How to tell between carbon powder or graphite powder?

Short answer: You can't reliably tell the "difference" (by any simple means)

Longer version:

First of all, the graphite you were taught about at school is a pedagogical construct. Any macroscopic sample of "real" graphite would contain both the layered honeycomb (crystalline) bits as well as amorphous bits of carbon (in addition to God-knows-what bunch of other substances that managed to wiggle their way into your sample).

What about charcoal then? It's amorphous carbon with significant quantities of several kinds of organic compounds (you can't expect them all to decompose to carbon...), depending on the source.

Trying to class your sample (which clearly has other stuff in it... I can see white/gray specks on it in the first picture ;P) as "graphite" or "charcoal" is an effort in futility; neither "real" graphite nor charcoal have a sharply defined constitution.

Simple tests of the physical nature of the powder would not be able to accurately distinguish it from being "mostly graphite" or "mostly charcoal", mainly, because it's a powder. Had you had the sample before it was pulverized you could (easily) tell the difference; the qualities that most people attribute to graphite (luster, slippery-ness) aren't easily discernible when you turn it into a powder and then compare it with powdered charcoal.

About the tests you conducted:

Your attempt to subject the sample to various (physical) tests is laudable (more users ought to do that before posting "Identify this thingy"-type questions here... it may give us something to work with). However those tests, apart from sorely lacking in some sort of mechanism/procedure to cut down on sources of error, don't really provide any meaningful data that you or me or anyone could use to determine what your sample is (largely) composed of.

Let's break down what you did

  Adding it to water, it sinks to the bottom and leaves a slight oily looking film on the top. Upon stirring, it darkens the water and then gradually sinks back to the bottom after some time.

The film isn't really a giveaway. Even particles denser than water are known to "float" in these circumstances... well, they aren't technically "floating", it's the surface tension of the water that's holding it up. This is easily disrupted by stirring/agitation, and as you've noticed, it'll sink.

Graphites typically have have a much higher density (slight variation depending on purity and crystal defects) than water, so pieces of "graphite" would sink.

Blocks of charcoal typically have lower densities (varies depending on type/source of charcoal) than water and hence floats, but that changes when you pulverize them (and destroy most of the air cavities in the process).

Conclusion: No useful data.

  I tried writing some thick lines on a piece of paper with a graphite pencil and testing for conductivity (high resistance).
  Same high resistance with the powder.

That was a lot of effort... for nothing (you have my deepest sympathies).

Okay, I think I see your reasoning here:

"Graphite's pretty conductive (low-ish resistance) and charcoal is a poor conductor (very high resistance), so doesn't this test tell me something?"

Nope.

Thing about pure graphite (of the "Holy cow! A non-metal that conduct electricity!!!" fame) is that it only conducts (really well) along a particular plane; try another plane perpendicular to this and you'll find it to be a thousand times less conductive ( see answer to this question).

Even if your powder is (largely) graphite, surely the individual powder particles aren't going to align themselves into an extended plane between your terminals. They'll be oriented randomly, so expect the high resistance.

You first drew a line on a piece of paper using a pencil and measured its resistance to create a baseline to work up from. Smart. Just two things though:

1) Pencil lead isn't graphite (neither is it "lead" for that matter ;P); it's a mixture of graphite and clay. So unless you know for sure that the impurities in your sample are clay, this doesn't really serve as a good baseline. Besides, the graphite particles aren't all aligned in that plane I was talking about.
2) Your makeshift resistor ("thick line on a piece of paper") is a poor standard for another reason: dimensions. Your line has no uniform dimensions (length, breadth or even depth). If you need resistance from resistivity of a given material, or the other way around, you need the sample's dimensions. Neither your line nor your powder have fixed dimensions in this test.

Trying to figure out what your sample is by comparing its resistivity with that of another substance is a really neat idea, so don't get me wrong; the issue here is (1) the design of the experiment, (2) of the two substances you're trying to compare (graphite and charcoal), the latter's electrical properties are altered because it's been powdered and (3) your "control-test/sample" (the line-resistor) is neither graphite nor charcoal, making it a rather pointless "control sample".

Conclusion: No useful data.

  Rubbing it into paper it has a much blacker colour than the graphite (which to me looks like a really dark gray). However, I don't know if that is a reliable indicator.

I don't need to tell you how poor a test this is (you already seem to doubt the validity of the test yourself)

Conslusion: No useful data.

  EDIT: Ok, I've made a paste using Isopropyl Alcohol and the carbon powder and made a small "stick" or rod of the stuff. I have now been able to achieve solid continuity at around 2 centimeters in distance and the resistance dropped from the high 40 - 50 kOhms for the loose powder, to the low 10s (ie 40 Ohms). If this helps anyone.

Same argument as before: This test is pointless because your sample doesn't have fixed, uniform dimensions. Nor do the particles (if they're graphite; doesn't matter if it's charcoal) align themselves in extended planes.

As for the drop in resistance, I reckon it's because the sample isn't completely dry. The "isopropanol" you get at the drugstore is at least one-quarter water (not pure isopropanol). After you mixed it with your sample, a ton of electrolytes would've dissolved in the water (pretty sure your sample has loads of stuff that isn't carbon/ carbon-based). When you measured the resistance across this stick of yours, all that water hadn't evaporated by then. This is probably the (main) reason for the drop.

Conclusion: No useful data.

Interesting question by the way, you have my upvote ;)


I would appreciate learning whether such reasoning, for these two
  type of substituted phenols is plausible or whether furthermore
  solid arguments should be invoked.

Your reasoning seems generally well thought out.  You clearly have a good understanding of inductive effects and you mention resonance effects in the cresol series.  Structural effects don't play a role with these compounds, but let me mention them for completeness.  If we have a compound such as I with a carbonyl ortho to the hydroxyl substituent, then in addition to inductive (stabilizes the anion) and resonance (stabilizes the anion) effects, we should also consider hydrogen bonding; which, in this case, would stabilize the phenol making it less acidic.  On the other hand, in compound 2, the bulky t-butyl group would force the ester carbonyl to twist out of the plane of the benzene ring.  In II, the inductive effect of the carbonyl would still operate, but the resonance and hydrogen bonding effects on acidity would be diminished.

  I would appreciate learning why the difference in acidity between
  2- and 4-chlorophenols is much more pronounced than the same
  difference between o- and p-cresols

This link discusses electrophilic aromatic substitution, where again both inductive and resonance effects need to be considered.  About two-thirds of the way down the page we see that chlorobenzene undergoes electrophilic nitration about 30 times slower than benzene, but is still ortho-para directing.  We also see that toluene reacts about 25 times faster than benzene and is also ortho-para directing.  From these and other examples, we learn that chlorine donates electrons weakly by resonance and withdraws electrons strongly through its inductive effect; and that a methyl group is a mild electron donor through resonance (hyperconjugation), while its inductive effect is also electron-donating, but is small.

Let's apply these concepts from electrophilic aromatic substitution to your two phenol series.  In the chlorophenols, we would expect the chloro substituent to influence the series primarily through a strong electron-withdrawing inductive effect.  Indeed, as the chlorine moves closer to the hydroxyl substituent, the inductive electron withdrawal increases in power and the $\mathrm pK_\mathrm a$ increases dramatically.  On the other hand, in the methyl series we expect mild resonance (hyperconjugation) effects to dominate (even though the inductive effect is in the same direction), therefore in the ortho- and para-cresols we expect the anion to be weakly destabilized and consequently a bit less acidic than phenol itself; while the meta isomer, with minimal resonance and inductive effects affecting the anion, should have an acidity close to phenol.


Is there any substance that's a 0-0-0 on the NFPA diamond?

Water!

You can find plenty of others if you perform this search on Wikipedia (but then, go check the MSDS for the individual compounds before doing anything serious with them).


Does lead(II) chloride dissociate into Pb²⁺ and Cl⁻ ions in HCl?

Your analysis about $\ce{PbCl2}$ in four points is correct.
Then ($\ce{A}$ and ($\ce{B}$) are correct, if they are in aqueous solution. The following equation shows that $\ce{SnCl2}$ is a reducing agent that is easily oxidized by air : $$\ce{6 SnCl2 + O2 + H2O  -> SnCl4 + 4 Sn(OH)Cl}$$ and, in ($\ce{B}$),  $\ce{SnO2}$ is soluble in $\ce{KOH}$ solutions according to $$\ce{SnO2 + 2 KOH + 2 H2O -> K2[Sn(OH)6]}$$ (D) is wrong, because the reaction is the following $$\ce{Pb3O4 + 4 HNO3 -> PbO2 + 2 Pb(NO3)2 + 2 H2O}$$ This reaction can be deduced by dealing with $\ce{Pb3O4}$ as if it was a mixture of $\ce{PbO2 + 2 PbO}$ so that only the two $\ce{PbO}$ react with $\ce{HNO3}$, without changing its oxidation number $$\ce{2 PbO + 4 HNO3 -> 2 Pb(NO3)2 + 2 H2O}$$


How many alternative chemical ways to transport oxygen in living creatures are known?

Yes! A variety of metalloproteins are used for oxygen transport. These include proteins that bind to porphyrins with different central metal atoms, as well as proteins that bind the metal atoms directly, without a porphyrin ligand. Fun fact- hemocyanin, which uses two copper atoms as prosthetic groups to bind oxygen, is responsible for the blue color of horseshoe crab blood!
Here are some relevant links:
https://en.wikipedia.org/wiki/Respiratory_pigment
https://en.wikipedia.org/wiki/Hemocyanin
https://en.wikipedia.org/wiki/Hemoglobin (scroll to "other oxygen-binding proteins" near the bottom)


What stops intermediates of alcohol hydration and dehydration reacting in the same way?

As you have spotted, the reaction is reversible and can proceed both ways. Which way it takes depends on the conditions and by implication on Le Chatelier’s principle.

In concentrated $\ce{H2SO4}$, any water will be immediately ‘captured’ by the strongly hygroscopic sulphuric acid. Thus, water is actively drawn out of the equilibrium and the alkene is produced by elimination.

If the reaction is run in aquaeous solution (or at least with water as a cosolvent) and only catalytic quantities of acid, then the nucleophilic attack of water is more likely and the alcohol hydration product is produced.

(In modern synthetic lab-scale chemistry, neither of the reactions are used due to their rather poor selectivity (especially hydration) and harsh conditions (especially elimination). Instead, hydration is typically performed with boranes (anti-Markovnikov) or Markovnikov-selective reagents while elimination is typically achieved by transformation of the alcohol functionality into better leaving groups such as mesylate.)


Intensive and extensive properties of gases

To give you a more clear explanation of the definition of these properties, it's more easy to remember them this way :

-Extensive properties are those which *are additive * ( mass, volume ) for example 1 L + 1 L = 2 L. On the other hand, intensive properties are not additive ( temperature, pressure, concentration ) for example 2 mol/L + 3 mol/ L does not give you a 5 mol/L solution. These are properties which don't change at any circumstance.

Some other thing to be considered :
The ratio of two extensive properties, gives us an intensive property. For example density= mass/volume. Since it is a ratio it will always be the same regardless of how much of a sample is measured.


What happens to magnesium in water with acid?

In water Mg oxidizes and the half-reactions are:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
$$\ce{2H2O + 2e- → H2 + 2OH-}$$

As the E($\ce{Mg}$/$\ce{Mg^{2+}}$)>E($\ce{H2O}$/$\ce{H2}$), Mg oxidizes and water reduces.

Metal ions can change their speciation depending on the pH. In acidic conditions $\ce{Mg^{2+}}$ dissolves and associates with malate (which in solution is deprotonated) and end up precipitating because the malate is in high concentration and the solution is saturated. And, in neutral o basic conditions they precipitate as metal hydroxides which are quite insoluble.

What you have to take into account is that metal ions can precipitate in basic conditions and dissolve in acidic ones.
Also, water has a reduction potential and has to be taken into account in rusting processes with metals. You can look for more information with Pourbaix diagrams where you can see the changes in speciation of metals in front of pH and the reduction potential of several metals.

Here's the Pourbaix diagram of Mg:

I hope this helps!


Pros and cons of Cartesian vs. Z-matrix representations of molecules?

Cartesian Space

In Cartesian space, three variables (XYZ) are used to describe the position of a point in space, typically an atomic nucleus or a basis function.  To describe the locations of two atomic nuclei, a total of 6 variables must be written down and kept track of.  The general ruling is that for Cartesian space, 3N variables must be accounted for (where N is the number of points in space you wish to index).

Internal Coordinates

Z-matrices use a different approach.  When dealing with Z-matrices, we keep track of the relative positions of points in space.  Cartesian space is 'absolute' so to speak.  A point located at (0,0,1) is an absolute location for a coordinate space that extends to infinity.  However, consider a two atom system.  The translation of the molecule through space (assuming a vacuum) will have no affect on the properties of the molecule.  An H2 molecule centered around the origin (0,0,0) is no different from the same H2 molecule being centered around (1,1,1).  However, say we increase the distance between the hydrogen atoms.  We now have altered the molecule in such a way that the properties of that molecule has changed.  What did we change?  We simply changed the bond length, one variable.  We increased the distance between the two atoms by some length R.  With Z-matrices, we keep tabs on internal coordinates: bond length (R), bond angle (A), and torsional/dihedral angle (T/D).  Using internal coordinates reduces our 3N requirement set by the Cartesian space down to a 3N-6 requirement (for non-linear molecules).  For linear molecules we keep tabs on 3N-5 coordinates.  When performing complex computations, the less you have to keep track of, the less expensive the computation.

Symmetry

Consider the following molecule, H2O.  We know from experience that this molecule has C2V symmetry. The OH bond lengths should be equivalent.  When using some sort of optimizing routine, you may want to specify symmetry in your system.  With a Z-matrix, the process is very straightforward.  You would construct your Z-matrix to define the OH(1) bond as being equivalent to the OH(2) bond.  Whatever program you use should automatically recognize the constraint and will optimize your molecule accordingly giving you an answer based off a structure that is constrained to C2v symmetry.  With Cartesian space this is not guaranteed.  Rounding errors can cause your program to break symmetry, or your program may not be very good at guessing the point group of your molecule based on the Cartesian coordinates alone.

Picking the Right One

As a preface, programs like Gaussian convert your Cartesian coordinate space (or your pre-defined Z-matrix) into redundant internal coordinates before proceeding with an optimization routine unless you specify it to stick with Cartesians or your Z-matrix.  I warn you that specifying your program to optimize using Cartesian coordinates makes your calculation much more expensive.  I find that I will explicitly specify 'Z-matrix' when I know I'm dealing with high symmetry and when I know my Z-matrix is perfect.

You will want to use Z-matrices on systems that are rather small.  If dealing with systems with high symmetry, Z-matrices are almost essential.  They can be rather tricky to implement and you will likely spend some time figuring out the proper form of your Z-matrix through trial-and-error.  If you wish to scan a particular coordinate, Z-matrices are also very helpful as you can tell a program to scan across a bond length, angle or torsion with ease (as long as you've properly defined that coordinate in your Z-matrix).

I use Cartesian coordinates for large systems, systems with very little or no symmetry, or when I'm in a hurry.


Can nonpolar molecules exhibit dipole-dipole forces?

Short answer: there are many electrostatic interactions between two non-polar molecules.

Beyond monopole (full charges) and permanent dipole moments (polar molecules), there is a full multipole expansion for the electrostatic potential around any molecule. (This is technically true for atoms and ions too, but higher-order terms are really only useful for molecules.)

So there are electrostatic potential energy interaction terms for charge-dipole, dipole-dipole, dipole-quadrupole, quadrupole-quadrupole, etc.

These terms are important - the quadrupole-quadrupole interactions dictate the orientation of the benzene dimer and $\ce{CO2}$ dimer in your example.1

The problem is that most of these interactions die off very quickly. The quadrupole-quadrupole term is:1

$$E(r)=\frac{-\Theta_1\Theta_2}{4\pi\epsilon_0r^5}\times\Gamma(\theta_1,\theta_2,\phi)$$

So roughly $1/r^5$, compared to $1/r^3$ for dipole-dipole interactions, or $1/r^6$ for dispersion forces like induced-dipoles.

When such molecules are close, the quadrupole moments (and other multipole electrostatic terms) can dictate packing and distances, but are not as strong or as long-range as dipole-dipole or charge interactions.


First of all, we cannot achieve temperatures of 0 kelvin in experiments, that is forbidden by quantum thermodynamics.

Currently, experiments can go down to seriously low temperatures of nanokelvins (source: a friend is working in that field, but spontaneously the only reference I can give  you is Wikipedia.)

At those temperatures the phases of 'solid' or 'liquid' don't really exist anymore. Rather collective quantum effects play a role, that make new phases appear, namely Bose-Einstein condensates and Fermi gases. What phase appears depends on the particles spin, which allows or disallows them to occupy the lowermost energy level.

Those exotic phases of matter count however still as 'condensed matter', which all elements at cold temperatures will have. The classical solid phase, which you're asking about, can be circumvented this way, but not at all pressures, as others correctly pointed out.


Does nitrogen inversion affect the basicity of amines?

Background

Amine basicity correlates with, among other things, the hybridization of the nitrogen orbital that is holding the lone pair of electrons.  The less s-character in this orbital, the more basic the amine. So, as the following figure indicates, a trialkylamine (the lone pair is in an $\ce{sp^3}$ orbital) is more basic (less acidic or higher $pK_{a}$) than the lone pair in pyridine (the lone pair is in an $\ce{sp^2}$ orbital), which is less acidic then the lone pair in a nitrile (the lone pair is in an $\ce{sp}$ orbital).

(image source)

The rate of nitrogen inversion also correlates with, among other things, hybridzation. But in cases where these "other things", such as ring constraints, limit nitrogen inversion, then the rate of nitrogen inversion may not correlate with hybridization. So the rate of nitrogen inversion does not always correlate with hybridization and is therefore not a good indicator of lone pair "availability".

Answer

Looking at your two compounds, a reasonable first guess might be that the nitrogens in these compounds are both roughly $\ce{sp^3}$ hybridized.  Therefore, even though one compound can undergo nitrogen inversion and the other can't, their hybridzations are similar and we would expect similar basicities.

We could refine this prediction by noting that in the bicyclic compound, due to geometric constraints, the carbon-nitrogen bonds around nitrogen are slightly pulled back a bit from the tetrahedral angle.  This would increase the p-character in these bonds (as a bond angle moves from 109° to 90° the p-character increases with 90° being pure p).  As a result, the p-character in the lone pair orbital decreases and the s-character increases.  As mentioned above, more s-character in the lone pair orbital leads to lower basicity.

Therefore we might expect both compounds to have similar basicity with the bicyclic compound perhaps being a bit less basic.

This analysis is corroborated by the $\mathrm pK_\mathrm a$ values of the conjugate acids (a more positive number means that the amine is more basic). According to J. Org. Chem. 1987, 52 (10), 2091–2094, quinuclidine has a $\mathrm pK_\mathrm a$ of 10.90, and 1-azabicyclo[2.2.1]heptane a $\mathrm pK_\mathrm a$ of 10.53. The decrease in basicity can be attributed to the increased strain and greater s-character of the lone pair. Triethylamine itself has a $\mathrm pK_\mathrm a$ of 10.75 (Evans table), making it marginally more basic than 1-azabicyclo[2.2.1]heptane.


Why are square planar molecules drawn from the side plane as opposed to the top plane?

You can do that. For example, a Google Image search for "cisplatin structure" suggests that although the hashed + wedged depiction is more common, the four plain bonds depiction is also in regular use.
For more detail on what's "acceptable" or not, refer to: Graphical representation of stereochemical configuration (IUPAC Recommendations 2006). Pure Appl. Chem. 2006, 78 (10), 1897–1970, DOI: 10.1351/pac200678101897. They recommend the hashed + wedged depiction, but do not prohibit four plain bonds (section ST-2.6, pp 1942–3). There is some rationale for this, which may go some way towards explaining why the hashed/wedged depiction is more common: it explicitly differentiates the square planar structure from other tetrahedral centres in organic structures.

The first recommended depiction should be drawn with two solid wedged bonds and two hashed wedged bonds. The two solid wedged bonds should be positioned below the central atom, while the two hashed wedged bonds should be positioned above. [...] It is also acceptable to draw similar structures with plain bonds. There are some cases where this drawing style would even be preferred, particularly when all four bonds are required to be in the plane of the paper by virtue of other aspects of the structural diagram [...] The similar depiction with the bonds themselves positioned along the axes is acceptable but should be avoided as it is extremely commonly used to represent tetrahedral centers, either if the tetrahedral configuration is nonstereogenic (carbon tetrachloride) or in Fischer projections.


What is Y-aromaticity? Is the trinitromethanide anion aromatic?

Introduction

It is fairly obvious that the statement given by Wikipedia cannot be entirely correct.

  There is some evidence that the anion (which obeys the 4n+2 Hückel rule) is aromatic.

The trinitromethane anion cannot obey the Hückel (4n+2) rule because it is neither monocyclic nor planar. And assigning a number of π electrons is fuzzy at best, so the (4n+2) part is also questionable. Hückel's rule is a very strict set, but an aromatic compound does not necessarily have to fulfil it. The easiest compound to understand this is naphtalene, which is clearly aromatic, but does not fulfil the criteria of Hückel's rule, i.e. it is not monocyclic. In extension to this, helicene is an aromatic compound that fails two of the given conditions. Another case is thiophene, which due to the included sulfur certainly does not entirely consist of trigonally hybridised (sp2) atoms (see Bent's rule).[1]

But can the anion still be regarded as aromatic? In order to explain that, we need to look at the concept of Y-aromaticity.

What is Y-aromaticity?

The term and the concept is not without controversial discussion. It is an attempt to give the "traditional" concept of resonance energy an new, fancy and more catchy name. It is often used to explain the extraordinary stability of the guanidinium cation, which has tree equivalent resonance structures, i.e. the electronic ground state is heavily delocalised.
Conventional Lewis structures employ two electron two centre bonds only, which therefore cannot represent the highly delocalised electronic structures, e.g. of aromatic compounds. Hence the concept of resonance had to be introduced to account for such stabilising effects. Obviously the same holds true for any compounds, that exhibit excessively delocalised electronic structures.
The term Y-aromaticity has been used for compounds, that are Y-shaped, with a trigonal planar coordinated (carbon) centre. In addition to that the π orbitals hold six electrons.[2] These compounds have a similar orbital picture and could therefore be grouped.

The electronic structure of Y-delocalised π systems

The π orbitals of a trigonal planar coordinated centre can be generalised by the following orbital picture, which was generated for the nitrate anion at the DF-BP86/def2-SVP level of theory.

What we can clearly see is, that there is one totally symmetric bonding orbital. Two degenerate (orthogonal) slightly bonding (or non-bonding) orbitals and another totally symmetric anti-bonding orbital.

The remarkable stabilisation comes from the lowest lying π orbital, which is delocalised over all atoms, just like in the more obvious cases like benzene. Additionally, such compounds often retain their planarity by undergoing substitution reactions, rather than addition reactions.

Pro and contra of Y-aromaticity

The concept itself allows the discussion of certain electronic properties in quite simple terms. It is related enough to the general concept of aromaticity so that it can easily be understood at the same level of teaching. Like aromaticity, it allows predictions - to a certain degree - about the reactivity of such compounds.

Peter Grund makes the following assessment, arguing in favour of the concept of acyclic aromaticity.[3]

  Despite recent trends, it would appear useful to designate systems as "aromatic" if they represent a delocalized system exhibiting peculiar chemical stability, in order to focus attention on such systems even when they are buried in complex structures. Since guanidine and its derivatives appear to possess really exceptional stability, and since their physical and chemical properties appear to be dominated by the tendency to retain the closed-shell, Y-delocalized 6 pi-electron configuration, we suggest that it is profitable to consider such substances as possessing a special type of "aromatic character," despite their acyclic nature.

For a general understanding of this feature it certainly does not hurt to keep it in the back of ones head.[4]

However, the applicability of this concept is limited to a quite small subset of compound (the same holds true for Hückel's rule, too). It therefore seems to be quite a superfluous concept, since it is in a more general sense already covered by resonance stabilisation.
After all, electronic structures are often not well enough described with simple models and in some cases these simple models can lead to wrong assumptions. This is one of these cases, where one can only use this concept as long as its restrictions are well known.

Some people might argue, that the naming itself is the problem and it may lead to confusion. I can partly agree with that. That does not change the fact, that the description in resonance, valence bond and molecular orbital theory terms will produce this result. In general I think the term aromaticity is unfortunate right in the beginnings. The concept has changed and extended quite a few times already. I often see problems when the term "Hückel rule" is used synonymously with aromaticity.

There is no true theoretical and at the same time all general explanation for aromaticity. It is often determined on a case by case basis experimentally. This limitation also has to apply to any derivations of it.

Electronic structure of the trinitromethane anion (TNM⊖)

Trinitromethane is a quite acidic substance, which can in part be attributed to the presence of the well delocalised π system. Cioslowski et. al. have investigated the anion and came to this conclusion. In their paper they also use the term Y-aromaticity.[5]

  Our calculations, however, show that at the HF/6-31G* level, the carbon atom is coplanar with the nitrogens while the oxygens of the nitro groups are rotated out of plane. The result is a structure with $D_3$ symmetry. The $\ce{C(CN)3^-}$  anion is likewise found to be planar, while the $\ce{CF3^-}$ ion exhibits a
  pyramidal geometry. In addition, we observe in $\ce{C(CN)3^-}$ a shortening of the carbon-carbon bond and a lengthening of the carbon-nitrogen bond relative to $\ce{CH(CN)3}$. We also find an analogous shortening of the carbon-nitrogen bond and lengthening of the nitrogen-oxygen bond in $\ce{C(NO2)3^-}$ relative to $\ce{CH(NO2)3}$. In contrast, the $\ce{C-F}$ bond length increases upon going from $\ce{CHF3}$ to $\ce{CF3^-}$. The planarity of the $\ce{C(NO2)3^-}$ and $\ce{C(CN)3^-}$ ions and the changes in bond lengths are obvious consequences of Y-aromaticity in these systems.

Obviously the bonding situation in the trinitromethane anion is a little more complicated than in guanidine, as the nitro group itself cannot be described by a single Lewis structure. The out of plane rotation of these groups further complicate things, but since the perpendicular $C_2$ axis is retained it is still somewhat possible to compare the resulting orbitals to a planar species. Therefore when you look at the orbitals you still find a remarkable resemblance to the picture above. For this purpose I have included the same schematic drawings next to the orbitals that resemble these best.

It is noteworthy, that because of the propeller structure there is no strict separation between σ and π orbitals, since the latter require a horizontal mirror plane, which is absent here.
But that is not all. Try counting the number of π electrons. This is far from trivial and it is in no case that makes sense six.[6] Don't let your brain explode about that fact, we are clearly dealing with a borderline case at this point.

And we have not even started to talk about the possibility of bonds between the oxygens of the nitro group.[7]

Take home message for TNM⊖

The trinitromethane anion has a strongly delocalised electronic system. It is extraordinarily stable due to this. If you accept the concept of Y-aromaticity as valid, then it is still a borderline case, where additional considerations have to be taken into account.

Notes, References and further reading

Bent's Rule

Bent, H. A. Chem. Rev. 1961, 61 (3), 275–311. DOI: 10.1021/cr60211a005.
What is Bent's rule?
Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

The six π electrons are probably the reason for the statement in the Wikipedia article.
Gund, P. J. Chem. Educ. 1972, 49 (2), 100. DOI: 10.1021/ed049p100.
I have used this concept already various times to explain certain electronic structures. From a personal point of view I find it quite helpful.

Delocalization of pi electrons in nitrate ion
Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

Cioslowski, J.; Mixon, S. T.; Fleischmann, E. D. J. Am. Chem. Soc. 1991, 113 (13), 4751–4755. DOI: 10.1021/ja00013a007.
There are several possibilities, that can be considered to count electrons towards the number of π electrons.

The probably easiest way would be to flatten the molecule for this thought experiment, then count electrons perpendicular to the plane. This gives you two electrons for the central carbon (sp2), no electrons from the nitrogens (sp2) and two electrons each from the oxygens (sp). That gives you a total of 14 π electrons.
Maybe as simple as this, but somewhat the easy way out would be to disregard all contributions from the nitro groups, leaving only 2 π electrons.
Regard the nitro groups as π-backdonor. Since only one of the intrinsic π orbitals can effectively overlap with the central carbon, each of them would donate two electrons (maybe less, because backdonation). Plus two from the central carbon, makes up to eight electrons. This certainly is a very fuzzy approach.
Count four electrons per oxygen, which is kind of cheeky, but it could work. This would give 26 π electrons.
Write a Lewis structure and count only the double bonds (× 2), which gives a total of eight electrons.
I am almost certain, there are still more ways to count the electrons.

In [5] they analysed the anion with the quantum theory of atoms in molecules. They found bond critical points between the oxygen atoms. How these can be explained and what they mean is certainly a story for another time.


How to Determine the Rate Law knowing time and concentrations

You could plot the data to determine the order of the reaction.  I would use Microsoft Excel to do this.  Reproduce the data and then add columns that calculate the $\ln [A]$ and $\frac{1}{[A]}$.  Then plot $[A]$, $\ln [A]$ ,and $\frac{1}{[A]}$ vs. time (I would use separate plots).  The plot that gives the "best" linear fit (which you can assess using the $R^2$ value), corresponds to the appropriate integrated rate law.

If $[A]$ vs. time is linear, then the reaction is zero order.
If $\ln [A]$ vs. time is linear, then the reaction is first order.
If $\frac{1}{[A]}$ vs. time is linear, then the reaction is second
order.

Hope this helps.


Is there a flame which doesn't produce smoke?

Yes there are flames that don't produce smoke, but the reasons are complicated and depend on the conditions of the flame as well as the key fuel creating the flame.

The obvious example of a smokeless flame is hydrogen. Pure hydrogen burns in oxygen to generate water and there is no way the normal things that make up smoke can be created (the smoke in the Hindenburg disaster was the gas-bag covers and the content of the zeppelin so isn't a counter example).

Smoke is, technically, a suspension of particulate matter in the atmosphere. Some things look like smoke but are not (like the steam from a kettle or the steam from a power station cooler tower: these consist of liquid droplets that will evaporate).

The main reason smoke is created is because the reaction driving the flame is incomplete and can produce small particles of something. Flames involving hydrocarbons, for example, have side reactions that can generate small particles of solid carbon or a variety of long-chain or aromatic hydrocarbons. The amount of these side-products (and therefore smoke) depends not just on the fuel but also on the control of the combustion reaction.

Flames where the fuel-air mixture is well controlled may generate very few side reactions and no visible smoke (most gas cookers are set up carefully and don't generate visible smoke, but a badly controlled Bunsen burner using the same fuel may be very smoky.) Candles emit smoke by design as this is what generates most of the light in the flame (small particles of carbon become incandescent in the flame emitting yellowish light just because they are hot). Candles are, essentially, optimised (by the design of the fuel and wick) to burn this way. But small burners using wicks and fuels like ethanol or acetone burn with very clean flames with little smoke (they are optimised for heat not light).

Even fairly well optimised burning can produce some smoke. Modern vehicle engines, for example, need to be as efficient as possible but both petrol (gasoline) and Diesel engines still emit some small particulates (though there are often not very visible unlike old Diesel engines which were notoriously smoky). A badly controlled flame from burning hydrocarbon can be very smoky, but a well controlled flame will emit little visible smoke.

Some things are hard to burn efficiently and smoke a lot. Raw coal is full of a variety of things other than carbon (lots of S and N containing compounds, for example) when burning these tend to promote smoky emissions and also release nasty sulphur and nitrogen oxides into the atmosphere which may react further creating more smoky nasties. "Smokeless" coal is heated before use to drive off the worst offenders and results in something like a halfway house on the way to charcoal, which is almost pure carbon. But the burning of solids is always harder to control than gases or liquids and even charcoal will emit small amounts of smoke though it may not be visible.

In short, smoke is usually a product of the control of a flame. Badly controlled flames smoke because side reactions are common; well controlled flames mostly don't as burning goes to the final products (CO2 and H2O which are gases). Well controlled flames don't smoke visibly (but might emit small amounts that are hard to see for some fuels). Pure hydrogen can burn with no smoke at all as there are no possible side reactions.


How to judge if a reaction is exothermic or endothermic

This is one of the most commonly misunderstood principles in thermochemistry - if you can wrap your mind around it, you are most of the way there!

The good news is, there is an easy way to do that. In any physics problem (thermochemistry is really physics) you have to be very careful with how you define the system and the surroundings. This is important for two reasons.

It lets you write energy balance equations across the system boundaries (so we can use math to solve the problem)
By choosing the right system boundaries, we can make the math much, much easier.

Since thermochemistry at this level always involves the same types of problems, chemists have done the hard part for you and already selected the best system boundaries.

The system is whatever is in the reaction equation.

That's it. The entire rest of the universe is the "surroundings."

Looking back at your question:

  The thermometer is in the solution which means it measures the temp of the solution. It's the system. The temp of a system increases means heat flows in, how could this be an exothermic reaction?

If the reaction molecules are the only thing in the system, then that means the water, the calorimeter, the themometer, you, me, the room - everything else - counts as surroundings. Now, the water did heat up. We know this because the thermometer absorbed some heat and showed a temperature increase. Since they are all part of the surroundings, the heat must have come from the system. Since heat left the system, the sign of $q_{reaction}$ is negative, and the reaction is exothermic.

As a teacher, I have found that getting system boundaries and the signs right is one of the biggest problems students have. If you focus on getting that concept down early - and really understanding why they are that way, you are well on your way.


What word, or phrase, do chemists use to describe the color saturation of a solution?

Using a spectrometer, you could accurately measure the degree to which light is absorbed by a solution ("absorbance" or "optical density") at a certain wavelength, and in a cell of certain dimensions. So, if you had a blue dye dissolved in water (which might appear blue because it primarily absorbs orange light), the absorbance (measured using orange wavelengths of light) would be proportional to the concentration of dye in water.


A cheaper and easier to find solution than EDTA (permeakra's answer) is citric acid. Citric acid reacts with metal carbonates (bicarbonates, phosphates, chlorides, etc.) to produce the soluble citrates. The citrate ion keeps the metals in solution by chelation, the same process that EDTA (ethylene diamine tetra acetate) does.

Citric acid can be purchased in the grocery store or convenience store as one of the principle components for any sour powdered drink mix (like Tang).

You could also use any substance or cleaner designed to remove hard water deposits or lime scale, such as CLR, which contains lactic and gluconic acids as chelating agents.


Enthalpy of dissolution of NaOH in small amount of water

The dissolution enthalpy for concentrated final solutions highly depends on the dissociation and hydration degree, that varies a lot with the substance and concentration.
Therefore particular values for concentrated solutions have to be either measured experimentally, either calculated by some empirical formula, derived from experimental data ( if available ).


Why can't equations of state be replaced by machine learning models?

For n data points there exists an n-1 order polynomial that perfectly fits the data.

Therefore there is no basis for "a neural network or whatever" being better.

Furthermore, it simply isn't true that the Peng-Robinson equation "has no underlying physical meaning".

The Peng-Robinson equation (like Van der Waals) recognizes that atoms/molecules occupy space and that attractive forces exist between them.

As Peng and Robinson say in their article:

  Since two-constant equations have their inherent limitations,
  and the equation obtained in this study is no exception,
  the justification for the new equation is the compromise
  of its simplicity and accuracy.

That being said, neural networks have been used to find equations of states.  See for example Equation of state and artificial neural network to predict the thermodynamic properties of pure and mixture of liquid alkali metals

  Generally, ANN is powerful and successful method for complex non-linear systems due to unique advantages such as high speed, simplicity and large capacity which reduce engineering attempt. In recent years, ANN modeling has been successfully used for prediction of thermophysical properties of pure and mixture fluids [24], [25], [26] and [27].


Most endothermic room-temperature reaction

Don't know if these are quite in the spirit of what you are looking for since, with the exception of the first reaction, the entropy change alone is not sufficient to drive them  (i.e., at room temp., their equilibria lie far to the left) (and, for ozone formation, the entropy change is actually unfavorable).  Furthermore, ozone formation and photosynthesis aren't single reactions, but multi-step processes.  But they all do take place at room temperature (or below):

Sublimation of ice (yes, snow can evaporate!):  2.8 kJ/g

Formation of ozone (O$_3$) from O$_2$ (photochemical, occurs in stratosphere): 3.0 kJ/g

Photosynthesis: 7.5 kJ/g

Electrolysis of water: 15.9 kJ/g

The values I've listed are all enthalpy changes (at constant pressure, the heat flow is equal to the change in enthalpy; at constant volume, the heat flow equals the change in energy).  In addition, by convention, the sign of the enthlapy for endothermic reactions is positive, since heat flows into the system, so the enthalpy is increased (not negative, as you've used).


How does Le Chatelier's principle work in a gaseous system?

Le Chatalier's principle states that if a stress is applied to a system at equilibrium, then  the system, will shift its equilibrium position in order to bring the system back to its equilibrium.

So the stress applied to the system could be temperature, pressure and the concentration of one of the components in the system.

I will choose to explain about pressure because apparently you have the problem understanding that.

Pressure.
If you add pressure to a system then the stress will be to decrease the pressure by shifting the equilibrium position towards the side with less number of particles.
If you  remove pressure from a system, then the stress will be to increase the pressure of the system by shifting the equilibrium position towards the side with the higher number of particles.

So it is true that when you increase the increase the pressure in a system, then the system is stressed to reduce the stress. In order to do that, the system's equilibrium will shift to the side with less number of particles.

For example:

2Fe$_2$O$_3$ + 3C -----> 4Fe + 3CO$_2$

So if you add pressure to this above system, then the stress is to reduce the pressure. On your reactants side the sum of the coefficients is 5 and on your products side, the sum is 7. Since you need to bring it down then the shift will be to the side with the least sum which is to the left-hand side. Therefore, the shift will be to the left-hand side.

Hope this helped you with the problem.


Why is europium used in Euro bank notes?

As already noted by the OP, many rare earths fluoresce under ultraviolet light and there seems to be no particular reason why Europium is a better choice than any other element, purely based on its fluorescent properties. (See update at bottom)

The European Central Bank keeps the exact nature of the compounds used a secret. However, it was discovered in 2002 by Frank Suijver and Andries Meijerink [1] that europium, complexed with two $\beta$ diketone molecules provides the red light but the other colours were more difficult to identify.

However, the reason for the choice of Europium is unclear. According to Hugh Aldersey-Williams[2]:

  It [the European Central Bank] wilfully misunderstands my request to know who fought for Europium and drearily asks for my 'understanding that, for security reasons, we cannot comment on the chemical components of the euro banknote security features.'

Apparently, in the original paper by Suijver and Meijerink they say that they were visited several years beforehand by European Bank officials who were investigating potential luminescent materials so it may be that they inadvertently gave the bank the idea for using Europium in its banknotes.

Europium is one of the more expensive lanthanides and so this is probably a factor in its usage to prevent against forgeries, as the OP suggests.

I suspect there is also simply an element of irony in the fact that Europium is used in Euro bank notes and it may be that it was just some bank official tasked with finding a suitable compound to use who thought that it would be funny to use Europium.

Anyway, the bottom line appears to be that nobody knows the reason because the European Central Bank refuses to tell anyone anything about the compounds they use.

UPDATE: In response to @Nicolau Saker Neto's comment I did some more research into the fluorescence of Europium. Europium(III) (red) is regularly used alongside Europium(II) (blue) and Terbium(III) (green) in trichromatic lighting systems. It may therefore be the case that Europium was chosen because it is already well studied and in common use and that they created a $\beta$ diketone complex in order to fine tune the exact wavelength of the emissions and make it unique to their banknotes.

References:

[1] http://dspace.library.uu.nl/handle/1874/25831 (Paper in Dutch which I can't read. However, Williams refers to some of the content in his book and I have alluded to this above. If anyone can translate the original paper and add anything to the answer that would be very helpful.)

[2]Hugh Aldersey-Williams; 'Periodic Tales: The Curious Lives of the Elements'; ISBN 978-0-141-04145-2 http://www.amazon.co.uk/Periodic-Tales-Curious-Lives-Elements/dp/0141041455


Strain Energy in di-axial vs di-equatorial

You can't have both in a chair conformation!

If the methyl groups in 1,4-dimethylcyclohexane adopt a bis-axial orientation, they are trans and they stay trans after ring flip:

If your 1,4-dimethylcyclohexane is cis-substituted the orientation of the methyl groups is axial and equatorial, respectively:


Ambiguity in synthesis (crystallization)

No.  This could also mean applying a vacuum to the toluene-dichloroglyoxime solution, and letting the solvent evaporate-away at room temperature.


How does one recognized Td/Oh symmetry in molecules?

You can detect $O_{h}$, $I_{h}$ and $T_{d}$ symmetry by checking that a molecule has all of the subgroup symmetries of these point groups.

According to this untitled document, which I presume1 is by W.C. Trogler, the elements are as follows:

$T_{d}$: $E$, $4C_{3}$, $3C_{2}$, $3S_{4}$, $6\sigma{_d}$

$O_{h}$: $E$, $3C_4$, $4C_3$, $6C_2$, $4S_6$, $3S_4$, $i$, $3\sigma{_h}$, $6\sigma{_d}$

$I_{h}$: $E$, $6C_{5}$, $10C_{3}$, $15C_{2}$, $i$, $6S_{10}$, $10S_6$, $15\sigma$

Obviously, you don't need to check for the identity symmetry $E$.

If you're a visual sort of person a symmetry spot check is to mentally superimpose a tetrahedron, cube or dodecahedron over the molecule and see whether the view down the surface normal of each face is identical. Cubes and octahedra are duals of each other, as are dodecahedra and icosahedra. The tetrahedron is self-dual.

Interestingly, H&S does not list chiral forms of these point groups, probably because they are so rarely encountered, however researchers have come up with molecules that obligately satisfy $T$, $I$ and $O$ symmetry2 (I have not yet read the paper).

(1) I would be indebted to anyone who can furnish me with a full citation to this work, and confirm the authorship.

(2) Narasimhan, S. K., Lu, X. and Luk, Y.-Y. (2008), Chiral molecules with polyhedral T, O, or I symmetry: Theoretical solution to a difficult problem in stereochemistry. Chirality, 20: 878–884.


Is it possible to remove salt from food without changing the flavor?

Short answer: no, there is no easy (or even moderately hard) procedure, and certainly not boiling and freezing.

The thing is that water dissolves sodium ions better than just about anything else, so almost everything will come out of solution before the sodium does. You're much better off with a product that starts with less salt to begin with.


Whatever that lye calculator is doing it is either wrong by design or internally correct and using a very obscure definition that no other chemist would understand.

Percentage solutions are always (barring calculators programmed by laymen) defined as weight of the solute divided by weight of the solution times 100. So a $\pu{30\%}$ solution of $\ce{NaOH}$ means $\pu{30g}\ \ce{NaOH}/\pu{100g}\ \text{solution}$.


Ok to separate steel from aluminum with bronze? (Galvanic action)

If you must use an aluminum hinge with a steel pin, anodize the aluminum. Anodizing builds up a coat of very hard, non-conductive $\ce{Al2O3}$, sapphire.

There are commercial metal fishers that will perform this, if you do not want to do so. There is a bit of an art in the process of creating a uniform coat of the required thickness: too thick and it cracks off; too thin and it wears off.


When recovering Copper from Copper Chloride via Aluminum, why can't we use Sodium instead to make Sodium Chloride?

As your teacher mentioned, sodium is a very reactive material. She mentioned that it is hard to obtain it naturally as a result. However, this isn't quite the reason as to why sodium isn't used for recovery.

If we were to only consider the cost in energy, aluminum would be a poor choice.
The main process for production of aluminum metal requires a current of 400 kiloamperes and temperatures of 950 C. In contrast, the main process for sodium metal production requires a current of 30 kiloamperes and temperatures of 600 C.

The main reason would be because sodium is too reactive to be used. The sodium metal would react with the water and form sodium hydroxide. This reaction would be too fast and violent that you wouldn't be able to recover copper.
So your teacher was right in pointing out that sodium is too reactive.


Difference between ideal & ideal-dilute solution

Raoult's Law is for ideal solutions (all of it: solvent and solute):

  Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. "An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components." (all quoted from above link)

As stated here concerning ideal-dilute solutions:

  "Many liquid mixtures deviate from the ideal behaviour predicted by Raoult's law. Ideal-dilute solutions are those for which the solvent obeys Raoult's law and the solute Henry's law. The difference in behaviour is because the solvent molecules tend to be surrounded by other solvent molecules so that they behave in an almost ideal way. However, the solute molecules tend also to be surrounded by solvent molecules so that their environment and therefore their thermodynamic behaviour is very unlike that of the pure substance." (all quoted from above link)

In summary: Ideal solutions are those which obey Raoult's Law; ideal-dilute solutions are ones where the solvent obey's Raoult's Law and the solute obeys Henry's Law.


How can x-rays show how many protons are in a nucleus?

X-rays are not emitted by protons. They are emitted by atomic electrons when they fall down from an upper orbit (or level) to another one near the nucleus. Of course the electrons have first to be excited (= sent to this highly excited level). But whatever the way of exciting the electron, the energy of the transition corresponding to the return to the lowest level is proportional to the number of protons in the nucleus. If you are able to measure the energy of the emitted X-ray produced during this transition (or fall), you can calculate the number of protons of the nucleus. I am sorry. I have used a rather simple language. Pure theorists will prefer using a better vocabulary, which may not have been understood by Kantura. I may be wrong.


Henry's law and O2 concentration in alveoli blood

Henry's law states that the amount of a gas that dissolves in a liquid
is directly proportional to the partial pressure of that gas.
Oxygen has a larger partial pressure gradient to diffuse into the
bloodstream, so it's lower solubility in blood doesn't hinder it
during gas exchange.  Therefore, based on the properties of Henry's
law, both the partial pressure and solubility of the oxygen and carbon
dioxide determine how they will behave during gas exchange.

I encourage you to read the article below.
Source: Boundless. “Henry's Law.” Boundless Anatomy and Physiology. Boundless, 21 Sep. 2016. Retrieved 16 Oct. 2016


Multiplet shape in proton NMR of morpholines

It looks as if the NMR of morpholine is an AA′XX′ spectrum (the chemical shift difference is 0.80 ppm, or 320 Hz on your spectrometer, two orders of magnitude larger than the coupling constant).
Unlike linear AA′XX′ systems where the bonds can rotate, in the morpholine it very much has a fixed conformation.

In this fixed conformation, the protons are predominantly in a 'gauche' type arrangement due to the shape of the chair, this is where the appearance of the 'triplet' comes from in the NMR spectrum.
The image below is from Hans Reich's NMR notes, and shows some simulated NMR spectra depending on the gauche/anti ratio (Karpus equation), alone with a true spectrum of cyanomorpholine (which you can see looks basically like your spectrum above).

Source: Hans Reich (wisc.edu)

As a complete side-note. In TopSpin (which is what it looks like you used to process your NMR data) you can change the window function to use a Gaussian function, for strong spectra (which it looks like you have), this can help get a bit more resolution (at the expense of S:N- peaks on the baseline will disappear) such that what looks like a broad slightly messy spectrum actually gives resolved multiplets where the J values can be calculated.
In TopSpin, this can be done by entering the wm command; select "gaussian" from the drop-down list, and enter the two required parameters LB and GB (a decent starting choice is -2 and 0.2). The modified FID will now appear; this can be Fourier transformed using ft and automatically phase-corrected using apk to give the modified spectrum. [Even shorter is to sequentially type the commands lb -2, gb 0.2, and gfp into the command line.]
In MestreNova, the apodisation interface may be brought up from the Processing menu (or directly with the hotkey W) and the relevant parameters set in the same fashion.


Entropy increase of both system and surroundings

We don't need to go to nuclear fusion to find an example. Lots of chemical reactions meet your criteria of being both exothermic (releasing heat to the surroundings) and increasing the entropy of the system.
A reaction is spontaneous if the free energy change of the reaction is negative ($\Delta_rG <0)$.The free energy change of a reaction at constant temperature is equal to the sum of the enthalpy change ($\Delta_rH$) and the temperature multiplied by the negative entropy change ($-T\Delta_rS$):
$$\Delta_rG = \Delta_rH - T\Delta_rS$$
One way for $\Delta_rG$ to be negative is for $\Delta_rH$ to be negative and $\Delta_rS$ to be positive. In a reaction conducted at constant pressure, the enthalpy change $\Delta_rH$ is equal to the heat exchanged between the system (reactants) and the surroundings. When $\Delta_rH <0$, the reaction is exothermic.
To find examples that meen your criteria, you only need a source of standard entropy and ethalpy data so that you can calculate $\Delta_rH$ and $\Delta_rS$ for a reaction. There are many such examples of reactions that are exothermic and have a positive entropy change of the system. Below is a simple example:
$$\ce{C(s) + O2(g) -> CO2(g)}$$
Use the NIST Chemistry Webbook, here are the standard enthalpies of formation and standard entropies of formation for carbon, oxygen, and carbon dioxide:
$$\begin{array}{|l|c|c|} \hline
 & \Delta_fH^\circ\ \mathrm{(kJ/mol)} & \Delta S^\circ\ \mathrm{J\cdot K/mol} \\ \hline
\ce{C(g)} & 0 & 5.8 \\
\ce{O2(g)} & 0 & 205.15 \\
\ce{CO2(g)} & -383.52 & 213.79 \\ \hline
\Delta_r & -383.52 & 2.84 \\ \hline
\end{array}$$
At all temperatures, this reaction is exothermic and has a positive entropy change for the  system.


I will first enumerate the four points that need to be addressed.

The vapor and liquid phases have the same composition in an azeotrope.
The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fractional distillation.

I will assume the definition of an azeotrope to be a solution whose composition is such that it exists at a local extremum of a Pxy or Txy plot. I prefer to instead take 1. as a definition, in which case my original definition follows as a property of an azeotrope, but both approaches are equivalent.

"Proof" of 1., by  contradiction. Overlay the corresponding vapor pressure curve as a function of $y_A$ to make your plot a Pxy plot for species $A$. This new vapor pressure curve must touch the original curve at the azeotropic composition $x^*=y^*=z$, for otherwise a tie line would connect two vapor phases at some $P$, which is physically unreasonable.

Proof of 2. Refer to the Pxy plot from 1. The dew point and bubble point coincide at the azeotrope, so a phase transition happens only at one value of $P$ and hence only at one value of $T$.

Proof of 3. Refer to the Pxy plot from 1. The dew point and bubble point coincide at the azeotrope, and so must the compositions of each phase.

Proof of 4. Fractional distillation is based on the composition of the vapor and liquid phases being different. This is precluded by 3. in an azeotrope.


What is the hybridisation of hydrogen in methane?

Hydrogen does not hybridise, as it only has one $s$ orbital. Hybridisation is the "mixing of $s$ and $p$ orbitals; here is a good explanation (for ethane, but it explains the general theory as well).


How accurate is this polar mechanism for the Diels-Alder reaction?

Disclaimer: This is not a full answer. I was just too long for a comment.

I think the suggested mechanism is not entirely wrong, it might be a contributing factor of the whole process, but not a very important one. In the last fifty years, the mechanism has been discussed in almost countless publications. One of the more recent ones by Goldstein et. al. concluded that a concerted pathway is the most economical.[1]
They also included possible diradical pathways and the similar. Of course there is no uniformly applicable mechanism, it is very dependent on the substrates. Many of the reactions can be effectively described by frontier molecular orbital theory, so it is also possible to do so for the simplest Diels-Alder reaction.

There might be many more publications out there discussing the whole mechanism in pretty exhausting detail, but for a start I thought the cited one will do just fine. Just to name some more, see [2] through [4].

The electrostatic interpretation is tempting, as it is simple and it explains the simplest case very well. I believe it explains it too well for the wrong reasons:
 I would not expect the polarisation, i.e. induced dipole, in ethylene and butadiene be strong enough to make this happen at all. Both are colourless gasses at room temperature. If there were more induced dipoles, they would be liquid, which means that these molecules are basically unpolarised.

In a publication by Gian Franco Tantardini and Massimo Simonetta it was stated, that the contribution of the polarised Valence Bond Structure is very little, see [5]. If I read it correctly, the contribution is less than 1%. This alone should suffice to disregard a mechanism purely based on electrostatic considerations. In the same publication it is stated, that ethylene has quite a significant contribution from its polarised structure (about 25%). However, these consideration would in most cases not be able to describe streospecific reaction products, that are obtained from different substrates.

As another aspect, one must not forget, that the ground state of 1,3-butadiene is trans and it requires just a bit of energy to rotated it to the necessary cis formation, which should be about 10 times less present. So what are the odds having a well polarised ethylene hitting a polarised cis-butadiene in the right angle at the right speed. There has to be more to it...

E. Goldstein, Brett Beno, and K. N. Houk; J. Am. Chem. Soc., 1996, 118 (25), 6036–6043.
R.B. Woodward, and Thomas J. Katz; Tetrahedron; 1959, 5 (1), 70-89. Mirror at znu.ac.ir.
Michael J. S. Dewar, Santiago Olivella, and James J. P. Stewart; J. Am. Chem. Soc., 1986, 108 (19),  5771–5779.
J. Sauer; Angew. Chem. Int. Ed., 1967, 6 (1), 16-33.
Gian Franco Tantardini and Massimo Simonetta; Int. J. Quant. Chem., 1978, 14 (S12), 131-141.


Is it possible to have an allotrope of carbon as a long unbranched chain of itself?

Well, this is a rather obvious development (in retrospect, that is). We had an $sp^3$ carbon with tetrahedral bonds, and it made diamond. We had an $sp^2$ carbon with trigonal bonds, and it made graphite. What if we just had a linear $sp$ carbon?

Such thoughts have been around for more than quite a while. Theorists swarmed around that non-existent tree. No wonder that people started trying to actually obtain that linear carbon.

Not knowing anything about you, I still can claim with a good deal of confidence that the first reports of its synthesis are quite probably somewhat older than you (or me, for that matter). Then they were proved... not quite reliable. Some groups agreed with that, some disagreed, a great deal of confusion abounded, then more reports came and things became hairy. So they stayed for the past few decades and will stay for a while, I guess.

Now to the Q&A.

Does this compound exist?

· Define "exist", otherwise no meaningful answer would be possible.

Is it stable?

· Define "stable", otherwise no meaningful answer would be possible.

Is it thermodynamically stable at any P and T?

· Most researchers agree that it isn't.

Is it electrically conductive along the strands?

· Most researchers agree that it is.

Is it made of double bonds or of alternating single and triple bonds?

· The difference is less than you seem to think, to the point that it might not be a question worth pursuing.

What is its crystal system?

· Nobody knows. Chances are there are polymorphs with different crystal systems.

Is it super strong?

· Nobody knows, as with any non-existent compound.

Should it be called a polymer?

· About as much so as graphite and diamond, which means: No.


Corrosion of metals and observations in the agar medium?

You have correctly identified zinc and iron as the anodes, since they are both less noble than copper. In the corrosion reaction, they are both oxidized and dissolved in the aqueous gel.

\begin{aligned}
\ce{Zn ->& Zn^2+ + 2e-} \\
\ce{Fe ->& Fe^2+ + 2e-} \\
\end{aligned}

The electrons flow to the copper cathode, where oxygen, which is dissolved in the liquid medium, is reduced to hydroxide.

$$\ce{O2 +2H2O + 4e- -> 4OH-}$$

As a consequence, the pH of the solution around the copper cathode rises, and the phenolphthalein indicator changes its color from colorless to violet-red in this area.

$\ce{Fe^2+}$ reacts with $\ce{K3[Fe(CN6)]}$ to form a deep blue iron(II,III) hexacyanoferrate(II,III).

$$\ce{K+ + Fe^2+ + [Fe(CN6)]^3- -> KFe[Fe(CN6)]}$$

Therefore, the solution near the iron surface turns blue. Zinc does not react with $\ce{K3[Fe(CN6)]}$ and the solution around the zinc anode remains colorless (source).


Are there any compound lists searchable by elemental content?

After some more searching, followed by asking reddit, I got a few good suggestions.
ChemSpider's advanced search method allows you to, among other things, search for elements containing specific elements and lacking other elements.

Also, for large lists of compounds I can use python tools to extract the tables from the CRC Handbook pdfs and then, search that data.


Relationship between rate equation and equilibrium constant

At equilibrium it does not matter if the reaction has multiple steps; your equilibrium constant only has to involve the initial reactants and final products.

For instance, in an S$_\mathrm{N}$1 reaction, obviously the RDS is the departure of the leaving group, but you don't have to (and usually are not in a position to) know the concentration of the intermediate carbocation in order to calculate the equilibrium of the overall reaction.  You can simply write an  equilibrium equation for the initial reactants and final products: $$K=\frac{[\ce{LG}][\ce{product}]}{[\ce{nucleophile}][\ce{substrate}]}.$$

Of course finding $K$ may be not be easy, even if you have access to good thermodynamic data for the reactants and products, because $\Delta G_0$ (and therefore $K$) depends on solvation, temperature and other conditions. But given identical conditions, $K$  is constant over a wide range of concentrations, so you might be able to find it in the literature or measure it with a single experiment.

All this assumes that the RDS is fast enough that you're able to reach  equilibrium before the solvent evaporates, or your reactants and products decompose via other pathways, or you die of boredom!

Note that you can also write an equilibrium equation for the RDS alone:
$$
K'=\frac{[\ce{LG}][\ce{carbocation}]}{[\ce{substrate}]}
$$
But since you seldom know the concentration of the carbocation, this equation isn't as useful in practice (it is of course extremely useful for mechanistic studies).

Edit: This page has a more complete and very nice answer!


If two electrons move up to a higher energy level in an atom, is it considered excited?

http://physics.nist.gov/PhysRefData/Handbook/Tables/germaniumtable5.htm
Energy levels of a neutral germanium atom
https://www.webelements.com/germanium/orbital_properties.html
Middle, electron binding energies for germanium

It will certainly be in one of its excited states, yes, probably followed by an Auger cascade.

Label   Orbital   eV
K       1s       11103
L I     2s       1414.6
L II    2p1/2    1248.1
L III   2p3/2    1217
M I     3s       180.1
M II    3p1/2    124.9
M III   3p3/2    120.8
M IV    3d3/2    29.8
M V     3d5/2    29.2


For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)?

Chemguide is simplified for A-levels and therefore in this case is strictly speaking incorrect.

The equilibrium constant $K$ is defined as a product of activities. I described this in a previous answer here.

The crux of the matter is that the activity of a pure solid or pure liquid is equal to 1, which means that it can be omitted from the expression for $K$ without affecting the value.

In your first reaction

$$\ce{H2O(g) + C(s) <=> H2(g) + CO(g)}$$

the chunks of carbon in the reaction are necessarily pure because they don't mix with the gases.

In the second reaction (yes, I am lazy, please feel free to edit for me)

$$\ce{EtOAc(l) + H2O(l) <=> AcOH(l) + EtOH(l)}$$

none of the liquids are pure, hence their activities deviate from unity.

Just as a final example, in the dissociation of a weak acid

$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A-(aq)}$$

water is omitted from the expression for $K_\mathrm{a}$ because water, as the solvent, is in large excess over $\ce{HA}$ and is therefore effectively "pure".


Difference between Sodium Bicarbonate and Sodium Carbonate reacting with HCl

It seems to me that the purpose of the 3 experiments with the HCl was not to identify the dry product. You can just write the reactions you observed, that was probably the point.


How does Melatonin put you to sleep? How is this compound radio-protective?

Melatonin is not "radio-protective" in the sense that it blocks radiation or anything. As you note, it's an antioxidant. One of the ways in which ionizing radiation can interact with the body is by providing enough energy to break chemical bonds, producing highly reactive radical species that can cause biological damage by reacting with various things in the body. Melatonin can easily be oxidized by many of these reactive species, preventing them from reacting with anything else. This paper shows the products for the reaction with radio-generated hydroxyl radicals.

The role melatonin plays in sleep is fairly well known (though the effects of exogenous melatonin, less so). Without going into too much detail, there are melatonin receptors in the body, found in cell membranes in the brain and a few other places (G protein-coupled receptors). When these bind melatonin on the outside of the cell, their conformation changes, further activating signalling processes on the inside of the cell. The effects of almost all signalling molecules in the body work in this way. The molecules activate some kind of receptor, which then activates a sequence of other signals, eventually resulting in biological effects.


Will an acyl chloride be attacked by water molecule to form carboxylic acid?

You need to lose a proton, then chloride becomes a better leaving group than hydroxide. In general, proton transfers are incredibly fast; in this scenario there is tons of water around to accept the proton, and the tetrahedral intermediate with a positively charged oxygen is fairly acidic anyway.

Mechanism from Clayden 2nd ed. (p 199).

In the absence of an external base (e.g. pyridine), water will act as the base itself here. Another way of approaching it is to note that hydrolysis of an acyl chloride will release one equivalent of HCl:

$$\ce{RCO\color{red}{Cl} + \color{red}{H}OH -> RCOOH + \color{red}{HCl}}$$

so in the absence of a second base, you will necessarily have to produce $\ce{H3O+}$.


Maltose is 4-O-α-D-glucopyranosyl-D-glucose.  The "O" in "4-O" indicates the attachment of atom of the substituent.  The substituent in this case is "-α-D-glucopyranosyl".  It is attached to the oxygen atom that is attached to carbon 4 in the base molecule.  The base molecule in this case is "-D-glucose".  "4-α-D-glucopyranosyl-D-glucose" could be interpreted as a different structure where the α-D-glucopyranosyl substituent is bound directly to carbon atom number 4.

The "base molecule" of "-D-glucose" still has an unsubstituted anomeric OH group.  Thus, in aqueous solution, maltose will be an equilibrium mixture of at least these three isomers:

4-O-α-D-glucopyranosyl-α-D-glucopyranose
4-O-α-D-glucopyranosyl-β-D-glucopyranose
4-O-α-D-glucopyranosyl-D-glucose (open chain on the right-most "glucose" moiety in the name)

Thus, it doesn't really make sense to disambiguate α vs. β vs. open-chain forms of the "right-most" glucose ring in maltose, unless you need to refer to a particular conformation of maltose that matters for a certain reaction or condition.  However, the "left" ring is locked (in maltose) to an α-D-glucopyranosyl conformation because it does not have an anomeric OH group and rapid equilibration in water does not take place.


About nitrogen acidity in function groups

First of all, a terminology correction. While a nitrogen atom may be basic, a nitrogen atom itself won't be acidic, rather, the hydrogen atom attached to that nitrogen atom would be acidic, if at all.

Now, whenever given an arbitrary compound, and this also holds true for atoms like carbon, oxygen, etc., here's what you need to do to determine the acidic and the basic atoms.

What does acidity mean? Acidity means the ability to lose a proton from the compound. The higher the ability to lose a proton, the higher the acidity. If a compound, on losing a proton, forms a more stable conjugate base, it is considered more able to lose a proton. Hence, it is also more acidic (than other similar protons in that compound).

What does basicity mean? Basicity means the ability to donate a lone pair of electrons to a proton. The higher the ability to donate a lone pair to a proton, the higher the basicity. If a compound, on donating a lone pair, forms a more stable conjugate acid, it is considered more able to donate a lone pair. Hence, it is also more basic (than other similar lone pairs in that compound).

They both sound similar, don't they? That's because they're on the same underlying principle: a chemical species would want to accept a proton or donate a lone pair if the resulting species is more stable.

Now, you can simply check for this in any given compound by deprotonating every H-atom and protonating every lone pair, while applying the simple rules of aromaticity, resonance, hyperconjugation, and inductive effects individually to each.


How are there only 8 structural isomers of C5H11Br?

You'll probably need to draw all the possible isomers to see it, but with only 5 carbons to work with the ethyl isomers are redundant with the methyl isomers: i.e. 1-bromo-2-ethyl-propane is the same thing as 1-bromo-2-methyl-butane.

Remember we are only talking about structural isomers, and ignoring stereoisomers.


In an acidic solution, the protonation methyl orange takes place, as shown below

The deprotanted form depicted on the left is basic, and is yellow in colour. The red form is acidic and is depicted on the right.

In general, a lot of acid base-indicators rely on the following mechanism of action

$$\ce{In^-}+  \ce{<=>[\ce{H^+}][\ce{OH^-}]} \ce{HIn}$$

At low pH, i.e high $\ce{H^+}$ concentration the $\ce{HIn}$ form is dominant since the equilibrium lies to the right, however, at higher pH $\ce{HIn}$ (which is weakly acidic) dissasociates to give $\ce{In^-}$ (its conjugate base).

Herein, $\ce{HIn}$ (the acid), and $\ce{In^-}$ (its corresponding conjugate base) have different colours. The colour changes are due to changes in absorption, due to change in the length of the $\pi$ conjugated system (as can be seen in the case of methyl orange pictured above).


$\ce{C2H4O2}$ is the molecular formula of acetic acid. As such, it's a correct method to write the molecule. However, you should note that, as Ivan Neretin mentioned, molecular formulae are not necessarily unique for a given molecule.


What will happen to bottle of cola after 50 years?

I'm going to make a time capsule for 25 or 50 years, and I don't know whether it's safe to put food, e.g. bottle of cola there.

Depends on the food and the way it's packed, but food is not something I'd suggest you pack for such a long period..

  Will it explode and ruin the capsule ?

Food or drink are very unlikely to explode (maybe alcohol would in the wrong circumstances).

But food can be unsafe to eat in quite a short time, depending on the food, preparation and packing and how you store it.  Bacteria could be a problem.

Most food will go bad over a period as long as you're describing without very specific storage (like refrigeration, and even then I'd say most things won't work).

The safest ones would be :

popcorn (kernels)
honey (jar of)
sugar (most forms)
alcohol (spirits like vodka, whiskey)
dried beans, dried lentils

I would not be planning to eat any of these stored for 25 years myself.  And in general I'd suggest testing the items before trying them after 25 years or more (if you feel you must).

I would not expect cans or glass or plastic bottles of soda to be in good shape after anything like 25 years.  The plastic might not survive without degrading.  The can and plastic might react with the liquid over that timescale and the glass would survive but I'd be less optimistic about a sugar laced chemical soup like soda or cola not undergoing some changes.  Hard to say.

If you want more info on this try this website.

  Will it be ok to drink it, if it won't explode?

I would not try it.  At best it soda would be flat and possibly not taste the same (chemical changes over that timescale ?) and at worst it could actually be harmful.

Exploding seems very unlikely.

  Also, what about Snickers or a hamburger in a ziploc package with air sucked out of it with vacuum cleaner?

Air isn't the issue.  There are bacteria that will happily live (and increase in numbers) on what's in the food.  Well, it is food, after all.  There are bacteria that will survive refrigeration as well.  Over the timescale you're talking about I'd say it's all bets are off territory.

  So: will Snickers, Hamburger in a ziploc, Bottle (or can) of Cola, all not opened, go crazy in 25 years? In 50 years?

All of those could be dangerous over such a long time period, IMO.  At the very least they'd taste bad and at worst they'd kill you if you consumed them.

  If so, can they go out of their packages and ruin the contents of the time capsule? If not, will it be safe to consume one of them?

Depends on the packaging.  Glass would last indefinitely baring physical force or extreme of hot and cold (which might possibly cause fatigue cracking).  The other wrappers would last pretty well (structurally), but 25 years is way past their design intentions.  It would be a dice throw.


What is the effect of Helium-based surroundings (99% Helium) on electrochemical gas sensors?

Aiming for dilution of the exhaust gases, you need a gas that is inert, i.e. does not react at these conditions with neither the gases present, nor with the metals present in in the catalytic converter (if there is such device installed) and electrodes of the sensors.

Under this restraint, using a nobel gas was a good idea.  Yet instead of helium, that you still may use to spot leaks in the tubings, I would opt for argon, which is much cheaper per litre.  Note that at ambient conditions, its density as gas is higher than the one of helium and of air and has a slightly lesser thermic conductance.


Yes, cuprates do react with carbon dioxide to give access to carboxylates / carboxylic acids.

Running a SciFinder search, I do not see any examples of traditional Gilman reagents (alkyl cuprates) reacting directly with carbon dioxide. This shouldn't be surprising though, since the alkyl cuprates are generated from the corresponding alkyl lithium compound. If you have the alkyl lithium compound, then you can just react that with carbon dioxide directly.

A more synthetically useful situation is when the alkyl cuprate is reacted with an alkyne, resulting in a syn-addition of the alkyl group and copper across the alkyne. The alkenyl (vinyl) cuprate can then be trapped with carbon dioxide to give a carboxylate.

A couple examples from the literature:

Z,Z-Dienes via acetylene carbocupration:  synthesis of navel orangeworm pheromone

The silylcupration of acetylenes: a synthesis of vinylsilanes

Also check the write up in Organic Reactions by Lipshutz (the world's expert in organocopper chemistry).

More complicated (non-Gilman) organocopper compounds have been reacted with carbon dioxide to give copper carboxylates:

Carbon-carbon coupling of activated alkanes with carbon dioxide by [(phen)(Ph3P)Cu(O2COH)]

Synthesis, Structure, and CO2 Reactivity of a Two-Coordinate (Carbene)copper(I) Methyl Complex


How do steric effects affect inductive effect in compounds with tetravalent nitrogen?

Since you are asking for a theoretical explanation, I believe a theoretical calculation might be helpful. It is possible to calculate the partial charges on atoms using electronic structure programs. I have calculated the partial charges at $\text{B3LYP/pcseg-1//GFN2-xTB}$ level (with GAMESS and xTB).
$\ce{Me-\overset{+}{N}(^iPr)3}$
We only need to look at the partial charges on the Me group to know the electron-withdrawing capacity of $\ce{-N+(^iPr)_3}$ group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.4367\\
\ce{H} & \;\;\, 0.1954\\
\ce{H} & \;\;\, 0.1957\\
\ce{H} & \;\;\, 0.1966\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.151$.
$\ce{Me-\overset{+}{N}(^tBu)3}$
Again, we only need to look at the Me group.
\begin{array}{cc} \hline
\mathrm{atom} & \mathrm{partial\;charge\;(e)} \\ \hline
\ce{C} & -0.3087\\
\ce{H} & \;\;\, 0.1436\\
\ce{H} & \;\;\, 0.1430\\
\ce{H} & \;\;\, 0.1414\\ \hline
\end{array}
Total charge on $\ce{Me} = 0.1193$
So, the partial charges would indicate that $\ce{-N+(^iPr)3}$ is more electron withdrawing than $\ce{-N+(^tBu)3}$.
Explanation?
I am not sure I have a definitive explanation for this. But it's possible to look at the differences between $\ce{^iPr}$ and $\ce{^tBu}$—

The cone angle formed by the three R groups for $\ce{-N+R3}$ would be larger for $\ce{R=^tBu}$ obviously, due to the steric bulk of that group. When the angles are the ideal $109.5^\circ$ each bond can be assumed to form from $\mathrm{N\;sp^3}$ orbitals. When N is $\mathrm{sp^2}$, the three bonds would be $120^\circ$ to each other. So, when the cone angle increases, the amound of p-character in those three N-R bonds can be assumed to go down. Conversely the amount of p-character in the 4th bond goes up which would mean less electron withdrawing capacity (p-orbitals are further away from nucleus). So, $\ce{-N+(^tBu)3}$ should be weaker EWG.

As there is no empty p-orbitals on N, so there should be no hyperconjugative effects for the compounds drawn. However, if instead of $\ce{Me}$, there is some other group that has a $\pi$ system attached to N, then there is a possibility of hyperconjugation.

The $\ce{CH3}$ groups have a very weak electron donating effect, so $\ce{-N+(^tBu)3}$ will be slightly weaker EWG than $\ce{-N+(^iPr)3}$. *

Notice that points 1 and 3 predict the same trend, which matches with the calculated trend. However, it is difficult to tell which factor is the most important (and whether these factors actually mean anything).

* Curiously, the partial charge on $\ce{C}$ of the side chain goes up when its $\ce{H}$ is replaced by $\ce{CH3}$, which would suggest that $\ce{CH3}$ is actually electron-withdrawing. This does not match with the standard notion of methyl groups being electron donating. I am not really sure if there is any explanation for this discrepancy. But note that there have been some evidence that $\ce{-CH3}$ groups can act as EWG's in some circumstances. The linked paper specifically mentions in the abstract—"That methyl groups attached to carbon atoms are electron donors must not be generally assumed."


Equilibrium Constant of a Salt knowing the solubility

The molar solubility $s$ equals:
$$s= \frac{15.9\times 10^{-3}}{78}= 0.2038\times10^{-3} \ce{mol/L}$$

According to the equilibrium: $\ce{CaF2 \leftrightarrows Ca^{2+} + 2F-}$, the  equilibrium constant is:

$$K_{sp}=4s^3$$

By substituting the value of $s$ in the above equation, we find:
$$K_{sp}=0.034\times10^{-9}$$


what's the ratio of atoms of these two isotopes?

You need more information.  They probably expect you to look in a book and see that the atomic weight (which is a weighted average over all isotopes) of $\ce{Tl}$ is 204.383.  Now if x is the amount of $\ce{^203Tl}$ with atomic weight 202.97 and (1-x) is....

Does that get you started?

$$\ce{(x*202_{.}97) + ((1-x)*204_{.}97)~=~?}$$

Can you get to the answer from here?


What to say 'one mole of C atoms or one mole of C molecules?

What you need to remember when using terminology involving moles is that moles are just a way of counting things but that the terminology only makes sense if you unambiguously specify the thing you are counting. It is entirely possible to use moles to count non chemical things (for an amusing example of the consequences of having a mole of moles (the small burrowing mammals) see this.) Most confusion arises because the way the thing is specified is ambiguous.

Some examples:

a mole of CO molecules is perfectly correct (the units being counted are single molecules of CO)
A mole of C atoms is perfectly correct (and, if you are counting C atoms in CO molecules that is also fine and the answer will be the same)
a mole of C molecules in CO isn't correct but only because the molecules is CO which consists of atoms of C and O.
if you were counting the atoms in CO2, then it would be correct to say there are twice as many moles of oxygen atoms than of carbon atoms or carbon dioxide molecules

The use of moles to count doesn't imply anything about what you are counting, you have to specify that as clearly as possible. When talking about the parts of a molecule we usually talk about the atoms that make up the molecule. CO2 doesn't consist of molecules of oxygen and carbon, but atoms of oxygen and carbon. But this is independent of the use of moles as a way of counting those components.


You can indeed "harden" glass by exchanging ions on the surface, but not because it turns into quartz glass.

Quartz glass is mostly so robust because it has an extremely small thermal expansion coefficient, and is therefore mostly free of internal stress due to uneven cooling. Ordinary glass already gets a lot more durable when you temper it, thereby relieving those internal stresses.

Generally, glass breaks under a very small extensional strain. It has a huge $E$ modulus and can take a lot of stress, but when you bend it, so the extensional stress and deformation is largest at the surface, where also you always have small scratches etc.), it shatters.

The trick is to put the glass surface under a large compressive stress. That way it can take a lot more pull before the critical strain (deformation) is reached and breakage occurs.

There are basically three ways to do that: Firstly, you can coat the hot glass with another glass that has a lower thermal expansion coefficient. That way the surface gets compressed during cooling. Tricky, and expensive.

Then, the most common way, is to rapidly cool the still nearly molten surface by e.g. blowing cold air on it. The outer shell is already cold, the inner part takes longer to cool and still shrinks, thereby compressing the surface.

The last, chemical trick (making supremely solid glassware) is to exchange sodium ions in the surface layer (100nm or so i think) by potassium ions. This was done in the eighties in the GDR for household glassware, under the brand name "SUPERFEST", by soaking the glasses in molten potassium nitrate (at 400 °C) for twenty minutes or so. The ions are larger, but the silica network lattice does not yield below at least 500°C, so the surface gets compressed by the exchange.

You know you had a glass that was treated in one of these ways when it does not break into large shards, but basically crumbles into small pieces or even dust. The classic example for the latter is Prince Rupert's drop, of the former any car windshield. (It seems the Na/K exchange does not (necessarily) do that, see your mobile phones "Gorilla glas".)


Let me venture an answer as well as I can.

I would point out that the article you are referencing is speaking of the arrangement of molecules in bulk liquid water when it is talking about this "fifth water" being present. The key is to note that they say a "fifth water molecule in the first coordination shell (23)." That means they are looking at how water molecules are coordinated with each other in liquid water. They are studying a massive (compared to a single water) amount of water, so the question of its liquidity is no question at all.

The "fifth water" thing is curious because it is common belief that waters arrange themselves in a tetrahedral arrangement. This is known to be true for ice, but is in question for liquid water. For instance, this article about the liquid water hydrogen-bonding network indicates that water, as it is heated, begins participating in only two hydrogen bonds rather than four.

Now, to answer the question of what makes liquid water be liquid water (when it is indeed liquid water). I think from the outset, it should be clear that this is a very difficult question to answer and any answer will likely be debated. I say that because you must define how many molecules it takes to make a liquid which is a very difficult thing to do.

I will tell you, however, that I have some useful information from research I am doing on water clusters and eventually liquid water (as a computational chemistry research assistant). The largest cluster we have studied is called DD*(20,1) which means (distorted dodecahedron with a water inside the cluster).

An idea of what $\ce{(H2O)21}$ can look is this: (taken from this very interesting article) water 21 http://pubs.acs.org.ezproxy.spu.edu/appl/literatum/publisher/achs/journals/content/jpcbfk/2006/jpcbfk.2006.110.issue-38/jp056416m/production/images/medium/jp056416mf00001.gif

The DD*(20,1) arrangement is image C.

As you can see from these different isomers, they are very distinct and no doubt each monomer behaves dramatically different from the next monomer. 21 monomers is quite a lot for a cluster, but even this does not mimic the properties of liquid water at all particularly because one would still consider this a gas-phase cluster. The fact that we even have isomers is indicative of not being near the point where something is considered liquid water. There are water clusters whose energy minima have been identified theoretically all the way from n=2 to (I've seen an article that says) n=60. Read more about some of that here.

One point I would like to make is that in structure C, for the research I'm doing, we have identified the vibrational properties of many of the monomers as being significantly different from those of molecules in liquid water.

So, all that being said about water clusters, we are certain that many more than five or 21 waters must be present to truthfully call something a liquid. We should then start looking for a system which is just large enough to accurately predict the properties of liquid water which are found experimentally.

As it happens, it is very difficult to do this on any scale, theoretically, at least, and identifying how many water molecules one is dealing with during an experiment to a quantifiable amount of precision is quite difficult. There are, however, good approximations of liquid water which are generated using Molecular Dynamics simulation software (here's a wikipedia page about MD software). The various models used for approximating the behavior of water do quite well at mimicking water's properties from an MD perspective. You can read about the TIP4P model of water if you want.

From the paper I just linked about TIP4P, they say that they approximate the properties of liquid water using 360 water molecules in their MD simulations, so that can act as a reasonable baseline for how many monomers are necessary before a system behaves like liquid water.

I say baseline for an important reason. The results found in that paper use MD simulations which pull a clever trick. Rather than having 360 waters that essentially float around in a box with nothing at all beyond the edges of the box, the software will actually mirror the behavior of a molecule beyond the walls of the box in which you are simulating. That means if you pick a single water, you would find that exact same water in the same place a full "simulation box" away from where you originally found it. So, if this says 360 waters, the number of waters for which interactions are being calculated is actually much larger than the amount of interactions 360 waters alone would have.

All that leads me to conclude that water behaves like water probably somewhere around 500 monomers. It must be noted, however, that the monomers behaving like liquid water as we know it will be the monomers on the inside of those 500 (but likely more) waters. Once you get towards a boundary, things start getting weird.

Hope that helps.


When concentrated sulfuric acid is heated, then it decomposes via the reaction
$$\ce{H2SO4 -> H2O + SO3}$$

I tried to determine under what circumstances the nascent oxygen (a free single atom of oxygen) is formed which would react with copper to create $\ce{CuO}$ but couldn't find the reaction in any modern source. I did find the proposed reaction in Essentials of Modern Chemistry
 By Charles Elwood Dull (1918) but I'm guessing that the reaction to give nascent oxygen is just wrong.

The gist here is that you can't piece reactions together from balanced equations but you must also define the conditions under which the reaction takes place. At a deeper level not only can the chemical equation be specified but also the mechanism of the reaction can be detailed.

Hydrochloric acid, $\ce{HCl}$, is typically in aqueous solution. Although copper won't react with pure hydrochloric acid, it can react with dissolved oxygen in hydrochloric acid.
$$\ce{Cu (s) + 2HCl (aq) + 1/2O2 (aq) -> CuCl2 (aq) + H2O}$$

Oxygen can also be formed in chlorine water:
$$\begin{align}
  \ce{Cl2 (aq) &-> HCl + HOCl}\\
  \ce{HOCl &-> HCl + 1/2O2}
\end{align}$$

But the concept of nacent oxygen from this reaction is wrong.


How can I calculate the number of branches in an alkane isomer?

Each instance of a comma-delimited number in the prefix is directly indicating a branch on the "backbone" or base-chain. It's quite literally the number of numbers.

The most complicated example you have can be dissected like this:

$2,2,3,3-$ tells us to anticipate four branching groups from the base-chain on the 2nd, 2nd, 3rd, and 3rd carbons.

$Tetra$ tells us that all four branching groups are identical (since there are four total)

$methyl$ tells us the name of the branches. The $methyl$ group is a $\ce{-CH3}$ attached to the backbone.

$butane$ tells us what the backbone is; in this case, a 4-carbon ($but-$) fully-saturated carbohydrate ($-ane$).

So the molecule will look like this, with the 4 branches indicated:


What is the mechanism for this intramolecular SN1 reaction?

You are correct on your initial assumption. This hydrolysis is assisted by the neighboring thiomethoxy group ($\ce{SCH3}$) as depicted in the following scheme:

The cyclic methylsulfonium intermediate, which is achiral in this case, is a mimic of bromonium ion in bromination of alkenes (also see Chemistry of Sulfur Mustard on releasing chlorine through cyclic alkylsulfonium ion). Realistically, this intermediate is mimic of bromonium ion of of cis-2-butene (also achiral), and hence would give a racemic mixture. Similar to bromonium ion intermediate, the upcoming nucleophile ($\ce{H2O}$) can attack at two places: through path $a$ or through path $b$. Keep track on the (R,S)-configurations of two stereo centers. The original compound has (S,S)-configuration. The (S)-configuration on carbon attached to $\ce{SCH3}$ is unchanged but carbon attached to $\ce{OTs}$ has been changed to (S)-configuration in cyclic intermediate, because it is necessary to have an anti-periplanar conformation for attacking group ($\ce{SCH3}$) and leaving group ($\ce{OTs}$), similar to $\mathrm{E2}$ mechanism.
During the path $a$ mechanism, $\ce{H2O}$ attack the carbon where $\ce{OTs}$ originally was. Therefore, the carbon bearing $\ce{SCH3}$ group does not change its (S)-configuration. However, the $\ce{H2O}$ attack happens in the opposite face of $\ce{SCH3}$ group, the (S)-configuration retain in the carbon bearing $\ce{OH}$ (as in the original compound).
However, both stereochemistry would be reversed during the path $b$ mechanism (see the scheme). Accordingly, $\ce{H2O}$ attack the carbon where $\ce{SCH3}$ originally was, which has (S)-configuration. Since the $\ce{H2O}$ attack happens in the opposite face of $\ce{SCH3}$ group, the (S)-configuration would be changed to the (R)-configuration in this carbon, now bearing $\ce{OH}$. (as in the original compound). The second carbon, now bearing $\ce{SCH3}$ group, would not change its (R)-configuration (as in the cyclic intermediate).
As a result, you get racemic mixture.


What chemical could dissolve low-density polyethylene?

The safest solvent might be supercritical carbon dioxide, $\ce{CO2}$. Supercritical carbon dioxide is also used in "green" dry-cleaning. Use for dissolving polymers requires a pressure vessel for ~5 to 50 MPa, able to work at ~40 to 100°C. There are (paywalled) articles at ACS and NCBI, if you want to investigate further.
According to Agilent, 1,2,4-Trichlorobenzene (TCB) is effective.
Some other solvents mentioned at Research Gate include toluene and xylene.

N.B. The aromatic solvents all are toxic, some quite so, and are volatile and flammable. All of them require temperatures near the melting point of polyethylene, ~100°C, to dissolve much of the polymer, so a reflux or pressurized system is required.

Supercritical $\ce{CO2}$, while not very toxic, can cause asphyxiation above ~5%, and the container is, effectively, a bomb containing a warm or hot pressurized fluid.