What is the IUPAC name of the given compound?

As for your other question, the simplified criteria for the numbering are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, you have correctly identified the principal characteristic group as ketone. It is expressed as suffix “one”. Thus, a low locant is assigned first to the ketone according to Rule (c), which yields “…1-one”.
Next, a low locant is given to the double bond according to Rule (e), which gives “3-en-1-one”.
Finally, a low locant is assigned to the remaining $\ce{-OH}$ substituent according to Rule (f).
Thus, the complete name is 5-hydroxycyclohex-3-en-1-one.
Rule P-14.4 applies to chains as well as rings.


The barrier to rotation in the allyl radical is low (~15-16 kcal/m) and it can undergo cis-trans isomerization.

As to question (c)(i), you're right, there are only two resonance structures.

What you show above with three resonance structures is the trimethylenemethane diradical, isobutene with 2 allylic hydrogens removed.


Combining Bond Enthalpies and Enthalpies of Formation

You can combine this data. You just need to find a sensible reaction cycle, and you have to make sure the temperatures (usually 298 K) match.

To give a simple example, consider $\ce{CO2(g)}$.

What is the $\ce{CO}$ bond enthalpy $\mathrm{\epsilon_{C=O}}$in $\ce{CO2(g)}$?

It is one half of the enthalpy for the reaction

$$\ce{ CO2(g)->C(g) + 2O(g)}~~~~~\mathrm{\Delta  H^\circ = 2 \epsilon_{C=O}}$$

The enthalpy of formation of $\ce{CO2(g)}$ is equal to the enthalpy of combustion of graphite:

$$\ce{C(s) + O2(g) -> CO2(g)} ~~~~~~\mathrm{\Delta _f H^\circ = -393.51}\pu{kJ mol^{-1}}$$

The heat of formation of $\ce{C(g)}$ is:

$$\ce{C(s) -> C(g) }~~~~~~\mathrm{\Delta _f H^\circ = 716.68}\pu{kJ mol^{-1}}$$

We also need a value of the $\ce{OO}$ bond enthalpy in $\ce{O2}$:

$$\ce{O_2(g) -> 2O(g) }~~~~\mathrm{\Delta  H^\circ = \epsilon_{O=O} =  498.3}\pu{kJ mol^{-1}}$$

Adding things up we obtain $\mathrm{\epsilon_{C=O}}$:

$$\mathrm{\epsilon_{C=O}} = \frac12(393.51+716.68+498.3)\pu{kJ mol^{-1}= 804.2 kJ mol^{-1}}$$


does it make any difference writing NO2 instead of O2N

The Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), include recommendations for contracted labels in structure diagrams. In particular

  GR-2.3.1 General guidelines
  Atom labels representing more than one non-hydrogen atom—also sometimes known as “contracted” labels—rely on the fact that many elements have consistent and well-understood bonding patterns. (…)

  GR-2.3.3 Orientation of symbols within contracted labels
  Contracted atom labels attached to only one bond should be read outwards from that bond, usually from left to right if the bond is on the left of the label. If the bond is instead attached to the right of the label, the label will normally be read from right to left, but ambiguities can result. Accordingly, contracted labels with a bond on the right should be avoided except for simple cases, usually limited to relatively small labels containing four or fewer combined element symbols and abbreviations. (…)

  GR-2.3.4 Interpretation of contracted labels
  In general, contracted labels are interpreted to fill as many valences as possible, as quickly as possible. (…)

Contracted labels for nitro groups cannot be interpreted with a simple application of valence rules, but also need some implicit charges to be added. It is equally possible to avoid entirely the issue of depiction and instead display the functional group using the $\ce{NO2}$ abbreviation. According to Table III in the Graphical Representation Standards, the left-to-right form of the label is $\ce{-NO2}$ and the corresponding right-to-left form is $\ce{O2N-{}}$.


Can an exothermic (endothermic) process increase (decrease) internal energy?

A primary reason for using enthalpy is the fact that heat transferred ($q$) is a fairly straightforward quantity to measure, whereas change in internal energy ($\Delta U$) is often not.

Under conditions where volume is held constant and no form of work other than pressure-volume work is possible, $\Delta U = q$, so we can get direct information about the change in energy from the amount of heat transferred.

But under the much more common laboratory condition of constant pressure, $q \neq \Delta U$. For this case, it is useful to equate $q$ with the change in a state function, so we have defined enthalpy such that $\Delta H = q$ when pressure is constant. Because enthalpy is a state function, we can make general conclusions about other paths between states based on our measurements of $q$, even though $q$ is path-dependent.


What is the seventh plane of symmetry in benzene?

Have a look at this picture which shows some of the symmetry axes and planes. You should be able to see other similar ones. Taken from www.molecule-viewer.com.(You can also use this site to practice working out point groups.)


Did the finding of the electron from cathode rays/ oil drop experiment at least temporarily throw Dalton's Atomic theory "out the window"?

Dalton's theory was about how atoms behaved: the observation of Cathode rays provided part of the explanation of why they behaved like this

The key elements of Dalton's theory were observational. He explained how the idea of atoms could explain chemical reactions and many observational facts about them. This was an important breakthrough and explained far more than previous ideas. But it didn't provide much of an explanation of why they behaved that way.

The observation of cathode rays was one of the first steps in providing the why not a fundamental challenge to Dalton's actual theory. Chemists didn't get a (mostly) complete explanation of what atoms behave the way they do or of some of the characteristic features of atoms until the discover of the neutron in the 1930s. Why, for example, do atomic weights not always follow the atomic numbers (or the pattern of chemical behaviour observed across the periodic table)?

We now understand that chemical properties are determined by the electrons in atoms and the number of electrons depends on the number of nuclear protons. But the mass also includes neutrons which occasionally alters the apparent periodic relationship between atomic mass and chemical properties.

None of this is a fundamental challenge to Dalton's ideas. All of it is a deeper explanation of why the observed properties and behaviours of atoms in Dalton's theory work as they do.

So we didn't throw Dalton's ideas out. We refined them with a more fundamental theory that explained why they behave like that.

This contrasts with other changes in our understanding of chemistry. For example, we discarded the old phlogiston theory of the explanation for combustion because observations didn't refine it but completely refuted it.


What is the mathematical meaning of the plus sign (+) in chemical reaction equations?

I think we are over-interpreting the poor $+$ sign. In older books you will also see an $=$ sign instead of an arrow just like what Lavoisier did. We have to get hold of La Traité Élémentaire de la Chimie (1789).
We should look at the historical perspective and I always want to see the earliest usage in chemistry just to see what was the original person thinking. I think we can credit Lavoisier for the "+" sign in chemical equations and let us see what was going on in his mind. I quote from the History of Chemistry, by F.J. Moore 1910 (Full text in Google Books).

The cumulative effect of these researches led to the sudden conversion of most French chemists about $1785,$ and the new ideas were firmly fixed by the publication of Lavoisier's great text-book La Traité Élémentaire de la Chimie in 1789. The change was well called the Chemical Revolution, for it inverted completely the chemical point of view. The mysterious hypothetical substance, phlogiston, which did not obey the law of gravitation, and changed its properties arbitrarily as theoretical considerations dictated, was banished from the science and the law of conservation of mass vindicated once for all. This made possible quantitative analysis and the chemical equation. The latter is a convenience in form of expression which we owe to Lavoisier, and of whose assistance in stating and solving chemical problems he made immediate use. He writes: $^{1}$
If I distill an unknown salt with vitriolic acid and find nitric acid
in the receiver and vitriolated tartar in the residue, I conclude that
the original salt was nitre, and I reach this conclusion by mentally
writing the following equation based upon the supposition that the
total weight is the same before and after the operation.
"If $x$ is the acid of the unknown salt, and $y$ is the unknown base
$I$ write: $x+y+$ vitriolic acid $=$ nitric acid $+$ vitriolated
tartar $=$ nitric acid + vitriolic acid + fixed alkali. Hence I conclude: $x=$ nitric acid, $y=$ fixed alkali, and the original salt was nitre."
Nothing like this had ever appeared before in chemical literature and
what a fog of mystery and superstition it removed!

Update Surprisingly, I cannot find this quote in Lavoisier' book (can't read French) but + sign is obvious! The only equation in his 382-paged book is
moût de raisin = acide carbonique +  alkool.
grape wort = carbonic acid + alcohol
Anyway, plus sign is quite old. In the book titled Historical studies in the language of chemistry by Crosland, Maurice P, a 1744 text is cited showing

Readers should consult this book (Internet Archive, need to make an account for borrowing it) if the want to dig deeper. The author discusses plus sign at several places at length.


Feasibility of using Magnesium, Beryllium, Lithium as light sources?

Berillium is toxic and quite hard to obtain, so it's off.

Lithium is more common but still quite rare. It still is used in some pyrotechnic compositions, but rarely in form of metal.

Magnesium IS used in some composition that are intended is light sources. At the dawn of photography photoflashes used magnesium-based compositions. But it is rarely used nowdays, both thanks to electrical flashes and because it generates too much UV light, so it may harm eyes.

BTW more recent photoflash compositions use aluminium to avoid avoid the bane of magnesium-based composition: extremely fine smoke.

BTW, to obtain any metal above a lot of electricity is required, and in a setting with such metals Yablochkov candle should be available and it has quite vintage/steampunk look.


Compressibility Factor Graph - Which gas attains a deeper minimum?

There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.

If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,\; P_r=P/P_c,\;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'

$Z$ vs $P_r$ at different reduced temperatures

For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume  and $T_C=8a/(27Rb), \; P_c=a/(27b^2) ,\; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.


Is it possible to have a (free-standing) single atom of carbon?

It's definitely possible, but such an atom would be highly unstable and likely would have a short lifetime. An example of this happening is if you blasted some carbon material with a laser. Here's a really cool paper where researchers blasted carbon nanotubes and graphene with lasers and observed single-atom chains (not an individual atom in this case, but the same principle applies) of carbon exhibiting some very interesting electronic properties:

http://www.pa.msu.edu/cmp/csc/eprint/DT086.pdf

  If not, what is the minimum size of a carbon particle (diamond or
  graphite network)?

Buckyballs (60 carbon atoms) come to mind.

Tangentially related....

A few years ago, a team of Australian scientists fabricated a single-atom transistor (the basic building block of integrated circuits). The atom was Phosphorus. Granted, the atom was not free-standing, but still very impressive. Such a device is extremely difficult to fabricate, and requires cryogenic temperatures, so it's doubtful that this would ever become a viable technology.

http://www.nature.com/nnano/journal/v7/n4/full/nnano.2012.21.html


Is there a liquid crystal or similar material that gets cloudy when an electric field is applied, but is otherwise clear?

Electrochromic materials are produced for this purpose and are typically sold as “smart glass”.


Can I compare DFT calculations with different grids?

Yes, your error analysis is valid for energy differences. However, I believe it is also valid for the absolute error $\epsilon_i$ of any calculated quantity, not just the error $\Delta\epsilon$ of an energy difference.

In any numerical computation, the key thing for situations like this is to ensure that these "structural" sources of error have been reduced to a magnitude that don't affect your results. In this case, this is achieved when you reach a grid quality $Q$ where the numerical results no longer change when you further increase to $Q+\delta Q$.

To be clear: in this case elimination of the imaginary frequencies is not a good criterion for a sufficiently large $Q$.  The proper stopping point for refinement of $Q$ is when the results no longer change appreciably.

Note that, as you allude to implicitly in the question, this situation is different than the one of trying to compare, e.g., absolute energies from computations run with different methods or at different levels of theory.  In this case, there are fundamental/theoretical reasons why those comparisons cannot be meaningfully made.


It does indeeed.
This reference describes the F-C reaction of anisole under AlCl3 or BF3 catalysis in good yields with the expected o/p direction.
J. Am. Chem. Soc. 106(1984) 5284

There are numerous other references

It should be noted that Anisole + AlCl3 is a method for demethylation to the phenol.


Comparing acidic strength in oxan-2-one and oxan-4-one

This has a simpler answer than you expect. First of all, your reasoning is wrong because inductive effect is much less dominant than conjugation of electron pairs. So, if the latter is unequal in both cases, we should first prioritize it over inductive effects.

  Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation

It seems you already know that the beta hydrogen of carbonyls is most acidic. Both molecules have carbonyl groups, and both have beta acidic hydrogens. So, you're on the right track!

Whichever molecule is more acidic, it will have a more stable conjugate base.

Note that in the second anion, the excess negative charge is completely delocalised over to the carbonyl group. Compare that with the first anion, where both the excess negative charge and the oxygen's lone pair are delocalising with the same carbonyl group. Thus, the excess negative charge is unable to completely delocalise with the carbonyl group.

Can you now identify why the first molecule is less acidic?


difference between glyceraldehyde-3-phosphate and 3-phosphoglycerate

The difference between the two molecules is highlighted in red.

The functional group in 3-phosphoglycerate is a carboxylic acid. That in glyceraldehyde-3-phosphate is an aldehyde.

Sugars have the general formula $\ce{C_$n$H_{$2n$}O_$n$}$. To accomplish this specific ratio of carbon to hydrogen to oxygen, the requirement is that one of the carbons must have a carbonyl group (i.e. forms an aldehyde or ketone), whereas the remainder of the carbons have one hydroxyl group each. For example, the open-chain structure of glucose is:

The carbonyl group is in red and the hydroxyl groups are in blue.

Therefore, glyceraldehyde-3-phosphate is simply a phosphorylated sugar: C-1 is the aldehydic carbon and C-2 and C-3 have hydroxyl groups, the latter of which is phosphorylated.

3-Phosphoglycerate is one oxidation state too high.


what is the exact meaning of oxidation state?

Oxidation state is a formal number ascribed to atoms, to tell if it has more or less electrons than in its natural state.

For single atoms it is fairly straightforward, as electrons can only be removed or added in discrete units. Therefore, an atom or ion has an oxidation state that equals it electric charge.

But electrons are not small particles, orbiting the molecule. In many ways an electron is better described as a cloud of negative charge hovering around the molecule and chemical bonds.

When different atoms are connected in a molecule, this cloud is not evenly distributed between the atoms. Example: In Carbonmonoxide (CO) the Oxygen attracts most of the electron-cloud. But not all, if so the molecule would part into ions.

In reality, the oxidation state of the atoms in CO should be calculated with decimals instead of whole numbers, but for practical reasons we ascribe the whole of an electron in a bond to one of the atoms.

In Carbonmonoxide the atoms are bonded by a double-bond (4 electrons) and if all of these “belongs” to oxygen it will have two more than the basic atom and thus the oxidation state is -2.


Yes, this is a subtle thing.

Using the Pauling electronegativities, one would expect CsF to have the larger electronegativity difference (3.2). So in principal, it should be "more ionic."

Unfortunately, an ionic bond requires separating charge, so $\ce{Cs+F-}$. The problem is that $\ce{Cs+}$ is much larger than $\ce{K+}$ and so the dipole moment for a fully ionic $\ce{Cs+F-}$ would be much larger.

We find that the charge isn't fully separated, likely because of the need to stabilize the large dipole moment.

So some people plot a curve of "% ionic character" as determined by the actual dipole moment vs. the expected dipole moment (i.e, full charge separation). We find that this curve reaches a limit ~75-85% ionic character.

IIRC, I think on that basis LiF is more Ionic than KF.

For more, I really like Bruce Robinson's lecture notes:
http://courses.washington.edu/bhrchem/c152/Lec23.pdf via the Internet Archive


How to get the freeze point of mixed liquids?

In general it is extremely hard to get the freezing point of a mixture purely by calculation. The reason is that the specific molecular interactions of the two substances you have mixed cause pretty much every mixture to need a unique formula for the calculation.

That said, there are some ways to get a rough estimate but these estimates mainly work for (linear) hydrocarbons because those are (fairly) simple to generalize. (google for "freezing point hydrocarbon mixture")

In your specific case, because it is ethanol-water your best way to find out the freezing point is to look up the experimental results of someone else. The engineering toolbox has a dataset on ethanol-water which you could use to get a reasonable estimate for your case (they don't list 95v%, but they do have the whole range from 0-100v% in steps of 10v%).


Calculating internal energy change of a reaction

An isolated system is one that can exchange no mass, heat, or work with the surroundings.  So, from the first law, the change in internal energy of an isolated system in any process within the system is $\Delta U=0$.
Now for my analysis of this problem in greater detail to determine the enthalpy change.  I'm taking as the initial state of the system one mole of CO and 1/2 mole of O2 at 298 K and 1 bar in a rigid adiabatic container, and the final state as 1 mole of reaction product CO2 in this same container at whatever temperature and pressure result when the reaction occurs spontaneously in this same adiabatic rigid container.
My game plan is to examine the internal energy change for the process implemented in 3 successive steps, based on Hess' law to determine the overall internal energy change:

Reactants reacting at constant pressure and removing whatever heat is necessary to hold the temperature constant at 298 K.  So the final state in step 1 is one mole of CO2 product at 298 K and 1 bar, at 2/3 the original volume (since 3/2 moles goes to 1 mole).

Expanding the 1 mole of CO2 isothermally from 2/3 the original volume back up to the full original volume, while decreasing the pressure from 1 bar to 2/3 bar.  So the final state of step 2 is 1 mole of CO2 at 298 K and 2/3 bar pressure.

Adding heat to the 1 mole of CO2 at constant volume to increase its temperature to whatever value is necessary for the overall change in the internal energy for combination of the 3 steps to be equal to zero.

Analysis of internal energy change for step 1:
The process described for step 1 corresponds to the conditions for the standard heat of the reaction CO + 1/2 O2--> CO2 at 298 K and 1 bar, $\Delta H^0=-283000\ Joules$.  So the change in internal energy for this step is the standard change in internal energy $$\Delta U^0=\Delta H^0-RT^0\Delta n=\Delta H^0+\frac{RT^0}{2}=-283000-(8.314)(298)(-1/2)=-282000\ Joules$$
Analysis of internal energy change for step 2:
Step 2 involves an isothermal change for an ideal gas (without chemical reaction), so the internal energy change in step 2 for the system in this step is zero.
Analysis off internal energy change for step 3:
Step 3 involves the constant volume heating of one mole of the ideal gas O2 from $T^0=298 K$ to the final temperature T, so the change in internal energy from this step is $$(\Delta U)_3=C_v(T-T^0)$$ where $C_v$ is the molar heat capacity of CO2.
Overall combined internal energy change
It follows from all this that the overall internal energy change for this combined 3 step process is $$\Delta U=\Delta H^0+\frac{RT^0}{2}+C_v(T-T^0)$$But we know that this overall internal energy change must be equal to zero.  This enables us to calculate the temperature rise of the system in the adiabatic rigid container:  $$(T-T^0)=-\frac{\left(\Delta H^0+\frac{RT^0}{2}\right)}{C_v}$$The corresponding enthalpy change for the combined process is $$\Delta H=\Delta H^0+C_p(T-T^0)$$After some algebraic manipulation, this reduces to $$\Delta H=-(\gamma-1)\left(\Delta H^0+\frac{C_pT^0}{2}\right)$$Substituting literature values for the molar heat capacities and heat capacity ratio into this equation then yields the enthalpy change for the overall process:  $$\Delta H=+83000\ Joules$$ This indicates that answers B and D are both correct.
So, for this problem, we find that $$\Delta H>\Delta U=0$$
The main error you made in analyzing this (difficult) problem was failing to account for the heat of reaction (where the internal energy and enthalpy can change at constant temperature).


Is it safe to lyophilize/Freeze-dry a chelated compound?

Lyophilization is regarded safer as normal drying because it is performed at considerably lower temperature; however in order to achieve this, you have to lower the pressure above the condensed phase, too.  Naturally volatile compounds (including some organic compounds solid at room temperature) may sublime under these conditions, too; which may be hold for hours, even days to remove the water gradually.  This simultaneous process may lower the yield of isolated product.

Other light organic molecules (like methanol, ethanol, THF) equally may leave the vessel, too.  The complexation may be affected if leaving molecules were involved as part of the no-further specified chelate, too.


Materials with solid volume greater than liquid volume

There is a bunch of such materials; among the elemental compounds, they include silicon, gallium, germanium, and bismuth. As for the properties... well, it just so happens that their crystal structure contains a little too much of empty space. Crystal packing is a tricky thing, it can't be predicted with certainty even with the most sophisticated methods.


As @Mithoron said, $\ce{NH2+}$ ion is called parent nitrenium ion. The following text is an abstract from a paper1 which discussed its generation and various trapping pathways:

[...] (It) has been generated in various solutions through photolysis of 1-amino-2,4,6-triphenylpyridinium perchlorate. It was
found that the hydrocarbons, toluene and triphenylmethane trap
$\ce{NH2+}$ through two independent pathways. (1) free radical pathway
which leads to bibenzyl (in the case of toluene) or the trityl radical
(in the case of triphenylmethane). (2) hydride transfer
pathway which leads to benzyltoluene (in the case of toluene) or
triphenylmethanol (in the case of triphenylmethane). Results from dilution, triplet sensitization, and EPR experiments lead to the assignment of the hydride transfer pathway  to the singlet state of $\ce{NH2+}$ and the radical pathway to the triplet state of $\ce{NH2+}$ [...]
$$\ce{PhCH2+ + NH3 <-[PhCH3] ^{1}NH2+ -> ^{3}NH2+ ->[PhCH3] NH3+^{.} + PhCH2. }$$

There is also another paper2 which discuss about the reaction of $\ce{NH+, NH2+}$, and $\ce{NH3+}$ ions with $\ce{H2}$ at low temperatures which leads to formation of ammonium ions (the general pathway of $\ce{NH4+}$ generation in interstellar environment):
$$\ce{N+ ->[H2] NH+ ->[H2] NH2+ ->[H2] NH3+ ->[H2] NH4+}$$
This reaction scheme is very much temperature dependent as it can lead to different pathways. For instance, reaction between $\ce{NH+}$ and $\ce{H2}$ has two pathways:
$$
\begin{align}
\ce{NH+ + H2 &->[k^{a}_{NH^{+}}] NH2+ + H} &\quad \Delta H^{\circ} = \pu{-1.9 eV}  & \tag{1}\\
\ce{NH+ + H2 &->[k^{b}_{NH^{+}}] H3+ + N} &\quad \Delta H^{\circ} = \pu{-0.8 eV} & \tag{2}\\
\end{align}
$$
The former pathway occurs at 300 K while the latter one occurs at 15 K.
Reference

On the Solution Chemistry of Parent Nitrenium Ion NH2+:  The Role of
the Singlet and Triplet States in Its Reactions with Water,
Methanol, and Hydrocarbons Sanjay Srivastava, Michael Kercher, and
Daniel E. Falvey The Journal of Organic Chemistry 1999 64 (16),
5853-5857 DOI: 10.1021/jo990324+
Reaction of NH+, NH2+, and NH3+ ions with H2 at low temperatures
The pathway to ammonia production in the interstellar medium
S. Rednyk, Š. Roučka, A. Kovalenko, T. D. Tran, P. Dohnal, R. Plašil and J. Glosík, A&A
Volume 625, 2019, DOI: 10.1051/0004-6361/201834149


reaction of diketone with ammonium carbonate at 100-115 degree celsius

1,4-diketones are likely to give pyrroles in a Paal-Knorr reaction.


Why isn't there a third buffering zone in the pH curve of this titration?

According to this, we have three pKas, as you observed: 2.12, 7.21 and 12.67 (at 25°C).
Your values may be different due to temperature.

The third value (12.67) can't show up in your graph because it doesn't go beyond pH 12.


How does the composition of a mixed solvent affect solubility?

In general, solubility is determined by the intermolecular interaction between a solvent and a solute. Electrostatic forces, hydrogen bonding, electron-pair donor/electron-pair acceptor interactions and solvophobic interactions are examples. The sum of the interaction forces can be related to the polarity of the molecule.

The dissolution of a substance requires that the interaction energy of the solute molecules and the interaction energy of the solvent molecules be overcome. In other words, for the solvation of A in a solvent B, we expect a stronger interaction A-B than A-A or B-B. Considering this, you will have the trend of solubility of polar and non-polar solutes in solvents: the old "like dissolves like" rule.

Now, if you have a solvent mixture, the interaction between unlike solvent molecules is important and leads to deviations from the ideal behaviour from Raoult's law, for example the water/alcohol mixture. Solvent mixtures can exhibit different physical properties compared with their pure components. Coming back to intermolecular forces, it has been found that the ratio of the solvent in the solvent shell of solvation can be different from that in the bulk solution, leading to a preferential solvation. This is caused by the variation of free energy of solvation of the solute in different solvents.

The acid-base equilibria is affected by a change in the 1) acidity or basicity of the solvent, 2) the relative permittivity of the solvent, $\epsilon_{r}$, and 3) its ability to solvate the species in the equilibria. There exists some formulas that relates the equilibrium constant and the relative permittivity (Born, Onsager, etc). In the case of the acetic acid, for example, the relative permittivity will have a strong influence in the solvation of the anionic and cationic species of the equilibria
$$\text{CH}_{3}\text{CO}_{2}\text{H}\rightleftarrows\text{CH}_{3}\text{CO}_{2}^{-}+\text{H}^{+}$$
since the two charges attract each other. The good solvation of the ions in a high relative permittivity solvent shift to the right the ionization. Consider that for water ($\epsilon_{r}\approx80$) and ethanol ($\epsilon_{r}\approx33$) the miscibility is similar, but also have distinct pKa's. Therefore, taking into account only the electrical interaction, between solute and solvent, is a poor approximation. The other problem is that it is practically impossible to find solvents that differ only in their relative permittivities and not in their acidity or basicity.

So, for instance, data on solubility in mixed solvents may be very difficult to obtain, but qualitatively the whole thing works in the form of intermolecular interactions.


Can Iron(III) oxide oxidize elemental iron in a nitrogen environment under room temperature without any catalysis?

Whether this qualifies as "without catalysis" is a matter of perspective, but it certainly does happen under ambient conditions.  One mechanism of atmospheric rusting involves the $\ce{Fe(III)}$ bearing rust reacting with fresh iron to form $\ce{Fe3O4}$, which then oxidizes in air to form more rust.  See Section 1.6 here.

We can capture the $\ce{Fe3O4}$ stage by merely placing the rusty material in water with a little electrolyte, such as rinse water after an acid pickling process, where air cannot easily penetrate.  The rust, being converted to $\ce{Fe3O4}$, turns black.  This emphasizes the fact that the iron reacts with the $\ce{Fe(III)}$ in the rust without oxygen, once the rust has first formed.

I've actually seen the $\ce{Fe3O4}$ formation happen on steel products during rinsing after pickling, giving black stains or stains bearing a black component.  To prevent it, we had to design and maintain the rinsing (and drying) processes to prevent the rusting.


There is a "readme" note from the author that says:

  The answers are not guaranteed, in the sense that not only
  might I make a mistake here and there, but I might have used different sources of
  data, or the data in a database may have changed. I have not been able to check all
  the answers, or to make them all consistent with the data in Appendix B, although I
  believe that most are.

So I think that this problem is an example of one of those situations.

The book itself works out the problem on page 267 and gets the same answer as you: −12306 J/mol.


Is the heat capacity infinite for a process at constant temperature?

If you assume ideal behaviour and a closed system, then since the ideal gas law holds, with pressure and temperature remaining constant volume must also be constant.So that means that nothing is happening to the system, everything is as it is with $dq=0$ and $dw=0$.Since there is no process involved there is no meaning for the term heat capacity.


Why is the combination of hydrated aluminium oxide and magnesium hydroxide used to treat heartburn?

Hydrated $\ce{Mg(OH)2}$, aka milk of magnesia, is an osmotic (saline) laxative, as well as antacid.

$\ce{Al(OH)3}$ is also an antacid, but has the opposite effect on the bowels: it's costive, i.e. a binding agent. Aluminum has also been linked (with much uncertainty!) with Alzheimer's disease, so it has become less popular as an antacid.

The combined effects of $\ce{Mg(OH)2}$ as laxative and $\ce{Al(OH)3}$ as a costive might tend to counteract each other, so as to avoid interfering with colonic function.

However, it would be far simpler to use $\ce{CaCO3}$ as an antacid, unless the release of $\ce{CO2}$ bubbles or excess calcium in the diet is an issue.


Excitation of electron in Niels Bohr's atomic model

Quantum mechanics received its name from the concept of the "quantum" which is Latin for "how much" or "amount" and is used to mean "small discrete piece or fragment". Classical physics prior to the development of QM described the energy of bodies as a continuum of possibilities. The novelty introduced by QM was the idea that within the continuum only a subset of the possible energies are in fact allowed, corresponding to allowed states. This means by extension that the energies of a body in different possible states differ by discrete rather than continuous amounts. This discreteness becomes increasingly evident for small and confined systems, where the wave-like properties of matter become more evident, as demonstrated for instance by the self-interference of the electron wavefunction in the double-slit scattering experiment.
Electrons in atoms fulfill all of the conditions for observation of QM phenomena. The particles involved are subatomic (very small) and are confined (by a Coulombic potential). The result is that electrons in atoms exist in a discrete number of allowed states, each state defined by a fixed energy and wavefunction that can be interpreted as the probability distribution of the electron.
When an atom is excited by a photon (a discrete light particle with a fixed amount of energy) it absorbs the entire photon. All of the photon's energy is acquired by the atom. However, such an excitation implies a transition between possible states of the atom. It is necessary that the energy of the photon match the difference in energy of the atomic states. In the case where the two states are bound (if an electron is not excited into the continuum of ionized free states) the result is that only a subset of photon energies can be absorbed by the atom.

As noted by Shoubhik R Maiti, photons can also be scattered by atoms and in the process transfer only part of their energy to the atoms. However the principle and meaning of quantization of electronic states remains the same. Electronic states are quantized (discrete) and so only particular amounts of energy can be absorbed (quanta corresponding to differences in energies of discrete levels).


What should be the reaction between lead oxide and dilute sulfuric acid?

YOur logic is wrong.

  I know for a fact that dilute $\ce{H2SO4}$ is a non-oxidizing acid...

sort of, but so far so good.

  ...which makes it a reducing agent.

This is where you are wrong, just because it is non-oxidizing does not make it reducing. Concentrated $\ce{H2SO4}$ is an oxidizer, diluting it does not make it less of an oxidizer, but limits the strength/mechanism.

A similar example is mixing nitroglycerin and diatomaceous earth doesn't make nitroglycerin less explosive, it just makes it harder for it to explode (dynamite).


I think phosphate buffer is the solution here. Reason:

$K_\mathrm{sp}$ of $\ce{CrPO4}$ at $\pu{25 ^\circ C}$ is $2.4 \times 10^{-23}$ while that of $\ce{Cu3(PO4)2}$ at $\pu{25 ^\circ C}$ is $1.4 \times 10^{-37}$, thus both ions will precipitate since concentration of each ion is $\pu{0.02 M}$.


Typically smoke point has to do with the composition and amount of free fatty acids in the fat/oil. Most fats and oils that you see on the market are mostly composed of triacylglycerols, with only small amounts of free fatty acids. The more free fatty acids you have the lower the smoke point. Similarly unrefined oils have a low smoke point. If you have fats/oils that contain high levels of unsaturated fatty acids, then that product will also have a lower smoke point than say a fat that contains significant amounts of saturated fat. In regards to the chemical mechanism, upon heating the triacylglcyerol structure gets broken down into free fatty acids and glycerol via a hydrolysis reaction. The free fatty acids volatilize and cause the visible smoke. Unsaturated fatty acid are more sensitive to heat and will be hydrolyzed quicker than saturated fatty acids. For more info read this article: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4424769/


When adding values together, you should add together their absolute uncertainties. For example, if you started with $500\pm5\mathrm{mL}$ and add $100\pm1\mathrm{mL}$, your new volume value is $600\pm6\mathrm{mL}$.

You would add the relative uncertainties if you are multiplying values. For example, if you started with $1\pm.06\text{ moles}$ and want to find the molarity in a solution with a volume of $500\pm5\mathrm{mL}$, your molarity will be $2\pm.14\text{M}$. The uncertainty was obtained by finding the sum of the relative uncertainties and multiplying it by the concentration to obtain an absolute uncertainty, as shown below.  $$(\frac{.06}{1}+\frac{5}{500})(2\text{M})=\frac{7}{100}(2\text{M})=.14\text{M}$$

For more details on uncertainty rules, this page from the University of Victoria gives some of the basics.


Calculating atomic mass of unknown element in given compound (gas) at STP

Assuming it is given that $\ce{X2O5}$ behaves like an ideal gas, the best approach to do this problem is manipulating $pv = nRT$ to get the equation, which fits the given data. Suppose $m$ is the mass of the gas, $M$ is the molar mass of the gas, and $\rho$ is the density of the gas. Then $n =  \frac{m}{M}$:
$$ pv = nRT = \frac{m}{M}RT \ \Rightarrow \ M =  \frac{m}{v} \ \cdot \frac{RT}{p} = \rho \frac{RT}{p} \tag1$$
Now, you can apply $\rho = \pu{5 g L-1}$ at STP to the equation $(1)$ to find $M$. Then apply $2M_\ce{X} + 16 \times 5 = M$ to find $M_\ce{X}$, which is the atomic mass of $\ce{X}$.


Why don't polar and non-polar compounds dissolve each other?

Very simply, you explain the reason for this solubility rule by taking in consideration the energy requirements for the breaking of intermolecular forces between the molecules in the solute and the solvent.

Note: this is only a simplified explanation as it also depends on other factors such as change in entropy

Here is some background information on intermolecular forces. In non-polar substances, there are dispersion forces between each molecule. These dispersion forces are relatively weak and hence only require little energy to break them. In polar substances, there are dipole dipole and hydrogen bonding (depending on the substance) between each molecule. These forces are much stronger than dispersion forces and require more energy to break.

Now lets consider the following cases:

Non-polar Solute and Solvent

For the solute to dissolve, the dispersion forces between the molecules in the solute and solvent need to break. This only requires very little energy. However when the solute dissolves into the solvent, they are able to be make dispersion forces with each other. The making of these forces releases very little energy. So simply put, very little energy is required to break the forces and very little energy released when making the forces. Hence overall everything balances out and the process occurs.

Non-polar Solute and Polar Solvent (and vice versa)

For the solute to dissolve into the solvent, both dispersion forces and dipole dipole forces are broken which require large amount of energy. However the molecules in the solute and solvent are only able to make dispersion forces with each other (as they aren't both polar). This only releases very little energy. Therefore overall, more energy is required than released and hence the process won't happen.

Polar Solute and Solvent

For the solute to dissolve into the solvent, dipole dipole forces are broken which require large amount of energy. However when they do dissolve, the molecules in the solute and solvent are able to form dipole dipole forces which releases large amount of energy. Therefore overall, everything balances out and the process occurs.


NaX (X = Cl, Br) is insoluble in acetone, but is soluble in water. So, if you have water/acetone system, the solubility of NaX is increased comparing to acetone. Solvation effect doesn't have a big role on iodide ions and does not greatly impact it's nucleophilicity. So, the mechanism is still going to follow SN2 path. So, after one halogen is displaced by I-, you have introduced a better leaving group in the substrate. Then, another iodide is going to attack the newly formed R-I, so the result is: 2 x inversion = configuration retention. As no NaX precipitate forms, it is better for I- to attack RI than RX. It turns out that after equilibrium is reached, you end up with racemic mixture, or something close to racemic mixture. Note that free iodide ions keep attacking R-I and that this is a dynamic process which constantly repeats in the solution.


How can we combine all the three gas laws into a single ideal gas equation?

It is indeed the case that the rearrangement of the different gas laws is not very satisfactory as it employs the different variables without any constraints. This issue has been addressed by S. Levine in J. Chem. Educ., 1985, 62 (5), p 399, but in the case you do not have access to this reference, I'll shortly repeat his arguments here.

As $P$, $V$ and $T$ are thermodynamic sate functions, we may write the total differential of $T$ as

$$\mathrm{d}T=\left (\frac{\partial T}{\partial V} \right )_P \mathrm{d}V + \left (\frac{\partial T}{\partial P} \right )_V\mathrm{d}P \equiv X\mathrm{d}V+ Y\mathrm{d}P$$

since this differential must be exact, the following relation must hold

$$\left (\frac{\partial X}{\partial P} \right )_V=\left (\frac{\partial Y}{\partial V} \right )_P$$

We can find simple functions satisfying this condition
$$X=cP\text{ and }Y=cV$$
where $c$ is a constant that we may choose to be $c=1/R$ for later convenience.  Substitution of these functions in the total differential gives
$$\mathrm{d}T=\frac{1}{R}\left (P\mathrm{d}V +  V\mathrm{d}P\right )$$
and integrating on both sides from 0K to T yields
$$T-T_0=\frac{1}{R}\left [(PV)_T-(PV)_{T_0} \right ]$$
and since $T_0=0$ and $PV=0$ at $\pu{0K}$ we arrive at the ideal gas law.


You've correctly identified the chiral center in your molecule (the carbon with the asterisk), that is the first step.  When doing work with chiral centers, it is very important to use solid lines to denote bonds in the plane of the screen, bold wedges to denote bonds coming from the plane of the screen towards the viewer and dashed wedges to denote bonds from the plane of the screen away from the viewer.  Drawing all 4 bonds as simple solid lines, as you did, will often lead to confusion.  Look at the figures below, both enantiomers of 1,1,2-trimethylcyclohexane are shown.  The next step in assigning R/S configuration is to rank the 4 groups around the chiral carbon by molecular weight, with "1" being assigned to the highest molecular weight group, and "4" to the lowest (you also did this correctly).  Now, grab the "4" group in your hand so that groups "1" through "3" form a circular pattern between your eye and your hand.  If moving from 1->2->3 involves clockwise movement, then this is the "R" configuration.  If the movement is counterclockwise, then it is the "S" compound you are observing.

Take a look at this link for a nice picture of what I'm describing.

You cannot determine the R/S configuration of the chiral carbon in the compound you've drawn on paper at the top of your post because it is unclear which group 3 or 4 is projecting towards the viewer and which is projecting away from the viewer.  Finally, note that my "R" drawing matches the "R" drawing at the bottom of your post.


Is volume expansion possible upon dissolution of solute?

Yes. In thermodynamic jargon, the onset of a change in volume under dissolution is known as an excess volume. If the solution takes up less volume than its constituent parts it is said to have a negative excess volume. Otherwise we have a positive excess volume. You might also find the term excess molar volume.
By googling each term we find many results concerning this, which seems to mainly manifest in metal alloys.


This molecule doesn't have any carbon to hydrogen bonds, so it likely wouldn't even be properly named using organic nomenclature. Tetrachloromethane, for example, is correctly named carbon tetrachloride. If you wanted to throw technicalities to the wind, however, along with the fact that this compound would likely be so incredibly unstable that no one would even bother attempting to synthesize and isolate it, it would probably be named tetratrioxidanylmethane.

You might notice the name bears some similarities to tetrachloromethane. That's because it's four of the same group, the trioxidanyl group (from trioxidane), bonded to a single carbon.


Why is periodicity seen in these certain properties?

For 2. I would add, on top of what you already have, that the elements start as metallic elements, hence we have metallic bonding. Across the period, the number of electrons increase, meaning there are more delocalised electrons among the metal cations so that electrostatic attraction increases along the period. For period 2, that's the trend.

For period 3 however, it is a bit more tricky because of the transition metal elements and I think Nicolau explains it better than I could in this answer. But I don't think that you will be required to know those for your level, but it's still interesting for your own knowledge. :)

For 3. I would add that the number of electron shells containing electrons remains constant. If it increased instead, the atomic radius would increase as well.

And you could use 2. and 3. to explain 1:

First, we observe that the atomic mass increases along the period, and that the atomic radius decreases. For giant structures, this generally means that the density increases ($\rho = \dfrac{m}{V}$ and while $m$ increases, $V$ is decreasing slightly, so $\rho$ increases)

When we hit the simple molecular structures, the density sharply falls again, because of the average distance between the atoms of the element has drastically risen, so the volume increases. With $m$ increasing gradually and a sharp rise in $V$, you get a sharp decrease in $\rho$.

These observations change down the groups though. For instance, Astatine is a solid, so you won't expect a huge fall in density across period 6 around group 5. But for the purpose of your syllabus coverage (I'm assuming High School), those should be enough.


What are the physical properties affected in ZnO when there is a sulfur impurity?

Of course, if $\ce{ZnO}$ has a very small $\ce{S}$ impurity level, this could be the reason of the yellow colour of products.
As for the effects of very little sulfur on tons of $\ce{ZnO}$ dust, it all depends on the end use of $\ce{ZnO}$:The applications of zinc oxide powder are numerous. Most applications exploit the reactivity of the oxide as a precursor to other zinc compounds. For material science applications, zinc oxide has high refractive index, high thermal conductivity, binding, antibacterial and UV-protection properties. Consequently, it is added into materials and products including plastics, ceramics, glass, cement, rubber, lubricants, paints, ointments, adhesive, sealants, concrete manufacturing, pigments, foods, batteries, ferrites, fire retardants, etc...

Clearly, if $\ce{ZnO}$ is used in rubber industry, sulfur impurities have no harm effect.
If it is used in electrical applications (as a semiconductor), sulfur will affect the electrical conductivity.
If it is used in coating and paints applications, sulfur will affect the the color ....


Which has the greater N–O bond length, NO₂⁻ or NO₃⁻?

There are two ways to think of it.  First: draw out the (major) contributing Lewis structures for both ions.  Each iin has one pi bond, but it's shared between two linkages in $\ce{NO2^-}$ ion versus three linkages in the $\ce{NO3^-}$ ion.  So the $\ce{NO3^-}$ ion has less pi bonding in each linkage making those bonds weaker overall.

Second: In all of those Lewis structuresthere are nonbinding pairs of electrons on the oxygen atoms, plus a pair on the nitrogen in $\ce{NO2^-}$.  There are more such pairs in $\ce{NO3^-}$ than in $\ce{NO2^-}$.  These pairs tend to repel each other and the bonding pairs, and $\ce{NO3^-}$ requires more room, thus longer bonds, to accommodate the greater number of closely packed nonbonding electron pairs.

These two factors tend to occur together when the molecules have nonnetalluc atoms that fill their votes, as here.


Coordination number of chromium(III) oxalate complex

This seems to be the case of "show, don't tell".

IUPAC's definition of coordination number (C.N.) which applies to inorganic complexes:

  In an inorganic coordination entity, the number of σ-bonds between ligands and the central atom. π-bonds are not considered in determining the coordination number.

Let's have a look at the one of the first determined structures of trioxalatochromate(III) inorganic salts [1]:

Figure 1. Crystal structure of $\ce{[Cr(C2O4)3]^3-}$ anion (ICSD-109607). Color scheme: $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#8A99C7}{\Large\bullet}~\ce{Cr}$.

Here, there are three oxalate $\ce{C2O4^2-}$ ligands arranged in distorted octahedral environment (depicted with semi-transparent coordination polyhedron).
Each oxalate is a bidentate chelating ligand, resulting in C.N. 6 of chromium center.

References

van Niekerk, J. N.; Schoening, F. R. L. The Crystal Structure of Ammonium Trioxalatochromate(III), $\ce{(NH4)3[Cr(C2O4)3].2H2O}.$ Acta Cryst 1952, 5 (4), 499–505. DOI: 10.1107/S0365110X52001428.


Finding OH- concentration given the H+ concentration - diverse methods

It is because you assumed the number 14 as a constant while it is actually $-\log(K_\mathrm w)$.
It is $-\log(1 \times 10^{-14}) = 14$ at 25 degrees Celsius but changed to $-\log(9.61 \times10^{-14}) = 13.02$ at 60 degrees Celsius.


Is there any actually observed excited state of a "hypervalent" main-group molecule that actually involves d-orbitals?

One candidate involves calcium, of all things. This paper by Krieck et al. [1] reports the synthesis of a dimeric calcium(I) complex $\ce{[(thf)3Ca\{μ-C6H3-1,3,5-Ph3\}Ca(thf)3]}$ in which the calcium atoms are bonded to each other and there are two unpaired electrons in antibonding orbitals of the aromatic ring system.

DFT calculations show that the orbitals containing the unpaired electrons have a significant contribution from calcium 3d orbitals which have the same symmetry as the ring antibonding orbitals.  A model of the compound is given below.

References

Krieck, S.; Görls, H.; Yu, L.; Reiher, M.; Westerhausen, M. Stable “Inverse” Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(Thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(Thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(Thf)3]. J. Am. Chem. Soc. 2009, 131 (8), 2977–2985. DOI: 10.1021/ja808524y.


Is the nomenclature "primary", "secondary", "tertiary", and "quaternary" carbon atoms limited to sp3 carbon atoms?

To best of my knowledge, the classification primary to quaternary is used for discussing reactivity of hydrocarbons, i.e. alkanes. There the number of substituents matters, as it governs carbocation stability, steric hindrance, etc...

In all other cases, the electronic structure of the molecule rules the reactivity. As you mentioned alkenes, alkynes or aromates, there it makes no sense to discuss such subtle effect as number of neighbouring carbons, when you have $\pi$-system at hand.

The IUPAC Gold Book entry for alkyl groups supports this distinction:

  The groups $\ce{RCH2}$, $\ce{R2CH}\ (\ce{R} ≠ \ce{H})$, and $\ce{R3C}\ (\ce{R} ≠ \ce{H})$ are primary, secondary and tertiary alkyl groups, respectively.


Do enolates get protonated at the carbon or oxygen atom?

Like all ions in solution, the enolate does not exist on its own in free form. It will be solvated with maximum efficiency. Depending on the solvent itself, the counter ion might be further or closer away, but charge balance needs to be maintained in the bigger picture.

Contrary to popular belief, the enolate ion is nothing that truly exists in non-gaseous phase. Therefore the question what gets protonated first turns into the question: Does the oxgen or the carbon have the stronger interaction with the solvent? This is quite hard to judge and very dependent on the whole system. there is certainly no general rule for it and as previously stated by bon, it does not really matter, because keto-enol tautomerism is fast and the ratio is entirely determined by the relative stabilities to the involved species.

There is another issue when talking about protonation. The proton itself is not free, either. In an aqueous solution the smallest model would be the hydronium ion. But this is far from being complete either. A whole branch of computational and theoretical chemists work on describing water clusters with models. It is extensive. One of the reasons for it being hard to comprehend is the high flexibility and mobility of protons even when bound to other species. The high conductivity of an acidic or basic solution is mainly governed by the Grotthuss mechanism. This kind of mechanism can in first approximation also be applied to keto-enol tautomerism.

But before we dig deeper into any mechanism, I would like to look at the properties bon outlined for the enolate. The following results were obtained at the DF-BP86/def2-SVP level of theory for the deprotonated butan-2-one resonance hybrid.
Let's look at the obvious first: resonance. With the help of the natural bond orbital theory (NBO) I performed the analysis. The first and most obvious thing is that it finds a so called hyper-bond between the oxygen and the two carbons with an occupancy of 3.95 electrons. As we will see in the refined analysis, this is in favour of the carbon-carbon bond (59.5/40.5 considering only these three atoms).
A more refined analysis, I chose a cut-off of 5%, include secondary resonance structures, seven in total, gives us the following weights for the major contributors:

From these structures we can support the highlighted arguments in the statement by bon:

  Attack on an enolate by an electrophile (in this case just a proton) is governed by the balance between electrostatic interactions, which are strongest at the oxygen atom due to its greater negative charge, and orbital interactions, which tend to be stronger at the carbon atom because the HOMO of the enolate has a greatest contribution at this atom [carbon].

The predominant contributor is is the enolate, hence in the overall description also its negative charge should be predominant. The π-system is certainly part of the HOMO in any case. Since it is mainly localised over the carbons, the highest contribution should be there.

We can further support that by looking at the natural charges of this model. More prominent is it looking at the charges obtained in terms of the quantum theory of atoms (QTAIM) in parenthesis.

The second part can be cleared up by looking at the HOMO; on this level of theory, the major contribution is C3-p of 56% and C2-p of 32%. The contribution of O1-p is below the threshold of printing, hence it is <10%. The major contribution of O1-p is found in HOMO -2 with 44% and a contribution of C2-p of 35%. It certainly looks a bit different in the picture. The contour value is set to 0.04.

All these calculations are only evaluations of the ion in the gas phase, so they can serve as an example to understand the true nature a bit more. It is in no way intended to be complete, it shall only provide a more concise educated guess of the general nature.

Now that we have a lot of information on the enolate itself, it's time to look towards a mechanism of a solvent-mediated keto-enol tautomerism. I have chosen the most simplest model possible, with one water (hydronium) explicitly included. The result is a six-membered cyclic transition state, where a proton is transferred from the oxygen of the enol to the oxygen of water, and one proton is transferred from water to the carbon of the double bond.
In theory there might well be many more water molecules involved. In principle any protic solvent is capable of doing this. The principle reaction path is the same for a base mediated mechanism, for example with ammonia.

If you look closely enough, I have borrowed heavily from the Grotthuss mechanism. A somewhat simultaneous transfer of protons. It needs to be acknowledged, that this is only an approximation of a stationary system. The animation only shows the transitional vibration mode (imaginary mode). It is not the calculated structural change or reaction path.
Unfortunately the calculations even on this low level of theory take quite some time. Currently I have no barrier height to report as an estimate, how fast this reaction would be approximately. I'll update this as soon as I can.

There is of course the possibility of a non-mediated transition state. This one, however, proceeds through a four-membered cyclic transition state. This one is highly strained, if you like the concept of ring strain, and seems unlikely compared to the above in any other medium than the gas phase.
If you are repulsed by the concept of ring strain, then you might want to look at it from a different angle. The deformation energy is high, since the carbon-carbon-oxygen is forced into a near 90° angle, hence increasing the p-content of the bonding orbitals. Higher p-orbital content means higher energy of the molecule. Therefore the activation energy will also be higher.

For a deeper insight into this, I recommend “Ab Initio Molecular Dynamics Study of the Keto–Enol Tautomerism of Acetone in Solution” (Clotilde S. Cucinotta, Alice Ruini, Alessandra Catellani, András Stirling, Chem. Phys. Chem. 2006, 7 (6), 1229-1234), which is available at researchgate.net.


ChemSpider versus SciFinder for finding chemicals without CAS number

SciFinder is a product of Chemical Abstracts Service (CAS), which is a division of the American Chemical Society. Its registry contains information on more than 111 million organic and inorganic chemical substances, and is updated with 15,000 new substances daily. Additionally, it has more than 66 million protein and nucleic acid sequences.[1] The chemical information is produced by CAS, and is prepared by the CAS Registry System, which identifies each compound with a specific CAS registry number, index name, and graphic representation of its chemical structure.[2]

ChemSpider is a free chemical database from the Royal Society of Chemistry. Its registry contains information on 49 million chemical structures, coming from 513 sources. You can search the database by systematic names, synonyms, trade names, database identifiers, and also using structure-based queries, with the option to draw structures in the web page or use structure files from your computer.[3]

Both registries allow you to find literature references, experimental and predicted property data (boiling and melting points, etc.), commercial availability, and spectra,[3] but the CAS registry also offers preparative methods and regulatory information from international sources.[1]

SciFinder may offer more, but it's extremely expensive. In fact, I'm not even sure if CAS offers SciFinder to individuals. You can see their pricing options here. Taking into consideration everything ChemSpider can do and the fact that it's free, it's an excellent resource, and I definitely recommend it.

References

CAS, Chemical Substances
Wikipedia, CAS Registry
ChemSpider


Balancing the equation for the preparation of dichromate from chromite ore

As of someone mentioned here, balancing most chemical equations is a a case of trial and error. Your equation is easy to balance by this way than go to extremes.

1) A relatively complicated formula within this equation is $\ce{FeCr2O4}$ on the reactants' side (LHS) of the equation. Since $\ce{Fe}$ atom in it is converted to the resultant product, $\ce{Fe2O3}$ and no other $\ce{Fe}$ bearing products on products' side (RHS), you should start with putting coefficient $2$ in front of $\ce{FeCr2O4}$ to balance $\ce{Fe}$ in LHS.

2) It would make four chromium atoms in LHS. Thus, you need to balance that by putting coefficient $4$ in front of $\ce{Na2CrO4}$ on RHS because it is the only $\ce{Cr}$ bearing molecule among products.

3) That action would make eight sodium atoms in RHS. Thus, you need to balance that extra $\ce{Na}$ by putting a coefficient $4$ in front of $\ce{Na2CO3}$ on LHS because it is the only $\ce{Na}$ bearing molecule among reactants.

3) Now, you would find four carbon atoms in RHS. To balance that extra $\ce{C}$ you may need to put a coefficient $4$ in front of $\ce{CO2}$ in LHS because it is the only $\ce{C}$ bearing molecule among products.

4) Now you have balance all atoms except for oxygen atoms. Thus, count the number of oxygen atoms in RHS and LHS, separately. If you count correctly there are 27 $\ce{O}$ atoms in RHS and only 20 of them in LHS. To balance that extra 7 $\ce{O}$ atoms, you should put a coefficient $\frac{7}{2}$ in front of $\ce{O2}$ in LHS. This would complete the balancing your equation:
$$\ce{2FeCr2O4 + 4Na2CO3 + 7/2 O2 -> 4Na2CrO4 + Fe2O3 + 4CO2}$$

In addition, some people do not like to keep fractions as coefficients in their balance equations. So, you may multify whole equation by 2 to get integers as a coefficients:
$$\ce{4FeCr2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$$


Calculating the biological standard potential

Unfortunately, the OP's Nernst equation is incorrect for several reasons. The balanced net redox equation is

$$\ce{5 Sn^{4+} (aq) + 2 Mn^{2+} (aq) + 8 H2O <=> 5 Sn^{2+} (aq) + 2 MnO4^- (aq) + 16 H+ (aq) \tag 1}$$
and n = 10 electrons for the reaction as written. The reaction quotient, Q, is

$$\ce{Q = \frac{[Sn^{2+}]^5 [MnO4^-]^2 [H+]^16}{[Sn^{4+}]^5 [Mn^{2+}]^2}              \tag 2}$$

and the Nernst equation is

$$\displaystyle{E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln(Q) = E_{cell}^\circ - \frac{RTln(10)}{nF} \log(Q)         \tag 3}    $$

The OP's question states that "all species except $\ce{H+}$ are in their standard states, so their activities are all equal to 1". Hence, in this very contrived example, Q reduces dramatically:

$$\displaystyle{Q = [H^+]^{16}             \tag 4}$$

The OP notes that $\displaystyle{E_{cell}^\circ }$ = 1.36 V (from tables). With T = 298.15 K, $R = \pu{8.31446261815324 J mol-1 K-1}$ and $F = \pu{96485.3321233100184 C mol-1}$, then

$$\ce{\displaystyle{ \frac{RTln(10)}{nF} \approx \frac{0.05916 \ V}{n} }        \tag 5}$$

So the Nernst equation is simply

$$\displaystyle{E_{cell} \approx 1.36 \ V - \frac{0.05916 \ V}{n} \log(Q) = 1.36 \ V - \frac{0.05916 \ V}{10} \log([H^+]^{16})      \tag 6}    $$

Hence, using $\ce{pH = -log[H^+] = +log(1/[H^+])}$, the Nernst equation can be expressed as

$$\displaystyle{E_{cell} \approx 1.36 \ V - 16 \times \frac{0.05916 \ V}{10} \log[H^+] =  1.36 \ V + 16 \times \frac{0.05916 \ V}{10} pH      \tag 7}    $$

If pH = 7, then $\ce{E_{cell} \approx 2.02 \ V}.$

It is better to use expression (3) in order to avoid getting too tricky. This is where the OP went wrong: forgetting the "16" exponent on the hydrogen ion concentration and also missing the fact that the negative sign before the pre-log correction term must be subsumed into the pH definition. In a realistic example, it will not happen that everything is in standard state except the hydrogen ion.

What about dividing the net ionic equation, i.e., equation (1), by 16? Then n = 10/16 = 5/8 and all the exponents in the Q expression also get divided by 16. This is no problem if everything is in standard state except the hydrogen ion. Otherwise, it is just facilitating making a silly mistake.


Ring expansion takes place in the step $\ce{D->E}$.
You were going right till the intermediate. Your intermediate will be attacked by $\ce{Br-}$ anions (as temp. is normal, therefore $\ce{S_N1}$ mechanism). Product will be B.
Then, it will form grignard reagent as C, which will react with aldehyde to produce alcohol D.
After that, it will react with HI to form a carbocation intermediate, which will rearrange to give product E.
The whole mechanism is as follows,


Which resonance structure to use for writing curved arrow mechanisms?

Any resonance structure can be used. Why? Resonance structures represent multiple views of the same compound, so it must be possible to start with any valid resonance structure. Because resonance structures consider different positions of electrons, the sets of arrows needed to describe the mechanisms of different resonance structures will be different.

While it's possible to use any resonance structure in a mechanism, selecting one resonance structure can be used to improve communication or emphasize particular property/reactivity of a compound.

As an example, below is a figure describing the formation of an enolate and its attack on an aldehyde (aldol reaction) using different resonance structures. For clarity, all of the curved arrows are describing reactions, not the relationships between resonance structures.

We know that carbonyls such as 1 have an important resonance structure which places a negative charge on oxygen and leaves a positive charge on carbon (2). While 1 is the most important contributor, 2 emphasizes the carbocationic nature of the carbonyl carbon. I like this structure, because I know some reactions of carbocations: SN1 and E1. This helps explain to me why the alpha-position is acidic. Deprotonation of 2 next to the carbocation, just as in E1, gives enolate 3. Thinking of the carbon-oxygen pi-bond as a leaving group, deprotonating 1 at the alpha- position gives the same enolate 3 but by a different set of arrows compared to 2 to 3.

Note that there is a third way (not shown) to describe the same deprotonation where a base deprotonates the alpha-hydrogen and the electrons are simply localized on the alpha-carbon to give structure 4.

Structures 3 and 4 are resonance structures, and again these can both be used to describe further chemistry. The blue set of curved arrows from 3 to 5 contains one additional electron pair movement to the red set from 4 to 5. Both are valid descriptions of the attack on an aldehyde, but emphasize different aspects of enolate reactivity.

Keep in mind that bond-line drawings, hybrid drawings, and curved arrows are just simple representations of molecules. Thankfully they work very well and allow us to reduce the complexity of electronic structure to a manageable level.

Finally, as @orthocresol suggests in the comments, using hybrid drawings is difficult. This is because bond-line drawings represent electron pairs as localized and curved arrows work best when showing the action of electron pairs.


The fact that manganese has a stable oxidation state in which it has (formally) lost all its $d$ electrons does not mean that it cannot be considered a transition metal, because it still has a lot of other oxidation states (like +II in $\ce{Mn^2+}$ or +VI in $\ce{MnO4^2-}$) with partially filled $d$ orbitals. Moreover, a naked $\ce{Mn^7+}$ ion does not exist in solution, it is only stable in the form of the oxocomplex $\ce{MnO4-}$.


Why does calcium chloride cause a greater freezing point depression in water than magnesium chloride?

As quantified in Fig. 17 of Manual of Practice for An Effective Anti-Icing Program (FHWA-RD-95-202):

By weight percent, MgCl2 depresses the freezing point of water to a greater degree than CaCl2, up to about 24%.  This is consistent with the freezing point depression formula, as Mg has less mass than Ca.

Beyond 24%, CaCl2 is superior.

This is because the eutectic point of MgCl2-water is at 21.6 percent, whereas the eutectic point of CaCl2-water is at 30%.

The freezing point depression formula is only valid at reasonably low concentrations.  It is certainly not valid beyond the eutectic point.


Monobromination of napthalene derivative

You have electronically asymmetric rings. The right ring is deactivated by the carbonyl group that acts like an acyl group (remember that acyl groups are strong meta-deactivators). The left ring is activated by both $\ce{-CH3}$ and the nitrogen substituent acting as an amine (activating ortho/para directing), so bromination will occur on the left ring (since it is more nucleophilic).

Monobromination is not guaranteed experimentally. Although the right ring is deactivated under the right condition (at higher temperatures) you can end up with more than one $\ce{Br}$ being added. Maybe this question is just showing that a bromination will occur probably in that particular position.


OK, for what it's worth:
Let $x_f$ be the final solution mass and $x$ be the mass of the 25% solution.
The dissolved mass can be defined by
$$0.3x_f~=~40+0.25x$$

The solution mass is:
$$x_f=100+x$$

Substitute $x_f$ into the first equation and solve. You should get the 25% solution mass to be added and then solve the easy solution mass equation to check your answer.


What exactly is happening when sodium, potassium, or other alkali metal explodes in water?

Until recently the answer was unknown, but a short time ago it was discovered that the reaction is in fact a Coulombic explosion. The rapid exchange of electrons between the sodium and the water causes the surface of the sodium droplet to become positively charged, and the ions repel each other. This behaves very like a negative surface tension, and the surface of the droplet increases in area rapidly forming a spiny, porcupine like shape as fingers of the molten metal are shot into the liquid at astonishing speed. The larger the surface area gets the faster the reaction occurs, leading to a runaway effect. The study was published in Nature Chem. (Ref.1).
References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, "Coulomb explosion during the early stages of the reaction of alkali metals with water," Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).


the molecular formula of an unknown hydrocarbon

$43\ \mathrm g$ of carbon for $7.2\ \mathrm g$ hydrogen.
Molar mass of carbon is $12\ \mathrm{g/mol}$ and that of hydrogen is $1\ \mathrm{g/mol}$.

Hence there are $\frac{43\ \mathrm g}{12\ \mathrm{g/mol}} = 3.58\ \mathrm{mol} \approx 3.6\ \mathrm{mol}$ of carbon for every $\frac{7.2\ \mathrm g}{1\ \mathrm{g/mol}}=7.2\ \mathrm {mol}$ of hydrogen.

Hence the empirical formula of the compound is $\ce{CH2}$.

Relative molecular mass of $\ce{CH2}$ is $12+2\times1 = 14$. Given relative molecular mass is $42$.

$\displaystyle\frac{42}{14}=3$

Hence the formula of the compound is $\ce{C3H6}$


Yes, amine inversion does effect(reduce) basicity and nucleophilic character of amines

In amine inversion a nitrogen which is sp3(tetrahedral) converts into sp2(planar) and then again coverts into sp3(tetrahedral).

This happens because nitrogen has a lone pair and this lone pair moves through nucleus towards the opposite side. This is known as quantum tunneling. While the electron pair is moving through the nucleus nitrogen has sp2 hybridisation. So, if nitrogen inversion is effective availability of lone pair decreases.

Now let's come to the case of quanidine. In quanidine, nitrogen inversion doesn't happen occur because of the impossible geometry it would have to form to do so.
However, in tri-methyl amine there is no such hindrance and hence achieve planar state relatively easily.


Can't you form a ring from almost any number of sides from carbon?

Carbon bonds have certain preferred angles. For example, the tetrahedrally symmetrical methane, which has sp3 hybridization,, $\ce{CH4}$, has an angle of 109.5 0. In benzene, where there is sp2 hybridization, the angle is 120 0.

These are the preferred angles for bonds. In some molecules, the geometry forces them to odd angles, such as cyclopropene, with a 60 0 bend. This imposes stress, much like making Tinker toy models where you have to bend the sticks a bit. Overdo this and the stick (bond) snaps.

So, here are some cyclic (ring) compounds with the following number of carbons, and at least one alkene bond:

3: cyclopropene

4: cyclobutene

5: cyclopentene

6: cyclohexene

7: cycloheptene

cyclooctene

Notice in the drawings of the molecular structure that larger rings are increasingly asymmetrical, degenerating to a twisted mess. So yes, very large carbon "rings" are possible, but eventually cannot hold their shapes. Some compounds contain more than on C=C bond, such as 1,3-cyclohexadiene, but I know of no cyclic compounds where all carbons contain a double (alkene) bond.

You can extend this, though, to build up three-dimensional polyhedra, such as cubene or the elegant (though not double-bonded) buckminsterfullerene.


How to determine if a complex has a crystal field stabilization energy of zero?

In the given example it is pretty simple to chose B as the correct answer, as it is the only option that could possibly create a spin forbidden transition $d^5$ or a $d^{10}$ configuration. For most situatons I have encountered it is sufficient to assume high-spin filling, until you encounter very stable configurations: completely full or half full orbitals (like $d^6$)


Name of the compund: Dimethyl nitroxide
Lewis structure:

The properly way to determine the Lewis structure, based on this example, is:

Total valence electrons: $4\cdot2 + 5 + 6 + 1\cdot6 = 25$
Total electrons needed for octets/doublets: $8\cdot4+2\cdot6 = 44$
Total shared/bonding electrons: $44 - 25 = 19$ (In other words, there is only nine single bonds and one electron left.)
Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 25 - 19 = 6$ (In other words, the are only 3 pairs of lone electrons.)

Furhter information on how to draw Lewis structures, please follow the link
Further information of related Lewis structures and how this knowledge is applied on real top science, I recommend seeing this reference: Moreira, G.; Charles, L.; Major, M.; Vacandio, F.; Guillaneuf, Y.; Lefay, C. and Gigmes, D. Beilstein J. Org. Chem., 2013, 9, 1589 - 1600.


Is Van der Waal's force the real source of friction?

While van-der Wall forces contribute to the frictional force, they are not the only reasons. Other secondary inter-molecular forces like electrostatic attractions, polar interactions, hydrogen bonding and more also play a significant part. Apart from this features on the surface of materials, present as abrasions are also responsible for locking together and resisting change.
The case of a gecko sticking to wall is not just about these forces and interlocking, but more about how the wrinkles on feet of a gecko act as several small vaccum cups which provide it the stability of sticking upside down as well(think how friction plays almost no role here)


Impossible pH For Aqueous Weak Acid Solution

Your reasoning is fine. But a major caveat (or additional factor) is that the pH scale itself no longer applies to concentrated solutions (say, more than 1 molar in acid) where the molarity of water falls significantly below 55M (pure water).

You can extend the pH scale down below 0 by empirically measuring how effective an acid solution is at protonating a weak base, but the scale deviates from the straight line predicted by the concentration of acid present.


Discrepancies between specific heat of vaporization of water and enthalpy of formation of water and steam

According to “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL, the standard molar enthalpy of formation at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\mathrm fH_\mathrm m^\circ=-285.8\ \mathrm{kJ\ mol^{-1}}$ for liquid water,

and

$\Delta_\mathrm fH_\mathrm m^\circ=-241.8\ \mathrm{kJ\ mol^{-1}}$ for gaseous water.

(These values are in agreement with the values that you have found on Wikipedia.)

Both values refer to a temperature of $T=298.15\ \mathrm K$; therefore, you do not have to raise the temperature of the liquid water to its boiling point. Nevertheless, you may do this; however, then you would have to cool down the steam back to $T=298.15\ \mathrm K$ later.

Instead, you can directly use the enthalpy of vaporization at the given temperature. Note that the enthalpy of vaporization depends on temperature, and that your value for the specifc enthalpy of vaporization of $\Delta_\text{vap}h=2258\ \mathrm{kJ\ kg^{-1}}$ refers to the boiling point at normal pressure, i.e. to a temperature of approximately $T=100\ \mathrm{^\circ C}$.

According to REFPROP – NIST Standard Reference Database 23, Version 9.0, the molar enthalpy of vaporization at a temperature of $T=298.15\ \mathrm K$ is

$\Delta_\text{vap}H_\mathrm m=43.987\ \mathrm{kJ\ mol^{-1}}$,

which is in good agreement with the difference of the values for the molar enthalpy of formation:

$-241.8\ \mathrm{kJ\ mol^{-1}}-\left(-285.8\ \mathrm{kJ\ mol^{-1}}\right)=44.0\ \mathrm{kJ\ mol^{-1}}$


(Computationally) finding similarity between two organic compounds

Using graphs to represent molecules, would it be reasonable to compute the distance between graphs A and B?

In a graph representation of a molecule, nodes are atoms and edges are bonds.

Example of an algorithm to compute distance between graphs: http://www.xavierdupre.fr/app/mlstatpy/helpsphinx/c_graph/graph_distance.html


Fehling's Test meaning for partial formation of a precipitate?

The reason is thus:

For aldoses (like glucose), the copper sulfate in the reagent easily oxidizes it into the carboxylic acid forming copper oxide (a precipitate).

For ketoses (like fructose), the ketone cannot be oxidized, so there is no color change. Over time at elevated temperature, the α-hydroxy ketone isomerizes to a α-hydroxy aldehyde. This aldehyde is then oxidized to the carboxylic acid, producing a red precipitate.


How to go about deciding the acidic strength of a functional group in an aromatic compound?

This question is, at best, flawed. The most acidic protons on the structure provided are not any of the labeled OH protons. Rather, the most acidic position is the one I have drawn a rectangle around below.

Deprotonation at this position produces an anion with a resonance contributor in which both rings are aromatic. The circa 150 kJ/mol resonance stabilization for each ring will vastly outweigh any delocalization that might occur for the oxyanions formed by deprotonating any of the OH groups.


Help understanding the following conversion

You can work backwards here. Step 3 converts cyclohexanol to the product by substitution reaction of methyl iodide and alkoxide. Sodium hydride is the base that deprotonates the alcohol.

Step 2 is a bit dubious. They want to substitute a secondary chloride with hydroxide. In principle this will work, though I am suspicious about whether or not it will actually take place. For cyclohexyl chloride (product from step 1), the substitution is probably only going to happen from the axial conformer; the equatorial conformer has the flag pole hydrogens blocking the approach to the $\sigma^{*}$ orbital. But from the axial conformer, I suspect that elimination is also a competitive pathway. Hard for me to say exactly what the relative rates are for sure.

Step 1 you've likely figured out. It's radical chlorination of an alkane.


Source of the no. of isomers in alkanes, and other organic molecules

Not counting stereoisomers and enantiomers, the number of isomers of the first ten alkanes is as follows:

Methane: 1
Ethane: 1
Propane: 1
Butane: 2
Pentane: 3
Hexane: 5
Heptane: 9
Octane: 18
Nonane: 35
Decane: 75

Source: http://elib.mi.sanu.ac.rs/files/journals/tm/21/tm1121.pdf and Wikipedia articles for these alkanes. The PDF has the number of isomers for the first 30 n-alkanes.

The number of isomers possible with an n-carbon alkane can be calculated by a recursive combinatorial function; however, as the linked paper states, this math isn't for the weak. The number of structural isomers of a 30-carbon alkane is over 4 billion, and a 50-carbon alkane has over 1031 possible isomers. Even organic chemists and petrochemists usually don't bother to identify any one isomer beyond about the pentane mark. Some of the more symmetrical isomers are interesting on their own, since a few of the more compact ones (like the n-methyl types) have markedly different behaviors such as boiling point from longer isomers (for example, tetramethylbutane, an isomer of octane notable for its relatively narrow liquid window). The 2,2-dimethylpropane isomer of pentane is visually interesting because of its isohedral geometry, and likewise the tetraethylmethane/diethylpentane isomer might find interest among a few supremacist groups but is otherwise just another 9-carbon alkane.


Do materials exist that absorb Visible light and emit Near InfraRed light?

Yes, there are some minerals that shows VIS-NIR luminescence here a couple:

Synthetic ruby $\ce{Al2O3:Cr}$
Kyanite ($\ce{Al2SiO5}$)

(Reference here)

A more sophisticated material is $\ce{LiSrPO4:Eu2+, Pr3+}$ you can read the full article in 11 February 2013 / Vol. 21, No. 3 / OPTICS EXPRESS 3161.

There are also some pigments like egyptian blue or cadmium yellow (CI pigment yellow 37, that is cadmium sulfide) that were also used in the past. Egyptian blue (calcium copper silicate ($\ce{CaCuSi4O10}$))  is safe, other pigments with cadmium are more toxic.

One of the most spread pigment shows VIS-NIR luminescence: Chlorophyll!

Have a look to this site for a nice overview.

All the luminesce processes have not a high quantum yield, so to record them, most of the times, you have to avoid reflection. Tungsten, halogen lamps and sun emit a lot of NIR radiation so is possible that reflected light overhang the luminescence. When I performed NIR luminescence I used VIS LEDs and a long pass filter at 850 nm.  If you don't have result, first thing you have to try is to remove the ICF (IR-Cut off-Filter) filter from your camera so NIR light can reach the sensor more easily.


Imagine a sky scraper where everyone wants to live on as low floor as possible, but landlord prohibits then to live more than 2 people per floor. These people will occupy lowest floors in pairs and the highest occupied level might have a single person or a pair. If someone moves out of the apartment on the 1st or 2nd floor, then someone from higher floors will move down to the vacant state. Electrons do the same. Floors are energy levels and the evil landlord is the Pauli exclusion principle.

Electrons are fermions. An electron's spin can be either +1/2 or -1/2. Two options. Two fermions (and, hence, electrons) cannot be the same. Electrons are the same if their spins are the same and they occupy the same orbital (Pauli Exclusion Principle). At lower temperatures (below red heat temperatures) electrons occupy the lowest energy level possible.

As a result, each new electron occupies the lowest allowed level. Because their spins can take two values 2 electrons are occupying each level. If you populate each energy level by a single electron you will get a high energy excited state that well emit light and go to lower energy two electrons per level state. If you try to squeeze more than two electrons it will not work because of Pauli principle. So, since they want to stay in lower energy levels and have to obey the exclusion principle we have electron pairs.

Bonus: here is why radicals recombine. For instance, why $\ce{2 H^. -> H2:}$

If an electron is the last one it is occupying the energy level alone.
So, electrons would like to all occupy the lowest level, but only two per level is allowed. In radicals only one electron occupies a level. When radicals A$^.$ and b$^.$ meet each of them sit on half empty orbital attracted to a single nucleus.  It is energetically favorable to share the same orbital (form a pair, which they are OK with) and orient fragments so that they "fly" (have the highest probability) between two nucleus (A and B). Staying between two opposed charged ions is better than staying next to just one charged ion. And electrons are allowed to stay in pairs (Pauli excluded 3+ electrons staying together).


There are two distinct forms of $\ce{HSO3-}$ (see picture below; all horizontal bars are formal $-$ signs).

Only the leftmost of these two forms can be called hydrogen sulphite (or ‘bisulphite’, if you really insist); the correct name of the rightmost form is sulphonate. The two differ in their chemical characteristics; hydrogen sulphite is, for example still acidic and has a lone pair on sulphur, while sulphonate is not.

Johansson, Lindqvist and Vannerberg from Göteborg solved the crystal structure of $\ce{CsHSO3}$ in 1980 and concluded that:

the $\ce{S-O}$ bond lengths are much closer to those in sulphate than those in sulphite ($\ce{Na2SO3}$)
the $\ce{O-S-O}$ angle is $113°$ and thus larger than the tetrahedral angle $\approx 109°$ (expected in sulphate) and much larger than the angle in $\ce{SO3^2-}$
the ion’s point group is $\mathrm{C_{3v}}$, so all three oxygens are equal
there is IR evidence of a $\ce{S-H}$ bond.

From this they deduced that $\ce{CsHSO3}$ be, in fact, cesium sulphonate, not cesium hydrogen sulphite.[1]

The first reference I found for the structure of a true hydrogen sulphite was the crystal structure of cis-$\ce{[Ru(bpy)2(HSO3)2]}$ by Allen, Jeter, Cordes and Durham from Arkansas in 1988. In their structure, $\ce{HSO3-}$ coordinates to the central $\ce{Ru(II)}$ ion via the sulphur atom: $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$. This coordination is, of course, only possible if there is a lone pair on sulphur, and thus the structure must be that of a hydrogen sulphite.[2]

(In my introductory chemistry class, Prof. Klüfers stated that structures for both forms be known and published.[3] Alas, I didn’t find the crystal structure of a simple metal hydrogen sulphite anywhere quickly.)

It gets easier when dealing with the corresponding acids’ esters. The hydrogen on sulphonic acid can be replaced with organic residues creating the vast class of organosulphonic acids. Sulphonic esters are also possible; they would have an organic residue attached to both the sulphur directly (replacing the hydrogen atom) and one of the oxygens. Every mesylate or tosylate (in formulae: $\ce{OMs}$ or $\ce{OTs}$) is, in fact, a sulphonic ester. Sulphurous esters are less prevalent in organic chemistry but do exist (e.g. dimethyl sulphite, $\ce{OS(OCH3)2}$). Of course, the chemical properties of methyl mesylate ($\ce{H3C-SO2-OCH3}$) — a sulphonate — and dimethyl sulphite are substantially different.

After discussing $\ce{H2SO3}$ and its salts, it is important to mention the wrong assumption you proposed in your question:

  we see that the $\ce{H}$ is bonded directly to the $\ce{S}$, not to the $\ce{O}$’s, as would be the case in other oxyacids.

This is not generally true. Phosphorous at least displays a similar chemistry, even to the extent that ‘hydrogen phosphite salts’ cannot be isolated and phosphonate salts are isolated instead. (One can synthesise both trimethyl phosphite and dimethyl methylphosphonate, the corresponding phosphorous and phosphonic esters, though.) I am not firm in the oxyacid chemistry of heavier homologues of phosphorous and sulphur, but it would deeply astound me if this chemistry is not observed there at least to some degree.

Note that the halogens do indeed follow your proposed rule: all their oxyacids are monoprotic and contain only a single, oxygen-bound hydrogen in neutral state.

References:
[1] L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2] L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.
[3] Refer to the internet scriptum of general and inorganic chemistry 1 (in German) as taught 2006 to 2008 and likely will be taught again in 2015.

(Note that the articles are most likely behind a paywall)


Thiourea ($\ce{(H2N)2C=S}$) is a very important component of tinning solution because tinning is a redox process. That's because thiourea is a powerful reducing agent (Ref.1), chemically and biochemically used in redox reactions in vitro or in vivo:

$$\ce{(H2N)2C=S <=> (H2N)2C-S-S-C(NH2)2 + 2e-} \quad E^\circ = \pu{0.418 V} \tag1$$

The oxidizing agent in this process is $\ce{Sn^2+}$ ion, which reduces it self during the process:

$$\ce{Sn^2+ + 2e- <=> Sn} \quad E^\circ = \pu{-0.1375 V} \tag2$$

As a result, following redox reaction happens:

$$\ce{(H2N)2C=S + Sn^2+ <=> (H2N)2C-S-S-C(NH2)2 + Sn} \quad E_\mathrm{rxn}^\circ = \pu{0.2805 V} \tag3$$

Since $E_\mathrm{rxn}^\circ $ is positive, the reaction is spontanious in standard conditions and no electric current is necessary. When you dip your circuit in the solution (or touch surface to surface of the solution), metallic tin would deposit as a thin layer.

As an alternative to $\ce{(H2N)2C=S}$, dimethylated or diethylated thiourea can be used in those reactions, which require electrode potential within the range of $\pu{0.418 V}$ or higher (Ref.2):

$$\ce{(CH3NH)2C=S <=> (CH3NH)2C-S-S-C(NHCH3)2 + 2e-} \quad E^\circ = \pu{0.460 V} \tag4$$

$$\ce{(C2H5NH)2C=S <=> (C2H5NH)2C-S-S-C(NHC2H5)2 + 2e-} \quad E^\circ = \pu{0.450 V} \tag5$$

In tanning process, theoretically, any component with oxidative potential bigger than  $\pu{-0.1375 V}$ can be used as the reluctant. Thus, best alternative come to my mind is ascorbic acid (Vitamin C). Recently, natural source of ascorbic acid source, extract of fruits of Averrhoa bilimbi (Sinhala name: Billin), has been used to prepare silver and gold nanoparticles from corresponding salts (Ref.3):

In any event ascorbic acid didn't work, there is a way to synthesize thiourea derivatives without using thiocyanate or isocyanate. The scheme is illustrated in the bottom box of above diagram (Ref.4). The $E^\circ$ of these derivatives should be closer to that depicted in equations $(4)$ and  $(4)$. However, any attemp to synthesize these compound should be done in certified chemical laboratory under the observation of an experienced chemist. Keep in mind that thiourea is a potent carcinogen.

Note: It is well known that silver polishing reagents contain thiourea ($50\%$ by weight; Ref.5). You may able to extract thiourea from such solutions.

References:

Paul W. Preisler, Louis Berger, “Oxidation-Reduction Potentials of Thiol-Dithio Systems: Thiourea-Formamidine Disulfide,” J. Am. Chem. Soc. 1947, 69(2), 322–325  (https://doi.org/10.1021/ja01194a048).
Paul W. Preisler, “Oxidation--Reduction Potentials of Thiol--Disulfide Systems. III. Substituted Thioureas and Compounds Containing Two Thioureido Groups,” J. Am. Chem. Soc. 1949, 71(8), 2849–2852  (https://doi.org/10.1021/ja01176a076).
R. S. Rimal Isaac, G. Sakthivel, Ch. Murthy, “Green Synthesis of Gold and Silver Nanoparticles Using Averrhoa bilimbi Fruit Extract,” Journal of Nanotechnology 2013, Article ID 906592, 6 pages (https://doi.org/10.1155/2013/906592).
Wei Tan, Jianpeng Wei, Xuefeng Jiang, “Thiocarbonyl Surrogate via Combination of Sulfur and Chloroform for Thiocarbamide and Oxazolidinethione Construction,” Org. Lett. 2017, 19(8), 2166–2169 (https://doi.org/10.1021/acs.orglett.7b00819).
Mark Laing, Mark Beeston, "Silver polish formulation containing thiourea," US Patent 5,958,854, 1999 (Application No.: 08/869,685 in 1997).


The short answer

The standard pressure is not an experimental condition, while the temperature is.

The long answer

The main reason the standard state is defined is because it leads to this very useful equation
$$K_p = e^{-\Delta G^\circ/RT}$$
Say you have this reaction: $A \rightleftharpoons B + C$ One way to use this equation is to compute the free energy of 1 mol of $A$, $B$, and $C$ at 1 bar using equations derived for an ideal gas, compute $\Delta G^\circ = G^\circ (B) + G^\circ (C) - G^\circ (A)$, and use that value to predict $K_p$.

If the gasses behave like ideal gasses "in real life" then the measured  $K_p$ will match the $K_p$ computed from $\Delta G^\circ$.  You can do the measurement at any pressure you want, not just at 1 bar.* The standard state refers to the pressure you use when computing $\Delta G^\circ$.  The only thing it has to do with the experimental measurement is that it defines the units you should use for your partial pressures when computing $K_p$

$\Delta G^\circ$ does also depend on temperature, but the temperature you chose should be the same as the experimental conditions.  So the temperature is not part of the standard state definition.

But what about "Standard temperature and pressure (STP)?"

Standard temperature and pressure (STP) refers to STP conditions under which $K_p$ is measured, not the pressure used to compute $\Delta G^\circ$. I know, they couldn't have made it more confusing if they tried when they named these things.

*Of course if you do the measurements at very high pressures or low temperatures, then the assumption that the gasses behave ideally will be less valid and the measured $K_p$ will differ more from the $K_p$ computed from Equation 1.  However, that is a separate issue unrelated to the standard state because the $K_p$ in Equation 1 refers to the $K_p$ you would measure if the gasses behaved ideally at the pressure and temperature used in the experiment.


Is diamond an allotrope or polymorph of carbon?

Allotropy is a special case of polymorphism. Polymorphism is the phenomenon of a substance exhibiting different crystal structures. Allotropism is the same phenomenon limited to the subset of all substances that contains only the chemical elements. According to the Wikipedia article on polymorphism:

  Polymorphism ... is the ability of a solid material to exist in more than one form or crystal structure. Polymorphism can potentially be found in any crystalline material ...

The allotropy article defines allotropy:

  Allotropy or allotropism (coined from Greek "other" + "form") is the property of some chemical elements to exist in two or more different forms ... The term allotropy is used for elements only, not for compounds. The more general term, used for any crystalline material, is polymorphism.

Thus, diamond and graphite represent two different allotropes of carbon, which also makes them different polymorphs of carbon.

Water is well-known for having multiple solid phases. These phases are polymorphs, but, because water is not an element, they are not allotropes.


Weak buffers and accuracy when mixing acid and base

What is the weakest buffer possible - that is, what has the lowest buffer capacity possible?

You've conflated two different aspects of a buffer. First there is the $K_\mathrm{a}$, or $\mathrm{p}K_\mathrm{a}$ for the species.

$$K_\mathrm{a} = \frac{\ce{[H^+][A^-]}}{\ce{[HA]}}$$

$$\mathrm{p}K_\mathrm{a} = -\log{K_\mathrm{a}}$$

So weaker acids would have a high $\mathrm{p}K_\mathrm{a}$, close to 7 or greater.

Then there is the buffer capacity which is a function of the absolute concentrations of the various species ($\ce{[H^+]}$, $\ce{[A^-]}$, and $\ce{[HA]}$). So if there is relatively much more $\ce{[H^+]}$ or $\ce{[A^-]}$ in the solution than acid or base being added, then the pH will change by only a small amounts.

  Also, how precise do you have to be when mixing a strong acid and base to get a pH value of 7 and not 6?

If the strong acid is $10^{-6}$ greater than the strong base then the pH will be 6.

$\mathrm{pH} = -\log\ce{[H^+]}$


Transparency and partially deuterated polymers

Deuteration lowers the vibrational frequencies C-H to C-D, but does not affect C-C etc vibrations.  Thus some lowering of vibrational frequencies will make the ir region slightly more transparent by bunching more frequencies together. In the UV deuteration hardy affects the spectrum and so here the polymer should be pretty much the same. Unless there is something in the processing of polymers that is improved by deuteration I can't think of anyhthing else.

EDIT

Having read your comments & looked at the data and thought more about it I now think that its the Franck-Condon factors that are responsible. These measure the strength of a transition and for electronic transitions in visible and uv are all proportional to $exp(-\alpha (\Delta R^2/2))$ where $\alpha = \sqrt(\mu k)/\hbar$ where $k$ is the force constant, (same for CH and CD bonds) and $\mu$  is reduced mass 1.7u for CD and 0.9u for CH so the deuterated bonds will be less intense as the exponential will be smaller for CD than CH. ($\Delta R$ is the displacement between levels in ground state and excited state). In the ir the intensity is proportional to the transition frequency multiplied by the square of $ \int \psi_i x \psi _{i+1} dx$ where x is the displacement from equilibrium bond length and $\psi$ a vibrational wavefunction with i quanta. Evaluating the integral shows that a transition is proportional to $\nu ^3$ where $\nu $ is the transition frequency between levels with i to i+1 quanta. The intensity is thus reduced by substituting D for H $via$ $\nu = \frac {1}{2\pi}\sqrt{\frac{k}{\mu}}$.

As CH/CD vibrations are of higher frequency than CC and the effect should be most pronounced at higher ir frequencies and proportional to the amount of deuterated species. Typically deuterated transition frequencies are 0.715 of the protonated values compared to the expected $1/\sqrt(2)$ =0.707, considered to be due to effects off cubic and quartic terms in the potential. This means that the protonated transitions are 2.7 times more intense than deuterated at approx 2900 cm$^{-1}$, the CH frequency in methane.