Why is the ionic product for water $10^{-14}$?

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

$$K_w = \frac{a(\ce{H3O+})a(\ce{OH-})}{[a(\ce{H2O})]^2}$$

where $a$ is "activity of".

Then you can approximate activity of water as 1 if appropriate and activity of the ions as concentration of the ions.

Just like any other equilibrium constant, the relative chemical potential (Gibbs energy) of the products and reactant determine the constant.

Consider the relative strengths of the bonds of $\ce{H2O}$, compare to those of $\ce{OH-}$ and $\ce{H3O+}$, as well as the number and strength of intermolecular hydrogen bonds to each of these.

For $\ce{D2O}$, the equilibrium constant is 8 times small.


Where is the mirror plane in 1-bromo-1-chloroethene?

For all symmetry operations, you should assume the atoms reduced to a point size. Then, if you have a perfectly flat molecule (like in the case of your disubstituted ethene), all atoms will be in the same mathematical plane meaning there is nothing above or below the plane.

Therefore, the plane is an element of symmetry that puts the nothing above the plane on the nothing below the plane and leaves the atoms themselves (in the plane of symmetry) untouched.


Ramachandran plot Phi(ϕ) Psi(ψ) dihedral angle Convention for Zero, Positive and Negative value- old and new

Evidently the old $\phi$ is measured from an angle oriented $180^\circ$ relative to the new convention. This implies that the amide H ($H_N$) rather than carbonyl carbon was used as the reference fourth atom used to define the dihedral, since these atoms are related by a $180^\circ$ rotation. The old choice of atom is opposite on the chain to the modern convention of using the heavy backbone atom (carbonyl carbon). You can obtain angles in the new convention by subtracting $180^\circ$ ie $\phi \rightarrow \phi - 180^\circ$.

The meaning of angles with value of zero for $\phi$' aka $\psi$ is the same. However the values between $180^\circ$ and $360^\circ$ in the original Ramachandran plot can be shifted below zero, as $180^\circ\rightarrow -180^\circ$ and $360^\circ\rightarrow 0^\circ$, or more generally $\psi \rightarrow \psi - 360^\circ$.
By the way inspection shows that this is the correct interpretation. See for instance the following diagram. The extended (beta/turn/pII) region is usually larger than the alpha region. The beta region sits at large values (below  $180^\circ$) of $\psi$, whereas the narrower alpha region is close  to  $0^\circ$ (at negative angles):

In Ref. 1 Ramakrishnan and Ramachandran explain the differences in the  conventions:

At a recent conference of some of the representative workers in this field held in Bethesda, it was decided to denote the two dihedral angles about the bonds N-Ca and C.-C' by $\phi$ and $\psi$ respectively, the sense of rotation being the same as that adopted here. The fully extended chain, with N-H and C'=O trans with respect to one another, is to be taken as the standard conformation with $\phi$=$\psi$=0.
It is readily seen that the new $\phi$ is the same as the old $\phi$, but that $\psi$=$180^\circ$+$\phi'$ (and $\phi'$=$180^\circ$+$\psi$). All the data reported here are thus readily converted into the ($\phi$,$\psi$) coordinates. The diagrams in Figs. 2, 3, and 6 have to be shifted up by $180^\circ$, or half the total length, along the vertical direction. In particular, the right-and left-handed alpha helices will have ($\phi$,$\psi$) equal to ($133^\circ$,$123^\circ$) and ($227^\circ$,$123^\circ$). As before, a helix with ($-\phi$,$-\psi$) will be inverse to one with ($\phi$,$\psi$); i.e., it will  be of opposite sense, but having the same number of turns per unit.
This paper was finalized well before this meeting, and so the older conventions are adopted here. However, it is proposed to use the new notation in the following papers in this series.

References
1.Ramakrishnan C, Ramachandran GN. Stereochemical criteria for polypeptide and protein chain conformations II. Allowed conformations for a pair of peptide units.  Biophysical Journal, 01 Nov 1965, 5(6):909-933. DOI: 10.1016/s0006-3495(65)86759-5.


Halogen-substituted aliphatic hydrocarbons are usually miscible with oils. Pick the compound by its density, price, safety, boiling point and other characteristics.

For example, trichloroethylene (trico) has a density of ~1.6 g/cm3, and a Material Safety Data Sheet shows a vapor pressure of ~60mm Hg at room temperature and a time weighted average (TWA) toxicity limit of 50 ppm (269 mg/m3) from ACGIH and OSHA. Though it's readily available, trico might be too toxic and volatile for your use. A similar chemical is tetrachloroethylene, sold as "perc", with the same toxicity issue.

In general, going down the halogen group increases density, so bromine or iodine substituted compounds would be denser than those with fluorine or chlorine.


Why can anthracene, but not phenanthrene, take part in Diels–Alder reactions?

Have a look at the reaction mechanism of the Diels-Alder Reaction, e.g. at Wikipedia.
We have talked about this reaction before on this site, and concluded that a sufficient explanation is only possible with the help of molecular orbitals: How accurate is this polar mechanism for the Diels-Alder reaction?
Borrowing from Wikipedia,[1] here is the most important part of the mechanism, the HOMO of the diene and the dienophile:

Now the most important part of this picture is the symmetry of the HOMO/LUMO of the diene, since this is the subject of your question. We need either a HOMO or a LUMO of these symmetries and then could match a suitable dienophile.

In anthracene we find the appropriate symmetry of the HOMO:

The reason for the regioselectivity has also been discussed before on this site: Diels Alder with Anthracene and Maleic anhydride.
We could also tune our reaction to be of inverse demand, as the LUMO has also appropriate symmetry:

In phenanthracene we do not have the appropriate orbital symmetry in the HOMO:

The LUMO is also not suitable for an inverse Diels-Alder reaction.

Reference and Notes

"FMO of Diels-Alder reaction" by Chrito23 - Own work. Licensed under CC BY-SA 3.0 via Commons.
Calculations on the DF-BP86/def2-SVP level of theory employing Gaussian09 rev. D. Contour values for the orbitals are at 0.05 e/rBohr. Occupied orbitals are depicted in blue and orange and unoccupied orbitals are depicted in yellow and red. Phases are chosen at random.


$K$, the standard thermodynamic equilibrium constant, is computed from $\Delta G^\circ$, using

$$\Delta G^\circ = -RT\log K \tag1$$

Generally speaking, $K$ in equation (1) is unitless. Its value depends on the specified reference standard states and $T$ (and obviously on the equilibrium activities of reactants and products).  Both $K$ and $\Delta G^\circ$ change (and are certainly allowed to) if you change the choice of standard states for any species.

You can also compute an equilibrium constant from a value of $\Delta G^\circ$. To do this reverse operation you need to know the standard state ($p^\circ=\pu{1 bar}$, $c^\circ=\pu{1 M}$, or $m^\circ=\pu{1 molal}$, $\chi^\circ=\pu{1}$) associated with each species in the equilibrium equation. At equilibrium each product/reactant is at the same temperature and in the same phase as in its associated reference standard state (if not at the same partial pressure, concentration, or mole fraction).

When you use either $K$ or $\Delta G^\circ$ in practice, you need to know the standard state of each species (this is often apparent from additional information that is provided with these parameters). If you are working with a gas-phase reaction then you can refer to $K$ as $K_p$. If you are using standard states then all reference states are $\pu{1 bar}$ $^\dagger$, and partial pressures computed directly from $K_p$ are in $\pu{bar}$ units (these can obviously be converted later).

To compute $K_c$ from say $K_p$ you have to convert units with an appropriate conversion factor. This is described in the linked post. Same applies to problems where $K_\chi$ is provided, where concentrations are described in mole fraction units. Note that if you change units, for instance to compute $K_c$, then you have (perhaps implicitly) changed the reference state (perhaps to a nonstandard state), and in that case $\Delta G^\circ$ also changes. This is also explained in an answer to the linked post.

$\dagger$ Components whose activity is constant can usually be ignored


What is the mechanism by which sugars glycate proteins?

One mechanism of glycation is the Maillard Reaction.

The mechanism starts with nucleophilic attack of the sugar carbonyl group by a side chain amino group from the protein.

The Milliard Reaction occurs in the human body and has been implicated in various adverse health conditions.

The cornea, lens and retina are known to be damaged by this reaction.

See The Maillard reaction in the human body. The main discoveries and factors that affect glycation Pathologie Biologie, Volume 58, Issue 3, Pages 214-219 for a recent review.


How to determine the bond order using the Lewis structure?

Determining the bond order starting from a Lewis structure is a task that can range from very easy to rather difficult. Luckily, most cases you will encounter are the easy ones. The first step should always be to draw out your molecules. For carbon monoxide and carbonate, this is what you should initially arrive at:

The second step is to check if you are missing any resonance structures. This is most obvious in carbonate whose three oxygens are all equal and whose bonds can be pushed around as shown below to give these additional two resonance structures:

However, also carbon monoxide can be drawn in a different resonance structure. These two resonance structures are obviously not equal since one equips all atoms with an octet while the second leaves carbon with a sextet:

Therefore, in a third step, you must weight the resonance structures. As stated above, the three of carbonate are all equal, we must consider them contributing the same to the overall structure. Those of carbon monoxide, however, are not equal; the triple-bonded structure will contribute far more than the double-bonded structure.
Depending on the level of your exam, you may be expected to completely ignore the lesser resonance structure (introductory level) or to assume weightings for both structures (advanced level) — or, anything in-between. Here, let us assume that the double-bonded resonance structure contributes almost nothing to the final result so we may ignore it (which is close enough to the truth).
Having done all this preparative work, we are now able to take a look at the actual bond whose bond order we want to determine. For each resonance structure:

Count the number of electron pairs in one specific bond

Multiply that with a weighting fraction

Sum up the values obtained in this way
$\displaystyle \text{B. O.} = \sum_i \frac{n_i(\ce{e-})}2 \times x_i$

For carbon monoxide, of which we have a single non-negligible resonance structure, we have 3 electron pairs in the resonance structure which contribute 1 (or the entirety) to the final structure so our bond order is 3.
For carbonate, we have three resonance structures that we must consider separately and we must multiply the individual structure’s electron pair count with $\frac13$ because each contributes a third to the overall picture. If we consider the upwards-pointing $\ce{C\bond{...}O}$ bond of the second figure, from the first resonance structure we have a contribution of $\frac23$, from the second we have $\frac13$ and from the third we also have $\frac13$. Adding these three values we have an overall bond order of $\mathbf{\frac43}$.

Finally, although this is already well into the advanced level I wish to point out why the weighting step is important. Consider the structure of a carboxylic ester as shown below. At first sight, this might look like ‘half a carbonate’, which would mean that both shown resonance structures are equal, contribute $0.5$ to the overall picture and therefore both bond orders would be $1.5$. However, this is only the case if we are talking about the carboxylate anion, not about the ester. In the ester’s case, as can be seen below, one of the resonance structures includes charge separation. Therefore, we must weight the non-charge separated resonance structure stronger than the charge separated one. We might choose weighting factors of $0.75$ and $0.25$ (without any additional data the choice of factors is entirely arbitrary!). This leads us to two different bond orders for the two $\ce{C-O}$ bonds, one being something like $1.75$ the other corresponding to $1.25$.

It is important to note that in the ester’s case we cannot ignore the contribution of the second, minor resonance structure since comparing the reactivity of an ester to that of a ketone shows a clear difference (the ester being less reactive, i.e. the double bond not being a full double bond). Likewise, replacing an ester e.g. with a thioester or selenoester further reduces the contribution of the second resonance structure closer to where it may safely be ignored again.


Why would the 2+ molecular ion of butane have a mass spectrum peak at 29?

The mass spectrometer observes the mass to charge ratio (m/z). Most of the time, small organic molecules like butane will be singly charged by the ionizer, which means that the mass can be read off directly since z =1.

If butane was somehow ionized to the +2 state, then the mass spec would observe 58/2=29, which is the mass of butane (58 g/mol) divided by the charge (+2).

This becomes particularly important for larger compounds with multiple easily ionizable groups, such as peptides. A series of peaks corresponding to different charge states can be used to identify the compound of interest.


Question during radius of gyration calculation from XYZ trajectoris of atoms

You don't say exactly what the difference is between your calculation and the LAMMPS calculation, but I'm going to make a guess that you have neglected to account for the periodic boundary conditions properly. The LAMMPS documentation for the compute gyration command says

  The coordinates of an atom contribute to Rg in “unwrapped” form, by using the image flags associated with each atom. See the dump custom command for a discussion of “unwrapped” coordinates. See the Atoms section of the read_data command for a discussion of image flags and how they are set for each atom. You can reset the image flags (e.g. to 0) before invoking this compute by using the set image command.

If you are trying to duplicate this calculation "by hand", you need to "unwrap" the atomic coordinates. This means that, for every molecule, all the atoms in the molecule need to belong to the same "copy" of the molecule, i.e. not to different periodic images of the molecule. You need to ensure this before you even calculate the position of the centre of mass of each molecule.

I'll illustrate one way to do this for a simple linear chain molecule of $N$ atoms. Something similar should work for a more complicated molecule, but you'll need to think about it yourself. I'm going to assume that the atoms are labelled in order $1, 2, 3, \ldots N$, and that $1$ is bonded to $2$, $2$ to $3$, and in general $j$ bonded to $j+1$. Let the atomic positions be $\mathbf{r}_j$. You may want to copy these to a separate working array, to avoid changing the originals.

We'll take the position of atom $1$ as a reference point. In the following loop, start by setting $j=2$.

Compute $\mathbf{d}_{j}=\mathbf{r}_{j}-\mathbf{r}_{j-1}$.
Apply the minimum image correction to $\mathbf{d}_{j}$.
Redefine $\mathbf{r}_{j}=\mathbf{r}_{j-1}+\mathbf{d}_{j}$.
Increment $j\rightarrow j+1$ and return to step 1, unless $j=N$ in which case stop.

Now you should be safe to compute the centre of mass of that molecule, and work out the radius of gyration, in the way you describe in your question. You can repeat the calculation for every molecule independently.

A further point about the LAMMPS compute gyration command is that it not only returns the scalar value $R_g$ (which should correspond to what you describe calculating), it also returns the separate $xx$, $yy$, and $zz$ terms, and also the off-diagonal $xy$, $yz$ and $zx$ terms. This is because the gyration tensor is, like the inertia tensor, a second-rank symmetric tensor. This may be of interest if your polymer configurations are significantly non-spherical in shape. Using these values, you can determine the principal axes of the gyration tensor for each molecule, and the corresponding three principal radii of gyration. It's basically a question of diagonalizing the $3\times3$ symmetric matrix whose components are returned by the LAMMPS command, along with $R_g$. I'm assuming, though, that this is not what you are worried about.
[EDIT: I've just noticed that @porphyrin made this same point in a comment while I was typing in my answer].

EDIT following OP comment.

It seems that "unwrapped" coordinates are already being used for this calculation, in post-processing. So it is possible that my answer does not  explain the difference between the LAMMPS calculation and the one implemented by the OP.

However, different definitions of "unwrapped" may be used for different purposes. In calculating atomic mean-squared displacements as a function of time, to get the diffusion coefficient, it is sufficient to make sure that periodic boundary corrections are never applied to the atomic positions as they evolve in time. However, for that purpose, there is no need to consider which molecule an atom belongs to. In order to compute the molecular centre-of-mass, and from that the radius of gyration $R_g$, the atoms must be grouped together by molecule. This means that, at some stage, a calculation similar to the one described above must be carried out. Maybe this has been done, in producing the "unwrapped" trajectories, maybe not.

So, it is still worth either carrying out the correction described above, or at least checking that bonds between atoms have reasonable lengths in the "unwrapped" trajectories. For example, for a linear chain of $N$ atoms, compute $d_j^2=|\mathbf{r}_j-\mathbf{r}_{j-1}|^2$ for $j=2,3,\ldots N$, and just store the maximum $d_j^2$ in the molecule. Extend this calculation to cover every polymer molecule, and print out the maximum value of $d_j^2$ discovered for the whole set of molecules. How does this compare with the square of the bond length between adjacent atoms, defined by the molecular force field? If there has been no error with the periodic boundaries, the two numbers should at least be similar.


A semi-transparent material that is a solid

There are other plastics that have better mechanical properties, but probably nothing is easier to find than polyester casting resin. It works really well in silicone moulds and there are various pigments you can buy to change its colour and opacity. It isn't as optically clear as some polyurethane or epoxy resins (you won't be making lenses out of polyester), but it's not bad and you can find it in a lot of art supply stores.

As for something with aerogel's smokey translucency, I don't know. Aerogel's translucency (its colour, as well) is due to scattering off its nanoporous structure. It may be possible to mimic the look by blending silica particles or something into a plastic to cause scattering, but I can't recommend anything specific.


Polystyrene properties after dissolved by acetone

What is happening is not actually the polystyrene melting but rather it dissolving in the acetone solvent. This is a physical change and not chemically altering the polymer. It should still retain its chemical properties as before.

However, other properties have changed. It is now a poorer insulator since the air pockets have been removed. It will also be more rigid for similar reasons.

The acetone ought to be able to be driven off by heating it in a ventilated space as acetone is quite volatile.


Negative voltage means the same as positive, except the electric current caused by it would flow in the opposite direction.

As for your another question, that's just what it is: having negative potential means being more reactive than H. See, that electrode potential is not an absolute value, it can be measured only relative to something. And the general agreement is to take the standard hydrogen electrode for that "something".

Suppose you'd make an electrode out of your metal immersed in a solution of some salt of itself, couple it to standard hydrogen electrode and see what happens. If a metal is more reactive, it will dissolve, the released electrons will run to the hydrogen electrode and reduce hydrogen there. With less reactive metals, it is the other way around: hydrogen will dissolve and send the electrons running in the opposite direction, to eventually reduce the metal.


Why does mercury have low melting and boiling points?

Mercury is different!  It is not as reactive as its neighbors in the Periodic Table, it doesn't conduct heat and electricity as well as other metals, and it is a liquid unlike other metals.  The electronic structure of mercury is $\ce{[Xe] 4f^{14} 5d^{10} 6s^2}$, so the first thing we notice is that all of its orbitals are full, there are no unpaired electrons in mercury.  This helps explain mercury's reluctance to react with other materials or form bonds with other mercury atoms.  Most metal atoms share their outer electrons with other metal atoms, in fact, all of these "shared" metal electrons exist as a diffuse "sea" of electrons.  It is the extended nature of this sharing that makes metals good conductors of heat and electricity.  The fact that the metal electrons are shared provides a strong bonding interaction between metal atoms that gives metals their solid structure.  None of this applies to mercury.  With its filled shell electron configuration it is very reluctant to form bonds even with other mercury atoms.

One additional, interesting effect makes mercury even more reluctant to share its electrons.  Special relativity suggests that for larger nuclei (and mercury with atomic number 80 is large enough for the effect to apply) electrons will begin to travel closer to the speed of light.  In turn, their mass will increase and as mass increases, orbital radius decreases.  The electron is closer to the nucleus, more attracted to the nucleus and less available for bonding.  Here is a nice reference that explains this relativistic effect and how it applies to mercury in plain English and a bit more detail.

These two effects, filled outer electron shell and contraction of the outer orbital closer to the nucleus, combine to make mercury reluctant to form bonds, even with other mercury atoms.  It is this lack of electron bonding between mercury atoms that makes it melt and boil at such low temperatures.


Change in pressure or volume of molten salt

Venting initiates from blockage - anywhere.  All your lines must be hot-traced.

Nitrate melts are powerful oxidizing agents.  If containment fails onto organics (e.g., wood) you've got a catastrophe.  Nitrate melts significantly shrink on cooling and solidifying, and they bond to polar surfaces (glass, metal).  If you have  a half-liter of solidified salt and heat the bottom of containment, the melt will expand against the plug of solid salt above it, possibly deforming or rupturing containment.  The very hot and oxidizing liquid shoots out.  Heat a stripe up the side to start.

Nitrate melts thermally decompose.  Even if the average temp is OK, heating hot spots may give you oxidizing, corrosive, and toxic by inhalation nitrogen oxides.

http://link.springer.com/article/10.1361%2F10599490417542
"Corrosion of stainless steels and carbon steel by molten mixtures of commercial nitrate salts"
http://prod.sandia.gov/techlib/access-control.cgi/2001/018518.pdf
http://www.electrochem.org/dl/ma/197/pdfs/0209.pdf

The articles say your hot end is out of the safety zone.  Welds.  Chloride.  Chloride corrosion, especially pitting, can be aggressive. Halide in general may destabilize nitrate.  Ammonium nitrate is rather tame.  Ammonium nitrate plus chloride contamination is Texas City, Texas 16 April 1947. 2300 tons.

http://www.energystoragenews.com/Molten%20Nitrate%20Salt%20for%20Solar%20Energy%20Storage.html
1600 tonnes of molten nitrate eutectic.

You might put it behind grout-filled cement block or cinder block with #5 rebar weaving vertical and horizontal, and into the foundation, when you build the wall.  Have a large area vent direction.


What does it mean when something is said to "contain ions"?

Your teacher is almost right but some terminology needs to be clarified.

Pure substances will almost always be electrically neutral (and the exceptions involve temporary local transfer of electrons from one substance to another as when you rub a rubber balloon on a wall surface). Beyond a certain level of charge separation you get enough electrical potential to cause sparks which allows the charges to equalise again. So persistent charge imbalance isn't easy to get. I think that is what your teacher means.

But if you look inside some pure substances, they do contain ions, just an equal number of each to avoid the bulk material having an overall charge. Common salt, for example, consists of ions (Na+ and Cl- in equal numbers). The substance is neutral, but the individual atoms making it up are all ions.

So your teacher is wrong if she meant that ions don't exist in nature but correct if she meant that bulk substances are electrically neutral.


Calculating the ph of an unknown solution by measuring the precipitate

You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate $\mathrm{pOH}$ of the solution as $\mathrm{pOH} = -\log \ce{[OH-]}$. Then, $\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}$. Now, you see, since $\mathrm{p}K_\mathrm{w} = 14$, only unknown is $\ce{[OH-]}$.

Now look at the reaction of your unknown hydroxide with lead (II) nitrate:
$$\ce{2 OH- (aq) + Pb(NO3)2 (aq) -> Pb(OH)2 (s) + 2 NO3 (aq)}$$

Since you know the precipitated amount of $\ce{Pb(OH)2}$ isolated from $\pu{1.0 L}$ of unknown hydroxide solution, you can calculate original $\ce{[OH-]}$:

$$\text{Amount of } \ce{OH-}\text{ in original solution} \\
= \pu {3.81 g}\text{ of } \ce{Pb(OH)2} \times \frac{\pu {1 mol}\text{ of } \ce{Pb(OH)2}}{\pu {241.21 g}\text{ of } \ce{Pb(OH)2}} \times \frac{\pu {2 mol}\text{ of } \ce{OH-}}{\pu {1 mol}\text{ of } \ce{Pb(OH)2}} \\
= \pu {0.0316 mol}\text{ of } \ce{OH-} $$
Thus, $ \ce{[OH-]} = \pu{0.0316 M}$, and hence,
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}= 14 + \log(0.0316) = 12.5 $$

Asking is approximate $\mathrm{pH}$ of orignal solution. Thus, this answer is acceptable.  However, keep in mind that trace of $\ce{OH-}$ will remain in solution due to the solubility of  $\ce{Pb(OH)2}$ in water $\left(K_\mathrm{sp} = 1.42 \times 10^{-20}\right)$.


Thermodynamics isn't quite that general.  It will tell you about the possibility of going from one well-defined state to another well-defined state, but as has been said, there is no way to do it in general.  What you are asking is akin to "is it hard to get there from here?" without specifying where both here and there are.

So you need to specify enough variables such as temperature or composition to unambiguously determine both the initial and final states.  How many variables do you need?  The Gibbs phase rule provides the answer.  Once that has been done, your question can be answered.


Fundamental forces behind covalent bonding

I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other?

The role of Pauli Exclusion in bonding

It is an unfortunate accident of history that because chemistry has a very convenient and predictive set of approximations for understanding bonding, some of the details of why those bonds exist can become a bit hard to discern. It's not that they aren't there -- they most emphatically are! -- but you often have to dig a bit deeper to find them. They are found in physics, in particular in the concept of Pauli exclusion.

Chemistry as avoiding black holes

Let's take your attraction question first. What causes that? Well, in one sense that question is easy: it's electrostatic attraction, the interplay of pulls between positively charged nuclei and negatively charged electrons.

But even in saying that, something is wrong. Here's the question that points that out: If nothing else was involved except electrostatic attraction, what would be the most stable configuration of two or more atoms with a mix of positive and negative charges?

The answer to that is a bit surprising. If the charges are balanced, the only stable, non-decaying answer for conventional (classical) particles is always the same: "a very, very small black hole." Of course you could modify that a bit by assuming that the strong force is for some reason stable, in which case the answer becomes "a bigger atomic nucleus," one with no electrons around it.

Or maybe atoms as Get Fuzzy?

At this point some of you reading this should be thinking loudly "Now wait a minute! Electrons don't behave like point particles in atoms, because quantum uncertainty makes them 'fuzz out' as they get close to the nucleus." And that is exactly correct -- I'm fond of quoting that point myself in other contexts!

However, the issue here is a bit different, since even "fuzzed out" electrons provide a poor barrier for keeping other electrons away by electrostatic repulsion alone, precisely because their charge is so diffuse. The case of electrons that lack Pauli exclusion is nicely captured by Richard Feynman in his Lectures on Physics, in Volume III, Chapter 4, page 4-13, Figure 4-11 at the top of the page. The outcome Feynman describes is pretty boring, since atoms would remain simple, smoothly spherical, and about the same size as more and more protons and electrons get added in.

While Feynman does not get into atoms how such atoms would interact, there's a problem there too. Because the electron charges would be so diffuse in comparison to the nuclei, the atoms would pose no real barrier to each other until the nuclei themselves begin to repel each other. The result would be a very dense material that would have more in common with [neutronium[(http://en.wikipedia.org/wiki/Neutronium) than with conventional matter.

For now I'll just forge ahead with a more classical description, and capture the idea of the electron cloud simply by asserting that each electron is selfish and likes to capture as much "address space" (see below) as possible.

Charge-only is boring!

So, while you can finagle with funny configurations of charges that might prevent the inevitable for a while by pitting positive against positive and negative against negative, positively charged nuclei and negatively charged electrons with nothing much else in play will always wind up in the same bad spot: either as very puny black holes, or as tiny boring atoms that lack anything resembling chemistry.

A universe full of nothing but various sizes of black holes or simple homogenous neutronium is not very interesting!

Preventing the collapse

So, to understand atomic electrostatic attraction properly, you must start with the inverse issue: What in the world is keeping these things from simply collapsing down to zero size -- that is, where is the repulsion coming from?

And that is your next question:

  Also, am I right to think that "repulsion occurs when atoms are too close together" comes from electrostatic interaction?

No; that is simply wrong. In the absence of "something else," the charges will wiggle about and radiate until any temporary barrier posed by identical charges simply becomes irrelevant... meaning that once again you will wind up with those puny black holes.

What keeps atoms, bonds, and molecules stable is always something else entirely, a "force" that is not traditionally thought of as being a force at all, even thought it is unbelievably powerful and can prevent even two nearby opposite electrical charges from merging. The electrostatic force is enormously powerful at the tiny separation distances within atoms, so anything that can stop charged particles from merging is impressive!

The "repulsive force that is not a force" is the Pauli exclusion I mentioned earlier. A simple way to think of Pauli exclusion is that identical material particles (electrons, protons, and neutrons in particular) all insist on having completely unique "addresses" to tell them apart from other particles of the same type. For an electron this address includes: where the electron is located in space, how fast and in what direction it is moving (momentum), and one last item called spin, which can only have on of two values that are usually called "up" or "down."

You can force such material particles (called fermions) into nearby addresses, but with the exception of that up-down spin part of the address, doing so always increases the energy of at least one of the electrons. That required increase in energy is a nutshell is why material objects push back when you try to squeeze them. Squeezing them requires minutely reducing the available space of many of the electrons in the object, and those electrons respond by capturing the energy of the squeeze and using it to push right back at you.

Now, take that thought and bring it back to the question about where repulsion comes from when to atoms bond at a certain distance, but no closer. They are the same mechanism!

That is, two atoms can "touch" (move so close, but no closer) only because they both have a lot of electrons that require separate space, velocity, and spin addresses. Push them together and they start hissing like cats from two households who have suddenly been forced to share the same house. (If you own multiple cats, you'll know exactly what I mean by that.)

So, what happens is that the overall set of plus-and-minus forces of the two atoms is trying really hard to crush all of the charges down into a single very tiny black hole -- not into some stable state! It is only the hissing and spitting of the overcrowded and very unhappy electrons that keeps this event from happening.

Orbitals as juggling acts

But just how does that work?

It's sort of a juggling act, frankly. Electrons are allowed to "sort of" occupy many different spots, speeds, and spins (mnemonic $s^3$, and no, that is not standard, I'm just using it for convenience in this answer only) at the same time, due to quantum uncertainty. However, it's not necessary to get into that here beyond recognizing that every electron tries to occupy as much of its local $s^3$ address space as possible.

Juggling between spots and speeds requires energy. So, since only so much energy is available, this is the part of the juggling act that gives atoms size and shapes. When all the jockeying around wraps up, the lowest energy situations keep the electrons stationed in various ways around the nucleus, not quite touching each other. We call those special solutions to the crowding problem orbitals, and they are very convenient for understanding and estimating how atoms and molecules will combine.

Orbitals as specialized solutions

However, it's still a good idea to keep in mind that orbitals are not exactly fundamental concepts, but rather outcomes of the much deeper interplay of Pauli exclusion with the unique masses, charges,  and configurations of nuclei and electrons. So, if you toss in some weird electron-like particle such as a muon or positron, standard orbital models have to be modified significantly and applied only with great care. Standard orbitals can also get pretty weird just from having unusual geometries of fully conventional atomic nuclei, with the unusual dual hydrogen bonding found in boron hydrides such as diborane probably being the best example. Such bonding is odd if viewed in terms of conventional hydrogen bonds, but less so if viewed simply as the best possible "electron juggle" for these compact cases.

"Jake! The bond!"

Now on to the part that I find delightful, something that underlies the whole concept of chemical bonding.

Recall that it takes energy to squeeze electrons together in terms of the main two parts of their "addresses," the spots (locations) and speeds (momenta)? I also mentioned that spin is different in this way: the only energy cost for adding two electrons with different spin addresses is that of conventional electrostatic repulsion. That is, there is no "forcing them closer" Pauli exclusion cost like you get for locations and velocities.

Now you might think "but electrostatic repulsion is huge!", and you would be exactly correct. However, compared to the Pauli exclusion "non-force force" cost, the energy cost of this electrostatic repulsion is actually quite small -- so small that it can usually be ignored for small atoms. So when I say that Pauli exclusion is powerful, I mean it, since it even makes the enormous repulsion of two electrons stuck inside the same tiny sector of a single atom look so insignificant that you can usually ignore its impact!

But that's secondary, because the real point is this: When two atoms approach each other closely, the electrons start fighting fierce energy-escalation battles that keep both atoms from collapsing all the way down into a black hole. But there is one exception to that energetic infighting: spin! For spin and spin alone, it become possible to get significantly closer to that final point-like collapse that all the charges want to do.

Spin thus becomes a major "hole" -- the only such major hole -- in the ferocious armor of repulsion produced by Pauli exclusion. If you interpret atomic repulsion due to Pauli exclusion as the norm, then spin-pairing two electrons becomes another example of a "force that is not a force," or a pseudo force. In this case, however, the result is a net attraction. That is, spin-pairing allows two atoms (or an atom and an electron) to approach each other more closely that Pauli exclusion would otherwise permit. The result is a significant release of electrostatic attraction energy. That release of energy in turn creates a stable bond, since it cannot be broken unless that same energy is returned.

Sharing (and stealing) is cheaper

So, if two atoms (e.g. two hydrogen atoms) each have an outer orbital that contains only one electron, those two electrons can sort of look each other over and say, "you know, if you spin downwards and I spin upwards, we could both share this space for almost no energy cost at all!" And so they do, with a net release of energy, producing a covalent bond if the resulting spin-pair cancels out positive nuclear charges equally on both atoms.

However, in some cases the "attractive force" of spin-pairing is so overwhelming greater for one of the two atoms that it can pretty much fully overcome (!) the powerful electrostatic attraction of the other atom for its own electron. When that happens, the electron is simply ripped away from the other atom. We call that an ionic bond, and we act as it if it's no big deal. But it is truly an amazing thing, one that is possible only because the pseudo force of spin-pairing.

Bottom line: Pseudo forces are important!

My apologies for having given such a long answer, but you happened to ask a question that cannot be answered correctly without adding in some version of Pauli "repulsion" and spin-pair "attraction." For that matter, the size of an atom, the shape of its orbitals, and its ability to form bonds similarly all depend on pseudo forces.


How do the electronic properties of a catalyst affect the rate of oxidative addition?

To quote Hartwig:

Oxidative addition to more electron-rich metal centres tends to be more favourable than oxidative addition at metal poor centres [1]

This is generally a thermodynamic issue, and can be rationalised quite easily if one considers that the change in electron density at the metal centre during the oxidative addition process (note, theres a difference between formal oxidation state and the 'actual' electronics about the metal centre).
As you point out, most catalysts systems used are either Pd0 to begin with, or generate Pd0 via a Pd2 pre catalyst (the palladium needs to be in its 0 oxidation state for oxidative addition to occur, usually, though 2 -> 4 is also possible in certain circumstances).
Your confusion seems to have came from the actual nature of the active catalyst, although you'll always end up with Pd0 , the metal centre has ligands attached, and these ligands are able to tune various properties at the metal centre.
The issue with trying to optimise a reaction ( as you seem to be doing) is that there is an almost infinite number of combinations of catalyst + ligand that you could consider, and indeed chemical companies such as Sigma-Aldrich even sell screening kits which provide many of the most common.
My honest advice in your situation would be to have a thorough search of the literature and try a few combinations of ligand + catalyst that you have available in your lab. Doing the optimisation rationally can be extremely interesting (doing it properly requires changing catalysts, ligands, solvents, temperatures, additives etc and is all time consuming, though often done in industry during optimisation before scale-up), but is of little use synthetically if you simply need to drag material through.

[1]: Hartwig, J. Organotransition Metal Chemistry- From Bonding to Catalysis; University Science Books:Herndon , 2010


Why is the initial carbocation not formed adjacent to the phenyl moiety?

I agree with your sentiment, so I run a very quick calculation with a semi-empirical method (GFN-xTB) to find the protomers. This has been done using water as solvent. Note that I did not choose to include any deuterium.

First I calculated the conformational space of the plain molecule (NIST: 1,1'-(3-methyl-1-propene-1,3-diyl)bis-benzene, InChI=1S/C16H16/c1-14(16-10-6-3-7-11-16)12-13-15-8-4-2-5-9-15/h2-14H,1H3/b13-12+, Google search InChIKey: GNQWHYWLSGTMSL-OUKQBFOZSA-N) to obtain the lowest lying conformation and show you the HOMO:

As you can see the double bond is a good protonation cite, with both carbon atoms having about equal contribution; one would therefore expect that the most stable intermediate will form. Undoubtedly this should be the one where the charge can be delocalised into the aromatic π-system.

The automatic generation of the protomers agrees here, and the lowest lying conformation (with partial π-stacking) is shown below, the protonated carbon indicated in bright white:

There is a tentative pathway via a four-membered (green) transition state to rearrange to the product shown in the exam question, but I do not have the time to investigate. Therefore here is my proposed mechanistic scheme:


Behavior of Natural Abundance and Mass Spec Ratios

You are probably mixing natural abundance (NA) and relative abundance (RA).
In mass spectrometry RA is a more valuable parameter as it can be directly obtained as the $y$-coordinate of a plotted mass spectra: the most abundant ion (isotope) corresponds to the base peak, which is always $100\%.$
In other words, RA reflect isotope ratio, not NA.
For the isotopes of the elements RAs can easily be derived from NAs via normalization; however, the problem of finding RAs of the various isotopic molecular species is a bit less trivial [1].
The following table contains compiled data for NAs [2, p. 1-12] and RAs [3, p. 89] for both elements you've mentioned:
$$
\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lllrlrlr}
\hline
Z & \text{Isotope} & & &\text{Mass}/\pu{u} & \text{NA}&\hspace{-1em}/\% & \text{RA}&\hspace{-1em}/\%\\
\hline
16 & \ce{^{32}S} & \ce{[E]}      & \d{31}{9720711744(14)} & \d{94}{99(26)} & \d{100}{000} &\hspace{-1em} \\
   & \ce{^{33}S} & \ce{[E + 1]}  & \d{32}{9714589098(15)} & \d{0}{75(2)}   & \d{0}{789} \\
   & \ce{^{34}S} & \ce{[E + 2]}  & \d{33}{96786700(5)}    & \d{4}{25(24)}  & \d{4}{433} \\
\hline
17 & \ce{^{35}Cl} & \ce{[E]}     & \d{34}{96885268(4)} & \d{75}{76(10)}    & \d{100}{000} \\
   & \ce{^{37}Cl} & \ce{[E + 2]} & \d{36}{96590260(6)} & \d{24}{24(10)}    & \d{32}{399} \\
\hline
\end{array}
$$
References

Margrave, J. L.; Polansky, R. B. Relative Abundance Calculations for Isotopic Molecular Species. J. Chem. Educ. 1962, 39 (7), 335. DOI: 10.1021/ed039p335.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.


In the Haworth projections of D and L glucose, is the stereochemistry at every carbon except the anomeric carbon just inverted?

Glucose is the hexose that projects its hydroxyl groups at positions C2, C3 and C4 as well as its CH2OH off C6 in the equatorial direction (in its low energy conformer that is). In the β and α configurations, its anomeric carbon C1 projects the hydroxyl in the equatorial and axial direction, respectively.
The D and L forms are mirror images in which all anomeric carbons have the opposite configuration between the two forms as:

Here are the equivalent structures using the Haworth projection.


Since it's the cathode where the leak is I presume it's the hydrogen you can't hold. Hydrogen is very hard to contain because of it's small molecules. I use plastic glass as its bubbler and a small hose to move it elsewhere. If it's your bubbler where the leakage is insulate it using silicon around the connections/openings.


Why is there a comparative lack of variety in positive counterions in drugs?

Here is what I managed to find:

In various researches, cationic counterions were associated with a high prevalence of unexpected incompatibilities in drug development processes. Owing to their low solubilities, (typically  <0.1% ) cationic counterions are infrequently used in many pharmaceutical preparations, however may be preferred to provide prolonged slow absorption while achieving effective plasma concentrations for some therapeutic uses, which is the case for benzathine and procaine penicillin G intramuscular injections.1 Furthermore cationic counterions, to a greater extent are susceptible to salting. Salting out results when highly hydrated inorganic ions (e.g $\ce{K+}$, $\ce{Na+}$) deprive organic ions and molecules of adequate water molecules to remain dissolved.

In addition another important factor influencing pharmaceutical salt selection is compatibility with excipients (diluents, binders, disintergrants, glidants, preservatives etc). Most amine derived counterions are incompatible with excipients. For example, the counterion of tromethamine (a primary amine) should not be used with reducing sugars as excipients because of a possible Maillard reaction, consequently limiting the available entities in the cationic counterions' pool .2

The trend can also be explained as a consequence of the changes in research techniques employed by the pharmaceutical industry. It can also be attributed to small absolute numbers of approved drug products containing salts formed from acidic entities.3

Overally, anionic counterions are favoured in drug development due to the wide availability of  weakly  basic pharmaceutically  active principles (approximately 75% of drugs are estimated to be weak bases) 4, consequently leading to correspondingly low availability of cationic counterions for acidic drugs which comprise only about 20%, with remaining percentage being neutral.

References

Newton D, W. Drug Incompatibility chemistry (American Journal of Health-System Pharmacy) Available from: http://www.medscape.com/viewarticle/590261.
Wu, Y., Levons, J., Narang, A. S., Raghavan, K., & Rao, V. M. (2011). Reactive impurities in excipients: Profiling, identification and mitigation of drug-excipient incompatibility. AAPS PharmSciTech, 12(4), 1248-1263.
Paulekuhn, G. S., Dressman, J. B., & Saal, C. (2007). Trends in active pharmaceutical ingredient salt selection based on analysis of the orange book database. Journal of Medicinal Chemistry, 50(26), 6665-6672.
Manallack, D. T. (2007). The pKa Distribution of Drugs: Application to Drug Discovery. Perspectives in Medicinal Chemistry 1( )25–38.


What is the order of the catalytic decomposition of hydrogen peroxide?

Check out this paper. There they studied the decomposition of hydrogen peroxide by iron hydroxide and found that the reaction is a second order reaction for low concentrations of hydrogen peroxide.

In your case, you are working with a soluble iron species which will change the reaction kinetics because the hydrogen peroxide does not need to adsorb to the surface of insoluble iron hydroxide particles. This is assuming you worked under neutral or better slightly acidic conditions, otherwise the iron will precipitate as iron hydroxide.

Therefore, in your case the kinetics should depend on the relative concentration of iron chloride and hydrogen peroxide. As long as the hydrogen peroxide is in excess, I would expect a zero order reaction because the iron will limit the reaction speed. If the iron is in excess, I would expect a second order reaction because you need two molecules of hydrogen peroxide to form molecular oxygen.

How you experimentally determine the reaction order is described on Wikipedia. Briefly, you need to perform the experiment at different concentrations of hydrogen peroxide and measure the initial reaction speed.  Then you plot the logarithm of the initial rates against the logarithm of the concentrations. The slope of the resulting line should be your reaction order.


Which d orbitals of sulfur take part in the pi bonds of SO3?

Unfortunately this is an issue that is, even though thoroughly disproved, still taught in many books and schools.

As Philip already pointed out in the comments: There is little to no contribution of the $d$ orbitals to the bonding in $\ce{SO3}$, as modern quantum chemical calculations prove (compare Wikipedia). This bonding scheme is a relic from a time where quantum theory was not yet widespread available. The little contribution of the $d$ orbitals (usually below 1%) we find is due to polarisation functions, that are added for a more accurate description of the molecule.

There are partial answers to the overall bonding in $\ce{SO3}$ in those two question:

Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

But I will go a little further here. In the second post linked, you will find the result of a quantum chemical calculation for $\ce{SO3}$ on the BP86/cc-pVTZ level of theory. Here is the part of the output, containing the major contributions to the molecular orbitals (including the LUMO):

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-88.394 is S1-s=1.00
 Alpha occ 2 OE=-18.855 is O2-s=0.67 O3-s=0.17 O4-s=0.17
 Alpha occ 3 OE=-18.855 is O3-s=0.50 O4-s=0.50
 Alpha occ 4 OE=-18.855 is O3-s=0.33 O4-s=0.33 O2-s=0.33
 Alpha occ 5 OE=-7.922 is S1-s=1.00
 Alpha occ 6 OE=-5.945 is S1-p=1.00
 Alpha occ 7 OE=-5.941 is S1-p=1.00
 Alpha occ 8 OE=-5.941 is S1-p=1.00
 Alpha occ 9 OE=-1.113 is S1-s=0.39 O2-s=0.17 O3-s=0.17 O4-s=0.17
 Alpha occ 10 OE=-0.993 is O3-s=0.37 O4-s=0.37 S1-p=0.18
 Alpha occ 11 OE=-0.993 is O2-s=0.49 S1-p=0.18 O4-s=0.12 O3-s=0.12
 Alpha occ 12 OE=-0.585 is S1-s=0.24 O3-s=0.14 O4-s=0.14 O2-s=0.14 O2-p=0.11 O4-p=0.11 O3-p=0.11
 Alpha occ 13 OE=-0.502 is S1-p=0.55 O3-p=0.14 O4-p=0.14 O2-p=0.14
 Alpha occ 14 OE=-0.482 is O2-p=0.31 S1-p=0.20 O4-p=0.15 O3-p=0.15 O2-s=0.12
 Alpha occ 15 OE=-0.482 is O4-p=0.26 O3-p=0.26 S1-p=0.20
 Alpha occ 16 OE=-0.374 is O4-p=0.36 O3-p=0.36 O2-p=0.13
 Alpha occ 17 OE=-0.374 is O2-p=0.43 O3-p=0.21 O4-p=0.21
 Alpha occ 18 OE=-0.358 is O2-p=0.60 O4-p=0.15 O3-p=0.15
 Alpha occ 19 OE=-0.358 is O3-p=0.45 O4-p=0.45
 Alpha occ 20 OE=-0.305 is O4-p=0.32 O3-p=0.32 O2-p=0.32
 Alpha vir 21 OE=-0.141 is S1-p=0.45 O2-p=0.18 O4-p=0.18 O3-p=0.18

Now it is important to realise, that these numbers have to add up to 1 (accounting for rounding issues 0.99-1.01). So the reminding contributions are those of polarisation functions. Most notably missing is in orbital 15 the contribution of the O2-p orbital. The missing parts in 16-18 can be attributed to the diffuse and polarisation functions, i.e. higher $p$ and $d$-functions, 0.15; 0.15; 0.10; 0.10. As you can see, the involvement of $d$ orbitals in the total bonding is very little, almost negligible.

If you, however, consider $d$ orbitals to play a role in the bonding, then you have to consider, that all will participate.


Can you discriminate a monooxygenase from other enzymatic mechanisms by the requirement of NADPH?

I would say this is not a correct statement.  The main reason being that just because an enzyme is not a monooxygenase doesn't mean it doesn't require a regenerating system for NADPH.
Examples would be many dehydrogenase, hydroxylase and synthase enzymes. They also require NADPH regenerating systems to operate efficiently.  So, one of the CYP 450 enzymes may fall into any of these categories requiring NADPH/NADP yet not be a monooxygenase.


Using solubility guidelines to predict a reaction of two ionic, water-soluble salts

Of course $\ce{NH4+}$ can join $\ce{Cl-}$ and $\ce{Zn^2+}$ can join $\ce{SO4^2-}$, but the formed salts are completely soluble in water, and they completely dissociate in water to gives the corresponding ions.

On the other hand, the starting salts:  $\ce{(NH4)2SO4}$ and  $\ce{ZnCl2}$ are completely soluble in water, and they completely dissociate in water to gives the corresponding ions.

So, all of the ions remain in solution, as if no reaction occurs.


Stereoisomers of octahedral co-ordination compounds of the form [Ma3b3]

According to me $\ce{Co}$ should be $\mathrm{sp^3d^2}$ hybridised giving an octahedral geometry.

No, please don’t do this. The most accurate description of cobalt here would be $\mathrm{d^5}$, maybe $\mathrm{d^5s}$. But don’t extend the hybridisation concept from main group elements to the transition metals; it doesn’t belong and it doesn’t help here. Rather, there are six ligands around cobalt, so you have an extremely high chance for octahedral geometry. Which is also the case.

  I know it can show geometrical isomerism, fac and mer.

That’s a good start.

  But can it show optical isomerism?

When asking about optical isomerism, ask yourself, if the compound is superimposable on its mirror image. One method to check this is to draw the compound, draw its mirror image and rotate either until they overlap — or until you think they cannot. The second method is much more efficient though: Still draw them, but check whether you can find a centre of inversion symmetry, a plane of symmetry or an axis of improper rotation (turn by $\frac{360^\circ}{n}$, where $n$ is a natural number and then mirror the result through a plane perpendicular to the rotation axis). If you can find any of these, the compound is automatically identical with its mirror image, and if there are none of these, then it displays optical isomerism. Remember to consider 3D structures.

It should be rather simple to determine this for the mer isomer. For the fac, consider a view where three ligands of the same type are above the paper plane, the central cobalt occupies the paper plane and the three ligands of the other type are below it. For both considerations, you can ignore the two hydrogens on each oxygen.

The results are given in a spoiler (mouse over to see).

   The mer isomer has planes of symmetry that transect three meridional atoms of one kind and the middle atom of the other three.

 The fac isomer has planes of symmetry that transect the cobalt, one ligand of each kind and the middle between the other two ligands of each type.

 Therefore, neither displays optical isomerism.


There are a lot of factors at play here:

The curie and becquerel only tell you the number of disintegrations per unit time—they convey no information of the type or energy of radiation produced.
How dangerous a radioactive source is depends on the type of radiation produced and how one is exposed to it. External exposure to alpha particles is not very dangerous because alpha particles are easily stopped by skin, but inhaling alpha emitters can be very dangerous because there is direct access to sensitive alveolar cells.
How the radioisotopes are dispersed will greatly affect individual exposure to the radiation. A large amount of radioisotopes spread over a large area may increase the incidence of cancer slightly for a large number of people, but a smaller amount concentrated in one place may be very rapidly fatal to a single person.

20 MCi (740 PBq) is indeed a very large amount of radiation (~1/7 that released by the Chernobyl accident), but in a big cloud spread over thousands of square kilometres may not be acutely dangerous to people. That many radioisotopes is equivalent to 231,000 kg of cesium-137 and all that (magically) concentrated into a point source would produce a dose of .62 Gy/s for a person 1 m away—fatal in seconds.


Does Zinc become black & dull after reacting with dil. HCl?

Look at it under a microscope.  If the surface is microporous, light goes into the reflecting cavities, bounces around, and is absorbed before it can escape.  Wilkinson Sword sold bolted stacks of double-edged razor blades as low power laser beam dumps,  Each sharp edge was mirror bright with a slight taper toward the edge.  Light bounced in and could not escape.  The sharp sides of the stack were astoundingly black.  375 blades x .004" = 1.5" square target.

Or, you have impure zinc and crud is accumulating at the dissolving surface.  Alloy 8111 aluminum foil contains about 2% or so of strengthening dispersed intermetallics.  Dissolving aluminum foil in acid or base leaves black residues.


Calculating Ka values from std. state thermodynamic data

The definition of $K_a$ for $\ce{HNO3}$ is $$\frac{[\ce{H3O+}][\ce{NO3-}]}{[\ce{HNO3}]}$$ (concentrations in mol/L), which is precisely the $K$ that you get from the $\Delta G_r^0 = -RT\ln K$ formula, assuming you use the right free energies (which is not a trivial matter since they must take into account the interaction of each species with surrounding water molecules, an effect known as solvation).

You may be wondering why the concentration of water does not appear in the equation for $K_a$ (or more generally the equation for $K$ in a reaction in aqueous solution).  Because that concentration remains essentially unchanged throughout, it is kept out of the defining expression for $K$, for simplicity. From the point of view of $G^0$ this is correct since the standard state to which the $G^0$ of water refers is pure liquid water, while the standard state for the other three species is 1M concentration.

  There's already a 1 molar concentration of hydronium ion IN solution under standard state conditions.

You're correct that for specifying the $G^0$ of $\ce{H3O+}$ the standard state is 1M. But the standard state of $\ce{H3O+}$ is not where the reaction takes place. The same $K$ will be found whether you add the $\ce{HNO3}$ to pure water or to a solution of $\ce{HCl}$ of pH zero -- the final concentrations of $\ce{H+}$, $\ce{NO3-}$ and $\ce{HNO3}$ will be different in the two cases, but $K$ will be the same.

You can find a good discussion of the "standard state" for ionic species in solution here:
http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set10/lecture10.doc

I'll quote a couple of paragraphs.  You can see from this discussion that the calculation of these standard free energies is based exactly on the equilibrium constants, so the formula your professor used works almost by definition!

  As always, we can talk about the Gibbs free energy, enthalpy and entropy of such reactions. However, if we try to break these reactions down into the Gibbs energy, enthalpy and entropy of formation of each species in the reaction, we end up defining things like the Gibbs energy of a sodium cation in solution. Yet there is no way to generate just a sodium cation in solution (that would violate charge balance). Thus, we can only determine the Gibbs energy of formation of pairs of ions. This does not cause any real physical chemistry problems, but it causes a book keeping problem: how do we write down the Gibbs energy of formation of a Sodium cation in a book so that we can use it in any reaction where sodium cations are formed?

  The answer is that we come up with a sort of standard ion, $\ce{H+}$ (really something more like $\ce{ H3O+}$) and we define the free energy of formation of all other ions relative to this one. For example, I can measure the Gibbs energy change when I dissolve HCl in water to form $\ce{H+}$ and $\ce{Cl-}$.
  $$\ce{HCl}\text{(g)}\to\ce{H+}\text{(aq)}+\ce{Cl-}\text{(aq)}
 \quad\Delta G^0{}_1
$$
  I can now just define the Gibbs energy of formation of $\ce{H+}$ as zero and then determine the relative Gibbs energy of formation of $\ce{Cl-}$.
  $$\begin{align}&
\Delta G^0{}_1=\Delta G^0_{\mathrm{F}}(\ce{H+})+\Delta G^0_{\mathrm{F}}(\ce{Cl-})-\Delta G^0_{\mathrm{F}}(\ce{HCl})
\\&
\Delta G^0_{\mathrm{F}}(\ce{H+})=0\quad\text{(definition)}\\&
\Delta G^0_{\mathrm{F}}(\ce{Cl-})= \Delta G^0{}_1+\Delta G^0_{\mathrm{F}}(\ce{HCl})
\end{align}
$$


Are there metals that react with sulfuric acid but not hydrochloric (or vice versa)?

OK, let's clean things up.

Reaction of an acid with a metal is generally a redox reaction. The oxidizing agent are protons generated in the process of dissociation of the acid (let's forget about anions for a sec). The property that decides whether a metallic element can be oxidized by protons is its standard reduction potential. In general, any metal with a negative reduction potential will react with any Brønsted acid (one that produces protons), no matter whether it is a weak (e.g., acetic) or strong acid (e.g., hydrochloric). And if a metal has a positive reduction potential, it won't react with acids. Therefore, if we do not consider other factors, there are no examples to make the point that different acids react differently with a given metal, in fact they all should behave in the same way.

But what makes Chemistry fun are those 'other factors'. The anions produced in the dissociation process of an acid can also act as an oxidizer. For example, nitrate ions produced from nitric acid will oxidize copper, so the metal will dissolve in nitric acid. Sulfate ions are also oxidizing agents, but much weaker than nitrate ions, so copper won't dissolve in diluted sulfuric acid, but a hot and concentrated sulfuric acid will do the trick. And hydrochloric acid in the absence of some other oxidants will not attack copper at all. Another factor to consider is the passivation of a metal. A thin, invisible layer of an oxide or salt can protect a metal from an acid attack. Aluminum, which is a very reactive metal, will not dissolve in oxidizing acids (e.g., nitric), because it will develop a protective layer of aluminum oxide on its surface. Non-oxidizing acids, on the other hand, will easily do the job. Metal salts that are insoluble in water can also create a passivation layer.

Regarding the original question, there are some examples for sulfuric and hydrochloric acid, where they will behave differently due to the formation of insoluble salts. One example is thallium - it dissolves in sulfuric acid but not in hydrochloric acid. An opposite is barium - it dissolves in hydrochloric acid but not in sulfuric acid.


How can I make a lot of microscopic rods of metal?

Here is a relevant 2013 paper: Synthesis of pure iron magnetic nanoparticles in large quantity, to quote:

Free nanoparticles of iron (Fe) and their colloids with high saturation magnetization are in demand for medical and microfluidic applications. However, the oxide layer that forms during processing has made such synthesis a formidable challenge. Lowering the synthesis temperature decreases rate of oxidation and hence provides a new way of producing pure metallic nanoparticles prone to oxidation in bulk amount (large quantity). In this paper we have proposed a methodology that is designed with the knowledge of thermodynamic imperatives of oxidation to obtain almost oxygen-free iron nanoparticles, with or without any organic capping by controlled milling at low temperatures in a specially designed high-energy ball mill with the possibility of bulk production. The particles can be ultrasonicated to produce colloids and can be bio-capped to produce transparent solution. The magnetic properties of these nanoparticles confirm their superiority for possible biomedical and other applications.

So, to answer your question, it actually may be feasible based on more recent work. Note: the article appears in the applied physics section of Journal of Physics, reflecting the likely significance of this topic in this area.


The theory for a muon-antimoun "atom" is essentially the same as any hydrogen-like atom, just with masses changed.  See this Wikipedia page for the relevant formulae for relativistic (Dirac) and non-relativistic treatments.

The non-relativistic energy levels are given by :

$$E_n =  \frac { \mu c^2 \alpha^2}{2n^2}$$

where $alpha$ is the fine structure constant and $\mu$ is the reduced mass, in this case because the masses are equal, it's $\mu=\frac m 2$ where $m$ is the muon mass.

Note that the muon mass is about $206.8$ times that of the electron and the reduced mass is then $103.4$ times than of the electron making the energy levels for the non-relativistic state about $103.4$ times that of the Hydrogen energy levels.

As the Wikipedia page explain (in more detail that you asked for), a more detailed relativistic calculation using the Dirac equation includes corrections for the spin-orbit interaction.

  The core is lighter for about 100 times but the muon is 100 times heavier than the electron

This is wrong.

Both the muon and antimoun have the same mass, hence the effect of reduced mass being a factor of one-half.

The reduced mass of two particles is given by :

$$\mu = \frac {m_1m_2}{m_1+m_2}$$

Strictly speaking there is no "core" as both the muon and antimoun are of the same mass whereas in Hydrogen the electron is of much smaller mass than the proton.


Comparison of organic and inorganic acid strengths

You will be likely be expected to memorize approximate pKa values for different functional groups. Molecules containing protons with lower pKa values are more acidic.

In this case, these are the functional groups you have:

a. alcohol (pKa ~16)
b. alkyl halide and carboxylic acid (pKa ~4-5)
c. hydrobromic acid (pKa ~-9)
d. carboxylic acid (pKa ~4-5)
e. ammonia (pKa ~35)

I would have to suspect that b is slightly more acidic than d, but they are going to be in similar ball parks. What you're looking for is the most and the least acidic compounds, and the pKas for those compounds are distinct in this question. Since the acidity of a group is inversely proportional to its pKa, the compound with the lowest pKa among its groups is the most acidic and the compound with the highest pKa among its groups is the least acidic. Things get a bit more complicated than that when multiple groups are involved, but this rule usually holds for simple cases.

Hydrobromic acid has the lowest pKa, so it is the most acidic. Ammonia has the highest pKa, so it is the least acidic.


The active centre is the site where the substrate binds to the catalyst in order for catalysis to take place. On a molecular level, most catalysts (or most substances really) are not just flat surfaces, they have complex three dimensional shapes. The substrate often has to bind to the catalyst in a very specific way in order for the catalysed reaction to take place - this is the active centre.


Why is the notation for reactions not explicit?

The reaction, described in the form given in the original question, i.e.

$$\ce{C + O2 -> CO2}$$

simply conveys that each formula unit of carbon reacts with one formula unit of oxygen to yield one formula unit of carbon dioxyde. As such, for example, useful to balance how many grams of carbon may be burnt with $\pu{22.7 L}$ of oxygen.  As such, yes indeed, it is a limited representation.  In an analogy, it is like book keeping; (within reason) is does not matter how you paid at a garage (cash, debit or credit) to get your car back -- but to satisfy both sides, it has to be balanced.

This view contrasts with many organic reaction equations as reaction mechanisms which are intentionally drawn with further detail like this:

(source)

Here, the how is important, and consequently outlined with more attention.  As if you would need to explain an apprentice how the teeth of a gear have to be adjusted in face, pitch, flank, etc. to transmit the force from the engine to the wheels.


Small amount of acid is required for hydrolysis.
As a result ammonium salt and formaldehyde are produced.

Update:
It can be explained with enthalpy of reaction Wiki Enthalpy of reaction.
To calculate it, enthalpy of each  reactant required.
From wiki: ΔH(C6H12N4)=-99.2kJ/mol, ΔH(H2O)=-285.8kJ/mol, ΔH(NH3)=-45.94kJ/mol, ΔH(CH2O)=-115.9kJ/mol;
ΔH(reaction)=6*ΔH(CH2O)+4*ΔH(NH3)-ΔH(C6H12N4)-6*ΔH(H2O)=918kJ/mol;

As ΔH(reaction) is large, positive number, a lot of heat required for this reaction, as a result, equilibrium is far on the side of the reactants.


Imagine viewing it from right side:

H is upwards and CH3CH2 is downwards.
OH is leftwards and CH3 rightwards.
H and CH3CH2 are backwards(from right) so they must be on vertical line of Fischer projection and similarly for OH and CH3.

Alternatively:

Put the dotted wedge CH3 in any vertical position let it be bottom.
Then imagine all other groups to be planar, we see in clockwise order:

CH3CH2 , H , OH

Draw them similarly on the Fischer projection in clockwise sense.!

Note that both are (S)


Why is NH3 a nucleophile in NH3+CH3 -> H3NCH3

When working with nucleophiles and electrophiles, it is easier to think of them as "positive-ophiles" and "negative-ophiles" respectively.

With that said, NH3 (ammonia) has a lone pair of electrons that is attracted to positive charges.  The carbon atom with three bonds is electron deficient and positively charged (which your image does not show).  Thus, as a "positive-ophile"/nucleophile, the ammonia through its lone pair is attracted to and forms a bond with the positively charged CH3+ ion.

Yes, nitrogen does have a full octet, but its lone pair of electrons is nonetheless still attracted to the electron deficient and positively charged CH3+ ion.  This is similar to how water molecules (also fully satisfied octets) can bind with positively charged sodium ions in solvation process of salt.


1,4 addition to α,β-unsaturated carbonyls?

It is called 1,4 - addition because when you go through the mechanism of this reaction, then actually you will find that the carbonyl bond will break and oxygen will get negative charge.
Then nucleophilic will attack on beta carbon, i.e. carbon at 4th position and proton will migrate to negative oxygen which is at first position. Double bond will form between the carbonyl carbon and  carbon at 3rd position.

Thus, 1,4-addition takes place. But in the last step, keto-enol tautomerism takes place. And proton transfer takes place from oxygen to the carbon alpha to carbonyl. Therefore, final product is again saturated carbonyl compound.


how to add a methyl group to a straight carbon chain

Why would you want an expensive process to convert cheap molecules into other cheap molecules?  Brominate ethane, make the organo-lithium, displace methyl iodide. Epoxidize ethylene, react with methyl lithium, eliminate to the olefin. Silly.

Doing something cheap, easy, and fast on a huge scale with unwanted methane from oil wells to form a high energy density liquid for pipeline conveyance will make you rich. Methanol is marginal - and still can't be done.  Perhaps methane plus air into titanium-silicalite (plus hydrogen peroxide, or cobalt- or vanadium-doped plus air) then that mixed product through ZSM-5 into aromatic gasoline.  That could be exciting in various ways.

http://dx.doi.org/10.1016/j.apcata.2008.04.020
http://chemelab.ucsd.edu/methanol/memos/final.html


Entropies of Solvation for Chlorate and Nitrate Anions

Look at the graph on p. 65, it puts the entropy of solvation (298 K) for chlorate and nitrate at about 25 and 21 kJ/mol respectively.  This reference suggests that the entropy term for nitrate is 116 J/(K-mol), multiplying that by 298 K yields 34.6 kJ/mol.


Stability of MnO4- in basic vs acidic conditions

The obviously feasible and spontaneous disproportionation reaction can be explained by considering the standard electrode potentials (standard reduction potential) involved (quoted as half–cell reductions, as is the convention).

$\displaystyle \ce{MnO4- (aq) + e- -> MnO4^2- (aq)} \quad\quad (E^0 = +0.56~\mathrm{V})$
$\displaystyle \ce{MnO4^2- (aq) + 4H+(aq) + 2e-  -> MnO2(s) + 2H2O(l)} \quad\quad  (E^0 = +1.70~\mathrm{V}, \text{in acid solution})$

2. has the more positive potential, so this will be the reduction half–cell reaction.

1. has the less positive potential, so this will be (reversed) the oxidation half–cell reaction.

$$E^0_\text{reaction} = E^0_\text{reduction} – E^0_\text{oxidation} = (+1.70~\mathrm{V}) - (+0.56~\mathrm{V}) = +1.14~\mathrm{V} \gg 0~\mathrm{V}$$

Therefore, the reaction is very feasible.

Incidentally: Given the two half–cell reactions, you get the complete balanced equation by adding 2. plus 2 x 1. reversed.

The greater stability of the manganate(VI) ion in alkali can also be explained by considering the electrode potential for 2. in an alkaline media.

$\displaystyle \ce{MnO4^2- (aq) + 2H2O(l) + 2e-  -> MnO2(s) + 4OH- (aq)}\quad \quad  (E^0 = +0.59~\mathrm{V}, \text{in alkaline solution})$

so, re–calculating gives

$$E^0_\text{reaction} = E^0_\text{reduction} - E^0_\text{oxidation} = (+0.59~\mathrm{V}) - (+0.56~\mathrm{V}) = +0.03~\mathrm{V} > 0~\mathrm{V}$$

Therefore just feasible! but on the basis of an equilibrium argument, here, the far lower $E^0_\text{reaction}$, suggests the $\ce{MnO4^2-}$ ion is far more likely to exist, i.e. more stable, in a very high pH solution and in practice it is stable for a few hours in alkali.


Mimic a volcano eruption with household chemicals?

The 'elephant's toothpaste' experiment with hydrogen peroxide decomposition works well to produce large amounts of foam and might be suitable for what you are looking for.
Low concentration hydrogen peroxide solutions are readily available and not dangerous provided you don't get it in your eyes or drink it or anything silly.

Fill the bottle with some hydrogen peroxide solution, add a few drops of orange food colouring and some dish washing liquid and then add something to catalyse the decomposition. You could use something like potassium iodide or manganese dioxide to give you a very rapid decomposition but if you want a slightly less violent and more sustained reaction then apparently dried yeast dissolved in warm water works quite well (yeast contains good amounts of catalase, an enzyme which catalyses the decomposition of hydrogen peroxide).

Having never done this experiment, I can't tell you exactly what quantities to use so you should probably test it before doing it as a demonstration. However, there are some descriptions of the experiment on the internet to give you a good starting point.

http://www.sciencebob.com/experiments/toothpaste.php
http://www.stevespanglerscience.com/lab/experiments/elephants-toothpaste


The box in my kitchen is labelled "reine Zitronensäure" (= pure citric acid) and "lebensmittelsauber".

According to the manufacturer, the latter means that citric acid in this quality can be used to descale (decalcify) kitchen items, but it is not intended to be used as a food supplement and should not be digested!

This may have real technical reasons, i.e. the product may in fact contain impurities resulting from production, processing and packaging. The manufacturer however does not specify the kind or amount of impurities.

On the other hand, this kind of labelling may just reflect different legal treatment (certification, monitoring, etc.) of products intended as food supplement and those only used to clean kitchen equipment.

Concerning waterfree vs monohydrate, neither the package in my kitchen, nor the website of the manufacturer provide any information.

If you know a chemist, this question is easily solved: Let him measure the melting point of a sample. Waterfree citric acid melts at 153 °C, the monohydrate at 100 °C.


What is the exact definition of the radial distribution function?

The atomic orbitals (wavefunctions) $\psi(r,\theta,\phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(\theta,\phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.

As a concrete example, the $2\mathrm p_z$ orbital ($\{n,l,m\} = \{2,1,0\}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:

$$R_{2,1} = \sqrt{\frac{1}{24}}\left(\frac{Z}{a}\right)^{3/2}\rho\mathrm e^{-\rho/2} \qquad Y_{1,0} = \sqrt{\frac{3}{4\pi}}\cos\theta \tag{1}$$

where $a$ is a collection of constants ($a \equiv 4\pi\varepsilon_0\hbar^2/\mu e^2$), and $\rho$ is related to $r$ by another collection of constants ($\rho \equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2\mathrm p_z$ orbital.

The atomic orbitals themselves are normalised in the sense that

$$\int_{\text{all space}} |\psi(r,\theta,\phi)|^2 \mathrm dV = \int_0^{2\pi} \!\!\!\!\int_0^\pi\!\!\!\int_0^\infty |R(r)|^2|Y(\theta,\phi)|^2 r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi = 1  \tag{2}$$

Conventionally, both the radial and angular wavefunctions are separately normalised. This means that

$$\int_0^\infty |R(r)|^2 r^2\mathrm dr = 1; \qquad \int_0^{2\pi}\!\!\!\!\int_0^\pi |Y(\theta,\phi)|^2 \sin\theta\,\mathrm d\theta\,\mathrm d\phi = 1  \tag{3}$$

(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)

Normally, since $\psi = \psi(r,\theta,\phi)$, the probability density $|\psi|^2$ depends on all three coordinates. The probability density (technically, $|\psi|^2\,\mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube frustum-like shape,† located at the point $(r,\theta,\phi)$. By summing these $|\psi|^2\,\mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.

However, often we are not interested in the angular dependency, and so what we can do is to sum up all these bits which have one value of $r$, but different values of $\theta$ and $\phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $\theta$ and $\phi$. By virtue of the angular wavefunction being normalised, the integral over $\theta$ and $\phi$ evaluates to $1$, and hence we have

$$|\psi|^2\,\mathrm dV = (|R|^2r^2\,\mathrm dr)\cdot (|Y|^2\sin\theta\,\mathrm d\theta\,\mathrm d\phi) \overset{\text{integrate over }\theta,\phi}{\longrightarrow} |R|^2r^2\,\mathrm dr \tag{4}$$

So, this quantity $|R|^2r^2\,\mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.

This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $\theta$ and $\phi$, it comes out to be $1$. The widely-cited alternative definition $4\pi r^2 |\psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have $\{l, m\} = \{0,0\}$ and

$$Y_{0,0} = \sqrt{\frac{1}{4\pi}}  \tag{5}$$

(you may check that this is normalised according to equation $(3)$.) Hence

$$|\psi|^2 = |R|^2|Y^2| = \frac{1}{4\pi}|R^2| \tag{6}$$

and it therefore follows that $|R|^2r^2 = 4\pi r^2|\psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4\pi r^2|\psi|^2$ is used (or $4\pi r^2 \rho$, where the probability density $\rho$ is simply $|\psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.

Based on the above arguments this is my take on the matter:

$|R^2|r^2$ is the best definition of the RDF;
$4\pi r^2|\psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);
I still don't fully understand why Housecroft has an additional factor of $4\pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4\pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $\sqrt{1/(4\pi)}$.

† Thanks to Ruslan for pointing this out! If you are curious as to what this looks like, there is a good diagram on LibreTexts, which I have re-uploaded to SE here. Note, however, that the convention in this diagram is $(\rho,\phi,\theta)$ instead of $(r,\theta,\phi)$.


Many software, both online and downloadable executables, are available for identifying a molecular structure - either by giving the official IUPAC name, or matching up the structure to a large database and returning IUPAC/trade names/properties/etc...

Online: NIST, Pubchem, ChemSpyder, Reaxys

Download: ChemBioOffice Software

A variety of options are available to search for structures such as by molecular formula, structure, identification values, etc... Easiest for most approaches is using any of the software's "Draw" features, to draw the exact structure and then search the database for it. Online options may have difficulty with the exact structure searches on molecules as large as yours above, so as much detail as possible if useful (stereochemistry, for instance, using dashes and wedges).

ChemBioOffice (specifically ChemDraw) is able to give you the IUPAC name, which would then let you look it up in any of the online databases.

If the drawing of the structures is causing you difficulty, you can also search with molecular formulas, but it may take some trial and error to identify the actual structure you want as you look for larger or more complex molecules this way. NIST is a great choice for this approach.

Outside these options, if you have any interest or experience coding, you may be interested in using RDkit (Python module), which has a large array of functions for molecular data interaction. For instance, if you use those above structure "Draw" features online, most will also show a SMILES code which is a way of shorthanding a molecular structure's features; place a SMILES code into a script that accesses RDkit and you can request it to draw the structure, look up information related to it (if it can find the SMILES code in its databases), etc... Warning - it is mostly just development manuals, no tutorials last I checked, so perusing the module can be rather arduous.

As an aside, your molecule in particular gave those online database structure searches a bit of difficulty when I attempted it, so you may need to find a way of giving it more structural information, or perhaps searching for sub structures first (such as just the fused and linked cyclohexanes without the Chlorines or amine group).

However, I found one very similar to your structure, but with aromatics instead for two of the rings: 1,1-dichloroheptadecane


How can we separate a mixture of two sulphide ores using froth floatation method?

In the froth floatation process, the role of the depressants is to separate two sulphide ores. They do so by  preventing one ore from forming froth.

In this process, we want some substances to settle down and some to float. Now ideally, we want the sulphide of the substance that we wish to extract tonfloat and the gangue to sink. However, sometimes the ore we are using also has other sulphides present in it as impurities. For example, lead sulphide generally has zinc sulphide also present as impurity which will also float along with the lead sulphide. To make sure that lead sulphide floats but zinc sulphide do not float, we add depressants.
Commonly used depressant is $\ce{NaCN}$.
To separate two sulphide ores ($\ce{ZnS and Pbs}$), $\ce{NaCN}$ selectively allows $\ce{PbS}$ to come with froth, but prevents ZnS. This happens because of the reaction of $\ce{NaCN}$ with $\ce{ZnS}$ to form $\ce{Na2[Zn(CN)4]}$.
The complete reaction is,
$$\ce{4NaCN + ZnS  →  Na2[Zn(CN)4] + Na2S}$$


What does "the magnesium salt of EDTA" mean?

The authors have some strange kind of humor... it's a riddle.

Below you might see

  2) If the magnesium salt of EDTA is unavailable, dissolve 1.179 g disodium salt of
  ethylenediaminetetraacetic acid dihydrate (analytical reagent grade) and 780 mg magnesium
  sulfate (MgSO4⋅7H2O) or 644 mg magnesium chloride (MgCl2⋅6H2O) in 50 mL distilled water.
  Add this solution to 16.9 g NH4Cl and 143 mL conc NH4OH with mixing and dilute to 250 mL
  with distilled water.

From the molar mass of $\ce{MgSO4 * 7 H2O}$ ($\pu{246.47 g/mol}$) you can calculate the number of moles of magnesium the "magnesium salt of EDTA" should contain. With the help of the data sheet of your purchased magnesium salt of EDTA you can calculate the required quantity in grams.

You also could calculate possible stoichiometries of the imaginary "magnesium salt of EDTA" if you like. But be aware that not all commercially available magnesium salts of EDTA have a well defined stoichiometry.


The term you are looking for is "allotropes", and carbon has a whole bunch of them, but the three most readily spotted in nature are graphite, diamond and fullerene, however, none of them are simple chains such as those postulated in the answer.

Graphite exists in large honeycomb flakes held together on two axes by covalent bonds, and on the third by weaker Van der Waals forces, which is why they come off readily under mechanical shear, leading to their popularity in pencil lead.

Diamond are huge, multiply interlocked complexes of crystalline carbon, although often with impurities locked into their structure, perverting the crystal structure and yielding interesting colors as a byproduct.

Fullerene is a molecular allotrope, consisting (usually) of 80 carbon atoms arranged in a buckyball. Initially thought to be a lab-grown species, it has since been identified as a natural result of certain combustion processes.

Now, none of these are chains like what you describe; while some of the compounds you sketch out are likely possible and present, they are also likely to be highly reactive, perhaps even to the point of decomposing spontaneously.

While amorphous carbon chains may very well exist, without a crystalline structure to shield their electrons, their reactivity would likely be such that they would be unable to exist for very long at all.

Simplistically speaking, the long stable chains of carbon found in hydrocarbons aren't entirely characteristic of carbon, as such, but rather, characteristic of hydrogenated carbon.


Which melting point would indicate a metallic structure?

Now, before we get to the "expected theoretical" answer, I should make one point clear: predicting the solid state just based on the melting point data can not be perfect! Browsing the data on melting points on Wikipedia, you'll find that vanadium has a melting point of $\pu{1910 ^\circ C}$, while zinc has a melting point of $\pu{420^\circ C}$. Both these values are very close to the options (1) and (2), hence, one can very well argue that both the options represent a metallic structure.

But this is definitely not the answer your teacher expects. A theoretical (and again, practically wrong) approach instead is to realize that the values of intermolecular forces increase in the order of molecular solid < metallic solid < ionic solid < covalent networked solid. Thus, the option (3) is probably a molecular solid, option (1) can be either an ionic solid or a covalent networked solid, while option (2) must be the metallic solid as expected.

  I understand that the third option would most likely be covalent.

Just get rid that misconception already. Covalently bonded networked solids have higher melting points than several metals and even ionic solids! Diamond melts at $\ce{3800 K}$, while boron melts at $\ce{2349K}$. Our beloved ionic solid sodium chloride melts at a poor $\ce{1073K}$, while all the 3d series metals melt under $\ce{2200K}$! Hence, the assumption, that low melting point indicates a covalent solid, is false.


Suppose you want to design an industrial scale continuous flow chemical reactor to produce some product stream from a set of feed streams.  You need to determine how big to make the reactor so that enough product is produced, given the reactant flow rates that you are feeding.  To figure this out, it's pretty important to know the rates of the reactions occurring within the reactor.  It's also important to know the amount of heat you have to add or remove from the reactor to keep it operating properly.  This is also determined by the reaction rates.

Another example is in atmospheric science where you need to know the rates of chemical reactions (including photolytic) occurring in the atmosphere so that you can quantitatively evaluate problems such as fluorocarbon/ozone and global warming.


Changing Temperature of a Galvanic/Voltaic Cell: Should I wait for the electrodes to also change in temperature?

In general the higher the temperature of the electrolyte, the more power you can draw from the cell.

The reason for this is, because the ions of the electrolyte can move more freely when the temperature is higher and chemical reactions do in general happen quicker with increased temperature.

On the other hand side the hotter a metal gets, the bigger its resistance will become. This is because electrons can’t move as freely when the temperature is increased.

However this effect usually is of importance when temperatures are quite high, like in an light bulb.

If you just heat the electrodes for 30-40°C, I doubt you would be able to even measure the difference in resistance (even though it does exist).

So to sum it up, the output power of a cell in relation to the temperature will mainly be altered by the temperature of the electrolyte.


Are there more positive azeotropes than negative azeotropes?

Summary. Positive-deviation azeotropes mean that like interactions are much stronger than unlike interactions, whereas negative-deviation azeotropes mean the opposite. The former case is more common because most substances interact better with themselves than with other substances---like dissolves like.

(So I've only just rephrased the question at a lower level: one could equally well ask why like should dissolve like, but believing this statement on faith is perhaps more palatable than just accepting that there are more positive-deviation azeotropes.)

First, a few definitions.

A binary solution showing positive deviations (negative deviations) to Raoult's law is one in which the dew-point curve in a pressure vs. mole fraction graph lies above (below) the line connecting the pure vapor pressure of the two species.

A binary azeotrope is positive deviation (negative deviation) or minimum boiling (maximum boiling) if it shows positive deviations (negative deviations) to Raoult's law and exhibits a local maximum (minimum) in the dew-point curve in a pressure vs. mole fraction graph. From solution thermodynamics, one can prove that the vapor and liquid mole fraction of either species is equal at these extrema.

Physically speaking, positive deviations (negative deviations) from Raoult's law imply that like interactions are stronger than (weaker than) unlike interactions in the solution, for the solution's dew point pressure is greater than (less than) that of an ideal solution.

Azeotropes are extreme forms of these positive deviations (negative deviations), when the dew-point curve further exhibits a maximum (minimum), such that at some composition the solution's dew point pressure is actually greater (lesser) than the pure vapor pressure of either species.

We expect, therefore, that positive-deviation azeotropes are more common than negative-deviation ones because it is more likely for two species to have stronger like than unlike interactions.


As you have not provided a source for your definition of $s_i$, I shall redefine it in terms of Raoult's Law and the Ideal Gas Law. To do this, we shall first take Raoult's Law and solve for $P$.

$$\begin{align}y_iP&=x_iP_i^{sat}(T)\\P&=\frac{x_iP_i^{sat}(T)}{y_i}\end{align}$$

we shall then take this form, and plug it into the ideal gas law, considering only the gas phase of the equilibrium system. Note that because the system is at equilibrium, $P_{bulk}=P_{liq}=P_{vap}$ and $T_{bulk}=T_{liq}=T_{vap}$. If the volume of the liquid phase is assumed to be significantly lower than that of the vapor, then $V_{bluk}=V_{vap}$. After plugging in this, we will then work towards the form of $s_i$ as shown.

$$\begin{align}P_{vap}V_{vap}&=n_{vap}RT_{vap}\\PV&=n_{vap}RT\\n_{vap}&=\frac{PV}{RT}\\n_{vap}&=\frac{x_iP_i^{sat}(T)}{y_i}\frac{V}{RT}\\\frac{n_{vap}y_i}{x_i}&=\frac{P_i^{sat}(T)V}{RT}\end{align}$$

From this, and the definition of $s_i$ as you stated, we can show that

$$s_i=\frac{P_i^{sat}(T)V}{RT}=\frac{n_{vap}y_i}{x_i}$$

which is subject only to the reasonable assumptions of ideality as stated above, and the bounds per the definition of each term. Namely, $\left\{P_i^{sat}(T),V,T\right\}$ must be positive, real numbers and that $\left\{x_i,y_i\right\}$ must be positive, real numbers with bounds of $\left\lbrack0,1\right\rbrack$.

No further inequalities are required to describe the system.

To solve such a system, you would then set up a system of equations and numerically backsolve for an answer, in terms $n_{vap},x_{i},y_{i}$. For example you might feed a set of equations the seed values of $n_{vap}=1$ and $x_i=.5=y_i$ such as seen below, then optimize your values to minimize the error between the calculated and target values.

The above seed values could then optimize to a particular solution that looks like. Note that this does not represent the whole of the solution space.

If you wish to further control where your particular solution ends up, you can restrict the solution space by hold one of the values to be a known constant. Below is one such example as described per your comments.

In this case, it is not possible to find a reasonable solution, as $n_{vap}=0.25~\text{mol}$ lies outside the solution space for this system of equations. Without going too heavily into the derivation, it can generally be stated that this value is a bit too low to garuntee that you will be able to find a solution. This more says that if you had this system, it would likely evolve off more of its liquid into the gas phase at in order to reach equilibrium than it says the system could not exist, though yours is not an unreasonable reading of the result.

A good rule of thumb for this would be that you can expect to find a reasonable solution for $\left\{x_i,y_i\right\}$ for any system in which $s_1 \leq n_{vap} \leq s_2$ holds, however this is by far the most complicated way to solve out this system of equations. A simpler approach to this would be to hold either $x_i$ or $y_i$ constant to restrict the solution space, as any constant value on the bounds of $\left\lbrack0,1\right\rbrack$, as already assumed will garuntee a unique solution to the system of equations.


Can you push a liquid and a gas into a container at different pressures?

Pressure is due to the random impact of molecules on the surface, regardless of whether water or carbon dioxide or anything else. Once equilibrium is reached, all components are at the same pressure.

That said, there's no problem using a moving stream of $\ce{CO2}$ to propel and to mix with water at much lower pressure, using Bernoulli's principle. The automotive carburetor and spray atomizer operate on this basis.

The alternative, of course, is to add the water to the container through a check-valve, then add $\ce{CO2}$ at a higher pressure.

The choice is a continuous process vs. a batch process.


Compound having highest standard reduction potential

Top 5 oxidants from the CRC Handbook of Chemistry and Physics [1, p. 5-83]:

TABLE 2. Reduction Reactions Having $E^\circ$ Values More Positive Than That of the Standard Hydrogen Electrode
$$
\begin{array}{ll}
\hline
\text{Reaction} & E^\circ/\pu{V}\\
\hline
\ce{F2 + 2 H+ + 2 e- ⇌ 2 HF} & 3.053\\
\ce{Tb^4+ + e- ⇌ Tb^3+} & 3.1\\
\ce{Pr^4+ + e- ⇌ Pr^3+} & 3.2\\
\ce{Cf^4+ + e- ⇌ Cf^3+} & 3.3\\
\ce{XeF + e- ⇌ Xe + F-} & 3.4\\
\end{array}
$$

Note that standard reduction potential is not characteristic for a single compound. Instead, it is only makes sense for a given redox pair – of course, assuming standard conditions ($\pu{298.15 K}$, $\ce{101.325 kPa}$), activity of all soluble species $\pu{1 .000 mol L-1}$ and measured against the standard hydrogen electrode.
For example, iodate $\ce{IO3-}$ has very rich redox chemistry and many SRP values [1, p. 5-79]:

$$
\begin{array}{ll}
\hline
\text{Reaction} & E^\circ/\pu{V}\\
\hline
\ce{2 IO3– + 12 H+ + 10 e- ⇌ I2 + 6 H2O} &1.195\\
\ce{IO3– + 6 H+ + 6 e- ⇌ I– + 3 H2O} &1.085\\
\ce{IO3– + 2 H2O + 4 e- ⇌ IO– + 4 OH–} &0.15\\
\ce{IO3– + 3 H2O + 6 e- ⇌ IO– + 6 OH–}& 0.26\\
\end{array}
$$

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.


How do I write the deterministic rate equation for bimolecular reaction with similar particles?

There is really only one independent equation here because

$\frac{dB}{dt} = -2\frac{dA}{dt}$

The "second equation" in the OP is just -2 times the first equation.

Every molecule of A destroyed creates 2 molecules of B.

Every 2 molecules of B destroyed creates 1 molecule of A.