Finding mass of MnF3 is produced in a reaction

Here we have an already balanced equation given in the question-

$$\ce{2 MnI2 + 13 F2⟶ 2 MnF3 + 4 IF5}$$

We find the given number of moles of each reactant by dividing their given masses by their respective molar masses

Given mass of $\ce{MnI2}=\pu{1.23g}$
Molar mass of $\ce{MnI2}= \pu{308.74 g/mol}$

Amount of $\ce{MnI2} = \frac{1.23}{308.74}= \pu{0.0039mol}$

Given mass of $\ce{F2}=\pu{25g}$
Molar mass of $\ce{F2}= \pu{38 g/mol}$

Amount of $\ce{F2} = \frac{25}{38}= \pu{0.657mol}$

Now we find the mole ratio from the given information and compare the calculated ratio to the actual ratio.
If more than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, the $\ce{F2}$ is in excess and $\ce{MnI2}$ is the limiting reactant. If less than $\pu{13 moles}$ of $\ce{F2}$ are available per $\pu{2 moles}$ of $\ce{MnI2}$, $\ce{F2}$ is the limiting reactant.

Let's say that all of the $\pu{0.657 moles}$ of $\ce{F2}$ were to be used up, there would need to be

$\frac{2}{13}\times0.657=\pu{0.101  moles} \text{ of } \ce{MnI2}$.

But there is only $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ available which makes it the limiting reactant.

If all of the $\pu{0.0039 moles} \text{ of }\ce{ MnI2}$ were used up, there would need to be

$0.0039\times \frac{13}{2} = \pu{0.0254 moles} \text{ of } \ce{F2}$.

Because there is an excess of $\ce{F2}$, the $\ce{ MnI2}$ amount is used to calculate the amount of the products in the reaction i.e. amount of $\ce{MnF3}$ produced.

From the equation given its easy to observe that the stoichiometric ratio of $\ce{MnI2}$ and $\ce{MnF3} = 1:1$

Thus amount of $\ce{MnF3}$ produced = $\pu{0.0039 mol}$
Molar mass of $\ce{MnF3} = \pu{111.94 g/mol}$

Mass of $\ce{MnF3} = 111.94 \times 0.00398 = \pu{0.445g}$


There are varying definitions of"acid" (or "base") depending on solvent system; some do not involve water at all. For example, there are Brønsted–Lowry acids (proton donors) and Lewis acids (electron acceptors), though the Arrhenius definition is based on aqueous solutions dissociating into $\ce{H+}$ (acid) or $\ce{OH-}$ (base) ions.

Of course, many acids are not liquids, such as solid citric acid, which may be anhydrous or the monohdrate.


The rules for systematic nomenclature in organic chemistry have changed several times. They were first issued in Geneva in 1892, followed by the Liége Rules in 1930. The IUPAC rules were issued in 1957 (Sections A and B), 1969 (Sections A, B, and C), and 1979 (Sections A, B, C, D, E, F, and H). The rules were revised in the 1993 Guide to IUPAC Nomenclature of Organic Compounds and amended by official publications from 1993 through 2002. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) was published in December 2013.
According to the current IUPAC recommendations, the preferred IUPAC name (PIN) of cinnamaldehyde is (2E)-3-phenylprop-2-enal.

However, many compounds can have two or more systematic names in accordance with several methods recommended by IUPAC. Other names (that are not preferred IUPAC names but that are generated in accordance with IUPAC recommendations) may be used in general nomenclature. In addition, various traditional names are retained for use in IUPAC nomenclature, though the number of retained names has been reduced with each succeeding edition of the IUPAC recommendations.
According to the following subsections of the current IUPAC recommendations, the traditional name cinnamaldehyde may still be used in general nomenclature.

P-65.1.1.2 Retained names only for general nomenclature
(…)
P-65.1.1.2.2 The following names are retained for general nomenclature with
functionalization but no substitution is allowed. (…)
(…)
cinnamic acid
(…)

P-66.6.1.2 Names of aldehydes derived from retained names of carboxylic acids
Names of aldehydes derived from retained names are formed by changing the ‘ic acid’ or ‘oic acid’ ending of the retained names of carboxylic acids to ‘aldehyde’. (…)
(…)
P-66.6.1.2.3 Aldehydes derived from retained acid names given in P-65.1.1.2 are only used in general nomenclature; no substitution is allowed. Preferred IUPAC names are systematic names (…).


Chemical bonding and hybridization of organic compounds

You talk about hybridisation but you haven't actually looked at any hybrid orbitals in your description. Since the geometry around the carbons is trigonal planar, each carbon hybridises its $\ce{2s}$, $\ce{2p}_x$ and $\ce{2p}_y$ orbitals to form three $\ce{sp^2}$ orbitals which lie in the $xy$ plane at $120^\circ$ angles to each other. One $\ce{sp^2}$ orbital on each carbon points towards the other carbon and so these orbitals overlap to form a $\ce{C-C}$ $\sigma$ bond. The other two $\ce{sp^2}$ orbitals on each carbon overlap with the $\ce{1s}$ orbitals on the hydrogens forming $\ce{C-H}$ $\unicode[Times]{x3C3}$ bonds. The remaining unhybridised $\ce{2p}_z$ orbitals on each carbon overlap above and below the plane of the $\unicode[Times]{x3C3}$ bonds to form a $\ce{C-C}$ $\unicode[Times]{x3C0}$ bond.

While $\ce{sp^2}$ orbitals still have two lobes, one lobe is substantially larger than the other and only that takes part in bonding. Note that this is the usual case: even in molecules that solely use p-orbitals, such as chlorine, only one lobe of a p-orbital will form a $\unicode[Times]{x3C3}$ bond; the other of the same orbital will not.

This link provides a nice visualisation of the different orbitals involved in ethene bonding.


Formation of toluene from (cyclobutane-1,2,3-triyl)trimethanol using conc. sulphuric acid

Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated.

The mechanism (on paper) seems to be as follows
Dehydration $1$
For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows:

Dehydration $2$
Similar to the first dehydration, a ring expansion occurs to stabilize the carbocation by forming a 6-member ring:

Dehydration $3$
In the third dehydration, a hydride shift takes place concomitantly forming a tertiary carbocation stabilized by conjugation with the double bonds. The double bond is then formed giving the compound aromaticity.


What's the name of the problem of finding equilibrium concentrations?

I don't think the phrase "direct equilibrium problem" is used in this context.

"Multiple-equilibrium problem" would be appropriate.

As in the phrase "It involves solution of the multiple equilibrium problem" J. Molecular Biology vol. 247 page 774.

Or "chemical equilibrium problem" if you don't want to imply that there is more than one equilibrium.


What are the products of the reaction between methanamine and nitrous acid?

No chance to get alkenes from methanamine because there's no $\beta$-hydrogen for E2 elimination. And E1 elimination isn't going to work because you're not going to generate a methyl cation from the diazonium. The mixture of alcohols is probably because the alkyl cation could rearrange, but you're not creating the cation intermediate in this reaction.

The diazonium is a very good leaving group, so this complex is certainly going to be susceptible to SN2 reactions. The best and most available nucleophile you have in your mixture is water (the solvent), so you're going to get methanol.

Now, methanol could be a nucleophile that reacts with diazonium, but you're in aqueous solution. That means that the amount of water is going to dwarf any methanol that is produced. A negligible amount of the dimethyl ether will be produced.


Does electron mass decrease when it changes its orbit?

I'd like to copy the answer by John Rennie to a similar question at Physics.SE, since it's much better than current answers here IMHO. Although it mostly speaks about binding energy, the same principle applies to the quanta of excitation energy.

The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron together then they come to $1.67353272 \times 10^{-27}$ kg. The difference is about 13.6eV, which is the ionisation energy of hydrogen (though note that the experimental error in the masses isn't much less than the difference so this is only approximate).
This shouldn't surprise you because you have to add energy (in the form of a 13.6eV photon) to dissociate a hydrogen atom into a free proton and electron, and this increases the mass in accordance with Einstein's famous equation $E = mc^2$. So this is a direct example of the sort of mass increase you describe.
However you can't say this is an increase of mass of the electron or the proton. It's an increase in mass of the combined system. The invariant masses of the electron and proton are constants and not affected by whether they're in atoms or roaming freely. The change in mass is coming from a change in the binding energy of the system.


Low viscosity, low surface tension liquid for volume measurement

Air would be a convenient fluid. Fill your sensor with air at a known (atmospheric) pressure - well, it is probably filled with air already.
Then connect it to a vacuum chamber tube of known volume with a pressure sensor, i.e., a tube with two valves and a pressure sensor. Attach a vacuum pump to one valve and evacuate the tube; attach the $CO_2$ sensor to the other valve and then open it and measure the new pressure in the closed system. All the air will have been supplied by the sensor volume, and the new system volume will be the sensor volume plus the (known) tube volume. Doing it this way is likely to be less messy and more accurate than using liquids, and it won't sacrifice a sensor - and you can check every sensor!
A problem with liquids in a blind experiment (you cannot see the internals) is that there could be blind alleys that would be hard to detect and hard to fill with liquid (or which would not empty easily).


IUPAC (2013) rule

  P-44.1.1 The senior parent structure has the maximum number
  of substituents corresponding to the principal characteristic
  group (suffix) or senior parent hydride …

(more e.g. here) is applied in the phase of selecting "parent structure", before than the "longest chain rule" (P-44.3.2, e.g. here) in the phase of selecting the "principal chain".

Therfore the latter name, 3-propylpentanedial, is correct.


Do we get high on "methampetamine" naturally?

All the drugs I know of are closely related to a molecule we produce naturally or are a molecule that gets produced by our body. This is kind of the point of drugs. But the main issue is the dose. We all have uranium in our bodies, does that mean consuming it isn't bad for you? Or to put it differently: We all have bones, don't we? But putting another bone into your body isn't really considered healthy.

First of all: Chemically, those two molecules of course are not the same. Let's get the obvious out of the way.

I think this is the wrong forum again because really what you are asking is more of a legal/ethical question than a scientific one (both questions). But I'll try to answer the 2nd question "scientifically" with a light esoteric touch: Consuming drugs means messing with a delicate balance in our brains that is so complicated we do not even begin to understand it. I'm not saying don't use drugs, I'm just saying maybe you want it to be not as bad as it really is.


Calculating the pH of a solution of Ca(OH)2

You did not account for the fact that $\ce{Ca(OH)2}$ is partly precipitated. Firstly, to find out the moles of $\ce{Ca(OH)2}$ precipitated:-
The number of moles in solution is $\frac{1.48}{74}M$ which is $0.02M$. The total moles precipitated is just the moles of $\ce{CaO}$ minus $0.02M$ since the initial number of moles of
$\ce{Ca(OH)2}$ is equal to the moles of $\ce{CaO}$ taken and hence the amount precipitated turns out to be $0.03M$. Nevertheless, this is not required. With $0.02M$ $\ce{Ca(OH)2}$ in solution, the $\ce{OH-}$ concentration turns out to be $0.04M$ which, assuming no contribution of $\ce{OH-}$ from water and complete ionization, gives a $p\ce H$of $12.6$.


What is the reaction in a child's 'add water and watch the egg hatch' toy?

The egg hatch toys are composed of a superabsorbent polymer. These polymers are able to absorb up to 30 to 60 time their volume in water by forming hydrogen bonds with the water molecules.

Sodium polyacrylate ($\ce{[-CH2-CH(CO2Na){}-]}_n%edit$) is the most common superabsorbent polymer and according to its wikipedia page, is the one used in the egg hatch toys.


Method of deriving some properties or possible method of synthesis of a specific monomer

I doubt that your "monomer" is very stable. Orthoesters, $\ce{R'-C(OR)3}$, are hydrolyzed under acidic conditions: $\ce{R'-C(OR)3 + 2 H+ -> R'-COOR + 2 ROH}$. The fate of your monomer is:


How To Scan "improper dihedrals" in Gaussian 09?

After I asked this question, I found answer. I am putting the solution here so maybe somebody else use this.
The solution is this: if gaussian exit from the job (for example) at step 6; change starting geometry and take it 6 step back. How? For example in my question every step size is 18 degree and step 6 is 90 degree. Open your *.gjf or *.com file select the same dihedral and decrease angle 90 by degree (don't forget 90 degree is just for this example, you wil change it according to your step and the size of step) *sometimes you may have to fix one group . And then start job again.

you can see my "wrong gjf file" in the question. And here is the my "right gjf file": I only changed the starting dihedral angles.

%nprocshared=4
# opt=modredundant am1 geom=connectivity

Title Card Required

0 1
 B                 -0.54526231    0.17208210    0.44752404
 H                  0.54967689    0.02475247    0.03303383
 H                 -1.22032311    1.04666245    0.03303383
 H                 -0.96514071   -0.55516861   -1.21043678

 1 2 1.0 3 1.0 4 1.0
 2
 3
 4

D 1 2 3 4 S 9 18.000000

NOTE: it seems I defined different dihedrals at first glance.
in question I defined:
D 4 2 3 1 S 10 18.000000 and here I defined:
D 1 2 3 4 S 9 18.000000 but out of the symmetry and all H bonds are equal two dihedrals are the same indeed. Here is the article about this:  https://pdfs.semanticscholar.org/ab92/5f43824e59dd9e3be3474ebbac57a64998c3.pdf (DOI 10.1002/jcc.21425  - "Symmetrization of the AMBER and CHARMM
Force Fields", 2009, EDYTA M. et all)
in this article in figure 5 shows only 3 different improper can exist for  I–J–K–L but in my question I=L=J and thus actually 1234 dihedral and 2314 dihedral are the same and gave the same results for the same technics.


Why do chromyl bromide and chromyl iodide not exist?

Chromyl Bromide does exist.  See Halogen Chemistry at page 246:

  Deep violet needles of chromyl bromide are obtained by treating chromium(VI) oxide with $\ce{CF3COBr}$ and HBr in the presence of phosphorus pentoxide. A less pure sample is made by treating chromyl chloride with a large excess of hydrogen bromide. The compound, which is hygroscopic, is purified by vacuum distillation at —135° (Zellner, 1949; Krauss and Starke, 1962).

The book continues:

  Chromyl iodide has not been prepared and may not exist because of the powerful oxidizing property of the chromium (VI) oxidation state.

See similarly The Chemistry Of Chromyl Compounds Chem. Rev., 1958, 58 (1), pp 1–61:

  Chromyl bromide has likewise been isolated within the past few years (273,
  274). The modern work corrects earlier reports of the compound (77).

  While there has been one report of the formation of chromyl iodide as a deep
  red liquid from a mixture of dichromate, iodide, and sulfuric acid (87), a majority of observers (77, 96, 128, 196, 255) deny this observation and the existence of the compound. There is little likelihood that the iodide can exist, in view of the large free-energy change attendant on the reaction:

  $\ce{2Cr^{+6} + I- -> I^{+5} + 2Cr^{+3}}$


What is the convention of numbering carbon atoms in organic molecules?

Glucose occurs as the open chain form aldehyde that determines atom numbering, and as the epimeric cyclic hemiacetal(s) you provided,

The "ether" is a hemiacetal.  C-1's -OH can be up or down.  Either way, it opens and reversibly dehydrates to the aldehyde R-C(=O)H.  Number one atom in a chain is the end that bears the highest total mass.  Double bonds count double their other end atom.  The cyclic form's C=1 also has the largest summed mass attached, two oxygens plus a carbon and a hydrogen.


Chlorine is a very large atom compared to hydrogen, which means the overlap between the hydrogen $1s$-orbital and the chlorine $sp^3$ orbital is poor, which translates into a low bond energy.

Additionally, chlorine is very electronegative, which helps pull the H-Cl bonding pair onto the chlorine atom. The chloride anion solvation in water also plays a big part.

It's simply because it's energetically favourable. Maybe someone knows in greater detail than I do.


What is the mechanism for this reaction?

In your question, you have mentioned:

The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

You have recognized the reaction correctly but the Lossen rearrangement is the conversion of a hydroxamate ester to an isocyanate (only the first step of given mechanism). Although the original Lossen rearrangement needed hydroxamate ester $\left(\ce{R(C=O)NH-O(C=O)R'}\right)$ to rearrange,  recent research has shown that the rearrangement is possible with free hydroxamic acids $\left(\ce{R(C=O)NH-OH}\right)$ as well (Ref.1):

Without using activating agents, a variety of free aromatichydroxamic acids could be rearranged to aromatic amines in the presence of base alone:

Source of the image: Ref.1
According to $\bf{(a)}$, free aromatichydroxamic acid $\left(\bf{1}; \ce{R(C=O)NH-OH}\right)$ react with a base to give  the corresponding carboxylate $\left(\ce{R(C=O)-O-}\right)$ as intermediate and hydroxyamine $\left(\ce{NH2-OH}\right)$ byproduct. The intermediate $\left(\ce{R(C=O)-O-}\right)$ reacts with another aromatichydroxamic acid $(\bf{1})$ to give necessary hydroxamate ester $\bf{A}$, which undergoes Lossen rearrangement in basic conditions to give the isocyanate $\bf{B}$. Mechanism for aromatic amine from isocyanate is depicted in $\bf{(b)}$.
For instance, the base-mediated rearrangement of free 4-methylphenylhydroxamic acid $\bf{(1a)}$ to  4-methylaniline $\bf{(2a)}$ is given as follows:

A mixture of N-hydroxy-4-methylbenzamide $\bf{(1a)}$ $(\pu{0.363 g}, \  \pu{2.4 mmol}),$ $\ce{K2CO3}$ $(\pu{0.332 g},$  $\pu{2.4 mmol})$, and DMSO $(\pu{2 mL})$ was heated to $\pu{90 ^\circ C}$ and stirred at that temperature for $\pu{2 h}$. The mixture was cooled to rt, and then treated with $\pu{2 M} \ \ce{ HCl}$ $(ca. \pu{3 mL})$. After the mixture became the clear solution, $\pu{2 M} \ \ce{NaOH}$ $(ca. \pu{3 mL})$ was added and extracted with $\ce{Et2O}$ $(\pu{15 mL} \times 3)$. The combined organic layers were dried over anhydrous $\ce{Na2SO4}$, filtered and evaporated under reduced pressure. The residue was purified by silica gel column chromatography $(\ce{hexane/Et2O}, 1:1)$ to yield the pure 4-methylaniline $(\pu{0.253 g}, \ 98\%)$ as a white crystalline solid.

The effects of the amount of base, reaction time, and presence of water are summarized in following diagram (Ref.1):

Thus, as shown in above diagram, presence of water may have direct effect on the conversion of the substrate to corresponding amine (even in DMSO). Further, Ref.1 clearly shows that this reaction requires polar aprotic solvent to proceed through Lossen rearrangement to give thiocyanate intermediate, which eventually converted to corresponding amine in the presence of aqueous base. For instance, if $\bf{(1a)}$ is reacted with $\ce{K2CO3}$ in methanol instead of in DMSO or DMF at for $\pu{90 ^\circ C}$ for $\pu{2 h}$, the yield was found to be only $10\%$ while it was $98\%$ in DMSO and $92\%$ in DMF under the same conditions. The authors reported $86\%$ of the starting compound has been recovered unreacted (none of it recovered in DMSO or DMF).
Note that the first reaction given in the question is in aqueous medium. Thus, according to the results found in Ref.1, the resulting product in the given reaction would be a trace of corresponding amine $\left(\ce{R-NH2}\right)$ and most of substrate would be recovered unreacted. Consequently, there won't be a reactive isocyanate $\left(\ce{R-N=C=O}\right)$, thus the reaction would not proceed to give the anticipated product in methanol as indicated.
Suppose this first reaction has been done in $\ce{OH-/DMSO}$ in place of $\ce{OH-/H2O}$. That would give you the anticipated isocyanate intermediate $\left(\ce{R-N=C=O}\right)$. It is well known that isocyanates reacts very differently in the presence of water, alcohol, and amine. In the presence of water, it would give you corresponding amine with evolution of $\ce{CO2}$. However, in the presence of alcohol or amine, it would give you corresponding carbamate or urea derivative, respectively$:

Since the given second reaction is in methanol (an alcohol), the product would be a methyl carbamate as shown in the scheme above (Ref.2). For example, polyurethane is prepared from diisocyantes:

References:

Yujiro Hoshino, Moriaki Okuno, Eri Kawamura, Kiyoshi Hondab and Seiichi Inoue, "Base-mediated rearrangement of free aromatichydroxamic acids (ArCO–NHOH) to anilines," Chem. Commun. 2009, (17), 2281-2283 (DOI: https://doi.org/10.1039/B822806J).
Greet Raspoet, Minh Tho Nguyen, Michelle McGarraghy, and Anthony Frank Hegarty, "The Alcoholysis Reaction of Isocyanates Giving Urethanes:  Evidence for a Multimolecular Mechanism," J. Org. Chem. 1998, 63(20), 6878–6885 (DOI: https://doi.org/10.1021/jo9806411).


Finding the number of orbitals on a central atom

According to the question, it is totally not clear that whether you want to know filled orbitals or total orbital (filled + vacant). Well, but after seeing the examples you gave, it can be concluded that you are talking only about filled orbitals. As I'm not clear and assuming that you are only talking about filled orbitals, so if you are talking about total, leave a comment.

As each orbital contains 2 electrons, therefore it is easy to find number of filled orbitals.

The number of filled orbitals in central atom will be equal to the sum of number of bonds and number of lone pairs present on central atom.

Example

As in case of $\ce{BeCl2}$, $\ce{Be}$ makes 2 bonds (both with $\ce{Cl}$) and have no lone pair, therefore,

$\text{No. of filled orbitals} = 2$

Let us take another example of $\ce{NH3}$.

In case of $\ce{NH3}$, $\ce{N}$ makes 3 bonds (each with $\ce{H}$) and have 1 lone pair, therefore,

$\text{No. of filled orbitals} = 4$

Take another example of $\ce{ClF3}$

In this case, the central atom $\ce{Cl}$ makes 3 bonds with $\ce{F}$ by sharing its one electron in each bond and have 2 lone pairs on it. (As $\ce{Cl}$ have 7 electrons, 3 shared in bonding and 4 left = 2 lone pairs.) Therefore,

$\text{No. of filled orbitals} = 5$

Hope, now everything is clear to you!


Could you be more clear on what 'burning in' something means? The chemical reactions require both reactants available, so they need to mix - in what sense is Hydrogen mixing and reacting with Oxygen any different that Oxygen mixing and reacting with Hydrogen?
In some sense, we do tend to control these reactions via the fuel, not the oxidant (oxygen free environments are difficult to reproduce on Earth's surface). However there may be some examples of controlling the mixture via restricting the oxidant in a fuel-rich environment... even though I can't even get through the sentence without that sounding wasteful.
Nitrous Oxide
Would you consider the example of Nitrous Oxide injection in an internal combustion engine an example of this? Pre-nitrous, there is compressed atmospheric oxygen and hydrocarbons at a set ratio, and to further increase the burn a nitrous system will inject either NO alone (dry system) or NO and more hydrocarbon fuel (wet system) to add more oxidants that would be available through the 21% Oxygen in the atmosphere.

The Chemical: https://en.wikipedia.org/wiki/Nitrous_oxide
Internal Combustion application: https://en.wikipedia.org/wiki/Nitrous

Candles
How about a candle? There's an abundant fuel source (wax), and we frequently control the shape and characteristics of the burn with shaped glass, not to mention put it out via starving it of oxidant.
In either case, if you were to, say take a lump of solid oxygen (nevermind the practical difficulties of dealing with temperatures below 54 Kelvin) and maintained a burn with Hydrogen, what you'd be likely to find would be combustion only happening at the edges, where the oxygen was evaporating and mixing with the reduction agent, similar to lighting a can of gasoline on fire (only the vapors are flammable).


Applying Raoult's Law in volatile solids–liquid solutions

Raoult's Law is only valid when the solute is non-volatile. Therefore, you cannot apply it.


Which of the following reactions doesn't represent an oxidation-reduction

If there is an element in the reactants and it is not an element in the products then yes, oxidation or reduction has occurred.
Acid-base reactions are pretty much non-redox reactions. But it is always handy to learn the oxidations numbers.
Extra sources
Do non-redox reactions exist?


A misunderstanding about the energy profile of reactions with a catalyst involved

The upper graph is just the most simple way to visualize the effect of a catalyst on a reaction $\ce{S -> P}$:

The activation energy is lowered.

The activation energy for the reaction in that direction is the difference of the energies of the starting material $S$ and a transition state $TS^\#$.

Since it is the same starting marterial in the presence or absence of the catalyst, the energy of the transition state is different.

Can the same transition state have two different energies - just through the remote distance magical action of a catalyst located somewhere? Probably not!

It is much more plausible that - in the absence and presence of a catalyst - two completely different (different in structure and different in energy) transition states are involved.

Exactly this situation is described in the second graph!

The catalyst "reacts" with the starting material, either by forming a covalent bond, by hydrogen bonding, etc. and thus opens a different reaction channel with different intermediates and a transition state not found in in the non-catalyzed reaction.

In the end, the same product $P$ is formed and the catalyst is regenerated, but this doesn't mean that the catalyst wasn't heavily involved in the reaction going on.


While you are correct that the first ionization energy should have a larger value for all the group 15 elements, this has a minuscule effect on the ionization energy compared to the energy that is required to separate charges. Second ionization energies are almost always higher than first ionization energies because the nucleus already has a positive charge, creating a larger coulomb force that needs to be overcome as part of the ionization process. In addition, additional ionization increases the charge on the nucleus of the ion, which progressively makes the ion less and less stable.


Are there examples of crystals with random but static mixtures of two unit cell configurations?

Possibly retracted soon, because I'm not aware how to include a picture in a comment to clarity your question.  Do you mean instances like the following, where sites seemingly are occupied by two atoms simultanously -- as in analogues of Prussian blue -- just because the model deposit in the .cif is the average of the unit cells' content?

(Chen et al. in Angewandte Chemie,  129, June 2017, 9151-9156, doi 10.1002/ange.201702955)


This is discussed pretty thoroughly in materials engineering textbooks when melting and freezing of (mostly) metals in general is considered. A good reference is Gaskell’s “Introduction to thermodynamics of materials”. But you will find helpful explnations to fill in the gaps of what uou know by reading up on undercooling in Wikipedia etc. Feezing and melting are related.

And if you read my answer to highest melting point materials, I mention some related issues.

One issue is that you cannot consider materials as “pure” or “impure”: a pure material of any amount large enough to test is in practice almost impossible to make. In fact any real world material is impure once you consider a large enough scale: which is on the order of anything bigger than (say) millions of atoms. You cannot actually make a “pure” material of appreciable size, but can only approach it once you get larger than a relatively small size of sample. In addition, a lot of materials have only been made as thin coatings, or as individual small crystals. And their behaviour is not the same as larger amounts of what you might think is the same material. A single crystal behaves very differently from a polycrystalline mass, and the surface of any material is very different from its interior, so a thin film is dominated by surface effects, as are very fine particles.

So you are looking at degrees of impurity, in terms of orders of magnitude. And the general effect of purity level is not linear but more like exponential in terms of its effect on material behaviour.

A related concept is to look at the theoretical strength of a material. From a theoretical point, a pure and defect free material would be incredibly strong, but in practice this is impossible to achieve, you can only approach it.


IUPAC of heterocyclic compounds terminology

This way of naming heterocycles is known as the Extended Hantzsch-Widman system of nomenclature (the above link is a free reproduction of a publication that revised this system : W.H. Powell, Pure Appl. Chem., 1983,55, 409-416).

The stems do actually come mainly from numeric prefixes. " 'ir' from tri, 'et' from tetra, 'ep' from hepta, 'oc' from octa, 'on' from nona, and 'ec' from deca."Source. They are mainly taken from the middle letters of the numeric prefixes. As for "ol" and "in" for 5 and 6, these are holdovers from the intial system from 1887 that was developed for nitrogen containing rings, so my guess is they were kept to maintain the connection to the endings of pyrrole and pyridine.


Large difference in boiling points of tetrafluoromethane and tetrachloromethane

The vapour pressure can be estimated by equating the rate of leaving the liquid surface with that returning to it from the vapour. This leads to an expression of the form $\displaystyle p\sim n_{l}RTe^{-\Delta E/RT}$ where $\Delta E$ is the energy needed to remove a molecule from the liquid to the vapour phase and $n_{l}$ is the number density (number / unit volume) of the liquid.

The energy $\Delta E$ will clearly depend on the type of molecules involved. CCl$_4$ has large easily polarisable chlorine atoms. This means that these electrons are easily influenced by nearby atoms and this leads to an increased intermolecular interaction compared to CF$_4$ because the F atoms, having fewer electrons, and ones close to the nucleus, are only slightly polarisable. The increased $\Delta E$ (CCl$_4$ vs CF$_4$) then leads to a lower vapour pressure for CCl$_4$ and hence a higher boiling point which is defined at the temperature when the vapour pressure reaches one bar, or one atm. if you prefer.

(Sometimes the polarisable interaction is called induced-dipole, induced dipole or London or in general van der Waals interaction.)


Para effects on substituents in aromatic compounds

At meta position the M effect is not considered. The reason behind is that during resonance of a six membered cyclic compound charges never reaches at meta position that why it is so try to proof it by drawing resonance structure and if you didn't get us then tell me in the comments below.


Why are the axial bonds of PF5 longer than those of the equatorial bonds? (Hybridization)

Older textbooks decided to use a d-orbital formalism for elements of the third and lower (in the periodic table) periods to explain hyperconjugation. Nowadays, most of these compounds are explained with multi-centre bonds.

I could go stealing pictures and explanations from ron’s great answer, but I’m not going to. Rather, I’m just going to quickly explain the gist and redirect you to that answer.

If you take away two of the fluorines, then you can imagine the $\ce{PF3}$ fragment to be sp²-hybridised with a remaining p-orbital containing two electrons. (This description is wrong, as can be seen from the pyramidial structure of $\ce{PCl3}$ and its analogues, but we’ll keep it as a working basis.) The p-orbital can now interact with two surrounding further fluorine atoms that approach it from above and below with an electron each in the way that is shown in the image in ron’s answer. The resulting three orbitals will thereby create a four-electron three-centre bond. This bond can also be described mesomerically, by assuming a $\ce{P-F}$ single bond, a positive charge on phosphorous and a negative charge on the other fluorine, with a second mesomeric structure as a mirror image of the first.

From this mesomeric consideration you can see that the bond order of the axial $\ce{P-F}$ bonds is not 1 but rather closer to 0.5. The greater the order the shorter the bond so the smaller the order the longer the bond — and there you have your result.


Question about liquid-phase in P-T Diagram

The gas phase is unique in its ability to expand to fill space, so if there is space above or around a solid or liquid, molecules will eventually enter the gas phase in order to fill that space. That's the vapor-liquid equilibrium you described.

The liquid region of the phase diagram is the region where, in a closed space, some of the material will remain indefinitely in the liquid phase, even though some is also in the vapor phase. For example, in a sealed but expandable container (like a piston) maintained at an external pressure of 1 atm, the equilibrium state at 80 C has some water and some water vapor. But at 110 C, it will be all vapor.

The "coexistence curve" is a misnomer for the curve you are referring to. It does not mean that the liquid and vapor simply coexist (which as you pointed out can happen throughout the liquid region). It means that the equilibrium vapor pressure is equal to the external pressure. At that point, you cannot increase the temp of the liquid any further if the pressure remains constant. All heat added will go towards converting liquid to gas. Only after it is all gas will the temp increase. In order to get liquid at a higher temp, you have to increase the pressure. We define this as the boiling point.


What does the small, steep curve at the very beginning of a weak acid strong base titration curve come from?

This is part of estsblishing the buffer associated with the weak acid. To establish a buffer you need relatively large amounts of acid and conjugate base, and a weak acid can't produce the necessary conjugate base by itself.  You need to push it along with a little strong base to get into the buffering condition.

Say your solution is 0.1 molar acetic acid, with $K_\mathrm{a}=1.8×10^{-5}$. Before titration the pH is given by the Henderson-Hasselbach equation.  The logarithm of the above $K_\mathrm{a}$ value is about -4.74, therefore:

$\mathrm{pH}=4.74-\log_{10}\left(\frac{[\ce{HC2H3O2}]}{[\ce{C2H3O2-}]}\right)$

Now the amount of $\ce{C2H3O2-}$ in the solution is just what you get from dissociation of the acid, but acetic acid is only a weak acid and the actual amount of dissociation is small, in this case only about 1.3% of the acid you dissolved.  Merely 0.01 molar strong base reacting with the acid is enough to make a lot more acetate ion than just the acid dissociation alone, so the ratio in the H-H equation makes a big change and the $\mathrm{pH}$ follows suit.  Only when you have large amounts of both acid and conjugate base does the $\mathrm{pH}$ become buffered against slight additions of base or acid.


This is a very broad and very interesting question, but the basic point is: polymorphs exist either because

their formation is controlled by kinetics (rather than thermodynamics, which would favor the most stable phase), or
the thermodynamic equilibrium between the various polymorphs is influenced by the synthesis conditions (temperature, pressure, presence of templating agents and other species in solutions…), and once the material has crystallized in a given form, it does not spontaneously undergo a transition to the most stable phase (again, because of kinetics)


Will the water at the bottom of the tube boil at the same temperature as the water at the top of the tube?

Here's a cool example that should help illustrate the correct answer to your question.  As for why it works that way, you might think about it in terms of bubble formation .. in order for a liquid to boil, it needs to form bubbles.  Is it harder or easier to do that when the ambient pressure is high?

As for the calculation, you can use the Clausius-Claperyon equation to calculate the correction as described here.  You will need some values for $P_0$, $T_0$ and $\Delta H_{vap}$; you can use 1 atm, 373 K and 40.68 kJ/mol.


Relative reactivities of alkyl halides in an E2 reaction

I believe you might be mixing up E1 and E2 elimination reactions. In the first case the reaction first proceeds by formation of a carbocation from the alkyl halide and then a proton extractio. In this case, the order of reactivity is indeed $3^∘>2^∘>1^∘$ as such is the order of carbocation stability - in order the E1 reaction to proceed the carbocation must be formed.

But the situation is completely different in an E2 elimination. In this case the mechanism is the abstraction of the proton that is oriented anti to the halide with the simultaneous formation of the double bond and halide leaving the molecule as anion. That means that in this case the stability of carbocation plays no role and the significant factor are the sterical considerations. If the carbon atoms around the halide make it hard for the base to get to the anti proton then the reaction is less likely to proceed via E2 mechanism.

But bear in mind that usually both E1 and E2 elimination can take place and because of that tertiary halides would usually undergo elimination more readily than primary halides - just not via E2 mechanism.

More details on the E2 mechanism, with pictures: Master Organic Chemistry


What does "correct orientation" in a chemical reaction mean conceptually?

What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example?
  $$\ce{2H2 + O2 -> 2H2O}$$

This is a good example, but a complicated one. It is a chain reaction with three steps that have radical species as reactants and products (chain propagation), and one step that has radical species as reactants but not as products (chain termination). If you add up reactions 1 through 4, you will get the net reaction.

$$\ce{ O + H2 -> OH + H [1]}$$
$$\ce{ H2 + OH -> H2O + H [2]}$$
$$\ce{ O2 + H -> O + OH [3]}$$
$$\ce{ H + OH -> H2O [4]}$$

In order for the chain reaction to start, you need an initiation reaction starting with stable molecules making radicals. This is a slow reaction requiring a lot of energy, and there is some debate what it is. Here are two suggestions discussed in the literature.

$$\ce{ O2 + H2 -> OH + OH [5]}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{ O2 + H2 -> H2O + O [6]}$$

These two reactions require different orientations of the dioxygen and the dihydrogen molecules when they collide. This is because different bonds are broken and made when the different products are made. The picture below shows a "good" orientation for reaction [5] on the left and for reaction [6] on the right (The circles show the reactant molecules, and the dashed lines show the bonds that have to be made to get the products).

With the orientation on the left, it is easier to make two OH radicals and more difficult to make a water (with an O atom remaining). With the orientation on the right, it is easier to make a water molecule (with an O atom remaining) and much more difficult to make two OH radicals. Depending what the products are, some orientations upon collision are productive (lead to product) and some orientations are non-productive (no reaction occurs).

There is a third suggestion for the initiation step in the literature, and it requires an orientation for the collision different from the two orientations shown above. Can you figure out the best orientation for reaction 7?

$$\ce{ H2 + O2 -> H + HO2 [7]}$$

Source for reaction mechanism: NASA report


Using the right equipment for successful detection of Novichok

Lacking additional information, answers likely will be speculation-based only.   To mention a few factors to consider:
Stating GC-MS analysis only states, that there is technique to separate compounds by different rate of retention over a stationary phase (gas chromatography, GC) combined (a.k.a. hyphenated) with an other technique to analyze fractions sorting at different times by mass spectroscopy (MS). Mass spectroscopy typically shatters the molecules into charged fragments which, by their mass/charge ratio, behave differently in electromagnetic fields.  However, depending on the setup, the separation of the fractions by GC may be better, or worse.  On the side of mass spectroscopy, there are multiple methods to ionize and fragment molecules and to record these.  It is a problem that the parameters of running a mass spectroscopy may alter the shape of mass spectra quite a lot; signals may vary in their relative intensity, or sometimes be below the detection limit altogether.  It need not be the engineering advantage among corporations, of Agilent's technology over the by  Bruker (if there is?).
In the line of smaller benchtop mass spectrometers typically mounted behind a GC, larger sector field mass spectrometers, and Fourier mass spectrometers, there is a significant increase of resolution to record tiny differences in the masses of the fragments  the experiment of mass spectroscopy generates.  This may help to attribute the signals to the correct analyte, especially if there already is a suspicion to search for a selection of compounds only.
Routine GC-MS used for forensics and IOC doping screenings, for example compare the mass spectra recorded for a sample against previously recorded trustful references.  (Different to IOC doping screening, it is not evident if the Russian government shares / shared such standards here, though.)   Depending on the compound to identify, though, the analysis does not need to stop at the level of a GC-MS.  Both other types of chromatography may have been used now (e.g., HPLC) to separate the compounds; or multiple mass spectrometers coupled into a tandem (MS/MS) to observe not only the initial fragments obtained by mass spectroscopy, but equally their further fragmented «daughter radicals», too which equally rises the chance to identify your compound in question.
Maybe the Germans did not trace the poison itself, but metabolites judged as sufficient evidence for earlier reported, related compounds.  Especially with small amounts of sample material, the success of an analysis equally depends on every step of sample preparation; here likely (incomplete listing): what was the delay between Nawalny's intake of compound X and obtaining a sample by him (metabolism), the nature of the sample (tissue, blood, hair, etc.), delay and conditions of storage of the specimen to get it to the lab (unwanted sample decomposition), concentration and digestion protocols in the lab to eventually submit the sample to the mass spectrometer; the people's experience and availability reference data / techniques.


What is the effect of pressure upon the melting and boiling points of a substance?

When we apply pressure on a substance, its melting point increases or decreases depending upon the change of volume which occurs during the phase change of that substance. To be specific, if the volume of the substance's liquid phase is greater than the volume of the solid phase, its melting point will increase upon the increase of volume. Again, if the volume of the substance's liquid phase is less than the volume of the solid phase, its melting point will decrease upon the increase of volume.
The boiling point of liquids always increases when pressure is applied on that liquid. This is because the molecules of the liquid will require comparatively more energy to turn into gaseous state when pressure is applied on that substance.


Why is NAD+ regioselectively reduced at C-4 to form NADH?

Living organisms utilize protein based enzymes to catalyze otherwise unfavorable reactions under very mild conditions. They achieve this by bringing the substrates together in a very specific manner. Dehydrogenases are a class of oxidoreductases that oxidize substrates using NAD$^+$. Here is the active site of lactate dehydrogenase:

As you can see, carbon-4 of the nicotinamide ring is oriented towards the substrate. The orientation of the molecules is why carbon-4 is selectively reduced.


This is a misprint. Here is a similar problem (OpenStax Chemistry, retrieved from https://opentextbc.ca/chemistry/chapter/7-4-formal-charges-and-resonance/) that makes sense and has a correct answer:

  As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: CNS–, NCS–, or CSN–. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:

  Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).

The statement made rationalizing the incorrect structure is also flawed:

  "is more stable because each atom has non-zero formal charge in the lowest energy state"

This is an ion. Some atom has to have a non-zero formal charge because the net charge should equal the sum of formal charges. Why having each atom have a non-zero formal charge should result in a more stable structure is not clear to me. Mentioning the lowest energy state is curious as well. Maybe they meant most relevant resonance structure.

One should also keep in mind that formal charges are mostly an accounting method and do not reflect actual charge distribution (see https://chemistry.stackexchange.com/a/119771).


Relationship between the water ionization constant and ionic strength

The equilibrium constant is given though $$K_x = \prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}},$$
where $x$ is a specific quantity. We can choose that to be the activity $a$.

For the auto-ionisation
$$\ce{H2O (aq) <=>[H2O] H+ (aq) + OH- (aq)}\tag{1}\label{water}$$
we find therefore
$$K_a = \frac{a(\ce{H+})\cdot a(\ce{OH-})}{a(\ce{H2O})}.$$

Assuming the equilibrium $\eqref{water}$ is largely on the left side and therefore there is an excess of water, a reasonable simplification is to treat $a(\ce{H2O})$ as constant.
$$K_a' = K_a\cdot a(\ce{H2O}) = a(\ce{H+})\cdot a(\ce{OH-})$$

In ideally diluted solutions we also find
$$\lim_{c\to0}\gamma_i = \frac{a_i}{c_i}=1,$$
and therefore
$$K_c = c(\ce{H+})\cdot c(\ce{OH-}) \overset{\mathrm{def}}{=} K_\mathrm{w}.$$

The ionic strength is defined as
$$I_c = \frac12\sum_{\ce{B}}c_{\ce{B}}z_{\ce{B}}^2,$$
or for our case
$$I_{c,\mathrm{w}} = \frac12 \bigg[c(\ce{H+}) + c(\ce{OH-})\bigg].$$

Because of $\eqref{water}$ we can write
$$I_{c,\mathrm{w}} = c(\ce{H+})$$
and
$$K_\mathrm{w} = c(\ce{H+})^2, $$
therefore
$$I_{c,\mathrm{w}} = \sqrt{K_\mathrm{w}}.$$

From this relation you can see, that when $K_\mathrm{w}$ increases, the ionic strength also increases, vice versa.

  If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH−}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.

The first part is correct, the activity of each ion would decrease. However, there are more ions, so the activity of the ensemble increases more (given ideal dilution), therefore also leading to an increase in $K_\mathrm{w}$.


Why isn't pure ortho hydrogen possible to obtain?

Ortho and para hydrogen differ in the alignment of their nuclear spins. The two hydrogen atoms each have a nuclear spin of 1/2 which results in a total nuclear spin of 0 or 1. The states with zero spin are called para and those with 1 ortho. The combination of the nuclear spins also results in the formation of a new spin wave function. There are 3 possibilities to get a nuclear spin of one with an even wave function and there is one possibilitie to get a spin of zero with an odd wave function. The overall wave function of the hydrogen molecule is given by the product of the electronic, vibrational, rotational and spin wave functions and should we antisymmetric with respect to exchange of the fermionic hydrogen nuclei. Because the electronic ground state of molecular hydrogen is even and the vibrational wave function of a diatomic molecule is always even, the wave function of para hydrogen has to be combined with the even rotational states and ortho with the odd rotational states. The absolute ground state of molecular hydrogen is thus the rotational state with zero rotation and is therefore para. The next state, with one quantum of rotation, is ortho.

You can only convert ortho and para hydrogen into each other using collisions or chemical reactions. To make pure para hydrogen, people use a catalyst at very low temperature. Because of the low temperature, all produced hydrogen will eventually end up in the absolute ground state, that is, para hydrogen. You cannot make ortho hydrogen this way because it does not represent the absolute ground state of the system. The 75% comes from the degeneracy of the Boltzmann factor: at low temperature (with no reaction catalyst) there are three times more ortho molecules with a spin of 1 than para molecules of zero spin.

Although you cannot make pure ortho hydrogen, you can study it by using narrow lasers that only excite odd rotational states.


If you specifically mean a hot plasma, I think it might be possible but it's a tough request. To get combustion, you'll need to start off with form of bonding present between atoms in the plasma. This requires finding a molecule with atoms so strongly bound that the bond dissociation energy is higher than the molecule's ionization energy.

Given that the lowest ionization energy known (3.51 eV or about 340 kJ/mol for ditungsten tetra(hpp)) is comparable to relatively sturdy covalent single bonds, there's not much room for negotiation. Ionization energies easily become larger than even strong covalent double bond dissociation energies, so there's a large tendency for any substance to be stripped into its constituent atoms before getting a chance to ionize.

The situation is easier if you allow the usage of "cold" plasmas. Plasma generators are commonly used to scour surfaces completely clean of any organic residues. I don't see why you couldn't just insert oxygen and some organic gas into such a generator, ionize them both with the strong and rapidly varying electric field inside, and cause the organic gas to combust.


How does one export a PDB structure to SVG?

If you haven't found answer till now, you can use VESTA (Visualization for Electronic and STructural Analysis) package. It is a freeware and can be downloaded from
http://jp-minerals.org/vesta/en/

It is available for both Windows and Linux. It can be used to import any many structural files (pdb included) and get an output in any supported format(which includes SVG). From their website:

  VESTA is a 3D visualization program for structural models, volumetric data such as electron/nuclear densities, and crystal morphologies.

However, if you do not strictly require SVG format them my personal favorite is "QuteMol". It gives way too many options to adjust visual representation
It is an open source (citeware, i.e. you have to cite their paper for any academic publishing) molecular rendering software available from:

http://qutemol.sourceforge.net/

It can generate PNG, JPEG anf GIF(animated) images of maximum dimensions up to 8192x8192 pixels so you can use them for high resolution printing/posters etc. too. One small flaw is that it is rather ill maintained and considerable amount of bugs are there.For e.g. using default options is fine but in case you want to use customized colour scheme then their example ART(atomic rendering table, just a table of atomic symbols, their radii, and colour in RGB codes) file gives following example:

  X    0       0.853  204    0    0

  H    1    0.930  199  199  199

  He    2  1.085  107  115  140

  Li    3  3.100    0  128  128

However in reality you will require 2 colums of radii for programme to recognize it (core and vander waal radii). Similarly atomic symbols with single alphabet are not rendered properly!!! But all these problems have work around and results are worth it (specially for large protein molecules).
sample this:

(source: sourceforge.net)


Why is the resonance concept not required in molecular orbital theory?

Because MO is closer to reality.

Resonance is a convenient fiction used to simplify the visualization of some molecules, and rationalize their actions in the frame work of Lewis structures. But every resonance structure you have ever seen in your life is a lie.

Well, this is not exactly true. It's more precise to say that resonance structures are like blind men trying to describe an elephant: they each give a partial snapshot of truth while being wholly inadequate to describing it all. The "true" structure is a weighted average of all the resonance structures based on how stable they would be if they were the actual molecule.

MO cuts to the chase and attempts to directly describe the molecule in terms of its molecular orbitals: another convenient fiction, mind, but one that is closer to the truth. The reason you don't need "resonance" MO structures is because molecular orbitals are inherently a probabilistic construct: they already take into account the uncertainty in the electrons' positions.


How does the application of Schrödinger equation to model a particle in a box explain the origin of degeneracy of atomic orbitals?

Is the particle in a box concept analogous to an electron in an orbital?

No, but there is an analogy to the $\pi$ systems of dye molecules, and (for the 3D box) to the band structure of nanodots. Some physical chemistry courses have a lab that explore this relationship, e.g. at Saarland university (sorry, in German, but the figures and the math are universally understandable, I hope).

Finally, if this equation can be used to explain the allowed energies of orbitals/electrons, what would L be in this instance?

It would be the length of the conjugated system (for the dyes) or the radius (for the nanoparticles)

Are the orbitals degenerate because this equation only includes the quantum number n and not the others and therefore all orbitals in a subshell will roughly be degenerate?

For a 1D case with degeneracy, you can explore the particle in a ring, Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems?.


Reason for the differences in third ionisation energy between elements (magnesium and fluorine)

The ionisation energy, as a rough guide, can be related to the effective nuclear charge and the principal quantum number:

$$\mathrm{IE} \propto \left(\frac{Z_\mathrm{eff}}{n}\right)^2$$

In the ionisation of $\ce{Mg^2+}$ and $\ce{F^2+}$, both electrons are being ionised from the second principal quantum shell, so $n = 2$ in both cases. However, the main reason why the 3rd IE of magnesium is larger is because the nuclear charge $Z$ is larger, and not because of any so-called stability of half-filled or fully filled shells.

Contrary to what you thought, the nuclear charge is not the charge on the species. It is the charge of the nucleus - hence "nuclear charge". The $\ce{Mg}$ nucleus has 12 protons, and the $\ce{F}$ nucleus has 9 protons. So, there is a greater electrostatic force pulling the electron towards the nucleus in $\ce{Mg^2+}$ than in $\ce{F^2+}$, and consequently it is harder to remove the electron from $\ce{Mg^2+}$.

In truth, the raw nuclear charge must be modified by the shielding constant $\sigma$ to obtain the effective nuclear charge, $Z_\mathrm{eff}$. However, the variation in $\sigma$ between the two compounds is really negligible compared to the variation in the actual nuclear charge.


Why are there 3N-6 degrees of freedom in molecules?

You have N atoms, each atom can be described by a set of cartesian coordinates (x,y,z). So for each atom we have 3 degrees of freedom which means we end up with 3N degrees of freedom. If we look at an isolated structure without any external field or anything like that then only the relative position of each atom to the others matters. So we can get rid of some degrees of freedom. Translating every atom in the same way into x, y or z direction won't change anything, so we can reduce the degrees of freedom by 3. Rotating it around x, y or z doesn't change anything either, so we are reducing by another 3, so we end up with 3N-6.


Regarding the coordination isomerism question, the reason why it's not considered as an isomer in the case of two uncharged complexes, is that there is no ionic attraction. The two coordination spheres should act as individual ions and hence participate in the ionic sphere. But in this case, where the two spheres carry no charge, there isn't a force, namely the electrostatic force, which binds the two spheres.

Your question on CrCl3.6H2O is a very debatable topic and there hasnt been a unanimous decision that has been put forth by all the authors. Some say there are 3 or rather most of them say 3, only one author, JD Lee says that there are 4.

The reason given by the authors to support there claim of only 3 isomers is not very concrete, but it still majority of the authors believe that it is true. The H2O that exists outside the coordination sphere is the water of crystallization. The water of crystallization is believed to be present within the voids created by an anion and a cation. Now if the isomer [CrCl3(H2O)3] 3H2O should exist then it must have these voids. But here there isn't any anion in the ionic sphere, in fact it is believed that there isn't any ionic sphere. Therefore this isomer doesn't exist and hence cannot be the 4th isomer.

Now, it's also believed that the H2O is actually held to the central metal atom (Cr) with the help of dipole forces and that the voids are not needed. So that would mean that this exists and since it is theoretically an isomer, it should also be an isomer practically.

That brings us to to this very important question, what is the significance of the water in the ionic sphere? Now most books say that it's the water of crystallization of that compound. So if the water of crystallization must be present in the voids so that they can be removed by physical processes like heating, the 3H2O wont be an isomer anymore, but just another compound that exists with the same molecular formula.


You're right in thinking that it's an alkane. Now I think I shouldn't blame you for getting confused by

  $\ldots$ as a molecular formula which is mainly used in cars as fuel $\ldots$

Because

First off, we don't know if $x$ in the molecular formula is an integer or a decimal number.
Second, we don't know what fuel is meant. It's just supposed to be a fuel used in cars. And what are the fuels used in motor vehicles again?

  Currently, the majority of motor vehicles worldwide are powered by gasoline or diesel. Other energy sources include ethanol, biodiesel, propane, compressed natural gas (CNG), electric batteries charged from an external source, and hydrogen.

What are these?

  The bulk of a typical gasoline consists of $\mathrm{\color{red}{hydrocarbons~with~between~4~and~12~carbon~atoms~per~molecule}}$ (commonly referred to as C4-C12).

  Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including n, iso, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\color{red}{\ce{C12H23}}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Ethanol is an $\color{red}{alcohol}$.

  Biodiesel refers to a vegetable oil - or animal fat-based diesel fuel $\mathrm{\color{red}{consisting~of~long-chain~alkyl~(methyl,~ethyl,~ or~ propyl)~ esters}}$.

Propane is an $\color{navy}{alkane}$.

  CNG is made by compressing natural gas (which is mainly composed of $\color{red}{methane}$, $\ce{CH4}$), to less than 1 percent of the volume it occupies at standard atmospheric pressure.

Electric batteries and hydrogen are $\ldots$ a bit irrelevant. Let's not make this answer longer.

Thus, the answer would be propane, since it's the only alkane among the common car fuels. But is it? Of course not; since propane cracking isn't in high school syllabuses anywhere in the world. There is a very very good chance the querent had gasoline mind.
Now that we trust our natural instincts (!) that they meant gasoline when they were referring to the fuel, we should remain perplexed and run in circles. Because gasoline isn't consisted of one or two distinctive and well-known substances, and it's a mixture of  It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes), where the usage of the terms paraffin and olefin is particular to the oil industry. To make matters worse, gasoline composition is obviously not the same around the world.

$\hspace{25ex}$
$\hspace{25ex}~\small{\mathrm{Some~of~the~compounds~commonly~found~in~gasoline}}$
$$\small{source}$$

So, let me rephrase your question into something actually meaningful:

  $\ldots$ as a molecular formula which is mainly known to be a really smooth-burning species in gasoline $\ldots$

Now we're talking. The best known gasoline would be one that consists of 100% isooctane, and that's why we have octane rating. It's not a real fuel, just the material in an imaginarily ideal mixture that would burn "real smooth". That's why your question is problematic and scientifically incorrectly asked.

Back to the question, 2,2,4-Trimethylpentane must have been the compound the querent had in mind. Now we have the answer for A.

$\hspace{25ex}$
$\hspace{28ex}$ Isooctane (PIN: 2,2,4-Trimethylpentane)
$\hspace{43ex}$ $\small{source}$

But what would the products of the cracking be?

According to the Wikipedia article on cracking the cracking process in entropy-driven and the radicals formed in certain conditions usually move the reaction forward. But the final products (and even which radicals form) highly depend on the methods of cracking (thermal, steam etc.) and/or the catalysts. There are numerous catalysts which demonstrate the diversity in the mechanism of cracking.

For example, take a look at this paper (It's just an abstract, but it gives you an idea)

  $\ldots$ The main reaction route on FAU is hydride transfer followed by β-scission leading to mainly C4 species, while on MFI protolytic scission is responsible for the formation of high amounts of C1−C3 species. $\ldots$

So, we're left with speculations. This isn't a real question. The only answer the examiners want is B's most probably butane, as it must be an alkane.

As this also demonstrates, a very simplistic view is that an octane can break into two ethene molecules and one butane molecule. So, the reaction they must have had in mind is
$$\ce{C8H18 -> C4H10 + 2C2H4}$$
Pfft.

TL;DR:

This isn't a good or correct question to ask at all; at the very least, it's very poorly worded. The answers are supposedly isooctane for A and butane for B, but the OP's confusion was innocent.


Can carbon dioxide be reduced to carbon monoxide and oxygen to produce energy?

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction:
$$\ce{CO2(g) -> CO(g) + O(g)}$$
This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the row for carbon dioxide that this reaction has an enthalpy change of $\mathrm{+532\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. The proposed reaction is therefore heavily endothermic. That is to say, it must absorb energy. Interestingly, it's true that the extreme strength of the bond in carbon monoxide has a measurable effect, making this process more favourable than expected. However, it is still overall extremely unfavourable, and therefore requires a large input of energy. I stress that this is unavoidable, no matter how fancy your machine - if the end result is the reaction stated above, then you must pay the energy cost somehow.
Part of the problem, though, is that we have monoatomic oxygen as a product, which is a very reactive, high energy species - it doesn't actually exist except in special conditions. A simple adjustment therefore is to have molecular dioxygen, $\ce{O2}$ (the kind in the atmosphere that you breathe). The reaction then becomes:
$$\ce{2 CO2(g) -> 2 CO(g) + O2(g)}$$
So what's the enthalpy change associated with this reaction? Looking up another table, this turns out to be $\mathrm{+283\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. Again, this reaction is endothermic, though much less so than the first one. Regardless, once more this reaction is an energy sink.
If you want more visceral confirmation of this fact, consider the following. It is well known that pure carbon monoxide burns in an oxygen atmosphere. The reaction is self-sustaining and releases considerable heat. If you pay close attention, the reaction in the video is the exact inverse of the second equation. By chemical thermodynamics, if the combustion of $\ce{CO}$ to $\ce{CO2}$ releases heat, then it is necessarily true that cleaving $\ce{CO2}$ to form $\ce{CO}$ and $\ce{O2}$ will consume energy.
As a last point, there are ways to make the production of $\ce{CO}$ from $\ce{CO2}$ feasible, but it requires changing the products. For example, if hydrogen gas is used as a reagent, the following becomes possible:
$$\ce{CO2(g) + H2(g) -> CO(g) + H2O(g)}$$
The enthalpy change for this reaction is $\mathrm{+41\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$ , which is still endothermic, but approaching the break-even point. This is not too surprising, as hydrogen gas can behave as a reductant, and the bonds in water molecules are strong, pushing the reaction forwards. Let us make one last tiny modification:
$$\ce{CO2(g) + H2(g) -> CO(g) + H2O\color{red}{(l)}}$$
By assuming the water produced is in the liquid state as opposed to a gas, the reaction surrenders a little more energy, and the calculated reaction enthalpy becomes $\mathrm{-3\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. This reaction is very mildly exothermic, which is to say it releases heat (admittedly, so little that it's within margin of error, and slightly different conditions could make the reaction overall endothermic).
If you're not dead-set on having carbon monoxide as a product, then there are further options still. For example, here is the complete reduction of $\ce{CO2}$ to methane ($\ce{CH4}$), a considerably exothermic process with a reaction enthalpy of $\mathrm{-253\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$:
$$\ce{CO2(g) + 4H2(g) -> CH4(g) + 2 H2O(l)}$$
Methane is not an ideal product, as it too is a greenhouse gas and is a low value chemical feedstock due to its abundance and relative lack of useful chemistry. There is much more interest in conversion of $\ce{CO2}$ to compounds such as methanol $\ce{CH3OH}$ and formic acid $\ce{HCOOH}$. These two particular reactions are also exothermic.
There are several issues with using hydrogen reduction of $\ce{CO2}$ as a carbon capture strategy to combat climate change, but perhaps the main one is a real-world factor: most hydrogen we currently produce is derived from fossil fuels, notably from the partial combustion of fossil methane (natural gas) with water at high temperatures, known as steam reforming. Therefore, while alternative sources of hydrogen gas using renewable low-carbon intensity energy are not available, this is a poor strategy to remove anthropogenic $\ce{CO2}$ from the atmosphere.


Is it possible to oxidize ether to ester?

Yes, it is possible.  One reagent to do so is acidic bromate salts.  This paper discusses a number of optimizations and characterizations of the bromate-powered ether oxidation reaction.

$$\ce{R-CH2-O-CH2-R + BrO3- ->[H+][H2O; {r.t.}] R-CO-O-CH2-R + Br^-}$$

Hydrolysis of the formed ester leads to an undesired byproduct: the free acid $\ce{RCOOH}$.  In addition to that byproduct, overoxidation of the substrate can also happen, leading to anhydrides or even $\ce{CO2}$.

There may well be plenty of other oxidants and catalysts than can achieve the same reaction, but I thought this one was notable because it works in water at room temperature.  Yield depends strongly on reaction conditions and the vagaries of the particular substrates you use, though.


Typical convention for organic molecules is to list carbon first, than hydrogen, and then other elements in alphabetical order. The Merck Index, for example, does it this way.


How is the process of creating PVC fabrics harmful for the environment?

I wouldn't think that the process of producing PVC fabrics is that much of a problem, for reasons:

PVC is produced by experts that have taken all the necessary precautions for all possible reductions of safety risks it may cause in the lab or in mass scale production.
The process of synthesizing PVC is not a secret. If it was too harmful (rather than its profitability), then companies would have been banned from producing such a thing.

Phew, I thought there was something serious about PVC fabrics.

There is! The problem is not the process, but VCM and phthalates.
Issue 1: VCM
With a chemical formula of $\ce{C2H3Cl}$, a boiling point of $-13.9°C$ and an approximate molecular weight of $62.5 \frac{g}{mol}$, VCM is the raw material for PVC. According to here, the death of four workers had been approved due to VCM in 1974. (Though wikipedia says it was three workers dead) It's reported that in 1975, an annual 5-billion-pounds rate of production of VCM was confirmed.
The industry had the worst ways of producing VCM, and has suffered for decades until better ways of producing it have come around. According to here, a traditional way of producing VCM was the reaction of acetylene with Hydrochloric acid, as following: $$\ce{C2H2} + \ce{HCl} ~->~ \ce{C2H3Cl}$$
It was in the presence of Mercury(II) Chloride as a catalyst! Nothing has been done to reverse the harmful effects done to environment because of this very reaction.
In the modern process, "polymerization" is known to be the most dangerous stage and the peak of the overexposure.
VCM is believed to be carcinogen. Severe symptoms have been reported due to the overexposure of VCM; such as: Hepatotoxicity (liver damage and tumor growth caused by chemicals and more precisely, angiosarcoma).
Issue 2: phthalates
VCM is the fundamental raw material for the synthesis of all kinds of PVC; rigid or soft. But Plasticizers are being added to rigid PVC to "soften it up"; so it would be applicable for the future use as "PVC fabrics". Phthalates are common plasticizers in the production of soft PVCs.

They are divided into two distinct groups, with very different applications, toxicological properties, and classification, based on the number of carbon atoms in their alcohol chain.

Phthalates aren't covalently bonded to the matrix of PVC. Therefore, the lighter ones can leach relatively very easy, usually due to heat or strong solvents.
The most common phthalate is diethylhexyl phthalate(DEHP). Studies show that Americans are exposed to DEHP. DEHP is believed to cause health complications in a variety of ages.
Oh, wait: Issue 3: PVC itself! But when it degrades
PVC is known for its strong linear polymer; but when it degrades (usually due to careless use or disposal) it degrades! Degradation is excessive reduction in a plastic's average molar weight. PVC will "microcrack" when it's degraded, into macroparticles that are a big threat to the environment. These particles "soak up" POPs and afterwards are usually ingested by living things, making it a big deal of our everyday plastic pollution exposure. (as wikipedia puts it)

What should I do now?! Panic?

Considering the advancement of technology and our anxiety to preserve our health we humans aren't sitting around to watch ourselves getting sick! I don't mean the hard-to-come-by sites that say overexposure to PVC isn't carcinogenic, but real efforts into solving the problem. News has been spread that REACH attempts to authorize DEHP and two other phthalates and ban their use in specific materials. DINP is another plasticizer, approved by the EU risk assesment, that will eventually replace the former plasticizers, hopefully.
Scientific researches for over ten years have proven that the amount of exposure to the chemicals mentioned is far from dangerous in current conditions and uses. So sit back relaxed; it's true that these dangers exist, but they're not serious threats to a healthy person, at least not yet.


Why is the maximum adiabatic temperature found for an equivalence ratio a bit higher than 1?

I actually found the answer in this 2011 paper, here is a quote from the abstract:

  The characteristic rich shifting of the maximum adiabatic flame temperature from the stoichiometric value for
  hydrocarbon/air mixtures is demonstrated to be caused by product dissociation and hence reduced amount of heat release.
  Since the extent of dissociation is greater on the lean side as a result of the stoichiometry of dissociated products, the peaking
  occurs on the rich side.


Why does pasta become sticky after being cooked?

After cooking, pasta has a lot of free surface starch ready to bond with things. This is why we rinse well after cooking to remove extra starch and cool down the pasta to prevent it from producing more surface starch.


I believe methanol does protonate the alkoxide. The resulting methoxide coordinates to $\ce{BH3}$ forming $\ce{(MeO)BH3-}.$ The electron donating character of the methoxy group makes the remaining protons more hydride which results in a more reactive hydride source.


Why does the stability of ascorbic acid decrease (increased rate of oxidation to dehydroascorbic acid) when pH increases?

The decrease of ascorbic acid is probably mostly due to oxidation.

Oxidation is loss of electrons.

Increased pH means more ascorbate, thus negative charge and higher electron density.

Higher electron density means easier loss off electrons, thus more oxidation and less ascorbic acid at the end.


In A Novel Example of Molecular Hydrogen Generation from Formic Acid at
Visible-Light-Responsive Photocatalyst, the authors describe the formation of carbon dioxide and hydrogen from formic acid using visible light from a halogen lamp and freshly prepared $\ce{Cu2O}$ (or platinum-covered $\ce{Cu2O}$) as a catalyst.

The preparation of the $\ce{Cu2O}$ is described as follows:

  Ten cubic centimeters of $\mathrm{1.0~mol~dm^{-3}}$
  $\ce{CuSO4*9H2O}$ aqueous solution was dropped
  into $\mathrm{100~cm^3}$ of $\mathrm{1.0~mol~dm^{-3}}$ $\ce{Na2S2O3}$ aqueous solution, and
  subsequently, the mixture solution was dropped into $\mathrm{200~cm^3}$
  of $\mathrm{1.0~mol~dm^{-3}}$ $\ce{NaOH}$ solution at 80 °C.

  The resulting reddish-brown precipitate ($\ce{Cu2O}$) was separated by filtration and washed with deionized water, after which the precipitation was dried at 40 °C in an $\ce{Ar}$ atmosphere.

The conversion of formic acid is described as follows:

  The photocatalytic experiment was conducted in a glass vessel (volume: $\mathrm{20~cm^3}$) containing $\mathrm{10~cm^3}$ of $\mathrm{2.5~mol~dm^{-3}}$ $\ce{HCOOH}$ and a known amount of photocatalyst (10 mg (ca. 70 μmol)) [...] in
  $\ce{Ar}$ atmosphere with stirring. The pH of the $\ce{HCOOH}$ solution was adjusted to 5.0 by the addition of $\ce{NaOH}$ solution.

  A halogen lamp was used as the light source (light intensity: $\mathrm{280~mW~cm^{-2}}$ ) along with a sharp-cut filter (S76-L42, Suruga Seiki Co. Ltd.), using which the samples were irradiated with light of wavelength $\lambda > 420~\mathrm{nm}$. [...]

In summary, it seems possible to perform the experiment at school. UV filters are available from different manufacturers, a Schott N-WG320 might work.


Mass number, (relative) atomic mass, average mass

It’s pretty simple:

The mass number is the integer you get if you count (and add up) the neutrons and the protons of a given element. Thus, a hydrogen atom of the $\ce{^1_1H}$ isotope has a mass number of $1$ (only proton), $\ce{^12_6C}$ has $12$ ($6$ protons, $6$ neutrons) and $\ce{^81_35Br}$ has $81$ (of which $35$ are protons, the remaining $46$ neutrons).
The atomic mass is what these atoms actually weigh in atomic mass units. For reasons that boil down to $E = mc^2$ (or so I believe) and the nonzero mass of an electron, this is not an integer except for one exception:

$\ce{^1H}$’s atomic mass is $1.007825032 1(4)$
$\ce{^12C}$’s atomic mass is exactly $12$. This is because $1~\mathrm{u}$ was defined as exactly $1/12$th of the mass of a carbon-12 atom.
$\ce{^81Br}$’s atomic mass is $80.916289(6)$

The average mass takes into account an elements different isotopes and their natural abundance and calculates an overall average. Thus, this is no longer defined on an isotopal basis but on an elemental one. The average masses of the elements discussed above are:

$\ce{H}$ has $1.00794 (7)$ (this is larger than the atomic mass of $\ce{^1H}$ due to the heavier isotopes.)
$\ce{C}$ has $12.0107 (8)$
$\ce{Br}$ has $79.904 (1)$ (this is lower than the atomic mass of $\ce{^81Br}$, because about half of the naturally occuring bromine atoms are $\ce{^79Br}$.)


I would say that acids are better.

Why? Bases and acids dissolve flesh pretty easily (I'm thinking of $\ce{NaOH}$ and $\ce{HCl}$ here, weaker acids/bases -- not so much). Most of our body is made up of proteins, which are overall pretty much neutral (generally). Both acids and bases can dissolve protein.

On the other hand, there is one major component of our body which is easier to dissolve with acid. Bones. The major component of bone is $\ce{CaCO3}$. The reaction $\ce{2HCl +CaCO3 -> H2O + CO2 ^ + CaCl2}$ is much more efficient/faster than $\ce{2NaOH + CaCO3 -> Na2CO3 + Ca(OH)2}$. Why is this? Well, for one, in the acid reaction, a gas is being released. This speeds up the reaction, by Le Châtelier's principle.

Apparently, NaOH is not able to dissolve bone completely, you are left with a fragile shell. Either way, it will be much slower than $\ce{HCl}$ while dissolving bone.

Here's an interesting article (no claims on validity) of why lime shouldn't be used to dissolve a corpse.

While (as far as I can tell) acids may be the better way to dissolve a body, I feel that filmmakers use acids because the term is known to most of the general public (unlike "bases", which isn't)


Methanol isn't particularly toxic in and of itself, although it's no walk in the park.

If methanol flowed through the body without being broken down, it would cause roughly the same kind of harm as ethanol, i.e. intoxication.

The real culprit is one of its metabolic products, methanoic acid, also known as formic acid.

To understand how formic acid, present as the formate ion, is toxic, we look to Wikipedia:

  Formate is toxic because it inhibits mitochondrial cytochrome c oxidase, causing the symptoms of hypoxia at the cellular level, and also causing metabolic acidosis, among a variety of other metabolic disturbances.

Edit: As DavePhD points out, an intermediate product in this process is formaldehyde, or methanal. While formaldehyde is also toxic, it is rapidly metabolized to methanoic acid.

Reedit: The deeper, more historical reason that this happens is that methanol isn't readily available in nature, meaning that few species have developed biochemical tools to deal with it. There simply hasn't existed an evolutionary pressure to deal with methanol.


Do acids and bases have the same coefficients if they neutralize one another?

Look at the chemical equation. One mole of $\ce{HCl}$ will neutralize one mole of $\ce{NH3}$. $\ce{HCl}$ is an acid so it loses a hydrogen ion and ammonia is a base so it gains a hydrogen ion.

$\ce{HCl + NH3 -> NH4Cl}$

So, you found the number of moles $\ce{HCl}$ added, and this equals the number of moles in the 5.00 mL sample of $\ce{NH3}$. Divide that by the volume and you get concentration, which is moles per liter. You don't need to find the molar mass of $\ce{HCl}$ and you didn't find it as described in your calculations above.


What is meant by "basis" in "... its basis in particles leads to a material comprising crystallites or grains tightly bound together, ..."?

The author is attempting to indicate that the "paste" is a suspension of particles that aggregate as it dries (ie the liquid portion evaporates), in contrast to a solution of a dissolved solid that recrystallizes as the solvent evaporates. Because the particles never dissolve, there is no possibility for a single crystal to form that is any larger than an already existing particle. Therefore, the resulting solid is necessarily composed of many small crystals/particles with boundary zones between them, rather than having a large scale uniform structure.


Effect of halogen atom on Markovnikov's rule

Markovnikov's rule is just a simple rule for predicting the structures of  alkenes in some reactions.
It is not advisable to apply this rule to all cases,it is often helpful to check the relative stability of carbocation intermediates.

In your question the carbocation will be stabilised by +M
effect of halogen as halogens have lone pair of electrons.


Relation of pressure and temperature for an ideal gas at constant volume

When using the value of $R=8.314$, the units are $\mathrm{\frac{J}{K\cdot mol}}$. One joule is equal to:

$$\mathrm{J=\dfrac{kg\cdot m^2}{s^2}=N\cdot m = Pa\cdot m^3 = W\cdot s = C\cdot V}$$

If you want to use volume in liters, you need to use a different value for R.


Here is a way to make the lactone:

To your aldehyde in dry THF at 0 °C, add the commercially available 3-butenylmagnesium bromide. Work up and collect the resulting secondary alcohol.
React the alcohol with dihydropyran in dry $\ce{CH2Cl2}$ to form the THP ether.
React this THP ether with $\ce{BH3.THF}$ as you did before, but do an oxidative work up with PDC to get the carboxylic acid according to Brown et al. [1], or do the basic peroxide as before to get the aldehyde, then oxidise that.
Treat the product acid with PTSA in MeOH to remove the pyran ether, the lactone will form spontaneously.

Your choice of alcohol protecting group will be driven by the oxidation conditions you use for the borane. THPO is known to be stable to PDC/$\ce{CrO3}$ in pyridine.

Reference

Brown, H. C.; Kulkarni, S. V.; Khanna, V. V.; Patil, V. D.; Racherla, U. S. Organoboranes for Synthesis. 14. Convenient Procedures for the Direct Oxidation of Organoboranes from Terminal Alkenes to Carboxylic Acids. J. Org. Chem. 1992, 57 (23), 6173–6177. DOI: 10.1021/jo00049a024.


How is $\ce{HCl}$, which is not even a reactant, 'displaced' from
  $\ce{NaCl}$, as there is no $\ce{HCl}$ to be displaced?

Well, there will be some $\ce{HCl}$ due to a well-known chemical reaction which is used both in the lab as well as in production to get the hydrogen chloride:

$$\ce{NaCl(s) + H2SO4(s) → NaHSO4(s) + HCl(g)}$$

The reaction proceeds at room temperature, but note, that the reagents should be dry. If you heat it up above 200 °C reaction proceeds even further:

$$\ce{NaCl(s) + NaHSO4(s) → HCl(g) + Na2SO4(s)}$$

Hydrochloric acid is then prepared by dissolving hydrogen chloride in water.


Why does there need to be a planar intermediate in 2,2'-dibromobiphenyl when one of the phenyl groups rotates?

I think that your professor was just mentioning a rotational angle vs relative energy plot. It is really important to understand relative energy difference between different points in such plots. The situation when two bromines are in the same plane and close to each other will represent the highest energy spot, this actually means that the probability of the molecule to be in such spot is extremely low, but technically not zero, that is why some reaction that require such spot may happen. In the picture below you can see the plot for biphenyl, in case of the dibromobiphenyl one of the peaks will be much higher then the others. But you are actually right, molecule itself would prefer to rotate the other way.


How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?

Unfortunately, nothing in the bonding situation in carbon monoxide is easily explained, especially not the dipole moment. According to the electronegativities of the elements, you would expect the partial positive charge to be at the carbon and a partial negative charge at oxygen. However, this is not the case, which can only be explained by molecular orbital theory. A complete analysis of this can be found in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.)
Responsible for the dipole moment is the highest occupied molecular orbital, a $\pmb{\sigma}$ orbital, which has its largest coefficient at the carbon atom. In first order approximation, this orbital can be considered the lone pair of carbon. All other valence orbitals are more strongly polarised towards the oxygen. The orbital that can in first order approximation be considered as the oxygen lone pair has almost only s character and therefore contributes only little to the dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow
\end{align}

I have reproduced the MO scheme of carbon monoxide for you below. Please note, that the blue/orange coloured orbitals are virtual (unoccupied) orbitals, which should be taken with a grain of salt.

There are two possible decomposition schemes to explain the bonding, both of them involve donor-acceptor interactions. The term "dative bonding" should be avoided here, it is better to use it only for bonds, that consist purely of donor-acceptor interactions, as for example in $\ce{H3N\bond{->}BH3}$.

Below, the two decomposition schemes are reproduced from figure 6 (b & c) in the linked paper. Please note, that this decomposition does not include hybridised orbitals.

The left decomposition is a better description, since it retains the $C_{\infty{}v}$ symmetry of the molecule. We can see a donor-acceptor $\sigma$ bond and two electrons sharing $\pi$ bonds.
In the right configuration we assume an electron sharing $\sigma$ bond, an electron sharing $\pi$ bond and a donor-acceptor $\pi$ bond.

It is very important to understand, that the concept of a dative bond, that you are trying to employ here is only right by coincidence. The reason that the dipole moment is oriented towards the carbon is only to find in the weakly bonding HOMO, the lone pair of carbon.


Origin of the names “thionyl” and “sulfuryl”

Different nomenclatures by IUPAC
For the convenience of the reader, here are the names of the compounds as mentioned in the IUPAC's 2005 Red Book[1, p 129]:

$\ce{SO2Cl2 = [SCl2O2]}$ – sulfuryl dichloride or sulfur dichloride dioxide[a], sulfuryl dichloride[b], dichloridodioxidosulfur[c];
$\ce{SOCl2 = [SCl2O]}$ – thionyl dichloride or sulfur dichloride oxide[a], sulfurous dichloride[b], dichloridooxidosulfur[c]

[a] – acceptable common name, [b] – functional replacement name, [c] – systematic (additive) name
The names which will be pertinent to this discussion are marked with italics, they should otherwise still be upright.
Also note that $\ce{H2SO4 = [SO2(OH)2]}$ is sulfuric acid[a], dihydroxidodioxidosulfur[c]. The molecular entity $\ce{H2SO3 = [SO(OH)2]}$ is sulfurous acid[a], dihydroxidooxidosulfur[c].
TL; DR: The general scheme of things is acid $\to$ radical, and the suffix used in this purpose is -yl. The difference comes from overlap as "properly" both radicals would have the same result: sulfuryl.
Etymology and history
Historically, the terms 'sulfureous' (sulphureous) and 'sulfurous' (sulphurous) date back to the 16th century. Lavosier was first to introduce the adjectives 'sulfuric' and 'sulfurous' to name acids containing sulfur in 1787[2]:

SULPHURIC, SULPHUROUS
The adjectives sulphureous and sulphurous meaning "pertaining to, or of the nature of, sulphur", date from the 16th century. The well-known oxy-acid was then known as the oil of vitriol (q.v.). In 1787, in the Méthode de Nomenclature Chimique, Lavosier Proposed the name sulphuric acid (Fr. sulfurique) for oil of vitriol and sulphurous acid (Fr. sulfureux) for the related acid containing proportionately less oxygen.[2, p 208]

The suffix -yl designates (hypothetically) as having derived from something.[2]

-YL
The suffix -yl, adapted from Fr. -yle, is derived from Gk. hylē which, in addition to meaning "wood", also means "the stuff or raw material (of any kind) from which a thing is made.[2, p 231]
[---]
(2) In the name of a radical formed by the removal of a hydroxyl group $\ce{-OH}$ from an acid.[2, p 232]
[---]
(b) Inorganic acids. A number of oxygenated inorganic radicals may be regarded as derived from acids (or from hypothetical acids) by the removal of one or more $\ce{-OH}$ groups. Such a radical is normally named by replacing the suffix -ic of the name ofthe acid by -yl. The more important of these radicals are sulphuryl $\ce{SO2\!\!:}$ (cf. sulphuric acid $\ce{SO2(OH)2}$) ... nitrosyl $\ce{NO}-$ (cf. nitrous acid $\ce{NO$.$\!OH}$)[2, p 232] [---]

As we can see, the -yl name does not care about the oxidation state of the acid. For sulfur, we see that both sulfuric + -yl and + sulfurous + -yl would give the same result: sulfuryl. One of them had to be different. For this purpose, the Greek language was used for the radical of sulfurous acid, and a Latin (possibly Sanskrit[2, p 20]) form remained for the radical of sulfuric acid.[2] This is an historic accident, it could have easily been the other way round.

THIONYL
The name given by Schiff in 1857 to the radical $\ce{SO\!\!:}$ as in thionyl chloride $\ce{SOCl2}$. It was formed from Gk. theion, sulphur, + -yl (q.v.). The radical may be regarded as the characteristic group of sulphurous acid $\ce{SO(OH2)2}$; compare sulphuryl $\ce{SO2\!\!:}$, the characteristic group of sulfuric acid.
(It is fortunate that the L. and Gk. names for sulphur (sulfur and theion) are quite different so that two different names could be formed for the two radicals.)[2, p 214]
SULPHURYL
The radical $\ce{SO2\!\!:}$, the characteristic group of sulphuric acid $\ce{SO2(OH)2}$ (see -yl), e.g. sulphuryl chloride $\ce{SO2Cl2}$ (1867)[2, p 208]

For the functional replacement name, $\ce{SOCl2}$ is named sulfurous dichloride to tie the radical more directly to its acid as well, sulfurous acid. Historically, as noted, the approach here was to use different languages of origin for the main element. This was possible because the words were very different.
If you wish to operate with common names, there is always some historic memorisation. Since this is a binary system, you could remember one and automatically derive the other. But as stated, this will not work for other similar cases: nitrosyl is for nitrous acid radical, not nitric acid.

References and bibliography
[1] Connelly, N. G.; Damhus, T.; Hartshorn, R. M.; Hutton, A. T. Nomenclature of Inorganic Chemistry; The Royal Society of Chemistry, 2005. ('IUPAC Red Book')
[2] Flood, W. E. The Origins of Chemical Names, 1st Edition.; Oldbourne, 1963.


Figuring out formula of a hydrocarbon from amount of products obtained and reactants used

From your interpretation I have to say that giving the stochiometric coefficent of $13$ is wrong because this number is not related to the molecules of the species (therefore the suspect that you multiplied the moles obtained by two is where lies the main error).

Since we are talking about a combustion, we can safely assume that every reagent is completely oxidised, so we can write the complete reaction with the generic coefficients:
$$
\ce{C}_{\color{blue}{x}}\ce{H}_{\color{blue}{y}(g)}+\left[\color{blue}{\frac{y}{4}}+\color{blue}{x}\right]\ce{O2_{($g$)}}\to\color{blue}{\frac{y}{2}}\ce{H2O}_{(g)}+\color{blue}{x}\ce{CO2_{($g$)}}
$$
Note that the rule of given saturated hydrocarbons implies that $y=2x+2$, but this is left intentionally generalized because the presence of specific double/triple bonds is unknown.

The purpose of writing this equation is to specify the relation between the reactants, so now we can assume the molar relation between the oxygen needed to the first hydrocarbon to obtain a simple equation in terms of $x$ and $y$:
$$
n_{\ce{O2}}=\left[\frac{y}{4}+x\right]n_{\ce{C}_{x}\ce{H}_{y}}
$$
Since, as you calculated, there are needed $6.5\:\mathrm{mol}$ of oxygen to oxidise $1\:\mathrm{mol}$ of the unknown hydrocarbon; we can substitute those values inside the above equation, giving:
$$
6.5=\frac{y}{4}+x\tag{1}
$$
After the complete combustion, the total amount of moles is indicated as you derived correctly, and there can be expressed in relation to the unknown reagent, and then added together to obtain the second equation as follows:
$$
n_{\ce{H2O}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}\quad n_{\ce{CO2}}=x\:n_{\ce{C}_{x}\ce{H}_{y}}
$$
Since the total volume is additive, then (because $\ce{C}_{x}\ce{H}_{y}=1:\mathrm{mol}$):
$$
4\:\mathrm{mol}=n_{\ce{H2O}}+n_{\ce{CO2}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}+x\:n_{\ce{C}_{x}\ce{H}_{y}}=\frac{y}{2}+x
$$
It now produces the second needed equation:
$$
4=\frac{y}{2}+x\tag{2}
$$
Solving the equations $(1)$ and $(2)$ gives the values of $x$ and $y$ needed to derive the brute formula for the unknown hydrocarbon:
$$
y=-10,\quad x=9
$$
Since the result is negative for one term, we can assume that the water produced is not in the vapor phase, hence we eliminate water term inside the equation $(2)$, $y/2\:n_{\ce{C}_{x}\ce{H}_{y}}$, giving the immediate value of $x$ because there is no effort given by the water:
$$
x=4,\quad y=10\to \ce{C}_{4}\ce{H}_{10}\quad\mathrm{butane}
$$
This result is describing a saturated hydrocarbon, because it follows the rule $\ce{C}_{n}\ce{H}_{2n+2}$.


The problem begins with your first statement:

$$\left(\frac{\partial S}{\partial V}\right)_U=0$$

Because in fact:

$$\left(\frac{\partial S}{\partial V}\right)_U=\frac{p}{T},$$

which is clearly non-zero (except in limit of zero pressure or infinite temperature).

More generally, being at equilibrium does not equate to a partial derivative being zero!  Rather, being at equilibrium equates to a partial derviative being defined (i.e., they are only defined at equilibrium).  Thus, whether or not a partial derivative equals zero has nothing to do with equilibrium, but rather with the specific nature of the system. And, most typically, partial derivatives are non-zero; it's only for certain specialized cases that a partial would be zero.

Consider, for instance, the total differential for $U=U(T, V,n)$:

$$\mathrm{d}U =  \left(\frac{\partial U}{\partial T}\right)_{\!V,\, n_i}\,\mathrm{d}T+\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}\,\mathrm{d}V +\sum_i \left(\frac{\partial U}{\partial n_i}\right)_{\!T,\, V,\, n_{j\ne i}}\,\mathrm{d}n_i,$$

where the last term allows for chemical reactions and/or matter flow into or out of the system.

Note that state functions ($U, H, A, G, S...$) are only defined at equilibrium, and thus these partial derivatives are likewise only defined at equilibrium.  What the above partial derivatives tell us, then, is how the state function (in this case, $U$) changes as we infinitesimally change the system from one equilibrium state to a new equilibrium state by varying $T$, $V$, or one of the $n_i$.

Thus, far from being zero at equilibrium, these partials typically need to be non-zero at equilibrium, unless the state function is independent of the change in the corresponding independent variable, which only applies in special cases.

One such special case is the internal energy of an ideal gas, which depends only on $T$ and $n$. Hence, for a single ideal gas, or a mixture of ideal gases:

$$\left(\frac{\partial U}{\partial V}\right)_{\!T,\, n_i}=0$$


Maximum pressure that can be produced by a combustion reaction

According to New Energetic Materials, table 1.2, a list of explosives ranges from 19-48 GPa in pressure achieved.  This is two orders of magnitude greater than the 60,000 psi figure.

As far as maximum theoretical pressure, the question isn't really clear whether the maximum absolute pressure at which combustion can occur or the maximum change in pressure due to combustion is the value of interest.  The answer could be related to the maximum pressure at which chemical bonds exist.


Liquid which neither dissolves in water nor oil

Fluorocarbons are compounds that are hydrophobe and lipophobe. It's a special property of perfluorinated compounds: They are non-polar and thus hydrophobic but in addition as the high electronegativity of fluorine reduces the polarizability of the atom, fluorocarbons are only weakly susceptible to the fleeting dipoles that form the basis of the London dispersion force. As a result, fluorocarbons have low intermolecular attractive forces and are lipophobic.
I'm not sure about their toxicity, but this seems to be an issue, e.g. here.