text:
How can I systematically solve a system of non-linear equations using a standard graphing calculator?
In analytical chemistry, when we have to consider more than one equilibrium, the problem can be reduced to solving a system of non-linear simultaneous equations by applying the mass and charge balance equations.
Conventionally, the equations are converted to a polynomial equation through simple substitution and then subsequently solved using an online equation solver. So, I was wondering if there's an easier way to solve the system of equations directly (i.e., without manipulating to a polynomial) using a graphing calculator.
I know this just sounds like I'm being lazy, but it would definitely save a lot of time considering that I'm used to dealing with 6 or more equations at a time. For reference, the graphing calculator I'm referring to is the TI-nspire CX II.
Any insight into this would be greatly appreciated. Thanks for reading!

You could try making a program in your calculator using Newton's method (Jacobian matrix iteration) with an initial guess. I made one for a TI-84 calculator some years ago (older version of your calculator), so it should be a little easier in your case.

@Sam202 The Newton method is very efficient when it converges, but it's convergence may be even for single independent variable tricky at unlucky choice of starting point and therefore of combinations of respective derivatives. (convergebce condition  |f.f''| < (f')^2 )

@Poutnik That is correct. Although not completely practical and far from perfect, one could manipulate convergence condition within the iteration loop to obtain an approximate answer, or perhaps re-run the program with a different initial guess if no convergence is obtained. I am guessing OP is reluctant to use Excel or alike because it is not allowed in the exam, whereas his calculator is.

Have you considered conversion of the set of nonlinear equations to an optimization problem?
If we have set of equations:
\begin{align}
L_1 &= R_1\\
L_2 &= R_2\\
..&.. \\
L_n &= R_n
\end{align}
then solve numerical optimization job = minimization of the value of the custom function:
$$U = \sum_{i=1}^{n}{\left(L_i - R_i \right)^2} = \mathrm{min}$$
$L_i$ and $R_i$ are the respective left and right sides of the equation No $i$.

I assume a set of 3 and more equations cannot be solved easily by a graphic way. OTOH many such calculators can be tasked by some simple automation program.

With handiness advantage, one can use the MS Excel optional add-in Solver, designed exactly for similar cases. It finds values of input parameters for minimization of the dependent target value. In case of problem with relative scaling of variables across many orders, there is option of scaling adaptation.

Good answer but please define the symbols.

@AChem I guessed it was not needed but then I have realized I used for right sides Czech symbols  P (pravá) instead of R(right).  :-)


text:
At the atomic or molecular level, why are some macroscopic materials more flexible, compared to others?
Before answering this question, keep in mind that I am a 2 year Biology student, with limited experience in studying chemistry. I am going to make assumptions which I explain, feel free to correct them.

Firstly, it would be great to have a scientific definition of material. The Wikipedia definition is as follows,

Material is a substance or mixture of substances that constitutes an
object.

It is worth knowing the Google dictionary definition of substance, so that we are all on the same page,

A particular kind of matter with uniform properties.

From these definitions I suspect that a material can be classified as being composed of one or more atoms (correct me if I am wrong), if the later is true, those atoms may be of the same or of a different element. If a molecule is considered a substance, it would follow that a material may be composed of one or more molecules, again, of the same type, or a different type. Тhus, I assume that the flexibility of a material, consisting of one molecule, may depend on the molecule's flexibility. On the other hand, I would expect the flexibility of a material consisting of two or more molecules of the same or different type, to depend on the weakness of intramolecular and intermolecular attractions, respectively.
Initially (thinking too hard on the atomic scale), I believed that in order for a material to have a different flexibility to another material, the chemical bonds of its constituent atoms would need to be more flexible. I had thought 'bond flexibility' could differ between molecules, perhaps, because the atoms of certain elements might have their energy dispersed over a greater distance (e.g., an a atom could have the positive charge of its proton spread over a greater distance). Thus, allowing two atoms to be separated over a great distance without 'losing' their bond. This was utter speculation and I now am nearly certain this is incorrect.
According to this source,

Free rotation around C—C bonds allows long polymer molecules to curl
up and and tangle very much like spaghetti. Branching - straight,
unbranched chains can pack together more closely than highly branched
chains, giving polymers that have higher density, are more crystalline
and therefore stronger. The spaghetti-like entanglements of polymer
molecules tend to produce amorphous solids, but it often happens that
some parts can become sufficiently aligned to produce a region
exhibiting crystal-like order, so it is not uncommon for some
polymeric solids to consist of a random mixture of amorphous and
crystalline regions. As might be expected, shorter and less-branched
polymer chains can more easily organize themselves into ordered layers
than have can long chains.

Therefore, I would expect materials consisting of several long and branched polymers to be more flexible compared to materials composed of several short, and unbranched polymers, because long, and branched polymers could not form as many intermolecular (and perhaps intramolecular) bonds or interactions with other molecules.
Anyway, from what I have read in this post,

Flexibility in polymers at the molecular level is determined by the
flexibility of the individual polymer chains and the interactions
between the chains.

Sachan goes on to explain that polymer flexibility is explained by the chemical nature of the monomers that make up the polymer chain. For example, monomers with polar groups increase intramolecular hydrogen bonding, or in other words, interchain interactions, thus, I imagine these interactions facilitate bonding between monomers of the polymer, by placing chain monomers closer together, allowing bonds between parts of the chain to occur. They also may facilitate hydrogen bonding between other polymers and thus, cross-linking between molecules, preventing each molecule from rotating independently.
Lower from the first link explains this concept similarly,

Side groups - polar side groups (including those that lead to hydrogen
bonding) give stronger attraction between polymer chains, making the
polymer stronger.

I would expect bonds between monomers of the polymer chain to make the polymer rigid to a higher degree than, weaker hydrogen bonds by preventing parts of the molecule from independently rotating (i.e., I suppose, two bonded monomers would make the polymer less flexible as a whole, compared to the two same unbonded monomers). Hydrogen bonds would also add resistance to parts of the molecule from separating and thus, rotating independently. In my head, I am thinking of two separate cysteine monomers, apart of a linear polymer chain, both containing sulfhydryl groups, where the sulphur is bonded to a hydrogen. If both hydrogens were removed and the sulphurs of two cysteine monomer bonded, the monomers could not rotate as before, because they would now be bonded to each other and the other monomers as before. It is really hard to visualise why this is the reason, but intuitively the atoms in a folded chain would be less free to rotate compared to the atoms in a linear chain. Adding atoms to a molecule does not necessarily decrease its flexibility, which I would have expected. If anyone can explain the logic behind this more clearly I would appreciate that.
Lastly, I envisage a material stretching in different directions due to atoms rotating around single bonds in the molecule/s that compose that material.
My final assumptions are that: non-polar polymers, composed of predominantly non-polar monomers are more flexible since cross-linking and hydrogen bonding do not impede the rotation of parts of the molecule, by allowing more atoms to be single bonded. Furthermore, Are polymers consisting of more double bonds, less flexible than polymer consisting of more single bonds, on the macroscopic scale? Are branched polymers more flexible compared to unbranched polymers?

Unfortunately, your post is very bloated. What even is the real question here? The title is quite misleading, but well... if it wasn't, it would be so broad it should definitely get closed. When you get to the specifics another issues arise and I don't know how this could be fixed. Certainly cutting the length down at least twice would be a good start. And bold for actual question not some speculation.

early here you make an assumption that makes understanding material properties much harder. You quote the googe definition "A particular kind of matter with uniform properties." but this is too narrow. Many steels, for example, look "uniform" but on microscopic examination have small crystal grains of different substances. And this is often very important in understanding the bulk properties of the material.

"non-polar polymers, composed of predominantly non-polar monomers are more flexible" not necessarily, HDPE is rigid.

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text:
Is there any amateur way to process fabric into cellulose?
This article interests me.
It says they can decompose fabric into more pure materials - cellulose, then splitting that polymer/polysaccharide into glucose - by soaking it in sulfuric acid.
I read that sulfuric acid is highly dangerous to work with, because it harms human tissue and also is harmful if inhaled.
So, acids are positively charged, so they attract electrons from nearby molecules, which weakens those molecule’s molecular bonds?
Is there a different naturally occurring acid that can also dissolve cellulose like this, for example, acetic acid distilled from vinegar?
Is there a common microorganism that digests fiber? Perhaps composting or enhanced biodegradation/bioprocessing could also convert fabric into usable new compounds?
What’s a feasible way for an amateur to try to break down or process fiber like these researchers have?

Cuprammonium rayon does not require $\ce{H2SO4}$. However, the solution is still somewhat toxic, disposal is difficult, and it readily stains things -- careless rayon chemists have blue-black hands and clothes.
It does make a good classroom demo, though, when performed with care: rayon from filter paper!

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text:
Reaction between iron oxide and metallic zinc or aluminum at 500C
High temperature corrosion of iron occurs on passenger automobiles in areas such as  exhaust headers, pipes and bolts that reach temperatures of about 500°C. The iron is in the form of steel or stainless steel and the corrosion is in the form of rust or scale.
What would occur if 99.9% pure metallic zinc (or aluminum) were brought into tight contact with iron in either its clean, rusted or scaled state and heated to normal exhaust temperatures in the presence of moisture and salts?
For example, exhaust flanges are clamped together using nuts and bolts and the bolts are long enough to allow a second nut to be installed. A zinc (or aluminum) washer is installed on the bolt and a second nut is used to crush it into the already-installed nut. The iron-oxide bolt / zinc or aluminum washer / iron bolt sandwich is heated in the presence of salts and moisture.
Would the zinc melt and drip away?  Would such an assembly provide any cathodic protection against rust or scale?  Would the preexisting flaky scale be converted into a blob of solid iron. Would a violent thermal reaction occur between the zinc and iron oxide and the exhaust burst into flames?
CLARIFICATION: the purpose of this question is to understand the likely interaction between Zn/Al and iron oxide/iron and uses exhaust pipes as a practical example.  Specifically:

At the moment the Zn melts, would it be expected to wick into rusted areas and reduce the rust back into iron and produce a coating of zinc oxide?
Would an Al washer crushed between two rust-prone nuts provide any protection against rusting, especially at high temperatures, in the small area around the Al washer?

Zinc and aluminum both melt. Aluminized steels area available. But the standard 13 % chrome SS does so well it is not used much any more. I have never seen corrosion problems on SS exhausts with well over 100,000 miles.


text:
How can we say that p in ΔU= q_p-pΔV and p in ΔH=ΔU+pΔV are equal? About the derivation of ΔH=q_p
I was studying about thermodynamics and everything was fine until I studied about enthalpy $H$. The first thing I wanna know is what is $p$ in eq. $H=U+pV$. Is it internal pressure or external pressure? Since $p_\text{in}$ and $p_\text{ext}$ are not necessary to be same and it is important to distinguish both.
At constant pressure we get $\mathrm dH=\mathrm dU+p\,\mathrm dV$ (eq. 1)
Another eq. we know at constant pressure is that , $q_p=\mathrm dU+p\,\mathrm dV$ (eq. 2).
I know that this pressure should be external pressure but I am not sure.
We derived $q_p=\mathrm dH$ (eq. 3) thinking $p$ used in both eq. is same.
I have read somewhere that $p$ in eq. 2 is external pressure. And $pV$ in eq. 1 tells the energy required to estabilish system. So I thought that this pressure would be internal pressure.
If this is true then $p$ in both equations eq.1 and eq. 2 would be different and the eq. 3 would get wrong.
So clear my confusion. I am totally confused to say if both are external pressure or internal. We can't use internal pressure in eq. 2. So $p$ should be external pressure. But on the other hand $p_\text{ext}V$ in eq. 1 looks wrong.

You should distinguish between equilibrium states of the system and transitions (processes) between these equilibrium states. Heat and work are exchanged during the transitions and are not equilibrium properties of the system, they are properties of the transition.
When in an equilibrium state (the endpoint of a transition), and when the system is subject to an external pressure, then the internal pressure of the system and applied pressure are the same: $p=p_{\mathrm{ext}}$. This condition is called mechanical equilibrium. When the system undergoes a transition for which the initial and final states have equal pressure, then if the system allows heat exchange with surroundings $\Delta U = -p \Delta V + q_p $ and $\Delta H = \Delta U+ p \Delta V = q_p$, assuming also that only pressure-volume work is done. Here $p$ is equal to the pressure of the system and surroundings, since the system and surroundings are in mechanical equilibrium.
Edit: Reversibility is also a property of processes, not of systems. For a process to be reversible the system and surroundings have to be in equilibrium throughout. That also includes satisfying the condition of mechanical equilibrium ($p=p_{\mathrm{ext}}$).
When considering an irreversible compression the initial equilibrium pressure is $p=p_{\mathrm{ext,initial}}$ and the final equilibrium pressure is $p=p_{\mathrm{ext,final}}$, where $p$ refers to the pressure of the system. To trigger compression the initial equilibrium condition is broken and the external (applied) pressure is increased above that of the system, from $p_{\mathrm{ext,initial}}$ to $p_{\mathrm{ext,final}}>p_{\mathrm{ext,initial}}$. The condition of constant $p$ of the system is not satisfied for the case of a compression against a constant external pressure since there is a change going from the initial to final equilibrium pressure. The external pressure is constant, not that of the system. The condition $\Delta H = q_p$ does not hold, and the change in enthalpy is given more generally by $\Delta H = \Delta U +p_2V_2 - p_1V_1$.
Regarding the other additional example:

Lets us consider an irreversible process with equilibrium at initial and final state. The external pressure decreases and then again becomes equal to internal pressure. Throughout this expansion internal pressure remains same due to heat gained by system.

The last sentence is a false statement. The internal pressure (that of the system) doesn't just "stay the same". It follows the external pressure (that of the surroundings) and therefore has to change accordingly. Initially there is a mismatch (mechanical disequilibrium) until heat has entered the system and the equilibrium is reestablished, so the internal pressure is not constant.

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In a Cu/Al galvanic cell, why does decreasing the aluminum concentration decrease the cell potential?
When I originally learned the Nernst equation, it was just conceptual without any hands on experience. I am now a chemistry teacher, and wanted to provide some hands on experience with it for my students. We did a lab where we set up a galvanic cell as follows:
Anode: Aluminum electrode in 1 M AlCl3 solution
Cathode: Copper electrode in 1 M CuSO4 solution
Then, we varied the concentration of each solution. The overall reaction should be:
2 Al + 3 Cu+2  -->  2 Al+3 + 3 Cu
So, we expected that diluting the CuSO4 solution would decrease the cell potential, while diluting the AlCl3 solution would increase the cell potential.
When we tried it, diluting CuSO4 decreased the cell potential as expected. However, diluting AlCl3 also decreased the cell potential! In fact, it seemed to decrease the cell potential MORE than diluting the CuSO4 did. All 10 labs groups independently got this same result, and their measured voltage values were all remarkably consistent, so I have ruled out procedural flukes. It is clear that there is something about this reaction I do not understand. What am I missing?
Here is some of our data in case it helps:
1 M AlCl3 and 1 M CuSO4: 0.83 V
0.01 M AlCl3 and 1 M CuSO4: 0.66 V
0.0001 M AlCl3 and 1 M CuSO4: 0.54 V
1 M AlCl3 and 0.01 M CuSO4: 0.78 V
1 M AlCl3 and 0.0001 M CuSO4: 0.73 V

I have always thought teachers verify the intended experiments themselves before deploying them in the lab class, unless overtaking already verified ones. Unexpected troubles confuse students a lot, if the teacher does not have answers for their questions. But the pro is you at least try to get answers a posteriori.

Aluminium electrode is not a good choice, as there are several side effects fighting against theoretical expectations.
If $\ce{Al/Al^3+}$ had been well behaving electrochemical system, $\ce{Al}$ would have created hydrogen bubbles in contact with the solution, with its strongly negative potential $E^\circ = \pu{-1.66 V}$.
With aluminium salt more and more diluted, it converges to water and metallic aluminium becomes more inert, with decreasing cell voltage.

The overall reaction is not $\ce{2 Al + 3 Cu^{2+} -> 3 Cu + 2 Al^{3+}}$ as expected, because, in contact with air, aluminium plates are always covered by a thin, continuous and colorless layer of aluminium oxide $\ce{Al2O3}$ (or alumina) that prevents any aluminum atom to get in touch and react with the aqueous solution it is dipped in. If a scratch happens on this outer surface, the $\ce{O2}$ molecules from air will immediately react and regenerate this protecting alumina layer. So in your $\ce{Al/Cu}$ cell, the metallic $\ce{Al}$ atom does not react : the overall reaction is due to water, as will be explained later. So please never try to make and measure an electrolytic cell with Aluminium plates in a classroom.
You may remove this alumina layer by dipping any aluminium plate into a mercury chloride solution for a couple of seconds. The $\ce{Hg^{2+}}$ ions have the strange property of going through the aluminium oxide layer, and more or less destroy it. The aluminum metal surface is transformed into an amalgam, which reacts immediately with air. You could see white alumina filaments growing at a visible rate on the unprotected metal. It stops when all aluminium amalgam has been oxidized. It is a surprising phenomenon, which was popular in classrooms when mercury compounds were allowed.
To go back to your cell, what happens can be described so : first a few molecules water are crossing the alumina layer through pores or by diffusion and react with aluminium metal according to : $$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$ This reaction happens as soon as the aluminium plate is dipped into water, and stops quickly. So a rather small amount (some nanomoles ?) of $\ce{H2}$ is adsorbed on the metallic surface under the oxide layer. It does not produce any bubbles. Later on, when the plate is coupled with another electrode, this adsorbed hydrogen produces the anodic half-equation $$\ce{H2 -> 2 H+ + 2 e-}$$ $$
\ce{E = E° + \frac{0.059 V}{2} log\frac{[H+]}{p(H2)}= \frac{0.059 V}{2} log\frac{[H+]}{p(H2)}}$$ But its potential is difficult to calculate with Nernst law, because both the pressure $\ce{p(H2)}$ is unknown and must be rather small; and the concentration [$\ce{H^+}$] must also be small (with an unknown pH < $5$). On the other electrode, made of copper, there's no problem: the half-reaction is as expected $$\ce{Cu^{2+} + 2 e^- -> Cu}$$ Its potential can be calculated by Nernst law.
So the cell is not $\ce{Al/Cu}$ but $\ce{H2/Cu}$ and the overall cell tension is not easy to calculate in a class room by ordinary students. So please don't make students measure and discuss an aluminium-copper cell. Replace the aluminium by another metal like zinc $\ce{Zn}$ or lead $\ce{Pb}$. This is an old chemistry high school teacher's advice !
However, let me analyze some of your results. First let's compare the cells with $1$ M $\ce{AlCl3}$ solution, when they are coupled with different copper solutions. When both $\ce{Cu}$ and $\ce{Al}$ solutions are $1$ M, the overall tension is $0.83$ V. As the $1$ M copper solution (from Tables) is at $\ce{E_{Cu, 1M} = +0.34}$ V, it means that the redox potential of the $1$ M $\ce{Al}$ electrode is $$\ce{E_{Al, 1M} = 0.83 \mathrm{V} - 0.34 \mathrm{V} = + 0.49 \mathrm{V} = \ce{E_{H_2,nearly zero M}}}$$ Doing the same calculation with $0.01$ M and $0.0001$ M $\ce{Cu^{2+}}$ solutions and their respective overall tensions $0.78$ V and $0.73$ V, one gets the following $\ce{Cu}$ redox potential $$\ce{E_{Cu, 0.01M}} = 0.78 \mathrm{V} - 0.49 \mathrm{V} = + 0.29 \mathrm{V}$$ $$\ce{E_{Cu, 0.0001M}} = 0.73 \mathrm{V} - 0.49 \mathrm{V} = + 0.24 \mathrm{V}$$ As a consequence, the three Cu potentials derived from measurements show a difference ($0.05$ V), which is nearly exactly what is derived from Nernst law : the theoretical difference should be $0.059$ V. It shows that the copper electrode behaves logically.
Similar calculations made on the aluminium electrode can hardly be carried out, for lack of knowledge about $p$H and pressure of $\ce{H2}$ on the electrode.


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Creating resilient (non-brittle non-water-soluble) verdigris (blue-green copper patina)
First off, I checked and there are indeed a number of questions on copper patinas, but none which deal with resilience.
Context
I own a number of copper writing utensils, which I'd like to get a nice patina on, preferably verdigris.
What I came up with thus far
Looking up the topic here and on wikipedia it seems that copper can patina either in brown or in blue. According to an unsourced claim from wikipedia, exposure to chlorides leads to blue-green coloring, while exposure to “sulfur compounds” [sic] leads to black-brown coloring. I find this highly doubtful, since copper sulfate forms bright blue crystals... so it might be more complex.
What I can conclude from natural experiments (e.g. old copper pennies), is that “normally” copper objects will patina black-brown (Fig. 1), and sometimes will develop some blue-green spots on them, which however bloom out of the penny, appear corrosive rather passivating (i.e. more like iron oxide than aluminium oxide), and are rough and brittle (Fig. 2). This stands in stark contrast with the patina I see on statues and on copper roofs (Fig. 3), which is indeed somewhat rough, but is never brittle, and looks more like proper passivation than corrosion.
I also consulted youtube for verdigris how-to's, and they all end up producing very brittle looking crystals, which are only preserved by adding some sealant to the metal after treatment — for what it's worth the color also doesn't really match what I see from statues/buildings (the architectural patina is more green, less blue, and less vibrant).
Someone brought up in the comments, that the passivating light turquoise patina is specific to bronze and not copper. Comparing Fig. 3 to Fig. 4, it seems it can be obtained on either, but I'm not asking about bronze in this question.
Figure 1, brown patina, as naturally develops with all handled copper objects.

Figure 2, brittle blue patina, as I've seen in isolated spots on pennies and analogous to what a lot of youtube videos show how to create — this is specifically what I don't want.

Figure 3, the glorious coating-like verdigris patina I'd like to get.

Figure 4, this shows the same coating-like verdigris, but on bronze rather than copper.

So I guess my questions are:

Is it possible to obtain a patina on objects which I handle which looks more like turquoise copper roofing and less like old pennies?
If so, how?
Should I be concerned about it being (significantly) toxic?

Addendum (not a self-answer since I still wasn't able to figure this out)
I tried a few experiments, basically taking a copper plate, cutting it to pieces which fit into a falcon tube, and sealing them in with tissue paper soaked in either vinegar (left) or ammonia (right). I left them there for about 1 month at ~10°C.

The first row is what they look like straight out of the tube, the second row is after handling them with dry gloves, the third is after handling them under water for a while. The Vinegar-treated plate basically had all the colour wash off immediately, not sure if the ammonia treated plate has truly resilient colouring either. This can't be what's on statues... Some of the colour indeed bleeds, but there's some layer that forms that's not water-soluble.

Many of the cast objects that have attractive patinas are not copper but bronze (a copper alloy with tin and other components which is much stronger than copper): these won't have exactly the same patina as copper. // Also exposure to "sulfur compounds" will usually create sulfides not sulphates so your doubt about the wikipedia claim is wrong.

@matt_black indeed “The Spirit of Detroit” is bronze — corrected that. But the patina seems indistinguishable from what is seen on copper statues, including the Vercingétorix Monument or the statue of liberty, neither of which look crusty or corroded. Also, copper shingles are made out of copper, bronze shingles seem much rarer.

When the US Statue of Liberty was repaired in the 1980's, there was considerable discussion of how to restore the green patina. Sodium carbonate was used to "sandblast" the statue, followed by ammonium compounds such as NH4Cl to restore the verdigris. See https://www.quora.com/Why-did-the-U-S-decide-to-not-to-continuously-clean-up-the-oxidation-on-Statue-of-Liberty-to-keep-the-copper-color-showing and https://en.wikipedia.org/wiki/Conservation-restoration_of_the_Statue_of_Liberty, for some info. Durability on your copper or bronze material is another question.

I it contains copper , it has some level of toxicity.

I assume artists involved in art restoration or aging may have much more practical, empirical experience with this topic. Most chemist may understand better the underlying chemical principles and processes. But they cannot easily guess the result from basic principles, unless they have by chance experience in this area too.

I'm not sure this would be helpful or not, but just for curiosity, when I check for Verdigris patina, this page came up. Accordingly:

Subtle Copper Patina Recipe: Make a solution with 3 tablespoons of 5% White Vinegar ($\pu{45 mL}$) and 1 teaspoon of table salt ($\pu{5 mL}$) Make sure to stir the solution very well until the salt dissolved completely. Let the copper soak for $\pu{30 min}$. Removed the copper and let it dry.
Green Copper Patina Recipe: Make a solution of $\pu{45 mL}$ of water and $\pu{15 mL}$ of Miracle Grow plant food in exact ratio (3:1). Make sure to stir the solution very well until all  Miracle Grow dissolved completely. Let the copper soak for $\pu{30 min}$. Removed the copper and let it dry. The colour will develop gradually while it dries.
Blue Copper Patina Recipe: Make a solution of $\pu{45 mL}$ of Red Wine Vinegar and $\pu{15 mL}$ of Miracle Grow plant food in exact ratio (3:1). Make sure to stir the solution very well until all  Miracle Grow dissolved completely. Let the copper soak for $\pu{30 min}$. Removed the copper and let it dry. The colour will develop gradually while it dries.

(+1) Nice! Now I have to go out to my shed and look at the ingredients on my box of Miracle Grow, which, sadly, never seems to save any of my plants. I think this is actually my fault: I have the opposite of a green thumb. ;-)

Tried something similar (updated the initial question) sadly it's just like a brittle sheen that forms, even after a month. Breaks apart from just touch, water dissolves most of it :/

I am not surprised. It was from an internet post, most of the time just fabrications. :-)


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Even in an acidic solution with a pH equal to 0, there are only 1 mol/L H3O+ ions, while there are approx. 54 mol/L water molecules
I find this sentence in my high school chemistry book to be confusing:
"Even in an acidic solution with a pH equal to 0, there are only 1 mol/L H3O+ ions, while there are approx. 54 mol/L water molecules."
How did they come to that conclusion and what were their calculations? I understand that the concentration of H3O+ ions is going to be 1 mol/L in a solution with a pH of 0, but how did they find that there are approximately 54 mol/L water molecules in this solution?

Does this answer your question? Why is the concentration of pure water is 55.5 mol/L?

From the solution composition and density, it is easy to determine the water mass in $\pu{1 L}$ of the solution. As the molar mass of water is known, it is easy to calculate the molar amount of water in $\ce{1 L}$ too.
For $\mathrm{pH}=0$, instead of $[\ce{H3O+}]=\pu{1 mol/L}$, rather $a(\ce{H3O+})=1$.  E.g. $\pu{2 M}\ \ce{H2SO4}$ has $\mathrm{pH}=0$ with just a small error, as activity coefficient of $\ce{H3O+}$ is near 0.5 ( and the 2nd hydrogen of $\ce{H2SO4}$ is mostly not dissociated, with $\ce{HSO4-}$ being a weak acid.)


text:
Water dissociation constant $K_w$ in vapour, or are there charges entities ($\ce{H3O+}$, $\ce{HO-}$,...) in water vapour?
When water is in vapour state, to which degree is there still some dissociation with the existence of charged species like $\ce{H3O+}$ and $\ce{HO-}$? (I would be interested by any reference to experimental studies that attempted to measure the amount of such charged species in water vapour.)
Alternatively, would you know what experimental techniques could be used to make a direct experimental measurement of the amount of charges species in a gas?

Eventual thermodynamic equilibrium is highly disturbed by ion creation due ongoing ionization by radon-222 decay and cosmic radiation.

Saying that there are no ions in gas would be a good initial approximation.


text:
Finding concentration of ascorbic acid
Question:
A person applied following procedure for the determination of ascorbic acid (Vitamin C) in a sample solution.
A volume of 20.00 cm3 of the ascorbic acid sample was reacted with 25.00 cm3 of 0.04 M KIO3
solution and excess KI in acidic medium. The remaining I2 is titrated against a 0.1 M Na2S2O3 solution until the purple color of I2 disappears.
If the burette reading (volume of Na2S2O3) is 40.00 cm3 then what is the concentration of ascorbic acid in the sample solution? In the presence of Iodine (I2), ascorbic acid gets oxidized to dehydroascorbic acid.
Use the following equations to support your answer.

(a) 0.10 M
(b) 0.02 M
(c) 0.025 M
(d) 0.04 M
(e) 0.05 M
My attempt:
Moles of Na2S2O3 = 0.1 x (40/1000) = 4 x 10^-3
Moles of I2 remaining = 2 x 10^-3
Moles of KIO3 = 0.04 x (25/1000) = 1 x 10^-3
Moles of I2 formed = 3 x 10^-3
Therfore moles of I2 reacted with ascorbic acid = (3 x 10^-3)-(2 x 10^-3) = 1 x 10^-3 mol = moles of ascorbic acid
Concentration of ascorbic acid = (1 x 10^-3)/(20/1000) = 0.05 mol/dm3
The answer according to the marks scheme is D) 0.04. Where did I go wrong?

Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger  permanent value.
You may find useful  formatting mathematical/chemical expressions/formulas.

I got the same answer. You get the answer in the key if you mistakenly think that the ascorbate was in 25 mL, not in 20 mL. Sorry about the inconsistent significant figures in the calculation below.
$V_{\mathrm{asc}}= 20.00\  \mathrm{mL}$
$n_{\mathrm{\ce{KIO3}}}= 25.00\  \mathrm{mL} \cdot 0.04\  \mathrm{M}$
$\ \ \ =1.0\times 10^{-3}\  \mathrm{mol}$
$n_{\mathrm{\ce{I2}}}= 3 \cdot n_{\mathrm{\ce{KIO3}}}$
$\ \ \ =3.0\times 10^{-3}\  \mathrm{mol}$
$n_{\mathrm{\ce{Na2S2O3}}}= 40.00\  \mathrm{mL} \cdot 0.1000\  \mathrm{M}$
$\ \ \ =4.000\times 10^{-3}\  \mathrm{mol}$
$n_{\mathrm{\ce{I2},remain}}= \dfrac{n_{\mathrm{\ce{Na2S2O3}}}}{2}$
$\ \ \ =2.000\times 10^{-3}\  \mathrm{mol}$
$n_{\mathrm{\ce{I2},asc}}= n_{\mathrm{\ce{I2}}} - n_{\mathrm{\ce{I2},remain}}$
$\ \ \ =1.0\times 10^{-3}\  \mathrm{mol}$
$n_{\mathrm{asc}}= n_{\mathrm{\ce{I2},asc}}$
$\ \ \ =1.0\times 10^{-3}\  \mathrm{mol}$
$[\ce{asc}]= \dfrac{n_{\mathrm{asc}}}{V_{\mathrm{asc}}}$
$\ \ \ =0.05\  \mathrm{M}$

Answer calculated with online calculator.

Now, why is this answer down voted? What is wrong?


text:
Why is collagen fibre autofluorescent?
Why is collagen fibre autofluorescent? Proteins with increased amount of trp, tyr, phe tend to fluoresce but I don't think collagen fibre has increased percentage of any of them. Some say collagen autofluoresces because of the cross links between the fibrils but how?

Any intrinsic trp , tyr, phe will fluoresce under uv excitation, and the emission wavelengths and fluorescence lifetimes in very many proteins is well studied. If the emission is at a different wavelength from these then some cross linking or other photo product could be being formed. Crystallin in your cornea, for example, undergoes a light induced chemistry.

Two proteins, collagen and elastin, are giving autofluorescence and both have a Pyridinoline compound as a cross linker, deoxypyridinoline in collagen and desmosine in elastin:

Thus, these two compounds should be responsible for the autofluorescence shown by these two proteins. Ref.1 has studied deoxypyridinoline, which is one of the two pyridinium cross-links that provide structural rigidity to type I collagen in bone. Ref.2 has given a structural model for desmosine cross-linked peptides. Likewise to collagen, this lysine-derived cross-links in elastin in large measure, contribute to the protein's elastic mechanical properties. The simplified crosslinking patten (Ref.2) is given in Wikipedia:

I have fallen in finding the crosslinking patten in collegen by deoxypyridinoline, but can attributed to the similar pattern displayed by desmosine because closely related structures of two. The crosslinking of collagen increases with age, and it is interesting to mention that the blue autofluorescence intensity of collagen was increased after the fifth decade of life (Ref.3).
In addition, hydroxyproline, which is derived from the post-translational hydroxylation of proline, is a major component of fibrillar collagen of all types, comprising ~14% of the total amino and imino acid content. Hydroxyproline increases collagen elasticity whereas proline reduces it. The 4-hydroxy group in hydroxyproline may have a role in crosslinking as well.
References:

Alessandro Rubinacci, Raffaella Melzi, Maria Zampino, Armando Soldarini, Isabella Villa, "Total and Free Deoxypyridinoline after Acute Osteoclast Activity Inhibition," Clinical Chemistry 1999, 45(9), 1510–1516 (DOI: https://doi.org/10.1093/clinchem/45.9.1510).
Robert P. Mecham and Judith A. Foster, "Total and Free Deoxypyridinoline after Acute Osteoclast Activity Inhibition," Biochem. J. 1978, 173(2), 617–625 (DOI: https://doi.org/10.1042/bj1730617).
D. Dayan, M. Wolman, and l. Hamme1, "Histochemical study of the blue autofluorescence of collagen in oral irritation fibroma: Effects of age of patients and of the duration of lesions," Histol. Histopathol. 1994, 9(1), 11–13 (PDF).


text:
Drano and Bleach reaction?
I’m not sure if anyone can answer this but I might’ve messed up, or I’m just overreacting.
My bathtub has been clogged for about a month now and I used a splash of Drano to see if it would do anything, which I know is less then what’s instructed. However, since it’s a super slow draining tub because of said clog, is it possible that the drano would react to bleach used to clean my tub from a week ago? I’m asking out of pure anxiety. Yes there’s been water moving through the drain overtime but I wasn’t sure if any of the chemicals could be sitting in with the clog and I didn’t know at the time that bleach and drano are a huge no, and I wasn’t sure if they could meet and mix over time.
Bleach used: https://stopandshop.com/groceries/laundry-paper-cleaning/household-cleaners/bathroom-cleaners/bathroom-tub-shower-cleaners/clorox-foamer-bathroom-bleach-trigger-spray-30-oz-btl.html Active Ingredient: Sodium Hypochlorite - 2.40%. Other Ingredients - 97.60% Draino used is mostly Mostly sodium hydroxide (NaOH). https://www.whatsinsidescjohnson.com/us/en/brands/drano/drano-liquid-clog-remover
Before anyone asks, yes I did call poison control. The person I spoke with didn’t seem too concerned and just told me to ventilate the room just in case but I’d like a second opinion because of the possibility of the chemicals sitting I the drain. I wasn’t sure if the bleach would be diluted by now or not.

It is always good to list composition of commercial products, so readers need not to search for it

Bleach used: https://stopandshop.com/groceries/laundry-paper-cleaning/household-cleaners/bathroom-cleaners/bathroom-tub-shower-cleaners/clorox-foamer-bathroom-bleach-trigger-spray-30-oz-btl.html Active Ingredient: Sodium Hypochlorite - 2.40%. Other Ingredients - 97.60% Draino used is mostly Mostly sodium hydroxide (NaOH). https://www.whatsinsidescjohnson.com/us/en/brands/drano/drano-liquid-clog-remover

Both bleach and Drano would coexist  well, in fact bleach solutions contain some hydroxide to stabilize them.
Both are washed away by water, so a week interval would separate them, even if they would react with each other.
Bleach must not come in contact with acids. That would lead to production of poisonous gaseous chlorine. But again, a week apart would separate them. The former would be long time away when the latter arrives.

So even with a clogged drain, there wouldn’t be a chance of anything being left behind? I’m mostly worried about the gases coming out of nowhere, and I’ve read there’s a possibility of explosions.
I’m certainly no chemist, my anxiety has just been racing for the past few hours.

No, these 2 products cannot evolve poisonous gas. Bleach does not have much use for clogged drain. Sodium hydroxide based products do (best if solid, poured by hot water and staying there at least overnight do).


text:
Do all bases release OH-?
Do all bases release OH-? Do all bases have the ability to receive an H+ ion? As an example, CaCO3 is a base but it doesn't release OH- ions.
PS. I am a high school student

What definition of acidity are you using? There are more than one.

Depends on what solvent they are in.

$\ce{CaCO3}$  can react with acids, but it is not a base

Be aware that laconic, not elaborated questions without hints of solving attempt are usually frown upon and often closed on the StackExchange network.  OTOH, well elaborated questions have higher probability to  receive acceptable answers.  How do I ask a good question?.

@Maurice CaCO3 is an Usanovich base, providing negative anion for the reaction.

@Poutnik. $\ce{CaCO3}$ contains a base, namely the ion $\ce{CO3^{2-}}$. But it is not a base itself.

@Maurice It depends on the base definition which depends on the applied acid-base theory. I do agree it is not a base according to the 3 usual acid-base theories. I also agree Usanovich theory is rather obscure one and I have not seen it used in practice. E.g. NaOH(s) is a base by Arrhenius and Usanovich theories, but not by Brønsted-Lowry nor Lewis theories (OH- is). For Usanovich theory,  CaCO3 or Na2O are bases too. There is also a Lux–Flood theory, used in geology or for molten salts, considering as a base the source of O^2- anion.

@Poutnik. Never heard of this Usanovitch theory. Is it important to develop it here ? Why do you mention a nearly unknown theory to a high school student, who is probably in his first year of chemistry ? I am not sure it will help him a lot.

Let us continue this discussion in chat.

Not all bases directly release $\ce{OH-}$ ion, but in case of water solutions, their chemical interactions lead finally to increasing of $\ce{OH-}$ concentration.
It was originally thought all bases are hydroxides, releasing hydroxide ion.(Arrhenius)
It was later realized not only hydroxides cause alkalinity, so acids and bases were redefined in sense of proton ($\ce{H+(aq)}$ in water) exchange (Brønsted-Lowry, B-L). It is not limited to water solvent nor even to liquids, being applicable to solids or gases as well.
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
$$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$
So ammonia does not release hydroxide ion, but captures proton from water. $\ce{CaCO3(s)}$ either releases carbonate anion that is a B-L base, or is able to capture $\ce{H+}$ itself.
Similarly for solid $\ce{NaOH}$.
$$\ce{CaCO3(s) + 2 H+(aq) -> Ca^2+(aq) + CO2(g) + H2O(l)}$$
This should be enough for you for now, the rest is left for optional further reading.

There are several developed theories of acids and bases, each being more general and involving older ones.
Bases by the Arrhenius acid-base theory release $\ce{OH-}$ ion. Like solid $\ce{NaOH}$ when dissolved.
Bases by the Brønsted-Lowry acid-base theory capture a proton like $\ce{NH3}$ or $\ce{OH-}$ or $\ce{H2O}$ or $\ce{CO3^2-}$. It is applicable for all gaseous, solid and liquid phases, not limited to water solutions.
Bases by the Lewis acid-base theory donate an electron pair, like in prior paragraph, or $\ce{NH3}$ reacting with  $\ce{BF}$ (a Lewis acid, accepting an electron pair). This remove the prior limitation to the proton exchange.

(image credit to Wikimedia Commons User Riptide360).
There are therefore often used terms "Arrhenius base", "Brønsted-Lowry base", "Lewis base".

There are even more general and complex acid-base theories in the link above, but that would be out of the needed knowledge scope

No, NH3 is a base. NH3 contains no oxygen - no hydroxide ions - so it cannot release hydroxide ions when dissolved in aqueous solution. I think you are getting the terms alkali and base mixed up - all alkalis contain hydroxide ions - they release them in aqueous solution - so all alkalis are bases but not all bases are alkalis.
Do all bases have the ability to receive H+ ions? Yes - the definition of a base is

a substance that can accept hydrogen ions in water and can neutralize an acid

So all bases have the ability to receive hydrogen cations (H+ ions) - or "protons" as some chemists call them - you may have heard the term "acid donates a proton" - that's speak for H+ ions leaving the dissolved acid compound and joining onto a base as part of a neutralisation reaction (acid + base --> salt - plus H2O if the base is an alkali OR plus H2O + CO2 if base is a carbonate).

There are TWO general types of chemical reactions. Those involving electron pairs or bonds, the electron pairs change their positions in or between molecules, are Acid-Base reactions. The base furnishes the electron pair the acid accepts it. It can be a subtle interaction or a brute force reaction between a strong acid and strong base in water. This means that a myriad of explanations has been presented to try to describe every nuance. Ignore them and learn Chemistry and the mechanisms for reactions. Acidity and basicity depend on the reaction.
The second general class of reactions involves the actual transfer of electrons usually one at a time. These are known as oxidation-reduction or REDOX reactions. The two usually happen together but can be separated. Unfortunately, these also have a wide range of explanations but usually involve neutral atoms, metals, and elements of differing electronegativity. A further complication is that, with imagination, an acid-base reaction can be written or thought of as a redox reaction and vice-versa.
The solution is to learn the definitions to pass the test, but to look into the overall reactions to try to understand more about what is going on and not be bound to restrictive definitions.


text:
Examples of stable organic cations
What are examples of organic cations that are stable at room temperature, can be isolated and that cannot be easily deprotonated by $\ce{NaOH}$? My only guess are various compounds with tetravalent nitrogen (tetraalkylammonium, N-alkylpyridinium, N-alkylimidazolium, etc.).
Are there some other examples?

Peralkylated ammonium is being often dealkylated by NaOH. Tetraalkylammonium salts, or anion exchanging resins based on them, are said not be with OH- ions too long .

Check out organic superacids and proton sponge

'Can be isolated' is rather vague. Cations obviously can't be isolated on their own, so I guess you're asking about isolating it as a salt—but salt stability will probably depend on the counterion. Does any counterion count, or do you specifically have something in mind?

Various trityl salts are commercially available e.g. trityl tetrafluoroborate https://www.sigmaaldrich.com/GB/en/product/aldrich/164577

There are a whole range of ionic liquids available https://www.tcichemicals.com/assets/brochure-pdfs/Brochure_R5058_E.pdf

I agree with orthocresol. And Waylander gave the commertially available compound with  simple carbocation. If you need a exotic one try this, which is also commertially available.

The conjugate bases of some other stable cations, called proton sponges or superbases, is discussed here[1]. Broadly, the compounds discussed have one or both of the following functions:
1,8-diaminonaphthalenes. In these compounds two nitrogen atoms are fixed into positions that allow a proton to be simultaneously covalent-bonded to one and hydrogen-bonded to the other (or symmetrically bonded between the two nitrogens), giving a conjugate-acid cation that may be weaker at redissociating the proton than water would be.
Bis(dialkylamino)cyclopropenimines. These have a structure similar to cyclopropenones, except the ligands on the ring carbon are all nitrogen atoms. The pi-donor character of the nitrogen atoms combined with the aromaticity of the cationic cyclopropenyl ring render the protonated molecule a strongly stabilized cation.
Reference

Lee Belding (2016). Bis(cyclopropenimine) Substituted Proton Sponge Derivatives: Theory, Synthesis, and Applications, Ph.D. thesis, Brock University, St. Catherine's, Ontario, Canada.


text:
Calculate volume of hydrogen gas evolved
MCQ Question:
Ammonia borane in water can release hydrogen upon hydrolysis while forming NH4BO2 as the other
product.
The volume (cm3) of hydrogen gas which will be evolved from the hydrolysis of 309 mg of
ammonia borane (NH3BH3) with a purity of 90 % w/w at 27 °C and at atmospheric pressure 1.0 bar is
approximately,
(a) 66.5 cm3
(b) 224.5 cm3
(c) 673.4 cm3
(d) 893.2 cm3
(e) none of the above
The answer from the mark scheme is C) 673.4 cm3
I took the reaction equation as
NH3BH3 + 2 H2O -> NH4BO2 + 3 H2
However I cannot figure out how to attempt this question. What method should I follow to get 673.4 cm3 of hydrogen gas?

Have you tried the ideal gas state equation pV=nRT?

Transform the initial data into moles. Multiply by 3 to get the amount of H2. Hopefully you know why 3. Then apply pV = nRT to get the final volume V.

Consider ammonia borane is the limiting reagent in the balanced equation. Thant gives you the $n$ in $pV = nRT$. What you don't know is only $V$.

NH3BH3 + 2H2 -> NH4BO3 + 3H2
309mg = 0.309g - use moles = mass/molar mass equation:
mol(NH3BH3) = 0.309/14+6(1)+10.8 = 0.309/30.8 = 0.01mol
Use stochiometric ratio 1:3 to recognize that you multiply the moles of NH3BH3 by 3 to get moles of H2, 0.01mol * 3 = 0.03mol = mol(H2)
Prepare variables for pV = nRT (pressure in Pa * volume in m^3 = 8.314 * moles of gas (H2) * temperature in kelvin) , rearranging equation to give V = nRT/p
To convert degrees C into Kelvin, add 273: 27 + 273 = 300K
To convert "bar" into Pa - one "bar" = 100000Pa
R is ideal gas constant 8.314
Substitute values into equation V = (0.03) * (8.314) * (300) / (100000) = 7.48 * 10^-4 m^3
Convert m^3 into cm^3 by multiplying by 1 million 7.48 * 10^-4 m^3 = 748.26cm^3
The 'purity' thing implies that only 90% of the calculated value is the answer - 784.26 * 0.9 = 673.4cm^-3 which is the answer.


text:
How come in biuret test the copper sulfate doesn't react with the sodium hydroxide?
As I understand the color change in the biuret test is due to the formation of a complex between the peptide (with more than 2 peptide bonds) and the $\ce{Cu^2+}$ ion, from $\ce{CuSO4}$. But before adding the $\ce{CuSO4}$ we add a base to our solution in this case $\ce{NaOH}$, my book says that's because we want an alkaline environment in order for the reaction to happen.
Shouldn't $\ce{CuSO4}$ react with the $\ce{NaOH}$  and then their product $\ce{Cu(OH)2}$ will precipitate?
Or at least both should happen. Every YouTube video I've seen, of the Biuret test, shows both tubes being clear and only a color change, and no percipitation.
In another case for example a Fehling's test for reducing sugars we add an extra compound (sodium potassium tartrate) in order to prevent $\ce{Cu^2+}$ reacting with the $\ce{OH-}$, why not in the biuret test?

Don't trust random YouTube videos. Internet is very ueful but it is also full of incorect information. You have to seek reliable sources from the internet for example, pages maintained by universities ending in edu in the address. See this one for example, https://www.clear.rice.edu/bioc111/bios111_biuret.htm

It does add potassium tartarate to avoid problems, which you just mentioned in the question.

What you need for this test is $\ce{Cu^2+}$ cation. Hope answer here give you insight.

@MathewMahindaratne I've already reviewed that  question but still got no answer to mine. Why the Cu2+ ions don't react with the NaOH as well?

@AChem the videos were just the same Biuret test demonstration, i watched them to see if other people had different findings. I was trying to see if percipitation did happen in any of them.

Fehling's  and Biuret reagents are complex mixtures, formation of which cannot explain by simple reactions. Here explaing why.

@MathewMahindaratne thank you for taking the time to reply and sharing this paper. It'll take me some time reading and understanding this one, when I figure this out I'm gonna post it here. Thanks once again

A similar effect is in the formation of a copper-ammonia complex in the cuprammonium reagent used in making  one type of rayon (https://en.wikipedia.org/wiki/Cuprammonium_rayon). Though it is in a highly alkaline solution, copper doesn't precipitate because it is bound to ammonia.


text:
How do solutions act as solvents?
For example:

Silver chloride, $\ce{AgCl}$, dissolves in dilute aqueous ammonia, so this can be used in qualitative analysis, after reacting $\ce{Ag+}$ ions with $\ce{Cl-}$ ions, to differentiate the white chloride-containing precipitate from a similarly colored cream bromide-containing precipitate ($\ce{AgBr}$) or a yellow iodide-containing precipitate ($\ce{AgI}$).

The ammonia is already dissolved in a solvent (presumably water), therefore the ammonia molecules are sparse and separated in the solution (because low concentration, hence "dilute"), how does the $\ce{AgCl}$ dissolve in an already dissolved solution - where the ammonia is not concentrated-in-one-space-enough to have something 'dissolve' in it (shouldn't it dissolve in the original solvent instead?)
Or do the polarities of the molecules (eg. $\ce{H2O}$ and $\ce{NH3}$) in the new solvent 'combine' somehow to make an even more polar solvent and therefore overcome the dipoles in the polar solute, making it dissolve? If this were case, why is more concentrated ammonia needed to dissolve $\ce{AgBr}$ (less polar than $\ce{AgCl}$ - hence shouldn't need as much concentration of ammonia (extra polarity) to dissolve it)? How do polarities of molecules in solvents combine?

In case of AgCl and ammonia solution, it is not true dissolution by solvent, but chemical reaction, forming soluble [Ag(NH3)2]Cl. AgBr has very limited "solubility" in ammonia and AgI is not soluble
.

For future reference: chemstry.stackexchange allows you to edit chemical formulae with mhchem.  Use it in questions, answers, comments; but because it is something special not all browsers understand well enough, do not use it in the title of a question.


text:
How is NADH or NADPH produced in industrial chemical synsthesis?
I am trying to find out how NADH or NADPH are produced in bulk. These molecules are very important biologically, however, they are also sold as chemical reagents and produced in large quantities. How are these chemically produced industrially? Is that through an organic synthesis (which synthesis?) or do they use engineered bacteria or enzymes and then harvest the NADH produced?
I tried to find this information online, but could not find it anywhere.

search the patent literature. A quick search suggests that state of the art is enzymatic synthesis starting from nicotinamide riboside which is chemically synthesized, but that might not be the dominant method.

Andrew might be on the right track today, and maybe for the past few decades.  However I'd bet that at one time (maybe as recently as 30 or even 20 years ago), the preferred method would have been purifying NAD+ or NADP+ from yeast extract, and if desired, converting to NADH or NADPH enzymatically.

The nicotinamide cofactors, $\ce{NAD+, NADP+, NADH,}$ and $\ce{NADPH,}$ are useful in a variety of enzyme-catalyzed oxidation and reduction reactions in organic synthesis. As Curt F. suggested in the comment,  $\ce{NAD+}$ as known as Coenzyme 1 (which is the longest-known coenzyme), is manufactured by a yeast based fermentation process (at least until 1999; Ref.1).
An early chemical synthesis of $\ce{NAD+}$ was reported in 1957 (Ref.2). In that approach, dicyclohexylcarbodiimide (DCC) was used for the coupling of adenosine monophosphate (AMP) and $\beta$-nicotinamide mononucleotide ($\beta$-NMN). This semi-synthesis was envisioned in 1950 (Ref.3) and achieved by 1957 (Ref.2). Finally, in 1999, Jaemoon Lee & coworkers have achieved the complete chemical synthesis of $\ce{NAD+}$ (Ref.4):

Where conditions are: (i) $\ce{HBr(g)/CH2Cl2}$; (ii) nicotinamide/$\ce{SO2},$ $\pu{210 ^\circ C},$ 90% from $\bf{7}$; (iii) $\ce{NH3/MeOH},$ $\pu{25 ^\circ C},$ $\pu{20 h},$ 80%; (iv) $\ce{POCl3}$ (4 equiv.), $\ce{PO(OMe)3},$ $25$ to $\pu{0 ^\circ C}, \pu{7 h},$ 80%; and (v) $\ce{MnCl2–MgSO4, HCONH2,}$
room temp., $\pu{16 h},$ 78% (58% isolated yield, > 99% purity).
Note: There is a peer-reviewed wore of enzymatic procedure capable of atomic labelling (Ref.5). The most recent published procedure is also semi-enzymatic while using $\ce{H2}$ gas to reduce and $\ce{D2O}$ to lable (Ref. 6).

References:

S. Budavari, editor, The Merck Index, 11th edition; Merck & Co., Inc.: Rahway, NJ, 1989, p. 6259 (ISBN 10: 091191028X).
N. H. Hughes, G. W. Kenner, and Sir. Alexander Todd, "735. Codehydrogenases. Part 111. A Synthesis of Diphosphopyridine Nucleotide (Coxymase), and Some Observations on the Synthesis of Triphosphopyridine Nucleotide," J. Chem. Soc. 1957, 3733-3738 (DOI: https://doi.org/10.1039/JR9570003733).
L. J. Haynes and A. R. Todd, "66. Codehydrogenases. Part 1. The synthesis of dihydronicotinamide-D-ribofuranoside [N-D-ribofuranosidyl-1 : 2(or 6)-dihydronicotinamide]," J. Chem. Soc. 1950, 303-308 (DOI: https://doi.org/10.1039/JR9500000303).
Jaemoon Lee, Hywyn Churchil, Woo-Baeg Choi, Joseph E. Lynch, F. E. Roberts, R. P. Volante, and Paul J. Reider, "A chemical synthesis of nicotinamide adenine dinucleotide ($\ce{NAD+}$)," Chem. Commun. 1999, (8), 729-730 (DOI: https://doi.org/10.1039/A809930H).
Kathleen A. Rising and Vern L. Schramm, "Enzymic synthesis of $\ce{NAD+}$ with the specific incorporation of atomic labels," J. Am. Chem. Soc. 1994, 116(15), 6531-6536 (DOI: https://doi.org/10.1021/ja00094a006).
Jack S. Rowbotham, Adam P. Hardy, Holly A. Reeve, and Kylie A. Vincent, "Synthesis of $[4S-\ce{^2H]NADH,}$ $[4R-\ce{^2H]NADH,}$ $[4-\ce{^2H2]NADH}$ and $[4-\ce{^2H]NAD+}$ cofactors through heterogeneous biocatalysis in heavy water," J. Labelled Compounds & Radiopharmaceuticals 2021, 64(15), 181–186 (DOI: https://doi.org/10.1002/jlcr.3899).

I have included 2021 paper as well.

According to a recent patent[1], the chemical synthesis approach is being replaced by an chemoenzymatic synthesis:
$$\text{nicotinamide riboside}\ce{->[NRK; ATP]}\text{nicotinamide mononucleotide}\ce{->[NMNAT; ATP]}\text{NAD}$$
where NRK = nicotinamide ribokinase and NMNAT = nicotinamide mononucleotide adenyl transferase.
The starting material nicotinamide riboside can be chemically synthesized, as described in another patent[2] (though I'm not sure what synthetic approach is used most for current commercial production).
[1] CN112437813A, Method for industrially producing NAD (nicotinamide adenine dinucleotide) by enzyme method
[2] US11242364 Efficient and Scalable Syntheses of Nicotinoyl Ribosides and Reduced Nicotinoyl Ribosides, Modified Derivatives Thereof, Phosphorylated Analogs Thereof, Adenylyl Dinucleotide Conjugates Thereof, and Novel Crystalline Forms Thereof


text:
Activated carbon filter desorb smells with dropping temperature?
I have a HEPA filter air cleaner with 16 pounds of activated carbon.  It's suppose to last up to years, but my experience is that after a few months (if not weeks) of operation under a light load like a bedroom, the air starts to exude a sweetish smell that is sometimes like honey and sometimes like citrus fruit.  This has been explained by the vendor as possibly due to the fact that the activated carbon is derived from coconut hust, and I've verified from Wikipedia that this is typical.
I noticed repeatedly over the years, however, that as the temperature drops a few degrees (e.g., from 22 to 19 Celsius), the honey smell seems like an unpleasant sour smell.  It only seems that way during downward transitions in temperature.  Once the temperature is at its low, it starts to smell like fruit again.
What mechanisms in activated carbon could explain this?
2023-02-09 afternote: I wonder if the cause could be related to humidity.  Small swings in temperature indoors are correlated with high swings outdoors, which correlate with relative humidity.

Could your air filter produce ozone which smells sweet? Does the device have an ionizator or filter based on high electric fields? Could the ozone pass more or less through the activated carbon layer at certain conditions, assuming the ozone is going through the activated carbon? Or the availability of the ozone for your nose changes by other means with the conditions.

The vendor is apparently handwaving to satisfy a customer with a natural reason. Activated carbon is usually prepared at 1000 C! No fruity smells or odors should survive at this temperature from any natural sources whether it is coconut or even if it is made from bones. In short fresh activated carbons do not have any odor.
Activated carbons have very large surface areas, this means they can adsorb anything, and since the filter is designed to "trap" bacteria, fungi, mold  and spores mechanically, these organisms can decompose/react with other adsorbed stuff collected over a period of time (other domestic odors) to generate volatile molecules. By luck, some of them might smell nice and some don't. Indoor air is full of man-made chemicals at very very low but measurable concentrations. They are called volatile organic carbons (VOCs). Imagine countless detergent/soap odors, flame retardants, creams, shampoos, perfumes, decomposing glues (on the flooring, carpets), and countless other things. So unfortunately, nobody can pin-point the exact identity of those fragrances and mal-odors until and unless the air sample is analyzed by professional services. What you are describing is not a repeatable scientific phenomenon, it must be very specific to your indoor air and building air circulation.
If this is a concern and if bad odors persist in indoor air, seek professional assistance for indoor air quality and/or medical advice.

I appreciate your informative response.  It confirms what I suspected.  But I was wondering, would you be aware of an absorption/desorption mechanism that might explain why the exuded smell becomes more pronounced as the temperature decreases, but this does not persist after the temperature reaches its low?  Yes, smell changes, but it is also less once the temperature stabilizes.  I've seen this with replacement filters over the years.  Thanks.

OK, thanks again.  It's just that the smell is noticeably reduced at slight higher temperatures.  I hardly notice it at above 20 Celsius, but it's quite noticeable below that, even though it's greatest during the downward transition rather than at the low end.


text:
When styrofoam shrinks from pressure, why doesn't it go back to normal after decreasing the pressure?
In this article (https://oceanexplorer.noaa.gov/explorations/16carolina/logs/sep3-2/sep3-2.html) and articles like this one, Styrofoam cups were taken to the depths of the ocean in an AUV to demonstrate the extreme pressure of deep water. Once the Styrofoam is brought back up to normal atmospheric pressure, why doesn't the air rush in and the Styrofoam expand again?

The structure of styrofoam isn't as perfectly elastic as you think
Most styrofoam is created by various mechanisms that blow gases into the polystyrene matrix as it cools. One common process involves creating small beads saturated with pentane which are then expanded by treatment with steam (the pentane expands creating bubbles in the beads). The resulting beads can be extruded to create specific shapes (polystyrene is a thermoplastic). This process also creates many air-filled voids.
It might be thought that the closed cells would simply compress when subject to high external pressure and would then bounce back when the pressure is released. But this assumes two things: the cells are perfectly closed and the physical properties of the polystyrene matrix is perfectly elastic (like a rubber balloon). Neither are true. Polystyrene is not perfectly elastic and the cells are not perfectly closed. External pressure can easily be enough to damage the walls of the closed cells so they don't "bounce" back or they don't stay closed. And the overall structure is often easily penetrated by external stuff like water or gases other than the original pentane or air.
So external pressure will often alter the structure in many ways that mean it does not recover its original form.

For one,

Foams are materials formed by trapping pockets of gas in a liquid or solid.

(Wikipedia's definition of foam)
and styrofoam is a form of extruded polystyrene consisting of "95-98% air" which yields a "normal density range of $11$ to $\pu{32 kg/m^3}$" (for comparison: distilled water has a density of $\pu{1000 kg/m^3}$, the ocean's water, containing minerals exceeds this value).  During the dive into the deep sea, the small cavities are irreversibly compressed by the high pressure while air escapes the material. As the pressure acts from all directions on the cups, the deformation isn't the same as e.g., accidentally stepping on one of the cups.  The forces are much larger than a manual compression of the sponge next to the kitchen sink or/and the one in the bathroom.

Assuming the cups are indeed made from expanded polystyrene (styrofoam), it is useful to look at the glass transition temperature of polystyrene. Wikipedia claims it is around 100 °C. So below that, polystyrene is a glassy solid.
In fact, polystyrene at ambient conditions or at temperatures in the deep sea is a rather brittle material that cannot be deformed to high strains without failure. You might know it from the lab in the form of disposable petri dishes, or otherwise from CD cases. These also tend to be cracked easily. The same is true for expanded polystyrene. The inner structure of the material is basically irreversibly shattered at high pressure. It is analogous to a shattered glass, only that in that case the damage is more obvious, and you would not wonder why the original shape is not recovered after removing the force.


text:
Could you tell me the name of this compound please?
Could you please tell me make of this compound?

That is a branched polypropylene glycol terminated with amine groups, a type of polyetheramine, available from BASF as Baxxodur® EC 310, or trimethylolpropane tris[poly(propylene glycol), amine terminated] ether (polyetheramine T 403) with CAS No. 39423-51-3. BASF markets it as a curing agent primarily for epoxy systems.
There is only one branching point, so it could be considered a single-generation dendrimer, except that it lacks perfect symmetry.
The image seems to be from the Sigma-Aldrich website (Merck):

There you can purchase the polymer with $M_n=440$ which has a degree of polymerization on each branch of ~n=1.8.


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Do whey proteins precipitate with heat?
I am currently studying about the biochemistry of yoghurt and when reading about the effect of heat in milk proteins specifically whey, my book says that heat can denature whey proteins.
So after the denaturation will the whey proteins coagulate and then peripitate?
If that is the case why in yoghurt that hasn't been homogenized, a layer of whey protein and fat forms on top. For example traditional greek yoghurts all form this top layer because they are left to cool down without any homogenization.
Does the denatured whey proteins go up or down?

Yogurt that hasn't been homogenized is far more complicated than just whey proteins.  Usually Greek yogurts will have up to $10^8$ "colony-forming units" or "CFUs" of bacteria per gram.  Practically, each CFU is a cell that weighs on the order of 1 pg, so each gram of yogurt has about 0.1 mg of live bacteria.  Many of those cells have flagella and other protrusions that can interact with denatured whey proteins and change their apparent buoyancy...

And then there's coagulation: cheese, labne, etc.


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What is the inconsistency in calculating the differences in Na2O and NaOH in solids (v2)?
(Note that this question has been fully edited for clarity, SE help advised not to delete the question and post another).
Background:
I have a computer model in which I feed it input data from various feedstocks usually containing sodium. My calculations input NaOH data (because it's an industry standard and easy to use for the user) but calculate Na2O for output quantities for the sake of consistency as everything else is in oxide form (e.g. SiO2, Al2O3, H2O etc.). Note that the full model incorporates the following metrics from each inputted feedstocks:

Feedstock Elemental Mol Quantities: (e.g. mols Si, mols Al and mols Na (etc.) from every feedstock, note that mols H instead of mols H2O is used for consistency in using elemental mols instead of changing to an oxide just for water).
Feedstock Oxide Weight Quantities: (e.g. wt% SiO2, wt% Al2O3, wt% Na2O (etc.) from every feedstock).
Feedstocks Solid wt%s: (i.e. $Solids_{i,wt\%s} = Feedstock_{i,Total,wt\%} - H_2O_{i,wt\%}$; the summed wt%s of all non-H20wt% quantities in a given feedstock)
Feedstock Liquids wt%: (i.e. $Liquids_{i,wt\%s} = Feedstock_{i,Total,wt\%} - Solids_{i,wt\%s} = H_2O_{i,wt\%}$

One such feedstock which is commonly used in the model are the standard reagent grade 97 wt% NaOH pellets. Knowing the reaction between NaOH and Na2O (i.e. 2NaOH -> Na2O + H2O), if I wanted to calculate the composition of a 1 g NaOH pellets I get the following (note that I did try to put everything in Chem SE table format but it was too difficult to do, format and the help page didn't provide too much clarity from my POV):
 1. First, the pellet's composition with respect to NaOHwt%:
Calculations are as follows:

NaOHwt% = 97 wt% NaOH as per the specification sheet mentioned in the link above.

Note that it has been assumed that $H_2O_{wt} = 100_{wt\%} - Solids_{wt\%} = 100_{wt\%} - NaOH_{wt\%} = 3_{wt\%} H_2O$

$mols Na = \frac{97 wt\% NaOH}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g NaOH}{100 wt\% NaOH} \times \frac{1 mol NaOH}{40 g NaOH} \times \frac{1 mol Na}{1 mol NaOH} = 0.0243 \frac{mols Na}{1 g pellets}$

$mols H = \frac{3 wt\% H2O}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g H_2O}{100 wt\% H_2O} \times \frac{1 mol H_2O}{18.02 g H_2O} \times \frac{2 mols H}{1 mol H_2O} = 0.0033 \frac{mols H}{g pellets}$

$Solids_{i,wt\%s} = 100_{wt\%} - 3_{wt\%,H_2O,i} = 97 \%$

$Liquids_{i,wt\%s} = 3_{wt\%,H_2O,i} = 3 \%$

2. Secondly, the pellet's composition with respect to Na2Owt%:
Calculations are as follows:

Note that Na2Owt = $ \frac{97 wt\% NaOH}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g NaOH}{100 wt\% NaOH}  \times \frac{61.98 g Na2O / mol Na2O}{40 g NaOH /mol NaOH} \times \frac{1mol Na/ mols NaOH}{2mols Na/mols Na2O}  = \frac{0.752gNa2O}{1 g Pellets} = 0.752 \frac{g Na_2O}{1 g pellets} \times 100\% = 75.2 wt\% Na_2O$

The resultant H2Owt% as per the above 3wt%, I presume instead needs to be  of $H_2O_{wt} = 100_{wt\%} - Na_2O_{wt\%} = 24.8_{wt\%} H_2O$ which seems ludicrous...

$mols Na = \frac{75.2 wt\% Na_2O}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g Na_2O}{100 wt\% Na_2O} \times \frac{1 mol Na_2O}{62 g Na_2O} \times \frac{2 mols Na}{1 mol Na_2O} = 0.0243 \frac{mols Na}{1 g pellets}$

$mols H = \frac{24.8 wt\% H2O}{100 wt\% Pellets} \times \frac{100wt\% pellets}{1 g pellets} \times \frac{1 g H_2O}{100 wt\% H_2O} \times \frac{1 mol H_2O}{18.02 g H_2O} \times \frac{2 mols H}{1 mol H_2O} = 0.028 \frac{mols H}{g pellets}$

$Solids_{i,wt\%s} = 100_{wt\%} - 24.8_{wt\%,H_2O,i} = 75.2 \%$

$Liquids_{i,wt\%s} = 24.8_{wt\%,H_2O,i} = 24.8 \%$

In can be seen that, although the mols Na are the same, neither the mols H nor the solids or liquids parameter match up. Which doesn't seem to make sense to me?
 3. Finally, the pellet's nonsensical composition with respect to Na2O:

The only solution I've managed to balance with the initial NaOH composition with respect to the Na2Owt doesn't make sense as I've had to add a nonsensical "Other solids?" compound which accounts for the missing solids to ensure that the mols of Na, H and solids and liquids quantities remain consistent.

My questions from the above are as following:

What is the missing component (maybe the "other solids" quantity) which resolves the inconsistency between the NaOH pellet's composition with respect to the NaOHwt and Na2Owt?

Is the H2O wt% assumption a good one to make due to the pellet's hygroscopic property?

Is it better to only assume the 2 NaOH <> Na2O + H2O reaction for solids and therefore only use it in aqueous solutions with NaOH or Na2O?

Should NaOH pellets only be quantified by their NaOH composition and not Na2O for better representation and accuracy?

Did you think about what is the last 3% of NaOH? I think it is water because NaOH is hydroscopic.

I had thought of that but assumed instead it was just trace elements of something else that exist from manufacturing.

There is near always some Na2CO3(s), as NaOH(s) reacts with aerial CO2. You obviously cannot consider Na2O taking 97% of NaOH. // As 2 NaOH <> Na2O + H2O is formal conversion, there is fixed mass ratio $\frac{2*M(\ce{NaOH})}{M(\ce{Na2O})}$ between them, and their molar ratio is 2 : 1.

I have just made a complete edit to the question so it's (hopefully) more easier and direct :) Also, Currently I'm leaning towards throwing out the 2 NaOH <> Na2O + H2O reaction for solids and only using it in aqueous solutions.

The first table is wrong, regarding molar amounts of Na versus H. If it were pure NaOH, the molar ratio would be 1 : 1. // Remember the 2 NaOH <> Na2O + H2O is purely formal reaction, describing NaOH/Na2O equivalence. So about 75% Na2O(s) + 25 H2O(l) in solid NaOH is purely formal. // Similarly, P in fertilizers is expressed conventionally as P2O5 in spite of there is no P2O5 either in solid ones either in liquid ones.

That's a bit confusing, thank you for the reply but could you please elaborate more?

FIRST THERE IS NO REACTION $\ce{2NaOH -> Na2O + H2O}$ NO ! This transformation from $\ce{NaOH}$ to $\ce{Na2O}$ never happens. As Poutnik says, it is purely formal. It dates from the $19$th century where chemists thought that all oxygenated mineral compounds were made of "jointed" oxides. For example $\ce{Na2CO3}$ was thought to be $\ce{Na2O + CO2}$. Then $\ce{H2SO4}$ was thought to be made of $\ce{H2O + SO3}$. And of course $\ce{NaOH}$ was thought to be made of $\ce{Na2O + H2O}$.  But there are no oxides in $\ce{Na2CO3, Na2SO4, H2SO4}$ and of course in $\ce{NaOH}$.

There are two mistakes in this post.

The formula used for calculate the amount of $\ce{Na2O}$ has a wrong unit. It does not give a weight as it should. It gives a percentage. This mistake is easy to prove by simply reasoning with units. In the first numerical example, a weight ($16.7$ g) is successively multiplied by a ratio of grams, and then by a ratio of moles. The final result must be expressed in grams, and not in percent. This observation is valid for both the first silicate solution and for the final pills.
The second mistake will be shown with the second calculation.
Here, one starts from $97$ g pure $\ce{NaOH}$, containing $97/40 = 2.425$ mol $\ce{NaOH}$.
These $2.425$ mol NaOH are not formed, but may be considered as formed by hydratation of $1.2125$ mol $\ce{Na2O}$, which weighs $1.2125 · 62$ g = $75.175$ g
The difference in weight $97 - 75.175$ g = $21.825$ g is water, but it is not the adsorbed water fixed at the outside of the $\ce{NaOH}$ pills. It is the amount of water that is theoretically necessary to add to $75.175$ g $\ce{Na2O}$ to get the final mass ($97$ g) $\ce{NaOH}$ after hydratation reaction. This theoretical water is not present in the pills. The amount of adsorbed water is only $100$ g - $97$ g = $3$ g.

The same mistakes have also been done in the first numerical example. But it is not so easily discovered, as water is used both as solvant and as theoretical water necessary to transform $\ce{Na2O}$ into $\ce{NaOH}$. These two sorts of water are not really separated in the calculations of the first numerical example.

1: I had simplified the formula but have added the exact steps in the newest edit just then, the resulting number is unchanged. 2: I'm still a bit confused here. From a mass balance POV, when I calculate the mols of Na for each scenario (i.e. using Na2O and NaOH) I get 0.0243 mols Na for both scenarios, however, when I do the same for water I get 0.0138 mols H2O and 0.0017 mols H2O respectively supposedly with the same system? How does that make sense with a different total number of mols? Should the reaction between Na2O and NaOH be ignored for the solid NaOH pellets?

I have just made a complete edit to the question so it's (hopefully) more easier and direct :)

NaOH does not react like you think and does not loose water and produce Na2O by heating.

Let me now comment your blue tables. They are particularly ridiculous.
The first one is called "NaOH pellets (NaOH)". The $3$ first lines give the composition of the pellets in weight percent. OK. Then there is a subtitle called "Elements (moles). This information has no meaning. Or it would have a meaning if some more information was given, saying that the given values in the two next lines are related to $1$ g of substance. One gram ! Let me study these values to convince you about this information.
$1$ gram of your pellets contains $0.97$ g $\ce{NaOH}$ and also $0.97/40 = 0.024$ mole $\ce{NaOH}$. There is also $0.03$ g $\ce{H2O}$ and $0.03/18 = 0.00166$ mol $\ce{H2O}$, which correspond is to $0.00333$ mole H. These values are the same as in your table. But I don't understand what is the use of calculating the amount of $\ce{H}$ atoms in the $3$ % of adsorbed water. Why ?
The second table is called "NaOH Pellets (Na2O)". The same observation can be made about the percent and mass units. It should be specified that the given amounts in mole are only valid for a $1$ gram sample.
Now let's comment about the moles of $\ce{Na2O}$. I need now to recopy the comment I wrote half an hour ago about the correspondance $\ce{NaOH - Na2O}$. Here is my comment :
In the $19$th century chemists thought that all oxygenated mineral compounds were made of "jointed" oxides. The nature of this joint was a mystery. For example $\ce{Na2CO3}$ was thought to be made of $\ce{Na2O + CO2}$. Then $\ce{H2SO4}$ was thought to be made of $\ce{H2O + SO3}$. And of course $\ce{NaOH}$ was thought to be made of $\ce{Na2O + H2O}$. But there are no oxides in $\ce{Na2CO3, Na2SO4, H2SO4}$ and of course in $\ce{NaOH}$. This old theory is not stupid, but it is no more useful today in chemistry; and it is still in use in geology, because it respects the oxidation numbers and the valences.
Let's go back to the second table "NaOH Pellets (Na2O)". It mentions that the $97$ % $\ce{NaOH}$ "corresponds" to $75.2$% $\ce{Na2O}$. This is right, and you have done this calculation yourself. But this is of no interest any more, except for geologists. The rest ($24.8$ %) is the sum of the really adsorbed water at the surface of the pellet + the theoretical amount of water necessary to transform $\ce{Na2O}$ in $\ce{NaOH}$. Why doing this sum which is of no interest ?
Later on, in the middle of the blue table, under "Elements", there is $0.024$ mol $\ce{Na}$ and $0.028$ mol $\ce{H}$. It should be stated that these amounts are present in $1$ gram pellets. These numerical values are only true if the $0.028$ mole $\ce{H}$ atoms are coming from the theoretical water necessary to transform $\ce{Na2O}$ into $\ce{NaOH}$. Has it any meaning ?
The last lines of this second blue table are specially ridiculous. They state that the pellets are made of $24.8$% liquid, as if constitution water was liquid in $\ce{NaOH}$ as analyzed by a 19th century technique.
I begin to get tired tomato disagreeable comments about your tables.
The worst is in the last blue table, where the author believes that some unknown solid is necessary to explain the composition of the pellets. The author has simply given the weight percent ($75.2$%) of $\ce{Na2O}$ necessary to give the real $\ce{NaOH}$ believing that it was the weight percent of $\ce{NaOH}$. The "missing solid" is not a solid. It is simply the theoretical amount of water necessary to transform the non-existant $\ce{Na2O}$ into $\ce{NaOH}$. I will not go any further today. How silly !

I had thought I was clear about the 'mols sections' due to their names but apparently not, my mistake, I have provided all the extended equations in the newest edit to be certain along with an explanation/context as to why mols H and not mols H2O was used in the background (it was because of other things in the model). I was also previously unaware of the whole NaOH and Na2O mix-up from 19th century geologists and am surprised that it's still a thing today as I regularly (and now confusingly) see some research literature with Na coming from Na2O instead of NaOH in some solutions (etc.).

@Hendrix13 Consider all NaOH/Na2O thing as equivalent masses or molar amounts, without any reaction ever happening. Like, if I have 1 kg or 1 mol of NaOH, how much it would have been, if it had happened to be Na2O. // Similarly, table salt NaCl food income can be evaluated as Na, without thinking you swallow metallic explosive sodium.

Yes good explanation @Poutnik, and I see what you mean about the NaOH/Na2O. I just didn't understand fully how the relationship worked with the H2O. I knew something was wrong and didn't make sense which is why I specifically referred to the "nonsensical composition" in my final blue table.


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What compound was drawn on this sidewalk?
A friend found this in the sidewalk of Kent town of Canterbury:

What compound is this, and why would its picture be there?

TNT aka trinitrotoluene.

Looks like a granite board with poorly drawn molecule of what would be TNT if not for weird $\ce{ON2}$ group. What symbol are you referring to, exactly?

@andselisk The entire compound: NO2, CH3, NO2, ON2, surrounding an O.  My last chemistry class was in 2009.

What is the connection with Kent town?

@BuckThorn No idea, really; there were several cryptic images like this in the sidewalk there of which he shared images.  For all I know it could be a Guy Fawkes reference, but I'm over the pond in "New England" so that's a guess out of the blue.

@andselisk: Nice catch. I'm sure caver didn't understand difference between $\ce{-NO2}$ and  $\ce{-ON2}$. Caver probably meant to go $\ce{O2N -}$, but seemingly lost in the transition! :-)

TNT had not been discovered around the time of Guy Fawkes, but looks like there's an answer posted with a good explanation....

Either this post should be fixed or closehammered. I mean seriously, calling chemical formula a "symbol"? Also "Simple question: what is it ?" isn't making it any favors.

@Mithoron Being much more familiar with mathematics and physics, my concept of a formula is a bit different, and I wasn't sure how to refer to what I know now is a "chemical formula".  I didn't want to confuse it with a "chemical equation", which I know now is something else different, indeed.  Anyway, not sure what "closehammered" is on StackExchange, or how to "fix" the post, but I do appreciate the feedback I've received, which has provided the insight I'd hoped when I added Chemistry to my Stack account.  Thanks, all !

Certainly, it's meant to be trinitrotoluene, TNT, with a "typo". Perhaps a craftsman tried to copy an image, such as that from Wikipedia, below, which shows the oxygen atoms outside the ring:

As to the reason for that sidewalk plaque:
Faversham, ~16 km from Canterbury, produced TNT, and on April 2, 1916, some 200 tons of TNT exploded, killing 115 people (apparently set off by other munitions -- it is rather difficult to detonate TNT alone). "News of the blast was kept secret due to the wartime sensitivities at the time. It is now recognised as the worst ever disaster in the UK explosives industry."
While there's a monument at the mass burial in Faversham, below, the plaque might be an additional memorial.

This is reminiscent of the Halifax, NS, CA munitions explosion, December 6 1917. with "equivalent energy of roughly 2.9 kilotons of TNT", killing about 2,000.
However, it might also be a crafty advertisement, if in somewhat poor taste, for either:

TNT Improv and Sketch Comedy Society in Canterbury at University  of Kent
TNT Delivery

There is an entire book about the disaster.

The inner of the structure formula neither is a character, nor a number.  It is a Thiele ring (suggested by Johannes Thiele 1865-1918 by 1899) to describe aromaticity, after Kekulé's approach already was disseminated.  It was an old practice (say up to about 1960s) you still may encounter in the electronic age, e.g.

when you neither want to draw one structure instead of the resonance structure on the very left, and the one on the very right, or because your sketcher program (ChemDraw, Chemdoodle, etc.) falls in this form.  Or, because you encounter such a drawing you want to assign a (kekulized) SMILES string:

(credit to both drawings: Noel O'Boyle, John W. Mayfield. We need to talk about... Kekulization, Aromaticity, and SMILES.  254th ACS National Meeting Washington, August 2017.  Archived on Noel O'Blog).
However, because it represents recurrent potential pitfall when counting how many electrons actually are present, especially if chemists from different experience/with different convention in their mind meet each other:

(Robert Grossmann The Art of Writing Reasonable Organic Reaction Mechanisms., already used earlier here)
especially in charged states (e.g., the assumed intermediate anions of Birch reductions), the safer (hence better) approach is to use mesomeric resonance formulae instead (and to recall that benzene isn't 1,3,6-cyclohexatriene with an alternating pattern of single and double bonds).

The Thiele ring is awfully small in the picture posted by the OP, and almost looks like an oxygen atom.

@Karsten, yes, it appears a craftsman without a technical background tried to copy the image. BTW, I've had difficulties getting a grave-stone properly engraved -- accuracy is not a priority for some commercial operations.

Probably the craftsman had observed that the whole image contains $4$ groups of $2$ capital letters with one number ($2$) in index. In three of these groups the index was at the end of the group. As it was not the case in the left group, he probably decided that the "internal index" was a mistake, that had to be corrected.


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General safety of aromatic vs. aliphatic hydrocarbon solvents
I am not a chemist and I could use some help from those who know. I am wondering about some properties of hydrocarbon solvents that are commonly used for degreasing in other fields. It seems many of the commonly available aliphatics with >40degC flashpoints have Kauri-Butanol values in the 30's. Aromatics (like those advertised as high flash aromatic napthas) in the C9-C10 range with >40degC flashpoints have Kauri-Butanol values in the 90's. Aliphatics are much more commonly used in this role.
My questions:
Assuming there is minimal benzene and VOC's are not a consideration (closed system)
Are the C9-C10 range aromatics (high flash aromatic naphthas) inherently more dangerous than the aliphatics of similar flash point range (mineral spirits)? I know they are more irritating short term in terms of odor and the like, but are they actually more toxic?
Would you expect greater performance of aromatics vs. aliphatics in degreasing tasks based on the Kauri-Butanol values?
Thank you!

Generally, yes.  But it's something that can be minimized with safety precautions such as good ventilation.  A xylene dip, followed by a warm caustic bath works very well for alkyd resin synthesis glassware.  (Wear long gloves and safety goggles).

Thank you for the response Robert. Advice taken.


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Orientation of spin polarization in O2
Oxygen gas in ground state is spin polarized. Is the direction of its spin (significantly) correlated to the orientation of the double bond?
I googled "orientation of oxygen magnetic moment" and couldn't find relevant result.
My guess is that there is no significant correlation between the orientation of the double bond and the orientation of the magnetic moment because the spin polarization arises from spin-spin coupling, not spin-orbit coupling. The lack of spin-orbit coupling (of relevant strength) means the spin "doesn't know" the orientation of the molecule and the different energy state that corresponds to different spin direction is essentially degenerate.
Is my thinking here correct?

Look for angular momentum coupling cases, such as Hund's rules, case (b) should be appropriate here. In this case the spin angular momentum is fixed in space (S does not couple with the intermolecular axis) and the molecule rotates underneath it.

got it. thanks.


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Why do Mn and Tc have lower melting points with respect to their immediate neighbors but Re doesn't?
Manganese (Mn) and technetium (Tc) show a sharp decline in their melting point in comparison to the neighboring transition elements, which is unexpected while following the general trend which says the extent of metallic bonding increases with the availability of unpaired electrons. rhenium (Re) however does not show the same tendency despite belonging to the same group.
Upon revisiting their electronic configurations, we can notice technetium is actually d⁶ which could've explained the disparity had technetium shown it instead of rhenium. Somehow manganese and rhenium are both d⁵ yet the sharp decrement is observed in case of manganese (d⁵) and technetium (d⁶) instead. This should suggest melting point isn't directly related to unpaired electrons or unpaired electrons isn't related to extent of metallic bonding, both of which sounds contradictory to the phenomena shown by the rest of the transition elements? What am I missing here?

Note that prediction of melting points is tricky business, with many variables, sometimes involving even Special relativity. ( Non SR prediction for Hg is 120 °C off, SR one within 1 °C error )

See: https://chemistry.stackexchange.com/q/4766/17368

https://chemistry.stackexchange.com/questions/4766/melting-and-boiling-points-of-transition-elements https://chemistry.stackexchange.com/questions/39786/what-is-the-lattice-structure-of-manganese

The melting point depends on how available the valence electrons (which are mostly paired in metallic bonding) are for metallic bonding, and that is not just a function of the number of such electrons formally existing in the atoms.
Most metals aside from alkali metals have at least two valence electrons that are relatively easy to ionize and thus also to delocalize into metallic bonds. Getting three or more metallically bonded electrons per atom is where you begin to run into some resistance, and thus you begin to see differences not only in melting point but also in density.
Let's look at third ionization energies of the relevant elements in Periods 4 and 5. All data are from Wikipedia and are in Joules per mole.
$\ce{Cr}=2987, \color{blue}{\ce{Mn}=3248}, \ce{Fe}=2957$
$\ce{Mo}=2618, \color{blue}{\ce{Tc}=2850}, \ce{Ru}=2747$
The higher ionization energy of the blue elements implies that getting more than two metallically bonding electrons per atom is more difficult for those elements than for their neighbors. With that come the lower melting points and also reduced densities. Manganese at $7.21$ g/cm³ is well under the average of chromium ($7.15$) and iron ($7.87$).
So why do we see this crest in third ionization energies? Despite the different electron configurations given to the atoms, the $+2$ ions all have the common configuration $\text{[noble gas]}(n-1)d^m$. When $m=5$ for the dication, as with manganese and technetium, the next stage of ionization requires breaking a configuration that has relatively strong exchange-enetgy stabilization.
We cannot extend this comparison to rhenium because the relevant ionization energy data are not available for all the elements in that part of the Periodic Table. But we should expect the $5d$ electrons to be more readily available for metallic bonding with rhenium on relativistic grounds (we do know that rhenium has a lower third ionization energy, $2510$ J/mol, than either manganese or technetium, per the table cited above). The contraction of the $s$ orbitals due to the increased mass-energy of those electrons in heavy elements makes them better at shielding other orbitals. The electrons in those other orbitals are therefore less strongly bound to the nucleus. This "reverse inert-pair" effect is prominent with valence $d$ orbitals in the sixth-period transition elements.


text:
Why didn't copper act as a reducing agent
In lab, we took $\ce{CuCl_2(aq)}$ and poured it into a cup. Then we took some aluminum foil and put it inside and started to stir. I could see the aluminum getting eaten by the copper because it was a better reducing agent. However, when I put silver instead of aluminum, nothing happened. I know it's because the standard cell potential of silver is greater than that of copper, but why didn't the copper act as a reducing agent?

What would the copper reduce?

Cu++ from CuCl2 has already given its electrons to 2 Cl-.  Aluminum metal is the reducing agent, giving them back to Cu++ --> Cu metal.  Similarly, Copper metal would reduce Ag+.

Copper is already oxidized as $\ce{Cu^2+(aq)}$ and silver reduced as $\ce{Ag(s)}$. Copper would act as a reducing agent in $\ce{Cu(s)}$ and $\ce{AgNO3(aq)}$ scenario.
Using $\ce{CuCl2}$ instead of $\ce{CuSO4}$ is a mistake, if $\ce{Ag}$ is involved, as forming insoluble $\ce{AgCl(s)}$ would complicate things. Eventually, $\ce{Cl-(aq)}$ may keep $\ce{Ag/Ag+}$ potential below or near $\ce{Cu/Cu^2+}$ potential:
$$E = E_{\ce{Ag/Ag+}}^\circ + 0.059 \log{[\ce{Ag+}]} =  E_{\ce{Ag/Ag+}}^\circ + 0.059 \log{\frac{K_{\mathrm{s},\ce{AgCl}}}{[\ce{Cl-}]}}$$
(simplified replacing activities by concentrations)
But, forming $\ce{AgCl(s)}$ layer would freeze it kinetically.

The answer lies in the redox potential chart:
Reactions:
$\ce{2 Alumimum metal + 3 Cu++ -> 2 Al+++ + 3 Copper metal}$
$\ce{2 Silver metal + Cu++ -> 2 Ag+ + Copper metal no reaction}$
Drawing the half reactions:
$\ce{Al+++ -> Aluminum metal   - 1.66 V}$
$\ce{Cu++ -> Copper metal       + 0.34 V}$
$\ce{Ag+ -> Silver metal        + 0.80 V}$
We can see why Copper ions react with Aluminum metal but not Silver metal.
Silver metal has a higher voltage potential to keep its electrons than both Copper and Aluminum metals. Aluminum metal (and Sodium) are strong electron donors.
In this case Copper metal sits right between Silver and Aluminum metal.  Copper metal will give electrons to Silver ions, and Copper ions will take electrons from Aluminum metal.  Silver metal and Copper ions will not react, nor will Silver metal and Aluminum ions.