How can you explain the energy change in this phenomenon?

Initially suppose we have a mixture with mole fractions $x_1$ and $x_2$.   Let $x_1$ be the solute and $x_2$ the  water and assume for simplicity that  deviations from  Raoult’s law behaviour are insignificant.  In this case the heat of mixing is zero and entropy alone drives the mixing process.

In this case the free energy of mixing per mole at constant temperature is
$\Delta G =RTx_1\ln(x_1)+RTx_2\ln(x_2)$
and in as there are two components $x_2=1-x_1$ (and changing $x_1 \rightarrow x$  for simplicity ) gives
$$\Delta G =RTx\ln(x)+RT(1-x)\ln(1-x)$$
This energy is negative over the whole range, except at $x=0,1$ where it is zero. The minimum value at $x=1/2$ is $-RT\ln(2)$

After the pure water is added in the left hand compartment and equilibrium is established again, the initial mole fraction can only drop to half its value by mixing equal volumes of water which is a mixing energy of
$$\Delta G\left(\frac{x}{2}\right) = RT\frac{x}{2}\ln\left(\frac{x}{2}\right)+RT(1-\frac{x}{2})\ln\left(1-\frac{x}{2}\right)$$

Again this quantity is negative over the whole range, thus the solution will continue to be diluted if nothing prevents this.

Dilution of the mixture could, in principle, be prevented by the gravitational energy $mgh$ where m is the (molar) mass and h the height the liquid has to be raised to. To prevent further dilution
$mgh  - \Delta G(x_f)  \gt 0 $
where $x_f$ is the final mole fraction. If $mgh \rightarrow 0$ then $x_f=x/2$.

Using values for water gives $mgh = 0.176h$ Joule for height h whereas thermal energy $RT=2494$ J.  If, for example the height is $h=9800 $ m then $mgh \approx RTln(2)$ which is the minimum value of $\Delta G$ then no dilution can take place as $\Delta G \gt 0 $ for all x.

At smaller heights whether or not  dilution occurs depends on the mole fraction of the initial mixture but at a height of $1000$ m as long as the initial mole fraction is greater than $\approx 0.035$ then complete dilution will occur. Below $1000 $ m the final mole fraction is effectively half the initial value, $x/2$.

Of course the heights involved in effecting any change are quite ridiculous, but this is just because gravity is a weak force; you may not find it so if you ever go hill running :)

The figure shows different values of $\Delta G/RT$ per mole.

(a) shows the normal mixing free energy $\Delta G/RT$ vs. mole fraction solute x (b) the change upon no dilution and a potential of $1000$ m .(c) maximal dilution (doubling volume) at $1000$ m, thus dilution can occur for all initial values of molec fraction above $\approx 0.035$.(d) $\Delta G/RT$ plus $mgh = RT\ln(2)$ for water $h \approx 9800$ m. No transfer of water can occur in this case as $\Delta G \gt 0$


Based on your previous question and your comment above in lines of «NIST's policy to use published data may be fine» I question if your literature search included a visit at NIST's Handbook of Basic Atomic Spectroscopic Data.
In the case of sodium, recorded lines for either air or vacuum are listed next to the data about persistent lines, and analogue data about the single ionized form.  While not yet in a .csv format, copy-paste into an editor is eased because you may toggle between the output to be table-like formatted form, or ASCII.
Addition:
An alternative could be to digitize emission spectra like the following:

(reference, final spectrum about the hot high pressure sodium lamp)


How to measure a distance along the surface of a protein in a PDB

I would use linear distances between a suitable amount of fixed points as an approximation. You cannot measure exact distances along a rough surface, as the distance will vary depending on minimum measuring size. This is known as the coastline length paradox in math. https://en.wikipedia.org/wiki/Coastline_paradox

In theory the length would approach infinity as the measuring size approaches 0.

In practice you can get around the issue if you use a step size compatible with the smallest stable feature.

A suitable measuring size would probaly be in the order of ~10^-11 meters. (1/10 the size of an carbon atom).


Why does cobalt have no negative charge?

Cobalt can be found in various oxidation states including negative ones such as Co(-I) in NaCo(CO)4. In this instance however the excess electron density on the cobalt center is relieved by backbonding with the CO ligands and therefore such a low formal oxidation state of the Co atom is stabilized. On the other hand the free Co$^-$$^1$ ion cannot hold the extra electron first due to the low electronegativity of Co as TAR86 hinted out in his comment and second due to the fact that the extra electron is placed in a d orbital which is not very  penetrating and thus shielded by all other electrons. Hence, it is unlikely to encounter compounds such as [Na]$^+$[Co]$^-$ where there are no ligands on the cobalt to stabilize the excess electron density. The opossite applies when you remove electrons from Co (and any other transition metal): the atom becomes positively charged and the nucleus attracts the remaining electrons even more effectively (Higher Zeff). Therefore, positive oxidation states are more common.


Solubility of Iodine in Some Organic Solvents

Iodine as a halogen has the strong electron affinity, what is the best met by the free electron pair of diethylether, then by $\pi$ electron density of toluene.

Note that iodine is also well soluble in ethanol, commercially used in medical treatment of skin injuries.


Not really. Overshooting doesn't happen, rather the concentrations gradually approaches the equilibrium concentration as an asymptote.

You'll soon see what I mean.

As an example, for simplicity, lets consider:

$$\ce{P <=>[k_1][k_2] Q}$$

I will assume here that the forward and backward reactions are first order, so we get the rate to be:

$$\text{Rate}_{\text{forward}}=k_1\ce{[P]}$$

and

$$\text{Rate}_{\text{backward}}=k_2\ce{[Q]}$$

As a result the net rate would be:

$$\frac{d\ce{[P]}}{dt}=-k_1\ce{[P]}+k_2\ce{[Q]}$$

Let's take the initial concentration of $\ce{P}$ to be $\ce{[P]_0}$, and the initial concentration of $\ce{Q}$ to be zero.

Suppose at any time $t$, the concentration of the reactant is $\ce{[P]}$. Thus, $\ce{[Q]} = \ce{[P]_0} - \ce{[P]}$. Now, we get the rate law to be:

$$\frac{d\ce{[P]}}{dt} = k_2\ce{[P]_0} - (k_1 + k_2)\ce{[P]}$$

Integrating this from $\ce{[P]_0}$ to $\ce{[P]}$:

$$\int_{\ce{[P]_0}}^{\ce{[P]}}\frac{d\ce{[P]}}{k_2{[P]_0}-(k_1 + k_2)\ce{[P]}}=\int_0^tdt$$

Solving the integral, we get:

$$\frac{-1}{(k_1 + k_2)}\log_e{\left(\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}\right)}=t$$

$$\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}=e^{-(k_1 + k_2)t}$$

Finally, on simplifying it, we get our integrated rate equation as:

$$\ce{[P]}=\ce{[P]_0}\frac{k_2 + k_1e^{-(k_1+k_2)t}}{k_1+k_2}$$

Let's plot this on a graph to understand what this is actually saying:

(purple-$\ce{[P]}$ and orange-$\ce{[Q]}$)

You can clearly observe that the concentrations never overshoot the asymptotes (black dotted line). The asymptotes denote the equilibrium concentrations, so it's evident that the reaction tends toward the equilibrium concentration.


Do I have to consider the water in the hydrate of a salt in a mass percent calculation?

As stated in one of the comments to your question, the waters of hydration constitute part of the crystal structure itself, just as much so as the rest of the molecule. You can somewhat think of the waters as being functional groups of the compound. For example, the dehydrated form of a compound can have very different chemical and physical properties than the hydrated form.  The presence of waters of hydration in no way means that the compound is aqueous or dissolved in water.

Thus, when calculating the mass percentage of hydrogen of the hydrated compound, you do include the hydrogens from both the waters of hydration and the rest of the molecule.

The common name of  $\ce{NH4MgPO4⋅6H2O}$ is magnesium ammonium phosphate or struvite. According to this Wikipedia article:

  Struvite (magnesium ammonium phosphate) is a phosphate mineral with
  formula: $\ce{NH4MgPO4·6H2O}$. Struvite crystallizes in the
  orthorhombic system as white to yellowish or brownish-white pyramidal
  crystals or in platey mica-like forms. It is a soft mineral with Mohs
  hardness of $1.5$ to $2$ and has a low specific gravity of $1.7$. It is
  sparingly soluble in neutral and alkaline conditions, but readily
  soluble in acid. Struvite urinary stones and crystals form readily in
  the urine of animals and humans that are infected with
  ammonia-producing organisms.

Here is a photograph of struvite crystals from the same article:


What, if any, hydration states does sodium bicarbonate have?

To my knowledge sodium hydrogen carbonate hydrates do not exist.

However, there exists a mixed hydrate of sodium carbonate and sodium hydrogen carbonate with the formula $\ce{Na2CO3 \cdot NaHCO3 \cdot 2H2O}$ (dihydrate of trisodium hydrogendicarbonate, $\ce{Na3H(CO3)2}$) that occurs naturally as the mineral trona.

The region labeled trona in the phase diagram below shows the conditions under which the hydrate can be crystallized from an aqueous solution of $\ce{Na2CO3}$ and $\ce{NaHCO3}$.

The picture is taken from US patent US7638109B2.


Photochemical rearrangement of 4,4‐diphenylcyclohexa‐2,5‐dien‐1‐one

The formation of structure 6 (your D) occurs as follows. Photolysis of dienone 1 affords excited state 2 that forms cyclopropane 3. In turn, 3 is capable of breaking a different cyclopropane bond to form 5-membered ring 4. Structure 5 is a resonance structure of 4. Structure 5 closes to the [3.1.0] product 6 (your D). I don't know how you formed A (12 here) but further photolysis of 6 can form 12 via a bridged phenyl migration as shown in structure 9. A similar migration in structure 2 leads to your B. Of course, structure C has only five carbons and was added to the choices as a trap. See reference 1 for an example of a dienone phenol rearrangement.

1) A. G. Schultz and S. A. Hardinger, J. Org. Chem., 1991, 56, 1105.


Substance of higher latent heat of melting than water, similar melting point

Indeed, it will be hard to find a substitute for water. You might want to look into how ice packs work.

Some of the things that are added to water in such packs include substances to prevent bacterial growth and to prevent the water to become fully liquid. The water thus become a gel.

  Gel packs may be made by adding hydroxyethyl cellulose (Cellusize) or vinyl-coated silica gel.

One factor that you might consider, is that if there is more mass (for example, dissolving something in the water) will provide more substance for the any heat to go to, so that the temperature rise will be slower.

Another, is hydrogen bonds, which are very present in water molecules and which explains its high specific heats (latent heat and heat capacity).

Unfortunately, adding in stuff to the water will tend to disrupt the hydrogen bonds. I haven't tried it, nor can find something to support my next sentence, but I would expect ammonia to be a possible additive. It has a higher specific heat capacity than water and dissolves easily as well since it does not disrupt hydrogen bonds that much due to its molecular structure.

Glucose is also a potential additive since it doesn't disrupt the hydrogen bonds much, but its specific heat is apparently a bit lower than ammonia. I found a site where I could input some parameters and it outputs the heat capacity of the sugar water and the outputs often say that the specific heat capacity is lower than that of ammonia alone (from the same site that michielm mentioned, ammonia has a specific heat capacity of a bit less than 4.6 at $0^oC$).


Transformation of virial coefficients from molar volume to pressure

We want to take this:

$$Z = \frac{PV_m}{RT} = 1 + \frac{B}{V_m} + \frac{C}{V_m^2}+…\tag{1}$$

And transform it into this:

$$Z = \frac{PV_m}{RT} = 1 + B'P+ C'P^2+…\tag{2}$$

We begin by expressing (1) in terms of $\rho_m = 1/V_m$:

$$Z=1+B\rho_m+C\rho_m^2+…\tag{3}$$

We then multiply (2) and (3) by $RT\rho_m$ to get (4) and (5), respectively. Remember that $ZRT\rho=ZRT/V_m =P$, because $Z=PV_m/RT$.

$$RT\rho_m Z = P = RT\rho_m + RT\rho_mB'P+ RT\rho_mC'P^2+…\tag{4}$$
$$RT\rho_m Z= P = RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…\tag{5}$$

Next, we use (5) to substitute for $P$ in (4):

$$P = RT\rho_m + RT\rho_mB'(RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…) +$$
$$ RT\rho_mC'(RT\rho_m + RTB\rho_m^2 + RTC\rho_m^3+…)^2+…\tag{6}$$

Collecting all the $\rho_m^2$ terms from (5) and (6):

$$RTB=(RTB')(RT)$$
Rearranging leads to the desired $$B'=\frac{B}{RT} \tag{7}$$

Next, collecting all the $\rho_m^3$ terms from (5) and (6):

$$RTC=(RTB')(RTB)+(RTC')(RT)^2$$

Simplifying, rearranging, and making use of (7) leads to the desired
$$C' = \frac{C-B^2}{(RT)^2}\tag{8}$$


What is the reaction mechanism of a 5-endo-trig cyclization?

Your mechanism to the 5-endo-trig product (4) is very close.  Notice however that when you do the initial cyclization, the intermediate formed has a positive charge on nitrogen along with the negative charge on oxygen (as you showed).  This cyclized intermediate then just needs to move a proton from the nitrogen to oxygen to form the neutral enol which then converts to the (ester) carbonyl - which is your desired product.  No other reagents are required.


Newman projections: how to tell which groups are bulkiest to determine what is the most stable conformation?

Well, I think your book is wrong. Read the extract:

  A-Values are numerical values used in the determination of the most stable orientation of atoms in a molecule (conformational analysis), as well as a general representation of steric bulk. The utility of A-values can be generalized for use outside of cyclohexane conformations. A-values can help predict the steric effect of a substituent. In general, the larger a substituent's A-value, the larger the steric effect of that substituent. A methyl group has an A-value of 1.74 while tert-butyl group has an A-value of ~5. Because the A-value of tert-butyl is higher, tert-butyl has a larger steric effect than methyl. This difference in steric effects can be used to help predict reactivity in chemical reactions.

I guess what you're looking for are A-values. The higher the A-value, the more "bulky" the group is considered.

So, the $\ce{-SH}$,$\ce{-Cl}$ and $\ce{-Br}$ groups seem to be less hindering than the $\ce{-iPr}$ group.

References:      A-value, Wikipedia


pH value of a buffer solution of HClO and NaClO

You're close.  Everything is correct, except that when you take the ratio of concentrations in the H-H equation that ratio is not in moles.   It's just a number, because you divide moles by moles .  Rule of thumb: logarithms and exponential should never involve anything with units.  So don't include the molar unit under the logarithm and you're good.


Why can't phosphorus trifluoride be used as a Lewis acid catalyst in electrophilic substitution of benzene?

The best answer I can give is that there is enough space on $\ce{Sb}$ in $\ce{SbCl3}$ for an additional Lewis Base to interact with a nonbonding orbital on the $\ce{Sb}$ atom. While $\ce{P}$ can be hypervalent, the LUMO in this case would be $\ce{P-F}$ $\sigma^\ast$, which is really high in energy (meaning that it doesn't easily accept electron density from other species in solution).


What is the hybridisation of titanium in the hexaaqua- and hexaamminetitanium(III) complex?

Immediately stop using the hybridisation concept for transition metals. Don’t even take it a step further. It is not helpful here at all.

Both complexes have essentially identical electronic structures except that the relative orbital energies differ slightly. The following image — originally taken from Professor Klüfers’ internet scriptum for his coordination chemistry 1 course (LMU Munich; in German) — shows a typical octahedral coordination compound and only includes the ligand–metal σ interactions.

You can see the metal’s orbitals on the left-hand side; from bottom to top they are 3d, 4s, 4p. 4d would be even higher in energy, possibly even higher up than 5s. These virtual energies can be really hard to calculate or measure accurately. On the right-hand side, you can see the six donor orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. (For some reason, the u is missing.) The ligands’ p-orbitals that can take part in π bonding would transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, and would therefore form another bonding-antibonding interaction with the metal’s $\mathrm{t_{2g}}$ orbitals. The ligand orbitals would again be bonding and the metal’s antibonding. If you check out the link and look for the image you’re seeing, you can also see pictures of the metal orbitals, and further down the scheme that includes the π interactions (only the relevant part highlighted).

The metal orbitals that interact with ligand orbitals in a σ fashion are two 3d, 4s and three 4p. The remaining three 3d orbitals interact in a π fashion. So if you really, really, really must use hybridisation in this case (did I say it is not recommended?) then you should report titantium’s hybridisation as $\mathrm{d^2sp^3}$ and not the other way round: 4d orbitals do not play any role.

The only difference between the hexaaquacomplex and the hexaamminecomplex is that the former has lower-lying ligand orbitals than the latter so the energy difference between $\mathrm{t_{2g}^*}$ and $\mathrm{e_g^*}$ will be slightly different. That’s all.


Why is a sodium chloride molecule stable?

What you have calculated is the energy required to make two separated ions out of two separated $\ce{Na}$ ad $\ce{Cl}$ atoms. When they are separated, these ions do not make a molecule or a crystal of salt. Separated ions attract one another and are able to release a huge amount of Coulomb energy when approaching each other.
Anyhow your calculation should still take into account the fact that chlorine atoms do not actually exist. They must be produced by breaking a $\ce{Cl2}$ molecule. And this requires some energy. You may also introduce the sublimation energy of sodium atoms


Determination of pKb of a mono acidic base

I appreciate your effort you showed in your question, but you missed one important thing that when a weak base is reacted with a strong acid a buffer solution is formed and the $\mathrm{pH}$ is not only contributed by base but also by the acidic salt ($\ce{BCl}$ in this case).

So, simply applying the Henderson–Hasselbach equation (derivation can be found here)

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log{\frac{[\ce{B-}]}{\ce{[BOH]}}}$$

we get

$$4 = \mathrm{p}K_\mathrm{b} + \log{\frac{1.2}{2.4}},$$

which gives us

$$\mathrm{p}K_\mathrm{b} = 4.3010$$


Can imines be catalytically hydrogenated?

Several methods on the reduction of imines are outlined on Wikipedia. Another method using catalytic hydrogenation is given here.


Carbocation rearrangement in alkene addition reaction

The mechanism shown is not a carbocation rearrangement. Those two structures are resonance structures, which means that the chloride anion can attack at either of the carbons that bear a positive charge. Because there are two different alkene products that are possible, the more substituted alkene will be favored. Strictly speaking, you're right, Zaitsev's Rule doesn't apply here, because this isn't an elimination reaction. However, the reasoning is the same: the most substituted, most stable alkene is the favored product.


How do I calculate the color of a compound?

Since this is quite a broad question, I will also only be able to provide an answer in very broad terms.

The short answer is: You cannot. At least not with (a real) pen and (real) paper.

The slightly longer answer is:
With a simple pen and paper approach this is virtually impossible. Even the use of more elaborate methods like MO theory can lead you to completely wrong results. Electronic transitions involve excited states, which can be incredibly hard to calculate. In most cases you can get an approximate spectrum by using time-dependent density functional theory. This can be right for the wrong reasons - it is still somewhat a crude theory. The best results you will get with multi-reference methods or big automated computations, which may take a long time on a supercomputer.

In other words: There is not easy, general, rigorous way to determine the colour of a compound.


What does the label of stability: 0 mean on an MSDS sheet?

According to the Wikipedia site for the NFPA 704   identification system, a Stability rating of 0 means:

  Normally stable, even under fire exposure conditions, and is not
  reactive with water.

The low concentration of the bleach, in other words it's mostly water, is the reason that it fits the "0" category.

Note that the bleaching component of the bleach solution is sodium hypochlorite (see your MSDS for bleach).  Pure solid sodium hypochlorite has a NFPA reactivity rating of "1".


Molecular orbitals and "generalized" aufbau principle for heteronuclear molecular configurations

Do the molecular orbital configurations and their order depends on the scheme and the linear combination of atoms (homo and heteronuclear, di and poliatomic)

Not only orbitals arising depends on atom configuration, but also on atomic orbitals available (s,p,d) and their ordering depends on geometry (i.e. size) and relative energy of atomic orbitals available. Unfortunately, there is no simple rule here.

Lucky enough, there is some amount of free computational packages you can use to calculate and then visualize molecular orbitals.


Lewis dot structures and exceeding the octet

For the first part of your question, you are correct that Mg has 12 electrons, which are arranged in shells in the pattern 2 - 8 - 2. But magnesium is a typical metal in that it tends to lose electrons to form a cation that then forms solids by ionic bonding. The octet rule predicts that the resulting cation will be stablest if its outermost shell contains eight electrons. And indeed the magnesium ion is $\mathrm{Mg}^{2+}$, two electrons less than the atom, with an electron configuration of 2 - 8.

What would happen if you had asked about a non-metal atom in the third period, say Cl? This has an electron configuration of 2 - 8 - 7, and tends to form a single covalent bond (e.g., to form $\mathrm{Cl}_2$ or $\mathrm{HCl}$), giving a configuration of 2 - 8 - 8 rather than 2 - 8 - 18. The reason for this is that the first eight valence electrons occupy the $s$ (2 electrons) and $p$ (6 electrons) subshells, while any further electrons in the third shell must go into the $d$ subshell. These orbitals are both higher in energy than those from the first two subshells (which is why, as you go through the periodic table, the $4s$ shell is filled before the $3d$) and extend less far from the nucleus, so that they are much less likely to overlap other atoms' orbitals to form a covalent bond. So it is often energetically preferable not to put electrons into this subshell if possible, which is why the octet rule can still apply to elements in this period.

Of course, it is possible to put electrons into this subshell, which is why it is possible for the central atom to have more than an octet in compounds like $\mathrm{SF_6}$. This leads into the second part of your question, which, as Manishearth suggested, looks like a duplicate of Can an atom have more than 8 valence electrons? If not, why is 8 the limit?.


Does energy come from fuel or oxygen? (surprise?)

[OP:] It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions.

An enzyme is a catalyst, so it does not change the enthalpy of a reaction. That part is correct. "Oxygen only enables the release of energy" is incorrect. Oxygen is one of the reactants, and the oxygen atoms also appear as products.

[Dr. Schmidt-Rohr in JChemEd paper] Generally, our analysis indicates that it is not the organic fuel but rather O2 that is “energy-rich”.

This might be true in the context the combustion reaction with the set of organic molecules he chose to include. The organic fuels exclusively had C-H, C-C, C=C, O-H, C-O and C=O bonds, if I understand correctly from the abstract and the figure. He is saying that the bond energy of most bonds in reactants and products are similar (e.g. C-H vs O-H, C=C vs C=O, 2 C-C vs C=O), and that the O=O "double bond" is an outlier.
I think it is worthwhile to state the amazing fact that we have a biradical (see https://chemistry.stackexchange.com/a/15061) in our atmosphere and are breathing in a very reactive substance. The only reason that our red blood cells aren't constantly undergoing mutations from oxidative damage is that they lack DNA once they are released into the bloodstream.
If I were to study the combustion reactions with elemental metals, I might come to a different conclusion than the paper. Try comparing $\ce{Na + O2}$ with $\ce{Hg + O2}$.
Final remark (quoting myself):

When a raw egg rolls off a countertop and falls on the floor, does it crash because the countertop is so high or because the floor is so low?


Yes, $\ce{IO4-}$ does exist and it is called metaperiodate anion. It can combine with a number of counter ions to form periodates, which may also be regarded as the salts of periodic acid.

  Metaperiodates are typically prepared by the dehydration of sodium
  hydrogen periodate with nitric acid.

  $$\ce{Na3H2IO6 + 2 HNO3 → NaIO4 + 2 NaNO3 + 2 H2O}$$

  They can also be generated directly from iodates by treatment with
  other strong oxidizing agents such as hypochlorites:

  $$\ce{NaIO3 + NaOCl → NaIO4 + NaCl}$$

  Metaperiodates adopt a distorted tetrahedral geometry with an average
  $\ce{I–O}$ distance of 1.78 Å

  The ortho- and metaperiodate forms exist in equilibrium.

  $\ce{H4IO6− ⇌ IO4- + 2H2O,      K = 29}$

  For this reason orthoperiodate is sometimes referred to as the
  dihydrate of metaperiodate, written $\ce{IO4^-·2H2O}$

One example of salt having metaperiodate ion is sodium periodate(@Gert). Its preparation is mentioned above. In fact, all other alkali metals and alkaline earth metals forms metaperiodate salts. Some examples include potassium periodate, rubidium periodate, calcium periodate etc.


How to calculate the mass of an element in a sample of a compound?

As suggested in the comments, keep track of the units!  If your answer doesn’t make physical sense, you made a mistake.

A thorough approach to this problem is to:

  calculate how much glucose you have.
  calculate the fraction of oxygen in glucose.
  multiply the (unitless) fraction found in (3) by the total glucose calculated in (1).

Using the values (and associated units) given in your problem:

  $0.5 \pu{\frac{mol}{L}} \text{ glucose}\times \pu{1 L} \text{ glucose}\times \pu{180.16 \frac{g}{mol}}$ glucose = $\pu{90.08 g}$ glucose.
  $\frac{\pu{96 g mol^-1} \ce{O}}{\pu{180.16 g mol^-1} \text{glucose}} = 0.533$
  mole fraction of $\ce{O}$ / glucose.
  $0.533 \times \pu{90.08 g}$ glucose = $\pu{48 g} ~\ce{O}$

Of course there are many shortcuts you could take or even different approaches altogether, but I like to parse a problem like this down stepwise in a way that shows what you are doing and facilitates keeping track of units.  This makes troubleshooting much easier and systematic if you find you've made a mistake or want to walk through it a second time to verify what you've done (as you definitely should!).


Assuming you fill the balloon only with liquid butane, the answer is very simple - the gas in the balloon at room temperature IS the liquid in the balloon initially. If you put some amount of liquid butane into the balloon and seal it at t=0, then allow it to warm until the liquid has evaporated, there is essentially no transport of other matter into or out of the balloon during that time. The only thing that has happened is that the initial charge of liquid has absorbed thermal energy from the surroundings and undergone a phase change from liquid to gas.

Because there is no change in the amount of material in the balloon - there are the same number of butane molecules in the gas phase as there were initially in the liquid phase - there can be no change in mass. Liquid and gas phase contain the same number of molecules, at some fixed mass per molecule.

Note that while the mass does not change, the density does.


Why are thiolates more nucleophilic than alkoxides?

It's true that thiolate ion is more nucleophilic than alkoxide ion. You are going in a right way. The order of nucleophilities is according to polarisability of atom. As we know, the size of atoms goes on increasing down the group.

atomic size: $\ce{S>O}$

Due to increase in size of atoms, the electrons are more loosely bound to nucleus. Hence, the atoms are now easy polarise. As a result, electrons are easily polarisable down a group.

Thus, $\ce{RS-}$ of thiolate ion is more polarisable than $\ce{RO-}$ of alkoxide ion. Therefore, thiolate ion is more nucleophilic than alkoxide ion.

The polarisability of nucleophile is more important in case of nucleophilic substitution reaction where the nucleophile electrons are polarised by the positive charge of electrophile and bond is formed during the reaction.

As nucleophilicity is a kinetic factor. More stronger the nucleophile, greater is its rate constant. The following is relative rate constant of both nucleophile on reaction with iodomethane by using methanol as reference whose rate constant is taken as unity:

$\ce{CH3-O-};~k  =  2\cdot 10^6$

$\ce{CH3-S-};~k  =  1\cdot 10^9$


Which chemical properties make a substance explosive?

I'm going to assume by 'good explosive' you mean 'high energy to mass ratio', which by no means implies anything about whether the explosive is 'good' in the 'won't suddenly explode when thought about' sense.

This is not intended to be a full answer as it doesn't touch on the terrifying explosivity of organic peroxides, however it's a start. The presence of nitro ($\ce{R-NO2}$) or azide ($\ce{R-N=N=N}$) groups in a molecule is strongly suggestive of explosivity. This has a lot to do with the large negative free energy change of formation of dinitrogen and (I guess) oxides of nitrogen upon decomposition of these molecules. $\ce{N2}$ has one of the strongest bonds known and releasing a lot of $\ce{N2_{(g)}}$ entails a large increase in entropy. Inorganic nitrates and azides are also likely to be explosive.

The state of the art in explosive research seems to involve cramming as many consecutive nitrogen atoms as possible into a molecule, as showcased on Derek Lowe's blog:

$\ce{C4H4N8}$

$\ce{C2N14}$ (what?!)

Alternatively, you can take a highly strained polyhedrane of some description and stick nitro groups all over it. The ur example of this sort of philosophy is octanitrocubane.

Bonus: tetranitrotetrahedrane

Disclaimer: Explosives are deadly and people who prepare them are either highly trained experts with exquisite safety infrastructure, or morons. Nothing in my answer should be taken by any person as an endorsement to attempt to synthesise anything described.


Priority of carbons attached to chiral group, one with double bonded oxygen, one with two oxygens

Given the same atom, those bond via triple bonds have higher precedence over double and double bonds have higher precedence over single. This is exemplified here and, a bit more formal covering all CIP-rules, here.


Alternative python modules to RDKit to convert SMILES to structure

I am not aware of any other Python tool not mentioned in the question @Mithoron linked to. Openbabel is available via pip by the way.

That RDKit is not available on PyPi should not hinder you from using it in a virtual environment. RDKit is available via the conda package manager which is easy to use and combinable with pip.

If you are on Linux, the probably best solution for managing different versions of Python (also Anaconda Python) and virtual environments is pyenv. I highly recommend to check it out.


How can you extract palladium from palladium chloride

Yes, there are several methods of reducing palladium chloride to elemental chloride.

By converting Pd(II) salt to palladium acetate ($\ce{[Pd(CH3CO2)2]3}$) and decomposing it at 200-300 $\pu{^\circ C}$. (link)

By using reducing agents

carbon monoxide
$$\ce{CO + PdCl2 + H2O → Pd + CO2 + 2HCl}$$
This is a qualitative reaction to determine the presence of carbon monoxide in the
air. A paper moistened with a solution of palladium chloride turns black in the presence of carbon monoxide due to the formation of finely dispersed metallic palladium. The reaction proceeds at r.t. The yield is high (84%).

Hydrogen
$$\ce{PdCl2 + H2 -> Pd + 2HCl}$$
Pd(II) salts can be reduced to finely divided elemental palladium by using
hydrogen. The reaction proceeds at r.t. but the reaction rate can be increased by
increasing the temperature. See this paper for more details.

Using $\ce{SnCl2}$

The reaction proceeds by formation of a complex. Yield is also high (90%). See here for more details.

Other reducing agents ($\ce{TiCl3, Fe, Cu, Zn}$)

Decomposition of palladium chloride. But the reaction is unreliable. The reaction proceeds at high temperature (590-740 $\pu{^\circ C}$). May release fumes of chlorine and hydrogen chloride which might be dangerous. Palladium obtained may be impure because of formation of various types of oxides. (link)

Decomposition of potassium tetra/hexachloropallade

Calcination in air. Dissolve the salt in a suitable solvent (usually water), "paint" the solution on a substrate, allow the solvent to evaporate and then start calcination. Pd(II) would become Pd(0), leaving a thin metal film on the substrate.

Binding of Pd-salt into an organic matrix (e.g., a zeolitic inorganic-organic polymer electrolyte), and then calcinate the product. However, using this approach, calcination should be done under vacuum or under an inert atmosphere (otherwise, oxygen might yield Pd oxide systems). In this case, Pd ends up as metal nanoparticles into a carbonaceous matrix.

Solution phase reduction. Impregnate silica with your metal salt solution, add a reducing agent like formic acid, ethylene glycol, $\ce{NaBH4}$ or hydrazine, heat at $\ce{150 ^\circ C}$ for ethylene glycol and at $\ce{80 ^\circ C}$  for the rest of the reducing agents, you will get Pd decorated $\ce{SiO2}$ particles. Nano particles are prone to oxidation at higher temperatures, so performing the reaction in inert atmosphere like $\ce{N2}$ or Argon will ensure the Pd(0) particles.

by Emeraldine base to form palladium nanoparticles. See here

Details about the points can be found in this researchgate article. But since this is a school project, points 3-7 are out of scope. The best method would be the "reduction through carbon monoxide".


Elimination of hydroxyl group while deprotecting acetal

As I said in the comments, it entirely depends on the conditions. It is easy to select mild conditions to cleave the dioxolane. Elimination would require very strong forcing conditions (concentrated acid and heating). The first site of protonation is the carbonyl, so you would form the hemi-acetal of the product long before elimination or substitution occured at the primary alcohol. If you were foolish enough to heat this with cH2SO4 then you could get elimination of the primary alcohol and generate quite a mess.


Initial guess for Unrestricted Hartree-Fock calculation

When $N_{\alpha} = N_{\beta}$, a restricted solution of Roothaan equations is a solution to the unrestricted Pople-Nesbet equations. This restricted solution always exists and necessarily results if an initial guess $P_{\alpha} = P_{\beta}$ is used. [...] In seeking this second solution [the unrestricted one], it is imperative that an initial guess $P_{\alpha} ≠ P_{\beta}$ be used or the equations will necessarily yield the restricted solution.

The quote you state from Szabo and Ostlund is not wrong, but isn't a practical quote in the sense that it helps to think about the different spin cases that one would want to handle. You already know how RHF/closed-shell calculations work in theory and in practice. I won't talk about restricted open-shell Hartree-Fock (ROHF), because it isn't relevant to this problem.

There are three UHF/open-shell scenarios, which will be considered in order:

Odd number of electrons
Even number of electrons, high spin
Even number of electrons, low spin

Odd number of electrons

For this case, take the copper(II) atom, which has a 3d$^{9}$ valence electron configuration; it is a doublet, so it has 1 more $\alpha$ electron than $\beta$ electron. There's no trouble here, because the indices for occupied orbitals run over different ranges; in the simplest case, the range for $\alpha$ spin will be 1 larger than for $\beta$ spin.

Even number of electrons, high spin

For this case, let's take triplet oxygen. In practice, this is actually identical to the above case, except we have 2 more $\alpha$ electrons than $\beta$ electrons. Again, we have differently-sized occupied blocks for each set of MO coefficients which will lead to different densities.

Even number of electrons, low spin

For this case, let's take singlet oxygen.

This is the difficult case, because the orbital degeneracy means for a proper qualitative description of the molecular orbitals, two determinants must be used, requiring a CAS(2,2) (at a minimum). Because this valence electron degeneracy isn't possible in a single determinant picture, the degenerate MOs will "split" into one lower in energy, the other higher. Requesting a singlet spin multiplicity means the two electrons will now occupy the same orbital; there's no reason for them to occupy different orbitals, at least for the initial guess. Hopefully it's (visually) clear that if the spatial components of $\alpha$ and $\beta$ spins are the same in this case, our guess is equivalent to the RHF guess. This means that for an open-shell singlet, one must break spatial symmetry in the initial guess, just as Szabo and Ostlund states. Since the RHF solution always exists as a local minimum on the "determinental" potential energy surface, whatever you choose as your SCF converger is most likely going to get stuck in this minimum.

This is a little hand-wavy, since we know that UHF calculations approximate two real determinants at the cost of spin contamination, but that isn't our goal.

Breaking spatial symmetry

How do we break spatial symmetry? Consider the initial steps of a typical RHF calculation that diagonalizes the one-electron (core) Hamiltonian:

$$\begin{align}
\mathbf{H}^{\text{core}} &= \mathbf{T}_{\text{e}} + \mathbf{V}_{\text{ne}} \\
\mathbf{F}^{0'} &= \tilde{\mathbf{S}}^{-1/2} \mathbf{H}^{\text{core}} \mathbf{S}^{-1/2} \\
\mathbf{F}^{0'} \mathbf{C}^{0'} &= \mathbf{\epsilon}^{0'} \mathbf{C}^{0'} \\
\mathbf{C}^{0} &= \mathbf{S}^{-1/2} \mathbf{C}^{0'} \\
P_{\mu\nu} &= \sum_{i}^{\text{occ}} C_{\mu i} C_{\nu i}
\end{align}$$

where the tick mark signifies a quantity is in the orthogonal AO basis, as opposed to the original (non-orthogonal) AO basis. Now assume we're doing a UHF calculation:

$$\begin{align}
\mathbf{H}^{\text{core}} &= \mathbf{T}_{\text{e}} + \mathbf{V}_{\text{ne}} \\
\mathbf{F}^{0'} &= \tilde{\mathbf{S}}^{-1/2} \mathbf{H}^{\text{core}} \mathbf{S}^{-1/2} \\
\mathbf{F}^{0'} \mathbf{C}^{\alpha, 0'} &= \mathbf{\epsilon}^{\alpha, 0'} \mathbf{C}^{\alpha, 0'} \\
\mathbf{F}^{0'} \mathbf{C}^{\beta, 0'} &= \mathbf{\epsilon}^{\beta, 0'} \mathbf{C}^{\beta, 0'} \\
\mathbf{C}^{\alpha, 0} &= \mathbf{S}^{-1/2} \mathbf{C}^{\alpha, 0'} \\
\mathbf{C}^{\beta, 0} &= \mathbf{S}^{-1/2} \mathbf{C}^{\beta, 0'} \\
P_{\mu\nu}^{\alpha} &= \sum_{i}^{\text{occ }\alpha} C_{\mu i}^{\alpha} C_{\nu i}^{\alpha} \\
P_{\mu\nu}^{\beta} &= \sum_{i}^{\text{occ }\beta} C_{\mu i}^{\beta} C_{\nu i}^{\beta}
\end{align}$$

This is misleading! Because we only have one initial Fock matrix, it will only have one set of eigenvectors, so $\mathbf{C}^{\alpha, 0'} \equiv \mathbf{C}^{\beta, 0'}$! If the ranges for forming the two densities are different (as it would be when $N_{\alpha} \neq N_{\beta}$), then having an initial pair of identical MO coefficient matrices is not an issue. For your Huckel guess (called the Generalized Wolfsberg-Helmholtz or GWH guess in Q-Chem, scf_guess = gwh), the initial MO coefficient matrices would still be identical, because you modify a quantity they have in common (the core Hamiltonian).

So, we must modify the MO coefficients (coefficient matrices) themselves. There are three options that I know of:

mixing
rotation
switching

These are variants of the same thing; rotation is the most general, since you can define switching or mixing of two orbitals via rotations.

Mixing

The most straightforward approach of the three that requires no prior knowledge of your system is mixing. The usual choice is to "mix" some amount of the HOMO with the LUMO of the same spin:

$$\begin{align}
C_{\mu,\text{HOMO}}^{\text{new}} &= \frac{1}{\sqrt{1 + k^{2}}} \left( C_{\mu,\text{HOMO}}^{\text{old}} + k C_{\mu,\text{LUMO}}^{\text{old}} \right) \\
C_{\mu,\text{LUMO}}^{\text{new}} &= \frac{1}{\sqrt{1 + k^{2}}} \left( -k C_{\mu,\text{HOMO}}^{\text{old}} + C_{\mu,\text{LUMO}}^{\text{old}} \right)
\end{align}$$

where $k = [0, 1]$. If $k = 0$, nothing happens; if $k = 1$, the HOMO and LUMO become the symmetric and antisymmetric linear combinations of the two, respectively. In theory, this doesn't need to be the HOMO and LUMO; any two MO indices could be chosen, as long as they will result in different densities. This is also only for a single $\mu$; because we want to work with matrices, we would be working with whole rows of the MO coefficient matrix. For Q-Chem, this is $\alpha$ spin only (scf_guess_mix = k); the Gaussian manual doesn't say which spin it takes.

Rotation

The most general form for operating on pairs of MOs is rotation. Recall the rotation matrix in two dimensions:

$$
R(\theta) = \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
$$

This can be used to "rotate" two arbitrary molecular orbitals $i,j$ by an arbitrary angle:

$$
\begin{bmatrix}
C_{i}^{'} \\
C_{j}^{'}
\end{bmatrix}
= \begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
C_{i} \\
C_{j}
\end{bmatrix}$$

where again each $C$ is a row vector. This can be expended into a set of 2 equations:

$$\begin{align}
C_{i}^{'} &= C_{i} \cos(\theta) - C_{j} \sin(\theta) \\
C_{j}^{'} &= C_{i} \sin(\theta) + C_{j} \cos(\theta)
\end{align}$$

There are a few interesting cases.

180 degree rotation: change phase

$$
R(180^{\circ}) = \begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix}
$$

leads to

$$\begin{align}
C_{i}^{'} &= -C_{i} \\
C_{j}^{'} &= -C_{j}
\end{align}$$

This is kind of pointless, since the energy should be invariant to a phase swap.

90 degree rotation: interchange

$$
R(90^{\circ}) = \begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
$$

leads to

$$\begin{align}
C_{i}^{'} &= - C_{j} \\
C_{j}^{'} &= C_{i}
\end{align}$$

This is the usual case when "rotate orbitals" is seen. This is equivalent to switching orbitals. I don't know about other programs, but if you do this in ORCA, the orbital energies will be swapped as well.

45 degree rotation: symmetric and antisymmetric linear combinations

$$
R(45^{\circ}) = \begin{bmatrix}
1/\sqrt{2} & -1/\sqrt{2} \\
1/\sqrt{2} & 1/\sqrt{2}
\end{bmatrix}
$$

leads to

$$\begin{align}
C_{i}^{'} &= \frac{1}{\sqrt{2}} \left( C_{i} - C_{j} \right) \\
C_{j}^{'} &= \frac{1}{\sqrt{2}} \left( C_{i} + C_{j} \right)
\end{align}$$

This last example is identical to the $k = 1$ (100%) case from the mixing example shown earlier.

Switching

We could also alter occupation numbers, such as by placing the $\alpha$ (or $\beta$) HOMO electron into the same-spin LUMO, or some other occupied orbital, but this isn't an automated procedure like orbital rotation, because you need to know your occupied/virtual orbital indices prior to starting the calculation. This is equivalent to switching orbitals.

A short summary of program capabilities (to the best of my knowledge):

Q-Chem: mix, switch
Gaussian: mix, switch
GAMESS: mix, switch
DALTON: switch
MOLPRO: rotate
ORCA: rotate

An important note about mixing: it doesn't matter that choosing the HOMO and LUMO to permute may be incorrect. Programs that have baked-in support for symmetry (DALTON, MOLPRO, unsure about GAMESS) will only mix/switch/rotate orbitals within a symmetry unless symmetry is disabled, but the point is only to break symmetry in the initial guess so that the local RHF minimum is escaped. If you know beforehand that your system has particular symmetries that need to be preserved, then you would probably be doing something more complex than just mixing the HOMO and LUMO anyway.


Writing a organic reaction in different form

It is depend on the reaction. For example, the given two reactions you have sighted are related to each other. It is a example of alkene to alcohol conversion using oxymercuration followed by demercuration with $\ce{NaBH4}$. The overall reaction is electrophilic addition of water to the alkene following Markovnikov's rule (when the nucleophile is water). The first reaction gives an organomercury intermediate, which is not important here for authors to mention. Yet, it is necessary to have a second reaction to convert carbon-mercury bond of the intermediate to carbon-hydrogen bond to get the product of interest. However, it is important to know (and is a common knowledge) that this intermediate should be isolated before treating it with $\ce{NaBH4}$ for following reasons:

One of the byproduct of reaction (1) is acetic acid, which violently react with $\ce{NaBH4}$.
One of the reactant of reaction (1) is water, which also react with $\ce{NaBH4}$.
The other reactant of reaction (1) is $\ce{Hg(OAc)2}$, which also react with $\ce{NaBH4}$.

Nonetheless, any chemist following this reaction sequence knows what hidden behind the sequence. It is also important to know that most organic reactions are written without byproducts, which are not important to the next step. To my understanding, it is okay to write two reaction sequence (even three or more) in a single reaction using steps (1) and (2) and more, as long as consequent intermediates are not important. For example, one of your intermediates $\ce{C}$ and $\ce{D}$ has to be a byproduct. But, if $\ce{C}$ and $\ce{D}$ give $\ce{E}$ and $\ce{F}$, respectively, reacting with $\ce{Y}$, you should show that in your sequence for readers benefits. I you are interesting only on $\ce{E}$ and $\ce{F}$ (together or you have a better separating method for the mixture), you can write it however you want it. The ball is always at your corner, but more explanatory way is always better.

Notes:
Your first reaction doesnt make sense unless $\ce{Y}$ is a catalyst. If not, you should draw it as tri-component reaction:
$$\ce{A + B + X -> C + D}$$
If it is a catalyst, then you can write it as:
$$\ce{A + B ->[\text{cat. } \ce{X}] C + D}$$
An common example for this is:
$$\ce{CH3CO2H + CH3CH2OH ->[\text{cat. } \ce{H2SO4}] CH3COOCH2CH3 + H2O}$$


What's the physicochemical difference between a chromophore and a fluorophore?

The physical chemistry of fluorescence is a little bit complicated, and so you should consult a text book for "the reason" that things fluoresce.

One of the things that I have noticed is that fluorescent compounds often have several "heavy halogens" Br and I attached to some poly-aromatic hydrocarbon.

Disclaimer: this is not to say that the halogens are the fluorescing portion of the molecule (it's probably almost the whole molecule, conjugation + halogen), and it does not mean that all compounds that contain Br and I will be fluorescent. It also does not mean that Br and I are required for fluorescence.

I think that a lot of fluorescent molecules were discovered by chance, or even brute-force changing substituents until the molecule exhibits the desired fluorescence.  (It's really too hard to predict with qualitative theory, and electronic structure theory descriptions of fluorescence are also very difficult.)


What should be the name of this substituted cyclopentane?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred suffix for the principal characteristic group $\ce{{}-CN}$ is ‘carbonitrile’. Therefore, the name of the parent structure of the compound given in the question is ‘cyclopentane-1-carbonitrile’ (note that the locant ‘1’ would be omitted in monosubstituted homogeneous monocyclic rings, i.e. ‘cyclopentanecarbonitrile’).

The relevant rules for the numbering of locants for the four substituent prefixes (‘amino’, ‘bromo’, ‘chloro’, and ‘hydroxy’) are:

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

and

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule (f) takes precedence over Rule (g).

In accordance with Rule (f), the compound given in the question could be named as 2-amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitrile as well as 5-amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile since both names correspond to the locant set ‘2,3,4,5’. However, according to Rule (g), this example is named as 2-amino-4-bromo-3-chloro-5-hydroxycyclopentane-1-carbonitrile rather than 5-amino-3-bromo-4-chloro-2-hydroxycyclopentane-1-carbonitrile since ‘amino’ is cited first as a prefix in the name.


Why is sulfuric acid used in crystallization

The sulfuric acid will dissolve any residual copper metal or any copper oxides that exist as impurities in the copper sulfate, converting those impurities into copper sulfate.

$$\ce{H2SO4(aq) + Cu(s) -> CuSO4(aq) + H2(g)}$$
$$\ce{H2SO4(aq) + CuO(s) -> CuSO4(aq) + H2O(l)}$$
$$\ce{2H2SO4(aq) + Cu2O(s) -> 2CuSO4(aq) + H2O(l) +H2(g)}$$


Question about reactions that are always spontaneous and non spontaneous

I was thinking about the boiling of water

That is a good example of a process that goes forward at higher temperature (water boils when heated above the normal boiling point) and goes in the opposite direction at lower temperature (steam condenses below the normal boiling point). This process also depends on the pressure, but my statements are valid at atmospheric pressure.

For this process, the entropy change is positive (water has more freedom to move in the gas phase compared to the liquid phase) and the enthalpy change is positive (endothermic, it takes energy to break the hydrogen bonds between water molecules).

  Reactions that are always spontaneous

How about combustion of cyclohexane? The entropy change is positive (lots of gas molecules are formed - six carbon dioxide molecules per hexane) and the enthalpy change is negative (exothermic).

  Reactions that are always non-spontaneous

Once you are certain that a reaction is always spontaneous, it is easy to find one that is always non-spontaneous - just swap reactants and products. So according to this argument, cyclohexane and oxygen will never form from carbon dioxide and water spontaneously. That sounds reasonable.


Could anyone suggest some catchy chemistry experiments or demonstrations, to be performed in front of a young audience?

To talk to kids about chemistry I don't think you just need some exotic experiments.

I visited the Sterling Hill Mining Museum and one of founders had a neat gimmick that got kids attention. He asked them to name some product that had nothing to do with mining. He'd then rattle off a number of things that are mined to make that product. So for a truck the oil and gas are from an oil well. The steel comes from an iron mine and the copper in the electrical wires comes from a copper mine. So have the kids try to name a product that doesn't use chemistry.

Another thing that has always bugged me is the perception of the general public about "chemicals." It seems that the public believes that only things created in some chemical processing plant are chemicals. Well table salt is a chemical. Sugar is a chemical. A strawberry is a bag of water with a zillion chemicals in it. Take a vitamin tablet? The tablet is full of chemicals that our bodies need. Our bodies run on chemistry. So chemicals are all around us.


Where does Oxygen come from in KO2 reaction with CO2?

Yes, the oxygen indeed comes from the decomposition of potassium superoxide ($\ce{K2O}$) but the other decomposition product is potassium peroxide ($\ce{K2O2}$) and not potassium monoxide($\ce{K2O}$).(Here)

$$\ce{2KO2 ->[290°C, vacuum] K2O2 + O2~~~~~~~~~~(1)}$$
Now, this potassium peroxide reacts with carbon dioxide to form potassium carbonate and oxygen.(Here)

$$\ce{2K2O2 + 2CO2 → 2K2CO3 + O2~~~~~~~(2)}$$

Adding $\ce{(1)\times2}$ and $\ce{(2)}$, we get full reaction. (Here)

  $$\ce{4KO2 + 2CO2 → 2K2CO3 + 3O2}$$

If reaction proceeds in aqueous medium, it gives potassium hydrocarbonate and oxygen.(Here)

$$\ce{4KO2 + 4CO2 + 2H2O → 4KHCO3 + 3O2}$$


The easiest way to measure the pH of a solution is to use an acid-base indicator. Limited to household supplies only, I would suggest a solution made from red cabbage, which should turn blue-greenish in the alkaline region.

The ions you actually get from your reaction is a higher concentration of $\ce{Na+}$ and $\ce{OH-}$, due to the fact that you are removing $\ce{H2}$ and $\ce{Cl2}$ (from $\ce{H+}$ and $\ce{Cl-}$ ions respectively) via the performed electrolysis. Under no circumstances would you generate a $\ce{Na2}$ species, you would rather write it as $\ce{2Na}$. Also, in aqueous solutions the $\ce{NaOH}$ most certainly dissociates. So speaking of the formation of solid $\ce{NaOH}$ from $\ce{Na}$ in water is quite a stretch.

Bottom line: Since the salt is already dissolved when you start the electrolysis, all you get is a change in concentration of the ions involved. The thereby induced change in pH can be measured observed using an acid-base indicator such as the juice of the red cabbage.


Not glass per se, but silica is often produced chemically by silicon alkoxide gelation to create aerogels.

The problem with creating glass by some cold process would be that the reaction side-products, such as water, would have to be removed, leaving holes in a froth of glass. This is fine for creating an aerogel, but is not helpful in making solid objects, such as test tubes or lenses. Of course, you could sinter the glass frit or froth, but that still requires heating.


Why not simulate every particle properties and interactions?

Forty electrons is tiny. Even if we limit ourselves to just the valence electons, cyclohexane already has 36 electrons. Anything drug-like has way more electrons that 40. For example, viagra has 178 valence electrons, and that's not necessarily a "large" drug. (Compare with vancomycin, for example.)

Even if you're dealing with things inorganic compounds, where the total number of atoms in the formula unit is small, the properties of the material don't come from a single formula unit, but come from the interaction of a large number of atoms. -- That's an example of a more general principle. The important properties for most materials you use (including drugs) don't come the molecule in isolation, but come from the interactions of the molecule with other molecules, either of the same chemical or of different chemicals. To be accurate, all of those interactions need a system with much more than 40 electrons.

The 40 electron limit comes from the implict assumption here that you're talking about quantum mechanical calculations. QM calculations are rather computationally expensive, as you have to account for all the interactions of all the electrons with each other at all positions in their delocalized superposition. There's various tricks (like DFT) which make the calculations for large numbers of electrons easier, but note that "easier" doesn't mean "easy". Even with DFT and other approaches, large systems take a lot of computer time to calculate accurately.

There are other approaches which don't suffer from the same limit as QM does, but they are able to make their gains in efficiency because they make approximations. For example, molecular mechanics approaches are able to simulate systems in the hundreds of thousands of atoms region. But they're able to do so because they don't actually calculate the position of electrons. Instead they treat the system "classically", and experimentally fit interaction potentials which approximate the underlying quantum effects. (For example, they don't exactly calculate the bond stretching potential, but instead approximate it as a harmonic one. That's "close enough" to the true bond stretching potential for the range of bond lengths typically seen in such simulations, but not 100% quantum mechanically accurate.)

There's many groups and companies which do use molecular mechanics and other similar approaches to inform their drug and material development process. The issue is that because the energetic potentials being used are only approximate the results from the simulation are also only approximate. Depending on what you're trying to simulate, the results of the simulation may or may not be accurate. As such, these simulations are treated mostly as a first step, to find potential leads/hypotheses, and then the scientists actually have to go into the lab and test the results to confirm.


dipole–dipole interactions: OCR Sample Paper wrong?

Permanent Dipoles occur when two atoms bonded to each other have significantly different electronegativities. When the atomic number of an element increases the number of electrons present in them also increases. These electrons are constantly moving around in the atom. At some point, there are more electrons on one side of the atom than the other. This creates an imbalance of charge, creating a small charge of $\delta ^+$ and $\delta ^-$. These small charges induce its neighbors to create additional charges of $\delta ^+$ and $\delta ^-$, thus creating a chain of induced dipoles. These induced dipoles are temporary as the electrons can move, and hence can be reverted. They are, therefore, known as induced dipole forces, or better, known as London Dispersion Forces. In reality, they are not really forces, but rather attraction between molecules.

Naturally, more the number of electrons in the atom, more the amount of London dispersion forces, and hence greater the attraction. The greater the attraction, the more energy you have to put to break the bonds apart. Between the considered hydrogen halides ($\ce{HBr}$ and $\ce{HI}$), iodine has more electrons than bromine, and hence $\ce{HI}$ has a greater attraction force between molecules than $\ce{HBr}$. As a consequence, $\ce{HI}$ has a greater boiling point between them. Appropriate values are listed in following table:

$$\begin{array}{cccccc}
\hline
\ce{H-X} & \text{BP}^\text{Ref.1} & \ce{X}\text{'s EN} & \mu ^\text{Ref.2} & \text{Polarizability }(\pu{10^{-24} cm^3})  & (r_\ce{H-X})^\text{Ref.3} \\ \hline
\ce{HCl} & \pu{-85.1 ^\circ C} & 3.16 & \pu{1.09 D} & 2.63 & \pu{1.27 Å} \\
\ce{HBr} & \pu{-67.1 ^\circ C} & 2.96 & \pu{0.82 D} & 3.61 & \pu{1.42 Å} \\
\ce{HI} & \pu{-35.1 ^\circ C} & 2.66 & \pu{0.45 D} & 5.44 & \pu{1.60 Å} \\
\hline
\end{array}$$

Moreover, a related question you can find here. Orthocresol has given very reliable answer, which included two equations to calculate the strength of permanent dipole interactions ($U_\mathrm{dipole}$) and the strength of dispersion interactions ($U_\mathrm{dispersion}$). The author has stated that:

  If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.

For example, for $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.

References:

J. F. Ogilvie, W. R. Rodwell, R. H. Tipping, “Dipole moment functions of the hydrogen halides,” J. Chem. Phys. 1980, 73(10), 5221 – 5229 (https://doi.org/10.1063/1.439950).
N. N. Greenwood, A. Earnshaw, “Chapter 17 – The Halogens: Fluorines, Chlorinee, Bromine, Iodine, and Astatine,” In Chemistry of the Elements; 2nd Edn.; Elsevier: Boston, MA, 1997, pp. 789–887.
K. Fajans, N. Bauer, “Electronic Structure and Stability of Hydrogen Halides and of Complex Ions $\ce{XO4}$,” J. Chem. Phys. 1942, 10(7), 410 – 415 (https://doi.org/10.1063/1.1723742).


Yes, "to the left" refers to the left side of an equilibrium expression.

Traditionally, the autoionization of water is written as

$$\ce{2H2O <=> H3O+ + OH-}$$

When we talk about equilibrium lying "to the left", it means that the educt/reactant is favored, i.e. more $\ce{H2O}$ than $\ce{H3O+}$ or $\ce{OH-}$.

Conversely, an equilibrium that lies "to the right" is one that favors products. In this case, it would favor the production of $\ce{H3O+}$ and $\ce{OH-}$ over $\ce{H2O}$,


protecting contact between aluminium and copper wires

This may depend on your location and local electrical codes.  I have the same issue in my house.  In the USA, there are NEC (National Electrical Code) compliant wire nuts rated for Al to Cu connections.  These come pre-packed with a protective goop. The goop can also be purchased separately in a tube for use with ordinary wire nuts. You may have to ask at your local electrical supply store.  In my area the pre-packed wire nuts are a distinctive purple color not found on regular wire nuts (for example: Home Depot Al/Cu nuts).

For code compliance, use approved solutions only, even if you think you know more chemistry than the electricians. Having your house burn down from a wiring issue is no fun.


What determines the electronegativity of an element?

There are various slightly different electronegativity scales.  The two most common are the Pauling scale and the Mulliken scale.  Of the two the Mulliken is the simplest.

The Mulliken scale is based on two quantities, both measurable.  One is the ionization energy $(E_i)$, the energy required to remove an electron from a neutral atom.  The other term is the electron affinity $(E_a)$.  That is the energy released when the neutral atom gains an electron.

The Mulliken electronegativity $\chi$ is then given by

$$
\chi = (E_a + E_i)/2
$$

It is that simple!

For more details and explanation see

http://en.wikipedia.org/wiki/Electronegativity


The reaction is $\ce{NaOH + CH3CH2COOH -> Na+ + H2O + CH3CH2COO-},$ so adding $\pu{0.04}$ moles $\ce{NaOH}$ will result in the solution having $\pu{0.06 +0.04 = 0.1 mol CH_{3}CH_{2}COO^-}$ and $0.05 - 0.04 = 0.01\, \pu{mol} \, \ce{CH3CH2COOH}.$ Then, you can use the Henderson-Hesselbach equation to find the final pH is equal to $-\log(1.3*10^{-4})+\log\left(\frac{0.1}{0.01}\right) \approx 4.886.$


What you have is classified as a Mg-air battery. You can read more about it here : https://en.wikipedia.org/wiki/Category:Metal-air_batteries

These are a class of primary(i.e. not rechargeable) batteries, where the anodic reaction is a metal being oxidised and the cathodic reaction is oxygen reduction.

The reactions are:

$\ce{Mg → Mg^{2+} + 2 e^{-}}$ on the anode

$\ce{O2 + 4 e^{-} + 4 H+ → 2 H2O }$ on the cathode, if the reduction goes to completion. Most of the time, it is only reduced to the peroxide or the superoxide which is considered incomplete reduction.

For it to be classified as a fuel cell, the fuel for the anodic reaction would need be brought from the outside(source: anectodal, personal thought).


Understanding the setup of a chemistry lab about redox

I'm wondering why is it not necessary to mix a metal like Fe into its own solution $\ce{Fe(NO3)2}$ (like you would do when setting up an electrochemical cell)?

First remember what the purpose of your experiment was:

  To determine the relative activity of several metals we observed reactions between the metals and aqueous solutions of dissolved metals.

In an electrochemical cell setup, $\ce{Fe}$ would be in two-way equilibrium with $\ce{Fe^2+}$:

$$\ce{Fe + Fe^2+ <=> Fe + Fe^2+}$$

When power was drawn from the cell you then get:

$$\ce{Fe -> 2 e- + Fe^2+}$$

Where as with another salt such as copper which is less reactive, the copper will oxidize the iron in one direction with no reverse reaction.

$$\ce{Fe + Cu^2+ -> Cu + Fe^2+}$$

or with zinc which is more reactive no reaction will occur at all:

$$\require{cancel} \ce{Fe + Zn^2+ \cancel{\ce{<=>}} Zn + Fe^2+ ->Fe + Zn^2+}$$


Should bromine water be called a solution?

Much like with chlorine or iodine, the disproportionation of bromine can occur in pure water, but it is not fast. As you can see, the reaction generates acids, so without any base present the equilibrium is pretty close to the bromine side. You do need bases to disproportion bromine in any meaningful amount — much like chlorine which is generally quenched with $\ce{NaOH}$ solution to generate hypochlorite solutions.

So if you just put bromine and water together, they may react very slightly, but not much, and most of the bromine will sink to the bottom (assuming you used a lot) while the lesser part will dissolve in the water first. Since the concentration of $\ce{Br2}$ is much higher (and we’re talking orders of magnitude) that those of $\ce{Br-}$ and $\ce{OBr-}$, it is best to consider this a bromine solution and not a bromide/hypobromite one.


What happen to the Salt in Water Electrolysis

When a battery is connected to an electrolytic cell one electrode becomes positively charged the other negatively charged. The ions in the liquid are attracted to the electrode with the opposite charge and move through solution. Ions with a negative charge move to the positive anode (anions), Ions with a positive charge migrate to the negative cathode (cations).

Ionic solids produce electrolytes. When an ionic solid is dissolved in water the aqueous ions are free to move and they carry the current. The anions of the salt and water give up electrons which flow through the circuit. The anions are oxidized at the anode while the cations of the salt and water recieve the electrons and are reduced at the cathode.

Cations will cluster around the cathode and anions will cluster at the anode. It is usually easier to displace hydrogen atoms from water molecules in the aqua ion than to displace the cation from the system  so electrons from the cathode react with water molecules , and hydrogen is displaced.

The result is decomposition of water molecules:

2H-OH(l) + 2e- → H2(g) + 2OH-(aq)

The solution becomes alkaline around the cathode due to the hydroxide ions being formed. If a metal is above hydrogen in the electrochemical series, hydrogen will always be produced at the cathode during electrolysis of solutions of its salts. If the ionic salt consists of a metal below hydrogen in the electrochemical series, electrolysis of its salt solution will always produce metal at the cathode.

Both ion flow and electron flow are occurring.

The intensity of an electromagnetic field reduces significantly the further away from the source.


Solubility of Aluminium Hydroxide in Sodium Hydroxide Solution

That's because $\ce{Al(OH)3}$ is amphoteric in nature, it acts as both acid and base. When you drop a chunk of $\ce{Al(OH)3}$ into a solution of $\ce{NaOH}$, you would get a soluble salt, which is sodium aluminate, and some extra water.

$$\ce{Al(OH)3 + NaOH -> Na[Al(OH)4]}$$


Are decomposition and dissociation reactions the same?

Decomposition and disassociation refer to similar processes, however they are different. The main difference between them is that decomposition reactions are usually irreversible while disassociation reactions are reversible and exist in equilibrium.

A decomposition reaction is where a compound breaks up into smaller elements or compounds. For example: $$\ce{2H2O -> 2H2 + O2}$$ is an example of a decomposition reaction. Note that in this case that this is actually a reversible reaction.

A disassociation reaction is the general process where molecules split into smaller particles which can include elements, compounds, ions and radicals. For example: $$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$ is a disassociation reaction.


What is the cause of decrease in volume of gaseous mixture in the following problem?

I think the fact you are missing is that after combustion, water vapors will be given away and the volume lost during combustion will be equal to the amount of water lost. $\ce{KOH}$ will absorb $\ce{CO2}$ and the volume absorbed by $\ce{KOH}$ will be equal to volume of $\ce{CO2}$ in the solution. Thus, you will have two equations and you will be able to proceed with the solution.


Determining number of electrons transferred

When determining the number of electrons transferred in a redox reaction is it the total in both half equations?

In the total redox reaction that is properly balanced, the number of electrons transferred is equal to the stoichiometric coefficient of the $\ce{e-}$ species in either balanced half-reaction.

  So would the number of electrons transferred in the redox reaction be two or four?

Your half-reactions and total redox reaction happen to be balanced, to the number of electrons transferred in this case is two.


Comparing basic strengths between pyridine and 1,2-dihydropyrazine

What I would do is I would compare their acid strengths. Due to inductive forces, deprotonating the 1,2-dihydropyrazine would result in a more stable compound than that in pyridine--the nearby nitrogen (electronegative) will share the burden of the negative charge. Therefore, 1,2-dihydropyrazine is more acidic and thus is less basic.


This question probably cannot be answered correctly without conducting any experiment.

However, I would not expect that any of the rocks given would be dissolved by highly concentrated sulfuric acid. None of them has a composition that may completely result in the formation of soluble salts when reacting with sulfuric acid.

Presumably (hot?) dilute sulfuric acid may be able to dissolve parts of the rocks if they are powdered. E.g. $\ce{Na2O}$, $\ce{K2O}$, $\ce{MgO}$, and up to a small amount $\ce{CaO}$ and $\ce{Al}$.

Hydrochloric acid would do better since it forms soluble salts not only with Na, K, Mg, Fe, and Al but also with Ca. E.g. it could almost completely dissolve limestone, dolomite, and marble.


Rusting of an Iron nail at oil water boundary surface

According to your description, this looks like localized corrosion. Due to differences in the availability of oxygen in the solution, the two half-reactions of the redox process take place on different areas of the surface of the corroding metal. The oxygen reduction reaction preferably takes place on the nail surface in the medium with higher concentration of dissolved $\ce{O2}$ (water).

$$\ce{O2 + 4e- + 2H2O -> 4 OH-}$$

The metal dissolution (oxidation) is then limited to the more oxygen-starved surface area, which is presumably right at the the phase border, where consequently the deposit of corrosion products is formed.

\begin{aligned}
\ce{Fe &->~ Fe^2+ + 2e-}\\
\ce{Fe^2+ + 2OH- &->~ Fe(OH)2 \rightleftharpoons FeO + H2O}\\
\end{aligned}

Since these two half-reactions are coupled, the amount of corrosion products formed depends on the oxygen concentration. When there is sufficiently dissolved oxygen available, $\ce{Fe^2+}$ is easily oxidized by to $\ce{Fe^3+}$, and thus the final product, which is commonly called rust, is a mixture of more or less hydrated iron(III) oxides and hydroxides.

$$\ce{Fe^3+ + 3OH- -> Fe(OH)3 \rightleftharpoons FeO(OH) + H2O}$$
$$\ce{2FeO(OH) \rightleftharpoons Fe2O3 + H2O}$$


Is there any software to simulate a laboratory?

There is a product called Yenka Chemistry which simulates basic chemicals/apparatus, and allows basic manipulations (for instance to show the pH change as more acid is added to base, or the amount of gas evolved).

The program evolved from another piece of software called Crocodile Clips, which was aimed at GCSE science in the United Kingdom. For this reason it is incredibly basic, mainly focusing on inorganic reactions (acid/base, titrations, reactivity of group 1 and 2 metals), with no real capacity for demonstrating organic reactions.


Aromaticity of 1,1'-bi(cycloprop-2-en-1-ylidene) and 1,4-dioxine

Antiaromaticity is a fictional state with real consequences,  that any molecule will distort to avoid.

In the case of molecule 1, the primary state will be a triplet (unpaired pi-electrons), with the central bond lengthening to decrease the pairing of the pi-electrons.  This will lead to 3 electrons in each pi-system, and hence preventing antiaromaticity, because there are no longer 4n electrons in the rings.  A fictional molecule, with the constituency shown for molecule 1 that is in a singlet state would be antiaromatic.

In the case of molecule 2, the molecule will distort itself to make the at least one of the oxygens non-coplanaer with the pi-system, and no longer being co-planar, the system will not be antiaormatic.  A fictional molecule with the constituency shown for molecule 2 that is co-planar, would be antiaromatic.

It is unclear what you are asking about for pyrene.


Yes, you are indeed right in the comment section that the reaction between thionyl chloride and ammonia will give trace amount of tetrasulfur tetranitride(1) and the intended product is thionyl imide ($\ce{HNSO}$)(2).

From the 'thionyl chloride' entry from the same site you used
in the question:

  Thionyl chloride reacts with ammonia giving rise to various complex
  products, the nature of which varies with the conditions.

As said, the main product from the reaction in gas phase is thionyl imide and trace amount of thiazyl chloride($\ce{NSCl}$) and sulfur dioxide. The reaction are as follows:

$$\ce{SOCl2(g) + 3NH3(g) -> HNSO(g) + 2NH4Cl(s)}$$
$$\ce{HNSO + H2O -> SO2 + NH3}$$

Ref. : Inorg. Chem., 1977, 16 (1), pp 216–217
DOI: 10.1021/ic50167a050


Why is dinitrogen tetroxide a planar molecule?

I think that one would expect that a stable form of the molecule might be planar and have D2h symmetry.

Some support for the claim:

Early crystal structures suggested that the ONO bonds on either side
of the N-N bond are easily modeled by low level DFT
(doi.org/10.1039/C19660000082).
More recently, simulations of the trans and cis N2O4 structure show
planarity when optimized at B3LYP/ aug-cc-pVTZ (aka, pretty good not
amazing) See here.

The simulations also suggest that the barrier to isomerization is low, meaning that while the stable structures are planar, there are low barrier (low temperature) out of plane rotations possible.

Summary:

Just because something has sigma bonds doesn't imply that it will immediately be non-planar
yes the molecule can form a conformer


Colorimetric test to measure dissolved oxygen in salt water

Unfortunatly I don't think there is an easy colorimetric technique available.

There are reagents that are strongly colored and oxygen-sensitive.  One exampleis reduced methylviologen radical.  That species can be prepared by mixing methylviologen with sodium dithionite in aqueous solution.  The one-electron reduced species is intensely blue.  It is rapidly oxidized by oxygen, leading to a loss in color.  In theory, you could use the degree of this color loss to assay DO.

The problem in real life with applying this reagent to DO determination is that oxygen in the air will also react, very quickly.  So you would have to put your seawater sample under an inert atmosphere (and also prepare the reduced MV radical solution under an inert atmosphere).  That is very inconvenient, and could create problems where your seawater sample outgases oxygen into the anaerobic environment before or during measurement.


Why H2O2's molecular geometry is what it is...?

As stated in the link provided by @Shadock, hydrogen peroxide, like many more other molecules, is subject to hindered internal rotation. When you rotate the two OH groups with respect to each other around the O-O bond, this requires a certain energy depending on the relative angle. In the eclipsed configuration (OH groups pointing in the same direction), it's easy to see that there must be a maximum in the potentiall due to the Paulli repulsion. In the anti-eclipsed configuration (both OH groups pointing in opposite directions) there is another maximum as you anti align the two dipoles of the groups. In between these maxima, you'll find a minimum and the potential can be expanded as a Fourier series in the relative bond angle

$$V(\gamma)=\frac{V_2}{2}\cos(2\gamma)+\frac{V_4}{2}\cos(4\gamma)+\ldots$$

where $V_2$ and $V_4$ are related to the heights of the two barriers. If the barriers were infinitely high, the OH groups would act as a quantum harmonic oscillator in one of the potential minima. On the other hand, if the barrier were very small, the OH groups would rotate freely around each other. In the case of hydrogen peroxide, the eclipsed barrier is so high that the OH groups can hardly go over/through it. The lower barrier is a different story. It is high enough to support a few vibrational levels, but not high enough to prevent the OH groups from tunnelling through the barrier and as a consequence, these energy levels are split in two. (In fact, these two levels are again split by the tunnelling through the higher barrier, but since this is so high the splitting is very small). Mathematically, the Schroedinger equation for this periodic potential is equivalent to the Mathieu equation (or more general Hill's equation).

In the case of H$_2$O$_2$ the tunnelling is so fast, that the molecule vibrates around the lower barrier with a period of ~100 fs. As a consequence, to determine the polarity, you have to average over this motion, resulting in a net polarity.

Note that, if the barrier would be so high that tunnelling would be sufficiently slow, H$_2$O$_2$ would be a chiral molecule!


Does the weird 4s and 3d rule apply for much of the p-block from Ga-Rn?

Well the only exception in the orbitals is for $s$ and $d$ orbitals. Meaning you will first remove electrons from $(n+1)s$ and then onto $nd$ orbital. It is only because the $(n+1)s$ sublevel is only lower in energy if there are no electrons in the d sublevel. As per the Aufbau principle, the the s orbital is therefore filled before. Then as the $d$ becomes filled with electrons, then the energy starts to flunctuate relative to one another and the s orbital ends up higher in energy. That is why when electrons are removed, they are first removed from s and then $d$. Luckily this only applies to $s$ and $d$ orbitals. When are you working with $p$ orbitals, well then the $np$ orbital will always be higher than the $(n-1)d$ orbital. Therefore for making ions, you will have to first removed electrons from $p$ then $s$ and then $d$ and so on.


Error in formula for the solution to a chemical problem

The equation is a little dubious, as I mentioned in the comments.

  How do we come up with coefficients of this sort?

Let's go from scratch. We don't know the coefficients at this point so we'll just call them $a$, $b$, and $c$. Note that the coefficient of $\ce{S$_x$H$_y$O$_z$}$ is taken to be 1. It doesn't have to be 1; however, for the sake of simplicity you might as well take it to be 1. If you take it to be 2, then you just have to double all the other coefficients.

$$\ce{S$_x$H$_y$O$_z$ + $a\,$O2 -> $b\,$SO2 + $c\,$H2O}$$

In the reactants, we have $x$ sulfur, $y$ hydrogen, and $z + 2a$ oxygen. In the products, we have $b$ sulfur, $2c$ hydrogen, and $2b + c$ oxygen.

Our aim is to express $a$, $b$, and $c$ in terms of $x$, $y$, and $z$. Equating all three, we get

$$\begin{align}
x &= b \\
y &= 2c \\
z + 2a &= 2b + c
\end{align}$$

which immediately tells you that $b$ and $c$ are equal to $x$ and $y/2$ respectively. Substituting these into the last equation,

$$\begin{align}
z + 2a &= 2x + \frac{y}{2} \\
a &= \frac{4x + y - 2z}{4}
\end{align}$$

So the full equation is:

$$\ce{S$_x$H$_y$O$_z$ + $\frac{4x + y - 2z}{4}$O2 -> $x\,$SO2 + $\frac{y}{2}\,$H2O}$$


Bubbles need nucleation points to form.
The growth of bubbles is limited by the surface tension of the liquid. The surface tension acts stronger on smaller bubbles and less strong on bigger bubbles.
Adding a solid salt to sparkling water means adding a lot of new nucleation points, which subsequently will lead to bubbling.
The small bubbles will combine to bigger bubbles, these bigger bubbles are not so much effected by surface tension and so even more gas is produced.
In other words, adding the salt will trigger a kind of avalanche effect, which weakens either if the dissolved CO2 gets towards used up or the salt gets dissolved completely.

It is not an reaction of dissolved potassium ions with the CO2, nor is it a reaction of dissolved citrate ions with the CO2.
What happens here is a physical reaction and not an chemical reaction.


No, the paper will not burn without oxygen being present.

Paper is made primarily of cellulose which is a polymer of glucose.

If you heat paper in a vacuum the cellulose simply decomposes to $\ce{H2O}$, $\ce{CO2}$, $\ce{CO}$ and carbon.  As the paper decomposes it will "char" or turn brown to black as the cellulose polymer degrades.  Here is a link to an abstract describing the process in more detail.


Can the compounds NOF3 and POF3 both exist?Why?

Both molecules can be considered the acid (hydrate) trifluoride of the respective acid (phosphoric acid or nitric acid). In the case of phosphoric acid, this is immediately obvious: Just replace the three hydroxide groups with a fluorine atom, each. For the nitrogen compound, you initially need to (formally) add water before you can do the substitution:

$$\ce{O=N+(OH)-O- + H2O -> [(HO)3-N+-O- ] -> F3N+-O-}$$

(Note: this is not a proper chemical reaction, merely an illustration of the thought processes.)

You could also approach this compound from the ammonium cation $\ce{NH4+}$ by replacing three hydrogen atoms with fluorine and the fourth with oxygen. Or you start off with $\ce{NF3}$ (isoelectronic to $\ce{PF3}$ and then oxidise the nitrogen by adding a single oxygen atom much in the same way you would oxidise from $\ce{HNO2}$ to $\ce{HNO3}$.

In fact, for many of the compounds traditionally explained with d-orbital participation for main-group elements, an explanation adhering to the octet rule (no more than eight valence electrons per main-group atom; two for hydrogen/helium) is often closer to the truth. This includes, but is not limited to $\ce{OPF3, SO4^2-}$, sulphonic acids and many more.


Your analysis was a good start, but then it dropped. You need to consider four electrons on each carbon atom. Each of the two carbons can then form a single bond to one hydrogen atom using one of its electrons. Each of the carbon atoms then uses two of the remaining electrons to form the double bond.

Thus, each of the carbon atoms has one extraneous electrons; the structure would be a diradical structure. It should better be represented e.g. like this: $\ce{H-C^.=C^.-H}$.

It is possible for these two radicals to be alongside each other — see for example the structurally strongly related dioxygen molecule which is a triplet (diradical) in ground state and not adequately described in simple Lewis structure formalisms. However, it is highly unlikely for two carbons without additional constraints to form a diradical like that; forming a triple bond ($\ce{H-C#C-H}$) would be much better energetically (for carbon!).


Why is the square of the wavefunction a probability density?

First, I would not call the Born rule a law as it is done in the Wikipedia article linked by OP in the question. It is, of course, the question of the definitions, but I would call the Born rule rather a postulate. A law in physics is a principle (usually universal) deduced from observations, while a postulate is an assumption, a statement that we can not prove or derive but which is assumed to be right and tested to be consistent with experimental observations.

The Born rule is a postulate, thus, there is no any proof of it; it is assumed to be right. The only thing we can do is to justify a postulate: we can make experimentally verifiable predictions on the basis of it and test them. And it happens so that so far the experimental outcomes are always in full agreement with the predictions done using the Born rule. Thus, we know it provides a correct description of physical systems.


Why should we be interested in catalyst for electrolysis?

Catalysts generally decrease the electrode overpotential, which leads to decreasing the needed voltage to perform electrolysis at given current densities toward its minimal theoretical value.

If higher current is possible at the same voltage due the catalyst, the lower voltage is needed for the same current, both leading to better efficiency.

As the energy needed to produce 1 mol of molecular hydrogen is

$$E=2 \cdot U \cdot F$$

where
$U$ is the used voltage
$F$ is the Faraday's constant cca 96490 C/mol

As the consequence, it has direct impact on efficiency of conversion of electricity to hydrogen for hydrogen based engines or just fuel.


What do the six polymorphs of chocolate look like?

The actual polymorphic compound is the fat (cocoa butter). They pure phases will all look like white powders and the 'bloom' referred to in your diagram is white cocoa butter extruded by the solid chocolate, and probably the most obvious  visual difference.

The melt-in-your mouth aspect of Form V is going to be because of its appropriate melting point w.r.t your mouth temperature. The decrease in mechanical stability as the melting point decreases is fairly typical of solids (think of plastics softening as you heat them up).

Afraid I don't know why Form V is shinier.

C.f. https://www.sciencedirect.com/science/article/pii/S0969806X04003019


Reason for the generation of London Dispersion Forces

You are correct that simply stating that the Schrodinger equation is time
dependent does not really answer the question.
In fact, time-dependence is not a component of the derivation of the dispersion forces. At a surface level, they just represent how much the ground state energy of a system of polarizable molecules is lowered by their interactions.
It's the same idea as any of the other intermolecular forces (coulomb, dipole-dipole, etc.) where some interaction between particles leads to a lower energy state than the free particles.

Generally, the London Dispersion interaction potential is not derived in a general chemistry class because it is purely quantum mechanical and rather involved. These slides provide a clear and thorough (albeit advanced) derivation, but I will explain the main points below.

Derivation Explanation

Let's imagine for this model that a polarizable atoms can be thought of as a simple harmonic oscillator with a positive charge ($+q$) on one end of a spring and a negative charge ($-q$) on the other end. The spring has an equilibrium length of 0, so that the atom has zero average dipole.
This model of an polarizable atom captures the physics we want because applying a force of some kind to the spring will induce a dipole in this atom  ($\vec{p} = q \vec{d}$).

So far we have only considered one of these atoms. What if we move to consider multiple of them together?
Remember we're really interested in the energy difference between two non-interacting atoms and two interacting atoms, so let's start by finding the non-interacting ground state energy of two atoms. We'll let the two atoms have two different spring frequencies $\omega_1$ and $\omega_2$ respectively.

Setting $z_1$ and $z_2$
as the displacements of the springs from equilibrium for atoms 1 and 2 respectively and $m$ as the mass of an atom,
we can write the total potential energy of the system as:
$$
U_{\text{non-interact}} = \frac{1}{2}\left(m \omega_1^2z_1^2 + m \omega_2^2z_2^2\right)
$$

Solving the Schrodinger equation for this potential yields the ground state energy for the two non-interacting harmonic oscillators:
$$
E = \frac{\hbar}{2} (\omega_1 + \omega_2)
$$

Now, we turn on the interaction and the problem becomes harder more interesting because the positions of the springs affect the magnitude of the interaction. Physically, because the dipole-dipole interaction can lower the energy, we expect that the interacting ground state of the two atoms will be a balance between minimizing the stretching of the springs (which increases the energy) and maximizing the amount of dipole-dipole interaction (which lowers the energy).

Setting $z$ as the distance between the two atoms, with some algebra one can show that the interaction energy between these two dipoles is given by
$$
U_{\text{interact}}(z,z_1,z_2) = \frac{q^2}{2\pi\epsilon_0}\frac{z_1 z_2}{z^3}
$$

Now, our total potential energy for the system is given by:
$$
U_{\text{total}}(z,z_1,z_2) = U_{\text{non-interact}}(z_1,z_2) + U_{\text{interact}}(z,z_1,z_2)
$$
While intimidating looking, this potential can also be solved exactly by changing variables to $z_+ = (z_1 + z_2)$ and $z_- = (z_1 - z_2$) and rearranging terms to make it look like the non-interacting potential energy. Once that is done the ground state energy can be found just as easily as in the non-interacting case.

With this done, we can find that the (one dimensional) energy stabilization is given by:
$$
U(z) = \frac{-\alpha^2 I}{2}\frac{q^2}{(4\pi\epsilon_0)^2}\frac{1}{z^6}
$$
Where the final step was to replace some of our less measurable parameters with their atomic counterparts: polarizability ($\alpha$) and ionization potential ($I$)


Effect of removing the product from equilibrium

First, let’s analyse the situation the graph presents. Obviously the reaction is (just restating for clarity):

$$\ce{PCl3 + Cl2 <=> PCl5}\tag{1}$$

We have already removed some $\ce{PCl5}$ from the mixture and are now looking at the self-reestablishing equilibrium. Obviously, for each molecule of $\ce{PCl5}$ we create we lose one molecule of $\ce{PCl3}$ and one of $\ce{Cl2}$ — this is the information we gain from the reaction coefficients (they are all $1$). This can then be expanded to the amount of substance reacting, namely $$\Delta n(\ce{PCl5}) = -\Delta n (\ce{PCl3}) = -\Delta n(\ce{Cl2})\tag{2}$$

Your graph uses concentration $c$ as $y$ axis. Concentration is defined as:

$$c = \frac nV\tag{3}$$

Since we can assume the volume $V$ to be constant, we see that concentration and amount are simply related by a constant factor; thus any differences in concentration will be analogous to differences in amount. Therefore, the labelled $\Delta x$ in the graph are indeed all identical as the represent the same concentration difference and thus the same difference in amount; and we have shown in equations $(1)$ and $(2)$ that the amounts are identical.

What about a case such as your hypothetical following case:

$$\ce{2A + B <=> 3 C}\tag{4}$$

Well, in general the same idea applies except that you must now take care of the different coefficients. Say we removed some $\ce{C}$ from the mixture. New $\ce{C}$ will be created and the creation of three molecules of $\ce{C}$ will use up two molecules of $\ce{A}$ and one of $\ce{B}$. Thus:

For each mole of $\ce{C}$ created we lose $\frac 13~\mathrm{mol}\ \ce{B}$
For each mole of $\ce{C}$ created we lose $\frac 23~\mathrm{mol}\ \ce{A}$

Therefore the relationship is:

$$\Delta n(\ce{C}) = -3\Delta n(\ce{B}) = -\frac 32\Delta n(\ce{A})\tag{5}$$


How do you stop a Pharoah's Serpent reaction?

Just put water on it.  A temperature of $\pu{165^\circ C}$ is required to sustain the decomposition. Adding water will remove heat from the reaction and will dissolve a small amount of the mercury thiocyanate. The heat removed by heating and boiling the water will cool the contents below the decomposition temperature.  Adding soap will help the water penetrate into the powder and more quickly stop the reaction.


Why do most character tables include an $(x^2 + y^2)$ term?

The functions on the right hand side of the table are the 'basis functions' for the irreducible representation, which are the rows of character, say $A_g$ or $B_g$ in C$_{2h}$. The basis functions have the same symmetry properties as the atomic orbitals with the same name (but other functions in $x, x^2, x^3$ etc and similarly for  $y,z$, could be used).
To determine where  an $xy$ function goes, look at a $d_{xy}$ orbital shape and subject it to each of the symmetry operations in turn,$E, C_{2}, i$ etc depending on point group.  (Assume z is the principal axis). Count +1 if the orbital (or function) is unchanged and -1 if not. So in C$_{2h}$, a $C_2$ operation leaves $xy$ unchanged as does $\sigma_h$ and $i$ so $xy$ is a 'basis' for A$_g$. The $xz$ transforms as B$_g$; a $C_2$ operation gives -1, as does $\sigma_h$, but $i$ produces +1.
(note in cases such as $z^2 -(x^2+y^2)/r^2$ the $x^2+y^2$ part is totally symmetric in the $xy$ plane and does not change anything so can be ignored; think of the donut around d$_{z^2}$.)


Klaus's answer has all the right explanations but I always find these things hard to follow without pictures, so here are some to clarify.

Note that in the pictures below we don't explicitly draw carbons or the hydrogens attached to them (we just mentally add enough hydrogens to make the number of bonds to each carbon equal 4).

First what is benzopyran?

The two numbers on the first structure tell us where the oxygen appears in the pyran ring and where the saturated carbon (with an extra hydrogen) appears. Both could be in different places. The second picture showns the numbering convention for each atom in the ring so we can keep track of where anything else is attached. The third picture shows what happens if we notionally add extra hydrogens on carbons 3 and 4 to give a fully saturated ring.

Now we add an OH (hydroxy) in certain positions:

The picture on the right shows how we number the parts of trihydroxy phenyl when it is attached to something else (it will be stuck on the 2-carbon on the benzopyran in the final compound).

Finally we have the epigallocatechin:

The "cis" indicates that the two hydrogens (mentally fill them in in positions 2 and 3) are on the same side of the benzopyran (the opposite side to the two wedge-like bonds on the OH and phenyl groups).

I hope that all makes some sort of sense. Note that none of the above bears any relation to how the molecules are made: this is all notional deconstruction for the purpose of explaining how the molecule is named.


In order for a substance to dissolve, the free energy change for the dissolution must be negative.  For the enthalpy change to be negative the heat of solvation (the solvent solvating the solute) must be greater in magnitude than the heat needed to break up the forces holding the solid solute together (in the case of NaCl lattice energy).  The entropy change of the solution process is accounted for by the gain in entropy from breaking up the solid and the loss of entropy from organizing solvent around the solute.  In forming most solutions, the entropy change is positive.

In this case the ability of carbon tet to solvate sodium and chloride ions is too low to make up for the lattice energy.  Carbon tet is nonpolar.  A good solvent for NaCl would be polar and the negative end of the solvent would surround sodium ions and the positive end would surround chloride ions.

http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/nacl2.gif


Predict the species formed by the reaction of mercury(II) chloride and iodine

You are right about the formation of $\ce{I3-}$ ion (from the reaction of $\ce{I2}$ and solution containing iodide ion, let's say $\ce{KI}$ solution). Reaction between $\ce{HgCl2}$ and $\ce{KI}$ solution at $\pu{70 ^\circ C}$ produces mercury(II) iodide ($\ce{HgI2}$):
$$\ce{HgCl2 + 2 KI → HgI2 + KCl}$$
But the story doesn't end here. This mercury(II) iodide reacts with $\ce{KI}$ solution to form potassium tetraiodomercurate(II)($\ce{K2HgI4}$):

Crystallizing from a concentrated aqueous solution of mercuric iodide
with potassium iodide is the monohydrate $\ce{KHgI3.H2O}$, which is pale
orange. In aqueous solution, this triodido complex adds iodide to
give the tetrahedral tetraiodo dianion.

So, the answer is  C. but there is a typo. It should be $\ce{[HgI4]^2-}$. The overall reaction is:
$$\ce{HgCl2 + 5KI + I2 -> K2HgI4 + KI3 + 2KCl}$$


What dictates cathode vs anode nomenclature use?

Who was responsible for this naming system and how can we change it?

Michael Faraday was responsible for the terms anode and cathode more than hundred years ago.

All the confusion regarding the nomenclature will vanish if you do not associate electrostatic signs with these two terms. One should identify the electrode labels with the redox processes rather than the signs.

Cathode: The electrode where reduction occurs
Anode: The electrode where oxidation occurs

These definitions are true whether you have a electrolytic or galvanic cell. This eliminates the need to call the negatively charged electrode as a cathode in an electrolytic cell or the positively charged electrode as the cathode in the galvanic cell. These definitions are self-consistent and hence nobody bothered to changed them.

Isoelectric focusing is "sort" of an an electrolytic "cell" hence its cathode is has a electrostatic negative sign because if we were to remove this gel and replace it with just salt water, we will see hydrogen evolution (= reduction of water) at this electrode.