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How should the reactivity of 1-propanesulfonyl chloride be expected to compare to tosyl chloride for sulfonylation?
I've come across some scientific literature (e.g., https://doi.org/10.1021/jm400751p) that notably feature 1-propanesulfonyl chloride (PsCl) in the selective activation of hydroxy groups, such as a primary alcohol at the expense of a secondary one. Since tosyl chloride (TsCl) is so much more commonly used than structural homologues of mesyl chloride (MsCl) in general alcohol activation, this has prompted me to wonder: Starting from and explaining with organic chemistry concepts, how should the reactivity of PsCl for alcohol activation (or amino-group sulfonylation) be expected to compare to TsCl?
Obviously, like MsCl, PsCl has access to alpha-carbon protons (though at least somewhat less acidic due to the longer alkyl chain of EDGs), so sulfene intermediates would be expected to form in the case of a sufficiently strong base such as triethylamine or DIPEA. Then, generating less of a reactive sulfene intermediate at a given time (compared to MsCl, see https://doi.org/10.1021/jo981532p for reference) should favor PsCl's use in ensuring greater selectivity in sulfonylation (specifically, when one substrate site is more nucleophilic than another). From a structural perspective, I am inclined to think of PsCl as being TsCl minus the methyl group and the "bottom" (from a 2-D perspective) three C-atoms of TsCl's benzene ring (along with, of course, no C-C double bonds), so sterically-speaking PsCl's S-atom should be at-least-slightly more available for a nucleophilic attack (e.g., courtesy of a weaker base like pyridine). Though, from the electronics-perspective I'm not sure what impact TsCl's aromaticity would have on the S-atom's electrophilic character (charge delocalization perhaps leading to greater electrophilicity?). TsCl also naturally doesn't have an alpha-carbon with protons, which would seem to preclude its having an E1cB or E2-like reaction capacity.
Here is a paper (https://doi.org/10.1021/jo00821a030) in which one can see that the rate of secondary hydroxy group activation in anhydrous pyridine (serving as solvent, in addition to catalyzing hydroxy-group activation) clearly proceeded faster for PsCl than TsCl. Since pyridine would be catalyzing hydroxy-group activation through an SN2-like attack on the PsCl/TsCl S-atom, this would seem to indicate that steric factors play some role in PsCl being more reactive than TsCl.

In your first citation there is apparently no amine base present, which would preclude initial salt formation with PsCl or deprotonation, leaving direct displacement by $\ce{RCH2OH}$. Hence, a source of selectivity. I recall from the distant past that MsCl reacts by a different mechanism in the presence of pyridine vs. triethylamine. In the former case, salt formation occurs; In the latter case, sulfene formation occurs. Sorry, but I don't have a reference.

@user55119 For the 1st citation, see Scheme 4 on the 3rd page and particularly the 6th page (which is in the Experimental Section). 2,6-lutidine, a pyridine derivative with di-ortho-substituted -CH3 groups, was used as the base (i.e., nucleophilic catalyst). 100% yield for the di-propanesulfonate (as crude material) was attained.

I only looked at reagent "b" in scheme 4 that has no mention of 2,6-lutidine while the experimental section does. IMO, 2,6-lutidine is only going to act as a scavenger for generated HCl. Certainly too hindered for attack at sulfur and unlikely to deprotonate PsCl. This still consistent with my thoughts above.


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How to synthesize 1,1,1-trichloro-2,4-dimethylpentan-2-ol from methyl isobutyl ketone?
Chlorobutanol can be made with chloroform and acetone in the presence of potassium or sodium hydroxide. Would it work with other ketones such as methyl isobutyl ketone to yield 1,1,1-trichloro-2,4-dimethylpentan-2-ol?

Chloroform anion generated with powdered KOH reacts with ketones to give addition resulting in the trichloromethyl methyl carbinol as reported by Willgerodt in 1881  (Chem. Berichte vol 14 245-60) as referenced here source which mentions that it has been extended to a wide range of ketones and aldehydes. So the direct reaction of methylisobutyl ketone with chloroform/KOH will give you your target compound.

source: Stack Exchange

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why an interstitial defect increases density in covalent solids and do not increases in ionic solid?
The interstitial point defect increases the density of a solid. Frenkel defect is also a type of this defect but it do not increases the density.
Also, the mass / volume concept of density tells us that density should be same in both cases, as they are basically interstitial defect. Then what happened to a covalent solid that density increased , and why it didn't happened to ionic solid?
(Please understand that it is not a duplicate of
Why Frenkel defect does not impact the density of the solid? )

In the standard Fresnel case, the small ion is displaced. It fits easily in a position between the larger ions without disturbing them much, if at all. For other materials and structures that may not be the case.

Does that means that in an interstitial defect in covalent solids , the particle in void disturbed other particles due to greater size? Is this is the only reason for differeneces of observed density? thank you

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How does iodic(VII) acid oxidise amines?
What happens when 3‐aminobutan‐2‐one undergoes oxidative cleavage with $\ce{HIO4}?$

I'm particularly interested in the amine part. Would the hydroxy and amino groups on the same carbon be stable? Or would it undergo some sort of rearrangement?

The primary amine will get oxidised to the hydroxylamine which will readily hydrolyse to the ketone giving the 2,3-diketone, see https://www.thermofisher.com/uk/en/home/references/molecular-probes-the-handbook/reagents-for-modifying-groups-other-than-thiols-or-amines/hydrazines-hydroxylamines-and-aromatic-amines-for-modifying-aldehydes-and-ketones.html

So does that mean there's no oxidative cleavage?

Correct - no oxidative cleavage


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Reason behind the existence of radioactive substances till now
If radioactive substances continuously radiates alpha particles, then that element would never be seen by us because it would had been converted into some stable atom a long time ago. Am I right?

The question is better suited for Physics.SE, where it has already been asked and answered: Since radioactive material decays how is it possible that there is any left after 4.5 billion years?; Does the half-life of an element mean it will never decay completely?

Have a seat and wait some $10^{30}$ years - by then there's a reasonable chance the second half the periodic table won't exist anymore. Enjoy it while you can.

You are right but the key point is your perception of "long time ago." Naturally occurring elements which are unstable, like U (Z=92) or Th (Z=90), have half lives on the order of the age of the Earth, i.e., they decay very very slowly. How old is the Earth by current estimates? About 4.5 billion years. So, these two elements, once they were formed at the time of the creation of this Earth (or the Sun?), are still present today. They are indeed continuously forming less stable elements, for example radon gas is formed in the decay series of U/Th. Their half-life are very short, on the order of a few days, but the gas is continuously being formed in the Earth's crust from the decay of U and Th daughters.
This is a very interesting topic. Paul Karoda, a Japanese American chemist, in the 1950s investigated all these questions which are coming to your mind and wrote a beautiful book (only 150 pages) titled "The Origin of the Chemical Elements"...buy it if you really think seriously about the origins of the elements on this Earth. Makes you wonder about the beauty of this universe and its Creator.


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Why is my battery of galvanic cells not enough to power a lamp rated for a lower voltage than the battery provides?
I made a series of 12 galvanic cells with saltwater as an electrolyte and zinc and copper as electrodes. It produces 8.8 V, but it can't even light a 2.5 V bulb (the bulb can be lit by a 1.5 V source).
Please assist in understanding what could be the problem here.
Update: the galvanic cells almost hit 10 volts (9.99 peak voltage) with 0.4 amps. However, it still can't light the bulb.

Assuming the bulb is working properly, the cell battery does not provide required voltage under the load, i.e. to the closed circuit. Measure the voltage when the bulb is attached. If it is enough then the bulb or the circuit are broken. If it is not enough then the bulb obviously cannot shine and then try changes suggested by Maurice.

It is not a problem of voltage. It is a problem of current and power. Whatever its nominal voltage, your bulb has a power to be known in watts. If your $\ce{Zn/Cu}$ cells are not able to produce the required current, the bulb will never be able to emit light. And the current is proportional to the surface of the zinc plate.
So first, determine the power of your bulb in watts. Divide it by 2.5 V. You will obtain the nominal current in Amperes. Measure your current with one of your arrangements. You will probably measure less than the nominal current. Then try to increase the dimension of your zinc and copper plates. Or make up a second arrangement of cells, parallel to the first one.
Also try to put the two plates as near to one another as possible. This will decrease internal resistance, and increase the current.

How can I determine the wattage of the bulb?

@Vince Try reading, what is written on the bulb or on its packaging. Alternative is to measure the current at nominal voltage as P = U.I

The seller said the bulb is 2.5 watts. @Poutnik

I see it now. 0.3 amps is imprinted on the bulb. @Poutnik

Maurice points out at the critical problem and answers the old classical problem. Why can't a 12 V ordinary battery be used in a car when a "large" lead-acid batery also generates 12 V? Your current set-up is not able to generate enough current to light the bulb. You can try the following:
(i) Decrease the internal resistance of the battery by adding more salt. You did not mention how much salt you added
(ii) Increase the surface area of the plates
(iii) Add more "cells" or use a bulb / LED that requires less electrical power.

Yes, this is another point, decrease the distance between the plates to reduce resistance.

@AChem Do you mean the decrease of the wires that connect the plates?

@EdV May you please say how to make saltwater wetted paper in simple steps? Thank you very much.

@EdV The copper plates are 10 cm long and 2.5 cm wide. The same goes with the zinc plates. The thickness of the copper plates is 0.05 cm and 0.02 cm on the zinc plates.

Should I decrease the length of the wires that connect the plates? @AChem

@Vince, if those wires are made of copper, then it should not be a proble. Increase the surface area of the plates and decrease the distance between the plates.


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Alternate electronegativity measures for exotic atoms?

An exotic atom is an otherwise normal atom in which one or more sub-atomic particles have been replaced by other particles of the same charge. For example, electrons may be replaced by other negatively charged particles such as muons (muonic atoms) or pions (pionic atoms).
On normal atom, electronegativity, symbolized as χ, is the tendency for an atom of a given chemical element to attract shared electrons (or electron density) when forming a chemical bond.

So for an exotic atom, will there be a new term beside electronegativity?
Such as:

Muonnegativity

Taunegativity

Antielectronpositivity (Postitronpositivity)

Antimuonpositivity

Antitaupositivity

Any others?

For a chemical compound made from exotic atoms and normal atoms, there will not be just an electronegativity within the chemistry bonds. Could we called it leptonnegativity scale, matrix, vector, tensors, or whatever it might be to measure its tendency to form chemical bonds and also antileptonpositivity for its counterpart?
Does it makes sense?

I think the "electro" in electronegativity has to do with electric charge, and is not intended as a direct reference to the electron itself as a subatomic particle. Contrast with the complementary term electropositivity. Not to mention that these are just rough concepts anyway, and anyone seriously working with exotic atoms would certainly make use of all the machinery of quantum electrodynamics at their disposal to make any predictions.

@NicolauSakerNeto But we also have terms: Electron affinity and Electron configuration. If electronegativity has to do with electric charge. Will those terms also change into electric charge affinity and electric charge configuration?

Funnily enough, those two additional concepts you mention are in fact directly tied to electrons, and you could readily define, for example, muon affinity and muon configuration. Unfortunately, I can't exactly make a list of which ones work and which ones don't. You need to be aware of the subtle difference between electro/electric and electron/electronic.

@NicolauSakerNeto Yes I'm aware, electron is a concrete entity. It could exist independently of mind. One of entity that hides behind it that we can draw from it, is charge, an abstract entity. The status of an entity that has an abstraction of charge is called electric. It's just that, I wonder if electron or muon affinity and electron or muon configuration, could be unified into let's say, charged particle affinity and charged particle configuration.

I do not think there is any sense in which "charged particle affinity" and "charged particle configuration" is useful, other than simply being a list of the individual properties. For example, in normal matter, muon affinities are wildly greater than electron affinities, with typical energies being around 200 times larger (around 1 kiloelectron-volt, or dozens of megajoules per mole). With regards to configuration, muons populate muonic orbitals completely independently of electrons, which populate electronic orbitals.

@NicolauSakerNeto Now that makes sense, I'm satisfied. Thank you very much.

https://chemistry.stackexchange.com/questions/80203/what-is-a-hydrogen-like-or-hydrogenic-atom https://chemistry.stackexchange.com/questions/125171/how-do-molecules-with-muons-and-electrons-bond https://chemistry.stackexchange.com/questions/44802/alternative-to-a-proton


text:
Water getting hotter once frozen?
I packed a fridge with temperature sensors and took over the temperature control in order to have extra cold beer.
One of those sensors is submerged in the middle of a container containing a 4.5% seltzer so I can get an idea of the temperature inside the surrounding beer cans.
This submerged sensor is the yellow curve at the bottom:

I'm trying to explain what happened around 6:25.
I was maintaining the seltzer at -3°C and I confirmed it was not frozen. However I think I was maintaining it slightly below it's freezing point.
This morning I saw that it somehow raised in temperature abruptly and it's now in this jelly/half frozen state.
I'd like to know why this really sharp raise in temperature.
Is this how freezing works?

When you wrote 4.5% seltzer, what does the 4.5% refer to?  Is it the level of carbonation, or something else (e.g., alcohol)?  More generally, what are the ingredients of your seltzer?  Does it contain only water and CO2, or are their other things in it as well?

It's 4.5% alcohol/vol. and the rest of the ingredients are carbonated water, sugars (glucose-fructose), Malted barley, Yeast, Natural flavors, Citric acid, Potassium sorbate.

I'm speculating, but it's possible this is what happened:
I looked up the freezing point of seltzer water (aka club soda), and it's $\pu{-10 ^{\circ}C}$. Plus your product also contains alcohol and sugars, which would further lower the freezing point. Since you have a temperature sensor in it, I assume it's open. Thus the seltzer will gradually lose its dissolved $\ce{CO2}$, and the alcohol will gradually evaporate.  As this happens, its freezing point will go up.
I'm guessing that your seltzer lost enough $\ce{CO2}$ and alcohol for its freezing point to exceed $\pu{-3 ^{\circ}C}$.  When that happened, the seltzer, which was at  $\pu{-3 ^{\circ}C}$, entered a "supercooled" state (meaning it was now a liquid below its freezing point), and thus started to freeze. When a substance freezes, it releases thermal energy (you have to add thermal energy to melt something; freezing is simply going in the opposite direction). That could be what caused the temperature jump.
It sounds like its freezing point is now $\pu{-1 ^{\circ}C}$, and the "half-frozen state" you are seeing is phase coexistence between ice and the remaining sugar-water solution.
I don't know why you're seeing the "jelly-like" consistency, but it could be from the dissolved sugars and malted barley left behind in the water as the ice forms.

I'm guessing sugar is a potential "contaminant." Water could freeze, leaving a liquid phase of concentrated sugar water.

@LevelRiverSt  Seltzer water doesn't contain sugar (it's nothing but water with dissolved CO2 and some mineral), unless some got in there by accident.

@theorist the OP called it a "4.5% seltzer" so I'm assuming its some kind of seltzer based cocktail-ish drink, not pure seltzer water.

When a supercooled liquid flash-freezes, it often turns into a very fine slush, which I suppose could seem like jelly in a way. If the slush is stirred aggressively, it returns to a more fluid state with compacted ice settling at the top.

@mbrig I thought the OP was referring to the level of carbonation.  But you're right, it could be something else. E.g., it could be hard seltzer, where the 4.5% is the concentration of alcohol. I will ask the OP!


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Orbitals in benzene and other delocalised structures
I've read that molecular orbital theory demands that if there are $n$ orbitals prior to some event like bond formation, there should be $n$ orbitals after this event.
Does this hold even in structures with conjugation like benzene, benzoic acid and so on? If so, then does the formation of benzene itself result in a chromic shift?

Chromic shift from what?

Yes, if there are a total of N atomic orbitals in the carbon and hydrogen atoms, then there should be a total of N molecular orbitals in benzene, benzoic acid, etc. As @Andrew mentioned, I'm not sure what you mean by "chromic shift."

@Andrew from the formation of benzene from say 1,3-cyclohexa diene similar to the seen when a benzenediazonium cation and, say for instance, phenol react to form 4-(phenyldiazenyl)phenol

Apart from the Chromic shift thing, which I don't understand, this is close to a duplicate of https://chemistry.stackexchange.com/questions/164346/how-do-computational-programs-decide-how-many-virtual-orbitals-to-compute/164352


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How to explain mesylate structural homologues (e.g., "esylate") being weaker leaving groups than mesylate itself?
I came across a piece of scientific literature (https://patents.google.com/patent/US20080045602A1/en) that would seem to imply leaving groups formed by reacting hydroxy groups with, e.g., ethanesulfonyl chloride (EsCl) or 1-propanesulfonyl chloride (PsCl) are, or at the very least can be, weaker than -OMs (from mesyl chloride). This has me wondering: How should I rationalize these longer-alkyl-chained leaving groups being weaker? Due to their structural similarity, would a comparison of estimated pKa values suffice, or would, for instance, a ranking of their leaving group abilities be more situational (e.g., solvent-dependent)? When I run estimated pKa calculations in MarvinSketch, the pKa goes up with increasing linear alkyl chain length (indicating that they would be "stronger" weak bases) in the homologous series starting from mesylate. However, I am also wondering if I should consider that, just as trifluoromethyl acts to stabilize the triflate leaving group through acting as a strong EWG, an extended (compared to the methyl portion of mesylate) alkyl EDG chain would have a destabilizing effect on leaving group ability.
Here are two quotations of note from the citation I linked above:

Although more reactive esters such as toluenesulfonyloxy or methylsulfonyloxy provide for quicker reaction times..., the use of less reactive esters such as benzenesulfonyloxy or ethylsulfonyloxy is advantageous because less and lower amounts of side products are formed.

Compounds of Formula III may be upon isolation and purification used in the next step. Alternatively compounds of Formula III can be used in [the] next step without purification i.e. a single pot synthesis from compounds of Formula II to V or from III to V is envisaged. In case those and/or subsequent intermediates are not isolated and purified, it is advantageous to make and use less reactive compounds, such as benzenesulfonyloxy or ethanesulfonyloxy or propanesulfonyloxy or butanesulfonyloxy.

Do note that the "compounds of Formula III" are the ones that contain sulfonate ester leaving groups. Additionally, page 17 of a thesis (https://irep.ntu.ac.uk/id/eprint/13/1/211217_ISMAIL'S%20thesis.pdf) mentions that in SN2 substitution reactions t-butylsulfonates (which could visually be thought of as mesylate but with three methyl groups in place of the C-atom's three H-atoms) have been found to be weak nucleofuges compared to mesylates.


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Details of what actually happens in cold sulfuric acid between 80 and 90%; what molecular changes cause the viscosity to skyrocket?
AChem's answer to Why does the graph of the electrical conductivity of sulfuric acid/water solutions have this knee in the ~85%-~92% range? includes this plot from Horace E. Darling in "Conductivity of sulfuric acid solutions" (Journal of Chemical & Engineering Data 9.3 (1964): 421-426.) and mentions:

There is a sharp increase in viscosity at 85%, which indicates there is a major structural change in sulfuric acid solution in the range 85-92%. Sulfuric acid forms a hydrate in this range. When the viscosity is high, the conductance goes down, there is a depression in the curve. This viscosity jump is causing the double hump. Once we are past the high viscosity range, conductance goes up again.
It is amazing how simple molecules do not stop from surprising us!

Das et al. (1997) Electrical Conductance and Viscosity of Concentrated H2SO4/H2O Binary Systems at Low Temperatures: Correlation with Phase Transitions (J. Phys. Chem. B 1997, 101, 4166-4170) do a thorough analysis and mention phase transitions and hydrate formation, but 25 years later are there more detailed molecular models of what is happening that might address the exact form of the hydrate(s) formed or any long-range correlations or ordering behind the peak in viscosity at a certain high concentration?
Question: What details are known or theorized as to what actually happens in cold sulfuric acid between 80 and 90%, what molecular changes cause the viscosity to skyrocket?

The composition where we see this rapid rise in viscosity corresponds to one of several sulfuric acid hydrates, specifically the monohydrate formally rendered as $\ce{H2SO4•H2O}$. As shown in this colorful diagram from Maynard-Casely et al.1, this monohydrate corresponds to a maximum in the melting curve, indicating that in the cold acid this composition is organizing itself into a solid structure.

Figure 1. The binary phase diagram of sulfuric acid and water (after Beyer et al., 2003). This shows the region of solid sulfuric acid hydrates (red) and the regions where liquid sulfuric acid and solid water ice coexist (blue). The labelled vertical lines indicate the stoichiometric compositions for the observed hydrates in this system; sulfuric acid monohydrate (SAM) 84.5 wt%, sulfuric acid dihydrate (SAD) 73.1 wt%, sulfuric acid trihydrate (SATri) 64.5 wt%, sulfuric acid tetrahydrate (SAT) 57.6 wt%, sulfuric acid hemitriskaidecahydrate (SAH) 45.6 wt% and sulfuric acid octahydrate (SAO) 40.5 wt%.

Sulfuric acid is, of course a strong acid towards water with respect to its first deprotonation, forming $\ce{H3O^+}$ and $\ce{HSO4^-}$ ions (the second deprotonation, forming $\ce{SO4^{2-}}$ ions, predominates only in very dilute water solutions). As we see with some other strong acids such as perchloric acid, these ions can form a salt, whose composition matches the monohydrate ($\ce{H2SO4•H2O}=\ce{(H3O^+)(HSO4^-)}$). Taesler and Olovsson2 find this ionic structure from single-crystal XRD, with the additional feature that the bisulfate anions are polymerized by hydrogen bonding. Hydrates formed with more water are also known, including the octahydrate studied in reference 1. But (with the additional water in their formulations) these involve either a more bulky, irregularly shaped cation than the monohydrate or have water molecules in their structures, both of which would favor a less stable and lower-melting solid than the monohydrate salt would be.
Reference
1.
H. E. Maynard-Casely, H. E. A. Brand and K. S. Wallwork (2012). "Structure and thermal expansion of sulfuric acid octahydrate". J. Appl. Cryst. 45, 1198-1207.
https://doi.org/10.1107/S0021889812037752. (also accessible in researchgate)
2.
I. Taesler and I. Olavsson (1968). "Hydrogen bond studies. XXI. The crystal structure of sulfuric acid monohydrate." Acta Cryst. B24, 299-304. https://doi.org/10.1107/S056774086800227X.

Thank fo for such a thorough and definitive answer! Are the conclusions based on X-ray diffractions of solid hydrate crystals and simply assumed to apply to the liquid phase, or are there some measurements of the liquid that confirm this. I don't doubt the conclusion, but like the block quote in my question mentions: "It is amazing how simple molecules do not stop from surprising us!"

The monohydrate has the expected ionic structure according to singke-crystal XRD,l, and there us also evidrnce that the anions are polymerized by hydrogen bonding. See Taesler and Olavsson (1968), _Acta Cryst.  B24, 299-304,
https://doi.org/10.1107/S056774086800227X.

I'm curious about work that examines aqueous solutions in the liquid form. I understand that crystallographic studies of pure solid forms of various hydrate structures are incredibly helpful, but as a liquid there's no guarantee that what is present is not a mixture of several different ones in equilibrium. It's possible that there's no data on this but there might be, and certainly it's something that can be addressed theoretically.

Liquids with high ion concentrations tend to have much higher viscosities than the pure solvent as they become saturated (which is what is basically happening here with the hydronium sulfate salt). We see this more commonly with sodium hydroxide solutions (https://www.protank.com/sodium-hydroxide for this effect with sodium hydroxide solutions); everything on either the ice or the solid sodium hydroxide saturation line is 20 cp or more versus 1 cp for water alone.


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Creating custom spectra graphs
I wanted to make a few custom spectra graphs for some homework questions I'm assigning. The issue is that the Education Board issued datasheets do not reflect online images. They often say that bonds absorb different wavelengths to images I can find like this one:

I was hoping that I could make my own graphs similar to this. I want to manually generate spectra that would match the Government Sheets' data on what bonds absorb what.
What is some software that could do this?

I'm not sure this is a chemistry question per se; but if you have the underlying data, this sort of graph is fairly easy to generate in any (reasonably high-level) programming language you may want to do. Python (with matplotlib), R, or Matlab are probably the most common choices in science. The tougher part of this is getting the numbers to plot. If you have those numbers, then it's simple: plotting an array of intensities (or logs or whatever) against an array of wavelengths, changing the axis labels, and adding a title will be covered in basic plotting tutorials.

@orthocresol The problem is that I don't have the numbers. Ideally some kind of drag and drop of points would be nice. Currently I'm just roughly hand drawing the curves but it'd be nice to make it look more professional

"drag and drop of points" -- then you are probably looking for graphic design software like Inkscape which will let you draw Bezier curves. I don't know if that's simpler, or whether it's in fact easier to use a pencil & ruler to extract the data from the graph you're trying to reproduce so that you can re-plot it in your language of choice.

@JohnHon If «I don't have these numbers» equates to «but I have the relevant graphs as illustrations (from different sources)», then you need a digitizer (e.g. webplotdigitizer). The subsequent creation of the data can be automated / scripted with a template for gnuplot into bitmap or vector format.

You can generate such custom spectra graphs using an open source python library : orca-uv and the ORCA tool
The orca-uv tool takes an .out file generated by ORCA for UVVis spectroscopy calculation and saves a .PNG file of the spectral graph.
Using a sample .out file obtained from the ORCA UVVis tutorial electronic spectrum,
a minimum reproducible example is shown below on a Google Colab Notebook:
!git clone https://github.com/radi0sus/orca_uv

sample_orca_output = '''
-----------------------------------------------------------------------------
         ABSORPTION SPECTRUM VIA TRANSITION ELECTRIC DIPOLE MOMENTS
-----------------------------------------------------------------------------
State   Energy    Wavelength  fosc         T2        TX        TY        TZ
        (cm-1)      (nm)                 (au**2)    (au)      (au)      (au)
-----------------------------------------------------------------------------
   1   19699.5    507.6   0.047917124   0.80078   0.87904  -0.16736   0.00724
   2   30354.1    329.4   0.012327447   0.13370   0.05769   0.36102  -0.00597
   3   31373.4    318.7   0.023725133   0.24896  -0.48626   0.08573  -0.07183
   4   32906.2    303.9   0.002354014   0.02355  -0.04012  -0.14813   0.00027
   5   34099.1    293.3   0.176508288   1.70411   1.27793  -0.24569  -0.10310
'''

with open("sample_orca.out", "w") as text_file:
    text_file.write(sample_orca_output)

!python3 /content/orca_uv/orca-uv.py /content/sample_orca.out

This saves an image called: sample_orca-abs.png in the root directory.
And the spectral plot in the image sample_orca-abs.png looks like this:

And to get the .out file for your data you can follow the ORCA UVVis tutorial


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Where does the energy required to initially begin the reaction of iron and oxygen come from?
When iron is left outside, over time it reacts with oxygen to form iron oxide or rust. But how does the oxygen react with iron, isn't iron held together by strong metallic bonds? Where does this required energy to initially begin the reaction of iron and oxygen come from?
Do all reactions initially require a input of energy?

Yes, that's why the fjords in Norway are popular for storing ships when they are not in use.

Regarding your last question see https://chemistry.stackexchange.com/questions/18924/are-there-any-reactions-with-no-activation-energies

Do all reactions initially require a input of energy?

Yes, it is called activation energy. When you say "initially required", it does not mean just when the overall reaction starts. Each time another atom or molecule reacts, you need that activation energy again.

But how does the oxygen react with Iron, isn't Iron held together by strong metallic bonds.

Yes, for every oxygen that reacts, there has to be some activation energy to start prying apart the bonds. For every iron atom that reacts, there has to be some activation energy to start prying it from the metallic bonds. In the case of oxygen reacting with iron, you get that energy back (and more) once the bonds between iron and oxygen form.

When Iron is left outside, overtime it reacts with oxygen to form Iron Oxide or rust.

If you leave iron inside, this also happens because there is as much oxygen inside as outside. However, when it is wet and salty, the reaction proceeds at a higher rate, and it might be more humid and saltier outside (especially in the ocean). You can avoid this by providing a barrier between iron and oxygen, for example by painting your boat.

Where does this required energy to initially begin the reaction of Iron and Oxygen come from?

It comes from the thermal energy. Thermal energy is not equally distributed, there are always some particles that have higher energy. They might react, or give that thermal energy to the next particle, which might react. We put our food in the refrigerator or freezer to slow down the reactions (and the metabolism of microbes) that would spoil our food.

Why is it that "wet" and "salty" speed up the reaction? I also read somewhere that pure Iron is very hard to rust. Why might that be?

Check out the answer by @AChem and the linked video.

When Iron is left outside, overtime it reacts with oxygen to form Iron
Oxide or rust. But how does the oxygen react with Iron, isn't Iron
held together by strong metallic bonds.

Let's forget about iron for now. Allow me to mention a very general trend. Any metal in the periodic table that is generally not found in its native form in the Earth's crust will eventually corrode when exposed to the open atmosphere. Using chemical reactions, humans input a lot of energy in the form of heat and electricity to extract metals out of the ore, but the metals would like to return to their original states, as they existed "naturally" in the combined state. Isolated metals such as sodium or potassium are extreme cases. They instantly corrode to sodium and potassium oxides, the instant air and moisture touches them, because these metals never occur naturally. The same holds true for iron. In the Earth's crust it occurs in a combined state with oxygen or sulfur. Thus pure iron would eventually combine with oxygen (as a matter of time). Copper reacts with the atmosphere, as does aluminum. However, gold which occurs in its native state does not corrode in the air. A crude example is that of a living human body. We come from the dust and eventually become dust. It is a matter of time.
Corrosion is not a simple process. People do PhDs in this subject trying to understand the process or slow it down. Corrosion is an spontaneous electrochemical reaction, i.e., in the case of iron and oxygen there is a redox reaction in the presence of moisture. This is beautifully demonstrated with the help of a galvanic corrosion demonstration.
Royal Society of Chemistry Website
As you state in the comments, impurities do help in forming local electrochemical cells on the metal surface. Sometimes pure metals can form a uniform and non-porous protective oxide layer. That aspect slows downs corrosion drastically even though the metal itself is very reactive. A common example is that of aluminum otherwise we would not be able to see aluminum pans in the kitchen.

Of course, free oxygen is not exactly a "natural" thing either. Plants and algae spend a lot of energy (coming from the sun) to split carbon dioxide, releasing waste oxygen in the process. The reason iron "naturally" rusts into iron oxides (etc.) is tied to the fact that right now, the Earth has a heavily oxidative atmosphere. That wasn't always the case.

@Luaan To add to that, isn't there a theory that oxygen levels in the atmosphere were kept artificially low for an extended geological period after the origin or photosynthesis as all the iron on the planet rusted?

@Dannie Yup; most of the available iron was dissolved in the oceans - iron oxides are no longer soluble in water and precipitate out, which is where the massive banded iron formations we heavily rely on come from. The "bands" tell us a lot about how this proceeded, much like growth rings in trees. In fact, it's thought that this was a very lucky thing for the algae - early on, they couldn't live in oxygenated environments themselves, and the iron-rich waters got rid of the waste oxygen periodically. When algae developed oxygen resistance, the oxygen production really took off.

Rust is a punctual phenomena. It happens on the surface of an iron piece, but only if there is some conducting impurity (carbon grain for example) on this surface. If a drop of water falls on this contact $\ce{Fe-C}$, the following electrochemical phenomena start.
The first $\ce{Fe}$ atom touching the grain yields two electrons according to $$\ce{Fe ->  Fe^{2+} + 2 e-}$$ These electrons are moving through the carbon grain up to the dissolved $\ce{O2}$ in the water drop, according to : $$\ce{O2  +  2 H2O  +  4 e- -> 4 OH-}$$ Then both ions $\ce{Fe^{2+}}$ and $\ce{OH-}$ are attracted to one another and produce a precipitate $\ce{Fe(OH)2}$. This precipitate is quickly oxidized by oxygen form the air according to :  $$\ce{4 Fe(OH)2 + O2 + 2H2O -> 4 Fe(OH)3}$$ Rust is mainly made of the brown precipitate $\ce{Fe(OH)3}$, which is later on partially dehydrated and even partially carbonated.
The main point here is the fact that this corrosion starts and continues at the very contact $\ce{Fe-C}$ and not in the bulk of the metal.


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Why is the value of the heat capacity ratio γ the same for molecules with the same atomicity?
The heat capacity ratio is defined as the ratio of the heat capacities at constant pressure and volume respectively:
$$\gamma = \frac {C_p}{C_V}\tag{1}$$
Then obviously
$$\gamma = \frac{n\,C_p\,\Delta T}{n\,C_V\,\Delta T} = \frac{\Delta Q}{\Delta U}\tag{2}$$
This means that $1/\gamma$ is the fraction of the heat $(\Delta Q)$ given or taken from the gas, which is spent in increasing or decreasing the internal energy $(\Delta U)$ of the gas, respectively.
I saw online that the value of $\gamma$ was the same for every monoatomic gas $(5/3).$ The same goes for diatomic $(7/5)$ and triatomic $(4/3).$ Why is it that this ratio is the same for these gases?
I think it has something to do with how only a certain fraction of the heat given or taken is contributed to by the change in internal energy, but I can't exactly find the right reason.
Another way this question can be asked is: why is the fraction of the heat given or taken that goes into the change in internal energy same for molecules with the same atomicity?

See also Heat capacity of gases

Note that those values of $\gamma$ are only approximations, theoretical predictions from statistic mechanics. Two gases with the same number of atoms can have significantly different heat capacities in reality.

So, for two real gases with the same atomicity, since there is a presence of potential energy, which may differ with change in number of electrons or protons, the value of internal energy may not be the same and thus the heat capacity ratio. Did I understand correctly, @electronpusher?

It is rather due different quantization steps of vibration energies due different atomic masses. They lead to different temperature ranges of progressive involving of vibration contribution to heat capacity. The same for rotation contribution for very low temperatures ( see also the page bottom on the referred hyperphysics link).

If we deal with an ideal gas then $C_p=C_V+R $ then $\gamma=1+R/C_V$ which then begs the question if $C_V$ is different for different molecules why is $\gamma$ constant and the same for molecules with the same number of atoms.
The answer is that for this to be true the temperature has to be sufficiently high that the slope of the internal energy with temperature, which is $C_V$ has become constant, i.e. $C_V=\left(\frac{dU}{dT}\right)_V$. As energy levels are discrete for $C_V$ to be constant with temperature energy levels have to be populated at a constant rate, i.e. the slope of $U$ vs. $T$ is constant. Only then will $\gamma$ be the same for molecules with equal number of atoms. (In practice  it may not always be possible to reach this temperature as the molecule may decompose first.)
At v. low temperatures as the average energy can be less than a quantum
then $C_V\to 0$ but as $T$ increases, by the Boltzmann distribution the average internal energy $U$ increases as more rotational and vibrational levels are populated and so $C_V$ increases. Eventually $C_V$ becomes constant but for many molecule this may be far above room temperature. So this means that experimentally $\gamma$ is usually greater than predicted from classical thermodynamics, but instead statistical mechanics has to be used. For Nitrogen, which has widely spaced quanta, at room temperature $\gamma \approx 1.4$ but classical prediction is 1.29, but for iodine with far more closely spaced levels the classical value is predicted.
In the classical limit (i.e. many levels populated) the heat capacity has a value $R/2$ for each degree of freedom, so is $C_V=3R/2$ for (x,y,z) translation and so $\gamma=1+2/3=1.667$ for atoms, and for a diatomic $C_V(trans) = 3R/2$,  $C_V(vib) = RT$,  $ C_V(rotation)=RT$ making $\gamma =1.29$. You can extend this for $n$ vibrations  where $C_V=(3+n)RT$. (Many Phys. Chem. textbook will give a fuller explanation and detailed calculations.)


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Which atom carries charge in polyatomic ions?
I have a basic confusion regarding  polyatomic ions:
Is it meaningful to say "which atom" in a polyatomic ion is charged?
Consider for instance hydroxide OH$^-$:
The oxygen has greater electronegativity than the hydrogen, so it is "more negative" than the hydorgen.
However, I am not sure: Is it wrong to say that the hydroxide is composed of a neutral hydrogen together with a negatively charged oxygen?
I guess this won't be an accurate statement/description, right?

In particular, If I understand correctly, then the formulas OH$^-$ and HO$^-$ describe exactly the same diatomic anion, so in particular on cannot deduce from the formula which is the more negatively charged atom between $H$ and $O$. (For a moment I wasn't sure whether the presence of the minus sign above the $H$ in OH$^-$ means that it's the $H$ that is more negative).

Your understanding is correct. $\ce{OH-}$ and $\ce{HO-}$ are exactly the same thing. The position of minus does not mean anything.

See: https://chemistry.stackexchange.com/questions/53568/what-is-the-structural-formula-of-alkali-hypohalite-mox-or-mxo

All atoms in such a ion are charged, effectively not in integer multiples of the elementary charge. Some are more charged than others. The charge is not limited to a single atom. The sum of atomic charges is equal to the ion charge.

And charges are not strictly localized, as you state, so think of an electron spending some time in the company of one atom, and sometimes with another. To confuse things more, there may be resonant electron layouts...


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Reduction potential of water
I can't understand what is meant by the reduction potential of water. In Electrolysis of water decomposition of water takes place as a result of which
$H+$ and $OH-$ are furnished. They migrate towards cathode and anode respectively where they get reduced and oxidised. Each of them have their respective reduction potentials. Same in case of $\ce{AgCl}$ what is meant by reduction potential of $\ce{AgCl}$ as it contains both cationic and anionic part.

See if that addresses your question, https://chemistry.stackexchange.com/questions/112840/why-is-reduction-potential-of-water-2h%E2%82%82o-2e%E2%81%BB-%E2%86%92-h%E2%82%82-2oh%E2%81%BB-is-0-83-insted-o

@AChem the link you shared doesn't actually answer my question. It is based on calculations and I have done many problems but I have doubt regarding the intuition behind terms like reduction potential of water

Okay. I will try to address it.

In the $\ce{AgCl/Ag}$ half-cell, the chlorine atom does not change its oxidation number. Only $\ce{Ag}$ is reduced from +$1$ to zero.

When electrochemists talk about the reduction potential of water, it specifically refers to an isolated half-cell. They are referring to the cathodic reaction only. When water gains electrons (from a source), it forms hydrogen and hydroxide ions.
$$\ce{2 H2O + 2 e- <=> H2 + 2 OH-}\label{rxn:2}\tag{1}$$
so there no cations here.
As you know, half-cells cannot exist alone, you need another half-cell to complete the circuit. The other half-cell is the hydrogen electrode whose value is arbitrarily set to zero by the German electrochemists more than a century ago.
You can see here how it is practically measured.
https://www.nemi.gov/methods/method_summary/5456/

Reduction of water is not set to zero. Look at the equation carefully, there are no protons there! https://chemistry.stackexchange.com/questions/112840/why-is-reduction-potential-of-water-2h%E2%82%82o-2e%E2%81%BB-%E2%86%92-h%E2%82%82-2oh%E2%81%BB-is-0-83-insted-o

SHE is set to zero at pH=0.

Reduction potential is defined as the tendency of a species to get reduced by gaining an electron and the value is standardized with a reference to the Standard Electron Potential (SHE) of Hydrogen having a value of 0.
Now, the reduction potentials are defined for isolated half cells.
For instance, in case of electrolysis of water, the cations and anions generated are H+ and OH- respectively. These ions migrate towards cathode and anode where they get oxidised and reduced. Here,the reduction potentials can be found out individually!
Same in case of AgCl. AgCl after ionisation gives Ag+ and Cl- ions which move towards cathode and anode, and then we can find out the reduction potentials individually. (We can also find the oxidation potentials of the above-mentioned species, but reduction is generally used as a convention, so we use that)
                      AgCl ⇌ Ag+ + Cl¯

The species which undergoes reduction (here, Cl-), has a positive reduction potential, as it is easier to be reduced whereas, the species which undergoes oxidation (here, Ag+) can also be defined in terms of reduction potential, but in this case, it will have a negative reduction potential as it is easier to oxidise Ag+ than to reduce it! (The negative sign means that the tendency of the species to get reduced is less than hydrogen and the tendency to oxidize is more than hydrogen.)
However, the reduction potential of a compound (AgCl) on the whole doesn't exist!


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Atomization enthalpy trends. The case of Phosphorus
I was looking again at the periodic trends like Ionization Energy, Electron Affinity, Radius and so on. I understood all the general trends and the exeptions for these ones, but I can't fully understand the trends for Atomization Enthalpy/Energy.
Quoting my book ( Inorganic Chemistry - Atkins ) : "[...] enthalpies of atomization first increase and then decrease across Periods 2 and 3, reaching a maximum at C in Period 2 and Si in Period 3. The values decrease between C and N, and Si and P: even though N and P each have five valence electrons, two of these electrons form a lone pair and only three are involved in bonding. A similar effect is seen between N and O, where O has six valence electrons of which four form lone pairs and only two are involved in bonding."
Now, I understand why the enthalphy decreases from C ( a covalent solid in its standard state ) to N ( a biatomic gas in its standard state ), because of course it's easier to "atomize" the gaseous molecule. But I don't understand why this happens from Si ( covalent solid ) to P ( molecular solid, P4, in its standard state ): I'd think that for "atomizing" P4 we would need much more energy because firstly we need to break the intermolecular forces and then the covalent bonds of the tetratomic molecules (P4). Even if I considered P4 as a single molecular species, not a solid, I would expect that ( as the book implies ) P4 would be harder to "atomize" than Si, because every P  has 3 unpaired electrons and it forms 3 bonds with other P atoms, while Si has only 2 unpaired electrons.
The trends in the d-block are clear to me. But this one bothers me.
Please excuse my english and thank you in advance.

Consider that the silicon crystal structure has 4 bonds per silicon atom.

@GeoffHutchison yeah you're right. How stupid I was for not considering that. But isn't the bond energy of P-P in P4 bigger than the one of Si-Si in silicon? I'd say so, because of the greater Zeff, in fact Si-Si distance is around 2,5A° and P-P distance is around 1.99A° so I would expect a bigger Atomization Enthalpy for P(s). Do I need to look deeper at the solid structure?

I don't think I can give a canonical answer, but here are some quick thoughts:

Breaking intermolecular interactions almost always requires significantly less energy than breaking covalent or ionic bonds. Consider that a typical hydrogen bond is ~20 kJ/mol while the O-H covalent bond dissociation is ~400 kJ/mol.

The silicon crystal structure is a network solid with four Si-Si bonds per atom.

The $\ce{P4}$ tetrahedra as you note only have three (strained) P-P bonds per atom.

Edit:
I previously mentioned the BDE for P-P and Si-Si, but the standard CRC tables mention offhand that the values are for diatomic molecules, even though they use the P-P single bond notation.
Using the canonical Comprehensive Handbook of Chemical Bond Energies:

$\ce{H3Si-SiH3}$ has a BDE of 321±4 kJ/mol
$\ce{H2P-PH2}$ has a BDE of 256.1 kJ/mol

The P-P dissociation energy is for an average P-P bond, not counting the high ring strain inherent in the $\ce{P4}$ tetrahedra, in which the P-P-P bond angle is 60°, not ~100° as you might expect in some sort of hypothetical P-P-P chain compound.

Consequently, it's this angle strain that makes $\ce{P4}$ fairly reactive.
So you have more bonds for silicon and they're a bit stronger, and the $\ce{P4}$ tetrahedra are strained.
On the other hand, to quote "The Thermodynamic Properties of Phosphorus and Solid Binary Phosphides." by Schlesinger:

Phosphorus is available in a bewildering variety of physical states, and the thermodynamic stabilities of most are understood poorly at best.

Most importantly:

Because of the difficulty in obtaining and performing experimental work with red and black phosphorus, the most reliable thermodynamic data has historically been obtained working with white phosphorus and the metastable liquid. That in turn has led to the choice of R-white/liquid phosphorus as the reference state for the element, despite experimental evidence showing that red (I), red (IV), red (V), and both black forms are all more thermodynamically stable.

In short, if you're confused by phosphorus enthalpies, you're not alone. It's been tricky to measure and understand as well.
When I perform a search for P-P bond dissociation energies, I find a huge range of values.

The 1., 2. and 3. points that you mentioned are all clear to me. The fact that the P-P bond dissociation energy is higher than that of a Si-Si bond makes me more confuse, but as you suggest I should consider the fact that the P-P bond in the P4 tetrahedra is surely a weaker bond than other types of P-P bonds. That might be the reason of the Atomization Enthalpy's values.
Thanks for sharing your reasoning and spending your time for a stranger.
Anyways, I was entitled to be confused if "the thermodynamic stabilities of most are understood poorly at best." ahahah

P-P bond has only 183 kJ/mol I think you perhaps wrote the value for triple bond?

I've seen a wide range of P-P single bond dissociation energies. I've now revised, based on a reliable BDE for $\ce{H2P-PH2}$ which seems comparable.

@GeoffHutchison and what about the lower atomization enthalpy for Sulfur ( standard state S8 (s) ). In the cyclo-octasulfur every S atom forms 2 bonds, with a lower bond distance ( S-S: 2.05 A°, P-P: 2.21 A°) but with a lower BDE ( S-S: 226 kJ/mol, P-P: 256 kj/mol). I'd say that the reason is that every S atom forms less bonds with a lower BDE. I considered Intermolecular forces as negligible.

But S(s) melts and boils at a higher temperature, I'm confused ahah

Boiling and melting temperatures don't involve breaking bonds, only intermolecular interactions. Consider ice, for example. Trends in atomization energy, however, do.

@GeoffHutchison  S has a higher melting and boiling point than P because S8 molecules are bigger than P4 molecules, therefore Van der Walls' forces are bigger. I understand that. But why standard atomization enthalpy of S is lower than P? Is that for the reason I mentioned?  Hope this will be my last question and that I won't bother you anymore ahah


text:
Why do different places on Earth have different concentrations of particulate matter in the atmosphere?
From the second law of thermodynamics, I would expect the pollutants to diffuse from a region of higher concentration to a region of lower concentration. The rates of diffusion in liquids and solids are significantly lesser than that of gases, however this must eventually lead to pollutant concentrations becoming the same everywhere on Earth. But this isn't what we observe empirically. Different PM2.5 levels have been measured even in cities that are adjacent to each other. I fail to understand how this could happen despite the local atmospheres being interconnected.
I suppose that the variation in pollutant concentrations is determined by certain factors more significant than diffusion. What are these factors and what is their mechanism?

Earth isn't a big balloon and "particulate matter" doesn't diffuse.

Vertical and horizontal convection are much more significant phenomena for the spreading pollutants than diffusion, even for gaseous ones.

It might help to consider a simpler case. Why do some cities have more humid air than others?

Another version of Mithoron's view: Earth is not made by solidifying a homogeneous solution!

Particulates are not dissolved in air, but suspended. Gravity operates on them. That said, sand from Africa does reach the Americas.
Also, the atmosphere is not in equilibrium -- there are sources of particulates and gases, such as water vapor from evaporation, and sinks for these, such as rain and snow. There is energy input from insolation and geothermal processes (think: Krakatoa), and energy output via radiational cooling. Even Pluto has weather!

Because the atmosphere is not in equilibrium
If the atmosphere were a small closed vessel, it might have a chance of reaching an equilibrium where everything in it would be evenly distributed. But it obviously isn't. Water vapour, for example, forms local concentrations called clouds and they sometimes produce local rain. In an equilibrium, water vapour would be even distributed equally everywhere.
This hints at the reason for particulates not being evenly distributed. More specifically, particulates are produced by a variety of specific local sources. Traffic is a major source of many types of particulate, for example, and traffic is highly concentrated in cities. Moreover, the processes that remove particulates from the atmosphere (one, for example, is rain which is highly local) happen far faster than the processes that would otherwise distribute particulates evenly throughout the atmosphere.
In a system as far from equilibrium as this, the second law does not apply. But the key observation is that the sources are local and the processes of removing the particulates are far faster than the processes that would otherwise drive them to an equilibrium concentration.


text:
Linearity of electrochemical cells and electrical conductivity measurement
I've measured the output signal of a few commercially available conductivity meters and was surprised to see that the signal was typically a square wave of amplitude in the range of 1.5 to 6 Vpp (peak-to-peak) and a frequency ranging from 4 to 20 kHz. One of the meters even had a dc (direct current) offset of about 200 mV.
My understanding is that for an electrochemical cell to behave as a linear system, and to avoid creating harmonic distortions, we typically use a sine-wave of around 10 mVrms (root-mean-square), but not much higher.
So, I'm a bit surprised with the characteristics of these meters.
Does anyone have an explanation for these meter characteristics? Mainly:

Why they use amplitudes higher than 10 mVrms?
Why the square wave instead of a sine wave?
Why this variability in the used frequency and how can one possibly determine the correct value to use?
What could explain the use of a dc offset although we know it could change the properties of our cell!

Any help is appreciated.

What you describe may be the standard design, but it would be interesting (but is not absolutely necessary) to know what quality of meter, maybe the brand.

@BuckThorn, Not sure this helps but I've tested one meter from Walchem and two different meters from Hanna Instruments and another no-brand cheap one. Interestingly this latter only sends a single period of a 20 kHz square wave repeating every 60 ms and it gives quite different readings whereas for the rest the readings tend to be similar.

@BuckThorn, What would you say is the best approach to use if you want to measure electrical conductivity of water using two solid metalic electrodes (e.g., platinum, hastelloy, stainless steel, etc.)? What would be the appropriate signal to send across the electrodes and how would one go about exploiting the measured signal across the electrodes to get conductivity values?

If you want to make your own device I would think you can't do worse than attempting to duplicate what a commercial device does, provided it is of the desired quality. These instruments are usually calibrated, which may circumvent some of the biases. If you have a highly specialized application you may want to search the literature, for instance for high precision conductivity meters, or make your question here more focused so it is clear what you are attempting to do.

You may also want to look into impedance spectroscopy, for instance: https://www.researchgate.net/publication/317622367_Electric_impedance_of_aqueous_KCl_and_NaCl_solutions_Salt_concentration_dependence_on_components_of_the_equivalent_electric_circuit


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What is the energy ordering of molecular orbitals in diatomic molecules of third and fourth period elements?
For diatomic molecules with less than 16 electrons, the list of molecular orbitals in order of increasing energy is:
σ1s < σ*1s < σ2s < σ*2s < (π2py = π2pz) < σ2px < (π*2py = π*2pz) < σ*2px
From O2 onwards, because of s-p mixing, the order of energies of the molecular orbitals changes to:
σ1s < σ*1s < σ2s < σ*2s < σ2px < (π2py = π2pz) < (π*2py = π*2pz) < σ*2px
I now want to know how the molecular orbital theory can be extended to diatomic molecules of heavier elements, such as Cl2 and Br2.
More precisely, what is the order of energies of molecular orbitals formed by the combination of atomic orbitals from the 3s, 3p, 4s, 3d and 4p subshells?

Heavier elements do not tend to form diatomic molecules, with $\ce{Cl2}$ and $\ce{Br2}$ being an exception.

@IvanNeretin Diatomic molecules are pretty tough to make for most elements, but in gas phase they fare perfectly well. For ex. making W2 vapor can be challenging, but the bond in there is rather impressive.


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How much analyte should I use for paper chromatography?
I am looking into using paper chromatography to distinguish between standard drugs and counterfeit drugs for my high school research project. I have decided on using aspirin as the "standard" drug and I plan to make my own counterfeit versions of aspirin by preparing a solution of aspirin and adding in other chemicals (eg. salicylic acid). I've tried searching online for "how much" analyte I should be using for the paper chromatography, but all I can find is the instruction of "spot the paper with the solution".
Two questions:

Does anyone know resources for me to get highly specific instructions on paper chromatography (most sources only talk about TLC)
What concentration of aspirin or the "counterfeit aspirin" should I use to spot the filter paper
Would placing an iodine crystal beside my filter paper do anything (I read that iodine is highly reactive with aromatic compounds, and can produce stains, but using iodine was suggested for TLC, so I'm not sure if it works for paper chromatography too)

I'm not sure if the following affects the concentration of analyte I should be using, but I will list them just in case they do:

I am using 95% ethanol as my mobile phase
I plan to use UV light (roughly 229 nm) to visualize the developed filter paper

And if you're wondering, I can't perform column chromatography or TLC because my school doesn't have a column or silica gel (so no plates either). I also have to get the experiment done in 3 weeks at max, so I don't think I will have the time to order the required instruments.
I just began researching chromatography last night, so I'm not the most confident about my experimental design. So, if you find something odd about my experimental design anywhere, please point it out. Thanks.

Have you watched Youtube videos on paper chromatography just for a general idea? Don't worry about the exact concentration. What I would suggest is that you make a relatively concentrated solution (1/3 of a spatula) in 4-5 mL of your mobile phase, ethanol in your case. Using a narrow capillary spot the paper. Using a too dilute solution of aspirin may make the visualization difficult.

Ideally use the same concentrations as suggested for TLC. https://chembam.com/online-resources/experiments/painkiller-chromatography/

You also need to try different solvents not just one of different polarities. There will be trial and error involved here.

You will have to experiment and go to the lowest convenient amount with good enough detection result. Too big amounts negatively affect the spot separation, cause strong diffusion and large spots. // Not from paper chrom, but as a rough idea of a first shot: When I used to work with CAMAG HPTLC UV scanners, mostly on silica plates, we used progressively spraying/drying typically several microliters of 1% solution, to determine "chromatographical purity" in order of 0.x%, using extinguishing of fluorescence of fluorescent indicator in the plate.


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Evaporation of sample solvent and redissolving
When I conduct a synthesis, after some workups, I usually evaporate large amount of solvent using rotovap, recover compounds sticked to the wall of the flask to the glass vial using small amount of solvent, and finally evaporate all residual solvent using vacuum oven (usually at low temperature about 20~30 ℃ and apply vacuum gradually overnight). After the solvent is completely evaporated, I usually get gooey gel-like compound that is sticked to the bottom of the vial.

I have two questions:

I have never obtained powder-like sample after evaporating the
solvent. But I have a few experience in organic synthesis, so does
the formation of gel-like form rather than powder form depend on the
characteristics of the compounds, not the evaporating method?
Actually, I have once obtained gel-like sample containing large amount of
unreacted starting materials, even though that material was weighed as powder
form at the beginning. How can I get the powder form rather than gel-like form when I evaporate the solvent? Does the evaporating method matter?

After I evaporate all the solvent, I try to redissolve the sample
with the same solvent, but it is not readily redissolved although I
use similar amount. Usually it takes some time and heat. But why? It was readily dissolved in the workup stages. Is this
because that the sample is sticked to the bottom of the vial?

Hard to tell without knowing what exactly is being dried. Unless you are dealing with ionic liquids, I'd expect powder from the salt, not necessarily the organic substance itself. You might want to try alternative methods such as lyophilization (takes talking with biochem people), or use vapor diffusion crystallization (takes hours (if very lucky)/days/weeks).

@andselisk I wondered if my evaporating method is generally not appropriate for obtaining powder form of the sample. Solvent I use is common organic solvents like toluene, dichloromethane, ethyl acetate, etc.

@andselisk Do you have any opinion about question 2? I have almost always faced that problem.

One of the common reasons for a lack of crystallization is impurities / trace impurities, including water, salts, can cause these problems.
Again for crystallizations for solids is to dissolve in a warm solvent with a minimum volume. Once it is dissolved, very slowly adding a solvent which does not dissolve it (usually polarity based choices) and slowly cool it. In such cases, often pure crystals can be obtained.

If your compound is brand new, then you have no clue whether the compound is supposed to be a solid or a liquid and its color is not known either. Let us assume that you were expecting a solid, based on known and similarly structured compounds.
The meaning of purity is quite different in different branches. What is pure for an organic chemist (on the basis of NMR etc.) is quite impure for an analytical/physical chemist. So purity is the key to successful crystallization.

The most obvious factors are poor solubility and low area of contact of viscous goo with the solvent. Also, the substance might get oxidized with air and you are dissolving a layer of something else that might be less soluble. Storing your substances under argon/nitrogen won't hurt. You can try using sonic bath to speed up dissolving if your substance is stable enough.

@AChem and andselisk, Thank you for the advices.

It would be most helpful if the compound in question were identified. Crystals of some complex compound with a low melting point will naturally tend to oil out when evaporated to dryness from a powerful solvent - especially with contaminating ingredients, like unreacted starting materials. And a high mw non-crystalline goo will be resistant to redissolution because each molecule will be immersed in the gooiness of other molecules and won't be exposing much of itself - it'll be like a herd of buffaloes protecting its calves from a pack of wolves by circling up.

@JamesGaidis Its structure has thiophene and aza-crown. These two compounds have low melting point, so your explanation makes sense. In fact it is not purified yet, I was investigating which eluent should I use for column. Thank you.

Also, looking at your vial, I'd say it is micro scale synthesis. So, tiny solvent trap with in matrix may not be easily removed so always seems oily. Ask some natural product chemists. :-)

@MathewMahindaratne Thank you for the advice.


text:
A durable rubber type material that can flex easily but return to its original shape very fast
I'm looking for a durable rubber type of material for a guitar pick holder. The material I'm looking for will need to have flexibility to pick individual notes of a guitar, but needs to return to its original shape/position as fast as possible for picking faster notes.
I've tried vulcanized rubber sheets of different thicknesses (1 mm to 4 mm), it has good flexibility, but returning to its original shape is very slow.
The material from a bouncy ball is the closest material I can imagine for what I need, but I need something in a sheet form.

I think this question is better asked on the Music (or Engineering) stackexchange since this question isn't about Chemistry.

I think "rubber type" and "very fast return" exclude each other. It would need some hard flexible plastics or metal. The softer and more compressible material is (like rubber), the weaker are the forces returning it to its original shape, causing slow return.


text:
What endothermic reactions can be effectively reversed by electrolysis?
Not only newbie, but total layman here; please be patient with me.
Given are either two different elements, two different molecules, or one element and one molecule (chemical A and chemical B). A and B are preferably not solid in the temperature range from approx. −20 °C to approx. +40 °C.
They are placed in a container in the aforementioned temperature range (−20 °C to +40 °C). Once the heat increases above +40 °C, A and B engage in an endothermic reaction preferably resulting in a single chemical C, a liquid (potentially a gas, but I'm not sure if the physics or chemistry will add up further down) that will preferably remain in liquid (or potentially gas) state even if its temperature after the endothermic process goes down drops below +40 °C and down to, and not necessarily below −20 °C.
Assume C is placed in a different container where it is subjected to electrolysis or other procedure requiring no mechanical procedure, but electricity to reverse C to original A and B.

If any, what element(s) and/or molecule(s) may be A and B?

If any, what chemical is C resulting from A and B's endothermic reaction above +40 °C?

Is it possible to revert C to A and B in any of the above described manners without the reaction producing thermal energy?

Note that there reportedly exists functional outdoorsy fabric that releases some amount of heat at low temperature and absorbs heat at high one, to help keeping the person temperature confort. Not sure about the nature of this process nor about values of the specific latent heat capacity.

Any elemental, i.e. single step, reaction and many more complicated reactions that can reach equilibrium, can be reversed. The forward and reverse reactions in an equilibrium situation follow the same reaction path. The requirements are straight forward. At equilibrium there must be products, reactants, proper mixing, sufficient energy so enough molecules have the activation energy and an appropriate mechanism or catalyst for the reaction to proceed at a reasonable rate. If products are missing the reaction will proceed to the right to give products, if the reactants are missing the reaction will proceed to the left to give the reactants. When equilibrium is reached the activities of the products and reactants satisfy the equilibrium constant. It is important to include energy as a reactant and product in the reaction.
I will give one example that you can look up on the web to get more info, the lithium-ion battery in your cellphone. When it is fully charged there is an anode and cathode in a high energy state both of which involve Li+ ions; there is a potential difference between the anode and cathode, between + and -, and electrons cannot flow because of construction. When the battery is used electrons flow thru the device doing work and the circuit is completed by Li+ ion flow in the battery. after all the ions have flown and the high energy states of the cathode and anode have shrunk to zero the cell voltage is zero, no more cell phone. Now the charger is plugged in and applies a higher reverse voltage hopefully bringing very thing back to original [well almost].
The same thing happens in simple chemical reactions.  If O2 and H2 are mixed nothing happens until a catalyst or initiating energy is introduced; then the reaction starts the temperature rises. If the heat is contained there will be a very hot mix of H2, O2, and H2O at equilibrium. If the energy is removed by radiation, or doing work, or generating electricity, more and more water will form until at room T only the slightest trace of hydrogen and oxygen remains. Conversely if we start with pure water at room T, add a catalyst and slowly add energy either by heat or electricity water will break down into H2 and O2. Done carefully there can be equilibrium every step of the way for the reaction in both directions. Any reaction that is removed from equilibrium, provided the conditions are there to reach equilibrium, is spontaneous. In one direction it is exothermic driven by the loss of energy; in the other direction it is entropy driven by the increase of entropy. At equilibrium Delta G = 0 and the change in energy is balanced by the change in entropy.
Careful observation of every day events shows many examples of steady state events that involve [almost] chemical equilibria: rain, snow, clouds, a steaming tea kettle your breath on a cool morning, putting on a sweater if cold, removing it if warm, global warming for a biggie.

Thank you, your answer pointed at a major flaw of my thought process. I was trying to workout a solid state cooling system that induces absorption of heat be the introduction of electricity into the system. I added a third question, but I believe it is a flat no, and my curiosity ended in a dead end. To my great delight, I find that such systems do exist. They are also called thermoelectric coolers or batteries. Really fascinating stuff. Cell phones could make great use of them.

@KortellyZamatosh See also results of googling of thermoelectric effects site:libretexts.org

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

@andselisk thank you! I’ll try to wrap my head around them!


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Does the reaction of solid Sodium Chlorate and Hydrochloric acid generate Chlorine Dioxide?
I did this reaction on a very small scale, swirled it around to mix and react, and saw a yellow green gas come out, as Chlorine gas is one of the products. However, based on what I found online about the reaction I found it ambiguous whether or not it also produced Chlorine dioxide, as it has a similar appearance. Is it maybe dependent on the concentration of HCl, temperature, or pH? (I used 6 M HCl at around room temperature)

ClO2 is a red-brown gas [lambda max 36o nm] very different from chlorine. It can be produced by reduction of Chlorate by oxalate in acid and supposedly by disproportionation of chlorate in acid. The latter reaction is questionable as you seem to have found.  It is definitely formed by chemical or electrolytic oxidation of chlorite or disproportionation of chlorite with acid especially HCl. Most of published equations involving ClO2 are at best only partially correct because the various chlorine oxidants interconvert, and yields are variable. Check Wikipedia for examples.

@jimchmst Wikipedia:  ....ClO2 that exists as yellowish-green gas above 11 °C, a reddish-brown liquid between 11 °C and −59 °C, and as bright orange crystals below −59 °C

I have done this experiment once. I have dropped about $10$ mL concentrated $\ce{HCl}$ in a $4$ liter cylindrical glass container, containing some potassium chlorate. This produced a heavy green gas which progressively filled the container. When half filled, I took some steel wool with a pair of pliers, put it in a Bunsen burner, and dipped it quickly into the green gas. It produced a small explosion. So the gas was not pure chlorine. I don't know what has happened, but I did not repeat this experiment

Long ago in college lab a friend used concentrated sulfuric to clean some fused permangate  + chlorates mess from a test tube . He dumped the result into a large stone sink - explosion. Lab instructor ran over and threw the tube into the sink- explosion.  Next day the professor lectured about the dangers of chlorine dioxide. (  No, I don't know the chemistry . My friend got to stay in school).

Sodium chlorate/hydrochloric acid mixture has been actively used as chlorinating agent and Wikipedia did mention that it generates $\ce{HOCl}$ or $\ce{Cl2}$ depending on pH which act as active agent in the chlorination reaction, the reaction does also generate chlorine dioxide. Again, it depends on the kinetics and pH of the reaction system. Following is the abstract from the paper[1]:

Based on the mechanism of $\ce{ClO3-/Cl-}$ reaction system, the kinetics for reaction of sodium chlorate and hydrochloric acid to generate $\ce{ClO2}$ was studied. The rate equation of this reaction system was deduced and simplified as a formula with mixed-order (combination of first-order and second-order) towards $\ce{ClO3-}$. This rate formula indicates that the initial rate of the reaction is the first-order with respect to $\ce{ClO3-}$, and the reaction rate is the second-order with respect to $\ce{ClO3-}$ when $\ce{[ClO3-]}$ becomes close to zero. The rate constants of the first-order were determined as $\pu{0.0168s^{-1}( 30 °C)}$, $\pu{0.0221s-1( 40°C)}$, and $\pu{0.0279s-1( 50°C)}$, respectively, and that of the second-order were obtained for $\pu{0.0019Lmol-1s-1( 30°C)}$, $\pu{0.0028Lmol-1s-1( 40°C)}$, and $\pu{0.0060Lmol-1s-1( 50°C)}$, respectively. The results of statistic test prove that the rate formula obtained in this work is credible.

Reference

Li, X.J., Jiang, D.D., Zhang, Y.J., 2013. Kinetics of the Reaction for Generation of Chlorine Dioxide from Sodium Chlorate and Hydrochloric Acid. AMR 634–638, 546–550. DOI: 10.4028/www.scientific.net/amr.634-638.546


text:
What are the available cheap proteins that are much less soluble than BSA
Currently, as a beginner, I'm performing solubility analysis (with UV based spectrometry) for practices.
The protein I'm using now is BSA. This is widely used and cheap option.
I'm now looking for a cheap negative control protein which I can use to compare with BSA.
It has to be much less soluble than BSA.
What are the good choice of proteins for that?

Protein solubility is not a simple concept. For instance proteins are colloidal and may aggregate to form a new highly hydrated phase that may slowly form and precipitate. One way to encourage aggregation is to use a buffer at the pI of the protein, so you may be able to do this with BSA.

@BuckThorn Thanks so much. Can you advice the simple protocol to go about it?
What type of buffer, how much etc.

@BuckThorn and with your method, how can we calculate the molecular weight of the new protein we create?

One technique to determine oligomerization state and distribution is size exclusion chromatography. Whether this is the right technique depends somewhat on the size and stability of the protein aggregates.

Can you explain how you would use an insoluble protein as a negative control? Is it just a precipitate at the bottom of the cuvette? And why is that a negative control rather than just zero protein as the negative control?

@Andrew I simply want to practice making two standard curves to show that one protein is less soluble than another.

How do you plan to make a standard curve of an insoluble protein?

@Andrew By insoluble  I don't mean completely insoluble. You still can measure the little of it that is soluble. So, in standard curve this much less soluble peptide will have lower absorbance score than those that are soluble.


text:
Variation of conductance of saturated aqueous solution with temperature
The following question was asked in JEE Mains 2020:

Let $C_{\ce{NaCl}}$ and $C_{\ce{BaSO4}}$ be the conductances (in S) measured for saturated aqueous solutions of $\ce{NaCl}$ and $\ce{BaSO4}$, respectively, at a temperature $T$. Which of the following is false?
(A) Ionic mobilities of ions from both salts increase with $T$
(B) $C_{\ce{NaCl}} >> C_{\ce{BaSO4}}$ at a given $T$
(C) $C_{\ce{NaCl}}$($T_2$) > $C_{\ce{NaCl}}$($T_1$) for $T_2 > T_1$
(D) $C_{\ce{BaSO4}}$($T_2$) > $C_{\ce{BaSO4}}$($T_1$) for $T_2 > T_1$

My attempt:
(A) As temperature increases viscosity decreases leading to higher mobility, hence this is true.
(B) $\ce{BaSO4}$ is almost insoluble in water, hence its saturated solution will have a much smaller quantity of ions compared to an $\ce{NaCl}$ saturated solution, hence this is true.
(C) As can be seen from the graph here, the solublity of $\ce{NaCl}$ does increase with temperature, but only slightly; nevertheless this option is also true because it does increase and also ionic mobility increases.
(D) Dissolution of $\ce{BaSO4}$ is an endothermic process, hence I would expect solubility (hence conductance) to increase with temperature (although I was not able to find any data), therefore this option is also true.
Where am I going wrong? This is a single-correct question. The given answer is:

 (A), but there are varied options found all over the web for this question

The answers are self-contradictory. If C and D are correct, then A must be correct. It cannot be false. Ion mobilities increase with temperature in solutions. I don't know what the examiner was thinking when writing this question.

Barium sulfate solubility increases with temperature. Here values from Kohlrausch : $\pu{3°C - 0.207 mg/L ; 18°C - 0.23 mg/L ; 26°C - 0.266 mg/L ; 34°C - 0.29 mg/L}$ These values are cited in a recent report : Arbaoui M. Ali, Energy Procedia 157 (2019) 879 - 891.

There can be catch if they mean NaCl solution is saturated at all T ( with excess of solid salt) or if just saturated solution is taken and heated up, becoming unsaturated. For the former, there can be competition of mobility increase due T versus effective mobility decrease for big electrolyte concentrations ( due ion atmosphere and ionic pair formation) .

But as NaCl solubility does not change much, I guess positive T effect on ion mobility would dominate.

Okay, there is actually a trick way to solve the question. If any of A, C or D are false the one of the others also becomes false. (As mentioned by @AChem).
So the only option left is (B), which should (ideally) be the correct answer.
Now, coming to the actual reason, as per what I’ve found, conductivity ($k$) of salts in $aqueous$ solution depends on their concentration and number of ions generated on dissociation.
$k$ in aqueous solution or water : $Ba^{2+}$ > $Na^+$.
$k$ in molten state : $Ba^{2+}$ > $Na^+$.

Please don't use MathJax for emphasis. Note that chemical element symbols are upright. Taglines such as "Hope it helps" are unnecessary noise. There are several IUPAC standard symbols for different conductivities, but none of them include $k$ (you can look them up in IUPAC Green Book). It would probably make sense to back up your answer by an equation or some literature data.

@andselisk Ah thanks again ! Actually I have trouble typing super scripts so use mathjax, also could you please tell me how I can italicise text ?

@andselisk also, “ The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by k, k, λ “, from Wikipedia. k(kappa) is used for conductivity. I don’t have any equation or data as such for the question, only reasoning from various sites I extracted.

One can use $\kappa$ to get $\kappa$. // Useful links for text and formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic // Use plain texts in CH SE titles.

You are welcome. //  For more, see Math SE MathJax tutorial.


text:
Can uranium be used to color aluminium oxide based gemstones, and/or diamonds?
Re-hosting this question from Physics to Chemistry, lightly edited.
Rubies and sapphires are chemically equivalent to the mineral corundum, with the exception of transition metal impurities that impart different colors to the aluminium oxide. Uranium is not a transition metal, but it can be used as a coloration admixture with silicon dioxide, and exists in various oxide forms.
Can uranium be used as a coloration impurity for 1) aluminium oxide based gemstones and/or 2) diamonds? If uranium cannot chemically bond with the substrate, then the ability to produce a consistent heterogeneous mixture, e.g. by suspending uranium oxide dust in molten aluminium oxide during the process of flame fusion would be a satisfactory answer in the affirmative.

Cross-posting is frowned upon, though...

@Mithoron A mod over at physics suggested here might be better. I see no reason, a priori, it would not be possible to dope the alumina with urania, in moderation. Maybe the result would not be transparent, like uranium glass?

@EdV Yeah, but migration would be proper way to solve it. BTW Uranium is way bigger then Al and that may be a problem.

This actually sounds like an interesting concept. Even if it isn't as easy as it would be for some smaller transition metals, at low concentration is might be interesting. And easy to test.

@Mithoron Agreed on both points. Still, it would be interesting to see what happens with a low doping level.

It's not quite alumina, but uranium(VI) oxide compounds have been used to color glass:

Oxides and uranates of uranium(VI) have been used in the past as yellow ceramic glazes as in Fiesta and to make yellow-green uranium glass.1

Source
But beware:

Both of these applications are abandoned due to concern regarding radioactivity of the uranium.

Cited Reference

Skelcher, Barrie (2002). The Big Book of Vaseline Glass. Atglen, PA: Schiffer Publishing. ISBN 978-0-7643-1474-2.

This reminds me of a video made by NileRed (cc @jpt4)


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Aqua Regia 0.1M solution
I was reading a method for total digestion, and the method uses 0.1 M aqua regia. I know that aqua regia has 3 parts Hcl and 1 part HNO3, by volume. How would I make a 0.1 M aqua regia solution, is the 0.1 M referring to the HCl or the HNO3?
Thanks in advance!

How can be aqua regia 0.1 M, when it is mixture of concentrated acids? It is like having 0.4% whisky.

There's no such thing as "Aqua Regia 0.1M"

You need to consult the source what they mean.  A asking B what C means by X is bad concept, if X does not makes sense or is unclear without more details or context.

As the comments imply, this isn't aqua regia at all. What makes something aqua regia is its content of molecules of nitric acid and whatever other stuff is produced when these molecules react with the hydrochloric acid. (Hydrochloric acid itself is stronger and may still be dissociated.) Put this stuff into enough water to make the concentration 0.1 molar, you dissociate those nitric avid molecules and hydrolyze the aqua Regia reaction products noted above, and all you're left with is dilute-acid ion soup.
"Aqua regia" is classically a 3:1 (respective) mixture of hydrochloric acid with nitric acid. I've never seen what that 3:1 is based on, probably it's by volume in which case the molar ratio is probably closer to 1:1. I would make 0.05 molar HCl and 0.05 molar HNO3 (labeling it like that, not an aqua regia solution) and call it a day.


text:
How to draw the structural formula for 3-ethyl-5-methyl-3-heptene?
This is the kind of structural formula we need to draw:

I took help from Reddit. Here is the attempt:

I have done one more attempt:

Though the 4th carbon of the left condensed version needs to have a hydrogen added, did not take a picture of that.

We do not need any unnecessary rants. We only need the question and the attempt you have made. That's it.

Sorry to say but the question is inappropriate! It is supposed to be 3-ethyl-5-methylhept-3-ene. Here's the solution:

First of all, thank you so very much! Now I know my answer is correct. (Got some help from Reddit before seeing your solution) The thing is, I really am innocent, just copied the question from the worksheet, you can look at the added picture, but I am still glad I was able to learn the correct naming structure for branched hydrocarbons. P.S. Nice handwriting.

It's not the question that's inappropriate, it's the name doesn't follow current nomenclature guidelines. Please familiarize yourself with How to ask and answer nomenclature questions? Here it would more make sense to cite a related rule or set of rules from the IUPAC Blue Book to back up your answer.


text:
How do you determine the charge of the metal ion in a single replacement reaction?
Say we have a simple single replacement reaction between a salt and a metal
$$\ce{2 AgNO3 + Cu -> Cu(NO3)2 + 2 Ag}$$
We know in normal circumstances, silver always has a $+1$ charge, and in normal circumstances for copper, it typically varies between a $+1$ or a $+2$.
Now I have been told that in copper nitrate, the copper ion will (almost?) always be in a $+2$ state, but this raises the question of why we would know that. Why can't the reaction be
$$\ce{AgNO3 + Cu -> CuNO3 + Ag}$$
instead?
Say we were to have a reaction between silver sulfate and copper. What charge would the copper take on? How do we know?

There are a few ways to make an educated guess if you don't know it already, from textbook - that's a generic reaction, you'd stumble on it sooner or later. 1) You should know that Cu is pretty easily oxidised to +2, so if you use pretty strong oxidiser like Ag+... 2) Another option is that Cu+ actually tends to disproportionate in aqueous solutions to Cu and Cu 2+

Hmm, now that I think of it you mean actual experimental reinventing the wheel? I.e. we just ignore few hundred years of chemistry and find it out?

Why copper(I) is unstable in aqueous medium?; Why do Cu⁺ ions spontaneously form copper metal and Cu²⁺ ions in solution?

You cannot know from basic principles. You must have background knowledge of element chemistry, eventually involving also concept of redox potentials. It can be also verified via element molar masses and experimentally determined equivalent mass ratios.

Say we were to have a reaction between silver sulfate and copper. What
charge would the copper take on? How do we know?

Silver sulfate is not very water-soluble. Let us say that we use a generic soluble Ag(I) salt and we wish to know what will be the resulting charge on the copper ion after the reaction. If we were discovering/ studying this reaction for the first time, we might do the following:
One is the pure observational way: We make a solution of a silver salt in water and place a copper plate in the solution. After some time, the solution turns blue. Using previous knowledge that Cu(II) solutions in water are typically blue, one can make an educated guess that the resulting ion is Cu(II) not Cu(I). Cu(I) is colorless.
Second option is pure analytical way: We let copper and silver nitrate react completely (i.e., excess copper), and after a reaction, we physically separate the silver particles and remaining copper by filtration. We evaporate the solution and analyze the blue crystals chemically. The resulting formula after analysis will suggest that copper must be in Cu(II) form, otherwise the elemental analysis data would not agree with Cu(I) salt.
Third approach is the electrochemical approach which you might not be familiar as yet: One can start with half-cells and look at electrode potential values. The question is, if Cu is given a choice to react with silver ions in water, what will be the most favorable product?
Ag(I)/ Ag  half-cell has an electrode potential of +0.80 V
Cu(I)/Cu half-cell has an electrode potential of +0.52 V
Cu(II)/Cu half-cell has an electrode potential of +0.34 V
Which half-cell of copper will give a larger positive potential difference with silver half-cell? It is the Cu(II)/Cu half cell, which means that if Cu(s) reacts with Ag(I), the thermodynamically favored product will be Cu(II) not Cu(I).


text:
Why is the rubber tip of my capacitive stylus no longer conductive?
In 2014, so eight years ago, I bought five styluses from China. Cheap stuff obviously. They worked fine while I used them, but soon after they were just left in a drawer. Last night I realized that they all don't work anymore. I cant even make a single click on my phone. I also tried with other devices.
The pen, all of it except for the rubber tip, is made from stainless steel.
The tip is made like a small bladder from black rubber. When you press it against a surface, from any angle, it presses against that surface with an are similar to a circle with a 5mm diameter.
The rubber does not appear to be cracked, deteriorated, discolored or anything like that.
From what I can read, the rubber tip is most likely made from so called conductive polymers. Is it normal that these age / deteriorate in such a way, that they lose their conductivity?
My friend suggested, that if it was some kind of conductive carbon / graphite, that might simply "fall of" the rubber as dust.
I am asking out of simple curiosity :)

Are you still using them with the same device?

@No: In 2014 I used it with an Acer tablet and a Samsung Galaxy Ace, but I don't have any of those anymore. Yesterday I tested it with two different phone models from Huawei and OnePlus, and a tablet from Apple.

I cannot imagine what should happen to an unbent rubber in a dark closet (under stress, rubber undergoes ozonolysis and develops cracks). My best guess would be the new phones want a stylus that is somehow different.

@Karl: I cant totally dismiss that. And unfortunately I dont see a way of finding out. However, my personal experience indicates that newer phones are more sensitive than older ones; for instance my newer smart phones work more reliably with gloves than my first ones did.


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Number of bridge oxygen atoms in a silicate mineral

Total number of oxygen atom(s) which act(s) as bridge between any two silicon atoms in a mineral with composition $\ce{MM’Si3O_x}$ $(\ce{M}$ is the divalent metal ion and $\ce{M’}$ is the tetravalent metal ion).

This is the question I have been stuck on. I do not understand how to approach it. Are we supposed to used the concepts of silicates? Like sheet silicate, chain silicate etc. or interstitial salts concepts?
Also, I don’t know whether it is possible to draw the structure as $x$ is a variable. Any hint would be appreciated.
P.S. Answer given is 3.

I am answering it to let others know that I have got the satisfactory answer, thanks to andselisk.
Since $\ce{M}$ is divalent and $\ce{M’}$ tetravalent, they exist as $\ce{\overset{2+}{M}}$ and $\ce{\overset{4+}{M’}},$ respectively. The anionic part is $\ce{[\overset{+4}{Si}_3\overset{-2}{O}_x]^6-},$ and taking oxidation numbers into consideration, $x = 9.$ $\ce{[Si3O9]^6-}$ moiety implies cyclic metasilicates:

There are three bridging oxygen atoms per ring or per formula unit. A real life example for the mineral from the problem could be benitoite $\ce{BaTiSi3O9}$.


text:
Comparison of enthalpy change for the dissolution of alkaline earth metal oxides
Compare the heat liberated per mole for the following reactions:
$\ce{MgO + H2O -> Mg(OH)2}$
$\ce{CaO + H2O -> Ca(OH)2}$
$\ce{SrO + H2O -> Sr(OH)2}$
$\ce{BaO + H2O -> Ba(OH)2}$
Data for lattice enthalpy; $\ce{MO (s)->M^2+ (g) + O^2- (g)}$

$^1$Metal Oxide
Lattice Enthalpy (kJ/mol)

$\ce{MgO}$
+3923

$\ce{CaO}$
+3517

$\ce{SrO}$
+3312

$\ce{BaO}$
+3120

Data for hydration enthalpy; $\ce{M^2+ (g)->M^2+ (aq)}$

$^1$Metal Ion
Hydration Enthalpy (kJ/mol)

$\ce{Mg^2+}$
-1921

$\ce{Ca^2+}$
-1577

$\ce{Sr^2+}$
-1443

$\ce{Ba^2+}$
-1305

Data for enthalpy of formation; $\ce{M (s) + O2 (g) + H2 (g)->M(OH)2 (s)}$

$^2$Metal Hydroxide
Formation Enthalpy (kJ/mol)

$\ce{Mg(OH)2}$
-925

$\ce{Ca(OH)2}$
-985

$\ce{Sr(OH)2}$
-959

$\ce{Ba(OH)2}$
-945

Energy required to break the lattice is positive, enthalpy of hydration is negative and enthalpy of formation is also negative. Hence, the net energy will sum up to a positive value. But these reactions are highly exothermic. Thus, there must be other energies to be considered.
Apart from lattice enthalpy, hydration enthalpy and enthalpy of formation what other factors to be considered?
P.S.: The highest value is for $\ce{Ca(OH)2}$.
References:

Concise Inorganic Chemistry- 4E by J. D. Lee & Sudarshan Guha (ISBN 9788126566495)

Table 6.3, Lange's Handbook of Chemistry- 15E

I miss the enthalpy of $\ce{O^2-}$ hydration and $\ce{M^2+(aq) + 2 OH-(aq)->M(OH)2(s)}$ precipitation, if the solid state is the final state. M(OH)2 formation enthalpy is for $\ce{M(s) + O2(g) + H2(g) -> M(OH)2(s)}$.

@Poutnik Now I think, it's very difficult to calculate by this method. Is there any short way?

Rather, it is difficult to obtain all needed input data. Shorter way is obtaining the results anywhere, or measuring it. BTW why the 1/2? 1/2 of O2 or H2 would not give 2 O or 2 H for M(OH)2.

I meant, by applying logic rather than calculating from data. My bad for 1/2.

As Poutnic said, you gotta account for O hydration energy too and its inevitable conversion to OH ion as aswell and its energy too. Frankly, if you need to measure any of these, don't get into theory. Just slake some of each in a calorimeter.

Some of the information pertains to the solid hydroxide and some pertains to the hydroxide in solution. As magnesium and calcium hydroxide actually have only limited solubility, I will assume the former applies.
You have enthalpies of formation for the solid hydroxides. From the Wikipedia articles for the respective compounds we can also get these enthalpies of formation, all in kilohoules per mole at 25°C:
$\ce{H2O}: -285$ (liquid)
$\ce{MgO}: -602$
$\ce{CaO}: -635$
$\ce{SrO}: -592$
$\ce{BaO}: -582$*
*There is surprising disagreeement over the heat of firmation of barium oxide, for instance NIST gives $-548$ kJ/mol.
From these values we may render all the hydroxide formations exothermic:
$\ce{MO + H2O -> M(OH)2(s)}$
$\ce{Mg}: \Delta H = -38$ kJ/mol
$\ce{Ca}: \Delta H = -65$ kJ/mol
$\ce{Sr}: \Delta H = -82$ kJ/mol
$\ce{Ba}: \Delta H = -78$ kJ/mol ($-112$ using NIST value for $\ce{BaO}$)
We see that on a molar basis, strontium or barium evolves the most heat, depending on which source is used for the barium data. But if we divide each of these results by the molecular weight of the product, we find that because of its lighter weight and still relatively high heat release calcium has the strongest heat effect per unit mass.

By using my second reference, the heat of formation of BaO = -548 kJ/mol from which the enthalpy of dissolution for BaO = -111 kJ/mol. So, the required enthalpy per mol is increasing from MgO<CaO<SrO<BaO. Can we explain this due to the increasing basic character from MgO to BaO?
The order of liberated heat per gram is: BaO<SrO<MgO<CaO

That seems odd with barium oxide, perhaps another question is why there is such a difference. I tried to be consistent by using the current Wikipedia data.

Except BaO, wiki data for other oxides matches with Lange's.