What is the major product on chlorination of 2-methylbutane?

Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.

According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:

22 % 2-chloro-2-methylbutane
33 % 2-chloro-3-methylbutane
30 % 1-chloro-2-methylbutane
15 % 1-chloro-3-methylbutane

The statistical factor is responsible for this.
If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be

1 hydrogen atom (8 %) → 2-chloro-2-methylbutane
2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane
6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane
3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane

In terms per hydrogen atom, the experimental yield is

22 % / 1 = 22 % → 2-chloro-2-methylbutane
33 % / 2 = 16.5 % → 2-chloro-3-methylbutane
30 % / 6 = 5 % → 1-chloro-2-methylbutane
15 % / 3 = 5 % → 1-chloro-3-methylbutane

Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.


Bisphenol-A is very inexpensive for its considerable formula weight - from benzene, propylene, air, acid; acetone (or other ketone or aldehyde) and acid.

1) Polycarbonate, as above.  If you don't like HCl, dimethyl carbonate will do.
2) Epoxy resin, by bis-etherifying with glycidyl chloride.  That is a huge market, including paints and coatings.
3) Bis-etherify with propargyl chloride.  That thermally B-stages (removes the shrinkage upon cure) then thermosets to an astounding thermally resistant super-crosslinked resin that adheres to any polar surface.
4) Bis-esterify with methacrylic acid to get into heavily cross-linked free radical or UV cures.
5) And if you really want something resistant, instead of acetone use fluorenone or a 2,7-disubstituted fluorenone.  Or be fiendishly clever and use 4,4'-dihydroxybenzophenone instead of acetone.  Now there is a tetrahedral tecton for synthesis!

Bisphenol-A is a claimed  endrocrine disruptor.  Eastman Kodak has immense production capacity for acetic anhydride (cellulose acetate photographic film).  It offshot into ketene dimer chemistry then mixed cis- and trans-2,2,4,4-tetramethyl-1,3-cyclobutanediol as a BPA replacement.  If you do not mind paying more and getting less, stuff as well as performance, the problem is solved.


Wurtz Reaction with acidic alkyl halides

The Wurtz reaction involves the reaction of alkyl halide with sodium in dry ether. In this reaction, two molecules of alkyl halide react to form a alkyl-alkyl chain and sodium halide is also produced -

$$\ce{R-X + 2Na + X-R ->R-R + 2NaX} \tag{$\ce{X}$ = halogen}$$

Now, sodium is a strongly electropositive metal, and is above hydrogen in the electrochemical series. Thus it replaces acidic hydrogen wherever it occurs. For example,

$$\ce{CH3COOH + Na ->CH3COO- Na+ + 1/2 H2}$$

So, sodium reacts with the acidic hydrogen present in the starting alkyl halide.

Also, acid-base reactions are, in general, faster than a substitution reaction. If there was no acidic hydrogen, alkyl sodium intermediate would have formed and an SN2 reaction would have taken place.

However, in this case, if any alkyl sodium intermediate is formed, it would accept the acidic hydrogen and convert into an alkane, stopping the reaction.

$$\ce{R- Na+ + H-M -> R-H + NaM}$$

(R = alkyl group, M represents the functional group or chain attached to the acidic hydrogen)


You have asked quite a number of questions in your post so let's tackle it one by one.

Why does the oxidation take place at that particular nitrogen atom?

To understand this, we first need to acknowledge that peroxyacids $\ce {RCO3H}$ are electrophilic oxidising agents. Specifically, the oxygen atom attached to the $\ce {H}$ is the electrophilic atom. From the frontier MO perspective, the reactive LUMO of the peroxyacid is the $\ce {O-O \sigma^*}$. This implies that the most nucleophilic nitrogen atom is likely to be favourably oxidised.

In fact, the oxidation of pyridine, which is a chemical species containing a nucleophilic $\ce {N}$ atom, to form pyridine $\ce {N}$-oxide using peroxyacids is well-known. It is documented by Clayden, Warren & Greeves (2012) on p. 730 as a method of increasing the reactivity of pyridine towards both electrophilic and nucleophilic aromatic substitution reactions. Notice that the bottom $\ce {N}$-containing ring actually resembles a pyridine fused with another benzene ring. However, note that the lone pair on $\ce {N}$ is not delocalised into the ring(s). Thus, there is no significant decrease in nucleophilicity of the bottom $\ce {N}$ in this compound relative to pyridine's $\ce {N}$ atom, suggesting that oxidation of the peroxyacid can very likely take place there.

Of course, we must be thorough in our analysis and also take a look at the relative nucleophilicities of the other two $\ce {N}$ atoms. Immediately, we will notice that for the $\ce {N}$ atom that appears to be making 3 single bonds has a lone pair that is heavily delocalised, in both the benzene ring below and the 5-membered ring it is in. Thus, we can eliminate the possibility of the the oxidation taking place at that nitrogen atom.

For the final top-most $\ce {N}$ atom, the lone pair may appear to be of the same nature of that of pyridine, i.e. held in an $\ce {sp^2}$ hybrid orbital. However, note that the presence of the adjacent $\ce {N}$ atom may result in inductive withdrawal of electron density from that $\ce {N}$ atom, decreasing its nucleophilicity.

Ultimately, I believe this first part is trying to test students if they are able to relate the structure and reactivity of pyridine towards oxidation by peroxyacids to the bottom pyridine-like ring. If students can make the connection, then they would instinctively know that the oxidation has taken place at that $\ce {N}$ atom.

From A to B what happened?

Well... There are a number of things which happened here. The most obvious being the first step: Basic hydrolysis occurs, with the leaving of $\ce {EtO^-}$. The next step is an electrophilic aromatic substitution, with the acidic deuterium ions as the electrophile. For pyridine $\ce {N}$-oxide, the 2- and 4-positions are activated towards EAS reactions. This can be shown simply by drawing the resonance structures of the compound with an $\ce {N=O}$ double bond. Similarly, the EAS takes place at the 2-position, relative to the $\ce {N}$-oxide group. It is difficult to explain why it doesn't take place at the 4-position as well. In general, for this type of structure elucidation questions, we would need to predict structures by looking at the resultant structures as well, especially if they are given. In this case, we see that only the 2-position has deuterium substitution taking place and we should simply leave it as that. Of course, if anybody has a good explanation for why that is the case, please propose it. Also, the other positions on the other rings may not be activated enough for the substitution to take place. Thus, we do not see substitutions taking place at those positions. Again, I am not able to provide a rigorous explanation for why that is the case.

Finally, the third step with $\ce {H2O}$ effectively makes the pH of the medium neutral again and also changes the carboxylic acid group from $\ce {COOD}$ to $\ce {COOH}$. This is simply due to a concentration effect. The high concentration of $\ce {H2O}$ simply pushes the equilibrium towards species containing $\ce {COOH}$.

What about from B to C?

From Wikipedia, it seems that $\ce {B}$ to $\ce {C}$ is simply a reduction step. The ammonium formate decomposes to give $\ce {NH3}$, $\ce {H2}$ and $\ce {CO2}$ in the presence of $\ce {Pd/C}$. The hydrogen gas evolved then gets adsorbed onto the palladium catalyst, where it can reduce the $\ce {N}$-oxide to give back the original $\ce {N}$-heterocycle.

And the last step involving conversion of C to the final product is as you have proposed.

Reference

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry (2nd ed.). Oxford University Press: New York, 2012.


Is the change in enthalpy (ΔH) for dissolution of urea in water positive or negative?

The sign of Q depends on the perspective. The water temperature decreased because it "lost" heat. The process of dissolving urea required energy, it "gained" energy. If I give you a penny, should that be +1 or -1 penny? Well, it depends who you ask.

In your answer, you are missing a negative sign in $\Delta H=−Q$ the way you start out with $Q$ from the perspective of the water.


The 3s orbitals are far too high in energy to participate in chemical bonding interactions. Lower in energy are the empty σ* antibonding orbitals of the N–H bonds, which can still be energetically relevant in further reaction chemistry. As a thought experiment, it might be possible for a hydride anion to donate electron density into an empty N–H σ* orbital in a sort of Lewis acid-Lewis base interaction. In fact, this is a hallmark of so-called `hypervalent' bonding of heavier main group elements. Rather than using d-orbitals (as is often incorrectly taught to intro chem students), these acid-base interactions are how the phosphorus(V) center of PF$_5$ bonds to the fifth fluoride ion, forming what is known as a three-center four-electron bond.

Why isn't this type of bonding seen with NH$_3$ and H$^–$ to form NH${_4}^-$? These `hypervalent'-type interactions involving 3c-4e bonds depend on the outer atoms having a high-enough electronegativity to stabilize a positive charge at the central atom. In the case of the hypothetical NH${_4}^-$ anion, this just isn't the case. Hydrogen is in fact far less electronegative than nitrogen.


(Z) is the same like cis

(E) is the same like trans

BUT cis and trans is usually only for double bonds with 2 different substituents. Confusion can arise when, for instance, you have 2-halobut-2-enes, because some take the highest priority when determining if this is cis or trans whereas others may take the identical groups.

You can use (Z) and (E) for more than 2 substituents, for those with higher priority. Like in your example.

In general it is better to use (Z) and (E) nomenclature to avoid these ambiguities.


Do orbital models derive from the solution of the stationary Schrödinger equation?

I know that hydrogen-like atomic orbitals can be seen as a solution of
  the stationary Schrödinger equation - but what about atoms with more electrons?

Orbitals are one-electron wave functions, i.e. they are wave functions that describe the state of only one electron. For a hydrogen-like atom, which is a one-electrons system, if treating its nucleus as a classical positively charged point particle, the wave function is indeed an orbital, since there is just one quantum particle out there and it is the single electron. So, right, orbitals are the solutions of the (non-relativistic) time-independent Schrödinger equation if the nucleus is assumed to be a classical point particle.

For any many-electron atom, though, even when its nucleus is though of as a classical point particle, the wave function describes more than one electron, so by the very definition it is not an orbital. It can be written in terms of the one-electron wave functions (orbitals) as their product, but this can be done only approximately due to interactions between the electrons. It is in this sense we say that orbitals aren't real.

Note though that, strictly speaking, even for a hydrogen-like atom orbitals aren't real since they arise in the non-relativistic treatment which is also an approximation. The reality is even more weird than its non-relativistic description, but for atoms from the first and second rows of the periodic table the non-relativistic description is usually a good one in a sense that it introduces relatively small errors that can be neglected for the most of the chemical purposes. The orbital approximation on the other hand might be the source of much larger errors. For instance, for a $\ce{He}$ atom in its ground electronic state the orbital approximation introduces the error of the $10^{-2} E_{\mathrm{h}}$ order while the relativistic correction is about two orders of the magnitude smaller, $10^{-4} E_{\mathrm{h}}$.


What makes cola efficient as a rust remover?

Rust consists of insoluble iron oxides and hydroxides, $\ce {Fe2O3·nH2O}$ and $\ce{FeO(OH)·Fe(OH)3}$. Acidifying these compounds protonates the oxide and hydroxide anions to for water:

$$\ce{O^{2-} + 2H^+ \rightarrow H_2O}$$
$$\ce{OH^{-} + H^+ \rightarrow H_2O}$$

This leaves the soluble iron $\ce{Fe(III)}$, which can be washed away.

Many colas contain phosphoric acid, and are quite acidic (Coke has a pH of ~2.5 according to this science fair project), thus they can dissolve the oxides.


Entropy is a physical property of a material that, for a single phase substance of constant composition, can be expressed as a function of temperature and one other intensive physical property such as pressure or specific volume.  The equation that you wrote down is a relationship that we can use to determine the change in entropy of a substance from one thermodynamic equilibrium state to another, by devising a reversible process from the first state to the second state, and calculating the value of the integral of dq/T for that process.  All reversible processes between these same two states will give exactly the same value for the integral.


How to determine the chemical state of a reactant or product in a chemical equation?

According to these basic solubility rules:

Calcium chloride (CaCl2) is soluble. Hence, you write CaCl2(aq) in the equation.


After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this?

Below is a basic example.

If percent yield is based on a mass, then a >100% yield can result from a product contains materials that should not be there.  For instance, a product that should be anhydrous may not be completely dry and water would add mass to the product increasing your yield.  Other present contaminants can also do this and can even appear from a product that reacted with the surrounding atmosphere.


Polar compounds that doesn't dissolve in water?

You may accidentally have chosen an awkward example because I'm not sure that $\ce{Al2O3}$ is ever properly molecular. It is usually crystalline and may even be in the form of nano-crystals, in which case there will be considerable lattice binding energy to overcome. In its doped form, with the odd Cr$^{3+}$ ion, it is the mineral Ruby which is clearly not soluble in water.

In the general case polarity is not the only consideration for solubility. There is entropy change of mixing which is always favourable when two species are mixed and also enthalpy change when placing molecules of one type into the other. The enthalpy is where the polarity part comes in.

In water in particular there are other  entropy  changes to consider because water is so hydrogen bonded. The H-bonding structure is broken up when a solute is added and water has  then to 'bridge the gap' as it were round the solute. This forms a more structured water arrangement than before adding the solute and so entropy is reduced. This is primarily responsible for insolubility in water.

We measure the process by considering the free energy change as $\Delta G = \Delta H - T\Delta S$ where $\Delta H$ is the change in enthalpy, $\Delta S$ that of the entropy. When the process if favourable $\Delta G$ will be negative.

Surprisingly adding, for example, non-polar  cyclohexane to water has a change in enthalpy ($\Delta H$) that is close to zero, whereas for a polar molecule $\Delta H$ will be negative. The H-bonding bridging with cyclohexane will make the change in entropy negative so making the free energy positive, which is unfavourable.

With a polar molecule in water H bonding bridging is not necessary to such an extent as for a non-polar molecule and the change in entropy is smaller and so with its negative $\Delta H$ the polar molecule has a negative $\Delta G$ and can dissolve.


Fragmentations and rearrangements in n-(4-alkyl phenyl) alkanoic acids in mass spectrometry

In your update you nearly got to the point: the McLafferty rearrangement from the carboxylic acid can lead to two products, either the ionized acetic acid (ML product) or the ionized alkene (ML'). In this case, ionization energy of the enol tautomer of acetic acid is probably much larger than that of alkenyl substituted benzenes (for instance the closest I could find in structure is o-Allyl-Toluene for which the IE is 7.78 eV). Thus from application of the Stevenson-Audier rule, the charge should remain on the moiety with the lowest ionization energy, leading to the formation of the m/z 188 moiety (ionized p-pentyl-allylbenzene).

This hypothesis is further reinforced as it also explains the formation of m/z 117 from a oss of 71 ($C_5H_{11}^\bullet$) from fragmentation of the m/z 188 radical cation, through a pathway similar to that leading to m/z 177 from the molecular ion.

(This is likely oversimplified mechanism interpretation: I strongly suspect that some other rearrangements or cyclisation likely occur in the process. I did not run a complete database search for other EI spectra of similar molecules, but for instance 10-phenyldecanoic acid leads to only a very minor loss of 60. Thus I would suggest that the m/z 188 probably undergoes cyclization and / or hydrogen migration processes leading other products.)


How to programmatically identify a chemical compound in my program?

Either way, you need to define what it means for a chemical to be 'in' your database. Do you care about the 3D coordinates (given that you have XYZ) or just the atom types and connectivity?

Short answer. use a cheminformatics library that provides structure matching or can encode your molecules (query and targets) as canonical SMILES, InChI, etc.

See Is there a way to use free software to convert SMILES strings to structures?

Long answer. Use a series of filters (or questions):

Are there any molecues in my database with the same number of atoms?
Do any of these have the same atom types?
Finally, are these atoms connected to the same neighbours?

The third one is a little trickier, but is still a quick way to determine isomorphism. If you  need true structure matching, see the short answer! If you need 3D matching, then again it's going to be much quicker to use a library method.


Using a modified Arrhenius plot to find activation energy

The use of $t$ only makes sense if $t$ is defined as the time it takes for the reaction to progress a certain relative amount (and if we are talking about first-order reactions).  For example if $t$ is the time it takes for half (or 90% or 99% or whatever, as long as it is specified) the reactant to be consumed, then the idea makes sense.  In that case it would be more helpful and clear to say $t_{1/2}$ (or $t_{90\%}$ or whatever) instead of just $t$.

Assuming $t_{1/2}$ is what is meant by $t$:

Arrhenius equation:

$k = A_0 e^{\frac{-E_a}{RT}}$

$k = A_0 e^{\frac{-E_a}{R}\frac{1}{T}}$

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T}$

First-order kinetics:

$C = C_0 e^{-k t}$

$\ln \frac{C_0/2}{C_0} = - k t_{1/2}$

$k = \frac{1}{t_{1/2}} \ln 2$

$\ln k = - \ln t_{1/2} + \ln \ln 2 = -\ln{t_{1/2}} + \ln \ln 2$

Combining:

$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T} = -\ln{t_{1/2}} + \ln \ln 2$

$\ln t_{1/2} - \ln \ln 2 = \frac{E_a}{R} T^{-1} - \ln A_0$

$\ln t_{1/2} = \frac{E_a}{R} T^{-1} - \ln A_0 + \ln \ln 2$

So if $t$ means $t_{1/2}$, then a plot of $\ln t_{1/2}$ vs. inverse temperature has a slope of $+\frac{E_a}{R}$, instead of $-\frac{E_a}{R}$ if you plot $\ln k$ vs. inverse temperature.

N.B. If $t$ just means time, then the idea doesn't make any sense, because then instead of $\ln \ln 2$ you would have $\ln \ln \frac{C(t)}{C_0}$, which is a term that depends on time, so instead of being an explicit equation that is clearly a line, you would have an implicit equation, non-linear in $t$, that would need to be solved before the plot was made, and the result would not be linear and would be impossible to use to determine $E_a$.


You need to protonate the oxygen in order to actívate the molecule towards nucleophilic attack by the nitrogen of Ethyl amine. pH plays a very important rule here. Too low pH gives an efficient protonation of the oxygen, but it also protonates the nitrogen of the Ethylamine(As ethylamine is weak base due to the lone pair of electrons on nitrogen in it), so the production of the imine results inefficient due to less amount of the desired reagent. On the other hand, at higher pH values, the oxygen does not get protonated enough to proceed the reaction as the electronegativity of oxygen dominates it from giving electrons to a low concentration of protons( As pH is high). So there is an optimum pH between pH 4.5 and 5, for this reaction to take place. So dil H2SO4 can give this pH level while weaker acids cannot. Also when you look at the mechanism for we need to dehydrate to get imine. Dil H2SO4 is good dehydrating agent.

For all the factors above diluted strong acids are better than weak acids.


What is the difference between a 500W high pressure mercury lamp and a 400W high pressure mercury lamp

I would say all other things being equal, i. e. Spectral distribution,  quantum efficiency,  that more wattage = more photons per second. And yes, more heat.


G-zero and G-prime.

G-zero has also in my presence been referred to as "G-ought", "G-nought", "standard-G" or "G at standard state". I am particular towards "G-zero" myself. (Mostly because of a witty limerick I heard during my studies. Zero rhymes with hero.)


The terminal bonds are stronger than the internal bonds.

In simple hydrocarbons where strain or electronegative substituents are absent, the bond strengths will correlate with the stability of the resultant radicals that would be formed when the bond breaks.

Because radicals are electron deficient species (they are one electron short of an octet) they are stabilized by electron donating substituents (alkyl groups for example).  Hence, a tertiary radical is more stable than a secondary radical which is more stable than a primary radical which is more stable than a methyl radical.

Using this information we can see that
$$\ce{R-CH2. + .CH2-R}$$
is more stable than
$$\ce{CH3. + .CH2-R}$$
Hence the internal bond will have a lower bond dissociation energy and break more readily.


Why must the rate law be experimentally determined?

Since rate of a reaction depends actually upon the way reactants react, there is not surety that the quoted stoichiometric coefficients may be the same for rate of a reaction, the rate depends on the way or mechanism of the reaction, and since all reactions have a bit different mechanism no strict formula could be quoted. It is only experimentally derived and can be theoretically approached if we know that it is an elementary (simple step) reaction or it falls under a already classified or studied reaction.

Zero Order, what if a reactant is in very large excess, now the rate is not limited or dependent on its concentration, also it's concentration is almost constant, so it is also merged into the constant and hence we have zero order, some adsorption reactions are maybe zero order too, it is because it actually depends on the surface area of the substance and it's constant.


Why does ammonia accept a hydrogen to form ammonium when its octet is already complete?

"Hydrogen" means different things at different times. It can mean $\ce{H+}$, $\ce{H.}$, or $\ce{H2}$. (Not to mention deuterium and tritium!)

A bare proton ($\ce{H+}$) does not bring an electron to the filled octet on nitrogen, but attaches to the lone pair. The positive charge is now shared over 5 atoms, and the energy of formation of a new bond stabilizes the ammonium cation. The octet on nitrogen is still only an octet.

Interestingly, a hydrogen atom ($\ce{H.}$) can also attach to an ammonia molecule to form ammonium radical. Electrolysis of ammonium salts with a mercury cathode  yields ammonium amalgam, which rapidly decomposes to $\ce{NH3}$ plus $\ce{H2}$. In this case, the extra electron in $\ce{NH4.}$ is also shared among the 5 atoms, and is more stable than when it is attached to just one proton. But not that much more stable, and the bond between two hydrogens is even stronger.


Sodium Polysulfide production from NaOH and Sulfur

If you calculate the oxidation numbers for everything before and after, you will see that the oxidation number of the oxygen atoms will always be $-2$. However, sulfur $\ce{S},\ (0)$ is being both reduced to $\ce{S4^2-},\ (\frac{1}{2})$ and oxidized to $\ce{SO3},\ (+6)$. Sulfur is both the oxidant and reductant in what might be called a disproportionation. Sodium hydroxide is thus acting as both a source of oxygen atoms in the formation of $\ce{SO3}$ and probably as the electrolyte if this is occurring in aqueous solution.

  Oxidation numbers:

Element    Reactant     Product
S           0            -1/2 in polysulfide
S           0            +6 in sulfur trioxide
Na         +1            +1
O          -2            -2 in sulfur trioxide and water
H          +1            +1

The half reactions (unbalanced) are:

Oxidation: $\ce{S -> SO3}$
Reduction: $\ce{S -> S4^2-}$

You might get some sodium hydrogen sulfate forming through the followin;
$$\ce{NaOH + SO3 <=> NaHSO4}$$


There are two types of control, that governs reaction path:

kinetic control (path with least activation energy) and
thermodynamic control (path with least energy of products).

Given temperature favors kinetic control.

Ozonolysis starts as 2+3 electrocyclic addition (with later regroup of the product), similar to Diels-Alder reaction . Please, read about this reaction in good organic chemistry textbook, especially about what controls geometry of products, it will help to understand what I'm going to say next.

Ozone has low energy of LUMO, so alkene part here provides HOMO, that is still lower then Ozone's LUMO. Oxygen from $OMe$ group donates some electronic density to $C-C$ double bound, so this bond becomes more active, then opposite one.


Collision that governs kinetic model of gas

There will be many molecules collide with the walls and the molecules will hit another molecules after its collision with wall (right?), thus the pressure of the system should be low. How often the collision of what kind happens under what condition?

The collision condition doesn't need to be strictly in Boyle's law condition, that the elastic collision is happened between anything and under any pressure condition. The requirement of large number of molecules but only collision interactions are applied implies that it is just not applicable to apply Boyle's law in kinetic model of gas.


Find product of the oxidation of 2-methylnaphthalene with chromium trioxide

Fun fact: this synthesis is/was the commercial method used by BASF to make the product, menadione, which is often found as a supplement in animal feeds (patent DE092952709). Apparently the use of stoichiometric chromium is worth it.

Why are alpha positions more reactive? There are two potential questions here, I'll try to answer both.

1) Why does methylnaphthalene react differently from toluene?
The naphthalene ring is generally more reactive than benzene because reaction intermediates do not fully destroy aromaticity (see below image, left)

2) Why is the alpha position more reactive than the methyl group?
This question of reactivity is tricky because other oxidants and reaction conditions will indeed oxidize 2-methylnaphthalene into 2-naphthoic acid. In general, electron-rich naphthalenes are able to oxidize into naphthoquinones, even just by exposure to $\ce{O2}$. If we assume (big assumption!!!) that these reactions to go through radical mechanisms, then we are either oxidizing the naphthalene pi system (left) or abstracting an H* (right). There's various arguments to be made about which pathway is more stable based on the extent of conjugation, and obviously that will depend on the reagent and solvents in use as well.

Alpha positions of which functional group have they mentioned?

The alpha nomenclature they use here is actually referring to the two different positions in naphthalene:


Nope, as wiki tells us, only outermost electron shell is considered
valence shell, though d-electrons for transition metals are
considered as valence electrons. No logic here, only tradition.
as you can see, there are three electron subshells above obviously inert Xe core. 5d-electons, that most of the time are far from the nuclei because of the form of their orbitals, feel dramatic increase in effective core charge through the third d-row. For example, for Ti effective charge, keeping its d-electrons is 4. For gold it is 11, so the electrons are well-held.

For s-electrons the form of the orbital puts very significant part of electronic density near the nuclei. This means, that for heavy elements (staring from as far as Ga) a so-known 'inert electron pair effect' is observed: in comparison with p- and d-electrons, due the form of the s-orbital, s-electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. As effect, Bi is highly unstable in valence state +5, perbromates are much stronger oxidizing agent than perchlorates, but same is not true for bromates and chlorates, and Ag, Au and Hg have very tightly bound s-electons, leading to relative inertness of said metals.

For Au and Ag, however, strongly bound s-electrons participate in extremely powerful (due to relative small size of s-orbitals) s-s interactions, while for Hg only relatively weak s-p and d-p interactions are possible because of filled d- and s- subshells. Thus, metallic Ag and especially Au in metallic state have s-electrons involved into effective covalent interactions, leading to relatively inert nature. Beware, however, that they are far from unreactive: Ag have high affinity to sulfur, and both Ag and Au dissolve in cyanide solutions when oxygen is present, not saying about such agressive reagents as chlorine and bromine.

Still, Au, Ag and Cu share one trait with Ia (I main subgroup) elements. Due to only spherical s-orbitals involvement in atom bonding in metal state, all three metals have high plasticity. Other d-elements have significant d-d bonding in metal states, and d-orbitals are not spherically symmetrical, leading to much less plastic behavior.


According to https://pubchem.ncbi.nlm.nih.gov, propylthiouracil has the following solubility properties:

  1 part dissolves in about 900 parts water at 20 deg C, in 100 parts
  boiling water, in 60 parts ethyl alc, in 60 parts acetone; practically
  insol in ether, chloroform, benzene (emphasis mine); freely sol in aq soln of ammonia
  and alkali hydroxides.

I would start with a small amount of your "sludgy" product. Stir it with ether as a first choice, chloroform as a second, for a while and see first of all whether you really get multiple phases or if just forms a less viscous sludge.  In a perfect world, after letting the solution set you would see the white propylthiouracil solid settle out at the bottom, with a sludge-discolored solvent layer on top.  If you see anything resembling this, then I would pursue this technique with more thorough stirring/shaking, maybe centrifuging to get better settling of your product, then further rinsing your product on filter paper with a vacuum flask, etc.  Try the other solvents if available if ether doesn't work.  The success/failure of this approach depends on the solubility of the sludgy stuff in the solvent.  It may fail, it may not give the purity you need, it may work wonderfully, but it is a cheap and simple way to test a method with a reasonable probability of success.


Why does the mole-fraction equilibrium constant depend on pressure and volume?

TLDR; Because only the concentration equilibrium constant is directly related to the rate constants themselves.

Given that pressure has a minimal effect on equilibria in solution, I'm going to answer assuming you're interested in gas-phase equilibria.  The first step in showing why pressure/volume matter to $K_x$ is to move to the partial-pressure equilibrium constant, $K_p$. For a system composed entirely of ideal gases, the partial pressure of each species is related to its concentration as:

$$
{P_i\over RT} = {n_i\over V} = C_i
$$

Substituting this into the expression for $K_c$ for your example reaction gives:

$$
K_c = {C_P C_Q\over C_A} = {1\over RT}{P_P P_Q\over P_A} = {1\over RT}K_p
$$

Thus, for this reaction:

$$
K_p = RT\,K_c \tag{1}\label{Kp}
$$

For other reactions, the power of $RT$ on the RHS of Eq. $\eqref{Kp}$ can differ, depending on the change in the sum of the stoichiometric coefficients between the reactants and the products.

Since $K_c = f(T)$  and $K_p = (RT)^nK_c$, then $K_p$ is also dependent only on temperature (with ideal gases only!), not pressure or volume.

Where pressure/volume enter the picture is when one goes from $K_p$ to $K_x$.  Again assuming ideal behavior, the mole fraction of a species is defined as:

$$
P_i = x_i P
$$

Substituting this into the $K_P$ expression:

$$
K_p = {P_P P_Q\over P_A} = P {x_P x_Q\over x_A} = PK_x
$$

Thus, for this reaction:

$$
K_x = {1\over P}K_p \tag{2}\label{Kx}
$$

Similar to the above, the power of $1\over P$ in Eq. $\eqref{Kx}$ can vary, depending on the change in the sums of the stoichiometric coefficients between the reactants and the products.

In any event, this is where the dependence on $P$ enters the expression for $K_x$.

This dependence on pressure implies a related dependence on volume, since by the ideal gas law:

$$
V = {nRT \over P}
$$

and, equivalently:

$$
P = {nRT \over V}\tag{3}\label{igl-P}
$$

If one is working with a system where the volume and temperature are the control variables but the pressure is not, then in general the pressure will vary implicitly according to Eq. $\eqref{igl-P}$. If the temperature and/or volume change in any way other than proportionally $($i.e., ${T\over V} = \text{constant})$, then the pressure dependence of Eq. $\eqref{Kx}$ will manifest itself in a fashion that looks like a volume dependence.


The electron configuration of  $\ce{Fe(II)}$ is $1s^22s^22p^6 3s^23p^6 3d^6$.

The electron configuration of  $\ce{Fe(III)}$ is $1s^22s^22p^6 3s^23p^6 3d^5$.

The half filled  orbital $3d$ is more stable than the same orbital filled with 6 electrons. So,  $\ce{Fe(III)} $ ion is more common than $\ce{Fe(II)}$.


What is the dipole moment direction of an aliphatic amine, i.e., R-NH2?

I really do not see an easy and straight forward way of determining the dipole moment's direction and quantity. It is very much dependent on how the molecule is currently shaped, i.e. rotations of the alkyl chain and also the electron cloud. Of course you could go ahead and calculate the dipole moment from tabulated values of certain bonds (for example), but this can be really tedious and it might now be always right.

You can probably find tabulated values for most common compounds online. Try http://cccbdb.nist.gov/ or maybe just google will do. If you are lucky enough to have access, the handbook of chemistry and physics has a large number of values. There is an online verison, but you can often find it in libraries, too.

For the meantime, I will try to illustrate something. In the series $\ce{R-NH2}$ the simplest one is $\ce{R~=~H}$, ammonia. Here the dipole moment is straight forward isn't it? I believe it is a little more complicated than that. Unfortunately you cannot just take the charges of the atoms in the molecule, since they are not point charges. Have a look at the total electron density, you will notice, that the isosurface is more extended on top of the Nitrogen, which is expectable, since it is the area where the lone pair is supposed to be. This also coincides with the highest occupied molecular orbital (HOMO). In this special case, the dipole moment is almost a direct correlation of that. (If you would add up all dipole moments of all occupied molecular orbitals, you will also obtain the total dipole moment.)
Below you see an isosurface of the total electron density in blue on the left, the isosurface of the homo in the middle and natural charges with the resulting dipole moment on the right. The values are taken from a DF-BP86/def2-TZVPP calculation with Gaussian09.

What we can learn from this is, that the lone pair has a huge effect on the dipole moment.

Let's go a bit further and look at $\ce{R~=~CH3}$, methyl amine. It can be expected, that the dipole moment is somewhat smaller than in ammonia, due to the methyl group. What you can also observe, that it is dependent on how the methyl group is rotated. Of course in room temperature, the group is freely rotatable and it will result in a mean (pulsating) value for the dipole moment. However, what you can observe here is, that the direction of the dipole moment is aligned with the direction of the lone pair. As you go further down the series this generally will hold, However, the magnitude will be decreasing since your "quite unpolar" moiety is becoming bigger.

For ethyl amine, $\ce{R~=~C2H5}$, the trend holds. But to not extend this answer too much, I only pick one representative amongst all possible conformations.

And for propyl amine, $\ce{R~=~(CH2)2CH3}$, we can observe it analogously, with decreasing magnitude of the dipole moment.

As the last example I have chosen heptanyl amine, $\ce{R~=~(CH2)6CH3}$, to example the trend. I removed some of the charges to not overcrowd the image. If you have a close look at the HOMO, you will notice, that it only significantly extends to the first carbon. The situation for any aliphatic amine can be expected to be very similar.

(I have no idea, why the vector for the dipole got bigger and bigger, while the magnitude was decreasing.)

TL;DR:
The dipole moment in aliphatic amines is about 1 Debye and the direction is through the nitrogen directed to the lone pair.


Caffeine is not acidic. In fact, it is basic. It will raise the pH of water to some degree. It is a weak base, so the free base will exist in equilibrium with its protonated counterpart.

Due to its basicity, it will be more soluble in low pH aqueous solutions than in high pH solutions. This is because in low pH solutions, caffeine would exist completely as a protonated salt, making it more hydrophilic. Despite this, caffeine is very polar and is still quite soluble in water at higher pH levels.

The six-membered pyrimidinedione can exist in an aromatic, zwitterionic form where both amide nitrogens have formed double bonds to respective adjacent carbonyls. This six-membered ring had been observed as being planar. Therefore, it has been concluded that caffeine does indeed exist primarily as its zwitterionic resonance form, with two positive charges and two negative charges. The negative charges exist on the two carbonyl oxygens. These may be slightly basic, too, but significantly less so than a structurally similar hydroxyphenol (e.g. resorcinol) due the electronegative nature of the attached aromatic ring. Therefore, the basic imidazole nitrogen likely has largest effect on the pH.

The fact that there is possibly more that one weakly basic atom on the caffeine molecule complicates calculations as far as determining a theoretical pH. Also, the calculation would require the quantity of caffeine you will add. If you need an exact number, you will have to measure it. You can be sure, however, that the pH of the water will rise as caffeine is added.


Why is the carboxyl proton in salicylic acid more acidic than the phenol proton?

More resonance structures does not necessarily mean more stabilization.

Phenolate has a negative charge at an oxygen atom which cannot be shared without "disturbing" the aromaticity of the ring. This is in part because in the resonance structures drawn, there are negative charged carbons within the aromatic ring, which either implies that:

The ring is still aromatic, so there must be a free electron pair in a p orbital (very high in energy).

Or:

The free electron pair is in an sp³ orbital, which means the ring is no longer aromatic.

In the case of the carboxylic acid, the charge is distributed between two atoms more electronegative than carbon and the ring retains its aromaticity.


What are energies in computational methods relative to?

Ab initio calculations give the absolute energy of the system of fixed nuclei and moving electrons (all considerations on the type of calculations are implicit such as the Born-Oppenheimer approx.). There isn't any distinction of concepts of absolute energies between softwares when ab initio methods are employed (no ambiguity is allowed as pointed out by Pentavalentcarbon). Maybe the confusion can be illustrated with an example: you can't obtain heats of formation, as an absolute energy, from an ab initio calculation but it serves as a good approximation depending on the method.


German Wiki article contains more use-cases for ethyl phthalate:

Formulation auxiliaries for plant protection products
Fixer for fragrances;
In the process of ethanol denaturing (in cosmetics this product is denoted as Alcohol denat.);
In manufacturing of cellulose esters as a very well-gelling and light-resistant plasticizer;
A plasticizer for cellulose-based varnish binders and as enteric coating for medicaments;
A part of cosmetic products due to its film-forming, softening and hair-conditioning properties;
Explosives phlegmatizer;
A carrier in the dyeing process of cellulose acetate or triacetate fibers.


The effect of pressure on boiling point?

Your flaw in the logic lies in that you think that the volume of the system includes the liquid. Le Chatelier's principle on pressure equilibrium only applies to when the compound involved is in gaseous phase, in an ideal sense. Therefore if you increase the pressure (decreasing the volume), then the reaction will progress towards the direction where gas volume DECREASES, which is towards liquid phase. Overall, the entire apparatus increases the vapor pressure on the gas (gas pressure on the external pressure), which increases the boiling point.


Relative Atomic Mass vs Atomic Weight
Relative Atomic mass ($m_a$) is the measurement of the mass of an atom relative to carbon-12. It's dimension is M.[1] It can be measured in Da/u, kg or g.* [3]
$$m_a \approx \mathrm{Mass\ Number} \times {1\ \mathrm{u}}$$
except for carbon-12 where they are the same.
Atomic weight is:
Relative atomic mass vs Standard atomic mass
Relative atomic mass or Relative atomic weight is the weighted average of the isotopes of an element according to their abundance in a given sample compared to the atomic mass of carbon-12. It's dimension is M/M or 1. [2] [4]
Standard atomic mass is the same except that the samples are restricted to terrestrial samples normally considering/averaging results from many different samples or results given by the CIAAW.
Relative Isotopic Mass vs Relative Atomic Mass
Relative Isotopic mass considers the mass of an isotope compared to the atomic mass of carbon-12. This is also dimensionless.
*Universal mass constant vs kg , g ,etc
The universal mass constant ($m_u$) is defined as $\frac{1}{12}$ of the mass of carbon-12 which defines 1 Da or 1 u. All measurements in Da are comparisons to the mass of a carbon-12 atom. Since the discovery of Avogadro's constant, we know: [3]
$$ 1\ {\rm{Da}} = 1.660\ 539\ 066\ 60(50)\times 10^{-27}\ \mathrm{kg}$$
Molar Mass vs Molecular Mass (or Formula Mass)
Molar Mass is the grams of a substance in 1 mole of it measured in gram per mole. It's dimension is $\mathbf{M N^{-1}}$.
Molecular mass or molecular weight is the sum of the atomic masses of all the component atoms of each element measured in u. It's dimension is M. Formula mass or formula weight is the sum of the atomic masses of component ions (not sure of the difference between this and molecular mass - in terms of quantity).[2]
[1] see the answer to this Q by M Farooq
[2] What is the difference between "molecular mass", "average atomic mass" and "molar mass"?
[3] Mass number, (relative) atomic mass, average mass
[4] Quick and simple explanation of molar mass, molecular mass and atomic mass


For a given configuration there is a corresponding and fixed rotatory effect. However, the actual rotatory effect - that seen by a polarimeter - can either be + or −, independently of the D- or L-configuration of the stereogenic centre that shows up in the compound name (in the case of sugars, it refers to the carbon furthest away from the =O group).

In brief, D-glucose is always D-(+)-glucose.


IUPAC name for 1,2,3-trichlorocyclopropane?

Achiral cyclic compounds like 1,2,3-trichlorocyclopropane may contain pseudoasymmetric centers.
Pseudoasymmetric centers have distinguishable ligands (“a”, “b”, “c”, “d”), two of which are nonsuperposable mirror images of each other (enantiomorphic). The lower case stereodescriptors “r” and “s” are used to designate the absolute configuration of pseudoasymmetric centers, for example (2R,3r,4S)‑2,3,4‑trichloropentane:

Note that the stereodescriptors “r” and “s” describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example “r” remains “r”, and “s” remains “s”), but are reversed by the exchange of any two ligands (“r” becomes “s”, and “s” becomes “r”).
The configurations (“r” or “s”) at pseudoasymmetric stereogenic units are determined by application of the Cahn-Ingold-Prelog (CIP) priority system and the sequence rules. In particular, note that an atom or group with descriptor “R” has priority over its enantiomorph “S”. In order to establish the order of precedence of ligands in a stereogenic unit, the atoms of the stereogenic unit are rearranged in a hierarchical diagram, called a “digraph”. Descriptors specified in digraphs may correspond to the final descriptors or to temporary auxiliary descriptors used only for ranking ligands and never appearing as final descriptors.
Similar to the above-mentioned case of (2R,3r,4S)‑2,3,4‑trichloropentane, the structure of (1r)‑1,2,3‑trichlorocyclopropane contains one pseudoasymmetric center, which has two ligands that are nonsuperposable mirror images of each other:

The other isomer of 1,2,3‑trichlorocyclopropane is (1s,2s,3s)‑1,2,3‑trichlorocyclopropane, which contains three pseudoasymmetric centers. Each one has two ligands that are nonsuperposable mirror images of each other:

It may not be obvious how to choose the stereodescriptors ‘r’ or ‘s’ for 1,2,3-trisubstituted cyclopropanes. The ranking is a complex process and requires the use of auxiliary descriptors.
Generally, the stereodescriptors ‘r’ and ‘s’ are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors ‘R’ and ‘S’. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.
The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) and which has been recommended by Prelog and Helmchen.
In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1r)‑1,2,3‑trichlorocyclopropane:

where $\ce{(C)}$ is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.
As the digraph is acyclic, it is now possible to specify the stereodescriptor ‘R’ or ‘S’ for the stereogenic unit at the atoms numbered ‘2’ and ‘3’ in each branch. These are auxiliary descriptors that are different in each branch.

Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that ‘R’ has priority over ‘S’.

P-92.1.3.5 Sequence Rule 5

An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.

Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1r)‑1,2,3‑trichlorocyclopropane can be determined to be ‘r’.

In the same way, the configuration of each pseudoasymmetric stereogenic unit of (1s,2s,3s)‑1,2,3‑trichlorocyclopropane can be determined as ‘s’.


In order to solve for this equation for heat capacity, do I need to treat it as a system of equations?

In the first equation, what does C designate.

Since the final value is energy in joules, then it should be the the unit for temperature. She should have used °C to be clear.

  If it designates the heat capacity of the system, do I solve for q by solving a system of equations?

No, because you don't have the correct equations. You need the amount of heat released from the hot water minus the amount of heat absorbed by the cold water to give you the heat absorbed by the calorimeter.


Acidity order of acetone, methyl acetate and N,N-dimethylacetamide

The increased acidity of the protons on the alpha-carbon in these compounds, compared to the acidity of the protons in methane, is due to stabilization of the resultant negative charge through resonance with the adjacent carbonyl.  In the top line of the following figure we show a ketone where the proton has been removed.  We can draw 2 resonance structures to describe this carbanion.  In structure II the negative charge has been delocalized onto the electronegative oxygen atom - a very favorable and stabilizing situation.

To whatever degree we remove the carbonyl's ability to resonance stabilize that adjacent charge we will decrease the acidity of the alpha-protons.

In the case of both the ester (III) and amide (VI) we can draw a third resonance structure (V and VIII respectively).  To whatever extent these resonance structures contribute to the overall description of the ester and the amide, the carbonyl will be less available to stabilize the negative charge through structures IV and VII.

It turns out that resonance structures V and VIII are significant contributors, with the amide structure VIII having a positive charge on nitrogen being more important than resonance structure V which places the positive charge on the more electronegative oxygen (less favorable).  In fact, resonance structure VIII is such a significant contributor that restricted rotation occurs about the $\ce{C-N}$ bond and we can typically observe two different R group signals in the proton-nmr spectrum of amides.

So as stated earlier, as the contribution of the third resonance structure increases, the contribution of the second structure (the structure that stabilizes our carbanion) will decrease making our compound less acidic.  The ketone doesn't have a third structure so it will be the most acidic structure. Structure VIII is more important than structure V, so the ester will be more acidic than the amide.  Therefore, the overall order of decreasing acidity is
$$\ce{CH3COCH3 > CH3CO2CH3 > CH3CON(CH3)2}$$
just as you noted.


Standard reference for thermodynamic properties for octane

The NIST Chemistry Webbook maintained by the National Institute of Standards and Technology is a freely searchable database of chemical and physical data on a vast array of substances.

Everything you want for octane can be found there.


Discrepancy between the apparent volume of the solution and the volume of the solute arising from the definition of solubility

I have solubility s given in g/mL. I understand this as mass of solute m divided by volume of solvent V.

Your confusion is arising from mixing the concept of solubility and concentration. Unfortunately, both are expressed the same way. You have to see the context. Most authors (like good scientists) try to write their experiment explicitly so that others can understand and repeat their experiment.

See the examples in this link: Solubility Concepts

Let us say, we find from the literature that the solubility of NaCl in water is 360 g / 1L. It means that if we take 1 L of pure water, we should be able to dissolve 360 g of salt in it. Note by adding 360 g of NaCl, the volume of water may not be exactly what 1 liter anymore. I cannot calculate concentration out of this information.

Now some authors may refer to concentration when they talk about solubility. For very dilute solutions it does not make a difference. So it is always good to see the context.

  I read online I should pour the solute into a measuring cylinder/cup first, and then add solvent until the volume fits the calculated value (V=m/s), but this seems to be at odds with the definition of solubility.

Don't trust the webpages made by amateurs or pages from non-reliable sources. With time you will develop the sense which sources are more reliable. There is a saying that paper never refuses ink. It means one can write whatever on paper, and the paper will never say no, even if it is wrong. The same is true for the web. If we search the web for corona virus, you will find a lot of non-sense. Only reliable sources are to be trusted.


How to translate the mobile phase from thin layer chromatography to column?

I never heard about the "10% rule" and I never applied it. Transferring TLC (silica on aluminium plates, Merck) results to preparative column chromatography on similar silica usually worked without changes, although the situation is a bit different in the column: the upper part of the TLC plate is rather dry when letting it run, while the silica in the column is not. Note that I was never a friend of the dry packing method and always prefered the slurry method to prepare the columns.

Anyway, I never thought about increasing the polarity.

Bad separation typically was a result of column overload, bad packing or cracking of the stationary phase, e.g., when the diethylether evaporates in the column.

+1 For citing the classic article on flash chromatography!

I can really recommend the technique and I used it ... more than once ;)


Deducing compounds from spectroscopic data for isomers of formula C4H8O

Your two proposed structures are correct, although your rationalisation of the 13C spectrum for Cpd D is not quite correct around the double bond. In a nutshell, carbon chemical shifts can be calculated as a function of their α, β and γ contributions. α for what is directly bonded to that carbon, β for substituents one carbon away, and γ for substituents two carbons away. Almost without exception, alkene carbons exhibit a downfield β shift and upfield γ shift for substituted alkenes. So, for both compounds C and D, the carbon closest to the substituents will be shifted furthest downfield. The β and γ contributions for the following substituents are shown (someone more tech savvy than me might tidy this up) :

[Substituent; β ;γ]
[-H; 0 ; 0]
[-CH2O-; +14.2 ; -8.2]
[-CH3 ; +9.0 ; -7.0]

The electronics contributing to β and γ shifts is not well understood, but is more complex than a consideration of just electronegativities/electron withdrawing potential.

Below are the spectra for the two compounds:

Your confusion with the description of the quintet label arises from how people report splitting patterns in the literature. I have discussed this on another question, but essentially the two methods of reporting splittings are to (a) report the observed splitting pattern (here a quintet) and (b) report the calculated/expected splitting pattern (here a doublet of quartets). As you can see, (or least I can), reporting the observed pattern can lead to some confusion, and gives no real information about how this splitting pattern arises.

Further, reporting couplings to the nearest Hz (here 6Hz) might be fine for some occasions where the difference in coupling for the two different partners is less or approximate to the observed linewidth, however, with a good sample and a good operator on a good magnet, what is reported as a quintet may very well appear different, especially once some apodization is applied. For example, below is a expansion of a simulated spectrum for compound C with Jbc=5.8 and Jbe=6.2 (smaller couplings from d and f ignored), and observed linewidth of 0.5Hz. The top spectrum is the normal spectrum, and the bottom spectrum is what it looks like with a gaussian linewidth function applied (gb 0.1, lb -1). They look quite different, and the bottom spectrum is very hard to rationalise as a quintet, but easily recognizable as a doublet of quartets.


Chemical kinetics with the reaction of tert-butyl bromide with azide ion

You really don't have to write all that differential equations.

Note: I'll write all my steps on a paper, since MathJax formatting is a little time-taking at the moment.

Consider the rate laws for the two reactions:

Since the $\ce{[(CH3)3C+]}$ isn't changing with time, we get

From which, you get

Plugging these in the Rate2 which I mentioned, you will get,

Which is the same as what's given in option (D).

I hope I have satisfied your queries.


The chelate effect is very simple. Consider the addition of $\ce{Na2H2edta}$ to a random transition metal (e.g. manganese(II)). Your starting complex is a hydrate complex; we shall assume it is $\ce{[Mn(H2O)6]^2+}$. The reaction equation is given below:

$$\ce{[Mn(H2O)6]^2+ + edta^4- <=>> [Mn(edta)]^2- + 6 H2O}\tag{1}$$

You started with two molecular entities, you end up with seven. Since $2<7$, the product side has a higher entropy (more independent particles). The same is true for any chelating ligand that displaces more than one monodentate ligand.

Yes, upon chelation the chelating ligand does lose degrees of freedom concerning bond rotation. However, the number of particles greatly outweighs that.


How to calculate the activation energy of diffusion of carbon in iron?

Maybe I'm wrong but my calculation has given another result:
\begin{align}
D_{298} &= D_0\cdot\exp\left\{\frac{-Q}{298~\mathrm{K}\cdot\mathcal{R}}\right\}\\
D_{573} &= D_0\cdot\exp\left\{\frac{-Q}{573~\mathrm{K}\cdot\mathcal{R}}\right\}\\
\end{align}

Thus, going through the $\ln$ form, you get
\begin{align}
\ln\{D_{298}\} &= \ln\{D_0\} - \frac{Q}{298~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{573}\} &= \ln\{D_0\} - \frac{Q}{573~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{298}\} - \ln\{D_{573}\} &= -\frac{Q}{\mathcal{R}}\cdot\left(\frac{1}{298}-\frac{1}{573}\right)
\end{align}

And then I obtained $Q\approx 38~\mathrm{kJ/mol}$.


What does negative current reading in galvanic couple mean?

The negative sign is simply indicative of how you've inserted your multimeter probes (which are labelled for polarity, or should be) into the galvanic battery circuit. In one case, the multimeter "sees" electrons flowing from the positive terminal to the negative one (hence the negative sign); in the other case, the opposite is true.


How does a shock, such as the snap of the disk in a reusable hand warmer, trigger the re-crystalization of a supersaturated solution?

Edit: Contribution of enthalpy and bond formation
Edit 2: Changed the description of entropy
After thinking for sometime, I realized that I neglected to mention three things:

The Driving force of precipitation reactions depends on the charge and size of the ions which corresponds directly to the magnitude and order of the hydration shells formed upon dissolution of the salt.

The enthalpy as a driving force of the precipitation of the salt. (Remember the Gibbs Free energy equation has two terms, entropy and enthalpy)

Additionally, the formation of ionic bonds can be lower in energy (more stable) than the electrostatic interactions present in solution. The release of energy during salt formation is transferred to the surroundings as heat. This would be a negative enthalpy (heat flowing out of the system) and if large enough, would overcome any decreases in entropy stemming from the formation of highly order crystalline structures and drive the precipitation.

In this case it is probably the enthalpy term that is driving the reaction, since you are putting a ton of heat into the system to dissolve the salt, once the solution cools, there is a lot of potential energy that is stored in the position of the atoms in the system. This is then released as heat upon the formation of the crystal structures.
Entropy changes in precipitation reactions:
In order to make a supersaturated solution, you have to add energy to form hydration shells around the ions (entropy in the system is decreasing). The thing to remember is that these hydration shells decrease the dispersal of the ions relative to that in the crystalline structure for some salts. When you heat and mix the solution with excess solute, you are providing sufficient energy to form hydration shells. When the solution cools (Slowly), the hydration shells remain intact, however, the system is unstable and applying enough energy to the solution will disrupt the hydration shells and crystallization of the salt will occur. The large amount of heat that is released is coming from the increase in the micro-states available for the ions (gain in entropy) in removing the hydration shells around the ions. The hydration shells are less dispersed than the crystalline structures of the salt and that is why entropy increases during some crystallization reactions, making this process exothermic.
Here is why the disc snapping is important:
Even though this process is very exothermic, it still requires activation energy. The mechanical force you are referring to (i.e. snapping the disc or whacking the heating pad) is doing just that by simply transferring energy to the molecules, through vibrations, which provides the activation energy for crystallization.
Just as walking can cause energy transfers from the mechanical forces of your legs to the molecules of the floor (i.e. through friction) and increase the temperature of the surface, some of the force applied on the disc transfers to the solution and if it is enough, will overcome the activation energy barrier of the process and crystallization will begin.


Does potential difference or current drive electrolysis?

For any electrochemcial reaction, electrons are transfer from the oxidant to the reductant. By definition, any flow of charge per unit time period is an electric current.

However, in the case of electrolysis, the electron transfer is not spontaneous. An external energy source is required for the reaction to take place. The energy (work) provided per unit charge is called the voltage.

For any electrolysis reaction to occur therefore requires a supply of electrical energy which means both a voltage and electric current are needed.

Specifically, the amount of work (energy) resulting from a voltage, $V$ and a current, $I$ applied for a time $t$ is given by:

Work produced by electrical energy, $W = V\cdot I\cdot t$

For a given electrolytic reaction to proceed, the required minimum voltage which must be applied is determined from the Gibb's function,

$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}$

Where $E^\circ_\text{cell}$ is the minimum voltage needed for the electrolysis reaction to occur, $\Delta G^\circ$ is the change in Gibb's free energy under standard conditions, $n$ is the number of electrons transferred and $F$ is Faraday's constant (96,485 coulombs per mole).

For example, if pure water is placed in an 'electrolytic cell' with two non-reactive electrodes (eg: platinum), electrons forced into the 'cell' by an electrical source (eg: battery) will react with water molecules at the cathode forcing them to lyze (split) into hydrogen ions and hydroxide ions.

$\ce{H2O(aq) ->[\text{elect}] H+(aq) + OH- (aq)}$

At the surface of the 'anode', hydroxide ion will be oxidised (dontate electrons) to form oxygen gas.

$\ce{4OH- (aq) -> 2H2O + O2 + 4e- }$ (anode half-cell reaction)

Meanwhile, at the surface of the 'cathode', hydrogen ions will accept electrons to form hydrogen gas.

$\ce{2H+(aq) + 2e- -> H2(g)}$ (cathode half-cell reaction)

The overall cell reaction is therefore:

$\ce{2H2O(aq) -> 2H2(g) + O2(g)}$ ; $\Delta G^\circ=237.2\ \mathrm{kJ/mol}$

The positive Gibb's function for this reaction indicates that the reaction will not occur sponteneously but requires an external energy source (eg: electrical energy), which is reflected in the negative cell potential.

The minimum (theoretical) voltage required is:

$E^\circ_\text{cell}=\frac{-\Delta G^\circ}{nF}=\frac{-237.2 \times 10^{3}}{2\times 96485}=-1.23\ \mathrm{V}$

However, in practice, a slightly larger voltage of 1.48 V is required since the  enthalpy (heating) of the products results in slightly lower efficiency which is manifested as an overpotential of about 0.25 V.

Once this critical voltage level is exceeded, the electrolysis reaction proceeds at a rate determined largely by the the current, since the current represents the rate at which charge is delivered to the system. Basically, the higher the current, the more molecules will react (electrolyze) and the more products (hydrogen gas) will form per unit time.


IUPAC defines "Solubility", but not "soluble". There is no single definition because there is no single meaning. The reason there isn't a single meaning is because it is useful to have different meanings in different contexts. (This is true of almost all words.) According to Wikipedia, 0.1g per deciliter is a common ceiling for insolubility. I've seen various values for the lower value for labeling a compound "soluble". I've seen 1g, 2g, 3g, 5g and possibly more claimed to be the (arbitrary) floor. I don't see any reason to call for standardization, because the use for such a property is varied and diverse.


Do reactions carried out in light always follow free radical mechanisms?

Often, yes; but sometimes not, and generalisations like that often don't work in organic chemistry.
It's a better idea to understand why there's a correlation between "light" and "radical reactions". One way of generating radicals is by breaking bonds, and one way of breaking bonds is to use light. Of course, this only really works for relatively weak bonds, such as the Br–Br bond, or basically any kind of bond between Group 16 and/or Group 17 elements (O–O, N–Cl, ...). Many of the classic radical reactions involve this kind of pattern, and this is probably why you're drawing this link.* Examples include: free-radical halogenation, the Hofmann–Löffler–Freytag reaction, and the Barton reaction.
Alternatively, if there aren't any weak bonds to break, it could be that you're trying to use light to promote an electron to an excited state. Now, because the vast majority of organic chemicals are closed-shell molecules (all electrons are paired), when you excite something you're going to get a radical species. A simple example is the Norrish type II reaction. This radical generation can be indirect as well, occurring via excitation of a photocatalyst, which then oxidises / reduces an organic substrate to form an organic radical. This is the idea behind photoredox catalysis, which has been a very, very hot topic in recent years.
It is not wise to make a sweeping generalisation, though. You could excite a molecule and have it react immediately, without generating an intermediate radical species. This is the case in photochemical electrocyclic reactions, for example. And there are many photoisomerisations which (as far as I am aware) occur in a single step, without going through a true radical pathway. A very important one is the cis-trans photoisomerisation of retinal. The mechanisms of these reactions are not well-described by the curly arrows of organic chemistry: you'd need to learn some photochemistry to get a good idea of how these work.
And not every chemical relevant to organic chemistry is closed-shell. Dioxygen, O2, is a diradical species; and when you excite it, you get an excited state where the electrons are paired up, usually referred to as singlet oxygen.† Singlet oxygen reacts with other organic molecules in a decidedly non-radical fashion: for example, it does classic Diels–Alder reactions and ene reactions.

* Note that many of these can be also accomplished by simple heating, which can break weak bonds too.
† Usually dioxygen is not excited directly. A photosensitiser like rose bengal is used.


Why do gases only transmit sound waves whose wavelength is long relative to the mean free path?

Since sound waves are longitudinal waves (generally in air) there needs to be some kind of interaction between neighboring "air" molecules (a mixture consisting mostly of $\ce{N2}$, $\ce{O2}$, $\ce{CO2}$ and $\ce{Ar}$). You can see that this is just not possible if the mean free path of the "air" molecules is larger than the desired wavelength. How would the "air" molecules interact on such distance if there are no "air" molecules to interact with? This interaction (i.e. sound of a particular wavelength) will only be possible if there are sufficiently many "air" molecules nearby to interact with, or formulated otherwise: if the mean free path is sufficiently much smaller than the desired wavelength of the sound. Of course, it is possible to make sound waves with a larger wavelength than the length of the mean free path, just not the converse.

One application I can think of lies more in Physics but whatever. In the days before Einsteins theory of special relativity was developed/accepted, there was a common consensus, which was supported by Maxwell himself, that light was travelling in a special reference frame called "The Aether". In this reference frame the speed of light equaled $c$. We now know of course that light can travel through vacuum and that there exists no such special reference frame as The Aether. There were various problems with this Aether theory, most famously the Michelson-Morley experiment that suggested that either the Aether did not exist or was moving for a weird reason with the exact same trajectory and speed as earth in space. Another problem, related to your question, was that the speed of waves in a medium is dependent on the particle density of that medium. And since (at least some) electromagnetic waves have an incredible small wavelength (the $\lambda$ of visible light for example was known), the density of this Aether medium had to be extremely high, yet it had to also weight almost nothing and be able to penetrate trough everything!

From this and other facts Einstein must have concluded that the Aether theory was in fact invalid. This idea ultimately lead to the theory of special and general relativity, both of which have many spectacular and far reaching consequences.


Is there any difference between thermal energy at room temperature in fluids(water) vs air?

First of all, just a minor thing, the thermal energy itself is always the same. You can't grab it, you can just measure it. What is different is the amount of energy "stored" in air or water, but its no different "energy".

The retainment of heat in the sea is due to the heat capacity.
https://en.wikipedia.org/wiki/Heat_capacity
This is a specific property of each molecule.
To increase the temperature of 1 g water by 1 K you need more Energy than for 1 g of air. That is the reason why the temperature near the sea is more moderate (absorbing heat in the summer and releasing heat in winter).

This is also a reason, why you can go into a sauna with 90 °C without burning yourself, but you can't grab into 90 °C hot water. This is also due to thermal conductivity but also due to the capacity. When your skin touches the 90 °C hot air, it is absorbing the heat providing you with cooler air around your skin. The same happens in water, but first, your skin has to absorb much more energy to cool down the water, second of all the surrounding water heats up the cooler water layer around your skin much quicker, than hot air does. That's why air is considered an isolation.


Common flame-retardant plastic film materials?

What have you found so far? In a situation like this honestly I would go ahead and research what hot air balloons are commonly made with already. Find somebody more experienced in the hobby and ask them if you can, or even an experienced fire fighter you can get in touch with.

From wikipedia: "For modern sport balloons the envelope is generally made from nylon fabric and the inlet of the balloon (closest to the burner flame) is made from fire resistant material such as Nomex"

The type of measurement you are asking for seems a bit more complicated than something like measuring mass or volume, it seems to be commonly represented with images, videos of tests, graphs, etc. You may feel more at ease by testing how useful materials are before and after flame testing them yourself, that is what I would do anyways. I hope this helps you.


If liquid hydrogen fluoride autoionises, how can it be an insulator?

Unlike metals which can conduct with free electrons, conduction in non-metallic liquids (ionic and covalent) is dependent on the ion concentration to carry charges. Pure water has a autoionization constant of $\rm K_w = 1\!\times\!10^{-14}$, which corresponds to a total ion concentration of $\rm 2\!\times\!10^{-7} \pu M$. At this concentration the resistivity of water is $\pu{18.2 M\Omega~ cm}$ which corresponds to a conductivity of $\pu{5.5 \mu S/cm}$ and is considered to be a poor conductor. Sea water on the other hand has about $\pu{3.4\%}$ sodium chloride and a conductivity of about $\pu{55,000 \mu S /cm}$ which is a million times improvement in conductivity which is considered conductive.

Hydrogen fluoride has an autoionization constant of $\rm 10^{-12.5} ~~(3.162\!\times\!10^-13)$ which corresponds to a total ion concentration of $\rm 1.125\!\times\!10^{-6}$ which is about five-times that of water. Five times the ions gives us five times the conductivity of $\pu{275\mu S/cm}$ or a resitivity of $\pu{2.28M\Omega ~cm}$ which is still a quite poor conductor. Thus it is necessary to add an ionic substance to increase conductivity to have appreciable conduction of the hydrogen fluoride similar to sea water.


How to determine the edge length of the aluminium unit cell?

Correct approach is to do two things:

1)  Put four atomic masses into one unit cell.

2)  Properly extract the cube root of the volume you obtain.  That volume should now be about $6.6×10^{-23}\text{cm}^3$ from part 1.  To take the cube root properly, render the exponent on 10 as a multiple of 3, thus  $6.6×10^{-23}$ = $66×10^{-24}$ where $10^{-24}$, not $10^{-23}$, is $(10^{-8})^3$.

Then your numbers should work out.


Does CO2 form carbonic acid in the atmosphere?

Short answer

It does form carbonic acid, as rain drops are water and carbon dioxide is soluble in water.
But the content of carbonic acid is minimal as you can guess from $\mathrm{pH=5.6}$

Longer answer

Majority of dissolved $\ce{CO2}$ does not form carbonic acid, 99.83% stays as hydrated oxide.

The equilibrium $\ce{CO2}$ concentration in water follows the Henry's law and is proportional to the $\ce{CO2(g)}$ partial pressure.

Therefore, as a drop falls, the partial pressure of $\ce{CO2(g)}$ grows together with the atmospheric pressure and  $\ce{CO2}$ solubility in water would increase as well, if temperature did not raise with decreasing altitude, typically $\mathrm{ 0.0065 K/m}$.
The total effect is general decreasing $\ce{CO2}$ solubility when a drop is falling.

Sure, some drops may evaporate during their fall ( seen as precipitation patterns without raining ) and the dissolved $\ce{CO2}$ becomes gas again.

Natural rain has $\mathrm{pH=5.6}$  because of dissolved $\ce{CO2}$ and equilibrium concentration of $\ce{H2CO3}$.

If all  CO2 converted to H2CO3, the natural rain and soda water would be much more acidic, because carbonic acid is quite strong acid with pKa=3.6, instead of formal 6.3 if 100% conversion is supposed.

In carst areas, there is ongoing reversible reaction of dissolving calcium carbonate by surface rain water going down the soil and precipitating it again in caves :

$$\ce{CaCO3(s) + H2O + CO2(aq) <=> Ca(HCO3)2}(aq)$$

The calcium bicarbonate comes also to surface water and makes frequently the majority of the calcium content.( and similarly for magnesium l


Why is the Kelvin Temperature Scale Based on the Triple Point of Water?

In reality all the "primary" scientific values have changed over time as experimental techniques have improved. Science has evolved to the point where a set of seven "SI base units" (ampere, kelvin, second, meter, kilogram, candela, and mole) have been selected and defined so that better measurements won't cause disruptive changes in scientific data.

So the gist is that the "temperature of ice water" depends on the atmospheric pressure. So it is necessary to define that 0 °C is the temperature of ice water at "standard atmospheric pressure." That means that if the "standard atmospheric pressure" changes, then so does the "temperature of ice water at standard pressure."

However the triple point of water doesn't change and has a unique value. So by defining the triple point of water as exactly 0.01 °C in 1954, then the Celsius scale is anchored as much as the Kelvin scale. By also defining absolute zero as 0 K and −273.15 °C then the temperature scale itself won't change if better ways to measure temperature are devised.

Thus by the time the whole notion of the primary base units was sorted out there was a massive amount of scientific data that had been accumulated using the old definition of the Celsius scale that was based on ice water at a standard atmosphere. Letting the triple point be exactly 0.01 °C kept all of that "old" temperature data "reasonably accurate" since 0.01° is only a 0.01/273.15=+0.0037% change.


Is there any scenario in which the size of a molecule increases due to an increase in temperature?

First, let me remark to “the size of a molecule” is not particularly well-defined. I assume in the following that you are interested about bond lengths (which are averages over time of distances between bonded atoms), as seems to be the case in your example. In the simplest case, you can consider an isolated diatomic molecule (think: gas phase N2), and we suppose we're not going to temperature high enough to ionize it or otherwise break its bond. Because the interatomic potential is asymmetric:

you can see that, as the temperature increases and the system is allowed to explore more of the potential energy curve (climb higher up the well), the average bond length will increase and not stay constant. So, for a single bond in an isolated molecule, it is expected that the bond lenghtens with temperature.

Now, one case where the overall “size expansion” can be clearly studied is that of crystalline materials. In solids, the evolution of volume as a function of temperature is named thermal expansion, and is characterized by the parameter of the same name, defined as: $$\large \alpha = \frac{1}{V} \frac{\partial V}{\partial T}$$ Because bonds grow longer with increasing temperature, most materials display a positive thermal expansion, i.e. they expand with temperature. This is, for example, the case for diamond, with $\alpha \approx +10^{-6} \text K^{-1}$.

However, some materials display the oppositive behavior, and shrink upon temperature increase. This phenomenon is called negative thermal expansion (NTE), and is rather uncommon.


Separating nickel coin into copper and nickel components

I recommend to dissolve the copper-nickel alloy in nitric acid and determine the dissolved $\ce{Cu(II)}$ by iodometric titration.

Start by weighing the coin, then dissolve it by placing it in nitric acid solution (under a fume hood). Continue to boil the solution for 20 min after complete dissolution of the coin (to remove any traces of remaining $\ce{NO2}$ in soution. The volume of the solution is about $40 \,\mathrm{mL}$.
$$\ce{Cu(s) + 4HNO3(aq)->Cu^{2+}(aq) + 2NO3^- (aq) + 2NO2(g) + 2H2O(l)}$$
$$\ce{Ni(s) + 2H3O^+(aq)->Ni^{2+}(aq) + H2 (g)  + 2H2O(l)}$$
Prepare a fresh solution of sodium thiosulfate of precise concentration $\approx 0.04\,\mathrm{M}$
Add $1.00 \,\mathrm{mL}$ of the diluted copper-containing solution to $15 \,\mathrm{mL}$ of $10\% (W/V)$ potassium iodide solution.
$$\ce{Cu^{2+}(aq) +  5I^- (aq)->CuI(s) + I3^- (aq) }$$
Titrate the yellow-orange suspension with sodium thiosulfate solution until the solution color has become yellow. To detect the end-point add about $1 \,\mathrm{mL}$ of starch solution.
$$\ce{I3^- (aq)  +  2S2O3^{2-} (aq)->S4O6^{2-} (aq) + 3I^- (aq) }$$
Repeat the titration to precisely determine the end-point.


How can I measure the internal temperature of a solid metal?

You can use ultrasound thermometry. It uses change in speed of sound under heat. AFAIK, there are industrial-grade solutions on the market.


Understanding the partial derivative of temperature with respect to thermodynamic beta (coldness)

You know that:

  $$\beta = \frac{1}{k_\mathrm BT}$$

Rearranging yields:
$$\color{red}{\beta} = \frac{1}{k_\mathrm B\color{blue}{T}}\implies \color{blue}{T} = \frac{1}{k_\mathrm B \color{red}{\beta}}$$

Thus:

$$\frac{\partial \color{blue}{T}}{\partial\beta} = \frac{\partial}{\partial\beta}\left(\frac{1}{k_\mathrm B \beta}\right) = \color{red}{-}\frac{1}{k_\mathrm B \beta^\color{red}{2}}$$

Substituting $\beta = \color{red}{\frac{1}{k_\mathrm BT}}$ yields:

$$\frac{\partial T}{\partial\beta} =  -\frac{1}{k_\mathrm B \color{red}{\left(\frac{1}{k_\mathrm BT}\right)}^2} = -\frac{k_\mathrm B^2T^2}{k_\mathrm B} = -k_\mathrm BT^2$$

$$-\frac{\partial T}{\partial\beta} = k_\mathrm BT^2$$
Q.E.D.


Copper can form a stable DMG complex (if it is there in the mixture in high concentration), it may also effect the color of the resultant Ni-DMG complex. Just go through Vogel's quantitative inorganic analysis book once.


How does one calculate oxidation states?

For compounds where you have a non-metal with a metal, there is likely an ionic bond. The oxidation states are just the charges on the ion. PbS has Pb2+ and S2-, so those charges are simply the oxidation states.

For PbSO4, you know sulfate is a 2- ion, so it must be bonded to a Pb2+. Therefore, again, Pb has a +2 oxidation number. When you look at the sulfate ion, you know that the sum of the oxidation states of the elements must equal the ion's overall charge. Oxygens are -2, so set up the equation: (X = oxidation number of sulfur)

X + 4(–2) = -2
X = +6

Calculating the oxidation state of carbon can be more complex. If you draw a dot and cross diagram showing the covalent bonds made by carbon, this will help. Look at the electronegativities of the elements in the bonds. Carbon takes the electron pair if it is the more electronegative, but this is rarely the case. For every electron pair carbon takes, it has a -1 added to the oxidation state, and +1 added if it loses the electron pair. For example, if you draw the methane diagram you find that the oxidation state of carbon is +4. Using this method, you can find the oxidation state of carbon in compounds.


How to perform molecular dynamics simulations of charged systems?

This is intrinsic to Ewald summation methods, not software implementations. The uniform charge arises from neglect of a reciprocal sum term. It does not directly affect the dynamics and may be a reasonable model of a spatially homogeneous system. See https://mailman-1.sys.kth.se/pipermail/gromacs.org_gmx-users/2015-October/101544.html for further details and https://www.mpibpc.mpg.de/14063977/Hub_2014_JCTC.pdf for a description and explanation of the likely artefacts.


How to remove paraffin wax drop from cam lense glass?

The first thing I would try is to re-heat the phone then try to wipe the paraffin off with your lens cloth. Paraffin wax typically melts between 46 and 68 °C (115 and 154 °F)$^1$. Assuming you don't have a good way to accurately heat the camera to this temperature, I would just repeat what you did to melt the wax in the first place; let the camera discharge it's battery and then recharge it. Once it has heated up, try wiping it off with the lens cloth.

Unfortunately, I have a feeling the above procedure alone may not be sufficient to remove all of the wax. It's worth trying because it's simple and safe for your phone. If that procedure does not remove all of the wax, the next thing I would try is using a solvent in addition to the heating. The wax should dissolve in diethyl ether$^2$, which is the primary component of most engine starting fluid products. You should be able to go to any auto parts store and purchase a can of starting fluid who's primary ingredient is diethyl ether (sometimes just called ether).

The first thing you will want to do with the ether is to verify that it does not attack the material holding/surrounding your lens. If the area around the lens is all metal and glass, you are safe and can skip this step. If the area around the lens is plastic (unlikely) then you will want to test a small area of the plastic to be sure it will not be dissolved or distorted in any way by the ether. Simply soak a corner of the lens cloth with the ether and touch it to the plastic. You have to be fast because ether evaporates very quickly. You might want to use a magnifying glass to look at the area that you whetted with the ether. If there is any change in the color or texture (you should feel the area with your finger), then you have to decide whether you want to proceed and risk ruining your camera. Again, it is unlikely that this will be a problem. I looked at some images of your phone model, and while there are some plastic parts to the construction, the lens assembly appears to be just metal and glass.

If you decide to proceed with cleaning the lens with ether, it should be a pretty simple, straight forward process. I would heat the phone by charging it as discussed above, soak a portion of the lens cloth with the ether-based starter fluid, and gently wipe the remaining wax from the lens. The lens glass and any metal surrounding it are impervious to the ether (and any other ingredients of the starter fluid), but avoid contact with any plastic components.

Given your problem as you've stated it and the make of your phone, I'm pretty confident that your camera lens can be thoroughly freed of the paraffin wax. Don't hesitate to ask for clarifications of anything I've said in the comments, and best of luck!

1) Kaye, George William Clarkson; Laby,Thomas Howell. "Mechanical properties of materials". Kaye and Laby Tables of Physical and Chemical Constants. National Physical Laboratory. Retrieved 25 October 2013.

2) Seager, Spencer L.; Slabaugh, Michael. "Alkane reactions". Chemistry for Today: General, Organic, and Biochemistry. Belmont, CA: Cengage. p. 364. ISBN 978-0-538-73332-8.


Is the electron configuration redundant?

The configuration you describe is the ground state configuration, the configuration of oxygen in which all electrons are in their lowest allowed energy states.

Substances can also be their excited states, where electrons are not in the lowest allowed energy states. Excited states happen when atoms or molecules absorb energy, usually electromagnetic radiation (but not always).

An excited state for oxygen may involve one of the $2s$ electrons being promoted to one of the remaining $2p$ spots:

$$\begin{aligned}
\ce{O} + E &\ce{-> O}^* \\
1s^2 2s^2 2p^4 &\ce{-> }1s^2 2s^1 2p^5
\end{aligned}$$

This configuration still describes an oxygen atom, but now the $2s$ subshell is partially empty.

The other reason you can't always assume that the lower subshells are always filled is that there are some exceptions to the aufbau principle. Some elements do not follow it, especially the lanthanides and actinides.

Chromium: $\ce{Cr} - \ce{[Ar]}4s^1 3d^5$ ($s$ is not filled so that $d$ can be half-full)
Copper: $\ce{Cu} - \ce{[Ar]}4s^1 3d^{10}$ ($s$ is not filled so that $d$ can be full)
Palladium: $\ce{Pd} - \ce{[Kr]}4d^{10}$ ($s$ is empty so that $d$ can be full)
Cerium: $\ce{Ce} - \ce{[Xe]} 6s^2 5d^1 4f^1$ ($f$-block elements are weird)


The most straightforward answer is that water in the form of ice I (or $\ce{I_h}$) has an open hexagonal structure (see figure) whereas liquid water does not. In ice the oxygens are precisely tetrahedrally positioned where each oxygen is hydrogen bonded by four neighbouring oxygens with an O..O distance of approx 0.28 nm. The H atoms lie very closely along the O-O axes. Two of the H bonds are short, as in liquid water, and two are long. This is expected from residual entropy calculations and also from thermal and neutron diffraction experiments.

As this ice structure has a rather open lattice this is the reason its density is less at its freezing point than that of liquid water, rather than being due to the high density of water. One could argue that the reason that water density is a maximum at 4 $\ce{^0C}$ (277 K) is that the structure of  liquid water is changing from that of ice $\ce{I_h}$ to the short range structure of the liquid.

Liquid water is, of course, also highly hydrogen bonded but the extra energy it contains means that its structure is far less regular and so becomes slightly denser than ice. As the water molecules can move about more in the liquid than in ice, the hydrogen bonds will bend a little or break as molecules rotate. This will lead to a shorter O - O distance and hence slightly higher density in liquid than in the solid.

see also answers to this post Why really is ice less dense than water?

This figure shows the hexagonal ice structure; The O atoms are shown as dots and the lines are the (almost linear) O-H-O bonds.(Image from Murrell & Jenkins 'Properties of liquids & solutions')

(Ice also exhibits eight different phases depending on temperature  and pressure of which $\ce{I_h}$ is the one found at normal atmospheric pressure. In these other phases, angles are distorted from being tetrahedral)


How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile?

The Darzens condensation can be carried out directly from the alpha-chloronitrile. The exact reaction needed here (benzaldehyde + chloroacetonitrile) is listed 15 times in SciFinder using different base catalysts (hydroxide or an alkoxide) and conditions (sometimes with phase-transfer catalysis). The epoxide will be formed as a mixture of stereoisomers, but often chemical yields are high (reportedly upto 100%).

With the epoxide in place, the nitrile can be reduced with DIBAL to give an intermediate imine, which is converted to the aldehyde after hydrolysis. This is pretty much the same strategy you outlined, but with a couple fewer steps.

As for sources of additional synthesis problems, old UC-Berkeley exams are usually very good. Check Chem 3A and 3B at this archive.


Does a mixture of gases contain equal volume to that of the container they're in? Why?

Well, the word "container" means a fixed volume. Let's look at your room for example (don't take into account any openings, like doors or windows, just the walls). It has some amount of gas in it at this moment. Now, if someone decided to "push in" let's say 100 extra moles of O2 gas, then that would not change the volume of your room. The volume of the containter remains constant, no matter how much gas we put into in.

This is also true for each component of the mixture. The oxygen in the original air of your room had the volume of your whole room, and even when we "push in" that 100 extra moles, it still has the same volume - because your room did not change. What can change then? Obviously, the pressure or the temperature can change.


How much time for a water molecule in pure water to reach the top of a glass of water?

I am posting here the answer since it has been quickly found thanks to the useful comments. To give an estimate the best way is to look at the self-diffusion coefficient.
This is defined as the diffusion coefficient when the chemical potential gradient equals zero, or more simply, it is the diffusion coefficient in a medium of the same molecules.
According to Ref.1, the self-diffusion coefficient of water at $\pu{25^\circ C}$ is $D = \pu{2.3 \times 10^{−9} m^2 s^{−1}}$.  The diffusion time can be calculated as: $$t \approx \frac{x^2}{2D}   $$
As pointed out by @Karsten Theis, you can use the calculator and insert the water diffusion coefficient D.
For example, if the distance x is $\pu{8cm}$ like a typical glass of water and we are at $\pu{25 ^\circ C}$, then $t \approx \pu{16 days}$!
Instead if the temperature is $\pu{90 ^\circ C}$ then it's around $\pu{5 days}$.
I have drawn a graph that shows how much time to reach the top of the glass depending on temperature:

References:

Manfred Holz, Stefan R. Heila, Antonio Sacco, "Temperature-dependent self-diffusion coefficients of water and six selected molecular liquids for calibration in accurate $\ce{^1H}$ NMRPFG measurements," Phys. Chem. Chem. Phys. 2000, 2(20), 4740-4742 (https://doi.org/10.1039/B005319H).


What exactly does it mean for the reaction of superoxide with
  non-radicals to be spin-forbidden?

For clarity, let's start with a simpler case.  If we take a ground state singlet molecule (singlet means all electrons are paired) and we irradiate it with photons, photon absorption will excite an electron to an excited state.  Will the excited state be a singlet or triplet immediately after it is produced?  For the transition to the excited state to occur readily it must be spin-allowed.  That means that spin must be conserved.  If we started with a singlet, then our excited state must still be a singlet.  In that excited state the excited electron will still have its spin paired with the electron that was its partner that remains in a lower energy orbital.  Creation of a triplet state, where the excited electron and its ground state partner no longer have their spins paired is spin-forbidden and is therefor much less likely to occur.  Spin must be conserved.

Well rules are made to be broken.  What can we do to increase the probability of a singlet -> triplet conversion that is formally spin-forbidden. Spin-Orbit coupling (the coupling of the electron spin with the orbital angular momentum of non-circular orbits) is our way out. We need to find a spot where the Morse potential curves overlap for both the singlet and triplet. Then, if there is a large spin-orbit coupling interaction we can readily cross over from singlet to triplet, or back.

Large atoms such as bromine and iodine, as well as transition metals with unpaired electrons have large spin-orbit couplings.  If such groups are attached to our molecule of interest, then spin need not be conserved and formally "spin-forbidden" reactions become more likely.

  Does the unpaired one have to bond with an unpaired one from another
  radical?

Yes, if spin is to be conserved.

  What does this also say about reactions of the excited state
  of singlet oxygen (shown in the middle below) with non-radicals?

The singlet oxygen shown in the middle of your figure ia a singlet species and will react readily with other singlet (non-radical) molecules.

  is superoxide a doublet
  Yes superoxide is a doublet.  Ground state $\ce{O2}$ is a triplet - it has two unpaired electrons in energetically degenerate orbitals, if we add one electron to form superoxide radical anion, that electron will pair with one of those two electrons leaving only one unpaired electron.


As adjectives, not in a math context: 'different' vs 'differential'

"Differential" is a compact way of saying "difference between the" (or its equivalent).  For instance, in your first example, where the author writes "Chromatographic separations result from the differential interactions of the enantiomers...", what's meant is  "Chromatographic separations result from the difference between the interactions of the [two] enantiomers...".

I.e., chromatographic separation results from the fact that there is a difference between how the two different enantiomers react with the stationary phase (the solid column packing material).

The first definition given in your link applies here: "Of, showing, or depending on a difference; varying according to circumstances or relevant factors."


How can I be safe and protect myself using trimethyl tin chloride?

In addition to the safety precautions mentioned in the question, the following can increase the safety for the experimentator:

Keep the fume hood as closed as possible with the head outside and the arms inside. Usually this keeps fumes from entering the body, since the fume hood airflow goes the other direction.
Work quickly. While this not only minimizes side reactions of the reactant with the air and other things, it also ensures less exposure time, minimizing vaporization.
As Aesin mentioned in the comment, if you have concerns about the infrastructure, you could always perform the experiment in a glovebox.

In the end, you have to feel confident that the safety measures are effective. If you don't, you will have fear of the reaction, and inevitably be less concentrated and more prone to errors. In combination with "work quickly" this may well become a disaster.

Maintain a healthy portion of respect, keep quenching agents close by and execute the reagent transfer with a steady hand.