How to calculate the number of atoms in a gas molecule?

Reason for rewrite: refer to edit summary.

From Wikipedia on Disulfur ($\ce{S2}$)

  This violet gas is commonly generated by heating sulfur above 720 °C

Since the sulfur is well above $720^\circ\pu{C}$, it would most likely be in the diatomic state, more like an analogous structure to dioxygen ($\ce{O2}$).

We can also judge this from the molar ratios too, as rose22 has pointed out. Since it is given that $40~\pu{mL}$ of $\ce{SO2}$ is formed from $20~\pu{mL}$ of $\ce{S_x}$ (assuming the atomicity of $\ce{S}$ to be $x$ at this temperature) and $20~\pu{mL}$ of $\ce{O2}$, they must come to terms with a simple whole number ratio.

$$\ce{\underset{20~\pu{mL}}{S_x}+\underset{40~\pu{mL}}{xO2}->\underset{40~\pu{mL}}{xSO2}}$$

Statement 1: Since the ratio of volumes of gases at a particular temperature and pressure is equal to the ratio of amount of substance in moles, it is clear that:

$$1:x:x=20:40:40$$

That is to say, $x=2$, so the atomicity of sulfur here is $2$.

In the question, the OP said:

  However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?

Statement 1 is a  result from the universal gas law.

$$PV=n\text{R}T$$

When $P$ and $T$ are constant,

$$\frac{V_1}{V_2}=\frac{n_1}{n_2}$$

Some may argue that this relationship is called Avogadro law, but again, this law too is a gas law, evidently.


This thesis, (Texture and Hydration of Expanded Rice, Clive Norton, September 1998), has a vast amount of information on rice, including puffed rice.

Rice is mostly starch, which when heated is gelatinised. This causes "the breaking of hydrogen bonds between poly-α-1,4-glucan chains, leading to loss of crystallinity" allowing those hydrogen bonding sites to absorb more water. The loss of crstallinity also leads to air pockets making puffed rice sponge like. The starches become "interspersed with gas pockets" which are filled by water through capillary action.

So, puffing makes the starch more hydrophillic, exposing hydrogen bonding sites, and the air pockets act as capillaries. The puffing also increases the ratio of surface area to weight. These factors cause the rice to be liftable with a "water chain".


Is the recipe for my shaving cream complete nonsense?

As @Mithoron commented, these are ingredients, not a chemical breakdown of the result.

And, as @Karl commented, traditionally you mix together some rendered fat or oil -- that's a mixture of fats and oils with the fatty acids and glycerin that make them up -- with potassium or sodium hydroxide to make the potassium or sodium salt (conjugate base) of the fatty acids, which is soap.

Here, you've got:

Solvents:

Water
Limonene

Fats, glycerin, and fatty acids:

Stearic Acid
Myristic Acid
Coconut Acid
Glycerin
Theobroma Cacao (Cocoa) Seed Butter
Melaleuca Alternifolia (Tea Tree) Leaf Oil

Base:

Potassium Hydroxide
Sodium Hydroxide
Triethanolamine

Extra smells

Parfum (Fragrance)
Also Limonene
Linalool
Also tea tree oil

The only reason it has so many is that they've mixed together either a few different soaps or a few different fats before saponification (soap-becoming), and each soap will have been (mostly) neutralised.

Edit: note that the cocoa butter and tea tree oil are almost certainly added after saponification, for fragrance, to alter the material properties of the product, and to be left on the skin.


According to this patent, alkaline catalyst of composition sodium and potassium hydroxides, carbonates, formates, and acetates can be used to speed up formation of paraformaldehyde from formalin solution (aq. formaldehyde).

  This invention relates to a new and improved process for the
  production of formaldehyde polymers. More particularly it relates to a
  process whereby high strength aqueous formaldehyde solutions can be
  concentrated to paraformaldehyde at a greatly accelerated rate.

Moreover, the patent describes the process of this reaction in details. Have a good read.


How do we know which compounds have higher solubility?

It depends on the solution and the solute. Usually, the rule is "like" dissolves "like". A polar solution will dissolve polar compounds and a non-polar solution will dissolve non-polar compounds.

https://www.youtube.com/watch?v=PVL24HAesnc

As far as I know, to find the solubility you have to determine it experimentally. This can be done by "feel" first. If you have a compound that doesn't have many polar groups on it, chances are it can be dissolved by a non-polar solution. If you know what the structure looks like try to match it with a solvent that looks similar to it.


Why is energy released as electrons move?

The previous answer is correct. I just wanted to add that atoms, electrons and generally matter, tries to ''organise'' itself in order to adopt a state, where it would have the lowest energy possible.
An electronegative atom tends to attract electrons and has the ability of handling their charge better. Thus when an electron moves from a less electronegative atom to a more electronegative one, it is like moving from a state of high energy to a state of low energy. This is also described as realease of energy.


If the thiolate bounds to the Au nanoparticle, both the electronic, as well as the mechanical properties of the thiolate and the Au nanoparticle change.  So you are going to monitor these changes with a spectroscopy susceptible to this, comparing the spectra of the pure starting materials with the one(s) of the decorated NP.

One of the easier available techniques is to rely on FT-IR spectroscopy, as by the established bond between the two partners, some absorption bands newly appear, and others change their position.  The former are the ones more easily taken as a quantitative reference, too (because yes, IR spectroscopy allows both qualitative, as well as quantitative characterisations, too).

In the instance of Au NP, to check for a bonding of the NPs to the mercapto decoration is possible both by IR spectroscopy, as well as by (surface enhanced) Raman spectroscopy.  In a work deploying $\omega$-mercaptoundecanoic acid (MUA) by Tripathy and  Yu, the later was the preferred check to monitor the S-Ag interaction:

(source, doi 10.1016/j.saa.2008.12.004, Spectrochimica A, 72, 2009, 841-844).

Similar as in the case of the Au NP (vide infra), the key feature is the disappearance of the S-H vibration.  To quote from the section about their characterisation by Raman spectroscopy:

  The most interesting feature was the disappearance of S–H stretching
  ($\pu{2547 cm^{−1}}$) peak. The disappearance of S–H stretching indicated the covalent interaction of sulphur and silver, which is in good agreement with the FT-IR data [16]. However some of the expected peaks such as C–Sgauche stretching could not be identified which may be due to their low intensity values.

the earlier answer, erronusly about Au NP

As an example, were gold nanoparticles were decorated by dodecanthiol (DT) chains:

(source, Sharma et. al. in Indian Journal of Pure & Applied Physics
52, 2014, 93-100.)

Other than the absorption band at $694$ and $\pu{728 cm^{-1}}$, the authors equally draw attention that

  the "S-H vibration band at 2576 cm−1 for pure DT is observed while it is almost absent for AuNPs. This suggests that as the thiol gets bound onto the  surface of gold, the S-H bond of thiol breaks in order to form the S-Au bond." (loc. cit., p. 97).

It may be helpful to mention that short wavelengths of light ($\pu{< 400 nm}$) may break the $\ce{Au-S}$ bond between nanoparticle, and thiol, hence if you doubt the presence of thiolate decoration on the NP, select an absorption band and monitor its attenuation while shining the short wavelength light on the sample.

And complementary to IR spectroscopy, is Raman spectroscopy.


Does the decay of Carbon-14 effect health in biological systems?

I mean take a glucose molecule C6H12O6, suddenly one of the Carbon atoms randomly decays to a Nitrogen atom. Does this molecule continue to function as a glucose molecule ? This question extends to every molecule in the body that contains Carbon atoms.

To answer this part of the question, obviously a glucose molecule which has had a carbon-14 atom decay isn't a glucose molecule anymore. What the glucose molecule turns into is anybody's guess, but no doubt there would be a wide range of molecular fragments from the multiple decays of glucose molecules. Some of the molecular fragments would be highly reactive and cause unusual reactions within our bodies.
Worse, the cabon-14 emits a beta particle when it decays. It would be sort of like a bullet ripping through the body. Thus the beta particle would disrupt the bonding in other molecules, not just in the original glucose molecule.


Does anyone have information about Piquet and Calitet (chemists)?

Can you exclude spelling errors? If not, there is

Amé Pictet, known from the Pictet-Spengler cyclization.
Claude Piguet, an inorganic chemist at the University of Geneva.

Both were/are is Swiss, not French.

However, the second name does not ring a bell for me.


¹H NMR and MS analysis of products obtained by Claisen condensation reaction between methyl 2‐methylpropanoate and sodium methoxide

Looking at your comment, I believe the main source of confusion may be understanding the underlying reason why the Claisen condensation requires the two α protons.
The first one, obviously, is required for the enolate to form. The second one is arguably equally important because it is the removal of this proton in the condensed product that prevents a retro-Claisen. The Claisen condensation is equilibrium driven to the product by that last deprotonation. Under the conditions listed here, you cannot deprotonate the product to generate the "real" product, so a retro-Claisen will quickly take place to provide the original starting material.
The "desired" product is the bis-methylated β-keto ester. You cannot use the conditions for a Claisen, but, as mentioned in comments, alkylation of the unmethylated β-keto ester should work. If you use a strong electrophile, like methyl iodide, you shouldn't need any harsh basic conditions, especially given the fact that a β-keto ester is quite acidic.


Why does electronegativity create a stronger acid, but so does greater electron distribution?

The main point about strong acids is ‘how stable is the conjugate base?’

The better stabilised the negative charge on a conjugate base of a neutral acid is, the happier the species will be to loose a proton and thus the more acidic the proton is. All of the points you noted just emphasise that.

Therefore, the only wrong assumption you are making is:

  more stable base = weaker acid.

The correct way is:

  more stable base = stronger acid.


I risk not directly answering the question as I have not perfectly understood it.

In your first example, you consider the reaction between two neutral atoms, $\ce{Na}$ and $\ce{Cl}$. An electron is transferred generating $\ce{NaCl}$ where the species are now charged ($\ce{Na^+ \  Cl^-}$). Then you follow up with a hypothetical reaction between a neutral atom ($\ce{Ca}$) and an already charged species ($\ce{OH^-}$). The two reactions are fundamentally different, but not because only the latter has polyatomic ions; the difference is in the very nature of the reactions you constructed (two neutral species reacting in the former, compared to one neutral species and an ion reacting in the latter).

The most direct way to obtain $\ce{NaCl}$ is the reaction between sodium metal (composed of neutral atoms, $\ce{Na^0}$) and chlorine gas (composed of neutral diatomic molecules, $\ce{Cl2}$). One then has:

$$\ce{2Na^0_{(s)} + Cl2_{(g)} → 2NaCl_{(s)}}$$

To obtain $\ce{Ca(OH)2}$, a similarly simple reaction would be that between calcium metal (composed of neutral atoms, $\ce{Ca^0}$) and water (composed of neutral molecules, $\ce{H_2O}$), where the water serves as a source of hydroxyl ions according to the equation:

$$\ce{Ca^0_{(s)} + 2H2O_{(l)} → Ca(OH)2_{(s)} + H2_{(g)}}$$

Now imagine you want to force a reaction between a neutral atom and an ion. There are many cases of this, but we can stick to something similar to what you wanted. In order to expose a substance directly to ions, a conceptually easy way is to add it to a molten salt. For now, let us consider pumping sodium vapour into molten caesium hydroxide ($\ce{CsOH}$). You seem to expect some interaction between the neutral $\ce{Na^0}$ and charged $\ce{OH^-}$, but in fact it is $\ce{Cs^+}$ that reacts with the sodium, according to the following equation:

$$\ce{Na^0_{(g)} + CsOH_{(l)} → NaOH_{(l)} + Cs^0_{(g)}}$$

Here we have simple exchange reaction. The hydroxide ion is actually only a spectator, and can be omitted without loss of understanding in an ionic equation:

$$\ce{Na^0_{(g)} + Cs^+_{(l)} → Na^+_{(l)} + Cs^0_{(g)}}$$

If we consider the reaction between sodium and molten caesium chloride instead, where chloride ions would be present, we have again a simple exchange reaction:

$$\ce{Na^0_{(g)} + CsCl_{(l)} → NaCl_{(l)} + Cs^0_{(g)}}$$

Indeed, it can be described by the same ionic equation above.

(Disclaimer: I actually am not sure the reactions between sodium and caesium compounds works as above. I don't have any understanding of the relative electrochemical potentials in molten salts, though I believe I've read that the reaction with molten chloride is a legitimate way to obtain metallic caesium. But that is mostly beside the point, either way. I avoided exemplifying with calcium and calcium hydroxide because I think the reaction wouldn't work, and the hydroxide is thermally unstable)


How do I find the expiration time of drug in the following example?

Your general approach to the calculation looks great.  My only question is, what is the meaning of $x$ in your calculation?

Here's how I would do the problem.

$$\ce{A->B}$$

$$A = A_0 e^{-kt}$$

$$\frac{A}{A_0} = e^{-kt}$$

$$0.35 = e^{-(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t}$$

$$ \ln{0.35} = -(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})t$$

$$ t = \frac{-\ln{0.35}}{(~{\mathrm 1 \times 10^{-8} {\mathrm s^{-1}}})}$$

If I did the units conversion from $\mathrm{s^{-1}}$ right, then the rate constant in units of $\mathrm{{yr}^{-1}}$ is 0.316.  (I just multiplied by 60×60×24×365.25).

$$ t = \frac{-\ln{0.35}}{(~{\mathrm 0.316~{\mathrm {yr}^{-1}}})}$$

When I plug in the numbers, I get 3.2 years.  It's possible I did something wrong, but my answer is way closer to yours than to the given answer.

Here are some possible mistakes that the original writer of the question could have made to give the wrong, lower, answer.

The question has a bunch of superfluous information about the chirality of the drug.  Mistakenly thinking that this is important to the question, perhaps somebody divided by 2 to "adjust" for chirality.  But this is not the correct approach.  We're given the activity of the drug and the decay rate of activity, so the structure of the molecule is completely irrelevant to the problem.
Another possible error is misinterpreting "is reduced to 35 % of original" mistakenly as "is reduced by 35%".  That leads to the calculation for $t$ being $\frac{-\ln{0.65}}{0.316}$ and gives the expected (but wrong) answer.

The bottom line is I think you mostly did the calculation correctly and the given answer seems incorrect, at least given the wording of the question as you supplied it.


I never really learned the method of solving this stuff with Avogadro's Law, but here's an almost identical way.

Your balanced equation is as follows...

$\ce{2H2S + 3O2 -> 2SO2 + 2H2O}$, and I think you agree on that by looking at your calculations.

At a given temperature and pressure, $1 mol = x L$. For example, at STP (standard temperature and pressure) this is $1 mol = 22.4 L$. It wasn't exactly defined what temperature and pressure are, but they are constant, so we can use a variable $x$ to denote some conversion (which cancels out anyway, I assume giving rise to Avogadro's Law).

$12.0L_{\ce{SO2}} * \frac{1mol_{\ce{SO2}}}{xL_{\ce{SO2}}} * \frac{2mol_{\ce{H2S}}}{2mol_{\ce{SO2}}} * \frac{xL_{\ce{H2S}}}{1mol_{\ce{H2S}}} = 12.0L_{\ce{H2S}}$

But this is assuming $100\%$ yield. I see you divided by $0.885$ above, which is correct, but I've gotten a metric ton of these wrong before, so I prefer the proportions way.

Regardless, $\frac{12L_{\ce{H2S}}}{0.885} = 13.6L_{\ce{H2S}}$

I think you definitely overcomplicated things by assuming a density of $1g/mL$ among other things. Your dimensional analysis also seems strange near the end of the left hand side where you somehow cancel $g$ with $mol_{\ce{SO2}}$.

If I were to do it with Avogadro's Law (taking a guess here), I'd still get $12.0L$ because the mole ratio is the same. And then I'd divide like you did.


Should an entire compound or just the redox-active element be written in redox pairs?

You generally use whichever compound is actually present in solution.

For manganese(IV), it will practically always be $\ce{MnO2}$ so use that. Manganese(VII) will always be $\ce{MnO4-}$ so use that. Manganese(II) salts dissolved in water dissociate and coordination complexes of manganese(II) are very labile. Thus, it doesn’t make sense to assume any $\ce{[MnL_n]^2+}$ complex to predominate and you should simply use $\ce{Mn^2+}$.

It does, for example, make a difference if your iron redox pair is $\ce{Fe^2+/Fe^3+}$ or $\ce{[Fe(CN)6]^3-/[Fe(CN)6]^4-}$. Likewise, a reduction using tin(II) should better be written $\ce{Sn^2+/[SnCl6]^2-}$.

If you are in doubt whether something dissociates or whether it should be considered a single entity, the cleaner solution is probably to write the fully dissociated salt, so you can get away with using $\ce{Sn^2+/Sn^4+}$. However, as soon as molecules or oxyanions are involved you cannot resort to this; oxalate reductions have to be $\ce{C2O4^2-/CO2}$ and permanganate has to be $\ce{MnO4-/product}$.


Relative molecular mass of an unknown gas in a mixture

After getting relative molecular mass ie $39.6$,

$$\mathrm{M_{net} = \frac{(n_1\times M_1 + n_2\times M_2)}{(n_1 + n_2)}} \tag{i}$$

Given that the molar ratio of gases is $4:1$, the molar ratio of oxygen is $\mathrm{4x}$ and that of the unknown gas is $\mathrm x$. Plug $\mathrm{n_1 = 4x,\ n_2 = x,\ M_1= 32,\ M_2 = M,\ M_{net} = 39.6}$ in equation (i) and get $\mathrm M$.


It appears to be an e1cb mechanism, where formation of the carbanion on the dichloro end is the rate limiting step. This appears to be a far more acidic proton than the ones on the fluorine-bearing carbon. Once the carbanion is formed, it rapidly eliminates fluoride ion to give the stable dichloroethylene product, rather than eliminating chloride (admittedly a better leaving group) to give the carbene, a higher energy product.


How can I determine the size of tropolone molecule?

You wanted to measure the distance "across tropolone" but you didn't specify from where to where. The distance from a remote hydrogen to each of the oxygens and to the hydroxyl hydrogen are shown below. You can do the measurements you want yourself here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ESPotential/ESPmisc2.html


Different formulas for copper pyrites and bauxite

Both $\ce{Cu2S.Fe2S3}$ and $\ce{CuFeS2}$ are the equivalent means to denote chalcopyrite. The first notation, $\ce{Cu2S.Fe2S3}$, commonly used a few decades ago, shows that two sulfides are not just a mechanical mix, but form a chemical compound (same as for crystallohydrates, e.g. $\ce{CuSO4 * 5 H2O}$). The second one, $\ce{CuFeS2}$, is a formula unit, a more universal and modern representation. Chalcopyrite is a mineral of $\ce{ABX2}$ type, crystallizes in $I\bar{4}2d$ space group.

Figure 1. Unit cell of chalcopyrite $\ce{CuFeS2}$. Color code: $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.

On the other hand, $\ce{Cu2FeS2}$ is a reduced formula of $\ce{Cu8Fe4S8}$, a superstructured bornite [1]. A compound of $\ce{AB2X2}$ type, crystallizes in $F\bar{4}3m$ space group.

Figure 2. Unit cell of superstructured bornite $\ce{Cu8Fe4S8}$. Color code: $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.

Structurally, chalcopyrite and superstructured bornite have very little in common. Unless there is a specific context given, I'd rather say that $\ce{Cu2FeS2}$ is an outlier among the three and is probably a typographic issue. Also, it's not a good practice to mix dot-notated formulas with formula unit representations unless one wants to underline some structural aspects (e.g. molecular assemblies/coordination polyhedra/domains etc.)

References

Ding, Y.; Veblen, D. R.; Prewitt, C. T. Possible $\ce{Fe/Cu}$ Ordering Schemes in the 2a Superstructure of Bornite ($\ce{Cu5FeS4}$). American Mineralogist 2005, 90 (8–9), 1265–1269. https://doi.org/10.2138/am.2005.1518.


Can reversible reactions proceed by different mechanisms in the forward and reverse directions?

Yes, you can. An ideal Carnot cycle (admittedly not a reaction) illustrates how this is possible. In such a cycle each step is reversible. To return to a state you can either reverse the direction of the process (running say a heat engine as a refrigerator) or follow the full path of the cycle until the state is again encountered.


Is there a standard symbol for molar density?

To my knowledge, n/V is known as molar concentration and uses symbol $C$ and it seems okay for me to use $C_t$ i.e. total concentration to mark sum of concentrations of all particles in system.


This looks like the differential of the ideal gas equation. I'm going to use $V_m$ for the molar volume.
$$PV_m=RT$$
Now we convert each side to the differential. We keep $P$ and $V_m$ together since they are both variable. We can move $R$ out of the differential because it is a constant.
$$d\left(PV_m\right)=RdT$$
We can evaluate $d\left(PV_m\right)$ using the product rule:
$$d\left(PV_m\right) = PdV_m + V_mdP$$
Now:
$$PdV_m + V_mdP = RdT$$
Which is equivalent to your first equation.
**However, at constant temperature $dT=0$, so at constant temperature, the equation simplifies to:
$$PdV_m + V_mdP=0$$
$$V_mdP = - PdV_m$$
$$\frac{dP}{dV_m}=-\frac{P}{V_m}$$
Which is not equal to your second equation. The closest I can get is the following:
$$\frac{dP}{dV_m}=-\frac{P}{V_m}\cdot\frac{V_m}{V_m}=-\frac{PV_m}{V^2_m}=-\frac{RT}{V^2_m}$$
Which is what we would get by evaulating:
$$\left(\frac{\partial}{\partial V_m}\right)_T \left(P=\frac{RT}{V_m}\right)$$


Why does a bonding molecular orbtial increase stability?

The forces involved are electrostatic, nuclei repelling one another and electrons similarly, but also being attracted to the opposite nuclear charges. Also there is kinetic energy of the electrons (for simplicity we consider the nuclei to be stationary) which form the total energy of the molecule.

The simplest molecule to consider is $H_2^+$. The interactions have potential energy arising from Coulomb interactions $$\displaystyle \displaystyle V(r) = -\frac{e^2}{r_1}-\frac{e^2}{r_2}+\frac{e^2}{R}$$

where $r_1,r_2$ are the separation of the electron from the nuclei and $R$ the separation of the nuclei themselves. In addition there is kinetic energy to consider and this has the form $p^2/2m$ where $p$ is the momentum (operator). Combining these terms the Schroedinger equation is formed and in doing so some integrals have to be solved.

Some of these have classical counterparts and are Coulomb type integrals, but one type, the 'resonance' or 'exchange' integral, is purely quantum mechanical. It is difficult to describe in words but it expresses the possibility of the electron moving from the vicinity of one proton to that of the other. All these integrals are positive but add together according to the signs in the potential.  The upshot of this is that the total energy reaches a (negative) minimum as the separation between nuclei is increased from a very small value, and becomes zero when the nuclei are well separated.

The electron has kinetic energy as well as potential energy so besides the potential energy effect of reducing the repulsive interaction between nuclei, by virtue of having increased density between them, the positive kinetic energy can counteract this effect.

To find out which is most important the Virial Theorem (as result from classical mechanics) can be used. In the special case of a diatomic molecule the average kinetic energy is given by $\langle T\rangle=-E-R(dE/dR)$ and the potential energy $\langle V\rangle=2E+R(dE/dR)$ where $E$ is the total energy at $R$ the internuclear separation.

The plot below shows that it is the minimum in the potential energy that caused the chemical bond to be formed not the smaller minimum in the kinetic energy which is positive anyway. Thus in forming a bond the kinetic energy increases (destabilising) but the potential energy decreases by more and this stabilises the bond.  This is shown below where $\rho$ is bond length in reduced units and $E_I$ is the ionisation energy. $E_0$ is the minimum total energy.


Do unsaturated fatty acids with an odd number of carbons exist?

C21:3 is a stable fatty acid that could indicate a variety of double-bond isomers.  It is "unusual" in biology because of the odd number of carbons, but that doesn't preclude its stability and accessibility to organic synthesis.  If someone really wanted to make this acid, they could do so easily.
Such a compound, 5Z,14Z,17Z-heneicosatrienoic acid,  has been reported in the natural environment.  It was found in a deep-sea clam, Calyptogena phaseoliformis, that lives in or near hydrothermal vents and lives off of the organic matter produced by chemosynthetic bacteria that live in close proximity to it.  So in addition to being synthetically accessible, it is also made in living organisms.

Thus, for all reasonable interpretations of "exist" in your question, does this fatty acid even exist?, the answer is yes.


Physical Interpretation of Gibbs Free Energy

Gibbs free energy is a  measure of the total non-expansionary work that a system can do. So in the case of your example, that system has the potential to do 20 joules of non-expansionary work per a mole of substance. Non-expansionary work refers to work other than pressure-volume work, such as electrical work. Hence not only is Gibbs free energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non-expansionary work a system, such as motor engine, can do.

Gibbs free energy is really derived is from the Clausius inequality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets assume that the pressure is constant so we can simplify this relation.

Constant Pressure

Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous


When simulating spectral line broadening, which convolution is preferred?

I'm not familiar with the computational packages -- I'm an experimentalist, not a theoretician.

As an example for why both might be present, we can turn to gaseous infrared spectroscopy:
one will quickly find that the line-widths in vibration-rotation spectra depend on pressure. At lower pressures, say less than a torr, Doppler-broadening is the main mechanism. This is Gaussian in nature. While at higher pressures the broadening is instead due to collisions, which are Lorentzian. You'll find combinations of the two and so forth depending on the specifics of your case.


What are some ways to predict that a double replacement reaction will occur?

To determine if a double replacement reaction will occur, you need to look at the products. You correctly balanced the equation and flipped the anions and cations, so let's take a look. A double replacement reaction will occur if one or both of the products is an insoluble precipitate, a liquid, or a gas. To check each of the products you use the solubility rules. You check each rule in order for each of the products. Here is a link to the rules I found online:

http://www.dorettaagostine.com/Solubility%20Rules.pdf

Look at the double replacement reaction chart. The rules are followed in order. Let's look at $\ce{KNO3}$ first. The first rule states that alkali metal salts are soluble. Potassium is an alkali metal, therefore potassium nitrate is soluble. The second product is $\ce{Pb(OH)2}$. Looking through the rules, none of them apply until the 5th rule which states that hydroxides are insoluble. Therefore lead hydroxide is insoluble.

As you can see, 1 of the products is insoluble, therefore the double replacement reaction will occur. I hope that helps! If you have anymore questions or clarifications just ask.

EDIT: How to find the net ionic equation.

First reduce the equation into it's respective ions.
$$\ce{2KOH(aq) + Pb(NO3)2(aq) -> 2KNO3(aq) + Pb(OH)2(s)}$$
...becomes...
$$\ce{2K+(aq) + 2OH-(aq) + Pb^2+(aq) + 2NO3-(aq) -> 2K+(aq) + 2NO3-(aq) + Pb(OH)2(s)}$$
However, keep in mind that $\ce{Pb(OH)2}$ does not break down into ions because it is insoluble, as we discovered earlier. Therefore it stays in it's full formula. At this point you do the "cancelling out" by removing duplicate ions on each side of the equation. The cancelled ions are also known as spectator ions because they do not actively participate in the synthesis of the product.
$$\ce{2OH-(aq) + Pb^2+(aq) -> Pb(OH)2(s)}$$
The potassium and nitrate ions are removed because they appear on each side of the equation in the same quantity. These cancel out leaving the net ionic equation show above.

The net ionic equation will show the creation of the insoluble, liquid, or gas precipitate that forms as a result of the reaction. I hope that helped!


What happens when butane and methane both absorb 100 J of heat?

First, consider this from a nice write-up on heat capacity:

  The amount of energy needed to raise the temperature of a substance by one degree in temperature will depend upon how many different ways the atom or molecule can store the energy in the form of these thermal motions and how much energy each kind of thermal motion involves.In general, for molecules with similar bonding the bigger the molecule the more heat has to be added to raise its temperature by one degree.

Now, examine the degrees of freedom for each molecule: there are more ways to store energy in the larger butane molecule (with 36 vibrational modes in addition to the translational and rotational modes) than there are in methane (9 vibrational modes plus translational and rotational ones).

Therefore, we'd expect a greater increase in temperature for the smaller molecule than the larger one for an equivalent amount of heat added to the system.


Why thermodynamic state variables of a sytem are only defined at thermodynamic equilibrium and not during any other instant?

Variables such as T and p are not well-defined when a system is fluctuating. Thermodynamic state variables are defined for systems that are static or quasi-static (in the latter case, transitioning very slowly - and therefore reversibly - through a series of equilibrium points). The main reason is that macroscopic fluctuations are absent in true equilibrium states. Each phase in such states is homogeneous. The argument is thus somewhat circular: thermodynamic states are homogeneous states that have fixed properties. The only way to guarantee fixed properties is by being at equilibrium.

Note that it is possible to obtain thermodynamic parameters by imposing restraints that trap a system in a local free energy minimum, and thereby prevent the system from transitioning to a global free energy minimum.


Yes, it's possible but as pointed out in Bamboo: an overlooked biomass resource? Biomass and Bioenergy, Vol. 19, pages 229-244:

  [bamboo] shares a number of desirable fuel characteristics with certain other bioenergy feedstocks, such as low ash content and alkali index

Potassium oxides are between 33-50% of the ash depending upon species and age of the plant.

Potassium oxides were between 0.16-0.58% of dry matter.

Sodium oxides were about a factor of 100 lower.


Does a red hot metal reflect light in the same way as a cold metal?

The light that you see from hot metals is incandescence.
Incandescence is caused by high temperature. The high temperature promotes electrons to higher energy levels. These electrons then relax and emit light.

On the other hand, the "light" (shiny) that you see from room temperature metals  is caused by re-emission/reflection of other light. Light from another source is absorbed by the metal and re-emitted, or the light from another source is reflected from the surface of a metal.

The reason as to why metals look shiny have to with plasmons. If I recall correctly, most materials will look "shiny" at a certain wavelength range. It just happens that metals are shiny in the visible wavelength range.

What determines what type of light a material will reflect is determined by its plasma/plasmon frequency. Temperature does seem affect the plasmon/plasma frequency. Several studies confirm this.

In general, it appears that the plasmon/plasma frequency decreases with increasing temperature. In other words, the material is able to reflect a narrower range of light as the temperature is increased. This would mean that if you had a material that is shiny in the visible light range, you could make it become dull by heating it. Of course, this is a simplification that doesn't take into account other factors.

However, I wasn't able to find an equation that directly relates temperature and plasma frequency. How the plasma frequency changes with temperature seems to depend on the material and the size of the material (nanoscale, bulk, etc). Furthermore, there are ab initio studies that suggest that the opposite must happen: the plasmon frequency increases with temperature.


Mathematically, how does the exchange integral for a closed-shell system reduce to zero?

Let's do the math as explicit as possible. So, the exchange integral is defined as follows,
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
:=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2)
r_{12}^{-1}
\chi_2(\vec{x}_1) \chi_1(\vec{x}_2)
\mathrm{d} \vec{r}_1 \mathrm{d} \vec{r}_2 \, ,
$$
where I used a more appropriate label for the spin coordinate ($m_{s}$) as well as distinguished between summation over discrete variables $m_{s1}$ and $m_{s2}$ and integration over continuous $\vec{r}_1$ and $\vec{r}_2$ ones.

Then, for the restricted spin orbitals,
$$
  \chi_1(\vec{x}_1) = \psi_1(\vec{r}_1) \alpha(m_{s1}) \, , \\
  \chi_2(\vec{x}_2) = \psi_1(\vec{r}_2) \beta(m_{s2}) \, ,
$$
where $\alpha$ and $\beta$ are the so-called "spin up" and "spin down" spin functions defined as follows,
$$
    \alpha(m_{s}) =
    \begin{cases}
        1, & m_{s} = +1/2 \\
        0, & m_{s} = -1/2
    \end{cases} \, ,
    \quad
    \beta(m_{s}) =
    \begin{cases}
        0, & m_{s} = +1/2 \\
        1, & m_{s} = -1/2
    \end{cases} \, ,
$$
we get then
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle \\
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}_1) \alpha(m_{s1})
\bar{\psi}_1(\vec{r}_2) \beta(m_{s2})
r_{12}^{-1}
\psi_1(\vec{r}_1) \beta(m_{s1})
\psi_1(\vec{r}_2) \alpha(m_{s2}) \\
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2})
\iint\limits_{-\infty}^{+\infty}
\bar{\psi}_1(\vec{r}_1)
\bar{\psi}_1(\vec{r}_2)
r_{12}^{-1}
\psi_1(\vec{r}_1)
\psi_1(\vec{r}_2) \, ,
$$
where we used the fact that spin functions $\alpha$ and $\beta$ are real-valued, so that $\bar{\alpha} = \alpha$ and $\bar{\beta} = \beta$.

At this point we could concentrate our attention exclusively on the coefficient in front of the double integral above, since we could show that it is equal to zero, so that the whole expression vanishes irregardless of the value of the integral. So, the coefficient can be written as follows,
$$
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2}) \\
=
\alpha(-1/2) \beta(-1/2) \beta(-1/2) \alpha(-1/2)
+
\alpha(-1/2) \beta(+1/2) \beta(-1/2) \alpha(+1/2) \\
+
\alpha(+1/2) \beta(-1/2) \beta(+1/2) \alpha(-1/2)
+
\alpha(+1/2) \beta(+1/2) \beta(+1/2) \alpha(+1/2) \\
= 0 + 0 + 0 + 0 \\
= 0 \, ,
$$
where all the terms vanish due to at least either $\alpha(-1/2) = 0$ or $\beta(+1/2) = 0$, so that we indeed get
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
=
0 \, .
$$

It can be noted that there is a shortcut in proving that the coefficient above is zero that uses the defining properties of spin functions. First, we could rearrange spin functions as follows,
$$
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s2}) \beta(m_{s1}) \alpha(m_{s2})
=
\sum\limits_{m_{s1}=-1/2}^{+1/2}
\alpha(m_{s1}) \beta(m_{s1})
\sum\limits_{m_{s2}=-1/2}^{+1/2}
\beta(m_{s2}) \alpha(m_{s2}) \, .
$$
Secondly we note that by the very definition of $\alpha$ and $\beta$ presented above,
$$
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \alpha(m_{s}) \beta(m_{s})
  =
  \alpha(-1/2) \beta(-1/2) + \alpha(+1/2) \beta(+1/2)
  =
  0 \cdot 1 + 1 \cdot 0
  =
  0 \, ,
$$
which is already enough to establish that the whole exchange integral vanishes. The same is, of course, true for the second factor as well,
$$
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \beta(m_{s}) \alpha(m_{s})
  =
  \beta(-1/2) \alpha(-1/2)
  +
  \beta(+1/2) \alpha(+1/2)
  =
  1 \cdot 0 + 0 \cdot 1
  =
  0 \, .
$$
It is also quite customary to use Dirac bracket notation for such expressions over spin orbitals, so that one could write these findings concisely as
$$
  \langle \alpha \lvert \beta \rangle
  =
  \langle \beta \lvert \alpha \rangle
  =
  0 \, ,
$$
where $\langle \alpha \lvert \beta \rangle$, for instance, is defined as follows,
$$
  \langle \alpha \lvert \beta \rangle
  :=
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \bar{\alpha}(m_{s}) \beta(m_{s}) \, .
$$
In general, for two spin functions $\gamma_1$ and $\gamma_2$, expression $\langle \gamma_1 \lvert \gamma_2 \rangle$ defined as follows,
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  :=
  \sum\limits_{m_{s}=-1/2}^{+1/2}
  \bar{\gamma_1}(m_{s}) \gamma_2(m_{s}) \, ,
$$
is zero if the functions correspond to different spin states and one otherwise which can be expressed as follows,
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  =
  \delta_{\gamma_1 \gamma_2} \, .
$$

Finally, I would like to make a short comment on "integrating" over discrete spin coordinates rather then summing up over it. One could indeed define
$$
  \langle \gamma_1 \lvert \gamma_2 \rangle
  :=
  \int \bar{\gamma_1}(m_s) \gamma_2(m_s) \mathrm{d} m_s \, ,
$$
and then think of integration over discrete $m_s$ variable being reduced to summation. But the integration here is just a "symbolic shorthand" since, strictly speaking, we have a summation over discrete $m_s$ variable from the get go. We could just symbolically write it down as an integration as well, if we think it looks more cute for some reason.

In exact same way the exchange integral can also be symbolically defined as follows,
$$
\langle \chi_1(1) \chi_2(2) \lvert r_{12}^{-1} \rvert \chi_2(1) \chi_1(2) \rangle
:=
\iint
\bar{\chi}_1(\vec{x}_1) \bar{\chi}_2(\vec{x}_2)
r_{12}^{-1}
\chi_2(\vec{x}_1) \chi_1(\vec{x}_2)
\mathrm{d} \vec{x}_1 \mathrm{d} \vec{x}_2 \, ,
$$
where the "integration" is informally done over joint spin-spatial coordinates of the electrons. And again, strictly speaking, a summation is done over the spin coordinates and an integration over spatial ones, as it is written explicitly in the very beginning of the answer.


Would molybdenyl chloride or tungstenyl chloride work as an alternative to chromyl chloride test?

In the chromyl chloride test, chromyl chloride is not a reagent that can be added to the sample to detect any chloride present. It is made in situ through a reaction of chloride containing sample, potassium dichromate and sulfuric acid.

$$\ce{NaCl + K2Cr2O7 + 3H2SO4 -> 2Na2SO4 + K2SO4 + 2CrO2Cl2 ^ (red fumes) + 3H2O}$$

Now in order to make molybdenum chloride work like chromyl chloride, you have to actually make molybdenyl chloride. For this, you need potassium dimolybdate. But it is quite unstable and is prone to decomposition. From here:

  Potassium Dimolybdate, $\ce{K2Mo2O7}$, is formed by fusion of suitable quantities of $\ce{K2CO3}$ and $\ce{MoO3}$, but is unstable; the aqueous extract of the fused mass yields large crystals which immediately begin to decompose.

Making tungstyl chloride is not feasible because in order to make it, you need potassium ditungstate which is costly, rarely used reagent and can be hardly found in chemical vendor or supplies. So, this method is out of scope and reach.

Potassium ditungstate or paratungstate has a unique formula that does not follow the general formula of potassium dichromate or potassium molybdate. Its formula is given as $\ce{5K2O.12WO3.11H2O}$ or $\ce{3K2O.7WO3.6H2O}$


Treatment of 2-vinylbicyclo[2.2.2]oct-5-en-2-ol with potassium hydride

The reaction in question is one of the early examples of an oxy-Cope rearrangement [1] in which a 3-hydroxy-hexadiene undergoes a [3,3]-sigmatropic rearrangement to afford a ketone.

The general mechanism is shown below:

Source: Organic Chemistry, Oxford University Press, 2 ed. Clayden, Warren and Greeves

In a general sense, the reaction is a [3,3]-sigmatropic rearrangement (if you look at the curly arrows, you'll see that we're essentially transposing a sigma bond from one end of the pi system to another).

The presence of the oxygen provides a driving force for the reaction, since the enol(ate) formed during the rearrangement is able to tautomerise, yielding the ketone product.

In the specific case you've given, the presence of a base means that during the sigmatropic rearrangement, the oxygen is deprotonated and therefore anionic. This specific case, using an alkoxide, is a specific kind of reaction known as an anionic oxy-Cope. The anion is thought to help speed up the reaction, to the extent that minimal heat is usually required.

References:

[1] J. Am. Chem. Soc., 1964, 86,5019.  DOI: 10.1021/ja01076a067


I'm not sure exactly what you mean by normalising as it can mean more than one thing. At each probe wavelength ideally you measure two signals simultaneously from the same probe pulse which is split into two parts before entering the sample. One part probes the sample at the same place that it is excited and the other probes unexcited sample. You then record the transmission of these two signals and calculate optical density change or work with transmission depending upon choice.  It is clearly necessary to normalise to the pump laser intensity since a more intense pulse will produce a bigger effect. Second the probe pulses have to be monitored and adjusted as necessary (without exciting sample) so that these signals are kept equal or at least in a constant ratio so they can be normalised later.
Pump-probe is just like any other transient absorption measurement (used to be called Flash-Photolysis) and a difference signal is measured. This produces a transient spectrum which might be positive or negative. If you have the exact optical density the you can construct the spectrum of a transient species knowing the ground state spectrum. If there is no ground state absorption where you have a transient spectrum then this must be positive and is the transient spectrum. If it is negative it might be due to transient gain, i.e. the probe causing stimulated emission from the sample. The schematic shows one way the experiment can be set up. The absorption change at t = 0 is exaggerated for clarity; a 1% change is a huge signal.


Is a spectrophotometric method of measuring the rate of a complex reaction valid?

This is a perfectly valid method of determining rates of chemical reactions. It is obviously limited to those reactions where either the product or the initial reactant can be monitored easily using spectroscopy as well as reactions which are slow enough to get an absorbance measurement at set intervals. I have used this method multiple times in undergraduate chemistry labs and it works, and in principle it does not matter if there are intermediates.

The intermediates should not matter because the rate of a reaction is defined as the rate at which product is formed. That is, the rate of a first-order reaction is,
$$
\text{rate}=k_{tot}[A]
$$
If the reaction involves only one elementary step then the prefactor $k_{tot}=k_1$, the rate constant of the one step (assuming no backwards reaction). Note that $[A]$ can be the concentration of either a product or reactant, but you might need a stoichiometric coefficient in there depending on the reaction. On the other hand, if there are intermediate steps, the prefactor can become arbitrarily messy. Maybe there's an equilibrium between two intermediates so you have a term in there $k_2/k_{-2}$ and whatever else you can imagine. This doesn't matter though because your experiment measures the change in concentration at either end, and the slope is then the effective rate constant for the reaction which has all of the intermediate stuff lumped together.

In fact, this method you use is probably the most direct means of measuring a rate in the sense that a rate is defined as a change in concentration over time. You are measuring a change in absorbance over time, so there is only step (and one assumption) in going from your data straight to the reaction rate. That assumption is simply the validity of Beer's law, which is usually not a problem for a large range of concentrations.

As a sanity check, look at this website which has a description of the same experimental setup and what appears to be real data. Note that they are measuring change in absorbance of a reactant while you say you are measuring the change in absorbance of a product, $\ce{CuCl2}$


Calculate the moles of acid and conjugate base needed

Since it is a buffer solution, the volume is constant ($V =\pu{0.1 L}$), hence
$$
\frac{c(b)}{c(a)} = \frac{n(b)}{n(a)}\overset{calc.}{=}0.72
$$
 and with
$$
n_{\text{tot}}=n(b)+n(a) \overset{\text{def}.}{=}\pu{0.05 mol}\\
n(a)=\frac1{1.72}\cdot\pu{0.05 mol}\\
n(b)=\frac{0.72}{1.72}\cdot\pu{0.05 mol}
$$
Then $$m=M\cdot n$$ and you should be fine.


Why doesn't NASA or SpaceX use ozone as an oxidizer for rocket fuels?

As is usual with rocket fuels, the problems of ozone are practicality not performance
Almost every answer for why a specific rocket fuel component is used or not will end up referring to John D Clarke's magnificent and sparklingly written book: Ignition: An informal history of liquid rocket propellants (a rare technical book worth reading for the brilliant and humorous style in addition to the technical content).
His summary of why ozone, despite its apparent functional advantages, is not more widely used is fairly simple: the practical problems outweigh the apparent advantages.
He points out the advantages:

What makes it attractive as a propellant is that (1) its liquid density is considerably higher than that of liquid oxygen, and (2) when a mole of it decomposes to oxygen during combustion it gives off 34 kilocalories of energy, which will boost your performance correspondingly. Sänger was interested in it in the 30’s, and the interest has endured to the present. In the face of considerable disillusionment.

But every available way of creating a liquid with a high proportion of ozone is dangerous. Ozone is extremely toxic and unstable:

For it has its drawbacks. The least of these is that it’s at least as toxic as fluorine. ... Much more important is the fact that it’s unstable—murderously so. At the slightest provocation and sometimes for no apparent reason, it may revert explosively to oxygen. And this reversion is catalyzed by water, chlorine, metal oxides, alkalis—and by, apparently, certain substances which have not been identified. Compared to ozone, hydrogen peroxide has the sensitivity of a heavyweight wrestler.

Work was done on solutions of ozone in liquid oxygen which is somehow more stable. But has the disadvantage that ozone/oxygen mixtures separate into two phases the ozone rich version of which is difficult to prevent in feed tubes after firing and is extremely unstable. Another mixture considered to make handling easier was with liquid fluorine (!!!).
Ultimately he concludes the known work on ozone mixtures of any sort:

For ozone still explodes. Some investigators believe that the explosions are initiated by traces of organic peroxides in the stuff, which come from traces, say, of oil in the oxygen it was made of. Other workers are convinced that it’s just the nature of ozone to explode, and still others are sure that original sin has something to do with it. So although ozone research has been continuing in a desultory fashion, there are very few true believers left, who are still convinced that ozone will somehow, someday, come into its own. I’m not one of them.

Maybe there are theoretical advantages, but they are outweighed by the practical and safety problems. In a profession used to testing things like FOOF (fluorine dioxide) and chlorine trifluoride, this is some admission to make.


How can there be concentration and pressure terms simultaneously in the Nernst equation?

This topic often causes confusion.

The most correct form of the Nernst equation uses thermodynamic activity of the substances:

$$E = E^{\circ} - \frac{RT}{zF} \ln\frac{a_\mathrm{red}}{a_\mathrm{ox}}$$

For diluted solutions, we can replace the activity of a substance with its concentration in the solution. But how do we express the activity of a gas?

This is basically the question of how to formulate the equilibrium constant of the redox reaction. There are actually multiple solutions to this. The problem is that all of them are correct.

Solution 1: Express activity of the gas as a function of partial pressure.
\begin{align}
pV &= nRT \\
\Rightarrow p &= \frac{n}{V}RT = cRT\\
a(\mathrm{gas}) &\approx c(\mathrm{gas}) = \frac{p(\mathrm{gas})}{RT}
\end{align}

Solution 2: Just use the partial pressure itself:
$$a(\mathrm{gas}) = p(\mathrm{gas})$$

Solution 3: Use the partial pressure divided by a reference pressure:
$$a(\mathrm{gas}) = \frac{p(\mathrm{gas})}{p(\mathrm{reference})}$$


Solution of the Roothaan Equations of H2 by Symmetry Arguments

After speaking to a coworker and reading some more I think I have a better understanding.
The symmetry of the wave function needs to follow the symmetry of the molecule (unless we are studying a broken symmetry solution, e.g. charge transfer, but that doesn't apply in this case). We are using a minimal basis, and hence we have only two orbitals. From atomic theory, these are s functions. Hence, they cannot form $\pi$ orbitals, and we are left with $\sigma$ orbitals. Between $\sigma_g$ and $\sigma_u$, $\sigma_g$ has the lower energy because it has no nodes, whereas $\sigma_u$ contains a node. From QM we know that this generally implies greater energy, but there's a nice explanation on this here (at the end he also specifically discusses a homonuclear diatomic molecule).
The specifics of how to get the symmetry representation for the molecule and match those with wave functions is still not entirely clear to me, but for this I will read a book on group theory and its chemical applications. It is, however, clear that $\phi_1+\phi_2$ and $\phi_1-\phi_2$ respect $\sigma_g$ and $\sigma_u$ symmetry respectively.


Why is my NaCl solution seemingly saturated, when I followed the recipe for an isotonic solution?

Salt added to food could be sea salt, common table salt or iodized table salt.

Common Table salt typically contains anti-caking agents (desiccants) such as magnesium carbonate, calcium silicate, dicalcium phosphate, tricalcium phosphate, calcium carbonate and sodium alumina silicates.

Iodised table salt may contains added potassium iodide for people who may otherwise have low consumption of this nutrient which is essential for proper thyroid function. Since iodide will slowly oxidize in moist air, stabilizing agents such as sodium thiosulphate or dextrose are also added. Some will also contain a base such as sodium bicarbonate, disodium phosphate or sodium pyrophosphate to also inhibit oxidation of iodide. There may also be added sequestering agents which bond to trace metal ions present as impurities to prevent them oxidizing the iodide.$^{[1]}$

The presence of magnesium and calcium will add "hardness" to the water which could cause precipitates such as calcium carbonate and magnesium carbonate to form.


Calculating concentration and volume in acid metal reactions

Your equation uses molar ratios, rather than weights or volumes.
If you know the weight of zinc used, you know the number of moles. From this, you can calculate the number of moles of zinc phosphate generated.
Supposed that the phosphoric acid is sufficiently concentrated to react all zinc, the number of moles calculated above together with the volume of the phosphoric acid used allows to calculate the concentration of zinc phosphate. Note: Do not omit units in your calculations!
As far as the hydrogen is concerned: The number of moles produced can be calculated using the amount of zinc used. Additional knowledge of the molar volume of gases under standard conditions is helpful to figure out the volume.


Is Nuclear Spin Isomer a kind of allotrope?

Some peer review journal articles and an India high school textbook and various study guides do refer to the spin isomers as allotropes.

There are physical differences such as different heat capacities.  See Orthohydrogen, Parahydrogen and Heavy Hydrogen for comprehensive information.

However, aside from the spin isomers, there are genuine allotropes of hydrogen.  Solid hydrogen has two normal pressure allotropes, a face centered cubic and a hexagonal closest packed allotrope.

Also, at high pressure, there is metallic hydrogen.


Strength of hydrogen bonding in phenol or methanol

Let me answer the second question first, because the answer is so much easier: Of course the hydrogen bond strength depends among others upon the carbon oxygen is attached to. Consider phenol and methanol or methanol and formic acid.

The guess as to which hydrogen bond is stronger is — I have to admit it — my speculation. But I think that methanol should be able to undergo stronger hydrogen bonding.

Phenol, of course, is aromatic and capable of delocalising negative (or positive) charges across its entire ring system — the reason why its $\mathrm{p}K_{\mathrm{a}} = 10$, a good five units lower than methanol’s. That means that at the same pH value of the solution, there should be less strong a negative charge on phenol’s oxygen than on methanol’s.

But hydrogen bonds are, to a certain extent, electrostatic interaction. So the system which has a stronger negative charge on the oxygen (methanol) should be able to build up stronger hydrogen bonds by attracting hydrogen more strongly.

One could also picture it differently: Phenol gets deprotonated at a lower pH, so the $\ce{O-H}$-bond must be a lot weaker than methanol’s. But if the $\ce{O-H}$-bond is weak, the oxygen has a larger tendency to simply loose the hydrogen to another electronegative partner (deprotonation/protonation) rather than share it (hydrogen bond).


Why do different alcohols have different effects on the brain?

As the chain length of the alcohol increases, it becomes more hydrophobic. This would make it easier for them to insert into membranes and hydrophobic pockets of proteins. This could interfere with the protein function. Hydrophobic molecules also tend to be harder to eliminate than shorter molecules and can accumulate in fatty tissues, including the brain. Longer chains can also be metabolized into a greater number of products simply because there is more to metabolize. And sometimes seemingly innocuous compounds can be metabolized into very toxic substances, such as n-Hexane, which can be converted to a molecule with an epoxide on each end. The epoxides can cross-link proteins, particularly in neuron axons.


No.

To answer your question we will assume certain things here.

Assume  $C_p = C_v$ for most liquids, where $C_p$ is the heat capacity at constant pressure, and $C_v$ is the heat capacity at constant volume.
Assume that $C$ is constant with respect to temperature, which is not the case in most real fluids. Here, in order to leave out the integration, we assume that it is a constant.

Then,

$Q = C\Delta{}T$

As different liquids have different heat capacities, the same amount of heat supplied to the liquid would result in a different $\Delta{}T$, the temperature change.

The higher the heat capacity, the more heat is required to raise a unit mass(or mole) of substance  by a unit temperature. Likewise, more heat has to be removed from the unit mass of substance in order to decrease the temperature by a unit temperature.

In other words, the higher the heat capacity, the more "reluctant" it is to change its temperature, for the same amount of heat supplied.


The second law of thermodynamics states that the entropy of the universe always increases.

$$\mathrm{d}S > 0$$

In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in entropy of the system.

$$\Delta S_\mathrm{sys} < 0 \qquad \Delta S_\mathrm{surr} > 0$$
$$|{\Delta S_\mathrm{surr}}| > |{\Delta S_\mathrm{sys}}|$$

This is because adsorption is an exothermic process and so the surroundings are heated up and therefore increase in entropy. If you consider the Gibbs energy change for adsorption it will be negative because the negative $\Delta H$ term is larger than the positive $-T\Delta S$ term. This chem.uic.edu page goes through the maths of this very nicely.


What makes polyimides good polymers for fibers?

Polyimides (PIs) are high-temperature engineering polymers originally developed by the DuPont™ Company. Polyimides exhibit an exceptional combination of thermal stability (>500 °C), mechanical toughness, and chemical resistance.It was based on pyromellitic dianhydride and 4,4’diaminodiphenyl ether.

image source

The presence of n-π conjugation between non-pair electron of nitrogen atom and π electrons of the carbonyl group makes them resistant to chemical agents and moisture.

image source

The linearity and stiffness of the cyclic backbone allow for molecular ordering. This phenomenon results in lower coefficients of thermal expansion (CTE) than those found for thermoplastic polymers having coiled, flexible chains.

Additionally, the morphology of long, linear ordered chains provides solvent resistance to the aromatic polyimides.

The comparatively rigid structure of polyimides provides high glass transition temperature (Tg > 300°C) and imparts good mechanical strength and high modulus.

The thermal stability of polymer can further be improved by incorporating aromatic rings on the backbone and/or side groups. In addition to such high thermal stability, the nature of the chemical structure consisting of rigid imide and aromatic rings always provides:

Excellent mechanical toughness
Excellent dielectric properties
High chemical resistance

These result in  thermoxidative stability, high mechanical strength, high modulus, excellent insulating properties, and superior chemical resistance.

Polyimide layers are suitable matrix for incorporation of metal, salts, chromophores as nanoscale particles to obtain of nanocomposite materials. It was discussed the possibility of use polyimides in materials chemistry and nanomaterials, one of these applications is the use for making biomedical implants for neurology, ophthalmology, biosensor device and chips which are a powerful tool in clinical diagnostics.

Another important trend is use in electronics and optoelectronics such as dielectric substrates and intermediate barrier layers, creating nanocomposite films with various nanosized particles such as dyes, metal, dielectric and other clusters.

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I am wondering if there are surfactants that exist in a gas state that can be used to reduce the surface tension of a liquid.

The most direct answer is "no".  Gases do not have surface tension and gas-phase molecules do not chemically affect the surface tension of liquids.

  I know its a broad question, so for example, are there any gases that could reduce the surface tension of water when it is exposed to it?

Gases could, however, dissolve in the liquid in question.  Then, they aren't really gases, but solutes in the liquid.  And solutes can affect surface tension.  You can see this earlier question on volatile surfactants for some examples of water-soluble volatile surfactants for some examples.  A paper entitled Surfactant in the gaseous phase was brought up by MAR in the comments and is another good example.


Why does the surface of an acid appear to have a stronger "dissolving power"?

Besides mechanical aspects

bubbles might form a sort of layer at the immersed part while at the air interface they evolve and detache easily renewing the metal/acid (oxidizer) contact

a more fundamental reason can be° surface excess. Normally we treat interfaces as very thin and therefore negligible (in mass or volume) as compared to the bulk.

However at an interface the chemical potential (and all related thermodynamic properties) differs from that of the bulk owing to the fact that a layer can be assumed to have the same number of entities per unit of surface.

While such a layer at depth is surrounded by the very same, at the interface it sees two different environments.

This aspect is quite familiar in physical phenomena, such as surfactivity, the shape of droplets, and even the tip effect in electrostatic. Another example  possibly familiar to many chemists is the difference in product of solubility / solubility for solids at very different sizes (digestion of precipitates to attain bigger grains/crystals is based on this, tough intermixed with the merely geometric fact that small particles dissolve faster).

As for in water the interface has higher potential and this results in spherical droplets to minimize it, in your specific case you can see the aqua regia doing the same by being more reactive. It is like the oxidizer trying to escape the air interface (tough a new one is continually formed, of course) by reacting more energically than the layers below, if this pictorial helps you.

If you dig for "potential at acid/air interfaces" or "pH of an acid at the air interface" and so on, you can find some material dealing with example somehow related to your example. For instance I have found study of the dissociation of weak acids (ie pKa at the interface vs bulk), The vast majority is however studies in which the very interface is at focus, like in monolayer LB films, and the surface excess comes naturally at play whatever explicitly considered or not.

(For instance a title such as "On the dissociation constant of carboxylic acids in monolayer thin films" implicitly assumes that some difference exist as compared to the bulk of the solution).

° can be as for I do not know if the evident effect you have spotted is mainly due to the "stirring effect" of evolving bubbles or to the higher activity of the interface, or their mix.


Identify structure of compounds from their reactions and molecular formula

You weren't far from resolution. The substrate is benzaldehyde, first reaction benzoin condensation, so Q is benzoin. It is later oxidated to benzil - R ($\ce{C14H10O2}$) (not ($\ce{H12}$) - in oxidation formula must change). Finally benzil undergoes benzilic acid rearrangement, so S is benzilic acid.

As for how one could get to the solution... Compounds having 6-8 carbons and highly unsaturated are usually benzene derivatives. In this case benzaldehyde is only common thing with this formula (it was unlikely to be for example tropone). First reaction must be dimerisation, next oxidation, and hydration.


Why do molecules have to first transform to gas before doing a reaction?

Not all reactions require vaporization. This process looks like the Born-Haber cycle for cesium iodide, an exercise in Hess's Law usually used to fill in one of the missing values in the cycle that might be hard to measure (usually lattice energy, but not always). The Born-Haber cycle can also be used to drive general chemistry students crazy.

The $E_N$ "energy of neutralization" looks like the lattice energy, which is defined as the enthalpy/energy change accompanying the formation of a solid crystal lattice from the gas phase ions. In your case, the lattice energy of $\ce{CsI}$ is defined as the energy change accompanying the following reaction:

$$\ce{Cs+ (g) + I^- (g) -> CsI(s)}$$

Because the lattice energy is defined for gas phase ions, you need the vaporization enthalpies for this process. Not all reactions require this vaporization step.


Lone pair electrons on secondary carbon vs primary carbon

I am not sure if you are referring to carbanions or carbenes (I suspect the latter), because both contain a carbon with a lone pair, so I will give explanations on both in the following.

Carbanions are destabilized by the exact same effect that stabilizes carbocations and radicals: hyperconjugation. Donation of electron density from an adjacent $\sigma$ bond into the empty/half-filled $p_z$ orbital stabilizes the cation/radical, but not a carbanion center with its negative charge in the fully-filled $sp^3$ orbital lobe. Therefore, a primary carbanion is more stable than a secondary one.

The inverse is true for carbenes: Secondary carbenes are way more stable than primary ones. This is because carbenes are per se a very reactive species, and because of this property they are often used in organic synthesis as intermediates which are generated in situ. To obtain a reasonably stable carbene, the reactivity of the lone pair needs to be somewhat reduced. Two substituents at the carbene carbon are needed which stabilize the carbene either by steric hindrance (i.e. making the lone pair difficult to reach for other reactants) or mesomeric effects (for example, lone pair donors which alleviate electron lack of the "sextet" carbon, delocalization of the carbene lone pair into an aromatic system). The Arduengo carbene (named after its discoverer) is a good example for this. The stability of this compound is such that it can be isolated as a crystalline solid at room temperature. The carbene lone pair can be delocalized into the five-membered ring which then contains 6 $\pi$ electrons (2 double bonds + lone pair of uncharged nitrogen) and is aromatic. Bulky $\mathrm{R}^1$ groups like adamantyl further contribute to stability by shielding the lone pair from potential reactants (source).

Arduengo carbenes are not the only persistent (especially stable) carbenes, and the understanding of their stability is still a subject of ongoing research, because structural elements (bulky substituents, aromaticity) which had been regarded as necessary for stability were often disproved as such by the synthesis of carbenes which did not contain them. The compounds do not need to be cyclic, as shown by bis(diisopropylamino) carbene, and even heteroatoms next to the carbene center do not need to be present, like in the case of bis(diisopropylamino)cyclopropenylidene.


Does tetramine copper dissolve cellulose?

As @iad22agp states, it is not sufficiently alkaline. Schweizer's reagent is tetraamminediaquacopper dihydroxide, $\ce{[Cu(NH3)4(H2O)2](OH)2}$. See Schweizer's reagent.

That is not Doktor Schweizer, I presume.


Normally we don't discuss quantum mechanics in this context, but an answer to your question requires it.

Taken from:

http://chemistry.umeche.maine.edu/CHY252/HOMO-LUMO.html

What we see here is the LUMO for CH3Cl.  This is where an incoming nucleophile will 'stick' its electrons to form the new bond.  Since the orbital is antibonding, electrons in the orbital will destabilize a bond, in this case the C-Cl bond.  The actual meaning of the size of this orbital, rigorously, is a little unclear.  However you can roughly infer that if a nuclephile gets its electrons here, a reaction is 'likely'.  The reddish region to the left encompasses the hydrogens.  That is also where the nucleophile has to go.

Compare the next from:

http://chemistry.umeche.maine.edu/CHY252/EtCl.jpg

This is for ethyl chloride.  There is an extra minor node on one of the hydrogens.  Because it is small, it is less important than the other regions.  Notice the methyl group is partly obscuring the backside of the orbital lobe near the electrophilic carbon.  There are less ways a nucleophile can approach this to cause a reaction.

I'm no longer practicing so I do not have software to produce analagous images, so you will have to imagine the extrapolation.  1-chloroopropane would not add much more to the steric hindrance, because the 'extra' methyl group would be dangling on the same of the molecule as the methylene group.  However, if we had two methyl groups, such as 2-chloropropane, the hinderance would be much more.  Three groups, as in t-butyl chloride?  Give it up, to much hinderance for a backside attack reaction.


What is the significance of enthalpy definition not under constant pressure?

There are many processes that take place at constant pressure, such as the change of phase from a liquid to a vapor for a pure substance at constant temperature.  Also, the heats of chemical reactions and heats of formation of compounds are typically cataloged at constant temperature and pressure.  So the enthalpy comes into play in both these instances.

The first law of thermodynamics for a closed system is expressed conveniently in terms of the change in internal energy of the system.  However, for an open system (in which mass is flowing in and out) operating at steady state, the first law is more conveniently expressed in terms of the enthalpy, not the internal energy.  And, of course, all industrial applications of thermodynamics involving continuous processes are operating at steady state.

Chet


The aldol condensation needs $\alpha$-hydrogens next to the carbonyl group in order to proceed. Check out this reaction equation from the Wikipedia article:

One of these hydrogen atoms is removed to form the enolate. See this mechanism (also from the Wikipedia article).

Now, look at the structure of p-methylbenzaldehyde. This molecule does not have any $\alpha$-enolizable hydrogen atoms. The $\alpha$-position is an $sp^2$-hybridized carbon atom in the aromatic ring. The aldol condensation is thus not possible.

We can also rule out the haloform reaction because there are no halogen compounds given. There are also no hydrolyzable functional groups (halides, ethers, esters, anhydrides, amides, nitriles, etc.).

The only remaining possibility is the Cannizzarro reaction, which requires a benzaldehyde derivative or other aldehyde lacking $\alpha$-enolizable hydrogen atoms.


It is not that surprising that you got this answer: the existence of azeotropes is due to non-ideal behaviour of mixtures (see for instance the Wikipedia page and the associated diagram). It is not surprising that when you impose an ideal behaviour of the solvent that follows Raoults law, you get some unexpected results. Basically, what your calculation leads to it that for ideal solvent mixture, the only situation in which the gas phase molar ratio can be equal to the solution molar ratio is when both solvent have equal vapour pressures.


What is the mechanism of APTES mono-layer formation on glass substrates?

Here's two links (1, 2) that show the simple scheme on forming monolayers. The essential figure is also shown here.

For the second part of your question... I can only point out this article that describes which properties change, if you make the molecules more bulky or add chloride, etc. I would guess that APTES has well known properties and they are "neutral" regarding common applications.


Activity of Materials in Equilibrium constant

While calculating concentrations is trivial (divide the amount of substance by the volume), activity is a complete different story. Activity is usually calculated by multiplying the concentration by a dimensionless number called activity coefficient($\gamma$):
$$a = c \times \gamma$$

The problem here is that $\gamma$ is not only different for every substance, it also varies with temperature, concentration of the substance itself, the diluent used, and so on. If the substance is a salt that dissociates then the calculation gets even more complex.

Since having the values of the activities would be quite a challenge, equilibrium constants are often written using concentrations, which are more readily available. It's not the most correct way of represent it, but it makes your life easier.


Bromination Pathways with alkane, alkene, and alkyne substrates

I wanted to offer some commentary on the points you've enumerated:

You mentioned halogenation of alkenes in the presence of $\ce{H2O}$. This reaction is known as halohydrin formation. It's worth mentioning that this reaction can also be performed in the presence of other nucleophiles, perhaps most commonly alcohols and salts of other halogens to give mixed dihalogenated products. (Other types of nucleophiles may also successfully react, but I can't recall seeing any specific examples.) The reaction isn't always efficacious. Frequently, addition of both halogen atoms (i.e., ordinary halogen addition) will occur; the proximity of the halide leaving group to the halonium bridge will often result in a rapid reaction before any other nucleophile can intercept.
Radical halogenation can be very difficult to control, even when dealing with substrates that should exhibit high regioselectivity. NBS (and its chlorinated counterpart, NCS) is actually an extremely useful and versatile reagent which mitigates some of the practical pitfalls of radical halogenation reactions. Under photocatalytic conditions, or with sufficient heating and a radical initiator, it can react by free radical pathways. It's often preferable to using the diatomic halogen directly (unless exhaustive halogenation is really the goal) because it generates lower concentrations of $\ce{Br2}$ in situ, so the reactions are somewhat more controllable and side-products from direct addition to $\pi$-bonds are minimized. It can also react analogously to $\ce{Br2}$ with alkenes to give a halonium ion, which makes it preferable for halohydrin formation (since there is no $\ce{Br-}$ leaving group present to immediately attack the bromonium bridge).

Overall, my opinion is that you have a good core knowledge base re: halogenation reactions within the confines of introductory organic chemistry. Of course, there are a multitude of other halogenation reactions operating on myriad functional groups, but I presume those aren't germane to this particular question.


In water, HCl reacts to form $\ce{H3O+}$ and Cl-.

Then, water molecules surround the respective ions in a stabilizing orientation.
The partially negatively charged O atoms of water interact with the $\ce{H3O+}$.
The partially positively charged H atoms of water interact with the Cl-.

None of the above can occur in benzene because benzene is very reluctant to accept a proton and because benzene is non-polar.

Generally speaking, polar solvents have an ability to stabilize ions that nonpolar solvents do not.


Why gives NaBH4 more Felkin-Anh product than LiAlH4 gives?

Lithium aluminum hydride is such a reactive nucleophile that it reacts readily with any conformation of the ketone. The reduced nucleophilicity of sodium borohydride allows for some discrimination of the conformations.

Note that the difference in selectivity between aluminum hydride and borohydride small. A large boost in stereoselectivity can be achieved by increasing steric bulk of the reducing agent (L-selectride gives a 10-fold increase in selectivity).


Why is H bonded to C5 drawn below the plane in Haworth formula for D-glucose?

The Haworth projection depicts the pyranose (or furanose, as the case may be) as a planar ring. The Fischer projection depicts the open chain carbohydrate with the carbon backbone in a single plane. Importantly, the groups on the horizontal bonds are coming out of the plane and the vertical bonds are going into the plane. The result of this is that the carbon backbone is curving toward itself, and the compound is partially poised to close a ring. However, the ring of the pyranose is not all carbon. In glucose, it's the oxygen at C5 that adds to the aldehyde to close the ring. We can manipulate the Fischer projection so that this oxygen at C5 is coplanar with the rest of the atoms that will make up the ring. Rotating around the C4-C5 bond places this oxygen on the Fischer projection's vertical axis. The Haworth projection is essentially a closed version of the Fischer projection tipped on its side with C1 at the right. Forming the hemi-acetal gives the pyranose form of glucose.

You imply in the question that groups that appear on the right side of the Fischer projection should appear on the bottom of the Haworth projection and groups that appear on the left side of the Fischer projection should appear on the top of the Haworth projection. That idea is true. However in the process above the -CH2OH is rotated to the the left side of the Fischer projection. Consequently, it appears on the top of the Haworth projection.


What's the purpose of a Grignard reagent initiation?

The first step in the Grignard reaction involves transfer of an electron from the surface of the metal to the alkyl halide to form a radical anion.  The radical anion then disproportionates into a free radical and halide anion.
\begin{aligned}
\ce{R−X + Mg &-> R−X^{•−} + Mg^{•+}}\\
\ce{R−X^{•−} &-> R^{•} + X^{−}}\\
\ce{R^{•} + Mg^{•+} &-> RMg^{+}}\\
\ce{RMg^{+} + X^{−} &-> RMgX}\\
\end{aligned}

When you take the magnesium out of the jar, the surface has a thin oxidation layer on it that prevents the initial electron transfer from occurring.  We used to take one piece of magnesium and scratch it good with a knife to expose a clean, fresh magnesium surface.  We would place this piece of magnesium in the flask (dry ether and the rest of the magnesium already added) and drip in a few drops of the alkyl halide.  Without fail, within a minute or two, mild bubbling would be noticed (local ether boiling), which became vigorous as the initiation phase of the reaction "cleaned" the surface of other pieces of oxidized magnesium.  After the vigorous bubbling started to subside we would drip in the remaining alkyl halide plus ether solution at a rate to maintain ether reflux.  So the reaction does involve a free radical but the initiation phase really has more to do with removing the oxidation layer from all of the magnesium and allowing that initial electron transfer step to occur.


What is the energy package released to the outside world?

Essentially yes - if you consider the two particles colliding into a potential well, the kinetic energy associated with each will (depending on the angular specifics) be apportioned into rotational, vibrational, and even occasionally electronic excitations. So, the two particles collide, and are then vibrating and/or rotating about each other with all the translational energy converted in the collision.

Non-reaction collisions with other ambient particles allow relaxation into lower-energy states, transmitting energy to those particles. After some collisions the new species formed is in equilibrium, transferring between low-energy rotational and vibrational states as it collides further, and the change in potential energy has been dissipated as rotational, vibrational, and translational energy in other particles.

(I guess I could really use some sort of animation for this.)


What is the difference between hand wash and body wash soap?

(My sister-in-law works at Unilever, and she makes these soaps)

These are all detergents.  The primary differences are harshness (hand soaps can be harsher, as hands are often more soiled), lathering, concentration, perfuming/scents, and marketing.

(Emphasis on the marketing.)


Is there anything like "Saturated Fullerene"?

I'm limiting my answer to $\ce{C60}$-fullerene, but the same idea applies to the higher fullerenes as well.

In short, it's not possible to fully hydrogenate $\ce{C60}$ to $\ce{C60H60}$. This is a result of the bond strain that would result upon hydrogenation, as the addition of hydrogen across an alkene in $\ce{C60}$ produces carbon atoms that would optimally be tetrahedral with angles near $109.5^\circ$, while the bond angles in $\ce{C60}$ are $108^\circ$ and $120^\circ$ for the pentagonal and hexagonal rings, respectively.

Partial hydrogenation of $\ce{C60}$ with zinc and hydrochloric acid has been done to form $\ce{C60H36}$,1 but this compound degrades quickly in the presence of oxygen to form hydroxylated groups; degradation of the fullerene cage and formation of ketones also occurs. Efforts to hydrogenate further under high pressure hydrogenation conditions results in complete breakdown of the fullerene cage, with pretty interesting results, including ejection of carbon atoms and $\ce{C2}$ loss, producing hydrogenated species with fewer than 60 carbon atoms.2 Collapse of the fullerene cage was also noted, forming hydrogenated compounds with 25 - 45 carbon atoms.

Cataldo, Franco. Fullerenes, Nanotubes and Carbon Nanostructures 2003, 11 (4), 295-316. DOI: 10.1081/FST-120025852
Talyzin, Alexandr V., Yury O. Tsybin, Jeremiah M. Purcell, Tanner M. Schaub, Yury M. Shulga, Dag Noréus, Toyoto Sato, Andrzej Dzwilewski, Bertil Sundqvist, and Alan G. Marshall. J. Phys. Chem. A 2006, 110 (27), 8528–8534. DOI: 10.1021/jp0557971


How to find the second order perturbation to wave function?

Disclaimer This post is some kind of a legacy post. Find the notation used in the question in the other answer. I added this proof as I was not entirely certain I understood the notation correctly. As it turned out, I did not. As a result I posted the complete derivation of RSPT up to second order, trying to guide anyone through it using a different notation.$\require{cancel}$
Obviously, I will come back to the initial problem only at the end, if you are impatient, just skip ahead.
Please also note, that I removed the parts that became redundant when I posted the second answer. If you are interested in legacy versions, just look them up in the history.
Also, this page is heavily loaded with $\mathcal{M}^{ath}\mathrm{J_{a}X}$, give it time to load completely.

In Rayleigh Schrödinger perturbation theory, we set up the Hamiltonian as $$\mathbf{H}=\mathbf{H_0}+\lambda\mathbf{H'}.\tag1$$

The perturbed Schrödinger equation is therefore $$\mathbf{H}\Psi=W\Psi_\mathrm{ST}.\tag2$$
You can develop this into Taylor series:
\begin{align}
W                &=
     \lambda^0W_0    + \lambda^1W_1    + \lambda^2W_2    + \cdots
     &&= \sum_{i=0}\lambda^i W_i\tag3\\
\Psi_\mathrm{ST} &=
     \lambda^0\Psi_0 + \lambda^1\Psi_1 + \lambda^2\Psi_2 + \cdots
     &&= \sum_{i=0}\lambda^i \Psi_i\tag4\\
\end{align}

Your unperturbed reference system is a complete solution denoted as
$$\mathbf{H}_0\Phi_i = E_0\Phi_i,\tag5$$
where $i\in\mathbb{N}$.

You chose your system to be intermediate normalised.
$$\langle\Psi_\mathrm{ST}|\Phi_0\rangle = 1\tag6$$
Since all your solutions of the unperturbed system are orthogonal,
$\langle\Phi_i|\Phi_j\rangle=\delta_{ij},$
the same holds for your perturbed system, hence
$$\langle\Psi_{i\neq0}|\Phi_0\rangle = 0.\tag7$$

Now we can combine $(1)$ through $(4)$ and gather them according to their order.
\begin{align}
\lambda^0&:&
  \mathbf{H}_0\Psi_0
    &= W_0\Psi_0\tag{8a}\\
\lambda^1&:&
  \mathbf{H}_0\Psi_1 + \mathbf{H'}\Psi_0
    &= W_0\Psi_1 + W_1\Psi_0\tag{8b}\\
\lambda^2&:&
  \mathbf{H}_0\Psi_2 + \mathbf{H'}\Psi_1
    &= W_0\Psi_2 + W_1\Psi_1 + W_2\Psi_0\tag{8c}\\
\lambda^n&:&
  \mathbf{H}_0\Psi_n + \mathbf{H'}\Psi_{n-1}
    &= \sum_{i=0}^n W_i\Psi_{n-i}\tag{8d}\\
\end{align}

We do have a complete set of solutions, hence we can express perturbations as linear combinations of these.
\begin{aligned}
\Psi_0 &= \Phi_0 &
\Psi_1 &= \sum_i c_i \Phi_i &
\Psi_2 &=\sum_i d_i \Phi_i &
\dots
\end{aligned}

Now we project equations $8$ onto $\Psi_0$. For $\mathrm{8a}$ this is trivial and it leads back to the expectation value of the energy for the unperturbed system. (In the last transformation we use the linear combinations.)
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
    &= \langle\Phi_0|W_0|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_0\rangle
    &= W_0\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9a}\\
W_0 &= \langle\Phi_0|\mathbf{H}|\Phi_0\rangle = E_0\tag{9a'}
\end{align}
The next step is similar, and everything else will follow from there analogously. For $\mathrm{8b}$ we obtain:
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &=   \langle\Phi_0|W_0|\Psi_1\rangle
       + \langle\Phi_0|W_1|\Psi_0\rangle\\
\langle\Phi_0|\mathbf{H}_0|\Psi_1\rangle +
\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &= W_0\cancelto{c_0}{\langle\Phi_0|\Psi_1\rangle} +
       W_1\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\\
\cancelto{c_o\cdot E_0}{\sum_i c_i \langle\Phi_0|\mathbf{H}_0|\Phi_i\rangle}
       + \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
    &= W_0\cdot c_0 + W_1\tag{9b}\\
W_1 &= \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle\tag{9a'}
\end{align}

Similarly we can obtain the coefficients by projecting on some other function than $\Phi_0$. I will cut this a little bit shorter.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle
     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
    &= \langle\Phi_j|W_0|\Psi_1\rangle +
       \langle\Phi_j|W_1|\Psi_0\rangle\\
\cancelto{c_j\cdot E_j}{\langle\Phi_j|\mathbf{H}_0|\Psi_1\rangle}
     + \langle\Phi_j|\mathbf{H'}|\Psi_0\rangle
    &= \cancelto{c_j\cdot E_0}{\langle\Phi_j|W_0|\Psi_1\rangle}
     + \cancelto{0}{\langle\Phi_j|W_1|\Psi_0\rangle}\tag{10b}\\
c_j &= \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{E_0-E_j}\tag{10b'}
\end{align}

Now this is impossible to solve for $c_0$, but we still have the intermediate normalisation.
\begin{align}
0 &= \langle\Phi_0|\Psi_1\rangle
   = \sum_i c_i \langle\Phi_0|\Phi_i\rangle
   = c_0\cancelto{1}{\langle\Phi_0|\Phi_i\rangle} +
     \sum_{i\neq0} c_i \cancelto{0}{\langle\Phi_0|\Phi_i\rangle} = c_0\\
\end{align}

Similarly this applies to all zeroth coefficients, $c_0 = d_0 = \dots = 0$.

Let's get back to business and continue with $\mathrm{8c}$.
\begin{align}
\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle
  + \langle\Phi_0|\mathbf{H'}|\Psi_1\rangle
  &= \langle\Phi_0|W_0|\Psi_2\rangle
   + \langle\Phi_0|W_1|\Psi_1\rangle
   + \langle\Phi_0|W_2|\Psi_0\rangle\\
\scriptsize\cancelto{d_0\cdot E_0}{\langle\Phi_0|\mathbf{H}_0|\Psi_2\rangle}
   + \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
  &= \scriptsize W_0\cancelto{d_0=0}{\langle\Phi_0|\Psi_2\rangle}
   + W_1\cancelto{c_0=0}{\langle\Phi_0|\Psi_1\rangle}
   + W_2\cancelto{1}{\langle\Phi_0|\Psi_0\rangle}\tag{9c}\\
W_2 &= \sum_i c_i\langle\Phi_0|\mathbf{H'}|\Phi_i\rangle\tag{9c'}
\end{align}

We do now insert $\mathrm{10b'}$ and get the second order correction:
\begin{align}
W_2 &= \sum_{i\neq0}\frac{
             \langle\Phi_0|\mathbf{H'}|\Phi_i\rangle
             \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle
            }{E_0-E_i}\tag{9c''}
\end{align}

Now we have the expression for the second order energy, let's do the messy stuff for the second order contribution to the wave function.

Skip here for the short version, which is still very long, but that's how we roll for the second order correction to the wave function

We start by multiplying $\mathrm{8c}$ with $\Phi_j$ from the left and integrating (projection). I will include a few more steps for this problem, hopefully that will help transferring it to your notation.
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
\end{align}

I will go through this term by term this time, as it is quite a long equation.
Let's start with the most simplest term, the third one on the right hand side. As you noted, it cancels.
\begin{align}
\langle\Phi_j|W_2|\Psi_0\rangle
   &= W_2 \cancelto{0}{\langle\Phi_j|\Psi_0\rangle }\\
   &= 0
\end{align}

Let's continue on the right hand side:
\begin{align}
\langle\Phi_j|W_1|\Psi_1\rangle
   &= W_1 \langle\Phi_j|\Psi_1\rangle\\
   &= W_1 \sum_i c_i \langle\Phi_j|\Phi_i\rangle\\
   &= W_1 \left(c_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle} +
         \sum_{i\neq j} c_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
   &= c_j W_1
\end{align}

And the same for the last contribution:
\begin{align}
\langle\Phi_j|W_0|\Psi_2\rangle
   &= W_0 \langle\Phi_j|\Psi_2\rangle\\
   &= W_0 \sum_i d_i \langle\Phi_j|\Phi_i\rangle\\
   &= W_0 \left(d_j\cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
        + \sum_{i\neq j} d_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\right)\\
   &= d_j E_0 &\mathrm{9a':}&(W_0 = E_0)
\end{align}

Let's go to the left side:
\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle
   &= \sum_i d_i \langle\Phi_j|\mathbf{H}_0|\Phi_i\rangle \\
   &= \sum_i d_i E_i \langle\Phi_j|\Phi_i\rangle\\
   &= d_j E_j \cancelto{1}{\langle\Phi_j|\Phi_j\rangle}
      + \sum_{i\neq j} d_i E_i \cancelto{0}{\langle\Phi_j|\Phi_i\rangle}\\
   &= d_j E_j
\end{align}

And the last one can barely be simplified:
\begin{align}
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle
  &= \sum_i c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
  &= \cancelto{0}{c_0 \langle\Phi_0|\mathbf{H'}|\Phi_0\rangle}
   + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
\end{align}

Now, Avengers reassemble, rearrange, and insert the expression for $W_1$:

\begin{align}
\langle\Phi_j|\mathbf{H}_0|\Psi_2\rangle +
\langle\Phi_j|\mathbf{H'}|\Psi_1\rangle &=
\langle\Phi_j|W_0|\Psi_2\rangle +
\langle\Phi_j|W_1|\Psi_1\rangle +
\langle\Phi_j|W_2|\Psi_0\rangle\\
d_j E_j + \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle
 &= c_j W_1 + d_j E_0\\
d_j (E_j - E_0)
 &= c_j \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle
   - \sum_{i\neq 0} c_i \langle\Phi_j|\mathbf{H'}|\Phi_i\rangle\\
d_j &=
   c_j \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
   - \sum_{i\neq 0} c_i \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}
\end{align}

Now introduce the first order base, $c_i$ and $c_j$ from $\mathrm{10b'}$, there will be some minus shuffling as well.
\begin{align}
d_j
  &=
   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)}
   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_j - E_0)}
   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(E_j - E_0)}\\
  &=
   \frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)} \cdot
   \frac{\langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(-1)(E_0 - E_j)}
   - \sum_{i\neq 0} \frac{\langle\Phi_i|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_i)}
     \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle}{(-1)(E_0 - E_j)}\\
 d_j &=
   \sum_{i\neq 0}
   \frac{\langle\Phi_j|\mathbf{H'}|\Phi_i\rangle%
         \langle\Phi_i|\mathbf{H'}|\Psi_0\rangle%
        }{(E_0 - E_j)(E_0 - E_i)}
   -\frac{\langle\Phi_j|\mathbf{H'}|\Psi_0\rangle%
          \langle\Phi_0|\mathbf{H'}|\Psi_0\rangle}{(E_0 - E_j)^2} \tag{10c'}
\end{align}

Well, I am officially exhausted right now.

And lastly, a small remark on your notation (I keep this paragraph for nostalgic reasons). It took me quite some time understanding it, and as we saw (in some earlier version of this answer), I was mistaken.
While I do not mind the hat notation for operators, indicating order on top of the quantities is problematic in my opinion. To me this place is reserved for powers. I suggest you change it to one of the following: $E_{n,0},\ {}^{(0)}\!E_n,\ E^{(0)}_n$. When you have a look at the other answer, you will notice, I have decided to stick to the last option. But as I do not know which book your are using it is also not easy to recommend. I believe there are as many different notations as there are books out there. I personally preferred the book "Introduction to Computational Chemistry, 2nd Edition"by Frank Jensen.
As we settled in the comments, I appreciate that you stick to Dirac's original notation. And also, notation is like Marvel and DC, you pick one and stick with it, you can still appreciate the other, but just not as much. Or a football (Americans pronounce this as "soccer") club, or ...
As long as you understand it and can explain it to someone else, you are fine.
And now we all deserve some chocolate and some coffee and cake. Thank you for reading this post until the end.


Does solid aluminum oxide conduct oxide ions at high temperatures?

Alumina is one of the most abundant compounds on Earth. If it would be also a solid electrolyte for oxygen ions, why in the world would anyone go into the trouble of dealing with the whole $\ce{Y2O3/ZrO2}$ thing? No, alumina doesn't and can't behave like that. Its oxygen ions are held in place pretty tightly.

Some forms of alumina may serve as conductors of $\ce{Na+}$ and similar cations, but that's another story.


It's all about the energy levels of electrons. Electrons exist in different energy levels and each energy level can only be occupied by a specific number of electrons (google quantum theory).

Example: Neon has an electron configuration of $1s^2 2s^2 1p^6$, which simply means that it has a full $1s$ orbital (the maximum is 2 electrons as denoted by the superscript), and has a full valence shell which is the $2s$ and $1p$ orbitals (which is the valence shell for molecules in row 2).

Now for neon to lose an electron (to bond) it would have to take an electron out of a full orbital ($1p$), which is what makes molecules reactive (a non-full valence shell), and would just take the electron right back.

Think of Ne$^+$ as more electronegative (wanting electrons) than elemental Fluorine, so it will pull an electron from almost anything and go back to its elemental form.

If this answer is not satisfactory then I can explain further, without diving head first into quantum theory.


"Hamiltonian operator has no effect on the spin function" what does it mean?

Because the operator $\mathrm d/\mathrm dx$ only acts on $x$ and not on $y$, we can write

$$\frac{\mathrm d}{\mathrm dx}[f(x)g(y)] = g(y) \left[\frac{\mathrm d}{\mathrm dx}f(x)\right]$$

Likewise, in this context, the Hamiltonian operator $\hat{H}$ only acts on $(x,y,z)$ and not $m_s$. This is explained by Levine1 in the paragraph immediately preceding this

  To a very good approximation, the Hamiltonian operator for a system of electrons does not involve the spin variables but is a function only of spatial coordinates and derivatives with respect to spatial coordinates [...]

Is there something about this paragraph which you don't understand? This is pretty much all there is to be said on the topic.

Levine, I. N. Quantum Chemistry (7th ed.), p 268


Why is Trouton's rule not valid for liquids with boiling points less than 150 K?

When a liquid vaporises its entropy goes from a modest value to a significantly larger one.  The ratio of $\Delta H_{vap}/T = \Delta S_{vap}$ is found to be approximately constant at the boiling point. Some authors give a value of $80~ \pu{Jmol^{-1}K^{-1}}$ others $90~ \pu{Jmol^{-1}K^{-1}}$.

Experimental values vary rather more than this and for gases such as neon, nitrogen, oxygen and methane whose liquids all boil below $150$ K, have values that are in the range $65 - 75$, benzene, many 'normal' liquids and liquid sodium, lithium and iodine, in the range $80-90$ and ethanol, water, hydrogen fluoride in the range $105 - 115 ~ \pu{Jmol^{-1}K^{-1}}$. Thus there seems to be nothing unusual about $150$ K but rather an influence from  intermolecular interactions.

The value of $ \approx 80~ \pu{Jmol^{-1}K^{-1}}$ corresponds to a cohesive energy of $\approx 9.5 kT$ per molecule and so the boiling point gives an indication of the strength of the cohesive energy holding molecules together in the condensed phase. When the cohesive energy exceeds this value, as in water, then the ratio $\Delta H/T$ is larger and conversely the ratio is smaller when the cohesive energy is less as in Neon or methane.

The $\approx 9.5 kT$ minimum energy per molecule is quite a modest energy; if a molecule has $6$ near neighbours this corresponds to about $3kT/2$ per interaction between two molecules, roughly the average thermal energy.


Why the H+ ion remains with the molecule even though there is no bond between it and hyperconjugative structure?

Resonance structures do not exist on their own! Resonance is not a rapid equilibrium between different structures. It is a description of a single delocalised structure using contributing forms which only involve 2-centre-2-electron bonds. It is therefore meaningless to think about what would happen to an atom in a resonance contributor if it were left around.

In your example the two structures are resonance contributors to the single delocalised structure and so the hydrogen ion cannot move away from the molecule because it never exists on its own as a hydrogen ion. It is always bonded to the rest of the molecule.

Hyperconjugation can be explained through resonance but it is much more clearly explained through molecular orbital theory. In this picture you have donation of electron density from the $\ce{C-H~\sigma}$ orbitals into the empty $p$ orbital of the positively charged carbon. This is possible because the two orbitals can be aligned in a plane due to rotation around the C-C bonds.