The problems with reacting pure nylon salt are as follows:

The molecules of hexamethylene diamine (HMD) and adipic acid will have a very hard time accessing each other's reacting groups, because of limited mobility within the solid structures.
The reaction is a condensation, so it will produce a substantial amount of water anyway, so pretty much any reason for keeping the reactants dry is irrelevant, because the final polymer will end up wet anyway.

It might be possible, but because of (1), your product might be impure, containing a lot of small unreacted adipic acid and HMD particles.

Also, If you don't want to deal with the tedious procedures above, considering that they were probably written for industrial applications, consider using this procedure. It's a lot more lab friendly.


Intoxication can be acute (short time of exposure, effects appear fast), or chronic (longer exposures, the onset toxic effects will take days, months or years). For Iron Man it seems clear that this would be a chronic intoxication.

I found this safety guideline from the EMA, which states:

  Mice given $\ce{PdCl2}$ in drinking water (5 mg palladium/l) for a lifetime
  showed suppression of body weight and an increase in amyloidosis of
  several inner organs. In another study of questionable value
  (chloropalladosamine was given enterally to rats for about 6 months) a
  dose of 0.08 mg/kg body weight (corresponding to 0.041 mg $\ce{Pd^2+}$/kg) was
  determined as the no-observed-effect-level (NOEL), above which changes
  in body weight, hemoglobin content, blood serum parameters as well as
  functional and morphological changes have been reported. There is
  insufficient histological information available from this study.

Extrapolating human toxicity from mice data is a long shot, but since this is all we have it will have to do. Also, bear with me for pretty inaccurate math from here on (it's fiction anyway).

It's been a while since I watched the movie, but if I recall correctly each palladium piece was roughly 7cm x 4cm x 1cm. That's a volume of 28cm$^3$. The density of palladium is 12.023g/cm$^3$, therefore this piece weighs a bit under 350g.

I'm going to assume that he burns through a palladium piece every two days. Good times.

The NOEL (see above) is 0.08mg/kg/day. For an 85kg human, that's 6.8mg/day,  roughly 0.02% of the mass in the reactor. I recall that the palladium was heavily corroded when he extracted it, so a loss of 0.5-1% of palladium seems pretty reasonable. Being conservative, the reactor would release about 0.9g a day. If more than one tenth of a thousandth of this palladium is absorbed by Tony's bloodstream (which doesn't sound like much), then he will be in trouble.

To sum up, yes, I do believe it possible to suffer from palladium poisoning if you have a chunk of that thing dissolving inside your chest day and night.

Answering the other half of your question, Tony Stark used palladium for his first arc reactor in the first Iron Man. I don't think it was planned, they just went with it.


Is the %V/V composition defined for homogeneous mixtures?

Let's be clear on what percent by volume means.  If I say component $X$ is 25% by volume of a mixture of $X$ and $Y$, with a total volume of 1 L, it does not mean there is some 250 mL portion of the mixture that is solely occupied by $X$.  That wouldn't be a mixture.  It doesn't even mean that if I mentally subdivide that 1 L mixture into tiny identical cubes, one for each atom or molecule, then 1/4 of those cubes are occupied by atoms or molecules of $X$.

Instead, all it means is that I can prepare 1 L of the mixture by mixing some volume $V_X$ of $X$ and 3 times that volume ($V_Y = 3 V_X$) of Y.  There's not even any guarantee that $V_X = \pu{250 mL}$, since sometimes volumes are non-additive, that is, $V_X + V_Y \neq \pu{1 L}$.

Now as it happens, for ideal gases, which the atmosphere is close to being, you can mentally subdivide the volume into tiny little equal-sized cubes, one for each atom or molecule in the mixture, and on average there would be one atom/molecule of $X$ in each fourth cube if $X$ is 25% by volume of the mixture.  (I emphasize "on average" because of course since the gas atoms or molecules are continuously moving around randomly, they will bunch up and spread out momentarily all the time, so at any given instant there may be zero or many more than one $X$ atom/molecule per tiny cube.)  Real mixtures in the liquid state often have at least some nonideality, e.g. even in a dilute solution of $\ce{NaCl}$ in water, you will find "structure" in the solution, with the $\ce{Na+}$ cations and $\ce{Cl-}$ anions surrounded by a fairly fixed arrangement of $\ce{H2O}$ molecules, almost like a tiny piece of clathrate, so a mental arrangement of tiny boxes is an even less accurate atomic-scale description of the solution.

In short, the percent by volume description of a mixture is only a way to characterize the amount of material that goes into it, it is not intended as any kind of implication of what the  mixture looks like at the atomic scale.

As for why NASA reports the composition of the atmosphere by percent by volume: probably because it's closest to the actual experiments done to measure it.  You would typically measure the composition of a gas mixture by physically separating it (e.g. by lowering the temperature until each gas liquefied) and then measuring the volume of each component.  You certainly could from those measurements easily calculate a percent by moles, but – why?  You introduce a calculational step between the measurement and what you report, and scientists tend to prefer getting original data, right what comes from the instrument, if possible, as a way of avoiding even the smallest risk of some error introduced in calculation.


Why does potential chemical energy decrease with increasing oxidation of carbon?

is there an intuitive explanation of why increasing oxidation state
  correlates with lower potential energy, at least for carbon?

The heat of formation of methane, carbon dioxide and carbon tetrafluoride are -74.9, -393.1 and -927.2 kJ/mol respectively (the trend remains unchanged even if you divide by MW and compare per unit mass).  As we shift electron density from electropositive elements to electronegative elements we stabilize electrons, their energy is lowered.

  Does increasing oxidation state correlate with lower potential energy
  for all elements?

I haven't checked extensively, but the trend seems to hold up.  I'd expect it to hold up.

$\ce{BeH_2}$, $\ce{BeF_2}$  127.2, -796.7 kJ/mol

$\ce{CaO}$, $\ce{CaF_2~~}$      39.1, -789.1 kJ/mol

$\ce{NaH}$, $\ce{NaF~~}$      126.2, -288.0 kJ/mol

  Why does it seem that oxygen forms rather strong bonds with other
  elements in many cases?

I'm not sure what you mean by "strong bonds".  The C-O bond in carbon dioxide is much more reactive than the C-H bond in methane.  Perhaps you're alluding to molecular stability which takes us back to the top, e.g. compounds with polar bonds typically have stabilized electrons, electrons in lower energy orbitals, as discussed above.


Alkene double bond is an area of high electron density and hence high (partial) negative charge. Bromine molecule, approaching double bond, shifts its electrons of covalent bond towards farther Br atom, it becomes polarized and an induced dipole.

That's the simplest explanation in terms of electron movement I would use.


Why does a glass tube show white bands after being heated with a torch?

It is sublimed silica according to the following reference at page 8.

http://www.public.asu.edu/~aomdw/GLASS/book.pdf


Naming a synthetic molecule from QM9 dataset

(note: The structure representation is suspicious and confusing. Linear nitrile $\ce{-C#N}$ group is "broken". But I see some software products generate such representation automatically.)

If the structure is really meant saturated, no stereochemistry is indicated.

3-aminooxolane-2-carbonitrile

(Watch the typo in your proposal …oxalane.) But it's not present at PubChem. (SMILES: N#CC1OCCC1N)

Maybe, if unsatured, furan derivative is meant:

3-aminofuran-2-carbonitrile


Schottky defect occurs when equal number of positive and negative ions are missing from the lattice thereby maintaining net electrical neautrality and leading to a decreased density in the lattice (since there are now less number of ions) .
It is observed to be occurring in cations and anions of similar sizes.

NOW CONSIDER THAT AN ION IS AT THE LATTICE POSITION AND A DEFECT IS TO BE OCCURRED.

There are 2 ways stoichiometric conditions can be satisfied. - either by loosing this ion as well as another of opposite charge in equal ratio(SCHOTTKY) ,
OR not losing this ion at all , but just shifting it's position from lattice position to interstitial one.(FRENKEL)----
If we think logically ,

Since it's a stoichiometric defect it should maintain neautrality as well as loose cations and anions in a certain ratio.

If the sizes were very different , there are chances the smaller ion would occupy an interstitial space causing it to create a hole and still maintain neutrality and ratio without losing any ions. (Which is what happens in Frenkel defect) . Since however ions are of similar sizes , one can't occupy an interstitial space .. and therefore losing equal no. Of anions and cations is the only way to go.
You can think of these as 2 cases of defect... Yes it's not necessary the ions are of similar sizes for a defect to occur BUT IF THEY ARE...They are most likely to cause the Schottky defect and if not , then Frenkel Defect.
As for your NaCl doubt , I would suggest try finding the critical radius ratio for one ion to occupy the interstitial space . This way you can verify whether it can perform Frenkel defect or not ..Or why it's performing Schottky.
Hope this helps.

EDIT:- I can't find any examples of ${\ce{AB2}} $ type as such , but you can again try finding critical ratio of ions in ${\ce{AB2}} $ lattices to decide what type of defect they are likely to occur.


name of a pharmaceutical drug used in dentistry that has Ionic bonding? and how this bounding is made?

The fluorocarbons that you mentioned do have polarized $\ce{C-F}$ bonds, but no ionic bonds. If you think ionic, think salts.

I'm not a pharmacologist, but what about the local anestethics used at the dentist? Lidocaine (1), mepivacaine (2), and articaine (3) are frequently used.

All these compounds have tertiary and secondary amino groups. (I've marked the relevant nitrogen atoms blue). When these compounds are treated with hydrogen chloride, the amines get protonated and the neutral compounds are converted to ionic hydrochlorides, $\ce{R3N ->[\textrm{HCl}] R3NH+ Cl-}$.


Hybridisation of Mn in potassium permanganate

This is formally a manganese(VII) compound and hence there are no 3d electrons.

The four $\ce{O^2-}$ ions are considered to be donating two electrons each to the atomic orbitals. Tetrahedral "hybridisation" can be achieved by using the $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$, and $\mathrm{3d}_{xz}$ AOs. Since in Mn(VII) the $\mathrm{3d}$ AOs are lower in energy than the $\mathrm{4s}$ and $\mathrm{4p}$ AOs, the "hybridisation" is best considered as $\mathrm{d^3s}$.

I have put "hybridisation" in quotes because it is not used in advanced chemistry (it is an artifact!) but it is correct in saying that the bonding MOs contain mostly $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$ character but they'll also probably mix some $\mathrm{4p}$ character in there for good measure!


In general, you can expect an epichlorohydrin to be attacked nucleophilically only on the outer epoxide carbon. First, let me present the synthetic argument: In their total synthesis of RK-397, Burova and McDonald used epichlorohydrin as the starting point for the synthesis of their C10–C16 fragment. The fragment synthesis is shown in the scheme below which is taken directly from their publication.

They start with an enantiopure epichlorohydrin which they open using vinylmagnesiumbromide. The resulting alcohol is recyclised under elimination of chloride by distilling from $\ce{KOH}$. Starting from (R)-epichlorohydrin, their sequence exclusively generates (R)-33. Epoxidation gives 35 as a mixture of diastereomers which they separate using hydrolytic kinetic resolution with the Jacobsen cobalt catalyst.[1] This procedure acts only on the epoxide and selectively cleaves the (R)-epoxide to a diol. If there had been substantial amounts of (S)-33, 36 and 37 would each have been isolated as a mixture of diastereomers; however, both were isolated as single diastereomers.[2]

If there had been any direct attack of the vinyl Grignard on the $\ce{C-Cl}$ bond with direct displacement of chloride, the opposide enantiomer (S)-33 would have been formed (see scheme 1 below; 33 would have been inert to $\ce{KOH}$ and thus simply been co-distilled). This would have led to the aforementioned mixture of diastereomers further down the synthesis. Since no mixtures were observed, we can safely conclude that the attack only happens under cleavage of the $\ce{C-O}$ bond.

Scheme 1:  Hypothetical attack on $\ce{C-Cl}$ gives the opposite enantiomer; would lead to 36 and 37 being isolated as a mixture of diastereomers.

The synthetic argument already establishes that epoxide-cleavage must be orders of magnitude faster than chloride displacement. This does allow the question: why? One general argument is that the epoxide is highly strained, so a ring-opening attack will release the ring strain. Furthermore, $\ce{C-Cl}$ bonds are rather strong compared to other halogens (especially $\ce{C-I}$) and thus not as easily cleaved. Furthermore, we are likely looking at some hyperconjugation effects: the orbitals of the $\ce{C-Cl}$ bond and the $\ce{CHR2-O}$ bond can overlap as shown in scheme 2; this interaction leads to the chloride displacement attack being further disfavoured.

Scheme 2: Possible overlap of the $\ce{C-O}$ and $\ce{C-Cl}$ orbitals shown here as $\sigma_{\ce{C-O}}$ and $\sigma^*_{\ce{C-Cl}}$

Coming back to your question, since the outer epoxide carbon is marked and since that is where ethanolate will attack, only (D) is correct and your proposed mechanism checks out.

References:

[1]: S. E. Schaus, B. D. Brandes, J. F. Larrow, M. Tokunaga, K. B. Hansen, A. E. Gould, M. E. Furrow, E. N. Jacobsen, J. Am. Chem. Soc. 2002, 124, 1307–1315. DOI: 10.1021/ja016737l.

[2]: S. A. Burova, F. E. McDonald, J. Am. Chem. Soc. 2004, 126, 2495–2500. DOI: 10.1021/ja039618+.


Can the light reaction of photosynthesis happen spontaneous in pools of water?

Photosynthesis is a lot more complex than just breaking down water

Yes, photosynthesis involves a reaction which partly involves absorbing photons and breaking down water. But overall it is far more complex than that as it also involves absorbing carbon dioxide and turning that into sugars and other molecules that plant can use a fuel. Breaking down water is a byproduct of that larger process as is the oxygen released by it.

The detail of the reaction involved is complex and consists of dozens of individual steps. It isn't as simple as absorbing a photon and splitting water. In fact, there are two separate and individually complex systems of proteins involved in the water-splitting part of the process: photosystem I absorbs light and generates higher energy electrons; photosystem II also absorbs light, breaks water molecules part and generates higher energy electrons. But these processes are so complex that we don't fully understand how they work (and our understanding of the structure of photosystem II was updated as recently as 2014). And that machinery is very complex involving dozens of individual proteins and components.

Given the complexity of the processes behind water-splitting in photosynthesis (which involves multiple photons and electrons) it should be clear that there is no easy way to perform the reaction.

You don't just need photons and water molecules. You need some of the most complex molecular machinery we know of and multiple photons. Sunlight + water isn't going to achieve it.


Compound $a$ is stabler than compound $c$ because the alkene in $a$ has $7$ hyperconjugative structures due to $7 \alpha$-Hydrogen, whereas $c$ has only $5$ hyperconjugative structures.
Hyperconjugation or Nathan-Baker Effect is a stabilizing phenomenon due to resonance-like charge delocalisation. Also, the reason why here hydride shift takes place is because it leads to a stabler carbocation.
Hence the product should be $a$.


Can cis and trans isomers have widely different flash points?

It is not totally surprising giving that they are stereoisomers. Think about maleic and fumaric acid, which due to their very different properties were given different names.
Moreover, but perhaps you are aware of, there is not in general a relationship between boiling point and flash point. In the first case factors which depends on the actual configuration - as molecular symmetry, polarity - should be considered for how they affect intermolecular forces. For the flash point, their influence to the intrinsic stability of the molecule and to the overall shape and accessibility for oxygen, should be considered instead.

For the less experienced ones: in few words volatility is required to reach the concentration (in air) at which the vapour can be ignite at a given temperature (flash point). More a liquid is volatile, easily that concentration is reached. But this does not imply a lower flash point, which depends on the thermodynamics and the kinetics of the reaction between the vapour molecules and oxygen in air.


Peak at m/z = 39 in mass spectrum of 3-chloropropene

It would indeed be $\ce{C3H3^+}$ assuming predominant isotopic species.  Although there can be other isomers, the cyclic isomer, the cyclopropenyl cation is strongly stabilized by aromaticity.

As PLD comments, there are really only two steps, not three, to forming this cation.  The allyl cation ($\ce{C3H5^+},m/z=41)$ is first obtained by loss of the chlorine atom plus an electron.  The aromatic cyclopropenyl cation noted above is then obtained by loss of $\ce{H2}$.


Kb values of N,N-dimethylaniline derivatives

chemicalize.com is  a cheminformatics website which is the kind of resource you are looking for.

You can use the search tool to find your compounds. The input is the structure itself, which you can draw in-situ.

Here is the $\mathrm{p}K_\mathrm{a}$ data I retrieved from the aforementioned site. Note that these are calculated / estimated, not measured $\mathrm{p}K_\mathrm{a}$ values.

Table 1. $\mathrm{p}K_\mathrm{a}$ values (of the conjugate acid)

$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & 3.71 & 5.19 & -0.98 & 1.44 & 2.66 \\
\textit{meta-} & 3.89 & 5.15 & 1.73 & 4.77 & 3.45 \\
\textit{para-} & 5.55 & 5.46 & 0.65 & 4.68 & 3.65 \\
\hline
\end{array}
$$

Since the question is asking for $K_\mathrm{b}$ values, I converted them too.

Table 2. $K_\mathrm{b}$ values

$$
\begin{array}{lccccc}
\hline
& 1 & 2 & 3 & 4 & 5 \\
\hline
\textit{ortho-} & \pu{5.13e-11} & \pu{1.55e-9} & \pu{1.05e-15} & \pu{2.75e-13} & \pu{4.57e-12} \\
\textit{meta-} & \pu{7.76e-11} & \pu{1.41e-9} & \pu{5.37e-13} & \pu{5.89e-10} & \pu{2.82e-11} \\
\textit{para-} & \pu{3.54e-9} & \pu{2.88e-9} & \pu{4.47e-14} & \pu{4.79e-10} & \pu{4.47e-11} \\
\hline
\end{array}
$$


Why did my glass pot explode when I boiled sugar in it and added cold milk?

Assuming initially your glassware contained a cup of sugar and a little of water, you basically prepared caramel.  Depending on the sugar employed, temperatures needed to trigger caramelization vary, but $\pu{160 ^\circ{}C}$ ($\pu{320 F}$) mentioned succrose, as example (which is the normal household sugar) is quite high. Most materials, including glass, dilate upon heating, which is fine as long as this is evenly done.

Now if you add cold milk, say of $\pu{20 ^\circ{}C}$, the glass aims to contract back to its initial state. Because of the quantity of milk added, and the large heat capacity of water as a major constituent of milk, milk served momentarily as considerable heat sink.  Now taking into consideration that the glass ware's walls are unevenly heated -- at the outside still heated by gas or stove to more than $\pu{160 ^\circ{}C}$, and the inner of $\pu{20 ^\circ{}C}$, this suddenly generates a lot of mechanical strain and stress on the material.  The more the walls of your glass ware are thick, the more easy these may then crack and shatter into pieces just by the sudden temperature change.  (For the same token, you place hot glass ware on a plank of wood to allow slow cooling to room temperature.)

In addition, if the glass ware were closed tightly just after addition of the cold milk, the hot ($\pu{160 ^\circ{}C}$) is able to boil off the water in the milk, too; generating steam that likes to expand, or -- if confined in volume -- will build up pressure.  Under normal circumstances, for each litre of (liquid) water, up to $\pu{1.7 m^3}$ of steam may be generated (at normal pressure). This represents an additional stress for the material, and standard kitchen glass ware is not designed to withstand such pressures.


From this paper,  it seems Electrochemical recovery of Neodymium is possible from the Magnet.


As I mentioned in my comment, the Wikipedia article on resonance provides a rather nice description of the subject where the Resonance in quantum mechanics section is of particular interest for the question asked.

However, I decided to somewhat clarify the matter emphasising few important points. And I will start by quoting the OP.

  We know that in reality benzene has no pure single or double bond.

Now please, take a deep breathe and repeat with me: there are no bonds. In reality we do have electrons and nuclei that constitutes a molecule, but a chemical bond is merely a theoretical concept introduced in the framework of a scientific theory. In other words, chemical bonds do not exist in the same sense as electrons and nuclei do.

For instance, the concept of resonance was introduced by Linus Pauling in the framework of the so-called valence bond theory, according to which

  a covalent bond is formed between the two atoms by the overlap of half
  filled valence atomic orbitals of each atom containing one unpaired
  electron.

So, this is the way the concept of chemical bonding enters the scene in the valence bond theory: a chemical bond is used to describe overlapped atomic orbitals of two atoms in a molecule. Note that atomic orbitals also do not exist in the same sense as electrons and nuclei do, but an orbital has at least some strict mathematical and physical meaning, while a chemical bond, as introduced above, does not. Read the above quote carefully one more time: it is not even a definition! What we just say is that whenever half filled valence atomic orbitals of two atoms overlap a covalent bond is formed, but we do not say what exactly is a covalent bond. It is just something; something which is formed whenever orbitals overlap. It is a concept, purely theoretical construction. It is practically helpful in description of a reality, but no one will say that it is actually a part of it.

Do not be afraid of all this business, because it is the way science generally works. Scientific theories basically tell us how nature could be described, and they do so in the language of mathematics, since it is the only language nature can unambiguously "speak" with us. However, it is often so tempting to force it to speak with us in a different language, in a language of our theoretical concepts (such as, for instance, the concept of a chemical bonding), no matter how fuzzy they are. This is a very dangerous road, because, after all, who are we to tell nature which language to speak? Nature will likely refuse to "speak" with us in that language, and that would be it.

That is what actually happened with the valence bond theory as it was originally formulated. Indeed, all carbon-carbon bond lengths are equal in benzene, but the valence bond theory were not able to account for that simple fact. Nature refused to "speak" with us in the language of chemical bonding. At that point chemists had two choices: either abandon the valence bond theory or introduce another theoretical concept into it, introduce a new word in our language making it possible to communicate with nature in this language. The concept of a chemical bonding was already so tremendously useful in chemistry that the choice was actually predetermined, and in 1928 Linus Pauling introduced the concept of resonance between several valence-bond structures of a molecule.

What is resonance? Is it a physical process? No way! It is just a concept, a word, used to describe

  delocalized electrons within certain molecules or polyatomic ions
  where the bonding cannot be expressed by a single Lewis formula.

The word "resonance" was probably not the best choice by Pauling, since it gives an unwanted feeling that the molecule actually do oscillate back and forth between the structures, which is, as I said, not the case. But putting aside its somewhat misleading name, the concept of resonance is as good as the concept of chemical bonding and many other concepts in chemistry and other sciences.


Will ferrocene undergo electrophilic aromatic substitution?

Due to the aromatic character of the cyclopentadienyl ligands, ferrocene can undergo electrophilic aromatic substitution reactions typical of aromatic compounds such as benzene. Ferrocene shows very high reactivity towards electrophiles, making it more closely comparable to phenol than to benzene, references and examples here and here.


How does cullet reduce the production temperature in glass production?

No expert here, but cullet would react faster than sand with the alkali, developing into a more uniform fluid that can be mixed at a (slightly) lower temperature so as to eventually dissolve the sand.

It is like beginning the reaction to produce glass at 25% (or however much cullet is used), rather than 0%, so that raising the temperature is not needed to to speed up the reaction. Lower temperatures can be used without needing to increase the time, and thus wear and tear on the equipment can be reduced.

So it's not that the cullet causes a reduction in temperature, but rather allows it.


Is sodium acetate strong or a weak electrolyte?

Sodium acetate is a strong electrolyte in water.  The status of sodium hydroxide as a strong base triumphs; it makes even a weak acid such as acetic acid dissociate essesitally 100% into ions (by stealing the protons).  And when something is essentially 100% dissociated into ions, by definition it's a strong electrolyte.


Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)?

In,Thermodynamics,Work is defined as
$W$= $-\int_{V_1}^{V_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V_1=V_2$.

Coming to the enthalpy,you are right.
$\Delta$H=$\Delta$U + $\Delta$($PV$).But if you apply the product rule then,
$\Delta$H=$\Delta$U + $P$$\Delta$V + $V$$\Delta$P.
At, constanst presurre,$\Delta$P=$0$. so at constant pressure,$\Delta$H=$\Delta$U + $P$$\Delta$V.But,according,to first law of Thermodynamics,$Q$= $\Delta$U - $W$.At,constant pressure, the $p$ in the integral comes out.So,$W$= $-p\int_{V_1}^{V_2}dV$= -$p$($V_2-V_1$) = -$p$$\Delta$V.So, first law in constant presuure becomes,$Q_p$= $\Delta$U - (-$p$$\Delta$V)= $\Delta$U+$p$$\Delta$V= $\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.


Chemical tests given by glucose (biomolecule)

This are two different types of reactions.

Tollens is a redox reaction in which silver is reduced and glucose is oxidized.
This reaction depends on the redox potential of glucose only.

Schiff and bisulphite tests are addition reactions. The reagents condense with the aldehyde group.
This occurs also in the case of glucose but it is not sufficient to give a positive result.
In the case of bisulphite, the adduct must precipitate, but it is too soluble.
In the case of the Schiff test, the adduct must dehydrate, but it forms the glucosamine derivative instead.


How to know about radical character of a diradical molecule?

Take a look at this is the article, which provides quantitative characterization of radical character:
M. Head-Gordon. Characterizing unpaired electrons from the one-particle density matrix. Chem. Phys. Lett., Vol. 372, Issues 3–4, 2003, p. 508-511

Based on natural orbitals, the analysis in this paper is applicable to any correlated method - it needs only one-particle state density matrix.


What is the simplest structural unit of an element or compound?

For a solid lattice (metal, ionic solid, crystal) I would say the answer is unit cell, for a non-charged compound  that isn't a lattice (including solution/gas phase) I would say molecule, and for a charged one I would say ion-pair / ternary ion-associates / quaternary-ion associates for two / three / four ion clusters. For polymers, I would refer to the basic unit(s) being the component monomer(s).

There is, as has been identified in the comments, an issue with allotropes, Carbon (Diamond, Graphite etc) can be encompassed within 'unit cell', but for elements like phosphorus and sulfur, it's a little more complicated. There I would contend that the basic unit of e.g. P4 (white Phosphorus) is the molecule, and for the rest, I would return to referring to the unit cell, or 'monomer' in the case of amorphous networks.


How to find pH in the following titration/buffer problem given concentration of base and acid

This is a titration problem - not necessarily a buffer problem. While you have $0.0055$ moles of $\ce{NH3}$, you have $0.01$ moles of $\ce{H}$ from $\ce{HCl}$ dissociating. $0.01 - 0.0055 = 0.0045$ moles of $\ce{H}$ left. Take H/total L to find M. Take the $-\log$ of $\ce{[H]}$.


Why is more of a catalyst added, when it is not consumed?

In short, a catalyst does participate in a chemical reaction the rate of which it increases, it just do so in a somewhat tricky way so that it does not appear in the overall reaction equation.

Wikipedia article on catalysis provides a concise explanation of why this is the case:

  Catalysts generally react with one or more reactants to form
  intermediates that subsequently give the final reaction product, in
  the process regenerating the catalyst. The following is a typical
  reaction scheme, where $\ce{C}$ represents the catalyst, $\ce{X}$ and $\ce{Y}$ are
  reactants, and $\ce{Z}$ is the product of the reaction of $\ce{X}$ and $\ce{Y}$:
  $$
     \ce{X + C → XC} \tag{1}
$$
  $$
     \ce{Y + XC → XYC} \tag{2}
$$
  $$
     \ce{XYC → CZ} \tag{3}
$$
  $$
     \ce{CZ → C + Z} \tag{4}
$$
  Although the catalyst is consumed by reaction 1, it is subsequently
  produced by reaction 4, so it does not occur in the overall reaction
  equation:
  $$
    \ce{X + Y → Z} \, .
$$
  As a catalyst is regenerated in a reaction, often only small amounts
  are needed to increase the rate of the reaction. In practice, however,
  catalysts are sometimes consumed in secondary processes.


Chemical reaction between dichloromethane and solid carbon dioxide

In brine cooling happens only on the surface of the solid carbon dioxide and causes the water to freeze there, forming an insulating layer (the same happens with ethylene glycol). In dichloromethane, which freezes below $-97$ degrees, no freezing occurs. Moreover, carbon dioxide is well soluble in dichloromethane, which makes the heat transfer more efficient. The latter property may cause you problems if you want to use your coolant in closed installations, because after warming the solution may release the excess of carbon dioxide.


Slope ≠ 1 in Henderson-Hasselbalch graph for determination of pKa

OP's question: From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ axis in this situation, why is this?

Answer for this question is easy. According to the graphs plotted using the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$ (Henderson-Hasslebalch equation), when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$ (meaning $[\ce{In-}] = [\ce{HIn}]$), then $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $. Therefore, the $x$-value where the straight-line crosses the $x$-axis is the best value for $\mathrm{p}K_\mathrm{a} $.


Which ultra-high temperature ceramics make a good electrical insulator?

Thorium dioxide gets you 3390°C.  Have fun getting it, $$$ and also the thorium is radioactive.  You also risk some nonstoichiometry, due to thorium having variable oxidation states, and that could make the compound somewhat conductive.  Magnesium oxide weighs in at 2800°C; but is more stable, cheaper, nonradioactive and having no variation in stoichiometry.


Can a metallic bond with a sea of electrons form inside the human body?

We may not produce nails from our bodies, but certain enzymes and proteins do involve metal clusters, especially clusters involving iron [1].  In the natural biomolecules iron-iron and iron-nickel compounds serve as functional groups for electron transfer.  Also discussed are iron-sulfur clusters, which have characteristics similar to all-metal clusters despite sulfur usually being considered a nonmetal.  Chemists have sought to mimic these natural cluster compounds with artificial molecules, not only using iron and nickel but also moving on to less common metals such as molybdenum and ruthenium.
Reference
1.
C. Fehl C and B.G. Davis, "Proteins as templates for complex synthetic metalloclusters: towards biologically
programmed heterogeneous catalysis.", Proc. R.
Soc. A 472: 20160078. (2016)
Link


Typcially, the terms saturated and unsaturated only apply to the carbon chain in itself. A saturated compound would be one whose systematic name ends in -ane while unsaturated ones would end in -ene or -yne — grossly simplifying. Going by this rather simplistically strict definition, at first sight acetone (propanone) is saturated. But.

You may have heard of the keto-enol tautomery. That is, a ketone is always in equilibrium with its corresponding enol form, for acetone that would be propen-2-ol — an unsaturated compound. It is shown in the scheme below. The first indication of the distinction being difficult.

Furthermore, acetone can be reduced by molecular hydrogen to isopropanol. That means that given the correct conditions, it can basically react much like an unsaturated compound would. It also reacts with elemental bromine similarly to how an unsaturated hydrocarbon would, although only the first mechanistic step is identical; only one carbon is brominated in the product.

There is also the measure of double bond equivalents, that is often used in the process of structure elucitaion. Taking the molecular formula of a compound (which can be deduced well from high-resolution mass spectrometry) and performing a simple calculation allows one to arrive at a number of double bond equivalents for a compound. The formula is:

$$\tag{1} \text{DBE} = \frac{2 \ce{C} - \ce{H} + \ce{N} + 1}{2}$$

Where $\ce{C}$ is the number of carbon atoms (including other tetravalent atoms such as silicon), $\ce{H}$ is the number of hydrogen atoms (including other monovalent atoms such as chlorine) and $\ce{N}$ is the number of nitrogen atoms (including other trivalent atoms such as phosphorus).

However, having 1 double bond equivalent or more does not automatically mean that a molecule is unsaturated, no matter by which definition. Take cyclohexane, $\ce{C6H12}$: It has one double bond equivalent which is caused by the fact that it is cyclic. It will not react with hydrogen or with bromine and no sane chemist would call it unsaturated. So this measure is not of much help.

Concluding, I can only say that both sides of the discussion are wrong. It is wrong that acetone will not react with hydrogen to give isopropanol. It is equally wrong that that makes acetone an unsaturated compound. The concept of unsaturation simply does not extend well to $\ce{C=O}$ double bonds.

The exam question should not be graded.


Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?

I think you are asking what does the HOMO and LUMO of the 1,3-dioxabiradical look like.

Let's start by using cyclopropane as a model compound.  In cyclopropane the $\ce{C-H}$ bonds are approximately  $\ce{sp^{2.46}}$ hybridized and the $\ce{C-C}$ bond is approximately $\ce{sp^{3.74}}$ hybridized (see here).  That's a lot of p-character in the $\ce{C-C}$ sigma bond. Dioxirane is likely similarly hybridized.  As soon as the 1,3-biradical forms (as soon as we break the $\ce{O-O}$ bond), the HOMO and LUMO will look like what is shown in the "initial" drawing below.  If the biradical lives long enough to undergo rotation about the $\ce{C-O}$ bond, then the HOMO and LUMO will look like what is shown in the "rotated" drawing.

These drawings just point out the conformational dependence of the HOMO and LUMO.  Note that the two lone pairs on each oxygen have not entered the discussion.  They are orthogonal to, and have no effect upon, the HOMO and LUMO orbitals.

Imagine bringing two oxygen atoms closer to each other.  If we push them closer and the orbitals containing the odd electrons are pointing towards each other as in the "initial" geometry, we will form a sigma bond and we know what the HOMO and LUMO look like for a sigma bond (just like what is pictured in the "initial" geometry, only the orbitals are closer together).  If we push them closer and the odd electrons are pointing parallel to each other as in the "rotated" geometry, we will form a pi bond and we know what the HOMO and LUMO look like for a pi bond (just like what is pictured in the "rotated" geometry, only the orbitals are closer together). In the 1,3-biradical the carbon atom holds the 2 oxygen atoms together, they are closer than they would be at infinity, but further apart then they would be if there was a bond between them.  This tells us that the HOMO-LUMO energy separation is greater than it would be at infinity (0), but less than if they were closer together.

We could refine our model by noting that, according to Bent's rule, we would expect the orbitals containing the oxygen lone pairs to have a bit more s-character that the carbon $\ce{C-H}$ orbital in cyclopropane.  Therefore, in dioxirane and the corresponding 1,3-dioxabiradical, we would expect both the $\ce{O-O}$ bond and the orbital containing the single electron in the biradical to have even more p-character than $\ce{sp^{3.74}}$.

Again, however, to re-emphasize an earlier point, the lone pairs on oxygen have no bearing on our discussion.  That said, if we would like to speculate on the hybridization of the orbitals containing the lone pairs, we may do so.  Most texts tell us that the oxygen lone pairs in water, ethers and alcohols are in (roughly) $\ce{sp^3}$ orbitals.  However data from photoelectron spectroscopy experiments seems to suggest that one lone pair is in (roughly) a p-orbital while the other lone pair is in an $\ce{sp}$ orbital.


The oxidation state of iodine in iodoxybenzoic acid

This is a written reply from Professor Zhdankin*, author of Hypervalent Iodine Chemistry:

Indeed, many literature sources classify pentavalent iodine compounds as iodine in the oxidation state +5. It makes sense because the electronegativity of iodine is almost the same as carbon, or even less than carbon. According to different studies, electronegativity of iodine varies from 2.21 [A. L. Allred and E.G. Rochow, J. Inorg. Nucl. Chem., 5, 264 (1958).] to 2.66 [A. L. Allred , J. Inorg. Nucl. Chem., 17, 215 (1961)]. The original Pauling's value of iodine electronegativity is 2.5, exactly the same as carbon. This terminology reflects the similarity of iodine in IBX to iodine in $\ce {I2O5}$ or $\ce {HIO3}$ by chemical and structural properties.
Personally, I prefer not to use the "oxidation state" terminology regarding organoiodine compounds. Pentavalent iodine is a perfectly correct classification for IBX.

*He has given his permission for his reply to be shared here.


What are the half-equations of manganese and oxygen in the following redox reaction?

The half-equations should look like this:

$$\ce{2Mn^2+(aq) \to 2Mn^4+(aq) + 4e^-}$$

and

$$\ce{4OH^-(aq) + O2(aq) + 4e^- -> 4O- + 2H2O(l)}$$


Why does temperature affect the dissolved oxygen concentration?

The solubility of a many gases in a variety of liquids is observed to decrease with increase in temperature, for example oxygen, nitrogen and methane in water.  This follows the qualitative predictions of Le Chatelier’s principle, which in thermodynamics is supplanted by the Clapeyron equation. As one of the two phases involved is a gas the Clausius–Clapeyron equation can be used.

This assumes that the molar volume of any gas/vapour is far greater than that of the solution. Assuming that all values correspond to one mole the Clausius–Clapeyron equation is

$$\frac{\mathrm d\ln(p)}{\mathrm dT}=\frac{\Delta H}{RT^2}$$
which in its integrated form between pressure and temperature $p_1, T_1$ to $p_2, T_2$, and by assuming that $\Delta H$ is independent of temperature gives

$$\ln\left( \frac{p_2}{p_1}\right)=- \frac{\Delta H}{R}\left( \frac{1}{T_2} - \frac{1}{T_1}   \right)$$

Despite the approximations involved in its derivation this equation does describe well the measurements of the vapour pressure over many solids and liquids, for example depression of freezing point, elevation of boiling point and solubility of solids and gases in liquids, although the latter is not as frequently described as the other two.

In the elevation of the boiling point the solution at a temperature $T_1$ has a lower vapour pressure $p_1$ than the pure solvent. The solution only boils when the temperature is increased to $T_2$  where its vapour pressure has increased to  $p_2$ (normally 1 atm ), which is the pressure at which the pure solvent boils.  By Raoult’s law  $p_1=p_2x_{\mathrm s}$, where $x_{\mathrm s}$ is the mole fraction of solvent
$\ln(p_2/p_1)=\ln(1/x_s)$
As $x_\mathrm s=1-x_\mathrm t$ where $x_\mathrm t$
is the mole fraction of solute,
$$\ln\left( \frac{1}{1-x_t}\right)= -\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$
and as $x_\mathrm t$ is far smaller than unity
$\ln(1-x_t) \approx -x_t -x_t^2/2 \ldots$
from which the second term can be ignored as it is insignificant compared to the first then,
$$x_\mathrm t= \frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$

Thermodynamically the elevation of boiling point and solubility are very similar but we must swap solute and solvent over so what we formerly called solvent has now to be solute and vice versa.

Considering now solubility, the variation of the mole fraction dissolved $x_\mathrm t$ can be written as
$$x_\mathrm t=\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_\mathrm B}   \right)$$

where what was $T_2$ is now the normal boiling temperature of the liquefied gas. Also $\Delta H_\text{vap}$ (a positive quantity) is now the (molar) heat of vaporisation of the liquefied gas and we approximate this to be the heat of vaporisation from solution.

This last equation indicates that the solubility decreases with increase in temperature and should be the same in all solvents in which it forms ideal solutions.

In the solution a gas molecule has an attractive interaction energy between itself and the solvent. (Experiment shows that $\Delta H_\text{vap}$ is positive thus the heat to dissolve is negative). The solvent molecules also have an attractive interaction between themselves otherwise they would not form a liquid.  Thus for a gas-solvent interaction a solvent-solvent one has to be destroyed. Overall the difference in energy is probably small.  However, there is an entropic effect to consider. Clearly this will have increased when a gas dissolves as now there is another species in the solvent and thus the number of ways the molecules can be arranged has increased and this will have increased the entropy. Together these effect make the free energy negative as expected for a spontaneous process.

As the temperature is increased the molecules gain in average energy and so are more able to overcome the attractive potential that keeps them in solution and thus the vapour pressure of the solvent increases and the solubility of the gas decreases. It is also worth noting that the average kinetic energy $(3/2)kT$ is the same in the solution and in the gas phase, in solution it is merely restricted to a smaller range on motion. Thus the notion that it is the kinetic energy that is acquired that causes a molecule to enter the gas phase in incorrect.


Creating ethane through electrolysis of vinegar

Indeed that is quite possible. The reaction was first described by Hermann Kolbe and is hence called "Kolbe's Electrolysis"

It is a decarboxylative dimerisation of two carboxylate ions (carboxylate salts--either sodium, or potassium are used instead of carboxylic acids for precisely this reason as they disassociate more readily to furnish carboxylate ions).

The general reaction scheme is is a two-stage radical process: first, electrochemical decarboxylation gives a radical intermediate; then two such intermediates combine to form a covalent bond.

For sodium acetate (again would furnish acetate ions more readily than acetic acid) one would see the following reaction:

$\ce{2CH_3COO^-}\ce{Na^+} \rightleftharpoons  \ce{2CH_3COO^-}+ \ce{Na^+}$

At the anode: a) $\ce{2CH_3COO^-} \stackrel{-2e-}\rightarrow \ce{2CH_3COO.}$

Electrons shuffle around a bit in the acetate radical and we get

b)$ \ce{2CH_3COO.} \rightarrow \ce{2CH_3.} + \ce{CO_2} $

c)$ \ce{2CH_3.} \rightarrow \ce{CH_3CH_3} $

At the cathode: a) $\ce{H_2O} + \ce{e-}\rightarrow \ce{OH^-} + \ce{H.}$

and, finally $ 2\ce{H.} \rightarrow \ce{H_2}$

Now, here we observe that the solution is becomes more basic as the reaction proceeds since hydroxide ions are being produced. If we make the conditions mildly acidic, and draw on Le Chatelier's principle, we can shift the reaction to the right.

So, it would make sense to use sodium acetate, in the presence of ethanoic acid to perform this reaction.


Which of these reactants will not generate this racemic amine?

This is an example of reductive amination, which generally proceeds through two main steps: (i) condensation reaction between the amine and carbonyl to generate an imine; and (ii) reduction of the imine to the corresponding amine.

Clearly (e) can't do this reaction since there is no carbonyl. The reason why (a) is also an answer comes down to the wording of the question - it specifically states that the product is racemic. The stereocentre in (a) is (R) in the starting material, and is not involved in the reaction so the product will exist as a single (R) enantiomer.


Voltaic cell: why does the rate of rusting increase when NaCl is added?

If your diagram is correct (the notation of the polarity of the electrodes) then the anode should be the one that is not being pumped with oxygen, because the negative end of any electro-chemical cell is the anode.

Let's see how this happens.

First of all let us consider the ions and molecules at two electrodes the system.

We have $\ce{O2}$ at electrode A and $\ce{Fe}$ at B   .

Now since the reduction potential of $\ce{O2}$ is greater than $\ce{Fe}$ a reaction takes place at the $\ce{O2}$ pumping electrode. This is a reduction:

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

Now at the other electrode, an oxidising reaction takes place:

$$\ce{Fe -> Fe^3+ + 3e-}$$

So you can see there's an excess of electrons at the $\ce{Fe}$ electrode. Now these electrons travel across to the $\ce{O2}$ containing electrode since the cell reaction there needs electrons. This travel of electrons give rise to an electric current. You might know from your previous lessons that current flows from negative terminal to the positive terminal inside a cell. Hence the $\ce{Fe}$ electrode becomes the negative anode.

As to your second question why does $\ce{NaCl}$ increase corrosion, for all these reactions to take place electrons should be moved about. $\ce{NaCl}$, being a strong electrolyte, increases the conductivity of water. That means electrons are moved rapidly giving a rise in the rate of corrosion.
And you have asked in the comments "How is just water doing the trick?". You should keep in mind that water exists as $\ce{H+}$ and $\ce{OH-}$ in the aqueous liquid state.


Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent?

Rigidity is definitely an important factor. At $295\ \mathrm{K}$, indeno[2,1-a] has a fluorescence quantum yield $\Phi_F$ of 0.94, (DOI). In an organic glass at $77\ \mathrm{K}$, even (E)-stilbene, which hardly shows any fluorescence at room temperature, exhibits  a $\Phi_F$ of 0.75.

Vibrational stretching of $\ce{C-C}$ bonds in substituents can significantly contribute to radiationless deactivation of an $S_1$ state and therefore reduce the fluorescence quantum yield $\Phi_F$. Admittedly, toluene isn't famous for its fluorescence ($\Phi_F = 0.15$), but the fluorescence of tert-butyl benzene is even lower ($\Phi_F = 0.03$).

Apart from radiationless deactivation through internal conversion (IC), an excited singlet state can be depopulated by intersystem crossing (ISC) to a triplet state. Under certain conditions, this is even possible for compounds that usually have very low triplet quantum yields.

Halobenzenes added to a solution of fluorescing anthracenes in dioxane (DOI) or ethyl iodide added to solutions of pyrene in alkanes (DOI) have shown to quench the fluorescence normally observed in the absence of the halogenated compounds. The latter exhibit a external heavy atom effect.

The lack of $S_1 \rightarrow T_1$ intersystem crossing is probably the most important factor that makes rigid compound show large $\Phi_F$ values.

It is striking that strongly fluorescent compounds typically have an extended $\pi$-system. Here, the $S_0\rightarrow S_1$ excitation is a $(\pi,\pi^*)$ transition.

On the other side of the fence, ketones, such as acetophenone or benzophenone, hardly show fluorescence. For these compounds, $S_0\rightarrow S_1$ excitation of the $\ce{C=O}$ group is a $(n,\pi^*)$ transition. From a $^1(n,\pi^*)$ state, spin-orbit coupling can both flip the spin and move one electron to a more distant and geometrically different orbital. $S_1 \rightarrow T_1$ thus is in fact a $^1(n,\pi^*) \rightarrow {}^3(\pi,\pi^*)$ transition.

For compounds like rhodamin B in their $^1(\pi,\pi^*)$ states, the  corresponding transition to a $^3(\pi,\pi^*)$ state is largely spin-forbidden ans does not occur. Provided that no external effects come into play, these molecules exhibit strong fluorescence.


Why are equations of state for a non-ideal gas so elusive?

It's been known since 1941 that the answer to your question is in the negative, i.e. that there will never be a closed form equation of state for a nonideal gas.

In 1941 Mayer and Montroll developed what is now known as the cluster expansion for the partition function of a nonideal gas whose particles have pairwise interactions. This cluster expansion provides a 1-1 correspondence between various integrals over the interaction potential and virial coefficients in the Kamerlingh Ohnes equation of state which has in principle an infinite number of terms in it. Therefore, it would not be considered a closed form equation of state. In practice the virial coefficients are known to decay so it is usually safe to truncate the expansion for practical calculations.

One could extend the cluster expansion to three-body and higher-order interaction potentials but this will not change the fundamental argument above.


Calculating the heat of reaction for Mg metal and HCl

The equations you listed can be used to partially solve the problem.  However, with the equations you used, you have made one error which was in your choice of "moles."  Your moles value must match up with your $\ce{C_{p}}$, so you must find the moles of Water in $\pu{95.00mL}$!  Alternatively, you can use $4.184 \pu{\frac{J}{g\ ^\circ C}}$ and $\pu{1 \frac{g}{mL}}$ and $\pu{95.00mL}$ of water along with the equation $\mathrm{Q = n \times C_p \times \Delta T}$.

On the other hand, the premise of your solution is close but wrong, which is obvious if you examine the units:

  $\pu{0.1773 mol} \times 75.3 \frac{\pu J}{\pu{mol K}} \times \pu{9.2 K} = \pu{122.8 J}$

The units for Heat of Reaction are $\pu{\frac{J}{mol}}$, not $\pu{J}$!

To finish off your solution, you must divide by the moles of reaction that occurred (given by the limiting reactant).  $\pu{-44.4 kJ}$ is the value calculated based on the moles of Mg reacted.  However, I suspect that the given answer is wrong because the balanced equation does not yield a 1:1 reaction ratio between $\ce{Mg}$ and $\ce{HCl}$, which means that $\ce{HCl}$ is the limiting reactant, which would make the Heat of Reaction Larger.


Identical compounds are not considered to be isomers. However, the following explanation should clear up any confusion from your question.

A and B are both isopentane and C is n-pentane (normal pentane). These make up two of the three isomers of pentane, the third being neopentane ($\ce{CH3CH(CH3)2CH3}$). The question is referring to pentane as $\ce{C5H12}$, a group of isomers in which n-pentane ($\ce{CH3CH2CH2CH2CH3}$) is included and not identical to.


Stereochemical aspect of heating of cyclohexane-1,2-diol

The reaction you are thinking of is likely the thermal decomposition of the cyclic sulfites prepared from the cis- and trans-diols.  Here is a link to the abstract.

In both reactions $\ce{SO2}$ is extruded.  Whichever group on the vicinal carbon that is lined up in an anti-periplanar fashion to the departing oxygen is the group that migrates.  This is because the antiperiplanar geometry allows for good orbital overlap during the migration; if a choice is available then hydrogen migrates in preference to an alkyl group just as with carbocations. In the cis-isomer, the vicinal hydrogen is properly aligned (I darkened the affected bonds to highlight the geometric arrangement), its migration produces cyclohexanone.  In the trans-isomer, only a $\ce{C-C}$ bond in the cyclohexane ring is properly aligned, its migration produces cyclopentanecarbaldehyde.


What total energy (in kJ) is contained in 1.0 mol of photons, all with a frequency of 2.75 • 10¹⁴ Hz?

The formula $E=h\nu$ is for the energy of one photon. You have a mole of photons. You need to use a slightly modified form:

$$E=Nh\nu$$

where $N$ is the number of photons, in this case

$$\begin{align}
N&=n\cdot N_\mathrm A\\[6pt]
&=1\ \mathrm{mol}\times6.02\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]
&=6.02\times10^{23}
\end{align}$$

Note that you are being asked to report energy in kilojoules not kilojoules per mole.

If I use $E=nN_\mathrm A h \nu$, I get the correct answer you cite of $110~\mathrm{kJ}$. Do you?


Why do non-polar covalent compounds dissolve in non-polar covalent compounds only?

There is the oft quoted adage in chemistry, which goes something like:

  like dissolves like

The reason for this can be understood on the basis of polar/ionic compounds have some degree of charge separation. In polar solvents, the partial charges on on the solvent molecules can interact with the partial charges on the solute molecules. These interactions lower the overall energy of the system, and thus dissolution takes place. This part you get, so it's fine.

First let's address why non-polar solutes don't dissolve in polar solvents. Consider the simple case, of a long chain hydrocarbon (solute) in water (solvent, present in larger excess). Typically, water molecules will form (and break) hydrogen bonds with each other. At any given time, a water molecule would like to maximise the number of hydrogen bonds it can form, simply because it is energetically favourable.

Remember that the water molecule, cannot form hydrogen bonds with the hydrophobic molecule; it can only do so with other water molecules.

So when a hydrophobic is dropped in an aqueous medium, hydrogen bonds between water molecules will be broken to make room for the hydrophobe; however, water molecules do not react with hydrophobe. This is bad energetically. So, water molecules would now try to adopt an arrangement that maximises the hydrogen bonding network; to do so they will arrange themselves in a "cage" like structure surrounding the hydrocarbon droplet. While, this allows the water to make more hydrogen bonds, it leads to overral "ordering" and thus decreases entropy. Again, bad. Thus to minimise this hydrocarbon droplets will clump together, and will be exluded by the solvent--this is why oil and water don't mix. (Read more here)

So what happens when non-polar compounds are tossed into non-polar solvents. True, there aren't any dipole-dipole, ion-dipole, etcetera kinds of interaction that you have with polar molecules, but you still have something called London dispersion, and non-polar entities often interact via these weak interactions.


Why nature favours lower potential energy?

It doesn't.  Nature favors higher entropy, and in many situations higher entropy can be achieved by turning potential energy into heat, so it looks like the general proposition is that systems evolve to states of lower energy. In fact, that was considered as a possible general law of nature by early philosophers.

However, there are plenty of counter-examples.  For example, consider the smoke from a fire.  It rises spontaneously, so the atoms and molecules in it are spontaneously ascending to a higher state of gravitational potential energy.  Or a steam engine, in which the flow of heat from a fire can easily do work (compress springs, move objects uphill, run a generator to charge a battery, and otherwise increase the potential energy of things).  Very early philosophers rationalized these distinctions by saying that things tended to "seek their own level," with certain things seeking states of low energy and others seeking states of higher.  This isn't very satisfactory, of course, because there's no way to determine ahead of time which thing seeks which state.  One of the greatest triumphs of 19th century science was to discover the underlying principle of entropy, and that all spontaneous changes can be understood as tending to increase the entropy of the universe.

The question of why nature favors states of higher entropy is much harder to answer, because such a preference is not built into the laws of physics.  Basic physics tells us states of higher entropy are favored only when you generally begin in states of low entropy.  So far as anyone can tell, the universe generally began in a state of very low entropy, and it has been increasing ever since, and that's the only reason why we observe an evolution to state of higher entropy.  Why the universe started off in a state of low entropy, nobody knows.


This picture is from How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?:

Let's say the red circles represent a layer of atoms above, and the pink circles a layer of atoms below. The T's and O's represent voids with tetrahedral and octahedral coordination located between layers.

  My visualisation: In the image, 3 inverted and 3 upright tetrahedron can be formed. Is this correct?

It is correct that there are inverted and upright tetrahedra. In the picture above, the T's squeezed between red circles are inverted, and the T's between pink circles are upright.

I can see eight tetrahedra with their center in the unit cell reposted below. Six have one of the orange atoms (in the image posted by the OP) as their tip, and two have all orange atoms as their base.

Complicating things, there are also tetrahedra with their center at the edge of the unit cell, making counting them difficult (see below).

  It is hard to visualise the octahedrons for an atom. And so, I can't figure out the number of octahedral voids in HCP.

The atoms forming the octahedral coordination sphers are hard to visualize in the picture of the question because the unit cell does not fully contain a full set of six. Here is the view from the top again, with the base of the unit cell shaded (red circles correspond to orange atoms in the unit cell shown form the side):

Because we are only showing two layers of atoms, this reflects only half of the unit cell. In this half of the unit cell, there are three voids with octahedral coordination. To visualize them, you need three "pink" atoms at the base of the unit cell (the center one and two vertices of the hexagon) and three "red" atoms (located at c = 1/2, i.e. the second hexagonal layer). Of these three, two are in the unit cell and one is in a neighboring one. If you complete the unit cell by adding another layer, you get a total of six octahedrally coordinated voids in the cell. Because none are on the edge of the cell, this is the total number.

  Also, I want to calculate how much part of the octahedron (see figure) is in each HCP structure.

I'm not sure how to do that. I can say that two of the coordinating atoms are in our unit cell, one is in the neighboring cell, and three are on the border of the unit cell. So the larger part of the volume is probably within the unit cell.

  Number of octahedral and tetrahedral voids present in a HCP structure

We already know that there are 6 octahedrally coordinated voids per unit cell. Let's get the ratio of the two types of voids. If you look at the "red/pink" pictures, you notice that T's and O's come in horizontal rows, with one row of O's next to two rows of T's. So there are twice as many T's than O's, or 12 tetrahedrally coordinated voids per unit cell.

You can also figure this out in a more complicated way. There are 8 T-voids with center inside the unit cell, and 12 T-voids with center on the edge. These are shared among three unit cells touching at these centers (3-fold axis). So the math is 8 plus (12 divided by 3) is a total of 12, as we already figured out.

Or you can make an argument how many tetrahedral coordination spheres an atom partakes in. Each atom forms the tip of two (up and down) and the base of six (three directions, up and down), that is eight. A tetrahedron has four vertices, so two tetrahedra per atom. The unit cell contains six atoms, so twelve tetrahedra per unit cell.

Guide to related questions

Same question for cubic-close packing: How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?
Counting atoms and voids is much easier in that case, but the answer is the same.

Shape of the voids and visualization: Distinguishing between octahedral and tetrahedral holes


Connection between irreducible representations and electronic states in diatomic molecules

In the linear $C_{\infty \mathrm{v}}$ and $D_{\infty \mathrm{h}}$ point groups there are two notations for the irreducible representations which are equivalent, in that

$$\begin{align}
\mathrm{A_1} &\equiv \Sigma^+ \\
\mathrm{A_2} &\equiv \Sigma^- \\
\mathrm{E_1} &\equiv \Pi \\
\mathrm{E_2} &\equiv \Delta \\
\mathrm{E_3} &\equiv \Phi \\
&\,\,\vdots
\end{align}$$

For $C_{\infty \mathrm{v}}$, these are the actual irreps; for $D_{\infty \mathrm{h}}$ you simply need to tack on a g or u label to denote the symmetry with respect to inversion. See e.g. the $D_{\infty \mathrm{h}}$ character table found at http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=1001&option=4 which contains both notations.


Why is a theoretical titration curve higher than a practical one?

One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.

The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on the pH of your aqeous system. As long as it is sufficiently acidic no carbonic acid can/will react to form hydrogen carbonate or even further to carbonate. But from $\ce{pH}\approx4$ the possibility that hydrogen carbonate will be formed rises. This means for your titration, that you are not only titrating your acid anymore but you are also titrating the dissolved $\ce{CO2}$. The longer you wait for equlibration near the the equlibrium point the more dissolved $\ce{CO2}$ reacts to hydrogen carbonate and distorts your ideal $\ce{CO2}$-free titration curve.

To show you the impact more clearly I created the following figures showing both of your systems with varying carbonic acid concentrations $\left(0.00\ldots0.05~\ce{mol~L^{-1}}\right)$.

At first your system with hydrochloric acid:

And secondly the system containing acetic acid:

There are two remarking differences between both titration curves.

You can determine the equlibrium point of the $\ce{HCl}$-titration (though it's not around pH 7 but at pH 4-5) no matter how many hydrogen carbonate will be formed during your titration.
You have to pay attention during the titration of acetic acid because the inflection points gets distorted above any detectable other inflection point. However it is not that important as it seems because titrations of acetic acid are fast and there is usually not much time to dissolve enough $\ce{CO2}$.

The other points are already mentioned before. Your pH electrode could not be calibrated correctly and your concentrations might not be as precise as you think they are. The titration curve of hydrochloric acid looks like your acid was less concentrated which results in a higher starting pH and in a lower ending pH. The titration of acetic acid indeed looks like the electrode was not properly calibrated because it looks quite good if you just raise the pH values per measuring point and it suffers from dissolved $\ce{CO2}$ which is why the inflection point is shifted to a higher volume.

The mentioned increase of volume that is most times neglected during the calculation of titration curves has no big impact unless the concentration of the base is much lower than that of the acid that has to be titrated.

On the one hand the first four images show that different base concentrations around your ideal concentration don't effect the curve itself very much but only the needed volume to reach the point of inflection.

On the other hand you can see that there is a visible influence on the titration curve if the base is 10 times as diluted as the acid.


Does sodium produce water when it reacts with water?

The reaction of sodium metal and water could (and usually does) indirectly produce more water, if enough heat is produced to spontaneously ignite the hydrogen gas which is produced by the initial reaction. This cyclic type of reaction is often found in explosive or extremely exothermic reactions, e.g. $\ce{K+ + I- + H2O2}$ will also react violently in a cyclic way.

Combustion of hydrogen in air is this:

$$\ce{2H2 + O2 -> 2H2O}$$

So, in a (well) closed system with excess oxygen, water could re-form and collect. It's worth noting that dissolution of $\ce{NaOH}$ in water ($\ce{NaOH -> Na+ + OH-}$) is also exothermic and adds even more heat to the system.


Can sodium carbonate-sodium bicarbonate buffer be used for EDTA titration of calcium?

Sodium hydroxide $(\ce{KOH})$ works as $\mathrm{pH}$ modifier (measure the $\mathrm{pH}$ by adding drop by drop of hydroxide until $\mathrm{pH}$ is approx. $10$ to $12).$

Carbonate ion will be combined with calcium(2+) cation to form calcium carbonate. You can measure the calcium content by precipitating the calcium using soda ash light solution. Good luck.


The problem gives that the solubility of Silver dichromate is $\mathrm{8.3\cdot10^{-3} g/100\,mL}$.
[…]
I changed the $8.3\cdot10^{-3}$ from g/mL to mol/L and got $1.9\cdot10^{-8}$.
[…] Were am I going wrong here?

Was the solubility given in $\mathrm{g/100\,mL}$?
$$\mathrm{8.3\cdot10^{-3} g/100\,mL = 8.3\cdot10^{-2}\,g/L = \frac{8.3}{431.72}\,mol/L = 1.92\cdot10^{-4}\,mol/L}$$


This source explains it well. It looks like part of class material, but it clearly explains the dimensionlessness of $K_{eq}$.

The resolution of this apparent paradox is that the above equation, while perfectly satisfactory for everyday use, is not technically correct.  A more correct version is:
$$K_{eq} = \frac{\frac{\lvert B \rvert_{eq}}{\lvert B \rvert_{ss}} \frac{\lvert C \rvert_{eq}}{\lvert C \rvert_{ss}}}{\frac{\lvert A\rvert_{eq}}{\lvert A\rvert_{ss}}}$$
where the "ss" subscripts refer to the concentration of that species in the standard state.  (By this definition, Keq is always unitless.)

It then goes on to state:

Strictly speaking, division by the standard state concentrations is also necessary in every thermodynamics equation in which you take the log of a concentration product, otherwise the units don’t come out right.)  We will NEVER use this "correct" version of the equation in this class (well, never except in one problem on this week’s problem set...), and $K_{eq}$ for a reaction with unequal numbers of reactants and products is ALWAYS given with units,
even in published papers.

[1] http://www.bio.brandeis.edu/classes/archives/biochem102/102-02stand.pdf


Thought Experiment: Is it possible to preserve the charge imbalance present during molten lead(II) bromide electrolysis?

An electret is a dielectric (non-conducting) material that has a semi-permanent electric charge or polarization, the electrostatic equivalent of a permanent magnet. Electrets were made by first melting a polymer or wax containing polar molecules, and then allowing it to re-solidify in a powerful electrostatic field (several kilovolts/cm). The polar molecules in the dielectric align to the direction of the electrostatic field, producing a dipole with a permanent electrostatic bias when the melt solidifies.
Although electrets are only in a metastable state, those fashioned from very low leakage materials can retain excess charge or polarization for many years. The Japanese used microphones polarized by electrets in WWII.
https://en.wikipedia.org/wiki/Electret
So, using the electret as an example, the likely charge imbalance in the solidified lead bromide melt would be very small because the electric field across the molten salt is nowhere near kilovolts per centimeter. It might be measurable, but very small, and probably not very long-lived.


About how much carbon monoxide is produced in these two different reations?

About the best that can done is simply to look at the ratio of carbon in the two cartridges.

    $\dfrac{56.7*0.09}{144.6*0.28} = 12.6$%

The ratio of carbon dioxide to carbon monoxide formed on burning would depend on particle size, how well mixed an so on. The second formulation with sulfur is akin to black gun powder. The sulfur burns first at a lower temperature to ignite the charcoal. The sulfur dioxide would be toxic as well, but not as fast acting as carbon monoxide.


How can change in Enthalpy for a reaction be written like this?

Your claim that $\Delta H = \Delta U + P_{ext} \Delta V$ is incorrect. It is only correct when the system is always in mechanical equilibrium with the external pressure, or in other words irreversible expansion against constant pressure. The more general statement is $\Delta H = \Delta U + \Delta (PV) $. Here as you can see even when you compare it with $\Delta H = \Delta U + \Delta n_gRT$ there is no problem as we don't have to replace $P_{in} $ with $P_{ext} $.


Is it better to use a smaller, more accurate measuring cylinder several times or a larger, less accurate one for the same volume?

You have three types of errors:

The errors in the accuracy of your measurement.  Assuming you are reading the water levels consistently (at eye level at the bottom of the meniscus), these should be random.
To add random, uncorrelated errors, the standard method is to sum their squares and take the square root.  So, for the 0.1 mL average error for the 10 mL cylinder:
$$\text{average error for seven combined measurements} = \sqrt{7  \times (0.1 \text{ mL})^2} = 0.26 \text{ mL}$$
As you correctly intuited, it wouldn't make sense to simply multiply the 0.1 mL by 7, since this would be saying you have either a +0.1 mL error every time, or a –0.1 mL error every time, but that's not how random errors work—they are just as likely to be positive as negative. In addition, they don't all have a magnitude of 0.1 mL; rather, their magnitudes follow a distribution. Thus essentially what you have is a random walk, with a random distribution of step sizes, where the average step size is 0.1. The square-root-of-the-sum-of-the-squares method gives you the average distance you'd end up from your starting point with 7 of these steps

Error in the calibration of the cylinder.   This would be a systematic error, and it would add to the error of each measurement.  Higher qualities of cylinders (e.g., Class A glassware) have lower errors.

Error in the consistency with which the cylinder delivers the measured amount.  These cylinders are probably calibrated as "TD", which means "to deliver".  That is, it is assumed not all the water will drain out, so the markings on the cylinder are calibrated to account for that.  However, there will be some random error in how closely the amount that drains out corresponds to the assumed amount that drains out.  The only way to determine this error would be to contact the manufacturer (or to test it yourself using a scale).  In addition, if your cylinder is not clean, more water may adhere to the walls, which could cause a systematic error in the amount of water delivered.

Note also the comment by Andrew Morton that there's an additional important practical consideration, which is that if you have to use seven measurements, you could miscount and thus end up being off by  $\pm$10 mL! And this error is much easier to make when measuring with a graduated cylinder than, say, a tablespoon, because each measurement with the cylinder takes focus to get the total volume to 10.0 mL, and that focus can cause you to lose count.  There are of course ways to address this, e.g., making a mark on a piece of paper for each 10.0 mL added.


Why is the less substituted alkene favoured in Hofmann elimination?

There are 3 main types of elimination reactions, the E1, E2 and E1cb.  Here is a generalized picture of the transition state for the E2 elimination reaction.

Breaking of the $\ce{C-H}$ bond and the $\ce{C-LeavingGroup}$ bond are roughly equally advanced, as is formation of the double bond. Because the 2 bonds are broken to roughly the same degree, it can be said that the E2 reaction is in the "middle" of the possible elimination mechanisms.

If we change this pattern and make the $\ce{C-H}$ bond breaking more advanced than the $\ce{C-LeavingGroup}$ bond breaking (we can do this by making the leaving group a poor leaving group [i.e. groups like ammonium, $\ce{-NR3^+}$, sulfonium, $\ce{-SR2^+}$, etc.]), then we shift the mechanism towards the E1cb side of the elimination spectrum.

In the E1cb case, because breaking of the $\ce{C-H}$ bond is advanced, partial negative charge develops on the carbon where the proton is leaving from.  Now primary carbanions are more stable than secondary carbanions, which, in turn, are more stable than tertiary carbanions (electron releasing alkyl groups will destabilize a negative charge).  Therefore, when we have a choice of which proton to remove, we remove the proton on the least substituted carbon as this carbon will better stabilize the developing negative charge.  As the LeavingGroup departs and the reaction completes, this will produce the olefin with the least substituted double bond, the Hofmann elimination product.

Sometimes it is argued that base size is responsible for elimination patterns and that when a large base is used the proton on the least hindered carbon is abstracted.  This is true but the effect is relatively small.  Changing the base from methoxide to t-butoxide only increases the amount of Hoffmann product by a modest 10-20% in simple haloalkanes.

So while sterics play a role, it is usually a small role with the electronic factors discussed above being more important.

For completeness, let me briefly mention the E1 elimination which is on the "other side" of the elimination spectrum of mechanisms. In this case, breaking of the $\ce{C-LeavingGroup}$ bond is advanced and partial positive charge develops on the carbon where the LeavingGroup is departing from.

Elimination of a proton from this intermediate can often proceed in two directions.  Since more highly substituted double bonds are more stable than less substituted double bonds and since the transition state for proton elimination involves partial double bond character, the transition state  leading to the more stable double bond (e.g. the more highly substituted double bond) will be lower in energy.  Hence, the product with the more highly substituted double bond (Zaitsev product) will form in E1 reactions.

Edit: Response to OP's comment

  I meant the pyrolytic elimination of quaternary ammonium hydroxide,
  which follows E2 mechanism

I wasn't saying that this reaction (the thermal Hofmann elimination) goes through an E1cb mechanism, rather I said "we shift the mechanism towards the E1cb side of the elimination spectrum."  In other words, it is an E2 reaction but $\ce{C-H}$ bond breaking is somewhat more advanced than $\ce{C-LeavingGroup}$ bond breaking. Therefore the primary $\ce{C-H}$ bond is more likely to be involved than the secondary or tertiary $\ce{C-H}$ bond, for the reasons outlined above.

Here's another approach.  Instead of electronic factors, let's look at steric factors - in this reaction they both lead to the same conclusion.

E2 eliminations prefer a trans or antiperiplanar geometry.  Below are Newman projections of the 2 conformations with this geometrical arrangement of the substituents.

Conformation A is set up to eliminate towards the least substituted carbon, the methyl group (Hofmann), while conformation B depicts elimination towards the more highly substituted carbon (Zaitsev).  The trialkyammonium group is extremely large and generates an unfavorcbale steric interaction in conformation B.  This makes conformation B higher in energy, hence elimination from conformation A to produce the less-substituted double bond will be preferred.


Why is difluoride a stronger oxidizing agent than dichloride?

If you think about it, strong oxidizing agents must have a high "desire" or propensity to be reduced themselves (as the nature of redox reactions is such). We'll build up on the explanation from the ground up.

Comparing both fluorine and chlorine gas, both are halogens and both only need two electrons each to be reduced to fluoride and chloride ions respectively:

$$\ce{X2 + 2e- <=> 2 X-}$$

Since it appears that the two are similar in most respects, we should look at the clearest differences between Fluorine and Chlorine instead: the number of protons and the number of electrons.

Fluorine is a row 2 element with 9 protons an electronic configuration 1s2 2s2 2p5, whereas chlorine is a row 3 element with 17 protons and electronic configuration 1s2 2s2 2p6 3s2 3p5. Although chlorine has more protons to attract electrons (and therefore be reduced / act as an oxidizing agent), it also has two inner shells worth of electrons that create a shielding effect to reduce the attraction. Fluorine, however, only has one inner shell to do this.

From a simplistic physics point of view, electronic attraction is governed by the charge and the inverse square of the distance. You can approximate that the shielding effect will make the effective charge of the atomic nucleus the same in both halogens, so distance becomes the key deciding factor. Since chlorine is larger, and therefore the attracted electron further than it would be if it was instead bound to fluorine, the attraction (and therefore oxidizing power) is greater in fluorine than in chlorine.

However, looking at the electron affinity (EA) values you'd find that fluorine doesn't follow the group trend of decreasing EA and there's a very good discussion on that over at ChemLibre. Long story short, it's because fluorine is so small that you have to account for the electronic repulsion as well which reduces the bond strength of fluorine vs chlorine (36.6 vs 58 kcal/mol)

There is another measurement which seems to be consistent with this whole fluorine being more oxidizing than chlorine though - and that is reduction potential. The reason that there is a difference between the values for reduction potential and electron affinity even though they are both supposed to be measuring the same thing (that is the energy change related to accepting an electron), the key is the environment in which they are measured. Electron affinities are measured in the gas phase where it is purely the energy related to the electron transfer whereas reduction potentials are measured in aqueous solution and so factors such as hydration energy start to play a role.

As your textbook correctly points out, the high hydration energy is due to fluorine's higher charge density compared to that of chlorine.


Experiment using hydroxides of alkaline earth metals

I would attribute this two 2 reasons:

Reason 1: Insolubility of Magnesium hydroxide.

Magnesium hydroxide is almost insoluble in water and will not easily dissolve in water.

Reason 2: Safety is one of the reason of the choice of compounds.

The formation of hydroxides of alkaline earth metals are generally exothermic and this intensity increases down this group.

Reaction of magnesium oxide with water will produce the
corresponding hydroxide:

$$\ce{MgO + H2O → Mg(OH)2  (very slight reaction)}$$

(The reaction quickly stops because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction).
The reaction of calcium oxide to produce its hydroxide is
exothermic:

$$\ce{CaO (s) + H2O (l) ⇌ Ca(OH)2 (aq) (ΔH_r = −63.7 kJ/mol of CaO)}$$

Barium oxide is  harmful to human skin and may cause irritation on exposure for prolonged periods or in large quantity.

Its reaction with water is also know to be violent.

So in my opinion it is much better to user hydroxides of Ba and Ca.


Synthesis of Pseudopelletierine, Mannich reaction base catalysed

You are very close. The only issue I see is that the Mannich reaction takes place between an enol(ate) and an imine (iminium). Your mechanism goes to the hemiaminal, and then the enolate must displace a hydroxide leaving group. Essentially that's an SN2 with a hydroxide as a leaving group, which does not occur. Instead, from the hemiaminal, the nitrogen kicks out the hydroxide to form an iminium, which is a much more reactive electrophile. The enolate attacks readily. Repeat that process, decarboxylate & tautomerize twice, and you're there.


Determining solubility of silver sulfate in its saturated solution

Solubility constants
You correctly state that they are defined as:
$$K_\mathrm{sp} = \frac{[\text{products}]}{[\text{rectants}]}$$
And in your case ($\ce{Ag2SO4}$) this would be:
$$K_\mathrm{sp} = \frac{[\ce{Ag+}]^2 \cdot [\ce{SO4^2-}]}{[\ce{Ag2SO4}]}$$
And concerning the reactand, the ‘concentration’ of a solid is always defined as $1$, so it is removed from the equation (technically this is an activity not a concentration but that doesn’t matter in this case).
The problem itself
What do you have? Let’s take another look at it:

Calculate the molar sulfate concentration in a solution that is $0.200\,\mathrm{M}$ in:
a) $\ce{Ag+}$ and saturated with $\ce{Ag2SO4}$

So we have dissolved $\ce{Ag+}$ with a concentration of $[\ce{Ag+}] = 0.200\,\mathrm{M}$ in aquaeous solution and also a solid $\ce{Ag2SO4}$ precipitate. You also state you are given the value of $K_\mathrm{sp} = 1.6 \times 10^{-5}$. And you are supposed to calculate $[\ce{SO4^2-}]$.
You have all the information you need and all you need to do is solve a single equation.


Control the pressure of Hydrogen in a small container without the use of a pressure regulator

More like a suggestion than a solution.

First, understand what you are trying to achieve.
You are generating hydrogen at some rate and storing it at high pressure. From your previous question I understand that the pressure of your stored $H_{2}$ is only a very general approximation.

Your objective is now to extract $H_{2}$ in a controlled way without using regulator. This sounds absurd.

My complaints:

I am very sure that you do not have a pressure gauge to measure the pressure of your $H_{2}$ container. This is a serious design flaw.
When you will try to extract $H_{2}$ the flow rate will be really high.

Now to the possible solution:

Incorporate atleast one dual valve regulator. This will take care of both issues mentioned above. You will be able to measure container pressure, and  regulate flow-rate.

(But you do not want a such a regulator ! okay. What you can do is to get the smallest one possible and connect the regulator using 1/8 inch or 1/4 inch PTFE tubes and keep it separate. The regulator will then need to have custom input connector. It is not difficult to design one. Regulator to tube can be connected using all metal nut-ferrule connection which are safe upto 2000 psi or more. )

Well, your safety should be prime concern than the size of device I guess. In case oxygen leaks in and temperature at some local point is enough there is a chance of SERIOUS explosion amplified by the high pressure.

Some things about which I am concerned:

Do you evacuate the container before filling $H_{2}$ in ? If not, you are making a mixture of $H_{2}$ and $O_{2}$. Be careful !
The container of $H_{2}$ can handle high pressure or not. Have you tested it ? You can test it with $N_{2}$ or just air before using $H_{2}$.


When can I use concentrations instead of activities in equilibrium calculations?

To calculate any equilibrum constant we need to be sure of the fugacity of each species. Note that fugacities are defined for every common state of matter (gas, liquid and solid), although they are most commonly used for gases. Look at the end of this answer for more details.

The activity of the coumpound $i$ written $a_i$ is defined as:

$$a_i=\exp\left(\frac{\mu_i -\mu_i^{\circ}}{RT}\right)$$

where $\mu_i$ is the chemical potential, which depends on the fugacity.

However this expression can be hard to use if you have not a strong idea of what you are doing because the are a lot of different ways to calculate things, depending on the formulas and laws or theories used.

What you must remember all the time

Activity can also be defined as $$a_i=\gamma_i \cdot \frac{X_i}{X_i^\circ}$$

where $X_i$ can be a concentration, a partial pressure, a molar fraction, a molality, and so on, and where $\gamma_i$ is the activity coefficient. Be careful; this coefficient depends on what $X_i$ is.

The role of $\gamma_i$ is to correct for the interactions between each constituent of the mixture.

If you are in the ideal regime, which means that the amount of your solute is almost negligible with respect to the entire mixture, then $\gamma_i \approx 1$. Only then can you use $X_i$ in place of $a_i$.

In general, the determination of this coefficient is not quite simple, especially for ions: see the example here for hydrogen ion.

Ionic solutions

Electrolyte solutions (e.g. a solution of $\ce{NaCl}$) can be modelled by Debye–Hückel theory. There are different versions of this model depending on the precision you want. One of the most used is given in the paper about hydrogen ion:

$$\log(\gamma_i)=-\frac{A_m z_i^2\sqrt{I_m}}{1+\sqrt{I_m}}$$

where $I_m$ is the molar ionic strength, given by:

$$I_m=\frac{1}{2}\sum_i z_i^2m_i$$

where $m_i$ is the molality of $i$.

To answer the other part of your question, whether your system is behaving "ideally" or not ultimately depends on the precision you want.

For example if you have a reaction constant expressed as:

$$K=\frac{a_\ce{A}^2 a_\ce{B}^3}{a_\ce{C}}=\frac{\gamma_\ce{A}^2 \gamma_\ce{B}^3}{\gamma_\ce{C}} \cdot \frac{(c_\ce{A}/c^\circ)^2 (c_\ce{B}/c^\circ)^3}{c_\ce{C}/c^\circ}=\frac{\gamma_\ce{A}^2 \gamma_\ce{B}^3}{\gamma_\ce{C}} K^\circ$$

which corresponds to the reaction $\ce{C -> 2A + 3B}$, you will have a large error if you assume you are under ideal conditions when you aren't.

For example if you assume ideality in calculating the $\mathrm{pH}$ of $\pu{5 mol/L}$ of $\ce{HCl}$ solution, like I did here, the answer will not be the same as if you were to consider the activity of the ion, like this.

So at least for strong acid, if the concentration is much than $\pu{1 mol/L}$, you must use activity coefficients. Depending on your problem you'll need to think by yourself and do calculations with and without assuming ideality. If you are unsure as to whether your solution can be considered as being ideal, it's good to do a check (and feel free to use a computer in hard cases).

I would add, as an example, if one of your compounds is 10% of the solution in mass, the solution must no longer be considered as ideal. But, as said earlier, the limit of what is ideal is fixed by you, depending on the precision you want.


Chemical hardener to cast breastmilk into gemstones

Referring to real milk (bovine, human breast milk; but not coconut milk) contains casein which may be polymerized.  You may end up with edible cheese, or green adhesives, or gemstone-like galalith, for example.

Frankly, I have no hands-on experience to experiment with breast milk.  And I recommend to refrain from experimenting with formaline at home.  However, the procedure "Sculpted Science: Turn Milk into Plastic!" in the open access section Bring Science to Home by the Scientific American, which triggers the process just by vinegar and gentle heat, sounds much safer and may be worth to follow.


Why does no Robinson Annulation occur after the Michael addition?

As illustrated in the following figure, the Robinson annulation reaction involves two steps.  The first step involves the Michael addition of a ketone enolate (the enolate derived from cyclohexanone in the example pictured below) to an $\alpha,\beta$-unsaturated ketone (methyl vinyl ketone below).  In a second step the methyl group adjacent to the carbonyl is deprotonated and undergoes an intramolecular aldol condensation to form the final product. This last step is very favorable since the reaction occurs intramolecularly and involves a favorable 6-membered transition state to form a stable 6-membered ring.

Often in Robinson annulations methyl vinyl ketone is used. It is important that  the $\alpha,\beta$-unsaturated ketone have an alkyl group attached to the carbonyl so that the enolate required in the second step can be formed.

  Explain why no Robinson annulation cyclic product is observed?

In your example, the phenyl vinyl ketone has no alkyl group attached to the carbonyl carbon - only a phenyl group. Hence, the enolizable hydrogens present in your Product C can only react with the other carbonyl through a highly-strained 4-membered transition state preventing any ring formation through an intramolecular aldol from taking place.


Why is the boiling point of polonium less than that of tellurium?

It's mainly about relativistic effects. Relativistic affects mainly affect heavy nuclei, such as polonium etc, so makes their properties slightly different to the trend.[ref] The outer electron configuration of Polonium could differ so much from its lighter family member tellurium due to the electron constriction, causing an outer electron to fall into a lower shell, that the intramolecular forces are weaker. Thus a lower boiling point.

It explains a lot of things that do not seem realistic with very heavy nuclei. Such as the "golden" colour of Gold. It should be silvery, but due to relativistic effects it is a "golden" colour. It's also like Copernicium behaving more like a noble metal, unlike the other metals of group 12, even being gaseous in room temperature.

Reference: Wikipedia article on relativistic quantum chemistry


Which has the largest bond angle between water, oxygen difluoride and dichlorine oxide?

There is practically only one effect that determines the angle, but there is an underlying mechanism that you also should understand.

Ideally, a central atom such as oxygen would want to use its p-orbitals exclusively for bonding and have the s-orbital electrons sitting comfortably as a lone pair close to its nucleus. However, that requires a $90^\circ$ angle which can create rather strained molecules — remember that each atom is surrounded by an electron cloud with negative charge that repulses other atoms’ electron clouds. There are now two possibilities of how to relieve this steric strain:

enlargen the bond lengths, so that different substituents are further away from each other while keeping a bond angle as perfect as possible;
enlargen the bond angles so the substituents move further apart from each other while retaining a close distance to their central atom with a more perfect bond length.

In practice, almost all systems stabilise themselves by modifying the angles. Increasing bond lengths decreases bond strengths and thus energy gained much more strongly than tinkering around with angles, because the overlap will inevitably get significantly smaller in absolute value number no matter which way you look at it.

An average bond length can be thought of as the distance which is just large enough for those orbitals that interact to do so favourably while minimising nonfavourable interactions. For certain atom pairs, bond length can essentially be considered almost a constant.

All of this taken together means that the very small hydrogen atoms have a much easier time of crowding together closer to the $90^\circ$ angle than the larger fluorine or huge (in this context) chlorine atoms. Even though the latters’ bond lengths are longer in absolute values their larger radii makes the effective bond lengths shorter and thus makes them require more space. Since chlorine is the largest of the lot, the $\ce{OCl2}$ bond angle is largest.