text:
What is the name of the glyoxal radical after removing two hydrogen atoms?
What is the name of the radical that results from removing two hydrogen atoms from glyoxal ($\ce{CHO-CHO}$)? It has a roughly this structure: $$\ce{O=C^.-C^.=O}$$
It is an important intermediate in heterogeneous electrocatalysis (for instance, J. Phys. Chem. Lett. 2015, 6, 11, 2032-2037). For simplicity, in most articles it is called "carbon monoxide dimer". However, it should have a systematic name like other crucial intermediates like $\ce{-CHO}$ (formyl) or $\ce{-OCH3}$ (methoxy).
Please note that the answer is probably not ethenedione, which is the hypotetic dimer of carbon monoxide in gas phase.

It is ethenedione even if it's in excited state.

The relevant rule for the systematic names of polyradicals in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-71.2.3 Multiple radical centers (Polyradicals)
Polyradicals containing two or more radicals centers, formally derived by the removal of two or more hydrogen atoms from each of two or more different skeletal atoms of a parent hydride, are named by adding to the name of the parent hydride combinations of the suffix ‘yl’ for a monovalent radical center, ‘ylidene’ for a divalent radical center, and ‘ylidyne’ for a trivalent radical center, together with the appropriate numerical prefixes indicating the number of each kind of radical center. The final letter ‘e’ of the name of the parent hydride, if present, is elided when followed by ‘y’. All substituents, including characteristic groups, when present, are cited as prefixes. Preferred IUPAC names result from the application of this rule.

Thus, the name of the parent structure is ethane-1,2-diyl.
The characteristic aldehyde groups of the original glyoxal (preferred IUPAC name: oxaldehyde; systematic IUPAC name: ethanedial) can no longer be expressed using the suffix ‘al’. They have to be cited as prefixes now. Therefore, the complete name of the compound that is given in the question is 1,2-dioxoethane-1,2-diyl.


text:
What are the chemical Tests and/or reactions to distinguish propanone from butanone?
Is there any series of chemical reactions that will help in distinguishing propanone and butanone? A simpler series of reactions or tests would be optimal.
I know it's easy to distinguish and possibly separate by physical properties like boiling point but I need a chemical test or reaction.
So far I have thought up using NBS (N-Bromosuccinimide) to brominate both ketones. And possibly converting products into alcohols and then using Lucas Reagent to distinguish between the different alcohols. In the case of propanone the $\ce{Br}$ and hence the $\ce{OH}$ group is attached to the methyl group whereas in butanone it is attached to the ethyl group. I am not sure if this will work.

What about physical properties such as boiling points?

You may also need to edit your posy to show your work to find the difference. Otherwise, this may get on hold tag as homework question.

The boiling points are $\pu{25°C}$ apart, it should be easy. Failing that, NMR or refractive index.

I appreciate the answers and I know we can practically use physical properties but this was asked in a test and I'm not sure my answer is the simplest.

Haloform reaction? Acetic acid from acetone; propanoic acid from 2-butanone. But now you have to distinguish between the 2 acids. Back where you started, in a sense.

Do analytic methods count? E.g burning and determining the elemental composition from the CO2 / H2O content would be a chemical solution to the problem.

Huh? Then I'd be interested in seeing what's in answer sheet for this test, because as far as I can tell, there's no chemical test that will tell them apart. Their chemical properties are virtually identical and neither Lucas no haloform or any simple text would work...

Different carbonyl compounds were identified in the past by making stable crystalline imines, which have sharp melting points and these melting points differ greatly from compound to compound.
This was accomplished by making imines out of semicarbazide or 2,4-dinitrophenylhydrazine.

They're hydrazones or semicarbazones, but your approach is correct

Carbonyl compounds such as aldehydes and ketones reacts readily with derivatives of ammonia containing a hetero atom attached to the ammonia nitrogen to give stable solid derivatives with sharp melting points. A well known examples of those derivatives of ammonia containing a hetero atom attached to the ammonia are hodroxyl amine ($\ce{H2N-OH}$), semicarbazide ($\ce{H2N-NH-C(=O)-NH2}$), and phenylhydrazine ($\ce{Ph-NH-NH2}$).
The most common phenylhydrazine used is 2,4-dinitrophenylhydrazine (2,4-DNP), because it gives hydrazone derivative, which can be used as qualitative test for these carbonyl compounds.
Traditionally, carbonyl compounds can be identified using sharp melting points of two of their nitrogen derivatives, e.g., corresponding semicarbozones and 2,4-dinitrophenylhydrazones. For example, one of the laboratory experiment for sophomore undergraduates taking Organic Chemistry II course at Texas A&M University is Reactions of Aldehydes and Ketones. During this lab experiments, students needs to make two derivatives (semicarbozones and 2,4-dinitrophenylhydrazones) of their unknown compound, and identify them by their melting points and other classification tests (to judge the unknown is aldehyde or ketone, and is the type of $\ce{CH3C(=O)-R}$). I listed the given values of several melting point values in the corresponding Laboratory Manual:
$$
\begin{array}{ccc} \\\hline
\text{Name of Unknown} & \text{Boiling point} & \text{M.p. of semicarbazone} & \text{M.p. of 2,4-DNP}\\\hline
\text{2-Propanone} & \pu{56 ^\circ C} & \pu{190 ^\circ C} & \pu{126 ^\circ C} \\
\text{2-Butanone} & \pu{80 ^\circ C} & \pu{145 ^\circ C} & \pu{118 ^\circ C} \\
\text{3-Methyl-2-butanone} & \pu{94 ^\circ C} & \pu{113 ^\circ C} & \pu{124 ^\circ C} \\
\text{2-Pentanone} & \pu{102 ^\circ C} & \pu{112 ^\circ C} & \pu{144 ^\circ C} \\
\text{3-Pentanone} & \pu{102 ^\circ C} & \pu{139 ^\circ C} & \pu{156 ^\circ C} \\
\text{2,4-Dimethyl-3-pentanone} & \pu{124 ^\circ C} & \pu{160 ^\circ C} & \pu{88 ^\circ C} \\
\text{2-Hexanone} & \pu{128 ^\circ C} & \pu{125 ^\circ C} & \pu{110 ^\circ C} \\
\text{4-Methyl-3-penten-2-one} & \pu{130 ^\circ C} & \pu{164 ^\circ C} & \pu{205 ^\circ C} \\\hline
\end{array}
$$
The first two is two compounds you want to identify: 2-propanone and 2-butanone. By the values given,, you'd be able to identify them exclusively by chemical and physical means. Needless to say that preparation of corresponding semicarbozones and 2,4-dinitrophenylhydrazones of given carbonyl compounds are so simple (simplest possible) that even undergraduate students with no prior experience in organic chemistry can also do it.
Then again, we have now introduced simple analytical method to identify unknown carbonyl compounds: $\ce{^1H}$-$\mathrm{NMR}$, aiming students to have well understand of spectroscopy. Following is two $\ce{^1H}$-$\mathrm{NMR}$ spectra of propanone (acetone) and butanone (ethylmethylketone). As you see, you can easily distinguish the two compounds in minutes: Propanone gives only one resonance for the methyl group, a singlet. Butanone, on the other hand, gives three signals - a quartet ($\ce{-CH2 -}$), a singlet ($\ce{-CH3}$), and a triplet ($\ce{-CH3}$).

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text:
What is tetrasulfur like? Why is there so little experimental information about it?
Phys.org's How sulfur helped make Earth habitable before the rise of oxygen says

Mapping the bonds and vibrational modes of molecules containing sulfur isotopes is helping to shed light on the chemical reactions that took place in Earth's atmosphere during the Archean era, before the atmosphere became oxygenated about 2.5 billion years ago.

and

The sulfur in the Archean Earth's atmosphere was emitted by volcanic activity, and through a process called mass independent fractionation, sulfur's various isotopes... became enriched in a manner that does not correlate with their mass.

It links to Computational analysis of vibrational modes in tetra-sulfur using dimensionally reduced potential energy surface and says that this paper

...explores some of the molecular bonds of a sulfur-4 (S4) molecule, and how these bonds affect the vibrational modes of the molecule, which in turn may influence the mass independent fractionation process.

Question: If the formation of tetra-sulfur is expected to be an important aspect of atmospheric chemistry in Earth's past, why is there so little experimental information on this molecule? Is it profoundly difficult to make, or to measure? Pardon the simple language, but What is tetrasulfur like? Why is there so little experimental information about it?
Wikipedia says only:

Tetrasulfur, S4
S4 has been detected in the vapour phase, but it has not been well characterized. Diverse structures (e.g. chains, branched chains and rings) have been proposed.[citation needed] Theoretical calculations suggest that S4 adopts a cyclic structure.

Source

While sulfur vapor might have been important  (I'm afraid is a bit speculatory though) who says this form was? It was only "detected" in vapor, so it's much less important part of it then S3 or S2. It is somewhat interesting why it's marginal.

Not all good quality information is online. Please consult Elemental Sulfur and Sulfur-Rich Compounds I By Bodo Eckert. He almost has a chapter on this topic.

@M.Farooq yes indeed, books! I'll find a copy this weekend. Thanks!

Have a look at this paper: https://www.cfa.harvard.edu/mmw/mmwlab/websitepdf/113.pdf

@NilayGhosh oh that's very cool, thank you! Figure 3 reminds me of my more recent question Does the nitrogen atom move with in the molecule in an ammonia maser, or does the molecule flip?

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text:
Are all stable Xenes (graphene-like 2D honeycomb sheets) buckled?
background: Per this abstract

Silicene, germanene and stanene are part of a monoelemental class of two-dimensional (2D) crystals termed 2D-Xenes (X = Si, Ge, Sn and so on) which, together with their ligand-functionalized derivatives referred to as Xanes, are comprised of group IVA atoms arranged in a honeycomb lattice — similar to graphene but with varying degrees of buckling.

question: This comment notes

Graphene is not flat, either. Cannot be, because only 3D crystallites have perfect long-range order. Landau-Peierls instability this is called.

How general is this? Can we expect all stable 2D crystals to be buckled, and any that would be flat and not buckled to "self-destruct" due to instabilities related to long-range order?

"X"enes? You need a better word. Also you didn't get the comment right. Graphene isn't "buckled", it's just that it's much like a scroll - without "support" or "gluing" to something it won't stay in plane.

"bucked" sounds like a rigid feature, wobbly is perhaps a better word. Oriented smectic liquid crystals are another well know example for this

@Mithoron perhaps you can tell these authors as well that they need a better word? 1, 2, 3, 4 Also, I'm pretty sure I have the word buckled misspelled (a missing "l" now included) but otherwise correct. See Nature paper (1st ref). Buckling refers to the periodic displacement, some unit cell atoms above/below the plane.

@Mithoron for graphene buckling see Buckling Analysis of Single-Layer Graphene Sheets Using Molecular Mechanics I've restored the title because I'm confident that it is clear, and correctly uses its terms.

Well, if even in paper title there was explanation what's X then your clarity of title is dubious...

@Mithoron Feel free to leave an expanded explanation as an answer to Is the clarity of my question's title really "dubious"? Is that even a good adjective to use in this way?

According to the authors of Ref. 1 this is a general property. They provide an explanation for the stability of such sheets based on formation of particular buckled geometries:

The discovery of a flat two-dimensional crystal known as graphene has contradicted Landau−Peierls−Mermin−Wagner arguments that there is no stable flat form of such crystals. Here, we show that the “flat” shape of graphene arises due to a microscopic buckling at the smallest possible interatomic scale. We show that the graphene, silicene, and other two-dimensional crystals are stable due to transverse short-range displacements of appropriate atoms.

This explanation is corroborated by experimental observations:

On the other hand, the generality of our prediction of the buckled universal shape has been revealed in recent experiments on silicene [...]

The authors note in the introduction that their explanation is only one among three that at that time were being seriously considered. The two other arguments are that (1) the sheets are trapped in a quenched metastable state that is not subject to large thermal perturbations compared to the lattice strength and (2) the sheets are slightly warped in the third dimension, the deformation suppressing thermal excursions.
The reference cited above is from 2012. Low-buckled 2D lattices have been reported more recently, for instance for germanene in Ref. 2.
Reference

A. O’Hare, F. V. Kusmartsev, and K. I. Kugel .  A Stable “Flat″ Form of Two-Dimensional Crystals: Could Graphene, Silicene, Germanene Be Minigap Semiconductors?
Nano Letters 2012 12 (2), 1045-1052. DOI: 10.1021/nl204283q
Jincheng Zhuang, Nan Gao, Zhi Li, Xun Xu, Jiaou Wang, Jijun Zhao, Shi Xue Dou, and Yi Du . Cooperative Electron–Phonon Coupling and Buckled Structure in Germanene on Au(111). ACS Nano 2017, 11 (4) , 3553-3559. DOI: 10.1021/acsnano.7b00687.

Thanks to Karl for the comment - a good lead. Also, evidently Mithoron's comment is one of the proposed alternative explanations justifying the stability of the nanosheets.

Thank you for your answer! It will take some time to dig in to these references. In the mean time, do I understand correctly that what's often termed buckling in Xene parlance (the periodic out-of-plane displacement with each unit cell) does need to be present to preserve stability?

I have not followed developments in this field closely enough to know which among the various theories is preferred to explain imperfections (deviations from planarity) or whether there is a universal structure, but there is agreement that they exist and are necessary (expected?). I suspect you could explain the necessity of deviations from planarity based on the presence of ZPE associated with out-of-plane vibrations and entropic considerations. I will have to look further to see what they suggest is limiting behavior is at 0 K.

I just wrote this answer at the edge of my comfort zone; how does it look? I can finally accept now that I've learned enough to understand your answer sufficiently :-)


text:
High errors in frontier orbital energy calculation using DFT
In a recent paper by Faber et . al., Mean Absolute Error and DFT (B3LYP) error (relative to the experiments) for 9 properties of QM9 dataset molecules has been reported.
\begin{equation}
\begin{array}{l|ccccccccc}
{} & {\mu} & {\alpha} & {\varepsilon_{\text {HOMO }}}& {\varepsilon_{\text {LUMO }}} & {\Delta \varepsilon} & {\left\langle\mathrm{R}^{2}\right\rangle} & {\text { ZPVE }} & { U_{0}} & {C_{v}} \\
 \hline & {\text { Debye }} & {\text { Bohr }^{3}} & {\text { eV }} & {\text { eV }} &{\text { eV }} & {\text { Bohr }^{2}} & {\text { eV }} & {\text { eV }} &{\text {cal/molK}}\\ \hline
 {\mathrm{MAD}} & {1.17} & {6.29}& {0.439} & {1.05} & {1.07} & {203} & {0.717} & {8.19} & {3.21} \\
 {\mathrm{DFT}} & {0.10} & {0.4}& {2.0} & {2.6} & {1.2} &{-} & {0.0097} & {0.10} & {0.34} \\ \hline\end{array}
\end{equation}

I was wondering why the DFT results for frontier orbital energies
have such high errors in comparison with atomization energy $U_0$.
For instance, the error reported for $\varepsilon_{\text {HOMO }}$
is $2.0$ while the breadth of values is $0.439$. Now compare this
with the error $0.1$ for $U_0$ with breadth of values for the
dataset reported as $8.19$, which is much more diverse.
A study on different DFT methods has been conducted by Zhang and
    Musgrave, which discusses inaccurate computation of frontier
    orbital energies. Some works have proposed the use ML methods to
    predict these values with very low errors. Since DFT is Ab initio,
    would suggesting to use ML methods even make sense?

Edits:

In the first paper (by Faber et al.) we read

Frontier molecular orbital energies (HOMO, LUMO and HOMO-LUMO gap) can
  not be measured directly. However, for the exact (yet unknown)
  exchange-correlation potential, HOMO and LUMO eigenvalues correspond
  to the negative of the vertical ionization potential (IP) and
  electron affinity (EA), respectively.

Based on this, I was wondering if DFT is the main approach to compute frontier orbital energies or this paper is just trying to find it using DFT in order to include the errors for DFT method.

Since all the methods are tested and optimized for atomization energies, it is not surprising that those are the best-reproduced numbers whatever methods you use while other properties (eg electron density, orbital energies...) can be much worse.

I'd like to add the small comment that DFT is not really considered an ab initio method, so you should not really be surprised that it is less accurate in cases than ML.

It is known (and some people have given analysis why) that range-separated hybrid density functionals (such as CAM-B3LYP and $\omega$B97 and related) give better results for orbital energies as measured by excited states performance.

@TAR86 Are these methods computationally more expensive than the simpler DFT methods?

@Blade Range-separated hybrids have the same formal scaling behavior as hybrid DFs such as B3LYP. Implementation details and different behavior in the estimation of negligible terms in practice mean additional costs on the order of 20% (in my estimation).

@Greg - in this particular paper, the authors implemented a range of ML methods - training each on different properties. So while the paper focuses on atomization energies, it does represent the best that particular implementations can do at predicting different properties (e.g., ~0.05 eV for orbital eigenvalues, etc.) on the QM9 set.

@GeoffHutchison Sorry, I haven't read the paper, but if you have garbage data, then your ML models will show all kind of things which are not there, so most probably the problem is still the same: DFT methods less reliable to calculate (especially unoccupied) orbital energies or densities than atomization energies. Just doing statistics on a dataset that contains false correlations or lacks real correlations, good because cancellation of errors etc will not fix the dataset, and different approaches can lead to different results due to how the biases plays out differently.

@Greg - I certainly wouldn't argue with you about using ML to fit garbage data. I'm just clarifying that the ML methods are tailored for fitting each property. I think you were commenting that the DFT methods are tested and optimized mostly for atomization energies (thermochemical properties) which is true, but not clear from your initial comment.

@GeoffHutchison You are correct, I was saying that the DFT (and any other QM method) is primarily optimized for atomization energies. That can be a very common source of bias in these data. Sorry that it was half-beaked and therefore not well articulated in the first comment.

IMHO, this particular comparison is a strange one.
You reference "Comparison of DFT Methods for Molecular Orbital Eigenvalue Calculations" by Zhang and Musgrave, and there are other similar comparisons between density functional orbital methods and ionization potentials and electron affinities.
In general, these Kohn-Sham DFT benchmark papers show that most functionals don't compute orbital energies that directly compare with ionization potentials and electron affinities. That is, in Hartree Fock, Koopmans' theorem suggests that the orbital energy of the HOMO should correspond to the negative of the ionization potential (-IP). In principal, for the exact density functional, a similar property should hold.
What does hold true (e.g., the Zhang and Musgrave paper) is that you can easily calibrate orbital eigenvalues from Kohn-Sham DFT methods to counteract systematic issues. Usually a linear regression is sufficient. (Computing the actual vertical ionization potential using two DFT calculations is generally even more accurate.)
Coming back to your question, I wouldn't be surprised if the absolute errors in orbital eigenvalues are mixed from DFT methods. But that's not how most research uses them - they either perform a calibration (e.g., linear regression) or are comparing multiple different molecules. In the latter case, the relative errors are much smaller (usually ~0.2 eV, IIRC - I'll look for some sources).
I believe the overall argument in the Faber et. al paper is that modern ML methods are very good. In some properties that's probably correct.

In reference to @Greg 's comment above, the other issue is that density functional methods are often optimized and tested for ground-state thermochemical properties like atomization energies, so other properties (like orbital eigenvalues) may not be as reliable.


text:
Reaction with ethyl chloroacetate and potassium carbonate
I am carrying out a reaction using ethyl chloroacetate, potassium carbonate in acetone for a compound containing $\ce{-NH}$ group. So, the substitution will be at $\ce{-NH}$ position.
I have tried stirring at room temperature, but no product obtained.
Refluxing has given a liquid product. Reported product should be white crystals.
Any suggestions for how to get the product?

What is the nature of the NH you are trying to substitute? You could add a bit of KI to make the iodoacetate in situ, this will be more reactive. Swap acetone for acetonitrile if there is a possibility the NH is reacting with the acetone.

Your target is a $N$-substituted glysine derivative?

Are you sure that obtained liquid product is not desired product as liquid? Maybe some solvent is remaining and is preventing crystallization.


text:
Sigma bond by combining two p orbitals

If I combine two p orbitals in such a way that they form σ bonds, which linear combination will represent bonding and antibonding?

My work on this problem:
$$
\begin{align}
&\text{Bonding:} &(\phi_1 - \phi_2)\\
&\text{Antibonding:} &(\phi_1 + \phi_2)
\end{align}
$$
where $\phi_1$ and $\phi_2$ are the wavefunctions of 1st and 2nd p orbitals.
Reason: While forming a bond, in bonding combination two lobes of same phase will overlap so they are actually mirror images of each other, and therefore they are negative of each other.
While in antibonding combination opposite phase combine, so the wavefunctions are actually added.
Is this correct?

There is no point in tracking individual orbital signs, for they are all arbitrary anyway. Other than that, you are right.

Hi thanks for your reply! U are saying that what I have written ( Y1-Y2) for bonding and (Y1+Y2) for antibonding is correct?

Provided that both orbital signs are chosen in the same way with respect to the axis directed along the bond, yes.

I want to share a picture with you how can I do that

There is no need for a picture, I think you got it right. Then again, you may edit your question and attach the image to it.

My teacher gave me wrong :(

Then they must have had a different picture in mind, and I would be glad to see both, but that's impossible unless you can read minds.

No our diagrams were same....ok imagine this axis of overlapping is z axis and 2 Pz orbitals are overlapping one coming from -infinity other coming from +infinity...now tell

That's the picture I had in mind from the very beginning.


text:
How can you use indicator titration to determine the pH of a complex solution like ocean water if at all
Could you measure the pH of ocean water by using a strong acid-weak base titration? I was thinking that you could set up a standard titration with ocean water as the "unknown", and drip the known HCl solution into it until the equivalence point but I don't know you would go about figuring out the equivalence point and consequently the pH of ocean water using this method. Is it even possible?

Titration doesn't measure pH in the first place, much like a tape-line doesn't measure weight.

I realize that titration isn't a direct method by which to measure pH but can't you use it to determine OH-/H+ concentrations and calculate pH using that?

No. Well, unless you know for sure there is one single fully dissociated acid (or base) in your solution. In that case it will work like you said. Otherwise it won't.

This might help: https://www.differencebetween.com/difference-between-ph-and-titratable-acidity/

Titration does not determine content of H+ nor OH-, as sea water is a kind of a diluted pH  CO2/HCO3- buffer.


text:
Is it possible to convert carbon soot/ black carbon in mg to CO2-equivalent? If so, how?
Since global warming is expressed in terms of carbon emissions or CO2 equivalent, can the short lived carbon soot/ black carbon that is produced from incomplete combustion be expressed in terms of CO2 equivalent? I learnt Carbon soot is a major health risk. The purpose of doing this is to calculate how much of heating effect can be reduced by controlling the emissions of these carbon particles.

Though both $\ce{CO2}$ and particulate carbon (soot) cause warming, a direct mathematical relationship of the effects would be specious for a few reasons:

The life-times of atmospheric $\ce{CO2}$ and soot differ, and vary with amount of sunlight (enabling plants to absorb $\ce{CO2}$, but not affecting soot appreciably), with amount of precipitation (affecting both), etc.
The heating effects differ: $\ce{CO2}$ remains in the atmosphere, absorbing and re-emitting infrared radiation and thereby raising atmospheric temperatures more than surface temperatures. Soot eventually settles out, warming the surface more than the atmosphere when radiation is absorbed.
There are other differences, such as effect of solar angle (seasonal and latitudinal variation), albedo of the surface (e.g. is the soot falling on snow or on a dark pine forest), etc.

BTW, particulate carbon and products of incomplete combustion also have direct impact on health, such as in burning fuels indoors or urban smog while the physoiological impact of $\ce{CO2}$ (at current levels) is less severe.

Your insights are great. Thank you so much for sharing. I happened to read https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/jgrd.50171
13.3.2 talked about the soot to co2 equivalent conversion idea and the uncertainty that revolves around it. Can you please go through and share your thoughts?


text:
Why polyethylene (PE), wool (polyamide) and cotton (cellulose) burnt with different smells, while they just product CO2 (and N2)?
Both PE and cotton produce CO2, but they distinctly have different smells while being burnt (Burning PE has smell of candle wax, while burning cotton’s smell is like burning leaves). Wool produces CO2 and N2 which is odorless, but yet has smell of hair burnt.
  Moreover, they also differ in burning process (PE and wool burns slowly, while cotton burns rapidly). So why there are such differences when their products are similar?

The reason is simple: incomplete combustion.

I don’t think about this case. Thanks for your quick reply!

@Valerie, why are you rejecting a valid answer??

I mean I have not thought about this case before so now I know it, it doesn’t mean I reject it. Maybe my words are not accurate, so now I fix it. Thanks!

Things are more complex than you suppose
There are two key reasons why these compounds burn with different smells.
The first is that your assumption about what the substances contain is wrong. Biological fibres contain plenty other than carbon and hydrogen. Often they will have lots of nitrogen and sulphur and various other elements and these often produce a variety of pungent chemicals that provide strong odours even in low concentrations. Plus artificial polymers will contain small amounts of other ingredients to modify their properties and these will often contain elements that give more complex combustion products some of which may be smelly.
The second is that burning these compounds in normal conditions is far from a perfect reaction yielding just the obvious oxidation products like carbon dioxide (which would be odourless). Even pure PE (and most of the PE you get will not be pure as it will contain various fillers and additives with other elements in it) will not burn cleanly to produce water and carbon dioxide). In normal conditions the burning will produce a variety of hydrocarbons and particles some of which will have odours. This is fairly obvious from the soot produced (which you would not get at all if the burning was completely efficient in producing carbon dioxide). The variety of those "imperfect" combustion products will be very different for different starting materials. So even compounds with roughly the same chemical composition (but different structure) may burn to give different combustion side products and, hence, will smell different.

What are your burning conditions?
PE fibers are indeed only carbon and hydrogen. But don't forget that wool is a protein (main component keratin) fiber and does not only contain carbon, oxygen and hydrogen, but also sulpher. That will also have an influence on your smell ;)
Whereas cotton is pure cellulose, if you do not have any waxes remaining.

They are burnt in air without any specific condition.

Cotton is taken from a dyed cotton towel, so it may affect a bit.


text:
Is the reaction of anhydrous aluminium(III) chloride and anhydrous ethanol reversible?
I'm planning an experiment that I'll do in silico in the near future, and one of the steps involves the extraction of the $\ce{AlCl3 · C2H5OH}$ adduct without any water in it.
Can the ethanol be simply boiled off to obtain anhydrous aluminum chloride, or does it end up as, say, a mixture of ethyl chloride and aluminum hydroxide or something?

You can buy solutions of Aluminium Chloride in Ethanol so it is reasonably stable. this suggests that the EtOH can be evaporated off to give you back the AlCl3

I'm just doing it in silico, partly since I'm not an adult yet (an 18 year old is still a minor where I live) and therefore might not be able to buy official reagents, and partly since I don't have physical equipment.


text:
How does solution which electrodes are immersed in effect electrode potential?
When we make an electrochemical cell, we dip the electrodes in the salt solution. How exactly does the electrode potential change as we dip the electrodes in the salt solutions?  Would different salt solutions change the effect of the electrode potential?

There is no electrode potential if the electrode is not dipped in the salt solution

@Maurice There is an electrode potential of not dipped in electrode, but in electrostatic sense, not electrochemical sense.

@ Poutnik. Right. There is an electrostatic potential. But this potential is constantly fluctuating, changing if you approach your hand or any other capacity effect.

@Maurice Surely it does. I remember an old, vacuum tube based pH meter in early 80s at a high school lab, where you had to stay 2 meters away and being still while reading pH value. The glass electrode and pH meter input had probably so high impedance it was like not being connected at all.

The short answer is that we want a cathodic reaction so we can use the metals with a positive standard reduction potential (because 0 is for $\ce{H2} $ reduction).
In reality however to really find a suitable electrolyte cyclic voltammetry is typically used and electrolytes with suitable electrochemical windows for the cell reaction are chosen.

The electrochemical window is the voltages where the current is zero in the picture above. If the wrong electrolyte is chosen then a full redox reaction happens(with the electrolyte participating) rather than half reactions at each electrode negating the need for the external circuit and creating heat

what is this? electrolyte cyclic voltammetry
and if current is there that means transfer of ions hence redox right?

We know that*Nernst eqn $$ \Delta G=-nFE $$ so a spontaneous reaction occurs at negative gibbs free energy or positive cell potentials(E). The cell potential can be calculated using a standard reduction table Ecell=Eoxidation+ Ereduction. With water hydrogen ions are reduced at zero volts so if a metal with a +ve oxidation potential (aka a metal with a negative reduction potential) were in the presence of water it would be oxidized and hydrogen ion would be reduced (+ve Ecell). So with WATER we use positive reduction potentials but with other electrolytes it depends on its reduction potentials


text:
Can a gas have a boiling point?

For each of the following pairs, choose the member with the lower boiling point. Explain your reason in each case.
HI(g) or HCl(g)

If they were both aqueous solutions, I would know that HCl has a lower boiling point. But the question specifically mentions gas. I believe that gas can't have a boiling point.
Can anyone help me with this?

When you make it cold enough, a gas will become liquid; that is also the boiling point.

Every gas has the boling point.  Substances considered as a (permanent) gas have their boiling point below the room temperature. The lowest boiling poist has helium-4, near 4.2 K.

I believe that gas can't have a boiling point.

The boiling point of a substance refers to the phase transition between pure liquid and pure gas. It depends on the pressure of the system, and the normal boiling point is taken at standard pressure, 100 kPa.
All pure substances that occur in a liquid and a gas state have boiling points. Some substances skip the liquid state and go directly from solid to gas (e.g. carbon dioxide at standard pressure). Some substances decompose before they boil when you heat up the liquid.
So the substance has the boiling point, not the gas. You could argue whether a pure liquid has a boiling point (you can't measure the boiling point without having gas present as well).

If they were both aqueous solutions, I would know that HCl has a lower boiling point.

Solutions (homogeneous mixtures) are more complicated than pure substances when it comes to boiling. Usually, the composition changes because one substance transitions preferentially into the gas phase. For example, if you boil maple sap (an aqueous solution of sugars and other solutes), the temperature at which the solution boils will increase during the process as the concentration of sugar increases. In this case, there is no fixed boiling point because it depends on composition.

Since 1982, standard state pressure is 100 kPa.


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Oxidation state of chlorine in ClO
Today, I had a friendly debate with my friend on what should be the oxidation state of $\ce {Cl}$ in the chlorine monoxide radical. I took the view that it should be $\ce {+2}$ while he took the view that it should be $\ce {+1}$. My reason for taking this view is simply that conventional state rules tell us that $\ce {O}$ should have a $\ce {-2}$ oxidation state assigned to it, when it is not part of a peroxide or in the superoxide ion, and when its bonding partner is less electronegative than it. His reason, on the other hand, is that one of the Lewis structures that can be drawn of this radical, features a $\ce {Cl-O}$ single bond. Hence, considering this single bond to be an ionic bond, we would thus observe charge separation to produce $\ce {Cl^+}$ and $\ce {O^-}$.
I would think that his perspective of the issue is flawed. Firstly, it is because there is more than one way of representing the covalent bonding between the two atoms. In fact, a simplistic MO diagram would allow us to deduce a bond order of $\ce {+1.5}$. Furthermore, a Lewis structure with a $\ce {Cl=O}$ double bond can also be drawn, although with $\ce {Cl}$ carrying a total of nine electrons. Hence, I believe that the approach of assigning states based on one's interpretation of the covalent bonding between the two atoms is inaccurate in this case. By following the conventional oxidation states, there would thus be no ambiguity in the assigning of oxidation states in this case.
Which view do you think is correct and why?

I'd say it would be correct not to apply the very idea of oxidation states here.

@IvanNeretin I agree with that the use of oxidation states to understand the bonding within this molecule would not be correct. However, there is some usefulness of this assignment when we consider the redox reactions of the this radical such as: $\ce {4ClO + 2H2O -> 3HClO2 + HCl}$. Clearly, if we take the oxidation state of $\ce {Cl}$ to be +2 in $\ce {ClO}$, then this reaction would just be a mere disproportionation of $\ce {ClO}$. We would have to consider reduction of $\ce {O}$ in $\ce {ClO}$ if we took the alternative view.

For the same reason, I'm reluctant to use the word "redox" when speaking about the reactions of this radical. Besides, the radical in question is not even a compound; it must have been obtained a microsecond or so ago from some real compounds, and it is their reaction in the first place.

@IvanNeretin The context is the photolysis of $\ce {HOCl}$ into $\ce {Cl}$ and $\ce {OH}$ and then, $\ce {Cl}$ goes on to abstract $\ce {H}$ from another $\ce {HOCl}$ to produce $\ce {OCl}$ radical.

A similar problem is assigning an oxidation state to iron in hemoglobin, when it is bound to oxygen. Depending on different resonance structures, you could take it to be +III, +IV or maybe even +V. Therefore, it is usually not assigned - and everything still works fine.


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Identify the structure of the compound (C3H5NO) given its NMR chemical shifts
I am trying to figure out the structure of the compound $\ce{C3H5NO}$ given just the following data from the 1H NMR spectrum:

$\delta 5.8 ~ \text{(singlet, 1H)}; \delta 6.2 ~ \text{(triplet, 2H)}; \delta 7.4~ \text{(triplet, 2H)}$

From 2-2-1 combination I have the following candidates:

Only the first and second one satisfy the multiplet data but I can't understand how their chemical shifts are so high. For a $\ce{-CH2}$ adjacent to $\ce{-C#N}$, the chemical shift reaches the value around $1.3 + 1 = 2.3$ but the given values are almost thrice of it. The $\ce{-OH}$'s proton's value seems fine nevertheless. Can anyone explain the reasoning behind the two other values? Or is the compound I have predicted wrong?

Was the molecular formula given, or could it be $\ce{C6H10N2O4}$?

You considered weird heterocycles and not acrylamide?

@Mithoron: Acrylamide is not the answer. It'd have 4 chemically and magnetically different protons, and hence gives 4 resonance. Coupling would not match either since there are $cis$- and $trans$-couplings. Besides, oxazole and isoxazole rings exist, so they are not weird.

@KarstenTheis Yes, molecular formula was given. #MathewMahindaratne yup that was my reasoning too.

Could you provide more information about the origin of the question, so as to confirm that there is indeed a structure that satisfies the constraints.

@BuckThorn it's from a uni chemistry assignment. I tried searching for it in my OC books but it wasn't present there.

This assignment looks similar to the one from Wade's Organic Chemistry where acrylamide is obviously the answer, but the multiplicities are mixed up and third shift isn't 7.4, more like 4.7 or 4.8.

Dhruba Charan Dash published the problem in his analytical chemistry textbook (ISBN 9788120353008). This is the picture that goes with the question:

Notice that the scale of the chemical shift increases with the strength of the magnetic field $B$. This means they are using the $\tau$ scale instead of the $\delta$ scale. The modern description of the signals would be $\delta = 10 - \tau$, where $\tau$ is the value given in the figure.
So we have a singlet at 4.30 ppm (B = broad?) and triplets at 3.83 ppm (typo in the figure, should be 6.17 instead of 5.17) and 2.63 ppm.
$$\delta\ 2.63 \text{ (triplet, 2 H)}$$
$$\delta\ 3.83 \text{ (triplet, 2 H)}$$
$$\delta\ 4.30 \text{ (broad, 1 H)}$$
For completeness, I will post the answer given by the author Dhruba Charan Dash:

PROBLEM 13.17 PMR spectrum of the compound $\ce{C3H5ON}$ in an acidic medium is shown below. Suggest its structure. Given that the peak 1 disappears in the presence of $\ce{D2O}$.
Solution Analysis of the spectrum:
(i) Since the peak 1 disappears in the presence of $\ce{D2O}$, it may represent a group containing active hydrogen. This group may be a hydroxyl group.
(ii) The peaks 2 and 3 must represent the hydrogen atoms attached to the methylene group.
(iii) The analysis of integral trace indicates the presence of 1, 2, and 2 protons within the peaks 1, 2 and 3 respectively.
The point (iii) suggests the presence of two methylene groups. These must be attached each other (point (ii)). The only structure consistent with the above facts is $$\ce{HOCH2CH2CN}$$

It is clear why the question posed by the OP (and by the teacher of the OP, as I learned in chat) was hard to solve. A single problem transcribing the question (replacing $\tau$ with $\delta$) obfuscated the intended chemical shift values.
For reference, here is the problem as presented to the OP:

At first I wanted to complain about why haven't you edited your existing answer, but after looking at this impressive detective work, I happily upvote one more time. Nice research and explanation!

I was sure OP's questuion was misleading and can't solve with the data given. Nice work to explain it correctly. You got $A^+$ from me! :-)

I'm writing this to show there is no solution. I'm hoping there is a flaw in my argument leading the way to a solution.
The key question, as the OP already stated, is what the two 2H triplets at 6.2 and 7.4 ppm arise from. As they are the only multiplets in the spectrum, it makes sense to assume they couple with each other. If we had an image of the spectrum, we could get the coupling constants for more information, but we don't.
Multiplets from $\ce{X-CH2-CH2-Y}$ fragments
This fragment occurs in the first two of the OPs three suggestions, and gives two triplets as seen in the spectrum. As the OP discusses, the problem is the chemical shift. No matter what the fragment is connected to (even directly to O or N), chemical shifts will not reach 6.2 and 7.4 ppm.
Multiplets from $\ce{=CH-CH2-CH=}$ fragment
One way to get higher chemical shift is to have olefinic hydrogens. The problem is that we need two equivalent hydrogens (same chemical shift at least). If the fragment shown above is built into a symmetric molecule without coupling to anything else, it would give two triplets. However, the chemical shift of the central two hydrogens is nowhere near 6.2 ppm. Also, for the specific molecular formula, the only atoms left are one H, O and N each, so you run out of carbon atoms.
Other ideas to get the chemical shift near 7 ppm
You could have a fragment like $\ce{-CH2-NH2}$, but usually no coupling is observed for structures such as those. You could invoke some co-solvent effect that increases chemical shift (lanthanide shift reagents?). You could consider a pair of cation and anion, or very small rings to open up the range of compounds, but none I tried solve the fundamental problem.
So my tentative answer is that there is no compound with a spectrum such as the one described.

Thanks to the fine work by Karsten Theis the remaining task is much easier (well, it's actually possible; please read the answers by Karsten Theis which explains that the shifts reported in the OP are incorrectly referenced). The corrected chemical shifts (from his answer) are:
$$\begin{align} &\delta\ 2.63 \text{ (triplet, 2 H)}\\
&\delta\ 3.83 \text{ (triplet, 2 H)}\\
&\delta\ 4.30 \text{ (broad, 1 H)} \end{align}$$
You can use chemspider to generate a library of plausible compounds:

and screen the compounds based on expected shift and multiplicity patterns or a spectral simulator:

Note that the triplets are apparent (the geminal protons are not magnetically equivalent). Couplings from NH to the neighbouring aliphatic protons are not resolved due to exchange or relaxation, resulting in an effective broadened singlet (the exchangeable NH proton is not included in the simulation by the program). The protons vicinal to nitrogen, resonating at $\pu{3.8 ppm}$ in the OPs spectrum, are presumably shifted downfield due to deshielding caused by H-bonding to NH by the solvent (water, at most partly deuterated in the OP spectrum, for otherwise the NH signal would not be visible).
The tricky compound appears to be azetidin-2-one.

I'm having trouble with the answer being azetidin-2-one because the predicted chemical shifts are much lower than the reported values. Additionally, the reported data were collected "in acidic media", so any exchangeable H would already be masked. Thus, the singlet is not likely exchangeable.

@BenNorris  Please read Karsten Theis' answers first, it might make clearer the difference in shifts. The simulation software does not provide predictions for exchangeable protons.

Ah. I see. I missed the tau scale issue. If understanding your answer requires catching one line out of another long answer, it is worth updating your answer to summarize that key point from the other answer.


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Validity of Maxwell-Boltzmann distribution for solids
Suppose there is a solid made of avogadro number of, say, aluminium atoms. This solid is kept at 273 K. According to Maxwell-Boltzmann distribution, all the particles or atoms will not have the same thermal energy. Some atoms are cooler than 273 K, some are much warmer than that etc. But is this true?
I mean, according to thermodynamics, heat must flow from warmer to cooler places (or particles) and must arrive at thermal equilibrium so that both places or particles will have same thermal energy or temperature.
So, atleast after some time, all the particles or atoms in a solid as I mentioned above must exchange energy with each other and must arrive at thermal equilibrium so that all the particles will have the same temperature (in the case I mentioned, 273 K), isn't it? But according to Maxwell-Boltzmann distribution, this will not happen.
-Why is this so?
-Or is Maxwell-Boltzmann distribution not valid for solids, so that the atoms will infact have same thermal energy or temperature atleast after some time?
-If it is valid for solids, why don't the atoms all arrive at a state where all of them have same temperature (which should happen from the point of view of thermodynamics)?
Edits:
I could clarify most of my questions from the link of another similar question provided. But, I couldn't find an answer to my last question:
-What might be the approximate value of the thermal energy or the temperature of the particle having maximum energy in the solid I mentioned as predicted by Maxwell-Boltzmann distribution?

The concept of temperature applies to an ensemble of atoms, not individual atoms. From a solid state physics perspective you have phonons (lattice vibrations) zinging around, just like you have vibrational/rotational states in polyatomic molecules.

@Ron That's a wrong way to think of it. Consider this: molecules in a gas collide all the time and exchange energy. How come they don't all have the same energy?

Thank you all for helping me. In gases, we can say that a single atom is collided by many atoms, thus the atom aquire more energy than others. This could happen with some other atoms too in a gas. So in gases even if atoms continuously pass on energy by collision, there will be some atoms with more energy because some atoms will be collided by MANY others. But in solids, especially, not many atoms are colliding with a single atom. Only surrounding atoms can vibrate with respect to that single atom. So gradually, in a solid, all atoms will distribute their energy and have same energy, isn't it?

Thermal vibrations wander through the solid much in the same way as sound does.

But these thermal vibrations must, over time, become uniformly distributed to all the atoms, must they not? If not, why is it so? Could someone explain it's mechanism? (Like in gases, as I mentioned above, the mechanism is that some atoms may be collided by many others, so that these will have a higher energy). Sorry if I'm asking too complicated questions.

Thank you for helping. But I have specified the part of my question for which I couldn't find an answer.

I suppose what you are wondering about is "how do you determine the optical mode": https://dspace.mit.edu/bitstream/handle/1721.1/46331/6-720JFall-2002/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-720JIntegrated-Microelectronic-DevicesFall2002/583B7E11-AA3D-408F-91E9-7527E94F1983/0/lecture1.pdf


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Question about the setting of DFT (Density Function Theory)-NEB (Nudged Elastic Band) calculation
I'm attempting to calculate Mg ion vacancy migration inside Pyrope unit cell. System has total 159 atoms, beginning and end geometry is optimized. For intermediate image, I'm using one coordinate that I acquired from crashed NEB case. This is not perfect intermediate geometry, but I thought this should be enough one to guide NEB.
1) Is there any rule of thumb to decide "num_of_images" based on moving distance of target atom?
In my script, I wish to move one Mg atom from one position to next vacancy which is about 3.5 angstrom away. In this case, how many images for NEB would proper? For example, if I choose 1 image per 0.5 angstrom, then 8 images should be chosen including the beginning and the end. But I'm not sure if 1 image per 0.5 angstrom would be the proper value. How do you decide "number of images" per given distance?
2) How many core number should be chosen for NEB?
I have some sense of core number decision based on my experience, but this is for relaxation. For NEB, I'm not sure how much core number is proper for given atom number or given number of images setting. I wish to learn if there are rule of thumb to decide core number for given NEB conditions.
3) How to decide intermediate image?
First, I tried my best guess of Mg ion location between two vacancies as an intermediate image. Then, I'm using one from one of the crashed NEB simulation result. Is this good approach?
4) Can freezing atom helps convergence? (freezing = fixing x, y, z coordinate of atom)
Is this better to freeze (put 1 1 1 beside the atom coordinate) the target atom for the beginning, intermediate, and end geometry? Or surrounding atoms around target atom? For now I'm not inducing any freeze to any atom in my NEB setting, but I'm curious if freezing helps convergence of NEB calculation.
Thank you

It is completely unclear which program you're using. It is also unclear which level of theory you are aiming at.

@Martin-マーチン Hello Martin, it is quantum espresso, and I'm using PBEsol PP.

There is not a general rule for choosing the right displacement between images. For very big systems, I sometimes launch two NEB calculations: one with shallow parameters (eg: steps of 0.5 angstrom on the reaction coordinate), and later I use the almost converged geometries near the TS for another NEB calculation, with finer (less than 0.1 angstrom) steps. Of course the nature of the system might help you choose: sometimes things behave funny, and you have to choose a strategy for each system. Anticipating your last question, freezing or constraining sometime helps: while it doesn't improve convergence, it can speed up the calculation (by telling the program that some movements are forbidden, so that the NEB algorithm doesn't need to try them before discarding them). The intermediate image should not be mandatory for a NEB calculation (if not given, it is generated by interpolation and optimized at every cycle), but it can speed up a calculation, if provided. It can be taken, for instance, from a geometrical scan, or from an approximate TS geometry.
As for the core number, if you are talking about processors, then generally speaking as much as possibile is a good choice (edit: provided that the software you are using is capable of parallelizing efficiently the slowest step of the calculation: your choice could be the same you use for geometrical optimizations at the same level of theory)

Thank you for your answers, those were greatly helpful for me. Yeah, maybe I could use NEB results from 'shallow' setting as an intermediate coordinates for further NEB simulation using 'deeper' setting calculations. I'm using QE and I find that QE has some I/O bottleneck issue, which makes calculation unstable with too much memory so I'm still searching for optimal core number. I will try freezing atom to help convergence. Thank  you!!!


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Hydronium ion concentration to compare pH values
I'm having a bit of trouble with this question:

Which has a higher $\mathrm{pH}$ value, $\pu{0.001 M}$ $\ce{HCl}$ or $\pu{0.1 M}$ $\ce{CH3COOH}$ $(K_\mathrm{a} = \pu{1.8E-5})?$

I have calculated that for the $\ce{HCl}$ solution (since it's a strong acid) the $[\ce{H3O+}] = \pu{0.001 M}.$
Then, for the $\ce{CH3COOH}$ solution I got $[\ce{H3O+}] = \pu{1.33E-3 M}$ (by solving a quadratic equation).
Since the $\ce{CH3COOH}$ has a higher concentration of hydronium ions, doesn't this mean that it will have a lower $\mathrm{pH}$ than the other solution? I'm confused because the answer says it's the other way around.

The concentrations look correct (next time please show your calculations and note that there is no zero in $\ce{H3O+}$), so you end up with pH 3.00 for HCl and pH 2.88 for acetic acid. What does the answer say, exactly? Could there be a typo?

The HCl solution has the higher pH value. The acetic acid solution has the higher concentration of hydrogen ions. What exactly did the answer key say?


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Reference request on formate potash process
Wikipedia article on Potassium Formate mentions a so-called "formate potash process" for production of potassium, presumably referring to elemental metal.
Concise encyclopedia chemistry by Mary Eagleson, page 888 is a given as a reference, but I can't find a PDF, hence this request.

The reference is correct, from [1, p. 888]:

Potassium formate: $\ce{HCOOK},$ colorless, deliquescent, rhombic crystals; $M_\mathrm{r}~84.12,$ density $1.91$, m.p. $\pu{167.5 °C}.$ P. is formed by neutralization of aqueous formic acid with potassium hydroxide or potassium carbonate. It is an industrial intermediate in the formate-potash process (see Potassium carbonate).

However, formate-potash process doesn't yield in elemental potassium, rather potassium carbonate [1, p. 886–887]:

Potassium carbonate, potash:
[…]
P. is also produced by the formate-potash process, in which potassium sulfate and lime are treated under pressure with carbon monoxide:
$$\ce{K2SO4 + Ca(OH)2 + 2 CO -> 2 HCOOK + CaSO4}.$$
The resulting potassium formate is oxidized to P. by heating it in air:
$$\ce{2 HCOOK + O2 -> K2CO3 + CO2 + H2O}.$$

English Wikipedia page claims formate-potash process is for “production of potassium” probably because the author attributed “P.” from the encyclopedia entry to potassium.
Whereas in reality it refers to the entry title, “Potassium carbonate”, which is a standard referencing method in data sources such as a handbook or an encyclopedia.
German Wikipedia page doesn't have this flaw and informs that “Formiat-Pottasche-Verfahren” is a technically insignificant method of producing $\ce{K2CO3}.$
References

Concise Encyclopedia Chemistry, English language ed., 2nd ed.; Scott, T., Eagleson, M., Eds.; de Gruyter: Berlin; New York, 1994. ISBN 978-0-89925-457-9.


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Why is 1,2-diiodoethane unstable?
Diiodoethane is a product of the reaction between glycol and $\ce{PI3}$. Why is 1,2-diiodoethane unstable?

https://chemistry.stackexchange.com/questions/68546/reaction-of-ethylene-glycol-with-pi%e2%82%83-p-i%e2%82%82/111098#111098

It isn't unstable at all.  1,2-diiodoethane is characterized with a melting point of 80–82 °C, and a mass spectrum with five major peaks.  It even serves as the iodine source in some rare earth iodide syntheses.  Read about it on Wikipedia.

Although 1,2-diiodoethane is commercially available (e.g., Sigma-Aldrich), it has been recommended that it should be stored in $\pu{2-7 ^\circ C}$. Isn't that means it may tend to decompose?

Please don't post links to the mobile versions of the websites (unless there is a really good reason to do so).

Missed that, sorry.  Was having tea at the time.  Don't drink and derive ;-) .


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R/S chirality question
Consider [(1S,2S,4S)-bicyclo[2.2.1]hept-5-en-2-yl]acetic acid:

(Source: http://www.chemspider.com/Chemical-Structure.22823922.html)
According to the chemspider nomenclature, this has three S stereocenters.  However, when I perform the CIP both manually and computationally, the center labeled ? comes up as R.  I am trying to figure out if I am misunderstanding CIP (particularly the expansion part for handling double-bonds) or if the literature is wrong in this case. I would greatly appreciate if some chemist here could work out this example to give me another data point.

Related: Using CIP rules in a bicyclic compound with carbonyl subsituent

According to the Cahn–Ingold–Prelog priority rules, a double bond considered as ($\ce{CCX}$), where $\ce{X}$ depends on the substitution on the double bond. In your case, $\ce{X}$ is hydrogen ($\ce{H}$). Thus, I arranged the priority as depicted in following diagram (chiral carbon in consideration is marked with asterisk):

Hydrogen is the least priority group on the chiral carbon. Among other three groups, the bridgehead carbon is a $\ce{-CH2 -}$ group, hence can be marked as ($\ce{CHH}$) on priority checking. The final two groups can be marked as ($\ce{CCH}$). I used red and blue color on one group (methine on the ring), and black on the other (double bond) to distinguish these two group. Since both has identical substitutions on first carbon, we need to moved to second carbon. On double bond, it is again ($\ce{CCH}$), but bothe red and blue carbon give ($\ce{CHH}$) priority checking. Thus, double bond has the highest priority as indicated. Therefore, assignment for this chiral center is ($S$)- as given in chemspider website.

Thank you very much!  I was misunderstanding the algorithm.  I thought that the idea was to merge the atoms at each successive level away from the chiral center, so that in the second round we would be comparing CCH vs CHH + CHH = CCHHHH.  Now a followup question is, in a hypothetical case where we have something like CCH vs CCH + CHH, does the second group take priority due to having an extra branch, or do we move to the next level of comparison?  I assume the former.

It's not clear to me what your question on hypothetical question. Yet, for your question in hand, I think you still confused how to go by carbon to carbon (since it is a ring). Well, first carbons from chiral atom here have: CCH, CHH, CCH, therefore, $\ce{CH2}$ group is the priority #3. Between other two, we have to look at second carbons: carbon with double bond still has CCH (twice) while other carbon has CCH twice. Therefore, carbon with double bond has the highest priority between them so, chiral center has a ($S$)-nomenclature..

Thanks again.  I understand how this works now.  Can't upvote you since I don't have the points but much appreciated!

I appreciate your concern. Just for your info, since it's your own question, you are allowed to upvote or downvote, and even accept any answer as you wish. :-)


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Schrödinger wave function for 3s orbital of hydrogen atom
I am unable to understand what these terms in the wave function represent and how to solve it.

For 3s orbital of hydrogen atom, the normalised wave function is
$$Ψ_\mathrm{3s} = \frac{1}{81\sqrt{3π}}\left(\frac{1}{a_0}\right)^{3/2}\left[27 - \frac{18r}{a_0} + \frac{2r^2}{a_0^2}\right]\mathrm e^{\displaystyle\frac{-r}{3a_0}}$$
If distance between the radial nodes is $d,$ calculate the value of $\displaystyle\frac{d}{1.73a_0}.$

The individual terms do not represent anything. As for the solution, you might want to find the first radial node, then the second one, then the distance between them.

To solve the problem, think about what is the condition for a radial node- what happens to the density $|\Psi_{\textrm{3s}}|^2$ at a node? First some definitions: $r$ is the distance from the nucleus and $a_0$ is a constant distance called the Bohr radius (=0.5291 Å).

 The density, given by $$|\Psi_{\textrm{3s}}|^2 = k\left[27 - \frac{18r}{a_0} + \frac{2r^2}{a_0^2}\right]^2 \mathrm{e}^{\displaystyle\frac{-2r}{3a_0}}$$ goes to zero (k is some constant which as we'll soon find out is not important) .

How can this go to zero?

 There are two ways this can happen: for large r the exponential term will go to zero. At intermediate r the polynomial in brackets can go to zero. The latter is the condition to solve for.

Now what?

 We can solve for the zeroes of the expression in brackets, it is a quadratic polynomial in r:$$ 27 - \frac{18r}{a_0} + \frac{2r^2}{a_0^2} = 0$$$$ r^2  - 9ra_0 + \left(\frac{27}{2}\right)a_0^2 = 0$$that we know how to solve for two roots using a closed form expression.

The solution for the position of the nodes is

$$ \begin{align} r_{node} &= \frac{9}{2}a_0 \pm \sqrt{\frac{81}{4}a_0^2 - \frac{54}{4}a_0^2 } \\ &= (4.5 \pm \sqrt{27/4})a_0  \\ &= (1.01, 3.76)Å \end{align}$$

The distance between nodes is then

 $d=\sqrt{27}a_0=5.20a_0$

so that

$\frac{d}{1.73a_0}=\frac{5.20}{1.73}=3.00$

Obviously a homework problem. . . .


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What is a material more easily cut than shattered?
I'm currently developing a sci-fi concept for animation which partially involves a typical "the earth has been overrun by XYZ" scenario.
One of the ideas for XYZ we're thinking of is a species of intelligent crystalline beings made of diamond -- or at least a diamond like material.
These are large, arachnoid in shape and fast. They kill by piercing their pray with sharp, lance-like protrusions on the ends of their forelegs.
The problem is, thus far we've been thinking that the trick to killing these things has been 'shattering' what essentially equates to a CPU or central core suspended like the 'body' of your typical daddy long legs. Which makes sense if the material were diamond -- however we realised it might be more interesting if kinetic force like cannonballs and explosive shockwaves were less effective, and that the key was 'cutting them.'
Maybe we're bad writers, and we just can't drop the idea of our protagonist being forced to use a 'chain-blade' up close to combat one of these things, but visually we're very keen on just such a scenario. So if those among you more educated than us (i.e. all of you) could point us towards a material where 'cutting' or 'sawing' with what could be called 'super-chainsaws' is more feasible than explosives, large projectiles, or other methods of blunt force, it would be greatly appreciated.
We Do understand just such a material will probably not exist in crystalline form, thus losing the overall aesthetic we're aiming for -- but nonetheless we'd love to hear what you guys have to say in general regarding the question and context.
Thank you very much for your time.

Maybe rubber or ballistic gel? Though I think that it would still be destroyed by an explosion.

Maybe go with an exoskeleton made of chitin (just like terrestrial insects).

The crystalline property that would allow to cut the "diamond creature" (or whatever those things are in your universe) you are looking at is cleavage (a writer might also use a term in a wordplay).
In crystallography, cleavage is the ability of crystals to crack or split in certain crystallographic directions.
This mechanical property of crystalline media is associated with their internal structure (the ratio of adhesion forces in the crystal lattice) and does not depend on the external shape of the crystals.
This property is also often utilized in the beginning of the process of fashioning the gemstones.
There is a fascinating demonstration with explanation: Crystals - Alan Holden 1958 (YouTube).
With diamonds you can do that too, just not always as easily as with mica: Diamond Cleavage (YouTube).
Diamond (as well as ﬂuorite, galena, topaz and some other minerals) possesses a so-called "perfect" cleavage, and the cleavage planes in diamond are parallel to the octahedral crystal faces [1, p. 30]:

Figure 4.1 Two examples of directional-dependant properties in diamond. The cleavage planes are parallel to the octahedral faces, which are also planes of near
maximum hardness. The easiest direction for sawing is at right-angles to the dodecahedral plane, which is also the plane of minimum hardness.

References

Read, P. G. Gemmology, 3rd ed.; Elsevier Butterworth Heinemann: Amsterdam; Boston, 2005. ISBN 978-0-7506-6449-3.

Based on further google searches into diamond cleaving and sawing, I just want to further ask if anything screams out to you as especially ludicrous (apart from the entire concept of course) if our protagonists wore visual aids that allowed them to view a 3D 'wire-frame' of the diamond organisms (akin to something like the Advisor or Sarine programs used by craftsmen today) in order to identify in real-time the directional weaknesses of the beasts -- and thus the best 'way' to swing their weapons at them.

@Delirium I guess the protagonist should be up-close and either have an optical device measuring light interference, or have an x-ray source and a properly placed detector. The latter would indeed reveal the structure and not just cleavage directions, but such an apparatus is at least of a size of a small fridge and requires cooling and powering units. However, if the creatures grow naturally, made on a conveyor or share a hive-mind:), these directions will be the same among all of them.


text:
Which chemical is used in drunken driver test?

Which chemical is used in drunken driver test?

Acidified K2Cr2O7
KMnO4
Chromyl Chloride
H2O2

On searching on Google, I got the answer as option 1, acidified potassium dichromate, but my questions are:

How is the test administered? How much time does it take for the results?
On searching, I got a hunch that it had something to do with oxidation of alcohol, so, why can't we use the chemicals of the other options mentioned?
Is my hunch correct?

The test is administered by breathing through a tube, and the results become apparent immediately. Indeed, this has something to do with oxidation of alcohol. As soon as the alcohol in the exhale reaches the contents of the test tube, the latter changes its color, and the police guys put you in the handcuffs. Or it doesn't, and they don't. That's a wonderful example of chemistry improving our life.
Now to the point of comparing different reagents.

Permanganate itself is way too dark. When looking through its solution, you won't be able to detect any trace of any other color, let alone the products of reduction. Besides, the solution is rather sensitive and won't last long.
Chromyl chloride is too dangerous. It is liquid, and the suspect is going to breath in some tiny droplets (and/or possibly vapors) of it, and that's not something you would want to happen to a human being.
Hydrogen peroxide... have you ever seen it, really? It has the almost same color as water (that is, no color at all), and the reaction products have the same color as well. You won't be able to see a thing. Besides, this one is also too sensitive. Besides, it won't oxidize ethanol right away.

So it goes.

Besides, it won't oxidize ethanol right away Why not ?

Er... Because chemistry.


text:
How to handle CO2 being evolved during vacuum distillation?
Setup: vacuum aspirator pump, distillation apparatus with stirring (instead of a bubbler). Partial vacuum around 5" Hg.
What would you suggest to hold the vacuum set-point?

Somewhere there is a vacuum pump, isn’t there? Actually controlling the vacuum level (as opposed to just pumping on it) depends on the set up and your equipment.

Is it possible to degas the solution before beginning the distillation?

A fine hand with the faucet to which your pump is attached to, or a vacuum controller.


text:
Why London force dominates in Alkyl Halides
Why London Force is greater than dipole-dipole interactions in Alkyl Halides?

As we know that most of the time London force is weaker than other forces. Then why in this case it's not so?
According to me, due to high charge separation (in C & X) the dipole-dipole interaction must me stronger than the London force which depends more on the surface area of molecules.
Source:Leroy G.Wade Organic chemistry

That depends on halide, for bromides and iodides textbook is totally right. For chlorides and fluorides that somewhat depends,

@scientist omega please refer ro this http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/substitution/alkylhalogenide/einfuehrung.vlu/Page/vsc/en/ch/12/oc/substitution/alkylhalogenide/eigenschaften_der_halogenalkane/eigenschaften.vscml.html

Actually size of halogen is so big that it kind of shields C atoms which makes it harder for dipole dipole interaction.Plus large surface area increases London force.


text:
Giving too much energy to an exothermic reaction
What will happen in an exothermic reaction if more than required energy is given? I mean, will it produce even more energy which will be highly dangerous or nothing will happen according to Le Chatelier's principle (as we are trying to carry the reaction in an opposite direction)?

Looks like you are wondering how thermal runaway (https://en.wikipedia.org/wiki/Thermal_runaway) can happen if exothermic reactions become less favorable with increasing T (given relation between K and $\Delta H$ according to the vant Hoff equation

There are two separate effects we need to consider here: thermodynamic and kinetic.
Let's assume you are only providing the energy thermally. So providing more energy means increasing the temperature (T).  Thermodynamically, if a reaction is exothermic, and you increase T, the reaction becomes less favorable (assuming it stays exothermic over that temperature range) according to, as you mentioned, Le Chatelier's principle.  I.e., the equilibrium shifts to the left.
However, we also need to consider kinetic effects.  Here, as you increase T, the rate of reaction increases, and thus the rate of thermal energy production increases, thus further increasing T, thus further increasing the rate, and so on, leading to the thermal runaway effect that Buck Thorn mentioned in his comment.  I.e., we get a positive feedback loop.

The equilibrium constant depends on temperature, the only thing it depends on. For an exothermic reaction the equilibrium constant will decrease ie shift towards the reactants side
Also consider kinetics as described above


text:
Determining structure from C-NMR spectrum, is my answer correct?

Determine the structure of the compound whose spectral data is given below and explain your reasoning (5 marks for structure, 15 marks for reasoning). The peak at 7.27 ppm does not belong to the compound. What is the source of this peak?
  (2.5 marks) Looking at the full spectrum you will see small peaks at about 4.2 and 3.7 ppm. What is the origin of these peaks? (2.5 marks)
Formula: $\ce{C9H10O3}$
IR: small peaks observed between 3010 and 2853. The largest peak is at 1693. All peaks are in $\pu{cm-1}.$
$\ce{^{13}C}$ NMR (DEPT-135 result in brackets): δ 190 (+), 153 (no peak), 149 (no peak), 131 (no peak), 127 (+), 111 (+), 109 (+), 58 (+), 57 (+) ppm.
$\ce{^1H}$ NMR (300 MHz. See expansions on next page):

I mainly looked at the $\ce{^{13}C}$-$\mathrm{NMR}$ aspect of this question. Given that there is a peak at $\pu{190ppm}$, it is probably an aldehyde with a $\ce{CH}$ (which explains the +DEPT sign). The molecular formula also showed a degree of unsaturation of 5, which could be explained by a benzene ring and a double bond (which links up nicely with the carbonyl group). But in order to come up with no peaks in an aromatic ring I added some methyl groups. I eventually came up with the structure 2,3-dimethylbenzaldehyde. The +signs indicated $\ce{CH}$ or $\ce{CH3}$, the 5 $\ce{CH3/CH}$ groups could be explained by the methyl groups and the rest of the carbons in the benzene ring. Do you think this structure is correct?

I'm trying to be nice here to pay respect to the comment of @orthocresol on your now deleted post. :-) Well, you structure is close to actual one, but not quite correct. For example, substitutions on benzene nucleus are not just two methyl groups. Did you pay attention to the chemical shifts of methyl carbons?

Oops, I forgot the oxygens. Is it 2,3-dimethoxybenzaldehyde instead?

That part is correct, but it is not 2,3-dimethoxybenzaldehyde. Would you look at $\ce{^1H}$-$\mathrm{NMR}$ instead? The coupling pattern in aromatic protons would tell the story.

Which part is correct? I'm looking at the coupling pattern and I can't seem to figure out which way to arrange the methyl groups to get 3 sets of doublets...

Your substitution pattern is not correct. The compound is 3,4-dimethoxybenzaldehyde.

@Mohamed Since this is not the first or the second time you are asking a similar question, please use StackExchange image uploader (icon on the toolbar). Don't upload screenshots to imgur.com. Make sure image URLs are https://i.stack.imgur.com/* and not https://imgur.com/*, this will assure that hosting won't delete your pictures later rendering the question obsolete.

@Mohamed Also, please avoid posting screenshots of text, spend some time and type it in whenever possible, and crop images appropriately — we don't need to know that you are running Chrome on Windows 10 in a VM. There are free tools to assist you with taking screenshots and running OCR such as ShareX or Capture2Text.

It is not always a good idea to focus only on $\ce{^{13}C}$-$\mathrm{NMR}$ when you are doing structure elucidation on unknown compound using spectral data. Yes, $\ce{^{13}C}$-$\mathrm{NMR}$ gives you valuable info, so does FTIR to some extent. Yet, $\ce{^{1}H}$-$\mathrm{NMR}$ gives the most info about the compound as I depicted in following diagram:

All your analysis is correct except for the final conclusion on the compound. It is not 2,3-dimethoxybenzaldehyde as you concluded. But it is 3,4-dimethoxybenzaldehyde. How do I know that? Look at the diagram and analyze the coupling pattern (disregard the small but significant meta-coupling, which give close to $\pu{2 ppm}$ here).
If the compound is 2,3-dimethoxybenzaldehyde, the $\ce{^{1}H}$-$\mathrm{NMR}$ spectrum of it should displays a doublet, a triplet (or dd), and a doublet in aromatic region (~$\pu{6.7-8.5 ppm}$), in that order going from upfield to downfield. However, what given $\ce{^{1}H}$-$\mathrm{NMR}$ exhibit is a doublet, a singlet, and a doublet in the aromatic region (disregarding the meta-coupling). The singlet indicates the presence of isolated aromatic ring proton, and two separate doubelets also indicate the presence of two isolated but neighboring aromatic ring protons. Since three substitution groups are one formaldehylde (-$\ce{CHO}$) and two methoxy groups (-$\ce{OCH3}$), there are three possibilities: The compound would be 2,5-dimethoxybenzaldehyde, or 2,4-dimethoxybenzaldehyde, or 3,4-dimethoxybenzaldehyde:

The possibility of been 2,4-dimethoxybenzaldehyde can easily be ruled out, because it would show the upfield shift of the chemical shift of the singlet corresponding to the proton on position 3 of the ring (been ortho to two -$\ce{OCH3}$ groups), even shielded than the doublet at the $\pu{6.99 ppm}$. This is justifiable conclusion since the resonance structures of 2,4-dimethoxybenzaldehyde (using only 2-methoxy group) indicate that the ortho ($\ce{C}$-3) and para ($\ce{C}$-5) positions will be more shielded than meta positions. Also, with respect to 4-methoxy group, the ortho ($\ce{C}$-3 and $\ce{C}$-5) positions will be even more shielded than corresponding ortho positions  in 3,4-dimethoxybenzaldehyde.
However, distingushing 3,4-dimethoxybenzaldehyde from 2,5-dimethoxybenzaldehyde is tricky. Both would have a doublet at $\pu{6.99 ppm}$ (for the proton at $\ce{C}$-5 for the former and the proton at $\ce{C}$-3 for the latter) since substitution effects are the same for both. Same argument is valid for the singlet (with meta-coupling) at $\pu{7.43 ppm}$ (for the proton at $\ce{C}$-2 for 3,4-dimethoxybenzaldehyde and the proton at $\ce{C}$-6 for 2,5-dimethoxybenzaldehyde) since, again, the substitution effects are identical for both. Yet, chemical shift for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde and that of the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde would not be identical. That is because ortho-effect of a substitution group is somewhat larger than para-effect of the same group in general (there are few examples you may find here), regardless of its sheilding or deshielding effect. Thus, the chemical shift for the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be more shielded than that for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde. My reasoning:

The proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be deshielded by para-$\ce{CHO}$ group (blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
The proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde should be deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by para-$\ce{OCH3}$ group (red shift).
The both protons at $\ce{C}$-6 position of 2,5-dimethoxybenzaldehyde and at $\ce{C}$-2 position of 3,4-dimethoxybenzaldehyde are deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
Comparition of reason 1 & 3 clearly shows these two chemical shifts should be switched (instead of singlet followed by doublet, it should be doublet followed by singlet).

Thus correct resonance order would be given by 3,4-dimethoxybenzaldehyde.


text:
Displacement by more electropositive metal in non aqueous environment
The more electropositive element should displace metal from it's salts' solution. This can be seen in reaction of copper salts with iron. But, if you try using for example potassium metal as the more electropositive metal in a displacement reaction you don't get a result since potassium will react with water before it has a chance to displace much of the other metal.

What makes dissolution in water favour displacement by the more electropositive metal, even if calculated gibbs free energy for the reaction would be positive? Is it the higher strength of hydrogen bonding of the more electropositive metal, or is it the lack of lattice energy of the salt that otherwise needs to be overcome?
This will depend on the answer to the first question, but is it possible to actually displace, say, Calcium with Potassium metal from any of it's salts provided that this salt is soluble and the solvent is inert towards Calcium and Potassium metal? If you had to do this in reality, what would you use, perhaps a phase transfer catalyst?

I imagine that if it's mainly the lattice energy making the difference one should be able to just use a different solvent, as it would be only important to dissociate the salt

Some of this was already answered when you asked earlier https://chemistry.stackexchange.com/questions/118031/what-would-be-the-concentration-of-products-in-this-calcium-producing-system

I realize (and mentioned it in the post) that water makes all the difference, now im asking how does it work? I can write the same equation for the reaction dry and in aqueous solution, and obtain the same gibbs free energy change, which would make it seem like it wouldn't happen even in solution. Yet, something makes the reaction favourable in water (or would make it favourable if not the mentioned reaction of potassium with the solvent). Is it because of stronger hydrogen bonding of the K+ ion? Or is the lattice energy the culprit and it's only important that the salt is dissociated?

I feel I explained what im asking better in the comment, so im going to edit the post to match.


text:
Characteristics of Isotopes
I'm currently applying information theory to physics, and have begun considering the nature of isotopes, and in particular, how much information is contributed to the properties of an element by its subatomic components.
One initial conclusion I've reached is that there's a window of ionization for most materials within which the properties of the material don't change in any macroscopically meaningful way.
For example, if I rub a balloon on my sweater, I will change the local electrostatic charge of my sweater, but it won't change in appearance in any meaningful way. This is obviously also true of conductive wires, and everyday compounds and elements generally.
We can say, therefore, that electrons don't contribute much information to the properties of a typical substance, within some window of ionization.
My question is whether this is true of neutron isotopes, and in particular, Gold 197. I understand that this is a naturally occurring, stable isotope, and I'm wondering if it's any different than ordinary gold in terms of appearance, conductivity, etc.
As a general matter, it seems possible that neutron isotopes are actually not physically similar to their elements, since, for example, a given isotope might not be stable. This suggests that neutrons generally contribute more information to the behavior and properties of a substance than electrons.

"electrons don't contribute much information to the properties of a typical substance" makes no sense... Charges on everyday stuff are so small in comparison with with their respective amounts of atoms that from chemical point of view they're as good as non-existent.

Given there is only one naturally occurring isotope of gold, you have picked an odd element to focus on. Au-197 IS naturally occurring gold.

@Mithoron - add electrons to a surface. Nothing about the surface really changes, other than its net charge. Change its proton number. You have a element. That is the different between contributing a lot of information and not.

@JonCuster Thanks, but that didn't answer my question. Is the neutron isotope any different than the element.

Well, you said "I'm wondering if it's any different than ordinary gold" - and it is ordinary gold. Your best bet for a difference is protium vs deuterium vs tritium.

Understood @JonCuster, is it different than the element.

As a general matter, it seems possible that neutron isotopes are actually not physically similar to their elements, since, for example, a given isotope might not be stable.

It was very well known (pre-1940s) that the number of protons (Z) distinguish one element from the other. If we talk about the simplest element with Z=1, whether it contains 0, 1 or 2 neutrons it is still a colorless gas which burns in air and forms a colorless liquid- water. Yet some of the physical properties of that "water" with 0,1 or 2 neutrons change such as refractive index, boiling or melting point etc, their chemical properties remain almost the same. The chemical properties are determined by the number of valence electrons. Hence if bromine is red, no matter how many neutrons its isotope has- they will all appear red.
Your observations about the number of neutrons are a classical example of casual relationship "A causal relation between two events exists if the occurrence of the first causes the other. The first event is called the cause and the second event is called the effect. A correlation between two variables does not imply causation."
https://www.varsitytutors.com/hotmath/hotmath_help/topics/correlation-and-causal-relation

This is very helpful thank you. So proton number is obviously the most important, and determines the element. Then, neutron number can determine some characteristics such as refractive index, boiling point, etc., but do not change the element. Finally, electrons seem to not change much at all about a given element.

Well it is difficult to say neutron number determines the refractive index, boiling & melting point, we should rather say that it having an additional neutron for an element with Z protons "affects" some of the physical properties even reaction rates. The chemical properties of the elements with Z protons is determined by the distribution of Z electrons around the nucleus.

That's an interesting point, so you're saying that the electrons affect the reactivity of certain elements. But for everyday materials, charge has absolutely not affect at all on a substance. Where do these electrons live in the material? For example, if I rub a balloon on my sweater, where are the electrons associated with that additional charge?

How many molecules are there in a drop of water? About 500000000000000000000 molecules. Now count the number of electrons (1 from hydrogen and 8 from O each). Remove a few electrons from this number. This would a highly charged droplet. What fraction of electrons have we removed? The key message is that we have to differentiate between bulk and molecular scale. The electrons on a balloon/sweater indeed come the atoms making up the polymer chains of your sweater and the polymer of your balloon. A physicist might be able to explain better as how dielectrics (=insulators) develop a charge.

On the contrary if we had a single water molecule in gas phase and you removed an electron, it would be positively charged. It would behave just like a positively charged particle and easily manipulated in electric and magnetic fields. This type of ionization is routinely done in mass spectrometers. What is your background in science and what is meant by applying information theory to physics?

I'm a mathematician and self-taught physicist. You can read my work here if you're interested: https://www.researchgate.net/profile/Charles_Davi

Good to know but I would not understand anything from information theory.


text:
Reduction of amides with lithium aluminium hydride
Since $\mathrm{p}K_\mathrm{a}$ of primary amides are close to alcohols, will there be an acid-base reaction between amides and $\ce{LiAlH4}$ besides reduction?

Not according to the commonly accepted mechanism here
The first step is

The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester.

Note: the pKa of primary amides is in the 22-25 range so not close to alcohols.

For one of the question in Stack exchange "Whether Grignard Reagent react with amides?", It was given that RMgX would deprotonate the amides. Now my doubt is that when alkyl carbanion from RMgX deprotonates the amide at Nitrogen atom, why not the hydride ion(a stronger base than carbanion, i suppose) deprotonates the same. Kinetically an acid base reaction is more feasible than a nucleophilic addition of hydride ion, if iam not confused.

Hydride from LiAlH4 is a very different species than R- from Grignard. What you need to consider is the role of both Al and Li in complexing to the carbonyl group and whether the hydride delivered is truly free H-. I do not think it is.

@Waylander: Does it really matter if the anion of the amide is formed? Aluminum complexes with negative oxygen and hydride is delivered to C=N, etc.

I don't think it does, but the mechanisms I could find do not involve anion formation


text:
Acidic strength of phenol derivatives
While comparing the strengths of o-nitrophenol and p-nitrophenol, p-nitrophenol is found to be more acidic. The reason I studied for this is $\ce{H}$-bonding in o-nitrophenol which makes the acid more stable, making it a weak acid.
But between o-methoxyphenol and p-methoxyphenol, o-methoxyphenol is more acidic because of more -I effect at o-position. Doesn't $\ce{H}$-bonding happen here?
Can someone please clarify the exact reason for opposite orders?

Possible duplicate of Arranging the phenol derivative in decreasing order of acidity

https://chemistry.stackexchange.com/questions/71859/is-there-an-intramolecular-hydrogen-bond-in-guaiacol

That's a really good question. The difference arises because -NO2 is -M group but -OCH3 is +M
So even if we think that the hydrogen of -OH group is doing Hydrogen bonding with OCH3 in the end the acidic strength will be decided by inductive effect of -OCH3 group at ortho and para position.
Since ortho is close its -I is dominating.
In NO2 it is a +I group so we didn't look at inductive


text:
Calculating the Ionic strength of sodium acetylsalicylate
With the addition of aqueous NaOH to an aspirin solutions in water the reaction will yield products of sodium acetylsalicylate and water. I would like to measure the amount of sodium apsirin formed that is disolved in water right after the NaOH reaction.
Attempt for the solution:
Given that sodium acetylsalicylate dissociates in water to form sodium ions and salicylate ions, I can consider finding the ionic strength i.e. ion concentration. However in order to calculate the ionic strength I need the molality which require the amount of solutes disolved, which is basically what I am trying to find in the first place.
Is there a recorded value for the Ionic strength of of sodium acetylsalicylate, which would intern allow me to calculate the mass of the solute or I'm going the wrong way about this?
Is there any other way I can use to algebraic calculate the mass of the sodium acetylsalicylate dissolved?
Any help will be greatly appreciated!

why not to calculate it from amount of added NaOH ?

Then it would be the amount of solute added not dissolved

As it is added to aspirin solution and as sodium aspirin as much more soluble..... it is still dissolved.

ahhh due to sodium aspirin being more soluble in water, we can assume the amount of sodium aspirin formed due to the REACTION with NaOH...as the amount of sodium aspirin dissolved. Is that right?

It is a simple acid base reaction.

After doing my calculation I find that the maximum amount of mass that can dissolve water is more than the mass of the solvent. Is that normal? Can the dissolved mass be more than the mass of the solvent.

It is common.. Many compounds are able to dissolve in several times smaller mass of water, and some even in any ratio.

Aspirin has a solubility of ~3 g/L. In addition from a patent by Galat [1]:

Sodium aspirin is almost 1000 times as soluble in water as aspirin itself.

All of the $\ce{NaOH}$  can be assumed to react with acetylsalicylic acid (aspirin) to form the sodium salt.  This means the amount of solubilized sodium aspirin is roughly equal to the amount of added $\ce{NaOH}$ (up to the equivalence point).
When computing the ionic strength you can therefore assume that the concentration of solubilized sodium aspirin is roughly equal to the amount of added $\ce{NaOH}$.
References

Galat, A. Stable Sodium Acetylsalicylate and Method for Its Manufacture. US3985792A, October 12, 1976. (PDF)

But we can assume that the amount of sodium aspirin formed by its reaction with NaOH, is the amount dissolved

@EPICTubeHD I thought you wanted to compute the ionic strength? In that case if the concentration of added NaOH is significantly greater than the concentration of dissociated acetylsalicylic acid then you can assume most of the ions are from Na aspirin.


text:
How can one determine an atomic mass with high accuracy using gravimetric analysis?
In the Wikipedia article on gravimetric analysis it is said that gravimetric analysis was used to determine the atomic masses of many elements in the periodic table to six figure accuracy. There are, however, no references to that claim.
Could someone please tell me, the molar masses of which elements can be determined using gravimetric analysis and what the exact procedure is?

Why, average analytical balance should be capable of doing that. "Six figure accuracy" doesn't mean it's "six figures after the decimal point".

@andselisk Ok, that's a good point, I haven't thought of that. But you cannot measure molar mass on a balance, so the questions about the specific procedure still stands.

Well, interesting question, I would re-word your question as "How can one determine an atomic mass with high accuracy using gravimetric analysis? Precision simply implies that you repeated a measurement and it agrees very well with itself even though it may be wrong. What we desire is high precision and high accuracy.
Almost all the naturally occurring elements had their atomic masses (used synonymously as atomic weight in older books) determined by gravimetry. For that you need to read Richard Theodore's Nobel Prize winning work who meticulously measured atomic weights for many many elements using gravimetry at Harvard in the early 1900s.
Do you know what was the experimental definition of atomic mass which made gravimetry possible? It is weight of an element which combined with 1 part of hydrogen or 16 parts of oxygen (we have to choose one of these but not both). Very sadly, nobody teaches this today so most chemistry students are oblivious to it and yet they can regurgitate the 1/12 of C-12 definition. This definition was invented for some other reasons rather than experimentally measuring atomic weights in the 1960s to resolve disputes among physicists and chemists.
So if I assign a weight of 16 units to oxygen and 65.4 parts of zinc combine with one oxygen in a 1:1 ratio (ZnO), we would assign an atomic weight of Zn as 65.4. Arbitrary as it may sound, this was the experimental way. What Richard did is that he used precipitation reactions using silver chloride and correlated that with oxygen 16.
I provide you with an exercise from Pauling's General Chemistry (forget about moles, Avogadro's, g/mol, molar mass etc.) to solve this problem. Let us go to 1940s, we have the above definition that atomic weight is weight of an element which combines with 16 parts of oxygen.

In a determination of the atomic weight of iron, 7.59712 g of
  carefully purified ferric oxide, $\ce{Fe2O3}$ , was reduced by heating
  in a stream of hydrogen, and found to yield 5.31364 g of metallic
  iron. Given the atomic weight of oxygen as 15.9994, to what value of
  the atomic weight of iron does this result lead?

Hint: Try to find what weight of iron would combine with 15.9994 parts of oxygen and note that there are two iron atoms and three oxygen atoms.

Very nice answer, thank you. Alas, I couldn’t find Theodore Richard’s work on the determination of molar masses. Could you point me to a reference, or describe his methods in more detail?

What grade are you in? Sorry, there is no single paper. It was element by element. Go to Google Scholar, set the year limit to 1930, and search Theodore +atomic weights. You will have search by elements name. https://scholar.google.com/scholar?hl=en&as_sdt=0%2C44&as_yhi=1930&q=Theodore+atomic+weights&btnG=

Don't search the word molar mass. Search atomic weight determination.


text:
Confused on how to draw cis-[Ni(H2O)4(NH3)2]Cl2
I got a question on my practice exam telling me to draw cis-diamminetetraaquanickel(II) chloride, but I'm not sure on how to place the two $\ce{Cl}$ atoms.

A structure of a coordination complex with properly assign wedged bonds for depicting stereochemical features should be be either fully enclosed in square brackets or denoted with an L-shaped brace on the top right corner with charge being superscripted.
Since here both chloride anions are not ligands and don't belong to coordination sphere, they should be placed outside.
There are no strict rules for the exact placement, however, according to IUPAC recommendations on Graphical Representation Standards [1], there are several limitations such as distance between ions and its alignment.
By taking these into account the following depiction would be appropriate:

If you are writing with $\mathrm\LaTeX,$ have a look at Counter ions and charge of a complex.
From [1, pp. 378–379]:

GR-5.7 Polyatomic ions
[…]
  If it is important to depict atoms and bonds explicitly, the ion should be drawn in full without concern for the delocalized charges. That structure should then be enclosed in large square brackets, and again any charge should be reported as a superscript following the rightmost bracket. For single fragments, it is equally preferred to forgo the leftmost bracket and depict simply a “corner” at the top right of the diagram, again with the charge as a superscript following the corner. The full paired brackets should be used in any situation where there is a possibility that the charge may be incorrectly associated with an additional structure diagram to the left of the intended one.

From[1, pp. 384–385]

GR-7.3 Salts drawn in unspecified form
In addition to the styles discussed above, an alternative style depicts the ions in structural form, but without any implied association between the ions. This is the preferred form for mixed salts where there is more than one positively charged atom or more than one negatively charged atom, and where the author wishes to treat them separately. In this style, the largest ion(s) should be enclosed in brackets, to emphasize that there is no structural association between it (them) and the other ion(s). The selection of ions to bracket is performed solely based on size, and so the bracketed ion(s) may be positively or negatively charged. The bracketed ions are placed on the left, and the other ions are placed on the right.
The various ions should be clearly separated from each other, but not separated so widely as to seem wholly unrelated. A separation distance roughly equivalent to the structure’s average bond length is usually reasonable for most purposes.

When an ion appears more than once, it may be depicted a single time, with an additional numeral present to indicate the repeat count.
  If the repeated ion is represented by a molecular formula, the repeat count is placed to the left of the formula in regular-style (non-superscripted, non-subscripted text) and separated from the formula by a space.

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277.