text:
Why do laundry detergents go bad?
Common lore on the internet is that liquid laundry detergents go bad 1-2 years after they are purchased. Is this true? I'd like to understand the chemistry behind this. This is the laundry detergent I am researching for o-chem.

A year or two? Mine gets exhausted long before then.

Hanging out with the wrong crowd will do it.

"Common lore" is not exactly evidence. Can you cite anything to back that up the lifetime of laundry detergents? What happens after that time -- are they no longer effective?
That said, many laundry (and dishwasher) detergents now contain enzymes, e.g., fungal alkaline proteases, to break down protein-containing dirt and stains(foods, blood, etc.). Since proteases are enzymes that break down proteins (cleavage),and since these enzymes are themselves proteins, they might shred themselves over time. In organisms, "Proteases... are synthesized in the inactive form so that they may be safely stored in cells... " Postponing self-cleavage could extend the shelf life of proteases.
Perhaps laundry detergents contain inhibitors to extend shelf life; some companies use dry granules for that purpose. Others are designing proteases more resistant to self-cleavage.
However, I've not noticed a difference in cleaning effect with old detergent; perhaps the issue of self-cleavage has been effectively ameliorated.


text:
Why does Euler's Number appear in so many equations?
Is there an intuitive explanation for $e$ in those randomly chosen equations from my theoretical chemistry book:
$$\Phi(x,t)=(2\pi \hbar)^{-1/2}\int_{-\infty}^\infty C(p){\color{red}e}^{i(px-E(p)t)/\hbar} \, \mathrm{d}p$$
$$\psi(x,t) = \phi(x){\color{red}e}^{-iEt/\hbar}$$
$$u_{ij}\equiv u_{ij|z}=(\overline{e}/2)|\mathcal E|\int \Phi_i^\star \cdot z \cdot {\color{red}e}^{ikx}\cdot \Phi_j \, \mathrm{d}V$$
I am aware that Euler's number is the base of an exponential function whose rate of growth is always proportionate to its present value, but it always bothers me that I can't really tell how it finds its way into such equations.

What makes you think it is more suited for the chemistry SE site and not the mathematical one?

@Poutnik Since I am concerned here with comtextual application, I think that one knows here more intuitive understanding in the chemical context.

Chemists seldom care about the Euler number, they have rather  applied approach to exp and ln functions.

@Poutnik what exactly do you mean with applied approach? I mean don't you have to understand something to apply it in new situations, I mean I somehow need to have a understanding to let Euler's number emerge in those equations.

Well, the Euler number is in none of those equations explicitly. And for using of the respective functions, you need not to know it either.

@Poutnik Hmm ok I see, maybe I just still have a problem with thinking more functionally when learning chemistry instead of trying to analyze every part.

Hello! This is a question that exceeds chemistry and science. However, you partly answered it. $\exp(x)$ is a particular solution to the differential equation $\mathrm{d}y/\mathrm{d}x = y$, so Euler's number appears. The second equation of your image is when you apply separation of variables to the time-dependent Schrodinger's equation (a PDE). The differential equation with respect to time, has the mentiond form.

Thus, from an epistemological point of view, the question may be formulated: if you consider that reality $=$ our conceptualization, why so many chemical and physical processes follow this simple differential equation? Other examples are first-order irreversible kinetics, radioactive decay, and the list goes on...

Further e is closely related to sin and cos, and hence periodic functions; before you know it you're into crystallography and anything that requires a Fourier transform, I.e. huge areas of spectroscopy

A better on-topic title of the question would be: "What is the use of Euler number in chemistry?"

Euler's number $e$ as such is less prominent than the exponential function $f(x)=\exp(x)$. It is this function, not $e$, that appears widely. For instance, the exponential function appears ubiquitously in assessing the asymptotic growth or decay of functions or populations.
In the real domain we certainly have $\exp(x)=e^x$, which is a consequence of the familar limit for $e$ combined with the algebraic properties of the exponential function. But that equality does not carry over into complex analysis. The function $\exp(z)$ is defined as a single-valued Taylor series that converges for all $z$, or equivalently as the unique solution to $df/dz=f, f(0)=1$; whereas $e^z$ is a branched function just like $a^z$ for other constants $a$. Mathematicians who deal with complex variables are well aware of the dustinction between the exponential function and $e$.

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text:
Can Spectator Ions Affect Reaction Kinetics?
Specifically, I am interested in the reaction of different hydroxides (in aqueous form) with $\ce{CO_2}$. I would like to determine which hydroxide would conduct the reaction the fastest. My initial thought would be that spectator ions would have no effect, and since the concentration of the hydroxide has a direct relationship with the reaction rate, therefore the most soluble hydroxides would react the fastest.
However, after seeing this article, I am confused as to whether spectator ions affect reaction rate or not. The abstract mentions "halide-acceleration effect," so I don't think the principle would apply to hydroxides, but I'm not entirely sure.

I don't know an answer here, but I just wanted to say that if "spectator" ions do have a big effect, then "spectator ion" isn't a good name for them.  (That doesn't mean the name isn't widely used; it just means, like many things in science, that the name doesn't make much sense.)

Without more specific information on the exact experimental setup it is impossible to speculate. However in "book problems" the real chemistry is oft ignored because the math is miserable. So to simplify the math concentrations are used in rate and equilibrium formulas instead of activities.  Concentrations should only be exchangeable for activities in very dilute solutions.

The initial premise "spectator ions would have no effect" is wrong.  Spectator ions do have effect. The term "spectator ion" means it is not involved in a reaction directly. But it has indirect effects, affecting e.g. solvation of other ions, ionic strength, relative permeability, viscosity, ion mobility, solubility, reaction kinetics etc, etc.

Everything present has an effect. The question is, is it distinguishable and measurable? Your proposal has many variables to consider; here are some. The amount of hydroxide, analysis by formulation or by titration, Activity of OH-, a good pH meter can help here, solubility of the hydroxide, carbonate, and bicarbonate; even the viscosity of the solutions. You will think up some more.
You intend to study the effect of different hydroxides in reacting with CO2. Since the differences are the counter-ions they will most likely have an effect. You will determine what the effect is and its importance. [avoid adding any halides]
There is much in the literature about this, check some out, a fresh look can be important.;

It does not have to be ions. Think of the familiar ammonia/hydrogen chloride experiment where you introducecthese gases at opposolite ends of the tube, and they diffuse towards each other until they meet and react somewhere in the "middle" (actually towards the HCl end due to the lighter ammonia molecules diffusing faster). This process is slowed down if you put "inert" nitrogen gas into the tube because of the impact of the obstructing nitrogen molecules on the diffusion process.

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text:
Energy Levels in Bohr's Diagram
When drawing the Bohr Diagram of Potassium with 19 electrons, the valence electron goes in the fourth concentric circle, but third energy level can hold up to 18 electrons? Does it mean that subsequent elements like Scandium will go back to the third energy level?

Bohr energy levels (quantum number n) are valid for atoms or ions with a single electron only, otherwise they split up and differ for different orbital types (quantum number l).
So the 4th, 5th levels start to fill before the prior level is filled, as 3d(4d) orbitals have higher energy than 4s(5s) orbitals.  Similarly, the 6th, 7th levels start to fill before prior 2 levels are filled, as described in Aufbau principle.
The general (with exceptions) orbital filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.  The respective orbital groups ns, np, nd, nf may contain up to 2, 6, 10, 14 electrons.
See also atomic orbital

ok that's what I suspected that subsequent energy levels get filled before prior level. Do you used "Bohr energy level" and "electron shell" interchangeably?

It can be used that way, but I haven't used the term electron shell as I rather avoid it. It creates false impression electrons are organized in some spheres. Bohr model is obsolete for almost 100 years and is taught mainly for its simplicity and for historical reasons.


text:
What kind of hybridisation is there around the central carbon atoms?

I am currently working on an assignment and I am confused about what kind of hybridisation exists for the first condensed structure CH3CN. Does the CH3 have sp2 hybridisation? I suspect its that because it has 3 hydrogens (1s^2) and 1 carbon (1s^2 2s^2 2p^2), allowing for 3 sigma bonds and 1 pi bond. I believe that the C standing alone would have sp3 because it looks like a bent shape? I am not too sure. Thank you.

CH3CN is not a bent molecule as drawn.

Should be straight.

In $\ce{CH3CN}$, the three atoms $\ce{C,C}$ and $\ce{N}$ are alined. The terminal $\ce{C}$ atom (in $\ce{CH3}$) is hybridized sp3. The central $\ce{C}$ atom is hybridized sp. The bond $\ce{C-C}$ between $\ce{CH3}$ and $\ce{CN}$ is sigma, and the bond $\ce{CN}$ has one sigma and two pi bonds.

Hi! Is the terminal C atom hybridised sp3 becuse its a tetrahedral shape? Or is there another way to understand why? Like how the hydrogen atoms are all single bonded to the carbon to create 3 sp3 hybrid orbitals? I'm trying to understand why. Thank you!

When $\ce{C}$ is linked to 4 atoms, it is hybridized sp3, and the bonds are all sigma. When $\ce{C}$ is linked to only 3 atoms, it is hybridized sp2, with three sigma bonds, and there is somewhere a pi bond. When $\ce{C}$  is linked to 2 atoms, it is hybridized sp and there are two pi bonds.

As Maurice said, the first carbon is sp3 and the second one is sp.
The easiest way to understand why is not to look at the shape (but the other way around - by knowing hybridization you determine shape), but at the number of atoms bonded on carbon. To bond 2 atoms together you need a sigma bond. If carbon has 4 atoms bonded on it, then it has 4 sigma bonds and is thus sp3 hybridized. If it has 3 atoms bonded, then it has 3 sigma bonds and one pi bond, and is sp2 hybridized. If it has only 2 atoms bound then it has 2 sigma bonds and 2 pi bonds, and is sp hybridized.
When you know the hybridization it is easy to determine the bond angles and shapes. Tetrahedral shape has 109.5° angles is typical for sp3 hybridization, sp2 has 120° angles and sp has 180° angles and linear shape.


text:
Why is sodium p-toluenesulfonate used as a supporting electrolyte in polypyrrole synthesis?
First, some preamble: I'm currently working on my thesis, which is about the electropolymerization of polypyrrole onto carbon fiber felt. My methodology, for the most part, follows a paper by one of my professors and a dissertation by one of his students. Both of them use sodium p-toluenesulfonate (NaPTS) as a supporting electrolyte during synthesis, which is the standard for other papers I've read on the topic.
Why is NaPTS the standard supporting electrolyte in PPy synthesis? Is it because pyrrole has limited solubility in water and p-toluenesulfonic acid increases its solubility? Are there any other electrolytes that would be suitable or even better for this application?
Thank you!


text:
Can Silicon hinder the precipitation of AgCl?
In my lab, I am having trouble with getting correct silver percentages.
For example, let's say a 99.99% silver sample. Digested in 10 mL of a 30% HNO3, fill up to 100 mL with DI, then concentrated (12M) HCl is added to precipitate the silver (around 10 - 15 drops of HCl around 1.50-2 mL). I was wondering if silicon can hinder the precipitation of silver because I recently had some contamination of silicon in my nitric acid and have not been able to get consistent 99.99% on this alloy.

Is the supposed accuracy that great?

Yes the accuracy of that alloy is 4 9's. Pure silver alloy

I mean the accuracy of the method to determine 99.99% Ag. What is the standard deviation of the measurement average or the respective confidence intervals?

Well the alloys measurement average is 99.99-100%. Most time I assay this alloy it usually comes out to around 99.97%. The accuracy of the method is 100% its just a simple gravimetric analysis.

Matt - RE: *The accuracy of the method is 100% its just a simple gravimetric analysis. * The method isn't 100% according to your information. You state that the sample is 99.99% Ag yet you average 99.97%. so you have a low bias which I would expect from a good gravimetric analysis. //  If you know you have silicon in the nitric acid then use that acid for something else and get a better grade of nitric acid. // Are you following some standard procedure?

The method of gravimetric analysis to my knowledge has a 100% accuracy. Not talking about my specific method, I do not know the measured accuracy of my procedure. My lab buys the highest grade nitric acid there is. I am following a standard procedure.

@Matt, "The method of gravimetric analysis to my knowledge has a 100% accuracy." In the real world, nothing has 100% accuracy. There is always some loss, or adsorption or contamination, or variability of atomic mass. Even were one to count atoms, there could be loss.

Okay I understand that nothing has 100% accuracy but no one has answered my question. Can the presence of silicon hinder the precipitation of silver chloride?

I can, as multiple other things that were not eliminated.

Gravimetric and coulometric methods are among the most accurate techniques known to analytical chemists despite their presence in dated textbooks. However, it's important to remember that in order to obtain highly precise results—such as when measuring atomic weights or ultrapure substances like the one you have—one must exercise high level of caution!
Dissolved silica should not interfere in the precipitation of silver chloride. If you are concerned about silicon, it is an insoluble solid. You might be concerned about dissolved silicon...if any. It may bias the results upon evaporation towards the higher side (just a conjecture). However, classical gravimetry books such as Hildebrand's Applied Inorganic Analysis, do suggest that silver cannot be accurately analyzed in the presence of silicate rocks. I think it will be rare to have the expertise and patience today to reach that level of accuracy as was done by the 19th century chemist with silver chloride precipitations/determination by Theodore W. Richards (Nobel Laureate).
Many questions for the apparently lower results for not getting 99.99 % purity or higher silver content from an ultrapure starting material.

Is your balance calibrated to the highest grade weights traceable to some standard organization?
Are you doing buoyancy corrections during weighing?
What is the solubility of silver chloride in that solution?
Are you using the right porosity filter for filtration
Is the method a standard method from a reputable organization?

In short if you are getting 99.97% for something labeled as 99.99%, then your analytical skills are highly commendable.

I assume dissolved silicate could  cause trapping part of Ag+ or AgCl in colloid, lost for precipitation.


text:
Confirmatory Test for Chitosan
Okay heres my problem. I cannot find this procedure online, can someone please help me?
Confirmatory Test for Chitosan Reagents and materials: Iodine/Potassium iodide (0.2 g iodine are added to 100 ml of a potassium iodide solution, w(KI) = 5%) and sulfuric acid = 1%
Small amount of chitosan powder was dissolved in water using a test tube.
Distilled water was placed in half of another test tube.
Then mix the dissolved chitosan solution in the distilled water test tube.
2-3 drops of Iodine/Potassium iodide was added.
The mixture was then acidified with 3-6 drops of sulfuric acid.
The solution was mixed using a stirring rod.
After the addition of Iodine/Potassium iodide, the precipitate turned dark brown. On the addition of sulfuric acid, the dark brown color turned to dark purple. This indicates the presence of chitosan (Lande et al., 2015).

You might be looking for the Van Wisselingh test for chitosan.
"Van Wisselingh test. This test is also called the 'chitosan-iodine test for chitin'. It is based on the hydrolysis of chitin to chitosan in the deacetylation process. While natural chitin does not react in this reaction, chitosan reacts with iodine, giving brown colored adduct. For chitin, the color seems black, but after a more thorough examination,
it is reddish-purple."
See also ResearchGate, Complex Formation of Chitosan with Iodine and Its Structure and Spectroscopic Properties--Molecular Assembly and Thermal Hysteresis Behavior, which is available for download sans paywall.


text:
Degeneracy of term symbols
Hello I was watching this video where term symbols are explained in detail. Moreover, he gives following illustration of a 2p1 state of hydrogen:

His calculation for the degeneracy is correct, however, how can I know which of the 6 electronic configurations corresponds to which term symbol?
For example, I don't understand why the $^{2}\text{P}_{1/2}$ state only has 2 states of the six. For example, look at the last state of the left row. If you add L=-1 to s=+1/2 you also get J=1/2. If you calculate every state through, you end with four states with J=1/2 and two J=3/2.
Where is my mistake? Or does one take the absolute value of L before adding it to S?

I have a feeling I know what you are asking, and I would like to answer if I can (if I'm right, I found this problem equally confusing), but the way you've written it makes it hard to tell. Could you edit your question to include explicit superscripts and subscripts? (Also, shouldn't one of those term symbols be $^2P_{\frac{3}{2}}$?)

You are correct, I typed one term symbol wrong. Edited accordingly

You can write $^{2}\text{P}{1/2}$ or $^{2}\text{P}{\frac{1}{2}}$ as $^{2}\text{P}_{1/2}$ or $^{2}\text{P}_{\frac{1}{2}}$

Thanks a lot :)

What do you think about this reference from the University of Santa Barbara in California? (Term symbols: why bother?)[https://web.chem.ucsb.edu/~devries/chem218/Term%20symbols.pdf];  Also what do you think of this reference? Term symbol?

First link is dead:(

A hydrogen atom was probably not a very good way to explain term symbols, because it doesn't show how the different quantum numbers relate to each other, and doesn't show the degeneracy well... If you can find it, I would highly recommend you look up Herzberg's Atomic Spectra and Atomic Structure; it explains everything very clearly.

If you add L=-1 to s=+1/2 you also get J=1/2

The main issue is that you are not being careful in distinguishing between $L$, which corresponds to the magnitude of angular momentum, and $M_L$, which denotes the orientation of the angular momentum (strictly its projection onto the $z$-axis).
When you write $+1, 0, -1$ on the p-orbitals these correspond to $M_L$, not $L$. (You might have seen the notation $m_l$; the small letters correspond to quantum numbers for a single electron, whereas $M_L$ signifies the sum of $m_l$'s. In this case there is only one electron, so they are the same thing.)
On the other hand, $L$ for a $\mathrm{2p^1}$ configuration is always $+1$, no more, no less. An electron in a p-orbital carries one unit of orbital angular momentum. (Again, you may have seen this denoted as $l$. In the one-electron case, $L$ is the same as $l$. With two electrons you have $L = |l_1 - l_2|, \ldots, l_1 + l_2$.)
The relationship between $L$ and $M_L$ is that for each value of $L$, the allowed value of $M_L$ ranges from $-L$ to $L$ in steps of $1$, which is exactly what you see in the statement $M_L = -1, 0, 1$.
The same is true of $S$ and $M_S$: the spin-up electron has $M_S = +1/2$ and the spin-down has $M_S = -1/2$, but both of these have $S = 1/2$.
Once you have $L$ and $S$, you obtain $J$ by the formula:
$$J = |L - S|, \ldots, L+S$$
in steps of $1$, which gives us $1/2$ and $3/2$. However, each $J$ term comprises several states which (again) have $M_J = -J, \ldots, J$. So:

The $^2\!P_{3/2}$ term comprises four states with $M_J = -3/2, -1/2, +1/2, +3/2$.
The $^2\!P_{3/2}$ term comprises two states with $M_J = -1/2, +1/2$.

How do you identify these states?
Because $M_J$ is a projection of the orbital angular momentum, as opposed to a magnitude, it can be obtained by simply summing up the projections of the two sources. In other words,
$$M_J = M_S + M_L.$$
You can now identify the state that has $M_S = +1/2$ and $M_L = +1$: it's the top-left one in your diagram. This state has $M_J = +3/2$ and thus must belong to the $J = 3/2$ term.
Likewise, you can find the state with $M_J = -3/2$ (bottom right); that must also belong to the $J = 3/2$ term.
The rest are slightly more complicated. You can't use a similar argument to assign the $M_J = \pm 1/2$ states to terms because both $J = 1/2$ and $J = 3/2$ can admit $M_J = \pm 1/2$. I'm fairly sure that you need to take linear combinations of the remaining states before you can map them to terms.
If you need to know this, it's probably worth a separate question (assuming it doesn't already exist somewhere here). However, I would strongly advise you to be comfortable with the various forms of angular momenta (magnitude / projection) before moving on to that.


text:
Combustion of organic materials in oxygen-18
Suppose I have some material that came from or was made from a plant or animal. I will assume that its molecular structure, along with carbon, will contain some oxygen as well as other elements. If I combust that materiel in pure oxygen-18 ($\ce{^{18}O2}$),the products will, of course, be $\ce{CO2}$, $\ce{H2O}$ and others.
Question: Will the $\ce{CO2}$ contain only the oxygen-18 or will it contain some of the oxygen-16 that was part of the molecular structure of the material?

Of course, even if the combustion is made in pure $\ce{(O-18)2}$ the $\ce{CO2}$ produced by this combustion contains some of the oxygen-$16$ atoms coming from the material to be burned.

To substantiate this answer: sources of oxygen by the organism can be all the carbohydrates (short sugars, glycogen, chitin), or carboxylic acids and their derivatives (lipids, proteins) to mention a couple of examples.


text:
Why doesn't ClO2 dimerise?
My teacher has told us that ClO2 does not dimerise due to delocalization of unpaired electron, but didn't elaborate on this. I didn't understand it.

The same question could be asked about the molecule NO. The reason may be that dimerization would produce a non-bonding molecular orbital. So Nature is reluctant to produce non-bonding orbitals.

The unpaired electron goes in a non-bonding orbital - as above. In the single molecule this weakens the bond - i.e. the 3 electron bond is weaker than the C=O bond. As mentioned above - as it's non-bonding - and therefore higher energy - presumably the dimer would be rather unstable.

All radicals dimerise, or even polymerise, to some extent, at least in low temperature. The resulting bonds may be so weak the equilibrium of thermal decomposition is far to the right at room temp., though.

@Maurice $\ce{NO}$ foes dimerize when condensed, and this plays some roke in its relatively high boilimg point for a diamotic molecule.

@Oscar Lanzi. Yes. NO dimerizes at rather low temperatures, well under $0$°C. I was speaking of the chemistry at usual temperatures, above $0$°C. I agree I should have mentioned it previously.

@Wallflower06 See my comments in https://chemistry.stackexchange.com/questions/81761/why-the-tendency-of-clo3-to-form-dimer-is-high

Thanks to all of you !


text:
How do I determine temperature and pressure rise whenever mols of gases are added in a isolated room?
Let's consider a fully isolated room, no gas or heat flowing from it.
We consider gases ($\ce{O2,N2,CO2}$) to follow the ideal gas law, and no chemical reaction occurs between them.
Pressure $p$, temperature $T$, volume $V$ and the amount of gas $n$ are known, and only $V$ is held constant.
What happens when $n$ suddenly changes?
You can imagine it as compressed gas container leaking $\mathrm{d}n$ moles of gas.
According to my intuition, both $p$ and $T$ would increase.
I can use the ideal gas law if I consider either pressure or temperature to be constant in order to obtain the other one, but I have no clue about what would happen in reality to both temperature and pressure.
The purpose of this question is to build "kind of" a simulation.

Even though you will say temperature and pressure would change, for a given arbitrary set of ($P,V$), volume should change with change in $n$ according to Avogadro's law.

I lack the knowledge but I assumed that Avogadro's law was only to be considered when setting temperature and pressure as constants.
Could you help me understanding what is wrong in my reasoning ?

Are you familiar with the open system version of the 1st law of thermodynamics?

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

The question can be answered by solving the material and energy balance for this case.

The tank is the control volume CV, of a constant volumne $V_\mathrm{CV}$, and its variables will be referred with the index $\text{CV}$. A constant molar flow rate in $\pu{mol s^-1}$ is exiting the CV.
The outlet won't have any index.

The process is depicted as follows:

1. Conservation of mass The rate of change of the amount of substance in the CV can only decrease due to the leaking
\begin{align}
  \color{blue}{\frac{\mathrm{d}n_\mathrm{CV}}{\mathrm{d}t}} = -\dot{n} \tag{1} \\
\end{align}
we have a decrease, hence the minus sign. The conservation of amount of substance here is valid, because we don't have a chemical reaction, so mass and amount are conserved quantities.

2. First law In its full form, where the index $i$ denote inlets and $o$ outlets, we have
\begin{align}
   \frac{\mathrm{d}\left[n_\mathrm{CV}\left(u_\mathrm{CV} +
   \dfrac{v_\mathrm{CV}^2}{2} + gz_\mathrm{CV}\right)\right]}{\mathrm{d}t}
    =&  \dot{Q} + \dot{W} \\ &+
   \sum_{i} \dot{n}_\mathrm{i}\left(h_i + \frac{v_i^2}{2} + gz_i\right)
   \\ &-
   \sum_{o} \dot{n}_\mathrm{o}\left(h_o + \frac{v_o^2}{2} + gz_o\right) \tag{2}
\end{align}
where our assumptions are:

We disregard the kinetic and gravitational potential energy vs the enthalpy.
There is no transfer of work.
There is no transfer of heat.

Thus, Eq. (2) with only one outlet of enthalpy yields
\begin{align}
  \frac{\mathrm{d}(n_\mathrm{CV}u_\mathrm{CV})}{\mathrm{d}t} &=
  \color{blue}{-\dot{n}}h
  \qquad\qquad \text{[Use Eq. (1)]} \\
  n_\mathrm{CV}\frac{\mathrm{d}u_\mathrm{CV}}{\mathrm{d}t} +
  u_\mathrm{CV}\frac{\mathrm{d}n_\mathrm{CV}}{\mathrm{d}t} &=
  h\frac{\mathrm{d}n_\mathrm{CV}}{\mathrm{d}t} \\
  n_\mathrm{CV}\mathrm{d}u_\mathrm{CV} &=
  (h - u_\mathrm{CV})\mathrm{d}n_\mathrm{CV} \\
  \frac{\mathrm{d}u_\mathrm{CV}}{h - u_\mathrm{CV}} &=
  \frac{\mathrm{d}n_\mathrm{CV}}{n_\mathrm{CV}} \tag{3} \\
\end{align}
Eq. (3) is as far as we can go. We will make two more assumptions:

The fluid inside the CV has the same variables as the fluid leaving the CV. Thus, $P_\mathrm{CV} = P$ and $T_\mathrm{CV} = T$, so that $h_\mathrm{CV} = h$.
The fluid behaves as an ideal gas. Thus, the differential of internal energy can be written as
$$ du_\mathrm{CV} = C_\mathrm{V}^\mathrm{ig} \, \mathrm{d}T_\mathrm{CV}
   \tag{4} $$

Combining both points we can simplify the nasty denominator in Eq. (3)
\begin{align}
  \require{cancel}
  h - u_\mathrm{CV} &= h_\mathrm{CV} - u_\mathrm{CV} \\
  h - u_\mathrm{CV} &=
 (\cancel{u_\mathrm{CV}} + P_\mathrm{CV}v_\mathrm{CV}) - \cancel{u_\mathrm{CV}} \\
  h - u_\mathrm{CV} &= P_\mathrm{CV}v_\mathrm{CV} \\
  h - u_\mathrm{CV} &= RT_\mathrm{CV} \tag{5} \\
\end{align}
Combining Eqs. (3), (4), and (5) and integrating
\begin{align}
  \frac{C_\mathrm{V}^\mathrm{ig}\mathrm{d}T_\mathrm{CV}}
  {RT_\mathrm{CV}} =
  \frac{\mathrm{d}n_\mathrm{CV}}{n_\mathrm{CV}} \\
  \frac{1}{R}\int_{T_\mathrm{CV1}}^{T_\mathrm{CV2}}
  \frac{C_\mathrm{V}^\mathrm{ig}\mathrm{d}T_\mathrm{CV}}{T_\mathrm{CV}} &=
  \int_{n_\mathrm{CV1}}^{n_\mathrm{CV2}}
  \frac{\mathrm{d}n_\mathrm{CV}}{n_\mathrm{CV}} \tag{6} \\
\end{align}
Strictly, even for an ideal gas, $C_\mathrm{V}^\mathrm{ig}$ is a function of temperature. We will analyze, for simplicity, the case where this magnitude is constant. The integration is not that hard if we consider a function like a polynomial, but we will get a result that is not analytic in terms of the final temperature $T_\mathrm{CV2}$.
Continuing with Eq. (6)
\begin{align}
  \frac{C_\mathrm{V}^\mathrm{ig}}{R}\int_{T_\mathrm{CV1}}^{T_\mathrm{CV2}}
  \frac{\mathrm{d}T_\mathrm{CV}}{T_\mathrm{CV}} &=
  \int_{n_\mathrm{CV1}}^{n_\mathrm{CV2}}
  \frac{\mathrm{d}n_\mathrm{CV}}{n_\mathrm{CV}} \\
  \frac{C_\mathrm{V}^\mathrm{ig}}{R}
  \ln\left(\frac{T_\mathrm{CV2}}{T_\mathrm{CV1}}\right) &=
  \ln\left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right) \rightarrow
  \boxed{\frac{T_\mathrm{CV2}}{T_\mathrm{CV1}} =
  \left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
  ^{R/C_\mathrm{V}^\mathrm{ig}}} \tag{7}
\end{align}
Now consider the ideal gas law in two situations
\begin{align}
  \require{cancel}
  \frac{P_\mathrm{CV2}\cancel{V_\mathrm{CV}}}
  {n_\mathrm{CV2}\cancel{R}T_\mathrm{CV2}} &=
  \frac{P_\mathrm{CV1}\cancel{V_\mathrm{CV}}}
  {n_\mathrm{CV1}\cancel{R}T_\mathrm{CV1}} \\
  \frac{P_\mathrm{CV2}}{n_\mathrm{CV2}T_\mathrm{CV2}} &=
  \frac{P_\mathrm{CV1}}{n_\mathrm{CV1}T_\mathrm{CV1}} \\
  \frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} &=
  \frac{n_\mathrm{CV2}T_\mathrm{CV2}}{n_\mathrm{CV1}T_\mathrm{CV1}}
  \qquad \text{[Use Eq. (7)]} \\
  \frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} &=
  \frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}
  \left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
  ^{R/C_\mathrm{V}^\mathrm{ig}} \\
  \frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} &=
  \left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
  ^{(R/C_\mathrm{V}^\mathrm{ig}) + 1} \\
  \frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} &=
  \left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
  ^{(R + C_\mathrm{V}^\mathrm{ig})/C_\mathrm{V}^\mathrm{ig}} \rightarrow
  \boxed{\frac{P_\mathrm{CV2}}{P_\mathrm{CV1}} =
  \left(\frac{n_\mathrm{CV2}}{n_\mathrm{CV1}}\right)
  ^{C_\mathrm{p}^\mathrm{ig}/C_\mathrm{V}^\mathrm{ig}}} \tag{8}
\end{align}

3. Remarks Eqs. (7) and (8) are the end. Final observations:

According to Eq. (7), as the tank is depleted, the temperature decreases as a power function.
According to Eq. (8), as the tank is depleted, the pressure decreases as a power function.
The pressure decreases, for the same level of depletion of the CV, more than the temperature does.
The total depletion of the tank, i.e. $n_\mathrm{CV2} \approx 0$, should not be evaluated with these equations. As temperature and pressure go down, a phase transition may occur in the fluid, so the equations are no longer valid because we don't have an ideal gas anymore.

Thank you SO MUCH for your answer ! It must have taken you some time to deliver that demonstration !
So methodic and detailed..! I feel priviledged that you put so much of your time for my question !
I'm really astonished by the community of this StackExchange!
About the complexity, it could have been too much if I didn't have enough algebra background.
I admit I wasn't expecting a complete demonstration but I suppose this is a more common kind of answer in the scientist world ahah !
Thanks again !

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

Suppose you are adding dn moles of gas to the tank.  From the open system version of the first law of thermodynamics, we have $$dU=d(nu)=h_{in}dn$$where U is the internal energy of the tank contents, u is the internal energy per mole, and $h_{in}$ is the enthalpy of the inlet stream.  (There is no heat transfer and no shaft work).  $$u=C_v(T-T_{ref})$$and $$h_{in}=u_{in}+P_{in}v_{in}=C_v(T_{in}-T_{ref})+RT_{in}$$where $T_{ref}$ is a reference temperature (of zero molar internal energy).  Substituting, we have $$nC_vdT+[C_v(T-T_{in})-RT_{in}]dn=0$$So, $$\frac{dT}{(T-\gamma T_{in})}=-\frac{dn}{n}$$Integraton gives:  $$\frac{T-\gamma T_{in}}{T_0-\gamma T_{in}}=\frac{n_0}{n}$$or$$T=T_0\frac{n_0}{n}+\gamma T_{in}\left(1-\frac{n_0}{n}\right)=
\frac{T_{0}n_0+(\gamma T_{in})\Delta n}{n_0+\Delta n}$$The pressure follows from application of the ideal gas law to the initial and final states:  $$\frac{P}{P_0}=1+\frac{\gamma T_{in}}{T_0}\left(\frac{n}{n_0}-1\right)$$If $(\gamma T_{in})\gt T_0$, both temperature and pressure increase with increase in the number of moles.

I gave the final answers for the temperature and pressure in terms of the number of moles at the end and at the beginning.  Isn't that what you asked for?

Yes thank ! I'm sorry, I didn't have time to study the answer yesterday !
And since I have very detailed answers it's difficult to pick ONE solution "winner".
Thank you very much for you answer !!

The question asks: "You can imagine it as compressed gas container leaking"... versus the answer has to do "with increase in the number of moles";  so for those applying these results from the question, just be careful with the sign of $n$, which should be negative for "leaking" and positive for "addition".  Also, I am not sure what assumption is made to come up with $\gamma$.  Is it reasonable just to symbolically reference whatever it is, or even assume an ideal monotonic gas where $C_v=\frac{5}{2}\times R$ and $C_p-C_v=R$ giving $\gamma=\frac{C_p}{C_v}=7/5$ when there is  $\ce{O2,N2,CO2}$?

@StephenElliottThe title says "whenever mols of gas are added..."

@StephenElliott Your equation for Cv is for a diatomic gas

The question states in the text "$\ce{O2,N2,CO2}$" , where $\ce[O2,N2]$ are diatomic in dry gas, and $\ce{CO2}$ is a small proportion.  Dry air has a gamma of 1.4 in this reference;  now on the question about that gas is being leaked into the room, the room gas pressure is given, but not significant.  Rather the gas pressure or temperature of the leaking gas canister is needed to calculate kinetic energy added to the room.

@StephenElliott That's your interpretation of the question, but it differs from mine.  Also, you used the word "monoatomic" in your first comment , not me.

These exact same results can be obtained based on using the closed system version of the first law of thermodynamics.

OP:  Did you mean that mols of gas are being added to the chamber of interest, or did you mean that mols of gas are being leaked out of the chamber of interest?

In fact, I asked for gas being introduced in the chamber of interest (from a bottle of gas leaking inside as an example).
But my sentence wasn't clear enough. It's my bad. I'll make sure my next posts are clearer.
On the other hand, though, Stephen Elliot's answer is still useful because at some point I'm gonna need to simulate bottle of gas also ^^
Thanks again for your answer !

So you are really interested in the more general situation of a container of high pressure leaking into a larger container of lower initial pressure (i.e., the changes in both containers) after a certain number of moles has leaked out of the higher pressure container?

IF the system is isolated and the source of the gas part of the room at the same T and V of the room constant the expanding gas can do no work and there will be no change in T from the simple mixing. The massive Kinetic energy from the directed air stream will eventually turn into heat. This can be eliminated by constraining the gas to diffuse randomly. There will be a Joule-Thompson effect either releasing or absorbing heat depending on the temperature. Finally, the entropy of the expanding, released, gas will increase, this means the entropy of the Universe will increase.
I imagine a pressurized helium balloon with a non-plastic wall but porous to helium, so no polymer crystal effects, slowly releasing the helium until the partial pressures equalize increasing the overall pressure at constant volume.
A correction: The compressed gas has potential energy equal to the work done in filling the tank. Upon release this is converted to KE, the wind generated. Eventually, as the wind is stilled, this becomes heat. This cannot be eliminated by a diffusion process if there is a pressure difference.

From the problem statement, we can have an initial $P_1, V_1, \text{ and } T_1$ with no gas or heat heat flowing from it, and the model of the Ideal gas Law and the Universal Gas Constant so that:
$$P_1 \times V_1 = n_1 \times R \times T_1$$
And quoting from the Reference 2 Universal Gas Constant, then

The universal gas constant, also known as the molar or ideal gas constant, is:

$$ R^* = 8.3144621\text{ }(75)\text{ } \frac{J}{\text{mol} \times K} $$

The gas constant for a particular gas is

$$ R = \frac{R^*}{m} $$

where m is the molecular weight of the gas. For a mixture, the "molecular weight" is a weighted mean of the molecular weights of the components:

$$m=\left( \frac{f_1}{m_1} +  · · · + \frac{f_n}{m_n} \right)^{-1}$$
$$\text{where } m_1, · · ·, m_n$$

are the molecular weights of the n gases,

$$\text{and } f_1, · · ·, f_n$$

are their masses relative to the total mass of the mixture.
The gas constant for dry air is

$$R_d = 287 \frac{J}{K \times kg}$$
The problem statement, "They don't create any chemical reaction (it would be O², N² and CO²)" is not precisely stated dry air.  This is a difficulty with the problem statement which may affect the answer quality, because when $R_d$ is used in the ideal gas law it may different from $R_\text{1 for the question}$ in the context of the question presentation.
Technically we do not have everything required to calculate the R value for the gas mixture in the room.  Hence, the answer needs to be conceptual about how to proceed (and I am amending the question a little with more detail to make it solvable), and then the actual Gas Constant for the room can perhaps be obtained later to follow the similar procedure to arrive at the desired answer, starting with the assumption that all the gasses are dry air and using the corresponding gas constant $R_d$ as:
$$R_d = 287 \frac{J}{K \times kg}$$
$$\text{and initially } P_1 \times V_1 = n_1 \times R_{1\text{ d}} \times T_1$$
$$\text{and finally } P_2 \times V_2 = n_2 \times R_{2\text{ d}} \times T_2$$
Now what happens with the next event in the question statement, "What happens when n suddenly changes?"  The usual meaning of "suddenly changes" is a model of an Adiabatic Process, and from this reference:

If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature
does not change; this is also called a Joule expansion.

The question here states: "You can picture a compressed gas container leaking n moles of gas".  When I imagine this, I see a picture of a gas container at very high pressure compared to its environment, and just a little bit of gas leaking.  This is not the scenario where the gas container is almost empty and at equilibrium with the atmosphere that it leaks to since that entity is completely outside of the problem statement.
Ideal Gas Thermodynamics: Specific Heats, Isotherms, Adiabats has the following illustration which shows what I envision from the problem statement.  Some gas is let out of the gas container quickly resulting is a situation similar to that pictured here.  When the high pressure gas from the container is let out, the volume of the gas inside the container and that outside of the container suddenly increases without additional significant release of heat (beyond that already in the gas that is leaked) to the environment.
Also, since this is a small and sudden leak, the work can be estimated as a relatively constant pressure times a small change in volume as:
$$\delta W_{\text{work from the container}}= P \times \delta V_{\text{volume of container and environment}}$$
$$=\delta n \times R_d \times T_{\text{Adiabatic constant in the gas container}}$$

Keep in mind in the above picture, that there is a sudden motion of the piston, dividing the molecules below the initial piston position and the molecules above the initial piston position finally into two separate regions (not shown in the picture).  With Adiabatic expansion, those molecules above do not have enough time to influence the molecules below.  And thus, those with a certain average energy before the sudden piston movement, below the piston, retain their average energies, independent of the molecules above the sudden piston movement.

From this equation one can see that the average internal energy of each molecule does not change, and rather it is the loss of molecules $\delta n$ during the gas release from the container that changes the complete energy stored in the container.  The temperature $T$ is related to the average energy of each ideal gas molecule in the gas container, not to how many gas molecules there are.  Thus, $T$ is constant, and total energy loss in the container is just the average energy per molecule times the number of molecules.
The initial temperature $T_1$ of course needs to be independent of the environment temperature that the gas is leaking to $T_\text{environment}$ because this is envisioned as a high-pressure gas container with nominal influence of the environment that it is leaking to (where the details of the environment's temperature and so forth are not given in the question statement).
So finally,
$$\delta W_{\text{work from the container to the environment}}= $$
$$= P \times \delta V_{\text{volume of container and environment}}$$
$$=\delta n \times R_d \times T_{\text{Adiabatic constant in the gas container}}$$
implies:
$$\delta \left(P \times V_{\text{container}}\right) = \delta P \times V_{\text{fixed size gas container}}$$
$$=\delta \left( n \times R_d \times T_{\text{Adiabatic constant for Joule heating in the gas container}}\right)$$
$$=\delta \left( n_2-n \right) \times R_d \times T_{\text{Adiabatic constant in the gas container}}$$
Just like a natural gas container of high pressure gas, the pressure starts out high and slowly decreases according to how much gas
$$\left( n_2-  n \right)$$
is suddenly used, while the natural gas or $CO_2$ container volume remains constant, like in the Adiabatic Expansion of Carbon Dioxide, where work is done by the gas after expansion, where its pressure is lowered to atmospheric pressure, and the pressure inside the canister of gas is proportional to the amount of $CO_2$, with the temperature of the canister being nearly constant, according to Adiabatic expansion.

Thus, contrary to the initial perception that: "According to my intuition both Pressure P and Temperature T would increase." It is quite different.  Think like it is a metal high pressure gas canister for a fixed volume of gas, like for soda production.  A release of gas decreases pressure in the tank and the volume remains fixed.  Of course the room is large and its temperature is not changed significantly with a small leak of gas from the canister.  Rather the pressure inside the room increases if the room is sealed (aside from the gas leak of the gas canister) in a proportional manner to how many moles of gas are added to the room from the gas canister.
Also, Temperature does not increase. Rather, because this is Joule work that the gas performs, and each molecule inside the container retains its average energy, the Temperature remains the same.
Finally then, for compressed dry air (with no liquid component inside the tank, we have):
$$R_d = 287 \frac{J}{K \times kg}$$
$$P \times V_{\text{in the container}} = $$
$$=n_{\text{inside the container}} \times R_{1\text{ d}} \times T_{\text{essentially constant due to Joule expansion from the container}}$$
Thus, taking account that $n_2=n_1-\delta n$ and $V$ is essentially constant and also $T$ is essentially constant, we have:
$$P_{\text{inside the gas container}} \times V_{\text{essentially constant in the container}} = $$
$$=\left(n_1 - \delta n \right)_{\text{inside the container}} \times R_{1\text{ d}} \times T_{\text{essentially constant in the container due to Joule expansion from the container}}$$
I am only calculating the parameters of the gas inside the tank because the relatively low pressure of the room are not deemed by the question to be significant enough to include there within.  And for most gas tanks at high pressure, the atmospheric pressure and temperature only have a nominal influence, like in the case of gas (only gas, no liquid) for cooking food, for instance.

Same as for Metal Storm, thank you VERY MUCH for the detailed answer !
This is amazing, you quoted every part of my question to justify every step.
Even though you both took my question under different angles (because I clumsily mentioned compressed gas container ^^'), you have a different conclusion being :

MetalStorm conclusion is that temperature vary along pressure.
You consider temperature remains the same.

Nonetheless, the only reason I don't upvote your answer is because my account is too recent for voting. ^^'

How can 2 answers with diametric results be correct? Your isolated constant volume room allows no work and Q and U do not change. It is essentially a free expansion of the added gas into the open space between molecules. You specify ideal gas behavior, so the only additional energy is the potential to kinetic energy of the compressed gas that is released.  The change in entropy is important. What you have described is a gas chamber.

@jimchmst The problem statement says:  "You can imagine it as compressed gas container leaking dn moles of gas."  From the problem statement I understand that the question is essentially similar to that of a gas chamber, which leaks after its initial conditions of as you stated "Your isolated constant volume room allows no work and Q and U do not change".  It is only later that $\delta n$ moles escape, they take with later their kinetic energy with them, lowering the total energy in the container, but not significantly the average energy from the problem statement of Adiabatic expansion.


text:
Is heat treatment the only way to increase grain size in steel?
Heat treatment is a well-established method for increasing grain size in steel. Is there any other approach known to man? I cannot seem to find anything via web search.

Chemists are not the best group to ask this. Metallurgists may help, but not sure about it in this specific case. Both domains overlap just marginally  and chemists may have the needed experience rather incidentally.

Second possibility: physics.se, because the site maintains a dedicated tag solid-state-physics (examples of questions).

Could be helping.  But as SSP is a huge and complex domain of science, I cross my fingers for the artist.

The examples are already more specific to grain within the tag of solid-state-physics.  Some posts' titles about grain size (examples) are encouraging to address the question there. Though, how to overcome activation in a slab of metal without heat by fire, or radio (contactless zone melting with a high-frequency RF coil) ...

Okay I will try the physics forum -- thank you both for the pointer!

@artist_and_not_EE_by_training, high magnetic fields have been investigated for controlling grain size: https://www.cambridge.org/core/journals/journal-of-materials-research/article/abs/effects-of-high-magnetic-field-on-austenite-formation-and-grain-size-in-a-hypereutectoid-carbon-steel/F7ADA09CCDF3B3C67D03A5D274765410 . But of course, that still involved heat treatment.


text:
Isochoric heating of a mixture of ideal gases with different heat capacities
I was thinking about the following situation:
Suppose we have an equimolar mixture of 1 mol argon gas (C=1.5 R) and 1 mol of nitrogen (C=2.5 R) and we heat it isochorically with 10 kJ from room temperature.
How can one calculate the final temperature then? I know we need to use the relation dU=C*dT, because the heat energy is converted to internal energy, but also argon should heat more quickly than nitrogen.
Does one need to use just the average of the heat capacity (2R) to calculate the temperature change or how can we do it? I couldn’t find practice problems with a similar question.

Note that in the case of gaseous mixtures, particular gases are not being heated separately as macroscopic entities, all is done on the molecular level. It does not matter how much of thermal energy a particular molecule of either of gases obtains during heat conduction. As this energy is extremely quickly (locally) evenly redistributed across all respective degrees of freedom by mutual collisions with the typical collision rate $\pu{E10 s−1}$). This still allows the macroscopic temperature gradient.

Work with the system heat capacity
$$C_\mathrm{V,sys} = n_1 \cdot C_\mathrm{1,V,mol} + n_2 \cdot C_\mathrm{2,V,mol}.$$
Then $$q = C_\mathrm{V,sys} \Delta T$$

The alternative approach is to formally use the effective molar heat capacity as the weighted average of molar heat capacities of involved gases:
$$C_\mathrm{V,sys, mol} = \frac{n_1 \cdot C_\mathrm{1,V,mol} + n_2 \cdot C_\mathrm{2,V,mol}}{n_1+n_2},$$
and then to use the molar amount of mixture molecules, regardless of the molecule types:
$$q = n_\mathrm{mix} \cdot C_\mathrm{V,sys,mol} \cdot \Delta T$$

Note that, generally, there is nonzero heat of mixing, as ChetMiller correctly pointed out.  But, for gaseous mixtures like $\ce{Ar}$ + $\ce{N2}$, at usual conditions, is this deviation negligible.  The heat of mixing is an issue rather for calorimetry in evaluation of heat capacities of mixed condensed phases.
There is more important effect of the temperature dependency of molar heat capacities of polyatomic gases, as quantized vibration modes become more active in higher temperatures.
Wikipedia, molar heat capacity of chosen gases as function of temperature:

Ah so it’s easier than expected, just sum the molar weighted heat capacities and calculate dT.

Q=CvDeltaT;  deltaT = Q/Cv ;  Cv is the weighted average by moles.

@jimchmst Does it mean the average we are using here is not actually the algebraic average but instead average by moles? I think this makes the difference.

@Proscionexium It is called a weighted average, where the weights are here molar fractions.

This applies only to an ideal gas mixture, where the heat of mixing is known to be zero.

Sure, but I thought a caveat would be worth mentioning.

Added mixing notes to the answer.


text:
Why are source and system at same temperature in a Carnot cycle
I am learning Carnot cycle and Carnot theorem from my textbook. I am also familiar with the graph of Carnot cycle featuring a pair of adiabatic and isothermal processes.
Say that the source is at a temperature $T_2$ and the sink at a temperature $T_1$. My query is only regarding the isothermal processes in it so I would take only isothermal compression or expansion.
It undergoes isothermal expansion from $V_1$ to $V_2$ being in contact with the sink at  temperature $T_2$. Then the work done ($n=1$) would be $$w =-RT_2 ln\frac{V_2}{V_1}$$
Since the process is isothermal and the work done is negative due to expansion, a $q$ amount of heat is absorbed by the system from the source. We also know that heat flows only due to a difference in temperature, therefore the system must be at some temperature lower than that of the source.
But the formula above asserts the temperature of the system as $T_2$ which is same as source's temperature. This means that the source and the system must be at the same temperature $T_2$ which shouldn't be right either as in this case there would be no flow of heat and the process will be no longer isothermal. Could anyone figure out what has gone wrong?

One ancient Greek dude said something along the lines that he's not going to be present at his own death, so why bother about it. Same thing here. When the system is at $T_2$, there can't be and isn't any process anymore. The process was before that.

We depict Carnot's process in the $T-S$ diagram:

Steps $2$ and $4$ are adiabatically reversible, also called isentropic, so we don't need to worry about those two. We focus in steps $1$ and $3$. Since they are isothermal, for an ideal gas, there is no change in the internal energy. Thus, $q = -w$ or
\begin{align}
  w_{12} = -RT_1\ln\left(\frac{v_2}{v_1}\right)<0 \qquad
  q_{12} =  RT_1\ln\left(\frac{v_2}{v_1}\right)>0 \tag{1/2} \\
  w_{34} = -RT_2\ln\left(\frac{v_4}{v_3}\right)>0 \qquad
  q_{34} =  RT_2\ln\left(\frac{v_4}{v_3}\right)<0 \tag{3/4} \\
\end{align}
For the surroundings, Carnot introduces the concept of a heat reservoir. This means "something" that can exchange an infinite amount of heat without a change of temperature. You may laugh at this, but they exist for practical purposes on earth. The most simple example is the ocean. We can dump a lot of high-energy toxic residuals.... Do you think the temperature of the ocean is going to change? No.
The entropy change for the a reservoir, since $T$ is constant, is just the famous formula $\Delta S = Q/T$.
We also remember the entropy change for an ideal gas as a function of temperature and volume
$$ \Delta S = \int_{T_1}^{T_2} \frac{C_V^\mathrm{ig}(T)}{T} \,
   \mathrm{d}T + R \ln\left(\frac{v_2}{v_1}\right) \tag{5} $$
First we will prove that the best outcome for the step $1-2$ is that the temperature of the system is equal to that of the surroundings. We will name this unknown temperature $\color{magenta}{T_\mathrm{sur,I}}$, as in the picture, in pink. Then we will discuss it.

Step $1-2$ Here, the fluid expands so the system does work. In order to maintain the temperature constant, the surroundings must give heat to the system. The entropy change of the system, by Eq. (5) since $T$ is constant, yields
\begin{equation}
  \Delta S_\mathrm{sys,12} = R\ln\left(\frac{v_2}{v_1}\right) > 0 \tag{6}
\end{equation}
For the entropy change of the surroundings, the heat that it loses is exactly Eq. (2), but with its sign reversed
\begin{equation}
  \Delta S_\mathrm{sur,12} = \frac{-q_{12}}{T_\mathrm{sur,I}} \rightarrow
  \Delta S_\mathrm{sur,12} = -\frac{RT_1\ln(v_2/v_1)}{T_\mathrm{sur,I}} \tag{7}
\end{equation}
Thus, the total entropy change in this step is the sum of Eq. (6) and (7)
\begin{align}
  \Delta S_\mathrm{12} &= \Delta S_\mathrm{sys,12} + \Delta S_\mathrm{sur,12} \\
  &= R\ln\left(\frac{v_2}{v_1}\right) -
  \frac{RT_1\ln(v_2/v_1)}{T_\mathrm{surr,I}} \\
  &= R\ln\left(\frac{v_2}{v_1}\right) \left(1 - \frac{T_1}{T_\mathrm{surr,I}}\right)
  \tag{8}
\end{align}
What can be the best possible outcome for the heat engine? We know the answer, and is that outcome that minimizes the entropy. Not only that, we can achieve its limit given by the second law, and by Eq. (8) this means that
\begin{equation}
  R\ln\left(\frac{v_2}{v_1}\right)\left(1 - \frac{T_1}{T_\mathrm{sur,I}}\right) = 0
  \therefore \boxed{T_1 = T_\mathrm{sur,I}} \tag{9}
\end{equation}
I leave to you the proof for step $3-4$, which is analogous, and derive that $T_2 = T_\mathrm{sur,II}$.

The idealization of Carnot engine stems from equaling Eq. (8) to zero. There is no need to do that, in fact, we can perfectly use a reservoir that is at a higher temperature than $T_1$. However, this inevitably brings some increase of entropy.
You are correct to affirm that to bring a change of temperature we need a driving force. This is what every phenomenological law about a flux tells us. Fourier's law states that a temperature gradient gives rise to a heat flux, Ohm's law that a density current flux gives rise to an electric potential difference, etc. Smaller is the temperature gradient, smaller will be the flux, and more time we will need to bring desired temperature changes to the system. This is an inevitable consequence of nature. We would like things to behave like Eq. (9) states, but things cannot move with Eq. (9). In real life, we need finite fluxes in order to move the system out of equilibrium, with the cost of an increase of entropy.
Eq. (9) represents a limit, in the sense that it is the best step for this heat engine. For practical purposes, to approach this in real life is not that crazy. Imagine your fluid in step $1-2$ is at $\pu{99.8 °C}$. It needs to recieve heat from a heat reservoir at a higher temperature. Boling water at $\pu{100 °C}$ can be chosen, and this is very near an isentropic operation, by applying Eq. (8). We will need a lot of boiling water, because as the reservoir gives heat, its temperature will go down. If the temperature of the system and the surroundings are equaled, beyond here, no temperature changes can be obtained.

Hmm, the arrows or the legends of the diagram must be reversed, as reversible adiabatic compression/expansion does not cause temperature drop/raise, but vice versa.

@Poutnik Hello! Wow, yes,  totally correct. Thank you very much for the comment, I have changed the labels now.


text:
What is happening in experiment coating a copper coin with zinc and turning it to brass?
Today in class we did the "gold penny" experiment, where we put metallic zinc and a penny in a zinc chloride $\pu{1 M}$ solution at $\pu{100 °C}$, which caused the penny to turn silver as the zinc coated it. Then, we held the penny to a flame to form brass.
I know this is a redox reaction where the zimc is being reduced, but I am confused about the details. What is the exact reaction that goes on? I think the zinc is being reduced into metallic form, and the copper is oxidized. However, I am confused about the specifics - do the $\ce{Cl-}$ ions re-form $\ce{ZnCl2}$ with the metallic $\ce{Zn}$? Or do they stay with the copper after oxidizing it?
Sorry if this is simple or obvious - I am 14 years old and just started my first chemistry class.

I found this video on a similar reaction today:
https://youtu.be/EaGDdTKFTHY?si=111nRXewSfsPLwZM
It seems to state that a voltage of about 1.1V is created between the zinc and copper, which is enough to drive the electroplating reaction.

Was the metallic zinc piece connected to the penny ? Could they form a galvanic cell  $\ce{Zn(s,zinc) + Zn^2+(aq) <-> Zn^2+(aq) + Zn(s, penny)}$? As Cu reducing Zn seems strange.

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To the person attempting to edit this and replace the word "penny" with the word "cent": in the US, it's called the penny; you would use the word "cent" to refer to the decimal portion of a price of something less than $1 (which we would equate as 100 cents).
Also, if you're going to edit, go all-in and fix all the things.

@Poutnik The zinc was granular, and it was sitting at the bottom of the beaker along with the pennies. Therefore, the pennies & zinc definitely had contact.

@Poutnik when doing research about the procedure, I found this:
“The trick in this step is to make sure all the pennies are touching zinc pieces, but not touching other pennies.
This works due to the zinc and copper starting to share electrons with each other and they start to chemically bond.“

The heat is needed to diffuse the zinc into the copper to make brass; an alloy.  Also, the flame must completely envelope the penny. And, be reducing to prevent oxygen from oxidizing the zinc when hot.  Cupper and zinc will form a broad range of alloys , to over 50 % zinc with various colors.  The most "gold" composition contains 30 % zinc.

Both the question and the other answer suggest reduction of zinc by metallic copper, which is impossible to any significant extent. Both the $\ce{Cu^{2+}/Cu}$ and $\ce{CuCl/Cu + Cl^-}$ couples have standard reduction potentials about 1 V greater than the $\ce{Zn^{2+}/Zn}$ couple, which would preclude any significant degree of reaction at equilibrium according to the Nernst Equation.
A more chemically plausible hypothesis is that a zinc(I) species is formed from the zinc metal and aqueous $\ce{ZnCl2}$. Wikipedia, citing Ref. [1], reports that molten zinc chloride dissolves metallic zinc and the solid formed by rapid cooling of this solution contains $\ce{Zn2^{2+}}$ ions:
$\ce{Zn^{2+} + Zn(s) <=> Zn2^{2+}}$
If such a reaction were to occur reversibly in aqueous solution, even to a minor extent, the copper could serve as a favored site for the reverse reaction with deposition of zinc. On this model the metallic zinc added to the solution as granules is effectively being redistributed onto the penny.
Once the zinc is deposited, the heating stage is simply an alloying of the zinc with the copper.
Modern-day US cents already contain zinc, so it may be possible to turn the surface to brass without the pre-deposition of a zinc coating from the solution.
Reference

Holleman, A. F.; Wiberg, E. (2001). Inorganic Chemistry. San Diego: Academic Press. ISBN 978-0-12-352651-9.

Thanks for the answer. On the tangent of the heating part, I was able to make a gold-ish penny just with a flame (albeit with colored bands of oxidization that weren’t present when melting the zinc onto the copper).

And it won’t work with a penny minted before 1982. It will work with some pennies minted in 1982, but not all, and it works with all pennies (except very rare U.S. Mint planchette errors) minted after 1982. (I was the first upvote, BTW.)

Modern US pennies are zinc with copper plating.

Starting some time in 1982 and continuing on, U.S. pennies are copper plated planchettes, with the planchettes being 99.2% zinc and 0.2% copper. The overall composition is 97.5% zinc and 2.5% copper and their clean uncirculated weight is 2.50g. Pennies in, e.g., 1975, were 95% copper and 5% zinc and their clean uncirculated weight was 3.11g.


text:
Reducing vaporization of concentrated hydrochloric acid solution
I have this bottle of hydrochloric acid solution. Although I do not know its concentration but I know it is quite concentrated. Every time I open it the fumes would get to all the places.
I have an idea of letting the bottle sit in cold water for a while before opening it. Would this reduce the amount of acid vaporizing? What are some other ways to do this?

Well, yeah, it could work... to some extent, but using fridge seems simpler. It's natural for concentrated HCl to fume, though, so use fume hood either way.

SAFETY - Do NOT put a bottle of HCl in your home fridge. I wouldn't store it in a lab fridge either. The out gassing of HCl will corrode any metal that it comes in contact with. The sophisticated solution is a vented acid supply cabinet.

Maybe a condensing column attached above the bottle?

HCl is very volatile, so even when reducing temperature will reduce it's vapor pressure I do not think it is enough to avoid the fumes. What we do in our lab is to use an acid dispenser. It is similar to the one in the picture below, although not so sophisticated. It works by suction and it will dispense the acid in the order of milliliters, through a small hole thus reducing the amount of fumes. This will allow you to control it better.
If you require to handle large amounts and you have no other option but to pour it and open the original bottle, then you have to use a hood. You will also require the hood to transfer the acid to the dispensing bottle.


text:
Difference between two forms of benzothiadiazole
Can someone please advise / point me to information regarding the difference between these two compounds:
https://www.combi-blocks.com/cgi-bin/find.cgi?HC-3087

https://www.combi-blocks.com/cgi-bin/find.cgi?JH-8440

The reason for this question is that rdkit assigns them the same 'inchikey', as if they were identical, or spontaneously interconverting substances (like tautomers).
But they aren't tautomers, as there is no atom being transferred, only electrons; and they are distinct commercial substances, sold in fairly large quantities, so I would guess they are individually sufficiently stable.
Many variously decorated commercial compounds containing either of these moieties are also found, reinforcing the above conclusion.
So it would seem that the machinery that calculates the inchikey is wrong in treating these two molecules as identical.
Any ideas?

EDIT summary of insights from the comments
According to both the people who commented so far, these two species are resonance structures of the same molecule.
The fact that different cas numbers, names, catalogue numbers from a single vendor are found for the two, would therefore seem to be incorrect, misleading and unjustified.
On the contrary, the fact that rdkit returns the same inchikey for the two seems to be correct.

Just learn to draw mesomeric structures, instead of using some "combi-blocks" - the second one has one bond too many.

Both structures have the same atom-to-atom connectivity (including H atoms) and the same formula of C6H4N2S.  Both rings of 2,1,3-benzathiadiazole are aromatic.  I think these are two resonance structures of the same compound.

Do you really think the two sulfur atoms in https://www.dougdiscovery.com/4-8-dibromobenzo-1-2-c-4-5-c-bis-1-2-5-thiadiazole.html are chemically distinguishable?

@Mithoron - surprising though it may sound to you, I do know how to draw chemical structures. I used the pictures from the vendor to show how they represented these molecules. And FYI, Scifinder (CAS) also draws these two molecules in the exact same way as Combi-Blocks does. If you have more information to provide, please do so. It is not really clear to me how making such snide comments or insinuations about people's intelligence or skills helps answering the question I am asking.

@CurtF. - that's what I thought, too, based on the principle that when only electrons are moving, we are talking about resonance, not tautomerism or anything that changes the identity of the molecule. But then how is it possible that the two species (and many derivatives) are represented as distinct, and sold with different catalogue numbers in vendors websites, described with different synthesis in papers, etc? BTW, we also need to establish if the representations here are both at least formally correct, because apparently someone (see note above) thinks one of the two is wrong.

Sorry for sounding harsh, but it boils down just to the knowledge about resonance (not intelligence, as you mentioned), Some structure drawing tools are just tools, users need to discern if results are good or not (they frequently aren't). In this case a slightly alleviating factor could be people buying into deprecated hypervalency - this second mesomeric structure is the "hypervalent" one. Also how some vendors draw it and sell, isn't exactly relevant, apparently can lead rather to confusion than to learning.

Could you link to some of those distinct syntheses from papers?  And also, just because there are two resonance structures, doesn't mean that both are equally prevalent.  I like the structure without the hypervalent sulfur better.

@Mithoron - well, I came here to discuss a scientific matter. You can choose to help or to make assumptions about my knowledge. On topic: 1) you still have not explained why you stated that the second structure is not correct. It is represented like that in Scifinder, as I mentioned. 2) as I wrote in my reply to CurtF., I too was initially inclined to think that these were equivalent molecules, given the identical inchikey etc, but came to to doubt my conclusion when vendors, scientific papers and Scifinder seemed to differentiate them. I cannot think of many examples where that happens.

I did say this structure is hypervalent, more precisely sulfur atom in there is, and hypervalency is a deprecated concept. Either way it's definitely just an additional mesomeric structure, definitely not a separate compound. Now I wonder if it would be even physically possible to be otherwise...

"the second one has one bond too many" - has it, really?

@Curt F. : Scifinder reports 125 references where the 1st molecule is a product. Example: "1,2,5-Thiadiazole 2-oxides: selective synthesis, structural characterization, and electrochemical properties" Konstantinova, Lidia S.; et al
Tetrahedron (2014), 70(35), 5558-5568. And it reports 8 references where the 2nd molecule is a product. Example: "Reactions of arylthiazylamides with internal and external fluoro electrophiles - formation of products with unusual structures" Lork, Enno; et al
European Journal of Inorganic Chemistry (2001), (8), 2123-2134.

Overall, the information you guys provided was useful and made me conclude what I mentioned in the edit. In essence, I should have trusted the fundamental principles of chemistry more than registry numbers, catalogue numbers, etc. I see people are voting to close this question. Please do so; or I can I will do it myself. Let's put everyone out of their misery. General comment on how this went: it would just be nice if people learned to exchange ideas without resorting to insults and condescendence. It does not reflect well on scientist when the only mode of communication they know is spite.

For the record: "Homework questions must demonstrate some effort to understand the underlying concepts." was the reason people provided for closing the question.
Wrong. The very first thing that I did was research these two substances, draw myself the resonance structures to see how they were conceptually the same (one electron pair from the S atom of the first one can be moved to the ring, all other pairs shift around until the last one reconnects to S, forming the N=S=N moiety).
I should have stopped at that and ignored the other misleading evidence pointing to them being different.

In practice, there is no difference between the two structures - they represent the same compound if you were to purchase them. That said, in terms of "correctness", the non-hypervalent structure is preferred, even though its Lewis structure still does not accurately represent the full electronic structure of the molecule. You could add some non-hypervalent charge-separated resonance forms which improve the description, as shown below.

The overarching truth is that occasionally you can find compounds with unusual Lewis structures available for purchase because they weren't sanity-checked. This is especially likely with combinatorial chemistry suppliers as they will often automatically populate their catalogues with thousands of structures. For example, sometimes you can find a cyclohexa-2,4-diene-1-one on offering, though on further inspection a chemist would realize these are actually a phenol tautomer, which typically only exist in part-per-trillion amounts.
But with benzothiadiazole specifically, the problem goes a bit deeper. While a trivial and satisfying Lewis structure can be assigned to it (or more precisely, benzo[c][1,2,5]thiadiazole), what about the related structure benzobis(thiadiazole) (more precisely, benzo[1,2-c:4,5-c']bis([1,2,5]thiadiazole)), which contains another fused benzothiadiazole ring opposite to the first? This is a well-known compound, of which hundreds of variants have been explored in organic electronics.
It turns out in this case that there is no "trivial" Lewis structure which can be assigned to the compound, at least not without a hypervalent sulfur atom. And curiously, that's what the literature has settled on - a quick Google image search of benzobisthiadiazole will turn up dozens of articles using the hypervalent structure shown below. It has become the standard in the field, and nobody particularly cares. It is possible to draw a non-hypervalent charge-separated resonance form, but it is ignored. The true electronic structure is even more complex, apparently including non-negligible biradicaloid character, which can be shown on a non-hypervalent Lewis structure.

As a last point, x-ray crystallographic data of benzobisthiadiazole compounds show that the two thiadiazole rings are typically completely equivalent (there is no single hypervalent ring, as the widely-drawn structures suggest), indicating that it really all boils down to Lewis structures just being limited tools at representing reality. You always have a trade-off between simplicity of representation and accuracy, and sometimes this trade-off is particularly unsatisfying.

Excellent, thank you very much for this detailed explanation! "occasionally you can find compounds with unusual Lewis structures available for purchase because they weren't sanity-checked" - that's quite bad indeed, but while one may forgive vendors for not being so strict with their checks, it remains shocking (to me) to find distinct cas numbers for species that do not even exist individually. I created a post about that. The phenol vs cyclohexadienone example is not strictly applicable: tautomers are indeed distinct species with, as you point out, measurable proportions in mixtures.


text:
Seeking Online Resources and Textbook suggestions
I want to learn chemistry, and I want to know if there is an online resource that would provide me with a list of textbooks, telling me the order to read them to bring me from a complete beginner to a college level. I have looked to see if any colleges released their course syllabus so I could follow, but I was unable to find any that also listed the textbooks. I also looked to see if the government has set something in place for self-education, but I was still unable to find anything. If something like what I have described does not exist, and if you still remember what textbooks you read, could you please tell me? Thank you in advance.


text:
How are enzyme inhibition constants assayed for "multi-inhibitors"?
Suppose there is a suicide inhibitor of an enzyme that reacts with the enzyme to form an inactive enzyme and another product. This "Another product", however, is capable of acting as a competitive inhibitor of the enzyme. An example of such an inhibitor is neostigmine, which reacts with acetylcholinesterase to give the carbamylated enzyme and a phenolic product that can also competitively inhibit AChE. If one wanted to assay the reaction rate constant for the formation of the inactive enzyme, how would one determine to what extent an observed decrease in enzyme activity is due to the suicide inhibition versus competitive inhibition by one of the products of the suicide inhibition?
Furthermore, if a studied inhibitor is thought to be able to deactivate multiple molecules of an enzyme (EA-3990 vs. AChE as an example), in a reaction scheme like this:
Initial Inhibitor + Enzyme-->Inactive enzyme+ Generated inhibitor
Generated inhibitor+ Enzyme--> Inactive enzyme+ competitive inhibitor
how could the reaction rate constant for the initial inhibition reaction be reliably assayed?

If it is true that the formation of the side product requires suicidal inactivation of the enzyme, seems like all you need to do is measure the rate of increase of the side product, e.g. by LC-MS.  Its presence would be directly proportional to enzyme inhibited.

I find the use of the word "assay" a little ambiguous here. An assay is a test, but here you seem to mean "measure". So your question seems to be how to measure the individual rate constants, right?

Experimental details would vary depending on the specific system. For example, if the inactivation is a stable covalent adduct, you could quench the reaction and use MS or some other relevant method to determine the fraction of enzyme with adduct at different time points. Or do a dilution/wash to remove the original inhibitor and the competitive derivative and measure activity to determine active/inactive enzyme.

@BuckThorn Correct.


text:
How can two separate phases interact with each other in an separatory funnel
Hello I was thinking about the following process.
I was working in the lab, and synthesized a chelator for various metals, iron, copper etc.; however, when extracting the mixture with water and ethyl acetate, the organic layer remained green. This made sense as the procedure for the chelator called for a copper (I) catalyst.
So to get rid of the copper (I) ions, I thought it would be a good idea to wash the organic layer with dipotassium EDTA. It actually got rid of the copper (I) and turned the organic layer brown (my chelator’s own color).
My question is now, how did this actually work? Water and ethyl acetate (to my knowledge) are in a good approximation not really soluble in each other. But how did the EDTA get the copper ions out of my chelator which was in the organic layer? I get that EDTA probably has a higher complexation constant than my chelator, but the EDTA solution couldn’t have diffused into the ethyl acetate (because it was charged, 2-).
But the resulting EDTA-Cu complex should be better soluble in water that’s for sure.
So by what mechanism does this work in general?

Did you shake the sample?

You can research interfacial diffusion in liquid-liquid extractions, finding papers like this one. Well, abstracts.

And here is a book chapter from a chemical engineering lecture introducing some quantitative models.

Just a sidenote, but water and ethyl acetate are substantially soluble in each other, just not infinitely. There are immiscible ratios, but roughly 5% of each solvent goes into the other phase. Compare with water and hexane, where the values are ~0.01%.

I did shake the sample.


text:
How do aqueous nanoparticle suspensions behave?
I have found many pictures of nanoparticle suspensions on the web, but I can find no indication of how these suspensions behave, e.g. do the nanoparticles remain in suspension indefinitely, or do they precipitate out after a while, and if they precipitate out, how long does that take?
Regarding appearance, if one had a suspension of iron oxide nanoparticles in water at neutral pH and an iron concentration of about 20 micro moles per litre, could this give the water a uniform brown tint when viewed through 20cm of water?
I believe that I may have produced a suspension of iron oxide nanoparticles via biological means, which is why I have included the biochemistry tag.
Update:
I understand that nanoparticles are considered to exist in the size range 1nm to 100nm.  Perhaps particles as small as 1nm will be kept in suspension by Brownian motion, and perhaps larger particles will eventually settle to the bottom of the container, but I don't know, that's why I'm asking!

Before the buzzword nanoparticles came into being, this science was called colloidal chemistry. In other words, nanoparticles were colloid particles. See Faraday's work on colloids. Gold colloids are stable suspensions for years. All colloids are prone to settling. It is just a matter of time, some colloid may take years to settle.

@AChem        Faraday’s colloidal gold is still unsettled after 150 years!

Note that even pollen particles in  micrometer range are visibly affected by Brownian motion, as that is how it was discovered. And the mass to be disturbed by random hits decreases with r^3.

Chances are that e.g. the Journal of Chemical Education holds a publication which can be both introduction into theory and application for a lab class (2018JChemEduc121 as an example).  In addition, nanoparticles and nanoparticle formation are suggested keywords in ACS' search mask to cover all their journals after typing just a couple of characters. Not specific to NP of iron, but frequently posters mention sodium dodecylsulfate (SDS) as detergent added to keep suspended NPs separate from each other (again because of energies).

Thanks to all who responded, but particularly to AChem whose input I found the most helpful.


text:
Can two resonance structures of a molecule have different cas numbers?
From this Wikipedia article :

Specifically, when a molecular structure is said to be represented by a resonance hybrid, it does not mean that electrons of the molecule are "resonating" or shifting back and forth between several sets of positions, each one represented by a Lewis structure. Rather, it means that the set of contributing structures represents an intermediate structure (a weighted average of the contributors), with a single, well-defined geometry and distribution of electrons. It is incorrect to regard resonance hybrids as rapidly interconverting isomers, even though the term "resonance" might evoke such an image.

Therefore, one should never treat two resonance structures of a molecule as distinct substances, right?
Well, here is an example of (what appear to be) two resonance structures of a molecule, which however have each a distinct cas number, different name, and are even sold separately (different catalogue number) by a single vendor:
https://www.combi-blocks.com/cgi-bin/find.cgi?HC-3087
273-13-2

https://www.combi-blocks.com/cgi-bin/find.cgi?JH-8440
22706-22-5

How can this be explained?
Is the attribution of distinct cas numbers to resonance structures of a molecule 'allowed' or 'correct', for some reason, despite the fact that they point to the exact same chemical entity? Or have they just got it wrong in this case?
Do you know of any other examples of resonance structures of a molecule being assigned distinct cas numbers?


text:
Refrence on shielding effect vs inter electronic repulsion
Recently I came across a very interesting concept , some scholars were saying that -
Shielding accounts just one component (radial) of interelectronic repulsion not complete repulsion
Who they are -
Shielding vs electron-electron repulsion
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=49293#p177071
https://chemistry.stackexchange.com/a/175176/129242
When I searched for this concept on internet, in books and articles I couldn't find there exists such a concept. Just as I was on the verge of giving up all hope and sinking into despair, I stumbled upon something that rekindled my hope - Answer  to problem 7.98 in the book "Chemistry the central science by Theodore L. Brown"

Although this book doesn't say anything about shielding and interelectronic repulsion is same or not except this answer which says that shielding and interelectronic repulsion are same for S orbital which concludes that  those may be different for P or D orbital.
That's all I found. Is there someone who could discuss about this concept in detail?
Or
could you just help me to find out some books or articles where can I find this type of language (radial and tangential component of repulsion, shielding is a component of repulsion etc) ?

You should rather add it to the original question as elaboration, not to fragment effort of yourself and others.  I am going to write/entend the answer somewhat later, not by a phone.

to which question  should  i add it ? i will definitely add

There is just one your original question.

The answers in the second link are clear enough, the shielding and the electronic repulsion are two different concepts, a quantitative analysis may be necessary the understand that they have a different energy impact.
The ion $\ce{He+}$ has an analytical ground state with one electron in the $1s$ orbital creating a shield for another incoming electron before the singlet state is formed, there is no repulsion but there is screening. The shielding and repulsion effects are related, but are different processes described as follows:
In the case of $\ce{He+}$, the ground state energy is $\epsilon_{1s}=-\frac{Z^2}{2n^2}$ ($Z=2$, $n=1$) a.u = $-54.42$ eV is also the orbital energy. To describe the repulsion, $\ce{He}$ is the simplest example, but the ground state has no analytical expression. The repulsion between the electrons at the  positions $r_1$, $r_2$, assuming that that the state remains a hydrogenic solution $\psi^2_{1s}(r)=e^{- Zr}$ is given by a perturbation $\Delta V_{ee} = \int\psi^2_{1s}(r_1)\frac{1}{|\bf{r}_1-\bf{r}_2|}\psi^2_{1s}(r_2)r_1^2r_2^2d\Omega_1d\Omega_2$, the value of this integral is $\Delta V_{ee} = \frac{5}{8}Z$ a. u. $= 34.01$ eV, the ground state energy $\epsilon_{1s}=-\frac{Z^2}{2n^2}+\Delta V_{ee} = -74.83$ eV, the accuracy is poor, there is a variationnal method leading to a better accuracy. The correct value of the orbital energy is given by the first ionization energy $-I = 24.59 $ eV. Obviously there is no shielding in this singlet state, the shielding starts with the lithium atom. There is no exact approximation : $\ce{Li+}$ is similar to the Helium atom, $\Delta V_{ee}(\ce{Li+}) = \frac{5}{8}(Z=3) = 51.20 $ eV, the repulsion is higher because the $1s$ orbital is more localized due to an additional nucleus charge.
The third elcction of $\ce{Li}$ has a $2s$ state orthogonal to $1s$, if the inner electrons $1s^2$ reduce the charge to an effective charge of $Z_{eff}=1$ on the third electron, the $2s$ level will have $\epsilon_{2s}^{approx}\frac{(Z=1)^2}{2\times 2^2} = -3.40$ eV higher in energy compared to the experimental value :  $\epsilon_{2s}=-I = - 5.39$ eV, if we use $\epsilon_{2s}=\frac{Z_{eff}}{2n^2}$, $Z_{eff} = 1.26$, smaller than the charge of the nucleus ($Z=3$).  The shielding of the nucleus $1s$ is $s = |Z_eff - 1(3-2) |= 0.26$, in agreement with the Slater's rules as an answer to the third link.
The shielding and the electronic repulsion are different and have different contributions, the repulsion has a significant contribution of about $51$ eV, but the shielding related to the nuclear charge has a correction $\frac{Z-Z^{eff}}{2n^2}$ of about $6$ eV, but they are correlated. The angular dependence of the orbitals for $\ell > 0$ gives spatial vacancies where the electronic density is very week, the shielding is less efficient compared to the example above and again not directly related to the electronic repulsion.

How can you say that -"there is no repulsion but there is screening". Repulsion without screening makes sense when we talk about same subshell or orbital but how does screening without repulsion makes sense, when screening is also a
part or component of repulsion. https://chemistry.stackexchange.com/a/175176/135928


text:
Why do we consider one elementary reaction to be in equilibrium while calculating rate law for complex reactions?
When we calculate rate of reaction pfor a complex reaction, if the rate determining step have intermediates, we calculate the rate of reaction using the step which is at equilibrium (and doesn't have any intermediate).
For example:-
Let the given reaction be a complex reaction$$\ce {A +B->C}$$
If this reaction takes place as follows:-
$$\ce{A +B<=> P + Q}.....(fast)$$
$$\ce{P +Q-> C}.......(slow)$$
So here rate determining step is second one. But as it consists of intermediate we say that the first elementary reaction is at equilibrium and solve.
But why do we assume that the first reaction is at equilibrium? Maybe it is still not at equilibrium. Also what if they can never reach equilibrium because of the conditions? Then how can we calculate rate of reaction?

Be sure you have searched this site and libretexts.org for related keywords, as this and similar topics are evergreens  and are asked regularly. // One of many links:  https://chem.libretexts.org/Courses/Purdue/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)/Chapter_05%3A_The_Study_of_Chemical_Reactions/5.7.%09Potential_Energy_Diagrams_for_Chemical_Reactions/5.7.1._Rate_of_Multistep_Reactions/Reaction_Mechanisms_and_Multistep_Reactions

I read the suggested article and it helped me. But I still didn't get what if there is no equilibrium in elementary reaction?

There is many more. Do a proper review before asking. As example,  searching and review search results for site:chemistry.stackexchange.com OR site:chem.libretexts.org reaction kinetics fast slow  step or for similar keywords. // Then enumerate what you learnt and know what you do not understand yet. No explicitly listed effort = no effort, as nobody can distinguish the cases.

I read many articles before. If I found my answer I wouldn't post the question at first place.

You should include in questions a short summary of what you already know or found (with eventual quotes/links), so others will not write what you are already aware of, or what you should know if you searched about it. If nothing was found, write at least what you have searched for.

The concentration (alternatively partial pressure) of P and Q would asymptotically raise until the steady state when $k_1[\ce{A}][\ce{B}]=k_2[\ce{P}][\ce{Q}]$

But why do we assume that the first reaction is at equilibrium?

It makes it easier to solve.

Maybe it is still not at equilibrium.

At the very beginning of the reaction, it will be far from equilibrium. So the model (fast equilibrium between reactants and intermediates) will not be a good description of the initial phase of the reaction.

Also what if they can never reach equilibrium because of the conditions?

Technically, no matter what the conditions, it does not reach equilibrium until all reactions are at equilibrium. However, in some cases, it is a pretty good approximation for most of the reaction time.

Then how can we calculate rate of reaction?

You can always break time into discrete steps and do a numeric approximation (step-wise integration). Their might be some special cases where you can get a closed-form integral, but that is probably rare.


text:
Does someone know a unique testable questions surrounding 1 of the 5 variables that affect the rate of a chemical reaction?
I am a year 11 student studying high school chemistry and we are currently studying catalysts.
I am working on an assessment and need to come up with a unique but testable hypothesis surrounding one of the 5 variables of chemical reaction rate.
The 5 variables that affect the rate of a chemical reaction that my teacher have given to the class are as follows:

The concentration of each reactant (pressure of gases)
The temperature
The surface area of reacting solids
The nature of catalytic material
The quantity of catalytic material

So far I have come up with one testable question that I think is unique but need some help from others who will most definitely be more knowledgable than me.
My current question is, “Does the physical condition of a catalyst change its effect on a chemical reaction rate?”
To make this make any sort of sense, I intend on comparing the effect a powdered solid has as a catalyst, compared to a chunk of something, for example - zinc crushed into a powder, if that’s possible, versus a solid chunk of zinc of the same mass in the reaction between zinc and sulphuric acid.
If anyone knows any other good questions that are unique, or if my question seems worthy of putting into a hypothesis, I would greatly appreciate any input. Thanks in advance to anyone who reads all of this and helps me, you are a legend!

The greater the surface area, the more accessible the reagent in a heterogeneous reaction.  Hence, assuming the same chemical purity, Zn (metal) is less reactive than Zn (shot), and this is less reactive as Zn (dust) e.g. in the course of a Barbier reaction (e.g., THF/aqueous $\ce{NH4Cl}$, Zn, allyl bromide and carbonyl compound; it carries some similarities to the Grignard reaction).  It equally «works» the less obvious way, i.e. old turnings of Mg (exposed for long to air) frequently have to be etched by a little of iodine to get past the oxidative layer to initiate the wanted Grignard reaction; finer shot is more prone to this, than coarse fresh turnings.
Though not used as a catalyst but as suspended reagent at lab scale, the efficiency of $\ce{MnO2}$ to oxidize organic compounds (e.g., allyl alcohols to the corresponding aldehydes and ketones) equally depends greatly from how this powder is synthesized in first place.*  Polymorphism (different spatial arrangement of molecules in the solid/crystalline state; you possibly know about the allotropes of sulfur) and different frequency of defects in the solid state's structure seem to contribute to this observation.  (The $\ce{MnO2}$ in dry cell batteries often is mangese dioxide prepared chemical synthesis (CMD), or electrolysis (EMD); this sum formula then can be a great simplification of the true composition.)
* See e.g. an earlier post here.

Careful using the word unique. The 5 conditions put too much emphasis on catalysts. It is basically 3 conditions: Concentrations, temperature, and adequate mixing. A catalyst changes the mechanism. The new mechanism responds to the same three conditions; there is just an additional component to consider: the catalyst amount and its molecular size ie. concentration. Mixing is next. There are two main methods, stirring or shaking and flow over a surface.
If you want a simple system to consider try electrolysis of water. a simple equation involving two reactions with all the components.  There is probably a kit in the school lab to set up an experiment, set up a cell in series with a battery, ammeter and light bulb. Current flow is the reaction rate; measure it by the ammmeter or a digital camera or Crookes radiometer to measure light intensity.

My current question is, “Does the physical condition of a catalyst change its effect on a chemical reaction rate?”

That is a good question.

To make this make any sort of sense, I intend on comparing the effect a powdered solid has as a catalyst, compared to a chunk of something, for example - zinc crushed into a powder, if that’s possible, versus a solid chunk of zinc of the same mass in the reaction between zinc and sulphuric acid.

If this is your question, you might want to ask "How does the surface area of a catalyst affect reaction rate". It is a common question, but becomes unique as you choose the reaction to study, and the specific set of experiments to test it.
Unfortunately, in the reaction you suggest, the zinc is a reactant, not a catalyst. However, you could take a different reaction, and test how the rate of reaction changes when you switch from a chunk of catalyst to a powder.

I am working on an assessment and need to come up with a unique but testable hypothesis surrounding one of the 5 variables of chemical reaction rate.

At the high school level, it is hard to come up with a unique but testable hypothesis. On the one hand, it makes sense to propose to work on a reaction that is well-studied (for safety reasons), on the other hand, that makes it less unique. Maybe "unique" in this assignment just means you thought about it carefully rather than taking an experiment from the internet (like getting a complete answer here).

If anyone knows any other good questions that are unique, or if my question seems worthy of putting into a hypothesis, I would greatly appreciate any input.

Your question is worthy. You just have to find a good reaction to test it on.
You could research the catalytic decomposition of aqueous hydrogen peroxide catalyzed by manganese oxide. It is a well-known reaction, and it is easy to observe the reaction because one product is elemental oxygen that bubbles out of the reaction vessel. You would have to think about how to make your question a bit different from the questions you encounter in your research.
For academic honesty's sake, it would be appropriate to cite all the resources you used to formulate your hypothesis. This is exactly how scientific progress works, researching and citing prior work that is relevant to your own work.


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D block elements and the cation of cobalt
The configuration of cobalt in its ground state is $\ce{[Ar] 4s^2 3d^7}$. when it loses 2 electrons, it supposedly leaves from s orbital making it $\ce{[Ar] 4s^0 3d^7}$. But since the s and d orbital have almost similar energy, they tend to exchange electrons from d to s and vice versa to attain a stable configuration. Wouldn't it make sense that when cobalt loses 2 electrons, 2 electrons are lost from s orbital and then 2 electrons are donated from d orbital to the s orbital, making it $\ce {Co^{2+} = [Ar] 4s^2 3d^5}$. since 3d orbital is half filled and it is more stable than partially filled orbital?

This is certainly a misconception and therefore an important question. There is a difference between unstable free ions that have a small mean free path before getting electrons somewhere and ions due to oxidation states in a compound. Let us assume that we have cobalt atoms in the air that do not interact with each other before they form the well known hexagonal solid. The ground state of a cobalt atom has $4s$ orbitals for two electrons and five $3d$ orbitals for seven electrons. The latter are assumed to be degenerate. The radial extent of the $3d$ orbitals is not large compared to the $4s$ orbitals, but the $s$ orbitals have the great advantage of being closer to the nucleus ($\ell=0$) and therefore always lower in energy compared to the $3d$ orbitals. The first and second ionization energies for cobalt are 760.4 kJ/mol and 1648 kJ/mol, respectively. When these free atoms are doubly ionized, the ionized electrons will come from the $3d$ orbitals because they are higher in energy and the configuration will be [$\ce{Ar}$] $3\mathrm{d}^5$ $4\mathrm{s}^2$, two electrons remain in the $4s$ orbitals.
The misunderstanding comes from the fact that without interacting (free atom), a transition of an electron from $4s$ $\rightarrow$ $3d$ is unlikely to occur. This occurs when the $\ce{Co^2+}$ ion interacts with the environment or is in a compound and have an oxidation state due to this interaction, the $4s$ state will be delocalized due to its radial extent and the orthogonality is almost broken, a partial $4s$ charge will fill the $3d$ orbitals. This process is exactly the same for hybridization, where a charge is transferred to higher $\ell$ for interacting with neighboring atoms, creating bonds. Without such interaction, the $4s$ orbitals should always be filled in free $\ce{Co}$ atoms.