond like the S->O, where the S still has a lone pair on it after forming the S-O sigma bond. That's why acetone is planar, but DMSO is pyramidal. Also look at the HOMO, which clearly has a nonbonding lone pair on S.  I'm sure the charges aren't fully -1 and +1, but certainly the O is negative and the S is positive because of the dative nature of the bond.

@Andrew Ok, that makes sense I think. Would you be willing to make it into an answer? If you do I will accept it.

The confusion seems to arise from the incorrect statement that the S-O bond has pi character and that there exist $\pi$ and $\pi^*$ molecular orbitals in DMSO. Instead, DMSO (despite the way it is often drawn) is most accurately depicted as having three single bonds and one lone pair on the the S atom (clearly visible as a dominant element in the HOMO shown in the question). The trigonal pyramidal geometry (also shown in the question) is consistent with this description.
The S-O bond is a dative bond, meaning that the S atom provides both electrons rather than having one electron come from each atom. That results in a formal positive charge on S and negative charge on the O. The O therefore is best represented as having one bond (to S) and three lone pairs, similar to the oxygen atom in an alkoxide ion. Because of the charge separation, the S-O bond also has a strong ionic component.
[Aside: so-called hypervalent sulfur compounds are typically drawn with double bonds that are actually combined sigma + ionic bonds rather than sigma + pi bonds. DMSO is not the only one. Keep this in mind in the future when trying to understand properties of sulfur-containing molecules.]
We now consider what happens when DMSO interacts with a transition metal. Similar to the O of an alkoxide ion, the O atom in DMSO can interact with metals as both a $\sigma$ and $\pi$ donor because of the multiple lone pairs. Both the $\sigma$ and $\pi$ donations result in decreased electron density on O, reducing its charge and weakening the S+/O- ionic interaction. The weaker ionic interaction means that the bond overall weakens, consistent with the decrease in the observed stretching frequency. There is no need for an S-O $\pi$ interaction to exist in order for this to happen.
If instead the S atom interacts with the metal, it does so primarily as a $\sigma$ donor via its one lone pair. There may be capacity for a $\pi$ acceptor interaction with $\sigma^*$ orbitals, as has been proposed for ligands such as triphenylphosphine, but if this occurs, it is with the $\sigma^*$ orbital associated with the S-C bonds, not the S-O bond, consistent with the LUMO shown in the question, which has a large contribution from the py orbital on S (ie the p orbital in the plane of the C atoms). Thus there is no weakening of the S-O $\sigma$ interaction due to $\pi$ acceptance.
As long as the $\sigma$ donation is stronger than the $\pi$ acceptance (which is what we expect), there is a net loss of electron density from S, which further increases the positive charge on the already electron deficient S atom. Increased charge means an increased strength in the ionic S+/O- interaction, so the S-O bond overall is stronger, consistent with the observed increase in the S-O stretching frequency.

The IR comparison between neat liquid DMSO and coordinated DMSO, according to above statements may be a little oversimplified. You would have to correlate the S-O distance of liquid DMSO with that (S-O distance) in Co(DMSO)6 or PdCl2(DMSO)2. The distances and TM_O-S angles should shed some light to stretching frequency of S=O or S-O bond.
Angular Overlap Model (AOM), "The core concept of the AOM is that different ligand orbitals will overlap with metal d orbitals to different extents because of the variety of angles at which these orbitals would approach each other. Stronger overlap leads to greater interaction", so does the IR stretching frequency.
https://www.youtube.com/watch?v=pS41j0P0Cjc
MO representation of DMSO as shown above, are those of the idealized vacuum structure of DMSO, not the solution of DMSO in the liquid nor the TM---O=S-Me2 MOs in the solid state. Care should be taken to correlate solvated DMSO with coordinated TM-DMSO. I tell you this because The DMSO-Cu-Benzoate XRD data is really hard to correlate with FTIR data. I'll keep you posted if I ever get satisfied with DFT Molecular Orbital Interpretation.


text:
Nernst Equation for Lithium ion battery
For electrochemical cells we know
$$
\Delta G = \Delta G^{\ominus} + RT\ln Q=- RT\ln Keq + RT\ln Q=-nFE
 $$
So for lithium ion batteries the reactions are
Cathode
$$ \ce{CoO2 + Li+ + e- <=> LiCoO2} $$
Anode
$$ \ce{LiC6 <=> C6 + Li+ + e^-}$$
Overall
$$\ce{ LiC6 + CoO2 <=> C6 + LiCoO2} $$
Is it safe to say the cobalt oxide is a solid as is the Lithium and can be ignored in the reaction quotient? As in a Zinc copper battery i do not think the mass of the zinc and copper affect the cell potential only the aqueous concentrations. For a lithium ion battery the cell potential is a function of the state of charge and temperature. but what are the concentrations in the reaction quotient for a lithium ion battery as most of the products and reactants are solids, is it not accurate to ignore them due to intercalation and are not exactly solids? What is nernst equation for lithium ion batteries

Imagine what the equation for Zn/Cu battery would be, if there were used respective amalgams instead of pure metals.

The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are  defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in the solid phase and the concentration $[C_6]_s$ increases.
The same reasoning may be done on the Nernst equation at the cathode :  $$E_c = E°_c + \frac{RT}{F}· ln{\frac{[Li^+]_e[CoO_2]_{s'}}{[LiCoO_2]_{s'}}}$$ where $[CoO_2]_{s'}$ is the concentration of $CoO_2$ in the solid phase of the cathode, and $[LiCoO_2]_{s'}$ is the concentration of $LiCoO_2$ in the solid phase of the cathode.
The overall voltage of the battery $\Delta E$ can be written :  $$\Delta E = E°_c - E°_a + \frac{RT}{F} · ln\frac{[LiC_6]_s [CoO_2]_{s'}}{[C_6]_s [LiCoO_2]_{s'}}$$ As a consequence, both $[LiC_6]_s$ and $ [CoO_2]_{s'}$ decrease when the battery is working. So the overall voltage decreases.

After discharge and charge the voltage jumps or drops immediately due to the activation and ohmic overpotentials. But the change due to the mass transport loss/concentration loss takes time. This change is due to the concentration diffusion but the Li+ ions concentration at the electrodes are not in the overall voltage. Are you saying the solid concentration is changing with time while no current is flowing?


text:
Hydrogen peroxide / sodium or potassium permanganate: reaction type
I have some people on youtube stating the steam generating reaction of $$\ce{3H2O2 + 2KMnO4 → 3O2 + 2MnO2 + 2KOH + 2H2O}$$
is catalytic, from where I'm sitting it is purely stoichiometric. Can someone please enlighten me as to why some people would consider this reaction catalytic with the presence of the permanganate the catalyst?
If this reaction really is regarded as catalytic how is this so, it flies in the face of everything I've learnt about reaction types. I'm only an amateur, can someone who is proficient in their chemistry please clarify this.

Practically everything catalyzes the decomposition of hydrogen peroxide and manganese dioxide does an excellent job in that regard: just add some to hydrogen peroxide and watch it go. So decomposition of the potassium permanganate produces manganese dioxide and that does the job. Admittedly, this is not the simple case of adding a catalyst and having it neither be produced nor consumed in the reaction.

BTW, because the MnO2 is in the form of freshly generated nanoparticles, it is particularly active. Just adding powdered MnO2 to 3% peroxide evolves oxygen only slowly. Using 90% or stronger H2O2 produces a finely divided chemist.

Learning from the comments and from looking at this video (https://www.youtube.com/watch?v=9S6v-NMMnB4), there are two reactions:
$$\ce{3H2O2 +2KMnO4 -> 3O2 + 2MnO2 + 2K+ + 2OH- + 2H2O}\tag{1}$$
$$\ce{2H2O2 ->[MnO2] O2 + 2H2O}\tag{2}$$
The first reaction creates the catalyst $\ce{MnO2}$, the second generates oxygen gas even after the permanganate is used up. If you look closely at the video, you will see bubbles (oxygen) coming out of the solution even after the solution is colorless (i.e. permanganate has completely reacted).
For some more chemistry, I also recommend this video by Prof. Martyn: https://www.youtube.com/watch?v=OLUyeCC-2Ko

Can someone please enlighten me as to why some people would consider this reaction catalytic with the presence of the permanganate the catalyst?

This is not accurate. What is true is that you need much less permanganate than hydrogen peroxide because only a little is required to make the catalyst, which then allows the reaction to proceed without being used up.


text:
What is the difference between equivalence and half-equivalence points?
I learned in class that the equivalence point is the point of neutralization where the amounts of acid and base are equivalent.
I was also told that the half-equivalence point is when the concentration of a weak acid equals concentration of conjugate base: $[\ce{HA}] = [\ce{A-}].$
I did a research and found similar definitions that don’t really shed any light on the differences between them and both terms seem identical to me. What is the difference between a half-equivalence and an equivalence point, exactly?

Look at the benzoic acid titration curve here: https://chemistry.stackexchange.com/a/117033/79678 . Halfway to the equivalence point, the concentration of benzoic acid equals that of benzoate ion. The is the middle of the buffer region. At the equivalence point, you just have a solution of sodium benzoate. There are lots of pH questions and answers here, so please use the search function.

In short then the difference is striking.

Start with exactly 1 M HCl solution. Now add exactly 0.5 moles of solid NaOH. This is the half-equivalence point: half of the total HCl has been neutralized, resulting in a solution that is 0.5 M HCl (aq) and 0.5 M NaCl (aq). Note that chloride ion is the (pathetically weak) conjugate base of HCl (aq). Now add another 0.5 moles (exactly) of solid NaOH. This is the equivalence point: all of the acid and base have exactly neutralized and you have a 1 M NaCl solution with pH 7. Hope this helps.

I assumed exactly 1 liter of the exactly 1 M HCl solution. Activities and auto-ionization of water are neglected. At the half-equivalence point, the molar concentrations of hydronium and sodium ions are each 0.5 M and the chloride concentration is 1 M. This is charge neutrality. At the equivalence point, the sodium and chloride ion concentrations are each 1M. The hydronium and hydroxide are the values at pH 7, i.e., 100 nM each.

The equivalence point is where the amount of moles of acid and base are equal, resulting a solution of only salt and water. If you are titrating an acid against a base, the half equivalence point will be the point at which half the acid has been neutralised by the base. For instance, if you have 1 mole of acid and you add 0.5 mole of base, exactly half of the acid will have been neutralised. The solution remaining will be half salt and half acid.

Now I was also told that Half equivalence point is when [HA] = [A−] , the concentration of a weak acid = concentration of conjugate base.

That is correct. You make the weak acid in situ when you titrate a weak base with a strong acid, or when you titrate a weak acid with a strong base. At the half equivalence point, the pH is roughly equal to the pKa of the weak acid.

What is the difference between a half equivalence and an equivalence point?

The equivalence point is what it always is. The point when you added stoichiometric amounts of titrant to the analyte. In the weak acid / strong base case, it means that you have quantitatively turned the weak acid into weak base, and don't have any excess of strong base (all was neutralized). The pH would be that of the weak base. In the weak base / strong acid case, it means that you have quantitatively turned the weak base into weak acid, and don't have any excess of strong acid (all was neutralized). The pH would be that of the weak acid.

Based on what i have been told, they sound the same to me.

Compared to the equivalence point, you have to add half the titrant to reach the half equivalence point, so they are different.


text:
Why is the pH of rainfall much lower than pure water?
Like it says on the tin. Several sites say that the pH of rainfall is somewhere between 5 and 5.5, and a few seem to suggest that it averages around 5.7. But why is the pH of rain much lower than that if pure water, which is 7?

This is because of the various oxides present in the air. Oxides of sulphur, nitrogen and carbon are always present in the air. They react with the water during rainfall to form their respective acids as shown below :
SO3+ H2O -> H2SO4
2NO2 + H2O -> HNO3 + HNO2
SO2 + H20 -> H2SO3
CO2 + H2O -> H2CO3
Hence, due to the presence of these acids the pH of rain is less than 7. When the levels of these oxides in the air increase due to excessive pollution it leads to further decreasing of the pH of rain which causes to acid rain.

Room temperature water saturated with normal levels of atmospheric CO2 would have a pH about 5.7.


text:
Electron pairing in atoms
A question asks:

This question refers to isolated gaseous atoms. In which atom are all electrons paired? Ba, Br, S or Si?

I thought it was Ba as its electron configuration is [Xe] 6s2.
However, the answer says Si, which seems incorrect to me.
Could someone please confirm the answer for me?
Thanks

Si seems incorrect to me as well.

The configuration for the free atoms is:

Ba - $\mathrm{[Xe]\ 6s^2}$
Br - $\mathrm{[Ar]\ 3d^{10} 4s^2 4p^5}$
S - $\mathrm{[Ne]\ 3s^2 3p^4}$
Si - $\mathrm{[Ne]\ 3s^2 3p^2}$

Thus:

Barium has no unpaired electrons.
Bromine has one unpaired electron in 4p subshell.
Sulfur has two unpaired electrons in the 3p subshell.
Silicon has two unpaired electrons in the 3p subshell.


text:
Why is absorbance in UV-VIS spectrometer negative even after calibration?
I was collecting data to plot a calibration curve using pure beta-carotene sample. Prepared 5 different concentrations of beta-carotene dissolved in methanol. Pure methanol was used as blank. However, data collected for all 5 concentrations observed was negative (between -0.005 and -0.010) at 610nm ( orange region along the spectrum). Please advice.

Beta carotene does not absorb at 610 nm. Are you speaking of a slightly displaced baseline?

I am aware that beta carotene does not absorb at 610nm, which gives rise to my confusion. however, the only peak in absorbtion was found at 610nm. it is also negative. hence i am checking what could have gone wrong.

Even not absorption features? Add this info to the question. But I guess no one can speculate of what went wrong. It can be everithing including peaking up the wrong files, for what we know. One can only point to a typical error of novices, ie absorption and scattering exceedingly high, so the instruments handle a kind of 0 as numerator...

Do the samples look orange? Are the solutions transparent or turbid?

What kind of spectrophotometer? What size cuvette/smple volume? As much detail as you can provide will be helpful. ALso, do you have a sample of B-carotene that is concentrated enough that you are able to see the peaks you are expecting?

A negative absorbance is the sign that the substance is slightly fluorescent.

Possible causes include:

If it is a dispersive instrument (simultaneous recording of blank and sample in two cells): is the logic about which one is sample and blank set correctly?  At least some spectrometers by PerkinElmer and Shimadzu allow to swap how front and rear path are managed by the software.
How long did you run the spectrometer to warm up?  Maybe the instrument wasn't yet stable, elder spectrometers need 10mn+ to warm up all their internals.
Check the lamp for its run time and compare with the specifications by the manufacture.  Depending on the setup, the instrument's software may tell you this, or / and the lamps may have a counter (the little box between the plug and the mount of the lamp shown in the picture below).  If the lamp's intensity over time is not stable any more while recording sequentially blanks and samples, you will have a hard time to record faithfully spectra.  This cause typically remains unnoticed with mono-beam instruments less well maintained and run until the lamp «breaks» (multiple times older than the recommended run time).

(image credit)

Check your instrument with a known standard.  This either may be a liquid solution of a known concentration (e.g., by SigmaAldrich) to purchase, or already sealed into a cell, e.g.

(image credit)
or solid samples of metals deposit on glass or quartz

(image credit)
Don't forget that groups active in materials science / spectroscpy / optics may have such references to calibrate their instruments regularly and may be happy to share with you how to use these accurately, too.


text:
What exactly is the 'shuttle effect' in sulfur batteries?
Ref :
https://www.electropages.com/blog/2020/01/why-lithium-sulphur-batteries-are-taking-so-long-be-used-commercially

Because they continually shuttle between the electrodes, they never change back into a neutral state (which the surface polysulphide ions do)

I don't think this explains the 'shuttle effect' clearly. I still don't understand what it is.
During discharging, lithium ions and sulfur form Li2S8 at the cathode, then it becomes Li2S4, Li2S2, Li2S with continued discharging. During charging, the Li2S move towards the cathode, now, why can't it be reversely reduced to Li2S4, Li2S8. Why can't they ever change back into a neutral state?

OP's title indicated OP is looking for the real mechanism of  the Shuttle Effect in $\ce{Li-S}$ batteries. Yet, OP is asking different questions in the text body. I'd try to address as much as possible in this answer.
A most recent review article (Ref.1) states that:

Lithium-sulfur ($\ce{Li-S}$) batteries are one of the most promising batteries in the future due to its high theoretical specific capacity ($\pu{1675 mAh g−1}$) and energy density ($\pu{2600 Wh kg−1}$). However, the severe capacity fading caused by shuttle effect of polysulfide needs to be addressed before the practical application of $\ce{Li-S}$ batteries.

To understand what is the Shuttle Effect, OP need to first understand the working chemical mechanism of $\ce{Li-S}$ batteries, and can take a head stat in that direction with this article about Lithium–Sulfur Battery, which gives a fair description of the chemistry involve in the battery. The same Wikipedia article simply described the Shuttle Effect as follows:

The lithium polysulfide $\ce{Li2S}_x$ ($6 < x < 8$) is highly soluble in the common electrolytes used for $\ce{Li-S}$  batteries. They are formed and leaked from the cathode and they diffuse to the anode, where they are reduced to short-chain polysulfide and diffuse back to the cathode where long-chain polysulfide are formed again. This process results in the continuous leakage of active material from the cathode, lithium corrosion, low coulombic efficiency, and low battery life.

The formation of polysulfide shuttle effect is classified into five steps in Ref.1:

Formation of long-chain polysulfide,
Detaching of polysulfide from sulfur host,
Dissolution of polysulfide into electrolyte,
Migration of polysulfide toward lithium anode side, and
Reaction between polysulfide and lithium anode.

If interested, OP may read Ref.1 and Ref.2, which thoroughly discussed the mechanism and possibilities of how to prevent the Shuttle Effect in depth. For example, one such techniques is modifying the separator, which has been done successfully recently (Ref.3):

References:

Wenchen Ren, Wei Ma, Shufen Zhang, Bingtao Tang, "Recent advances in shuttle effect inhibition for lithium sulfur batteries," Energy Storage Materials 2019, 23, 707-732 (https://doi.org/10.1016/j.ensm.2019.02.022).
Chao Deng, Zhuowen Wang, Shengping Wang, Jingxian Yu, "Inhibition of polysulfide diffusion in lithium–sulfur batteries: mechanism and improvement strategies," J. Mater. Chem. A 2019, 7(20), 12381-12413 (https://doi.org/10.1039/C9TA00535H).
Yanpeng Fan, Zhihui Niu, Fei Zhang, Rui Zhang, Yu Zhao, Guang Lu, "Suppressing the Shuttle Effect in Lithium–Sulfur Batteries by a UiO-66-Modified Polypropylene Separator," ACS Omega 2019, 4(6), 10328-10335 (https://doi.org/10.1021/acsomega.9b00884).

I've read this paper before asking my question. But it doesn't say, why does leakage of cathode hurts the performance, why the sulfur deposited on the anode destroys the battery, or why is corrosion of anode a problem. Specifically, of the 12345 items in the quote, why does each pose a problem?

This battery is based on lithium and sulfur. Any time you loose one of them hurts the performance (similar to basic dry battery, which normally loose zinc electrode). Here, when lithium polysulfide ($\ce{Li2S}_x$, $x \gt 6$) is formed it is dissoved in organic electrolyte, making it mobile.

I think the cell works BECAUSE the polysulfides dissolve in the electrolyte and become mobile. If it never dissolves, only the lithium atoms will attach to the the sulfur electrode and the battery will stop working.
When you say you "lose" electrode materials, how come you can't regain it by charging? In Zn/Cu battery, when discharging, Zn dissolves and Cu deposits on the Cu electrode, when charging, Zn deposits on the Zn electrode and Cu dissolves. It's reversible.
Why isn't Li-S battery similarly reversible?

Apparently, the main drawback of the Lithium-Sulfur battery is that it should work so : At the anode made of lithium, $Li^+$ ions are emitted and go to the cathode made of sulfur. This sulfur should accept the electrons, to make $S^{2-}$ ions which neutralize the arriving $Li^+$ ions. And it should produce a deposit of insoluble $Li_2S$ (or $Li_2S_2$, or $Li_2S_4$) on the cathode. The trouble is that sometimes, the sulfur does accept electrons, but the $S$ atom is part of a big molecule $S_6$ or $S_8$, and that this molecule fixes an electron producing $S_6^-$ or $S_8^-$ ions, which do not neutralize $Li^+$ ions. On the contrary, they cross the electrolyte in the "wrong" sense, and get deposited on the anode. This destroys the cell. Nobody has yet found the way of getting rid of this stupid problem.

But why does the sulfur deposited on the anode destroy the battery? Why can't the deposited sulfur continue to react during a subsequent charge/discharge cycle? I think this is my major confusion.


text:
Intuition behind errors
I want to ask a question about the multiplication of two variables and the error associated with the process.
I was learning about calculating errors in the physical labs at school and was taught that the following formula was used to calculate errors:

$\frac{\triangle Z}{Z} = \frac{\triangle A}{A} + \frac{\triangle B}{B}$

This made sense, given that we are in essence calculating percentage uncertainties and these are added when calculating the uncertainty of Z.
However, in my current undergraduate lab, we square each term:

$(\frac{\triangle Z}{Z})^2 = (\frac{\triangle A}{A})^2 + (\frac{\triangle B}{B})^2$

and I cannot find any reference to explain the purpose of squaring the values.
Why do we square each term as above?

Do you assume that the uncertainties of $A$ and $B$ are correlated or independent?

It is complicated. See the examples in the Wikipedia article on "Propagation of uncertainty."

We are told the measurements of A and B are independent.

In principle, the second formula with the squares is better, according to the Gaussian theory of error propagation. But in the lower classes, we accept the formula with the non-squared errors. Anyway the difference between the two calculations is not severe. And the beginning student hate both squares, and then extracting square roots. This is also true for students at higher levels...

@vik1245  Possibly you stumble over «variance».  For the later, if you have a function $z$ which is the result of two mutually independent functions $a$ and $b$, you access the variance of $z$ via the partial derivatives: $\mathrm{var}(z) = (\frac{\partial{}z}{\partial{}a})^2 \cdot \mathrm{var}(a) + (\frac{\partial{}z}{\partial{}b})^2 \cdot \mathrm{var}(b)$. Using the squares here ensures that the variances will add, regardless of the sign of either summand, hence for either $z = a + b$, or $z = a - b$.  So the second equation comes closer to the principle of determining the variance.

@vik1245 ... as the $\Delta{}b$, for example, could be negative, hence rendering the fraction $\frac{\Delta{}b}{b}$ negative -- just depending on the order of minuend and subtrahend.  This could lead to the (erroneous) result that the error in $z$ decreases.

@Buttonwood nailed it! Never use the first expression, period. For lots of relevant information, including worked examples, download this (free) on the web: EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012. It is often just referred to as QUAM.

@EdV With my accidental find of the «Breakthroughs in Statistics» (https://link.springer.com/book/10.1007/978-1-4612-0919-5,  https://link.springer.com/book/10.1007/978-1-4612-4380-9, https://link.springer.com/book/10.1007/978-1-4612-0667-5; especially volume 2) and now your suggestion on top, the initial anticipation of a reading weekend is at least exchanged by a study weekend ahead.  Thank you for the hint.

@Buttonwood Thanks for links to those amazing books!

@EdV Again an unexpected find indeed (at least, for a chemist).  An additional example of potentially missed knowledge for each Joe who thinks «just email the library to xerox me pages x--z of the book indicated by google and Scifinder» (helas, just missing to peek into the ones in the same shelf).


text:
Can the direction of electron flow in a voltaic cell be reversed?
I am wondering if the direction of electron flow in a voltaic cell be reversed such that the cathode becomes the anode and vice versa. I understand this is not possible for two standard half-cells, but what about non-standard half-cells at STP ?

Electrolysis perhaps

A voltaic cell ( or a galvanic cell) is made up of two half cells between which a current will flow when they are connected. In other words, a battery. The Nernst equation gives the half cell voltage under non-standard conditions. However the Nernst equation assumes a infinitesimally small current flow. A simple model of real battery uses an internal resistance to account for the fact that the battery's voltage drops as more current passes. // Some batteries are rechargeable, and some are not.

@MaxW is the x intercept for a graph of the Nernst equation, where x represents the concentration of the products, equal to concentration after which the reaction become non-spontaneous and the reverse reaction becomes spontaneous ?

@Physics - Sorry, I'm not a mind reader. I have no idea what graph you mean. Please explain in detail...

@MaxW Sure. Say you have the Daniell cell at non-standard conditions.
E = 1.1  -  (R(298)/2F) x ln(x/1.0)  where E is the emf of the cell at non-standard conditions and x is the concentration of Zn(2+) ions. So, is the x-intercept, the concentration beyond which the reaction becomes non-spontaneous and hence the reverse reaction becomes spontaneous ? In the case of this cell, the x intercept is physically impossible (due to the solubility limit), but may be it is possible for other cells. In case : The reaction is :  Zn(s) + Cu(2+) --> Zn(2+) + Cu(s)

It is not necessary to draw graphs and lines, if you want to reverse the equation accusing in a cell. You only have to apply a tension which is higher that the voltage given by the cell under study. If you want to reverse the chemistry of a Daniell cell (delivering 1.1 V), you just connect the positive pole of a stronger battery like the Lead battery (delivering 2 V) to the positive pole of the Daniell cell. And you also connect both negative poles. The Daniell cell will then works in the electrolysis mode, and deposit Zinc metal on the negative pole, which is then a cathode.

There are two half-cells in the Daniel cell. So if the two half cells were connected then the current flow would stop when the cell voltage (sum of both half-cells) drops to zero. In other words each half-cell won't individually go to zero. // There are a couple of gotchas with the Nernst equation. It is really based on activities, not concentrations. Also for the Zn half cell you could reach the solubility limit of whatever Zn salt is being created in the cell.

@Maurice but is it possible to do the same by changing concentrations only without using any batteries ?

@Physics - I'll point out that for a Daniel cell the cell voltage is: $$E = 1.10 - \dfrac{0.0591}{2}\log{\dfrac{\ce{[Zn^{2+}]}}{\ce{[Cu^{2+}]}}}$$

Can going to non-standard conditions reverse the cell potential of a voltaic cell?

I have already answered that question, in the affirmative, for a simple tin and lead galvanic cell, here: https://chemistry.stackexchange.com/a/116734/79678.
What follows is an elaboration with some specific values and illustrative figures. In fig. 1, the tin and lead voltaic cell is shown:

In fig. 1, the tin and lead ion concentrations are each exactly $\pu{1 M}$ and the cell potential is $\pu{+0.014 V}$, with tin as the anode. All necessary equations are shown in the figure and the usual simplifying assumptions are made: unity activity coefficients, zero cell resistance, etc. The electron flow is from $\ce{Sn}$ to $\ce{Pb}$.
In the next figure, the lead ion concentration is $\pu{0.3363 M}$ while the tin ion concentration remains at exactly $\pu{1 M}$. This results in zero cell potential, as shown in fig. 2:

Since the cell potential is zero, it is a moot point to call tin the anode: the cell is not doing (net) electrochemistry. There is no net electron flow.
In the final figure, the lead ion concentration is exactly $\pu{0.1 M}$ while the tin ion concentration remains at exactly $\pu{1 M}$. This results in a cell potential of $\pu{+0.016 V}$, as shown in the figure:

But now lead is the anode and tin is the cathode. The electron flow is from $\ce{Pb}$ to $\ce{Sn}$.
So changing the ion concentration to non-standard values can reverse the electron flow direction. But this was an especially easy case because the two metal electrodes had standard potentials that were relatively close. For the Daniell cell, with $\ce{Zn}$ and $\ce{Cu}$, it would not be feasible.

Tangible and plausible answer

Wonderful answer! But please don't make it a habit to write EDIT whenever you add something to your post, we have the edit history to keep a record of that, further you could use a bit of MathJax whenever you feel so, though its not always compulsory. For a comprehensive read refer to this meta post


text:
Are the entropy values calculated with average heat capacities
I asked myself whether the entropy values in an ordinary table in thermochemistry are calculated with exact specific heat capacities for each little temperature change or if they are averaged the capacities and assumed to be constant.

Which ordinary table in thermochemistry? Each source should cite the origin of the data and, if it is a primary source (the one that computed the entropies in the first place, not one duplicating the values) then provide a description of the computations involved.

This one https://www.engineeringtoolbox.com/standard-enthalpy-formation-value-Gibbs-free-energy-entropy-heat-capacity-organic-d_1979.html
There stands "internationally agreed values".

Not sure where that site gets its data, might be from CRC or NIST. For the later, you may start looking in https://webbook.nist.gov/chemistry/thermo/


text:
Why can't strong bases cleave ethers?
I'm imagining a strong base like NaNH2 attacking the adjacent carbon of the ether, giving the oxygen a lone pair and forming an alkoxide. NH3 has a pKa of 36 while an alcohols (conjugate acid of alkoxides) have a pKa of around 15 so shouldn't this be favorable?

Do you mean to say that after abstracting a hydrogen from the alpha carbon, you would conjugate that and end up with an alkene+ an alkoxide?

They can, it just so happens that in many cases a really strong base is needed. Alcoholate-anion is a very shitty leaving group, so something mild like sodium amide won't work normally. Diethyl ether, for example, requires alkyllithium to be cleaved. The products are lithium ethilate and ethylene.

Yes, like permeakra said, basic cleavage be a whole lot of trouble. It's much easier to stick to our usual acidic ethereal cleavage with HI

@permeakra I meant attacking the carbon directly. So if NaNH2 attacked diethyl ether, then why can't you get sodium ethoxide and ethyl amine?

Sodamide is not actually a great nucleophile but is a very strong base. It is fair more likely to remove a beta proton causing alkene formation by elimination of alkoxide

Adding on to Waylander's point as to why it is so, the antibonding orbital of the C-H sigma bond is a hard electrophilic centre, while the antibonding of the C-O sigma bond is comparatively softer. NaNH2 would be categorized as a chemically hard species due to it's high charge density, hence it will go for the C-H bond more than the C-O one


text:
Converting Lactic Acid to Potassium Lactate
I am looking to try an experiment incorporating potassium lactate into soap making. This seems to be new territory for soap makers in that sodium lactate is usually used. I found a bottle of "Lactic Acid 88%" at a beer making supplier and already have 90% potassium hydroxide but do not know how much of one is needed to neutralize the other. Can anyone point me to the right tables or resources to do the calculations?

OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $M_\ce{LA} = \pu{90.08 g mol-1}$) and potassium hydroxide (MW: $M_\ce{NaOH} = \pu{50.11 g mol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:
$$\ce{CH3CH(OH)COOH + KOH -> CH3CH(OH)COO^-K+ + H2O}$$
This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.
Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%\mathrm{(w/w)} \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$.  Thus, mass ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$:
$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$
Similarly, since OP has only $88\%\mathrm{(w/w)}$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid.  Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:
$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$
Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:
\begin{align}
m_{\ce{KOH}}
  &= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90}
 = \pu{311.7 g}\\
m_{\mathrm{LA}}
  &= \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88}
 = \pu{511.8 g}
\end{align}
Keep in mind that these calculations are valid only if purity of both reagents is given in $\%\mathrm{(w/w)}$. Otherwise, additional information such as density is needed.

You should be aware, that in concentrated solutions lactic acid polymerizes into mainly lactyllactic acid Watson (1940), which reduces the amount of “free” acid by around 20%. The polymers are sufficiently stable at room temperatures to create inconsistencies between what is actually necessary for neutralisation (titration) and the theoretical value. Therefore, analyses for titration of concentrated lactic acid prescribe boiling with excess of Na/K-hydroxide for about 20 minutes and a back-titration of the remaining OH.

90 g pure lactic acid reacts with 56 g pure potassium hydroxyde. As your substances are not pure, you must use 102 g of your impure lactic acid (88%), and make it react with 62.2 g of your impure (90%) potassium hydroxyde.

It is unnecessary to post a new answer. Please edit an older one; if it is deleted and you cannot undelete it, please flag it for moderator assistance.


text:
Subtracting standard electrode potentials
As far as I'm aware, if we construct a half cell ($\ce{A + e- <=> A-}$) under standard conditions then the standard electrode potential $E^⦵$ is the potential of the $A/A^{+}$ couple wrt. SHE, which doesn't change depending on whether the reaction is an oxidation or reduction in this half cell. This is because potential is a physical, electrostatic concept.
Consequently, the cell EMF is then also defined to be the potential of the cathode subtract that of the anode, $E^⦵_{cell} = E^⦵_{cat} - E^⦵_{an}$. All of the above I have obtained from IUPAC recommendations.
However, some people say that reversing a half equation reverses the sign of $E^⦵$, and then proceed to add the electrode potentials to find the cell EMF. Although this method is algebraically equivalent, it doesn't make any sense - since the electrode potential has nothing to do with the chemical concept of a reaction direction!
The only justification I have seen is that if you write $\Delta G = -nFE^⦵$ for each half equation, and sum them, then the EMF comes out as expected only if we do change the sign. However this seems inherently flawed as well, since although Gibbs free energies are additive, $\Delta G = -nFE^⦵$ only applies to a cell as a whole (because it is derived by considering the energy change if charge flows from one terminal to the other) and is meaningless in the context of a half equation.
Consequently, why is the notion that reversing the half cell equation negates the electrode potential still taught - since to me at least it seems fundamentally incorrect!
Thanks a bunch!

Let's take a look at what you're doing specifically for computing the potential difference. The potentials at the cathode and the anode are all relative to some standard. Importantly, it must be the same standard. Let's refer to this as ground.
$$E_{\mathrm{cathode}} = V_{\mathrm{cathode}} - V_{\mathrm{ground}}$$
$$E_{\mathrm{anode}} = V_{\mathrm{anode}} - V_{\mathrm{ground}}$$
The $V$'s here represent absolute potentials on some arbitrary scale.
What's the change in potential when I move a test charge from the anode to the cathode? Since potential differences are path independent, we can look at this via states: moving the charge from anode to cathode is the same as moving the charge from anode to ground and then from ground to cathode.
The potential difference is expressed as:
$$E = (V_{\mathrm{cathode}} - V_{\mathrm{ground}}) - (V_{\mathrm{anode}} - V_{\mathrm{ground}}) = E_{\mathrm{cathode}} - E_{\mathrm{anode}}$$
The key is the cancellation of the ground potential, so that we have a potential difference.
This is exactly what are doing when working with half-reactions. And it works because both the reduction and oxidation half reactions are referenced to the standard hydrogen electrode.
Let's look at two half-reactions:
$$\ce{A -> A+ + e-}\ ,\ \ E^{⦵} = -E_{A}$$
$$\ce{B+ + e- -> B}\ ,\ \ E^{⦵} = E_{B}$$
As a general rule, we tabulate reduction potentials, so let's treat the $\ce{A}$ reaction with a reduction potential:
$$\ce{A+ + e- -> A}\ ,\ \ E^{⦵} = E_{A}$$
Now, both of these half reactions are relative to the standard hydrogen electrode.
In the reaction
$$\ce{A + B+ -> A+ + B}$$
we can look at the equivalent process of moving an electron from $\ce{A}$ to a proton in a standard hydrogen electrode, and then taking a electron from dihydrogen in the standard hydrogen electrode and moving it to $\ce{B}$.
The potential difference is:
$$E^{⦵}_{\mathrm{cell}} = E_{B} + (-E_{A}) = E_{B}-E_{A}$$
This is exactly the same process because the anode potential relative to ground is the reduction potential of the oxidation half-reaction. The potential difference is computed by subtracting the potential from the other one.
Alternatively, if you already expressed the anode potential as a oxidation potential (inverse of the reduction potential), then the minus sign is implicitly included, and you can just add the two potentials.

Thanks for your answer, but I'm finding it difficult to agree. You say "How else would you know to subtract a potential...", though this is quite standard in circuit theory, i.e. the potential difference across a cell is the difference in electric potential between the cathode and the anode. Potentials definitely do not always sum.

I'm drawing mainly from this provisional recommendation from IUPAC: https://iupac.org/wp-content/uploads/2019/07/PAC-REC-18-01-09R2_PR190703MC.pdf

@jameswirth Poorly worded on my part. Will need to edit. You're taking the potential difference of two things that are referenced to the same value. If they're not referenced to the same value, you cannot take the difference. That's the argument I'm trying to make here. If potentials have the same reference, you can take the difference. That's the equivalent of summing the oxidation and reduction potentials, since those two approaches are related by an implicit negation.

Thanks a bunch for clarifying, you've just relieved me of a lot of anxiety!

@JamesWirth Re-written. Thanks for your feedback.

Many thanks, a great answer. What was throwing me off is that there are actually several different reference systems in use - the most common being that of a SHE in the same cell. Though I've also seen others such as the "single electrode potential" (which is actually not experimentally amenable!) - where the reference is defined as the electrolyte! At the end of the day, the differences will all agree, the distinction between different "flavours" of electrode potential isn't usually made clear!


text:
Parallel 1st Order Reaction Rates
Consider some general parallel first order process, where $\ce{A -> B}$ and $\ce{A -> C}$ (rate constants $k_1$ and $k_2$ respectively)
The overall rate equation of this process is said to be $-\frac{\mathrm d[A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$
I get the impression that this is derived from the addition of each individual rate equation, i.e. the addition of $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A]$ and $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_2[\mathrm A]$ would result in the overall rate equation shown above.
However, if this is the case, then why isn’t the rate equation actually $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$? I mean, you’re adding two $-\frac{\mathrm d[\mathrm A]}{\mathrm dt}$ terms together; and so considering basic algebra, surely this would result in $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt}$?
Or do differentials just not work in this way, and I’m simply confusing myself?

You just have two independent parallel decay pathways, so the faster one, with larger k, dominates. Just add the two k values: you do not have two separate pots of “A” to drain away.

@Ed V I suppose the overall rate can therefore be seen as the rate constants of both individual pathways complementing each other? In other words, it’s as though the rate constants for both pathways combine to make one whole, ‘big’ rate constant and this is what dictates the overall rate?

That is what I think. Imagine a silo full of water and two holes shot into it right at ground level. The water will drain from both holes, with faster drainage through the larger one (if one hole is larger). But this is equivalent to just having one hole that is somewhat larger than either of the other two.

@EdV Yes, very aptly put: my frame of thought was something along those lines too, but more in the sense that the overall rate is simply the combined rates of each individual process, i.e. one ‘big’ rate... I understand it now, though!

As you have worked out, species A can only decay  as the sum of rate constants $k_1+k_2$. (It cannot sometimes decay with rate constant $k_1$ and at other times $k_2$). B appears with the total rate constant ($k_1+k_2$)as does C, however, the fraction of B produced is $k_1/(k_1+k_2)$ and the fraction of C $k_2/(k_1+k_2)$.

There is no need adding the number $2$ in front of the definition of the rate. So the rate equation is :
$$-d[A]/dt = k_1 [A] + k_2[A] = (k_1 + k_2)[A]$$

Some of the comments posted already are more elaborate than this answer, which does not even address the main question.


text:
How to dissolve magnesium malate in water?
When I mix magnesium citrate with hot water, it quickly dissolves and the water becomes transparent. When I do the same with magnesim malate, it does not dissolve,  it sinks to the bottom as sand.
The question is how to dissolve the magnesium malate so that the water becomes transparent like with magnesium citrate?

The term you are looking for is "solubility".

According to this publication from the EFSA https://efsa.onlinelibrary.wiley.com/doi/full/10.2903/j.efsa.2018.5292. Magnesium malate is freely soluble in water so if your material does not dissolve are you sure of its identity?

@Waylander I have tried couple of different brands and all behave the same.

@waylander this MSDS begs to differ on solubility.  Whassup?

I cannot reconcile the difference between the 2 sources but the OPs experience suggests yours is correct.

You don't.  It's insoluble in water.
Being a carboxylic acid salt, it may be introduced into the body by reaction with hydrochloric acid in gastric juice.
Similar data in magnesium citrate, shown here for the dibasic salt $\ce{MgH(C6H5O7)}$, indicate more solubility for the citrate.  Possibly the citrate ion, with more oxygen atoms, can complex magnesium and other metal ions more efficiently through chelate formation.


text:
How would one favor the formation of p-TsCl over o-TsCl in the chlorosuflonation of toluene?
I found an old synthesis of $\ce{o-TsCl}$ that purportedly gives a yield of $85\%$ here and was wondering how, still using chlorosulfonic acid, I could favor the formation of $\ce{p-TsCl}$ instead.

Chlorosulphonation using $\ce{ClSO3H}$ is an irreversible, electrophilic aromatic substitution reaction.
In the book Chlorosulfonic Acid, On taking Toluene as the aromatic substrate, it is seen to yield two products -

$para$-tosylchloride i.e. the thermodynamic product (because it is devoid of the steric hindrance offered by the methyl group)

$ortho$-tosylchloride i.e. the kinetic product (because of more activated ortho position due to proximity with methyl $(+I)$ group, which is favourable for the electrophile to attack on despite facing steric hindrance)

At low temperatures like $0-5°C$, the kinetic product is favoured and thus the ortho-isomer is formed. This reaction is exploited in the manufacture of the non-fattening sweetener $Saccharin$.
At higher temperatures like $75-80°C$, the thermodynamic product is favoured i.e. the para-isomer.
According to Chlorosulfonic Acid, The yield of para-tosyl chloride can be increased by

heating, or

addition of ammonium chloride or other catalysts

However, Clayden instead goes on to suggest on Pages 486 and 566, that the preference of ortho-isomer in chlorosulphonation and para-isomer in sulphonation is due to the fact that the latter is reversible unlike the former.

If the reaction is irreversible (as in chlorosulphonation), the alpha-product is usually formed. (with respect to naphthalene, here analogous to ortho position on toluene)
Sulphonation at low temperatures give the alpha-product by kinetic control and sulphonation at high temperatures gives the beta-product (corresponding to para-position on toluene) by thermodynamic control. The beta product is formed more slowly but it is more stable as there is less steric hindrance.... (as the case in toluene)

...which suggests that chlorosulphonation may not be temperature-dependent at all and shall always yield the ortho-isomer, which I am a bit doubtful about. (help needed)
However, the thermodynamic-kinetic explanation seems very satisfactory to me.


text:
Since a Voltmeter does not account for the internal resistance of a voltaic cell, how can the actual EMF of the cell be measured?
Since a voltaic cell has some internal resistance the voltage measured by the voltmeter connecting the electrodes, is actually lower than the emf of the cell, so how can I measure the EMF of the cell exactly?

The voltage drop across the internal resistance will be negligible if negligible current is drawn from the voltaic cell. So we typically use a high input impedance voltmeter to approximate true open circuit cell conditions. An inexpensive DMM/DVM typically has 10 megohm input resistance and an expensive electrometer system, such as used with pH electrodes, may be up around 1 trillion ohms. So current draws are typically 100 nA or so for the DMM and much lower for the electrometer system. Of course, high electrolyte concentrations also help.

Another thing that helps is to have all conductive pathways, including the salt bridge and solution reservoirs, be relatively short and have relatively large cross section. A long, narrow bore salt bridge would be a poor choice.

@EdV Since you mentioned salt bridges, what type of salt bridge is the best because I have seen people do them several ways. Is the glass tube with cotton at the edges good or should I use soaked filter paper ?

There are lots of variants, but most are glass tubing having ends capped with fritted glass, glass wool, cotton, agar or whatever. Search on that well-known 4 letter auction site for “salt bridge” and note the construction with  filler tube at the top of the inverted “U” shape. Personally, I hate the paper towel, filter paper and porous cup work-arounds.

If you're stuck using a low impedance voltmeter, e.g. a D'Arsonval moving-coil meter, you could determine the cell resistance by drawing more current with a known resistance and noting the voltage drop.
From that, and the known resistance of your meter, you can then calculate the open-circuiit potential more accurately.
That said, high-impedance digital voltmeters are available for less than US$7, and the reading would be within the accuracy of the meter itself, sans calculation.

What, no wheatstone bridge and galvanometer? :) These modern times ...

How about an infinite-impedance electrostatic VM? They're still in some dosimeters. BTW, use the Triboro Bridge rather than Wheastone. ;-)


text:
How are unstable transition states determined?
The Wikipedia article about the Grignard Reaction says that

The addition of the Grignard reagent to the carbonyl typically proceeds through a six-membered ring transition state.

This got me thinking: How was it determined that the Grignard reaction has the ring as a transition state? In fact, how are unstable transition states in general determined?

It is clear from the text that you are curious about transition states. It should be noted that transition states and reaction intermediates are two different things. I would advise you to correct your question.

It´s called educated guessing. ;)

Experimentally, and surprisingly,  there is only circumstantial evidence for most short lived species in any type of reaction, i.e. no one has spectral or structural evidence for a transition state, such as often portrayed in textbooks, other than for example NaI dissociation or atom-diatomic molecule reactions e.g. H+OH  etc. done in ultra-low vacuum.
For longer lived intermediates (picoseconds to longer) short pulse laser pump-probe (flash photolysis) methods have to be used to obtain spectral signatures but give no structure, and this method only works for light triggered reactions which are rare. An alternative is stopped flow, but the time scale is slow, many microseconds so usually too slow for most intermediates, and again does not give structure. NMR is immensely slow also. Time resolved x-ray only works on crystals. So far light driven reactions in proteins have been studied down to a few tens of femtoseconds.  There is a huge and exciting field here to be explored.


text:
What will be the number of molecules per mole in a set of different gases when compared with hydrogen?
The source of my question is due the nature of a problem which I found misleading in the way how it was presented.
The problem is as follows:

The molecular weight of hydrogen ($\ce{H2}$), helium ($\ce{He}$), nitrogen ($\ce{N2}$) and oxygen ($\ce{O2}$) are $2$, $4$, $28$ and $32$, respectively. If $N$ is the number of molecules of $\ce{H2}$ per mole. Then find the number of molecules per mole of $\ce{He}$, $\ce{N2}$ and $\ce{O2}$ respectively.

The alternatives given are as follows:
$\begin{array}{ll}
1.&\textrm{2N, 14N, 16N}\\
2.&\frac{N}{2},\,\frac{N}{12},\,\frac{N}{16}\\
3.&\textrm{N, N, N}\\
4.&\textrm{4N, 28N, 32N}\\
5.&\textrm{N, 7N, 8N}\\
\end{array}$
This problem in particular has left me confused on what should be the answer? as in my opinion is not very clear or precise what the author is exactly intending to mean. First of there isn't any indication regarding the weight of any of the gases.
From the way how this question is stated (as giving references of the molar masses of each gas) I think its implying that in all of them there is one mole, if such is the case then there's one mole of helium, one mole of nitrogen and one mole of oxygen, and therefore there are $N$, $N$ and $N$ in all cases.
But the other interpretation which I thought was:
$\textrm{N= number of molecules}$
$2\frac{g}{mol}=N$
Therefore for: helium
$4\frac{g}{mol}\times\frac{N}{2\frac{g}{mol}}=2N$
for nitrogen:
$28\frac{g}{mol}\times\frac{N}{2\frac{g}{mol}}=14N$
for oxygen:
$32\frac{g}{mol}\times\frac{N}{2\frac{g}{mol}}=16N$
and it will be:
$2N$, $14N$, $16N$.
Respectively. But what will be the answer for this question?. I am still doubtful due the ambiguity. Will it be $N$, $N$ and $N$ or $2N$, $14N$ and $16N$?. What sort of interpretation should be made here?.

Actually none of the answers are correct. The IUPAC definition of a molecule notes that a molecule must have two or more atoms. Thus the He atom is not a molecule. The correct answer would be: 0,N,N.

@MaxW That's exactly what I was thinking but I suspect that the author "forced" beyond the boundaries of such definition to include He. But I agree with your interpretation.

Option $(3)  N, N, N$ is correct.

According to IUPAC, the definition of one mole is as follows:

The mole, symbol $mol$, is the SI unit of amount of substance. One mole contains exactly $6.02214076 × 10^{23}$ elementary entities. This number is the fixed numerical value of the Avogadro constant, $N_A$, when expressed in $mol^{−1}$, and is called the Avogadro number.

Thus, irrespective of the type of gas, 1 mole of its molecules will have the same number of molecules i.e. $N$.
As far as your second interpretation goes, I can't quite understand it. Here are my thoughts.

You assigned $N=2 gmol^{-1}$ which seems dimensionally incorrect. Number of molecules would be a dimensionless quantity.
Judging by your interpretation, I think you're trying to apply unitary method of some sorts (like if 2 grams of a gas is 1 mole of it, then 4 grams of it must equal 2 moles for all types of gases).
But this is incorrect as $2g$ of not  every gas equals 1 mole of it. You should go by its definition.

The source of confusion was what moles were for each gas? Because whoever proposed this question to my understanding did not provided the mass of the gases from which to find the moles. So I assumed that they were intending to say that there was $\textrm{1 mol}$ of each gas. From it, there can be concluded that there was $N$ molecules of each gas which of course is the Avogadro's number.

Because of that "missing" grams, I attempted to use the unitary method was to bypass that ambiguity by assuming that the molar masses could be interpreted as a "floating" proportion hence the method.

The question is referring to moles of molecules of gas. Now you might be thinking about Helium - because the term 'moles of molecules' won't be valid for Helium because it is one atom. However, the definition for 'molecule' in the Merriam Webster dictionary is: "the smallest particle of a substance that retains all the properties of the substance and is composed of one or more atoms."  A molecule has to also exist freely and stably, which Helium does. So we would call He(g) a monoatomic molecule. Now the question seems valid, I guess.

I haven't had concrete proof for this statement, calling He(g) a molecule, except the formal definition of a molecule, but I and my professor (just asked him about this) have personally always treated 'He' as such.

Okay, interesting. The IUPAC goldbook defines a molecule to be of more than one atom without equality. Seems like there is a conflict of knowledge. In that case, the question must refer to the amount of smallest entity of the standard form of each gas, be it molecule or atom. That does call for more information to be provided in the question, you're right.

There is a fixed number in this universe called the Avogadro's constant.
One mole of a substance always contains this fixed number of entities (in this case: molecules). That is the very definition of "amount of substance" in moles.
Your answer would be $\mathrm {N, N, N}$, without ambiguity.

P.S. I believe you should consider going through the Concept of moles and Avogadro's number. Here's a decent page to get you started.

I am well aware of the meaning of mole. But in this given context there was no specification of a mass from which to "find" the number of moles. They were only giving the molar mass that's it, and from there I believed that the intended meaning was that there were $\textrm{1 mol}$ of each gas. That was the source of confusion or ambiguity.

"and from there I believed that the intended meaning was that there were 1 mol of each gas." isn't this true?


text:
Batch Process engineering
For a fast chemical reaction where reactant A is fed into an agitated solution of reactant B.
Can the selectivity of the reaction be strongly dependent on the location of the feed pipe outlet?
I am not too sure if specific power input has anything to do with it. I can't seem to understand the principles behind the selectivity. If anyone could help, that would be great.
Thanks

Is the system allowed to equilibrate?

If the reaction is potentiated when there's support, the walls could be a way for that species to react more.
Now, you could check if the agitation is right and check for selectivity feeding from the border and from the center, then comparing results.


text:
Why doesn't an aliphatic acyl chloride react with cyanide ions in aqueous sulphuric acid undergo nucleophilic addition elimination?
I'm asking this because im proving a friend wrong. I get how cyanide ions won't react in an ethanol solvent because the solvent is too reactive but why won't this?

How is your chloride turning into a methyl? Or did it just get cropped out of the picture? In general, compounds that look like (R)(R')C(OH)(X) are pretty unstable, where X is any kind of half-decent leaving group.

Thank you, I said the thing about it being unstable and tbf the chloride going to methyl went right over my head thank you. Going to go prove him wrong now.

Cyanide ions in aq. sulfuric acid don't stay as ions very long, they get protonated to give HCN so cannot act as nucleophiles. Also aliphatic acyl cyanides are about as reactive as aliphatic acyl chlorides, both will readily hydolyse to the carboxylic acid in aq. soln.


text:
How to access implicit bond angles that don't appear in Gaussian's z-matrix input?
Suppose I have a Gaussian script to calculate the energy of a relatively large molecule like hexamethyldisiloxane:
# method ...

Title
 0 1
O
Si      1        1.63776
Si      1        1.63776     2      148.48211
C       2        1.87584     1      107.42571     3      198.44765
H       4        1.08758     2      111.25460     1      180.71860
H       4        1.08686     2      110.86273     1      300.80346
H       4        1.08691     2      110.96626     1       60.56110
C       2        1.88065     1      110.11571     3      318.27904
H       8        1.08696     2      111.18628     1      301.17666
H       8        1.08749     2      111.53142     1       61.67217
H       8        1.08768     2      110.77221     1      181.40186
C       2        1.87887     1      109.39896     3       78.96340
H      12        1.08704     2      110.94415     1       57.63532
H      12        1.08764     2      111.00260     1      177.47274
H      12        1.08699     2      111.36071     1      297.48941
C       3        1.87887     1      109.39896     2       78.96340
H      16        1.08707     3      111.36057     1      297.48596
H      16        1.08696     3      110.94715     1       57.63340
H      16        1.08766     3      110.99845     1      177.47927
C       3        1.88071     1      110.11501     2      318.27662
H      20        1.08768     3      110.76810     1      181.39759
H      20        1.08698     3      111.18376     1      301.17588
H      20        1.08741     3      111.52706     1       61.67191
C       3        1.87581     1      107.42600     2      198.44461
H      24        1.08749     3      111.25560     1      180.71260
H      24        1.08679     3      110.86218     1      300.80333
H      24        1.08694     3      110.96749     1       60.55900

The idea I have is to modify the values (bond length, angles, dihedral) of the script to make a rigid scan and plot energy curves.
However, there are angles in the molecule that are not explicit in Gaussian script, such as HCH. This also occurs with some dihedrals.
Would I be able to add a line to the script to characterize the HCH angle?
What I would like was to have an input for the HCH angle in the Gaussian script and to vary it.
Should I use dummy / ghost atoms?
Any suggestions are welcome.

For such large number of atoms, you might be interested in supporting the creation of a Stack Exchange site for large-scale molecular modeling: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ZmExZTg5NWU3NDk2OGI1Y2ZhNDYxMmNlMjAwOTBlNTY3MjFmNjhjZmM5NDJkNWNmMzM2OTU1YTAxMDQ2YWUzOF0BhlVdtMJOmY0ql8ZBP3z0cCmJx6fZnuQ47jsgdHZa0. You could ask here, but that site is going to be exclusively modeling questions, and a higher concentration of experts. Make sure to click "sign-up" before you click commit.

You can reorder the Z-matrix with a free program like Molden. For example open the molecule in Molden and press the ZMat editor. From there you should see a button that says reorder Z-matrix. Click the atoms one by one, making sure to select the HCH atoms sequentially so they have an explicit angle. Dummy atoms are another good option.

Very good answer. I hadn't used MOLDEN's ZMAT editor before, but I foresee it being very useful in the future!


text:
Using Maxwell Relations or any other mathematical or thermodynamic relations, prove that (∂Cp/∂P)=−T(∂^2V/∂T^2)
Prove that:
 $$\left(\frac{\partial Cp}{\partial P}\right)_T = -T \left(\frac{\partial^2 V}{\partial T^2}\right)_P$$
Since $Cp = (\frac{\partial H}{\partial T})_P$
$$\left(\frac{\partial Cp}{\partial P}\right)_T = \frac{\partial}{\partial P} \left(\left(\frac{\partial H}{\partial T}\right)_P\right)_T$$
Using the chain rule:
$$\frac{\partial}{\partial P} \left(\left(\frac{\partial H}{\partial T}\right)_P\right)_T = \frac{\partial}{\partial P} \left(\left(\frac{\partial H}{\partial S}\right)_P\left(\frac{\partial S}{\partial T} \right)_P\right)_T$$
From:
$dH = TdS + VdP$
$(\frac{\partial H}{\partial S})_P = T$
$$\frac{\partial}{\partial P} \left(\left(\frac{\partial H}{\partial S}\right)_P\left(\frac{\partial S}{\partial T} \right)_P\right)_T = \frac{\partial}{\partial P} \left(T\left(\frac{\partial S}{\partial T} \right)_P\right)_T$$
Expanding the derivative:
$$\frac{\partial}{\partial P} \left(T\left(\frac{\partial S}{\partial T} \right)_P\right)_T = \left(\frac{\partial T}{\partial P}\right)_S \left(\frac{\partial S}{\partial T} \right)_P + T\frac{\partial}{\partial T}\left(\left(\frac{\partial S}{\partial P}\right)_T\right)_P$$
By the cyclic rule and subsequently Maxwell Relations:
$$-\left(\frac{\partial S}{\partial P}\right)_T = \left(\frac{\partial T}{\partial P}\right)_S \left(\frac{\partial S}{\partial T} \right)_P = \left(\frac{\partial V}{\partial T}\right)_P$$
Substituting these in, I get an extra term:
$$\left(\frac{\partial V}{\partial T}\right)_P-T \left(\frac{\partial^2 V}{\partial T^2}\right)_P$$
Above was my first approach to the question. Why do I get an additional term in this version of the proof?
I was able to later prove it by finding $\frac{\partial H}{\partial P} = T\frac{\partial S}{\partial P} + V$ and solving using similar steps to get the correct answer.

The first part of the demonstration is not necessary since by definition the specific heat at constant pressure is the derivative of the entropy with respect to temperature multiplied by the temperature and you can get it easily by just taking the diffential of entropy with (T, P) as variables
Concerning your problem, by quickly looking I believe you don't need to expend the derivative since T is constant

A good equation to start with is $$dH=C_pdT+\left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right)dP$$

Thank you very much Tom! I guess it would be correct to assume that there is no expansion of the derivative.


text:
Understanding the term 'heat of hydrogenation'
From Chemistry Libretexts:

Heat of hydrogenation (symbol: $ΔH_\mathrm{hydro}$) of an alkene is the standard enthalpy of catalytic hydrogenation of an alkene.

Does the heat of hydrogenation include only the double bonds in an alkene or does it also include other forms of unsaturation, such as rings?
For an example, while discussing the heat of hydrogenation of cyclooctatetraene (which is non-aromatic), would we say that the heat of hydrogenation is equal to just the energy required to hydrogenate the four double bonds, or do we also account for the energy required to open the ring and convert it into octane?

It is not that easy to open a ring.

Would you say that cyclohexane is an alkene?

No it is the enthalpy of reaction for every reaction A + nH2 - - - > AH2n.

@Alchimista the reaction you have stated would imply the ring being broken, which is exactly what my question is about.

The point is simply that first there is a reaction then you can name it and refer to it. If I would be able to hydrogenate cycloexAne I will certainly speak of its enthalpy of hydrogenation.

"Heat of hydrogenation" is a subset of "heat of reaction" which is nothing but the standard enthalpy change ($\Delta H^{\circ}$) in the system as reactants are converted to products. In catalytic hydrogenation, as hydrogen molecules are added to atoms bearing double/triple bonds, the enthalpy change which accompanies the reaction is called as "heat of hydrogenation".
Now coming to your question:

Does the heat of hydrogenation include only the double bonds in an alkene or does it also include other forms of unsaturation, such as rings?

Heat of hydrogenation just refers to the enthalpy change required to convert double and triple bonds to single bonds in a molecule. It doesn't account for ring opening. This is because opening a ring involves breaking at least one carbon-carbon covalent bond which requires a great deal of energy. But that doesn't mean rings cannot be broken. If they could be broken, then the enthalpy change is plainly called as "heat of reaction" but not "heat of hydrogenation" since it would lead to ambiguities.
Thus, heat of hydrogenation of cyclooctatetraene is just the enthalpy change associated with those four double bonds, even though cyclooctane has a Index of hydrogen deficiency (degree of unsaturation) as one.


text:
Is Tkatchenko's MBD-NL method self-consistent?
I have a question regarding Tkatchenko's [1] new nonlocal many-body dispersion method (MBD-NL). Does anyone know if it's a self-consistent method? I read this paper many times but I'm still not sure.
References

Hermann, J.; Tkatchenko, A. Density-Functional Model for van Der Waals Interactions: Unifying Atomic Approaches with Nonlocal Functionals. arXiv:1910.03073 [cond-mat, physics:physics] 2019.

This question might get more attention on the new Materials Modeling SE


text:
How are the parameters of a force field obtained for the bond angle HCH?
In my head, I can't see molecules that I could vary the HCH bond angle without varying other bond lengths, angles and dihedrals to build an energy vs. HCH bond angle curve.
If I vary a single HCH angle of a simple methane molecule, for example, I would automatically be varying the other HCH angles of that same molecule, which would affect the construction of the actual energy vs. HCH bond angle curve.
I may be talking nonsense.

You don't necessarily need a potential energy curve to fit a force field. I answered a related question about fitting a force field from quantum calculations.
Let's assume you have a harmonic angle term. You could either do this in terms of the angle bending or the distance between the end atoms, but let's take the angle for our example:
$E_{angle} = k (\theta - \theta_0)^2$
The first parameter you need is the optimal angle $\theta_0$ which you can get from a geometry optimization, experiment, etc.
Your question stems from the second parameter, $k$ - the force constant for whatever this angle type happens to be (e.g. H-C-H in methane for your example).
There are a few ways to get the force constant, but it comes from the second derivative, right? So you can get that from a Hessian or vibrational calculation. My favorite discussions of this center around the "Badger's rule for angle force constants, e.g."

"Maximally diagonal force constants in dependent angle-bending coordinates. II. Implications for the design of empirical force fields" J. Am. Chem. Soc. 1990, 112, 12, 4710-4723

Unfortunately, most such force fields are defined in well-determined
  sets of internal coordinates, whereas empirical potentials use larger
  sets of dependent coordinates. This paper illustrates a unique
  “localized” representation of the angle-deformation potential in
  dependent coordinates which is exactly diagonal for in-plane bending
  at trigonal-planar centers and is nearly diagonal for angle bending at
  tetra coordinate centers.

Modern force field fitting methods typically use scripts that minimize the differences between an in-development parameter set and a set of Hessians and/or experimental data.
As I mentioned in the other question, there's a huge pile of such papers, many with code to derive force fields from quantum chemical data.

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

Thank you very much for your comments, Geoff Hutchison! I am very grateful for your help. As this area is new to me, it has been difficult for me to understand some things for now.
The first idea I had was to derive the parameters of the force field through potential energy curves. Some things are explicit and I am managing to solve it in Gaussian, for example. Others don't.

Hutcison I was thinking about one thing ... Every scan that we perform to obtain the energy curve (varying bond length, bond angle and dihedral), indirectly we would be changing the non-bonded interaction parameters.
That would be a limitation, right? It would be practically impossible to obtain the isolated parameters of bond length, bond angle and dihedral.
Did I say any nonsense?

@EmersonPL - please read the 1990 Halgren paper. It goes into the issues of 'separability' between QM / experiment and the typical design of force fields.


text:
Condensation reaction between catechol and acetone
I was thinking of ways to synthesize 2,2-bis(dihydroxyphenyl) propane and I thought it could be made by a condensation reaction between dihydroxybenzene (catechol) and acetone in strongly acidic conditions.
At first, the catechol would probably form the acetal (even though the former compound could also take part in the reaction). This would then react with acetone, form a cation and this would react with another molecule of the catechol. At last, the diacetal would be hydrolyzed to make the desired compound.
My question is, whether this makes sense. I am especially worried about self-condensation of acetone, which might ruin the reaction.

If you replace your diphenol by ordinary phenol, the reaction works perfectly. It is even a usual method for  making 2.2-bis(p-hydroxyphenyl)propane, which the starting point for producing epoxy resins.

Baker (J. Chem. Soc., 1678 (1934)) describes the condensation of catechol with acetone and it went down another path. Indeed the acetone self-condensation interferes with your reaction plan: acetone self-condensation eventually leads to phorone, which condensed with two molecules of catechol to form "5:6:5':6'-tetrahydroxy-3:3:3':3'-tetramethylbis-1:1'-spirohydrindene".
The author gives a very good explanation and analysis of multiple derivatives of this interesting compound. Even without mass spectrometry or NMR results, I completely believe the structural interpretation. Top notch science!


text:
Reaction rate ambiguity
Let’s suppose the following reaction:
$$\ce{ 2 H2O2 —> 2 H2O + O2}\tag{1}$$ with reaction rate $v$
Now let’s suppose the same equation with different balance:
$$\ce{ H2O2 —> H2O + 1/2O2 }\tag{2}$$
I assume that the reaction rate must be the same ($v$), right? It should not change with stoichometry.
However, depending on the balancing, I get different rates for each substance as shown in the equations below, different $k$ constant and so on.
$$\begin{align}v_1 &= -\frac12 \frac{\mathrm d [\ce{H2O2}]}{\mathrm dt} = \frac12 \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\\[1em]
v_2 &= \frac{\mathrm d [\ce{H2O2}]}{\mathrm dt} = \frac{\mathrm d[\ce{H2O}]}{\mathrm dt} = 2\cdot \frac{\mathrm d[\ce{O2}]}{\mathrm dt}\end{align}$$
Any help?

Logically, they must proceed in sequence: first nascent oxygen is released, then O2 forms from the atomic O.

According the picture , the first relation represent the rate of appearing $\ce{O2}$ , the second relation represent the rate of disappearing $\ce{H2O2}$ or the rate of appearing $\ce{H2O}$

$k$ constant of disappearing $\ce{H2O2}$= $k$ constant of appearing$\ce{H2O}$=2($k$ constant of appearing$\ce{O2})$

I assume that the reaction rate must be the same (v), right? It should not change with stoichometry.

No, the reaction rates changes when you double all coefficients. Nothing real changes though (i.e. the rate of disappearing $\ce{H2O2}$ etc.).
You have a similar case with equilibrium constants: when you double or halve all the coefficients, the new equilibrium constant will be the square or the square root of the original one. The molar reaction enthalpy will also change when you change coefficients. It is a choice, and it still correctly describes the reaction.

Any help?

There is no need for help. If you use either chemical equation format to calculate a measurable quantity (changes in concentration, equilibrium concentrations, enthalpy change for a specific amount that reacts, for the three topics respectively), you will get the same result. It is almost like the choice of describing a volume using liters or milliliters - it does not change the result even though it might look different to someone not used to the milli prefix..


text:
For group 13 elements M³⁺, why is the formation of MCl₄⁻ more favorable than for group 12 elements M²⁺?
I'm not a chemist and I'm trying to understand a chemistry point from a talk I went to recently. Any help would be much appreciated.
The speaker was talking about $+3$ (Group 13) elements in the periodic table. He found out that one of them forms $\ce{MCl4-}$ complexes, where $\ce{M}$ is a $+3$ element.
He made a comment that I didn't understand, that the formation of $\ce{MCl4-}$ is electrostatically more reasonable than $\ce{MCl4^2-}$, where $\ce{M}$ would be a $+2$ metal.
Why would this be true?  I am hanging my head in shame that I don't know this.

Let's take an example. M can be Aluminum $Al$. In order for $Al$ to have an outer electronic layer equal to 0 or 8, it can either loose 3 electrons and become an ion similar to Neon. But it can also use its 3 electrons to make covalent bonds with three $Cl$ atoms, and still add a 4th $Cl^-$ ion, which brings one doublet. In this option, the Aluminum is surrounded by 3+3+2 = 8 electrons, and it looks like Argon. This is the most favorable choice, because it would cost much more energy to remove three electrons. Three !
It is quite different for metals having 2 electrons in their outer shell, like $M$g. If $Mg$ wants to look like a noble gas, it may loose 2 electrons, and become $Mg^{2+}$ which is similar to $Ne$. Of course it costs energy to remove 2 electrons, but not that much. And it would be much more difficult to make covalent bonds. With its 2 electrons, $Mg$ could make two covalent bonds with $Cl$ atoms. But 4 electrons are still missing to look like Argon. These 4 electrons may be brought by 2 $Cl^-$ ions, making $MgCl_4^{2-}$. But this structure is not favorized, because the two approaching $Cl^-$ ions repell one another.
All that to explain why, combined with Chlorine,  $Al$ does not make $Al^{3+}$ and prefers covalent structures like $AlCl_4^-$ions, and why $Mg$ prefers to make $Mg^{2+}$ ions

Thank you, Maurice.  I'm not sure why you got a downvote but this is starting to stir some memories from a long-ago chemistry course!  Something I still don't understand is where you say that Al can make covalent bonds with three Cl (I understand that) and then still add a fourth Cl- ion, which brings one doublet.  The part I don't understand is, why would adding a fourth Cl- be a good idea and how would it bring a "doublet"?  What is a doublet here?  Maybe I'm not being clear about where my confusion is.  Please feel free to let me know.  Thank you.