text:
Effect of addition of inert gas on concentration in an equilibrium
We have the following equilibrium established in a container with movable piston at atmospheric pressure.
$$\ce{N_2 (g) + 3 H2 (g) ⇌ 2NH_3 (g)}$$
Now how does the concentration of the reacting species change upon addition of $\ce{He(g)}$ to the container?
I know that the reaction would shift towards left because it has larger number of moles than the right side due to Le-Chateliers principle. So the moles of the reacting species increases. Also, the volume of the container increases. So how to determine whether concentration, which is $\ce{\frac{Moles}{Volume}}$ increases or decreases as both Moles and Volume increases.

No,it doesn't talk abt the problem I mentioned.

First, pretend the reaction is super-slow and just consider what happens to the concentrations. Then and only then consider in which direction the reaction will have to go to attain equilibrium. This reaction will counteract the change imposed, but not completely to back to the original concentrations. https://chemistry.stackexchange.com/a/18717/72973 is relevant here.

Suppose initially you had 1 mole of $\ce{N2}$ and 3 moles of $\ce{H2}$, and that $n$ moles of $\ce{N2}$ had reacted.  Then you would have, at equilibrium, $1-n$ moles of $\ce{N2}$, $3(1-n)$ moles of $\ce{H2}$, and $2n$ moles of $\ce{NH_3}$.  So the total moles at equilibrium would be $4-2n$, and the mole fractions (equal to the partial pressures in atm) would be $$x_{\ce{N2}}=\frac{1-n}{4-2n}$$$$x_{\ce{H2}}=\frac{3(1-n)}{4-2n}$$ and $$x_{\ce{NH3}}=\frac{2n}{4-2n}$$So the equilibrium relation would be $$\frac{(2n)^2(4-2n)^2}{3(1-n)^4}=K_p$$If we had added $\delta$ moles of inert to the mix, we would have obtained:$$\frac{(2n)^2(4+\delta-2n)^2}{3(1-n)^4}=K_p$$or equivalently $$2n(4+\delta-2n)=\sqrt{3K_p}(1-n)^2\tag{1}$$
ADDENDUM
If we solve Eqn. 1 analytically for $n=n_0$ in the base case $\delta=0$, we obtain:
$$n_0=1-\frac{2}{\sqrt{4+\sqrt{3K_p}}}\tag{2}$$Furthermore, if we differentiate Eqn. 1 with respect to $\delta$ and take the limit as $\delta\rightarrow 0$, we obtain:  $$\left(\frac{dn}{d\delta}\right)_{\delta \rightarrow 0}=-\frac{n_0}{2\sqrt{4+\sqrt{3K_p}}}\tag{3}$$Therefore, the conversion decreases as we add inert gas to the reaction mixture.
The overall molar concentration $C_0$ in the cylinder is constant (irrespective) of addition of inert to the mixture), and is given by:  $$C_0=\frac{P}{RT}$$where P is the total pressure.  Therefore, the concentration of NH3 is equal to the overall molar concentration times its mole fraction:  $$C_{NH3}=\frac{2n}{4-2n+\delta}C_0$$If we employ Eqn. 3 to evaluate the derivative of $C_{NH3}$ with respect to $\delta$ at $\delta\rightarrow 0$ we obtain:
$$\left(\frac{dC_{NH3}}{d\delta}\right)_{\delta\rightarrow 0}=\frac{-n_0}{2\left(1+\frac{2}{\sqrt{4+\sqrt{3K_p}}}\right)}C_0$$Therefore, as the number of moles of inert gas increases, not only does the number of moles of ammonia decrease, but so also does the ammonia concentration in the gas mixture.

I agree that n is decreased which means moles of $\ce{N_2}$, which is $1-n$ is increased. but what happens to the concentration, which is $n/V$? Increase of moles alone doesn't imply that concentration is increased right? As there is a volume term.

SEE THE ADDENDUM

You can see this graphically as well, as $\delta=\sqrt{3^3K_p}(1-x)^2/2n+2n-4$ plot the rhs on the x axis and $n=0\to 1$ on y. As $\delta$ increases $n$ decreases.

@Chet Miller: Can you please explain why you made $\delta \to 0$ after differentiating?

The analytic solution I got for $n = n(K_p, \delta) $ was complicated and it wasn't obvious from it if n and C increased or decreased if inert was added.  Also, the question implied that the main interest was the direction (sign) of the effect.  So I judged that it would be easier if I just evaluated the incremental effect on n and C of adding a small incremental amount of inert.  This would be by expanding n in a Taylor series about $\delta = 0$.  This gave a clearcut analytical answer.


text:
Why do chemists work with common logarithms (base 10)?
In my understanding, every operation we do in $\log_{10}$ can be done in the natural logarithm itself and it should be better because mathematical integrals naturally give out expressions involving the natural logarithm function.
However, it seems that in my chemistry books there are elaborate schemes set up to avoid this and use the base ten logarithm instead. For example, we define $\mathrm{pH}$ as the negative of log base ten of hydrogen concentration instead of choosing the natural logarithm.
What are the historical reasons in the study of chemistry which caused this preference?

Largely due to comparing things that span many orders of magnitude. pH of 6 and pH of 8 is otherwise a factor of 100: try plotting that. As a geologist, I'll direct you to the inconvenience of base 2 for sorting grain sizes in sediments. It's all about who did it first AND how convenient it is for all parties concerned.

If we had 8 digits in our hands instead of the usual 10, we would use a base-8 system (most likely). Consider also the odd situation of computer scientists who choose to speak of kB, MB, GB, TB .... when in fact they mean the nearest binary number. It's ingrained in our way of counting learned from childhood. Perhaps kindergartens should teach to count using the binary system (base-2). Base-e (natural log) is then something else. You are not even using rational numbers. Much harder to rationalize its use.

As MFarooqs answer explains, a unit pH difference is easy to conceptualize: a factor of 10 difference in concentration. pH 2 difference = factor of 100, etc. Easy! Now think about ln scale:  -lnH difference of 1 means difference of e in concentration; a difference of 2 means $e^2$... uhm, what ? Note we already use a base-10 (decimal) system. Why deviate from that (like the Imperialists)? My favorite unit in this context has to be the decibel.

@BuckThorn Though in French, e.g., shopping for a thumb drive, you often read ko, Mo, Go, To where «o» stands for «octet» in lieu of kB, MB, GB (example) recalling that there is some base of eight in play ...

@Buttonwood that makes some sense since one byte  is 8 bits, but the meaning of kilo remains ambiguous (1024 bytes).

@Buttonwood AFAIK, In context of IT, there are officially decimal based units kB, MB, GB....  and binary based units kiB, MiB, GiB... ( i as binary) with 1000 resp. 1024 multiples. Unfortunately, the former units are unofficially used for both cases, even more for the latter, what brings sometimes inconsistency. E.g.hard disk capacity is provided in decimal units as GB/TB, while IT specialists use them in binary context. The latter XiB units are seldom used.

There are cases where 2.303 (ln(10)) coefficient has to be used if log10 is used instead of ln.  There are cases where 1/2.303 coefficient would be used if ln was used instead of log10. All depends if the scenario has "natural" or decimal nature.

When you use slide rules, you use log10 quite naturally. In then end it is simple to convert from one base (log or not) to the other, but for better or worse humans like thinking in multiples of 10.

I am sure you are mathematically gifted, and that's why you sensed this trend in an experimental science like chemistry. It seems that your textbook author(s) put an unnecessary emphasis on avoiding natural logarithms. The quick answer is that depending on the experimental situation, one can use either the natural logarithm or Briggsian logarithm. Chemists do not avoid natural logarithms. For example, in chemical rate laws or radioactive decay, the natural logarithm is the natural way to go.
However, when it comes to solution preparation, base 10 is experimentally feasible and of course very convenient even for mental calculations. When chemists like to prepare solutions, often the concentration is a simple multiple of 10.
Our South Asian readers will immediately recognize the normality notation often labelled on acids or bases as 1 N, N/10, N/100 solutions. Normality is equal to gram equivalents/ L. In the 18th and 19th century, this unit was in vogue. It makes solution preparation and dilution easier. Recall that burettes or graduate pipets are all scaled like 0.1, 0.2, 0.3 mL volumes and so on rather than multiples of natural logarithm base $\mathrm e$. You will notice natural logarithm avoidance in the pH scale as well as the Nernst equation, because concentration terms are involved. As one reader pointed out Nernst equation is typically found with natural logs on Wikipedia. I looked the original 100 year old text of Nernst, and he indeed used natural logs. My bias is toward analytical chemistry and then consulted a classic on electroanalytical chemistry text (Lingane: Electroanalytical Chemistry), there the author after doing all the derivations switches to Briggsian log. The reason that one of the most reliable methods to determine pH was electrochemical methods. All your pH meters rely on that equation i.e., Nernst equation with Briggsian log.
So this choice, is again dependent on the needs of the researcher.
Sørensen is popularly called the originator of pH concept. However, the hydrogen ion concentration idea was well known in his time. I checked Sørensen's original essay (Biochemische Zeitschrift. v.21 1909, freely available from Hathi Trust), and all his concentration units are in normality. As you can see for the pH, the scale goes like the picture below (Ref: Italian Journal of Medicine (2011) 5, 147—155). Now imagine setting up a natural logarithm concentration scale. This is not experimentally feasible. The numbers look "friendlier" with base 10 logarithms.
$$
\begin{array}{rll}
\hline
\mathrm{pH} & [\ce{H+}]/\pu{g Eq L^-1} & [\ce{H+}]/\pu{g Eq L^-1} \\
\hline
0 & 1 & 1.0 \\
1 & 0.1 & \pu{1.0E-1} \\
2 & 0.01 & \pu{1.0E-2} \\
3 & 0.001 & \pu{1.0E-3} \\
4 & 0.0001 & \pu{1.0E-4} \\
5 & 0.00001 & \pu{1.0E-5} \\
6 & 0.000001 & \pu{1.0E-6} \\
7 & 0.0000001 & \pu{1.0E-7} \\
8 & 0.00000001 & \pu{1.0E-8} \\
9 & 0.000000001 & \pu{1.0E-9} \\
10 & 0.0000000001 & \pu{1.0E-10} \\
11 & 0.00000000001 & \pu{1.0E-11} \\
12 & 0.000000000001 & \pu{1.0E-12} \\
13 & 0.0000000000001 & \pu{1.0E-13} \\
14 & 0.00000000000001 & \pu{1.0E-14} \\
\hline
\end{array}
$$
And students of the 1970 and early 1990s era will recall that base 10 logarithm tables or a slide rule were part and parcel of every student who studied mathematics and natural sciences. We used to have printed copies of base 10 logarithm tables in school in mathematics classes.

Re "And students of the 1970 and early 1990s era will recall that base 10 log tables or a slide rule were part and parcel of every student who studied mathematics and natural sciences". Where? I have never encountered them (only outside of the education system - my parent's generation knew them - I have my father's slide rule).

I doubt the "early 1990" era for slide rules. I was born in 1966, and at my school, we were the first 7-graders - so this would have been 1979 - who got pocket calculators, a friend who was 1 year older had still been taught to use slide rules.

But the Wikipedia article on the Nernst equation has way more $\ln$'s than $\log_{10}$'s; and the $\log_{10}$-based equations require the awkward conversion factor of $\ln 10 = 2.303$. After all at the end of the day it boils down to thermodynamics which uses $\ln$: $\Delta G = \Delta G^\circ + RT\ln Q$, and then to $\mu = \mu^\circ + RT\ln a_i$. In general I agree, but I think a better example could be chosen.

@PeterMortensen, The log table were purely academic excercises in schools (not colleges or universities) in South Asia until the late 90s.

@orthocresol, Good observation, I updated the post.

Uhm sorry if i seem like dumb , but i dont think we require mathematical gifts to be able to sense the base 10 trend in chemistry:) its quite obvious if one studies Physical Chemistry :)

@RishiKirti, Well, I have not been this question. The profile of the OP says that she/he wants to do a PhD in mathematics so certainly this is an above-average student's wish list. :-)

@M. Farooq that's why you sensed this trend in an experimental science like chemistry. I disagree with the idea there is a base-10 trend chemistry. I've not seen it in any of the publications I read, or the presentations I attend, or in any of  textbooks I teach from.   And I disagree with the implication that, because chemistry is an experimental science, chemists are  mathematically unsophisticated.  Also, what the OP presents as an example of a 'natural log avoidance scheme' by his textbook isn't that at all, since he's referring to the pH scale, where base-10 makes sense

@orthocresol In my school textbook on mathematics all equations are introduced in lns then converted into base 10 for presentation, so it seemed to me there was such an aversion.

see nernst equation for example in page-73 here

@Buraian, Both forms of Nernst equation are common (especially in some undergraduate/college textbooks, I have also noticed log).  The pre-log factor is 0.0591/n so it is easy to memorize, if you use log base 10. There is nothing special.

All fields of science (biology, chemistry, physics, and geology) use both natural and base-10 logarithms.
Is there a distinguishing characteristic that (typically) determines when one is used vs. the other? Yes: There is a class of applications where the natural log is used, and a qualtitatively different class of applications where the base-10 log is used.
Natural logarithms are used when describing physical processes whose underlying mathematics are exponential (specifically, base-$e$ exponential, which is commonly referred to simply as "exponential").  Examples:
Biology:  Population growth
Chemistry: First-order rate laws
Chemistry and Physics: Nuclear decay
Exponentials are commonplace because they result from a frequently-seen type of physical process: Those where the rate at which the quantity of an item changes is proportional to how much of that item you have (i.e., where the derviative of a function equals its value):
$$\frac{dy}{dx} = y \implies \frac{dy}{y} = dx \implies \ln y = x \implies y = e^x$$
Natural logs are used to linearize these functions.
In summary, natural logs appear when mathematically characterizing physical processes that follow a very common functional form: the exponential.
[In the comments, Peter Mortensen noted that base-2 may be more natural for nuclear decay than base-$e$.  The underlying physics is exponential, but base-2 becomes convenient specifically if you want to work in terms of half-life—which is, of course, a standard way of specifying nuclear decay rates:  $N(t) = N_o e^{-\lambda t} \implies N(t) = N_o e^{-t\ln 2/\tau}  \implies N(t) = N_o 2^{-t/\tau}$, where $\tau$ = half-life.]
By contrast, base-10 logs are used when deploying a measurement scale that needs to cover a wide range of values. Examples include the pH scale in chemistry, and the Richter scale in geology. None of these scales are mathematically tied to any underlying physical process. They are purely descriptive.  E.g., the Richter scale is based on the base-10 logarithm of the amplitude of seismogram waveforms.
When creating such purely descriptive scales it's typically considered much more convenient and intuitive to base them on powers of 10 rather than powers of $e$.  Hence in converting the values of these phenomena to their respective scales, the base-10 logarithm is typically used rather than the base-e ("natural") logarithm.
So no, it's not the case that "every operation...should be better [with the natural log]"; nor are chemists concocting "elaborate schemes" to avoid using it.  Rather, chemists (and scientists in all other fields) readily use both the natural log and $\log_{10}$, depending on what makes sense.
[Note: I'm not saying that $\ce{[H^+]}$ or seismogram waveform amplitude couldn't be tied to some underlying physical process; I'm simply saying these scales don't do that.]
The base-10 logarithm is also often used purely for presentational purposes, when showing data covering a wide range. Here's an example from ATLAS Collaboration, Probing Dark Matter with the Higgs boson, April 2020, https://atlas.cern/updates/briefing/probing-dark-matter-higgs-boson:

To sum it up: Decimal numbers use base 10, which naturally propagates where there are no (significant) other concerns.

Isn't base 2 more natural (no pun intended) for nuclear decay (half-life, no need for ln(2))? (Not a rhetorical question.)

@PeterMortensen  Good point. The exponential function corresponds directly to the physics of nuclear decay.  However, yes, it is natural (and usual) to think in terms of half-lives and, specifically for that reason, it is not uncommon to see nuclear decay re-expressed in base-2, i.e.:  $N(t) = N_o e^{-\lambda t} \implies N(t) = N_o e^{-\ln 2 t/\tau}  \implies N(t) = N_o 2^{-t/\tau}$, where $\tau$ = half-life.   I'll edit my post to add that. But my broader point remains: those working in the field use what makes sense for the intended use and, if applicable, the underlying physical process.

I'm guessing, of course, but my answer is: graph paper.
Every proportionality produces a plot on cartesian graph paper
that is a straight line.   And, every power law produces a plot
on log-log paper that is a straight line.   Every exponential
(exponentials are faster-rising than any power law) can be made to describe a
straight line on a plot using semilog paper.
So, part of yesteryear's equipment any experimenter needed was a selection of
graph papers to hand-enter data points onto; that was the way to make
a curve fit for those (fairly common) sorts of relationships.
I'm sure Pierre Curie used log-log paper to determine the power
law for magnetization diminution near the Curie temperature, for instance.
How else, before easy computer access, would he have found that
constant (about 1/2)?
And, to enter a decimal  number onto logarithmic scale, you are
using axes for your graph that have a small-integer-ratio that
makes the log(base 10) trivial to read off.   One still  hears
lots of '3 dB per octave' chatter, in some circles, which is
a reference to the 'octave', a  unit interval of a time or frequency
scale, which 'unit' is in the logarithm base 2.
While some use  is found for base 10 and base 2 scales, I've never
seen a graph paper log 'cycle' which was a factor of 'e'.

In addition to the other answers:
In the old days, log calculations were not simply a click or press on a computer / calculator, but needed to be looked up in paper tables - which in turn took man-years of manual calculations to produce, and therefore were expensive. Why buy another such booklet for ln, if you already had the one for log10 (and knew a lot of logs by heart)?

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text:
Considering Gibbs energy or electrode potential to find disproportionation tendency
The following data was given (without the asterisks and footnote) asking us to find the oxidation states that would have a tendency to disproportionate -

By tendency to disproportionate, I understood that the disproportionation reaction must be spontaneous, implying change in Gibbs free energy should be negative. As it is related to the given electrode potentials as $\Delta G = -nFE$, I proceeded by solving $\Sigma nE$ should be positive. (I didn't forget to consider the fact that all potentials are reduction potentials, and for disproportionation the oxidation half would have its sign reversed).
However, my teacher later told that in such questions one simply considers the sum $\Sigma E$ and not $\Sigma nE$, for tendency to disporportionate. From the given data, for each such species both $\Sigma E$ and $\Sigma nE$ are positive, so I cannot infer from it whether my approach was correct or my teacher's. Please clarify which method is correct, and in case $\Sigma E$ is correct, then why $\Sigma nE$ doesn't work.
An example of my approach - Consider $\ce{MnO4^2-}$
For its disproportionation, reduction potential for reducing to $\ce{MnO2}$ = $\ce{E1}$ = 2.26 and number of electrons =  $\ce{n1}$ = 2. Oxidation potential = $\ce{E2}$ = -0.56 and number of electrons =  $\ce{n2}$ = 1. Then the sum $\Sigma nE$ = $\ce{E1n1 + E2n2}$ = 3.96 which is positive. So $\Delta G$ is negative, hence the disproportionation is spontaneous.
My teacher told that $\Sigma E$ = $\ce{E1 + E2}$ = 1.70 is positive, would suffice.
(Normally I would never doubt my teacher, but she didn't reply when I messaged her regarding why my method was wrong, so I still need an explanation.)

Comments are not for extended discussion; this conversation has been moved to chat.

Your teacher is correct. I will take the example that you gave in the question. You wish to analyze the reaction:
$$3\ce{MnO4^{2-}}\to 2\ce{MnO4^{-}}+\ce{MnO2}...(1)$$
This reaction can be split into two parts, the oxidation half reaction, which forms the anode:
$$\ce{MnO4^{2-}}\to \ce{MnO4^{-}}+\ce{e-}...(2)$$
And the reduction half-cell, which forms cathode:
$$\ce{MnO4^{2-}} +2\ce{e-}\to \ce{MnO2}...(3)$$
Notice that Gibb's free energy is an extensive property, and hence is additive. So, if original reaction has $\Delta G_1$, since:
$$(1)=2×(2)+(3)$$
We have:
$$\Delta G_1=2×\Delta G_2+\Delta G_3$$
Which means:
$$-\Delta G_1=2×(1×F×E_2)+1×(2×F×E_3)$$
This means that checking the sign of $E_2+E_3$ is sufficient to conclude if $\Delta G_1$ is positive or negative, taking the sign of $E_2$ into consideration, of course.
In general, we have the formula:
$$E^{°}_{cell}=SRP_{cathode}-SRP_{anode}$$ which could have been used to conclude the same.

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What is the n-factor of ferric sulfide when it reacts with oxygen to give ferrous sulfate and sulfur dioxide?

What is the n-factor of $\ce{Fe2S3}$ in this reaction?
$$\ce{Fe2S3 + O2 -> FeSO4 + SO2}$$

I broke this reaction down into two parts
$$\ce{Fe2S3 -> FeSO4}$$
$$\ce{Fe2S3 -> 3 SO2}$$
And in the second case the change in electrons is 18. But how do we find it in the first reaction?
I think this is a hypothetical reaction. The answer is given to be 20.

Even if we assume that Fe2S3 is a normal thing and not a typo, I don't believe the reaction will produce FeSO4. Also, 18 looks rather cryptic to me. Why 18?

@IvanNeretin i guess it produces FeO; but anyways, how do we still calculate overall n-factor here for Fe2S3 here? These types of questions do occur in our UG exams like JEE. So I wish this question gets an answer!

I don't believe in FeO either.

@IvanNeretin Well, disregarding what the product should have been and restricting the relevance of the reaction to 'hypothetical purpose only' ,now let us expect an answer!

Then I guess $n$ is the number of electrons transferred.

@Raviraj Bhosale this is the question of fiitjee aits 2020

in the 2nd case i think the answer must be 6 and not 18 as only 1 S atom from Fe2S3 contributes in SO2. the rest of the 2 S atoms could be seen in FeSO4 once the equation is balanced

@Raviraj Bhosale but we have to balance the reaction first of all.

@Ivan Neretin but we have to balance 2nd reaction first of all

@IvanNeretin supposedly

Please note that n-factor is an outdated concept. See: The definition of n-factor, what was it when it was introduced?

We first need to find out which definition of n-factor we need to use here. This is simple, considering that the reaction is a redox reaction. Thus, the definition we use is the number of electrons lost or gained by the compound.
Therefore, we find the need to balance the equation.
Upon balancing, we see that the equation transforms into
$$\ce{Fe2S3 + 5 O2 -> 2 FeSO4 +SO2}$$
This can now be done from two points of views.
POV 1: From $\ce{Fe2S3}$
When you look at the reaction from the point of view of $\ce{Fe2S3}$, we see that the main reaction is actually the combination of two reactions
$$\ce{Fe^3+ + S^2- ->  2Fe^2+ + 2S^6+}\tag{1}$$
$$\ce{ S^2- ->  S^4+ }\tag{2}$$
Note: I do not care about the balancing of these equations since that has already been done and is not the focus of this question. Here only $\ce{Fe2S3}$ is important and so we tunnel vision.
Now, in half-reaction (1), Fe gains two electrons, and sulfur loses $(8 \times 2)$. Therefore net change in half-reaction(1) = -16 + 2 = -14.
Now, in half-reaction (2), S loses 6 electrons. Therefore net change in half-reaction (2) = -6
Therefore net change from the POV of $\ce{Fe2S3}$ is 20 electrons.

Our second POV is much simpler and also uses the charge neutrality concept - electrons lost by one compound are gained by the other in the reaction.
POV 2: From $\ce{O2}$
From the POV of $\ce{O2}$, we only have one reaction.
$$\ce{5O2 -> 10 O^2-}$$
Therefore the net change in electrons from the POV of $\ce{O2}$ is 20 electrons gained.
By the principle of charge neutrality, this means that $\ce{Fe2S3}$ loses 20 electrons.

Therefore from both POVs, we get the same result. The n-factor of $\ce{Fe2S3}$ is 20.

The written equation is impossible. I explain.
The reagents contain the compound $\ce{Fe2S3}$, where the initial iron is ferric, i.e. at the oxidation number +$3$. In the final compound $\ce{FeSO4}$ the iron is ferrous, i.e. at the oxidation number +$2$. In this equation, the iron atom is reduced from +$3$ to +$2$. This is impossible in an oxidation with $\ce{O2}$. The oxidative action of $\ce{O2}$ cannot lead to a decrease of the oxidation number of an atom present in the reaction. The oxidation of $\ce{Fe2S3}$ must produce a ferric compound, like $\ce{Fe2(SO4)3}$ instead.


text:
What is the IUPAC name of perchloroisobutane?

What is the IUPAC name of $\ce{(CCl3)3CCl}?$

The answer I got was 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane. But when I looked up the question online, most sources say the answer is 2-(trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane, even though ‘chloro’ comes before ‘trichloromethyl’ alphabetically. What is the reason ‘chloro’ is written after ‘trichloromethyl’?

According to Chemdraw JS' Structure to Name feature which generally gives the PIN of the molecule, the PIN for $\ce{(CCl3)3CCl}$ is 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane

From Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book),

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)

P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Here we have two main groups attached to $\ce{C-2}$, the (trichloromethyl) group and the (chloro) group.

P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Taking the example of 7-(2,4-dimethylpentyl)-5-ethyltridecane from the Blue Book, we see that in such a case, the di- is considered to be a part of the complete name and so needs to be considered as well.
Therefore the two names in the nomenclature of our compound would be 1,1,1,2,3,3,3-heptachloro and 2-trichloromethyl. Our parent chain is propane and so the PIN for the given compound becomes 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl)propane which is the same name as suggested by Chemdraw JS.
Therefore, your proposed answer is correct and the source online is wrong when it comes to the PIN.


text:
Resonance of Tosylate ion

For the Tosylate ion, the negatively charged oxygen is stabilized via resonance with the oxygen atoms. But what are the orbitals of sulfur that are involved in the overlap? Is the sulfur $\ce{sp^2}$ hybridized and thus have a p-orbital for resonance?
If it is $\ce{sp^2}$, would that mean that it is possible for the electrons in oxygen to delocalise into the pi-electron cloud of benzene via the continuous overlap of p-orbitals in sulfur, carbon and oxygen atoms?

The sulphur has 4 $\sigma$ bonds and no lone pairs, so it is most likely $sp^3$ hybridized. Not written as an answer as I'm not completely sure.

Yes, @TRC is right, the sulfur centre is tetrahedral, so the oxygen is not in the same plane as the ring.

Thanks to @Shoubhik R Maiti for confirming that my reasoning was correct
I hope you know the meanings of the terms $\sigma$ bond, $\pi$ bond and how to find hybridization of an atom. If not, you can get numerous good resources on Google.
To find the number of hybrid orbitals, we take the sum number of $\sigma$ bonds + number of lone pairs. Here, sulphur has six valence electrons, all bonded. So there are totally 4 $\sigma$ bonds and zero lone pairs. Resulting in hybridization $sp^3$, giving it a near-tetrahedral geometry (near because the aryl groups and oxygen atoms different, slightly distorting it from perfect tetrahedral). Since it is tetrahedral, the oxygen and phenyl group are not in the same plane, so the continuous overlap which you suggested is not possible.

What you are asking is if the electrons of the oxygen atom can delocalize into the benzene ring. Firstly I need to mention that the negative charge is equally distributed on all three oxygen atoms, due to equivalent resonance. Now, from the canonical structure you have given in the question, there can be two cases of oxygen's electrons delocalizing into the benzene ring:

The electrons of negatively charged oxygen bond with the sulphur atom. In that case the $\sigma$ bond between sulphur and benzene must completely break to accommodate the fresh electron pair from oxygen. You will be left with $\ce{SO3}$ molecule and a toluene with negative charge. This obviously can't happen.

The electrons from one of the double bonded oxygen atoms delocalizes into the ring. Then you shall have an oxygen atom with a positive charge on it, singly bonded to the sulphur, and the benzene ring double bonded to the sulphur with a negative charge on one of the carbons. This is energetically unfavourable because it involves unlike charges' separation (which takes energy), and has + charge on electronegative oxygen and vice-versa, making it an extremely unstable resonance structure. Hence even if it is a contributing canonical structure, its contribution to the final resonance hybrid is very, very small, almost negligible.

Hence in no case is it possible to have electrons from oxygen atoms delocalizing into the benzene ring.

+1 I would also add that even though direct delocalisation cannot take place, hyperconjugation/ negative hyperconjugation can take place to some extent, but since oxygen is electronegative, the electrons from benzene ring are withdrawn. This is why SO3H is a meta-directing group.

The $\ce{-SO3-}$ group ist tetrahedral, that is the only reason why it can be described with sp³ hybrid orbitals. This is disregarding whether or not hybrid orbitals are indeed a good description in this case. Then there is the problem of double bonded oxygen, which is a stretch. If ever, this is a zero order approximation. I'm sorry, but this answer falls short. It is one of those 'good resources' that unfortunately feed into the myths that hybridisation could be a cause for anything. It is a mere mathematical interpretation, making that clear from the start would go a long way for science.

It doesn't matter whether you are aware or that or not; your answer uses reverse reasoning, it is wrong. Unfortunately it is employing those hand-wavy models which have been popular and continue to be so as these are fed by answers like this. I wholeheartedly disagree that this does not lead to wrong results. Apart from this, nobody who would need to read the correct answer will ever do so, and I do not need to post an answer for people who already know that. || Also, Newton's laws are a valid and good approximation; the comparison is moot and frankly quite insulting.

@Martin-マーチン The hybridization approach is something that is taught in high school in chemical bonding everywhere in India (I do not know of other countries' curriculum). I have found it in a number of introductory textbooks too, and since all of it is wrong, it is necessary first to amend those textbooks first so that people like me don't learn and use the wrong concept right from the start. The myth is largely fed and popularized by the people who include it in the curriculum, much more so than answers of Chem SE. I don't see any point in carrying ahead this discussion in any case.

Simply put, the oxygens' lone pairs are NOT delocalised into the benzene ring, and the stabilisation of the sulfonate group must be allocated more to ionic interactions(or loss of such; see below) than to resonance effects(the latter of which does not really matter that much, due to the instability of the "toluenide anion"-containing resonance canonical).
As many people have pointed out in many different answers posted on this site, there are no double bonds in sulfate groups and the like- the sulfur has a +2 charge while the oxygens have -1 charges each. These charges attract each other, resulting in the "double-bond-like" sulfur-oxygen distances. If an electrophile attacks and bonds to the oxygen, then the oxygen would not be able to have a negative charge anymore, resulting in destabilisation from the loss of an electrostatic attraction interaction.
Now, both nitric and toluenesulfonic acids are strong acids. In the nitrate case, there are two factors (pi-resonance and  (+1 charge nitrogen)-(-2/3 charge oxygen) electrostatic attractions at work here, while in the sulfonate case, there is only one factor at work here, namely the (+2 charge sulfur)-(-1 charge oxygen) electrostatic attractions. Note that the elctrostatic attractions in the sulfonate anion are much stronger than those in the nitrate anion.
The destabilisation of the loss of electrostatic attractions upon protonation of the former is nearly of the same order of magnitude of the destabilisation from the disturbing of the pi-system upon protonation of the latter(if any, since the proton certainly would not want to attack the pi-system and violate VSEPR theory), and hence both acids become nearly equally strong acids.
Note: all of the above is true because the "toluenide anion" is unstable. In situations like the trifluoromethanesulfonate anion, the three oxygens' sp³ lone pairs, in the same direction as the carbon-sulfur bond, CAN delocalise into the sulfur-carbon antibonding orbital(which is mainly of sulfur 3p character; remember that one side of a 3p orbital has one node, just like the whole of a 2sp³ orbital), since the "trifluoromethanide anion" is relatively stable due to the inductively electron withdrawing fluorines. In support of this argument, optimisation calculations done on Mopac using the PM7 level of theory indicate that, while there is a charge <-1 on each oxygen in the toluenesulfonate anion(i.e. no delocalisation of the oxygen lone pairs), while there is a charge slightly >-1 on each oxygen in the trifluoromethanesulfonate anion(i.e. some delocalisation of the oxygen lone pairs).


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Why is Electron Gain Enthalpy of only Be and Mg positive, in group 2?
While going through my book, I came across the following observation:

$\Delta H_{eg1}$ is positive for Be and Mg

Where $\Delta H_{eg1}$ is the first electron gain enthalpy.
It seemed quite reasonable as Be and Mg have fully filled $\ce{ns^2}$ configuration and also that the incoming electron would enter the higher energy p orbital which is well shielded by the s sub shell (we know that screening of s is maximum).
But why don't the other elements follow this? I thought maybe because they have d orbitals and thus get stabilised but Mg also has a d orbital right? So that doesn't seem a reasonable explanation.

All atoms have $d$ orbitals, but in general occupying them is energetically favorable only if all lower energy orbitals are filled.  So you have to fill orbitals up to $3p$ first, therefore at least an argon core.

Oh. So it is indeed because of the presence of... sort of 'accessible' d orbitals which help in stabilising the formed anion?

The presence of $d$ orbitals can indeed provide a means for otherwise closed-shell atoms of alkaline-earth metals to accept electrons.  Wu et al. [1] describe carbonyl complexes of Ca, Sr, and Ba in which the valence $d$ orbitals of these metal atoms are indeed engaged in (covalent) bonding.
In principle, all atoms actually have $d$ orbitals. Getting them to accept electrons in the ground state, however, is a different matter. With two units of orbital angular momentum and thus at least two nodes (corresponding to a principal quantum number of at least 3), all such orbitals are relatively high in energy and require a relatively deep potential well to make a bound state with the additional electrons. With 20 positive charges in the nucleus in a calcium atom the well is deep enough as the $3s$ and $3p$ orbitals privide only partial shielding from the nuclear charge. Magnesium with only 12 positive nuclear charges does not get there. It does have $3s$ electrons and perhaps bonding with $3p$ orbitals, these orbitals having two total nodes in a magnesium atom; but these have less angular momentum than the $3d$ orbitals we seek to occupy.
References
1.
X. Wu, L. Zhao, J. Jin, G. Wang, M. Zhou, G. Frenking, "Observation of alkaline earth complexes M(CO)8 (M = Ca, Sr, or Ba) that mimic transition metals", Science  31 Aug 2018: Vol. 361, Issue 6405, pp. 912-916.
DOI: 10.1126/science.aau0839

A simple, intuitive answer based solely on high-school chemistry and available emperical data; 1s < 2s < 2p < 3s < 3p < 4s < 3d
For Be, all the orbitals up to 2s are filled, so the "new" elctron has to go into the 2p orbital. Since there is a large gap between the 2s/2p energy levels of Be(it is one of the only elements in the 2nd period that cannot easily form pi bonds; BeF2 is polymeric while BF3 is monomeric with meaningful B-F pi bonds), the first EG enthalpy is positive. Same goes for Mg using its 3s/3p orbitals.
However, the situation changes when one goes to Ca. All the orbitals up to 4s are filled, but the next orbital is 3d instead of 4p. According to several papers, including this one, the 3d orbitals on Ca nearly act as frontier orbitals in well-defined compounds- exactly like the transition metals. Since the 4s/3d gap in the early 3d transition series is small, one can assume the same for Ca, and hence the first EG enthalpy of Ca is negative, just like the 3d transition metals. Again, the same goes for Sr using its 5s/4d orbitals.


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HCP and hexagonal primitive cell
There are 14 types of Bravais lattices one of them is hexagonal which is just present as primitive cell ie the basis are at its corners , now when I read about hexagonal closed packing I came to know that it is a part of hexagonal system (from here
) But it confused me that if it belongs to hexagonal system then it should be present in primitive cell form ,but there are atoms in the hexagonal closed packing prism which are present inside it. Why does this happen , can anyone clear my doubt?

Bravais lattice type does not tell you anything about where the atoms are. True, on the typical pictures they just sit at the corners and centered positions (if any), leading the students to believe there is a law to that effect. No, there isn't.

Green is a simple hexagonal Bravais lattice. Adding the orange atoms makes it a hexagonal lattice with a two-atom basis, resulting in a close-packed structure called hexagonal close packing (hcp).

You may do better than pasting a screen photo of your cell phone showing a part of a web page.  All the information about ad by your mobile carrier are not relevant to the question.  Instead, format your question, and add a accessible reference to the illustration and quote.


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Nanocomposite hydrogel for wet adehsion in biomedical applications
My name‘s Tristan and I‘m a bachelor‘s student in Germany. I‘m trying to synthesize an injectable nanocomposite hydrogel for wound closure and wound healing applications. The problem is that, next to bioactive glass nanoparticles, I‘d like to incorporate chitosan or carboxymethyl chitosan (https://doi.org/10.1016/j.ijbiomac.2018.09.131) as a biopolymer matrix but I don‘t know how to get it to adhere to tissue in wet environments. I‘ve been researching catechol/quinone chemistry as an option for functionalization, since it is said that those groups could adhere to virtually anything (https://doi.org/10.1016/j.ijbiomac.2021.02.091). Therefoe I want to try to synthesize some adhesive polydopamine or tannic acid (because of its gallol moieties) hydrogels. I‘ve already tried an approach with tannic acid (https://doi.org/10.1021/acsami.0c02037), which has failed to adhere in wet environments and I‘m now looking for a viable approach inculding polydopamine.
However, I now don‘t really know how to specifically functionalize chitosan to get it to properly adhere to tissue in wet environments. Does anyone of you have an idea about a method or substance that could do the job?
Greetins from Germany:)


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Do the differential expressions of second law of thermodynamics imply a smooth increase in entropy?
Another statement of second law of thermodynamics can be formulated in terms of system properties and not properties + surroundings (isolated system). For a closed system at constant temperature and pressure:
$$dG_{T,p,n_i} \leq 0$$
Are $n_i$ constant during a process?
We know that during a chemical reaction Gibbs free energy take its minimum value at equilibrium. But during the reaction the number of molecules $n_i$ change. So would it better the inequality to take the form of:
$$dG_{T,p} \leq 0$$
?
Interpretation of the differential
In mathematics the differential of a function $f(x)$ can be interpreted as the change in $f(x)$ for an infinitesimal change in the argument $x$, that is:
$$df(x)= f(x+dx)-f(x)$$
I have trouble in the interpretation of $dG \leq 0$. When I was introduced in thermodynamics I always thought of the spontaneity inequalities in the following manner.
Suppose that the system is at equilibrium state (state $1$) at time $t$. Something happens (e.g. a chemical reaction takes place) so the system will reach a new equilibrium state (state $2$). In that case $G$ can be thought only as function of $ξ$ (the extent of reaction). Because the Gibbs energy must take its minimum value at equilibrium I interpreted the $dG \leq 0$ as:
$$dG(t)= G(t+dt)- G(t) \leq 0$$
In other words Gibbs free energy decreases until it takes its minimum value. And because we can plot $G$ as function of $ξ$ the reaction will proceed in the direction where $G$ is minimized as shown in the following figure (the left figure):

Although the above interpretation seems intuitive I have some problems.
$1$st Problem
With the above interpretation is not clear if the equality should be interpreted as that the function have a constant value over an inteval or that it has reached a minimum value. For example if we were monitoring $G$ over time and we noticed that $G$ has a constant value would this imply that the system have reached equilibrium? Look at the following figure:

In the mid region of the function (where it is constant) the differential is zero. So how $dG \leq 0$ should be interpreted? In general we can't find a function $φ$ so that $G=φ(t)$ and this should be expected as Thermodynamics doesn't deal with how the state of the system varies with time.
$2$nd Problem In the first scheme I have drawn an alternative path (right figure) that the system could take in order to minimize $G$. In that case there is an abrupt change in the extent of the reaction but still the systems keeps decreasing its $G$. To be honest I haven't seen any reaction reaching equilibrium in that way but I can't find a reason why it should not behave this way. It should be noted that the line representing the path depicts the change in $ξ$ in an abrupt manner and it should not be confused with a continuous change (think it like a $5$ step process).
In summary I want to make clear how to interpret the differential expression of the second law. Is a smooth incease of entropy (of surroundings + system) over time assumed?

Usually, it is the derivative of G with respect to reaction conversion is equal to zero at equilibrium.  For multiple reactions, it would be minimum of G with respect to all species perturbations under constraint of conservation of atomic species.

Point 2 it is interesting. I do not think anything goes exactly like that but I think to see a similarity with oscillating reactions. I also would like to know how to discard it by reasons/principle rather than by feeling.

Yet, the trick of thermodynamics is to swap the real dynamics under the carpet.... At the end what you are following are state functions, and their construction does require smooth continuous (and slow) changes. But, then, you have state functions. So between left & right the observation at equilibrium would be the same.

I can't understand why the question is closed.


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diprotonated vs monoprotonated inorganic phosphate
We have made confounding observations in the physiology lab working with inorganic phosphate. The cell processes we study rely upon the transport of diprotonated inorganic phosphate. But we get different results whether the extracellular buffers are bicarbonate or HEPES. We wonder if the two buffering systems, bicarbonate vs HEPES, differ in their making available diprotonated vs monoprotonated inorganic phosphate.
If we buffer our solutions with bicarbonate (25mM), pH to 7.4, and maintain with 95% O2/5% CO2, we get expected physiological responses over a range of 0-5 mM inorganic phosphate (Pi), added in the form of sodium monobasic phosphate.
However, if we try adding sodium dibasic phosphate, it is supersaturated at ~>2 mM Pi and pH 7.4 and unuseable.
If we buffer our solutions with HEPES (20mM), pH to 7.4, and bubble with 100% O2, the solution looks fine but we do not get expected physiological responses over the range of 0-5 mM inorganic phosphate (Pi), added in the form of sodium monobasic phosphate.
Sodium dibasic phosphate also dissolves fine into the HEPES buffer at pH 7.4. However, we still do not get the expected physiological responses.
It seems as though there is no diprotonated inorganic phosphate in the HEPES buffer no matter how Pi was originally added to the solution.
Other confounding issues could include whether bicarbonate (does) or HEPES (some argue does not) get into the intracellular space. Before getting into that question, we'd like to know:
Do the two buffering systems, bicarbonate vs HEPES, differ in their making available diprotonated vs monoprotonated inorganic phosphate at pH 7.4, room temperature?
Many thanks in advance.

Do the two buffering systems, bicarbonate vs HEPES, differ in their making available diprotonated vs monoprotonated inorganic phosphate at $\mathrm{pH \ 7.4}$, room temperature?

I do not have any physiology lab experience. However, my general knowledge is telling me that you have not completely disclosed what physiological response you are looking for and what components in the buffer would make that response (it is true that you have said that the study rely upon the transport of diprotonated inorganic phosphate, but should the cell enzymes taking monoprotonated inorganic phosphate and converting it to diprotonated inorganic phosphate?). If we know those little details, if would have been easy to answer your question.
Well, $\mathrm{p}K_\mathrm{a2}$ of HEPES buffer is 7.5 at $\pu{25 ^\circ C}$ while that of diprotonated phosphate $(\ce{H2PO4-})$ is 7.2 at $\pu{25 ^\circ C}$. Therefore, at $\mathrm{pH} \ 7.4$, monoprotonated/diprotonated phosphate ions would exists in HEPES buffer, if you added some $(\ce{NaH2PO4})$ to the buffer. However, the amount of monoprotonated phosphate $(\ce{HPO4^2-})$ would be more compared to that of diprotonated phosphate ions $(\ce{H2PO4-})$ in HEPES buffer $(\mathrm{p}K_\mathrm{a2, \ HEPES} = 7.5)$ during your physiological response checking.
On the other hand, $\mathrm{p}K_\mathrm{a}$ of carbonic acid is 6.1. The bicarbonate buffer system is an acid-base homeostatic mechanism involving the balance of carbonic acid $(\ce{H2CO3})$, bicarbonate ion $(\ce{HCO3-})$, and carbon dioxide $(\ce{CO2})$ in order to maintain $\mathrm{pH}$ in the blood and duodenum, among other tissues, to support proper metabolic function. That's why you need to maintain the partial pressure of $\ce{CO2}$ in the buffer system when you are using bicarbonate buffer. What I understand according to the process you have described in the question is when you added $\ce{NaH2PO4}$, $\mathrm{pH}$ might move from 7.4 towards 7.2 giving equal or higher amount of $\ce{NaH2PO4}$ compared to that of $\ce{Na2HPO4}$ $(\mathrm{p}K_\mathrm{a, \ \ce{H2CO3}} = 6.1)$ during your physiological response checking.
Although you get better response in bicarbonate buffer compared to that in HEPES buffer in this particular case probably based on the reasons given above, keep in mind that HEPES is widely used in cell culture when compared to bicarbonate buffers (which are also commonly used in cell culture), largely because it is better at maintaining physiological $\mathrm{pH}$ despite changes in carbon dioxide concentration, usually produced by aerobic respiration (Ref.1 & 2). I have attached the PDF of Ref.2 so that you can read and learn an specific case of comparison of HEAPS and bicarbonate.

References:

Simona C. Baicu and Michael J. Taylor, "Acid–base buffering in organ preservation solutions as a function of temperature: new parameters for comparing buffer capacity and efficiency," Cryobiology 2002, 45(1), 33-48 (DOI: https://doi.org/10.1016/S0011-2240(02)00104-9).
Dayle H. Geroski, Roberts Grosserode, and Henry F. Edelhauser, "A Comparison of HEPES and Bicarbonate Buffered Intraocular Irrigating Solutions: Effects on Endothelial Function in Human and Rabbit Corneas," Journal of Toxicology: Cutaneous and Ocular Toxicology 1982-83, 1(4), 299-309 (DOI: https://doi.org/10.3109/15569528209079878)(PDF).


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Find atomic radius from angle, wavelength, (311) plane and FCC structure
I've worked this problem and seem to be off by a factor of $2$ somehow.  From Callister, 6th edition, problem 3.56W (but I don't have access to the "W" web material that actually explains Bragg diffraction).  Using other texts, I found that plane spacing $d$ satisfies
$$\frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2},$$
which for the $(311)$ plane in an FCC lattice (therefore cubic) gives $a = \sqrt{11}d$.
The problem gives $\theta = 36.12^\circ$ and $\lambda = \pu{0.0711 mm}$.  Bragg's Law says
$$n\lambda = 2 d \sin \theta,$$
so for first order diffraction (the problem says reflection but I assume that is a mistake) $n=1$ and I get $d = \pu{0.0603 mm}$, which gives $a = \pu{0.2000 mm}$ and using FCC, $r = \pu{0.0707 nm}$, which is about half the correct answer according to Wikipedia $(\pu{142\pm 7 pm}$).
Can anyone tell me where I went wrong?

Nobody answered this and I figured it out, so I'll explain here.
It turns out that glancing angle is $\theta$, the angle the incoming beam makes with the surface (not the normal to the surface, as is usual in optics).
Meanwhile the diffraction angle is $2\theta$, the total amount the incoming beam gets "turned" to become the outgoing beam.
The problem stated the diffraction angles.  They need to be divided by $2$ before being plugged into the Bragg Law.

As mentioned by RobertTheTutor, the angle the incoming beam makes with the surface is $\theta$, therefore the diffraction angle is $2\theta$ (the total angle between the incoming and outgoing beam).
So the angle given of 36.12° needs to be divided by 2 before being plugged into the Bragg's Law.
$$2\theta = 36.12^\circ \rightarrow  \theta = 18.06^\circ$$
The interplanar spacing for the (311) is calculated,
$$d_{311}= \frac{n \lambda}{2 \sin \theta} = \frac{1 \times \pu{0.0707 nm}}{2 \times \sin(18.06^\circ)} = \pu{0.1147 nm}$$
For a cubic lattice,
$$ a = d \sqrt{h^2+k^2+l^2})$$
thus for the (311),
$$a = 0.1147 \sqrt{3^2+1^2+1^2} = 0.1147 \sqrt{11} = \pu{0.3804 nm}$$
Finally, using Pythagorean theorem to relate the atomic radius, $r$, to lattice parameter, $a$, for the face centered cubic structure:
For the front face,
$$a^2 + a^2 = (4r)^2 \rightarrow \sqrt{2} a = 4 r$$
So,
$$r = \frac{\sqrt{2}a}{4} = \frac{\sqrt{2}\times \pu{0.3804 nm}}{4} = \pu{0.1345 nm}$$


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Hydrogen Peroxide & Vinegar Film reversal, can someone help me understand this reaction?
I'm interested in DIY photo development and I came across a reversal process that uses chemicals easily available at the grocery store. The purpose of a reversal process is to take a partially developed negative, which is a suspension of silver and silver bromide in gelatin, and turn it into a positive by removing the silver and leaving the silver bromide behind.
The two in question are 3% hydrogen peroxide and 5% acetic acid. I have tried this and it works but I don't understand why.
The process is as follows:

30ml of 5% acetic acid solution is added to 500ml of 3% hydrogen
peroxide, the rest is just water.
The mixture is then heated to 60°C.
In complete darkness the partially developed film is placed in the
liquid for 6-10 minutes.
After this the silver is gone, but a yellow tint is left behind.
After this the silver bromide is then exposed and redeveloped leaving a positive
image.

I believe the tint is silver oxide but I am unsure as I don't know enough about what is actually taking place in this reaction. I'm left with all sorts of questions:

What reaction is taking place to remove the silver?
What other reactions might be happening?
What is the fate of the removed silver after this reaction?
Why does it need to be hot?
Why is there a yellow tint left behind? Is it silver oxide or
acetate, and how might it be removed/avoided?

The slow reaction dissolving silver must be : $$\ce{2Ag + H2O2 + 2CH3COOH -> 2CH3COOAg + 2 H2O}$$ The mixture has to be hot because the rection is extremely slow. Its rate is negligible at room temperature. Even at $60$°C, about $10$ minutes are required to dissolve nanoparticles of silver  on the milligram scale. Silver acetate $\ce{CH3COOAg}$ is moderately soluble in water ($\pu{10 g/L}$).

The acetic acid stops the developer, and the hydrogen peroxide oxidizes the silver particles that were selectively reduced by the developer. More detail in this thread enter link description here. Probably needs to be hot just to force an acceptable reaction rate from the peroxide, or to get any reaction at all. The yellow tint is likely just the natural color of the silver halide in the original film. I doubt the tint can be removed, but the remaining silver halide, having been so treated in the dark, should still be light sensitive, so you may try exposing it to light, then developing normally to see if the silver halide gets reduced to silver, leaving the normal black.
Using a fixer would probably complex & remove the silver halide, leaving blank film.
If you want a positive, you could develop the negative film normally, then "contact print" to a fresh piece of negative film, which you then develop normally.

Thanks, but I don't think the yellow is Silver Halide. The image you see above is actually after re-exposure & redevelopment and then fixed to remove any possible residual halides. The yellow is actually in areas where the silver has been removed, and where the halides have been converted by the first developer. This leads me to believe that the yellow color comes from a compound produced during the bleaching process.

Don't you think that this yellow color may be due to the gelatin or the film itself getting old and somewhat oxidized without any silver compound involved ?


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Why is it technically inaccurate to say that the decrease in reactivity of halogens is due to decreasing electronegativity?
I came across the following information in this post.

Below the infographic there is a paragraph with a disclaimer:

As another disclaimer, the reactivity of the halogens decreasing down the group isn’t actually due to the fact that it gets harder to add an electron. However, this is the explanation peddled by most exam boards up to A Level in England, so despite it being a technically inaccurate explanation, it’s the one that makes it on to the graphic.

However, I cannot find relevant information to support this. If possible, please clarify the alternate explanation if the disclaimer is true.
My understanding on reactivity for chemical nature here is using electronegativity as explanation. The statement given above contradicts my understanding. I know that halogens show anomaly in trend in terms of electron affinity and bond energy. Are these two the correct explanations? However, this anomaly begs other explanations which are inter electronic repulsions and pd hybridization. Perhaps, other factors play here, so I am asking for answers. Thank you very much.

Well fluorine is more reactive than chlorine,but as you say fluorine shouldn't be more electronegative than chlorine? Halogens get less reactive as the get heavier,and more electropositive.

I think the answer depends on which system you are looking for. For example, if you are looking at aqueous phase reactions, then fluroine suddenly becomes less reactive, because it is highly solvated by water. If you are asking about metal complexes, the fluorine is a strong pi-donor but weak sigma-donor, so it really depends on the d-electron count, and the bonding properties of other ligands.

It doesn't get harder to add an electron to a halogen atom as you go down the group! It just gets less easy (Ref 1). Big distinction, since the addition of an electron to the radical is exothermic. However, other than this imprecision in the graphic, it is generally true that reactivity drops as you go down the column, as electronegativity does also, so there is a correlation. The electron orbit gets farther from the nucleus as the atom gets bigger, heavier; the electron has less to gain from bonding. Generally, except that there is the lower electron affinity of F compared to Cl.

I believe Pandora's box has been opened. First to fly out was the electron affinity imprecision already mentioned. Next is the vagueness of electronegativity. How is that measured? With an electronegativity meter? No, it's guestimated from a bunch of data that someone selects, and several other people have come up with useful guestimates, but electronegativity is still a fuzzy numerical measure. And I think the results are smoothed out as much as possible. Isn't that an exceptionally smooth decline? (also Ref 1)

Next to fly out is reactivity: what is reactivity? Perhaps the total energy released in spontaneous reaction(s)? Obviously, you want to compare reactivity - but how completely? Would you require data for all possible reactions of each halogen with all other reagents and then average the data? Probably not, just  get a "more than" or "less than" comparison for a few reactions of the halogens.
The hydrogen-halogen and carbon-halogen bond energies do drop smoothly down the column, as does the electronegativity (the columns of data are for F, Cl, Br and I from left to right) (Ref 2): And the last line is another set of electronegativity numbers, from a different set of data points.

You might wonder why the reactivity of one chlorine radical for another (the Cl-Cl bond energy) is so much greater than the reactivity of any other halogen radical for its twin; the Cl-Cl bond energy is highest of all the halogens (top line)! Adding that last electron into the chlorine (the electron affinity) is definitely preferred over addition to any other halogen, so when you do it twice in one bond, I guess you get a boost.
The problem is that a few words can oversimplify a picture that is itself worth 1000 words, but a few numbers can clarify the situation. Sometimes I think chemistry is too complicated to describe briefly - and near-impossible without numbers.
Ref 1. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/0Group_17%3A_Physical_Properties_of_the_Halogens/Group_17%3A_General_Properties_of_Halogens
Ref 2. http://wwwchem.uwimona.edu.jm/courses/CHEM3101/Halogens2.html


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Hydrogen permeation from low to high absolute pressure area due to difference in partial pressure?
I need to get help with a question regarding hydrogen permeation which I do not really how to get around.
The statement is as following:
A steel container contains helium gas at high pressure (100 bara). Surrounding it is hydrogen gas at a lower pressure (approx. 3 bara) and at a great volume. Hydrogen can permeate the steel walls of the helium container but helium is impermeable through steel. After infinite time, will the pressure increase, decrease or stay the same within the helium container?
I am a bit puzzled about this. There is a concentration gradient and a gradient in partial pressure of hydrogen, but on the other hand, there is a much higher pressure inside the helium container. Is permeation driven by concentration gradient or "mechanical" flux/mass flow?
Hope someone got an easy answer for me, because this is keeping me up at night. Thank you all in advance!

The He cannot escape, so the pressure cannot go down, so it is either constant or increases. The partial pressure of H2 is initially zero inside so diffusion will occur until the H2 pressures are equal. Consider how could the He prevent the H2 from diffusing.

Porphyrin is right. But does it make sense to make a statement that steel is permeable to hydrogen and not to helium ? Both have about the same dimension. I would have been less dubious if the inner gas was $\ce{CO2}$ or $\ce{CH4}$

Hydrogen can permeate a steel barrier, which helium cannot. As hydrogen can move as free ions through the lattice structure.

Related to a manufacturer who filled tennis balls with sulfur hexafluoride, so the balls would stay “fresh” on the store shelves. Then no need for pressurized tennis ball cans, so less expense. But air diffused into the balls faster than the sulfur hexafluoride diffused out of them, so the pressure kept increasing and tennis balls randomly exploded (“loudly burst”) on the store shelves.

@Ed V, great comment really made me laugh!


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In the mechanism of Cannizzaro reaction, how do we rationalize hydrogen acting as leaving group?

Mechanism of Cannizzaro from wikipedia

I am trying to understand how to rationalize in the first step of third row that the hydrogen act as a leaving group and a nucleophile. After googling, I found this quora post, from which the only notable information I gathered was this following paragraph

It is unlikely that the hydride ion being transferred in step two is actually a discrete separated hydride ion H- floating around in solution. Rather it is just an H- sliding across from one carbon atom to another, without every becoming completely detached. This will require a high concentration of the starting materials so that they are close to each other.

I can understand the above, but still how would one rationalize hydrogen , a poor leaving group, acting as a leaving group in the above scheme?

This paper asked the same question as well :D https://pubs.rsc.org/en/content/articlelanding/1985/p2/p29850000717#!divAbstract

https://chemistry.stackexchange.com/questions/25202/cannizaro-reaction-hydride-shift-reason


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Is sulphonamide group prone to undergo diazotization?
I wonder if it is not a problem that there is another $\ce{-NH2}$ group in the molecule not connected to aromatic ring and if it can get protonated and react with $\ce{HNO2}$ too, in this case the one sulphonamide group is containing. Is the first reaction below correct?
Should it be essential not to have such group to have this reaction nicely going? - I mean getting only aromatic amine undergo diazotization.

Sulphonamide is very different then amine - won't get protonated, much less nitrosated.

The -NH2 of the sulfonamide group is more acidic than basic so it is not going to protonate in the way you have drawn.

So it won't protonate even if there is great excess of acid? I guess it will in THIS case. I mean in excess

Think about how electron-withdrawing the sulfonyl group is.

Ok, there is no problem because -NH2 from sulfonamid won't get easily protonated, so it won't react with HNO2


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Entropy change during electrolysis of water
Consider the electrolysis of water reaction:
$$\ce{H_2O -> H_2 + 1/2O_2}$$
At room temperature and atmospheric pressure, the change in enthalpy of this reaction is $\Delta H = \pu{+286 kJ mol-1}$. Now at constant pressure, $\Delta H = Q + W_\text{other}$, so in order for this reaction to occur, at the bare minimum, we must provide an amount of energy equal to $\pu{286 kJ mol-1}$. The question then becomes, of the $\pu{286 kJ mol-1}$ required for this reaction to occur, how much can we supply as heat, and how much can we supply as work such that the 2nd law isn't violated.
Now according to standard reference tables (all values being per mol):
$$S_\ce{H2O} = \pu{70 J/K} \qquad S_\ce{H2} = \pu{131 J/K} \qquad  S_\ce{O2} = \pu{205 J/K}$$
So if this reaction occurs, the increase in the systems entropy according to these values will be:
$$\Delta S_\text{sys} = (131 + \frac{1}{2}\cdot 205-70) = \pu{+163 J/K}$$
My textbook (Schroeder's Introduction to thermal physics) then says that because $\Delta S_\text{sys} = \pu{+163 J/K}$: "the maximum amount of heat that can enter the system is therefore $T\Delta S = (\pu{298 K})(\pu{163 J/K}) = \pu{49 kJ}$". This fact is bugging me a bit. I can see that because the systems entropy is increasing by $\Delta S_\text{sys} = \pu{+163 J/K}$, we can afford for the entropy of the surroundings $(S_\text{surr})$ to decrease by a maximum of $\Delta S_\text{surr} = \pu{-163 J/K}$ without violating the requirement $\Delta S_\text{universe} \gt 0$. Now since the entropy of the surroundings is reduced when heat is provided to the system $(\Delta S_\text{surr} = -\frac{Q_\text{to sys}}{T})$, the maximum heat the surroundings can supply is simply the aforementioned $T\Delta S_\text{surr} = (\pu{298 K})(\pu{163 J/K}) = \pu{49 kJ}$. We can always have the surroundings supply an amount of heat less than $\pu{49 kJ}$ provided we increase $W_\text{other}$ but we may never have the surroundings supply heat more than $\pu{49 kJ}$.
My problem is the fact that when an amount of heat $Q \lt \pu{49 kJ}$ is supplied to the system, we have $\Delta S_\text{surr} \gt \pu{-163 J/K}$. But we also have that $\Delta S_\text{sys} \lt \pu{163 J/K}$ because $\Delta S_\text{sys} = \frac{Q_\text{to sys}}{T}$. But this change in the systems entropy is clearly less than the required change of $\Delta S_\text{sys} = \pu{+163 J/K}$ given by the tables. So surely it is the case that the amount of heat supplied by the surroundings must always be equal to $Q = \pu{49 kJ}$, for if $Q \gt \pu{49 kJ}$, then we violate the second law, but if $Q \gt \pu{49 kJ}$ then we fail to create enough entropy for the system to meet its entropic requirement of $\Delta S_\text{sys} = \pu{+163 J/K}$.
What am I missing here? Is entropy created within the system by other means in just the right amount? If so, how?

I would suggest to make the Q smaller , simpler. Otherwise , no one would be interested to read the Q. Also , make speakers for paragraphs

As you truly mentioned, Q cannot be more than 49 KJ. Q=49 KJ happens when the system works under reversible conditions. Q<49 KJ happens when system is not working reversibly (i.e., the reaction rate is fast and not enough heat can be supplied by the surrounding). In such a condition, some entropy is generated because of irreversibility which adds to thermal entropy.

Do you know why $\Delta S^0\neq\frac{\Delta H^0}{T}$ for a reaction of ideal gas species?

$2^{nd}$ law will not be violated anyways.

$\Delta S_\text{sys} = \frac{Q_\text{to sys}}{T}$ Is incorrect. Entropy is a state function, so it is independent on the path that determines work and heat exchange. The surrounding entropy change is related to the heat exchange because nothing else happens in the surrounding.

The amount of heat that must be added is $Q=T\Delta S^0=49$ kJ.  The amount of electrical work that must be done on the system is $-\Delta G^0=-\Delta H^0+T\Delta S^0=286-49=237\ kJ$.  The real question for you is "why isn't Q equal to $\Delta H^0?"$

The answers to all your questions can be provided by using the open-system (control volume) version of the 1st and 2nd laws of thermodynamics to analyze a specific continuous flow system featuring negligible irreversibility.  The control volume contains an ideal Van't Hoff equilibrium box reactor, followed by ideal isothermal flow compressors to compress the pure hydrogen and pure oxygen streams coming out of the reactor.
The reactor receives a pure feed stream of liquid water in the standard state, and discharges pure oxygen and hydrogen through semipermeable membranes; the pressures of the pure oxygen and hydrogen exiting the reactor through the membranes are the same as the partial pressures within the reactor; so they are essentially at equilibrium with the gas mixture in the reactor.  The reactor is operating at steady state, so the amounts of pure oxygen and hydrogen streams exiting the reactor correspond stoichiometrically to the amounts present in the inlet water stream.
Once the pure oxygen and hydrogen streams exit the reactor (at pressures << 1bar), they are fed to isothermal continuous flow compresors which raise their pressure up to the stand state pressure of 1 bar.  Since the compressors are operating isothermally, the enthalpy change of these streams in the compressors is zero; therefore, all the enthalpy change of the overall combined process takes place in the reactor.
ANALYSIS OF REACTOR
From the open system version of the 1st law applied to the reactor, we have:  $$0=Q_R-W_{SR}-\Delta H_R=0$$or $$Q_R-W_{SR}=\Delta H=+\pu{286 kJ/mol}$$where $W_{SR}$ is the shaft work done by the reactor on the surroundings.
From the open system version of the 2nd law applied to the reactor, we have $$Q_R=T\Delta S_R=\pu{298 K}\cdot \Delta S_R$$If we combine these two equations, we obtain $$\Delta G_R=-W_{SR}=\Delta  H_R-T\Delta S_R=0$$where $\Delta G_R=-W_{SR}=0$ because the reactor is operating at quasi-static  equilibrium.  Therefore, we have $$Q_R=\Delta H=\pu{285 kJ/mol}=T\Delta S_R$$$$\Delta S_R=\frac{\Delta H_R}{298}=\pu{959 J mol-1 K-1}$$
ANALYSIS OF THE ISOTHERMAL FLOW COMPRESSORS
From the open system version of the 1st law applied to the isothermal;, we have:  $$0=Q_C-W_{C}-\Delta H_C=0$$Since the compressors are operating isothermally, $\Delta H_C=0$ or $$Q_C=W_{SC}$$ where $W_{SC}$ is the shaft work done by the compressors on the surroundings.
From the open system version of the 2nd law applied to the compressors, we have $$Q_C=T\Delta S_C=\pu{298 K}\cdot \Delta S_C$$
Since the Gibbs free energy change of the reactor is zero and the Gibbs free energy change of the overall process in the control volume is 237 kJ/mol, the Gibbs free energy change of the compressors is 237 kJ/mol.  It therefore follows that $$-T\Delta S_C=\pu{237 kJ/mol}$$ $$\Delta S_C=\pu{-795 J mol-1 K-1}$$and$$Q_C=\pu{-237 kJ/mol}$$
RESULTS FOR OVERALL COMBINED OPERATION
$$\Delta H=\Delta H_R+\Delta H_C=(286+0) \pu{kJ/mol}=\pu{286 kJ/mol}$$
$$Q=Q_R+Q_C=(286 -237)\pu{kJ/mol} =\pu{49 kJ/mol }$$
$$W_S=W_{SR}-W_(SC)=(0-237) \pu{kJ/mol}=\pu{-237 kJ/mol}$$
$$\Delta S=\Delta S_R+\Delta S_C=(959-795) \pu{J mol-1 K-1}=\pu{164 J mol-1 K-1}$$
$$\Delta G=\Delta G_R+\Delta G_C=(0+237) \pu{kJ/mol}=\pu{237 kJ/mol}$$


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What is the radial probability distribution function and what is its significance?
Below I have given three graphs for the $\ce{1s}$ orbital.
$R(r)$ is the radial part of the wave function of the electron. $R^2(r)$ is the radial part of the wave function of the electron multiplied by its complex conjugate which gives us the probability of finding the electron at some point. $R^2(r)4πr^{2}dr$ is what is called the 'radial probability distribution function'. It gives us the probability of finding the electron in a thin spherical shell of thickness $dr$.
Clearly from the second graph we can see that as $r$ increases  $R^2(r)$ decreases which means that as we move away from the nucleus, the probability of finding the electron decreases. Now, let us look at the third graph. It is telling us that as we move away from the nucleus, the probability of finding the electron in a thin spherical shell of thickness $dr$ increases and it reaches a maximum value at some $r$ which we will call $r_{max}$. $r_{max}$ is the distance from the nucleus at which the probability of finding the electron is maximum. Now I have a couple of questions.
From the second graph, we know for sure that as we move away from the nucleus the probability of finding the electron decreases. But what the third graph is telling us is the probability of finding the electron increases and reaches a maximum value. But how is this possible? We just saw that the probability decreases with increase in distance from the nucleus. How can the probability attain a maximum value at some $r$ if it is decreasing with $r$? Aren't the two graphs contradictory to each other? Moreover it is a $\ce{1s}$ orbital and so by the electron density diagram we can tell for sure that the probability of finding the electron decreases with increase in distance from the nucleus. As it turns out, $r_{max}$ is equal to $0.529×10^{-10} m$. It means that at this distance from the nucleus probability of finding the electron is maximum. But we clearly know that as we move away from the nucleus the probability of finding the electron decreases. So how can the probability of finding the electron be maximum at $r_{max}$? The probability of finding the electron should be maximum just outside the nucleus and not at $r_{max}$, right?

Be aware that $R^2(r)$ is probability density function, with probability $P_{r_1,r_2}=\int_{V}^{}{R^2(r) \cdot \mathrm{d}V}=\int_{r_1}^{r_2}{R^2(r) \cdot 4 \pi r^2 \cdot \mathrm{d}r}$ Therefore, even if probability density has maximum for r=0, probability for r,r+dr for r going to 0 converges to 0 as well.

@Poutnik, but the probability of finding the electron should be maximum just outside the nucleus and not at $r_{max}$, right?

What case has higher probability ? probability density=100 and volume = 1, or probability density 1 and volume = 1000 ? Maximal density for r=0 does not help if sphere surface converges  to 0 faster than the desnity rises ?

Calling two different graphs the same name ("probability") is a great way to get yourself confused, especially when you know for sure that they are different, and moreover, how exactly are they different.

Imagine a tango party with a large dance floor and a single porta-potty (one square meter floor area). There is a larger chance of finding people on the dance floor than in the restroom. On the other hand, there might be a larger chance of finding someone in the restroom than on a specific square meter area on the dance floor.
In other words, the two graphs address two different questions. The second shows you how likely it is to find an electron in a box of given volume closer or further from the nucleus. Perhaps surprisingly, the highest probability for the electron in a hydrogen atoms ground state is right at the nucleus. The third graph shows how likely it is to find an electron at a given distance. Because the volume available at a given distance increases with the square of the distance, you get a different shape of the curve. The third curve is relevant for calculating the mean or the average distance of the electron from the nucleus (or a surface within the electron is located inside with a probability of 90%).
Below are all three functions in the same graph (the y-axis is arbitrary). You can see that the radial part of the wavefunction (green) is different from its square (purple), which again is different from the radial probability density (orange). The x-axis should read $r / a_0$.

(source)

Nice answer! But I have a question; (1)So, mathematically it does make sense why radial probability is maximum at a certain distance for (say) 1s orbital. But is there a physical explanation for this, shouldn't the electron be more stabilised just outside the nucleus?

Yes @Rishi even I have the same question.

@Rishi The potential is strongest near the nucleus, this is why the density is highest near the nucleus. The population density in New York State is highest in New York City (Manhattan, probably), but a majority of people live outside of the city. The average distance of people to New York Times Square places them outside of Manhattan as well. The conceptual hiccup comes with the math, not the physics or the chemistry.

Brilliant analogy @Karsten Theis! I have one last question : If someone asks where is the probability of finding the electron in a $\ce{1s}$ orbital maximum - at a distance of $0.529 × 10^{-10} m$ $(r_{max})$ from the nucleus or just outside the nucleus? Then what would we answer? It would mean the world to me if you could answer the above question.

@user281837 It depends on the nuances in the question. If they ask "at what distance from the nucleus would I observe the electron, I would say at around $r_{max}$. If they are interested in scattering by electrons (like in crystallography), I would say the closer to the nucleus, the more scattering because the electron density gets higher as you approach the nucleus. Of course, that is specifically for the 1s, where we don't have any nodes..


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buffer "recipe" needed
I have a question, that I hope can be answered without the respondant showing off how much they know about chemistry and talking in "words of art". It's been 50 years since I tool chemistry in college, and an answer in grams, rather than moles, etc, would be appreciated. In that regard, here's the question. Starting with 4 liters of an aqueous solution of water plus 200ppm of hypochlorous acid, with a pH of between 6 & 6.2, what amounts, in grams, of NON-TOXIC buffers can be added to stabilize the pH in that range. Becoming more acid is not good for the skin, and becoming more alkaline will cause the HClO to become hypochlorite (in effect, bleach). A solution in which the buffers are solids that dissolve or precipitate out, to maintain a constant pH, would not be acceptable. Thanks, in advance, for your help.

Honestly, I stopped reading after two sentences.

A pH buffer at pH $6.2$ can be made in $4$ Liters water, by dissolving $\ce{104.8 g KH2PO4}$ plus $\ce{5.44 g NaOH}$. Adding $\ce{NaOH}$ will simply transform part of the dihydrogenophosphate ion into monohydrogenophosphate ion. And these two ions are not toxic.


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Combustion and oxidation
What is the difference between combusting and oxidizing? Oxidation of alcohols leads to formation of aldehydes and carboxylic acid whereas the combustion of alcohols leads to the formation of carbon dioxide and water. What is going on if oxygen is reacting with the same molecule?

The same reagents may give different products at different conditions. Surely you must have heard this before.

Apart from the fact that not all combustion involves oxygen (virtually everything combusts in chlorine trifluoride and oxygen doesn't come into it) the major difference is how far the reaction goes.
In normal combustion (involving oxygen) the reaction is uncontrolled and tends to go mostly to completion (that is most things end up in their most thermodynamically stable state: hydrocarbons are fully reacted to water and carbon dioxide).
When chemists talk of other oxidation reactions they tend to talk of much more specific, much more controlled reactions. So specific conditions are used to convert an alcohol to an aldehyde or an aldehyde to a carboxylic acid. These are "oxidations", but the chemist uses conditions to control the reaction so they get a specific product.
Combustion is oxidation, but uncontrolled. Most oxidations that chemists talk about are far more specific and controlled. That's the difference.

Fire is a kind of combustion and most of fires are controlled.

@Poutnik You are using "controlled" in a very different sense to me. From a chemical perspective fire is very uncontrolled. Sure, most coal fires in your house don't burn down your house, but that doesn't mean the fire is as "controlled" as an alcohol to aldehyde conversion in a chemist's reaction vessel.

Well, oxidation is a superset to combustion, it is not a non-overlapping different set. // Combustion still can be controlled and partial, as well as non combustive oxidation.

Combustion versus oxidation is not about what products are created, but about the way how they are produced.
Combustion is the kind of oxidation, usually using gaseous oxidant, sufficiently exothermic, with a sufficiently high reaction rate. Characteristic combustion behavious is high temperature raise with thermal IR and light emission, frequently with flame, if combustible gases are present/created.
Both combustive and non combustive oxidation may or may not involve reaction with oxygen.
E.g. during fermentation of wine to vinegar, ethanol is not combusted, but is enzymatically oxidized. No flame, no light, no significant temperature raise. Ascorbic acid and other antioxidants in food are progressively oxidized, not combusted.
In ethanol burners, ethanol is combusted, forming heat, light and flame.


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Is diethyl and methyl preferred over an ethyl and an isopropyl group as substituents when both give the same longest chain?

According to what I took (look for the longest chain, start close to the branches, etc etc) I see two paths I could follow to give me a hexane.
One where you go through the carbon in the methyl, and the other by going through two carbons in the ethyl.
In one case I'm left with a diethyl and a methyl and in the other I'm left with an isopropyl and one ethyl. Which of these is correct?

In such cases (when 2 or more possible routes give the same number of carbons in the parent chain) we look for which of the possibilities has the highest branching. The one with the highest branching is considered as the parent chain. In your case, the chain you must consider as the parent chain is the one having diethyl and methyl as its branches.

Thank you! Must've missed that somewhere. any good simplified sources on IUPAC naming rules?

Taken from the answer to In Parent chain choosing should one consider longest chain or chain with more substituents? by Loong.
From the Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)

P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

Therefore in this case, the diethyl and methyl make up 3 substituent groups whereas the ethyl and isopropyl group only make up 2 substituents. Therefore the chain that has the diethyl and methyl group needs to be taken as the parent chain here.

Therefore, the PIN of this compound is 3,3-diethyl-2-methylhexane.
As a side note: A brief set of the IUPAC rules can be found here


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Is equilibrium point for a coupled equilibria of arbitary number of coupled reactions is a unique point?
I was solving a question of salt hydrolysis for a mixture of Weak Acid-Base Salts, given the initial amount of salt and Equilibrium constants. I was trying to approach the problem in two ways, Assuming initially that we have all reactants and no products and move the reaction forward OR assuming all the reactants have been converted into product instantaneously and moving the reaction backward for equilibrium.
As concentrations of various species in equilibrium depends on equilibrium constant powers of their coefficients, which does not seem like any linear or straightforward relation. I wonder if both approaches are equivalent.
I.e. What guarantees the uniqueness of an equilibrium point (that is, specific concentrations of various species at eq.), given just the specific values of initial variables (initial concentrations)?
The paper https://aip.scitation.org/doi/10.1063/1.1669753 mentions something like that but the math went over my head. I wonder if there is any simple explanation based on the first principles of thermodynamics, like the curve for free energy for the whole system, does have some restrictions (it should  be concave or something like that).

Both approaches are equivalent if you do not apply any simplifying assumption. E.g. the notorious formula pH=0.5*(pKa - log c ) for weak acid pH applies 2 such assumptions ( c >> [H+] and [H+] >> [OH-]).

My question is more on theoretical side...how can we prove that both approaches are equivalent.

Such task is rather for mathematicians. Ask yourself, does the solution of the given set of equations for tthe given set of variables depend on the way how you solve it ? The existence of multiple solutions and the proof of it is matter on mathematics. Fortunately, usually just one of solutions makes scientific sense, e.g. concentrations cannot be negative.

Usually, resolving the equation set leads to a polynomial function f([H+])=0 and only one value makes sense. Then, given [H+] leads to unique concentrations of various compounds forms.

One of the first experimental demonstrations that the same equilibrium point is reached if the reaction starts with all reactants or all products was $\ce{2HI<=>I_2 + H_2}$ made by Bodenstein in $\approx 1894$. I have redrawn the data.

figure.  $\ce{ 2HI=I_2 + H_2}$. $x$ is the fraction of HI at $721$ K. $x=[HI]/([HI]+2[I_2])$

In the reaction $\ce{A<=>B}$ the equilibrium constant is determined by the ratio of rate constants forward and back, $k_1,\;k_{-1}$ respectively. If another reaction is coupled to the first one $\ce{B<=>C}$ then this has a different equilibrium constant with rate constants $k_2,\;k_{-2}$ but these are separate reactions and so the rate constants and hence equilibrium constants are unchanged when A, B, C are together in solution.  The concentrations are changed, of course, because when C is added it changes the concentration of B and hence A.
From a thermodynamic perspective the free energy determines the equilibrium constant and adding C does not change the enthalpy or entropy leading to reaction between A and B.
The difficulty is to show (as in your reference) that this is true for all combinations of coupled reactions, linear as well as those forming rings e.g. A-B-C-D-A etc. It is the case, however, that only one set of concentrations is found otherwise the system of reactions could flip between one and the other and a perpetual motion machine could be constructed which is impossible.


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Why is the bond angle of trifluoromethane higher than that of difluoromethane?
The bond angle $\ce{\angle(F-C-F)}$ of $\ce{CHF_3 > CH_2F_2}$ whereas that of $\ce{\angle(Cl-C-Cl)}$ is $\ce{CH_2Cl_2 > CHCl_3}$ why is this the case? Some explanations I read suggest the Bent's rule but I am not able to understand the idea well.

Where did you see this result?

"Wiley Physical Chemistry for JEE"

@Ashish This may help. https://chemistry.stackexchange.com/questions/152367/why-does-a-larger-bond-angle-correspond-to-greater-s-character

According to NIST CCCBDB, the $\angle(\ce{F-C-F})$ bond angles of $\ce{CH2F2}$ and $\ce{CHF3}$ are ${108.421 ^\circ}$ and $108.099 ^\circ$ respectively as calculated using CCSD=FULL/aug-cc-pVTZ.
Similarly, the $\angle(\ce{Cl-C-Cl})$ bond angles of $\ce{CH2Cl2}$ and $\ce{CHCl3}$ are ${112.329 ^\circ}$ and $109.992 ^\circ$ respectively as calculated using CCSD/aug-cc-pVTZ.
Both show a similar trend for bond angles unlike as mentioned in the question.
Now, this trend can be explained using Bent's rule which states:

Atomic s character concentrates in orbitals directed toward electropositive substituents

As the number of electronegative substituents increases, the p-character in each bonding orbital between $\ce{C-Cl}$ increases. This increase in p character leads to a decrease of the bond angle as observed here.
There is a similar decrease for $\ce{C-F}$ as seen above. The reason for the chloromethanes having a higher bond angle is due to steric effects. Chlorine has an A-value of 0.43 where fluorine has an A-value of 0.15. Therefore, the steric effect is more pronounced in the chloromethanes.

The book mentions $108^\circ6'$ for CHF3 and $108^\circ3'$ for CH2F2 whereas it is $111^\circ3'$ for CHCL3 and $111^\circ8'$ for CH2CL2 which is just so off that given on the website. Not sure where they might have taken the data from

@Ashish Convert the degrees into decimal or vice versa and you will find that both are nothing but equal.


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Does vinyl chloride deactivate the benzene ring?
I came across the following option of a multiple choice question (The options are in the form of assertion and reasoning side-by-side and the incorrectly matched statements are to be marked):

According to the answer key, this is a correct option (i.e., an incorrect relation). Now this is only possible if either the first(left-side) statement or second(right-side) statement, or both statements are wrong. If a group is activating, it will be ortho-para directing. So, this means that (based on the answer key) vinyl chloride is a deactivating group (and the given styrene derivative is meta-directing). But I think it should be an activating (precisely, weakly activating) group due to hyperconjugation with the benzene ring ("Weakly" activating because there is only one α-H available and it belongs to an alkene).
So, could anyone please explain why is vinyl chloride a deactivating group?

There is a double bond, so there is conjugation, no need to complicate matters by bringing in hyperconjugation, whose effect is much weaker than conjugation, and does not apply in this molecule in any case. Vinyl chloride is a deactivating group because the Cl is electron withdrawing. Draw the resonance forms from conjugation of the benzene ring with the vinyl double bond, and you will see that some resonance forms where the vinyl double bond takes away electrons from the benzene, are stabilized by Cl, so contribution of those forms are higher.

@ShoubhikRMaiti I understood it now, thanks a lot for your explanation

@Pal If you got your answer, feel free to self-answer it. Self-answers are always appreciated and it will also help future readers.

@ShoubhikRMaiti, you could post this as an answer and I would upvote :D

While it's possible, that it's deactivating, I wouldn't buy it too easily - some proof is needed to answer it properly.


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Which temperature would allow a molecule to vaporize at ultra-low pressure?
I must vaporize a molecule with a normal boiling point of 320°C. Now I work with ultra-high vacuum and I need to bring my molecule to the gas phase. Has anyone a rough idea how high the temperature has to be, to bring my molecule to the gas phase with a pressure of $\pu{5E-8 mbar}$.

The vapor pressure increases with temperature as described by the Clausius–Clapeyron Equation. One parameter you need to know is the heat of vaporization, which correlates with the strength of intermolecular interactions. Without knowing the type or size of the molecule, there is no way to make the estimate.

I am afraid there are no precise answer to this question. It only depends on the nature of the interaction between your molecule and the walls.

Have a look for the method called MALDI used in mass spectrometry to do what you want.

Molecules don't have boiling points, substances do. BTW melting point is more important - it's not like you need you boil something to vaporise.

Like @porphyrin said, please look at MALDI! The wikipedia article is here. Heating is only one way to get big molecules up and going!

This strikes me as a hwk problem. What are your attempts to solve this? KarstenTheis's approach is your safest bet from a simple theoretical point of view. Otherwise recommend experimenting and following other advice (MALDI).


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Limitations of Quantum Chemistry
I have a physics background and I am interested in the applications of quantum mechanics in predicting chemical reactions, and I’m wondering where should I start?
I am particularly interested in the Quantum chemistry of species dissolved in water. How could one compute the time evolution of a chemical system taking into account a carrier medium such as water?
Could one use quantum chemistry to predict the affinity of a certain compound of a particular chemical?
How could the existence of catalysts, temperature and pressure be taken into account when predicting reactions?
Or are these well beyond the scope of current quantum chemistry?

So... you're asking how big and complex system can be practically modelled with taking account of QM? Well, that quite depends.  Even that would be probably too broad, and you're asking more then that.

Since you are considering modeling the system you might want to post your question at Matter Modeling SE. It's all within the scope of quantum chemistry, though. However simulation of quantum systems is very computationally costly. This is the main limitation.


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What causes horizontal lines on an excitation-emission matrix in fluorometry?
I am doing excitation-emission matrices to test for CDOM, and we just got some brand new quarts cuvettes. When I hold them in front of a light, they're totally clear and clean.
I ran a sample, and in addition to the normal diagonal scattering lines, there are some intense horizontal lines in the lower left of the EEM diagram. We were getting these before with old cuvettes and figured they were just scratched up and that was the cause, but now that they're happening again I'm a little concerned. Is this a normal phenomena or something hardware or sample-specific?
The horizontal lines are part of the excitation signal (y axis) which should be independent of the sample. I was able to run a few samples today before it happened, though.

Just some speculation. Are you sure it is not Raman scattering? Those horizontal stripes stop before the Rayleigh line and the gaps get bigger as wavelength increases. Any way to have the software use or plot wavenumbers instead of wavelengths? Also, are the cuvettes Suprasil? The fact that you got several samples run before the problem suggests contamination.

I am not sure it's not Raman scattering, but since it doesn't always appear I'm inclined to think it's not. The cuvettes are quartz from Hellma; I'm runnign a scan now without the built-in correction factors in case the factory-supplied correction factors are the cause of the problem. Again, since the problem originates in the excitation, I'm assuming it's hardware-related.

@EdV I reran the same scan but without the "mask scatter" boxes checked. The horizontal lines had disappeared, but so had my data (the big white blob). Looking at this link here, there's no kind of scattering I should get that looks like what's occurring in my data. http://dreem.openfluor.org/tutorials/smoothing/.

Yeah, this does not make sense to me. The tutorial has nothing similar. The uniform intensities of those artifact bands makes me think the problem is in the software or hardware, but I have no clue how. Do you have a known standard that has worked well in the past?

@Edv I just started using this machine a couple of months ago, and so far there's no pattern to what gives this signature and what doesn't. The lamp in the machine is brand new as well. Running the same scan and sample without the corrected signals produces a profile that is simply a white to black vertical gradient, but not in the same way those stripes exist in the OP.

The fact that they’re perfectly flat suggests it’s a baseline correction issue or something like that in the software. Does it even collect emission data at wavelengths well below excitation? Looks like it might be just filling that part of the spectrum in with a flat line

It does look like a fault to me, in the spectrum you show, the signal is absolutely constant up to 350nm then goes down then rises with scattered or Raman light. It is just non-physical. Whether the fault is in software of mechanical only you can determine. Have you asked the manufacturer about this?

Not yet porphyrin, I'll ask my PI about it - last time we went to Horiba with a problem, they broke our machine even further.