What is the difference between "vapour" and "gas"?

Vapor is a much older word alluding to dampness and it was not coined by scientists. It is in use since the 1300s. The actual meaning of meaning of vapor is "Matter in the form of a steamy or imperceptible exhalation; esp. the form into which liquids are naturally converted by the action of a sufficient degree of heat. This is the original 13th century meaning of vapor. The German equivalent of vapor is Dampf (which is relatively clearer in terms of feeling- imagine entering a basement which is damp from water vapor). Gas is relatively a new word as per the unabridged Oxford English Dictionary, and it entered English in the 1600s.
In fact, the OED mentions the distinction "Gas: A substance in a state in which it expands freely to fill the whole of a container, having no fixed shape (unlike a solid) and no fixed volume (unlike a liquid); spec. (distinguished from a vapour) such a substance above a critical temperature such that it cannot be liquefied by the application of pressure alone"
You can also think of the general meanings of vapor in scientific usage:
A substance which is a liquid at room temperature or under ambient conditions, and if it is heated enough so that the liquid evaporates (same root of word), you can call that as vapor of that substance.
Examples: Mercury is a liquid metal, when it is heated enough in a an electric bulb, it becomes a gas, hence the name mercury vapor lamp. It is not called a mercury gas lamp.
It is not necessary that heat be involved. You will find the usage "cold vapor method for Hg analysis" again for mercury which means that Hg was a liquid in a sample but by blowing a gas over it, it has been evaporated.
The recent meaning of vapor from OED, "Vapor: In modern scientific use, a fluid that fills a space like a gas but, being below its critical temperature, can be liquefied by pressure alone."
This is clear enough. Now apply this distinction to your phase diagram and notice the position of the critical point. The vapor "state" is below the critical point.


How does sulfuric acid catalyze the reaction of potassium chlorate and sucrose?

I think the sulfuric acid isn't really a catalyst here -- I think you use the heat from the reaction between sulfuric acid and sucrose (which will start happily at room temperature) to start the reaction between the potassium chlorate and the sucrose (which usually won't).

ETA: Once you've done that, it's pretty much a straight-up messy combustion reaction, only with the oxygen being provided from a solid source (thermal decomposition of chlorate to chloride and oxygen as the reaction you mentioned shows) rather than anything mechanistically clean. Compare, for example, this fairly typical demonstration just using hot chlorate and a sugar source: Oxidation of Sugar or Gummi Bear with Potassium Chlorate.


How many p-orbitals are there in each period?

For the p subshell, the azimuthal quantum number, which describes the angular momentum and shape of the orbitals, is $l=1$. For the associated magnetic quantum number, discrete values in the range $-l, -l+1, ... 0 ..., l-1, l$ are allowed, so the number of different $m_l$ values is $2l+1$.

With $l=1$, it follows that there are $2 \cdot 1+1=3$ p orbitals with the $m_l$ values of -1, 0 and +1, respectively. Each orbital can hold 2 electrons, so the p subshell can contain a maximum of 6 electrons. This also explains why there are 6 elements in the p block of each period. With increasing atomic number $Z$, one p electron is added per element in this block.


Selective etchant for Al2O3 (Does not etch Al, Cu, Zn, Sn)

In Fabrication of Microstructures by Wet Etching of Anodic
Aluminum Oxide Substrates Chem. Mater. 2005, 17, 4049-4052,

5% phosphoric acid is used and is selective relative to aluminum, but you need to check how it would do with respect to the other metals you mention

  Finally, the AAO substrate was wet-etched in a 5 wt % H3PO4 solution
  at 20 °C for 8 h


What is the energy released in chemical reactions?

Let's start off on common footing by making sure we understand the definition of energy:

Energy is the capacity to do work.

An object may have many types of energy, we'll get to those in a second, but the total energy of an object will be the sum of its kinetic energy and its potential energy.  Kinetic energy is the energy associated with the movement of an object and potential energy is the energy associated with an object's position.

To take a physical example, hold out a book (preferably your heavy, expensive general chemistry textbook) and drop it on the floor (preferably, not on your little brother's toes).  The book falls to the ground and a loud noise is heard.  As the book moved, the potential energy stored in its original position was converted to kinetic energy as it fell to the ground.  Upon hitting the ground the individual atoms that make up the book crashed in to the individual atoms that make up the ground, causing them all to move.  This movement of atoms is thermal energy.

Let's focus on thermal energy for a moment, the majority of chemical reactions, including the combustion reactions you refer to, involve the transfer of thermal energy.  Hold a cup of coffee in your hand; what happens to your hand?  Assuming the coffee was hot, and your hand was not, your hand starts to feel warm.  We might say your hand is heating up.  This thing we refer to as *heat is the transfer of thermal energy; the molecules in the cup of coffee were bouncing around, crashing in to the cup, which made the molecules in the cup crash around hitting your hand.

So now let's turn to chemical energy which is the energy associated with the position of nuclei and electrons in atoms and molecules. Since we are talking about (relative) positions of objects, this energy is a form of potential energy.  Some molecules, such as methane, have nuclei and electrons that are positioned in such a way that they store a large amount of potential energy.  When the positions of these nuclei and electrons are moved, potential energy is converted into kinetic energy.

So in summary, the energy of chemical reactions is best described in terms of potential and kinetic energy, or in other words the relative position and movement of nuclei and electrons.  Often, the energy we experience in chemical reactions can be adequately described by the transfer of heat, and therefore thermochemistry and thermodynamics take up a large portion of our instruction in Chemistry.  Other types of energy, such as electromagnetic radiation, electrochemical, and nuclear, can all be described in terms of potential and kinetic energy, namely the position and movement of objects.


What is the intuition behind 'mol' as a unit 'symbol'

The symbol mol is due to Ostwald who was a very influential and respected physical chemist more than a century ago. In German, "mole" is "Mol". It is a shortform of Molekül.
I believe the Internet Archive has an English translation of his historically famous book of 19th century. This is where I recall reading this. The original reference is W. Ostwald Hand- und Hilfsbuch zur Ausführung physikochemischer Messungen. This book is available on Google Books (pg 278). A footnote appears there:

Um nicht stets den schleppenden Ausdruck Gramm -Molekulargewicht brauchen zu müssen, werde ich diese Größe in der Folge kurzweg Mol nennen, und erlaube mir, diesen Vorschlag zur allgemeinen Annahme zu empfehlen. Man schreibt ohnedies diese Größe häufig Mol., und braucht daher nur den Punkt fortzulassen.

In order not to always have to use the sluggish expression gram - molecular weight, I will simply call this quantity Mol in the following, and allow me to recommend this proposal for general acceptance. Anyway, this quantity is often written as Mol., so you only have to omit the point.
Translated via (www.deepl.com) with some manual changes.


What does the volume of a gas depend on?

I must clear one thing up. gases always fill their container. If you take away gas particles, it will always fill its container. The volume of gas is proportional to many different factors, though. Based on the ideal gas equation, volume is proportional to moles and temperature. It is inversely proportional to pressure. when the text say that volume is proportional to amount, it is assuming all other variables are constant, and not changed. So yes, in general, as there is an increase in volume, there is an increase in number of moles of gas.


A radical, according to the IUPAC, has an unpaired electron. Radicals may or may not be ions.

  A molecular entity such as $\ce{.CH3}$, $\ce{.SnH3}$, $\ce{Cl.}$ possessing an unpaired electron.

http://goldbook.iupac.org/R05066.html

Contrary to your book's claim, the sodium ion is not considered a radical. It actually has all its electrons paired. It is isoelectronic with helium.

Regarding your book's usage of "radical:" your book's terminology is outdated. Here is the IUPAC statement on your book's usage of the term:

  In the past, the term 'radical' was used to designate a substituent
  group bound to a molecular entity, as opposed to 'free radical', which
  nowadays is simply called radical. The bound entities may be called
  groups or substituents, but should no longer be called radicals.


Is there an established procedure for titanium tetrachloride titration?

$\ce{TiCl4}$ is amenable to distillation if done carefully so as to exclude moisture. That doesn't mean it isn't a pain, however.

We do a lot of chemistry using $\ce{TiCl4}$ and other Ti Lewis acids (including some Ti aldol reactions on v. small scale), so end up having to distill it quite often (roughly once a quarter). Once distilled, it lasts in the fridge until it runs out so long as the flask is well sealed and everyone remembers to flush with argon before they're finished using it.

I'll quote from Purification of Laboratory chemicals for general information, but also note that we don't follow this procedure. (Mercury makes me uncomfortable)

  Ref: 'Purification of Laboratory Chemicals', 6 ed.

In practice, we simply distill the $\ce{TiCl4}$ under reduced pressure (high vacuum pump), discarding the first few mL as would be common procedure. This usually yields $\ce{TiCl4}$ that is clear in colour and sufficiently pure for our chemistry. If some colour remains, a second distillation from copper turnings is occasionally done, but this is rarely needed.

The key bit of information is to make sure that you quench all of the glassware and tubing used with a weak base (sodium hydrogen carbonate is sufficient), and to ensure that the pump exhausts into a good fume cupboard incase any fumes do escape. All of the glassware used should also be heavily dried prior to use.


What are the units of P and V when R is expressed in terms of J/(mol K)?

Would you agree that pressure is expressed as a force acting on a surface? If so, let's juggle with units and start with the pascal.
$1\ \mathrm{Pa} = 1\ \mathrm{Nm^{-2}}$
will certainly be familiar to you ;)

Would you further agree that we can express the joule as $1\ \mathrm J = 1\ \mathrm{Nm}$?


How did they assign absolute configuration to these cis and trans 2-methylcyclohexanols?

You have somehow decided that the front side of the ring is actually the rear side … or something. So your drawings show exactly the wrong stereochemistry. For the hydroxy-carbon, you thus would, in theory, get the incorrect assignment (S) but you use the rule you reference to reverse that into (R). Naturally, because your initial assertion of which atoms are in which plane was incorrect, your application of the rule was also incorrect. For the methyl carbon, your inverted ring reproducibly gives you the inverted assignment (R) and you do not apply any incorrect correction.

Figure 1: the molecule in question.

In figure 1, I have reproduced the molecule in the more ‘standard’ chair configuration which gives a better approximation of the bond angles. The wedged/bold portion of the ring is in front of the paper plane, the thin solid portion behind. I have also rotated the ring by $60°$ so that the hydroxy function is on the rightmost carbon atom for easier discussion.

It should be a rather simple exercise to determine that the hydroxy-carbon (carbon 1) indeed does have an (R) configuration: the hydroxy group has highest priority, the front side of the ring second-highest, the back side third and the hydrogen pointing downwards fourth. Therefore, we must look from above the ring plane to put the lowest substituent to the back; the rotation from hydroxy to methyl side to non-methyl side is thus clockwise, therefore (R).

The same analysis performed on carbon 2, the methylated one, should give you:

carbon 1 ($\ce{CHOH}$) is highest priority; going to the back right and slightly downwards
carbon 3 ($\ce{CH2}$) is second; going to the left, slightly downwards while staying in the plane
the methyl group is third; going up while staying in the plane
the unmarked hydrogen is fourth; this goes to the front right and slightly downwards.

Thus, we should look from inside the ring as the smallest substituent is pointing towards us. However, we can also look from the smallest substituent and invert the assignment; the way we see the three groups is again clockwise, so if we were to look from inside the ring (or behind it) we would see them anticlockwise, thus (S).

Note that the wedged and bold bonds — both in the paper’s drawing and in my figure — are not there to indicate a positioning from the stereocentre but to indicate which half of the ring is in front of the paper plane and which is behind. Thus, you cannot — like you did — assume the bolded bond to be in front; it actually is in your paper plane when you are looking at the two stereocentres on either side of it.


Carbocation stability + determining position of attack on a carbocation in SN1 reaction

It is important to note that the two structures shown aren't actually different. That's what resonance implies that in reality, it exists in between both of the as a 'hybrid'.
So it is not right to state that the nucleophile attacks on the left structure 'or' the right.

Coming to the carbon, (again, it's not right to take the structures to be different)

"I just don't know why the methanol attacks the $\ce{sp2}$ carbon in the right structure instead of the carbon with the $+1$ formal charge on the left structure"

It's actually the same carbon! and the hybridisation is still $\ce{sp2}$ no matter what, as the carbocation is also a planar species($\ce{sp2}$). And this is where the attack occurs.

"I'm also not sure why the MeOH attacks the carbon next to the carbocation and not the oxygen itself, where it has the positive charge"

'Carbon next to the carbocation' ? I believe you are confusing the +1 formal charge on oxygen to be a carbocation which is not true( Carbocation has the positive charge exclusively on Carbon) .
The attack is ocuring at the carbocation.

Our nucleophile here, is $\ce{OH-}$ while oxygen only has a +1 formal charge and also, the positive charge is more localised on the carbon as Zhe mentioned.
This is because oxygen being a more electronegative species, doesn't really 'like' the positive formal charge on it.
But still,

"doesn't the nucleophile attack the carbocation, or whichever atom bears the positive charge?"

Well, a nucleophile is a species that donates electrons at electron deficient sites. But oxygen still has it's lone pair! and the $\ce{OH-}$ will face repulsions from them if it tried to attack here.
The real electron deficient species here is only the carbocation and thus it is the preferred site for attack by the nucleophile.


Do microwaves heat polar molecules or molecules with polar bonds?

It depends on the strength of the incident EMF. Pure $\ce{H2}$ can be ionized by microwave radiation in multi-photon collisions, as well as being heated. Even nonpolar molecules have quantum fluctuations in the electron cloud, so they interact to some extent with n electromagnetic field.


Pressure at which graphite and diamond are in equilibrium

For each phase (graphite or diamond) you can write

$$ \mu = \mu^\circ+\int_{P^\circ}^{P} V_mdP $$
at constant T.

We are asked to find the pressure $P=P_{eq}$ at which carbon coexists in the two phases (the Gibbs free energy is equal in both) so that

$$ \mu(diamond) = \mu(graphite)$$

This leads to

$$\Delta \mu^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

and, since $\Delta \mu^{\circ}= \Delta_f G_m^{\circ}$, ultimately to the expression you provided:

$$\Delta_f G_m^{\circ}=-\int_{P^\circ}^{P_{eq}}\Delta V_mdP$$

The approximation you are allowed to make at this point is that the solids are incompressible, such that their $V_m$ are constant with change of pressure. It follows that

$$\Delta_f G_m^{\circ}=-\Delta V_m\int_{P^\circ}^{P_{eq}}dP=-\Delta V_m(P_{eq}-P^\circ)$$

which can be solved for $P_{eq}$:

$$P_{eq}=-\frac{\Delta_f G_m^{\circ}}{\Delta V_m}+P^\circ$$


What is parachemistry? What does a parachemist do?

I'll try to present the best I could find. The titles "para", "meta" and "ortho" were perhaps given by Hermann Kolbe (1818-1884) way back in the nineteenth century. For one thing, Kolbe had the reputation of being a well recognized teacher and a skilled and productive chemist. He is also well known today (partly) as the author of a critique of the foundation of stereo-chemistry, with was established by the Dutch chemist, Dr. J.H. vant Hoff. He had directed bitter remarks not only towards vant Hoff for his hypotheses (which was perhaps the boldest), but at nearly the entire realm of contemporary organic chemistry.
Background

$\ldots$ Once he gave his amanuensis the empirical formulas for three novel compounds for which their discoverer had just assigned structural formulas. The man reported back in a half hour with several more candidate structures, some of which looked more probable to Kolbe. Kolbe concluded that structure theory is a dangerous toy, especially for inexperienced chemists, and that structures are often assigned "in one's sleep," with little or no empirical warrant. $\ldots$

Privately, Frankland "entirely dissent[ed]" from this judgment and Volhard also objected privately; he joined Frankland in expressing regret over the personal character of many of Kolbe's remarks. Kolbe's response was pretty straightforward. As he remarked he must criticize and contest the chemical ideas that he considered false and worthless, he would be happy if his views were contested. He later wrote to Varrentrapp that he would rather be considered sharp-tongued than cowardly.
The crusade

In Kolbe's third and fourth retrospects, those for 1873 and 1874, he went after his favorite example of structuralist excess: Kekulé's benzene theory. He absolved Kekulé himself of much of the blame, for he was convinced that Kekulé regarded the theory merely as an intriguing and useful hypothesis. However, most chemists by this time viewed the hexagon as "infallible dogma," as "the Pope is for Catholics." They were true fanatics, Kolbe wrote, and viewed him (Kolbe) as a rank beginner, of weak understanding. They were right, he thought, for he could not understand arguments built "in the air" or "on loose shifting sand."

The end of their sand castles was not distant. Kolbe continued,

The modern chemist, who knows exactly what a chemical compound looks like in its middle and its end, how the six carbon atoms of benzene are symmetrically linked together in a plane, who then further purports . . . to have a clear conception of the spatial arrangement of the atoms, of their ortho, meta, and para positions, who determines the positions of all of the atoms in the compound, has long since abandoned the solid ground of exact science; the scientist has become a metaphysician.

He then named true physicists as "orthophysicists" as opposed to "paraphysicists" who construct hypotheses on the basis of experimental work of others and "metaphysicists" who believed that they can dispense entirely with empirical evidence.
In his own view, he was an "orthochemist" and Kekulé was a "parachemist. But as the  years went by, he increasingly believed that even the term "parachemist" was too kind; his opponents were "metachemists", to whom "chemical theories are as cheap as blackberries".

References:

Kolbe Versus the “Transcendental Chemists”: the Emergence of Classical Organic Chemistry, A. J. Rocke
The Quiet Revolution Hermann Kolbe and the Science of Organic Chemistry, Alan J. Rocke
Image and Reality: Kekulé, Kopp, and the Scientific Imagination, A.J. Rocke


Confusion in hybridisation of coordination compound

Related

Basically, it isn't really wise to use hybridization on coordination complexes to begin with. However, the hybridization of $d^2sp^3$ makes sense if you think of it as just describing the shape of the molecule. This should be octahedral, due to its 6 equivalent groups around the metal center, and $d^2sp^3$ is the valence bond description of the hybridization of an octahedron.

Using Ligand Field theory and MO diagrams is probably a better way to understand how the orbitals overlap and where electrons are in the final structure. This diagram is for $\ce{[Ni(H2O)6]^2+}$, but the diagram will look very similar for the $\ce{V}$ complex, just with fewer electrons. (Image source: dx.doi.org/10.1021/jp4100813 | J. Phys. Chem. B 2013, 117, 16512−16521)


Which white powder first "boils off" and then catches fire at high temperatures?

The fact that it turned black when heating tells you that:

It was an organic substance and did not burn completely leaving some carbon behind, or
It was a transition metal that had a lower oxidation state and was oxidised during burning.

Let's remove the obvious ones:

Not a nitrate. These will release toxic brown gas, which is not something you give to high school students.
Not an alkali (Na, K, etc) or alkali earth (Ca, Mg, etc) element because they're not pink nor they turn black.
I doubt it's a sulfur compound of any kind. You should have detected a sulfur smell when burning.

Maybe this can help you a bit.


Is the density of a two-component mixture always between the densities of its individual components?

No, there is no law that requires the density of a mixture to fall between the densities of pure components. It does so most of the time, but the exceptions are not unheard of. Here's one.
Water: density $1.00\ \rm{g/cm^3}$
Hydrazine: $1.02\ \rm{g/cm^3}$
Hydrazine hydrate: $1.03\ \rm{g/cm^3}$
So it goes.


How do I calculate the concentration of sulphuric acid by a titration experiment with sodium hydroxide?

Molarity of diluted $\ce {H2SO4}$ (solution 2): $\pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M}$
  (I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)

No, you used the correct volume since you want to know the concentration of the $\pu{25 mL}$ that you added to the flask.

  $\pu{0.54172 M}$ - Molarity of original solution of $\ce {H2SO4}$.

Correct.


Since a destructive method is allowed, the other suggestions did not work for you (despite being excellent, especially the Archimede's scales suggested by @porphyrin) and you have hot caustic soda (aka $\ce{NaOH}$), I suggest this plausible method (which, thankfully, I have never had to use). First, crush a small amount of the suspect lead glass, trying to get it a finely crushed as you reasonably can. Second, mix that crushed glass with solid  $\ce{NaOH}$. You do not need much of the mixture: a teaspoon (or a couple of grams) should suffice. Third, do a fusion of that mixture. A crucible would be used in a laboratory, but you can improvise using a small cast iron skillet. They sell very small ones (4 or 5 inch diameter) at some hardware stores. Absolutely do not use aluminum because the $\ce{NaOH}$ will destroy it!

You need to do the fusion at moderately high temperature, such as can be obtained with a charcoal grill, gas grill or a campfire. You may need to be a bit inventive on this and take care to avoid having anything spatter onto yourself! This is where a grill cover is very helpful. Fourth, let the fusion cool down and add some white vinegar to neutralize the $\ce{NaOH}$. You can add an excess, but don't get carried away: dilution is not your friend. If lead was present, you should now have some 'sugar of lead' (lead acetate) in solution. Fifth, hardware stores usually sell inexpensive test kits for lead (from leaded paints). So use one of those to test your neutralized solution for lead.

What about false positives? Do the whole thing over with crushed glass that definitely does not contain lead. It would be best to do this 'blank' first, to avoid contaminating the new little skillet (which, by the way, can be cleaned, seasoned anew, and used for its orthodox purpose).

I have not tried this and I hand-wash my lead 'crystal' single malt glasses, so best of success!


Effect of incorrect volume reading on calculated density value

It's asking for the calculated density, not the calculated volume. If the volume, as measured by the student, is too small, then the density (= mass / volume) will be too large. Of course, that's assuming that the mass was measured correctly.


Reaction of white phosphorus with aqueous sodium hydroxide

Actually, there are two reaction involving:

$$\ce{P4 + 3NaOH + 3H2O -> PH3 + 3NaH2PO2}$$

  White phosphorus react with sodium hydroxide and water to produce
  phosphine and sodium hypophosphite . Sodium hydroxide - cold,
  concentrated solution. This reaction takes place slowly. (Chemiday 1)

The O.S. of phosphorus in sodium hypophosphite is calculated in @Sayantan.G's answer.

$$\ce{P4 + 8NaOH + 4H2O ->[\Delta]  4Na2HPO3 + 6H2}$$

  White phosphorus react with sodium hydroxide and water to produce
  disodium hydrogen phosphite and hydrogen. Sodium hydroxide -
  concentrated solution. The reaction takes place in a boiling solution.(Chemiday 2)

The O.S. of phosphorus in disodium hydrogen phosphite is:

$$\ce{ +2*1 + 1 + x + 3*(-2) = 0}$$

$$\ce{ x = +3}$$


What is the mechanism for the double bond migration in 4-methylcyclopent-2-en-1-one?

After Step 2, re-protonate alpha to the carbonyl (the double bond has now migrated to the beta-gamma position).  Now enolize the carbonyl - remove the proton next to the methyl group (the top one), then push the electrons all the way back, protonating the gamma position.  Job done!


Assuming you mean NaCl - the common "salt" (chemists call lots of things salt!).

If you extract the salt by physical means, it's a physics question.

Assuming that the "dirt" is not (or poorly) soluble in water, I would simply dissolve the salt in water, filter the liquid, then recrystallize (by evaporation of the liquid) and weigh the resulting crystals. Not sure if you call that chemistry or physics... it's physical chemistry. It is well described in text books and online - see for example this Scientific American page

Depending on how accurate you want to get, there are a slew of precautions to take. That would be the realm of the chemistry.stackexchange sister site.


Difference between lattice energy and thermal stability

1) Lattice energy is the measure of the stability of an ionic solid.

It is the energy required to separate the ions in one mole of a solid ionic compound into gaseous ions.

The weaker the bond between the ions, the easier it will be to break that bond and separate the ions into gaseous ions and the lesser will be the lattice energy.

Therefore, the stronger the bond between the ions, the more is the lattice energy of the ionic compound.

2 )Thermal stability depends upon the amount of heat required to break the ionic compound. Clearly, the stronger the bond between the ions is, the more heat will be required to break the bond and as a result, the thermal stability of the compound increases.

Therefore, stronger the bond, more is the thermal stability of the ionic solid.

In general, we can say that as thermal stability increases, the lattice energy also increases, but they also also depend upon the crystal structure of the solid and other factors.

So, you cannot say with certainty that if the thermal stability is high, then lattice energy will be high too.

Hence, the answer is no. If MX has a higher thermal stability than MY, then it may or may not have higher lattice energy than MY.


Deriving rate law from a reaction mechanism for a chain reaction using steady-state approximation

In the book that Wikipedia cites (Advanced Organic Chemistry: Reaction Mechanisms
By Reinhard Bruckner, ISBN 9780080498805), they have a different set of assumptions. They are saying the formation of chlorine radicals proceeds by a fast equilibrium. Then, they say that the organic radical is at steady state, ignoring reaction (4) given by the OP (the termination reaction).

Where am I going wrong?

I'm not sure why your steady state approximation is less appropriate than that of the textbook. I'm sure that there is some experimental data (and some fitted rate constants) that would speak to that. Also, the appropriate approximation depends on the initial conditions.

[Comment from Poutnik] The initiation rate may not depend on [Cl2]( for high enough [Cl2]), but on irradiation level only. i.e being a reaction of zero order.

The derivation in the book is for the thermal reaction, but the OP asked about the photocatalyzed reaction, so there is a discrepancy there.


Is it possible to electroplate from a gallium anode?

Yes, it is possible to electrodeposit gallium. The difficulties are: efficiency due to the need for low electrodeposition potentials, formation of extremely fine hydrogen bubbles which spoils the surface finish, the requirement to use extremely low temperatures and the melting/alloying of gallium with the metal it's plated on, and poor repeatability of the process. There doesn't seem to be a patent describing using a pure gallium anode.

The difficulties are described in these patents:

US20110226630A1 - "Gallium electroplating methods and electrolytes employing mixed solvents":

  ... Gallium electrodeposition electrolytes and electrodeposition methods for solar cell manufacturing processes have many more stringent and special requirements than the electrodeposition methods and solutions employed for many other commonly plated metals such as Cu, Ni, Co, Pb, Sn, Ag, Au, Pt, and their alloys, etc. This stems from the facts that;

  i) Ga is one of the lowest melting point metals in existence, with a melting point of about 30° C.,

  ii) Ga has a high negative electrodeposition potential and thus Ga electrodeposition efficiency is naturally low since high electrodeposition potentials cause hydrogen generation, in addition to Ga deposition, at the cathode surface in aqueous electrolytes,

  iii) hydrogen bubbles generated on a cathode surface form defects such as un-deposited regions unless such bubbles could immediately be removed from the surface,

  iv) Ga has a tendency to form low temperature melting alloys with many alloy-partner materials such as In, Cu, Ag, Pb, Sn, etc. Furthermore, such alloys may form during electrodeposition of Ga onto surfaces comprising any of such alloy-partner materials.

  Electrodeposition solutions employing glycerol are very viscous and difficult to handle. The viscosity of glycerol at room temperature is 1500 centipoise (cP) compared to the viscosity of water, which is 1 cP. Gas bubbles such as hydrogen bubbles formed on the electroplated (cathode) surface during Ga plating in viscous electrolytes cannot be easily removed from that surface and therefore cause voids and other defects in the electrodeposited films. Such defects may be acceptable for some applications of thick electrodeposited Ga globules. However, they cannot be tolerated in electronic device applications such as solar cell absorber formation applications where they cause compositional non-uniformities, morphological non-uniformities, and pinholes etc., all of which negatively impact the device performance.

  Glycerol based plating solutions become more viscous as their temperature is lowered and therefore the problems cited above may get worse at lower temperatures. One other important point about the electrodeposition process for Ga is its sensitivity to the nature of the substrate surface on which the electrodeposition is performed. For example, to form a Cu/In/Ga precursor stack, the Ga film needs to be electrodeposited on an In surface. To form a Cu/Ga/In precursor stack, on the other hand, Ga plating needs to be performed on a Cu surface. One Ga electrodeposition solution that performs well for plating Ga on a Cu surface may not perform well for electrodepositing Ga on an In surface because the electrodeposition efficiency of Ga on one surface may be very different from its electrodeposition efficiency on another surface.

  As mentioned above, gallium is a low melting point material with a melting temperature of around 30° C. As a result, when electrodeposited out of aqueous electrodeposition solutions kept at about room temperature (20-25° C.), it often forms rough films comprising molten surface features, especially at high electrodeposition current densities such as current densities greater than about 5 mA/cm$^2$. This is because even though the electrodeposition solution may be at a temperature lower than the melting point of Ga, the local temperature on the cathode surface may actually exceed this melting point due to the heat generated by the electrodeposition current. As further mentioned above, when Ga is electrodeposited on surfaces of materials that easily form alloys with Ga, molten droplets of Ga alloys with low melting temperatures may be formed on such surfaces. If the Ga film is electrodeposited over In and/or Cu, the local heating and Ga melting may actually promote alloying between the plated Ga film and the underlying In and/or Cu because there are low melting alloy phases between Ga and these materials such as In—Ga alloy phases and CuGa$_2$ alloy phase. As a result, the surface roughness of the deposit may further be increased due to the above mentioned reaction and the formation of molten alloy phases. For example, Mehlin et al. (Z. Naturforsch, vol. 49b, p.1597 (1994)) attributed the rough morphology of their electroplated Ga layers to the alloying of the electrodeposited Ga with the underlying Cu surface of the cathode and the formation of a molten CuGa$_2$ alloy.

  Gallium may be electrodeposited from the electrodeposition solution at temperatures below −10° C., preferably below −20° C., most preferably below −30° C., so that local melting of the deposited Ga and its possible reaction with the materials on the cathode surface are avoided. Furthermore, at these low temperatures, the electrodeposition current densities may be increased to levels above 5 mA/cm$^2$, preferably above 10 mA/cm$^2$ and even above 20 mA/cm$^2$ without causing melting and/or alloying on the cathode surface. As a result, the electrodeposition rate and therefore the process throughput may be increased while, at the same time, the deposited film roughness is reduced. All of these benefits are important for the successful use of electrodeposited Ga layers in thin film solar cell manufacturing. For example, the melting point of methanol is −97° C. and the freezing point of a methanol/water mixture is a function of the ratio of methanol to water in the electrodeposition solution. A mixture of 75% methanol and 25% water, for instance, has a freezing point of −82° C. (−115° F.). This means that a Ga plating electrodeposition solution comprising 75% methanol and 25% water may be operated at a temperature as low as about −70-80° C., thus avoiding the melting, reaction and surface roughness problems described above.

  ...

US7507321B2 - "Efficient gallium thin film electroplating methods and chemistries":

  ... Gallium is a difficult metal to deposit without excessive hydrogen generation on the cathode because Ga plating potential is high. Hydrogen generation on the cathode causes the deposition efficiency to be less than 100% because some of the deposition current gets used on forming the hydrogen gas, rather than the Ga film on the substrate or cathode. Hydrogen generation and evolution also causes poor morphology and micro defects on the depositing films due to the tiny hydrogen bubbles sticking to the surface of the depositing film, masking the micro-area under them, and therefore impeding deposit on that micro-area. This causes micro-regions with less than optimum amount of Ga in the film stack.

  Poor plating efficiencies inherently reduce the repeatability of an electrodeposition process because hydrogen generation phenomenon itself is a strong function of many factors including impurities in the electrolyte, deposition current densities, small changes on the morphology or chemistry of the substrate surface, temperature, mass transfer etc. As at least one of these factors may change from run to run, hydrogen generation rate may also change, changing the deposition efficiency.

  Electrodeposition of Ga out of low pH aqueous electrolytes or solutions may suffer from low cathodic efficiencies arising from the presence of a large concentration of H+ species in such electrolytes. Therefore, hydrogen gas generation may be expected to lessen at higher pH values. However, as the pH is increased in the solution, Ga forms oxides and hydroxides which may precipitate as reported in the literature. Only at extremely alkaline pH values these oxides/hydroxides dissolve as soluble Ga species. Therefore, it becomes possible to electrodeposit Ga in a bath of pH>14 containing Ga salts as was done in prior-art techniques using high concentrations of KOH and NaOH in the bath formulation.

  High concentrations of alkaline species, however, cause corrosion problems for the equipment as well as the cathode material itself. There is also a limit of the Ga amount that can be dissolved in the form of acidic Ga salts (GaCl3, Ga(NO3)3 etc) in such solutions before Ga starts to precipitate. Therefore, the pH needs to be adjusted again by further addition of alkaline species such as NaOH and KOH.

  As pointed out above, solutions comprising a large molar amount of caustics are difficult to handle and they also have high viscosity. High viscosity makes the hydrogen bubbles formed on the cathode stick more to the cathode making it very difficult to remove them by stirring or other means of mass transfer. As explained above, such gas bubbles on the cathode surface increase defectivity of the deposited Ga layer.

  ...

US8545689B2 - "Gallium electrodeposition processes and chemistries" and
WO2012028416A2 - "Gallium and indium electrodeposition processes":

  "... Current processes for electroplating metal containing thin films such as gallium and gallium alloys present numerous problems. These problems include, among others, low cathodic deposition efficiency due to excessive hydrogen generation, poor repeatability of the process, partly due to the poor cathodic efficiency, and the poor quality of the deposited films such as their high surface roughness and poor morphology.

  Gallium is generally considered a difficult metal to deposit without excessive hydrogen generation on the cathode because gallium plating potential is relatively high. Hydrogen generation on the cathode causes the deposition efficiency to be less than 100% because some of the deposition current gets used on forming the hydrogen gas rather than for forming the gallium film on the substrate or cathode. Moreover, hydrogen generation and evolution is a causal factor for introducing porosity into the deposited film, thereby contributing to increased surface roughness and microdefects.

  The plating efficiencies inherently reduce the repeatability of an electrodeposition process because hydrogen generation itself is a strong function of many factors including impurities in the electrolyte, deposition current density, small changes on the morphology or chemistry of the substrate surface, temperature, mass transfer, and the like.

  ...

  Summary:

  The present invention is generally directed to solutions and methods for electrodeposition of a substrate. In one aspect, the solution for electrodeposition of the metal containing film comprises a metal salt, wherein the metal is selected from the group consisting of gallium, indium, a combination thereof, and a combination of any of the preceding with copper; an acid selected from the group consisting of an alkane acid, an alkene acid, an aryl acid, a heterocyclic acid, an aromatic sulfonic acid, an aromatic sulfuric acid, hydrochloric acid, perchloric acid, and nitric acid; optionally a metalloid compound wherein the metalloid compound is selected from the group consisting of arsenic, antimony, bismuth, a combination thereof; and a solvent to dissolve said metal salt, wherein the pH of the solution is in a range selected from the group consisting of from about zero to about 2.6 and greater than about 12.6 to about 14, and wherein the solution is free of a complexing agent.

  The method for electrodepositing a substrate comprises contacting

  (i) a solution comprising a metal salt, wherein the metal is selected from the group consisting of gallium, indium, a combination thereof, and a combination of any of the preceding with copper; optionally an inorganic metalloid compound; further optionally an organic additive having at least one sulfur atom; and a solvent to dissolve said metal salt, wherein the solution is free of a complexing agent; and

  (ii) a substrate; adjusting pH of the solution to a range selected from the group consisting of from about zero to about 2.6 and greater than about 12.6 to about 14; and applying a current to electroplate the substrate with a metal containing film.

  ...


Ground state term symbol for Cerium (Ce)

Iván, you are right on your determination using Hund's rules. However, for heavy elements, with strong spin-orbit coupling, the LS coupling assumed by Hund's rules may not apply. In this case, the j-j coupling may give better agreement with experiment.

There are extensions of Hund's rules for j-j coupling [1].

If you check NIST Atomic Spectra Database Levels Data, you will see that the wavefunction for the ground state of the Ce atom is 55% $^1\text{G}^\circ$ and 29% $^3\text{H}^\circ$. This mixture between singlet and triplet is due to the spin-orbit coupling.

[1] T. E. H. Walker, J. T. Waber, Modified Hund’s Rule for jj Coupling. Physical Review A. 7, 1218–1224 (1973). https://dx.doi.org/10.1103/PhysRevA.7.1218


In crystal field theory, why do all the metal electrons enter the d orbitals and not the s orbital?

Your example is a little questionable. I would think you are talking about $\ce{[Fe(H2O)6]^2+}$, in which case you are actually supposed to have an $\ce{Fe^2+}$ ion with a $[\ce{Ar}]\mathrm{(3d)^6}$ configuration. There are no $\mathrm{4s}$ electrons in the $\ce{Fe^2+}$ ion.

Nevertheless, it is true that in crystal field theory, electrons are assumed to be entirely in the d-orbitals, even for low-oxidation state complexes. For example, nickel(0) in $\ce{[Ni(CO)4]}$ is assumed to have a "$\mathrm{d^{10}}$" configuration. There is a reason for this, the explanation of which requires a molecular orbital treatment. Consider a generic octahedral complex, $\ce{[MX6]^{n-6}}$, which is formed from a $\ce{M^n+}$ ion and six $\ce{X-}$ ligands. We could construct an MO diagram:

(Source: Miessler & Tarr, Inorganic Chemistry 3rd ed., p 347)

Each $\ce{X-}$ ligand contributes one pair of electrons, which gives rise to a total of six pairs; these six pairs of electrons enter the low-energy bonding MOs $\mathrm{a_{1g} + e_g + t_{1u}}$. These MOs are similar in energy to the ligand AOs, and therefore are likely to be mostly centred on the ligand, which means we can essentially ignore these MOs when we are describing the properties of the complex. (Of course, these MOs are important in the sense that they are responsible for the bonding between metal and ligand. If these bonding MOs did not exist, the complex would not form. However, we are usually much more interested in the higher-energy orbitals.)

The electrons in the $\ce{M^n+}$ ion can be in any of the AOs ($\mathrm{3d}$, $\mathrm{4s}$, or even $\mathrm{4p}$) - it does not matter. The only important thing is that, once the complex is formed, these electrons will enter the lowest-energy MOs according to the aufbau principle. Since those low-energy bonding MOs are already occupied by ligand electrons, the metal electrons will have to go into the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ MOs, which in the simplistic treatment of crystal-field theory are called the "d-orbitals".

The truth is that those two MOs have quite a lot of metal d-orbital character (as you can see from the dotted lines), so crystal-field theory is not entirely wrong. Nevertheless, you can see from the MO diagram why it would be wrong to assume that the metal $\mathrm{4s}$ orbital is filled: what CFT calls the $\mathrm{4s}$ orbital is actually the $\mathrm{a_{1g}^*}$ orbital in the MO scheme. This orbital can only be filled if the $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ MOs are already filled, which is not the case unless we are talking about some kind of zinc(0) complex (very unlikely!).

Similar diagrams can be constructed for other geometries (tetrahedral, square planar etc.) and the same point will hold. Therefore, even though $\ce{Ni}$ itself has a $[\ce{Ar}]\mathrm{(3d)^8(4s)^2}$ configuration, the tetrahedral $\ce{[Ni(CO)4]}$ complex will have a $\mathrm{(e)^4(t_2)^6}$ configuration (as you are probably familiar with, in tetrahedral complexes, the d-orbitals are split into $\mathrm{e + t_2}$). Again, in the language of crystal-field theory, both the $\mathrm{e}$ and $\mathrm{t_2}$ orbitals are simply called "d-orbitals", so this is equivalent to having a "$\mathrm{d^{10}}$" configuration.


Effect of impurities on melting and boiling points

All generalities are false (as is this one). Though impurities usually lower the melting point (m.p.) by disrupting crystallization on the atomic order, consider the phase diagram of the binary alloy (amalgam) HgxAg1- x: Pure Hg melts ~-39°C, and adding even a little bit of impurity raises the melting point considerably!
And without resorting to entropic explanations, consider that a high boiling point (b.p.) solute would "hold onto" the solvent, decreasing vapor pressure. $\ce{CaCl2}$ will absorb $\ce{H2O}$ from the air, driving down the vapor pressure (and humidity) even after the $\ce{CaCl2}$ liquefies.
That said, the answer they expect on the exam is that impurities decrease the m.p. and increase b.p.


Does trans 1,3-dichlorocyclobutane have zero dipole moment?

Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.

The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid.  The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.

Only molecules that belong to symmetry classes (point groups)

$\ce{C_{n}}$ (the molecule only contains a $\ce{C_{n}}$ axis)
$\ce{C_{nv}}$ (the molecule contains a $\ce{C_{n}}$ axis and a $\ce{\sigma}_{v}$ plane)
$\ce{C_{s}}$ (the molecule only has a plane of symmetry)

can have a dipole moment.

Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.

Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.


Is protein folding thermodynamically favourable? If yes.. why?

What determines whether a process is spontaneous or not, is the change in free energy, not the increase of entropy. Entropy is one important factor, as is enthalpy. Entropy in this case is not increasing since the system is becoming more organized, however,enthalpy is greatly favoured since protein folding creates many Hydrogen bonds between its polar aminoacid fragments and surrounding water and also many hydrophobic interactions between their non-polar aminoacids constituents. Take a look at this page to have a better understanding about protein folding thermodynamics (introductory text, though it´s a good starting point).
http://www.chembio.uoguelph.ca/educmat/phy456/456lec02.htm


Latent Heat, Vapor pressure and Equilibrium

All is about  3 basic principles.

Liquid evaporation consumes thermal energy, taken from liquid itself or surrounding. Vapour condensation releases this energy.
Liquid saturated vapour pressure (about) exponentially raises with temperature.
If saturated vapour pressure exceeds external pressure, liquid boils. If vapour pressure exceeds the saturated vapour pressure, vapour condensates.


Can the van der Waals coefficients be negative in the van der Waals equation for real gases?

This question requires a simplistic notion of real gas behavior.

The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is

$$(P + a/V_\mathrm{m}^2)(V_\mathrm{m} -b) = RT$$

If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.

The rest of the story...

The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)


Meaning of Partial Half-Life for Two Parallel Reactions

The partial reaction half-life relates to the reaction speed constant by the same way as for the "normal" reaction half-life.
If there are 2 parallel reactions of the first order, $\ce{A -> B}$ and $\ce{A -> C}$, and if there is the reaction rate for the former:
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=k_{\ce{B}} \cdot [\ce{A}]$$
then the partial half-life w.r.t. $\ce{A -> B}$ is:
$$t_{1/2,\mathrm{B}}=\frac{\ln{2}}{k_{\ce{B}}}$$
Analogically the similar for the other reaction:
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_{\ce{C}} \cdot [\ce{A}]$$
$$t_{1/2,\ce{C}}=\frac{\ln{2}}{k_{\ce{C}}}$$
The overall half-life w.r.t. $\ce{A -> X}$ is then
$$t_{1/2}=\frac{\ln{2}}{k_{\ce{B}}+k_{\ce{C}}}=\frac{1}{\frac 1{t_{1/2,\ce{B}}}+\frac 1{ t_{1/2,\ce{C}}}}$$
----  Responses to comments
The partial half-life is the extrapolated time after which all $\ce{A}$ would have decayed, if it had been decaying by the current and constant rate of given reaction and only by this reaction. But the main meaning is as a kind of reciprocal value of the reaction constant.
If we draw the chart of the partial reaction rate, then it's tangenta at $t=0$ will cross $x$-axis at the reaction partial half time. It is the same as if it was the only reaction. It does not say half of $\ce{A}$ decays in this partial half time.
If $\Delta t \ll \min{(t_{1/2})}$, then it does not matter it is just partial halftime, as we can neglect decay of the other parallel reactions.


How the Azimuthal quantum number describes 3d shape of the orbital?

what they call 'orbital' of an electron in chemistry is called wave function in physics. It determines how the electron moves and how it behaves. In fact it determines everything of the electrons. The azimuthal quantum number is just the quantum number used to describe angular part of the wave function, so it determines the shape of the orbital. This is a very rough explanation. You need to learn something like quantum chemistry or quantum mechanics in order to get a better understanding on that.


From SDS:

Special hazards arising from the substance or mixture: carbon oxides ($\ce{CO}$, $\ce{CO2}$) nitrogen oxides ($\ce{NO}$, $\ce{NO2}$ etc.) hydrocarbons, isocyanate vapours and hydrogen cyanide can be released in case of fire.
The polyurethane polymer formulation greatly changes so the fumes that can form at high temperatures (above the melting temperature). A suction and exhaust system is required. To measure the exact composition of the fumes very specific analysis are required.


What is the mechanism for demercuration?

Apparently, the current assumption implies formation of $\ce{R-Hg-H}$ intermediate, followed by radical chain decomposition This, however, should be taken with a grain of salt, i.e. very skeptically. Proving mechanism is an incredible amount of work, and considering relative unimportance of organomercury chemistry in modern organic synthesis, it is unlikely any considerable resources will be allocated to the task.


Does a strong acid's conjugate base act as Brønsted-Lowry base?

In theory, a Brønsted-Lowry base is a chemical substance which can accept a proton.

But in the case of a strong acid, the acid will lose a proton and form a weak conjugate base. Say there is a strong acid HA.

$$\ce{HA + H2O <=> A- +H3O+}$$

So if we take a certain quantity of HA, most of it will undergo the forward reaction since HA is a strong acid. A- is a weak base and will not want to accept the proton and thus only a small quantity of the conjugate base will do undergo the backward reaction. Hence practically, you could say that the conjugate base of a strong acid is not a Brønsted-Lowry base because of its reluctancy to accept the proton.


In non-nuclear chemistry, everything is electrostatic interactions. This is why you can learn and predict so much just by "following the electrons"

Covalent bonds are also formed because of electrostatic interactions - they are just more complicated conceptually than ionic (actually, ionic bonds are more accurately described by wavefunctions, we just try to keep things simple in the beginning). Since electrons exist as waves, when you confine them they start to do weird stuff (or what seems weird to us). For example, the shapes of atomic orbitals are all the way they are because the electrons act like standing, 3-dimensional waves trapped between a nucleus with a positive charge and a "zero-point" at infinite distance. When you put two atoms close together, the electrons on different atoms interact with each other, and the wavefunction becomes much more complicated. The result is that in some pairs of atoms, the wavefunctions combine to form bonding orbitals.

So the short answer to your first question is: "Molecular orbitals hold atoms together in covalent bonds, and those are a result of electrostatic interactions and the quantum nature of electrons."

Yes, ionic compounds are large collections of ions, and you can't really define "molecules" for them - instead we talk about "formula units" which are the lowest possible whole-number ratio of elements that represent the compound. Groups of covalently bonded atoms are also held together by electrostatic interactions, but since the covalent bonds are so much stronger, a molecular compound can exist "on its own" as a single molecule. Collectively, the forces that hold collections of molecules together are called van der Waals forces if they don't involve ions. In any atom or molecule, there is never a completely uniform charge density on the surface. For some molecules, this is extreme (water is a good example) and we say it is very polar, or that it has a large dipole moment. This is just another way of saying that one part has a negative charge and the other has a positive charge. In water it looks like this (from wikipedia):

In this picture, red means "more electrons" and blue means "less electrons." Water can form hydrogen bonds, which are very strong electrostatic interactions. Some atoms and molecules have an almost uniform charge density on the surface. We call these "non-polar" molecules - noble gases are good examples. However, even noble gases have what is called an induced dipole due to statistically correlated fluctuations in electron density when the atoms are near each other. As a result, even noble gases can be cooled to the point where they become liquid - the very, very weak electrostatic interactions will hold them together at low temperature, when they are not moving very fast. These forces are called London Dispersion Forces - after the guy who first described them. London dispersion forces are important, because they are found in all molecules - polar or not. In fact, this is what makes most plastics solid. Polyethylene, for example, is made of very long chains of essentially non-polar molecules (from wikipedia):

Each chain is attracted to the others through weak London dispersion forces, however, since each chain has tens of thousands of atoms, those tiny forces quickly add up to make large forces holding the polymer together. This is why polyethylene is solid at room temperature, and can be used to make things like shopping bags!


Physical meaning of (negative) pressure in molecular dynamics simulations

Looks fine. 2. I'd phrase that more generally as the simulation volume would tend to expand. 3. But can only do so if the ensemble permits it (pressure can be negative in NVT).

The fluctuations are a consequence of making an observation of a macroscopic quantity over small numbers of degrees of freedom, or time. See http://www.gromacs.org/Documentation/Terminology/Pressure


Certainly, the most important organization for chemical nomenclature is the International Union of Pure and Applied Chemistry (IUPAC). In most cases, the various IUPAC recommendations represent the best references.

However, not all IUPAC recommendations are universally accepted. For example:

The correct internationally standardized name of element 111 is ‘röntgenium’; IUPAC recommendations spell it ‘roentgenium’, which is explicitly considered wrong by international standards.
According to current IUPAC recommendations, the formula of ions like sulfate is written as $\mathrm{SO_4{}^{2-}}$, which has not yet gained much acceptance. According to ISO and various national standards, the correct internationally standardized notation is $\mathrm{SO_4^{2-}}$.
In the 1970 IUPAC recommendations, a chelating ligand with two coordinating atoms was indicated by the term ‘bidentate’. In the 1990 IUPAC recommendations, the term ‘didentate’ was used rather than ‘bidentate’, for reasons of linguistic consistency. However, the term ‘bidentate’ remained in common usage, which is reflected by the reversion to the previously accepted term ‘bidentate’ in the 2005 IUPAC recommendations.
The @ notation for “fullerenes with metals inside” was proposed in J. Phys. Chem. 1991, 95 (20), 7564–7568 and became generally accepted. The corresponding IUPAC proposal, which was presented in Pure Appl. Chem. 1997, 69 (7), 1411–1434, did not use the @ symbol, but it was not as successful as the first one.

Many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. In addition, various traditional names, which have been coined and used over the years, are retained for use in IUPAC nomenclature. Therefore, current IUPAC recommendations include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names. Nevertheless, other names that are generated in accordance with IUPAC recommendations may be used in general nomenclature.

Apart from the large and universally active IUPAC, further smaller and specialized organizations exists, for example the International Union of Biochemistry and Molecular Biology (IUBMB). Such organizations may issue their own recommendations concerning their field of expertise. However, many of these recommendations are actually harmonized with IUPAC recommendations.

Another large and universally active organization that prescribes independent naming rules is the Chemical Abstracts Service (CAS). CAS names may differ from IUPAC names. For example, the preferred IUPAC name of Buckminsterfullerene is (C60-Ih)[5,6]fullerene, whereas the corresponding CAS name is [5,6]fullerene-C60-Ih.


Mechanism of Friedel-Crafts acylation with succinic anhydride

The reactivity of acid chlorides and acid anhydrides are actually quite similar. It may be helpful for you to think of an acid anhydride like this:

When we do, the acylation reaction is almost identical to that of an acid chloride, where an acylium cation $\ce{R-C#O+}$ is generated:

The subsequent proton exchange restores aromaticity:


Aside from carbon, what other substances can be made "superhard"?

Interestingly, nobody addressed the reason why diamonds are hard in the first place. The pressure (and temperature) are not the reason why they're hard, only the reason why they are formed. The diamonds are hard because the carbon atoms are bonded together by sigma ($sp^3$) bonds, which are the strongest chemical bonds. Other materials exhibiting the same kind of bonding are potentially also very hard. But the small atomic number of carbon (Z=6) means that the bonds will be stronger than in higher atomic number atoms, such as silicon.

However, amorphous (non-crystalline, like glass) materials can also be extremely hard. There is even an example that is even stronger than diamond (although not naturally occurring). And this material is made of... carbon! Exactly like diamond. This material, dubbed Q-carbon, was discovered in 2015, and is even created at atmospheric pressure, and around 4000 K with the help of a laser pulse. And it is estimated to be 10-20 % harder than diamond.

Bottomline: gigantic pressures are not required to create extremely hard materials, and diamonds is no longer the strongest known material.


Calculating the pH of a buffer made by a diprotic acid and its double salt

Can I multiply Ka1 and Ka1 to eliminate [$\ce{C4H5O6−}$], and then get the concentration of C4H4O62− necessary by plugging in 0.1 M for [C4H6O6] and the target pH in the appropriate form in [H+]

No, because $\ce{C4H5O6−}$ is a one of the major species. In fact, if you add the tartaric acid and its double salt at equimolar ratios, $\ce{C4H5O6−}$ will be the only major species. As a first approximation, you can let this reaction go to completion. If you have an excess of the acid, there will be a mixture of $\ce{C4H5O6−}$ and acid as major species. If you have the double salt in excess, there will be a mixture of the double salt and $\ce{C4H5O6−}$ as major species.

  Do I have to use the Henderson-Hasselbalch equation? Make certain assumptions?

If your ratio of reactants and products is such that you have a conjugate acid: base pair as the major species, you can use Henderson-Hasselbalch as the next step to estimate the pH. (If your major species are $\ce{C4H6O6}$ and $\ce{C4H5O6−}$, use $K_{a1}$, if they are $\ce{C4H5O6-}$ and $\ce{C4H4O6^2−}$ use $K_{a2}$.) In some cases, that's a good approximation, in others, not. You can test that by calculating the concentration of all species (based on the approximate pH) and then calculate Q for the two acid base equilibria to see how well they match K.


Hydrogen bonding relating to molecular orbital theory

Yes you can think of hydrogen bonding in terms of MO theory. Consider a hydrogen bond between two water molecules as a simple example.

The donor hydrogen donates electron density into the $\pi^*$ orbital of oxygen which is the LUMO of the acceptor oxygen. This thus weakens the donor $\ce{O-H}$ bond, but shortens the distance between the donor hydrogen and acceptor oxygen.

Thus, there are two components to be considered: the pure electrostatic attraction and the stabilization due to covalent interactions.

Because all you're asking about is the MO side of things, as explained above, hydrogen bonding can be viewed as donation of electrons into the $\pi*$ orbital of some acceptor atom. This could indeed be called a Lewis acid-base interaction, but I don't hear this very often.

I'm not sure what your textbook is talking about with hydrogen bonding being a Brønsted-Lowry reaction. Hydrogen bonding is not a reaction (in the traditional sense) so I would disagree with your textbook. The acceptor oxygens are accepting electrons much more than they are accepting the whole proton which would make them Lewis acids. It's certainly possible for water/alcohols to behave as either Lewis acids or bases and Brønsted-Lowry acids or bases, but the former is much more common based on the fact that auto-ionization is quite a small effect but hydrogen-bonding is a very large (and common) one.

Regarding the MO view of things, the exact same logic can be applied in inorganic chemistry when considering carbonyl compounds or dihydrogen/dihydride compounds. In that case the donation of electrons into a $\pi*$ orbital (or $\sigma*$ orbital in the case of dihydrogen compounds) is internal to the molecule however and we call it backbonding.

Here's an MO and orbital diagram showing that interaction.


How to extract pure caffeine from tablets?

The best option would be an organic extraction with dichloromethane. All the other ingredients (except possibly D&C yellow #10 aluminum lake) will be insoluble and therefore unaffected. Filter, let the dichloromethane evaporate and you should have your caffeine.

EDIT: I agree with J M. This caffeine IS NOT EDIBLE. You could easily become intoxicated by the small amounts of dichloromethane that will remain. Also be careful when handling the dichloromethane. Vapors are also toxic and if the liquid is heated or shaken in a closed recipient it might explode. Use caution.


With the $\ce{C=C}$ double bond of the stilbenes as geometric reference, you may think of the phenyl rings as substitutents increasing the electron density of the former.  This creates a small, locally constraint, dipole moment.  Dipoles like dipoles differ from a scalar (like e.g., temperature) that they have a direction, which is why I drew little arrows in the simplified representation of either isomer in the illustration below:

In the case of (E)-stilbene, these vectors oppose each other; for symmetry reasons, they may cancel out each other as if there were no vector present.  This contrasts to the case of (Z)-stilbene where the sum of the two vectors does not vanish; symbolized (the doodle is a concept, not at exact scale) by the dashed arrow: there is a non-zero molecular dipole moment.
The drawing of (Z)-stilbene's configuration however does not state (much) about the molecule's energetically preferred conformation(s).  To lessen intramolecular steric hindrance of the hydrogens of the two phenyl rings, the following spatial arrangement is more likely to happen:

Note, regardless how the phenyl rings rotate around the bond between $\ce{C_{ar}-C_{sp^2}}$, the overall (molecular) dipole moment will not be affected from these local variations.
Equally note that to achieve an efficient recrystallization of the three constituents of your mixture, the isomers of stilbene and TPPO, you use a mixture of ethanol/water heated to reflux then allowed to cool slowly to room temperature.  The addition of water is to attenuate the solution capacity of ethanol for each of the three compounds so that one preferentially crystallizes out.  This is why this purification method is called fractional crystallization.


Unfortunately, there is no caliper-like tool to measure ionic radii directly.
They are determined based on experimental data such as crystal structure or crystal lattice energy using empirical relationships.
The exact value would inevitably deviate from one method to another.

Pauling [1, pp. 150–152] used a hard-sphere model which doesn't account for overlap in crystal lattice and was predicting higher interatomic distances compared to the crystallographic data, hence an overestimated value of the anionic radii.

Several limitations of the hard-sphere model can be resolved by applying a soft-sphere model.
According to the recent work by Lang and Smith [2], who also compared applicability of both hard- and soft-sphere models, $r_\mathrm{i}(\ce{H-}) = \pu{139.9 pm}.$

CRC Handbook of Chemistry and Physics [3, p. 12-22] lists the value of thermochemical radius determined using Kapustinskii equation from a set of known lattice energies: $r_\mathrm{i}(\ce{H-}) = \pu{(148 ± 19) pm}.$

References

Pauling, L. The Chemical Bond; Cornell University: Ithaca NY, 1960.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.
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.


How to choose a UV lamp for photoisomerization of 4,4'-azobenzene dicarboxylic acid?

I'd probably try a more powerful lamp. Plain low-pressure mercury tubes have their emission maximum at 254 nm, which isn't what you want here. But these tubes also come in various coated variants with emission maxima around 300 or 365 nm.

A reasonable choice would be a Rayonet RPR-3000A or a Luzchem LZC-UVB lamp in a proper housing.


Does this reduction mechanism of an diazonium via stannic chloride sense?

Agreeing with Jan's comment, there are a few problems, and they all have to do with your reducing agent. That's not tin!

Tin is $\ce{Sn}$ and not $\ce{Zn}$ (zinc). Your mechanism is using $\ce{ZnCl2}$, zinc chloride. Let's say you made a typo and replaced $\ce{Sn}$ with $\ce{Zn}$. Your electron configurations match tin and not zinc. Then your mechanism is good. $\ce{SnCl2}$ is a reducing agent and could do this reaction. $\ce{SnCl2}$ is not stannic chloride, but stannous chloride. Stannic chloride is $\ce{SnCl4}$. Regardless $\ce{SnCl2}$ would work. $\ce{ZnCl2}$ would not work. The zinc in $\ce{ZnCl2}$ is already in zinc's highest common oxidation state. $\ce{ZnCl4}$ does not exist.


Is pyrene aromatic despite failing Hückel's rule?

To clarify @ron's point, the general "cheminformatics" rule for deducing aromatic fused-ring systems is whether there is one path (usually the peripheral one) that satisfies $4n+2$:

So for naphthalene:

The bold bonds give you a ring path of $\mathrm{sp^2}$ carbons with 10 π electrons, so it's aromatic. There are some arguments about whether the central bond is aromatic or not. (I say baloney, but it's fair to say aromaticity generates lots of arguments.)

So yes, the bold path gives you 14 π electrons for a $4n+2$ path. Again, I've heard people say the central two carbons aren't aromatic. (IIRC, the $\ce{^13C}$ NMR disagrees.)

I wrote the code for Open Babel and aside from some tricky nitrogen-heteroaromatic rings, this rule generally matches chemical reality. (Setting aside the question of the interior carbons.)

I think another fairly safe rule is "if a fused aromatic system is made up of individually aromatic rings, aromaticity is retained."


How can over grinding lead to reduced API extraction in HPLC analysis?

I sort of feel this is actually more of a mathematical/statistical/modeling related question than traditional chemistry.
In essence, one argument is that the sample derived effective API concentration is impacted by possible changes in the particle size distribution due to grinding. Now, there are, in effect, known statistical distributions that address particle size distributions (PSDs), and associated methods to assess parameters of said distributions relating to sample means and variability, and changes thereto.  This would seem to be a possible component of the analysis.
As background, I reference Wikipedia on the topic, which importantly notes, to quote:

The PSD of a material can be important in understanding its physical and chemical properties...It affects the reactivity of solids participating in chemical reactions, and needs to be tightly controlled in many industrial products such as the manufacture of printer toner, cosmetics, and pharmaceutical products.

as both surface area (a function of particle size) and concentration influence a reaction.
Also, some detail on PSD distributions:

The log-normal distribution is often used to approximate the particle size distribution of aerosols, aquatic particles and pulverized material.
The Weibull distribution or Rosin–Rammler distribution is a useful distribution for representing particle size distributions generated by grinding, milling and crushing operations.
The log-hyperbolic distribution was proposed by Bagnold and Barndorff-Nielsen[9] to model the particle-size distribution of naturally occurring sediments. This model suffers from having non-unique solutions for a range of probability coefficients.
The skew log-Laplace model was proposed by Fieller, Gilbertson and Olbricht[10] as a simpler alternative to the log-hyperbolic distribution.

Now, for example, in the case of the Log-normal distribution, the best estimate of the parent population mean particle size (which here can be used to estimate total concentration) is obtained by working with the log values of the sample particle sizes. Wikipedia on the Log-normal distribution notes that the sample mean (say 'm') and variance (${s^2}$) of these log-transformed values can be employed to more accurately specify the parent population (aka, the Log-normal) mean, which turns out to be ${Exp(m + 0.5*s^2)}$. Note, simply taking the straight average of the observed Log-normals size values is not recommended as a Log-normal random deviate can be very noisy being the ${Exp()}$ of a Normal random deviate, hence the use of a log-transform. The Weibull distribution is another example of where the parent population mean is a function of two of the distribution's to-be-estimated parameters.
Further, it may indeed be the case that excessive grinding affects particle size distribution and reaction rate. However, this should be verified experimentally to determine the potential impact. Perhaps using a proxy, that produces a more readily measurable product (as in gas evolution, or strongly colored compound) may be of assistance.
Finally, this educational source comments on the inter-connection between reaction rate, mass and concentration, noting an exception, to quote:

The concentration of the reactants determines the speed of the reaction. In simple reactions, an increase in the concentration of reactants accelerates the reaction. The more collisions over time, the faster the reaction can advance. The small particles have less mass and more surface area available for the collisions of other particles. However, in other more complex reaction mechanisms, this may not always hold true. This is often observed in reactions involving huge protein molecules with large masses and convoluted structures with reaction sites buried deep within them that are not easily approached by collision particles.

So, still some experimenting required here in an obviously specialty discipline with a possibly large molecule.


Is this an exception to the rule that strong acids/bases always replace weak acids/bases?

Every reaction is considered reversible to some extent and feasibility of a reaction is mainly dealt in the terms of favourability of either products or reactants. When we say that creating a weaker acid from a stronger base is feasible, we actually mean that the equilibrium favours the formation of the products in such a case. However some other factors can be at play and can well influence an equilibrium. One such governing principle is Le Chatalier's.

$$ \ce{H2SO4 + 2NaCl <=> 2HCl + Na2SO4 }$$

Here's a youtube video showing the production of hydrochloric acid under laboratory conditions. Notice the $\ce{HCl}$ produced is a gaseous product, and hence there's an imbalance in our equilibrium; $\ce{HCl}$ escapes out as vapors from the solution, lowering $Q_c$ of the reaction thus making the reaction go forward.


Can I completely dry air with calcium chloride?

"Completely dry"? No, because molecules are always moving and there will be some exchange between desiccant and air. If you mean "below 10% relative humidity (RH)", there are a number of ways to do so:

Consider that another way to discuss humidity is the dew point. Use a cold trap: after drying the air with a desiccant, pass the air slowly through a U-tube (or better, a through a spiral condenser) cooled with ethanol/dry ice or similar bath. As the air emerges, the dew point, ~200 K for dry ice, remains that of the bath, but as it warms to room temperature, the RH would fall.
Dry the air with desiccant at room temperature, then heat the air (not desiccant) and the RH drops.
Synthesize air from ~80% liquid nitrogen and ~20% liquid oxygen. If you're a stickler for accuracy in the simulation, add a dash of liquid argon, and a smidgen of liquefied CO2 (not dry ice, which, being open to air, quickly condenses water). These liquefied gases can probably be obtained from a welding supply shop.

BTW, don't forget to bake your apparatus before use! Metal, and particularly glass, adsorb water quite well, and slowly release it at low RH or pressure.


No, the reaction does not produce hydroxide ions. But the solution is already basic (probably between pH 12 and 14).


How to name an unsaturated spiro compound with many substituents?

The parent structure is a spiro compound, which has a fixed numbering starting in the smaller ring.
The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-24.2.1 Monospiro alicyclic ring systems
Mono spiro parent hydrides consisting of two saturated cycloalkane rings are named by placing the nondetachable prefix ‘spiro’ before the name of the unbranched acyclic hydrocarbon with the same total number of skeletal atoms. The number of skeletal atoms linked to the spiro atom in each ring is indicated by arabic numbers separated by a full stop, cited in ascending order and enclosed in square brackets; this descriptor (called in these recommendations the ‘von Baeyer spiro descriptor’) is placed between the spiro prefix and the name of the acyclic alkane. Numbering starts in the smaller ring, if one is smaller, at a ring atom next to the spiro atom and proceeds first around that ring, then through the spiro atom and around the second ring.

In this case, the name of the parent spiro hydrocarbon ring structure is spiro[6.9]hexadecane.
The fixed numbering of the spiro hydrocarbon ring is not modified by the introduction of the double bonds and functional groups. However, low locants must be attributed if there is a choice.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(a) fixed numbering in chains, rings, or ring systems, i.e., when the numbering of a system is fixed, (…), this numbering must be used, both in PINs and in general nomenclature;
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Thus, low locants are first assigned to the principal characteristic group, which is carboxylic acid. This leads to the name spiro[6.9]hexadecane-8,14-dicarboxylic acid.
Next, locants are assigned to the double bonds according to Rule (e), which yields (8E,11Z,15Z)-spiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid.
Finally, locants are assigned to all other substituents. Therefore, the correct name for the complete compound is (8E,11Z,15Z)-4-amino-11-carbamoyl-3-chloro-2-(chlorocarbonyl)-9-cyano-5-formyl-4-hydroxy-13-methoxy-12-(methoxycarbonyl)-10-oxospiro[6.9]hexadeca-2,8,11,15-tetraene-8,14-dicarboxylic acid.


What is the heaviest possible water using oxygen and hydrogen isotopes, and what might it's physical properties be?

The heaviest water obtained with stable isotopes would be $\ce{D_2^{ 18}O}$.  This compound is not well-charactetized itself, but Ref. [1] indicates that the melting point is only about 0.28°C above that of ordinary water.  The implication is that overall physical properties of $\ce{D_2^{ 18}O}$ are close to those of heavy water with oxygen-16($\ce{D_2^{ 16}O}$); apart from density the isotopic effect of $\ce{^{18}O}$ versus $\ce{^{16}O}$ is much less than that of deuterium versus protium/"hydrogen".
References

F. Steckel and S. Shapiro, "Physical Properties of Heavy Oxygen Water.  Part 1.  Density and Thermal Expansion", Trans. Faraday Soc. 1963, 59, 331-343.


Electronic energy is indeed an eigenvalue of electronic Hamiltonian, and, as you correctly pointed out, it doesn't include the contribution from repulsions of the fixed nuclei (which has nothing to do with electrons). But sometimes the total energy for a particular nuclear configuration (one that includes the above mentioned interactions contribution) is still (incorrectly) termed "electronic energy" while just an eigenvalue of electronic Hamiltonian is called "pure electronic energy". That is essentially wrong, but it is sort of a custom.

And I would not say that "all the electronic structure packages quote the Total energy as the electronic energy". Did you check all of them? For instance, I quickly checked one of my DALTON calculations, and found this:

@    Final DFT energy:           -210.248305383291
@    Nuclear repulsion:           158.997839473485
@    Electronic energy:          -369.246144856776

So the first one is the (total) energy, then come its two components with the right names.

Same for ORCA:

Total Energy       :         -210.25104538 Eh           -5721.22181 eV

Components:
Nuclear Repulsion  :          158.99783872 Eh            4326.55115 eV
Electronic Energy  :         -369.24888411 Eh          -10047.77296 eV

GAMESS-US (here first two terms = electronic energy)

        ONE ELECTRON ENERGY =    -596.6619030866
        TWO ELECTRON ENERGY =     228.7817683952
   NUCLEAR REPULSION ENERGY =     158.9978510234
                              ------------------
               TOTAL ENERGY =    -208.8822836680

NWChem (middle three terms = electronic energy)

         Total DFT energy =     -210.248032406037
      One electron energy =     -597.377034042581
           Coulomb energy =      258.596296569204
    Exchange-Corr. energy =      -30.465145956066
 Nuclear repulsion energy =      158.997851023405

And I don't want to continue, since I'm pretty sure that most of the programs use the right terms.


Why is my conclusion inconsistent with the van't Hoff equation?

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $\Delta G^\circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -\frac{\Delta H}{R} + \frac{T \Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-\frac{\Delta H}{R} \times \frac{1}{T}$ tends to zero and $\frac{T \Delta S}{R} \times \frac{1}{T}$ cancels to be just $\frac{\Delta S}{R}$.


Yes, you are exactly right.  Anthracene is better described by this structure (where there are no fixed single and double bonds, but rather a delocalized aromatic structure)

than a structure like this (where the single and double bonds are fixed and distinct from one another).

So in anthracene, the aromatic character must be disrupted to some degree in order to undergo the Diels-Alder reaction.  In cyclopentadiene, there is no aromaticity to disrupt, the double bonds are basically fixed like in buta-1,3-diene (and, as you point out, locked in the required s-cis conformation).

A final note, benzene won't undergo a Diels-Alder under normal conditions because its aromaticity is completely disrupted and lost during the process.  In anthracene the aromaticity remains in two of the rings, a much smaller price to pay.


Is pyrite (FeS₂) an ionic or a covalent compound?

You seem to have fallen into the trap of thinking that ionic and covalent bonds are fundamentally different. They are not - they are just two ends of a spectrum, which has an arbitrary division somewhere in the middle into an ionic and covalent regime. This is explained in the answers to this question.

In the case of pyrite we have a relatively hard cation, with a small ionic radius and charge of +2, and a rather large anion, with a charge of -2. Therefore there will be a significant degree of covalent character in the Fe-S bonds due to the polarising effect of the cation on the anion. This is confirmed by experimental results and theoretical calculations which suggest that the charge on Fe is about +2/3 and the charge on S is about -1/3. This is significantly less than the expected charges of +2 and -1 from a purely ionic model and so indicates that there is significant sharing of electrons. This is supported by measurements and calculations of the electron density, which show significant electron density between the atoms.

Reference: http://pubs.rsc.org/en/content/articlepdf/2014/sc/c3sc52977k


You ask:

  Are common rocks (e.g. rock from mountains or rivers, bricks, or even concrete) actually silicon oxides? Or it is just one of compositions?

Rocks can contain silicon oxides, but that's not part of what the definition of a "rock" implies. From an introductory course in geology at the University of Texas at Austin, a rock is defined as:

  Any coherent, naturally occurring substance generally composed of minerals.

That's all there is to it. Whether or not the minerals include silicon oxides has nothing to do with a rock being called a rock.


What is heat of vaporization? How can it be used at temperature as low as 25 °C?

Think of vaporization as the removal of the fastest molecules. The separation of the gaseous form from the more closely-bound liquid therefore removes energy from the liquid. The heat of vaporization can apply at much lower temperatures than the BP of water at atmospheric pressure. Even on a day when the temperature is well blow freezing, laundry will dry on a clothesline because ice evaporates (well, sublimes). Also, water at skin temperature soon feels cold as it loses heat through evaporation.

BTW, the boiling point of water is 100 °C only at 1 atmosphere; see this video of water boiling at room temperature, ~25 °C.


Less maintenance-dependent continuous pH testing

The most common approach is to measure periodically.

pH glass electrodes (the most common ones) need to be "wet" in order to be used, so if you are going to remove it from the solution every hour you'll need to keep it submerged in the appropriate solution. Not only that, you need to wash it (preferably with the most pure water you can get) between measurements. Read section 2.2 of this monograph to understand why it needs to be "wet" and what solution to store it in. There's information on a lot of types of electrodes. This explains how to wash it. Another problem you might face with continuous measuring is not the electrode, but the meter. From that monograph:

  The condition of the pH measuring assembly is subject to fluctuations over the course of time. Depending on the accuracy demands, a pH calibration should therefore be performed periodically, e.g. once a day.

You'd have to test and see what accuracy you get with your equipment.


A chemical change is defined as when your substance reacts with another substance to form some products, or decomposition into other products. In a chemical change, there will be formation of new chemicals that are different in terms of chemical formula.

Thus, breaking of intramolecular bonds will not be a chemical change in the case of C-C bonds in diamond/graphite, because there is no formation of new chemical substances with an alternative formula. Similarly, ionization (without formation of complex ions) would be physical change


Electrolysis of aqueous copper (II) nitrate

There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since

$$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$

Your first half-reaction is an oxidation (as written). You have the wrong sign on that $E$ value.
\begin{align}
  \ce{Cu (s) &-> Cu^2+ (aq) + 2e-}&
  E^\circ_\mathrm{ox} &= \pu{-0.34 V}
\end{align}

Your second half-reaction is a reduction (as written)
\begin{align}
  \ce{NO3- (aq) + 4H+ (aq) +3e- &-> NO (g) + 2H2O (l)}&
  E^\circ_\mathrm{red} &= \pu{+0.96 V}
\end{align}

The combination of these two half-reactions produces a positive $E^\circ_\mathrm{cell}$ which is a spontaneous reaction. You do not need to use the oxidation of water as your oxidation half-reaction. Copper is a much better reducing agent than water:

$$
\ce{3Cu (s) + 2NO3- (aq) +8H+ (aq) -> 3Cu^2+ (aq) + 3NO (g) +4H2O (l)}\\
E^\circ_\mathrm{cell} = \pu{+0.96 V} +(\pu{-0.34 V}) = \pu{+0.62 V}
$$

How does this change your analysis?


You are approaching this question from the wrong angle. One does not simply ‘add a gas to a drink’ for no reason. If one wanted to add any gas, one would probably use boring old nitrogen since it is the most abundant in the atmosphere (See the Guiness example as pointed out by Ivan in a comment).

Carbon dioxide is present in fizzy drinks such as soda to give them their fizz. The reason lies in the equilibrium between $\ce{CO2 + H2O}$, $\ce{H2CO3}$, $\ce{HCO3-}$ and $\ce{CO3^2-}$. No other gas would display this behaviour and give a slight acidic taste along with the tingle on the tongue.

Note that oxygen is sometimes added to drinks to improve the assumed value. It doesn’t really do anything but sells for a higher price. I don’t think that would be the case for nitrogen- or argon-enriched water (but I am likely underestimating the powers of marketing here). Many other gases would simply make no sense to add at all because they are either toxic ($\ce{Cl2, H2S}$) or disgusting ($\ce{H2S}$) or simply way too expensive ($\ce{Xe}$).


Will the concentration of 3% w/v H2O2 be lowered after 3 weeks?

Hydrogen peroxide is somewhat unstable even when not exposed to air and sunlight. It decomposes to water and oxygen gas over time.
According to this Wikipedia article:

Hydrogen peroxide is thermodynamically unstable and decomposes to form water and oxygen with a $\mathrm{\Delta H}$ of $\mathrm{-98.2}~\pu{kJ/mol}$ and a $\mathrm{\Delta S}$ of $\mathrm{70.5~\pu{kJ/mol}\ce.\pu K}$:
$$\ce{2H2O2 -> 2H2O + O2}$$
The rate of decomposition increases with rising temperature, concentration and pH, with cool, dilute, acidic solutions showing the best stability.

So, not knowing the exact conditions during the 3-week period or other factors, it's not possible to say for certain whether the $\ce{H2O2}$ degraded or whether it just was a little low to begin with. But the potential for degradation over that period of time is certainly there.


How do I solve this problem on mass balance from blast furnace iron making?

"The Mn in iron ore should be equal to Mn in slag." vs
"the solution in my book reads MnO from ore equals Mn in hot metal plus MnO in slag."

It seems that the calculation desired involves breaking down the MnO in ore to Mn + O, and putting some of the Mn into the metal; the rest of the Mn goes into slag, as MnO. The hot metal would contain only metallic Mn, not MnO. It shouldn't be a huge adjustment in the answers.


Any molecule that cannot be superimposed on its mirror image is described as being chiral. If the molecule is superimposable on its mirror image the object is called achiral.

For simple alkenes, the carbon atoms of the double bond and the four atoms bonded to them all lie in the same plane, which divides the molecule into two halves, which are mirror images of each other. This plane of symmetry ensures that the mirror image forms of the molecule are superimposable. Therefore, simple alkenes like 2-chloroprop-1-ene and 1-bromo-2-chloro-2-fluoro-1-iodoethene are achiral.

Nevertheless, owing to the barrier to rotation in alkenes, 1-bromo-2-chloro-2-fluoro-1-iodoethene exists as two (achiral) stereoisomers: (E)-1-bromo-2-chloro-2-fluoro-1-iodoethene and (Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene.


How does expansion of electron shells work?

You're right! The 4s subshell is filled before 3d. Electrons fill the shell   with least energy first and the 4s shell has a lower energy than the 3d subshell.
Use the Aufbau Principle to find the shell with least energy.

Basically it says that shells with a lower n+l (n and l are the principal and azimuthal quantum numbers) value are filled first.
If two shells have the same n+l value, then the shell with lower n is filled first.
There are a few exceptions to this rule - Copper, Chromium, Palladium, etc.

Reason: n+l gives the number of nodes in an orbital. The more the number of nodes, the less stable the orbital and so the higher the energy.
Valence shells are filled in the s and p blocks. The inner shells are filled in the d and f blocks.


This link has a good summary of hydrogenation. Syn addition is very much preferred because of the way the molecule 'adheres' to the surface. As you note, however, there are cases where anti-addition is observed - particularly for sterically hindered molecules such as 1,2-dimethylcyclohex-1-ene which shows up to 30% of the trans/anti-addition product (which also so happens to be the thermodynamic product).

The reason we see the syn-addition more often is because the Hydrogen atom attack happens really quickly (it's the kinetic product). If you pump more heat energy into the reaction though, you could expect to see more trans product - but this is dangerous as hydrogen is explosive.


Expanded structure formula and Lewis dot structure, whats the difference?

Yes. Extended molecular structures replace dots with line emphasising electron pairs that are shared between two atoms. In other words each line does represent a bond. Most often the EMS omit electrons that are not directly involved in bonds. As such the depiction of electrons between C and O in, e.g., a carbonyl group is simply sketched as C=O. However, whenever the presence of lone pairs is relevant to the discussion, those are explicitly shown.

In case in which the structure does not correspond to a single Lewis structure, the most important one is depicted and the situation is considered to be known to the reader. An archetype example is benzene. It can be sketched as having three double bonds but their delocalization is assumed. Those six delocalized electrons can be depicted as an inner ring, an expedient which is more evocative but must be abandoned when fine details are to be discussed (mechanisms of reaction or mesomeric effects). Moreover a circle in aromatic rings doesn't always represent six electrons.

Finally you can find depiction of delocalized electrons as variously dotted lines enveloping the group of atoms over which they do delocalize, for instance a kind of "cloud" around the two oxigens of a carboxylate anion. Their meaning is usually clear.

To the second question: on the light of the above yes, formal charge are to be assigned. They are usually important while discussing the relative stability of limiting forms which enter the description of the actual hybrid of resonance.

https://en.m.wikipedia.org/wiki/Lewis_structure. Particularly the subsections Resonance and Usage and limitations.

Edit: For sake of clarity I must add that unpaired electrons are always depicted. Indeed they are always of chemical relevance!