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What are the preferred IUPAC names of these two compounds?
In a recent homework assignment we are asked to give IUPAC name to the following two compounds:

I am not asking this question for homework help -- for that purpose I got both names correct, to the extent of the standard given in my orgo textbook. What intrigues me is that compound naming softwares seem unable to agree upon the names of these compounds.
For structure I: (MarvinSketch is consistent with homework answer, because bromo comes alphabetically before bromobutyl)

ACD/ChemSketch (freeware) 2017.1.2: 4-(2-bromobutyl)-6-methyloctan-3-ol
MarvinSketch 18.4: 6-bromo-4-(2-methylbutyl)octan-3-ol

For structure II: (ACD/ChemSketch is consistent with homework answer, because this name recognizes 3 substituents, while the other recognizes only 2)

ACD/ChemSketch (freeware) 2017.1.2: 4-(2-bromobutyl)-6,6-dimethyloctan-3-ol
MarvinSketch 18.4: 6-bromo-4-(2,2-dimethylbutyl)octan-3-ol

Is it because IUPAC rules are so complicated they can be only followed by hand, or because the tie-breaking rules presented in textbooks are too simplistic to give preferred names in every case?

If you can follow the rules by hand then a computer can certainly be programmed to follow them. So either the programs were using the wrong algorithm, the rules changed, or the textbook cherry picked problems for which its subset of rules yields the right answer.

Related: IUPAC naming – choosing the main branch

We can only speculate about the various automatic algorithms that are available to generate names for chemical structures. Anyway, all known programs have limitations or shortcomings. In particular, the algorithms may not include all rules or they may use obsolete rules. Note that the IUPAC recommendations have changed several times and that the current (2013) recommendations have more than 1500 pages.
In my experience and also in this case, ACD/ChemSketch is more reliable than MarvinSketch. However, such programs are typically updated frequently, and some issues might be solved in a new version.
The compounds that are given in the question are not trivial cases. In order to find the preferred IUPAC names, we have to dig a bit deeper and consider the rules given in Section P-45 of the current recommendations. These rules are applied if two or more names may result according to the basic nomenclature rules, because of different substitution patterns or multiple occurrences of the same senior parent structure. Apparently, these rules are taken into account in the present version of ACD/ChemSketch but not in the present version of MarvinSketch.
Thus, ACD/ChemSketch correctly gives the preferred IUPAC name for the first compound as 4-(2-bromobutyl)-6-methyloctan-3-ol.

The name ‘6-bromo-4-(2-methylbutyl)octan-3-ol’, which was proposed by MarvinSketch, is not wrong, it is just not the preferred name. Remarkably, this name was in accordance with obsolete IUPAC recommendations from 1979 (retained in 1993). Therefore, some homework exercises might still expect this answer. However, this does not necessarily mean that MarvinSketch is using old IUPAC recommendations from 1979 or 1993; it could be a mere coincidence as the second example below shows.
ACD/ChemSketch also correctly gives the preferred IUPAC name for the second compound as 4-(2-bromobutyl)-6,6-dimethyloctan-3-ol.

The name ‘6-bromo-4-(2,2-dimethylbutyl)octan-3-ol’, which was proposed by MarvinSketch, is not in accordance with IUPAC recommendations (neither 1979 nor 1993 nor 2013). Apparently, the relevant (past or present) tie-breaking rules for such cases are not included at all in the present version of MarvinSketch.


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Why is hypophosphorous acid (H3PO2) a strong reducing agent?
Why is hypophosphorous acid $(\ce{H3PO2})$ a strong reducing agent? the oxidation state of phosphorous in $\ce{H3PO2}$ is $+5$, as hydrogen and oxygen both are more electronegative than phosphorous, and phosphorous can never increase its oxidation state to more than $+5$, so how can it be a strong reducing agent?

If you render the oxidation state of hypophosphorous as +5, then the electrons in the phosphorous-hydrogen bonds are attributed to the hydrogen atoms in those bonds.  You can do that, but then the hydrogen atoms involved have oxidation state -1 not +1.  So the reducing action of hypophosphorous acid is attributed to the negative oxidation state hydrogens.

so its all because of the negative partial charge on hydrogen that will be responsible for the reducing ability of the acid?

Hyrogen in a negative oxidation state (-1) is reducing.

can u plss explain how negative oxidation state of hydrogen is reducing?
everytime i encountered with oxidation problems i always used to see the oxidation state of the central atom and then judge the oxidizing and reducing nature of the compound in a reaction..

The central atom is often decisive, but not always.  You really have to consider the oxidation state of all atoms.

The oxidation state of phosphorus in H3PO4 is indeed +5, but in this case you have only 2 and not 4 oxygens. This means that your P is only +1. The reason it's a reducing is because that P really wants to "become" +5.
Oscar Lanzi in his answer suggests that you can assume the H is H-, but this introduces a complication of this being a hydride, and these are not usually stable in solution. Note that this distinction is artificial. Eventually, this is all covalently bonded and the use oxidation numbers suggests ionic bonding.

If you assume phosphorous is +5, then hydrogen attached to it is -1.  You can assume instead that all the H is +1 but must then accept P being +1.  No matter how you slice  it, you render something in a state that makes the molecule strongly reducing.

$\ce{H3PO2}$ not $\ce{H3PO4}$

@A.K. yea I know, that's what I said "only 2 and not 4 oxygens".

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Using acetone instead of acetonitrile on reverse-phase prep-HPLC
I've been using reverse phase prep-HPLC a LOT lately, and I'm growing increasingly concerned about the amount of acetonitrile needed. In an attempt to use a cheaper and more environmentally friendly solvent I had a read around and discovered you can substitute MeCN with acetone.
Has anyone ever tried this? How did it compare in a broad sense?

If you want to go completely green, you should use ethanol.
In many sectors it is considered the only green solvent (except for water of course).
Then, to be really green, you should make sure that it comes from fermentation, not from synthesis.
This quality exists.
The only problem with ethanol is the same as with acetonitrile, the price.
In this case it is due to the taxes that it normally has.
However, if you are working for some institution, it is probable that you can buy it tax free.
Another alternative is to use methanol.
It is not so green, but you can call it wood alcohol.
Methanol works very well in HPLC.
The only difference with acetonitrile is that the pressure in the system will be higher.
This is normally not a problem because the equipment is designed for it.
I prefer not to use acetone because it can give reaction with some compounds, or with itself to form isophorone or other condensation products.
As Waylander correctly points out in his comment, the UV cut off of acetone is also a problem in case you are using UV detection.
                 UV cut off (nm)
 acetonitrile        190
 methanol            205
 ethanol             210
 acetone             330

To change from one solvent to another you will have to change the gradient.

You will also need to look at your detection system. If you are using UV then be aware that acetone absorbs across a different range than MeCN

@Waylander Very true. Thank you for the hint. I have edited.

Yes - reports are easily located using a search engine and keywords such as "HPLC", "acetone", "green chemistry", and "acetonitrile":
One abstract here:
The use of acetone as a substitute for acetonitrile in analysis of peptides by liquid chromatography/electrospray ionization mass spectrometry. Keppel et al. Rapid Comm. Mass Spect. December 2, 2009.
and another write-up (non-refereed):
HPLC going green: Generic replacement of acetonitrile with acetone
and another abstract here:
Acetone as a greener alternative to acetonitrile in liquid chromatographic fingerprinting. Funari et al. J. Separation Science. April 2, 2015.
Comparisons are favorable and generally require modest changes in what to expect in terms of retention times and peak broadening.

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Why are conformational isomers considered as isomers?
I was taught that geometric (cis-trans) isomers are considered isomers because there is a high energy barrier to breaking the double bond for rotation (to convert into the other isomer).
 Given that there is a low energy barrier to rotation about a single bond, why are conformational isomers considered as isomers?

Gaurang's answer deals well with the theory part I just wanted to give a visual aid.

(Taken from Atropisomerism of naphthyl alcohol)
Let's call the rotating substituent on the naphtahlene a fan.
The fan has two black fins one red fin.
Now let us consider two conformations

Red fin at 3'o clock

And

Red fin at 9'o clock

If you were to take snapshots of both the conformations they look different.
Now these two isomers are different and naming them as 9'o clock and 3'o clock  is obviously very inefficient.
That's why conformational isomers and their nomenclature exists, for us to point out specific configurations of a compound.

I was taught that geometric (cis-trans) isomers are considered isomers because there is a high energy barrier to breaking the double bond...

There's a problem here. Geometrical isomers are isomers because the arrangement of different groups around the pi bond is different. Consider the following molecule:

It also has a pi bond, we might be tempted to say that they are geometrical isomers. But, they are not. In fact, they are identical.
Also, geometrical isomerism can occur without a pi bond, as in 1,2-dimethylcyclohexane.
The actual condition for geometrical isomerism is the restricted rotation of different groups, leading to different isomers that can be isolated from each other. These isolated isomers usually have very different physical and chemical properties from each other, and the Wikipedia page does a pretty good job of detailing that.

Now, recall that geometrical isomerism and conformational isomerism are both forms of stereoisomerism. That is, they arise because the relative configuration of their atoms in space is different.
You are correct that conformational isomerism arises because of unrestricted rotation of a molecule around its sigma bond. In fact, there are infinite conformations for any given molecule! Notice that I said conformations, because conformational isomers are finite, and refer to only those conformations that have specific energy minima or maximas. Here are, for example, the finite conformational isomers of butane:
 (source)
The reason why these conformational isomers are given special study ("conformational analysis") is because (to quote) they "can be used to predict and explain product selectivity, mechanisms, and rates of reactions". Again, the Wikipedia page details this nicely with examples.

To sum this up:

Both geometrical isomerism and conformational isomerism are a type of stereoisomerism
Geometrical isomers are considered so because of restricted rotation (about a pi bond or a ring), wile conformational isomers have unrestricted rotation (about a sigma bond)
Geometrical isomers are distinguished from each other because of their very different chemical and physical properties. Hence, they are considered as isomers and separately studied.
Conformational isomers usually have similar properties, and rapidly interconvert. Yet, they are sometimes helpful in explaining the mechanisms of certain reactions. Again, they are considered as isomers and need to be studied.

I hope it helps!

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Why is the following reaction not considered a redox reaction?
The following question appeared in the Cambridge International AS and A level Chemistry Coursebook:

Choose from redox, substitution, elimination, addition and hydrolysis to give the type of reaction for the following:
  $$\ce{C2H6(g) + 2Cl2(g) -> C2H4Cl2(l) + 2HCl(g)}$$

The answer given by the answer key was substitution since $\ce{Cl2}$ replaces $\ce{H2}$.
However, why is redox not a suitable word to describe the reaction? After all, the oxidation number of $\ce{C}$ in the reactants is $-3$, while it is oxidised to become $-1$ in the products. Furthermore, the oxidation number of $\ce{Cl}$ in the reactants is $0$ while it is reduced to $-1$ in the products.
Hence, by that logic, isn't the reaction also a redox reaction since $\ce{Cl}$ is reduced and $\ce{C}$ is oxidised?
Thanks in advance for any help.

You are right. The question is definitely misleading. If it would ask for the mechanism, (radical) substitution would be the correct term, as a reduction does not say anything about the mechanism. But "type of reaction" can basically mean anything, and in this case both answers are correct.

Electrophilic substitution? This mechanism is likely free radical substitution. An alkane is reacting.

Sorry, my bad. Edited the answer. The main intention of the answer of course still holds.


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Tetragonal (z-in/z-out) in tetrahedral complexes?
Is it possible to have tetragonal distortion in tetrahedral complexes? Because distortions in octahedral complexes take place along the C4 axis to reduce the symmetry to D4h, but there is no C4 operation in the Td point group, I was inclined to think there cannot be z-in/z-out as such?

Since $D_{h4}$ is not a subgroup of $T_d$, the z-in/z-out distortion you talk about does not give rise to a tetragonal distortion for a tetrahedral structure. It is possible to have tetragonal distortions in tetrahedral complexes, and has been observed in some cases (Bates and Chandler 1975) and (Sharnoff 1965). There is an illustration in Figure 1 of (Virot, Hayn, and Boukortt) which shows what turns out to be the easiest and most common tetragonal distortion for a tetrahedron, the one to the point group $D_{2d}$. So while z-in/z-out is associated with octohedral groups (and is usually the easiest way for them to have a tetragonal distortion), there are other distortions which occur in tetrahedrons.

So, wouldn't it be kind of wrong to call it a tetragonal distortion? Shouldn't it simply be a J-T distortion which of course would arise for any non-linear electronically degenerate molecule? I'll look up the first two references as soon as I can. Thank you so much!

I think it depends on what you mean by "tetragonal distortion." I guess the way I am thinking about it that a Jahn Teller distortion that lowers the point group symmetry to a tetragonal point group is a "tetragonal distortion", even if it is not z-in/z-out and I think that is how some of the references talk about it. However, the vast majority of the time, "z-in/z-out" and "tetragonal distortion" are synonymous. At least this is my interpretation . . .

I see. What would be the defining criterion for a tetragonal point group? A fourfold rotational axis?

Ashcroft and Mermin has a good explanation of this in Chapter 7. But, the short answer is, you take your seven Bravais lattice crystal systems and reduce the symmetry to derive your 32 crystallographic point groups. Then, a crystallographic point group is in the category (of the Bravais lattice point group) it was originally derived from until all its symmetry elements can be found in a Bravais lattice crystal system group of lower symmetry. A fourfold axis means a point group is at least tetragonal, though it could be cubic. Since $D_{2d}$ has S4, it is not orthorhombic so it is tetragonal.


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Are coefficients in a reaction taken as the exponents in the rate law expression?
In my book it is written that if we have an reaction equation such that
$$\ce{a A (g) + b B(g) + c C(g) -> product}$$
The rate expression of the reaction will be
$$R = k[\ce{A}]^a [\ce{B}]^b [\ce{C}]^c.$$
But, when I researched on expression of rate law, I understood that $a$, $b$ and $c$ are not coefficients of the reaction. But, in most part of my book, the coefficients are taken as the powers $a$, $b$ and $c$.
Is my book wrong or is there something that I don't know about rate expression?

But when i researched on rate of expression i undertood that a, b and c are not coefficients of the reaction

You are correct. $a,b$ and $c$ summed together represent the order of the reaction. Note that this is very different from the molecularity of a reaction.

But in most part of my book the coefficients are taken as the powers a b and c

Without knowing the reactions for which this is done, it is impossible to comment further. However, assuming that the book is correct, we can say, that in those specific cases, your book was dealing with elementary reactions.
Recall that molecularity of an elementary reaction is the same as its order. However, most natural reactions occur through complex mechanisms, and in those cases, molecularity is rarely equal to its order.
For more details on the difference between molecularity and order of a reaction, have a look at this question.

The reaction is $Pb(aq)+2I(aq)⟹PbI₂(s)$ will as this reaction occur in one step we can say that it is elementary so we can write rate of reaction as          $R=[Pb][I]^2$ but if the reaction occur in several step it's Rate expression would be the rate expression of its slowest step am I right?

@MarvaJami Yes you are absolutely correct Marva. Well done! :)


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Does a non-terminal alkyne react with sodamide?

Hydrocarbon (A), $\ce{C6H10}$, on treatment with $\ce{H2/Ni}$, $\ce{H2}$/Lindlar catalyst and $\ce{Na}$/liquid $\ce{NH3}$ forms three different reduction products (B),(C) and (D) respectively. (A) does not form any salt with ammoniacal $\ce{AgNO3}$ solution but forms a salt (E) on heating with $\ce{NaNH2}$  in an inert solvent. Compund (E) reacts with $\ce{CH3I}$ to give (F). Compound (D) on oxidative ozonolysis gives n-butanoic acid along with other other products. Give structures of (A) to (F) with proper reasoning.

From the formula of (A) and the details given, it is clear that (A) is an alkyne. I could make out that (B) would be an alkane while (C) and (D) would be cis and trans alkene respectively. Now, as (A) does not react with ammoniacal $\ce{AgNO3}$ solution, it is not a terminal alkyne. But, then I got confused as the question told, (A) reacts with $\ce{NaNH2}$ to give a salt. Till now, I knew that only terminal alkynes react with $\ce{NaNH2}$. What is it that am I missing?

It is a part of the assignments that I had received, so i don't know as to which book it was but a quick google search, and I found this link : link

Use of a strong base at elevated temperature allows for alkyne-1,2diene equilibrium. Among other things, it allows migration  of triple bonds into terminal position.

@permeakra That means one can use $\ce{NaNH2}$ to convert an internal alkyne into a terminal alkyne ?

@RajdeepDutta and vice versa, it's an equilibrium. The product is guarded by thermodynamics.

The ozonolysis question is a duplicate:  https://chemistry.stackexchange.com/questions/90256/ozonolysis-reaction/90259#90259

By the given question conditions you should easily guess that $(A)$ is $\ce {Hex-2-yne}$. When it is reacted with $\ce{NaNH_2}$ in presence of heat it can isomerise to terminal alkyne by following this mechanism and then $\ce{CH_3I}$ addition will simply follow an $\ce{S_N2}$ mechanism.

Secondly,Ozonolysis will yield an aldehyde which can get oxidised to acid in oxidative medium.


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How can a decomposition reaction be exothermic?
Decomposition reactions involve breaking of bonds which requires energy. Therefore, they are generally endothermic.
But how can decomposition reactions release energy? Examples are the respiration reaction or the decomposition of vegetable matter.

Pure decomposition reactions are exceedingly rare. Typically, some bonds are broken, and some other bonds are formed.

Yes it's the bonds that are formed that release that huge amount of energy, which is more than the energy absorbed when bonds are broken in the case of your compost pile.

The famous decomposition of ATP is exothermic. The explosion of nitroglycerine, or of TNT are exothermic decompositions. All these exothermic decompositions started by endothermic phenomena, followed by the much more exothermic formation of new bonds

It just can be so.  Look for something that is favored by cooling instead of heating.
In the case of decomposition of something to pure elements that is exceedingly rare, but decomposition involving other compounds offers more possibilities.  During hot processing of steel (in the hot rolled coil, actually) we can get this reaction which is favored by cooling below 570°C:
Wustite --> Magnetite + Iron
$4 \ce{Fe_{1-x}O} \rightarrow \ce{Fe_3O_4}+(1-4x)\ce{Fe}$
$x$ typically 0.05-0.10.
See the phase diagram here noting what happens to the $\ce{Fe_{1-x}O}$ phase on cooling.

Note that Fe-Fe metallic bonds are formed, which is crucial to driving the reaction forward. We don't have free gaseous iron atoms flying off!

You have to consider the whole reaction not just the initial step
The problem with your first assumption is that, while often partially correct, it is an incomplete view of any chemical reaction. Yes, many–if not most–reactions start with some bonds breaking (which requires energy) but that is not the whole picture.
Consider the breakdown of nitrogen tri-iodide. This decomposition involves breaking a bunch of N-I bonds each of which requires some energy input. But, as many chemical students have witnessed, the input energy to kick of the explosive decomposition is low. A touch from a feather will trigger it. But the overall reaction releases a fair amount of energy, which is why it goes BANG and is an incredibly dangerous substance to work with.
Why this occurs is that the ultimate products that form (dinitrogen and molecular iodine as I2) release a lot of energy when they form, especially the N2. The reaction is not over when the N-I bond breaks but only when the intermediate products have reformed into stable compounds. And, if we tot up the net movement of energy we find that energy has been released.
The problem with looking at just the initial step is that this only tells us about the energy barrier to starting the reaction, not the overall amount of energy involved. In simple reactions this would be the activation energy barrier (the energy input required to kick things off) not the overall amount of energy involved once the reaction goes to completion.
Plenty of real world reactions need a lot of energy input to start but release even more once they get going. Burning dirty carbon in the form of coal, for example, takes a lot of effort to get going but was the primary way of providing domestic and industrial heat for many people for a century.
In short, for any reaction, you need to consider the whole reaction not just the first step to understand how much energy is involved.


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Why do transition metals make colored compounds despite the remaining d orbitals being hybridized?
I know that transition metals make colored compounds when they forms a bond with a ligand, as the ligand increases the energy level of $\ce{d_{x^2-y^2}}$ and $\ce{d_{z^2}}$ orbital by repelling these orbitals with the lone pair electrons of the ligand. So, the electrons move to the lower energy orbitals and when they absorb energy, they again move to the higher orbitals and that absorbed energy creates color.
Now my question is, $\ce{d_{x^2-y^2}}$ and $\ce{d_{z^2}}$ orbital become hybridized while forming a bond with the ligands and those orbitals get filled up with the lone pair electrons of ligands. Then, how are the electrons excited from lower energy orbitals into the higher energy orbital where there is no place left for them?

There are many theories in coordination compounds. VBT, CFT, LFT, <more that I probably don't know of>. Which one are you using?

@GaurangTandon Seems you're rather overoptimistic here. I'm afraid not one sentence here is correct.

I am talking about CFT(Crystal Field Theory).

I agree with Mithoron here. Not one single sentence is correct from its first capital letter to its final full stop. However, some contain correct fragments in-between. Many of OP’s misconceptions may clear up once they go back to their books and fully understand what is written there.


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Is volume expansion possible upon dissolution of solute?
I am aware that there is volume reduction upon dissolution of ionic solutes, due to the formation of hydrated spheres of aqueous ions where water molecules aggregate around the solute ions.
But similarly could there be a volume expansion?

Positive deviation from Raoult's law comes to mind, though they aren't necessarily ionic solutes only.

Duplicate question (https://chemistry.stackexchange.com/questions/90960/effect-on-volume-of-solution-due-to-dissolution-of-solute) within a couple hours. Please don't

Yes. In thermodynamic jargon, the onset of a change in volume under dissolution is known as an excess volume. If the solution takes up less volume than its constituent parts it is said to have a negative excess volume. Otherwise we have a positive excess volume. You might also find the term excess molar volume.
By googling each term we find many results concerning this, which seems to mainly manifest in metal alloys.


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Are there any (simple) molecules with very different absorption and emission dipole directions?
When a single molecule absorbs or emits light, it does so perpendicular to the direction of the respective transition dipole. In principle, the directions of the dipoles for absorption and emission can be different, due to different electronic configurations of the ground and excited states of the molecule, but in practice the difference is very small. Photons are thus preferentially reemitted into the same plane in which absorption took place.
A large angle between these two dipoles would probably require a significant change in the molecular structure when the molecule is excited. I have heard of some fluorescent proteins which exhibit large angles between absorption and emission directions, but I do not know any details. (Edit: I found this interesting paper, which reports negative fluorescence anisotropy in YFP and a large angle between the absorption and emission dipoles.)
I would be interested in specific compounds which have a large angle between the two transition dipoles. Particularly, are there any simple compounds (less than ~100 atoms) that show such a behavior?
I have read about twisted intra-molecular charge transfer (TICT), in which a segment of the molecule rotates by ~90 degrees upon excitation. I could imagine that in such a case the emission might take place in a plane perpendicular to the original plane of absorption, but I haven't found any confirmation of this either.

Update: to illustrate, I have created a small sketch that outlines what I have in mind. The molecule I drew is just for illustration and does not necessarily show that behavior; I also don't know how exactly such a restructuring process actually takes place (the one I drew would correspond to a rotation of the rings on the right around the single bond). I further assumed that the dipole directions coincide roughly with the positions of the two rings, which might not actually be the case.
That said, the process I have in mind looks as follows:

A photon with a polarization vector parallel to the dipole direction arrives (its direction of propagation is necessarily perpendicular to the dipole direction)
The photon is absorbed and the molecule is left in an excited state
In the excited state some internal rearrangement of the molecular structure takes place, such that the dipole direction changes
A fluorescence photon is reemitted perpendicular to the new dipole axis. The photon has been effectively "redirected" (and shifted to higher wavelengths)

"Very" different may be the issue here, but did you take a look at optical rotation? You can find all sizes of compounds that are optically active.

I think you are right. Moreover ICT plus twisting should also lead to a new "ground" to decay to.

@TAR86: thanks, but this is not exactly what I was looking for. Optical rotation changes the direction of the polarization, but not the actual direction of propagation.

@Alchimista: is this because of the energy that is consumed by the rotation? I imagine the new ground state to be lower than the original one, right?

Not necessarily. Actually the molecule will finally decay to the original one else is photochemistry  Like a state for the emission is formed in the gap.  Emission could be even more red shifted. But this is what seems to me.  I am not sure/ don't know the situation you ask for. ...

I do share your interest and I actually calculate spectra of compounds with this property. I never realized that the direction of emission dipole could be of any importance. I will dig through my data to find this out.
Meanwhile, have a look at the most conscious list of fluorescent molecules under a name I would never guess: http://www.chem.ucla.edu/~craigim/pdfmanuals/catalogs/Lamdachrome-laser-dyes.pdf

OK, then I would suggest to take a look at molecular motors, some are driven by light absorption and I suspect that some of those also emit after rearranging.

Great first question!

In part of the naphthalene molecule you draw you indicate a transition moment. This is the direction for absorption into the lowest excited state $S_1$. As naphthalene is highly symmetric the fluorescence is also emitted from a transition moment aligned in the same direction. The second excited state $S_2$ has a transition dipole set perpendicularly to the long axis and so to the $S_1$ direction.  Again excitation and emission would be from similarly aligned transition dipoles if emission was from $S_2$. However, in solution $S_2$ decays very rapidly ( a few picoseconds) to $S_1$ thus the observed absorption and emission dipoles would be at right angles to one another.
The wavefunction (molecule orbital) in an excited state is different to that in the ground state since an electron is excited to an anti-bonding orbital. This means that bond lengths and angles can change a little. If the molecule has polarisable substituents that are asymmetrical placed then the excited state dipole may differ in direction from that of the ground state. However, in molecules such as rose bengal, rhodamine 6G and many dicarbocyanines, which are not strictly symmetric ($C_1$ point group), the absorption and emission dipoles are parallel as determined by experiment.  (Presumably this is because substituents are not sufficiently close in energy to affect the electronic state in a significant way)
How can we know what the angle between absorption and emission is? The probability of absorption is proportional to the cosine squared of the angle $\theta$ between the electric field direction of the light and the transition moment direction, $\approx (\bar e \cdot \bar\mu)^2= e^2\mu^2\cos^2(\theta)$. So we can describe the molecules are being photo-selected, the probability of absorption being greatest when $\theta=0$, i.e. absorption and excitation directions are parallel.
If we use polarised light to excite the molecules, and assume that it is polarised along the $z$ direction, then the  emission intensity observed through a polariser parallel to the $z$ (or vertical) direction is $I_v\approx (\bar e_z \cdot \bar\mu_z)^2$  and observed at right angles to this ($x$ or horizontal direction)  $I_h\approx (\bar e_z \cdot \bar\mu_x)^2$ where $\bar e_z$ is a unit vector aligned along the electric field and  $\bar \mu_z$ the unit vector component of the fluorescence dipole along the $z$ and $\bar \mu_x$ along the  $x$ direction. The amount of polarisation of the fluorescence is usually measured via the  anisotropy which is defined as
$$r=\frac{I_v-I_h}{I_v+2I_h}$$
where the denominator is the total emission intensity.
In fluid solution molecules undergo rotational diffusion and if the molecule is regarded as a sphere, in fact not such a bad approximation in many cases then, the anisotropy becomes $\displaystyle r(t)=r_0e^{-6Dt}$ where $D$ is the rotational diffusion constant. The anisotropy can be measured by recording the fluorescence decay such as by time-correlated single photon counting. Taking the decay back to time zero produces $r_0$, this is the anisotropy that would be observed if the molecules were in rigid solution and effectively measures the angle between absorption and emission dipoles.
Now this relates directly to your question. The value of $r_0$ can be also be calculated as well as being measured and, again, considering the angle between absorption and emission dipoles $r_0=(3\cos^2(\alpha)-1)/5$ where $\alpha$ is the angle between absorption and emission dipoles. When the angle is zero then $r_0=0.4$ which is easily measured and when at right angles $r_0=-0.2$ so this limits the range of anisotropy.
(references. Text. M. Daune. Molecular Biophysics publ. OUP 1999. Some original literature; T Tao, Biopolymers v8,p609, 1969, Y. Yguerabide Meth. Enzymology, v26, p498, 172, Chuang & Eisenthal, J. Chem. Phys. v57, p5094, 1972)

Thanks for this detailed review. Do you happen to know any substances for which $\alpha$ is large, i.e. $r_0$ is very small or negative?


text:
Why doesn't TEMPO react with itself?
TEMPO ((2,2,6,6-tetramethylpiperidin-1-yl)oxyl, or (2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl) has the following structure:

Given that it's a free radical, why doesn't it react with itself, despite being capable of reacting with other radicals in the system itself)?
In addition, according to Wikipedia, it's a red-orange, sublimable solid. How can it be a solid despite being a radical?

Everything can form a solid. Being or not being a radical is irrelevant here.

1,1-Diphenyl-2-picrylhydrazide is a stable, purple, solid free radical.

You don't need a complicated or sterically hindered molecule.  Oxygen and the $\ce{NO_x}$ oxides are quite stable free radicals under typical ambient conditions.  Credit goes to molecular orbital structure, in both the radical and any supposed dimer.

I believe the first part of your question is asking why the oxygen with the unpaired electron doesn't find a second one to form a peroxide.
The answer is actually quite simple. It is by design. Next to the nitrogen, you see the 4 methyl groups that give TEMPO the "TEM" part of it's name. These methyl groups give TEMPO a lot of steric bulk around the oxygen, and this bulk makes it very difficult for two TEMPO radicals to form an oxygen-oxygen bond with each other. There is just not enough space. You can build a model to convince yourself of this.

https://chemapps.stolaf.edu/jmol/jmol.php?model=CC1(C)CCCC(C)(C)N1OON2C(C)(C)CCCC2(C)C is a link to the 3D structure of the molecule formed from two TEMPO radicals.

It is definitely not only steric repulsion but also a lone pair effect. It is mostly that the formation of a NO-ON bond is quite unstable due to the lone pair repulsion which is larger than the energy gain of forming a sigma bond. (The same reason that peroxides are relatively unstable)

Steric hindrance is the way to make them stable.
I have never worked with TEMPO, but I have seen this one in our lab:

Here the steric hindrance is very evident.
This is also a nice stable solid.

You don't even need the chlorine substitution.

@OscarLanzi Right, but the one I have seen making is with chlorine. Surprisingly easy: trichlorobenzene, chloroform and aluminium trichloride.

The stability of TEMPO and hence is formation is thought to result from resonance involving the nitrogen lone pair as well as hyperconjugation from the numerous methyl groups. See also the Wikipedia article on 3 center 2 electron bond.
The resonance involves oxoammonium contributors as those postulates in oxidation catalyse by TEMPO and carbanions, hence invoking sacrificial hyperconjugation.

As stressed in other answers the steric hindrance of the methyl groups also plays a major role "caging" the radical center and leading to longer than usual bonds when TEMPO reacts.
Consider that coupling two TEMPO would lead to a peroxide. Peroxides are, with few exceptions, unstable and reactive.
The above effects makes a hypothetical $\ce{O-O}$ bond in TEMPO even less stable than that one in other peroxides,  where it is already characterised by energy of about $\pu{50 Kcal/mol}$, i.e. less than the already unstable $\ce{O-H}$ bond in the same molecule and generally speaking less than a half of $\ce{C-C}$, $\ce{C-H}$ and $\ce{C-O}$.

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


text:
empty 30 Gal. barrel of Potassium Hydroxide and Sodium Hypochlorite - safe for collecting Maple syrup?
I bought an empty 30 Gal. barrel of Potassium Hydroxide and Sodium Hypochlorite - CIP 150 - Wondering if it is SAFE to use for the collection of Maple Tree Syrup?  (Label say the white plastic barrel had been rinsed 3 times - however it does say that the previous contents (above) were corrosive. - AND, I plan on rinsing it a few more times) Answers? Suggestions?

As an iron clad rule chemical containers should never be reused for foodstuffs. It is an bit overkill but guessing that reuse would be ok is just nuts.

A more detailed explanation of your answer, as a possible chemist or whatever, would be appreciated. Otherwise - sounds like a anti-maple-syrup-corporate-lackey answer. LOL

Maple syrup lover here, no affiliation with any producers or anti-maple syrup organizations. The safe bet is to err on the side of caution and buy a cheap alternative without the dodgy chemical pedigree. However, the stuff that was allegedly in the barrel is water soluble, and several rinses should clear it, but don't name me in your lawsuit if you singe your trachea: I'd want to be the guy doing the rinsing, scrubbing, and lightly neutralizing residual base with an acid wash.

I'm voting to close this question as off-topic because on-site experts should be consulted for questions like these.

@Øle I don't know where you get the lackey feeling from MaxW, but what he wrote is the answer from any chemist to such a question. Don't ever use containers which were used for chemical appliances or

While I also consider the question somewhat off-topic, I would still like to answer here. In a time in which people eat tide pods for fun it seems kind of necessary...
So, for general information, regardless of what chemical substances were inside the container or what you plan to put into, you should never (re)use containers for chemicals to store food!
Here are some reasons why:

Standard food containers are readily available, do not cost much, and are much better suited to store and conserve food
you are sending a disastrous message to children if you are taking food from a container with chemical warning symbols on it
there is a chance that you mix up your food container with one that contains what is actually on the label
you might not be able to remove all the previous content from the container by rinsing it with water, e.g. if the previous substance is non-soluble in water or if it is greasy. Minuscule quantities of the remaining substance may be toxic or hazardous
the substance may have changed over time if it was stored for a prolonged time or under inadequate conditions, like exposure to heat, moisture or sunlight. Products may precipitate or form films on the inside of the container.
you do not know the history of the container. Maybe the previous owner used up the original content and then used the empty container to store chemical waste. This is also not a good practice, but still more common than to store food!


text:
Is it possible to solve the equation for the diffusion of gas through a sphere analytically by applying e.g. "combination of variables"?
In the case of diffusion through a thin film it is possible to combine the variables in the following way:
$\sigma = \frac{x}{\sqrt{4Dt}}$
where

$x$=length
$D$=diffusion coefficient
$t$=time
$\sigma$=a function of x and t

Thereby, x and t is substituted by the single variable sigma
Is there a similar expression for a sphere?
E. L. Cussler discribes in detail how to solve the equation for a thin film by applying "combination of variables".

Good, but still unclear. You can combine (say, multiply) any variable with any other variable. What is the significance of this particular combination?

@Sigils while not off topic here, you might get a better answer on Physics SE.

Are you looking for a solution which gives you for example in how much time a sphere of sugar will dissolve in water by only diffusion?

Given the diffusion equation
$D(\frac{d^2y}{dr^2}+(\frac{2}{r})\frac{dy}{dr})=\frac{dy}{dt}$
You can put in
$y=z/r$
And then
$\frac{dy}{dr}=(\frac{1}{r})(\frac{dz}{dr}-\frac{z}{r})$
$\frac{d^2y}{dr^2}=(\frac{1}{r})(\frac{d^2z}{dr^2}-(\frac{2}{r})\frac{dz}{dr}+\frac{2z}{r^2})$
$\frac{dy}{dt}=(\frac{1}{r})\frac{dz}{dt}$
Plug these into your diffusion equation and a lot of terms cancel leaving the same equation you would have in rectangular coordinates:
$D\frac{d^2z}{dr^2}=\frac{dz}{dt}$
So you then use the same combination of variables as in the rectangular case except you put $z=ry$ as your dependent variable.

Then I get a result of the type:
z(sigma) = C1+erf(sigma) * C2
Which leads to an equation of the following type with respect to c:
c(r,t) = (C1 + erf(x/(sqrt(4Dt))) * C2)/(r)
Is this correct?

Your boundary conditions on $z$ will not be analogous to those on $y$ in the rectangular case.  You can reduce your problem to one independent variable but you might get different functional forms depending on the form of the boundary conditions for $z$.  So treat that carefully.  Also remember to use $r$ for your coordinate throughout.

Yes. Let's improve the wording here: that is not just an expression, it's the definition of a dimensionless variable that eases the understanding of the solution of the transient diffusion is a semi-infinite medium. Because there is no clear characteristic length as in a finite diffusion path problem, Buckingham π theorem will tell us the two dimensionless quantities we need are
$$\Pi_1 = \frac{z}{\sqrt{D t}}$$ (Or any function of it, like squaring it or dividing it by 4, such as your $\sigma$)
And
$$\Pi_2 = \frac{y-y_{A\infty}}{y_{As} - y_{A\infty}}$$
With $s$ being the surface index and $\infty$ indicating a position "at infinity" (often we use $y_{A\infty} = 0$).
Then we conclude $\Pi_2 =\textrm{function}(\Pi_1)$. This particular function can found by solving Fick's second law with the appropriate boundary and initial conditions. In the most common case, it yields a Gauss error function.
Had we be initially interested in a sphere instead of a planar surface, the only major difference would be the use of spherical coordinates instead of a Cartesian frame, which would result in a different functional solution. But the problem of defining the dimensionless variables would be the same, given we assume perfect angular symmetry. The $z$ position would just be substituted by a radial position $r$.


text:
Electrochemical manufacturing of small metal components
Metal crystals easily can be grown via electrochemical pathways. Copper is a great example of this. I recently had an idea of growing those crystals in controlled ways (i.e.: surrounding the electrode on which the crystallization occurs with a "mold" of the desired shape with holes in it) but these crystals are tending to grow towards the another electrode in the shortest way possible. Let's say that I would like to grow a copper tetrahedron on a triangular copper plate surrounded by plastic sheets which are controlling the crystal growth.

The problems which I came across:
     1. How should I arrange my equipment to get the crystals fill the mold?
     2. Shold I worry of liquids being trapped in the crystal?

Nice idea, but difficult to do.
The copper cations must continuously enter the "mold" electrode to grow there, but they must not be able to grow outside.
If there are holes in the "mold" the crystals will grow towards them because there is where the cations will come from.
A permeable "mold" would let cations in over all the surface, but nothing would stop them to continue growing outside the membrane.
If the crystals tend to grow to the other electrode in the shortest possible path, you could place a very small electrode just above the center of the triangle of the other electrode.
The shape will, however, not be very defined.
Maybe in combination with a permeable "mold"...
Trapping of liquids inside the piece will be almost certainly a problem.
You are not growing a single crystal.
Many crystals will start growing all at the same time in different places of the electrode.
When they join somewhere, they will not fit nicely together, so some liquid will be trapped.
This occurs also when big crystals are made in a normal crystallization, especially when there is no stirring.


text:
Why are silicon and germanium used as semiconductors but not other metalloids?
The semiconductor materials are either elementary such as silicon and germanium or compound such as gallium arsenide. Silicon is the most used semiconductor for discrete devices and integrated circuits.
However, other metalloids like antimony and tellurium are not used as semiconductors. Is it because they are too expensive to extract or is there another reason for this?

Astatine is highly radioactive and unstable, most stable isotope half life about 8hrs. Not a desirable characteristic for a semiconductor

It's not important if sth is metalloid, important is its bandgap.

Alright, Astatine is out of the picture, but what about the others? @Waylander

The ones you mention aren’t semiconductors. Carbon is a wide band gap semiconductor, with limited doping options. Tin and lead are semi metals and metals. Not much else to say...

The statement, "metalloids like [a]ntimony and [t]ellurium are not used as semiconductors", is untrue:

Tellurium thin-film transistors have been fabricated.
Bismuth nanowires have been used as electronic gates.
Carbon, in both diamond and graphene allotropes, is used.
Phosphorus, in the form phosphorene, structurally similar to graphene, has been used as a semiconducting photodetector.

That said, there are various reasons why these semiconductors have not replaced silicon, which is plentiful, usable to beyond 400 K, resistant to degradation by air, and has an established fabrication technology, for the majority of electronic uses. GaAs, GaN, SiC and other semiconducting compounds are being used increasingly, though, for high-speed electronics, high-temperature devices and lasers.
See the Wikipedia list of semiconducting elements and compounds... it's longer than you might expect.

While bismuth telluride is indeed used in thermoelectrics, it is a compound. Neither Bi nor Te is itself a semiconductor (like neither Ga nor As are, individually, semiconductors).

A recent paper on the thermodynamics of the Bi-Te system is https://www.sciencedirect.com/science/article/pii/S0364591617301219?via%3Dihub - three line compounds and a broad phase field in addition to the elements.

Te-Bi, as @JonCuster states, is not a single element. However, various literature cites it as either an alloy or an intermetallic compound of definite proportions. In either case, it shouldn't be in the answer and has been removed.  BTW, Bi and Te can be used for semiconductor devices.

What about other elements in the C Si Ge priodic table column, like Sn Pb? I've seen no usage of such elements as semi-conductors. Are there some?

AFAIK, they conduct too well to be used as semiconductor (though they're superconductors), but compounds such as PbS and PbSe are used in IR photodetectors. https://www.lasercomponents.com/de-en/ir-components/ir-detectors/pbs-and-pbse-detectors/, as well as Sn compunds: https://onlinelibrary.wiley.com/doi/10.1002/advs.201900751


text:
Why is Red Phosphorous used in the Nagai Method?
The most common method of producing Dextromethamphetamine (Crystal Meth) is supposedly the Nagai Method, where Pseudoephedrine, Hydriodic acid and Red Phosphorous are refluxed overnight, producing a mixture of Dextromethamphetamine, water and Red Phosphorous which may be reused.
The thing is, I don't quite understand how the Red Phosphorous ends up in the mixture as though it were acting as a catalyst.
The best mechanism I could think up was the following:
Pseudoephedrine + Hydriodic acid --> Iodoephedrine + Water
Iodoephedrine + Hydriodic acid --> Methamphetamine + Iodine
Using no Red Phosphorous at all.
The only use of Red Phosphorous I can see here would be to convert the Iodine back to Hydriodic acid, which would give Phosphoric acid as a byproduct, not Red Phosphorous.
What exactly is the Red Phosphorous supposed to be doing here?
EDIT: I notice that this question is very similar to mine, however whilst the answer to that question does address the use of the Red Phosphorous (Phosphoric acid is produced as I had assumed), it unfortunately doesn't explain how it ends up back as Red Phosphorous in the end.
Disclaimer: I am merely interested from a theoretical standpoint and have no intention of making Dextromethamphetamine. Just saying.

@Mithoron I may have accidentally clicked on the 'Possible Duplicate' banner's "This answers my question" button... Can that be undone by any chance?

Your question has entered a reopen queue. If 5 people consider it reopenable, it'll be opened.


text:
How to calculate the pH of a sodium bicarbonate-sodium hydroxide buffer solution?
This website claims that if you add $\pu{50 mL}$ of $\pu{0.05 M}$ sodium bicarbonate, and $\pu{5 mL}$ of $\pu{0.1 M}$ sodium hydroxide (and dilute to $\pu{100 mL}$), you should create a solution with $\mathrm{pH }= 9.6$.
I'm struggling a bit getting this result.
My attempt:
The sodium bicarbonate reaction would be:
$$\ce{NaHCO3 + H2O -> H2CO3 + OH- + Na+}$$
We know the $\mathrm{p}K_\mathrm a$ of $\ce{H2CO3}$ is 6.4, so the $\mathrm{p}K_\mathrm{b}$ of $\ce{HCO3-}$ is:
$$14 - 6.4 = 7.6$$
Furthermore, $\ce{pOH} = \mathrm{p}K_\mathrm b + \log\frac{[\ce{B+}]}{[\ce{BOH}]} = 7.6 + \log\frac{[\ce{H2CO3}]}{[\ce{HCO3-}]}$
However, how do we find $[\ce{H2CO3}]$ ?
I'm not entirely sure this is the right approach, but none of the other resources I've used to determine this have helped.

Your idea is good, but you have thought of the wrong reaction.
You have written that $\ce{HCO_3^-}$ hydrolyses to give $\ce{H_2CO_3}$ and $\ce{ OH^-}$, but think carefully when you add base, $\ce{OH^-}$ is in excess in this medium. So, shouldn't your proposed reaction proceed in backward direction instead? Moreover, your reaction is actually not going to occur.
When you are adding strong base, $\ce{HCO_3^-}$ acts an acid to give $\ce{CO_3^2-}$ like:
$$\ce{NaHCO_3 + NaOH -> Na_2CO_3 +H_2O}$$
 or, more simply
$$\ce{HCO_3^- + OH^- -> CO_3^2- + H_2O}$$
Initially, number of millimoles of $\ce{HCO_3^-}$ were $=50\times0.05 = 2.5$ and number of millimoles of base added $=5\times0.1 = 0.5$
So, $\ce{CO_3^2-}$ produced will be also 0.5 millimoles as the base added was a limiting reagent, and $\ce{HCO_3^-}$ left $= 2.5-0.5 = 2$ millimoles.
As, you can see now, the solution acts an acid buffer as, $\ce{HCO_3^-}$ is a weak acid and $\ce{CO_3^2-}$ is the salt after reacting with a strong base. According to Henderson-Haselbach equation,
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{[\text{salt}]}/[\text{acid}]$$
$\mathrm{p}K_\mathrm{a}$ of $\ce{HCO_3^-}=\mathrm{p}K_{\mathrm{a2}}$ of $\ce{H_2CO_3^}=10.3$ . Now [$\ce{HCO_3^-}$] = $\pu{0.02 M}$, and [$\ce{CO_3^2-}$] = $\pu{0.005 M}$. Now if you calculate $$\mathrm{pH}= 10.3 +\log\frac{0.5}2 = 10.3- \log(4) = 9.69$$ So, the final $\ce{pH}$ of the medium has come out to be $= 9.69$.
The difference in digits after decimal may be due to some given $\mathrm{p}K_\mathrm{a2}$ which is experimental and you should do calculation according to that.


text:
Thermal oxidation of gold(I) sulfide
As far as I know the chemistry of $\ce{Ag}$ and $\ce{Au}$ are almost the same; because $\ce{Ag}$ belongs to the 4d series and $\ce{Au}$ belongs to the 5d series, we know the size of 4d and 5d are similar, so their chemistry remains similar.
Thermal study and mechanism of $\ce{Ag2S}$ oxidation in air
In this paper, it is given that $\ce{Ag2S}$ can be oxidised thermally in the presence of $\ce{O2}$ to give $\ce{Ag}$ and $\ce{SO2}$ over 510 °C. I am not able to find any references where a similar experiment is done for $\ce{Au2S}$. Can somebody please help me ?

One aspect:  gold-sulfur binary compounds are thermodynamically unstable to decomposition.

Gold is generally extracted by pressure oxidizing a refractory ore such as pyrite or chalcopyrite. They are common sulfide ores where finely divided nanosized gold are present. They are roasted to break the sulfide matrix and recover the metallic gold. These refractory ores are considered low grade ores (where gold% is less) and are therefore roasted to extract the gold.
Critically, the size at which one can efficiently liberate gold from the host mineral of the particle will determine the process to use for optimal gold recovery. Gold(I) sulfide is not considered a refractory ore as because gold is not present in microscopic level and hence are generally not roasted.
But it is possible to roast gold(I) sulfide but the yield will be less as compared to the gold  yielded when it is leached.
$$\ce{Au2S + O2 ->[200-500°C] 2Au + SO2}$$
References

Chen, T.T., Cabri, L.J. & Dutrizac, J.E. Characterizing gold in refractory sulfide gold ores and residues. JOM 54, 20–22 (2002). https://doi.org/10.1007/BF02709181
http://mine-engineer.com/mining/minproc/gold-in-pyrite.htm
Leaching of a gold bearing partially roasted sulphide. Laboratory scale studies by M.F. Almeida,  Mat. Res. 4 (4), Oct 2001,  https://doi.org/10.1590/S1516-14392001000400014
https://www.911metallurgist.com/blog/gold-leaching-flotation-process

A similar question was asked in IIT-JAM 2018, I knew that Cu and Pb can be obtained by self-reduction.I also knew that least electropositive metals can be reduced via O2. So I chose Au also. Now in their answer key, Au isn't given. Should I challenge their answer?


text:
Why do gases liquify under pressure?
To my understanding, an increase in pressure (of gases in a container) is proportional to an increase in gases' kinetic energy, and hence temperature. However I read that gases liquify under pressure... However, gases liquify in colder temperatures, so how come gases liquify under pressure, where the temperature increases?
Thanks to all responses.

Do you know what a phase diagram is?

Related: https://chemistry.stackexchange.com/q/80492/31775

@Raditz_35 A phase diagram describes the phenomenon but does not explain it.

@owjburnham I'm aware. The comments are not to give answers but e.g. to ask for information to clarify the question, so asking if the OP knows about a certain thing or is completely new to the field offers valuable information to people trying to answer - imo. You can skip the explanation of phase diagrams for example if the OP knows about them

https://chemistry.stackexchange.com/questions/31508/why-do-some-gases-liquefy-easily-under-pressure-and-normal-room-temperature

https://chemistry.stackexchange.com/questions/86154/why-butane-exist-as-a-liquid-in-a-lighter https://chemistry.stackexchange.com/questions/81744/why-do-gases-need-to-be-cooled-to-liquefy-solidify

https://chemistry.stackexchange.com/questions/40325/london-forces-and-liquifying-noble-gases

Think of it simply as spheres in a box. The spheres are your molecules. The higher the temperature, the higher the energy the molecules have and the more they move around. So when they have high enough energy for the molecules to move around a lot and stay away from each other they are a gas. (This is a simplified answer). If you decrease the energy they have (cool the system down), they move about less and there's a certain point the molecules aren't moving far away enough from each other to stay as a gas, so they liquify. As Raditz_35 said, look at a phase diagram. So when you put a gas under pressure, the molecules are getting "squished" together and they don't have a lot of room to move around, so they can't move away from each other enough and thus liquify. This is how they store Hydrogen as fuel. It's too dangerous as a gas and they put the Hydrogen under pressure to store it as a liquid. Hope this helps.

Both statements are correct: Gases tend to liquefy under pressure and  gases liquefy in colder temperatures. However, in both cases it is implicitly assumed that the other quantity remains unchanged. This is, if you increase pressure slowly enough to give your gas time to transport the increase in heat to the surroundings, then it will eventually liquefy.
To illustrate, look at the working principle of a chiller. In a first step, a working gas is compressed; it increases in pressure and in temperature. The compressed gas is then lead through a heat exchanger on the outside of the machine, where it cools down to ambient temperature. The heat is transported to the surroundings (this is why your fridge is hot in the back). Then the dense gas is allowed to expand again on the inside of the machine and its temperature decreases below the original temperature. The temperature decrease during expansion is just the inverse the temperature increase during compression. With this mechanism way you can cool down gases and eventually liquefy them. The important thing is to dissipate the heat to another medium.


text:
How to account for the interface between two different phases in a discretized diffusion model?
I have tried to set up a model for the diffusion of a gas into a liquid. The two media are next to each other and the geometry is spherical because the system should simulate the diffusion out of a gas bubble.
I have made a diffusion model for the gas phase (the left compartment) and a corresponding diffusion model for the liquid phase (the right compartment).
The two diffusion models are as follows:
\begin{align}
  \frac{\partial c_\mathrm{left}}{\partial t}
  &= D_\mathrm{left} \cdot \frac{1}{r^2} \frac{\partial}{\partial r}
     \left(r^2 \frac{\partial c_{left}}{\partial r} \right)\\
  \frac{\partial c_\mathrm{right}}{\partial t}
  &= D_\mathrm{right} \cdot \frac{1}{r^2} \frac{\partial}{\partial r}
  \left(r^2 \frac{\partial c_{right}}{\partial r} \right)\\
\end{align}
Then I have discretized the volume. So the PDEs are turned into ODEs of the form:
\begin{align}
  \frac{\mathrm{d}c_{i,\mathrm{left}}}{\mathrm{d}t}
  &= D \left(\frac{c_{i+1,\mathrm{left}}
     - 2 \cdot c_i + c_{i-1,\mathrm{left}}}{\Delta r^2}
     + \frac{2}{r_{i,\mathrm{left}}}
     \cdot \frac{c_{i+1,\mathrm{left}}
     - c_{i-1,left}}{2 \cdot \Delta r} \right)\\
  \frac{\mathrm{d}c_{i,\mathrm{right}}}{\mathrm{d}t}
  &= D \left(\frac{c_{i+1,\mathrm{right}}
     - 2 \cdot c_i + c_{i-1,\mathrm{right}}}{\Delta r^2}
     + \frac{2}{r_{i,\mathrm{right}}}
     \cdot \frac{c_{i+1,\mathrm{right}}
     - c_{i-1,\mathrm{right}}}{2 \cdot \Delta r} \right)
\end{align}
When discretizing I got the following expressions:
\begin{align}
  \frac{\mathrm{d}^2c_i}{\mathrm{d}r^2}
  &= \frac{c_{i+1} - 2 \cdot c_i + c_{i-1}}{\Delta r^2}\\
  \frac{\mathrm{d}c_i}{\mathrm{d}r}
  &= \frac{c_{i+1} - c_{i-1}}{2 \cdot \Delta r}
\end{align}
But how does one incorporate the following two boundary conditions at the gas-liquid-interface, where $H$ is Henry's constant?
\begin{align}
  c_\mathrm{left,interface} &= H \cdot c_\mathrm{right,interface}\tag1\\
  J_\mathrm{left,interface} &= J_\mathrm{right,interface}\tag2
\end{align}
Meaning that there is equal flux across the interface between the two compartments.
The concentration in the last compartment of the gas phase (before the interface) is approximated by linear extrapolation.
This means that the concentration just before the interface is given by:
$$c_{3} = \frac{c2-c1}{r2-r1} \cdot r_3 + c_1 - r_1 \frac{c_2-c_1}{r_2 - r_1}$$
In this example it is assumed that only three compartments are present on each side of the interface.

How are you discretizing the time derivatives?

I am only discretizing the spatial derivatives. So I divide the spherical volume into smaller compartments. Then I carry out Taylor approximations to find approximations for the second order derivative and the first order derivative in the general diffusion equation. So I substitute the results from the Taylor approximations into the general diffusion equation thereby generating a system of ODE's.

You may have more luck getting an answer for this on Computational Science.

I have to agree with hBy2Py. I like the question, but I imagine a computationalist or physicist is going to be able to give a better answer.

I posted it on  Computational Science and Math, thank you


text:
Why is the boiling point of polonium less than that of tellurium?
My textbook states that polonium has a lower boiling point than tellurium because it has weaker intermolecular forces of attractions (van der Waals forces).
Why are van der Waals forces of attraction less in polonium as compared to tellurium?
Polonium's atomic size is greater than tellurium, and I believe van der Waals forces are directly proportional to the surface area. Is this the reason?

Strongly related: https://chemistry.stackexchange.com/questions/33815/trend-in-the-melting-point-down-the-group

I was told by my friend that, "Due to inert pair effect valence electrons in Po will be less availably and hence van der Walls forces will be less".... Seems dubious, please someone confirm me if this is right/wrong.

@Zenix It's wrong. Also it's not reasonable to think that that crude rules of thumb would be helpful here.

It's mainly about relativistic effects. Relativistic affects mainly affect heavy nuclei, such as polonium etc, so makes their properties slightly different to the trend.[ref] The outer electron configuration of Polonium could differ so much from its lighter family member tellurium due to the electron constriction, causing an outer electron to fall into a lower shell, that the intramolecular forces are weaker. Thus a lower boiling point.
It explains a lot of things that do not seem realistic with very heavy nuclei. Such as the "golden" colour of Gold. It should be silvery, but due to relativistic effects it is a "golden" colour. It's also like Copernicium behaving more like a noble metal, unlike the other metals of group 12, even being gaseous in room temperature.

Reference: Wikipedia article on relativistic quantum chemistry


text:
How to determine the edge length of the aluminium unit cell?
I'm having trouble understanding why I can't get the correct answer from this homework problem:

Solid aluminum forms a face-centered cubic unit cell.  Aluminum has a density of $\pu{2.70 g/cm^3}$. Determine the edge length of the $\ce{Al(s)}$ unit cell in $\pu{cm}$.

Usually if I enter the wrong answer the it will show all the calculations to determine the correct answer. However this one does not show the calculations.
Here is the work I've done:

$\ce{Al }= \pu{26.98 g/mol}$
$\text{Mass} = \frac{26.98}{6.022\times10^{23}} = \pu{4.480E-23 g//atom}$
$\text{Mass of one atom} = \text{mass of one unit cell}$
$\text{Mass of one unit cell} = \pu{4.480E-23 g//unit cell}$
$D = \frac{M}{V}$, thus $V = \frac{M}{D}$
$V = \frac{\pu{4.480E-23}}{2.70} = \pu{1.66E-23 cm^3//unit cell}$
For a cube $V = l^3$, thus $l = V^{1/3}$
$\text{Unit cell edge length} = V^{1/3} = {\pu{1.66E-23 cm^3}}^{1/3} = \pu{1.18E-8 cm}$

The correct answer that the book gives is $\pu{4.049E-8 cm}$.
The only example I have to go off of in my book is for a simple cubic unit cell. Since $\ce{Al}$ is a face-centered cubic unit cell and has 4 particles per unit cell I tried multiplying my final answer by $4$, but that gives me $4.72$. I also tried multiplying the unit cell mass by $4$ which gave me a final answer of $1.87$. I just can't seem to get $4.049$. Any help understanding this would be greatly appreciated.

Who said that mass of one atom equals mass of one unit cell?

My book said that in the explanation for the simple cubic unit cell.

True, but your cell is not simple. It is fcc.

Hey @haruku, check this out. Your step "I also tried multiplying the unit cell mass by 4 which gave me a final answer of 1.87." should give you the correct answer instead. Are you making a calculation mistake?

You're right, I went back and redid the calculations and it turns out my calculator wasn't grouping numbers properly.

WRONG $4.049\times10^{-8}$ is wrong. You only know the density to 3 significant figures so the unit cell edge shouldn't have any more than that. So the answer should be $4.05\times10^{-8}$. // Also don't keep rounding intermediate results so much. Since you know the answer will have 3 significant figures, carry 2 extra (so 5) in the intermediate calculations. Then round the final result.

Correct approach is to do two things:
1)  Put four atomic masses into one unit cell.
2)  Properly extract the cube root of the volume you obtain.  That volume should now be about $6.6×10^{-23}\text{cm}^3$ from part 1.  To take the cube root properly, render the exponent on 10 as a multiple of 3, thus  $6.6×10^{-23}$ = $66×10^{-24}$ where $10^{-24}$, not $10^{-23}$, is $(10^{-8})^3$.
Then your numbers should work out.


text:
"The centre of positive and the centre of negative coincide" What does that mean?
I'm doing AS level chemistry right now and my teacher told us that when a molecule is non polar , the centre of positive and the centre of negative coincide , I don't really get what he means by that .

Hi Lia, did your teacher follow up with a diagram of what he meant? Or an example molecule? Please add that to your question. Thanks!

Well, he meant just that. Molecules consists of atoms, and atoms consists of even smaller things, many of them charged. Take all positive charges, find the center...

Each molecule has negatively charged electrons, and positively charged nuclei.
If the centroid of the electrons' charges (electron probability density) is different than the centroid of the nuclear charges, then the molecule will have a permanent electric dipole moment.

Several concepts matter to what your teacher meant (but didn't seem to explain well).
One is that bonds are formed when electrons are shared between atoms.
When the atoms involved in the bond are not the same, the electrons are not distributed entirely evenly. When this happens there is, effectively, a mismatch between the distribution of positive charges (from the protons in the atoms' nuclei) and the negative charges in the distribution of the electrons in the bond. The extent of the mismatch is dependent on the relative electronegativity of the atoms.
So a bond between carbon and oxygen, for example, has the distribution of electrons biased towards the oxygen. So in a compound like CO the electrons tend to be closer to the oxygen. The mismatch of positive and negative charges gives the molecule a small electrical dipole. This matters for its chemical behaviour and how it interacts with other molecules.
But the shape of the molecule matters as well, not just the polarity of the bonds. So CO2, a linear molecule with two relatively polar bonds is not a polar molecule. This is because the two bonds are pointing in opposite directions so the net distribution of charge in each cancels out. The overall distribution of electrons and protons in the molecule now coincide with the centre of the molecule, giving it no net dipole moment. This, I think, is what your teacher means.

Basically, the molecules have two atoms of different electronegativities (ability to attract the bond pair of electrons); therefore this causes one atom nuclei to become surrounded by more electrons than the other atoms nuclei. This causes both the atoms positive and negative charges to not coincide.


text:
What chemical can cause such a fire?
A friend of mine who thinks that she is under influence of black magic met an astrologer. The astrologer asked her to wear a t-shirt for three nights and then bring that to him. The astrologer then cut some lemons and rubbed them throughout the t-shirt. Then he poured some water on her hands and face; and asked her to keep the t-shirt on her palm. After about 15 minutes, the y-shirt caught fire and the astrologer dropped the rest of water to put the fire down.
I'm now wondering what chemical could cause this fire. I know that the water looked muddy. Would you please help me to find out what material was potentially used by the astrologer? I'm suspicious of the acid in the lemons and water which has been used, as well as body fat or sweat perhaps was on the t-shirt too. She was telling me that at first, smoke color was white but then it turned to black after the t-shirt caught fire. The smell of the fire was as bad as rotten meat.

Was an alcoholic smell around?

Not sure about alcoholic smell. I been told that was very bad smell seems a dead body smell

And here my  inner sceptic activates and a question arises: I don't see any proof here so how am I supposed know if you or your colleague didn't made it up? Even if it's not the case, then it was a trick and these are made to be hard to analyse and fool people. Therefore I don't see how it's supposed to be valid question.

Is this just a curiosity, where you believe the events can be explained by chemistry, but don't yet know the chemicals?  Or do you plan to accomplish something with this information?  Or are you worried about how long you can wear a shirt before drinking margaritas becomes hazardous? ;-)

Thanks any way for your attention. I don't know how shall I answer your curiosity to make you sure what I said is true and real. All I can tell you, even when I heard what's happened there, at the first I believed that should be some kind of mysterious power behind it, but soon after I come to my senses that should be explaining by chemistry. Chemistry ain't my field, and I just looking for some answer. Any way the astrologer asked for money now to complete the practice. That's why I'm looking for some logical answer to feed my own curiosity.

Magic tricks don't usually rely on exotic physics or chemistry, they just rely on the fact that humans have limited perceptive abilities. I'd personally guess that the guy had a hidden lighter somewhere before assuming there's some secret chemistry at play.

White phosphorus can spontaneously ignite in air, and produces white smoke.
Some oils, like linseed oil, on bunched-up cloth can also spontaneously combust.

The problem is we don't know what exactly he used. All I know is that after he rubbed cut lemons on cloths, and after about 15 minutes the cloths ignited. Perhaps it make sense to you that whatever trick it was, he done it perfectly without she realises what's going on.

All the clue that I have, the time of around 15 minutes and white smoke at the first of ignition as well as bad smell like a dead body smell. That's all I know. I'm not sure how long does it take for white phosphorus or linseed oil to ignite.

In meantime, in my point of view, this should be an old style alchemy that used by such people and passed generation to generation. I don't feel it's a new chemistry science.

@user59309 linseed oil soaked rag took 3 hours and 10 minutes in this experiment: http://abcnews.go.com/GMA/ConsumerNews/summer-safety-diy-dangers-linseed-oil/story?id=11328546 however, if the ambient temperature is high, it would take less time.

@user59309 "a small chunk of white phosphorus has been left exposed to air for about 15 minutes (which is not shown); the white phosphorus then spontaneously ignites, producing a brilliant" https://www.angelo.edu/faculty/kboudrea/demos/burning_phosphorus/burning_phosphorus.htm

@user59309 see "The Alchymist, In Search of the Philosopher’s Stone, Discovers Phosphorus, and prays for the successful Conclusion of his operation, as was the custom of the Ancient Chymical Astrologers" https://en.wikipedia.org/wiki/The_Alchemist_Discovering_Phosphorus#/media/File:Joseph_Wright_of_Derby_The_Alchemist.jpg

I also heard that, after the cloths ignited, the astrologer  dropped the muddy water on that and then put it in a plastic jar. She told me the jar was shaking afterward while it still producing smoke.


text:
Alternatives to EDTA for calcium ion chelation
Are there any $\ce{Ca^2+}$ chelating agents (other than EDTA) that are as good as EDTA in terms of binding calcium ions and effective in a pH range of $5$ to $9$?

Hi Smruthi, welcome to Chem.SE! Have you done research on this topic? If so, please share your research on this topic! It will help potential answerers on the extent of your knowledge.

I don't have time to write an answer now, but this looks interesting, since crown ethers are known for their ion selectivity: https://dx.doi.org/10.1021/ac00098a016

EGTA.  https://en.wikipedia.org/wiki/EGTA_(chemical)


text:
What is the most "important" resonance structure of SCN⁻?
Numerous online references say that $\ce{SCN-}$ has two resonance structures:

I am wondering why this structure is not also possible?

I expect structure 3 to be rare because of the high formal charges, but shouldn't it be included as a possible resonance structure?
Furthermore, there is disagreement about whether structure 1 or structure 2 is more common. I would expect structure 2 to be more common because the negative charge is on the more electronegative N atom. However, this worksheet says that structure 1 is more common. On the other hand, this video says that structure 2 is more common. Which should it be?

What do you mean my "more common"?

First I must note the improper use of the terms common and rare as we are not to answer which structure most frequently occurs. I consider this as due to not carefully chosen words.
We have to predict which of the above sketched limiting structure is the more stable or, precisely, the most important one,e.g. that entering the molecular orbital with higher weight.
The one you proposed is indeed possible and you also know why is not the major contributor, and even not a major one.
Usually, as you said, discerning between structures with a formal charge is done by placing it according to the elements electronegativity.
In our case this rule points to structure 2, with the negative charge on nitrogen.
However, examining the energy of the corresponding bonds we note that 2 is a cumulene, which is not a particularly stable configuration around a carbon atom.
Opposite in 1, a stable C-N triple bond is attained, with the big sulphur atom still capable to spread electron density over itself.
We are therefore facing a case in which is not very easy to answer, and myself I will have to doubt.
As matter of the fact I remember that 1 is indeed the major contributor. In SCN anion the negative charge is about 50 % on sulphur and 30 % on the nitrogen side. But on the values I can be wrong.

I performed a quick calculation on the DF-BP86/def2-SVP level of theory and analysed it with Natural Resonance Theory (from the Natural Bond Orbital Theory).
This results in the following major contributors to this wave function:
$$\left[
\underset{(1)}{\overset{67.49\%}{\ce{^-S-C#N}}} \ce{<->}
\underset{(2)}{\overset{21.25\%}{\ce{S=C=N^-}}}
\right]$$
The third contributor is a weird structure with a 'long-distance-bond' between the sulfur and the nitrogen with $7.22\%$. All other contribution are neglected/discarded.
While your structure 3 is indeed a valid contributor, its actual contribution will be very small. Forcing the program to use it as a structure, it resulted in an error, as it was unable to match the orbitals to that structure. The reason for that is likely that the overlap between the sulfur and the carbon is too poor to actually be considered a good contributor. This would also explain the smaller contribution of the second structure.
Here are the localised (according to NBO) molecular orbitals:

(Colour code: blue/orange - occupied molecular orbital [Lewis];
red/yellow - virtual molecular orbital [unorccupied, non-Lewis])
The total contributions in terms of atomic orbitals to the above:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.98209) LP ( 1) S  1            s( 79.86%)p 0.25( 20.13%)d 0.00(  0.01%)
   9. (1.77474) LP ( 2) S  1            s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
  10. (1.77474) LP ( 3) S  1            s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
  11. (1.96342) LP ( 1) N  3            s( 52.42%)p 0.91( 47.54%)d 0.00(  0.04%)
  12. (1.99743) BD ( 1) S  1- C  2
               ( 44.86%)   0.6698* S  1 s( 20.64%)p 3.81( 78.61%)d 0.04(  0.75%)
               ( 55.14%)   0.7425* C  2 s( 51.41%)p 0.94( 48.47%)d 0.00(  0.12%)
  13. (1.99846) BD ( 1) C  2- N  3
               ( 41.02%)   0.6404* C  2 s( 48.44%)p 1.06( 51.48%)d 0.00(  0.08%)
               ( 58.98%)   0.7680* N  3 s( 48.02%)p 1.08( 51.75%)d 0.00(  0.23%)
  14. (1.99735) BD ( 2) C  2- N  3
               ( 43.98%)   0.6632* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 56.02%)   0.7485* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  15. (1.99735) BD ( 3) C  2- N  3
               ( 43.98%)   0.6632* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 56.02%)   0.7485* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
 ---------------- non-Lewis ----------------------------------------------------
  16. (0.01904) BD*( 1) S  1- C  2
               ( 55.14%)   0.7425* S  1 s( 20.64%)p 3.81( 78.61%)d 0.04(  0.75%)
               ( 44.86%)  -0.6698* C  2 s( 51.41%)p 0.94( 48.47%)d 0.00(  0.12%)
  17. (0.01384) BD*( 1) C  2- N  3
               ( 58.98%)   0.7680* C  2 s( 48.44%)p 1.06( 51.48%)d 0.00(  0.08%)
               ( 41.02%)  -0.6404* N  3 s( 48.02%)p 1.08( 51.75%)d 0.00(  0.23%)
  18. (0.22011) BD*( 2) C  2- N  3
               ( 56.02%)   0.7485* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 43.98%)  -0.6632* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  19. (0.22011) BD*( 3) C  2- N  3
               ( 56.02%)   0.7485* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)

               ( 43.98%)  -0.6632* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)

On the terminology.
Alchimista already explained most of this, however, I cannot stress enough: There is no such thing as a most stable resonance structure. Therefore when you say common, you probably mean large contribution to the wave function, and when you say rare, you probably mean little contribution. None of the resonance structures can be independent from each other, as they are all hypothetical.
Please read more about that here: What is resonance, and are resonance structures real?

One simple way to rationalize this at the student level is that third row elements like S generally form weak pi bonds relative to second row elements like C and N.

@Andrew I think that may be an okay rule of thumb, but ultimately it is too simple, and when they come across thioketones you have to start making exceptions to these rules. There's really no winning with simplicity here.

I don’t see why thioketones are an exception. This rule explains why the enethiol tautomer is usually preferred if one is accessible and why they polymerize so readily

@Andrew I don't wish to explain the reactivity of thioketones, from a bonding perspective they are still best explained with a double bond and that would include a pi bond. I reject your statement as it does not rationalize any of what I have written, but instead condenses it down in a rule of thumb. If anything your statement could somewhat be rationalised through this example, but not the other way around. You're welcome to contribute an answer, but if your statement is the takeaway of my post, then it, my answer, clearly doesn't get the point across and should rather be deleted.

My comment was not intended as a "take-away" from your answer nor was it criticism. You've given a great answer to the specific question. I was adding to it by offering a rule of thumb with which one can intuit the qualitative answer without resorting to advanced computation. With regard to thioketones, I still do not understand why they are problematic. I didn't say S could not form pi bonds, just that they are weak compared to second row elements, a fact which is consistent with the behavior of thioketones and also consistent with your analysis here of SCN-

Off topic but just wanted to say this answer helped me (successfully) challenge the answer key in an Indian Olympiad exam, hence the bounty (after all who can refuse to believe actual computation chemistry calculations :D)

This should be marked the correct answer, OP should update.


text:
Reaction rate constants for methane combustion
I can't find a rate constant or activation energy for the methane combustion reaction given by:
$$\ce{ CH4 + 2O2 -> CO2 + 2H2O}$$
Do all reactions have a rate constant? The NIST web site lists several rate constants for reactions of methane that yield other products, but none for the reaction above. Is there a known rate constant for the reaction given above?

Very good question. The reaction you have is the sum of several intermediate reactions, and many of them have been quantified in terms of their kinetics. This publication is a bit dated but offers a wide overview of the type of problem you have. Additionally, and sentimentally for me, it is co-authored by Bill Gardiner, with whom I spent countless hours in graduate school, as his TA for physical chemistry and as his friend.

This looks pretty good.


text:
Solubility of oxygen in liquid hydrogen; is anything near 1:2 (molar ratio) possible?
In a question on another Stack Exchange site a proposal was forwarded to premix oxygen and hydrogen in the stoichiometric molar ratio of 1:2 to be used as rocket fuel, as opposed to the standard method of storing each liquified gas (LOX, LH2) separately.
According to Wikipedia (here and here) the freezing and boiling points of oxygen (presumably at standard pressure) are around 54 and 90 Kelvin, while those of hydrogen are around 14 and 20 Kelvin, respectively, suggesting that oxygen would freeze well before gaseous hydrogen would condense into a liquid.
However, one might (naively) wonder if oxygen might still exist as a solution in liquid hydrogen at some useful fraction, and at a pressure of roughly one atmosphere or less for practicality of storage on a weight-conscious rocket.
Even the wording of this abstract makes this sound fairly unlikely, but without some confirmation from someone more knowledgable it's hard to address the other linked question.

It sounds like new famous last words. Interest in liquefaction of such mixture isn't particularly reasonable.

@Lex I've edited the question, trying to word it a bit more carefully in hopes that it could be reopened. Still waiting for the edit to be accepted...

@uhoh You took quite some liberty here, but I see no harm in doing that. However, If you, Lex, don't like these edits, please [edit] it yourself.

@Martin-マーチン indeed, but since we're both looking at the same original question elsewhere I'm pretty sure this is the same basic question, or at least close enough so that the answer will be helpful there.  I'd already left a comment for Lex there as well. Thanks!

Looks good to me.

It would be the other way round.
Since the oxygen is the first to liquify, it would be hydrogen dissolved into liquid oxygen.
In this article they show the following calculated vapour liquid equilibrium diagram.

The 2:1 mixture, $x_{\ce{H2}}=0.67$, would exist at around 85 K and 50 bar.

This is great and extremely helpful! Any conclusions or available info that might address "...and at a pressure of roughly one atmosphere or less for practicality of storage on a weight-conscious rocket."

From what I could get from the paper, I think the x-axis in your graph is a mass ratio not molar ratio. This leads to x_H2 = 0.11 which is great since it gives a less extreme temp.


text:
Electrophilic substitution of an indole as opposed to benzofuran and benzothiophene
Regioselectivity in electrophilic aromatic substitution of benzofuran and indole
This response as to the regioselectivity of the electrophilic aromatic substitution of indole left me with a few more questions than it did answers.

In figure 3b, does the pi bond in the bottom half of the benzene on the left get pushed to the nitrogen, allowing for stabilization across the benzene ring in addition to the resonance structures of the nitrogen cation? And if so, and this is the case of a viable nitrogen cation contributing to the stabilization of a valid resonance structure, is it true that all (or most) polycyclic compounds that undergo electrophilic substitution at a location on the nitrogen-containing ring that result in the temporary resonance structure of a nitrogen cation along with stabilization across a benzene ring will be more stable than substitution at a location that would only provide the resonance across the benzene (e.g. without the possible nitrogen cation, had electrophilic substitution occurred elsewhere)?
Furthermore, does this imply that in the electrophilic aromatic substitution pattern of benzothiophene, seeing as how sulfur is less electrophilic than both oxygen and nitrogen, that we would see a similar substitution pattern at the 3-position, as opposed to the substitution at the 2-position that we would see (And that is, to my puny little brain, comprehensible) on benzofuran?

The positively charged nitrogen here isn't electron deficient: it has a double bond to carbon next to it and hence has a complete octet. It is not like, say, $\ce{PhCH2+}$ where the positive carbon has an empty p orbital.

So that lack of electron deficiency - that results in it being more stable than the other resonance structure afforded by substitution at the 2-position? And does electrophilic substitution occur at the 3-position on benzothiophene?

because when electrophile react on 3 position more resonance structures are posible than the 2 position and thus it show stability.Thats why we say that position 3 is more good for indol to attach electrophile.


text:
How are polarizabilities measured experimentally?
Most common spectroscopies that produce either a full spectrum, a tensor, or a scalar value have a specific instrument associated with them that is relatively self-contained and not custom. For example, (linear) IR spectra come from benchtop FT-IR spectrometers, NMR chemical shifts use a FT-NMR spectrometer (which is capable of many experiments), and so on. One spectroscopic property that serves as a benchmark for electronic structure theory and is important for designing optical materials is the (electric dipole) polarizability, which "is defined as the ratio of the induced dipole moment $\mathbf{p}$ of (a molecule) to the electric field $\mathbf{E}$ that produces this dipole moment":
$$
\mathbf{p} = \alpha \cdot \mathbf{E}.
$$
Physically, this corresponds to how easy it is to shift a system's charge density or electron cloud around. More explicitly, it is a symmetric rank-2 tensor:
$$
  \begin{bmatrix}
    p_{1} \\ p_{2} \\ p_{3}
  \end{bmatrix}
  =
  \begin{bmatrix}
    \alpha_{11} & \alpha_{12} & \alpha_{13} \\
    \alpha_{21} & \alpha_{22} & \alpha_{23} \\
    \alpha_{31} & \alpha_{32} & \alpha_{33}
  \end{bmatrix}
  \begin{bmatrix}
    E_{1} \\ E_{2} \\ E_{3}
  \end{bmatrix},
$$
as the applied electric field may have 3 independent Cartesian coordinates, and the system being measured may have some anisotropic response to this field. In some cases, calculating the tensor may be time-consuming, but formulation is well-understood and there are standardized methods for doing so. However, it is not obvious to me how it is measured experimentally. There is the Lorentz-Lorenz equation (rearranged),
$$
\alpha = \frac{n^2 - 1}{n^2 + 2} \frac{3}{4\pi N},
$$
which relates the refractive index $n$ of a system or medium to its polarizability. There is also the Clausius-Mossoti relation, which relates the dielectric constant $\epsilon$ to the polarizability, as the refractive index and (complex) dielectric constant are related:
$$
n \propto \sqrt{\epsilon}.
$$
This implies to me that if you can measure the refractive index with a refractometer, then that is the preferred way of measuring the polarizability.

Is the refractive index the most common method of experimentally measuring the polarizability? Can it be measured directly, or must be through a relationship such as those above?
Does this relationship hold for oriented (anisotropic) polarizabilities, or only averaged (isotropic) ones?
The refractive index and the polarizability are frequency-dependent properties. How is the static (frequency-independent) polarizability measured? Or, is it even possible without extrapolation from multiple frequency-dependent data points?

Good question. There are several ways depending on the state of the molecule. For  gas phase molecules Rayleigh scattering is used. Raman scattering also works. In these cases polarization of light is also important.  A more detailed exposition on this will take time for me to compile.

Here is a starting point: https://pubs.acs.org/doi/full/10.1021/jp9845322

@GeoffHutchison It's a bit far from my comfort zone but I gave it a shot.

Related: https://chemistry.stackexchange.com/q/51292

The polarizability of molecules determines how a molecule interacts with other molecules (intermolecular interactions and chemical reactions), how it interacts with electromagnetic radiation (optical properties), and how it interacts with electrical fields (electrical properties). Thus, there are multiple ways to compare the polarizability of two molecules or samples, and to quantify it.

[OP] Is the refractive index the most common method of experimentally measuring the polarizability?

That is hard to say without a systematic search of all the polarizability data. A starting point for such a search might be the polarizability data provided by NIST, which includes references that describe methods used. In general, there are electrical techniques (static electrical field or slowly oscillating) and optical techniques (electromagnetic field). Light scattering depends on the electric polarizability of the medium, so any technique that is based on light scattering (Raman spectroscopy, X-ray diffraction, refraction) in principle could be used to determine polarizability.

[OP] Can it be measured directly, or must be through a relationship such as those above?

The refractive index can be measured directly. The polarizability, while it has direct impact on molecular properties like the strength of dispersion forces, is determined from other measurements.

[OP] Does this relationship hold for oriented (anisotropic) polarizabilities, or only averaged (isotropic) ones?

As written in the question, the Lorentz-Lorenz equation is for the isotropic case. The literature linked to below also shows the more general equation.

[OP] The refractive index and the polarizability are frequency-dependent properties. How is the static (frequency-independent) polarizability measured? Or, is it even possible without extrapolation from multiple frequency-dependent data points?

The refractive index is frequency-dependent while the polarizability is not. You measure the refractive index at multiple wavelengths and extrapolate to infinitive wavelength. Alternatively, you can do a non-linear fit to the respective theoretical dispersion function.
Measurements of molecular polarizability are done at low concentrations in the gas phase, and yield the isotropic polarizability. Here is one paper that compares the experimental data to computational results for the iodine molecule. More complicated experiments involving polarized light are required to figure out the anisotropy, if any. For example, this paper from the same two labs compares the anisotropy of the polarizability of butadiene to computational results. An example of a quantumchemical computation of polarizability is here, for the hydrogen atom. In the solid phase with oriented molecules, the anisotropy of polarization is easier to measure (for example with cross-polarized filters as in the example below showing a plastic lens under mechanical stress):

Source: https://commons.wikimedia.org/wiki/File:Stress_analysis_of_plastic_lens.png


text:
Does this C17H28 isomer exist?
Can anyone tell me what this $\ce{C17H28}$ molecule is? Has it been synthesized?  Does it have a name?
I was just interested in the molecule because it looks interesting.

This compound (formally called "hexadecahydronaphtho[1,8-de]naphthalene", but in common parlance also "[6.6.6.6]fenestrane" - thanks Mithoron!) has never been actually synthesised before, but some of its properties have been studied computationally.
It does not particularly matter, but it is worth pointing out that this compound possesses stereochemistry which was not indicated in your original drawing. Hence your question is essentially asking about a collection of stereoisomers.

Gund and Gund1 investigated the strain energies of a few $\ce{C17H28}$ isomers, shown above, using empirical force field methods. The compound you are interested in is 4b in the scheme (the various 4a are constitutional - as opposed to configurational isomers - of 4b). The numbers next to the names are the strain energies in kcal/mol. The higher the energy, the more strained; hence it seems that 4b, if synthesised, would be generally thermodynamically unstable with respect to 4a (kinetic stability is of course entirely separate). The letters 'c' (chair), 't' (twist boat), and 'b' (boat) refer to the conformations of each six-membered ring in the structure. The labels cis and trans refer to the relationship between each neighbouring pair of fused six-membered rings.
The potential for systems containing fused six-membered rings to undergo "ring flips" was also investigated computationally (using DFT) in a recent paper by Alder et al.2 In particular, the authors were interested in the all-cis isomer of your compound, i.e. ccct/4b6 in the screenshot above - which incidentally has a rare $S_4$ symmetry. They found that inversion has a whopping barrier of +88.9 kJ/mol (B3LYP-gCP-D3/6-31G*//B3LYP/6-
31G*) - almost double that of cyclohexane (~45 kJ/mol). The authors conclude that this isomer of [6.6.6.6]fenestrane "probably has the highest inversion barrier for any nonpolymeric multi-ring structure built from cyclohexane rings".
Although the compound has never been made, the authors propose that at high temperature the inversion process (which interconverts equatorial and axial hydrogens) may be studied by NMR spectroscopy. However, due to the significantly slower rate of inversion relative to cyclohexane, temperatures of 100–200 °C would be needed (as compared to cyclohexane which requires low temperatures for NMR studies).

References

Gund, P.; Gund, T. M. How many rings can share a quaternary atom? J. Am. Chem. Soc. 1981, 103 (15), 4458–4465. DOI: 10.1021/ja00405a028.
Alder, R. W.; Butts, C. P.; Sessions, R. B. Perhydrohelicenes and other diamond-lattice based hydrocarbons: the choreography of inversion. Chem. Sci. 2017, 8 (9), 6389–6399. DOI: 10.1039/C7SC01759F.