Which of the following carbanion is most stable?

Some points to check while comparing stability of a carbanion:

Inductive effect: preferentially $\ce{-I}$ effect
Resonance: preferential $\ce{-M}$ effect
Hybridization: preferentially attached to carbon atom having more s-character

... many more: stability
Now the problem is option 1 and 3 are in itself very 'nice' (w.r.t. stability) possible cases of Hybridization-stabilization and Resonance-stabilization respectively.
So the natural (what I thought) way out here is comparing $\ce{pKa}$ 's of acetylene and toluene respectively for option 1 and 3.
from here :

$\ce{pKa}$ of acetylene is about 25.
$\ce{pKa}$ of toluene is about 43.

Less $\ce{pKa}$ denotes how (much more) feasible is for the acid to dissociate by deprotonating itself.
So, acetylene carbanion is more stable than toluene carbanion


On the exact definitions of Frenkel and Schottky defects

Well, you can say that they are kind of similar...the only major difference is that Schottky defects are only for ionic compounds, and the number of missing cations and anions must be equal. Vacancy and interstitial defects can be shown by non-ionic solids. Ionic solids must always maintain electrical neutrality. Rather than simple vacancy or interstitial defects, they show these defects as Frenkel and Schottky defects.
Vacancy Defect: When some of the lattice sites are vacant, the crystal is said to have a vacancy defect. This results in a decrease in the density of the substance. This defect can also develop when a substance is heated.

Schottky Defect: It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal.
Like a simple vacancy defect, the Schottky defect also decreases the density of the substance. The number of such defects in ionic solids is quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. In 1 cm3 there are about 10^22 ions. Thus, there is one Schottky defect per 10^16 ions. The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For example, NaCl, KCl, CsCl and AgBr. It may be noted that AgBr shows both, Frenkel as well as Schottky defects.

Also, with respect to your comment: "I thought Frenkel defects weren't vacancy defects but classified separately under interstitial defects."
Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. Frenkel defect is also called dislocation defect. It does not change the density of the solid. Frenkel defect is shown by ionic substances in which there is a large difference in the size of ions, for example, ZnS, AgCl, AgBr and AgI due to the small size of Zn2+ and Ag+ ions.

Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial defect. This defect increases the density of the substance.

Also, here's why AgBr shows both Schottky and Frenkel defects, even though Frenkel defects are shown by ionic substances in which there is a large difference in the size of ions, and Schottky defects are shown by ionic substances in which the
cation and anion are of almost similar sizes.

Schottky defect in AgBr is exhibited due to precipitation of both cations and anions. In AgBr, Ag+ ion is small in size and when removed from lattice point they can occupy interstitial sites and hence show both Frenkel and Schottky defects.


Why doesn't the atomic number change by two in the process of production of Americium?

What might be confusing is that it's actually two reactions. Two neutrons are captured followed by a single beta decay:

$$\ce{^239_94Pu ->[2(n,\gamma)]^241_94Pu->[\beta^-]^241_95Am}$$

The first reaction leaves the new nucleus in an excited state, so a gamma ray is emitted. The second reaction is actually quite slow with a halflife of ~14 years. It's a beta minus decay so that's why an electron is emitted.


The name of this compound is carbonyl(η5-cyclopentadienyl)(trimethylphosphane)cobalt. Note that, generally, the prefixes denoting the organic groups and any other ligands are placed in alphabetical order before the name of the metal.
The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) presents the rules for formulating and naming coordination compounds (Chapter 9) and organometallic compounds (Chapter 10). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
In particular, the ‘η5-cyclopentadienyl’ term used here is explained in Section IR-10.2.5.1. as follows:

The special nature of the bonding of unsaturated hydrocarbons to metals via their π-electrons has led to the development of the ‘hapto’ nomenclature to designate unambiguously the unique bonding modes of the compounds so formed. (…) The Greek symbol η (eta) provides a topological description by indicating the connectivity between the ligand and the central atom. The number of contiguous atoms in the ligand coordinated to the metal is indicated by a right superscript numeral, e.g. η3 (‘eta three’ or ‘trihapto’), η4 (‘eta four’ or ‘tetrahapto’), η5 (‘eta five’ or ‘pentahapto’), etc. The symbol η is added as a prefix to the ligand name, or to that portion of the ligand name most appropriate to indicate the connectivity, (…)
The ligand name η5-cyclopentadienyl, although strictly speaking ambiguous, is acceptable as a short form of η5-cyclopenta-2,4-dien-1-yl, due to common usage.
These ligand names are enclosed in parentheses in the full name of a complex. Note the importance of making rigorous use of enclosing marks, etc. to distinguish the above bonding modes from the other four cases below. (…)


What is the exact definition of state functions? How to determine state functions?

Where am i going wrong in my understanding?

In short: everywhere. I mean all possible ways. Trying to cheat by randomly throwing symbols and words out there without deep thinking is a way to nowhere. Thermodynamics is a tough subject because it requires a strong mental discipline, so to start from, you should know and understand the definitions. Funky pictures won't help unless you understand the actual meaning of the analogies used there.

State function

A state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.

Using this definition we could answer the question

  How do you know if something is a state function?

trivially: if a quantity is uniquely determined by the equilibrium state then it is a state function, otherwise it is not. $\Delta U$ is clearly not a state function, since by its very definition ($\Delta U = U_\mathrm{final} - U_\mathrm{initial}$) it is not determined uniquely by the state of the system: neither by the final, nor by the initial one. $\Delta U$ is not even a characteristic of the state (since it clearly refers to two of them, not just one), rather it is the characteristic of a thermodynamic process of transition from one state into another.

Internal energy

  I know Internal Energy ($U$) is a state function since it is the sum of
  energies, which is constant at a given time.

Sorry, but this sounds like a nonsense to me. What do you mean by constancy of internal energy at a given time due to it being sum of some energies? From a microscopic point of view, internal energy of a system can indeed be subdivided into kinetic energy of motion of all the system's particles and potential energy due to all kinds of interaction between particles in the system,
\begin{equation}
 U = E_{\mathrm{micro,k}} + E_{\mathrm{micro,p}} \, .
\end{equation}
But what kind of constancy follows from this subdivision? Do you want to say that $\Delta U$ is always zero or what?

Note also that such definition of the total energy belongs already to the realm of statistical thermodynamics, not the classical one. In classical thermodynamics the total energy is postulated to be a state function. Essentially, the first law of thermodynamics tells us that there exist a state function called internal energy, $U$, the increase in which for a closed system during a thermodynamic process is equal to the heat supplied to the system plus the work done on the system,
\begin{equation}
 \Delta U = q + w \, ,
\end{equation}
or in infinitesimals for a quasistatic process,
\begin{equation}
    \newcommand{\dif}{\mathrm{d}}
    \newcommand{\difbar}{\text{đ}}
 \dif U = \difbar q + \difbar w \, .
\end{equation}
Note that infinitesimal heat and work transfers in a quasistatic process are denoted by $\difbar$ rather than $\dif$, used to denote infinitesimal change in internal energy, to emphasize the fact that heat and work in contrast to internal energy are not state functions. Mathematically speaking, $\dif U$ is an exact differential, while $\difbar q$ and $\difbar w$ are inexact differentials. For more on that story, check this answer of mine. What is really important for our current purposes is the very final sentence there: essentially it was experimentally found that for a closed system $\oint \difbar q \neq 0$ and $\oint \difbar w \neq 0$, but $\oint (\difbar q + \difbar w) = 0$. This finding implies that there exist a state function $U$ such that $\dif U = \difbar q + \difbar w$. So, the law that postulates that $U$ defined by $\Delta U = q + w$ is a state function is based on experimental evidences.


Are 6-hydroxyhexan-2-one and hexanoic acid isomers?

Isomers are compounds having same molecular formula, but a different structural formula.

As they satisfy the above condition, they are isomers and they are structural isomers. Structural isomers have a few variety based on connectivity of constituent atoms.

Chain : which has a different arrangement of carbon molecules
Position : different position of same functional group
Functional : different functional group for the same atoms
the others (tautomerism, metamerism and ring-chain isomerism)

They are functional isomers as they have different functional groups. First one has a ketone and an alcoholic group, while the second one has a carboxylic acid functional group.

You will come to see that it is easier to recognise structural isomers compared to the rest (stereoisomerism).

When speaking about isomerism, a picture is worth a thousand words:


For a copper/copper sulfate half reaction, do sulfate anions move to the anode and lose electrons?

I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.

This is one way of achieving charge balance, but there are many other ways.

Charge balance of half-reactions

Half-reactions have electrons either as reactants (reduction half-reaction at the cathode) or as products (oxidation half-reaction at the anode). For the cell in question, the oxidation half-reaction at the anode is:

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

There is no physical state associated with the electrons, but they are removed through the wire. So if electrons appear in the half-reaction, you need some ions to balance it out. In a oxidation, we produce electrons, so either we need to produce cations at the same time or consume anions (or have ions on both sides of the equation, but of unequal charge). The ions (in solution) are also responsible for the charge transport between anode and cathode.

Oxidation state imbalance of half-reactions

At least one atom (as element, simple ion, or part of a compound or complex ion) has to undergo a change in oxidation state. The species that balance the charges, however, might or might not undergo a change in oxidation state. In the half-reaction in question, copper changes oxidation states, and the copper ions balance out the charge of the electrons so that both sides of the half-reaction have equal charge (zero, in this case).

To give a different example, here is a half-reaction involving lead:

$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + H+(aq) + 2e-}$$

Here, the lead gets oxidized (from element to +II), but remains at the electrode. The charge balance is achieved by hydrogen sulfate ions traveling to the electrode and remaining at the electrode as part of the solid lead(II)sulfate, and the released hydrogen ion.

Direction of ion flow

In the reaction in question, copper cations travel from anode to cathode in the solution to balance the charge transport due to electrons traveling from anode to cathode via the wire. We can write the equation showing this explicitly by combining the half-reactions and keeping anode and cathode species labeled.

$$\ce{Cu(ano) + Cu^2+(cat) + 2e-(cat) -> Cu^2+(ano) + 2e-(ano) + Cu(cat)}$$

Because we can't have electrons or copper ions accumulate in or at the electrodes, we need a transport mechanism for both species (through the wire and through the solution, respectively).

Here are two more examples with different ion transport.

Water electrolysis

$$\ce{4H+(aq) + 4e- -> 2H2(g)}$$
$$\ce{2H2O -> O2 + 4H+(aq) + 4e-}$$

At low pH, the cathode consumes hydrogen ions while the anode produces hydrogen ions. At neutral pH, you would say the cathode consumes hydrogen ions and the anode consumes hydroxide ions produced from auto-dissociation of water (i.e. anions travel to the anode and cations travel to the cathode - the classic case giving anions and cations their name).

Lead-acid battery

$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + 2e- + H+}$$
$$\ce{PbO2(s) + 3H+(aq) + HSO4^-(aq) + 2e- -> PbSO4(s) + 2H2O}$$

The element that changes oxidation state is lead in various solid compounds (elemental, lead(IV) in lead oxide and lead(II) in lead sulfate). The ion transport is due to hydrogen sulfate going from solution to both electrodes, forming lead sulfate, and hydrogen ions going to the cathode to combine with oxygen to form water (and hydrogen ions being released from hydrogen sulfate at the anode). So here, ion transport is accomplished with anions going to both electrodes, but an excess of cations (hydrogen ions) going to the cathode. To emphasize, hydrogen sulfate anions are traveling to the cathode. This is an counter example to "anions travel to the anode".

Cells with salt bridge

None of the cells in my three examples had salt bridges. If half-reactions have separate solutions connected by a salt bridge, ions don't travel the entire way between electrodes; instead, redox-inert ions such as potassium cations and chloride anions negotiate the charge transport between solutions of the half cells.


How do nylon and polypropylene blend together in the production of a toothbrush?

A toothbrush need not be made of the two materials you mention.  The two materials are choosen because of their low price of production*), ease of manufacture of the intended tooth brush (heat molding of the frame), moderate / tuneable flexibility of the material for the bristles (nylon).  In comparsion to bristles of natural origin, nylon equally is easier to cleanse by rinsing with water, and dries rapidly ‒ offering less ground for bacteria to grow on them.

*) Waste disposal is not accounted here. While the brush, consisting of polymers, may be incarcerated, why do we throw away the whole toothbrush away if apparently only the heads / bristles wear out?


What do the brackets mean in this inline structure representation?

When you see a formula like the one in the question in organic chemistry, the person writing it down either already assumes a lot of chemical knowledge on the reader’s side or wants to teach them how to interpret this. So to a practising organic chemist, one inline formula may mean one thing while a similar one (with brackets in the same positions etc) may mean something completely different. Sometimes even, inline formulae are (and must be, due to their nature) potentially ambiguous. Therefore, organic chemistry relies a lot on actually drawing out the structures, whether it be each atom explicitly or in the sum formula formalism.

In the case of $\ce{Br(CH2)4Br}$, organic chemists will know that bromine will typically only form one single bond. Additionally, each $\ce{CH2}$ unit is known to be able to form two bonds — either two single bonds or one double bond. Therefore, an organic chemist will automatically expand the formula to $\ce{Br-CH2-CH2-CH2-CH2-Br}$ or 1,4-dibromobutane.

However, you might also sooner or later encounter inline formulae such as $\ce{BrC(CH3)2CH2Br}$ or $\ce{(CH3)3COH}$. Here, the brackets do actually mean repeated addition of said group to a single common atom. so the first of these two formulae describes 1,2-dibromo-2-methylpropane and could also be written $\ce{H3C-C(CH3)(Br)-CH2Br}$ while the second one represents tert-butanol or 2-methylpropan-2-ol.

Immediately realising which type of brackets is used requires practice and experience.


Hexane is the solvent for the two halogens. It does not react with either so leave it out of your answer.


Designing synthetic route to dicarbonyl compound

Everything you said is correct except for the first step.  Free radical bromination is reasonably selective, tertiary positions are about 20 time more reactive than secondary carbons, so the 1-bromo, 1,4-dimethyl compound is formed.  Everything's a go from there!


Which is a dye: Ferric Acetate or Ferrous Acetate?

The gist is in the solubility products for the hydroxide precipitates of $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$.

  $\ce{Fe(OH)3}$ doesn't really exist but is rather $\ce{FeO(OH)}\cdot x \ce{H2O}$. However the $K_\mathrm{sp}$ is calculated as $\ce{Fe(OH)3}$.

Now for the vinegar solution let's consider a somewhat "spent" solution where a lot of the acid has reacted. So consider a nominal pH of 5.5, so pOH = 8.5 and $\ce{[OH] = 3.16\times 10^{-9}}$. Let's calculate the equilibrium molarity of Fe species as "Mol(Fe) @ pH=5.5" in the table below.

\begin{array}{|c|c|c|}\hline
 & K_\mathrm{sp} & \text{Mol(Fe) @ pH=5.5}\\ \hline
\ce{Fe^{2+}} & 8.0 \times 10^{−16} & 80.12 \\ \hline
\ce{Fe^{3+}} & 2.79 \times10^{−39} & 8.84\times 10^{-14}\\ \hline
\end{array}

A $\ce{Fe^{2+}}$ solution can't be 80 molar of course, but what that does mean is that all the $\ce{Fe^{2+}}$ formed will stay in solution.

The calculations also show that a substantial amount, if not all, of the $\ce{Fe^{3+}}$ created will precipitate. The precipitation does lower the pH, so the equilibrium values are indeterminate without knowing the concentrations of the acetate anion and the nominal concentration of $\ce{Fe^{3+}}$ (amount dissolved plus amount in the precipitate). However looking at the overall reactions, it wouldn't seem that the pH could be lowered much.

\begin{array}{c c c c}
 & \ce{2Fe + 6H+} & \ce{<=>} & \ce{2Fe^{3+} + 3H2} \\
 & \ce{2Fe^{3+} + 6OH- } &\ce{<=>} & \ce{2Fe(OH)3} \\
 & \ce{6H2O} & \ce{<=>} & \ce{6H+ + 6OH-} \\ \hline
\text{Overall} &\ce{2Fe + 6H2O} & \ce{<=>} & \ce{2Fe(OH)3 + 3H2}\\
\end{array}

A further consideration here is that oxygen from the air, which dissolves into the solution, can oxidize $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$.

Now I'd assume that it is the $\ce{Fe^{2+}}$ salt that you'd really want since that salt could diffuse into (soak into) whatever you're coating the solution with. The $\ce{Fe^{2+}}$ would then be oxidized to $\ce{Fe^{3+}}$ in the interior of the item.  The $\ce{FeO(OH)}\cdot x \ce{H2O}$ on the other hand would just sit on the surface of the item like a paint.


Is Hexafluorocyclohexane the "Most Polar" Small Molecule?

At first, I thought, that those 6.2 D of O’Hagan et al. have been measured somehow but it as later described, they calculated it (1):

  A molecular dipole value of 6.2 D for 1 was calculated at the M11/6-311G(2d,p) theory level using natural bond orbital (NBO) analysis with NBO 6.0.

I was using your given database for a little search, starting with nitroaniline, and came up with a few small molecules with higher dipole moments. Those seem to be PM7 calculated values and are shown in the following table:

\begin{array}{lll}
\hline
\text{Name} & \text{Dipole moment in D} & \text{Source}\\
\hline
\text{4-Nitroacetanilide} & 8.36 \pm 1.08 & (2)\\
\textit N\text,\textit N\text{-Dimethyl-4-nitroaniline} & 9.53 \pm 1.08 & (3)\\
\textit P\text ,\textit P\text{-Dimethyl-}\textit N\text{-(4-nitrophenyl)phosphinic amide} & 10.89 \pm 1.08 & (4)\\
\hline
\end{array}

As there were some comments, “it seems that this level of calculation does not provide accurate estimates of dipole moments”, I did some further calculations on levels that might provide better estimates.

There is a quite new paper of Hickey and Rowley (5) which says:

  CCSD, MP2, or hybrid DFT methods using the aug-cc-pVTZ basis set are all able to predict dipole moments with RMSD errors in the 0.12–0.13 D range and polarizabilities with RMSD errors in the 0.30–0.38 Å³ range.

Within the paper, there is later on a comparison between cc-pVTZ, Sadlej-pVTZ and aug-cc-pVTZ:

  The Sadlej basis set yields results that are improved over the cc-pVTZ basis set and have effectively the same accuracy as the aug-cc-pVTZ basis set.

That’s why I chose Sadlej’s basis set at first … but I wanted to look if it could not be easier hence faster. Therefore I also chose the def2-SVPD and looked, what an improvement a def2-TZVPD single point calculation on this geometry would yield.

I use ORCA 3.0.3 for calculations – the command line is like the following:

! B3LYP Sadlej-pVTZ TightSCF TightOpt

Table 1: "Pure" dipole moments (in Debye) straight from the calculations
\begin{array}{lllll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide} & 9.18261 & 9.00380 & 9.07922 & 8.6738\\
\textit N\text ,\textit N\text{-Dimethyl…} & 7.39418 & 7.29188 & 7.30862 & 9.4209\\
\textit P\text ,\textit P\text{-Dimethyl…} & 8.58913 & 8.44430 & 8.51591 & 11.8352\\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.17422 & 5.97013 & 6.04794 & 5.6298\\
\hline
\end{array}

What I did then, was to think of how to produce values that could be properly compared to the experimental values from the paper. As they said, with their methods they can yield RMSDs of around 0.12 D. So I took their values for the Sadlej basis set and made my own calculations on their test molecules for the two def2 basis sets (and also for PM6 with Gaussian 09 Rev. A.02).

Table 2: Some analytical values to compare with the publication values; the units are in Debye
\begin{array}{lllll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{MAE} & 0.08 & 0.08 & 0.08 & 0.40\\
\text{RMSD} & 0.13 & 0.12 & 0.12 & 0.61\\
\hline
\end{array}

So for my two ways, the MAE and the RMSD are not too far away from the triple zeta values but are calculated way faster. Now I have made a linear regression between the experimental and the calculated dipole moments from Hickey and Rowley’s test molecules to be able – as said before – to calculate proper error ranges at a given probability.

Figure 1: Regression functions for the four tested methods (blue), prediction intervals at 95% (gray); from top left to bottom right: def2-SVPD, def2-TZVPD//def2-SVPD, Sadlej-pVTZ, PM6

Now let’s turn the “pure” values from Table 1 into somehow more real values:

\begin{align}
\text{def2-SVPD} \to \text{exp}&: -0.0097439128 + 0.9763275\ x\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: -0.0045050428 + 0.99095691\ x\\
\text{Sadlej-pVTZ} \to \text{exp}&: -0.012922547 + 0.98812009\ x\\
\text{PM6} \to \text{exp}&: -0.16057241 + 0.89545049\ x\\
\end{align}

Prediction Intervals (for the error):

$$\small\begin{align}
\text{def2-SVPD} \to \text{exp}&: 0.24473369 \sqrt{1.0217391 + 0.010348758 (x-1.4590217)^2}\\
\text{def2-TZVPD//def2-SVPD} \to \text{exp}&: 0.23719452 \sqrt{1.0217391 + 0.010656662 (x-1.4321957)^2}\\
\text{Sadlej-pVTZ} \to \text{exp}&: 0.23333979 \sqrt{1.0217391 + 0.010593477 (x-1.4448261)^2}\\
\text{PM6} \to \text{exp}&: 0.98446665 \sqrt{1.0217391 + 0.0097476423 (x-1.7592391)^2}
\end{align}$$

Table 3: Converted dipole moments (in Debye) from Table 1 including error ranges at 95% probability

\begin{array}{llll}
\hline
 & \text{def2-SVPD} & \text{def2-TZVPD} & \text{Sadlej-pVTZ} & \text{PM6}\\
\hline
\text{4-Nitroacetanilide}  & 8.96 \pm 0.31 & 8.92 \pm 0.30 & 8.96 \pm 0.30 & 7.61 \pm 1.20 \\
\textit N\text ,\textit N\text{-Dimethyl…}     & 7.21 \pm 0.29 & 7.22 \pm 0.28 & 7.21 \pm 0.27 & 8.28 \pm 1.24 \\
\textit P\text ,\textit P\text{-Dimethyl…}     & 8.38 \pm 0.30 & 8.36 \pm 0.29 & 8.40 \pm 0.29 & 10.44 \pm 1.40 \\
\text{all-}\textit{cis }\ce{C6H6F6} & 6.02 \pm 0.27 & 5.91 \pm 0.26 & 5.96 \pm 0.26 & 4.88 \pm 1.06 \\
\hline
\end{array}

What can be seen from this now? The smaller B3LYP/def2-SVPD method yields nearly the same values as the more expensive Sadlej version. Putting the def2-TZVPD single point on top of them, yields slightly deeper results with actually no accuracy improvements. PM6 is no real good option as the error ranges are way higher, with values of about $1-1.4$ D.

In the end, the dipole moment of all-cis 1,2,3,4,5,6-hexafluorocyclohexane is probably somewhere around $6$ D but there are probably many other molecules with higher dipole moments … “my” three molecules shown above go at least up to somewhere around $9$ D.

As long as no experimental values are available, everything here is only a good guess – but you will know this too good.

Sources:

O’Hagan et al., Nature Chemistry, 2015, 7, 483–488
http://pqr.pitt.edu/mol/NQRLPDFELNCFHW-UHFFFAOYSA-N
http://pqr.pitt.edu/mol/QJAIOCKFIORVFU-UHFFFAOYSA-N
http://pqr.pitt.edu/mol/FICBIFYYTQXSEY-UHFFFAOYSA-N
A. L. Hickey, C. N. Rowley, J. Phys. Chem. A, 2014, 118 (20), 3678–3687


How does pH affect the spontaneity of biochemical processes?

Note that pH effect is reaction specific.
If a reaction is e.g. $\ce{A + B + H+ <=> C + D+}$ , or $\ce{A + B- <=> C + D + OH−}$,
then lowering pH supports the spontaneity.
If OTOH, the reaction is $\ce{A + B + OH-  <=> C + D−}$ , or $\ce{A + B+ <=> C + D + H+}$,
then lowering pH goes against the spontaneity.
It can be generalized by applying the Le Chatelier's principle, saying systems adapts to external changes by such equilibrium shifts that partially decreases the system chage caused by the external factor.
Lowering pH supports spontaneity of reactions consuming H+ and vice versa.
Increasing temperature prefers endothermic reactions consuming heat and vice versa.
Increasing pressure prefers reactions decreasing the total molar amount of gaseous components and vice versa.

Note that biological systems are much more complex than chemical systems.
Lowering pH may support thermodynamic spontaneity of some particular reaction. But an enzyme catalyzing its kinetic rate may start loosing its activity, get denatured, cells may stop producing the enzym, cells may stop overall life functions, or may die.


Why is FORTRAN so commonly used in computational chemistry?

I don't think that's really true anymore.

Some Fortran use is historical (i.e., early codes were developed in FORTRAN because that was the best programming language for number crunching in the 70s and 80s). Heck, the name stands for "formula translation."

Some Fortran use is because of performance. The language was designed to be:

  especially suited to numeric computation and scientific computing.

Many times, I find chemistry coders sticking to Fortran because they know it and have existing highly optimized numeric code-bases.

I think the performance side isn't necessarily true anymore when using modern, highly optimizing C and C++ compilers.

I write a lot of code in C and C++ for performance and glue a lot of things with Python. I know some quantum programs are written exclusively or primarily in C++. Here are a few open source examples:

Psi4 - Written in C++ and Python
MPQC - Written in C++
LibInt - Written in C++ for efficient quantum integrals.
LibXC - Written in C with Fortran "bindings" for DFT exchange-correlation functionals

This is my opinion, but my recommendation for faster performance in chemistry would be Python with some C or C++ mixed in.

I find I'm more efficient coding in Python, partly because of the language, partly because of the many packages, partly since I don't have to compile, and that's all important.

Also, you can run Python scripts and functions in parallel, on the GPU, and even compile them, e.g. with Numba. As I said, if I think performance is crucial, I'll write pieces in C or usually C++ and link to Python as needed.


Is there a definition of rotatable bond?

Is there an operational definition, or a published study where a
general definition is given, which would allow me to recognize in a
generic molecule (or at least an organic molecule) whether a given
bond is rotatable or not?

It depends on your purpose. Why do you want to identify some bonds and call the rotatable? For instance, PubChem, one of the biggest databases of molecules, defines rotatable bond as follows:

Rotatable bond is defined as any single non-ring bond, bounded to
nonterminal heavy (i.e., non-hydrogen) atom.

To some extent all this boils down to the following IUPAC definition of free rotation:

free rotation (hindered rotation, restricted rotation)
In a stereochemical context the rotation about a bond is called 'free'
when the rotational barrier is so low that different conformations are
not perceptible as different chemical species on the time scale of the
experiment. The inhibition of rotation of groups about a bond due to
the presence of a sufficiently large rotational barrier to make the
phenomenon observable on the time scale of the experiment is termed
hindered rotation or restricted rotation.

At usual lab conditions a rotation around a single non-ring bond, bounded to nonterminal heavy atom will be indeed almost free, thus, PubChem definition can be used as a very simple rule-of-thumb to identify such bonds. But in general it depends on conditions. And purposes.

With respect to protein backbones being rotatable or not: I'm not a speacialist in this area, but Wikipedia article on peptide bond claims that (emphasis mine)

Significant delocalisation of the lone pair of electrons on the
nitrogen atom gives the group a partial double bond character. The
partial double bond renders the amide group planar, occurring in
either the cis or trans isomers.

So, this partial double bond character might contribute in protein backbones being not quite rotatable. Besides, the Wikipedia article also claims that (emphasis again mine):

In the unfolded state of proteins, the peptide groups are free to
isomerize and adopt both isomers; however, in the folded state, only a
single isomer is adopted at each position (with rare exceptions).

So, at least in the unfolded state protein backbones are rotatable. The fact that only two configurations (cis and trans) are possible is irrelevant. The notion of rotatable bond does not imply having few specific possible configurations.


How many electrons are there in the pi-system of cyanogen?

Here is the structure of cyanogen

Each carbon has 4 valence electrons and each nitrogen has 5, for a total of ((4x2)+(5x2)) 18 electrons as you said.  Again, as you noted, 6 electrons are involved in sigma bonds, the 2 C-N sigma bonds and the one C-C sigma bond. Each nitrogen also has a lone pair of electrons and that accounts for (2x2) 4 more electrons.  That leaves 8 electrons to fill the 4 pi bonds.  Just like acetylene with 2 pi bonds, the two carbon-nitrogen triple bonds in cyanogen also have two pi bonds, with each pi bond holding 2 electrons.


As the dipolar moment is a vectorial property, you have to take into account both the number of chlorine atoms around the carbon atom, as well as their relative arrangement -- among them, as well in respect to the carbon atom.

Lacking access to experimentally determined data about the molecular geometries,you may sketch the molecules in question, for example with Avogadro, and optimize the geometry (here, MMFF94 was deployed). You are literally able "to walk around" the molecule and its dipolar moment.

You easily will recognize a feature in common -- the similar direction of the resulting dipolar moment.  Simultaneously, you will recognize the different angle defined by Cl-C-Cl, too.  However, beside the angle, distances like C-Cl and C-H are affected by the different steric demand and electronic properties, too; equally influencing the overall dipole moment.  Hence the molecule of dichloromethane is more polar than the one of chloroform.

Sidenote: a force field like MMFF94 is a computationally cheap approach to optimize molecular structures.  There are better methods, but more costly; for the level of the question addressed, however, I thought it already provides the insight necessary here.


Consequences of ortho effect on resonance

If they are off the plane due to steric effect,  then resonance wouldn't occur, hence the (-)R effect of - COOH group vanishes and hence "generally" acidity of the respective acid increases.


Radial Probability Distribution Curve versus ψ² versus r curve for 1s orbitals

It is very important to note here that $R(r)$ is just the radial part of wavefunction. Wavefunction does not describe any observable. The Born interpretation says that the probability density of finding electron between any two points $x_1$ and $x_2$ is given by
$$\rho=\psi\psi^*.$$
While the probability is
$$P =\int_{x_1}^{x_2}{\psi\psi^* \mathrm{d}x}.$$
(This is only in one dimension. In spherical coordinates you must integrate over the volume).
Why is there a difference between these two quantities? I think it will be easier to understand using an analogy. Consider a rod of unit length in which the charge varies as a function of $x$. Say the function is
$$q(x)=x\mathrm{e}^x-1$$
The average charge of the rod is $0$. While the average charge density of this rod is $\mathrm{e}-2$. Where average is given by
$$\langle f \rangle=\frac{\int^{b}_{a}{f \mathrm{d}x}}{b-a}.$$
I hope you have understood the difference between function density and the function itself. The radial probability and probability density are something similar. By seeing the graphs you can to a certain extent predict the average of the probability density and probability. In the second graph it is $0$ while in the second graph it is close to the peak which is $a_0$.


UV Absorption of iodine with some solvent

Iodine is highly polarisable and will form a (Mulliken) charge-transfer complex with many aromatics. Very many pairs of donor - acceptor pairs do this and have been studied for approx $100$ years, see for example tetracyanoethylene–pyrene (J. Chem. Phys. 105,1996 p2287) for an ultrafast time resolved study.  In  absorbing a photon the ground state complex DA transfers  an electron directly to a charge transfer state $DA \overset{h\nu} \to \mathrm D^+A^-$. The interaction is thus that of all the orbitals in one molecule with all those in another.

Books

R. S. Mulliken and W. B. Person, 'Molecular Complexes' Wiley, New
York, 1969. and G. Briegleb, 'Elektronen Donator Acceptor Komplexe' Springer, Berlin, 1961.


Why is there an upper limit to the photon energy that can be observed in the discrete emission spectrum of the hydrogen atom?

The closest an e- can get to nucleus is the 1st, it can't get closer and so can't get to lower E. If an e- is beyond the influence of the nucleus and falls to 1s it can't fall any further  (drop in E any more ). This represents the largest E emission.  Once an e- is beyond the influence of the nucleus it doesn't matter how much further it goes it is still beyond its influence and hence the energy of the transition doesn't increase


Is the water in an aqueous solution undergoing a reaction considered part of the system or surroundings?

The boundary between system and surroundings is defined by you the scientist. Still, some choices make more sense than others, depending on the question you want to answer. When you perform a calorimetric experiment, for instance, you monitor the heat generated or absorbed during a reaction (such as formation of a precipitate). The heat originates or is consumed in the local combination of reagents to form the bonds between precipitate atoms and during segregation of solvent molecules. That heat then dissipates into the rest of the vessel and is registered, for instance as a temperature change. In the final analysis, what you call the system here would probably include only what has changed and you are specifically interested in, namely the molecules involved in the actual reaction. This would include solvent molecules associated with the solubilized reagents (the involvement of solvent is made explicit when reagents are labelled as aqueous, for instance). The behavior of the bulk solvent, which serves an "auxiliary" role, for instance by transferring heat across the vessel, is important during calibrations, but is not the question of interest, and would usefully be regarded as the surroundings.

Water can be regarded as a catalyst in many aqueous reactions, but when a reaction occurs in a solvent that associates strongly "catalyst" is not generally a proper description for the solvent, as the energy of both reagents and products may be stabilized/destabilized by the solvent, and to different degrees (and not just the energy of an intermediate in the reaction).


Is there a way to liquefy a super absorbent polymer without using salt?

change the pH to less than 3.   absorbents do not like low pH.
since absorbents  are a polymer of sodium acrylate.  lowering the
pH covert  sodium acrylate to the acid form (acrylic acid) the acid form is not a super absorbent.


When volume decreases and the product has less moles than reactants, why is there no increase in the reaction quotient?

Keep in mind that for reaction

$$\ce{r_{1}R_{1} + r_{2}R_{2} + \ldots + r_{n}R_{n} <=> p_{1}P_{1} +\ldots + p_{m}P_{m}}$$

the reaction quotient is:

$$Q = \frac{\prod_{i = 1}^{m} [P_{i}]^{p_{i}}}{\prod_{j = 1}^{n} [R_{j}]^{r_{j}}}$$

Specifically, here:
$$Q = \frac{[\ce{COCl2}]}{[\ce{CO}][\ce{Cl2}]}$$

For your specific reaction, the reaction quotient is getting smaller because the denominator has a quadratic dependence on the pressure where as the numerator only has a linear dependence on pressure.

The goal as usual is to make $Q = K_{\mathrm{eq}}$, which is equilibrium.

Since now, $Q < K_{\mathrm{eq}}$, you want to push the reaction forwards. That will increase the concentrations of product and decrease the concentrations of reactant, making the reaction quotient bigger.

Intuitively, that makes sense in your example. There are two molecules combining to form one molecule. By decreasing the volume and increasing the pressure, you're squeezing them together into the product.


Look this as:

$$\ce{2Ca + O2 -> 2CaO}\qquad\text{(complete reaction)}$$

then we can this split into two sub-reactions:

$$\ce{Ca - 2 e- -> \overset{+2}{Ca}}\tag{1}$$

Here "$+2$" means that $\ce{Ca}$ lost $2$ electrons (or two negative units) and therefore it was oxidized. Since it lost something negative it has a higher number of protons (positive particles) remained. In other words the atoms is now positively charged. The initial state was $0$ (the same numbers of protons and electrons, or positive and negative particles/units) and it was increased by loosing two electrons (the number increased).

$$\ce{O + 2 e- -> \overset{-2}{O}}\tag{2}$$

The oxygen got two electrons or two negative particles. Now oxygen has two negative particles more than positive particles (protons) and therefore it is $-2$. In other words, the number (oxydation state) decreased ($0 \to -2$).


"Tie-breaker" rules for IUPAC nomenclature of organic compounds

For Question 1

Example 1

You should select the (green + blue) chain because, no matter which among the blue and yellow you pick, the numbering would not start from the green chain rather the yellow or blue.
Now if you pick the blue chain the substituent ($\ce{methyl}$) would get number 4, whereas if you picked the green chain the substituent($\ce{isopropyl\ or\ methylethyl}$) would get the number 5, so in accordance to the First Point of Difference Rule you should select the green + blue chain.
Also Worth noticing is the fact that no matter which chain you pick, the other chain which becomes the substituent would always get the number 7

So locants (4,7) are better than (5,7).
$$\begin{matrix}
\text{Better name:}& \text{7-(2-isopropylhexyl)-4-methylheptadecane}&\\
\text{Rather than:}& \text{5-isopropyl-7-(3-methylhexyl)heptadecane}&
\end{matrix}$$

Example 2

In this case according to the above mentioned rule, you should check the position of the double bond, now since there is a double bond already in the green chain at the first position it would get number 1 and the numbering would start from there.
If you choose the green + blue chain the other double bond would get number 5 and for the green + yellow chain the other double bond would get number 6. So you should go for the green + blue chain in this case.

So locants (1,5) are better than (1,6) with reference to double bonds.
$$\begin{matrix}
\text{Better name:}& \text{(5}\textit{Z}\text{)-5-ethenylhepta-1,5-diene}\\
\text{Rather than:}& \text{(3}\textit{Z}\text{)-3-ethylidenehepta-1,6-diene}
\end{matrix}$$

For Question 2

If the chain has to be numbered and two substituents have the same locant the n you see the alphabetical order of the names of the groups/substituents and if they are also the same it only means that they are the same substituent and hence are written together with there locants separated by a (comma),.

For example in you would name the below compound as
$\text{2,4-dichloropentane}$

For Question 3

Example 1

The substituent in this example would be named $\ce{ethylidene}$.
$$
\begin{matrix}
\text{Final name:}& \text{5-ethylidenenona-1,8-diene}
\end{matrix}
$$
Similar substituents can be named as

$$
\begin{array}{|c|c|c|c|}
\hline
\ce{Substituent& Saturated\ form& Name\ of& Name\ of \\
&& saturated\ form& substituent \\
\hline
R=CH2& R-CH3& Methyl& Methylidene\\
\hline
R=CH-CH3& R-CH2-CH3& Ethyl& Ethylidene\\
\hline
R-CH=CH2& R-CH2-CH3& Ethyl& Ethenyl\\
\hline
R=CH-CH2-CH3& R-CH2-CH2-CH3& Propyl& Propylidene\\}
\hline
\begin{align}
\ce{R=C&-CH3\\
|& \\
C&H3}
\end{align}&
\begin{array}{rl}
\ce{R-C&-CH3\\
|\ &\\
C&H3}
\end{array}&
\ce{Isopropyl& Isopropylidene}\\
\hline
\end{array}$$


Why is gum so sticky and what chemicals make for easy removal of gum?

The International Chewing Gum Association tells us:

Gum base is what gives chewing gum its “chew.” It is made of a combination of food-grade polymers, waxes, and softeners that give it the texture desired by consumers and enable it to effectively deliver sweetness, flavor, and various other benefits, including dental benefits.

A more complete description of what makes chewing gum sticky states:

Terence Cosgrove, Professor of Chemistry at the University of Bristol and Chief Scientific Officer of Revolymer, explains that the chemical bonds between the molecules in a polymer-based chewing gum make it difficult to remove from a surface. When you pull a piece of used gum off a surface, most of the energy goes into stretching the polymer bonds in the gum, rather than actually breaking the bonds between the surface and the gum. Since polymers are long chains of covalently bonded molecules that give the gum its elasticity and "chewiness," the attractive forces between the atoms are formed by the sharing of electrons [8]. Thus, the bonds between the repeating units tend to remain intact as they lengthen and contract from an applied external force.
The temperature at which the polymer base is exposed to the air also affects both the elasticity and adhesiveness of the chewing gum. When a piece of gum is heated by saliva and deformed by the grinding of human teeth, its polymer chains align in the direction of these forces. The degree of alignment is a function of the magnitude of applied stress, which explains why the gum becomes tougher and less elastic the longer or more rigorously you chew [8]. After the gum is removed from your mouth and placed in a cooler environment, the drop in temperature causes the orientation of the polymer chain to freeze, resulting in a hardened piece of used gum.
Most commercial polymer gum bases are hydrophobic (water-insoluble), which is the reason they stick easily to oily surfaces and are difficult to remove, even with the help of household cleaning solutions. A hydrophobic substance tends to repel polarized molecules like water and attract non-polar compounds, such as the grease and grime on streets. A hydrophilic substance, on the other hand, behaves in the opposite manner by attracting water and repelling fats and oils [8]. A compound that exhibits both hydrophobic and hydrophilic traits is known as "amphiphilic" [9]. These substances, which attract both water and oil to some degree, can be used to create synthetic polymers that could help eliminate the sticky mess of chewing gum pollution.

Note: The numbered references in the above block quote can be pulled from the linked article. The article also details efforts aimed at producing a new type of chewing gum that doesn't stick to things as much as normal/currently-manufactured ones seem to. It's a good read.
But to your main point, I've linked the MSDS for a popular product for removing gum and related compounds (it's called Goof Off), which consists almost entirely of acetone (also commonly found in nail polish remover). Given the deleterious health effects of acetone (and its high vapor pressure), it's not a good candidate for your needs.
Something non-stick and more permanent, such as a Teflon spray might be ideal for your needs.


What is the name of this organic compound?

The longest chain has 7 carbon atoms. So you would have an ethyl group in carbon 3 and the amino group will be a ramification in carbon 5. I would stand for:

  5-(aminomethyl)-3-ethylheptan-2-ol


What you are calling "taste" is actually produced by the olfactory sense – "smell."  True taste, which requires contact of a substance with your tongue, is limited to sensations of sweetness, sourness, saltiness, bitterness, and umami.  The chemistry of taste is certainly quite interesting, but anything beyond those "flavors" is produced by smell (via far more complex physiological processes).  You have certainly smelled rust (or, more accurately, various oxidation states of iron that exist in blood, likely from wet or weathered steel).  So you are, in fact, correctly associating the smell of iron oxides from two different sources: steel and hemoglobin.


$\ce{(C6H10S2O3)3}$ implies that there are $3 ~ \ce{C6H10S2O3}$ molecules bonded together instead of $1~ \ce{C18H30S6O3}$ molecule. Although they are the same thing, the correct notation is your teachers way because it shows the isometric formula for the molecule. Also if you were using that notation you should have put $\ce{(C6H10S2O)3}$ because $\ce{(C6H10S2O3)3}$ has 6 too many oxygens.


Can 25mL of 10% phosphoric acid substitute for 10mL of 25% solution?

Not reliably. Acid strengths measured in percentages are usually given as percent by weight, and if that's the case just swapping the volume numbers around will not provide you an accurate quantity of acid.

You first need to be sure of how the percentages are expressed.  Assuming they are percents by weight, you would then need to look up the density of phosphoric acid at the two percentages and work out the calculation so that you were adding the same number of moles of $\ce{H3PO4}$.


Is it more economical to liquify natural gas to ship overseas or convert it to propane (C3H8) and ship it overseas?

Gas to liquid technology is known, but propane isn't a good choice of end product

The idea of converting natural gas (which is mostly methane) into liquids is widely used not least because the earth has, inconveniently, located the best natural gas fields some way from the places that most want to use natural gas.

Right now (in 2016), though, it seems it is still not desperately uneconomical to convert the methane into a liquid by refrigeration and ship it to customers in the largest known thermos flasks (LNG tankers). Both the ships for transporting liquefied natural gas (LNG) and the plants to create it are very expensive (tens of billions of dollars for the minimal economic scale). There are something like 410 of these ships and there seem to be dozens of new units being built every year (see this report on the industry for more detail: warning pdf).

But there are some examples of plants that convert the local gas supply directly to more transposable liquids. Shell, for example, have developed a process that converts gas directly into clean liquid fuels (the initial plants produce very clean diesel, but other fuels are possible). See this description of the Shell process. Some other firms are operating similar processes.

So it seems that, right now, LNG is the best economical option. But if you are going to use a technology to go directly to something more transportable than methane, it is better to go the whole hog to a widely used fuel (diesel and petrol are both basically longer chain hydrocarbons than propane).

This may, of course, change.


Why does reducing pressure cause the equilibrium to shift towards the side with less moles?

Actually, the shift of reaction towards left on decreasing pressure (and towards right on increasing pressure) is due to Le Chatelier's Principle, which states that if a change is brought in the equilibrium conditions of a reaction, the reaction will proceed in such a manner that it counteracts the change.

In case of increasing pressure the reaction shifts to right due to lesser number of moles on right. And according to gas equation, lesser moles means lesser pressure. The opposite happens when the pressure is decreased.


Are primary, secondary, and tertiary amines considered functional group or chain isomers?

When we talk about chain isomers we mean isomers with different carbon skeleton.

If the carbon skeleton is changing as well as the functional group we report it as functional group isomerism.

For the given compound in your question, primary, secondary and tertiary amines will show functional group isomerism as they are forming different types of amines itself.


Pressure of two flasks connected to each other at different temperatures

The easiest way to prove the equality in pressure is by taking a cross-section of the narrow tube and finding the forces acting on it. By extending this, we can prove that the pressures are equal.
Take a cross-section area of the narrow tube. We assume this is at steady state and there is no average motion of particles in the entire system.
Now, since there is no net average motion, this means that there is no pressure gradient across any cross-section. Expanding this cross-section into the tube, we see that the ends of the tube have to be at the same pressure since there is no net average motion of the gas within the two flasks at steady state (which is where the parameters are asked to be measured).
Therefore $$p_\mathrm{left} = p_\mathrm{right}$$


Ammonia for cleaning - what is the interaction between ammonia and oils?

Common food fats and oils usually contain some free fatty acid1. The alkalinity of household ammonia serves to form a soap which will emulsify the main body of the fat/oil, if enough elbow grease (i.e. vigorous agitation with a scrub brush or wash rag) is applied. Other alkalis do as well - better, if you consider breathable ambient air to be desirable.
Some brands of commercial household ammonia already contain fatty acid soaps and exhibit a cloudy appearance.
It also seems to me that a dilute solution of washing soda ($\ce{Na2CO3}$) would clean quite well, and ammonia should do no better. But when the odor of $\ce{NH3}$ disappears, and the oily/fatty surface is now clean, it is as if there are now two indicators of cleanliness: no odor and no oiliness, therefore the surface must certainly be clean, because all evidence of the cleaning process has been removed. It must be a psychological thing, like foaming with detergents.
Reference:

Analytical approaches for the assessment of free fatty acids in oils and fats
S. A. Mahesar, S. T. H. Sherazi, Abdul Rauf Khaskheli, Aftab A. Kandhro, Siraj uddina, Anal. Methods, 2014 ,6, 4956-4963, DOI: 10.1039/C4AY00344F


Is this stoichiometric question answerable? (unknown variable)

With a lot of assumptions, 0.5 M is a reasonable answer.
Here are the holes you have to fill in.

If it is a titration of NaOH and HCl, it is probably an acid/base titration where we stop the titration when the pH is neutral.

Titrations are often performed to figure out the concentration of a solution. You would titrate a smaller sample of the solution to figure out the concentration. Technically, the concentration of HCl after the titration is zero because all of it is supposed to react as you reach the endpoint of the titration. So it makes sense to ask about the concentration in the original solution.

The stoichiometry of the neutralization of HCl and NaOH is one-to-one:
$$\ce{NaOH(aq) + HCl(aq) -> H2O(l) + Na+(aq) + Cl-(aq)}$$

Of course, it is silly not to give any context when posing such a question on a quiz. I won't comment on the statements you attribute to your teacher other than to say that hydrochloric acid always contains water (pure HCl is a gas under ambient conditions).


Can sodium acetate anhydrous be used as a desiccant?

No, not in most applications. While it may act as a dessicant, it is fairly volatile, so you'll get acetic acid evolving from it in the presence of water. Generally, you don't want to expose what ever you're keeping dry to acid. The two obvious issues when it comes to food uses (ie toxicity concerns) is smell (and taste) and acceleration of spoilage (although in the cases of acidic foods, it might not be an issue). You don't generally want your dessicant mixed into the product so keeping them separate is optimal but with the volatile acetic acid it could be a problem. If you wanted to use its acidity to enhance stability or taste (mixed into the food), then you'd have problems with changes in consistency as water was absorbed.


I assume you mean tetrakis(acetonitrile)copper(I) hexafluorophosphate. This compound contains a tetrahedral complex of $\ce{Cu+}$ with 4 $\ce{CH3CN}$ ligands which stabilize the copper(I) center. They are, however, easily exchanged in the presence of other potential ligands, and with water, disproportionation to $\ce{Cu^2+}$ and elemental $\ce{Cu}$ will occur.

$$\ce{2[Cu(CH3CN)4]^+ + 6H2O ->~ [Cu(H2O)6]^2+ + Cu\downarrow + 8 CH3CN }$$

The pale blue color of the hexaaquacopper(II) complex might not be visible at low concentrations, and the pink color that you have observed arises from fine particles of the elemental $\ce{Cu}$ precipitate.


How does adding sodium chloride to a solution of copper sulfate change the equilibrium?

When you mix $\ce{CuSO4}$ and $\ce{NaCl}$ solutions, the following equilibrium forms:
\begin{align}
  \ce{[Cu(H2O)6]^2+ + 4Cl^- &<=> [CuCl4]^2- + 6H2O},&
  \Delta Η &= \text{positive}\\
\end{align}

Upon heating the mixture, the equilibrium shifts forward (following Le Chateliers principle) to produce more of the yellowish coloured $\ce{[CuCl4]^2-}$ ions, also known as tetrachlorocuprate (II) ion. As the amount of yellow colour in the mixture increases the solution becomes greener due to subtractive mixing of colours, i.e. blue and yellow solutions mix to give a green coloured solution.

The experiment can be found on YouTube, for example: Chemical Equilibrium ($\ce{CuSO4 + NaCl}$)


How to change color of water based on its acidity?

I'm assuming that you're doing this at home. A popular (and SAFE!!) experiment at home is to use red cabbage, vinegar, and baking soda.

The indicator comes from the cabbage.

Vinegar turns the solution acidic.

Baking soda turns the solution more alkaline.


Do the units of the law of mass actions convey any meaning?

Well, you can calculate difference in number of moles of gases and active liquid species on reactant and product side species using units of $K_c$ and $K_p$, respectively, as these are mostly used for homogeneous reactions.

For example, the unit of $K_c$ is $\text{(concentration)}^n$ if $n > 0$, more liquid or aqueous (ionic) species on product side.

Then, you can predict whether the product is formed or reactant when you increase or decrease external pressure or volume of a container containing a mixture in equilibrium. Temperature's effect would be only heat of reaction dependent.

Hint: write $K_c$ in terms of moles and volume to check effect of volume, write $K_p$ in terms of mole fractions and total pressure to check effect of external pressure.


Why is nylon durable and how does it have such exceptional strength?

Polyamides (such as Nylon) are molecules containing a large number of amide functions:

In those functions, the $\ce{C=O}$ bond is polarized (due to the difference in electronegativity between C and O) : $\ce{^{\delta+}C=O^\delta-}$.
The N-H bond is polarized, too: $\ce{^{\delta-}N-H^\delta+}$.
This leads to H-bonds between the chains:

While H-bonds are weak interactions, their high number (as there are a lot of amides in polyamide chains) leads to strongly bonded chains, which results in increased strength and durability.


Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?

Excerpts from my answer here.

The mechanism proposed for the cleavage reaction involves an initial attack by the donor on one boron atom in diborane, leading to cleavage of one $\ce{B-H-B}$ bridge.

This is followed by the attack of a second donor molecule, cleaving the remaining $\ce{B-H-B}$ bridge. If the boron atom in the pendant $\ce{BH3}$ group is attacked, two moles of adduct result, whereas attack on the atom already carrying a donor produces cation and borohydride:

Support for this view comes from the fact that the intermediate ($\text{I}$) is detectable by tensimetric titration of amine-borane with diborane and by $\ce{^{11}B}$ nmr.

It can then be interpreted that hard bases like ammonia don't experience the steric effects which softer bases like trimethylamine do. Sterics inhibit the attachment of the second donor at the same boron atom.
Source: Boron Hydride Chemistry, Earl Muetterties


Libraries which support Water model Simulation:

Assisted Model Building with Energy Refinement(AMBER)
Chemistry at HARvard Macromolecular Mechanics(CHARMM)
The Molecular Modelling Toolkit

Cool Site for Water Model Generation:

Water Models

Image Source: http://biomodel.uah.es/en/water/index.htm


Do the concepts of Newtonian and non-Newtonian fluids still hold at a nanoscopic level?

Generally Fick's laws of diffusion have been found to apply pretty well to molecular translational and rotational diffusion in solution. This means that we assume that the solvent is a continuous structureless fluid, which in practice means that the diffusing molecule should be far bigger than that of the solvent so that the solvent's molecular nature can be ignored. Surprisingly this works well with 'slip' and 'stick' boundary conditions down to situations where the solute is not that much bigger than the solvent, e.g.  rhodamine and similar dye molecules in solvents such as acetonitrile or ethanol. In your example of a protein in water then you might expect that it would behave 'normally' as just described.

However, in the cytoplasm the close packing of many similar sized and large molecules means that these can easily block the diffusion of any other (except for far, far smaller ones such as water) and so the situation is totally different and will be more like a percolation process where a slight reduction in packing can lead to a large, almost stepwise increase in diffusion coefficient. The idea of a single diffusion coefficient is therefore not helpful.


Are different enclosing mark types in chemical names needed to differentiate between substances?

There is no real difference between the types of bracket - indeed, in online searching in the CA Registry file on STN all brackets are replaced by round () brackets. However, there are conventions with regards to square brackets in display in some names and these should be kept.


Mechanism for the conversion of an unsaturated alkyl halide to an alcohol

For what it's worth, this reaction does not appear in the literature. So, whether it actually forms the product or not is anybody's guess.

However, if I had to suggest a mechanism... I would note first that an SN1 reaction on a primary centre, as you have proposed, is difficult. I think it more likely that the alkene assists in the expulsion of the chloride.

The resulting cyclopropylmethyl carbocation is stabilised and is electrophilic at multiple sites. I have only drawn one resonance form, but there are a few more. Attack of water at the tertiary carbon leads to the desired product after deprotonation.


Can the Smirnov-Zamkow reaction be described as a pericyclic process?

I was going to write a comment, but I think this is good enough for an answer.

The issue that symmetry rules forbid a [2+2] thermal cycloaddition if the addition is suprafacial-suprafacial.

However, you can achieve a suprafacial-antarafacial [2+2] thermal cycloaddition. This is symmetry allowed because the antarafacial symmetry is exactly opposite suprafacial symmetry.

I'm not familiar with your exact example, but I am familiar with [2+2] thermal cycloadditions of ketene. It is thought to proceed via suprafacial-antarafacial additions.

See for example: Burke, L. A. J. Org. Chem., 1985, 50, 3149.


Why does NF3 have a smaller bond angle than NH3?

This can be argued on the basis of Bent's rule; concisely stated

  Atomic s character concentrates in orbitals directed toward electropositive substituents

What follows below is a crude explanation. Before that, I'll note that we concern ourselves with the hybridisation of the orbitals at the central atom.

Since s orbitals are lower in energy than p orbitals, they are better at stabilising electrons. Which is why molecules like to use them (s orbitals) in bonds where there is "more" electron density to stabilise.

Thus, Bent's rule can just as easily be formulated as follows:  p character concentrates in orbitals directed towards electronegative substituents; (the central atom doesn't have to waste its low energy s orbitals, the orbitals on the electronegative atom, can (possibly) take care of it.

What has all of this got to do with bond angles? Well, more s character leads to larger bond angles. You can think of how s orbitals are spherical in shape, and a large s character would lead to more "spherical" hybrid orbitals, and that would lead to larger bond angles.

Going back to your question, we are supposed to compare $\ce{NF3}$ and $\ce{NH3}$.

Fluorine is more electronegative that hydrogen, and the $\ce{N-F}$ bond would have greater p character than the $\ce{N-H}$ bond, (or $\ce{N-H}$ bond has greater s character, whichever you prefer). Thus $\ce{H-N-H}$ angle is greater in $\ce{NH3}$ than it is in $\ce{NF3}$

P.S I'm sure you can learn more about what I just tried to explain (rather poorly) by reading the Wikipedia article, and/or a textbook and/or using the search function on this site to find related questions/answers.


Represent bonds spanning a unit cell boundary plane

As the developer of both Open Babel and Avogadro I can say the answer is "not yet."

The main thing isn't the file format. It's that Avogadro at the moment doesn't have support at the moment to show bonds across a unit cell.

I believe there was a patch for that, but it was too slow for typical use:
https://github.com/dlonie/avogadro/commits/ENH_intercell_bonds


There is no need adding the number $2$ in front of the definition of the rate. So the rate equation is :
$$-d[A]/dt = k_1 [A] + k_2[A] = (k_1 + k_2)[A]$$


Correct stability order of resonance structures

Compound II and III have 2 aromatic rings whereas I and IV have only 1 aromatic ring, and more the number of aromatic rings, the more the compound is stable.

  Aromatic molecules are very stable, and do not break apart easily to
  react with other substances. Organic compounds that are not aromatic
  are classified as aliphatic compounds—they might be cyclic, but only
  aromatic rings have special stability (low reactivity).

Thus stability of II= stability of III and stability of I = stability of IV.

And since II and III have greater number of aromatic rings than I and IV,

Stability(II,III) > Stability(I,IV)

Thus the answer would be [Option D] and not [Option C].


Is there a substance I can use to coat a projectile to eliminate friction heat?

Try using carbon fiber in a uniquely woven shape. Maybe apply a dual layer with the top layer having a unique shape for heat dissipation. There have been a few carbon fiber helix designs that maintained structure and supported weight with a 2000 F acetylene/butane flame directed on it.

Another option is go with layers of powdered graphite. See if you can convert thermochemical properties with blends. Maybe you can add a coolant system for your beer can with the excess energy stored.

If you are really trying to have your beer survive extreme heat,"Thermo-electric Reactions and Currents Between Metals in Fused Salts" by Thomas Andrews is probably a good start for prototype blends. I think google has a free .pdf available through Google books.

Here is a graph that might help with selecting your surface metals, if you ever take this project on seriously:

If nothing works, at least your efforts should land you a pretty hardcore beer-koozie.

Final deciding factors come down to the degree of effort you are wishing to exert and material expense.


Do chemists use molecular orbital theory to explain physical characteristics of a compound (i.e. melting point, boiling point, etc)

Certainly the electronic structure of compounds are of interest when predicting or interpreting the physical properties - but not necessarily the molecular orbital theory in and by itself. The symmetry of compounds (and by extension this also means;) the symmetry of the bonds in the compound are vital to many physical properties. However, MO theory is mainly used to describe bonding in more detail, not to infer "macro"-properties FROM bonding - so I guess the answer is... "no and yes". Yes, chemists think about these things, at least the ones that work close to the quantum realm. No, MO theory is not really useful in the way you are describing - with one giant exception:

When we are talking about "possible" physical properties of not-yet-synthesized compounds, the answer is resoundingly "yes". Graphene, for instance, has more theoretically predicted qualities than have been measured. Super conductance, tensile strengths, diffusion rates... there is just so much exiting being suggested all a result of computer modeling on the quantum level from MO theory and the likes.


Pourbaix diagram and the order of the species

Pourbaix Diagram In Environmental Chemistry
Pourbaix diagrams are very useful to identify the nature chemical species in different pH environment.
The figure contains additional information on the stability of water. The lines at 1.23 V and 0.00 V represent the redox chemistry of water. At 1.23 V water is being oxidised to oxygen. At 0.00 V the water is reduced to hydrogen. As labelled in Fig.1, there are 3 regions where behaviour of water is represented. The middle region is where the water in "stable state".
Note: These voltages are obtained at 298 K. In order to derive these lines, you can use Nernst Equation (where gradient is: -RT/nF, as the both reactions involve 2 electrons the gradient is the same in both cases.)

Figure 1: Stability of water.
Now moving on to iron. Overlapping water behaviour and iron as in Fig.2. (The dashed lines - redox chemistry of water). In the case of water, knowing the pH given you can predict the behaviour of iron. For the sake of understanding how to utilise the Pourbaix diagrams.
Assuming you are analysing alkaline lake (high pH: 10-12) and assume there is not much oxygen content (to utilise also the y-axis, low oxygen level - E decreases). In Fig.2, the region of your interest is pH 10-12 and since low oxygen, and region around 0.00 V. The Pourbaix indidicates that at these conditions you can have some amounts of Fe(OH)2, Fe(OH)3.

Figure 2: Analysis of Iron species in 1 M of water (298 K).
To see different behaviour of other species:
http://www.crct.polymtl.ca/ephweb.php


How can I solvent exchange from water to alcohol?

Since it's a colloid, rather than a true solution, put the water-based mixture in a dialysis membrane of appropriate pore size and immerse it in concentrated alcohol. The water will dissolve into the alcohol, while alcohol slowly seeps in to displace the water, but the colloidal particles cannot get through the pores. Depending on how little water is desired in the final product, this step may be repeated multiple times.

For some materials, inexpensive cellophane sausage casing may be suitable.


$\mu_i=\mu_i^o+RT ln(a_i)$ is the definition of activity ($a_i$).

Therefore, the standard state for standard chemical potential ($\mu_i^o$) is necessarily part of the definition of activity.

For a pure liquid or solid, the standard state is the pure substance at 1 bar.

For a liquid solvent, the standard state is also the pure liquid at 1 bar.

For a gas, the standard state is the fictitious state of an ideal gas at 1 bar.

For solutes, there are multiple conventions, for example the fictitious state of the solute having a molality of 1 mole/kilogram and exhibiting the behavior it does at infinite dilution.

$\Delta G^o = \sum_i \nu_i \mu_i^o$ by definition.

So there is no one standard state for $\Delta G^o$, because the various substances may have different standard states.


Reactivity is a vague concept in chemistry. It is rather confusing than useful. Reactivity involves both thermodynamic and kinetic factors, it predicts whether a chemical compound will undergo a chemical reaction (either by itself or with other compounds) or not and if it does how fast is the reaction going to completion i.e., rate of reaction.

Reactivity is also used to compare or study rates of two or more chemical reaction. For example when we say reactivity of HBr is more than reactivity of HCl - here we refer to how fast the compund will give acid-base reaction.

Reactivity is sometimes referred to bulk and sometimes to a single molecule but mostly it is referred as bulk, but it is not predictable.

In pure compounds, reactivity is regulated by the physical properties of the sample. For instance, grinding a sample to a higher specific surface area increases its reactivity. In impure compounds, the reactivity is also affected by the inclusion of contaminants. In crystalline compounds, the crystalline form can also affect reactivity. However, in all cases, reactivity is primarily due to the sub-atomic properties of the compound.


Relative rates of two acyclic dienes in the Diels-Alder reaction

You are correct that the having the diene in an s-cis conformation is extremely important.  Only from this conformation can the Diels-Alder occur. If there is a high concentration of the s-cis conformer the Diels-Alder will likely occur rapidly; if there is a low concentration the reaction will occur more slowly.

In cyclopentadiene, the diene component has no choice, it is locked in an s-cis conformation by the ring.  However with acyclic 1,3-dienes an equilibrium exists between the s-cis and s-trans conformations.

In 1,3-butadiene the amount of the s-cis conformation is relatively small (~2%) because of the steric interaction between the "internal" hydrogens at the ends of the chain in this conformation. In accord with this expectation, 1,3-butadiene is not as reactive as cyclopentadiene in the Diels-Alder reaction.

If we replace one of these "internal" hydrogens with a methyl group (penta-1,cis-3-diene), the steric interaction between the "internal" methyl and hydrogen in the s-cis conformation is even greater, further destabilizing the s-cis conformation and reducing the amount present at equilibrium.  On the other hand, replacing one of the "exterior" hydrogens with a methyl group (penta-1,trans-3-diene) doesn't cause any increased steric interference in the s-cis conformation and the amount of the s-cis conformer is about the same as 1,3-butadiene. Again, in accord with expectations, penta-1,trans-3-diene reacts at a rate similar to 1,3-butadiene, whereas penta-1,cis-3-diene reacts slower.

(image source)

Using this information we can make predictions for your compounds. We would expect cyclopentadiene to react the fastest because it is held in an s-cis conformation. In the case of cis, cis-2,4-hexadiene we now have an internal methyl-methyl interaction, which severely destabilizes the s-cis conformation, so there is very little present at equilibrium and this compound reacts the slowest in the Diels-Alder reaction. trans, trans-2,4-hexadiene only has an "internal" hydrogen-hydrogen interaction when in the s-cis conformation, therefore it should react about as fast as 1,3-butadiene - in between the extreme rates of the other two compounds.


Is ruthenium really invulnerable to aqua regia?

Page 807 of A Textbook of Inorganic Chemistry says that iridium, rhodium. and ruthenium are "almost unaffected" by aqua regia.  So "completely invulnerable" = no, but "resistant" = definitely yes.


Finding equilibrium temperature when two things are mixed

In this type of question, you should bring all the matter into one state and calculate net heat.
In this solution, we will consider heat released as positive heat, absorbed as negative, and the state to which we will bring all the matter be water at 0 °C.

Here we were left with 20 g of water at 0 °C and zero heat if heat was some positive quantity we would use the heat to raise 20 g water to a specific temperature.
By equation $20\times T\times1=\text{heat left}$
$T$ is temperature to which 20 g water is raised
If heat was negative quantity, we would convert part of water at 0 °C to ice at 0 °C by equation
$M\times80=\text{heat left}$
$M$ mass of water which is converted to ice


Your protocol gives you the answer: to make sure your sediment is neutral. After all, you titrate with NaOH afterwards, so any wrong starting point in terms of pH would skew your result.


Why do transition elements make colored compounds?

You are absolutely correct, it all about the metal's electrons and also about their d orbitals.

Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level.

However when the metal starts bonding with other ligands, this changes. Due to the different symmetries of the d orbitals and the inductive effects of the ligands on the electrons, the d orbitals split apart and become non-degenerate (have different energy levels).

This forms the basis of Crystal Field Theory. How these d orbitals split depend on the geometry  of the compound that is formed. For example if an octahedral metal complex is formed, the energy of the d orbitals will look like this:

As you can see, previously the d orbitals were of the same energy, but now 2 of the orbitals are higher in energy. Now what does this have to do with its colour?

Well, electrons are able to absorb certain frequencies of electromagnetic radiation to get promoted to higher energy orbitals. These frequencies have a certain energy which correspond to the energy difference between different orbitals. Now most substances are only able to absorb frequencies of radiation which are outside the visible light spectrum, for example they might be able to absorb radiation which has a frequency of $300$GhZ (that is infrared radiation). This means that it reflects all other types of radiation, including the full spectrum of visible light. So our eyes see a mixture of all the colours; red, green, blue, violet, etc. This is seen as white (this is why several organic compounds are white).

However transition metals are special in that the energy difference between the non-degenerate d orbitals correspond to the energy of radiation of the visible light spectrum. This means that when we look at the metal complex, we don't see the entire visible light spectrum, but only a part of it.

So for example, if the electrons in an octahedral metal complex are able to absorb green light and get promoted from the $d_{yz}$ orbital to the $d_{z^2}$ orbital, the compound will reflect all other colours except for green. Therefore by using the colour wheel, we can find the complementary colour of green which will be the colour of the compound, which is magneta.

This explains why not all transition metal complexes are colourful. For example copper sulfate is a bright blue compound, however zinc sulfate on the hand is a white compound despite being a transition metal. The reason behind this is because zinc's d orbitals are completely filled up with electrons, meaning that it is not possible for any electron to make a d-> d transition as they are all filled up. Hence you might sometimes see zinc referred as not being a transition metal.


As discussed in the comments... The two component chemiluminescence reaction:
(Solution 1) To 50 ml of deionized water add 10-50 mg of luminol, 10-20 ml of 10% ammonia solution and 1-5 ml of 3% hydrogen peroxide. (Solution 2) To 100 ml of deionized water add few milligrams of copper salt (e.g., sulfate) or few drops of blood or fresh horseradish juice. These are approximate quantities (they should work), but you can experiment with the concentrations, after all experimenting is the most fun part of chemistry. Now take the solution 2 (whole) and add few milliliters of solution 1 to it and shake it. You should be able to observe a faint glow. A brighter emission requires organic solvents (DMSO, DMF) and corrosive chemicals (KOH). Adding a third liquid that will separate the two other (as the OP suggested) is not a simple thing to do. I would do this otherwise: add some neutral ingredient to solution 2 to increase its density (possibly sugar, but I'm not sure if that will not quench the chemiluminescence) and place the solution in a container (preferably a long tube). Then carefully pour the solution 1 on the surface of the solution 2 (so that the solutions won't mix). When you shake the container the solutions will mix and produce glow. Another idea: place one solution in a container and layer it with a hot paraffin from a candle and allow it to cool down. Then you can pour the second solution on the solid paraffin. To start the reaction, simply break the thin paraffin layer by shaking (you can place some glass marbles in the container).


Why is phosphorous acid more acidic than phosphoric acid?

I'm pretty sure it is because phosphorus acid is more polar than phosphoric acid. This is because in the phosphoric acid there are more OHs and an O so it kind of balances out more towards the center. In phosphorus acid, there is a H on the main carbon. Hydrogen doesn't pull as much on electrons as Oxygens do and hence that part of the molecule becomes more positive causing a stronger dipole moment than that in phosphoric acid.


Provide all the data that are necessary to reproduce the experiment:

TLC material (silica gel or aluminium oxide)
composition of the mobile phase
special tricks, such as running the TLC in methanol first for 5 mm to squeeze the initial spot to a very narrow band
detection of the separated bands/spots:

$\mathrm{R_f}$ values
do the spots have a colour
do the show fluorescence
which reagents did you use for staining

UPDATE

If you report on a series of measurements, it's usually good practise describe the conditions that didn't change (type of TLC plate, staining solutions, etc.) in the general section of the experimental part of the report/article/thesis.


Will the following complex show optical isomerism

As already commented by Ivan Neretin, the cis form also does not show optical isomerism. Elaborating a bit further on the comments:

Complexes show optical isomerism when they are asymmetric (also known as chiral). They should not have any elements of symmetry, such as planes/axes/centres of symmetry. The trans form has a clearly visible plane of symmetry. It's a bit trickier to find the plane for the cis form. Here's a diagram showing the planes:

(source: Chemistry libretexts)


Explain why the noble gas Kr does follow the general trend for covalent radii

Since the increased atomic radius causes a decrease in the electric interaction between electrons and nucleus, the heavier noble gas elements, such as Kr and Xe can form compounds with other elements. (KrF2, XeF4, ... &c.)

The covalent radius is quite meaningless in for elements such as Ar and Ne since its compounds seldom exist, but Kr and Xe compounds can occur with relatively high stability, so its 'covalent radius' can be meaningful.

That 0.109nm is probably a covalent radius.

Something like measuring a Kr-F bold length in KrF2 and then subtracting the covalent radius of a fluorine atom.

Trivially, the Van der Waals radius tends to be bigger than the covalent one, the noble gas atom radius seems as if it does not follow the trend, but from Krypton which the covalent radius is measurable, the data follows the general trend in periodic table.


Why do all atomic masses have decimals if there are some elements that don't have isotopes?

The answer to your question has to do with Einstein's most famous formula
$$
E=mc^2
$$
which relates energy to mass, but what does it mean? Let's consider a simple example first. The simplest atom is of course the hydrogen atom which consists of one electron and of one proton. If you add the mass of the proton $1.672621898\times 10^{-27}$kg, to the mass of the electron $9.10938356\times 10^{-31}$ kg, you find a value of $1.67353284\times 10^{-27}$ kg. If you compare this to the mass of the $^1$H atom of $1.6735326915926\times 10^{-27}$ kg, you see that the hydrogen atom is lighter than the combined mass of the proton and the electron by $2.44017148\times 10^{-35}$ kg. What happened to the mass when the electron and proton got bounded together? Well, according to Einstein the difference in mass is converted to the binding energy of the electron! If you use Einsteins formula you find a binding energy of $-2.19311675\times 10^{-18}$ J or $-13.7$ eV which is very close to the Rydberg constant!

Let's now get back to other atoms. In the same way that the formation of a hydrogen atom from an electron and a proton lowers the mass of the atom by the corresponding binding energy, the binding energy between the neutrons, protons and electrons that make up heavier atoms also results in a reduction of the mass of the atom. Because the binding of the neutrons and protons is much stronger than the binding energy of the electrons to the core, the effect of the mass is also much larger. See for instance here for further details.


Check out this 1953 article from ACS on this exact reaction: https://pubs.acs.org/doi/abs/10.1021/ja01103a521

Does a variety of systems and calculates rate constants for comparison. I'd then recommend using similar terms they use to locate further resources on particular molecule hydrolysis, and even Google Scholar's (or Web of Science, Scopus, JSTOR, etc...) "cited by" option to work your way forward in time to papers that deal with this type of work. Using search terms from any articles you come across in databases (ACS, ScienceDirect, Scifinder, etc...) should get you going with plenty of benchmarking material as well.


Why are 'table salt and sugar are both terrible candidates for recrystallization'?

The solubility of $\ce{NaCl}$ has very poor temperature dependency. If you dissolve it at $100^\circ\rm C$ and precipitate at $20^\circ\rm C$, you'll essentially lose about 35 g to purify 3 g. Even using a fridge, you'll still lose more than half of the product, which is quite a bad business. Industrial process relies on evaporation, I think.

Sugar is another story. It does have a nice temperature dependence, but it dissolves too well, even when cold. Just imagine dealing with that awfully viscous syrup, like molasses. It would take forever to filter, and quite a while to crystallize. Industry can handle that, but for home chemistry I'd suggest something more pleasant.


Why does depurination happen at a higher pH (or under less harsh conditions) than depyrimidation?

Really interesting question!
Mechanism of acid catalyzed hydrolysis of purine nucleosides is known to proceed through unimolecular departure of base moiety(DOI: 10.1021/ja00709a055).

It seems that someone else has also asked the same question: "Why reaction rate with pyrimidines is so much slower?" -and published paper of the reaction mechanism (DOI:10.1039/P29940000309).

In the paper it is suggested that pyrimidines follow different reaction mechanims. This would explain huge differences in reaction rates.

  In other words,
  the structure-reactivity correlations lend additional support to
  the previous suggestion, according to which the acid-catalysed
  hydrolysis of 4-oxopyrimidine nucleosides follows an openchain
  mechanism

Open chain mechanims:


Configuration of chiral center in cocaine

You are actually right, that stereocenter does have R stereochemistry. The four substituents to the chiral center in question are:

1 - $\ce{CO(O)}$ = $\ce{C(O)3}$
2 - $\ce{C(O)(C)(H)}$
3 - $\ce{C(N)(C)(H)}$
4 - $\ce{H}$

These are ranked and numbered according to the Cahn-Ingold-Prelog priority rules. Adding the numbering to the structure:

Orienting the lowest priority group away from us and following the direction of the numbering, we confirm that it does indeed have R stereochemistry:


How do you create primary amines from alcohols?

In the conversion of primary alcohols to primary amines

$$\ce{R-CH2OH -> R-CH2NH2}$$

direct alkylation of ammonia normally is the last thing you want to do in the lab.

Under conditions where $\ce{OH}$ is a good leaving group, i.e. in acidic medium, the nucleophilicity of ammonia is reduced due to protonation. Moreover, there's little chance to prevent further alkylation of the initial product.

More promising strategies are:

Gabriel synthesis: Hydroxy-halogen-exchange, followed by reaction of the primary halide with phthalimide and subsequent hydrazinolysis
Mitsunobu reaction ($\ce{Ph3P, DEAD}$) of the primary alcohol using hydrazoic acid $\ce{HN3}$ as a nucleophile, followed by reduction of the alkyl azide
An (almost-one-pot) combination of the Appel and the Staudinger reaction

Heat the primary alcohol ($\ce{R-CH2OH}$) with $\ce{Ph3P}$  and bromotrichloromethane $\ce{CBrCl3}$ in benzene, then cool down to RT (room temperature) to yield $\ce{R-CH2Hal}$
Add $\ce{NaN3}$,tetrabutylammonium bromide and N,N-dimethylformamide and reflux again to obtain the azide $\ce{R-CH2N3}$, then cool down to RT, pour on water, extract with benzene and dry the organic layer
Add triethylphosphite $\ce{(EtO)3P}$ to the organic layer, stir at RT to yield the iminophosphorane $\ce{R-CH2N=P(OEt)3}$
Hydrolyze the iminophosphorane by refluxing with hydrochloric acid, then neutralize to release the amine


Why does a chiral carbon affect some hydrogen's hydrogen environment?

There is a big difference between the two hydrogens and the hydrogens in the two methyl groups.

The issue is that the methyl hydrogens are part of a group that can rotate quickly and so they see an averaged environment which is the same for all the hydrogens. If the bonds were "stuck" and the methyl groups could not rotate your argument might be correct, but the is not the case.

The two hydrogens on the carbon next to the chiral centre are also connected to bonds that can rotate. But they never see the same environment even on average because they will always have a different relationship to the hydrogen and hydroxyl on the next carbon. Whatever way you rotate the bond connecting their carbon, the two hydrogens will be in a different environment, so they appear with distinct shifts.


Is there a reaction between ethanol and aluminum?

The reaction of aluminum and ethanol forms aluminum triethoxide,  $\ce{ Al(OC2H5)3}$, but the reaction is fairly slow at room temperature, particularly if the aluminum is anodized. In the presence of any moisture it decomposes into aluminum hydroxide and ethanol.

The Safety Data Sheet indicates that aluminum triethoxide is both flammable and can cause "severe skin burns and eye damage", but in atmospheric conditions it likely decomposes as quickly as it forms, so the main adverse effect will be the corrosion of the aluminum, covering it with a white crust of $\ce{Al2O3}$ (I've seen this happen to an aluminum lid on a container used for ethanol).

UNH has posted a chart that shows that anodized aluminum is resistant to ethanol attack, but does not state at what temperature, so you may have to experiment to see if it suitable in your application.

BTW, many other alcohols react in a similar manner, such as isopropanol, which forms aluminium isopropoxide with aluminum.


Is the equation dU= TdS - PdV always valid (even for irreversible processes?)

The differential statement of the first law, $$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V,$$ is valid for all states---but not all processes---for the reasons you have described. When we apply it, we are always assuming that a reversible process connects the two states of interest; irreversible processes are outside the range of applicability of this equation. States and processes get muddled up a lot, because we can always find a reversible process that connects any two states.

The generalization of this equation for all processes is given by $$\mathrm{d}U = \delta q + \delta w,$$ which is just the statement of energy conservation. We cannot, however, write $$\mathrm{d}U \leq T\mathrm{d}S - p\mathrm{d}V,$$ for this would imply that $T$ and $p$ are defined for a system at all points during an irreversible process, which is not true.


Obtaining mercury from mercury(II) nitrate

The process you described would be more appropriately called "reduction of mercury(II) to elemental mercury".
Unfortunately, the trick with iron likely won't work (something more inert like copper would be a better choice though).

Mercury(II) oxide is weakly basic, so mercury salts in general would easily undergo hydrolysis and form basic oxosalts in aqueous solution unless acidified.
Mercury(II) nitrate quickly hydrolyses to poorly soluble yellow mercury(II) oxide upon dilution or addition of alkali:

$$\ce{Hg(NO3)2(s) + H2O(l) -> HgO(s) + 2 HNO3(aq)}\label{rxn:R1}\tag{R1}$$

which can be converted back to nitrate by adding excessive amount of nitric acid, which in turn won't leave any chance for iron not to be oxidized, so you end up with both metals in solution.

I didn't want to dive deep into the discussion of how mercury(II) salts hydrolyze and leave it simple, but after receiving critique from Maurice claiming $\ce{Hg(OH)NO3}$ to be the “real” product, I think I allow myself to add a paragraph or two.
Raman spectroscopy and x-ray scattering studies in late 1960 demonstrated that hydrolysis of mercury(II) salts produces series of polynuclear oxo-bridged species of types $\ce{Hg2OH(H2O)2^3+},$ $\ce{Hg3O(H2O)3^4+}$ or $\ce{Hg4O(OH)(H2O)3^5+}$ [1, 2].
Formation of mercury(II) hydroxide nitrate $\ce{Hg(OH)NO3}$ as a hydrolysis product was taught in 1940–1950s era, and it stuck in several textbooks published later probably because it's listed in all editions of Pauling's General chemistry up to 1988.
However, not only it's oversimplified (granted, reaction \eqref{rxn:R1} is also an oversimplification in a sense it's a boundary case), but is also an incorrect concept.

The most up-to-date summary of what is really happening when $\ce{Hg(NO3)2}$ is dissolved in water can be found in Mercury handbook [3, p. 115]:

  $\ce{Hg(NO3)2}$ solutions are only stable in the presence of a certain amount of nitric acid, which prevents hydrolysis.
  $\ce{Hg(NO3)2}$ quickly hydrolyzes in excess water and produces a precipitate of
  $\ce{Hg3O2(NO3)2 · H2O}$ or, when boiled in dilute solutions, forms mercury(II) oxide $(\ce{HgO}).$

Regarding the reduction of mercury(II), there are two pathways: $\ce{Hg^0}$ may be obtained from mercury(II) nitrate using a dry or a wet method.

Speaking of a dry method, the most straightforward way to obtain metallic mercury from nitrate is to heat $\ce{Hg(NO3)}$ in a distillation apparatus (b.p. of mercury is 357 °C).
Above $\pu{400 °C}$ it readily decomposes:

$$\ce{Hg(NO3)2(l) ->[\pu{400 °C}] Hg(g) + 2 NO2(g) + O2(g)}\label{rxn:R2}\tag{R2}$$

\eqref{rxn:R2} is a brutto-reaction; the nitrate first decomposes to the red mercury(II) oxide at lower temperatures (which, in turn, decomposes to the elements):

$$\ce{2 Hg(NO3)2(l) ->[\pu{350 °C}] 2 HgO(s) + 4 NO2(g) + O2(g)}\tag{R3}$$

Wet method suggest mild conditions and a reaction in solution.
For example, formic acid (also used in silver refining) being a strong reducing agent would cause a precipitate from ammoniacal mercury(II) nitrate solution.

Since nitrogen dioxide, mercury vapors as well as mercury salts and oxides are highly toxic, the reactions must be carried in a fume hood which makes it a poor fit for a talent show.
Considering the possible risks and your level or preparation (no offence), I'd strongly advise to take extreme caution doing mercury chemistry and avoid public demonstrations until you become more experienced.

Note: chemical reactions are adopted from [4, p. 310–312]

References

Cooney, R.; Hall, J. Raman Spectra of Mercury(II) Nitrate in Aqueous Solution and as the Crystalline Hydrate. Aust. J. Chem. 1969, 22 (2), 337. https://doi.org/10/b6t3h2.
Johansson, G.; Haugsten, K.; Rasmussen, S. E.; Svensson, S.; Koskikallio, J.; Kachi, S. An X-Ray Investigation of the Hydrolysis Products of Mercury(II) in Solution. Acta Chem. Scand. 1971, 25, 2787–2798. https://doi.org/10/bn5j2g. (PDF)
Kozin, L. F.; Hansen, S.; Kit, M. Mercury Handbook: Chemistry, Applications and Environmental Impact; RSC Publ: Cambridge, 2013. ISBN 978-1-84973-409-7.
R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Amines do not react with hydroxide to give amide ions - not significantly at least.

The typical $\mathrm pK_\mathrm a$ of an amine is $\sim 30$ (see Evans pKa table) and that of $\ce{H2O}$ is $\sim 15$. The equilibrium constant for the second equation you wrote is therefore on the order of $10^{-15}$.

In order to generate any amide ions you need a much stronger base than simply hydroxide. A typical choice is an alkyllithium (LDA is commonly made by reacting diisopropylamine with BuLi).

Back to the question whether amines are amphoteric: If your definition of an amphoteric compound is simply "can react either as an acid or a base", then yes, it technically is, since it reacts both with acids ($\ce{HCl}$) or bases ($\ce{RLi}$).

However, it may be useful to draw the line at reacting with aqueous acid and aqueous base. Otherwise, nearly everything with a hydrogen can be considered to be amphoteric (for example, $\ce{HF}$ can technically act as a base to form $\ce{H2F+}$). Under this definition amines are not amphoteric.


Calculating how many grams will I get from a reaction

Here, $\ce{H2O}$ is a product and you are assuming that the reaction goes to completion, so it does not affect the yield of the reaction in any way. Maybe a suitable analogy is to think of the reaction system as a car - you give the car a certain amount of petrol (say 2 gallons) and it converts it to a certain number of miles (say 70 miles), as well as a lot of exhaust gases. Now the amount of exhaust gases that your car is putting out doesn't affect the fact that if you put in 2 gallons of petrol, you get 70 miles out of it.
If you were to treat the reaction as a reversible one, where you have an equilibrium between reactants and products, then in general, the amount of products present would also matter. But don't worry about this for now, I'm sure you will come across this in your study of chemistry at some point in time.
Good question - if the reaction is not balanced, then yes, you do have to balance it by finding the appropriate stoichiometric coefficients. If you don't know exactly how many miles per gallon (35 in this case) your car consumes, then you won't be able to work out how many miles you can drive if you feed your car 2 gallons of petrol. Not all reactions can be balanced, but if a reaction can't be balanced in one way or another, it can't happen in real life. For example, you can't balance the reaction $\ce{Pb -> Au}$, but as we all know, alchemy was (mostly) a load of rubbish. For this reason, any reaction that you come across in your study of chemistry, as well as in real life, can be balanced.


Why is one lobe of an sp3 hybridized orbital smaller than its other half?

When combined at a given atomic center, any two atomic orbitals add in a vectorial
manner. For example, consider the orbital $\phi$ defined by $\ce{p_{x}}$ and $\ce{p_{y}}$ atomic orbitals as

\begin{align}
\phi = c_1 \ce{p_{x}} + c_2 \ce{p_{y}}
\end{align}

The orbital addition can be pictured like this

for the two cases $c_1 = c_2 > 0$ and $c_l = -c_2 > 0$, respectively. The ratio $c_1 / c_2$ controls how much the orbital $\phi$ is tilted away from the $x$ (or $y$) axis.

The mixing of atomic orbitals with different angular momentum quantum number is
also controlled by a vectorial addition and leads to various types of hybrid orbitals
some examples of which are shown here:

The example a) would be an $\ce{sp}$ hybrid orbital. It gets its sense of direction from the $\ce{p}$ orbital used to construct it because the $\ce{s}$ orbital is isotropic. Its shape comes about because the parts of the orbitals that have an opposite sign (this is hinted at by the shading in the pictures) cancel each other out to some extent. So, if the left lobe of the $\ce{p}$ orbital is negativ and the right one is positive and the $\ce{s}$ orbital is also positive then the left half of the $\ce{s}$ orbital overlaps with the left lobe of the $\ce{p}$ orbital in a destructive manner and they cancel each other out, such that only a small portion of the initial $\ce{p}$ orbital lobe remains. But the right half of the  $\ce{s}$ orbital overlaps with the right lobe of the $\ce{p}$ orbital in a constructive manner and so the right lobe of the initial $\ce{p}$ orbital becomes larger.