Metallic bonds depend on the Pauli exclusion principle to cause a single energy level to be separated into a sea of very closely spaced levels.

Consider metallic Ca, with its s2 filled with two electrons of opposing spin state. You cannot have another s2 electron in the same atom. Now, add a second Ca atom that can freely interchange electrons with the first (e.g. metallic conductivity). Quantum mechanically, the electrons are indistinguishable, and an electron of the first atom might be found on the second one, so the combination acts as a "super-atom" with two nuclei but a single electron cloud. To allow this without violating the Pauli exclusion principle, one s2 level "moves up" and the other "moves down", splitting the level into two bands.

As more atoms are added, the Fermi "sea" becomes populated to the extent that the energy levels merge into a continuum.

See http://fog.ccsf.cc.ca.us/~wkaufmyn/ENGN45/Course%20Handouts/15_ElectricalProps/02_ConductivityConductor.html and http://www.chemexplore.net/sodium.htm for further explanation.


How is bond angle of a molecule measured?

The most obvious method is single crystal xray crystallography, of course, and for simple molecules, IR and microwave spectroscopy in the gas phase can be used.
For many complex (e.g. bioorganic) molecules, this fails, but there are two (afaik) last hopes to help solve/prove structures if they don´t crystallise and the usual NMR spectroscopy with added computational chemistry is not enough.

Single molecule xray diffraction, a very new approach that can only just be done on the one most brilliant xray "laser" source in the world, the XFEL in Hamburg. It cost 1.2 billion euros to build, and this is the "killer" application it is supposed to prove feasible.

RDC (residual dipolar coupling) measurements with NMR, a very fancy method where the interesting compound is trapped in an oriented (sort of "stretched") gel. If the molecules of the compound have a preferred orientation in that gel (they usually do, because very few biomolecules are spherical), you observe dipolar splittings in the spectra, which would normally be averaged out due to the free tumbling of the molecules. These couplings can be used to calculate relations between distances and angles between the magnetic field and the direct connecting line between the atoms involved.

No 2 of course relies heavily on additional input from conventional (spectroscopic) methods and computer simulations.


How to form 6-31G basis set; specific instructions

I believe you have not correctly interpreted the first basis set, which should read:

O1 = 0.061 f(172) + 0.358 f(25) + 0.7 f(5.5)
O2 = -.39 f(3.66) + 1.21 f(.77)
O3 = 1 f(.195)

Note that these are only the s-orbitals. Using the coefficients in the right-most column, two triplets of p-orbitals are also present. By doing it this way for the second basis set, which is 6-31G, you would obtain 3 s-orbitals and two triplets of p-orbitals, making it nine AOs in total. Note that all orbitals in a tuple must have the same contraction scheme, otherwise, the calculation would not be rotationally invariant (to within integration accuracy).

Aside: Why does this basis specify s- and p-orbitals this way?
Depending on the integral scheme, one can reuse certain intermediates of some integrals for others and save some computational time. However, modern programs tend to work better with segmented basis sets (i.e. basis sets that do not allow for reuse, because the $\alpha$ exponents are all different). The results are also of higher quality. IMHO, the Pople basis sets are only of historical interest or for hobbyists with very limited resources.


Products of Reimer–Tiemann reaction of 4-methylphenol

The Reimer-Tiemann reaction is known as the ortho-formylation of phenols under strongly alkaline conditions.

It's not an addition of a carbene to a double bond, like the Simmons-Smith reaction, the pyrrole to pyridine ring enlargement or the Buchner expansion, which is catalyzed by transition metals.

As ron already pointed out, pH is one crucial requirement, having a phenol (and not just any other arene) is the other.

What we're actually facing here is the addition of a carbanion to a strongly electron-deficient carbene.

As again already described by ron, there is significant charge density at the 4-position of the phenolate, and the initial addition step might indeed happen here too.

The resulting intermediate can't rearomatize other than in a back reaction and that's what it will mainly do! Only a minor part will grab a proton and yield C.

That was close, but no cigar! Let's have a look at the addition at the 2-position.

Through the interplay of dienone, phenolate and phenol we get rid of one chloride and all further steps to the final benzaldehyde are likewise assisted by this tautomerism.

We still haven't answered the question why no 1,3-dialdehyde is formed. Let's have a look at the major product B under alkaline conditions and its resonance structures.

One of them looks like the enolate of a 1,3-dicarbonyl conpound, with additional charge distribution over the $\pi$ system.

This species has a significantly lower localized electron density than the anion of the starting material. So, the party stops here.

Edit In a comment, it was asked whether the geminal dichloride will convert to the corresponding aldehyde under alkaline conditions.

Such a transformation could involve a nucleophilic substitution, leading to a geminal halohydrine which then eliminates HCl to furnish the aldehyde.

In the case of benzylic dihalides, this is possible even without the intramolecular assistance of a phenolate (cf. this example) using secondary amines as a base and in the case of diphenyldichloromethane, the transformation to benzophenone is achieved by heating in the presence of water.

In the case of C, the simple deprotonation of the dihalide might be a dead-end alternative pathway to substitution.


How to calculate the probability of a particular value of an observable being measured

$$
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}\rvert\,}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
\newcommand{\op}[1]{\hat{#1}}
$$

Solution should be a pretty straightforward application of the Born rule.

In this particular case, we have a self-adjoint operator $\op{L_z}$ with eigenfunctions and eigenvalues satisfying
$$
  \op{L_z} \psi_l(\theta) = l \hbar \psi_l(\theta) \, ,
  \quad \text{where} \quad
  \psi_l(\theta) = \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{\mathrm{i} l \theta}
  \quad \text{and} \quad
  l = 0, \pm 1, \pm 2, \dotsc
$$
such that any wave function $\psi$ can be expanded over orthonormal basis $\{ \psi_l(\theta) \}_{l=-\infty}^{\infty}$ of the corresponding eigenfunctions of $\op{L_z}$ as follows,
$$
  \psi(\theta)
  =
  \sum\limits_{l=-\infty}^{\infty} c_l \psi_l(\theta) \, ,
  \quad \text{where} \quad
  c_l
  =
  \braket{\psi_l}{\psi}
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta \, .
$$
And a measurement on a system in arbitrary state $\psi$ can yield any of eigenvalues $l \hbar$ with the probability given as follows
$$
  \Pr(l \hbar) = |c_l|^{2} = c_l^* c_l \, .
$$

Now, since $\psi(\theta)$ is defined to be
$$
  \psi(\theta)
  =
  \begin{cases}
  \frac{1}{\sqrt{\pi}} & \text{if} \quad \theta \in [0, \pi] \\
  0            & \text{if} \quad \theta \in [\pi, 2 \pi]
  \end{cases} \, ,
$$
the coefficient $c_l$ is equal to
$$
  c_l
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi} \psi_l^*(\theta) 1/\sqrt{\pi} \mathrm{d} \theta
  +
  \int\limits_{\pi}^{2 \pi} \psi_l^*(\theta) 0 \mathrm{d} \theta \, ,
$$
where the second term trivially vanishes leading to
$$
  c_l
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-\mathrm{i} l \theta} \frac{1}{\sqrt{\pi}}
  \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2} \pi} \mathrm{e}^{-\mathrm{i} l \theta}
  \mathrm{d} \theta \, .
$$
And then some simple algebra leads to the final answer, which I leave to the OP, since it is far into the night for me and I'm afraid I will make a lot of stupid mistakes.


This is a question has been one of interest since 1850, and I found a paper that nicely summarizes some of the hypotheses that you yourself listed, as well as what is considered to be true. I've paraphrased it below.

  Michael Faraday was the first to propose that there exists a film of water on ice that freezes when in contact with other pieces of ice, but remains liquid by itself. Though he performed several experiments that supported his claims, most of which involved sticking two pieces of ice together, his ideas were largely dismissed until C. Gurney also suggested that this film of water on play a role in ice's slipperiness.

  Gurney hypothesized that molecules, inherently unstable at the surface due to the lack of other molecules above them, migrate into the bulk of the solid until the surface becomes unstable, which prompts the formation of the liquid phase. This came to be accepted, and Charles Hosler compiled very compelling evidence on the the forces required to pull ice spheres apart. He discovered that at vapor pressures below ice saturation, no adhesion occurred below $-4\:^{\circ} \mathrm{C}$. From this they inferred that the expected roughness of the surfaces was removed by the presence of a liquid film whose thickness was sufficient to provide a smooth surface of contact.

  More experiments were performed trying to determine the exact thickness of the water film at different temperatures, and the phenomenon was found to occur at temperatures even below $-22\:^{\circ} \mathrm{C}$.

  Here is a depiction of what the surface of ice looks like at a molecular level.

Rosenberg, R. Why Is Ice Slippery? Phys. Today Physics Today. 2005, 58, 50–54.


A picture is worth a thousand words (see the two tetrahedra heights making up the vertical c-axis?):

The HCP packing has layers of closest-packed spheres in an ABAB... pattern. The distance between layers is one height of a tetrahedron, and you need two layers to get a repeating structure along c, i.e. the unit cell parameter.
(picture found at https://chemistry.stackexchange.com/a/111921)


It would be wrong to say cyclotriene, rather cyclohexatriene would be an appropriate name. And it would also be a highly systematic IUPAC name for said compound, following the rules of e.g. the cyclopentadienyl anion, the cycloheptatrienyl cation, cyclooctatetrene and more.‡

Let’s look at benzene’s history. It was discovered as early as 1825 by Michael Faraday who called it bicarburet of hydrogen. In 1833, Eilhard Mitscherlich isolated it and named it benzin. In 1836 it was isolated by Auguste Laurent who called it phène. The name benzene also dates back to the first half of the 19th century and, like benzin, derives from the gum benzoin. Note that this already predates the founding of IUPAC by about 80 years. The first IUPAC nomenclature was not established until 1921.

At the time when it was discovered, chemists had no clue as to what structure it had. It wasn’t until 1865 when Friedrich Kekulé published his ground-breaking paper[1] on the structure of benzene which in the end turned out to be correct (-ish). At that time, nomenclature of chemicals was still the Wild West, since there was hardly any way for structure elucidation and thus most chemicals acquired trivial names — some of which survive up to date (acetic acid being a prime example).

When IUPAC finally derived comprehensive nomenclatures for organic (and inorganic) compounds, and when structure elucidation was finally good enough to prove structures, the names benzene, benzin and possibly also phène had been in use for many decades, much like many other trivial names. Had IUPAC decided to obsolete them all altogether, they could have just as well trashed their own recommendations, because nobody would have followed them — even today, many chemists are too traditional to take up newer nomenclature schemes even if they have been around for two decades.

Probably to help gain acceptance, IUPAC has always allowed a certain set of retained names to persist under official nomenclature. While the list has gotten smaller over the years, it is extremely unlikely that benzene, phenol, phenyl etc. will ever get removed from it, since calling those compounds that way is so prevalent — and short.

Similar things happened to carboxylic acid. Hardly anybody says methanoic acid or butanoic diacid in place of formic acid and succinic acid. And sometimes, trivial names even become part of the official nomenclature, as was the case with fullerene because it provided a much shorter way to label a certain class of compounds.

So the short answer to your question: IUPAC has retained a common name. The reasons are tradition, prevalence and simplicity.

Note:

‡: Yes, as a chemist I know that the structure of benzene is incorrectly represented by drawing three pure double bonds and three pure single bonds, that in reality the molecule has a very high $D_\mathrm{6h}$ symmetry rendering all bonds equal and the electrons in the π system are delocalised all across the molecule. However, that is not what the question is asking about, and not something IUPAC nomenclature cares about.

Reference:

[1]: F. A. Kekulé Bull. Soc. Chim. Paris 1865, 3, 98.


How much sulphuric acid can be obtained from 5 kg of sulfur

$\ce{S6}$, $\ce{S8}$, $\ce{S12}$ – does it make a difference or is it just a trick to make the question more complicated than it is?

What you know for sure is:

$M(\ce{S}) = 32.065\ \mathrm{g}\cdot \mathrm{mol}^{-1}$
$M(\ce{H2SO4}) = 98.079\ \mathrm{g}\cdot \mathrm{mol}^{-1}$

$\frac{5000\,\mathrm{g}}{32.065\,\mathrm{g}\cdot \mathrm{mol}^{-1}} = 155.93\,\mathrm{mol}$

$155.93\,\mathrm{mol} \cdot 98.097\,\mathrm{g}\cdot \mathrm{mol}^{-1} = 15.29\,\mathrm{kg}$, which is close enough to the answer.


How can I identify and neutralize the chemical leaching out of new nitrile gloves?

Pack a glove into a glass container, let it sit for a few hours or days, get the container to your institute's mass spectroscopy and gas chromatography departments, and have them have a go at the gas phase in the container. I think that might be your best bet at finding out what is leaking out of these gloves. An IR or UV/Vis of the gas phase might also yield results, as Spießbürger suggested above.

Actually, if you got an analytics department at your university/institute, why not go to them, they might be able to have someone do a B.Sc. or M.Sc. thesis on that. Even if you end up trashing the gloves, at least then it won't have been a complete waste.


Why does the potential difference of a current drop, when silver nitrate is added to deionised water?

The potential of an electrode in the absence or paucity of redox reactions at the interface is not predictable using the Nernst equation. This situation is known as an unpoised electrode and the electrode's measured potential is related to the formation of an electrical double layer at the interface, which behaves very much like a capacitor.

If you try to measure the potential of an electrode that won't easily oxidize or reduce in deionized water, due to the very low concentration of ions, establishing the double layer is slow and easily disrupted. You can think of it as having a large solution resistance in series with the double layer capacitance. When there are few ions, the resistance is very high. On that same note, how good the voltage measurement instrument is becomes important. Any current that a voltmeter draws will induce a large voltage at the electrode (as there is a large resistance to bringing ions to the interface, excess charge at the electrode is not easily balanced)—a voltmeter with a 10 nA input bias current may induce 100–200 mV of error in a very resistive solution, so it takes very good instruments to deal with these types of measurements.

This error is proportional to the solution resistance and even adding a tiny amount of ions to deionized water will dramatically reduce the resistance. I'm guessing that repeating the experiment with a supporting electrolyte will prevent this effect.

The other trouble here is that neither of your electrodes can be assumed to maintain a fixed potential in your solution. The potential of the silver electrode should vary according to the Nernst equation once you have an appreciable quantity of silver ions in solution, but the graphite electrode is not in equilibrium with any components of your solution and its potential is free to fluctuate as the solution composition changes. Without a supporting electrolyte, adding the silver nitrate may have a large effect on the potential of the graphite electrode.


Reduction of a ketone in the presence of an aldehyde

The best strategy is probably to chemoselectively protect the aldehyde before the borohydride reduction.  Protecting groups are derivatives of certain moieties in a molecule that are far less reactive under most conditions, but which can be easily removed under very specific conditions to regenerate the original moiety.

For carbonyl compounds -- this includes both aldehydes and ketones -- the most common protecting groups are all introduced by ketalization/acetalization.   Common examples are dimethyl acetals, 1,3-dioxolanes, and 1,3-dioxanes.

However, in general, aldehydes are far more prone to acetalization than ketones are to ketalization.  That offers a means to chemoselectively protect aldehydes.  Acetals such as 1,3-dioxanes are stable to borohydride reductions, but, as you note in your question, ketones are sensitive.

Poking around reveals several potential strategies for selective chemoprotection of an aldehyde.  Here are two examples.

Tetrabutylammonium Tribromide (TBATB) as An Efficient Generator of HBr for an Efficient Chemoselective Reagent for Acetalization of Carbonyl Compounds:

  This convenient, mild, chemoselective method allows acetalization of an aldehyde in the presence of ketone...

Chemoselective Protection of Aldehydes in the Presence of Ketones Using RuPVP Complex as a Heterogeneous Catalyst

Putting it all together, a recipe for selective borohydride reduction of a ketone in the presence of an aldehyde would be something like:

Protect the aldehyde.  The best choice is probably as a 1,3-dioxane acetal by reacting your compound with 4 equivalents of 1,3-propanediol, 1 equivalent of triethyl orthoformate, and 0.01 equivalent of tetrabutylammonium tribromide.  Under suitable conditions, the ratio of protected aldehyde (desired) to protected ketone (undesired) should be >50.
Borohydride reduction of the aldehyde-protected compound.  Only the unprotected ketone will react.
Deprotection.  Acetals can be removed in acidic conditions.  A number of catalysts that will effect deprotection without requiring acid have also been developed.


At what temperature does tautomerization stop?

Tautomerisation, as an observable process by NMR, is entirely system dependent, and there are many cases where non-exchanging or very slow-exchanging systems are observed at room temperature and higher. As always, equilibrium is both sterically and electronically controlled. Some systems require very low temperatures (liquid nitrogen and liquid helium temperatures) to slow the exchange processes, and of course, going to higher magnetic fields will also improve the case of observing individual exchanging species.

Acetaldehyde is a very small molecule; this would require extremely low temperatures; as you have said 11K allows for observation without interconversion.

A lot of this sounds like a homework question .... you might want to be more specific about what your own research is contributing towards answering your own questions.


Determine the equilibrium constant for this reaction

You made a slight error in calculations for B and C. While those concentrations are correct, that is 1 M at 15 mL. When doing RICE tables and equilibrium calculations, we can use concentrations, however I prefer to convert concentration into moles.

We are trying to find $K_c$, and we know that equal moles of the reactant react to give us 0.0045 mol of BC.

$$\ce{B + C <=> BC}$$

\begin{array} {|c|c|c|c|}
\hline
\text{Initial conc.} & \text{0.015 mol} & \text{0.015 mol} & \text{0 mol}\\
\hline
\text{Change conc.} & -x & -x & +x\\
\hline
\text{End conc.} & \text{0.015 mol} - x & \text{0.015 mol} - x &  0 + x \\
\hline
\end{array}

We know that our change in equilibrium is 0.00450 mol, as that is the end amount of product that we have in our equilibrium calculations.

Accounting for the concentrations, we divide our number of moles by the total volume to get mol/L. 0.5 mol/L, 0.5 mol/L, .15 mol/L.

Assuming that our reaction is at equilibrium:

$$K_c = \frac{[\text{0} + \text{.15 mol/L}]}{[\text{0.5 mol/L} - \text{.15 mol/L}][\text{0.5 mol/L} - \text{.15 mol/L}]} $$

$$K_c = \frac{[\text{.15 mol/L}]}{\text{[0.35 mol/L]}^2} $$

$$ K_c = 1.22 $$

Thanks @Nicolau for pointing out the error in the question and reviewing my work.


V, you are mixing terms a little.

Vapor pressure is a quantity indirectly describing the amount of a substance must be in the gas phase for the substance to be in equilibrium between the liquid and gaseous states.  If the partial pressure of the gas (say water vapor) is equal to the vapor pressure of the liquid (must be the same substance, so water) than the liquid will have no net flow either from the gas phase to the liquid phase or vice versa.

There is no need for the substances to have the same density. This is only true for super-critical fluids.  Super-critical fluids exist once you have raised the temperature above a critical point after which there is no phase transition between liquid or gas.


Role of calcium oxide in soda lime decarboxylation

Calcium oxide is not necessary for carrying out this reaction. It only prevents NaOH from touching the glass of the container where the reaction is done. If a solid mixture of sodium acetate and NaOH is heated to high temperature, some NaOH will attack the glass before reacting with acetate, and make the glass melt down. If NaOH is intimately mixed with CaO, this trouble is avoided. The glassware is saved.


Would the End Point and Equivalence Point be Labelled the Same?

The endpoint and equivalence point are different.

According to this Wikipedia article , "The endpoint (related to, but not the same as the equivalence point) refers to the point at which the indicator changes colour in a colourimetric titration."

Essentially, the endpoint is when the titration visually ends in an experimental setting, but the equivalence point is when the titration actually ends, or when the moles of acid are equilalent to the moles of base.

Since you didn't use an indicator, you didn't have an endpoint.

Also, if you did use an indicator, you wouldn't really know what pH the titration ended at because you don't have a pH meter to measure the pH with.

Unless you had both a pH meter AND indicator (which seems kind of useless), you shouldn't have to label the endpoint on any titration curve.


What is the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate?

Plain hydrolysis would just replace trifluorexthoxide with hydroxide.

But you might want to consider that your staring material in equilibrium with another tautomer:

Deprotonation of the $\ce{OH}$ in the right structure, followed by an intramolecular reaction (transesterification/lactonization) in which $\ce{CF3CH2O-}$ is kicked out should give rise to an oxazolone.


Can I use fluorescence microscopy with a solution of hydrogen peroxide?

The Hydrogen peroxide will definitively accelerate bleaching of your fluorescent microspheres. In fact, we routinely add glucose oxidase, glucose and catalase to our single molecule fluorescence experiments in order to get rid of oxygen and hydrogen peroxide in our solution and thus reduce photobleaching. However, Fluorescence microspheres are very photostable and whether or not they work for your application depends on several factors:

how long the microspheres have been in the solution before you start imaging (the sooner you start the better, because of less reaction time between dye and hydrogen peroxide).
How long you need to image (the shorter the better, because of less photobleaching)
How large the microspheres can be (the larger the better because large microsphere contain more dye that is less accessible to the hydrogen peroxide in solution).

Basically, I suggest you try (and let us know how it went :) )


Do there exist metals that are composed of molecules?

According to A Conducting Crystal Based on A Single-Component Paramagnetic Molecule,
[Cu(dmdt)2] (dmdt ) Dimethyltetrathiafulvalenedithiolate) J. Am. Chem. Soc., 2002, 124 (34), pp 10002–10003 :

  Single-component molecular metals should greatly extend the
  development of new types of molecular conductors. For example,
  while the first metallic molecule-based magnet has been reported
  recently, [reference 8] it may be possible to create such metallic molecular
  magnets from single-component molecules containing paramagnetic
  ions such as Cu2+ and Co2+.

Reference 8 is Coexistence of ferromagnetism and metallic conductivity in a molecule-based layered compound Nature 408, 447-449

  The temperature dependence of the resistance exhibits a metallic behaviour over the entire temperature range that we investigated (down to 2 K).


Thermal Decomposition of Sulfur Hexafluoride

From an IEEE paper1:

$\ce{SF6}$ begins to decompose at $\pu{300 °C}$ and the main decomposition components contain $\ce{CO2 , SO2F2 , SOF2 , SOF4 , SO2}$ and $\ce{H2S}$; [...] $\ce{H2S}$ is the special component only appears when thermal fault proceeds to some degree (above $\pu{360°C}$).

For more information about the decomposition mechanism and product yiels, see here.
References

F. Zeng et al., "Decomposition characteristics of SF6 under thermal fault for temperatures below 400°C," in IEEE Transactions on Dielectrics and Electrical Insulation, vol. 21, no. 3, pp. 995-1004, June 2014, doi: 10.1109/TDEI.2014.6832242.


How are non-Stoichiometric compounds determined?

You can measure non-stoichiometry by determining how many atoms are present.  There are a number of ways to do this.

You can dissolve your unknown material in an acid and use a reagent to precipitate away one element but not another.  By keeping careful track of the number of moles of reagent you used, you can know how many atoms were precipitated and thus determine the stoichiometry of the original sample.  Once upon a time, this approach was used to demonstrate the fact that some chemicals ARE stoichiometric!

Today, another commonly used method is to count the number of X-rays emitted by a sample in an electron microscope.  Each atom emits X-rays with different amounts of energy so you can determine not only what atoms are present, but how many relative to other atoms.

For example, you can measure iron sulfide which stoichiometrically should have one Fe atom for each S.  Fe emits X-rays with about 2.5 times the energy of S, so they are easy to tell apart.  You then place a known stoichiometric iron sulfide (for example, using chemistry as above) in the microscope and measure how many Fe and S X-rays you get.  Let's say, hypothetically, you get one X-ray from Fe for each X-ray from S (just to keep the math simple).  Then you place a non-stoichiometric iron sulfide in and suddenly you get about 0.9 Fe X-rays for each S.  Presto, you know you have about 9 Fe atoms for each 10 S.

Note: For simplicity sake, I left out all the intricacies of the experimental method which would have to be considered in a real measurement.  If you are interested you can google: Energy Dispersive Spectroscopy (EDS), wavelength dispersive spectroscopy (WDS), and wet chemical analysis.


Why are NaCl and NaBr not soluble, whereas NaI is soluble in acetone (in the Finkelstein reaction)?

The solubility of a salt in a solvent is for a good part determined by the difference of stability in the dissolved state (solvated ions) vs. the solid state (ions in crystal lattice). The different solubility of the sodium halides in acetone can be explained by the crystal structures.

With iodide, the solvate $\ce{NaI \cdot 3(CH3)2CO}$ can be formed, in which each $\ce{Na+}$ is coordinated by 6 acetone ligands via oxygen lone pairs, and the $\ce{I-}$ ions fill in the gaps between these octahedral units, being surrounded by the methyl groups of the acetones. $\ce{I-}$ has a larger ionic radius compared to $\ce{Cl-}$ and $\ce{Br-}$ and thus fills the gap completely, thereby minimizing empty space in the lattice and its distance to $\ce{Na+}$, which is, however, still larger than in solvent-free sodium iodide. This larger distance between the ions decreases the lattice energy, which in turn makes the compound readily soluble in acetone.

$\ce{Cl-}$ and $\ce{Br-}$ are too small to fit in these gaps, and as the cation-anion distance becomes too large, the only way to minimize the lattice energy is the formation of a solvent-free structure $\ce{NaX}$ (X = $\ce{Cl-}$, $\ce{Br-}$). When the lattice energy is larger than the energy released by solvation of the ions, namely that of $\ce{Na+}$ by complexation with acetone, the compound becomes insoluble and precipitates from the solution (source).


What does it actually mean by stability of an atom

First you need to understand what stability is. Normally, octet rule is based on noble gas configuration. Since, noble gases hardly react with other atoms, and they have 8 electrons as their shell configuration, thus we consider them "stable".

So, taking noble gas 8 electrons shell configuration, we tend to generalise all other atoms/elements to have 8 electron filled orbit to gain stability.

You must understand that scientist create models to explain natural phenomena. Rules are  designed to explain already existing phenomena.  Stability truly means "less energy" or inert behaviour where the atom neither gains nor lose electrons to other atoms/molecules. The lesser the energy of the atom/system, more the stability.

Thus your doubt on "does not explain why an atom is stable when it obeys those rules" is very vague. Think of inert gases, their configuration, their inert nature and you will understand.


What makes many water soluble fertilizers blue?

What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes?

I have some Miracle Gro on hand, the water soluble kind: The blue stuff is copper sulfate, or ${\rm CuSO}_{4}$. Here is the MSDS for Miracle Gro Water Soluble All Purpose Plant Food, which lists the chemicals in the mixture.

  Is this necessary or for aesthetic purposes?

It's necessary in these formulations for purposes detailed (in part) here and here. In short, it is included in these formulations because plants need copper for biosynthesis; and because it acts as a fungicide, presumably leading to a nicer ecosystem in one's lawn/garden.


Why doesn't the SeO2 oxidation of ketones stop at the hydroxyketone stage?

As noted in the cited paper by Sharpless and Gordon$^{[1]}$, the problem with the Corey$-$Schaefer mechanism is the existence of the selenous ester (7) which, due to its instability, would undergo hydrolysis prior to any elimination reaction.
A study by Qingjiang Li and Tochtrop$^{[2]}$ supports Sharpless and Gordon's counterargument. The former pair give a good overview of the hypotheses put forward up to that point (i.e., 2014).
$\hskip 1 in$

$\beta$-hydrogen transfer (Speyer mechanism)

Based on the logic of the proposed mechanism, we reasoned that replacement of $\beta$-hydrogens of the substrate with deuterium would result in a ﬁrst-order
isotope eﬀect in the reaction rate.

Consequently, a $\beta$-deuterated fatty acid was prepared ... and converted to triﬂuoromethyl ketone to react with selenium dioxide. The reaction rate was monitored by NMR and compared with that of the analogous fatty acid without deuterium.$^{[2]}$

Based on this experiment, Li and Tochtrop conclude$^{[2]}$,

No signiﬁcant diﬀerence in reaction rates was observed, which suggested that the removal of $\beta$-hydrogen to form a carbon−carbon double bond is not likely to be the rate-limiting step, and we therefore excluded the $\beta$-hydrogen transfer mechanism.

Verdict: not the best explanation.

Sharpless$-$Gordon

While solving the selenous ester controversy, Li and Tochtrop found other problems introduced by Sharpless and Gordon.

[A]dding acid should accelerate the Sharpless mechanism by accelerating the enol−ketone tautomerization, as the rate limiting step is electrophilic attack of selenium dioxide to the enol. Since no signiﬁcant increase of the reaction rate was observed when adding acetic acid to the reaction mixture, the Corey−Schaefer mechanism seemed more reasonable.$^{[2]}$

To additionally verify this result,

$1.2$ equivalents of $\ce{Ti(O^iPr)4}$, a strong Lewis acid to coordinate with the carbonyl group, was added to the triﬂuoromethyl ketone for $1\ \mathrm{h}$ prior to selenium dioxide addition. This experiment resulted in no reaction, which subsequently served as further evidence that the reaction initiates with the ketone instead of the enol.$^{[2]}$

The Sharpless$-$Gordon mechanism also fails to explain some products detected in Li and Tochtrop's work with carboxylic acids.$^{[2]}$
Verdict: not the best explanation.

Corey$-$Schaefer

Verdict: not the best explanation due to the selenous ester intermediate.$^{[1]\ [2]}$
New mechanism proposed by Li and Tochtrop
Li and Tochtrop came up with a fourth mechanism to account for experimental evidence. Here, the reaction initiates via the ketone. The further pathway is dictated by whether $1$,$4$-elimination is restricted.$^{[2]}$
Li and Tochtrop examplify their mechanism with a trifluoromethylketone that has varying radicals attached to the non-fluorinated $\alpha$-carbon.$^{[2]}$
$\hskip 1 in$

As shown in Scheme $5$, we propose that high valence selenium reacts with
triﬂuoromethyl ketone to form the enol ester $(\bf{17})$. The enol ester undergoes $1$,$4$-elimination directly instead of nucleophilic attack of the carbon−carbon double bond to form selenium ester.$^{[2]}$

[---] When $1$,$4$-elimination is restricted, nucleophilic attack at the $\alpha$-position occurs and the subsequent hydrolysis of selenium($\mathrm{II}$) ester $(\bf{18})$ aﬀords alcohol as the product instead of enone.$^{[2]}$

In the context of the Li$-$Tochtrop mechanism, we can thus answer OP's original question.

The $\alpha$-hydroxyketone (10) is not observed. Why is that so?

The intermediate (4) has a small hydrogen attached to the original $\alpha$-carbon. In other words, $\ce{R2 = H}$. The $1$,$4$-elimination is not restricted. Hence, nucleophilic attack of the $\ce{C=C}$-bond does not take place.
The mechanism by Li and Tochtrop is also useful because it explains, to an extent, the dependence of the end product upon the starting material. Solvent and $\mathrm{pH}$ effects still have to be considered, and the pathway that yields diketones is not covered.

The nature of the starting material seems to have a great influence on the course of the reaction but usually one of the two types of compounds can be specifically obtained by using the right solvent. For example, dioxane favors the formation of the dione whereas t-butanol favors the unsaturated carbonyl compound.$^{[3]}$

The amount of diketone (or even triketone) will further depend on $\mathrm{pH}$ and the quantity of $\ce{SeO2}$.$^{[2]\ [3]}$ Depending on the mixture (e.g., with $\ce{H2O2}$), carboxylic acids and $\alpha$-ketocarboxylic acids are also possible$^{[3]\ [4]}$ with various degrees of saturation$^{[2]}$.

Some information on how to eliminate sidereactions that lead to diketones can be found here:
0 Hans J. Reich, James M. Renga, Ieva L. Reich. 'Organoselenium chemistry. Conversion of ketones to enones by selenoxide syn elimination'. Journal of the American Chemical Society 1975, 97(19), 5434$-$5447. DOI: 10.1021/ja00852a019.

1 K. Barry Sharpless, Kenneth M. Gordon, 'Selenium Dioxide Oxidation of Ketones and Aldehydes. Evidence for the Intermediacy of 8-Ketoseleninic Acids,' Journal of the American Chemical Society 1976, 98(1), 300$-$301. DOI: 10.1021/ja00417a083.
2 Qingjiang Li, Gregory P. Tochtrop, 'New Methodology toward $\alpha,\beta$-Unsaturated Carboxylic Acids from Saturated Acids,' Organic Letters 2014, 16(5), 1382$-$1385. DOI: 10.1021/ol500127u.
3 Alain Krief, Laszlo Hevesi (1988), Organoselenium Chemistry I. Functional Group Transformations, pp 115$-$180. ISBN: 978-3-642-73243-0 (print) 978-3-642-73241-6 (online).
4 Thomas Wirth (2012), Organoselenium Chemistry: Synthesis and Reactions, p 212. DOI: 10.1002/9783527641949.


What's the reason that alkanolamines change their color at storage?

Oxidative degradation by oxygen certainly occurs and has been studied 1, 2. I can't access the papers so I have no information on possible mechanisms. I've seen no mention of UV degradation, but that doesn't mean it does not happen.


How to come up with overall equation of a galvanic cell

Since $ \ce{Zn} $ never touches any $ \ce{Cu^2+} $ ions in a galvanic cell - the salt bridge/porous plate is preventing this - it would not be helpful to take this chemical equation as a starting point. As the aim of such cells is to separate the reduction and oxidation process, this is where the equations should originate.

Direction of the Reaction

To start with, one should find out which of the 2 electrode materials is more noble, hence will be the oxidizing agent aka cathode material, and will get itself reduced during the process through receiving electrons from the anode over the external circuit. A higher standard electrode potential ($E^\circ$) correlates with higher nobility. Information on different $E^\circ$ values can be found in galvanic series (Wikipedia: Standard electrode potential (data page)). A lookup of $\ce{Cu}$ and $\ce{Zn}$ gives:

$$
\begin{array}{rclc}
\hline
\text{Oxidant} & & \text{Reductant} & E^\circ/\mathrm{V}\\
\hline
\ce{Zn^2+(aq) + 2 e− & <=> & Zn(s)} & −0.7618\\
\ce{Cu^2+(aq) + 2 e− & <=> & Cu(s)} & +0.337\\
\hline
\end{array}
$$

Since $\ce{Cu}$ has the higher $E^\circ$ (is more noble), $\ce{Cu}$ will act as cathode, and $\ce{Zn}$ as anode, to provide the reactions spontaneity.

Setting up the Reaction

Looking at the electrochemical process, first the $ \ce{Zn} $ from the zink electrode gets oxidized according to:

$$ \text{Ox - Anode half equation:}\qquad\ce{Zn(s) -> Zn^2+(aq) + 2 e-} $$

After passing through the external circuit the electrons get to reduce the $ \ce{Cu^2+} $ ions according to:

$$ \text{Red - Cathode half equation:}\qquad\ce{Cu^2+(aq) + 2 e- -> Cu(s)} $$

Which shows, that the anode is going to lose mass through corrosion, while the cathode is going to gain mass through the formation of solid copper on its surface.

After having these two half equation we can sum them up and cancel out the $ \ce{2 e-} $ respectively to yield:

$$ \text{Redox - Total equation:}\qquad\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)} $$

While these processes happen, the salt bridge/porous plate is allowing the balancing of charges in both half cells.

Note: Of course someone could begin with the reduction, rather than with the oxidation, but it would anyway end in the same total equation, and it's harder then to imagine the electrons wandering through the external circuit.

Edit: As requested in the comments: The reactions with spectator ions:

$$\begin{align}
\text{Ox - Anode:}&& \ce{Zn(s) + SO4^{2-}(aq) &-> Zn^2+(aq) + SO4^{2-}(aq) + 2 e-}\\
\text{Red - Cathode:}&& \ce{Cu^2+(aq) + SO4^{2-}(aq) + 2 e- &-> Cu(s) + SO4^{2-}(aq)}\\
\text{Redox - Cell:}&& \ce{Cu^2+(aq) + Zn(s) + 2SO4^{2-}(aq) &-> Cu(s) + Zn^2+(aq) + 2SO4^{2-}(aq)}
\end{align}$$

Edit 2: Added Direction of the Reaction


What is the molecular mechanism behind oncotic pressure?

The molecular basis of osmotic pressure (oncotic pressure is a special type of osmotic pressure) is similar to filtration but on a very small scale.

Some membranes are partially permeable: they let small molecules pass but block larger ones. This asymmetry means that a flow of small molecules is created. Imagine things at the molecular level where a lot of individual motions are essentially random. The big molecules can't pass through the membrane but small molecules like water can pass through. If one side of the membrane consists of small molecules only and the other side has small and large molecules, then random motion will tend to lead to a net flow from of small molecules into the side containing large ones just because the concentration of small molecules is lower on the large molecule side. There are fewer opportunities for random small molecules to move from that side to the small molecule only side because there are fewer small molecules per unit volume on that side.

Obviously I'm simplifying a little, but this is the essential reason behind osmotic pressure.


Why do the halides run in reverse orders in the spectrochemical series and the trans-effect series?

The two features you cite — trans-effect and spectrochemical series, i.e. ligand-induced crystal field split — are based on two different phenomena, two different structures and two entirely different observations that warrant two different discussions. Thus, the probability for the two trends to be inverse is $0.5$.

Spectrochemical series

The spectrochemical series in itself is already a convolute of multiple different effects and it is defined for octahedral complexes. However, if we stick to small regions of it, we can make out trends based on general observations. Here, we are dealing with four very similar ligands which are all weak-field ligands, σ bases and π bases. To understand the effect of their π basicity and to relate them to each other, we need to take a look at the molecular orbital diagram of a typical $\ce{[ML6]}$ complex accounting for π effects. Figure 1 shows such an MO scheme.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.

We see that the $\mathrm p$ orbitals (or more precisely: that $\mathrm p$ group orbital with $\mathrm{t_{2g}}$ symmetry) interact with the metal’s $\mathrm{t_{2g}}$ orbitals in a bonding/antibonding manner of π symmetry — thus the interaction is weaker than the $\mathrm{e_g}$ interaction which features σ symmetry. In the case of all halido ligands, the ligand orbitals are fully populated meaning that the metal electrons end up in the $\mathrm{t_{2g}^*}$ orbitals which have been raised with respect to the neutral (or π-effect absent) state. This reduces the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ and thus the value of $\pu{10Dq}$ or the field split.

The different halides differ only in their electronegativity and thus in the relative positioning of their orbitals. The higher the electronegativity the lower the corresponding orbitals — i.e. fluoride will have the lowest-lying orbitals. Since these are all non-metals with relatively high ionisation energies (even for the anions that we are looking at here), they will all be below the metal orbitals. However, a larger interaction — and thereby a lower value of $\pu{10Dq}$ — is typically obtained if the interacting orbitals are close in energy.

Iodide is the one that is the least far away, meaning that the &#3c0; interaction is largest in an $\ce{[MI6]^n-}$ complex, meaning that $\pu{10Dq}$ is lowest meaning that it is on the very weak end of the spectrochemical series. The higher up we go through the group, the lower the halide’s $\mathrm p$ orbital energies will be, thus the weaker the interaction and the higher $\pu{10Dq}$. Naturally, the orbital energy decreases monotonously and therefore the ligand series of the halides will be as observed.

Trans-effect

The trans-effect is typically observed for complexes of platinum or related elements and is typically referred to for square planar geometries. It concerns ligand exchange which has been shown by a number of experiments to proceed via an associative mechanism:

$$\ce{[MX4]^n- + L -> [MX4L]^n- -> [MX3L]^${(n-1)-}$ + X-}\tag{1}$$

The initial and final complexes are square planar, the intermediate is trigonal bipyramidal. Going by the principle of least movement, the additional ligand $\ce L$ should approach via the distinguished axis (I’m calling it $x$) towards the inital complex and displace two ligands which will now simultaneously occupy the $xy$ plane (meaning that $z$ is the axis of the two non-moving ligands). Then, one of the three $xy$ ligands will get displaced; it has to be the one that is trans to the ligand which exercises a trans-effect and it has to be one of the two in the $xy$ plane — i.e. those that were originally on the $y$ axis.

Since the pentacoordinated transition state is obviously most important here, we should draw up its molecular orbital scheme. This is a scheme much less known than the octahedral scheme shown above so I will point out a few things. The trigonal bipyramidal complex’ point group is $D_\mathrm{3h}$. The metal’s $\mathrm d$ orbitals transform as $\mathrm{a_1' + e' + e''}$ with the orbitals within each irreducible representation being $\mathrm{d}_{z^2}, (\mathrm{d}_{xy}, \mathrm{d}_{x^2-y^2})$ and $(\mathrm{d}_{xz}, \mathrm{d}_{yz})$, respectively. The ligands’ σ-symmetric orbitals transform as $\mathrm{2a_1' + e' + a_2''}$. Broken down, the $z$ ligands are $\mathrm{a_1' + a_2''}$ and the $xy$ ones are $\mathrm{a_1' + e'}$. The π-symmetric $\mathrm p$ orbitals of the $z$-ligands should transform as $\mathrm{e' + e''}$ and the π-symmetric $\mathrm p$ orbitals of the ligands in the $xy$ plane are $\mathrm{a_2' + a_2'' + e' + e''}$. We can see a lot of $\mathrm e$ irreducibles which is where all the fun takes place. For completeness, the central metal’s $\mathrm s$ orbital transforms as $\mathrm{a_1'}$ and its $\mathrm p$ orbitals as $\mathrm{a_2'' + e'}$.

Figure 2: Qualitative molecular orbital scheme of a trigonal-bipyramidal $\ce{[ML5]}$ complex including both σ and π interactions.

The MO scheme is given in figure 2 above. The key to the analysis is to find out why certain ligands preferentially occupy the $xy$ plane but are not displaced themselves. The key lies within the orbitals that are labelled $\mathrm{e'}$: they bond in a σ fashion will all equatorial ligands in the intermediate being degenerate while in the reactant and product state they have different energies — one corresponding to a σ bond, one to a π bond. Thus, in moving from the reactant to the intermediate, we are generating a new interaction of σ symmetry which — again according to symmetry — affects all three equatorial ligands.

As I pointed out before, the lower a halide’s electronegativity the higher its orbitals energy levels and thus the greater the interaction. Therefore, by having the less electronegative halide participate in the new σ bond (i.e. by having it as part of the equatorial ligand sphere and thus by displacing the ligand trans with respect to it) will generate a more stable intermediate and will thus be a more favourable path.

Thus it is the same property of the halogens that create the two line-ups but the final effect is opposite in the two cases.


Why is salt needed when using vinegar to clean pennies?

The copper surface is oxidized, which slows down the formation of copper acetate. When you add chlorides to the solution, the chloride penetrates the oxide layer and creates a hole in that layer that allows acetate to react with the copper. To find more information on the effects of chlorides on the surface of copper look up copper corrosion and chlorides.


What chemical explosives detonate when exposed to alpha particles or nuclear fission products?

(not really an answer here, but anyway ...)
I would suspect any highly sensitive explosive is theoretically capable to do that, possibly depending on how close you are to the point where it would just thermally explode. They checked  "Crystals of lead azide, silver azide, cadmium azide, silver acetylyde and nitrogen iodide" in the paper cited above. The four with heavy metal ions don't work, I guess it makes some sense that those can (better) absorb a very local energy surge like from an alpha particle.

My list of suspects would thus include organic ozonides, peroxides, azides, tetrazoles and other compounds that only contain CNOH. Number 1: solid $\ce{H2O2}$.

And basically anything that is known to sometimes explode without an obvious external triggering event.


One of the goals of the InChI project was to ensure uniqueness: [1]

Strict uniqueness of identifier
The same label always means the same substance, and the same substance always receives the same label (under the same labelling conditions). This is achieved through a well-defined procedure of obtaining canonical numbering of atoms.

Whilst this is often the case for InChI strings, there are some (complex) examples where the above goal has not been met, for example in the natural product Spongistatin (below) where two isomers incidentally have the same InChI key: [2]

InChI=1S/C63H95ClO21/c1-33(19-42(67)18-17-35(3)64)20-53-55(72)57-39(7)58(79-53)59(73)63(75)31-51(70)37(5)52(85-63)16-14-12-13-15-44-22-43(68)27-61(81-44)29-47(76-11)23-45(82-61)25-50(69)38(6)56(78-41(9)66)36(4)34(2)21-49-28-60(10,74)32-62(84-49)30-48(77-40(8)65)24-46(83-62)26-54(71)80-57/h13,15,17-18,36-39,42-49,51-53,55-59,67-68,70,72-75H,1-3,12,14,16,19-32H2,4-11H3/b15-13-,18-17+/t36-,37+,38+,39+,42+,43+,44+,45-,46+,47+,48-,49+,51-,52-,53-,55+,56+,57-,58+,59+,60+,61+,62+,63-/m0/s1
Although InChI can sometimes break (as per above) most of the issues with InChI strings are with implementation (how the structure is parsed to the thing generating the string in the first place). Several (common) features are at present un-supported by the InChI implementation:

Polymers
Complex organometallics
Markush structures
Mixtures
Conformers
Excited state and spin isomers
Local stereochemistry/chirality
Topological isomers
Cluster molecules
Polymorphs
Unspecific isotopic enrichment
Reactions

Generation of InChI strings
InChI is, by nature, an algorithm designed to be ran by a computer. Whilst strings can be parsed by humans (with ennough effort), the complexity of the strings is such that it is challenging to ensure they're correct. The InChI FAQ specifically deals with this:

You should not do so (though you of course can). This may give apparently reasonable answers but it is error-prone and may break relations in the InChI.

The most recent implementation of InChI is provided by the InChI trust, open source and free of charge.[3]
As alluded to above, a common source of InChI errors is the way in which the structure of interest is passed to the algorithm rather than with a fundamental flaw with the process used to generate it. Sh*t in, sh*t out, so to speak.
To give a concrete example, consider the heterocyclic system below:

Clearly, the two tautomeric forms cannot be distinguished chemically, but depending on how the InChI string is generated, they may end up having the same or different strings. In this case, we need to specify to the InChI algorithm whether we want to fix the hydrogens (to show a single tautomer, each of which would have a unique InChI string), or not fix them (such that both tautomers have the same InChI string)
Validation
The real question you're possibly interested in isn't How to determine the correct InChI for a certain compound?, but rather How to validate an InChI string for a certain compound?.
Given the complexity of the InChI strings, this is a challenging thing, and to my knowledge there is no tool which allows a string to be provided and says whether it is valid or not (similarly you can't provide a proposed IUPAC name to any tool which will say whether it is the IUPAC preferred name).
One thing you can do is use the InChI string to generate a structure (ChemDraw does this). Using your strings from the question, it's evident that they actually refer to different forms of the molecule (cyclic vs acyclic, just imagine the terminal primary alcohol attacking the ketone).

Chemically, it may be that the cyclic and acyclic forms are in equilibrium- in this case there is no way for InChI to represent the mixture (and hence which structure used to generate the string is ambiguous). It could also be that they are separable chemically and not interconverting, in which case quite rightly they should have different InChI strings.

[1]: Journal of Cheminformatics, 2015, 7, 23
[2]: http://www-jmg.ch.cam.ac.uk/data/inchi/ Accessed 3-Sept-2017
[3]: http://www.inchi-trust.org/about-the-inchi-standard/
Accessed 3-Sept-2017


In electrochemistry, the rate of electrolysis depends on rate of charge entering the cell. Second most important point is that one can either control potential or current but not both during electrolysis i.e. you cannot have both values set. If you are fixing potential at 2.5 V, the value of current is not in your hand. I don't think Ohm's law remains valid during electrolysis in general.

So the main question one should ask is: What rate of electrolysis do I need? For example, you may want 200 mL of $\ce{O2}$ in 60 min. From here you calculate the charge Q needed to collect this volume of oxygen. Current is charge per unit time (3600 s) and this is your required current above the decomposition voltage of current. See second figure: http://www1.lsbu.ac.uk/water/electrolysis.html

See the sample calculation here: https://www.quora.com/What-amount-of-current-is-required-to-produce-0-224-cc-STP-of-O_2-at-anode-during-the-electrolysis-of-water


How many cis-trans isomers does 1-O-stearoyl-2-O-arachidonyl-phosphatidyl-inositol-4',5'-bisphosphate have?

Perhaps the answer is correct. If by cis/trans isomers the question includes the four stereogenic double bonds of the arachadonic acid chain and the six stereogenic carbons of the inositol ring, then there is a total of ten or 210=1024 isomers. But the answer is 1023 stereoisomers. That is because the question says "cis-trans-isomers OF...", which means NOT including the one shown. [210-1=1023.] Had the question asked "How many stereoisomers OF ...", then the lone stereogenic center of the glycerol moiety would have to be included. Then the answer would be 211-1=2047.


Why is the dipole moment of ethyl chloride larger than that of vinyl chloride?

I see two explanations for why the dipole in vinyl chloride is smaller than ethyl chloride.

Consider a resonance structure where the chlorine donates electrons to the alkene. This puts a formal positive charge on chlorine:

This would be a very minor contributor given the electronegativities and the mismatch in orbital size. However, this is not possible in ethyl chloride at all.
The more likely explanation is that an sp2-hybridized carbon is more electronegative than an sp3-hybridized carbon. Since the electronegativity of chlorine will be closer to the electronegativity of the carbons in vinyl chloride than the carbons of ethyl chloride, the C–Cl bond will be less polarized in vinyl chloride, giving a smaller dipole.


This is a partial answer to your query.  According to the paper Degradation and Recyclability
of Poly (Ethylene Terephthalate)
 (Venkatachalam, 2012), there are a few degradation products, from the article:

  Apart from aldehyde, $\ce{CO}$, $\ce{CO2}$, ethylene, benzene, biphenyl are also identified as
  degradation products

(The article also shows diagrms of the chemical processes of PTE degradation).

an important aspect to note is experiments/observations have

  shown that rate of thermal degradation in the closed system is
  about three times higher than that in open system.

In regards to burning of PET, the products can vary according to the page EPB 433- Health and Environmental Effects of Burning Waste
Plastics
 (Saskatchewan Ministry of Environment) stating that

  the composition of byproducts of plastic combustion as to their type and concentration depends
  on the combustion temperature and the flame residence time.

According to the Peiyu Plastics page (a reference from Wikipedia), if complete combustion occurs, then the products would be $\ce{CO2}$ and $\ce{H2O}$, incomplete combustion could produce soot and carbon monoxide.

Further reference:

Here is the MSDS for PET

A study of thermal decomposition and combustion products
of disposable polyethylene terephthalate (PET) plastic using
the High Resolution Fourier Transform Infrared
Spectroscopy, Selected Ion flow Tube Mass Spectrometry
and Gas Chromatography Mass Spectrometry  (Sonova et al. 2008).


Heat Capacity temperature dependence and Gibbs energy

At constant pressure, $dG=-SdT$.  Also, the definition of G is $G=H-TS$, so $S=\frac{H-G}{T}$.  Substituting,
$$dG=\frac{G-H}{T}dT$$
This equation can be manipulated into the form:
$$\frac{d(G/T)}{dT}=-\frac{H}{T^2}$$
So,
$$\frac{d(\Delta G/T)}{dT}=-\frac{\Delta H}{T^2}$$
I leave the rest up to you.

A math error was corrected on 10/16/15


What are the predominant acids in sphagnum peat moss?

This 1992 article by Gagnon and Glime suggests that the ability of Sphagnum to lower pH depends on the presence of cations in the surrounding environment and is not (contrary to prevalent belief) due to organic acids in the plant itself. Here is a more recent article from New Phytologist amplifying that cation exchange is responsible for lowering of pH.

If you want to acidify soil using sphagnum/peat--assuming the authors are right--it's more important what cations you put into the soil. The sphagnum will then exchange them for $H^+,$ lowering the pH of the soil.

This does not mean there aren't organic acids in moss. You would have to look in specialized journals to determine what acids are found in the plant, but they are not necessarily very acidic and are, according to both articles, not the cause of acidity of the soil associated with the plants.


Replacement for Centrifuge in Calcium Content Determination

If your "source" claimed that a centrifuge increased the reaction rate, then you need to find a better source. (I admit that it is possible that the compressive work a centrifuge does causes heating which for high speed centrifuges can be significant, but it isn't likely to be a factor here with small analytical amounts at relatively low speeds, <300 g's, imho) Without some calibration, you really need to find a method you can follow closely. If you have some knowns, say some calcium phosphate and some dietary calcium supplements, then you may be able to validate any deviations from the given method by showing that both highly insoluble salts and highly bioabsorbable calcium (usually calcium carbonate, acidified with HCl to the Ca++ ion) both can be analyzed with the modified method. (I also would guess that the centrifuge wouldn't "push down" other suspended (colloidal) material in the milk (which might either also contain calcium or improve the precipitation rate or amount of the calcium oxalate).)


What alkali and alkaline earth metal oxides will turn moist red litmus blue and finally white?

The standard oxides $\ce{Li2O, MgO}$ and $\ce{SrO}$ are water soluble and hydrolyse to form hydroxides and a basic solution. $\ce{BeO}$ is insoluble in water.

$\ce{RbO2}$ is a superoxide which in aqueous solution undergoes disproportionation to $\ce{O2}$ and $\ce{OH-}$ so will give a blue litmus test.

The peroxides $\ce{Cs2O2}$ and $\ce{BaO2}$ react with water to give hydrogen peroxide and the hydroxides, so again a basic solution(source). The hydrogen peroxide produced will bleach the litmus paper.

$\ce{KO3}$(Potassium Ozonide) is highly unstable, decomposes in water to $\ce{O2}$ and $\ce{KOH}$

The oxides which will NOT decolorize litmus are $\ce{Li2O, MgO, BeO}$ and $\ce{SrO}$. The peroxides ($\ce{Cs2O2}$ and $\ce{BaO2}$), the superoxide($\ce{RbO2}$) and the ozonide($\ce{KO3}$) will decolorize litmus.


What is known about the phase of matter at very high temperature and enough pressure?

There is a point in which it is too hot for a molecule to remain bonded and will break apart, no matter how much pressure is applied. Now with more pressure the electrons and nuclei will start free flowing in what is known as a plasma. And add a little more heat and fission happens.

So basically the reason they don't show you any more of the curve because your molecule under enough pressure and heat is no longer a molecule.


Comparing the reactivity towards aromatic electrophilic substitution reaction

We can answer your question by first looking at each individual, unsubstituted position in the benzene ring and determining the relative reactivity at each of these positions. Next we can compare the molecules and see which one has the most activated positions.

I've carried out the first step of this exercise and show the results in the following figure.

Looking at the top molecule, m-dinitrobenzene, we see that 3 positions are deactivated by the resonance effects of both nitro groups (that is, these positions are either ortho or para to each nitro group).  One position is not deactivated since it is meta to both nitro groups. We can analyze the para-dinitro isomer in a similar manner.  In this case we find that each position is only deactivated by one of the nitro groups.  Clearly, a position that is doubly deactivated will react much slower than a position that is singly deactivated.  Now, if we try to estimate the relative molecular reactivities by summing up and averaging the reactivities at each position, we see that the meta isomer is roughly deactivated by an average of (0+2+2+2/4) 1.5 at each position, whereas the para isomer is roughly deactivated by an average of (1+1+1+1/4) 1.0 at each position.  Therefore our analysis suggests that the para isomer will be the least deactivated and consequently react faster.

In the dihydroxybenzene case, hydroxyl groups are ortho-para activating.  Applying the same type of analysis leads us to conclude that the various positions in the meta isomer will, on average, be more reactive than the positions in the para isomer.  Hence the meta isomer will react faster in an EAS reaction.

Similarly, in the cresol case both substituents are again ortho-para activating but with the hydroxyl group being more activating than the methyl substituent. In the meta isomer, 3 positions are doubly activated and one position is not activated (or deactivated).  In the para isomer, each position is singly activated, and only 2 of these positions are activated by the strongly activating hydroxyl group.  This would lead us to suspect that the meta isomer will react the fastest.


Migratory aptitude in Baeyer-Villiger reaction

The migratory aptitude list for the Baeyer-Villager oxidation is as it is because that is how the molecules undergoing the reaction behave. When we are learning about new reactions from a textbook, we often read about the mechanism and theoretical explanations of chemo-, regio-, and stereoselectivity. It is easy to forget that the experimental observations came first.
The migratory aptitude list $(\ce{H>R3C>R2HC}\approx\ce{Ph(>H)>RCH2>Me})$ is the result of many experiments. Nor is it a linear list as is usually presented. $\ce{H}$ migrates more readily than tertiary groups, which migrate more readily than secondary groups, which migrate more readily than phenyl groups (which migrate more readily than $\ce{H}$).
The mechanism and theoretical explanations came later. The partially complete mechanism from the Wikipedia page is shown below. The key step (center) is a concerted collapse of the tetrahedral intermediate with a [1,2]-sigmatropic shift of $\ce{R}^1$ and ejection of the carboxylate leaving group. Note that this mechanism shows none of the proton transfer steps.

Since the whole thing is supposedly concerted, I can understand your objection to the build-up of $\delta^+$ on $\ce{R}^1$. However, with the exceptions of the anomalous behavior of $\ce{H}$ and $\ce{Ph}$, the migratory aptitude list is the same as the carbocation stability list. Thus, researchers looked for a link between the two phenomena and a way to relate one to the other. The link may not be real, but the trends are there. To a student learning organic chemistry, it is much friendlier at first to apply an existing trend in reactivity to a new reaction (with an exception), than to learn a whole new trend that is unrelated but cosmetically similar.


I have encountered the same problem with plant extracts. I think the term is hydro/sol when the extract is suspended in aqueous solution. Anyway, I've had some success with heavy sodium chloride brine then chilling the solution to near freezing for an extended time (a week or so). Even then the return hasn't been that impressive. In most cases making a tincture (or gas)extraction then distilling the product (if you still need to) produces a much better yield.


Aromaticity of two rings connected by double bond

Actually, the breakage of double bond doesn't happen in the way you are thinking it does. Let me illustrate the phenomena what happens in such rings connected by double bonds.

If you have two rings connected by double bonds where the breakage of that will produce anti aromatic rings, the molecule itself becomes very inert towards that breakage. The molecule always tries to gain stability through any perturbations to it. That's why production of anti-aromatic rings becomes extremely difficult in those cases.

In your given case, the molecule becomes highly polar due to unsymmetrical distribution of electron cloud (i.e. more negative charge on cyclopentadienyl ring and more positive charge on the other ring). But in a case, where you are assuming that breakage of double bond gives anti-aromatic cyclopropenyl anion or cyclopentadienyl cation, that breakage becomes extremely impossible. Even if you would subject it to extreme conditions, the molecule will never go into that transition state as that will have immensely high energy. Instead it will always shift towards that transition state by which it can gain some stability, by forming non aromatic or aromatic rings.

If it can't stabilize by that also, it will then become almost inert towards any electrophile or any external reagent which tries to make it anti-aromatic. Even if you perform ozonolysis, it will remain as a perfect keto-group, no unsymmetrical distribution will be observed for these classes of rings.

Considering the number of π-electrons in continuous conjugation of the whole molecule, we can judge if the molecule is aromatic or anti-aromatic. But if the molecules breakage of one double bond gives it anti-aromaticity, it will try to stay in its own form, whatever it may be.


Calculating heat capacity of a calorimeter

First, is this a question from your book, and do you have the correct answer to check your work.

Do you know how to use dimensional analysis?

Start with 50 mL. You have 2 Molarity of some subtance, how do you write that as a quotient and what are the units of the denominator and numerator? You will also need to know how many mL are in a liter and you should get the correct moles of that substance. It is very important that you write this down on paper correctly and make sure your units will cancel leaving only moles. It is hard to be wrong when your units cancel correctly.

Do you mean the heat of neutralization is $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

The more information you give us, the more help you will get.


You are not going to dissolve 304 and leave the concrete. Use sandpaper to remove the surface rust as needed : Or mechanically ( cut/grind) the 304 to remove it.


An alternative basis set for analytical computation of two electron integrals

A few things come to mind: 1) my understanding is that (ab|cd) in a gaussian basis is exactly solved for arbitrary "angular momenta" using recurrence relations such as the Obara-Saika method. The O-S method also has ways of generating the requisite derivatives with respect to coefficients, without which it is difficult to minimize the energy or do anything practical.

2) On the question of alternative basis sets to the Gaussian, one simply needs a complete basis set (of which there are many), helps to be orthogonal (almost as many), has to describe the boundary conditions of vanishing at infinity or being periodic with unit cell (fewer), hopefully is cusp-like near the nuclei (very few), and integration has to be tractable without resorting to quadrature, since there are four indices (each with 3 dimensions) but also millions of these integrals to compute -- and that leaves you with the Gaussian.


Is there Any Advantage to Using Multiple Steel Plates in an Electrolyzer?

As the reactions in electrolysis take place on the electrodes surface, it can therefore be concluded, that the more electrode surface (contact surface) exists, the more water molecules can be electrolyzed in one instance of time. Therefore, more steelplates result in more surface, which increases the productivity of the electrolyzer.


What are the products of the hydrolysis of ICl3?

$$\ce{3ICl3 + 6H2O → 9HCl + HI + 2HIO3}$$

  Iodine(III) chloride react with water to produce hydrogen chloride,
  hydrogen iodide and iodic acid. The reaction is called Hydrolysis of
  iodine trichloride.(source)

The reaction is same as @xavier_fakerat noted in the comment section.


Assuming monochlorination of 2-Methyl Butane ,the following products are possible as shown in the figure below.


Am I limited by the pKa value of an weak acid when preparing a buffer solution of it (and its salt)?

If I was to make an acidic buffer system with the ph of, for instance, 4, would I be limited with my choice of weak acids?

Yes you are limited to weak acids that have a pKa of +/-1 of your desired buffers pH value. This is a rule of thumb used in many labs.Some theroetical background is quite nicely explained on wikipedia wikipedia#buffering  capcacity.

This reference also includes some useful equations related to buffering capacity.


Are there any safety guidelines for mixing sulfate with chloride?

Overall, always wear disposable gloves, goggles, and a dust mask.  Add a lab coat, smock, or apron so as not to contaminate your clothing.  Avoid skin contact, don't ingest and don't inhale (this includes the dust).

Manganese heptoxide is a special case with concentrated sulfuric acid.  Avoid mixing strong oxidizers (permanganate, chromate/dichromate, perchlorate) with organics.  NEVER screw around with chlorate salts.  They are powerful oxidizers with very low kinetic barriers.  Finely divided metals (especially titanium) can go to white-hot burning in open air.  Titanium burns under air, carbon dioxide, nitrogen, water...

Palladium is a heavy metal, poison and tissue corrodant.  A microgram of tellurium by any contact gives "tellurium breath"

Local contact with chromates can cause ulceration.  Neodymium gives you purples, cerium yellow to orange.  I seem to remember that cobalt blue (cobalt aluminate after firing) in tin oxide applied to white ware gives bubblegum purple.

Tin Glazing

Gold via the Purple of Cassius is very nice.

Here's a pretty!  Suppose you mixed clay with hollow glass microbubbles, then fired.  Would the clay foam to an interesting surface or whole body?  This is the stuff, sold generally as lightweight resin filler, including hobby stores.  Most prices are outrageous when sold by volume.  Density of 3M Scotchlite K20 is 0.2 g/cm^3.  Goggles and dust mask.  Be careful or the stuff goes everywhere.

http://www.aircraftspruce.com/catalog/cmpages/bubbles.php
Good price.
http://solutions.3m.com/wps/portal/3M/en_US/Energy-Advanced/Materials/Product/?N=5002440+7570097&rt=r3


What is the difference between hydrogen chloride and hydrochloric acid?

Hydrogen chloride is a gas. It is the molecule made up of a hydrogen and a chlorine atom bonded by a single bond. It comes in gas cylinders for use in chemistry.

Hydrochloric acid is a solution (no, not to your problem of a stubborn teacher). It is composed of water molecules and hydrogen chloride molecules. In terms of mass percentage, concentrated hydrochloric acid is often $37~\%\ \ce{HCl}$ in water. It comes in glass bottles (often brown glass for ease of bottle use; the chemical company does not have to distinguish between compounds that can live in clear bottles and those that must live in brown ones) for use in chemistry.

Hydrogen chloride can also be dissolved in other solvents such as ether. In this case too, it may be termed hydrochloric acid, although a specification of the solvent should be added.

Whether or not either of the two names is a misnomer depends on what you are addressing. Using hydrogen chloride for the liquid solution is as wrong as using hydrochloric acid for the gas.


Water molecule bent shape and electron configuration

According to VSEPR theory (there are other models, but they give the same qualitative results), since there are four electron pairs around oxygen they will be in a tetrahedral configuration. Assume the hydrogen atoms are at A and B as you've described them in your question; C and D (the lone pairs) are in a plane perpendicular to the one containing A and B. Below is a ball-and-stick model of water, showing the lone-pair electrons as yellow "atoms".
 (Public domain image from Wikimedia Commons)

It sounds as if you're thinking in two dimensions with the hydrogen atoms and lone pairs in the same plane. If water existed in only two dimensions it could conceivably be linear, but it really is in three dimensions. (The molecule itself is flat because the lone pairs affect the shape but are not counted as part of it.)

All of the above applies to water under normal conditions. I'm not aware of any excited or exotic states of water in which the VSEPR prediction doesn't apply, but I can't guarantee that there aren't any.


Does minimizing ANY type of energy ALWAYS predict a state of equilibrium?

When you find a minimum in the thermodynamic potential energy of a system, you will have found an equilibrium point.

The key is the thermodynamic potential part. Depending on the constraints of the system (i.e. which thermodynamic state variables are held constant) the thermodynamic potential energy function will be different.

For example, in a system of a fixed number of particles at constant temperature and pressure (we would call it NPT), the Gibbs free energy is the thermodynamic potential, and it will be minimized at equilibrium.

For NVT (constant volume and temperature) the Helmholtz free energy is the thermodynamic potential, and that is what will be minimized at equilibrium.

In a system consisting of a rock dropped from a height above the ground, gravitational potential energy is the thermodynamic potential (I am abusing the language a little bit here) and it will be minimized at equilibrium (when the rock hits the ground).


Use methylcellulose(MC) as a binder and Ammonium Polyacrylate (NHPA) as your dispersant or deflocculant. The concentration of methylcellulose can be set according to your requirement as its non-ionic (if you want liquid slurry use MC in large amount and if you don't use less amount). For NHPA you have to make different samples of fixed alumina concentration and NHPA of different concentrations. Check viscosity of each sample and the one which gives you least viscosity means the best-dispersed sample and hence your operating concentration of NHPA.


Why can't we calculate the atomic radii of noble gases?

Covalent radius is measured as the distance ($r_\text{cov} = d/2$) between the nuclei of two bonded atoms (covalent). But, if you try to do the same for noble gases/inert gases (good luck!), as they have fully filled $np$ orbitals, they will repel each other, hence the closest distance between the two atoms is taken (high pressure, low temperature) as the van der Waals radius.


Single Point Energy in Gaussian software

The energies you're seeing are those resulting from solving for the eigenfunctions of the Coulomb Hamiltonian under the Born-Oppenheimer approximation, where the Hamiltonian can be broken down into the following 5 components:

The most important thing to take away from these expressions is that they all contain summations over nuclei, electrons, or both; combined with the presence of the set of nuclear charges $\{Z\}$, there is a non-linear dependence of the molecular energy on more than just the nuclear masses.

This is one reason why it usually isn't instructive to compare absolute energies of unrelated molecules.


Why isn't iron(III) oxide an effective catalyst for decomposing hydrogen peroxide?

Ah but iron(III) does act as a catalyst in peroxide decomposition (YouTube video).

It can be improved by the addition of some copper.

Some hydrogen peroxide is mixed with inhibitors (radical scavengers or chelating agents) to suppress the decomposition.


Why is a strong base needed for an E2 reaction and not a an E1 reaction?

E2 is a concerted mechanism. The alpha proton in an E2 reaction substrate is also only weakly acidic because the bond between the alpha proton and the alpha carbon is relatively strong; there is only some inductive withdrawal of electron density from the C-H bond by the halogen.

Therefore, we need a strong base to rip the proton away from the alpha carbon.

E1 however involves two steps. In E1 the leaving group leaves, and then the base attacks.

When the leaving group leaves, a carbon lacking an octet is formed. This carbon is highly electronegative and will withdraw electron density. This makes any alpha proton more acidic as hydrogen is less electronegative than plain old carbon - much less a carbon deficient of an octet in a covalent framework.

In other words, electron withdrawal by the carbocation weakens the bond holding on to the alpha proton, and a weaker base suffices for taking away the proton.

Note that I don't particularly like the below picture because it doesn't show anything abstracting the proton. Still the general idea is correct; electron density from the alpha C-H bond flows toward the hypovalent carbon. Heck, we can draw valid, hyperconjugative resonance structures for this baby.


Solubility of benzoic acid(mw 122.12):

1.7 g/L (0 °C) - 0.00139 moles/100 ml
2.7 g/L (18 °C)
3.44 g/L (25 °C)
5.51 g/L (40 °C)
21.45 g/L (75 °C)
56.31 g/L (100 °C) - 0.04611 moles/100 ml

Solubility of sodium benzoate (mw 144.10):

62.69 g/100 mL (0 °C) - 0.4350 moles/100 ml
62.78 g/100 mL (15 °C)
62.87 g/100 mL (30 °C)
71.11 g/100 mL (100 °C) - 0.4935 moles/100 ml

So yes, sodium benzoate is more soluble than benzoic acid, but consider calcium benzoate which is less soluble.

Calcium benzoate (mw 282.31):

2.32 g/100 mL (0 °C) - 0.00822 moles/100 ml
2.72 g/100 mL (20 °C)
8.7 g/100 mL (100 °C) - 0.03081 moles/100 ml

So the "real" answer is - it depends...


Which is a better oxidising agent: concentrated nitric acid or dilute nitric acid?

I think your assignments for the equations need to be inspected closely. Let's look at those equation with their redox potentials:

In dilute conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 8:3):
$$
\begin{align}
\ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)} &E^\circ = \pu{0.957 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\
\ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.615 V}
\end{align}$$

In concentration conditions (say, ratio of $\ce{HNO3}$ to $\ce{Cu}$ is 4:1):
$$
\begin{align}
\ce{2NO3-(aq) + 4H3O+(aq) + 2e- &<=> N2O4(g) + 6H2O(l)} &&E^\circ = \pu{0.803 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &&E^\circ = \pu{-0.342 V}\\
\ce{2NO3-(aq) + 4H3O+(aq) + Cu (s) &-> Cu^2+(aq) + N2O4(g) + 6H2O(l)}&&E^\circ_{\text{rxn}} = \pu{0.461 V}
\end{align}$$

Because  this concentrated reaction is exotermic, the colorless $\ce{N2O4}$, gas decomposes to reddish brown $\ce{NO2}$ gas according to $\ce{N2O4 (g) <=> 2NO2 (g)},~\Delta H^\circ = \pu{+57.2 kJ mol^{-1}}$

Based on $E^\circ_{\text{rxn}}$ values, one can assume that the concentrated reaction is less spontaneous. However, some dissolved $\ce{N2O4}$ can still undergo following redox reaction which is the most spontaneous reaction among three:

$$
\begin{align}
\ce{N2O4(aq) + 4H3O+(aq) + 4e- &<=> 2NO(g) + 6H2O(l)} &E^\circ = \pu{1.035 V}\\
\ce{Cu(s) &<=> Cu^2+ (aq) + 2e-} &E^\circ = \pu{-0.342 V}\\
\ce{N2O4(aq) + 4H3O+(aq) + 2Cu(s) &-> 2Cu^2+(aq) + 2NO(g) + 6H2O(l)}&E^\circ_{\text{rxn}} = \pu{0.693 V}
\end{align}$$

If we can consider the loss of $\ce{N2O4}$ to $\ce{NO2}$ is negligible, then total oxidation of $\ce{Cu}$ with both conc. $\ce{HNO3}$ and $\ce{N2O4}$ can be written as:
$$
\begin{align}
\ce{2NO3-(aq) + 8H3O+(aq) + 3Cu (s) &-> 3Cu^2+(aq) + 2NO(g) + 12H2O(l)}&E^\circ_{\text{rxn}} = \pu{1.154 V}
\end{align}$$

Although it is identical reaction to that in dilute conditions, the reaction proceeds much more favorably (compare $E^\circ_{\text{rxn}}$ values in both conditions).

Late edition:

I have tried to avoid criticizing literature values on this subject, but my answer so far arose questions from concerned people. Thus, I'd try to do my best to handle it as delicate as possible.

To my knowledge, there shouldn't be a question about whether conc. $\ce{HNO3}$ is better oxidizing agent than dilute $\ce{HNO3}$. The oxidizing reagent here is the $\ce{NO3-}$ ion, as my equations described above. For example, any $\ce{NO3-}$ ion (say $\ce{KNO3}$) will oxidize $\ce{Cu}$ in an acid medium (say, $\ce{HCl}$). In using dilute or concentrated $\ce{HNO3}$, we conveniently provide both $\ce{NO3-}$ ions and $\ce{H3O+}$ for our redox reaction. This reaction is highly based on reaction conditions. I think nobody knows exact process, but everybody speculate by the results. That's the reason you are seeing too many different equations in different sources. I'm using the comprehensive table in: http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrochemical%20series.pdf, which I believe more realistic. Now, consider following 2 questions:

  The question 1: If concentrated nitric acid is a better oxidizing agent, then why does it oxidizes only 2 moles of Cu per 8 moles of nitric acid against the 3:8 ratio in dilute one?

Answer: Conc. $\ce{HNO3}$ has more $\ce{NO3-}$ available to react. When available, two moles of it react with 1 mole of $\ce{Cu}$ to give more stable product, $\ce{N2O4}$ ($\Delta H^\circ_f = \pu{+9.16 kJ/mol}$) compared to $\ce{NO}$ ($\Delta H^\circ_f = \pu{+90.25 kJ/mol}$).

  The question 2: Why is there change of -1 in oxidation state of $\ce{N}$ in concentrated solution as opposed to -3 in dilute solution?

Answer: Look at the last series of reactions involving $\ce{N2O4}$ given in my original answer. Eventual change of oxidation state of $\ce{N}$ is same in both cases. This reaction is not just a matter of power. I don't think anybody studied this process in details yet. I'd like to make my point by giving some known redox processes of of $\ce{NO3-}$.

$$
\begin{align}
\ce{NO3-(aq) + 3H3O+(aq) + 2e- &<=> HNO2(aq) + 4H2O(l)} &E^\circ = \pu{0.934 V}\\
\ce{NO3-(aq) + 4H3O+(aq) + 3e- &<=> NO(g) + 6H2O(l)}&E^\circ = \pu{0.957 V}
\end{align}$$

Both processes are under relatively similar conditions (say, $\pu{2M}$ vs $\pu{3M}~\ce{HNO3}$). In process (1), the oxidation state of $\ce{N}$ changes from +5 to +3 while in (2), it was +5 to +2. Which process dominate when they were subjected to oxidize appropriate amount of $\ce{Cu}$?


I guess that it actually was a silver cleaner, not one for gold or precious stones.

The latter mostly contain nonionic surfactants, which are made by the reaction of water-insoluble long chain alkanols (C12 - C15, from natural fatty acids) with oxirane. These washing agents dissolve well in water and help to remove grease and fingerprints from the jewelery. Their action isn't much different from that of a dishwasher for manual cleaning.

These cleaners don't smell the way you described and the car retailer probably used similar cleaners for the car.

Silver cleaners are a different league. The do contain the same surfactants (washing agents), but in order to actually remove the tarnish (silver sulfide) from the silverware, they do also contain a bit of acid (typically sulfuric acid) and thiourea, which probably is responsible for the smell.

Repeated shampoeing probably is the best method to get rid of the thiourea without ruining the carpet.


How to predict the effusion/diffusion for a mixture of gases?

Sort of, but not quite.

If our hole is large (say, as if pierced by a toothpick), then the gases will go out in pretty much the same composition that is found inside, so their ratio in the mixture will stay nearly the same till the very end.

If the hole is small, about the size of free path of a molecule in a gas, then it's a different story. BTW, these conditions are called Knudsen flow. Methane weighs 16, and helium weighs 4 (I know you've said "hydrogen", but let's assume it was helium, just to make the calculations easier). So helium particles are 4 times lighter than those of methane, hence they fly 2 times faster. So there would be two times as many of them in the outgoing stream - that is, if the mixture inside was 1:1. See, it's not like all helium flows outside, while methane molecules stand politely in the line, saying "After you".

As the story progresses (which is not going to be fast with just one tiny hole), the mixture inside gets enriched with methane. So does the outgoing flow, but still it remains richer in helium relative to the mixture inside, and hence continues to shift its composition in the same direction. Yet still, pure methane is a limit which is never actually attained.

That being said, pressing gas mixtures through porous membranes is a valid way to separate the components by molecular mass regardless of their chemical properties, and as such, was used for isotope enrichment.


Why are p-type solar cells more prone to degradation in space than the n-type ones?

The space radiation environment is a quite complex subject. Generally speaking one needs to be careful about just where in space you are concerned about - near earth orbit (within the Van Allen Belts), outside the belts, regularly passing through the belts, interplanetary, etc.

That being said, the three main areas of concern are protons from the solar wind, electrons from the solar wind, and cosmic rays from the rest of the universe. For testing purposes the standards for the solar wind are 10MeV protons and 1MeV electrons. I will focus on these for the purposes of this answer.

As the protons or electrons hit a solar cell, the will go in some ways and lose energy through interactions with electrons and the nuclei (electronic and nuclear stopping, respectively). Electronic excitations at best make a little more current in the depletion region. The interactions with nuclei are more problematic - either particle can transfer enough energy to a nucleus to displace it off a crystalline lattice site (protons are about 1000 times more effective at it then electrons because their mass is much higher). So, as the cell accumulates dose, more and more atoms are off their sites.

OK, what does that do? Well, point defects in semiconductors often create mid-gap electronic states that allow carriers to recombine - your cell is now less efficient at making current since some gets lost. Further, point defects (vacancies and interstitials) can combine with either other point defects or impurities to form more complex defects, and these may be even better recombination centers.

Of particular relevance for your question is the impact of boron. To quote from
T. Hisamatsu et al. Radiation degradation of large fluence irradiated space silicon solar cells (1998):

  Radiation-induced defects associated with boron atoms are said to form
  recombination centers for minority carrier electron and this phenomenon is said to
  cause the degradation in low-fluence regions. Therefore, using highly boron-doped
  substrates in order to avoid the sudden drop of cell performance should not be
  a proper means to provide cells with higher radiation tolerance.


Pyrite can actually come in a range of crystalline morphologies, as seen below:

Image source: Modelling nanoscale $\ce{FeS2}$ formation in sulfur rich conditions (J. Mater. Chem. 2009, 19 (21), 3389), who state that the shape is dependent on the concentration of sulphur during its formation.

Modelling of cubic pyrite crystals, as in the picture in your question, are explained in the article Modeling the Shape of Ions in Pyrite-Type Crystals (Crystals 2014, 4 (3), 390–403.), modelling the following structure:

From the caption, the blue spheres represent $\ce{Fe}$ and the ellipsoids represent $\ce{S}$. The shapes of the constituent chemicals used in this diagram is based on their modelling, specifically one of the conclusions the author makes is that:

  It turns out for $p$ valence shell ions that an isotropic ionic radius only occurs on cubic lattice sites. For all other site symmetries, however, two or three radial parameters will apply. Appropriate geometric shapes for $p$ valence shell ions are thus given by ellipsoids instead of spheres.

The $p$ valence ion in this case being sulphide.  With this model in mind, the explanation provided in the caption for this diagram states:

  The mesh inscribed into the sulfur ions now reveals the ellipsoidal compression along <111> directions. In this model, the number of contact points of $\ce{S}$ and $\ce{Fe}$ ions become four and six, respectively, and the ellipsoidal deformation is concluded to enable a stable packing.

The molecular shape does not necessarily mean that the pyrite crystal has a cubic shape, however if the conditions are right (pressure and temperature), and the composition is 'correct' - if the formation is in a iron rich environment (as per the first reference), then this molecular shape could express as a cubic crystal.


how to speed up Ferric Chloride and copper reaction

Chemicals are probably not the good way to go. What is the temperature of the solution? Increasing the temperature speeds up the reaction. Btw: take the proper safety precautions when etching pcb's!


The questions has at least two aspects to address.

The first thing is, there is no guarantee that the reaction is first-order with respect to each of the reactants in first place. In other words, without further investigation you can not just take $r=k[\ce{A}][\ce{B}]$ because $r=k[\ce{A}][\ce{B}]^2$ or $r=[\ce{A}]^2[\ce{B}]$ or something more exotic may happen, depending on the nature of the reaction.

OK. So let's say for simplicity we take $r=k[\ce{A}][\ce{B}]$ when the system has only $\ce{A}$ and $\ce{B}$, as we don't need to be quite educationally correct. Then yes, after mixing many reactants, the equation still holds for each pair of the reactants, provided they are elementary reactions. It will be a completely different story if, for example, neither of two of the reactions are elementary and they share the same reaction intermediates.

I also have a reminder that when stepping from the expression of the rate to the expression of concentration with respect to time, you have to be careful and do the integration properly instead of just naively extrapolating the concentration expression when only two reactants exists. Let's say that we don't consider chemical equilibrium for the moment (i.e. consider all the reverse reactions to be prohibitively difficult to happen), then as far as I can see, the $r-t$ curve will not be smooth throughout the time and will have some turning point every time a reactant is fully consumed.

Side note: I suggest that you do your formatting properly in your program for the superscripts and subscripts.


Why is phenolphthalein an appropriate indicator for titration of a strong acid with a strong base?

If you look at the titration curve, which plots the volume of base added vs pH (source):

you can see that the equivalence point occurs at pH = 7.

Phenolphthalein is fuchsia in pH's roughly between 8.2 and 12, and is colorless below pH 8.2. When the number of moles of added base is equal to the number of moles of added acid (or vice versa; example valid for strong monoprotic acids/bases assuming 100% dissolution), the pH is equal to 7.

You might say, if the pH needed is 7, and phenonphthalein changes at pH's around 8.2, how can you use this indicator?

Well, again looking at the curve, from pH = 11 to about pH = 4, pH changes very rapidly with from an infinitesimally small change in the volume of base added. Since one drop of added titrant will cause this large change, even though the change in color of phenolphthalein does not occur right on the equivalence point, it is within approximately one drop. This kind of uncertainty is "acceptable uncertainty" in using titration to volumetrically determine concentrations.

To clarify what I mean by "acceptable uncertainty", you should realise that each of your measurements has some kind of uncertainty to them:

When you weighed out the primary standard to titrate against, was the balance perfect?
Was your solution made up precisely to the graduation in the volumetric flask?
Did you pipette the exact volume of the aliquot or were you off by a drop or two?
Are you able to state the volume added from the burette to arbitrary precision or is there some uncertainty beyond the two decimal places given by the graduated lines?

In the scheme of things, ~1-2 drops will not be a significant factor in getting an accurate result, but you should most definitely acknowledge that there is uncertainty in your answer.


Why does the second electron affinity has an opposite sign of the first one?

Many non-metals like those in Group 16 and 17 have first electron affinities that are negative.  In other words, the following reaction is exothermic and energy is released.

$$\ce{X + e^{−} -> X^{−} + ~energy}$$

By gaining an electron the group 17 atoms attain a complete octet of electrons and become stabilized.  By gaining an electron the Group 16 atoms are just one electron away from having a filled shell.  Further in both the Group 16 and 17 cases, the protons in the nucleus are not fully screened by the outer electrons, so adding one more electron to form an anion is electrostatically favorable.

However, if we try to add a second electron the reaction becomes endothermic; adding a second electron is unfavorable and energy must be added (endothermic).

$$\ce{X^{-} + e^{−} + ~energy~ -> X^{2−}}$$

It is difficult to add a second electron primarily because of the increased electron-electron repulsion that occurs with the second electron.

Since adding the first electron is exothermic, while adding the second electron is endothermic, the sign reverses when we measure the energy added or released by these two processes.


Balancing the reaction of potassium permanganate and sulfuric acid with hydrogen peroxide

It seems to me that this is simply a redox reaction in the presence of acid.

If you write the 2 half reactions, and balance in the presence of acid (sulfuric acid), you end up with the same answer (with 5 O2 instead of 10 O on the product side).

This seems to be more of a mathematical manipulation rather than a chemistry problem, IMHO.


What is the difference between organosilyl, alkoxysilane and silane compounds?

Silane is the equivalent of methane for silicon, $\ce{SiH4}$.
Silicon containing organic derivatives of many kinds exist.
Generally speaking, organosilanes are those in which $\ce{Si-C}$ bonds exist, like for example, tetramethylsilane ($\ce{TMS}$, used as a reference on $\ce{{}^{1}H~NMR}$).

Alkoxysilanes are those compounds that contains an organic group bonded to a silicon but not directly, through an oxygen atom instead, like so: $\ce{R-OSi(Me)3}$, common structure on silyl enol ethers.

A surface is hydrophobic if it's repulsions towards water overcome its attractions, and this is the case for many silicon compounds, but not all of them. For example, glass (silicon dioxide, $\ce{SiO2}$) is hydrophilic, due to the possibility of formation of hydrogen bonds between the glass oxygen and water molecules.


Mistake in finding bond angles in VSEPR theory

Looking at compound D, one of the first things we notice is the carbonyl group. The carbon atom in a carbonyl is $\ce{sp^2}$ hybridized, so angle 6 involves an $\ce{sp^2}$ hybridized carbon. In such cases the $\ce{H-C-O}$ bond angle is ~ 120 degrees.

Just to elaborate a bit more:

If you have four objects arranged in space about a central carbon atom (the $\ce{sp^3}$ hybridization case), the four bonds from carbon are going to arrange themselves spatially so that the electron - electron repulsions between the four bonds are minimized.  This is achieved by arranging the bonds such that they point towards the vertices of a tetrahedron.  In a case like $\ce{CH_4}$, where the 4 atoms bound to carbon are identical, the C-H bonds adopt the perfect tetrahedral angle of 109.5 degrees. In the case of $\ce{CH_2Cl_2}$ where the four bonds are not equivalent, the bond angles will differ somewhat from 109.5 degrees (can you guess which angle will be greater then 109.5 degrees and which angle smaller) - but still, the tetrahedral angle is a reasonable first approximation.
In the case of an $\ce{sp^2}$ hybridized carbon, the 3 sigma bonds will arrange themselves in a plane with an approximate angle of 120 degrees between the bonds.  This arrangement again minimizes the electron - electron repulsion between the 3 sigma bonds in this geometry.
Finally, what geometry will minimize electron - electron repulsion between the two sigma bonds in the $\ce{sp}$ hybridized case?

(mouse over for answer)

   Yes, the two bonds will be in a plane 180 degrees from one another, just like the $\ce{H-C-C}$ bond angle in acetylene.

Take home message:

$\ce{sp^3}$ hybridized carbon has angles ~109 degrees
$\ce{sp^2}$ ~ 120 degrees
and $\ce{sp}$ ~180 degrees.


How to obtain sodium oxide from sodium chloride?

There is no need (or possibility, really, in terms of standard lab capabilities) to oxidize sodium(I).
In fact, one method relies on sodium(I) reduction to metal as a method of eliminating unwanted chloride.

Method 1

Electrolysis of molten sodium chloride:
$$\ce{2 NaCl(l) -> 2 Na(l) + Cl2(g)}$$
Oxidation of sodium metal to oxide by burning:
$$\ce{4 Na + O2 ->[>\pu{250 °C}] 2 Na2O}$$
Drawback: pure sodium oxide cannot be obtained by direct oxidation of sodium.
Instead, a mixture of sodium peroxide and sodium oxide is formed.
In order to suppress the formation of peroxide, sodium metal or sodium nitrate is added in excess to the mixture in inert atmosphere:
$$\ce{Na2O2(s) + 2 Na(l) ->[\pu{150 °C}] 2 Na2O(s)}$$

Method 2

Convert sodium chloride to sodium bicarbonate using Solvay process:
$$\ce{NaCl(aq, conc) + H2O(l) + NH3(g) + CO2(g) -> NaHCO3(s) + NH4Cl(aq)}$$
Thermal decomposition of bicarbonate first yields in sodium carbonate:
$$\ce{2 NaHCO3(s) ->[\pu{250 - 300 °C}] Na2CO3(s) + CO2(g) + H2O(g)}$$
which is subsequently calcined to form an oxide:
$$\ce{Na2CO3(l) ->[>\pu{1000 °C}] Na2O(s) + CO2(g)}$$
This appears to be a preferred method as it is less energy- and resources-consuming and allows to obtain $\ce{Na2O}$ selectively.