text:
What is this copper compound I made?
I started with copper (II) sulfate solution left over from a recrystallization. I added sodium carbonate (definitely an excess) and about 200 of water and got a lot of bubbling. I heated it, for no particular reason. Few hours later I filtered the solution, which now looked dingy gray. I got a deep blue solution with a pH of around 11, and what was in the filter was black...
The filtrate is still deep blue with no precipitate, I'm washing and drying the residue from the filter, but I'm still curious what I made. I should have gotten basic copper carbonate, I've done this before to get copper carbonate, I can't remember if I've used sodium carbonate or bicarbonate last times though. I didn't think it made a difference.

https://chemistry.stackexchange.com/questions/34605/what-are-the-products-of-a-reaction-between-copper-sulfate-and-sodium-bicarbonat/63778

You made basic copper carbonate.

I though copper carbonate was 1: insoluble in water and 2: green-blue powder. I have a dark blue solution and black power... I've done this before, I have copper carbonate from previous runs, idk this time is different.

It is probably copper oxide, that you get by heating copper hydroxide and possibly basic copper carbonates as well.

Any idea why the filtrate is deep blue? That, to me, indicates dissolved copper but I shouldn't have anything soluble after adding the sodium carbonate, right?

There are two surprising phenomena in this experiment.
First : $\ce{CuSO_4}$ reacts in a unusual way with $\ce{Na_2CO_3}$. The equation is :  $$\ce{2 CuSO_4 + 2 Na_2CO_3 + H_2O -> Cu(OH)_2·CuCO_3 + CO_2 + 2 Na_2SO_4}$$ This forms some bubbles of $\ce{CO_2}$ as you have seen. And it produces a strange precipitate of the so-called basic copper carbonate, $\ce{Cu(OH)_2·CuCO_3}$, which is well known in geology as the mineral called malachite. This formula is sometimes written as $\ce{Cu_2CO_3(OH)_2}$. This is what you have obtained as your insoluble deposit.
Second : Basic Copper carbonate gets redissolved in a big excess of carbonate ions, producing a deep blue solution containing an anion which is probably $\ce{[Cu(CO_3)_2]^{2-}}$ This is what you have obtained in your blue filtrate.

Wow, thank you for the very specific answer! I'm wondering, is there a way to somehow recover the copper carbonate/anion from the filtrate? After filtering I've let it sit in a wide mouth jar covered with a coffee filter and there's some white precipitate at the bottom and fractal-tree looking crystals crawling up the sides.

Your white precipitate may be sodium hydrogenocarbonate $\ce{NaHCO_3}$ which is not very soluble in water : $6.9$ g in $100$ mL, and less in $\ce{Na_2CO_3}$ solutions. This substance appears in $\ce{Na_2CO_3}$ solution exposed to air and/or to $\ce{CO_2}$

To recover copper carbonate, the excess carbonate must be destroyed, by adding an acid. The trouble is : first the excess carbonate ions will  produce huge amounts of CO2. Then the anion $\ce{[Cu(CO_3)_2]^{2-}}$ will be decomposed and the malachite precipitate will soon appear again. But it is difficult to know with some precision the amount of acid necessary to obtain malachite quantitively, because if an excess of acid is added, the malachite will be dissolved and at the end you will obtain a  $\ce{Cu^{2+}}$ solution like in the very beginning of your experiments.

I guess I'll add sulfuric acid little by little with stirring. This really is just an experiment with leftover chemicals so...


text:
What is the name and formula of this structure?
Could anyone please let me know the formula and name of this structure? Any answer is helpful. Could you also summarize how to deduce the formula?
Here is an image:

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.Please see our guidelines on how to ask nomenclature questions.

Do some research on statins before asking for an answer to what looks like a homework assignment.

That's Atorvastatin, a cholesterol lowering drug (works by inhibiting the enzyme HMG-CoA reductase).
https://en.wikipedia.org/wiki/Atorvastatin
For more data, see e.g. https://www.trc-canada.com/product-detail/?CatNum=A791725

source: Stack Exchange

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

sort: votes


text:
Stacking two square (2D) layers to make a 3D close packed structure
My textbook states these ways of stacking 2D layers to make 3D close packed structures:

Square close packing layer over Square close packing layer (though not written explicitly, the illustration imply stacking done in a non staggering manner), generating Simple primitive cubic unit cell lattice.
Hexagonal (2D) close packing layer over Hexagonal (2D) close packing layer (here, again not stated explicitly, but the illustrations imply it is done in staggering manner so that one layer just fits right in the depressions of other layer) giving two different close packed structures: hcp and ccp

However, they did not mention the two cases arising naturally (out of curiosity  maybe):
a) Square close packing layer over Square close packing layer done so that the second layer fits just right in the depressions of the first layer, i.e. in a staggering manner so the layer pattern is of 'ABAB...' type.
b) Hexagonal close packing layer over Hexagonal close packing layer in a non-staggered manner, which might generate a lattice with the primitive hexagonal unit cell, layer put in the pattern 'AA...' type.
The case (b) was easier to work out its unit cell's type, however I am unable to comprehend the 3D structure of what type of unit cell case (a) can generate.
My question is, are these cases of lattice arrangements possible? If yes, then what are their respective unit cells?

Case (a) gIves12 neighbors for each atom. A diagonal slice through unit cell formed by atoms from consecutive A layers shows that this is hcp.

I object. Case (a) is actually ccp in disguise.

@IvanNeretin Yes, I agree. That was a typo.

@IvanNeretin how shall I choose the unit cell to come to that conclusion? my paper drawings suggest that the unit cell is either fcc with two diametrically opposite face particles removed or body centered cell which seems monoclinic, but both are impossible,  it will be helpful if you could please provide a full answer with an illustration ~thanks

although i could figure out the coordination number easily and i tried 3D modelling software for finding its unit cell, but I was still unsuccessful in finding a right unit cell

It's actually a sneakily disguised version of a lattice you already know.  Read on ... .
Let's try out the staggered square packing. The distance unit used here is the center to center distance between adjacent spheres the diameter of a single sphere.
Start with the A layer with spheres A1, A2, A3, A4 in rotational order at the corners ofca unit square.  On top of the center of this suuaqre place sphere B1, then C1 through C4 eclipsed over A1 through A4 respectively.
You have made a tetragonal structure.  The base of this structure is a unit square, the height obtained by drawing right triangle A1-C1-C3 is tge square root of 2.  Curiously, this height equals the diagonal of the square base.
Let's count the number of spheres touching B1.  There are four apiece in the A and C layers, plus spheres B2, B3, B4, B5 (in rotational order) you can draw touching B1 within the B layer.  Thus 12 spheres, suggesting a close-packed structure.  The distance between the B2 and B3 centers is the diagonal of a square, so again that square root of 2 is coming up.
Hmmm, a tetragonal close packing?  Aren't close-packed lattices supposed to be cubic or hexagonal?  And what does the height of the cell have to do with the diagonal of its base when these two quantities seem to be equal?
The Big Reveal
Add layer D on top of layer C.  The D spheres D1 through D5 are eclipsed over B1 through B5 respectively.  Thus B2 through B5 and D2 through D5 form a square prism with B1, D1, and the C spheres at the face centers.  And because of all the lengths we rendered above as the square root of 2, this prism has all its edges equalling that length and therefore it's a cube.
You have a face centered cubic, therefore cubic close-packed lattice.  It's just rotated from the one you got with the apropriate packing of the hexagonal layers.
There is a more subtle lesson here.  If you come across a "tetragonal" structure and the height is close to 1.41 times the side of the square base, look again -- it could actually be (face-centered) cubic.

In the picture, the cubic face-centered packing is shown, with layers of identical atoms colored in red and blue for better contrast. The unit cell shown with thin lines is the conventional face-centered setting. It has red spheres on the corner, and on the center of the bottom face (the faces not normal to the plain of the paper have blue spheres in their centers).
The unit cell shown with thick dashed lines is what you were asking about. It has square close packing of the red spheres, with square close packing of the blue spheres shifted below. (The third unit cell vector is not shown, and is the same for both, from red to red layer).

So to answer your question, just tilt your head by 45 degrees, and you have your "close square" packing rather than face-centered cubic. As you can see, the three unit cell vectors will not be of equal length anymore, so the unit cell is no longer cubic, and it is body-centered instead of face-centered. The volume is smaller by a factor 2, and the number of atoms per unit cell is smaller by a factor of 2 (2 atom per cell instead of 4 atoms per cell).
In exploring this question, I made use of this site: https://scilearn.sydney.edu.au/fychemistry/iChem/solids.cfm?page=ccp&type=cubic

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


text:
What are possible anode materials for copper(II) sulfate electrolysis?
I'm a student doing a research paper about copper sulfate electrolysis with an experiment by varying anode material, and aside from Pb and graphite (and pure Ti which I think wouldn't work without further alloying) I haven't been able to find or source feasible anode materials as most studies I've found called for Ti-alloys and Pt. I'm also trying to see if Fe, Al, and Zn could be used but I haven't found any literature which even mentions the possibility of using them.
Edit:
My research paper was originally focusing on electrowinning copper, but because of circumstances being as they are I've had to simplify my methodology to using copper sulfate with added sulphuric acid and iron salts to at least somewhat replicate electrowinning electrolyte conditions.

If it's a research paper, I'd try to test as many metals and alloys (e.g. bronze, brass) as possible (also think of various stainless steels, maybe some screws or alike will do, just grab what you can get) and check what will happen when the metals are used as anode or cathode.
For testing cathode materials, it might a good idea to use a copper anode and v/v.
Also, are you supposed to use copper sulphate alone or should you add some sulphuric acid to increase conductivity (which might change the behoviour of some metals)?

Why don't you test Copper ? With a copper anode, you will produce exactly the sane amount of Copper ions in the anode the you reduce at the cathode. So the composition of the solution will not change with the time. Anyway don't try with Fe, Zn, Al, or any other reactive metals, as they will react with copper ions even without electrolysis : $$\ce{Fe + Cu^{2+} -> Cu + Fe^{2+}}$$

My research paper was originally focusing on electrowinning copper, but because of circumstances being as they are I've had to simplify my methodology to using copper sulfate with added sulphuric acid and iron salts to at least somewhat replicate electrowinning electrolyte conditions. So using copper anodes would be somewhat redundant, but thank you anyways I should have specified.

Iron, Aluminum and Zinc are all significantly more reactive than Copper, so they will react with the Copper Sulfate and displace the Copper. I'm assuming that's not what you're trying to do, so you'll likely have to use an inert metal (Platinum, Gold, Silver) that conducts electricity well as your anode.
Edit: When I say that Iron, Aluminium and Zinc are more reactive, what I mean is they have a greater electropositivity compared to Copper and are, therefore, more likely to donate/lose electrons to form positive ions.

When you mean more reactive, is it mainly because their standard potential is higher than Copper's?

The first step should be to ask a question so that you can get an answer which gives you data that you can analyze and discuss. Research generally tries to ask a question that results in some improvement for the world. How to better electrowin copper would be a good question. It seems that you are not just looking to move copper from one electrode to the other, but to understand what might be going on within the solution, because, after all, impure copper in one place is not that different from pure copper in another place.
What is the question that changing anode materials will answer? Different anode materials may dissolve (or not) and change the composition of the solution, which could be desirable, or it might be considered a contamination. Different anode materials will give different voltage vs current curves, because active metals will assist the electrolysis.
In a commercial environment, purifying copper should be routine until the solution becomes so loaded with active metal contaminants that voltage adjustments have to be made to continue plating purest copper, and now a decision has to be made about disposing of the solution. More noble contaminants, like silver and gold would precipitate.
So what could you do with a solution of copper sulfate and ferrous sulfate? Zinc has been mentioned as a frequent impurity, but in large operations, nickel is the soluble impurity of considerable value https://metalrecyclingmachines.com/metal-recycling/metal-recovery-systems/copper-electrolytic-refining-plant.html Lower pH keeps some metal hydroxides in solution rather than precipitating, but will increase the electricity needed to transport a given amount of copper. Nickel sulfate may not be available for you, so I suppose ferrous sulfate could be used.
Perhaps you could pick one anode material and vary the solution to mimic the end of life solution loaded with impurities (in your case, ferrous sulfate). In this case, an iron anode would seem best. Then with variable ratios of copper to iron (and perhaps at a couple of pHs, since this could be changed to assist the process), what voltages, or probably better, what current densities (amps/cm^2) are optimal for squeezing the last bit of copper out of the solution? So the anode could remain the same composition, but vary in area. (You will also have to try just dumping in a wad of steel wool - skip the electrolysis. This is an out-of-the-box shot to just see what will happen. No guarantees.) I suspect that too low a current takes too long but will give pure copper output (as seen on a stainless wire cathode), while a high current density will get the copper out of the solution faster, but it might be contaminated (how? air, iron, solution - I don't know).
Good luck. Be safe.


text:
Comparison of basicity of Aniline, p-Anisidine, and o-Anisidine
I have a question where I am required to compare the basicity of Aniline, o-Anisidine, and p-Anisidine.

Now here's what I do know:-
In ortho substituted aniline, an ortho effect acts which reduces the basicity of the ortho-substituted aniline. But, my text book (Solomons and Fryhle) states that this effect doesn't act if the ortho substituent is $\ce{-OH}$ or $\ce{-OCH3}$. So I should be able to do the comparison here without taking into account this ortho effect.
Now here's my try:
In Aniline, clearly the lone pair is quite a bit delocalised in the benzene ring which reduces its availability. Now, $\ce{-OCH3}$ is a group that has both $\ce{-I}$ and $\ce{+M}$.
Taking a look at the para variant, clearly there is $\ce{+M}$ acting which reduces the extent of delocalization of the Nitrogen lone pair, thus making it more available. Also, the $\ce{-I}$ effect in the para is lesser than that in the ortho variant, so in the para variant, much more of the electron density of the Nitrogen lone pair is available which helps in easier donation, and thus I find the para variant to be the most basic which is all well and good.
The problem arises in the next part of the comparison, between Aniline, and o-Anisidine.
If the ortho effect isn't acting, due to the $\ce{+M}$ effect of the $\ce{-OCH3}$ group,  the lone pair of the Nitrogen should be more available for donation than that of Aniline, (but less than the availability in the para case, due to stronger $\ce{-I}$ in the ortho case), and thus the Ortho variant should be more basic.
But my book states that the basicity order is : p-Anisidine>Aniline>o-Anisidine
I cannot really understand why the basicity order of the last two (Aniline>o-Anisidine) is the way that it is, if no ortho effect is acting and $\ce{-OCH3}$ has a $\ce{+M}$ effect.
PS: Just took a look at Peter Sykes and the order given is same as that in my book. This this due to the fact that in ortho, the molecule goes slightly out of plane, so $\ce{+M}$ effect may not be as effective?

@Mithoron No it doesn't. It says that the inductive effect in the Ortho variant is stronger than the resonance but doesn't really explain why or how.

Do believe the basicity order is : p-Anisidine>Aniline>o-Anisidine ? That is easily looked up.

@MaxW I know it can be LOOKED UP. I need to be able to explain the order during my examinations


text:
How do we obtain the π molecular orbitals for allene via Hückel theory?
For a conjugated system, for instance 1,3-butadiene, we can try to obtain the π molecular orbital energies and wavefunctions via Hückel theory. In our given molecule, there are 4 interacting Pz orbitals. By creating a 4x4 Hückel matrix and solving for the energy eigenvalues, we can arrive at 4 possible linear combinations (of p-orbitals) to get 4 molecular orbitals.
However, how can we use Hückel theory to deal with a molecule like 1,2-propadiene (allene)?
My attempt: I thought that since this molecule has essentially 2 separate π-systems:

we may be able to apply Hückel theory to each one individually. Then, we can take all the possible combinations of the individual energies to arrive at the possible total energies. But I am not sure if this is correct. What would be the correct way of approaching this problem?

For p-orbitals $i$ and $j$ which are orthogonal to each other (i.e. different colour in your diagram), the off-diagonal matrix elements are simply zero:
$$\langle i | H | j \rangle = 0$$
If we ignore the overlap matrix,* the eigenvalue equation to solve therefore has the form:
$$\mathbf{Hc} = E\mathbf{c},$$
or in explicit form, using the fact that the matrix elements of $\mathbf{H}$ are $\mathbf{H}_{ij} = \langle i | H | j \rangle$,
$$\begin{pmatrix}
\alpha & \beta & 0 & 0 \\
\beta & \alpha & 0 & 0 \\
0 & 0 & \alpha & \beta \\
0 & 0 & \beta & \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}
=
E
\begin{pmatrix}
c_1 \\ c_2 \\ c_3 \\ c_4
\end{pmatrix}$$
This matrix has block-diagonal form, so solving for the eigenvalues of the entire matrix simply reduces to solving for the eigenvalues of each individual $2\times 2$ block (lots of discussion can be found on Google). For a physical interpretation, this means that both π systems have no interaction with each other, so you can indeed solve them individually.
Each block gives rise to one MO with energy $\alpha + \beta$ ("bonding") and one with energy $\alpha - \beta$ ("antibonding"), so all in all you will have two degenerate MOs with energy $\alpha + \beta$ and two degenerate MOs with energy $\alpha - \beta$.

* If you don't ignore the overlap matrix $\mathbf{S}$, whose matrix elements are defined by $\mathbf{S}_{ij} = \langle i | j \rangle$, this becomes a "generalised eigenvalue equation"
$$\mathbf{Hc} = E\mathbf{Sc}$$
However, because the overlap matrix also has the same block-diagonal form, ignoring the overlap matrix doesn't change the fact that the problem is separable into two $2\times 2$ blocks. It merely changes the exact eigenvalues and eigenstates by a small amount.

Does this mean that our OVERALL molecule has 4 MOs? I am having trouble understanding how each MO could represent the entire molecule, since each come from a single 2x2 (representing one of the π-systems) .

@JamesBond yes; well, four π-type MOs, at least. And yes, MOs are typically more delocalised than the conventional Lewis bonds, but the keyword is "typically". There is no requirement that they must stretch over the entire molecule. In fact, even localised lone pairs on an atom can be perfectly valid MOs sometimes. (Think e.g. the 2px and 2py lone pairs in HF, which cannot mix with the other orbitals due to symmetry mismatch.)

That makes sense. I completely forgot that there isn't a strict requirement for MOs to always be delocalized throughout the entire molecule. Like some sigma MOs: formed via overlap of orbitals between 2 atoms (so mainly localized), but nonetheless considered a MO.

Indeed. In this case, if you were to go beyond simple Huckel theory, there is probably going to be some overlap between the π-type orbitals and the C–H σ-type orbitals on the opposite side of the molecule. The overlap will not be amazing anyway, though, because they are relatively far apart. Huckel theory assumes complete separation between the π and σ frameworks, so that overlap is set to zero, which allows you to treat the π system separately (this is completely true in systems like ethene or benzene, where the π and σ orbitals are always orthogonal because of symmetry).


text:
Purification of primary amines using Schiff base immobilization
I want to purify all primary amines containing compounds in a complex mixture (cell-extracted metabolites). I'm unaware of any published methods to do so. So I am thinking of making my own using a reversible Schiff base immobilization.
My input is a 1 mL metabolite extract that contains hundreds of compounds with a total molarity of ~1 mM. Of this ~50% contain a primary amine. Glyoxal agarose typically has 40–50 μmol aldehydes per mL, so capacity is not an issue.
Impurities that I want to get rid of are, among others, organic acids like malate, pyruvate and succinate, sugars like glucose and glucose-phosphate, lipids and all non-volatile salts.
The idea is to use Schiff base chemistry:
$$\ce{R^1CHO + R^2NH2 <=>[$k_1$][$k_2$] R^1CH=NR^2 + H2O}$$
An immobilized aldehyde is used to capture all compounds containing primary amines. A Schiff base is generally stabilized under high pH and destabilized at low pH. It can be further stabilized by dehydration to shift the equilibrium towards the Schiff base side. Therefore, I would add my input at pH = 10 in 95% MeOH, wash under dehydrated conditions e.g. using 100% MeOH, and elude at pH = 2 using 0.1 M aqueous formic acid. Alternatively, adding glycine in the elution buffer to compete off remaining amines.
For immobilized aldehydes I'm thinking about using aldehyde activated beads e.g. glyoxal agarose beads:

My questions are:

Are there alternative ways to purify primary amines from a complex mixture which I'm unaware of?

Did anyone publish a paper that used Schiff base chemistry to purify primary amines?

What would be your odds that this could work? Is it worth pursuing?

Any ideas for improvements? Eluate has to be run on mass-spectrometer, so my requirements are:
a. no non-volatile salts;
b. no solvents with MW above 150 Da.

Might be easier for a first pass to pass the solution through an SCX cartridge which will retain all basic species then elute them off with ammonia soln.

Thank you @Waylander. I have been thinking along those lines too but it's great to be reminded. My two biggest reservations are: 1) many metabolites are charged but not containing the primary amines that I am looking to purify (e.g. a slew of phosphatidylcholines), and 2) very low pH would be required to push all primary amine containing metabolites into positive charge (e.g. glutathione with pK1 = 2.12, and probably even lower for S-Succinylglutathione).

I think it would be better to reduce this post to 1 and 2. Then hope for an answer and ask a followup question. I think it currently is too open-ended, and especially point 3 invites for too much speculation and is primarily opinion based.

Let's take your questions one at a time.

Are there any alternative ways to purify primary amines from a complex mixture that I am unaware of?

Your end application is LC-MS.  There are plenty of papers on amine-specific derivatization methods for metabolome analysis. For example, there is one by Zhao et al. [1]. Instead of purifying the primary amine metabolites and isolating the purified mixture as a separate step, they just rely on the analytical instrumentation to be able to identify primary amines from their derivatization pattern.
Alternately, as suggested in the comments by @Waylander, you could acidify the sample to form ammonium ions, and then pass the sample through a cation exchange column, where only positively charged ions will bind.  As you noted, this will not be perfectly selective for primary amines, because species like quaternary ammonium salts etc. will also be retained.
You could combine the derivatization approach with the SCX approach, of course.  If you chose a primary-amine-selective derivatization, you might get a selective method.  At some point, however, the complexity of the method will become a drawback.

Have anybody else, published or unpublished, used Schiff base chemistry to purify primary amines?

Well this question is impossible to answer for the unpublished case.  I couldn't find any published methods in a quick search, which makes me suspect it is not the best approach to your problem.  But I could be wrong of course!

What would be your odds that this could work? Is it worth pursuing?

Unanswerable without data.  I suspect that secondary amines, at least some times, will also react with the aldehydes to form Schiff-base-like iminium cations, and may also be selected by your proposed method.

Any ideas for improvements?

I think derivatization, without separation or isolation of the amines, would be an attractive direction for you to consider.  The paper I linked, its references, and the papers that reference it might be good starting places.
Reference

Zhao, S.; Li, H.; Han, W.; Chan, W.; Li, L. Metabolomic Coverage of Chemical-Group-Submetabolome Analysis: Group Classification and Four-Channel Chemical Isotope Labeling LC-MS. Anal. Chem. 2019, 91 (18), 12108–12115. DOI: 10.1021/acs.analchem.9b03431.


text:
Can one exclude calibration points using a combination of permutation testing and a non-parametric test statistic?
I work in bioanalytical chemistry. I have used an analytical instrument (LC-MS) to measure the response (peak area) of several compounds spiked in to equivalent samples at known concentrations. Each sample, at each concentration, was analysed three times over three separate days. Spike concentrations, for this example, are: 0 (blank), 1, 2, 5, 10, 50, 125, 250, 500, 1000 ng/L. I now wish to construct sensible calibration curves using linear least squares regression (hence, assuming homoscedacity of data, but still acknowledging the potential for outliers) of the response (peak area) versus spike concentration (i.e. the spike concentration).
From visual inspection of data files, it appears that for some compounds there are samples that were spiked at > 0 ng/L that are indistinguishable from blank samples (i.e. those at 0 ng/L). I’d like to test my suspicions statistically and, if significant at a reasonable threshold, would exclude these samples prior to performing the least squares regression.
Given that, for many compounds, the data are often non-‘normal’ around the lower-end of the calibration curve in such dataset, I thought to use permutation testing to generate a sampling distribution. Here, I’d define a null hypothesis that the median peak area of blank samples (0 ng/L) is equal to that of non-blank samples at a defined concentration (e.g. 1 ng/L), i.e. a statistical test of location. Drawing from the world of parametric tests, I’d hoped to use: [median(peak areas of non-blank samples at single concentration) - median(peak areas of blank)] / median absolute deviation(peak areas of non-blank samples at single concentration).
Here, the sampling distribution would be generated by repeatedly randomising class labels (i.e. concentration labels for 0 ng/L and the calibrant level being tested) and calculating the test statistic. The p-value would be derived as the proportion of the permuted test statistics that were equal-to or greater-than the actual measured statistic, i.e. as calculated before randomising labels. If the p-value turns out to be greater than some pre-determined threshold (probably going to be 0.05 or 0.01), I accept the null hypothesis. Given that I would repeat this test for each concentration level, I’d also plan to apply a Benjamini-Hochberg correction to the final list of p-values and reject then exclude concentration levels with adjusted p-values above the cutoff (I’d aim for either 0.05 or 0.01 – yet to decide).
So, feedback on the following would be great:

Does the proposed test statistic seem reasonable as a non-parametric replacement to the t-statistic?
Is it appropriate to exclude samples from the regression, for which the p-value for the test statistic exceeds the defined threshold?

Personally, I oppose excluding samples unless there are strongly compelling reasons to do so. I do not know what your detection limit may be, or which definition you may use, but I recommend reading reference 29 in my answer here.

Or you might try these two papers: H. Evard, A. Kruve, I. Leito, “Tutorial on estimating the limit of detection using LC-MS analysis, Part I: theoretical review”, Anal. Chim. Acta 942 (2016) 23-39 and
H. Evard, A. Kruve, I. Leito, “Tutorial on estimating the limit of detection using LC-MS analysis, Part II: practical aspects”, Anal. Chim. Acta 942 (2016) 40-49. The paper I suggested in my first comment also addresses weighted regression, if the measurement noise is heteroscedastic.

This is more of a comment, but a typical comment doesn't allow a figure...
You have a very wide range (1-1000) for the independent variable. Using a simple linear least squares analysis isn't appropriate.
I grabbed the image below from webpage https://www.mathworks.com/matlabcentral/answers/392636-confidence-band-around-linear-least-squares-line

The point I want to make is shown by the shaded area. Assuming homoscedasticity, you wouldn't weight the errors. Thus the confidence interval for the line is narrowest at the center of the fitted range and gets much wider at either end of the range. So your low samples ( 1, 2, 5, and 10) will essentially be irrelevant.
So you need to assume hetroscedasticity and weight the sample errors somehow.

1. Does the proposed test statistic seem reasonable as a non-parametric replacement to the t-statistic?

Just because in general non-parametric tests are less powerful than parametric tests, I would try hard to avoid using a non-parametric test on parametric data.

2. Is it appropriate to exclude samples from the regression, for which the p-value for the test statistic exceeds the defined threshold?

I'll assume that you mean to exclude some of the standards not samples. It is possible to exclude standards, but don't use three standards and then drop the one that you "know" isn't on the line. :-)

Hope this helps a little. I just don't know enough about S/N for LC/MS or the detector system that you're using to tell you what you should do.
Your statement "Given that, for many compounds, the data are often non-‘normal’ around the lower-end of the calibration curve in such dataset,..." is disconcerting.

Before running a nonparametric analysis, try viewing the problem from a parametric vantage of Generalized Least-Squares. To quote a source:

In that regard, GLM repeated measure is a statistical technique that takes a dependent, or criterion variable, measured as correlated, non-independent data.  Commonly used when measuring the effect of a treatment at different time points.  The independent variables may be categorical or continuous.  GLM repeated measure can be used to test the main effects within and between the subjects, interaction effects between factors, covariate effects and effects of interactions between covariates and between subject factors.  GLM repeated measures in SPSS is done by selecting “general linear model” from the “analyze” menu.  From general linear model, select “repeated measures” and then preform “GLM repeated measures.

So, one of the important aspects of your data is that you have repeated observations, likely with correlated error structure.
Also, you measured your explanatory variable concentration amounts. Depending on how easy this was accomplished, there could be some associated measurement error also in your explanatory variable. This may lead to bias regression coefficients (see this source).

+1 for some way other than ordinary least squares to analyze the data. However without knowing a lot more about the instrumental technique (LC-MS wasn't my thing...) and the data set, just jumping on some statistical technique seems inappropriate.

Actually, much more inappropriate is the possible employment of an ad hoc non-parametric technique as suggested by MRJ.  Now, GLM involves a covariance matrix that incorporates a weighting of data as you, yourself, suggested, which I agree is likely appropriate, so exactly how is this remiss?

AJKOER - I'd agree that some type of weighting is almost certainly needed. However (1) GLM is only one of many techniques (2) GLM will need variances determined from the data. The sample size to get good precision on variances is much larger than the sample size to get good precision on expected values. // Who knows it may be enough to assume that the variance is proportional to the concentration of the standard.

MaxW: Your comments are not precisely accurate. For example, a simple version of a GLM model could contain a user-specified diagonal covariance matrix. It is also not restricted to the current data only, aka apriori weights can be specified in what is known as Ridge Regression (a form of Bayesian Linear  Least-Squares).


text:
Constructing phase diagram given temperature-concentration-phase data
This question may look straightforward to be answered but somehow I can not figure it.
I received these papers [1-2]. Here, the papers (both papers study same system) provide complete information of phases in every point. What I am still mulling on is that usually I was given the phase diagram and locate compositions and phases in the phase regions. Suppose that the phase diagrams are not shown, how can I draw the phase regions from these points? How can I know that at some points outside the data makes transition to other phases when the temperature or composition is varied?
I would be very happy if someone can explain me to draw the phase regions as shown in the papers.
Thank you.
References

Campbell, A. N., Campbell, A. J. R. J. Am. Chem. Soc. 59, 2481-2488 (1937).
Rhodes, F. H., Markley, A. L. J. Phys. Chem. 25, 527-534 (1921).

Images
From ref. 1

https://pasteboard.co/J54SVM6.png

https://pasteboard.co/J54T2yI.png
From ref. 2

https://pasteboard.co/J54S1VH.png

https://pasteboard.co/J54SqoS.png

https://pasteboard.co/J54SDhZ.png

https://pasteboard.co/J54SM7y.png
Text about above figure

"Concentrations, Total and Partial Vapor Pressures, Surface Tensions and Viscosities, in the Systems Phenol—Water and Phenol—Water—4% Succinic Acid", Campbell, A. N., Campbell, A. J. R. J. Am. Chem. Soc. 59, 2481-2488 (1937) https://doi.org/10.1021/ja01291a001 // Paper behind ACS firewall...

"The Freezing–Point Diagram of the System: Phenol–Water," Rhodes, F. H., Markley, A. L. J. Phys. Chem. 25, 527-534 (1921). https://doi.org/10.1021/j150214a003 // again behind ACS firewall

@MaxW Wait, I'm going to put the pictures of the data there

@MaxW Done. I uploaded the pictures on other website than Imgur because Imgur is blocked in my country.

I added pictures to Imgur so that they will stay with the question.

Figure 1 in last image is the phase diagram for the phenol/water system. Presumably the letters on the diagram would be described in the article's text.

@MaxW Here is the description of the letters https://pasteboard.co/J5517x1.png

@MaxW Other pages in the paper describes other figures (viscosity, vapour pressure, so on)

@MaxW I can share the pdf to somewhere on website. Is it alright to put it there for a moment? (I will delete it later).

some of the user's here will have access to the paper. I've reached my curiosity limit.

@MaxW Alright :)


text:
Why is the strontium ion smaller than the potassium ion?
The ionic radius of the $\ce{Sr^2+}$ ion is $\mathrm{132\,pm}$, while the ionic radius of the $\ce{K^+}$ ion is $\mathrm{152\,pm}$. Why is this the case? I would have thought that since $\ce{K^+}$ has one less shell of electrons than $\ce{Sr^2+}$, its radius would be significantly smaller.

See also Libretexts.org - Periodic_Trends_in_Ionic_Radii

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Some quick analysis: $\ce {Rb^{+}}$ has an ionic radius of 166 pm. What this means is that we cannot attribute the effect to d-orbital contraction, i.e. poorer shielding effect provided by the 3d orbitals. Going from $\ce {K^{+}}$ to $\ce {Ca^{2+}}$, there is a significant drop from 152 to 114 pm. Similarly, there is a big drop from $\ce {Rb^{+}}$ to $\ce {Sr^{2+}}$, from 166 to 132 pm. It may well be the case that the addition of another proton cause a significant reduction of the ionic radius, such that it outweighs the increase in radius due to presence of an additional electron shell.

Related: https://chemistry.stackexchange.com/questions/64335/why-is-the-radius-of-the-caesium-cation-smaller-than-that-of-the-chloride-anion.


text:
Why is it important to do the nitrosation of aniline to diazonium salts at low temperatures (0-5°C)?
The reaction the formation of diazonium salts starting from aniline (nitrosation of aniline), which should be performed at low temperature.

Diazonium salts are unstable and decompose readily at higher temperatures, sometimes explosively

If you work at higher temperatures, it's an expensive way to produce phenol.

The diazonium salt solution contains the ion $\ce{C_6H_5NN^+}$. At room temperature it is decomposed into phenol $\ce{C_6H_5OH}$ according to the equation $$\ce{C_6H_5NN^+ + 2 H_2O -> C_6H_5OH + N_2 + H_3O^+}$$


text:
Number of optically active isomers of [Co(NH3)2Cl2(NO)(OH)]?
My Attempt:
Since Compound is of the form M[a2b2cd]
I got three optically active isomers of the form M[(ab)(ab)(cd)] , M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)]
and since NO is a bidentate ligand, total number of optically active isomers should be multiplied by 2 due to linkage isomerism.
Hence i was getting the answer as 2*3=6 but answer given was 4.
Can you please help me with this?

NO isn't bidentate ligand, is it? Neither ambidentate...

@Zenix then i am still getting only 3 isomers but according to answer it is 4. Also doesnt NO donate through both oxygen and nitrogen?

I would write that as (ON)...

@RajeshKoothrapalli NO might be able in general to donate through both N and O, but it's sterically incapable of donating from both N and O to the same metal ion center.

Try drawing out [M(ab)(ab)(cd)]: you'll see that it's actually optically inactive!

So, there are 2 optically active isomers, which along with their respective enantiomers adds up to 4 (M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)],  as you've mentioned).

if a and b are drawn on the vertical lines then there is no plane of symmetry so isnt it optically active?

The plane of symmetry should pass through every type of ligand that occurs an odd number of times, here that means is must be vertical passing through the c and d ligands.  It exists in this isomer but you drew it wrong.  Also to get a superposition you need to combine refkection with a twofold rotation, which is allowed for an achiral molecule.

@Oscar Lanzi Right, my bad. I'll change it.

You have two choices really. (1) use the plane of tge paper and add the twofold ritation. (2) use a plane perpendicular to tge paper and passing through c and d, bisecting the a-M-a and b-M-b angles; this does not need the additional rotation.

@Oscar Lanzi can you explain what you mean by two fold rotation? i was not able to superimpose the mirror image on the molecule.

Since there are two of $\ce{NH3}$ and $\ce{Cl}$, therefore in trans-form, we can always draw a plane of symmetry (in figure) to which the trans-line (in figure) is normal as depicted below for $\ce{Cl}$,

So, only that configuration will be optically active which has both the groups in cis-form as follows,

which gives us two optical isomers (this and its mirror image). The other two optical isomers can be drawn by swapping the positions of $\ce{NO}$ and $\ce{OH}$ in the above figure.
Hence, we get a total of four.


text:
Which of the following compound will undergo rearangement?
Question

Attempt
I believe the answer should be B,C and D. In C and D the oxyge(or Nitrogen in D) can rearange by donating lone pair to carbocation.
But the books answer says B. Where am I wrong. Please help.

In (C) and (D) the carbocation is quite stabilized by conjugation with the lone pairs of O and N respectively. Rearrangement here refers to one caused by a Wagner–Meerwein rearrangement , as resonance is not a form of rearrangement. Only in B does a rearrangement lead to a more stable intermediate with more alpha hydrogens

@YusufHasan in (b) better will be ring expansion after rearrangement..

@Zenix I would say that ring expansion isn't absolutely necessary in this case, because 5-membered rings aren't strained a lot as compared to 3 or 4 membered ones. They will only go for 6 membered if there isn't a huge loss in hyperconjugation. Here, ring expansion after rearrangement would lead to a loss of alpha hydrogens


text:
What happens in an exothermic reaction at the atomic level?
Exothermic Reactions are those chemical reactions in which heat is released.
How does this happen?

What I mean is where does  the heat energy come from? Which form of energy is converted to heat energy? What makes some reactions release heat? What affects the amount of heat energy released?
Thanks!

Exothermic reactions are the reactions in which relesed energy is more than that which was absorbed. It comes from the internal energy of the reacting species Eg.vibrarational energy.

Absorbed energy? I'm sorry, I don't understand quite what you mean by absorbed energy here...

To initiate a reaction we need to supply energy in any form which makes the reacting species overcome the activation energy

Here is a nice description from Richard Feynman.

What is the energy supplied when water reacts with quick lime to form slaked lime. $CaO$ + $H_{2}O$ $\longrightarrow$ $Ca(OH)_{2}$. This is an example of exothermic reaction

It would be the dissociation energy of water and lattice energy of CaO. Of course the reaction is self sustained at room temperature  because at this temperature the reacting molecules have sufficient energy. As the temperature is higher than the equilibrium temperature the reaction occurs spontaneously. As the temperature comparison is relative even if we don't supply energy (raise the temperature) the reaction occurs. This does not mean that there was no energy absorbed.

@NicolauSakerNeto That's a very nice video, but I believe Feynman's description of photosynthesis contains an error! AFAIK carbon fixation does not separate the carbon from the oxygen in CO2, but incorporated the whole molecule into the sugar it's creating. The oxygen plants release comes from water-splitting.

@gardenhead You are correct, he got that part wrong. Fortunately, he's going for a "feeling" of what happens, rather than a rigorous description, so it's acceptable for a lay audience.

What happens in an exothermic reaction at the atomic level?

In a chemical reaction, bonds are broken and bonds are formed. For example, when elemental oxygen and hydrogen form water, we are breaking bonds in the elements and making heteroatomic bonds connecting hydrogen and oxygen atoms:
$$\ce{O2 + 2H2-> 2H2O}$$
To emphasize the bonds, we could also write:
$$\ce{O=O + H-H + H-H -> H-O-H + H-O-H}$$
In a thought experiment, we could first separate the elements into "isolated" atoms, and then recombine them to form water. Bond dissociation energies estimate how much energy it takes to separate bound atoms into "isolated" atoms. In the case of the example reaction, it takes less energy to break all the bonds of the reactants (one double bond connecting oxygen atoms and two single bonds connecting hydrogen atoms) than to break all the bonds of the products (four single bonds between oxygen and hydrogen atoms). As a result, there is an excess of energy that becomes available, resulting in an exothermic reaction.

What affects the amount of heat energy released?

The reaction follows a more subtle path than completely ripping molecules apart into atoms and then making new molecules. However, because enthalpy is a state function, the result in terms of reaction enthalpy is the same. So the only thing that affects the reaction enthalpy is the set of reactants and products.
The amount of heat depends on how you set up the reaction. If the temperature before and after the reaction is the same, if the pressure is constant and there is no non-PV work, it is equal to the reaction enthalpy. If the reaction is used to do work (electrochemical cell, steam engine etc), if it is a photochemical reaction, if it gives off light, or if any other of the conditions given in the previous sentence are not met, that will affect the amount of heat transferred.

Which form of energy is converted to heat energy?

The energy "comes from" the different energy of electronic states in the reactants and products. You can call this chemical energy, or a combination of kinetic and potential energy of the electrons. We see that electrons going from a higher energy state to a lower energy state give off energy when considering atomic spectra (sodium giving off a yellow light in the Bunsen burner, for example). In the case of reactions, electronic states are different between reactants and products not because they are in excited states, but because of the different configuration of nuclei giving rise to different electronic states.

How does this happen?

Unless the reaction gives off energy directly (by emitting electromagnetic waves - rather rare), the available energy gets transformed into kinetic energy of the nuclei (e.g. molecular vibration), and then is transferred to the surrounding by the common ways of heat transfer until the system is at thermal equilibrium again. If the reaction is endothermic, some of the activation energy (provided by molecular collisions) is not "returned to the pool" but instead used to get the electrons into their higher energy states in the products (compared to the reactants), leading to a net heat transfer from surroundings to the chemical reaction.

The dissociation energy curves of O2 or H2 could be useful to illustrate how the potential and kinetic energies are reorganized. Here is a generic curve https://www.britannica.com/science/potential-energy-curve.

@Karsten on a different note, would you allow any user of the these sites to ask questions, comments despite their reputation numbers? (because nobody is an expert in every area, and even the experts do have the freedom to ask infant questions without being barred for rep-numbers when in doubt or when learning, understanding things)

@bonCodigo This would be a good question for meta. From the help pages: "The three most important activities on Chemistry Stack Exchange are Asking, Answering and Editing - none of which require any reputation at all!". https://chemistry.stackexchange.com/help/whats-reputation

@Karsten well, context of my comment was more to do with that rep (whether 10 or 10,000) should not be taken as a measure to assumed a user knows everything or stop them from asking questions in their own learning process. There was a person who was practically throwing tantrums at me (often others too) hideously pointing at my reps in comparison to a question I was trying to solve;

@Karsten that sorta behaviour in the community sites like this clearly not welcoming, and the arrogant went on to take their revenge by downvoting practically most of my questions and answers on this site at the expense of his own reps. That was sick! if I was a chem expert I would be teaching in an university and asking my colleagues, wouldn't bother asking questions here...

@bonCodigo We try to be welcoming while also allowing each single user to decide on what they want to upvote, downvote, comment on or ignore. This freedom allows for community-based evaluation of answers, but also allows for occasional frustration, disagreement or disappointment. You are welcome to ask questions at any level, and I expect everyone to treat you with respect.

@Karsten Been here for awhile, I can appreciate the nature of the various. This wasn't just my experience. We get to see a thing or two off here. I have only experienced it once last year and yes ignored it. I guess a human frustration is perhaps still tolerable compared to what it may be with darn AI chatbots...if it comes to conquer the sites/communities. Good day!

An exothermic reaction is one which results in heat or light being released by the system and absorbed into the surroundings, usually referred to as a change in enthalpy. Enthalpy $H$ is the energy contained within the bonds of a molecule.  This amount, the bond energy, is roughly the amount of enthalpy necessary to break the bond.
Important Concept!
An input of energy is always required to break a bond ($\Delta H\gt0$) which means the process is endothermic.
Anytime a bond is formed energy is released ($\Delta H\lt0$), so forming bonds is exothermic.
Try not to get the above $\Delta H$'s mixed up with any $\Delta H_\mathrm r$. The above is relevant to breaking/creating single bonds; reactions are dynamic processes that are much more complicated that formulas make them appear to be.
In an exothermic reaction there is less energy in the products' bonds than there was in the reactants' and when that energy difference is absorbed the the surrounds it causes a positive increase of heat in the surroundings. The difference in enthalpy can be calculated with the following equation:
$$\Delta H_\mathrm r=\Sigma H_\mathrm{f(\text{products})}-\Sigma H_\mathrm{f(\text{reactants})}$$
$H_\mathrm f$ is the enthalpy of formation, defined as the change in enthalpy when one mole of a substance is formed from its pure elements. You will most likely see it used along with a degree symbol, $H^\circ_\mathrm f$ indicates this reaction takes place at STP. You will usually be able to look up the necessary $H^\circ_\mathrm f$ in a reference table, but the amount is related to bonds strength, length, charge imbalance, and some other attributes.
Exothermic reactions have a negative change in enthalpy $(\Delta H_\mathrm r\lt0)$ The energy released during an exothermic reaction is the difference in bond energy between the reactants and the products. The energy lost by the reaction becomes energy gained by the system which will result in an increase of heat (or light). The difference in signs for the change in enthalpy of the reaction and change in energy of the system is a common point of confusion, so be sure to remember this:
Exothermic:

Energy flows from the system into the surroundings, $(\Delta H_\mathrm r\lt0)$

Endothermic:

Energy flows from the surroundings into the system. $(\Delta H_\mathrm r\gt0)$

Thanks, by the way, what level Chemistry is this?

This is the level I was taught when I first took chemistry in 10th grade; I didn't take any intro level courses at undergrad but I imagine this would be covered in the second semester of the first course, along with a lab component.
If anything is unclear or you want to know more feel free to ask; I'm extremely rusty and the summary I wrote is pretty rough by I can definitely explain more if you have any specific questions.

@Rajdeep_Sindhu hey I didn't realize we weren't supposed to have casual conversations in the comments; looks like some got deleted. If you ever wanna chat about chemistry you can find me on Twitter or telegram, same name as here. Cheers

There is also [chat] for casual conversation, and yes, obsolete comments will get cleaned up occasionally. This may be somewhat related: Is Bond Formation “Strictly” Exothermic?


text:
2-Methylpropan-2-ol and Iodoform test
Can 2-methylpropan-2-ol be converted into methyl ketone the action of $\ce{I2}$ in the presence of alkali base and thus give iodoform test?
Answer given to this question is that it can't give iodoform test. Please could someone confirm this with an explaination.

$\ce{C^2}$ of "2-methylpropan-2-one" will be pentavalent...

No it cannot. 2-methyl-propan-2-ol cannot be oxidised further and no rearrangement is likely to occur to give a methyl ketone

No, 2-methylpropan-2-ol cannot give iodoform test, as it cannot be converted into a methyl ketone.
Generally, alpha hydroxy methyl type (this is not the actual nomenclature) groups can give iodoform on oxidation.


text:
cheap anti-fog agent!
What are some safe and widely available alternatives to specifically formulated anti-fog sprays/treatments? Specifically for use on glasses, goggles, and face shields.

If the humidity from breath reaches the goggles, it indicates that the mask is not fitted properly. The exhaled air is supposed to go through the fabric, filtering out some of the droplets and aerosol, not escape into the goggles. Otherwise, the mask is not providing any benefits.

Due to COVID-19, we've widely deployed face shields at work. Initially, there were a lot of employee complaints over fogging, and at the time it was difficult to find a steady supply of commercial products. We ended up testing a number of different alternatives, and finally settled on baby shampoo as it was both effective and unlikely to cause eye irritation if it got mixed with sweat and dripped into the employees eyes.
Get a couple drops on the surface, then use a paper towel (or preferably a microfiber rag or lint-free paper) to wipe it over the entire surface being protected. Once dry, it works very well.

Baby shampoo works well to prevent fog on swim goggles too.

Here is an excellent review article, which seems unbiased:
To quote the summary of favorites:

Among those who were able to fulfill their promise, causing them to earn the trust of many anti-fog spray users, are Optix 55, Just Add Water, Splaqua, Z Clear, SURF BUTTA, Cressi, Hendlex, iGK, Scott Sports, and Atlan.

My personal recommendation is the most expensive one, Optix 55, as it lasts for days, while many others require frequent re-applications, which can apparently be not particularly trivial.
DISCLAIMER: I have no affiliations with any product and limited personal use.

I bought some anti mist sprays to find out more about them. there is no information about the ingredients of the spays on the bottle. I want to know what is the main antifog substance inside them.


text:
Find order of reaction 2A → B + C

Derive the integrated rate expression for a reaction of type:
$$\ce{2A -> B + C}$$

I first need to determine the order of the reaction in order to find integrated rate law of that particular order. How to find the order of above reaction? Please note that there is no extra data is provided, and the rate law expression is also not given.
It appears there is not enough data in the question to come to a conclusion. Normally, experimental data would be provided or at least the rate expression. I tried to search a lot, but it was all in vain.

Order of reaction is an experimental data.

The question demands integrated rate law for the above reaction which is only possibly to derive after knowing the order...hence need to find the order...

Basically the question is wrong.Please try to understand order is an experimental data

You can't simply tell the order from an equation like this..

If
$$
2\ce{A ->[$k$] B + C}
$$
is assumed to be an elementary step then the reaction is second order
$$
-\frac{1}{2}\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]^2
$$
because the sum of the exponents on the right-side of the equation is $2.$
The integrated rate expression is found by taking the rate of change, rearranging and integrating from an initial concentration $[\ce{A}]_0$ at $t  =0$ to $[\ce{A}]$ at time $t$:
$$
\int_{[\ce{A}]_0}^{[\ce{A}]}\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = -2k \int_0^t\mathrm dt
$$
to yield
$$
\frac{1}{[\ce{A}]} = \frac{1}{[\ce{A}]_0} + 2kt
$$

This is almost a complete answer. What is missing is to emphasize that you can't answer the original question without making the crucial assumption that the reaction is an elementary step - the question the OP cites is not well posed.


text:
Why does high blood gas partition coefficient of a inhalational anaesthetic drug mean slower induction of anesthesia?
High blood gas partition coefficient of a drug corresponds to higher solubility of the drug in blood leading to high concentration of the drug in blood with respect to gas (air in alveoli). If the drug is present at high concentration in the blood, wouldn't it correspond to faster delivery to target tissues-leading to faster anaesthetic effects or induction? What is meant by "more time is required to attain partial pressure in blood for induction" if the drug has high blood gas coefficient"?

I don't have much knowledge on pharmacology, but I remember the mechanism of action of gaseous anesthetics has something to do with the gas dissolving in the isolating lipid sheath around axons, messing with nerve impulse transmission. Even inert gases like nitrogen can do that at high enough pressures. The blood is mostly water, a polar solvent. I guess if the gas has high affinity to polar blood, it will have lower affinity for nonpolar lipids. But I'm not sure.


text:
Why does my metal phone ring stand becomes iridescent, and turns back to normal after a while?
I got this metal phone ring stand from online. When I was meddling with it, I notice the ring started to produce iridescence. (See photo-only the centre portion, the moveable ring is not affected) I believe the mild heat from my hand caused this to happen. I left it alone on a table overnight, and the next morning, the iridescence disappeared. Te ring is back to normal.
Whenever I applied mild heat again (I placed it at the back of my vacuum cleaner for 5seconds), the iridescent will appear, but will then disappear again after I leave it overnight.
The iridescent consists of greenish, pinkish and purplish hues. It doesn’t seem to be forming at the surface, but rather it seems to be forming in the interior.
The seller claims that the ring is made up of zinc alloy, but it is magnetic. Some guesses of the material that might be in the product: zinc alloy, nickel, steel, chrome plating (the centre portion is very reflective like a mirror) Please correct me if my guesses are wrong.
I am just very perturbed by this phenomenon occurring in such simple object and am curious to find out what causes this!!
Thank you!!

This is an interference phenomenon, more specifically interference of light between two thin layers with an air gap. Recall the beautiful soap bubble as an example. Whatever coating your have on this object gets very very slightly separated from the surface upon mild heating. This generates an air gap between two layers.
Iridescence is a different phenomenon, which is angle dependent in the sense that you see different colors at different angles.
If you have over heated an empty stainless steel frying pan by mistake, that also forms a interference pattern (purple tinge). This is because of an oxide coating on the surface but it is not reversible. You have to scrub it away.

I see, thank you for your clarification. However, this pattern on the ring does not change colour/shape when I look from it a different angle. Hence, is this observation still consistent with the proposed thin film interference theory?

yes that is why this term iridescence is not correct in the context of your query. Thin film interference is more appropriate.

Elaborating on M. Farooq answer, if you have a surface covered in a thin transparent coating (oil, grease, oxide, et cetera), depending on the angle of incidence and wavelenght of light, the upper reflection may be either boosted or hampered by light coming from the lower reflection, so you see different colors depending on the angle. I did a lame drawing:

This fact is used in powerful techniques like X-ray diffraction. See the wikipedia article on Bragg's law.


text:
CNT Raman ratio of G band to 2D band?
I believe that the G band (~1600 cm-1) shows the stretching mode of the graphene in a carbon nanotube (e.g. chiral, zigzag, armchair) and the 2D band (~2680 cm-1) shows the number of walls in the CNT. I vaguely remember there is some information that can be gleaned from the ratio of the two. Is this the case?

I think there might be a mistake in your question...
Around $\pu{2680 cm-1}$, there is the $S_1$ band (attributed to overtone or combination scattering). I think the $2D$ band you are referring to is the band shouldering the $G$ band (also called $D'$) which is around $\pu{1680 cm-1}$.
About the ratio, there are two ratios that are often used to characterise $\mathrm{sp^2}$ carbon based materials such as the CNT.
$$\text{Intensity ratio} =  R_1 = \frac{I(G)}{I(D)}$$
$$\text{Area ratio} = R_2 = \frac{A(G)}{A(G)+A(D)+A(D')}$$
Both these ratios help characterise the degree of organization of your sample.
For more information or more details, I would recommend checking out Spectrochim. Acta, Part A 2003, 59 (10), 2267–2276


text:
Will increasing temperature have an effect to the pH of honey to be able to carry out an acid-base titration?
After doing some research on how to decrease the viscosity of honey to undergo acid-base titration with NaOH, one of the ways was by increasing the temperature of honey but the problem is that the pH of honey will decrease so is this true or not? and what is the best way to decrease the viscosity of honey to undergo acid-base titration? (I am comparing acidity of honey from different regions)

o.O Only reasonable way to do something related to pH, titration and honey would be dissolve a bit it water and titrate the solution. You also seem to be missing the point of titration and that's not the first misguided question on this topic.

So you mean if I add the same amount of water to all the different kinds of honey would not affect viscosity? also, can you please more specific about the point I'm missing? Thank you very much!!!

Only diluting it may times over would give viscosity  and sugar concentration low enough to talk about titrating it. The point is why to do this? I imagine estimating content of carboxylic acids, but for eg. aminoacids may complicate things. Another thing is I'm pretty sure there's some standard procedure to be followed, and you're reinventing wheel...

Measure the volume or mass (depending on the titration approach you are using, you can do it by mass) of undiluted honey that you are titrating. After that, provided that you don't splash any out or loose some by transferring to another container, you can add as much distilled water as you like. It will make no difference to the titration result. Yes, the pH of the sample will be nearer 7, but with a titration you are measuring the amount of the acid or base not the concentration. Once you have the amount, you calculate the original concentration from the volume (the most usual) or mass of the sample, measured before you added any water, indicator or titrant.
You can also do a precise dilution beforehand and measure aliquots out of this. This is quite normal and works for the same reason as above: you are measuring the amount by titration and know what volume or mass of the original sample is contained in your aliquot. I usually prefer to use a good balance rather than volumetric glassware, so pre-diluting does not offer me much advantage, though I think it would generally be the most common non-automated technique.
Adding water will certainly decrease the viscosity, but that is good in that it will allow good mixing of your titrant. I don't think it is at all important that the viscosity is adjusted by the same amount for each honey. That's not what you're measuring by titration.
High heat would wreak all kinds of havoc on your honey, but slight heating would lower the viscosity and increase the reaction rate without negatively affecting your result. I doubt it's worth it though as you can just dilute with distilled water to achieve good mixing. Adding water will also give a clearer solution, which is useful if you are using an indicator not a meter.


text:
Non-conducting substance with a glass transition temperature around 400 to 600 °C?
Am looking for a substance to seel two pieces of ceramic that will have molten plastics flowing between at 300+ °C however silver solder circuit traces inside the device prevent me from going much above 600 °C without destroying them – otherwise I would be using some form of pottery glaze but the lowest melting one I have found reference to is around 1000 °C.
The ideal would be something I could bring back up to the glass point and separate the pieces for servicing.
Can any one either suggest a compound or recommend a resource that would allow me to search based on constraints like I outlined?

That´s very tricky.  One trick could be high-temp 2K epoxy glue. They are usually filled with iron powder, but maybe you can fill a regular 2K epoxy with something ceramic. I´ve used "JB Weld", it survives 250°C for a while. At 500°C it will surely die after a while, so you should be able to take it apart.

There are a lot of special (and somewhat expensive) cement mixtures on the market, which come with low shrinkage and defined thermal expansion ratios. Those are a one-way street of course.

Had thought about JB Weld also but that 250°C is a definite deal breaker since the device will be operating for hours on end. Am now thinking of going with sodium silicate (water glass) as an adhesive. Instead of raising the temperature to melt the seal (1000+C) it will be dissolved using an ultrasonic cleaner with plain water.

Water glass I have also tried at the time, it´s not bad except that it has a high shrinkage. Don´t try to fill space with it, only use like a real glue, as thin as possible between a flat contact. I wouldn´t bet on getting it off via ultrasonication after it was as 300°C for a while.

But I´m curious: What are you building there?

New type of filament extruder for a 3d printer.


text:
Change in concentration when acids are diluted
I came across this formula to calculate the resultant [H]+ concentration when two weak acids are mixed (Both acids of equal volumes are mixed together).
The only problem i have with this is that, we are completely ignoring the fact that the concentration of both the acids are going to be halved after they are mixed together since, the volume is doubling. Is there anything i'm missing? The only way, this could be correct is if the concentrations that are given to me in the question are those after the acids have been mixed.
EDIT: This is the question i was talking about

This is the solution i was given

Comments are not for extended discussion; this conversation has been moved to chat.

Let suppose $c, c_1, c_2$ are concentrations of weak acids, and $K_\mathrm{a}, K_\mathrm{a1}, K_\mathrm{a2}$ are their respective acidity(dissociation) constants.
The equations $\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$ for a weak acid, respectively for 2 acids $\ce{[H+]}=\sqrt{K_\mathrm{a1} \cdot c_1 + K_\mathrm{a2} \cdot c_2}$
are approximately valid, if $c, (c_1, c_2) \gg K_\mathrm{a}(, K_\mathrm{a1}, K_\mathrm{a2}) \gg 10^-7 = \sqrt{K_\mathrm{w} }$, what also implies $\ce{[H+]} \gg 10^{-7}$.
Then we can approximate all $\ce{H+}$ comes from the acid dissociation, but OTOH, all acid is in non-dissociated form, leading to $$K_\mathrm{a} = \frac{\ce{[H+][A-]} }{\ce{[HA]}} \simeq \frac{{[H+]}^2 }{ c}$$ and $$\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$$
If, OTOH, $c \ll 10^{-7}=\sqrt{K_\mathrm{w}}$  and $c \ll K_\mathrm{a}$ we can approximate all acid is dissociated.
Then $$K_\mathrm{w} =  \ce{[H+][OH-]}$$
$$\ce{[H+]} - \ce{[OH-]} = c_1 + c_2$$
$$\ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}} = c_1 + c_2$$
$${\ce{[H+]}}^2 - (c_1 + c_2) \cdot \ce{[H+]} - K_\mathrm{w} = 0 $$
$$\ce{[H+]} = \frac{c_1 + c_2 \pm \sqrt {{(c_1 + c_2)}^2 + 4 \cdot K_\mathrm{w} }}{2}  $$
$$\ce{[H+]} = \frac{\pu{2.1e-9} \pm \sqrt {{(\pu{2.1e-9})}^2 + \pu{4e-14} }}{2} = \pu{1.0105e-7 mol/L }$$
$$\mathrm{pH} = 6.995$$
It is obvious, that if with full dissociation $\mathrm{pH}$ does not reach even the value $6.99$, the result of partial dissociation cannot be $6.7$

The pH of a mixture of two weak acids can be obtained as:
$$\mathrm{pH} = \sqrt{K_\mathrm{\alpha 1}\times c_1 + K_\mathrm{\alpha 2}\times c_2}$$

Ok, that's what we are trying to prove, but the equation is wrong from the start. The right hand side calculates $\ce{[H+]}$, not pH.  Note that $c_1$ and $c_2$ are the nominal concentrations of the acids in the mixture.

Consider two weak acids say HOCN and HCOOH
For HOCN: $K_\alpha = \dfrac{\ce{[H+][OCN-]}}{\ce{[HOCN]}}$
For HCOOH: $K_\alpha = \dfrac{\ce{[H+][HCOO-]}}{\ce{HCOOH}}$

Both of the above $K_\alpha$ expressions would be valid for any concentrations of the weak acids and any values of the $K_\alpha$ constants.

Also, $\ce{[H+]}$ in solution will be same and thus:
$$\ce{[H+] = [H+] by HOCN + [H+] by HCOOH}$$
$$= \ce{[OCN-] + [HCOO-]}$$

This assumes that $ \ce{[OCN-] + [HCOO-] \gg [OH-]}$

$$\ce{[H+]} = \dfrac{K_{\alpha,\ \ce{HOCN}}\times \ce{[HOCN]}}{\ce{[H+]}} + \dfrac{K_{\alpha,\ \ce{HCOOH}}\times \ce{[HCOOH]}}{\ce{[H+]}}$$
$$\therefore \ce{[H+]^2} = K_{\alpha,\ \ce{HOCN}}\times \ce{[HOCN]} + K_{\alpha,\ \ce{HCOOH}}\times \ce{[HCOOH]}$$
$$\text{or}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
$$\ce{[H+]} = \sqrt{ K_{\alpha,\ \ce{HOCN}}\times \ce{[HOCN]} + K_{\alpha,\ \ce{HCOOH}}\times \ce{[HCOOH]}}$$

Now we know that $C_\ce{HOCN} = \ce{[HOCN] + [OCN-]}$. If $\ce{[HOCN] \gg [OCN-]}$ then $C_\ce{HOCN} \approx \ce{[HOCN]}$ and we can substitute $C_\ce{HOCN}$ for $\ce{[HOCN]}$.
Likewise for $C_\ce{HCOOH}$ and $\ce{[HCOOH]}$
Substituting we get the desired result.
$$\therefore \ce{[H+]^*} = \sqrt{K_\mathrm{\alpha,\ \ce{HOCN}}\times C_\ce{HOCN} + K_\mathrm{\alpha,\ \ce{HCOOH}}\times C_\ce{HCOOH}}$$
But the hydrogen ion concentration has been notated as $\ce{[H+]^*}$ to denote that the value is conditional. The condition is that
$\ce{[H+]^* \gg 1\times 10^{-7}}$ for the value to be valid.
$\ce{[H+]^* \gg 1\times 10^{-7}}$ is somewhat fuzzy.

If $\ce{[H+]^* \gt 1\times 10^{-6}}$ then we can have 1 significant figure in the answer.

If $\ce{[H+]^* \gt 1\times 10^{-5}}$ then we can have 2 significant figures in the answer.

If $\ce{[H+]^* \gt 1\times 10^{-4}}$ then we can have 3 significant figures in the answer. (3 significant figures for such calculations is more wishful thinking about the chemistry than reality.)

I've been scratching my head on this part, but I believe that $\ce{[H+]^* \gg 1\times 10^{-7}}$ encompasses any other assumptions.


text:
How can randomly oriented molecules in a liquid solution create a certain polarization pattern (d or l)?
How can a large number of moving, randomly-oriented chiral molecules in a watery solution twist light only to the left or only to the right?
Wouldn't a levorotary molecule become dextrorotary if I turned it around?

Please see here: https://chemistry.stackexchange.com/a/6277/79678.

Does this answer your question? Molecular chirality and optical rotation

In that sense, you have to think of it like a screw. A screw has the same twist regardless of how you look at it. That's what we mean by left or right handed.


text:
Which elimination is this?
This is the question:
$$\ce{Cl-CH2-CH2(Cl)2->[alc KOH][Δ]?}$$
Now, I got the answer right but I think that the reaction should go by the $\ce{E1}$ mechanism because of the high temperature/heat, however, my teacher says that the mechanism will be $\ce{E2}$.  I don't understand why it will be like that. Everyone seemed to agree, so I didn't have the guts to ask in class. Please tell me why it's $\ce{E2}$? And then, what's the use of high temperature/heat? It was there to make ionization easy, right?

I don't think high temperature facilitates E1, I also think it should go by E2... It's better to ask your teacher such conceptual doubts. Don't be shy in such cases....

I guess this link about alpha elimination will be useful, it woyuld be similar to E1cb I dare say

@Zenix E1 is generally more favorable under heating just because you need to promote dissociation of the electrophile. That doesn't mean of course that if you need heating, it's an E1...

@Zhe won't any elimination reaction be more favourable at high temperature, why specifically E1? I read your answer's last para... And first para.... I guess for E1 to occur, dominating factor will be stability of carbocation, which isn't stable (due to reasons you mentioned)

Sorry, i was a bit misled about the temperature/heat factor. While, solving problems I gave preference to first order reactions if there was heat involved. It usually worked, and unknowingly I made it a rule of thumb to predict first order reactions if heat was there. Clearly , it was a mistake. @Zenix E1 mechanism has two steps (if we look at kinetics), formation of carbocation is the first one. Obviously heat would favour ionization.

@Zenix If that was your claim, I don't really have any complaints. Frequently E2 reactions are spontaneous enough that no heating is required. In that case, if you heat it, yes, it will go faster, but you're also more likely to have less selective reactions or side reactions.

An E1 mechanism is highly unlikely. I don't know what level you're at, but one could invoke the Hammond Postulate in this case, and look at the high energy cation intermediate. There are three chloride leaving groups that could generate one of two intermediates. Both of these intermediates are primary cations. A chloride substituent isn't likely to be able to stabilize this much via resonance but more likely destabilize due to inductive effects.
The other potential mechanisms to consider (that do not involve a cation intermediate) are E2 and E1cb. E1cb might work for the first step depending on where the anion is formed, especially if the inductive effect of the chlorines is enough to help stabilize the carbanion. E2 just requires an anti-periplanar arrangement, which isn't hard here with free rotation.
The reason I'd favor E2 and the reason the reaction likely requires heat is the fact that hydroxide is just an OK base. Add to that the fact that it's in a polar protic solvent that might further dampen basicity. If you tried this reaction with an amide base (e.g., LDA) in toluene, I would guess no heating is necessary.
Heating is generally helpful for E1 because the dissociation of the leaving group is slow. However, the need to heat a reaction is by no means indicative that it is E1 (it might not even be an elimination!). Any number of factors could conspire to make the reaction slow, in which case, heating might help make the reaction proceed at a faster, more practical rate.

Ok, thanks. But i still don't understand "it's in a polar protic solvent that might further dampen basicity". I understand that polar protic solvent will dampen nucleophilicity but why basic strength ? Being in polar protic solvent(alcohol here) should increase basicity because $\ce{C2H5O-}$ ion is more basic than $\ce{OH-}$ , you can verify. Being a strong base is the reason why elimination occurs instead of substitution. Correct me if i'm wrong.

No, I'm saying that hydroxide would be a stronger base in something like DMSO which is aprotic. The hydrogen bonding in a protic solvent will make basicity weaker. It doesn't matter if you're looking at an alkoxide or hydroxide.

Cf. the pKa values for water in water and in DMSO: https://www.studocu.com/en-ca/document/university-of-victoria/advanced-organic-synthesis/other/evans-pka-tables/689204/view

can you please provide a reason why hydrogen bonding would decrease basicity. I just can't understand why.

The same reason hydroxide is more basic in a nonpolar solvent. Generally, bases have high energy lone pairs; that's what makes them a base. In our case, the base is also negatively charged. Any stabilization of the base, for example, via charge-dipole interactions in a polar solvent or hydrogen bonding in a protic solvent, make it less basic.


text:
Explanation of Specific Chemical Shifts on an H-NMR Spectrum
Recently, I've been looking into chemical shifts on H-NMR Spectrums and I came across a graph I didn't quite understand...
As you can see from the spectrum attached, the signal emitted from the protons near the Cl (colour coded orange) is further downfield than the signal emitted by the protons near the O (colour coded dark blue). However, shouldn't the protons near the Oxygen be further downfield due to their proximity towards the more electronegative atom? How can this be explained?

Actually, more interesting question is how $3^\circ \  \ce{-OCH}$ is more shielded than  $1^\circ \  \ce{-OCH3}$.


text:
Main industrial method for producing ethyl ethanoate (acetate)?
I am aware there are various methods, however, which is most prevalent in industry? Researching on Google has indicated Fischer esterification or Tishchenko reaction and other versions. Which is most common?

This source (though 10 yrs old)  says it is the condensation of ethanol and acetic acid with cat. cH2SO4  https://www.icis.com/explore/resources/news/2007/11/02/9075322/ethyl-acetate-production-and-manufacturing-process/  The Avada process which makes it from ethlyene and acetic acid is also significant

According to ACS Molecule of the Week:

As you might expect, ethyl acetate was first synthesized from ethanol and acetic acid. The reaction was the classic acid-catalyzed Fischer esterification, which dates back to 1895. This is still the most widely used commercial synthesis. An alternative method is the Tishchenko reaction in which acetaldehyde disproportionates in the presence of base to the alcohol and the acid that then esterify in situ.

The mechanism of  Tishchenko reaction is depicted in following diagram where $\ce{R = CH3}$:


text:
Overlap integrals in Hückel theory
In Hückel theory we are only interested in π systems, where $\mathrm p_z$ orbitals overlap. One of the approximations in Hückel theory is that the overlapping $\mathrm p_z$ orbitals are orthonormal:
$$
S_{ij} =
\begin{cases}
1\quad i = j \\
0\quad i \neq j \\
\end{cases}
$$
If orbital overlap leads to bond formation, how can the overlap integrals, between $\mathrm p_z$ orbitals, be equal to 0? Maybe I am missing something obvious, but doesn't a π-bond demand non-zero overlap between p-orbitals?

You have the right idea, see here: "This neglect of orbital overlap is an especially severe approximation. In actuality, orbital overlap is a prerequisite for orbital interaction, and it is impossible to have $H_{ij}=\beta$ while $S_{ij}=0$." So its just to make the math easier and you are correct that it makes the result less physically realistic.

The overlap of an orbital with itself is 1 for normalised orbitals and approx 0.25 for adjacent orbitals and less for those further away. Huckel made a dramatic assumption that the overlap is zero except for $i=j$. The simplicity and application to a range of important compounds has given this theory a surprising success. However, the expectation of accurate results should not be high.

I see, so in reality, there IS overlap. Just the calculations are simplified with this approximation. I wonder though, is this approximation absolutely mandatory? (i.e. can I evaluate the overlap integrals precisely, if I wanted to, while still using Hückel theory?)

@JamesBond There is the Extended Hückel Theory, which includes overlaps (J. Chem. Phys. 39, 1397 (1963)).

@FelipeS.S.Schneider I initially thought Extended Hückel theory introduced overlap between the pi and sigma frameworks. I was not aware  that it also included the overlap between the p-orbitals!

Hückel theory would still work if there would be overlap.
In fact, it is possible to create a completely new basis by applying a transformation to the basis set by which a new set of orthogonal basis functions is created. The advantage of setting $\hat{S}$ (the overlap matrix) to $\hat{I}$ (the identity matrix) is that the eigenvectors of the matrix equation can be directly interpreted in terms of the $\pi$ orbitals in the system.
There is the opposite view in Extended Huckel Methods whereby the assumption is that the overlap entries depend upon the off-diagonal elements. In ordinary Huckel theory, the diagonal elements are represented as
\begin{align}
\alpha_{ii} &= \langle \psi_i | \hat{H} | \psi_i \rangle \\
\hat{H} &= -\frac{\hbar^2}{2}\nabla^2 + V
\end{align}
But in the extended thoery, the offdiagonal elements are not denoted by $\beta_{ij} = \delta_{ij}$ but more as
\begin{align}
\beta_{ij} &= - \frac{C}{2}\Big \langle \psi_i \Big |-\frac{\hbar^2}{2}\nabla^2 + V \Big | \psi_j \Big \rangle\\
&= - \frac{C}{2} \left(\alpha_{ii} + \alpha_{jj} \right)S_{ij}
\end{align}
Where $S_{ij}$ is the usual notation for the overlap between the corresponding atomic orbitals.
Here $C$ is a constant, usually taken equal to $7/4$ and has units of energy.


text:
Why is CuI2 not stable in solution while other copper(II) halide salts are?
Why is $\ce{CuI2}$ (copper (II) iodide) not stable in solution while other copper(II) halide salts are? And how can you tell this from the redox potentials?

https://chemistry.stackexchange.com/questions/53768/what-happens-when-potassium-iodide-is-added-to-a-copper-ii-salt?rq=1

FWIW, $\ce{CuI2}$ has never been isolated as solid state but it is made as a greenish-blue solution by reacting cuprous iodide and iodide. It is extremely unstable and readily decomposes. It forms stable cuproammonium salt, $\ce{CuI2.4NH3.H2O}$ on reacting with ammonia/ammoniated cuprous iodide. The compound $\ce{2CuI.CuI2.4NH3}$ is also known. It exist as brilliant-green needles, insoluble in water, but soluble in ammonia solution.

It is common knowledge that copper (II) iodide ($\ce{CuI2}$) does not exist. Thus, your question "why is $\ce{CuI2}$ not stable in solution?" is clear on that, but unlike this previous question, you wasn't clear about what happens (or what you have observed). I have no knowledge of what would happen when some one add $\ce{KI}$ solution to $\ce{CuSO4}$ solution. Yet, I can speculate what would happen by considering few redox half reactions:
$$\ce{I2 + 2e- <=> 2I-} \qquad E^\circ = \pu{0.536 V} \tag{1}$$
$$\ce{Cu^2+ + 2e- <=> Cu} \qquad E^\circ = \pu{0.342 V} \tag{2}$$
$$\ce{Cu^2+ + e- <=> Cu+} \qquad E^\circ = \pu{0.153 V} \tag{3}$$
For convenience, let;s rewrite equation $(1)$ in oxidation format:
$$\ce{2I- <=> I2 + 2e-} \qquad E^\circ = \pu{-0.536 V} \tag{4}$$
Sum of $(2)$ and $(4)$ would gives the appropriate redox equation:
$$\ce{Cu^2+ + 2I- <=> I2 + Cu_{(s)}} \qquad E_\mathrm{EMF}^\circ = \pu{-0.194 V} \tag{5}$$
Similarly, sum of $(3)\times 2$ and $(4)$ would gives the appropriate redox equation:
$$\ce{2Cu^2+ + 2I- <=> I2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.383 V} \tag{6}$$
The excess $\ce{I-}$ would react with $\ce{Cu+}$ and $\ce{CuI}$ would precipitate as a result:
$$\ce{Cu+ + I- -> CuI_{(s)}} \qquad K_\mathrm{sp} = 1 \times 10^{-12} \tag{7}$$
Therefore, albeit forward reactions of equations $(5)$ and $(6)$ are not thermodynamically favorable (e.g., for equation $(5)$, $K_\mathrm{eq} \approx 2.7 \times 10^{-7}$), both forward reactions persist and they would be driven to completion because of one product separate as a solid (Le chatelier principle).
Let's now consider oxidation half reactions of other three halides:
$$\ce{2F- <=> F2 + 2e-} \qquad E^\circ = \pu{-2.866 V} \tag{8}$$
$$\ce{2Cl- <=> Cl2 + 2e-} \qquad E^\circ = \pu{-1.358 V} \tag{9}$$
$$\ce{2Br- <=> Br2 + 2e-} \qquad E^\circ = \pu{-1.087 V} \tag{10}$$
If $\ce{Cu^2+ -> Cu+}$ reduction happens, the respective three redox reactions are:
$$\ce{2Cu^2+ + 2F- <=> F2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-2.713 V} \tag{11}$$
$$\ce{2Cu^2+ + 2Cl- <=> Cl2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-1.205 V} \tag{12}$$
$$\ce{2Cu^2+ + 2Br- <=> Br2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.934 V} \tag{13}$$
And if $\ce{Cu^2+ -> Cu}$ reduction happens, the respective three redox reactions are:
$$\ce{Cu^2+ + 2F- <=> F2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-2.524 V} \tag{14}$$
$$\ce{Cu^2+ + 2Cl- <=> Cl2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-1.016 V} \tag{15}$$
$$\ce{Cu^2+ + 2Br- <=> Br2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-0.745 V} \tag{16}$$
Since values of $E_\mathrm{EMF}^\circ$s are so high negatives, these reactions are practically not happen. For example, $K_\mathrm{eq}$s are in the range of $6.14 \times 10^{-26}$ (for eq. $(16)$) and $1.55 \times 10^{-92}$ (for eq. $(11)$). Also keep in mind that, contrary to $\ce{CuI}$, $\ce{CuBr}$ and $\ce{CuCl}$ are also water soluble and $\ce{CuCl}$ is water soluble as well as hygroscopic.


text:
Ripening of bananas?
So, obviously the ripening of bananas involves changing it’s chemical composition. For example, ripe bananas have a much higher amount of sugar.
It is also known that reaction rates are generally increased with increasing temperature, yet bananas ripen faster in a refrigerator.
Why do bananas ripen faster in the fridge, even though it’s colder? And is there any difference (except ripening time) between ripening a banana in the fridge, and having it ripen in room temperature?

One word: ethylene

It would then apply to all fruits. But it seems to apply just to fruits from warm regions, that do not like low temperature. It may be rather alternative process to ripening.

Here is a paper which discuss the science behind ripening of bananas.

That paper doesn't seem to have any induced ripening with changing temperature.

Similarly, even tomatoes are not recommended to be stored in fridge, as it affects their taste when they are (warmed up ) used later.

I like bananas, but I almost went bananas doing some research for this question!
Let me post some quotes - too much info to post the sources:
1) Is it safe to store my bananas in the refrigerator?
Dr. Brecht: “Never, never, never put bananas in the refrigerator!
2) The answer, far from dissuading me from keeping bananas in the fridge, only confirms my habit. Banana-eaters are divided into two fairly equal camps: those who like the delicate flavour of the green (unripe?) banana and those who prefer the quite different strong flavour of the yellow (over?)ripe banana. Using cold to inhibit the enzymes which aid the ripening process is just what is needed by the 'greens' but should be avoided by the other camp.
3) If they're kept at a cold temperature, the enzymes that enable them to ripen are inhibited. And as those enzymes become inactive, other enzymes operate more efficiently. Some cause cell damage, while others (browning enzymes) cause the skin to blacken. Bananas do best if stored at around 10°C.

4) Bananas require gas, that is ethylene, to trigger the ripening process.
5) Cooler temperatures also slow down the chemical ripening process. Once bananas are ripe to your likening, store them in the fridge.
6) De gustibus non est disputandum
So it seems that bananas do change because of the cold, but the process is not exactly ripening. It does involve a chemical change, which is so complex that it can be predicted and analyzed, but is not useful as a guide to developing a tasty banana for a specific person. Too much variation in the people who eat bananas! I did an experiment some time ago, wondering if the economy of buying green-yellow bananas (with noticeably thick, firm skins) was significantly less than buying bananas with exactly the color for immediate eating (deep yellow for me), where the skins were noticeably thinner, softer. How much weight loss do I incur and pay for, but don't get the benefit of? Not much, it seems, about 10% of the weight of a banana, maybe 15% if you got a green one and let it go almost brown.
So banana color may not be the whole story: enzyme activity occurring at different temperatures may affect taste without being "ripening" - a banana may go from underripe to overripe without ever being exactly ripe! The process of marketing, from green to yellow, is described in some detail, and interestingly, ripening bananas develop heat (and I would guess, ethylene also), so the recommendation was to unpack the banana boxes quickly when they get to the grocery store to reduce the temperature (and C2H4) so the bananas keep longer.


text:
What type of FDA process does a 222nm Kr-Cl excimer UV-C light have to go through
There are plenty of articles and preliminary research about this type of light being able to kill virus as effectively as 254nm UVC light, but not harmful to human skin and eye. This is wonderful news to the current mess created by COVID-19. Not sure whether there are companies already applying for FDA approval, and what kind of process does it need to go through which will basically translate to how long we have to wait to see this available?

No wonder there are saying that this type of lamps are quite expensive. Not as much probably as in 87. But for this to be available to general public, it for sure has to be affordable.


text:
What is heterogeneous catalysis in gas phase?

Give examples of heterogeneous catalysis (i) in gas phase and (ii) in liquid phase.

Now, I came to know that heterogeneous catalysis involves different phases between the catalyst used and the reactants. That way, either the catalyst can be in solid phase where the reactants can be in gas/liquid phase. But in the above question, what does "in gas phase" mean? Is it about the gas phase of the catalyst or the gas phase of the reactants?
Also, it would be of great help if you could provide me with examples of the above question.

Look up the “Haber process” in wiki and scroll down to “Catalysts”. One of the most important catalyzed processes in the world.

"in gas phase" refers just to the phase of the reactants and products. The catalyst in these heterogeneous systems is usually solid phase.

"Gas phase" means that the reactants are gases
The basic answer is that heterogeneous catalysis "in the gas phase" just means that the materials being reacted are gases. They catalysts can be liquids or solids (more common).
A simple, but very, very important example is the reaction used to convert ammonia into nitric acid (vital for fertiliser production, important for many organic chemistry building blocks and explosives). In this reaction gaseous mixtures of air and ammonia are passed over platinum gauze and, initially, produce NO (nitric oxide) which is further oxidised homogeneously to dinitrogen tetroxide which ultimately reacts with water to give nitric acid. A good summary of the reaction and its history is given in this Johnson Matthey article.
So gas-phase reactants + solid catalyst -> product. Gas phase heterogeneous catalysis.

Here is a liquid phase example, 'Iron–cobalt mixed oxide nanocatalysts: Heterogeneous peroxymonosulfate activation, cobalt leaching, and ferromagnetic properties for environmental applications, to quote, in part:

(i) need of heterogeneous activation of sulfate salts using transition metal oxides, (ii) nanoscaling of the metal oxide catalysts for high catalytic activity and promising properties with respect to leaching, and (iii) easy removal and recovery of the catalytic materials after their applications for water and wastewater treatments. In this study, we report a novel approach of using Fe–Co mixed oxide nanocatalysts for the heterogeneous activation of peroxymonosulfate (PMS) to generate SRs targeting the decomposition of 2,4-dichlorophenol...

where SRs refers to sulfate radicals.
For gas phase, I cite: 'The importance of heterogeneous and homogeneous reactions in oxidative coupling of methane over chloride promoted oxide catalysts'. To quote:

The formation of ethene from ethane and methane in a silica reactor has been studied both in the presence and in the absence of chloride-containing catalysts.

Key to these mixed catalysts is surface chemistry activation reactions.


text:
Double Replacement Reaction in which everything cancels out
We are currently trying to work out the double replacement solubility reaction for Nickel (III) Nitrate and Rubidium Sulfate.
$$\ce{Ni(NO3)3 + Rb2SO4 -> Ni2(SO4)3 + RbNO3}$$
Balancing gives:
$$\ce{2Ni(NO3)3 + 3Rb2SO4 -> Ni2(SO4)3 + 6RbNO3}$$
We believe all four compounds are salts and therefore split all four into their cations and anions:
$$\ce{2Ni^{3+}(aq) + 6NO3-(aq) + 6Rb+(aq) + 3SO4^{2-}(aq) -> 2Ni^{3*}(aq) + 3SO4^{2-}(aq)  + 6Rb+(aq) + 6NO3-(aq)}$$
At this point, we want to eliminate the matching spectator ions. But they are the same on both sides of the equation and we're left with nothing and no precipitate.
We suspect there is something or things that we are misunderstanding. We don't even know if it's ok for these types of reactions to not have a precipitate and if it is ok, if that is actually the case here with these two compounds.

"We believe all four compounds are salts and therefore split all four into their cations and anions:
2Ni+6NO3+6Rb+3SO4⟶2Ni+3SO4+6Rb+6NO3"
Charges are missing. In solution, you'll just have a mix of ions. If you want to isolate any species, check the solubility of all compounds (take in account that there might be several forms with different water content) and try fractionated crystallisation.

We did look up the charges, but I omitted them in the question because we had pretty low confidence in those as well. We did check the solubility of all and all four compounds "seemed" to be soluble according to the limited solubility chart we were given.

Regarding solubility, do you have access to the CRC Handbook of Chemistry and Physics? Another source to check out is PubChem (https://pubchem.ncbi.nlm.nih.gov/).
Regarding charges, as there is no redox involved, give it a try. Hint: Rb behaves like sodium (e.g. NaCl), sulphate is 2-

Most importantly, do $\ce{Ni(NO3)3}$ and $\ce{Ni3(SO4)3}$ exist? Most of nickel compounds are in +2 oxidation state. For instance, there is no solubility date for $\ce{Ni(NO3)3}$ and $\ce{Ni3(SO4)3}$ to be found.

If it's a true double displacement reaction, not everything will dissociate. Otherwise, it's just two dissociation reactions mashed together.

This problem has no solution, because Nickel(III)  nitrate and Nickel(III) sulfate do not exist.

This problem is impossible, because $\ce{Nickel(III) nitrate}$ and $\ce{Nickel(III) sulfate}$ do not exist. You loose your time looking for their solubilities. The only known compounds where Nickel has an oxidation number $+3$ are Nickel oxide $\ce{Ni_2O_3}$ (which is insoluble) and the complex ion $\ce{[NiF_6]^{3-}}$. Simple $\ce{Ni^{3+}}$ ions do not exist in water. As a consequence, no corresponding nitrate or sulfate exist. And no double replacement may occur with these non-existing substances.
See N. N. Greenwood and A. Earnshaw, Chemistry of the Elements, Pergamon Press, Oxford, England, 1986, p. 1340

Thanks for your help. It did turn out that that was what we were meant to find. And basically the fact that everything cancels out is an indicator of that.


text:
What is the edge length of ZnS, given the radii of Zn and S ions?
What is the edge length of $\ce{ZnS}$, given the radii of $\ce{Zn^2+}$ and $\ce{S^2-}$ ions?

You need to know the crystal structure too.

In ZnS, $Zn^{2+}$ is located in alternate tetrahedral voids and $S^{2-}$ at the vertices of cube of ZnS crystal. If $a$ is the length of cube, then the tetrahedral void is present at a distance $√3a/4$ from the ion at the nearest vertex on that diagonal of cube,i.e.,$S^{2-}$.
So, $Zn^{2+}$
 is at a distance
 $√3a/4$ from $ S^{2-}$.
Tetrahedral void is present between centre of cube and it's vertex on the diagonal of the cube.
So $√3a/4$ = radius of $Zn^{2+}$+ radius of $S^{2-}$=$d$ say.
Hence, $a$=($4/√3a$)×$d$

Is this for sphalerite or wurtzite?

It is for sphalerite (ZnS in FCC crystal structure).


text:
Are the reactions in a simple cell overall exothermic? Why?
For context, I'm in high school. I'm trying to mentally connect electrolysis with the concept of energy changes and kinda having trouble.
I understand the bottom line is that a simple cell converts chemical energy to electrical energy in a spontaneous redox reaction, and in an electrolytic cell, electrical energy is used to cause a non-spontaneous chemical reaction to occur.
I've found resources that vaguely say the overall reaction (?) in a simple cell is exothermic. I find this unclear and confusing.

If the reaction is exothermic then the products should have lower energy than the reactants. What are the products and reactants involved in the system, to derive the enthalpy change from?
Where did the energy lost go?
Don't both bond breaking (oxidation @ anode) and bond making (reduction @ anode) occur? Does the bond making somehow involve more    energy? How? (does this have something to do with the cation getting    discharged at the cathode being of a less reactive element than the cation formed at the anode? must the electrolyte have a cation that fulfils this requirement?)

For the sake of simplicity assume a simple cell setup with just electrodes, external circuit, and electrolyte. Not connected to a lightbulb or something.
Thank you

Your second paragraph is exactly correct. But then you wander off a bit. Please search for “Lead-acid battery” in wiki and look at the electrode reactions and net reaction for this extremely common and useful rechargeable battery. Then recharging is electrolysis: input electrical energy results in re-formation of the substances present in the fully charged battery.