What causes positive and negative deviations when weighing analytical samples?

There can be plenty. There is no definite list. I would suggest to look up the following terms in Skoog or Daniel Harris' Analytical Chemistry. Worth a trip to the library now :-) The following concepts give rise to systematic errors

(i) buoyancy error
(ii) temperature differences
(iii) electrostatic charge build up on the sample
(v) moisture
(iv) above all loss of calibration of the balance.


How could you dissolve a coating of CaCO3(s) from a metal plate?

It might take some time, but I would use vinegar.

The acid in vinegar is strong enough to dissolve the calcium carbonate, but it is not strong enough to dissolve metals (unless the metal reacts with water).


Degree of dissociation of acetic acid at infinite dilution

As the dilution becomes infinite for the amount of acetic acid, then the acetic acid's contribution to the solution's acidity falls, and the pH of the water is controlled by the acid dissociation of water. So the pH falls to 7.00. Thus,

$$\begin{align}
\frac{(10^{-7})[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 1.85 \times 10^{-5} \\
[\ce{CH3COO-}] &= 185 [\ce{CH3COOH}]
\end{align}$$

The fraction of unionised acid, $x$, is therefore given by

$$\begin{align}
x &= \frac{[\ce{CH3COOH}]}{[\ce{CH3COOH}] + [\ce{CH3COO-}]} \\
&= \frac{1}{1+185} \\
&= 0.54\%.
\end{align}$$


Chemical energy is really nothing but electric potential energy.

The crucial difference is that chemical energy involves energy that can be extracted by breaking/making bonds.

For example, we say that the reaction $\ce{2H2 + O2->H2O}$ is accompanied by a release of energy. This energy is the difference of potential energies of the bonds in the reactants and products. Which potential energies? The bond potential energies. Bonds are, at the most basic level, formed due to attraction and repulsion of electrons and nuclei (coupled with some quantum mechanical effects). This carries some potential energy. The change in this potential energy achieved by breaking the bond is the "bond energy", and forms part of your "chemical energy". Some other parts are lattice energy, hydration energy etc, which are (in the end), due to electromagnetic forces/electromagnetic potential.

So yeah, chemical energy is just a "subset" of potential energy,

  If chemical energy is an "own" form of energy, why does it not have a specific equation?

Since chemical energy is an amalgamation of various types of electric potential energy in atoms, it cannot be given a formula (though you can calculate it for some compounds if you know some standard values).

  Is chemical energy an energy that is "conserved" and if so, why is chemical energy not just potential energy?

Chemical energy isn't really conserved anywhere unless there is no change to the chemical composition of the system. However, in a chemical reaction, most of the time any changes in energy are compensated for by absorption/release of heat. But it is just potential energy, as mentioned.

  Did we have chemical energy before the big bang or when did it start?

Meaningless question, there was no "before" the big bang. Also, energy really isn't something that you should classify into compartmentalized types.

But, since atoms formed a few minutes after the Big Bang, there wasn't anything to facilitate the existence of any chemical energy -- so the first occurrence of chemical energy was then.

  Can you have chemical energy at or close enough to absolute zero?

Yep. Bonds exist at absolute zero. Note that we can never reach absolute zero, but bonds exist both close to and hypothetically "at" absolute zero.

  How much of the total energy in the universe is chemical energy?

I can't answer that, and I doubt it can be answered to any reasonable approximation. I think the percentage would be rather low -- our universe is mainly stars (besides being empty space). The average star doesn't have many chemical bonds--it contains hydrogen/helium plasma, and plasma can't exactly form bonds.

  What's the point of many different definitions of energy?

In certain situations, you know that there is no method to facilitate conversion of one form of energy into another. In such a case, one can separately conserve that form of energy. So the forms of energy help in analyzing/solving a situation, but not much more.


What is the perfect definition for chirality?

The correct definition of chirality is given in the IUPAC gold book as follows:

  chirality
  The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

From this we can deduce, that all organic compounds, which have tetrahedral carbons with four different ligands, are chiral. This is a rule of thumb and an easy way to recognise chirality.
Glycine is optically inactive as it has an internal mirror plane:

Often we teach this as a rule or a definition, because it is so easy to comprehend and to see.

In inorganic chemistry, we usually deal with more complex structures, where it is not necessary, that all ligands have to be different. One example for this is the cation in tris(ethylenediamine)cobalt(III) chloride:

Another interesting example is (Λ/Δ)-cis-dichlorobis(ethylenediamine)cobalt(III) chloride.


Effective chemical for protecting hide glue from microbial attack

Silver salts are used to kill bacteria, algae and fungi.  Different types of clothing are 'laced' with silver to kill the bacteria that cause body odor.  Silver spoons were suspected historically to have self-sanitizing properties.  I ran a small analytical chemistry lab that was exempt from most emission oversight, but we could not dispose of silver waste into the sewer as it would destroy the 'good' bacteria at the water-treatment plant.  Different silver compounds, often phosphates, are used to strike a balance between required toxicity and durability/longevity.  If I were starting your project, I would at least start with a good literature search along these lines and some small scale proof-of-concept experiments if it seemed at all feasible.

I do want to be clear however, that I know of no reason to expect silver to be toxic to beetle larvae, at least not at the concentrations that would provide anti-microbial protection. Maybe you would need a second additive for that purpose


How does mCPBA convert an amine to an N-oxide?

Firstly, any stereoselectivity in this reaction is not affected by the second step you drew (which is a simple proton transfer). This is because your second step neither creates nor destroys any stereochemistry present in the reactants.

However, mechanistically it is probably easier to deprotonate in the same step using the carbonyl oxygen of mCPBA.


What chemicals are going to be used to make the smoke black or white?

I think Titanium Tetrachloride would be a good option for white smoke (I've heard it's used quite often in smoke bombs). But it releases quite a lot of $HCl$ gas and so it might not be very safe.
A quick Google search comes up with a lot of answers to your question though. This site even answers your question in the same context (The elections going on at the Vatican):
http://thehappyscientist.com/science-experiment/black-and-white-smoke
Apparently you can make white smoke using Magnesium powder as well. (The white colour is due to fine particles of Magnesium Oxide $MgO$ )
It seems the Vatican has experimented using many sources before (Military flares, Smoke bombs and so on) though apparently they didn't turn out too well...
EDIT
The Wikipedia article on Coloured smoke gives another possibility. You can add Titanium Dioxide ($TiO_2$) to a mixture of Potassium Chlorate ($KClO_3$) and Lactose for white smoke. $KClO_3$ is a powerful oxidizing agent which reacts with Lactose (which is a reducing sugar) to create an explosion. $TiO_2$ is safe and is used as a pigment in most white paints and as a food colouring as well.
EDIT2
The official recipes have been revealed by the Vatican press office. See this story on the New York Times website.
The previous edit was partially correct. They use a mixture of $KClO_3$, lactose and pine rosin for white smoke. For black smoke, they use Potassium perchlorate ($KClO_4$), anthracene (an aromatic compound made of three fused benzene rings) and sulfur.
The process (quoted from the site):
"The chemicals are electrically ignited in a special stove first used for the conclave of 2005, the statement said. The stove sits in the Sistine Chapel next to an older stove in which the ballots are burned; the colored smoke and the smoke from the ballots mix and travel up a long copper flue to the chapel roof, where the smoke is visible from St. Peter’s Square. A resistance wire is used to preheat the flue so it draws properly, and the flue has a fan as a backup to ensure that no smoke enters the chapel."


The dissolution of calcium carbonates in water

Regarding the slurry statement, you can only make slurries with insoluble compounds or they would dissolve and your slurry would go away leaving you with a clear solution!

Your attempt to alter the eggshell composition by extracting with the surfactant should have had little effect.  You didn't give any quantitative results from your ICP analysis, but I think one or both of two things happened:
1) There was residual Ca from the surfactant remaining on your eggshell
 and/or
2) You may have extracted the few percent of protein that makes up eggshells, leaving a more "concentrated" $\ce{CaCO3}$ residue as compared to the untreated shells.


The method is only outlined below, the full details should be in your textbook or lecture notes.

Your $H^{(1)}$ is just the (small) perturbing potential energy added to the original hamiltonian so you could write $H^{(0)}+\lambda V$.  ($\lambda$ is dimensionless and varies from 0, no perturbation, to 1, full perturbation ).

The Schroedinger equation is now $(H^{(0)}+\lambda V)\varphi_n=E_n\varphi_n$  where $\varphi_n,\; E_n$ are the new wavefunctions and energy. These are expanded as a Taylor series in $\lambda$, e.g.  $\displaystyle \varphi_n \approx \varphi_n^0 + \lambda \frac{d\varphi_n}{d\lambda}+\frac{\lambda^2}{2!} \frac{d\varphi_n^2}{d\lambda^2}+ ..\equiv \varphi_n^{(0)}+\lambda\varphi^{(1)}_n +\lambda^2\varphi^{(2)}_n+..$ and similarly for $E_n$.

Next this result and that for  $E_n$ are substituted back into the Schroedinger eqn and powers of $\lambda$ factored out. These remaining equations are solved but none now contain $\lambda$; for example the first order term is $\lambda\left((V-E^{(1)}_n)\varphi^{(0)}_n+(H^{(0)}-E_n^{(0)})\varphi^{(1)}\right)$ where in this new notation  $\varphi^{(0)}$ is the original unperturbed wavefunction and similarly for the energy.

The equation to solve is now $(V-E^{(1)}_n)\varphi^{(0)}_n+(H^{(0)}-E_n^{(0)})\varphi^{(1)}=0$.

After (formally) integrating the equation some fiddling around you find that $\varphi^{(1)}$ is needed. The important step now is to expand this wavefunction in terms of the original basis set of unperturbed wavefunctions $\phi$. This can be done because these form a complete orthonormal set, thus $\varphi^{(1)}= \sum_k a_k\phi_k$ where $a_k$ are the expansion coefficients and these have to be found.  Again after some more fiddling with equations and using the orthonormality of the $\phi_n$ the perturbation energy $E^{(1)}_n = \int \varphi^{(0)*}_nV\varphi^{(0)}_nd\tau = \int\phi^{0*}V\phi^0 d\tau$.

Notice how only the unperturbed wavefunctions contribute to the energy.  The new wavefunction is $$ \varphi^{(1)}_n=\sum_{n\ne k} \frac{E_n^{(1)}}{E_n^{(0)}-E_k^{(0)}}\varphi^{(0)}$$

which is a linear combination of the unperturbed wavefunctions with the $a$ coefficients as ratio of energies.


Why is there no effective alternative to metal salts in antiperspirants?

Which biological and chemical processes underlie the effect of    antiperspirants that make metal salts so effective?

Aluminium-based complexes present in the antiperspirants react with the electrolytes in the sweat to form a gel plug in the duct of the sweat gland. The plugs prevent the gland from excreting liquid and are removed over time by the natural sloughing of the skin.  From a biological point of view, it interact with the keratin fibrils in the sweat ducts and form a physical plug that prevents sweat from reaching the skin’s surface.

Why are aluminium salts used most commonly?

Aluminium salts have a  astringent effect on the pores causing them to contract, further preventing sweat from reaching the surface of the skin. Aluminium salts are also natural antimicrobial agents, so they control bacteria on the skin, reducing unpleasant odors.
Some of the aluminium salts used in antiperspirants:

Aluminium chloride
Aluminium chlorohydrate
Aluminium zirconium tetrachlorohydrex gly (an aluminium-zirconium compound)


Martin - マーチン has given excellent answer here. I'll give here an supporting review article for your benefit, which also gives some mathematical overview:

Louis S. Kassel, “The Mechanism of the Combustion of Hydrogen,” Chem. Rev. 1937, 21(2), 331–345 (DOI: 10.1021/cr60069a010).

Overall, mechanism can be summerized as follows:
\begin{align}
\ce{H2 + \text{heat} &-> 2 H^.}\\
\ce{H^. + O2 &-> HO-O^.}\\
\ce{HO-O^. + H2 &-> HO-OH + H^.}\\
\ce{H^. + HO-OH &-> H2O + HO^.}\\
\ce{HO^. + H2 &-> H2O + H^.}
\end{align}

The propagation would continue with having $\ce{H^.}$ at the end and the presence of excess $\ce{O2}$.


How to calculate the equivalent weight of a metal in a metal sulphate?

I assume that if you are dealing in equivalent weights then you must be well associated with the concepts of equivalents.
Now,
$$ \textrm{equivalent} = \textrm{weight} / \textrm{eq. weight}$$
and since compounds combine in equal amounts of equivalents we have
$$\frac{\textrm{weight of metal}}{ \textrm{eq weight}} = \frac{\textrm{weight of sulphate}}{ \textrm{its eq weight}}$$
$$\rightarrow 2/x = 2.6/48$$
Hence $x= 39.92$.


Is there a way to calculate bond energies using computational chemistry?

Is there a way to calculate bond energies using GAMESS?

Yes. You would normally model the bound and unbound states

$$\ce{ML_{n - 1} + L -> ML_{n}}$$

The definition of bond energy is based on the enthalpy of reaction in the gas phase (thanks DSVA!), so you would calculate it as follows:

$$\Delta H^\circ = H_{\ce{ML_{n}}} - (H_{\ce{ML_{n - 1}}} + H_{\ce{L}})\text{.}$$

While not strictly bond energy by definition, a value for the reaction Gibb's free energy might also be useful in some contexts:

$$\Delta G^\circ = G_{\ce{ML_{n}}} - (G_{\ce{ML_{n - 1}}} + G_{\ce{L}})\text{.}$$

This takes entropy into account.
For the calculation of Gibb's free energies, most codes calculate the molecular partition function from translational, rotational and vibrational contributions and get enthalpy and entropy from it.
Due to the vibrational contributions, you'll need to diagonalize the Hessian matrix of the system. This will give you an association (minus dissociation) energy value.
(You might take solvation into account through explicit/implicit solvation, use decent level of theory, validate/benchmark, etc.)
There are a lot of approximations and considerations in the process. More can be found here (some equations) and here (more practical).

Observe that

$$K = e^{- \frac{\Delta G^\circ}{R T}}$$

is the equilibrium constant associated with the process above, which is called the stability constant, formation constant or binding constant.
This is widely used to "measure the strength of the interaction between the reagents that come together to form a complex" (source).
Bear in mind that $\Delta G^\circ$ goes inside an exponential, i.e., small errors in $\Delta G^\circ$ become (exponentially) large errors in $K$, so $\Delta G^\circ$ might be more useful in a computational context than $K$ itself.

  Is there a way to determine if the complexes would form?

This is complicated.

Your reaction might be too slow though, i.e., actual formation might depend on transition states. Furthermore, there might be faster processes in competition that you might miss.


What is the relation between Rate of reaction and rate of consumption of reactants?

For your first part you're supposed to multiply it by -3 instead if $\frac{-1}{3}$  as the rate of reaction is given for one mole. That is if $R$ is the rate of reaction it is defined as .
$$\frac{-d[A]}{dt} = \frac{-d[B]}{3dt} = \frac{d[C]}{2dt} = R$$
So the rate of consumption of B is $-3R$.

As for your second part it's a tad unclear for me if you could further edit your question.I could take a shot at answering your second part as well


Experimental verification of the field effect

The inductive effect is transmitted through bonds; the field effect is transmitted through space.

The $\ce{C-Cl}$ bond is polarized due to the electronegativity difference between the two atoms (carbon 2.55; chlorine 3.16), the chlorine end being more negative than the carbon end. We could note this excess of electron density (partial negative charge) around the chlorines in your structures by placing a ${\delta^{-}}$ by the chlorine and a ${\delta^{+}}$ by the carbon.

Now, when the acid ionizes there is also a negative charge associated with the carboxylate group.  When the chlorines are on the same side of the bridge as the carboxylate (syn isomer) the (through space) field effect acts to destabilize the situation because we have placed the partial negative charge around chlorine closer to the negative charge on the carboxylate (placing the negative charges closer together is destabilizing since like charges repel). The destabilization results in a higher activation energy being needed to reach the transition state for ionization.

On the other hand, the anti isomer with the chlorines further from the carboxylate will be less destabilized because the negative charges are further apart.  This means that the anti isomer will have a lower activation energy for ionization compared to the syn isomer.  Therefore, the anti isomer will be more acidic (display a lower pKa) than the syn isomer due to the field effect.


Are clathrates and interstitial compounds the same thing?

Clathrate compound :

  A clathrate is a chemical substance consisting of a lattice that traps or contains molecules. The word clathrate is derived from the Latin clathratus (clatratus), meaning ‘with bars, latticed’. Traditionally, clathrate compounds are polymeric and completely envelop the guest molecule, but in modern usage clathrates also include host–guest complexes and inclusion compounds. According to IUPAC, clathrates are inclusion compounds "in which the guest molecule is in a cage formed by the host molecule or by a lattice of host molecules."

The famous is the methane in ice clathrate $\ce{4 CH4 \cdot 23 H2O}$
Another case are clathrates of noble gases, found in hunting for noble gas compounds.

Many clathrates are stable just at low temperatures and/or high pressures.

Interstitial compound :

  An interstitial compound, or interstitial alloy, is a compound that is formed when an atom with a small enough radius sits in an interstitial “hole” in a metal lattice. Examples of small atoms are hydrogen, boron, carbon and nitrogen. The compounds are industrially important.

So in some loose sense, the former has more or less stable non metallic lattice, the latter has stable, metallic lattice.


Temperature dependence of reaction rates

Well mostly it does but there can be exceptions.

Eg Consider the reaction $$\ce{2NO + O_2 -> 2NO_2}$$

It has a negative temperature dependence. The rate decrease in increasing the temperature.

There is a simple reason to why this happens.
We have a RDS with an intermediate species. The rate determining step must have a positive temperature dependence.  But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence , that affects the overall reaction.

The Mechanism

$\ce{NO + O2 <=> ONOO}$ Fast but reversible

$\ce{ONOO + NO -> 2NO2}$ Rate determining step

If Step 1 is exothermic, as a bond is being formed between the reactants.
This has a greater impact on the overall reaction and by le Chatlier's principle the equilibrium shift backward . We get a negative temperature dependence!


Why do strong polyprotic acids not dissociate completely, whereas strong bases with multiple (OH)- groups do?

In simplistic terms, Arrhenius acids are molecules while Arrhenius bases are ionic compounds. Much like $\ce{BaCl2}$ – another ionic compound – it is a good first rule of thumb to assume it completely dissociates into its respective ions: one $\ce{Ba^2+}$ and two single-charge anions. You could say that barium were never really connected to either hydroxide; ionic compounds are often shown as collections of balls simply in a certain arrangement (crystal structure) but easily separable.

Arrhenius acids, which are all molecules, are different. There is no inherent separation point; dissiociation in water actually requires the transfer of a proton, i.e. the breaking of an $\ce{X-H}$ covalent bond. Two such bonds don’t get broken simultaneously, instead in a first reaction one proton is removed – but then you end up with an anion. Removing another $\ce{H+}$ has to go against electrostatic attraction and that is a harder job to do so it happens less and the $\mathrm pK_\mathrm a$ values differ.

In actual fact, the picture I have drawn for Arrhenius bases is simplistic. Ionic bonds also involve orbital interactions; no bond is purely ionic or purely covalent. Cations and anions obviously also electrostatically attract each other. As barium is twice positively charged, it does tend to bind the hydroxide fragments rather tightly. Overall, upon dissolution it is much more likely to end up with what effectively is a coordination compound of the $\ce{[Ba(OH)(H2O)_n]+ + OH-}$ type.


Why does LiAlH(O-tert-Bu)3 not reduce acyl chlorides to alcohols?

Without being very specific it looks that $\ce{LiAlH(O-^{t}Bu)_3}$ is a lot less reactive than $\ce{LiAlH4}$, mostly because the $\ce{^{-}O-^{t}Bu}$ is a much weaker base than $\ce{H-}$. Also its much bulkier, further reducing reactivity.

The aldehyde being less reactive than the acyl chloride the reducing agent first eats up all the acyl chloride, and then starts reducing the aldehyde. If you were careful to add stoichiometric amounts, you can stop right when you run out of the acyl chloride.

See also this link.


Calculating the pH of a highly dilute solution of HCl

At first you have to think of where the protons come from. That is as you mentioned it, hydrochloric acid (for every chloride ion in solution there has to be a proton) and water (for every hydroxide ion from the autoprotolysis there also has to be a proton). $$\ce{[H+] = [OH-] + [Cl-]}$$

The concentration for the hydroxide ions can be derived from the autoprotolysis of water, as I've said before.
$$k_\mathrm w = \ce{[H+]~[OH-]}
\Rightarrow \ce{[OH-]}=\frac{k_\mathrm w}{\ce{[H+]}}$$

Using some funny math, one can find out, that the concentration of the chloride ions in solution can be described the following eqation. Recognizing shortly after that the hard way was not necessary yields to the simplification that hydrochloric acid is completely dissociated.
$$\ce{[Cl-]} = \frac{\ce{[HCl]0}~k_\mathrm a}{\ce{[H+]} + k_\mathrm a} \xrightarrow{k_\mathrm a \gg \ce{[H+]~~}} \frac{\ce{[HCl]0}~k_\mathrm a}{k_\mathrm a} = \ce{[HCl]0}~(=c_0)$$

This changes the first equation to be like
$$\ce{[H+]} = \frac{k_\mathrm w}{\ce{[H+]}} + c_0$$

Rearranging everything gives you a second order equation in $\ce{[H+]}~(=x)$
$$x^2-c_0~x-k_\mathrm w = 0$$
Wherefrom you get two solutions (from which only the positive one makes sense):
$$x = \frac{c_0}{2} + \sqrt{\frac{c_0^2}{4}+k_\mathrm w}$$

Let's compare this solution to your mentioned version. The following graph shows you the difference between both assumptions as $$\operatorname{abs}(f_1(c)-f_2(c))$$

$\hskip3.5cm$

A maximum difference of 0.1 pH between the easier and the "more complex" version is acceptable.


OK, here goes. Chemists use a variety of representations of molecules to try to predict their behavior and reactivity, but none of them come close to preparing you for what happens in the laboratory. In that sense, none of them are "safe" to use. Lewis structures, and the curly arrow formalisms, are good for organic chemists to quickly explore possible avenues of reactivity and synthetic pathways. But they are not very good at dealing with chemistry south of the second row of the periodic table, nor at phenomena such as aromaticity, electrocyclic reactions, bulk properties, solvation, steric effects, polarity, polarizability, photochemistry, free radicals, etc. So we add in to the mix some "rules", try to think of molecules as being best described as "multinuclear atoms", and try to improve and promulgate our best tools for computational chemistry. For all that, we still need to approach the laboratory with a degree of humility--things that look good on paper usually do not look so good when you try them out in the real world.


Plasma globes do indeed produce small quantities of ozone when they are touched.
This is because the high-voltage within the globe is at very high frequency and this can drive high kinetic-energy electrons through the glass to produce ozone generating micro-discharges at your fingertips.
Nox generation is highly unlikely at the low energies used.


It's not really the same vanadium dioxide.  The 67°C temperature corresponds to a phase change.  Compare with the phase change in solid tin.


Which isomers of C6H4(OH)2 has a higher melting point?why?

Polarity rules the day if there is no other interactions involved. There is quite a lot of potential interactions in organics, that may overwhelm polarity by orders of magnitude.

To name a few.

There is a well known effect of halogen-halogen interactions, that may raise bp/mp dramatically. This is due to so-known dispersion forces, caused by correlation of electron movement in two atoms. Sometimes it is described as interaction of 'momentary dipoles'. To produce strong dispersion forces, atoms involved must be large and have a lot of electrons. Naturally, heavier atoms in a column are typically larger and have more electrons, thus we have strong increase of mp/bp for most haloalcones whith shift from chlorine to iodine.

Another well known effect is formation of intermolecular hydrogen bonds. Hydrogen bonds are common in compounds containg $\ce{O-H}$ or $\ce{N-H}$ bonds, but sometimes may pop-up in other, usually quite exotic settings. The most common form of hydrogen bonds is a bond of form $\ce{R-O-H\cdot\cdot  \cdot O(R)-R}$ . Such a bond may be stronger of weaker. It is stronger if the second oxygen atom has strong negative charge and the hydrogen involved has strong positive charge. Hydrogen bond is stronger if it is linear, i.e. the entier $\ce{O-H\cdot\cdot  \cdot O}$ fragment is linear. Intermolecular hydrogen bonds have competition : intramoleculear hydrogen bonds. Intramolecular hydrogen bonds prevents two potential sweet spots from interacting with other molecules, and thus a compound capable of formation of intramolecular bonds typically has lower mp/bp than compound that cannont, but have same amount of potential hydrogen bonding sites.

And, finally, come electrostatic interactions. They are not limited to just polarity, some compounds are ionic and this typically results in high bp/mp. Ionic bonding is common in inorganics, but may occur in organics as well, say, $\ce{N(CH3)4 PF6}$

Now, about example at hand. All three molecules are capable of formation of hydrogen bonds. So, the one that is better in it wins the contest of bp/mp.

The 1,2-disubstituted compound is capable of forming intramolecular bonds, and the oxygen atoms are a bit crouded. This makes bonding a bit weaker.

The 1,3-compound is better in this regard, but has $\ce{OH}$ groups in meta-position. It is known, that oxygen in such group is capable of donating some of its electron density into aromatic ring. However, if we put two oxygens in 1,3-pattern,  the two will suppress donation of each other thanks to resonance effects. The last compounds would have the most polar hydrogen bond. It also has a high symmetry, which helps. It is well known, that slight asymmetry mar lower mp considerably. Consider, for example, toluene. It has higher bp than benzene but significantly lower mp.

Consequently, compound 3 (hydroquinone) wins the prize.


Does adding ice to sparkling mineral water extend the 'fizziness' of the solution before the drink goes flat?

Let me lightly gloss over the issue of nucleation sites and ice. While one observes bubbles forming on ice in the glass, they continue to form on the glass as well.

Instead, one could well ask the impact of temperature on release of the carbon dioxide, and thermodynamically that does come down to the solubility of carbon dioxide in water. Since this is of great importance to the soda industry, the study goes back some ways. R. Wieb and V.L. Gaddy, "The solubility of carbon dioxide in water at various temperatures from 12 to 40 degress and at pressures to 500 atmospheres - Critical phenomena', Journal of the American Chemical Society 62 815-817 (1940) shows measured solubility curves close to the temperatures of interest. One should note the increase in solubility as the temperature is dropped from room temperature down towards zero C.

While the carbon dioxide is supersaturated in the soda water, the driving force relates to the difference of the supersaturation to the solubility limit. As the temperature decreases, the thermodynamics favor retention of more carbon dioxide, which will then come out in your (warmer) mouth, giving you that 'fizz'.

Further, ignoring nucleation sites on the ice being much better than on the glass, as the temperature is decreased, the kinetics of nucleation will be at least partially suppressed as well.

Plus, I like my soda (and beer) cold, not warm!


Is activation energy related to reactivity?

If by reactivity you mean, "does the reaction occur quickly", then yes, a highly reactive molecules do have a low activation barrier. The fastest molecules can react bimolecularly is if they do so on first encounter with one another. This effectively means that  there is a very small potential barrier between reactants and products so that they have enough energy, at the prevailing temperature, to always overcome this. In solution the rate constant is determined by how fast the molecules can diffuse together and in mobile solvents such as water or acetone the bimolecular rate constant is around $\ce{10^{10} dm^3mol^{-1}s^{-1}}$ depending on exact solution viscosity and far less in viscous solvents such as glycerol.

If by reactivity you mean that the reaction is very exothermic, i.e. free energy is large and negative, $\Delta G < 0$ then there is generally no way to precisely predict how big the activation barrier will be. It is usually measure using the Arrhenius equation and plotting log(rate constant) vs. reciprocal temperature. I'm not sure if ab initio quantum calculations can predict this in general. In fact many exothermic, thermodynamically spontaneous reactions occur slowly. Most organic substances react with oxygen exothermically, but are quite stable, because the activation energy to do so is so large. You and me for example :)


Why is chlorine and not fluorine labelled as the most reactive halogen for halogenation?

The reactivity of halogens in sunlight

OK, this tells us that we are dealing with a free radical reaction.  Sunlight breaks the halogen bond ($\ce{X-X}$) to generate halogen free radicals.

$$\ce{X2 ->C[{h\nu}]\  2X.}$$

In the next step the halogen radical abstracts a hydrogen atom from any hydrocarbon present to generate $\ce{HX}$ and a hydrocarbon free radical (note that multiple hydrocarbon radicals may be generated depending on the structure of our hydrocarbon).

$$\ce{X. + R-H -> HX + R.}$$

Finally, the hydrocarbon radical can react with the halogen free radical to produce a halogenated product.

$$\ce{R. + X. -> R-X}$$

Let's now consider the energetics of these steps.  The first step, generation of the halogen free radical, uses the sunlight to break the $\ce{X-X}$ bond, so no additional energy is required for this step. Then we break a $\ce{C-H}$ bond (so we must add energy) and make an $\ce{H-X}$ bond and a $\ce{C-X}$ bond (so energy is given off).  The various bond strengths are contained in the following Table.

\begin{array}{|c|c|c|c|c|c|} \hline
\ce{C-H}~ \text{Bond Strength} & \ce{H-X} & \text{Bond Strength}& \ce{C-X} & \text{Bond Strength} & \text{Overall}  \\ \text{(kcal/mol)}\ & & \text{(kcal/mol)} & & \text{(kcal/mol)} & \text{(kcal/mol)} \\ \hline
\ 99 & \ce{H-F} & -135 & \ce{C-F} & -116 & -152\\ \hline
\ 99 & \ce{H-Cl} & -103 & \ce{C-Cl} & -81 & -85\\ \hline
\ 99 & \ce{H-Br} & -88 & \ce{C-Br} & -68 & -57\\ \hline
\ 99 & \ce{H-I} & -71 & \ce{C-I} & -51 & -23\\ \hline
\end{array}

(bond strengths taken from Bill Reusch's table)

We can clearly see that all of these free radical reactions will occur (e.g. they are exothermic) in the order you suggested, with fluorine being, by far, the most reactive.

$$\ce{F2 >> Cl2 > Br2 > I2}$$

Saying that chlorine is the most reactive is incorrect.

Note: 99 kcal/mol was used as a typical $\ce{C-H}$ bond strength.  We could refine our model by using more exact $\ce{C-H}$ bond strengths (see this Wikipedia table) and calculate the relative rates of halogenation at the various carbon atoms in a complex hydrocarbon.


What is the systematic name of a "double benzene"?

As @ron already said in his comment, the systematic name of the naphtalene is Bicyclo[4.4.0]deca-1,3,5,7,9-pentaene or Bicyclo[4.4.0]deca-2,4,6,8,10-pentaene since you can start the carbon numerotation from 2 chemically equivalent carbon.

You can find all the references here and here


will the autoionization of water reduce due to common ion effect?

Adding acid to the water will not stop the auto-ionization of the water, so: the auto-ionization will continue to give equal concentrations of hydroxide and hydronium. But the addition of acid to the water increases the concentration of the hydronium ions in the solution, pushing water balance with its ions to the left to resist the increase in the concentration of hydronium and reach a new equilibrium. The concentrations of hydronium and hydroxide produced by auto-ionization of water are reduced, where the concentrations of $\ce{H_3  O^+}$ or $\ce{OH^-}$ from water ionization is less than$\pu{ 10^{-7}}$.

$$\ce{{[H_3O^+]}_\text{(after acid  addition)}=\ce{[H_3O^+]}_\text{(produced by acid dissociation)} +\ce{[OH^-]}_\text{(produced by water auto-ionization )}}$$


If vibrational modes are quantised, why is infrared spectroscopy smooth looking?

You should first have a look at the gas phase spectrum of water to convince yourself that vibrational levels are indeed "discrete". The textbook treatment of vibrating molecules assumes that there is no molecule nearby and each and every unit is independent. Which state of matter is closest to these assumptions? The gas phase, the lower the pressure, the further the molecules are apart (large mean free paths). Therefore if you wish to see quantitization visually, it is better to look at the spectrum of gases (rotations, vibrations and electronic transitions). Recall you study atomic spectrum of hydrogen gas not of solid hydrogen or liquid hydrogen.

Now, you may ask, that water is a triatomic molecules with 3N-6= 3 vibrational modes. What are the rest of those lines? Recall that water can also rotate in the gas phase. Those very thin lines represent rotational energy transitions.

The story changes when you deal with liquids or solids. Now the vibrating molecules are so close to each other that they are no longer independent from each other's influence. The rotational structure is completely smeared. The vibrations are still quantized but now you see a collective behavior millions of close-by water molecules which are strongly interacting by hydrogen bond. If we had a very fancy ultra ultra high resolution infrared spectrometer, we might be able to see some structure in each broad band, but the smearing is so bad in solids and liquids that we all see a overall wide peak, which is just an envelope of all possible vobrational transitions by millions of molecules in different environments on the liquid and solid state.


Luminol (IUPAC name: 5-Amino-2,3-dihydrophthalazine-1,4-dione) displays chemiluminescence when it is oxidised, and glows a blue colour. It is used in forensic science to detect iron (in haemoglobin) and also to detect copper in biological material.

The process leading to luminol emitting light is explained in this image.

As for whether luminol can be "infused" in fabric, this is difficult to answer, as you have not explained what you mean by "infused". As it is soluble in water, it will easily be washed out, and is thus not suitable for long-term use. Moreover, it will react with a variety of substances (apart from blood), for example bleach.


Do light and heavy water form an azeotrope?

This is a good question, but no, there is no azeotrope in any isotopic mixture, as an azeotrope requires non-ideal mixing of two components to form a  minimum or maximum critical point on the plot of total vapor pressure as a function of composition as shown in Figure 1 below. Isotopic Mixtures have a near-zero enthalpy of mixing, making isotopic mixtures the closest thing to an ideal liquid mixture. Because of this ideality of mixing the boiling of protium-deuterium oxide mixtures looks much like the phase diagram in Figure 2 which follows Raoult's law, though this is actually quite exaggerated compared to semi-heavy water mixtures.  Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of $\ce{D2O}$ or $\ce{H2O}$ and be unobservable due to the "doped" species existing as semi-heavy water ($\ce{DHO}$).  This is further evidenced by the fact that other physical properties of isotopic mixtures follow a linear curve when measured as a function of concentration.

Figure 1. Generic Plot of vapor pressure as a function of composition

Figure 2. Generic Boiling phase diagram for ideal solutions

To answer your question about distillation only enrichment, it is possible though impractical.  Commercial enrichment of $\ce{D2O}$ first uses hydrogen sulfide in the Gridler-Sulfide process, once enriched to around 30%, distillation is used to finish the process to the desired isotopic purity.


Work done in Irreversible Isothermal Expansion

There is some confusion about terminology. The figure shows two processes. The curved path represents a reversible expansion in which there is pressure balance between system and surroundings throughout, so that the applied pressure can be described by the ideal gas law (or Boyle's law $p=\frac{p_1V_1}{V}$). This process is clearly isothermal, since $pV=nRT$.
The second process shown as two straight paths consists of a pressure drop and and an expansion against constant external pressure and is also isothermal (in thermodynamic lingo). The fact that the initial and final points are at equal T makes it isothermal. The temperature (and other properties except presumably composition) of the system during the process is on the other hand not well-defined. This process is irreversible.


What do short-range and long-range corrections mean in DFT methods?

There are many good review articles, but I really like the freely available "The ABC of DFT" by Kieron Burke (and friends) and a recent review article that tries to keep to understandable language.

The problem is treating the Coulomb interactions. Conventional DFT is not "asymptotically correct." That is, at long range, the 1/r behavior is not met, and dispersion interactions (e.g., van der Waals forces) are not properly handled either.

For example, in Neon atom (from ABC of DFT by Burke):

The problem of course is that most DFT methods start from the local density approximation (LDA) which only considers the electron density at a particular (local) point. LDA indicates that the long-range behavior falls off exponentially instead of 1/r:

$$v_{xc}^{\mathrm{LDA}}(\mathbf{r}) = \frac{\delta E^{\mathrm{LDA}}}{\delta\rho(\mathbf{r})} = \epsilon_{xc}(\rho(\mathbf{r})) + \rho(\mathbf{r})\frac{\partial \epsilon_{xc}(\rho(\mathbf{r}))}{\partial\rho(\mathbf{r})}\ $$

So some recent efforts have used "range separated" functionals, e.g.,

$$ \frac{1}{r} = \frac{1-g(r)}{r} + \frac{g(r)}{r}$$

The first (short-range) term can be handled by the generalized gradient approximation (GGA) used by basically all modern functionals. The second term is usually handled by Hartree-Fock (i.e., exact exchange), and there's some function g(r) which smoothly scales between the two.

I think modern range-separated hybrid functionals are quite good, but choosing a functional depends a bit on the property that you want to predict. Thermodynamics? Excitation energies? Oxidation/reduction potentials from orbital eigenvalues? etc.


Molecular explanation of the Hofmeister Series

The molecular origin of the Hofmeister series is not well understood and still a very active area of research. For example, see the upcoming Faraday discussion conference. The syllabus for the conference says:

  Although it is now clear that the Hofmeister series is intimately connected with ion hydration in homogeneous and heterogeneous environments and with ion pairing, the molecular origin of these effects has been poorly understood. Biochemists and physical chemists have been typically using the term Hofmeister series to put a label on ion specific behaviour in various environments, rather than to reach a molecular level understanding and, consequently, an ability to predict a particular effect of a specific salt ion.

(So, this is not an answer, but rather too long for a comment…)


What exactly does a "system" include in chemistry?

Note: This was not originally written by me. This was first written by @airhuff. I just thought it was relevant in this question as well.
According to this Wikipedia article:

...a physical system is a portion of the physical universe chosen for analysis. Everything outside the system is known as the environment. The environment is ignored except for its effects on the system.
The split between system and environment is the analyst's choice, generally made to simplify the analysis. For example, the water in a lake, the water in half of a lake, or an individual molecule of water in the lake can each be considered a physical system.

Every system has some relationship to it’s surroundings. As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:

Isolated system: Neither matter nor energy may pass into or out of the system.

Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.

Open system: Both energy and matter can pass into and out of the system.


Finding Pv in Pv=nRT for molecular hydrogen

You can solve for either $P$ or $V$.  Choose one, say $P$, and then divide that into the right hand side $nRT$ to get $V$; i.e. $V=\frac{nRT}{P}$.

More seriously, you need to check your units!  You seem to be using Kelvins for $T$, which is right, and $n$ is just a count in moles.  But I do not recognize your value of $R$.  The SI value for $R$ is $8.3144621(75)$, in units of joules per Kelvin-mole.  Joules, in turn, are Pascal-meters$^3$. Personally, I think most people find Pascals too small a unit for pressure.  Far more practical would be Torr-liters, and then $R=62.36367(11)$.  You can examine lots of different unit combinations for $R$ at the ideal gas constant page on Wiki: Ideal Gas Constant


The colours of fireworks are caused by the electrons emitting energy when dropping orbital levels. Metal oxides are solids, mainly dull colours (even the yellows and reds are not in the same class as firework colours), and could not cause bright light in the air. The temperature of fireworks is far below that needed to form a plasma.


When we ask, for example, "how many moles of Iodine (or Oxygen or Hydrogen) is contained in 100g?" It ALWAYS refers to the molecule (e.g. I2, O2, H2 etc). Therefore, the number of 6.02 x 10^23 refers to the number of molecules, not atomes of Iodine and therefore you actually have 12.04 x 10^23 atomes of Iodine contained in one mole of molecular Iodine. However, when you look at the periodic system, they always refer to atomic mass not molecular. In case of Oxygen, the atomic mass of 16 g/mol refers to 1 mol of O, not O2 as molecular form. Further, when you want to know mass of the Oxygen (gas form, O2) then: 2 (number of atoms) x 16 g/mol = 32 g/mol - this is the mass of 1 mol of O2 (Oxygen-diatomic form - gas present in the atmosphere).


How can we produce gasoline from electricity without fossil fuels?

Historically, this sort of process has been done with the Fischer–Tropsch process.

The core of this process is to use "syngas" (a mixture of $\ce{CO}$ and $\ce{H2}$) in the presence of a catalyst to make higher molecular weight hydrocarbons:

$$\ce{(2n + 1) H2 + n CO → C_{n}H_{2+2n} + n H2O}$$

Under normal conditions, this will actually make mostly straight-chained hydrocarbons, which are great for diesel fuel, but not so great for gasoline/petrol engines. However, there are standard catalytic cracking techniques in the petrochemical industry which can adjust the branchedness of hydrocarbons, making them more "gasoline like".

Now, the only question is where to get the $\ce{CO}$ and $\ce{H2}$ from? The $\ce{H2}$ is easy, as it is readily obtained from the hydrolysis of water.

The $\ce{CO}$ can be obtained from $\ce{H2}$ and $\ce{CO2}$ by the water-gas shift reaction (here technically run in reverse):

$$\ce{CO2 + H2 \rightleftharpoons CO + H2O  }$$

As I mentioned, the Fischer–Tropsch process was actually used during WWII by the Germans to produce motor fuel, as conventional sources were blocked off by the Allies. Their syngas didn't come from $\ce{CO2}$ and water, though, but from reformed coal, which they had a ready supply of.

This process is not done much because it's hugely inefficient, compared to just digging oil out of the ground. But if you have a "too cheap to meter" supply of electricity, the efficiency concerns go away. The main issue is getting a ready supply of $\ce{CO2}$, and there are a number of techniques being proposed at the moment for scrubbing $\ce{CO2}$ from the atmosphere - either by capturing it from point-source generators like power plants and cement manufacturers, or from "artificial trees" which capture $\ce{CO2}$ directly from standard atmospheric concentrations.

These "artificial trees" tend not to use normal photosynthesis for carbon capture, but instead often use something like a (strong) base in aqeous solution or $\ce{CaO}$. In these systems the base or the $\ce{CaO}$ reacts with dissolved $\ce{CO2}$ to form a carbonate salt, which either stays in solution or precipitates out. This carbonate salt formation is favorable, and happens spontaneously. The energy intensive portion of the process is the regeneration of the basic solution/$\ce{CaO}$. This normally happens by heating the carbonate salt, which causes $\ce{CO2}$ release. If done correctly, this can produce a stream of relatively pure carbon dioxide. But again, the regeneration step is quite energy intensive, and so would not be feasible without a cheap source of power.

Other environmental considerations would be minimal, aside from the land needed for the $\ce{CO2}$ devices (you would need a fair number to get a suitable steady volume of $\ce{CO2}$), and any disposal concerns about the process catalysts (which shouldn't be any more onerous than dealing with the catalysts now used in the petrochemical industry).


What is the difference between balancing a redox reaction in acidic solution vs basic solution?

You basically just have to add hydroxide ions to either sides of your equilibrium.

$$\ce{2MnO4 + 5H2O2 + 6H+ + \color{blue}{6OH-} -> 2Mn^2+ +5O2 +8H2O + \color{blue}{6OH-}}$$

$\ce{H+}$ and $\ce{OH-}$ ions can be collectively re-written as $\ce{H2O}$, so we have:

$$\ce{2MnO4 + 5H2O2 + \color{blue}{6H2O} -> 2Mn^2+ +5O2 +8H2O + 6OH-}$$

Finally, we can cancel out the excess water molecules which leaves us with:

$$\ce{2MnO4 + 5H2O2  -> 2Mn^2+ +5O2 +2H2O + 6OH-}$$


Truly Understanding the Second Law of Thermodynamics and Entropy

Be careful in citing equilibrium as an example, as it fully integrates entropy in its definition. See this article for the details, but in a spontaneous reaction the entropy is increasing, and part of reaching equilibrium is that the energy gain and loss among products and reactants is balanced. Also, most experimental reactions are (nearly) closed systems for short periods of time, and if no energy is gained or lost (again, for short periods), you can have an equilibrium where there is no change in entropy. If you let your experiment sit and the heat of the system changes (for example), the equilibrium will change as well. And eventually you end your experiment, clean out your flask, and thus increase the entropy of the universe again!

@MARamezani's comment pointing to the article on Biology looks like a good answer to your biology questions. Earth is most definitely not a close system with that giant energy source 93 million miles away.


Does NaOH react with O3 to produce NaO3?

My own Google search does show that sodium ozonide can form [1].  But it requires low temperature:

Sodium ozonide was produced by the ozonization of sodium hydroxide at -80 to -100°[C] followed by extraction with liquid ammonia and removal of the latter under vacuum at-50° [2]. Composition of the product obtained: $\ce{NaO3}=80.1, \ce{NaOH•H2O}=11.2, \ce{H2O}=8.6$ [%].

Upon warming it decomposes to give the superoxide, and then in the presence of water back to sodium hydroxide:

The process of thermal decomposition of sodium ozonide to $\ce{NaO2}$ occurs at −20 and −10°.

At temperatures above 0°, the process of thermal decomposition of sodium ozonide is accompanied by the process of interaction of the $\ce{NaO2}$ formed with $\ce{H2O}$, and decomposition proceeds to the formation of $\ce{NaOH}$.

References
1.
Tokareva, S.A., Pilipenko, G.P. Thermal decomposition of sodium ozonide. Russ Chem Bull 13, 686–688 (1964). https://doi.org/10.1007/BF00845322.
2.
S. A. Tokareva and M. S. Dobrolyubova, Izv. AN SSSR, Ser. Khim. 1964, No. 4.


Given this liquid-liquid phase diagram, is the green line the nitrobenzene-rich phase?

Here is the same diagram with more annotation from the physical chemistry textbook by Atkins:

It shows that above the temperature $T_{uc}$, there is only one phase (the substances mix in any ratio). Below that temperature, there are two phases, with compositions shown by the solid lines. The horizontal line shows that if you try to make a mixture with a mole fraction of $a$ at the given temperature, you will get two solutions with mole fraction $a'$ and $a''$ instead. You can calculate the approximate ratio of the volumes using this technique.

  Am I right in assuming that the right side (green line) is the nitrobenzene rich phase?

Yes, because the x-axis is the mole fraction of nitrobenzene, the phase with higher fraction of nitrobenzene will be that on the right, as you say. It does not necessarily mean that the fraction is higher higher than 50% on the right and lower than 50% on the left, though.


Why is it only electrons that take part in chemical reactions?

The nature of an electron is more complex than a single charged particle. Once you get down to the quantum level, our intuition about how matter interacts with other matter fails.

A simple answer to your question is that electrons occupy a much larger volume than nuclei. When two atoms get in close proximity to one another, their electrons will "bump into one another" and interact far before their nuclei will.

You are also correct that the nucleus of a neighboring atom is essential for the electrons to interact with one another. If there was no positive charge to attract another electron, a bond would not be formed. The point your instructor was trying to get across is that in most (excluding nuclear reactions) chemical reactions, electrons are the species moving around and forming/breaking bonds.

If you would like a more in-depth discussion on the true quantum nature of electrons and their interactions, you should check out Mulliken's Nobel Lecture from 1966. It is a great introduction.


Soil-solution cations in competition for exchange sites

Only half as many Ca2+ are needed as Na+ to neutralize the charge of the ion exchanger (here: clay).

The affinity of clay for bivalent ions is larger than that for monovalent ions:
Ca2+ > Na+

For most clays, the affinity for equally charged ions increases with the nonhydrated radius of the ions:
Cs+ > K+ > Na+ > Li+
Ba2+ > Sr2+ > Ca2+ > Mg2+


A microbial fuel cell (MFC) is an electrochemical cell that converts chemical into electrical energy, and in which one redox half-reaction is catalyzed by enzymes of microorganisms. The fuel cell consists of anode and cathode compartments which are separated by a membrane that is only permeable for specific cations. At the anode, bioorganic fuel is oxidized to $\ce{CO2}$ and $\ce{H3O+}$. The gained electrons flow from the anode through an external electrical circuit to the cathode, where they are consumed together with protons by the reduction of $\ce{O2}$ to $\ce{H2O}$. The $\ce{H3O+}$ ions are transferred from the anode to the cathode compartment through the membrane.

The fuel is a medium rich in nutrients which can be oxidatively degraded by the microbes, like, for example, sugars (in the following formula, $\ce{C12H22O11}$ stands for a variety of disaccharides):

$$\ce{C12H22O11 + 13H2O -> 12CO2 + 48H+ + 48e-}$$

The electrons from this half-reaction are either transferred directly to the anode (mediator-free MFC) or via a mediator (mediator MFC), which integrates into the electron transport chain of the cell and transfers electrons to the anode by passing the cell membrane. The mediator gains and loses electrons in every transport cycle, and switches between its reduced form (laden with electrons from the cell) and oxidized form (has given up electrons at the anode).

The scheme below (image source: MFCGuy2010/Wikimedia Commons) shows a MFC which uses soil as biofuel. Nutrients from the soil, like acetic acid, are oxidized by microbes in the anodic biofilm to $\ce{CO2}$ and $\ce{H3O+}$. Protons are then consumed in the reduction of supplied oxygen at the cathode, and electrons flow through the external circuit from anode to cathode. In this case, the soil acts simultaneously as nutrient medium and proton exchange membrane.


Relative magnitudes of 2- and 3-bond coupling constants

Be careful when generalizing from a single example.  There are many situations where the geminal coupling is much larger than vicinal coupling.  Pick pretty much any molecule that has an aliphatic CH2 where the two H's are diastereotopic and you'll find that the geminal coupling for those two H's can be 12-16 Hz vs. the typical ~7 Hz vicinal couplings.  However, if you're familiar with 6-membered rings in chair conformations, you've also seen vicinal coupling that range from ~2-15 Hz which leads us to the real answer to your question: the Karplus relationship.

Martin Karplus (who got a nobel prize for completely different work) is most well known in the NMR community for his theoretical work relating the values of J coupling to bond angles, bond lengths, electronegativity of substituents, etc.  In other words, there's no reason that 2 bond coupling is necessarily larger (or smaller) than a 3 bond coupling – the exact values will depend on the functional group and local geometry of the molecule.  Most organic chemistry texts describe a highly simplified Karplus relationship (ignoring all effects except for bond angles) for 2-bond and 3-bond J values.  I've attached two common examples.

You can see that for a typical aliphatic CH2, a bond angle of ~109° would predict a 2J ~ 14 Hz.  Whereas, a terminal alkene with a bond angle of ~120° would predict a 2J ~ 3 Hz.  Comparing this with the 3-bond Karplus relationship, the vicinal coupling for alkenes should be 8 Hz or 10 Hz depending on the cis/trans relationship.  For an aliphatic system with free rotation about a C-C bond, you have to average the Karplus relationship over all angles and you end up predicting ~ 7 Hz.  For a cyclohexane derivative in a chair conformation, you can see that an axial-axial coupling (180°) is predicted to be ~10 Hz, while an axial-equatorial coupling (~60°) is predicted to be < 2 Hz!

Qualitatively, this matches what we observe quite well.  Quantitatively, we quickly run into a lot of problems.  Many axial-axial couplings are 12-15 Hz and the J coupling observed in alkenes typically deviates quite a bit from the predictions given by these graphs.  Again, these graphs are grossly oversimplified - the real theory would fold in information about nearby substituants and steric effects, etc.  However, you'd then need run a bunch electron structure calculations and average these over many different conformations, etc.  So most people just live with the qualitative nature of these graphs to rationalize the patterns that they see.


I'm not sure whether $\ce{PCl3}$ can lead to the installation of a chlorine atom at C-2. I performed a cursory Reaxys search and in the first page saw several examples of deoxygenation without chlorination, but no examples of concomitant chlorination.

As far as I know, treatment of pyridine N-oxides with $\ce{PR3}$ simply leads to deoxygenation, with $\ce{P(III)}$ being oxidised to $\ce{P(V)}$.

However, if you were to treat pyridine N-oxide with phosphorus oxychloride $\ce{POCl3}$, then you would see chlorination at C-2.

If you have a source that says that $\ce{PCl3}$ can deoxygenate and chlorinate, please do inform us. Otherwise as far as I can tell, you need $\ce{POCl3}$ to effect a chlorination.

References

Joule, J. A.; Mills, K. Heterocyclic Chemistry, 5th ed.; Wiley: Chichester, U.K., 2010; pp 121, 155.
Weickgenannt, A. Pyridine N-Oxides, Baran Group Meeting, June 9, 2012; p 3.


You are using the Heisenberg uncertainty principle to relate the uncertainty in position $x$ to the uncertainty in velocity $v$.

However, the quantitative version of this principle actually is

$$\Delta x\cdot\Delta p\geqslant\tfrac12\hbar $$

where $\Delta x$ is the uncertainty in position $x$ and $\Delta p$ is the uncertainty in momentum $p$.

Certainly, since
$$p=m\cdot v$$
or
$$v=\frac pm$$
where $m$ is mass and $v$ is velocity, you may also use the uncertainty principle to compare the uncertainty in position $x$ to the uncertainty in velocity $v$.

However, the mass of a proton is $m_\mathrm p=1.672\,621\,923\,69(51)\times10^{-27}\ \mathrm{kg}$[CODATA2018], whereas the mass of an electron is only $m_\mathrm e=9.109\,383\,7015(28)\times10^{-31}\ \mathrm{kg}$[CODATA2018]; i.e. a proton is over a thousand times heavier than an electron.

Therefore, a given uncertainty in position $x$ and the corresponding uncertainty in momentum $p$ result in a smaller uncertainty in velocity $v$ for the proton.


Atomic number ranges in "integral units"?

This is only a way of saying that the atomic number will never be a decimal number like 1.5 or 16.8

One consequence is that if you carry out an experiment which gives you a decimal number as an atomic number, then you must have a mixture or something is wrong.


Is sugar in mead different from "regular sugar"?

Honey usually contains mainly naturally occuring glucose + fructose, with ratio typically 100:104-170. Sacharose occurs as well, but in minority.

Fermentation consumes just part of sugars, but that is not limited to mead. It applies to beer,wine, and bases for spirits as well.

E.g. typical "future beer" before fermentation has sugar content  $\pu{100-120 g/L}$.


Why is there an orbital angular momentum if the electron isn't properly revolving around the nucleus?

In quantum mechanics, the classical concept of trajectory of a particle is abandoned. The quantum state of a particle i.e. the electron is characterized by a wave function $$\psi(\pmb{r}, t)$$ that contains all information about the particle and that extends over space. The simplest case where angular momentum appears in the context of quantum mechanics is the one of a particle in a “ring”. Now you have a wave function which extends over a circular region of space which implies that your particle has angular momentum. When you apply  the same idea to the Hydrogen atom you will find the orbital angular momentum.

The energy levels of the Hydrogen atom is given by
$E_{n} = E_{0}/n^{2} $,
where $E_{0} = 13.6 eV$ and n = 1, 2, 3, ... is the principal quantum number and describes the size of the orbital. Therefore, we see that $E_{n}$ is independent of the orbital angular momentum. If the atom is in the presence of a magnetic field then things change.


What happens when (+) 2-iodobutane is treated with NaI in acetone?

There is no equilibrium in the second reaction because the sodium bromide formed by the iodide displacement is not soluble in acetone so there is no Br- to displace iodide. In contrast with the first reaction, NaI is soluble in acetone so there is always excess I- around to continue the reaction. This is called the Finkelstein reaction, further information here finkelstein reaction


These two weak acids have similar disassociation constants in water.  A low Ka indicates a relatively small number of acid molecules will form an anion stabilized by water.  Two weak acids with similar disassociation constants will coexist in a solution without interacting.  If two weak acids have very different Ka then there can be a reaction.  In this case nothing would happen.   For more generalized information search for mixture of weak acids.

  Carbonic Acid

$$\ce{H2CO3 + H2O <=> HCO3- + H3O+}$$
$$K_a = 4.3 \times 10^{-7} = \frac{[\ce{H3O+}][\ce{HCO3^-}]}{[\ce{H2CO3}]}$$

  Citric Acid

$$\ce{H3C6H5O7 + H2O <=> H2C6H5O7- + H3O+}$$
$$K_a = 1.8 \times 10^{-6}  = \frac{[\ce{H3O+}][\ce{H2C6H5O7^-}]}{[\ce{H3C6H5O7}]}$$


According to this thesis, the polymerization is determined by thermal initiator that is used in PMMA mixture. The polymerization (corsslinking) starts when the initator is added to the mixture, but it is really slow at low temperature. Higher temperature lowers hardening time (speeds up the crosslinking). Since the dustsheet does not provide the information about the initiator, I would suggest to determine the cross-linking temperature experimentally. Some material related to the polymerization process:http://www.chemistry2011.org/,

BULK POLYMERIZATION OF METHYL METHACRYLATE IN A KNEADER
REACTOR
P.-A. Fleury
LIST AG
CH-4422 Arisdorf,

High conversion polymerization of methyl methacrylate
in the presence of poly(methyl methacrylate) of different relative
molecular mass
J. BARTOŇ, V. PASTUŠÁKOVÁ, V. VAŠKOVÁ, and P. AMBROVIČ
Polymer Institute, Centre of Chemical Research, ChemicalPapers


The graphic from the book is correct, those two molecules are mirror images.  Same for your drawing, those are the same two mirror images as in the book.  Remember that the electrons in the aromatic ring are delocalized.  The double bonds in the aromatic ring are not fixed.  Where you choose to draw the double bonds in aromatic rings is your choice and does not play any role in determining if molecules are enantiomers.  Bottom line, the following three structures all represent the same thing.


In a multi equilibrium system, how does change in one equilibrium affect the other?

Increasing $\ce{[E]}$ lowers $\ce{[C]}$ by a simple application of Le Chatelier's Principle. Regarding the effect of increasing $\ce{[E]}$ on $\ce{[A]}$, your understanding is correct. To make your observation more obvious, we may reverse reaction (ii) and add it to reaction (i): $$\ce{A(g) + E(g)<=>B(g) + D(g)}$$
So we can, once again, simply apply Le Chatelier's Principle and see that increasing $\ce{[E]}$ lowers $\ce{[A]}$.


Unsure whether this classification of Volatile Organic Compound and/or Particulate Matter is correct

2-Acetylaminofluorene is a solid, mp 192-196 C. It satisfies the definition of being a VOC by virtue of being a compound of carbon, not excluded by the definition. It could be fine enough to be classified as PM 2.5 or 10 if it is powdered; more information would be necessary about the source of the compound. It is also carcinogenic, but that does not appear to be involved in the questionnaire.
Chloramben can be looked up and is also a solid with a high melting point: a VOC and possibly a PM.
Hydrazine is only dangerous, but not a VOC, and a liquid with a bp of 114 C. Not a PM. I don't think a spray of droplets would qualify as a PM; they would just evaporate.
Diisocyanates: VOCs for sure. Hazardous for sure. If you allow a pot of diisocyanates to evaporate, that's all: no PM. However, I can imagine a source of sprayed diisocyanates (in a paint?) that would result in polymerized particulate matter: then it could be a PM. Thus the classification would depend on the source of the material - the environment and the application.
Polycyclic aromatic compounds: most often encountered in smoke from various sources: VOC for sure, probably also PM because they could be larger agglomerates than single molecules.
This could be a fun exercise for a chemist to look up and play with mentally, to get to know and understand a large number of chemicals. For a non-chemist, this could be very difficult.


Ketone/aldehyde synthesis from N-acylazetidines or aziridines

I think your suggestion about the stability of the tetrahedral intermediate is correct, but it's not that the tetrahedral intermediate itself has stability. I think it's because the collapse of the tetrahedral intermediate leads to a less stable intermediate.

My key insight for rationalizing this is that lithium hydride reduction of amides generally affords amines. We generally prefer to eliminate the lithium-oxygen adduct from the tetrahedral intermediate, forming an iminium ion.

With the two substrates you've provided, the iminium ion would be very highly strained because of the geometry of the small ring. I think collapse of the intermediate is thus disfavored. On work up, we actually kick out the amine to collapse the intermediate, providing an aldehyde or ketone.


Verify my expected product of two acid/base reactions and does chosen solvent change reaction?

You will not get toluene in either reaction, you will get the potassium and sodium salts of p-toluene sulfonic acid for the respective reactions. This is because 1.) the toluene sulfonic acid is not a salt of toluene and sulfuric acid and 2.) aromatic rings (unless special conditions are applied) undergo electrophilic aromatic substitution. i.e. positive groups cause substitution on the ring, which sodium and potassium are not electrophilic ("electron loving") enough to effect such reaction. If you wanted toluene from these reactions you would have to use a dilute mineral acid in water and the $\ce{H+}$ ions from the acid would cause a substitution, removing the sulfonyl group and yielding toluene and sulfuric acid.

If you really need to neutralize this material, just add the mentioned bases as solids and allow them time to react. Bases don't need to be in solution to react with acids.


Part of the color change might be due to creation of F-centers, crystal defects where there are electrons filling the place of an atom. A corresponding H-center (for halogen interstitial) may also be created for each F-center.

That said, such color changes tend to be rather pale, and as you state, $\ce{NaCl}$ heated to melting in an iron crucible may appear dark or even black. This is likely due to dissolution and oxidation of the crucible. The molten salt attacks the metal at grain boundaries and also dissolves protective oxide coatings, allowing oxygen from the air to attack the underlying metal. Even in a closed system, such as a molten (mixed) salt thermal energy storage system, corrosion is a problem.

The color change could also be due to trace impurities in the "table salt". For example, sodium or potassium ferrocyanide, $\ce{Na4[Fe(CN)6]·10H2O}$, is intentionally added to keep the salt free-flowing despite humidity changes. This might react at high temperatures to form a deeply-colored compound.

To find out what is causing the discoloration, you might try heating a small sample of high purity $\ce{NaCl}$ in a ceramic or platinum (a bit pricey) crucible, and you could analyze heated table salt for dissolved iron compounds.

BTW, F-centers and other crystal defects can be created intentionally with optical lasers and x-rays. If memory serves, there was an article in Scientific American's Amateur Scientist about creating F-centers using an electrostatic generator.


The nuclear shell model is a useful "first approach" to determining nuclear spin.  It doesn't always work, but it is a relatively simple way to make a first attempt.

Here is a nuclear shell energy diagram.  As you can see it is somewhat analogous to an electron orbital energy diagram.

A set of shells are filled by neutrons and a separate set is filled by protons.  The protons will pair when possible, same applies to the neutrons.  Notice that shells with J=1/2 hold 2 nucleons (a nucleon can be either a proton or a neutron), shells with J=3/2 can hold 4 nucleons, shells with J=5/2 can hold 6 nucleons, and so on.  This is because the shell with (for example) J=5/2 has 3 sub-shells: J=5/2, J=3/2 and J=1/2, each capable of holding 2 nucleons.  Higher J sub-shells are "usually" filled first.

Let's apply this model to a few examples.

$\ce{^2H}$ (deuterium): 1 neutron, 1 proton; the neutron goes into the neutron $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; the proton goes into the proton $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; total nuclear spin = 1/2 + 1/2 = 1

$\ce{^{17}O}$: 9 neutrons, 8 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the 9th neutron goes into the neutron $\ce{1d_{5/2}}$ shell and is unpaired, spin = 5/2; all 8 protons are paired in the 1s and two 1p shells, spin = 0; total nuclear spin = 5/2 + 0 = 5/2

$\ce{^{23}Na}$: 12 neutrons, 11 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the last 4 neutrons go into the neutron $\ce{1d_{5/2}}$ shell and each set of two is paired, spin = 0; the first 8 protons fill the proton 1s and two 1p shells and are all paired, the next 2 protons go into the $\ce{1d_{5/2}}$ J=5/2 sub-shell and are paired, spin = 0, the last proton goes into the $\ce{1d_{5/2}}$ J=3/2 sub-shell and has spin 3/2; total nuclear spin = 0 + 3/2 = 3/2


Effect of side reactions on galvanic cell using iron rod as anode and sodium chloride as salt bridge

Since you state, "We measured a voltage way below the expected value and it was dropping quite fast," it indicates a few things that could lead to those observations.

How did you measure voltage? If you used a low impedance voltmeter, it would show much lower voltage because the impedance of the meter is low compared to that of the cell. Voltage is distributed in proportion to the impedance (here, the DC resistance) of the meter and the cell. For example a 1 volt cell with internal resistance of 20 kilohms, measured by a voltmeter with resistance of 20 kilohms, would show as only 0.5 VDC, because half the voltage is dropped inside the cell.
As a cell delivers current, the surface of the electrodes change. Corrosion, surface deposits and bubbles can cause current to drop. While the cell is being measured, observe the electrodes closely. Look for any change in appearance that progresses during measurement.
How well were the two half-cells isolated? Did the sponge effectively block liquid (as opposed to ion) flow between them? You can test that by putting some dye, such as food coloring or bromothymol blue, on one side and seeing if it migrates to the other. Flow might indicate the cell was "short circuited".


(Source: chem.libretexts.org)
In the image, we see that for a "real gas" (indicated by non-ideal solution), Henry's law is applicable for $\ce{\chi} << 1$.
As, $$\ce{p = K_H \cdot \chi}$$ hence the pressure should be low.
Limitations of Henry's law:

The pressure should be low and the temperature should be high i.e the gas should behave like an ideal gas.

The gas should not undergo compound formation with the solvent or association or dissociation in the solvent.


how is the relation between Gibbs free energy and cell potential?

The Gibbs energy is just a way to measure the thermodynamically favoured direction of a reaction. It doesn't tell you anything about the kinetics, so it might well be that the activation energy is too high for a reaction to happen, although it is thermodynamically favored.

The Gibbs free energy is related to the cell potential $\Delta E$ via
$$ \Delta G = -nF\Delta E\; ,$$
where $n$ is the number of Electrons transferred per mole product and $F$ is the Faraday constant. It follows that the larger the cell potential is, the more negative the Gibbs free energy of the system.

The cell potential can be calculated using the Nernst equation:
$$ \Delta E = \Delta E^0 - \frac{RT}{nF}\ln\left( \frac{[\text{Red}]}{[\text{Ox}]}\right) $$

$[\text{Red}]$ and $[\text{Ox}]$ refer to the concentrations of the reduced and oxidized species respectively.

Using this information allows you to determine in which direction a given redox reaction will develop.


How does a carbocation have a positive charge?

I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge?

You need to count ALL the protons and electrons in the ion. What about the atoms other than the central carbon?


In a heating curve, when it is plateau, where does all the energy go since the temperature does not get higher?

You may have learnt at some point that a change of phase is endothermic (i.e. needs heat to happen). So, when you transition from a phase to another and keep adding heat, the temperature will remain the same, as all the heat you'll add will be used for the change of phase. Hence, the temperature will remain the same until the transition is over (i.e. you gave your system enough heat for the transition to be complete), then will vary again.


Explaining boiling using internal energy

You are wrong when you say the potential energy of particles decrease when they move  apart.

The forces between the particles are attractive not repulsive.

Recall the definition of potential energy. It is the work done by external force on a charged particle under influence of electric field.

Therefore, the potential energy for particles, which attract each other, increases when they move apart.

So when to particles get close their potential energy decreases, hence their stability increases. This also happens to be one of the reasons why they come closer.


How much precipitation occurs for a given ionic product?

Solubility product is the product of concentration of ions that can be soluble / in aqueous ionic state under given set of conditions. When ionic product is more means, the remaining is not soluble. Only that amount precipitates out. Still there are ions in water, product concentration of which is exactly equal to solubility product.

For basic understanding of equilibrium constant, you can refer: http://www.adichemistry.com/physical/equilibrium/introduction/chemical-equilibrium.html


What factors other than Gibbs free energy determine formation of a compound?

Thermodynamics only dictates whether or not the process is favorable. It gives no indication as to whether the process will actually take place or how quickly if it does.

You're asking about chemical kinetics which measures how quickly reactions proceed. In this case, the formation of the more stable product under ambient conditions is slow because the barrier to reaction is high.


Enantiomers have different physical arrangement of atoms. This difference plays a crucial role in biology.

  But, why do enantiomers have different biological activities?

  Recognition of chiral drugs by specific drug receptors is explained by
  a three-point interaction of the drug with the receptor site, as
  proposed by Easson and Stedman. The difference between the interaction
  of the two enantiomers of a chiral drug with its receptor is
  illustrated below.

  Source: https://www.khanacademy.org/test-prep/mcat/chemical-processes/stereochemistry/a/chiral-drugs


Olfactory Receptors Bonding with Organic Material

The so-called "bond" you are talking about is relatively weak. Binding between most odiferous compounds (most of them are organic compounds, but there are some inorganic ones like ammonia and hydrogen sulfide) and olfactory receptors (think proteins with a groove that fits the shape of the odor chemical) is rapid and reversible. There are no covalent bonds formed- just weak intermolecular forces like London forces and maybe hydrogen bonding. You need that rapid reversibility in order to tell which way a smell is coming from  or to track something by smell.

For sure, high doses of odiferous compounds can temporarily overload your olfactory receptors - the result being that your sense of smell is degraded or qualitatively changed for minutes or perhaps hours. I do not know for sure but I suspect that some chemical substances (probably only if they are designed to do so) can irreversibly bind to receptors (similar to so-called "suicide" substrates that inactivate an enzyme) and permanently inactivate them, in which case the olfactory cells would have to produce more receptors (hours to days?) in order to restore your sense of smell.

The question I think you need to consider is: is there a chemical substance that would purge the olfactory receptors of old odor compounds and render them freshly able to detect new odors? I think the answer may be no. Even if you could place a zeolite or activated carbon particle next to an olfactory receptor, the odor chemical that is bound to it would still have to diffuse out (on its own) in order to bind with the particle.

But lets see if others out there have any ideas about this. And give my regards to your grandmother!


Each of the terms you generate has a different energy, and may have different spin multiplicity. The spin multiplicity $(2S+1)$ is given by the left superscript (6 and 4) . The quartet (right) it seems has the lowest energy with spin of a triplet (sym e) x doublet ($b_2) \to$  triplet + quartet.  To do this add spin vectors triplet $S_T=1/2+1/2$ , doublet $S_D= 1/2$ and  make series $S_T+S_D \to |S_T-S_D|$ in unit steps, (Clebsh-Gordon series) this gives total spin 3/2 & 1/2 or quartet($\uparrow\uparrow\uparrow$) + doublet($\uparrow\uparrow\downarrow$). The symmetry has next to be associated with the spin multiplicity and this is complicated. It is necessary to find the character ($\chi$) under, say, quartet multiplicity for each symmetry operation $R$ for each symmetry species $A_1,A_2 ..$  etc. $R^2$  means square the operation. In this case this is

$$\chi(quartet) = (1/6)\left([\chi(R)]^3-3\chi(R)\chi(R^2) +2\chi(R^3)\right)$$

The result has to be reduced into symmetric $\chi^+$ and antisymmetric components $\chi^-$  as $\chi^+=\left( [\chi(R)]^2+\chi(R^2)\right)/2 $ and   as $\chi^-=\left( [\chi(R)]^2-\chi(R^2)\right)/2 $ and the set of characters produced reduced to find the symmetry species.

Chapter 5 of 'Symmetry & Spectroscopy' by Harris & Bertolucci gives a detailed account of how to do this.


The pH of a solution is dependent on acid/base equilibria of the substances in the solution. Each of these equilibria will be affected differently by temperature changes. If reactions are endothermic, the equilibrium will shift towards the products upon increasing temperature, otherwise towards the reactants. How much the equilibrium shifts depends on how exo/endothermic the reaction is.

One of the equilibria is the autodissociation of water, which is endothermic. Increasing the temperature will increase product (H+ and OH-) concentration in pure water. Per definition, the pH will decrease as the H+ concentration (or activity) increases.

More complex solutions will either show an increase or a decrease in pH when temperature is raised. For example, some buffers are notorious for changing pH a lot while others are pretty stable, see this table.


What does the amount of acetic acid in vinegar depend on?

Vinegar comes from French "vin aigre", aka sour wine. The transformation of wine into vinegar happens when some (possibly unwanted) aerobic bacteria transform the alcohol into acetic acid by oxidation. So any alcoholic mixture which allows growth of those bacteria can be turned into vinegar, like cider, sake...

The amount of acetic acid depends on the alcohol content which in turn depends on the initial concentration of sugar. It is also dependent on the time allowed for fermentation.

In fact it is much more complicated than that. Have a look at Wikipedia for more info.


Percentage error. How to find percentage error?

Skipping to your question about

  $$\frac{0.1\alpha^3\alpha^3}{( 0.1 - 0.1\alpha)^2}$$

I'll point out two things.

(1) Significant figures come into play here. The Kc is only given to one significant figure as is the molarity of the solution. So each of those quantities has a +/- 50% error. Remember that in Chemistry 0.1 m isn't the same as 0.1000 m.

So the molarity of the solution is 0.1 +/- 0.05. If $\alpha < 0.1$ then the assumption that $0.1\alpha$ is negligible is reasonable.

(2) You can solve the equation for alpha exactly by trial and error. Solve as you did for $\alpha$, then use a factor 10 times larger and one 10 times smaller. One number should be too big and the other too small as compared to Kc. Calculate the average of the two values and test that number. That average will be too big or two small. Discard one side of the bracket. So you're creating an ever shrinking bracket about the "true" value. 6-8 iterations should get you very close.

Now with the "true value" and the estimate from the assumption that $0.1\alpha$ was negligible you can calculate the "real" percentage error for $\alpha$.


Why lactols being tetrahedral complexes do not break?

You may notice that B is an intermediate to eventually reach D.  Indeed, it is quite possible, performing the reaction in the lab, B would spontaneously yield D with only little chance to isolate this intermediate as such.
$\ce{NaBH4}$ reduces the aldehyde to an alcohol.  Then, mediated by the acid, the actetal is converted into an aldehyde to yield B.

The aldehyde in B already is more reactive than a ketone would be, but the presence of acid (still present from converting the acetal) may increase the reactivity of the carbonyl group even further.  The subsequent cyclization is favored

because the chance that alcohol and aldehyde are close enough to form a new bond is much higher for an intramolecular reaction, than for an intermolecular one.  This is a general rule worth to retain. In addition, the formation of five- and six-membered rings equally is energetically favorable.


To your first question: Consider what happens if you put an iron nail or a zinc pellet into a copper(II) solution: the less noble metals will oxidize and reduce the copper ions to copper(0). So even if some of the less noble metal ions are reduced and deposited at the cathode, they would react with the copper ions (that are the most numerous in this scenario).

To your second question: if ions of more noble metals than copper were in the solution, they would be deposited on the cathode. However, by carefully tuning the voltage and current, no more noble metal ions are produced at the anode. As stated by Ivan Neretin in the comments, the cathode will reduce what is available to be reduced: copper ions (which are most numerous and the potential fits).


Building on @Philipp and @chipbuster's answers-
What is Activity?
When we have dilute solutions or small quantities of a gas or other substance, the substance behaves as if the particles are just points, rather than having a size. In these settings, we often use a simplified set of equations.
For example, with the hydrogen ion, we use $\mathrm{pH} = -\log_{10}[\ce{H+}]$ where $[\ce{H+}]$ is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).
Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).
In these circumstances, we talk about ‘activity’, which is a way to correct for this. Basically, it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words, it uses a correction factor for the simplified laws.
Activity, Concentration and Activity Coefficient
The thermodynamic definition of activity is:

$a_{i} = \operatorname{e}^{\frac{\mu_i - \mu_i^⦵ }{RT}}$

Which in low partial pressures and dilute solutions is simplified to:

$a = \gamma\cdot\frac{c}{c^⦵}$ for a solution
$a = \gamma\cdot\frac{p}{p^⦵}$ for a gas

where:

$a$ is the activity
$\gamma$ is the activity coefficient (a constant number based on what the substance is, what the solvent is if there is one, the temperature and other factors)
$c$ is the concentration (for something dissolved); $p$ is the partial pressure (for a gas)
$c^⦵$ is the standard state concentration; $p^⦵$ is the standard state pressure

pH and Activity
The original version of pH as defined by Sørensen is:

$\mathrm{pH} = -\log_{10}[\ce{H+}]$ (hereafter referred to as $\mathrm{p}[\ce{H+}]$)

However, the modern definition that takes into account activity is:

$\mathrm{pH} = -\log_{10}a_{\ce{H+}}$

$\gamma$ is found experimentally. In the case of the hydrogen ion, it is found using electrochemical methods.
Activity of Hydrogen Ion in Dilute Solutions
Standard thermodynamic state for $\ce{H+}$ is considered to be 1 mol/L ($\mathrm{p}[\ce{H+}]$ 0); thus:

$a = \gamma\cdot\frac{c}{c^⦵}$ (as a dilute solution)
$a_{\ce{H+}} = \gamma\cdot\frac{c}{c^⦵} = \gamma\cdot\frac{[\ce{H+}]}{1\ \mathrm{mol/L}} = \gamma\cdot [\ce{H+}]$

Also, in dilute solutions, activity can be approximated/calculated using an extension of the Debye-Hückel equation, the Güntelberg equation, which applies to concentrations of < 0.1 mol/L:

$\log_{10} \gamma_i=\frac{-Az_{i}^{2}\sqrt{I}}{1+\sqrt{I}}$

where:

$A$ and $B$ are constants that depend on solvent and temperature (0.51 and $0.33\times10^8$ respectively for water at 25 °C)
$z$ is the charge of the ion (1 for $H^+$)
$I$ is concentration of the ion

For $\ce{H+}$ in a solution of water at 25 °C this would be:

$\log_{10}\gamma=\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

Substituting, we get:

$a_{\ce{H+}}=\gamma[\ce{H+}]$
$\log_{10}a_{\ce{H+}}=\log_{10}[\ce{H+}]+\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$
$-\log_{10}a_{\ce{H+}}=-\log_{10}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

and so for dilute solutions only:

$\mathrm{pH} = \mathrm{p}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$


Do you break the metallic bond when you break a metal into two pieces?

In addition to ManishEarth’s answer, I'll add that this is only strictly true if you take a pure single crystal of metal and break it. If you take a real-life piece of iron, it is not a pure crystal but contains a large number of crystalline defects. The most common (and easiest to think about) in a polycrystalline material is the presence of grain boundaries between the many crystallites that make up the metal particle:

  Micrograph of a polycrystalline metal; grain boundaries evidenced by acid etching (from Wikipedia)

So, your metal piece is not a single crystal, and crystalline defects are key in the understanding of mechanical response to applied stress, including the case of fracture in which you are interested. So, by breaking a piece of metal in two, you may not actually break so many metallic bonds :)


On the unit of standard reduction potential and its independence of elements involved

A an electric potential (Volt) is equivalent to the unit of J/C. This is because if you move a charge through a potential you gain energy. Volt is kind of a mysterious unit and by the time people reach it the sub-units of J and C are really well understood so it often helps to make sense out of potential. In the same way you can understand speed by thinking of it as distance/time.

The standard reduction potential can be thought of as the potential produced if the reaction happens (if positive) or required to make it happen (if negative). It is determined usually by measuring the total potential (of both reactions) with one half reaction being the one you are interested in and the other which has a zero potential so the standard reduction you are looking for is then just the net potential.

When you have two of those reactions which are both non-zero, the net potential does depend on both.

There is an excellent article on it at CLT :

Standard Reduction Potential


Why are osmoles not considered SI units?

The unit osmole (symbol: $\mathrm{osmol}$) is a coherent derived unit with a special name and symbol like the similar unit equivalent (symbol: $\mathrm{eq}$). Both units have not been accepted for use with the International System of Units (SI).

The main reason why these units have not been adopted is probably that such use is neither necessary nor convenient. All concerned quantities have the same dimension as the amount of substance $n$ $(\dim n = \mathsf{N})$. Therefore, all values can be expressed in terms of the base unit mole (symbol: $\mathrm{mol}$).

Note that, according to the definition of the mole,

  The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is “mol”.
  When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles.

Thus, the amount of substance can always be expressed in terms of the unit mole, regardless of the considered elementary entities. For example,

amount of mercury atoms: $n(\ce{Hg})=1\ \mathrm{mol}$
amount of hydrogen molecules: $n(\ce{H2})=1\ \mathrm{mol}$
amount of chloride ions: $n(\ce{Cl^-})=1\ \mathrm{mol}$
amount of electrons: $n(\ce{e-})=1\ \mathrm{mol}$
amount of equivalent entities $\ce{1/2H2SO4}$ corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction: $n(\ce{1/2H2SO4})=1\ \mathrm{mol}$
amount of equivalent entities $\ce{1/5MnO4-}$ corresponding to the transfer of an electron in a redox reaction: $n(\ce{1/5MnO4-})=1\ \mathrm{mol}$
amount of solute that contributes to the osmotic pressure of a solution: $n_\text{solute}=1\ \mathrm{mol}$

Nevertheless, certain coherent derived units with special names and symbols have been adopted for use with the SI. The main reason is convenience. The special names and symbols are simply a compact form for the expression of combinations of base units that are used frequently, for example “one kilogram metre squared per second squared” may be written “one joule” $(1\ \mathrm{kg\ m^2\ s^{-2}}=1\ \mathrm J)$. However, his reason does not apply to the units osmole and equivalent, since “osmole” and “equivalent” are not shorter than “mole” and $1\ \mathrm{osmol}$ and $1\ \mathrm{eq}$ are not more compact than $1\ \mathrm{mol}$.

In many cases, coherent derived units with special names and symbols may also serve to remind the reader of the quantity involved. For example, a value expressed in the unit newton may help to remind the reader that the considered quantity is a force. However, the unit symbol should not be used to provide specific information about the quantity, and should never be the sole source of information on the quantity. Therefore, even if the special name “osmole” or “equivalent” is used, the special nature of the considered quantity would still have to be explained in the text.

There are, however, a few important exceptions. For the quantity plane angle, the unit one is given the special name “radian” (symbol: rad), and for the quantity solid angle, the unit one is given the special name “steradian” (symbol sr). For the quantity frequency, the unit reciprocal second is given the special name “hertz” (symbol: Hz), and for the quantity activity, the unit reciprocal second is given the special name “becquerel” (symbol: Bq) in order to emphasize the different nature of the quantities. For the quantity absorbed dose $D$, the unit joule per kilogram is given the special name “gray” (symbol: Gy), and for the quantity equivalent dose $H$, the unit joule per kilogram is given the special name “sievert” (symbol: Sv) to avoid the grave risks of confusion between these quantities (e.g. in therapeutic work). Apparently, the concept of “osmotic concentration” (formerly known as “osmolarity”) is simply not important enough to justify a similar exception for the unit osmole.


What would make guanidinium ions and urea more effective denaturants of proteins in some cases, but not in others?

Chaotropic agents increase the solubility of nonpolar substances in
  water. Consequently, their effectiveness as denaturing agents stems
  from their ability to disrupt hydrophobic interactions, although the
  manner in which they do so is not well understood. Conversely, those
  substances  that stabilise proteins strengthen hydrophobic forces,
  thus increasing the tendency of water to expel proteins. This accounts
  for the correlation between the abilities of an ion to stabilize
  proteins and to salt them out.
  1

Among other well established chaotropic agents, guanidinium ion ($\ce{Gu+}$) and the nonionic urea, which, in concentrations in the range 5 to 10 M, are the most commonly used protein denaturants. The effect of the various ions on proteins is largely cumulative: $\ce{GuSCN}$ is a much more potent denaturant than the often used $\ce{GuCl}$, whereas $\ce{Gu2SO4}$ stabilizes protein structures.

There are factors which have been investigated to explain why these two chaotropes appear to be effective in different cases (i.e they tend to be effective in some cases but not others). In one experimental investigation, the mechanistic basis of thermodynamic destabilisation due to urea and guanidinium was tested in order to find possibility that they exert their denaturing effect through peptide group hydrogen bonding (done experimentally by measurement of acid and base catalysed hydrogen exchange (HX) in a small peptide model)[2]

Urea

It has been established that urea increases the acid catalysed rate and decreases the base catalysed rate.

The acid catalysed HX rate in urea solution is higher than in the absence of urea. This can be expected when it is understood that acid catalysed HX proceeds dominantly through transient protonation of the peptide carbonyl group and deprotonation of the peptide NH by water or another base like urea and thus the proposed explanation is that urea forms an H-bond to the peptide carbonyl.[2]
In the case of base catalysed HX rate, which strongly decreases upon urea, the peptide NH group transfers its proton to a transiently H-bonded hydroxide. If urea is H-bonded to the peptide N-H, hydroxide can’t access the proton and HX is blocked.[2]

Guanidinium

With gaunidinium, as explained above, the effect of the various ions on proteins varies by some margin as they exert differently on the ion. It doesn’t selectively H-bonds to the peptide group (base catalysis) because guanidinium cannot H-bond to the peptide NH. Instead studies seem to suggest that guanidinium tends to engage in transient interactions with itself.[2]
Guanidium is often found to be approximately twice as efficient as urea in denaturing proteins but this varies with protein target. Helical peptide studies show that gaunidinium can be up to four times more efficient than urea when planar amino acids are major contributors to the helical stability. In contrast, guanidinium is barely more efficient if stabilisation is due to salt bridges. Thus protein targets are major contributing factor to the varying efficiency of guanidinium in denaturing proteins.[2]

Tl;dr

From above discussion, the varying effeciences depend on things like:

acid/base environment
nature of ions (the case of guanidinium)
protein target (e.g planar amino acids in helical peptides)

References

Voet and Voet Biochemistry. Section 7-1. Primary Structure Determination of Proteins.
Urea, but not guanidinium, destabilises proteins by forming hydrogen bonds to the peptide group. Woon Ki Lim 2595-2600 vol 106 no 8.


Why does electrode potential depend upon enthalpy of hydration and sublimation?

In a thought experiment, you can substitute the electrochemical half reaction by three steps:

Removing atoms from the electrode into the gas phase (sublimation)
Removing electrons from the atoms (ionization)
Solvating the resulting ions

If you have quantitative descriptions of these processes, you get a quantitative description of what is actually going on. This is because many thermodynamic functions (such a enthalpy) are functions of state, i.e. they don't depend on the path from A to B, just on the state of A and B.

I would have said these quantities are related rather than one depends on the other. There is really no causality here.