text:
Common name of CH2Cl2
My book says geminal dihalide or alkylidene halide is the common name for dihaloalkanes where both halogen atoms(if of same element) are bonded to the same carbon atom and alkylene dihalide if 2 halogens of same element bonded to adjacent Carbon atoms.

But in a list of examples in the same book, $\ce{CH2Cl2}$ is named as Methylene Chloride. Shouldn't it be Methylidene Chloride?

Here is text from NCERT Chemistry Class 12 Part 2 Chapter 10:

The dihaloalkanes having the same type of halogen atoms are named
as alkylidene or alkylene dihalides. The dihalo-compounds having both
the halogen atoms are further classified as geminal halides or gem-dihalides
when both the halogen atoms are present on the same carbon atom of the chain and vicinal halides or vic-dihalides when halogen atoms are present
on adjacent carbon atoms. In common name system, gem-dihalides are
named as alkylidene halides and vic-dihalides are named as alkylene
dihalides. In IUPAC system, they are named as dihaloalkanes.

If your book calls C2H2 methylene chloride you need to get a better book

@Waylander sorry I made a typo!

Even after correction, the formula of methylene chloride is not $\ce{C2H2Cl2}$. Methylene chloride is $\ce{CH2Cl2}$, with $1$ carbon atom.

@Maurice sorry again, I made a typo in the edit too. The title was correct only

No ! $\ce{C2H2}$ (from $4$th line) is named ethyne. It has also a more usual and older name : acetylene. But today the author wants to discuss about $\ce{CH2Cl2}$
$\ce{CH2Cl2}$ is usually called dichloromethane. But it has an older name : methylene chloride, which can be justified by stating that $\ce{CH2Cl2}$ looks as it was a chloride (and even a dichloride) of the group $\ce{CH2}$ who was earlier called methylene. This nomenclature was coming from before IUPAC (created $1918$). Before $1918$, all compounds of chlorine were called "chlorides".
In the $19$th century, the alkanes had an end in -ane (like today). With one $\ce{H}$ removed, the molecule becomes a radical in -yl (like today). But with two $\ce{H}$ lost, it becomes -ene or -ylene. Today, we specify on which atom the $\ce{H}$ atoms have been removed, in molecules containing $2$ or more Carbon atoms. Strangely enough, removing $3$ $\ce{H}$ atoms from an alkane gave the group $\ce{CH}$ which was sometimes called form- (from formic acid).
Let's go back to all chloro- derivates of methane: Removing $1$ or several $\ce{H}$ atoms from methane $\ce{CH4}$ yields first the radical $\ce{CH3}$ which was and is called methyl; then $\ce{CH2}$ is obtained which was called methylene. So, before IUPAC, $\ce{CH3Cl}$ was called methyl chloride (today chloromethane), and $\ce{CH2Cl2}$ was called methylene chloride (today dichloromethane). Furthermore, trichloromethane $\ce{CHCl3}$ was called "chloroform", and tetrachloromethane $\ce{CCl4}$ was called "carbon tetrachloride".

Methylene chloride is still widely used among lab chemists so students of chemistry should be aware of this name

Nice history lesson in there. Chemistry retains many of these older names.

So, is this an exception to the "Alkylidene halide & Alkylene dihalide" thing ?

What's the meaning of :  *Is this an exception to ... ?" Which "this" is it ?

@Maurice the name of CH2Cl2

Why call $\ce{CH2Cl2}$  a methylidene chloride ? Why insert this "id" inside the word methylidene ? Why "id" ?

@Maurice my book says that if 2 halogens are present on same carbon, it should be called alkylidene halide. Alkylene dihalide is when both halogens are on adjacent Carbon atoms.

Maybe your book is right. I don't know. But what I know is the fact that $\ce{CH2Cl2}$ is called and sold under the name methylene chloride or dichloromethane. Never methylidene chloride. Anyway this ending in -idene is never used. It is better to specify the number of the C atom in the alkane chain  where the chlorine atoms are fixed.


text:
Why is silver(I) phosphate so unexpectedly thermally stable?

Silver(I) oxide is completely decomposed in air already at $\pu{300 °C}.$
Formulate the decomposition reaction and the corresponding mass action expression.
$\ce{Ag3PO4}$ can be melted at $\pu{800°C}$ without decomposition.
Explain the unexpected thermal stability.

The decomposition reaction and the corresponding mass action from the first part are:
$$\ce{Ag2O(s) <=> 2 Ag(s) + 1/2 O2(g)} \qquad K'_p = \sqrt{p(\ce{O2})}$$
But how can the thermal stability of $\ce{Ag3PO4}$ be justified?

Hint: oxygen combines to form O2. What about PO4?

@DrMoishePippik Polyphosphate structures?

Is PO4 found by itself?

@DrMoishePippik is it something like P4O10?

PO4 ≠ P4O10. What are you trying to calculate?

@DrMoishePippik I don't know what you mean with your first comment. "oxygen combines to form O2. What about PO4?" Didn't you asked to what PO4 combines?

Consider entropy changes in the reactions, the total energy balance.

Fun fact: Although silver(I) phosphate is thermally stable, it is very photosensitive and can photolytically decompose: http://www.sciencemadness.org/talk/viewthread.php?tid=66463

@ Spinell. $\ce{PO4}$ does not exist. It is not a stuff. It is an ion. If silver phosphate decomposes, it may produce metallic silver $\ce{Ag}$. Why not ? But it cannot produce $\ce{PO4}$, as neutral $\ce{PO4}$ does not exist, even by dimerizing or being transformed into some new molecule. This is the main difference between the pyrolysis of $\ce{Ag2O}$ and the pyrolysis of $\ce{Ag3PO4}$. $\ce{Ag2O}$ may produce $\ce{Ag}$ and $\ce{O2}$. But $\ce{Ag3PO4}$ cannot produce anything with $\ce{PO4}$. This is why it resists pyrolysis.

@Maurice "..even by dimerizing or being transformed into some new molecul." But what happens at idk. 4000°C (or whatever temperature is finnaly enough for pyrolysis of Ag3PO4)? What would the decomposition products look like? I mean, some temperature must always be sufficient for pyrolysis.

I don't know what happens to $\ce{Ag3PO4}$ at  very high temperatures (> $2000$°C). But I think that this salt is decomposed  into oxides then into elements, by a reaction like : $$\ce{2 Ag3PO4 -> 3 Ag2O + P2O5}$$ which is immediately followed by $$\ce{2 Ag2O -> 4 Ag + O2}$$ and $\ce{O2}$ is partially decomposed into $2$ oxygen atoms. And maybe a fraction of $\ce{P2O5}$ is decomposed into elementary phosphorous and oxygen.

Since silver oxide decomposes at 300C it will not be an intermediate compound at 2000C. A more likely scenerio would be transfer of electrons from phosphate oxygens to silver giving P2O3:  2Ag3PO4  =  6Ag + 2PO4[!]; [2PO4 = P2O8 = p2O3 +5/2O2] The equation is!!  2Ag3PO4  =  6Ag  + P2O3 + 5/2O2. This balances for mass and redox. It is possible to write equations that balance for mass and not for redox.

Compare it with silver sulfate with similar behavior. (m.p. 652.2–660 °C, b.p(decomp)  1,085 °C)

Silver orthophosphate just like other orthophosphates are thermally stable because they a) they have a strong ionic bond b)evolution of phosphorus pentoxide requires huge energy [See: Do (some or all) phosphates thermally decompose?]
It has been noted in the past how stable silver(I) phosphate is and even after prolonged heating, it refuses to convert into a different species and only converting to pyrophosphate at white heat kept for long time:

Pyrophosphate of silver fuses with extreme facility, even at a temperature below that of redness, yielding a dark brown colored liquid, which, without suffering the least appreciable decomposition, becomes a radiated crystalline mass on cooling. By the first impression of heat, long before fusion, it changes to brownish yellow, and retains when cold a shade of the same color. Phosphate of silver, on the contrary, is very infusible; so that it may be heated in a glass tube, or on a platinum foil, to full redness, without being fused. Its color changes to a reddish-brown by very slight elevation of temperature but, as it cools, the original yellow tint is restored. At a white heat it is fused; and then, like the pyrophosphate, it assumes a dark brown color, but regains the yellow on cooling. If kept for a long any time in a fused state, a portion of pyrophosphate is generated; and consequently, the mass becomes more fusible, and acquires a paler tint. A very minute quantity of pyrophosphate greatly increase the fusibility of the phosphate of silver.

Source: The Edinburgh Journal of Science, Volume 3 by Thomas Clark, 1830

Phosphate to pyrophosphate is a condensation reaction involving loss of water. You are proposing an oxidative mechanism involving Ag+.  How about giving an appropriate balanced equation.

The nonexistant ionic bond has little to do with it. To decompose silver phosphate silver must be reduced an something must be oxidized. Since phosphate cannot be oxidized an O from the phosphate is the only candidate. If it is balanced out there are not enough electrons to go around so phosphate must also oxidize some O, this is not energetically favorable so phosphates are stable beyond expectations whatever they might be. A simpler way to say this is that the P-O bonds in phosphate are very strong.

The stability of oxides and phosphates in general should be compared. I have always wondered about the stability of the O= ion because it is endothermic and many oxides are colored indicating loose electrons somewhere. Stable oxides have high lattice energies as in Al2O3 or MgO or strong covalent bonds as in PO4[3-] or SO4= or even CO2. The comparison is between the oxide ion and the phosphate ion. Also the charge transfer or covalency between the silver ion [or other +ion] should be considered.
Why is the stability of a phosphate "unexpected" phosphates are usually stable compounds. The question is biased and misleading.
I am adding a comment I posted above giving a fictionalized mechanism for phosphate decomposition the actual mechanism is complicated but pondering or researching it might give an idea why phosphates are stable.
Since silver oxide decomposes at 300C it will not be an intermediate compound at 2000C. A more likely scenario would be transfer of electrons from phosphate oxygens to silver giving P2O3:  2Ag3PO4  =  6Ag + 2PO4[!]; [2PO4 = P2O8 = p2O3 +5/2O2] The equation is!!  2Ag3PO4  =  6Ag  + P2O3 + 5/2O2. This balances for mass and redox. It is possible to write equations that balance for mass and not for redox.

Thank you for your answere but I wasn't really able to extract a fitting answere. Should I argument with ionic character? Your last part doesn't really offer an explanation (at least for me), it seems more like a characterisation but in an exam I couldnt write it like that.

Phosphates are stable because the P-O bonds are strong.  Metallic oxides with lower lattice energies are more unstable because the O-2 ion is endothermic.

Isn't the bond between the silver and the phosphate/ oxygen more relevant in the pyrolysis?

With 2 down votes my answer is questionable. Researching P2O3 it disproportionates to P4O10 and P. Ionic compounds do not have specific metal ion bonds. If anything an Ag-O attraction should aid phosphate decomposition. This subject seems to be inadequately researched.  My opinion is still that phosphate is just a very stable ion.

After reading the above references They are low T studies not decomposition of phosphates and not relevant.  The MP is a function of the lattice energy. decomposition the stabilityof the anion. The colors observed possibly support my fictionalized mechanism Sine any P formed in an O2 or air atmosphere would be oxidized to phosphate

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text:
f-f transitions and their cause
I was looking at this question:
How can f-f transitions happen?
And got some answers that answered some of the questions I had, but not all. I am still trying to conceptually grasp why f-f transitions happen at all because a) there should be no crystal field splitting because of the lanthanide contraction (orbitals are closer to the nucleus, are not perturbed by the ligand sphere as there is next to no interaction) and b) then all f-electrons should be degenerate.
This is in contrast to d-elements, where d-d interactions can be explained by crystal field splitting and, more precisely, by looking at Tanabe-Sugano-diagrams (because not every d-d transition is the same due to the symmetry of the orbitals involved and spin-orbit-coupling).
So how to best explain this?
Edit:
I am asking because I was confused by this diagram:

In this diagram, one excited Eu(III) state of 5D(0) can decay to 7F(6) state. Is this a d-f transition or a f-f transition?
Sorry if this is obvious, but even though the term symbol is a D, this not always coincides with the orbital; for example, Sm(III) has the ground state term symbol 6H(5/2), but obviously the outer electrons don't reside in a "h-orbital".

To say that there is no crystal-fiekd splitting is to overstate the fact. There is a small amount of splitting and $f-f$ transitions can be seen, but they are much weaker than $d-d$ transitions in transition-metal complexes.
Both $d-d$ and $f-f$ transitions are Laporte-forbidden if we have pure $f$ orpure $d$ orbitals. In the real case, molecular vibrations and the formation of molecular orbitals between metal and ligands break the symmetry that causes the Laporte-forbidden condition, allowing these transitions to take place. However, the orbital mixing process is less effective for $f$ orbitals than for $d$ orbitals because the former are relatively isolated from the ligands and undergo only weak metal-ligand bonding interactions. Therefore $f-f$ transitions are much weaker than $d-d$ transitions.
In contrast transitions heyween $f$ and $d$ orbitals ($f-d$ transitions)  are not Laporte-forbidden because of the different inhetent symmetry of $f$ and $d$ orbitals. These $f-d$ transitions are therefore quite stron. It is $f-d$ transitions, not $f-f$ transitions, that predominate in the UV and visible spectra of lanthanide complexes.
Experimental spectra are given fir some lanthanide(III) complexes by Purohit and Bhojak[1].
Reference

Suresh Purohit and N. Bhojak (2013). "The Absorption Spectra of Some Lanthanide (III) Ions". Research & Reviews: Journal of Chemistry. 2(2), e-ISSN: 2319-9849.

Could you maybe take a look at my edit? Seems to incorporate the f-f or d-f transition question

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text:
Operating a chemostat at the optimum dilution rate
I am wondering what precautions should one take to operate a chemostat at the optimal dilution rate. This is a question on one of the exercise sheets from a class I took two semesters ago and that I am currently revising, but I cannot really make much progress on this. I tried looking it up in the literature, but with no result, so I would really appreciate some help with this.
EDIT: Here dilution refers to the dilution with a sterile broth medium and it equals the ratio of the flow divided by the volume of the chemostat.

What sort of "dilution rate" are you speaking of ?

@Maurice I added some explanations, is it clear now?

Paradoxically, the optimum operation is not recommended for a chemostat, so typically it should be avoided. This is because the optimum dilution rate, more often than we would like, is near the critical dilution rate. Therefore, small variations of the volumetric flow rate $F$ will lead to a washing of the biomass. I will present to you the proof of this behaviour, and then use some data that I personally used in my biochemistry classes, that illustrates this problem.
The sketch of an academic chemostat is on the left, while reality is on the right:

1. Biomass balance With a zero inlet flow of biomass, i.e. $X_0 = 0$, at steady state we have
\begin{align}
  \require{cancel}
  \cancel{FX_0} - FX + \mu XV &= \frac{\mathrm{d}X}{\mathrm{d}t} = 0 \\
  -FX + \mu XV &= 0 \\
  -F + \mu V &= 0 \rightarrow \boxed{\mu = \frac{F}{V} = D} \tag{1} \\
\end{align}
2. Substrate balance Let $Y_\mathrm{X/S}$ denote the stoichiometric/maximum yield of biomass/substrate. At steady state, we have
\begin{align}
  \require{cancel}
  FS_0 - FS - \frac{\mu X}{Y_\mathrm{X/S}}V &=
  \frac{\mathrm{d}S}{\mathrm{d}t} = 0 \\
  F(S_0 - S) - \frac{\mu X}{Y_\mathrm{X/S}}V &= 0
  \hspace{1 cm} (\text{divide by $V$)} \\
  D(S_0 - S) - \frac{\mu X}{Y_\mathrm{X/S}} &= 0 \hspace{1 cm} (\text{use Eq. (1))} \\
  \cancel{D}(S_0 - S) - \frac{\cancel{D} X}{Y_\mathrm{X/S}} &= 0 \\
  (S_0 - S) - \frac{X}{Y_\mathrm{X/S}} &= 0 \rightarrow
  X = Y_\mathrm{X/S}(S_0 - S) \tag{2}
\end{align}
Assuming the specific growth rate follows the Monod's equation
\begin{align}
  \mu &= \frac{\mu_\mathrm{max}S}{K_\mathrm{S} + S} \\
  S &= \frac{\mu K_\mathrm{S}}{\mu_\mathrm{max} - \mu}
  \hspace{1 cm} (\text{use Eq. (1))} \\
  S &= \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D} \tag{3} \\
\end{align}
and combining Eq. (2) and (3)
$$ X = Y_\mathrm{X/S}\left(S_0 - \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D}\right)
   \tag{4} $$
Eq. (4) shows a feature of the chemostat at steady state. If you don't choose well the volumetric flow rate, the biomass at the exit will be zero (the washing I mentioned at the beginning). Setting Eq. (4) to zero gives the critical dilution rate
$$ \boxed{D_\mathrm{crit} = \frac{\mu_\mathrm{max} S_0}{K_\mathrm{S} + S_0}} \tag{5} $$
3. Optimum condition The biomass productivity is defined as $P_\mathrm{X} := DX$. Thus, combining Eqs. (1) and (4) and differentiating with respect to the dilution rate $D$ (I let that derivative to you) gives the optimum dilution rate
\begin{align}
  P_\mathrm{X} &= D Y_\mathrm{X/S}\left(S_0 - \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D}\right) \\
  \frac{\mathrm{d}P_\mathrm{X}}{\mathrm{d}D} &=
  \frac{\mathrm{d}}{\mathrm{d}D}\left[D Y_\mathrm{X/S}\left(S_0 - \frac{D K_\mathrm{S}}{\mu_\mathrm{max} - D}\right)\right] = 0 \rightarrow
  \boxed{D_\mathrm{opt} = \mu_\mathrm{max}
  \left(1 - \sqrt{\frac{K_\mathrm{S}}{K_\mathrm{S} + S_0}}\right)} \tag{6}
\end{align}
4. Example In my class we used saccharomyces cerevisiae at $T = 30 ^\circ \pu{C}$ and the limiting nutrient was glucose. The important data is:
\begin{array}{|c|c|}
  \text{Magnitude} & \text{Value} \\ \hline
  \mu_\mathrm{max} & \pu{0.5 h^-1} \\
  K_\mathrm{S}     & \pu{0.025 g glucose/L} \\
  S_0              & \pu{12 g glucose/L} \\
  Y_\mathrm{X/S}   & \pu{0.5 g biomass/g glucose} \\
\end{array}
Replacing this values in Eqs. (5) and (6) give
\begin{equation}
  D_\mathrm{opt} = \pu{0.4772 h^-1} \hspace{1 cm} D_\mathrm{crit} = \pu{0.4990 h^-1}
\end{equation}
This goes back when I started the post, both values are pretty near to each other. The evolution of $X$, $S$, and $P_\mathrm{X}$ is shown below:

If the dilution rate goes up a bit, you abruptly wash all the biomass. If it lowers a bit, the consequences are not so serious. The golden characteristic of the chemostat is depicted in Eq. (1), because we can manipulate specific growth rate by changing $D$. Unfortunately, by the same cause, changes in $D$ will bring up a new set of steady-state values (after a transient period), that may be vastly different.
Fortunately, it can be proven that the chemostat is an autoregulator, and a new steady state will always be settled. I invite you to search for the Monod parameters for various microorganisms, apply these equations, and see that it is generally the case.


text:
Why is there inconsistency about integral numbers of protons in NMR in the Clayden: Organic Chemistry 2nd ed.?

I was studying the 1H NMR chapter of Clayden: Organic Chemistry 2nd ed. until I came across this case and became curious about integral numbers of protons because it didn't seem right. Then I measured them and the ratios from left to right were 1:1:2.16:3.28. However it should have been 1:1:4:6 according to the structure of compound. Why is it?
I am also curious about the chemical shift of red and green protons. both should be around 5.7 according to additive guidelines for determining chemical shift provided in the book. Basically saying for CH proton we should start from 1.7 and add 2 for each of Halogens or ethers present. The red proton seems ok but the green one is too far from expected value. Can you provide an explanation?

Iirc 100 MHz doesn't give you a very good spectrum. But I'm out of the game for too long.

@Martin-マーチン it is not very good in terms of resolution. But a 100 is not very different from a 300 in terms of accuracy of integrals (or should not be, at least).

The error in the integrals in the 100 MHz 1H NMR spectrum of the diethyl acetal of dichloroacetaldehyde is only one of the problems. The spectrum is simulated and it is incorrect. ChemDraw produces a similar spectrum and it is also incorrect (Fig. 1). The methylene protons at δ 3.70 are not chemically nor magnetically equivalent but rather they are non-equivalent and diastereotopic.
 Fig. 1
Consider the 90 MHz spectrum of acetaldehyde diethyl acetal from the SDBS Spectral Data Base (Fig. 2). Notice that the pattern in the region of δ 3.5 is more complex than a simple quartet. Each methylene carbon bears two protons, Ha and Hb, that are coupled to each other as well as to the methyl group. This issue has been addressed previously on ChemSE.
Fig.2a
Fig.2b

In general, you should not expect NMR integrals to be perfect. There are several reasons for this. Two main ones are:

The integral in the spectrum depends on the proportion of the spins that are aligned with the magnetic field before the experiment—this is called the polarisation. In principle, you would like this quantity to be the same for all the spins in the molecule. In practice, it isn't, because different spins relax at different rates.

Different spins are affected differently by the radiofrequency used. For a given pulse length and power, you will excite a different proportion of one spin than another.

Avoiding these and getting quantitative results requires setting up your NMR experiment carefully. This isn't done routinely because it takes time (in particular, you need to use a longer relaxation delay, i.e. a larger gap between experiments to allow spins to regain their polarisation). Furthermore, super-accurate integrals are not often needed for routine analysis.
For more information see section 4.1.2 of Claridge, High-Resolution NMR Techniques in Organic Chemistry (3rd ed.).


text:
Does Hydrogen Gas need to be bubbled in during Reduction of the electrode?
I have read in textbooks and websites that hydrogen gas needs to be bubbled into the SHE (Standard Hydrogen Electrode). I understand why for oxidation of the half cell - the gas is oxidised into H+ (aq) and the electrons move into the other half cell. Anions from the salt bridge then move to maintain cell neutrality.
However, I fail to understand why hydrogen gas needs to be bubbled into the cell during reduction, as hydrogen intake is not needed. As H+ (aq) gets reduced into gas, it will just bubble out whereever it is formed, not necessarily through the tube. Cations from the salt bridge move to maintain neutrality.
I suspect that it is something to do to maintain stability/neutrality. Can anyone please help?

What potential would SHE have if you did not bubble H2 through it?

You use a salt bridge only to close the electric circuit (i.e., charge transport).  It is filled e.g. with diluted aq. $\ce{KCl}$,* and plugged by diaphragms to hinder the travel of $\ce{K+}$ / $\ce{Cl-}$ into either one half cell across this boundary.  You do not want a travel of matter (e.g., $\ce{Cu^{2+}}$) across this connection into the other half-cell.

* An example of a strong electrolyte (i.e., complete dissociation) of ions with similar mobility which are not subject to electrochemical reaction in the difference of electrode potentials of your intended half-cell reaction(s).

The SHE is not used as a working electrode with passing current, it would affect its potential.
It is used as the third, reference electrode with negligible current. The potentials of the working anode and/or cathode with eventually passing significant current are then measured against SHE.
Similarly are used other, more practical calomel or argentchloride reference electrodes $\ce{Hg|Hg2Cl2|KCl}$ or $\ce{Ag|AgCl|KCl}$. The latter is often integrated with pH glass electrode, forming the compact 2-electrode pH sensor.
Note that the real potential of SHE with deviations of $a_{\ce{H+}}$  or $p_{\ce{H2}}$ is:
$$E_\mathrm{SHE} = E^{\circ}_\mathrm{SHE} + \frac{RT}{2F} \ln{\left(\frac{a_{\ce{H+}}^2}{p_{\ce{H2}}}\right)}$$
The SHE requires $a_{\ce{H+}}=1$ and $p_{\ce{H2}}=\pu{1 atm}$, with the solution saturated by hydrogen. This can be hardly done other way than by bubbling hydrogen through the solution, what is BTW the part of SHE definition.


text:
Can glycerol in substituted natural fats be a monomer according to European REACH Regulation?
European REACH regulation for registration of chemical substances has an own definition of a polymer according to this regulation.
I've noticed the following problem.
I want to discuss here only the chemical aspects.
Can glycerol in substituted natural fats be a monomer according to European REACH Regulation?
Let's assume a fat obtained from animals, plants or fishes as industrial raw material (means a mixture of different triglycerides) that is substituted with a functional group (e. g. acetylated, hydroxylated, oxidated, sulfated or sulfited fat), e.g. at the double bonds of the unsaturated triglycerides or at the hydroxyl groups of castor oil. (See e.g. Wikipedia for the structure formulas of such triglycerides.)
REACH Regulation (1907/2006/EU), May 8 2023, states:
"For the purposes of this Regulation:

polymer: means a substance consisting of molecules characterised by the sequence of one or more types of monomer units. Such molecules must be distributed over a range of molecular weights wherein differences in the molecular weight are primarily attributable to differences in the number of monomer units. A polymer comprises the following:

(a) a simple weight majority of molecules containing at least three monomer units which are cova-lently bound to at least one other monomer unit or other reactant;
(b) less than a simple weight majority of molecules of the same molecular weight.
In the context of this definition a ‘monomer unit’ means the reacted form of a monomer substance in a polymer;

monomer: means a substance which is capable of forming covalent bonds with a sequence of additional like or unlike molecules under the conditions of the relevant polymer-forming reaction used for the particular process"

Let's assume paragraph 5 is fulfilled for the mixture of differently substituted different triglycerides of the considered natural fat.
My question is:
Can glycerol meet the definition of a monomer according to paragraph 6?
If we synthesize the triglycerides starting with the reaction between the fatty acids and the substituent, glycerol is a monomer according to paragraph 6: glycerol is capable of forming covalent bonds with the sequence that consists of the substituent and the fatty acid.
But if we synthesize the triglycerides starting with the reaction between glycerol and the fatty acids, glycerol isn't capable of forming covalent bonds with a sequence of further molecules in the polymer formation reaction.
That means, if the mixture is a polymer according to REACH or not depends on the order of the synthesis steps in this case.
Am I right?
I found that there are gaps in the REACH polymer definition. I want to see if I'm right or looking for counterarguments.

polymer: means a substance consisting of molecules characterised by the sequence of one or more types of monomer units. Such molecules must be distributed over a range of molecular weights wherein differences in the molecular weight are primarily attributable to differences in the number of monomer units. …

This

… are primarily attributable to differences in the number of monomer units …

rules out triglycerides – substituted or not – from being a polymer.

monomer: means a substance which is capable of forming covalent bonds with a sequence of additional like or unlike molecules under the conditions of the relevant polymer-forming reaction used for the particular process"

And this

… under the conditions of the relevant polymer-forming reaction used for the particular process …

rules out glycerol from being a monomer as there is no polymer.

REACH aside, purely from a chemical point of view and in context of eventually substituted triglycerides, very few chemists would consider such triglyceride as a polymer and glycerol as a monomer.

As you consider glycerol as a monomer and as there is just a single glycerol unit.... // You would not consider as a PE/PET polymer a compound with just 1 ethylene, ethyleneglycol or TF acid unit.

You would not consider as a PE/PET polymer a compound with just 1 ethylene, ethyleneglycol or TF acid unit. // Ethene, or ethylene diradical are not PE.  1 Ethyleneglycol esterified with 2 terephthalic acids, or vice versa, are not PET.

Not substituted triglycerides are not REACH polymers, as I understand the wording. // The best choice would be asking REACH about interpretation.  A asking B "What does C mean by X?" may not provide a satisfactory nor reliable answer. A should ask C.

Non-substitued triglyceride mixtures and your short-PE/PET example with only 3 molecule units aren't REACH polymers because at least 3 monomer units are necessary which are covalently bound to at least one other molecule unit. We need at least 4 molecule units. Let's assume your PE/PET copolymer with 4 molecule units. The problem remains that a monomer must be capable of forming covalent bonds with a sequence of further molecule units. If a molecule unit is a monomer depends on the order of the polymer forming steps. A molecule is a monomer only if there is already a molecule sequence.

Could you please correct your answer to be suitable and remove some of your many comments?

You seem repeatedly trying to manipulate contributors to write what you want to be written and to remove what you do not want to see. // Another closure reason for this question may be "opinion based" as it is just opinion how REACH may interpret it.

Hm. Please, please excuse me. Sharing scientific arguments is opinion-based and manipulating? I would expect suitable counterarguments, not arguments that are'nt suitable.


text:
Why is prismane explosive but cubane is not?
So I was reading about polyhedral hydrocarbons, because I dig the topic, and I read (admittedly on wikipedia entries... But they cited papers, I swear!) that prismane is explosive due to the massive ring strains of its faces, but cubane isn't, even though it has ring strain as well in its 90-degree edges.
The explanation that I read (this time on the appropriate paper) was that the symmetry break required for cubane to react leads to high activation energies for the transition states, which kinetically traps the molecule and prevents spontaneous reaction with the oxygen in the air. The result is that cubane, despite its ridiculous energy content, is remarkably stable.
Now... Prismane looks pretty symmetrical to me as well and I guess it has less symmetry than cubane, because prisms belong to lower-symmetry point groups than cubes. But is that the true explanation for why prismane is explosive? Higher energy content but lower symmetry?

Presumably you mean the C6H6 prismane not one of the larger prismanes. in which case the more obvious explanation is the much higher ring strain of carbons with 60° angles compared to the 90° angles in cubane. Symmetry doesn't seem like a good explanation for the barriers to reaction.

According to Wikipedia, "Once formed, cubane is quite kinetically stable, due to a lack of readily available decomposition paths." In other words, the first step in decomposition requires a fair amount of energy in all available paths -- it's trapped in a "well" at the top of a hill, with no easy path out of that well.
Consider $\ce{H2 + Cl2 -> 2HCl}$. The $\ce{H2}$ and $\ce{Cl2}$ require initial energy to break their bonds to react.
Think, perhaps, not in more ways cubane is symmetrical, but rather that there is no easy path through the activation barrier. All it takes is one easy path out, and decomosition could take place. In the same way, diamond is metastable at STP... but there's no quick way out of that state.

Yes, I understand that. The reason why these paths are blocked is, according to the literature, a large loss of symmetry from the electron density of the cubane. Now, prismane is also very symmetrical but, unlike cubane, it is aggressively explosive. I am puzzled by that. Shouldn't prismane also benefit from high-symmetry decomposition blockages?

Cubane is symmetric in more number of ways than the number of ways in which prismane is symmetric. I suppose that is why.

-1 You're basically repeating what OP said.

@Mithoron The OP has asked for confirmation: "But is that the true explanation for why prismane is explosive? Higher energy content but lower symmetry?" Unless you feel this answer is incorrect, it's a great answer. (+1)

@urquiza what Proscionexium suggests is interesting - "number of ways in which prismane is symmetric' could be the basis of a follow-up question.

@uhoh Gosh, OP somehow doesn't sound satisfied in the comment. Primary question is "Why is prismane explosive but cubane is not?" If it was just a matter of confirming of what OP says, than it would be "amirite" q. where OP's actually right - which isn't a valid question at all. And, no I don't intend to discuss it further.

@Mithoron When asked well, "amirite" questions can certainly be perfectly valid Stack Exchange questions. You do some reading, then some thinking, try to work it out, come to a tentative conclusion, then vet your work in Stack Exchange. Happens - and is well received - all the time. Sometimes they're so right that there is not much left to answer. Do you see anything actually missing here that needs to be said?

@uhoh That's not "amirite" question - you need to go further than OP to answer - how does symmetry or whatever give cubane high Ea of decomposition and it doesn't happen for prismane.

I suppose, the main reason prismane explosive, is bond angles, less natural for carbon.
As about symmetry, think about it that way: in cubane all C-C bonds and their energy levels are equal. So any disturbance distributes through the structure freely, as in model of a cube made of springs.

You mean to say that all C-C bonds are equivalent and degenerate in cubane which is not so in the concerned prismane?

I don't specifically meaned the word, but probably. I had in mind rather classical physics picture, consider two strings:
|-----------------------|  ,
|-------------=====|
First one is symmetric, second not. Waves (of energy) could travel one string freely, other not.


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Help with exchange integral interactions!
A single determinant configuration constructed in a basis with 7 spatial orbitals has the following exchange integral interaction energy terms:
-2K12 -K13 -K23 -K14 -K24
(i) Draw a configuration that has exactly these exchange energy terms.
attempt:
1st orbital -> 2e-
2nd orbital -> 2e-
3rd orbital -> 1e-
4th orbital -> 1e-
(ii) Write down the full energy expression using the one-electron (hii),
Coulomb (Jij), and exchange (Kij) integral contributions.
attempt:

Your question looks like a homework or exam question. Could you elaborate your question so we can help you with the answer? I think there is a guide to submit this kind of questions but I can not point you to it right now.

Hint.  Remember that the exchange integral $K_{ij}$ are cero if both spin-orbitals have different spin functions.


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What reaction products are formed if an unsaturated fat is reacted at 20 - 50 °C with H2O2 of 35 %?
My question is:
What reaction products are formed if an unsaturated fat is reacted at ca. 20 - 50 °C with aqueous hydrogen peroxide of ca. 35 %?
To find the answer, we can subdivide the question into the questions about the elementary reactions:
a) Which reaction products are formed in the oxidation step?
b) Which reaction products are formed in the further reaction steps?
I already know:

Blowing with compressed air at higher temperatures yields a hydroperoxy group in $\alpha$-position to the double bond.
Epoxidation with hydrogen peroxide or alkyl hydroperoxides can be improved by transition metal complexes.
Peroxycarboxylic acids (e.g. performic acid or peracetic acid formed in situ from the aqueous acid and aqueous hydrogen peroxide), which are more electrophilic, convert the double bonds to epoxides without metal catalysts.

But what about aqueous hydrogen peroxide alone?
We know hydrogen peroxide is a hydroperoxide and a weak acid. Does the hydrogen peroxide react according to ionic or radical mechanism under my reaction conditions?
Are the hydroperoxides formed, the epoxides, the monoalcohols, or the dialcohols?
The references contradict each other in these points or don't contain any information on the reaction conditions.


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Can there be two transition states (first order saddle point) connecting the same pair of minimas on a potential energy surface?
This is a purely hypothetical question. Consider a pair of local minimas A and B on a multi dimensional potential energy surface. Can there be two first saddle points (IE two different configurations) which would connect A and B ?
If the answer is yes, does this situation happen in a physical system ? Is there any example of a reaction proceeding in this format.

That sounds quite possible.

Yes. It sounds posible. But the situation between the two saddle points is an intermediate product.

I am asking about a situation where say you start an IRC scan from two transition states, they would connect the same pair of minimas.

By symmetry you could easily have this situation, but in general I would expect that one saddle point would be at a different height to the other and so not accessible.

@porphyrin Could you give an example (or a reference) of a reaction where you see this? In case you have come across a situation like this before before.

Sorry, no but I was thinking rather simply of something such as a metal complex, say, MX3  where there is 3 fold symmetry.


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Why would iron found in a nanoparticle not technically be a mineral?
An quote from an article at https://newatlas.com/science/antarctica-blood-falls-iron-red-water/ states, "Now a new study has examined samples of the water and found that the iron appears in an unexpected form. It’s not technically a mineral – instead it takes the form of nanospheres, 100 times smaller than human red blood cells."
Why does being in nanoparticle form cause iron to not technically be a mineral?  If not a mineral then what would the iron be classified as?

From the article you linked: “In order to be a mineral, atoms must be arranged in a very specific, crystalline, structure. These nanospheres aren't crystalline, so the methods previously used to examine the solids did not detect them.”

The article is not forthcoming with information.  It seems iron salts are involved and an ion analysis of the colorless water would be interesting. What anions are present?

Where can non-crystalline iron be found?  How is it produced?  I am suspecting some biological process?  Is the iron in hemoglobin crystalline?


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RMS Speed of Gases as Standard Deviation
In kinetic molecular theory, the average velocity of gas particle is zero since the molecule move in different directions, and the overall effect is zero. Howeever, you can calculate different speeds.
Specifically, RMS (Root Mean Square) speed is the speed of a particle in a gas with the average kinetic energy. This is
$$u_{rms}=\sqrt{\frac{3RT}{M}}$$
Conceptually, if $v_1, v_2,...,v_n$ are the speed of $n$ particles in a gas, the RMS is
$$u_{rms}=\sqrt{\sum\frac{v^2_i}{n}}$$
As the Wikipedia page on RMS points out, RMS is a special case of standard deviation when the mean is zero. Hence:

Is it correct to say that $u_{rms}$ is the standard deviation of the magnitudes of the velocity (not speed).
If it is correct, is there any significance/use of this interpretation?

Speed IS the magnitude of velocity. // As there are no negative values, the mean cannot be zero.// The formal similarity with formula of s as sigma estimation with zero mean is coincidental, as any square root of the arithmetic mean of quadrates.

Don't confuse the vector of velocity (which averages to zero) with the magnitude of the velocity (which clearly isn't). And the speed distribution is not gaussian.

I'll answer in reverse.
(2) There is so-called Central limit theorem in math, which basically says that under certain precise conditions (which I won't discuss here) the sum of many independent random values behaves according to the Gaussian, or normal, distribution. It is enlightening to know that the gas molecule velocity, being a result of many independent random collisions, behaves somewhat similarly, in that each component ($v_x,\,v_y,\,v_z$) is distributed by Gauss. Of course the average values is 0 (since the whole body of gas is not moving anywhere), hence the RMS is the same as the standard deviation. Actually, the latter would be true for any other kind of distribution, so I don't even know why I brought up all that Gaussian stuff.
(1) One may say that the velocity has a 3D normal distribution with average being the zero vector and RMS being our $u_{rms}$. The magnitude of velocity, AKA speed, is another thing entirely. To begin with, it is always positive (except when 0), hence its average is positive and not 0. In fact, it is distributed by Maxwell-Boltzmann, and its variance (which you may see by following the link) is, naturally, the difference between the mean square speed and the square of the mean speed.
So it goes.

Velocity components have the Bolzmann distribution $p(v_x) \propto \exp{(\frac{-mv_x^2}{2kT})}$, which is the Gauss one just in different colour. Integrated over spherical surface, one gets the Maxwell-Boltmann one.


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Gallium & Electrolysis
Preface this with I'm a chemistry newb. Ive finished chem I and II and did extremely well, but I think that means I know just enough to get myself in trouble....Also, sorry if this is a little long. I've been wanting to do this for awhile now and I finally have the time, but I really only have the time to try it once or twice and get it right. I definitely don't have the time to start from scratch and make a new sample.
I've been working with gallium over the summer after becoming fascinated with it over the last couple of semesters. I have two samples of what I think is gallium chloride, but over the last few days of reading I'm not so sure.
The 1st sample was prepared from the little tube that the gallium was shipped in that had a thin film of gallium I wasn't able to remove. That took about 7 days and I didn't use any heat or stirring, but I did use concentrated HCl. 7.5mL concentrated HCl and based on the mass of the tube before and after removing the gallium, it had 0.71g of gallium. After it was fully dissolved, I set it on a hotplate in a hood and evaporated off all the liquid and was left with some small white crystals. I filled it up with water, the evaporated it again, the added about 30 mL of water and put it in a little bottle. Today I ran it through a filter because it had some white solid in it. It clogged a glass and gave off a little gas so I set up another buchner with a paper filter and was able to pull it all through. This sample is currently clear.
The 2nd sample was prepared by dissolving 2.33g of gallium in 60mL of 6M HCl. I could have used concentrated HCl but I grabbed 6M off the shelf. Took about 5 days. I let it sit overnight but while I was on campus I had it on a hotplate with light heat and moderate stirring. After it was completely dissolved, I let it cool and poured it into a storage container as is. It ran through the glass buchner fine. This solution is currently yellow.
After filtering each, the pH on sample 1 was about 2, and the pH on sample 2 was about 1. I did the math on the 2nd sample and I think I have about 3.61g excess HCl in solution, I didn't do the math on the 1st sample but I was surprised at the pH considering I evaporated off the solution until crystals formed, twice.
What I'm attempting to do with all of this is put the gallium chloride solution into a small dish and run electrolysis to pull out the gallium metal. The only info I've really been able to find on electrolysis of gallium has been a couple of youtube videos. In both videos they mention that electrolysis needs to be performed under basic or highly basic conditions. After they make their gallium chloride, they add sodium hydroxide to make sodium gallium hydroxide. I guess I could have just dissolved the gallium in a concentrated solution of sodium hydroxide, but the one example I saw of this seemed like it would take way to long.
So, a couple of questions:

Why can't I/shouldn't I just put the gallium chloride I currently have into a beaker and run electrolysis?

I did the math on it and it looks like gallium metal should be able to be pulled out of this solution, but I believe chlorine gas would form at the anode. The videos I saw didn't look like they were done in a lab or under a hood so I was thinking maybe this was why they made the sodium gallium hydroxide, the the chloride ions are still in solution so this explanation doesn't really make sense. I don't really understand why electrolysis needs to be performed under acidic or basic conditions. We only touched on electrolysis in chem II, mostly just how to tell what's going on at the anode or cathode.

Can I use graphite for both electrodes?

I understand why the pH is close to 1 for solution 2 that had so much excess HCl, but I don't understand why solution 2 is so acidic. After evaporating off the solution twice, I dissolved the crystals in about 100mL of water. I guess I just thought the pH would a little higher. I had an organic professor help me today with the filtration and he didn't really have an explanation for this either, just thought it was interesting after the two evaporations.

I'm assuming solution 2 is yellow because I didn't do anything with it after the gallium dissolved. Is that correct?

If I need to or should add sodium hydroxide to make the sodium gallium hydroxide, can I add sodium hydroxide to each solution as is, or should I evaporate off solution 2 until it's clear? I understand this process is very exothermic and I saw it boil over in one video. I was going to basically add 1 pellet of sodium hydroxide at a time from our stock bottles and let the solution cool back to room temp and do this until it's finished... how do I tell if it's finished? Can I/should I do this in an ice bath or chill the solution first?

Ok last one.... IF I did all my math correctly, I should be able to run solution 2 as is at 2.7amps for 1 hour to pull out the 2.33g of gallium. I don't think that would change if I use the sodium gallium hydroxide.

The two examples I've seen of this are done with heat. One mentioned it was being performed at 40*C, so they got liquid metal. I wanted to try it once this way and then I'd like to try it in an ice bath and see if I can get gallium crystals. As mentioned earlier, we got a very brief into to electrolysis in our class. How does the math change for time and amps between a run with heated vs cooled solution?
Ok that was a lot. Thank you in advance!

TL: DR $  $...

Gallium does not readily crystallize. https://en.wikipedia.org/wiki/Gallium "Ga nanoparticles can be kept in the liquid state below 90 K." Can you electrolyze it at liquid N2 temp?

The school has liquid N2, I'm pretty sure one of my professors would let me try it. They've been pretty helpful with this essentially pointless project so far. Not sure they'll be as accommodating when the fall semester starts and they have to start teaching again.  What about the electrolysis set up at room temp forming molten gallium?

To preface, I can only answer the electrochemistry questions.
The pH at which you do electrodeposition is dependent on a range of factors, in general the first thing to do is to check a Pourbaix diagram (from: Chung, Yonghwa & Lee, Chi-Woo. (2013). Electrochemistry of Gallium. Journal of Electrochemical Science and Technology. 4. 1-18. 10.33961/JECST.2013.4.1.1.)

As you can see if you do not want to form a passivating film of Gallium oxide you need to deposit from either acidic or alkaline solutions in aqueous electrolytes. Next you see the deposition voltage is below the dotted line, these lines signify the electrochemical limits of water. You will partially be forming hydrogen gas during deposition, depending on your substrate and the catalytic activity of Gallium it can be that in acidic or alkaline conditions you're more likely to form more hydrogen gas (or even form hydrides) which limits the current efficiency (how much is going to Gallium deposition vs to some other processes). Sometimes depositing from different pHs makes you get a smoother finish or other results that are preferred, now for your educational exercise this is not super relevant.
Next, on the question of graphite. You can use a graphite cathode, but I would advise against a graphite anode. Because, as you've mentioned, you have chloride in solution. Looking into a good oxygen evolution catalyst as an anode might prevent this.
This is on the sixth question, 2.7A is a crazy high current for lab scale deposition. You will not be able to draw this amount of current without just making a bunch of hydrogen. You have to note that near your electrode surface, locally, the Gallium in solution will be depleted and more needs to come to the electrode by stirring and diffusion. So if you apply 2.7A, a small fraction will be Gallium and all the other current will be your electrolyte (forming hydrogen, or other stuff) Extracting all the Gallium from the solution by electrodeposition will be a very long process and it is beneficial to work in as high concentrated solution of Gallium as possible to not run into mass transport issues. Maybe it would be better to do the experiment in a potentiostatic way (apply a constant voltage).
Finally, I think it might be better to attempt to coat something with Ga electrochemically instead of extracting all of it. Bulk extraction is way easier with other methods. I have attempted Ga electrodeposition before, but from organic electrolytes. I can refer you to this paper, here you can have way higher current efficiencies as your electrolyte is not unstable in the potential range needed to deposition Gallium: https://pubs.rsc.org/en/content/articlelanding/2021/cp/d1cp01074c

THANK YOU! That article is a huge help, at least what I can understand. I also found the article that the chart you shared came from which is also helpful. For the 2.7amps, I just did my calculations based on only having one hour. I think I can get a spot in the lab to let it run for longer, but I was seeing what it would look like if I only had 1 hour.

https://www.youtube.com/watch?v=6HBMMgvqVY0&list=PLdW7n7SKdpHng1dl39_DfX4xovwVgk92Q&index=28&t=600s   <- this is what I am wanting to recreate. I didn't know if it would be possible to form crystals of Ga rather than a molten solution. I was looking at graphite because I'm not sure if I can convince a professor to give me a titanium and the only nickel I've seen in lab is a jar of nickel balls.


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Regarding the standard reduction potential and electrolyte/solid interface
I am modelling lithium-sulfur batteries ($\ce{LiS}$ batteries) and a strange phenomenom happens during charge, named infinite charging, which I show below:

The factor $f_\mathrm{C}$ that appears in the description doesn't matter, and the compound is one of the electrolytes that are used in the $\ce{LiS}$ battery. As we can see, while the current $I$ goes down, it takes more time to recharge. If low enough, the battery doesn't charge at all, as the voltage enters a plateau and never gets out.
The reaction mechanism in $\ce{LiS}$ batteries is challenging, due to the many species during the electrochemical reactions. We also have disproportionations, precipitations, and dissolutions. However, from a modelling perspective, we can summarize it as follows during discharge.

Positive electrode (cathode): solid sulfur $\ce{S8(s)}$ is dissolved into liquid $\ce{S8(l)}$. After that, a huge number of reductions take place to $\ce{S8^2-}$, $\ce{S7^2-}$, $\ce{S6^2-}$, up to $\ce{S^2-}$. This last one is insoluble in the electrolyte, so $\ce{Li2S(s)}$ appears at the end of the discharge. The charge transfer occurs at the carbon/electrolyte interface. We add carbon, or similar forms, because $\ce{S8(s)}$ is a horrible conductor and charge transfer will not occur over a sulfur/electrolyte interface.
Negative electrode (anode): the lithium is just oxidized
$$ \ce{Li(s)} \rightarrow \ce{Li^+(l)} + \ce{e^-}$$
and the charge transfer occurs at the lithium/electrolyte interface. The symbol $\ce{(l)}$ here denotes always the electrolyte in this post, for all species.

During charge, there is a consensus in the community that the reaction mechanism changes. However, we may simplify matters and just say that the reactions are inverted. In addition, we are sure that the shuttle effect takes place, where the infinite charging is one of its consequences. This shuttle effect I also summarize below:

$\ce{Li2S(s)}$ starts dissolving into $\ce{S^2-}$ and $\ce{Li+}$. After that, the oxidations start to take place at the positive electrode, so we start getting $\ce{S_2^2-}$, $\ce{S_3^2-}$, $\ce{S_4^2-}$, $\ce{S_5^2-}$, etc.
Many of these guys, called the polysulfides, suffer electromigration to the negative electrode.
These tranported guys suffer reduction. For example, $\ce{S_5^2-}$ reduces to $\ce{S_4^2-}$, or $\ce{S_7^2-}$ to $\ce{S_6^2-}$, etc.
These reduced polysulfides suffer electromigration to the positive electrode and are re-oxidized.

Funnily, the same electrochemical reactions take place in both electrodes, but in the opposite direction. If we combine these points, it is easy to understand why we obtain the infinite charging in the upper image. The compounds that are being oxidized at the positive electrode, are then reduced in the negative electrode. The net effect is that the charge is just circulating back and forth, and we cannot go back to the starting place. Again, there isn't a consensus about the exact behavior of the whole shuttle effect in the community, but this will suffice.
From a modelling perspective, however, a simple approach may function to capture the infinite charging: once charging initiates, we just add another reaction in the negative electrode, that mimics its counterpart in the positive electrode. For example, while the following happens in the positive electrode (oxidation)
$$ \ce{\frac{1}{2} S8^2-(l) \rightarrow \frac{1}{2} S8(l) + e^- (carbon/elyte interface)} $$
we just include the opposite in the negative electrode (reduction)
$$ \ce{\frac{1}{2} S8(l) + e^-(lithium/elyte interface) \rightarrow \frac{1}{2} S8^2-(l)} $$
Question It is not important the details of how to mathematically model this. Enough is to say that, in order to implement it, we need to know the standard reduction potential $E^\circ$ of these reactions in order to use Nernst's equation. Since they are the same reaction, we may assume that $E^\circ$ is the same. However, both happen at different interfaces, since one is carbon and the other is lithium. Is there any evidence that a same reaction happening at a different interface may have a different standard reduction potential, or different $E^\circ$? Is it a conceptual mistake to assume that they are different?
If the $E^\circ$ was different in both reactions, that would allow to implement this computationally. If they are equal, it can be shown that many problems happen. I am looking forward to either experimental or theoretical proofs, if possible.
References Image from:

Mikhaylik, Yuriy & Akridge, James. (2004). Polysulfide Shuttle Study in the Li/S Battery System. Journal of The Electrochemical Society. 151. A1969-A1976. 10.1149/1.1806394.

I was curious when you will pull up this topic. :-) Interesting stuff.

You were curious? Oh thanks, Poutnik. Unfortunately, it is heavy stuff for me. Do you think the post is clear?


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Comparing Succesive Ionisation Energies
I am trying to understand successive ionisation energies. In particular, an explanation for why the second ionisation energy is greater than the first ionisation energy. I'm looking for a clear and concise explanation.
The common explanation is that it's harder for an electron to be removed from a cation than the atom or that the second electron feels a greater nuclear charge. However, these explanations do not satisfy me.
For starters, How can the nuclear charge be greater for the second electron when the number of protons within the nucleas remains unchanged. Assuming both the first and second electrons are on the same energy level, they experience the same amount of shielding from inner electrons. So it seems to me, both electrons feel the same nuclear charge. I can see the second electron, perhaps feeling repulsion from the first that results in less nuclear charge. Would this then be correct?
I don't particularly understand the explanation about it being harder to remove an electron from a cation. The way I see it, is that the atomic radius decreases when the first electron is removed, meaning the second electrons feels more electrostatic attraction from being closer. Is this explanation sufficient? Could someone please provide a clear explanation for this?

In fact, all this is well explainable by the rules of classical electrostatics, one need not even apply quantum mechanics.

Sure is - that's a first approximation, but you can just treat it as separating -1 charge from +1 vs -1 from +2 - should be pretty obvious.

I will give a classical and a quantum mechanical consideration for a potential answer of your question. The ionization is only dependent on the potential as purely kinetic particles are considered unbound. You are right by stating that it depends on the repulsion between the particles. In the classical case of $N$ electrons as particles occupying the outermost shell, the effective potential before the ionization (1) and after the ionization (2) is :
$$
V_{eff,1} = - \sum_k^{N} \frac{Z_{eff}e^2}{r_k} + \sum_i^{N}\sum_j^{N} \frac{e^2}{|r_i-r_j|} \text{ (1)}\\
V_{eff,2} = - \sum_k^{N-1} \frac{Z_{eff}e^2}{r_k} + \sum_i^{N-1}\sum_j^{N-1} \frac{e^2}{|r_i-r_j|}\text{ (2)}
$$
Assuming that the electrons in this shell are closed enough without taking into account the orthogonality and the symmetry of orbitals, the average distance between the electrons in the shell and the nucleus is approximately $r_k \approx R$,. The average distance between the electron is assumed to be  $|r_i-r_j| \approx R'$ :
$$
 \Delta V = V_{eff,2} - V_{eff,1} = \frac{Z_{eff}e^2}{R} - (2N-1)
 \frac{e^2}{R'}
$$
As $R \gg |r_i-r_j|$ ($R \gg R'$), when an electron is ionized, we obtain $\Delta V \approx - 2N \frac{e^2}{R'}$ which is purely the repulsive part giving $V_{eff,2} < V_{eff,1}$ . Therefore in this simple case the ionization affects mainly the repulsive part of the potential. This means that the potential of the nucleus is more significant after ionization compared to the repulsive part, the bonds are stronger and the second ionization needs more energy.
A quantum mechanical treatment is quite similar, there is no defined position $R$ or a distance $R'$ for the electrons in this case. But the repulsive part has a significant value because $|r_i-r_j| \ll $ especially for singlet states where there is an overlap. This distance is also small for triplet states without an overlap compared to the mean distance to the nucleus (where the probability density is higher) $|r_i-r_j| \ll  R \gt a_0$ giving the same result more justified in this case.

Assuming your knowledge level, I will try to focus to easy to understand verbose description. Let assume for the illustration the case of a helium atom.

If we start with the helium nucleus $\ce{He^2+}$, it has the charge $+2e$.
If we add 1 electron to have $\ce{He+}$, the electron is attracted by the charge $+2e$.
If we add another electron to have $\ce{He}$, things get more complicated.

If this second electron had been hypothetically somewhere very far and weakly bonded, we would have neglected it and still assumed the first electron had been effectively attracted by $+2e$ of the helium nucleus.
If this second electron had been hypothetically inside the nucleus, we could have assumed the first electron had been effectively attracted by the charge $+2e - e = +1e$.
As it is none of these opposite extreme cases, it is clear the effective charge of nucleus will be somewhere from the interval $(+e, +2e)$.

If we qualitatively evaluate the mutual interaction of both electrons, the electrons effectively:

Partly repulse the other electron out of the atom like if they were in the nucleus, pushing the effective nuclear charge toward $+e$ instead of $+2e$.
Partly increase the nucleus attraction of the other electron. As they are further from the nucleus, pushing the other electron toward the nucleus and the effective nuclear charge toward $+3e$ instead of $+2e$.
Due mutual electron repulsion is the mean distance of electrons in $\ce{He}$ atom bigger than for the single electron in $\ce{He+}$

Overall effect is expressed by semi-empirical Slater's rules for the effective nucleus charge $Z_{\mathrm{eff}}= Z - s$, where for a helium atom $Z=2$ and for $\ce{1s}$ electrons, $s = 0.30$.
Therefore, for $\ce{He}$ atom, $Z_{\mathrm{eff}} \ne 2$, $Z_{\mathrm{eff}}= 1.7$
Combination of

the smaller effective nuclear charge (= smaller net force)
the larger mean distance from the nucleus

leads to the lower ionization energy for the first electron and higher one for the second one.
Helium ionization energies:
$$\begin{array}{|c|c|}
\hline \text{IE1} & \text{IE2} \\
\hline \pu{24.58738 eV} & \pu{54.41776 eV} \\
\hline
\end{array}$$

Fur further reading, see

Wikipedia: Effective nuclear charge
Wikipedia: Slater's rules

Whatever the structure of the atom, it is more difficult to remove a (negative) electron from a positive structure than to remove an electron from a neutral structure. And it is even more difficult to remove an electron from a doubly charged atom that has already lost two electrons. And if the atoms has already lost three electrons, it is immensely more difficult to remove a $4$th electron. Etc.


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Why do lipids decrease in viscosity with increasing temperature?
Is the answer simply that the carbon bonds break up and that the molecules get smaller?

Generally, all liquids  have decreasing viscosity with T, for water based liquids there is a rule of thumb 3% per 1 deg C.// there are special exceptions due internal effects.

NO ! The carbon bonds do break up when increasing the temperature up to very high temperatures. But this effect, called pyrolysis, is negligible around room temperature. It starts at temperatures much higher than $100$°C, and it is not reversible, as the viscosity does change but the chemical composition also changes. So the viscosity of heated and pyrolyzed liquids (and then cooled down) is not comparable to the viscosity of the non-heated liquid.
To close the discussion, the decrease of viscosity due to increasing the temperature is a general phenomena, observed with all liquids, lipids or not lipids.


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Which of the given samples will have the largest number of atoms?
The samples that we have in question are as follows:

1g Au
1g Na
1g Li
1g Cl2

I know that the sample of element which has the largest number of moles of atoms of the element must have the largest number of atoms.
I am also aware of the following formula:
n = m/M
where n stands for the number of moles of the substance, m stands for its mass and M denotes its molar mass.
Can I argue that the number of moles of atoms in a given mass of an element is inversely proportional to its molar mass, and use this fact in the solution of this problem?

Using "mass/amount of substance" is preferred to using "number of kilograms/moles of substance".

Yes. your reasoning is correct. One mass unit (gram or kilogram) of small objects contains more objects than the same mass of heavy objects.

Basically, the question is like If there is more stones in 10 kg of smaller stones or in 10 kg of bigger stones.

Yes, your analogy is correct. The amount of substance $n$ is inversely proportional to molar mass $M$ $(n = m/M).$ Because the mass $m$ is fixed, increasing molar mass will lead to heavier atoms, reducing the number of atoms present in that fixed mass.
So, the answer to the question is lithium since it has the smallest molar mass among the listed compounds.


text:
How can i preserve a coating of baked in powder on activated carbon pellets whenever it rains?
I'm trying to create pellets of activated carbon that have a coating of a soluble white powder. To make this I first dissolve a significant amount of this powder in boiling water to the point where the activated carbon is saturated with the powder and cannot absorb any more. Then I add some more and let the boiling water evaporate completely. This leaves a coating around all of the pellets.
However this coating will wash off within 10 minutes when exposed to falling water droplets like from the shower. I'm looking for biodegradable binders that are soluble in boiling water and could help to preserve the coating around the pellets whenever its raining.
According to chatgpt i could use starch but i don't know if that's very water resistant.
What cheap water resistant and biodegradable binders could i add to the coating to prevent it from falling apart whenever its exposed to falling water droplets?

If I understand correctly, you are looking for a water soluble substance which is not soluble in water.

@Maurice I am looking for a way to preserve the coating.

But the coating is soluble in water.

@Maurice Yes but so are a lot of other things like water based paints. Water based paints has binders to retain the pigment. I need a biodegradable binder that can prevent the coating from falling apart.

Water based paints are not solutions in water. They are emulsions, or suspensions. The painting operation is irreversible.

Water-based paints, and many water-based glues, are not dissolved in water, but rather are suspensions. When the water evaporates, the suspended particles come together to make a continuous, water resistant film. Depending on the film-forming binder, such as acrylic, polyvinyl acetate, milk or egg protein (in tempera paint) and additives, the paint may be made less or more water resistant.
If you have a specific schedule for release of the underlying "soluble white powder", whether hours or years, that should have been in the question. As stated, you would have to experiment for your use.
If such a paint, as you have suggested, meets your needs, why not use it?


text:
What is the correct IUPAC nomenclature for this compound?

I recently came across this type of compound, and normally, I would follow the rules I've previously learnt, which is to find the longest carbon chain that contains the highest priority functional group. As Wikipedia states:

It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest group precedence should be used.

This aligns with a lot of other material I've read. Following this, the double bond takes precedence, and therefore, with the ethyl group as a substituent, the IUPAC systematic name should be: 3-propyloct-1-ene.
However, I've also found material that suggests otherwise. The Brief Guide to the Nomenclature of Organic Chemistry published by IUPAC and available online as a pdf states:

SENIORITY OF PARENT COMPOUNDS
The systematic name is based on the name of the senior parent compound, which is chosen by applying the following criteria in the order described below, until a decision is reached.
f.1: Contains more atoms:
A nine-atom chain is senior to an eight-atom chain (even if it has fewer double bonds)

If this is the case, then the correct name of the compound above should be: 4-ethenylnonane.
Which name then, is the correct one?

In the present case, more atoms is a criterion to be considered earlier than  the one of maximum of multiple, than double bonds.  IUPAC's brief guide illustrates this by the following flow chart:

(credit to the Brief Guide by Hellwich et al., p.534)
Later, the brief guide illustrates the application of the rules; example f (p. 536) where the vinyl (or systematically: ethenyl group) is attached to the nonane chain:

This is why your second suggestion is the correct one.
(Side note: This and additional references to chemical nomenclature are compiled in the site's dedicated section Resources for learning Chemistry.)
Hellwich, K.-H.; Hartshorn, R. M.; Yerin, A.; Damhus, T.; Hutton, A. T. Brief Guide to the Nomenclature of Organic Chemistry (IUPAC Technical Report). Pure Appl.Chem. 2020, 92, 527–539. https://doi.org/10.1515/pac-2019-0104. (open access)

Addendum: Consultation of the open-access excerpt by Moss (April 2022) of IUPAC's 2013 Blue Book as whole pdf includes a noteworthy section P-44 about the seniority order for parent structures.  The introduction (P-44.0) reads

"The selection of a preferred parent structure is based on the  seniority of classes (see P-41), which gives priority first to characteristic groups expressed as suffixes and then to parent hydrides when different classes are present. Section P-44.1 covers the selection of a preferred parent structure when different classes are involved and the selection between rings and chains in the same class. When there is a choice among cyclic parent hydrides, the senior ring or ring system is chosen in accord with the seniority order of rings and ring systems (see P-44.2). When there is a choice among acyclic parent hydrides a principal chain must be chosen (see P-44.3). The three seniority orders, for classes, rings and ring systems, and the principal chain, are expressed in a general seniority order called ‘seniority order for parent structures’. Section P-44.4 covers criteria for selection of a senior parent structure applicable to either rings, ring systems, or acyclic chains.
A thorough revision and extension of the seniority order of classes, of rings and ring systems, and for selecting the principal chain was needed in the context of preferred names. This revision incorporates a major change from earlier recommendations given in the 1979 edition (ref. 1) and the 1993 Guide (ref. 2).
In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier
recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain
is the length of the chain; unsaturation is now the second criterion."

The emphases in italics/bold are additions by me. In the original (both Moss' version, as well as RSC's e-book), the last paragraph is set in a framed box I'm not aware how to replicate here.

"In earlier recommendations unsaturation was senior to chain length" deserves elaboration. (Was this previously different? When was it changed? Any additional explanations?) -- Different sources using different versions of the rule goes a long way to explaining why there's a discrepancy in what the resources call for.

@Buttonwood Thanks so much. It all makes sense now!

@R.M. Thank you for your input; there is an amendment to the answer to highlight the change and at least the date of transition.


text:
Four year old opened but rarely used bottle of 99,9% Isopropanol: Risk of peroxides?
Let me begin this post by stating that I do not know basically anything about chemistry, which is why I am asking for your help. Please be kind :)
Today, I saw the following post on Twitter and Reddit and it somewhat scared me:

I had no idea that Isopropanol, stored for too long, can potentially turn into something quite dangerous (and according to the comments on Twitter / Reddit many people are in the same boat).
In the beginning of 2019, so over four years ago, I bought a 1 liter bottle of Isopropanol 99,9% on Amazon. The product page on Amazon explicitly stated that it contains no additives. I bought it for cleaning purposes, for example cleaning heat paste from CPUs and so on, but I don't do that too often, so it was not used a lot and there is still quite a lot of Isopropanol in it. I probably have not used it for several years now, but I thought that nothing would happen to it.
The bottle is a white, somewhat opaque, "soft" plastic bottle. This bottle is still stored on my shelf. One thing I have noticed with the bottle is that it somewhat compressed by itself, but I guess that does not indicate anything dangerous (?).
The room it is in usually does get some light, but not very bright light, and sun light is rare. That might have been slightly different in the years 2019 and 2020, I cannot remember for sure.
I am now worried that this might have happened to my bottle of Isopropanol, but other than the time that has passed since I opened it, I have no indication for it. I am not really sure how to assess the risk.
At the moment, I would not even move the bottle, but I might be overcautious here and fueled by unnecessary anxiety. Due to the opaqueness of the bottle, I probably would not even be able to tell for sure if there are crystals in the bottle, most likely I would also have to move it to be able to check it properly.
What is your risk assessment and what are your suggestions on how to deal with it? I would love to just get rid of it, but that would probably mean bringing it to a waste disposal site, and I assume that could already be dangerous, if something of that sort had happened.
Thank you all for your help, suggestions and comments.

Well, isopropyl is listed as a peroxide former. Up to you to decide what to do. I’ve had bottles last longer than that, but that is no guarantee yours isn’t going south.

Since you state, "the bottle is... somewhat compressed," it indicates that the $\ce{(CH3)2CHOH)}$ solvent has slowly evaporated through the plastic container, but air has not entered. In essence, the plastic bottle is a semipermeable membrane. The pressure decrease is cause by by osmosis of a vapor. Not harmful, and since it indicates more air is not entering to form peroxides, likely the solvent is still safe, though it might be possible a small amount has formed on the bottle the cap thread.
Note that isopropanol is on this list. A similar question has also been asked.
Since this is a personal safety question, it probably does not belong on this site, but I felt this non-authoritative answer worth the effort, though it should soon be removed.

The document cited by DrMoishe Pippik is called "this list", The very first lines of this document are repeated here : Peroxide formation in solvents and reagents have been the cause of occasional accidents. Even the relatively innocuous solvent, isopropyl alcohol is capable of peroxide formation. This can occur upon exposure to air, heat, light, or simply with passage of time. .

Thanks, @Maurice, I fixed that glaring error!

It is a pity that solution of FeSO4 or Na2S2O5 cannot be used as they are for diethylether or diisopropylether.

What about the solid in the picture the OP posted. Is that fake?

Don’t distill it: https://pubs.acs.org/doi/10.1021/acs.oprd.2c00112


text:
Why is there no neutral [Cr(OC)6] or anionic [Fe(NC)6] - isomers of cyanide and carbonyl complexes?
I was just reading about linkage isomerism, that usually arise due to the fact that some ligands are ambidentate (i.e. $\ce{SCN}$ and $\ce{NCS}$).
I then think to myself, considering only the ligand structure, $\ce{CN-}$ and $\ce{CO}$ ligand also should also be ambidentate ligands (I'm not including bridging). However, I haven't found any reference to that. Here's now, my questions:

Are $\ce{CN-}$ and $\ce{CO}$ really ambidentate? why or why not?
How different are the complexes? (i.e. $\ce{Cr(OC)6}$ vs $\ce{Cr(CO)6}$)

Also, if anyone found a complex with $\ce{NC-}$ or $\ce{OC}$ ligand, please do put the reference to that.

And why not include bridging? That's the main situation you get O-coordinated carbonyls.

Mostly for personal preference. However, I felt like bridging force the ligand to use more than one electrons pair at the same time, instead of having 2 (or more) electrons pair that can be used in multiple ways.

Well, in the answer, there sure is bridging...

In the case of cyanide we actually do have iron coordinated to the nitrogen ends of the cyanide ligands in Prussian blue. In fact  iron atoms are coordinated to both ends of the cyanide ion ligands; the iron forms a simple cubic lattice with the cyanide ions laid out along the edges of the cube like an expanded perovskite structure:

From Wikimedia Commons user Ben Mills

From Wikipedia user Smokefoot
In general, the pi-acceptor ligand that cyanide ion is will bond first through the less electronegative carbon atom, where the $\pi^*$ orbitals that accept electrons from the metal have higher amplitudes (and thus potentially better overlap). But when that bond is already taken up, as in the Prussian blue structure or with nitrile ligands where the cyanide carbon is bonded to another carbon atom, then the nitrogen end of the cyanide can from a second bond.


text:
Please explain the order of Lewis acidic strength in trichlorides of boron group
The order of decreasing Lewis acid character as per online sources is as follows:
BCl3 > AlCl3 > GaCl3 > InCl3
Reason cited is: As we move down the group, the size of atom increases, and as a result, the tendency to attract electrons decreases. This leads to a decrease in Lewis acid nature down the group.
However, it is also known that BF3 is the weakest Lewis acid among the boron trihalides on account of back bonding effect (BI3 > BBr3 > BCl3 > BF3) as the energy gap between boron and fluorine (2p-2p) is less resulting in a stronger bond and the energy gap between boron and iodine (2p-5p) is the most, resulting in the weakest bond.
Why can't we apply the above explanation to BCl3, AlCl3, GaCl3, InCl3 as well, saying that Al and Cl should have the least energy gap (n=3 for both) and thus have the strongest bond and weakest Lewis acid character.
Please feel free to point out any fundamental mistakes in my assumptions and reasoning as I am still figuring things out and may have a limited/incorrect understanding of back bonding.

Related: https://chemistry.stackexchange.com/questions/135554/lewis-acidity-order-of-aluminum-halides/135643#135643, https://chemistry.stackexchange.com/questions/153768/towards-a-hard-lewis-base-like-water-which-is-the-stronger-lewis-acid-silicon-t, https://chemistry.stackexchange.com/questions/129558/why-is-phosphorus-trifluoride-the-strongest-lewis-acid-among-the-phosphorus-trih


text:
When a 7-membered ring is formed during GC/MS, does the substituent need to start with a carbon?
When looking at the GC/MS for Allylthiobenzene aka allyl phenyl sulphide, two clear peaks are seen at 91 m/z and 65 m/z, most common for 7 membered and 5 membered carbon rings respectively. The problem is that the substituent group associated to this compound begins with a sulphur, which, when integrated into the ring, does not produce a signal at 91. Is it possible for this to be cleaved and a different carbon (possibly from the tail swinging around) to be integrated?

Welcome to chemSE. The likely cations mz=65 and m/z=91 are $\ce{C5H5+}$ and $\ce{C7H7+}$ (tropylium?), respectively. There is the possibility that the initial event is a Claisen rearrangement. This would form a $\ce{C10}$ framework from which the fragments arise.

It is also true that the MS of toluene, thioanisole and allylbenzene all show the presence of m/z=65 and m/z=91 fragments. Look here.


text:
Organic solvents distillation for purification
I've realised that the majority of the solvent that we use in my lab are contaminated with grease (peaks at 1.26 and 0.86 ppm in $^1H$ NMR in $CDCl_3$), since they come in plastic containers (I guess that's the reason).
One solution would be to buy better grade solvents, but we have a large amount of these solvents in the plastic containers. I was thinking about distilling them to purify them and keeping them in glass bottles, but I thought of some issues that could appear.
Some organic solvent present stabilizers. For example, dichloromethane is often added amylene (like in my case), so I'm afraid of losing the stabilizer when distilling. Amylene has a boiling point of 38.5 ºC, so it should distill with the DCM. However, I was thinking of using the rotavap for practical reasons, but I think that maybe the amylene would not condense at the conditions used for the DCM. Also, I don't know if more stabilizers are added, since sometimes I get peaks that resemble BHT but I don't use any solvent that has BHT in it.
Could I use the rotavap for distilling the DCM or should I perform a simple distillation? Would the simple distillation work for keeping the stabilizer and preserving the DCM for long periods? Also, in the bottles of ethyl acetate and n-hexanes it doesn't say anything about stabilizers, is it alright to assume that they don't need any and I can distill them without problem?

If your solvents come from the producer in plastic bottles, the impurity is probably not grease. It is probably some impurity coming from the plastic itself. You can distill DCM in a Rotavap : the amylene will distill together with the dichloromethane. I have never heard that ethyl acetate and hexane contain stabilizers.

I'd argue that, if a supplier has used plastic bottles, they are not the most likely source of contamination. They will be designed to transport pure solvents. More likely, other lab practices are creating impurities. We used to use cheap PVC piping for inert gas transport (when doing air sensitive chemistry). The plasticisers (often alkyl phthalates) and preservatives in those pipes were a big source of contaminants.

Can you filter thru activated charcoal? Or thru a silica column?

Using the rotovap to distill the DCM proved useful, as it left a yellow oil as a residue. After that, the solvent was clean by NMR, only with some peaks that seem to be from amylene (the stabilizer present in DCM).
As far as I know, n-hexane and ethyl acetate don't need stabilizers. The ones we have in our lab are clean, although they also come in a plastic container.
I don't know if the contaminated DCM that we have is due to a bad batch, the plastic being dissolved or the actual quality of the DCM. It could be easily tested, but I haven't done it.

Purifying solvents can be difficult. I sort of solved it by encountering a local glassblower and devising a minimal holdup system with a Vigreux column. I used a dedicated distilling flask for each solvent. Collection was into a glass bottle saved from purchases of HPLC grade solvent. I tested by evaporating about 10 mL in a CLEAN [washed, Oven dried, pyrolyzed, at high T], Erlenmeyer flask and letting the last bit dry on a 4-bounce atr cell No detectable residue. This worked for methanol, methylene chloride, n-pentane, chloroform, cyclohexane, benzene, DMF and heptafloro-2-propanol[used to extract fluorinated oils]. Ethanol and iso-propanol were distilled from activated magnesium to remove water. The alcohols and DMF were checked by infrared or the nmr OH peak.
The removal of stabilizer is not a problem if purified solvents are used deliberately and not stockpiled.
A story about stabilizers: A colleague synthesized a rather large molecule with an ethyl group. He had an nmr with a weak quartet and came to us to corroborate his assessment. the chemical shift looked suspicious. It turned out that the deuterochloroform was stabilized with regular ethanol; we never heard back.


text:
How to neutralize 60L of a 2% Aqua Regia with Various Metals suspended
I am encountering an issue in my laboratory. My supervisor and I are currently faced with the task of neutralizing a volume of 60 liters of Aqua Regia, which contains suspended metals, at a concentration of 2% or lower.
Our current approach involves using Granular Sodium Hydroxide (NaOH) as a neutralizing agent. However, we have observed that we often end up using excessive amounts of NaOH, resulting in an overly basic solution. Therefore, I seek guidance on the appropriate quantity of NaOH required to neutralize the Aqua Regia effectively.
Furthermore, I would like to inquire about the viability of preparing an aqueous solution of NaOH prior to its addition to the 60-liter drum. Will this method enhance the neutralization process?
Any advice is greatly appreciated.
EDIT: At my job we digest silver alloys and we use 10 mL of 30% HNO3 and 5 mL Concentrated HCl. When we filter to collect AgCl we use DI H2O and the amount that we use is not the same through out all of the filtering. So the concentration of the Aqua Regia is most like below 2%. I just used 2% as a guess really.

Seems curious that you do not have an environmental health and safety (EHS) outfit, or the like, that can tell you what you can and cannot do with hazardous waste, particularly when you have substantial volumes. Where I was, there were lots of rules and regulations, yearly online safety training and processing hazardous waste was a no-no. Everything went into properly labeled and appropriate containers, for later pick up by the EHS people.

At my job we like to do things our selves because I work for a family run business and It can get costly to get EHS to come and pick up all of our waste. We use alot of acid if you can see my edit and we would need them to come almost every other day. We do follow OSHA Guidelines and do have the correct PPE. (safety glasses, lab coats, gloves etc...)

Thanks for the clarification! I know from experience that the EHS folks can be weirdly pedantic: one time I saw them bring a big plastic tub, on wheels and loaded with baking soda (just in case of a spill in transit), to pick up a single bottle of chemical. That chemical was sodium bicarbonate. ;-(.  I have not voted, by the way.

Wow that's just a bit over the top haha. I started my job last September and I remember my supervisor and I talked to the President of our company for EHS and he told us it was some ridiculous price to get our waste picked up every week. We would need it more than once a week as well.

There's no "2% Aqua Regia", you can have aqua regia (small letters!) diluted with water.

Instead of adding sodium hydroxyde $\ce{NaOH}$ to your solution, it is advisable to replace it by sodium bicarbonate ($\ce{NaHCO3}$) which is also named sodium hydrogen carbonate. This powder is not corrosive (like $\ce{NaOH}$). But it will produce plenty of gas ($\ce{CO2}$) when in the contact with acidic solutions. As aqua regia is made by mixing $\ce{HCl}$ and $\ce{HNO3}$, their reactions with $\ce{NaHCO3}$ are :  $$\ce{NaHCO3 + HCl -> NaCl + H2O + CO2}$$
$$\ce{NaHCO3 + HNO3 -> NaNO3 + H2O + CO2}$$ And $\ce{NaHCO3}$ will also react similarly with $\ce{NOCl + Cl2}$ which are the products of the reaction of $\ce{HCl}$ plus $\ce{HNO3}$

I do not have access to Sodium Bicarbonate, we could buy it but my facility already has large amounts of NaOH.

You do not say, what exactly you mean by 2% aqua regia ($\pu{1.2 L}$ diluted to $\ce{60 L}$?), nor its state after reaction with metals, so no precise numbers are possible.
Take small aliquote and realize how much hydroxide, carbonate or bicarbonate you need to use.
Then scale it up.

Alright I did not specify, the aqua regia is about 2% because of the volume being at 60L.

Aqua regia cannot be $2$%. If you mix of $3 \ce{HCl}$ + $1 \ce{HNO3}$, both being concentrated, it produces aqua regia. It can dissolve gold. If the mixture is diluted more than $2$ or $3$ times, it is not aqua regia any more. It is a mixture of diluted nitric  + hydrochloric acids. And this mixture does not dissolve gold.


text:
Conductive ink on silicon to bridge two isolated regions of the wafer
Can I use a thin layer of conductive ink on silicon to bridge two isolated regions of the wafer (p-type)? I know that using PVD methods would be the best option, but I don't have access to these expensive tools. How would these conventional conductive inks adhere to silicon? Or would silicon be "inkphobic"?

What is the scale of the device, and what is the allowable circuit resistance in your use? Most conductive inks are made of metal particles in a non-conductive binder. Given a large enough trace width and thickness, enough metal particles contact each other to complete a circuit. A thin trace would likely have high resistance.

This is a really interesting question! Local interconnects on silicon (as opposed to going up into the metal layers and then back down again) come and go in VLSI design. For "inkaphobicity" note that absolutely clean silicon (which doesn't last long in air) is not particularly hydrophilic, but it quickly grows a native oxide layer in air; once you get monolayer or two of SiO2 on top it's just as hydrophilic as glass. You can delay that by passivating the dangling Si bonds on the surface with things like fluorine or hydrogen (the "HF dip" step)

Also, since silicon is a semiconductor and not a metal, it is a challenge to make good electrical contact with the conduction band. You need to think about how to make an ohmic contact between each end of the ink path and the semiconductor. Using very highly doped silicon helps, but only on one end or the other. Thus for a given direction of current flow you'd need a heavily p-doped silicon pad under one end and heavily n-doped silicon pad under the other.

potentially helpful would be some reading on printed electronics


text:
Preparation of cycloalkanes
Is it possible to get cyclooctane from calcium suberate? Is this reaction similar to one when getting cyclohexane from calcium adipitate?

Consider to substantiate your question by provision of details what reaction(s) you have in mind to create cyclohexane/cyclooctane.  Don't hesitate to include a reaction scheme (e.g. with the help of ChemDraw's sample page) to prepare an illustration. For one, it documents you invested some effort to solve the question by your own. For two, it improves the quality of your question for future readers potentially interested in this method.

Synthesis is done in opposite direction. Why would anyone synthesise cyclohexane?

Kolbe's Electrolysis
It is a decarboxylation dimerization reaction that follows a free-radical pathway and, formally, occurs by passing current through an alcoholic $\ce{KOH}$ solution to ensure presence of carboxylate ion instead of carboxylic acid. In your case, the base is $\ce{Ca(OH)2}$.

Intermolecular
Intramolecular


text:
How many aldehydes and ketones can be formed from molecular formula C5H10O?
Question:

What are the possible aldehydes and ketones with molecular formula $\ce{C5H10O$ (including the stereoisomers)?

Answer: 5 aldehydes and 3 ketones
The degree of unsaturation of the compound is 1, so the ketone part is understandable since you take a carbon chain of length 5 and we have 3 possibilities for attaching double bond oxygen.
For the aldehyde part, I'm having trouble drawing the isomers.

Aldehydes

Structure
IUPAC Name
Common Name(s)

Pentanal
Valeraldehyde, n-butyl aldehyde

3-methylbutanal
Iso-butyl aldehyde,  isovaleraldehyde

Pivalaldehyde
Neopentanal, tert-butyl aldehyde

(R)-2-methylbutanal
R-sec-butyl aldehyde

(S)-2-methylbutanal
S-sec-butyl aldehyde

Ketones

Structure
IUPAC Name
Common Name(s)

Pentane-2-one
Methyl n-propyl ketone, methyl propyl ketone

Pentane-3-one
Diethylketone (DEK), dimethyl acetone, metacetone

3-methylbutan-2-one
Iso-propyl methyl ketone, MIPK


text:
Franck-Condon principle and shift of wavelength
I was revisiting my notes for spectroscopy, and I came across the Franck-Condon-principle. I understand it well, but want to clear some doubts regarding this.
I know that the Franck-Condon-principle states that during an electronic transition from the ground state vibrational state (according to Boltzmann, the most populated), a transition happens, but not to the ground vibrational state of the electronically higher state, but to a vibrationally excited state of the electronically higher state, as seen here:

So, as seen by the excitation (highlighted in blue), the excited state of the electronically higher state decays through collision with other molecules to the ground vibrational state and fluorescences to a vibrationally excited state of the electronic ground state. That is the reason why the Emission spectrum is lower in energy than the absorption spectrum (Stokes shift):

For me there remains a doubt:
So basically for the transition to happen, the electronic energy difference needs to be overcome plus the vibrational difference, all by one photon that matches the energy difference as seen in picture 1. However, in statistical thermodynamics it is always assumed that electronic and vibrational transitions are uncoupled, so one photon should only ever excite an electronic transition, or a vibrational transition, but one photon can’t change both modes (electronic + vibrational) at the same time. Where did I think wrong?

No problem ! One photon can easily change both modes (electronic + vibration) at the same time. This is even what happens in nearly all photon absorptions in the visible or ultra-violet region.

Okay that makes sense then, but is the premise of statistical thermodynamics then wrong?

Wrong ? What sort of "premise of statistical thermodynamics" are you speaking of ?

The geometry  of the ground state and excited state are different so that transitions from any ground state vibrational level to any excited state one are allowed (supposing also spin allowed). The intensity of these vibronic transitions are determined by the wavefunction overlap which directly produces each FC factor. btw in the gas phase at low pressures there can be no collisions (during several excited state lifetimes) so that emission occurs from the initially excited level.

As @Maurice says, you are messing up your Statistical Thermodynamics.  When describing the population of the molecule in the ground state, it is usually assumed that the different molecular degrees of freedom are uncoupled so the electronic, vibrational, rotational and translational molecular degrees of freedom can be treated separately. Normally, the energy of the excited electronic excited electronic states is much higher than that of the ground state, in a good approximation the upper state is empty, and this state can only be accessed through photon absorption.

The absorption intensity is related to the initial population of the initial state, so there is no absorption from an empty state.
However, the absorption intensity also depends on the transition probability which depends on the wave functions of the excited state and the ground state and their transition moment. There is no need for Statistical Thermodynamics here.

A bit of a necro here, but I think one key thing missing in these comments is that the vibrational and electronic degrees of freedom being decoupled (approximately) does not mean that they are separate states. The states of the system would have to specify all of the energetic contributions, no matter how you actually found them. So transitioning between different electronic and vibrational levels simultaneously is not only allowed, it is what you would expect to happen by default.