text:
Enrichment in phase diagram
After my post at UF4-ZrF4 phase diagram (the phase diagram is given below)

Question: Consider the coordination (40%; 900oC), the cooling to 850oC would form x(ZrF4) = 0,22. When the cooling progresses further, both liquid solution and solid solution are enriched by ZrF4 (Comparing the solution at point A and B). Is it true in this case?
Thought: I was thinking that at liquid solution of composition ZrF4 = 0,4 which is enriched in UF4, The cooling should result in higher UF4 solid and ZrF4 liquid. The diagram indicates otherwise.

The initial solid is higher in UF4, being to the left on the solidus line.

@JonCuster just to be clear, I added the blue line which represents my first liquid solution being cooled down, the cooling progresses with the enrichment in ZrF4 at both liquid solution and solid solution. Is that true?

I guess I have had the answer to my problem. To be more specific, when we start from the solidus line, the liquid composition is mainly UF4. If the temperature is increased a bit more, both ZrF4 and UF4 melts to liquid form. The composition of both solution should change.
It is a bit tricky to understand the solid-liquid phase diagram that is browsed in my question. So let just take liquid-vapor phase diagram of methanol (MeOH) and ethanol (EtOH) as another example.

This answer merely points out the composition changes in the LS region.
I guess "phase diagram nerd" like me should wonder about the extent of the vaporization which could be rephrased "As the methanol is the more volatile component, its composition in the gas phase should be enriched and the opposite trend should be seen in the vapor phase." It turns out to be a total mistake.
To put it simply, let's take the example of the starting solution composing of $\pu{60 g}$ MeOH and $\pu{40 g}$ EtOH. At the bubble point line, suppose we have $\pu{10 g}$ of MeOH vaporizes (nearly 100% in the vapor phase). A slight increase in temperature would result to the vaporization of both MeOH and EtOH.
Let's consider the vapor phase first, after the increasing in temperature, suppose MeOH vaporizes additional $\pu{10 g}$, EtOH vaporizes $\pu{5 g}$ simultaneously. Then the composition in the vapor phase now is MeOH ($\pu{20 g}$ or 80%) and EtOH ($\pu{5 g}$ or 20%). So we see the decrease in composition fraction of MeOH (even though MeOH is the more volatile component)
Next, let's consider the liquid phase, after the increasing in temperature, MeOH composition is $60 - 20 = \pu{40 g}$ and EtOH composition is $40 - 5 = \pu{35 g}$. Calculating the fraction, we should see that for MeOH is 53,33% and 46,67% for EtOH.
Comparing two results, the MeOH composition decreases in two phases and it is opposite for EtOH composition. I hope this explanation is useful to other people.

At the bubble point line, suppose we have 10g of MeOH vaporizes (100% fraction in the vapor phase). - This is not true, as the gaseous phase always contains both, if both are present in liquid.

If $p_\text{M0}$, $p_\text{E0}$ are vapor tension of methanol/ethanol at given $T$, and if $x_\text{l}$ and $x_\text{g}$ are molar fractions of methanol in liquid and gas, then $p_\text{M}=p_\text{M0} . x_\text{l}$, $p_\text{E}=p_\text{E0} .( 1 - x_\text{l})$ and $x_\text{g} = p_\text{M} / ( p_\text{M} + p_\text{E} ) = p_\text{M0} . x_\text{l} / ( p_\text{M0} . x_\text{l} + p_\text{E0} .( 1 - x_\text{l}) )$. All with assumption with zero deviation from Raoult's law.

I edited. The phrase was to show how the simultaneous and proportional change, so it was a bit unclear.

Try  $\ce{UF4}$  // Chem+Math formatting reference:
MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

In phase diagram, both $x,y$ denoted as MeOH. Please correct which one is $x$ and which one is belong to EtOH.

Putting points A and B between the solid and liquid curves is misleading and may confuse for you what is happening. As for the given scenario, the total composition of the whole system is rather a formal parameter after the solidifying starts.
The solid, enriched by $\ce{UF4}$, solidified earlier at higher temperature, is effectively taken away from the equilibrium system and does not take part in further phase exchange.
Therefore, the effective total composition of the system in equilibrium is not constant, but is converging toward the composition of eutectic point.
The liquid phase is enriched by $\ce{ZrF4}$, compared to the new "total composition", which is itself enriched by $\ce{ZrF4}$, compared to initial composition.
The newly being formed solid phase is enriched by $\ce{UF4}$, compared to the new "total composition".
Therefore, in total numbers of fraction composition, both the liquid phase and the newly being formed solid phase are being progressively enriched by $\ce{ZrF4}$. As the newly being formed solid phase is always relatively enriched by $\ce{UF4}$, compared to current liquid phase.
By other words, the total composition of the "equilibrium part of the system", being progressively solidified while cooling, does not reach the inside area between curves. It rather crawls along the upper liquid curve of the diagram toward the eutectic point.
Similarly, the composition of being formed solid phase is crawling along the lower "solid" curve toward the meeting point at the eutectic point.


text:
How and when do redox indicators work?
In redox titrations, for example cerimetry, cerium(IV) is reduced to cerium(III) with a reducing analyte. The Ce(IV)/Ce(III) redox potential is around $\pu{1.76 V}$. We keep track of the solutions' potential using an electrode or using a redox indicator.
For the latter, the indicators' standard redox potential $E^\circ$ needs to be in between the endpoints' potential and the potential right before the equivalence point. But why is that the case? If we added the redox indicator right at the start of the experiment, wouldn't it immediately be oxidized as well, for example using ferroin $(E^\circ = \pu{1.06 V})?$
Or is the reason that it does not react instantly because the potential difference from cerium(IV) to the reducing analyte (lets say iodide, around $E^\circ = \pu{0.35 V})$ is greater than to the ferroin? And because a bigger potential difference usually results in a faster reaction (according to Marcus theory) due to the higher overpotential?

It is fully analogical to pH indicators and acid-base titrations.
The potential of the equivalence point should lay within the potential range of indicator transition. The optimal case is when the potential of the first sign of the visual change is as close as possible to the potential of the equivalence, preferably just after.
Both analyte and indicator are kept at the same potential.
As the standard reduction potential of ferroin is much higher than of iodine, near all iodide is oxidized when ferroin starts to be oxidized (except temporary effects out of equilibrium).
It is similar as when much stronger acid (like acetic acid) reacts with NaOH first and much weaker one (like phenolphthaleinum) starts to react when the former has practically fully reacted.
Even if the indicator ferroin is temporarily oxidized by cerium(IV), it is kept reduced by iodide until it cannot be kept reduced any more.

source: Stack Exchange

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text:
What can I do if a peptide won't go in solution in a biological assay?
I have working on the realisation of on assay, that is intended to examine the activity of a protein. The assay works in a way, in which the product of the target enzyme is transfered by a support protein to a peptide. The peptide-product complex that is formed after the secondary reaction becomes fluorescently active and this is my activity/inhibition readout.
The unpolar peptide has a cysteine residue and therefore a DTT reduction must take place beforehand. Concentration of the peptide in the assay is 2 micromolar.
My plan was to create a peptide stock solution 200 fold concentrated to weigh-in the peptide more accurately (3mg on 10 ml buffer) and than dilute 1 part of the concentrated stock with 199 parts of the buffer to give the final concentration. Unfortunately, it was not possible to bring the peptide in solution in the stock. I tried warming it a little, vortexing and ultrasonic treatment. Adding 10 percent of DMSO to the stock did not clear the precipitate.
The peptide seams to be soluble in DMSO, but adding the peptide in DMSO to the buffer seems to result also in precipitation.....
Can you give me additional Tipps?  Would it be advisable to do the DTT treatment directly in DMSO and than try to get this into aquous solution? O would assume, that after the SS double bonds have cleaved, solubility should go up?

I maybe could try the following: Dissolve 3 mg in 1 ml of DMF, add the DTT and incubate. Than transfer to the buffer slowly, and hope that at this step no precipitation occurs. DTT is maybe better suited than DMSO because i of the cysteine? Final DMF concentration in the assay would be fairly low,  under 0,1 % if the concentrations will work.

Aside from solubility issues related to crosslinking, have you checked the solubility of a cysteine free peptide? (eg C->S). Can you use detergents? What data do you have on predicted solubility?

So, it is a very unpolar peptide, and it did not go into the buffer solution. What I tried, and what finally worked was using pure DMF as a solvent. To this I added the DTT and did the reduction entirely the organic solvent. Than I could transfer to the buffer without precipitation.

Dealing with insoluble peptides is a pain. If you have some time you can write up what you did into an answer. Ideally you might provide more information about the length, predicted solubility or any other properties of the peptide to make the answer more useful (you might not want to provide the sequence if its important to keep under wraps).

So, it is a very unpolar peptide, and it did not go into the buffer solution, neither vortexing, ultrasonic treatment or gentle heating worked. The 3 mg of peptide was attempted to be solved with 10 ml of buffer and a tensid was already included.
What I tried, and what finally worked was using pure DMF as a solvent to create a stock solution. To this I added the DTT and I performed the reduction entirely in the organic solvent. Than I could transfer to the buffer in a 1:2000 ratio without precipitation, resulting in a concentration
of roughly 5 micromolar. The assay system worked than as expected. When you are facing a similar problem, try to use DMSO to bring the peptide in solution. If the peptide contains cysteine, I would prefer DMF, because DMSO can oxidize cysteine and I don't know if DTT can fully compensate for this. DMSO is used up to 5 % in cell assays, but DMF is cytotoxic and might only be applicable for protein based assays. Other possible solvents might be acetic acid, acetonitrile, methanol, propanol, or isopropanol, pure or in mixture with water or buffer. If your peptide is acidic or basic instead try to use the corrosponding aqueous conditions first (PBS, acetic acid).
Find more information here:

https://www.bachem.com/knowledge-center/technical-notes/peptide-solubility/
https://www.lifetein.com/handling_and_storage_of_synthetic_peptides.html

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How to indicate that a substance is in a standard state?
An superscript circle or Plimsoll  is typically used to indicate that a thermodynamic property is that of a substance in its standard state. On the other hand, when writing a chemical equation we typically use abbreviations to designate the state or phase of the substance, e.g. "g" for gas, "aq" for aqueous. Is there a common or standard way of describing symbolically that a substance is in its standard state, e.g. $\ce{Cl2(g,^\circ)}$ or $\ce{Cl2(\mathrm{std})}$, or is it necessary to spell out those conditions explicitly and allow the reader to infer that those are standard conditions, e.g. $\ce{Cl2(g,\pu{1 bar})}$?

What about the notation $M^\circ(\ce{compound,state})$ where $M$ is the thermodynamic property? I wouldn't add the pressure since the $^\circ$ already implies $p^\circ \equiv \pu{1 bar}$. If I need the temperature going on during the algebra, I use $M^\circ(\ce{compound,state};T)$. E.g., $G^\circ(\ce{Cl2,g};T_2) - G^\circ(\ce{Cl2,g};T_1)$

@MetalStorm That would be the way to do this if I was referring to the property, but I am attempting to describe the substance. The thought occurred to write "$\ce{Cl2^\circ}$" but rather than invent my own notation I thought I'd ask around. There is likely a convention, I haven't seen this asked here before though.

You don't react Cl2 at 1 atm with H2 at 2 atm. The conditions refer to the reaction as a whole and should be marked as such.

Ohh, my bad. I haven't either, but I would use something like ${\ce{substance}}^\circ$ (e.g., ${\ce{MnO4^-}}^\circ$) as not to conflate with symbols relating to charge, radical, transition state, and so on.

@IvanNeretin But substances in solution in their standard state are (commonly) each at 1 M concentration. In the gas phase, standard state means each component at 1 bar. That does not refer to the pressure of the vessel under reaction conditions.

Regardless, the question refers to a substance in its standard state. Whether a reaction occurs is not relevant. What abbreviated notation is used, if any?

I'm not aware of a standard notation, so in case of need I'd probably invent my own.

On my browser (Chrome) using a circle in MathJax gives a silly small symbol, using o  instead  I think is much clearer, i.e G^\circ,  vs G^\text{o}, or $G^\circ$,  vs. $G^\text{o}$

@porphyrin my question is not really about aesthetics but ok. Since on the subject, an advantage of using the o is that it is ascii and readily transferred between editors, whereas the mathjax \circ symbol translates into nonsense when copy-pasted into a basic ascii editor. Which is related to my question, which is about using notation that is widely adopted and not confusing. As regards aesthetics I choose a balance between formatting effort and appearance. As mod I sometimes have to force myself to follow formatting guidelines more attentively.


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Is a microplastic saturated ferrofluid polar or nonpolar?
In my experiment (Junior science fair) I submerged a magnetic chip collector holding my ferrofluid (magnetite and grape seed oil) with polypropylene beads (attracted as part of the experiment) into various other fluids to see which fluid would do the best at removing the beads. Isopropyl alcohol and vegetable oil worked the best. I've been looking into why and have been researching solubility and polarity, 'like dissolves like', but the alcohol is polar and the vegetable oil is nonpolar. Should I be considering the nonpolar beads and grapeseed oil being 'like' the vegetable oil? Isopropyl alcohol did better than vinegar which is also polar (and the acid that I hypothesized would work best), why did they work at all? Any help is appreciated, I have not had any chemistry classes and only have a basic knowledge.

"Remove" is an awfully broad term. What exactly happened to the beads (and fluids, for that matter)?

"attracted as part of the experiment" <- Are the beads dispersed in the ferrofluid or attached to the surface of the liquid? A guess is that there may also be interfacial or electrostatic issues at play which do not involve solubility as such (somewhat along the lines of what is suggested in the current answer) // Vinegar is a solution of acetic acid (and other stuff) in water.

The question title is not meaningful: a mixture is not a pure substance. Each component may be more or less polar. One can measure relative permittivity (dielectric constant) of a mixture, which is based on the polar nature of the components.
That said, the amount of polarity is a matter of degree. In a sufficiently strong field, any substance can exhibit some amount of polarity. Compare the extent to which each component is polar, and to which each solvent is polar. Also consider that molecular polarity is not the only factor in solvent effectiveness.


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How to remove color from bromostyrene/dibromostyrene
Distillation under low heat can remove color (excess bromine) but this is time consuming and there is a danger of bromostyrene polymerization even with added TBC.
In aqueous, sodium thiosulphate can react with bromine to produce sodium bromide, which is water clear, but I don't know if this reaction will take place in pure bromostyrene/dibromostyrene. Additionally, since I am later polymerizing the bromostyrene, any additives that will interfere with polymerization will need to be removed.
Any comments or ideas on how to 'remove' typical yellow-orange-browm color from bromostyrene/dibromostyrene would be much appreciated.

Out of curiosity I wonder why you selected the "everyday-chemistry" tag?

If the organic mixture is treated by sodium thiosulfate to remove $\ce{Br2}$, the created ions bromide and sulfate will remain in the aqueous phase. Practically no thiosulfate will remain in the organic mixture that may interfere with future polymerization.

I second a thiosulfate/metabisulfite wash


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How would you balance this equation by half reaction
$\ce{Na + Hg2Cl2 —> NaCl + Hg}$
I was told to balance this reaction by using the half reaction method. I am confused to as how you would do that since the $\ce{Cl}$ is with both the $\ce{Na}$ and the $\ce{Hg}$.
I did with the inspection method and got $\ce{2Na + Hg2Cl2 —> 2NaCl + 2Hg}$ as my answer.

Half reaction method? It is like rubbing behind the right ear by the left hand.

Agree with @Poutnik. This could be easily done by hit and trial.

AS $\ce{Hg2Cl2}$ is a covalent compound, you can write those half reactions as:
\begin{align}
\ce{2 Na &-> 2 Na+ + 2 e-}\\
\ce{Hg2Cl2 + 2e- &-> 2 Hg + 2 Cl-}\\
\ce{2 Na+ + 2 Cl- &-> 2 NaCl(s)}
\end{align}
Therefore
$$\ce{2 Na + Hg2Cl2 -> 2 Hg + 2 NaCl}$$


text:
Multiple equilibria
I was reading about multiple equilibria reactions and can't quite comprehend some things.
Suppose we have two reactions:
\begin{align}
\ce{A + B &<=> C + D},& K_\mathrm{eq} &= 10^{20}\tag1\label1\\
\ce{A + E &<=> F + D},& K_\mathrm{eq} &= 10^{10}\tag2\label2
\end{align}
Both share the common components $\ce{A}$ and $\ce{D}$.
Due to the large difference in equilibrium constants, is it fair to say that in a mixture that initially contains only $\ce{A}$, $\ce{B}$ and $\ce{E}$, reaction $\eqref1$ will basically take place to completion while reaction $\eqref2$ won’t really happen at all? Will the difference of concentrations of $[\ce{C}]$ to $[\ce{F}]$ be $10^{10}$?
Or will both reactions happen to a significant degree in equilibrium as both equilibrium constants are large?
And how would one go about calculating the exact concentrations for every component given starting values?

K is practically more of a thermodynamic thing then kinetic. It doesn't really say about speed of reactions.

The answer depends on initial conditions. E.g. if A is in excess and  B is a limiting reagent then it does not matter much how big is the R1 eq. constant. // Mathematically, it is a set of 2 nonlinear equations of 2 variables. Well, in 4 variants, depending on reactant initial ratios.

Let’s say A,B,E are used stoechiometrically. Like initial conditions for each of them is 1mol/l.

Then reaction 1 should be favored heavily right?

No ! The reaction (1) is not necessarily favored. It depends on the activation energy. If (1) has a rather high activation energy, much higher than (2), the yield of (1) would be negligible.

I meant in a thermodynamic sense

Ignoring kinetic considerations such as mentioned in comments, and limiting discussion to thermodynamic ones, you might consider what happens if you mix A, B and E in one pot. You can then combine the two reactions into one:
$\ce{F + B <=> C + E},~~~~~~~~~~ K_\mathrm{eq} = 10^{10}\tag3\label3$
Clearly product C and reactant E are strongly favored. If however the reactions you describe are not coupled (you have separate reaction vessels) this line of argument does not hold.
To put this on a quantitative footing you can solve the quadratic equation that follows when the initial conditions are that the reactants are present in equimolar amounts. This leads to the solution for uncoupled (1) or (2):
$[\mathrm{product}] = A_0\frac{(K-K^{1/2})}{K-1}$
When K>>1 this simplifies to
$[\mathrm{product}] = A_0(1-K^{-1/2})$
Therefore for $K\geq 10^{10}$, $[\mathrm{product}] \approx A_0$.
In addition for the two uncoupled reactions the ratio of products is
$$\begin{align}\frac{[C]}{[F]}&=\frac{(1-K_1^{-1/2})}{(1-K_2^{-1/2})}\\&\approx(1-K_1^{-1/2})(1+K_2^{-1/2})\\&\approx(1-K_1^{-1/2}+K_2^{-1/2})\\&\approx(1+K_2^{-1/2})\end{align}$$
In other words, the product concentrations for both uncoupled reactions differ by a tiny amount and are approximately $A_0$.

I think that the equations are coupled because each is an equilibrium. So if we start with A+B and let these equilibrate then add E all the concentrations will adjust.

@porphyrin the point in my answer is that F is hardly produced if you mix everything together (A,B and E). Instead if only A and E are mixed nearly all A is converted to F.

Yes, when the equilibrium is reached, there will be significantly more $\ce{C}$ than $\ce{D}$ present.
To calculate the equilibrium concentration, you just go through the standard method of using the ICE table. You obtain a system of two nonlinear equations for two variables, which you have to solve using a numerical solver.
Let's denote initial concentrations with $[\ce{A}]_0$, $[\ce{B}]_0$ and $[\ce{E}]_0$, and let $x$ and $y$ be the extents of the reaction (1) and (2), respectively. In the equilibrium, the following two equations hold:
$$
\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]} = \frac{x(x+y)}{\left([\ce{A}]_0 - x - y\right)\left([\ce{B}]_0 - x\right)} = K_{\rm eq, 1}, \\
\frac{[\ce{F}][\ce{D}]}{[\ce{A}][\ce{E}]} =\frac{y(x+y)}{\left([\ce{A}]_0 - x - y\right)\left([\ce{E}]_0 - x\right)} = K_{\rm eq, 2}.
$$
I will use Wolfram Mathematica* (but you can use any other software with a numerical solver) to get the solutions of these two equations, namely the $x$ and $y$, with $[\ce{A}]_0 = \pu{1 M}$, $[\ce{B}]_0 = \pu{1 M}$, $[\ce{E}]_0 = \pu{1 M}$, $K_{\rm eq, 1}=10^{20}$ and $K_{\rm eq, 2}=10^{10}$:
$$\begin{align}x &\approx \pu{1.0 M}, \\ y &\approx \pu{1.0e-5 M}.\end{align}$$
Therefore, the ratio of products is:
$$\frac{[\ce{C}]}{[\ce{F}]} = \frac{x}{y} \approx 10^5.$$

* This is the code I've used:
N[Solve[{
    (x (x + y))/((A0 - x - y) (B0 - x)) == Keq1,
    (y (x + y))/((A0 - x - y) (E0 - y)) == Keq2,
    x < 1, y < 1
    } /. {A0 -> 1, B0 -> 1, E0 -> 1, Keq1 -> 10^20,
    Keq2 -> 10^10}, {x, y}, PositiveReals], 4]
(* {{x -> 1.000, y -> 0.00001000}} *)


text:
Temperature dependence on the feasibility of an equilibrium
If we consider a reaction A + B ⟶ C + D, with $\Delta$S $>0$, then regardless of whether $\Delta$H is positive or negative, increasing the temperature will make $\Delta$H - T$\Delta$S more negative, and the reaction will be more feasible
If we now consider a reaction A + B ⇌ C + D, still with $\Delta$S $>0$ for the forward reaction, if we express $ln(K)= -\frac{\Delta H}{RT}+\frac{\Delta S}{R}$, now increasing temperature only makes $ln(K)$ larger if $\Delta H >0$ and is independent of $\Delta S$
Why has the dependence of the feasibility and equilibrium constant of the reaction switched from  entropy to ethalpy. Am I confusing a relationship between the feasibility of a reaction itself and the equilibrium constant, i.e. a higher equilibrium constant does not mean the same thing as the forward reaction being more feasible?

That's not true. Dividing by T doesn't really change anything. If T makes S component larger, or H one smaller, doesn't matter.

See also https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation


text:
Surface treatment and current density: electrodeposition of Cd and Ag
I'm currently working on a cadmium and silver electrodeposition project.
Let me explain:
I'm doing electrolytic surface treatment on low-alloy steels as well as stainless steels. At present, we have no mathematical way of defining the intensity and therefore the current density to be applied to the bath, as each colleague does this according to their own experience.
I'd like to set up a calculation program or some other tool that would enable us to work in the same way, and make it easier to train newcomers.
I've tried looking for mathematical relationships (Faraday's law, etc.), but it's complicated because there are too many parameters.
Does anyone know anything about this? I need some light!
Thanks in advance for your comments

Can you clarify your question further? It seems very unclear what the actual problem is. You have some deposition setup and you do not know what current you apply, or rather the current density you apply? Current density is as simple as current divided by area. If you can measure the area of your electrode and look at your power source what current it applies you can simply calculate current density.


text:
Negative absorption coefficient
What does a negative absorption coefficient mean?
In detail - I participated in a Synchrotron virtual laboratory. I have to analyze the results and understand what a negative absorption coefficient means, as you can see in the pictures. The area where the absorption coefficient is negative is the area before XANES.
In the first picture you can see the graph from the theoretical material, and in the second picture the graph from the analysis of the results.

Usually it means that there is a little fluorescence effect, or some radiation lost by diffusion.

you could look at https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/X-ray_Spectroscopy/XANES/XANES_-_Theory_II

Thank you!
Fluorescence effect caused by absorption of radiation, so the absorption coefficient should be positive. How is it possible ?


text:
Dependence of elasticity on temperature and degree of vulcanization of typical commercial rubber band
I am doing a school experiment that tries to find how the elasticity of rubber band depends on temperature.
From all the research I have done, theory suggests stress to increase linearly with temperature at a given strain, this I understand can be derived by modeling rubber as ideal chains and consider entropic force.

However, my data seems to show the opposite:

Now I have digged up the paper that gave the first graph, I believe the method they used is the same as mine (stretching the rubber in water bath), so the only key difference has to be the material we used. The sample they claimed to use is 'eight part sulfur and one hundred parts pale crepe, vulcanized at 147 C, for 180 min', whereas the rubber band I used is just one of those really standard rubber band you can buy in a stationary store (link for store: https://item.jd.com/5326713.html#none) (Another link for a typical commercial rubber band brand: https://www.amazon.com/Rubber-Bands-Band-Depot-Pound/dp/B082WMN7PF/ref=sr_1_2_sspa?keywords=rubber%2Bbands&qid=1704758500&sr=8-2-spons&sp_csd=d2lkZ2V0TmFtZT1zcF9hdGY&th=1 ). The store's webpage contains no detail about the exact material of the band, and I couldn't find any articles showing the exact material or degree of vulcanization of ANY typical rubber band. Now my current explanation for my opposite trend is that in order to increase performance and strength, commercial rubber band are highly vulcanized and otherwise treated, causing severe steric hindrance, so that the degree of tangling and entropy increase in contracting becomes insignificant, so that the mechanism for rubber band contraction are now like most other material.
So what I would appreciate is any rough estimate of the degree of vulcanization of a typical rubber band, approximately how different is that compared to the 8% sample, any link that might lead to the detailed manufacturing of a specific rubber band brand, or critisism / confirmation of my current ideas.

This looks like a wild goose chase, although I'm no expert on vulcanization. It might help to look for a similar product in another website, maybe one that is more accessible to the anglophone audience at chem SE, or to provide a link to a translated page.

If I interpret this correctly strain is how far you stretched the band, and stress is the force you needed to apply. I'm surprised the experiment was performed like this (constant strain). I would think it'd be easier to apply a constant stress (say by hanging a weight from the band) and measure the strain response over T. How did you do the experiment?

Perhaps make your own rubber test samples, so they can be standardized. Vulcanization is fairly straightforward: https://www.theslingshotforum.com/threads/homemade-rubber-vulcanisation-experiment.30091/. You might also test artificial rubbers, such as used in inner tubes.

@BuckThorn, I have provided a link from amazon, what I did is that I first did force over extension test at a fixed temperature, so I initially get a bunch of stress over force over extension graph at different temperature, but because I chose a fixed set of extension each time I do the experiment, I could make the force over temperature graph I showed here.

One thing that stands out when comparing the graphs is that the strain in yours was much greater, even if the forces aren't, for instance 175% vs 22% at 2 kg/cm2. This suggests you used a much softer rubber band (smaller force constant), or that the units in your plot are different, or that you calculated % extension some other way. Would be good if you look into this. // As for commerical rubber band manufacturing info (vulcanization recipe) that is probably a trade secret. You might also consult diy sites like  DrMoishe Pippik suggested.


text:
Can multispectral imaging be used for identification of materials, substituting for hyperspectral imaging?
Nowadays a lot of research and applications are using Hyperspectral Imaging for identification and segregation of plastics by material. But the cost of a HSI system is quite high.
Could a multispectral solution yield good enough results, to substitute for HSI?

There is an explanation /comparison here https://www.edmundoptics.co.uk/knowledge-center/application-notes/imaging/hyperspectral-and-multispectral-imaging/ . Notice that this company has a commercial interest in these methods.

You might be more specific regarding the application involved. The link provided by porphyrin is very informative.


text:
Are there pigments that will only show their colour after applying water?
I am looking for pigments that at first are not showing their colour (and appear colourless), but after adding water or another 'component' like oil/alcohol etc. will show their colour.
My excuses for not knowing the right terminology and terms. I have little experience with Chemistry

All kinds of "magic inks" have that property (typically, you add something other than water, like acid, base, or heat).

First, do you definitely mean "pigment", rather than "stain" or "dye"?
Second, there are many dyes, stains and pigments that change color when dampened. The do so in a number of ways.

Finely-ground transparent substances, such as powdered colored glass, often look white or just a pale version of the bulk color because of internal reflection and refraction of light. Dampen them, and the water lowers the change in refractive difference between the substance and air. As soon as water evaporates, the powder pales.
Chemicals may change color when water of hydration is added. Anhydrous copper sulfate, $\ce{CuSO4}$, is white. As it's commonly sold, the copper sulfate pentahydrate, $\ce{CuSO4.5H2O}$, is blue. Heat it to dry it, and it reverts to white.
Chemicals may combine irreversibly, changing color when wet, and staying that color even when dried. For example, ferric ammonium sulfate, $\ce{NH4Fe(SO4)2.12H2O}$ is grayish-white, and potassium thiocyanate, and potassium thiocyanate, $\ce{KSCN}$ is colorless. If the two are kept dry, the mixed powders are white, but a drop of water turn them red, permanently.

Glazes for pottery are powered glass. It is difficult to anticipate the vibrant colors achieved after firing from the mute colors when applying the glaze.

@Karsten, exactly! The powdered frit merges into one surface.


text:
How the tertiary carbons of cyclohexane are assigned the stereo-descriptors (r)?

We assign the stereodescriptors R and S for chiral centres - carbons and other atoms, as per the set of sequence rules (CIP rules).
The pseudochiral centres are given the descriptors (r) & (s) whenever there are stereoisomeric groups present on a pseudochiral centre with different configurations R/S or E/Z.
In this compound, how are the priority 2 & 3 being decided for the tertiary carbon of the cyclohexane ring?
[The stereochemistry has been software generated using ChemDraw]

In "Search on Chemistry" at the top of your screen, type "user:55119 descriptors". You will find useful links here on ChemSE. Also, look here and here. That is, E>R>S>H = r.

The diene 1a not only has a plane of symmetry passing through $\ce{C1-C4}$ but it also has a two-fold axis of symmetry passing through the bonds $\ce{C2-C3}$ and $\ce{C5-C6}$. What is true for $\ce{C1}$ is also true for $\ce{C4}$.
The first step is to modify the double bonds and add the "phantom" carbon atoms shown in red in 1b. Clearly, the branched side chain has top priority while the hydrogen has the lowest priority. But what are the priorities of the methylene groups $\ce{C2 and C6}$? To answer this question, one turns to diagram 1c.
The compound is deconstructed into two pathways around the ring: $\ce{C1 -> C2 -> -> (C1)}$ and $\ce{C1 -> C6 -> ->  (C1)}$. The red dots, $\ce{(C1)}$, are carbons of atomic number six and they are bound to three atoms of atomic number zero. The temporary assignments R0 and S0 at $\ce{C4}$ are made as you would for any stereogenic center. Because R precedes S (CIP Rule 5), the priorities at $\ce{C1}$ are (E)-chain>R0 >S0 >H. The configurations at $\ce{C1}$ and $\ce{C4}$ are "r".

That was too deep to realize that even achiral compounds (at least in cyclic rings) would get the stereo-descriptors (r and s).
So, in the process we have treated the cyclohexyl ring as the two separate groups to decide its descriptor.
In that sense, should the tertiary carbons be called pseudo-chiral or not?

I avoid "pseudo" terminology. I prefer the terms of Mislow and Siegel. For a link and example go here. R/S, stereogenic and chirotopic; r/s, stereogenic and achirotopic; no descriptor, non-stereogenic and chirotopic. If you are compelled to use "pseudo", use it.


text:
Does polyamide react with hydrocarbons (group III engine oil) in its thermoplastic state?
I need to form (bend) thermoplastic polyamide hose in rubber jacket to a desired shape.
I can use heat gun but I thought for better results I could use heated to 150C group III engine oil (your standard hydrocracked oil -> so called full-synthetic) to evenly distribute heat.
My question is does polyamide in its soft state reacts with petroleum product such as engine oil?

It certainly doesn't react, but perhaps could start to absorb it, or even dissolve a bit in it (if it's Nylon 11 you mention in newer question).


text:
Why is my Briggs-Rauscher reaction not oscillating?
I have been trying to run the Briggs-Rauscher reaction following this guide, which is pretty similar to other guides out there, yet the reaction is failing to not only oscillate, but display any color change whatsoever (although the hydrogen peroxide is visibly reacting and decomposing). I have perfectly followed the guide, so that cannot be the issue. I ran an iodine test on the starch I was using and it reacted, so the issue is not with that. I do not have any silver nitrate to run a chloride test, but all the chemicals come from reputable sellers and do not have any evidence of contamination. (And yes, I used distilled water)
Because of the cold weather and where the chemicals are being stored, the solutions are significantly below room temperature. Could this be the issue? If not, what could I go about doing to try to get the reaction to work?

Try tweaking the amounts of malonic acid and other reactants, i did this reaction for my high school chemistry project, and it took a bit of trial and error to get it to work. (I used the measurements from NileRed's video.)

Where i live it's pretty hot, so im also interested as to if temperature messes with things

Is your source of distilled water reliable? I had an issue with silver nitrate reacting with water in a glass bulb and I had no idea why—turned out the filter we used for our distilled water had stopped working. Also, oscillating reactions are extremely sensitive to impurities—perhaps your glassware isn't clean enough?


text:
Activation energy depends on the form of the rate law
I was going through the following example, on p. 73, in a textbook called Chemical Reaction Engineering (3rd ed.) by Octave Levenspiel:

Experimental studies of a specific decomposition of A in a batch reactor using
pressure units show exactly the same rate at two different temperatures:

At $\ce {400K}$, $\ce {-r_A = 2.3 p_A^{2}}$, while at $\ce {500K}$, $\ce {-r_A = 2.3 p_A^{2}}$, where the reaction rate is measured in $\ce {mol m^{-3} s^{-1}}$ and the partial pressure of A is measured in $\ce {atm}$. Right away, we can conclude that the activation energy of this reaction, based on the fact that the rate constants determined at two different temperatures are the same, is zero. Activation energy is a measure of temperature dependence of the reaction rate, so if reaction rate does not change with temperature, then activation energy is zero. The same conclusion can be arrived at using the Arrhenius' equation for the activation energy.
Then, the example proceeds to change the rate laws to be based on concentration instead. The ideal gas law is made use of and after some unit cancellations, we obtain the following rate laws at $\ce {400 K}$ and $\ce {500 K}$:
At $\ce {400K}$, $\ce {-r_A = 0.0025 C_A^{2}}$, while at $\ce {500K}$, $\ce {-r_A = 0.0039 C_A^{2}}$, where the reaction rate is measured in $\ce {mol m^{-3} s^{-1}}$ and the concentration of A is measured in $\ce {mol m^{-3}}$. This would suggest that the activation energy is no longer zero and in fact, by the Arrhenius' equation, we obtain a value of $\ce {7394 J mol^{-1}}$.
The textbook does not explain why this change in the value of $\ce {E_a}$ occurs but my best guess is that in the original rate laws based on pressure, the temperature dependence of reaction rate was hidden within the partial pressure of A. On the other hand, after converting to concentrations, which are independent of temperature, the temperature dependence of the reaction would then be transferred to the rate constants. Is my explanation for why this happens correct?
Also, the textbook claims that the activation energies obtained using the rate laws based on pressure are incorrect and that it is only correct to derive activation energies based on the rate law involving concentrations.
Edit: From what I recall learning in high school, if a rate law is written in terms of pressure, the rate will be expressed in units of pressure per unit time (e.g. $\ce {atm s^{-1}, Pa s^{-1}}$), so in the example above, reporting rate as $\ce {mol m^{-3} s^{-1}}$ while writing the rate law based on pressures, seems a bit strange.

@porphyrin It's not a typo. In fact, you can see that the rate constants based on the rate laws for concentration differ exactly by the ratio of the square of the temperatures: $\frac{0.0039}{0.0025} = (\frac{500}{400})^2$

The frequency factor A from the Arrhenius equation is function of temperature too. The function A(T) is different for concentrations and pressures, leading to frequency of geometrically suitable collisions, By other words, it is A(T) what differs for p and c, not Ea.

The textbook does not explain why this change in the value of $\ce {E_a}$ occurs but my best guess is that in the original rate laws based on pressure, the temperature dependence of reaction rate was hidden within the partial pressure of A. On the other hand, after converting to concentrations, which are independent of temperature, the temperature dependence of the reaction would then be transferred to the rate constants. Is my explanation for why this happens correct?

I will do an example to show one way to arrive at this situation.
Imagine we have the following data for a constant-volume batch reactor at two different temperatures:
\begin{array}{c|c|c}
 t \, (\pu{s}) & c_\mathrm{A} \, (\pu{M}) \quad (T = \pu{350 K}) &
 c_\mathrm{A} \, (\pu{M}) \quad (T = \pu{495 K}) \\ \hline
 0   & 1000 & 500 \\
 100 & 500  & 250 \\
 200 & 333  & 166 \\
 300 & 250  & 125 \\
 400 & 200  & 100 \\
\end{array}
We will explore if the data responds to a second-order reaction. The amount balance for species $A$ leads to the well-known formula
\begin{equation}
  \frac{\mathrm{d}c_A}{\mathrm{d}t} = -kc_A^2 \rightarrow
  \frac{1}{c_A} = \frac{1}{c_{A0}} + kt \tag{1}
\end{equation}
So, we check the linearity for $1/c_A$ vs $t$ and obtain the following ($m$ is the slope and $b$ the y-intercept):
\begin{array}{c|c|c|c}
 T \, (\pu{K}) & m \, (\pu{M^-1 s-1}) & b \, (\pu{M^-1}) & R^2 \\ \hline
 300 & \pu{1.00 \times 10^{-5}} & \pu{1.00 \times 10^{-3}} & \approx 1 \\
 495 & \pu{2.00 \times 10^{-5}} & \pu{2.00 \times 10^{-5}} & \approx 1 \\
\end{array}
Thus, the rate constants are
\begin{equation}
  k(\pu{350 K}) = \pu{1.00 \times 10^{-5} M^-1 s^-1} \hspace{2 cm}
  k(\pu{495 K}) = \pu{2.00 \times 10^{-5} M^-1 s^-1}
\end{equation}
Now, we put the rate law in terms of the partial pressure of species $A$
\begin{equation}
  -r_\mathrm{A} = kc_\mathrm{A}^2 = k\left(\frac{p_A}{RT}\right)^2 =
  \frac{k}{(RT)^2}p_A^2 \rightarrow -r_\mathrm{A} =k'p_\mathrm{A}^2 \tag{2}
\end{equation}
where $k' \equiv k/(RT)^2$. If we calculate this quantity we get
\begin{equation}
  k'(\pu{350 K}) =  \pu{1.21 \times 10^{-8} M s^-1 atm^-2} \hspace{2 cm}
  k'(\pu{495 K}) =  \pu{1.21 \times 10^{-8} M s^-1 atm^-2}
\end{equation}
and you will report the following rate laws
\begin{align}
  -r_\mathrm{A} &= (\pu{1.21 \times 10^{-8} M s^-1 atm^-2}) p_A^2 \quad
  \pu{(350 K)} \\
  -r_\mathrm{A} &= (\pu{1.21 \times 10^{-8} M s^-1 atm^-2}) p_A^2 \quad
  \pu{(495 K)} \\
\end{align}
Thus, if we calculated the activation energy with these two rate constants, it will lead to $E_a \approx 0$. According to the author, since the transition theory is derived using concentration units, the rate constants must be reported with the rate law put in terms of the concentration. This was rather a surprise to me, and it seems to be true.
Nonetheless, if you just do the whole derivation you will quickly see that you are searching for the 'real' $k$ and no confusion will arise.


text:
When Should you use grease to lubricate ground glass joints?
I am getting ready to try steam distillation following this guide which references an earlier chapter which says "If using grease as recommended by your instructor, lightly grease all joints." Since i am just doing this on my own when is grease recommended in a general case, and what would my specific case (orange peels with possible use in candy) change those general recommendations.

The greasing is not needed for the specific cases of steam distillation  of volatile essential oils, perhaps should be even avoided.
Using light greasing(*) of glass joints (not limited to distillation) is recommended:

When there is expected significant pressure difference, compared to ambient pressure, like vacuum distilations.
When there are present harmful substances that could escape via loosen joints.
When the assembly is being  manipulated, mainly before or after use, grease somewhat suppresses its tendency to accidentally disassemble. (not a reason to use it, rather a side effect)
When eventual presence of traces of grease does not affect the quality of product.
When there is danger or experience of joints getting stuck together.

Avoiding greasing is recommended:

When there is not expected significant pressure difference, compared to ambient pressure.
When eventual presence of traces of grease could affect the quality of product.

(*) - just enough to be able to wet both surfaces of the joint by mutual rotation,   with visual control. Better repetitive adding small amounts until enough than using too much.

+1 "When there is expected significant pressure difference," that's a bit woolly. Should be "when you are going to run the glass apparatus at a lower than ambient pressure, i.e. with a vacuum regulator." Because it's extremely difficult to make it air-tight without grease.  And if it's "mechanically challenged": fix that, or don't assemble it at all. ;)

As I say - significant pressure difference. // You have misinterpreted that. There is nothing to fix.

But what do you mean by "mechanically challenged"?

@Karl It might not be the best fitting expression. I as in the update, I mean there are scenarios, not limitied to distillation,  that involve occasional manipulation and grease little bit helpts to hold it together. // I remember my high school class graduation lab work, with distillation of ammonia for later titration in 1980s. The lab assistent was unhappy due large amount of broken glass.

OK, I see what you mean. That works, although in my own (painful) experience, the flask with your product of three days of work then still falls off when you least expect it. Perhaps you had better grease than we did in the late 2000s. ;)


text:
Vapor / liquid balance in a closed container?
How is the vapor-liquid balance in a closed container computed as a function of temperature? In other words, if I have a container half full of water and I heat the container from 20-degrees C to 30-degrees C, then how would that affect the amount of water vapor in the container?
I know there is the equation PV=nRT, where "n" is the amount of something, but I am not sure how to use this equation to compute the amount of vapor as a function of temperature. For example, if T goes up and V remains the same, then obviously P will go up also, but the n I guess refers to the total amount of water vapor in the container (assuming no air is present) which, what?, stays the same?

Search for water vapor tension (partial pressure of saturated water vapor) as function of temperature. There are many data in form of tables, charts or empirical equations.

@Poutnik Those tell me the PRESSURE of the vapor, not how much vapor there is (molar quantity).

But that is with negligible error idealized as $n = \dfrac{pV}{RT}$

Say you have an initial amount of water $n$ in a closed container at pressure $p_1$ and temperature $T_1$. The closed container has a volume $V$. We know that the final state is at vapour-liquid equilibrium.

You give some heat $Q$ so that the temperature increases to $T_2$. The energy balance disregarding kinetic energy, potential energy, and any work done is simply
$$ Q = \Delta U \rightarrow \boxed{U_2^\mathrm{sat}(T_2) = Q + U_1(p_1,T_1)} \tag{1} $$
Note that Eq. (1) gives us the value of the final internal energy. Since the final state is a VLE equilibrium, its thermodynamic properties depend only on one intensive variable. One way to apply this abstract equation is to go to the water tables. Once you know $p_1$ and $T_1$ read the value of $U_1(p_1,T_1)$ in $\pu{kJ/kg}$ or any other unit. Then, sum to that value $Q$. Finally, go the the saturated part of the table to identify where $U_2$ is located, and you should also read $T_2$.
The volume of the container cannot change. Let $v$ be the molar volume, then
$$ V = n^\mathrm{vap} v_2^\mathrm{vap}(T_2) +
       n^\mathrm{liq} v_2^\mathrm{liq}(T_2) \tag{2} $$
Since you know $T_2$, you can also locate the saturated liquid and vapour molar volumes in the table. Another obvious equation comes from the conservation of the amount of substance
$$ n = n^\mathrm{vap} + n^\mathrm{liq} \tag{3} $$
Eqs. (2) and (3) form a set of two equations with two unknowns, $n^\mathrm{vap}$ and $n^\mathrm{liq}$, which are the ones you are interested in finding.
You can repeat this procedure for several $Q$'s given, so that you end up at higher temperatures. Repeat the procedure and plot the amounts $n^\mathrm{vap}$ and $n^\mathrm{liq}$ as a function of the final temperature $T_2$.


text:
False positive test for chloride ions during lab practical
In my highschool practical examination yesterday, I was given an inorganic salt to identify using chemical tests. The test I performed for the presence of chloride was to heat salt in a test tube with $\pu{2-3 mL}$ of mildly concentrated sulphuric acid. I observed white fumes of $\ce{HCl}$ (or so I thought), so the procedure was to keep a glass rod dipped in $\ce{NH4OH}$ over the mouth of the test tube. I observed (denser?) white fumes of what I thought was $\ce{NH4Cl}$, but apparently not as my salt was actually aluminium nitrate as I confirmed later (brown $\ce{NO2}$ is unmistakable, and aluminium confirmed via cobalt nitrate (blue) ash test).

What exactly did I observe and why?
Could I prepare $\ce{H2SO4}$ by heating $\ce{NaHSO4}$ and $\ce{HCl}$? If not, why only the other way?

Please note that $\ce{NH4OH}$ is very, very outdated. https://chemistry.stackexchange.com/q/10230/4945

I agree with Maurice's answer, here I add some details:

Moisture in the air may stabilize nitric acid against decomposition, which would keep the fumes white or at least more white.

Ammonia can react to produce ammonium nitrate, which would form more intense fumes than the acid alone. The ammonium nitrate fumes would be more stable against decomposition and woukd remain white.

A drop of silver nitrate solution (or a rod that has been dipped in such a solution) held over the reaction mixture would form a precipitate with a hydrochloric (or hydrobromic or hydriodic) acid fumes but not with nitric acid.

I live in one of the most humid places in the world, just a few kilometres from the ocean. This makes a lot of sense, thank you.

I swore I could smell ammonia, is this produced by decomposition of ammonium hydroxide?

This is indeed the confirmatory test, i shouldve done it but my lab teacher just told me it's nitrate lol

"Ammonium hydroxide" is not really so. It's just ammonia solution that has a few hydroxide ions in it because ammonia is a weak base. So your ammonia smell is directly from the solute.

If your powder, mixed with "mildly" concentrated sulfuric acid, produces white fumes that react with ammonia, the white fumes may be $\ce{HCl}$ , $\ce{HBr}$, or $\ce{HNO3}$. But if it is $\ce{HNO3}$, the fumes are usually not white, as nitric acid is often partly decomposed into brown $\ce{NO2}$. Have you ever seen any slightly brownish fumes ?
Second question. $\ce{H2SO4}$ cannot be obtained  from $\ce{NaHSO4 + HCl}$, because the reaction proceeds in the reverse direction. Heating a mixture of solid $\ce{NaCl + H2SO4}$ produce gaseous $\ce{HCl}$ plus $\ce{NaHSO4}$. So the only thing that can be made by heating a mixture $\ce{NaHSO4 + HCl}$ are vapors of $\ce{HCl}$.

Thank you for the answer. Actually, the manual recommends concentrated H2SO4 but i doubt the one provided is very concentrated hence the 'mildly'.

No, I don't remember seeing brownish fumes, it was mostly white/offwhite. Its probably just my inexperience and I should've done the test again, but unfortunately in the exam writing matters more than performing.

Why does the reaction proceed in the reverse direction? Does it have anything to do with electrode potentials?

The reaction proceeds according to H2SO4 + NaCl -> NaHSO4 + HCl and not in the reverse direction, because HCl is the most volatile substance, and gets first out of the mixture, especially when heating a bit. This behavior has nothing to do with potentials, as the oxidation numbers of the atoms do not change during the reaction.


text:
Extended resonance and colour
I was pretty sure that this would've been asked before but I couldn't find anything here. I've noticed that system with multiple benzene rings and pi donating/accepting groups tend to be quite colourful compared to their less conjugated counterparts, like azo dyes or indophenol or methylene blue. My guess is that the extended conjugation somehow pushes down the energy gap between the ground state and excited state, which  kicks up the wavelength of light absorption out of the UV spectrum perhaps by molecular orbital magic.

Is there any rigorous explanation for this observation?
(Answered)
Naphthalene and anthracene are both colourless, so do the nitrogen/oxygen substituents play an important role in imparting colour ie why do heteroatoms seem to create more colourful molecules?

https://chemistry.stackexchange.com/questions/84720/explaining-color-differences-using-bond-conjugation

https://chemistry.stackexchange.com/questions/109670/finding-the-color-of-crystal-violet-theoretically https://chemistry.stackexchange.com/a/10648/9961

https://chemistry.stackexchange.com/questions/67770/simple-way-to-understand-why-chromophores-often-have-double-bonds

In your defence, starting search with resonance and colour proved to be a surprisingly bad approach, while it shouldn't be. Your question might fix it.

This answer by Philipp is delightful, thank you very much.

There's more in follow-up https://chemistry.stackexchange.com/questions/8925/the-energy-gap-between-a-pi-conjugated-system-with-2-bonding-and-1-anti-bonding

Umm, you closed it yourself, and then started reopening :/ Some of the links^ talk about substituents.

My mistake, I didn't know that answering yes would close the question.

@Mithoron do i delete this question then? i don't know the procedure here

No, it may be useful to have a duplicate like this one, like I mentioned earlier. I didn't put all these links just for you.


text:
Searching for a phase transition dataset for substances
Disclaimer: I'm not a chemist, just a curious engineer.
As per the title, I'm searching for a dataset (or relevant Python library) with phase transition information for different substances. Eg: water, hydrogen, methane, ethylene... I'm looking to plot a phase transition graph for each substance (temperature vs pressure) so that I can compare them in different environments (by moving the bounds for each axis).
I'm currently using CoolProp, but this doesn't seem to include information to form the solid/gas and solid/liquid curves.
Maybe I'm approaching this the wrong way, or maybe I've not looked hard enough, but I would have thought this information would be relatively easy to find.

That link doesn't work. Apparently they don't have https capabilities.

Thanks @Martin-マーチン, fixed,  but as you said no https...

Oh, alright. That works. I wasn't sure you can use links to http sites anymore. Anyway. You're looking for phase diagrams of pure substances. I'd expect the handbook of chemistry and physics to have that. But I cannot check.

Note you don't have phase transitions for molecules. A phase transition is a property of the whole macroscopic system, not an individual molecule, so it would be better to say something like "substance" rather than molecule. And an upvote for the avoidance of non-https sites

Thanks @IanBush, edited accordingly.


text:
How long would it take fluoroantimonic acid to melt through a tiger's body?
I'm writing a story in which a character throws a knife containing fluoroantimonic acid contained within hydrofluoric acid, which is itself contained within fluorinated plastic, into a tiger's mouth. Upon hitting the tongue of the tiger, the plastic shatters, causing the acid to be released into the tiger. I want to be accurate in this story, so could someone explain how long it would take this fluoroantimonic acid to melt through the tiger's body?

Use butter of antimony, and the tiger would run around in anger, and melt into butter.

Related: https://chemistry.stackexchange.com/questions/4877/is-it-actually-possible-to-dispose-of-a-body-with-hydrofluoric-acid

This is not Chemistry. It can take as long or as little time as your plot allows, no one would be any the wiser.

Who needs fluoroantimonic acid anyway? The hydrogen fluoride would mess up the tiger all by itself.

https://chemistry.meta.stackexchange.com/questions/2874/how-many-shots-will-it-take-me-to-kill-you-with-a-neutron-gun

It would have to be tested, e.g. on fresh meat, tissues, leather of dead animals. It cannot be derived from basic principles.
It would be also matter of mass ratio. With low ratio of mass of a superacid and mass of tissues to get through, it does not matter much how super the acid is. It would get spent along the path.
Comparably, if you read about chlorine trifluoride accident, a liquid putting in fire near everything, the damage was in proportion to its amount. It did not dig a new coal mine.


text:
What determines the pressure in the system when the gas tanks are opened?
I have two gas tanks of nitrogen and oxygen. Mixture of gases is created at specified composition and used in the system of interest.
Let's say I want to create an atmospheric conditions in my system, mass ratio of oxygen to nitrogen is 0.3 and total pressure of 1 bar (oxygen partial pressure 0.21 bar). Therefore, mass flow ratio of the gases needs to be 0.3 to create a gas mixture of atmospheric composition. Gas tanks are equipped with volumetric flowmeters. This isn't a problem as mass and volumetric flowrates can be easily related by the ideal gas law. I'm planning using air at atmospheric pressure and temperatures 500-800 $^\circ$C which I think justifies ideal gas law.
What I'm not clear about is when the tanks are opened and the flowrates are specified, what will determine the pressure in the system? This is important as I need to know and change the oxygen partial pressure during this experiment (electrical conductivity relaxation).
Schematic is provided to give a better insight into the process:

Hi! I may help you but I have some doubts. You start with two tanks, open their valves, and then both streams enter a third tank/vessel? Is the third vessel closed?

I have understood it as an open system at ambient pressure, being flushed by some flow of artificial air. So N and O  pressure are such that maintain the required respective flows or flow ratio against atmospheric pressure. Right or wrong ?

@MetalStorm The system (vessel) is opened as gases flow in and out of it. I added schematic of the process in the main post. Flowrates of gases can be controlled.

Flow in vs flow out. I would use mass flow controllers and a butterfly valve on the outlet controlled by a barotron.

To get mass flow rate from volumetric, pressure and temperature are required. These are constants in the experiment, but would like to know what sets pressure in this setup.

@DarioMirić Great, just one last question. The vessel is attached to an electrical furnace, so we have an exchange of energy between it and the air within the vessel. Is that correct?

@MetalStorm Yes.


text:
The Role of Acidified Potassium Manganate(VII) in Oxidation of Alcohols
The chemical equation for the oxidation of ethanol is given as:
$$\ce{C2H5OH + 2[O] -> CH3COOH + H2O}.$$
Acidified potassium manganate (VII) solution $\ce{KMnO4}$ is added as the oxidising agent. My question is how does the $\ce{KMnO4}$ help in the oxidation? My guess is the reaction between acid and $\ce{KMnO4}$ releases oxygen that will be used in the oxidation of alcohol. Is that correct?

You may find the reactions starting from acidified permanganate solution and producing $\ce{HMnO4}$, then this acid being decomposed and producing oxygen atoms, and finally oxydation of ethanol with these oxygen atoms. But this problem is more easily solved by using half-equations, like all redox reactions. You should know this general approach. It gives, with ions of permanganate, instead of $\ce{KMnO4}$  :
$$\ce{MnO4^- + 8 H+ + 5 e- ->  Mn^{2+} + 4 H2O }$$ and $$\ce{C2H5OH + H2O -> CH3COOH + 4 H+ + 4 e-}$$ The only specialty is the high total number of electrons involved, $20$ in total. The sum of these two half-equations gives : $$\ce{4 MnO4^- + 5 C2H5OH + 12 H+ -> 4 Mn^{2+} + 5 CH3COOH + 11 H2O}$$.

The mechanism involving [O] is specious. 20 electrons is not a problem they are necessary.

So is that true that the equation of the oxidation of alkene to produce alcohol with KMnO4 is : 2MnO4^- + 6H^+ + 5C2H4 + 2H2O -> 2Mn^2+ + 5C2H4(OH)2

@Zhang Jin. Your equation produces glycol, which can be considered as a diol.

Ok got it... so does that means the equation is correct but the product is glycol, not alcohol ?

Writing an equation does not guarantee a reaction. A strong oxidizing agent such a permanganate will give different products under different conditions. Under acidic conditions with total reduction to Mn++ ethylene would be oxidized to carbon dioxide and water. Cold neutral dilute permanganate will oxidize ethylene to ethylene glycol while being reduced to MnO2. Buy an older edition of Morrison and Boyd and read it cover to cover to get an idea of what is going on.

The equation you quote is not a chemical equation. It is a symbolic equation stating that an oxidizing chemical is needed to effect oxidation of an alcohol to an acid. A chemical equation shows the reactants and the products and is properly balanced to show the stoichiometry. Ethanol can be oxidized to acetic acid by many different oxidants such as permanganate, $\ce{MnO4-}$ [btw. $\ce{Mn(VII)}$ not $\ce{Mn(IV)}$], dichromate, $\ce{Cr2O7^2-}$, both in acidic solutions, molecular $\ce{O2}$ usually with the help of enzymes as in the conversion of wine into vinegar by microorganisms or the removal of ethanol from our bodies. The mechanisms of these reactions do not involve the formation of oxygen atoms. They involve the transfer of electrons from the alcohol to the oxidizing chemical with the aid of acid and water.
Your homework assignment is to find the reactions of ethanol with permanganate, dichromate and $\ce{O2}$  and to determine the properly balanced equation for each oxidizing agent.

So like what you have said oxygen atom don't really exist in the reaction, the equation only indicates that an oxidizing chemical is needed and the reactant that involve in the reaction is only water,  ions of permanganate and ethanol. Am I right?

Write out the balanced chemical equations. That is a minimum of what is involved. A detailed mechanism is needed to determine everything involved.  One is never "right" in understanding anything.


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History of Azeotrope (azeotropic mixtures)
I'm currently researching the history of azeotropic mixtures for an essay, and I've come across some interesting information. The term "azeotrope" was coined in 1911 by English chemists John Wade and Richard William Merriman. This term is derived from the Greek words ζέειν (boil) and τρόπος (turning) with the prefix α- (no), creating the overall meaning of "no change on boiling."
The earliest documented evidence of azeotropes dates back to 1911, as found in the Journal of Chemical Society. However, it is believed that azeotropic mixtures were known before this time, possibly before 1910, although concrete historical records or publications from that period are scarce.
While there is a general acknowledgment of azeotropic mixtures existing prior to 1910, it is challenging to find specific information on their discovery. It's worth noting that the term itself was only formally introduced in 1911, which may explain the limited historical documentation.
I would appreciate any insights or references related to the discovery of azeotropic mixtures before 1910 if you have come across any reliable sources. Thank you!
References :
https://www.oed.com/dictionary/azeotropic_adj?tl=true#:~:text=The%20earliest%20known%20use%20of,formed%20within%20English%2C%20by%20derivation
https://en.wikipedia.org/wiki/Azeotrope

Whether someone had a name for it, I’d say the concept has been known for as long as distilling liquor has been done.

You might try instead History of Science and Mathematics SE. I can migrate this there if you like.


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What is a degenerate orbital
I don't really get what degenerate orbitals are, I mean I know that orbitals are degenerate when they have the same energy level, and I know that the energy levels depend on n, but does that mean that 1s is never degenerate and 2s and 2p are always degenerate or am I misunderstanding something, because 2s and 2p have the same principal quantum number n=2, thus same energy levels? Is this correct or am I missing something

It seems that you are mixing two ideas. If you solve the Schrödinger equation for your system $\hat{H}\Phi_i =E\Phi_i$ you would obtain two kind of solutions. The scalar (eigenvalues) $E$, the energy, and the wavefunction $\Phi$ (eigenfunctions). In the case of the hydrogen Hamiltonian solutions, you will find that there are wavefunctions that have the same energy (they are degenerate), but when other observables are considered (e.g. angular momentum), they render different values. Check your Physical Chemistry textbook.

See also: https://chemistry.stackexchange.com/questions/49209/are-electron-orbitals-in-a-helium-atom-degenerate?rq=1

The orbits $2$s and $2$p are degenerate in $\ce{H}$ atom. In heavier atoms, these orbitals have different energies. They are not degenerate any more.

And that's because with heavier atoms the energy levels also depend on $l$ and then f.e. 2p is degenerate but not 2s?

yes, the three 2p orbitals are degenerate

@Maurice I think we're glossing over an important concept here which is that in an atom with a single electron, the degeneracy is across the level. Adding additional electrons breaks the degeneracy because nuclear penetration is different.

Note that ns and np orbitals are not exactly degenerate in H atom either, but the involved effects of later developed quantum electrodynamics are not addressed by Schroedinger equation.

There is multiplicity and degeneracy. In the H atom the n=2 levels the multiplicity is four one 2s, and three 2p, and they are also degenerate in the simplest models. This means that these levels are equal in energy. However, even in the H atom experiments show that this is not strictly true because in the 2p orbital the electron has some orbital angular momentum and as the electron also has angular momentum (due to its spin) these interact by a tiny but measurable amount and the multiplicity remains but the degeneracy is broken. (The interaction is called spin-orbit coupling). The three p orbitals scan also be split in energy, thus removing their degeneracy, by an electric field (Stark effect) or by a magnetic field (Zeeman effect).

Concerning your doubts, I suggest you to approach the problem in a more mathematical fashion.
Orbitals for hydrogenoid atoms (or even for molecules) are stationary wavefunctions $\phi_i$ that satisfy the time-independent Schroedinger's equation:
$$
\hat{H}\phi_i = E_i\phi_i
$$
Now, if the set of orbitals ${\phi_i}$ is such that it is possible to have k different $\phi_ij$ such that $E_i$ is always the same for different j.
$$ \hat{H}\phi_{ij} = E_i\phi_{ij}\ \ \forall j={1,...,k}$$
We call those orbitals "k-fold degenerate orbitals", i.e. k orbitals with the same energy value (due to the particular simmetry of the molecule/atom under study).
Now: for the hydrogen atom there is no difference in the energy of any orbital with the same quantum number $n$.
The solution of Schroedinger's equation for the hydrogenoid atom, however, gives us different energies for the orbitals $\phi_{n,l,m,s}$ associated with the usual quantum numbers n, l, m, s.
For the same n, there exist n-1 orbitals with the secondary quantum number l ranging from 1 to n-1, each set of orbitals with a different l has a different shape and a different energy (in general this holds for approximate orbitals of non-hydrogenoid atoms).
For the same n there exist a range of quantum numbers m={-l,..., 0,..., +l} of orbitals with the same shape but different orientation in space, these have the same energy and are therefore $degenerate$.
The spin number s is not of our concern here, it can assume only half-integer values for any atomic orbital.
In this context:
In the energy levels with n=1 there is only one spin-orbital with $l=0$, $m=0$ and $s=\pm\frac{1}{2}$, meaning that there is one orbital that we commonly call 1s.
For levels with n=2 there are two set of possible orbitals:
$$n=2,\ l=0,\ m=0,\ s=\pm\frac{1}{2} \implies \ 2s$$
$$n=2,\ l=1,\ m=\{-1,0,+1\},\ s=\pm\frac{1}{2}\ \implies 2p_x,2p_y, 2p_z$$
So 1s and 2s orbitals are nondegenerate while 2p orbitals are 3-fold degenerate.
In the same fashion 3p orbitals are 3-fold degenerate and 3d orbitals are 5-fold degenerate ($l=\{-2,\ -1,\ \ 0,\ +1,\ +2\}$).

The energy levels for orbitals in mono-electronic species (i.e. atoms or ions with one electron only; for example, hydrogen) are dependent on just the principal quantum number $n$. But, for multielectronic species (i.e. atoms or ions with more than one electron), the energy level depends on both the principal quantum number $n$ and the azimuthal quantum number $l$.
That is,

For monoelectronic-species:
$1s<2s=2p<3s=3p=3d<4s=4p=4d=4f\ (\text{Energy comparison})$

(Image Source)

For multielectronic-species:
$1s<2s<2p<3s<3p<4s<3d<4p<4d<4f\ (\text{Energy comparison})$

(Image Source)

This energy order for multielectronic species, by the way, is exactly what you derive from the Aufbau principle.

(Image Source)

Notice that in the orbital energy level diagram for monoelectronic species (each box corresponds to one orbital), all the orbitals in the $2s$ and $2p$ subshells, for example, have the same energy, and are thus degenerate. In the orbital energy level diagram for multi-electronic species, however, $2s$ and $2p$ subshells have different energies and thus all their orbitals "together" are not degenerate.
But, (irrespective of whether the species in question is mono-electronic or multi-electronic) the three orbitals in the $2p$ subshell, namely $2p_x$, $2p_y$ and $2p_z$, do have the same energy, and thus we can say that they ARE degenerate, as Gaurav also pointed out. Hopefully, that alleviates your confusion.


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How was it known certain materials were elemental (Hydrogen 1766, Phosphorous 1669, ...) before Dalton showed that matter was made of atoms in 1803?
Dalton showed in 1803 that matter was made of atoms rather than continuously distributed. But phosphorous, hydrogen, and other elements were identified before this discovery. How could they have known that these were elements before it was shown that matter was made of atoms?

For the ancient chemist, an element was a substance that had never been decomposed into smaller fragments. So if nobody had been able to decompose a substance by electricity, by heat, or any other process, this substance was considered as an element. Phosphorous and Hydrogen had never been decomposed in Dalton times. So they were put in this category.

"Dalton showed" is quite a bit of an overstatement. Atomic theory was around for a long time before that, and was not firmly proven for a long time after.

https://chemistry.stackexchange.com/questions/4989/when-did-the-atom-theory-become-indisputable https://chemistry.stackexchange.com/questions/10855/how-were-the-elements-discovered-before-modern-technology?rq=1

As elements were considered such substances that could not be dissembled to more substances.
Before oxygen and hydrogen discovery and observation they form water together, water was considered as an element.
Before realising air major composition, air was considered as an element.
Before praseodymium and neodymium were separated, their mixture was erroneously considered as an element didymium.
At some point, most substances being considered as elements were verified to be real chemical elements at atomic level.


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Does dative bonding cause polarity?
When two atoms form a dative non-polar bond, does the one that shares the couple of electrons acquire a +δ charge and the other a -δ charge?
I have been told that in a dative polar bond the charge fraction is twice as much as in a normal covalent bond, but, since the electrons behave right the same way, shouldn't that imply the dative bond itself creates a charge?

Generally dative bonds form to reduce charge densities. Examples such as carbon monoxide and zwitterionic aminoacids are of lower polarity.

Unequal sharing of electrons can indeed induce polarity. A clear-cut example comes from ozone, the triatomic form of oxygen. The referenced Wikipedia article reports a dipole moment of 0.53 D (citing [1]) despite all atoms being identical, an unusual feature for a molecule with only one element.
The three atoms in the ozone molecule form a bent chain in which the central atom shares three electron pairs and each of the end atoms shares only a pair and a half (there are two equivalent contributing structures in which one end atom shares two electron pairs and the other shares one). The molecular orbital structure of ozone also shows the central atom giving more electronic charge to the end atoms than vice versa in the three-center, four-electron pi bond. And, indeed, this unequal sharing of electrons creates a charge separation and a dipole moment as documented by Wikipedia.
Cited Reference

Kenneth M. Mack, J. S. Muenter; "Stark and Zeeman properties of ozone from molecular beam spectroscopy". J. Chem. Phys. 15 June 1977; *66 (12): 5278–5283. https://doi.org/10.1063/1.433909


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Amalgam Reaction
I have some doubt about Amalgams.
On google it says that amalgams are mixtures. However, when I mixed Na and Hg it 'reacted' to form the amalgam. If they are just mixtures, why do they react to form?
Also, two states of Na-Hg form when you reduce the concentration of sodium.
If you take less sodium, is the sodium and Hg completely amalgated or does sodium act as a 'limiting agent'? Also, it is in liqiud form and also is uniform, so I think it is completely amalgated.
Can anyone please explain to me why amalgam is a mixture and not a compound if the above metals react?

Wikipedia describes amalgams broadly as "alloys, intermetallic compounds, and solutions." All three may apply. Certainly, as "intermetallic compounds", one expects chemical interaction. It is a matter of degree and of definition, as in "polar vs. nonpolar", rather than a simple true/false statement.
For example, "Metallic sodium dissolves in mercury exothermically, i.e. with the release of heat... generating sparks [and causing} localised boiling of the mercury." This is more than simple mixing [and exceedingly hazardous to the experimenter and environment].
What I find amazing is the mercury can stabilize the ammonium radical, forming mercury-ammonium amalgam, and even quaternary ammonium amalgams!

Apparently there are a number of phases in the Na-Hg binary system. The OP should look at the phase diagram.

The Hg-Na binary phase diagram is rich in phases. An article in the Journal of Alloys and Compounds (Volume 246, Issues 1–2, 15 January 1997, Pages 80-90) shows the many different compositions and crystal structures that may form. The phase diagram given there is (the available pdf is a horrible scan, sorry):

As to why they 'react' - well, those phases are thermodynamically stable, so they would have a lower Gibbs Free Energy compared with the pure Na and Hg phases. At the very least a 'mixture' would have a reduction in $G$ through the entropy of mixing.


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Why is the melting point of meta-diiodobenzene higher than that of ortho-diiodobenzene?
I'm currently trying to reason for as why the melting point of meta-diiodobenzene higher than that of ortho-diiodobenzene and I'm unsure of what to say. I understand the para isomer has the highest melting point due to symmetry and packing efficiency and would also understand the reasoning if the ortho isomer melting point was higher than the meta melting point (due to possible greater dipole moment and dipole dipole interactions), but the only reasoning I currently have is that there is some orbital overlap or hinderence between the iodine atoms which negate these effects? Please can someone help me... maybe it's something to do with the mesmoric or resonance effects (which I have little understanding of), many thanks

If you have little understanding of mesomeric effects then maybe concentrate on that. Trying to analyse melting points (and other random values) may be enticing, but in practice quickly becomes a waste of time.

Melting point is not a thing to be explained by such qualitative arguments. (Boiling point is a different story altogether.)

First off, what actually are the melting points? They may not both be known; Sigma-Aldrich fails to give one in the SDS for its ortho compound.

According to https://www.stenutz.eu/chem/solv6.php?name=m-diiodobenzene and https://www.stenutz.eu/chem/solv6.php?name=o-diiodobenzene the MPs are 40C and 27C respectively. So the same at this level of handwaving


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Is my derivation for the reaction orders of a single-step reaction valid?
Recently in my chemistry class I learned about reaction orders. For some reaction $\ce{r_1A1}+\ce{r_2A2}+\dots+\ce{r_nA_n}\rightarrow\ce{B}$,   $m_1,\dots{,m_n}$ are the orders of that reaction. The meaning of the order $m_i$ is as follows: If we increase the concentration of reactant $\ce{A_i}$ by a factor $\lambda$ we know that the reaction rate increases by a factor of $\lambda^{m_i}$.
In general, I learned, the orders must be determined experimentally. However, for reactions without intermediate steps, they coincide with the coefficients of their respective reactant.
I might've found a way to derive this fact, which is why I'd like to ask whether my derivation is valid. Moreover, in my opinion, this is a rather lengthy derivation, which is why I'd also like to ask whether there is an easier one.
Here we go:
My idea is to show that the reaction rate $v$ in fact increases by approximately $\lambda^{r_i}$ if we increase the amount of $\ce{A_i}$ by $\lambda$.
The number of possible collisions $C$ of  $r_1, \dots, r_n$ particles from substances $\ce{A1},\dots, \ce{A_n}$ which can cause a reaction, is proportional to ${N_i\choose r_i}$, where $N_i$ is the number of particles of $\ce{A_i}$. Now see what happens if we increase the amount of $\ce{A1}$ by $\lambda$: For the binomial coefficient we get
$$
{\lambda N_i\choose r_i} = \lambda N_i(\lambda N_i-1)\dots(\lambda N_i-r_i+1)\\
\approx \lambda^{r_i} N_i(N_i-1)\dots(N_i-r_i+1)\\
= \lambda^{r_i} {N_i\choose r_i},
$$
so $C$ also increases by $\lambda^{r_i}$. And because $C$ is itself proportional to the reaction rate $v$, $v$ also increases by the same factor and we are done.


text:
How to name an ester with an alkene or alkyne as the R group?
More specifically, see the image down below:

Would it be pentan-2-ene-3-yl propanoate? Do you just add the 2-ene to specify the position of the double bond?

Prefix the name with (Z).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the names of esters are generally formed by placing the alcoholic component in front of the name of the acid component as a separate word (e.g. ethyl acetate).

P-65.6.1 General methodology
Neutral salts and esters are both named using the name of the anion derived from the name of the acid. Anion names are formed by changing an ‘-ic acid’ ending of an acid name to ‘-ate’ and an ‘-ous acid’ ending of an acid name to ‘-ite’. Then, salts are named using the names of cations, and esters the names of organyl groups, cited as separate words in front of the name of the anion.

The organyl group is named using the using the usual methodology.

P-29.2 GENERAL METHODOLOGY FOR NAMING SUBSTITUENT GROUPS
The presence of free valences formally derived from the loss of one or more hydrogen atoms from a parent hydride is denoted by suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’, together with multiplying prefixes indicating the number of free valences; lowest locants are assigned to all free valences as a set, then in the order ‘yl’, ‘ylidene’, ‘ylidyne’. In names, the suffixes are cited in the order ‘yl’, ‘ylidene’, ‘ylidyne’. The suffixes ‘ylidene’ and ‘ylidyne’ are used only to indicate the attachment of a substituent to a parent hydride or parent substituent by a double or triple bond, respectively.
(…)
Systematic names are formed by using the suffixes ‘yl’, ‘ylidene’ and ‘ylidyne’, with elision of the final letter ‘e’ of parent hydrides, when present, according to two methods:

The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups and for the mononuclear hydrides of silicon, germanium, tin, and lead. Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;
The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ are added to the name of the parent hydride with elision of the terminal letter ‘e’, if present, when followed immediately by the letter ‘y’. The locants for the atoms of free valences are as low as is consistent with any established numbering of the parent hydride and, except for mononuclear parent hydrides or the suffix ‘ylidyne’, the locant ‘1’ must be cited. This method is used to generate names of ‘alkanyl-type substituent groups’ that are simple substituent groups with free valences at positions other than ‘1’.

(…)

In this case, Method (2) is used because the free valence is not at position ‘1’. The name of the parent hydride (2E)-pent-2-ene is changed to the alkanyl-type substituent group pentenyl by addition of the suffix ‘yl’ and elision of the terminal letter ‘e’. A low locant is given first to the free valence, which gives penten-3-yl. After adding the other locants we finally get the complete name (2Z)-pent-2-en-3-yl propanoate.


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Use of Gibbs-Helmholtz Equation to Calculate Electrochemical Work
When we derive the expression $\mathrm{d}G=V\mathrm{d}P-S\mathrm{d}T$, if my understanding is correct, we do not consider the term representing useful (non-PV) work. Based on this expression, we conclude $$\left(\frac{\delta G}{\delta P}\right)_T=V$$ and $$\left(\frac{\delta G}{\delta T}\right)_P=-S$$
Further, we use these functions when deriving a relation for the EMF of a cell with temperature, obtaining the result
$$E_{cell}=\frac{-\Delta H}{nF}+T\left(\frac{\delta E_{cell}}{\delta T}\right)_P $$
My confusion arises from the fact that while deriving the first two results, we neglected the useful work term, but used the result in a process where we are concerned with the non-PV work. Where am I going wrong in my understanding?

Chem+Math formatting reference:
MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles.

The equation you wrote for dG is for a pure component, not a mixture of reacting components.

@Chet Miller could you please elaborate a bit on this? I would be very grateful as I'm really struggling with this concept, so if you could write out an answer it would really help me out

For a mixture, $$dG=-SdT+VdP+\sum \mu_jdN_j{}$$ where $\mu_j$ is the chemical potential (Partial molar free energy) of the j'ta species and $N_j$ is change in the number of moles of the j'ta species

Thanks a lot, it's clear now


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What are the building blocks of matter?
Question 1: If atoms are made up of fundamental particles, then why we say that atoms are the building blocks of matter?

Atom, the basic building block of all matter. (1)

Atoms are the basic units of matter. (2)

The atom is considered the basic building block of matter. (3)

Question 2: Is it right to say that fundamental particles are the building blocks of matter?

Gosh, golly, I thought building blocks are what kids play ;> Oh, it must be one of these, whatchamcallit, metaphors! But better get to the source. Building blocks of kids' constructions are... building blocks, aren't they? And they are certainly made from smaller stuff, yet they are building blocks, so what's the problem?

@Mithoron Your chagrin is understandable but this seems a reasonable and researched question. The focus is on definitions that play a bit too loosely with language. The OP is right to question them. What you write as a comment is part of the answer. A complete answer would address the history of seeking (the why) and the identity (the what) of the smallest division  of matter.

Elementary particles are building blocks on physical level, atoms on chemical level.
The smallest particle of a chemical element is an element atom. The smallest  particle of a chemical compound is a compound molecule.
But the currently smallest particles of baryonic matter are quarks, electrons and virtual particles of force interactions.

Force carrier aren't necessarily virtual.

@Mithoron They are not, but such are not necessarily present.


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How salt helps to melt ice even when there is no dynamic equilibrium between water and ice?
I am a high school student and I am confused about freezing point depression.
We say that when in winter's it gets even colder than normal freezing point Temperature then we use solute to melt ice that is there on the floor but we also say that the "dynamic equilibrium" between water and ice exists only at freezing point Temperature. These two statements are contradictory because the process by which the solute helps to melt ice is by disturbing the dynamic equilibrium between water and ice(as it doesn't let water molecules to for ice crystals , as a result more ice transition into water than water to ice). But at temperatures below freezing point, there was no water to begin with and hence no dynamic equilibrium between water and ice. So how the solute here can melt ice? The explanation they give in textbooks is "because it lowers the freezing point" but it doesn't make sense if we talk about the mechanism.

Have a look at the answer here https://chemistry.stackexchange.com/questions/116302/where-does-the-energy-come-from-to-lower-the-temperature-of-a-brine-solution/116316#116316

I think my question is somewhat different from that one

If it was not, it would be closed as a duplicate. // Consider ice vapor tension and hygroscopicity of salt at high relative humidity, leading to saturated salt solution.

Does this answer your question? Melting ice with salt - how does it start?

While there may not be dynamic ‘equilibrium’ initially, there are dynamics going on all the time. Add salt, and those dynamics will lead the system towards a new equilibrium.

@Mithoron. The answer to that question explains it from a completely different POV. It creates more confusion from me because the answer on the internet and from the textbooks are always like "the salt no let let water to freeze, but the ice continues to transition into water, thereby it melts or we say the freezing point gets depressed. But according to that answer all of this is incorrect?

@Poutnik in that answer , it is being explained that how the temperature of whole system drops down. I have watched some videos on it and it states that if we are using Nack, it takes more energy to dissolve it than the hydration energy, thereby it's an endothermic process which takes away energy from water and hence decrease it's temperature. But for salts like Cacl2, this is actually exothermic. Anyways my question is different. I am taking about ideal solute and the mechanism by which the ice melts on adding it even though there was no dynamic equilibrium (of ice and water) to begin with

Endothermic is in the first place ice melting. For comparison, to melt 1 kg of 0 deg C ice needs the same heat as warming up 1 kg of water from 0 to 80 deg C. // NaCl dissolving in water is thermally almost neutral. // Dissolving of CaCl2 is exothermic, of CaCl2 . 6 H2O is endothermic.

There always is a dynamic equilibrium — because ice can evaporate, with measurable vapor pressure (between 100 and 600 Pa in the temperature range where saltwater is in equilibrium with ice: "salt" is taken here to mean sodium chloride). When the vapor recondenses onto salt particles, it creates a salt solution. This solution then becomes the liquid phase into which ice melts.
The implication there is a kinetic aspect to the melting of ice by salt. As the temperature drops so does the amount of vapor pressure (note the wide range quoted above) and thus the rate at which the dynamic equilibrium can operate. Thus we find salt losing effectiveness even at temperatures somewhat above the nominal thermodynamic limit. Those of us using Fahrenheit temperatures have a rule that sodium chloride is not really effective below $\pu{0 °F}$ (about $\pu{-18 ^\circ{}C}$).

So the equilibrium between ice and water exists at all temperatures? And no matter how low the temperature gets there will be some water over it you trying to say?? But that is more problematic ,one can ask if that water (the whole ice +water system) is at that low temperature why it is not freezing

Ice is always in equilibrium with water vapor. There is, however, a lower temperature bound for a liquid phase also to be in equilibrium.

The dynamic equilibrium is between ice and water vapor [perturbed by atmospheric or mechanical pressure]. Addition of a salt or any solute dissolves a small amount of water.  The resulting solution has a lower chemical activity and vapor condenses lowering the pressure. The heat of condensation melts some ice and some ice evaporates to attempt to raise the VP. Since the heat of sublimination is greater than the heat of condensation the temperature lowers until the vapor pressure of the solution and the ice are the same. In an adiabatic situation the result is freezing point lowering; in real life on the roads heat is added from the environment so there is no equilibrium. There is a slushy mess until the water and ice evaporate leaving the solute. Since addition of a solute lowers the vapor pressure of both solid and solution adding salt actually increases the time of slushy roads over simple plowing.
[This is happening on the right side of the water-ice-salt phase diagram not the left side. To examine the process without the atmosphere use the water-ice vertical line in the water phase diagram and imagine the results of drastically lowering the water activity.]

The mechanism of ice melting in the presence of salt can be explained to high school students by using a human analogy (Sorry for the analogy). It may be stated that salt "wants" or "is happy" if it can make a solution in the presence of water. When salt touches ice, it "feels" that it is nearly water. It just needs a little bit of heat to make water and get dissolved. But there is no heat around. So salt feels a way of producing heat when there is no heat available. Each grain, each piece of ice is transformed into two parts : the outer layer and the core. The core (or inner part) gives heat to the outer layer, that melts, dissolves the salt and flows away. Of course the inner core gets colder, and soon becomes the outer layer. The same phenomena is repeated until the remaining inner part becomes too cold to melt, at about -$20$°C. Sorry for the "human" interpretation of the phenomena. It is wrong, I know. It is even contrary to the 1st principle of thermodynamics. Salt is never "happy" or never "wants" something. But in my personal experience, this explanation can be accepted or understood by high school students.


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Using biochemical standard state versus chemical standard state
Can I use the table below to calculate the standard cell potential, $E^{\circ}$, of ethanol oxidation to acetaldehyde (EtOH + 1/2O2 $\rightarrow$ Acetaldehyde + H2O)? Does it matter that the half-reactions provided are in the biochemical standard state (ie at pH 7)? I'm asking because the H+ terms in the half-reactions cancel out, and H+ isn't in the net reaction.

(x + 0.421) - (y + 0.421) = x - y


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How are these green, blue, and purple markers fluorescing under monochromatic 589nm (yellow) light?
I have some green, blue, and purple markers which clearly fluoresce under monochromatic light with a lower wavelength. Here they are illuminated by 589nm light emitted by a low pressure-sodium lamp (SOX light):

And here lit by the diffuse light from a a 532nm +/- 10 green laser:

This is surprising because fluorescence is defined as:

the visible or invisible radiation produced from certain substances as a result of incident radiation of a shorter wavelength such as X-rays or ultraviolet light.

By my understanding, that means it shouldn't be possible for either the blue or purple marker to be fluorescing under either light source because blue and purple light has a shorter wavelength.
What is happening?
More Photos
Markers lit by 650nm +/- 10 red laser light. The blue marker looks like it might be slightly blue to my eye, but it's hard to say.

Markers lit by sunlight:

Markers lit by incandescent light:

Low Pressure Sodium Lamps
The yellow light is a low-pressure sodium lamp (also known as a SOX lamp). As wikipedia describes:

These lamps produce a virtually monochromatic light averaging a 589.3 nm wavelength (actually two dominant spectral lines very close together at 589.0 and 589.6 nm). The colors of objects illuminated by only this narrow bandwidth are difficult to distinguish.

Viewing the SOX lamp through diffraction grating with 100 lines/mm:

In contrast to an incandescent light viewed through the same grating:

In contrast, "high-pressure sodium lamps emit a broader spectrum of light than the low-pressure lamps, but they still have poorer color rendering than other types of lamps" per wikipedia. High pressure sodium lamps are more common in streetlights.
The green laser shined at the wall through the diffraction grating:

Unlikely Explanations
Both prior research and answers/comments here make these seem like unlikely explanations. I have retained them for context.
I found a chemistry question here which explains that:

[Emitting light at a shorter wavelength] is certainly possible to make in a fluorescent / phosphorescent light scenario, but the engineering challenges (or, more specifically - the cost) involved in such a solution would be steep.

Another answer notes:

This method is called two-photon microscopy. Two photons of longer wavelength are absorbed by a dye molecule which then emits one photon of shorter wavelength. As mentioned before, this process is highly unlikely, but this property can be used to reduce the size of the fluorescent spot in confocal mocroscopy.

These answers provide a mechanism by which emission of green, blue, and purple light is possible, but make it sound like that only happens in exotic, expensive materials. Noting the process is "highly unlikely" also makes me expect that the blue and purple fluorescence would be much dimmer than they are.
These markers were bought in Thailand for just a couple dollars. While they glow the brightest of any green-purple objects I have, they are far from the only thing I own that exhibits this phenomenon.

Not two photon absorption, you would need an very bright  laser source to do this.  If these are dyes then just normal absorbance/fluorescence will explain observation, but if semiconductors  or nano particles then upconversion is just possible.

This is an interesting observation and one can appreciate that the OP thought about two photon absorption as a. A couple of observations first that challenge the hypothesis. Ordinary sodium lamps or even street sodium lamps are not monochromatic at all. If you have diffraction grating, view this lamp via narrow cardboard slit and you will see an entire line spectrum which is very beautiful. The Na-D 589 nm line is intense but other shorter wavelengths are present too. This is why you can see the markers, but they are not fluorescing at all in my opinion. If it were pure 589 nm, the markers would be appearing black just like the two on the right. You can test the fluorescence rather in solution with a yellow highlight. Dip the tip in water and view it in sunlight and in this low pressure lamp. You will see the difference.
Two photon absorption has a very very low probability. For that one needs an intense flux of photons and this is achieved with lasers as already commented above.

I added a picture of the light from the low-pressure sodium lamp to the original post. I also added photos of the markers lit by monochromatic laser light.

@HenryMarshall, Nice work and hopefully now you see that was not fluorescence or two photon fluorescence.

Thanks. I am even more confused now. The crisp lines from the diffraction grating has me convinced that both the SOX lamp and green laser are in fact monochromatic. Does this seem true to you? If the light is monochromatic, how is it possible that the markers are showing blue and purple?

@HenryMarshall, The SOX is not monochromatic. It has green to green. Otherwise the markers would be invisible (black).


text:
Is it not possible to prepare LAH using aluminium hydride?
Wikipedia lists this reaction for preparation of lithium aluminium hydride:
$\ce{4 LiH + AlCl_3 -> Li[AlH_4] + 3 LiCl}$
But is it possible to prepare it by reacting aluminium hydride with lithium chloride?
$\ce{Al_2H_6 + LiCl ->Li[AlH_4]}$
(I have not written all the products which may form)
If no, why?

First, at least try to enumerate the formal reaction, before asking if it is possible to do so.

Could you expand this comment into an answer if you are trying to hint that it won't be possible? I couldn't guess what products may form.

First off, aluminum hydride is not a molecular compound. It has a macromolecular structure with empirical formula $\ce{AlH3}$. Since the compound is not molecular, the empirical formula is the one we use.
You should, indeed, include all the products in the proposed reaction in order to ubderstand it properly. There must be a chloride product, and we can reasonably assume it's lithium chloroaluminate. Upon adding this product and balancing the equation the following is obtained:
$\ce{4AlH3 + 4LiCl -> 3LiAlH4 + LiAlCl4}$
The problem here is you are requiring the aluminum hydride to "disproportionate" its hydride ligands, with one mole surrendering all the hydride ligands to the other hree moles. In other words, you require the same material to act as both a hydride ion donor and a hydride ion acceptor. This generally does not happen.
Instead, we need to be adding hydride ions from an external source that provides them more readily than aluminum hydride (whose metal atom is accepting the hydride ions). Lithium hydride is predominantly ionic (thus a good hydride ion source) and relatively easy to dissolve into polar organic solvents (which solvate lithium ions well). Hence the reaction between lithium hydride and aluminum chloride us the one to use.