Can 1 gram hyaluronic acid really hold 6L of water?

Various figures can be found “on the internet”. E.g. ‘Hyaluronic Hydro Gel Has 1000 Times Water Holding Capacity Perfect for Hair to Heel’ (commercial link intentionally omitted).

Gel liquid phase absorption ability measure seems to be a ‘swelling capacity’ (ratio).

For hyaluronic acid (HA), it can be increased by additional crosslinking. In one paper from 1992, usual swelling ratio was found to be in the order of magnitude 1 to 10, the maximum figure was 120 (for crosslinking ratio much higher than 2.1; 25–30 for the ratio 2.1).[1]

In another paper, another guys are using biphasic nanoparticles (some composite of crosslinked and non-crosslinked HA), they state they have achieved for two different particle types ‘… swelling ratio 1270, 1000 %, respectively’[2] (note the percent sign). Which is, in both cases, approximately, not 1000× capacity, but rather 10× capacity, I'd say.

However, there are polymers with the claimed property – superabsorbents:[3]

  3.27 superabsorbent polymer

  Polymer that can absorb and retain extremely large amounts of a liquid relative to its own mass.

  Note 1: The liquid absorbed can be water or an organic liquid.
  Note 2: The swelling ratio of a superabsorbent polymer can reach the order of 1000:1.
  Note 3: Superabsorbent polymers for water are frequently polyelectrolytes.

References:

Shah, Chirag B., and Stanley M. Barnett. "Swelling behavior of hyaluronic acid gels." Journal of applied polymer science 45.2 (1992): 293-298.
Chun, Cheolbyong, et al. "Effect of molecular weight of hyaluronic acid (HA) on viscoelasticity and particle texturing feel of HA dermal biphasic fillers." Biomaterials research 20.1 (2016): 24.
Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008)


Explanation of certain chemical bond diagrams

The second structure you posted is a line angle diagram. In this diagram, methyl groups (-CH3) are implied using sticks with nothing at the end.

Carbon atoms are also implied; carbons exist at every "corner" of the polygon which doesn't have an element explicitly identified. Hydrogens attached to carbons are also implied. So at the intersection of two lines, if there are no other attachments and no charge, you can expect there to be two hydrogens (carbon is generally tetravalent).

The third structure you posted is a wedge-dash diagram. Wedges depict substituents angled toward you; dashes are substituents pointing away from you; the lines are coplanar. The advantage of this diagram is that it shows the 3D nature of the molecule. Your first picture of the fluoro-carbon compound implies 90 degree bond angles. However, you should realize that the carbon atoms are sp3 hybridized and thus should not have 90 degree bond angles.

Further reading and exploration of other depictions of molecular structure:

http://www.masterorganicchemistry.com/2010/10/15/the-many-many-ways-to-draw-butane/


Why are peroxides unstable but disulfide bridges considered stable? Why are esters stable but thiolesters are unstable?

Some hard data: bond enthalpies (in $\pu{kJ mol-1}$)

$$\begin{array}{c|c|c|c} \text{Bond} & \text{Enthalpy} &
     \text{Bond} & \text{Enthalpy}  \\ \hline
\ce{C-C} & 350 & \ce{Si-Si} & 226\\
\ce{N-N} & 163 & \ce{P-P} & 201\\
\ce{O-O} & 146 & \ce{S-S} & 226\\
\ce{F-F} & 155 & \ce{Cl-Cl} & 240 \\
\end{array}$$

We observe a decrease in bond energy from carbon to nitrogen and from silicon to phosphorus. It is clear and probably have same reason: appearance of lone electron pair of the atom. There is a repulsion between lone pairs, that is in partially compensated by stronger bonds formed by smaller atoms (the atomic radii falls down to the end of the row). When moving from second to third row, this repulsion is reduced thanks to larger size of atoms of the elements of the third row.

The general decrease in energy from carbon to silicon is thanks to increased size of the orbitals of valence level, making them more diffuse with less efficient overlap and longer bond length.

The stability of esters vs thioesters is ruled by different reasons. Sulfur is not as electronegative as oxygen, but when you consider an anion, sulfur anions have larger radius and are less prone to grabbing first positive charge they find. So sulfur anions are more stable intermediates (kinetically), and thus polar dissociation of thioester bonds is easier.


Keto-enol equilibria reflect a delicate thermodynamic balance between the two forms.

The bonding differences between the keto and enol structures shown above are:

keto: C=O double bond, C-C single bond, C-H bond

enol: C=C double bond, C-O single bond, O-H bond

If we look up the bond energies for these bonds we find

keto: 745 + 347 + 413 = 1505 kJ/mol

enol: 614 + 358 + 467 = 1439 kJ/mol

This rough calculation tells us that, generally, the keto form will predominate.  However, the energy difference between the two forms is only 66 kJ/mol, a relatively small number; so it is likely that small differences can cause a dramatic shift in the relative concentrations of the two species.

Some general guidelines for predicting shifts in keto-enol equilibria

Aromaticity and conjugation - The keto-enol equilibrium for phenol (I) lies entirely on the enol side due to thermodynamic stabilization provided by aromaticity. Compound III would be expected to have a higher enol content than compound II due to the extended conjugation present in the enol of III

Substitution - Replacing hydrogen with alkyl groups on a double bond stabilizes the double bond. Therefore, we would expect compound V to have a higher enol content than compound IV

Dipolar repulsion - The carbonyl dipoles in butane-2,3-dione (VI) can reduce their electrostatic repulsion by adopting the geometry pictured with the carbonyl groups oriented away from one another.  Due to the constraints of the 5-membered ring, cyclopentane-1,2-dione (VII) cannot adopt a similar geometry.  In this case, the dipolar repulsion is lessened by increasing the enol content in the equilibrium to a point where the enol predominates.

Hydrogen Bonding - Hydrogen bonding can stabilize the enol form.  If the hydrogen bond is strong enough, and particularly if other factors also stabilize the enol form, the enol form can predominate.  Your molecule, 2,4-pentanedione (IX), is a good example.  There are two possible enol forms, VIII and X. X is by far the predominant enol as it has 1) extended conjugation and 2) substitution on the carbon-carbon double bond.  In addition, X can form a very stable hydrogen bond involving a 6-membered structure between the two oxygen atoms.

Solvent Effects - Especially in cases where hydrogen bonding is involved in stabilizing the enol, the solvent can have a dramatic effect. In benzene, where intramolecular hydrogen bonding predominates, the IX:X ratio is approximately 5:95. In water, where the intramolecular hydrogen bond is replaced by hydrogen bonding to the solvent, the ratio is roughly reversed.

Answer to your question

Cyclopent-2-enone can form two different enols, XI and XIII. Neither enol has any features that would be expected to provide significant enol stabilization.  There is no more conjugation than what is found in the ketone, there is no special hydrogen bonding, and no substitution of the carbon-carbon double bonds.  Indeed, the proton nmr of cyclopentenone is just as expected, no evidence for a significant amount of enol.  On the other hand, 2,4-pentanedione exists primarily as the enol form, at least in non-polar solvents, for the 3 reasons (conjugation, hydrogen bond, double bond substitution) discussed above.

As an aside, you mentioned that a 5-membered ring with two double bonds (a cyclopentadiene) might not be very stable; actually, they are quite stable and quite common, 2 double bonds in a 5-membered ring do not create a lot of strain.


How are partial molar Gibbs excess functions correctly defined?

The confusion is perhaps not uncommon. However, whichever book it is has jumped from discussing the free energy of pure materials, and their variation with various things (T, P, V, ...), to discussing mixtures.  In this case, all variation of the 'pure' substances is present in the ideal terms - that is because those variation exist without regard for any mixing terms.  What you are focused on here are the impacts of making a solution (or binary alloy, ...).

So, to make the mixture, you take a certain amount of substance A, and a certain amount of substance B, and mix them together. Before mixing, the total free energy of the system is $x_A G_{A} + x_B G_{B}$ where $G_{A,B}$ are the Gibbs free energies of those substances at whatever the conditions you specify before mixing. So, what does mixing do? Well, at the very least you get the increase in entropy, $RT(x_{A}\ln{x_{A}} +x_{B}\ln{x_{B}}$. If there is any interaction between A and B, you now have non-ideal terms which can be represented in any number of different ways. However, the point is that they are "all the rest that happens" during the mixing, and those things may be functions of concentration, temperature, pressure, ...

But, to come back to your question - no, this is not an error in the textbook (whichever it is). But, the did not make it clear that they were leaving behind the explicit dependencies of free energy on the various parameters - they figured you had that part down.

I would also note that, when one calculates a phase diagram (ultimately one hopes the book might get to that, and miscibility gaps and whatnot), you are free to choose the reference points or phases of A and B yourself - that choice makes no difference in the phase boundaries and tie lines. And, this comes back to just what $G_{A,B}$ means.


As TAR86 mentioned, you seem to understand where deuteration occurs in crotonaldehyde. The process is usually conducted under alkaline conditions. All of the steps are reversible and need not occur in the same order as in the diagram. For instance, enolate 2 can deuterate at the α-position prior to γ-deuteration. Hope this helps.


How does VB theory explain the Si-O-Si bond angles in SiO2?

Tl;dr version: the O-Si bonds utilize sp hybrid orbitals so that “lone pairs” are in nearly pure p orbitals and, rather than being true lone pairs, engage in some sort of $\pi$ bonding, the exact nature of which is unknown.  Other hypotheses are that the bond is highly ionic or that it is favored due to oxygen-oxygen repulsion.
Long version: First, let me point out a slight inaccuracy in the posted question:

Since the oxygen bonds to two silicon atoms, one would expect a sp3 hybridization, with two orbitals filled with lone pairs —

This is an incorrect interpretation of VB theory. Although we might approximate a molecule such as $\ce{H2O}$ as having an sp3-hybridized O atom, quantitative approaches to VB theory account for the fact that the contribution of s and p atomic orbitals to localized orbitals differs if the group differs. That is, lone pairs and bond pairs can have (and usually do have) different contributions from the s and p orbitals, ie different hybridization. A commonly cited example of this is $\ce{SH2}$, where the S-H bond orbitals are nearly entirely comprised of p orbitals, so the lone pairs are necessarily essentially sp hybrids.
Furthermore, one quantitative approach calculates the p orbital contribution based on the bond angle (Coulson's Theorem). In silicate, the Si-O-Si bond angle varies between ~140$^\circ$ and 180$^\circ$, depending on polymorph. We can calculate that the hybridization of these bond orbitals thus varies from sp (180 degree angle) to sp1.3 (140 degree angle), which is actually a relatively small variation in the s orbital fraction (43-50%). The lone pairs, therefore, are nearly pure p (93-100%) orbitals.
Although it is known that the $\alpha$-quartz polymorph (bond angle 144 degrees) is the most stable at low temperature, the observation of many polymorphs suggests that there is not much of an energetic penalty for the variation in bond angle within this range.
For simplicity, let’s consider the extreme case of the 180 degree bond angle, so that the O-Si bonds are formed from sp orbitals on O, and the lone pairs are in pure p orbitals. Both VB theory and MO theory indicate that this arrangement is only energetically favorable relative to a bent conformation if either the p orbitals are empty (as in $\ce{BeH2}$ for example) or if the electrons in the p orbitals are involved in $\pi$ bonding, for example the carbon atom of $\ce{CO2}$ or $\ce{HCN}$.
COnsistent with this idea, researchers have argued since at least as far back as the 1960's that the “lone pairs” on oxygen in silicate are in fact not lone pairs, but are involved in $\pi$ bonding to the silicon. This is supported by the short Si-O bond lengths and high bond strength which suggest a bond order greater than 1.
Historically, this interaction has been proposed to involve empty d orbitals on silicon, but more recently there has been a shift away from d orbital involvement, for similar reasons that d orbital involvement in phosphorus bonding has generally been disregarded. Interaction with empty $\sigma^*$ antibonding orbitals on Si, however, also does not appear satisfactory, as that would weaken the Si-O bonds (assuming that all Si-O bonds are equivalent, so all would have increased bond order from $\pi$ bonding, but decreased from addition of density to $\sigma^*$). Essentially, there would be a significant contribution of a resonance structure in which each Si has two Si-O bonds and one Si=O bond, similar to a carbonate.
Nonetheless, the idea that a large bond angle is favored due to some sort of $\pi$ bonding seems to be the prevalent view. Two other hypotheses, however, are still supported by some.
Other hypotheses
Ionic interactions
One alternative explanation is that there is very little difference in bond orbitals and lone pairs, which could result from a strongly ionic structure of $\ce{O^2-}$ anions and $\ce{Si^4+}$ cations[1]. On the oxygen, all of the electrons then are lone pairs, and the “bond angle” (which isn’t between any true bonds) is simply optimized to minimize repulsion between negatively charged oxygen atoms and between the positively charged silicon atoms. This would result in a tetrahedral arrangement of the negative charges around Si (as is observed) and a linear arrangement of the two Si atoms around each oxygen (the occasionally observed 180 degree angle). Since there aren’t any bonds, VB theory doesn’t really apply, and we consider the oxyanion as a lone atom with completely filled valence shell.
The problem with this argument is that free $\ce{Si^4+}$ is not observed under any circumstances, and most computational approaches suggest a high degree of covalency in the Si-O bond.
Oxygen-oxygen repulsion
A third hypothesis is that VB theory does not apply here because the minimization of energy of the Si-O-Si bond angle is not the only factor. Specifically, oxygen-oxygen repulsion favors a larger angle so as to increase the distance between silicon atoms and, therefore, the oxygens attached to adjacent silicon atoms.
Even at the small scale of a single molecule of $\ce{(HO)3Si-O-Si(OH)3}$, it has been argued that oxygen-oxygen repulsion is sufficient to cause a large Si-O-Si bond angle without any need for $\pi$ bonding[2]. The observed angle is a compromise between repulsion, which favors a 180 degree angle, and bond orbital energy, which favors a bent structure with bond angle closer to 110 degrees.
A problem with this hypothesis is that the Si-O-Si bond angle in disiloxane ($\ce{H3Si-O-SiH3}$) is also observed to be 144 degrees, even though there should be no oxygen atom repulsion effects and general steric effects should be much less than in silicates.
[1]Gillespie, RJ and Johnson, SA (1997) Inorg Chem 36:3031-3039. doi:10.1021/ic961381d
[2]Noritake F (2019) J Comput Chem Jpn 5:1. doi:10.2477/jccjie.2018-0016


In the original German paper [1] Adolf Strecker used Aldehyd-Ammoniak or aldehyde-ammonia as a precursor, that's where the name derives from:

  Vor einigen Jahren habe ich gezeigt, daſs Aldehyd-Ammoniak und Blausäure beim Erwärmen mit verdünnter Chlorwasserstoffsäure sich zu einer schwachen Basis, Alanin genannt, vereinigen [...]:
  $$\ce{\underset{\text{Aldehyd-Ammoniak}}{C4H4O2 * NH3} + HCl + \underset{Blausäure}{C2NH} +2 HO = \underset{Alanin}{C6H7NO4} + NH4Cl}$$

As David Richerby mentioned in the comments, Strecker's brutto-formula ($\ce{C6H7NO4}$) deviates from the moden one ($\ce{C3H7NO2}$), also the reaction scheme is a bit different.

Strecker, A. Annalen der Chemie und Pharmacie 1854, 91 (3), 349–351. DOI 10.1002/jlac.18540910309.


Disproportionation is a redox reaction. Since
$$\Delta G = -zF\Delta E$$
you may compare the redox potentials for both half-reactions in order to determine whether disproportionation occurs spontaneously ($\Delta G < 0$).

Note that nitrogen dioxide ($\ce{NO2}$) exists in equilibrium with dinitrogen tetroxide ($\ce{N2O4}$)
$$\ce{2 NO2 <=> N2O4}$$
and literature values for redox potentials are often given for $\ce{N2O4}$ (values at $\mathrm{pH}=14$):

$$\begin{alignat}{2}
\ce{N2O4 + 2 e- \;&<=> 2 NO2- }\quad &&E_1^\circ = +0.867\ \mathrm{V}\\
\ce{2 NO3- + 2 H2O + 2 e- \;&<=> N2O4 + 4 OH-}\quad &&E_2^\circ = -0.86\ \mathrm{V}
\end{alignat}$$

The potential $E_1^\circ = +0.867\ \mathrm{V}$ for the reduction of $\ce{N2O4}$ is larger than the potential $E_2^\circ = -0.86\ \mathrm{V}$ for the oxidation of $\ce{N2O4}$; therefore, disproportionation should occur.

In equlibrium:

$\begin{align}
E_1^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},1}}{a_{\text{red},1}} &= E_2^\circ + \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}}\\
E_1^\circ - E_2^\circ &= \frac{0.059\ \mathrm{V}}{z}\log\frac{a_{\text{ox},2}}{a_{\text{red},2}} - \frac{0.059\ \mathrm{V}}{z}\log \frac{a_{\text{ox},1}}{a_{\text{red},1}} \\
\left(E_1^\circ - E_2^\circ\right) \cdot \frac{z}{0.059\ \mathrm{V}} &= \log \frac{a_{\text{ox},2} \cdot a_{\text{red},1}}{a_{\text{red},2} \cdot a_{\text{ox},1}} \\
\left(0.867\ \mathrm{V} + 0.86\ \mathrm{V}\right) \cdot \frac{2}{0.059\ \mathrm{V}} &= \log\frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{\left[\ce{N2O4}\right] \cdot \left[\ce{N2O4}\right]} \\
{10^{58.5}} &= \frac{\left[\ce{NO3-}\right]^2 \cdot \left[\ce{NO2-}\right]^2}{{\left[\ce{N2O4}\right]}^2} \\
{10^{29}} &= \frac{\left[\ce{NO3-}\right] \cdot \left[\ce{NO2-}\right]}{{\left[\ce{N2O4}\right]}}
\end{align} $

Therefore, $\ce{N2O4}$ is unstable with respect to disproportionation:
$$\ce{N2O4 + 2 OH- -> NO2- + NO3- + H2O}$$


What happens to the water inside the egg when it is boiled?

The water present in the egg is a colloidal solution of mainly proteins in water.  On boiling proteins get coagulate. The coagulate proteins absorb water (solvent and it disappear).


Mechanism for the conversion of dimethyl cyclohexa-1,4-diene-1,2-dicarboxylate to dimethyl benzene-1,3-dicarboxylate using bromine

Firstly, I didn't get which oxygen you are talking about to break the ring. If you are considering the oxygens that are part of esters, that is not possible. Secondly, $\ce{Br^-}$ is the conjugate base of s very strong acid $\ce{HBr}$. So, $\ce{Br^-}$ is a very weak base. Therefore, it is very very difficult for that $\ce{Br^-}$ to abstract the hydrogen atom and act as a base.

Instead what you can simply do, you can first perform bromination i.e.  treating the substrate with $\ce{Br_2/ CHCl_3}$. You will first get a bromonium intermidiate, like this:

The intermidiate will form on the farther double bond as the $\pi$ -electron cloud of the other one is pulled by the electron withdrawing ester groups, and also to form that ring, there will be huge steric crowding. The other $\ce{Br^-}$ will act as nucleophile to break this ring, (normal mechanism) and you will get this product:

Now, the two $\ce{Br}$ are in anti position. So, if you now treat the compound with excess $\ce{EtO^-}$ in $\ce{EtOH}$ and heat it, anti-periplanar E2 elimination will occur, and finally you will get your desired product as the one mentioned above.

The elimination will occur from two different carbons as the bromines are in anti position. So, to follow anti elimination of $\ce{Br}$ with $\ce{H}$, the elimination has to be from two different non adjacent carbons.


How can endothermic reactions be spontaneous?

A spontaneous endothermic reaction can occur when the changes in enthalpy and entropy yield a negative Gibbs free energy. An endothermic reaction can be spontaneous if entropy increases by more than the change in enthalpy.

$\Delta G = \Delta H - T\Delta S$

For endothermic r̥eactions, $\Delta H$> 0 and thus if the value of $-T\Delta S$ overshadows the $\Delta H$ value (usually at very high temperatures) then the overall value of $\Delta G$ becomes negative and thus the reaction becomes spontaneous.


How does precipitate affect cell potential?

The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to
$$E_\ce{Pb} = \pu{-0.13 V} + \pu{0.0296 V}\times\log{10^{-8}} = \pu{-0.37 V}\tag{1}$$
As a consequence, the $\ce{Pb}$ potential decreases. But the cell potential increases and goes from the initial value (without sulfate)
$$E^\circ_\mathrm{cell, i} = \pu{0.34 V} - (\pu{-0.13 V}) = \pu{0.47 V}\tag{2}$$
to the final value (after precipitation of lead sulfate)
$$E_\mathrm{cell, f} = \pu{0.34 V} - (\pu{-0.37 V}) = \pu{0.71 V}\tag{3}$$


What's a good resource/book on pH measurements?

Looks like you understand pH measurement, but you are trying to go deeper into the details of potentiometry and open circuit potential measurements.

There I would recommend basic electrochemistry texts. Here is one simple book that is cheap, easy to read and very informational:

Electrode Potentials; Richard G. Compton and Giles H. W. Sanders


From the point of atom efficiency, methylsulfonate is advantageous compared to the p-toluenesulfonate.
On the reagent itself, the methanesulfonate is more polar than the p-TsOH; so the later dissolves better in less polar organic solvents than the former.  Methanesulfonate even may be used to dissolve some metal salts.
If used to generate a better leaving group, on workup, a methanesulfonate enters the aqueous phase easier than the tosylate.
p-TsOH is a cheap, widely available solid (sometimes sold as hydrate) you may prepare by your own in large quantities; methanesulfonate however is a liquid, hence potentially easier to add continuously, and more expensive (even more so, per mL, the triflates).
Methyl sulfonates are able to alkylate DNA, hence potentially carcinogen.  For tosylates, currently there is not such evidence (but obviously the dust of this strong acid is considered harmful).

As a personal note, among other reasons I found consultation of journals like Organic Process Research & Development just under this aspect (including green chemistry) as useful in two ways: i) publications describing how running a particular reaction within a chain of others works well / better than an other -- especially on larger scales of application--, and ii) the sections in line of "Some Items of Interest to Process R&D Chemists and Engineers".

Addendum:  There are dedicated compilations to figure out scope and limitations of chemical reagents.  To mention only two, Fieser & Fieser and EROS edited by Leo Paquette (there is an e-EROS as electronic ressource, too) are multivolume references.


How could I produce/rig this flame-colour lottery?

Smokeless and quick burning isn't hard to achieve, just use flash paper. For white smoke, I would use flash paper again but with a white oxide powder such as titanium oxide with a flame retardant to slow the burn. For red, again flash paper but, you need a compound that releases bromine when burned. I think copper(i) bromide or silver(i) bromide would be ideal as both are easy to decompose and tolerant to moisture in air, though silver bromide is not tolerant to prolonged light exposure. These salts may discolor the paper and you may need to color the non-red smoke papers to make them all look the same.


How do we determine if some metal will undergo passivation under suitable environment?

There is not much we can do. It is certainly not a property that is explained by a rigorously established single measurable parameter like oxidation potential. Passivation is a delicate interplay of multiple factors, some of which are not even related to the metal itself (remember, a metal may be passive in some environments, but not in the others).

However some trends can be observed. Surely, having a hard, high-melting, and chemically inert oxide does help a lot.


Reaction of 1-(1-chloroethyl)-1-methylcyclobutane under E1 Conditions

You are correct that an alkyl shift will take place but there are two possible options for which group moves. Your option is possible but there is another option which involves a ring expansion to generate product A.

This mechanism is favoured because it relieves the ring strain in the four membered ring. Since the reaction is run under reversible conditions, the thermodynamic product (with the least ring strain) will dominate.


I would bet activation energy for this sort of reaction is quite high, and since primary and tertiary carbons aren't hugely different in stability the sort of H for Li exchange reaction would be very slow. For nBuLi, the reaction would practically not happen, and for sBuLi and tBuLi intramolecular rearrangement (proton shift) is more likely than reaction with solvent, resulting in nBuLi and iBuLi respectively.

Agreed that elimination and precipitation of LiH will be faster than alkyl exchanges in these solution, though I've never checked an old bottle of tBuLi for any precipitate so this is entirely speculative.

Investigating this would be quite challenging, as loss of base power in practice usually happens from slow exposure to O2 and H2O,which would resolve Li2O and LiOH respectively, also basic precipitates.


What property of zinc chromate makes it a good anticorrosion paint

Long story short, if it's scratched in the presence of water(humidity) the chromium ion inside the chromate will get reduced to Cr 3+ from Cr 6+. This 3+ ion will then form a Chromium hydroxide compound,which is insoluble in water and corrosion resistant, inside the scratch. This reduction of the chromium ion first is what inhibits the oxidation of the underlying metal.

Here's the full article http://www.electrochem.org/dl/interface/wtr/wtr01/IF12-01-Pages34-38.pdf

The answer: Being water resistant in the first place and being capable of producing ions that can be reduced forming another corrosion resistant compound( chromium(III)oxy-hydroxide.)


Yes, resonance should never stop at the most likely resonance structure. If you’re really willing to play the game, you can take it even further and consider singlet diradical structures where the bond has been broken into a spin up and a spin down electron, each one of which resides in a different orbital. And more, and more.

(Side note: this is why it is useless to ‘count’ resonance structures. The only correct answer to ‘How many resonance structures does $\ce{X}$ have?’ is ‘Always one more.’.)


Hybridization of carbon atoms present in the smallest ester?

You are right. Methyl formate has one $\mathrm{sp}^2$ and one $\mathrm{sp}^3$ hybridized orbital.

Your book probably has a misprint.


Are the following expressions for base dissociation constant equivalent?

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$$
This expression is infact $1/K_\mathrm a$ which is related with $K_\mathrm a K_\mathrm b=K_\mathrm w$

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$$
here $K_\mathrm b$ is correct but nowadays any constant concentration (here water) is not written in the expression. As it is assumed that concentration of water is not changing when reaction proceed.


What is the standard molar weight of Carbon, and why?

As you stated the molar mass is a weighted average based on isotopic abundance, specifically abundance on Earth. We can obtain the molar mass of carbon by averaging the mass of $\ce{^{12}C}$ (about $98.938\%$ abundance) and $\ce{^{13}C}$ (about $1.078\%$ abundance). $$(\mathrm{mass\space of\space \ce{^{12}C}})*(.9889)+(\mathrm{mass\space of\space \ce{^{13}C}})*(.0111)=(12)*(.9898)+(13.00335)*(.0111)=12.0206\text{g}$$ This deviates from your value above because the values I used for the abundance aren't very precise.

In terms of where these abundances comes from and how the error bars are determined, the short answer is that its complicated. The long answers is here in a technical report by IUPAC, particularly sections 1.2-1.4.


Extended Hückel method: what are the atomic orbital energies usually used?

Please see the following reference:

  A. Herman, Modelling Simul. Mater. Sci. Eng. 2004, 12, 21.

It has $H_{vv}$ terms for different orbitals of all elements of the table.


Quoting Greenwood & Earnshaw's Chemistry of the Elements, discussing the triatomic polyhalide anions:

  [I]nteratomic distances [...] are individually always substantially greater than for the corresponding diatomic interhalogen.

Unfortunately, the following data table does not contain an actual value for the bond length, though it does indicate that the two $\ce{Br-Cl}$ bonds are of equal length (which we probably could've safely assumed from symmetry considerations and the energetic degeneracy of the orbitals of the two chlorine atoms, if nothing else), and that the $\ce{Cl-Br-Cl}$ bond angle is (approximately) $180^{\circ}$. Bond length can often be correlated to both bond strength and bond multiplicity, especially when comparing compounds of the same atoms. Since the bonds in $\ce{BrCl2-}$ are longer than those in $\ce{BrCl}$, it's reasonable to infer that they are also weaker, and do not have greater multiplicity (i.e., there is no additional $\pi$-bonding). I would further reason that any $\pi$-orbital overlap would have to be very poor (judging by the mismatch in both size and energy of the $3p$ and $4p$ orbitals of chlorine and bromine, respectively), which makes $\pi$-bonding unlikely. (For the sake of completeness, I would note that I could be mistaken here, since $4p$ orbitals are somewhat smaller and lower in energy than expected, given the poor shielding of the nucleus by $3d$ electrons. That said, I don't think the magnitude of this effect is large enough to make a major difference here.)

A full quantitative molecular orbital theory treatment of the molecule would require computation, and I wasn't able to find many useful literature references. However, as the molecule is relatively simple, we can probably make reasonable progress with even a very naive analysis. On the valence level(s), we have a total of $22$ electrons in $12$ twelve atomic orbitals, all $s$ or $p$. These will combine to form twelve molecular orbitals, eleven of which are going to be doubly occupied. Note that for any bonding MO generated, an anti-bonding MO will also be generated. Hence, even if we had six bonding MOs (and, in reality, some would almost certainly be non-bonding), we would have five anti-bonding MOs occupied as well. This would yield an overall bond order of $1$, indicating a bond order of $\frac{1}{2}$ for each individual bond. By analogy to similar (but more common and hence extensively studied) polyhalogen anions (e.g., $\ce{I3-}$), this conclusion seems justified. $\ce{I3-}$, and numerous similar isoelectronic molecules, are often described using the three-center four-electron (3c4e) bond, and I think this model is probably applicable to $\ce{BrCl2-}$ as well.

The upshot of all this is that, while it's true that you can draw certain valid resonance contributors, they may not accurately reflect the structure and electron density of the actual molecule. The structure with the negative formal charge on the bromine is best, which agrees with your intuition regarding its size and polarizability being able to stabilize the charge. Due to the disparity in electronegativity, however, the bonds are nevertheless polarized towards the chlorine atoms, so the formal charge alone doesn't fully describe the electron density correctly.


Why don't we see these lanthanide species?

This is really the exact same question but just in a different context of transition metals: Cr(II) and Mn(III) - their oxidizing and reducing properties?

The answer is because exchange energy (which is the reason behind the stability of half-filled and fully-filled shells) is not the main factor that decides the predominance of an oxidation state.

Whether an oxidation state can be reached depends on the balance between two things:

An energetic cost of having to ionise the element, i.e. the ionisation energy
The energetic payback obtained by forming covalent bonds / ionic bonds / solvation in solution / etc.

Exchange energy effects only give rise to minor variations in ionisation energies. The increase going from the nth to (n+1)th ionisation energy is far greater than any exchange effects.

Just like the magnesium example I gave in the other question, the lanthanides have very small second ionisation energies, and these can also be easily compensated for by bonding or ion-dipole interactions in solution. Take your example of $\ce{Sm+}$. The fact that it has a $\mathrm{f^7}$ configuration means that the second IE is marginally larger than would be expected. However, this marginal increase hardly changes the fact that the cost of ionising Sm a second time can be easily recouped.

Likewise, your example of $\ce{Pr^5+}$ is totally unrealistic because those 5th IEs are so huge that there is absolutely zero chance that they will give it up, no matter what exchange stabilisation there is.

The predominance of the +3 ionisation state for the lanthanides is therefore simply a consequence of IE1 + IE2 + IE3 being sufficiently small, and IE4 being way too large. That's all there is to it. Note how the variations in the graph below (due to exchange energy effects) are so small that all the IE graphs never cross over each other. IE5 is so huge that they didn't even bother plotting it!

(source: Inorganic Chemistry 6ed, Weller et al., p 630)


The deuteration of phenol and aniline hydrochloride has been investigated by Werstiuk and Kadai by the High Temperature/Dilute Acid (HTDA) method. For phenol, ortho/para substitution is faster than meta (2.7 D/3.0 H vs. 1.3 D/2.0 H). OD/OH exchange was conducted at the end of the reaction. The presence of conc. HCl undoubtedly lowers the efficiency of incorporation of deuterium.
Incorporation of deuterium without the presence of conc. HCl allows for efficient deuteration of aniline hydrochloride. [RND3+ to RNH3+ exchange was conducted.] The authors note that conversion of aniline hydrochloride-d5 to phenol-d5 is efficient via diazotization.

Werstiuk, N. H.; Kadai, T., Can. J. Chem., 1974, 52, 2169.


Why is the test for sodium as dihydrogen antimonate so specific?

It has no (direct) relation to the electrochemical series nor to acido-basic properties.
It is not a true chemical reaction, rather just crystallization of the least soluble salt from the available ions.
There are no distinguished molecules $\ce{NaCl}$, $\ce{KH2SbO4}$, $\ce{NaH2SbO4}$, $\ce{KCl}$ in the solution, but mutually independent hydrated ions $\ce{Na+}$,  $\ce{Cl-}$, $\ce{K+}$, $\ce{H2SbO4-}$ instead (plus some minor hydrolysis products). Which salt precipitates first depends on their solubility products. $\ce{NaH2SbO4}$ has smaller solubility product than $\ce{KH2SbO4}$.
Similar things would happen if you mix solutions of potassium  chloride and sodium nitrate ( chlorate, perchlorate ). If original solutions are concentrated enough, and especially if you cool the mixture below ambient temperature, there will precipitate the respective potassium salt with the oxoanion.
Another case is the $\ce{NaCl}$ solution  with dissolved $\ce{NH3}$ And $\ce{ CO2}$, with ions $\ce{Na+, NH4+, Cl^-, HCO3-}$, leading to precipitation of $\ce{NaHCO3}$ and solution of $\ce{NH4Cl}$. It is used commercially in the Solvay process
producing sodium carbonate from sodium chloride.


Why does an element's chemical properties rely only on its valence electrons and not on anything else?

Technically, other electrons could play a part in determining chemistry. However, there is a significant energy barrier between the p-orbitals of one period and the s-orbital of the following period.

Consider copper. The characteristic deep blue colour of $\ce{[Cu(NH3)4(H2O)2]^2+}$ derives from the absorption of photons whose energy corresponds to the difference of the $\mathrm{3d}_{z^2}$ and $\mathrm{3d}_{x^2 - y^2}$ orbitals. This is spin-allowed because copper(II) is a d⁹ system. The wavelength corresponds to approximately $610\:\mathrm{nm}$.

On the other hand, the energy between a $\ce{2p}$- and a $\ce{3s}$-orbital can be measured using x-ray transition — it corresponds to a photon of a wavelength of approximately $1\:\mathrm{nm}$. This is about six-hundred times the energy difference compared to inside the d-orbitals.

Of course, this is a slightly unfair comparison. The energies between $\ce{3d}$ and $\ce{4s}$ are noticeably larger than inside of the $\ce{3d}$ orbitals, and the energy difference between $\ce{4s}$ and $\ce{4p}$ should again be well larger — but they will remain significantly smaller than energy differences across shells. The former are typically visible or UV, the latter are typically x-ray.


What is the direction of the net electric dipole moment in hydronium H3O+?

In aqueous medium, the proton ($\ce{H+}$) is thought to exist as hydronium, $\ce{H3O+}$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be difficult to measure the dipole moment of $\ce{H3O+}$ experimentally. I haven't found any reference that did this, and I suspect it is impossible.
So we have to turn to computational chemistry to calculate the dipole moment of $\ce{H3O+}$. For one isolated $\ce{H3O+}$ (i.e. in gas phase):

The z-axis is set along the $C_3$ axis.
$\text{M06-2X/aug-pcseg-2}$ gives a dipole moment of $\text{1.515 D}$ in z-direction. In x and y direction, the dipole values are negligible:
         DX          DY          DZ         /D/  (DEBYE)
    -0.007834    0.000072    1.515546    1.515566

This means the dipole points along +z-axis in the picture. Partial charges(ESP) fitted to the dipole moment are:
 NET CHARGES:
 -------------------------------------
 ATOM                CHARGE    E.S.D.
 -------------------------------------
 O                  -0.3668    0.0000
 H                   0.4561   -0.0000
 H                   0.4553   -0.0000
 H                   0.4554    0.0000
 -------------------------------------

As you can see, the oxygen atoms has a partial negative charge and hydrogen atoms have partial positive charges. So, the formal charges that we assign to the Lewis structure does not have much significance in this case.
(IUPAC convention of dipole moment is that the arrow points from the negative charge towards the direction of the positive charge. This is exactly opposite of what you wrote in the question)
Another method, $\text{MP2/aug-pcseg-2}$ gives $\text{1.507 D}$ along z-axis, which is close to what we got from the DFT calculation.
(Note that I have not enforced symmetry for the calculation, which is why the partial charges are slightly different for each hydrogen. This is a numerical error, because in reality all the hydrogens are equivalent)


Why is the only H NMR resonance of 1,4-dibromobenzene a singlet?

All four hydrogens are equivalent and therefore they do not display any coupling to each other. (They do couple, but the coupling cannot be seen in the spectrum.) See: Why don't equivalent hydrogens cause splitting in NMR? for more discussion of this topic.

Just to be extra nitpicky, I wouldn't write "all four hydrogens show up as singlets", since that implies that there are four different singlets. I'd write something like "one singlet is observed in the spectrum, corresponding to all four hydrogens".


Why radioactive elements emit alpha beta and gamma rays

The stability of nuclei is really a sophisticated topic in theoretical quantum mechanics. But there is a simple way to think about what is happening that doesn't get too intense with the quantum mechanical theory.

Nuclei are made from two particles: protons and neutrons. But protons are positively charged and repel each other. The electromagnetic force is very strong and therefore this force is very large. So the first mystery is why all nuclei don't just fly apart.

The reason they don't is that there are two very short-range but very strong forces that bind the nucleus together: the strong and weak nuclear forces. Without getting into mind-bending topics in theoretical physics we can understand something about their net effect like this.

The interaction of the electromagnetic force and the two nuclear forces has some structure (it's quantum stuff, just accept it). Some combinations of protons and neutrons are more stable than others. Each combination has an energy level and some combinations have lower energy than others. Nuclei with even numbers of protons and neutrons are more stable than odd-odd combinations and nuclei with wildly unbalanced neutron to proton ratios are less stable. Neutrons act a little like a glue, helping protons stick together (this is an oversimplification as too many neutrons is also a source of instability: this is a consequence of a complicated interplay of several forces). But bigger nuclei are less stable and need a higher ratio of neutrons. And some large nuclei are just too large for the forces to keep them together so beyond a certain point all nuclei are unstable.

Some nuclei can be transformed into a more stable (lower energy) nucleus by various forms of radioactive decay. Nuclei with too many neutrons can emit a beta particle (this decay mode converts a neutron into a proton); elements with too many protons can emit a positron (converting a proton into a neutron). Bigger nuclei can become more stable by kicking out an alpha particle (which makes the nucleus significantly smaller, moving it towards the stable zone). Gamma radiation is associated with some of these modes: the high energy photons "mop up" the excess energy (I'm simplifying a lot).

This picture (more detail on wikipedia) sums up what we know about nuclear stability:

To sum up: nuclei are held together by the interaction of several forces but the forces not equally effective for all possible combinations of protons and neutrons. Some nuclei have so much energy that they are unstable and can decay into more stable nuclei in various ways depending on the mix on neutrons and protons they have. In some cases the forces hiding the nuclei together are strong enough that they never decay.


Although fluorine is more electronegative than chlorine it has less electron gain enthalpy because of its small size.

The electrons in the outer shell of fluorine repel the incoming electron more effectively than the electrons in chlorine. More closer the electron comes to the nucleus more energy is released. In case of chlorine, the electron comes more close to the nucleus.

Thus, chlorine has more electron gain enthalpy.

Size matters in chemistry. It is more dominating factor than electronegativity in most cases.

You would not expect enormous difference in enthalpy of formation of $\ce{CsF}$ and enthalpy of formation of other alkali halides. In case of $\ce{CsF}$, The bond length of the resultant salt will be larger than other alkali halides due to large size of cesium atom and small size of fluorine atom. The bond between them would not be very strong. Consequently, the reaction will not be strongly exothermic.

Putting the theory part aside, here is the data :

$\Delta{H^{\circ}_f}$  -  $\ce{CsF}$ $\rightarrow$ $-550\frac{KJ}{mole}$,

$\Delta{H^{\circ}_f}$ - $\ce{NaF}$ $\rightarrow$ $-575\frac{KJ}{mole}$,

$Electron\space gain\space enthalpy\space - Chlorine$ $\rightarrow$ $-349 \frac{KJ}{mole}$

The comparisons in your question are very different and they can only be compared by looking at the data. The prediction will be very inaccurate. If you want to compare two properties, you will at least need some similarity between the two properties. Like, you can compare enthalpy of formation of two different alkali halides or electron gain enthalpies of Halogens or chalcogens. Electron gain enthalpy and enthalpy of formation of alkali halides are very different things.


The lowest energy conformer for a cyclohexane ring with multiple substituents will be the one in which the largest group occupies an equatorial position.

Steric parameters are tabulated in many organic chemistry textbooks, the most common of which is the A-value. The A-value represents the difference in energy (ΔG) between a group being axial/equatorial on a cyclohexane ring, thus, as a group gets bigger, its A-value increases, reflecting the increasingly high energy needed to flip the ring.

Bulky substituents such as tert-butyl are often said to "lock the conformation", that is, with a tert-butyl group on a ring, the energy to flip is sufficiently high that ring is "locked" in the conformation with the tert-butyl group equatorial.

In your example, building a simple model with a kit will show you that having the iso-propyl group axial isn't terribly favourable (interestingly the iso-propyl group has a similar A-value to ethyl, due to the ability of the group to rotate, avoiding having a methyl group pointing into the ring).

I ran a few simple calculations (molecular mechanics followed by DFT optimisation - gas phase) on the axial and equatorial conformations, which showed that the two conformations are separated by around 3.4 kcal/mol.  The minimum energy conformation of your example is shown below:


Why isn't dry CaCl₂ used to dry HCl gas instead of concentrated sulfuric acid?

There are several reasons:

Efficiency
When a stream of wet hydrogen chloride is run through a tube filled with coarse calcium chloride "rocks", interaction with the drying agent takes place at at its surface.
When the same wet $\ce{HCl}$ ist passed through sulfuric acid, the contact area is given by the surface area of the bubbles. Moreover, bubbles mean mixing. The trapped water is distributed over the drying agent.
Safety
Wet calcium chloride will clog, which might result in a pressure buildup in the drying tube.


Derivation of formula used in calculating molecular mass from a mixture

Hint: $$v_{rms}=\sqrt{\frac{3RT}M}$$

And weighted average of speeds for mixture.

EDIT:

$$ v_{mix} ^{rms}= x_1 v_1^{rms}+ x_2 v_2^{rms}$$

This will complete the derivation of the formula you gave.


Rate law of the reaction of propane with oxygen

The oxidation of hydrocarbons ($\ce{RH}$) such as this one proceed via a chain mechanism. The initiation step is

$$\ce{RH + O2 -> HO2^. + R^.}$$

and the propagation

$$\ce{ R^.+O2 -> RO2^. \\ RO2^. +RH -> ROOH +R^.      }$$

But it does not stop there as the hydroperoxide decomposes $\ce{ ROOH -> RO^. + OH^.}$ and there are other reactions such as

$$\begin{align}
\ce{ R^. + O2 -> RO^. + O^. \\ RH + OH^. -> H2O + R^. \\ RCH2OO^. + O2 -> RCH2O + O3  \\ RH + O3 -> RO^. +HO2^.}\end{align}$$

and chain branching $\ce{ \quad O^. + RH -> OH^. + R^.}$.

The exact mechanism depends on temperature and pressure, and in a plot of $T$ vs $p$ there are regions of slow reactions, explosion and hot and cold flames. (An original paper is by Newitt & Thorne, J. Chem. Soc 1937, p1656). Thus you can appreciate that sorting out the mechanism is no simple task and will need much specialised equipment.


Is having a lone pair sufficient and necessary for a Lewis base? What about empty orbitals and Lewis acids?

No, not every molecule that has a valence lone pair is automatically a Lewis base.

For example, tin forms divalent salts such as $\ce{SnCl2}$ wherein tin is in the $\mathrm{+II}$ oxidation state and thus its electronic configuration is $\ce{[Kr]}\ \mathrm{4d^{10}\,5s^2\,5p^0}$. Even if we say that the d-electrons are not really valence electrons, since they are not really chemically accessable, you cannot deny that tin(II) has a lone pair in its 5s orbital. However, this lone pair does not exhibit Lewis basic properties; instead, tin(II) functions as a Lewis acid. The reason for that is the spherical symmetry of the s-orbital along with its notable dispersion, which means it cannot be reasonably donated in a Lewis-basic matter.

Note that the lone pair is accessable chemically: tin(II) can be oxidised to tin(IV) e.g. by $\ce{HgCl2}$.

A somewhat similar (but much more questionable!) case can be made for electron sextets and Lewis acids. NHCs are a pretty common substance class. The most generic structure is that of imidazole, but with both nitrogens substituted so that one of them bears a positive charge (the one with a double bond to the carbon sandwiched between the two). If a base is added to remove that sandwiched carbon’s proton, a carbene is generated in which said carbon is only bonded to the nitrogen, has an $\mathrm{sp^2}$ configured lone pair and is considered to have an empty $\mathrm{p}$ orbital. However, this carbene is nucleophilic, which is why they are also called nucleophilic heterocyclic carbenes. They react as Lewis bases with metal centres; the second generation of Grubbs’ metathesis catalysts made use of NHCs.

Of course, it is slightly debatable whether the p-orbital on that NHC carbon is truly empty. One could argue that significant electron pair donation from both nitrogens should happen.

(As a side note: radicals, while not having full octets, are typically never considered Lewis acids — but then, their orbital is already occupied by a single electron, so they don’t have an empty orbital that would make them a potential Lewis acid, anyway. Thus, your question should better be worded differently to explicitly exclude radicals.


It's not a catalyst here. It speeds up the reaction, which is why it acts like a catalyst.

Here is the reaction without sulfur:

$\ce{6 KNO3 + C7H8O -> 3 K2CO3 + 4 CO2 + 2 H2O + 3 N2}$

And here it is with sulfur:

$\ce{4 KNO3 + C7H4O + 2 S —> 2 K2S + 4 CO2 + 3 CO + 2 H2O + 2 N2}$

(Source)

Sulfur decreases the ignition temperature, as well as probably contributing a bit to the rate. Here, it is effectively turning $\ce{CO2}$ and $\ce{K2CO3}$ to $\ce{CO}$ and $\ce{K2S}$. If you take into account Le Chatelier's principle, the conversion of the products of the basic reactants speeds it up. (The decrease in concentrations/activities of the products due to their conversion means that the equilibrium shifts in the forward direction to compensate for it.)

So it is speeding up the reaction and making it easier to carry out. Not a catalyst.

A note: There's no reason why any element/compound can't act as a catalyst. Transition metals are generally catalysts in a certain type of reaction, and they generally catalyze the reaction by adsorbing the gases. Catalysts can work in many other ways—generally, they form intermediates in the reaction and present an alternative path via the intermediate. ($\ce{V2O5}$, for example, is a catalyst in many reactions involving organic compounds since it forms cyclic intermediates with various groups).


Yellow precipitate in the presence of ammonium molybdate?

[...] diluted $\ce{H2SO4}$ and $\ce{HNO3}$ [...] $\ce{NH4OH}$ [...] and ammonium molybdate.

The yellow precipitate indicates that phosphate was present too. Under these conditions a yellow ammonium molybdatophosphate is formed:

$\ce{H2PO4^- + 22 H+ + 12 MoO4^2- +3 NH4+ -> \underset{yellow}{(NH4)3[P(Mo3O10)4]} +12H2O}$


Thermochemistry Problem - How did they get that volume?

I think most of your issues regarding this question will be solved if you actually put the states in your chemical equations, so it becomes: $$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$ It is important to note that all the substances are in the gaseous state except for water which is a liquid. Therefore your application of Gay-Laussac's law is incorrect here as the number of moles of gaseous reactants is not equal to number of moles of gaseous products. Thus the initial and final volume shouldn't be the same. This is also why in the question they have stated to assume that the volume of water is negligible as water is a liquid which has very little volume compared to a gas.

The change in volume can be calculated by noting that 3 moles of gas reacts to form only 1 mole of gas. Since there is 2L of methane, that means that the volume is going to decrease by 4L.

Therefore: $\mathrm{V_f = V_i - 2.0L}$


(b) does not show resonance because the second resonance structure (with negative charge at oxygen and double bond in the ring) violates Bredt's rule. The carbon $\ce{sp^3}$ orbital with the negative charge and the $\ce{p_z}$ orbital of the adjacent carbonyl C cannot overlap because they are not in plane, so the charge cannot be delocalized into the carbonyl group.


Does Le Chatelier's principle apply to the gas inside an air mattress?

No, the air we breathe is about 78% nitrogen, 21% oxygen, 0.5% carbon dioxide and other gases. Nitrogen doesn't react with anything at room temperature and there is nothing to burn inside the air mattress.

What probably happened was that there is a very small hole that the air leaked out. About the only way to find the hole would be to submerge the air mattress in a pool.


How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value?

$$\ce{NaOH + H2O-> Na+  +OH- + H2O}$$
\begin{align}
\mathrm{pH}_1 = 7.00 &\implies\mathrm{pOH}_1 = 7 \\
\mathrm{pH}_2 = 12.42 &\implies\mathrm{pOH}_2 = 1.58
\end{align}
so $[\ce{OH-}]_1 = 10^{-7}$ and $[\ce{OH-}]_2 = 10^{-1.58}$
So, initially, you have $16^{-3}\times 10^{-7}$ mol of $\ce{OH-}$ in your water. Your $\ce{NaOH}$ solution is $\pu{0.3 M}$ so you'll bring $\pu{0.3\times 10^{-3} mol}$ each $\mathrm{mL}$.
Then, the resulting concentration will be $$C = \frac{n_\text{initial} + (0.3\times 10^{-3})\times x}{V_\text{initial} + x\times \pu{1mL}}$$
You know $V_\text{initial}$ to be $\pu{16 mL}$
You know $n_\text{initial}$ to be $\pu{1.6\times 10^{-9} mol}$
And you know the target concentration to be $10^{-1.58}$
I let you do the solving for $x$ where $x$ is the volume of $\ce{NaOH}$ solution to be added ;) be careful with the units.


Why are oxime geometrical isomers stable?

It is the same reasoning that is used to predict relative barrier heights in amines.

In the example pictured below, the imine starts with the nitrogen lone pair in "more or less" an $\ce{sp^2}$ orbital.  In the transition state for interconversion, the lone pair is in a $\ce{p}$ orbital.

The term "more or less" is important.  If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit more p character than $\ce{sp^2}$ (say $\ce{sp^{2.2}}$ for example), then it is already closer to the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be reduced.  On the other hand, If we start with a molecule that has the nitrogen lone pair in an orbital that has a little bit less p character (more $\ce{s}$ character) than $\ce{sp^2}$ (say $\ce{sp^{1.8}}$ for example), then it is further from the $\ce{p}$ hybridized transition state and the activation energy required to reach the transition state will be greater.

In the case of oximes ($\ce{X=OH}$) or other molecules where the substituent on nitrogen is electronegative (electron withdrawing), the electron density in the $\ce{N-X}$ bond is shifted towards the electronegative atom.  Consequently there is less electron density in the nitrogen portion of this bond.  As a result the nitrogen rehybridizes. Since $\ce{s}$ orbitals are lower in energy than $\ce{p}$ orbitals, orbitals with increased $\ce{s}$ character are better able to stabilize electrons. Since there is less electron density in the orbital nitrogen is using for this bond, $\ce{s}$ character is removed from the orbital.  This "extra" $\ce{s}$ character is used to stabilize the orbital containing the lone pair of electrons.  The lone pair orbital now has more $\ce{s}$ character and is further away from the $\ce{p}$ hybridized transition state - the barrier to reach this transition state increases.  If our substituent is an alkyl group (X=R, an imine), the carbon is less electronegative then nitrogen, there will be increased electron density in the nitrogen orbital used to form the $\ce{N-R}$ bond.  The nitrogen will hybridize such that there is more $\ce{s}$ character in this orbital so as to better stabilize these electrons.  As a consequence, there will less $\ce{s}$ character (more $\ce{p}$ character) in the lone pair orbital.  The lone pair orbital is therefore closer to the $\ce{p}$ hybridized transition state and less energy will be required to reach the transition state.

As a result of this reasoning $\ce{NF3}$ has a higher inversion barrier then $\ce{NH3}$ and oximes have a higher barrier to inversion than imines.


Strongest chloride complex as predicted by HSAB theory

The link to your handout explains why $\ce{Fe^{3+}}$ (and it is $\ce{Fe^{3+}}$ in $\ce{FeCl4^{-}}$) is a harder acid than $\ce{Co^{3+}}$ or $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$.

  Hard acids (in context, HA) are characterized by (s,f blocks, left side of d block in higher OS's)

Iron is the furthest to the left of the four metals you mention. Thus, it has fewer electrons in its $d$ shell to balance the positive charge:

$\ce{Fe^{3+}}$ is $\ce{[Ar]}3d^5$
$\ce{Co^{3+}}$ is $\ce{[Ar]}3d^6$
$\ce{Ni^{2+}}$ is $\ce{[Ar]}3d^8$
$\ce{Cu^{2+}}$ is $\ce{[Ar]}3d^9$

and

  Low electronegativity (χ) of the acidic atom. A value in the range 0.7-1.6 is typical of hard acids;

Iron has the lowest electronegativity of the four metals:

$\ce{Fe}$ 1.83
$\ce{Co}$ 1.88
$\ce{Ni}$ 1.91
$\ce{Cu}$ 1.90

Iron is the hardest Lewis acid of the metal cations you list.

As for $\ce{Ni^{2+}}$, it does form a complex $\ce{NiCl4^{2-}}$ with $\ce{Cl-}$. Here is a Google search for it.


How to use dispersion correction with TPSSh in Gaussian 16?

The answer is quite simple and already given in the manual:

  You can use this empirical dispersion method with other functionals via the IOps(3/174-178) (S6 should be 1.0).

Parameters and the standalone program can be found can be found on the DFTD3 homepage at Mulliken Center for Theoretical Chemistry. Here are links via The Internet archive to the tables for zero-damping and BJ-damping.

The Overlay3 manual can be found on gaussian.com, or The Internet Archive:

  IOp(3/174)

  S6 scale factor in Grimme’s D2/D3/D3BJ dispersion.

0          Default (see subroutine R6DS6).
-1         Set S6 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

  IOp(3/175)

  S8 scale factor in Grimme’s D2/D3/D3BJ dispersion.

0          Default (see subroutine R6DS8).
-1         Set S8 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

  IOp(3/176)

  SR6 scale factor in Grimme’s D2/D3/D3BJ dispersion.

0          Default (see subroutine R6DSR6).
-1         Set SR6 to 0.
NNNNNNNN   A value of NNNNNNNN/1000000.

  IOp(3/177)

  A1 parameter in Becke-Johnson damping for D3BJ and XDM.

0          Default (see subroutine R6DABJ/XDMABJ).
-1         Set A1 to 0.
NNNNNNNN   A value of NNNNNN/1000000.

  IOp(3/178)

  A2 parameter in Becke-Johnson damping for D3BJ and XDM.

0          Default (see subroutine R6DABJ/XDMABJ).
-1         Set A2 to 0.
NNNNNNNN   A value of NNNNNN/1000000 Ang.

For TPSSh the route section should therefore look similar to the following:

#P TPSSh/def2TZVP           ! verbose, method/ basis set
EmpiricalDispersion(GD3BJ)  ! Switch on D3 with Becke-Johnson damping
IOp(3/174=1000000)          ! Set s6 to 1.000
IOp(3/175=2238200)          ! Set s8 to 2.2382
IOp(3/177=452900)           ! Set a1 to 0.4529
IOp(3/178=4655000)          ! Set a2 to 4.6550
opt(MaxCycle=100,VeryTight) ! Run optimisation
int(ultrafinegrid)          ! use finer grid for dfa


Difference between conformational, constitutional, and structural isomers and the same and different molecules

I'm going to assume that you don't need an explanation for the checkboxes the same molecule and difference molecules, so that leaves the following:

The terms constitutional isomers and stereoisomers make up two broad categories of isomers (molecules with the same chemical formula).  Constitutional isomers typically have different connectivities and stereoisomers have the same connectivities but differ in spatial arrangements.

enantiomers are stereoisomers that are non-superimposeable mirror images of one another.  Your hands are (roughly) enantiomers.  both are optically active is true of enantiomers.

diastereomers also known as geometric isomers are also stereoisomers that are not mirror images of one another. cis- and trans-dichloroethene are diastereomers.

If you do a web image search of isomer flowchart you'll see a lot of informative information on isomers.  Linked below is the public domain version from Wikipedia which is an example (although not the best in my opinion.)


Is there a more direct way to show this identity for the projection operator?

Before I go into how this result can be derived, I want to point out a mistake in your attempt. $|\psi\rangle\langle\psi|\neq1$, but rather $\sum_i|i\rangle\langle i|=1$, where the sum of is over a complete set of states. Basically, this result only holds when you sum over all the basis vectors that span the space you are studying, not just for any single vector in the space. Its easiest to understand this in analogy to more conventional vectors. Assuming a column vector $v$ is normalized, than $v^\dagger v=1$, which is the inner/dot product of the vector with itself. This is analogous to $\langle\psi|\psi\rangle=1$ in bra-ket notation. If we reverse the order of the product, $vv^\dagger=\mathbf A$ where $\mathbf A$ is some matrix. This operation is called an outer product and is akin to $|\psi\rangle\langle\psi|$. As it turns out, for a complete orthonormal basis, the sum of these outer products has to equal the identity matrix $\mathbf1$. We can make sense of this by recognizing that the sum is a sum of projectors onto all the basis vectors, which in QM might be all of the possible eigenstates of the system. Your wavefunction has to be some linear combination of these basis vectors, so if your project your wavefunction onto all the eigenstates and then sum those projections, you should get your original wavefunction back, hence the sum of the all the projectors is the identity matrix.

For the actual derivation, the result proceeds directly from the time dependent Schrodinger equation (TDSE) $$i\hbar\frac{\partial\psi}{\partial t}=H\psi\to i\hbar\frac{\partial|{j}\rangle}{\partial t}=H|j\rangle$$

where in the second half of the equation, I have written the equation for the state $j$ that the projection operator references. Writing the projection operator explicitly as $P_j=|j\rangle\langle j|$, we can rewrite your original
expression as $$i\hbar\frac{\partial(|{j}\rangle\langle j|)}{\partial t}=i\hbar\bigg[\frac{\partial|{j}\rangle}{\partial t}\langle j|+|j\rangle\frac{\partial\langle{j}|}{\partial t}\bigg]$$ where the equality is due to the product rule. The first term contains $i\hbar\frac{\partial|{j}\rangle}{\partial t}$ which we know from the TDSE equals $H|j\rangle$, so the first term becomes $H|j\rangle\langle j|=HP_j$. To find the other term, we take the complex conjugate transpose of the TDSE to obtain $-i\hbar\frac{\partial\langle j|}{\partial t}=\langle j|H$, which leads to the second term being $-P_jH$ and thus $$i\hbar\frac{\partial P_j}{dt}=[H,P_j]$$


Why don't the solute molecules in a volatile solvent also escape to contribute to vapor pressure?

It is true that "non-volatile" is a relative term and that virtually all compounds have some, possibly infinitesimal, vapor pressure, but the idea is that this is negligible. And you are also correct that "...the molecules of the solute that are at the surface between the solution and the vapor and should not be held by the strong intermolecular bonds of the solid."

Concerning that latter statement however, there are many reasons that a compound might maintain very low volatility as a dilute solute, even though the intermolecular forces within the bulk solid are not at play. For example, compounds having a large molecular weight tend to be non-volatile, and these will still require much more kinetic energy to be "bumped out" of the solvent, meaning this will be a statistically rare event compared to the transfer of solvent molecules to the vapor phase.

Ionic solids are another type of solute that are typically quite non-volatile. In order for an ionic species occupying a solute surface site to escape into the gas phase, in the case of an aqueous solvent for example, a rather large amount of energy would be required to dissociate the ion from it's solvation sphere and kick it out into the gas phase as a free ion.

I hope this answers your question "where did I go wrong in my reasoning". Don't hesitate to ask for clarification in the comments if I missed something.


Syrup is just a very concentrated solution of sugar.
Because of the high concentration (for instance 65% w/w for USP Syrup, the concentration that avoids bacterial growth) the viscosity increases, and the solution gets syrupy.

In the case of sucralose, the solubility is much lower than that of sugar.
That is normally not an issue, since the sweetening potency is about 600 times higher, at moderate concentrations (80 ppm sucralose have the same potency as 5% sugar).
But to make a syrup, you need a very high concentration.
With a solubility of 28.3 g/100 mL, the viscosity will, probably, be much lower than that of sucrose syrup.

To compare, note that the viscosity of a 30% sugar solution is 3.21 mPa•s, while that is a 60% sugar solution is 56.76 mPa•s.
Take into account that sucralose differs from sucrose only in the exchange of three $\ce{-OH}$ for $\ce{-Cl}$.


It does not exist in nature, but you could make it in the lab by adding pressure.

From a traditional pencil and paper calculation I found the pressure needed for

Al2/3O (s) + CO2 (g) = Al2/3CO3 (s)

is about 10 GPa...

Or you could make an Al2(CO3)3 (aq) solution with concentration of $10^{-5}$ M.

Well, still not impossible. And there should be some special solvents and/or complex structures that easily allow this reaction (postulation).


Is the P-Cl bond energy same in PCl5 and PCl3?

The $\ce{P-Cl}$ bond in $\ce{PCl3}$ is stronger ($\pu{322 kJ/mol}$) than that in $\ce{PCl5}\ (\pu{260 kJ/mol})$. However the $\pu{326 kJ/mol}$ value comes from the average of all $\ce{P-Cl}$ bonds found in all known molecules containing $\ce{P-Cl}$ bonds (including but not limited to $\ce{PCl3}$ and $\ce{PCl5}$).


Could an ester react with magnesium oxide?

Magnesium oxide can be an effective catalyst in transesterification reaction of vegetable oil samples mostly containing high carbon esters:

$\ce{MgO}$ is used as catalyst to increase the rate of transesterification reaction of sunflower oil in supercritical methanol.
The reaction in case of soybean oil has also been reported. This method was developed using metal-chitosan complex leading to higher yield of fatty acids and ethyl ester (biodiesel).
Sometimes alkaline supported catalyst like $\ce{MgO-K2CO3}$ ($\ce{K2CO3}$ fused $\ce{MgO}$) has also been used in the transesterification reaction of canola oil.

Another reaction using magnesium oxide as catalyst is the carbon-acylation of malonates with either acyl chlorides or chloroformates. Various malonic esters (methyl, ethyl, isopropyl and isobutyl) were easily alkoxycarbonylated to give the corresponding methanetricarboxylic esters.

References (in the order of mentioned points)

A. Demirbas, "Biodiesel from Vegetable Oils with MgO Catalytic Transesterification in Supercritical Methanol" Pages 1645-1651, 22 Jul 2008 DOI: https://doi.org/10.1080/15567030701268401
Gizelle I. Almerindoa, Luiz F., D.Probsta Carlos E. ,M. Campos Rusiene , M.de Almeidac , Simoni M. P.Meneghettic, Mario R.Meneghettic , Jean-Marc Clacens, Humberto V.Fajardoe "Magnesium oxide prepared via metal–chitosan complexation method: Application as catalyst for transesterification of soybean oil and catalyst deactivation studies" Journal of Power Sources Volume 196, Issue 19, 1 October 2011, Pages 8057-8063 DOI: https://doi.org/10.1016/j.jpowsour.2011.05.030
http://www.asianjournalofchemistry.co.in/User/ViewFreeArticle.aspx?ArticleID=28_2_6
Jacek Skarżewski "Carbon-acylations in the presence of magnesium oxide. A simple synthesis of methanetricarboxylic esters"Tetrahedron Volume 45, Issue 14, 1989, Pages 4593-4598 DOI: https://doi.org/10.1016/S0040-4020(01)89094-3


19F NMR of hexafluorobromine(VII) cation

$\ce{BrF6^+}$ does have an octahedral structure consequently all of the fluorines are equivalent.  As you point out, bromine is a spin = 3/2 nucleus, therefore, it is also a quadrupolar nucleus (all nuclei with spin > 1/2 are also quadrupolar). A spin 3/2 nucleus that is coupled to another nucleus (fluorine in this case) will split the nmr signal for the other nucleus into a 1:1:1:1 quartet.  In the case of bromine we don't normally see this coupling because bromine typically undergoes rapid quadrupolar relaxation (some quadrupolar nuclei do, some don't) which washes the coupling out (e.g. rapid quadrupolar relaxation effectively decouples the nuclei, that's why the protons in $\ce{CH3Br}$ are a singlet, for example).  However, when the quadrupolar nucleus exists in a highly symmetric environment (octahedral, tetrahedral geometry) the electric filed gradient at the nucleus is near zero and quadrupolar relaxation becomes slow (on the nmr timescale) and coupling can often be observed in such cases.

Finally, bromine occurs naturally as a mixture of two isotopes, $\ce{^{79}Br}$ (50.7%) and $\ce{^{81}Br}$ (49.3%).  Both isotopes are quadrupolar with spin = 3/2.  Because of the slight difference in electronegativity of the two bromine isotopes, they will produce a slight difference in chemical shift for the fluorines bonded to them. Hence, this mixture of $\ce{^{79}BrF6^+}$ and $\ce{^{81}BrF6^+}$ will produce two 1:1:1:1 quartets in the $\ce{^{19}F}$ nmr spectrum.


Please explain which polarity is correct, and why.

In a fuel cell, the anode does sit at a more negative potential than the cathode. This polarization is opposite that of an electrolytic cell, where the anode is positive relative to the cathode.

The chemical composition of the fuel $(\ce{H2})$ and oxidant $(\ce{O2})$ feeds at the anode and cathode, respectively, cause this.  Thermodynamically, in this system the product of the fuel cell reaction, $\ce{H2O}$, is much more stable than the reactants.  Thus, the system biases itself to move in favor of the forward reaction:

$$
\ce{2H2 + O2 <=> 2H2O}
$$

As the $\ce{H2}$ is oxidized, the electrons from this half-reaction are passed into the anode:

$$
\ce{2H2 -> 4H+ + 4e-}
$$

These electrons have to move through the circuit in order to be able to supply the electrons needed for the oxygen reduction reaction:

$$
\ce{O2 + 4H+ + 4e- -> 2H2O}
$$

Here's the key: In order for the electrons to move through the circuit in the correct direction, the potential has to be more-positive at the cathode: the negatively charged electrons are attracted toward the positively charged cathode, along the conductive pathway of the circuit.

  However, why do $\ce{H+}$ ions move away from the anode, when they should by attracting because of negative charges?

For the most part, the circuit is going to be composed of (semi)conducting materials. Conductive materials in a circuit cannot sustain a very substantial charge polarization at all--as soon as they become very slightly polarized, they allow current to pass that equalizes the charge imbalance. (This is why they are used to carry current!)

As a result, the actual charging of the anode is negligible, even though it develops a finite electrical potential.  Thus, the electrostatic attraction of the $\ce{H+}$ ions to the anode can in general be ignored, as it is orders of magnitude too small to keep the protons from diffusing away, toward the cathode where they are being consumed to produce water.


Are "gas phase" calculations the same as "vacuum" calculations?

I don't know the etymology of the phrase, but yes, "gas phase" calculations are the same as "vacuum" calculations. There is no difference.

Think of it this way: the only option other than a "gas phase" calculation is one where solvent is included, either as explicit atoms or implicitly via a mathematical model or even a solid-state/bulk-model calculation. This indicates that the solute is in the gas phase, and hypothetically not interacting with the environment in any way.


Which mechanism is not seen during hydrolysis of esters?

The reaction mechanism said actually corresponds to the following mechanisms,

$A_{AC^2} $ which means acid catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
$A_{AL^2} $ which means acid catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.
$B_{AC^2} $ which means base catalysed acyl carbon oxygen bond breaking bimolecular mechanism.
$B_{ALC^2} $ which means base catalysed alkyl carbon oxygen bond breaking bimolecular mechanism.

In the last one it is unimolecular, and acid catalysed and acyl carbon oxygen bond breaking mechanism.

In esters, Hydrolysis can occur via breaking of two possible bonds. If the ester is $\ce{R'-CO-O-R}$, then the $\ce{CO-O}$ bond is the acyl carbon oxygen bond, whereas the $\ce{O-R}$ bond is alkyl carbon oxygen bond. Either of these two breaks in the process of hydrolysis.

If you take usual hydrolysis of esters, they can occur in both acid and base catalysed reaction medium condition, and in that case the $\ce{R'-CO}$ part of the ester goes to the acid, and the $\ce{-O-R}$ goes to alcohol. So, this is the acyl carbon oxygen bond breaking and bimolecular as the $\ce{OH-}$(in base catalysed) and $\ce{H2O}$ (in acid actalysed) attacks at the rate determining step. So, $A_{AC^2}$ and $B_{AC^2}$ is easily seen.

In an ester, where the attack of $\ce{OH-}$ or $\ce{H2O}$ at the carbonyl carbon is highly sterically hindered (e.g. in mesitoic ester) the reaction passes through alkyl carbon oxygen bond breaking, and thus the $\ce{R'-CO-O}$ part of ester goes to acid, whereas only the R part goes to alcohol, and they are also bimolecular generally, as in the slow step, both the nucleophile attacking at the alkyl carbon remains present along with the molecule. That's why $A_{AL^2}$ and $B_{AL^2}$ are also seen but it is slightly rare.

But for showing $A_{AC^1}$, the molecule doesn't have any other chance rather than forming an acyl carbocation(as, the carbocation intermidiate follows unimolecular reactions). But if it forms acyl acrbocation in acid medium, the lone pair on the adjacent oxygen is already donated to $\ce{H+}$, so, there is no extra stabilising factor rather than any highly electron donating group, like tert butyl or phenyl attached to carbonyl carbon. But if you have those, the molecule generally prefers the alkyl bond breaking to avoid steric hindrance.So, $A_{AC^1}$ is very very rare to happen.

But $A_{AL^1}$ can be seen where R is tert butyl group. because, it can get stabilisation by forming a carbocation of its own, and the remaining portion of molecule can get stabilisation through resonance.


A gas occupies the volume of whatever the container it is in. Take a balloon filled with Helium for example. Now the volume of Helium gas is the volume of the ballon.

If you increase the temperature by 1 °C at constant pressure the volume of the balloon increase by 1/273th of the original volume of the balloon. Hope this might clear things up.


Why does aqeous fluorescein shine when exposed to UV light?

The process is called fluorescence. Some substances, such as calcium sulfide, $\ce{CaS}$, absorb light and release it slowly in a process called phosphorescence, glowing for some time after the source of light is removed.


Is there any difference between a completely filled orbital and an half-filled one?

While what ssavec says about not taking orbitals away from a structure is true, it's possible to make some generalisations, particularly for lone atoms where the symmetry constraints make the orbital divisions more clear.

When you generate orbitals for an atom, you get the set of s, p, d, f (and so on) orbitals separated out, and you can generally say that for the same principal quantum number (usually denoted n), a doubly-filled orbital will produce more diffuse orbitals than a singly-filled one. This is an obvious consequence of Coulombic repulsion. But, again, all the other orbitals you generate for the altered wave-function will be affected by the change in occupancy and overall charge.

It would be possible to construct a somewhat artificial pair of calculations in which you kept fixed all but one of the occupied orbitals, and then optimised the wavefunction using one electron in one calculation and two in the other, and I would expect you to find that in the calculation with two electrons, the varied orbital would be more diffuse.

Because the two orbitals would be different in spatial form, you wouldn't expect the value at a given point to be a multiple of the other.


Hand sanitizers are generally ~50% - 70% ethanol or isopropanol. When these alcohols mix with water, they may release heat. Starting at room temperature, mixing ~100% ethanol with an equal volume of water will raise the temperature of the mixture ~8 0C.


Stereoselectivity differences in Lindlar's catalyst and Na/NH3?

In Lindlar's catalyst, it is a matter of the physical way in which the molecules react. Firstly, $\ce{H2}$ dissociates into atomic hydrogen and binds to the catalyst, and the alkyne coordinates to the metal surface. Since the surface of the catalyst can be thought of as a vast plane, all carbon to metal bonds must be facing down onto the plane. The hydrogen atoms then get inserted in place of these carbon to metal bonds, and the result is syn-addition.

In the $\ce{Na/NH3}$ reduction, when the electrons are first added to the triple bond and the $\mathrm{sp}$ hybridization is broken, the groups bonds on either side the triple bond repel each other into the lowest energy conformation, the trans position. This is why the $\ce{H}$ atoms being donated by the $\ce{NH3}$ add through anti-addition.


Is it possible to prevent sacrificial anode contaminate the soil?

I wonder if a sacrificial anode near the surface could prevent corrosion at the deepest part of your pipe, where the water and wet soil are. Usually, sacrificial anodes are placed fairly close to the item being protected.

It would be worth testing some of the water (that comes up thru the pipe) for pH and corrosivity. If the water is alkaline (hard water), it may not be very corrosive. The oxygen level far below ground would be low, perhaps not enough for rapid corrosion.

Maybe a sacrificial anode is unnecessary.


What is the mechanism of the reaction of a ketone with sodium nitrite/ hydrogen chloride?

$\ce{NaNO2/HCl}$ is as you suggest a source of $\ce{HONO}$. This will be protonated by $\ce{HCl}$ to give the nitrosonium ion $\ce{NO+}$ (in equilibrium with nitrosyl chloride). This strongly electrophilic species will react with the enol of the ketone to nitrosate the alpha position. This nitroso compound then rearranges to give the alpha oxime.

Image from: Emilia Iglesias and Isabel Brandariz, A further study of acetylacetone nitrosation. Org. Biomol. Chem. 2013, 11, 1059-1064. DOI: 10.1039/C2OB26073E; available as pdf from pdfs.semanticscholar.org.


Compare the ionization energies of C and Cl?

It's the other way around,

moving left to right across a row increases the IP due to increasing nuclear charge, and
moving down a column decreases the IP, due to increasing shielding of the nuclear charge by core electrons.

Chlorine is only one row below carbon, but it is three columns to the right.  As the following chart illustrates, moving to the right by one position generally has a similar or larger effect than moving down one position.

(image source)

Therefore, in this case moving further to the right by 3 positions (and increasing the IP) strongly outweighs moving down by only one position (and decreasing the IP).  Hence, in this case the IP of chlorine would be predicted to be greater than the IP of carbon.  Indeed, the IP of chlorine is reported to be 13.0 eV, while that of carbon is only 11.3 eV.


The ionic equation for this reaction is:

$\ce{Fe + 2H+ ~(+\;SO4^2- ) -> Fe^2+ + H2 ~(+\;SO4^2- )}$

As two half-reactions, it is seen that the iron is oxidized:

$\ce{Fe -> Fe^2+ + 2e- }$

The reductant (iron) transfers electrons to hydrogen, and is thus itself oxidized.

And the hydrogen is reduced:

$\ce{2H+ + 2e- -> H2}$

The oxidant (hydrogen) removes electrons from iron, and is thus itself reduced.

To calculate the reduction potential you have to solve the following equation:

$\Delta E = E_\text{acceptor} - E_\text{donor} = E_{\ce{2H+/H2}}- E_{\ce{Fe/Fe^2+}}$

$\Delta E = 0\ \mathrm{V} - (-0.41\ \mathrm{V}) = 0.41\ \mathrm{V}$

The reactions runs voluntarily if $\Delta E > 0$.

You can do the same for b).


Are pesticides used on fruit easily washed off?

Pesticides never were my main field of research, so you might want to take the following with a grain of salt:

Most pesticides i'm aware of are insoluble in water, but you might want to cross-check that for particular substances in the huge Pesticide database of the EC.

This insolubility seems reasonable, otherwise the substances would get washed off after the next rain.

Concerning your second question, the answer unfortunately is yes. To my knowledge, different pesticides are typically applied at the various state of plant growth. In order to suppress weeds, one might spray fields in an early, pre-emergent state. Later, fungicides or other substances might be applied during blossom. In order to protect the bees, it is however not recommended to apply insecticides during that time. Reasonable and minimized use of pesticides will keep the load in the fruit close to the limits of detection or beyond. Depending on the origin of the fruit, this might however not be the norm. In these cases, high concentrations of pesticides, even those banned in your home country, are detected both in the fruit and its products.

This leaves us with the question what we can do as consumers. If the fruit is highly contaminated, we can't do anything. Therefore, it might be reasonable to avoid products from countries infamous for breaking environment protection rules and go "organic" instead. Honestly, i never peel apples, but i wipe the surface with a piece of cloth. But i avoid industrial sorts, buy from local farmers, or pick them from the ground of meadow orchards.

I hope that this contribution, although more personal and less scientific, was helpful.


Does the "extent of reaction" have a physical meaning?

The stoichiometric coefficients $\nu_i$ are arbitrary anyway; the only thing that matters is the ratio of them. $\ce{3A -> 2B}$ and $\ce{6A -> 4B}$ both describe the same physical process occurring (unless it is implying something about mechanisms, but in this context of thermodynamics, that's not the case). Clearly, even for a single unchanging system, its value of $\xi$ will vary depending on how you define the stoichiometric coefficients. So, I wouldn't expect $\xi$ to have any physical meaning. Personally (in my limited knowledge of thermodynamics) I just view it as a useful, but ultimately arbitrary, definition.

[In the same vein, I would say that the magnitude of the equilibrium constant $K$ doesn't have any physical meaning either, since it also depends on how you define the stoichiometric coefficients. But I digress.]


What is the major product formed on addition of Hydrogen Iodide to Pent-2-ene?

I would say the reason that the iodine ends up in the 2-position is because at the 2 and 3 positions the charge of the carbo-cation is on a secondary position and has no thermodynamic driving force. The adjacent secondary hydrides may shift without much energy loss leaving the major product to be kinetically determined. Since there are two positions for an addition to the 2-position carbon and one for the 3-position carbon, your 2-position addition has a statistical advantage. Combine this with the fact that two adjacent ethyl groups would be more stericly hindered to attack than a methyl and propyl group and it makes reasonable sense that the addition occurs on the 2-position.


According to Uranium partitioning between liquid iron and silicate melt at
high pressures: implications for uranium solubility in
planetary cores

  The solubility of U in
  liquid Fe is in the range of 0.6 to 800 ppm and increases with temperature (T) and
  pressure (P).

See also U solubility in Earth’s core which finds a solubility of 2 ppm at zero pressure (run number 198).


Heats of combustion and stability of rings

Compounds A-D all have the same molecular formula, $\ce{C6H12}$.  We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation

$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$

Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).

This link provides the heats of combustion for some useful model compounds.  Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained.  The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.

Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".


Why does nitrogen have a maximum covalency of 4?

Recall that covalency is the number of shared electron pairs formed by an atom of that element. Nitrogen's maximum covalency is indeed $4$. And no, it does not break up its lone pair.

I'll give you a simple example. Have a look at the Lewis structure of the ammonium ion:

(source)

Notice that nitrogen's octet is complete as soon it bonds with three $\ce{H}$ atoms (aka forms ammonia). The fourth covalent bond is actually a coordinate covalent bond, formed when that nitrogen atom's lone pair gets donated to a proton.

This is also the maximum covalency for the nitrogen atom, since it has no more unpaired electrons that could be paired up with other atoms, to form more covalent bonds.

Homework exercise: Can you now deduce the maximum covalency of nitrogen's elder brother, oxygen?

Caution: extra stuff ahead this will help those with knowledge beyond high school level. If you're at or below high school level, go back home and play with your cat.

After-thought: Valency is a useless term that doesn't help you do any chemistry. Different sources define it differently (two definitions on Wikipedia). It just helps fill high school textbooks with more pages, but it becomes irrelevant with the introduction of better terms like coordination number, which actually tells you something about the structure of the molecule. While "valency" maybe  useful in basic beginners course at getting a grip, there are limitations to this point of view.

Here's an example of such a contradiction. You may expect elements of period $3$ and above to display higher covalencies. One such example is phosphorus. Though it belongs to the same group as nitrogen, it can form compounds like $\ce{PCl5}$, (apparently) increasing its maximum covalency to $5$ instead. The reasons for these are usually attributed to hypervalency/octet-expansion, but they are wrong and obsolete concepts, having been superseded by newer concepts. In fact, $\ce{P}$ still has a covalency of four in $\ce{PCl5}$, since there are only four shared pair of electrons (the non-bonding electrons don't count). This reinforces the idea that coordination numbers are a better and more useful term than valency. ($\ce{P}$ now has a coordination number of $5$ in $\ce{PCl5}$)


The only given concentration is the initial concentration [...]

Unless there is a major misunderstanding on my side, you have all the data you need.

Given the initial concentration $c_0$ and knowing the path length $d$ of your cuvette, you measure the extinction $E$ at this point and use - as you have correctly stated in your question - the Lambert-Beer law to calculate the molar extinction coefficient $\epsilon$.

$$E = \epsilon\cdot c\cdot d$$
$$\epsilon = \frac{E}{c\cdot d}$$

Did I mention units? Units are your friend!

Write the Lambert-Beer law with units!
Percent (%) is NOT a proper unit for the initial concentration of your protein.

  but I'm supposed to be able to calculate the final concentration and maximum reaction rate

Having determined $\epsilon$, you can now correlate at any point along your reaction the measured extinction with the actual concentration of your sample, including the final concentration.

From the slope of the curve of extinction $E\ $ vs concentration $c$, the rate of your reaction can be obtained.

If the relation is not a straight line, logarithms are your friend.


Can you determine whether a compound can form given particular elements?

Answer on first question is no, on second is yes.

Consider this short list C, H, O, N this elements can produce unlimited number of compounds.
If you have short formula like C6H6 and you assuming only covalent bonds in molecule, define valence 4 for Carbon and 1 for Hydrogen you can make mathematical calculation according to graph theory and you will get 217 structures. But most of them doesn't exist due to bonds strain and intersection. If you want to obtain existing molecules you can filter this 217 models using for instance quantum mechanics calculation.


Chiavarino, et al. [1] report  that where electrophilic substitution occurs with carbocations (borazole more often undergoes addition), it does so on nitrogen.  Nucleophiles such as methanol prefer boron.
We can explain that result in terms of both molecular orbitals and the Wheland intermediate.  In the molecular orbital explanation, recall the familiar delocalized pi orbitals in the benzene ring.  With the polar bonds of borazole the occupied bonding orbitals whose electrons will add to an electrophile are shifted towards the nitrogen atoms, so the electrophile will tend to bond there rather than on the boron (the latter has instead mostly the empty antibonding pi orbitals and is thus favored in nucleophilic substitution).
In terms of the intermediate, adding the electrophile to the ring leaves the remaining pi electrons in a deficient condition -- only four electrons cover the remaining five conjugated atoms, with most of the electron deficiency falling on atoms ortho and para to the attack site.  Attacking on nitrogen puts the less electronegative boron atoms at these deficient sites leading to a more stable EAS intermediate.
Reference

Barbara Chiavarino, Maria Elisa Crestoni, Simonetta Fornarini, "Electrophilic Substitution of Gaseous Borazine," J. Am. Chem. Soc. 1999, 121(11), 2619-2620 (https://doi.org/10.1021/ja983799b).


How to determine the iron content of a screw?

There is another problem with this.

Potassium permanganate is a very strong oxidizing agent, which is capable of oxidizing Chloride ions to Chlorine. This will result in additional consumption of $\ce{KMnO4}$ solution, which makes you think there is more iron in your screw than there actually is.

To solve this, you should look up the titration of Iron using the Reinhardt-Zimmermann solution. It contains sulfuric acid, Manganese (II) sulfate and phosphoric acid.

The $\ce{MnSO4}$ changes the redox potential so the acidic $\ce{KMnO4}$ solution can no longer oxidize Chloride ions. Because Iron(III) gives an intense yellow solution, the addition of Phosphoric acid is necessary since it will react with Iron(III) to colorless Iron phosphates. That makes it easier to detect the end point.

So the way I did this back in my first semester was to make sure we have all Iron as Fe(II) by adding an excess of $\ce{SnCl2}$. The excessive $\ce{SnCl2}$ was quenched using $\ce{HgCl2}$. Then the Reinhard-Zimmermann solution was added and titration with $\ce{KMnO4}$ could start.


Is the elimination reaction between 1 1 dibromo cyclopropane and phenyl lithium feasible in alkaline medium (and some heat)

There are two pathways that phenyllithium could follow - it could deprotonate, but there really not any reactive protons, or it could do lithium-halogen exchange with one of gem-bromines. This second pathway will lead to a carbene by loss of $\ce{Br-}$ and the carbene (a highly reactive intermediate) will rearrange to give allene $\ce{H2C=C=CH2}$.

This method is documented:

In Page 275: H. Maskill, "Chapter 1: Three- and Four-Membered Carbocyclic Rings," In Specialist Periodical Reports - Aliphatic, Alicyclic and Saturated Heterocyclic Chemistry, Volume 1, Part II: Three‐ and four‐membered rings (carbocyclic and saturated heterocyclic); W. Parker, Ed. (Senior Reporter); The Chemical Society: Burlington House, London, England, 1973, pp. 3-292 (ISBN: 0-85186-502-X).
In Pages 1492-1493: Armin de Meijere, Editor, Carbocyclic Three-Membered Ring Compounds, Houben-Weyl Methods of Organic Chemistry, Volume E 17 b - Cyclopropanes: Synthesis; Georg Thieme Verlag: Stuttgart, Germany, 1997,  p. 1492 (ISBN: 978-3-13-798204-3).


Equation for reaction between chlorine fluoride and potassium bromide

Let's subtract the two reactions from each other to simplify things a bit:

$$\ce{2ClF + 2KBr -> 2KF + Cl2 + Br2}\ \ \ \ \ \ \ \ \ \ [1]$$

$$\ce{ClF + 2KBr -> KCl + KF + Br2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2]$$

$$\ce{ClF + KCl -> KF + Cl2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] = [1]-[2]]$$

Now we can discuss [2] and [3] separately by splitting them up into half reactions.

Reaction 2

$$\ce{ClF + 2KBr -> KCl + KF + Br2}$$
Potassium is a spectator ion. The reduction half reaction is:

$$\ce{ClF + 2e- -> Cl- + F-}$$

The oxidation half reaction is:

$$\ce{2Br- -> Br2 + 2e-}$$

Reaction 3

$$\ce{ClF + KCl -> KF + Cl2}$$

Potassium is a spectator ion. Fluorine does not change oxidation state. Chlorine undergoes comproportionation (oxidation states +1 and -1 to oxidation state of 0). The reduction half reaction is:

$$\ce{ClF + e- -> 1/2 Cl2 + F-}$$

The oxidation half reaction is:

$$\ce{Cl- -> 1/2 Cl2 + e-}$$

Which reaction happens?

That depends on the reduction potentials, the kinetics of the two reactions and the concentration of species present. I'm not sure how you would figure it out without experimental (or computational) data.

Comparing the oxidation half reactions, you could make an argument about electronegativities of bromine vs chlorine, but I have no insight to compare the two reducation half reactions, which are very different.

The Wikipedia article on chlorine monofluoride states:

  Many of its properties are intermediate between its parent halogens, $\ce{Cl2}$ and $\ce{F2}$.

If you substitute $\ce{Cl2}$ for $\ce{ClF}$ in reactions 2 and 3, they both look reasonable (reaction 3 becomes a self-exchange, i.e. equilibrium constant = 1, whereas reaction 2 reduces chlorine and oxidizes bromine, products are favored).


Good resource for stereochemistry of more than 4 substituents?

I'm not sure how easy this is to implement algorithmically, but for what it's worth, the official IUPAC recommendations are as follows.
Section P-93.3 in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) describes the guidelines for assigning configuration at non-tetrahedral atoms. The portions of this answer will be quoted from there.
These guidelines, and further details on certain aspects, are also discussed in Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) (link), section IR-9.3.

Stereodescriptors for nontetrahedral configurations are composed of three parts.
(1) A symbol indicating overall geometry called the 'polyhedral symbol';
(2) A symbol called the 'configuration index' that describes the configuration of ligands around the central atom;
(3) A symbol indicating the 'absolute configuration' associated with the central atom called the 'chirality symbol'.

For the first example you provided:

which is a trigonal bipyramid, the appropriate 'polyhedral symbol' is TBPY-5 (P-93.3.2, Blue Book). The numbers in red are the priorities according to the Cahn–Ingold–Prelog system, which will soon be required.
The 'configuration index' is

a series of numbers to identify where each atom or group is located. It is based on the CIP priority order of the atoms attached. The atom or atoms with highest priority are given priority number one '1'; the next priority number two '2', and so on. [...] (P-93.3.3.1)

For a trigonal bipyramid, this "consists of the priority numbers of the two apical atoms (lower number first if different), representing the reference axis of the system" (P-93.3.3.5.1). In this case, the two apical atoms have priority numbers 1 and 2, so the configuration index is simply 12.
The 'chirality symbols' available for a trigonal bipyramidal system are 'A' and 'C'.

The atoms in the plane perpendicular to the reference axis are viewed from the side with the atom or group of highest priority on the reference axis. If the direction from the atom in this plane with highest priority to the atom with the next higher priority is clockwise the chirality symbol is 'C', if anticlockwise 'A' (P-93.3.4.2.1).

Looking along the reference axis (which is the axis containing the axial ligands), from the direction of the higher priority atom (oxygen), the equatorial ligands decrease in priority going in a clockwise order. So, the chirality symbol is C. (Essentially, this is analogous to assigning an R-configuration for an ordinary chiral carbon.)

The three parts of the stereodescriptor are enclosed in parentheses and separated from each other by hyphens and cited at the front of the name separated from it by a hyphen. (P-93.3.1)

The full stereodescriptor is therefore (TBPY-5-12-C).