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How does an electron identify whether to absorb a photon of a certain wavelength or not in absorption spectrum?
How would an electron identify the specific photons of a certain wavelength which are to be absorbed in a Hydrogen atom while creating an absorption spectrum? Well, I was thinking of it like this-
Way-01: The electron will absorb any photon coming towards it regardless of its wavelength and jump. Then if the wavelength is the exact amount required to jump from one shell to another, then the electron simply jumps to the higher shell. Otherwise, it cannot find a shell and since electrons cannot exist in between shells, they simply re-emit that photon and we perceive the incident as if the photon weren't absorbed at all.
Way-02: The explanation given above is false and electrons choose the photons via some other mechanism
Now, my question is- is the "Way-01" correct? Or do electrons identify the photons via some other way? If so, then what is it?
Note: I'm only a high-schooler, so any relatively simple qualitative explanation will be greatly appreciated.

“relatively simple qualitative explanation” There isn't one. You can come up with stories that seem to make intuitive sense to yourself, like your explanation above, but I doubt any of them will be close to reality, i.e. quantum mechanics and possibly quantum electrodynamics. Electrons can't exist in limbo and "choose" whether to re-emit a photon or not; they don't have a brain, they aren't sentient. It's fun, but incorrect, to anthropomorphise them.

[It's perhaps not the answer you want, but I do not mean it in a condescending manner at all, as I don't understand it myself. See also Feynman's famous quote: "If you think you understand quantum mechanics, you don't understand quantum mechanics.".]

I like Way-01. Slight modification: the electron gets vibrated, rather than absorbed, by any incoming photon and may just shift into a cavity (shell) where the vibration echoes just right. Coupled pendulums illustrate the transfer of energy from one item to an identical item, so with the substitution of photon energy for electron energy, you have a similar picture. https://www.youtube.com/watch?v=3c73d5GJ2V0

@JamesGaidis Don't confuse him further.

Well, actually, maybe some kind of resonance analogy might work...? It would quickly degenerate into a very circular description if I tried to explain it, but someone better than me could do it. I'll bow out...

I must reiterate what orthocresol said: don't even start making up stories about quantum stuff. You don't yet have the right words for that, and won't have for a while. The explanations you come up with may be witty and colorful and all that, but they won't lead you in the right direction. For example, contrary to your "Way-01", the photon is decidedly not re-emitted. We know that because there are cases when a photon is genuinely re-emitted, and it behaves very differently then. To begin with, it "forgets" where it came from and has no reason to continue in the previous direction.

@Habib I agree with the others that some sort of intuitive classical will ultimately be misleading. However, if I were to offer one, it would be this: If you toss a handful of differently shaped objects such as spheres, cubes, prisms, etc. at circular hole the width of the sphere, how does the hold identify whether to allow an object through or not? I suppose you could say the hole will absorb the objects of the right size, and not interact with the others. For electrons, the measurement by which it will or won't interact with photons is energy.

The Bohr model of the atom is wrong. It's primitive and wrong. "However, because of its simplicity, and its correct results for selected systems, the Bohr model is still commonly taught to introduce students to quantum mechanics." https://en.wikipedia.org/wiki/Bohr_model

[OP..] How would an electron identify the specific photons of a certain wavelength which are to be absorbed in a Hydrogen atom while creating an absorption spectrum?

As discussed in the comments, this is an observed phenomenon. We can try to describe this observation in a way that we can make predictions, but that does not answer the questions "how?".

[OP..] Otherwise, it cannot find a shell and since electrons cannot exist in between shells, they simply re-emit that photon and we perceive the incident as if the photon weren't absorbed at all.

There are processes that do something similar as the one you describe, but they involve observable events. When an atom scatters light, it scatters light of the same wavelength as the incoming light, but in all directions. There are also processes where the light is absorbed and then emitted again after a while (sometimes at different wavelength). What you suggest is an explanation for observing no interaction between the electron and the photon, and explaining it with an interaction (absorption and instant re-emission at the same wavelength in the same direction) that we can't observe.
You could have explanation models of the same flavor for many things that don't happen. For example, on a day it does not rain, I could say drop form in the clouds but instantly are destroyed again, so we can't observe them ever.

Way-02: The explanation given above is false and electrons choose the photons via some other mechanism

As you can read in the comments, using the term "choose" for something electrons do is not so helpful. We know under what conditions the photon might get absorbed (not all of them are), but to go beyond that and explain the "how" or even "why" is difficult to impossible (and not taught in e.g. the 4-year college chemistry curriculum, as far as I know).

Since yesterday I was thinking on how to decently write this. The important part is to point out that something similar to way 2 exist but in an observable way. Plus 1. To OP: Still, one can put it even in more pictorial way. For instance, taking the photon as the anthropomorph does already a better job than way 1. It simply goes where it can, like a marble goes into hole in which it fits. The marble and the hole don't need to know it a priori (of course a marble falls also in bigger holes, the photon not.. This open the way to some of the scattering process mentioned above).


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Compressibility Factor and Generalized Compressibility Chart
From my understanding, the compressibility factor is defined by
$$Z=\frac{pv}{RT}=\frac{p\bar{v}}{\bar{R}T}$$
It can also be defined by the ratio of the real molar volume of a gas to the ideal molar volume of a gas at the same temperature and pressure. Essentially it corrects for the deviation of a real gas from an ideal gas.
On a generalized compressibility chart, the compressibility $Z$ is plotted as a function $f=f(p_R,T_R)$ of the reduced pressure and temperature. I don't understand why exactly; it would be nice if someone could explain that a little more.
Another thing I'm confused about is the psuedoreduced specific volume, given by
$$v'_{R}=\frac{\bar{v}}{\bar{R}T_c/p_c}$$
Why don't we use the reduced specific volume as opposed to the psuedoreduced specific volume?
In my thermodynamics class, I was presented with the following question:
Determine the temperature, in °C, of air at 30 bar and a specific volume of 0.013 $m^3/kg$. Use compressibility chart.
I was confused about how to figure out this problem since I can't find the reduced temperature without the actual temperature of the air.

This is a trial and error calculation.

@ChetMiller How so? Is there not one answer?

There is one answer.  Do you know what a trial and error calculation is?

@ChetMiller No, I'm not sure how to go about this.

You start out with an initial guess for the temperature.  I suggest starting out using the value that you would calculate using the ideal gas law.  You then,

Calculate the reduced temperature and pressure
Get z at this reduced temperature and pressure
Use these to calculate a value for the specific volume.
If the value for the calculated specific volume does not match 0.013, adjust your guess for the temperature and go back to step 1.

Continue doing this until the calculated specific volume matches the prescribed value of 0.013

source: Stack Exchange

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Why does the pH of the endpoint decrease when temperature of the titration is increased?
I was doing an experiment in chemistry class involving titrations. I would heat up different samples of vinegar and titrate them using NaOH. The trend that I found was how has temperature increase, the pH of the endpoint decreases. Can someone explain this to me please?

If you were heating the solutions without covering the flasks, the evaporating water left the flask concentrating the acid, hence why you obtained a lower pH reading. Commercial vinegar is usually a 5% solution of acetic acid (CH3COOH).

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How can you increase the pH of an acidic solution by 1?
In order to increase the pH of a solution of some acid $\ce{HA}$ by 1, how many times must it be diluted by? The pH log scale suggests that the required dilution factor is 10. Is that generally valid for strong and/or weak acids, or do we need to take into account the dissociation constant for weak acids?

You can't tell without more data.

Since you have not given any values some of the important values need to be considered as variables,
Also assuming its a monobasic acid $\ce{HA}$
$$\ce{HA_{aq} <=> H^+_{aq} + A^-_{aq}}$$
The important values are the

Acid dissociation constent $K_\mathrm{a}$
Initial concentration $c$

Writing the equation for the acid dissociation constant
$$K_\mathrm{a} = \ce{\frac{[H^{+}][A^{-}]}{[HA]}}$$
Initially a $x$ amount of acid gets dissociated equation simplifies to
$$K_\mathrm{a} = \frac{x^{2}}{(c-x)}$$
This is a $ax^2+bx+c=0$ type equation which can be solved easily simplified,
$$x^2 + K_\mathrm{a}x - K_\mathrm{a}c = 0$$
Take the accepted value for $x$ then (Not thinking that you will get complex number for real $K_\mathrm{a}$ and $c$ values),

Optionally you can consider that $c \gg x$ so $c - x \approx c$ simplified $x = \sqrt{K_\mathrm{a}c}$

Since $x$ is found now, we now know the initial $\mathrm{pH}$ value or $\mathrm{pH}_1$
$$\mathrm{pH}=-\log(\ce{[H^+]})$$
so,
$$\mathrm{pH}_1=-\log([x])$$
Final $\mathrm{pH}$ value $\mathrm{pH}_2$ goes up by 1 so,
$$\mathrm{pH}_2=-\log([x]) + 1$$
So the new dissociated $x_2$ value,
$$x_2 = 10^{-(1-\log([x]))}$$
Since $\ce{[H^+]=[A^-]}$ new equation,
$$K_\mathrm{a} = \frac{x_{2}^{2}}{(c_{2}-x_{2})} = \frac{10^{-2(1-\log(x))}}{(c_{2}-10^{-(1-\log(x))})}$$
Now you can solve this equation and get $c_2$ new concentration now the answer is almost complete just need to get the dilution factor
$D = \dfrac{c}{c_{2}}$
$D$ is how many times you need to dilute
But in the non monobasic acids this is a bit more complex since it has more than one dissociation constant $K_{\mathrm{a}_{1}}$ $K_{\mathrm{a}_{2}}$ $K_{\mathrm{a}_{3}}$ etc. need to check if the ratios of the dissociation constants if so that they are effective in the $\mathrm{pH}$ calculation.

I assume the task author goes for the simplified formula, based on $c_0 \gg [\ce{H+}] \gg [\ce{OH-}]$

Hmm yeah I have now added that in a block just now

why does 2=10^−(1−([]))  and not x2=10^-log[x]+1

Its just $\ce{pH = -\log{([H^{+}])}}$ substitute the new $\ce{pH_2}$ to the equation and try to get the new $\ce{H^+}$ concentration you made a mathematical mistake watch out there is a minus sign there


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Why do 4d and 5d metals have more orbital overlap than 3d because of size?
When considering the electrical conductivity of metal oxides, they will conduct better if the metals orbitals can reach each other to overlap. Whether this can happen is contributed to by a couple of things such as Z-eff charge (electronegativity), the anion, oxidation state etc.  We're told that one of those properties is that as you down a group, the size of the orbitals increases, and hence more overlap.
But isn't making the atoms bigger just going to mean that the distances between the atoms get bigger proportionally? So switching a 3d for a 4d, but keeping the same structure would just look like "zooming in", but not change any proportional overlap?
As I understand f orbitals are inside (having an electron density maximum closer to the nucleus), and so don't participate in bonding.

Just to be clear, it is not just "zooming in" and everything cancelling out neatly in the end when we you move downwards in the same group. We wouldn't need the modern relations if things were so nice and tidy :)


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Accidentally sprayed the wrong product in my mouth
I sprayed the wrong product in my mouth the other day. It was a spray hand cleaner. Noticed the mistake right away and immediately spit it out and then rinsed my mouth. But even so, my tongue and throat burned for a good 10 to 15 minutes afterwards. The label on the product lists the following ingredients but no % of concentrations is given:

List item
benzethonium chloride
myristamine oxide
polyquaternium 7
deionised water
fragrance

Again, the amount that I actually swallowed was infinitely small but I was surprised it burned that much. I would like to know your opinion if I should consult a doctor just the same. This happened 3 days ago and now feel pretty good, except for perhaps the occasional burning still in my tongue and throat. But the tongue looks good, no burning marks or anything like that that I can see.
Thanks for your time.


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Organic Based Photocathode
What organic molecules have a relatively small work function, preferably in the visible spectrum? Polycyclic aromatic hydrocarbons are in the low UV spectrum, making it unsuitable for the photocathode requirement. A work function in the green-cyan spectrum, or anything below3 eV is the main focus. Efficiency, thermal stability, etc is not the main focus for the time being

Look online for conjugated polymers. (I don't embark in a discussion or rant on the cons of using solid state physics concepts and words because it is a lost battle).

Also it will be useful to have a frame.

A work function in the green-cyan spectrum, or anything below3 eV is the main focus. Efficiency, thermal stability, etc is not the main focus for the time being

I don't know any. An organic with a very low oxidation potential & availability is PEDOT. In it oxidised form it can be cast in film having a work function similar to gold. I am not aware of HOMO levels lying higher than that in readily handled organics. But shall be the cathode be photoactive? Your question isn't that clear.

PEDOT has a work function of 5 eV, well into the UV spectrum. The main goal is a work function of 3 eV of below, all other factors are not of importance for the time being.

Therefore I said PEDOT has already a very low oxidation potential. It seems you are looking for an organic comparable to alkali metals!!!!


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Stereochemistry of the product of crotonic acid with bromine
Is there a a simple rule for determining the stereochemistry for the product of a reaction between an unsymmetrical alkene like (2E)-but-2-enoic acid and bromine (anti reaction)?

If it's an anti reaction, does that mean that I just need to put each bromine atom on a carbon, which makes the two carbons of the double bond different configurations, like carbon 2 R and carbon 3 S, or the other way around?
And if it's a syn reaction, does that mean that the two carbons in the product need to be same configuration? If that's the case, I get it. But if it's not, how would you decide how would the bromines go in without models?
Last but not least, in the second part I need to do an elimination (I made this question by the way to understand this concept), how would I know if the product is E/Z?

As far as I remember, bromine water is used instead of just bromine though I am not sure if water has any effect on the reaction.

it's just Br2/CCl4 , no water. but same logic.


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Photoelectric cells
I've read about photoelectric cells, and most of the sources say that Potassium and Cesium can be used as electrodes in the same. But why not Rubidium?
The abundant isotope of Rb (72.2%) is not radioactive either, and is actually quite stable.
The ionization energy of Rb lies in between that of Cs and K, so its pretty easy to remove electrons from Rb too, and does not require much energy.
So, is Rb used in photoelectric cells or not? If not, why?

All of the alkali elements can be used in photocells, and they are suitable because of the comparatively low first ionization potential.
Consider the photoelectric effect, though; one of the first clues to quantum mechanics. An incident photon must have sufficient energy, E = hν, to knock out that electron. A red light photon is ~1.5 electron volts (ev), and violet ~3 ev, so photocells using cesium are insensitive to red light, and potassium photocells would respond only to midrange UV. Rubidium would work for long-wave UV and shorter wavelengths.
So, it depends on the wavelength for which the photocell is intended (as well as the volatility of the metal at the temperature of use, cost, etc.).

yes, im aware of the wavelength argument. but i was wondering if theres any specific reason why only Cs and K are used. the others may be used, but thats not usually the case. why?

Misapprehension "Rubidium is also used in the manufacture of photocells." https://www.azom.com/article.aspx?ArticleID=6063


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Why are Methyl vibr. modes called r modes and methylene vibr. modes called d modes
I am coming from a chemical background and currently work in the field of vibrational spectroscopy. When I talk to fellow physicists or chemists I always encounter the same question. "Why are the vibrational modes of methyl- and methylenegroups are called r and d modes".
Does this have a group theory background? Can someone elaborate on that?

Can you show a reference/ webpage which shows r and d modes? What is the context?

Vibrations In this picture i illustrated the methyl and methylene stretching vibrations.

Where is the reference to this picture?


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Why is the ester carbonyl peak in benzocaine IR spectrum so low?
Having looked at multiple IR spectra for benzocaine, I am struggling to understand why the carbonyl C=O peak is so low in the spectra compared to where an ester carbonyl group usually is on other IR spectra

What do you mean by "low"? Could you provide any numbers or visual cues?

I think it might be due to conjugation with the benzene (or rather, aniline) ring.

Let's first have a look at the structure of benzocaine, to see how it is different from an usual aliphatic ester:

One thing that should be immediately obvious is that the ester is attached to an aromatic ring (which has an amine group). The aromatic ring would participate in conjugation with the ester oxygen, particularly because the amine group is strongly electron-donating (via conjugation).
This is the IR spectrum of benzocaine (from Chemical book):

The peak at $\mathrm{1682\;cm^{-1}}$ can be assigned to the ester C=O stretch. Usually aliphatic esters have IR stretches around $\mathrm{1730-1750\;cm^{-1}}$. Esters attached to unsaturated systems or aromatic groups have lower stretches, as the conjugation weakens the C=O bond.
Compare the $\ce{C=O}$ IR stretches of some esters:

Ethyl acetate  : $\mathrm{1752 cm^{-1}}$
Ethyl benzoate  : $\mathrm{1726 cm^{-1}}$

The presence of the benzene group has reduced the stretch wavenumber by $\mathrm{\sim30 cm^{-1}}$.

methylparaben  : $\mathrm{1682 cm^{-1}}$

Here the $\ce{-OH}$ group is quite strongly electron-donating, and it brings down the ester C=O stretch wavenumber even further, and now the IR stretch resembles that of benzocaine. $\ce{-NH2}$ is also a stronlgy electron-donating group like $\ce{-OH}$.


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Why does whey protein powder make porridge much less viscous?
When I mix whey protein powder into warm porridge (that I cooked in the microwave), the thick porridge becomes much less thick.
I searched this, and I found several forum threads discussing this too so it seems to be a common phenomenon. Happens when cooking porridge on the hob too, not just from mixing protein in after cooking.
I make the porridge through simply heating porridge oats + cow's milk. It doesn't matter if the oats are steel-cut or rolled, part of a fast-cook microwave pack, or completely natural.
How come the porridge becomes more liquidy when whey is stirred into it?

Seasoned Advice SE often plays with disparate metallurgy and rheology stuff.

I suspect that the thickness of the porridge is in fact, due to it being a non-Newtonian, shear thickening fluid. When the protein powder is mixed, perhaps the mixture becomes less non-Newtonian. That's just my guess though.

@ShoubhikRMaiti according to this podcast I just found, porridge is shear-thinning

I found that whey proteins are used as thickeners, at least in yogurt.

@Alchimista that's why this behaviour in porridge is very surprising

@theonlygusti you have a complicated matrix, so one has to identify and study really in details. What is thickening a fluid yogurt can reduce the viscosity of a thick porridge. In the latter case it suffices, but I am inventing, that the proteins in the powder are smaller than those of milk. Voilà, now the ibterchain interactions are between long and shorter chains, disrupting gels or anyway acting as "lubricants". It is just a possibile mechanism. Or there can be more cysteine, etc.

Does it contain lecithin as an emulsifier? https://www.renegadedad.net/why-lecithin-is-used-in-whey-protein-powder Also, if hydrolyzed, that also could include chemicals breaking down the starch.


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Does increasing one of the electrolytes in a galvanic cell increase the voltage?
I would like to know how you could increase the voltage in zinc and copper galvanic cell:
$$\ce{Zn(s) + Cu^2+ → Zn^2+ + Cu(s)}$$
I know that temperature and surface area is a factor that affects the voltage, but I would like to specifically know about how concentration changes the voltage.
If I want to increase the voltage, which electrolyte do I change? The anode or cathode? And what's the reason for that?

Surface area does not affect the voltage of a cell. Temperature does affect the voltage but not much. The main effect is the concentration, according to the Nernst law. Apparently you do not know this law. Or do you ? In two words, in the zinc/copper cell, you may increase the cell voltage by increasing the copper concentration $\ce{[Cu^{2+}]}$ or decreasing the zinc concentration $\ce{[Zn^{2+}]}$

It looks like you ask for explanation of electrochemistry basics.  For fundaments, studying is better than asking. The latter is better for just filling the gaps.
See electrochemistry basics on Hyperphysics, Libretexts and  Wikipedia

As suggested in the comments, the key is Nernst equation. The fundation is the Gibbs free energy given by:
$\Delta G = \Delta G^0 + RT \ln Q_r$
where $\Delta G^0$ is the standard free energy for the process, while $Q_r$ is the reaction ratio. Remember that the reaction ratio is the analogous quantity of the equilibrium constant, but with the non equilibrium concentration. Remember also that the Gibbs free energy is proportional to the voltage. Then you need to increase the reaction ratio.

If I want to increase the voltage, which electrolyte do I change?

Hope you are familiar with the term electrolyte which is the solution where ions flow in the electrochemical cell

The electrolyte solutions used here are $\ce{ZnSO4(aq)}$ solution and $\ce{CuSO4(aq)}$ solution

The anode or cathode?

Yes you can chose both or one of them
Now lets just consider the anode
The half cell reaction is,
$$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$
But more correctly it is,
$$\ce{Zn(s) + 6H2O(l) -> [Zn(H_{2}O)_6]^{2+}(aq) + 2e^-}$$
This reaction happens in acidic medium or may be neutral medium but since Zn is a amphoteric metal it likes to get dissolved in basic medium rather than acidic medium.
The new reaction,
$$\ce{Zn(s) + 4OH^-(aq) -> [Zn(OH)_4]^2-(aq) + 2e^- }$$
Standard reduction potentials for acidic and basic status are $\ce{E^° -0.762 V, -1.190 V }$ receptivity
Thus for the anode try changing the medium to basic by adding $\ce{NaOH}$ in excess
But for the cathode there is not much solution easy as above but you can try adding $\ce{NH3}$ or $\ce{CH3COOH}$ maybe cream of tartar $\ce{KC4H5O6}$ in excess these will make complex ions in cathode side of $\ce{Cu^2+}$ ions which will either increase or decrease the half cell value for anode. But in order to have a higher voltage of the final cell you need a lower half cell reduction potential for cathode and for the anode it is higher cell potential value. (I have given only few most of them are house hold chemicals you can test)
Here Check this link it has different $\ce{Cu}$ complexes and reduction potentials!
$\ce{[Cu(dmphen)_{2}]^2+}$ looks good with $\ce{E 0.590 V}$ default is $\ce{0.339 V}$ but not verified.
I suggest you to look for nerst equation for understanding how to increase the voltage  other than changing electrolyte (except concentration) it is defined for only for one reaction equation. Look for standard reduction potential table for changing the metal.


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Does cinnabar react with iodine?
Does cinnabar react with bromine and iodine to form elemental sulphur? This reaction was described in "Explorations in the history of science and technology in China" but I have doubts...

"…reports$^{22,23}$" — you sure have checked these reports, haven't you? Also, what is "Explorations in the history of science and technology in China", an article, a book, a report or a blogpost? How are we supposed to check these references?

I did not verify the possibility of the above mentioned reaction experimentally.

Explorations in the History of Science and Technology in China . A Special Number of the " Collections of Essays on Chinese Literature and History . ” Shanghai : Chinese Classics Publications House, 1982.

I did a cursory search and found some information particularly from books of late 19th century(1,2):
Cinnabar and iodine

On heating, it (cinnabar) is readily decomposed with formation of small scarlet crystals of mercuric iodide.

Cinnabar and bromine

[...] After digesting the cinnabar with excess of bromine liquid (or bromine water containing 3% of bromine or solution of hydrochloric acid containing 13% of bromine) for a few days, the whole is dissolved, and also any metallic mercury present [...] The amount of bromine required is very considerable, 1 part of mercury is obtained and 3.5 parts of hydrobromic acid remains in the solution.

References

Chemical News and Journal of Industrial Science, Volumes 37-38, Chemical news office., 1878
Dictionary of Arts, Manufactures and Mines by Robert Hunt, Andrew Ure, 1878


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Do the cathodes of all batteries increase in mass as the battery is used?
I understand the mechanism behind the cathode's increasing mass in many batteries. As the cathode receives electrons, ions are reduced to form atoms. In short, more electrons equals more mass in the cathode. what I was wondering is whether or not there are batteries where the cathode does not increase in mass for whatever reason. Are there batteries where electrons, or some amount of mass is able to return to the anode? Thanks to anyone who responds to this question!

Electrodes of vanadium energy storage cell batteries have constant mass.VO2^+/VO^2+// V^3+/V^2+

Fuel cells? https://en.wikipedia.org/wiki/Fuel_cell

The original zinc batteries have cathodes made of $\ce{MnO2}$ in contact of a solution containing $\ce{NH4Cl}$. When getting electrons from the zinc anode, the following reaction happens :$$\ce{2MnO2(s) + 2NH_4^+ + 2 e- -> Mn2O3(s)  +  H2O + 2 NH3}$$ The reaction produces $\ce{H2O + NH3}$ which both pass into solution. The cathode looses one oxygen atom when 2 electrons are coming. As $\ce{MnO2}$ molar weight is $\pu{87 g/mol}$, the cathode becomes lighter by about $\frac{16}{2·87} = 0.092 = 9.2$ %


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What is the anodic reaction of aluminum in water
Standard electrode potentials are reported for materials that are not usable as actual electrodes in water because they react spontaneously and rapidly. https://www.engineeringtoolbox.com/electrode-potential-d_482.html
Some metals, like zinc, are useful as cell electrodes if pure, but if not so pure, exhibit cathodic areas that cause the metal to corrode and produce hydrogen without delivering useful external current. One way to counteract that was to coat the surface with a tiny bit of mercury, which apparently coated the cathodic sites but permitted the zinc to migrate thru as if the mercury were not there.
Other metals, like aluminum, develop a film which inhibits corrosion by developing a resistance to atom/ion movement. It has been suggested to me that the actual electrode reaction on aluminum in water is production of hydrogen, so that the half-cell using an aluminum electrode in water is the hydrogen electrode, because bare aluminum reacts immediately with water. This would be analogous to the zinc anode with surface impurities (cathodic sites), except that we do not consider zinc to be producing a hydrogen electrode, just corroding.
it seems that there is considerable usage of metals like zinc, aluminum and magnesium for cathodic protection under conditions where no hydrogen is expected, so it is reasonable to visualize the protection to be developed by reaction of the metal, possibly thru a partially protective film which is equivalent to a cell resistance rather than changing the chemistry to a hydrogen electrode. https://performancemetals.com/pages/sacrificial-anodes-faqs
Is there any evidence that metals like aluminum (or zinc or any other metals) could support a hydrogen electrode like platinum or palladium do? The standard half-cell with Pt or Pd is in 1 M HCl, which would not be appropriate for Al, Zn or Mg. Would any other conditions be better?

Aluminum cannot be used easily as support for a hydrogen electrode, because it is always covered by a thin, continuous and insulating layer of aluminum oxide. As this layer is an electrical insulator, its resistance is high, and ions like $\ce{H+}$ have some difficulty in crossing it. So the measured potential of hydrogen measured on such a protected electrode is not a good measurement of the redox hydrogen potential. It corresponds to an important overpotential, which is to be avoided if you want to measure or determine a redox potential.


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How can I find out how much caffeine is in my decaf tea and coffee?
I know that the decaffeination process is not perfect and that it leaves some caffeine. I've looked online and there was a company in 2009 that offered caffeine test strips but they seem to be no longer offered.
Another approach would be to use a GCMS, but used GCMS devices are expensive and it takes alot of effort to keep them in calibration (at least it did when I was in college).
Are there services where I can send a sample off and they will tell me how much caffeine they contain? Failing that, how would I reasonably test how much caffeine is in my tea and coffee?

Why do you want to do that?

Yes, there are many labs that do food testing, and include caffeine testing among their services.  But it's expensive; the pricing is designed for businesses rather than consumers.  E.g., https://www.mydrinkbeverages.com/services/new-beverage-development/laboratory-services does caffeine testing by LC/UV, but it costs $150 per sample (search this document for "caffeine": https://4557656.app.netsuite.com/core/media/media.nl?id=1271913&c=4557656&h=df3d7a19fb96962fe6e7&_xt=.pdf )

Why do it yourself when others have already done the analysis? As this report shows.

I want to do it for myself because every cup is different. And every batch of tea is different. And I want to know what’s going on with my food, not other food.

I originally thought that UV/vis on tea samples (the liquid from contacting tea with hot water) would not be a good method as the tea is a complex mixture. I think that with a lot of clean up that it would be possible to make the measurement. I know that HPLC with a UV detector would be the best method of making the measurement.
But after reading the work of a famous tea and coffee chemist, I have come to the view that direct measurement of the brewed tea would work using measurement at 273 nm.
There was an electrochemist at Imperial College named Michael Spiro, he did a lot of work on tea and coffee. One of the things he did was to model and measure the brewing process which occurs when you make a cup of tea.
See for example
Kinetics and equilibria of tea infusion, Part 15. Transport of caffeine across a teabag membrane in a modified rotating diffusion cell, Michael Spiro and Deogratius Jaganyi, Food Chemistry, Volume 69, Issue 2, 1 May 2000, Pages 119-124
I would not want to use GCMS for caffine as the caffine is so very polar.

Thanks. I wasn’t aware that GCMS wouldn’t work for this.

Also a great reference!

You may try to extract the caffeine with DCM and determine its concentration by spectrophotometry.

So the idea would be to get a used UV spectrometer?

I would suggest so. See https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4641934/. The spectrum of caffeine has a peak of absorbance at 274 nm (UV).


text:
Betaine HCl solution at neutral pH
I was trying to make a $2\text{M}$ solution at $\text{pH}\, 7.5$ of Betaine $\ce {HCl}$.
I started by weighing out the appropriate amount of Betaine $\ce {HCl}$, added some water and started stirring. At this point, the $\text{pH}$ of the solution was $\sim1-2$.
I went on to titrate the solution with $\ce {NaOH}$.
At about $\text{pH}$ $3-4$ the betaine $\ce {HCl}$ was completely dissolved.
as I added more base, I noticed that the solution began to warm up and give off a some sort of a "plasticy" smell and foamed a little bit. This began to happen at around $\text{pH}$ $5.5-5.7$.
It seems to me that the Betaine $\ce {HCl}$ is somehow reacting at high $\text{pH}$.
I was wondering if someone can explain what may be happening to the Betaine $\ce {HCl}$.

Trimethylamine is quite a good leaving group

@Waylander That has nothing to do here. Helloo, this is hydrochloride!


text:
Calcium hydroxide questions (as it relates to corn and nixtamalization)
Please pardon my chemistry ignorance! I actually have done a lot research on these specific topics, but can't find anything definitive--anywhere!

When adding calcium hydroxide to water and corn for nixtamalization, does the heat of the water have any effect on the "activation" of the calcium hydroxide (specific for this process)? How so?

1a. Specifically, what causes the change in water color (from clear to yellow)?

Does the ratio of calcium hydroxide to water have any effect on the process, or only calcium hydroxide to volume of material (i.e. corn)? Can you change the pH levels in water alone, with the more or less cal you add to the water? This may be obvious, but not to a non-chemist.

Since the solubility of calcium hydroxide decreases with increasing temperature, how does this affect the nixtamalization process?

I am by no means an expert in food processing, so just some general chemistry point of view:

Higher water temperature increases speed of all chemical reactions. The typical increase is in the range 2-4 times by every $\pu{10^{\circ}C}.$. The reason behind it is that molecules at higher temperature move in average faster, have higher energy and more easily overcome the energetic barrier for the reaction to happen.

The corn seeds are yellow. The process partially dissolves surface layers and the dye dissolves partially in water. Additionally, the color can be partially caused by oxidation of dissolved organic matter.

There must be enough calcium hydroxide to form saturated solution, and to keep saturated solution during processing. This makes pH more or less constant, with extra hydroxide not having effect on this. BUT - with too low solution to corn ratio, solution can be partials exhausted, with additional calcium hydroxide eventually not readily supplied by dissolving.

Effect of decreased solubility is much weaker than effect of increased reaction rate at higher temperature.


text:
Number of π electrons in all trans-2,4,6-octatrienoic acid

Sketch the Lewis structure of all trans-2,4,6-octatrienoic acid.
(a) Consider the delocalized π electrons. Employ the particle-in-a-box model. Drawing on your knowledge about the occupation of orbitals in atoms what is the highest occupied state? Sketch the occupation of the energy levels with electrons using ↑ and ↓.

I wondered if the electron lone pairs on the oxygen atoms also counted as π electrons since resonance structures exist where these lone pairs can be delocalised over the molecule.

If so, how would this correspond to using them to find the length of the conjugated system because wouldn’t that give extra length where there was no double bond?

No, discard them. Also, are you sure that such resonance structures exist?

Which oxygen do you mean? The carbonyl oxygen (=O) lone pairs are orthogonal and cannot participate in the resonance. The carboxyl oxygen (-OH) can participate in resonance to some extent, but since the question asks about particle in a box model, you can't really consider that.

In UV-Vis spectroscopy, these two electron pairs on carbonyl-O and alcohol-O are named n electron pairs because these are not shared with the extended $\pi$ electron system.


text:
Find gas pressures of injection knowing mixture pressure without knowing volume
Let's say we have a gas 1 moving inside a pipe (we don't know the geometry) at mass flow rate $Q_1$ and temperature $T_1.$ We inject a gas 2 at some point inside the pipe with gas 1 at mass flow rate $Q_2$ and at temperature $T_2.$
We know that the resulting mixture is moving at the mass flow rate $Q_\mathrm{mix} = Q_1 + Q_2$ at the temperature $T_\mathrm{mix}$ and is at pressure $p_\mathrm{mix}.$ All gases are ideal gases.
Is it possible to find the gas 1 pressure $p_1$ and the gas 2 pressure $p_2?$ And if so, what is the relation?
So far I haven't found any clues to solve this problem.
I thought about partial pressure, but I feel wrong to say that $p_1$ and $p_2$ are respectively equal the partial pressures of gas 1 and gas 2 in the gas mixture because there is no mention of volume in this problem.

Can parameter $\ce{M_w}$ molecular weight of the gas used to answer this question?

@Avon97 Yes it can be used. But I don't see where I can use molecular weight of each gases to answer the question

Of course it is a pipe and it has a infinite volume hypotactically,
To use ideal gas equations is a bit tricky with mass flow rate need to convert to molar flow rate of each gas thus need to divide by $M$ molecular weight respective to each gas.
Consider $1$ time unit where inside of pipe gets
$$\frac{Q_1}{M_1}$$
$$\frac{Q_2}{M_2}$$
moles of gas to one volume section of pipe
At one time unit according to the ideal gas equation
$$pV=nRT$$
Gas 1 will consume $V_1$ amount of volume
$$V_1=\frac{Q_1RT_1}{M_1p_1}$$
Gas 2 will consume $V_2$ amount of volume
$$V_2=\frac{Q_2RT_2}{M_2p_2}$$
Total of $V$ volume per each time unit
$$V=V_1+V_2$$
But $V$ is just a section of the pipe it just shifts to whatever direction the gas is flowing and keeps going forever in the infinite pipe.
But as further the gases go away just like you said turbulence will stabilize and final temperature and pressure inside pipe will be almost equal to $T_\text{mix}$ and $p_\text{mix}$
When it is stabilized,
Gas 1 will consume $V'_1$ amount of volume
$$V'_1=\frac{Q_1RT_\text{mix}}{M_1p'_1}$$
Gas 2 will consume $V'_2$ amount of volume
$$V'_2=\frac{Q_2RT_\text{mix}}{M_2p'_2}$$
Total of $V'$ volume per each time unit
$$V'=V'_1+V'_2$$
The important factor here is that number moles of gas will shift to next stage will be the same thus molar fraction for each gas is a constant. Think as $V'$ stage in pipe.
Molar fraction can be easily calculated
$$X_1 = \frac{\left(\frac{Q_1}{M_1}\right)}{\left(\frac{Q_1}{M_1}\right)+\left(\frac{Q_2}{M_2}\right)}$$
$$X_2=\frac{\left(\frac{Q_2}{M_2}\right)}{\left(\frac{Q_1}{M_1}\right)+\left(\frac{Q_2}{M_2}\right)}$$
Using the partial pressure equation
$$p_\mathrm A=X_\mathrm Ap_\mathrm{tot}$$

what is the relation?

Refer here!
$$p'_1=\frac{\left(\frac{Q_1}{M_1}\right)}{\left(\frac{Q_1}{M_1}\right)+\left(\frac{Q_2}{M_2}\right)}p_\text{mix}$$
$$p'_2=\frac{\left(\frac{Q_2}{M_2}\right)}{\left(\frac{Q_1}{M_1}\right)+\left(\frac{Q_2}{M_2}\right)}p_\text{mix}$$
There you have it final answer but consider that its one time unit thus take only the magnitude with respective unit of mass flow rate * unit time $\ce{|Q1| |Q2|}$ symbol ` is there to show gases are stabilized.

Thank you. So does it mean that $p_1$ and $p_2$ I am looking for are equal to $p'_1$ and to $p'_2$ (or at least very close) ?

Well that depends on $\ce{T_{1} and T_{2}}$ difference the closer the values are faster the partial pressure of gas gets to $\ce{p_{1} p_{2}}$ you mean stable value yes

So for instance, if $T_1 =T_2$, then we can say  $p_1=p'_1$ and $p_2 = p'_2$ . But when $T_1 >> T_2$ it is impossible to find $p_1$ and $p_2$. But what if we consider both $X_2 << X_1$ and $T_1>> T_2$ ? Can we say in that specific case that $p_1=p'_1$ and $p_2 = p'_2$ ?

Yes if you say $\ce{ X_{2} \ll X_{1}}$ there is hardly any gas 2 in the pipe so yes $\ce{p1 = p1^' and p2 = p2^'} $ also $\ce{T_1 \approx T_{mix} p_1 \approx p_{mix}}$


text:
do electrons in different orbitasl in a same subshell have same energy
In valence bond theory, I suppose that electrons in say, 2px and 2py have the same energy because of the same structure of orbitals. But what about 3dxy and 3dz^2? Do they also have the same energy? I would say, no for the difference in their shape. But I am not quite sure since they are in the same subshell. could anyone plz clarify?

Yes they also have the same energy because of the same structure of orbitals. Their apparently different shape is not a fact of any consequence.

But then how would one describe the colour of transition metals? Since I have learnt that they are colourful because of the energy emitted while electrons jump in between 3dxy and 3dz^2

That's a different story altogether. They have the same energy in a lone atom, but not necessarily so in various compounds.

In the isolated atom, yes, they will have the same energy. Dealing with d orbitals, you are in the field of transition metal chemistry. As correctly said by Buck Thorn, the degeneracy can be removed by external electromagnetic field. In the case of transition metals, it can be extensively seen in the crystal field theory. The ligands act as a source of electric field, then the spherical symmetry of the metal center is removed. Therefore an electronic configuration like d$^2$ will split into terms, which are a collection of microstates. You have to see in advanced inorganic chemistry textbooks.

Degeneracy (=equal energy) may be broken by external electric or magnetic fields.
For instance, in NMR the application of an external magnetic field can break the degeneracy of possible nuclear spin states (in $\ce{^1H}$, antiparallel $\alpha$ and $\beta$). This is known as the Zeeman effect and its influence on atomic (electronic) spectra lead to the postulation of angular momentum quantum numbers by Arnold Sommerfeld (for which he should have received a Nobel Prize) and to the discovery of quantum spin.


text:
Molar conductivity of a strong electrolyte
In 1910, Kohlrausch showed that the dependence of the molar conductivity of strong electrolytes is well described by the following equation:
$$\Lambda_m = \Lambda_m^\circ-K\sqrt{c}$$
This equation, known as Kohlrausch's law, states that the molar conductivity of strong electrolytes decreases with the square root of the concentration, with slope given by the quantity K (Kohlrausch's parameter). It is also found that this slope of the stoichiometry of the electrolyte.
I need to find for a paper the slope value for magnesium chloride, MgCl2, and the molar conductivity for different concentrations. But, I can't find this experimental data anywhere. I hope you can help me.


text:
Do water filters expire even if underused?
This filter rates a 3 month lifespan with 1210 liter capacity, but I expect to filter only 300 liters in 3 months. Is it still expired after 3 months - if so, why? What degrades a filter while it's unused?

What does "expire" mean for you.

@Karl Unsuitable for consumption (health hazard etc).

Many things are aging even if not used, like lithium ion power cells.
There is filter aging independent of the passed water volume. There is eventual growth of chemical and microbial stuff. E.g. chemical sediments on the active surface of antimicrobial silver causes deterioration of antimicrobial effect. Active carbon for organic traces and ionex resin of ion exchange can be affected by this as well.
The life span with 2 criteria is very generally –  not limited to filters nor chemistry – always meant  "whichever one is reached sooner".
In a funny analogy, 1.5 L of bottled water have criteria e.g. ( illustratively ) 2 days after opening, or 1.5 L of provided water, whatever comes first. If you drink just 500 mL, the left 1 L does not have an eternal lifespan either but 2 days are still valid.

Good answer and I've upvoted. Tough reasonable guess or measurements could be made case by case.

So filters do have post-opening expirable components - sense enough. A trick would be to use up the entire filter and store results somewhere... in a cubic meter container.


text:
Will changing the salt in the salt bridge change the voltage produced in a galvanic cell?
I have used zinc sulfate and copper sulfate as the solutions, and used zinc and copper electrodes.
For salt bridge I took lithium, potassium, calcium, sodium and zinc nitrates. The results are as below:
$$
\begin{array}{lr}
\hline
\text{Compound} & E_\mathrm{cell}/\pu{V} \\
\hline
\ce{LiNO3}    & 1.081 \\
\ce{NaNO3}    & 1.057 \\
\ce{KNO3}     & 1.071 \\
\ce{Ca(NO3)2} & 1.069 \\
\ce{Zn(NO3)2} & 1.071 \\
\hline
\end{array}
$$
As it shows, there is very insignificant difference. Is the results supposed to be like this or have I done something wrong in my experiment or something so that the voltage was constant?
If I have done something wrong, then should I have changed the anions instead of the cations? Or would it have been a mechanical problem?

The voltage should not depend on the composition of the salt bridge. What were the concentrations of your solutions ? Were your measurements constant and stable in time ? Or were they slightly and slowly moving ?

@Maurice I took the same concentrations for all the solutions, and I made sure that my measurements were stable most of the times. Also, I have done many other trials as well, which all look like this.

How did you make your salt bridge ? With agar ? With another gel ? With a membrane ?

@Maurice my teacher told me to use filter paper. It wasn't like there was no current.

Well ! With a filter paper, you cannot prevent the anode and cathode solutions from intermixing. So the measured voltages are not reliable, because some copper ions can get into the zinc compartment. This will short the circuit, as the reaction $\ce{Zn + Cu^{2+}->Cu + Zn^{2+}}$ happens in the anode compartment and will not produce any current. A salt bridge must be made in such a way that only negative ions from the copper side can cross it, and no ions in the opposite direction.


text:
What is the difference between atom and nucleus?
Atomic nucleus-The atomic nucleus is the small dense region consisting of protons and neutrons at the center of an atom discovered in 1911 by EARNEST RUTHERFORD. Atoms are made up of a positively charged nucleus surrounded by a cloud of negatively charged atoms. Nuclei are very dense and extremely small. They contain more than 999/ of the mass of an atom and are 10000 times smaller than an atom.
My question is what is the role of the nucleus in an atom? Does the nucleus of an atom move?

What is the difference between atom and nucleus? Electrons.  what is the role of the nucleus in an atom? Without nucleus, there would be nothing left to kep electrons together. Does the nucleus of an atom move? How would you breathe if they did not ?

A nucleus has the following roles in an atom.

It contains almost all the mass of the atom

It is positively charged and hence keep the electrons in orbit via electromagnetic force (attractive)

A nucleus contains protons which determine which element the atom is of

That's all. And, no, it is safe to assume that nuclei don't move within the atom, they are stationary, mostly for their large mass. But if the atom itself goes somewhere, so does the nucleus. Quite like you sitting on a bus, which can move but you will not get up from your seat

I was thinking of it with the atom itself being the reference frame. A nucleus surely doesn't go anywhere within the atom. Anyways, I will point the ref frame out


text:
Why are shells softening in the ocean?
I learnt about how ocean acidification can be described in terms of equilibrium. This is summarized in the textbook diagram below:

However, this diagram doesn't explain to me why the shells should be softening. As more CO2 is put into the atmosphere, this will increase $\ce{CO2_{(aq)}}$. By the second equation this should create more H+ and HCO3-.
People say that the increase in H+ concentration will eat up the CO3 ions in solution and thus the loss of CO3 will shift the dissolution equilibrium right and make shells soft. However, although there is an increase in H+ which eats up the CO3, there is also an increase in the amount of HCO3 so shouldn't that prevent a shift in equilibrium?

If we play a little with all the equilibrium systems shown in the diagram we can get:
$$
\require{cancel}
\begin{align}
\ce{\cancel{\ce{CO2(aq)}} &<=> CO2(g)} \tag{R1} \\
\ce{HCO3^-(aq) + \cancel{\ce{H^+(aq)}} &<=> \cancel{\ce{CO2(aq)}} + H2O(l)} \tag{R2} \\
\ce{HCO3-(aq) &<=> \cancel{\ce{CO3^2-(aq)}} + \cancel{\ce{H^+(aq)}}} \tag{R3} \\
\ce{Ca^2+(aq) + \cancel{\ce{CO3^2-(aq)}} &<=> CaCO3(s)} \tag{R4} \\
\hline
\ce{Ca^{2+}(aq) + 2 HCO3^-(aq) &<=> CaCO3(s) + CO2(g) + H2O(l)} \tag{R5}
\end{align}
$$
From this overall reaction, we can see that increasing the concentration of $\ce{CO2}$ will result in an equilibrium shift to the left according to Le Chatelier, which translates to the decomposition of $\ce{CaCO3}$.
Another explanation is that the solubility of $\ce{CaCO3}$ increases with decreasing pH, and vice-versa. In other words, if the ocean becomes acidic enough, more $\ce{CaCO3}$ will dissolve in the water.


text:
Why does denaturation of albumen protein cause pH of solution to increase?
I'm interested in the denaturation of proteins by alcohols, specifically by ethanol.
I have devised a simple experiment in which I inject solutions of ethanol (of variable concentrations) into solutions of egg albumen (containing proteins such as conalbumin). I have observed a marked pH change in solution after injection of ethanol. The rate of change of pH is in fact proportional to the concentration of ethanol injected (as expected).
However, I am having trouble explaining the theory behind why ethanol's disruption of tertiary protein structure increases pH. As I understand, ethanol disrupts proteins such as conalbumin by disrupting hydrogen bonds in the tertiary structure of the molecule.
But why does this cause the pH to increase (in effect, the concentration of H+ ions to decrease)?

Just to be sure, is the pH change compared to pH change of a blank sample without albumine ? Is pH change caused by ethanol addition eliminated ?

@Poutnik I’m a chemistry student, so forgive me if I make mistakes with nomenclature and such… but I basically have a pretty-made solution of albumen protein into which I inject ethanol. pH is measured first of the albumen without ethanol and then the change is measured when ethanol is injected.

Be aware the ethanol presence itself affects the measured pH value even without albumine. This effect has to be eliminated. I would try simulation with  some aminoacid with comparable mass concentration as albumine, with the same initial pH.

Yes, you misunderstood it - change in pH is a result of that you have a different solvent - any pH in water-ethanol mixture is different then in pure water. I'm also disappointed that answerers didn't consider it.

@Mithoron Good point:https://pubs.acs.org/doi/10.1021/j100803a022 Well, that's why the site exists I suppose, to aggregate and pass judgement over answers :).

@Poutnik This is an excellent point which I completely neglected. However, the below answers clearly indicate that there exists a mechanism which raises pH. How would I go about quantifying the relative weights of that mechanism vs. the flaw that you pointed out?

Aside of mentioned aminoacid simulation, what about alternative ways of denaturation ? Like heating ?

The main component of egg white, ovalbumin, contains 93 negatively charged residues (Asp + Glu) and 68 positively charged residues (Arg + Lys), as well as 14 His. If you make the solvent less polar by adding ethanol to water, they will tend to form the neutral species (Arg+ and Lys+ will lose a proton, and Glu- and Asp- will pick up a proton).
If you had the protein unfolded already, you would expect that the solution gets more basic if the effect on all charged amino acids is the same because there are more negatively charged ones.
To test whether the observed effect has to do with denaturation of the proteins or with changes in protonation state of side chains already exposed to water, you could measure the pH of amino acids (preferably with a modification of the main chain carboxylic acid and amino functions so only the side chains show acid/base reactions) at different ethanol concentrations.

Most proteins are polyelectrolytes because the constituent aminoacids may have ionizable groups, typically carboxylic and basic (imino, guanidino) groups on the sidechains, in addition to the terminating amino and carboxylic groups (when not modified). The degree of ionization depends on the fold of the protein and the pH of the solution. Denaturation of a protein can expose to the solvent buried sidechains with ionizable groups, resulting in acid/base reactions with water.
If the solution is buffered and the protein is not present at high concentrations, then the degree of ionization of the protein will depend on the solution pH, but the solution pH will not be particularly sensitive to the presence of the protein. On the other hand, if there is a lot of protein in the solution and/or the solution is not buffered so as to resist changes in the degree of ionization of the protein, then such changes will translate in changes in pH. Concretely, changes in secondary or higher structure of the protein will translate into changes in pH, as it appears you observe.
Unfortunately (as noted in the comments), the measured pH of the sample may vary strongly with the concentration of ethanol even in the absence of protein and in the presence of buffer. This is in addition to the effect which the organic solvent will have on the dissociation constant of ionizable groups on the protein noted by Karsten Theis. It is quite likely that those effects will exceed in importance that of denaturation, since charged groups are most often exposed on the protein surface. The extent of exposure and the dissociation constant may increase with denaturation, of course, and disentangling the different effects is not simple, requiring accurate measurements.
To start with you might check the extent to which the addition of alcohol may be responsible for the increase in pH. A lot of work has already been done in measuring such effects, and the following table compiles some results reported in the IUPAC Compendium on Analytical Chemistry (Ref. 1) for the  0.05  m  potassium  hydrogen  phthalate  (KHPh)  buffer.
Other posts on this site address measurement of pH in organic media.
$\begin{array}{c|c} \hline
\text{mole fraction ethanol} \\
\hline  \begin{array}{c|c|c|c|c}
    T (°C)      &0.0416     & 0.0891        & 0.2068 & 0.4771  \\ \hline
   -5.0000  &  4.2660 &   4.5700 &   5.1120  &  5.5270 \\
    0.0000      &   4.2490 &   4.5440 &   5.0760  &  5.5000\\
  10.0000   &   4.2350   & 4.5130 &   5.0260  &  5.4690\\
   25.0000  &   4.2360  &  4.5080 &   4.9760  &  5.4720\\
 40.0000   &    4.2600 &   4.5340  &  4.9780 &   5.4930 \\\hline
\end{array}\end{array}$
Reference

Web edition of the Compendium on Analytical Nomenclature (The Orange Book) (2002). https://media.iupac.org/publications/analytical_compendium/

Thank you so much for your input. I hugely appreciate the recent edit, in which you added methodology for mitigating ethanol's own effect. There are little (if any?) straightforward resources on the net that answer this question, so this page will surely come in handy for many people. I am always surprised at the sheer quality of answers on the Stack Exchange network :)


text:
How to clean out a contaminated aluminum automotive radiator
I have a very expensive custom made automotive aluminum radiator that has had a mouse or mice nesting inside the top tank.  I have removed most of the nesting material through the top inlet but I want to clear the radiator tubes that may be plugged up with the nesting material (looks like cotton of some sort) and be sure the radiator is flowing water through it as it should. As it is an aluminum radiator I am told they cannot remove the tank and "rod out" the radiator tubes.  The only solution offered by the company is replacement.  The radiator cost about $1200.
I have thought about an enzymatic drain cleaner but am not sure that it will not harm the aluminum radiator internally.  What ideas do you have?

I think an enzymatic cleaner should be OK. DO NOT USE DRAIN CLEANER or anything based on hydroxides.

Well I actually tried a small jar of the Enzymatic product with a sample of the nesting material I already removed as well as small piece of .023 6061 aluminum... Overnight the aluminum was completely decomposed.  The nesting material seemed unfazed.  So much for that idea!


text:
pKb of a strong acid?
What does the pKb (or $K_b$) of a strong acid mean?
For instance this link reports the Kb of HI, an extremely strong acid. Are they reporting the reaction constant of the reaction:
$HI + H_2O \implies  H_2I^+  + OH^- $
Is the $K_b$ for the equation?
That reaction seems so bizarre; it seems like the products are so unstable and reactive.

No, it's $K_\mathrm{b}$ of the conjugate base. The table header tells you this, although I guess it could be a bit more explicit: "Values of Ka, pKa, Kb, and pKb for Selected Acids HA  and Their Conjugate Bases (A−)".

https://chemistry.stackexchange.com/questions/24342/what-is-the-ka-of-oh-and-kb-of-h3o

Ah, I see. And the stronger the acid, the weaker the conjugate base. In this case, the conjugate base is $I^-$; however, it seems like it should have a good ability to accept a proton? Therefore, wouldn't it be a strong base?


text:
Catalyst and Inhibitors home experiments
My teacher assigned me a science experience involving catalysts and inhibitors at home.
I think I understand the principle of it, but I fail to find possible experiments that I can do at home with a thermostat and some home substances.
The experience:
3 reactions...
1 with the catalyst,
1 with the inhibitor,
1 without anything.
I will compare them by measuring the reaction speed with a factor (preferably the heat variation).
EDIT: I can also measure the volume of the product.
Thank you.

You might look up the decomposition of hydrogen peroxide, $\ce{H2O2}$. The 3% solution sold as antiseptic should be safe to use at home, as would be some readily available catalysts and inhibitors.
Horseradish might be used for a catalyst... but don't eat it in the lab.
To measure speed of reaction, measure gas evolution with an inverted, liquid-filled graduated cylinder or small measuring cup over time.
You can have fun burning things (carefully) in the evolved $\ce{O2}$, too.

Thank you, I am not really good in chemistry, but what I understand is that I need mix the solution with some horseradish powder so that the decomposition of H2O2 happens faster? In this experience, what will the inhibitor be? I don't think I will burn something in the O2; I will breath it :-)

Actually, fresh horseradish probably has more peroxidase than powdered. But there are many other catalysts you can use. Use the 'net, young master... there is a plethora of info on peroxide catalysts and inhibitors.


text:
Selectivity in Thermodynamically-Driven Silyl Enol Ether Formation, and Optimistic Avoidance of Protecting Group Chemistry
I am aware of the regioselectivity of silyl enol ether formation at an unsymmetrical ketone under 'thermodynamic conditions'.
Does this selectivity extend to systems containing more than one ketone group?
In other words, if some starting material were to contain two ketone groups, under 'thermodynamic conditions' with one equivalent of TMSCl, would the overall most substituted double bond product be formed in majority?
Context:
I am particularly curious about the specific example below as it would form part of a solution to a synthesis problem I've been working on. I fear that here, assuming the answer to the above is yes, the somewhat conjugation-supported product might be formed between both carbonyls instead? If this is the case I would protect the lower ketone as a ketal which wouldn't be unhelpful as my next intended step is a Mukaiyama-Michael, which might pose another selectivity issue due to possible competition with Mukaiyama-Aldol (?). Naturally, I just want to avoid using protecting groups.
I apologise for the crude appearance of my ChemDraw-ings.

My intuition is that you will get the lower product as the most acidic proton is on the methylene between the carbonyls, and none of the upper product. There will be little selectivity in which of the oxygens gets silylated.

@Waylander Thanks! Do you think it's possible to protect the lower ketone selectively given it's a little less hindered, or is that unlikely to be successful?

Possible. These papers offer examples  https://cdnsciencepub.com/doi/pdf/10.1139/v83-245     https://www.researchgate.net/publication/216711546_Solventless_Rapid_Synthesis_of_3-Alkoxy-2-cyclohexen-1-ones_from_1_3-cyclohexanedione_Promoted_by_Indium_III_ChlorideSilica_Gel

That's great, thanks very much for your help.


text:
IUPAC Nomenclature using Iso alkoxy groups
Is the name of the compound    4-isopropoxy-pentan-2-one or is it 4-(1-methylethoxy)pentan-2-one... Are both the correct "systematic" names or am I messing up:'.

Note that the prefix names ‘1-methylethyl’ or ‘isopropyl’ may be used in general nomenclature; the preferred prefix, however, is propan-2-yl.
The relevant rule for the systematic names of ethers in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

63.2.2.1 Systematic names
P-63.2.2.1.1 Substituent prefix names for $\ce{R'-O-{}}$ groups are formed by concatenation, i.e., by adding the prefix ‘oxy’ to the substituent prefix name for the group $\ce{R'}$. These compound prefixes require the numerical multiplying prefixes ‘bis’, ‘tris’, etc.

However, a problem with this rule is that many examples in the Blue Book were wrong and had to be corrected because they ignored the rules for parentheses. Note that parentheses are used around compound prefixes according to Rule P-16.5.1.1 and around simple substituent prefixes to separate locants of the same type referring to different structural elements according to Rule P-16.5.1.2.
Therefore, the systematic name for the compound that is given in the question is
4-[(propan-2-yl)oxy]pentan-2-one
since the simple substituent prefix ‘propan-2-yl’ includes a locant that does not refer to the same structural element as the initial locant ‘4-’ and ‘(propan-2-yl)oxy’ is a compound prefix.


text:
Trans/Cis Alkene with mCPBA products?
What is the different in stereochemistry between the product of cis alkene with mCPBA and trans alkene with mCPBA, I drew the products, they do look different but the configurations are the exact same, does that mean that they give the same products?

The configurations of the products of the trans alkene are wrong!

can you please elaborate why?

The first is RR (not RS) and the second is SS (not SR). Remember the CIP rules. You should look at each carbon atom in such a way that the hydrogen atom points backwards.

Sorry Im not following, in the first molecule, the carbon on right has already hydrogen going inside, the O has the highest priority and it goes counterclockwise to the carbon on the left (priority 2)

See this diagram: https://imgur.com/a/38EkRS5

In your diagram, why cant i just draw an arrow from 1 to 3 counterclock wise?

Because oxygen is in pointing outwards the screen. An arrow from 1 to 3 (passing through 2) is clockwise.

Let us continue this discussion in chat.


text:
Solute and Solvent classification in a Solution
When I look up the definition of solute and solvent, the definition by the Oxford English Dictionary is
solute

/ˈsɒljuːt,sɒˈljuːt/

noun

the ***minor*** component in a solution, dissolved in the solvent.

What does minor imply here? Does it classify them as solute and solvent as per mass/volume comparison? Or does it refer to their concentration?

These general definitions focus rather on usual cases and have troubles with extremes which put them in a question.

Not that it clarifies much as compared to the above dictionary but here it is https://goldbook.iupac.org/terms/view/S05746 which becomes a bit clearer togheter with https://goldbook.iupac.org/terms/view/S05746. Also upvoted the answer by P. M. below.

@Alchimista You posted the same link twice.

@Martin-マーチン thanks. The first link above should have been https://goldbook.iupac.org/terms/view/S05744.


text:
Conversion factor for conductivity of an aqueous solution
I am not 100% sure if this is the right place to ask, but please point me in the right direction if not.
I am designing a conductivity meter, and when it comes to displaying the results in mS/cm or µS/cm, I also need to convert to EC. The first place I found was THIS website which seems great as it has all the units that I can convert here and there, no problem. This site states that 1EC is equal to 1µS/cm.
I had started to do most of my calculations based on the conversions from this website. However, when getting a reference calibration liquid, I noticed it was set to 2.76EC/2.8mS, which goes against what was stated in the website. I then found THIS website which shows 1EC as equal to 1mS/cm. I have now seen another place online (closed the tab and can't find it!) that also agrees with this.
Another website I have found HERE gives a bit more information on the conversion, and states that EC is measured in deciSiemens per meter (dS/m), and 1EC = 1dS/m, which would be 1mS/cm. Then later, it has a conversion thing at the bottom which has a statement saying:

Conversions:

1 deciSiemens per meter (dS/m) = 1 mmho/cm = 1
milliSiemens per cm (mS/cm) = 1000 microSiemens per cm (uS/cm) 1
mhos/cm = 1 S/cm = 1000 mS/cm

Electrical Conductivity units (EC's) IF
measured in uS/cm: 1 EC= 1uS/cm = 0.001 dS/m = 0.001 mS/cm = 1 mS/m

(Example 1.5 dS/m = 1.5 mS/cm = 1500 uS/cm = 1500 EC)

And surely that contradicts itself where it says 1EC = 1µS/cm, then the example after says that is not the case?
I just wanted to ask and be sure on this.... When converting, is 1EC equal to 1mS/cm, or 1µS/cm?

Your question is not clear, because you have forgotten to mention the unit meter or centimeter in you last sentence.

@Maurice edited to sort that out. I don't think it really made the question unclear. It is pretty obvious what it meant so the downvote was probably a bit harsh for that. I assume that was yourself. If so, I hope the edit is enough to reverse it

For the 2 guys downvoting, it would be nice to leave a comment to explain why you feel the downvotes were deserved so I can improve the question.

Wow, it's amazing what can happen if you interpret something literally. In one of the documents you link to it says "EC units". This could be interpreted to mean either "the set of units EC" or "the units for the property EC".   It seems at some point (and IMHO somewhat bizarrely) "EC" has been adopted by some as a convenient shorthand for a particular set of conductivity units (mS/cm or $\mu$ s/cm). Another link does state

An electrical conductivity unit is a decimal unit of electrical conductivity, which by definition is equal to 1 microsiemens per centimeter (μS/cm). 1 EC = 1 μS/cm. The unit is often used on conductivity meters made in the USA.

However this is highly unorthodox, if not absolutely horrendous abuse of nomenclature, and it is not surprising you are confused. Good job by M Farooq in pointing this out.
My recommendation is to find the manual for the instrument you are using and read it. Some instruments allow display of measurements in different units. Compare those if available to determine the appropriate conversion factors.
In addition, when you do convert from electrical conductivity units to TDS, keep in mind that TDS is not a unit of conductivity. The result depends on different scales that assume different proportionality constants between TDS and conductivity based on different calibrations. Use the conversion that makes sense for your situation.

I'm fine with the conversion from ED to TDS, that's not a problem. The issue is we want to sell this product to the US market where they do tend to use EC as a unit. The manuals I have seen just tell you how to change the display units, not what they mean. It does seem to be that the majority of places tend to convert 1EC to being 1mS/cm, so I may have to go along with that. Unfortunately, I get told to design to the brief supplied by the sales team....

No serious researcher should buy a meter with EC written on it without providing any units in the manual.

@M.Farooq well, then it's lucky I'm not a serious researcher in chemistry then.

I think you have badly confused yourself by mixing the concept of a property of a solution with the units used to express this property.
Total dissolved solids (TDS) is a property of a solution. Theoretically, it can have any units kg/ kg of a solution, part per million, percentage, or whatever you like. However, people have standardized what they mean by TDS is usually mg solids/L of solution or even ppm.
In the same way, EC is electrical conductivity. It is a property of a solution. It can be expressed in units of microSiemens/cm or any multiple of it. EC itself is not a unit.

When you look online, there are places that use EC as a unit in some conductivity meters. I believe it happens often in America mainly. I understand what TDS and ED are, but I have seen places using EC as a unit, and it has been a requirement of this project

Do not trust everything online.

It is in multiple places online about this subject, plus a lot of conductivity/TDS meters include EC units as one of the units to display measurements in. So there is an answer to this question.

Have you checked the manuals of those meters? EC unit has to be defined. It does not mean anything on its own. You have to find authentic information from the manual because EC is not a unit, it is a property. Did you see this website "EC units: Electrical conductivity can be expressed using a number of different units, but the typical unit is siemens per meter2 per mole (S/m2/mole) or millisiemens per centimetre (mS/cm). The mS/cm unit is generally used in Europe as a guide to the concentration of nutrients in water. "?

https://www.canna-uk.com/electrical-conductivity#:~:text=EC%20units,concentration%20of%20nutrients%20in%20water.

I haven't looked at the manuals of all of them. The one I have looked at just mentions how to switch between displaying mS/cm, EC and ppm. The more I have looked, the more it is looking like 1EC is equal to 1mS/cm in most places


text:
Cinnabar decomposition by heating optional method?
Is it possible to decompose cinnabar into $\ce{Hg}$ and $\ce{S}$ instead of $\ce{Hg + SO2}$ with roasting and heating. What conditions need to be satisfied in the new method?

New method for what? Sulfur is inexpensive and would never be made from HgS decomposition. Mercury is obtained this way, by roasting cinnabar. But there is already a surplus of mercury world-wide. So what is the point?

Even though the chemistry is often not about why, but rather about why not, I agree with @EdV that obtaining elements from mercury(II) sulfide is not economically feasible. By the way, is you read Russian (judging from your nickname), feel free to have a look at the reactions for HgS listed in Лидин, Р. А.; Молочко, В. А.; Андреева, Л. Л. Химические Свойства Неорганических Веществ, 3rd ed.; Химия: Москва, 2000. This should give an idea what can be done easily and what not so.

Yes, it is possible to fully decompose cinnabar to elemental mercury and sulfur.

What conditions need to be satisfied in the new method?

Before reaching to decomposition temperature, cinnabar phase transitions to a different polymorph. Data from this paper1 indicate that α-HgS (trigonal) → β-HgS (cubic) phase transition occurs at 673 K and is completed at 698 K. The thermal expansion of cinnabar is fairly isotropic. At 744 K, the thermal decomposition of HgS occurs to elemental mercury and sulfur accompanied by a change in mechanism2.

So, is this decomposition technique feasible for producing mercury?

Not really, because you have to achieve very high temperature for cinnabar to fully decompose. Then, you have to separate mercury and sulfur from the mixture which is again additional work.
So, basically roasting process is applied. In this case, crushed cinnabar ore is roasted in rotary furnaces. Pure mercury separates from sulfur in this process and easily evaporates. A condensing column is used to collect the liquid metal, which is most often shipped in iron flasks. If the ore is heated in an atmosphere of oxygen, then mercuric oxide is formed which gets decomposed to mercury (which is deposited below while the gases escape). The other advantages is that the roasting temperature is lower than the decomposition temperature, so it is easy to achieve.
$$\ce{2HgS + 3O2 -> 2HgO + 2SO2 ^}$$
$$\ce{2HgO + HgS -> 3Hg + SO2 ^}$$
$$\ce{2HgO ->[\Delta] 2Hg + O2 ^}$$
Another technique is to heat cinnabar in an atmosphere of hydrogen. The sulfur that is forming quickly reacts with hydrogen to form hydrogen sulfide which escape as a gas. Simultaneously, hydrogen helps in the reduction of cinnabar and the process quickly accelerates to completion.
$$\ce{HgS + H2 ->[\Delta] Hg + H2S ^}$$
References

Ballirano, Paolo & Botticelli, Michela & Maras, Adriana. (2013). Thermal behaviour of cinnabar, α-HgS, and the kinetics of the β-HgS (metacinnabar) → α HgS conversion at room temperature. European Journal of Mineralogy. 25. 957-965. 10.1127/0935-1221/2013/0025-2341.
Baláž, P., Godočiková, E. Thermal Reduction of Mechanically Activated Cinnabar (HgS) and Stibnite (Sb2S3). Journal of Thermal Analysis and Calorimetry 65, 51–57 (2001). https://doi.org/10.1023/A:1011564232088

(+1) Your usual thorough and nice work!


text:
Reduction of alcohol to alkane
I have been looking everywhere for hours now, and can't find anything on a simple lab reduction of an alcohol to an alkane that doesn't involve any "fancy" reagents such as tosylates, metal catalysts etc.
I would have expected that under acidic circumstances, the oxygen donates an electron to the proton from the acid, and a hydride from a general reducing agent such as LAH or NaBH4 then attaches itself to the carbon, pushing out a molecule of water, and alkane.
However, I have found no proof or mention of the existence of such a reaction. If this is is impossible, then why, and what would be the best general purpose way to reduce an OH to a H

Of course it's impossible. For ex. https://chemistry.stackexchange.com/questions/15586/hi-p-reduction-mechanism is an option

If you regard tosylates as fancy reagents then your career in synthetic chemistry is going to be very limited.

May I introduce you to the Barton-McCombie reaction for the reduction of alcohols to alkanes.

image and further information here

Perhaps too "fancy". ;)


text:
What pH value of 1M sodium crotonate solution supposed to be?
I need to prepare the $\pu{1M}$ solution of sodium crotonate, however, I don't have one in lab, but I do have sodium hydroxide and crotonic acid.
Thus, I calculated that for $\pu{100 mL}$ of $\pu{1M}$ solution I need: ~$\pu{4g}$ of $\ce{NaOH}$ and ~$\pu{8.6g}$ of crotonic acid. After preparation of solution (approx $\pu{80 mL}$ of DI water to dissolve the crotonic acid in a beaker, add $\ce{NaOH}$, mix until dissolving, let it cool down, quantitatively transfer a solution into a $\pu{100.0 mL}$ volumetric flask, adjust the volume), the $\mathrm{pH}$ of the final solution is 6.1 (which also stays the same after purging with $N_2$ gas and autoclaving).
To my understanding, the solution of sodium crotonate supposed to have a basic $\mathrm{pH} \ (\gt 7)$ since it is formed by weak crotonic acid (with a $K_\mathrm{a} = 1.97 \times 10^{-5} $, reference) and a strong base. Thus, the rough estimation of $\mathrm{pH}$ should be done according to formula:
$$
\mathrm{pH} = 14 + \lg \left(\sqrt {\frac{K_\mathrm{W} \times C_\text{sodium crotonate}}{K_\mathrm{a}}} \right)\\= 7 + \lg \left(\sqrt {\frac{C_\text{sodium crotonate}}{K_\mathrm{a}}} \right) \\= 7 + \lg \left (\sqrt\frac{1}{1.97 \times 10^{-5}} \right) = 9.35
$$
Which results in a far higher value.
On the other hand, if I suspect the formation of buffer solutions as sodium crotonate + crotonic acid (if there was not enough hydroxide), then the final $pH$ should be estimated according to:
$$
pH = pK_a + \lg \frac{C_{salt}}{C_{acid}}
$$
Which roughly estimates 3.8% of unreacted acid in solution. However, even if I arbitrally assume only 0.1% of unreacted acid in solution, the formula results in $pH$ of 7.7, which is anyway higher than I obtained, but still not a 9.35 as should be.
So, my questions are:

What is the real value of $\pu{1M}$ solution of sodium crotonate supposed to be (like if prepared from salt reagent)?
I think I should adjust the $\mathrm{pH}$ with $NaOH$, but for what final value? 9.35 or 7.7?
Are there any verified protocols for preparation such solution (USP / EuP / etc.)?

P.S. I also found paper where sodium crotonate solution was prepared by reacting sodium bicarbonate and crotonic acid followed by adjusting $\mathrm{pH}$ to a final range of 6.7 ... 7.0, but I still don't get why they stopped at $\mathrm{pH}$ of 7, not higher.
P.P.S. As also stated here, the pH "without neglecting the water autoionization" can be estimated by solving this equation:
$$
\left\{\begin{matrix}
\left [ H^+ \right ] ^3 +\left ( K_a + C_{salt} \right ) \times \left [ H^+ \right ] ^2 - K_w \times \left [ H^+ \right ] - K_w \times K_a = 0
\\
\left [ H^ + \right ] > 0
\end{matrix}\right.
$$
However, it results into
$
\left [ H^ + \right ] = 4.43847 \times 10 ^{-10}
$
and $pH = 9.35$ which is not much different from the previous estimation.

Recall buffers. What do you think aqueous crotonic acid/sodium crotonate soulution would be?

Solutions of salts of weak acids are always mildly alkaline. pH = 14 - 0.5 ( pKb - log ) = 7 + 0.5 ( pKa + log c ). Acidic pH in the case using bicarbonate was due dissolved CO2. Similar thing in lesser extend can happen, if hydroxide contains carbonate.

@Poutnik I wasn't considering the influence of $CO_2$ due to: 1) the solution is autoclaving, which should stimulate the release of $CO_2$ from solution, but pH still remains about 6.1 after autoclaving; 2) the crotonic acid is stronger than carbonic acid, and no carbonates should be in solution. Do you think it is still a factor after autoclaving?

@MathewMahindaratne I was considering the buffer system (see updates to the post), but even those calculations are higher than 6.1. Which still remains the question, what the final pH should be?

1/ Autoclaving is essential info, that should be provided in the question. How should we know ? But even at high temperature, with aventual reached CO2 partial pressure, there would be still significant CO2 dissolved, affectingn pH. 2/ The stronger the acid is, the weaker base is its salt. And it is weak acid ( CO2 ) versus weaker base ( NaA ) pH clash. I have said carbonate in hydroxide. Reaction of the acid with such hydroxide releses CO2. Another important factor is the accuracy of reached stoichiometry ratio, due calculation and purity deviation errors.

I would rather not to rely on calculations and would adjust the final pH by adding NaOH. As NaOH ( not sure about the acid ) does not have well reproducible purity due Na2CO3 content and its hygroscopicity. Small deviations from stochiometric acid : NaOH ratio causes big pH differences. I would use an intermediate step of boiling up and cooling down at weakly acidic or neutral pH stage

@Poutnik I agree on adjusting the pH with NaOH, but for what final value? Just neutral or higher?

Let us continue this discussion in chat.


text:
Why is the Suzuki coupling done under inert atmosphere?
There are plenty of sources saying that the Suzuki coupling is air-sensitive (even though there are some versions that require oxygen). However I'd like to know why a typical Suzuki coupling is air-sensitive. I cannot find what the oxygen actually does.  Pd(0) doesn't seem like it would get oxidised in air that easily, does it?

A Suzuki coupling isn't oxidative; it shouldn't require oxygen. What schemes are these? Can you attach them?

Sure, here https://pubs.rsc.org/en/content/articlelanding/2011/ob/c0ob00524j#!divAbstract

However I'm asking about the normal version where it's done under N2 or Ar, I'd like to know why.

Ah, I now see that perhaps my question was ambiguous.  I will try to amend it.

Hmm, I'm pretty sure that the standard one is oxygen-sensitive because Pd(0) is easily oxidised. That oxygen Suzuki is a bit underexplained, there is only a speculative suggestion in an earlier paper by the authors...

Yes, it is to keep Pd at (0)


text:
Why does iron oxide form hydrous iron oxide while chromium oxide does not?
Both Cr2O3 and Fe2O3 share a similar structure, and both can form some sort of passivating layer, but only one of them hydrates, thus turning into rust and providing a pathway for more water and O2 to further oxidize the iron. I believe the same applies to aluminum and nickel as well. iron seems to be the only one to do this, but I cannot figure out why.

I object. Iron oxide doesn't hydrate either. Whenever you end up with hydroxide, you've never had the Fe2O3 in the first place.

Iron iii oxide doesnt hydrate? I thought that adding H2O molecules was hydration, and I definetly thought that Fe2O3.H2O was a thing, though I may be mistaken

Fe2O3.H2O is a thing all right, but it is not produced by putting together Fe2O3 and H2O. Ditto for Cr2O3.H2O.

hmm, okay, thankyou. It appears you are correct. Is there anyway you can tell me how Cr2O3, and well as Al2O3 are formed then? I was told that Cr2O3 "does not readily hydrate" like Fe2O3... do they form via the same mechanism as Fe2O3 and then loose the H2O?

No, unlike the iron rust, they don't form hydrated. That's where the difference originates.

ahh, i see. so is the general mechanism the same, with the exception being that they simply don't bond to that water molecule?

I'd say the general mechanism is different in that Fe2O3 is produced hydrated, and Cr2O3 is produced non-hydrated.

TL;DR: Guignet's green is a green pigment with composition $\ce{Cr2O3.2H2O}$.
Long answer
Chromium oxide is largely used as a green pigment in various industrial and commercial purpose as it is the only green pigment which can withstand both acidic as well as alkaline environment. It is also used as permanent paints for cements, distempers and paints. It can also withstand high temperatures, hence used for coloring glasses, enamels and pottery glazes.
Guignet's green or $\ce{Cr2O3.2H2O}$ is the hydrous form of the oxide which is used for similar purpose as the anhydrous form. This pigment becomes anhydrous when heated at 200 °C, hence not suitable for glass and ceramic industries.
It is prepared by mixing 3 parts of boric acid and 1 parts of potassium dichromate which is heated in reverberatory furnace. It becomes a dull red mass initially but swells up releasing oxygen to give a nice green color. It is then boiled with water whereby chromium borate decomposes to hydrated chromium oxide and boric acid. The precipitate is dried and separated while the mixture is washed with mother liquor to evaporate the boric acid.
$$\ce{2K2Cr2O7 + 4H2B2O4 -> 2Cr2B2O6 + 2K2B2O4 + 3O2 + 4H2O}$$
$$\ce{Cr2B2O6 + 3H2O -> Cr2O3.2H2O + H2B2O4}$$
Reference

S. P. Das Gupta & H. N. Bose (1948) A Note on Preparation of Chromic Oxide Green, Transactions of the Indian Ceramic Society, 7:2, 60-63, DOI: 10.1080/0371750X.1948.10877850


text:
Entropy problem

A tank is divided into two equal chambers by an internal diaphragm. One chamber contains methane at a pressure of $\pu{500 bar}$ and a temperature of $\pu{20 °C},$ and the other chamber is evacuated. Suddenly, the diaphragm bursts. Compute the final temperature and pressure of the gas in the tank after sufficient time has passed for equilibrium to be attained. Assume that there is no heat transfer between the tank and the gas and that methane obeys the van der Waals equation of state. Data: $C_p = \pu{35.56 J mol^-1 K^-1}$

You should post your thoughts on how to solve this. This is not a HWK solving site.

We are given the initial and final volumes, the amount of gas, and the initial temperature. The expansion is adiabatic and free so no heat is exchanged and no work is done, meaning that $\Delta U = 0$. If it were an ideal gas, this would mean that the temperature is unchanged and that the final pressure is given by Boyle's law. However it is a van der Waals gas. We therefore make use of the thermodynamic equation of state:
$$dU = \left[ T\left(\frac{\partial P}{\partial T}\right)_V - P  \right]dV + C_V dT $$
For the vdW gas
$$\left[ T\left(\frac{\partial P}{\partial T}\right)_V - P  \right] =\frac{a}{V_m^2} $$
so that
$$\Delta U = 0 = a\left( \frac{1}{V_{m,ini}}-\frac{1}{V_{m,fin}} \right) + C_V \Delta T$$
or
$$T_{fin} = T_{ini} + \frac{a}{C_V}\left( \frac{1}{V_{m,fin}}-\frac{1}{V_{m,ini}} \right) $$
From this last expression you can determine the final T, and knowing the final volume you can compute the final pressure using the vdW EOS.

Actually, we are given the density or molar volume of the gas, but the answer is the same.

To integrate the temperature term, one must first verify that Cv is not a function of molar volume for a VDW gas.

@ChetMiller Good point, I overlooked that and assumed it was constant.


text:
How do we know the natural abundance of isotopes on Earth?
Without knowing the Average Atomic Mass or the percent abundance, how do we know that Protium is the most prevalent hydrogen isotope? What methods did scientists use to come to this conclusion? How did they reach such a conclusion?
I'm still starting out in learning chemistry and I felt like I hit a boulder. I know the formulas and how to get the natural abundance, but I haven't heard a concrete explanation of the concept. I know that it's impossible to survey all the hydrogen elements on Earth to see which isotope is abundant, so what method did scientists use?

Same as with everything else in life. Do we have to bite every apple in a crate to find out they aren't rotten?

Sampling and lab measurements. // The mass of individual isotopes is known with far greater precision than the averaged mass of a particular element because of isotope variation. Look at copper for instance.

Side note: Equally see IUPAC's committee working on this (example, their biennial reports example), or more recently, https://www.isotopesmatter.com/.

Isotopic ratios in rhino horns. Trade and diet. http://luna.cas.usf.edu/~rtykot/NPR2%20-%20Rhino%20Horn.pdf

Because we can sample the environment with great precision
You seem to assume that we can't know the exact makeup of a sample unless we already know the components. This is wrong. Mass spectroscopy can reliably measure the exact mass of every isotope in a sample and the relative abundance of each. If, for example, a sample of sea-water is measured, we can reliably tell the exact proportion of all the hydrogen isotopes and all the oxygen isotopes in it.
Given that his is possible, estimating the true natural abundance is a matter of statistical sampling. Ocean and river water is fairly well mixed so this is easier than it sounds (at least when there are no processes that radically alter the isotopic composition). So we have a lot of samples, for example, from different oceans and different locations in those oceans. Given that, we can be fairly confident that the isotopic composition doesn't vary a lot and, therefore, we can get a good estimate of the "natural" abundance of each isotope.
But the techniques we use are precise enough to notice that there are some (small) variations in that composition. Some processes do (very slightly) alter the isotopic ratios. Evaporation tends to happen slightly faster for lighter isotopes, for example, as do some biological processes. Some of those processes are temperature dependent leading to the ability to identify the ambient temperature preset when some fossils were alive by measuring the ratio of oxygen isotopes in them.
The key point is that we do have tools that can measure isotope ratios very precisely. So we know the ratios in any sample very precisely. With a very large number of samples, we can generate a reliable picture of the typical abundance of each isotope which is what we refer to as the natural abundance.

These numerical values have been determined by mass spectrometry.
In brief, it starts form a box like a match box in the vacuum. No air in the room around the box. There are many plates and tubes in this box. First a small tube is introduced for delivering some hydrogen gas in the box. Second two thin metallic points have been introduced in advance one through the roof and one through the bottom of the box. Then a high tension is applied between the two points : a spark is produced that breaks the $\ce{H2}$ molecules into parts, producing $\ce{H}$ atoms, electrons, $\ce{H2^+}$ and $\ce{H+}$ ions. These fragments will soon recombined as soon as they get out of the spark. Except if they are prevented to do it by applying an outer tension. And indeed there are two vertical metallic plates on the lefthand and righthand sides of the box.
One of these vertical plates is charged positively and the other negatively. This last one attracts positive ions like $\ce{H+}$ and $\ce{H2+}$. They will move and fall on this plate. Not all ! Because this plate plate is pierced in its middle, and just on the other side there is another plate which is charged at a higher voltage. So the positive ions arriving in the middle of this hole are more strongly attracted by the second plate and they cross the hole.
Well ! This second plate is also pierced in its center, and it is followed by a third pierced plate, etc. The last plate is not pierced but it is far away from the last but one plate. This whole system of plates produces a thin beam of charged particules crossing all holes before being discharged when hitting the last plate. But between the last plate and the last but one plate, a big magnet is approached that is strong enough to deflect the trajectories of the ions in the beam. Small ions like $\ce{H+}$ are more deviated than heavier ions like $\ce{H2+}$. So by dividing the last plate into many independent small parts, you may know whether and how many ions have been slightly or strongly deviated. So you may know the proportion of isotopes in the beam.
OK ?


text:
pH of an aqueous solution of Mg2+ kept rising above that of clean water. What's going on?
In a recent inorganic chemistry lab session, we tested the pH of various ions in water. The water turned out to be on the acidic side (5.5 with $\mathrm{pH}$-paper, 5.59 with $\mathrm{pH}$-meter). A solution of $\ce{Mg^{2+}}$ gave some strange results. The $\mathrm{pH}$ kept rising when using a $\mathrm{pH}$-meter, going from around 4.2 to 5.83 (higher than that of the pure water). We stopped at this point, thinking something must be wrong. A classmate experienced the exact same thing, and stopped at 6.5 or something.
I expected $\ce{Mg^{2+}}$ to act as a very weak cation acid, making the solution at least somewhat more acidic than the water. In my limited knowledge, I would think that the only thing $\ce{Mg^{2+}}$ can do in this scenario is to slightly polarize the water ligands, thus making them deprotonate.
Was this a fault in the equipment, or was there some other reaction going on?
I'm having a hard time finding expected $\mathrm{pH}$ of $\ce{Mg^{2+}}$ in aqueous solution, so any data on that would also be welcome.

Read this article.

Thanks, but this doesn't explain my results. I know what cation acids are, and Mg$^{2+}$ should be a very weak one, but still lower the pH of the water.

The pH of pure water in contact with air is never $7$. It is about $5.5$, due to the action of  $\ce{CO2}$ in air, which is slightly soluble in water and produces some ions $\ce{H3O+}$ . As the amount of $\ce{CO2}$ in air is not constant and changes from time to time, the pH of pure water is also changing in time.

That's fine, but the water pH was tested straight before the water+Mg${^2+}$ mixture was tested, and the latter still showed a higher pH. That shouldn't make any sense, am I right?

pH meters are not very accurate, and don't give consistent results. I have used one in my lab, and the pH value kept changing even though it was dipped in a buffer solution, which is supposed to have a constant pH, regardless of CO2 from air etc. I had to recalibrate it every 5 minutes or so, therefore, you shouldn't place too much importance on the pH meter reading changing from 4.2 to 5.83, that may well be machine error.

What magnesium salt did you use?  Magnesium hydroxide is  sufficiently strong so that its salts with weak acids could be basic.

Mg(NO$_3$)$_2$. I guess I have gotten my answer from all the comments above. It doesn't make sense that Mg$^{2+}$ solution has higher pH than water.


text:
What color would be produced by the burning of the listed chemicals of the projectile weapon the carcass?
The carcass was a projectile weapon developed by the French in the second half of the 17th Century. It was "named carcass because the circles which pass from one ring, or plate, to the other, were thought to resemble the ribs of a human carcass."

illustration from the article "The British Fire Bombing of Quebec City in 1759"
The internal chemical constituents are described as
"10 parts of pounded gunpowder, 2 of nitre (potassium nitrate, KNO₃), 1 of sulfur, and 1 of colophony (rosin)"
OR
"6 of gunpowder, 4 of nitre, 4 of sulfur, 1 of beaten glass, 0.5 of antimony 0.5 of camphor, 1 of sal armoniac, and 0.25 of common salt"
and the exterior was covered in
"4 parts of melted pitch, 20 of rosin, 1 of oil of turpentine, and as much ground gunpowder as was needed to reduce it to the consistency of a paste. After immersion, the shell was to be covered with tow, and immersed again"
Is there any way to determine or estimate what color the exterior would burn during flight, or what would be the color when the interior ignited? What color would be the smoke either during the flight or during the burning of the interior?

The smoke is usually made of soot (impure charcoal), which is black. And when it burns, soot produces a yellow color

But the flames might be coloured by the potassium in the nitre and gunpowder.


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Why do dihaloalkanes lose both halogen atoms during mass spectrometry?
I was looking through mass spectrums of haloalkanes and found the one for 1,2-dibromobutane confusing:

The base peak is at 55 m/z, and the only fragment I can think of with that ration will be $\ce{C4H7+}$. The two bromine atoms must be absent as the relative molecular mass of one bromine atom is 80. I can't think of how this will happen. Does the removal of one bromine atom force the other off? Does it form an alkene? It is present in all positional isomers (1,3-dibromobutane, 1,4-dibromobutane etc.) I initially thought that the lone pair of the bromine will be donated to the positively charged carbon, but if the carbon is 2 carbons away from the bromine, I can't imagine this happening.
I feel like this is somehow related to the addition of bromine to alkenes, maybe the bromonium ion is an intermediate and bromine gas is formed? The chemical formula might be that of an butene radical. But since the peak still shows in the spectrum of 1,4-dibromobutane, I don't think that is the case.
If anyone has any theories, please do share!

That's hardly surprising. C-Br bonds are weak and Br radicals pretty stable.

It isn't just both bromine atoms that are lost, although that might be expected in mass spec given the relative weakness of the carbon-bromine bond.  You also lose a hydrogen atom.  And that is key.
We might suppose that a single bromide ion (or bromine atom plus electron) comes off first, forming a $\ce{C4H8Br^+}$ ion.  That is the peaks at 135 and 137, recalling that in mass spec the mass of a bromine atom is both 79 and 81 which are the most common isotopes.  There is nothing spectacular about such an ion per se, but now it can eliminate $\ce{HBr}$, or $\ce{H + Br}$ to give a resonance-stabilized cation:
$\ce{CH2=CH - \overset{+}{C}H - CH3}\leftrightarrow\ce{\overset{+}{C}H2 - CH=CH - CH3}$
Thus the formation of this ion is energetically favored, and it would account for $\ce{C4H7^+}$.
In the case of 1,4-dibromobutane you should expect the initial (primary) $\ce{C4H8Br^+}$ carbocation to rearrange with a hydride transfer to the end carbon that first lost the bromine, making the ion secondary.  Then the ion will match that obtained with the other isomers.

I thought that an alkene cation was involved! However, I'm still quite confused on how it can eliminate the H and Br like you say. I'm assuming it involves the positive charge travelling through resonance? But there are no double bonds, and the bromine atom isn't necessarily beside the charge...

Once the second bromine atom comes off with that weak bond, the extra hydrogen also gets shoved out of the way to complete formation of the stabilized cation.

So the bromine just comes off spontaneously without any electron transfer?

Why not?  There can be neutral fragments in a mass spec chamber.  We just miss them in the spectrum.