Electronegativity of zinc vs copper in galvanic cell

There are two common confusions in your question. I hope I can clear them up.

  Electrons are not flowing from zinc to copper. Not exactly.

In the Zinc-Copper galvanic cell, the net reaction is:

$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}$$

Zinc is being oxidized:

$$\ce{Zn(s)->Zn}^{2+}(aq) \ce{+2e-} $$

Copper is being reduced:

$$\ce{Cu}^{2+}(aq)\ce{+2e- ->Cu(s)}$$

Electrons, therefore are being transferred from zinc not to copper, but copper (II) ions, which are different than the copper atoms in Cu(s). We need to be comparing something about zinc atoms and copper (II) ions.

  Electronegativity has nothing to do with electrochemistry.

Electronegativity is the attraction an atom has for a pair of electrons in a covalent bond.

Electrochemistry is about controlling the electron transfer that accompanies a redox reaction. Usually this involves ionic species, and so electronegativity is not the relevant quantity to compare. In this case, we care about the energy change that occurs when copper and zinc lose electrons (ionization energy) and the stability of the resulting ions in solution. Ionization data mined from this NIST publication.

For zinc:
$$\ce{Zn->Zn+ + e-} \ \ \ \ \ \Delta H= 9.585\times 10^2 \text{  kJ/mol}$$
$$\ce{Zn+->Zn}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.733\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$

For copper:
$$\ce{Cu->Cu+ + e-} \ \ \ \ \ \Delta H= 7.454\times 10^2 \text{  kJ/mol}$$
$$\ce{Cu+->Cu}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.958\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.703\times 10^3 \text{  kJ/mol}$$

  Why is zinc oxidized and copper reduced?

Because doing so leads to a lower energy state. It takes more energy to remove two electrons from copper than to remove two electrons from zinc. Thus, to get a negative energy change (spontaneous), when need to add the zinc reaction as written and the copper reaction in reverse:

$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{  kJ/mol}$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ \ \ \Delta H= -2.703\times 10^3 \text{  kJ/mol}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ \ \Delta H= -11 \text{  kJ/mol}$$

The estimate above assumes that the solvation energies of both ions are similar and their standard entropies are similar. I could not find a good source of those data.

In electrochemistry, we care more about the potential of the cell, $E_{cell}$, which is the potential for the cell to do electrical work. Potential, $E$, is related to free energy change $\Delta G$ by the following equation, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant $F=96,485 \frac{\text{J}}{\text{V mol}}$.

$$\Delta G=-nFE$$

For a process to be spontaneous, $\Delta G$ must be negative. $\Delta G$ is negative when $E$ is positive. Thus, in a galvanic cell, electrons will flow in the direction that produces a positive cell potential.

How can we determine which direction that is? First, we could set up the zinc-copper cell, and connect it into a circuit and let it run. In this cell, a piece of copper metal serves as one electrode and a piece of zinc metal serves as the other. Whichever electrode increased in mass is the electrode at which reduction occurred: metals cations are converted to metal atoms, coming out of solution and increasing the mass of the electrode. The other electrode must be the one responsible for oxidation (metal atoms become cations and dissolve away); the mass of this electrode must decrease.

Or, we could do a simpler experiment. Dip a piece of copper metal into a solution of Zn2+. Nothing exciting happens (no reaction). Then, dip a piece of zinc metal into a solution of Cu2+. Zinc reduces the Cu2+ ions and copper metal plates onto the zinc.

Or, we could rely on tables of standard reduction potentials for half cells, measured against the standard hydrogen electrode SHE (for which E is defined to be ZERO). Standard reduction potentials come from Wikipedia, which mines them from a number of legitimate sources.

SHE:
$$\ce{H2(1 atm) + 2e- -> 2H+(1 M)} \ \ \ E^o =0 \text{ V}$$
For zinc:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=-0.762 \text{ V}$$
For copper:
$$\ce{Cu->Cu}^{2+} \ce{+ 2e-}  \ \ \ E^o =+0.34 \text{ V}$$
To generate a spontaneous reaction, these half cells are combined in such a way so that the overall cell potential is positive:
$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=+0.762 \text{ V}+$$
$$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ E^o=+0.34 \text{ V}$$
$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ E^o=+1.10 \text{ V}$$
This analysis only applies to standard conditions. Under nonstandard conditions (for example [Zn2+]>>[Cu2+], then the reverse reaction could be spontaneous.


Book reference for electronic occupation of TM

Element Name and Symbol + Atomic Number + Electron Configuration

NIST Standard Reference Database 111 provides an up-to date reference for the electron configuration of neutral atoms in the ground state [1] in both tabulated and periodic table formats.

Common Oxidation States

An overview of elements' oxidation numbers is provided in practically every inorganic chemistry textbook. Arguably, the most extensive one is given throughout Greenwood and Earnshaw's Chemistry of the Elements [2].

More compressed information is provided in CRC handbook of chemistry and physics (Section 4. Properties of the Elements and Inorganic Compounds) [3, pp. 4-1 — 4-92].
There is also a periodic table at the end which includes oxidation numbers of the elements in the upper right corner of each cell:

References

Kramida, A., Ralchenko, Yu., Reader, J., and NIST ASD Team (2018). NIST Atomic Spectra Database (ver. 5.6.1), [Online]. Available: https://physics.nist.gov/asd. National Institute of Standards and Technology, Gaithersburg, MD. DOI: https://doi.org/10.18434/T4W30F
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; CRC Press, 2017; Vol. 97. ISBN 978-1-4987-5429-3


Rationalising the order of reactivity of carbonyl compounds towards nucleophiles

Here is my "old school" explanation.

Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left.  As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well representing the tetrahedral intermediate.  $\ce{E_{act}}$ is the activation energy required to pass over the transition state.  To whatever extent a substituent might stabilize the starting carbonyl compound, the energy of the carbonyl compound will be lowered, and consequently $\ce{E_{act}}$ will increase.  The central carbon (it was the carbonyl carbon) in the tetrahedral intermediate is fully saturated (just $\ce{sp^3}$ bonds) and without charge, so the substituent X cannot interact through resonance to stabilize the tetrahedral intermediate.  In other words, while substituent X can stabilize or destabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

Below are the resonance structures we can draw for our carbonyl compound.

If structure III contributes significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structure III contributes little.

Let's start by comparing the ester to the amide.  In both cases structure III involves a 2p-2p pi bond, but the resultant positive charge is placed on oxygen when we start with the ester and on nitrogen when we start with the amide.  Clearly, for electronegativity reasons, the amide case is better with the positive charge on the less electronegative amide nitrogen.  Our prediction: the amide is more stabilized than the ester and therefor should be the slower reacting of the two.

$$\ce{ester > amide}$$

In the case of the carboxylate anion, resonance structure III is equivalent to resonance structure I (note: there is no positive charge in structure III for the carboxylate case).  Since they are equivalent, they carry high weight stabilizing and in describing the carboxylate anion.  The carboxylate anion is so stabilized by resonance structures I and III that it is basically unreactive to nucleophiles

$$\ce{ester > amide >> carboxylate}$$

There's really no difference between the ester and carboxylic acid in terms of resonance structure III contribution so we would expect them to have similar reactivity towards a nucleophile.

$$\ce{ester \sim acid > amide >> carboxylate}$$

Let's compare the ester to the anhydride.  Resonance structure III is equally effective for both compounds, BUT, there are two carbonyls in the anhydride for the oxygen to interact with.  Therefor, the oxygen will only be half as effective at stabilizing any one anhydride carbonyl as it was for the ester carbonyl.

$$\ce{anhydride > ester \sim acid > amide >> carboxylate}$$

In the case of the acid chloride, resonance structure III involves a 2p-3p pi bond, not very effective overlap due to the size difference (also chlorine is electronegative so placing a positive charge on it is not desirable, but the poor pi overlap is what limits the contribution from structure III in the acid chloride case).

$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$

For the ketone, our resonance structure III has a slightly different look, it involves hyperconjugation as shown in the following figure.

Since there are 3 hydrogens, there are three such hyperconjugated structures,  They are not super significant resonance structures, but there are 3 of them, they do place the positive charge on a proton, so they do count for some stabilization.  To be honest, I would have guessed that it creates more stabilization than resonance structure III in the acid chloride case, but I wouldn't have known where it fits between the acid chloride and the ester.  Wherever I put it, I'd put the aldehyde to its left.  There is no resonance structure III for the aldehyde.

I wind up with
$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$
and $\ce{aldehyde > ketone}$ fitting in somewhere between the acid chloride and the ester


Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered

I think you are asking about preferred conformations for $\ce{sp^3-sp^3}$ single bonds and $\ce{sp^3-sp^2}$ single bonds.

You are correct that the staggered (or gauche) conformation is preferred in simple alkanes.  This is generally explained by noting that

adverse steric interactions are minimized in the staggered conformation, and
stabilizing hyperconjugative interactions are maximized

(From: “Isomers of Organic Compounds”, section 24.2 from the book Principles of General Chemistry (v. 1.0).)

In the case of the $\ce{sp^3-sp^2}$ single bond as found in propene, toluene, acetaldehyde, etc., the eclipsed conformation is now the more stable.

(Michael Di Maso, Jared Shaw, the UC Davis ChemWiki by University of California)

Again both steric and electronic arguments can be advanced to explain this preference.  The steric argument says let's use the bent model to describe the double bond.

Now replace one of those ethylene hydrogens with a methyl group and note that in the eclipsed conformation (build a model if you can) none of the methyl hydrogens eclipse the hydrogen on the adjacent olefinic carbon or any of the bent bonds.  In the staggered conformation all 3 of the methyl hydrogens are eclipsing something, the 2 bent bonds and the $\ce{C-H}$ bond on the adjacent olefinic carbon.  The basis of the electronic argument can be found here.  Personally, for me, the steric argument is easier to understand.


First of all, the difference between ionic and covalent bonds is not sharp. As electronegativity differences increase, you move away from covalent and towards ionic bonds. There are "in between" states like polar covalent, where one side of the bond is stronger but not fully ionic. And this I think is the main reason: hydrogen has fairly high Pauling electronegativity (2.20), rather close to oxygen (3.44), which seems polar covalent overall (and why we get hydrogen bonding with water). In contrast, the alkali metals all have electronegativity less than 1.00, a much bigger difference versus oxygen and thus a more ionic bond.


What is the product of a copper and vinegar reaction?

If it looks green/blue on paper, you likely have copper (II) acetate. You likely have the anhydrous version $\ce{Cu(OAc)2}$ which is darker than the hydrous $\ce{Cu(OAc)4(H2O)2}$

Wikipedia Images include:

The rate of formation is a property inherent to the reactants and the products, but it also depends on temperature.


Alkenes v/s Alkynes in two different reactions

The first stage of electophilic addition is addition of a cation to alkene/alkyne with formation of a carbocations. Vinyl carbocations are much less stable then alkyl cations, so the path is strongly hindered for alkynes.

The first stage of catalitic hydrogenation is addtion of hydride-ion on the metal-surface to unsaturated hydrocarbon, forming a carbanion bound to metal surface. Vinyl-anion are more stable than alkyl-anyons for the same reasons vinyl-cations are less stable than alklyl-cation. Acetylenes also are much less hindered sterically than alkenes and adsorb on a metal surface much easier.

Please, note, than nearby substituents may affect these properties dramatically. Deactivation of alkenes in electrophilic addition by strong $\pi$-electron-withdrawing groups is well known.


Why does a molecular sieve work for gas separation?

I have seen finer zeolites but this one does not seem as coarse as you sounded it. Consider it again, likewise in fractional distillation, there will be some molecules with never being an equilibrium with another phase. However, much more of them will go through zeolites and with the passage of each unit distance and time, you will have more of the compound you want to get and less of the redundant one. If your column is enough long and zeolites are sufficiently fine and ordered, it will be a good purifier.

Try to visualize it in more detail, while some molecules try to pass through the cavities between zeolites, other molecules will be forced to collide and travel through directly zeolites.


Material that changes solubility when electric potential is applied

This is a tentative that perhaps does not fit with restrictions posed by the setup of the question, but does offer an example of a class of materials whose solubility (integrity, to be more precise) is easily altered by an electrochemical process.

A number of conjugated polymers can be deposited atop an electrode by
A: electrochemical polymerisation ( this resulting in a film of the polymer in a more or less conductive state ) or

B: as pristine material if available and soluble in some solvent ( spin coating, dipping, dr blading. ..).

The state of the polymer can be electrochemically controlled by the applied potential. ( however the complete neutralisation to its insulating/semiconducting state is generally difficult ).

Overoxidation of these films results in chemical damage associated with removal of the material, directly or by subsequent washing.

3-methylthiophene and poly(3-hexylthiophene) are well known examples for an electrochemical depositable polymer and for a polymer which can be directly cast from its solutions.

In both cases the poly(3-alkylthiophene) (P3AT) film is insoluble in polar solvents and electrolyte solutions. Applying an electrochemical potentially higher than, say, 2 V, overoxidize it damaging the skeleton to such an extent that the material fall out the electrode and / or can be washed out.

Specifically to the question: only case B is viable.

Looking at the figure in the Q, bottom, the orange part will be consisting of neutral P3AT.

Resolution will be not too precise as overoxidation will take place at the edge of the electrode, too.


What are the consequences of having an element with an atomic number above 137?

The critical value "137" occurs in calculations based upon relativistic theories that treat the nucleus as a point, such and Dirac theory or Sommerfeld theory.
However, when the nucleus is correctly treated as having a non-zero size, a significantly higher value is calculated.
The most recent calculations I see are in New method for solving the $ \bf Z > 137$ problem and determining hydrogen-like energy levels (2014) Phys.-Usp. 57, 189.  (alternative link to pre-print):

Following the development of the Dirac theory in 1928, Dirac [reference 2], Darwin [reference 3] and Gordon [reference 4] obtained expression (1) as a result of exact solution of the Dirac equation in the Coulomb field of a point charge (-Ze). ...  Formula (1) [results in a complex number] if Z [is greater than 137]...the complexity of energy levels in (1) is often called the “Z>137 catastrophe”.
In 1945, Pomeranchuk and Smorodinsky [6] considered
an atomic system with [the nucleus being a sphere of radius $0.8 \times 10^{-12}$cm and having uniform charge within the nucleus and found 175 to be the critical value].

The article goes on to explain that since the 1945 calculation, various calculations, with various nuclear charge distribution models, have calculated critical values in the 170-175 range.  The article's own calculations find 178.


Why is there no salt bridge in electrolytic cell?

Electrolytic and galvanic cells have separation of electrolytes for different reasons. In galvanic cell you wouldn't have current if you mix electrolytes from both cells. In electrolytic cell you can dip both anode and cathode in the same solution. Then yes, you don't have a salt bridge because you use the same electrolyte.

Salt bridge passes ions at a certain rate that is not too high in order to prevent electrolyte mixing. In electrolysis you can allow mixing so you don't need a bridge.

In electrolytic cell you have two transports. Lets say you have Zn/ZnSO4 and Cu/CuSO4

(1) Electrons are flowing from more active (Zn) to more passive metal (Cu). This is essential for creating the electric electric current in the cell. In solution without a salt bridge you have constant concentrations of anion (SO4$^2-$). However, concentration of Cu2$^+$ is decreasing (Cu deposits on metal) and concentration of Zn${2+}$ increases (Zn dissolves). This creates charge imbalance that will quickly compensate difference in potentials of Zn/Zn${2+}$ and Cu/Cu${2+}$.

(2) You cave a salt bridge. It has to pass ions at a right speed. Not to slow and not too fast. If ions can travel too fast (no wall) then diffusion will mix electrolytes in two cells, copper will precipitate on both sides and there will be no current. If ions travel to slow (solid wall) then the charge difference on both sides of the electrolyte will prevent further current.

historically, it was just a piece of fabric between two compartments. Diffusion through it is slow, but drift along the electric potential is fast.


Greater entropy: an atom or macromolecule?

There are many equations that describe the entropy of a system, one is
$$\ce{S = k~ ln ~W}$$
where S is the entropy of the system, k is Boltzmann's constant and W is the number of states a system can exist in.  If we take an atom, it can only move in 3 directions x, y and z - that's it.  If we take a small molecule like ethane, it can also move in these same 3 dimensions but it can also rotate about any one of the atoms, the various bonds can stretch and compress and the various bond angles can expand and collapse.  Clearly, there are a lot more states that ethane can exist in as compared to a single atom, clearly W is larger for ethane than a single atom, clearly S is larger for ethane than a single atom.  As we increase the complexity of the molecule, W and S increase exponentially.  A macromolecule will have a much larger entropy than an atom.

Edit (based on @Gdgames Gamers comment below)

Entropy is a measure of the disorder in a system.  It will always be true that the entropy of substance in the gas phase will always be greater than the entropy of the same substance in the liquid phase, which, in turn, will be always greater than the entropy of the same substance in the solid phase.

Perhaps the person who wrote the exam is assuming that the atom is in the gas phase and the macromolecule is in the solid phase and that based on the phase difference alone the atom has a larger entropy than the macromolecule.  I don't think this is always correct.  First off, the temperature is not specified, nor the state of the macromolecule (is it in solution?).  The macromolecule, depending upon the temperature and state, may or may not translate and rotate through space, but it will vibrate (even at 0K).  I'm just not sure how to compare the entropy contribution from the potential 3n-6 vibrational modes in the macromolecule to the entropy from the 3 translational modes of the atom.  Very poor question by the examiner.


Why did my vinegar not create the traditional volcano?

Acetic acid can be decomposed by methane-producing bacteria. Note that the citation is for soil bacteria, but these methanogens can be found almost anywhere.  Indeed, there is an entire microbial community in vinegar, including nematodes (worms). The more "natural" the product, the more organic material to nourish the microbes and to speed decomposition.
That said, check the labels for acetic acid concentration when purchased. The nominal concentration for distilled vinegar is between 5% and 10%, but some types of vinegar may be only 4% acetic acid.


Oxidation of conjugated dienes into carboxylic acids by potassium permanganate

Here is a relevant extract from a PDF (Dr. P.G. Hultin, 28 July 2016) depicting oxidative cleavage.  For the compound in hand (your question) the process will be done twice. $\ce{H2O2}$ will serve as the source of $\ce{H3O+}$.

  When alkenes are treated with $\ce{KMnO4}$ in acidic solutions, the diol is not formed. Instead, the alkene is cleaved. The reaction proceeds by the same mechanism at the start, forming a cyclic manganate ester (although since the reaction is under acidic conditions the structure is protonated). In the schemes below, the alkene carbons are highlighted throughout, so you can see where they end up in the product.

  Now, under these conditions the manganate ester is not very stable and it undergoes a cyclic fragmentation process, which results in breaking the C-C bond between the two oxygens. Notice that in this case, since there was a hydrogen atom attached to each of the alkene carbons in the starting material, there is a hydrogen attached to the carbonyl carbon in the product and therefore the product that is initially formed is an aldehyde

  Aldehydes are very easily oxidized to carboxylic acids, and thus the aldehydes formed in the cleavage reaction do not survive. They are rapidly transformed into carboxylic acid groups, by a complex reaction whose mechanism you need not worry about.

  Now, if the alkene had not had any hydrogens attached, the product in that case would have been a ketone
  rather than an aldehyde. Ketones are not easily oxidized further, and the reaction would have stopped at that
  stage.

  If one of the alkene carbons had a hydrogen substituent, while the other did not, then we would get both acid
  and ketone groups in our product, as shown below.


Pauling's equation on the relationship between Electronegativity and Percentage Ionic Character

'e' is a natural exponential function, so not an actual measurement.

The equation is correctly written as:

% Ionic character $= (1 − e^{−(\frac{Δχ}{2})^2})×100$

where the $−(\frac{Δχ}{2})^2$ term is raised as a power of the natural exponent 'e'.

Source: The University of the West Indies page Lecture 3. Polarizability, you'll need to scroll down the page a bit to the section labelled 'Bond character based on electronegativity differences'.


What does it mean for meth or any other compound to be 'pure'?

When a compound is "pure" it means the contents of that compound are exactly what we claim them to be.  If something is $99\%$ pure, that means $99\%$ of it is the expected material (in this case, meth), and 1% is other non-meth compounds.

Purity matters in chemistry and biology because the other compounds can be very nasty.  In theory, if you had a bag of 100% pure meth, and a bag of $50\%$ pure meth, you could simply double the dose with the second bag and end up with the same amount of methamphetamine in your system.  However, you would also have a double dose of $50\%$ "other" stuff.

Sometimes "other" stuff isn't a big deal.  It all depends what that "other" stuff is.  I don't know much about the methamphetamine trade, but the effect of purity is a big deal in the ethanol world.  Moonshiners know to discard the "head" and the "tail" of the distilation, the first and last parts of the distilation, because they contain fusel oils (which taste really bad), and methanol (which can make you go blind in large enough dosages).  They want very "pure" ethanol.

To you talk about the purity of prescription drugs, typically what matters more for the drug companies is that they fully understand what "other" things are in the pills or injections, and they are confident they are safe.  That's why, when you look at many pill vials they will list their "inactive ingredients." They want to make sure you are very confident that the only things you are putting in your body are the "right" things.  Going back to the alcohol industry, "pure" ethanol is actually $95.6\%$ ethanol by volume, and $4.4\%$ water.  That is the limit for how "pure" ethanol can get with fractional distillation.  To have ethanol in any higher percentages than that you have to go to exotic processes to create anhydrous alcohols.  However, that is deemed "pure" because we still know what's in it.  We know its $95.6\%$ ethanol and $4.4\%$ water.

Of course, $100\%$ purity isn't always the desire.  There are myriad alcoholic spirits out there.  All of them are sold at 80 proof ($40\%$ alcohol by volume), so they contain the same amount of ethanol.  Its the "other" stuff that makes each spirit unique.  You pay a great deal of money for the particular impurities that make a great well aged scotch.  You pay very little for the impurities in a cheap tequila that leave your head pounding the next morning.


This is a question worth debating and discussion. First of let us realize the fact that we are used to looking at relative atomic weights or masses (if someone wants to be too accurate). No Table lists the absolute atomic masses of atoms in the periodic table, i.e., through out the history we have to assumed either the weight of hydrogen atom is exactly 1.0000, or oxygen-16 had a weight of 16.0000 or lately the weight of carbon-12 atom is exactly 12.0000 units.  So let us forget the picture that someone actually made a measurement, took a single atom of hydrogen or carbon and weighed by some fancy means. These are all man made references and numbers. We could have taken the weight of C-12 as 100, and all the atomic masses would change proportionally.

  So far so good. Now hydrogen weights 1 g/mol and carbon weights 12 g/mol. Two questions to that: How did we actually measure the weight of one carbon atom in the first place?
  Why is the molar mass not precisely 1 and 12?

See above. You have to choose a reference. Fix hydrogen to 1 or carbon as 12 exactly but not both. Now you may ask how these numbers came up? You have to look at the question historically, when Avogadro's number was not used. If you understand German, have a look at Handbuch der Anorganischen Chemie by Abegg. There is long history how the number 12  got associated with carbon.

https://archive.org/details/a2handbuchderano03abeguoft/page/12;


What is the most acidic proton of 2-methylcyclopent-2-en-1-one?

There are two possible enolates that can be formed from your compound: the α deprotonation (left equilibrium arrow in the scheme below) and the γ deprotonation. Both protons have roughly the same $\mathrm{p}K_\mathrm{a}$ values, so deciding which one is more acidic is non-trivial.

However, experimental procedures give us a hint. Usually, the α deprotonation is achieved with strong bases at low temperatures — kinetic conditions. Typically conditions such as LDA in THF at $-78~\mathrm{^\circ C}$. This leads to the conclusion, that the α deprotonation is kinetically favoured. This is more evident in open chain compounds and especially if the α carbon is a methyl group.

The γ deprotonation is generally achieved under more thermodynamic conditions: Weaker bases, warmer temperatures. I didn’t do it myself, so I cannot state conditions off the top of my head, but I would guess conditions such as $\ce{$t$BuOK}$ and $0~\mathrm{^\circ C}$. Therefore, this system must be the thermodynamically favoured one and the γ proton must be slightly more acidic. (But only a tad, or the kinetic product would not be so easily accessable.)

I cannot provide a proper explanation, but it would seem to me like the non-branched π system achieved by γ deprotonation is more stable than the branched one following α deprotonation.


Difference of dipole moments of dichloromethane and trichloromethane?

Just to add some quantification to Ben Norris's answer. Consider each $\ce{C-Cl}$ bond, which has a bond dipole moment of magnitude $A$. The contribution from $\ce{C-H}$ is neglected here to simplify the calculations.

Now consider dichloromethane. The resultant of the two $\ce{C-Cl}$ dipoles will have a magnitude of $$2A\cos\frac\theta2$$
where $\theta\approx109.5^\circ$, the angle between the vectors from the ends of tetrahedron to the centre. This turns out to be $1.154A$.

For tricholoromethane, the three $\ce{C-Cl}$ dipoles will be at an angle $\theta$ such that $\cos\theta=1/3$ with a straight line symmetrically passing through its centre. Thus the resultant in this case will be $$3A\cos\theta$$
which will be $1A$, lesser than in the case of dichloromethane.

To visualise how can you get those angles and cosines, see this page.


This is a wild guess, so I am open to criticism.

Brownian motion could explain this phenomena.

Water was on the board, so it is possible a film of water condensed on the board some time in the night.  This means the humidity was high while the board was still relatively cool. This film may have allowed the oil particles to suspend.
The ink bound tighter to itself than the non-porous glass board, and the ink itself may be buoyant in water.
The ink binding to itself explains why the letters stayed semi-intact instead of diffusing into a mess.
The water molecules are moving constantly which would allow the bigger particles, which in this case would be the letters, to do the random walk characteristic of Brownian motion.
(As a note the random walk characteristic of Brownian motion tends in the direction of gravity.)
The diffusion coefficient of the letters would be temperature dependent, so the temperature in the room may have briefly spiked just warm enough for the oil to diffuse some then it came back down so the ink did not run completely off the board.


How fast a surface oxide layer is formed on bulk chromium?

First, a definition of the Langmuir unit, used in the graph in the question, which is not often run across except by vacuum (particularly UHV) folks. One Langmuir is the exposure of a surface to a gas pressure of $10^{-6}$ Torr for one second. It roughly corresponds to one monolayer equivalent of gas hitting the surface. At normal atmospheric pressure, with 20% oxygen in air, 1 Langmuir of oxygen hits the surface in about 6.6 nanoseconds (yes, nanoseconds).

Another paper that could be of interest is C. Palacio et al., "AES, XPS and EELS study of the initial oxidation of polycrystalline chromium", Surface Science 182(1-2) 41-55 (1987). They see chemisorption of oxygen in the first 1-2 L, nucleation and lateral growth of oxide islands up to 10 L, and continued growth of an oxide film above 10 L.

Overall, there is not a huge difference between the two papers. By ~10 L of exposure to oxygen there is a continuous oxide on the surface at 300K. Following from the numbers in the first paragraph, that occurs in less than 100 nanoseconds. Basically, it oxidizes as it is cut.


In a two electrode system, is the applied current density referring to just the working electrode or to the auxiliary electrode as well?

It is normal in many laboratory electrochemistry experiments for the working electrode to have a much smaller surface area than the counter (auxiliary) electrode. This when the current density at the working electrode is 0.5 mA per square cm, the current density on the counter electrode is much smaller.

It is more common that someone will express the potential and the current density on the working electrode.

If for example we have a reference electrode which is at 0 volts (standard hydrogen electrode) and the potential of the working electrode is + 400 mV. We could have a flow of 1 mA out of the working electrode. As no current flows through the reference electrode all of this current flows through the counter electrode.

The electrical potential between the counter and the reference electrodes is known as the compliance voltage. This is the voltage required to create a flow of current which is just right to make the right amount of current flow through the working electrode to keep its potential at + 400 mV.

This is done using a potentistat, a range of different designs of such devices exist. I think that the most simple design and one of the best is the general design used by PAR (Princetown Applied Research) in the classic 173. It has been joked that at one point almost all electrochemists have used a 173, I rather like the 173 it is a nice machine. But we are not here to discuss classic electrochem machines.

Back to the design of a machine like a 173, If we set the drive at -400 mV then the top op amp will impose a current on the cell via the secondary electrode which will bring the potential of the reference electrode to -400 mV.

The bottom op amp which is a I to V converter will hold the working electrode at zero volts (ground or GND) as far as the electronics is concerned. Thus for the chemist if they hold the view that the refernece electrode is the zero volts point then the working electrode will be at +400 mV.

If 1 mA flows into the working electrode and the value of resistor R is 1000 ohms then the output of the I to V converter (bottom op amp) will have to be -1 volt for the sum of the currents coming to the inverting (-) input of the op amp to be zero. We assume that the input resistance of an op amp is infinity. Thus if we attach a x-y plotter to the drive terminals (x to E or potential axis) and the (y axis to the current indication terminal) we will get the plotter to plot -400 mV, -1 volt.

A word of warning the Europeans and the Americans often follow different conventions on what sign to give a reducing current (cathodic) and what sign to give a reducing potential.

If I recall rightly the European convention is for bottom left to be reducing current and reducing potential.


Is there a relationship between loss of electrons and EM wave reflection?

Actually, I think you have it backwards. The color you see is due to the colors of light which are removed from white light when traveling past the iron. That is, if you go and find an absorption spectrum of iron cations in different oxidation states, you will find they primarily absorb the complementary colors of the color you actually see.

As far as the relationship between charge and the color, the ions with a smaller charge will likely have a larger HOMO-LUMO gap which means you will see colors of shorter wavelength, which is what you report. This property will depend very sensitively in the presence of ligands and other environmental changes, so it is not always possible to guess the color of a compound based only on charge.

This is typically how colors work for systems that absorb light. It is, however, possible for the we see to be due to reflection. Pure reflection, though, is independent of wavelength, so the phenomenon you describe is actually scattering, which does depend on wavelength. This type of phenomenon is responsible for the color of the sky.

Finally, you sometimes see the color which is emitted after absorption, as is the case in fluorescence and phosphorescence.

Hopefully that clears up your confusion


Propane Combustion in a Gas Grill with Irregular Flameouts

The brilliance of the orange in flares suggests suggests incomplete combustion of a carbon residue.  This can happen with a poor oxygen mix, but the blue in the core of the flame indicates that mix is fine.

Also, on the base of the grill there is some charred residue.

Based on the color, and the residues I expect somewhere in the flow you have some carbon deposits that are occasionally getting dislodged, causing the sputtering.


What is density in density functional theory?

The density in density functional theory (DFT) refers to electron density, which is the object of all DFT calculations. DFT differs from various other wavefunction-based methods which focuses on determining the ground-state $\ce {3N}$-dimensional wavefunction*, such as the Hartree-Fock (HF) method. Electron density, often denoted as $\rho$ or n(r), refers to the 3-dimensional function of 3 spatial coordinates describing the distribution of electrons in a given system. It is of interest in DFT as it allows us to determine all observable properties of the system without knowing the exact ground-state wavefunction, hence providing great computational ease. The following is the usual expression of electron density (for spin-restricted calculations, for a system of an even number of $\ce {N}$ electrons, which would have $\ce {N/2}$ different electronic energy levels since electrons in the same orbital are considered to have the energy, thus also meaning that we are summing only $\ce {N/2}$ of the different individual densities, and then multiplying by 2) that you would find in resources on DFT:

*$\ce{N}$ refers to the number of electrons in the given system. The wavefunction is a function of $\ce {3N}$ dimensions since each electron has 3 spatial coordinates. If spin is considered as a fourth coordinate, then the wavefunction becomes $\ce {4N}$-dimensional. Obviously, such a wavefunction would not be easy to solve for exactly.

  However, I find it difficult to understand how density is used to calculate the energy of the system.

This is where the functional in the name DFT comes in. The 1st theorem of Hohenberg and Kohn tells us that:

  Ground-state energy of the system is a unique functional of its electron density.

What exactly is a functional? A functional is a function of a function. A commonly used functional would be the definite integral. In Kohn-Sham DFT (which treats the system as comprising of non-interacting particles moving in an effective potential), notably the most widely-used version of DFT (another version being orbital-free DFT), the functional can be seen as comprising of three main contributions: 1) kinetic energy of electrons, 2) coulombic interactions between electrons and nuclei, and 3) coulombic interactions between pairs of electrons. However, there are some effects that we would need to take into account. For example, when we consider coulombic interactions of each electron with the electron density (in solving the Kohn-Sham equations), we also include a non-physical self-interaction (i.e. the interaction of the electron with itself). Particles in the system can also undergo exchange interactions and their motions are also correlated (i.e. they are not independent). Hence, another term called the XC (exchange-correlation) functional is also included into the overall density functional expression to account for all these additional effects that have not already been accounted for.

The following is an expression of the Kohn-Sham density functional taken from Wikipedia:

The first term is the electron kinetic energies, the second represents the electron-nuclei interaction energy, the third represents the electron-electron interaction energy, while the last term is the mysterious XC functional.

The site has more posts on the topic which you may find useful:

Are there any full worked examples of DFT calculations?

Density Functional Definition


The problem with NK Yu's diagram is that the bond rotation is unnecessary to obtain the correct conformation for epoxide formation. The epoxidation and the bromination/base give the same product and both processes are stereospecific. (Z)-Butene (1) gives the cis-oxirane 2 while (E)-butene affords the trans-oxirane (not shown). Peracid epoxidation adds oxygen to either face of the π-system of (Z)-butene to provide meso-oxirane 2. Bromination of 1 affords an intermediate meso-bromonium ion 3 by the same type of addition as was seen in the peroxidation. When hydroxide effects an SN2 (blue arrow) inversion of stereochemistry occurs (R --> S), the S,S-bromohydrin 3 is formed. With equal probability the path of the red arrow leads to the R,R-enantiomer 4. Deprotonation of the racemic bromohydrin leads to a second inversion of stereochemistry to form meso-oxirane 2, i.e., R,R --> R,S and S,S --> R,S. The commonality here is that direct, peracid epoxidation involves no inversions of stereochemistry while the bromination/base route involves two inversions of stereochemistry. The bottom line is that an even number of inversions, zero being considered even, produces the same stereochemistry.


What is the correlation between electronegativity and the sharing of electrons?

Is there a specific equation that can give us how the bonding electron pair is shared?

No. The electronegativity scale is based on bond strength, not on partial charges. There are quantum-chemical methods to calculate the electron density in a molecule, and you can derive partial charges from that.

  For example in a C-O bond, the electronegativity difference is 1.2. does this mean that the Oxygen has 1.2 of the shared electrons and Carbon 0.8 (or some other numbers)?

No, the electronegativity difference does not allow you to make a quantitative statement like that. It is useful, however, to predict how a bond would hydrolyze. It also gives you an estimate about how polar a bond is.


Nope, no rule, and again there is no difference, like your question about the names for iron oxidation states.  The "bi" form is the older form, and the "hydrogen" form is now preferred, again because it more explicitly describes the form of the compound.

EDIT: Some additional detail was requested.  When I say "preferred", I mean the name sanctioned by the IUPAC (International Union of Pure & Applied Chemistry), also known as the systematic name, so called because the rules to form the name are "systematic"; i.e. consistent and clear. Older names are called common names, which were in used before the IUPAC.  So "bisulfate" is a common anion name, while "hydrogen sulfate" is systematic.


Steam distillation- how it actually works?

You boil the water and the product co-distills.  There is an anomalous selective interaction between steam and many hydrophobes.  If you heat the pot you deplete the water contained and may pyrolyze contents on the hot wall.  Blowing in steam limits temp to near 100 C, moves a lot of energy into the flask and out (540 cal/g enthalpy of vaporization), and maintains the water content.

You could filter off the solids, acidify the aqueous raffinate,  extract non-basics, then alkalinize and extract the aniline.  Since aniline is air sensitive to oxidation and extremely toxic by skin contact, you can appreciate advantages of steam distillation.


Yes, spatial configuration of attached groups matters. Meaning different stereo configuration of $C_1$ and $C_3$ may render $C_2$ either chiral or not:

Also because you have more than 1 stereo center you're not limited to just enantiomers, you can get diastereomers - the molecules will be different even though they're not mirror images of each other:


How to determine the source of PVA: crude or plant-based?

It certainly would be possible to get an indication of some of the sources of carbon through mass spectrometry and through radiocarbon dating. For example, fossil carbon sources would have little 14C, sources predating the 20th century, e.g. older trees, would have more. Post-Atomic Age sources might have more yet, beyond that due to natural decay of 14C.

That said, it's a nice question, but are you prepared to pay for the analysis to have it answered? Perhaps your government might perform such testing to verify a claim of being "organically produced".


Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?

They can, they do and they color. The problem is, the corresponding transition requires so much energye, that the adsorption is well above upper range of visible spectrum in most cases. Linked article provides us with 3 optical spectra of $\ce{Cr}$, $\ce{Mo}$ and $\ce{W}$ carbonyls. For all adsoprtion mostly happens above 300 nm, with only small bits between 300 and 400 nm. Visible spectrum is between ~400 and ~700 nm. Since only bit of upper blue part of the spectrum is adsorbed, only a pale yellow coloring is observed.

This, of course, if you want so badly to apply crystal field theory to carbonyls, which does not work well in the case.


To obtain $Q_\mathrm{sp}$ for

$$\ce{Ag2CrO4 <=> 2 Ag+ + CrO4^2-}$$

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{CrO4^2-}]$$

$$Q_\mathrm{sp} = c(\ce{Ag+})^2\cdot c(\ce{CrO4^2-})\label{eqn:1}\tag{1}$$

you need to use new concentrations of ions in solution $c(\ce{Ag+})$ and $c(\ce{CrO4^2-})$, which can be found from the corresponding values for initial concentration $c_0$ and volume $V_0$:

$$
\begin{align}
c(\ce{Ag+}) &= \frac{c_0(\ce{AgNO3})\cdot V_0(\ce{AgNO3})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &= \frac{\pu{0.10 mol L-1} × \pu{105 mL}}{\pu{105 mL} + \pu{125 mL}}\\
  &= \pu{4.57E-2 mol L-1} \label{eqn:2}\tag{2}
\end{align}
$$

$$
\begin{align}
c(\ce{CrO4^2-}) &= \frac{c_0(\ce{K2CrO4})\cdot V_0(\ce{K2CrO4})}{V_0(\ce{AgNO3}) + V_0(\ce{K2CrO4})} \\
  &= \frac{\pu{0.35 mol L-1} × \pu{125 mL}}{\pu{105 mL} + \pu{125 mL}} \\
  &= \pu{1.90E-1 mol L-1} \label{eqn:3}\tag{3}
\end{align}
$$

Finally, plug in the concentrations of ions from \eqref{eqn:2} and \eqref{eqn:3} in \eqref{eqn:1} and you find a value for $Q_\mathrm{sp},$ which indeed exceeds $K_\mathrm{sp}$:

$$Q_\mathrm{sp} = (\pu{4.57E-2 mol L-1})^2 × \pu{1.90E-1 mol L-1} = \pu{4.00E-4 mol3 L-3}$$


Proving using ideal gas law that molar volume at NTP is 22.4 litres

The issue is that one mole of a gas at NTP should not be 22.4 L based on the ideal gas. This is only true for a gas at STP. The key difference between the two conditions is they are defined by different temperatures (STP is for 273 K while NTP is for 293 K).

They are also different in that the IUPAC now defines STP with respect to 1 bar while NTP is defined with respect to 1 atm.

See The Engineering ToolBox for more info.


How does the polarity of the eluent and sample affect the Rf value in thin layer chromatography?

Never in my years in the lab have I seen a more polar eluent that causes a lower $R_\mathrm{f}$ value than a less polar eluent. It also does not make sense theoretically.

TLC 2 is obviously the one with the least polar eluent and compound $z$ is the least polar compound.

I fully disagree with half of MaxW’s proposed answer.


How does gefitinib affect the EGFR(ecptor) and why is there relapse

How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.

No, EGF binds on the extracellular side (to the ectodomain), and gefitinib binds on the intracellular side (to the tyrosine kinase, purple in the figure). The inhibitor interferes with ATP binding (see below, middle panel - the part boxed in green has structural similarities to adenine in ATP). The ATP would otherwise phosphorylate the tyrosines of other signalling proteins. So even if EGF binds to the ectodomain, there is no downstream effect because the tyrosin kinase domain is inhibited.

Source of images: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6523254/

I recommend section 3.2 to learn about how various inhibitors work.

  It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?

Additional mutations might directly prevent the drug from binding, or change signalling pathways in a different fashion to render the drug inefficient.


Mechanism for palladium-catalysed spirocyclisation of conjugated diene

You don't have to wonder about which alkene coordinates to the Pd(II) centre. In this case, they are conjugated and therefore can coordinate together as a diene, which is a 4-electron donor. Many pi complexes are known ranging from alkenes (2e donor), to allyl (3e), to diene (4e), to cyclopentadienyl (5e), ... so this is not something really surprising.

Coordination of the diene to Pd(II) makes it more electrophilic. The pendant hydroxyl group simply does a 5-exo-trig cyclisation to form the C-O bond, generating a Pd(II)-allyl species. In terms of stereochemistry, the OH group has to attack anti to the metal. Deprotonation can happen with acetate, or the hydroquinone dianion that benzoquinone is later reduced to.

This Pd(II) allyl species is an intermediate in the well-known Tsuji–Trost allylation. In this case, the Pd-allyl species is being generated in a different way, but goes on to react in the same way. Chloride being a relatively soft nucleophile attacks the allyl moiety, again anti to the Pd, leading to a Pd(0)-alkene complex. The stereochemistry of the chlorine isn't indicated in your product but this is my guess at what's going on.

Dissociation of the alkene directly gives the desired organic product and reoxidation of Pd(0) to Pd(II) with benzoquinone regenerates the catalyst.

Looking this reaction up on Reaxys, it seems that it is taken from a 1991 article by Bäckvall and Andersson.1 The mechanism posited in the paper is identical to what I have written above, and on top of that my guess at the stereochemistry was correct: the authors indicate that the product obtained was 99% cis (referring to the chloride and ether oxygen in the product).

Interestingly, with $\ce{Li2CO3}$ instead of $\ce{LiCl}$, the product obtained has a trans acetate. Carbonate isn't a great nucleophile so it doesn't act as one, but it can deprotonate acetic acid. The acetate ion thus produced then forms a bond to Pd(II) before it migrates to the allyl system.

Reference

Bäckvall, J. E.; Andersson, P. G. Stereocontrolled oxaspirocyclization of conjugated dienes via palladium catalysis. J. Org. Chem. 1991, 56 (7), 2274–2276. DOI: 10.1021/jo00007a002.


Because the target structure contains the motif of a cyclohexane-1,2-diol, I suggest an approach including a Diels-Alder reaction (DA in the illustration below) of 1,3-butadiene and the commercially available diethyl acetal of propargyl aldehyde:

The hydrolysis of the acetal then sets the stage to install the methyl group with the cuprate and 1,4-selectivity in respect to the carbonyl group.  As an intermediate protection of the aldehyde, a cyclic acetal is installed. The Prévost reaction typically yields 1,2-(trans)-diols, thus possibly here, too (brief review).  Eventually, the deprotected aldehyde is transformed into the exocyclic alkene with the Lombardo reagent. Contrasting to the conditions of a Wittig reaction, the conditions of the Takai-Lombardo reaction are milder (absence of a strong Bronsted base).
Alternative installations of the (trans)-diol could be achieved e.g., by epoxidation and subsequent ring opening with a peracid like mCPBA or (in situ generated by) oxone.


Suppose an atom of element 'X'

Let's assume the charge on this atom is 0 - hence, the number of protons in the nucleus and the number of electrons revolving about the nucleus must be equal, which results in a net 0 charge on the atom.

Now, lets assume that another atom of 'X' has +1 charge over it.
Well, how could this happen?
There are two ways for this to happen.

1) Nucleus gains an addition proton (+1 charge)

2) Atom loses an electron(-(-1) charge)

Both of these cases result in the atom gaining +1 charge.

Case 1 is very difficult to achieve, hence its not a good idea to think about in any chemical reaction (in fact it is not one at all, see comments!), instead consider case 2.

Notice, losing an electron results in +1 charge over the atom - another way to think of this is that we have ruined the balance that existed between the number of electrons and number of protons.

So we could say the above sentence in your words, we subtract the total charge present on the atom from the atomic number (remember, originally when the atom had no net charge, atomic number = number of protons = number of electrons. This is no more the case now!) to get the number of electrons present in the atom.

This is a very basic form of conservation of charge.


Ranking gases by value of the van der Waals equation-of-state constant 'a'?

You look up the values of "a" in a table of constants for the van der Waals equation.

\begin{array}{lrr}
   \text{gas}                  &
    a / \mathrm{L^2bar\,mol^{-2}} &
    b / \mathrm{L\,mol^{-1}}      \\\hline
   \text{Hydrogen}             &    0.2476  &         0.02661 \\
   \text{Carbon dioxide}       &    3.640  &         0.04267 \\
   \text{Nitrogen dioxide}     &    5.354   &         0.04424 \\
   \text{Water}             &    5.536  &         0.03049 \\
   \text{Carbon tetrachloride} &  19.7483  &         0.1281  \\
\end{array}


Why is tetrafluoromethane non-polar and fluoroform polar?

Draw the structures in 3D and then you will see why one is polar and the other not.

$\ce{CF4}$:

As you can see this molecules adopts a tetrahedral geometry which is perfectly symmetrical in every direction and so the dipoles of the four $\ce{C-F}$ bonds cancel out, leaving no overall dipole.

$\ce{CHF3}$:

Although the molecule has some symmetries, it is not perfectly symmetrical. Since the dipoles of the $\ce{C-F}$ bonds are far larger than the basically non-existent dipole of the $\ce{C-H}$ bond, the dipoles do not cancel out and you are left with a molecule which has a notable net dipole moment of 1.649 D, with the negative end over the fluorines and the positive end over the hydrogen.

As general advice, when determining whether a molecule has a net dipole, always consider the molecule in 3D as opposed to just looking at individual bond dipoles.


What is cheap material that absorbs water in packaging?

What you are looking for is most probably silica gel.
http://en.wikipedia.org/wiki/Silica_gel

It is a highly water absorbent material with large surface area. It is often used in packaging as well as in laboratory in desiccators.

You can actually buy it packed or in bulk easily (let me not link explicit offers here). You can regenerate used (old) silica by heating / drying / pumping, so you don't need to buy every single time new one if the old gets less effective.


Treating a sample with water, how to deduce present compounds given precipitate and solution parameters?

You correctly point out that Cl "is soluble with all the given metals". This statement is a little unclear, though, which might be why you're confused. What this means, exactly, is that no precipitate that incorporates Cl can form. For example, the solid $\ce{KCl}$ cannot precipitate, because the combination of $\ce{K}$ and $\ce{Cl}$ is soluble.

This fact alone doesn't forbid precipitates of other species. For example, the solid $\ce{BaSO4}$ is insoluble in water. We know that $\ce{BaCl2}$ alone would be soluble in water, but the barium atoms in $\ce{BaSO4}$ don't care that there are $\ce{Cl-}$ ions in the solution; all that matters is that the combination of $\ce{Ba}$ and $\ce{SO4}$ prevents the two from separating. As a consequence, when $\ce{Ba^2+}$ and $\ce{SO4^2-}$ meet in solution, they merrily form the insoluble product $\ce{BaSO4}$, regardless of the other ions present in the solution.


Why are the masses of atoms less than the sum of their subatomic particles?

This is due to the mass-energy equivalence and a phenomenon called binding energy.
Forming a nucleus releases energy because the nucleons are falling into a potential energy well. Due to Einstein's mass energy equivalence this results in the mass of the new nucleus being less than that of the particles that formed it.
The binding energy of carbon-12 is quoted on Wikipedia as $\pu{92.162 MeV}$. Therefore we can estimate the mass defect of a carbon-12 atom, $\Delta m$, using $ E = (\Delta m)c^2$:
$$\Delta m = \frac{(\pu{92.162 \times 10^6 eV})(\pu{1.6022 \times 10^-19 J eV-1})}{(\pu{2.9979 \times 10^8 m s-1})^2} = \pu{1.6430 \times 10^-28 kg} = \pu{0.098943 u}$$
The difference in the mass of carbon-12 to the mass of its constituent particles is $\pu{0.08940 u}$, so we can see that our calculation is a reasonable estimate of the mass defect. The slight difference is due to other more complicated factors that I have not taken into account, but it still illustrates the main reason for the mass defect.


What happens when the amount given doesn't allow for equilbrium to be achieved in a reaction?

This takes the assumption that the water vapour present acts as an ideal gas. Therefore $PV=nRT$ is valid.

Flaws/Misconceptions:
(1) You directly used pressure as you have in the RICE table.

Using pressure in a RICE table is not very useful since the RICE table deals with the amount of reactants and products and not pressure.

(2) The reaction doesn't reach equilibrium

The reaction reaches equilibrium as shown below. This means that the question in title doesn't need to be considered.
However, in such a case where equilibrium isn't reached, then the reaction stops due to lack of reagents to go forward. When you add more reagent it keeps moving forward until equilibrium is reached after which addition of solid reactants wouldn't cause a reaction to take place.

Now, as OP has stated, the $P_\ce{H2O}= \pu{20 mm Hg}$ at equilibrium. and it is given that relative humidity is $80\%$ for a saturation vapor pressure of $\pu{85 mm Hg}$ and so that means the intial pressure exerted by $\ce{H2O(g)}$ is:
$$P_\ce{H2O}= \frac{8}{10}\cdot 85=\pu{68 mm Hg}$$
Now using the ideal gas law, for the initial condition, we get:
\begin{align}
n_i(\ce{H2O})&= \frac{68}{76}\cdot\frac{1}{0.0821 \times 300} \\
&\approx\pu{0.00036 mol}
\end{align}
Similarly for the final case, we get:
\begin{align}
n_f(\ce{H2O})&= \frac{20}{76}\cdot\frac{1}{0.0821 \times 300} \\
&\approx\pu{0.00011 mol}
\end{align}
Therefore, $\pu{0.00025 mol}$ of $\ce{H2O}$ reacts for the condition to reach equilibrium. Now drawing the RICE table for the reaction we get:
\begin{array}{lc} \hline
\ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline
i &1\times10^{-3} && 1 \times10^{-3} && 3.6 \times 10^{-4} \\
c &1.25\times10^{-4} && -1.25\times10^{-4} && -2.5\times10^{-4} \\
e &1.125\times10^{-3} && 8.75\times 10^{-4} && 1.1\times10^{-4} \\ \hline
\end{array}
So the final composition is:
\begin{array}{lc} \hline
\text{Component} &\text{Amount} \\ \hline
\ce{A.3H2O} & \pu{1.125 \times10^{-3} mol} \\
\ce{A.H2O} & \pu{0.875 \times10^{-3} mol} \\
\ce{H2O} & \pu{1.1 \times10^{-4} mol} \\ \hline
\end{array}


While solubilities are typically quoted in $\mathrm{g/l}$ or related units, the actual physicochemical idea behind it is an equilibrium constant known as the solubility product. In general, dissolution and precipitation can be considered two directions of the following equilibrium:

$$\ce{m A^n+ (aq) + n B^m- (aq) <<=> A_mB_n (aq) <=>> A_mB_n (s) v}\tag{1}$$

The outer sides are both preferred with respect to the central one; thus, the central compound is usually ignored, giving the following equilibrium constant:

$$K_\mathrm{sp}^{-1} = \frac{a(\ce{A_mB_n})}{a(\ce{A^n+})^m a(\ce{B^m-})^n}\tag{2}$$

Notes concerning equation $(2)$:

technically, this would be a precipitation equation; the solubility product is derived from the opposite equation hence the superscript $-1$.
activities should be used here, and not concentrations. This makes a difference as it allows for an activity of a solid (which has no concentration in solution — being solid outside of the solution).

Going the dissolution pathway, the activities of $\ce{A^n+}$ and $\ce{B^m-}$ are in the numerator and the solid’s activity is in the denominator. Thankfully, the activity of any solid compound is always $1$ per definition. That gives us $(2')$:

$$K_\mathrm{sp} = a(\ce{A^n+})^ma(\ce{B^m-})^n\tag{$2'$}$$

Which is the way the solubility product is typically represented. Now since this is an equilibrium process, Le Chatelier’s principle influcences it. Most importantly, you can add any one reactant to generate more of the product and vice-versa. If you take a look back at $(1)$, you’ll notice that the dissoluted ions enter the equation separately. Thus, we can use higher concentrations of one ion to reduce the overall solubility since then less of the other ion can remain dissolved.

This already tells us that all of options B through D result in a lower solubility since bromide ions (or silver ions) are dissolved. But how do we compare the three? The key here is that a $\pu{1M}$ solution of a salt does not necessarily mean that all its ions have a concentration of $\pu{1M}$. Instead, for $\ce{CaBr2}$ two moles of bromide are in solution for each mole of salt so if $c(\ce{CaBr2}) = \pu{1M}$, the concentration of bromide will be $c(\ce{Br-}) = \pu{2M}$.

Therefore, the solubility of $\ce{AgBr}$ will be lower in $\ce{CaBr2}$ (twice the concentration of bromide) and lower still in $\ce{FeBr3}$ (thrice the concentration of bromide). Hence B is your answer.


Why does the addition of oxygen render materials colourless?

Many organic dyes (arguably, all of them) produce color due to electron transitions in conjugated $\pi$ systems. Strong oxidants (not just the peroxide, mind you, but also hypochlorite bleach and others) snap the double bonds and thus destroy these conjugated systems. So the color vanishes.

This is not applicable to inorganic dyes, which usually would not get bleached by peroxide. On the contrary, some transition metals ($\ce{Ti^4+,\;Cr^6+}$, and some others) which have little or no color by themselves would produce intensely colored complexes when hydrogen peroxide is added. These reactions are sometimes used in analytic chemistry to discover their presence.


Here entropy change will be zero because resistor convert the electrical energy into heating so it take as work not heat so entropy change will be zero


Can the thermodynamic information of a compound be calculated or must it be found emperically?

For gaseous compounds heat of formation can be calculated using quantum chemistry packages as a difference between energies, assigned by the package, though inaccuracies are possible. Entropy of formation is a bit trickier, it requires calculation of vibrational and rotational states, but still pretty straightforward. For condensed phase (liquids, solids) the situation is harder, as knowledge of structure of solid state must be known beforehand, or full exhaustive search of possible crystal structures must be performed.

In practice, however, both heat and entropy of formation are found experimentally.


Does 2-chloro-2-methylpentane react with NaI faster in acetone or water?

It  is simple this is the famous finkelstein reaction.

As you can see in the image NaCl is formed which is insoluble in acetone hence it precipitates and the reaction cannot go reverse.

But if we take water the reaction will shift in backward direction as NaCl is soluble in water.

 That is the reason we take acetone.


pH probe bulb - what is happening within the glass?

My reference for all information and pictures is Harris' Quantitative Chemical Analysis, 9th ed., pp 347-9. I think it'll be worth your while to consult those pages, but I'll try to summarize the important points here.

An ion-selective electrode is characterized by a thin membrane that, well, selectively binds ions. The glass electrode is an ion-selective electrode for $\ce{H+}$ made of amorphous silicate glass, which consists of connected $\ce{SiO4}$ tetrahedra.

Presumably, no special preparative techniques are required for the glass, and the depth of the hydrated gel layer is mediated by the strength and range of the intermolecular interactions between water and the glass.

Protons are the main ions that bind to the layer, leading to the selectivity of the electrode. They diffuse between the solution and the hydrated gel layer and displace the metal ions originally present on the surface of the glass, which describes an ion-exchange process. Note that they cannot, however, diffuse through the inner glass layer.

A few side remarks.

Equilibrium is reached when the favorable binding of protons to the glass surface is balanced by the unfavorable electrostatic repulsion and chemical potential gradient that result from diffusion into the hydrated gel layer. This provides an equation relating the potential difference to the pH of the solution and allows for $\ce{pH}$ measurement.
Something has to be able to move through the inner layer of the glass membrane to conduct a current and hence allow for a measurement of the potential difference. It turns out that sodium ions can move through this inner layer, but only sluggishly---the resistance of the glass membrane is about $10^8\,\Omega$.


The salt in the oceans came originally from rocks. Rain can dissolve some of the minerals found in rocks, especially rain that is slightly acidic from dissolved $\ce{CO2}$. The chemicals in the minerals are broken down into ions in solution. Some of the ions will tend to precipitate out of solution, or be taken up by living things (for example calcium is removed from water by the shells of molluscs and diatoms). Other ions will tend to build up in the water. Two of the most soluble and common ions are sodium and chloride. Over time these built up in the ocean. The sodium chloride in sea water is what makes the seas salty.


Is ethanol an acid or a base or amphoteric?

Because we are talking about Brønsted–Lowry acids and bases, we are only concerned with whether or not a molecule has (or the degree to which it has) the ability to donate a proton (acid) or the ability to accept a proton (base) or to do either (amphoteric).

The case of ethanol is rather simple to evaluate. Firstly, the hydroxyl proton could be donated to a strong base, although this only happens to a minimal degree in an aqueous solution as water is a much stronger acid than ethanol. Still, ethanol has the ability to act as an acid because of the ability to donate it's hydroxyl proton.

However, aqueous solutions of ethanol are slightly basic. This is is because the oxygen in ethanol has lone electron pairs capable of accepting protons, and thus ethanol can act as a weak base.

Summary - TL/DR:
As ethanol has the ability to both donate and accept protons, it should be considered an amphoteric compound with respect to Brønsted–Lowry acid-base theory.


Can a part of a compound be a reducing agent?

For accounting purposes, we can assign oxidation numbers to all the atoms. Oxygen could be -II in all species here, then sulfur would be +IV in $\ce{SO4}$ and -II in $\ce{Cu2S}$ (oxygen and sulfur are in the same group and both are more electronegative than copper, so assigning the same oxidation number in $\ce{Cu2S}$ and $\ce{Cu2O}$ makes sense). Oxidation numbers of atoms in elements are zero, and the copper in both compounds would be +I to balance out the -II of the chalcogens.

Once you are in agreement with your discussion partner on these oxidation states, you can make a statement about which atoms undergo oxidation or reduction: Sulfur gets oxidized, and copper in both copper compounds gets reduced.

In cases where only one atom in a compound changes oxidation states, or when all changes are in the same direction, we do talk about the compound being reduced or oxidized. For example, it is common to say that when you oxidize an aldehyde, you get a carboxylic acid (and when you reduce it, you get an alcohol or an alkane).

On the other hand, if you hydrate an alkene, some folks don't consider that a redox reaction.


Electrolysis of copper sulphate using copper electrodes: Mass lost from anode is greater than mass deposited on cathode

$\pu{3 V}$ is too high. Kinetics of deposition of $\ce{Cu}$ is saturated, so another electroactive system kicks on, creation of hydrogen. Similarly for anode. Try to evaluate volt-ampere characteristic of the system and stay on the current plateau before gases start to be created.
Generally, is is more advantageous to use a power source keeping constant current rather than constant voltage.
If gases are being evolved, you cannot expect the dissolved and deposited mass of copper would match, as various portions of the current are spent on gases, instead of copper. The current density that Maurice suggests is a good one.
For low enough voltage, copper deposition + dissolution can absorb the current. At particular current density, one of these reactions is saturated and cannot take more current any more, being diffusion controlled. Then either hydrogen, either oxygen starts to be created, depending on the electrode where it occurs first. When it happens for the other reaction as well, the other gas gets created as well.

Each electrode has a voltampere characteristic given by I=f(E) function, that has a shape of irregular rounded step-wise slope. When a cathodic potential gets low enough value for a given compound/ion, current starts to raise. When reaction gets saturated, i becomes being controlled by ion diffusion and there is zero concentration at the electrode. Then the current reaches a plateou. The plateou ends at a potential where another redox system starts to interact. Similarly for anode.
When an external voltage is applied, electrodes gain such potentials to meet 2 conditions:

The potential difference equals to the external voltage.
Currents of both electrodes, according to the above function, are equal but the sign.

Lower current density means more positive cathode potential, where $\ce{Cu^2+}$ gets selectively reduced. Higher density means lower cathode potential where both $\ce{Cu^2+}$ and $\ce{H3O+}$ may get reduced. It is like if you used mildly reducing reagent, reducing just copper ions versus strongly reducing reagent, reducing also hydronium ions. Similarly for the anode.
The only necessery thing is the cathodic current matches the anodic current. Processes on both electrodes are independent. There is no "quantum entanglement" between these processes. If $\ce{Cu^2+}$ diffusion and $\ce{Cu}$ deposition can absorb all the cathodic current, there is no need of hydrogen creation.
Try to measure the steady current as a function of the applied voltage. You can use a small resistor is a serie with a cell, measuring voltage on it, which is equivalent to the passing current as $I = U_\mathrm{R} / R.$ Find the first current plateau. Or the "like to be a plateau, just if the 2 voltage steps were not so close together."
Choose the voltage from this plateau, but be aware it may change with the change of the cell geometry or electrolyte composition.

Even in cased hydronium ions do not participate on electrode reactions directly, they can participate indirectly by influencing I=f(U) characteristics of the cell via:

Keeping or affecting chemical state of involved ions. ( E.g. pushing equilibrium $\ce{[Cu(H2O)4]^2+ + H2O <=> [Cu(H2O)3(OH)]+ + H3O+}$ to the left.
Balancing the local gradient due charge disbalance
Increasing conductivity and therefore the solution potential difference, that subtracts from the applied voltage.


Magnetic and acidic properties of nitric oxide

$$\ce{2NO + O2 → 2NO2}$$
Nitric oxide reacts with oxygen to produce nitrogen dioxide. This reaction takes place quickly at room temperature in air. (chemiday)
The bond order in $\ce{NO}$ is 2.5
The oxidation state of $\ce N$ is +2, and this is a colourless, paramagnetic gas (ie. it is a radical, or contains an unpaired electron). (source)
Neutral oxides are those oxides which show neither basic nor acidic properties when they react with water. Examples include carbon monoxide $(\ce{CO})$ and nitrous oxide $(\ce{N2O})$ which are only slightly soluble in water, and nitric oxide $\mathbf{(NO)}$ which is appreciably soluble in cold water. (source)

So, option 3 is the correct answer as it is an inaccurate statement.


Prepare 1 liter of 28% NH3(aq) solution from 1 gal of %10 NH3(aq)

Disclaimer: ammonia is extremely hazardous including flammable in some vapor concentrations.

  How do I know when I have reach %28 NH4OH concentration?

You can measure the specific gravity and compare to the table on page 2 of
this document:

Next, it is important to understand that if you heat/boil/distill the 10% solution the vapor coming off is not 100% ammonia, it is a mixture of ammonia and water, but with increased ammonia concentration.

To predict the percent ammonia in the vapor phase, you need to look at the dew point line in a table like Fig. 1.2 here, but that one is for 16 bars or Fig. 2 of Ganesh et al. Journal of Mechanical Engineering Research Vol. 3. (1), pp. 25-39 that shows many pressures including 1 bar.

As the original solution is heated, the system progresses toward the upper left in Fig. 1.2/Fig.2, the liquid phase along the bubble point line and the vapor phase along the dew point line.  The concentration of ammonia in both phases progressively decreases.  Eventually, the concentration of ammonia in the vapor phase drops below the concentration of ammonia in the original solution.

From Fig. 2 of Ganesh et al. it can be seen that at 1 bar starting from 10 weight% ammonia liquid phase, the initial vapor phase concentration would be close to 28% ammonia.

This shows that even if you added no water to the condensate, there is no way you can get 1 liter of 28% ammonia from 1 gallon of 10% ammonia in one simple distillation step.

Only through fractional distillation could you get that much 28% ammonia from that little 10% ammonia.

Also, practically speaking, it will be difficult to avoid lossing ammonia vapors to the environment and safety needs to be considered.


How to eliminate electrostatic charges from titanium dioxide powder

The best way I've found is to use an anti-static gun (sold by chemical companies such as Sigma-Aldrich).

If this isn't available, taking gloves off helps enormously, as the gloves can easily become 'static' in the same way as rubbing a balloon against some fabric. Gently tapping the flask against a metal object also helps, as the metal 'takes' the static away.

I have also known chemists to wear the same anti-static/grounding bracelets used by people who work with computers so as not to damage the electrics internally, however for a bench-chemist only occasionally working with fine powders, this would probably be over kill.


One possible way to intuit the result is to remember that all of the orbitals are normalized. Thus, the contribution of a single AO to the three MOs must add up to 1, and the contribution of the three AOs in each MO must also add up to 1. (This is a bit of an oversimplification, since the normalization is of the square of the wave function, but conceptually it's the same idea). So if a particular AO contributes a lot to one MO, it is proportionally less represented in the others. Also, the other AO's have to contribute less to the first MO in order for it to have the right magnitude.
In MOs composed of AOs from atoms of differing electronegativity, the lower energy MOs have a larger contribution from the more electronegative atom (and therefore smaller contribution from the less electronegative atoms). The higher energy MOs must therefore have a smaller contribution from the electronegative atom.
As a very simple first example, consider the $\sigma$ and $\sigma^*$ MOs of a polar diatomic molecule such as HF. The valence F AO is much lower in energy than the hydrogen orbital, so the bonding MO has a large contribution from F and very small contribution from H. In fact, the MO is nearly indistinguishable from the fluorine AO, as evidenced by the bond being nearly completely ionic. In order for the normalization to be correct, the antibonding orbital must be the reverse - a very small contribution from F and large contribution from H.
A slightly more complicated example is a triatomic like H2O. Here is a drawing of the lowest three valence MO's for XH2, when X is more electronegative than H:

(This is a portion of Figure 7.3 of Albright et al. Orbital Interactions in Chemistry).
You can see that again the electronegative atom (O) is overrepresented in the lowest energy MOs, at the expense of the H atomic orbitals. Therefore, in the high energy orbitals, we have to compensate with the reverse, an overrepresentation of the H orbitals (as seen in the 3a1).
Now getting to your question of C-C-O, again the O is most electronegative, so it contributes most to the lowest energy orbital. That means it has less to contribute to the higher energy orbitals, where the carbon orbitals are the bigger contributors.
This overly simplistic approach is harder to apply to the difference between the two carbons, for which you really have to incorporate the actual math of the atomic orbital contributions, but maybe it at least helps with the concepts.


Why does pool chlorine smell only when wet?

According to americanchemistry.com:

  Pool smell is due, not to chlorine, but to chloramines, chemical
  compounds that build up in pool water when it is improperly treated.
  Chloramines result from the combination of two ingredients:  (a)
  chlorine disinfectants and (b) perspiration, oils and urine that enter
  pools on the bodies of swimmers. Chlorine disinfectants are added to
  pool water to destroy germs that can give swimmers diarrhea, ear aches
  and athlete's foot. Perspiration, oils and urine, however, are
  unwanted additions to pool water. By showering before entering the
  pool, and washing these substances from the skin, swimmers can help
  minimize pool smell.   When chlorine disinfectants are added to water,
  two chemicals are unleashed that destroy waterborne germs:
  hypochlorous acid, $\ce{HOCl}$, and hypochlorite ion, $\ce{OCl-}$.

Some chemical reactions that then lead to the formation of chloramines include:

$$\ce{HOCl + NH3 -> NH2Cl + H2O}$$
$$\ce{HOCl + NH2Cl -> NHCl2 + H2O}$$
$$\ce{HOCl + NHCl2 -> NCl3 + H2O}$$

These are fairly volatile compounds, but can also, to different degrees, "stick" to your skin.  When you take a hot shower, these are then liberated and you can smell it again.


First, let's tackle what an irreversible reaction is. Baking a cake, for example, is commonly considered "irreversible." The eggs, flour, milk, etc. which you started with are no longer present in their original forms after the cake has been baked.

Now, can you reconstitute the eggs, flour, milk, etc.? Of course you could take the molecules of the cake and piece them together to make eggs, flour, and milk, but doing so would be a very formidable task. A lot of energy will have to be expended to reconstitute the egg, for example, because eggs are made of protein and with heat (baking) proteins are denatured (lose their original 3D conformation and possibly their linear makeup or amino acid sequence as well).

Another example of an irreversible reaction is combustion of hydrocarbons - i.e. burning gasoline in your car. This reaction is definitely reversible in the long run - it just takes a few million years, a lot of heat, and a lot of pressure for the hydrocarbons to be formed again. But for semantic's sake we say that combustion is irreversible because it's not reversible on a really relevant time scale.

So in general, the difference between irreversible and reversible isn't exactly well defined - i.e. it might be hard to reverse a reaction (turn a cake back into eggs, etc.) or it might take a long time to reverse a reaction, but what's hard? What's long? Tellinging, the IUPAC (International Union of Pure and Applied Chemistry) does not try to define irreversible vs. reversible reaction.

A deeper treatment would involve discussion of equilibrium constants (symbolized by capital K) but I'll let another poster get into this if necessary.

  It is generally said that reactants react so that they can achieve a
  lower energy state. Then why does a reversible reaction occur in the
  first place?

Good question. Remember that we can always add energy to make an unfavorable reaction proceed. For example, the sodium ion, which is isoelectronic with neon, is stable with a full octet of electrons. However, we can still take away more electrons. It just takes a rather sizable application of energy.


This question is very good as it brings together many disciplines$-$astronomy, chemistry, atmospheric physics. Regretfully, as a consequence it is hard to compentently answer by a member of one field. I look forward to a healthy discussion in the comment section.

Hydrogen

A scientifically sound story ought to be such in most aspects. Specifically,

how does that planet possess hydrogen in high concentrations?

Hydrogen is a molecule with small molecular mass. As such it occupies the high end of Maxwell-Boltzmann speed distribution:

$$f(v)\begin{align}&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}.$$

$\ce{H2}$ would thus escape the atmosphere over time since they often reach escape velocity

$$v_e\approx\sqrt{\frac{2GM_{planet}}{r}}.$$

Possible explanations why this would not have occured include:

1. Relatively young planet

A newly-formed celestial body is very likely to be hot. As such, it will not contain liquid water, definitely not to the extent of forming oceans.

One study by Tian et al. 'A Hydrogen-Rich Early Atmosphere' ($2005$) suggest the possibility of oceans and a concurrent hydrogen mixing ratio of $30$%. However, this claim has been contested.

  Tian et al.’s argument that hydrogen escape was inefficient on early Earth assumes that the upper atmosphere would be cold as on Venus, because the atmosphere is composed mostly of CO2. Their claim is controversial because their model depends on gases other than hydrogen to provide the radiative cooling, but their model does not actually include gases other than hydrogen. In particular, the key assumption that the upper atmosphere was cold is not obvious and needs to be addressed quantitatively. [----]

  Visconti’s calculations contrast markedly with the cold thermospheres in Tian et al.’s pure hydrogen escape models.

  source: Zahne et al. 'Earth’s Earliest Atmospheres'. (2010) link

2. Massive size and a magnetoshpere

Massive size would mean it has not really lost any atmosphere. Unless it is close to a star, in which case hydrodynamic escape would become important. These planets are mostly gas giants. We do not really know what is going on at their center, but it is almost certainly not a liquid ocean.

Without a magnetosphere the planet's atmosphere would be washed away by solar wind, similarly to that of Mars.

3. Hydrogen-producing microorganisms, e.g., a hyperactive, modified Chlamydomonas reinhardtii

This would be your best  bet. I am not qualified to discuss the likelihood of simultaneous thriving of both hydrogen-producing and oxygen-producing organisms. At face value$-$and this is speculation$-$it is probably not viable. Especially with so high pressures of hydrogen (high solubility).

What if?

Whether the reaction between $\ce{O2}$ and $\ce{H2}$ actually takes place (with considerable speed) depends on numerous variables. The ratio of these gases, temperature, pressure.

  Can there be oxygen dissolved in water in a hydrogen atmosphere?

Short answer is yes.

  [Is it] a significant amount?

Probably not. There ought to be some equilibrium amount but negligible. Even leaving aside the reaction between hydrogen and oxygen, the oxygen would be used to oxidise minerals. After these are, in a sense, saturated and when aquatic organisms adapt to more oxygen-rich water will the concentration of $\ce{O2}$ grow. It is now that the high hydrogen content becomes an issue.

Extra discussion requires actual calculation with supplied parameters, and the definition of what constitutes a significant amount. I have provided some sources for the kinetics of $\ce{O2/H2}$ mixture if this a route you wish to undertake.

Marcus Ó Conaire et al.. 'A Comprehensive Modeling Study of Hydrogen Oxidation' link
Alekseev et al.. 'The effect of temperature on the adiabatic burning velocities of diluted hydrogen flames: A kinetic study using an updated mechanism' Combustion and Flame, 162, 5, 1884$-$1892. (2015). link
Martin Hersch. 'Hydrogen-Oxygen Chemical Reaction Kinetics in Rocket Engine Combustion'. NASA Technical Note. (1967). link

Rogue planets

In a paper by David J. Stevenson 'Life-sustaining planets in interstellar space' (Nature vol. 400, page 32, 1999, link) he proposes that planets which form sufficiently far from their star or are gravitationally accreted may be sent on a course outside their star system. These planets could retain their hydrogen because the effective temperature would be fairly low: estimated to be about $30 \ \mathrm{K}$. According to Stevenson,

  [P]ressure-induced far-infrared opacity of $\ce{H2}$ may prevent these bodies from eliminating internal radioactive heat except by developing an extensive adiabatic (with no loss or gain of heat) convective atmosphere.

Thus surface temperatures may be enough for liquid water to exist. Assuming the planet collided with enough asteroids and comets at a stage where water was retained we arrive at the conditions:

hydrogen-rich atmosphere, Stevenson estimates that the pressure could be $100$ to $10000\ \mathrm{atm}$, so your intuitive estimate was pretty adequeate;

Sidenote: Pressure at the bottom of oceans on Earth is (at the deepest) about $1000\ \mathrm{atm}$. So we would have the same amount of pressure on the surface of the ocean.

liquid oceans.

A possibility for life would be anaerobic chemoautotrophs: they do not need oxygen, and would be able to be their own source of organic molecules. They do require a geothermal energy source. Volcanic activity at the bottom of the ocean floor suffices. Through random genetic mutation these organisms might transition from being obligatory to being facultative anaerobs. Remember: if $\ce{O2}$ is your product, $\ce{CO2}$ is likely the raw material the organism is utilising.

I have no idea how things could proceed from here. We could have a cosy ecosystem in such an ocean (especially near vents) but how and why there would be evolutionary pressure toward oxygen-rich life at other depths... I simply do not know. Only way is to decrease hydrogen content but, well, we need that to keep warm :-). Perhaps a gradual enrichment or replacement of hydrogen by other gases without lowering temperature further is possible? (see suggestion below)

As Stevenson notes,

  If life can develop and be sustained without sunlight (but with other energysources, plausibly volcanism or lightning in this instance), these bodies may provide a long-lived, stable environment for life (albeit one where the temperatures slowly decline on a billion-year timescale). The complexity and biomass may be low because the energy source will be small, but it is conceivable that these are the most common sites of life in the Universe.

Suggestion

If you wish for more information, consider follow-up questions on our

biology,
physics,
world building,
astronomy site.

I suggest against asking in all sites at once. Consider taking a concrete point (for example the transition from anaerobic to aerobic on biology.SE). Read their guidelines to be sure it is on-topic, and check for possible duplicates. Link to this question when asking again.

Good luck!


Luminescent materials under IR radiation?

There are two types of materials I'm aware of that produce light in the way I think you want: two-photon fluorescence and photon upconversion. Two-photon fluorescence is basically the same as normal fluorescence but two photons are absorbed simultaneously, emitting a single photon when returning to the ground state. Photon upconversion similarly involves the absorption of multiple photons, but sequentially (by various mechanisms), followed by emission of a single photon.

Two-photon fluorescence is mostly seen used for two-photon excitation microscopy and useful dyes tend to be big conjugated organic systems. Materials used for photon upconversion are generally lanthanide-doped nanoparticles and they've found many uses in applications traditionally served by quantum dots and organic fluorophores. They're both particularly useful in medical imaging as IR can penetrate tissue deeply, but upconverting nanoparticles have a big advantage in that two-photon fluorescence requires very high excitation intensity (pretty much just lasers) as both photons must be absorbed simultaneously, but upconversion processes are sequential and can thus occur at lower intensity. Though, for some applications, the intensity requirements for two-photon absorption are useful as one can play tricks with focusing such that only a small region receives enough excitation light to emit, making imaging small areas possible.

As for a database of compounds, Thermo-Fisher has a catalogue of two-photon dyes, but I don't know that there are a lot of commercial manufacturers of upconverting nanoparticles, though Sigma-Aldrich formerly carried up converting nanocrystal under the Sunstone brand.


Elimination in α,β-dibromohydrocinnamic acid

Dibromohydrocinnamic acid can eliminate in a number of different ways as shown in the following diagram. I'm not sure what conditions lead to cinnamic acid, but I know it can happen.  You have recently been asking questions about E1 and E2 eliminations, so my guess is that you are running a well-known lab experiment that generates cis- and trans-bromostyrene and allows the student to study the effect of reaction conditions on the mechanism of an elimination reaction.

Usually, the experiment starts with the bromination of trans-cinnamic acid.  The bromination is run under conditions that lead to trans-addition of bromine so that only the erythro (see here for diagrams and mechanism) or (S,R) [and its mirror image (R,S) enantiomer] dibromocinnamic acid diastereomer is formed.  The other dibromohydrocinnamic acid diastereomer, the threo isomer [(R,R) and its (S,S) enantiomer], which can only be formed by cis-addition of bromine is not produced.

If you build a model or draw a Newman projection it will make it easier to understand what follows.

Using the erythro [(S,R) and (R,S)] diastereomer obtained from the above bromination, running the elimination reaction in a solvent like acetone (a solvent that won't stabilize a carbonium ion as much as protic solvents like water or methanol) you drive the mechanism towards the E2 elimination side.  With your model or drawing, put the carboxylate group (the base has removed the acidic carboxylic acid proton to produce the corresponding carboxylate anion) and the bromine on the next carbon in an anti-periplanar arrangement (you've discussed anti-periplanar in your earlier questions on the E2 mechanism) and see how in a one-step, concerted elimination of both carbon dioxide and bromide you wind up with the cis-bromostyrene isomer.

If you ran the reaction in a protic solvent (like water or methanol) that stabilizes carbonium ions, you push the reaction towards the E1 mechanism involving loss of the bromine adjacent to the benzene ring to produce a stabilized benzylic carbonium ion.  The stable carbocation ion has an appreciable lifetime and will rotate so 1) the soon-to-be-departing carboxylate group is lined up with the carbocation p-orbital on the adjacent carbon and 2) that the benzene ring and remaining (bulky) bromine are as far apart as possible (the lowest energy conformation).  Elimination of carbon dioxide from this conformation produces trans-bromostyrene.

Typically, the student is given several test tubes and the reaction is run under a variety of conditions where solvent and reagents are changed and then the amounts of cis- and trans-bromostyrene are measured by nmr.  From this information the student can see how solvent and reagent can shift the mechanism.  See this link for a bit more detail on the actual experiment along with some good Newman projections.

BTW, the lab should smell nice since the trans-bromostyrene produces a hyacinth-like odor.


Why do square pyramidal molecules have only two vertical mirror planes?

Either your book is wrong or you are misunderstanding what it is trying to say.

There are 4 vertical planes of symmetry.

See for example GROUP THEORY AND ITS APPLICATIONS IN CHEMISTRY, SECOND EDITION , section 1.5.2

See especially:

  Vertical planes are further classified into two, namely $\sigma _v$ and $\sigma _d$ The $\sigma _v$ planes fall along the bonds and the $\sigma _d$ planes bisect the bond angles in molecules. The subscripts v and d stand for vertical and diagonal. However, both $\sigma _v$ and $\sigma _d$ are vertical planes


PEG (polyethylene glycol) and ITO (indium tin oxide)

PEG "silane" is $\ce{HOCH2CH2O-PEG-CH2CH2CH2-Si(OCH3)3}$.  (Pt-catalyzed $\ce{HSi(OCH3)3}$ addition to the terminal allyl monoether.)  A metal oxide surface can be hydrated, (bulk)M-OH.  The TMS end hydrolyzes and dehydrates to tetrahedrally covalently bonded $\ce{HOCH2CH20-PEG-0Si(OM#bulk3)}$.  The alcohol polyethoxy end hydrated waves in the water, and the protein ignores the PEGylated surface.

An especially fine biocompatibilizer rendering nearly any organic surface except silicone rubber invisible to protein deposition and denaturation is poly(N-vinyl pyrrolidone).  It can be linearly radiation grafted into organic surfaces (visualize with Congo Red).  Plexiglass gives a 19 degree average contact angle in saline and implants with neither cellular or humoral reaction.  NVP plasma deposition gives you sludge.  There is no obvious (inexpensive) chemical attachment of its low polymer or oligomer to a surface.  Pity.


Could sodium in ethanol reduce amides to amines?

Yes, that is used as a pathway for synthesis of amines from amides, albeit rarely.

I quote from the abstract of a communication that reports the same: (emphasis mine)

  The reduction of amides to their corresponding amines has been used extensively in the synthesis of many drugs and their intermediates. This reaction has usually been carried out by means of metal hydrides (especially lithium aluminium hydride) or by catalytic hydrogenation over various catalysts. Metal and alcohol has been used for the reduction of ketones to their corresponding hydroxy compounds and oximes to their corresponding amines, but the reduction of amides under these conditions, although previously reported, is not well documented in major works of organic chemistry. This procedure is specially useful for industry since it is safer, less expensive, requires less time and provides higher yields (65-80 per cent) when compared with lithium aluminium hydride or catalytic reduction methods.

The emphasized text is written in a slightly confusing language but nonetheless confirms that amides can be reduced to amines using $\ce{Na/ROH}$, otherwise known as Bouveault-Blanc reduction.

References:

Bhandari, K. , Sharma, V. L. and Chatterjee, S. K. (1991), ChemInform Abstract: A Convenient Method for the Reduction of Amides to Their Corresponding Amines.. ChemInform, 22: no-no. DOI: 10.1002/chin.199118077


Why is up spin in the px orbital and down spin in the py orbital considered an excited state?

Hund's rule of maximum multiplicity states that for a given electron configuration, the term with the lowest energy has a maximum multiplicity $2S+1$. $S$ is the total spin angular momentum of all electrons in the partially-filled subshell.

For the application of the rule, we consider the 2 electrons in the $p$ subshell and add up their spin quantum numbers to obtain $S$ and the multiplicity. For the ground state configuration, with both electrons "spin-up", we get

$$S=+\frac{1}{2}+(+\frac{1}{2})=1$$
$$2S+1=3$$

For your proposed configuration with one $p$ electron "spin-up" and the other one "spin down":

$$S=+\frac{1}{2}+(-\frac{1}{2})=0$$
$$2S+1=1$$

Since the multiplicity value is smaller, this configuration is not the one which is lowest in energy, and therefore cannot be regarded as the ground state.