Because $\ce{CO_2}$ isn't exactly what's being dissolved. When it encounters water, it immediately encounters the following potential reaction:

$$\ce{CO2~(g) + H2O~(l) <=> H2CO3~(aq)}$$

Carbon dioxide is non polar, yes, but carbonic acid readily dissolves (and dissociates, but I didn't care to show that) in water.

The sound we hear is because when we open the bottle, the equilibrium above shifts to the left, creating more carbon dioxide that is being released. Why does the equilibrium shift? It has to do with the pressure in the bottle.

When the bottle is closed, it's under a certain set pressure. When it's closed at the same temperature and pressure for a long time, the reaction above reaches an equilibrium state, where the forward and backward versions of the reaction are happening at the same rate. However, when the bottle is opened, the pressure inside the bottle decreases, as the gas in the bottle suddenly has a lot more space to expand into. A decrease in pressure favors the side of an equilibrium that has more gaseous molecules, which in this case means the carbon dioxide. Ergo, when you open the bottle, the gas is formed, escapes from solution, and rapidly expands outward, making the infamous popping noise.


What facts are being used in this atomic structure question?

The only correct answer is #1.

In actuality, the ground state of the hydrogen atom has zero angular momentum.

Regarding #2 and #4, there is no preferred plane in which one point should orbit another point.  By symmetry, they are all equivalent.

Regarding #3, there is no angular momentum in the ground state, and in a state whether there is angular momentum, it doesn't need to be in the same direction as the electron spin angular momentum.


In what approximations, if any, can we assume a constant temperature in a Bomb calorimeter?

You have more or less answered your own question. Enthalpy is a function of state, so you can write an expression of the form $\Delta {H} = {H_f}-{H_i}$, and we know that this depends only on the initial and final states, and not the path taken between them.

You also have the relation, $H = U + PV$, using this and the previous equation, and the fact that $U$ is also a state function (so we can write $\Delta {U} = {U_f}-{U_i}$)

$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i \tag{1}$

One comment, in the link that you posted equation (6) namely,

$\Delta {H} = \Delta {U} + \Delta(PV) = \Delta {U} + P\Delta V + V\Delta P $ seems absurd, for finite, non-infinitesimal differences.

For infinitesimal changes, i.e if we write the differential

$\mathrm{d}H = \mathrm{d}U + \mathrm{d}(PV) = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P $ this holds true; (chain rule from mathematics)

Anyway, for solids in liquids, the $PV$ term can be neglected on account of their small molar volume. I guess this would be one approximation/assumption you are interested in.

Second, assumption/approximation would be that a gas behaves ideally, and thus follows the equation of state: $PV = nRT$

Plugging it into (1);

$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i = \Delta {U} + R(n_fT_f-n_iT_i) $

This expression is true for finite differences. Now, the finally approximation is indeed $T_f \approx T_i \equiv T$, i.e $\delta T$ is "sufficiently" small, then indeed we can write

$\Delta {H} = \Delta {U}+ RT\Delta n $

Note however, that we don't necessarily require $\Delta T \to 0$ in a strict, mathematical sense. What can be considered "sufficiently" small depends on your system, your problem, and the kind of solution you need.
What I mean to say is in some cases/applications $-4.96\ \mathrm{kJ}$ is just as good as $-4.88\ \mathrm{kJ}$

There are many other simplifying assumptions in calorimetry when it is taught such as heat-capacities are independent of temperature for example; this only, really true over narrow ranges of temperature and in general, heat capacity is in fact a function of temperature.


Why do different substances have different boiling points?

The boiling point of a liquid depends on the intermolecular forces present between the atoms or molecules in the liquid since you must disrupt those forces to change from a liquid to a gas. The stronger the intermolecular forces, the higher the boiling point.

Two oxygen molecules are attracted to each other through London dispersion forces (induced temporary dipoles between the molecules) while water molecules are attracted to each other by hydrogen bonding (attraction of the + dipole on H in one molecule to the – dipole on an oxygen in an adjacent molecule) that is relatively strong. (Hydrogen bonding is an important intermolecular force for molecules where H is directly covalently bonded to F, O or N, which are quite electronegative and thus form bond with H with a relatively strong dipole.) London dispersion forces become more important for atoms and molecules with more electrons. Dipole–dipole attractions are also important in some molecules.


Potential wells and symmetry in absorption spectroscopy

This is a bit tricky to explain, the way I imagine it is this: In a totally symmetric mode where, say, each bond extends in phase the potential minimum need not be the same in the excited state and in the ground state but it might be so 'accidentally'. All the forces are balanced. In a non-totally symmetric mode, one set of bonds stretches as the others compress so the potential has to be the same in both states otherwise there would seem to be an internal force being generated that would move the molecule, which is not possible.

edit:

Other points are

(a) The spectrum reflects the change in geometry between states and thus is that of totally symmetric vibrations only, but see (d) below.

(b) the excited and ground state have to have the same symmetry,

(c) the asymmetric stretch is a symmetric function of normal coordinates, which is what your lower picture shows.

(d) the non-totally symmetric excited state mode need not have the same frequency  as in the ground state, i.e. potential can be wider or narrower. In this case there are selection rules based on the wavefunction's symmetry; both must be the of the same symmetry in the FC factor. The symmetry of vibrational wavefunctions for non-totally sym. modes change with vibrational quantum numbers and it works out that these type of transitions can have $0 \pm2,\pm4..$ transitions only.


Rules to identify the most stable resonance structures

You will encounter situations such as these when the two rules seem to contradict each other, or they confuse you.
However, when this happens, you must use another criteria to determine the best resonance structure. Specifically, you must take electronegativity into account.

What this means is that the most electronegative element should have the lowest formal charge, and the most electropositive element should have the highest formal charge.
This is because the electronegativity "rules" explained above are what are likely to happen (that is how thiocyanate is likely to appear). This makes sense because the higher EN and element has, the more likely it is to "hog" the electrons and have a lower formal charge.
When you use these rules to evaluate the resonance structures of the $\ce{CNS^{-}}$ ion, you can figure out why the first resonance structure is preferable.


Maximum (Z) to (E) ratio in the photoisomerization of azobenzene?

Your question is not completely clear to me, so I am answering based on thinking that you are asking about why, after photoisomerization, the system can relax with roughly equal probability to either the E or Z minimum.

Strangely I know something about the photochemistry of azobenzene. One overarching theme is that azobenzene molecules undergo very complicated isomerization processes which are sensitive both to how you substitute them and to the solvent in which the experiment is done. Additionally, the photoisomerization and thermal isomerization processes don't necessarily share anything in common. There is a pretty in-depth interesting literature on different types of azobenzene molecules discussing the relative roles of inversion and rotation relaxation pathways, solvent effects, and so on... I just bring this up in case you are interested to read more.

The case you describe, I think has a relatively straightforward explanation actually. In ref. [1], a computational study is done on azobenzene to find which pathway, rotation or inversion, is dominanant upon photoisomerization of azobenzene. To clarify, rotation means rotation about the CNNC dihedral, and inversion means moving one of the rings along the CNC angle.

The paper shows that there is a conical intersection along the rotation pathway which is only 2 kcal/mol above the $\ce{S1}$ excited state minimum. Thus, the molecule will follow this conical intersection upon relaxation. Because the molecule is already above both states, and the molecule can essentially end up going toward either the E or Z minimum from this conical intersection, to first order, the two should end up equally populated on relaxation. Of course, from there, one would expect to move back to a Boltzmann distribution of the two states thermally.

[1]: Cembran, A., Bernardi, F., Garavelli, M., Gagliardi, L., & Orlandi, G. (2004). On the mechanism of the cis− trans isomerization in the lowest electronic states of azobenzene: S0, S1, and T1. Journal of the American Chemical Society, 126(10), 3234-3243.


Henry's Law is mathematically expressed by $$p=k_Hc$$ where $p$ is the partial pressure above the liquid, $k_H$ is the Henry constant for the specific gas at the given temperature, and $c$ is the concentration of gas in liquid. If you set up one equation for each gas and divide the two, you get $$\frac{p_{Ar}}{p_{Kr}}=\frac{k_{H, Ar}}{k_{H, Kr}}\frac{c_{Ar}}{c_{Kr}}.$$ You are given both the ratio of partial pressures and the ratio of the Henry's Law constants; the ration of concentrations should be easily calculated from that point.


You should know that the molecularity is not relevant in your case and has no significance for multi-step reactions, it is just applicable to elementary reactions. The relevant parameter for multi-step reactions is the, reaction order wich is originally an experimental parameter  indirectly related to the molecularity of all elementary reactions that makeup the reaction mechanism.
So for saponification the proposed mechanism includes 3 steps, the last one of them is irreversible and controls the reaction. The order of the reaction is considered 2, the molecularity for the step that you have mentioned is also 2 but it is just one step in the mechanism.
Also, there are two types of mechanisms through which this reaction takes place: a base catalyzed mechanism (with simple base or alkoxides) and an acid catalyzed mechanism.


The problem lies in the electron count. Each bond contains 2 electrons and the pair is counted twice, once for being around each atom in the bond. And don't forget the extra electron to make the ion.

The + and - are from the point of view of the atom nucleus: e.g., consider the nitrogen to share 2 electrons with the oxygen on the right, to give 2 to the oxygen above, and to give 1 to the oxygen on the left. The other electron in that bond is one that makes the anion negative overall.

Then nitrogen gains its octet of 8 electrons, but formally only shares 50% of those electrons, so it started out with 5 but only controls 4 - it now has a positive charge. The right oxygen shares, so no charge. The top oxygen got 2 from nitrogen, so is negative, and the left oxygen got 1 from nitrogen and one from the metal of which this is the anionic part. And then, draw up the other two diagrams that have the "neutral oxygen" in the other positions, add them all and average them. Then each oxygen winds up with a 2/3 negative charge (but the nitrogen is still +1), while the anion is -1 and planar.


Which reactant to use to calculate yield?

You have the right thought process - you need to find which is the limiting reagent.
Your calculation is correct, $N_2$ is limiting, so it can be used to calculate yield. Looks like we weren't really needed!


The problem can be solved using retrosythesis.

You look at the product and imagine how that was made.

It's like looking at Lego construction, trying to identify the pieces - or looking at a pair of jeans, trying to identify how it was stitched together.

Then you go back and do the same for the previous building blocks.

The final product apparently is an imine. I'm sure we have seen that before among one of the exercises you worked on - in that case it was an intramolecular reaction.

Imines are usually made from carbonyl compounds and primary amines under acidic conditions. This one is made from propanal and ethylamine.

Looks good here, I see $\ce{H2NCH2CH3,\ H3O+(cat)}$ among your reagents.

What about the aldehyde then? The number of carbon atoms in the aldehyde and the amide (propionamide is the starting material) are the same, just the (formal) oxidation of the carbonyl/carboxyl group are different.

Let's think about a direct reduction of the amide with $\ce{LiAlH4}$:

OK, that isn't it. Just two steps would not have been enough anyway ;-)

But we do remember that

esters can be reduced to alkanols with $\ce{LiAlH4}$
alkanols can be oxidized to aldehydes with PCC (pyridiniumchlorochromate)

All this is among the list of reagents. Nice, we know to make the propanal from a propionic ester!

How to turn propionamide into a propionic ester then?

Direct reaction with methanol under acidic conditions does not work, but after alkylation of the amide with an alkyl halide, we should be ok.

Lets's summarize our five steps then:


Computing reaction mechanisms using solvation effects

1) Yes, geometries must be calculated both in the gas phase and in the solvent. To analize your results in the solvent you need something to compare that's why the gas phase calculation is always needed and often is considered "the only source of truth".

Frequencies must be calculated both in the solvent and in the gas phase as well, on the respective converged geometries.

2) I never worked with explicit solvent but I feel that the use cases of two methods are not overlapping. When performing calculations on a multi step reaction usually you don't have enough computational resources to calculate all the solvent molecules needed. This is why people use implicit solvent calculations. If you want to benchmark your solvent calculations, compare it with the gas phase ones and evaluate wether there is some improvement in the replication of experimental data.


Effective attenuation coefficient for polychromatic light

The intensity transmitted at  wavelength $a$ where the extinction coefficient is $\epsilon_a$ is $I_a=I_{0a}\exp(-\epsilon_a L C)$ with path length $L$ and concentration $C$, and similarly for a second wavelength $b$. If, for clarity, we let $x=\epsilon l C $ then then $I_a=I_{0a}\exp(-x_a)$ and $I_b=I_{0a}\exp(-x_b)$. The total transmitted light is $I_t=I_a+I_b$ and the total initial amount $I_{0t}=I_{0a}+I_{0b}$. Thus the total transmittance ratio is

$$ \frac{I_t}{I_{0t}}=\frac{I_{0a}e^{-x_a}+I_{0b}e^{-x_b}}{I_{0a}+I_{0b}}$$

You want to make this equal to $I_{0ab}e^{-x_{ab}}$  but there seems no systematic way to do this and this confirms the fact that the simple  Beer-Lambert law only applies for monochromatic light.


What is the relationship between pressure and stoichiometry ratios

The actual reaction being discussed is $$\ce{3O2 -> 2O3}$$

We also know from the gas law that $$\ce{PV=nRT}$$

In the current experiment the only things changing are the pressure and the number of moles; the volume and temperature remaining constant.  In this case the gas law simplifies to $$\ce{\Delta P = \Delta n}$$

Therefor, we know the change in moles for the overall reaction (-3+2=-1) is $\ce{\Delta n}$ and by the above stoichiometry the change in the moles of oxygen will be three times this amount or $\ce{3\Delta n}$ or $\ce{3  \Delta P}$ or $\ce{3x}$.


Do cyclic oxygen molecules exist and their theoretical stability?

Unlike the likes of carbon, oxygen-oxygen single bonds are highly unstable. Because oxygen is such an electronegative compound, it can only become satisfied with bonds where it hoards a majority of the electrons. Because of this, compounds such as peroxides ($\ce{RO-OR}$) will readily decompose into two radicals. This is why hydrogen peroxide is used as a radical initiator for chemical reactions. Since a single $\ce{O-O}$ bond is so unstable, it would be extremely unlikely for a larger chain to exist. Peroxides by themselves are highly unstable. Not only do they decompose into radicals, but they also rapidly decompose into water. Peroxides have a single $\ce{O-O}$ bond, and are somewhat stabilized by donating groups on either side of the peroxide. Now, imagine a chain that looks like this: $\ce{O-O-O}$. The central oxygen is not only bonded one inductively withdrawing oxygen as in peroxides, but two of them. In addition, it is not stabilized by electron donation from any source. Despite this, it is predicted that small amounts of cyclic ozone might exist in the atmosphere. Cyclic ozone is so much less stable than regular ozone for the reasons I listed above, but it should exist. Any larger oxygen ring will most likely not form.

Carbon does not have this same problem. Since carbon has a mid range electro-negativity, it is stable forming bonds with itself. In addition, carbon can form four covalent bonds, normally two of which are with hydrogens (as in a cycloalkane). Carbon is perfectly stable with this electron density, leading to the stability of cyclic alkanes.


Is it possible to make synthetic oil without fossil fuel?

Yes, it is being theorized and it's being done:

Example A:
https://www.researchgate.net/publication/264810861_Engineering_cyanobacteria_for_production_of_a_cyclic_hydrocarbon_from_CO2_and_H2O

Example B:
https://www.sciencedirect.com/science/article/pii/S2211926416304982

Example B is more theoretical with economic considerations, and example A is more methodical. In these papers, we see cyanobacteria being used to make the aromatic hydrocarbon limonene with the limonene synthase gene derived from a plant.

As for can we make synthetic oil that is chemically proportional to fossil fuel, Probably not ... but by adjusting the ratios of aromatic hydrocarbons, we can measure some useful properties like the heat value of the fuel for which limonene has a heat value very close diesel fuel and the density is close as well as you can see if you read the aforementioned Life Cycle Analysis paper. If you want to explore the chemical composition of fossil fuel then you need to do some spectroscopy work.

But fossil fuel isn't perfect. There is residual incombustible material, and perhaps by adjusting the proportion of combustible aromatic hydrocarbons in fossil fuel maybe we can get more energy from fossil fuel. This is also a function of the funding that these research projects get and also industrial support which is also a function of money ... which is why you need high yield in these reactions, hence genetically engineered cyanobacteria.


What bases are fit for human consumption and can be used in foods?

Unfortunately Googling for "alkaline food" these days is a bad idea; results are dominated by nonsense sites espousing the "alkaline diet" nonsense / woo.

There aren't many foods that are naturally alkaline.  The two exceptions I know of are:

baking soda
egg white

Sometimes when I'm cooking and accidentally add too much acid (e.g. lemon juice or vinegar) to a broth or sauce, I can partially correct the mistake by adding small amounts of baking soda.  This converts the excess acid to the sodium salt and $\ce{CO2}$ is evolved.  For the example of vinegar (acetic acid):

$\ce{CH3COOH + NaHCO3 -> CO2(g) + CH3COONa + H2O}$

So sodium acetate salt is left behind, which limits the degree to which you can correct overaddition of acid.  (You don't want your food to be too salty.)$%edit$


Stability of carbocation directly attached to a halogen atom

some people say that its because of octet completion of the carbocation
and the NS reactivity that you mentioned is due to the -I effect of chlorine which makes it the most reactive and not the least reactive


Are difluorine and hydrogen chloride considered isoelectronic?

If we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have the same number of electrons, they are actually isoelectronic. Please see "Ultraviolet Spectroscopy And Uv Lasers"
 edited by Prabhakar Misra, Mark A. Dubinskii, page 35 (available in google books)

But if we take this definition of isoelectronicity: Two or more molecular entities (atoms, molecules, or ions) are described as being isoelectronic with each other if they have a similar electron configuration and the same structure, they are not.


Does aromaticity require a co-planar pi system?

The $\pi$ system is not obliged to be planar at all; the only limitation is to keep the angular twist between the adjacent atoms low enough so they still can be conjugated. (Biphenyl violates that, so it is out of luck.) The whole structure can be pretty much anything. It can be slightly out-of-plane, like corannulene or circulene; it may have cylindrical shape (see the example below), in which case the Hückel's rule still applies, or it may be all twisted to the topology of a Möbius strip, in which case the said rule is reversed.

Fullerenes are not aromatic, unless you stretch the definition of aromaticity to utter pointlessness. Some of their derivatives are, though. $\ce{C60Cl30}$ comes to mind: its 30 remaining $sp^2$ carbons are grouped in two 6-membered benzene-like rings and one 18-membered belt, which (judging by its equalized bond lengths) is an example of a cylindrical aromatic system.


How to name alicyclic compound with hydroxymethyl substituent?

The principal characteristic group of the compound that is given in the question is $\ce{-OH}$, which is expressed as suffix (‘-ol’). Thus, the senior parent structure of this compound is methanol. The problem in this case is that the compound contains two separate methanol structures.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a structure with two or more identical senior parent structures that are linked by di- or polyvalent groups can be named by multiplicative nomenclature if certain conditions are met (otherwise the usual substitutive nomenclature is used).

  P-45.1 MULTIPLICATION OF IDENTICAL SENIOR PARENT STRUCTURES

  P-45.1.1 Multiplicative nomenclature is senior to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical senior parent structures, other than alkanes, in the name of the parent structure (see P-51.3.1). In most cases multiplicative names are shorter than regular substitutive names. A preferred IUPAC name is generated by multiplicative nomenclature when the following criteria for its use are met (see P-51.2.3).
  (1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical; and
  (2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
  (3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

  When these conditions are not met, substitutive nomenclature generates preferred IUPAC names.

The given structure fulfils these requirements; therefore, the preferred IUPAC name is generated by multiplicative nomenclature as follows.

  P-15.3.1.3 Multiplicative name formation
  Multiplicative names are formed in accordance with the number of occurrences of identical structural units as defined in P-15.3.1.1 and the relationship of the linking multiplicative substituent group to the identical structural units.
  When a compound contains identical structural units as defined in P-15.3.1.1 linked by a symmetrical simple, compound, complex, or concatenated multiplicative group (a di- or polyvalent substituent group), it is named by stating successively:

  (a) the locants for the positions of substitution of the linking multiplicative substituent atom or group to the identical parent structural unit (the locant 1 is omitted when alone in the name of a mononuclear parent hydride);

Here: omitted (The locant 1 is the only available locant in methanol.)

  (b) the name of the linking multiplicative substituent atom or group;

Here: cyclohexane-1,2-diyl

  (c) the numerical prefix ‘di’, ‘tri’, etc.; and/or ‘bis-‘, ‘tris-‘, etc., with no elision of the final vowel before the name of the identical parent structural unit;

Here: di

  (d) the name of one of the identical structural units including the principal characteristic group and substituents, if any, enclosed in appropriate enclosing marks (see P-16.5).

Here: methanol

Therefore, the preferred IUPAC name is (cyclohexane-1,2-diyl)dimethanol.

By way of comparison, the (correct, but not preferred) name based on usual substitutive nomenclature is [2-(hydroxymethyl)cyclohexyl]methanol


Electrolysis of molten nitrates, sulphates and carbonates

A common theme is disproportionation of the anion, and potential reactivity with the electrodes.

Nitrates

Electrochemical behavior of nickel electrodes in molten potassium nitrate with increased basicity, Electrochimica Acta, Vol. 36, No. 3/4, pp. 65S-660, 1991:

  $\text{Ni}$ electrodes in a $\ce{KNO_3}$ melt [...] some ionic species of nickelite ($\ce{NiO_2^{2-}}$) and nickelate ($\ce{NiO_2^{-}}$) type

Voltammetric investigations of the oxygen electrochemical systems in molten sodium nitrate at 420°C, Electrochimica Acta, Vol. 33, No. 4, pp. 545-549, 1988:

  the potential range is limited anodically by the electrochemical oxidation of nitrate ions into nitrogen dioxide and oxygen: $\ce{NO3-} \to \ce{NO2} + \frac{1}{2}\ce{O2} + \ce{e-}$ and cathodically by the electrochemical reduction of nitrate ions into nitrite ions: $\ce{NO3-} + \ce{2e-} \to \ce{NO2-} + \ce{O^2-}$

Carbonates

Electrochemical formation of carbon nano-powders with various porosities in molten alkali carbonates, Electrochimica Acta, Vol. 54, pp. 4566–4573, 2009:

  reduction of carbonate ions: $\ce{CO3} + \ce{4e-} \to \ce{C} + \ce{3O^2-}$ [and reduction of metal ions, along with oxidation of carbonate ions:] $\ce{2CO3^2-} \to \ce{2CO2} + \ce{O2} + \ce{4e-}$

Sulfates

Polarization measurements on solid platinum-molten sodium sulphate-sodium chloride interfaces, EIectrochimica Acta, Vol. 23, pp. 381-388, 1978:

  the limiting reactions [...] are suggested to be the $\ce{SO3}$ reduction to $\ce{S^2-}$ and $\ce{O2-}$ and the $\ce{SO4^2-}$ oxidation to $\ce{SO2}$ and $\ce{O2}$


How is Langmuir adsorption isotherm different from Freundlich's?

There is a fundamental difference between both models.

Langmuir's model was a theoretical construct, while the Freundlich isotherm is empirical.

In the Langmuir model, it is assumed that at maximum coverage, there is only a monomolecular layer on the surface. This means that there is no stacking of adsorbed molecules. The Freundlich isotherm does not have this restriction.


There are several reasons why $\mathrm{^{13}C}$ experiments are far less sensitive than proton experiments.

The natural abundance for $\mathrm{^{13}C}$ is only 1.1%, while $\mathrm{^{1}H}$ has a natural abundance of 99,9%. This difference alone means that if you don't label your sample with $\mathrm{^{13}C}$, the spectra will be around 90 times less sensitive than proton spectra due to the lower natural abundance alone.

$\mathrm{^{13}C}$ also has a four times lower gyromagnetic ration than $\mathrm{^{1}H}$, which also lowers the sensitivity. But the way the experiment is usually measured, you get magnetization transfer from directly-attached protons to the carbons, which increases the sensitivity.

$\mathrm{^{13}C}$ also relaxes more slowly, which means that the time between experiments is longer and you can measure fewer scans than for proton experiments in the same time.

Another factor that is important is which kind of probe your spectrometer is using. For an inverse probe like e.g. a Bruker TXI, the inner coil is the proton coil, which means it's especially suited for proton experiments. For $\mathrm{^{13}C}$ experiments an observe probe like a BBO is more sensitive, as the carbon is measured on the inner coil. If you have multiple spectrometers, you should ask which one is better suited for $\mathrm{^{13}C}$ experiments.

If you have a very small amount of sample, a Shigemi tube allows you to measure in half the sample volume (200-250 $\mu$l) compared to standard tubes, you can achieve a higher concentration that way. They're pretty expensive though and they have to be matched to the solvent, so you would need a CDCl3 tube.

And there are various problems that could drastically reduce your sensitivity for $\mathrm{^{13}C}$, the most important ones if we disregard hardware problems, are proper tuning and matching for $\mathrm{^{13}C}$, and the correct pulse length for $\mathrm{^{13}C}$.

I've avoided your actual question, as the required concentration depends on quite a few factors. I'd recommend you to ask the person responsible for the spectrometers, as they should be able to give you a better estimate based on the actual hardware you have.


Phenolphthalein indicator spill onto skin

Didn't you receive the SDS for phenolphtalein (or its methanolic/ethanolic solutions)?
Weren't you given instructions on how to handle it?
Didn't you wear suitable gloves in addition to your safety goggles?

Back in the days, phenolphthalein was apparently used as a laxative. This is no longer recommended!

However, the current GHS classification is as follows:

  Germ cell mutagenicity, Category 2; H341
  Carcinogenicity, Category 1B; H350
  Reproductive toxicity, Category 2; H361f

(cited from the GESTIS database)

To my knowledge, phenolphtalein is on the REACH Candidate list of SVHC (substances of very high concern) since 2011.

My suggestion is:

Contact a physician.
Report the exposition to your supervisors.


How to determine if a reaction proceeds via SN1, SN2, E1 or E2 mechanism?

$\ce{EtO-}$ is a strong base, as well as a strong nucleophile, and the temperature is not too high. Therefore a mixture of products is most likely to be formed, where substitution will lead to the major and elimination will lead to the minor product.
Also, the reactant is a primary halide. So, in substitution, SN2 will be favoured, whereas in elimination E2 will be favoured.
So, final product will be,

Ethoxy hexane (via SN2) (major)
Hex-1-ene (via E2) (minor). (Maybe, a very little amount of hex-2-ene may also form by very very small fraction of E1).


Why are ionic reactions very fast in aqueous medium?

Water has a permanent dipole character, which means it has a slight positive and negative charge permanently in it (due to electronegativity difference). So, in case of an ionic reaction, it helps in easier solvation of ions by pulling them with its charged centres (positive centre attracts negative ions and negative centre attracts positive ions)
Aqueous is a term specially for water as a solvent. Any polar solvent can support ionic reactions.


Yes, amide bonds can fragment during electrospray ionization.

Intentional fragmentation in a collision cell

Amide bonds are intentionally fragmented in mass spectrometers all the time, during proteomics analysis.  In that case, peptides are ionized, particular ions are isolated (by the quadrupole of the mass spectrometer), and then the intact peptide ions are fragmented intentionally in a collision cell inside the instrument.  After collision with high-energy gas molecules, peptides fragment in mostly predictable ways giving rise to y, b, a, and z ions.

"In-source" fragmentation

The key step of the fragmentation process is collision with high-energy gas molecules.  That can happen accidentally during electrospray ionization in the ion source.  That's because the ion source, being at atmospheric pressure, has many many more gas molecules around than a collision cell, which might be at 10$^{-5}$ torr or lower in pressure.  So after your ions of your peptidic molecule form in the ion source, some of them might bump into these atmospheric gas molecules and be fragmented in the ion source, before they even make it into the mass spectrometer.

Controlling in-source fragmentation

The propensity of this "accidental" fragmentation to happen is somewhat tunable by ion source parameters -- nebulizer and capillary voltages, gas flows and temperatures, etc.  The details depend entirely on your exact model of MS.  If the carboxyl and amine fragments are both more intense than the "molecular ion", it could be a sign that the ion source is tuned too aggressively -- trying lowering a voltage or two and see if you boost the molecular ion signal.  (Alternately, for some applications, in-source fragmentation could be desired.)


How does dimethyl sulfoxide transport other molecules through skin?

DMSO interacts with the outer membrane of skin, the stratum corneum.  The stratum corneum is a typical lipid bilayer and DMSO can solvate the exposed hydrophilic lipid head.  This results in a weakening of the lateral forces between the individual phospholipid molecules.  The weakened lateral forces result in a widening of the channel between the phospholipid molecules and the widened channel allows small molecules to pass through the lipid layer.  After the DMSO is removed the lipid bilayer restores itself back to normal.

Although it is agreed that small molecules can pass through the DMSO-affected bilayer, there is some evidence that larger molecules such as viral agents might also pass through.  There is also some controversy over the effect of DMSO on the blood-brain barrier.  Here is a link to a nice review of DMSO's pharmacological effects, including transport of larger molecules and transport across the blood-brain barrier.


Between ethene and propene which is more acidic and why?

The difference between there acidity arises only due to the difference in inductive effect[or the electron donating capacity of the alkyl group] of the alkyl group attached to the $\ce{C=C}$ in both the cases.

By this site there exists a difference of total one unit

For more info on Inductive Effect

Relative acidity of a compound is checked by removing the acidic hydrogen and then comparing the stability of the anions.The more stable the anion is the faster it is formed and hence the acid is stronger.

The anions formed are $\ce{H3C-CH=CH-}$ and  $\ce{H2C=CH-}$ . In the former one the anion or the negative charge on the anion is destabilized by the electron donating ability or the inductive effect of $\ce{CH3-}$, hence it is less stable and less rapidly formed and hence less acidic.

  Why $\ce{H3C-}$ has electron donating effect?

In an alkyl group, for example, in methyl group carbon is linked to three hydrogen atoms. Carbon is more electronegative than hydrogen. So carbon pulls bonded electrons nearer to carbon and carbon acquires a slight extra negative charge (delta-) and correspondingly hydrogen lose hold over bonded electrons and acquires a slight positive charge (indicated by delta +). Thus carbon get a little more electronic charge which the carbon will push towards the group to which it is linked.When this electron rich carbon centre (methyl group) attached with other compounds it behaves as electron donor.
And thus methyl group acts as electron donating group.
Be in mind, the electronegativity difference between carbon and hydrogen is very less.

$\ce{H3C-CH=CH-}$ is less stable due to presence of above factor.

You can also use conjugate base method or charge holding capacity method to compare the acidity of compound to give the same result as above.


In NaCl electrolysis, why does chlorine ion get oxidized but not oxygen ion?

I am afraid that as stated in another answer there is no simultaneous evolution of oxygen or chlorine during controlled electrolysis. It all depends on the concentration of ions in the solution. Keep in mind that the concentration of water is 55 M, so the ion concentration is nowhere near it no matter what you do!

The key reason is that the electrode potential of chlorine/chloride couple is quite different from that of oxidation of water, so if you have a high concentration of chloride ion, chlorine is oxidized first, once chlorine falls below a certain threshold, water begins to oxidize. Nernst equation can help in predicting that.


If you are interested in making deutero-chloroform, there are better ways to do it.

From a purely mechanistic standpoint, chloroform isn't a very strong acid ($\ce{pKa \sim 15}$) so $\ce{D2O}$ is unlikely to deprotonate it at a useful rate.  Further, chloroform isn't very soluble in water making your desired reaction even less likely (but you could engineer your way around this problem by using a phase transfer catalyst).

Your suggestion to use a stronger base is reasonable, but the problem you run into here is that a different reaction is brought into play.  Specifically, once a stronger base like potassium t-butoxide (in t-butanol) is used, dichlorocarbene formation occurs.  Once the trichloromethylcarbanion is formed, it quickly ejects $\ce{Cl^-}$ to form dichlorocarbene.  Dichlorocarbene is an extremely energetic species and would quickly react with whatever else is around - perhaps the t-butanol to produce dichloromethyl-t-butyl ether.


Okay, let's tackle at least one problem here. Consider the rotation of 1,2-dichloroethane (BP86/cc-pVDZ):

These conformational changes can be further rationalised:

C and C' are the same conformation, since these are mirror images.
The same applies to B and B'.
A and C are local minima and can be referred to as conformes in the above given way
B and D are transition states - they are not stable.
If you consider the rotation A to C as a reaction, this state would mark the maximum energy needed for the conversion. I believe with reactions it is meant, that an accurate description of the transition state by InChI is not possible.

For the term non-local chirality I suggest you look at BINAP or BINOL.

In this case I have used a slightly abridged version of BINAP for educational purposes, where $\ce{Ph -> Me}$.

Other wonderful examples of non-local chirality include helicenes, which have helical chirality. Here is one example of a fancy azonia derivative of hexahelicene (parent compound and links). The structure is available at the Cambridge Crystallographic Data Centre (CCDC) Database. I removed the anion and converted it to xyz coordinates which I include in the appendix. Here are the structures, left P-chirality, right M-chirality:

On another account of non/local chirality, there is the planar chirality, as can be seen in trans-Cyclooctene (the bottom two). For further reading I suggest Neuenschwander et.al. "The Conformations of Cyclooctene: Consequences for Epoxidation Chemistry."

Markush Structures are a kind of summary structure for certain compound classes. A very simple example would be a monosubstituted biphenyl system, where the upper structure is a representation of all three below. However, it is noteworthy, that each instance of a Markush structure can be well explained by InChI.

More complicated structures are often used in patents and in other scientific publications, usually specifically referring to a very general procedure or mechanism. I found one on wikipedia commons, that could well represent a million molecules:

And another way to break the InChI are clusters. Although the definition is not very specific, the $\ce{[KO{}^{\mathit{t}}Bu]_4}$ tetramer (wikipedia, chemspider) certainly contains one. While the monomer can be described via InChI, the tetra- or Polymer should fail. It was published by Chisholm et.al. in Polyhedron and the *.cif file can be obtained from CCDB. I cleaned it up a little bit and included the *.xyz in the appendix. You can quite nicely see the $\ce{K4}$ tetrahedral cluster in the middle (hydrogens omitted for clarity).

Last but not least, I'd like to cover the matter of complicated organometallic molecules. As such I'd like to introduce the Grubbs' catalysts. Grubbs obtained the Nobel Prize in Chemistry 2005 for his research in conjunction with Yves Chauvin and Richard R. Schrock. One of the key publications was "Ring-opening metathesis polymerization (ROMP) of norbornene by a Group VIII carbene complex in protic media". The crystal structure can also be obtained at CCDC. Another example stems from "Synthesis and Applications of $\ce{RuCl2(=CHR')(PR3)2}$:  The Influence of the Alkylidene Moiety on Metathesis Activity" and is also available via CCDC. This structure can also be found in the appendix. It is generally known as a 1st generation Grubbs catalyst (Hydrogens omitted for clarity).

Appendix

The following section contains the nuclear coordinates in *.xyz(XMol) file format and angstroms as unit. (If not denoted otherwise, structures are obtained at BP86/cc-pVDZ.)

1,2-dichloroethane A

C        0.000000000      0.000000000      0.000000000
C        0.000000000      0.000000000      1.521307000
Cl       1.723641000      0.000000000     -0.593797000
H       -0.486731000     -0.904743000     -0.410753000
H       -0.486731000      0.904743000     -0.410753000
Cl      -1.723641000      0.000000000      2.115104000
H        0.486731000     -0.904743000      1.932060000
H        0.486731000      0.904743000      1.932060000

1,2-dichloroethane B

C        0.030242000      0.085780000     -0.007620000
C       -0.030242000      0.085780000      1.528927000
Cl       1.518312000     -0.758527000     -0.631286000
H       -0.837191000     -0.442465000     -0.444148000
H        0.070555000      1.115212000     -0.412109000
Cl      -1.518312000     -0.758527000      2.152593000
H        0.837191000     -0.442465000      1.965455000
H       -0.070555000      1.115212000      1.933416000

1,2-dichloroethane C

C        0.066377000      0.127741000      0.004558000
C       -0.066377000      0.127741000      1.516749000
Cl       1.083384000     -1.241726000     -0.615213000
H       -0.919786000      0.050600000     -0.492443000
H        0.568277000      1.063384000     -0.317636000
Cl      -1.083384000     -1.241726000      2.136520000
H        0.919786000      0.050600000      2.013750000
H       -0.568277000      1.063384000      1.838943000

1,2-dichloroethane D

C        0.164887000      0.141535000     -0.000240000
C       -0.164887000      0.141535000      1.521547000
Cl       0.337146000     -1.473219000     -0.795160000
H       -0.633932000      0.666019000     -0.558099000
H        1.123149000      0.665665000     -0.177529000
Cl      -0.337146000     -1.473219000      2.316468000
H        0.633932000      0.666019000      2.079406000
H       -1.123149000      0.665665000      1.698836000

R-BINAP

C       -0.086423000     -0.055821000     -0.031546000
C       -0.090885000     -0.047699000      1.480085000
C        1.148913000     -0.158515000      2.237313000
C        1.064686000     -0.139089000      3.676951000
C       -0.131097000     -0.014450000      4.357670000
C       -1.357362000      0.116010000      3.652970000
C       -1.345263000      0.107319000      2.211098000
P        2.768926000     -0.321256000      1.451350000
H        1.988600000     -0.233144000      4.265654000
H       -0.135982000     -0.008251000      5.459401000
C       -2.590782000      0.269421000      4.370415000
C       -2.623021000      0.305506000      1.559779000
C        0.437392000      1.073619000     -0.788457000
C        0.410661000      0.991852000     -2.228054000
C       -0.079670000     -0.105834000     -2.908992000
C       -0.579162000     -1.233570000     -2.204534000
C       -0.579999000     -1.219208000     -0.762676000
P        1.107602000      2.557133000     -0.001640000
H        0.790955000      1.839345000     -2.816335000
H       -0.077540000     -0.112552000     -4.010726000
C       -1.063317000     -2.378355000     -2.921905000
C       -1.047368000     -2.424579000     -0.111239000
C       -3.792005000      0.431032000      3.705083000
H       -2.558632000      0.263827000      5.471798000
H       -4.735477000      0.552772000      4.258523000
C       -3.797353000      0.457087000      2.279022000
H       -2.662047000      0.340912000      0.466043000
H       -4.750761000      0.602046000      1.748556000
C       -1.503463000     -3.517255000     -0.830391000
H       -1.036094000     -2.475958000      0.982488000
H       -1.854657000     -4.415344000     -0.299810000
C       -1.523650000     -3.499569000     -2.256520000
H       -1.052652000     -2.347534000     -4.023279000
H       -1.890198000     -4.377463000     -2.809877000
C        4.061075000     -0.509159000      2.730645000
C        2.844045000     -1.846744000      0.446017000
C        1.585856000      3.772639000     -1.280334000
C       -0.173490000      3.383258000      1.008127000
H        0.254332000      4.289560000      1.485098000
H       -1.067700000      3.703930000      0.418133000
H       -0.532498000      2.717241000      1.815764000
H        1.997887000      4.670081000     -0.777298000
H        2.377005000      3.362349000     -1.938832000
H        0.749130000      4.118465000     -1.941170000
H        5.044249000     -0.601334000      2.227532000
H        4.100503000      0.383637000      3.385840000
H        3.944572000     -1.404262000      3.395364000
H        3.842692000     -1.931312000     -0.030903000
H        2.674387000     -2.779210000      1.039384000
H        2.087489000     -1.826871000     -0.361580000

S-BINAP

C       -1.383946000     -3.448543000     -2.256996000
C       -1.065989000     -2.266144000     -2.910129000
C       -0.652376000     -1.110469000     -2.179491000
C       -0.563545000     -1.177237000     -0.738280000
C       -0.899026000     -2.413534000     -0.098700000
C       -1.299620000     -3.519323000     -0.838209000
C       -0.154302000     -0.008671000      0.004167000
C        0.166467000      1.182512000     -0.673969000
C        0.072046000      1.221328000     -2.098779000
C       -0.323252000      0.113168000     -2.831301000
C       -0.061735000     -0.093981000      1.507135000
C       -1.110430000      0.372113000      2.322674000
C       -0.989013000      0.267385000      3.742090000
C        0.132224000     -0.289631000      4.336601000
C        1.210137000     -0.777916000      3.542877000
C        1.113248000     -0.682564000      2.103804000
C        2.206553000     -1.174312000      1.320794000
C        3.329975000     -1.726929000      1.922868000
C        3.419059000     -1.820539000      3.340084000
C        2.377998000     -1.353835000      4.130094000
P       -2.590653000      1.186493000      1.502738000
C       -3.750658000     -0.293139000      1.402088000
P        0.636241000      2.692004000      0.339668000
C        2.511918000      2.555623000      0.253510000
C       -3.385046000      2.079953000      2.947654000
C        0.388676000      4.078351000     -0.898992000
H       -1.802841000      0.635802000      4.383638000
H        0.202504000     -0.360978000      5.434128000
H        0.319305000      2.150696000     -2.632672000
H       -0.386992000      0.166196000     -3.930250000
H        2.436377000     -1.419019000      5.228476000
H        4.313241000     -2.262708000      3.805535000
H        2.146133000     -1.110484000      0.224695000
H        4.157224000     -2.097887000      1.298347000
H       -0.834575000     -2.478936000      0.997042000
H       -1.552289000     -4.458551000     -0.322477000
H       -1.128190000     -2.199481000     -4.008215000
H       -1.700498000     -4.331984000     -2.832408000
H        2.950535000      3.473538000      0.698290000
H        2.842563000      1.687658000      0.858736000
H        2.884117000      2.439555000     -0.786611000
H        0.678948000      5.023670000     -0.395042000
H        0.990357000      3.982867000     -1.828352000
H       -0.686746000      4.150262000     -1.160750000
H       -4.302108000      2.573768000      2.564446000
H       -3.673500000      1.421270000      3.794842000
H       -2.700063000      2.872570000      3.312044000
H       -4.742112000      0.063129000      1.051704000
H       -3.359638000     -1.014277000      0.656570000
H       -3.871382000     -0.807607000      2.379049000

M-8a-azonia[6]helicene (from crystal structure)

N        3.565811000      0.259443000     13.724558000
C        7.445232000     -1.196589000     12.564039000
C        8.629416000     -1.762757000     12.151335000
C        8.885324000     -1.978555000     10.783139000
C        7.914184000     -1.694865000      9.864763000
C        6.665024000     -1.171130000     10.265448000
C        5.584378000     -1.088690000      9.330104000
C        4.339207000     -0.800151000      9.738213000
C        4.088732000     -0.406137000     11.083608000
C        2.763235000     -0.187914000     11.543327000
C        2.522579000      0.029096000     12.830042000
C        3.246403000      0.443720000     15.064121000
C        4.198045000      0.692252000     15.970654000
C        5.531025000      0.961394000     15.546639000
C        6.527506000      1.327523000     16.488699000
C        7.769629000      1.675468000     16.086777000
C        8.044342000      1.864594000     14.697195000
C        9.259232000      2.459858000     14.281133000
C        9.466841000      2.806590000     12.982045000
C        8.445715000      2.608977000     12.035743000
C        7.265466000      1.985829000     12.396483000
C        7.061105000      1.553020000     13.720492000
C        5.833567000      0.912900000     14.172787000
C        4.884651000      0.303088000     13.284635000
C        5.151668000     -0.288539000     12.002691000
C        6.444932000     -0.841371000     11.632053000
H        7.296301000     -1.066868000     13.466331000
H        9.304643000     -2.036748000     12.840647000
H        9.760406000     -2.388329000     10.496986000
H        8.038683000     -1.964007000      8.943383000
H        5.780686000     -1.333585000      8.395468000
H        3.604979000     -0.909262000      9.084780000
H        2.029733000     -0.315211000     10.922946000
H        1.672696000      0.060618000     13.273677000
H        2.290439000      0.327335000     15.253240000
H        3.954366000      0.824398000     16.914659000
H        6.255785000      1.333585000     17.427224000
H        8.447606000      1.891266000     16.702563000
H        9.925491000      2.630800000     14.988120000
H       10.286033000      3.200604000     12.619714000
H        8.559741000      2.909640000     11.135042000
H        6.570317000      1.891266000     11.711236000

M-8a-azonia[6]helicene (from BP86/STO-3G)

N        0.241946000      3.510759000     13.746203000
C       -1.207087000      7.492952000     12.512135000
C       -1.801472000      8.692275000     12.089683000
C       -2.102492000      8.912745000     10.707738000
C       -1.853647000      7.899258000      9.772330000
C       -1.278056000      6.643678000     10.182732000
C       -1.146616000      5.553783000      9.229249000
C       -0.770948000      4.283969000      9.646130000
C       -0.392477000      4.045979000     11.023979000
C       -0.161103000      2.688046000     11.484550000
C        0.042219000      2.428893000     12.821687000
C        0.400500000      3.183323000     15.136378000
C        0.639392000      4.176721000     16.059802000
C        0.949043000      5.527644000     15.625469000
C        1.361398000      6.519481000     16.597330000
C        1.808770000      7.766658000     16.182869000
C        1.981223000      8.066019000     14.770425000
C        2.627560000      9.287588000     14.362758000
C        2.913363000      9.530579000     13.012425000
C        2.578886000      8.546652000     12.028110000
C        1.916409000      7.363991000     12.392255000
C        1.565834000      7.097902000     13.763105000
C        0.907348000      5.851096000     14.211296000
C        0.281492000      4.885356000     13.292158000
C       -0.309154000      5.144896000     11.968134000
C       -0.894790000      6.445456000     11.575101000
H       -1.007504000      7.339193000     13.585554000
H       -2.048406000      9.472285000     12.834293000
H       -2.558542000      9.868716000     10.388465000
H       -2.123797000      8.035037000      8.708340000
H       -1.422274000      5.738729000      8.172836000
H       -0.763274000      3.430969000      8.943112000
H       -0.251900000      1.843110000     10.776492000
H        0.087281000      1.420761000     13.274365000
H        0.295196000      2.107892000     15.372408000
H        0.697253000      3.916258000     17.133249000
H        1.320711000      6.254708000     17.669775000
H        2.110372000      8.530075000     16.926174000
H        2.921715000     10.016221000     15.141242000
H        3.423497000     10.463796000     12.708003000
H        2.853573000      8.716110000     10.969950000
H        1.691114000      6.611596000     11.618453000

$\ce{KO{}^{\mathit{t}}Bu}$ tetramer (from crystal structure)

K       -1.309381000     -1.309381000     -1.309381000
O       -1.313567000      1.313567000     -1.313567000
C       -2.107232000      2.107232000     -2.107232000
C       -3.502008000      2.298114000     -1.478495000
H       -3.842748000      1.465100000     -1.465100000
H       -4.060420000      2.821364000     -2.059512000
H       -3.516240000      2.821364000     -0.619528000
O       -1.313567000     -1.313567000      1.313567000
O        1.313567000     -1.313567000     -1.313567000
K       -1.309381000      1.309381000      1.309381000
K        1.309381000      1.309381000     -1.309381000
C       -1.478495000      3.502008000     -2.298114000
C       -2.298114000      1.478495000     -3.502008000
K        1.309381000     -1.309381000      1.309381000
C       -2.107232000     -2.107232000      2.107232000
C        2.107232000     -2.107232000     -2.107232000
O        1.313567000      1.313567000      1.313567000
H       -1.465100000      3.842748000     -1.465100000
H       -2.059512000      4.060420000     -2.821364000
H       -0.619528000      3.516240000     -2.821364000
H       -1.465100000      1.465100000     -3.842748000
H       -2.821364000      2.059512000     -4.060420000
H       -2.821364000      0.619528000     -3.516240000
C       -1.478495000     -3.502008000      2.298114000
C       -3.502008000     -2.298114000      1.478495000
C       -2.298114000     -1.478495000      3.502008000
C        3.502008000     -1.478495000     -2.298114000
C        1.478495000     -2.298114000     -3.502008000
C        2.298114000     -3.502008000     -1.478495000
C        2.107232000      2.107232000      2.107232000
H       -1.465100000     -3.842748000      1.465100000
H       -2.059512000     -4.060420000      2.821364000
H       -0.619528000     -3.516240000      2.821364000
H       -3.842748000     -1.465100000      1.465100000
H       -4.060420000     -2.821364000      2.059512000
H       -3.516240000     -2.821364000      0.619528000
H       -1.465100000     -1.465100000      3.842748000
H       -2.821364000     -2.059512000      4.060420000
H       -2.821364000     -0.619528000      3.516240000
H        1.465100000     -1.465100000     -3.842748000
H        3.842748000     -1.465100000     -1.465100000
H        4.060420000     -2.059512000     -2.821364000
H        3.516240000     -0.619528000     -2.821364000
H        2.059512000     -2.821364000     -4.060420000
H        0.619528000     -2.821364000     -3.516240000
H        1.465100000     -3.842748000     -1.465100000
H        2.821364000     -4.060420000     -2.059512000
H        2.821364000     -3.516240000     -0.619528000
C        1.478495000      2.298114000      3.502008000
C        3.502008000      1.478495000      2.298114000
C        2.298114000      3.502008000      1.478495000
H        1.465100000      1.465100000      3.842748000
H        2.059512000      2.821364000      4.060420000
H        0.619528000      2.821364000      3.516240000
H        3.842748000      1.465100000      1.465100000
H        4.060420000      2.059512000      2.821364000
H        3.516240000      0.619528000      2.821364000
H        1.465100000      3.842748000      1.465100000
H        2.821364000      4.060420000      2.059512000
H        2.821364000      3.516240000      0.619528000

Grubbs I catalyst

Ru      1.923022    1.782458    4.279841
Cl      3.508375    2.054525    6.061616
P       0.620071    3.495015    5.336792
C       2.967497    2.881791    3.239789
Cl      0.101984    1.197998    2.838894
P       2.855052   -0.401312    3.741075
C       2.954465    3.343954    1.839333
Cl      3.310061    4.678711   -2.473972
C       1.880873    3.257547    0.960743
C       1.973325    3.667349   -0.352875
C       3.185197    4.168660   -0.812833
C       4.252303    4.292067    0.012884
C       4.145603    3.890748    1.341393
C      -0.637087    4.207002    4.175769
C      -1.979932    4.694567    4.692950
C      -2.932151    4.983356    3.538285
C      -2.339331    5.946324    2.517809
C      -0.957803    5.505405    2.056512
C      -0.041546    5.252113    3.235271
C       1.483967    4.856669    6.241316
C       0.518279    5.817396    6.934517
C       1.256971    6.738015    7.899778
C       2.346867    7.495106    7.161401
C       3.302557    6.547168    6.465690
C       2.565538    5.630984    5.497585
C      -0.333586    2.698072    6.719178
C      -1.014762    1.418485    6.241149
C      -1.859827    0.796208    7.358000
C      -1.043744    0.536540    8.601177
C      -0.361483    1.805588    9.075860
C       0.493782    2.427234    7.979254
C       2.628515   -0.975242    1.991593
C       3.236690   -0.059043    0.937820
C       2.687448   -0.416553   -0.435697
C       2.971677   -1.856587   -0.778868
C       2.466883   -2.799363    0.290130
C       2.999176   -2.423218    1.675026
C       4.622054   -0.540862    4.248552
C       5.523290    0.453308    3.517538
C       6.887074    0.518241    4.175936
C       7.549339   -0.842629    4.233326
C       6.649166   -1.859830    4.903939
C       5.271941   -1.927013    4.232992
C       1.995357   -1.737789    4.704830
C       2.010202   -1.452357    6.212537
C       1.292567   -2.535071    7.008639
C      -0.120796   -2.764155    6.507353
C      -0.110733   -3.109699    5.031603
C       0.567738   -2.010334    4.218268
H       3.716836    3.154747    3.670969
H       1.120535    2.917815    1.276640
H       1.189061    3.632874   -0.962082
H       5.017596    4.645026   -0.262690
H       4.792512    4.003484    1.897392
H      -0.815388    3.399122    3.615754
H      -1.902150    5.510235    5.223685
H      -2.378700    3.974732    5.228704
H      -3.726326    5.342882    3.821555
H      -3.181986    4.184074    3.125510
H      -2.984577    6.066660    1.748479
H      -2.250423    6.781889    2.909669
H      -1.054116    4.674447    1.537657
H      -0.576082    6.145292    1.467384
H       0.723372    4.965838    2.939787
H       0.125317    6.138803    3.761321
H       1.921040    4.375890    6.908583
H       1.676706    6.235725    8.636983
H      -0.145301    5.369077    7.433963
H       0.091950    6.422548    6.224250
H       0.676388    7.347621    8.295653
H       2.791319    8.070082    7.723424
H       1.896364    8.099751    6.522077
H       3.914233    7.048872    6.008409
H       3.698100    6.058287    7.055823
H       3.116098    5.021349    5.094850
H       2.158660    6.180195    4.786983
H      -1.006024    3.336226    6.935353
H      -1.559257    1.598311    5.456258
H      -0.325083    0.765200    6.003389
H      -2.608220    1.404611    7.524315
H      -2.214635    0.029761    7.044111
H      -0.418035   -0.117788    8.397718
H      -1.627822    0.182306    9.277813
H       0.176244    1.645756    9.829964
H      -1.053680    2.460437    9.333028
H       0.877440    3.225032    8.287287
H       1.235327    1.881817    7.763580
H       1.720632   -0.869158    1.853889
H       4.184273   -0.146643    0.965428
H       3.012517    0.820454    1.136093
H       1.847900   -0.200343   -0.409930
H       3.096540    0.151437   -1.100956
H       3.904789   -1.982340   -0.878422
H       2.584028   -2.105280   -1.617970
H       2.763571   -3.703670    0.105411
H       1.556168   -2.778368    0.317905
H       4.607903   -0.216537    5.166797
H       3.990062   -2.477871    1.686571
H       2.642313   -3.025155    2.339113
H       5.156005    1.289122    3.512016
H       5.590006    0.129866    2.598457
H       6.737696    0.838429    5.032942
H       7.431919    1.106030    3.739569
H       8.390108   -0.830548    4.679900
H       7.729962   -1.148361    3.309561
H       6.514020   -1.632968    5.797587
H       7.041226   -2.668142    4.887375
H       4.662063   -2.600521    4.676553
H       5.418404   -2.225128    3.326293
H       2.516573   -2.545817    4.572816
H       1.564734   -0.621934    6.319621
H       2.866067   -1.356878    6.528769
H       1.774146   -3.375710    6.968817
H       1.285356   -2.270001    7.937592
H      -0.502310   -3.449358    7.037418
H      -0.746615   -1.928105    6.670990
H       0.328488   -3.937948    4.917492
H      -1.013735   -3.235955    4.696631
H       0.558651   -2.294513    3.264385
H       0.034321   -1.214512    4.228140


The universe doesn't calculate anything. Energy and it's conservation and everything that follows from it is a human-made concept, a physical model. It describes the things that happen in nature quite well (in the right circumstances), but it is only a model. In how far this lends some kind of "reality" to the concept/model I couldn't say. It's more of a philosophical question. Physical laws only describe how nature works, but not why it works the way it does.

Edit:

Concerning the physical content of the collision you described: I think of it in the following way: The projectile hits the first atom of the molecule, let's call it A1, and transfers its momentum to it. Now, this atom moves in the direction along the bond axis directly towards the next atom, A2. While it does so, it has to work against the spring force of the bond between A1 and A2. Thus, kinetic energy of movement is converted to kinetic energy of vibration. A2 will now be accelerated in the direction of the next atom, A3, and it has to work against the spring force of the bond between A2 and A3 and(!) the restoring force brought about by the bond between A1 and A2. Again kinetic energy is converted to vibrational energy. But now you have a system of coupled oscillators. This process will be repeated for every atom in the molecule until every bond vibrates. So, it is not true that

  Each bond in the setup doesn't "know" about the existence of the other
  bond and each atom could be said to be unaware of all but the atoms
  bonded or colliding with it.

They are coupled through the spring forces of the bonds and thus each atom "feels" the mass of every other atom in the molecule as well as the bond strength of every bond in the molecule.


What is the correct reaction of sulfur dioxide with chlorine?

Sulfuryl chloride ($\ce{SO2Cl2}$) is prepared by the reaction of $\ce{SO2}$ and $\ce{Cl2}$ over activated charcoal, details of the preparation are here 1.

Thionyl chloride may be prepared from $\ce{SO2}$ and $\ce{Cl2}$ but the reaction requires an equivalent of $\ce{SCl2}$ to be present.


Does the potential energy increase when temperature is raised?

Yes, potential energy increases with increasing temperature for at least the following three reasons:

At a higher temperature, more atoms/molecules are in excited electronic states.  Higher electronic states correspond to greater potential energy. Potential Energy is -2 times Kinetic Energy. So actually, at higher temperature, when more atoms are in higher electronic states, there is more potential energy and less kinetic energy (just considering electronic energy).
At higher temperature, more molecules are in excited vibrational states.  Higher vibrational states correspond to greater potential energy. Half the energy of each vibrational state is kinetic, half potential.
At higher temperature, more molecules are in excited rotational states.  While for a rigid rotor, this would not involve increased potential energy, real molecules are not rigid.  The rotation of a diatomic molecule about an axis perpendicular to the bond stretches the bond as a centrifugal distortion, which represents an increase in potential energy.

Additionally, for intermolecular forces, to the extent that they involve potentials of the form r^x, the Virial Theorem requires that total energy is divided into potential and kinetic energy in a specific proportion, so total energy can not be increased without increasing potential energy.

Free translational motion is not subject to a potential, nor is rigid rotation.


Cut off as much of the "steel" as possible (hacksaw or angle grinder will work - use gloves & eye protection) then immerse the stub in 33% hydrochloric acid until the "steel" remnants are fully dissolved. Neutralise the acid with bicarbonate before disposal.


Geometrical isomers refers to isomerism in which the connectivity of atoms is preserved, and changes refer only to the spatial arrangement of the atoms. In other words, it only allows transformations around certain features that can be arranged in different ways in three dimensions (and which aren't just conformers that will easily transform into each other due to molecular motion).

I mention this because your reference to a "cycloalkene [...] having [8 or more] carbon atoms" suggest that you are looking for constitutional isomers (i.e. compounds with the same atoms but different connectivity) and not just geometrical isomers (i.e. compounds with the same connectivity but different spatial arrangement).

If we consider geometrical isomers of the compound you gave, 1) the cycloalkene ring, 2) the side chains and 3) the substitution sites in the ring are fixed (as changing them would produce constitutional isomers), so we are left with isomerism due to cis/trans double bonds.

You have three double bonds in the side chains that will produce $2^3=8$ isomers. Cyclooctenes are in principle large enough rings that their trans form can be isolated, so this leads to $16$ isomers; you can figure them out from your posted example (which is cis in the cycloalkene and trans in the side chains) by changing double bonds.

If you want to figure out constitutional isomers, that's a far more complex task; you'll have to account for ring size, number of side chains, ring substitution pattern, branching, position of insaturations and so on; with a hydrocarbon of this size, the number of possibilities will be huge.


Is a beta pleated parallel sheet always formed from more than one peptide?

While it is true that some antiparallel beta sheets have neighboring strands that are adjacent in primary sequence, other cases do exist. As for parallel beta sheets, the polypeptide chain between strands could be connected via a loop or a helix if the two strands are close in primary sequence.
The mapping of strands on the primary structure has been studied thoroughly, and there are two types of diagrams to quickly show what is going on. For the "TOPS" diagram, we pretend to look down the strands, and up and down triangles represent strands in one or the other direction:

Source: https://onlinelibrary.wiley.com/doi/pdf/10.1002/pro.3285
Circles represent helices. If you look closely at the connecting lines, you can see whether the connection is above or below the plane of the paper.
The other representation has the sheets in the plane of the paper, e.g.:

Source: https://image.slideserve.com/258316/beta-sheet-topology-diagrams-l.jpg
Take-home messages

Most sheets are mixed (combo of parallel and antiparallel).
Most neighboring strands are not adjacent in sequence.
When two neighboring parallel strands are adjacent in sequence, there is a loop or a helix between them (or a strand from a different sheet) to connect them in the proper orientation.


Check My Beginner's Guide to Shells, Energy Levels, Subshells, and Orbitals

All looks good at a school level, but I would suggest or convince you that we should not attach any physical pictures or meaning to all these atomic concepts. If you pursue science in future, you will have to unlearn a lot of these stories ingrained by school or even university teachers. It is okay to acknowledge and say that we do not fully understand everything as yet. Nature is very very complex. Our only effort is to make simple models of the phenomena we observe in the Nature.

The shell model is outdated and obsolete yet it is good for educational purposes. It was inspired by the planets rotating around the Sun, so learning about it from a historical perspective is good. For example, you wrote "The shells are farther away from the nucleus and thus the electrons in it are less attracted to the nucleus." It may not be a correct picture, because in reality there are no circular shells per se, like an onion ring around a nucleus. Brilliant physicists spent their life time in understanding the atom so don't be surprised if you find atomic structure overwhelming. Indeed it is.

More importantly just delete this interpretation from your mind "An orbital is an actual path (probability distribution function, PDF) that either 1 or 2 electrons can be whizzing through at a time."

An orbital is just a mathematical construct with no physical reality as textbooks show them as balloon models and so on. The rigorous and correction version of an orbital is "Wavefunction depending explicitly on the spatial coordinates of only one electron" (IUPAC)

Their shape depends on the coordinate system being used to solve the Schrodinger's equation. Can you believe that Schrodinger never showed or drew a single picture of orbitals in his four famous papers?

The typical shapes of s, p, d, f orbital are shown in Cartesian coordinate system, but physicists and mathematicians have plenty of other coordinate systems to solve differential equations. Do the electrons care what coordinate system a human chose to represent them or calculate their energies? The answer is no. Their energy turns out to be the same regardless of the coordinate system. Nobody teaches that and I also recently learned this from a physicist who has published a lot on orbital shapes. That gentleman is a big opponent to the teaching of orbitals and the myths associated with them in undergraduate curriculum. Not all agree with him including Linus Pauling.

The atomic structures "stories" as taught to chemistry students as facts is one of the worst things in modern educational system. In my humble opinion, chemists should take the help of modern physicists and ask them how to represent the latest views of atomic structure right from beginning classes. It does not need to have crazy mathematics, but good scientists can explain and elaborate things in words as well. As someone said, "If you can't explain it simply, you don't understand it well enough."

Book recommendation: Hydrogen: The essential element by J. Rigden (from Harvard). A thin book which is very illuminating. Chemical educators should adapt this style. The last chapter or so talks about anti-hydrogen.


Looking for an old (1950s) "chemical experiments" book

There are 2 books in German which inspired me to become Chemist. I had them in the late 70s.
"Chemische Experiemnte, die gelingen" (Chemical experiments which succeed) and "Organische Chemie im Probierglas" (Organic chemistry in a test tube). Both published originally in 1939 by Hermann Römpp and Hermann Raaf. Kosmos Verlag (Kosmos Publishers). No idea if there ever were English translations, but you might still find them in antiquary bookshops.


Nitrogen along with it's 3 bonds has a lone pair. That's what comes into play. The H+ ion has no electrons of it's own and therefore when it forms bond with N, nitrogen coordinates it's lone pair to H+ completing it's duplet and at the same time it's octet and forms a coordinate bond also known as dative-bond.

You can find more about coordinate bond here


Absorbtion of light by a coordination compound

The given wavelength of $\lambda=600\ \mathrm{nm}$ may be directly converted to various other quantities, namely wavenumber $\tilde\nu$, frequency $\nu$, photon energy $E_\text{p}$, and molar energy $E_\text{m}$:

$$\tilde\nu=\frac{1}{\lambda}=16\,700\ \mathrm{cm^{-1}}$$
$$\nu=\tilde\nu c=\frac{c}{\lambda}=5.00\times 10^{14}\ \mathrm{s^{-1}}=500\ \mathrm{THz}$$
$$E_\text{p}=h\nu=\frac{hc}{\lambda}=3.31\times 10^{-19}\ \mathrm{J}=2.07\ \mathrm{eV}$$
$$E_\text{m}=N_\text{A}h\nu=\frac{N_\text{A}hc}{\lambda}=199\ \mathrm{kJ\ mol^{-1}}$$

With regard to electronic spectra of coordination compounds, usually wavenumbers $\tilde\nu$ expressed in $\mathrm{cm^{-1}}$ are used.

The electronic configuration of $\ce{M^2+}$ in the given complex $\ce{[M(H2O)6]^2+}$ is unknown. Nevertheless, we may assume that the complex has octahedral geometry and that the observed absorption maximum probably corresponds to promoting one electron from the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$) to an $\mathrm{e_g}$ orbital ($\mathrm{d}_{x^2{-}y^2}$ or $\mathrm{d}_{z^2}$). Thus, it directly gives an estimate for $\Delta_\text{o}=10\ \mathrm{Dq}$ as $16\,700\ \mathrm{cm^{-1}}$.

The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{NH3}$, the value of $\Delta_\text{o}=10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are increased.

An empirical equation for estimating the value of $\Delta_\text{o}$ for any pair of metal and ligands is $$\Delta_\text{o}=f \cdot g$$ where $f$ is the ligand parameter and $g$ the contribution from the central atom. The parameter $f$ describes the ligand strength relative to water; i.e. the value for $\ce{H2O}$ is $f=1.00$. The empirical value for $\ce{NH3}$ is $f=1.25$.

Since the ligand field splitting for $\ce{[M(H2O)6]^2+}$ is $\Delta_\text{o} = 16\,700\ \mathrm{cm^{-1}}$, the corresponding ligand field splitting for $\ce{[M(NH3)6]^2+}$ may be estimated as $\Delta_\text{o} = 1.25 \times 16\,700\ \mathrm{cm^{-1}}= 20\,900\ \mathrm{cm^{-1}}$.

The resulting wavenumber $\tilde\nu=20\,900\ \mathrm{cm^{-1}}$ may be converted to a wavelength of approximately $\lambda=480\ \mathrm{nm}$.

Since $\lambda=\frac{1}{\tilde\nu}$, the new wavelength could have been directly calculated from the given original value of $600\ \mathrm{nm}$ using the parameter $f=1.25$ for $\ce{NH3}$:
$$\lambda=\frac{600\ \mathrm{nm}}{1.25}=480\ \mathrm{nm}$$


What is the product of the reaction of carbon with sulfuric acid?

Carbon reacts with sulfuric acid to produce carbon dioxide and sulfur dioxide gas along with water. Sulfuric acid should be a concentrated, heated solution.

$$\ce{C + 2H2SO4 ->[\Delta] CO2 + 2SO2 + 2H2O}$$

More insight about this reaction can be found here.

Copper(II) oxide, a black solid, and dilute sulfuric acid react to produce copper(II) sulfate, giving a characteristic blue colour to the solution. From this solution, blue copper(II) sulfate pentahydrate crystals may be obtained if desired.

$$\ce{CuO + H2SO4⟶CuSO4↓+H2O}$$

It is a general "insoluble metal oxide is reacted with a dilute acid to form a soluble salt" type of reaction. It is a standard class experiment.

More insight about this reaction can be found here.


What is the definition of current efficiency?

The theoretical definition is the ratio of electrons used to make the useful product out of the total electrons that were supplied. The problem in the chemistry literature with efficiency definitions, however, is that you will more than likely find that different people will use different definitions for one reason or another (easier to calculate, some type of comparison in their work, or simply because it sounds or looks better). The right way to deal with efficiencies, then, would be to define how the efficiency is calculated in order for the reader to know what you are talking about. If this is an academic setting, I would stick to the theoretical definition, but ask your professor as to what their favorite definition would be.


Trivial name of a compound vs its chemical formula

No you are absolutely right here, but even sulphate or nitrate is wrong. If you take a closer look at what a sulphate essentially is, you will see the following: It is a separate unit, it has a central atom and ligands and the whole unit is negatively charged. If this was a metal we would say it's a complex and name it with some sort of -ate in the name. And this is exactly what you see here. So tetraoxosulfate in short is sulfate. Whit this shorter notation you are just not able to guess it unless it is so common that you should know about it. For lithium aluminium hydride it's the same thing, it should be called lithium tetrahydroaluminate.

Now you may ask yourself how are you supposed to work with these problems? In all scientific cases just google it. That's what we do as well, if this is class or homework related, then your teacher will only give you certain examples. If he doesn't, talk to him about it.


Why does Calcium (specifically Calcium Carbonate) inhibit the Citrate buffer system?

"Inhibited" is not probably the right term, as there is no catalyst inhibition, affecting the reaction kinetics.

If we consider about neutral $\mathrm{pH}$ range, the citrate buffer keeps $\mathrm{pH}$ by equilibrium reactions

$$\ce{HA^2- <=> H+ + A^3-}$$
$$\ce{HA^2- + OH- <=> H2O + A^3-}$$

with $\mathrm{p}K_\mathrm{a3}=6.4 $

Carbonate as a base spends the buffer capacity

$$\ce{CO3^2- + 2 HA^2- <=> CO2(g) + H2O + 2 A^3-}$$

If it was sodium carbonate, sodium ions would be just bystander ions.

But calcium ions form chelate with citrate 3:2, decreasing concentration of free hydrogen citrate + citrate ions.

So the overall calcium carbonate effect is being base and chelating citrate.


Resonance stability: ester vs. thioester

Try to understand it in a layman's way, greater the extent of overlap between the atoms, the closer their nucleus will be. The stability of the molecule will thus increase as stability of a molecule is inversely proportional to the bond length of the atoms in the molecule.

However this is applicable only upto some extent like your situation but if the nucleus of two atoms get too close a stronger force which is nuclear force starts acting on them which repels them, it is a very short length force.


From many years as a trained firefighter, I can tell you that there is certainly science involved. There are a number of very heavy calculations for calculating things like solid combustible burn time and heat release rates for combustible materials, which might possibly be useful to predict the perfect shape, fuel size and arrangement for a campfire, but are probably well beyond the scope of producing the perfect sausage sizzle. However, a number of key factors that influence fire behaviour, and which must be considered in building the best campfire include:

Size of fuels. For a given fuel type, the larger the fuel piece, the greater the resident burn time, but the greater the energy input required to obtain ignition. So small fuels will ignite and burn first, and provide energy through preheating to ignite larger fuels.
Availability of fine fuels (typically those under 5mm diameter). For ignition to occur, you need to have fine fuels, especially in the absence of other freely burning ignition agents (fire starters, a cup full of petroleum etc). As heat rises, the fine fuels need to be placed at/near the base to preheat larger available fuels. Plenty of fine fuels is critical - you can't have too much.
Available air capacity. Compact fuels that do not have sufficient air flow are poor propagators of fire. Oxygen is one of the critical elements of the fire triangle (or quadrilateral if you choose), and if not being positively injected (bellows etc), there must be sufficient air flow between fuels to allow drawing in of fresh oxygen supplies (As fire burns, the hot gases rise, and fresh oxygen is drawn in at the base, and must be able to freely disperse.
Exposure to radiant and convection heat energy. Heat energy passes via radiant, convection and conductance pathways. Solid fuels need to be preheated to release the combustible gases, and this requires sufficent exposure to radiant and convection energy pathways, which is usually above the burning fuel source.
Fuel moisture. Critical factor. Many fuels with greater than about 15% fuel moisture will have difficulty igniting; around 25% most fuels have difficulty in sustaining burn. Fine fuels exchange moisture with the atmosphere far more readily, and so weather conditions must form a real factor in your calculations (temperature, relative humidity, wind). On hotter, drier days, larger diameter fine fuels can be used for ignition. Drier fuels also burn hotter and provide greater energy output. Moist fuels release lower energy flames, taking longer to ignite other fuel.
Resident burn time. Different materials (including different types of wood) have different resident burn times, and so your choice of fuels will be significant. Iron bark, for instance, has a greater resident burn time, and greater energy output than radiata pine. This derives from the density and composition of the fuels themselves. Fine fuels, of course, have a very short resident burn time, so it is important to have enough to ensure sufficient preheating of larger fuels. How much will depend on fuel moisture!
Surface area of fuels. Smooth, polished wood will take longer to ignite than rough sawn timber

In addition, you may have to stipulate some other limitations:
Availability of renewable fuels. Is you campfire fuels source static, or are you able to add more fuels as the campfire burns. This clearly influences how long it lasts, and heat generation
Stability of design. As the finer fuels burn, there is a point of critical collapse, where the top fuels collapse on the sub-layers. This will influence a number of factors, including availability of air and exposure to preheating.

My design for my favourite campfire is somewhat similar to your log cabin design. It is a crib design, which means it is layered, quite often with the bottom layer of fine fuels (tinder) sitting on a spaced grid of larger sticks to allow air flow underneath. The mix of large fuel and smaller fuels (kindling) is staggered to give stability and access to preheating. The overall shape gets smaller as the layers go up, so the final shape is somewhat square-pyramidal. Separation of fuels needs to be close, while still ensuring maximum air flow.


Are all exothermic reactions spontaneous?

No, it is not the case that all exothermic reactions are spontaneous.  More generally, it doesn't even make sense to make a broad statement that a reaction is or is not spontaneous without also specifying the concentrations of reactants and products, since for a certain relative concentration of reactants and products, the reaction might be spontaneous in the forward direction, while for another set of relative concentrations it might be spontaneous in the reverse direction. [See illustration below.]

One could, however, make the general statement that chemicals will react spontaneously if, in so doing, the entropy of the universe increases.  This, however, is not one of the options you are given.

Why, then, is "B" the correct choice?  Answer: It's because this Ga and As are forming a chemical bond, and bond formation is exothermic. In addition, there is a phase change here from a gas to a solid, and that is also exothermic.  Furthermore, since Ga and As are both monatomic gases in this reaction, no chemical bonds need to be broken to allow the Ga and As to combine. Thus, since we have chemical bonds being formed but not broken, and a phase change from gas to solid, the reaction is exothermic.

Here's a graphical illustration of the general concept articulated in the first paragraph:


what is the issue in evaluating two electron integral?

The main problem of 2-electron integrals is their amount. They are given by

\begin{equation}
  I_{ijkl} = \langle ij|r^{-1}_{12}|kl\rangle
\end{equation}

where $i$, $j$, $k$ and $l$ run over all $N$ atomic orbitals. So we need to calculate $\mathcal{O}(N^4)$ integrals, which makes this a bottleneck (in terms of memory and computing time).

For Gaussian Type Orbitals we can evaluate the integrals analytically, which is much faster than numerical approaches. But this does not help with the $N^4$ scaling.

Evaluation for Gaussian Type Orbitals is explained in Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund for s orbitals ($l=0$). More general evaluation schemes for higher angular momentum are explained in detail in Molecular Electronic-Structure Theory by Helgaker, Jorgenson and Olsen.


What is the viscosity of menstruation blood?

It is difficult to give a typical value as the composition and hence the physical properties of menstrual blood varies over the course of a woman's period. This reference has a good discussion which you may find useful; you can find on p. 159:

  It is fair to say that a large proportion of menses samples are considerably more viscous than blood or water - about four times as viscous as venous blood and 35 times as viscous as water.


Effect of nitro group basicity order of o,m and p nitroanilines?

First we have to understand what affects basic nature of a compound.

Tendency to donate lone pair
Electronic effects— Inductive, resonance etc
Factors stabilizing conjugate acid (or acidity of conjugate acid
Nature of solvent

Tendency to donate lone pair is affected by the acidity of conjugate acid and electronic effects.

Grab a pen and paper and draw conjugate acids of

$$2-nitroaniline$$
$$3-nitroanliline$$
$$4-nitroaniline$$

And you know that electron withdrawing inductive (–I) effect increases acidity and electron withdrawing resonance (–M or –R) also increases acidity.

In conjugate acid of

$$2-nitroaniline$$
You will find there is only –I inductive effect and there is no –M effect of $-NO_2$ group because of SIR (In SIR - STERIC INHIBITION OF RESONANCE - the nitro $-NO_2$ group shows steric hindrance to $-NH_3^+$ group and the p–orbital of Nitrogen (in $-NH_3^+$ group) moves out of plane of p–orbital of Carbon (in benzene ring) which inhibits resonance or does not allow resonance to happen). So in conjugate acid of $2-nitroanliline$ there is only –I effect which increases acidity.
$$3-nitroanliline$$
You will find that there is no –M effect obviously!! But there is –I effect of nitro ($-NO_2$) group which increase the acidity of conjugate acid of $3-nitroanliline$ but it is less than what we have seen in $2-nitroanliline$ (due to distance factors). Therefore conjugate acid of $2-nitroanliline$ is more acidic than conjugate acid of $3-nitroanliline$
$$4-nitroanliline$$
You will find there are both –M and –I effect which increase acidity of conjugate acid of $4-nitroanliline$. So conjugate acid of $4-nitroanliline$ is more acidic than conjugate acid of $3-nitroanliline$ or $2-nitroanliline$

Conclusion
Order of acidity of conjugate acids
$$conj.\ acid\ of\ 3-nitroanliline<conj.\ acid\ of\ 2-nitroanliline<conj.\ acid\ of\ 4-nitroanliline$$

Since the order of basicity of 'orginal' compounds is just opposite of order of acidity of conjugate acids, so answer to your question is as follows
$$3-nitroanliline>2-nitroanliline>4-nitroanliline$$
Or m>o>p


internal alkyne stability to strong alkali

According to A Textbook of Organic Chemistry by V. K. Ahluwalia and Madhuri Goyal

  1-Butyne may be isomerised to a more stable 2-butyne in presence of alcoholic potassium hydroxide and the reverse of this reaction may occur with sodamide

(see 17.2.1 of Best Synthetic Methods: Acetylenes, Allenes and Cumulenes for reaction conditions in the 2-alkyne to 1-alkyne direction, actually many relevant reactions are in this book)

So in the present of the strong base $\ce{NaNH2}$ the internal alkyne $\ce{H3CC#CCH3}$ rearranges to $\ce{HC#CCH2CH3}$

My thinking is that with just hydroxide the more stable 2-butyne is favored, while if an even-stronger base is used, that fact that the acetylenic-H can deprotonate pulls the equilibrium toward 1-butyne.

Additionally, see 17.2.11 Isomerisation of 1,4 bis(alkoxy)-2-butynes to the corresponding allenes, which is catalyzed by t-BuOK.

Add see table 17.1 "Base catalyzed isomerisations of acetylenic compounds" about 10 different internal acetylene reactions are listed in the table.


The Scatchard eqn can be written as $\displaystyle \frac{Y}{[L]}=\frac{n}{K}-\frac{Y}{K}$ with $[L]$ the free ligand concentration and $Y$ the fraction bound, (i.e. $[L]$ bound /total protein) so the x-axis intercept (when $Y/[L]=0$) is $n$ the number of ligands bound, the slope is $-1/K$ and the y-axis intercept is $n/K$. Your plot confirms this equation, the interpretation of intercepts and slope follow from the equation.
(Your plot seems to plot bound concentration not fraction bound so needs correcting with total protein to make $Y$ and this is presumably why $n$ comes out to be a small value)


does aluminum oxide float on molten aluminum and by what mechanism?

A chunk of $\ce{Al2O3}$ would sink in molten aluminum. However, $\ce{Al2O3}$ usually appears in smaller pieces (down to microns), with irregular surfaces, probably filled with air ($\ce{N2}$) from when it was cool. A mix of metallic aluminum and $\ce{Al2O3}$ could allow aluminum to sink thru the solid particles after melting, allowing the particles to rise above the surface of the liquid. $\ce{Al2O3}$ has a very high melting point.

The mass of solid particles would have some very weak structure, and some buoyancy by trapped air. The surface tension of aluminum is very high (~900 dyne/cm) , so it might very well account for the covering of the metal by a heavier substance and the non-wetting of the oxide.

This is reminiscent of floating a needle on water.

In the electrolytic production of aluminum from $\ce{Al2O3}$:

  The electrolyte is a molten bath of cryolite ($\ce{Na3AlF6}$) and
  dissolved $\ce{Al2O3}$. Cryolite is a good solvent for $\ce{Al2O3}$
  with low melting point, satisfactory viscosity, low vapour pressure,
  and density lower than that of liquid aluminum (2 vs 2.3 g/cm3), which
  allows natural separation of the product from the salt at the bottom
  of the cell.

References

https://link.springer.com/article/10.1007%2Fs11663-999-0108-4
https://en.wikipedia.org/wiki/Aluminium_smelting


Can 2‐chloropropane and acetyl chloride give the iodoform test?

As mentioned by user55119, ethanoyl chloride will not give iodoform test, because acid derivatives will show nucleophilic substitution reactions. Resulting will be acetate ion, which is no more a good nucleophile. For a more detailed answer check out here.

2-chloropropane will show iodoform test, since through bimolecular substitution reaction $(\mathrm{S_N2}),$ chlorine will be replaced by $\ce{-OH}$ group. Propan-2-ol so formed will be oxidised to acetone, which will show iodoform test.

Remember that only $\ce{CH3COR}$, (where $\ce{R}$ can be alkyl or aryl part) or which can be converted to it, give haloform tests, until steric hindrance prevent even some of these compounds to react.


Pressure Vs Volume plot for real and ideal gasses

When the pressure tends to infinity, the volume of the ideal gas tends to zero, and the volume of the real gas does not. It tends to the proper volume of the molecules, which is greater than zero. This is exactly what you see on the upper part of the picture.

And when the pressure is extremely low, the volume increases but not as much as calculated with the ideal law, because there is some small attraction between the molecules. This is exactly what you see on your picture if you draw a horizontal line at its lower part


I think the answer to your question can be described based primarily on thermodynamic considerations.  Ta has very high fusion (25 kJ/mol), vaporization (760 kJ/mol) and atomization (780 kJ/mol) enthalpies which are a result of strong metal-metal bonding in the solid (see Greenwood and Earnshaw, Chemistry of the Elements).

To further understand the protective nature of the native oxide that forms on Ta, it is fruitful to look at an Ellingham diagram, which describes the stability of a metal oxide as a function of temperature. For more authoritative sources than the Wikipedia page, see a comprehensive Inorganic Chemistry textbook such as those by Shriver and Atkins or Housecroft and Sharpe.  I found this site particularly interesting for purposes of visualizing Ellingham diagrams and used it to make the following plot:

(Apologies for the color scheme, which was not adjustable)

Here I've plotted the free energy for the oxide formation of tantalum oxide along with oxides of iron and aluminum.  The x-axis is temperature.  The large negative values for each of these oxides indicates that they are all thermodynamically favorable at the temperatures plotted.  Room temperature would be very slightly to the left of the intersection of the x and y-axes.  A complete analysis of the stability of the metal oxides would also require consideration of kinetics, but we can use thermodynamics to postulate that the very large negative values for the free energy of formation of the tantalum oxide are one factor in the stability of the oxide film.  I'll leave it to the OP to play around with thermochemical cycles, which should demonstrate why the oxide films are resistant to many types of acids.


Does vinegar neutralize the strongly basic compounds in cement?

If you look for Portland Cement safety issues, it is stated that cement is highly alkaline so it can cause alkaline burn and presumably since vinegar is acetic acid, it neutralises the alkaline effects of cement as tested here and the burns you suffered are probably related to alkaline burn. You can look at this article section 6.5 for some details on alkaline burns, which was suggested by @Cornelius or perhaps the mentioned chapter in this book (Harchelroad FP, Rottinghaus DM. Chemical burns. In: Tintinalli JE, Kelen GD, Stapczynski JS, Ma OJ, Cline DM, eds. Emergency Medicine: A Comprehensive Study Guide. 6th ed. New York, NY: McGraw-Hill; 2004: chap 200.).


Your textbook is correct, the $K_\mathrm{w}$ expression is as follows:
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}$$

Since for every $\ce{OH-}$ there must be one $\ce{H+}$ ($\ce{H+}$ in water is synonomous with $\ce{H3O+}$) the mol ratio is 1:1, if moles are equal the molarities are equal. I'm going to substitute $x$ in for both
$$K_\mathrm{w}=[x][x]=[x]^2=1.0\times10^{-14}; x= 1.0\times10^{-7}.$$

So the molarity of $[\ce{OH-}]$ and $[\ce{H3O+}]$ are both $1.0\times10^{-7}$ That's a really really low amount, small enough to ignore.

However, a lot of reactions (especially in Organic Chemistry) you'll see a negative ion, for example, $\ce{Cl-}$ pulling the $\ce{H+}$ off of a water. So when $[\ce{OH-}]$ or $[\ce{H3O+}]$ are needed, water can be torn apart to make it.


Does the entropy of a chemical system increase in an endothermic reaction?

Not necessarily.
Exo- and endothermic refer to the direction in which heat is exchanged between system and surroundings, not to changes in the entropy of the system, or of the temperature of either system or surroundings. Your logic though is partly correct: exothermic means the entropy of the surroundings increases, as it would do with an increase in T - the catch is that we usually assume that the surroundings has constant T, but that's because we say it has infinite heat capacity.
It is a recurrent subject of confusion to conflate changes in the entropy of the surroundings, of the system, and of the sum of both. Always keep the second law in mind. What determines whether a reaction will be spontaneous is the sum of the changes in entropy of the surroundings and the system.
When a reaction occurs reversibly, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}=0$.
When spontaneous, $\Delta S_{\mathrm{surr}} + \Delta S_{\mathrm{sys}}>0$.
When a reaction is exothermic,  $\Delta S_{\mathrm{surr}} >0$.
When a reaction is endothermic,  $\Delta S_{\mathrm{surr}} <0$.
Nothing about the thermicity (whether the reaction is endo or exothermic) tells you what happens to $\Delta S_{\mathrm{sys}}$, unless you already know whether the reaction is spontaneous or not (in which case the sum of the entropy changes is positive).


Does brewing tea with hard water reduce the amount of bioavailable fluorine?

$\ce{CaF2}$ has a relatively low solubility in water, about 15 mg/L, with a $K_{sp}$ of about $4\times 10^{-11}$.  According to the paper you linked, the concentrations of fluoride in tea are around 5 mg/L, which corresponds to about $1.3\times 10^{-4}$ M. Thus, the concentration of $\ce{Ca^2+}$ required to precipitate the fluoride out is about 2.5 mM, which corresponds to 0.1 g/L. Hard water is defined as having a calcium concentration above about 0.6 mM, with very hard water being 1.8 mM calcium. Thus, it is unlikely that there would be high enough concentrations for $\ce{CaF2}$ to precipitate out of solution.

Because of the large solvation energy of fluoride ion, the molecules in solution are likely to be nearly completely ionized. That is, the fluoride anions and the calcium cation will each have their own solvation shell of water molecules. That means that they will not interfere with each other until the concentrations are high enough that solid $\ce{CaF2}$ begins to precipitate.

As long as the concentrations are below the solubility limit, the presence of calcium ions will therefore not affect the availability of the fluoride ions.

Furthermore, since HF is a weak acid, in the acidic environment of the stomach some of the fluoride ions will be protonated to HF, decreasing the concentration of free fluoride and increasing the apparent solubility of $\ce{CaF2}$.


Reaction mechanism of aqueous ammonia with 3,4-dibromopyridine

The reaction between ammonia and 3,4-dibromopyridine is a nucleophilic aromatic substition reaction (SNAr). This is possible because the ring is rather deactivated due to the presence of the electron withdrawing nitrogen atom of the pyridine drawing electron density towards it. The ammonia attacks carbon at the 4-position preferably, rather than the carbon at the 3-position as this position is more deactivated than the latter. This attack results in the breakage of the double bond, resulting in a loss of aromaticity. This results in the formation of a carbanionic Meisenheimer intermediate. The bromine at the 4-position then leaves and the double bond reforms, resulting in the restoration of aromaticity in the ring. This results in a net substitution of $\ce {-Br}$ with an $\ce {-NH2}$ group.