text:
Why is everything confused up in thermodynamics

Is the formula application done properly when calculating work. I think the volume should be final minus the initial. So in first step it should be $V_1-V_i$ and in second step it should be $V_f-V_1$. But totally different things are given and I can't understand. There is also the graph based on this and conclusions too. I wonder whether they are correct. Please help me.why is it so confusing?

What in your opinion should the work be?

@ChesterMiller yeah I have written it in the question. My problem is with the volume term in the work

@ChesterMiller sir please help me

You have a two-step process.  In the first step, the mass m1 is removed from atop m2, and the system is allowed to re-equilibrate.  In this step, if I do a Newton's law force balance on the combination of the piston (assumed massless) and m2, I get:  $$F_g-m_2g-P_{atm}A=m_2\frac{dv}{dt}$$where $F_g(t)$ is the force exerted by the gas on the base of the piston during step 1, A is the area of the piston, and v is the velocity of the piston.
If I multiply this equation by the velocity of the piston $\frac{dx}{dt}=v$, I get:$$F_g\frac{dx}{dt}=m_2g\frac{dx}{dt}+P_{atm}A\frac{dx}{dt}+m_2v\frac{dv}{dt}=\left[\frac{m_2g}{A}+P_{atm}\right]\frac{dV}{dt}+\frac{1}{2}m_2\frac{dv^2}{dt}$$where V is the volume of gas at time t.
If I integrate this equation from t = 0 to arbitrary time t during the process, I obtain:
$$W_g(t)=\int{F_gdx}=\left[\frac{m_2g}{A}+P_{atm}\right][V(t)-V_i]+\frac{1}{2}m_2v^2(t)$$where $W_g(t)$ is the work done by the gas on the combination of piston and m2 up to time t, V(t) is the volume of gas at time t, $V_i$ is the initial volume of gas, and v(t) is the velocity of the combination of piston and m2 at time t.  The final term on the right hand side of the equation represents the kinetic energy of the combination of piston and m2 at time t.
What do you think will happen to the velocity of the combination of piston and m2 at very long times?
(a) the velocity will continue increasing forever
(b) the velocity will slow down gradually until the piston and m2 finally stop at the equilibrium position $V_1$
(c) the piston and mass will overshoot the equilibrium position (until their velocity slows to zero), reverse direction and overshoot the equilibrium position going in the other direction (until their velocity slows to zero again), reverse direction again, etc.  They will oscillate about the equilibrium position forever (like a mass/spring system in simple harmonic motion)
(d) same as (c), except that, even if the piston is frictionless, viscous damping stresses in the gas will ultimately damp out the oscillation until the combination of piston and m2 settle at the equilibrium position $V_1$
Actually, the correct answer is (d).  However, I suppose (b) would be possible if the gas were somehow very viscous.  Either way, however, the conclusion is the same.  At final equilibrium of step 1, the piston is no longer moving.  So the amount of work that the gas does on the combination of piston and m2 when final equilibrium is attained in step 1 is just:
$$W_{g1}=\left[\frac{m_2g}{A}+P_{atm}\right][V_1-V_i]$$This is the result they obtained in your reference (aside from the sign (since they are considering work done by the surroundings on the system, and I am considering work done by the system, in this case the gas, on its surroundings).

If you look closely sir they have written Vi-V1. That's what bothering me(leaving off signs) . Again in second step we must go from V1 to Vf in the integral but they have given Vi to Vf. Am I correct?

I'm okay with what you have explained me sir

Yes, you're correct.  That's a typo.

then what do i do with the graph sir. how is the work done in two step greater than that in one step?

I don’t understand this question.

sir I'm just asking how is the work more in two step process of than that in one step

In the two step process, first the gas does some work, and then in the 2nd step it does some more work.  So the total amount of work is more the if it just did the first step.

But when I look at the graph the shaded region is less in area than in one single step. In one step ,the integral would give me the whole are under the P-V curve. Am I correct?

No.  The gas re-equilibrates after removing m1, while doing as much work as it can on m2 and the external atmosphere.  For it to do more work, you remove m2.  It then does more work, but only on the external atmosphere, since m2 has now been removed.  So the total work is what they have shown in the figure.  In step 1, you are doing work on m2 and the external atmosphere, but in step 2 you are doing work only on the external atmosphere.

You can't do the whole thing in one step because, with m1 removed, the gas can't expand beyond V1, since this is the volume at which it reaches equilibrium.

yes sir now I understand this.

Since we are doing work against a pressure higher than the atmospheric pressure the work his higher . Whereas in case of reversible expansion we do work only against the atmosphere in the whole process

Let us continue this discussion in chat.


text:
Can a car's battery be used as a sacrificial anode?
The title of the question says it all. Can I hook the anode of my car's battery up to the car's shell and prevent it from rusting? Also would the car's battery get electrically charged from this?

If by anode you mean negative terminal, no it won't prevent rusting or charge the battery; the negative terminal is already connected bodies of most cars.

A sacrificial anode must be in direct contact to the corrosive medium and electrically connected the thing it's supposed to save. Your battery sits high and dry under the hood, for good reason. Doesn't work.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Nomenclature in case when same fucnctional group is present more than once

here since Only CN group is present, it is the major functional group. In this molec. we cannot have a chain which has carbons of all CN group. hence in such compounds we do not take any carbon of cn in the parent chain.so answer should be 3.
but few books and many reference texts are showing
name as option d.

The IUPAC rules do not seem to have an explicit rule for general cases when more than two cyano groups are attached to the (linear) alkane chain, like it does for aldehydes (see Is propane-1,2,3-tricarbaldehyde the correct IUPAC name of CHO-CH2-CH(CHO)-CH2-CHO?), but do have a related one:

P-66.5.1.1.2 If an unbranched alkane is linked to more than two terminal cyano groups, all cyano groups are named from the parent hydride by the substitutive suffix ‘carbonitrile’, preceded by appropriate multiplying prefix and locants.
  Example:
$\ce{\overset{4}{C}H3-[CH2]2-\overset{1}{C}(CN)3}$
  butane-1,1,1-tricarbonitrile (PIN)

I think that similar usage in general is implicitly allowed in the rule

P-66.5.1.1.4 When a group is present that has priority for citation as the principal characteristic group or When all $\ce{-CN}$ groups cannot be expressed as the principal characteristic group, the $\ce{-CN}$ group is designated by the preferred prefix ‘cyano’.
(…)

(Three examples follow, two of them have another higher-priority group $\ce{-COOH}$; the other one has three cyano groups, not on linear chain but one of them is attached via a chain branch, $\ce{NC[CH2]2CH(CH2CN)[CH2]2CN}$, 4-cyanomethyl(heptanedinitrile).)
I our case, we can express all the groups as principal characteristic group, so the correct name should be propane-1,2,3-tricarbonitrile.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Chloroantimonic vs fluoroantimonic acid
Does chloroantimonic acid exists? What do we get when we mix hydrochloric acid and antimony pentachloride and is the mixture stronger or weaker than fluoriantimonic acid. How about bromo and iodoantimonic acids?

Of course they are weaker, how fluoroantimonic would be strongest liquid acid otherwise?

Chlroantimonic acid has been synthesised by dissolving antimony trioxide in concentrated hydrochloric acid followed by saturation with chlorine. The relevant information can be found here[1]:

Complex compounds are known to which the formulae $\ce{HSbCl6}$, $\ce{H2SbCl7 }$ and
$\ce{H3SbCl8}$ have been ascribed; these may be regarded as meta-, pyro- and
ortho-chlorantimonic acids [...] Neither the ortho- nor the
pyro-chlorantimonic acid appears to have been isolated, although salts
are known. Meta-chlorantimonic acid has been obtained by dissolving
antimony trioxide in concentrated hydrochloric acid and saturating the
solution with chlorine. The solution darkens at first, afterwards
changing to a bright greenish-yellow colour. Crystals can be obtained
by concentrating slightly over a water-bath, adding hydrochloric acid
and leaving over sulphuric acid, the temperature not
exceeing 0° C. The crystals are very hygroscopic; they are soluble in
cold water, alcohol, acetone and glacial acetic acid. The solution in
water undergoes hydrolysis, especially on warming, hydrated antimony
pentoxide separating out. It is prevented by
hydrochloric acid and to some extent by nitric acid. The solutions in
organic solvents are more stable. Various salts of chloroantimonic acid are known [...]

Bromoantimonic acid is prepared from crystallization from antimony bromide/bromine and sulfuric acid. From Here[2]:

Of the three possible acids only one has been obtained in the free
state. By crystallisation from a solution containing antimony
tri-bromide, bromine and either hydrobromic or sulphuric acid the
compound meta-bromoantimonic acid, $\ce{HSbBr6.3H2O}$, has been
prepared in the form of hygroscopic, irregular, six-sided, black
tablets. It decomposes readily, with evolution of bromine, leaving a
residue of antimony tribromide.

Pure iodoantimonic acid is not known although a mixed acid, $\ce{HSbBrI3}$ is known. Here[3]:

Salts of a complex antimony iodohydrobromic acid, $\ce{HSbBrI3}$, have
been obtained by triturating equimolecular proportions of antimony
triiodide and metallic bromide with a non-aqueous substance such as
acetic acid or xylene.

Carefully note that in this acid, antimony is in the +3 rather than +5 oxidation state.  Iodide ion, more easily oxidized than lighter halide ions, would accordingly be less stable in combination with antimony(V).
References

http://antimony.atomistry.com/chloroantimonic_acids.html

http://antimony.atomistry.com/antimony_pentabromide.html

http://antimony.atomistry.com/antimony_triiodide.html

$$\ce{Sb2O5 + 12HCl → 2H[SbCl6] + 5H2O}$$
http://www.easychem.org/en/subst-ref/?id=3722

What about their acidic strength.

I indicated that the iodine-bearing acid has antimony in the oxidation state +3 rather than +5.  Feel free to roll back if inappropriate.


text:
How do we determine molecule chirality in these examples?

I understand that in the first example, the molecule is chiral as there is no symmetry. For the second and third example, I see no symmetry in the 3d conformation after twisting the right hand side. In the third example, I do see symmetry if we twist the right hand side so that the carbon molecules are in one plane. What am I missing?
Edit: typo corrected. Also I was not clear in that the picture I attached is the answers the professor provided. I see why the first one is right as there is no symmetry. But I did not see why the second and third ones were considered chiral and achiral, as I saw symmetry in the third example and not in the second. From the answers though, I am understanding that we consider if the molecule is super imposable with it's mirror image?

Are the answers you have circled correct?

Also, welcome to Chemistry.SE!

I think you have made a typo. You write the molecule is not chiral as there is no symmetry; the molecule is chiral. And in the third example is a meso compound, so you have to find symmetry. I don't really understand your question, it seems to me you have done it right.

For E and F, look here: https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415

Although you have arrived at the correct answers and seem to have followed a reasonable process for getting there, it's important to be precise with the use of terms like symmetry and chirality. For a molecule to be chiral, it must be non-superposable with its mirror image. It's important to really embed this definition before talking about symmetry and things like chiral centres. For example, it is possible for molecules to be chiral without any chiral centres, and it's possible for molecules with multiple chiral centres to be achiral.

the guaranteed way to determine if a molecule is chiral: draw (or build a model) of the mirror image and determine whether they are superposable

Now, symmetry is a different, but related, topic. A molecule with a point or plane of symmetry will be achiral. However, molecules with a rotational axis of symmetry can still be chiral, eg. many often-used chiral ligands are C2 symmetric.
So, in answer to your question, "What am I missing? - I would just say: the rigor in use of the terminology. Your approach of drawing different conformers of the molecules with two chiral centres and looking for a plane of symmetry (that shows the molecule is a meso compound, and therefore achiral) is sound and led you to the correct answers.

I would slightly augment Explained's fine answer with a caution that, after you build a model of the molecule in question, be sure to consider all of its easily accessible conformations to see if any of them are superimposable on its mirror image. If so, the molecule is achiral.


text:
Does hydrogen peroxide in basic medium only give oxygen gas on oxidation?
Does hydrogen peroxide on oxidation gives oxygen gas only in basic medium?
I thought that was true but I read about these reaction of $\ce{H2O2}$ in acidic medium:
$$\ce{2MnO4- + 6H+ + 5H2O2 -> 2Mn^{+2} + 8H2O + 5O2}$$
$$\ce{K2Cr2O7 + 4H2SO4 + 3H2O2 -> K2SO4 + Cr2(SO4)3 + 7H2O + 3O2}$$
What is the possible explanation of both the title and the reactions?

I don't get it, what would expect when H2O2 is oxidized?

In both reactions $\ce{H2O2}$ is oxidized by the more powerfull oxidizers permanganate ($\ce{KMnO4-}$) and dichromate ($\ce{Cr2O7^2-}$).
Whether or not oxygen is produced depends on the actual reaction and not necessarily on pH. A rule is that when hydrogen peroxide is oxidized by a stronger oxidant, free oxygen is produced. It is not always the case when hydrogen peroxide is the oxidizing agent itself.
See this discussion: Half equations for $\ce{H2O2}$


text:
Hybridization in ethene
I recently learned that there is such thing as a hybridization in chemistry, but I don't really get it.
For example, in $\ce{C2H2}$ (Ethene), they say the there are 3 electrons in three ${sp^2}$ and one electron in a normal $p$ orbital. How do we decide what orbitals get hybridized and what doesn't?

Welcome to ChemSE. Please keep in mind that the concept of hybridization you mention here applies on atomic, and eventually on molecular orbitals; and not on electrons.

We do not decide either which. Hybridization is an observational theory - it is a theory crafted to fit observation, it is not made from the bottom up. That doesn't make it any less valuable (well... maybe it does...) but your question becomes rather easy to answer. We decide which orbitals become hybridized based on how we measure the molecule to be. We see the symmetry.
Hybridization is just a model. But then again so are most things we learn in chemistry. I would say you are not missing out on anything if you just skip learning it and go directly to molecular orbital theory. Your teacher might argue though, do what (s)he says.

You have overall 4 valence electrons in the carbon atom. They can be involved either in sigma-bonds or in pi-bonds. All atomic orbitals involved in bonding do so as molecular orbitals (remember LCAO as mathematical approach).
When a carbon is involved in single bonds only, its ${s}$- and three ${p}$-orbitals are ${sp^3}$-hybridized. When a carbon is involved in a double bond (alkenes, carbonyls etc.), three of four orbitals are ${sp^2}$-hybridized, because the double bond consists of one $\pi$-bond and one $\sigma$-bond. When a carbon is involved in a triple bond (alkynes, nitriles etc.) two of four orbitals are ${sp}$-hybridized, because the triple bond consists of two $\pi$-bonds and one $\sigma$-bond.

Welcome SimonE to ChemSE.  I edited your answer mainly because the concept of hybridization applies to orbitals, rather than to electrons.


text:
What is the meaning of K in the thermodynamic equation for equilibrium constant?
I just studied the formula for equilibrium constant in thermodynamics. But when an equilibrium is established $\Delta G$ is always $0$ as told by my teacher. Then $k$ has to be always $1$ whenever an equilibrium is established.
But that does not happen always as equilibrium constant is not always $1$. I want to know where I am going wrong here.
And this $k$ is $k_\mathrm p$ or $k_\mathrm c$ ?
Any help will be really appreciated.
Thanks in advance.

You are confused between $\Delta G$ and $\Delta G^{\varnothing}$, the standard free energy change.
The equation you are describing is actually:
$$\Delta G^{\varnothing} = -RT \ln K$$
These numbers are fixed for some reference temperature and pressure.
The specific $\Delta G$ you are referring to is a function of $Q$, the reaction quotient, which describes the position of your equilibrium. Its value, relative to $K$, will determine whether $\Delta G$ is positive, negative, or zero.
Specifically,
$$\Delta G = \Delta G^{\varnothing} + RT \ln Q$$
$Q$ is computed the same as $K$ except that you plug in the actual concentrations you have, not the values at equilibrium.
There are three cases to consider:

$Q = K$. This corresponds to $\Delta G = 0$. We are equilibrium.
$Q > K$. $\Delta G > 0$. We are favoring products to much. We need to shift the reaction back to decrease the value of $Q$.
$Q < K$. $\Delta G < 0$. We are favoring reactants to much. We need to shift the reaction forward to increase the value of $Q$.

Thank you Sir. But I am still confused about this k . Is it kp or kc?

Technically, it is neither, but it's closer to $K_{\mathrm{c}}$ than $K_{\mathrm{p}}$.


text:
How a molecule electron cloud interact with polarised light to show optical activity and why only asymetrical molecules only do so?
I want to know how exactly light electrical component or anything else interact with e- cloud

I think this begs the question what do you know about quantum mechanics or basics of spectroscopy.


text:
How to identify whether an electrophile will attack benzene ring when a deactivating group is attached to it?
I was asked to find out the major product formed when this molecule is subjected to nitration using $\ce{HNO3/H2SO4}$ mixture

Here $\ce{NO2+}$ will be the electrophile. Lets call the benzene ring which is directly attached to the carbonyl group as ring 1 (benzylic ring) and the other as ring 2 (phenolic ring). Now ring 2 is attached to an activating group as $\ce O$ will show +M and will increase the relative electron density at o-p positions. But the electrophile will preferably attack para position due to steric hindrance at ortho position.
Benzene ring 1 is attached to a carbon atom which in turn is connected to a more electronegative atom. So it is partially positively charged and will show -M effect thereby deactivating the ring. But it is also meta directing group. So will the electrophile attack meta position of ring 1?
The answer to this question says, ring 1 won't be attacked by the electrophile as it is attached to a deactivating group. The the molecule after nitration is

Then how to identify whether an electrophile will attack benzene ring when a deactivating group is attached to it?

It's not a matter of if, but of on what time scale will the reaction proceed. minutes, hours, geological eras.

I agree that the phenolic ring reacts first, but if it stops there depends on the conditions because it is possible to nitrate benzene rings with deactivating groups, after all dinitrobenzene is fairly easy to make https://www.docsity.com/en/preparation-of-dinitrobenzene-organic-chemistry-lab-manual/405022/


text:
Color change of aqueous ferric salt when reacted with phosphate
I have a question about an equilibrium system:
$$\ce{Fe^3+ (aq) + SCN-(aq) <=> [Fe(SCN)]^2+ (aq)}\\
\text{(yellow)} \quad \text{(colorless)}\quad \qquad\text{(red)}\qquad\quad$$
The question is about what happens to  $\ce{Fe^3+}$ when you add $\ce{HPO4^2-}$ to the solution above.
I'm not sure (that's why I'm asking), but when you add $\ce{HPO4^2-}$ to the solution it reacts with $\ce{Fe^{3+} (aq)}$ and forms $\ce{Fe2(HPO4)3}$ (I’m guessing):
$$\ce{2Fe^3+ (aq) + 3HPO4^2-(aq) -> Fe2(HPO4)3}\\
\ \ \ \ \text{(yellow)} \quad \text{(colorless)}\qquad \qquad\text{(colorless)}\qquad$$
My hypothesis is: When you add $\ce{HPO4^2-}$, there is a "loss" of  $\ce{Fe^3+}$ because $\ce{HPO4^2-}$ reacts with $\ce{Fe^3+}$. This also means that there is a chemical shift to the left (The solution produces more  $\ce{Fe^3+}$), and because $\ce{Fe2(HPO4)3}$ is colorless the solution also turns colorless.

Please COMMENT if you also want me to elaborate on something.

Well, that's still reposting, from new account this time. You know you can edit your posts to improve them? And improved old, post may get reopened. Check out how A.K, improved equations, earlier they looked as bad in old posts. https://chemistry.stackexchange.com/questions/102629/how-are-fe3-affected-when-you-add-hpo42?rq=1

And what do you mean by saying that my equations look bad? They are just fine?

Well, compare before and after edited with mathjax

I know how to use "Mahtjax" now, so what's the problem?

My first point was that one question shouldn't be asked multiple times. So in the future, if question gets on hold, don't ask it again, but edit it. This post looks mostly OK, though isn't perfectly clear. You ask what's the reaction or why product is colorless, while substrate isn't?

Okay I know that now, thanks. I ask both questions.


text:
Why are the calculated NMR values on the DFT & GIAO level of cyclopentane-1,3-dione so far off?
I'm trying to simulate the 13C NMR spectrum of cyclopentane-1,3-dione (PubChem CID: 77466; CAS 3859-41-4; ChemSpider ID: 69875; SDBS No: 15258; SMILES: C1CC(=O)CC1=O). There are three equivalence classes of carbons: The two directly attached to the oxygens (≈197 ppm, call this class 1), the one carbon between those two (≈105 ppm, call this class 2) and the "other two" (≈31.3 ppm, class 3).
I've been using Gaussian 16 via the guidance provided by Dean Tantillo's group at UC Davis, the chemical shift repository (CHESHIRE) in particular first performing molecular structure optimization at some level (e.g. B3LYP/6-31+G(d,p)) and then calculating shieldings (at mPW1PW91/6-311+G(2d,p) with solvent corrections, scrf=(solvent=chloroform, smd)). But I keep getting shielding values that are radically off.
In particular, regardless of the settings I arrive at isotropic shielding values for classes 1, 2, 3, of ≈-34,  ≈136, ≈144. Now of course observed chemical shifts and shieldings are often related by a linear scaling, per the Tantillo group's work (and others), which means that it's very odd that classes 2 & 3 here (the non-Oxygen-attached carbons) have nearly identical shielding values.
I am new to chemistry and DFT methods, so I worry I might be something wrong, but it might also be the case that this is just a "hard" molecule for DFT methods to get right. I wanted to know:

Are there any higher levels of theory that I should try / better methods for shielding calculation that would be worth assessing?
Is there any way of analyzing the output from Gaussian16 to get a hint that things might be going awry?
Is there something that just makes this a "hard" molecule to calculate? I get the impression that the contributions in liquid state from various rotamers, etc. is relatively minor as it seems quite rigid (although I have not quantified this).

Not an expert in this area of computation, but it might be worth seeing if your solvent model is sufficient. You may want to try including some explicit solvent molecules, with DFT hopefully this won't make the calculation too costly. https://pubs.acs.org/doi/10.1021/ct900498b

Could you please add the in-put you are using to run the calculation, it might make things easier to reproduce and play around. Also it appears you wanted to include a warning you observe in Gaussian, but it is missing.

Also related to this: When using Gaussian to calculate NMR, what's the default solvent and frequency?

I have also experimented with this just now and got the same values. I then used their formula to calculate the shift $$\delta=\frac{\text{intersection}-\sigma}{-\text{slope}}= \frac{186.5242-\sigma}{1.0533}$$ and got $\delta_\ce{C^1} = 214.8 [197.6]$, $\delta_\ce{C^2} = 48.6 [105.0]$,  and $\delta_\ce{C^3} = 40.3 [31.3]$. Not too bad, but $\ce{C^2}$ is still way off. I checked against DMSO, but there is not a lot deviation. Even the change in G16 vs. G09 should not make a significant change. Not sure why...

@Martin-マーチン I've found that in G09 the default electrostatic scaling factor of 1.1 for PCM is problematic in some cases and has led to poor results (worse agreement with experimental values) in my experience.  Going to alpha=1.2 fixed many of those issues.  Granted, I was looking at NMR chemical shifts of a transition metal, not main group.  I haven't used G16 yet to see if the problem was fixed or not.  (Also, in my own personal experience, PCM made my computed NMR results worse overall w.r.t. solvent-free NMR comps with a non-relativistic Hamiltonian).

@Eric J Try IGLOII basis set for your NMR computation.  Do a solvent-free (i.e. no PCM) NMR computation, one with DMSO, and one with DMSO and alpha=1.2.  I want to see those results.  I don't have any experience with mPW1PW91 but PBE0 (i.e. PBE1PBE) should work well.

@Martin-マーチン Using your formula, I got the same results with various computational protocols.  $\mathrm{C}^2$ is way off by about 50-60 ppm. https://i.imgur.com/Rp6QQBT.png

@LordStryker I think you should make that an answer. It seems like this molecule really is problematic, and your values at least state that they are consistently garbage.

Did you observe enol form in the nmr spectra?

Does anyone have a primary reference for the experimental values?

@LordStryker Biorad has spectrum in DMSO online https://spectrabase.com/spectrum/1hO3D6ynkiq

Brief Summary
The computational results given below indicate that there is something inherently problematic with this seemingly simple system.
Computational Methodology
Full optimizations and corresponding harmonic vibrational frequency computations of cyclopentane-1,3-dione was were at two levels of theory:

B3LYP/6-31+G(d,p)
MP2/aug-cc-pVTZ

Each optimized geometry was subjected to GIAO-NMR computations at the following levels of theory:

mPW1PW91/6-311+G(2d,p)
mPW1PW91/IGLO-II
PBE0/IGLO-II
mPW1PW91-SMD/6-311+G(2d,p)
mPW1PW91-SMD/IGLO-II
PBE0-SMD/IGLO-II

The NMR levels of theory were chosen specifically to look at basis set effects as well as implicit solvation (SMD) effects.  The solvent chosen was chloroform.
All computations were carried out using the Gaussian 09 software package.  All DFT computations employed an ultrafine integration grid.  The frozen-core approximation was invoked for the MP2 computations.
Results
Table 1 presents the absolute isotropic shielding constants ($\sigma\ce{C}$) as well as the computed chemical shifts ($\delta\ce{C}$; determined using Equation 1 given by Martin in the comments) with all values given in ppm.  The carbons are labeled in figure 1 for cyclopentane-1,3-dione.  Despite the level of theory used for the optimization  or NMR computation, the $\delta\mathrm{C}^2$ values are consistently off by ≈50 ppm with respect to the 'experimental' values obtained from equation 1.

Figure 1: Cyclopentane-1,3-dione with labeled carbons

Equation 1:
$$\delta=\frac{\text{intersection}-\sigma}{-\text{slope}}= \frac{186.5242-\sigma}{1.0533}$$
I am actually quite surprised by these results as I did not expect this system to be difficult to characterize with straightforward NMR computations.

The slope and the intersection are a bit different for the messages, see here for the recommendations and would have to be calibrated against the test/probe sets. That being said, it does not change that 2 & 3 are so very close in calculation, but not in experiment. It would probably be quite interesting to see how the methods perform for the H-shifts. Btw. ChemDraw predicts 207, 56, 39, and I think they use increments. Or https://www.nmrdb.org/13c/index.shtml: 198, 105, 31 (I don't think they calculated that).

@Martin-マーチン this is incredible, thank you so much!

@EricJ Don't thank me, LordStryker did the hard work. If this answer helps you, you can mark it as accepted.

I am not a computational chemist, but I agree with permeakra's comment on the question that tautomerism could be a significant factor affecting the experimental shifts here.


text:
Synthesis of 3-(2-oxocyclohexyl)propanenitrile
Does this seem correct?
The point is to draw the mechanisnm of which 3-(2-oxocyclohexyl)propanenitrile is synthesized from 1-(cyclohex-1-en-1-yl)pyrrolidine.

You might want to Google "enamine". The enamine of your second structure in the first line will have the double bond in the less substituted position.

The first step is incorrect:

You take electrically neutral cyanoethylene, but on your mechanism it is negatively charged. So you create negative charge of nothing on zeroth step of the mechanism. The same is for positive charge on enamine
You cannot pluck the proton from enamine in this way.

===
At the first step the electrons of π-orbital of enamine interact with the π*-orbital of acrylonitrile(Michael acceptor) resulting in formation of the bond between alpha-carbon of enamine and beta-carbon of acrylonitrile.
The transformation of iminium to carbonyl occurs via tetrahedral intermediate and not via formation of vinylic carbocation.


text:
What is a "tri-salt polymer" as described by this product that eliminates and prevents mold?
This answer mentions an anti-mold product called Concrobium Mold Control. The FAQ linked on their website says:

Q: I see Sodium Carbonate listed on the label. Is Concrobium Mold Control just Sodium Carbonate (washing soda) and water?
A: No. Sodium Carbonate is the listed “active ingredient” on the Concrobium Mold Control label, but in fact there are two other ingredients in the solution which, when combined with the sodium carbonate and the water, result in this patented tri-salt polymer that eliminates and prevents mold. (Sodium Carbonate and water alone are not effective against mold.) The solution contains no bleach, ammonia or VOCs. Our Material Safety Data Sheet is posted on our website.

The MSDS lists two particular ingredients:
Chemical Name            CAS #          % by Wt

Trisodium phosphate   7601-54-9           1-5
Sodium Carbonate       497-19-8           < 1

I'm guessing the third might be water.
Question: What might the term "tri-salt polymer" refer to in general, and if possible to answer, what might it be in this particular case?
I did find a patent in a google search but I'm not sure of it's relevance and I don't understand it.

CA2504014C, Non-toxic water soluble inorganic anti-microbial polymer and related methods

Beside the three-step deprotonation  of (ortho) phosphoric acid, $\ce{H3PO4}$, along
$\ce{H3PO4 <=> H2PO4^- + H+ <=> HPO4^{2-} + 2H+  <=> PO4^{3-} + 3H+}$
where each of the oxygen-bound hydrogen atoms may be substituted by a cation, phosphoric acid may dehydrate partially to yield polyphosphates.  Which seems, reading a related Canadian patent CA2504014 (A1) (deposit in English), the working principle here:

(loc. cit., p. 6; cations generalized)
The know-how includes tuning the proccessing as such that the composition of these blend polyphosphates is obtained reliably, since the complete dehydration simply would yield the waxy polyphosphoric acid used as acidic catalyst in the lab.

Names of elements aren't capitalised.

Thank you for your answer. The Wikipedia subsection Polyphosphoric acids says ""backbone" chain of these types of molecules consists of alternating P and O atoms covalently bonded together. Polyphosphoric acid molecules can have dozens of such phosphoric units bonded in a row." Wow!

@Mithoron Agreed.  Better to adhere to the rules of IUPAC's Red Book.

@uhoh This pattern of smaller ions joining each other may occur with anions of other oxoacids, too.  Beside the smaller dichromate anion, the silicates for example represent a multitude of different ways the formal $\ce{[SiO4]^{4-}}$ tetrahedra may be attached to each other (vertex, edge, face), too.

This is really interesting, thank you! I'm going to retreat to a good book or three to read further; I'd never associated silicates and phosphates (I'm not a chemist) but will now.  It's never too late to learn Chemistry.


text:
Insoluble salts and Gibbs free energy
Is it true that according to gibbs free energy equation, insoluble salts such as AgCl don't dissolve because the process would be too endothermic? The right side of the equation, TΔS, would have to be positive because entropy increases when salts dissolve, which means that ΔH would have to be high enough to make the equation positive. Is this correct?

G = H - TS. Let G = 0. It is known empirically that H is -127 and S is .0963, therefore T must be -1318.8 K. This means the reaction occurs at all temperatures, and therefore insoluble salts must not dissolve for some other reason. Correct me if I'm wrong, but that's my understanding of it.

@ChristopherMarley aren't those numbers the standard enthalpy of formation of AgCl from the elements in their standard states, and the standard entropy of AgCl? Not related to solubility?

It is incorrect to say that "entropy increases when salts dissolve" as a general truth. That is sometimes correct, but it can also happen that the solvent becomes more ordered in order to create a solvation shell, for example, which can overcome the gain in entropy of breaking up the salt lattice, so the $\Delta S$ is negative. Such salts are only soluble if the $\Delta H$ is negative and greater in magnitude than the positive $-T\Delta S$ term.

Yes, this is correct. Assuming that you are talking about solubility in water, it is discussed by JF Liebman, Structural Chemistry, 15, 165 (2004) which needs a subscription to read in full, but much of the key information is given on the first couple of pages which are available as a preview. The relevant thermodynamic quantities (in kJ mol$^{-1}$) for dissolution of one mole in water are given as a table, with AgCl compared with NaCl and KCl:

$$ \begin{array}{cccc}  & \Delta H & \Delta G & T\Delta S \\ \hline \text{NaCl} & +4 & -9 & +13 \\ \text{KCl} & +17 & -5 & +22 \\ \text{AgCl} & +45 & +37 & +8 \end{array} $$

The author got the original data from NBS Tables of Chemical Thermodynamic Properties.
All the entropy terms are favourable.
All the enthalpy terms are unfavourable, but the AgCl has a much more unfavourable enthalpy of solution than the alkali metal chlorides, and the
entropy cannot compensate for this. The Ag$^+$ ion is poorly hydrated in solution
compared with Na$^+$ and K$^+$.


text:
Bonding metal pieces using electroplating - what's the process called?
So I think the title all most says it all - given say two pieces of copper can we make solid 'weld' using electroplating chemistry? If so whats the accepted terminology for this process so I can do some literature research - also am I correct in my assumption that the join would be as strong as the raw metal?

Just to clarify I'm imagining holding both sides of the join in position with the cathode attached to both pieces and the join area submerged in the plating bath.

You could electroplate Pb/Sn on the copper, then reflow the solder in a furnace to join the pieces. // I don't think that you could get a good mechanical joint by just trying to electroplate the two pieces together in a solution.

How would electroplating penetrate between sheets?

There's a 1974 Sandia labs paper on "Joining by Electroplating", abstract and citation available here.  They recommend tapering the joint and plating a triangular void closed, which brings up the important point that you need to let fresh electrolyte in to reach the joint as it builds up material.  For example, trying to electroplate two flat disks face-to-face would likely result in most of the deposit forming a ring around the disk edges, sealing a layer of corrosive electrolyte between the disks rather than securely welding them together.
The terminology for this sort of technology has rather permeable boundaries, but includes:

Electroplating, normally aiming to deposit a mostly uniform surface coating over an existing substrate.  Common in a variety of consumer goods, often for corrosion protection.
Electroforming, depositing material over a removable mandrel.  Common in micromanufacturing.
Electrowinning, extracting or purifying material.  Common in mining.


text:
Comparing thermal stabilities of MgCl2 and BaCl2
I read that to compare thermal stabilities of ionic solids:
if the solid has a monoatomic anion, we compare thermal stability on the basis of lattice energy which is inversely proportional to size.
if the solid has a polyatomic anion, we compare thermal stability on the basis of ionic character which can we done using Fajans' Rule.
Could someone please confirm that these are correct?
I have a doubt because I also read from somewhere else that MgCl2 is less thermally stable than BaCl2,
which contradicts the rule as Cl- is monoatomic, Ba2+ is larger in size, so BaCl2 has lesser lattice energy so it should be less thermally stable.
So what is the correct answer?

I also came across this order(without any proper sequence) for thermal stability of alkali metal chlorides: LiCl>KCl>NaCl>CsCl. What could be the possible reason?

In this case, I think lattice energy should be the primary consideration. Magnesium chloride has the most exothermic LE while barium chloride has the least exothermic one. Thus, MgCl2 with the most exothermic LE, is the most thermally stable.

Isn't the lattice energy of MgCl2 more than that of BaCl2 as Mg2+ is smaller than Ba2+. I googled the values I got MgCl2 is +2326 KJ/mol and BaCl2 is +2056KJ/mol.

Lattice energies are by definition exothermic. Where did you get your values from?

There are two definitions of lattice energy. One of formation other of disassociation(just opposite in sign but magnitude remains same).

Either way, I too think that MgCl2 should be more thermally stable but in my book(question) it is given otherwise.


text:
Stability order of Cr2+ ,Mn2+ , Fe2+ ,Co2+
Stability order is given as $\ce{Mn^2+ > Fe^2+ > Cr^2+> Co^2+}$
Of course I know that $\ce{Mn^2+}$ is the most stable form as it has half filled degenerate orbitals. But my query is regarding the stability of $\ce{Fe^2+}$ and $\ce{Cr^2+}$.
It is given in my book that $\ce{Fe^2+}$ has greater stability than $\ce{Cr^2+}$ through exchange energy. I want to know how exchange energy stabilizes $\ce{Fe^2+}$ more than $\ce{Cr^2+}$ .

What do you mean by stability here? Stability WRT reduction? Stability WRT oxidation? Stability of the complexes these ions form? Or something else?

@IanBush  it is a question from the medical entrance examination in India.

Well as given it is more or less meaningless

@IanBush it is not perfectly given in what terms the stability order is asked but they just asked the stability of M2+ ions. No specificity regarding this.

Formation of ion takes place by removing electron from outermost shell.
Electronic configuration for Fe2+ is [Ar]3d6 4s0
Electronic configuration of Cr2+ is [Ar]3d4 4s0
Cr2+ is a stronger reducing agent that Fe2+ in aqueous solution(d3 configuration more stabls than d5 in aqueous solution).
So Cr2+ has higher tendency on losing electron compared to Fe2+ so Fe2+ is more stable than Cr2+.
Mn2+ has most stable and Co2+ is least stable(least exchange energy due to only 3 unpaired electrons).

You have read from here http://www.answers.com/Q/Why_is_d3_orbital_more_stable_in_water_than_d5 why d3 is more stable than d5 in water.


text:
Why formic acid is stronger than benzoic acid? Don't explain through pKa values
Acidity in organic compunds, specially carboxylic acids.

One possible explanation:  solvation.  Formate ion is compact with essentially no hydrocarbon component, and so is more strongly solvated by water solvent than other carboxylate ions.  One could say the same about neutral formic acid versus other carboxylic acids, but solvation effects are stronger for ions than for neutral molecules.  So the favorable solvation for formic acid/formate tends to promote formation of the ion making the acid stronger.

Good explanation, but doesn't this only hold in the cases where the acid is aqueous?


text:
Diluting methanol in water: is it exothermic?
Can someone explain in simple terms why does temperature increase when mixing water with methanol? I do not have a strong background in chemistry, but I usually have to mix both liquids (sometimes also water + ethanol) (as part of my work in a biology lab) and have always wandered about the reason behind the evident increase of temperature upon mixing. I guess that the break/formation of H-bonds is involved but I haven't found a clear ("easy-to-understand") explanation in online searches.

Do you have any existing thoughts on this matter? It may help us explain the science if we understand where your thought process has taken you so far.

I do not have a strong background in chemistry, but I usually have to mix both liquids (sometimes also water + ethanol) (as part of my work in a biology lab) and have always wandered about the reason behind the evident increase of temperature upon mixing. I guess that the break/formation of H-bonds is involved but I haven't found a clear ("easy-to-understand") explanation in online searches.

When the methanol and water are separate, they both exhibit hydrogen bonding with themselves. When the water and methanol are mixed together, some of the existing hydrogen bonding (water-water or methanol-methanol) is disturbed and now there is hydrogen bonding between water and methanol. This new hydrogen bonding pair (water-methanol) releases energy when it forms as it is more favorable than maintaining only water-water and methanol-methanol hydrogen bonding; the temperature increase you observe is the result of the released energy warming the solution.


text:
Carbonyl bond vs alkene bond
Why are carbonyl bonds stronger than those of alkenes? I've read several times that carbonyl bonds are shorter and that you need to apply a higher amount of energy to break them. Why is that so?
Is it due to the electronegativity of oxygen, which pulls the bonding electrons near it, thus attracting the carbon as well?

Yes you are correct the electronegativity difference between C and O imparts partial ionic character to the bond. Consider that there is a partial negative charge on O and a partial positive charge on C and these charges attract each other.

I see. So the same can be said regarding the triple bond between carbon and nitrogen when compared to the alkyne bond?

Yes you are right. These charge separations may be slight in some cases but they do have significant consequences on bond strength.

Why are carbonyl bonds stronger than those of alkenes?

I will use homolytic bond dissociation energies here to quantify bond strength. Other ways to quantify it exist.

I've read several times that carbonyl bonds are shorter and that you need to apply a higher amount of energy to break them. Why is that so? Is it due to the electronegativity of oxygen, which pulls the bonding electrons near it, thus attracting the carbon as well?

There is a pattern of bonds between different atoms being stronger than bonds between identical atoms. If you take a table of bond dissociation energies to calculate the differences
$$\ce{2 BDE(X-Y) - BDE(X-X) - BDE(Y-Y)},$$
where BDE stands for bond dissociation energy in kJ/mol, you get the following picture for single bonds:

Single
H
C
N
O

C
44

N
183
101

O
344
224
93

F
539
469
248
79

So for example, two C-H bonds are approximately 44 kJ/mol stronger than a C-C bond and a H-H bond.
For double bonds, you get (taking the bond in dioxygen as a double bond):

Double
C
N

N
210

O
498
298

Bond dissociation energies are from http://ch301.cm.utexas.edu/thermo/#thermochemistry/enthalpy-bonds.html.
There is a trend as you turn two homo-atomic bonds into two "mixed" heteroatomic bonds: The higher the electronegativity difference, the higher the gain in energy. This is a rough calculation because it is based on average bond dissociation energies. To make a more rigorous case, you would have to go to specific examples, e.g. comparing a given amount of ethane and hydrazine with twice the amount amino methane:
$$\ce{H3C-CH3 + H2N-NH2 -> 2 H3C-NH2}$$
This only shows the correlation, not the cause. In the most direct way, the reason why two C=O double bond is more stable than a C=C bond and a "O=O" bond is because that's what we get when we calculate an approximation of the solution of the Schrödinger equation. And we can test it experimentally.
Trying to explain it with partial ionic interactions, partial charges, or electronegativity is a slippery slope because these are all constructs we have come up, rather than measurable quantities.

It's stronger because of a dipole moment. oxygen is more electronegative than carbon, so there is a stronger attraction between C and O than a C=C bond.

A more polar bond is not necessarily stronger than a less polar one.

@Martin I agree with your point. None of the answers here seem to establish the necessary causal relationship. Would you care to answer the question, or point me towards a helpful resource?

@Ray I don't think I could answer this in depth here. Get yourself a book on quantum chemistry and explore for yourself. It'll be more rewarding. I liked Jensen's book quite well: https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry/37304#37304

A very simplistic approach is that electrons in MOs are attracted to both nuclei and the added protons in the O atom with similar shielding lowers the energy of the bond a bit. This can be interpreted as increased bond polarity or additional resonance structures, but they are an invention also.

Basically there can be two reasons
•electronegativity of oxygen is more than carbon which creates partial negative charge on oxygen and partial positive on carbon and they attract each other resulting in a shorter bond length
• i consider the second reason more dominant- as oxygen is in same period as that of carbon and has greater atomic number too that is it is already having shorter size than carbon this means we will have C-C double bond longer than C-O double bond which will increse the bond strengh as bond length is shorter.....similarly the case of C-C triple bond and  C-N triple bond

Shorter bond length does not necessarily allow the conclusion that the bond is stronger.

authors: Αντώνιος Κελεσίδης


text:
Inorganic IUPAC for Compounds with bridged ligands?

Can we use  bis(tetraammine) instead of octaammine is the 2nd example ?
As in the 1st example they used  bis(pentaammine)  and also in the 3 one the used bis(carbonylcyclopentadienyliron)

Btw, IUPAC Red Book 2005 mentions probably the same complex ion 1. [{Cr(NH₃)₅}₂(μ-OH)]⁵⁺, but the name contains complex net charge instead of metal oxidation state: μ-hydroxido-bis(pentaamminechromium)(5+). Most importantly, the parentheses seem to be incorrect in the question.

@mykhal  Yes I think the parentheses is incorrect It should be like the 3rd example , but still I am confused .

Here is an example of a similar ligand and it can be named in three ways.
All you have to do is to put hydroxido/hydroxo in place of superoxido.

So yes, bridged ligands can be named in a few ways, combining any amount of recurring atoms separately, but most common used would be (if the LHS and RHS are the same) μ-amido-μ-hydroxidobis(tetraammine)dicobalt.
Reference

Crystal structure of .mu.-amido-.mu.-superoxo-bis[tetraamminecobalt(III)] tetranitrate, $\ce{[(NH3)4Co(NH2)(O2)Co(NH3)4](NO3)4}$ by
Gary G. Christoph, Richard E. Marsh, William P. Schaefer, Inorg. Chem. 1969, 8, 2, 291-297, DOI: https://doi.org/10.1021/ic50072a022


text:
Is there also a leveling effect for weak acids in water?
The leveling effect says that any acid stronger than the hydronium ion ($\ce{H3O+}$) will completely dissociate in water. Hence no differentiation of acid strength is possible between eg HBr and HCl in water.
I was wondering why no one ever talks about the leveling effect for acids weaker than water. Isn't it so that these acids would not dissociate at all and hence no differentiation is possible on this side either?
Also, what is the reasoning behind the fact that acids stronger than the hydronium ion will completely dissociate in water? Why can't it be an equilibrium (say if it's only slightly stronger than hydronium)?

"Dissociation" is somewhat incorrect term. When you dissolve, say, gaseous HBr in water, the water molecule plucks proton from HBr. So dissociation is, in fact, acid-base reaction. That means, if you dissolve an acid whose acidity is 2 units of pKa(or more) lower, than the acidity of hydronium, it will be, in practice, completely "dissociated".

Ok, I appreciate the comment, dissociate is indeed the wrong term. However, as you say, complete 'reaction' will only occur if  the pKa differs more than 2 units, so the statement that all acids stronger than H3O+ can be considered to react completely is not really correct? Also what about the weaker acids than H2O?

Your point about strong acid is entirely valid. Of course there's equilibrium, but in diluted acids it can be approximated as complete dissociation. On the other hand this talk about weak, doesn't seem to make sense.

For weak acids you have to think also of the conjugate base. So $\ce{HA}$ might not be a great proton donor, but $\ce{A-}$ can be a great proton acceptor.

First of all, I personally don't get why people are so eager to downvote questions of people who are genuinely interested, without argumentation too. If an acid is weaker than water it's pKa is larger than 14 and by default the pKb of the conjugate base smaller than 0. This means that virtually no acid will be transformed into it's conjugate base. And this is the same for all acids weaker than water, it's like the levelling for strong acids but to the other side of the equilibrium.

To answer your first question, weak acids don't get levelled in water. Each solvent has their acid-base equilibrium window, and water has one that is between 0-14, meaning that any solvent with a pKa within that range can be distinguished in water. There are other solvents with a ore narrow acid-base discrimination Window, such as methanol can acid, with a range between -10 to about 3, which will be unable to discriminate between weak acids. The reasoning behind the acid-base discrimination window is a bit long winded (and maybe a little confusing), so I will try to break it down as much as I can.
Say you have an acid in water, then the pKa + pKb = pKw
This is true for any acid or base in water. If pKa<0 or pKb<0 then then most of the compound will be dissociated in water, as an acid or base, respectively. This is undesirable, especially if the pKa/pKb<<0. As you have noted in your comments, when the pKa of a compound is less than 2 units compared to water, then in water only 1% of it will remain in the undissociated form. If pKa>pKw, you will obtain a negative pKb which will behave more like a base in water. It follows that compounds with a 0< pKa< pKw  will fulfill the condition where both pKa and pKb> 0. Now guess what pKw is? 14. Hopefully this might explain to you why water is able to distinguish between weak acids with a pKa between 0 to 14.

Ok I think now I totally got it. I know it's a matter of semantics, and not that important. But including pKa's and pKb's in my reasoning: If the pKa is smaller than 14, say at least 2 units below as well. Then in this case the weak acid would only dissociate around 1% right? And so below this pKa weak acids can not be distinguished as you said. So that's what I called a leveling effect for weak acids in my original question.


text:
Can we interpret the "extent of reaction" as the number of reactions that happened?
Consider this reaction:
$$\ce{\alpha\ A + \beta\ B -> \omega\ C + \delta\ D}$$
Where $\ce A$ and $\ce B$ are the reactants, $\ce C$ and $\ce D$ are the products, and $ \alpha, \beta, \omega, \delta$ their respective stoichiometric constants
We can then define the extent of reaction at instant $t$ as: $\xi (t)$.
My question is: can we interpret the extent of reaction as the number of reactions that happened at an instant $t$ considering that say, for reactant $\ce A$ at instant $t$: $$n_{t}(\ce A) = n_{i}(\ce A) - \alpha\times \xi(t)$$

If I understand this correctly you want to measure $\ce A$ to infer the extent of reaction. I think you should have started with mathmatically defining the extend of reaction:
$$\xi(t) = \left[\frac{n_i(\ce A) -n_t(\ce A)}{n_i(\ce A)}\right]\tag 1$$
Where $n_t$ is the concentration at a given time $t$. Here $n_i(\ce A) -n_t(\ce A)$ gives the consumed concentration of $\ce A$ after time $t$ which is normalized by the initial concentration to yield  an extent of conversion.

Assume $\mathrm{2\ a}$: If the stoichometric constants are normalized $({\alpha} + {\beta}= 1)$ to be taken as the initial concentrations then we can substitute: $\alpha = n_i(\ce A)$ for example $$\ce{0.33A + 0.67B -> 0.5C + 0.5D}$$
$$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right]\tag {2 a}$$
  otherwise:
  $$\left(\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1\right) \implies \frac{\alpha}{\alpha + \beta} = n_i(\ce A)\tag {2 b}$$

Which we can rearrange to :
$$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right] \implies \alpha-n_t(\ce A) = \alpha\ \xi(t) \tag{3} $$
Subtracting $\alpha$ and then negating gives:
$$n_t(\ce A) = \alpha - \alpha\ \xi(t) = n_i(\ce A) - \alpha\ \xi(t) \tag{4 a}$$
Which would make you correct, or in the case of assumption $\mathrm{2\ b}$:
$$n_t(\ce A) = n_i(\ce A) - \frac{\alpha}{\alpha + \beta}\ \xi(t) \tag{4 b}$$
Which would make you almost correct.

Thank you very much. And sorry for the late answer <3


text:
What about Miller Indices of planes passing through origin?
For Miller Indices of planes if we encounter a case where the plane passes through the origin then I have heard that the only remedy can be arrived by shifting the origin to a different point.However I do not understand whether or not I am supposed to change the index of the other planes whose origin was not passing through the previous origin.Please give me a bit of justification for the case.

Miller indices are not a property of planes at all. Rather, it is a property of systems of parallel planes.

@IvanNeretin Does that mean that I have to change the indexing of the planes that I have already named with a separate origin?

Or, put slightly differently, all lattice points are the same, so moving the origin to another point changes nothing.

What is meant by symmetry of planes/directions in Miller Indices?For cubic it says that the principal directions are symmetric while for tetragonal it excludes the one perpendicular to the basal plane...Something similar is for the planes as well.Please describe this as vividly as possible.

Symmetry is another thing altogether. You'd better ask a new question to that effect.

Since a comment does not allow the inclusion of a picture, perhaps this way is helpful:
Have a look at the following picture, derived from an illustration about the Miller Indices on wikipedia.  It depicts a simplified situation of a unit cell which may occur if your crystal is of the cubic lattice system, i.e., the three vectors defining the unit cell are orthogonal to each other, and all are of same length. In the drawing, the author choosed $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ for the unit cell's dimensions.

Not only the plane closest to the coordinate system's origin drawn is a valid Miller plane, equally so the second, farther one drawn.  For each set of Miller indices $(hkl)$, the corresponding Miller planes are parallel to each other.  The distance between these planes depends both on the indices, as well as on the symmetry of the crystal.
Since a crystal consists of many unit cells adjacent to each other, it is perfectly reasonable to draw Miller planes either in "positive direction" along vectors $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ -- as shown here -- or in opposite direction along $-\vec{a_1}$, $-\vec{a_2}$, and $-\vec{a_3}$ with the same set of $(hkl)$.


text:
POI is a bent molecule similar to water. How are the two molecules different it terms of bond angles and what causes this?
phosphorus oxyiodide ($\ce{POI}$) has the VSPER AXE formula of $\ce{AX2E}$ while water, $\ce{H2O}$ has the AXE code of $\ce{AX2E2}$. According to my work sheet $\ce{POI}$ should have a $120^\circ$ bond angle while $\ce{H2O}$ should have a bond angle of $104.5^\circ$.
Does this mean that due to the extra lone pair it creates less resistance resulting in a greater angle? Also, does it having three different elements in the equation do anything?

You mean the molecule O=P–I? Because to me, italicised $POI$ sounds like... a triangle, in geometry.

The worksheet says $POI$ multiple times. In those exact words. I can't seem to find it on the internet. Either there is a mistake (which was repeated multiple times) or it is made up to prove a point?

Well, I'm pretty sure the molecule doesn't exist. But it does fit into the VSEPR scheme of $\mathrm{AX_2E}$, which is why I suspected it was that. And if your only goal is to learn how to apply VSEPR theory, it doesn't particularly matter whether the compound exists or not; you just need to be aware of what you are saying, i.e. "if this molecule existed, VSEPR would predict a bond angle of x degrees" and not "this molecule has a bond angle of x degrees".

Does this mean that:

Due to the extra lone pair $\color{red}{\text{it}}$ $\color{limegreen}{\text{creates less resistance}}$ resulting in $\color{blue}{\text{a greater angle}}$?

More like:
Due to the extra lone pair $\color{red}{\text{in } \ce{H2O} \text{ the pendant atoms}}$ $\color{limegreen}{\text{experience more repulsion}}$ resulting in $\color{blue}{\text{a smaller angle}}$ $\color{magenta}{\text{compared to }\ce{POI}}$?

Also, does it having three different elements in the equation do anything?

No, I think the three different elements in the question were there to evaluate if you were:

estute to the fact that oxygen forms a double bond thus
phosphorous has 1 unbonded pair of electrons (not 2) and
if the double bond would hang you up in obtaining an answer.


text:
Trouble with Converting from Structural Formulas into Newman Projections
I have the following question from my Organic Chemistry textbook:
I'm asked to convert the following line diagram into a Newman Projection.

The arrow pointing to the carbon signifies "looking down the carbon".
My logic was the following:
At the carbon that is being pointed at by the arrow, there are 3 bonds to other carbons, all of which are in the same plane due to the absence of any 'dashed lines' or 'solid wedges' that would indicate otherwise.
I then concluded that the geometry of that carbon is trigonal planar and that it has a lone pair, since all the bonds to other carbons are in the same plane.
However, drawing a Newman Projection for an atom with trigonal planar geometry proves to be impossible.
I'm not sure how to go about converting this line diagram into a Newman projection.
How can I do so?
Thank you.

You don't have a realistic structural formula to begin with. It should have had some wedge/dash information, but it is lost.

Your assumption about the geometry of the arrowed carbon is wrong: it isn't trigonal planar. In structure pictures like this you need to add missing hydrogens to give each carbon its normal compliment.

this is how you can convert the str.s into each other. It should have had some wedge/dash or sawhorse information... for the conversion.


text:
Boiling temperature of salt water at low pressure
With a hydro aspirator with re-circulating water (see here), the maximum vacuum you can get is set by the water vapor pressure given the temperature.
Can you lower the vapor pressure measurably by adding salt to the water (I get −700 mmHg, at 25 °C, with regular tap water already)?

Note that “$-700\ \mathrm{mmHg}$” is a gauge pressure, which is unfortunately still widespread in engineering but not useful for thermodynamic calculations since it depends on the barometric reference pressure. The equilibrium pressure of water at $T=25\ \mathrm{^\circ C}$ is $p=23.776\ \mathrm{mmHg}$. For example, at a reference pressure of $p_0=760\ \mathrm{mmHg}$, the corresponding gauge pressure would be $p_\mathrm g=-736.22\ \mathrm{mmHg}$. Also note that the use of the unit $\mathrm{mmHg}$ is deprecated.

According to this 1927 paper, "THE VAPOR PRESSURE OF THE SATURATED AQUEOUS SOLUTIONS OF CERTAIN SALTS" at 25 °C, 6 mol per kg salt solution gives a pressure of 18.5 mmHg instead of 24.5 mmHg for pure water.
So yes there should be a measurable difference (unless salt destroys the diaphragm pump).
Edit: This recent paper "Water vapour pressure above saturated salt solutions at low temperatures", gives the same results


text:
Stabilization of hydrogen peroxide
It was given in a chemistry book that things like orthophosporic acid, acetanilide and sodium stannate stabilize hydrogen peroxide, while carbon, $\ce{MnO2}$ and $\ce{Pt}$ help decomposing it.
Why is this so?

Part of the explanation is that orthophosphate forms complexes -or even precipitate, traces of heavy metal ions present in the solutions. Many of these ions, like manganese and platin, will act as catalysts for the degradation of hydrogenperoxide. The phosphonic acid HEDP (Etidronic acid) is another example of a stabilizer.

Can u give complete equations for all of what u mentioned ?

I´d rather not. The chemistry of complexes in not my specialty so my answers would probabely either be wrong or to general to be of any use.


text:
Equalization of pressure in heat exchanger
I was solving numericals on Kinetic Theory of Gases when I came across this question

Two closed vessel of equal volume contain air at 105 kPa, 300 K and are connected through narrow tube. If one of the vessel is now maintained at 300 K and other at 400 K what will be the pressure in the vessel?

I don't want the solution of this question but I want to know why the pressure will be same in both the vessel when one is maintained 300 K and other 400 K. The molecules in  the vessel at 400 K will have greater kinetic energy and will exert greater pressure on the walls than the molecules of the other vessel.
Now one will say that some of the molecules of vessel at higher temperature will move to the other vessel so that the pressure remains same in both the vessels, but why is it so. Why can't the pressure in both the vessels be different?

The pressure can't be different precisely because the vessels are connected, and some molecules will move to the other vessel to equalize it. That's what pressure is all about.

That's what I m asking why will some molecules move to another vessel to equalise it. As motion of molecules is random, it might happen that some molecules move to vessel at higher temperature.

Of course they will. You are right, the motion of molecules is pretty random; also, there is quite a lot of them. How can we tell which way the net flow will be? We can't, unless we are able to find a measure. Luckily, some guys did that for us. The measure is called pressure.

@IvanNeretin pressure is definitely not a measure of flow. It's a measure of force per unit area.

Macroscopics
Imagine the tube is initially closed. Then one vessel is heated up. As $V$ and $m$ are constant:
$$p=k\,T$$
So if $T$ is increased, $p$ is increased. As $p$ is increased, there will be a net flow of mass, until they are balanced. That's what we observe. Now, we can think of it microscopically.
Microscopics
Initially, there are the same number of particles in each vessel. When heating up there is an increase in velocity. So, in the high-temperature vessel the same number of molecules are with higher velocity. Imagine a plane in the middle of the narrow tube, the number of collisions per time, because of velocity, will be higher in $400\ \mathrm K \to 300\ \mathrm K$ direction. It indicates a net flow of mass. This flow of mass will equal pressures on both sides.

I can't follow, so what if the collision rate is greater in the vessel at higher temperature? I want to know why will the pressure will be same in both the vessels

@santimirandarp Your second to last sentence is very, very misleading. It seems to draw an equivalence between the gas flow and the velocity of the molecules - this is very much not the case though! You should really be clearer as to how the greater velocity of the molecules exerts a greater force, which causes gas flow.

@santimirandarp Why does a higher number of collisions increase the flow?

This experiment divides the container so that pressure (a "flow" of gas molecules against the walls) can be equalized thru the narrow tube but net flow of heat from hot molecules passing thru the narrow tube is seriously impeded (essentially does not occur because each section is maintained at 300 K or 400 K).
This non-equilibrium condition is held static by the external devices maintaining the set temperatures and the narrow tube which connects the  pressures in the two "systems" but disconnects the heat flow.
To see that the pressures will eventually be the same, imagine connecting the two vessels at 300 K by the narrow tube, then quickly heating one to 400 K. The pressure will increase in the heated tube, and depending on the narrowness of the tube, will decrease as gas flows into the cooler tube (where the pressure will increase because there will be more molecules there, even tho still at 300 K). After some time, the pressures will equalize.

I'm going to answer this question in two parts: First, why will some of the molecules of the vessel at higher temperature move to the other vessel?
Pressure is simply the force per unit area that the gas exerts. When a gas is sitting in a vessel, it exerts a force on the walls - and the walls in turn exert an equal an opposite force. Thus, nothing moves (on a macroscopic level) in the system.
If you open a tube to another vessel, then the gas in each vessel will be exerting a force on the gas in the tube. When both vessels are at equal temperatures and pressures, they exert equal and opposite forces, and the gas remains static (on a macroscopic level), and there is no net flow.
Heating up one vessel would increase the pressure due to the ideal gas law:
$$pV=Nk_\mathrm BT.$$
Since the higher temperature vessel will have a higher pressure, it will exert more force on the gas in the tube, and the gas has to move away from it.
Secondly, why can't the pressure in both the vessels be different? The answer is that the system is closed, and the number of molecules is conserved - molecules cannot enter from outside the two vessels. Thus, molecules can't move from the hotter vessel to the cooler vessel forever - the hotter vessel would run out of molecules.
The critical part of the ideal gas law to keep in mind is the $N$, the number of molecules. In the hotter vessel, molecules will travel out, and $N$ will be reduced, as long as the pressure is greater than the cooler vessel. Since volume and temperature are constant, pressure must drop until the pressures are equal.


text:
Why are diagonal arrows disallowed in Jablonski diagrams?
My professor mentioned in one of our lectures that diagonal arrows in a Jablonski diagram are disallowed. As far as I know, horizontal arrows signify conservation of energy, while vertical arrows signify conservation of momentum. Is this true/connected to the issue?

Since nobody has responded to this yet, I'll offer a short answer.
In a Jablonski diagram, the electronic states are grouped together horizontally by spin multiplicity $2S+1$. So, in the simplest example, you see a column of singlet states $S=0$ and a column of triplet states $S=1$. (Quite often, the vibrational levels are also illustrated schematically). Diagonal lines, then, represent transitions between states of different multiplicity.
These are more usually termed "forbidden transitions" rather than "disallowed". An example is phosphorescence, the decay from an excited triplet to a lower-lying singlet, most commonly the ground state. There is a general selection rule, $\Delta S=0$, which in principle makes the transition moment vanish. However, the rule can be violated: one possible reason is that $S$ is not a perfectly good quantum number due to spin-orbit coupling. Typically, the consequence is that the transition occurs very infrequently (which is why phosphorescence occurs slowly, and weakly).
So, it is not "disallowed" to put diagonal lines on a Jablonski diagram,
but in most cases they correspond to weak, "forbidden", transitions.


text:
Advice on forming a 10% solution of Maltol
I’m having great trouble forming a 10% solution of Maltol. Specifically, I’m trying to dissolve 3 grams of Maltol in 27 grams of solvent.
I’ve tried everything from Propylene Glycol, to Deionized Water, to Ethanol, including some mixtures of these few solvents, but none of them can work to fully dissolve all the Maltol.
The best one so far appears to be hot Deionized Water, which appears to dissolve the most Maltol. Adding a small amount of Ethanol helps dissolve the Maltol further, but not completely.
Here I’ve attached a photo of the result. The entire mixture was immersed in a hot water bath. As you can see, the Maltol appears to form fluffy clumps that float to the surface, and are really stubborn to dissolve.
If anyone could advise me on how to achieve a complete dissolution of the Maltol, that would be very much appreciated!
Thank you!

Looks like you're trying to produce too concentrated a solution.  Have you tried Chloroform/MeOH?  One gram of maltol dissolves in 82 mL water, 80 mL glycerin, 21 mL alcohol, 28 mL propylene glycol; Freely soluble in hot water, chloroform; sparingly soluble in benzene, ether, petroleum ether; soluble in alkali hydroxides giving yellow solutions
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 1061

Maltol (3-Hydroxy-2-methyl-4-pyrone) appears to be most soluble in ethanol.
https://pubchem.ncbi.nlm.nih.gov/compound/maltol#section=Melting-Point gives solubility as
1 to 10 mg/mL at 72° F in water, 10900 mg/L at 15 deg C (about 1%); one gram dissolves in 82 mL water, 80 mL glycerin, 21 mL alcohol, 28 mL propylene glycol; Freely soluble in hot water, chloroform; sparingly soluble in benzene, ether, petroleum ether; soluble in alkali hydroxides giving yellow solutions
Your fluffy floating clumps suggest something like water, which would reduce the solubility of maltol and could clump it up. or maybe there is some other impurity that complicates things.
You could try is to make a 5% solution in warm dry ethanol (apparently no problem...) and then evaporate off excess solvent. If you can get 10% when hot, but the solution drops crystals on cooling, you will have a saturated solution. But maybe it won't precipitate...

Thanks! I guess I just have to accept that a 10% strength solution is not possible for Maltol and start looking at weaker solutions...

Only just come across this.
This (maltol) is a common material used in flavourings.
As a Flavourist, I am very aware the maltol has a maximum solubility in propylene glycol and ethanol of about 2.5-3.0%. Higher levels may be achieved if trying to dissolve in something like lactic acid.
Sometimes if using heat to dissolve the maltol you may think you can dissolve more than 2.5% however when the solution cools down, the excess maltol will crystalise out.
My advice would be to use a weaker solution, as mentioned above 2.5% is a good level.


text:
Calorific value of a gas in combustion
The Wikipedia article on heat of combustion says:

The calorific value [...] may be expressed with the quantities:

energy/mole of fuel
energy/mass of fuel
energy/volume of the fuel

Is the energy obtained by the combustion mainly proportional to the mass of the gas, being then values in $\pu{J/mol}$ and $\pu{J/kg}$ a characteristic of the gas composition and basically independent of the pressure and temperature?
That is, for a specific gas, values in $\pu{J/mol}$ and $\pu{J/kg}$ will be mainly independent of the pressure/temperature of the gas in combustion, while value in $\pu{J/m^3}$ will be strongly dependent due to the relation between volume, pressure and mass.
In the latter case, given the heating value, $f_0$ (in $\pu{J/m^3}$) for a gas at $\pu{°C}$ and pressure $p_0 = \pu{101 kPa} = \pu{1 atm}$, we could (?) find new heating value $f$ (in $\pu{J/m^3}$) at some other temperature $t$ (in $\pu{°C}$) and pressure $p$ (in $\pu{Pa}$) applying the expression:
$$f = f' \frac{p}{p_0} \frac{273}{273+t}$$

Are you familiar with Hess' Law?

@ChesterMiller: a few, but I do not see now the relation

The heat of combustion changes with temperature because the specific heat of the products is not equal to the specific heat of the reactants.  Google Hess' Law and see how it is applied to chemical reactions.  So, yes, the heat of reaction does vary with temperature.

“Energy/mole of fuel” is not a quantity; it’s a mixture of quantities and units. The correct quantity would be “energy per amount of fuel” or $E/n$, which could be expressed in J/mol.

Calorific values are usually quoted under standard conditions making the distinctions a matter of preference and convenience
The apparently different units used in quoting heat of combustion are not as different as they appear. We need to know heat of combustion for many reasons (eg how to bill domestic customers for the correct amount of heating or cooking their gas supply delivers). But the units used are all equivalent once we define a standard set of conditions when measuring them.
For example, a mole of pure methane at standard temperature and pressure (STP) occupies 22.4 L and will weigh about 16g and will burn in air releasing about 890kJ of energy. This means that, if we stick to the same conditions (STP in this case) we can interconvert the energy/mol, energy/mass and energy/volume with a simple calculation.
Which one we choose to use is a matter of convenience. Domestic gas meters might choose the energy/volume as they measure volume flow at a standard pressure and temperature. Liquefied gas ships may choose to measure by mass as the methane is mostly liquefied and the mass may be easier to measure than the volume, temperature or pressure (but, knowing the mass, we can work out the volume at STP to work out how much domestic gas can be produced). If a chemist is working on some reaction scheme, then they will probably prefer molar metrics as they make balancing the equations easier.
But the essential point is that a given amount of substance always has the same heat of combustion. How you measure the amount is a matter of convenience.


text:
Effective attenuation coefficient for polychromatic light
The Beer-Lambert law gives the optical intensity of collimated light as a function of depth $z$ as:
$$I(z)=I_{0}\mathrm e^{-\gamma z},$$
where $\gamma = \alpha + \beta$ is the wavelength-dependent attenuation coefficient, with $\alpha$ and $\beta$ being the absorption and scattering coefficients respectively.
Suppose two completely different wavelengths $\lambda_1$ and $\lambda_2$ are present in the light beam. Is it possible to define the "effective attenuation coefficient" of the material $\overline{\gamma}$ for two or more wavelengths as an addition of the different attenuation coefficients?
That is,
$$\overline{\gamma}=p_{\lambda_1}\gamma_{\lambda_1}+p_{\lambda_2}\gamma_{\lambda_2},$$
where $p$ is the fraction of the beam which is of a given wavelength (indicated by the subscript). The mathematics seems to work out, but I have never seen this used in literature before. Any explanation is greatly appreciated.

What's to explain? You invented this value, now find a use for it.

I don't see what is unclear about this question. How do I make two attenuation coefficients into one?

@IvanNeretin This could be very useful in the area of material processing by laser beams. Some lasers can have two wavelengths present simultaneously (for example, when you frequency-double a laser).

The intensity transmitted at  wavelength $a$ where the extinction coefficient is $\epsilon_a$ is $I_a=I_{0a}\exp(-\epsilon_a L C)$ with path length $L$ and concentration $C$, and similarly for a second wavelength $b$. If, for clarity, we let $x=\epsilon l C $ then then $I_a=I_{0a}\exp(-x_a)$ and $I_b=I_{0a}\exp(-x_b)$. The total transmitted light is $I_t=I_a+I_b$ and the total initial amount $I_{0t}=I_{0a}+I_{0b}$. Thus the total transmittance ratio is
$$ \frac{I_t}{I_{0t}}=\frac{I_{0a}e^{-x_a}+I_{0b}e^{-x_b}}{I_{0a}+I_{0b}}$$
You want to make this equal to $I_{0ab}e^{-x_{ab}}$  but there seems no systematic way to do this and this confirms the fact that the simple  Beer-Lambert law only applies for monochromatic light.


text:
How does ring size affect SN2 reactions?
Suppose we have to compare the $\mathrm{S_N2}$ reactivity order for cyclopropyl and cyclopentyl chloride. How can we do that?
I think that cyclopentyl should be more reactive towards $\mathrm{S_N2}$ because of larger bond angle and therefore lesser steric hindrance to the atttacking nucleophile but I am not sure about this.
Also, can the trend be generalised i.e. does increasing ring size lead to faster $\mathrm{S_N2}$?

TL;DR - Bond angle strain gets a lot worse in going from cyclpropyl chloride to the transition state for the 3-membered ring example, than for cyclopentyl chloride going to the corresponding transition state.
The rate for each reaction will be determined by the free energy barrier in each case. Cyclopropyl chloride is strained because the carbons "want" to be tetrahedral, with sp3-hybridized orbitals (109° bond angles), but are forced to have much more acute (60°) bond angles within the ring. However, at the transition state, the carbon undergoing nucleophilic attack "wants" to be trigonal planar, with sp2-hybridized orbitals (120°), but is still geometrically-constrained to 60° bond angles within the ring. This makes the free energy barrier for SN2 very large.

Cyclopentyl chloride has bond angles close to the ideal 109°. There will be some bond angle strain at the transition state, but the effect won't be nearly as bad as for the three-membered ring.

Now, can the effect be generalised? Mostly. As expected, cyclobutyl chloride reacts faster than cyclopropyl, but slower than 5 or 6-membered rings. Five vs six is a tricky one. Cyclopentyl chloride is actually slightly more reactive. If you look at cyclohexane, it is very stable in the chair conformation. All nice bond angles, no eclipsing interactions. At the transition state for SN2 with cyclohexyl chloride, bond angle strain is introduced and there is steric interaction between the incoming nucleophile and the adjacent axial hydrogens. Cyclopentyl chloride doesn't start off quite so perfect - there are some eclipsing interactions. Also, the transition state may actually even relieve some of these eclipsing interactions.
Beyond six-membered rings, things get very complicated quickly. You could apply the same approach, but the number of possible conformations quickly rises from 8-membered and above. It becomes difficult to take all the possible factors and conformations into account, so without high-level calculations, it is difficult to make predictions.
If you have access, the reactivity of 3-6 member cycloalkyl chlorides was included in a publication by Rablen et. al..[1] It includes calculated energy barriers for SN2 reactions of these cycloalkyl chlorides with cyanide ion as nucleophile.
Reference:

Rablen, P. R.; Mclarney, B. D.; Karlow, B. J.; Schneider, J. E. How Alkyl Halide Structure Affects E2 and SN2 Reaction Barriers: E2 Reactions Are as Sensitive as SN2 Reactions. J. Org. Chem. 2014, 79 (3), 867–879 DOI: 10.1021/jo4026644.

There is a literature study[1] on the rates of reaction of cycloalkyl bromides with sodium benzenethiolate [PhSNa] in the aprotic, polar solvent dimethylformamide [DMF] at 0 °C. Bromide is a good leaving group and the thiolate is an excellent nucleophile for an SN2 reaction. You are correct that the cyclopentyl halide is more reactive than the cyclopropyl halide. Indeed, their rates bracket the examples studied by over a million-fold! Why is the cyclopropyl halide the slowest of the group? Well, you said it — bond angle strain. Clearly, the idealized C-C-C bond angle of 120° in the transition state cannot be attained. The rate of the cyclobutyl halide displacement is sped up by the greater C-C-C bond angle by a factor of ~10,000 relative to the cyclopropyl system.
The cyclopentyl halide reacts about 10% more rapidly than the acyclic model, 3-bromopentane. This difference cannot be predicted a priori. While 3-bromopentane is not constrained by bond angle issues, the "floppy" methyl groups can inhibit the attack of the nucleophile. This problem is absent in the cyclopentyl  bromide, which is a "tied back" version of 3-bromopentane. The internal C-C(Br)-C bond angle of the cyclopentyl bromide ring is ≈104°, which is still not close to the  bond angle of 120° in an idealized SN2 transition state.[2]
The cyclopentyl bromide with its "envelope flap" shape does not present any severe steric interactions for the nucleophile.

At first sight cyclohexyl bromide might be expected to have the fastest rate because the C-C(Br)-C bond angle is ≈112°. However, steric factors now come into play. The chair conformation of bromocyclohexane having the equatorial C1-Br bond has a pair of axial hydrogens at C3 and C5 that impede the attack of benzenethiolate. The observation[3] that cis-4-tert-butyl-1-bromocyclohexane (axial C-Br) reacts with benzene thiolate in aqueous ethanol 58 times faster than with its trans-counterpart lends credence to the reaction proceding through the chair conformation having the axial C-Br bond, thereby avoiding steric issues incurred with the equatorial C-Br bond.
Bromocycloheptane has a C-C-C bond angles of ≈115° that facilitate a favorable TS but this benefit is countered by the onset of transannular steric factors. Nonetheless, this displacement is ≈100 times faster than the cyclohexyl example. Steric transannular effects come into serious play in the cyclooctyl, cyclodecyl and cyclododecyl ring systems. In summary, bromocyclopentane reacts the fastest of the cycloalkyl bromides having the best balance between angle strain and steric factors. With bromocyclohexane as some what of an outlier, steric factors dominate in rings greater than 5-membered, while angle strain prevails in rings smaller than 5-membered.
Notes and References

Masson, E.; Leroux, F. The Effect of Ring Size on Reactivity: The Diagnostic Value of ‘Rate Profiles’. Helv. Chim. Acta 2005, 88 (6), 1375–1386 DOI: 10.1002/hlca.200590110.
Note: Recall that the degree of bond-breaking and bond-making need not be equal in the transition state [TS]. The TS certainly would be expected to be early given the nucleophilicity of the thiolate and the electrophilcity of the C-Br bond. Therefore, an angle less than 120° may well suffice for this displacement.
Eliel, E. L.; Haber, R. G. Conformational Analysis. VII. Reaction of Alkylcyclohexyl Bromides with Thiophenolate. The Conformational Equilibrium Constant of Bromine. J. Am. Chem. Soc. 1959, 81 (5), 1249–1254 DOI: 10.1021/ja01514a058.

From a thermodynamic point of view, cyclopropyl ring would look to be more reactive because of higher ring strain. But the answer comes from looking at the transition state, which determines the Gibbs energy of activation. In transition state, for any general $\mathrm{S_N2}$ reaction, the non reacting ones substituents would be forced to be in a plane in this case the ring and another hydrogen, but the carbon atoms in cyclopentane ring are already at an angle close to $120^\circ$ and they would require much less "force" in order to get to the transition state, but for cyclopropane ring, should go through a change in bond angle, which would much more "force", and hence $\Delta G^\ddagger$ for cyclopentane ring would be lower than for cyclopropane ring and thereby, $k$ for cyclopentane ring would also be higher.
I am not an expert in chemistry, so please correct me if I am wrong.

Yep, cyclopropyl ringwill not be able to bear the partial positive charge generated on the ring due to the difference in the relative rates of bond breaking(for the leaving group) and bond formation(for the nucleophile) in accordance with Bayer's strain theory(which is quite valid uptil five membered rings). So it will be less reactive than cyclopentyl ring


text:
Reducible representations - choice of new basis
I was looking at this tutorial about reductions of representations. At first, the basis consists of four orbitals $(s_N, s_1, s_2, s_3)$ and the representations look like this

and we want to reduce the 3D $\Gamma^{(3)}(g)$ representation. Because its matrices aren't block-diagonal, we need to perform some similarity transform to get them into the convenient form. For that, we should choose a new appropriate basis, i.e. the linear combination of our orbitals.
Now the tutorial says that the appropriate new basis looks like this:

And yes, the basis is correct and the new block-diagonal matrices can be further derived using it.
But I have no idea, how did the author come with this specific basis.
So, how can I find the appropriated basis to reduce a reducible representation?

I suspect from a purely mathematical point of view it’s highly non-trivial (https://mathoverflow.net/questions/199911). However, I suppose you can find the irreducible representations (using the reduction formula) then use the projection formula to get the form of the MOs. See e.g. slide 13 of http://goicoechea.chem.ox.ac.uk/resources/s2_presentation.pdf onwards.

@orthocresol Thank you! Please, try to write down a more detailed answer, so I could accept it.


text:
Boiling of a liquid
Consider water as an example. I m really confused when does boiling takes place. When it's vapour pressure equals atmospheric pressure or when it's saturation atmospheric pressure equals atmospheric pressure.
Wait this is even more confusing. Either of two doesn't make sense. Consider this image

Here the water isn't covered by lid so there won't be any vapour pressure, as all the vapours will escape. This is the case where vapour pressure will exert
vapour pressure.

But in the first case irrespective of the vessel covered by a lid, boiling takes place. Neither vapour pressure equals atmospheric pressure nor saturation vapour pressure equals atmospheric pressure.

Related: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

Saturated atmospheric pressure is a phrase meaning that the vapor pressure is equal to 1 atmosphere. So it isn't just "atmospheric pressure" but a pressure of exactly one atmosphere. // For a open pot boiling imagine a thin layer above the liquid which is the pure vapor of the liquid. The vapor then diffuses into the atmosphere.

@IvanNeretin reading your post makes me even more confused. Firstly why are bubbles forming at the bottom of the liquid. Secondly, as you wrote that the bubbles that are formed at the bottom contains pure water vapour so why does boiling takes place when vapour pressure equals atmospheric pressure, because at the bottom there will also be and additional pressure of $h\rho g$ which would prevent the formation of bubble

$\rho gh$ will be there all right; I simply ignored it.

The red heat of the burner makes the bottom of the pot a little hotter and causes the highest vapor pressure. If "bubbles"  developed at the very top surface of the liquid, they might be invisible.

It's the first of the two options given. Vapour pressure is a property of the liquid, not of the vapour, and it is 1 bar at 100°C!

Water boils because the vapor pressure IN THE LIQUD reaches a level where the molecules can overcome the external pressure of the surrounding atmosphere and form bubbles. This level can be reached either by raising the temperature (the molecules gets more energy) or lowering the surrounding pressure (like in the mountains).
Actually, the pressure a water molecule has to overcome to turn into vapor, is the sum of the pressure from the water above plus the atmosphere. That is why water in hydrothermal vents can be several hundred degrees without boiling.
It makes absolutely no difference whether the atmospheric pressure comes from a bone dry gas or from pure water vapor.


text:
Referring to numbered atom (e.g. C-1 vs C1)
Is there a recommended style how to designate an atom by its number according to the structure numbering (given by nomenclature, or any other one in the text context)?
The are (at least) two different styles used by IUPAC itself. E.g.[1]

Diastereoisomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.

or[2]

Note: The configuration at C-4 (method 2) or C-1 (method 1) cannot be given because the attached substituent is unknown.

or[3]

Small substituents on the C-8 and C-9 atoms of steroids should be drawn within the ring system in this fashion even if there are no substituents on the adjacent fusion atoms C-10 and C-14.

vs[4]

The oxygen atom of the $\ce{CF3CO}$ portion of the ligand is attached
  to C2, while that of $\ce{MeCO}$ is attached to C4.

I can see some disadvantages for both of the styles.
With hyphen (dash):

Might be mistakenly thought, in favorable cases, that it is an isotope abbreviation. E.g. C-14 for carbon-14 ($\ce{^{14}C}$).
Not very likely, but eventually might be thought as a mathematical expression (e.g. when talking about 1st halogen atom X-1).

No hyphen:

When the text formatting was lost, it might be interconfused with molecular formula. Maybe not for carbon / organic chemistry, more likely for other elements in inorganic compounds, e.g. P4. Especially at chem.SE with Georgia font ☺.

References:

Basic terminology of stereochemistry (IUPAC Recommendations 1996), p. 2199 (anomers, via IUPAC Gold Book)
Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, p. 551
Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), p. 356
Nomenclature of Inorganic Chemistry (IUPAC Recommendations 2005), p. 156

If I'd be really worried about possible lost formatting and other noise immunity issues, I'd go with C#1.

@IvanNeretin … which then could be confused with some broken incomplete line formula notation (e.g. SMILES) with triple bond.

How about using the label plus the number encircled? Would be hard to confuse and misinterpret. I actually use this myself when writing on paper or typesetting on computers.


text:
Reaction rate coefficient constant multiplier for equation with multiple atoms consumed?
For the elementary reaction $\ce{2O -> O2}$, when calculating the rate of concentration change of atomic oxygen as $\frac{d[O]}{dt}$, would I use the standard rate coefficient $k_f$ or would I have to multiply by 2 since 2 oxygen molecules are being consumed?
$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -2 \cdot k_f \cdot [O] \cdot [O]$
or
$\ce{O + O -> O2}$ with $\frac{d[O]}{dt} = -k_f \cdot [O] \cdot [O]$
Or would the rate constant define the dependency on atomic oxygen?

Given that equation, the rate is for production of $\ce{O2}$. So you do need to factor for the stoichiometry of $\ce{O}$ relative to $\ce{O2}$.

In general if the reaction is $\ce{aA + bB \to cC +dD}$ the the rate is written as $$r=\displaystyle -\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$$.
If the reaction is indeed second order in oxygen atoms as stated by the OP, then the first equation given by the OP is correct, but unusual (you could just leave the factor two in the denominator on the left hand side instead of multiplying by two on the right hand side).

When we measure  the rate of the reactant concentration change (rate of disappearance of the reactant) or the rate of formation of the product experimentally, we neglect the coefficient and equation, but we take the balance equation into consideration when calculating the rate of disappearance or formation of each species in the reaction,
so: rate of disappearance of oxygen atoms = double the  rate of formation of  oxygen molecules.
$\ce{\frac{d[O]}{dt} = - k \cdot [O]^2}=2 \cdot{\ce{\frac{d[O2]}{dt}}}$


text:
How does MP2 level of theory affect the natural orbital picture of the system?
I have performed CASSCF calculations for a series of molecules with varying input orbitals. I have used UNO, UHF, ROHF and MP2 based natural orbitals to perform the CASSCF calculations. For my system I just wanted my input to be just pi-orbitals but the number of orbitals containing pi orbitals is different set of theory to form pi orbitals, with being less in number and least conjugated with MP2 based natural orbitals. When I am using MP2 based natural orbitals, the active space is turning out to be same for some molecules in the series. For example, I have an acene type of series and I am getting same active space say (8,8) for systems containing 6 benzene rings , 7 benzene rings and 8 benzene rings. Since the length of the system is increasing in a systematic fashion in terms of conjugation and the number of benzene rings, shouldn't the number of pi orbitals also increase in same manner ? But I am not able to see that, and therefore not able to justify the use of same active space for all. (For other cases like UNO, UHF the number of pi orbitals are increasing and also giving highly conjugated pi-orbitals in picture which is giving difficulty in convergence and for molecules big in the series exceeding the CASSCF limit.)


text:
Assign stereodiscriptor to 4th carbon of 3-ethyl-2,4-dimethly-5-(propan-2-yl)heptane-1,7-diol
Here is the molecule. Clearly the 4th carbon is chiral. I have drawn the diagraph for it, but the exploratory pathways of the left and right chains are the same. So, the problem is that the two chains are actually different, but same in the diagraph. How to assign stereodiscriptors in such cases?

If you can visually tell that they're different, then the graphs you build is different. If you're getting different graphs, you're doing something incorrect in that respect.

You mean "if you're getting same graphs"? The graphs are different, but the exploratory pathways are same, i.e. same set of atoms at any distance from the chiral carbon for both the chains

it's in how do you hierarchically evaluate the paths' nodes …

You just look at the sets of atoms at a distance of 1,2,... from the chiral carbon and wherever you find the first point of difference, then you can assign priorities. But, in this molecule the sets of atoms at all distances are same. So, you can't decide

No, the connectivity matters. 2-propyl < 1-hydroxy-2-propyl You don't just evaluate the nodes at each distance. You need to evaluate the whole group, but closer atoms have more influence than atoms that are farther away.

Not the proper, somewhat complicated, IUPAC ‘digraphs’… but might help.
Bold/colored bonds are the left and right branches' resulting paths taken when advancing to next sphere (because the decision could not be yet done) during determining a higher priority branch by finding a first difference of the compared atom set rankings. Sphere numbers a Roman. Grey digits are semi-conveniently arbitrarily chosen numbering just for reference, not nomenclature numbering nor priorities(!).

Following list shows chosen highest priority atoms in comparison in each nth sphere and left/right branch (bold; also comparison with other sibling atoms in the same branch, if they are not just hydrogen), chosen according to the next bonded atoms from the (n+1)th sphere (in 1st braces; atom number in 2nd brace).
sphere № : [left] vs [right]
I: [ C(CCH) (1) ]   =   [ C(CCH) (4) ]
II: [ C(CCH) (2) > C(CHH) (3) ]   =   [ C(CCH) (5) > C(CHH) (6) ]
III: [ C(HHH) (7) = C(HHH) (8) ]   <   [ C(OHH) (9) > C(HHH) (10) ]
Please note that in the II. sphere, left, the carbon (3) leading to terminal -OH group could not be chosen, because it's C(CHH), compared to higher priority, chosen C(CCH) (2).
Also note that in sphere III, left, any of the two methyls' carbons could be chosen (equivalent).
In this III. sphere, the decision finally could be made. The right branch (R1) has higher priority – finally only hydrogens bonded at left: C(HHH) (7), vs oxygen and hydrogens bonded at right: C(OHH) (9).
(Next priority is left branch R2, then methyl, then H.)

So the compound configuration/name is
(4S)-3-ethyl-2,4-dimethyl-5-(propan-2-yl)heptane-1,7-diol.

Oh! Ok, so you don't just keep on comparing sets of atoms for each of the chains at increasing distances from the chiral carbon and then try to get assign priorities. Instead, you see which branch to traverse in each chain, based on the priorities of atoms of each branch connected to the atom in the previous sphere. Am I right?

@Shashank This is what I did not want,… to confuse.  You just keep comparing, but your question looked like you thought that you traverse in each chain. I rather accidentally looks like it should be traversed, maybe the structure was intentionally designed such way (by our teacher?), or at least to be somewhat difficult. I'll try to improve the answer, mention some rules.

Yes, I thought each and every branch of both the chains have to be traversed

@Shashank … yes, it should be traversed (at least to the first atom in each branch☺) but with the evaluation, and finding the first difference. Your question looked like you are comparing sets of all possible traversal paths without evaluation/ranking of the attached atoms at every node.

Yes, the was the mistake I made. Just to clarify, you didn't use the chain starting from C-3, leading to the terminal OH group. I thought, that will also have to be traversed