The documentation for the Gaussian software is actually pretty complete, even if qualitative. On the subject of optimizations, I recommend you look up the freq and opt keywords.

The manual explains the procedure followed by the Berny optimization as involving construction of an initial analytical estimate of the Hessian using a simplified force field, with subsequent optimization steps improving on that estimate using computed gradients. A number of options are provided for improving the initial estimate of the force constants where the defaults derived from the estimated Hessian are not good enough.

For frequency calculations, accuracy requires a good estimate of the full Hessian, better than estimates that might suffice for geometry optimizations. Gaussian allows for analytical computation of force constants for some of the methods (e.g. RHF) and uses analytical gradients to compute a numerical Hessian with some other methods, or if requested. In optimization cases where convergence is an issue, Gaussian facilitates convergence to a real minimum by employing the Opt=CalcAll keyword, which results in analytical (where possible) computation of the full Hessian (at some cost though).


Hydrogen peroxide: catalyst changes colour during peroxide decomposition

Recently, I answered your question and relevant information can be found from that answer.

At first, I thought that the brown colour is due to ferrous chloride that formed during the course of reaction but since the reaction is occuring in a aqueous solution, it should be green in colour. Ferrous chloride is brown in anhydrous form. So I thought to be either due to formation of a complex, iron(II, III) oxide($\ce{Fe2O3}$), or iron(III) oxide-hydroxide($\ce{FeOOH.nH2O}$) or a mixed salt of oxide-chloride i.e. iron oxychloride($\ce{FeOCl}$).

The following information is from the paper 2 of that answer:

  According to the mechanism proposed by KREMER and STEIN, an
  intermediate oxygen complex of iron with oxidation number +V is
  primarily formed by the reaction of $\ce{Fe^3+}$ with $\ce{H2O2}$.
  This complex reacts with another $\ce{H2O2}$ molecule to water and
  oxygen thereby reforming $\ce{Fe^3+}$.

  $$\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> [Fe^{V}O]^3+ + H2O ->[H2O2] Fe^3+ + 2H2O + O2}$$

  According to the mechanism proposed by HABER and WEISS the $\ce{Fe^3+}$ ions
  initiate a radical reaction, after which the chain reaction consumes
  the hydrogen peroxide. This mechanism can explain the high reaction
  rate very well.

  Chain initiation:  $\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> Fe^2+ + HOO. + H+}$

  Chain propagation: $\ce{Fe^2+ + H2O2 -> Fe^3+ + 2OH.}$
  $\ce{Fe^3+ + H2O2 + OH. -> Fe^3+ + HOO. + H2O -> Fe^2+ + H+ + O2 + H2O }$

That intermediate oxygen complex of iron with oxidation number +V maybe responsible for the brown colour of solution.

As for the potassium iodide, the color you are seeing is basically food coloring. Various food coloring  is used to make the foam produced colorful. If food-color is not used, still color is seen. That color is basically color of pure soap foam. This elephant toothpaste experiment not uses food-color but still it has color that is color of soap-foam.


There would be no aromaticity in pyrrole after protonation since there is no lone pair to allow for Hückel's rule for aromaticity (the $4n+2$ rule). So, upon protonation, you are destroying aromaticity in pyrrole. Also the lone pair is completely delocalised in pyrrole and it cannot easily participate as a base.
However, in aniline, there is a partial loss of the lone pair due to resonance but the lone pair of nitrogen is still available for protonation. Aromaticity is not lost in the resonance structures of aniline.
Therefore aniline is more basic than pyrrole but is less basic compared to pyridine (where the lone pair is out of plane of resonance).


What I usually do when I start numbering is look for branching and start numbering at the end with the closest branch. Also, keep in mind that with branching you'll want to consider numbering from one of the branches (if it'll yield a long chain.)

I'm using the structure you made as an example:

Here I numbered from the side with the most branching. But then I numbered it in two different ways. One way was starting with the straight chain which gave me a 7 carbon chain. Then I renumbered it (the image on the right) starting from one of the branches. This gave me an 8 carbon chain. (You could also number starting from the other branch, but that gives you 7 as well.)

Choose the longest chain. Since stuff that looks like a side chain can actually be part of the main chain, seriously keep this in mind during exams. It's totally okay to number the same alkane in different ways and play around with it to find which is the longest chain.

4-ethyl-3-methyloctane

Now, if you're having trouble drawing them:

I would draw just a zig-zag pattern (like I drew above) containing the number of carbons you need. You need a heptane, so start out with a 7 carbon chain.

This will just be your bare, skeleton structure. Then number the carbons, then add the ethyls to carbons 2 and 3.

HOWEVER, the question is flawed. You actually can't have an ethyl on carbon number 2 here, because it just makes the chain longer.


Is it possible to save an insoluble snowflake?

With gas no absolutely. To preserve ice at room temperature using the pressure of a gas you should reach a pressure of one billion Pascals. Too dangerous for Penny safe...
In fact there are some imprecisions in the video Leonard cites two different methods to "preserve" snow flakes with an organic polymer. In fact these methods create a snow flakes replicas, they are called snow crystal fossils, because there is no more ice inside them. Both methods are describe here, have a look!
Formvar Replicas

Leonard: "it will last forever I preserve it in a 1% solution of
polyvinyl acetal resin"

This is the Schaefer and Day technique [1] the solution Leonard is talking about is made from 1% of polyvinyl acetal resin in ethylene dichloride.
Superglue Replicas

Leonard: "in fact is a pretty simple process, you see, Cyanoacrylate
are monomers which polymerize mmm... "

Cyanoacrylate are the main components of Super Glue  this is easy method propose by the chemist Tryggvi Emilsson that you can do at home.
You have to choose one of these methods, you can't do both of them as Leonard (it seems) did... Take in account that Leonard is a physicist he doesn't know much about this stuff...;-)
[1] V. J. Schaefer and J. A. Day, Peterson Field Guides: Atmosphere (Houghton Mifflin, 1981).


Should equiatomic iron–cobalt alloy be written as FeCo or CoFe?

Current IUPAC recommendations suggest following electronegativity ordering. The symbol for iron (a less electronegative element among the two) should be written first: $\ce{FeCo}.$
From the section IR-4.4 Sequence of citation of symbols in formulae [1, pp. 58–59]:

IR-4.4.2 Ordering principles
IR-4.4.2.1 Electronegativity
If electronegativity is taken as the ordering principle in a formula or a part of a formula, the atomic symbols are cited according to relative electronegativities, the least electronegative element being cited first. For this purpose, Table VI* is used as a guide. By convention, the later an element occurs when the table is traversed following the arrows, the more electropositive is the element.
[…]
IR-4.4.3. Formulae for specific classes of compounds
IR-4.4.3.1 Binary species
In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1) is most often used in binary species.2
[…]

For intermetallic compounds, earlier recommendations prescribed alphabetical ordering
rather than by electronegativity (see Section I-4.6.6 of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 1990, ed. G.J. Leigh, Blackwell Scientific Publications, Oxford, 1990).

Examples:
[…]

$\ce{Rb15Hg16}$
$\ce{Cu5Zn8}$ and $\ce{Cu5Cd8}$

Table VI as seen in [1, p. 260]:

Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


what is the difference between a redox reaction and an ionic bond?

An ionic bond is formed by electrostatic attraction of electrons between one atom and the other whereas in redox reactions the electrons are transferred from one atom to the other. Hope this helps.


Acetone - Can it already be anhydrous when purchased?

As correctly pointed out by @MaxW, the technical acetone still contains water.  In addition, it is better to quantify "dryness", so by passing a mark you may draw the line of "dry enough" for the application you aim for; especially as Karl Fischer analysis may provide you a reproducible figure of merit about the water content.

Now, if you need dry acetone, stirr the acetone with potassium carbonate or calcium chloride and subsequently distill it off into a dry container.  In the instance of acetone, however, I refrain from molecular sieves because they catalyze aldol reactions.


hydroxide ion in water -- How does the dynamic change?

It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions.

This is not quite right. Because of the hydrogen bonding a wide variety of $\ce{H3O+(H2O)$_n$}$ ions are in dynamic equilibrium, where $n$ can take values as large as few tens. It has been shown that $\ce{H3O+}$ can donate three hydrogen bonds, so that it's first solvation shell of 3 water molecules, and the hydrogen bonds between $\ce{H3O+}$ and $\ce{H2O}$ are quite strong. This means that the $\ce{H3O+}$ cation can actually be considered as $\ce{H9O4+}$ in solution. For more details, see here.

  Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion.

Reasoning by analogy is a very dangerous road. As mentioned by sheogorath there is no possibility for a coordinate covalent bond as in the case of $\ce{H3O+}$. But the hydrogen bonding is possible and the hydroxide ion is indeed strongly hydrated as well as $\ce{H3O+}$. The details of the hydration are not well known yet, but recent studies indicate $\ce{OH- (H2O)4}$ as the prominent complex in an aqueous solution. For details, see here.


The short answer is nuclear binding energy, which is the energy needed to disassemble an atom into its subatomic parts (or in some cases the energy released when this happens). The binding energy is a consequence of the strong and weak nuclear forces that hold atoms together.

Where does this energy come from? It comes from the mass of the nucleons! What? It is true; most atoms have less mass than the sum of their parts, and that mass defect is converted into the energy that holds them together. If you do not believe me, let us look at one atom of carbon-12, which is used as the definition of the atomic mass unit:

$$\mathrm{1\ u}=\dfrac{1}{12}\text{ of the mass of one }\ce{^12C}\text{ atom}$$

Thus, the atomic mass of an atom of $\ce{^12C}$ is by definition $\mathrm{12\ u}$. The atom is constructed from 6 protons, 6 neutrons, and 6 electrons, and the masses of those particles are:

particle   mass                     number    total mass
p          1.00727646681290     u   6         6.0436588008774    u
n          1.0086649160043      u   6         6.0519894960258    u
e          5.485799094622×10^−4 u   6         0.0032914794567732 u
                                     TOTAL:  12.0989397763600    u

The sum of the parts of the $\ce{^12C}$ atom are more than the mass of the atom! This mass defect, $\mathrm{0.0989397763600\ u}$ in this case, is the value of the nuclear binding energy when converted to energy using $E=mc^2$.

This handy graph from the Wikipedia article linked above shows that the binding energy per nucleon is not the same for all nuclei.

If it were, then we would have integer values for monoisotopic nominal masses. However, the consequences of this change in reality could be as drastic as the heavier elements (like iron) not having enough energy to hold themselves together. An additional consequence is that there would be no productive nuclear reactions (fission, fusion, or radioactivity). All nuclear transmutations are driven by conversion of nuclei with lower binding energy per nucleon to nuclei with higher binding energy per nucleon, resulting in a net conversion of mass to energy. Without a difference in binding energy per nucleon, we would not have nuclear bombs (and probably we would be better off as a civilization), but we would also not have nuclear power and radioisotopes for compound labeling, cancer therapy, and medical diagnostics. We would also not have a ready supply of deuterium for NMR solvents, and we might not be able to use variations in isotopic distributions of $\ce{H}$, $\ce{C}$, and $\ce{O}$ to determine all sorts of things like did that "locally grown" food from that restaurant you like actually come for your geographic region?


Why sodium sulphate has water absorbing capability?

Sodium sulphate reacts readily with water at room temperature to form hydrates up to sodium sulphate decahydrate, $\ce{NaSO4\cdot10H2O}$. $$\ce{NaSO4 + 10H2O -> NaSO4\cdot10H2O}$$ This means that $\ce{NaSO4}$ can absorb up to 10 mol of water for every 1 mol of salt that is used, making it one of the most effective drying agents in terms of sheer capacity.

Using data from this source we can calculate the Gibbs energy change for the above reaction at different temperatures. $$\mathrm{\Delta G_{20^\circ C} = -1.33~kJ~mol^{-1}}$$ $$\mathrm{\Delta G_{30^\circ C} = 1.28~kJ~mol^{-1}}$$

Since the entropy change for the reaction is negative, we can see that the cooler the solution you are trying to dry is, the more effective $\ce{NaSO4}$ will be as a drying agent.


What is the carbon content, by weight, of vegetable oil?

Vegetable oils are mainly made of glycerol esters of oleic acid, linoleic acid and palmitic acid. For example, the fatty acids extracted from olive oil are a mixture of $74$% oleic acid, $11$% palmitic acid, and $10$% linoleum acid, plus 5% other acids. These proportions are average values that may change from one sample to the next one.
The mass proportion of carbon atom is $77.2$ % in glycerol trioleate, $78.3$ % in glycerol trilinoleate, and $75.8$ % in glycerol tripalmitate


Why do we multiply two wavefunctions in valence bond theory instead of adding them?

Actually, the MO theory wavefunction can also be written as a product. While the individual molecular orbitals are just weighted sums of the atomic orbitals, the wavefunction can, to a first approximation, be written as a Slater determinant of the occupied orbitals.

For your example of the hydrogen molecule, this would mean taking the antisymmetrized product of the occupied $\sigma$ spin orbitals. (Note, if you haven't encountered this before, spin orbitals are a spatial and spin description of a single electron, rather than the commonly encountered spatial orbitals which describe two electrons of opposite spin.


What happens to chemical shifts of protons close to aromatic systems in ¹H-NMR?

The aromaticity found in certain ring systems (e.g. benzene, see Hückel's rule) causes a magnetic field because of the circular motion of the electrons in the ring (see picture).

(Source: Clayden et al. Organic Chemistry 2nd edition, p. 277)

This additional field enforces the magnetic field of the NMR machine inside the ring, while countering the field outside the ring, causing the signals of the outer hydrogens to shift downfield (i.e. to the larger shifts).

Now for your other questions:

  First, what does being near an aromatic system do to protons? If we
  look at the spectra of 1-napthalenemethanol and then
  9-anthracenemethanol, the peak of the two aliphatic protons of the
  benzylic methylene group get shifted way up. Can someone explain why
  adding in that third aromatic ring does this?

Lookig at the difference in shift between the benzylic methylene protons of 1-Naphthalenemethanol and 9-Anthracenemethanol we must conclude that the induced magnetic field of the second is stronger at that position than that of the former, as there is no other appreciable difference. Intuitively this seems to make sense as the field of the 9-Anthracenemethanol is more uniform at and closer to (with respect to the "right" side of the molecule as in the pictures) the discussed proton. Of course this could/should be confirmed by quantum mechanical calculations, but I do not have the tools or the knowledge to do that (if someone has, please feel free to expand this answer!)

(1-Naphthalenemethanol Proton NMR)

(9-Anthracenemethanol Proton NMR)

  I was also under the impression that protons near electronegative
  elements would be deshielded and be shifted higher. However in the
  spectrum of Benzil, the protons para to the carbonyl group are higher
  shifted than the protons meta to the carbonyl. Why is this? Is it to
  do with conjugation of the carbonyl with the aromatic ring? I'm having
  trouble finding a proton nmr in the literature, so there's nothing to
  compare it to.

Yes, conjugating effects and inductive effects can counter each other, but there is no easy general way to say which will win out in a particular situation, although usually it is the conjugating effect.


The complete, formal mathematical answer is a bit complex. It might be better to get that discussion from Math Stack Exchange.

An answer to the world of Chemistry is more straightforward: Yes, for any physically relevant operator and any relevant function, you always can. That’s a fundamental property of quantum mechanics: (functional) states are superpositions of eigenstates.


Enthalpy of Formation problems using Hess's Law

Find the $\Delta H^\circ_f$ of the following reaction:
$$\ce{C + O2 -> CO2}$$

We have our list of reactions with known enthalpies:
$\ce{SrO + CO2 -> SrCO3} \quad \Delta H^\circ_f = \pu{−234 kJ}$
$\ce{2SrO -> 2Sr + O2} \quad \Delta H^\circ_f = \pu{1184 kJ} $
$\ce{2SrCO3 -> 2Sr + 2C + 3O2} \quad \Delta H^\circ_f = \pu{2440 kJ} $
Hess's Law takes advantage of the fact that enthalpy is a state function, looking at our reactions, we need to find a way to derive what we want:
Carbon dioxide is on the reactants side, however, we want it on the products, so we can run that reaction reverse; however we need to change the sign of enthalpy if we do that:
$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
We need to get rid of $\ce{SrO}$, thus, we use our second equation:
$\ce{2SrO -> 2Sr + O2 } \quad \Delta H^\circ_f = \pu{1184 kJ} $
Since, there's only one strontium oxide in the first reaction, we divide by the integer 2 across the chemical equation including the enthalpy of formation:
$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
Now, with equation 3, we need carbon and oxygen on the reactants side, and need to remove $\ce{SrCO3}$ and also divide again by 2, since we only have one strontium.
$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $
Now we add up all our reactions and enthalpies, and cancel out common elements on both sides:
$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$
$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $
$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $

$\ce{C + O2 -> CO2} \quad \Delta H^\circ_f= (\pu{234 kJ}) + (\pu{592 kJ}) + (\pu{-1220 kJ}) = \pu{-394 kJ}$


Why is active mass of a pure solid or liquid always taken as unity?

Does the activity of a solid or liquid change over the course of a reaction?

The density of a solid or liquid reactant doesn't change over the course of a reaction. The mass and volume do as it is consumed, but the ratio of the two is constant. If the reaction causes a temperature change then there are small changes in density, but that would also alter the equilibrium constant and the concentrations of the species involved.

The equilibrium constant is based on activities, which you can think of as the ratio of the concentration (molar density) at standard state to the concentration in the reaction. Since solid/liquid density is constant at a given temperature, activity is too.  It takes quite a large change of temperature or pressure to make a substantial change in the density of a substance, assuming no phase changes occur.

What about gases, why don't they show the same phenomenon and have fixed activities? (Asked in comments.)

For solids or liquids, as mass decreases the volume also decreases by a proportional amount. For example, $\mathrm{100mL}$ $\ce{H2O}$ $\mathrm{= 100g}$ and half as much, $\mathrm{50mL}$ $\ce{H2O}$ $\mathrm{= 50g}$. Both have molar densities of $\mathrm{55.5 mol/L}$.) Gases keep the volume of their container, so if you remove some via a reaction, the mass of the gas decreases, but its volume remains the same, and its concentration decreases as a result.

Why is active mass given a value of 1 then?  Is this simply a convention?  Shouldn't its value depend on the compound? (Asked in comments.)

No, it's not a convention. Activities, active mass or any other activities, are really ratios. The ratio here is a comparison between the activity under experimental conditions and the activity at standard state under standard conditions.  For gases that is 1 bar of pressure at 25°C, for solutions that is 1 molar at 25°C and 1 bar, and for solids and liquids that is the molar density of the substance in its most stable phase and allotrope at 25°C and 1 bar.  The activities of real substances require some additional work and there are particular compounds that do not behave quite as nicely as expected, but the basic idea of activity being the ratio between experimental state and standard state is sound.

With that in mind, consider the activity of solids and liquids.  Their standard state is their density at standard conditions.  Their experimental state is their density at experimental conditions.  It takes significant pressure and temperature changes to cause substantial changes in the density of a solid or liquid, so their experimental state is approximately equal to their standard state, and the ratio of these two states is therefore quite close to 1.


Authors mention that pyrogallol (PG), among other standards, has been purchased from Sigma Aldrich US and used as is without further dilution. The amounts of polyphenols were calculated as PG equivalent by external standard method, and utilizing additivity of absorption coefficients the proportion has been established.

$62.5$ is just a dilution factor for the herb extract probe. From the protocol, first time the extract was diluted $5$ times (from $\pu{5 mL}$ to $\pu{25 mL}$), second time – $12.5$ times (from $\pu{2 mL}$ to $\pu{25 mL}$), resulting in total dilution factor of $5 \cdot 12.5 = 62.5$.


Nitrogen lone pair electrons on 2-acetyl-1-pyrroline

The nitrogen lone pair is in plane with the ring because it is in one of the three $sp^2$ hybrid orbitals of N. The other two hybrid orbitals form $\sigma$ bonds to the adjacent carbons, and all three are in plane, with an angle of 120° between them. The $p_z$ orbital of N which is perpendicular to the ring plain forms the $\pi$ bond of the C=N double bond. The acetyl substituent should also lie in plane with the ring because this would allow energetically favorable conjugation between the two double bonds (C=N and C=O).


Is there a sixth type of d-orbitals with index 2z²-x²-y²?

There are 5 linearly independent solutions to spherical harmonics for $\ell = 2$.

We generally pick $\mathrm{d}_{z^{2}}$, $\mathrm{d}_{x^{2}-y^{2}}$, $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{xz}$.

Ignoring, $\mathrm{d}_{z^{2}}$, notice that it's really easy to see what the $\mathrm{d}$-orbital looks like figuring out the nodal planes. The lobes of the $\mathrm{d}$-orbitals go in between, and the sign of the function tells you how to switch the sign of the wavefunction. So, in your diagram for (a), (b), and (c), I would take off points for the reason that there is no indication that the diametrically opposite lobes have the same shading.

Here's the question that you want to wrap your head around. Why is there no $\mathrm{d}_{x^{2}-z^{2}}$ and $\mathrm{d}_{y^{2}-z^{2}}$ in analogy to $\mathrm{d}_{x^{2}-y^{2}}$?

The answer is by choice because you can add orbitals together to get new orbitals. The orbitals we write are actually linear combinations of the complex solutions we got from solving the spherical harmonics for the Schrödinger equation. Now, the full name of $\mathrm{d}_{z^{2}}$ is $\mathrm{d}_{2z^{2}-x^{2}-y^{2}}$. This is not hard if you think about the where the node is (a nodal bi-cone) and the signs, the torus is different sign than the lobes. With this and $\mathrm{d}_{x^{2}-y^{2}}$, you can construct the two missing ones (being extremely loose with notation):

$$
-\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y{2}} + \mathrm{d}_{x^{2} - y^{2}})
= -\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2} + x^{2} - y^{2}})
= -\frac{1}{2}\mathrm{d}_{2z^{2} - 2x^{2}}
= \mathrm{d}_{x^{2} - z^{2}}\\
-\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2}} - \mathrm{d}_{x^{2} - y^{2}})
= -\frac{1}{2}(\mathrm{d}_{2z^{2}-x^{2}-y^{2} - x^{2} + y^{2}})
= -\frac{1}{2}\mathrm{d}_{2z^{2} - 2y^{2}}
= \mathrm{d}_{y^{2} - z^{2}}
$$

So, all you need is 5.


Is sulfide ion a stronger base than hydroxide ion?

Equilibrium Expressions Related to the Solubility of the Sour Corrosion Product
Mackinawite Ind. Eng. Chem. Res. 2008, 47, 1738-1742 compiles 18 different studies concerning the second Ka of H2S and shows that they vary over 7 orders of magnitude from $10^{-19}$ to $10^{-12}$

So firstly it isn't accepted whether or not Na2S is stronger than NaOH.

Putting that aside, the only way that 1M sulfide could yield 1 M hydroxide (and hence pH be equal to pH of 1 M hydroxide) is if sulfide was infinitely strong.  If they were equally strong, it would only yield 0.5M hydroxide.

Student 1: clearly wrong.

Student 2: correct (based upon the information given)

Student 3: can't say without seeing the table, but if the table actually says that SH- is weaker than water than the answer is true, but not correct in the sense that the instructions say "Based on this information".


I looked and there are a lot of questions on the site about units cells, many of which are pertinent. However I couldn't find one that answers your question directly.

First there is no reason to guess that $\ce{LiH}$ has a FCC structuree. You can look that up, for instance in Wikipedia.

Second $\ce{LiH}$ doesn't really have molecules in the solid. Think of $\ce{LiH}$ as  an empirical formula rather than a molecular formula.

UNIT CRYSTAL

I lifted the following from website https://opentextbc.ca/chemistry/chapter/10-6-lattice-structures-in-crystalline-solids/

Now the text says:

  The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula NaCl.

That is correct. The unit cell must have the correct stoichiometric ratio. However the section seems to imply that the middle drawing, as show below, is the proper unit cell. It isn't!

First realize that when looking at proper unit cells is that that the lattice points between which the edges are drawn are the nuclei, not points in space that would contain whole atoms. So consider the drawing below. For the cell as shown in the illustration, there are 8 corners with 1/8 atom each. Four corners are Na, and four are Cl. Thus the cell shown only has $\frac{1}{2}$ of a Na cation and $\frac{1}{2}$ of a Cl anion.

Now down the rabbit hole!

The proper unit cell for NaCl is shown below:

From Wikipedia article on Cubic crystal system:

  The face-centered cubic system (cF) has lattice points on the faces of the cube, that each gives exactly one half contribution, in addition to the corner lattice points, giving a total of 4 lattice points per unit cell (1⁄8 × 8 from the corners plus 1⁄2 × 6 from the faces). Each sphere in a cF lattice has coordination number 12. Coordination number is the number of nearest neighbors of a central atom in the structure.

So the unit cell contains $\ce{Na4Cl4}$ !?!

The reason is that the proper unit cell must also contain all the symmetry elements of the crystal. Look at the drawing below:

I won't get into all of how group theory applies to crystals and molecules, but the  red line shows a four-fold rotation axis. In other words all the atoms are identical. You can rotate the crystal 90 degrees about the red axis and the sodium and chlorine atoms are in identical spots. The axis is a four-fold axis because you have to rotate 90 degrees four times to get the identical atoms back to their starting position.


Naming of elements 113, 115, 117, and 118

IUPAC will certainly have a press release on their news page, given that they are officially responsible for the names.

Currently, that's at this address: http://www.iupac.org/news/latest-news.html

Sadly, they don't appear to have an RSS feed. (Not even an Atom feed [rimshot noise].)


The 'preferred' oxidation states of s-block metals can be explained by looking at their electron configurations:

Note that for the alkali metals, each element has one more electron than a noble-gas configuration.  This electron is very weakly bound to the nucleus as evidenced by the ver low ionization energy:

It is much harder to remove the 2nd electron, since it is contained within a completely filled shell.  (Please ignore the 2nd IE for Fr, as it doesn't exist in my data source.)

Therefore, the alkali metals have a tendency to lose just one electron, forming the +1 oxidation state.

The same thought process can be applied to the alkaline earth metals.  Here are the first three ionization eneriges for each:

Hopefully, it is clear that in the case of the alkaline earth metals, it is the 3rd electron that is in the completely filled shell, and that electron is extremely difficult to remove.  Thus, the metals in this group tend to stay in the +2 oxidation state.


For the given structure, the principal chain could be a 3-hydroxyhexanoic acid or a 3-methylhexanoic acid; i.e. the name of the structure could be 3-hydroxy-2-(pentan-2-yl)hexanoic acid or 2-(1-hydroxybutyl)-3-methylhexanoic acid.

The IUPAC recommendations on this matter have been changed.
According to the old IUPAC recommendations (1979, retained in 1993), the principal chain was the 3-hydroxyhexanoic acid. Thus, the preferred name was 3-hydroxy-2-(pentan-2-yl)hexanoic acid.
However, the relevant rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-45 SELECTION OF PREFERRED IUPAC NAMES
(…)
P-45.5 CRITERIA RELATED TO ALPHANUMERICAL ORDER OF NAMES
P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (…). Alphabetic letters are considered first in the order that they appear in the name; (…)

Therefore, the preferred name is 2-(1-hydroxybutyl)-3-methylhexanoic acid (not 3-hydroxy-2-(pentan-2-yl)hexanoic acid) since ‘hydroxybutyl…’ is earlier alphabetically than ‘hydroxy-pentan-yl…’.


Different force constant in ground vs excited state in spectroscopy

At the risk of providing a somewhat circular answer, in the harmonic approximation of an interatomic potential, the force constant is a measure of the force of attraction between the atoms (in a classical description they would obey Hooke's Law $|\vec{F}|=-k(x-x_\circ$)). The answer to the question is therefore that the attractive interatomic force in the electronically excited state of $\ce{I2}$ is weaker than in the ground state.

The following figure from Ref. 1 illustrates the potentials in the ground and excited states:

The potential in the excited (upper) state is broader. In the Morse functional approximation, the potential has the form $$V(r)=D_e\left(1-e^{-\beta(r-r_\circ)}\right)^2$$

where the width of the potential is proportional to $\beta$ and $$\beta = \left(\frac{k}{2D_e}\right)^{1/2}$$

A smaller force constant results in a smaller value of $\beta$ and a broader potential well. This fits nicely with the results noted in the OP and the above figure (the equations are also outlined in the accompanying text). The energy accompanying a vibrational transition is proportional to the square root of the force constant and therefore inversely proportional to the width of the potential well. This is analogous to the behavior observed for a QM particle in a box, wherein the spacing between energy levels narrows as the box size increases. Working backward from vibrational transition energies leads to the conclusion that smaller transition energies are associated with greater oscillatory amplitudes (broader potentials) and smaller harmonic force constant estimates.

References

Bayram, S.B. and Freamat, M.V., A Spectral Analysis of Laser Induced Fluorescence of Iodine. arXiv eprint 1507.02600, 2015.


Cp-Cv For a gas following Van Der Waal equation of state

You could start with $C_V=(\partial U/\partial T)_V=T(\partial S/\partial T)_V$  and H and p instead of U and V for $C_p$ as appropriate. Then generate an expansion for S as $dS=(\partial S/\partial V)_T dV+(\partial S/\partial T)_VdT$ and differentiate wrt T. You should then get an expression in $C_V$ and $C_p$ plus other terms that you can find using the vdw equation.


Alkenes are very capable of a similar carbometallation reaction as you can see in the scheme below (ref: R. M. Sultanov et al. / Tetrahedron Letters 54 (2013) 6619–6623):

I think the reasons why such reaction with alkenes is not so common is:

It is useless (correct me if i am wrong). Hence the reason why the work in the scheme above was published in a low impact journal. Why would anyone want to make alkanes in such a complicated way instead of just getting them from crude oil for example?
With alkenes, the reaction prefers to keep incorporating alkene to form really useful dimers or polymers. If one takes a look in the step wise mechanism of olefin polymerization (e.g. Ziegler-Natta) and dimerization the transmetallation/insertion step is equivalent to the one in the question posted:


Is light a reactant in photochemical reactions?

It is neither reactant nor catalyst, and equilibrium concepts do not apply
There are processes other than photochemistry that to go in a direction away from equilibrium. They require mechanical or electrical work, and there is no established way to incorporate these into a chemical equation.
If we write a conceptual equation for charging a battery, it could look like this:
$$\text{empty battery} \ce{->} \text{charged battery}$$
The external power source is neither a reactant nor a catalyst; it is work done on the system, which serves to move the system away from chemical equilibrium.
If we write a conceptual equation for a refrigerator (or more generally, a heat pump), it could look like this:
$$\text{warm body + warm body} \ce{->} \text{cold body + hot body}$$
Again, the mechanical work done by the compressor is neither a reactant nor a catalyst, and it serves to move the system away from thermal equilibrium.
For both processes, you could not say that in the presence of work, an equilibrium is reached. It is just the opposite - we are moving away from the equilibrium.

[OP] So is photon absorption just a physical process and therefore the chemical equilibrium concept doesn't apply?

Photochemical reactions can move a reaction away from equilibrium (as in the example of cis/trans isomerization the OP mentions). This makes it a non-equilibrium process, so equilibrium concepts need to be expanded. If you want to describe the situation under certain irradiation conditions, you might use the term photostationary cis:trans ratio (see section on stilbene in https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/photchem.htm).

[OP] I read somewhere about photoisomerization and that favors the less thermodynamic stable product and that make me think of how is it possible for the value of equilbrium constant to change (considering interoconversion as a "reaction").

The equilibrium constant does not change. The reaction just does not go to equilibrium (or in fact moves away from equilibrium).

[OP] First the notation hν can't be thought to be a catalyst as it doesn't make sense because isn't regenerated.

Things written on top of the reaction arrow are not necessarily catalysts. In general, that spot is used for reaction conditions, such as solvent, high temperature, or "reflux".

[OP] Then how one should think about light in a chemical reaction? It is a reactant?

Some textbooks write "heat" as reactant or product, which already does not make sense (but is used in combination with Le Chatelier principle to memorize how equilibrium constant change with temperature). For photochemical reactions, however, there is a stoichiometric aspect to the role of photons; for every reaction, one photon has to be captured.
A comprehensive published treatment
The abstract of this paper (Perspective – life and death of a photon: an intuitive non-equilibrium thermodynamic distinction between photochemistry and thermochemistry) addresses some of the misconceptions for photochemical reactions. It is a bit long, so I am breaking it up into smaller parts.
First, it states that photons are not chemical reactants in photochemical reactions:

Neither the thermodynamically determined probability isotherm nor its kinetically manifest rate isotherm can be applied to photo-absorptive reactions such that the participants, including photons, may be treated as if they were chemical reactants. Photons and chemical reactants differ from each other fundamentally: firstly, a photon's energy is absolute and, in all instances of practical relevance to the present paper, independent of its surrounding electrochemical field, while the energy of a chemical reactant is relative and defined by its surrounding field; secondly, while both photons and chemical reactants can and do engage in entropy creation, only chemical reactants can engage in entropy exchange.

It then goes on to address the mistake of treating photochemical processes using equilibrium concepts:

Clarification of these differences requires identification and abandonment of fundamental historical errors in photochemical thought deriving from inappropriate overreach of analogies drawn between light and ideal gases, and including: treatment of photo-absorption as a reversible chemical reaction; attribution to light of thermal potential, or temperature (as distinct from the idealised abstraction of a ‘temperature signature’); attribution to light of exchangeable entropy content.

Then, it addresses how entropy plays a role in these processes:

We begin by addressing widespread misapprehensions concerning the perennially misunderstood concept of entropy and the frequently overlooked distinction between entropy creation and entropy exchange. Armed with these clarifications, we arrive at a useful perspective for understanding energy absorption and transfer in photosynthetic processes which, through the chemical ‘kidnapping’ of metastable excited states within structured metabolic pathways, achieves outcomes which the Second Law denies to thermal chemical reactions


Are there ionic solids that conduct electricity?

Yes.

Another example is VO2. Over 67 °C its crystal structure changes to rutile with a band gap of ~0.1 eV. Dopants of other substances can reduce the temperature at which it becomes a conductor.


What is the difference in the effect of DTT (Dithiotreitol) and SDS (sodium dodecyl sulfate) for PAGE?

Proteins are thought of as long polyamide chains. This long chain of amino acids is known as the primary structure. Due to hydrogen bonding of the backbone amide groups, a protein chain will organise most of itself locally either as an α-helix or a β-sheet; which one depends on the angles to the α carbon. These secondary structure elements then cluster to form a tertiary structure in which hydrophobic residues typically cluster together, aromatic residues may form π stacks, positively and negatively charged residues come together to build up ionic interactions, cysteins come together in disulphide bridges and hydrogen bonds are also formed where applicable. This last step is repeated if a protein consists of multiple subunits to form the multi-protein complex’s quarternary structure.

Almost all of the structure elements I mentioned that explain tertiary structures are intermolecular, meaning that they do not constitute a chemical bond between the connected fragments. Therefore, these structure elements are easily broken by the action of SDS whose aliphatic chain inserts into these tertiary structures, interacts with them and practically captures them on its inside.

The only exception to the former paragraph are disulphide bridges between cysteins. These are actually a chemical bond conneting the two fragments. In a simplified reaction scheme, the synthesis of a disulphide bridge looks like this:

$$\ce{\underset{\text{cystein}}{2 R-S-H} ->[][- 2e- \ -2 H+] \underset {\text{disulphide bridge}}{\ce{R-S-S-R}}}$$

Chemically, it is an oxidation of two cysteins.
Therefore, these disulphide bridges cannot be broken by the action of SDS alone. It takes a reductive agent such as DTT to resupply the electrons to reform the two cysteins. If no reducing agent is present (but there are disulphides in the protein structure), the disulphide bridge will remain intact and the protein will move not through the PA gel as it should according to its size since it is cyclic.


What am I doing incorrectly with this IR/ Mass spec data?

What I was trying to say was that firstly the molecular formula has to be determined correctly. From the mass spectrum of the given molecule, the molecular ion peak ($\ce{M+}$) was determined as $\frac{m}{z}=142$, which corresponds to the molecular formula $\ce{C9H18O}$. Note that, it can also correspond to $\ce{C10H6O}$, but in your $^1\mathrm H$ NMR spectra, a total integration of at least $9$ is there, thus the latter consideration is anyway not possible to be the correct formula.

After determining the chemical formula, it's good to look for index of hydrogen deficiency(i.e. degree of unsaturation) of the molecule. If the formula is $\ce{C9H18O}$, the degree of unsaturation is $1$. From the IR spectra also, we can see a sharp distinct $\ce{C=O}$ stretch near $1700\ \mathrm{cm^{-1}}$, and also no aldehydic proton peak near $9.5{-}10\ \mathrm{ppm}$, therefore we can conclude there is a keto group present in the molecule.

Now, we come to analyse the $^1\mathrm H$ NMR spectrum, which shows only total relative integral of $9$, but remember the integrals in proton NMR are always relative, they are never absolute. So, you can scale them up or down relative to your choice. Here to match the total protons in our system, we have to scale it up by a factor of $2$. Therefore the integrals now correspond to $4,8$ and $6$ protons respectively, and also by this scaling we can also say that the molecule is symmetric in nature.

Now peak at $\delta_\mathrm H=2.5\ \mathrm{ppm}$ is splitted into a triplet and these correspond to the protons attached to your carbonyl, resulting in the most downfield shift, near $\delta_\mathrm H =1.5\ \mathrm{ppm}$, two peaks merge and therefore they are actually the methylene protons, and lastly near $\delta_\mathrm H\approx0.9\ \mathrm{ppm}$, the triplets correspond to methyl proton peaks and thus we arrive at the final most likely possible structure as,

Analysis of the splitting pattern to find out the $^3J$ values, I leave it to you...


Which is the most intuitive to perform Dynamic Monte Carlo calculations?

If you want to perform Kinetic Monte Carlo for a chemical reaction system, you can use ZACROS. I am not sure what your system is? You should follow some good text book like "Understanding Molecular Simulation" to identify the type of your problem and appropriate software for that. There are plenty of freely available code on the internet but they all are for some specific purpose, So basically it is your responsibility to choose the best available software that will suit your need.


Which factors determine the keto:enol ratio for aldehydes, ketones, and 1,3-dicarbonyl compounds?

Your reasoning is along the right lines, but it is incomplete.

enol of B is more stable because of more substituted alkene.

The methyl group in acetone does stabilize the carbon-carbon double bond in the enol form for the reasons you suggest. However, the methyl group also stabilizes the carbonyl double bond in the keto form (see here for example, this is why a carbonyl in a ketone is slightly stronger than a carbonyl in an aldehyde). In fact, in simple carbonyl compounds, the methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond. As a result, simple aldehydes generally have a higher enol content than simple ketones. Hence, the enol content in acetaldehyde is greater than the enol content in acetone.
$$\ce{carbonyl <=> enol}$$
$$K_\mathrm{eq}=\frac{[\text{enol}]}{[\text{carbonyl}]}$$
$$
\begin{array}{lc}
\hline
\text{compound} & K_\mathrm{eq} \\
\hline
\text{acetaldehyde} & 6 \times 10^{-7} \\
\text{acetone} & 5 \times 10^{-9} \\
\hline
\end{array}
$$
The two β-dicarbonyl compounds have a higher enol content than the two monocarbonyl compound because hydrogen bonding and conjugation stabilize their enols. The enol content in C (a mono aldehyde) is higher than D because of the reasons outlined above.
Therefore, the overall order of increasing enol content is C > D >> A > B.
Unless you've studied this before, you might not know that a "methyl group stabilizes the carbonyl double bond a bit more than it stabilizes the enolic carbon-carbon double bond". So, a general rule that might come in handy in these situations is that the enol content increases with the acidity of the enolic hydrogen. A more acidic α-hydrogen implies a weaker $\ce{C-H}$ bond. Since the position of a keto-enol equilibrium is dependent on the relative stabilities of the keto and enol forms (the compound with the highest bond strengths overall will be more stable and will predominate at equilibrium) a weak $\ce{C-H}$ (lower $\mathrm{p}K_\text{a})$ generally implies a higher enol content.
$$
\begin{array}{lr}
\hline
\text{compound} & \mathrm{p}K_\text{a} \\
\hline
\text{acetone} & 19.3 \\
\text{acetaldehyde} & 16.7 \\
\text{2,4-pentanedione} & 13.3 \\
\hline
\end{array}
$$
Finally, this earlier answer provides a review of the key factors controlling keto-enol equilibria.


Comparing Methods of Determination of Dissolved Oxygen Content in Water Sample

Let's abstract your task to determine dissolved oxygen (DO) to something more general:

You have to tie an observable to a concentration.
It would be nice if the method would work over a broad range of concentrations.
The method should be specific, or, alt least, very selective.
If possible, this relation should be linear.
It would be nice if the method could be packed into a sensor that works on its own (permanently or in cycles) and does not require a lot of personal interaction (= autonomous probe)

What properties of molecular oxygen dissolved in water can be measured?

In water, $\ce{O2}$ can be reduced electrochemically: $\ce{O2 + 4e- + 2 H2O -> 4 OH-}$→ Have a look for galvanic and polarographic oxygen sensors
Molecular oxygen is known to quench the luminescence of electronically excited states of dyes. The observed luminescence intensity (and lifetime) is reciprocally proportional to the the concentration of the quencher(Stern-Volmer equation)
→ Have a look for luminescence-based oxygen sensors


Calculating Raman Intensities from Static Raman Activities

You have two options in that case.

You can try to understand the frequency dependent (or dynamic)  property in the formulation for Raman activity. This property is a bit complex and is known as polarizability. For Raman spectra,  certain invariants of this tensor are involved in the intensity formulation. Be careful with this.

You can then find the formulae for activity first (by refering to the Gaussian manual and cheking referred literature there). After having a formula, then calculate the components required (including polarizability) individually, and finally obtain the dynamic activity.

You can calculate static activity very accurately and scale it using dispersion relations. Dispersion relations are relations which relate a dynamic property to static ones. (I am not sure if Gaussian has dispersion relation calculation.) In some cases you can calculate dispersion coefficients also and use them for scaling. Important is that you need to verify whether static activity can be scaled in such a manner. (I know that polarizability can be scaled but, not near absorption regions.)


How is antioxidant activity of carotenoids determined?

As Buck Thorn and I commented elsewhere, the antioxidant activity of carotenoids has been determined using $\ce{DPPH}$ radical ($\ce{DPPH^.}$)(Refs 1 & 2). According to Ref.1, antioxidant activities of the isolated α- and β-carotene have been analyzed in five different concentrations of 0.001, 0.005, 0.01, 0.05, and 0.1% ($w/v$) using several assays: The metal chelation, $\ce{DPPH}$ radical scavenging, and superoxide scavenging activities. All of these assays have been showed almost similar results in terms of high activity at lowest concentrations. Thus, it is not true that $\ce{DPPH}$ radical assay cannot be use to measure β-carotene free radical inhibition.

For latest information on $\ce{DPPH}$ radical assay, read Ref.3. In general, following is a typical procedure for the assay:

  The $\ce{DPPH}$ assaying of $\beta$-carotene can be performed using known antioxident (e.g., Trolox, ascorbic acid, etc.) as reference compound. The assay can be carried out in either alcoholic media (e.g., methanol or ethanol) or in an aprotic solvent (e.g., acetonitrile). To the degassed solutions ($\pu{2 mL}$) of antioxidant (with a final concentration from $0$ to $\pu{60 μM}$), were added $\pu{2 mL}$ of $\pu{140 μM}$ $\ce{DPPH^.}$ stock solution (70 μM $\ce{DPPH^.}$ final concentration). The solution was stirred at room temperature in the absence of light, and, after $\pu{60 min}$, the absorbance was measured at $\pu{517 nm}$ using an UV-vis spectrometer. The working $\ce{DPPH^.}$ concentration was calculated from the absorbance response at $\pu{517 nm}$ of the initial solution in the absence of antioxidant. The $\ce{DPPH^.}$ scavenging percentage or inhibition ($I\%$) was calculated from the measured absorbance using the following expression:

  $$I\% = \frac{A-A_0}{A_0}\times 100$$
  Where $A$ and $A_0$ represent the absorbance at a given antioxidant concentration and the absorbance in the absence of given antioxidant, respectively.

References:

Lars Müller, Kati Fröhlich, Volker Böhm, “Comparative antioxidant activities of carotenoids measured by ferric reducing antioxidant power (FRAP), ABTS bleaching assay (αTEAC), DPPH assay and peroxyl radical scavenging assay,” Food Chem. 2011, 129(1), 139−148 (https://doi.org/10.1016/j.foodchem.2011.04.045).
Surashree Sen Gupta1, Mahua Ghosh, “In Vitro Antioxidative Evaluation of $\alpha$- and $\beta$-Carotene, Isolated from Crude Palm Oil,” Journal of Analytical Methods in Chemistry 2013, Article ID 351671 (10 pages) (http://dx.doi.org/10.1155/2013/351671).
Om P. Sharma, Tej K. Bhat, “DPPH antioxidant assay revisited,” Food Chem. 2009, 113(4), 1202−1205 (https://doi.org/10.1016/j.foodchem.2008.08.008).
The most recent reference for $\ce{DPPH}$ assay: Magda M. Becker, Gilvanda S. Nunes, Danilo B. Ribeiro, Francisco E. P. S. Silva, Gaëlle Catanante, Jean-Louis Marty, “Determination of the Antioxidant Capacity of Red Fruits by Miniaturized Spectrophotometry Assays,” Journal of the Brazilian Chemical Society 2019, 30(5), 1108-1114 (http://dx.doi.org/10.21577/0103-5053.20190003).


Beryllium chloride - Highly hydrolysable?

The structures of $\ce{BeX_2}$ where $\ce{X = F}$ or $\ce{Cl}$ can be found in A.F. Wells, Structural Inorganic Chemistry 5th edition, 1984 pages 412 and 1047.  There you will see that while the predicted structures for $\ce{BeX_2}$ are linear (sometimes rationalized with Lewis dot structures displaying double bonds and ignoring formal charge considerations of the halogens).  However, the actual structure of $\ce{BeCl_2}$ in the vapor-phase is a 3-coordinate dimer that converts to the linear structure at higher temperatures.  This behavior supports your 'gut feeling' that the electron deficiency of $\ce{Be}$ should be compensated for by another $\ce{Cl}$ atom.  The dimer network formed in the solid state allows lone pairs from $\ce{Cl}$ to be donated to the $\ce{Be}$ in an attempt to fill the valence shell of beryllium.

The use of a double bond to describe the nature of the $\ce{Be-Cl}$ bond is supported experimental data which shows that the bond length is shorter than expected for a single bond.

The image below is taken from Inorganic Chemistry by Miessler, Fischer and Tarr, 5th edition.


Why CCl4 does not dissolve in water while SiCl4 does?

$\ce{SiCl_4}$ does not quite dissolve in water; rather, it reacts with water. So does $\ce{CCl_4}$, albeit extremely slowly, so for most purposes one may safely assume it doesn't. The difference is mostly due to the atomic size of C and Si. Smaller С atom is completely blocked off by four bulky chlorine atoms, so the water molecule can't reach it. Larger Si is easier to attack.

Another factor is that silicon is a less electronegative element than carbon, so that the Si-Cl bonds will have more polar character.

Whether or not the d-orbitals of silicon contribute significantly to the transition state, is (IMHO) a question of secondary importance.


What would oxygen or nitrogen snow look like?

I don't know if there would actually be oxygen and nitrogen snow. I would think under those circumstances, there might be carbon dioxide  snow, but the rest of the atmosphere would probably liquify first, there may be oxygen and nitrogen rain with some hail or sleet, but snow? Not sure it would happen.
If it did, oxygen and nitrogen snow would be different from our normal water based snow because of the absence of hydrogen bonds.
Individual snowflakes hexagonal in shape would not form, as this is a function of hydrogen bonds and this would not happen with oxygen or nitrogen. Water ice is also less dense than liquid water because of the way hydrogen bonds cause "voids" in the solid structure. This would not be the case with oxygen and nitrogen. So you'd probably get something more like needles than snowflakes.
Also, solid oxygen is blue as noted by Ed V below, so blue snow if it happens at all.
Qualifier:
If the earth is out there (beyond the orbit of Pluto) long enough, enough of the atmosphere would eventually freeze and the pressure may drop below the triple point of nitrogen. If that happens nitrogen would sublimate, then fall as snow as temperatures fluctuate.
But even Pluto's surface temperature and pressure is above the triple point of oxygen most of the time (temperature does sometimes drop below though) so any oxygen in the atmosphere would probably fall as rain (blue rain) or possibly hail/sleet rather than snow.
Another thing to remember is that if the atmospheric pressure is that low, there won't be much drag/resistance either, so no "softly drifting" snow. Those needles will reach terminal velocity pretty damn fast and shatter on the ground.


Examples from everyday life for diastereoisomers ?

Nowadays, it's really easy to find information on the Internet or in books. Being resourceful and being able to search for stuff yourself will help you greatly.

Nevertheless...

Carbohydrates

Glucose and galactose, two of the most common monosaccharides found in the body, are diastereomers of each other:

Lactose, which is found in milk, is formed by a condensation reaction between glucose and galactose.

Eyesight

Retinal is a molecule found in the photoreceptor cells of eyes that normally exists in the 11-cis form. When it absorbs light, it undergoes an isomerisation to form the all-trans form:

This forms part of the biochemical pathway by which light is converted into an electrical signal in the nervous system.


Zwitterion (amino acid) equilibrium and the equilibrium constant

If you assume that the groups react independently, you have two equilibria with equilibrium constants $K_a(\ce{-NH3+})$ and $K_a(\ce{-COOH})$:

$$\ce{^+(H3N)-R <=> (H2N)-R + H+}$$

and

$$\ce{R-COOH <=> R-COO- + H+}$$

You are interested in the overall reaction

$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COOH}.$$

You can take a two step route to achieve this:

$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COO- <=> (H2N)-CHR-COOH}$$

The equilibrium constant for the first step is $K_a(\ce{-NH3+})$, and for the second step is $1/K_a(\ce{-COOH})$. Combined, you get a result of

$$ K_\text{combined} = \frac{K_a(\ce{-NH3+})}{K_a(\ce{-COOH})}$$

Because the $K_a$ values of ammonium and carboxylic acid are different by about four orders of magnitude, the uncharged species occurs about ten thousand times less often than the zwitter ionic form in aqueous solution. In general, comparing p$K_a$ values of two groups competing for a proton tells you which group will usually have the proton (the one with the larger p$K_a$) and how unlikely a proton transfer to the other group would be ($10^{\Delta pK_a}$).

  What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?

I don't understand line 2 and 3 of your argument, and how you get from line 5 to 6. The result you get differs from mine in a square root.

  I thought zwitterions only consist a small portion of the mixture

This depends on the solvent. In aqueous solution, zwitterions are the common species, whereas in organic solvent, the uncharged species is the common one.


When calculating enthalpy changes of reaction using, why is only the amount of substance of one reactant used?

Consider the following reaction ($\Delta H_{\text{rxn}}$ is the enthalpy change for one mole of the reaction, or the amount of moles of each substance found in a balanced equation: in this case, 1 mole $\ce{A}$, 1 mole $\ce{B}$):

$$\ce{A->B}\qquad\Delta H_{\text{rxn}}=~?$$

You are given 5 moles of $\ce{A}$, and you measure the change in heat to be $\pu{-100 kJ}$. However, this $\pu{-100 kJ}$ is actually just the $\Delta H$ of this reaction: $$\ce{5A->5B}\qquad\Delta H=\pu{-100~kJ}$$

Looks the same, right? That's because it is the same. However, if you want to calculate the enthalpy change of the first reaction, you are going to have to divide the measured enthalpy change, $\pu{-100 kJ}$, by 5. Thus, the $\Delta H_{\text{rxn}}=\pu{-20kJ}$.

Now consider this:
$$\ce{A +B->C}\qquad\Delta H_{\text{rxn}}=~?$$
If you were again given 5 moles of $\ce{A}$, along with 5 moles of $\ce{B}$, and you measured the $\Delta H$ to be $\pu{-100 kJ}$, you would get something similar to the first scenario:
$$\ce{5A + 5B -> 5C}\qquad\Delta H=\pu{100kJ}$$

Looks familiar right? Now, if you wanted to calculate the $\Delta H_{\text{rxn}}$, you wouldn't divide 100 by 10; you would divide it by 5, since the second reaction is just $5x$ the first one.

Side-note: This same reaction would be written if we have more than 5 moles of $\ce{B}$. Since $\ce{A}$ and $\ce{B}$ react in a 1 : 1 ratio, adding more $\ce{B}$ than $\ce{A}$ would mean that there is excess $\ce{B}$.


If you really want to neutralize the acid of your mould removal product and don’t want to leave solid deposits, you can try a dilute aqueous solution of ammonia.

You can find typical solutions containing for example about 25 % ammonia (sometimes labelled “ammonium hydroxide”) in hardware stores. Such solutions are corrosive; they can cause severe skin burns and eye damage. You should dilute the solution to well below 5 %.

Ammonia is volatile; you will notice its characteristic pungent smell and you should thoroughly air the treated area. For the same reason, excess ammonia will evaporate and not leave solid deposits. However, it can form salts with the acid of your mould removal product and with carbon dioxide from the air. Such salts are water-soluble and, if necessary, they could be removed easily.

Ammonia is not corrosive to the metal parts of your car (ammonia is actually added to the feedwater of industrial boilers in order to keep the pH high and to reduce the risk of corrosion), except for open electrical contacts made of copper. However, you should test the ammonia solution on an unobtrusive spot to see whether it affects the colour of your carpet.


The non-stoichiometric form of nickel (II) oxide and copper (II) oxide are black, with the same square-planar (at metal) solid-state structure, (with crystal defects), so the metal centres are bridged by oxides. The assignment of oxidation state is +II; therefore, could consider a fragment of the structure as: -O-Cu(II)-O-Cu(II)-; but, with redox-active metals, this also:
-O-Cu(III)-O-Cu(I)- thus. Electrons can drift through the structure giving rise to intervalence electronic transitions, with strong involvement of oxygen atoms. The influence of pie-donating oxygen prevents the formation on Cu(0), the native metal.

Intervalence transitions can give rise to intense colours (e.g. black); normally, forbidden to colours arising from d-d electron transitions. (This, also seen with Prussian Blue, where Fe(II) & Fe(III) are bridged by cyanides.) A discrete Cu(II) centre would have a weak, less-intense, colour, arising from said d-d transitions.

Stabilisation of highly-polarising Cu(III) & Ni(III) is achieved by the oxide, which is a very electronegative and pi-donating ion, stimulating low activation-energy electron-flow.

Transitions between p and d orbitals (as in permanganate: metal and oxide) give rise to very intense colours. These extended oxide lattices are also susceptible to being defective, so missing ions; or, ions in the wrong location in the lattice. This also gives rise to colour. An example of this is seen in zinc (II) oxide, when heated. In this case it is not mixed valence; but, loss of oxygen from the lattice, that gives rise to defects that allow lower energy (than in "complete" ZnO) electronic transitions to occur and absorption at the UV-end of the visible region. On cooling the lattice picks up oxygen, from the air, becoming complete again, reverting to its original white from the, high-temperature, yellow colour.


We can share the women, we can share the wine
We can share what we got of yours, cause we done shared all of mine

"Jack Straw" Grateful Dead
But seriously now, the most important thing in the Princeton.edu article (which explains at the bottom that its content comes from Wikipedia) is
"The distinction from ordinary covalent bonding is artificial"
Does $\ce{H2}$ come from a hydride and a proton or from two hydrogen atoms?  The reason a molecule is stable is independent of a particular path by which it is made.
For $\ce{CO}$, it is only in your mind that oxygen brought more wine (electrons) to the party, but what really matters is that there is enough wine (electrons) at the party for everyone to be satisfied.
$\ce{CO}$ is isoelectronic with $\ce{N2}$ and $\ce{CN-}$; the reason for bonding is the same for each.


Adiabatic processes and the Ideal Gas Law

No, it doesn't contradict the ideal gas law at all. When you are saying that rise in volume means increase of temperature, you are inherently assuming that Pressure remains constant. But that is not the case at all in an adiabetic expansion.

In fact, you should realise that it is the product $PV$ which is decreasing. So, Volume might increase, but pressure drops even faster than volume, that's why the product $PV$ overall decreases. If some of the weights are removed, the gas expands slowly according to the equation, $PV^{\gamma} = C$. For most of the gases $\gamma > 1  .(\gamma = \frac{C_p}{C_V})$ Thus by simple observation, you can see if Volume increases $x$ times, Pressure will decrease $x^{\gamma}$ times, which is more dominating in the product.Thus, the product $PV$ will definitely decrease with increase in volume in case of an adiabetic process.

Also, your second argument is also not acceptable, because you already realise that pressure will decrease and by Kinetic theory of ideal gases $PV = \frac{1}{3}m\ n\ v^2_{rms}$. Thus if you understand the decrease in the product $PV$, you will also understand that $v^2_{rms}$ will also decrease, and hence the kinetic energy of gas molecules also.

Also you are thinking in a much more complicated way towards the situation. Think simply. For ideal gases, internal energy ($U$) is a measure of the translational kinetic energy of gas molecules and also $U = \frac{3}{2}RT$ depicting that for ideal gases, kinetic energy is only a function of temperature only. Also, by the first law of thermodynamics, $dU = \delta q + \delta W$ ( $\delta $ here represents inexact differentials). For adiabetic process, $\delta q =0$. So, $dU = \delta W$. Now, $\delta W$ is work done on the system which is negative for an adiabetic expansion, which makes $dU$ also negative, i.e. internal energy decreases and so does temperature.


tl;dr
The thermodynamic product of the dehydration reaction of 1-methylcyclobutan-1-ol is 1-methylcyclobut-1-ene, not methylenecyclobutane. This is supported by [1],[3] and my own calculations.

Some details
I was unable to find any reference to why the dehydration of 1-methylcyclobutan-1-ol should yield methylenecyclobutane (3) as the main product. From what I could gather up until now, the major product is 1-methylcyclobut-1-ene (1). I also tried finding suitable mechanisms and/or protonation (hydration) energies for these two molecules according to the following scheme, to compare their relative thermodynamic stability to each other.

Wiberg and Hao report the enthalpy difference between 3 and 1 to be only $\Delta H = -0.54~\mathrm{kcal/mol}$. They obtain this value from equilibrating 1 and 3 over sodium on alumina at various temperatures.[1]  This would translate to a ratio of 1:3 of about 1.0 : 0.4.[2]

Lajunen and Himottu quite similarly report, that 1 is more stable than 3 by about $4.3~\mathrm{kJ/mol} \approx 1~\mathrm{kcal/mol}$. They obtained this value via equilibrium constants measurements in perchloric acid.[3]  This would in return translate to a ratio of 1 : 3 of about 1.0 : 0.2. It is also reported, that the transition states of protonation are quite similar, but lower for 1.

I also conducted some quantum chemical calculations, using DF-BP86/def2-SVP and the more sophisticated MP2/def2-TZVPP. These are in reasonable agreement with the previous findings. The enthalpy differences, approximated at $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$, are $1.4~\mathrm{kcal/mol}$ and $1.3~\mathrm{kcal/mol}$ respectively. This gives a 1:3 ratio of about 1.0 : 0.1. (More values, as well as the proton affinities (PA),[4] yielding 2, are given in the table at the end.)

It is not surprising, that these structures are quite similar in energy. They both contain a four membered ring. In 1 two $\ce{sp^2}$ centres are present in the ring, forcing planarity. As a result of the shorter double bond, two angles are slightly larger than $90^\circ$, while two are forced smaller.

In 3, only one $\ce{sp^2}$ centre is present. However, the dihedral angle is a mere $18.4^\circ$, the out-of-plane angle is $26.8^\circ$, respectively. This is apparently not enough to significantly reduce ring strain. Because of symmetry constraints, i.e. $C_\mathrm{s}$, this arrangement forces three inner ring angles to be larger than $90^\circ$.

The concept of ring-strain, or strain in general, is not well defined and it cannot be measured directly. There are relative methods, to obtain this virtual quantity, sometimes involving hypothetical molecules. There are different reference systems and as far as I am aware none of them are considered the ultimative choice. In this sense, any arguments can be made. It is therefore also almost impossible to derive ring strain from just a pen-and-paper exercise, especially when the molecules are that similar.

References

K. B. Wiberg and S. Hao, J. Org. Chem. 1991, 56, 5108-5110.
How the ratio is determined
M. Lajunen and M. Himottu, Acta Chem. Scand. 1987, 41A, 252-258.
How proton affinity is defined

Appendix

\begin{array}{llrr}\hline
  & & \mathsf{\pmb1} & \mathsf{\pmb3} \\\hline
\text{DF-BP86/def2-SVP} & \Delta E_\mathrm{el}    & 0.0   & 1.3   \\
\text{DF-BP86/def2-SVP} & \Delta E_\mathrm{o}     & 0.0   & 1.6   \\
\text{DF-BP86/def2-SVP} & \Delta H                & 0.0   & 1.4   \\
\text{DF-BP86/def2-SVP} & \Delta G                & 0.0   & 1.6   \\
\text{DF-BP86/def2-SVP} & \text{Ratio} (\Delta H) & 1.0   & 0.1   \\
\text{DF-BP86/def2-SVP} & \text{PA}               & 200.2 & 198.8 \\\hline
\text{MP2/def2-TZVPP}   & \Delta E_\mathrm{el}    & 0.0   & 1.1   \\
\text{MP2/def2-TZVPP}   & \Delta E_\mathrm{o}     & 0.0   & 1.5   \\
\text{MP2/def2-TZVPP}   & \Delta H                & 0.0   & 1.3   \\
\text{MP2/def2-TZVPP}   & \Delta G                & 0.0   & 1.7   \\
\text{MP2/def2-TZVPP}   & \text{Ratio} (\Delta H) & 1.0   & 0.1   \\
\text{MP2/def2-TZVPP}   & \text{PA}               & 399.3 & 398.0 \\\hline
\end{array}


can magnesium chloride and iodine react together?

Magnesium chloride and Iodine with form an equilibrium.  This is a bit of a simplification of the reactions but the dissociated chlorides and free iodine will reversibly react to form iodide and triiodide ions and chlorine.
$$\ce{Cl- + 1/2 I2 <=> 1/2 Cl2 + I-} $$
$$\ce{MgCl2 + I2 <=> MgI2 + Cl2} $$
$$\ce{I- + I2 <=> I3-} $$
$$\ce{MgI2 + 2I <=> Mg(I3)2} $$
The amount of products of this reaction though will not be very appreciable, but yes a very literal sense magnesium chloride and iodine will react.


Is the quantum number $L$ a combination of the azimuthal and magnetic quantum numbers $(\ell, m_\ell)$?

The values of $L$ come from the individual $\ell$ values, according to the Clebsch-Gordan series:
$$L = |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2.$$
where the dots indicate all "in-between" values with a step size of 1. The values of $S$ come from the individual $s$ values, in a similar fashion. Note that it is $s$, not $m_s$.
Finally, the values of $J$ come from $L$ and $S$:
$$J = |L - S|, \ldots, L + S.$$
The principal quantum number $n$ does not play any role in this, since it has no direct relation with angular momentum.

Usually it helps to see an example. Consider the case of the ground-state carbon atom; the $\mathrm{1s}$ and $\mathrm{2s}$ orbitals can be neglected since those are closed-shell, so we only need the two highest-energy electrons, i.e. a $\mathrm{p^2}$ configuration.
We therefore have two electrons, both having $\ell = 1$ (p-orbital) and $s = 1/2$ (because they are electrons, and all electrons are spin-$1/2$). The permissible values of $L$ are
\begin{align}
L &= |\ell_1 - \ell_2|, \ldots , \ell_1 + \ell_2 \\
  &= | 1 - 1 |, \ldots, 1 + 1 \\
  &= 0 , \ldots, 2 \\
  &= 0, 1, 2
\end{align}
The permissible values of $S$ are
\begin{align}
S &= |s_1 - s_2|, \ldots , s_1 + s_2 \\
  &= | 1/2 - 1/2 |, \ldots, 1/2 + 1/2 \\
  &= 0 , \ldots, 1 \\
  &= 0, 1
\end{align}
The permissible values of $J$ depend on what $L$ and $S$ are, and must be evaluated on a case-by-case basis. So far, we have identified three possible values of $L$ and two possible values of $S$. Consider the case where $L = S = 0$: here,
\begin{align}
J &= |L - S|, \ldots , L + S \\
  &= |0 - 0|, \ldots, 0 + 0 \\
  &= 0 , \ldots, 0 \\
  &= 0
\end{align}
So when $L = S = 0$, the only permissible value of $J$ is also $0$. If you've seen term symbols before, the corresponding term symbol is denoted as $^1\!S_0$; if you haven't, then all you need to know is that this is a chemist's "shorthand" way of writing $(J, L, S) = (0, 0, 0)$, which is the combination we just found.
You will need to work through this for the other five combinations of $L$ and $S$. This will give you a full series of term symbols, except that there's a catch. The Pauli exclusion principle forbids some of those six combinations: to be precise, the combinations $(L, S) = (2, 1)$, $(1, 0)$, and $(0, 1)$ are not permissible, which leaves only three valid combinations, one of which I've already demonstrated. See Pauli-forbidden term symbols for atomic carbon for more details.

Addendum. As for the $m_\ell$ and $m_s$ quantum numbers: they do have a use, but in a rather more subtle way. I don't expect this very brief summary to be useful, but I'll try; feel free to ignore it at this stage of your learning if it doesn't make much sense.
Much like how (for one electron) $\ell = 1$ implies $m_\ell = -1, 0, 1$, for the entire atom each value of $L$ is associated with a range of $M_L$ values: for example, $L = 1$ implies $M_L = -1, 0, 1$. Therefore, the $L = 1$ term is actually not one single electronic state, but rather a collection of three different values of $M_L$, which is why we call it a "term" rather than a state. Likewise, the $L = 2$ term actually corresponds to five different values of $M_L$. How does this relate to $m_\ell$? Well, it turns out that $M_L$ is the sum of the individual electron $m_\ell$ values. A similar situation exists with $S$, $M_S$, and $m_s$. However, I think that's a topic for another question, if necessary.


Why does the photoelectric effect depend on the frequency of the light?

The photoelectric effect is described by the following equation

$$E_\mathrm{max} = h\nu - \mathrm{WF_M}$$

where $E_\mathrm{max}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\mathrm{WF_M}$ is the workfunction for the particular metal. The kinetic energies of all electrons emitted are distributed from $0$ to $E_\mathrm{max}$.

  Why was monochromatic light used in the experiment? Why is a proper
  frequency used and not any other frequency?

The experiment is typically performed by scanning through a continuous range of monochromatic wavelengths from lower to higher energy.  At some specific wavelength, the observer will notice that electrons start to be emitted (the threshold).  As the scan continues to wavelengths with even more energy, the emitted electrons will increase in kinetic energy.  From the threshold energy and the above equation, the experimenter can determine the workfunction of the metal.  If the experiment had been run with light containing many different wavelengths (non-monochromatic light), electrons would still be ejected, but you wouldn't know what the threshold wavelength was and you wouldn't be able to determine the workfunction of the metal being studied.

If you'd like more information on the photoelectric effect, here's a good, concise reference.


DFT calculation of solids with different periods

I have found a useful reference here

Basically the idea is to avoid the mismatch by best fitting their primitive cell unit vectors.


How to calculate the percentage yield for enantiomers in free radical halogenation?

I'll say that this is just a matter of convention, and not really worth worrying about. Your underlying method and concept is 100% correct. Your final question boils down to:

  Are enantiomers separate chemical compounds?

To elaborate, if the question was:

  Find the percentage of major product formed, counting the enantiomers separately.

Your approach would have without doubt been correct.

But since the question only says:

  Find the percentage of major product formed.

It is difficult to ascertain what the question wants. Enantiomers themselves have different chemical reactions - so they may be interpreted as different chemical products. But, your teacher has interpreted them as a single compound, presuming that optical isomers are just one compound.

Again, I guess this is just convention, because I haven't been yet able to find references to favor either definition.

Finally, to satisfy both the parties involved, and write a foolproof answer that guarantees a complete score, you can frame your results like so:

  Percentage of (R)-2-chloro-3-methylbutane = $x$
  Percentage of (S)-2-chloro-3-methylbutane = $x$
  Total percentage of 2-chloro-3-methylbutane = $2x$

where $x$ is your calculated value for a single enantiomer.


You can regard the reaction to be pseudo-first order, as the concentration of $\left[\ce{V}\right]$ is much higher than the concentration of $\left[\ce{AV}\right]$. Therefore, $k\cdot\left[\ce{V}\right]_0=\text{const.}=k_{2}$. The rate law of this reaction would change to $r=k_{2}\cdot\left[\ce{AV}\right]$. Integrating this would lead to the expression: $\ln\left[\ce{AV}\right]=\ln\left[\ce{AV}\right]_0-k_{2}\cdot t$. The dependence of $\ln\left[\ce{AV}\right]$ on time has the slope $-k_{2}=-0.32\ \mathrm s^{-1}$, so you can easily calculate $k$ from $k_{2}$ and $\left[\ce{V}\right]_0$.


Equlibrium for a non-spontaneous reaction

Note: This first part isn't actually that related to the specific scenario you are giving, but I think it's a misconception that you had that led you to ask this question. If you want a direct answer, go to the last paragraph.

Just because it is non-spontaneous, it doesn't mean that no reaction will occur. In fact, the reaction will always occur to some extent, even if only 0.001% of the particles react.

Consider this formula, which I'm sure you have seen before:  $$\Delta G=-RT\ce{ln}K$$
This makes sense because if the reaction is spontaneous ($\Delta G<0$), $K>0$, so the forward reaction, which is spontaneous is favored over the reverse reaction; vice versa.

If we, for example, say $\Delta G = +100$ kJ, at 298 K, $K$ would equal $3.4\cdot10^{-17}$. Thus, the forward reaction pretty much completely dominates the reverse reaction, so we say that the reaction is essentially irreversible. However, the reverse reaction is still happening, to a very small extent.

If we happened to make $\Delta G = +10$ kJ, $K$ would equal only about $56$, so the reverse reaction becomes visible. The terms spontaneous and non-spontaneous are only generalizations; it doesn't mean that a non-spontaneous reaction doesn't occur at all. There will always be some particles in that substance that will overcome the activation energy and react.

In terms of your question, since the reaction $\ce{A->B}$ is non-spontaneous, the reaction $\ce{B->A}$ is actually spontaneous (think about their equilibrium constants). Thus, the reverse reaction will be favored until the system reaches equilibrium, or when $Q=K$.


How can I get analytical form of GTO primitives for a atom?

Main differenece betweet STO's and GTO's is a $r$-expotetnt. Normalizing factor of GTO one can get from common procedure:
\begin{equation}
N^2 \int_0^\infty \left(r^{n-1}e^{-\zeta r^2}\right)^2 r^2 dr =1
\end{equation}
Of course, computation is not so simple for $n > 1$, but we can use online WolframAlpha (clickable calculation) and we get the formulas from the question post.
For a Lithium atom we have contraction (6s,3p) -> [2s,1p], which means
the first 3s of 6s GTO's cotract to one s "1s"-STO
$$\mathrm{STO}(1s) = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta_1 r},$$
second 3s of 6s GTO's contract to another s "2s"-STO
$$\mathrm{STO}(2s) = \sqrt{\frac{\zeta^5}{3\pi}} r e^{-\zeta_2 r},$$
3p GTO's contract to one p - "2p"-STO
$$\mathrm{STO}(2p) = \sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r}$$
(notation in orbitals in quotation marks "1s", "2s" means the same as for the hydrogen atom). So, as we can see, every atomic STO contract by only 3 GTO's (which justifies their name).

For a Litium atom we have for "1s" atomic orbital
\begin{equation}
\sqrt{\frac{\zeta_1^3}{\pi}} e^{-\zeta_1 r} \approx
C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} +
C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} +
C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_3 r^2}
\end{equation}
where
\begin{align}
\alpha_1 = 0.1611957475E+02; C_1 = 0.1543289673E+00;\\
\alpha_2 = 0.2936200663E+01; C_2 = 0.5353281423E+00;\\
\alpha_3 = 0.7946504870E+00; C_3 = 0.4446345422E+00;
\end{align}

For "2s" atomic orbital
\begin{equation}
\sqrt{\frac{\zeta_2^5}{3\pi}} r e^{-\zeta_2 r} \approx
C_4 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_4 r^2} +
C_5 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_5 r^2} +
C_6 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_6 r^2}
\end{equation}
where
\begin{align}
\alpha_4 = 0.6362897469E+00; C_4 = -0.9996722919E-01;\\
\alpha_5 = 0.1478600533E+00; C_5 = 0.3995128261E+00; \\
\alpha_6 = 0.4808867840E-01; C_6 = 0.7001154689E+00;
\end{align}

For "2p_x" atomic orbital
\begin{equation}
\sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r} \approx
D_1 \left(\frac{128\alpha_4^5}{\pi^3}\right)^{1/4} xe^{-\alpha_4 r^2} +
D_2 \left(\frac{128\alpha_5^5}{\pi^3}\right)^{1/4} xe^{-\alpha_5 r^2} +
D_3 \left(\frac{128\alpha_6^5}{\pi^3}\right)^{1/4} xe^{-\alpha_6 r^2}
\end{equation}
where
\begin{align}
\alpha_4 = 0.6362897469E+00; D_1 = 0.1559162750E+00;;\\
\alpha_5 = 0.1478600533E+00; D_2 = 0.6076837186E+00; \\
\alpha_6 = 0.4808867840E-01; D_3 = 0 0.3919573931E+00;
\end{align}

(For $p_y$ and $p_z$ --- do the same)
Also, for STO's best fit $\zeta_1 = 2.69$ and $\zeta_2 = 0.81$ are consistent with those given in the article Hehre, W. J., Ditchfield, R., Stewart, R. F., & Pople, J. A. (1970). Self‐Consistent Molecular Orbital Methods. IV. Use of Gaussian Expansions of Slater‐Type Orbitals. Extension to Second‐Row Molecules. The Journal of Chemical Physics, 52(5), 2769–2773. doi:10.1063/1.1673374


Cobalt and Iron ions are the only ones to produce colored complexes when mixed with thiocyanate ion $\ce{SCN^-}$. If you mix up a dilute acidic solution of Iron(III) Chloride $\ce{FeCl_3}$ and a similar solution of ammonium thiocyanate $\ce{NH_4CNS}$, a extremely dark red solution is obtained, due to a complex containing the ions $\ce{Fe^{3+}}$ and $\ce{SCN^-}$. If you replace $\ce{FeCl_3}$ by a solution of Cobalt Chloride $\ce{CoCl_2}$, you obtain a blue color due to a complex containing $\ce{Co^{2+}}$ and $\ce{SCN^-}$. No other metallic ions are known to produce such colored complexes with thiocyanate ions, at least among the usual ions studied in chemical analysis courses. The blue Cobalt complex has also the property of being soluble in pentanol, where it can be extracted if the initial solution is colored.

Any way it should be mentioned that :

1) The reaction with Iron is extremely sensitive. The red color may still be seen if the concentration of the complex is 1 - 2 ppm. The blue complex made with Cobalt ions is only visible at much higher concentrations.

2) If this reaction is made with a Cobalt solution containing Iron as an impurity, the solution will turn red, as the blue color of Cobalt will be hidden. In this situation, a couple of milligrams of Potassium fluoride $\ce{KF}$ should be added to the solution. This will produce a colorless complex of $\ce{FeF_6^{3-}}$ and the blue color of the cobalt complex is restored.


Will squeezing lemon juice on top of foods with ferrocyanide additive release hydrogen cyanide gas?

No. The hexacyanidoferrate(II) complex is a very inert complex that does not undergo ligand exchange easily. This is somewhat exemplified by the harsh conditions alchemists used to synthesise hexacyanidoferrate(II), which included high temperatures and strong bases.

Furthermore, whatever amount of lemon juice you add the resulting acidity will be much less than in your stomach. If you swallow a cyanide source such as potassium cyanide, this will get protonated by your stomach acid to liberate hydrogen cyanide and potentially kill you. Food additives — especially approved ones — undergo various testing procedures to make sure that does not happen.


What is the difference between reacting and dissolving?

CCl4 is a covalent compound (having no free ions). Thus it cannot enter into an ionic reaction/double displacement reaction.

Basic Requisites for checking reactivity

Reactivity Series (Elements higher up in the series replace the lower elements)
Checking whether bonds are covalent or ionic (You might just need to remember which compounds are which type. Some compounds can have both type of bonds)

Bonds

Ionic = "Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions. These ions represent atoms that have lost one or more electrons (known as cations) and atoms that have gained one or more electrons (known as anions)." Wikipedia
Thus the elements can take or give electrons. In a solution they exist as ions.
Covalent = "A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs and the stable balance of attractive and repulsive forces between atoms when they share electrons is known as covalent bonding." Wikipedia
Thus it has no free electrons to give or take.


Why do iron(II) ions and chromium have different electronic configurations?

For starters, it isn’t nearly as simple as it is usually put in chemistry classes to explain chromium’s ground state electron configuration.

$\ce{Cr(0)}$ and $\ce{Fe(II)}$ have the same number of valence electrons. You might have heard of the concept of isolobality, which, briefly explained, suggests that compounds with similar electronic configurations (isolobal configurations) will react similarly; so you could replace $\ce{F}$ with $\ce{OMe}$ or even $\ce{[Co(CO)4]}$. The concept has its values and its strong simplifications which I will not go into here.

However, even assuming the isolobality concept’s validity, it would not count $\ce{Cr(0)}$ and $\ce{Fe(II)}$ as isolobal to each other. The reasoning is rather simple: One is a neutral metal, the other is a cation with $+2$ charge. For every electron an atom looses, the atom’s orbitals will contract and the atomic (or ionic) radius will decrease. This also changes the energy levels between orbitals and will affect lower d-orbitals differently than higher s-orbitals.

The result is that the $\ce{4s}$ and $\ce{3d}$ orbitals are now so dissimilar in energy, that electrons hopping back or forth is now no longer an option.


What is natural bond orbital theory used for?

Yes, along with a family of related methods. I honestly don't fully understand the mechanics behind it, but the orbitals obtained with the NBO method can be used for a treatment called Natural Resonance Theory, which gives you an idea of what sort of Lewis-structure resonance components there are and in what weightings, as well as bond indices, and a measure of covalency.

There are three papers from the developers of the technique, covering theory, usage, and some more examples. (I'd probably start with the usage paper.)

Also, along with NBO, there are the Natural Atomic Orbitals which are often lumped in because they're also computed by the NBO program, and Natural Population Analysis which you can use the Natural Atomic Orbitals to do. These give you an idea of what sort of atomic orbital character is contributing to your molecular orbitals, and also atomic charges and net spins.


You're right in that 40% weight/volume means that you take 40 grams of ferric chloride and dissolve it in 100 mL of water. This is the definition of w/v(%) and there is no need to work out the molar concentration - the main benefit of this method of stating concentration.

The total volume will not be (volume of liquid + volume of solid) in most cases of dissolution. In most cases, you will observe very little change in the volume of the resulting solution as the solid molecules separate and fit into the spaces between the liquid molecules. You may even occasionally observe a decrease in volume as shown in this experiment.


Transmission, absorption, and reflection of light

While I can't explain the phenomenon of light transmission, I can say a few things. The energy levels correlating to physical states and the gaps between the states are quantized. This an all or nothing reaction. The frequency of light must exactly match the energy (E=hv) needed for the excitation event (electron is pushed into a higher orbital, a vibrational mode to expressed, etc) for absorption to occur. When the light frequency is mismatched and does not absorb, it will reflect. The frequency that induces excitation is theoretically the frequency of emission, but phenomena such as intersystem crossing may create multiple paths by which the excitation energy will disperse. Hope this helped.


Database for molecular dipole transition moments?

You can find some experimental information on dipole transition moments in the NIST Chemistry Webbook in the section on constants of diatomic molecules.

http://webbook.nist.gov/cgi/cbook.cgi?ID=B3000043&Units=SI&Mask=1000


Net ionic equation with only aqueous ions

If you have a balanced equation like the following:

$$\ce{NaCl_{(aq)} + KBr_{(aq)} -> NaBr_{(aq)} + KCl_{(aq)}}$$

and the solubility products for all the species are big enough that there's essentially no solid, there's really no distinction between the left and right sides of the equation. It's a balanced equation, but nothing has changed. We started with $\ce{Na+_{(aq)} + K+_{(aq)} + Br^{-}_{(aq)} + Cl^{-}_{(aq)}}$ in solution and we still have that on the right side. So there's really not a net ionic equation because there's no net change.


Comparing the rates of hydrolysis of different haloalkanes

This confusion arise because OP did not look at the energy profile of $\mathrm{S_N}$1 mechanism. The first order kinetics of $\mathrm{S_N}$1 reactions (e.g., reactions given in the question) suggests a two-step mechanism in which the rate-determining step consists of the ionization of the substrate (here, it is an alkyl halide), as shown in the following diagram:

The first step, $\ce{R3C-X <=> R3C+ + X-}$, which is the ionization of alkyl halide, is the slowest and the rate determining step. It is true that this step is reversible and is in equilibrium.
However, the next step(s), $\ce{R3C+ + H2O <=> R3C-O+H2 -> R3C-OH + H+}$, is fast and gives the intended product(s). As shown in the energy profile of this mechanism, a carbocation is formed as a high-energy intermediate in first step, and this intermediate bonds immediately to nearby nucleophiles (either $\ce{X-}$ or $\ce{H2O}$ in this case). The rate of this latter reactions would be depends on the values of $\mathrm{\Delta E_\ce{X}}$ (reversible reaction of first step) and $\mathrm{\Delta E_\ce{2}}$ (the second step, $\ce{R3C+ + H2O <=> R3C-O+H2}$), respectively. The value of $\mathrm{\Delta E_\ce{X}}$ may depend on how fast $\ce{X-}$ could take away from the electrophile, meaning the rate of solvation of the respective halide ions come to play here. However, since this is an example of hydrolysis reaction, protic solvent molecules are much more around the electrophile than halide ions, and this competitive reverse reaction is minimal. It is also worth noting that final step, $\ce{R3C-O+H2 ->[X-] R3C-OH + HX}$ is irreversible (not shown in the profile) and would have gone towards to give products eventually.
Therefore, the rate is only depend on the first step. Such reactions would be considered unimolecular and would follow a first-order kinetics. That is the reason to call it $\mathrm{S_N}$1 reactions and having undergo with relevant mechanism. The first step depends on the leaving group ability of the halide. As you learn in the subject, the leaving group ability increase in halogens when going down in the periodic table: $\ce{I- \gt Br- \gt Cl- \gt F-}$ similar to $\ce{C-X}$ bond dissociation energy. Thus, rate of hydrolysis of alkyl halides are:
$$\ce{R3C-I \gt R3C-Br \gt R3C-Cl \gt R3C-F}$$
Hence, $k_1 \gt k_2 \gt k_3$.

Note: Your teacher is correct of the hydration rates of halide ions: $\ce{I- \lt Br- \lt Cl- \lt F-}$ (Ref.1)

References:

Kenzo Hiraoka, Susumu Mizuse, and Shinichi Yamabe, "Solvation of halide ions with $\ce{H2O}$ and $\ce{CH3CN}$ in the gas phase," J. Phys. Chem. 1988, 92(13), 3943–3952 (https://doi.org/10.1021/j100324a051).


How to recast Henderson–Hasselbalch when fitting data on a chemical shift vs. pH graph for NMR

The derivation of these expressions begins from the premise that the observed value of the NMR parameter (chemical shift) is an average over values for two states weighted by their mole fractions: $$\sigma = \sigma_1 \chi_1 + \sigma_2 \chi_2= \sigma_1 \chi_1 + \sigma_2 (1-\chi_1)$$ This is a reasonable assumption in NMR experiments if there are two states (protonated and deprotonated) with distinct shifts and exchange between the two states is fast compared to the chemical shift difference such that the observable signal is a single time average. This allows us to write that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma_2)} = \chi_1 $$ Similarly we can write that $$\frac{(\sigma-\sigma_1)}{(\sigma_2-\sigma_1)} = \chi_2 $$ such that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{\chi_1}{\chi_2} $$ or, in terms of concentrations, that $$\frac{(\sigma-\sigma_2)}{(\sigma_1-\sigma)} = \frac{c_1}{c_2} $$ Introducing this last expression into the Henderson-Hasselbalch equation $$pK_a=pH+\log_{10}\left(\frac{c_{HA}}{c_{A}}\right)$$ gives upon rearrangement
$$10^{(pK_a-pH)}=\left(\frac{c_{HA}}{c_A}\right)=\frac{(\sigma-\sigma_A)}{(\sigma_{HA}-\sigma)}$$ which can be further rearranged into $$1+10^{(pK_a-pH)}=\frac{(\sigma_{HA}-\sigma_A)}{(\sigma_{HA}-\sigma)}=\frac{\Delta \sigma}{(\sigma-\sigma_{HA})}$$ and then further into the expressions in the articles (note the sign convention for $\Delta \sigma$).


What does it mean in a Ni(II)complex if the HNMR spectrum appears unbroadened and unshifted & what does this tell you about the coordination geometry?

If the proton spectrum is unbroadened and not shifted out of the ~0-10ppm range, then I would have to say it is a diamagnetic compound.

It could be low spin square planar.  Tetrahedral and octahedral d8 cannot be diamagnetic.

Image source: http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png

Could also be two nickel atoms in the complex that are antiferromagnetically
coupled for a net spin of zero.


Why is a temporary name given to an element with an atomic number above 100?

The systematic element names refer to a temporary name and symbol for hypothetical and newly synthesized elements. Because of naming controversies in the past, these names were introduced by the IUPAC to discuss the elements without ambiguity. After the discovery is confirmed, they receive a permanent name.

The IUPAC states in the red book this controversy as follows:

  In the past, some elements were given two names because two groups claimed to have discovered them. To avoid such confusion it was decided in 1947 that after the existence of a new element had been proved beyond reasonable doubt, discoverers had the right to suggest a name to IUPAC, but that only the Commission on Nomenclature of Inorganic Chemistry (CNIC) could make a recommendation to the IUPAC Council to make the final decision.

As this text suggests, this is a highly political process, that can be dragged out quite easily. The systematic nomenclature and symbols for new elements are stated in IR-3.1.1 as follows:

  Newly discovered elements may be referred to in the scientific literature but until they have received permanent names and symbols from IUPAC, temporary designators are required. Such elements may be referred to by their atomic numbers, as in ‘element 120’ for example, but IUPAC has approved a systematic nomenclature and series of three-letter symbols (see Table II).2 The name is derived directly from the atomic number of the element using the following numerical roots:
  0 = nil
  1 = un
  2 = bi
  3 = tri
  4 = quad
  5 = pent
  6 = hex
  7 = sept
  8 = oct
  9 = enn
  The roots are put together in the order of the digits which make up the atomic number and terminated by ‘ium’ to spell out the name. The final ‘n’ of ‘enn’ is elided when it occurs before ‘nil’, and the final ‘i’ of ‘bi’ and of ‘tri’ when it occurs before ‘ium’.
  The symbol for the element is composed of the initial letters of the numerical roots which make up the name.
  Example:
  1. element 113 = ununtrium, symbol Uut

As for the transfermium elements, their names have been finalised only in 1994 (Pure Appl. Chem., 1994, Vol. 66, No. 12, pp. 2419-2421).

However, for Mendelevium the first publication already suggested the name and a symbol. It was in use before officially recommended by the IUPAC (A. Ghiorso, B. G. Harvey, G. R. Choppin, S. G. Thompson, and G. T. Seaborg, Phys. Rev. 1955, 98, 1518).


Why is NH4OH used in Polarimetric analysis?

Do you see a potential corresponding to the sugars, or to their Schiff bases, or to said imines' trimers?  Perhaps ammonia gives you a nice pH buffer, as opposed to alkali hydroxides that will wildly swing pH and allow Cannizzaro disproportionation of reducing sugars' aldehyde groups at high pH. Ammonia does not reduce at the cathode.


Charged nucleophile and SN1/E1/SN2/E2 reactions

Strong nucleophiles tend to undergo SN1/E1 reactions whereas weak nucleophiles tend to undergo SN1/E1 reactions.

An increase in electron density will make a nucleophile stronger. This effect can be clearly seen when conjugate bases of common nucleophiles are stronger than the nuclephiles themselves. For example: $\ce{NH_2^- > NH_3} $.

In SN1/E1 reactions, the weak nucleophile can only attack the carbocation once the leaving group has left. This is precisely because it is a weak nucleophile.

IN SN2/E2 reactions, the strong nucleophile attacks the carbon before it can even form a carbocation. This forces the leaving group to leave. This is precisely because it is a strong nucleophile.


A gram of salt added to pure (distilled) water introduces around $2 \times 10^{22}$ ions?