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How will the degree of conjugation effect the light emitted? (glow stick reaction)
So regarding the chemiluminescence in a glow stick reaction, the glow is due to the unstable compound 1,2-dioxetanedione decomposing into carbon dioxide, releasing energy which is then absorbed by the electrons in the molecules of the dye. This promotes the electron to an ‘excited state’ and when the electrons fall back to their ‘ground state’, they emit photons of light, where the exact energy of the light given off is dependent on the structure of the molecule.
Since I learned in organic chemistry that the more conjugated the molecule, the less energy required to promote the electrons to an excited state, I assume that is the reason why more conjugated systems (in the photo, Rhodamine B) emit lower frequencies of light (red), since lower energy is required to excite the electrons.
Am I correct?
In addition, if that is so, wouldn't the amount of conjugation of dye affect the efficiency of the chemiluminescence reaction? So for example, since more conjugated systems require less energy for the electrons to be excited, they will emit more light with the same amount of energy compared to less conjugated systems that need higher energy (emits higher frequency light).
I have asked my teacher multiple times, but she does not provide me with a satisfying question, I pray for help.
Thank you.

To my knowledge, the substituents present on the conjugated system also effect colour through "push" and " pull" of electrons, arising from inductive and resonant effects e.g. CN, OMe. This was certainly the case in the early development of oligomeric and polymeric organic LEDs

While first part seems OK, then that's quantum efficiency that you're got into and I doubt it depends on the gap between HOMO and LUMO in any simple way.

@Beerhunter Can you give me more detail on this?

First part is straightforward. Second part I don't think the teacher can, see Mithoron comment (that I agree with).

The initial chemiluminescent (CL) species transfers its excited state energy to the dye. This depends on the overlap of the CL emission spectrum with the absorption spectrum of the dye and their spatial separation, i.e. better spectral overlap and higher concentration more transfer and more dye emission. (see Forster energy transfer for mechanism).  The dyes used have 80-100% fluorescence quantum yield, so v. high chance of emission. The intensity you experience will also depend on how your eyes respond to different colours.
The energy gap in the dyes does not greatly affect their fluorescence efficiency, this is primarily governed by interactions within the dye to produce triplet states which drain energy out of the fluorescent singlet state.


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Why work done Is 0 and why not a negative value?
If a gas expand in vacuum, $P_{ex.} =0$. Then why is the work done 0? Since the gas is expanding, it is doing some work on the piston so the work done must have some negative value.

How about a balloon bursting with the gas expanding into vacuum? Or opening a valve (you would do some work opening the valve, but the gas would not be involved)? There are many situations where the process described is an "ideal" one that can only be approximated in reality.

There is no work done on moving a piston against zero external pressure (not if it is ideal, ie lacks mass and there is no friction in its movement). You might consider the definition of $pV$ work
$$w = -\int_{V_\mathrm{ini}}^{V_\mathrm{fin}}p_\mathrm{ext}
\mathrm{d}V$$
Obviously if the external pressure $p_\mathrm{ext}=0$ then $w=0$.
Conceptually, the point is that a collision by a single gas molecule would suffice to drive the piston to its extended position. In addition, that gas molecule would expend no energy, it would not slow down or change course in pushing the piston, it wouldn't feel it at all (this can get complicated, as explained further below). As absurd as that sounds, this is accurate within the constraints imposed by this conceptual question.
The following schematic contrasts the effect of a gas molecule impinging upon an ideal versus a slightly more realistic piston.

In the real case forces during the collision transfer momentum to the piston and reduce or reverse the forward momentum of the gas molecule. In the case of the ideal piston the collision is best modeled as a perfectly inelastic collision between the massless piston and the gas molecule, the resulting united body having the same momentum (in the direction of the piston axis†) as the original momentum of the gas molecule. There are no forces however and therefore no work is done.
† Components of the molecule's momentum along the orthogonal direction are conserved as well, and the molecule must be reflected once it reaches the chamber wall.

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Selective reduction of nitro group to amine, in benzene ring containing nitrile?
I would like to know if there is a possible route of synthesis that may reduce 3-nitrobenzonitrile A to 3-aminobenzonitrile B.

I would normally reduce a nitro to an amine using $\ce{Pd/C}$ and $\ce{H2}$, but since the nitrile is sensitive to this method, I need an alternative.

Clarification: If there are easier methods to arrive at product "B" through alternative reactions, besides reduction from "A", then I would gladly hear them.

I think it would also works using Fe/HCl

So would Sn/HCl or Zn/HCl

Sure! There are some "active metals" that would work

I just did a Chem Abstracts search of this reaction with 3031 hits! Many of them involve hydrogenations. Hope you have access.

I was always a fan of Zn dust/NH4Cl (see e.g. Chem. Sci., 2012,3, 537)

My answer is intended to support Waylander's suggestion elsewhere. I thought it'd be better to give OP some insight of this reduction mentioned (Ref.1). To my knowledge, after doing thorough literature search, the aromatic nitro group reduction by $\ce{SnCl2.2H2O}$ is the best method so far in the presence of other sensitive  groups such as nitrile group on the aromatic nucleus. The abstract of the reference 1 stated that:

Aromatic nitro compounds are readily reduced by $\ce{SnCl2.2H2O}$ in alcohol or ethyl acetate or by anhydrous $\ce{SnCl2}$ in alcohol where other reducible or acid sensitive groups such as aldehyde, ketone, ester, cyano, halogen, and O-benzyl remain unaffected.

This point is depicted in following diagram as two examples given in Ref.1:

However, as shown in the separate box in the diagram (under the title, Stephen reduction of nitriles), if you use $\ce{HCl}$ (hydrogen chloride gas) with $\ce{SnCl2}$ as other suggestions, the result would be different (Ref.2). $\ce{HCl}$ is added to the nitrile group to give the imidoyl chloride salt first, which is then reduced by the tin(II) chloride. The initial product is the complex with $\ce{SnCl4}$ (which is often crystalline) that is then hydrolyzed to the corresponding aldehyde (see the insert in the diagram).
For the convenience of OP and other readers, I'd include a typical reduction procedure in Ref.1:

The reduction of p-nitrobenzoic acid: A mixture of $\pu{1.67 g}$ ($\pu{0.01 mol}$) of p-nitrobenzoic acid and $\pu{11.275 g}$ ($\pu{0.05 mol}$) of $\ce{SnCl2.2H2O}$ in $\pu{20 mL}$ of absolute ethanol is heated at $\pu{70 ^\circ C}$ under nitrogen. After $\pu{30 min}$ the starting material has disappeared and the solution is allowed to cool down and then poured into ice. The $\mathrm{pH}$ is made slightly basic ($\mathrm{pH}\ 7-8$) by addition of 5% aqueous sodium bicarbonate before being extracted with ethyl acetate. The organic phase is thoroughly washed with brine, treated with charcoal and dried over sodium sulfate. Evaporation of the solvent leaves $\pu{1.5 g}$ ($94.5\%$) of p-aminobenzoic acid, which gives one spot on TLC and melts over $\pu{300 ^\circ C}$.

References:

F. D. Bellamy, K. Ou, “Selective reduction of aromatic nitro compounds with stannous chloride in non acidic and non aqueous medium,” Tetrahedron Letters 1984, 25(8), 839-842 (https://doi.org/10.1016/S0040-4039(01)80041-1).
James A. Knight, Harry D. Zook, “Reduction of Aliphatic Nitriles by the Stephen Reaction,” J. Am. Chem. Soc. 1952, 74(18), 4560-4562 (https://doi.org/10.1021/ja01138a031).

I prefer basifying with 50% NaOH. The initial  basification produces a persistent emulsion of Sn(IV) salts. If you add enough NaOH, Sn(OH)4  forms which goes back into solution and the extraction becomes straightforward.

@Philip: You should accept Waylander's answer since it is the one prompt me to do some research. And also, it is the best method so far.

@MathewMahindaratne: Thank you for your very expounding answer <3

@Philip: I'd appreciate if you accept Waylander's answer, because my answer initiated by his reference. :-)

$\ce{SnCl2.2H2O}$ in refluxing $\ce{EtOAc}$ as reported by Bellamy & Ou here will do this very cleanly. Work up into ice then add concentrated $\ce{NaOH}$ until you get phase separation. I have done this transformation.

Reduce a nitro to an amine using 1% Pt on carbon  and $\ce{H2}$, since the nitrile is sensitive to Pd/C catalyst. I have done reduction of this type material using Selective 1% Pt on carbon  grade and low hydrogen pressure, at ton level.

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Determine bond length via rotational spectroscopy

Estimate the bond length of 12C16O (pure rotational spectrum) given J''=3 (15.356 cm-1) and J'=9 (38.356)

I understand that the first step is the calculate the rotational constant. So to do this I assume I would use one frequency value and one J value? And the equation to use would be
$$\nu = BJ(J+1)$$
If I do this with the numbers for $J=3$, then I get $1.28$ for $B$, which is apparently incorrect. I have seen someone else's attempt and they use
$$12B = (38.356 - 15.365)=1.916$$
And solved for B this way and they get the correct answer. Why is the way I thought not correct? Why is the second way correct and where did B come from?

The method they are using seems to be looking at the separation between two different transitions to determine $B$.
$J''$ denotes the lower energy state in a transition and $J'$ denotes the higher energy state. Here, you are looking at two different transitions: $$J_1'=4\to J_1''=3 \text{ with } \nu_{4\to 3}=\pu{15.365cm^-1}$$ $$J_2'=9\to J_2''=8 \text{ with } \nu_{9\to 8}=\pu{38.356cm^-1}$$
We can see in the figure below that $\Delta\nu=2B|J_2'-J_1''|$ or that the separation between transitions depends on difference between the $J$ of the higher state of the higher energy transition and the lower state of the lower energy transition. In your case, this gives that $\Delta\nu=2B|9-3|=12B$.
$\hspace{5cm}$
You could also get $B$ from the individual transitions, but you have used the wrong formula. $E_J=BJ(J+1)$ is the energy of the rotational state $J$, but you are given the energy of the transitions, which can be computed as $\nu=2B(J''+1)$. Checking with either transition should give the same result as using the spacing between the transitions.
One reason you might use the spacing to compute $B$, rather than individual transitions, is that if you have any systematic error in measuring transition frequencies, taking the difference will cancel out this error.


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Why is N- a better nucleophile than S-?

I can't rationalize why S- is a worse nucleophile than N-. I thought nucelophilicity was related to electronegativity, the more electronegative an atom is, the less of a nucleophile it is. Sulfur is less electronegative than Nitrogen. What concept am I not considering or misunderstanding?

Well, if that was so simple... In organic chemistry thinking in terms of electronegativity is hardly enough to guess even acidity.

I think it would be nice to consider the charge density. The S- is more stable than N- because it has more electron shells, so the negative charge density is better distributed in the atom. It makes that the N- is more likely to give up electrons and act as a nucleophile

Why wouldn't that apply to oxygen versus sulphur?

I would swear that the carboxylate can stabilize by resonance and thiolate not


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Why hyperconjugation can't stabilize carbanion?
If Resonance can stabilize both carbo cation and anion then why can't hyperconjugation stabilize carbanion ? Since both Resonance and Hyperconjugation are pretty much the same, i.e both involve delocalisation or diffusion of electrons( at least in model) ?

Unfavourable geometry.

Hyperconjugation involves the electrons from an adjacent bond to "donate" electrons to an empty p orbital. Now if we consider a filled p orbital, or a carbo-anion, then there will be a repulsive force according to Coulomb's law. This will destabilize the molecule in question. However, resonance involves the delocalization of electrons in a molecule; whether this stabilizes the molecule depends on what exactly is going on.

In carbanion the carbon carrying negative charge contains one $\mathrm{sp^3}$ hybrid orbital which contains a pair of electrons and therefore there is no possibility that it will undergo bonding interaction with the neighbouring $\ce{C-H}$ bond, due to repulsion between the electrons.


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How can we calculate the "degree" of conjugation?
So a conjugated system is a system of connected p orbitals or alternating single and multiple bonds.
I know that conventionally, we count the number of adjacent parallel p orbitals or alternating single-double bonds to see which molecule is "more conjugated" or not. However, when considering the following molecules:

I cannot seem to find out which molecule is "more" conjugated or not.
Is there a way to calculate the degree of conjugation? And in this case, would the rhodamine B be the most conjugated compound in the list?

For a deeper insight on the degree of conjugation, have a look at Clar's rule, for instance: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3982536/

Not a "degree" but you can calculate energy of conjugation,

There are several but this is a field itself. Note that in complex structure   counting how many double bonds are in alternate sequences does not suffice. The first example that comes to my mind is that of antiaromatic cycles.


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Starch Amylase Experiment using Iodine
The starch and iodide test is to detect amylose, a polysaccharide. We add an iodide solution to a sample and observe a color change with the starch; it turns dark blue. I am doing a bio/chem experiment on how pH changes Amylase digestion of Starch.
My recent trials have all failed, as the amylase starch solution is not showing any blue colour. Since I think its because that the amylase and starch solution is too concentrated and thats why they are digesting quickly, I am wondering, what percent should my amylase and starch solution be?


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Resonance structures of 4‐aminobut‐3‐en‐2‐ylium
Problem

How many resonance structures can be drawn for the following molecule?

(a) 1
(b) 4
(c) 3
(d) 2

Answer

 (d) 2

Question
I think that the molecule can be drawn in three ways, which contradicts the book's answer. Can anyone explain?

Asking is how many (more) structures can be drawn. First one is already here. You can draw two more, thus answer is D.

@MathewMahindaratne But that isn't what the question asks. The one shown can be drawn, too. They should specify "additional" if that is what they want.

@ SendersReagent: Compare Q1 & 2, to see how different they were written. Anyway, you and I can't read writer's mind. That's the bottom line.

Just shift the pi-bond from the double bond to the right side. The carbocation then shifts to the left. That's the second resonance structure. The first one is already shown, making a total of two different structures (D).
You might be wondering about the allylic lone pair on the Nitrogen atom. Well, that could have resulted in another resonance pair, but if you shift the lone pair to the adjacent C-C bond on the right, there will be a total of 3 charges on the resonance structure. This is a highly unstable form, so it's not a significant resonance structure. Thus, it can be neglected.


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Calculating partial pressure with 3 equilibrium equations
So my friend had sent this question to me some days ago and it seems I still am not able to solve it. I am confused whether the amount of A that dissociates in the first equation would account for the third one and how they inter relate.
The correct answer is 8.

I am confused whether the amount of A that dissociates in the first equation would account for the third one and how they inter relate.

All atoms of A are in the form of $\ce{A2}$ initially. After equilibrium is established, some are in the form of atomic $\ce{A}$, and some are part of $\ce{AB}$ while the rest remain as $\ce{A2}$. If you had a dial you could turn, increasing the dissociation constant of $\ce{A2}$ (leaving everything else constant), you would lower the concentration of the $\ce{A2}$, which would disturb the equilibrium of the third reaction, consequently lowering the concentration of $\ce{AB}$ (and increasing that of $\ce{B2}$. So it is all linked, and you have to solve the entire set of equations simultaneously.
Solution
It is possible to calculate all five partial pressures using the mass balance of A and B (two equations) and the third equilibrium constant (one equation) together with the given partial pressure of AB and the total pressure (two equations). Initially, the partial pressure of $\ce{A2}$ was 1 atm (same for $\ce{B2}$).
Let $x$ be the partial pressure of $\ce{A2}$ and $y$ the partial pressure of $\ce{B2}$ at equilibrium. From the stoichiometry of reactants and products, we get the following expressions of the partial pressure of A and B (2 atm is the partial pressure if all of A were present in atomic form, starting with 1 atm of $\ce{A2}$):
$$P_A = \pu{2 atm} - P_{AB} - 2x = \pu{1.5 atm} - 2x$$
$$P_B = \pu{2 atm} - P_{AB} - 2y = \pu{1.5 atm} - 2y$$
All five species together have to account for the equilibrium pressure:
$$ x + y + P_A + P_B + P_{AB} = \pu{2.75 atm}$$
Substituting the expressions for $P_A$ and $P_B$, we get:
$$x + y + (\pu{1.5 atm} - 2x) + (\pu{1.5 atm} - 2y) + \pu{0.5 atm} = \pu{2.75 atm}$$
Simplifying, we get:
$$x + y  = \pu{0.75 atm}\tag{*}$$
We can express $y$ by (using the equilibrium constant for the third reaction):
$$ y = \frac{\pu{0.125 atm^2}}{x}$$
and use that in equation (*), simplifying to this quadratic:
$$ x^2 + \pu{0.125 atm^2} = \pu{0.75 atm} \cdot x$$
The two solutions for $x$ are $\pu{0.5 atm}$ and $\pu{0.25 atm}$, where corresponding values for $y$ are $\pu{0.25 atm}$ and $\pu{0.5 atm}$. The two solutions reflect the symmetry of the problem (A and B are interchangeable without changing anything). From this, you can calculate partial pressures of A and B, and the equilibrium constants. Their ratio is 8 for one solution and 1/8 for the other. The last sentence in the problem breaks the symmetry, and the answer is 8, as stated.
This was a brute force answer - I'm sure there is a more elegant approach.

I understood the question, all thanks to you sir but I would like to ask a few certain doubts. I still did not get how you wrote the equation for the partial pressure of A and B the 1st equation. Could you elaborate a little more on that? also i'd give you all my upvotes but my votes don't count at the moment...

@theRock If all $\ce{A2}$ would dissociate and no AB forms, the partial pressure would be 2 atm. However, there is 0.5 atm of AB, so we have to subtract that. And there is $x$ of $\ce{A2}$ left over, so we have to subtract $2x$ (factor two because of the 2:1 stoichiometry of $\ce{A2 <=> 2A}$).

Finally got it! Thank you so much sir you are a saviour!


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If volatility depends on intermolecular forces, then why in Raoult's law, two substances have different volatility?
I have a doubt, Ideal solutions are solutions where intermolecular forces between solute-solute and solvent-solvent are nearly equal to solute-solvent. Since intermolecular forces between A-A = B-B, doesn't that mean, they both have same volatility? But in Raoult's law, the graph shows that one of the vapour pressure is lower than the other resulting in two substances  having different volatility. I'm confused, where did I go wrong?

Assume surface molecules are restrained by a QM harmonic potential, with energy levels described as $$E_n = \hbar \omega(n + \frac12)$$
where $$\omega=\sqrt{\frac{k}{\mu}}$$
is the frequency of the harmonic oscillator
and $$\mu=\frac{m_1m_2}{m_1 + m_2}$$
is its reduced mass.
The actual shape of the potential is determined by $k$, the force constant, ie classically we would write $$U_{pot}=\frac12 k x^2$$
Therefore we might have two particles with identical restraining potentials, but due to differences in reduced mass they might differ in the nature of the available QM energy levels and their thermal populations. This means you can in fact have solute and solvent interacting via identical potentials and yet having different volatility.
It can be readily shown that a condition satisfied by an ideal mixture that follows Raoult's law is that the heat of mixing
$$\Delta_{mix} H=H_{12}-\chi_1 H_{11} - (1-\chi_1)H_{22}$$
be equal to zero.  Note that $H_{12}$ depends on the composition (on $\chi_1$). However, identical potentials (force constants) does not guarantee that the condition is satisfied. Consider now a simple model that assumes that the oscillator consists of a dangling surface molecule and the rest of the bulk, such that $\mu \approx m_i$ (where $m_i$ is the mass of the dangling molecule). Then it can be shown that the ideality condition represented by the equation $\Delta_{mix} H=0$ holds for statistical mixing if the force constants are independent of the nature of the bulk (but may depend on the dangling molecule). This is probably not an unreasonable description of van der Waals bonding, which depends on the area of the interaction surface of the solute more than on the composition of the solvent (assuming chemically similar compounds eg hydrocarbons).


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Method to find out the amount of solvent replaced by another solvent
So I am performing a relatively simple coprecipitation experiment to synthesize ceramics. All my precursors are in aqueous solutions. But due to the high surface tension of water, drying the water is affecting my surface area and porosity. So I  have introduced a solvent washing step, where I try to replace water by ethanol using ethanol washing, as the surface tension of ethanol is quite low.
So I am hoping that when I replace water by ethanol, the morphology of my particles will be improved.
I need to measure how much water has been effectively replaced by ethanol.CHN analysis is one way to it, but we don't have that in our lab.
Is there any other relatively simple way in which I can do this?
Thanks for assisting

Someone more into analytical chemistry could provide a better solution, but a dirty way comes to my mind: if only water and ethanol are present, you could use molecular sieves 3A as selective media for absorbing water (some testing or literature search might be needed, to check if the method is indeed appropriate). My idea: place your compound with a measured amount of vacuum-dried molecular sieves in a pressure-resistant, closed container under inert gas.

Heat the container in a oven for some time, and let it cool. Weight the molecular sieves, which will contain mainly (only?) water. Vacuum dry your compound, and weight it. The difference should represent the weight of ethanol.

@The_Vinz but my system contains a lot other compounds mainly metal nitrates and also ammonium carbonate as the base for precipitating all the metals hydroxides and carbonates down.


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Preparation of compounds using acetoacetic ester synthesis
I've been given the following exercise in organic chemistry, and I'm at a complete loss. The text book doesn't cover the subject at all. I have some knowledge of melonic ester synthesis. As an undergraduate student in biology, I have a really hard time grasping the concepts, so an in-depth explanation would be much appreciated.
The acetoacetic ester synthesis is closely related to the melonic ester synthesis, but involves alkylation with the anion of ethyl acetoacetate rather than diethyl malonate. Treatment of the ethyl acetoacetate anion with an alkyl halide, followed by decarboxylation, yields a ketone product:

How would you prepare the following compound using an acetoatic ester synthesis?

Can you identify what R would give you this structure? X is normally a halogen (Cl, Br, I)

Could R be the aromatic ring with bromide attached?

better if it is the aromatic ring with CH2Br attached

Thank you. Would it possible to do the exercise without being given the contens of the first picture. And if so, how would one do it?

Diethyl malonate + sodium ethoxide in ethanol, add benzyl bromide. When that reaction is complete add 5N HCl to pH1. Heat to reflux until reaction complete. Concentrate, partition the residue between water and Ethyl Acetate. Product is in the ethyl acetate phase.


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Difference between Particle in a Box and Harmonic Oscillator approximations
Why in the particle in a box model do the values of n begin at 1 but in the harmonic oscillator they begin at 0? I understand what the wave-functions and their corresponding probabilities look like and that the PIB has 0 nodes for n=1 which means the number of nodes is n-1, so for n = 0 it would have -1 nodes which is physically unreasonable. But I do not understand why the PIB cannot have n=0 whereas the Harmonic oscillator can.

PIB cannot have n=0 precisely for the reason that you said yourself just a sentence ago: that the corresponding $\psi$ would not be physically meaningful. If that is not enough of a justification for you, then what is?

This still doesn't answer the question of why PIB cannot have an n=0 whereas the HO can. All you've stated is that we can change the rules by redefining things. But redefining things doesn't change the facts. PIB can never have n = 0 because this would mean the energy is zero, which means the particle is not moving so momentum is 0 and we can know the position, but this violates the uncertainty principle. But why can HO have an n=0??

Ivan Neretin, an explanation to WHY something is not physically meaningful. Thanks

$n=0$ is not a fact; it is a convention, and as such, can be changed, like Buck Thorn said. $n$ is not a physical value at all. Also, look at the HO state with $n=0$: its energy is not $0$.

Do you realize that there is a zero point energy associated with the harmonic oscillator? That is, when n=0 the energy is not zero?

Quantum numbers pop up when you solve Schrödingers wave equation for systems whose solutions are discrete due to the presence of a restraining potential. There is some degree of arbitrariness to what the range of quantum numbers might be, because it depends on how you define the quantum number.
You could just as well define the quantum number and associated energy levels for a 1D  pib as $$\begin{align} E_{n'}&=k(n'+1)^2 \\&=k n'^2 +2kn' + k\end{align} $$ where $$k=\frac{\pi^2 \hbar^2}{2mL^2}$$ Here $n'$ defines a "new" quantum number related to the original one as $n'=n-1$ and therefore ranges from 0 to $\infty$.
For the harmonic oscillator, you could similarly rewrite the energy as
$$E_n = \hbar \omega n'$$
where $n'$ ranges in integer increments from $\frac12$ to $\infty$.
Physically speaking, the non-zero ground-state energy of both the pib and harmonic oscillators is usually  attributed to the Heisenberg Uncertainty Principle, because a particle with no energy would have no velocity and therefore would be completely unrestrained in position. There are of course plane waves, but this would be a strange plane wave, with no amplitude.


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Reaction of alkyl halides with potassium cyanide
Its given in my book that alkyl halides react with alcoholic potassium cyanide to give alkyl cyanide through $\mathrm{S_N2}$ mechanism.
But why do we need to use alcoholic $\ce{KCN}$ (potassium cyanide dissolved in alcohol)? My understanding is that polar protic solvent favours $\mathrm{S_N1}$ over $\mathrm{S_N2}.$

Solubility. Potassium cyanide does not dissolve in non-polar aprotics

Does alcohol have any role in reaction. Can we have reaction using only KCN

As @Waylander said, alcohol does not participate in the reaction itself. It only dissolves KCN so that it can dissociate to give the $\ce{CN-}$ nucleophile

Then instead of alcoholic can we use aqueous KCN

No because then the alkyl halide will not be soluble

Could use aq KCN in a two phase system (water/dichloromethane is good) with the addition of a phase-transfer catalyst such as tetrabutylammonium bromide which will transport the CN- into the organic phase.

@Waylander Correct, that is the common method


text:
The relationship between the temperatures at different times and pressures at different times is?

At the instant the boiling point is reached, the vapour pressure of the solution should be equal to atm. pressure. And since the boiling is continued the pressure will keep on increasing. Thus the relationship between the pressures will be be $P_5  < P_{10} < P_{15}$.
And since at boiling the temperature of the liquid remains constant the temperature should be equal at all interval. That is Option C should be correct.
But the given answer is option A which doesn't make any sense to me. Can you help me with the question?

Boiling of a solution differs from boiling of a pure liquid a great deal. To begin with, there is no longer such thing as boiling point.

When making candy, the boiling temperature increases as water boils off and the sugar concentration in the liquid increases, see e.g. https://www.craftybaking.com/howto/candy-sugar-syrup-temperature-chart

As more of the solvent is boiled off, the molality of the solute increases, which increases the boiling point (as $\Delta T_B = mK_b$). Therefore, $T_5<T_{10}<T_{15}$ (Karsten Theis mentioned a good link in the comments giving a practical example)
Since the solution is boiled in an open vessel, the vapour pressure of the solvent will at all times be equal to the atmospheric pressure, as it is an isobaric process. Therefore, $P_5=P_{10}=P_{15}=P_{\text{atm}}$

I understand the case for temperature. However, the same can't be said for pressure. Anyways, Thanks a lot.

Your argument for the temperature relationship seems straight forward to me. However, I don't agree with your reasoning for the vapor pressure relationship (though the relationship itself is correct). The vapor pressure of the solution is by definition equal to atmospheric pressure, so the vapor pressure is equal to atmospheric pressure at every time and thus P5=P10=P15=Patm.

@airhuff didnt notice the 'open vessel' statement in the question. Edited accordingly.


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Tungsten-Rhenium alloy connections to Copper. Methods?
I want to use some fine WRe wire (50 micron diameter, 80/20) in a particular environment where it has to conduct a few millamps of current while supporting 100g of mass. The problem being the contact between the wire and its copper/bronze supports has to be extremely consistent. We are looking for changes in current in the nanoamps, so just clamping it in place will not work given oxide layers, surface electronic effects etc. not to mention mechanical tension.
We have done this before with a Platinum alloy which we could solder.
So, first question - can this be soldered or spot welded in place?
Second, if not can we perhaps alloy the surface with Gold by heating the wire to (say) 2500degC when coated in a Auric salt or Gold foil? And would this help us to find a solder that would "stick" to the surface?

Tough question. Neither W nor Re have any alloys with Cu, so while spot welding might physically work the joint might be weak or brittle. If you could plate the end of the wire with Ni that might be better - Ni has some solubility with both W and Re, and plays well with Cu.

@JonCuster Thanks. Might be worth trying before we go the Gold route.


text:
How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination?

How many distinct dichlorination products can result when isobutane is subjected to free radical chlorination?

I think that three distinct products may result namely
a) 1,1-Dichloro-2-methylpropane
b) 1,2-Dichloro-2-methylpropane
c) 1,3-Dichloro-2-methylpropane
But the book says only two distinct products may form. Can anyone explain how?

The following three products wil be formed:

Your book may be giving the answer to the number of monochloro products formed, as there are only two distinct monochloro products formed on chlorination of isobutane.


text:
XRF is "non-destructive," but what actually happens to the sample?
Editing for clarity and more details.
My colleague has been using XRF to quantitatively analyze 47-mm diameter Teflon membrane filters for metals, minerals, and rare earths. Much of their analysis has been focused on quantifying metals like iron, steel, and copper. It seems they are looking into replacing their large floor-standing WD-XRF with a benchtop ED-XRF or another floor-standing WD-XRF.
They are also planning to attempt quantification of trace metals that others report to be in the ppt range, so they are considering adding an ICP-MS downstream of the XRF. However, ICP-MS is completely destructive (the membrane filters go through solvent extraction).
My colleague and I discussed that if they were going to put an ICP-MS downstream anyway for trace metals analysis, perhaps it would be better to budget for a benchtop ED-XRF upstream and a more expensive ICP-MS downstream, so that the XRF acts more as a selection mechanism for which filters should be extracted for ICP-MS.
One important question we have is on our assumption that XRF is "non-destructive" to the sample. The membrane filter will certainly be intact after XRF, but will there be some amount of loss of particulate matter that will be significant for trace metals being analyzed in the ICP-MS step of this hypothetical measurement setup?
Thanks for your time and help.

If think that rather X-rays themselves, not heat are inherently destructive, it's just the less intensity an shorter time the smaller damage.

An ED-XRF will typically use a much less powerful x-ray tube. You can "cook" a sample with WD-XRF but that is due to chemical reactions caused by ionization created by xrays not really "heat" in the sample. // The two types of instruments are somewhat complementary. ED-XRF is usually better for a quick qualitative analysis, but WD-XRF is typically better for quantitative work. WD-XRF also is better for low energy XRF.

I did XRF for about two decades. To decide between the various  XRF instruments you really need to think about what  kind of samples you want to analyze, and if you want a semi-quantitative or the best quantitative analysis possible. // Frankly your question tells me that you don't have a clue about the technique...

This question is considered to be "broad." I agreed with @MaxW on his latest comment. You need to narrow this question by giving information on types of samples you need to analyze, and whether you are looking for quantitative or qualitative analysis (not just looking for sample heating).

I thought it was a pretty simple question, and it's certainly not my area of expertise, but it's also not my instrument. I thought maybe the forum would help my colleague, who is simply not a user of forums. Thanks for the efforts from the community, anyway, I'll just go talk to the techs, who have been helpful.

@MaxW Considering your expertise with XRF and numerous previous answers on this topic which were often very helpful, would you like to post an answer by expanding your comments a little? I personally think the question is OK as it is about characteristics (probe requirements) of the instrumental method, which should have a certain answer. In my opinion, despite the question is from a person who is not an expert in XRF, it appears literate and sensible.

@andselisk - I really can't give an answer with any more detail without knowing what kind of samples need to be analyzed. ED-XRF can yield a very good quantitative analysis for the right kind of sample.

For instance yes, XRF is generally considered a non-destructive technique. However that really means that the technique, in general, doesn't destroy the prepared samples. However most lab instruments require a 1.25 inch in diameter sample. So an art museum is unlikely to think that cutting 1.25 inch discs out of a painting is "non-destructive."

As stated, XRF was already being used and we're certainly not suddenly going to start cutting up paintings that we weren't already cutting up before. But I will edit the original post with some clarification to the question.

I edited the original post and appreciate everyone's patience and willingness to contribute.

The title of the question is about sample heating by ED-XRF instrument. Yet, OP is wondering its limitations and like to understand what kind of compounds analysts might lose to some sample heating if they switch to ED-XRFs (bench-top?). I hope this may be addressed by MaxW, when OP has said whay kind of samples they are analyzing. However, general knowledge on these instruments would predict both WD- and ED-XRF instruments would heat samples regardless of what kind of instrument is in use, but I'd suggest ED-XRF would heat less than that of WD-XRFs since they are low energy instruments.
Actually, to me, the bigger question is which kind is more sensitive to particular elemental analysis, regardless of sample kind (e.g., environmental or archaeological). My literature survey on the subject indicate that ED-XRF instruments is more qualitative than quantitative toward analysis, specifically towards analysis on low-$Z$ and medium-$Z$ elements (References 1 & 2).
The authors of ref.2 pointed out that the energy dispersive X-ray fluorescence analysis of low-Z-elements can generally be seen as a challenge. They have adjusted the instrument in several ways to get better data. The abstract of the article has well reflected the point:

Photoninduced energy dispersive X-ray fluorescence analysis in total reflection geometry has been applied to low-$Z$ elements. To avoid the absorption in the $\ce{Be}$-window a windowless $\ce{Si(Li)}$-detector has been used. Icing of the detector crystal is prevented by a thin $\ce{C}$-foil. Physical reasons of the low energy background are reduced using a high energy cut-off reflector. Measurements of low-$Z$ elements from $\ce{O}$ ($Z= 8$) onwards are possible and detection limits in the $\pu{ng}$ region are obtained.

Anybody interested in these adjustments should get the article, which has explained them with some depicted diagrams.
On low-$Z$ element analysis, authors of Ref.2 claimed that:

WD-XRF is the best XRF approach for measuring and quantifying low-$Z$ elements. ED-XRF spectrometers are subject to limitations, such as Bremsstrahlung radiation and escape peaks, which affect detection and quantification of low-$Z$ elements in particular. Although we will demonstrate in Section 2.3 that, under the right conditions, pXRF spectrometers can simulate the performance of bench-top ED-XRF instruments, there are two notable exceptions: $\ce{P}$ and $\ce{Na}$ (Note: the notation pXRF spectrometers are for portable-ED-XRFs).

The following diagram from Ref.2 well explained point, which shows exceptional correlation ($R^2=0.995$) of expected to measured concentrations of $\ce{P2O5}$ when WD-XRF is used:

References:

Alice M. W. Hunt, Robert J. Speakman, “Portable XRF analysis of archaeological sediments and ceramics,” Journal of Archaeological Science 2015, 53, 626-638 (https://doi.org/10.1016/j.jas.2014.11.031).
C. Streli, H. Aiginoer, P. Wobrauschek, “Total reflection X-ray fluorescence analysis of low-Z elements,” Spectrochimica Acta Part B: Atomic Spectroscopy 1989, 44(5), 491-497 (https://doi.org/10.1016/0584-8547(89)80055-2).

I agree with everything you stated. I'd add that typically with WD-XRF you could take a background reading above and below the peak of interest and use the average of those counts for a background. With ED-XRF you're going to almost certainly do some sort of curve/pattern fitting becuse the individual x-ray peaks won't be fully resolved. All instruments these days would have a computer attached so such fitting/data reduction isn't a major problem, but it does impact S/N.

OK, now with more information I can maybe help a bit more...

47-mm diameter Teflon membrane filters

This seems larger than a "typical" sample. I'd guess that your commercial instrument choices would be very limited by this sample size. (Note that you'd also want to spin the samples for greater uniformity...)
Typical way to analyze such a sample would be to suspend the filter between two thin sheets of polypropylene (or other plastic sheet) with a sample cup. I know such sample cups exist for 1.25 inch discs, not sure about 47 mm. (This might not be totally "non-destructive in that some particulate matter could transfer between the filter and the covering polypropylene. You could always cut both sheets of plastic out with the filter, and digest all three layers for ICP-MS.)
Having a film over the filter would obviously limit sensitivity to lower energy x-rays.

degradation of Teflon filters in x-ray beam

While at IBM there was an analysis of Al/Si films on teflon using a WD-XRF. There was a long term stability problem with the standards in that the x-rays created $\ce{F2}$ gas in the teflon which leached the Si to a volatile silicon fluoride gas.

rare earths

In general the "raw" samples the rare earths tend to behave very similarly chemically. Thus do you want to analyze a mixture of the rare earths, or a single rare earth that has been purified by processing?
Mixtures of rare earths would be hard with ED-XRF.

Si(Li) detector has poor efficiency for $K_\alpha$ lines.
L- Lines for mixtures rare earths overlap.

Mixtures of rare earths would be also be hard for WD-XRF.

For mixtures there isn't a lot of separation of L-lines.
For $K_\alpha$ lines I don't think that there is a really good crystal for dispersion. (I can't find $2\theta$ tables on web right now, but I think LiF 420 has smallest spacing. I'd guess that the $K_\alpha$ lines would be at very small $2\theta$ angles)

Filter loading

I'd guess that there is quite a bit of data of filter "loading" for air particulate analysis. Basically it would ideal if the amount of particulate matter was low enough so that the sample was infinitely thin. When analyzing dust samples from mines the filter canister was sometimes half full of particulate matter and loose material was sliding around the inside of the canister.
The overall point here is that the infinite thickness for low energy x-rays is very small. So particles might be infinitely tick for light elements and infinitely thin for heavier elements. Thus you might need to use the weight of the particulates on the filter to make a correction.

KI pellet

With WD-XRF one of the ways to see where xrays are hitting a sample is to use a KI pellet. Free $\ce{I2}$ is produced by the ionizing radiation turning the exposed area purple. The technique isn't terribly useful with ED-XRF since the tube power is so low...

"Inverted" Optics - Most XRF systems have the x-ray tubes and detectors below the sample. Hence the need to have a film to hold the particulates on the filter.


text:
Copper wire smoking underwater
I am doing an electrolysis experiment with two copper wires inside of test tubes that are both submerged underwater, with 20 V per wire. Test tube A is creating oxygen gas at a high rate, but test tube B is creating a smoke-like gas that is heavier than the water and coming out of the test tube that is submerged in water and settling on the bottom of the beaker.
Why is the wire creating the smoke-like gas and settling on the bottom of the beaker? The wires have been in the water for about a week and only turned on when we are working on the experiment. Is the smoke-like gas a reaction to the fresh tap water that has been sitting?

A gas that is heavier than water is quite an extraordinary claim.

A gas heavier than water is not a gas. Not for common substances and conditions.

It's not smoke, but Copper(II) oxide (cupric oxide, CuO), which is a black solid. It formed on the surface of your copper electrode and is falling off. Even after falling off, copper oxide could continue to form on the copper surface.


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What does ³D mean in energy level diagram for helium?
I'm coming from a Physics background and am trying to make sense of the energy level diagram for para-helium and ortho-helium.
From what I've gathered each column shows the different energy levels for different orbitals so that for example the leftmost column with $^1S$ written on top shows the energy levels of 1s, 2s, 3s etc.
What I don't get is what the different denotations on top mean, what does for example $^1P$ and $^3D$ mean?

Related:https://en.wikipedia.org/wiki/Term_symbol#LS_coupling_and_symbol

They refer to different electronic states which have different amounts of spin angular momentum and orbital angular momentum. A physical chemistry / spectroscopy textbook will have a full explanation...

The index 1 or 3 is used for saying that the spins of the two electrons are parallel or anti-parallel. One spin is up or +1/2, and the other one is down, or -1/2. This is the case in the ground state : the 2 electrons of Helium have two antiparallel spins (one +1/2 and one -1/2). The atom has no total spin, as the two electrons cancel the magnetic effect of their spins. The total spin of the atom is then 1/2 - 1/2 = 0. In such a configuration, the He atom is called para-helium. This remain true if one electron is excited and goes to an upper orbital like 2s, 3s, etc. provided that the two electrons maintain their spins (one up and one down). The total spin is zero, and the atom is an excited state of para-helium.
It is difficult, but not impossible, to change the spin of one electron, so that both electrons have the same spin. In this configuration, the helium is called ortho-helium. The total spin for the whole atom is 1/2 + 1/2 = 1. In that case, there are three possibilities for the projection of the total spin in an external magnetic field. +1 in the sense of the field, -1 in the opposite sens, and zero, if the total spin is perpendicular to the magnetic field. This number 3 is reported in upper index. The differences of the three projections must be an integer.


text:
Energy diagram of copper iodide complex
We did an experiment where copper iodide 4-methylpyridine and copper iodide imidazole was formed and now we are asked what orbital the electron is excited from and to.
After much research we still struggle to answer this question. We came to the conclusion that the structure of our two products are tetrahedral and thus the electron should be excited from $\mathrm{e_g}$ to $\mathrm{t_{2g}}$. However, since copper is a $\mathrm{d^{10}}$ metal, isn't the $\mathrm{t_{2g}}$ orbital full? So, to which orbital would the electron be excited? Have we misunderstood anything?
All help as well as links to helpful information is much appreciated!


text:
Why does thiocyanate bind to Fe³⁺ with the S side instead of the N side?
According to hard and soft acids and bases (HSAB) theory, "soft" Lewis bases bind to "soft" Lewis acids, and "hard" Lewis bases bind to "hard" Lewis acids.
$\ce{Fe^3+}$ is a hard Lewis acid. In $\ce{SCN-}$, either the $\ce{N}$ or the $\ce{S}$ side of the molecule can bind with $\ce{Fe^3+}$. Of the two, $\ce{N}$ is the harder Lewis base. But $\ce{[Fe(SCN)6]^3-}$ has a $\ce{Fe-S}$ bond and not a $\ce{Fe-N}$ bond. Why is this so?

Already asked https://chemistry.stackexchange.com/questions/93792/why-does-ironiii-form-the-trithiocyanate-complex-instead-of-the-triisothiocyan


text:
Remove stain from synthetic leather
I have a valuable soccer ball. It is not real leather, but a synthetic leather.
I used this cleaning agent to remove a few marks:
https://www.amazon.co.uk/Motsenbockers-Lift-Off-Graffiti-Remover-Trigger/dp/B00N3B25LU
However, I stupidly forgot to wait for the ball to dry and then placed it within an orange plastic carrier bag. Now the cleaning agent has reacted with the dye and there's a yellow-tint stain.
I have tried using acetone and these:
https://www.amazon.co.uk/Dr-Beckmann-Stain-Devils-Survival/dp/B07TKJPH7D/ref=sr_1_14?keywords=dr+beckmann&qid=1575488301&sr=8-14
but I cannot remove the stain.
Does anyone with chemistry knowledge know what else I could try?
UPDATE:
Got it out in the end. I used Dr Beckmann Stain Devils tea & wine, cut some cloth in to the shape of the area I wanted to clean, poured water, then the powder, then water on top and left for hours. Took multiple attempts.

That original cleaner did most likely not "react" with the orange dye, but simply dissolve it partially. Is it water based? Then I would wrap the ball in a wet (white ;)) cloth for a night, and see if the dye starts coming off. I dont want to write an answer, because frankly it is impossible to tell without knowing what exactly the dye and cleaners are.

@Karl This is the patent, it describes the product in high detail: https://patents.google.com/patent/US5227085A/en

"Water-based cleaning compositions for removing a variety of stains or deposits from a variety of substrates, the compositions consisting by weight essentially of tri-sodium phosphate (TSP) between about 0.85% and about 3.5%; ethylene diamine tetra acetic acid (EDTA) and/or phytic acid or a mixture of them between about 0.9% and about 9.0%; ethylene n-butyl ether (glycol EB) and/or butyl salicylate or a mixture of them between about 2.27% and about 27.00%; and acetone and/or a member of the group consisting of dimethyl adipate, dimethyl glutarate, dimethyl succinate or a mixture of them in wat

@Karl soak the cloth in cold water, not the original cleaning agent?

Start with plain water for the first night. I dont want to be resposible for damaging your valuable ball. ;) Next try adding ten percent isopropanol or acetone, if that doesnt help try the motsenbocker (what a name ;))))

@Karl it's fine. This was my own fault. It's typical these reactions will take hours to reverse? Are you sure water is worth trying?

It wont hurt. The (my) idea is to see when the dye starts to get off, i.e. stain a white cloth. If it rubs off with water already, or dilute alcohol, all the better for your ball. Pure acetone for a night might well damage it. Might. I am, like you, stabbing in the dark here.

Last idea: Try with a bit of hair bleach, 1% peroxide, 3% perhaps. Again: I dont know what that does to your ball, and the color it originally had. ;) Wash it off after 5 min, let the ball dry, if it looks OK, increase to 15 min, and so on. Good luck!

What about tooth paste (baking soda) or is that an urban myth?

Soda would be one of my later resorts. Tooth paste I would indeed file under "urban myth".

The dye has probably migrated a bit into your ball`s surface, so whatever (solvent) you try to get it out with again, it might take a while.
Remember the classic school (chromatography) experiment: You draw a line with a felt pen on a piece of paper, 2 cm from the edge, and stick the paper into a glass filled only 1cm high with acetone.
You can try to transport the dye so far into the ball you dont see it any more, or try to "fill the glass up to 3cm with acetone". ;)
And remember: Its hard to tell how resistant your ball and its original painting is to acetone or hair bleach or soda.

Got it out in the end. One of the products I tried initially worked when left for hours. It was the fact you emphasised leaving it overnight which really helped. Also I cut the cloth to the shape of the printed shapes on the ball, so as to not affect any of the intended design. The product which worked was Dr Beckmann Stain Devils tea & wine

Hi there (again). Got a small problem which occurred during cleaning. Given how knowledgeable you were before, could I see your help again? https://chemistry.stackexchange.com/questions/134990/synthetic-leather-football-has-blistered-during-cleaning


text:
Valency of each atom in a molecule with python
Consider ethanol, where each carbon atom has a valency of 4 and oxygen a valency of -2. I would like to determine these values for many compounds given their SMILES or InChI keys. (For the purpose of computing the degree of reduction, which in the case of ethanol is 6)
I would like to find a systematic method to calculate this for many molecules. I attempted to use Open Babel as described below, but I am open to any other programmatic solution.
Attempt with Open Babel on Python:
I attempted the following with openbabel.pybel (after installing the appropriate packages):
from openbabel import pybel

ethanol = 'CCO'  # SMILES

mol = pybel.readstring('smi', ethanol)

for atom in mol.atoms:
     print(f"type={atom.type} charge={atom.formalcharge} \
     degree(valence)={atom.degree} implicitvalence={atom.implicitvalence}")

This returns the error:
AttributeError: 'OBAtom' object has no attribute 'GetImplicitValence'

If I understand correctly the implicit valence is the value I am after. I am stuck and the documentation is not helping (is not even consistent with the behavior of the module)
openbabel.__version__ returns 3.0.0.
This is based on a 6 year old question

Possible duplicate of Counting valency of atoms, in a molecule with python

There's no 'implicit valence' in Open Babel 3 because we found it to be a very confusing concept. I think you should ask a more general question outside your specifics with Pybel.

@GeoffHutchison Thank you for noting that. I edited my question to make it more general, I am open to use any tool that would get the job done, it does not have to be limited to python or open babel.

If you consider valency to be the number of bonds a compound forms, then it cannot be negative and oxygen’s valency is also 2. If you consider it as some kind of oxidation state, then it can be negative, but then carbon’s oxidation states are $\mathrm{-III}$ and $\mathrm{-I}$.

I'm positive this would now be on topic at [mattermodeling.se]. However, from a chemical and definition point of view in regard to valence it does not make sense. Valence is a fixed number for each element. https://doi.org/10.1351/goldbook.V06588
This is 4 for carbon and 2 for oxygen and 1 for hydrogen… you need a table and probably not a program for this.

It is possible to extract implicit and explicit valence from a SMILES string using rdkit.
I am using rdkit version 2022.9.4.
The definitions of implicit and explicit valence according to rdkit are:
GetExplicitValence((Atom)arg1) → int :
Returns the explicit valence of the atom.

GetImplicitValence((Atom)arg1) → int :
Returns the number of implicit Hs on the atom.

Here is fully reproducible code to get implicit and explicit valence of atoms in a molecule:
from rdkit import Chem
from rdkit.Chem.Draw import IPythonConsole
IPythonConsole.ipython_useSVG=True

smiles = 'CCO'
mol = Chem.MolFromSmiles(smiles)
mol

implicit_valence = [(atom.GetSymbol(), atom.GetImplicitValence()) for atom in mol.GetAtoms()]
implicit_valence

# Outputs >> [('C', 3), ('C', 2), ('O', 1)]

explicit_valence = [(atom.GetSymbol(), atom.GetExplicitValence()) for atom in mol.GetAtoms()]
explicit_valence

# Outputs >> [('C', 1), ('C', 2), ('O', 1)]

Is there a word definition for implicit and explicit valence? Valence is a fixed number for each element, so they must use a much different definition.

I could not find a definition per se, and I am not expertised in chemistry, but as far as I understand from their documentation, explicit valence is the number of bonds that atom forms in the molecule. And implicit valence is the number of implicit H atoms attached to the atom.

@Martin-マーチン - some formats / representations (e.g., SMILES, ChemDraw, SD files) do not explicitly indicate hydrogen atoms - think a standard 2D organic molecule depiction. So these are "implicit" hydrogens vs. formats that require all atoms to be indicated (e.g., quantum chemistry).

@Geoff Thank you for clearing that up. I can only wonder, why we reuse these terms, when the original is already confusing enough…


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Rotational Energy Levels
Why does the rotational constant B decrease and transition spacings decrease as the mass of a particle increase?
I understand from a purely equation perspective that since
$$B = \frac{h} {8\pi ^2 cI}$$
that as $I$ increases the denominator increases and so $B$ decreases. But what is the physical reasoning behind this? Why or in what way is the rotational constant dependent on mass?

I is the moment of inertia of the molecule, which is given by
$$I = \mu R^2$$
R is the distance between the two atoms and $\mu$ is the reduced mass of a bimolecular system, given by
$$\mu = \frac {m_1 m_2} {m_1 + m_2}$$ If the mass of either particle increases, then the reduced mass increases, causing $I$ to increase, which then causes $B$ to decrease.

thank for simply restating my question and not answering it at all.

How is the energy related to B?

See, it is pretty much the same with any quantum system (think of PIB, think of HO). A heavier particle means more classic-like behavior, which means "less discrete" energy spectrum, which means smaller transition spacings.

First, take a look at classical physics. The angular momentum of a particle rotating in a plane is defined as
$$L = I \omega$$
and its kinetic energy is
$$E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}.$$
So if you formulate your energy in terms of the angular momentum of your rotating particle, you arrive at the inverse relation.
In analogy to the classical picture, the eigenvalues of the rotational Schrödinger equation,
$$ E = hcBJ(J+1),$$
likewise depend quadratically on the angular momentum quantum number, and thus have a similar inverse dependence on the moment of inertia through $B$.


text:
Why does the rusting process start from the bent/squeezed portion of iron?
I have noticed that the rusting of iron starts from the squeezed or bent parts of iron piece.
 I want to know what is the peculiar reason for the observation that rusting starts from bent or irregular surface than smooth part of a given iron piece and it spreads to everywhere.

Which direction is gravity acting in this photo? Maybe water settles in the troughs? Also, is the pure Fe or is it an alloy or plated? The bends will have more cracks and therefore more susceptible to corrosion.

That is an aluminum  roofing sheet and the photo was taken on its bottom part not the sun facing part.

Sorry for the mistake. It's not aluminum

I think this question is same as wondering as to why a certain cancer patient has a cancer in the kidney but not on the skin? Nobody knows. In the same way, rusting is a random process as well. It can start anywhere. This iron sheet is not ultrapure iron sheet of homogeneous nature. One can see a lot of spots in troughs and crests of this sheet. Secondly you do not know the history of this sheet. Is this galvanized or not galvanized? This rusted spot may have a local impurity which can initiate a local electrochemical cell.
In formal atmospheric corrosion studies, my colleague uses flat sheets of test material and rusting pattern is totally random. They also do a scratch test by drawing a cross X on the flat coated sheet, in case they wish to test a coating. The purpose is to expose the "fresh" metallic surface. The rusting then starts from the X pattern. This image is an example of flat sheet used in ASTM coating. The rusting pattern is nothing but random design on the uncoated sheet (left most).
https://hardide.com/technical/astm-b117-corrosion-resistance/


text:
How is the chirality of sidechains written in the IUPAC name?

I circled what I believe to be the chiral centers and I added the locants for the longest carbon chain. One of the chiral centers is in a side chain, so that carbon doesn't have a locant. How should its absolute configuration be included into the IUPAC name?

Side chain have their locants. Your question is rather how to number them and even more general, how to number/call a molecule and has little to do with chirality.

You have correctly ideentified and numbered the parent structure (the longest chain), which is undec-3-yne.
Including the substituent groups, the complete name is 10-bromo-5-(butan-2-yl)-2-methylundec-3-yne.
Note that names for simple substituent groups derived from acyclic hydrocarbons with the free valence not at the end of the chain are generically called ‘alkanyl’. A low locant is assigned to the free valence (i.e. the suffix ‘yl’) in accordance with the numbering of the chain. In this case, butan-2-yl is such a substituent group.
Also note that the locant ‘2’ in butan-2-yl refers to a different structural element compared to the other locants in 10-bromo-5-(butan-2-yl)-2-methylundec-3-yne. Therefore, parentheses are used around the substituent prefix to separate the locants.
If you need to indicate the stereochemical configuration, the same numbering and separation of locants is used, for example: (5R,10R)-10-bromo-5-[(2S)-butan-2-yl]-2-methylundec-3-yne.

Very well explained!


text:
Inductively Coupled Plasma and Mass Spectrometry
I was doing some homework and I was reading about how ICP is suitable for mass spectrometry and it said how its high temperature favours ionization which is good in MS. But also, because the temperature is so high in a plasma, wouldn't there be a very high collisional frequency which could cause side reactions/disruption of the signals, etc?


text:
Electrochemistry conditions for voltage to increase in a galvanic cell
I understand that increasing the concentration of reactants in a galvanic cell generally causes the voltage of the cell to increase. But is that really the case always?
Consider a standard cell of zinc and copper, 1 M and 1 L each solution of copper nitrate and zinc nitrate in each half cell. Another cell, increased the concentration of copper nitrate to 4 M, but added 0.25 L of it and kept the same zinc nitrate 1 M and 1 L.
This way I increased the concentration, but the amount of substance is the same. Does the voltage increase in the new cell? I would really appreciate someone to explain to me the specifics of what causes the voltage of a cell to increase based on the reactants

Hmm so are you referring to the Nernst equation then? https://en.wikipedia.org/wiki/Nernst_equation#Expression

Please don't use a zinc plate in a solution of zinc nitrate. Metallic Zinc reduces nitrate in neutral solutions and produce nitrous ions and then hydrazine.Better use zinc sulfate.

@Maurice, zinc does not reduce nitrate in neutral condition. Nothing happens if you dip Zn plate in 1 M zinc nitrate.

In an electrochemical cell, increasing the concentration of reactants will increase the voltage difference, as you have indicated. A higher concentration of reactant allows more reactions in the forward direction so it reacts faster, and the result is observed as a higher voltage.
If you have adjusted the cell volume to keep the total amount of reactants the same, the rate of electron flow thru the cell will be higher at first. Then the cell will run down faster, but the total amount of electron flow will be the same.
Electrical energy is the product of the voltage times the total amount of current (total number of electrons flowing). Ordinarily we say volts x amps = power (which is the rate of energy produced or used), but by energy here I mean volts x amps x time. You might think that there is a free source of energy here: higher concentrations give higher voltage at the same current. Can we make this into a perpetual motion battery?
Well, no, because the increase in cell voltage is rather minuscule (0.0178 V vs 1.12 V, according to ManRow). One way to explain this is that the more dilute solution has more entropy, but we did not extract the work associated with that when we made up the dilute solution. So if you find a source of dilute reactants and try to concentrate them to get more energy out, you have to put in work to concentrate the solutions, and it will cost you more to do that than you gain in voltage.

Did you by any chance mean voltage x current = power, not electrical energy? Also thanks for the thought-provoking idea regarding perpetual motion

Yes, thanks! I was thinking total electron flow (accumulated) and was imprecise. I will edit the original answer.

Yes, your voltage will change in accordance with the Nernst equation
$$\Delta E_\text{cell} = - \frac{RT}{zF}\ln Q_{\text{rxn}}$$
So, in your copper-zinc battery, where
$$\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)}$$
with a standard $E^\circ_\text{cell} \approx \pu{1.10 V},$ setting the $\left[\ce{Cu^2+}\right] = \pu{4 M}$ should raise your cell potential by
$$\Delta E_\text{cell} = -\frac{RT}{2F}\ln\frac{1}{4} = \pu{+0.0178 V}$$
for a new cell potential (at $\pu{25 °C})$ of
$$E_{\text{cell}} = \pu{1.10 V} + \pu{0.0178 V} \approx \pu{1.12 V}$$

So what is the reasoning for the increase in voltage? There is still the same number of moles of reactant. Why is it that only the concentration affects the cell potential rather than number of moles? Moreover, i want to know if there are any other factors that can affect voltage of a galvanic cell besides the reaction, and the reaction quotient. Finally, does the new cell last as long? Sorry I am self learning this subject and I don’t understand some concepts. Thanks.

By varying $\left[\ce{Cu^2+}\right]$ you are shifting the system either closer to or farther away from chemical equilibrium, often expressed as a ratio of product & reactant concentrations. This principle also applies to galvanic cells like in your example as well, and reflects in the EMF that we can measure from them as well. Namely, the closer the system is to equilibrium, the smaller the magnitude of the EMF (and vica versa).

The new cell will last as long as the system has not reached equilibrium. However the closer the system is to equilibrium the lower the cell potential will be. Finally, yes, there are additional factors that can influence the cell potential as well, such as temperature, for example. In your example if we set $\left[\ce{Cu^2+}\right] = 4\text{M}$ and also raised the temperature (to something warmer than the standard 25°C), the new cell potential would above the calculated 1.12V as well.

I suppose it would benefit this answer to make clear at what point the concentration will effect the potential. At least for me it is not obvious, why Q = 1/4 with the change in concentration.

At what point? The Nernst equation is technically relevant at "all points" and times of the reaction, not only initial conditions. The reaction quotient (see https://en.wikipedia.org/wiki/Reaction_quotient) I computed was for the initial conditions however (though, I thought that was obvious). An interesting side effect is that you can also use the Nernst equation to determine the equilibrium ("final") concentrations of the reactants, by finding the reaction quotient that would give a total, zero cell potential.

Also I've noticed the OP might also be asking about why concentrations are used for the terms of "activities" (https://en.wikipedia.org/wiki/Thermodynamic_activity) in the reaction quotient., but it looks like that relationship is already covered by Mark Wolfman's answer : )

As others have answered, yes the potential of the cell will still change. It sounds like you're looking for a more "physics-y" answer than the ones provided so here goes.
The Nernst equation (and equilibrium equations in general) work based on concentration and not total moles. To understand why, start with a homogeneous reaction, like $\ce{HCl + NaOH <-> NaCl + H2O}$, or more usefully $\ce{H+ + OH- <-> H2O}$. In order for one molecule of $\ce{H2O}$ to be produced, we need a proton and a hydroxide ion to collide. If there are lots of them, this reaction will happen often. However, at the same time, some number of $\ce{H2O}$ molecules will dissociate into protons and hydroxide ions (i.e. the reverse reaction). If the forward reaction happens quickly compared to the reverse reaction, then the equilibrium will favor the products. If you double the volume but keep the concentration the same then your increasing the number of collisions of protons and hydroxide ions but also increasing the number of dissociations of water molecules so overall the equilibrium stays the same. Only if you increase the number of protons and hydroxides relative to the number of water molecules will the equilibrium shift. This is what you do when you increase the concentration.
Now let's switch to your battery example and consider the half-reaction of $\ce{Cu^{2+}_{(aq)} + 2e- <-> Cu_{(s)}}$. This is a heterogeneous reaction so we have to think about the interactions at the surface of the electrode. In order for the forward reaction to occur, a solvated copper ion must shed its solvation layer and adsorb to an active site on the surface of the metal then receive two electrons from the copper. At the same time, some number of atoms on the copper metal are releasing two electrons, solvating and entering the bulk electrolyte. Let's say you have a cell in equilibrium. This means that the rate of the forward reaction and the rate of the backward reaction are the same. Increasing the concentration of the electrolyte means that now every second more ions will be hitting the copper metal and it increases the likelihood of a conversion to copper metal. This increase in the rate of the forward reaction pushes the equilibrium farther towards the products. However, increasing the total volume of the electrolyte without changing the concentration doesn't affect the rates of adsorption of ions to the surface of the electrode, since that's an inherently surface phenomenon.
To use an analogy: the number of cars per minute that get through a toll-booth is only determined by how quickly the very front cars pay and drive off. Adding a bunch of cars in a queue behind the booth doesn't change anything. You could make the queue a mile long and still only the very first row of cars actually paying will set the rate of traffic. Of course, having a long line means it will take longer for all the cars in the queue to get through, just as in your case the additional of volume of electrolyte will increase the capacity of the cell.

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. What is wrong on the other hand is using mesomeric arrows ($\ce{<->}$) instead or equilibrium arrows ($\ce{<=>}$).

Your link is broken, but sure I'll make sure to get the arrows right next time. Any thoughts on the actual chemistry, though?

Sorry for the old link, fixed: recommendations (Sec. 2.1.). You might want to have a look at the Grotthuss Mechanism. In aqueous solution, there not really are any of the ions you suppose there to be, i.e. $\ce{H+}$, $\ce{OH-}$; collisions isn't the right word to use there. I can't really follow the parts after that.


text:
What is the oxidation state of copper in Gilman's Reagent?
What is the oxidation state of copper in Gilman's reagent? In $\ce{R-Cu}$ bond the oxidation state is +1. Thus in $\ce{R-Cu-R}$ bond the oxidation state has to be +2. But if that so why positive $\ce{Li}$ act as counter ion in Gilman's reagent? And why is it named as dialkyl cuprate (-ate for anionic complexes)?

All of your questions can be answered by the fact that R-Cu-R in a Gilman reagent has a negative charge on it.

Why it has a negative charge

Because copper is not in the +2 oxidation state; it's +1.

Since it make bond with two alkyl group ,should it be +2

Oxidation state isn't determined by how many bonds to alkyl groups you have. By that logic the Mn in permanganate ion $\ce{MnO4-}$ would have an oxidation state of +8, since it's bonded to four oxygens.

Do the math. Two R- and one Cu+ = R2Cu with a negative one charge. Li+ makes R2CuLi neutral.

Don't count bonds.  Count electrons.
Copper has $11$ valence electrons as a neutral atom, as the $d$ orbitals are able to participate in reactions.  In the oxidation state formalism the alkyl groups, whose atoms are more electronegative than copper, are each considered to take an electron away from the copper to complete their carbon octets.  But then the lithium, bonded to copper (bonding lithium to the alkyl groups would give a separate organolithium molecule instead), adds an electron.  So the copper goes from $11$ valence electrons down to $10$ in the oxidation state formalism, losing one overall; the oxidation state of copper is thereby $+1$.


text:
Using calculus to find the rate of a bivariate reaction
Recently, I've been learning about rates and have found the concept fairly easy to understand in the case of a reaction with a single reactant, which is essentially solving $$f'(t) = kf(t)^n$$ where $f(t)$ gives the concentration of the reactant at time $t$. Now, for a reaction with two reactants, I'm not exactly sure how to represent the rate in terms of a function. My best guess was defining a function $r(t) = \left(x(t), y(t) \right)$ where $x(t)$ and $y(t)$ represent the concentration of both reactants respectively at time $t$, then solving $$\mid \nabla r(t) \mid=kx(t)^ay(t)^b $$ This boils down to mostly a math problem but I don't know much about multivariable calculus. I'd appreciate if anyone could provide an answer or a link to a source. Thanks.

Reactants don't react independently of each other, they are linked by stoichiometry. It is still a single variable problem.

Agree with Karsten. Also, this is first of all chemistry, not maths, so there is no 'bivariate' reaction. In fact, regardless of how many different reagents are present, the rate of the reaction depends on its mechanism. A reaction with one reagent can be second order. Some reactions are pseudo-zero order because the rate-determining step depends on species whose concentrations don't change much. So my main advice here is: look at this as a chemistry problem first, understand what is actually happening in the system, and only then apply the appropriate maths to solve it.

Do you mean something like this?

@Rodrigo: look at the comment by Ivan Neretin to the post you linked. Far too complicated for its own good. And again, while maths is necessary to solve some problems related to chemical systems, one must first understand what is happening chemically, then find the simplest mathematical model that addresses it. IMO, the main issue with the OP is that the 'two reactants' are treated as independent variables in some abstract mathematical world, whereas they are actual molecules colliding with each other, and their concentrations are linked by mass balances. It can't be right.

@user6376297 So, how can the model be fixed?

Look at the comment by orthocresol to the post you linked. That's what is usually taught in reaction kinetics courses (or at least, it was when I was a student; maybe now one has to be a nobel laureate in quantum physics just to describe an aldol condensation).

@Rodrigo - actually I see now that your own answer to that post does take into account the concentrations mass balance, so it's correct, and basically the same as orthocresol suggested. Maybe you should have linked your own answer https://chemistry.stackexchange.com/a/84059/41751 rather than the OP.


text:
What are the steps of Hartree-Fock Theory?

Outline the steps involved in the Hartree–Fock method for the calculation of molecular electronic structure.

I understand the basics of HF, in that it does not account for electron correlation (neither non-dynamical or dynamical) but I do not understand what the question means by steps.
Are the steps of the HF method just those of the self-consistent procedure or is this something difference?
In addition, does Hartree-Fock account for the antisymmetry requirement or not? I have notes that say it does, because it uses a Slater determinant, but I also have notes that say DFT accounts for the antisymmetry neglected by HF.

The SCF and HF are essentially the same, if you include specifying the nuclear coordinates/charges/#of e- and computing various integrals as part of the SCF. I have never heard of DFT including antisymmetry neglected by HF

Hope this would be helpful: http://vergil.chemistry.gatech.edu/notes/hf-intro/hf-intro.pdf

HF accounts for exchange correlation exactly. DFT is a whole other story, see some more here: Are there any full worked examples of DFT calculations?

Martin is correct here, though the terminology around this issue is easy to get confused. Exchange accounts for a certain type of correlation (same spin electrons avoid being in the same place). A lot of authors call the energy added by post-Hartree Fock methods the correlation energy, which is mostly true because it accounts for the rest of the correlation (electrons moving in response to each other electron rather than their average potential), but exchange is still a form of correlation.

For another phrasing of my previous comment, I would point you toward the second and third paragraphs of the "Atomic and Molecular Systems" section: https://en.m.wikipedia.org/wiki/Electronic_correlation

Yeah, I think it asks you to write the energy as expectation value, vary one orbital and get the Roothan equation.
The Hartree Fock model is the extension of the Hartree model where the wavefunction was schematized by the following guess:
\begin{equation}
\psi=\phi_{a}(1)\phi_{b}(2)...\phi_{n}(n)
\end{equation}
If you work out the equations, you'll see the equations look a lot like the HF ones, but you'll miss the exchange term. This is because the wavefunction is not in the correct fashion to describe a fermionic system not made by infinitely distanced parts.
So you need to plug in the antisymmetric form of the wavefunctions (Slater determinants are a basis in the space of antisymmetric functions). Also, the HF approximation requires you to just approximate the wavefunction with just one determinant, just one piece of the expansion. This is clearly a simplification but it can be correct enough if there is a significative HOMO-LUMO gap and the system is closed shell.
So the HF model gets some correlation (known as fermi hole) meaning that electrons having same spin do avoid each other (you enforced it using antisymmetric wavefunctions) but the electrons don't avoid themselves respecting the Coulomb repulsion (two electrons cannot be in the same place even if they have not the same spin) this is a first piece of correlation you lose, the other being the fact you simply are using one determinant. Unfortunately correlation is described as everything going beyond HF because of Quantum information theory, so you have to be careful when you talk of correlation of HF. As it regards DFT, you'll plug in it functionals, so you may include functionals in it that allow you to recover some more correlation, but remember that DFT functionals are, most of the time, semiempirical.


text:
Gibbs free energy for an ideal gas
You know that for a generic system which during a process exchanges heat with a reservoir at a constant temperature $T_{a}$ it holds:
$$\Delta G=\Delta H-T_{a}\Delta S$$
Where:
$\star)$$\Delta G$ is the Gibbs free energy change of the system during the process.
$\star)$$\Delta H$ is the enthalpy change of the system during the process.
$\star)$$\Delta S$ is the entropy change of the system during the process.
Note that during the process (unless it is quasistatic) the temperature of the system is not defined, so it is not constant.
Said this, if I want to simplify the previous equation for a ideal gas:
$$
\left\{
\begin{array}{c}
\Delta H=n c_{p} \Delta T \\
\Delta S=n c_{p} ln(\frac{T_{2}}{T_{1}})+n R ln(\frac{p_{1}}{p_{2}})\\
\end{array}
\right.
$$
$$\Delta G=[n c_{p} ln(\frac{T_{2}}{T_{1}})+n R ln(\frac{p_{1}}{p_{2}})] T_{a}+n c_{p} \Delta T$$
Where $T_{1}, p_{1}, T_{2}, p_{2}$ are referred to the system in its inital and final state.
On my book I find written:
$$\Delta G=n R T ln(\frac{p_{1}}{p_{2}})$$
Which is the equation I got by putting $T_{1}=T_{2}$. Someone could confirm that my formula is right and the book is just considering a particular case? (The book isn't very accurate in this explanation)

Please provide the source of your initial equation.

Atkins-Jones-Lavermann Chemical principles

What exactly do they say, because this equation does not seem correct.

Indeed it is suspicious you are using incorrect equations: under isothermic process, there is an enthalpy change, yet it would be zero according to your equation. The enthalpy expression you use is the isobar expression, isn't it?

They assume temperature and pressure constant @Greg

@Landau and how do you do that? Also, fix the question because it wasn’t what you asked.

It is not possible to confirm your derivation. If you describe the change in free energy of a system with the expression
$$\Delta G=\Delta H-T\Delta S$$
then you are implicitly assuming that the initial and final temperatures of the system are equal to $T$. The more general expression for $\Delta G$, from the definition $G=H-TS$, is
$$\Delta G = \Delta H - \Delta (TS)$$
From your expression
$$\Delta G=\Delta H-T_a\Delta S$$
it can be assumed that the initial and final temperatures of the system are equal to $T_a$, that of the surroundings. Given that you know that the initial and final temperatures are identical, and that for an ideal gas $V=nRT/p$, you can use a convenient reversible path to compute the change in $\Delta G$ by integrating $$dG=Vdp-SdT$$ at constant $T$, which yields the textbook expression $$\Delta G = nRT \ln \left(\frac{p_1}{p_2} \right)$$ (note here $p_1$ is the final pressure).


text:
Methylhydrazine Calorimetry
Below is a chemistry practice problem.  I am curious why the change in heat of the oxygen molecules doesn't affect the calculations.

Methylhydrazine ($\ce{CH6N2}$) is commonly used as a liquid rocket fuel. The combustion reaction is below. When 4.00 g of methylhydrazine is combusted in a calorimeter, the temperature increases from $\pu{25.00 ^\circ C}$ to $\pu{39.50 ^\circ C}$.  The calorimeter constant is $\pu{7.794 J/^\circ C}$ and the specific heat of methylhydrazine is $\pu{2.93 J/g ^\circ C}$. What is the overall heat of the reaction (kJ/mol methylhydrazine)?

$$\ce{2 CH6N2(l) +   5 O2(g) -> 2 CO2(g)+   2 N2(g) +   6 H2O(g)}$$
$$\begin{align} \mathrm{molecular\; mass\;\;} \ce{CH6N2} &= \pu{46.08 g/mol}\\
\rightarrow \pu{4 g} \cdot \pu{1 mol}/\pu{46.08g} &= \pu{0.0868 mol} \ce{CH6N2} \\ \rightarrow
 q &= (\pu{4 g})(\pu{39.5 ^\circ C} - \pu{25 ^\circ C})+(\pu{7.794 J/^\circ C})(\pu{39.5 ^\circ C} - \pu{25 ^\circ C}) \\  &= \pu{283 J}  \\ \rightarrow \Delta H &= \pu{-3.26 kJ/mol} \end{align}$$
The change in heat accounts for the $\Delta T$ and temperature change of the calorimeter and the temperature change and $C_p$ and mass of the methyl hydrazine, but why doesn't the oxygen also absorb heat in accordance with its mass and specific heat constant?
-3.26 kJ/mol is the "correct" answer in the answer key, meaning that the absorption of heat by methylhydrazine IS required to get the answer which I didn't expect-I though it would just be the heat absorbed by the calorimeter, but I understand why the methylhydrazine would absorb heat from the reaction, but then why doesn't the oxygen absorb heat since there is actually more mass of oxygen in the calorimeter as mass of methylhydrazine.

You are saying that the change in heat is the sum of the heat absorbed by the calorimeter and the heat absorbed by the initial compound methyl-hydrazine and that gives the answer that is considered to be correct, but why does the methyl-hydrazine absorb some of the heat according to its Cp and mass, but the oxygen that reacts with the methyl hydrazine doesn't absorb heat.  I'm not sure why.  Is it an error in the problem?

^^That's what I'm asking

There seems to be an error in your equation for q (doesn't affect the question otherwise).

You need to include the Cp of methylhydrazine in the equation for q.  I don't know why the "correct" answer is arrived at by including the heat change in mehtylhydrazine but not also in the oxygen.  I would think, neither or both.

I suggest adding the following to the end of your question:
-3.26 kJ/mol is the "correct" answer in the answer key, meaning that the absorption of heat by methylhydrazine IS required to get the answer which I didn't expect-I though it would just be the heat absorbed by the calorimeter, but I understand why the methylhydrazine would absorb heat from the reaction, but then why doesn't the oxygen absorb heat since there is actually more mass of oxygen in the calorimeter as mass of methylhydrazine.

Simple question.  When calculating the heat of reaction in a calorimeter, does the temperature (or state) change of the reactants and or products affect the calculation?


text:
What is the physical significance of the fact that spin operators do not commute?
So we looked at the Stern-Gerlach experiment and I understand the basic principle; when a stream of silver atoms is passed through an inhomogeneous magnetic field, and the Ag atoms are deflected either up or down depending on their spin.
What I don't understand is how this proves that spin operators don't commute. When it refers to spin operators I presume they are referring to S^2 and Sz. If these did commute why would we see a smeared blot?
And why, after the first pass through a field, do they electrons split a second time after a second pass through a magnetic field?

I think the reason you're having trouble is your assumption that the operators are $S^2$ and $S_z$. In fact, those two operators do commute. The operators that do not commute are the individual components $S_x$, $S_y$ and $S_z$.
You also seem to have misunderstood the significance of the lack of "smear" result. When the silver ions make one pass and we do not see a smear, that simply tells us that angular momentum is quantized. If it could take on a continuous range of values, we would see the smear.
For the question of commutation, when two operators commute, it means that we can know the values of both at the same time. This necessitates that measuring one does not affect the value of the other. If the operators do not commute, we cannot know both their values simultaneously. Experimentally, this means that measuring one affects the value of the other.
The way the S-G experiment demonstrates that, for example, $S_x$ and $S_y$ do not commute is by first separating particles according to the value of angular momentum on one axis, for example $S_x$. So the output is one stream of particles with positive $S_x$ and one with negative. We then pass only one of these streams (say +x) through a magnetic field aligned to separate based on $S_y$. If the measurement of $S_y$ does not affect the value of $S_x$, then all of the particles should still have +x. However, passing the output of the $S_y$ "filter" into an $S_x$ filter again yields two spots, indicating that the $x$ component of the angular momentum has been scrambled by the measurement of the $y$ component.
Importantly, passing the +x output through an $S_x$ filter without first passing through an $S_y$ filter gives only the expected +x spot, indicating that the $x$ component values do not spontaneously scramble in the absence of the $S_y$ measurement.
Because the $S_y$ measurement scrambles the $S_x$ value, we cannot know both values simultaneously, and this is consistent with the mathematical result that the operators do not commute.

Thank you very much! This was incredibly helpful!


text:
Tight binding model for nanoparticles as finite systems
I'm still trying to calculate the band structure ab initio for some hypothetical nanoparticles. It seems that I can directly build a tight-binding model by forming Bloch functions, numerically calculate the dispersion relation, and then apply Brus equation to get band gaps for nanoparticles.
Another strategy I've tried is just to treat the nanoparticle as a large molecule and use MO calculation; however, the problem is that the matrix grows too large when working in three dimensions and using basis sets with multiple orbitals (consider a $20 × 20 × 20$ cubic lattice with $\mathrm{sp^3d}$ basis, that should be a whooping $20^3 · 5^2 = 200\ 000 × 200\ 000$ matrix).
I'm asking if there's some modified version of tight-binding model that specifically work for finite systems. I'm aware that tight-binding model can be used to study nanowires, but the wires themselves are assumed to be infinitely long in one direction so that's not quite what I want.
My chemistry professor suggested modifying the Bloch functions using wavefunctions that are solutions to the particle in a box model. However, I'm not sure how to do this properly, or what's the justification for this.

I'm not sure I understood all of what you are writing. If you can treat nanoparticles as molecules, then the xtb program might be worth having a look at: https://github.com/grimme-lab/xtb To my knowledge, it works fairly well for extensive biomolecules.

@Martin-マーチン Thank you! I will look into this. It seems to be helpful for me.

If the software helped you, I would very much appreciate it, if you could write a short answer with an example how to do it for the future generations to come.


text:
Can I completely dry air with calcium chloride?
I'm trying to dry air by leading it through calcium chloride.
I'm measuring the resulting humidity with a simple capacitive humidity sensor. I do not see measurements below 10% relative humidity, but it is quite possible that it is a result of a calibration error (The sensor is not actually calibrated.)
But how dry can I get air by exposing it sufficiently long to calcium chloride dihydrate?

"Completely dry"? No, because molecules are always moving and there will be some exchange between desiccant and air. If you mean "below 10% relative humidity (RH)", there are a number of ways to do so:

Consider that another way to discuss humidity is the dew point. Use a cold trap: after drying the air with a desiccant, pass the air slowly through a U-tube (or better, a through a spiral condenser) cooled with ethanol/dry ice or similar bath. As the air emerges, the dew point, ~200 K for dry ice, remains that of the bath, but as it warms to room temperature, the RH would fall.
Dry the air with desiccant at room temperature, then heat the air (not desiccant) and the RH drops.
Synthesize air from ~80% liquid nitrogen and ~20% liquid oxygen. If you're a stickler for accuracy in the simulation, add a dash of liquid argon, and a smidgen of liquefied CO2 (not dry ice, which, being open to air, quickly condenses water). These liquefied gases can probably be obtained from a welding supply shop.

BTW, don't forget to bake your apparatus before use! Metal, and particularly glass, adsorb water quite well, and slowly release it at low RH or pressure.


text:
Is an aqueous solution of calcium chloride hygroscopic?
Calcium chloride is deliquescent, it is hygroscopic, so much so that it dissolves in the water it has attracted by hygroscopy.
Sulfuric acid is hygroscopic, even if not anhydrous. So a solution containing water can be hygroscopic.
Is a concentrated solution of calcium chloride that was created by the solid salt attracting water still hygroscopic, attracting more water?
If yes - down to which concentration is that the case?

It will attract more water until the vapor pressure over the solution matches the ambient vapor pressure.

Interesting.  My daughter in 5th grade tried to make crystals from magnesium chloride for a science fair project along with other salts.  The water never evaporated to yield crystals.  By adding MgCl2 to the liquid we were able to get crystals appear for a short time but then it absorbed more water.  That leaves a possibility that we could cool the solution to produce crystals by reducing the solubility which would also lower the vapor pressure of the solution so is is possible that there is a temperature at which the MgCl2 would not be dissolved in its water as well as some humidities.

Even the saturated solution of  table salt is hygroscopic at relative humidity >75%.


text:
In an isothermal process, how can the change in internal energy be 0?
It was written in my textbook,
$$
\mathrm{d}U =
  \left(\frac{\partial U}{\partial T}\right)_V \mathrm{d}T
 +\left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
$$
If the process is isothermal, $\mathrm{d}T = 0$. So, the equation reduces to:
$$
\mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
$$
I was told, $\mathrm{d}U = 0$ in an isothermal process, does that mean $\mathrm{d}V = 0$, or  $\left(\frac{\partial U}{\partial V}\right) = 0$? But how is it possible that $\mathrm{d}V=0$ in an isothermal process, or $\left(\frac{\partial U}{\partial V}\right) = 0$?
If someone could explain what am I doing wrong, that would be very helpful.

People like treating things like an ideal gas, and they should not.

The internal energy of an ideal gas is simply $$ U = \alpha nRT,$$
where $$\alpha = \frac{\text{degrees of freedom}}{2}$$
So, in an isothermal process,
$$
\Delta T = 0 \Longrightarrow \Delta U = \alpha nR\Delta T = 0,$$
and likewise any
$$\left(\frac{\partial U}{\partial P}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T = 0.$$

I was told, $\mathrm{d}U=0$ in isothermal process.

That is not generally true.  It is, however, true for ideal gases, which is probably what you were discussing.   No attractive or repulsive forces exist between ideal gas particles.  Hence the only type of internal energy an ideal gas can have is kinetic energy, i.e.,  energy due to the motion of its particles. And since kinetic energy depends only on temperature, an ideal gas's internal energy likewise only depends on its temperature.  As a consequence, $\mathrm{d}U = 0$ for ideal gases in isothermal processes, because the temperature doesn't change.

does that mean $\mathrm{d}V = 0$, or  $\left(\frac{\partial U}{\partial V}\right) = 0$? But how is possible that $\mathrm{d}V=0$ in isothermal process, or $\left(\frac{\partial U}{\partial V}\right)=0$?

Because the internal energy of an ideal gas depends only on its temperature, $\left(\frac{\partial U}{\partial V}\right)_T=0$, i.e., the internal energy doesn't change with volume at constant $T$.

For completeness, since there is already a well-explained answer addressing how and why $\mathrm{d}U=0$ for an isothermal process is a hallmark of an ideal gas, here is a short derivation of a general expression for the energy.
Start from the total differential for the free energy:
$$
\mathrm{d}U =
  \left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V
  +\left(\frac{\partial U}{\partial T}\right)_V \mathrm{d}T
  \tag{1}\label{eq:total-differential}
$$
Define the heat capacity at constant volume as
$$
C_V =
  \left(\frac{\partial U}{\partial T}\right)_V.
  \tag{2}\label{eq:heat-cap-const-v}
$$
Evaluate the partial derivative with respect to $V$ in \eqref{eq:total-differential} from the 1st law of thermodynamics:
\begin{align}
  \mathrm{d}U &=
    -P\mathrm{d}V + T\mathrm{d}S \\
  \rightarrow \left(\frac{\partial U}{\partial V}\right)_T &=
    -P + T\left(\frac{\partial S}{\partial V}\right)_T
    \tag{3}\label{eq:first-law}
\end{align}
Make use of the following Maxwell relation:
$$
\left(\frac{\partial S}{\partial V}\right)_T=
  \left(\frac{\partial P}{\partial T}\right)_V
$$
So \eqref{eq:first-law} becomes
\begin{align}
  \left(\frac{\partial U}{\partial V}\right)_T &=
    -P + T\left(\frac{\partial P}{\partial T}\right)_V
    \tag{4}\label{eq:with-Maxwell}
\end{align}
Then inserting the results of \eqref{eq:heat-cap-const-v} and \eqref{eq:with-Maxwell} into \eqref{eq:total-differential} we obtain the general result (when only $pV$ work is done):
$$
\mathrm{d}U =
  \left[ -P + T\left(\frac{\partial P}{\partial T}\right)_V \right]
  \mathrm{d}V + C_V \mathrm{d}T
  \tag{5}
$$
If you plug in the equation of state for an ideal gas you then obtain
$$\left(\frac{\partial U}{\partial V}\right)_T = 0$$
and the expected result
$$\mathrm{d}U = C_V \mathrm{d}T.$$


text:
What is the meaning of an oxide being 'Higher' or 'Lower'?
To put it simply, many books use the term 'Higher Oxide' or 'Lower Oxide'.
What does this mean?

I think it generally means that the average oxidation state of the metallic element in the oxide is higher or lower. You can also say that the mass percentage of oxygen in the oxide is higher or lower.


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Large difference in boiling points of tetrafluoromethane and tetrachloromethane
Why is carbon tetrachloride $\ce{CCl4}$ is seen to posses liquid state (b.p. $\pu{76.72 °C}),$ whereas carbon tetrafluoride $\ce{CF4}$ is in gaseous state at room temperature (b.p. $\pu{−127.8 °C})?$

Compare the bond strength of C-F and C-Cl bond.

@Nilay Ghosh should we be more concerned about the intermolecular forces than intramolecular ones?

The vapour pressure can be estimated by equating the rate of leaving the liquid surface with that returning to it from the vapour. This leads to an expression of the form $\displaystyle p\sim n_{l}RTe^{-\Delta E/RT}$ where $\Delta E$ is the energy needed to remove a molecule from the liquid to the vapour phase and $n_{l}$ is the number density (number / unit volume) of the liquid.
The energy $\Delta E$ will clearly depend on the type of molecules involved. CCl$_4$ has large easily polarisable chlorine atoms. This means that these electrons are easily influenced by nearby atoms and this leads to an increased intermolecular interaction compared to CF$_4$ because the F atoms, having fewer electrons, and ones close to the nucleus, are only slightly polarisable. The increased $\Delta E$ (CCl$_4$ vs CF$_4$) then leads to a lower vapour pressure for CCl$_4$ and hence a higher boiling point which is defined at the temperature when the vapour pressure reaches one bar, or one atm. if you prefer.
(Sometimes the polarisable interaction is called induced-dipole, induced dipole or London or in general van der Waals interaction.)

When you write the proportionality p~RT... I am sure you don't mean that p has units of energy. Maybe you could just write p~exp(...) and remove that RT factor, or be more specific.

Thank you, a good idea, then I realised that multiplying by number density makes the units pressure, so have made this change.

The answer is brilliant but you don't need to have a prefactor...

It is really just rewriting $\Delta G^o = -RT \ln K$ for the liquid-vapor equilibrium, where K=p. Since the $\Delta S$ term is approx a constant for the two compounds it can be ignored, and then if $\Delta U = \Delta H$ which it approx is for a liquid, you have your result.