When we say "the energy in a system" (in chemistry) do we mean the energy stored in the chemical bonds in the compounds present in that system?

The answer to this question depends on the nature of the system we have defined.

Firstly, to define the term system according to this Wikipedia article:

  “...a physical system is a portion of the physical universe chosen for
  analysis. Everything outside the system is known as the environment.
  The environment is ignored except for its effects on the system. The
  split between system and environment is the analyst's choice,
  generally made to simplify the analysis. For example, the water in a
  lake, the water in half of a lake, or an individual molecule of water
  in the lake can each be considered a physical system.”

Every system has some relationship to it’s surroundings.  As another way of stating the above, we can describe the surroundings, or environment, as all parts of the universe that is not part of the system. We can then generally categorize the system as being isolated, closed or open:

Isolated system: Neither matter nor energy may pass into or out of the system.
Closed system: The exchange of energy is allowed but matter may not pass into or out of the system.
Open system: Both energy and matter can pass into and out of the system.

In the context of your question (as I think I understand it), we are talking about isolated systems.  It wouldn't make much sense to discuss the energy of chemical bonds without constraining the system, even if hypothetically, in this way. To use the lake analogy, we talk about the chemical bonds within a single water molecule in a manner that excludes the matter or energy in the rest of the lake (and the rest of the universe).  If we are talking about hydrogen bonding between the water molecules for example, then we are considering there to be some hypothetical container within the lake that excludes all matter and energy from outside the container.  The contents, matter and energy, of the container, comprise the system.


Ahem, forget it.

Theoretically speaking, you can get $\ce{Zn}$ from alkaline batteries (see details here), but you have no chance to get pure $\ce{N2O}$ this way and reaction may easily go out of control.

Attention! Detonation hazard !!!

The usual and easy way to produce $\ce{N2O}$ is to heat very carefully widely available $\ce{NH4NO3}$, to $180-200~^\circ\mathrm{C}$. It is an absolute imperative to keep temperature of the reaction mixture below $240~^\circ\mathrm{C}$. Decomposition of ammonia nitrate ($\ce{NH4NO3}$) is exothermic and may proceed explosively. To be on the safe side, use sand bath to heat the reaction flask and put thermometer into the sand, so its temperature never risen above safe limits. Liquid bath with appropriate liquid may be better, but I'm not sure if a school lab would have proper liquid (silicone oil with high flash point). The produced gases should be cooled in Liebig condenser to separate water, and $\ce{N2O}$ may be further dried with $\ce{H2SO4}$. Of course, all joints must be sealed using proper oil. And keep a bath of cold water to cool the reaction mixture in case if temperature raises above safe limits.

It is advised to use lab grade, not fertilized grade ammonia nitrate.

Possible safer, but costlier route

It is reported that it is possible to produce $\ce{N2O}$ by heating mixture of   $5~\mathrm{mL}$ $6~\mathrm{M}~\ce{HNO3}$, $1~\mathrm{g}~\ce{NH4NO3}$ and $0.05~\mathrm{g}~\ce{NaCl}$ on water bath at $\approx 80~^\circ\mathrm{C}$. If this works, it is a safer route, as hot ammonia nitrate is diluted and autospeeding reaction cannot occur. Link


What is the difference between doxycycline HCl and doxycycline hyclate?

[...]they prefer hyclate because it crystallizes more readily than HCl salt

The principal component is doxycycline - in both cases. The active component is a metabolite.

Doxycycline is an antibiotic and belongs to the tetracyclines (= four rings).

In order to prepare nice ions that can traverse membranes and pumps by the body easily, pharmaceutics containing amino groups or 'alkaloids' (see the red nitrogen atom in the structure), are often converted to their hydrochloride, citrate, oxalate, bromide etc. salts.

These are salts with the (protonated) pharmaceutical as the cation and chloride as the anion. Salts are water soluble.

Here, the notation for this is doxycycline$\ce{.HCl}$.

The crystallisation is done in a solvent or a mixture of solvents.

When the doxycycline$\ce{.HCl}$ precipitates (=when it forms a solid), molecules of the solvent may get embedded, trapped, or entrained in the crystal structure (but the crystals gets pulverized anyway and re-crystallized).

Note that this is not just a wet solid! There would be little or no absorption in that case.

Exactly this happens when the hyclate is formed during the crystallization of doxycycline hydrochloride in a mixture of water and ethanol.

Hyclate is just an abbreviation for hydrochloride hemiethanolate hemihydrate, which means that each molecule of doxycycline comes with 1 $\ce{.HCl\ }$  and 0.5 $\ce{.CH3CH2OH}$ and 0.5 $\ce{.H2O}$ in the crystal.

Whether it is more concentrated (= more active component in one pill) only depends on the formulation of the respective pharmaceutical companies.


Why are calorimeters often made of materials with low specific heats?

Actually, a calorimeter with high specific heat would have less temperature change, which is harder to measure. Therefore, the error in the calculated enthalpy of the reaction would be higher.
Specific heat does not mean heat conductivity, as described by Fourier's Law.


Two points before we begin:

+ charge means deficiency in electrons.
- charge means excess of electrons.

Now lets come to the reaction you are talking about.

  Two chromium atoms gain 3 electrons each. They go from 0 to 3+

Moving from neutral to +3 charge means it just lost 3 electrons. (Electrons have negative charge.)

  Similarily, 6 iodine atoms go from -1 to 0, thus gaining one electron. However my textbook says that Iodine has lost electrons. How?

Your textbook is right and you're wrong here. Moving from -1 charge to 0 means that an atom just lost an electron (in other words, it lost the excess negative charge it had and now it is neutral).


Activated alumina, silica, thoriated silica gel, phosphorus pentoxide-silica gel, aluminum phosphate, titanium dioxide
Catalytic Dehydration of Formic Acid, E. G. Graeber and D. S. Cryder
Industrial & Engineering Chemistry 1935 27 (7), 828-831, DOI: 10.1021/ie50307a019
Mechanism of dehydration decomposition of formic acid over alumina and silica. Adsorption measurements during the catalytic reaction, Kenzo Fukuda, Yuko Noto, Takaharu Onishi and  Kenzi Tamaru, Trans. Faraday Soc., 1967,63, 3072-3080, DOI: 10.1039/TF9676303072

Mixed titanium-vanadium oxide
Sadovskaya, Ekaterina & Chesalov, Yuriy & Goncharov, V. & Sobolev, Vladimir & Andrushkevich, Tamara. (2017). Formic acid decomposition over V-Ti oxide catalyst: Mechanism and kinetics. Molecular Catalysis. 430. 52-64. 10.1016/j.molcata.2016.12.010

Various homogenous transition metal catalysts
Homogeneous Catalytic Dehydrogenation of Formic Acid: Progress Towards a Hydrogen-Based Economy, Gábor Laurenczy, Paul J. Dyson, J. Braz. Chem. Soc. vol.25 no.12 São Paulo Dec. 2014, DOI: 10.5935/0103-5053.20140235

Monoclinic bismuth chromate hydroxide proto-catalyst, $\ce{m-Bi(OH)CrO4}$
Two-step catalytic dehydrogenation of formic acid to CO2 via formaldehyde, Viasus et. al., International Journal of Hydrogen Energy, Volume 44, Issue 3, January 2019, Pages 1534-1543, DOI: 10.1016/j.ijhydene.2018.11.178


Why is methyl chloride a primary alkyl halide?

Yes, you are correct.

The IUPAC defines alkyl groups in the following manner:

  Univalent groups derived from alkanes by removal of a hydrogen atom from any carbon atom $\ce{–C_{$n$}H_{$2n+1$}}$. The groups derived by removal of a hydrogen atom from a terminal carbon atom of unbranched alkanes form a subclass of normal alkyl (n-alkyl) groups $\ce{H(CH2)_{$n$}}$. The groups $\ce{RCH2}$, $\ce{R2CH~(R \neq H)}$, and $\ce{R3C~(R \neq H)}$ are primary, secondary and tertiary alkyl groups, respectively.

Note that for primary alkyl groups hydrogen may be a remainder, i.e. $\ce{R $=$ H, Alkyl, Aryl}$, therefore any halogenoalkane with the structure $\ce{RH2C-X}$ is primary.


Supercooling is a general phenomenon and it is not limited to any particular substance, although the degree you can supercool something will vary. The idea in thermodynamics is that at the melting point the free energy of the solid and the liquid is equal and a lower temperature will favour the formation of the solid.

However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.

The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling.


Are the rules proposed by Slater for the calculation of effective Z right?

Slater's rules are pretty good, but they're not perfect. Consider that he published his work in 1930: Phys Rev 1930 v. 36, 57
There are a number of people who have modified Slater's scheme to better reflect shielding effects and correct for experimental ionization energies and electron affinities (and similar properties) in later rows of the periodic table.
If I remember correctly, one such modification is that d and f orbitals shouldn't have as much shielding as suggested by Slater, but the correction is small. There's another scheme that uses an exponential correction instead.
For example, Wikipedia gives effective nuclear charges from Clementi and Raimondi, who did full self-consistent quantum calculations and you can compare these to Slater's.
The main take-home from Slater's rules is that they work pretty well for a wide range of elements, including 1st, 2nd, and 3rd row. I think the corrections generally tweak things when we get into the 4d and 5d transition metals and some of the lanthanides and actinides. Not bad for an 84-year old paper!


Calculating standard enthalpy of reaction, which reactant to choose for the amount of substance

The difference between the Enthalpy of Reaction $\Delta_{\mathrm{r}} H$ and the Standard Enthalpy of Reaction $\Delta_{\mathrm{r}} H^{\ominus}$

The standard value is not related to a standard temperature (although Standard Enthalpies of Reaction are often given with respect to the standard temperature - they could have been given for any arbitrary temperature). The term "Standard" refers to standard pressure and the reactants and products being in their standard states. The standard states are usually those of the pure compounds. Furthermore gases are usually assumed to behave like ideal gases. This definition of standard states means that Standard Enthalpies of Reaction are often purely calculational quantities since the underlying reactions are practically not feasible. For example, the reactants usually have to be mixed in order to react with one another, meaning that they are not present in their pure form. Similarly, the products of a reaction are often formed as a mixture and not as pure substances. And real gases certainly don't show the same behavior as ideal gases.
So, enthalpically the difference between the Enthalpy of Reaction and the Standard Enthalpy of Reaction is that you would have to add the Enthalpy of Mixing of the reactants to $\Delta_{\mathrm{r}} H$ and subtract the Enthalpy of Mixing of the products from it in order to get $\Delta_{\mathrm{r}} H^{\ominus}$. Furthermore you would have to account for the enthalpic difference between real and ideal gases if there are any gases involved. Furtunately, those differences are usually quite small compared to the often very big Enthalpies of Reaction so that they can be ignored without introducing a large error. But you have to be mindful of the difference between $\Delta_{\mathrm{r}} H^{\ominus}$ and $\Delta_{\mathrm{r}} H$: for example, if $\Delta_{\mathrm{r}} H^{\ominus}$ is small and the reaction is conducted under very non-ideal conditions with respect to the behavior of gases and/or including liquid mixtures with a very large Enthalpy of Mixing, then you have to expect big differences between the (tabulated) standard values and the measured values.

So, in this sense you would have quite a hard time to calculate the exact Standard Enthalpy of Reaction from your measured $\Delta_{\mathrm{R}} H$. But, under the appropriate conditions your measured value is a good approximation.

How to calculate $\Delta_{\mathrm{r}} H$?

The key to solve your calculational problem is to look at the definition of $\Delta_{\mathrm{r}} H$:

\begin{equation}
  \Delta_{\mathrm{r}} H = \biggl(\frac{\partial H}{\partial \xi} \biggl)_{p, T} \ ,
\end{equation}

whereby the extent of reaction $\xi$ is defined as

\begin{equation}
  \mathrm{d} \xi = \frac{\mathrm{d}n_{i}}{\nu_{i}}
\end{equation}

with $n_{i}$ and $\nu_{i}$ being the amount and the stoichiometric coefficient of substance $i$, respectively.
Consider the reaction:

\begin{equation}
  \ce{\nu_{A} A + \nu_{B} B <=> \nu_{C} C + \nu_{D} D}
\end{equation}

From its definition you see that $\Delta_{\mathrm{r}} H$ is the amount of heat that is exchanged with the surroundings in a isothermal and isobaric process when $\ce{\nu_{A}}$ moles of A and $\ce{\nu_{B}}$ moles of B are converted to $\ce{\nu_{C}}$ moles of C and $\ce{\nu_{D}}$ D (= "conversion of one formula unit").
So, using the differential form of your calorimetric formula (under the assumptions you made) and "dividing" it by $\mathrm{d} \xi$ you get

\begin{equation}
  \mathrm{d} H = m C \mathrm{d}T \\
  \frac{\partial H}{\partial \xi} = m C \frac{\partial T}{\partial \xi} \\
  \Delta_{\mathrm{r}} H = m C \frac{\partial T}{\frac{\partial n_{i}}{\nu_{i}}} = m C \nu_{i} \frac{\partial T}{\partial n_{i}} \ ,
\end{equation}

where $m$ is the mass of the stuff you heat in the calorimeter (assumed to be constant) and $C$ its heat capacity.
Considering finite instead of infinitesimal changes of $n_{i}$ leads to

\begin{equation}
  \Delta_{\mathrm{r}} H = m C \nu_{i} \frac{\Delta T}{\Delta n_{i}} \ .
\end{equation}

So, in order to get your Enthalpy of Reaction you simply have to pick one substance from your reaction (it doesn't matter which) and substitute its stoichometric coefficient and the amount that was consumed/produced during the reaction into the equation above. Note that since $\xi$ is always positive the sign of $\Delta n_{i}$ is positive for products and negative for reactants (the same is true for $\nu_{i}$).

Side Note: Standard Enthalpy of Formation $\Delta_{\mathrm{f}} H^{\ominus}$

The Standard Enthalpy of Formation is a special Standard Enthalpy of Reaction for which further conditions are in effect:

the target compound is to be formed by reacting the (pure) elements
it consists of, whereby each element is expected to be in its most
stable modification for the given temperature.
The reaction equation is to be formulated in such a way that $1 \,
   \text{mol}$ of the compound in question is formed.


I didn’t find anything definite. However, I found a PDF from the State’s Institution for Protection of the Enviroment in Baden-Württemberg (Landesanstalt für Umweltschutz) which gives me two or three relevant clues. Along with my intuition and a chat with a labmate, I conclude:

There is nothing systematic or conventionalised in the names of the isomers. They were likely named after the order in which they were isolated or their structures were determined.

Clues I used:

The $\unicode[Times]{x3B6}$, $\unicode[Times]{x3B7}$ and $\unicode[Times]{x3B8}$ isomers were synthesised last.
$\unicode[Times]{x3B1}$, $\unicode[Times]{x3B2}$ $\unicode[Times]{x3B3}$ and $\unicode[Times]{x3B4}$ were characterised first by a guy called Teunis van der Linden, after whom the active insecticide $\unicode[Times]{x3B3}$-hexachlorocyclohexane was named lindane.
Thereby concluding that $\unicode[Times]{x3B5}$ was discovered after $\unicode[Times]{x3B1}$ to $\unicode[Times]{x3B4}$.


Qualitative analysis of the anionic part of a salt using it's soda extract

If a solution of $\ce{H2SO4}$ is added to a some carbonate, bicarbonate or sulfite, it will produce some $\ce{CO2}$ or $\ce{SO2}$ effervescence with gas bubbles filling the container where this test is done, and maybe cause it to overflow. With sodium bicarbonate the equation is :
$$\ce{H2SO4  + 2 NaHCO3 -> Na2SO4 + 2 H2O + 2 CO2}$$
With sodium sulfite, it is :
$$\ce{Na2SO3 + H2SO4 -> Na2SO4 + H2O + SO2}$$


What are the correct resonance structures of bromoethene?

Your resonance structure is perfectly alright. But the resonance structure with the negative and positive charges won't be very important and here is why: For one thing, it contains seperated charges and usually those resonance structures are most important that contain as few charges as possible. This is simply a matter of electrostatics - seperating charges and holding them apart from one another costs energy. A second reason has to do with the "electronic origin" of this resonance charge: It describes a conjugation between a free electron pair of $\ce{Br}$ and the $\pi$ system of the double bond - more accurately put the filled free-electron-pair-orbital can interact with the $\pi^{*}$ orbital of the $\ce{C=C}$ bond to form two new orbitals: an in-phase combination that is lower in energy and will contain the two electrons from the free electron pair and an out-of-phase combination that is higher in energy and will remain unoccupied. This stabilizes the system. The stronger the overlap between the free-electron-pair-orbital and the $\pi^{*}$ orbital of the $\ce{C=C}$ bond and the closer they are in energy the better the stabilization of the system through this conjugation is and the more important is the second resonance structure. But in the case at hand the $\pi$-overlap between the bromine orbital and the $\ce{C=C}$ bond is very weak (the reason for this you can find here) and so the conjugation is very weak, too. Thus, this resonance structure is not very important.


Valence bond theory and molecular orbital theory — how do they complement one another?

You are correct that Valence Bond Theory (VB) is still very active in the chemical bonding community along with MOs. I recommend this article/conversation by Roald Hoffmann about this topic (Acc. Chem. Res. 2003, 36 (10), 750–756). Succinctly, they both are capable of providing a description of bonding in molecules; however, VB theory is better at demonstrating the existence of localized bonds while MO theory is more generalizable to delocalized systems (and ultimately the foundation of predictive computational chemistry). If you have an organic molecule, for example, the VB picture is more helpful for understanding reactivity because these molecules (for some reason) usually behave according to the intuition accessible within such a local bonding framework. MO theory could give you a numerically exact answer, but the wavefunction is in general hard to interpret since it's delocalized across the entire molecule.

One topic that might be of interest to you in this field is MO Localization. This is the field of study which aims to take the MOs and the high accuracy wavefunction and transform them into orbitals that reflect chemistry as intuitively as VB theory. (One can take arbitrary rotations of MOs without changing the observables of the wavefunction). This problem is technically ill-defined, however, there are many approaches that have done this to great success. For example: Natural Bond Orbitals (NBOs), Adaptive Natural Density Partitioning (AdNDP), Intrinsic Bond Orbitals (IAOs), and many others. As the titles suggest, there is a belief that localized (VB-like) bonds are real things hidden in the true wavefunction.


What chemical process occurs when dead leaves make a mark on concrete?

The shape of the leaf is consistent with a type of maple tree leaf, and the brown coloring is characteristic of the water soluble tannins that would be extracted by water. For comparison, here are two photographs, one of tanic acid in water and one of a dried mixture of tannins. The photos were taken from this Wikipedia article.  Although the tannic acid in the photograph below was extracted from different parts of a redwood tree, it is a ubiquitous tannin found in most hardwood tree leaves and the color is characteristic of many such tannins.


Regioselectivity of epoxide hydrogenation using Pt versus Ni catalyst

First of all, I should clarify that hydrogenolysis of epoxides does not always proceed with such predictable regioselectivity simply based on a switch in catalyst. Often it depends on the substituents present, as well as their relative configuration (for example, cis/trans for 2,3-disubstituted epoxides).[1]

In Reaxys I find no records of the epoxide in the question (2,2,3-trimethyloxirane) being hydrogenated. However, I did manage to find one study in which the authors measured the rates of hydrogenolysis of propylene oxide over Pt and Ni, finding that Pt gives more 2-propanol and Ni gives more 1-propanol.[2]

On top of that, there are also a few more examples that show similar reversals in regioselectivity for Pd (not Pt) and Ni.[3,4]

Bartók and Notheisz have put forward an explanation for the difference in regioselectivity between Ni and Pd/Pt.[2] They suggested that the key step is an oxidative addition of the metal into the C–O bond in the epoxide. This is fairly logical, considering the weakness of the C–O bond in a strained three-membered ring. For a Pt catalyst, the regioselectivity is determined by steric effects and insertion into the less sterically hindered C–O bond is preferred:

On the other hand, the authors write that for Ni

  Because of the higher electron affinity of Ni, its insertion becomes ionic, formation of the surface species is controlled by electronic factors, and rupture of the C–O bond precedes formation of the Ni–C bond.

In modern parlance the way of expressing this is that the oxidative addition is asynchronous: the two new bonds are not being formed at exactly the same time. Because of the greater electron affinity of Ni compared to Pt, the Ni–O bond forms earlier than the Ni–C bond. If we were to draw a mechanism using curly arrows, it would be something like this: the flow of electrons depicted by the red arrow occurs earlier than the green arrow.

Consequently, there is a buildup of positive charge on the carbon undergoing bond breaking (hence "ionic"). This is not unlike the acid-catalysed opening of epoxides. The "electronic factors" that the authors refer to is simply the fact that the more substituted carbon is better able to carry this positive charge, which leads to the reverse in regioselectivity.

References

Murai, S.; Murai, T.; Kato, S. Reduction of Epoxides. In Comprehensive Organic Synthesis; Trost, B. M., Fleming, I., Eds.; Pergamon: Oxford, U.K., 1991; Vol. 8, pp 871–893. DOI: 10.1016/B978-0-08-052349-1.00249-3.
Bartók, M.; Notheisz, F. Stereochemistry of the hydrogenolysis of oxacycloalkanes on metal catalysts. J. Chem. Soc., Chem. Commun. 1980, 667–668. DOI: 10.1039/C39800000667.
Mitsui, S.; Kudo, Y.; Kobayashi, M. Stereospecific hydrogenolysis of benzyl-type alcohols. Tetrahedron 1969, 25 (9), 1921–1927. DOI: 10.1016/S0040-4020(01)82813-1.
Augustine, R. L. Organic Functional Group Hydrogenation. Catal. Rev.: Sci. Eng. 1976, 13 (1), 285–316. DOI: 10.1080/00087647608069940.


The term "active mass" is a historical term.

The concept of an equilibrium constant was developed by Cato Maximilian Guldberg and Peter Waage.  The Law of Mass Action has also been referred to as the Law of Guldberg and Waage, historically.

Guldberg and Waage defined the term "active mass" in the 1867 Études sur les affinitès chimiques.

  l'on peut choisir un volume arbitraire, par exemple un centimètre cube. Les
   quantités d'un corps qui se trouvent dans lcc du volume total se nomment la
  masse active du corps.

or roughly translated:

  One can choose an arbitrary volume, for example a cubic centimeter. The
  amounts of a body located in 1cc of total volume are called the active mass of the body.

And as stated in The Laws of Reaction Rates and
of Equilibrium J. Chem. Ed. 1956, 33 (4), p 178 :

  Guldberg and Waage in their study of chemical
  affinity formulated the law of the rate of chemical reactions
  as proportional to the "active mass" of the reagents.
  This they called the law of mass action and
  very clearly defined "active mass" as amount per unit
  volume. (emphasis added)

So, from a historical point of view, it is false that "active mass" is dimensionless.

However, in relating Gibbs energy to the equilibrium constant, it is necessary for the equilibrium constant to be dimensionless, because only the logarithm of a dimensionless quantity is logical.  For this reason, the text mentioned in the OP redefines "active mass" to be dimensionless.


Are we technically reducing electrodes/wires for the instant that an electron passes through them?

Reduction and oxidation are defined as gaining and donating electrons from the valence shell of an atom respectively.
Whereas in the case of electrons flowing in the electrodes and connecting wires the electrons are passing from the interatomic voids and not being released by the wires and the electrode.
(There will be some electrons which will interact with the conductor's atoms but what will happen is that they will get deflected. There will be a slim chance that a few electrons will replace an electron from the conductor's electron cloud but that doesn't change the situation as both the electrons are same in a classical sense. )
If your claim was true than the anode would have got some extra negative charge and the cathode extra positive which would have driven up the voltage higher than that expected from the cell alone.
In a similar way when we connect a wire it doesn't magically change the measured voltage.
Even when we perfom the same experiment with different inert electrodes and different kinds of wires the measured reading doesn't change much. (There are some changes in current flowing through the circuit due to the difference in resistances of different materials)


where can I find experimental compressibility factor values?

The NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) is a program that can calculate the thermodynamic and transport properties of some important fluids and their mixtures. It was developed by the National Institute of Standards and Technology (NIST) and is commercially available as NIST Standard Reference Database 23 here.

For example, the following selected results have been obtained with REFPROP Version 9.0. The number of digits is set to the default value (5 digits), which are not necessarily significant.

Methane
Temperature: 200.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99369
2nd Virial Coef.: −0.0065173 m³/kg
3rd Virial Coef.: 0.000015219 (m³/kg)²

Methane
Temperature: 287.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99802
2nd Virial Coef.: −0.0029520 m³/kg
3rd Virial Coef.: 0.000010232 (m³/kg)²

Carbon monoxide
Temperature: 200.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99739
2nd Virial Coef.: −0.0015501 m³/kg
3rd Virial Coef.: 0.0000035524 (m³/kg)²

Carbon monoxide
Temperature: 287.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99952
2nd Virial Coef.: −0.00040865 m³/kg
3rd Virial Coef.: 0.0000027781 (m³/kg)²

REFPROP can also be used to plot the results, for example for carbon monoxide:


Eigenvalues of Hessian matrix when the molecule is not at its equilibrium configuration

I am quite sure a much more thorough answer can be provided, but I will give a brief answer.

The answer is both yes and no. The answer is no in the sense that a harmonic analysis should always be performed at the true minimum on the potential energy surface (PES) because the eigenvalues have a physical interpretation as the force constant, and this must be a positive value. This is true because if one wishes to calculate a quadratic approximation to the PES, one will Taylor expand the potential about the minimum. In this case, the leading term is quadratic in the displacement and the coefficient is the second derivative with respect to the potential. This form of the potential is identical to the usual one for a harmonic oscillator, so the coefficient is a force constant and it is positive because the PES is strictly concave up at the minimum.

On the other hand, the answer is yes both approximately and exactly. First, I have seen people sometimes perform a harmonic analysis away from the minimum and simply trust the frequencies which they get out. This can be justified by noting that if the displacement is mostly along an eigenvector of the Hessian, then one can expect to basically still be at the minimum with respect to all other coordinates because these eigenvectors are mutually orthogonal. You will see this if you do a harmonic analysis slightly away from the minimum. That is, the modes whose eigenvectors do not describe a motion similar to the displacement you performed will have basically the same frequency as at the true minimum. However, if you make many displacements (i.e. randomly jiggle the atom positions) then you have no guarantees the answer you get out means anything.

On the other hand, these frequencies can be made rigorous by projecting out the eigenvectors which correspond to negative eigenvalues and re-diagonalizing the new Hessian. This guarantees all positive frequencies and can be interpreted physically as the instantaneous frequency when the system reaches a particular point on the PES. I think people have tried to use this type of procedure to determine linewidths of vibrational transitions. I am not sure how effective of a method it is. This kind of thing can at least give you some insight into how sensitive a particular frequency is to motion in other vibrational coordinates.


Why is the ionic product of water not divided by the concentration of water?

The short answer is: substances in their standard form are not included in equilibria expressions, and most commonly this includes pure liquids and solids.

The long answer is: the equilibrium expression relates all species to a standard activity (this is why they are all unitless). This question and answer detail that well.


You seem to approach the separation of alloys much like the separation of water-soluble salts, probably based on the fact that both involve the liquid/solid phase diagrams, which may even look alike. This is wrong; these systems are fundamentally different.

Say, you have a water solution of some well-soluble salt which you need to recrystallize. You cool it down, so the salt precipitates. You filter the crystals out of the solution, then you probably rinse them with something which would remove the traces of solution without damaging the crystals, or just let them dry.

Now suppose you have a metal alloy, maybe not even of the kind you mentioned above; let it have a simple eutectic diagram, like Bi-Cd.

You melt it, then carefully cool it down to a certain point, and end up with a mixture of some (supposedly) pure metal crystals with a still-liquid alloy. What good does it do? You can't really filter it. You can't rinse the crystals. You can't let them "dry". It is useless.

With metal alloys, your best bet is to dissolve both metals and separate them chemically, based on their different properties. In some relatively rare cases, you might be able to melt the alloy and perform chemical conversion of one component without affecting the other; examples include cupellation or Bessemer process referenced earlier in the comments. In yet more exotic cases, you might be lucky enough to chemically etch one component out of solid alloy; this is how Raney nickel is made. Unless you are that lucky, go back to the start of this paragraph.


What could be a high school experiment that involves changing a condition to get the highest yield of a certain product?

You could try the so-called Job's experiment. You prepare in advance 100 mL of two solutions of reactants which are supposed to have the same molar concentration. For example : CaCl2 0.1 M and Na2CO3 0.1 M. You take 9 test tubes in the same rack. Write 1 on the first, 2 on the second tube, etc. up to 9 on the last tube.

Now, with a plastic pipette, you put 1 mL of the CaCl2 solution in test tube 1, 2 mL CaCl2 in test tube 2, 3 mL of CaCl2 in test tube 3, etc. up to 10 mL in the tube 10. Then, you take another plastic pipette, and transfer 9 mL of NaCO3 solution in the tube 1, 8 mL of Na2CO3 in test tube 2, etc. up to 1 mL in tube 9. You mix thoroughly all tubes. You must see a white precipitate that settles in the bottom of  all tubes after a while. If you measure the height of the deposit, you will find that the highest amount is in the tube 5, if the stoichiometry is 1 + 1. If the mot important deposit is in another tube, it means that the stoichiometry is 1:2, oe 2:3, or any other ratio.

You may change and use any other combination of compounds. You will find that in the mixture CuSO4 plus NaOH, the stoichiometry is not 1:2, as you might suspect, as the precipitate is not Cu(OH)2, but 3Cu(OH)2·CuSO4.


1,3-Butadiene (Preferred IUPAC name: Buta-1,3-diene) is the organic compound, which is industrially important as a monomer in the production of synthetic rubber. It is also the simplest conjugated diene among organic compounds. Since all carbon atoms are $\mathrm{sp^2}$ hybridized, orbital overlap between double bonds is possible and, theoretically, two different conformer shows maximum overlapping, which are s-trans conformers with $\ce{C2-C3}$ dihedral angle of $180^\mathrm{o}$ and s-cis conformer with $\ce{C2-C3}$ dihedral angle of $0^\mathrm{o}$. However, due to steric hindrance, this s-cis conformer is approximately $\pu{16.5 kJ/mol}$ higher in energy than its s-trans counterpart, making this geometry a local energy maximum. Thus, in contrast to the s-trans geometry, which is global minimum, s-cis geometry is not a conformer (Wikipedia).

The double bonds of the s-cis geometry are twisted to give a dihedral angle of around $38^\mathrm{o}$, is considered to be a second conformer (called gauche-rotamer) that is around $\pu{12.0 kJ/mol}$ higher in energy than the s-trans conformer. The high level ab Initio calculations, utilizing coupled cluster theory with quasi-perturbative triple excitations and augmented quadruple $\zeta$ level basis sets, done by Feller and Craig in 2009 showed that the transition state separating the trans- and gauche-rotamers lies $\pu{26.8 kJ/mol}$ above the trans global minimum (Ref.1). These calculations further showed that the gauche rotamer lies $\pu{12.6 kJ/mol}$ above the trans-rotamer and the s-cis-form is a transition state $\pu{2.0 kJ/mol}$ higher than the gauche-rotamer. Wikipedia state that:

  This increased rotational barrier and strong overall preference for a near-planar geometry is evidence for a delocalized π system and a small degree of partial double bond character in the C-C single bond, in accord with resonance theory.

This is illustrated in the following scheme with the most stable conformer of 1,3-butadiene, which has the s-trans conformation (Ref.2):

Quantum chemical calculations on Π-Electron Delocalization in Butadiene has also done and published in 2006. The resultant bond lengths were illustrated in the following diagram and the total abstract of the paper was reproduced here for readers with interest:

  Abstract: Equilibrium structures have been determined for s-trans-1,3-butadiene and ethylene after adjusting the rotational constants obtained from rotational spectroscopy by vibration-rotation constants calculated from the results of quantum chemical calculations. For butadiene, the formal $\ce{C=C}$ bond length is $\pu{1.338 Å}$, and the formal $\ce{C−C}$ bond length is $\pu{1.454 Å}$. For ethylene, the $\ce{C=C}$ bond length is $\pu{1.3305 Å}$. These values appear to be good to $\pu{0.001 Å}$. It is shown for the first time that π-electron delocalization has the structural consequences of increasing the length of the formal double bond by $\pu{0.007 Å}$ and decreasing the length of the formal single bond by $\pu{0.016 Å}$. Comparisons are made with structures computed with several quantum chemical models. The MP2/cc-pVTZ results agree best with the new $r_e$ structure.

References:

D. Feller, N. C. Craig, “High Level ab Initio Energies and Structures for the Rotamers of 1,3-Butadiene,” J. Phys. Chem. A 2009, 113(8),  1601–1607 (DOI: 10.1021/jp8095709).
N. C. Craig, P. Groner, D. C. McKean, “Equilibrium Structures for Butadiene and Ethylene:  Compelling Evidence for Π-Electron Delocalization in Butadiene,” J. Phys. Chem. A 2006, 110(23), 7461–7469 (DOI: 10.1021/jp060695b).


Octet rule violation in Period 2 elements

Typically the octet rule becomes less likely to work, regardless of the period, when we move to a group where there are too few valence electrons to go around.  Prominent examples in Period 2 include compounds of beryllium in Group 2 (such as the fluoride, especially when gaseous or liquid) and boron in Group 13 (as in alkyl boranes like this one).  Carbon can also form species that are most stable with fewer electrons than the octet rule would predict, a "violation" most prominently seen in the cyclopropenyl cation, where another "rule" that is much more reliable when properly used comes into play.


What does the term 'by mass' mean with respect to different contexts?

In $\ce{H2O}$, the proportion of $\ce{H}$ counted by atoms is $\ce{2/3 = 66.7}$ %. In the same $\ce{H2O}$, the mass of $\ce{2 H}$ is $2$ u. As the mass of $\ce{1 O}$ is $16$ u, the proportion of H by mass in $\ce{H2O}$ is $2/18 = 11.1$%. This percentage $11.1$% is rather different from $66.7$%


What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD?

This isn't the most well-thought-out scenario, because if you dissolve $\ce{KOD}$ in $\ce{H2O}$ all the deuteriums will be lost immediately anyway.

$$\ce{KOD + H2O <=>> KOH + HOD}$$

Assuming that the amount of $\ce{KOD}$ is much less than the amount of $\ce{H2O}$ (after all $\ce{H2O}$ is the solvent) then the position of this equilibrium will lie very, very far to the right. So, at the end of the day, if you are using $\ce{H2O}$ as the solvent, the use of $\ce{KOD}$ won't have any effect. If you want to see any difference in deuteration, you have to at least use $\ce{KOD/D2O}$. However, there is a second problem with this:

If you dissolve an amine in $\ce{D2O}$, then all the amine hydrogens will be very rapidly replaced with deuteriums. It does not even matter mechanistically what the pathway is. Regardless of whether the initial product is $\ce{RNH2}$, $\ce{RNHD}$, or $\ce{RND2}$, as long as there is $\ce{D2O}$ in the mixture, all three will become $\ce{RND2}$. The same can be said of dissolving them in $\ce{H2O}$.

To answer your questions:

(1) No. In general, protonation can and will occur by any solvent molecule that happens to be close by. If you are running the reaction in $\ce{D2O}$ solvent, then there is much, much more $\ce{D2O}$ than there will ever be $\ce{H2O}$ produced. So, protonation by $\ce{H2O}$ is statistically close to impossible.

(2) None of us can answer that, as we cannot presume what your teacher was thinking of when they set the question.

(3) Depends on what your solvent is, see comments above on dissolving amines in $\ce{D2O}$ or $\ce{H2O}$.


How was the quantum three-body problem solved?

It never was solved, analytically (which means an exact mathematical solution)

It was solved approximately, using perturbation theories.  In the case of the helium atom, an estimate of the energy electron-electron repulsion is added to the calculation.  The appropriateness of the perturbation added is based on the comparison of the result to experimentally determined values.


Which substance(s) can solve gelatin (on film stock)?

Hot water, about 100 C, should do the trick. Also, this article has some information on the properties of gelatin layers in photographic films.

Properties and Stability of Gelatin Layers in Photographic Materials


Boiling point of ethanamide vs propanamide

The relatively high boiling points for such low molecular weight compounds suggests a primary effect from hydrogen bonding.

Using ethanamide as a reference point, propanamide has a higher molecular weight, which should raise the boiling point a little, but three relatively nonpolar carbons (plus hydrogens) vs two in ethanamide. A lower boiling point would then not be unexpected. Butanamide seems to support this, being a little higher in bp than propanamide by virtue of its higher mw, but still lower than ethanamide because of its lower overall polarity.

But to throw in some difficulty, formamide (methanamide) has a bp of only 210 °C, whereas its low content of nonpolar atoms suggest it should be much higher.

And then there are the melting points, which are difficult to relate to polarity: formamide 2 °C; ethanamide (acetamide) 80 °C; propanamide (propionamide) 80 °C; and butanamide (butyramide) 115 °C.

Formamide could be an outlier because of nitrogen flipping back and forth thru almost-planar nitrogen. Butanamide's high mp is a bit surprising. There could be packing factors in the solid that are favorable.

Maybe too many factors and too small overall effects to be able to make a lot of predictions, even tho we may be satisfied with some explanations after seeing the numbers.


You might view the relevant principle as follows: components whose activity changes during a reaction tend to put a stop to a reaction before they are completely consumed. Components with constant activity (independent of the amount of other reactants), on the other hand, can be completely consumed (exhausted): they can be limiting reagents, provided they are less abundant (stoichiometrically) than the other reagents. If a reagent with constant activity is exhausted before equilibrium is reached the reaction stops - it is then a limiting reagent. Reagents whose activities change during a reaction (because their concentration changes, for instance) are in principle not exhausted this way because their activity increases too much for the reaction to proceed any further. Reactions involving only reagents with variable activity tend to go to an equilibrium in which none of the species is completely exhausted. If reagents with constant activity are present in excess (such that their initial amount exceeds that at equilibrium), then they will not limit the extent of reaction. Note also that if the increase in activity with dilution of a reagent with variable activity is negligible, its concentration at equilibrium may be so low that it can be considered exhausted (and limiting).

A good example of the relevant principle is a mixture of two reactive solids or two immiscible liquids of constant activity that react at the interface between them. In either case, if the reaction is favorable ($Q\gt K_{eq}$) then it will proceed until one or both of the substances has been completely consumed (both being completely consumed if present in stoichiometric proportion). On the other hand, when a solid dissolves into a solvent, the activity of the solid remains constant and it will continue to dissolve provided the activity of the dissolved species does not increase above that of the solid. If it doesn't, then the solid is completely dissolved (exhausted).

If a reaction is already at equilibrium, adding more of a solid won't make a difference (provided there are no surface effects). But if you are not at equilibrium, it will if the amount present is limiting. A good example again is solubilization of a solid in a solvent. Added solid will continue dissolving until you reach the saturation point. Past this point the concentration of the compound in solution does not change any further. In addition, with saturation the activity of the compound in the solid and solution phases becomes equal.


Difficulty understanding redox in terms of hydrogen and oxygen transfer?

You are quite right, those rules of thumb are not always valid.

To be precise, the "adding hydrogen = reduction" rule only works for elements more electronegative than hydrogen, and the "adding oxygen = oxidation" rule only works for elements less electronegative than oxygen. For example, sodium is less electronegative than hydrogen, and as such adding hydrogen to sodium ($\ce{Na -> NaH}$) is in fact an oxidation of sodium from 0 to +1.

However, I wouldn't be so quick to limit their application to "organic chemistry". The elements that are less electronegative than hydrogen are just the metals, and the only element more electronegative than oxygen is fluorine.

Therefore, adding oxygen to chlorine is an oxidation, although the oxides or oxoanions of chlorine are classified under inorganic chemistry. Likewise, nearly the entirety of the p-block obeys these rules (except for Group 13). Think about germanium, arsenic, etc.. those are undoubtedly not "organic chemistry".


You didn't give a mechanism for your expected product, so I can only speculate as to how you intended to form the final product.

If you thought to use an Sn2 mechanism, I'd say that the reaction is highly improbable as the halide is attached to a trisubstituted carbon, i.e. a carbon with a lot steric hinderance, and therefore, it would be rather difficult for an Sn2 reaction to take place.

If you chose to use an Sn1 mechanism: with the first step, when iodine leaves, you would create a primary carbocation, which is unstable. Additionally, being that fluorine is an electron withdrawing group, the three fluorines would likely serve to further retract electrons from the already electron-poor carbon, and not serve to stabilize the primary carbocation. Therefore, I doubt that this reaction would occur as well.

In the textbook provided mechanism, the oxygen attacks the accesible iodine, which is next to a very psuedopositive carbon. And although this places a  negative charge  on the carbon atom, the electron-withdrawing fluorines serve to help stabilize the carbanion via their inductive properties. In the second step, the carbons rids itself of its negative charge, by taking a hydrogen atom from a nearby hypoiodous acid (pKa 11).

References:
McMurry, J. (2015). Organic chemistry: with biological applications. Stamford, CT: Cengage Learning.


What do final zeros to the right of decimal point indicate?

"$2.000$" does not mean $2.000 \pm 0.0005$.
"$2.000$" does not mean the interval [1.9995,2.0005].
"2.000" means that there is an unspecified amount of uncertainty in at least the last, and possibly the last two digits.
See the NIST Good Laboratory Practice for Rounding Expanded Uncertainties and Calibration Values:

Example 5
The correction for a weight is computed to be 285.41 mg and the uncertainty is 33.4875 mg.
First, round the uncertainty to two significant figures, that is 33 mg. Then, round the correction
to the same number of decimal places as the uncertainty statement, that is, 285 mg

So in other words, for people who follow the NIST standard of stating uncertainty to two significant digits:
2.000 can mean anything from 1.990-2.010 to 1.901-2.099
Others only express uncertainty using one significant digit, in which case:
2.000 can mean anything from 1.999-2.001 to 1.991-2.009


Does an increase in pressure always causes increase in molar concentration of each gaseous substance involved?

Your reasoning is correct as long as there is a decrease in volume. Note that pressure can also increase with temperature and the volume can stay constant. In this case, the molarity will not change. Molarity is not a very useful unit for expressing gas concentrations in equilibria because you have to quote pressure, volume, temperature and of course moles to quote molarity.


Realistic conditions for fuel container explosion

The topic of explosives in movie production is a specialty application as one does not generally employ classic explosives!
In movies, the aim is NOT to kill/injure anyone (especially, your famous lead actor). As such, one can select even a flameless explosive, which, for example, could include Ammonium nitrite ($\ce{NH4NO2}$). The latter is primarily a pressure explosive which, per a chain reaction, quickly releases a massive amount of gas. Such explosives find applications, for example, in blasting in coal mines. Apparently, mines can develop a potentially explosive mixture of methane and air, where a flame/spark could result in a massive secondary explosion!
Also, distinct from a battlefield setting, consider low brisance explosives, which do not produce lethal shockwaves.
In essence, your objective should differ significantly from the design/use of military-grade explosives. Instead, aim at much flash and dust, but less lethal. The sound level of the explosion can be addressed in the post-editing of the film.
To accomplish this, you need more research and the opinion of someone who actually specializes in this area.


In cracking, why does C form double bonds?

When we heat hydrocarbons to high temperatures (somewhat lower temperatures if we use a catalyst) they crack.  That means that carbon-carbon bonds are broken and smaller fragments are formed.  In the top part of the following drawing, I've picked n-heptane as an example.  You can see that there are 3

different bonds we can break and a number of different fragments will result.  Each of the fragments that result from breaking a C-C bond has an unpaired electron (there were two electrons in the bond we broke, each fragment retains one electron).  These fragments with an unpaired electron are called radicals.  The carbon with the unpaired electron does not have a stable octet configuration, therefor radicals are extremely reactive species and will react rapidly in a variety of ways to regain their stable octet configuration.

I've picked one of the possible radicals generated in the top drawing and show how it may react further in the bottom half of the figure.  Sometimes it will break a C-H bond and lose a hydrogen atom and form a stable compound with a double bond (an olefin), sometimes it will remove (abstract) a hydrogen from a hydrocarbon (heptane in our example) to generate a new alkane (pentane) and a new radical (perhaps a radical not formed in the initial cracking step).  Many of the radicals formed can rearrange to more stable radicals, this adds further complexity to the overall process.

So starting with just heptane we can get a large variety of saturated hydrocarbons and unsaturated hydrocarbons.  The situation becomes exponentially more complex if, as usual, we do have many more hydrocarbons present in addition to heptane.

The key here is that radicals are very reactive species - they want to stabilize by regaining the octet of electrons around the carbon atom that has the radical.  They can get that octet and become stable by adding a hydrogen atom (alkane formation) or breaking a C-H bond and eliminating a hydrogen atom to form an alkene.


What is the complete reaction for Cyanide ions reacting with Thiosulfate?

Source 1(If you consider $\ce{Na+}$ as counterions)

$$\ce{Na2S2O3 + NaCN -> Na2SO3 + NaSCN}$$

  Sodium thiosulfate reacts with sodium cyanide to produce sodium
  sulfite and sodium thiocyanate.

Source 2(If you consider $\ce{Ag+}$ and $\ce{Na+}$ as counterions for cyanide and thiosulfate ion respectively)

$$\ce{AgCN + 2Na2S2O3 → Na3[Ag(S2O3)2] + NaCN}$$

  Silver(I) cyanide react with sodium thiosulfate to produce sodium
  dithiosulfatoargentate(I) and sodium cyanide.

In this case, thiocyanate does not form, rather a complex forms.

Explanation of this reaction can be found by observing the related reactions given in the Wikipedia article of sodium thiosulfate.

  Silver halides, e.g., AgBr are typical components of photographic
  emulsions, dissolve upon treatment with aqueous thiosulfate:

  $$\ce{2S2O3^{2−} + AgBr → [Ag(S2O3)2]^{3−} + Br^{−}}$$

Since, $\ce{CN-}$  and $\ce{Br-}$ hold same charge,  you can easily replace it with silver bromide with silver cyanide  to get  cyanide ions as products.

Furthermore:

  Sodium thiosulfate is a component of an
  alternative lixiviant to cyanide for extraction of gold to form a
  strong soluble complex $\ce{[Au(S2O3)2]^{3−}}$. The advantage of this
  approach is that thiosulfate is essentially not toxic and that ore
  types that are refractory to gold
  cyanidation (e.g. carbonaceous or Carlin-type ores) can be leached by
  thiosulfate. Some problems  include the high consumption of
  thiosulfate, and the lack of a suitable recovery technique, since
  $\ce{[Au(S2O3)2]^{3−}}$ does not adsorb to activated carbon, which is
  the standard technique used in gold cyanidation to separate the gold
  complex from the ore slurry.

This process is also suitable for silver extraction using sodium thiosulfate and silver cyanide to form corresponding silver complex, $\ce{[Ag(S2O3)2]^{3-}}$. Just replace gold cyanide with silver cyanide and it will work pretty much well.


Why is fullerene 60 an insulator while graphite is a conductor?

To reiterate Ivan's comment fullerene is a bad conductor because that's what the measured properties produce as a result. The mechanism that makes it a bad conductor is that it has shorter range continuity than graphite. In graphite the carbon is made of sheets that can be as long as the sample. These sheets have fairly low conductivity, but when the electrons must jump between sheets to continue its path, this is when the conductivity value for the bulk material lowers. Extrapolate this to fullerene. If the fullerene molecule itself had high conductivity, it would still require the electrons to jump between molecules every few nanometers. This can produce a rather large resistivity for the bulk material much more so than observed in graphite.


Most acidic compound amongs the following unsaturated ketones?

Its not because of any ring expansion but because of formation of an aromatic compound.

Lets Start with (C) and (D). Both compounds after formation of carbanion are stabilized by extended conjugation:

But (A) forms an even more stable carbanion because of aromatic nature of:

Cyclopentadiene has a $p_{ka}$ of 16. Its high acidic strength is because of the above reason.

You can analyse (B) yourself. But it wont form a carbanion as stable as (A).


Confused Reading of Pressure on Vacuum Gauge

Pressure gauges typically measure pressure relative to ambient, so "0" on a gauge means equal to ambient (~ 1 bar absolute pressure depending on where you are). In some cases, the units have a "g" at the end, for example "psig" to indicate that the measure is "gauge pressure" rather than absolute, but that isn't always the case.


How many alkanes having a molecular weight of hundred, are chiral?

You are essentially correct.  Of the nine isomers of heptane, only two are chiral.  However, if you count each enantiomer as a separate molecule, then there are "four" chiral isomers:

(R)-3-methylhexane
(S)-3-methylhexane
(R)-2,3-dimethylpentane
(S)-2,3-dimethylpentane


Carbon number 4 is a pseudo chiral center because its chirality depends on the chirality of the second and sixth carbons. For example, if the second carbon is S and the sixth carbon is R, then carbon number 4 can be r or s, but if the chirality of the second and sixth carbons are the same, then carbon number 4 is not the chiral center, because it doesn't have 4 different groups around itself.


Bubbles consist of a gas trapped by a liquid, where the liquid has a surface tension high enough to encapsulate the gas. This gives bubbles three ingredients: the liquid, the gas, and surface tension.

The first ingredient, the liquid, is obviously the rainwater.

The second ingredient, the gas, is (also obviously) air. But what isn't obvious is how the gas becomes entrained in the liquid. Asphalt is porous, meaning the microstructure of asphalt consists of lots and lots of tiny random tubes through the solid material. Thanks to the capillary effect, water is sucked into those tiny tubes, and gas is forced out. This gas being forced out of the porous asphalt material and through the liquid on the surface creates bubbles.

Surface tension is the third ingredient, and controls the bubbles' shape, size, and strength. Surface tension can be enhanced by surfactants (big molecules which, when added to a liquid, agglomerate and increase the surface tension). The surface of a road accumulates lots of surfactants - mostly, oil. Oil consists of many long-chain hydrocarbons, which are perfect surfactants (most detergents are actually long-chain hydrocarbons that are polarized at one end). Surfactants tend to agglomerate (that's why, if you pour a little oil on top of water, it will tend to clump together as it spreads out over the surface of the water). Having oil act as a surfactant increases the surface tension of the rain water on the surface of the road, and makes any bubbles formed in the water stronger. That explains the large bubbles you saw.


Why is copper(II) carbonate not selected as an unknown compound for back titration? (Whereas another unknown carbonate is used instead?)

Copper is a transition metal. As such, its ion will be a Lewis acid. The effect will, as with iron, be stronger with increased charge--so Fe(III) salts of strong acids give acidic solutions, more so than Fe(II) salts of the same acids.

This is because lone pairs can be donated into the empty d or hybrid orbitals from water. The structure produced in this way tends to lose hydrogen ions. This means that the acidity in your experiment has two sources for hydrogen ions.


Is it true that under certain conditions, Mg can reduce SiO2?

Yes it is true.

Conditions are temperature in the 650 to 850 degrees C range.

There is a cool looking youtube video of the reaction, not that that makes it true.

For a more serious discussion see Production and Purification of Silicon by Magnesiothermic Reduction of Silica Fume

and Ordered Mesoporous Silicon through Magnesium Reduction of Polymer Templated Silica Thin Films


What's the chemical equation for the lithium iron phosphate battery

Close, the cathode reaction is:

$$ \ce{LiFePO4 -> FePO4 + Li^+ + e^-}$$

The anode is:

$$\ce{Li^+ + C6 + e^- -> LiC6}$$

The best description I can give is that the lithium ion intercalates into the graphite and "steals" some electron density from the six-membered conjugated ring.


Why is the amino acid cysteine classified as polar?

[...] we still put it in polar category

If you search for amino acid classification, you will find that there is no agreement on how to classify cysteine.
What you should know is that cysteine has different roles in proteins. In cysteine proteases, it acts as a nucleophile (and its surrounding often makes it more prone to deprotonation compared to a netural aqueous environment). Cysteines that form disulfide bonds often are at or near the surface of the protein; once the disulfide is formed, you have a quite hydrophobic group. Protonated cysteine is incapable of making conventional hydrogen bonds, and the electronegativity of carbon and sulfur are quite similar. This explains why methionine, the other sulfur-containing amino acid, is classified as hydrophobic.

[...] , why?

The classification of amino acids is sometimes used as proxy of the tendency of amino acid side chains to be located on the surface or in the interior of a protein. Similarly confusing, proline is sometimes classified as polar amino acid; however, its side chain is made exclusively of carbon and hydrogen atoms. If you study where proline is mostly located with a folded protein, you find it is on the surface. This is not because proline is polar but because it is a helix and strand "breaker", so it is often found in turns, which are mostly on the surface of the protein.
Also, you should know that there is no all-or-nothing classification of every amino acid side chain. Tyrosine, threonine, even lysine have hydrophobic parts while being capable of making hydrogen bonds. Often, the environment is just right, with hydrophobic parts interacting with other hydrophobic parts, and all hydrogen bond potential satisfied, either by interactions with other side chains, with main chain or with solvent.
Why is cysteine not classified in a more consistent manner?
Here is an excellent starting point to answer this question, starting with the abstract:

Cysteine (Cys) is an enigmatic amino acid residue. Although one of the least abundant, it often occurs in functional sites of proteins. Whereas free Cys is a polar amino acid, Cys in proteins is often buried and its classification on the hydrophobicity scale is ambiguous. We hypothesized that deviation of Cys residues from the properties of a free amino acid is due to their reactivity and addressed this possibility by examining Cys in large protein structure datasets. Compared to other amino acids, Cys was characterized by the most extreme conservation pattern, with the majority of Cys being either highly conserved or poorly conserved. In addition, clustering of Cys with another Cys residue was associated with high conservation, whereas exposure of Cys on protein surface with low conservation. Moreover, although clustered Cys behaved as polar residues, isolated Cys was the most buried residue of all, in disagreement with known physico-chemical properties of Cys. Thus, anomalous hydrophobic behavior and conservation pattern of Cys can be explained by elimination, during evolution, of isolated Cys from protein surface and clustering of other Cys residues. These findings indicate that Cys abundance is governed by Cys function in protein rather than by the sheer chemical and physical properties of the free amino acid, and suggest that high tendency of Cys to be functionally active can considerably limit its abundance on protein surface.


Regarding first of two questions:
Assuming duplex DNA (dsDNA), the molecules should repel each other and thus stay in solution at low ionic strength, such as in Tris/EDTA buffer. This is due to the abundance of negative - repulsive - charges. Increasing the ionic strength of the solution, such as high concentration of sodium acetate should neutralize the charges and allow the dsDNA fibers to sediment. In fact, a controlled sedimentation condition may help crystallize the nucleic acids.


Is it possible to have Zero order Elementary Reaction?

While zero order reactions are certainly possible and have considerable technological use, for example, in breathalysers used to determine alcohol content in ones breath, the question is what do you consider an 'elementary' reaction to be ?

In a pure sense the only 'elementary' reactions that occur are unimolecular and bimolecular ones (possibly also termolecular but exceptionally rare). A unimolecular reaction could be a bond dissociation or cis-trans isomerisation which can occur in an isolated molecule. Bimolecular reactions such as atom/molecule & molecule/molecule reactions are very common and strictly involve only two species.

So in this sense zero-order reactions have to have a more complex nature, such as involving an enzyme as in alcohol decomposition.

(Note also that the order of a reaction is measured experimentally as the sum of the powers to which reactants are raised in the rate expression, $rate= kA^aB^bC^c$ the order is $a+b+c$. Often it is the case that an apparently simple reaction such as $\ce{H2 + I2 }$ is only fully understood by a complex series of reactions, in this case the rate expression has fractional and reciprocal orders. The molecularity m refers to the coefficients in an elementary reaction, for example $m = a + b$ if the reaction actually occurs in a step $aA+bB \rightarrow$ so that m is a small positive number usually $1$ or $2$, for example $ m=1;\, A\rightarrow product$ or $ m=2;\, A+B\rightarrow product$)


Misconception? "That gas is heavier so it sinks."

Without significant mixing, diffusion takes a long time to mix gases.  Our understanding of entropy tells us that we will indeed finish with mixed layers, but that doesn't give us a time frame for that mixing, only an outcome.

Given a slow but steady production of a dense gas, layers absolutely will form due to density differences.  There are plenty of places where this can easily be seen, from your WWI example to the ability to pour carbon dioxide or hexane vapor into a container.  They're not stable over the long term, but they do exist for substantial periods of time, days in some cases.

If confined by an open container, huge areas can experience this effect, as at Lake Nyos.  Heck, even small temperature based density differences drive atmospheric layering that has major impacts on weather and pollution.

It's not an enormous effect, but you may be underestimating the strength of the forces involved here.  The same pressure differential that forces a helium balloon upward acts on an open volume of helium as well, and more strongly due to the missing mass of the balloon.  A body of a gas like $\ce{NO2}$ has about as much force pushing it downward (in air) than an equal volume of helium has pushing it upward.  Without outside mixing to cause it to disperse, it will happily form a short term layer until diffusion carries it away.


Wikipedia states

  The instability of $\ce{NI3}$ and $\ce{NI3 · NH3}$ can be attributed to the large steric strain caused by the three large iodine atoms being held in close proximity to each other around the relatively tiny nitrogen atom. This results in a very low activation energy for its decomposition, a reaction made even more favorable due to the great stability of $\ce{N2}$.

The steric strain between the atoms can be better visualized by the representations here.

The decomposition reaction is
$$\ce{2NI3\to N2 + 3I2}$$
Note that there is an increase in the number of moles (4 versus 2), which means that the reaction is favored by entropy, because there are more possible microstates.

More specifically, the enthalpy of formation is 154.4 kJ/mol, which is incredibly favorable for a reaction.

Reference: UC Davis ChemWiki


Can an atom bond with more than 8 other atoms?

Yes, there are coordination complexes of large elements which have coordination numbers greater than eight. Some examples are:

$\ce{[ReH9]^2-}$ with a tricapped trigonal prismatic structure. The nine hydride ligands are small enough to fit around the relatively large rhenium atom fairly easily. This ion can be isolated as a potassium salt $\ce{K2ReH9}$.

$\ce{[U(NO3)6]^2-}$ with an impressive coordination number of 12, consisting of six nitrate ligands each forming two coordinate bonds to the central uranium ion.


Specific Heat Ratio for a perfect gas mixture

In general for a perfect or ideal gas,
$$C_p=C_V + R'$$
(using your notation) where the heat capacities are molar quantities.
It follows that for a perfect gas mixture $(C_p)_\text{mix}=(C_V)_\text{mix} + R'$.


You are confused because the answer key is wrong.
The way to figure it out is to assign oxidation numbers to each species following the rules. Oxidation states are simply a convenient way to keep track of electrons without having to draw out Lewis structures, count electrons pairs and so on. The rules help make that happen.

Notice the position of the sign relative to when you indicate charge. It is not the same.

The correct answer is C though: $$\ce{CuO + CO -> Cu + CO2}$$
with oxidation states
$$\ce{\overset{(+2)}{Cu}\overset{(-2)}{O}
    + \overset{(+2)}{C}\overset{(-2)}{O}
   -> \overset{(0)}{Cu}
    + \overset{(+4)}{C}\overset{(2\times(-2))}{O2}}$$
Notice Cu went from +2 to 0. It gained 2 electrons to do that, thus it was reduced. What gave up the electrons? The other carbon, because it went from +2 to +4: its oxidation state increased by 2, thus it lost 2 electrons.

Recall OIL RIG: Oxidation is loss of electrons and reduction is gain of electrons.
Whatever is oxidized increases its oxidation state, and whatever is reduced decreases it.


What causes spray cans to get cooler when shaken?

TL;DR: Spray cans don't actually get colder when shaken.  However, shaking a can does increase heat conduction from your hand to the can, making it feel colder.

Humans don't actually sense external temperature directly; our thermoreceptors are located under the skin, and thus effectively measure the rate at which body heat is lost through the skin.  This is why, for example, touching a metal surface will feel noticeably colder than a wooden or plastic surface at the same ambient temperature.

When you pick up a spray can, it is typically at room temperature, i.e. significantly below body temperature.  The can itself is a thin metal cylinder, and thus conducts heat well, but has very little heat capacity in itself.  Thus, the rate of heat loss from your hands (and thus the sensation of coolness) mainly depends on how quickly whatever's on the inside of the can will absorb heat from the surface.

Now, a typical aerosol spray can contains a propellant (mixed with the actual payload fluid) which would be slightly above its boiling point (and thus in gaseous form) at room temperature and pressure, but is kept liquid by the elevated pressure inside the can.  As the spray valve is opened and the pressure reduced, some of the propellant fluid will boil, bringing the pressure back up until the liquid and gas phases are again at equilibrium.

What this means is that, in a half-empty spray can, the top half of the can will contain gaseous propellant (which, being a gas, will not absorb heat very effectively), while the bottom half will contain liquid propellant (mixed with the payload fluid) at just under its boiling point.  This liquid will absorb heat much better than the gas at the top, both because it's denser (and thus has higher heat conductance and capacity), and also because heating it will cause some of it to boil, converting part of the absorbed heat into its latent heat of vaporization.  (The boiling effect actually causes a half-empty spray can to act, to some extent, as a diode heat pipe.)

Thus, even without shaking the can, it's possible to observe that the bottom end of a half-empty spray can will feel cooler to your hand than the top end.

Now, when you hold the can and shake it, you cause the propellant liquid to slosh around inside, coating all of the inner surface of the can instead of just the bottom.  The sloshing also significantly increases the heat transfer from the can to the liquid, by causing the warmer liquid near the surface of the can to mix with the colder liquid in the interior.  (This is essentially the same reason as why moving air feels cooler on your skin than still air.)  Thus, a shaken can will generally feel even cooler to touch than either end of an unshaken can.


Write a Net Ionic Equation to Represent the Reaction of Aqueous Ammonia with Propionic acid

We know that ammonia is $\ce{NH3}$ and propanoic acid is $\ce{CH3CH2COOH}$.

Ammonia is a basic substance, and propanoic acid is... an acidic substance. More specifically, ammonia fits the definition of a Bronsted-Lowry base, so it will act as an $\ce{H+}$ accepter in solution.

Arrhenius acids are expected to increase the concentration of $\ce{H+}$ in solution, and propanoic acid is no exception. In water as propanoic acid dissociates...

$\ce{CH3CH2COOH ->[H2O] CH3CH2COO- + H+}$

This is where the $\ce{H+}$ comes (and is later bonded to the ammonia via)...

$\ce{NH3 + H+ -> NH4+}$

After those two steps occur, the products $\ce{NH4+ + CH3CH2COO-}$ are formed. Your ionic equation is as follows...

$\ce{NH3 + CH3CH2COOH -> NH4+ + CH3CH2COO-}$

There appear to be no spectator ions (ions that appear on both sides of the equation), so there is nothing to cancel out. As such, this is ALSO the net ionic equation. They both appear to be balanced.

So, yes, you had to do both. You were supposed to add an $\ce{H+}$ to the ammonia, but that happens if you dissociate the acid.

As for the equation with the calcium, carbonate ions sometimes create carbon dioxide and water (via carbonic acid readily decomposing, like in sodas).


The effect of reactant concentration on when equilibrium is established

I you have an equilibrium with significant amounts of $\ce{A}$ and $\ce{B}$ you need to consider the backwards reaction. If it is first order in $\ce{B}$, the rate law is

$$\frac{\mathrm d[\mathrm{A}]}{\mathrm dt}= -k_1 \cdot [\mathrm{A}] + k_2 \cdot [\mathrm{B}] \tag{1}$$

If the initial concentration of $\ce{A}$ is $\mathrm{[A]}_0$ and the initial concentration of $\ce{B}$ is $0$ the solution of the differential equation is

$$\frac{\mathrm{[A]}}{\mathrm{[A]}_0} = \frac{1}{k_1 + k_2} \cdot \left(k_2 + k_1 \cdot \mathrm e^{-\left(k_1 + k_2\right) t}\right) \tag{2}$$

The equilibrium concentration of $\ce{A}$ is obtained as the limit for $t \to \infty$:

$$\frac{\mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = \frac{k_2}{k_1 + k_2} \tag{3}$$

From $(2)$ and $(3)$ it follows that

$$\frac{\mathrm{[A]} - \mathrm{[A]_{eq}}}{\mathrm{[A]}_0} = k_1 \cdot \mathrm e^{-(k_1 + k_2) t} \tag{4}$$

or, logarithmically,

$$t = -\frac{1}{k_1 + k_2} \left(\ln\left(\mathrm{[A]} - \mathrm{[A]_{eq}}\right) - \ln\left(\mathrm{[A]}_0\right) - \ln\left(k_1\right)\right) \tag{6}$$

The criterion for equilibrium $\left|\mathrm{[A]} - \mathrm{[A]_{eq}}\right| \ll 1$ lets us neglect the term $\ln\left(\mathrm{[A]}_0\right)$ in equation $(6)$.

This shows that the time it takes to reach the equilibrium is independent of the initial concentration of $\ce{A}$.


What is the difference between hexanes and petroleum ether

Petroleum ether:
This is a mixture of a lot of different compounds (aliphatic hydrocarbons in this case) that as a mixture have a boiling point within a specified range. The contents are dealt with as having low amounts of aromatic compounds and high amounts of this and that, not one specific formula. It's like a blackbox to me and can be used for crystallisations that are usually easy to do.

Hexane:
In the synthesis laboratory I worked in, hexane was dealt with as we would deal with petroleum ether, since it was a mixture the many possible constitutional isomers and not pure n-hexane. We bought the technical grade hexane in our chemical store and purified it with a solvent purification system to have it dry and compatible for glove box use. Since they appeared so clean in our Schlenk flasks I thought in the beginning that it was pure n-hexane but a simple 1H NMR spectrum taught me otherwise.
The hexanes mixture was used for medium difficulty crystallisations.

Pentane:
The pentane we bought and purified, however, was n-pentane and was used for difficult crystallisations that were not easy to reproduce. Already among organometallic compounds there are examples in which obtaining crystals at all is an enormous struggle and having single crystal quality is simply a matter of hope.

Therefore, by going from petroleum ether to hexane to n-hexane (or n-pentane) is a matter of going from a wide mixture to a compound to a pure chemical.


if I have concentrated acetic acid (99%) what are the out put gases during electrolysis?

Upon electrolysis, organic acids will lose the carboxyl group and dimerize.
This is known as the Kolbe electrolysis:
https://en.wikipedia.org/wiki/Kolbe_electrolysis
In the case of electrolysis of acetic acid, ethane and $\ce{CO2}$ will be formed.


Is it possible to make a solution of zinc oxide in a high school chemistry lab?

Zinc oxide is not soluble. It does dissolve at very acidic or alkaline pH, but it is a chemical reaction, not true dissolution.
The role of zinc oxide ( or titanium dioxide ) is mainly to cause UV light dispersion, effectively prolonging a lot the ray paths to allow sufficient absorption.  Dissolving the sunscreen formula would create a thin suspension with very different UV properties, which would have tendency for sedimentation.
I see a sense rather in measurement of passing UV through a thin film of a cream with zinc oxide, prepared by some controlled way, e.g. controlling its mass per area.

You have to do some sunscreen solubility tests first.
The used formula would be an emulsion either "oil in water", either "water in oil". The solvent should dissolve well both.  I would perhaps dissolve sunscreen in acetone, performing filtration/centrifugation or sedimentation, whatever applicable, to get rid of eventual $\ce{ZnO}$ or $\ce{TiO2}$ particles.
Then I would evaporate the aceton and add defined $\ce{ZnO}$ amount.
Or, you can try your own sunscreen formula.


Given solubility of sodium carbonate, find mass of hydrate sodium carbonate dissolved?

To answer this kind of problem, it'd be better starting with elimination approach. It is given the solubility of sodium carbonate in $\pu{100 g}$ of water as $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$.  By that, you know, solubility of its hydrate should be more than $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$ because the mass of hydrate is including additional mass of water as well (as your calculation shows). Therefore, you can eliminate answers (a), (b), and (c) directly, because they are less than 25 and are not correct. Your calculations shows it must be greater than $\pu{67.5 g}$ (therefore, answer (d) is incorrect as well). It should be greater than $\pu{67.5 g}$, because you forgot to include additional amount of salt, which would dissolve in 10 equivalents of water in that added salt: Molar mass of $\ce{Na2CO3}$ is $\pu{106 gmol^{-1}}$ while Molar mass of $\ce{Na2CO3.10H2O}$ is $\pu{286 gmol^{-1}}$. This means, each $\pu{mol}$ of hydrate has $\pu{186 g}$ or $\pu{186 mL}$ of water. That is a lot and we need to consider that as well in this question. Now we have only two answers to be considered: (e) and (f).

To do the calculations, suppose $x\; \pu{g}$ of $\ce{Na2CO3.10H2O}$ would dissolve in $\pu{100 g}$ of water at $\pu{22 ^{\circ}C}$. Thus,
$$\text{amount of }\ce{Na2CO3}\text{ in the solution} = \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
$$\text{and, amount of }\ce{H2O}\text{ in the solution} = \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
From given data for anhydrous salt, we can conclude that upon saturation:
$$\text{The amount of }\ce{H2O}\text{ in the solution} = 4\times \text{amount of }\ce{Na2CO3}\text{ in the solution}$$

$$\therefore \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g} = 4 \times \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
Solve for $x$:
$$\left(4 \times 106 - 180\right)x = 28600 \; \text{and therfore, } x=\frac{28600}{244}=117.2$$
Therefore your answer is (f) ($\pu{117 g}$).


Esterfication of 2,2,2-triphenylethanoic acid

Disclaimer: The question has pretty much been answered in the comments. I'm just making it substantial.

The triphenylacetic acid is dehydrated in the presence of concentrated sulfuric acid to give an acylium ion, namely triphenylacetylium cation, which is stabilized by resonance.

Then, the triphenylacetylium cation undergoes decarbonylation (removal of carbon monoxide) to give triphenylmethyl cation, which is stabilized by the three phenyl groups.

The oxygen in methanol acts as a nucleophile to attack the carbocation, and is deprotonated to give our desired product triphenylmethyl methyl ether.

It is noted that this reaction should be done in a well-ventilated area because of the evolution of toxic carbon monoxide gas.


Is it possible to light a match by scratching it on hard surfaces, like they do in movies?

Most matches these days are safety matches: they're designed to need something more than ordinary levels of friction to ignite, by splitting the combustion materials between the match-head and the striking strip, i.e.  the brownish paint thing down the side of the matchbox or across the front of the matchbook.

Most "regular" matches now are safety matches. The brownish strip has red phosphorous in it. The match head has potassium chlorate $\ce{KClO3}$. Combine those two with friction, and you get fire.

There are still striking matches, with heads of phosphorus sesquisulfide $\ce{P4S3}$, still around. Though largely in countries where consumer protection is not yet well-developed.

With sufficient energy, you can still ignite a safety match without the striking strip. But you'll need more energy than the friction of scratching it on a wall or bit of stubble.

(PS Don't play with fire. Get a friendly neighbourhood chemist to play with fire on your behalf.)


you have both answers wrong.

(1) b - The volume of oxygen and nitrogen are both 1 liter at 1 bar, but they are put into a 2 liter container. So the pressure stays the same.

(2) c - The volume of oxygen and nitrogen are both 1 liter at 1 bar, but they are put into a 1 liter container. Since the volume is halved the pressure must double.


Le Chatelier's principle is not about pressure, it is about concentrations and temperature (Please also see the good article on Wikipedia for Le Chatelier's principle).

Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

Lets imagine that $\rm A,\,B,\,C,\,D$ are gases. So you wrote correctly that a pressure increase will cause the reaction equilibrium to shift to the "right" - more products $\rm C,\,D$ will be formed. When we increase the pressure in the system, it means that we reduce volume. We squeeze our system, and there is less volume for molecules to be in. They hit each other more frequently, and the probability of product formation increases. When you reduce the volume by increasing the pressure, you basically increase the concentration of reagents. The same amount of atoms or molecules occupy lesser volume.

I'll make a silly comparison. Let's take this reaction:
$$\rm Boy+Girl\longleftrightarrow Pair$$
If there is too little space on the dance-floor to dance on your own it could be a good idea to find a partner to reduce the volume you both occupied to feel much more comfortable. But when the other people start to leave the club you can unpair with your partner...

Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.

$P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$
$P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$
$P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$
$P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$

Our system is in equilibrium, so we can calculate Equilibrium constant: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$
$$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$

Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:

$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$
 where

$\upsilon'(A)=2\,mol-2x\,mol$
$\upsilon'(B)=3\,mol-3x\,mol$
$\upsilon'(C)=1\,mol+x\,mol$
$\upsilon'(D)=1\,mol+x\,mol$

Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$

$X'(A)=(2-2x)/(7-3x)$
$X'(B)=(3-3x)/(7-3x)$
$X'(C)=(1+x)/(7-3x)$
$X'(D)=(1+x)/(7-3x)$

Solving the eqution gives $x=0.325\,mol$, so now:

$\upsilon'(A)=1.35\,mol$
$\upsilon'(B)=2.03\,mol$
$\upsilon'(C)=1.33\,mol$
$\upsilon'(D)=1.33\,mol$

But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right.

$P'(A)=0.45\,atm$
$P'(B)=0.67\,atm$
$P'(C)=0.44\,atm$
$P'(D)=0.44\,atm$

Ok, but when we add inert gas, what happens:

$$\rm 2A+3B+inert\,gas \longleftrightarrow C+D+inert\,gas $$

It does not do anything because even though the pressure increased, the probabilty for 2 atoms $\rm A$ to hit 3 atoms $\rm B$ does not change. The partial pressures stay the same. Why is it so? It can be easily shown. For example we have this reaction:
$$\rm A_g+B_g\longleftrightarrow C_g$$

Lets say we put this reaction in a jar of 134.4 L volume at room temperature, we have 3 mol of $\rm A$, 2 mol $\rm B$, and one 1 mol $\rm C$. 6 mol of gases being put in 134.4 L at RT will give us 1 atm pressure.
$$6\,{\rm mol} \cdot 22.4\,{\rm L\cdot mol^{-1}}=134.4\,{\rm L}$$

The partial pressures are:

$ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$
$ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$
$P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$

Now we will add 1 mol of inert gas:
$$\rm A_g+B_g+inert\,gas\longleftrightarrow C_g+inert\,gas$$

Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same):

$P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$
$P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$
$P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$

As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.

Let's take this reaction:
$$\rm Boy+Girl+Cat\longleftrightarrow Pair+Cat$$
You cannot dance with a cat. Adding extra cats on the dance-floor will increase the pressure in the crowd, but will not increase the probability of formation of a dancing pair.

Let's imagine that $\rm A$ and $\rm B$ are liquids, and see what a pressure increase will cause. Usually it does not cause a lot, because it is very difficult to squeeze liquids. No matter how you try, the volume which molecules of liquid occupy stays the same. So increasing the pressure in reaction:

$$\rm A_\ell+B_\ell\longleftrightarrow C_\ell$$

will not do practically anything, because concentrations of reagents will not increase under the pressure.

When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as "product". If you remove this so called "product" (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.

If we increase temperature, it is like we added more of this so called "product" (heat) in the jar, so reaction is moved to the left.

If $\Delta H>0$, you need to think about heat not as "product", but as "reagent".

For better understanding try to "dive" into Equilibrium thermodynamics.


Is there a way to estimate the nutritional calorific value of a chemical

If carvone ($\ce{C10H14O1}$) is digested by the body using oxygen to generate carbon dioxide and water, you can estimate the heat of this chemical reaction:

$$\ce{C10H14O1 + aO2 -> xCO2 + yH2O}$$

After balancing the equation:

$$\ce{C10H14O1 + 13O2 -> 10CO2 + 7H2O}$$

The required enthalpy can be know from the formation heats of each molecule multiplied by the found factors.

A quick estimation

I just did this exercise using computational chemistry (B3LYP/6-31G(d) in Gaussian09) and found the following energies (in kcal/mol):

carvone: -291456.0217
oxygen: -94322.73333
carbon dioxide: -118326.709
water: -47947.30569

The change in energy of the reaction is then:

$$\begin{multline}10\times (\pu{-118327kcal/mol})+7\times (\pu{-47947kcal/mol})\\-1\times (\pu{-291456kcal/mol})-13\times (\pu{-94323kcal/mol}) = \pu{-1137 kcal/mol}\end{multline}$$

this is $\pu{-1137 kcal/mol}$ (exothermic, as expected).

Since molecular weight of carvone is $\pu{150.2 g/mol}$, we can convert to $\mathrm{kcal/g}$, the negative value indicates it releases heat:

$$\frac{\pu{-1137 kcal/mol}}{\pu{150.2 g/mol}} = \pu{-7.6 kcal/g}$$


Which d-block cation has the maximum hydration enthalpy?

Note that hydration energy is the energy released when water molecules are adsorbed on the surface of the cation. Hence the cation should have a high charge density, which means it should have a small size. So, the smaller, the better.

Here, $\ce{Ni}$ has the smallest atomic radii compared to all other options due to greater effective nuclear charge, so it's ionic radii would also follow the same trend. Hence, $\ce{Ni^2+}$ would have greater hydration enthalpy.


Is aqueous medium necessary for complex formation? if so, why?

It should be noted that ammonia is a mildly polar solvent and that (for instance) you can prepare $\ce{[Cu(NH3)6]^2+}$ by reaction with liquid ammonia1. The description of the geometry (two pages earlier) of $\ce{CuSO4.5H2O}$ (namely containing a square planar $\ce{[Cu(H2O)4]^2+}$ unit with two additional $\ce{[SO4]^2-}$ above and below the plane) suggests that ammine ligands could fulfill a similar role to water.

Bear in mind that passing ammonia gas over a solid will probably mean that if the reaction happens it will only happen near grain surfaces.

To answer your second suggestion, Housecroft and Sharpe1 identifies a reaction producing $\ce{[Ag(NH3)4]+}$ from $\ce{Ag2O}$ in liquid $\ce{NH3}$.

Hope this is informative.

[1] Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 2nd Edition; Pearson Prentice Hall, 2005; pp. 635–637, 693.


How to specify atomic carbon terms in the coupled and uncoupled representation?

I'll try to illustrate the procedure for two of the nine states that form the term symbol $^3P$. This term symbol has $S = 1$ and $L = 1$, which makes for three possible values of $M_S$ and three possible values of $M_L$, for a total of nine states. These states are "coupled" in both the spin and the spatial dimensions:

$$|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle, |1,0,1,+1\rangle, \cdots$$

The microstates in the table are states which are "uncoupled" in both dimensions:

$$|m_{s1},m_{s2},m_{l1},m_{l2}\rangle = |{+1/2}, {+1/2}, {+1}, {+1}\rangle, \cdots$$

To go from the term symbols ("doubly coupled" - don't use this terminology, I made it up) to the microstates ("doubly uncoupled"), we have to convert both spin and spatial components from coupled to uncoupled representations.

Thankfully, the spin and spatial components are separable, so we can write

$$|S,M_S,L,M_L\rangle = |S,M_S\rangle \cdot |L,M_L\rangle$$

Uncoupling the spin part is simple enough:

$$\begin{align}
|S=1,M_S=+1\rangle &= |\alpha(1)\alpha(2)\rangle \\
|S=1,M_S=0\rangle &= \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \\
|S=1,M_S=-1\rangle &= |\beta(1)\beta(2)\rangle \\
\end{align}$$

where $|\alpha\rangle$ denotes $m_s = +1/2$ and $|\beta\rangle$ denotes $m_s = -1/2$, as usual.

For the spatial parts, you will need to use the Clebsch–Gordan coefficients to expand $|L,M_L\rangle$ in terms of the "uncoupled representations" $|m_{l1},m_{l2}\rangle$. You already seem to understand this. For illustrative purposes we'll use

$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|m_{l1} = +1, m_{l2} = 0\rangle - |m_{l1} = 0, m_{l2} = +1\rangle)$$

and we'll simplify the notation by denoting $m_l = +1, 0, -1$ as $p^+, p^0, p^-$ respectively, so:

$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\}$$

Now, say we wish to find the microstate corresponding to $|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle$. What we need to do is to expand both spin and spatial components in terms of their respective uncoupled representations, then multiply them together again.

$$\begin{align}
&|S=1,M_S=+1\rangle \cdot |L = 1, M_L = +1\rangle \\
&\qquad = |\alpha(1)\alpha(2)\rangle \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\
&\qquad = \frac{1}{\sqrt 2} \left\{|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle - |p^0(1)\alpha(1)\,p^+(2)\alpha(2)\rangle\right\}
\end{align}$$

Notice how this is a (suitably antisymmetrised) state where one electron is in $p^+$ with spin up, and the other electron is in $p^0$ with spin up. This corresponds exactly to microstate 1.

Although the microstates are uncoupled states, it doesn't mean they aren't antisymmetrised. By virtue of quantum indistinguishability, they have to be antisymmetrised: that's why most of the microstates are linear combinations of "composite states" such as $|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle$ (such a state on its own is not physically permissible, whereas the microstates are physically permissible states).

For an uglier but more instructive case, let's look at $|S,M_S,L,M_L\rangle = |1,0,1,+1\rangle$.

$$\begin{align}
&|S=1,M_S=0\rangle \cdot |L = 1, M_L = +1\rangle \\
&\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\
&\qquad = \frac{1}{2} \left\{ |p^+(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^+(2)\alpha(2)\rangle + |p^+(1)\beta(1)p^0(2)\alpha(2)\rangle -  |p^0(1)\alpha(1)p^+(2)\beta(2)\rangle \right\}
\end{align}$$

Notice the first two terms in this expansion describe an (antisymmetrised) state where one spin-up electron is in $p^+$ and one spin-down electron is in $p^0$. This corresponds to microstate 7. Likewise, the second two terms correspond to microstate 9. All in all, we can write:

$$|S = 1, M_S = 0, L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|7\rangle + |9\rangle)$$

Using this procedure you can work your way through the rest of the states in $^3P$ (and indeed, $^1S$ and $^1D$ too). As a final example, let's look at one of the five $^1D$ states:

$$\begin{align}
&|S = 0, M_S = 0, L = 2, M_L = 0\rangle \\
&\qquad = |S = 0, M_S = 0\rangle \cdot |L = 2, M_L = 0\rangle \\
&\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle - |\beta(1)\alpha(2)\rangle\right\} \cdot \\
&\qquad\qquad\qquad \left[\frac{1}{\sqrt 6}\left\{|p^+(1)p^-(2)\rangle + |p^-(1)p^+(2)\rangle \right\} + \sqrt{\frac{2}{3}}|p^0(1)p^0(2)\rangle \right] \\
&\qquad = \frac{1}{\sqrt{12}}\left\{ |p^+(1)\alpha(1)p^-(2)\beta(2)\rangle - |p^-(1)\beta(1)p^+(2)\alpha(2)\rangle \right\} \\
&\qquad \qquad + \frac{1}{\sqrt{12}}\left\{ |p^-(1)\alpha(1)p^+(2)\beta(2)\rangle - |p^+(1)\beta(1)p^-(2)\alpha(2)\rangle \right\} \\
&\qquad \qquad + \frac{1}{\sqrt{3}}\left\{ |p^0(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^0(2)\alpha(2)\rangle \right\} \\
\end{align}$$

The first line of this expansion corresponds to $(\sqrt{1/6}|8\rangle)$; the second to $(\sqrt{1/6}|11\rangle)$; and the third to $\sqrt{2/3}|14\rangle$. So, your expansion was correct; however, be careful, because the uncoupled representations of the spatial wavefunctions

$$|m_{l_1} = +1, m_{l2} = -1\rangle$$

do not themselves correspond to the microstates. Instead, it is only by multiplying this with the spin wavefunction can you obtain a wavefunction that corresponds to the microstates.


NMR analysis regarding cis-1-isopropyl-3-phenylcyclohexane

The two methyl groups do not see the same average environment

The issue here is not whether the two methyl groups are freely rotating about the bond to the CH group: the issue is whether they see the same environment on average when the bond between the CH and the cyclohexyl ring rotates.

The easiest way to see that they don't is to build a model and play with it.

If you do you will see that their average environment can never be the same no matter how much rotation you do. Imagine rotating them so that the CH is facing the same way as the Ph with respect to the ring. The two methyls are different: one will be closer to the Ph than the other. Now rotate the CH-ring bond 180 degrees. The CH now points away to the opposite side of the ring to the Ph. But the environments of the methyl groups are not reversed: the other methyl is now closer to the Ph but is now on the opposite side of the ring so sees a different environment. So that rotation does not equalise the average environment for each methyl.

This is generally true in chiral compounds.


Sodium Perborate as a detergent ingredient

Connecting with water, sodium perborate releases hydrogen peroxide.
$$\ce{NaBO2*H2O2*3H2O->Na+ + BO2- + H2O2 + 3H2O}$$
$$\ce{H2O2->H+ + HOO-}$$
$$\ce{HOO- ->HO- + O}$$

Dissolution of sodium percarbonate in water at hand washing and the washing machine washing, leads to the formation of the perhydroxy anion. This anion is an effective bleach, but it is formed in sufficient quantities only at temperatures above 60 ° C.
From the perhydroxy anion obtain atomic oxygen, which has a strong oxidizing capacity.