text:
Fastest way to determine if molecule has resonance structures
I came across this old exam problem:

For which of the following choices, is every molecule a resonance hybrid with resonance structures?
A. $\ce{SO2},$ $\ce{NH3},$ benzene
B. $\ce{SO3^2-},$ $\ce{NH3},$ benzene
C. $\ce{SO2},$ $\ce{NO3-},$ benzene
D. $\ce{SO3^2-},$ $\ce{NO3-},$ benzene
E. $\ce{SO3},$ ethanol, benzene

So, in order to solve this problem I don't actually have to draw any Lewis structure and what not, I just have to determine which molecules have resonance structures. But what is the fastest way to do that?
How can I determine if a molecule has resonance without having to draw anything, if that's even possible? I hope someone can help me out.

In the end, you do have to draw the molecules.

Incidentally, the answer is C although old textbooks might give it as D :D

You don't have to do anything for benzene because it occurs in every choice.
You have to compare $\ce{SO2}$ and $\ce{SO3-}$ because they distinguish A) from B) and C) from D).
You have to compare $\ce{NH3}$ and $\ce{NO3-}$ because the distinguish A) and B) from C) and D).
If you use this strategy, you miss out on the chance to figure out  $\ce{SO3}$, ethanol and benzene, and you would learn less.

Concerning 4: not really. For example, it could be that e.g. both $\ce{NH3}$ and $\ce{NO3-}$ have none and it could be that more than one line is correct.

When you say "compare" do you indicate that I have to draw them? On a different note, are you danish? Your name seems danish.

@Jan Yes, I'm not saying it's a perfect strategy. In fact, I think it is a bad strategy for multiple reasons.

@Carl - I can visualize them in my head because I already drew them "a couple" of times. $\ce{NH3}$ has just one none-hydrogen atom, so any resonance would have to invoke something unusual. I don't understand the question - what does "one or more resonance structures" mean? How can you have resonance with a single structure?

You are right. I made a mistake copying the problem. I edited the question.


text:
Can carbon have electrons in a 5d orbital?

(a) Sketch the radial probability distribution for a 5d orbital in a carbon atom. You should label the axes, but do not need to include numbers. Use arrows to indicate the radial nodes.
(b) Label the most probable radius, $r_\mathrm{mp},$ on your 5d radial probability distribution with a *.

The question asks about a $\mathrm{5d}$ orbital in a carbon atom, but to my understanding a ground-state carbon atom should have a configuration of $(\mathrm{1s})^2 (\mathrm{2s})^2 (\mathrm{2p}_x)^1 (\mathrm{2p}_y)^1,$ and no electrons in a $\mathrm{5d}$ orbital.
I understand why there would be two radial nodes $(n - l - 1 = 5 - 2 - 1),$ but shouldn't that be irrelevant because a carbon atom doesn't have electrons in the 5d orbital? Or maybe I am missing something more fundamental and it is still possible for a ground-state carbon to have electrons in $\mathrm{5d}?$ Any suggestions are appreciated.

No one says it has to be ground state...

Empty orbitals are still orbitals, in meaning of the wave functions, allowed quantum states of electrons and respective electron occurence probabilities.
It is true that at the carbon atom ground state, quantum states with $n>2$ are not occupied. But it could happen 5d orbital is occupied in high temperature plasma or by non thermal high energy excitation ( like photons with energies little below the carbon ionization energy), as an electron in carbon atom or ion can get excited to 5d orbital.
There are known Rydberg atoms with very high principal quantum numbers, e.g. $\ce{H}$ atom with $n=137$ has the size about $\pu{0.001 mm}$ and $\ce{K}$ atom with $n=600$ has size almost $\pu{0.1 mm}$.

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text:
Second order reaction with different reactants

In a laser photolysis reaction $\ce{C6H5NH2}$ $(C_0 = \pu{60 µM})$ is oxidized to its radical cation. The disappearance of the cation can follow three pathways: (1) Dimerzing of the cation, (2) Reaction between the cation of the initial substance, (3) Reaction between the cation and the solvent. Decide which pathway is most likely to occur.
$$
\begin{array}{l|cccccccc}
\hline
t/\pu{µs}~(\text{efter puls}) & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\
\hline
[\text{Radikalkatjon}]/\pu{µM} & 20 & 11.12 & 6.86 & 4.47 & 3.00 & 2.06 & 1.43 & 1 \\
\hline
\end{array}
$$

I first checked for the dimerizing of the cation by assuming it to be a second order reaction with the same reactants and plotting $1/[\mathrm{cation}]$ against $t,$ which didn't give a good fit. So I wanted to plot it as a second order reaction with different reactants $(\ce{A + B -> products})$ but I don't understand how that is done.
I should plot $\displaystyle\ln\frac{a-x}{b-x}$ against $t,$ but what exactly is $(a-x)$ and $(b-x)?$
I know that I should set up a reaction like:
$$
\begin{array}{l|ccc}
\hline
& \text{A} & \text{B} & \text{product} \\
\hline
t=0 & a & b & 0 \\
t=t & a-x & b-x & x \\
\hline
\end{array}
$$
And I would assume that my $a = \pu{20 µM}$ and $b = \pu{60 µM}.$ But where do I go from there? All help is appreciated!

The general solution, if the reaction is $aA+bB \to product$, is to allow an amount $x$ to react at time $t$ then there will be concentrations $(A_0-ax)$ and $(B_0-bx)$ of species A and B respectively at time $t$ if $A_0,B_0$ are the concentrations at time zero.  The rate equation is $\displaystyle  -\frac{1}{a}\frac{d(A_0-ax)}{dt}=\frac{dx}{dt}=k(A_0-ax)(B_0-bx)$ which can be integrated ($0\to x, 0\to t)$ to give
$$ \ln\left(\frac{A_0-ax}{B_0-bx} \right)=k(A_0b-B_0a)t +\ln\left(\frac{A_0}{B_0} \right)$$
If the reaction follows this scheme plot the left hand side vs $t$ and this should produce a straight line of slope $k(A_0b-B_0a)$.
Should it happen that concentrations are initially $A_0/B_0 = a/b$ so that species are present in stoichiometric amounts then as $A_0b-B_0a=0$ the  solution above cannot be used. In this case the rate equation becomes $\displaystyle  \frac{dA}{dt}=-bkA^2$ which gives $\displaystyle \frac{1}{A}-\frac{1}{A_0}=bkt$.

Before jump into rate equation, I'd like to approach this in different direction.
A laser photolysis reaction of aniline, $\ce{C6H5-NH2}$ ($\ce{A}$), has created an $\ce{A^{.+}}$ radical cation, which has decayed with time to (unknown) products. No other additional data were given except for aniline concentration $(\ce{[A]} = \pu{60 \mu M})$. When plotted experimental values of $\ce{[A^{.+}]}$ against time $t$, you obtained a hyperbola ($xy = a$ type).  Thus, I plotted  $\ln \ce{[A^{.+}]}$ against time $t$ to see what happens. I got a very straightline with $R^2 = 0.993$ agreement:

Thus, you can conclude that the decay process follows first order kinetics:
$$\text{disappearance rate of } \ce{[A^{.+}]} = k\ce{[A^{.+}]}$$
That means:
$$-\frac{d\ce{[A^{.+}]}}{dt} = k\ce{[A^{.+}]}$$
$$\frac{d\ce{[A^{.+}]}}{\ce{[A^{.+}]}} = -kdt$$
Integrate both sides:
$$\int^\ce{A}_\ce{A_\circ}\frac{d\ce{[A^{.+}]}}{\ce{[A^{.+}]}} = -\int^t_0kdt$$
$$[\ln\ce{[A^{.+}]}]^\ce{A}_\ce{A_\circ} = -k[t]^t_0$$
$$\ln\ce{[A^{.+}]} = -kt + \ce{[A_\circ]} \tag1$$
The equation $(1)$ is a straight line. The equation of best trend line of the plot is:
$$y = -0.0839 x + 2.8439 \tag2$$
Hence, $k = \pu{0.0839 \mu s-1}$ and $\ln\ce{[A_\circ]} = 2.8439$ or $\ce{[A_\circ]} = e^{2.8439} = \pu{17.18 \mu M}$ (actual $\ce{[A_\circ]} = \pu{20 \mu M}$).
Based on this fact you may able to determine which path it takes to decay. Keep in mind that decay process only depends on the concentration of radical cation concentration.

@ Mathew Mahindaratne. Well. The alignment of the points ln A vs. t is not ideal. The points at the beginning (t = 0) and at the end (t = 35) are above the line. The points in the middle of the graph are under the line. It looks as if the order is higher than 1, but smaller than 2, as I state in my previous comment. It may be due to two simultaneous reactions, one 1st order and another one 2nd order.

A kinetics may be proven to be second-order when the initial concentrations of A and B are different, by using Frost and Pearson's treatment. In page $16$ of their manual, they say that the kinetics law is the following : $$\ce{\frac{1}{B_o - A_o}ln\frac{A_oB}{B_oA} = kt}$$ In order to simply prove that the reaction is second order, they recommend to simply plot log(B/A) against t.
Ref.: Arthur A. Frost, Ralph G. Pearson, Kinetics and Mechanism, A Study of Homogeneous Chemical Reactions, John Wiley & Sons, Inc. New York, 1965
The trouble with the numerical values is that the only known values are those of the radical cation $\ce{A}$. The corresponding values of $\ce{B (= C_6H_6NH_2)}$ are not known. They are $60$ µM at the beginning. But they remain unknown at t > $0$. I have made different hypothesis about these variations, with $\ce{\Delta A}$ = a $\ce{\Delta B}$, with $\ce{a}$ being $0, 1$ or $2$. The obtained values are never alined, as if the reaction is not of $2$nd order.
But the reaction should be "about" $2$nd order, since the half-life is growing. Interpolation of the given data shows that it is first $6$ µs, then $7$µs., then $9$ µs, then $11$ µs. This is typical for an order greater than $1$, but smaller than to $2$.

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text:
How to know whether a molecule is 'stabilized'
According to a reference book, in acid-base reactions, the position of equilibrium is shifted towards when the conjugate base is more 'stable' and the relative stabilities are compared by

different atoms
resonance
induction
orbital

What I don't understand is that in the first 3 of criteria. They are about how able for the electrons to move around (delocalize). But in the fourth criteria, presence of triple bonds make the molecule most stable since the electrons are confined in a small place (stronger attraction to the nucleus). Doesn't this mean electrons are 'localized' instead of 'delocalized' (i.e. less stable)? Do I get it wrong? Thank you very much.

I’m voting to close this question because you refer to a source without naming it.


text:
When does nucleophilic attack and proton transfer happen?

from the reaction, the 3rd molecule (carboxylic acid) formed undergo neutralization with OH- (proton transfer to a base).
However, is it possible for the 3rd molecule to behave as the 1st molecule (ester) and undergo nucleophilic attack by the OH- as they both have an electrophilic centre (carbon)?
If yes, will this reaction form a cycle (never forms carboxylic salt). If no, may I ask why? Is it related to stability or other effects? Thank you very much

What would you expect the product of such a reaction to be?

@LigninPauling I think it may be a mixture of first 3 molecules in equilibrium as the OH in 2nd molecule seems to be a leaving group (regenrated) Sorry that I am a total beginner

If you chose the carboxylic acid as the starting compound, as opposed to the ester, would you not find that the nucleophile and the leaving group are the same?

Oh I see so can this be one of the reason why the first 2 steps are reversible (i.e. has chance to undergo the attack? Also, does this means the proton transfer step occurs by higher chance (lower energy level)? Thank you for answering

Yes indeed. For the first step, either the hydroxy group or the alkoxy group may leave, giving reactants or products, respectively. For the second step, it is reversible as the generated alkoxide ion may acts as a nucleophile to attack the carboxylic acid and give rise to the intermediate structure once more.

Proton transfer reactions are very rapid. https://chemistry.stackexchange.com/questions/16592/why-is-proton-transfer-so-fast?rq=1

Because you are a beginner to organic chemistry, I'll tell you this: Both esterification (e.g., Fischer esterification) and de-esterification reactions are reversible. The question you had at the end is why the last step of base catalyzed de-esterification reaction is only forward reaction. As you have learned in general chemistry class, acid-base reaction is one of the fastest reactions in chemistry (not counting nuclear reactions). Thus, the last step is governed by preference for acid base reactions:
$$\ce{R-COOH + OH- -> R-CO2- + H2O}$$
The $\mathrm{p}K_\mathrm{a}$ of $\ce{R-COOH}$ is about $4.7$ ($\mathrm{p}K_\mathrm{a}$ of $\ce{CH3-COOH}$ is $4.75$) while $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is $15.7$. Therefore, forward reaction is about $10^{10}$ times faster than backward reaction.
Also remember, this reaction is in aqueous medium, therefore there are much more $\ce{H2O}$ molecule available than $\ce{R-OH}$ molecules. Thus, going back to ester reaction is much slower than becoming acid (governed by Le Chatelier principle).

This means in every step of reactions I should first consider position of equilibrium (pKa) of acid-base reactions other mechanisms (e.g. nucleophilic attack) since it occurs much easier? Thank you for answering

It is depends. Think, your reactant is a strong base as well as a good nucleophile. For example, consider Grignard reaction.Grignard reagent is a strong base, but under goes nucleophilic reactions with ketone, etc. But, if you hace trace of water or prtic solvent, it react with them acting as a base. But, even if $\alpha$-H in ketone is relatively acetic Grignard reaction prefers nucleophilic attack rether than abstracting $\alpha$-H. See, it's a little complicated. But, you'd be okay when you got enough experience. :-)


text:
Oxidation to allylic alcohols without stoichiometric metal reagents
Common methods of oxidation of alkenes to allylic alcohols in one step usually use stoichiometric metal oxidants, e.g., sodium dichromate.
Without using this type of reagent, are the other methods typically peroxide forming,  requiring further chemistry to adjust to the allylic alcohol, e.g., allylic hydroperoxides or t-butyl peroxides, or are there more "direct" methods?

Almost all allylic oxidation reactions have been using metal catalysts such as chromium, selenium, copper, etc. because it is relatively easy to abstract hydrogen radical from allylic position and metals are well known for doing so with or without single electron transfer (SET). This fact is well documented in the review published in 2013 (Ref.1), the abstract of which simply states that:

Although $\ce{C–H}$ oxidation of hydrocarbons is generally difficult, allylic $\ce{C–H}$ oxidation is relatively simple and predictable, even on a preparative scale, because active species generated at the allylic position are stabilized by the double bond. Therefore, allylic oxidation has been employed in natural product synthesis, and a variety of reagents and conditions for allylic oxidation have been reported...

Some of the specific catalysts the review referred are  selenium Dioxide, diphenyldiselenide–Iodoxybenzene, chromium Trioxide–3,5-Dimethylpyrazole, pyridinium chlorochromate (PCC), chromic acid, etc., all of which are toxic to humans (e.g., selenium and chromium compounds) and fairly expensive (e.g., palladium and rhodium compounds). Recently, a paper published in Nature journal describes an electrochemical allylic $\ce{C–H}$ oxidation strategy that does not need toxic and expensive metal catalysts, but uses inexpensive and readily available materials. The method exhibits broad substrate scope, operational simplicity, and high chemoselectivity. The notable advantage of this electrochemical allylic $\ce{C–H}$ oxidation strategy is It represents a scalable allylic $\ce{C–H}$ oxidation (results have demonstrated on $\pu{100 g}$ scale reactions).
The General Procedure for electrochemical allylic oxidation is described as follows in the paper:

With no precautions to exclude air or moisture, a test tube is charged with a stir bar, $\ce{Cl4NHPI}$ $(\pu{0.2 equiv})$ and $\ce{LiClO4}$ ($\pu{0.6 equiv}$, $\pu{0.1 M}$). Acetone ($\pu{6 mL/mmol}$ substrate) is added, and the resulting suspension is stirred until most of the solids dissolve. Substrate ($\pu{1 equiv}$) is added, followed by pyridine ($\pu{2 equiv}$) and $70\%$ aqueous tert-butyl hydroperoxide ($\pu{1.5 equiv}$). The anode and cathode (RVC, separated by a glass slide) are placed in the solution, and the reaction mixture is electrolyzed at a current of $\pu{10 mA/mmol}$ substrate until the reaction is complete as judged by TLC or GCMS analysis. The reaction mixture is transferred to a separatory funnel, and the electrodes are washed thoroughly with hexanes, diethyl ether, or ethyl acetate. The organic layer is washed twice with $\ce{H2O}$, once with brine, dried, and concentrated. The resulting crude product was then purified on $\ce{SiO2}$ to furnish the desired enone product. Note: Pyridinium perchlorate can be added to the reaction ($\pu{0.1–0.2 M}$) to improve solubility of the reaction components. This co-electrolyte has negligible effect on product yields and may improve the reaction if the substrate contains any easily reduced functional groups since this increases the concentration of protons available to be reduced at the cathode.

Note that the mediator, $\ce{Cl4NHPI}$ is N-Hydroxytetrachlorophthalimide, which was synthesized from tetrachlorophthalic anhydride and $\ce{H4N-OH/HCl}$ in pyridine using thermal or microwave technique for this research. However, it was available in Sigma-Aldrich for $\pu{62 USD}/\pu{10 g}$.

The schematic diagram of the general procedure is depicted in above diagram (Figure A).  The variety of mediators used during the progress of the process are listed in Figure B. Among them, N-Hydroxyphthalimide ($\ce{NHPI}$) and $\ce{Cl4NHPI}$ have given the best results (have given total yield of $56\%$ and $77\%$, respectively in otherwise optimized conditions). They are drastically efficient than others, e.g., next best has been TEMPO, which has given only $26\%$ total yield under identical conditions. Among the co-oxidants tried, $\ce{t-Bu-OOH}$ has given the optimal results ($51\%$). The next best is $\ce{PhC(CH3)2-OOH}$ with $43\%$ yield. Acetone ($56\%$) and $\ce{CH3CN}$ ($51\%$) have given equally impressive results as choice of solvent, and acetone was chosen for optimal conditions as the solvent for this research. Hence optimized electrochemical parameters were: $(\pu{0.16 M})$ of substrate with $\ce{Cl4NHPI}$ $(\pu{0.2 equiv})$, pyridine $(\pu{2.2 equiv})$, $\ce{t-Bu-OOH}$ $(\pu{1.5 equiv})$, $\ce{LiClO4}$ $(\pu{0.6 equiv})$ in acetone.
The tentative mechanism is illustrated in Figure C. The formation of tert-butyl-peroxy-intermediate (III) was confirmed. In a separated experiment, one of corresponding tert-butyl-peroxy-intermediate was synthesized in a different route and reacted in identical condition used in electrolytic procedure ($\ce{Cl4NHPI/t-BuOOH/LiClO4/pyridine/acetone}/RT$). The reaction has completed both with and without electrolysis within $\pu{6 h}$, confirming requirement of electrolysis during initial tert-butyl-peroxy-intermediate formation steps (Ref.2).
References:

Akihiko Nakamura, Masahisa Nakada, “Allylic Oxidations in Natural Product Synthesis,” Synthesis 2013, 45(11), 1421-1451 (DOI: 10.1055/s-0033-1338426).
Evan J. Horn, Brandon R. Rosen, Yong Chen, Jiaze Tang, Ke Chen, Martin D. Eastgate, Phil S. Baran, “Scalable and sustainable electrochemical allylic C–H oxidation,” Nature 2016, 533, 77–81 (https://doi.org/10.1038/nature17431).

I really like the answer, but the question related to allylic alcohols,  preferably in one step (albeit a step could occur during work-up).

The oxidation of alkenes with t-butyl peroxybenzoate in acetonitrile or benzene in the presence of copper compounds has been reported by Levina and Mazart, Tetrahedron Asymmetry 6 147 (1995) e.g. cyclohexene to the allylic benzoate in 59% yield.
Similar oxidations of alkenes using t-butyl peroxybenzoate with catalytic amounts of bisoxazoline copper complexes have been reported by MB Andrus et al, Tetrahedron Letts. 36 2945 (1995).


text:
How do you calculate the amount of substance from molar mass and volume
I am struggling with one calculation and it goes like this:

Calculate, how much of $\ce{CO2}$ is made during the reaction: $\ce{2CO + O2}$, if we use $\pu{33,6 dm^3}$ of $\ce{O2}$, also calculate the amount of $\ce{CO}$ used. M(CO)=28, M(CO2)=44.

The answer is $\pu{132 g}$ of $\ce{CO2}$ and $\pu{67,2 L}$. I especially don't know how I am supposed to get to the litres. Any help is appreciated :)

A cubic decimeter is one liter.

We do not do homework problems, or the like, here. You need to read the site’s homework policy and show your effort in solving the problem. We will help if that is done, but otherwise no.

How do I ask homework questions on Chemistry Stack Exchange?

Was there a temperature and pressure given for the gas? // As the problem is now presented you can solve for the volume of CO but not the grams.

R3vive - Hint The oxygen gas and the carbon monoxide must be at the same temperature and pressure. Assume them both to be ideal gases. So for a stoichiometric reaction, what must the volume of carbon monoxide be?

R3vive - The problem doesn't have enough information to solve for the grams of carbon dioxide. You could calculate a volume of carbon dioxide. However back calculating from the answer I can tell that the conditions used were 0 °C and 101.325 kPa. Thus you must be in India.

Hint. All the volumes are simple multiples of 22.4 liters

The given question is a bad question.

Calculate, how much of $\ce{CO2}$ is made during the reaction: $\ce{2CO + O2}$, if we use $\pu{33,6 dm^3}$ of $\ce{O2}$, also calculate the amount of $\ce{CO}$ used. M(CO)=28, M(CO2)=44.
The OP gives the answer as 132 g of CO2, and 67,2 L of $\ce{CO}$.

The issues with the problems are:

"...how much of $\ce{CO2}$ is made..."
For $\ce{CO2}$, this doesn't specify if it is the volume of gas, the number of moles, or the mass which is sought.
"...also calculate the amount of $\ce{CO}$ used."
Again, this doesn't specify if it is the volume of gas, the number of moles, or the mass which is sought.
The problem doesn't specify that $\ce{CO}$ and $\ce{O2}$ are present in stoichiometric proportions.
The problem uses units $\pu{dm^3}$ for the $\ce{O2}$ but gives the answer for $\ce{CO}$ in $\pu{Liters}$.
The problem doesn't specify that the reaction went to completion.

Utilizing only the information given, and the following assumptions, then the relative volumes of the gases could be found.

That $\ce{CO}$ and $\ce{O2}$ are present in stoichiometric proportions.
That the reaction went to completion.
That after the reaction the gas volume was measured under the same conditions, pressure and temperature, used initially.

With the assumptions the volume of $\ce{CO} = \pu{67.2 L}$ and the volume of $\ce{CO2} = \pu{67.2 L}$.
It is to be noted that $\pu{33.6 dm^3} = 1.5\times 22.4$ where 22.4 is the old molar volume of an ideal gas at STP. So adding the assumption that the gases were measured using the old STP conditions allows the calculation of the grams of $\ce{CO2}$.

Don't know why but it seems that all the posters from India are still using the old STP specifications (before 1982).

Here is what I think is a better formulation of the question. It would assume that the solver understand that the old specification is used for STP, but that would be consistent throughout whatever book was being used.

Given that the reaction $\ce{2CO + O2 -> 2CO2}$ goes to completion, calculate the grams of $\ce{CO2}$ produced when, at STP, a stoichiometric amount of $\ce{CO}$ is added to $\pu{33.6 dm^3}$ of $\ce{O2}$. Also calculate the volume of $\ce{CO}$ added. M(CO)=28, M(CO2)=44.


text:
Alternative python modules to RDKit to convert SMILES to structure
Python module RDKit is great, but it cannot be installed by pip, and therefore it is difficult to use it sometimes in a virtual environment.
I am wondering that if there are some alternative python modules that can convert smiles to chemical structures?

Does this answer your question? Is there a way to use free software to convert SMILES strings to structures?

@Mithoron, Thanks a lot for your kind help. These tools are not suitable for me. Except Pysmiles, the rest tools are not python-based. Pysmiles cannot convert smiles to chemical structures.

There is also the Indigo Toolkit. An example appears in an answer to this question.

https://chemistry.stackexchange.com/questions/61887/portable-library-to-render-2d-structural-formulas-as-vector-graphics-from-smiles

I am not aware of any other Python tool not mentioned in the question @Mithoron linked to. Openbabel is available via pip by the way.
That RDKit is not available on PyPi should not hinder you from using it in a virtual environment. RDKit is available via the conda package manager which is easy to use and combinable with pip.
If you are on Linux, the probably best solution for managing different versions of Python (also Anaconda Python) and virtual environments is pyenv. I highly recommend to check it out.


text:
Has the half-life of an isotope ever been accurately predicted before it could be measured?
The half lives of many radioactive isotopes have been measured directly. But are their half-lives known to be a function of their composition?
In other words, can someone say "given an atom with this many protons, neutrons and electrons, and given the known strengths and ranges of the various forces at play inside an atom, we can quantify how unstable such an atom would be by saying that a collection of them would have a half-life of X"?
Has such a prediction ever been made about an isotope before people had encountered it, then later borne out by experiment?

I think the answer is probably technically no: it was not until fairly recently that we knew bismuth is actually radioactive, albeit with a ludicrously long lifetime. But this is an excellent question, particularly if lifetimes are restricted to a billion years or so.

Actaully $^{209}\ce{Bi}$ would seem to be an example. Theorists had evidently long suspect that the isotope was radioactive. See for example: "A SEARCH FOR α-PARTICLES FROM THE DECAY OF Bi209" E. P. Hincks and , C. H. Millar, Canadian Journal of Physics, 1958, 36(2): 231-251, https://doi.org/10.1139/p58-027

@EdV What is supposed to be excellent about it?  Unless clarified, I don't think this question is OK, at all. There are theories of nuclear structure and all kinds of predictions. Maybe about only thing that would make this viable is the notion of accuracy. If clarified - how accurate is accurate enough.

@Mithoron My thought was that making an accurate prediction, and then finding experimental verification, is what science is about. I agree completely about the accuracy aspect. The bismuth example had a predicted range that was way too large and the experimental half-life was near the lower end. I am not impressed with that. If a new isotope were to be created and had an experimental half-life within 10 or 20% of an a priori prediction of the half-life, I would be impressed and say it was a significant achievement. But YMMV.

@Mithoron My apologies if my question doesn't meet your standards. I'm very much a beginner. My motivation: decay rates are used in radiometric dating, and young earth creationists sometimes allege that decay rates may have changed. I'd like to know if there exists a function that can accurately tell us the half life of any isotope based on its structure and the basic nuclear forces, such that half lives can't have changed unless nuclear forces have. An accurate prediction would be strong validation for such a function. I would leave quantifying the accuracy to the judgment of respondents.

Other options would be the relatively recently created elements, but there good accuracy of the predicted lifetimes is getting within an order of magnitude.

@NathanLong - Don't waste any effort trying to convince young earth creationists that their belief is wrong. They have seen "The Flintstones" on TV and know that people and dinosaurs roamed the earth together. Corollary:"Never try to teach a pig to sing. It frustrates you and annoys the pig."

@MaxW There are definitely those who can't be persuaded, but don't write people off categorically. Good arguments persuaded me away from YEC, and there are many personal stories on the BioLogos website of a similar nature. I continue to try to improve my understanding of how all the pieces of evidence fit together. The most helpful explanations to me are those that tell not only what is known, but how we know it. Those are the kinds I want to offer in my own conversations.

A quick perusal of the post-WWII literature (particularly Glenn Seaborg) pulls up papers like The New Element Californium (Atomic Number 98) where comparisons of the observed alpha decay half life are said to be in line with predictions. Such predictions follow along from papers such as Systematics of Alpha-Radioactivity.
So, yes, it would seem that predictions of radioactive half lives were part and parcel of the creation of new elements in the laboratory.


text:
Reaction between halogens and concentrated nitric acid
I have read that the reaction between $\ce{I2}$ and conc. $\ce{HNO3}$ gives $\ce{HIO3}$ and $\ce{NO2},$ but there is no mention of the other halogens' products. Do they all give halous acids and nitrogen dioxide? Specifically with chlorine, how does the reaction proceed?

@Waylander That website looks fishy to me. Is this journal even trustworthy? BTW, iMedPub is in the list of predatory publishers.

@andselisk I know nothing about this journal other than Google found me a reference to the oxidation of bromine by nitric acid. In all honesty the world does not need yet another method for the oxidation of benzylic acohols, but it appeared to demonstrate the reaction the OP asked about. There are other references to Br2/Br- oxidation by nitric acid

@ThenardRinmann From what Google is showing me, it appears that nitric acid will oxidise chloride to chlorine but no further.

Previous question in chem.SE: (fluorine + nitric acid) and (iodine + nitric acid).

I may want to have a look at the redox potential Wikipedia data page. $\ce{HNO3}$ does not have much chnace to oxidize chlorine.

@Poutnik The net standard electrode potential for the oxidation of chlorine to chlorate and simultaneous reduction of HNO3 to NO, which can further undergo aerobic oxidation to NO2, is in fact positive. However only god knows about the kinetics of the reaction.

@Thenard Rinmann Well, for acid environment, it is +0.958 V for NO3-/NO versus  +1.49 V ClO3-/Cl2, so I rather insist on my prior comment.

@Poutnik yeah I checked back with the data, ran sum numbers: you're right about the negative potential. I am new to this. Should I delete the question and/or previous comment or will you put that as an answer to the question?

This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:
$$
\begin{align}
\ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\
\ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &= \pu{0.934 V} \tag2\\
\ce{2NO3- + 4H+ + 2e- &<=> N2O4 + 2H2O} &\quad E^\circ &= \pu{0.803 V} \tag3
\end{align}
$$
Now, let's look at reduction potential of halides:
$$
\begin{align}
\ce{F2 + 2e- &<=> 2F-} &\quad E^\circ &= \pu{2.866 V} \tag4\\
\ce{Cl2 + 2e- &<=> 2Cl-} &\quad E^\circ &= \pu{1.3583 V} \tag5\\
\ce{Br2 + 2e- &<=> 2Br-} &\quad E^\circ &= \pu{1.066 V} \tag6\\
\ce{I2 + 2e- &<=> 2I-} &\quad E^\circ &= \pu{0.5355 V} \tag7\\
\end{align}
$$
Accordingly, we can list oxidizing power at standard conditions in decreasing order:
$$\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{NO3-} > \ce{I2}$$
Therefore, at standard condition, concentrated $\ce{HNO3}$ acid (meaning $\ce{NO3-}$ in acidic medium) can easily oxidize only $\ce{I2}.$ One such example is documented in Ref.1. According to the authors, the oxidation of iodine by hyperazeotropic nitric acid ($\pu{16–21 M}$) was shown to proceed in at least two steps:
$$
\begin{align}
\ce{I2 + 3HNO3 &<=>[$k_1$][$k_2$] 2I+ + 2NO3- + 2HNO2 + H2O}\\
\ce{I+ + 6HNO3 &<=>[$k_3$][$k_1$] I^5+ + 4NO3- + 2HNO2 + 2H2O}
\end{align}
$$
Since reduction potentials of $\ce{HNO3}$ and $\ce{Br2}$ are in such close range that each can oxidize the other. In fact, the kinetic studies of this phenomenon has been performed in literature (Ref.2).
In different aspect, there is a publication (Ref.3) claiming that the presence of bromine in fuming nitric acid makes it a brominating reagent (electrophilic aromatic substitution):

This bromination has worked well with deactivated aromatic nucleus. In absence of bromine, the reagent nitrate the ring. Thus, it is safe to suggest that $\ce{HNO3}$ might promote formation of '$\ce{Br+}$', meaning oxidation.
References:

J. C. Mailen, T. O. Tiffany, "The reaction of iodine with concentrated nitric acid," Journal of Inorganic and Nuclear Chemistry 1975, 37(1), 127-132 (https://doi.org/10.1016/0022-1902(75)80139-4).
Istvan Lengyel, Istvan Nagy, Gyorgy Bazsa, "Kinetic study of the autocatalytic nitric acid-bromide reaction and its reverse, the nitrous acid-bromine reaction," J. Phys. Chem. 1989, 93(7), 2801–2807 (https://doi.org/10.1021/j100344a021).
Alexander M. Andrievsky, Vera I. Lomzakova, Mikhail K. Grachev, Mikhail V. Gorelik, "Aromatic Bromination in Concentrated Nitric Acid," Open Journal of Synthesis Theory and Applications 2014, 3(2), Article ID:44941, 5 pages (doi: 10.4236/ojsta.2014.32003).

$\ce{HNO_3}$ does not react with $\ce{Cl_2}$, as $\ce{Cl_2}$ is produced in the famous "aqua regia", which is a mixture of nitric and chlorhydric acid, having reacted according to :$$\ce{HNO_3 + 3 HCl -> NOCl + Cl_2 + 2 H_2O}$$
This equation is given in Cotton and Wilkinson, Chapter 5-13. In Chapter 12-8, it is stated that NOCl is never pure. It is decomposed to $\ce{Cl_2 + NO}$ to the extent of $0.5$% at room temperature.
NOBr is produced in a mixture $\ce{NO + Br_2}$ and it is an equilibrium, where ~ $7$ % of NOBr is decomposed at room temperature and $1$ atm.
Ref.: F. Albert Cotton, Geoffrey Wilkinson, Advanced Inorganic Chemistry, A Comprehensive Text, 3rd Ed., 1972, Interscience Pub., John Wiley & Sons, Inc. New York, London.

I'm not sure why you decided to delete your previous answer and re-post it again with minor changes instead of editing it, but this is still not exactly the answer to the question IMO. I don't understand the logic behind "$\ce{HNO3}$ does not react with $\ce{Cl2},$ as $\ce{Cl2}$ is produced in the famous 'aqua regia'" statement, and how the paragraphs after the reaction are relevant to the question.

@ Andselisk. The question was : How does the reaction $\ce{HNO_3 + Cl_2}$ proceed ? My reasoning is that those two substances do not react at all. If $\ce{Cl_2}$  would react with the solvant containing  nitric acid, it would not be produced in the aqua regia. It would immediately be transformed into something else, and would not exist in aqua regia.  This reasoning is a little bit similar to the stupid equations that poor students sometimes establish "showing" that an acid reacting with something produces a hydroxide. Sorry for the silly comparison !

Took me a while to obtain the edition you are citing, but I haven't found the reaction you provided in chapter 5-13, only the statement on p. 179: "Aqua regia (~3 vol. of cone. $\ce{HCl}$ + 1 vol. of conc. $\ce{HNO3})$ contains free chlorine and $\ce{ClNO},$ and it attacks gold and platinum metals, its action being more effective than that of $\ce{HNO3}$ mainly because of the complexing function of chloride ion."

Keep in mind that the reaction at the room temperature is in fact reversible, and doesn't involve instant formation of chlorine gas, which IMO sort of defeats your reasoning: $$\ce{HNO3(conc) + 3 HCl(conc) <=> NOCl + 2 Cl^0 + 2 H2O}$$ The irreversible reaction occurs when the mixture is heated: $$\ce{HNO3(conc) + 3 HCl(conc) ->[\pu{100–150 °C}] NO(g) + 1.5 Cl2(g) + 2 H2O}$$ I'm not saying your answer is wrong, but to me, the explanation is not convincing and it also only covers the reaction with chlorine, ignoring other halogens (see the question's title).

The main thing is that Chlorine $\ce{Cl_2}$ cannot be oxidized by nitric acid to some higher oxidation number. The production of halous acid and nitrogen oxides, as suggested by Thenard Rinmann, simply does not happen with nitric acid.


text:
Openbabel is not converting Gaussian Z-matrix
I ran into an issue when OpenBabel (a great and extremely useful problem) could not convert my custom-generated Gaussian Z-matrix to XYZ format. The problem is that probably major flaw of OpenBabel are very terse error messages (mostly just "0 molecules converted", and even with ---errorlevel 25 option I got only:
==============================
*** Open Babel Audit Log  in ReadChemObjectImpl
  OpenBabel::Read molecule Gaussian Z-Matrix Input
==============================
*** Open Babel Audit Log  in Clear
  Ran OpenBabel::Clear Molecule
==============================
*** Open Babel Audit Log  in ReadChemObjectImpl
  OpenBabel::Read molecule Gaussian Z-Matrix Input
==============================
*** Open Babel Audit Log  in Clear
  Ran OpenBabel::Clear Molecule
0 molecules converted

). Some other file was converting just fine, with no obvious differences between their contents! I spent a long time trying to understand what is wrong with the second file, but to no avail.
The file which converts without any problem:
# This is a dummy route line, required for OpenBabel for the conversion

   This molecule contains 26 atoms.

0 1
C
C   1    1.4141
C   2    1.3856  1    117.29
C   1    1.4077  2    121.06  3    -1.09
C   4    1.4088  1    121.27  2    2.37
C   3    1.4075  2    121.95  1    -0.93
C   3    1.5118  2    128.77  1    -176.79
C   7    1.4378  3    105.16  2    175.55
C   6    1.5156  3    108.80  2    -176.32
N   8    1.4317  7    122.53  3    -177.58
C   10   1.3629  8    120.12  7    -146.43
C   11   1.3942  10   120.44  8    -178.97
C   12   1.4024  11   119.82  10   -3.12
C   13   1.4110  12   118.94  11   1.63
C   10   1.3704  8    119.11  7    32.69
O   7    1.2445  3    125.27  2    -1.31
O   9    1.2448  6    124.96  3    -179.37
H   2    1.0882  3    121.72  6    -181.70
H   4    1.1010  1    119.57  2    -178.73
H   5    1.0951  6    121.12  3    179.30
H   11   1.0928  10   115.87  8    1.70
H   12   1.0908  11   118.95  10   178.60
H   13   1.0963  12   120.77  11   182.82
H   14   1.0959  13   121.94  12   -180.61
H   15   1.0855  10   115.78  11   178.99
H   1    1.0923  2    119.78  18   0.45

The file which fails to convert:
#T HF/3-21G Opt Integral(Grid=SG1)

  H2O2 H2O2 GeometryOptimization GeomOpt GeometryOptimization 1

0 1
O
H   1    1.0
O   1    1.4  2    105.0
H   3    1.0  1    105.0  2    176.0
O   1    5.0  2    90.0   3    90.0
H   5    1.0  1    105.0  2    52.5
H   5    1.0  6    105.0  1    178.0

This is OpenBabel version 2.4.1 on Debian 10.
P.S. I sincerely don't know whether this question belongs better to the Chemistry StackExchange or to other forum, but I presume the question might appear to be useful for someone someday.

This is absolutely fine here and well asked. Could you please add the version of OpenBabel and the operating you are using it on.

OK, thank You. Added the information You requested.

Great, thanks. For what it's worth, and I'm positive this won't change any of the Q&A, Open Babel is now available as 3.1.1. (You can flag the comments if you want.)

Yes, but I have Debian 10, as I mentioned. The 'rock stable' one :-) That's why 2.4.1.

Turns out that OpenBabel is sensitive to whether blank lines are really blank lines or lines with space characters only. In my case, this was the blank line between the route and the title line. One can see them by pressing the button with a paragraph sign (¶) on it in Your favourite text editor (Notepad++, or even MS Word or  LibreOffice Writer), or by entering :set list in vim or M-x whitespace-mode in Emacs.
What does OpenBabel require for conversion:

First line of the file must begin with a hash sign (#) – a substitute for the Route line
Next line must be absolutely empty
Next line can contain anything You want
Next line again must be absolutely empty
Then comes the charge and multiplicity
And then comes the Gaussian Z-matrix in either compact or full form.

What doesn't OpenBabel require for conversion:

the file to have the extension GZMAT, it can also be GJF
title line not to begin with the hash sign (#) or any other character
route line to contain actual keywords (if You are not trying to convert them to any other input file format, that is)

When it comes to coding, "MS Word" or "LibreOffice Writer" should not be the in the same sentence with "your favourite text editor".

Probably so, but I am not acquainted to any other than those mentioned to know about the ¶ button (Bluefish doesn't have and also I do not use Emacs) :-)

I guess OpenBabel is as serious about its blank lines as Gaussian would be.

Actually not, I have just checked – Gaussian 16 didn't care at all when I put a single or multiple spaces or even tabs in any of these blank lines.


text:
Associated occupation numbers of frozen natural orbitals (FNOs) from MP2 method
The paper Frozen Virtual Natural Orbitals for Coupled-Cluster Linear-Response Theory gives the following (simplified) description of frozen natural orbitals (FNOs).
In the MP2 method, the unrelaxed one-electron density matrix can be written in terms of spin orbitals
$\gamma_{pq}=\langle\Psi^{1}|\{a_{p}^{\dagger}a_{q}\}|\Psi^{1}\rangle$
where $|\Psi^{1}\rangle$ is the first-order correction to the Hartree-Fock wavefunction.
In the MP2 based NO method, the virtual-virtual block is constructed:
$\gamma_{ab}=\frac{1}{2}\sum t_{ij}^{ac}t_{ij}^{bc}$
and then diagonalised:
$\gamma V=nV$
The eigenvectors V are the virtual NOs, and the eigenvalues n are the associated occupation numbers.
If the eigenvalues are found from a virtual-virtual (unoccupied-unoccupied) block, how can there be an associated occupation number?

Not sure about the right answer, maybe it is related to mixing the term "occupied" for Hartree Fock

@user1420303 I thinking along the lines that occupation numbers are associated with single, double, excitations into the virtual orbitals. Though I would need clarification on this.

I mean that when natural orbitals are constructed, their occupation number is not integer. Many of them are very close to zero or one. But they are not exactly zero or one (specially in multireference systems). Then you can classify nat orbs as occupied when they are essentially 1 or unoccupied when they are close to 0 (and eventually you can find orbitals with intermediate occupation numbers).

Virtual or unoccupied refers to the Hartree-Fock solution - the virtual orbitals are thus unoccupied in Hartree-Fock, i.e. have occupation number 0, whereas the occupied orbitals have occupation number 2 (spin-traced occupation in restricted HF) or 1 (for a given spin in unrestricted HF). The first-order MP2 correction to the wave-function will have different occupation numbers - the occupied orbitals will become slightly less occupied, whereas the virtual orbitals become slightly occupied by the same total amount as a result of electron correlation. This doesn't change the nomenclature: occupied and virtual orbital still refers to the HF reference wave-function.


text:
Why can alcohol undergo elimination just by sulphuric acid?

In both $\mathrm{E1}$ and $\mathrm{E2}$ reactions with alcohol, loss of a leaving group and proton transfer are required. And proton transfer requires a base to get that hydrogen in alcohol.
I know that $\ce{H_2SO_4}$ can make a bad $\ce{ROH}$ group into a good leaving group ($\ce{ROH_2^+}$) by protonation, but how can it (strong acid) serve as a base for proton transfer to happen (in this 1 step process)?
What I think:

$\ce{H_2SO_4}$ is a dehydrating agent, so it can 'suck out' the $\ce{OH}$ together with heat, like dehydration with sugar into pure carbon (from my high school textbook)? However, is this mechanism even called $\mathrm{E1/E2}$?

In E1, the way I understand it is that the resultant carbocation's protons become reasonably acidic as the conjugate base (an alkene) is very stable. They only require a weak base to be removed, so practically anything can remove them. It looks to me as if the likely weak base candidates are the water produced from the leaving group, or even another alcohol in the reaction mixture.
Primary alcohols will be less susceptible to dehydration by this method, as they follow an E2 mechanism which progresses much more slowly in acidic conditions due to the absence of a strong base (which would usually act to 'shove' the leaving group off, a weaker base gives a weaker 'push').
I have seen a few schemes for E1 and E2 in which the hydrogen sulphate ion acts as the weak base; the usual assumption is that there is a weak base that is able to capture the proton, but we don't necessarily care what it is because anything will do. As long as there isn't another half decent nucleophile in the reaction mixture, elimination will likely prevail.
As you can see from this image I found in Clayden Organic Chemistry, the proton in the reaction mechanism is often drawn as just 'falling off', emphasising that any weak base is capable, and abstraction of the proton is very likely by a large number of species.

So is it possible that the heating already provides the activation energy for 'falling off' of proton without even using a base? (Since activation energy required for proton transfer << loss of leaving group)

If the reaction is heated, water is removed, and the reaction is made reasonably irreversible in the final step of the mechanism. This will push the reaction to the right, effectively encouraging the protons to be pulled off despite the presence of only a very weak base (because of Chatelier's Principle).

As in case of SN1 and SN2 reactions 1 and 2 represents the molecularity of the reaction

(molecularity= no of molecules colliding in the RDS step).
  *In similar manner in case of elimination reactions E1,E2,E1CB the significance of 1,2 is the representation of the molecularity.

In case of E1 reaction  like dehydration of alchol >formation of carbonation  C+ from OH2+ is the RDS step .The reason for it being E1 is that there is involvement of only one substrate molecule in RDS ie molecularity =1 !

[]>(https://i.stack.imgur.com/Lq160.jpg)
  In case of E2 reaction like Alc KOH removal of H and LG takes place simultaneously.It is one step rxn with two molecules colliding in RDS step ie molecularity =2 .One is substrate and other is the Base .

  In case of EICB reaction the condition are same as of E2 reaction only but the molecularity is 1.
  In E1CB reaction removal of H and LG is not simultaneous .
  First of all Base traps H and there is a formation of Carbanion.Then this carbanion removes LG in the next step.The molecularity of the reaction is 1 as only one molecule is involved in the RDS step .Hence called E1CB reaction .


text:
What is the safe level of energy output for hydrogen combustion at home?
With confinement, my son would like to do some home experiments for a class. So far, he has a good setup that seems sound for the electrolysis of water. A later part of the experiment would combine back the $\ce{O2}$ and $\ce{H2}$ to get back water vapour.
What amount of atmospheric pressure $\ce{H2}$ and $\ce{O2}$ can be safely combined, at home, without it turning to a fireball and/or a news event?  From Googling, I found 12,749 joules per litre of hydrogen. What I'm asking is:
How does dissipating 12K joules look like in a plastic bucket? Is this a case of needing only gloves and eye protection, or does it need further protection? Will it be just some heat, just a small flame, or a McGyver-type explosion?
Also, will $\ce{H2}$ and $\ce{O2}$ at atmospheric conditions instantly react, or would this be a progressive reaction?

Certainly don't do it on a large scale. Small amounts of hydrogen oxygen mixtures from electrolysis bubbled through soapy water in a standard bucket will create a big enough bang when ignited to be troublesome indoors. Even perhaps 100mL of bubbles will be a big bang.

I'm not as familiar with this SE as some others. I understand my question might not be correct as is. Down-voters, if you would please comment, in addition to down-voting, that would give me the opportunity to fix the question.

I did not downvote, but I have heard the sharp retort from a small balloon filled with hydrogen and oxygen. The small soap bubbles method suggested by @matt_black is the way too go: no fast moving fragments of significance. As for downvotes, things that blow up are problematic for a variety of reasons, as you can well suppose, and safety must be paramount.

I remember a high school experiment presentation in early 80s with bubbles from O2+H2 mixture in a china bowl. It crashed the bowl, making from the fragments a kind of a geometric projection on the table.

First things first: hydrogen gas and oxygen gas do not spontaneously react but need activation. Activation can be provided by a flame, an electric spark or similar.
That said, here is the important part. The reaction will not turn into a fireball but even small volumes will explode with pretty loud bangs. It’s not the loudest bang known to chemistry instructors but it’s certainly something to be aware of – you don’t want your neighbours ringing the police because they think a shooting has been going on.
If you’re going to try this, I recommend starting with the smallest volumes possible, not bigger than a shot glass. My chemistry teacher in high school used a glass cylinder approximately the size of a small (200 ml) water glass. That gave an even stronger bang but the cylinder itself remained unharmed, probably because there was a large opening on one side.
However, the experiment is unlikely to show you much because the water vapour is obviously gaseous and will dissipate rather than being visible anywhere. It would almost be more instructive to identify the two gases as hydrogen (burns when brought to a flame) and oxygen (ignites glimming wood) by the standard procedures used in high school (at least in mine).
Tl;dr: I suggest you don’t try igniting a 2:1 mixture unless you’ve seen it performed recently and know what you’re doing.


text:
Is there a Python module for calculating Ovality of a molecule?
Ovality gives a measure of how the shape of a molecule approaches a sphere (at one extreme) or a cigar shape (at the other). Ovality is described by a ratio of volume to area (see https://en.wikipedia.org/wiki/Ovality).
So, it is basically "just" about intersecting Van-der-Waals spheres. Although this sounds simple, it can get pretty complicated.
There is a nice paper from M. Petitjean explaining how to get correct values of Van-der-Waals volumes and surfaces (see https://hal.archives-ouvertes.fr/hal-01955983).
According to this paper, inacceptable approximations have been used in the past.
I can follow this paper to a certain extent, but when it comes to implementing this in Python it would exceed my mathematical and programming knowledge.
Therefore my question:
Is there maybe already a Python module which can calculate Ovality from a given molecular structure?
Input would be: atom types, atom positions, and Van-der-Waals radii of the atoms.
I'm already using RDKit and maybe it can do it, and I just haven't found this feature.
Thank you for any hints.
I guess there was need for clarification:
The ovality (O), a parameter used as shape descriptor, is defined as the ratio between actual surface and minimum surface for the same volume (the surface of a sphere with a volume that equals the actual volume) (see
https://www.sciencedirect.com/topics/chemistry/molecular-size)
For this, the van-der-Waals surface and the van-der-Waals volume of the molecule need to be calculated, basically by intersecting spheres with different sizes at different positions.

It may be too crude a method, but why not start by calculating moments of inertia assuming all masses are the same. This will give you a prolate or oblate spheroid from which you can get area & volume etc. and is simple to calculate.

You are looking for the terms oblate (round & flat) and prolate (rugby ball). You usually quantify their shape with the aspect ratio.

@Karl and porphyrin, I would like to calculate the van-der-Waals-surface and the van-der-Waals-volume of a given molecule. How could "moments of inertia" with a crude approximation and the terms "oblate" and "prolate" help me achieving this?

Because I have never heard the term "ovality", and suspect that you may find other literature with more common search terms, which also don´t exclude the oblate part of molecules? Spheroid might also be a good search term.

@Karl, ok, I see. Sorry, I was not aware that "ovality" is not such a common term. I will update the question.

If I recall correctly, actually, I have found the term "ovality" used more in SAR analysis (in the bioinformatics/drug discovery/ligand design field) rather than in the pure chemical field. By the way I am not able to check it out, but maybe DataWarrior can compute such descriptor :)

yes, ovality it's a thing in chemistry! check this: https://www.pirika.com/ENG/TCPE/Den-H5.html

@The_Vinz, thank you for the hint. DataWarrior looks interesting, but I haven't yet found this feature. Even if it has it, it would be cumbersome to integrate it into a Python program.

Sphere vs cigar, but what about pancakes? (Pancakes are tastier and less damaging to health than cigars .)

I created a python library for estimating the ovality of molecules called pychemovality.
To estimate ovality for a molecule follow the below steps:
Clone the repo and cd into root dir:

git clone https://github.com/vandan-revanur/pychemovality
cd pychemovality

Three types of coordinate files can be used as input for calculating ovality.

XYZ file
PDB file
MOL file

To get the ovality of the molecule in your coordinate file run the following:
python pychemovality --coord-file-path <path-to-your-coord-file>


text:
Chemical change - decomposing single element?
If I had a molecule of a single element such as atmospheric oxygen O2, and I added energy to decompose it into two separate oxygen atoms (i.e. O2 + ENERGY –– 2 O ) is this a chemical or physical change?

Please see my response on chemical and physical changes https://chemistry.stackexchange.com/questions/131474/why-is-solubility-often-listed-as-a-chemical-property-isnt-the-action-of-disso/131477#131477

At an elementary level, your decomposition of O2 molecule into O atoms is a chemical change.

Thanks for the feedback. I assumed that this was a chemical reaction because the chemical formula is changing. It does seem the distinction between physical and chemical changes can get a bit nebulous though.

Yes and in the long term this labeling is useless.

Something like physical and chemical "change" has only notion in some introductory textbooks. Chemical reaction is the important term.

It is a chemical change
The aspect of this that is probably confusing you is that both the product and the result in the case of an element are still a form of the element.
But elements can exist in different chemical forms and moving between them is still a chemical process. A relevant example is the reaction occurring mostly in the upper atmosphere that converts dioxygen (O2) to ozone (O3). This is a specific example of what happens when you add energy (in this case from UV light) to an oxygen molecule in the presence of other oxygen molecules: the split-apart oxygen atoms react to form ozone (which is more stable than a free oxygen atom). This is clearly a chemical change as the molecular structure is different, though ozone is still elementary oxygen so the element has not changed.


text:
How is the ΔL = ±1 selection rule derived?
I understand there are selection rules in spectroscopy and quantum mechanics which governs whether a transition happens or not. For example, in atomic spectroscopy, the total orbital angular momentum quantum number $L$ can only change by plus or minus one:
$$\Delta L = \pm 1$$
I'd like to know the derivation of this selection rule, but I dont know where to start from.

I assume you mean atomic spectroscopy, but please edit if I have misinterpreted.

The photon is spin 1. Note that this is not an absolute rule.

yes i mean atomic spectroscopy, sorry! How could I continue on from this @JonCuster

It is due to the photon, whose spin is equal to $1$. If a photon is absorbed, this spin is not lost. It must change the quantum number of the target.

Angular momentum is always conserved. The photon has one unit and so the atoms's angular momentum must change by this amount also. When examining the equations, involving the dipole $E.x$ induced in the atom(or molecule) by the radiation and the wavefunctions, the selection rules are found so that a solution is not zero, i.e. there is a finite chance of absorbing a photon. In one dimension the equations have the form $E\int\psi_f x\psi_idx$ where $\psi_{i,j}$ are the initial and final wavefunctions and $x$ the displacement that causes the dipole, and $E$ the amplitude of the radiation.


text:
Optical activity of 3‐chlorobutan‐2‐ol
In the picture that I have uploaded, I have read that the given compound is chiral.
But on drawing the sawhorse structures (for better visualisation), I feel like the compound is symmetric about a plane that could bisect both OH and Cl.
What is it that I'm thinking incorrectly?

OH isn't a single point, to begin with, and who said the projection is eclipsed and not staggered (MolView)?

Since when methyl groups are identical with hydrogen atoms?

What you call a symmetry plane is not a symmetry plane. For example the H atoms above the plane are transformed into something else, namely CH3 by your "planar" symmetry. They should have been transformed into H atoms !

@Mithoron That's the elephant in the room I totally missed and started with nitpicking. Nicely noticed. Maybe you could post a brief answer?

Originally, both of us, andselisk and I, thought you are taking about a plain which goes through all four groups ($\ce{H-H-Cl-OH}$ squire). It is a forbidden plane when it come to symmetry of a compound, if it had been the case. Fortunately it is not the case. We later realized after Mithoron's comment that the plane you are talking about actually goes through $\ce{C^*-C^*-Cl-OH}$ squire instead (asterisks indicate chiral $\ce{C}$s). That is an allowed plane for symmetry concern. Yet again, you are wrong about it's been a plane of symmetry. For your convenience, I draw all four possible isomers of 3‐chlorobutan‐2‐ol and relevant sawhorse structures showing the plane of your concern:

I tried to put the plane as it is a mirror on a plane where $\ce{C^*-C^*-Cl-OH}$ squire overlap. The particular stereoisomer representing both Fischer projection and sawhorse structure (with your 'plane of symmetry') is $(2R,3S)$-3‐chlorobutan‐2‐ol, so I drew that particular isomer first. Now look at my sawhorse structure below that particular structure. The $\ce{C^*-C^*-Cl-OH}$ squire is on the plane so that those groups and atoms do not count on symmetry. The only countable groups are two methyl groups and two hydrogen residues. If you consider the plane as a mirror (which dissects each $\ce{H-C^*-CH3}$ angle), you would see both $\ce{CH3}$-groups are in the front of the mirror and hydrogen residues are inside of the mirror (resembling mirror images). As a result, we can conclude that the mirror is not a plane of symmetry because each $\ce{H}$ atom can't be the mirror image of adjacent $\ce{CH3}$ group. Therefore, both carbon atoms marked with $\color{red}{red}$ $\color{green}{green}$ circles are chiral centers. As a result, there are 4 stereoisomers for 3‐chlorobutan‐2‐ol: $(2R,3S)$-, $(2S,3R)$-, $(2R,3R)$-, and $(2S,3S)$- where first two and last two isomers are enanthiomers of each other (as indicated in the diagram).

Thank you so much i get it now. I have just another doubt. Can you add the sawhorse diagram of meso and chiral forms of tartaric acid too? Only if you are okay with it

Check here. If you don't understand, let meknow or post as a new question, but make sure to show your work.


text:
How to convert pCO2 (uatm) in seawater to grams
I have seawater pCO2 measurements, and I need to convert them from uatm to moles or grams for around a 50m depth integration.
I know the solubility coefficient and temperature among other data.
For example, I have:

445.87uatm seawater pCO2
2.77*10-5  (mol atm-1 m-3) solubility coefficient
50 m Hypothetical integration depth
12 (carbon to mole conversion)

The air-sea flux equation is F=delta pCO2 (atm) * solubility (mol atm-1 m-3)  * gas transfer velocity (m / yr)
Which gives an answer in mol / m2 / year.
To calculate the total volume of carbon in grams, can I use:
pCO2 (uatm) * 1 000 000 (atm) * solubility  (mol atm-1 m-3) * integration depth (m) * 12 (to convert to grams). And then if I had a time component, divide by the number of days to get (g/m2/day).
This gives:
(445.87*1 000 000)*2.77e-5*50*12 =  7410359.39 (gC / m2)  or 7410.36 kgC/m2.
Is this correct? Is the integration depth a valid assumption in replacing gas transfer velocity?
Is using the solubility the correct way to calculate this mass using Henrys' Law?
Can you point me in any resources to check these calculations?
Thanks very much.

The solubility coefficient seems to me to be an error. It is only 1.2 mg /m3 for pure CO2. BTW, It is better to stay in algebraic equation form as long as you can, plugging in numbers in the last minute. Correctness or mistakes are easier to be seen.

I'm a bit befuddled about what you are doing, but I'd expect the solubility coefficient to be affected by pH, temperature and salinity.


text:
Chemicals to increase surface tension of water
I want to increase the surface tension of water by 1.2 - 2 times. How about adding one or more chemical substances to the water that can possibily increase its surface tension by the desired amount? But I don't know what chemical substances to add. Do you what chemical substances can I add?

The surface tension of water is $\pu{72 dyne/cm}$; it is sometimes compared to mercury, which has a surface tension of $\pu{486 dyne/cm}$. The higher surface tension of the mercury is because the atoms of mercury bond much more tightly.
A molecule at the surface of a liquid experiences net inward cohesive forces. Usually, we try to reduce those forces in water so it can wet surfaces more easily. Water does not wet waxed surfaces because the cohesive forces within the drops are stronger than the adhesive forces between the drops and the wax (source).
It is claimed that most inorganic solutes increase the surface tension of water. Another source suggests that surface tension could be increased if you increase the polarity of water by adding a more polar substance. But there seems to be contradictory evidence both $\ce{NaOH}$ and $\ce{Na2S2O3}$ decrease the surface tension of water. And there is an article that shows $\pu{8M} \  \ce{NaCl}$ to have a surface tension of about $\pu{85 dyne/cm}$. So it would seem that if there is an effect, it is small.
So let's try another method to increase the net inward cohesive forces. We could add a thickener, like xanthan gum, which at very low concentrations (less than 1%), will thicken the water so that you could float not only a needle, but a penny or a nickel on the surface - for a long time, maybe not forever. In a way, the surface tension is increased, but it is time-dependent, because the water molecules are highly immobilized, yet still somewhat free to move. Other thickeners could behave similarly.
I will yield if someone claims that the thickener achieves not a surface tension increase, but rather a viscosity increase.


text:
Chemistry of heat and egg whites
I tried to make Italian buttercream icing the other day and followed this recipe.
Unfortunately, the icing was completely runny / liquid. So I know I made a mistake somewhere.
But since I don't understand the chemistry of egg whites I am not sure how to figure it out.
I think the key to the change from liquid to creamy/spreadable is when you pour hot syrup down the side of the bowl while keeping on mixing at high speed.
So what exactly happens there? I think this is where something went wrong but I am not sure what. I had my perfect egg whites beaten to nice firm froth and my no longer boiling syrup that I poured down the side of the bowl while furiously mixing and yet the result was close to water in consistency, even after adding the butter!
I appreciate any help in understanding how something liquid can be turned into something of high viscosity with this process! Thank you so much in advance for giving me insight into egg whites in the kitchen!

The recipe linked to the question is not very detailed. It does not even say to beat the egg whites (which you did and which needs to be done). Here is a snippet of another recipe:

You just need a sugar thermometer so you can cook the sugar until it reaches a temperature of 113°C/235°F. Once the sugar syrup reaches that temperature, immediately take it off the heat – or you’ll risk bringing the sugar up to ‘hard-ball’ stage – and carefully drizzle it into the foamy egg whites, mixing continuously. You do remember the foamy egg whites, right? I bet you do. Drizzle the hot (!) syrup slowly into the egg whites, making sure not to pour it directly onto the whisk attachment of your mixer, or the syrup could splatter against the sides of the bowl or into your face.

Source: https://thetoughcookie.com/2014/09/11/different-kinds-meringue-make-italian-meringue/
I think the key is "drizzle into, mixing continuously". And here is a picture of what it should look like at that stage (from the same source):

[OP} But since I don't understand the chemistry of egg whites I am not sure how to figure it out.

There is some chemistry and some physics here. You want to add a lot of sugar and little water. You need the sugar in solution so you can add it quickly, and if the syrup cools down, it is no longer a liquid. On the other hand, you don't want the egg white to heat up too much, so that's why you drizzle in the syrup while mixing vigorously. That way, the syrup cools down while being diluted with the water in the beaten egg whites, effectively removing water from the bubbles surrounded by the protein from the egg white, making the foam more stable.
If everything else was correct (water content of syrup, using bowls that are not greasy, correct ratio of egg whites to syrup), I think the key is to drizzle the syrup into the egg white where it gets mixed right away. You describe adding it to the side of the bowl, but the recipe does not say that. If I were to experiment with the recipe, this step would be my focus.

Thank you very much for your detailed answer. You wrote I want to add a lot of sugar and little water. So basically the water in the syrup is just for the purpose of making the sugar liquid so I could try with less water?


text:
Why doesn't core charge increase down a group?
Atomic radius increases down a group because the electrons feel a lesser attraction to the positive nucleus (due to shielding from inner shells). Why then, doesn't core charge decrease seeing as core charge is just the attraction felt by the valence electrons towards the nucleus (and that attraction is said to decrease down a group as per the reasoning for the increase in atomic radius)?

Technical note: Try to formulate shorter sentences. As a guidence for technical/scientific texts, it is adviced to write sentences with 20-25 words in average and discouraged writing sentences longer than 35 words. Your question has 66 words and is difficult to read and comprehend. Try to read your own question after several month. You may forget the beginning before you reach the end. Then you may need to read it again and again.

The felt attraction is proportional to $\frac{Z_\mathrm{eff}}{r^2}$. Ramember also electron repulsion and in the first place the quantization, which is the primary reason for atomic size getting bigger.


text:
Finding initial amount of ammonia using equilibrium constant and amount of nitrogen gas
Question

When an amount ammonia is added at $\pu{600 K}$ in a $\pu{1 L}$ container the following reaction takes place:
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
The equilibrium constant $K_c = 4.20$ at $\pu{600 K}.$
At equilibrium, it is known that $\pu{0.200 mol}$ of $\ce{N2}$ gas exist in the container. What amount of ammonia was added at the start of the reaction? Choose from the answers below:
$\pu{0.826 mol};$ $\pu{0.482 mol};$ $\pu{1.226 mol};$ $\pu{0.400 mol};$ $\pu{0.800 mol}.$

My attempt
I created an ICE table, but I think it's wrong:
$$
\begin{array}{lccc}
\ce{&N2(g) &+ &3 H2(g) &<=> &2 NH3(g)} \\
\text{I} & 0 && 0 && y \\
\text{C} & +x && +3x && -4x \\
\text{E} & 0.2 && 0.6 && y-0.8 \\
\end{array}
$$
$$K_c = \frac{(y - 0.8)^2}{0.200\times 0.6^3}$$
Solving the equation with Maple gives me none of the answers above. I must have messed up on the table. Can anyone tell me where I messed up?

The method is ok, but you made a slip, the term for ammonia is not -4x, correct this and you will get one of the answers. In an exam you can get the answer easily with a hand calculator.

Since the volume is given and it's a constant, the initial amount of ammonia $n_0(\ce{NH3})$ can be found from its initial concentration $c_0(\ce{NH3}):$
$$n_0(\ce{NH3}) = c_0(\ce{NH3})\times V\tag{1}$$
To find $c_0(\ce{NH3}),$ an ICE table might indeed come in handy; however, yours needs corrections.
First, I suggest to rewrite it according to the process occurring in the system, namely dissociation of ammonia:
$$
\begin{array}{lccc}
\ce{&2 NH3(g) &<=> &N2(g) &+ &3 H2(g)} \\
\text{I} & c_0 && 0 && 0 \\
\text{C} & -2x && +x && +3x \\
\text{E} & c_0 - 2x && x && 3x \\
\end{array}
$$
Second, note that the provided equilibrium constant $K_c$ is given for the synthesis of ammonia, hence in our case a reciprocal value is relevant:
$$
\begin{align}
\frac{1}{K_c} &= \frac{[\ce{N2}][\ce{H2}]^3}{[\ce{NH3}]^2}\tag{2.1}\\
\frac{1}{K_c} &= \frac{x (3x)^3}{(c_o - 2x)^2}\tag{2.2}\\
\frac{1}{K_c} &= \frac{27x^4}{(c_o - 2x)^2}\tag{2.3}\\
\frac{1}{\sqrt{K_c}} &= \frac{3\sqrt{3}x^2}{c_o - 2x}\tag{2.4}
\end{align}
$$
$$c_0 = 3\sqrt{3K_c}x^2 + 2x\tag{3}$$
At the equilibrium there is $\pu{0.200 mol}$ of nitrogen in the $\pu{1 L}$ vessel, so $x = \pu{0.200 mol L-1},$ and the initial amount of ammonia can be found as follows:
$$
\begin{align}
n_0(\ce{NH3}) &= (3\sqrt{3K_c}x^2 + 2x) × V\\
  &= (3\sqrt{3\times 4.20}\times (\pu{0.200 mol L-1})^2 + 2\times \pu{0.200 mol L-1})\times \pu{1 L} \\
  &= \pu{0.826 mol}\tag{4}
\end{align}
$$


text:
Does the first step in electrophilic aromatic substitution follow equilibrium? If so, can equilibrium constant be calculated?
Using chlorine (a Lewis base) and ferric chloride (a Lewis base) as an example, the first step I saw in the textbook looks like this:

The picture above depicts the Step 1 in the EAS mechanism using chlorine as an example.
I noticed that this step has two reversible arrows. I think it means chlorine, ferric chloride, the molecular complex and the ion pair will all be present in a system and their concentration values follow chemical equilibrium.
From high school knowledge (I am a freshman), we can calculate equilibrium constant using concentration. Can we still do it in this case? If so, how?

The intermediate molecular complex would only exist for a very brief period of time. It would be important in the kinetics of the overall reaction. However the equilibrium would probably be best expressed as an overall equilibrium.

If there’s an equilibrium, a constant can be calculated. Often, however, the results aren’t very meaningful especially in a case like this where the intermediate complex does not have a long lifetime.

Let me put it this way: Every compound exist in a 3N dimensional configuration space (may be simple XYZ coordinates of the system, may be internal coordinates, based on your choice), and based on the configuration, its potential energy is defined. The potential energy surface (PES) can be calculated quantum mechanically or using some force field classically. The probability of the system to be found in the given set of coordinates is given by the distribution function defined according to statistical mechanics. For a canonical ensemble, the formula is given as (Z is the partition function):
$$P(\vec{r}_{3N})= \frac{1}{Z}e^{-E(\vec{r}_{3N})/k_BT}$$
Now imagine this, a reaction between compounds containing M and N atoms is actually an exploration of the potential energy surface in 3(M+N) dimensions. Reactants, product and intermediates are given by local minima in the PES while transition state is given by maxima along one dimension and minima along all other dimensions. In other words, for a stable state (reactant, product, intermediate, etc.) all the frequencies (square root of the second derivate of the PES) is real, while for transition state one of the frequencies is imaginary.

Now coming to your question: You will require to calculate the energetics of each of the step of reaction considering the presence of solvent (this is probably happening in presence of solvent), plug in your values to the appropriate statistical ensemble (if it is inside solvent without release of gaseous products, canonical ensemble should be good enough) probability function and you will get an idea about the percentage of intermediates you get.
In most high school textbooks, the reaction marked with double arrows simple means the reactant, product (and in your case the intermediate) is not too much different in terms of potential energy. For a single headed arrow, it means that the difference between potential energies of the minimas (reactant and product) is significant and most of the compounds (in an ensemble) will be trapped in the lower minima.
In case you want to read more about it, I suggest the following books:

Statistical Mechanics by McQuarrie
Physical Chemistry by Atkins
Physical Chemistry by Levine

For more details about reaction dynamics, you may refer to:

Molecular Reaction Dynamics by R. D. Levine


text:
Is reduction of copper oxide a combustion reaction?

Is the following reaction a combustion reaction or just a single displacement? Could be both?
$$\ce{H2 + CuO -> Cu + H2O}$$

A friend of mine claims it is a combustion reaction because it is exothermic and a redox reaction.

No, combustion is a loose term which implies a visible flame.

Think about what all is required for a combustion reaction (think formulas)

@kskinnerx16, There is no formula for combustion if you are thinking of carbon dioxide and water as products. All you need is a (usually) visible flame to call it a combustion process. Sodium can "burn" in chlorine, right?

@M.Farooq Certainly. Perhaps we need more context from OP, as I took this as a fairly basic level question (identifying reaction types).

For the record, so nano-related thermites are not associated with 'combustion reaction happens quickly, producing heat, and usually light and fire. " My advice, do not even attempt one for safety reasons.

Kareem: If you want to impress your teacher with your knowledge by citing a 2019 article reporting, for the 1st time, successfully stabilized nano-CuO and possible thermal and reactive consequences, therefrom, do mention a nano perspective on this reaction.

By the question asked, I assume OP is in high school of freshman in college. Thus, I'd use this general description of combustion as a starting point:

Combustion reactions are common and very important. Combustion means burning, usually in oxygen but sometimes with other oxidants such as fluorine. A combustion reaction happens quickly, producing heat, and usually light and fire. Combustion describes how the reaction happens, not the reactants and products. Chemists as early as Lavoisier suggested that people get their energy from combustion-like reactions, but even though the products and reactants are the same when you burn food in a fire and in your body, the way it happens is different. In a combustion reaction, the thing that burns (the reactant that isn't $\ce{O2}$ or $\ce{F2}$) is called the fuel. Combustion reactions are a type of redox reaction.

As  M. Farooq pointed out a combustion reaction happens quickly, producing heat, and usually light and fire. For example, lets look at combustion reaction of an alkene (a hydrocarbon). If it is a complete combustion, the fire have a blue flame:
$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$
If it is a partial combustion, it can have a multiple $\ce{C}$ compounds as products, and have a yellow flame due to presence of elemental $\ce{C}$:
$$\ce{C_nH_{2n} + x O2 -> m C + p CO + $(n-m-p)$CO2 + n H2O}$$
where $x = \frac{2(n-p-m) +p}{2} = \frac{2n-2p-2m +p}{2} = \frac{2n-p-2m)}{2}$. In your reaction would not produce fire and it didn't use either oxygen or other oxidants ($\ce{CuO}$ is not that type of oxidant). It is true that the reaction is a redox reaction.

@KareemAli, he means CuO is not a gaseous oxidizer, which is typically needed in a combustion reaction.

LOL - The reaction is also a hard way to get hydrogen to oxidize.

@ M. Farooq: Man, you read my mind! :-)

@MaxW, this reaction was historically very important in analytical chemistry. "Copper oxide (CuO) is used extensively in elemental analysis for the oxidation of light hydrocarbons and carbon monoxide into carbon dioxide and of hydrogen into water"

Actually, with nano CuO, a remarkably exothermic reaction with hydrogen is expected per my cited Science Direct source.

So, nano related thermites are not associated with 'combustion reaction happens quickly, producing heat, and usually light and fire. " My advice, do not even attempt one for safety reasons.

Your friend is right, it 'can' produce actually an extraordinary combustion/thermite reaction.
In fact, for example, one can expect extreme temperatures when mixing fine metal powder (for example Aluminum dust) with nano-CuO. Per this June 2019 source from Science Direct, to quote from the abstract:

Nanothermites (metal oxide/metal) can offer tremendously exothermic self sustained reactions. CuO is one of the most effective oxidizers for naonothermite applications. This study reports on two prospectives for the manufacture of CuO nanoparticles....The superior performance of colloidal CuO particles was correlated to their steric stabilization with employed organic solvent. This is the first time ever to report on fabrication, isolation, and integration of stablilized colloidal nanothermite particles into energetic matrix where intimate mixing between oxidizer and metal fuel could be achieved

And from the body of the work:

CuO/Al is one of the most effective thermite mixtures in terms of combustion temperature, heat output, and gasous products [9,10]. CuO/Al can offer an adiabatic flame temperature of 5718 K with heat of reaction of about 4100 J/g, and a reasonable amount of gaseous products (0.345 g gas/g thermite). Furthermore the formed Cu metal (resulted from thermite reaction) can be transformed into gas due to its low boiling point (1000 °C) [11].

Working with gaseous hydrogen in place of Aluminum is expected to be less exothermic, but more likely, I suspect, an explosion hazard. For support, based on this 1940 article, for my supposition based on the known combination reaction of H2 with O2 in the catalytic presence of Cu metal (likely with a Cu2O layer). This likely implies a possible radical-induced hydrogen atom acting on oxygen (nano-CuO) pathway.

I do not know who has a problem with my cited source  (Science Direct) and is further not acquainted with the activity of  nano-oxides, but I will kindly provide educational references.

So, nano-related thermites are not associated with 'combustion reaction happens quickly, producing heat, and usually light and fire. " ? My advice, do not even attempt one for safety reasons.


text:
Will toluene vapors damage a mirror made from PETG?
I'm working on a project for work and trying to avoid using an acrylic mirror because I know that doesn't hold up well against toluene, but I'm wondering if a mirror made out of PETG would?

PETG is also not resistant against toluene.
A month-long contact under ambient conditions results a complete swollen/disintegrated PETG material (PDL rating 0), and an year-long exposure even causes whitening [1, p. 2660]:

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; PDL handbook series; William Andrew: Norwich, N.Y, 2012. ISBN 978-1-4557-7896-6.

Thanks, have you yourself ever heard of a plastic that's actually resistant to toluene that can be used for a mirror? I'm having trouble finding anything, I mostly see Teflon and Nylon..

@Zayas12 From the top of my head, not really, sorry.


text:
What exactly makes a carbon atom "α" in a protein residue?
I'm a computer scientist and have no biochemical background, but was working on a project recently that had me going through a lot of protein residues for one reason or other. Feel free to correct me from here on, but my understanding is that proteins, also called "peptides" also "subchains" when a part of a larger macromolecule are made of amino-acids(central dogma) which are also frequently referred to as "residues".
Upon closer (programmatic) inspection, each residue (amino-acid) turns out to be a bunch of atoms like so:

I was wondering what is the significance of that CA there in each of the residues, and whether i identified it correctly as "alpha-carbon". Protein Data Bank docs refer frequently to a residue's alpha-carbon. If it is indeed the alpha-carbon, i have two questions which are probably dead-simple to anyone with biochem background:

Why is that one(in space) considered alpha and not the carbon atom next to it?
Is it a given that each residue out there has only one alpha-carbon? Seems pretty obvious, but please.

Thanks a ton!

The $\alpha$-carbon is the carbon attached to the carboxyl (-COOH) functional group.

@MaxW In this case these are connected with amide groups (besides the C-end, where it's indeed COOH).

Huh, your output there seems to be indeed confusing. What's its source?

The protein backbone runs as  $\mathrm{\cdots C-N-C_\alpha -C-N-C_\alpha \cdots }$ but is more often written as $\mathrm{-C(=0)N(H)-C_\alpha(R_1)-C(=0)N(H)-C_\alpha(R_2)-}$ where the $R$ are the amino acid residues always attached to the carbon alpha, and the N atom always has a hydrogen and the C always an oxygen. The $C_\alpha$ are usually labelled $C_{\alpha 1},  C_{\alpha 2}$ etc. It would be easier to look up some pictures.

Question 1: Why is that one(in space) considered alpha and not the carbon atom next to it?

All human proteins consist of $\alpha$-amino acid residues. An $\alpha$-amino acid means the carboxylic acid group ($\ce{COOH}$) and amino group ($\ce{NH2}$) are separated by one $\ce{C}$ carbom atom, which is called $\alpha$-carbon ($\ce{C}_\alpha$; See the insert at bottom right of the diagram):

Usually, backbone of a protein ($\alpha$-helix) is written as: $$\ce{H2N-C_\alpha(R^1)-C(=O)-NH-C_\alpha(R^2)-C(=O)-NH-C_\alpha(R^3)-C(=O) -}\cdot \cdot \cdot \ce{-NH-C_\alpha(R^n) -COOH}$$
For example, the dipeptide in this diagram can be written as:
$$\ce{H2N-C_\alpha(R^1)-C(=O)-NH-C_\alpha(R^2) -COOH}$$
As demonstrated in these backbones, you'd see each $\ce{C_\alpha(R^1)}$ is in place between $\ce{NH}$ and $\ce{C(=O)}$ (the $\ce{R^1, R^2,}$ etc. are corresponding side chains of the particular amino acids). Thus, $\text{<Atom C>}$ next to $\text{<Atom CA>}$ ($\ce{C}_\alpha$) in the written chart of the program is referring to the carbonyl $\ce{C}$ of that particular amino acid. For example, let's consider three amino acids in the given chart: (10 Arginine; (2) Lysine; and (3) Leucine:

Arginine (arg) residue in the chart is can be written as follows (the side chain in parenthesis):
$$\ce{-HN-C_\alpha(CH2CH2CH2NHC(=NH)NH2)-C(=O) -}$$
According to the nomenclature off the program, you can rewrite it as:
$$\ce{-HN-C^A(C^BH2C^GH2C^DH2N^EHC^Z(=N^{H1}H)N^{H2}H2)-C(=O) -}$$
where $\ce{A #} \alpha$; $\ce{B #} \beta$; $\ce{G #} \gamma$; $\ce{D #} \delta$; $\ce{E #} \epsilon$;$\ce{Z #} \zeta$; and $\ce{H #} \eta$ (following Greek alphabet letters). Since two nitrogen atoms are attached to $\ce{C}_\zeta$, they are appropriately labelled as $\ce{N^{H1}}$ and $\ce{N^{H2}}$ after next Greek letter $\eta$.
Since the program has avoided hydrogen, let's rewrite it again without hydrogen atoms:
$$\ce{-N-C^A(C^BC^GC^DN^EC^Z(=N^{H1})N^{H2})-C(=O) -}$$
Thus, program writes it as $\ce{N -C_\alpha -C(=O) -}$ first and then the the atoms in side chain next. Hence, Residue ARG: $\text{<Atom N>}$, $\text{<Atom CA>}$, $\text{<Atom C>}$, $\text{<Atom O>}$, and then the side chain in parenthesis as: $\text{<Atom CB>}$, $\text{<Atom CG>}$, $\text{<Atom CD>}$, $\text{<Atom NE>}$, $\text{<Atom CZ>}$, $\text{<Atom NH1>}$, $\text{<Atom NH2>}$.
Similarly, lysine (lys) residue in the chart is (side chain in parenthesis):
$$\ce{-HN-C_\alpha(CH2CH2CH2CH2NH2)-C(=O) -}$$
You can rewrite it according to the nomenclature off the program (avoiding $\ce{H}$s):
$$\ce{-HN-C^A(C^BC^GC^DC^EN^Z)-C(=O) -}$$
Thus, program writes it as $\ce{N -C_\alpha -C(=O) -}$ first again, followed by the atoms in side chain. Hence, Residue LYS: $\text{<Atom N>}$, $\text{<Atom CA>}$, $\text{<Atom C>}$, $\text{<Atom O>}$, and then the side chain in parenthesis as: $\text{<Atom CB>}$, $\text{<Atom CG>}$, $\text{<Atom CD>}$, $\text{<Atom CE>}$, $\text{<Atom NZ>}$.
For leucine, $\ce{-HN-C_\alpha(CH2CH(CH3)CH3)-C(=O) -}$,you can again rewrite the formula according to the nomenclature off the program (avoiding $\ce{H}$s):
$$\ce{-N-C^A_\alpha(C^BC^G(C^{D1})C^{D2})-C(=O) -}$$
Note that since two carbon atoms are attached to $\ce{C}_\gamma$, they are appropriately labelled as $\ce{C^{D1}}$ and $\ce{C^{D2}}$ after next Greek letter $\delta$.
Thus, program writes it as $\ce{N -C_\alpha -C(=O) -}$ first again, followed by the atoms in side chain. Hence, Residue LEU: $\text{<Atom N>}$, $\text{<Atom CA>}$, $\text{<Atom C>}$, $\text{<Atom O>}$, and then the side chain in parenthesis as: $\text{<Atom CB>}$, $\text{<Atom CG>}$, $\text{<Atom CD1>}$, $\text{<Atom CD2>}$.

Question 2: Is it a given that each residue out there has only one alpha-carbon?

As explain in above backbone of the protein, you'd find only one $\ce{C_\alpha}$ fir each amino acid (which is chiral).

This is even more exhaustive and to-the-point than what i was looking to pick up. Thank you!

This nomenclature is due to the fact that amino acids are carboxylic acids.
Near the carboxylic acid moiety, the carbon chain is unbranched and simple, so the positions are named like an unbranched, simple aliphatic carboxylic acid.
The carboxylic acid ($\ce{-CO2H}$) is not indicated with a position. But the carbon immediately next to it is $\alpha$. The one next to that is $\beta$. That should be enough to explain the $\alpha$ in your context because that carbon is the one immediately next to the carboxylic acid. In a protein or polypeptide, the carboxylic acid is most frequently converted to an amide in a peptide bond, but the Greek letter nomenclature is unchanged.
This nomenclature is used in other cases for example, with an $\alpha$,$\beta$-unsaturated carbonyl.
The most publicly visible use of this nomenclature is when we use $\omega$ to denote the end of the carboxylic acid chain. This is the origin of the term $\omega$-3 [unsaturated] fatty acid.


text:
Verb for Sorption
Slightly silly question! Intuitively I want the present-simple verb for sorption to be "sorps", but saying it out loud, "sorbs" sounds much more natural. However, Google doesn't turn anything up for either of these, and they both make me sound like a redneck.
Since it definitely does not seem to be used often, it seems like the best choice would be to fall back to "adsorb" or "absorb", but I was wondering if there was a verb for the generalized sorption because it is a word I would like to use! Thanks.

Yes, not surprisingly, the verb for sorption is sorb. It has been used for > 100 years in physical chemistry.
Here are the early references from the unabridged Oxford English Dictionary (by subscription only) so I cannot give you a link.
. transitive. To collect by sorption. Also absol.
1909   J. W. McBain in London, Edinb. & Dublin Philos. Mag. 6th Ser. 18 918   An idea of the quality of the carbon employed may be obtained from the amount of gas sorbed by it in actual experiment.
1938   Proc. Royal Soc. A. 167 407   The two zeolites in the form of three-dimensional networks sorb ammonia copiously without ammoniate formation.
1954   P. Alexander & R. F. Hudson Wool viii. 261   When wool is immersed in hydrogen peroxide, some is initially sorbed by the amino and imino groups without reaction.
1970   New Scientist 2 July 9/3   Papers with inked designs sorb best on the inked areas.
1972   Physics Bull. Oct. 583/1   This has the advantage that exhausted water vapour is not sorbed by the trap on the fine side of the pump.

Thank you! For some reason I could not find any official reference to it.

The most authentic reference is the unabridged Oxford English Dictionary. It is a paid access. I am copying the references for you in the main answer.

Would be better to spell OED out. You did it in the comments, but they are not meant to last.


text:
Example of chemical reaction with a given form for the kinetic equations
I’m an applied mathematician looking for examples from various applications fields (right now, chemistry, obviously) to illustrate the following reaction-diffusion system
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\
\frac{\partial v}{\partial t} - d_v \Delta v = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4}
\end{cases}
$$
Above, all constants $d_u$, $d_v$, $k_i$, $p_i$, $q_i$ are positive, and $p_i,q_i\geq 1$. $k_2-k_3$ and $k_7-k_8$ might be of any sign. The system need not be self-contained — if you need to add a third equation in order to make these two exist, feel free to do so.
If the diffusion part is bothering you, then I’m also fine with
$$
\begin{cases}
\frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u^{p_1} v^{p_2} - k_5 u^{p_3+p_4} \\
\frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u^{q_1} v^{q_2} - k_{10} v^{q_3+q_4}
\end{cases}
$$
And if the $p_i$ and $q_i$ are bothering you, then one example where they are all equal to $1$ is
$$
\begin{cases}
\frac{\text{d} u}{\text{d} t} = k_1 v + (k_2-k_3) u - k_4 u v - k_5 u^2 \\
\frac{\text{d} v}{\text{d} t} = k_6 u + (k_7-k_8) v - k_9 u v - k_{10} v^2
\end{cases}
$$
One more precision: if $k_2-k_3\leq 0$, then $k_5=0$ is allowed, and similarly, if $k_7-k_8\leq 0$, then $k_{10}=0$ is allowed. Therefore the following system would be a satisfying answer:
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_2 u - \tilde{k}_3 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_2 u - \tilde{k}_1 v - \tilde{k}_4 uv
\end{cases}
$$
(where the constants $\tilde{k}_i$ differ a priori from the previous constants $k_i$).
I looked for actual examples of first-order reversible reactions. I found $\ce{Cu+ <-> Cu^{2+}}$ there but I do not know if having an irreversible reaction “$\ce{Cu^+ + Cu^{2+} ->}$ some product” in this configuration is possible.
Can you think of any “real” chemical system that would satisfy such kinetic equations?

This one seems to be a homework problem. Please show your efforts before you expect an answer here.

You need to have rate constants that are common between the two reactions.

@MitradipDas: I tried but I have to admit that I am no chemist. I think that the $k_1$ and $k_6$ part corresponds to a reversible monomolecular reaction. The $k_4$ and $k_9$ part could produce a third quantity, let’s say $U+V\to W$, and similarly we could have $2U\to X$ and $2V\to Y$. Is that more convincing?

@porphyrin: yes, some rates can be equal if needed, it’s fine by me. You can also have $k_2-k_3=k_7-k_8=0$ and forget about this part of the equations, if needed.

@MitradipDas: I added more details in the opening post. Hope this is satisfying.

A chemist friend found an answer for the simplest case:
$$
\begin{cases}
\frac{\partial u}{\partial t} - d_u \Delta u = \tilde{k}_1 v - \tilde{k}_{-1} u - \tilde{k}_2 uv \\
\frac{\partial v}{\partial t} - d_v \Delta v = \tilde{k}_{-1} u - \tilde{k}_1 v - \tilde{k}_2 uv
\end{cases}
$$

$u$ is the concentration of ethenol;
$v$ is the concentration of ethanal;
the tautomerization of ethenol and ethanal gives the first-order part of the reaction;
the aldol reaction of ethanal with ethenol, giving the 3-hydroxybutanal, gives the second order part.

This is definitely a satisfying answer.
Nevertheless I’m still interested in other answers, especially with autocatalysis of either $u$ or $v$ (that would decrease the first-order consumption of it) or with autoreactions (that would give the term $-k_i u^2$ or $-k_i v^2$). As a matter of fact, the French Wikipedia claims that an aldol reaction of ethanal with itself in presence of a base can also produce 3-hydroxybutanal, so that the same example would also give a term $-k v^2$? But maybe these two aldol reactions correspond in fact to the same elementary mechanism?


text:
Substitution and Elimination Reactions
Why strong bases only give elimination products as major while they could give substitution products too being good nucleophiles? (I am excluding those non nucleophilic bases)

This is an awfully general statement...

You might want to look into HSAB theory. Generally, a saturated carbon centre is a 'soft' electrophile, while an acidic proton is a 'hard' electrophile. This can be the difference in reactivity of certain nucleophiles.

As what Zhe mentioned, the statement you are making seems to be overly general. Substitution and elimination are in competition and the ratio of the two reactions depends on not just the strength of the reagent as a nucleophile or base but also, on the solvent and temperature. E.g. Hydroxides may seem like a good base and nucleophile but their reaction with alkyl halides differ based on other conditions. Generally, $\ce {KOH}$ in ethanol at high temperature would favour elimination for primary alkyl halides while aqueous $\ce {KOH}$ at lower temperatures may favour dominance of substitution.


text:
Total number of isomers formed on monobromination of 2-methylbutane

What is the total number of isomers formed on monobromination of 2-methylbutane (including configurational and stereoisomers)?

I tried working it out. However, I found only five:

According to the source, the answer is six. Which structure am I missing?

What’s the source?

One of the online practice questions i was doing

Here are the six structures that your source is probably referencing.

Oh yes! I overlooked the fact that the first structure had a chiral carbon too! Thank you very much!


text:
Distinguishing between different geometries of bromochlorobis(triphenylphosphine)platinum(II) using ¹⁹⁵Pt NMR
Question
The question off which I am basing my question assumes that $\ce{[PtClBr(PPh3)2]}$ can exist in either the square-planar or tetrahedral geometry. If one would use $\ce{^{195}Pt}$ NMR spectroscopy, how would the spectra of the possible molecules differ from each other? The effect of the $\ce{Cl}$ and $\ce{Br}$ can be disregarded.
My reasoning
For the square planar geometry, either the cis or trans isomer is possible. I am assuming that you won't be able to distinguish between the square planar trans molecule and the tetrahedral molecule since the $\ce{PPh3}$ ligands in each molecule are equivalent. The $\ce{^{195}Pt}$ NMR spectra for these two orientations will thus be the same, a single triplet.
Will the square planar cis molecule also produce the same spectrum or will there be second-order coupling present since the $\ce{PPh3}$ ligands are non-equivalent? How would the $\ce{^{195}Pt}$ NMR spectra then look for the cis isomer?


text:
Chemical that will dissolve the oils left over from broken down coffee grounds?
When coffee grounds breakdown in pipes there is a black sludge residue as a result. Is there a chemical that would dissolve this residue?

The obvious answer here is a good “professional strength” commercial drain cleaner: they are formulated to do the job without harming the pipes (whatever they happen to be made of). Maybe it will take multiple applications, but that is far better than risking real problems by using chemicals that are not approved for the purpose. There is also the environmental aspect of the waste stream after the cleaning.

@EdV Not whatever they're made of - Al pipes can be damaged, for example.

@Mithoron I was not implying that an arbitrary drain cleaner formulation would work with an arbitrary pipe material: the containers tell purchasers all the usage and safety details. My own house plumbing is a mix of copper pipes, cast iron stand pipe and several types of plastic, but no aluminum. Of course, there are always people who ignore all safety and usage information and mix various bathroom and plumbing chemicals, sometimes with fatal consequences. As well, I seriously doubt aluminum pipe would be legal in homes in the USA.

@Mithoron, I don't recall any place where aluminum drain pipes are used in kitchen sinks, or washrooms. Nobody would use a highly reactive metal to make pipes for kitchen sink waste which is always alkaline from detergents etc.