text:
Solubility of Ag2S in excess sodium thiosulphate solution
The question is pretty much self-explanatory,I wanted to know whether Ag2S dissolves in excess hypo solution.Given that we know AgBr dissolves in hypo forming a complex I expected the same complex to form here,but a question in a textbook I follow says it isnt.I'd like to know why.

The ion $\ce{Ag+}$ may react with $\ce{S2O3^{2-}}$ or with $\ce{S^{2-}}$ according to
$$
\begin{align}
\ce{Ag^+ + 2 S2O3^{2-} &-> [Ag(S2O3)_2]^{3-}} &\quad K_1 &= \pu{1.7E-13} \label{rxn:1}\tag{R1}\\
\ce{2 Ag^+ + S^{2-} &-> Ag2S} &\quad K_\mathrm{s} &= \pu{6.3E-51} \label{rxn:2}\tag{R2}
\end{align}
$$
So, if the reacting ions $\ce{S2O3^{2-}}$ and $\ce{S^{2-}}$ are both in concentration $\pu{1 M}$, the remaining ion $\ce{Ag+}$ would have a residual concentration equal to $\pu{6E-14 M}$ according to \eqref{rxn:1}, and $\pu{8E-25 M}$ according to \eqref{rxn:2}. So as soon as $[\ce{Ag+}] < 10^{-13},$ the ion $\ce{Ag+}$ does not react according to \eqref{rxn:1}, but it reacts according to \eqref{rxn:2}.
When dissolving pure $\ce{Ag2S}$ in water, the concentration of $\ce{Ag^+}$ is
$$c(\ce{Ag^+}) = \left(\frac{\pu{6.3E-51}}{4}\right)^{1/3} = \pu{1.1E-17 M},$$
which is much too low to produce any reaction according to \eqref{rxn:1}.

In aqueous solutions, $\ce{Ag2S }$ is practically insoluble:
$$\ce{Ag2S (s) + H2O (l) <=> 2 Ag^+  (aq) + S^{2-} (aq) }  \tag{1}$$
with
$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{S^2-}] = \pu{6.3E-50} \tag{2}$$
When excess of aqueous $\ce{S2O3^2-}$ is added, following complex formation reaction would taken place:
$$\ce{Ag+ (aq) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq)}   \tag{3}$$
with
$$K_\mathrm{f_1} = \frac{[\ce{[Ag(S2O3)2]^3-}]}{[\ce{Ag+}][\ce{S2O3^2-}]^2}= \pu{2.0E13} \tag{4}$$
The net reaction can be obtained by the sum of the equations $(1) + 2 \times (3)$:
$$\ce{Ag2S (aq) + 4 S2O3^2- (aq) <=> 2[Ag(S2O3)2]^3- (aq) + S^2- (aq) } \tag{5}$$
Thus,
$$K_\mathrm{net} = \frac{[\ce{[Ag(S2O3)2]^3-}]^2[\ce{S^2-}]}{[\ce{S2O3^2-}]^4}= \frac{[\ce{[Ag(S2O3)2]^3-}]^2}{[\ce{Ag+}]^2[\ce{S2O3^2-}]^4} \times [\ce{Ag+}]^2[\ce{S^2-}] \\ = K_\mathrm{f_1}^2 \times K_\mathrm{sp} = \left(\pu{2.0E13}\right)^2 \times \pu{6.3E-50} = \pu{2.5E-23}$$
Thus, adding an aqueous solution of $\ce{S2O3^2-}$ in excess will produce a higher concentration of the complex ion in solution according to the equation $(5)$ and thus the solubility of $\ce{Ag2S}$ will be much higher in an aqueous $\ce{S2O3^2-}$ solution $(K_\mathrm{net} = \pu{2.5E-23})$ than that in water $(K_\mathrm{sp} = \pu{6.3E-50})$.

Note: $(K_\mathrm{sp} = \pu{5.0E-13})$ for $\ce{AgBr}$:
$$\ce{AgBr (s) + H2O (l) <=> Ag^+  (aq) + Br- (aq) }  $$
Thus, when excess amount of aqueous $\ce{S2O3^2-}$ is added to $\ce{AgBr}$, same complex formation reaction would taken place:
$$\ce{Ag+ (aq) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq)}   $$
Similar to $\ce{AgS}$, The net reaction can be obtained by the sum of above equations:
$$\ce{AgBr (aq) + 2 S2O3^2- (aq) <=> [Ag(S2O3)2]^3- (aq) + Br- (aq) } \tag{5}$$
Thus,
$$K_\mathrm{net} = \frac{[\ce{[Ag(S2O3)2]^3-}][\ce{Br-}]}{[\ce{S2O3^2-}]^2}= \frac{[\ce{[Ag(S2O3)2]^3-}]}{[\ce{Ag+}][\ce{S2O3^2-}]^2} \times [\ce{Ag+}][\ce{Br-}] \\ = K_\mathrm{f_1} \times K_\mathrm{sp} = \pu{2.0E13} \times \pu{5.0E-13} = \pu{1.0E1}$$
Here, $K_\mathrm{net} = \pu{1.0E1} \gt 1$ favoring forward direction, and therefore almost all of $\ce{AgBr}$ would dissolve.
Source for $K_\mathrm{f_1}$: Source-1 and Source-2


text:
How to make standard sample for NMR signal-to-noise ratio
I want to make a sample for proton NMR signal-to-noise-ratio measurement. It is conventional to use 1% ethylbenzene in d-chloroform, as specified here.
I have two questions:

What is meant exactly, by 1% ethylbenzene? In some places, I see the notation (v/v), which I take to mean 1% by volume. However, I am still not clear what this means. Does it mean that an amount of ethylbenzene should be used that would fill 1% of the final volume when the solvent is not present? Or does it mean to fill 99% of the final volume with the solvent and then add ethylbenzene until I reach the total volume? Or something else?

What is a good procedure for making such a solution? I imagine a volumetric flask would be useful, but should the solute be measured out by weight? Or using a pipette? I only need about 10 mL of the final solution.

10 ml of final solution by 1% v/v should contain 0.1 ml Ethyl benzene.

My personal opinion is that it's better measure out the solute by weight, because common lab weight machines have more precision (4 decimal places i.e. 0.0001 g) than common pipettes (~0.1 mL)

Ask the manufacturer for a standard suitable for the experiment and spectral range in mind (example) or a befriended NMR group because eventually, these are flame sealed (no evaporation of the solvent across the cap). I saw strychnine in multiple places (equally to anticipate the time of recording necessary) to run a test according to ASTM E2977 and the MNova program.

It is often much easier to buy these standards - most major chemical/NMR suppliers have them. Note also that the S/N standards are field dependent: a 1% EB standard is for lower fields (<300MHz by memory). Higher fields use 0.1%EB. The preparation of such standards follows the same protocols as for the preparation of any quantitative solution, but it is always easier to make them on a larger scale and only use the volume you need.

Thanks for these comments. I will use the tabulated density of ethylbenzene and a balance to fill 1% of the volume of a volumetric flask with ethylbenzene, then add solvent to the graduation mark.

For small quantities of a liquid, you may fill your volumetric flask to about four fifths with with your solvent "X".  Then record the mass difference of a small syringe a) filled with toluene, b) after disposing the toluene in "X" as the mass of toluene added.  Eventually fill the flask to the mark with solvent "X".  This allows you to work rapidly, because otherwise quite some of toluene may evaporate from the flask / a Pasteur pipette until the stem solution is prepared.

Toluene <—> Ethylbenzene. This sounds like a good idea, to measure the mass change of a small syringe rather than the mass change of the flask. I don’t understand why the standard is not specified by molarity. It would give the same result but one would not have to know the density of the solute.

Oh I see my error. You are spot on to object that toluene is not ethylbenzene.  The practice to put a small syringe (e.g., the about 1 mL) on a balance is a technique borrowed from determining manually water concentrations by Karl-Fischer, or organometallic / Grignard reagents against e.g., diphenyl acetic acid.  Of course you may reach out to machines for the former, or record a no-D NMR for the later (e.g., this Hoye paper) if you may record the spectrum by yourself.

source: Stack Exchange

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

sort: votes


text:
When does a piston stop in an irreversible process?
Suppose we have an ideal gas filled inside a container fitted with a piston. The external pressure is equal to internal pressure.
If we suddenly decrease the external pressure, we say that the gas will expand until the internal pressure and external pressure become equal.
Considering an irreversible process, shouldn't the gas expand beyond that point because by that time the piston will have gained some velocity?

You're right.  The piston can gain kinetic energy and overshoot its equilibrium position, but then, the force of the gas on the piston will drop until the piston comes back the other way.  The piston will undergo an oscillation back and forth, but this oscillation will be damped as a result of viscous dissipation stresses in the gas.  Eventually, the oscillation will decay until the piston stops.  At this point, the gas pressure will be in equilibrium with the lower pressure outside.
If I do a force balance (Newton's 2nd law) on the piston, I get
$$F_\mathrm{g} - p_0 A = m \frac{\mathrm{d}v}{\mathrm{d}t},$$
where $F_\mathrm{g}$ is the force the gas exerts on the piston, $p_0$ is the outside pressure, $A$ is the piston cross sectional area, $m$ is the mass of the piston, and $v$ is the piston velocity.  If I multiply this equation by the velocity of the piston $v = \mathrm{d}x/\mathrm{d}t$, and integrate with respect to time, I obtain:
$$
W_\mathrm{g}(t)
  = \int\limits_0^{x(t)} F_\mathrm{g}\,\mathrm{d}x
  = p_0(V(t) - V(0)) + m\frac{v(t)^2}{2},
$$
where $W_\mathrm{g}(t)$ is the work done by the gas on the piston (its surroundings) up to time $t$, $V(t)$ is the gas volume at time $t$, and $m\frac{v(t)^2}{2}$ is the kinetic energy of the piston at time $t$. At infinite time, the piston is no longer moving, so its kinetic energy is zero at final equilibrium. Therefore, at final equilibrium, the work done by the gas is just
$$W_\mathrm{g}(\infty) = p_0\Delta V,
\text{ where } \Delta V = V(\infty)-V_0.$$

If we use damping factors to explain these points, then shouldn't we count them too while calculating work? We only count external pressure.

No.   That is internal.  The net work done on the surroundings when the final equilibrium state is attained is just the outside pressure times the volume change.

BTW, I can prove that if you want to see it.

Please show it, if it can be done using only high school calculus.

See the ADDENDUM

And what if we want to calculate work on the system? Can we neglect it then? Thanks.

The work done by the surroundings (the piston) on the gas (the system) is minus the work done by the gas on the surroundings.

Let us continue this discussion in chat.

@ChetMiller Is this description limited only for irreversible process? Could we use the same justification even for a reversible process with the only difference that the pressure difference is infinitesimal?

@ado sar Sure, as long as you recognize that, for the reversible case, the external pressure cannot be held constant, and has to be decreased gradually.

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


text:
Fluorochrome (fluorescein) emission decay when linked to macromolecule
I measure the intensity and polarization of light emitted from a blood sample that contains excited fluorescein (experiment known as fluorescence polarization assay). The excitation is done, as usual, with vertically polarized light.
In case the sample has certain macromolecules of interest (an antibody, in this case), the fluorescein links to the much-larger macromolecules, yielding a higher polarization factor, giving a positive- or "sick"- result. Alternatively, if there is no antibody, the excited fluorescein randomly rotates and then emits light, and the polarization factor is smaller giving a negative-or "healthy"- result.
The question:
I find that the light intensity (in both vertical and perpendicular polarizations) decreases for the positive case. Why is that so? Could it be that a non-radiative process like FRET acts stronger for fluorescein linked to macromolecules than for free fluorescein?
EDIT 1: I have used two different intensity detection systems, PMTs and SiPMs, and got similar behaviours.

There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission.  When bound to the protein fluorphores are in a different environment and this can affect their fluorescence yield, ANS (anilino naphthalene sulphonate) is a good example.  The same is probably happening in your case, you know the motion is restricted as the polarisation is greater so there has to be some interaction.

Think that many good fluorescent molecules cease to be so when in solid state or strongly interacting with a solvent. Non radiative channel are opened. To me, what you've described and taking aside fine details, should be a quite general behavior. In fact looking at polarisation and not only intensity should be just for better sensitivity.

@Alchimista Thank you. The fluorescein-solvent interaction for sure gets stronger with the sample positivity - the more antibodies, the more linking. I am not sure what are these "non-radiative channels"?

@nicolas005 this is far more general than in your biological oriented experiments. An excited species can collide and be de-excited. Now think of something adsorbed or solvated or anyway interacting. Decay can happen through vibrational levels instead that radiatively. At the end the energy goes off as heat.

Specifically to the point about FRET see below. I upvote that.

There should be no FRET for the free fluorescein unless you have specifically added an acceptor and in any case this would reduce the fluorescein emission.  When bound to the protein fluorphores are in a different environment (e.g. pH  change, hydrophobic pocket, tight binding, polar/non-polar nearby residues etc.)  and this can affect their fluorescence yield and lifetime up or down, ANS (anilino naphthalene sulphonate) is a good example.  The same type of thing is probably happening in your case, you know the motion is restricted as the polarisation is greater so there has to be some interaction. Fluorescein emits at a relatively long wavelength so aromatic protein residues cannot act via FRET as they absorb in the UV. You would need a chromophore (such as heam/chlorin/porphyrin) that absorbs in the green to red region to be a FRET acceptor from the fluorescein and so reduce its intensity.

Good to know that FRET should not be the cause of the intensity decay. Thank you for the answer.


text:
N-N bond length in Dinitrogen Tetroxide
The $\ce{N-N}$ bond length in $\ce{N2O4}$ is $\pu{1.64 Å}$ which is much longer than in the case of Hydrazine $\ce{N2H4}$ $\pu{(1.47 Å)}$. This means that the bond is quite weak. Another unusual case I see that $\ce{N-N}$ bond length in $\ce{N2O3}$ ( the blue asymmetric form ) is $\pu{1.86 Å}$ !!
That is very weak for a Nitrogen single bond. Is there a satisfactory explanation for the same? And amazingly molecules with such long bonds tend to have a nonplanar structure due to bond rotation but $\ce{N2O4}$ is planar.
I found a similar question on N2O4 but the answer to the planarity question wasn't so satisfactory. Also, I wanted the extend upon the weak bond question.

Related: https://chemistry.stackexchange.com/questions/133751/why-is-dinitrogen-tetroxide-a-planar-molecule

also https://chemistry.stackexchange.com/questions/91233/why-is-the-n-n-bond-length-greater-in-dinitrogen-trioxide-than-in-dinitrogen-tet

https://chemistry.stackexchange.com/questions/133085/the-source-for-the-n-n-bond-dissociation-energy-of-240-kj-mol-in-the-table

https://crawford.chem.vt.edu/wp-content/uploads/2019/11/The-weakly-bound-dinitrogen-tetroxide-molecule_-High-level-single-reference-wavefunctions-are-good-enough.pdf might be of interest

Images are from Wikipedia.
The first obvious thing that you can see is that N2O4 is planar but N2H4 is not. And the primary reason starts here.
N2O4 can be viewed as two nitrogen dioxide molecules joined together. If you draw a Lewis structure of NO2, you will recognize two things. Firstly, it has one unpaired electron. So imagine when two NO2 molecules come near together, they will share their two unpaired electrons (like how radicals react that you might be familiar with in organic chemistry) to form a single sigma bond. This is the very bond you are asking about.
And secondly, it is sp2 hybridized at the nitrogen atom. Thus, when you join the two NO2 molecules together, taking the relative orientation of the sp2 orbitals and how they should overlap to form a sigma bond in mind, you can see why N2O4 is planar. Because the N=O bonds (I write the "=" symbol for double bond for simplicity, it should be about 1.5 order due to resonance) heavily repulse each other when they are in the planar position, it weakens the N-N bond. Indeed this bond is so weak that it is homolytically dissociated at near below room temperature or above, creating basis for the popular equilibria between NO2 and N2O4.
The situation is even worse for N2O3. N2O3 can be viewed as formed from a NO (itself has an unpaired electron and a sp2 nitrogen) and a NO2 molecule, with the exact mechanism as above. And guess what? N2O3 has a lone electron pair on the nitrogen (in place of the missing oxygen compared to N2O4). Lone pairs repulse much, much more heavily than the bonding pairs do, and this further weakens the N2O3 N-N bond. This is illustrated by the easily disproportionation of N2O3 into NO and NO2.
Then why N2H4's bond is stronger? Because the nitrogen atoms are sp3 hybridized in this case (note that oxygen is divalent but hydrogen is only monovalent), thus they form a tetrahedral electronic structure. Thus they can comfortably adopt a skew conformation to minimize the repulsion. Either N2O3 or N2O4 could achieve this because the sp2 hybridization confines them to planar geometry.
Using Pauling's formula with the 1.14 A triple bond in N2 as a reference:
D(n) = D(m) - 0.6log(n/m)
With D(n, m) is the length of the bond with order n, m respectively.
The N-N bond order in N2O4, N2O3 and N2H4 can be estimated to be 0.44, 0.19 and 0.85 respectively.

"you can see why N2O4 is planar - Sorry, but no I can't, especially if the N-N bond is purely a sigma bond. Further the iso-electronic oxalate ion can be found in both planar and non-planar forms (https://en.wikipedia.org/wiki/Oxalate). I agree that iff N2O4 and similar are planar then repulsion between the oxygens is a good hand waving argument to explain the long bond, the problem is that very argument suggests to me that it should not be planar in the first place. So could you expand on why this explains the planarity of the species in the first place?

I'm seeing $D_{\rm 2h}$ symmetry for the $s$-$\ce{N2O4}$ dimer here - as to the why, that's another kettle of fish.

@Ian Bush this paper might help: https://onlinelibrary.wiley.com/doi/abs/10.1002/jcc.540020106

@ĐỨcLêHồng Unfortunately where I work doesn't let me through the paywall. But the abstract seems to confirm my own rationalisation for why N2O4 is planar: "Experiment says it is planar. Good level theory says it is planar. Accept it - Chemistry is difficult, hand waving doesn't solve everything"

@ĐỨcLêHồng Thanks for the opinion, that bond order thing makes it quite clear.


text:
How to make cyclohex-4-ene-1,2-diol from cyclohexa-1,4-diene?
As the question says, I'm supposed to make cyclohex-4-ene-1,2-diol using cyclohexa-1,4-diene:

I can use any inorganic reactant, any solvent, and any organic molecule with 3 or fewer carbons. I am not sure how to go about this question, as I do not know how to use regional-specific reactions. Any help, even just mentioning a reaction mechanism I can use for this would be greatly appreciated.
I could put the cyclohexa-1,4-diene on $\ce{KMnO4}$ for a catalyst and $\ce{H2O}$ with $\ce{HO-}$ as solvents in order to break the double bond and replace it with the $\ce{HO-}$ molecules. Problem is that this would affect the molecule as a whole – not just a specific side which is what I need.
I also thought of using bromine as something temporary, since they are easy to remove however, again, this affects the whole molecule.
I also thought about Diels-Alder reaction, but we haven't learned anything about that so I do not know how to use it- much less on a cyclic molecule.
I have also seen the osmium tetroxide oxidation of alkenes. However, I still have the problem that I can't control this reaction to only affect one side.

It doesn't matter you're not sure. Write some idea even if it wouldn't work.

I mean, I could put the cyclohexa-1,4-diene on $\ce{KMnO4}$ for a catalyst and $\ce{H2O}$ with $\ce{HO}$ as solvents in order to break the double bond and replace it with the OH molecules. Problem is that this would affect the molecule as a whole- not just a specific side which is what I need. I also thought of using Bromine as something temporary, since they are easy to remove however, again, this affects the whole molecule. I also thought about Diels-Alder, but we havent learned anything about that so I do not know how to use it- much less on a cyclic molecule.

You  should edit it into the question.

Ok, I added that to the question.

Why not just use one equivalent of $\ce{KMnO_4}$?

I think if you use 1eq of KMnO4 you will still get some tetra-ol, the diol can and will react. The mixture will be separable and maybe this is how you have to do it.

Use limited amount of KMnO4, it will give you a good enough yield of the diol.

I think the easier way of doing this is to epoxidise the diene with 1 eq of a peracid epoxidation with peracids . The standard reagent m-Chloroperoxybenzoic acid (MCPBA) breaks your rule of organic molecules with 3 carbon atoms or fewer, but you can also do it with peracetic or pertrifluoroacetic acid preparation of peracetic acid. You will inevitably get some di-epoxide, but this will be separable by chromatography. The mono-epoxide can then be reacted in mild aqueous acid to give the diol. You could also carry the mixture through to the aqueous acid treatment as the tetraol will be far more water soluble than the diol and can be removed by an aqueous wash.

You mentioned the use of $\ce{OsO4}$ to form the enediol. The secret is not to perform the osmylation catalytically ($\ce{OsO4/H2O2}$) but rather stoichiometrically in ether. The initially formed osmate ester often precipitates from solution thereby suppressing overoxidation. The osmate ester may be decomposed with $\ce{H2S}$. The downside --- prohibitively expensive.

In addition OSHA will be concerned about the toxicity of $\ce{OsO4}$ (reference).

@Buttonwood: The OSHA concern goes without saying. A one gram ampoule of OsO4 is ~ 4 mmol.

What is the desired stereochemical outcome of the product?

For cis-diol, use 1/2 eq KMnO4/NaOH
For trans-diol, use 1/2 eq MCPBA followed by catalytic H2SO4/H2O (excess H2SO4 will cause protonation of the π bond and lead to undesired polymerization and other unwanted side products)

Does this help? :)

Item 1: What prevents overoxidation. One is bound to get tetraol and recovered diene. also. Item2: Why wouldn't protonation of the double bond then lead to hydration?


text:
Why do galvanic cells need a salt bridge while electrolytic cells don't?
I see the questions of why a galvanic cell needs a salt bridge and why a electrolytic cell doesn't need a salt bridge being answered separately but I still can't grasp why you are not allowed to mix or interchange the reasons of the two. Can someone explain why both cells are set up in that way and provide a comparison?
From what I've read, it has to do with some of the following differences but I can't seem to put it together into a common picture.

A galvanic cell needs to prevent electrolyte mixing in order to produce current. Of course electrolytes need to be connected in order to maintain neutrality. But instead of a direct connection such as using the same electrolyte for both electrodes, the electrolytes for the anode and cathode are separated but connected by a salt bridge in order to maintain this neutrality but prevent the electrolyte mixing by having ions pass at a rate not too high. Could someone explain again what electrolyte mixing has to do with the current? Meanwhile since in an electrolytic cell, the chemical reactions are what is being produced the electrolytes can be mixed.
The electron supply in both are different. For a galvanic cell, it is supplied by the species getting oxidized. For an electrolytic cell, the external battery supplies the electrons.
I should also reiterate here how galvanic cells have spontaneous chemical reactions and do not need the external voltage supply while electrolytic cells have nonspontaneous chemical reactions and need the external voltage supply.
For a galvanic cell, the anode is negative while the cathode is positive. For an electrolytic cell, the anode is positive while the cathode is negative. However, it should be noted that by definition an anode always has oxidation while a cathode always has reduction.
It should be noted that the flow of electrons is the same when based on the charge of the electrode. Electrons always move from the negative electrode to the positive electrode. For galvanic cell it is anode to cathode while for an electrolytic cell it is cathode to anode.

It's not like I can't understand the two concepts of the galvanic and electrolytic cells separately. But I want to be able to fully understand the comparison and contrast on why you can't just interchange the two and how their many opposites and differences have something to do with that.

Look at my drawing of a Daniell cell: https://chemistry.stackexchange.com/a/118744/79678. If you get rid of the salt bridge and just use one beaker with a mixture of the two solutions, then zinc metal will spontaneously be oxidized by the copper ions, thereby wasting the available free energy as heat, instead of getting electrical energy via the external electrode pathway. This is trivially simple to demonstrate: place a zinc metal strip in a copper sulfate solution. Zinc gets oxidized and copper plates out, usually as a poorly adherent coating.

Think of a typical battery cell. If you discharge then it is a galvanic cell, if you are charging it then it is electrolytic cell. Thus your supposition isn't true.

Salt bridges or equivalents keep cells from more or less fast reaching the chemical equilibrium. While electrolysis can be done any time, even for systems in equilibrium, systems in equilibrium do not provide electromotoric power, their voltage is zero.


text:
What does it mean to "shear" an emulsion?
The following instructions are given for a preparation and I am wondering what it means to "shear an emulsion"?

The water phase components and the oil phase components are each mixed
in separate containers and filtered and the refractive index of each
is measured. The refractive index of the water phase is adjusted to
match the refractive index of the oil phase to within 0.0004 by
addition of water or propylene glycol as required. The water phase is
then slowly added to the oil phase at about 18.degree. C. with
sufficient mixing to form a clear emulsion with minimum aeration. This
emulsion is then sheared to form a clear gel with a viscosity of
about 130,000 to 160,000 cP.

Check what is meant by shear forces in fluid mechanics. Hint: also see shear thickening and shear thinning.

While fluid mechanics is part of curricula in chemical engineering, note that engineering.se may offer a more receptive audience for this question.  See this tag as an example.

Search for "shear mixer" and you'll get a good idea eg https://en.wikipedia.org/wiki/High-shear_mixer

You may also want to ask "why shear mix an emulsion?" Look into the Wikipedia article linked above.

An emulsion is made of small droplets in suspension in another liquid. Usually it is not stable. After some time, the two phases get separated. But if the mixture has been sheared, the emulsion is stable like milk and will not get separated.


text:
Does PubChem have stereocenter location data?
I have spent the last couple of days learning to navigate, find useful information and use the download service on the PubChem website. One of the pieces of information you can find is a stereocenter count, but I can't seem to find anything about the stereocenter location.
This might just be something that isn't available, but since I find parts of the PubChem website (specifically the download services), a bit unintuitive it might just be that I missed it. Therefore my question is this: is information about the stereocenters' location available in PubChem? Also, if it's just something that is not available in PubChem, are there any similar websites where it is available?
Note: the format doesn't really matter - it can be in terms of coordinates of the 2D-image or the number of the carbon counted from the highest priority functional group or whatever.

Doesn't InChI have stereocentre data?

First, let me be pedantic: it's 'stereocenter', not 'chiral center'. Chirality is a property of the whole molecule, not of individual atoms or groups. Then, see: https://pubchem.ncbi.nlm.nih.gov/compound/4369359#section=Isomeric-SMILES . SMILES strings contain '@' symbols on atoms with explicit stereochemistry. You would however miss stereocenters that are not specified. But I am not sure what you are actually after. What is the 'location' of a stereocenter? Depending on how you number the atoms, the list of atoms that are stereocenters could change, so what is the value of such a list?

See also: https://www.rdkit.org/docs/RDKit_Book.html#identification-of-potential-stereoatoms-stereobonds

And this: https://en.wikipedia.org/wiki/Chemical_table_file#Bond_block_specification


text:
What is the composition of diet coke?
I've been trying to find the values of any sort of percentage composition of diet coke and I haven't found any luck other than one interviewee (Dr. MAUREEN STOREY [Senior Vice President, Science Policy, American Beverage Association]) who states

Soda is comprised mostly of water. A full-calorie soft drink has 90 percent water, and a diet soft drink is 99 percent water. Water is the most important nutrient that we have...

And that was all the information I have.

Is not energy content published on any drink bottle/can ? 1 g of sugar is equivalent to about 4.3 kcal, or about 4.2 times more kJ. // Referring cal as calorie ( "small calorie") and kcal=1000 cal as Calorie ( "big calorie" ) is very old and deprecated way to do so. Calorie itself is a deprecated, but tolerable unit among old school chemists.

Soda being 99 percent water is high questionable. Sweet sodas are at about 30% sugars by mass (!) Diet zero soda might approach that as for the CO2 content was calculated here in SE to be a few gram per liter soda.

I edited your question. One would not "boil a soda" to find it's caloric content. You might not understand what is meant by calorie content of a soft drink. The caloric content refers to the energy content (in kcal) of nutrients that a person can use to derive energy, for instance fat, carbs and even protein. Diet soda is diet soda because the sweeteners have a much higher sweetness than sugars such as sucrose or glucose. That means you hardly need sweetener, and therefore energy content is very low.

It is a pity the question is closed. Because the formula of Coca-Cola contains 16 ingredients. They are described in page 422 of the book by Mark Pendergrast, "For God, Country and Coca-Cola", Phoenix, London 1994. ISBN 1 85799 180 X. There is not enough room in this "Comment" to be more precise and give more details.

@Maurice is the fact that the recipe is treated as industrial top secret a legend?

@Alchimista. The Coke formula is known by 3 persons. In Coca-Cola factories, the constituants are called "Material X1 or M2". So the workers mix unknown liquids. The discovery of the formula is a rather long story covering about 10 pages, where the author explains that its research took many years. He said at the end that he told the Head of the Company that he knows the formula. The President was not impressed. He said it is not easy to copy the formula, because for example one of the constituant is described by "oil of unripe orange flowers". But what does it mean : "unripe flower" ?

The water content was bracketed as 99.3% on the ingredients list for some cans being sold in the UK.

Water and bad stuff.

The major ingredient in Diet Coke is water (H$_2$O). The second ingredient is carbon dioxide (CO$_2$), about 30 psig, or three times atmospheric pressure. (I measured it with an automobile tire gauge.) Since CO$_2$ is soluble about 1.7 g/Kg at 20$^o$C at atmospheric pressure, Ref 1, this is about 5 g/L, or ~0.5%, using a density for Diet Coke of 1 g/L Ref 2.

So, in the list of ingredients for Diet Coke, Ref 3, everything will be between a maximum of 0.5%, down to about 0.01295% ("46 mg caffeine/12 fl oz" on the label of my 2L bottle).
And, as Matt Black noted, the recipe varies from country to country, and from time to time (and is noticeably different in cans vs fountain delivery, according to those with highly developed preferences), but the big impacts are provided by
#1. water
#2. carbon dioxide
#3. artificial sweetener
#4. caffeine and
#5. flavoring, which might include a little more acid, like citric or phosphoric.
Seltzer is just #1 plus #2; many caffeine-free beverages just leave out #4.
Ref 1. https://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html
Ref 2. http://web.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/36.34.html
Ref 3. https://www.coca-cola.ca/brands/diet-coke; see also https://www.huffpost.com/entry/coke-zero-sugar-taste-difference_n_597f71d7e4b00bb8ff3875d9 for varieties of Diet Coke

This needs a bit a MathJax love. On my phone the formatting breaks. But I'm not going to try to fix this using my phone... It'll only become worse.

The major ingredients in diet coke vary by country but none are major components compared to sugar in full-fat drinks
The exact ingredient list of Diet Coke is not publically available but most of the flavourings are present in very low quantities. Also, the exact recipe varies over time and across countries (this is claimed to be an adaptation for local tastes). So it is hard to be exact
The major difference between a ~330 mL can of Diet Coke and normal (sugary) Coke is huge. Sugary Coke has about 40g of sugar for reference (other ingredients will be present in amounts similar to Diet Coke). The major components are:

sweetener (usually aspartame but others are sometimes mixed in at similar levels): ~100mg
caffeine: ~40mg
citric acid (or citrate): ~50mg
phosphoric acid (or phosphate): ~100mg

The total mass of water, for comparison, is about 330g (330,000mg) so there really isn't much of any of those key ingredients relative to the water.
This is very different to the sugary version of the drink where the sweetener is more than 10% of the total mass of the drink.

Full fat or sugar rich?

@Alchimista In normal, casual conversation "full fat" is a synonym for normal, sugary Coke.

ah focus is on the side effects. Got it.


text:
Application of statement: extent of overlapping is always less than 50% because of electronic repulsion
Question from the online class:

Can two s-orbitals overlap as shown?

Dotted circles in diagram depict maximum probability region of s-orbitals in ideal situation when atoms A and B are not bonded. Black dots denote nuclei.

Answer by teacher: no because extent of overlapping should always be less than 50 %.
My answer and argument:

Practically the two spheres are not overlapping 50 % mathematically even if the nucleus are on radius of each atoms (as shown in diagram).

Van der Waals radius of hydrogen is $\pu{120 pm}.$ Bond length of $\ce{H-H}$ bond is $\pu{74 pm}$ in hydrogen molecule. The above two data clearly states that nucleus of hydrogen atom lies inside the sphere of another hydrogen atom (as bond length is less than two times the radius of an atom).

Second argument provides a supportive  example (for my answer) of hydrogen molecule that the overlapping shown in diagram (infact better than diagram) is possible. So, given diagram is correct or at least possible.
What is right at undergraduate level? Even a small answer (with authentic source if possible) is also welcome because question itself contains much opposing reasons.

To answer the question, you have to first define what those circles actually mean? s-orbitals are just wavefunctions i.e. a probability distribution, you cannot directly draw them. You can draw a surface that covers some percentage (like 95%) of probability of finding the electron.

Uhhh. The problem isn't merely geometric. In the macroscopic world you are asking if two football balls can superpose each other and be in compenrtration. Electrons comes with charge, nuclei too. This means there is a potential requisite to fulfil. That is what quantum mechanics do. At this stage better trust sources (though I have no idea about the 50% figure).

@Jay because in doing so you are again in a merely geometric scenario. The VdW radius of H isn't much of a help discussing H-H. It can be seen as a limit of closest approach. But it does not say much in case the two bond togheter. Take the discussion about overlap with a grain of salt. For what matters, the message should be that s orbitals can have good overlap independent of orientation. Moreover the figure 50% or whatever wouldn't be reflected by those distance. We are dealing with "balls" (assuming the draw means the percentage bla bla...) at the end.

You're confounding vdW radius with atomic radius. for vdW it's a distance to a neighboring nucleus

If you all want I shall delete this post (just asking to confirm)

What is the source of the idea that 50% overlap is max? The argument sounds electrostatic (Coulomb repulsion and attractions balancing). In general sounds like a crude approximation to real H2. Invoking vdW radii can lead to problems since these are imprecise. They might be from say crystal structures.


text:
Which oxidation states were used when Pauling developed his electronegativity scale?
Paulings electronegativity is a relative scale, based on the difference in electronegativity between X and Y, $\Delta EN = 0.102 \sqrt {\Delta}$, where $\Delta = (X-Y)_{measured}-(X-Y)_{theoretical}$ bond energies.
But what is the oxidation state of X and Y? Is it an average of the common oxidation states for each element, or is it simply the most common oxidation state? Or do I have it all wrong to begin with?

Wikipedia Electronegativity article has a section dedicated to this issue. A single valued EN, whatever defined, it is a simplification nevertheless good in many instances. It is expected that one comes here when something isn't clear or covered by very easy to find and more or less reputed sources.

As far as I can tell, the wikipedia article only mentions that EN changes with oxidation state, and that Allred Rochow calculated some EN-values based on averages of some transition metals. I can't find anything about what oxidation state Linus Pauling himself used for his EN-values..

Basically a value for each oxidation states he deal with. But the article should satisfy your question nevertheless. There is no a periodic table electronegativity in reality. Yet, that or other derived values proved useful.


text:
What is the dissociation percentage of a polybasic organic acid at a certain pH?
For my Master's Thesis I am working with the dicarboxylic acid muconic acid $(\ce{H2MA},$ $\mathrm{p}K_\mathrm{a1} = 2.9,$ $\mathrm{p}K_\mathrm{a2} = 4.0).$ It thus dissociates into $\ce{HMA-}$ and $\ce{MA^2-},$ respectively.
I am aware of how to calculate the $\mathrm{pH}$ and the subsequent dissociation percentage when dissolving a weak acid in water. But I am wondering how you can calculate it the other way around. Say I dissolve $\pu{1 g}$ of the acid in water and adjust the $\mathrm{pH}$ to 4 with a strong acid such as $\ce{H2SO4},$ how do I know what the concentration is of each species $(\ce{H2MA},$ $\ce{HMA-}$ and $\ce{MA^2-})$ at that given $\mathrm{pH}?$
In general, how can I calculate the dissociation percentage of a weak polybasic organic acid when the pH is given?

Hi, Welcome to Chem.SE! Please note the edits. About the question, if the pH is fixed, then you have $\ce{[H+]}$. You have the initial concentration of the acid which will be the total of $\ce{[H2MA] +[HMA-] +[MA^2-]}$ in the solution. Write the equations for protonation/deprotonation equilibria and then use the above, and there should be a system of equations you can solve.

Following on from the detailed comment by @ShoubhikRMaiti, you just compute the dissociation fractions at pH 4 or for a range of pH values. Here is what they look like for phosphoric acid: https://chemistry.stackexchange.com/a/118933/79678.

I remember posting such calculation formulas, perhaps few months ago. See the answer  https://chemistry.stackexchange.com/questions/149282/how-to-numerically-model-a-phosphoric-acid-titration-curve

I have read them few times somewhere and derived them once for myself many years ago. Since then I remember them. Just taylor it for just 2 dissociation degrees.

Reused my answer to how-to-numerically-model-a-phosphoric-acid-titration-curve, edited for diprotic cases:
First, calculate the common denominator $CD$:
$$a_1 = [\ce{H+}]^2$$
$$a_2 = [\ce{H+}] \cdot K_\mathrm{a1} = a_1 \cdot \frac {K_\mathrm{a1}}{[\ce{H+}]}$$
$$a_3 = K_\mathrm{a1} \cdot K_\mathrm{a2} = a_2 \cdot \frac {K_\mathrm{a2}}{[\ce{H+}]}$$
$$CD = a_1 + a_2 + a_3 $$
.. and then reuse it and its additive terms:
$$[\ce{H2A}]=c_0 \cdot \frac {a_1}{CD}= c_0 \cdot\frac{[\ce{H+}]^2 }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$[\ce{HA-}]=c_0 \cdot \frac {a_2}{CD}= c_0 \cdot\frac{ [\ce{H+}] \cdot K_\mathrm{a1} }{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$[\ce{A^2-}]=c_0 \cdot \frac {a_3}{CD}=c_0 \cdot \frac{K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + [\ce{H+}] \cdot K_\mathrm{a1} +  K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$c_0$ is the total concentration of respective particular acid forms.
For the general degree of "N-proticity", the expression terms span from $\ce{[H+]}^N$ to $K_1 \cdot ... \cdot K_\mathrm{aN}$


text:
Will Avogadro's number change if definition of atomic mass unit is changed?
I know that the relation between atomic mass unit and Avogadro's number is a reciprocal relation.
My question is if the definition of atomic mass unit is changed, suppose instead of 1/12th mass of carbon-12 isotope, we consider 1/24th mass of carbon-12 isotope, will the Avogadro's number change?
And if Avogadro's number changes will that change the mass of a mole of atoms of any element?

No , it will not. Avogadro number is related to mole as 12 is to dozen.

The Avogadro constant is defined exactly as $N_\mathrm{A} = \pu{6.02214076E23 mol−1}$

See the wikipedia article Avogadro constant which gives an interesting oversight on the constant. The gist is that IUAPC has defined some fundamental constants, including the Avogadro number,  as having exact values which are not expected to change ever again.

Simple and actual answer:
The current definition of Avogadro's number doesn't depend on the unit of mass. They are completely independent quantities. The Avogadro's number is currently simply defined as $\pu{6.02214076e23}$. It's just a number which is important enough to get a special name.
The answer you are probably looking for:
Historically the Avogadro's number was defined in such a manner that it did depend on the unit of mass.
At that time the Avogadro's number was defined as the number of atoms present in $\pu{12 g}$ of $\ce{^{12}C}$ isotope of carbon.
Now as you can see even if we defined an atomic mass unit as $1/24$ of the mass of an $\ce{^{12}C}$ atom instead of $1/12$, it won't make any difference to the Avogadro's number as it was defined on the basis of grams of carbon and not atomic mass units of carbon.
Now coming to your second question of whether this would have changed the mass of a mole of a substance. Yes it would have changed that, all the molar masses would have been double of as known currently.
Note:
All this discussion is just about how someone wrote a definition and the loopholes someone can detect from it using language.
For example if we really would have defined an atomic mass unit in this different way, we would have also defined Avogadro's number as number of atoms in $\pu{24 g}$ of $\ce{^{12}C}$. Even $\ce{^{12}C}$ would have been written as $\ce{^{24}C}$!
The takeaway is that the scientific idea and origin of all these units was(and is) always there  but it's just the nuance of defining it in such a way that it doesn't create such problems was the problem. That's the only reason all the standard units were restandardized recently. It's a sad reality that schools are still teaching the old definitions even after we have said them a good bye.

My understanding is that if we consider Avogadro number to be a constant then mass of 1 mole shouldn't be affected by changing the definition of 1 amu as it would affect only the relative mass of elements. The absolute mass would still remain the same right? So why should the mass of 1 mole change on changing definition of 1 amu?

@dumb We initially defined moles on the basis of gram-equivalents. $\pu{12 g}$ of carbon-12 was selected because we assumed that mass of $\ce{^{12}C}$ to be $\pu{12 amu}$. If it was $\pu{24 amu}$ we would call $\pu{24 g}$ of carbon-12(24) a mole. Nothing is special even with a gram. If they decided at that time to use imperial units than the mole would be defined in that way(eg. Calling 24lb of carbon-12(24) a mole) . A mole of substance really isn't something special, other than the fact we made it the standard one.


text:
Reaction of Tollen's reagent with 7‐oxabicyclo[2.2.1]heptan‐1‐ol

Identify whether the given compound will get oxidised by Tollen's reagent.

The problem here is with the bridged alpha carbon. It cannot pertain sp2 character in the middle of the reaction as it is not permittable according to Bredt's rule.

Is the answer given, yes? I too think you are right, and also, tollen's reagent oxidises only compounds with free CHO group, like aldehydes and alpha hydroxy ketones, so why should this give positive tollen's test?

I agree. I think this will not get oxidised by Tollens

The structure shown is 7-oxabicyclo[2.2.1]heptan-4-ol, which exists as 4-hydroxycyclohexanone, and not the title compound, 7-oxabicyclo[2.2.1]heptan-3-ol. Neither compound will be oxidized by Tollen's reagent (cc. @Waylander).

The answer given was incorrect and even I thought so. Now it's cleared

@AbhishekP If you got your answer, feel free to self-answer it. Self-answering is always encourages and it will help future readers.


text:
Noble gas configuration and valence shell electrons
I know that the maximum number of electrons within a shell is equal to 2n^2. I would think that the noble gases would reflect this but that isn’t always the case
n = 1
2(1^2) = 2, this is the atomic number of He ✅
n = 2
2(2^2) = 8, plus the 2 e from n = 1, this is 10, which is the atomic number for Ne ✅
n = 3
2*(3^2) = 18, this is the atomic number for Ar, but I realized that the 3d orbital is empty. The 3rd energy level should have a total of 18 e and the total amount of electrons until the 3rd energy level should include 10 e from the previous 2. I understand that this is because of the aufbau principle, but it got me confused about how to consider noble gases. I thought they were supposed to have the maximum number of valence electrons, but then I thought what about group 12, Lutetium, and Lawrencium? Just based on their electron configuration, every orbital they have is completely occupied. I know these are very different from noble gases. I’m just wondering why, especially because these elements can be very reactive (right?).
Edit: I forgot to mention that after n = 3, nothing matches up at all.
n = 4
2(4^2) = 32, For Kr, Z = 36
n = 5
2(5^2) = 50, For Xe, Z = 54
n = 6
2(6^2) = 72, For Rn, Z = 86
And lastly, Idk if Oganesson is even considered a noble gas but it’s in the same group so I’ll put it here.
n = 7
2(7^2) = 98, For Og, Z = 118

Noble gases were never about filling everything that is available. They were about filling everything until you need to increase the n.

By increasing n, do you mean in the order electrons occupy orbitals?

Like you keep adding electrons just before the electrons can occupy the next energy level

@lbby Yes, because that's when we say a period in the periods table is changed.

@NisargBhavsar thanks, but then why are they described as having the maximum number of valence electrons?

@lbby Because they do have the maximum number of valence electrons. Note that the electrons filled in d and f group are not filled in the "valence" shell.

Like a wise person once said: don't forget the Aufbau principle... you may want to look at Ben Norris' or Brian's answers in this answer (other answers may go into too much advanced detail). There are many linked posts on the site you may want to browse as well.

The number of orbitals per shell (n) is n2. At most there are two electrons per orbital, each with opposite spin. However, due to nuclear screening, the subshells (n and l), and as a result the orbitals (n, l, and ml) are not of the same energy.
When you get to more complicated situations, such as the third row transition metals, and you see exceptions such as chromium and copper (and their respective group members) having ns1 (n-1)d5 and ns1 (n-1)d10 configurations, the added stability from half-filled and fully filled sub shells outweighs pairing the ns electron as it is more distant from the nucleus and less stabilized by screening.
With lutetium, the configuration is [Xe] 4f14 5d1 6s2; for lawrencium, the configuration is [Rn] 5f14 7s2 7p1.
In terms of reactivity, valence electrons further from the nucleus are easier to remove, so reactivity can vary. With regard to these rare earth metals you mentioned, I am not as familiar and have limited knowledge on them. But, in terms of filling orbitals, you are correct to follow Aufbau. (f will come before d and be (n-1) lower than the respective d, and d will be (n-1) lower than the respective s/p.
Does this help? :)

Thanks for your answer. This helps with some other questions I have but there’s just one part I’m stuck on. The other person here told me that d and f group electrons are not part of the valence shell. I have verified this as true but Idk why this is. If it’s not too much trouble, could you explain please?

Also I forgot to mention that I thought Luterium and Lawerencium were the inner transition metals with f14s2 electron configuration but I was off by one in each case. I was trying to use them as examples to illustrate completely filled orbitals so thanks for pointing that out.

The valence (outer most) shell of lutetium is n = 6 and that of lawrencium is n = 7. Both the d and f orbitals for the respective elements have principal quantum numbers smaller than this. See above discussion.

Thanks for elaborating. That makes so much sense. Idk why I didn’t consider that


text:
Indacene pericyclic mechanism question

I am looking for any help in developing a mechanism for this transformation as well as determining the stereochemistry on the starred centers. I do not see any obvious connections. I was thinking that the second to last step would involve some sort of electrocyclic ring closure since the hydrogen atom orientation is a consideration for the product. Beyond this, I can't justify any further reactivity. There is no acid to lead to carbocation-mediated ring closure. I can't see any obvious sigmatropic rearrangements nor cycloadditions that would either directly or indirectly lead to this.
I had already taken the suggestion of rotating the single bond into consideration. Can someone please confer with me on the stereochemistry, though? Kindly see below.

Rotate the central single bond and see if there's no "obvious sigmatropic rearrangements nor cycloadditions" ;)

What you have proposed is a 4n (n=4) thermal, conrotatory electrocyclization where the hydrogens at the newly-formed sigma bond will be trans disposed. Some double bonds in the final two structures are missing or misplaced.

@user55119 - this is how it was listed in the source - since there are an even number of electron pairs (6), would this not following thermal/even/conrotatory Woodward-Hoffman rules? The double bond may be obscured by the orbitals in the drawings; that is a software issue and I apologize! Do you agree or disagree with trans disposition?

Yes, the hydrogens will be trans as predicted bt W-H rules. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/pericyclic/pericyclic.htm#Electrocyclic


text:
How can I mix methylene chloride, methanol and hydrofluric acid to make a paint stripper?
HF paint stripper on data sheets lists
Methylene chloride 70>85%
Methanol           10>25%
Hydrofluric acid.   5%
Previously I have found the chemical splits. Is there a particular method to properly blend all 3 parts?

Offhand, I can't think of a more dangerous DIY activity than what you're proposing, if in fact this is headed that way. If you're just curious, disregard my comment.

Sorry I’m fully aware of the dangers. I forgot to mention I work with this everyday in well extracted rooms. We use this for the paint removal of alloy/steel wheels which require a acid based stripper to remove the epoxy primers applied from factory. We currently use around 2000litres every 4-6 weeks.

We do have all the PPE and anti HF gels inc shower room. Getting back to the point and the reason for my question is costs across transportation of this blended stripper have increased massively. However we are able to order the raw materials in for half the cost and mix on site if the process was straight forward or is there a particular method?

Why do want to blend all "3 parts"?

The mix is available to purchase but costs have doubled for various reasons. The reason for mixing is the raw materials are half the cost. Paint strippers without the hf, in particular for alloy wheels do not remove the primer and with the hf we are able to strip wheels in less than 30mins. Our process requires 300plus wheels per week to be stripped and approximately 2000litres of chemical per month.

It is not clear how reliable the concentrations given in the SDS are. It is not uncommon to give a range, but actually manufacture product at one end of the range.
I would take 25 parts of methanol and 70 parts of dichloromethane and add hydrofluoric acid till it becomes cloudy - then back off (add some more CH$_3$OH + CH$_2$Cl$_2$). You may be able to add 4.6 parts of hydrofluoric acid (round up to 5 for the SDS).
The solubility of dichloromethane in water is 1.60% at 20$^o$C and the solubility of water in dichloromethane is 0.24% at 20$^o$C. Ref 1  The solubility of dichloromethane in water increases as the temperature decreases. Ref 2
One unknown is the concentration of hydrofluoric acid and whether higher concentrations of HF are more easily dissolved in this mixture than lower concentrations. The SDS may not give you useful information, but I suggest trying to use the most concentrated HF available - you can always dilute it, and H$_2$O is no friend of this combination.
Methanol and water (and thus, probably also hydrofluoric acid) are miscible and so are methanol and dichloromethane. A recent question on StackExchange (Ref 3) gave a ternary diagram of water-methanol-dichloromethane:

The line separating the one-phase (M) from the two-phase regions (I) does not go all the way to the top, but you can see that there are homogeneous mixtures with a lot of water, methanol and dichloromethane.
The Safety Data Sheet is required to list hazardous ingredients, but not non-hazardous materials. Some stabilizer might be added, hidden in the 15% range of dichloromethane and 10% range of methanol, that could keep the mixture homogeneous. One compound that comes to mind is propylene glycol. If the commercial product is cloudy, it might be stabilized with a thickener like guar or xanthan gum. Wax has been used as an evaporation inhibitor in paint removers containing dichloromethane; it might also act as a thickener and stabilizer.
Ref 1. https://macro.lsu.edu/HowTo/solvents/Dichloromethane.htm
Ref 2. https://en.wikipedia.org/wiki/Dichloromethane
Ref 3. Mixture of water, methanol and dichloromethane


text:
Qualitative analysis of the anionic part of a salt using it's soda extract
While testing for anions using dilute $\ce{H2SO4}$, we prefer using the soda extract for tests rather than the actual salt. But we don’t use it for tests of carbonates, bicarbonates and sulphites. Why is this so?
I get that soda extract itself is a carbonate so no point to use it for carbonates but why not bicarbonates and sulphites?
Also I’m told that we use soda extract because sodium salts are mostly soluble. Then why specifically sodium? Isn’t there any other metal we can use?

If a solution of $\ce{H2SO4}$ is added to a some carbonate, bicarbonate or sulfite, it will produce some $\ce{CO2}$ or $\ce{SO2}$ effervescence with gas bubbles filling the container where this test is done, and maybe cause it to overflow. With sodium bicarbonate the equation is :
$$\ce{H2SO4  + 2 NaHCO3 -> Na2SO4 + 2 H2O + 2 CO2}$$
With sodium sulfite, it is :
$$\ce{Na2SO3 + H2SO4 -> Na2SO4 + H2O + SO2}$$

I don’t understand why “Cause it to overflow” is a problem. Why can’t we just use a bigger container and prevent overflowing?

Also, is there any chemical aspect to this problem?

@Apoorv Sharma. Before  doing the experiment, you don't have any idea how far and high the foam may go up. You may have to give up your first experiment, and start again with a second sample. Apart from this aspect, there is no chemical difficulty or secondary reaction.


text:
Making a bridge on a cyclic molecule
So, my question is how would I make a bridge as seen on the following molecule (right side)?

As for the molecule on the left, I am supposed to make it using that molecule as a base. The only requirement is that every other molecule I use that uses carbon, can only have a maximum of 2 carbons.
To reiterate- I need to use the molecule in the left using any inorganic compound, and organic molecules with up to 2 carbons, and turn it into the molecule in the right.
My biggest problem here, is how to make it so that the molecule only reacts in that area. I was thinking of using a mixture of condensation and $\ce{HBr}$ in order to break the $\ce{-OH}$, and temporarily remove a hydrogen.
However, again, I don't know how to make it so it would only react in that specific area.

This is non-trivial.

This is most definitely not a homework question, so the close votes do not make sense.


text:
why doesn't alcohol react with sodium hydroxide but react with potassium hydroxide?
I have read that ROH+ NaOH doesn't form alkoxide ions because water is a stronger acid than all alcohols except methanol. However while reading xanthate test I found that alcohol reacts with KOH to form alkoxide ions. Why is it so?

The reaction of alcohol + NaOH producing $\ce{C2H5ONa}$ + water is an equilibrium which is usually strongly pushed to the left hand side. But in the presence of $\ce{CS2}$, the alkoxide $\ce{C2H5ONa}$  is transformed into xanthate. So the equilibrium is driven to the right hand side.

And the same goes with the elimination reactions in which alcoholic KOH is used for alkene formation from alkyl halide?


text:
Why does the dissolution of calcium hydroxide have a negative entropy?
We had an experiment on the dissolution of calcium hydroxide and we obtained the experimental entropy value $\pu{-203 J mol^-1 K^-1}$ with a percent error of $26.7\%.$ Doesn't dissolving a solid in a solvent increase entropy?

It usually does increase the entropy, but compound solubility usually does increase with temperature. It is not the case of $\ce{Ca(OH)2}$.

When ionic compounds dissolve in water, water molecules form a complex ion around the calcium cation.  The presence of the hydroxide anion has a similar effect, though the polarities of the water ligands is reversed.
Thus, dissolution of Ca(OH)2 in water is actually a chemical reaction.  In some cases, complex ions may be more structured than the crystals of undissolved ions.       In such a case, the entropy change of solution will be negative.

Entropy can be measured and aproached differently, (i) from an equation that relates entropy and Enthalpy to measure the heat of reaction, the remainder of the work done per degree of Temperature (°) is
or if the change in enthalpy is measured then simply.
In this case the Vector of entropy has a negative direction hence an endothermic process and the magnitude is 203.
(ii) Considering the amount of microstates, Boltzmann constant would relate the energy and the temperature, between solid and liquid dissolution; As a function of Omega (W).
For Omega (W) evaluate the microstates in your solid sample and the liquid volume using Boltzmann Relation.
The change in Entropy between solid and liquid state is measure by Math approach, and is proven by Covention I and II (Physical Chemistry; Ira N. Levine Pg.295 6th Ed) for the Activity Coefficients on Gibbs free energy of standard solid state (S* solid 83.36 J/mol*K) and ideal model for gas state (S°gas,1 bar 285.62).
An increase in Entropy would be, then proven, as a solid State entropy is of approx. 83 J/molK and raise to 203 J/molK, through an endothermic process.
NIST Ca(OH)2 https://webbook.nist.gov/cgi/cbook.cgi?ID=C1305620&Mask=FFF


text:
Keto-Enol tautomerism acidity of α Hydrogen vs stability of double bond
If I consider a ketone that has an ethyl group and a methyl group attached to it. The α-Hydrogen of the methyl group would be more acidic than that of the ethyl group due to -I effect of the adjacent carbon in ethyl group. However the double bond formed in the enol form would be more stable in case it is formed using the α-carbon on ethyl since it's more substituted. In this case which enol form should I choose to be the more stable form ?

Stability of transition state depends upon the double bond formed. The more substituted enol should be major product here.

How does the question above answer my question? It compares between aldehyde and ketone but now doesn't talk about this

Look at the linked answers within the question linked..


text:
Searching for low cost alternative for zirconia ceramic crown
I have seen one company call as you have seen below:

Call for "Development of Zircon Alternative Formulations in Ceramic Tile Glazing Industry"

So I have searched for finding the low-cost alternative for zirconia crowns by these terms:

diy ceramic crown
low-cost dental crowns

So as dental industry is more sophisticated on the ceramic and crown I searched the tooth crown and seen this result :

All-resin dental crowns are the least expensive compared to the other types of crowns listed. However, they can wear down more quickly and fracture more easily than the crowns made of porcelain and metal. For these reasons, resin crowns are rarely a long-term solution and are usually a temporary stand-in.

And found some alternative like porcelain, so I tried to find out how could create one resin crown (Do It Yourself).
So I want to find out some table or comparison, between the :

cost
strength
lifetime

Of the zirconia, porcelain and resin crown with laboratory instruments or if possible as Do It Yourself (DIY) projects.
Thanks.


text:
Synthesis of (1S,2S,4S)-4-ethenyl-5-methylcyclohexane-1,2-diol
I'm trying to devise a synthesis for the following question.

I am aware it involves retrosynthetic, Wittig, Grignard and diels alder reactions, but unfortunately I can't really grasp all concepts which make me stuck. If there's any help or further explanation would be great.
I have tried to do retrosynthetic for the alkyne compound, also the Wittig reaction. I think Grignard should be the last step where it will change the oxygen bond to $\ce{OH}$. But I can't get myself to there yet.

I would start with the commercially available methyl propiolate and do the D-A on that first

Because the target structure contains the motif of a cyclohexane-1,2-diol, I suggest an approach including a Diels-Alder reaction (DA in the illustration below) of 1,3-butadiene and the commercially available diethyl acetal of propargyl aldehyde:

The hydrolysis of the acetal then sets the stage to install the methyl group with the cuprate and 1,4-selectivity in respect to the carbonyl group.  As an intermediate protection of the aldehyde, a cyclic acetal is installed. The Prévost reaction typically yields 1,2-(trans)-diols, thus possibly here, too (brief review).  Eventually, the deprotected aldehyde is transformed into the exocyclic alkene with the Lombardo reagent. Contrasting to the conditions of a Wittig reaction, the conditions of the Takai-Lombardo reaction are milder (absence of a strong Bronsted base).
Alternative installations of the (trans)-diol could be achieved e.g., by epoxidation and subsequent ring opening with a peracid like mCPBA or (in situ generated by) oxone.

This looks good except for the use of propargyl aldehyde which is not commercially available and potentially tricky to handle. A preparation of it is here  http://www.orgsyn.org/demo.aspx?prep=CV4P0813

@Waylander I agree with your argument.  For the purpose of the DA, the dienophil may be an ester (e.g., methyl propiolate). Reduction with LAH, oxidation with TEMPO to generate the aldehyde chemically is possible, but renders overall the reaction lengthy to a total of then eight steps for such a little molecule. If each of them yields the next product by 90%, overall 0.90 ** 8 = 0.43 or 43% over all steps. I anyway redraw it for orthocresol's comment.

You could use the diethyl acetal of propargyl aldehyde as the dieneophile as it is commercially available.

Do you need to reprotect to do the prevost reaction?

@Waylander It may be that the reprotection of the aldehyde is not necessary (so far, I never performed a Prévost reaction).  On the other hand I consider the dihydroxylation as an oxidation once and because I was once surprised by an aldehyde quickly degrading into an acid I speculated a protection would be a useful safeguard here.  The Kürti/Czakó book about name reactions mentions an example (by Deslongchamps et al.) where at least a primary and a secondary alcohol remain intact. Crosslink.

One of those reactions where you don't know until you try it.

@Waylander «One can certainly plan research, but not the results! (Addressed to all those engaged in the distribution of research grants!)» is right in the beginning of D. Seebach's 48 pages long tour d'horizon  «Organic Synthesis -- where now» in Angew. Chem. Int. Ed. Engl. 29 (1990) 1320--1367; doi 10.1002/anie.199013201.


text:
Aromaticity of fused rings
I understood how to find aromatic nature of single aromatic rings (like benzene) or that of combination of rings which are independently aromatic, such that the electrons of one ring do not enter into another ring (like biphenyl).
However I don't have much clarity on finding the aromatic nature if electrons from 2 or more rings can delocalise into each other.
Like biphenylene

(Image Source:pubchem)
If I only look at the central ring it seems to be antiaromatic. However there can be resonance throughout the entire molecule that can cause it to be more anti aromatic. Also the top and bottom rings might have aromatic character on resonance in individual rings too. These are just my guesses so I don't know what is the actual nature of the compound, but I didn't understand why this compound would be stable if it is associated with some antiaromatic character? Please elaborate on this aspect

related https://chemistry.stackexchange.com/questions/82084/why-does-benzyne-dimerise-and-not-trimerise https://chemistry.stackexchange.com/questions/136917/is-benzocyclobutadiene-aromatic-or-anti-aromatic

You are much on a right track, at least on seeing the complexity of the matter.

You may want to look at this answer, in which two isomeric neutral polycycles are compared.  Even though they have the same formula and same count of conjugated pi electrons, only one of them is aromatic and to make the other one aromatic, you should firm the di-cation instead.  Topology enters the discussion -- the outcome depends on how the rings are connected.

Antiaromaticity doesn't exist in states that are observable. Its extended presence is a contradiction to its definition. Instead the double bond system exists as an independent system instead of being conjugated. This effect is known as the Jahn-Teller Distortion. A very famous example of said effect is cyclo-butadiene which exists in the form of a rectangular compound as explained in this answer
Proof that this is also the case in biphenylene is as follows:
Taken from the wikipedia page on biphenylene, (emphasis mine)

[...] Questions of bond alternation and ring currents have been investigated repeatedly. Both X-ray diffraction and electron diffraction studies show a considerable alternation of bond lengths, with the bridging bonds between the benzenoid rings having the unusually great length of $\pu{1.524 Å}$. The separation of the rings is also reflected by the absence of the transmission of NMR substituent effects through the central [4n] ring. However, more sensitive NMR evidence, and particularly the shifting of proton resonances to high field, does indicate the existence of electron delocalization in the central [4n] ring. [...]

The reference stated in the Wikipedia article is Tetrahedron 1964, 20 (5), 1179–1184, which states:

Taking into consideration the fact that the molecule is planar and that the two six-membered rings are known to be $\pu{1.52 A}$ apart and assuming that the charge density around the molecule is uniform, we calculated the chemical shifts between the ortho and meta hydrogens due only to ring currents in two cases, i.e., two ring currents (0.13 ppm) and three ring currents (0.34 ppm). The chemical shift for the two ring-current case is in good agreement with the experimentally determined value (0.105 ppm), but it should be borne in mind that ample physical chemical evidence exists to show that the six-membered rings are not entirely independent electronic systems.

Plus 1 for remembering that a molecule is antiaromatic when it cannot be so. But it does not seem to answer the question.

@Alchimista I noticed that the question was just not about biphenylene. I'm currently short of time but I will add to this answer in the near future.

make it general because a detailed answer takes a chapter at least.

The first paragraph is a very good start for antiaromaticity in general. The third sentence is a bit confusing though. In any case, I think this covers the question quite alright.


text:
Calculating the molecular mass of an element using relative molecular mass in several compounds

The relative molecular masses of A, B, and C were found to be $101.30$, $87.30$ and $144.90$ respectively, compared to an assigned value of exactly $2.00$ for $\ce{H2}$. Subsequently, these substances were subjected to elemental analysis and were found to contain $37.2\%$, $64.8\%$ and $78.0\%$ of $\ce{X}$ respectively by mass.
Find the molecular mass of element  X, found in A, B & C.

I found the relative molecular mass of X in each of the three compounds:
In A, $$101.3 \times 0.372 = 37.68$$
In B,
$$87.3 \times 0.648 = 56.57$$
In C,
$$144.9 \times 0.78 = 113.022$$
I calculated the HCF of these three values. However my answer doesn't match with the actual answer. What am I doing wrong here?

It would be good to hear your final answer.

Is the answer 18.84?

The twist is that there are one or more atoms of X in compounds A, B, and C.


text:
Why does NH3 cause pairing in case of some metal complexes and doesn't in case of others?
I was met with a question that required me to find which complex is an outer orbital complex, and two of the options included $\ce{ [Ni(NH3)6]^2+}$ and $\ce{[Co(NH3)6]^3+}$,
I want to know why $\ce {NH3}$ acts as a strong field ligand and cause pairing of electrons in d orbital in the case of $\ce{[Co(NH3)6]^3+}$ but acts as a weak field ligand and doesn't cause pairing in the case of $\ce{ [Ni(NH3)6]^2+}$ (hence making it an outer orbital complex)?
I know that $\ce{NH3}$ lies in the middle of spectrochemical series, so how do I identify if it will cause pairing in a complex or not?

The concept of "outer orbital complex" is way outdated and not taught in most places, though India seems to be an exception (maybe that's where you are?). It is now known that both high spin and low spin complexes use "inner orbitals" for bonding (that is (n-1)d orbitals, so 3d for cobalt rather than ever using 4d).

@Andrew For cobalt, it is easy to see that it will use 3d, 4s and 4p orbitals to form 6, d2sp3 hybridised orbitals (I know that it has been stressed many times in ChemSE that we should not use hybridisation in case for coordination compounds but use of MOT is beyond my knowledge). The problem is with nickel complex, if $\ce{NH3}$ causes pairing of electrons then we left with only one vacant 3d orbital and thus, cannot form d2sp3 orbitals.

So, can we say that "because $\ce{NH3}$ cannot form d2sp3 orbitals, it will not cause pairing of $\ce{e-}$ and use 4s, 4p and 4d orbitals to form 6, sp3d2 orbitals". As chemistry theory somehow tries to explain the experimental facts.

@Apurvium There’s no reason you can’t use 3d orbitals just because they’re full. It just means that you end up with mostly filled anti bonding orbitals. These anti bonding orbitals are the ones we look at when we talk about splitting in the spectrochemical series. The 6 bonding orbitals are filled by the six electron pairs from the NH3 ligands

Also, the actual answer here is trivial. As $\ce{Ni^2+}$ is $d^8$, there is only one possible d electron configuration for an octahedral complex. It will have completely filled $t_{2g}$ orbitals and unpaired electrons in the two $e_g$ orbitals regardless of the strength of the ligand.

recommended reading: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Theory

I think OP wants to ask about the possibility of eg electron pairing in octahedral complex for a d8 type configuration. Even if we use a very strong field ligand, d8 only shows high spin complexes. I think the reason (according to hybridization approach) lies in the orthogonality of 3d's and 4d's, which will require that either we use 2 d orbitals from 3d's as in the cobalt case above OR 2d orbitals from 4d's as in Nickel case. Never the third possibility of 1d orbital from 3d and other from 4d, which could give something like dsp3d.  Perhaps a theoretical chemist can help us out here.

In case of $\ce{[Ni(NH3)6]^2+}$, the configuration is $3\text d^8$. After splitting from spherical to octahedral field, the lower energy orbitals, i.e., $\text t_{2g}$ are already filled. Therefore, pairing of $\ce{e-}$ will not occur whether the ligand is strong or weak. So, the compound is a high-spin complex.

$\uparrow$ $\downarrow$
$\uparrow$ $\downarrow$
$\uparrow$ $\downarrow$
$\uparrow$
$\uparrow$

${|<------------\text t_{2g}---------->}|{<----\text e_g----->|}$
In case of $\ce{[Co(NH3)6]^3+}$, the configuration is $3\text d^6$. After splitting from spherical to octahedral field, the lower energy orbitals, i.e., $\text t_{2g}$ are not fully occupied. Therefore, pairing of $\ce{e-}$ can occur depending on the ligand. In this case, the compound is a low-spin complex.
$\text {CFSE}_\text {high spin}=-\frac{2}{5}\times 6 +\frac{3}{5}\times 2 + 0\cdot P$
Before pairing:

$\uparrow$ $\downarrow$
$\uparrow$
$\uparrow$
$\uparrow$
$\uparrow$

${|<-----------\text t_{2g}--------->}|{<-----\text e_g------->|}$
After pairing:

$\uparrow$ $\downarrow$
$\uparrow$ $\downarrow$
$\uparrow$ $\downarrow$
$~~$
$~~$

${|<-----------\text t_{2g}----------->}|{<----\text e_g----->|}$
$\text {CFSE}_\text {low spin}=-\frac{2}{5}\times 6 +\frac{3}{5}\times 0 + 2\cdot P$

I'm quite confident that the layout is completely broken on my device for these attempts at making an image. The site design is responsive, so it won't look the same for everyone that reads it. Unfortunately, if you need images, you should prepare them on your device and upload them.


text:
Siiloxane polymer experiment

I believe the first reaction would be a substitution of $\ce{SiMe3}$ for $\ce{I}$ and creating $\ce{SiMe3Cl}$.
The second process I think would involve the $\ce{Mg}$ removing the two $\ce{Cl}$ and forming diphenylsilane, which would likely be a repeating unit in a silicone polymer.
Could any explain the mechanism for these reactions?

You are correct about reaction 1, but I have yet to see a mechanism for this

@Waylander Wouldn't the first reaction just be an $\mathrm{S_E Ar}$ ?

@ShoubhikRMaiti  maybe, but I would like to see a reference that says so

A)SiMe3 Increases the inductivity and reduces Aromaticity,
so by free radical (uv termination) its likely to have p-iodideSi(Me)3Bencene.

B)Dimer crystal solution in Cl- ions.  (Ph)2Si(Cl)Mg+---ClSi(Ph)2
Or Solid crystals of (Ph)2Si(Cl)MgCl.

Er no. ICl iodinates with desilylation in the presence of silver salts http://www.xiuzhengrd.com/ejournals/pdf/synthesis/doi/10.1055/s-1993-25912.pdf


text:
will splitting a solution will keep the concentration values (and moles) the same?
I have this exercise from my H.W:

and I am asked: if we split the solution into 3 different containers equally -

how many moles will the solution have in each container?
will the Cl- concentration change?

for the first question, I answered 0.08/3 moles. my common sense say if I split something to 3 it will split equally.
for the second I wrote that the Cl- concentration will not change from the original solution, but im not sure how to explain this.
Am i right?

Using photos/screenshots of text ( especially if handwritten ) instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it  can be challenge to decipher.  Consider  copy/pasting or rewriting of the essential parts and using of MathJax for eventual formatting of mathematical/chemical formulas or equations.

Concentration is intensive and amount of substance is extensive.

I wonder if you really need that sketch to answer something obvious.

If the solution is Standarized, distributing equally the ions; then the concentration [C] is a ration of volumes for your complex formation.
[C]i*(Vf/Vi)=[C]f
When Vf1=Vf2=Vf3, then all 3 [C]f will have an approximate Concentration.
Ignoring solutions fugacity, Cl- ion concentration must be constant as well; it is likely to reach equilibrium way before complex formation.
The formation constant Kf for CuCl2- is 3e5, an amount will be in relation to the solubility product of CuCl Ksp 1.72e-7

Don't make things more complicated than they are. The answer doesn't change in whatever solution or if activity is considered. The only requisite is that of homogeneity. Nor the solute nature plays a role.

Ignoring the handwritten notes, which are difficult to read (you can use "Markdown" language, instead), your common-sense answers certainly seem correct!
To explain the thought-process behind stating the concentration stays the same when splitting a solution, consider that if the ratio of solvent:solute is X, then the ratio of solvent/3:solute/3 is still X. You've simply divided the top and bottom of the equation by the same amount. Without the math, you've split off the solute and solvent in the same proportions that they originally had.
One might quibble that when dealing with picoscopic quantities, i.e. just a few atoms of each, one sample might have only one ion while the other had two, because chemistry doesn't split ions, but that's splitting hairs.


text:
Why is tellurium-128 so stable compared to the other radioisotopes?
Tellurium-128 is the longest lived radioisotope so far with a half-life of 2.3 septillion ($2.3*10^{24}$) years. I know I've heard something about so called "magic numbers" that if you have the perfect amount of protons and neutrons in an atomic nucleus it usually is still unstable but has an absurdly long half-life (examples bismuth-209,xenon-124,). But why is tellurium-128 this stable? And why is this for the ones I have just said?
Wikipedia Link: https://en.wikipedia.org/wiki/Isotopes_of_tellurium

Somebody has to come out on top.

Tellurium-128 is in no way extraordinary or unique.
Cape Chelyuskin in Russia is the northernmost extreme point of continental Eurasia. Tellurium-128 is not an extreme point of anything. There are hundreds of isotopes which are less stable than it and hundreds of those which are more stable. The latter kind is called simply "stable".
Now you might think there is a clearcut boundary between the stable isotopes and the radioisotopes. No, there is none. As our detection methods improve, more and more isotopes previously thought to be stable are found to have some decay path, usually with an absurdly long half-life (otherwise they would have been found earlier). When I was a kid, the heaviest stable element had the number 83. This is no longer the case.
Magic numbers of protons and neutrons have no role in this. Well, I mean, they are related to the same field: if a nucleus has either of these, or better both, it is expected to be more stable than its peers. $\rm^{128}Te$ doesn't and isn't.
If anything, $\rm^{128}Te$ is (or was; I didn't check) a temporary winner in a perpetual race. In due time, longer half-lifes will be detected, and then longer yet, and so on.
So it goes.

We have $_{180}^{73m}Ta$ waiting in the wings for a predicted, but highly spin-forbidden, decay, no?

Apparently so. Then tellurium-128 is not even the record holder.

When it comes to double beta decay, the mode by which Te-128 decays, there are many nuclei which are predicted to be unstable but no decay has been observed as yet - see https://en.wikipedia.org/wiki/Double_beta_decay

And then there's always the projected decay of the proton, predicted half life > 10^30 years

There are unstable nuclides more stable than tellurium-128.
Multiple nuclides are considered thermodynamically unstable with predicted decay path, but they are observationally, kinetically stable. Some of them are eventually observed to decay, directly or indirectly, as the detection techniques improve over time.
Two of interesting examples are

calcium-48 ($\pu{6.4E19 y}$), the double-magic-number nuclide with the extra long half-life in spite of the highly disbalanced n/p ratio.
Metastable tantalum-180m, the only (just observationally) stable odd-odd nuclide heavier then nitrogen-14. It is stabilized by its high spin state and "forbidden" transition. The lower energy tantalum-180 has the half-life about 8 hours, beta decaying to hafnium-180 or tungsten-180.

Some long-life nuclides like calcium-48 and tellurium-128 perform double beta decay, like
$$\ce{^{48}Ca -> ^{48}Ti + 2 e- + 2 \bar{\nu_\mathrm{e}}},$$
or
$$\ce{^{128}Te -> ^{128}Xe + 2 e- + 2 \bar{\nu_\mathrm{e}}}.$$
They are quantum tunnelling  through less stable odd-odd nuclide ( scandium-48 or iodine-128, respectively ) with higher energy (even-even -> odd-odd -> even-even). These intermediate nuclides form an  energetic barrier analogical to very high activation energy of chemical reactions (like kinetically stable diamonds).


text:
R or S configuration?
The respective configutations of a and b are...?

I tried to find the answer for a (for me is: 2S, 3S) but I'm not sure of my results, can someone help me?

Here is my sage advice to you, Silvia95. Do not redraw structures!!! Use the method described here. Transcribing structures to place the lowest priority group in the rear will invariably lead to errors. Leave the structure where it is and use your hands as tools to determine chirality. Make a "thumbs up" gesture with either hand and point the thumb from the asymmetric carbon to the lowest priority group. Your fingers should point in descending order from highest priority to the next to lowest priority. If not, you are using the wrong hand. Right hand, R; left hand, S.C2 of dibromide a has the hydrogen in the rear. The fingers of your left hand show Br>C3>C1. C2 in dibromide b has the hydrogen in the front. Now the thumb of your left hand points toward you and Br>C3>C1. Using the same technique for C3 of a and b, your right hand fingers reveals Br>C2>C4. These two structures are identical, (2S,3R)-2,3-dibromopentane. If you rotate either structure by 180o about a vertical axis in the plane of the screen, the two structures will superimpose.

Of course, there's a shortcut for this particular problem if you've cultivated some skill in visualization.  It is possible to recognize directly that the two figures represent the same diastereomer: they are related to each other by a 180-degree rotation about an axis parallel to vertical direction of the page, and (pure) rotations preserve hand.  With that fact in mind, one can observe that only one of the answers offered has the needed property.

Clearly, I'm well-aware of their identity. My pet-peeve is the way textbooks teach the rules for R/S and the usual consequences of novices redrawing structures.

I can't speak much to the current crop of textbooks, but I will say that back in the day, it was the method you describe in this answer that I learned.


text:
About the Vilsmeier-Haack formylation of this compound
I was reading an article and found the following reaction sequence:

The second step involves a formylation of the hydrazone with the Vilsmeier reagent to, first synthesize the pyrazole ring and then, to formylate it to produce the final compound. However, when reviewing this reaction I found many examples in which it is used to formylate phenols and other electron-rich arenes. My question is why doesn't the formylation occur on the phenol of the example?
I'd appreciate if someone could provide an answer.

The hydrazone exists as an imine/enamine tautomeric equilibrium and attacks the Vilsmeier reagent when in enamine form. Could it be that the enamine is so electron-rich that it attacks the Vilsmeier reagent faster than the phenol ring could? Maybe the phenol ring acts as an electron-donating group to the enamine and this promotes the attack? Is the pyrazole ring much more reactive than the phenol so that it gets formylated before the phenol?


text:
Disambiguate molecule's bonds, angles or dihedral angles selection
I am working on a project where one of the goals is to store data from computational quantum chemistry calculations for later retrieval.
Currently, we are trying to store the outcomes of simple bond (or angle or dihedral angle) "scan jobs". The main problem we have is in finding a way to represent the scan coordinate that would be independent of the order of the atoms in a molecule.
As an example, given a simple $\ce{C-C}$ bond scan in $\ce{CH3CH2CH2CH3}$ molecule, we may define the scan coordinate using atom labels and their order in the molecule (where the order may be taken as the order of atoms in an xyz coordinate file) so that we know which $\ce{C-C}$ bond has been scanned.
However, that would only work if the order of atoms is kept the same. What if there are other scan jobs on the same molecule and the same $\ce{C-C}$ bond, but with different order of atoms. How could we define that $\ce{C-C}$ bond in an unambiguous way?
Our goal is to store the scan jobs in a distributed way and access them via a simple query that would define the molecule and the bond (or angle or dihedral angle) we want the scan data for.
I was thinking that maybe SMARTS strings would allow us to somehow describe the scan coordinate pattern we want to retrieve (e.g. two atoms for a bond or three atoms for an angle or four atoms for a dihedral angle). However, it is not yet clear how could I do that with SMARTS.
Another problem is that sometimes to disambiguate which bond (angle/dihedral) has been scanned one needs to provide extra context information about the bond chemical environment(certain functional groups near the bond atoms etc.). Any advice on how to define such scan coordinates will be welcome.

You definitely want to use SMARTS, likely with atom classes to define the atoms involved in the scan (e.g., c1c[cH:1]c:2C:3O). Take a look at QCArchive's Torsion Drive notes: https://qcarchivetutorials.readthedocs.io/en/latest/basic_examples/torsiondrive_datasets.html


text:
Why is the tropylium anion non-aromatic?
Recently I came across a question asking which compounds are anti-aromatic. This was one of the options:

I’m confused as this is not an answer. I think the lone pair electrons on the carbanion participate in resonance and so there are 8π electrons which, follows the 4nπ electrons rule for anti aromaticity.
So why isn't it anti-aromatic?

The rules in determining the aromaticity of a compound are as follows:

The system must have 4n+2 π electrons
The system must be planar
The system must be completely conjugated
The molecule must be cyclic.

The cycloheptatriene anion does not follow two out of these four rules. Namely,

The anion system has 4n electrons (whereas it should have had 4n+2 electrons in it π system)
The anion system has an sp3 carbon and is hence non-planar (where it should have been planar)

Due to these two issues (second issue mainly), there is no ring current in the complete ring and so no conditions need to be considered whether the compound is antiaromatic or aromatic in nature.
Cyclopentadienyl anion also has an sp3 carbon as an anion. However the anion formed is aromatic if planar and so this overcomes the ring strain and so retains aromaticity.


text:
Why doesn't an sp3s hybridisation exist?
Trying to prove that nitrogen doesn't split its lone pair to form 5 bonds, I thought of a situation that I couldn't rule out; the paired electron being excited to the 3s orbital, so that five bonds could be formed- in an "$\ce{sp3s}$" hybridisation.
I know hybridisation is just an observation-based explanation, but I'd like to know how to rule wacky situations like this out.
Can this - and all other nonstandard hybridisations you can come up with - be ruled out by reasoning that the electron excitations involved take up too  much energy?

Yes, you are correct, the hybridisation scheme has to make sense in terms of energy. There is also a symmetry factor to consider. Without looking up a character table, my guess is that sp3s has the right symmetry to form a 5-coordinate system. However, it doesn't work as 3s is too high in energy in N. (Also when does N form 5 bonds??)

@ShoubhikRMaiti: haha, N doesn't form 5 bonds anywhere; I was just trying to rule out ways in which it could. Namely, this.

@ShoubhikRMaiti: wait, what? Sp3s can exist? (You said "too high in energy for N", are there cases where it can work out? )

No sp3s cannot exist. Because once you go down the period you end up on P, where there is a 3d orbital between 3s,3p and 4s, so I doubt there is any case where sp3s exists. It has the right symmetry for N, but it doesn't work out due to energy differences.

@ShoubhikRMaiti: I see. Is there any list of hybridisations which can actually be applied to molecules, as I've commented below? Can all made-up hybridisations (ie., the ones not in such a list) be proven not to exist by the energy argument?

Hmm, I have not come across any such list. Well, most of the hybridisation schemes we use work as they match the energy differences. So, other schemes must not exist due to the same reason. Note that many of the common hybridisation schemes like sp3d, sp3d2 etc. for 3rd row elements were found not to be right due to 3d being too high in energy.

@ShoubhikRMaiti: kay, guess it's a matter of experience to know if a hybridisation is invalid right off the bat. This answers it anyway, thanks.

@ShoubhikRMaiti: 'kay, guess it's a matter of experience to know if a hybridisation is invalid right off the bat. This answers it anyway, thanks.

The counter question that might help you rule this out is: "Which 3s orbital?" Note that atomic Nitrogen doesn't have a 3s orbital as N itself belongs to the 2nd period(n=2).

Also note that in hybridization, Atomic orbitals of same or almost similar energies intermix to form same number of new Hybrid orbitals of same shap and equivalent energy.
That being said, you cant expect 2s or 2p orbitals to intermix with a 3s orbital
(if it was a case. Since this case isn't possible with N)

From what I understood from this answer, every atom does indeed have every orbital, it's just that they're unoccupied. Which would imply that there is an empty 3s orbital here. I just needed to know if the energy-not-enough explanation was correct, or if it's wrong, the correct explanation.

@harry You are correct, every atom has all the orbitals. But at the same time there are actually no orbitals. Orbitals aren't physical boxes in which we place electrons. But we tend to say that some orbitals aren't in an atom, what we mean by that is that a large amount of energy will be needed to move an electron to that orbitals. For a simple example we can excite the electron in hydrogen to higher states. But we never see this naturally because why would something like to horde energy when it has a way stabler option?

@NisargBhavsar: right, thanks. So can every non-standard hybridisation you can think of be proven not to exist using this?

@harry What exactly do you mean by non-standard hybridizations? Does $\mathrm{d^3s}$ count as one? If yes, than I am sorry to say but it exists.

@NisargBhavsar: I've heard of d3s. I meant ones like sp3s; hybridisations you can just make up using whatever explanation and that has never actually been used to describe a molecule before (so I guess they're "wrong" hybridisations). As far as I know, there isn't a comprehensive list of all the hybridisations that actually exist, so 'standard' and 'non-standard' are fuzzy terms. If anyone knows of such a list, linking it here would be great.


text:
How is it possible to electroplate using an Al3+ solution, when water is a stronger oxidant?
Suppose that I wanted to electroplate Al(s) onto a Cu(s) Cathode, (using an Aluminium Nitrate solution and Al(s) anode).
At standard conditions, Al(s) is the strongest reducing agent, which is good. However, H2O (l) is a stronger oxidising agent than Al3+, which should mean that no Al3+ can be reduced to form the coating on the cathode.
Despite this, I have seen several experiments achieve such an electroplating (with a solution, not molten).
How is this possible?

Can you link to those experiments because this is very unlikely, unless they have cracked the code of making a solution of Al3+ with very high activity.

@NisargBhavsar https://core.ac.uk/download/pdf/11725439.pdf.   Experiment F. I’m trying to replicate this experiment with oxidation state of 3+, and Al3+ is the only one that I can think of that can reduce directly to solid. Maybe there is another metal that could work

No ! It is not possible to deposit aluminum by electrolysis of a water solution. It may be done in a solution of $\ce{AlCl3}$ in an organic solvent like toluene, but not in water. And even in toluene, the conductivity is low. Metallic aluminum reacts with water, when this metal is not protected by an outer layer of aluminum oxide $\ce{Al2O3}$ as it is as soon as an aluminum piece is in contact with air.
For those who doubt about this reaction, let me add a few words. The reaction ion Al + H2O is not really violent, but it is visible. The only way of seeing this reaction is to prevent the alumina layer to cover the surface of the metal. And the only way of preventing this phenomena, is to dip a piece of metallic aluminum into an aqueous solution of mercuric chloride $\ce{HgCl2}$.  In this case, the ions $\ce{Hg^{2+}}$ are able to cross the alumina layer and react with aluminum according to : $$\ce{2 Al + 3 Hg^{2+} -> 3 Hg  + 2 Al^{3+}}$$ This reaction produces the appearance of some atoms of metallic mercury which form an amalgam with aluminum. Here the alumina layer is immediately formed but it does not cover entirely the metallic surface of aluminum. So the layer of alumina is not air proof or waterproof. So the corrosion continues. If the wet piece of aluminum is removed from the $\ce{HgCl2}$ solution, dried and deposited somewhere on a table, the alumina is formed steadily making sorts of white filaments that are growing before one's very eye. In a couple of minutes these filaments look like a delicate beard and measure several centimeters long. This phenomena stops when the fist layer of aluminum atoms is oxidized, because the mercury atoms are also eliminated. And the next oxidation of the aluminum surface is entirely covered by alumina.

I see. I’m currently doing research on oxidation states, and metals which ions of 3+ oxidation states seem few and far between. I suppose the report I read failed to mention any additional solutions used.

"Metallic aluminium reacts violently with water" I highly doubt that, I have seen aluminium chips from cutting aluminium pieces come into contact with water, and nothing has happened. At best, it will tarnish slowly.

You can buy gallium aluminum “amalgam” and use it to generate hydrogen gas by putting it into water. Very expensive hydrogen generation, but same principle as with mercury: breach the aluminum oxide coat and reactivity happens! I upvoted, by the way.

@EdV Thank you for for the information about the alloy Al-Ga, the I did not know.


text:
What is a suitable material for a cathode in electroplating experiment?
I am going to do an electroplating experiment, whereby I will be using a Cu(NO3)2 solution to plate onto a cathode.
What is an easily accessible cathode which I can use for this? Would a strip of Ag (s) be sufficient, since it is a weaker reducing agent than Cu(s), and should therefore not react with any of the solution? And furthermore, using a strip maximises surface area and there will be a more noticeable change in mass of the cathode?
I am performing these at standard conditions.

Certainly worth trying. Presumably you will be weighing the silver cathode before and after. If it works well, let us know!