Pubchem: list all compounds for which Kovats retention indices are available

Ha NCBI helpdesk just replied with this:

The easiest way to do this is to use the PubChem Classification Browser to access the list of Compound records with Kovats data.

You can find the PubChem Classification Browser here:  https://pubchem.ncbi.nlm.nih.gov/classification

1) Click on the "Select classification" pull-down and select "PubChem Compound TOC".
2) Under "Data type counts to display" click on "Compound".
3) Scroll down to the TOC listing and expand the "Chemical and Physical Properties" section and then "Experimental Properties".
4) Look for the entry "Kovats Retention Index" with the number of counts next to it.
5) Click on the number of counts (currently 79,656) to take you to PubChem Compound results list populated with those records shown. (https://www.ncbi.nlm.nih.gov/pccompound?DbFrom=pchierarchy&Cmd=Link&Db=pccompound&LinkName=pchierarchy_pccompound&IdsFromResult=1856976)

To save the list of CIDs for these records there are two options....

1) Click "Send to" (near the top).
2) Select "File".
3) Select "UI List"
4) Click "Create file"

OR

Click "Structure Download" (on the right)& follow the directions....to download the SDF or XML formats of these records.

If you don't need the 3D versions of the records/images this will work fine, HOWEVER if you want the "3D Records/Images" versions of these - there is a limit of 50,000.  So you'd need to download the CID list and split it in two to upload in this tool for downloading the file.

I hope this helps.

NCBI User Services


Is converting SMARTS to SMILES a "lossless" operation?

SMARTS is deliberately designed to be a superset of SMILES. That is, any valid SMILES depiction should also be a valid SMARTS query, one that will retrieve the very structure that the SMILES string depicts.

However, as a query language, SMARTS can be more general than SMILES is. For example, CC as a SMILES string depicts a single compound: ethane. As a SMARTS query, though, CC will match ethane, but will also match propane, acetic acid, cyclohexane, vancomycin, etc.

There's also SMARTS strings which are not valid SMILES strings. You list several in your question: [#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8] is not a valid SMILES representation, unless your SMILES parser is being particularly generous. Even if it is, [#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8,#7] is also a valid SMARTS query which should match your molecule, but not even a generous SMILES parser is likely to accept it.

Despite being similar, SMARTS and SMILES are intended for fundamentally different things - the purpose of SMILES is to represent particular compounds, whereas SMARTS represents a query against a range of possible molecules, or an abstract description of a set of possible molecules. As such, they're not inter-convertible, even if the strings are literally identical.

For your particular case, yes, CC1CC=CC(=O)C1 is both a valid SMILES and a valid SMARTS, but as a SMARTS query, it represents not just 5-methyl-2-cyclohexen-1-one, but also 5-propyl-2-cyclohexen-1-one and 3-hydroxy-5-butyl-6-amino-2-cyclohexen-1-one, as well as many others, all of which contain that substructure. The SMARTS viewer you link doesn't depict this explicitly, because it's implicit in the use of SMARTS that it's a substructure pattern for a broader class of compounds.


When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group...
Why amino acid? Not carboxyl acid, or some mix of the two, or something else?

You stated why it is called an amino acid, the name comes from the amine (amino) and the carboxylic acid (carboxyl group).
Amines are weak bases, so the acid portion of the name must come from the carboxylic acid.


How to find the number of valence electrons?

'p' and 'd'.

There is one type of 's' orbital. It is spherical. So an s orbital can hold 2 electrons only.
There are three types of 'p' orbitals. They are the same shape (like a dumbell) but are oriented in different directions - along the x, y and z axes.  So a 'p' orbital can hold $2$x$3$ = 6 electrons.
There are five types of 'd' orbitals (10 electrons) and seven types of 'f' orbitals (14 electrons).

Fill up the orbitals in the following order (following the red arrows), which is increasing order of energy. Lowest energy fills up first.

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p

So let's take the example of sulfur.
The total number of electrons is 16. We'll distribute them in the following manner.

$1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons in the last (3rd) orbital. Therefore the valence number is 6.

It is not simply the last 2 numbers while writing the electronic configuration.

For Phosphorous, we have:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^3$, which gives us 5 electrons in outermost (3rd) orbit.

For Lithium (3 electrons), we have:
$1s^2, 2s^1$, which gives us a total of 3. Outermost orbit is the 2nd orbit and the total number of electrons in it is just one. Therefore valency of Lithium is one.

However, it is much simpler to do it the way I've shown above as with Sodium. Don't need to bring in all the orbitals

For the spin, just remember that you have to fill up the entire orbital with up spin electrons before you start putting the down spin electrons. Don't just alternate up and down.

Let's talk about the 'p' orbital. Each 'type' of the p-orbital must have up-spin electrons before adding the down-spin. So there will be 3 up spin electrons before you start adding the down-spin ones. Also, you can not move on to the next orbital without completely filling the previous one with an EQUAL number of up and down spins.

For Phosphorous; let's say we've filled the 1s, 2s, and 2p orbitals. Now we add one up spin to the 3s, followed by a down-spin. Which leaves us with 3 more electrons to be added to the valence orbit (3rd). It is a 'p'-orbital, which needs to have 3 up spin electrons before any downspin ones can be accommodated. But we have only 3 remaining! So all will be up-spin electrons in the valence orbit. So we'll have a total spin of +3/2.

Hope I didn't complicate too much, but I've actually tried to make it as short as possible. I've still left out quite a bit.


What eco-friendly substance can I use to write and erase on a wall for UV glowing without detection in daylight?

I think that in a wall you can't write and erase so easily. You can write and then re-paint the surface in this case you can use Zinc White (make sure that what you use is $ZnO$). It was a common white pigment before $TiO_2$ discovery and has a strong fluorescence under UV light so if your wall is white (best $TiO_2$ painted, $CaCO_3$ may have some fluorescence) you will not notice $ZnO$ area in visible light but only under UV light.


Does nitrogen pressure in tyres fluctuate to much a lower degree than air with changes in temperature?

There can be no scientific explanation, because the argument itself is incorrect.

For the most part air is nitrogen.  To be specific, about 80% of dry air is made up from nitrogen.  In fact both nitrogen and oxygen, the second major components of air that takes up almost all the other 20%, behave extremely like ideal gas.   Pressures of dry air and pure nitrogen gas at different temperatures, from far below freezing point to above boiling point, have indeed been measured to incredibly high accuracy.  The difference between these two gases is below 0.1%.   In fact, near room temperature, many types of gas are close to ideal gas and have very similar temperature profiles.

I am not sure why people are comparing nitrogen with water in that question thread you mentioned.  Water is a minor component of air and its concentration varies so dramatically across places.  However, even in a summer day, at the maximum humidity of 100%, the amount of water in the air is only 10% of that of oxygen.

Basically, pressure of nitrogen gas do not fluctuate less with change of temperature compared to normal air.


Where do the bubbles in a glass of water come from?

Gases are less soluble in warm water than cold water.

The bubbles are likely from dissolved gas coming out of solution as the water warmed.

The bubbles are unrelated to self-ionization of water.  The self-ionization does vary somewhat with pressure, so at an extreme depth in the ocean you might need to consider what Kw is, but not in a glass of water.

The solubility of gases increases with pressure, so on this basis it is more likely that you would have bubble appear near the top.  However, the pressure is only 1-2% greater at the bottom of the glass compare to the top.  There may be fewer nucleation sites (such as tiny scratches) near the bottom of that particular glass.  Also, if the bubbles move up along the wall sometimes, it would be logical that there are less at the bottom.


Repeating unit of poly lactic acid (PLA)

In fact, both pictures show the same polylactic acid, only different disconnection points have been chosen:

The red disconnection (or your picture 1) is preferred because you can immediately see that it is a polyester.

Edit:

Your teacher is right in that, during the polymerization, the $\ce{-COOH}$ group loses $\ce {-OH}$ and the $\ce{-OH}$ group loses $\ce {-H}$ to form $\ce{H2O}$.
Therefore, your teacher prefers this arrangement:

If we draw it the other way, it are the wrong $\ce{-OH}$ and $\ce{-H}$ at the end of the chain, in the sense that it are not the ones that will give water:

However, note also that in either case, the part between brackets is not formed by identical fragments.
The atoms are indeed the same, but since not all are connected to the same atoms, they are not all chemically equivalent. There are different functional groups.
There are acid, alcohol and ester groups; mainly ester of course.
So in reality, we are writing $\ce{[...]_n}$, but those $n$ are not all the same and thus it is not correct.

If you want to depict the whole chain including the polymer chain and the ends that can continue polymerizing, the most correct way is to draw the end monomers explicitly.
At that point, you can place the brackets where you want, since the structures are absolutely equivalent.


Why is the nomenclature glucose 6-phosphate, not 6-phosphateglucose?

Reading this closely, technically, the compound should be 6-phosphonoglucose but glucose 6- phosphate is acceptable usage, probably derived from considering it as a salt analogue rather the ester that it is.


What should the Rehm-Weller equation look like?

Both expressions are identical. It's just juggling of the signs, since

$$E^{\mathrm{red}}(D^{\cdot+}/D) = - E^{\mathrm{ox}}(D/D^{\cdot+})$$

Relevant primary references would be:

Rehm, D.; Weller, A. H. Isr. J. Chem. 1970, 8, 259-271.
Rehm, D.; Weller, A. H. Ber. Bunsen-Ges. Phys. Chem. 1969, 73,
834-839.
Weller, A. H. Z. Phys. Chem. NF 1982, 133, 93-98.

A normative glossary of terms in photochemistry is available from the IUPAC.


reaction between Iron(ii) and Vanadium(v)

I think I realize (a part of) the answer, concerning the conditions of a redox titration of vanadium(V), for which a metavanadate, such as $\ce{NH4VO3}$ is typically used as starting material. Under acidic conditions, i.e. in the presence of 85% $\ce{H3PO4}$, the colourless $\ce{VO2+}$ is formed.

$\ce{VO3- + 2 H+ -> VO2+ + H2O} $

Again under acidic conditions, the latter species is reduced by Fe(II) to the blue $\ce{VO^{2+}}$.

$\ce{VO2+ + Fe^{2+} + 2 H+ -> VO^{2+} + Fe^{3+} + H2O} $

Is is conceivable that the peroxodisulfate is used to oxidize the excess of Fe(II), which otherwise would be oxidized by the permanganate?


How many decimal places should a volume measurement in a beaker include?

It may have to do with the fact that it's a beaker, rather than some other kind of glassware.

First, beakers are poor vessels for volumetric measurements due to their large diameter/height ratios.  You have to add a lot of liquid, relatively speaking, in order to increase the height of the fill level by (say) a millimeter, and that means the uncertainty in the fill volume is large.

As well, beakers are usually not graduated precisely. Thus, even if you very very carefully fill the beaker to exactly the $15.5~\mathrm{mL}$ line with liquid, there's a very real chance that the line doesn't actually correspond exactly to a $\mathit{15.5~mL}$ fill volume.

So, unlike with a piece of glassware specifically made for accurate volumetric measurements, such as the burette mentioned by MaxW in a comment, it is likely not appropriate to assign more significant figures to a beaker measurement than are "available" in the graduation increment.


What constituents in soft drinks are the cause their strong pungent smell?

Yes, that strong smell is mostly $\ce{CO2}$, possibly with a faint tint of volatile aromatizers from the beverage.

Our sense of smell works in surprisingly non-straightforward ways. Indeed, the smells of the same compound in high and low concentrations are sometimes perceived as quite different, to the point of being totally unrecognizable. A classic example is skatole, which may produce either a pleasant flowery aroma or a fecal stench, depending on the concentration alone. Looks like $\ce{CO2}$ is just a somewhat less extreme example of the same effect.

Air with 5% of $\ce{CO2}$ doesn't smell much of anything (not that it is OK to breathe it for long), which leads most people to conclude that $\ce{CO2}$ is odorless. This is quite wrong. If you try to inhale some pure $\ce{CO2}$ (which I do not recommend or endorse), that's when the smell kicks in. Your breathing just stops instinctively, as if you were trying to inhale water. By that moment, though, some $\ce{CO2}$ reaches your olfactory nerves. I am at a loss of words as to what it feels like. Being slapped across your face with a shovel seems like a moderately adequate approximation. Besides, you feel that familiar refreshing sharp odor of an open Coca-cola bottle.

So no, this is not due to some mysterious additive.


Are `dTDP-4-dehydro-6-deoxy-alpha-D-galactose` and `dTDP-4-keto-6-deoxy-D-glucose` the same thing?

There are four stereocentres in hexoaldoses giving rise to a set of $2^4 = 16$ different molecules with the same connectivity; you can group these to eight pairs of enantiomers (perfect mirror images of each other). The enantiomers are labelled D and L depending on the orientation of the stereocentre which is farthest from the aldehyde group. In linear orientation, these are the following molecules:

(All images taken from Wikipedia; all D-isomers.)

In the order of reading these are allose, altrose, glucose, mannose, gulose, idose, galactose and talose. If you compare glucose (3rd in the first row) with galactose (3rd in the second row) you will realise that they differ only in the orientation of the hydroxy group in 4-position. The posh terminology is epimerism — D-glucose is the C4 epimer of D-galactose and vice-versa.

You asked about two labels where the sugar is preceded either by 4-dehydro-6-deoxy or 4-keto-6-deoxy. The agree in the 6-deoxy part which means that the oxygen on C6 is missing; you will find a methyl group $\ce{CH3}$ there instead. The labels 4-keto and 4-dehydro may also be confusing. The first says that a keto group is there instead of the hydroxy group, while the second says that two hydrogens have been removed from that position. The result is the same, instead of having $\ce{CHOH}$ as fourth carbon we have $\ce{C=O}$. This also deprives us of the stereocentre (a carbonyl group is never stereogenic).

Putting all of this together: Yes, the two are identical. If you start from the C4-epimers and remove the stereocentre at C4 it does not matter which you started from. Both galactose and glucose are common sugars so both choices are understandable (even though glucose is probably known even a tad more). I would, however, strongly object agains 4-keto-D-talose in the place of 4-keto-D-mannose (mannose and talose being C4-epimers) since mannose is much more known than talose.


High end vodka is primarily good vodka+branding. You pay extra to get that designed bottle. Perhaps the manufacturer also does something extra that they claim makes the vodka special. Whether it actually does depends on what they have done.

Good vodka has most flavours removed. Some flavours - specifically those that both have a boiling point close to that of alcohol and the ability to pass through an active carbon filter will remain.


What pressure does liquid nitrous oxide need at −45 °C?

Using NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9, I found the saturation point of nitrous oxide at equilibrium for a temperature of $T=-45.000\ \mathrm{^\circ C}$ at a pressure of $p=783110\ \mathrm{Pa}=7.8311\ \mathrm{bar}$;
for $T=-78.000\ \mathrm{^\circ C}$, it is $p=182290\ \mathrm{Pa}=1.8229\ \mathrm{bar}$.


Procedure to assign the irreducible representation of a derivative operator

The transformation of a derivative operator under a linear transformation $\hat R $ is given by
$$
\frac{\partial}{\partial x_i} \rightarrow \sum_j R_{ji} \frac{\partial}{\partial x_j} = \sum R^T_{ij}\frac{\partial}{\partial  x_j}
$$
where $R_{nm}$ are the matrix elements of the matrix representation of the transformation operator in regard to cartesian coordiantes. The matrix is defined as the coordinate transformation matrix that takes the initial coordiantes to the transformed coordinates.
$$
\hat R : x \rightarrow  Rx
$$
The character of an representation with the derivative operators as basis is therefore the same as the character of the corresponding coordinate representation, since the trace of a matrix and a matrix transpose are identical. The derivative operators form thus a basis for the same irreducible representation as their coordinates and belong to the same irreducible representation.


A monocyclic 6 carbon ring with 6 double bonds

There are several monocyclic $\ce{C6}$ isomers.  Three of them are pictured below

A - the compound you've drawn above, cyclohexahexaene
B - cyclohexatriyne
C - cyclohexa-2,5-diyne-bis-1,4-ylidene

While all of the carbons in these molecules are (more or less) $\ce{sp}$ hybridized, the molecules are different from one another due to differences in bond lengths and bond angles.  For example, all of the bonds in A are exactly the same length due to symmetry, whereas the triple bonds in B will be shorter than the adjacent single bonds.  Therefore, A-C are isomers, not resonance structures.

Ideally the atoms attached to an $\ce{sp}$ hybridized carbon form a linear arrangement (180° bond angle) with the central $\ce{sp}$ carbon, acetylene and the central carbon in allene serve as examples.

(image source)

(image source)

However in the cyclic $\ce{C6}$ compounds A-C, the $\ce{sp}$ hybridized carbons are forced by the ring to deviate from the desired 180° arrangement to something down around 120°.  Therefore, quite a bit of angle strain is incorporated into A-C making them extremely strained and unavailable through laboratory preparation to date.

As mentioned above, A-C have different bond lengths and bond angles therefore they are isomers, not resonance structures.  A fascinating question is,

Do A-C each sit in a potential well, separated from one another by an energy barrier and if so, which one is the most stable (energy minimum)?

or

Is there another monocyclic species lower in energy than A-C? Perhaps a molecule that could be described by resonance structures A'-C', where the prime denotes that all of the bond lengths and angles have been adjusted and made equal in A, B and C.


From Second Law of Thermodynamics viz. $$\Delta S_\textrm{sys}+ \Delta S_\textrm{res}\geq 0\,,$$ it can be concluded that $$W_\textrm{sys}\leq (U_1- U_2)- T(S_1 -S_2) \tag{1}\;.$$

Now, $$W_\textrm{sys}= \underbrace{W_T}_\textrm{pressure-volume/compression work}+ \underbrace{A_T}_\textrm{non-compression work}\;.$$

Putting the the first term on the RHS of $(1)\,,$ it yields,

\begin{align}A_T &\leq (U_1- U_2)- T(S_1 -S_2) - W_T\\ &\leq (U_1- U_2)- T(S_1 -S_2)+ P(V_1 -V_2)\;.\end{align}

Now, this is equal to the negative change of Gibbs energy $G$ provided the thermodynamic process is done at constant pressure and temperature.

So, that means,  $$A_T \leq G_1 -G_2\;.$$ Note the $\leq$ sign; this implies $(G_1-G_2)$ is the maximum value $A_T$ can attain.

Since, $\Delta G \lt 0\;,$ this implies $A_T$ is positive. Now, if the process is reversible, then the non-compression work $A_T$ is maximum viz: $$A_T= G_1 -G_2\;.$$

This definitely supports the term free energy as the decrease in $G$ gives the amount of energy "that can be 'freed' and made available for work".

  That is, more energy is released by the system than the enthalpy of reaction.

Of course, the extra energy came from heat energy absorbed from the environment as the entropy of the products was greater than that of the reactants by $\Delta S\;.$


Think about the pKa of the carboxylic acid protons compared to the pKa of the enolisable proton (hint: the acid is more acidic).

Once the carboxylic acid is turned into the carboxylate, enolisation is not possible, as the carbonyl cannot simultaneously conjugate with the carbon and the oxygen.

The use of phosphorus tribromide is known as the Hell-Vollhard-Zelinski reaction, it allows carboxylic acids to be directly brominated at the alpha position without initial protection of the carboxylic acid via in situ formation of the acyl bromide which hydrolyses upon workup.


Boiling point elevation with multiple solutes

When there is more than $1$ solute in the solution, the total elevation in boiling point ($\Delta T_\mathrm{b,total}$) or the depression in freezing point ($\Delta T_\mathrm{f,total}$) is defined as the sum of each individual value for each type of solute.
What is Van't Hoff factor? According to Chem Libretexts, the Van't Hoff factor is defined as:

$$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \tag{1} \label{1}$$

Now, why does this work? Let's take an example to see how it works.
We take a simple case as what you have mentioned in the question:

Find the elevation in boiling point of a solution of $\pu{1 m}$ glucose and $\pu{2 m}$ $\ce{NaCl}$ [$100$% dissociated] in water.
$\mathrm k_\mathrm {b,water}= \pu{0.52 ^\circ C kg mol-1}$

We can do this in two ways. One, where we take $i = 2$ for $\ce{NaCl}$. Second, we state that $\ce{NaCl}$ dissociates into $\ce{Na+}$ and $\ce{Cl-}$ and take the three types of solutes to be present instead of two [going from basic principles].
Method $1$
Here we would have two different solute particles. Finding the value of $\Delta T_\mathrm{b}$ for both glucose and $\ce{NaCl}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,NaCl} \\
&= \mathrm{k_b}[1\cdot1 \,+2\cdot2] \\
&= 5\cdot\mathrm{k_b} \\
\end{align}
Method $2$
Here we would have three different solute particles. Here, $\ce{Na+}$and $\ce{Cl-}$ would have a value of $i=1$ and their individual molality would be identical to that of $\ce{NaCl}$ since the salt was $100$% dissociated in solution
Finding the value of $\Delta T_\mathrm{b}$ for glucose, $\ce{Na+}$ and $\ce{Cl-}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,Cl^-} +\Delta T_\mathrm{b,Na^+} \\
&= \mathrm{k_b}[1\cdot1 \,+1\cdot2 + 1\cdot2] \\
&= 5\cdot\mathrm{k_b} \\
\end{align}
Therefore, both methods yield the same answer. The second method is actually the derivation for value of $i$ for a dissociating salt.
Now, you can apply $i_\text{total}=i_1+i_2$ only when the concentrations of the given solutes are the same. The proof is similar to how the second method proves the first above. In such a scenario, you can assume them to be one solute since colligative properties don't care about how many types of solute there are but rather care only about the amount of solute particles present in the solution.


I think your notation of "Changing one atropisomer to another requires bond breaking (in some cases the removal and reattachement of steric groups, according to my understanding)" is not correct. To my knowledge, the free rotation of one single bond of an atropisomer is not restricted by connecting bond(s). I think you get confused by some literature explaining breaking a bond selectively directed to pair of atropisomers (e.g., Ref.1):

According to your post, IUPAC definition of atropisomers:

Atropisomers - A subclass of conformers which can be isolated as separate chemical species and which arise from restricted rotation about a single bond.

A variety of sources confirm this definition. For instance, Ref.2 and 3 described atropisomers as:

Atropisomers are stereoisomers (rotamers) resulting from hindered rotation about single bonds (having an energy barrier to rotation about a single $\sigma$ bond) usually due to steric hindrance. This energy barrier to rotation is high enough to allow the isolation of the conformers (rotamers).

Nonetheless, the name, coined by German biochemist Richard Kuhn in application to a theoretical concept 1n 1933 (Wikipedia) is derived from Greek (a = not and tropos = turn; thus, atropos meaning "without turn"), which supported the definition. However, the atropisomerism was first detected in 6,6'-dinitro-2,2'-diphenic acid by Cristie, et al. in 1922 (Ref.4):

It is also important to know the nomenclature of atropisomers (Wikipedia):

Determining the axial stereochemistry of biaryl atropisomers can be accomplished through the use of a Newman projection along the axis of hindered rotation. The ortho, and in some cases meta substituents are first assigned priority based on Cahn–Ingold–Prelog priority rules (Ref.5 & Ref.6). One scheme of nomenclature in based on envisioning the helicity defined by these groups (Ref.7). Starting with the substituent of highest priority in the closest ring and moving along the shortest path to the substituent of highest priority in the other ring, the absolute configuration is assigned $P$ or $\Delta$ for clockwise and $M$ or $\Lambda$ for counterclockwise. Alternately, all four groups can be ranked by Cahn–Ingold–Prelog priority rules, with overall priority given to the groups on the "front" atom of the Newman projection. The two configurations are termed $R_\mathrm{a}$ and $S_\mathrm{a}$ in analogy to the traditional $R/S$ for a traditional tetrahedral stereocenter (Ref.8}:

Atropisomers are also found in nature. The main pigment of cotton seeds, gossypol is an atropisomer, which exists in both form of $(R_\mathrm{a})$- and $(S_\mathrm{a})$-conformations. Most commercial Upland (Gossypium hirsutum) cottonseeds have an $(R_\mathrm{a})$- to $(S_\mathrm{a})$-gossypol ratio of ~$2:3$, but some Pima (Gossypium barbadense) seeds have an excess of $(R_\mathrm{a})$-gossypol. Between two isomers, $(R_\mathrm{a})$-gossypol is more toxic and exhibits significantly greater anticancer activity than its $(S_\mathrm{a})$-atropisomer:

Two fun facts about atropisomers are (KU.edu):

Atropisomers are detectable by $\mathrm{NMR}$ if half lives exceed $\pu{10^{-2} s}$.
Atropisomers are isolatable if the half-life is above $\pu{10^3 s}$.

References:

Gerhard Bringmann, Thomas Hartung, “Atropo-enantioselective biaryl synthesis by stereocontrolled cleavage of configuratively labile lactone-bridged precursors using chiral H-nucleophiles,” Tetrahedron 1993, 49(36), 7891-7902 (https://doi.org/10.1016/S0040-4020(01)88014-5).
Alan R. Katritzky, Christopher A. Ramsden, John A. Joule, Viktor V. Zhdankin, In Handbook of Heterocyclic Chemistry, Third Edition; Elsevier Limited: Amsterdam, The Netherlands, 2010 (ISBN: 978-0-08-095843-9).
Jonathan Clayden,  “Atropisomerism,” Tetrahedron 2004, 60(20), 4335 (https://doi.org/10.1016/j.tet.2004.03.002).
George Hallatt Christie, James Kenner, "LXXI.—The molecular configurations of polynuclear aromatic compounds. Part I. The resolution of $\gamma$-6 : 6′-dinitro- and 4 : 6 : 4′ : 6′-tetranitro-diphenic acids into optically active components," Journal of the Chemical Society, Transactions 1922, 121, 614–620 (https://doi.org/10.1039/CT9222100614).
R. S. Cahn, Christopher Ingold, V. Prelog, “Specification of Molecular Chirality,”  Angew. Chem. Internat. Ed. Engl. 1966, 5(4), 385-415 (https://doi.org/10.1002/anie.196603851) and Corrigendum: Angew. Chem. Internat. Ed. Engl. 1966, 5(5), 511-511 (https://doi.org/10.1002/anie.196605111).
G. P. Moss, "Basic terminology of stereochemistry: IUPAC Recomendations 1996," Pure and Applied Chemistry 1996, 68(12), 2193-2222 ()(PDF).
http://goldbook.iupac.org/terms/view/H02763: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
http://goldbook.iupac.org/terms/view/A00547: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


Radicals vs atoms having unpaired electrons

Yes, a chlorine atom is very much the same as chlorine radical.

With sodium, the two things are technically similar too, but you'll hardly ever encounter a mention of "sodium radical".

Why is that? Well, I guess there is a difference between Na and Cl. A chlorine radical exists in its own right (if only briefly), and is an important intermediate in many reactions. A sodium radical, on the other hand, is pretty much not a thing. Single sodium atoms just don't wander around alone. You may see reactions involving metallic sodium, or ionic salts of sodium, or maybe (not sure about that) some compounds which can pass for covalent, or (unlikely) even $\rm\bf Na^{\pmb-}$... but that would be all.

So it goes.


transported electrons in $\ce{NADH}$ have a higher energy than those in $\ce{FADH2}.$

This is jargon describing the redox potential of the electron carrier $\ce{NADH}/\ce{NAD+}$ vs the electron carrier $\ce{FADH2}/\ce{FAD}.$ So if you compare the two reduction half reactions
$$
\begin{align}
\ce{NAD+ + H+ + 2e- &-> NADH} &\quad E^{\circ '}_\mathrm{red} &= \pu{-0.320 V}\tag{1}\\
\ce{FAD + 2H+ + 2e- &-> FADH2} &\quad E^{\circ '}_\mathrm{red} &= \pu{0.031 V}\tag{2}
\end{align}
$$
$\ce{NADH}$ is the stronger reducing agent.

Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

Another way of saying this is that the reaction of $\ce{NADH}$ with dioxygen is more exergonic (the equilibrium lies further on the side of the products, more free energy is available from it) than the reaction of $\ce{FADH2}$ with dioxygen. For completeness, the reduction half reaction for dioxygen/water is:
$$\ce{1/2 O2 + 2 e- + 2H+ -> H2O} \quad E^{\circ '}_\mathrm{red} = \pu{0.816 V}\tag{3}$$

Do they mean that when these electrons are transferred to the electron carriers they are part of molecular structures in which they sit in high energy positions?

It has to do with how tightly the electrons are bound in one and the other electron carrier. The OP is correct to place the term "high-energy" in quotes when referring to electrons; electrons are indistinguishable, so their energy is with respect to the bound state that they are in.


Proof of buffers inability to maintain exact same pH

A buffer is a solution that can resist pH change upon the addition of
  an acidic or basic components. It is able to neutralize small amounts
  of added acid or base, thus maintaining the pH of the solution
  relatively stable.

Buffers do not maintain exact same pH, but keep it relatively stable i.e. close to the original value.
log(ratio) << pKa, therefore, pH changes will be minimal, unless substantial amount of acid has been added to the solution.

For instance, when the ratio between the conjugate base/acid is equal to 1, the pH = pKa.

If the ratio between the two is 0.10, the pH drops by 1 unit from pKa since log (0.10) = -1.

If a ratio increases to a value of 10, then the pH increases by 1 unit since log (10) = 1.

As you can see, changing ratio by 10 times, affects pH value only by 1 unit.


It is non-aromatic even though it has 4$\pi$ electrons because nitrogen cannot form five bonds. Hence delocalisation is not possible and so it is non aromatic.


Why aren't chaperones considered catalysts?

I don’t think you are wrong. Actually molecular chaperones have been named enzymes in some texts. The molecular chaperones comprise several unrelated classes of proteins that have rather different functions.

  Most molecular chaperones are ATPases (enzymes that catalyze ATP
  hydrolysis), which bind to unfolded polypeptides and apply the free
  energy of ATP hydrolysis to effect their release in a favourable
  manner.

It is also made evident in the way some of them function (they possess some ATPase activity), it suggests they’re kind of  catalysts  enzymes:

  They function in an ATP-driven process to reverse the denaturation and
  aggregation of proteins (processes that are accelerated at elevated
  temperatures), to facilitate the proper folding of newly synthesized
  polypeptides as they emerge from the ribosome, to unfold proteins in
  preparation for their transport through membranes.

Two classes of molecular chaperones have been well studied. Both are found in organisms ranging from bacteria to humans.

The heat shock proteins 70 (Hsp70)

The first class is a family of proteins called Hsp70. They function in an ATP-driven process to reverse the denaturation and aggregation of proteins (processes that are accelerated at elevated temperatures), to facilitate the proper folding of newly synthesized polypeptides as they emerge from the ribosome, to unfold proteins in preparation for their transport through membranes and to subsequently help them refold.

Chaperonins

Are also heat shock proteins that form large, multisubunit, cagelike assemblies that are universal components of prokaryotes and eukaryotes. They bind improperly folded globular proteins via their exposed hydrophobic surfaces and then, in an ATP-driven process, induce the protein to fold while enveloping it in an internal cavity, thereby protecting the folding protein from nonspecific aggregation with other unfolded proteins.

Other chaperones e.g. Hsp90 proteins (have a regulatory role in that they induce conformational changes in nativelike substrate proteins that result in their activation or stabilization) and Trigger factor (prevents the intra- and intermolecular aggregation of newly synthesized polypeptides as they emerge from the ribosome by shielding their hydrophobic segments) However, unlike most other chaperones, trigger factor does not bind ATP.

Enzymes are more or less catalysts in biological systems, so yes its right to synonimise them with catalysts.

References

Biochemistry (Voet and Voet) Section 9.2 Folding Accessory Proteins, Molecular Chaperones: The GroEL/ES System.
Lehninger Principle of Biochemistry Chapter 4, Protein Denaturation and Folding.
Biochemistry R. Garret, C. Grisham. Chapter 31 Completing the Protein Life Cycle: Folding, Processing, and Degradation


Why does everything go in the direction of positive entropy change?

This answer us quite long, but at each stage should be straight forward and will hopefully show you exactly where the second law comes from in classical thermodynamics. If you are going to spend the time learning something you might as well do it properly right?

Introduction

The first law of thermodynamics tells us that in any process energy is conserved, it can be transformed and converted into different forms, but the total amount is unchanged.

The second law imposes limits on the efficiency of energy conversion processes. For a robust understanding of the second law it is necessary first to understand Carnot cycles. As such I will divide my answer into sections. Many of the points in the following are intricate in their application and a good knowledge of the first law is required. At the time classical thermodynamics was developed the main interest of the research was into the efficiency of heat engines. Historically these are quite interesting, but that is why you need to understand a Carnot cycle first which in essence is just a really efficient engine (the most efficient).

Keep in mind that in any heat engine a supply of heat $Q$ can either be converted to work $W$ or internal energy $\Delta U$. Since an isothermal process means $\Delta U=0$ then the most efficient cycle will have two isotherms. However two isotherms do not constitute a cycle. The most efficient way to connect these two isotherms is with adiabatic processes (no heat exchange with the surroundings). With two adiabatic lines we finally have our cycle. Graphically to maximise the work you want to maximise the area under the cycle.

By cycle we mean that a system returns to exactly the same state as before the change or process occurred.

Carnot Cycles

Work can be obtained from an engine of there are heat sources at different temperatures, and heat will flow from a hot body to a cold body without work being performed until thermal equilibrium is attained. Carnot realised therefore that any temperature difference either produces work, or is wasted as spontaneous flows of heat.

Lets use an ideal gas as our substance. The figure shows a Carnot cycle. It is reversible and has two reservoirs. Where $ab$ is an isotherm at temperature $T_1$ and heat $Q_1$ enters from a reservoir at $T_1$. The line $cd$ is an isotherm at a lower temperature $T_2$ where heat is given to another reservoir at $T_2$. The line $bc$ and $da$ are adiabatic processes. The work done $W$ is the area enclosed by the cycle.

We might choose to represent the Carnot engine according to the figure below. Where $W$ is the work done by the engine. The efficiency $\eta$ is a measure of "what we get out for what we put in" so in this case the work output for the heat input.
\begin{equation}
\eta =\frac {W}{Q_1}
\end{equation}
If we remember the first law and that $\Delta U=0$ then we may write $W=Q_1-Q_2$ denoting the sign convention for heat into the system and work done on the system to be positive and vice versa.
\begin{equation}
\eta =1-\frac{Q_2}{Q_1}
\end{equation}
Kelvin-Planck statment

It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at a uniform temperature and the performance of an equivalent amount of work.

This basically says that some heat needs to be given to a body of colder temperature otherwise by the above equation we would have 100% efficiency. If you were to violate this you could just have perpetual motion machines or cars that just absorbed heat and produced work. It may be summarised as a process whose only effect is the complete conversion of heat into work is impossible. However the opposite process is allowed  where all the work is transferred to heat completely

The Clausius statement

*It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body. *

What this means is that work must be performed if heat is to be transferred form a colder to a hotter body. We might be more familiar with this statement if we called this engine a fridge. An engine that extracts heat from a cold body and delivers it to a hot body when work is performed.

The kelvin-Planck statement is entirely equivalent to the Clausius statement. We will skip any proof or mathematics of why this is so.

Carnot's Theorem

No engine operating between two reservoirs can be more efficient that a Carnot engine operating between those same two reservoirs.

The proof is as follows. Assume that such an engine did exist $E'$ with efficiency $\eta '$ that takes heat $Q_1'$ from the hot body and does work $W'$ delivering heat $Q_2'=(Q_1'-W')$ to the cold bath.

Let us now imagine a Carnot engine $C$ with heats $Q_1$ and $Q_2$, efficiency $\eta$ adjusted in such a way that it performs the exact amount of work $W$ as the "more efficient engine $E'$ so that $W=W'$, and $Q_2=Q_1-W$.

Iff,
\begin{equation}
\frac{W'}{Q_1'}>\frac{W}{Q_1}\ \ \ \ \ \ \ \ \ then \ \ \ \ \ \ \ \ Q_1>Q_1'
\end{equation}
The Carnot engine is reversible so it may be driven backwards, acting as a composite device with $E'$. It extracts positive heat $(Q_1-Q_1')$ form the cold reservoir and delivers the same heat to the hot reservoir with no external work being required.

However since the reservoir is large the temperature is unchanged with the addition of this heat. This means that the Clausius statement is violated. Therefore $\eta'>\eta $ is not true, and $E'$ can never exist. If $\eta '=\eta$ then $Q_1'=Q_1$ and no net heat transferred for no work. Therefore we say,
\begin{equation}
\eta' \leq \eta
\end{equation}
Entropy

Consider a substance that can do work under going a cycle so that at the end of its cycle its state is unchanged at $T_1$. Lets say we change the first state to infinitesimally close neighbouring state 2 at $T_2$ by adding a small amount of heat $\delta Q_1$ from . If we do this with a Carnot engine with the temperature baths of state 1 $T_1$ and some reservoir at $\bar T$ then $T_1$ supplies $\delta Q_q$ to the working substance while the Carnot engine supplies the same amount to the $T_1$ reservoir to leave it unchanged, in turn taking heat $\{\bar T/T_1\}\delta Q_1$ from a principle reservoir.

So in short we have performed a change from states 1 to 2 with external work $\delta W_1$ and  extraction of heat $\{\bar T/T_1\}\delta Q_1$ from the principle reservoir. This is one system of a composite process of identical systems. The diagram really helps here!

The composite system is assumed that $\Delta U=0$ and that the heat supplied to it is;
\begin{equation}
Q=\sum _i\frac{\bar T}{T_i}\delta Q_i
\end{equation}
Where the sum is over all the little systems making up the composite. The external work performed is;
\begin{equation}
W=\sum _i\delta W_i
\end{equation}
The first law is then $0=Q-W$. Looking at the second diagram below for the composite system we can see that this violates the Kelvin-Planck statement. The only way that this works is if work is done on the system and an equal quantity of work flows out. Or it works if both $W$ and $Q$ are zero.
\begin{equation}
W=Q\leq 0
\end{equation}
So
\begin{equation}
\bar T\sum _i\frac{\delta Q_i}{T_i}\leq 0 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \sum _i \frac{\delta Q_i}{T_i}\leq 0
\end{equation}
If you take the limit of this then over one complete cycle we arrive at the Clausius inequality!
\begin{equation}
\oint \frac{dQ}{T}\leq 0
\end{equation}
It is important to know that the $T$ appearing in the integral is the temperature of the heat supplying bath!

If the cycle is reversible then there is no reason why we couldn't do the proof in the other direction and hence obtain the inequality sign the other way around. This only holds for reversible cycles. The only way to satisfy both inequalities physically is if for reversible cycles we equal the integral to zero exactly. This means that for reversible cycles the integral is equal to zero!

The final step is to now use this Clausius inequality to prove the second law. Imagine a cycle that is completely reversible between two states $i$ and $f$ with two reversible paths $r_1$ and $r_2$ connecting them.
\begin{equation}
\oint \frac{dQ_r}{T}=_{r_1}\int^{f}_{i} \frac{dQ_r}{T}+_{r_2}\int_{f}^{i}\frac{dQ_r}{T}=0
\end{equation}
Thus,
\begin{equation}
_{r_1}\int ^{f}_{i}\frac{dQ_r}{T}=_{r_2}\int^{f}_{i}\frac{dQ_r}{T}\end{equation}
What all this means is that the integral is path independent SO THERE MUST BE A PATH INDEPENDENT FUNCTION !! We call this state function entropy.
\begin{equation}
S_f-S_i=_r\int ^f_i \frac{dQ_r}{T}
\end{equation}
For this equality to hold the heat exchanged must be defined over a reversible path! That is to say if the entire cycle is not reversible then we have to use the Clausius stamen again. Lets suppose that one off the paths is irreversible now.
\begin{equation}
 \int^f_i\frac{dQ}{T_0}+_r\int ^i_f\frac{dQ_r}{T}\leq 0
\end{equation}
 With a bit of interval switching,
\begin{equation}
\int ^f_i\frac{DQ}{T_0}\leq -_r\int ^i_f\frac{dQ_r}{T}=_r\int ^f_{i}\frac{dQ_r}{T}=S_f-S_i
\end{equation}
And finally the way we normally write this is in an infinitesimal form,
\begin{equation}
\frac{dQ}{T_0}\leq dS
\end{equation}
This is more general than before since it includes the previous equation too, equality holds if the cycle is reversible. So , in an infinitesimal irreversible process between a pair of states the entropy change is $dS$, if the system is thermally isolated then $dQ=0$ and we generate the principle of increasing or "maximum" entropy.
\begin{equation}
S_f-S_i \geq 0
\end{equation}
Or if you prefer $dS\geq 0$. The entropy of a thermally isolated system increases in any irreversible process and is unaltered in a reversible one.


Why does carbon monoxide have a greater affinity for hemoglobin than oxygen?

Excursion into simple coordination chemistry: Bonding, backbonding and simple orbital schemes

Please refer to Breaking Bioinformatic’s answer for the MO scheme of carbon monoxide, it is very helpful. You might also look at the orbital pictures in this answer by Martin. Carbon monoxide can bind to metal centres via a σ coordinative bond where the HOMO of $\ce{CO}$ interacts with metal orbitals and also by the π backbonding, Breaking Bioinformatics mentioned. I’ll start by touching the σ bond so we can later better understand the π bond. In figure 1 you can see the molecular orbital scheme of a complex composed of a central metal ion and six ligands that donate in a σ manner exclusively.

Figure 1: Molecular orbital scheme of an octahedral complex with six σ donors around a central metal. Copied from this site and first used in this answer of mine. Metal orbitals are 3d, 4s, 4p from bottom to top; ligand orbitals are of s-type.

You will notice that figure 1 contains the irreducible representations of the orbitals beneath them. Orbitals can only interact if they have identical irreducible representations; otherwise, their interactions will sum up to zero. We can see that the metal d orbitals are split up into a $\mathrm{t_{2g}}$ set and and $\mathrm{e_g}^*$ set. This is the effect of σ bonding and why it stabilises the entire entity.

π bonding can only occur if the ligands have available orbitals of π symmetry. The 2π orbitals of $\ce{CO}$ are a nice example, but one could also simply assume a halide with its p-orbitals for the same effect. Further down in the internet scriptum I originally copied the picture from, you can see a set of two pictures that introduce p-orbitals. Twelve ligand p-type orbitals will transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, thus a further stabilising/destabilising interaction is introduced for the $\mathrm{t_{2g}}$ orbitals. Due to the $2\unicode[Times]{x3c0}^*(\ce{CO})$ orbitals being similar in energy to the metal’s $\mathrm{t_{2g}}$ and empty, they can stabilise the former well, creating a possibly strong stabilisation for the overall system. Since this is an interaction of two orbitals creating both bonding and antibonding orbitals, and since the resulting molecular orbital is filled with two electrons that stem from the metal centre this is termed π-backbonding.

The binding mode of carbon monoxide to haemoglobin

After this extensive background discussion, it may be clear that $\ce{CO}$ is generally a good ligand. For reasons I didn’t go into, a carbonyl complex of metal ions isn’t that stable, but a single carbonyl ligand will almost always be beneficial. Such is the case for the porphyrin-iron(II) system that forms the heart of haemoglobin: The central iron(II) ion is coordinated well from five directions (four from the porphyrin ring and one histidine of haemoglobin) and has a weakly bound water atom in the ground state in its sixth coordination slot, sometimes displaced by a distal histidine. Carbon monoxide can diffuse in and bind very well to this system, displacing the weakly bound water and histidine. In fact, isolated haem can bind carbon monoxide $10^5$ times better that it can oxygen.

This is something that nature could not really inhibit, because it is based on fundamental properties, but also something nature didn’t really care for, because the natural concentration of carbon monoxide is very low, and nature rarely had to deal with it competitively inhibiting oxygen binding to haemoglobin.

The binding mode of oxygen to haemoglobin

Oxygen, the $\ce{O2}$ molecule, is an absolutely lowsy σ donor — especially when compared to $\ce{CO}$. Its molecular orbital scheme is generally that of carbon monoxide except that it is entirely symmetric and two more electrons are included: They populate the 2π orbitals to give a triplet ground state. These orbitals are now the HOMOs and they hardly extend into space in any meaningful way — plus the still existant lone pair on each oxygen atoms is now at a much lower energy and also does not extend into space and thus cannot bind to a metal centre in a σ manner.

What happens here is rather complex, and the last lecture I heard on the topic basically said that final conclusive evidence has not yet been provided. Orthocresol discusses the different viewpoints in detail in this question. The diamagnetic properties of the resulting complex are, however, unquestioned and thus one must assume a singlet ground state or one where antiferromagnetic coupling cancels any spins at molecular levels. Since the ground state of haemoglobin has a high-spin iron(II) centre and the ground state of oxygen is a paramagnetic triplet, it makes sense to assume those two to be the initial competitors.

Professor Klüfers now states the steps to be the following:

  Iron(II) is in a high-spin state with four unpaired electrons;
  Paramagnetic triplet oxygen approaches and one of its $\unicode[Times]{x3c0}^*$ orbitals coordinates to the iron(II) centre in a σ fashion.
  This induces a high-spin low-spin transition on iron and slightly reorganises the ligand sphere (pulling oxygen closer to the iron centre). We now have a low-spin singlet iron(II) centre and a coordinative σ half-bond from oxygen. We can attribute that electron to the iron centre.
  Through linear combination, we can adjust the spin-carrying σ orbital and the populated iron orbitals so that an orbital is generated which can interact with the other $\unicode[Times]{x3c0}^*$ of oxygen in a π fashion.
  We thus expect an antiferromagnetic coupling and an overall state that can be described best as $\ce{Fe^{III}-^2O2^{.-}}$ — a formal one-electron oxidation of iron to iron(III), reducing oxygen to superoxide ($\ce{O2-}$).
  If the spin-carrying orbitals are all treated as being metal-centred, we gain a $\ce{Fe^{II}-^1O2}$ state.

(Content in the link I quoted from are in German; translation mine and shortened from the original.)

Only due to the complexly tuned ligand sphere and also due to the stabilising high-spin low-spin transition (plus reorganising) is oxygen able to bind to iron at all. The distal histidine further stabilises the complex by a hydrogen bond, alleviating the charge slightly. It is to be assumed that nature did a great deal of tuning throughout the evolution since the entire process is rather complex and well-adjusted for collecting oxygen where it is plentiful (in the lungs) and liberating it in tissue where it is scarce.

Comparison of the oxygen and carbon monoxide binding modes

The simpler picture I drew above for carbon monoxide is not correct. Inside haemoglobin, carbon monoxide also binds in an angular fashion as if it were oxygen — see bonCodigo’s answer for space-filling atom models. This is because the entire binding pocket is made to allow oxygen to bind (as I stated) thus attempting everything to make oxygen a comfortable home. Carbon monoxide is rather strained in there, its binding affinity is reduced by a factor of $1000$. However, since we started with a binding affinity difference of $10^5$, carbon monoxide can still bind $100$ times better than oxygen can.

Summary

Carbon monoxide is generally a good ligand that can bind to metal centres well.
Oxygen is generally a poor ligand.
Nature did everything to make oxygen a comfortable home in haemoglobin.
In doing so, the binding pocket became substantially less comfortable for $\ce{CO}$.
But since carbon monoxide was so good and oxygen so poor to start with, the former still binds better than the latter.
Nature didn’t really care, because individuals seldomly come in contact with carbon monoxide so the collateral damage was taken into account.


Mechanism of Fischer esterification: Does the carboxylic acid gives off OH- or H+?

The accepted mechanism is far more likely than your proposed alternate mechanism for a number of reasons.

First, let’s consider an esterification reaction employing a chiral secondary alcohol such as butan-2-ol. Protonation of the alcoholic oxygen is no difficulty and will happen in equilibrium. To generate the corresponding ester from the conjugate acid of butan-2-ol, we would require some nucleophilic attack mechanism: the acid would either attack according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. If we assume an $\mathrm{S_N2}$-like mechanism, the asymmetric carbon atom would be inverted. If we assume an $\mathrm{S_N1}$ mechanism, it would racemise. Experimental evidence shows, however, that the alcoholic carbon atom strictly retains is stereochemistry — and this principle is used in many total synthesis in which asymmetric alcohols are first generated stereochemically and then transformed into esters of the same stereochemistry. To give you just one of many examples, consider the total synthesis of aetheramide by Gerstmann and Kalesse.[1]

Scheme 1: Example esterification of (R)-butan-2-ol

Scheme 2: Esterification performed during the synthesis of aetheramide by Gerstmann and Kalesse.[1]

Second, and immediately tied to the scheme as referenced above, many alternate esterification methods exist in the literature. In principle, they all work by replacing the acidic particle ($\ce{H+}$) of a Fischer esterification with a less reactive and/or more specific compound. Practically all methods — Steglich or Yamaguchi to name just two — work by adding a compound that is known to interact with the carboxylic acid and not as much with alcohols. However, this is only a weak indication since different methods could follow different mechanisms — even though the quantity of the evidence does seem overwhelming.

Third and finally, the acidic hydrolysis of esters — the reverse reaction of the Fischer esterification — has been studied by $\ce{^18 O}$ marked water. Polanyi and Szabo could show that upon hydrolysis the heavy oxygen atom remains bound to the acid fragment and not to the alcohol.[2] This offers intriguing evidence for the mechanism of nucleophilic attacks on a carbonyl group. I had thought that a corresponding isotope-marking experiment had been performed for the Fischer esterification, too. Unfortunately, I was unable to find a reference. Do note, however, that using isotopically marked oxygen on either the acid or the alcohol would, by careful analysis of the product, rule out or confirm one mechanistic pathway.

Scheme 3: Mechanism of hydrolysis as confirmed by Polanyi and Szabo using $\ce{^18O}$ marked water.[2]

References:

[1]: L. Gerstmann, M. Kalesse, Chem. Eur. J. 2016, 22, 11210–11212. DOI: 10.1002/chem.201602682.

[2]: M. Polanyi, A. L. Szabo, Trans. Faraday Soc. 1934, 30, 508–12. DOI: 10.1039/tf9343000508.


How to rationalise the increasing bond length order in the carbonate ion, carbon monoxide, and carbon dioxide?

Actually, in regard to bond length, you should be looking at the bond order of a single bond within the molecule. The bond order is 2, not 4, for $\ce{CO2}$, and the bond order for the carbonate ion is somewhere between 1 and 2 due to resonance. Thus, the carbonate ion has the longest bond length, followed by carbon dioxide, and finally carbon monoxide.


Why are yields of some chemical reactions so dependent on the stoichiometry of the reagents?

The proper stoichiometry may not be the only reason for the friend's success, but also the overall preciseness of their preparation procedure.

Chemical reactions are not generally that much critical, but there are lot of  cases, that are.

There are few aspects to consider:

Excessive reagents may affect the effectiveness of the product isolation.
Excessive reagents may alter chemical environment in final mixture, affecting formation of products.
Excessive reagent may not remain, but may form another product, instead of the desired one

Typical case for 3. is e.g reaction of $\ce{NaOH}$ with $\ce{H3PO4}$. Phosphoric acid forms 3 salts.

$$\ce{NaH2PO4 + Na2HPO4 + Na3PO4}$$

Let suppose you want to produce
$\ce{NaH2PO4 }$

$$\ce{NaOH + H3PO4 -> NaH2PO4 + H2O}$$

Extra  $\ce{NaOH}$ will create from the wanted salt  other salt...

$$\ce{NaOH + NaH2PO4 -> Na2HPO4 + H2O}$$

... and the amount of the final product decreases.

More detailed answer depends on providing your particular scenario.


Which acid/base does a strong base/acid react when added to a buffer solution?

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$

  My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?

You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.

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

If you add up the two reaction, you get a third one:

$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$

You can go either way (use the second or the third reaction) to explain that $\ce{OH-}$ is part of a neutralization reaction. As a consequence, the concentration of $\ce{CH3COO-}$ increases and that of $\ce{CH3COOH}$ and hydronium decreases.

Hydroxide reacts with hydronium

In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about $\pu{e-5}$ M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.

Hydroxide reacts with acetic acid

In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.

Which way is better?

It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.

  Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?

No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.


How does radiation facilitate the formation of tetravalent iron?

The ionization energy of $\ce{Fe^3+}$ to $\ce{Fe^4+}$ is 54.8 eV.
The energy of typical gamma radiation is much higher (in the order of roughly some keV to a few MeV). Depending on the energy, the interaction of average gamma radiation with the affected electron is mainly due to Compton scattering.


Reaction mechanism of 2-amino-4,6-dimethylpyrimidine as a nucleophile

Your mechanism is almost correct; this is essentially an aldol condensation. The only issue is with the protonation states, which (as you've drawn) aren't entirely feasible.

Let's start by establishing how acidic that methyl group actually is. You're right that it is more acidic than a typical methyl group, because the charge can be delocalised onto those two nitrogens. But note that this delocalisation is pretty similar to that in a typical ketone enolate:

so to a very, very rough first approximation, that proton you're removing has a $\mathrm{p}K_\mathrm{a}$ of $\sim 20$, roughly that of a ketone. Yes, there are two nitrogens in the ring which improves the delocalisation, but there is also an amino group sticking off the ring, which is also trying to delocalise electron density onto the ring nitrogens. The value of $4.8$ cited in the comments is far more likely to be the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid (where the amino group is protonated).

Secondly, if anything were to be deprotonated, the $\ce{NH}$ in the amino group is going to be more acidic than the $\ce{CH}$ in the methyl groups. This is exactly analogous to how ammonia is more acidic than methane, and how amide $\ce{NH}$'s are more acidic than ketone $\ce{CH}$'s.

And lastly, note that your reaction is being carried out using acetic acid as a solvent. The chances that a carbanion (and a strongly basic one, as previously established) can exist for any reasonable amount of time under these conditions are... low, shall we say!

Now, you've probably drawn mechanisms for aldol condensations in acid. In this case, you don't deprotonate to form an enolate; you just use the enol tautomer, which is still nucleophilic on carbon. So the strategy here is not to deprotonate the methyl group outright, but to use the tautomer. Basically, use the same mechanism you've drawn, but add a proton to everything:*

That includes the final elimination, too. Note that if you protonate the $\ce{OH}$ group and let it fall off, you can get a benzylic cation that is stabilised by the 4-hydroxy substituent.

* You could delocalise all the way onto the other ring nitrogen if you wanted.


According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), heteromonocyclic compounds with up to and including ten ring members are named by the extended Hantzsch–Widman system. According to P-68.1.1.3.2, this also applies to heterocyclic parent hydrides containing Group 13 (boron group) atoms.

The Hantzsch-Widman system stem for an unsatured 7-membered ring is “epine”. Therefore, the preferred name for the compound that is given in the question is 1H-borepine.
The name 1H-1-boracyclohepta-2,4,6-triene is not recommended.


Would a melted crystal recrystallize back into the same lattice structure when frozen?

If cooling is rapid enough, then crystals don't have time to form, so the substance may be amorphous or glassy (though there may be some localized structure). For example, there are metallic glasses and amorphous water ice. Metallic glasses may have improved wear and corrosion resistance over a crystalline alloy. One interesting intermediate form is a quasicrysal, which can have 5-fold symmetry. Roger Penrose proposed this geometrical tiling before it was found in metallic glasses.
It depends on the conditions in which the material freezes. For example, graphite can be forced into a diamond structure by heat and pressure. A pseudomorph could form if the substrate initializes crystallization in a different form.


What actually happens to the volume of a gas in an isolated system?

The volume of a gas is based on the pressure, temperature and number of moles.
According to the ideal gas law :

$PV = nRT$

Where P is the pressure, V is the volume, n is the number of moles, R is a gas constant and T is the temperature.

In your case, the volume of both containers is the same, thus the volume of both gases are equal. However, assuming the temperature of the system is constant, the pressure that the gases exist must be different :

V, T, R = Constants -> $ n \propto P$

Therefore, the pressure of the container with 5 moles is five times of that of the one with 1 mole.

To actually make a fair comparison of the volumes of both moles of gases, the pressure that they are in need to be the same. So, if 5 moles of the gas is placed in a container that has 5 times the volume of the one with one mole, the pressure would be the same but the volume would be different (5:1).

Here is a helpful video for you to understand the concept: https://www.youtube.com/watch?v=BxUS1K7xu30


Unknown yellow substance from mixing ascorbic acid, baking soda, and rubbing alcohol, then boiling

That's sort of the colour that vitamin-C tablets go if they are kept long beyond their expiry date. I'd go with it being oxidation of some sort but chemistry.oregonstate.edu/courses/ch130/old/VITCTEXT.htm indicated that the simple oxidation of ascorbic acid is colourless... so what you've got is a more complex mix of oxidation products.

Instead of boiling off in an open pan, boiling off in a stable atmosphere might work better. You'd probably need some distillation glassware and a source of non-oxidising gas to flush out the oxygen dissolved in the water / keep an oxygen free atmosphere. You could probably get away with any arc welding gas or even use carbonated water as a source of CO2 gas to flush the air out of your equipment. Or of course you could go the whole chemistry-department route and boil the water off in a low pressure system without adding much heat.


Why do pharmaceutical drugs have a expiration date?

It's a good question, and I cannot answer your specific example but can give some generic reasons why pharmaceuticals have an expiration date:

Degradation of the compound/formulation over time. This obviously comes to mind, but there is more to it than meets the eye. For example, you gave the example of an ointment and cited the active ingredient: however, the product you buy (and use) is more than just the active molecule. The whole formulation has to be stable, not only the benzydamine. The ointment is likely to be a complex emulsion, designed to target many different physical and chemical properties, including chemical stability, thermal stability, but also viscosity (so it's easy to apply), color, etc. I suppose in that particular case, the emulsion might be less stable than the active ingredient taken alone.
Also to note: many pharmaceuticals are sensitive to air, in many cases because they are unstable in the long term in presence of oxygen, or moisture. Such products are sold enclosed in a plastic, metallic or composite packaging that protects them. There, the limited stability in time might simply be due to the limit of stability of the (cheap) packaging itself: e.g., after some time of possible light exposure, plastic will become porous and let moisture in.

Being conservative, because we're talking about health.

Being conservative, because the company selling the product doesn't want to be liable.

Economics: it takes money to make absolutely sure the drugs are safe for a longer period of time. Are customers willing to pay more? Most people don't store drugs for use years later (nor is it recommended), so they won't see the point in paying more for longer expiration.


Is it possible to have same mass and volume for a solvent and its solution

As Klaus suggests above, you can't add mass to another mass without it increasing.  That would defy the first law of thermodynamics that energy cannot be destroyed since mass and energy are directly linked.

However, much of this depends on what degree of precision you're measuring to.  If you're working with a cubic metre of water and only measuring to the first two significant figures, mass would be 1000kg and volume 1000l, and you add 40kg of sodium chloride to make a 4% saline solution, you would still only have 1000kg and 1000l because you're not counting the third significant figure.

It's all about context ;-)


Are there types of glue that are designed to be dissolved easily?

"Wood glues" can have several different compositions, chances are if the ones you are working with respond to vinegar they are either protein based or PVA.  In the protein case the solvation is likely accomplished by the vinegar protonating the polymer chains, and therefore changing their attraction to each other. In the case of PVA based glues, vinegar is likely contributing to acid catalyzed hydrolysis.  The softening of cyanoacrylate (super glue) with acetone is a more traditional dissolving mechanic, as no chemical changes are being made.

That being said, all three mechanisms mentioned above can be pretty well represented in the glue world with different polymers.  Based on what you consider cheap solvents, you have a few options. Cyanoacrylate (superglues), some UV-cured adhesives (which are frequently very well crosslinked and strong), PVC, and many others are all softenable and removable by acetone

Additionally, anything that can be taken out by acetone will be solvated in MEK (methyl ethyl ketone).  MEK is less water miscible and more aggressive though, which may be relevant to your needs.

Silicone adhesives are great for flexible applications, but frequently need stronger organic solvents (like toluene) to swell it and then it can be mechanically removed.  There are also commercially available silicone removers that appear to use acidic or basic bond cleavage to break up the adhesive.

Rubber cement is another flexible and relatively strong bonding adhesive mainly composed of solvated polymers, drying and tackifying by solvent evaporation.  Reintroducing the solvents (many commercially available paint thinners should work) will soften the adhesive.

As an aside note, depending on the materials you are bonding together, heat and cold can also be used to weaken joints and destabilize the adhesive bond separately or in tandem with chemical treatments.


Difference between solubility and dissolution?

Solvation [...] is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. As ions dissolve in a solvent they spread out and become surrounded by solvent molecules.
  [...]

  Dissolution is a kinetic process, and is quantified by its rate. Solubility quantifies the dynamic equilibrium state achieved when the rate of dissolution equals the rate of precipitation.

Quote from Wikipedia.


Why is the formation of a peptide bond not spontaneous?

Based on a quite old reference 1 (which I'm using because it's available free by open access), peptide bond formation at 25 C is unfavorable only because of a large enthalpy change, on the order of 1.5 kcal/mol (6.3 kJ/mol). The entropy change is actually favorable, with $T\Delta S$ being about 1 kcal/mol (4kJ/mol), so the net free energy change is ~ + 500 kcal/mol (approx. +2 kJ/mol).
The reason for both values being what they are is because the free acid and amine groups are ionized at neutral pH. Thus, the reaction can be represented as
$\ce{R-CO2- + ^+H3N-R -> R-CONH-R + H2O}$
This can be broken into two steps conceptually:

$\ce{R-CO2- + ^+H3N-R -> R-CO2H + H2N-R }$

$\ce{R-CO2H + H2N-R -> R-CONH-R + H2O}$

As we know, proton transfer from an acid to a base typically involves a large negative enthalpy change, so reversing that to neutralize both has a positive enthalpy change. Since the products remain neutral, we never get that energy back. And the bond energies of a neutral carboxylic acid and amine are not much different from an amide and water, so the neutralization dominates.
For the entropy change, we start with two molecules and end with two, and the degrees of freedom are similar. The authors of the cited paper propose that the positive entropy change is also related to the loss of ionization. They believe that the more structured hydration shell of the ionic species vs neutral is sufficient to make the free amino acids lower in entropy than the neutral products, making bond formation entropically favorable, but not enough to overcome the unfavorable enthalpy change.
Later paper have further shown that the amide formation reaction becomes favorable around 60 C because the favorable entropy change finally overcomes the enthalpy change at that temp.
1 Dobry, A., Fruton, J.S. and Sturtevant, J.M. (1952)  J. Biol. Chem. 195, 149–154.
(Available by open access here)


Does ultrasonic bath cleaning damage laboratory volumetric glassware calibration?

According to technical information for volumetric measurement provided by Brand, it is acceptable to use an ultrasonic bath.

  Both glass and plastic labware may be cleaned in an ultrasonic bath. However, direct contact with the sonic membranes must be avoided.

Nevertheless,

  For gentle treatment of labware, clean immediately after use – at low temperatures, with brief soaking times, and at low alkalinity. Glass volumetric instruments should not be exposed to prolonged immersion times in alkaline media above 70 °C, as such treatment causes volume changes through glass corrosion, and destruction of graduations.


Do both iodine and potassium iodide turn dark in the presence of starch?

Starch is a long ribbon or a long filament made of a great number of glucose units attached to one another like the wagons in a train. But this filament is wound in a helicoidal way. And the inner part of this helix is a sort of long hole. And the dimension of this hole is just big enough to allow the iodine molecule $\ce{I2}$ to enter and stay there, maintained by van der Wals forces. But when the iodine molecule is inserted in the starch helix, it produces a "starch-iodine" complex which has an intense blue color.


Ionic reaction between lithium and chlorine

Maybe your book is wrong. It can't be. There has to be 8 electrons in the outermost occupied shell and to complete its octet. I searched over the google about it and found three links to cover my answer.

https://in.answers.yahoo.com/question/index?qid=20111018173152AAn9DHz
https://quizlet.com/38997670/chem-1305-exam-2-study-guide-flash-cards/ (question number 3)
http://slidegur.com/doc/1661637/ionic-bonds (slide number 18)


Free chemical database to search for physical properties?

As I've mentioned on other related questions, this is very much an open topic. There aren't many open-access databases in chemistry, and far fewer that have APIs.

As suggested in the other answer, there are multiple ways to query / scrape Wikipedia.
ChemSpider offers some property search features.
Reaxys offers some property searching, including spectra features. It's not open access.


Drawing the condensed structure of ethylcyclobutane

$\ce{CH3CH2CH(CH2)3}$ is the best you can do.

If you want to draw the cycle, then you should make it pendant, drwing a bond from the ring to the ethyl group.


Melting point of aspirin, contradicting sources

Wolfram company doesn't conduct any experimental determinations of physical constants for chemical compounds and uses literature data sources.
The webpage for ChemicalData Source Information lists numerous sources of chemical information used by the company's products, including Wolfram Alpha.

Wolfram Alpha Knowledge Database is linked with 87th ed. (2006) of CRC Handbook of Chemistry and Physics (m.p. $\pu{135 °C}$), whereas Wikipedia cites newer 92nd ed. (2011) of CRC Handbook of Chemistry and Physics (m.p. $\pu{136 °C}$).

Obviously, Wolfram's sorting algorithm preferred another data source despite it's quoting the CRC Handbook.
Why and how it did that cannot be answered as the server-side engine is proprietary.

Note that even newer 97th ed. (2017) of CRC Handbook of Chemistry and Physics adds measurement error of $\pu{4 °C}$ (m.p. $\pu{136(4) °C}$).
The difficulty of determining the exact value probably arises from the fact that at that temperature range thermal decomposition already begins, which is reflected in m.p. dependence on heating rate and atmosphere and melt solidifying at lower temperatures (down to $\pu{118 °C}$ [1, table 3.4]).

References

Aspirin and Related Drugs; Rainsford, K. D., Ed.; CRC Press; Taylor & Francis: London, 2004. ISBN 978-0-7484-0885-6.


Which hydrogen halides react with group 14 oxides?

All reactions taken from this site.
First some maths: 4 hydrogen halides can react with 5 Group 14 monoxides and dioxides. So, there should be 40 different chemical reaction combination possible. But not all reactions are feasible.
HF

With $\ce{SiO2}$ (already mentioned)

$$\ce{GeO2 + 4HF → GeF4 + 2H2O}$$

$$\ce{SnO + 2HF ->[333 K] SnF2 + H2O}$$

HCl

$$\ce{GeO2 + 4HCl(g) ->[450-500 C] GeCl4 + 2H2O}$$

$$\ce{GeO2 + 4HCl(l) ->[170-180 C] GeCl4 + 2H2O}$$

$$\ce{SnO + 3HCl → H[SnCl3] + H2O}$$
$$\ce{SnO2 + 6HCl → H2[SnCl6] + 2H2O}$$
$$\ce{PbO + 2HCl → PbCl2 + H2O}$$
$$\ce{PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O}$$

$\ce{HBr}$ and $\ce{HI}$ do not show such type of reaction.


I won't comment on B for the moment, but liquid-liquid interface electrochemistry is quite a vibrant field. There's a review of it here, if you're interested. The simplest interaction involves ions: if one applies a potential across the interface of sufficient strength and polarity, the energy barrier of transferring a given ion from one phase to the other can be overcome. If you do this with something like a solvatochromic dye, it will change colour as it moves from one phase to the other. (It would have to be something that doesn't partition into the second phase easily on its own, of course) One could also have some reagent in one phase that reacts with the ion being pushed across the interface that produces some coloured product. (iodide-starch, perhaps?) There are also voltage-sensitive dyes, though I don't have any experience with them.

One thing to note is that neither water nor oils and other immiscible solvents are generally very conductive. To keep the cell resistance down to a reasonable level, it will be necessary to dissolve electrolytes in both. Things like Tetrabutylammonium hexafluorophosphate are commonly used as electrolyte in non-aqueous electrochemistry.


How exactly does the reaction between alkyl halides and silver nitrite occur?

The concept of hard and soft acids and bases (HSAB) proved to be
  useful for rationalizing stability constants of metal complexes. Its application
  to organic reactions, particularly ambident reactivity, has led
  to exotic blossoms. By attempting to rationalize all the observed
  regioselectivities by favorable soft–soft and hard–hard as well as
  unfavorable hard–soft interactions, older treatments of ambident
  reactivity, which correctly differentiated between thermodynamic and
  kinetic control as well as between different coordination states of ionic
  substrates, have been replaced. By ignoring conflicting experimental
  results and even referring to untraceable experimental data, the HSAB
  treatment of ambident reactivity has gained undeserved popularity. In
  this Review we demonstrate that the HSAB as well as the related
  Klopman–Salem model do not even correctly predict the behavior of
  the prototypes of ambident nucleophiles and, therefore, are rather
  misleading instead of useful guides. An alternative treatment of
  ambident reactivity based on Marcus theory will be presented.

  — H. Mayr’s, M. Breugst’s and A. Ofial’s introductory remark in their review pulished in Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.

Within the review, in the section on the nitrite ion, the authors give two competing effects that control product distribution.

For electrophiles with an electrophilicity parameter $E < -3$ (the bis-(4-methoxyphenyl)carbenium ion has $E = 0$, lower values are more stabilised carbocations such as bis-(4-dimethylaminophenyl)carbenium) the reactions proceed at equilibrium under thermodynamic control and the product yields are equivalent to their relative stabilities. Since nitro compounds are more stable than nitrite esters, they predominate.

$$\ce{R-ONO <=> R+ + NO2- <=>> R-NO2}$$

For electrophiles with a parameter $E > 0$, the reactions are essentially diffusion controlled, i.e. the free electrophile will capture the first nucleophile it can find. This includes tertiary alkyl carbocations ($E \approx 7{-}8$). The reactions do not go through classical transition states, and any means of describing them via frontier orbitals must fail. That explicitly includes describing them according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. In these reactions, kinetic control determines that the $\ce{O}$-attack is slightly more likely with $k_\ce{O} / k_\ce{N} \approx 3.4$. (This corresponds to computational models that show a slightly smaller activation barrier for the oxygen attack.)

Reality is more complex than these simple models imply especially for the diffusion controlled reaction. While the attack of oxygen is favoured kinetically, the attack of nitrogen results in less required reordering of the molecule (the two $\ce{N-O}$ bonds would still be identical while they become $\ce{O-N=O}$ in the nitrite ester) which reduces the expected favourism for oxygen. Effects such as solvent and counterion slightly affect the product distribution, and the highest ratio of nitromethane : methyl nitrite can be achieved using $\ce{Bu4N+ NO2-}$ and $\ce{MeI}$ in $\ce{CDCl3}$ (a product ratio of $70\,:\,30$ in favour of nitromethane). Note that $\ce{AgNO2}$ and $\ce{MeI}$ in DMSO yield more methyl nitrate ($54\,:\,46$) although the same electrophile is used. The authors conclude the nitrite section with the statement:

  It should be noted, however, that the selectivities
  shown in [the table presented in the review] are also not related to the hardness of the electrophiles.

Note

Throughout this answer, electrophilicity parameters according to Mayr’s equation are used:

$$\lg k_{20~\mathrm{^\circ C}} = s(E + N)$$

$E$ is a nucleophile-independent electrophilicity parameter while $s$ and $N$ are electrophile-independent nucleophilicity parameters.

Bibliography

A paper related solely to studying the nitrite anion: A. A. Tishkov, U. Schmidhammer, S. Roth, E. Riedle, H. Mayr, Angew. Chem. Int. Ed. 2005, 44, 4623. DOI: 10.1002/anie.200501274.

The review mentioned in the opening, concerning all sorts of ambient nucleophiles often explained via the HSAB concept: H. Mayr, M. Breugst, A. Ofial, Angew. Chem. Int. Ed. 2011, 50, 6470. DOI: 10.1002/anie.201007100.


How are pure substances (elements, compounds, etc) produced?

No lab reagent is 100% pure. But you can get really close to that.

How 'pure' are they?

Read the label or do the analysis. It varies A LOT from one reagent to another.

Is it possible to create relatively pure substances with household goods?

Depends on the substance, depends how pure you want it. For instance, you can very easily distill water at home, but pure water is hard to get.

How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab)

You'll need to be more specific. There are entire classes you can take at the uni on the large scale production and purification methods of this and that type of reagents. I see you used the ionic-compounds tag, but that's far from specific. A more answerable question would be "How are compounds X and Y produced and purified?"


Is it possible to make deuterated organic compounds?

Deuterium occurs at around the 0.02% level in nature (natural abundance level).  So all compounds that contain hydrogen will already contain a low level concentration of the deuterated isomer(s).  So in fact, deuterated compounds do exist in nature.

Despite this low level of natural deuterium incorporation, the deuterium can still be detected by various spectroscopic techniques (mass spectrometry and nmr for example).

Further, organic compounds can be synthesized with deuterium in specific locations. These deuterated compounds are still considered organic compounds. Since deuterium is a stable nucleus (e.g. does not undergo radioactive decay), the compounds containing deuterium are also stable.  Because the $\ce{C-D}$ bond is a bit shorter and therefore a bit stronger than the corresponding $\ce{C-H}$ bond, there is an isotope effect in reactions where these bonds are broken. Because the $\ce{C-H}$ bond is weaker, it will break up to 7 times faster than the corresponding $\ce{C-D}$ bond. This can have a significant effect on biological processes.

$\ce{D2O}$ and $\ce{D2}$ gas are relatively common and inexpensive materials that can be used to introduce deuterium into other molecules. Here are 2 examples involving the use of these materials for deuterium incorporation.  In the first example $\ce{D2O}$ is used as the source of deuterium, in the second example deuterium gas is used to introduce the deuterium.  If you google "organic preparation of deuterated compounds" you'll find many more examples.

image source

image source

Specifically deuterated compounds are widely used in many areas of research and industry.


Why does the 31P{1H} NMR spectrum of cis-[Mo(CO)2(dppe)2] show two signals?

In this complex there are two different 31P environments which are not related by symmetry:

The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted.

As extra proof, consider that the green P is cis to both carbonyl ligands whereas the purple P is cis to one and trans to the other.

They therefore have different chemical shifts and in the spectrum you would expect to see two different peaks. Presumably they would show coupling to each other, so I would expect two triplets, if we ignore satellites from all other nuclei (e.g. 13C).


What are the products formed when phenol is reacted with benzyl bromide?

I believe the SN2 mechanism will predominate. The benzyl bromide is not sufficiently electrophilic to perform a Friedel-Crafts Alkylation, as that often requires the use of a metal catalyst.


What chemicals have been named after Star Wars characters?

I heard about one other Star Wars compound named after a character: Kinobeon A.[1]

There is also a supramolecular structure named after a Tie fighter.[2]

Apparently there is also a Yoda1 analogue (note the additional sulfur atom) called Dooku1.[3]

References:

Kanehira, T.; Takekoshi, S.; Nagata, H.; Matsuzaki, K.; Kambayashi, Y.; Osamura, R. Y.; Homma, T. A novel and potent biological antioxidant, Kinobeon A, from cell culture of safflower. Life Sci. 2003, 74 (1), 87–97 DOI: 10.1016/j.lfs.2003.06.033. Mirror: www.chm.bris.ac.uk
D'souza, F.; Gadde, S.; El-Khouly, M. E.; Zandler, M. E.; Araki, Y.; Ito, O. A supramolecular Star Wars Tie Fighter Ship: electron transfer in a self-assembled triad composed of two zinc naphthalocyanines and a fullerene. J. Porphyrins Phthalocyanines 2005, 09 (10), 698–705 DOI: 10.1142/S1088424605000812.
Evans, E. L.; Cuthbertson, K.; Endesh, N.; Rode, B.; Blythe, N. M.; Hyman, A. J.; Hall, S. J.; Gaunt, H. J.; Ludlow, M. J.; Foster, R.; Beech, D. J. Yoda1 analogue (Dooku1) which antagonizes Yoda1-evoked activation of Piezo1 and aortic relaxation. Br. J. Pharmacol. 2018, 175 (10), 1744–1759 DOI: 10.1111/bph.14188.


Why does ethanol have a health rating of [2] in NFPA 704 whereas methanol, which is far more toxic, has a rating of only [1]?

I believe that the rating is dependent on toxicity tests.

The acute oral toxicity LD50 value of acetone is 5800 mg/kg

The acute oral toxicity LD50 value of methanol is 5628 mg/kg

The acute oral toxicity LD50 value of ethanol is 3450 mg/kg

Low LD50 values means that only a small amount of chemical is needed to kill someone. According to these values, you would need to ingest a larger amount of acetone or methanol to die.I am not sure how valid these toxicology test are. Fun fact: fructose is deadlier than methanol according to LD50 values.

Furthermore, there appears to be some variation in the values as well. Wikipedia lists the LD50 value of ethanol as 7,060 mg/kg.
Maybe this question can be addressed at Biology SE.


Why can't unimolecular reactions be 2nd order?

A first order reaction involves the behavior of individual particles, usually their decomposition into smaller, more reactive species.  The rate of such a reaction relies solely on the number of particles present, so if we double the number of particles then the rate doubles simply because there are twice as many particles available to decompose.

A second order reaction involves the interaction of two particles with one another. (Even if both particles are the same it is still a bimolecular reaction.)  If we double the concentration of either species we double the likelihood of a collision between those species.  If both species are the same (ex: $\ce{2A -> A2}$) then the result of doubling the concentration is a quadrupling of the rate because we simultaneously double the number of particles that can interact and the number of interactions that can occur.

An metaphorical example involving people:

Unimolecular:
A couple can spontaneously break up into two separate people because of its internal properties.  (Assume there are no intercouple issues involved here.)

$\ce{C -> M + W}$

If we double the number of couples we would expect breakups to occur twice as often because there are twice as many couples to be broken up.  However, since there are no intercouple issues in this example, adding additional couples to our consideration does not increase the likelihood of existing couples breaking up.

Bimolecular with Two Different Species:
A man and a woman can join together to make a couple if they collide in the right orientation with enough energy. :-)

$\ce{M + W -> C}$

If I increase the number of either men or women, it is more likely that I will form a couple simply because there are more interactions between men and women occurring.  Each species' increase is first order though, since they require a partner of the other gender to couple with.  i.e. Doubling the number of women only doubles the rate at which couples are formed because they still have to connect with a man, and the number of men has not changed.

Bimolecular with a Single Species:
Homosexual men can join together to form a couple.
$\ce{M + M -> C'}$

If I double the number of men, then each additional man can interact with every other man in the room, greatly increasing the likelihood of a couple forming.  This means that not only are there more men available to make couples (the first doubling of the rate), but that the existing men are also more likely to be involved in a connection that results in a new couple being formed (the second doubling of the rate.)


Okay, so if we talk about the acidic strength of a compound, consider adding a base to a solution of all the above. Practically, a reaction is bound to take place-if possible-due to thermodynamic reasons. Hence, reaction in IV will take place since the base will try to find the most acidic Hydrogen, which is why one should consider the removal of H atom from which resonance can take place.

Now for the order of acidic strength, I, II and III are individually more acidic than IV, due to extensive conjugation in all three but the last. This consequently makes the conjugate bases stable and acids stronger. Now for the order among these, notice that Cross-conjugation takes place in III, as we have COO- group. Hence, III will consequently be less stable and a weaker acid.

The order and reason stated by you for II and I are correct, and hence the order for acidic strength is:

II>I>III>IV

i.e. option A.


Why is the alpha particle called a particle when it is made of four particles?

Actually, the alpha particle contains two protons and two neutrons. It is what is emitted from the nucleus under alpha decay, giving off what was first classified as alpha radiation by Rutherford at the end of the 19th century (very late 1800's).

In 1932 the neutron was discovered. Only then did it become clear that the nuclei of atoms contain more than just some positively charged particles (which we know as protons).

So there is this historic reason for the naming of the alpha particle.

"Particle"

The word particle does not only pertain to elementary particles (and in [quantum] chemistry, we stop at the level of nuclei as particles, everything that goes deeper is nuclear physics). There are smoke particles, dust particles, etc.

There is also the wave-particle duality. In principle, since we can observe this "alpha radiation" from radioactive processes, there must be a particle that belongs to the wave. With electromagnetic radiation, this is the photon, and in alpha radiation it is the alpha particle.


Can I change glass transition temperature (Tg) and impact resistance with additives?

The glass transition temperature depends on the cooling rate and molecular weight distribution and could be influenced by additives:

Most liquid plasticizers are low molecular weight organic materials with a glass transition temperature in the range of $125-225^0\text{C}$. When this kind of additives is added to an organic polymer, the result is a weight averaging of the glass transition temperature between that of the polymer and that of the plasticiser if they are miscible. The reduction of the the glass transition temperature of a polymer by the plasticiser is usually about   $5^0\text{C}$ per weight percent of plasticiser.

Impact modifiers ( a kind of additives) can affect the impact resistance of a polymer. Impact modifiers are key additives for
increasing flexibility and impact strength
to meet physical property requirements of
rigid polymers. All impact modifiers are elastomeric or
rubbery in nature, with a lower modulus
than the host polymer. The dispersed
rubber phase acts to absorb or dissipate the
energy of impact in order to stop craze or
crack propagation.


Why does the earth have more radon in some locations?

Radon nuclides are short-lived members of the natural radioactive decay chains of thorium and uranium. On a time scale that would be relevant for geology, you should better ask why some locations have more uranium or thorium. The distribution of radon concentrations in the ground air is a result of the distribution of uranium or thorium in the ground.

The main reason for the inhomogeneous distribution is igneous differentiation.  The original liquid magma might have been rather homogenous. But when the magma cooled down, fractional crystallization started. This removal of mineral precipitates also changed the composition of the remaining melt. Uranium and thorium ions are incompatible, i.e. they do not fit well in the crystal lattices of the silicate minerals that precipitate first. Therefore, they were accumulated in the remaining melt. Primary deposits of uranium and thorium minerals (such as pitchblende) were formed where the last parts of the cooled magma solidified.

Furthermore, secondary deposits were formed when uranium or thorium minerals were dissolved, transported and precipitated in ground waters. (This includes the nicely coloured and fluorescent uranyl ($\ce{UO2^2+}$) minerals.


Why doesn't the current stop in neon lamps when all the atoms have been ionized?

If current is the flow of charge why doesn't it stop once all the atoms have been ionized?

The short answer is that it is impossible to ionize all the gas at once. It just requires huge amount of energy to ionize the whole atoms at once. Just for the sake of discussion, even at 10000 °C, still a very small fraction of atoms is ionized! This temperature is hotter than the surface of the Sun! A neon discharge tube is not even close to those temperatures or let us say energies.

The red glow of neon tubes you see is mainly due to electronically excited neon atoms. The pressure of the gas is pretty low inside the tube. Once you start applying a voltage, electrons emit from the cathode and ionize the gas forming a a number neon ions. These ions and free electrons are responsible for carrying the current. Since you are constantly supplying current, it continues to flow through the discharge tube and the process of ionization and recombination continues. Remember that ionization is an equilibrium process in the gas phase as well. An additional process can also occur, that the electron is promoted to a higher energy orbital instead of leaving the atom. Once it jumps back to its original orbit, it emits its characteristic wavelength of light.

I think the recombination process stated above should yield a continuous spectrum so the red characteristic glow of neon tubes is not from recombination process.


How to make cyclohex-4-ene-1,2-diol from cyclohexa-1,4-diene?

I think the easier way of doing this is to epoxidise the diene with 1 eq of a peracid epoxidation with peracids . The standard reagent m-Chloroperoxybenzoic acid (MCPBA) breaks your rule of organic molecules with 3 carbon atoms or fewer, but you can also do it with peracetic or pertrifluoroacetic acid preparation of peracetic acid. You will inevitably get some di-epoxide, but this will be separable by chromatography. The mono-epoxide can then be reacted in mild aqueous acid to give the diol. You could also carry the mixture through to the aqueous acid treatment as the tetraol will be far more water soluble than the diol and can be removed by an aqueous wash.


How many d-eletrons are in the outer shell of following compounds:

Sulfur does not have any "d" electrons as you said.

Before losing two electrons from the 4s sublevel, Zinc's outer sublevel is 4s (not 4p as you mentioned).  After those two electrons are lost, the outermost sublevel becomes the 3d sublevel which contains 10 electrons.

Yes, it does matter whether the atom forms a compound or not, so we can understand where the electrons are allocated/transferred.

Here is the general rule for how to consider electrons in atoms.  Electrons fill up an atom via the Aufbau principle (image attached).  However, after filling, they arrange themselves in the energy according to the their principle quantum number (1, 2, 3,...) and their sublevel type (s, p, d, f).  For example, 4s fills before 3d in zinc, but after filling, 4s gains a higher energy than 3d.


According to “Structure of Free Molecules in the Gas Phase”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL., the bond angles in propane are

$\angle\ce{CCC}=112^\circ$

and

$\angle\ce{HCH}=107^\circ$

Therefore, the remaining bond angles should each be about

$\angle\ce{CCH}\approx109^\circ$